Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. Calculus is incredibly easy and trivial if you already know calculus
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  2. It is easy. The harder part is learning all the prerequisite material you need to know to start to learn calculus. But if you know algebra, trig, and geometry really well, calculus is incredibly easy.
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  3. Looking back on calculus, most of the things I actually had issues with were not the core concepts, but in fact was my ability to perform algebra without making small mistakes, remembering and applying trigonometric identities, and getting used to new notation. To anyone going through Calculus I urge you not to stress too much about it, just do your best it comes in time!
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  4. Division by 7 always produces 142857. Spare key to the Universe, if you lost the master copy
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  5. One of the first exams I had in physics the teacher gave us velocity over time graphs and we had to “be the car” and move in distance over time. Now that I’m 63 and still remember this tells me it was one of the best learning tasks I ever had.
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  6. In high school, I was surprised by how by far the hardest parts of Calc I and II simply involved a lot of steps of algebra. Things like partial fraction decomposition are a major pain, but actually integrating the resulting rational functions was very straightforward--once you did the necessary algebra (completing the square, etc.). Then in Calc III, I found it was much the same. Vector algebra is obnoxious, but the calc part really isn't so bad. I will say though that it gets much worse. Nonlinear differential equations are way harder than anything you have to deal with in a high school algebra class. I'd sooner factor ten solvable quintics than stare at a system of nonlinear PDEs until my brain melts.
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  7. In school whenever I noticed patterns such as these divisibility tests, my teachers discouraged me from pursuing them because they themselves were not sure if they'd always hold and were concerned they'd lead me astray. Another example that I recall is my noticing that each power of two is equal to one more than the sum of the lesser powers of two. That's well-established and taken for granted in computer science, yet was unknown to my teachers and regarded with skepticism. I remember also my mom pleading with my teachers to stop counting my work wrong for my daring to use techniques I developed myself from having explored the mathematical foundations of the rote mechanisms they taught. I understand that the pressures on elementary school math teachers drive them to stick with safe techniques, but for them to feel threatened by a student privately moving beyond that is frankly an indictment of the whole system of education.
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  8. As an Indian it is so disappointing that so far I wasn't aware about this Indian gem...... But I am grateful to you for revealing it before us🙏🙏🙏
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  9. I wish my childhood teachers had shown half the love and passion for knowledge that you do. Thanks for your work!
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  10. Do you really think im going to watch the whole 50 minutes? Because you are totally right
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  11. Your proof is decent, Mathologer, but Nathan's proof is by descent.
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  12. If it isn't Euler it's Gauss. Remember folks, if a murderer comes into your house and asks who proved a theorem you never heard of the Best survival strategy is to say "Euler or Gauss"
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  13. (HO)³ : A Christmas joke for mathematicians (HO)₃ : A Christmas joke for chemists
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  14. This demonstrates something non-math people don't get: Infinity is full of trap doors, subtleties, and other frustrations. The early infinity theorists like Cantor nearly lost their mind over this kind thing.
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  15. Mathologer: What does a partition have to do with a pentagon (aside from beginning with "p"). Me: blinding flash of insight They both end in "n"!!!
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  16. this man sort of comes across as a bond villain but is friendly enough so that I think he would be the assistant to the bond villain and would end up somehow disarming the nukes of the villain as a sort of double agent. these are the things I thought about in college. and I wonder why my degree didn't work out.
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  17. Ramanujan is the classic kid that doesn't listen in class, forgets to take notes, does no homework but then FREAKIN' ACES the test because he found his own way of doing things... He will never cease to amaze me!
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  18. The most memorable part was when you giggle, and my wife in the other room says "You're watching that math guy again?" As always, thank you for expanding my knowledge base.
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  19. Every part of calculus is mirrored in sequence calculus EXCEPT the chain rule. This is what makes infinitesimal calculus extraordinarily more powerful.
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  20. I was taught the concept in elementary school, under the name “casting out nines”. Sadly, it was presented as a trick or technique, without real explanation, which I had to discover for myself. So much is lost when mathematics is taught as a bag of techniques without the underlying beautiful patterns!
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  21. Shirt = To infinity and beyond?
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  22. The concepts of calculus are easy, and so is making programs to do it. The hard part of calculus is how they teach it in school. They want you to solve it with all of the rules to memorize. But memorizing all those rules- and the exact situations in which to use them- is the difficult part. The ideas of differentiation and integration can frankly be understood by anyone who can understand the area of a circle and how to graph a line; in other words, a late elementary school student or older. But for me, calculus was the first math class where suddenly there was no ability to look at a problem and know immediately how to solve it; you had to try different things on the same problem until it worked. And that does make it more difficult than any previous math class. Granted, it really doesn't have to be that way. Teachers could teach it differently and you wouldn't have that problem.
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  23. It seems that you are able to provide a geometric interpretation for any mathematical concept that exist. Simply amazing.
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  24. Fermat: "Hey, here's this cool thing about numbers." Mathematicians: "Amazing! Can you prove it?" Fermat: "I already did." Mathematicians: "Wow! Can we see it?" Fermat: "Hmmm... nah."
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  25. This proof was discovered by Roger Heath-Brown in 1971, and was later condensed into the one sentence version by Don Zagier. It's one of two proofs of this theorem found in the wonderful book "Proofs from THE BOOK" 6th ed by Martin Aigner and Günter M. Ziegler in chapter 4.
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  26. I tend to forget that Burkard is actually German (?), so when he read out the title from Moessner's thesis I was like: "Damn, his German is amazing." and felt silly a few moments later...
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  27. I remember seeing this video 3 years ago when I really was getting into maths and science. I could never make sense of how this could possibly work. What's nice about this video is that now I'm studying chemical physics and have the maths to fully understand this. Kind of nostalgic looking back on problems that I once was unable to grasp.
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  28. As easy as the derivate of e^x
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  29. All YouTube:trees Mathologer: let's sum quadrillion of powers
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  30. I'm convinced Tesla thought this was a neat mathematics trick, and someone picked it up as "mystical" because Tesla is smart.
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  31. I don’t think I can express how grateful I am to live in a time when I can be taught by one of the best math teachers in the world from my bedroom. Thank you so much Mathologer, we don’t take you for granted.
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  32. When he said “what a crazy, crazy year right?” I’ve been conditioned to expect him to say “Wrong!” 😂
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  33. When you showed the manuscripts(written in my mother tongue Malayalam, spoken over by only less than 40 million people in a small state known as Kerala in India) towards the end of the video, it gave me chills literally. I am glad that the efforts of these ancient Indian mathematicians are being appreciated by the mathematics fraternity all over the world.
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  34. "If you're ever captured by an evil toymaker, you'll be ready" And they say math has no real-world applications.
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  35. You're the Bob Ross of mathematics.
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  36. A new upload, great stuff <3
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  37. Do this? Pascal's triangle Do that? Pascal's triangle What time is it? Pascal's triangle
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  38. When I want to double a cube, I just pull out my complex ruler and e^i(pi) compass and get to work. Easy.
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  39. That moment when you realize the number line is just the 1d case of Moessner's Miracle...
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  40. This is mad! I am a PhD student who is actively studying number walls! Specifically, I have written a program to generate the number wall from a sequence and I have some nice formulas for the number of sequences that have a given window in their number wall. They have some beautiful relationships to diophantine approximation over function fields and linear complexity! It's so excititing to see them be popularised more.
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  41. The opposite of this is a vuvuzela, which is a horn with finite surface but infinite volume
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  42. I love this process. We learnt this before moving on to “conventional” calculus at my school and I took it for granted that everyone had too.
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  43. I'm Brazilian and I've studied the high school in a technical school. I recall learning about Heron's formula as a secondary topic, but no proof nor any thorough explanation, as the one presented, was given. Thanks for the lecture!
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  44. As a programmer, I sometimes wake up with "dream functions" lol
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  45. I have no idea who dislikes videos like this. The amount of effort put into it is just tremendous.
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  46. Ramanujan was such a great mathematician that on 22nd Dec.,as his birthday is celebrated as national mathematics day in India.
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  47. 7:17 Actually, I would argue that neither you nor the young man in the clip missed out on that oversight, because there was no oversight to miss. The coach never stipulated that all of the cards had to end up being face-up, only that the sequence of moves had to terminate. In other words, the task was simply to show that there is no possible infinitely long sequence of moves, or to put it in yet another way, once you apply a move to any arrangement of cards, it is impossible to apply any combination of moves to the resulting arrangement such that the result is the original arrangement. Therefore, the answer to the question of what to do if only the right most card is face down is to do nothing. Such an arrangement simply represents a terminating condition. Also, about the shorter video, I actually enjoy this format very much and would enjoy more shorter videos.
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  48. Madhava was truly one of the greatest mathematicians of all time
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  49. <3
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  50. Imagine discovering this in the 1980s, then learning that actually some geezers had gotten there not only before the invention of computers, but before the printing press...
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  51. This video is my best shot at animating and explaining my favourite proof that pi is irrational. It is due to the Swiss mathematician Johann Lambert who published it over 250 years ago. The original write-up by Lambert is 58 pages long and definitely not for the faint of heart (http://www.kuttaka.org/~JHL/L1768b.pdf). On the other hand, among all the proofs of the irrationality of pi, Lambert's proof is probably the most "natural" one, the one that's easiest to motivate and explain, and one that's ideally suited for the sort of animations that I do. Anyway it's been an absolute killer to put this video together and overall this is probably the most ambitious topic I've tackled so far. I really hope that a lot of you will get something out of it. If you do please let me know :) Also, as usual, please consider contributing subtitles in your native language (English and Russian are under control, but everything else goes). Today's main t-shirt I got from from Zazzle: https://www.zazzle.com.au/25_dec_31_oct_t_shirt-235809979886007646 (there are lots of places that sell "HO cubed" t-shirts) Merry Christmas, burkard
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  52. When you mentioned Calculus Made Easy I thought, "Hey, I have that book!" and ran to the bookshelf to retrieve it. As it turns out, no I don't. I have a book called Calculus the Easy Way by Douglas Downing of Yale University, © 1982. It's a fun little book wherein the protagonist is involved in a shipwreck and washes ashore in the land of Carmorra where he, in essence, helps its denizens invent calculus in order to answer burning questions involving the speeds of trains, the areas of fields, the simple harmonic motion of a spring-powered chicken scaring machine, etc.
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  53. Geoff Smith, who wrote the problems for this film (and some of the Olympiad questions) was my lecturer in first year. When the film came out they were extras for our problem sheets.
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  54. Despite serious competition, Mathologer remains the greatest math channel imo ^^ Thanks for another awesome video!
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  55. Nice explanation of calculus. However, I would have preferred the masterclass on Galois theory you promised three years ago 🙂
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  56. Good Stuff Burkard! :)
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  57. Every month when Mathologer comes out of his hiding, you know it's going to be a great video
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  58. One day I hope to answer your “did you see it?” with an equally enthusiastic “Yes!” 💕🐝
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  59. You have a gift for showing us what math is really about. It's pure amazement, wonder, curiosity and entertainment. Thank you for capturing the essence of math!
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  60. I have literally used a slide rule for a quiz in engineering school around 2014. The battery in my calculator died and I didn't have a backup. But the rules of the class said I could use any device for calculation with the professor's permission. So I dug the slide rule out of my bag, held it aloft and said "Excuse me, professor! May I use this for the quiz?" He - having grown up in the age when digital calculators were still uncommon, at best - laughed out loud and said "If you know how to use it, sure!" He was certainly proud that I aced the quiz with that slide rule.
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  61. I feel like Ramanujan was an alien that was sent to Earth to accelerate our knowledge in mathematics, and once he taught everything he could to the human race, he left Earth to teach another underdeveloped civilization.
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  62. "Are we there yet?" "No just 1+1/2+1/3+1/4+... more minutes."
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  63. I actually guessed Gauss! Take that Mathologer! Me: 1 Mathologer: 387ish
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  64. Thanks for making a video about my theorem!
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  65. Archimides was doing calculus without algebra. No wonder he's your favourite. That's pretty much the spirit of this channel if you think about it.
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  66. Surprised you didn't call good and bad numbers nice and naughty numbers respectively.
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  67. Calculus is only hard until the point where it clicks in your head and then you feel: "How could I not understand it?" Tangent 🤣: Archimedes was really close. I think he makes for an incredibly plausible "what if?" scenario. What if he discovered calculus? How much would it change the world?
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  68. Ramanujan was and is great.
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  69. This is perhaps the best math video I've seen. Clever, well-explained, and elegant. Keep up the great work. Stay safe amid the Covid.
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  70. 69 x 10k subs, nice :^D
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  71. When I started learning Calculus in High School, I began to realise that everything I had learned in maths before then from simple arithmetic, geometry, algebra and trigonometry was leading up to it. Does this mean that one reason we learn how to add and subtract is so that we can eventually do Calculus?
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  72. /w magic
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  73. (28 March) Really bizarre, this video was basically invisible for almost two weeks with hardly any recommendations going out to fans. Only now YouTube has decided to actually show it to people. Who knows, maybe it was a mistake to mention cat videos in previous videos and the YouTube AI is now under the impression that the target audience for these videos has changed :) The longest Mathologer video ever, just shy of an hour (eventually it's going to happen :) One video I've been meaning to make for a long, long time. A Mathologerization of the Law of Quadratic Reciprocity. This is another one of my MASTERCLASS videos. The slide show consists of 550 slides and the whole thing took forever to make. Just to give you an idea of the work involved in producing a video like this, preparing the subtitles for this video took me almost 4 hours. Why do anything as crazy as this? Well, just like many other mathematicians I consider the law of quadratic reciprocity as one of the most beautiful and surprising facts about prime numbers. While other mathematicians were inspired to come up with ingenious proofs of this theorem, over 200 different proofs so far and counting, I thought I contribute to it's illustrious history by actually trying me very best of getting one of those crazily complicated proofs within reach of non-mathematicians, to make the unaccessible accessible. Now let's see how many people are actually prepared to watch a (close to) one hour long math(s) video :). Have a look at the description for relevant links and more background info. The first teaching semester at the university where I teach just started last week and all my teaching and lots of other stuff will happen this semester. This means I won't have much time for any more crazily time-consuming projects like this. Galois theory will definitely has to wait until the second half of this year :( Still, quite a bit of beautiful doable stuff coming up. So stay tuned.
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  74. If you consider the right boundary of the Problem, where you interpreted the Rules to mean that you only turn over one card, then there should turn up an interesting result if the cards are layed out in a circle. That would mean that when turning over the rightmost card, you would also turn over the leftmost card. That way, the sequence would not need to terminate, i think. Because you could shift the Face Up cards infinitely to the left and looping around when they "hit the edge" of the deck.
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  75.  @plaierdifortnaiti9955  On the same level as an integral from 0 to 0
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  76. As a 50 year old man, I may have done better in maths if I had teachers like you! Thank you for your simplification of complex maths. :-)
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  77. Let's gooooooooooooooooooooooooo! :D
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  78. Most memorable part: me losing my life after failing the “no nines sum converges”
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  79. A "short" video for a change, "only" 30 minutes long :)
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  80. Hey. Compared to the amount of minutes in 2000 years, this video is remarkably short.
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  81. Minecraft villager be like: 5:30
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  82. WOW IT TOOK ME MY ENTIRE DEGREE TO MAKE SENSE OF THIS THEOREM AND HERE MATHOLOGER COMES SHUFFLING CARDS TO PROVE IT CLEARLY
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  83. Looks like I unknowingly introduced this to myself and my wife and daughters with a little game we used to play while travelling. We would add up the numbers on license plates and see who came up with the "digital root" the quickest, even though we didn't know that was the term to use. We saw very quickly that any combination of numbers that add up to 9 could be eliminated so 572 would be 5 without going through the process of adding. Later, as 3 or 4 number plates lost its challenge, we included letters. The letters "I" and "R" could automatically be eliminated since they corresponded to the number 9 and 18 respectively. This expanded the challenge because you had to figure out the numbers corresponding to the letters. As you played the game this became more intuitive when you could eliminate combinations of letters that added up to 9 for elimination. Example GSP562 would be 1. One of my daughters got so good at it that within seconds she could get the digital root of signs with just letters such as names of towns or short sentences.
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  84. Sometimes you wonder how mathematicians come up with things... Other times, you wonder how mathematicians don't come up with things...
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  85. Once you add both the triangles and squares, it looks like a 4D projection.
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  86. There's actually a sort-of-explanation for why e^π is roughly π+20. If you take the sum of (8πk^2-2)e^(-πk^2), it ends up being exactly 1 (using some Jacobi theta function identities). The first term is by far the largest, so that gives (8π-2)e^(-π)≈1, or e^π≈8π-2. Then using the estimate π≈22/7, we get e^π≈π+(7π-2)≈π+20.
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  87. Really appreciate how much effort is put into these videos. Very well structured and illustrated. YouTube really doesn't deserve this level of effort.
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  88. Ah yes, my favorite mathematicians, iron man and towel man!
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  89. This is just the discrete Fourier transform working in the background. Amazing video !
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  90. Two 3b1b AND a Mathologer upload? I am almost in heaven...
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  91. This is mathematical proof that cutting corners gives bad results ! ;-)
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  92. I get the feeling that Euler was a pretty smart guy.
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  93. My solution to the coin paradox: imagine the two coins are like gears with the centers fixed in space. Both turn once every time the other one does. Then imagine the camera rotates with one of them, so it looks fixed. This feels more intuitive since nothing "rolls away" or gets wound around.
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  94. I love how much of induction in this video even tho the word "induction" is not mentioned
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  95. The other paradox: how every Mathologer video manages to fit in so many interesting facts and beautiful demonstrations!
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  96. Any divergent series: exists Ramanujan: Allow me to make it convergent.
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  97. People will stay at home Time to make 1hour long video
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  98. Mathologer recently hit 500K subscribers and I would like to thank you all for your interest and your support over the years. Since I started the channel four years ago, it has pretty much been a one-man labour of love. However, maybe now is a good time to take Mathologer to the next level and hire someone to assist with editing the videos, preparing subtitles, etc. In preparation for this, I recently monetised the videos by switching on the least annoying ads on YouTube. I also just launched a Patreon page a couple of minutes ago: https://www.patreon.com/mathologer If you enjoy these videos and you can afford it, please consider taking out one of the Patreon memberships or making a one-time donation via PayPal: paypal.me/mathologer :) My plan is to also use this Patreon page as a platform to share more Mathologer materials with you, get into live chats, etc. Let’s see what’s possible and what makes sense here. Merry Christmas :)
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  99. At first I was sceptical as there is no "real magic". But then I saw an infinitely stretchable rubber band and now I'm convinced magic is real ;D
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  100. Always appreciate when a Rubik’s cube example “accidentally” gets you some group theory ;)
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  101. For me is impressive how almost everything can be explained geometrically but commonly isn't thaught like that, even when I find it more beautiful and comprehensible
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  102. When dealing with actual shoes what goes commonly overlooked is the order in which the laces overlap at the intersection. It matters immensely. For experiment tie your shoe with standard zigzag but make sure every overlap lays right over left. Walk around for a day and then try again with overlaps left over right now. You'll notice they pull vastly differently on the foot. The lace overlapping the other lace is free to adjust slightly while pinning down other down restricting its movement. I have a bony knuckle top cent of my feet so I make sure the overlaps at this spot are laced a certain way. It's so much more comfortable than overlapped the other way. All this while never changing the actual layout of the lacing.
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  103. "And mathematicians wonder why people think they're weird." My mother was a singer, actress, secretary, homemaker, and social butterfly. NOT a math person (that was my dad). One day, I was trying to explain a math problem to her and I needed to pick a small number to use in a simple example. So I said "How about 1?", and she starts laughing. "Why 1? It's so small! Why don't you pick a REAL number?" 😄
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  104. The last remarks hinting to asymptotic series had me on the verge of my chair while watching the video. I am a physics PhD student and your "masterclasses" are gold to me even if they have nothing to do with my research. The amount of care, precision, passion and entertainment that is present in your videos is outstanding; making such complex and rich subjects accessible to the amateurs and casuals as well as interesting and non trivial to the experts is a miracle made possible by your talent and commitment. I am not saying this to appear sappy, it is just that I believe that it is priceless the fact that I can have access to high-quality content like this for free. I just want to show appreciation for the creators that spend time and energy to put together a video like this one. Thank you very much and keep up the good work!
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  105. Donald E. Knuth calls this "finite calculus", as opposed to the usual "infinite calculus" that is commonly taught. His book, "Concrete Mathematics: A Foundation for Computer Science", uses this "finite calc" to tackle subjects such as hypergeometric functions, generating functions and asymptotics, and derives lots of analogies between the two variants. For example, establishing a power rule, an analogue to exponentional and logarithmic functions and even a "summation by parts" technique. Finite calc is a powerful tool that lets us work wonders, and makes lots of sums we usually cant tackle, easily reducible. I'd say check the book out!
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  106. I found calculus to be really easy when I first learned it, but it was always the algebra that held me back. Just as they say, people take calculus to finally fail algebra.
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  107. Wow… this whole finding differences of a set of points of a polynomial is something I figured out in 11th grade. I made this program on my calculator that you input a table of points and it will calculate the area under the curve, and I found as you say, for ‘well behaved nice functions’ it was super accurate. Kinda proud I was able to figure this all out with only knowing very basically calculus at that point in my life.
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  108. I'm here to brag about my bragging rights for choosing yellow arbitrarly
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  109. the rubber band explanation for logs clicks so naturally, in a way that few other explanations do!
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  110. As a German with a rather casual interest in mathematics, I just want to point out how nicely Moessner presented his idea in the original text. No overstatement, no attempt at making it seem more complex or grandiose than it is, and a straight to the point, easy to understand text. It's literally just "here is a generalisation that does not seem to have been noticed so far, have a look". Great text and an awesome video!
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  111. Unlike mathematicians, real-world painters know that the area they can paint with a given volume of paint depends on how thick they slather the paint on the surface. Volume = Area x Height. They also know that they cannot spread the paint infinitesimally thin—otherwise just one drop of paint would be more than sufficient to paint everything. So no paradox for them.
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  112. I was not that surprised by the square pattern, nor the linear one. But I was truly amazed when he pointed out that in the linear pattern the natural numbers appear exactly one by one 😇
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  113. 9:39 I love this thing about 2^n being equivalent to e^x. It's as if someone came along and very simplistically assumed that "we're dealing with integers, so we should round e down to 2". It's crazy! :D
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  114. A year ago I left a comment on one of these video's saying I was so inspired I was going to make my own math education you tube video's. I have something very special for everyone coming very soon, it's a free software project that I created while working on a tool to make animations for my video's and is almost ready to be released. I just published the first video on my channel, check it out!
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  115. There must be millions of people who have heard of the Tower of Hanoi puzzle and the simple algorithm that generates the simplest solution. But what happens when you are playing the game not with three pegs, as in the original puzzle, but with 4, 5, 6 etc. pegs? Hardly anybody seems to know that there are also really really beautiful solutions which are believed to be optimal but whose optimality has only been proved for four pegs. Even less people know that you can boil down all these optimal solutions into simple no-brainer recipes that allow you to effortless execute these solutions from scratch. Clearly a job for the Mathologer. Enjoy :)
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  116. Feedback: I've just finished a master's degree in mathematics, now starting with a PhD. Nevertheless, there are many things I'm learning from these videos. It's cool seeing recreational math videos reaching this level.
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  117. I think your visual derivations of the fundamental mathematical concepts is amazing. It helps the people who struggle with visualising the algebraic operations with something visually tangible and making connections which were forgotten by the maths teachers.
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  118. Well, after that one, I (probably irrationally) suppose mathematicians tend to avoid too much sun exposure cos tan is a sin.
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  119. "I made it to the very end."
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  120. Most memorable part: In university Mathologer apparently came up with an original finiteness proof for Kempner's series, and the grader failed the homework because they couldn't be bothered to check a solution that was different from the one on the answer sheet.
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  121. 40 minutes of Professor Polster + links to 400 hours of reading = 1 helluva education Every single time, thank you for all you're doing, Mathologer!
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  122. "Are there any other special numbers?" Well obviously 2 works :p
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  123. Truly the man who knew infinity.
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  124. I will never tire of 99999999 in a strong German accent
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  125. I have indeed been taught this at school. Not in the math class, but during Latin class. Our Latin teacher also knew ancient Greek and had found the Greek text about this interesting enough to teach us about it. I found it very enlightening.
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  126. Greetings from Melbourne. For a change I am posting this video at a reasonable time, 8:51 a.m. on a Sunday morning. We are still in lockdown around here, but things appear to be improving: 63 new cases. There is a very interesting footnote to what I am talking about today contained in the description of this video. Check it out :)
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  127. always something uncanny about that ai generated yassified sophie germain
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  128. This video makes a lot of cents.
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  129. The good, the bad and the even
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  130. 42:40 "To be continued" I see what you did there.
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  131. I have made a career of mathematics, but these videos make me feel that childhood joy of mathematics all over. Thank you so much for making these.
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  132. 17:40 "This recipe also solves another Hanoish puzzle." I think the correct adjective is "Hannoying".
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  133. For the record, this is also a path to discovering derivatives and calculus. For many reasons it's a shame this wasn't stared at harder a while back. I wonder what we're just a few steps from discovering
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  134. One neat fact that was left out is that when the four numbers in the box are viewed as fractions, the two fractions are equal to the tangents of half of each of the two acute angles of the triangle.
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  135. RIP conway, he will be missed
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  136. It seems a miracle indeed that anyone would do a half-hour video on the Moessner Miracle, until you read the name Mathologer.
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  137. I'm almost 40 and back in school to become an engineer, not having touched math since my teens. This type of visualization is so helpful getting my old brain back on track! I have a feeling I'm gonna be replaying this video a lot 🤓
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  138. as easy as proving fermats last theorem, walk in the park.
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  139. Mathologer often uses the expression “mathematical spidey sense”. He is right. Math is not invented or discovered,; it is sensed.
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  140. First challenge: no, you can cut out the angle of the board
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  141. This was a foundation subject taught in actuarial studies long before the advent of PCs and spreadsheets. Much of the early work of actuaries relied upon such techniques to analyse empirical mortality and morbidity data in life insurance. It’s actually a cornerstone of a broader field called numerical analysis. Another long forgotten subject that might merit your attention is spherical geometry, the basis for navigation and astronomical work. It’s parallels and extensions to Euclidean geometry are fascinating.
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  142. I wasn't even taught the divisibility test, let alone the remainder function! This is really cool!
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  143. Super interesting! Slight error: @6:11 you go to 3.18 on the diagram (each sub-sub-tick is .02).
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  144. Conway’s circle has the larger diameter. The “diameter” of the shape of constant width is exactly one of the chords we were discussing. But in Conway’s circle, those same chords are not diameters, so are less than a diameter, so the circle has a greater diameter.
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  145. When I saw Burkard with hair, my head turned into a black hole.
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  146. This channel has brought me intellectual ecstasy for years
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  147. <3
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  148. Cube time! I stopped the video around 5:00 where you said those slices don't fit nicely around the cube. Well I say they fit pretty good if you just grab a couple of little unit helpers. So we get two 1^3 helper cubes to add to 5^3 + 6^3, then a 5 cube plus six 6x6 slabs plus two unit helpers in the corners we get the 7 cube putting everything together! So 2*1^3 + 5^3 + 6^3 = 7^3. So lets go further. Following from our two previous families, the central pattern for cubes looks like its gonna be 6=6*1, 18=6*(1+2), 36=6*(1+2+3), etc. So lets look at the second one. 16^3 + 17^3 + 18^3 ~= 19^3 + 20^3. We're gonna need more helper cubes, this time two sets - one for each big cube on the right. Putting stuff together we get 2*(1^3 + 2^3) + 16^3 + 17^3 + 18^3 = 19^3 + 20^3 makes perfect fit! Now it looks like our helper cubes have a pattern very similar to the central number: 6*1 <-> 2*1^3 and 6*(1+2) <-> 2*(1^3+2^3). So now I'm gonna go right into the third cube equation with central number 36=6*(1+2+3) and say that we need 2*(1^3+2^3+3^3) helper cubes. So just jump right to the equation 2*(1^3+2^3+3^3) + 33^3 + 34^3 + 35^3 + 36^3 = 37^3 + 38^3 + 39^3, and it works! If you want to keep the Christmas tree aesthetic, we can call our helper cubes the ornaments :)
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  149. I'd be really interested in that "infinite soldiers and moves" solution
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  150. You know it's going to be a great video when you're only one minute in and you've already seen three excellent results.
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  151. Challenges: 1) We can guarantee at least 5 on one side. There are two ways to split the pigeons: 3-2 and 4-1. The max you can hit for both of them is 3 and 4 respectively. So we have at least 3 pigeons on one side. Add in the two pigeons on the equator and we have 5. 2) let the mystery value equal x. by multiplying by 10^some integer until the digits line up again, and then subtracting, we have 10^whatever*x - x = the non-repeating part you're left with. Factor out x on the left and that 10^whatever - 1 becomes a bumch of 9s. Then divide and we have the non-repeating remainder over a bunch of 9s. If ever the remainder isn't an integer just multiply both the numerator and denominator by 10 to knock out the decimal. Example: 0.4318181818... x = 0.4318181818... 100x = 43.18181818.... 99x = 42.75 (all the 18s afterward cancel out) x = 42.75/99 x = 4275/9900 x = 19/44 (simplify) 3) let's pretend we color the "didn't shake hands" white. We leave the "shook hands" black. So all we have to do is prove a triangle exists somewhere, and it doesn't matter what color. Note that each dot will always have at least three lines of the same color connecting to it, for the same reason the pigeons in challenge 1 can be split into a side with at least 3 and a side with less. Look at a point, and follow three lines of the same color (I'll choose black) to three other points. Any line connecting two of these three points would either be black or white. If it were black, that would make a triangle with the two points and the starting point. If it weren't, then the three dots all have white lines between them, which makes another different triangle. 4) 315. And no, I didn't count manually. Here's how I found out: The 7 edges UB, UR, RB, DB, DL, LB, and UL, all commute in one big cycle of length 7. The remaining 5 edges do the same. The 5 corners URF, ULF, DRF, DLF, and DRB go back in their positions but in the wrong orientation. So it takes 3 cycles to get back to their original orientation. The same thing happens with the other 3 corners in the back, only this time they take 9 cycles because they permute around each other too as well as reorient themselves. All that's left is to compute the lcm of 7, 5, 3, and 9, which turns out to be 315. 5) 1, 2, 4, 8, 16, 32, and 64. Any collection just results in the binary representation of the number. Since every number can be written in binary in only one way, every single collection will add to a different number. 6) Queen of Hearts. The 9 of Hearts at the start signals the suit of the missing card is Hearts. The remaining three kings can be sorted as MBT (Diamond, Club, Spade), which is assigned 3. 3 spots after 9 is the Queen, and putting it all together we have Queen of Hearts. Edit: I forgot one, challenge 5 at chapter 6
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  152. the chemist version of the joke is very "basic"..pun intended.
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  153. Would be a very natural thing to do but sadly hardly anybody ever gets to know about this beautiful topic (until now of course :)
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  154. My father was a carpenter. He built various buildings, and the last one was a cottage where I helped. We checked that the corner was at right angle using the 3-4-5 measurement.
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  155. Never commented here before... Burkard, you seem like the coolest person! Loved every one of your videos that I have watched. I wish I had the internet when I was a kid. Learning math with you as a kid would have been so much simpler and so much more fun. Thanks for everything! You rock!
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  156. Thanks Burkard for putting together such a nice presentation and filling in so many connections. It has been a load of fun working with you and Henry on this.
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  157. 4:45 "On the right, what do we have. Oh, well, whatever that is, it should be infinity. HMMM" "On the left, we are dividing by zero. Double HMMMM" DEAD
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  158. Always found it insanely impressive when ancient mathematicians used normal language and text instead of our modern formalism to transfer ideas and results. Imagine how hard it must be to do mathematics that way.
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  159. I already had a python interpreter open, so I did Bernoulli's sum in about 20 seconds! 😝
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  160. Those "complete the sequence" questions are my pet peave. The thing is, any number can continue any sequence, and there will be a formula (a polynomial; actually, infinitely many polynomials) to produce the resulting new sequence. That type of question is routinely used in school tests and intelligence tests, but what it really tests for is a kind of learned bias toward small integers.
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  161. When the rubber band snaps, do you get complex numbers? Maybe surreal! Great stuff as usual
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  162. Regarding the tree puzzle at 15:50: The bottom square's area is 1. The Pythagorean theorem states the two squares attached it share the same total area, so they are each 1/2. The total area so far is then 2, with 1 contributed from the big square and 1/2 + 1/2 = 1 contributed from the small squares. The next level down, the tinier squares attached to one of the small squares has to add up to 1/2, so they are each 1/4. There are 4 of them, so the total area of these squares is 1, and the total area is now 3. You continue down the line, adding 1 to the total area for each iteration of the tree. There are 5 iterations, so the total area is 5.
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  163. 6:28 - The reason taxes come out first isn't some kind of greed on the part of government. It's because only paying the taxes first ensures that nobody will go bankrupt to their friends as a means of avoiding the payment of taxes to the government. If my assets are only just enough to cover my back taxes, I can call my friend and say "Do you remember that money I borrowed from you 20 years ago? We kinda forgot about it, but I think it's time to formalize it with a document because I'm about to go bankrupt". The friend comes over and I sign the note, which is for 99 times the amount of my assets. If taxes are not given priority, then when I go bankrupt the government gets its penny on the dollar in proportion to all debts, while my friend gets 99 cents on the dollar. When the dust settles, my friend simply gives me back that 99% of my assets. I end up losing 1% to back taxes instead of 100%. But if everyone knows that taxes come first, then everyone sees the fruitlessness of such schemes and they won't be attempted. Of course, I can run this same scheme on any NON-governmental entity to whom my debt equals my assets, and get 99% of everything back while the debt to that creditor is marked "paid in full". That creditor will go to court and challenge the validity of a promissory note signed one day before a bankruptcy-filing, but the government, by taking taxes first, doesn't have to go through that hassle.
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  164. Wow. 50 Minutes of Mathloger. It is like Christmas already!
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  165. Madhava, Aryabhatta, Brahmagupta, Baraha- the list of such geniuses goes on and on. This talk was so beautiful it brought me immense joy. A flowers fragrance benefits all - no problem. The problem is the same people not recognizing, others walking nearby not recognizing. It is good to see western and other intellectuals finally recognizing. We must recognize where it is due - like the agricultural genius from South America, Math and philosophy from India, material science from China.
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  166. So elegant. At 19:17, I understood where this proof is going, that is the happiest moment of your video when I understand where the proof is going 😃
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  167. 20:00 Number of tilings of the Arctic Circle: 2^(-1/12). Got it. 😏
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  168. Oh, that's what an empty lunch box means. I just thought my mom was a little absent-minded. But this makes so much more sense.
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  169. Am i really watching a 29 min video at 3am about the best way to tie your shoe laces ? YES.
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  170. Related to the coin rolling paradox: Despite there being 365.25 days in a year, the earth actually rotates 366.25 times on its own axis in a year.
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  171. I'll have to watch 3 times. It always takes me at least 3 times because I constantly want to pause, and go explore something that was said...😂
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  172. so this is a kind of duality between two different triangles, neat
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  173. One ring for each integer above 1. And one ring to rule them all.
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  174. What's better than an "AHA moment"? Over 40 minutes of endless "AHA moment"s, of course! Loved the video, make more, Burkard! (And preferably faster :D )
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  175. Here's the thing. I just got home and am extremely tired and definitely not ready to get my brain working on math, or anything demanding that is. So, I got hooked in the first few seconds and my brain started working again. Your skills, sir, are of another world.
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  176. -He says something about rings -I think it will be another Mathologer analogy or something -He is actually talking about damn rings from abstract Algebra -I get excited
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  177. As we’re all home-schooling at the moment, the school is sending challenge sheets out to all the kids. The final question on my daughter’s maths one was “how many different ways can you make £3.10 using coins”. She’s 7 years old... I just changed the question to “think of a few ways you could” instead, because even I didn’t know how to solve it. But now I do! I don’t think I’ll be teaching this to her yet though.
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  178. 24:22 For the 3,4,5 triangle the line connecting the incenter and the Feuerbach point is parallel to the shortest side. Thus, the parent triangle of the 3,4,5 triangle is the degenerate 0,1,1 triangle (and its clear why this construction cannot be taken further).
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  179. A few years ago I used to binge watch this channel. You might imagine my surprise when I started my science degree (undergraduate biochemistry where maths is thankfully a necessary subject to take) and walked into a maths lecture and was confronted by this wonderful person in the flesh. That was about 4 or 5 years ago. Keep up the saintly work you do!
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  180. There's a mistake at the end. In Brahmagupta's Formula the third bracket should be (A+C+D-B) not (A+B+D-B). But great video. That are some beautiful equations and you explained all of it really nicely with all these brilliant animations. Love it!
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  181. 22:21 cos 120 = -1? wtf?
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  182. The big flaw of the Talmud method is that it disincentivises big loans (no one wants to be the biggest creditor), and can be gamed by splitting up big loans into smaller ones. In the extreme case, it converges to proportional splitting.
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  183. The most memorable thing is how ugly the optimal leaning tower is
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  184. Archimedes' claw, seriously? What a shame not to name it Archimedes' pumpkin 🥲 Or Archimedes' Kabocha to be even more accurate.
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  185. Completely understood the math behind it, but I would never be able to come up with that binary analogy on the spot if you asked me to do the same.
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  186. Ramanujan may have been The Man Who Knew Infinity, but Mathologer is the Man Who Made Infinity Long Videos About Them :)
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  187. I'll be honest, I don't know why, but the second problem of integration, about there not being elementary antiderivatives of all elementary functions, just crushed me psychologically when I was learning calculus. I know there are ways around the problem, and I memorized them enough to do decently well in calc, but somehow the trial and error nature of it just lost some of the luster off something I previously quite enjoyed, and turned me off of pursuing any higher forms of calculus. To this day I can handle most high school math up to that point with only minimal references, but all the various methods of integration slipped away from me like water once the class was over.
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  188. Wow just the first 2 minutes of introduction is just amazing, what a great mind Ramanujan had, and beautifully explained professor!
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  189. Dear Mathologer, Seeing your video this morning has brightened my day so incredibly much. Your videos allow me to transcend my body(have pain) and live in a world of pure mathematics. Please never stop. - Your fan and student.
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  190. His shirt doesn't have a 27
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  191. “But of course close doesn’t win the game in carnivals or mathematics” Analysts: “Allow us to introduce ourselves” ???: “Amateurs” Analysts: “What did you say?!” Numerical analyst: “AMATEURS”
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  192. When I got 4th rank in Kerala for the Madhava Mathematics competition, I realised that I had the potential to be a mathematician. Thank you for letting more people know about him!
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  193. A Parker heptagon!
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  194. Twenty minutes later I felt the power of Matrices.
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  195. Wow, I've watched this video four times over the past 1+ year and each time I understand the cubic formula better, seriously! Especially since I learnt about complex numbers last year only. Amazing video, one of my favourites on all of YouTube!
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  196. It is good to see that mathloger is back online...
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  197. The most suprising part for me was the "terrible aim" the fact that odd/even is never an integer is so simple yet i would have never thought about it
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  198. It seems algebra, geometry, modular arithmetic, and fractals are all the same thing wearing different hats.
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  199. Speaking only for myself I appreciate when you explain the same problem in three or more ways. I feel like I gain a little bit of understanding with each iteration.
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  200. When you said he solved this instantly I couldn’t help but feel small
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  201. That “hexagon exists in a cubic lattice” is why two sorts of crystal lattices in chemistry are identical. I can’t remember which ones, but I think it’s hexagonal and either face-centred-cubic or body-centred-cubic. Also there’s both tetrahedra and octohedra within a cubic lattice, which tesselate with each other in 3D space. The way I’d look at that initial problem, finding equilateral triangles in a cubic lattice, is that all points in a cubic lattice are either 1 or sqrt(2) from their neighbours, and an equilateral triangle needs a sqrt(3) in there. Plus or minus an inverting scale-factor. But on a cubic lattice, the distance between diagonally opposite points is sqrt(3). Not exactly rigorous, but intuitive to me.
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  202. 0:06 Just started and I already pressed like for the t-shirt
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  203. Woah, I was able to guess the shortcut rule! Do I get bragging rights? 🙃
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  204. I the 1950's I was taught to check math problems by something the teacher called "casting out nines." I didnt know why it worked but was intrigued by it. 60 years later I stubble across the answer.
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  205. "Don't worry. Be happy. And let's leave the demons for later." - Life advice from Mathologer, 2019
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  206. So excited when the notification popped out!
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  207. 3:55 omg I just realized that’s a pumpkin pi.
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  208. I learned of Heron's formula when I was about 8 years old from a table of mathematical formulae in the back of a dictionary, but I was never taught it in school, and I spent most of my life being curious about how to derive it in an elegant and geometrical manner. Your video is fantastic, it has unlocked some remaining pieces of the puzzle I had not figured out.
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  209. Wow that last part of the proof was really nice :)
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  210. Some years ago, a company made a perfect mathematical paint as described. Unfortunately they couldn't get it to market because the paint leaked past the molecules of every container they put it in.
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  211. Worth mentioning quadrilateral as a special case - we have 90 and 180 ears to add (180 are essentially midpoints). When we add 90 degrees first we get Van Aubel's theorem as a special case. When we build 180 ears first (i.e. taking midpoints, giving us a parallelogram) we get Thebault's theorem (which also follows from Van Aubel).
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  212. Adding 1 = 1 at the top would be very satisfying. Length of 1. Justified with the sum if a single number = product of a single number.
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  213. Most memorable: that the harmonic series narrowly misses all integers by ever shrinking margins
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  214. I'm doing my part! 😄
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  215. Everyone: maths is boring :( Mathsloger : let me take care of it. ;) Btw your videos are very interesting and full of knowledge...... Love from india 🇮🇳❤❤
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  216. First challenge: The chessboard cannot always be tiled after removing 2 black & 2 green. Suppose we remove the two black squares adjacent to the upper-left corner and the two green squares adjacent to the lower-left corner. Then the left corner squares are no longer adjacent to any other squares, so the board cannot be tiled. EDIT: Second challenge: if m and n are odd, then ⌈m/2⌉ = (m+1)/2 and ⌈n/2⌉ = (n+1)/2. Now the j=(m+1)/2, k=(n+1)/2 term of the product is 4cos²(π/2) + 4cos²(π/2) = 0, so the product is 0. Moreover, if m is not odd, then 0 ≤ j < (m+1)/2 in all terms of the product, hence 0 ≤ jπ/(m+1) < π/2 in all terms, so cos²(jπ/(m+1)) > 0 in all terms, so each term of the product is nonzero. This means the formula gives a nonzero answer whenever m is even -- symmetrically, the answer is nonzero whenever n is even. Thus, the formula returns 0 if and only if m and n are both odd.
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  217. I thought this vid was going to be more than 30 minutes long, but it's still nice to have an occasional short video from time to time! :) As for the puzzles: 5³ + 6³ ≈ 7³ (just off by two which just so happens to be the cube root of the next number 8) The answer to the second puzzle is 2 since 10² + 11² + 12² = 13² + 14² = 365 (I knew it was 365 earlier since when mentioning this equation some people like to point out that there are about 365 days in a year!) Funnily enough the sequence in the third puzzle goes 5, 6, 6.893..., 7.80557... which seem to round up to 5, 6, 7, and 8. Now I'm curious to know if this rounding pattern continues or if it's just a coincidence . . .
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  218. I prefer Flammable Math's avoid positivity shirt -|x|
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  219. That's it, I'm stealing that. I'm no longer a Software Engineer, I'm an Algorithmologist.
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  220. quick answer: a lot
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  221. Merry Christmas! I think the easiest way to show pi = lim N(r) / r^2 is to say that the squares centered in grid points <= r units away from (0; 0) must cover a circle with radius r - 1 and center (0; 0) but are contained in a circle with radius r + 1 and center (0; 0). Which gives us pi * (r + 1)^2 / r^2 >= N(r) / r^2 >= pi * (r - 1)^2 / r^2 and since both leftmost and rightmost side of this inequality approach pi as r->inf, we get what we wanted
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  222. There’s a reason that the all-time mathematical greats like Euler, Ramanujan, Laplace, and Fermat were fascinated by magic squares and other patterns! It’s the knack and habit of recognizing those underlying structures which led them to some of the greatest insights and advances in mathematical history. THIS is why it’s so important to teach our kids more than rote memorization of numbers, facts, tables, and theories; we need to teach them how to see them as patterns which lead to other patterns within even further patterns. People with the gift of innate intuition about patterns are the people who change the world.
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  223. Mathologer: explains rules about two colours added -> the third one me: ok, addition mod 3, easy Mathologer: two same colours -> that colour me: hmm me: something something group theory??? Mathologer: -(a+b) mod 3 me: takes maths degree certificate off wall
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  224. I had my worst two weeks lately, and this 30ish minutes was the first I was feeling joyful. Thank you!
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  225. On the rational main-angles in a goniometer. "At some point I'll do a whole Mathologer Video in German. Promised."
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  226. Glasses: +50 iq Nerdy math t-shirt: +80 iq German accent: +6000 iq
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  227. One of the coolest parts of the Gleich paper is that it leads very naturally to the question of how to express normal powers in terms of falling powers; e.g., n^3 = 1 + 7n + 6n(n-1) + n(n-1)(n-2). Equivalently, is there a pattern in the first numbers of the rows of the difference scheme for f(n) = n^k? Go read the paper in the description for the answer!
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  228. A day with a new Mathologer video is a better day
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  229. The disaster has been averted. Glad you’re back!
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  230. What amazes me is just how accessible the Mathologer videos are. I would rate myself as 'clever' up through High School mathematics (pre-calculus), but also a bit lazy and underperforming. I am not a 'math guy.' I remember the general idea of trig but only the most basic identities, a handful of geometric proofs, some of the definitions in calculus but none of the methods or shortcuts. And yet, excepting a very few, I almost always manage to watch these explanations and follow along - delighted and inspired the whole way through. The man is truly a very gifted teacher and a master of this "you-tube" way of presentation. Thank you, Dr. Polster, for leading us on this wonderful journey.
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  231. I really appreciate the "insane" videos. A lot of math content on YouTube is watered down or repeated, so it's nice to see this stuff that I've never heard of.
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  232. What a weekend! 3b1b + mythologer in the same weekend!
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  233. I have never seen any of these proofs - I graduated HS in '56 - got a BSEE in '60 - worked in Aerospace 38 years - retired and became a HS math teacher in 2001 - retired in 2014. Never offered any of these proofs to my students. I think they (and I) were shortchanged.  Love this stuff.
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  234. lmfao I tried to google "Niche method". Thinking this was some mathematician Then after seeing results for Nike, realized the joke. "Just Do It" facepalm wow I'm slower than I thought
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  235. Decades upon decades of maths: agonal spasm over sheets of paper Mathologer: so basically we use Pascal's triangle here and that's it
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  236. This looks highly related to the discrete derivative and discrete integral which is cool stuff..
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  237. I love your Ramanujan inspired TAXI 1729 T-shirt
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  238. Hell ... at around 23:50 I clearly understood (several times) the term 'the center of mess' until I realized that this has absolutely no mathematical meaning. But in that moment, it made perfect sense to me ... the point clearly was in the center of all that mess.
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  239. I find really wholesome this man's dedication to speak and explain so passionly for 50+ minutes straight. As a phisicist that I am, I love how matematicians like this one continiously inspire us all everytime they can. Keep on the good work, stay amazed and happy holidays!
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  240. German mathematician: "Here's another kitten, in a cube. Very cute. Feeling revived?" Quantum mechanics students: "NO ERWIN, PLEASE, NOT AGAIN!"
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  241. I've been tying the "Ian's knot" ever since 5th grade (roughly 11 years ago) and only today I just found out where the name came from.
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  242. "You're here for the mysterious iris in the thumbnail, aren't you?" Sorry, but no. I'm here because it's another Mathologer video :)
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  243. For the bug problem: An equivalent question would be to ask how long it takes for any one ant to reach the ant it is chasing. The chased ant always moves orthogonally to the chaser, so it is neither going towards it nor away from it. Thus the distance between the two ants simply decreases by the speed of the chaser. It thus takes 1 unit of time for the ants to meet.
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  244. I already know Heron's formula and Brahmagupta's formulas but I haven't seen a proof for them until now, also that general formula for the area of a quadrilateral is beautiful, thanks
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  245. Why isn't this in my subscription feed... I had to search up the channel name to find it
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  246. "If you're a 5 dimensional creature, that's a no brainer... Just start by visualizing 5D shells of 5D cubes" 🤯👽🤓
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  247. Wow. Such an amazing and carefully constructed video. Also, the important warning about ‘changing the rules’ shows (to me in my opinion) how seriously and solemnly the ethic of education is handled on this channel. You have a very satisfied and excited new subscriber!
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  248. As an American born-and-raised who was in the public system as both student and teacher… our math education is disgustingly deficient in number theory. High school graduates (even some going into stem fields) do not even know the Euclidean algorithm. They have almost no experience working with modular arithmetic. Too many decades of parents complaining about this “useless” math subject has led to them and their children being mystified by the simplest of number theory diagrams. Thank you mathologer for making so much explanatory content paced for victims of the US public school number theory book banning. (Inb4 some other American tells a story about their one teacher that taught them number theory)
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  249. I searched for Mathologer the other day and noticed he had missed some months of uploading, watched some videos, only to see another upload today. Lovely!
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  250. This seems very similar to second order automata - each cell's value is determined by the three cells over it, and the cell 2 over it.
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  251. A fun proof for the 4x4 case: make the colors prime numbers 2, 3, 5 and 7. The product of all numbers in the grid is (2*3*5*7)^4. Then divide by the numbers in the 2x2 squares between the corners (forming a cross that overcounts the middle). This gives 1. Since we divided by the middle 2x2 square one time too much, we can multiply by the middle square again to compensate, which gives 2*3*5*7. The number we are left with is the product of the numbers in the 4x4 grid, when you take away the middle cross, i.e. the corners. Since we got 2*3*5*7 for that number, the corners must be 2, 3, 5 and 7.
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  252. One ring to rule them all
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  253. The manuscript in the end was really enlightening. It makes you wonder how much past (and possibly present) mathematicians were (/are) influenced in their thinking by the writing conventions of their time.
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  254. Ramanujan has no formal education,he taught himself and made himself a genius
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  255. Pascal's triangle is inaccurately named as it only has two sides!
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  256. I believe a person in the Swiss Patent Office published a number of papers in 1905 that garnered some attention and may have elbowed out other interesting papers.
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  257. I said the first pattern should continue with 31. I didn't expect you to add the "evenly spaced" criterion.
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  258. It's neat how this feels like a sort of "hacky" way of approaching calculus, if that makes sense. Colleges should teach this imo. Just from this 40 min video, I could see some of these concepts plugging into all sorts of annoying classwork problems we've done in calc and differentials
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  259. There is another sequence that goes 2, 4, 8, 20, infinity: the number of faces of a regular polyhedron made of triangles. The first one is a degenerate case with 2 triangles sharing all vertices, then tetrahedron, octahedron, icosahedron, and finally an infinite tiling when you attempt to meet 6 triangles at each vertex.
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  260. I really love how there's subtitles for every video since I'm still learning English Thanks for the great content
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  261. Great video <3
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  262. They should have called it the 'Barely Divergent Series'
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  263. "A conspicuously simple and universal pattern is more likely a feature of the observer's perspective than the universe being observed." ... seems a more profound lesson than anything one could wring from an obsession over Tesla circles. Thanks!
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  264. Im a math failure. I'm here for your t-shirts 😅
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  265. The power of 2 coins can make any amount in exactly one way! (binary representations :))
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  266. Earlier I wondered: "When is mathologer posting his next video?" Couldn't have gotten a better answer :)
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  267. Pro tip: if you are manipulating equations without regard for units, you are doing mathematics. If instead you DO consider units, you are probably doing physics, engineering, or some other useful thing ;-)
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  268. Thank God someone's finally talking about these things! I stumbled upon one of these Tesla 3 6 9 videos ages ago, and I knew that everything they said definitely wasn't magic or anything and probably had a simple explanation in number theory, but I could never put it into words myself. I'm so glad someone finally put together an understandable and informative response to those things.
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  269. The weirdest thing you showed is definitely the unusual optimal brick stacking pattern.
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  270. I am from Austria and we never learned that the number, which remains actually is the remainder (9:45). When I learned about modular arithmetic in math Olympiad, I guessed that fact to be true while doing an example. Not even my highly invested teacher was sure, whether the solution was right. Infuriating, that you do not learn these deeper truths about mathematics at school.
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  271. This is very much nit-picking, but clickbaity tiles like "Why was X lost/forgotten/ignored/missed for Y years?" does the channel a disservice. Like question in headlines, this makes it look like poor journalism (despite it being the opposite). Furthermore, the interesting thing that the video focusses on is the maths itself, not the sociology around it, which makes the title somewhat inaccurate. Again, not a major issue, but such great educational content deserves better than a gutter-level, click bait title - no matter what a YouTube brand consultant might say.
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  272. "Mathematicians, artists, wizards, and a lot of crazies." So... just crazies, then?
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  273. My dad was a property assessor, and had many well made straight and circular slide rules. My grandpa was a brick mason, but never used one. I showed him how they work, and he thought it was interesting. It was in the mid-70's. I then showed my grandpa my six-place logarithm table. It's an Austrian book, from 1924 by Dr. Ludwig Schrön. The logarithms are to six figures, and the column/row lookup goes to 5 (but you can use proportions between answers to get a sixth). I showed him how to turn multiplication into addition, and division into subtraction, and how to back-reference to find the answer. He was quite amazed by the whole process. I gave him my table for a while as he was quite fascinated and spent time using it on many different occasions.
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  274. The number of math-related shirts Burkard owns, is somewhere between 12 and Graham's Number. Burkard, do you ever throw out old shirts?
    110
  275. Don't let other YouTubers doing the same topics stop you. Please! I like to see different takes on the same topic from my favorite YouTubers.
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  276. I liked the color proof better, because the swivel proof seemed like it relied more on visuals, which could be deceiving.
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  277. The fact that amyone on earth can connect to the web and have their mind expanded in this manner, for free, is wonderful. I am very grateful for this and other maths content on Youtube. Thank you!
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  278. This channel and 3Blue1Brown have fantastic visuals that are very helpful. Both also have excellent, clear presentation. No other mathematics channel I've found come close to your skill in communication, which obviously involves a great deal of work to create all the graphis.
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  279. Hey, Mathologer is back after a short interruption of service. Nice to see you again!
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  280. Really cool that you mentioned Think Twice. He's such an underrated youtuber, you really, really should check him out!
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  281. I have to say (and I am hard to impress ) this is one of the best high level explanations I have seen. The amount of work involved in putting this to air both at the animation level and the level of explanation are truly top drawer. But as at least one other viewer has pointed out, don't burn yourself out on this stuff at the expense of your own research.
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  282. That image of Madhava seriously looks like you. I could not believe it wasn't an AI generated image of you from 15th century India.
    107
  283. My answer to the question on why people focus on calculus is because calculus is indeed a simple solution and thinking with calculus is way helpful for cases where the "neat" solutions don't work
    107
  284. Most memorable: The harmonic series misses all integers up to infinity
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  285. I never never ever thought that I might see as strong visual proof for this amazing formula as this guy's👏🏻👏🏻👏🏻 this is so 100% enough
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  286. Continued fractions are truly amazing and, for most people, mysterious mathematical objects. This is really a pity, because just as the natural base for logarithms is e rather than 10, and the natural measure of an angle is radians rather than degrees, the most natural representation of a real number, in a sense, is a continued fraction.
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  287. Another fun but elementary observation: the sines of the "nice" angles 0⁰,30⁰,45⁰,60⁰,90⁰ are √0/4, √1/4, √2/4, √3/4, √4/4. I could never remember what their values were until I noticed that!
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  288. Archimedes video is in the pipeline :)
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  289. Finally, a mathematical proof that devils and angels are equally evil
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  290. The most memorable part for me might be the idea of that gamma value: especially the super quick visual proof that it had to be less than one by sliding over all the blue regions to the left
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  291. I liked the binnary solution because not only is it "elegant", it also shows intuitively why the number always gets smalller.
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  292. I think it's easy to get distracted by the fact that there is a matching negative for every positive term in the sequence. A similar paradox makes it more intuitive what's wrong with rearranging terms. ∞ = 1 + 1 + 1 +... ∞ = (2 - 1) + (2 - 1) + (2 - 1) +... ∞ = (1 + 1 - 1) + (1 + 1 - 1) +... ∞ = 1 + 1 - 1 + 1 + 1 - 1 +... Then we can pull out positive and negative terms. 1 + 1 + 1... - 1 - 1 - 1... So every +1 is canceled by a - 1. You can even create a mapping from the nth positive 1 to the n*2 negative term, so every positive term has a negative to cancel it. This, to me, intuitively shows why you can't add infinite sums by rearranging terms. You need to look at how it grows as you add terms.
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  293. Math: '*exists* Euler: "First!"
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  294. How odd...
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  295. For me, the "no integers" part was the most memorable, but honestly the whole video was of great quality (as expected).
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  296. Apparently the names of the hyperbolic functions are pronounced differently in different places! I had no idea. Perhaps this is a US vs. UK thing? In my schooling in the US, I have never heard anything other than COSH for cosh(x) and SINCH (homophonous with cinch) for sinh(x). The other functions I've heard more variation. For tanh(x) some people say TANJ like 'tangerine' and some TANCH. The most difficult one by far to incorporate, I think, is csch(x) wherein you can try to say COSEECH or instead bail out to '1 over SINCH.'
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  297. Solution to the problem at 16:14 Start with this equation, which is true for every value of k: 5^k * 2^k = 10^k The digital root of any power of 10 is 1, so DR(5^k * 2^k) = 1 Using the multiplication rule you explained earlier, DR(DR(5^k) * DR(2^k)) = 1 In other words, DR(5^k) and DR(2^k) have to be multiplicative inverses of each other. Taking the “digital root” of an integer is equivalent to modding it by 9. (The only difference is that if DR(n) = 9, then n mod 9 = 0.) In mod-9 arithmetic, every number except for 0, 3, and 6 has a unique multiplicative inverse. Since the digital root of a power of 2 is never 3, 6, or 9, this means that DR(2^k) completely determines DR(5^k). As k increases, the value of DR(2^k) cycles as follows: 2 4 8 7 5 1 2 4 8 7 5 1 … Taking the multiplicative inverse of each number above gives the values of DR(5^k). 5 7 8 4 2 1 5 7 8 4 2 1 … So DR(2^k) and DR(5^k) cycle through the same values, but in reverse.
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  298. I liked your evil mathematician back story, with the teacher refusing to grade the "wrong" proof.
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  299. This is a great length of a Mathologer video, nothing wrong with this! Thanks
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  300. Most memorable: being invited to take a moment and post why it might be obvious that gamma is greater than 0.5 and then doing it. Hmm... why is it obvious that gamma is greater than 0.5? Well it didn't seem obvious... But imagine the blue bits were triangular; then there would be equal parts blue and white in the unit square on the left i.e. a gamma of 0.5. But the blue parts are convex, they each take up more than half of their rectangles and together take up more than half of the square.
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  301. Yet another awesome topic that has me saying "how is this the first time I'm hearing about this?"! The mystery of why the number walls will always have integer entries would've been enough to hold my attention, let alone all the other cool properties! Also, fun fact: The characteristic polynomial is the denominator of the generating function of the sequence! The numerator will be a polynomial of degree less than the denominator, determined by the initial terms of the sequence.
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  302. The recursive formula for the squares translates to: x^2 = 3(x-1)^2 - 3(x-2)^2 + (x-3)^2 Expanding the right hand side gives us: (3x^2 - 6x + 3) - (3x^2 - 12x + 12) + (x^2 - 6x + 9) = 3x^2 - 3x^2 + x^2 - 6x +12x - 6x + 3 - 12 + 9 = x^2 The formula works :) The coefficients of the formulas look like the binomial coefficients, they are matching with the rows in pascals triangle. You can generate the formla by expanding the right side of the equation 0 = (x - 1)^n and replacing every x^k by the k-th coefficient c_k. Then solve for the highest power (x^n) and you have the formula for the sums of x^(n-1). Notice, the formula for the squares uses the coefficients of the 3rd row, expanding (x-1)^3 not (x-1)^2.
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  303. Somehow, my first reaction to the equilateral triangle version was "Huh? Surely there can't be any?"
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  304. I just watched the first 5 minutes and have to really compliment the way you present you material. It's inspiring how you structure it in a way that makes it engaging. The "tricking" shows how important it is to really check what's going and that's what math is all about :)
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  305. This proof is so beautiful that I wrote an entire essay about numbers as the sum of two squares. When the essay was "finished" (I admit that it wasn't), I sent it to the main competition for this type of math essays in the Netherlands, and it got third place. Also, because I heavily studied the subject in my spare time and Olympiad training, I got really good at this type of number theory. When I participated at the IMO in Oslo this year (second time), I solved question 3 with full points, which was about this type of number theory. I got a perfect score on the first day, and scored 7+5+4=16 points on the second day, for a total of 37 points! GOLD! 19th place worldwide! Relative best for my country ever! I really don't know if I would have gotten this score without this proof, so thank you so much for making this video. I hope that you are going to inspire lots of other people as well!
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  306. In 2013, I discovered this Gregory-Newton formula myself using insight from Pascal's triangle. I found the formula while trying to solve an arithmetic sequences and sums problem with 4th difference for a private high school student I taught. I had no idea how to answer the question using the standard arithmetic sequence and sum formulas taught in schools. I struggled with that one question for more than an hour before I found an insight in Pascal's triangle and then came up with this helpful formula. I tested the formula with couples of made up cases to make sure the formula works and indeed it worked! I then told my student to use this formula to solve that particular question. The next day, my student told me the answer was correct, but his teacher didn't give him full mark because he didn't use the standard formula. 💔 I didn't know the name of the formula I had discovered until today. But at that time, I was sure someone else should have found the formula. Imagine the humiliation I would get if I had named that formula by my name. 😅 Thank you for this video, sir. This brings back all the memories of that moment. 🙏🏼
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  307. Ramanujan is my favorite. He saw things in a way nobody else ever has. Maybe someday we will have another like him. I love his work it seems so natural yet is so deep.
    97
  308. We dont celebrate Christmas in my country but I celebrate any Mathologer video anytime
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  309. For those who are interested, contributions of Madhava's school are covered in the book titled, "A Passage to Infinity" by George Gheverghese Joseph. Another book, "Mathematics in India", by P.P. Divakaran covers 5000 yrs of history of mathematics in India starting from the Indus valley civilization. P.P. Divakaran also states that Madhava was a pioneer in the methods of calculus much earlier than it development by Newton and Leibniz.
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  310. Nuclear physicists play dominoes because they like starting chain reactions, according to my own interpretation of game theory.
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  311. The regular polygons you can fit inside a 3D grid are also the regular polygons you can use to tile a surface. Coincidence?
    95
  312. Imagine not getting a heart and reply from Mathologer.
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  313. He is German, more precisely from somewhere in Unterfranken. He said so in one of his videos. I think it was close to Würzburg or Würzburg itself.
    94
  314. If I'd lived in Classical Europe, I'd totally have worshiped geometry as sacred! Great video!
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  315. I am just amazed by how Ramanujan's mind worked.
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  316. Euler is quickly rising in the ranks of "potentially the smartest human ever lived" in my estimation.
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  317. "We started with an infinite sum so let's count our blessings" Well it does seem like there are countably many...
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  318. I've been fascinated with Fibonacci numbers and Pythagorean triples since I discovered them when I was about 8. 45 years later you taught me some new things and helped me understand the "why" behind some of what I already knew. Thank you.
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  319. I’d like to add to your two criteria for the “best” lacings: shortest and strongest, please. As I am a sneakerhead and lacing is an intrinsic aspect of one’s “stylishness”, that is a top criterion, though quantifying it might be a matter of subjectivity! Thank you for your studies and presentation 😄👏🏼👌🏼
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  320. 10:50 this is what bond villains see before they die
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  321. 10:12 just out of curiosity, how do we differentiate between higher dimension convex and concave polyhedra??
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  322. I really loved the binary representation, though my gut instinct would have been to invert the bits (starting state is 0). The proof would have still worked since we have a maximum upper bound, but having it descending is easier :P The other reason that I like the binary proof is that you don't have to know what happens in that right-most edge case-- even if you can't flip it, the proof still works. It doesn't reach 0, but it can't get smaller, therefore no moves can be made and therefore it terminates
    88
  323. The most unbelievable and beyond reason thing about this channel is that it does not have (yet) 1M subscribers! All the other assertions are left as an exercise for the reader...
    88
  324. The beauty of mathematics lies in the way how seemingly unrelated threads interweave to create the fabric of utmost mathematical elegance. And Mathologer .......you do a great job untangling those threads and making us see and appreciate the beautiful connections lying underneath. Thank you so much.
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  325. The best complex logics/math film I have ever seen. By “complex” I mean “consisting of many, sometimes, non-trivial elements”. If I confess I am awarded Best University Lecturer for many years, it is only to pay tribute to the quality of this film – to keep things so ordered and clear is SIMPLY AMAZING! I do appreciate the apologies for not explaining why complex numbers needed to be introduced (but no fully explained) when analytical functions were being talked about. It gives a lot of security to a lay listener that all vital things were introduced even if no all were fully developed. Yes, the content still can be completely wrong (I am not an expert to judge) but it is certainly “CONSISTENT and COMPLETE” – in contrast to the film it was commenting. The detailed and well paced debate with the statements of Numberphile content were excellent. Well, it was really impressive. I do not subscribe to any channels and social media but believe me, I will be watching you regularly!!! Well done (you know it 😊).
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  326. Most memorable: If my life depended on knowing if the sum of no nines series is finite I would not be alive
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  327. The Professor's workroom looks like a boy's playroom, and vice versa; an anti-shapeshifting space, home to a brilliantly playful and playfully brilliant mind.
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  328. I started playing with Meccano gears when I was 7 and rapidly discovered the anomalies that seem to happen when one gear rotates around another. It really bugged me til I got it figured out. Later, during my engineering degree, I couldn't understand why the lecturer made such a meal of it all, with complicated formulas, when to me it was (by now) all so obvious but then he lacked your clear insight and pretty animations. Many years ago Meccano issued a striking clock kit. A striking mechanism requires a wheel that goes round once in 12 hours in 1+2+3...+12=78 steps, one for each hour strike. Awkwardly 78=6*13. They had to use a 12:1 geared ratio with 13 given by mounting the 12:1 gears to go round with the big wheel epicyclically. For more puzzlement tie a loop of string into a trefoil knot and drop it over a post and discover that the string only goes around the post twice despite the three loops. Or count the number of times on a clock, the minute hand crosses the hour hand in 12 hours.
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  329. 33:14 Marty does not like the Fibonacci sequence, and neither does Matt Parker. From this, we can deduce that having a first name that starts with "Ma" and contains a "t" predisposes you to disliking the Fibonacci sequence. Mathologer doesn't count because, presumably, that's not his actual first name.
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  330. Most memorable part: derivation of γ. As in high school we learn about the approximation of the area under the 1/x curve but not many actually focus on the 'negligible part of the area' which in fact adds up to something trivial to the whole field of number series. 24:23 Sinple proof for γ>0.5: All the tiny little bits of that blue areas are a curved shape. By connecting the two ends of that curve line we can see each part is made up of a triangle and a curved shape. The total area of those infinitely many triangles equals to 0.5 so the total area of the blue sharpest be greater than 0.5.
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  331. 18:28 mentions the general rule of filling odd magic square we were taught in China in primary school. The rules was written as so: 一居上行正中央，依次斜填切莫忘 上出框时向下放，右出框时向左放 排重便在下格填，右上排重一个样 Translation: 1. put 1 in the middle of the first row 2. fill next consecutive numbers diagonally 3. when going out on the top, put into the bottom row 4. when going out on the right, put into the left column 5. if the square is occupied, put into the square below 6. if going diagonally on the top right, the same rules applies
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  332. That final example of constructing a lantern from a soda can was awesome.
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  333. Maybe you could do a follow-up video about the Risch algorithm to find anti-derivatives for elementary functions. Would be interesting to get an understanding how that works in principle.
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  334. Nobody who watches Mathologer videos is put off by them being long. Just ask 3 Blue 1 Brown.
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  335. For the second puzzle, my answer is 2 I already learned from the video that 10² + 11² + 12² = 13² + 14² so the numerator is just 2(13² + 14²). When (slowly) performing the addition in my head, I was surprised that 13² + 14² actually equals to 365, cancelling out the denominator and leaving 2. Knowing the result, it's fun to think about making an efficient one-page calendar where the front is a 13x13 square and the back is a 14x14 square, with each square containing a date :D
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  336. If you make a PDF I will buy it, and probably I will not be the only one. This really associates the formal math and the intuition in a striking way. Well designed, brilliant accomplishment.
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  337. there is no such thing as a coincidence
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  338. It is interesting to me that people find the 'infinite surface area and finite volume' combination to be so paradoxical. People seem much more willing to accept/reconcile the existence of 'infinite length/perimeter and finite area'.
    83
  339. Slide rules are not so out of fashion like most think. Several weeks before i saw a technician in a test center uses a slide rule on calculate some chemical rations to mix together. Asked her on it and she explained that its safer. The slide rule (made out of steel) you can throw into the autoclav for sterilization but calculator or phone not. And no battery can be empty. Sterilization chemical are very strong and calculator and phone display surfaces are destroyed if use them. Slide rule is the most esiest and safe option. But its expensive,. one cheap calculator maybe is 15 euro and a slide rule of medical steel about 250 euro. And you need many cause they go 8 hours into the autoclav after each use. Many test places buy cheap calculators and burn them. It seem cheaper but its not and waste resources.
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  340. Ramanujan was a freaking force. What a beast!
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  341. ​ @Mathologer  I wouldn't be surprised if it was already published somewhere, but I haven't been able to find it anywhere. I was working on some problems involving modular forms and I tried differentiating the theta function identity θ(-1/τ)=√(τ/i)*θ(τ). That gave a similar identity for the power series Σk^2 e^(πik^2τ). It turned out that setting τ=i allowed one to find the exact value of that sum.
    82
  342. I watched this a few months ago and understood very little, now after a semester of calculus I can really appreciate the beauty of it. Great video Mathologer!
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  343. For the demonstration of A(X) + A(Y) = A(XY) Definitions : - we define A’(x,y) as the area under the curve between x and y (we notice that A’(1,y) = A(y) and that A’(a,b) + A’(b,c) = A’(a,c)) - we define f_p(z) the transformation of an area z by a factor p Let’s find an expression for f_p: Given z=A’(a,b), the top left hand corner of the area is mapped to: 1/p * 1/a = 1/(ap) (squeezing) which corresponds to the inverse function for a value ap (shifting). Same goes for the top right hand corner. We can conclude that: f_p(A’(a,b)) = A’(pa,pb) = A’(a,b) Let now prove the theorem : A(x) + A(y) = A’(1,x) + A’(1,y) = A’(1,x) + f_x(A’(1,y)) = A’(1,x) + A’(x,xy) = A’(1,xy) = A(xy)
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  344. Dear Mathologer. I can see how much work goes into these video's, but please never stop doing them. When I was a child I remember asking myself why we didn't have a TV channel that just showed educational programs. You (and a handful of others) make youtube what it can and should be. Thank you.
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  345. ​ @Mathologer  It really is sad, colonialism extends to every aspect of society, even its perception of something as objective as science. You would mostly only hear Indians trying to show India's scientific achievements by misrepresenting them, if not straight up making things up, and its not because they are so enthusiastic about science and its history, but because they are insecure about the perception of India in the world, which ultimately is a result of colonialism. Its really sad how conquest and colonialism affect scientific progess. India went from one of the few bright spots in the scientific world to a mediocre country after centuries of destruction. Not to mention the texts that were literally destroyed during invasions. Its scary to think how much knowledge humanity has lost just because of wars and plundering.
    81
  346. Fun fact: Brazil is probably the only country who calls the quadratic formula "Bhaskara's formula". Really don't know why, but is really unusual the use of the regular one
    81
  347. The sum of angles: Imagine a point with a nose walking along the edges of the star. In every corner, it will rotate 180°-a, where a is any angle of the star. If you carefully watch its nosr, it does three rotations, so 360°×3 or 180°×6. We get (180°-a)×7=180°×6, so the sum of angles is equal to 7a, which from the equation is 180°
    81
  348. And as always the first person who prove ... Leonhard euler
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  349. Made it all the way to the end wondering what Steve Mould would think of this, he likes water level stuff!
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  350. As an Indian and a Keralite I'm ashamed of not knowing anything about this great Mathematician. Thanks a lot sir for enlightening us with this information. My country had to deal with lot of violence in the past for centuries. But I feel if we can move past them and try to rebuild the pieces to understand what was lost would bring us more peace. And what you're doing goes along that way. Salute.
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  351. I always liked to describe differentiation as just a bunch of rules you have to apply and it's usually straight forward how to do it. Integration on the other hand consists of either knowing the answer or trying to manipulate the function until you do.... with the optional third step of giving up and looking it up on a table. Also I like that the music got way more epic as soon as you got to the chain rule.
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  352. Instant classic. This just feels like the perfect “Mathologer-y” topic! It also had the possibility for some great animations, which I’m so happy you delivered on flawlessly. I’ll probably watch this a few times throughout the future to enjoy it again!
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  353. I'm homeschooling my children, and interestingly I just taught them Heron's formula yesterday as an example of where square roots can be very practical in real life problems. Specifically, we talked about how convenient it can be if you are trying to determine the size of a plot of land you want to buy when all you can measure are distances and no idea if the land is square. Just walk the perimeter, walk the diagonal, apply Heron's formula to the 2 triangles, and you know immediately how big the plot is. For 7th graders who don't yet know trigonometry, the example worked beautifully. It should absolutely be taught in schools. It is a crime that it is not considered essential these days.
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  354. I think the shirt in this might be one of my top five favorite shirts I've seen him wear so far. Impossible triangle made of rubik's cubes, perfect hexagram in the middle, and it almost looks 3D. This shirt is a winner.
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  355. Some carnival guy right now: scribbling furiously
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  356. Given that you can do the operation on the final card without having to turn a card to the right (since it doesn't exist). The largest possible number of turns to terminate the sequence is equal to [n*(n+1)]/2. edit: If you cannot perform the operation on the last card then the most turns possible to terminate the sequence is given by x=(n-1)*n/2 though the sequence may then terminate with one card left.
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  357. Another gem from Mathologer. It's because of Mathematicians like you out there, Maths is still beautiful and elegant.
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  358. Haha, was just wondering why you didn't cover partitions and then seen this. Very interesting and an intriguing topic with the contributions of several important people like Euler and of course Ramanujan .
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  359. I once performed a version of the trick at 26:18 where the AUDIENCE got to choose the card to keep, and the only choice the magicians had was the arrangements of the 4 cards. Of course that’s normally be impossible, so we cheated a little and used an extra bit, a right-hand vs left-hand choice. It goes pretty smoothly if you’re quick at mental math, and the audience never suspects a thing!
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  360. Nice :)
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  361. Once again I am rewarded for my procrastination: I never saw the original. "Hard work often pays off after time, but laziness always pays off now".
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  362. all values that have paradoxical meaning are just new planes to fly our imaginary numbers on.
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  363. I don‘t recall seeing any of these proofs in school. Most relevant formulas were simply introduced to us without ever being proven before college.
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  364. Some of us will recognise Donald Knuth as the creator of the typesetting language TeX. For that alone, he has been responsible for the professional quality appearance of thousands of PhD theses.
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  365. He might not be your typical name, which you could give and everyone would recognize it, such as its with Newton. Despite that however, Ramanujan is a genius, who sadly didn't live for long and would probably be one of the most important mathematicians, should he have lived and published more papers, perhaps even be an advisor to some future mathematicians :)
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  366. At 28:42, your generalization is wrong- it should be that cbrt(ABC)<=(A+B+C)/3 rather than that sqrt(ABC)<=(A+B+C)/2. As stated, (3,3,3) (for example) breaks your generalization. At 35:50, it’s because the bug that each bug is walking towards is always moving orthogonally to the first bug’s path, never getting closer or farther outside of the first bug’s motion. So it has to go exactly the same distance as it was at the beginning. At 36:30, it’s 1/5. You can rearrange the square pieces to get five squares of equal area, summing to area 1.
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  367. I studied the infinite generating function for high school math competitions. They are very useful in combinatorics problem!
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  368. The cicadas :) Nice and crisp video as always. And big thanks to Andrew!
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  369. Yet again, you manage in 30 mins to better explain something than my maths teachers could over a year. Bravo, sir!
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  370. 1:36 "circle gets turned inside out" *"smiles and frowns" flashbacks intensify*
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  371. "...The Euclidean Algorithem, an ancient mathematical superweapon" perfect description!
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  372. Let's all just forget about yesterday shall we? Great to see the videos are back and I really hope they work their situation out (IN PRIVATE)!
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  373. Red Cross solution: Align the centre of one of the smaller crosses to the centre the big cross, matching their orientation. Rotate the smaller cross until its vertices touch the edges of the big cross. The 4 segments produced form the 2nd smaller cross.
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  374. I remember watching 3B1B's video on Hanoi and feeling such awe, that a seemingly simple puzzle could weave recursion and binary counting together. I never thought anything could replicate that awe again! But damn you! Thank you soo much for making me experience that awe again, and this time much muchh more! I just can't wait to get on my computer and try to animate all the animations that you showed today (and perhaps even a generalised version for small n using Frame-Stewart algorithm?). But seriously lot's of love and respect man for making me fall in love with Maths over and over again!
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  375. I'd say the Axiom of Choice is true as long as you don't have an Electoral College :(
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  376. 9:45 12 million "and change"?? I'll take your change then, thank you!
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  377. 15:52 "There's no hidden trickery" I'm not complaining, but I'd say that counting faces is quite tricky, as it relies on topology. To see when we're removing a face and when we're not, is visually obvious and yet non-trivial. And also one needs to be careful not to disconnect the network (as you said, "starting from the outside" should guarantee this).
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  378. The most memorable proof is the original proof of the harmonic series' divergence simply for the fact that this probably the only proof I could present to my year 10 math class and most of them would understand it.
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  379. Nicolas Bourbaki will be furious. 😉
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  380. Ich würde wirklich gerne ein Mathologer Video auf deutsch sehen. Greetings to Australia! Great Video!
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  381. You had me going at the beginning. Because of the particular choice of original triangle, you briefly had me wondering whether the "folded" triangle might be (geometrically similar to) the mirror image of the original. But no, not in the general case.
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  382. I really love the graphical intuition added onto that one sentence proof. It makes it a lot clearer WHY that function is an involution and has exactly 1 fixed point. Also, you misspoke. At 28:54 you said "b squared" instead of "c squared." >.< Gotta be tough to get through that stuff without any mistakes. At least it's clear what you meant cause of the written equations.
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  383. So happy to see a new video. I was really upset when you didn't upload in February.
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  384. I am a simple man. I see mathologer upload Einstein, I click!
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  385. Engineers: Hold my Taylor Series
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  386. Mathologer’s Theorem: π is the sum of two squares. 21:19
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  387. As an Indian and a student of mathematics it is ridiculous that I came to know about Madhava and his discoverys through you sir, although I am familiar with that series but don't heard Madhava's namae before Thank you sir for this video .🙏🙏
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  388. Just like the Indian mathematician Madhava, there are other Indian mathematicians who did advanced mathematics way before western mathematicians
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  389. 21:15 "and therefore pi is a sum of two squares" 🤔 now that is some mathologer magic I missed in between the lines
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  390. One nit: the binary argument demonstrates that the sequence of moves must terminate, however, there is an additional step required to say that it terminates specifically in 00000 == all cards face up. Consider for example if one were in the state 00001, and we know every move decreases the value, but perhaps they decrease it by 10 or something. We'd overshoot 0 and be stuck at 00001. It's important to call out that as long as there is a 1 present, there always exists a legal move that does not overshoot 0
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  391. 8:30 You can still use the exponential function in the following way: Let y = ce^f(x) y' = f'(x)ce^f(x) Therefore f'(x) = x and f(x) = ½x² + const So in the case of y' = xy we have y = ce^½x²
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  392. What makes Quadratic Reciprocity so special, and in particular, why was it so important to Gauss? Why is squaring integers so important in Number Theory? To understand the answers to these questions, you need to appreciate the work Gauss did in the theory of quadratic forms and their application to differential geometry. Gauss's "fundamental theorem" of differential geometry is about how the 2nd order partial derivates of a function that defines a "surface", e.g. `f(x,y)=height`, depends on a specified quadratic form called the "metric tensor" of the surface (and its first derivates), which gives a rule for how to measure distances "on the surface". This shows a profound and deep insight into the nature of the relationship between a large number of otherwise seemingly unrelated abstract concepts. Squaring numbers plays a fundamental role in both Number Theory and Geometry. After watching this video twice, going on a few wild goose chases, and not being able to stop thinking about it: I finally think I understand why Gauss placed such a high degree of importance on this particular theorem, and feel like I have a lot of reading to do now!
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  393. The Auto Algebra Song™ makes me so happy.
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  394. Imagine having to use 360° because the formulas look weird with 2pi. If only there was a better constant to represent full circles...
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  395. log(1+2+3)=log(1)+log(2)+log(3)=log(1*2*3)
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  396. I think I'd honestly prefer the first proof, but I was too busy shouting at the screen about the second proof to enjoy it.
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  397. Wow, I had ventured into speedcubing back in 10th grade and I conjectured to my friend that no matter what repetitive steps you use you'll get back to solved state and we discussed about it at great length. This video has brought those memories back and how I never thought about it afterwards even after doing a course on group theory but now you have opened my third eye. Thank you!
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  398. I always get a 'WOW!' when I watch your videos. Thanks for bringing this stuff to YouTube.
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  399. Most memorable moment was the cat going "μ".
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  400. Sequence Calculus is just low resolution calculus.
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  401. Another video of Mathologising beauty. The 4D cube rotating in space was a delight.
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  402. for the 1/x is a anti-shapeshifter if you squish the graph by 2 along the y axis, then each value of y gets mapped to y/2, while values of x stay where they are. the function that describes this graph is f(x)=(1/x)/2 = 1/(2x) now let's stretch this graph by 2 along the x axis each value of x gets mapped to 2x while y doesn't change so f(2x) = 1/(2x) and so f(x)=1/x the function of this graph is 1/x; same as before
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  403. 8:24 The magic constant is sum of all numbers in the square divided by the number of rows. For those looking for a concise formula: it should be (n²(n²+1)/2)/n or simply n(n²+1)/2. Plugging in 33 will give us the magic constant of 17985. 18:48 Go up-right one space. If you exit the board, wrap around to the other side. If you run into a space already filled in, drop straight down one space instead.
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  404. Only 20 years too late! This is the calculus lesson I wish I had while I was an engineering student. Very well done!
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  405. love to see a video about Sophie Germain's work, she doesn't get nearly enough attention
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  406. Thanks! Excellent video as always. So interesting to learn that this type of math was being done in the 1400s.
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  407. Mathematicians doesn't believe in "coincidence", these are true mathematical Facts. 😌 3³+4³+5³=6³ but not possible for further case. Pls continue doing such 10-15 min videos also, Sir. 🙏🏼 Long vid are itself goldmine. ❣️❣️❣️
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  408. 15:19 Am I the only one who thinks it's funny how he dodges the triangle? By the way, it was a great video, very interesting!
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  409. i am but a simple man i see video about Ramanujan I click
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  410. The best thing about Reve's puzzle; "We're going to have some cheese."
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  411. I've been a fan ever since your humble beginning, and I'm more excited than ever for new Mathologer vids!
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  412. As long as you don't start to follow the YouTube "shorts" trend everything is fine;)
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  413. I thought I'd know all of this already, but was pleasantly surprised when you put the circle and hyperbola together and at the geometric proof of the e^α identity. Great video!
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  414. Your videos are really calming to the mind. Pleasant music during algebra autopilot and then fascinating math explained in a natural way!
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  415. 9:23 I'm a high-school maths teacher. I have taught that dumbed-down divisibility-by-9 test without going into digital-roots and the remainder property. I shall never do it again. Props to your deliciously made videos.
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  416. 3:20 there is a tiny mistake: after the four non coincident digits you have in one case10582 and in the other10581. The "2" and the "1" should be coloured as different digits. Amazing video, many thanks.
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  417. That sounds like it would be related to the amount of ways you can permute an infinite list, which sounds uncountable to me.
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  418. I would rather have expected the Spanish Inquisition than the 1800-year-old Talmudic bankruptcy problem. Professor Polster never ceases to amaze us.
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  419. "What comes next?" Me: Almighty Lagrange's interpolation
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  420. Clearly, the highlight of the Euler-Mascheroni constant is a splendid part of the video...the sum of no 9's animation is very impressive.
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  421. 16:00 I found the area of the tree to be 5, since we have a depth of 5 and for every iteration, the new squares sum up to 1, thus a a tree with infinite iterations has an infinite surface area (still counting overlapping surfaces)
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  422. as a teacher, I'd avoid using Blue because of how easy it is to confuse with side B of the triangle.
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  423. I feel like this sort of "discrete calculus" was something I was vaguely aware might exist, but I've never seen it formalised like this before.
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  424. I'm confused, what if I take a 5d hypercube, if I pick a random point won't the 5 points connected to it form a regular pentagon? Edit: Aha, these points are not even on the same plane.
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  425. Mathologer Nice “save” from a small mistake in the video.
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  426. You are damn right
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  427. So for that first puzzle... I think my mind just got blown I started with 5 since I was basing on the fact that the first equation of both patterns end and start with 3 (1 + 2 = 3 and 3² + 4² = 5²), so why not do the same for the cube pattern. What I ended up is that 5³ + 6³ = 341, which is almost close to 7³ = 343. I was curious enough, so I tried to expand this by utilizing the same trick from the previous patterns and used 6*(1 + 2) for the next pattern and got 16³ + 17³ + 18³ ≈ 19³ + 20³ (with the difference being 18) Expanding this pattern leaves me with a list of differences and... I don't know about you, but (2, 18, 72, 200, etc.) just screams "I have a pattern"... and it does! Each difference is just twice a triangular number *squared*, and knowing that just blew my goddamn mind... because THAT'S LITERALLY HOW I STARTED working on this pattern. For each sum that uses 6(1 + 2 + ... + n) cubed as a starting point, there is a difference of 2(1 + 2 + ... + n)². How cool is that!? In honor to document this amazing pattern, here's my christmas tree for the cube pattern, with the difference added in. You can think of them like they're ornaments or something xD (the formatting might only work with monitors) 5³ + 6³ = 7³ - 2 16³ + 17³ + 18³ = 19³ + 20³ - 18 33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - 72 56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - 200 85³ + 86³ + 87³ + 88³ + 89³ + 90³ = 91³ + 92³ + 93³ + 94³ + 95³ - 450 ... and here's another one with the sum expanded, just to see the beauty :D 5³ + [6(1)]³ = 7³ - 2(1)² 16³ + 17³ + [6(1 + 2)]³ = 19³ + 20³ - 2(1 + 2)² 33³ + 34³ + 35³ + [6(1 + 2 + 3)]³ = 37³ + 38³ + 39³ - 2(1 + 2 + 3)² 56³ + 57³ + 58³ + 59³ + [6(1 + 2 + 3 + 4)]³ = 61³ + 62³ + 63³ + 64³ - 2(1 + 2 + 3 + 4)² 85³ + 86³ + 87³ + 88³ + 89³ + [6(1 + 2 + 3 + 4 + 5)]³ = 91³ + 92³ + 93³ + 94³ + 95³ - 2(1 + 2 + 3 + 4 + 5)² ...
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  428. behold: new Amazon Prime service translating in 4k+1 resolution :)
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  429. "I actually never gave much thought to bankruptcy problems before this video, nor did I question the idea that proportional distribution is the best solution. But now, I actually see the Talmudic approach as superior in many real life situations." As an Orthodox Jew who has studied Talmud for years, both inside and outside of a yeshiva setting, and as someone obsessed with mathematics, I found this to be an incredibly touching statement. And, I'm incredibly flattered that you chose to make a video on this subject, and thrilled that you did it such justice and improved my understanding of my own religion! You rock, Mathologer!! ❤
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  430. 15:32, it does have a turning property but you must rotate around a cardinal axis. This does mean it is useless for finding other grid points though.
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  431. "Is the no 9 series finite? You life depends on this!" Me: suspicious, has to be finite! "Believe it or not, it is finite!" Me: YAY
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  432. It wouldn't be discrete if it was working in the foreground.
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  433. 11:54 “It’s always 2 pluses, followed by 2 minuses, followed by 2 pluses and so on” How do you just expect me to know where this sequence is going? Hahaha
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  434. Proof of the length of the wheel being ln(10) : When the number x reaches the wheel, the rubber band has stretched by a factor 1/x (we consider that the rubber band is numbered from 0 to 1 like in the video). If you now wind the band just a bit more so that you reach x+dx (with dx infinitesimal), then the length added on the wheel is dl=a(x)dx. So the total length of the wheel is the integral of dl for x=0.1 to 1 that is int(dx/x, x=0.1..1), which is equal to ln(10).
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  435. The off-by-one counter-intuitive logic reminded me of (and is clearly related to) how there are only 11 places on the face of a 12-hour clock where the hands point the same direction. One of my favourite weird facts is that if either one of the two hands on that clock went backwards (what a strange clock that would be!) the hands would point in the same direction in 13 places instead.
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  436.  @Mathologer  Rest assured that your efforts are greatly appreciated. This is one of the (if not the) best mathematics channels on youtube and the reason is that people recognize quality and hard work when they see it. Happy π-day!
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  437. This has to be a record - two minutes in and I'm already pausing the video to pick exploded bits of my mind out of the carpet. How did I never learn this before? In school, Heron's formula was presented as a side note, and I never really understood the derivation (meaning I'd have to look it up again any time I wanted to use it). Given a little bit of reflection it all seems crystal clear, now. Nice!
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  438. So the "2" in "x+2" came from the fact that an edge/vertex is defined by two points and a line/edge between them. Is there a mathematical entity where you would use "x+3" or any arbitrary "x+N"?
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  439. 1 - trivial 1 1 - still trivial 1 2 1 - the middle ones have two as their greatest common factor 1 3 3 1 - GCF = 3 1 4 6 4 1 - GCF = 2, NOT 4 1 5 10 10 5 1 - GCF = 5 1 6 15 20 15 6 1 - GCF = 1 1 7 21 35 35 21 7 1 - GCF = 7 1 8 28 56 70 56 28 8 1 - GCF = 2 So, when P is prime, doing Pascal mod P makes the triangles of size P^k + 1 special because the GCF of the middle numbers in the P'th row is P itself
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  440. Most memorable: the fact that gamma is the Ramanujan summation of the harmonic series.
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  441. My only regret is that I can never watch 19:57-23:03 for the first time again.
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  442. Euler's formula for polyhedra can easily reach #1 if you realise it's actually d0-d1+d2-d3+d4...dn=1 where di is the number of i-dimensional objects that form an n-dimensional polyhedron
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  443. After checking a feel cases, I think that to get the tower from A to B, you start by moving the smallest disc to: • C, if there are 2n discs; • B, if there are 2n+1 discs.
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  444. Your video is a treasure. I hope it is incorporated in early STEM learning. My experience watching, caused me dynamic insights in three separate domains. In my opinion, Mathologer is a "miracle" math educator.
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  445. I watched this video 5 times and I have to watch 10 times more in order to really get it, but I just wanted to tell you how much I love your videos and the way you teach. Your are exceptional!
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  446. In India we have a whole chapter by the name Herons Formula . After that I decided to derive my very own formula for finding area of a triangle, but I accidentally deriverd Herons Formula. I was so Happy at that time. Those were the days. 🙂🙂
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  447. Doing seperation of variables in Diff Eqs was the first time it really hit me how nice Leibniz notation is. It's really just the chain rule, but still, super nice.
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  448. One of Mathologer best videos. The great graphic design, The clear explanation, The general idea, all are simply perfect!
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  449. Didn't learn about Ptolemy's theorem until my late 20's, well after having learned calculus and other pretty sophisticated algebraic properties up through grade school and all. Should have been the first thing I learned.
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  450. That example at the end, where you are owed 1 dollar and the bank is owed a million, where in a "logical" proportional system you would get a millionth of a dollar (ie. nothing) and the bank gets everything, while in this system you get your dollar and the bank gets the rest, really clarified why this actually sounds a more fair system than the proportional system, even though the latter sounds a lot more logical at first.
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  451. How on earth could you resist to check what happens with the sequence of primes??? Thanks for the video.
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  452. This the the only strand-type math problem
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  453. I have a bachelor’s degree in mathematics. I was able to follow everything except for the formula at 40:45. Im sure that requires more thought/explanation but overall, great video.
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  454. I wonder if Neil Sloane might be interested in including this Moessner-Transform and reverse transform into his database of integer sequences.
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  455. The visualization for finding the base (14:30) to (16:30) is marvelous :D
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  456. I was watching one of your videos about infinite fractions and... WOW, NEW VIDEO, THAAAANKS
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  457. I feel like a similar version of this problem is in NCERT Ex. 5.4 (Optional) Class X Mathematics. Who else have tried that problem using AP?
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  458. When I saw the 1+1/2 case for figuring out the base (and mentally extrapolated the rest of the proof), I was shocked- such beauty!
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  459. For the m x n board with m and n being odd numbers: Since m and n are odd, the denominators (m + 1, n + 1) in the fractions inside cos will always be even. And, since we round up in the expression above the PIs, j and k will in one factor both be exactly half of m + 1 and n + 1 respectively. When this happens we get cos(Pi/2) for both terms. Squaring the cos of course changes nothing. And when the product has one zero factor the entire thing will equal zero. Much fun this one!
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  460. You make complicated-looking maths problems sound so easy! Please keep up the great work!
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  461. I'm pretty sure it is the case that debts are consolidated before shares are calculated. The only way to have multiple claims is to have multiple legal entities asking for compensation and the courts not noticing that they are same creditor. That means that splitting up loans into smaller claims is not a useful strategy.
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  462. It’s always interesting seeing how close genuine mathematics, particularly number theory, gets to numerology sometimes - the choice of base ten is essentially completely arbitrary. On a barely related topic, I would love to see a video which covers some of the bignum arithmetic algorithms, like karatsuba, toom-cook, and possibly even FFT based multiplication. We rely on these algorithms thousands of times per day for RSA in TLS, so I feel they deserve some love.
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  463. Pretty sure that wheels made of logs is a very old invention ... this one just added special Powers.
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  464. There is a special place in heaven for people like you who make math accessible and fun, who put the intermediate logical steps in their demonstrations, thank you.
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  465. Removing two black and two green cannot always work. By counterexample: you can isolate a corner square. But sometimes you *can*: example: just remove the squares occupied by any two dominoes of a domino-filled chessboard.
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  466. Like others, I'm an old engineer who used a slide rule during my exams at university: cheap calculators hadn't been invented yet. (I still have my slide rules.) I also noted that log graph paper has the scales exactly the same as that of a slide rule. So when I was travelling in the UK after graduation (and before calculators and phones), I cut out two of the axes on the graph paper and taped them to cardboard at the proper point so that I was permanently calculating the conversion between Canadian dollars and pounds Sterling. While shopping, I could easily convert an item's cost to familiar currrency.
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  467. Mr. Mathologer. Thank you for your good work. I understand every single thought, every single word in every single second in this video and it is an extreme pleasure to follow the steps of these old math geniuses. The perfectness and beauty of math amazes me for years now. And it never stops. In math we see ... god.
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  468. A teacher showed me this back in high school, and I thought it was really pretty. I had wanted to include it in my combinatorics series, but I had to cut it for time. Great to see it covered in Mathologer fashion.
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  469. 13:27 In 1,023 moves, moving the smallest disc every other move means we're moving it 512 times, which is 2 (mod 3). So we need to move it counter-clockwise (A -> C -> B -> A) to make sure it ends at B. In general, 2^(n-1) is 1 (mod 3) if n is odd and 2 (mod 3) if n is even, so we should move the smallest disc clockwise if n is odd and counter-clockwise if n is even.
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  470. I'm a finn born in 1960. When I got my groung schooling it was given that I need to learn germany as most of the books in the universities in Finland were written in germany. The time (1981) when I entered uni in Helsinki, Finland to study mathematics, all the books were written either in finnish or english.
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  471. ​ @Mathologer  that's the sad state of affairs today sadly 😔
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  472. I have always thought about the coin rotation “paradox” in a rather simple manner. The distance travelled by a coin with radius r around the outside of another circle with radius R is simply 2*pi*(R + r), or 2*pi*(R – r) if it travels on the inside. Given the constraint of no-slip and the correct ratios, the result is hardly surprising.
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  473. I'm a High Schooler, didn't understand some parts😁. But I enjoyed the video so much🥰🥰 . Whenever I watch these Mathologar videos everytime I learn a new thing beyond High School syllabus. I just love this channel, he teaches in a very creative,fun and funny way.
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  474. (OH)4: A Christmas joke for innumerate dyslexics
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  475. Another masterpiece as always. I am a math nut, have been my whole life. When I go for a walk I think about numbers and sequences and formulas and physics. I spent my career as an engineer but now that I am at retirement I spend large amounts of time on 2 of my loves - math and physics. This is am absolutely fantastic channel. Thank you!
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  476. An Italian Math and CS Senior Lecturer here. Just want to share that, as it happened to the Mathologer, when my professor did the Riemann rearrangements theorem in Real Analysis I, as a freshman, was totally upset and amazed by this counterintuitive result. Congrats to The @Mathologer whose videos always make us see the things we know under new and interesting perspectives.
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  477. Me: takes out ring, proposes GF: says yes, crying Me: starts talking about the number of vertices on the diamond of the ring GF: takes off ring
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  478. I'm amazed that the paraboloid & the parabola were even known and studied that long ago. How did they define it without coordinate geometry?
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  479. Is asking people to spam "I made it to the very end" a way to drive eNgAgEmEnT?
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  480. I just realized: for the hexagon, if you turned it into a 3D broken cube, then looked at it from any of the open faces (assuming you're looking directly at the face from a single point), you would see a complete square of one color. This would be the same from all three open sides, and there would be no hidden faces. Thus, the sides being equal, there will always be an equal number of each color domino
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  481. 13:00 My guess: a^2 = a×a, so I think the output sequence will be a^a.
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  482. My favorite fact about the Dürer magic square: it was completed in the year 1514, and in the middle of the bottom row you can see it says 15 14. The only similar example I know of is the Basel problem, posed in 1644 whose solution is pi^2/6 = 1.644... However, sources differ on whether it was first posed in 1644 or 1650.
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  483. I cant believe after all these years of calculus ive finally seen an intuitive argument for how lnx appears when integrating 1/x 😮
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  484. I was wondering what could possibly be interesting about the number of ways to tile the glasses. I couldn't have guessed it would be the number of the beast! I also didn't realise that the number of the beast could be written in a neat little expression involving only the first 4 primes: 666 = 3^2(5^2+7^2)
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  485. Mathematics is like an intellectual equivalent of Willy Wonka's Everlasting Gobstopper -- it is harder than anything else on Earth, it tastes wonderful and no matter how long you work at consuming it, it never diminishes.
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  486. 42:50 That time when you see Notch in the Paypals section
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  487. For the grid problem, sum up every the possible 2*2 suqare exactly once. Notice that the total amount of times red, green, yellow, purple appeared MUST BE EQUAL. The corner squares are counted once, the edge suqares are counted twice (it is calculated in 2 of the 2*2 suqare), and the squares in the middle are calculated 4 times(it is in 4 of the 2*2 squares). Since the total amount of 2*2 squares is ODD, every color must appear an ODD number of times. Ans since only the corner squares are counted an ODD number of times, each color MUST appear in the corner square. QED. By the same argument, we can show that every color appears on the corner of the 4n*4n*4n cube, if it is colored with 8 colors such that every 2*2*2 cube consist of 8 distinct colors. We can generalize this further to m dimentional cube with 4n as its sides.
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  488. Not as had as "what comes next in the following sequence: 1, 3, ? " Answer "TREE(3)", not acceptable, must give an exact integer value.
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  489. Very much so :(
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  490. It makes me happy to see someone as famous / well-known as you acknowledge a mistake, own it, and correct it. As a fellow mathematician / professor, I genuinely appreciate the humility you have shown!
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  491. Congratulations on 100 videos! Your channel is awesome ❤
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  492. When the rubber band snaps you get the so-called "broken numbers", which were studied in 1763 by mathematicians working in a causally-disconnected region of spacetime, so unfortunately we will never find out what they discovered.
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  493. 4:54 This actually isn't quite the origin of the name. From Wikipedia: Dirichlet published his works in both French and German, using either the German Schubfach or the French tiroir. The strict original meaning of these terms corresponds to the English drawer, that is, an open-topped box that can be slid in and out of the cabinet that contains it. (Dirichlet wrote about distributing pearls among drawers.) These terms were morphed to the word pigeonhole in the sense of a small open space in a desk, cabinet, or wall for keeping letters or papers, metaphorically rooted in structures that house pigeons. Because furniture with pigeonholes is commonly used for storing or sorting things into many categories (such as letters in a post office or room keys in a hotel), the translation pigeonhole may be a better rendering of Dirichlet's original drawer metaphor. That understanding of the term pigeonhole, referring to some furniture features, is fading—especially among those who do not speak English natively but as a lingua franca in the scientific world—in favour of the more pictorial interpretation, literally involving pigeons and holes. The suggestive (though not misleading) interpretation of "pigeonhole" as "dovecote" has lately found its way back to a German back-translation of the "pigeonhole principle" as the "Taubenschlagprinzip".
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  494. 24:30 It's "obvious" because 1/x is concave, meaning between any two points the graph is below the secant line connecting those two points. Dividing the 1x1 square into rectangles in the obvious way, the blue areas include more than half of each rectangle and hence more than half of the 1x1 square.
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  495. As you were finding the tiny areas and adding them up, I couldn't help thinking that essentially the Jacobian is involved. If you don't get the Jacobian right, your area won't be right. That is, the Jacobian is what tells you how to get areas right. And it is also related to your little spikeys.
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  496. Very subtle post dubbing in the last chapter ;)
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  497. Mathologer walks into a party "Hey wanna see a card trick?"
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  498. 2^2+ i^2=3
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  499. 👍
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  500. Almost number of the beast. In Dutch we called the slide rule a “gokstok” (gamble stick). In English this device could be the “wheel of fortune”.
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  501. Dear Mathologer, I would like to repeat my question I asked in the previous version of the video. Will you please please please ever publish that "insane" (your word! Your promise even) video about the Galois theory. I've really been waiting for it for a very long time now. You promised it after all. You did that in your video about the cubibal polynominals that they didn't teach for over 500 years. Thank you foe keeping your promise in advance. Happy 2025, A.
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  502. General nxn grid is: n^2 + SUM(i = 1 -> n-1) { 2 * (i)^2 }, so it's palindromic
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  503. Burkard, you asked specifically for feedback on how successfully you made these explanations accessible. I am an electronics engineer who has always been fascinated by mathematics, but struggled with the higher, more abstract elements of the subject. My way of learning is like constructing a building: laying foundations and fully understanding foundational material before laying the next layer of bricks. And I always felt uncomfortable just learning calculus identities without understanding their derivation. I think that's one of the big things that held me back. Your videos follow a logical progression, building layers of bricks on top of understanding I already have, and at last I can trust in my gut that these identities are correct because you have shown me, even if I cannot repeat the derivation myself. A few times in this video I had to stop, rewind and check some simplification or rearrangement you did, because it wasn't immediately clear to me, but I got there. Thank you for making this fascinating subject accessible in a way I have never seen before. You have rekindled my love of higher math. By the way, I much prefer the taurean version of Euler's identity: e to the i tau equals one, and I think your beautiful graphical demonstration could easily be adapted to demonstrate this form. I'd love to see your taurean conversion of other identities, which will be similar but I think lead to a far greater intuitive resonance than using pi.
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  504. Wow, damn, I sorta figured out a couple of these things for myself when I was in school (and later on in college) but I was never explicitly taught any of it! It's so cool to see that my thoughts about "wait, is 2^x a kind of discrete version of e^x?" were actually a thing that people had studied and wasn't just a weird quirk!
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  505. Best math channel on youtube by a total landslide, there's simply no competition. Keep up the amazing work Burkard! but I do hope that you don't work too much on youtube to the detriment of your own research (or worse that you burn yourself out); you're doing such a great and important service to math by inspiring so many future mathematicians and showing them ways to think mathematically that almost no degree would ever teach them, but it is crystal clear that you yourself are more than capable of contributing every-bit-as-important new results to mathematics - it is my opinion, and I'm sure also many others', that you are one of the best mathematical minds alive in the world today, and that you can achieve something big (yes, big big , but we know it's not winning a million dollars that you care about :P) Merry christmas! - we wish you the best combination of luck and continued improvement in skill, in choosing your mathematical battles; may you make major breakthroughs in the new year or beyond and push your field forward. I have zero doubt that you'll do us all (and most importantly, yourself) proud!
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  506. I love Zagier's sentence, even without the windmills. It serves as a great exercise in reading proofs. If I ever teach one of those "intro to proofs" class, I would assign the task of deciphering it as some sort of class discussion for the day.
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  507. 27:46 when he paused, I was already like 90% sure he was going to say Euler. If you're ever on a quiz show and one of the questions is "Who first proved X?" if you just guess Euler you're probably good.
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  508. I'm a simple math lover. Mathloger uploads. I click.
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  509. Maths.... is the most beautiful thing ever. So subtle, so abstract and so transcendental... it always existed and will always exist. What we know is a tiny fraction, the most will always be beyond our comprehension.
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  510. In recent years, when I buy shoes they tend to be laced in strange ways so that I always have to relace them (to criss-cross lacing). I wonder why this is. Are there lacings that can be done FASTER in the factory? Have you ever considered speed of lacing as a criterion?
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  511. As soon as you moved the negative fractions below the top line, my first instinct was "Wait...isn't the top part 'outpacing' the bottom part?" Then I lost confidence when you collapsed them, lining up all pos and neg, lol. I was like "but, but, but...." Anyway, I love that stuff!
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  512. I remember when I first noticed the derivative relationship between volume, surface area, and circumference of the sphere. It’s been over a decade since my first calculus course but it’s still so satisfying. When you consider that the derivative is a rate of change, the relationship begins to make perfect sense. I think what really made it click was when we started doing integrals. If you integrate the circumference from r=0 to r=R, you wind up with a sweep of concentric rings that cover the whole area, meaning that the rate of change of the area is given as the circumference as r is varied. Similarly, as you integrate the surface shells across the range of r- values, you wind up with the whole sphere! It’s brilliant once it all clicks and you can see those animations in the mind 😊
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  513. Pilot's computer (E-6B) - which has many other functions as well - including an analog vector adder. When using a slide rule, always estimate the value in your head first to avoid factor of 10 assumptions.
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  514. Typical Fermat. Claiming he has proofs but not delivering. Unlike Mathologer of course 😜
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  515. It also visually shows that log(0) doesn't make any sense, which is cool!
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  516. Hey, I reverse engineered the progression of XP required for each level in final fantasy 5, using this technique (I didn't know about the Gregory-Newton's formula but somehow I figured a formula).. I was 13 at the time or something. I wanted to make a game and for some reason the XP progression was going to be a big part of it. (I ended up making a small dungeon in rpg maker) It turns out that the XP progression is a polynomial, but the last row is a bit random - instead of being the same number (and thus the next one being all zeroes, and the other all zeroes too, etc), eventually it had a random noise of 0, 1 and -1. I figured out that this meant the coefficients of the polynomial weren't perfectly integer, and rounding messed up the differences.
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  517. For the hexagon puzzle: looking at the picture as a 3D stack of blocks, it is obvious that each tiling can be gotten by adding one block at a time. This corresponds to rotating a hexagon with side length 1 by 180 degrees. Thus the number of tiles in each color doesn't change. But the numbers are equal when there are no stacked blocks.
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  518. It always irks me when people teach the subtraction rule and the quotient rule as separate unique rules. The subtraction rule is the addition rule and the constant multiple rule combined. After all, f-g is simply f+c*g where c=-1. Same rules, no need to make a special subtraction rule. The quotient rule can be explained by the product and chain rules combined. f/g = f*h (product) where h = g^-1 (chain). I just dislike having to learn special cases when they're not at all special cases.
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  519. Best one Ramanujan's explanation I have never seen before . Awesome MATOHLOGER .
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  520. I made it to the very end, Burkard. And I don't regret it.
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  521. I can't believe it's actually finally here. I've been waiting for this since the joke at the end of pi is transcendental proof. Thank you, Mathologer. You've somehow outdone yourself yet again!
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  522. First the Mathvengers: Eulergame, then Numberphile hit pi million subs, and now a new Mathologer video?! Is it Christmas? Oh wait, it is.
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  523. "Things are even much hairier in Australia than they appear at first sight." At first sight of this video - it looks really rather un-hairy, I agree... ;)
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  524. A truly beautiful visual and mathematical feast. The complicated symmetry is an extra detail that (for those of us obsessed with symmetry) elevates this to astronomical heights. Thank you, Burkhard for reminding me how much I love Mathematics. A special round of applause to the crystal clear animations.
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  525. Another good one Mathologer. The closing moments were the most important IMO: aspiring mathematicians and those interested should always remember that they're free to redefine expressions and loosen axioms depending on their area of work - much new material can be discovered in this way (e.g. NE Geometry). Similar arguments made for defining what 0**0 should be, and different answers depending on who you ask of course. Perhaps even video worthy :) Thanks again, have a good one.
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  526. I'm surprised there aren't more comments about how your shirt literally says "To infinity and beyond" in math geek. At least, I think it does?
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  527. I cant believe I hadnt heard of Quadratic Reciprocity considering my honours dissertation was on finite fields. Granted, it was efficient computation of matrix opperations under GF2 on GPUs, but still, I cant believe I'd never come across any of this.
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  528. I'd love to see you do a video on dual quaternion skinning. It'd be combining topology, N-D transformations and general matrix/linear algebra. Alot of bad resources on it not really well explaining it's power.
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  529. I taught woodwork for years, in good selective schools and there were always pupils that thought they could build a hollow cube out of six equal squares of plywood. My boss would actually cut them the squares and watch then trying to assemble them. I regret that it did not occur to me to tell them it was impossible because of Fermat's last theorem, that would have been cool. Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube.
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  530. IMO Mathologer perfectly hits the balance between presenting topic in depth and in interesting way. There are other channels more attractive in form, there are channels discussing math deeper, but here I can follow up what's going on, and I want to know where it's going on.
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  531. Awesome
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  532. I remember translating this theorem's Wikipedia page to Hebrew a few years ago, I came up with the coloring proof. but when I saw the swivel proof I just couldn't help but smile. I liked it better than my original proof. this proof made me smile so much. this always happens in your videos, in what you call "aha moment" I experience euphoria. your videos just make me smile from ear to ear, every one of them.
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  533. Am I the only one to notice the infinite sum on the last slide is wrong? It has the product of natural numbers in the denominator instead of just odd numbers. And it includes even powers of x as well.
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  534. 1+2+3 = the amount of seconds between the release of the video and me clicking on it
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  535. Just solved the 4D cube for the first time thanks to your breakdown video! Thanks! It's been a dream of mine since I first learned about it nearly 10 years ago and never thought I'd ever solve one! Great video!
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  536. 7:00 yes it can. The number ends in 81. That's a multiple of 4 + 1.
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  537. "Coins will be gone soon!" Laughs in Zinc Lobbyists
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  538. You triggered my math competition PTSD 😂
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  539. I can hear the ancient Greeks yelling, "Mathematics is just geometry, man!"
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  540. Finally a math topic I’ve never heard about! Thank you Mathologer, you’re great
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  541. My wife called me a nerd when she caught me watching this! 🤓
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  542. Thank you from the heart. You have such kindness, generosity, and humor in your lectures. I trained to be a mathematician but realized that I didn't have any real talent, so I became an Engineer ( all three). And tutored math in my free time.
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  543. Actually, here in Russia we are in fact taught Heron's formula at school, but I'm not really used to it since it was a lockdown year when we were taught this. So if a problem requiring Heron's formula to solve it occurs, I'm always a bit perplexed as it doesn't come to mind at first.
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  544. I have to say that you're one of the few people I know who actually pronounces other mathematicians' names correctly. I appreciate that you take the time to get them correct - it not only shows respect for those mathematicians, but also saves me from having to hear someone say "Weierstrass" with a W or pronounce "Weil" as "Wile" again. Thanks for paying attention to those little details!
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  545. Brilliant
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  546. I rarely ever comment but my appreciation for these master classes merits a comment here. Other channels either prove easy results that look pretty in animation or mention crazy results without giving proofs. What I like about these videos is that they actually take the time to prove those crazy results. I would love to see any videos you may post on Galois Theory or any other weird topic only math insiders get to see and that other channels say "so and so proved this 100 years ago but the proof is really hard"
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  547.  @Mathologer  It does rhyme! and it is deliberately written in the form of seven-word-poem (much like the solution poem to the Chinese remainder theorem). On this same topic, there's another, much more ancient (and famous) "poem" on just the 3x3 magic square: 九宫之义，法以灵龟，二四为肩，六八为足，左三右七，戴九履一，五居中央 Translation: """ The way to fill a 9-square palace, is to imagine a turtle (back): 2 and 4 as the shoulder, 6 and 8 as feet, 3 on the left, 7 on the right, 9 as hat, 1 as shoe, and 5 in the middle """ This text came from an ancient manuscript, that says these numbers/pattern comes on a turtle, and it's a sign of miracle. I guess that's part of what you said in the video: some people think magic square is truly magical.
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  548. Just in case nobody else has pointed it out, may I add that the frictional force between the elastic band and the wheel increases exponentially. This is why cowboys in westerns sometimes do not tie their horses to the rail. A few turns around the rail makes the horse just as secure.
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  549. 28:12 The error is in the final term for the sum of ninth powers, which should be -3/20 nn.
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  550. Don’t…..blow my mind like that
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  551. I have almost no grasp of basic algebra. I watch these videos in complete aww of the innate problem solving potential of human beings. I feel like learning math is akin to finally being able to leave Plato’s allegory cave, in that math seems like a key to understanding the entire world around us.
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  552. I am not a studied mathematician by anyone's measure. Yet, I carry away so much from your videos! Thank you so much for your well-constructed presentations. This one was wonderfully startling!
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  553. Calculus is so beautiful and elegant theory....and also it is really easy if learned from the correct teacher or book
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  554. It was me, I discovered this proof back in grade school when making arts & crafts. I wrote a note in my journal of discovering the proof, but I had to also go back and watch Power Rangers.
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  555. As many people have already pointed out in the comments, a compromise between the Talmudic method and the proportional method would involve some kind of hourglass shaped containers. This would smooth out the payout curves, but I can't think of any other practical benefit. Just something fun to think about.
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  556. Legendre looks angry!
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  557. "I'd like to finish off the video" he says roughly half way through the video...
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  558. I use the subtitles on YouTube to compensate for my hearing loss and I kindly suggest that you watch this video with the subtitles to see the issue of having crucial information/animation in the lower parts of the video. I hope that this is seen as the constructive feedback that it's intended as.
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  559. Hydroxide, hydroxide, hydroxide!
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  560. I love the second "simpler" proof. It is intuitive and I can even explain it to members of the family who are not true maths lovers.
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  561. The most memorable part was me dying because I didn't know the no 9's series was convergent
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  562. We were taught Heron's formula in Bulgaria. And half the perimeter is denoted as p, not S. That's the area.
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  563. I fear most schools prefer not to promote "visual proofs" since it could have pupils do them when the visual is "close enough" yet still false. Meanwhile, the algebraic ones have "rigid" rules, that can be more easily explained and corrected.
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  564. A Strand Is A Part Of A Rope Or Bond, While Stranded Means Being Washed Up On The Shore, And Being Stranded Is When You Can't Go Home.
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  565. Now that I have watched this video it's been up for 50 minutes and 94 likes so at least two people liked it in the same minute. This is not interesting up to here, but this number will initially increase for a bit and then decrease later. And the peak is dependent on the number of notification bells and subscribers. So with this simple thing we can find out about the number of notification bells set on any channel. (Disclaimer, I don't know if you can actually see this number anywhere on YouTube or not)
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  566. Mouse flavoured cat food. How have they not thought of this?
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  567. Absolutely amazing. I did not want the video to end!
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  568. Best part of this was realizing how I can use the logic to solve 3 of my yet unsolved ProjectEuler problems! Awesome video!
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  569. I kept hearing "a 4k+1 prime" and wondered how or if the primality mattered. It's amazing how late, and how crucially, it finally comes into play.
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  570. I love this video. When you mentioned the problem, I first thought that it sounded familiar, then realised that I've heard of it before, including the fact that it is not possible to move 5 steps past the line. However, I have never seen a proof of it before, and I thoroughly enjoyed the proof.
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  571. Thanks a lot, Mathaloger! I have studied some works of Madhava but didn't know about these beautiful correction terms. Our textbooks in India still reflect a post-colonial mindset, resulting in a significant lack of awareness and ignorance. Unfortunately, many important contributions from Indian scholars are overlooked or not adequately highlighted. For instance, Baudhayana, who authored the world's first geometry textbook, including the Pythagorean theorem long before Pythagoras, is rarely mentioned. Similarly, Pingala's profound insights into permutations and combinations within the context of literature are often disregarded, despite their beauty and significance. Aryabhata's foundational work in trigonometry, which forms the basis of our modern understanding, also receives insufficient attention. Furthermore, there's no mention of Madhava, or any mathematician from the Nila school, in our great textbooks. They only perpetuate a subtle sense of inferiority complex by failing to acknowledge these remarkable achievements. It is high time for a change in our educational curriculum. One aspect I particularly admire is India's significant contributions to philosophy and spirituality. These traditions are not only highly logical but also explore the limitations of logic itself, as evident in the concept of non-dualism. Additionally, architectural marvels like the Kailash temple exemplify India's profound knowledge in the field of architecture. Regrettably, we have lost a great deal of knowledge due to constant invasions, including the destruction of the precious Nalanda University, the world's first residential university. There is so much more to be said about India's marvelous contributions to the world. It is my hope that our generation recognizes and appreciates the immense legacy left by our ancestors. Thank you so much again. Keep up the amazing work that you do!!
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  572. If we really want to go back to the beginning of calculus, there's strong evidence that Archimedes had something very very close to it (including rigorous definitions of limits), if not bona fide integral calculus itself. Which is some 1600 years before Madhava. Of course, I would only be half survived if some even older Indian, Chinese, Babylonian, etc... mathematician had invented something equivalent to calculus even before that. Scholars typically remember the start of still-living lineages, rather than the original discoveries that faded for centuries before being rediscovered.
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  573. And the neat thing is that you can generalize the discrete approach to several dimensions, and talk about heat equation on grids... But why stop there? Isn't a grid like a special case of a graph? Like the plane, that is a very particular case of a manifold? With some reasonable hypothesis, many theorems have a parallel discrete contrepart!
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  574. Heard of this before, it’s just mod 9 arithmetic with a coat of mysticism thrown over it.
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  575. nice :)
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  576. The animation in this video is just insane - about the best I've ever seen on Youtube in fact. I can't imagine all the time it must have required to get all that to flow so smoothly. Well done. I especially liked the part about the bugs, shows you how to solve such a problem in visual steps - a great problem solving technique. And the animation was just beautiful.
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  577. I'm not a fan of the transition slides with the slightly twitching eyes, they're unsettling to look at and in general the use of AI makes me uncomfortable.
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  578. The last time Mathologer posted a video the night before my exam, I got selected even though I didn't prepare much. I'm hoping that the trend continues. :D Seriously though, the timing couldn't have been any luckier.
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  579. If I could only have had 30 seconds in Ramanujan's brain
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  580. Second proof is more simple and easy to spot, first proof is a little more complicated but elegant, in my opinion
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  581. My daughter (ninth grade) just sent me this video with the message, “this is so interesting!” Thank you for the proud dad moment, Mathologer!
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  582. Also, the point of leaving the last card only down, if you consider either that one card only is flip, or that no move is possible it is terminated.
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  583. Of course, it's not taught, but as long as one thinks, not teaching this won't stop a few, very thoughtful students from discovering this method.
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  584. 13:00 Taking a = 1 and b = c = ... = 2, we see that the output sequence for {n^2} must be {n! * (n-1)!}. Neat!
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  585. I am a Ph.D physicist and working as a data scientist but never knew this simple intro to e. Thanks .
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  586. I would love it, if you would continue with this concept! 🙌
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  587. Re: the end questions, undergrad computer science, and would be fantastic to see more of the "insane stuff" :)
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  588. Can you sometime do a geometric proof like these that shows why: 1^3 = 1 2^3 = 3 + 5 = 8 3^3 = 7 + 9 + 11 = 27 4^3 = 13 + 15 + 17 + 19 = 64 Etc
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  589. Those are some pretty non-christmassy glasses if you ask me, you're a beast.
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  590. Your teaching style is just so good! I think it's a combination of the interesting topics, your smooth as heck animations, giggles, and the quick glances you give at the end of each chapter to summarize (it's especially nice for note-taking!). Not even to mention the fact that you don't give direct answers to questions you bring up, but instead direct the viewer to introductory terms and topics to look up and gain knowledge themselves. I wish I could attend one of your lectures, but until then this will have to do!
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  591. Awesome video (im a 33m into the past time traveler)
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  592. "Hey programmers" I see my work is requested.
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  593. 5:38 All primes that can be written as a sum of two squares are primes
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  594. Thanks for the video! When I first heard about this proof, I asked Alexander Spivak who invented the visual version. And he said that there was no other source, it was his own idea. Because we don't know anybody who came up with this before 2007, it's almost certainly that he was the first. Unbelievable, but the Zagier's proof (and the previous proof by Heath-Brown) had appeared without any connection to geometry.
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  595. This log wheel also provides a nice illustration of Benford's Law, the "observation that in many real-life sets of numerical data, the leading digit is likely to be small" (wikipedia). If you pick a random point around the perimeter of the wheel, it is very likely that the number it represents starts with the digit 1, less likely that it starts with 2, etc. all the way up to 9 which is very unlikely. This is independent of the magnitude (decimal shift).
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  596. The most memorable is the fact of the partial sums being non integer.
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  597. For the width 4 triangles, there are 10 "essentially" different triangles, if your symmetries are permuting colors and reflections across the vertical axis. I used Burnside's lemma twice.
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  598. and a cool T-Shirt :)
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  599. Videos like this make me marvel at the internet. Growing up I could never have access to content like this but now I can watch a brilliant mathematical mind explain fascinating concepts to me. this channel is an example that should give everyone faith in the future of humanity.
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  600. For a "change," I see what you did there.
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  601. with the last two videos this channel has outdone itself. I have seen and re-watched them several times and as an amateur and enthusiast I believe that they are the two best calculus lessons I have attended. So illuminating and profound, they hold together all those details that leave one confused in a school course and which here instead receive the right attention and are explained with incredible ease. Bravo! ❤
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  602. "A Tour of the Calculus" (Berlinski) was successful in terms of sales, but a pompous flop in terms of making calculus accessible. Martin Gardner prepared an edition of Silvanus Thompson's "Calculus Made Easy" that includes some additional chapters. Steven Strogatz's "Infinite Powers" is a fascinating contemporary take on how calculus works and what it can do. Highly recommended. I also took a crack at it in "A Stroll through Calculus," which is subtitled "A Guide for the Merely Curious" because it keeps the math as elementary as possible. (It's been used as a textbook for non-STEM majors who need a math class.)
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  603. I was learning about “discrete calculus” like this in my undergrad and I was blown away when I learned that 2^n is its own difference, much like e^x is its own derivative. It’s like 2 and e are analogs of each other. I’m still trying to grapple with the mathematical-philosophical implications of discrete calculus being epitomized by “2” and continuous calculus being epitomized by “e”. They aren’t that far apart on the number line, after all.
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  604. Odd factors of 2020: 1,5,101,505, all good. This tells us there are 16 pairs. 2020 = 16^2 + 42^2 = 38^2 + 24^2. The two permutations and four sign choices yield all cases.
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  605. How does it feel to never be able to make a video about something “barely anyone knows about” when you will inevitably popularise it? I think it’s CGP Grey that has to deal with people commenting on his “common misconceptions” videos saying how “everyone knows it” but the only reason the correct knowledge is so well known is partially because of said video
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  606. 23:33 there’s another lovely pattern that’s easy to miss in this table; look at the antidiagonals. See anything familiar?
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  607. I hadn't seen the theorem before. When you explained the phenomenon, I said to myself, 'wow, cool application of circulant matrices'. I didn't trouble to pause the video to try to work it through, but as soon as you showed the five components, is was "oh wow, that's just a Discrete Fourier Transform, summing roots of unity" and everything else was obvious in that perspective. (Seeing the title of Alvy Ray Smith's paper clinched it!) I'm curious whether that's the approach Petr, Douglas and Neumann all took, or whether one or another of them had some more painful route to the conclusion. But not curious enough to try to plow through a paper written in Czech.
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  608. Simplesmente maravilhoso
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  609. I already knew about generating functions, but seeing that automated algebra and the reason for all those 3s and 0s was seriously amazing
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  610. My physics background led me to guess the name eigenpentagons just before you revealed the eigenpolygon name. Cool!
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  611. @JoelNeely, I fully agree to your comment. I was also taught this at elementary school, for a later confusion as follows: Since these divisibility rules are Base-10 dependant, I had thought for many years that the divisibility of a number with another was Base dependant, and that perhaps on another base those same numbers were "conmensurable". A gross mistake that hindred developing intuition on numbers theory. I loved the explanation where Prof. Burkard decomposes a base 10 number in: a (9+1) + b (99+1) + c (999+1) ... seen it that way is so straightforward !
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  612. I can't take my eyes off of your epic Escher rubic cube shirt.
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  613. The coin rotation: 5. The stationary sircle is 3/2 times bigger than the rolling one, so it will be 3/2+1=5/2 reotations of a rolling sircle per a roll, so 5 for two rolls
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  614. "I'm not about to propose" You just broke my (math) heart!
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  615. That's new angle system, where 120=π
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  616. Yes! That was the first thing that came to my mind as well. Additionally, for the first part of the video, you get rewarded for claiming everything, which is just bad in practice.
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  617. I never made it past geometry in public school, and yet I was able to follow most of this well, and appreciate how beautiful this proof really is. I chalk that up not only to your ability to explain things in various ways, but also to just how clean and professionally edited this video was. Well done. You have yourself a new fan. (Or... a new windmill.)
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  618. Firstly, thanks. I'm looking forward to this. Before I get too far in and while I remember: Do you take requests? Can I suggest a Beginner's Guide to p-adic Numbers.
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  619.  @Mathologer  Bald +10IQ No need to scratch head to solve problems 😁
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  620. Funny story: After discovering an alternative to Faulhaber's formula about 1.5 years back I've (in search of validation) sent it to Marty and Burkard. Being the nice and generous guys they are, they took a look at it and confirmed that it appeared to be true and also provided me with some additional papers about the topic. If anyone here is interrested in my solution, I'll leave a link to one of my quora-posts about it (which in turn contains a link to the formula's proof) down below. https://www.quora.com/There-is-a-formula-for-the-sum-of-consecutive-natural-numbers-Is-there-a-formula-for-the-sum-of-their-squares-cubes-and-other-exponents/answer/Samuel-Gantner?ch=10&share=9e599769&srid=oJzsi
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  621. 6:11 He LIED! Three pigeons were hurt in the end of the video 😅
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  622. Ramanujan–Sato series is fastest known pi converging algorithm to calculate pi values using computers. The Chudnovsky algorithm is based on Ramanujan’s π formulae.
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  623. India is very rich in mathematical and metaphysical knowledge as well as philosophy. The mathematics is earlier times were in the form of shlokas . Respected sir , i am an IIT student currently pursuing my degree in aerospace engineering, but i love maths . Here is a request from me to please make a video on baudhyan theorem and Pythagoras theorem , baudhyan theorem came earlier than Pythagoras , but baudhyan stated it in a very different way , leading to the same result as Pythagoras . It would be intresting to know about both of them , their differences and similarities from a great teacher like you .
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  624.  @friedrichschumann740  Würzburg, according to Wikipedia.
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  625. Opposing angles subtend a partition of the whole circle. The measure of an angle is half the measure of the arc it subtends. Therefore, the sum of the opposing angles equals half the sum of opposing arcs, which is half of 360 degrees, which is 180 degrees.
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  626. What could go wrong at 28:16 ? The corner, where the three tetrahedrons meet, corresponds to a spherical triangle. The side lengths of wich are the angle wich meet there. These side length have to conform the triangle inequality, otherwise the corner can't be form. Lucky as we are, the angles are just the angles of the corner, where the lower case edges meet. So all is fine.
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  627. @ There's always a way to defeat any consolidation rules. For example, you can agree with an unconnected creditor to loan a sum to someone, and that creditor can sell some debt to you. When you're talking about banks and other big financial institutions who have many many loans and lend to each other, it quickly becomes a tangled mess that makes it impossible to tell that two loans "have the same creditor".
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  628. "If there are any parts of this video that you struggled with, just ask" Yes. Where do you get your T-shirts from? It took me a few seconds to figure this one out, but hey - you've got a friend :-) Edit: Greetings from AUstria to Australia
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  629. The duality relationship between the triangle and its folded form is simply beautiful. As a triangle lover, I absolutely love this video. I cannot believe this was not known.
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  630. I think I independently discovered the devil lacing when I was weaving my shoelaces so as to use up extra lace length and turn lace-up shoes into slip-ons. With the right goal in mind, it can be the best way to lace up your shoes!
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  631. Today is my birthday, this is not less than a gift, thanks!
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  632. As a chemist, I’d feel much better if there was a trivalent cation to accompany those hydroxides :/
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  633. I'm just realising we did too, but no emphasis was placed on it, nor was it's source explained
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  634. It's 12 : 00 AM in India Looking forward to the following 50 minutes And Merry Christmas!!
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  635. I have explained this as follows with no reference to the horn: Take any amount of ideal paint (that is, paint that can be spread as thin as you want), say 1 quart. Spread it over a surface of 1 square yard of an infinite plane. It still has some thickness so, spread it until it covers twice the area. Now it's only half as thick, but it stil has some thickness you can repeat this, over and over--each time doubling the area this quart of paint covers. There's no end to this procedure, so the paint will cover the entire infinite plane.
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  636. It is always a pleasure to watch your vids. Not only because these are great educational videos, but also because your voice and wordings make them even better
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  637. I wonder who is that one person who disliked this video ? Are they trying to balance ? 😉
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  638. I remember watching this video when I was younger, and it partially inspired me to get into math competitions. Glad more people will be able to see Nathan’s beautiful solution!
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  639. The dot proof is more emotionally satisfying. :)
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  640. "lots of 4's" - I'm a genius! "but that's not it" never mind...
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  641. I paused and worked out as much of the problem as I could on paper. Then I unpaused and he covered everything I did in 5 seconds. :~(
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  642. I'd guess it's because also eg. 13=13 or 423=423 ... so, it would defy any further pattern... 😅
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  643. that's why i always hate these what come next questions in "IQ tests"
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  644. The general formula for the sum of 1/x^(2n) is hidden in here too, and only requires a few extra steps. Start from the chapter 2 formula and take logs and derivatives to get the formula at 10:40: cot(x) = 1/x + 1/(x-pi) + 1/(x+pi) + 1/(x-2pi) + 1/(x+2pi) + ... Move the 1/x term to the left side, and take 2n-1 more derivatives. The k'th derivative of 1/x is (-1)^k * k!/x^(k+1), so when k = 2n-1 this is -(2n-1)!/x^(2n). So we have (d/dx)^(2n-1) (cot(x) - 1/x) = -(2n-1)! * (1/(x-pi)^(2n) + 1/(x+pi)^(2n) + ...) Now divide by -(2n-1)! and take the limit as x -> 0. 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1/(-pi)^(2n) + 1/pi^(2n) + 1/(-2pi)^(2n) + 1/(2pi)^(2n) + ... On the right side, all the negatives are squared away, and we end up with 2 copies of each term. So multiply by pi^(2n)/2, and we get this: pi^(2n)/2 * 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1 + 1/2^(2n) + 1/3^(2n) + ... = zeta(2n) This formula looks messy, but there's a trick: notice that on the left, we have something of the form 1/k! * k'th derivative of f(x) at x=0. These are just taylor series coefficients! The left side is really just pi^(2n)/2 times the coefficient of x^(2n-1) in the taylor series of 1/x - cot(x). There are a few other things we can do to make the formula easier to read. We can multiply the function by x to make the powers line up nicely (otherwise the 1/k^8 sum will be related to the coefficient of x^7, instead of x^8). This gives: zeta(2n) = pi^(2n)/2 * coefficient of x^(2n) in the taylor series of 1 - x cot(x) The next thing we can do is move the pi^(2n) "inside" the taylor series, by replacing x with pi x. We can also move the factor of 1/2 into the function. Then we get: zeta(2n) = coefficient of x^(2n) in the taylor series of (1 - pi x cot(pi x))/2, or equivalently, (1 - pi x cot(pi x))/2 = sum n=1..inf, zeta(2n)x^(2n) And indeed, if you ask wolframalpha to compute the taylor series of (1 - pi x cot(pi x))/2, you get pi^2/6 x^2 + pi^4/90 x^4 + pi^6/945 x^6 + pi^8/9450 x^8 + ... Finally, comparing this series to the standard taylor series for cot in terms of Bernoulli numbers gives Euler's general formula for zeta(2n)
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  645. The Toymaker is Hanoi'ing me again.
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  646. Most Memorable: getting the Mathologer seal of approval
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  647. Since I made it to the end of this, and since you asked nicely: I'm an enthusiastic amateur, never took much beyond AP calculus formally but I read & study the subject pretty broadly as a hobby. I was cruising along smoothly here right up until about chapter 4 or so, after which there was a lot of pausing, rewinding, workings-out on paper or Python, and a couple side trips into Wikipedia & Wolfram. The Pascal/Bernoulli thing was mind-blowing to see in action. It definitely made sense on the surface, though as it got deeper i started to feel like it was going in circles? Something like, "We can easily derive S using B. But how do we get B? Well by deriving it from S of course." At least that was my initial impression. Really enjoyed it even if it did start to outpace me towards the end (or more accurately, because it did). It'll be a few more viewings before everything clicks for sure, but I'm looking forward to the challenge!
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  648. Fun Fact: If that Hexagon tiling were blocks in Minecraft, then if that represented a sloped hill/mountain in Minecraft, it would be scalable as there is a path from the bottom to the top. Although it's not really obvious that that would be the case.
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  649. I did know the Euler McLaurin from Wigner-Weisskopf Theory in Quantum Optics, but for me as a Physicist ist was just a mathematical tool to solve a hard integral Thanks to Mathologer I now know the true beauty and significance of this formular To answer the question, all of it worked for me, but I'm studying Physics so my background in math isn't that bad Grüße aus Deutschland
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  650. Hi, I did find another way to "build" youre cubes. Instead of adding the hexagons, you can add 1+ (1+2+3), than add 7 + (3+4+5), and 19+ (5+6+7), ... If you look in the video at 8:39 you see the second cube. To go to the next cube you need to add one side, so 3 pieces, lets say on the left, then you add 4 pieces to the right side (3 pieces for the original cube and one piece for the just added row), and finaly you add the bottom with 5 pieces to get back to the cube. Now if you look at the numbers you add they are left and right from the yellow marked number in the top row (2 and 4 sandwidch the 3; 5 and 7 the 6,...) And these yellow numbers are diviceble by 3 (by construction) So you get a sequens of (3n-1)+(3n+1) and that is the same as 2(3n). We could write 2(3n) also as 3(2n). This we can convert in (2n-1)+(2n)+(2n+1). And this is the secuence you are adding to youre cube. (Note "n" is the base number from youre last cube, and the counting number of youre tripple (where 3 is the 1st; 6 the 2nd;9 the 3th;12 the 4th,..)) Notice that by every step you go up you start with the same number you stopt with the last time. (2n+1)=(2n+2-1)=(2(n+1)-1) So the cube you end up with has a sidelenght of 2n+1. And you start the next step just adding that side to youre cube I hope you can see how i find this a bit easyer to construct the cube. Also this is in a way the same you construct youre squares (if you start with "skipping" just one number at the top) if you notice that every odd number is the sum of its position + its position -1. (5 is the 3th odd number and it is 3+2; 9 is the 5th odd number and it is 5+4; ...). Also here you get that the last digit you added the last time you start to add the next time. (but you only get 2 numbers). Note; this methode does not work in the fourth or higher dimensions. Why? Well I'm not a mathematitian. So I don't know. And I have trouble imagining in four dimentions. Hope this is helpfull. Sorry for my bad English, I speak Dutch. Greetings from Belgium.
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  651. We summoning Satan with this one🗣️🗣️ 🔥🔥💯💯
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  652. SPOILERS SPOILERS SPOILERS: It's uncountable; here's a funny algorithm: given a sequence (n_k) of natural numbers, we start out just as normal, adding positive terms until we overshoot. But rather than turning around immediately, we keep going for n_0 many positive terms, and only then turn around. Next, once we undershoot, we don't immediately turn around, but keep adding n_1 many negative turns before we turn around. Then once we overshoot again, we keep adding n_2 many positive terms before we turn around, etc. Note that the original algorithm is what you get when you start with the constant sequence (0,0,0,...). Now of course, this isn't guaranteed to converge; you could end up over/undershooting too wildly. But - certainly if all the n_i are 0 or 1, it will converge. And there are already uncountably many sequences of just 0s and 1s.
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  653. There were/are mathematicians, who can smart formulas not only elaborate, but they really see and feel the underlaying deep relationship. Also Ramanujan was one of them.
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  654. Mathlogger: "Do you recognize this shape that is starting to materialize?" Me: "I do, it is an a-" Mathlogger: "The cardioid!" Me: "-ardiod. Of course that is what I was about to say"
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  655. My favorite thing about this rubberband stretching construction is that it doesn't matter how big the wheel is or what numerical base you want to use. Simply take a wheel of any size, stretch the band as done in the video for 1 revolution, mark the point on the band where the revolution is completed, unravel and add evenly spaced markings for whatever desired numerical base, then rewrap the rubberband around the wheel.
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  656. My brain cant keep up to this! 😭 Why is it im so slow at this field 😭😭😭
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  657. That was also my immediate thought after seeing that a polygon is the sum of some ideal polygons! I scrolled down to comment and what's the first thing I see?
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  658. This is like a grand unified theory for pi formulas. Amazing!
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  659. Thanks for another interesting video! One question though: In the 3rd charpter, about 28:30 min in the video, in the difference scheme of the squares, shoudn't the first differences (the second row) be 1,3,5,7 instead of 1,3,7,9?
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  660. Most memorable: The 700 year old proof by a bishop
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  661. Thanks for taking us on a stimulating mathematical journey! I love the animated proof at the end. It's almost like a poem.
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  662. I was hoping for a crazy and unexpected method that I could start using today but it seems the criss-cross method reigns supreme (probably should have seen that coming). At least I can check out Ian's site and see if there's some other weird tricks to use!
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  663. Simple. Get velcro instead.
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  664. Beautiful presentation. Euler's original book "Introductio in Analysin Infinitorum" is a treasure. It's easily readable, even if written in Latin! (there are translations, of course.) It's exactly the same spirit as your presentation.
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  665. That’s interesting that the addition and multiplication tables for 2 are the logic tables for XOR and AND gates 7:00
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  666. Congrats on your 100th video! Very special number, since it's the fourth octadecagonal number, amongst other things 😆
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  667. Most memorable: How strange the optimal leaning tower is
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  668. Ich freue mich schon auf das Mathologer-Video auf Deutsch! Schöne Grüße aus Marburg.
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  669. I didn't knew anything about it up to now.
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  670. The Arctic Circle with hexagons can also be called Q*Bert's Heaven.
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  671. More (fabric designers) should partner with mathematicians. Gorgeous animations!
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  672. In my time it was called the calculus of finite differences. Love the Mathologer videos.
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  673. 21:40 Decompose both expressions as products of fractions by pairing each term in the numerator with the term below it in the numerator. On the right, you have that each fraction is > 1, so their product will always grow. On the left, however, they alternate between > 1 and < 1, so it's at least possible for it to converge.
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  674. It would be fun if WolframAlpha implemented a "what comes next"-function which uses the Gregory Newton Formula along other algorithms or databases (like the OEIS) to predict the likelihood of the next upcoming integer :)
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  675.  @zacharyjoseph5522  It's a bit tricky to explain without diagrams, but I'll do my best. Let us call the whole shape G for glasses. First consider the pair of 2x3 rectangles at the extreme left and right of the glasses. If a domino is placed which crosses the border between the 2x3 and the rest of the glasses, then the 2x3 is left with a region of size 5, so cannot be covered. We therefore know that the dominoes covering the pair of 2x3 areas are necessarily wholly within the 2x3 areas, so the tilings would be the same if the 2x3 regions were actually disconnected from the glasses. So if we let G' be the glasses without this pair of 2x3 regions and if we let N be the function counting the number of tilings of a shape, then we have N(G) = N(2x3)^2*N(G'), as the tilings of the pair of 2x3 regions and the tiling of G' are independent. We'll need names for a few other things, so we'll call the boundary of a hole E for eye. So the eye is the 10 squares directly surrounding the hole. Consider the middle 2x2 in G'. Imagine a vertical line L cutting this 2x2 into two pieces. This line divides G' into two equal copies of the same shape, an eye with a 2x1 hanging off one side and a 2x2 hanging off the other. We'll call this shape E+2x2+2x1 Now if we consider tilings with no domino crossing L, then clearly the number is N(E+2x2+2x1)^2, as the tilings of the two shapes on either side of L are independent. If instead we have a domino crossing L, then we must in fact have 2 dominoes crossing L. This is because otherwise we'd create a region of odd size, which couldn't be tiled. In this arrangement with 2 dominoes crossing L, we again have two identical regions to tile, but this time they are E+2x2, following the naming convention for the previous shape. In this case there are N(E+2x2)^2 tilings. So we've shown that N(G') = N(E+2x2+2x1)^2+N(E+2x2)^2. Next we consider E+2x2+2x1. If the 2x1 area is covered by one domino, then we're left with tiling E+2x2. Otherwise, then the tiling is forced all the way to a remaining 2x2 region. Therefore N(E+2x2+2x1) = N(E+2x2)+N(2x2). Now we consider E+2x2. Imagine a line L' dividing the E from the 2x2. If no domino crosses L', then the E and 2x2 are tiled separately, so we get N(E)*N(2x2) tilings. If instead a domino crosses L', then the rest of tiling is forced, so we get 1 such tiling. Therefore N(E+2x2) = N(E)*N(2x2)+1. The remainder is not especially hard to check. N(E) = N(2x2) = 2, so N(E+2x2) = 2*2+1 = 5, N(E+2x2+2x1) = 5+2 = 7, N(G') = 7^2+5^2 = 74. Finally N(2x3) = 3, so N(G) = 3^2*74 = 666.
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  676. A joke for linguists: Carissimus Dei.
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  677. I was really worried about the kitten trapped inside the hypercube, but then beard-man appeared and used his Shadow-Squish Super Power and saved the day! What a great story!
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  678. In university I always struggled with this number theory problems and their proofs. Mathologer should be a movement for teachers all over the world
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  679. Yay! Another video - thank you. Can we get that one on Galois theory? Or did YouTube messed up again and didn't notify me and you already published it? Regardless - I love your videos!
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  680. Hooray I was sure that the videos would need to be reuploaded from scratch. Very glad to see the old comments were not lost.
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  681. Can't wait for them to hear about the rest of elementary number theory.
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  682. Absolutely! Masterpiece, one after the other!!!
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  683. The sophistication, simplicity and humor of these videos is always so delightful! Just wanted to say thank you!
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  684. I made it to the very end! And I actually followed everything you presented, cuz by the time you got to the p(n)(O-E) setup, I bursted aloud: "Some are zero, and the others will be pentagonal exceptions alternating between 1 and -1!!!" I felt sheepishly proud, but really, it was only obvious because the previous 47 minutes were presented so masterfully by you!
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  685. Draw the chord that joins the vertices of the two green segments (top). This gives an isosceles triangle, hence the axis of the chord is a bisector of the top angle of the original triangle. Hence this axis goes through the triangle's incenter, and this incenter is also equidistant from the two origial top points. The same is true for all three pairs of points. Now choose the top right and the most right point: both lie on a (green+blue) segment. Reason just the same as before and you get that these two points also are equidistant from the triangle's incenter. Rinse and repeat, and you get that the incenter is equidistant from all six points.
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  686. There's an extended formula for three dimensions which shows that meshes consisting entirely of triangles have a fixed volume even if they aren't rigid. Would love to see a video on that.
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  687. Having to write all your formulas as poetry sounds both amazing and horrifying at the same time. It must have made everything so much more difficult!
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  688. At first, the fact that the tree contained every irreducible fraction broke my mind. Then all of a sudden it seemed obvious. I can't explain why though.
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  689. 13:46 I was so annoyed that the pattern was that simple. I had worked out a completely different, more complicated pattern. The sums of the differences were always factors of the double position number they surrounded. (e.g. the position numbers surrounding 2 were 1 and 3, which sum to 4, which is double the original position number). Furthermore, the number needed to multiply the sum to get to double the position number went 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, etc. It was a pattern, just a way more complicated one. Now I'm going to have to try to prove that they are equivalent patterns, which may be quite difficult.
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  690. A fun problem to wake up to.
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  691. Tristan's proof is exactly multiplying by x+2. Wonderful. I wonder if there's a link between these generating functions and the genus of the figure they define.
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  692. Not even once Mathologer, can I make it through a video of yours without learning non stop, start to finish! People like you are truly some of humanity’s most valuable gifts. To think how many new physicists, engineers, programmers, mathematicians in the making that have been added to society, spear-headed by your influence, the value is clearly demonstrated as we are all benefactors of that reality.
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  693. Me: "Wait, am I missing something? It's not obvious to me that the "slap on ears" operation distributes over the components. Am I being dumb?" Video, almost immediately after I have that thought: "I should say something about that." :)
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  694. 11:56 challenge question: The Pythagorean triple (153, 104, 185) corresponding to the box [ 9, 4, 17, 13 ]. If you call the children A,B,C, the 153, 104, 185 is the A'th Child of 'CCC'
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  695. Watched the whole video, this stuff is absolutely amazing. I'm in 11th grade
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  696. 21:15 "Oh wow, this formula implies that the coefficients for 5n, 5n+1, 5n+2, 5n+3, 5n+4 must be the same!" thinks about original problem for a minute ... Oh.
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  697. is this a proof by the infinite descent? Nice. Also, 16:42 looks like a rope bridge.... perspective is crazy.
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  698. Awesome video. I learned late in life that this kind of math isn't something I'm naturally bad at, just something that requires more effort on my part than, say, writing an essay on Wittgenstein's late period thought. But then again, calculus is something that requires a lot of effort for MOST people. Anyway, it's great to have resources like this, which are obviously the product of a great deal of passionate labor on the part of Mathologer.
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  699. As an ex Maths teacher in UK the reason this doesn't get taught is simply because it is not in the curriculum, and to actually get through the curriculum leaves no time to play with Maths, or indulge the class in whatever the teacher finds interesting or stimulating (supposing he/she has a higher understanding of Maths in the first place!).
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  700. This didn't show up for me. Thought you might want to know that youtube seema to do some weird stuff again.
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  701. Most memorable part: the 100 zeros sum being larger than the no nines sum.
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  702. This was one your best videos ever Mathologer; thank you. I'm curious if anyone has the answer to the puzzle about the Red Cross at 2:18? Cheers.
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  703. Es hat wie immer sehr viel Spaß gemacht. Die Zusammenhänge der Funktionen geometrisch in dieser Form zu erleben ist wirklich der Hammer.
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  704. Great content, made my day! One curiosity about Horner form for polynomials: it is used in computers to actually compute the value of a polynomial, of degree say n, because it involves n multiplications and n sums, instead of computing all powers and summing up, that involves n sums but n(n+1)/2 multiplications. Horner is much more efficient, because of lower error propagation in numerical arithmetics at multiplications, and also requires less memory.
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  705. An alternate solution to the bugs problem: Let each of the bugs have velocity 1 unit/s. Split each velocity vector into a vector that points towards the center of the square and one that points 90 degrees to that one. Using the 45-45-90 triangle created with the original velocity vector and two component vectors, we find that the two new vectors have magnitude sqrt(2)/2. Thus the bugs are going towards the center at a rate of sqrt(2)/2 units/s. The center is sqrt(2)/2 units away so it will take the bugs one second to get to the center. Distance = rate*time so distance = 1 unit/s * 1 s = 1 unit.
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  706. 7:00 yes actually. In 9th grade (or, as we call it here, Class IX), there's an entire chapter in the book called Area of Triangle and it's simply filled with good old Heron. Respect from India
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  707. You're the only math youtuber that makes me audibly gasp every video, let alone multiple times. Keep up the amazing work!
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  708. I dislike, and object to, "What comes next?" questions on IQ exams. And so I appreciate your video as you elevate this to more of a science. I still think there should be no such questions on IQ tests. Or...here is my alternative approach...Give the sequence, ask what comes next, and give 4 answers, where each answer includes 3 or more terms of continuation. In other words, you have something to check against. Otherwise you are just testing a very obscure math thing, that has zero applicability and is proof of nothing. Rant over. Summary: "too many answers are possible"
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  709. Merry Christmas to Mathologer and to all!
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  710. I studied electrical engineering and at University we learned something called symmetrical components theorem, of wich we use the specific simpler case for triangles. In reality we're not taught it as a triangle, but rather, as the three components of our three phase electrical grid (be it current or voltage). You see, in the electrical grid, the 3-phase system is not always symmetrical (with 3 phasors of the same length and 120° apart) because of asymmetrical loads and asymmetrical faults and such. So it would be very complicated to model and visualize it as a 3 phase system because of the symmetries. So instead, we model it as a sum of 3 symmetrical systems (analogous to those 5 symmetrical base pentagons). And seeing them as triangles would be like one equilateral triangle, one "mirrored" equilateral triangle and one "triangle" with all 3 vertices in the same location. It's sad the in my university we didn't go into much depth with this theorem, we just saw a glimpse of it and were immediately learning the application. But it's really nice to now see a video about it explaining everything in detail in an easy to comprehend format.
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  711. Congratulations on 100 videos. Your videos are impressive, to say the least.
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  712. Exercising my bragging right - yellow.
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  713. I think instead of being a wheel, it's really the cross section of a log. I'll see myself out...
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  714. "Time for a cat video?" lol
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  715. I was so excited when I found this pattern with the exponents when I was younger Edit: yay you went over it
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  716. If the L-shaped part was called a "gnomon", I propose the term "kutos" (κύτος), for the 3D version. As in, the hull of a ship :-)
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  717.  @Mathologer  Fantastic to hear it. On subject of "easy" books. I have Polish book from 1946, written by one of the few survivors of Polish School of Mathematics. It is on Complex numbers. In less then 30 pages it takes you from "what is complex number" to "calculus on complex numbers". It is incredibly easy to follow and it's free. The only problem is that it is in Polish.
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  718. 200 hundred different proofs. Me, an intellectual: Sorry this margin is too small to contain a proof 🤷🏻‍♂️
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  719. The 2nd of the basic geometry theorems (that the inscribed angle is half the measure of the subtended arc) can also be used to easily prove the 1st: two opposite angles of a cyclic quadrilateral subtend arcs that clearly add to the entire circle, and their measures would then add to half of a circle, or 180°.
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  720. "Logarithm explode to infinity" What a violent explosion 😂
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  721. Nathan's digit proof was so much more elegant. The faces (mathologer's) proof was just based on observation.
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  722. This ruling is actually from the Mishna, although it was reprinted in the Talmud. In fact, most of the Talmud is essentially just a commentary on the Mishna. So on the page you showed (12:19), it essentially says "The Mishna says XYZ... Now here's various Talmudic rabbis' arguments over why the Mishna said that" You may notice two boldfaced words in the central column. The bit between the two boldfaced words is the Mishna quote, and the bit after it is the Talmud's discussion on it (Rabbi Shmuel said this, Rabbi Ravina said this, etc) You can look up Ketubot 93a for more. Sefaria has a nice translation
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  723. Hey I found 2019 in your π-shirt after the 244th digit after the decimal I am a man of culture
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  724. Ok, can we take a moment to appreciate the slide transition at 25:40? It's magnificent.
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  725. Nooooo!! It's a nice way to spend your time indoors... and humanity needs you right now... keep it up!
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  726. I was taught Heron's formula in high school, in Greece. I found it rather interesting, but considered it a curiosity, mostly, as it wasn't connected to anything else I was taught.
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  727. Awesome, Mathologer! Another ab-so-lute-ly delightful journey. Apart from its important and most beneficial applications, Mathologer never ceases to amaze with another revelation on how maths has just this amazing beauty and harmony in itsself. This channel is such a gem on YT. ✨Thank you very much, indeed, Sir. 🙏
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  728. "Why are manhole covers circular?" "Because manholes are circular." I wish I had the balls to give that answer.
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  729. I gasped out loud when he pointed out that the windmills pair up with each other. That was amazing
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  730. You remove all the tedium, replacing it with slick graphics, perfect music and your happy personality. Your love of math is the "answer" for us. Thank you!
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  731. oh just realised that's what you show at the end
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  732. nvm just found it in the description 😅
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  733. I learned Heron's formula in school in the US state of New Jersey in the 1990s. Unfortunately, it was not proven, though I did derive my own proof using Law of Cosines similar to the one in the video. I am always on the lookout for better proofs, and the one in the video is definitely beautiful. I love the way the exposition on the 345 triangle gently leads the viewer into the appropriate conceptual space. Well done.
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  734. Having the third part is silly. The second and third is just a log rule, but even worse the third gives up the joke
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  735. Given that there are 7! = 5040 possible orderings of the 7 proofs and a subscriber count in the hundreds of thousands, I can say with certainty that multiple viewers of this video will have the same ordering of most liked to least liked :)
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  736. Undergraduate mathematician here. The better I get at math, the more I appreciate your videos. These videos give a great visual experience which is generally not taught in proof courses. My favorite chapter was probably Chapter 5, reminded me of some of the concepts discussed in my analysis course.
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  737. "I can't be bothered to do the hard cases" in my day = "Left as an exercise." "I can't be bothered to do the hard cases" today = "Leave your thoughts in the comments."
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  738. The iris was interesting, but I'm here for Conway!
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  739. Made it to the very, very end - as usually.
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  740. I made it to the very end... ...and I liked it. I know a decent bit of recreational math and most Mathologer videos contain "something old, something new, something borrowed, something blue". But this one - apart from the concept of the partition numbers - open a new part of the math world. Thanks Burkard for coming up with these amazing and very followable adventures! 👍
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  741. I find it particularly interesting how a square grid can give rise to a circle... basically you get rotational symmetry out of something that is not... And why does it have to be L2 symmetry not some other Lp?
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  742. Speaking of Sophie Germain, my son and wife were looking up facts about the number 47, when I mentioned that 47 is the end of the Cunningham chain of the first kind with the smallest Sophie Germain prime as the initial number.
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  743. Shhhhh.... they don't want us knowing these things, hahaha. Thank goodness you are here!!!! What a time to be alive!!!
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  744. The reading at 6:14 is 3.18 not 3.14, but the magical thing really understood you were multiplying by pi :)
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  745. This is also why the two obvious ways to pack spheres are the same.
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  746. I feel like I watch this guy 20% for his amazing math demonstrations and 80% for him laughing at his own jokes <3 thanks for the incredible content! Math is an amazing thing!
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  747. Wunderbar, diese Wiederbelebung der fast in Vergessenheit geratenen Entdeckung von Steinbach! Wie immer, Vergnügen pur und besten Dank!
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  748. Yeah, ancient Indians loved converting everything and anything into Sanskrit verses. If you don't know how to properly decode it (like most people today), there is a high chance that you will mistake it to be a flowery story or poem. I think they did this so that the students could memorize easily. Edit: Oh, and being a Malayali myself I am surprised I couldn't completely read the manuscript. It is so fascinating that the script changed a lot with time!
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  749. Second challenge: it is obvious that in this hexagon there is the same number of dominoes of each color because transposing this 3D volume(from isometric axonometry) into projects on the XOYZ axes we obtain identical squares on OX , OY and OZ planes so an identical number at any scale of dominoes ;(and whenever we place the dominoes , the projections will always be squere).
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  750. A fun intuition for why the frozen sections show up with high probability: Imagine the left corner of the n'th Aztec Diamond has a horizontally-oriented block on it. If you draw it on paper, you will see that this completely determines that the entire left side of the diamond is only horizontally-oriented blocks, what remains undetermined is nothing more than the (n-1)'th Aztec Diamond. So the number of configurations with a horizontal domino in the left corner is equal to A(n-1), which is fairly intuitively a small fraction of A(n).
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  751. Whenever you feel marooned in a parallel universe and nothing seems to remind you of Kansas, five words will snap you back into familiar territory, the land of reason and kindness - "Welcome to another Mathologer video!" Or three notes for that matter.
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  752.  @friedrichschumann740  Selbstverstaendlich! 🙃
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  753. "Mathologerization" The act of explaining Math "Mathologerers" People that attempt to follow along, and never give up. "Matholgerized" The Math, explained
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  754. 2:15 Let (a, b), where a≥1 and b≥0, be the lowest side of a square written in vector notation. Precisely, it is the side containing the bottommost (and leftmost in case of a tie) vertex as its left endpoint. This enumerates all possible squares uniquely. There is room for (n - (a+b))^2 such squares in the grid, so the total number of squares is (A2415 on OEIS): sum[k = 1 to n-1] sum[a+b=k | a≥1, b≥0] (n - k)^2 = sum[k = 1 to n-1] k (n - k)^2 = n^2 (n^2 - 1) / 12
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  755. our monthly dose of underrated mathematics
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  756. I'm not a math historian but I believe that it is because the Greeks studied sections of the cone. If you cut the cone parallel to the base, you get a circle. If you cut the cone oblique to the base, you get an ellipse as long as the cut is not parallel with the sides. If you cut the cone parallel to the side, you get a parabola. If you cut it more oblique than the sides, you get a hyperbola. https://en.wikipedia.org/wiki/Apollonius_of_Perga
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  757. For the curious, r and s are solutions to cubic equations, so WolframAlpha present them with formulas involving complex numbers, even when the final result is real.But when asked to simplify, it found a closed form using trig functions: (1 + Sqrt[7] Cos[ArcTan[3 Sqrt[3]]/3] + Sqrt[21] Sin[ArcTan[3 Sqrt[3]]/3])/3
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  758. The pigeon that disliked this video got stuck in pigeonhole
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  759. maybe once explain us the Poincaré Theorem proof :D
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  760. This method likely originated due to lack of proof of claims which would have been very common in the past as documentation was not robust in the past, it assumes that all claims are not completely true and are likely with equal probability and we end up with this type of distribution.
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  761. I'm just simply amazed that it never occurred to me that solving "next element of sequence" puzzles on IQ tests is actually just applying differential calculus. I guess I just never gave it deeper thought... Amazing video, great explanation :)
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  762. 37:25 Proof: Let n be the number of cards, Let k be the number of cards bigger than the last one. By transferring the last card to the other side i eliminated k instances in which sometimes bigger than it is before it but also added (n-1)-k instances in which its bigger the other numbers (because the rest of the numbers are smaller then it). So eventually i just added n-1-2k . Since its a power of two i can ignore the -2k (its even). Now, i am left with just n-1 added to the power. If n is even then n-1 is odd and therefore the sine changes. If n is odd then n-1 is even and therefore the sine doesn't change. [|||]
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  763. Due to covid, I finally managed to have enough time to watch a whole mathologer video :)
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  764. The simple part: any odd number n that can be written as the sum of two squares must be the sum of an even square a^2 and an odd square b^2. Now a^2=0 (mod 4) and b^2=1 (mod 4), so that n must be 1 (mod 4).
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  765. The most memorable part was Tristan's fractal. Fractals are beautiful and they always show up when you expect them the least.
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  766. Bonus problem: Show that the sequence shown at the start of the video (x_k = 1, 2, 4, 8, 16, 31, ...) is the maximum number of pieces that can be formed by slicing any convex four dimensional polyhedroid using k–1 hyperplanes.
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  767. I just spent twenty or so minutes trying to make a number-wall for the partition function by hand. I may have given myself a mathematical concussion in the process😵Great stuff! Maybe next time I'll start with a simpler sequence though 😂😭
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  768. Actually you don't even need to know anything about binary numbers. The proof is equal even if we read the number as decimal.
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  769. I love watching your videos even though this is way out of my field. You make math very interesting and dare I say it? Fun. Thanks for the hard work.
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  770. ​ @Mathologer  Yeah but I heard his disciples never got the angles right...
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  771. i feel like 4k+1 and 4k-1 is much prettier
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  772. Definitely love the short form video option. I have about 50 of your videos in a list to watch later, but I rarely have time for a 45 minute video. (I'm slowly getting through the "mystery of the sums" one... lovely stuff).
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  773. Always nice when you upload a new video!
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  774. Merry Christmas! 6:11 - You could pick the two blacks in the top left corner. It would isolate the corner green square, so not every combination of 4 squares removed is tileable. 10:13 - say m = 2p-1 and n=2q-1. The denominators in the cosines will be 2p and 2q. Carrying out the product, when j=p and m=q, we will have a term (4cos²(π/2)+4cos²(π/2)), which is 0, cancelling out everything else 13:50 - Lets say T(n) is the number of ways to tile a 2xn rectangle. First two are obviously T(1)=1 and T(2)=2. For the nth one, lets look at it from left to right. We can start by placing a tile vertically, which will isolate a 2x(n-1) rect. - so T(n-1) ways of doing it in this case. If we instead place a tile horizontally on the top, we will be forced to place another one directly below, so we don't isolate the bottom left square, this then isolates a 2x(n-2) - so T(n-2) ways of doing it in this case. ---- We have T(n) = T(n-1) + T(n-2). Since 1 and 2 are fibonacci numbers, the sequence will keep spitting out fibonacci numbers 14:33 - It's 666. I did it by considering all possible ways the center square can be filled and carrying out the possibilities. It was also helpful to see that the 2x3 rectangles at the edges are always tiled independently. I was determined to do all the homework in this video, but hell no I wont calculate that determinant, sorry 30:12 - I'll leave this one in the back of my mind, but for now I'm not a real math master. I'm also not a programmer, but this feels like somehting fun to program 37:28 - Just look at the cube stack straight from one of the sides, all you'll see is an nxn wall, either blue, yellow, or gray
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  775. I have a simple solution to who found this proof: By the elimination of variables. As a variable, I eliminate myself. Good luck with the other 7 billion variables.
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  776. solution to there bricks with overhang of 2 units: place one brick with overhang of 1. then place another bring on top of this one all the way to the right with its own overhang of 1 unit. clearly this will fall. now place the third brick to the left of the second, making the top layer 4 units (2 bricks) long, and the bottom layer centered around it. Done.
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  777. Just remember to use multiplied triangle instead of measuring just up to five what ever units you're using. For example measure to 30, 40, and 50. The larger the triangle, the less chance there is for the inevitable measurement error when doing it haphazardly by hand.
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  778. Gotta say, as someone who knows a fair number of artists first-hand, seeing AI "art" being used is always a bit gut-wrenching knowing what they are going through right now because of AI.
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  779. im already a big fan of long-form content in general, but this channel has such a distinct and enjoyable format that it easily sits in the top of the list. instead of continually building to a large and complex problem/explanation, its like you meander through a selection of related small problems that can be appreciated on their own or as a collective. it makes them very nice to watch and rewatch
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  780. And turns out, there is a 3D equivalent for magic squares called a magic cube. Going beyond 3D, we have magic hypercube. How can I never hear of this before? I love that the well-known classic problem has a lot of different variations to tinker about. Especially when each has a unique approach to the problem. The wonder of math always amazes me.
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  781. was this the first STRAND type puzzle?
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  782. HELLO OUT THERE...OUt there...There...ere...
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  783. The level of video and animation is amazing. This is a huge and talented work! СПАСИБО БОЛЬШОЕ 👌
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  784. I don't understand why so many people are afraid of calculus. Calculus, geometry, analytical geometry and linear algebra were the easiest math disciplines I had. Anyway amazing video, I did maths years ago, I abandoned it for software development, but I love how I'm always learning something new with your videos.
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  785. Whenever I feel smart, I watch Mathologer to knock my ego down a few pegs.
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  786. Nice one! I have also done the versions of AM-GM and Cauchy Schwarz like you did here - love this diagram for those visual proofs! Used it to animate Priebe’s and Ramos’ sine of a sum too. Was gonna create a mashup of them to show they are all the same but here you’ve done it better. Thanks!
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  787. 19:32 Can't believe you missed using the color sequence Red, Black, Blue, Yellow, Pink. This would be a delightful easter egg to the Power Rangers fans.
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  788. If the tower has an odd number of discs, first move the smallest to the place you want the tower to end up. If even, move to the space you don't want the tower to end. Then just consistently follow the direction you moved the smaller disc to (clockwise/counterclockwise if arranged as a triangle, to the left/right with wraparound if arranged in a line).
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  789. Thank you the video was fantastic!! I never heard of number walls before, and the graphical presentation was very informative. I look forward to learning more from the links.
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  790. Simple solution for the sum of 7-star angles: 1. Without loss of generality, assume the 7 points lies on a circle 2. Those 7 angles are produced by 7 unique arcs that form a whole circle 3. For a certain arc, the angle at centre produce by that arc is always twice of the angle at circumference (Theorem "Angle at centre is twice angle at circumference") 4. Therefore, twice of the sum of all angles is 360 degrees. 5. Sum of required angles = 180 degrees
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  791. 50 minutes that didn't feel like 50 minutes. I was ready for you to get into the zeta function and gamma (both the function and the Euler-Mascheroni constant). Re: your request at the end: I'm coming up on 30 years out of high school and I haven't taken a formal mathematics course other than introductory statistics since 1992. I got into recreational mathematics thanks to the late Martin Gardner and a few other writers, and then I found channels like yours and 3blue1brown on YouTube a few years ago. Lately I've been pondering the Riemann Hypothesis and the Collatz Conjecture, and I'm currently re-reading Julian Havil's book Gamma, about the Euler-Mascheroni constant. My instinct is that it's transcendental, but I'd love for somebody to prove it. I'm really looking forward to seeing what you have to say about gamma and the zeta function... and take as long as you like on those topics! Edit: Yes, I've seen your previous videos on Riemann and related topics. Also, something I remember discovering while playing with numbers on my own in high school algebra class came to mind. I remember finding that the perfect squares had differences of consecutive odd numbers, and then when I took this to higher powers I found the factorial of the exponent at the bottom of the formula, but I was never able to complete the formula for the full general case. I still have my original spreadsheet file from 1990 with my calculations.
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  792. this is not the reason why this system was introduced actually, remember that those times written records of things were perishable or even not present at all, so the money that was claimed had no solid proof on paper but only testimonies from witnesses, there was a probability of claim being exaggerated so it tries to punish high claims without solid proof, imagine that someone claimed 90% of the money without solid proof, it would be unfair to just equally divide on that basis because the others would only get like 10% of the money without solid proof that this distribution is fair, the talmudic system assumes that claims are not 100% accurate like that in the case of the inheritance problem in the video, its a silly system anyways because claims without proof truly have not real fair way to be resolved.
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  793. That was a wonderfully powerful intro. So beautiful and fully deserving of the absolutely epic music.
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  794. In information theory there are two related problems where the solutions are similar to the topic of this video. One problem is like, you have three information channels with different noise levels, how should you split data to transmit in each channel to maximize the total data transfer speed. The solution is called “water filling algorithm”. There’s a dual problem with the solution as “reverse water filling problem“. Such problems don’t have any game theory flavor and the objective is very clear. The solutions are also analogous to communicating vessels.
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  795. Your concerns are well founded. I was a math and science teacher at two private schools in the United States (each time for two years). The textbook selections never aligned well with my ideals of making math and science relevant, useful and entertaining. For 3 of the four years, I was able to accomplish that in spite of the books. During the 4th year the academic police state finally caught up with me, insisting that the reasons we teach anything are foremost to increase test scores and grow our market share through test-based reputation. That authority banned all non-standard curricula and forced me out of the profession I loved. All the texts from which I taught were ostensibly aligned with their goals. They were also filled with endless drills, BS examples, incomplete history, and frankly serpentine reasoning far more likely to confuse that to convey any valuable understanding. I’ve lived 4 years in Germany (which was a bit better) and 4 more in South Korea (which seemed much better). Unfortunately, the Korean kids were under immense pressure to perform, making almost all of them profoundly depressed and/or stressed from perhaps 10 years of age. Most of the great contributors to the betterment of our world recognize the importance of conquering fun challenges. Both mindless, meaningless repetition and idiotic complication turn beautifully curious and malleable children into miserable adults. When teaching became primarily the indoctrination of future workers, it necessarily ceased teaching young people to think critically and creatively in favor of teaching them to do what those in authority tell them. This bodes terribly for the future and causes me to be deeply concerned for our posterity. Thank you for making learning the entertaining challenge it’s meant to be!
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  796. This is amazing. Everything truly is connected to everything. I could hardly have been more surprised if Pascal's triangle had also made an appearance. 🙂
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  797. It makes sense that the area is the derivative of the volume, if you think that the volume is created by adding all the surfaces of the spheres with smaller radiuses. Basically like blowing up a perfectly spherical balloon. That trick should work for any shape, not only spheres. For example a cube defined by the 8 points (+/-r,+/-r,+/-r) has a volume of 8r^3. If you drive that by r, you get 24r^2 and that is exactly the surface of such a cube with side length 2r.
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  798. It would be nice to see a follow up filling in the details that were left on the table at the end of the video. A video on the connection between continued fractions, cutting sequences, and trajectories of billiard tables could also be a fun "spiritual successor". ;)
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  799. Tesla's 3-6-9 and Vortex Math is the key to Youtube views.
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  800. It's weird to see you mentioning falling powers, and their derivative... a long time ago I was bored, and started writing down the math for such functions to pass the time. I came to it because of the combination formula! It's very cool to see it actually has a functional purpose, more than I thought! I definitely noticed it could be helpful, but I never thought it could be THIS. This is so basic, and yet so important for the rest of calculus we learn in school, it's really amazing. Teaching this first, at least a little, could really help with people's intuition of Calculus.
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  801. In 1967, I took an Engineering Measurements course at U.C. Davis. One of the measuring tools was a polar planimeter which allows you to find the area of any figure by tracing its perimeter. If you have a plot of land drawn on a map. you can easily find the enclosed area. This is a very practical application of Heron's Formula.
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  802. We were meditating over these principles in 1988, when I was in 7th grade and we demonstrated tons of problems around that. How was this discovered only 10 years ago?
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  803. If there is any mathematician here that wishes he or she could go back in time so they could discover proofs for fame consider this: All of those past mathematicians would trade places with you in a heartbeat to reach the level of understanding we have now of maths and the world. We have these proofs, because of their frustration of the ignorance of the population around them.
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  804. Look up "Kitab ul Hind". It's a book written by an Arabic scholar when he was visiting India (India under the muslim turkic occupation). He translated knowledge into Arabic and took it back to the Islamic world in the 11th century. In fact, the so called "Islamic golgen age" wouldn't have happened if it weren't for their findings in India. Europe took that knowledge from the Arabs in the coming centuries.
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  805.  @Mathologer  yes I do. Because I love to make experts correct when they do something wrong. But man this video just blew my head. But my wit and will to make experts correct is damn huge, so I will continue to work on these summation formulas even though It will take a Quadrillion year! BTW Congratulations to Mathologer for hitting 500K. You know I really don't get sick of watching your video till end. I enjoy it. And trust me I mean it from my heart. !!! LOVE YOU MATHOLOGER!!!
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  806. The movie "Fermat's Room" is indeed excellent, I'm glad it's getting a bit of publicity!
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  807. A very weird rabbit hole I went down related to the four bugs problem: The function that gives the path length in terms of the step size is f(x) = x/(1 - sqrt(x^2 + (1-x)^2)), and we saw that the limit of f(x) as x -> 0 is 1. What I did was look at the Taylor series of f(x) around x = 0 and what I noticed was the terms up to x^15 all had positive coefficients, but from x^16 on the signs of the coefficients are periodic with period 8 (starting from x^19 the pattern is 4 +'s then 4 -'s). I think I understand why it's periodic with period 8; it has to do with the roots of x^2 + (1-x)^2 being (1+i)/2 and its conjugate, where the angle corresponding to (1+i)/2 in the complex plane is 1/8 of a turn. And I also sort of have a formula for the coefficients: if the Taylor coefficients of 1/sqrt(x^2 + (1-x)^2) are a[n] and the Taylor coefficients of f(x) are b[n], then b[n+1] - b[n] = a[n-1] - a[n] + a[n+1]/2. But the a[n] sequence itself is probably not expressible in closed form so I'm not sure how to get more info on b[n].
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  808. The vid has been up for 15 minutes. 1000+ views. 100+ likes. Just an ordinary fantastic Mathologer vid.
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  809. I like your calendar idea :)
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  810. Same energy: why does a sidereal year last an extra day compared to a solar year? You have to count the extra rotation of earth that results from its revolution around the sun. Or, the earth doesn't have to spin quite one full turn around its axis per day, since it will move around its orbit and make up the rest of the rotation (about 4 minutes per day)
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  811. I think the proof that the rational numbers (or fractions, actually, and thus rational numbers) is countably infinite is one of the most beautiful. But it's not included in the paper! What a travesty.
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  812. this channel is absolutely fantastic!
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  813. Thompson's book is superb. It's like being led through the steps of calculus by a friendly Victorian uncle
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  814. I'll call it a Parker Anagram.
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  815. The toy maker himself wouldn't expect this video to come out one day during that time
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  816. 0:35: "We're always on the lookout for the mathematical sole of things"
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  817. The -1/24 doesn’t really do much of anything anyway so let’s throw it out! Ha ha Ha ha ha.
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  818. Once more a greatly enjoyable introduction to a topic that is tangentially relevant to a lot of my mathematical interests but I knew little about. I can't wait to watch what you Mathologerize next (maybe some simplification of Mihăilescu's theorem? A big ask, I know)
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  819. If you don't already know that 10² + 11² + 12² = 13² + 14² then you can still calculate it quickly by noticing that it's approximately 5 x 12² = 720, plus an "error term" of (13-12)(13+12) - (12-11)(12+11) + (14-12)(14+12) - (12-10)(12+10) = (13+12-12-11) + 2*(14+12-12-10) = 2+8 = 10. In fact, in general, (a-2)² + (a-1)² + a² + (a+1)² + (a+2)² = 5a² + 10.
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  820. I have had difficulty with arithmetic since I learned to count. As a result I looked for patterns in numbers to make life easier. I learned the adding digits to find out if a number was a multiple of 3 and that a multiple of 9 always has the digits add up to 9. In fact I used a similar trick to add up a series of numbers from 1 to n by noticing that 1+n is the same sum as 2+(n-1) and so on...... I learned later in life that Gauss figured out this trick at age 9. I figured it out at age 11.
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  821. This day just got a lot better 🥰
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  822. The variations on the Harmonic series were definitely my favourite - who even thought to ask such a strange question as "What's the Harmonic Series, but if you remove all the terms with a nine in them?" It would never have occurred to me to ask a question like that.
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  823. 9:40 I don't know if I was taught it, but it always seemed trivial that this could be done from the moment I first heard about all this.
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  824. 29:49 The width of the white rectangle is sqrt(2) - 1 . Its height is 1 - (sqrt(2) - 1) = 2 - sqrt(2) = sqrt(2) * (sqrt(2) - 1) . This height divided by the width is: sqrt(2) * (sqrt(2) - 1)/(sqrt(2) - 1) = sqrt(2) .
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  825. Hi, I learned it in school in the GDR, in an extended math class in the begin of the 1970ths. It was rather interesting and fascinating.
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  826. just starting to really enjoy maths and discovered it as one of my passions, (doing hours everyday), your videos are very fun and useful! much appreciated!
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  827. 15:59 What is sed zeven? :)
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  828. As year 10 students were explicitly mentioned:- First the horizontal strech by a factor of 2, means that the input x was changed to x/2, hence plugging that in gives 2/x. The vertical squish divides that by 2, leaving us with 1/x.
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  829. 2 minutes into the video amd here's what's on my mind: the rule is the same as the average modulo 3, meaning (x+y)/2. If you assign a number to each color, and use this average, the answer will be the number for the corresponding color. For example: (1+0)/2 =1/2 = 2 (inverse of 2 modulo 3 is 2). Average of x and x will be x. This also reminds me of tricolorability of knots. (Or whatever it's called. I forget)
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  830. The amount of effort you put into making these animations are truly astonishing.
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  831. ich liebe dich
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  832. Sir please make a video on collatz conjecture and staircase paradox
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  833. This is the most mind boggling video ever made on this equation. and definitely a masterclass. How Ramanujan thought this is simply impossible to fathom. It appears that when it comes to Maths first comes god and then comes Ramanujan.
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  834. 4k-1 would have been more elegant than 4k+3, IMO
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  835. Something I was waiting for but wasn't mentioned: the horizontal symmetry, which can be explained by simply numbering the points with negative numbers going anticlockwise.
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  836. If mathematics is one side of the video, the music is the other. Thank you for both astonishing mathematic topic and making me discover so great music. <3
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  837. Her hair isn't even curly 😭
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  838. Lol I stumbled into it because I on and off for a few years looked at varying degrees of triangular sums and their closed forms. Of course I didn't arrive anywhere from that other than just saying "oh well isn't this neat". I eventually, looking at matrix algebra, blundered into the idea of using systems of equations to determine coefficients for the general polynomial solution which disappointingly contented my former self. Faulhaber's formula is what you want to look at if you want all the closed form solutions for polynomial sums/series. It's interesting.
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  839. Proud to b Indian❤️
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  840. It's uncountable because for any arbitrarily large integer n, there are 2ⁿ ways to pick the first n terms (either positive or negative) and then the remaining terms are fully determined by the over/under method.
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  841. no one expects the spanish inquisition😂
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  842. Thanks for the video! One more bump you might want to mention is that we're assuming continuous functions and sometimes assuming continuous derivatives. Works great for the Mathologermobile, which I assume can't teleport or go from 0 to 60 in zero seconds.
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  843. Exactly! I am almost 50 years old, and I can distinctly remember playing with these 'folds' in the EXACT same way as shown in the video when I was around 10 years old with my own arts and craft projects at home. In fact, I might still have it laying around somewhere in some boxes at the attic. I suspect when he says 'discovered' he actually means either A) officially described in some math paper, and/or B) a proof was found. Which are VERY different things than 'known/discovered' .
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  844. Really enjoyed it the first time around and looking forward to a second. Is junior Mathologer a grad student or your son or maybe even both?
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  845. I am a simple man. More Mathloger is better than less Mathloger.
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  846. Hey, I won gold because of this channel (long story lol), and have a suggestion for a small addition to the part two. The correspondence between the map of Lambert and Kepert is done by taking circle segments between two points, and varying the "angle" of the circle segment. 0° is a line segment, and 90° is a halfcircle, like used in the video. One of my own proofs, of the cylic quadrilateral angle theorem (that has undoubtedly been found by someone else as well) is that given a quadrilateral, you can look at the line segments like they are circle segments. The "circle segment quadrilaterals" have invariant α–β+γ–δ. Since you can merge two pairs of circle segments, you basically directly get the theorem. This even works for hyperbolic geometry! It is a nice proof, using unorthodox techniques, so it probably has to show up on this channel eventually, although it likely won't fit. It also has a dual theorem, where the opposing sides of a quadrilateral sum to the same length if and only if the quadrilateral has an inscribed circle.
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  847. Never thought I'd see such a detailed video on this topic. I've heard a little bit about all of these concepts (Aztec squares, Kesteleyn's formula, rhombic tilings, etc.) while watching Federico Ardila's great lecture series on combinatorics on YouTube, and I really think the accessibility of this subject benefits from visual-oriented, thorough, and intuition-driven videos like these. As always, great video.
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  848.  @Mathologer  Sure, just checking us. There goes my hypothesis that you started writing 120 degrees, then realized you'd basically already written it since cos 120 = -cos 180 - 120 = -cos 60, but left a half-written entry which got merged with the entry for 180 degrees.
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  849. Once you mentioned the higher dimensions I expected the Pascal triangle will pop up somehow in this videos.
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  850. When I was a kid, I created my own lacing, where each eyelet pair had one string going across, out the first eyelet and in the corresponding one, and that was all that was visible - no diagonal lacings. Everything else was under the eyelets and did not cross, and were thus invisible. I was very proud of it. I still lace my shoes with this style.
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  851. Just to go a step further, the perimeter of the shape described by the wiper arcs is exactly pi times the perimeter of the triangle i.e. (pi * the chord length), whereas the circumference of the circle is (pi * 2 * the radius) and, as noted above, it is easily seen that (2 * the radius) is greater than (the chord length). To check that the perimeter of the SOCW is (pi * the chord length), consider each arc length. If we label the triangle sides A,B,C and their opposite angles a,b,c, then the arc lengths are given by: Aa + Bb + Cc + (B+C)a + (A+C)b + (A+B)c = (A+B+C)(a+b+c) = (chord length)(pi)
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  852. Infinitely mesmerised by the guitar music. Then I consider, is the tablature some kind of code?
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  853. I had worked on this problem in my pre-university days. Though I eventually had to see the solution out (couldn’t solve it), it was my first deep exposure to Ramanujam’s mind and mathematical thinking. (The first time mathematical connect was obviously the introduction to limits concept). That was when I truly understood why he is called “the man who knew infinity”. No wonder, one of the greatest son of Indian soil 🙇
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  854. You can get the non-recursive function from a recursive function using the z-transform. This is used often in Digital Signal Processing problems.
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  855. Don't you get half of your claim, so 50 cents? You only get your 1 dollar if the estate is enough for both creditors so 1,000,001 dollars.
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  856. 7:02 yes it can. It's sufficient to look at the last two digits of a number to check if it's divisible by 4 since 4 divides 100. The last two digits were 81 which is one above a multiple of four.
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  857. I love these visual proofs. The animation of the visual proof at 22:40 is extra fun because two of the triangles are re-scaled. But they are re-scaled instead of having a more complicated animation. Well done!
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  858.  @Noughtgate  the actual amount of elastic force doesn't matter, only the fact that the free portion elongates uniformly while the portion in contact with the wheel remains fixed and does not slip.
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  859. 49:53 the answer is the odd-numbered Fibonacci numbers!
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  860. It was probably only published by someone 10 years ago. People playing around with triangles ABSOLUTELY have discovered this many times over history.
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  861. But the real question is: is that a Math or a Moth shirt?
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  862. My supernatural psychic sense is telling me that your card is... 👸🏻♥️
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  863. One of your finest videos in my opinion, I really enjoyed it!
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  864. I can't believe you mentioned Tree With Deep Roots. Knowing next to nothing about Korean or Korean, I stumbled onto that drama around 2011 and was intrigued by the story of Hangul woven into it. I learned to read it and eventually became somewhat proficient in Korean, all because of a random K-drama! Funny how those things work.
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  865. Lemme check the Nobel prizes... the high point of that guy's career was the photoelectric effect?? Sorry, not impressed.
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  866. It is a riff on "The Claw of Archimedes", a super weapon created by Archimedes, also called the "Ship Shaker". Look it up.
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  867. As a slide rule user, engineering in the 1960s, I found this a fascinating explanation but the graphics are absolutely brilliant, Even knowing what was going to be shown, I was mesmerised by the animation. So beautifully done.
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  868. I want a heart pls 😭😭
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  869. This is why I love your channel, a machine gun of “Aha!” moments makes me feel like Baldrick with a cunning idea. Love it!
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  870. What a joy to see this topic covered with such clarity! I wanted to make my own video about it a few years ago, and I went so far as to create animations in Desmos and tease the topic at the end of my video on Napoleon's Theorem. I came across the idea of using DFT's to explain Petr's theorem in a paper by Gregoire Nicollier, "Some Theorems on Polygons with One-line Spectral Proofs" (Forum Geometricorum Volume 15 (2015) 267-273). It was a lot of fun learning about it and trying to animate it. I just couldn't quite simplify it enough to make it accessible to my audience. So, thanks for the gift of seeing "my video" realized by a real pro!
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  871. 4. The necklace has 12 parts of equal length between the big pearls (? idk), which can be streched into a 3-4-5 triangle to check if an angle is right. Ropes like these were used in ancient Egypt to make right angles, though they weren't so cool-looking, just ropes with knots
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  872. Why not 4k-1 For 3,7,11,19.....
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  873. This kind of single move algorithms often show up in computer science and an often overlooked follow up question is if their were "multithreading" where if two subsequent movements occurred to and from different pegs they could be done simultaneously as 1 move. Obviously this would only apply to larger n but I would be curious if the solutions for that were actually superior given how ordered the solutions seem to be already.
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  874. I was already... not comfortable, but let's say "resigned"... to the fact that there exist very slowly divergent series; but the fact that there are very slowly CONVERGENT series, whose sum is impossible to approximate computationally within a reasonable margin of error, like the no nines series... was a shock!
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  875. What's really ironic is that when I was studying engineering, after 3 years of teaching me how to solve differential equations, they finally admitted that real-world engineering problems rarely have closed-form solutions. So that's when they started teaching numerical methods like Finite Element Analysis and Computational Fluid Dynamics! This video—which drew a parallel between sequences of discrete numbers and continuous functions—reminded me of that! Calculus is soft, silky, continuous, and exact. Numerical methods are rough, gritty, discrete, and approximate. I think that's why math departments continue to teach contrived exercises that resolve to beautiful, closed-form solutions. Whereas engineering departments are forced to tackle real-world problems head-on. Engineers need to arrive at reasonable solutions that are practical to implement; they need not be perfect.
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  876. Fantastic video! A more computationally-efficient method than the determinant method, which also easily and transparently deals with single and multiple zeros: Step 1: After using k delayed copies of the original sequence, instead of repeatedly computing determinants (as in the video), take a rectangular “snapshot” — this is a rectangular matrix of size k-by-n, with n > k. Step 2: Transpose this matrix (so now it’s n-by-k with n > k) and compute its “economy” SVD (singular value decomposition). There are many existing software libraries to do this. This step is the only computationally-intensive step, whose complexity is O(n k^2), which is much less than the determinant method as stated in the video, which is O(n k^3), or even more if care is not taken… Step 3: Now, look at the singular values: If one or more of the smallest singular values is/are zero, then we’re sure this is what Burkard calls “Fibonacci-like sequence”, something usually called “linear recurrence with constant coefficients” (which he mentioned in the video). However, if none of the singular values is zero, then repeat with larger k, i.e., more delayed copies of the original sequence. Step 4: The linear recurrence with constant coefficients is easily determined from the column of the SVD result corresponding to the zero singular value (because this is the vector which can zero-out any set of k consecutive elements of the sequence). Specifically, if the “economy” SVD result of the original n-by-k matrix is U, s, Vt, then the last row of the square k-by-k matrix Vt will be the desired set of linear constant recurrence values. This last row has size 1-by-k, of course. These values will probably need to be scaled by a common multiple if nice integer values are desired! Move the 1st element of this set of k values to the other side of the equation to use as a prediction recurrence equation, i.e., the next value is determined linearly from the previous k-1 values. Remark: The idea of a singular value decomposition (SVD) is mathematically very closely related to a determinant, because it essentially also determines linear combinations of columns (and/or rows). But it is more computationally-efficient in this case, because it can be computed for a rectangular set of values simultaneously, rather than small squares of values one after the other. The second advantage of the SVD is that it can also give us the linear recurrence with constant coefficients “for free” (from the same computational result).
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  877. Don't forget to bring a towel...
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  878. 1:51 Challenge accepted ! (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 = 91409924241424243424241924242500 Pretty faster that way !
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  879. I'm quite happy as long as the longer content still exists. You're one of the few STEM channels that continues to carry videos of 40 minute length or longer. Some of my favorite channels have gone from a 40m to 8/15 minute formats for the sake of the algo and it broke my heart. It's difficult to find a topic you love, and then have to find a new video on that same topic after just 10 minutes. However...it's also nice to sprinkle in those shorter vids. These are the Tough questions that try men's minds. As Felix Unger once said..."Let it be on your head." 😁
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  880. Noticed another "super pattern" for the squares. Looking at the base of the last term of one row, you can derive the base of the first term of the next row. Multiply the base by (n+1)/n, where n = the row number. For example, the base of the last term of the 1st row is 5, and 5 * 2 / 1 = 10, which is the base of the first term of the second row. 14 * 3/2 = 21, 27 * 4/3 = 36, etc...
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  881. 0² + 1² = 1 1² + 2² = 5 0² + 3² = 9 2² + 3² = 13 1² + 4² = 17 ?² + ?² = 21 0² + 5² = 25 Is 21 really a "good" number?
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  882. I’m stunned. What a sublime concept, especially the animations that produce the cool little fractal trees
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  883.  @AlexanderQ689  A short video on change for a change :)
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  884. I am so happy to have access to such great content without any charge. I love mathematics so much and this satiates my curiosity! Looking forward to more of your amazing work ❤️
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  885.  @enderyu  If all the creditors divide their loans into equally-sized chunks, it is equivalent to proportional. Using the example in this video, if creditor A has 10 loans of $10 each, creditor B has 20 loans of $10 each, and creditor C has 30 loans of $10 each (so that's 60 loans of $10 each) and the estate has $200 left, then each $10 loan gets $3.33 paid back. Creditor A gets $33.33 in total, creditor B gets $66.67 in total, and creditor C gets $100 in total.
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  886. 1 cubed plus 2 cubed equals 3 squared. (10sq + 11sq + 12sq + 13sq + 14sq) /365 = 2 because 10 sq + 11sq + 12sq = 13sq + 14sq and 169 + 196 = 365. 3cubed + 4 cubed + 5 cubed = 6 cubed.
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  887. Mathematician looks at 3:10 - "hey, neat, the area proof of pi!" Engineer looks at 3:10 - "hey, look, nuclear reactor tiles!" ;)
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  888. Related, cryptographic hash collisions must exist. Finding them is a bear, so many damn pigeonholes 😆
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  889. Burkard, rather than heap additional praises regarding the dependable brilliance of your channel's content, I thought I might remark on your delivery: Your pacing, pitch, tone and inflections are such that, rather than lulling my mind to sleep in class, you have a special way of always educing and coaxing my brain forward (stubborn mule that it is).
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  890. Best thing about you, you always tell the real mathematician of the theorem. 😀👍🏻 BTW merry Christmas🥳🥳
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  891. This feels like a deliberate deep dive into the slightly shoddy limit taking at the end of the previous video. Brilliant!
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  892. This is mathematical proof that there's the right way and the wrong way of cutting corners!
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  893. Interesting, I noticed years ago that if you free hand draw the best pentagram that you can manage and keep connecting the points of the pentagon in its middle into another pentagram and keep doing this you will quickly end up with horribly crooked pentagrams. I hypothesized (🧐) that what was happening was basically an amplification of whatever small inaccuracies were already in the pentagram that I started with and that if I had drawn a perfect pentagram the inner pentagrams would also be perfect forever.
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  894. Let's collect some interesting comments and remarks as they come in (also check out the description of this video): VincentvanderN I don't know if this is already down here somewhere in the comments, but 2 x 2 = 2 + 2 can be used to show that there are infinitely many prime numbers. We generate a sequence a_1, a_2, a_3 etc by starting with a_1 = 3 and a_{n + 1} = a_n^2 - 2. Now we use our freak equation to show that if q is a prime divisor of some a_m, it cannot be a prime divisor of a_n for any n > m. (And then, as a result neither of an a_n with n < m, because otherwise we could repeat the proof with n in the role of m and get a contradiction.) Here is the argument. Look at the sequence a_m, a_{m+1}, ... modulo q. We have a_m = 0 mod q, a_{m+1} = - 2 mod q, a_{m + 2} = 2 mod q, and then, by the freak equation 2 x 2 = 2 + 2 (in the form 2 x 2 - 2 = 2) we get that a_{m + k} = 2 for all k >= 2. Neat, right? I believe I learned this from Proofs from the Book. charlesstpierre9502 Use 2x2=2+2 to make a formula for generating Pythagorean triples. Start with two positive integers m > n. Then (2×2)(mn)² = (2+2)(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² + (m⁴ - m⁴) + (n⁴ - n⁴) (2×2)(mn)² = m⁴ + n⁴ + 2(mn)² - m⁴ - n⁴ + 2(mn)² (2mn)² = (m² + n²)² - (m² - n²)² Set: a = m² - n² b = 2mn c = m² + n² Then: a² + b² = c² This will not work for: aᵏ + bᵏ = cᵏ; k > 2 franknijhoff6009 Hi Burkhard, the 3-variable equation plays a role in integrable systems. In fact, it appeared in Sklyanin's work on quadratic Poisson algebras (around 1982) providing solutions in terms of elliptic functions, and (with an extra constant) as the equation for the monodromy manifold of the Painleve II equation. Sklyanin's paper is: Some algebraic structures associated with the Yang-Baxter equation. Functional.Anal.i Prilozhen 1982 vol 16, issue 4,pp 27-34 ; look at equation (27). Furthermore, in L.O. Chekhov etc al., Painleve Monodromy Manifolds, Decorated Character Varieties and Cluster Algebras , IMRN vol 2017, pp 7639--7691 you can find in Table 1 a close variant of the 3variable equation with extra parameters. I think it would be interesting to explore this idea with exponents. For example 1+1+1+1+2+3=3^2^1^1^1^1 2¹×2¹=2¹+2¹ 3½×3½×3½=3½+3½+3½ 4⅓×4⅓×4⅓×4⅓=4⅓+4⅓+4⅓+4⅓ 5¼×5¼×5¼×5¼×5¼=5¼+5¼+5¼+5¼+5¼ ... tanA + tanB + tanC = tanA x tanB x tanC (that's the identity that features prominently in the Heron's formula video) For completeness sake and for fun let's mention 2⁴ = 4² 2 + 2 = 2 × 2 = 2² log(1+2+3)=log(1)+log(2)+log(3) log(2+2)=log(2)+log(2)=2log(2) https://www.mtai.org.in/wp-content/uploads/2023/09/IOQM_Sep_2023_Question-paper-with-answer-key.pdf ... 29th question in an Indian maths olympiad problem Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) http://www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem 5.1. says If there are exactly two sum-equals-product identitites , then one of the following two conditions is true: 1. n - 1 is a prime and 2n - 1 \in { p, p^2, p^3, pq } , 2. 2n - 1 is a prime and n - 1 \in { p, p^2, p^3, pq } , where p, q denote prime numbers. Why don't we start out the sequence of basic sum-equals-product identities with 1=1? Well, N=N for all N :) Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :) For this video I played around with generating some AI animated photos of Sophie Germain using https://www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz https://www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :) Not perfect, but will be interesting to find out what the machines can do in 6 months time (and at what point the machines can make better Mathologer videos than the Mathologer :( In general quite a bit of AI hate in this comment section. Strangely I've never had people complain about me using any of those generic white men with beards images of ancient mathematicians that museums are full of. Sophie Germain primes and safe primes (the 2n+1 primes) are important in cryptography, those involving prime fields. Because if the size field is a prime p, the size of its multiplicative group is p-1. For the group to be secure, the multiplicative group must have a large subgroup of prime size so p-1 must have a prime factor. So, having a prime q such that 2q+1 is also a prime is a good candidate. Even though it is not strictly required as long as p-1 has a big enough prime factor, this is what students are taught as a start. Kaprekar's constant is among the lengths corresponding to exactly two sum-equals-product identities (discussed at the end of this video :) https://en.wikipedia.org/wiki/6174 Cunningham chains are an interesting generalisation of Sophie Germain primes (chains of primes such that the next prime in the chain is always the previous one times 2 plus 1). https://en.wikipedia.org/wiki/Cunningham_chain If you are happy to also play this game with negative integers then you get more solutions. In particular, since (-1)+(-1)+1+1=0 and (-1)x(-1)x1x1=1, you can splice these two blocks into a/any sum-equals-product identity of length n to arrive at a sum-equals-product identity of length n+4. And for complex integers we've also got things like this: (1 - i) + (1 + i) = 1 - i + 1 + i = 1 + 1 -i + i = 2 + 0 = 2 = 1 + 1 = 1 - (-1) = 1 - i^2 = (1 - i)(1 + i) 2xy - (2+x+y) +1 = N-2 <=> 2xy - 2 - x - y +1 = N-2 <=> 2xy - x - y +1 = N <=> 4xy - 2x - 2y +2 = 2N <=> 2x(2y - 1) - 2y +2 = 2N <=> 2x(2y - 1) - (2y - 1) = 2N-1 <=> (2x - 1)(2y - 1) = 2N-1 Probably the easiest way to demonstrating this equivalence is to go backwards and start by expanding (2x-1)(2y-1)... Relevant Project Euler problem: 88 https://projecteuler.net/problem=88 Relevant Online Encyclopedia of integer sequences: A033178 @Frafour Tangentially related to 2x2 = 2+2: there is this video of Gromov (Gromov: 4 = 2 + 2 as "proof of Donaldson's theorem") https://youtu.be/bgePMb8wxv0?si=DmTf2wzbIr1F21f8 observing that 4 = 2+2 in 3 ways. That is, there are 3 ways to partition a 4-element set in two 2-element subsets. The fact that 3 is smaller than 4 is unique to 4, and produces a surjection from A_4 to A_3 - the alternating groups on 4 and 3 elements. This shows that A_4 is not simple, 4 is the only number where this happens, and this is responsible for many weird things in dimension 4. In another direction, I first encoutered Sophie Germain primes accidentally when learning group theory in my undergrad. There is an elementary proof of simplicity of the Mathieu groups M_11 and M_23, which actually gives a general simplicity criterion for subgroups of S_p, p prime https://www.jstor.org/stable/2974771?origin=crossref. When I read this I wondered if this ever works for proving simplicity of the alternating group A_p. If I remember correctly, it works precisely when p is a Sophie Germain prime! @YSCU261 The roots of a quadratic equation of the form x^2-bx+c=0 satisfy the following equations : r1+r2=b r1*r2=c So we have that the solutions of xy = x+y can be expressed as the solutions to the quadratic equation x^2-bx+b for all b this can be extended to x+y+z=xyz and so on with vieta's formulas
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  895.  @cheater00  Every positive integer is a some of 4 squares. https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem A positive integer is a some 3 squares if and only if it not of the form 4^a(8b+7). The "only if" direction is easy. https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem Finally, a positive integer is a some of 2 squares if and only if the exponent of each of its prime factors of the form 4k+3 in its prime factorization is even. (This could have been mentioned in the video.) https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem
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  896.  @Josh-dj1bl  You change the structure by permuting infinite terms, that doesn't guarantee the sum stays the same. And there are already uncountably many real numbers that must be covered by those permutations. My intuition is that it is countably infinite
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  897. I understand every word you say, but your sentences are mostly beyond my understanding! I am amazed at how you can explain these issues (I last studied Math over 50 years ago, but remain fascinated).
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  898. Ja bitte!
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  899. I remember going through a full and rigorous proof of the euler characteristic formula in graph theory, and all I recall was it being quite a doozy! Enjoyed your mathologerized version very much.
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  900. Fascinating. With the crushed cans and cylinders at 22:00, it's like the can or cylinder is trying to retain its original surface area while being crushed smaller, so it tends towards a Schwarz lantern shape. For what you said at the end, I didn't feel that anything necessary to understanding was left out. Perhaps because the "staircase diagonal = 2" idea is familiar to me, I actually felt there was more explanation than necessary, but perspectives may differ so don't take that as a criticism.
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  901. 12:00 Woah!! That was awesome!!! I just finished taking my Random Processes course as an Electrical Engineer last semester, and I'm so glad I watched this video after having taken that, cuz that was brilliant!
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  902. I am a retired Engineer old enough to have been trained in the art of using slide rules. I have owned several, including one with a Pi folded scale that avoided "falling off the end". However, my most treasured slide rule is one I inherited from my father who used it when he worked as an accountant. It is a cylindrical slide rule with the scale wrapped around two one inch diameter concentric cylinders. The upper cylinder slides within the bottom one and they are overridden by a "cursor" cylinder. On the upper cylinder there were two cycles of 1 to 10, on the lower cylinder just one. The cycles consist of 22.5 turns around the cylinders giving an equivalent linear scale length of nearly 1.8 metres. Given this length, four significant figures was easy and five could be reliably interpolated. Interesting that an accountant would only need answers to five significant figures!
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  903. I wish everyone talking about the harder sciences (physics, chemistry, etc.) and math spoke English in a German accent. It seems appropriate. Additionally, biologists should speak English with a British or American accent and Philosophers should speak English with a French accent. Am I crazy?
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  904. Fröhliche Weihnachten! You also can get this formula by using a function where pi is hiding in prime regularities, 3Blue1Brown did a similar video about it :D
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  905. The most memorable was the optimal towers, as I always thought that the leaning tower of lire was the best way to stack overhangs. It looked so perfect that I never questioned if there was a better way to do it!
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  906. Yeah, every time there's an attempt to teach more theory in American math classes, a lot of parents get angry because they don't know how to help their kids with their homework. It happened with "New Math" in the '60s, and it happened with Common Core in the 2010s.
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  907. For the bugs distance problem: Fix the frame of reference to one bug. The bug moving towards it, in that frame of reference, just travels along the shortest distance towards it, meaning the covered distance is exactly 1
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  908. Viewer in 2030: "what is "change"?" Viewer in 2040: "what are "dollars"?"
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  909. It's a pie chart of Benford's law. I have an Otis-King helical slide rule, the scale is about a yard and a half long but wrapped around six inch rods that slide telescopically.
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  910. I just discovered this channel a couple hours ago, and I have spent over 3 hours in a row watching your videos about sequence, derivatives, and of course trig and log .. it’s funny because I’m not really a math person because I feel like knowing something by just memorizing its efficient enough to apply it on a test but not into real life so that really doesn’t motivates me into doing maths. So it’s nice seeing a deep explanation of these concepts and not just something superficial.
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  911. Mathologer bingo: “This was shown to me by a friend of mine” “… that not even most experts are aware of!” “Why was this lost/forgotten/undiscovered for X years?” Or “why is this not taught” on the thumbnail Banter with Marty Shirts
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  912. It's gonna be yellow. Literally only took me a second (cause I looked at your previous examples and noticed something)
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  913. I checked your channel three times this week, I was expecting this, LET'S GOOOOO
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  914.  @Mathologer  thanks to you,now more people will know him.
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  915. If it is every month I think i missed some videos, going back to watch them
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  916. Wow, this is so amazing and beautiful! That's why I love maths and especially number theory, thank you for that video!
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  917. Knowing that Mathloger might expand was my best recieved Christmas gift so far
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  918. Thank you, Mathologer for your wonderful videos! David Wells's survey sadly omits Cantor's diagonalization, which, in my opinion, belongs no lower than position 2 on his list of most beautiful proofs. Cantor's proof is also the granddaddy (through Goedel) of Turing's proof of the undecidability of the halting problem (which also sends chills down my spine whenever I read it), and which ushered in the field of computer science.
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  919. A new mathologer video? Awesome :D
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  920. 1:27, now that's what I call "completing the square!"
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  921. Why do you always seem to post a video on these topics after I finish studying it
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  922. Most memorable: The most efficient overhanging structure being the weird configuration instead of an apparently more ordered one.
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  923. You lost me at "turn any one of the green circle into an orange circle", im colorblind, couldnt se any diference between that kind of green and that particular orange, a regular red should have been more noticeable 😅 But that didnt stop me for keep watching the demostration, amazing as ever, its nice to have you back. 👍🏼💪🏼
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  924. Videos are back :D
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  925. 8:54. I never applied myself in school at all but watching your videos the past year or so has been enlightening! I was a bit intimidated I don’t remember the first topic I saw but I noticed you are a great teacher! … Anyway the time stamp is because I knew it was going to have something to do with a circle lol.
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  926. The number of hairs in your head is, of course, your "Hairdős Number" (sorry...)
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  927. this is a great presentation. easy to understand and breaks down seemingly mysterious mathematical intuition. thank you!
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  928. First off, the video wasn't explicit about the condition of 'terminate.' This can confuse some, but you can infer it from the condition of choosing a facedown card, which infers that if there are no facedown cards there is nothing suitable to choose. From there, it's even simpler than math[s]. It's logic. If you have a finite set, and your task is to find one thing in a certain condition, change that condition and also change the condition [regardless of what that condition is] of a related thing, eventually you will run out of things in the [unchanged] condition, because the one definitive move of every action is from unchanged to changed.
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  929. I used to do the tower in the head on the way to school, three pegs and I managed up to 8 discs traveling in a tram. The difficult part was the nice looking girls. Just try that.
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  930. I think all proof papers would be better if instead of QED they ended with "ta-dah!"
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  931. I can’t get over how big the dude’s forehead in the thumbnail is.
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  932. Mathematics and its magic it's terrifyingly beautiful!
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  933. Truly a polite and precise explanation of Pi easy to understand youtube video of the day.. Great work Mathologer 👏
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  934. I can confirm that I've just watched the video for the first time today, even though I exclusively use subscriptions feed page and watch all of it thoroughly. It's very likely that it wasn't in the subscriptions feed at all at the moment of posting.
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  935. Always😀😀
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  936. I have a degree in Math. As most degrees needing math, there were Calc 1, 2, and 3. There is usually another one after that. After finishing those, I remember taking a Math 300, Into to Calc. This class made all the rest of the calc classes easy/obvious, I never did figure out if the 300 class was so enlightening since I had the other classes or if I could have taken the class earlier and the other classes would have been easier.
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  937. The fact that all the partial sums are nonintegral was quite memorable; especially once it was explained the graphic at the beginning of the chapter made so much more sense. Also the fact that the sum of the “100 zeros” series is greater than the sum of the “no 9s” series (assuming the approximations were of similar accuracy) despite being less dense is quite mind boggling - that one I’ll certainly not forget 🤯 Edit: skimmed the paper and I believe the term these days is “nice”
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  938. The hexagonal theorem is what just brought it all into perspective for me! I could not figure out why we were subtracting the “AB” and then the visual for the hex came with the “-6T” and it all clicked with the extra triangles! And then dividing the hex by six and getting the trithagorean buttoned it all up like a beautiful winter coat…
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  939. Interesting. Playing on a plane was fun, yet I wonder what would be the generalizations of this into multiple dimensions. I also wonder how it would change on a curved surface.
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  940. Legendre: >:[ Wherever he is now, I'm sure he is much happier knowing that there is such a great video discussing his work!
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  941. The best Christmas gift
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  942. Many years ago when it first came out I went for a job interview and the person interviewing me pulled out a Rubik's cube scrambled it and then said "I'll give you 10 minutes see if you can solve it" and then left the room. 10 minutes later he came back and asked how I had done. I said that I had not succeeded but I knew it was possible because I had seen my brother do it but that when I got one part solved it would mess up somewhere else. I asked him what the purpose of the exercise was and he said he just wanted to see how I approached a seemingly intractable problem whether I just gave up or persevered. Being willing to keep trying to solve the problem even if it did not work was what he was looking for.
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  943. It’s gonna be interesting
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  944. As a regular watcher of the Numberphile and Sixty Symbols channels, (both excellent, in my opinion), this sort of thing seems to pop up all over the place in math and get expressed in various ways in physical reality. This is a nice exploration of this particular instance. Now that I have discovered your channel, I will add it to my "interesting math channels" subscriptions. Thanks!
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  945. Really loved this video thanks. Pascal's triangle is the gift that just keeps on giving. Although strictly speaking the rule here is: " Twice the number above left plus the number above right"
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  946. He is missed, sorely. I'm disappointed I didn't get to know everything he'd done before he died, only knowing about his Game of Life.
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  947. I WATCHED THE WHOLE VIDEO AND I WANT TO WATCH MORE
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  948. This video reminded me of Grant Sandersen's (3Blue1Brown) video entitled: "Pi hiding in prime regularities" However I find this video more straightforward. Thanks for the present and Merry Christmas!
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  949. Great video! Another way to speed up the convergence of the Madhava series is to use the Shanks transformation; this makes the assumption that the difference between the nth partial sum and pi is roughly geometric, or at least that the ratio between successive errors changes very gradually. Repeatedly applying Shanks can give 7 digit accuracy from just the first 10 or so terms! I like the correction terms in the video better, though; they're more specific to this series and they help to explain the paradox.
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  950. Most memorable moment: the posture problems due to excessive obsession with mathematics
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  951. I for one really appreciate this short interlude type video. The smaller, more easily digestible content without needing to establish a large number of foundational knowledge is refreshing!
    18
  952. Pascals triangle seems to be everywhere
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  953. Music, "Chris Haugen - Fresh Fallen Snow" (I hear it on a lot of videos, love it)
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  954. Madhava is an Indian (Keralite) mathematician. Actually I am watching this video miles away from his native place (Irinjalakkoda, sangamagrama as you said). Nice work sir👏
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  955. 11:01 this moment is meme-potential. had me laugh out loud at his expressions
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  956. 65yo, haven't used calculus in over 40 years, still remember the heart of calculus and find it easy to follow this wonderful presentation. I see calculus as "depowering" or "powering" operations. Exponents become addition and back again. I imagine this ranking of increasingly more powerful operations, and calculus as the rules for going up and down in effect. Very similar to the way explained here by graphs and the table of elementary operations. Go up for slope, go down for area. From the first it seemed so intuitive back when I was young. Then again, so did dancing molecules which led to my career in medicine and biochemistry as an expert in single carbon metabolism ( the dance of the B vitamins). Mathologer is a wonderful channel because he loves this. There's an inherent beauty that simplicity brings; but you must first love knowing for the sake if knowing, not some other goal. Sorry, an old man reminiscing of when he still had a functional mind here. Soon again, I will think.
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  957. Merry Christmaths!!
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  958. Why there is 27 missing on your "I can't even" t-shirt?
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  959. It is impossible that they did Tailor's series (well, we have to use the same term for clarity) with verses. I think they did do some form of symbolic math, but it was the tradition to use only verses in the final copy. One reason for that may be that they thought it would be harder for future mathematicians not familiar with their symbolism to understand the symbolic math compared to reading verses.
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  960. This channel is mostly above my head but the quality of education here never ceases to amaze me <3
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  961. This was one of the most ah-ha-dense videos you've made. Long time fan; this was one of your best. I came away thinking about an entire mathematical space I hadn't thought about much before.
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  962. I remember noticing that the 2nd difference of the square numbers is constant and the same for the 3rd difference of the cubes when I was about 12 in school, so this is weirdly nostalgic. I've never learned about any of these details though! Very cool
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  963. This makes me want to crack open my old calculus books.
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  964. I couldn't wait for the video proof of the Christmas thoerem and found something AMAZING on mathoverflow that I wanted to give you a heads up about in case you had not seen it before. I think the famous one sentence proof can be Mathologerized using the incredible geometrical interpretation of the involution in the answer by Moritz Firsching to the MathOverflow question titled "Zagier's one-sentence proof of a theorem of Fermat". I have read many books/papers on polynomial automorphisms and never saw anything like that before in my life. After going through a bunch of proofs I think the one-sentence version is the best hope to do at the level of your videos using that geometry trick to explain where the involution comes from. There is also a wonderful numberphile video on this topic with an outline of the proof.
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  965. given that us Jews weren't allowed to own land or animals etc etc, we had to go into banking and trading and the like, so of course we know a lot about it :)
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  966. we need mythologer vs 3B1B teaching competition. they pick 5 random subscribers from the 2 channels and teach them something. then those teams have a competition to see who is a better teacher.
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  967. I love the feeling of having just about finished watching another great math video, and then I remember this is Matholologer and I've only just finished chapter 2 of 5! You spoil us with knowledge!
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  968. There are no ways to rearrange these terms to get exactly Pi, because this algorithm can never actually be run. Unlike other infinite series that sum to pi, this one can only generate up to the number of significant digits you already know. To produce Pi exactly, you have to already know the value of Pi exactly.
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  969. there's a small error in the red 'pentominoes' at 28:10 : one of the shapes (the 'W' shape) has 6 instead of 5 squares
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  970. my answer for the nxn points in a square is: sum of (i*i*(n-i)) from i=1 to n-1 so for 5x5: 1*4+4*3+9*2+16*1 wolframalpha says that can be simplified to (1/12)*(n-1)*(n^2)*(n+1)
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  971. A great math video to start the new year!
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  972. You know that it is going to be a great video when it is Mathologer, you have just started the first second of the video, and you've already seen three comments pointing out how exceptional the video is. I am confident that I am going to agree with your assessment.
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  973. Although technically, Euler's polyhedron formula also works perfectly for non-convex (concave) polyhedra, as long as they don't have any holes.
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  974. Most memorable: the optimized leaning tower! Although it was very messy, I think there's a lot of beauty in the fact that the most optimal arrangement of bricks is such a mess. It reminds me of how an extremely simple physical system (like a double pendulum) can result in chaos!
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  975. Ramanujan and Euler is everywhere ... And I love it ... ❤️
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  976. This is my boyfriend's favorite video. He watches it at least once a week.
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  977. Encore une vidéo prouvant le génie de Ramanujan. Démonstration hallucinante !!!! Bravo, comme d'habitude. Les vidéos mathématiques les plus géniales du net.
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  978. Math was very challenging until grade 10, honestly, in my case, I understood math very quickly afterward. Calculus was the easiest subject during university. Now after 10 years I still remember these rules. Math is all about understanding concepts and visualization of problems in your head, most students I came across learn math by doing examples and memorizing types of problems, at first glance, it seems the correct approach, but in reality, when the problem is slightly changed they struggle to solve it.
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  979. Hey Mathologer, My solution to the Strand Problem has a much more intuitive way to find the x^2+x = 2y^2 identity. It helps a lot to instead write in this way: x(x+1)/2 = y^2 x is the largest number house, and 1 is the 1st house. The sum of all house numbers is the average of the highest house x and 1, so (x+1)/2, multiplied by the number of houses there are, which is x. Ergo, sum of all houses is x(x+1)/2. If the sum of all houses is equal to a perfect square, y^2, then the occupant lives in the house y. After you reach this point in the puzzle, you can solve with wolfram to find all solutions. (1,1) (6,8) (35,49) (204,288) (1189,1681) Wolfram gives an equivalent formula to the one you reach at end of video. Would have done it myself but too lazy 🙂
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  980. The beauty of the interconnectedness of mathematics
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  981. Top paypal: markus persson, aka notch :)
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  982. Back in my early teens I was fascinated by this stuff. I'd spend hours re-discovering all these relations (nobody taught me that in school, but I stumbled on it myself by some chance). Thanks for the fun reminder 😃
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  983. There is a slight problem in the coding challenge: It can be implemented only on a device which has infinite storage and can do infinite precision arithmetic in finite time.
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  984. I was introduced to the Steinbach ratios 8 years ago. When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated: a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7) Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0 It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0 Also, to expand on your formulas at 43:00, A(m,n)=C(m,n+1), B(m,n)=C(m+1,n), and C(m,n)=(1*(r/1)^m*(s/1)^n+r²*(-s/r)^m*(1/r)^n+s²*(1/s)^m*(-r/s)^n)/(1+r²+s²)
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  985. 13:43 "Fun, no? ... Well I don't care. I think it's fun."
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  986. Hey Mathologer! What are the three chords at the beginning of your videos? Are they somehow important, or do you just like them?
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  987. I made it to the very end.
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  988. The optimal solution is very dependent on the exact happiness function used. It is not clear that getting equal parts of the contested estate rather than, say, proportional to the initial claim is a better happiness function.
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  989.  @Mathologer  High school maths (Algebra and Geometry) teacher here. The reason that we don't do classical geometry anymore is because school is compulsory. Back when classical geometry was taught at this level, school was an elective or privileged activity. The people who were taking classes did so because they wanted to and/or understood the value of it. Most people do not value math. Take a look at your subscriber count compared to mindless entertainment like Markiplier or something. I have good reason to believe that most of my students are not only uninterested in math (in fact, they hate it -- even the cool stuff), but are incapable of comprehending it. So how did the school curriculum respond? By boiling all of the math down so that the students with the lowest math performance could still pass the classes -- that is to say, teaching kids how to push buttons in a calculator, but not why the formulas work or what they mean. It's a waste of everyone's time, but that's what you get when teachers become glorified babysitters for families with working parents.
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  990. Mathologer making complicated math available for amateurs since 2016
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  991. If you use surreal numbers you can make the paradox "disappear". This summation is an infinite set of games which has a total game value of ON (infinite moves for Left Player) + OFF (infinite moves for Right Player) better known as DUD (Deathless Universal Draw). The winning move is not to play, because there is no winning move. Surreals make infinity easy and clean!
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  992. "This is the first Strand type puzzle"
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  993. Most memorable: fractal visualization of no-9's series being finite
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  994. A few years ago, a friend of mine participated in a popularization event for high school students (Maths en Jeans, Paris) and we offered an iPad to any student that would come up with a working solution for the 5th row. We had kids waiting in line to try, and they did not mind (nor go away) after we showed up that it was impossible. It’s one of my fondest memories as a teacher.
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  995. ALSO just to add, because (1 + 2 + ... + n)² is just 1³ + 2³ + ... + n³, we can replace each squared triangular sum with that so we can get rid of the pesky squares and get a purely cubic pattern :D
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  996. 2 minutes in and you've already blown my mind Mathologer! Thanks! Love your videos
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  997. Yeah! And as someone from the 16th Century who knows a fair number of medieval scribal monks, seeing the printing press being used is always a bit gut-wrenching knowing what they going through right now because of the printing press. This channel isn't going to commission an artist to draw something for a couple of seconds of transitions between chapters. It's not taking away anyone's job. It's the perfect use case for generative AI - to liven up and add moderate-quality art to something that would otherwise not have any.
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  998. What do I do with a quarter garment?
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  999. 19:00 Zagier's "One Sentence Proof". In case you want to investigate a bit I'll leave here a rough explanation of what an involution is. Imagine a process that pairs elements so if you have element A it returns element A' and vice versa: for element A' you get A. One example of this would be when you get a direction, say North (in Projective Geometry we called it a point at infinity). If we had a parabola tangent to the Y axis, drawing lines parallel to the North would give us either 2 crossing points, 1 unique double-crossing point (the tangent to the parabola through the Y axis) or 0 points. These crossing points form an involution since each relate to another and vice versa through some crossing and all involutions behave in a way that you always get a double point that is related to itself, which we can call a fixed point. This fixed point is what assures that we have an odd number of answers, that we can then interchange y and z in the involution and then get 4y^2 = (2y)^2 which summed to x^2 yields p. In the proof, the involution is crafted very delicately so it proves Fermat's Christmas Theorem, but I've not spent the time to look into how it exactly works. I hope this helps understand that part of the video.
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  1000. 36:29 there is always more to dig on every subject in math, this is a never ending quest.
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  1001. Most memorable: The intuitive geometric proof that γ is finite, hadn't seen that one before!
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  1002. Thank you, At 71 years of age I thought I would never see anything about slide rules again. When the video started I thought to myself - this could be about slide rules. It was surprisingly satisfying that my suspicion was correct. When I was a senior in HS calculators were just becoming available. True engineers keeping slide rules on hand is like current day sailors keeping and using sextants at sea. Young people are going to learn the hard way that all of this electronic technology is going to fail them at some point.
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  1003. Yet another Mathologer gem. It's definitively my preferred channel! Thx Mr. Polster, for give us such a wonderful presentation in all senses.
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  1004. 28:11 Hey! Euler is your Patreon patron! 28:25 And so is Mandelbrot!
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  1005. The locus of points with AB=A+B is the hyperbola XY=1 shifted up and right 1. This gives nice pairs like (3/2,3), (4/3,4), (1+1/n,n+1).
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  1006. Hi Mathologer, could you please explain why the perimeter of the ellipsis has no finite formula?
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  1007. ​​ @Mathologer  I went through a few minutes of the Hindi subtitles (another Indian language), it's not perfect but it's pretty decent in my opinion. Definitely better than no subtitles.
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  1008. Wow, this is a great application using math. I also like math modeling myself, and this video is in the same context of my interest so much. Great job and nice explanation!
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  1009. And I thought train spotters were strange. Now I'm aware there are triangle spotters, too.
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  1010. Uhh, a new Mathologer video! I'm soo excited!
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  1011. Most memorable: Tristin’s visual proof of the finiteness of the no-9 series.
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  1012. This approach is awesome! We covered the law of quadratic reciprocity in number theory class, but the proof was omitted, and it came down to uninspired manipulation and flipping of Legendre symbols. But now I'm 95% convinced that I understand why it all works :) I suppose something that could have been expanded upon for the benefit of other viewers is what the Legendre symbol is used for (which was briefly mentioned in the video), such as an example of solving a quadratic congruence over Z/pZ. Maybe that would make the whole LQR/Legendre business feel more motivated, but it's great as it is. And maybe another visualisation of the quadratic residues/non-residues mod a larger prime (like 17 maybe?) so we can get a feel for the numbers that appear along the diagonal, in particular what to expect (somehow it is easy to forget obvious things like (49/p) = 1 for any prime p, no need to find remainder first, as the symbol can lose attachment to its definition when purely evaluating by rule).
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  1013. On a scale of 1-10 for how beautiful that recursive argument is, because 10 is just a special case of 4, the rating I give is 4.
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  1014. Personally I find the limit definition of calculus very unwieldy and intuitive. Working with Leibniz notation like this is always much preferred in my eye. Turning your differential operations into statements about the convergence of sequences doesn't seem like a particularly natural step and is one that I don't tend to see people do when working with calculus in physics or dealing with them numerically. In physics and while working with numerical methods I feel like what we do most resembles non-standard calculus, especially ERNA and ERNA^A. Sam Sanders 2010 article "More infinity for a better finitism" gives a really nice write up of this. Being able to manipulate infinitesimals symbolically without fearing that one of your implied sequences stops converging is great peace of mind. In particular their example of pushing sums through integrals without fear (as long as your result is sensible) is one of those things I bet thousands of people do without thinking about whether it is justified.
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  1015. Witchcraft!
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  1016. Feedback: Mechanical engineering student with interest in math . Think I got almost everything. Will try to do the bits you left for the viewer. Thank you for awesome videos like this. :D
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  1017. Puzzle at the end: Conway's circle is larger. The weird curve's diameter is precisely the length of the chords since the chords are perpendicular to the curve's tangent at the intersection (this can be trivially shown from the curve's construction), so the diameter of the circle is necessarily larger (specifically, the square of the diameter of Conway's circle is the square of the incircle's diameter plus the square of the sum of the three sides of the triangle).
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  1018. 31:05 I believe I read, was mentioned in one of my Logic classes, or ???: For every Provable True Theorem, there are an infinite number of True Unprovable theorems. "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.".
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  1019. Every so often I think to myself, "It sure has been a while since Mathologer put out a new video..." and it seems that more often than not, you come through with something lovely in a day or two :)
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  1020. Programming challenge (did this around 13:41): because all of the coins evenly divide a dollar, it must be the case that a sum of coins to a multiple of a dollar can be separated into some groups of coins in which no single denomination adds up to a dollar, and the rest, in which each single denomination adds up to a multiple of a dollar. The former cases can be enumerated by doing the product trick without the exponent of 100, and taking the coefficient of each exponent divisible by 100, including 0. There are ways to make change like this for zero through four dollars inclusive. This gives us one part of the solution. The other part is to determine how many ways the remainder could be made. But this is a simpler problem, because it's equivalent to asking "How many ways are there to add five (number of denominations - 1) boundaries to a set of a given size" which is equivalent to asking for (size of set + 5) choose 5, which can be hard-coded as a sequence of multiplications followed by a division. (I could have tried expanding out the polynomial explicitly, but that sounds like effort). So, the final answer is to, for each amount of dollars that can be made without using a dollar's worth of a single denomination, multiply that by the number of combinations for the remaining dollars. I coded this up in Python, and it's pretty fast. I just started added zeroes to the exponent on the ten, and it was pretty acceptable performance up to 2 * 10 ^ 100,000. I'll post the value for 2 * 10 ^ 1,000,000 when it finishes, because that's much slower. Okay, yeah, that took a few minutes. One sec... Oh geez, it overflowed the buffer. I'm not pasting five million digits in here. See https://pastebin.com/MA2a9p3R EDIT: At 23:02 I'm seeing the same coefficients in the center row that I had my program calculate for "ways to make dollars without a single denomination adding up to a dollar" So I guess this is going in the same direction.
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  1021. The number of ‘toys’ in your office is astounding. Always amazed at your presentations.
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  1022.  @arikwolf3777  yes. Nowhere do any of the equations depend on the digits of the numbers
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  1023. These are some of your most visually satisfying animations to date, which is really saying something!
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  1024. I had to teach myself about digital roots when i was in middle school, although the deepest I came up with was 9 or 18...didnt even think about 1+8 equaling 9 because I only went one sequence in. To me, the fact that 99 equaled 18 was enough. Glad to be getting deeper 26 years later. Edit: wouldnt this also be considered modular number maths? Edit: aaaaand I got to 16:50
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  1025. This is almost click bait. Actually, I was curious if it was a female or just a male mathematician with long curly hair and feminine features. So I actually clicked to check the name. Does that constitute click bait?
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  1026. But since the horn is an ‘open’ surface as 1/x never ‘closes’ by crossing zero then wouldn’t your paint just run out the other end?
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  1027. I've now realized how easy it is to miss the edge cases when using visual proof. It's a crucial lesson in mathematical proof. I'll keep in mind when I'm dealing with visual proof in the future. Thank you for the beautiful insight professor. Yet another gem you've shared with us 🎉
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  1028. youtube: HERE’S SOME MATH me: okay
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  1029.  @Mathologer  I'm just picturing you stuck in a perfect sphere where the internal boundary is a blackboard.
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  1030.  @ravenecho2410  In fact some actuaries are fantastic at pure maths but most are what you would say “good at maths”. It’s prerequisite to study highest level at school to enter courses. As you’d know, but many wouldn’t, actuarial studies is a discipline of applied maths combining a diverse range of fields such as demography, finance, investment, commerce, marketing, accounting, risk analysis, modelling to problems in insurance, banking, health, pensions etc. So the emphasis is on applied maths to real world problem solving. Being brilliant at calculus or number theory is not of much use. if your friend chose another calling where such maths fields are useful, that’s fine. My son is an aeronautical engineer, who despite all the theoretical maths work at uni, discovered real world work generally didn’t require it.
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  1031.  @Macieks300  page 21 of the book I mentioned. As a source it cites: D. R. Heath-Brown: Fermat's two squares theorem, Invariant (1984), 2-5. latex version, with appendix on history, January 2008, at eprints.maths.ox.ac.uk/677/1/invariant.pdf The URL is archived at: https://web.archive.org/web/20110606154228/eprints.maths.ox.ac.uk/677/1/invariant.pdf
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  1032. I have the most excellent documentation of who came up with the windmill interpretation of this proof, but there isn't enough space to place it into this youtube comment.
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  1033. In 1983 I got a Commodore 64 for my 8th birthday. My dad got some 5-1/4" floppy disks full of public domain BASIC games and one of them was Tower of Hanoi. That was the last time I heard of Tower of Hanoi. 🙂
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  1034. Surprisingly fun for such a short video! Also, as a big fan of empty operations, I was quite happy to see the "0 =" and "0^2 =" at the tops of the trees :)
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  1035. This was an amazing video. Ptolemy theorem is one of my fav theorems now!
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  1036. TheOneThreeSeven. I love the fine structure of his name
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  1037. My brain ain't braining
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  1038. 20:58 All this time, I never knew that ln stands for logarithmus naturalis always thought that it's just *natural* to use ln (pun intended) Also, another beautiful visual explanation from Prof. Polster Thank you.
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  1039. For the challenge at 13:00, I got the sequence n!(n-1)!
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  1040. Is this it? Is this the hemisphere turned inside out? That wasn't easy to follow, was it?
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  1041. Pure mathematics at its finest Thanks for revealing this bag of gems
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  1042. Simple is King.
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  1043. 6:32 The beauty of the movement looks like a 4-dimensional hyper cube / Tesseract. So many great videos on this channel. Thank You!
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  1044. Mathologer Then your students are very lucky. I am an. 83 year old Maths lover, who got as far as calculus but would love to have had a greater understanding of Maths. But, which, with you help, I now have.
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  1045. For the trick to convert pascal's triangle to our game traingle, I originally thought of multiplication by 2. Which is the same as multiplication by (-1) anyway since we are doing modular arithmetic here :)
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  1046. I only realized it was a 50 min video when I read this comment after the video
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  1047. colouring proof is the one i came up with after you pointed out the incircle. also, the shape of constant width has width equal to one of the chords of the iris, definitely less than the diameter of the circle.
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  1048.  @JMairboeck  Yes. As he mentions at the end of the video, any odd number can be written as x^2-y^2. So any odd prime p has p=x^2-y^2=x^2+(iy)^2
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  1049. Would you like to know more?
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  1050. I'm mathematically challenged, but this was one of the most entertaining videos I've seen in a while! Job well done! 👍
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  1051. I did wonder if it would appear in any "discrete math" or other "math for computing" contexts but my more advanced algorithms texts (where it might be found) are elsewhere. It is not in the DM book I have here.
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  1052. 18:16 SPOILERS FOR ANSWER: (The method used) Start by placing 1 on the middle column and in the top row. Move: Place the next tile diagonally up 1 space and right 1 space. If the ‘move’ makes you go off the board, cycle back to the bottom row or to the left most column. (If you hit the corner do both) If the ‘move’ makes you end up on a space occupied by a tile placed previously, place the next tile the space directly below the previous one instead. (I.e. move 1 space down) This will ensure you fill the square with numbers in such a way that you will end up with a magic square.
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  1053. Squeezing and stretching the snake, that sounds like lots of fun, and the result is quite beautiful indeed.
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  1054. Thank you for educating us on Madhava...never taught in Indian school books..
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  1055. CHALLENGES! 2:21 "Why is everyone using calculus - or being scared of it - when it's not required?" The calculus explanation is probably the most popular and obvious one. What can I say, calculus is robust. 15:49 "I hope you can forgive me for my little lie" Honestly, I thought something was off when you advertised the volume to be 8. I went through the calculus at the beginning of the video and got an answer of pi, checked wolfram and also got pi. But your reasons are perfectly valid, plus you gave a fix anyway to actually make the volume 8. You are forgiven. 18:06 "Let me know how this calculus-free exposition fares for you" Ties in REALLY WELL with your trilogy, and is a really nice way to finish off the series of this magical shape. Bonus As noted by someone in the comments, real world painters understand the amount of paint they use depends on how thick they apply it. One drop of paint is technically enough to cover the entire planet earth if it's spread thin enough, so it comes as no surprise that only 8 liters of paint can cover an infinite horn. Surface area and volume just work like that.
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  1056. The most memorable is Kempners proof!
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  1057. What An hour? hand hovers indecisively over play button and it's a masterclass? hand drifts away from the play button and I need to do my flute practice hand retreats to lap (oh very funny, plz don't go there) OTOH it's a video with a funny bald German in it hand approaches keyboard but there are probably no cats in it being cute hand falls to side crickets more crickets reciprocity prime what the? wat IS dat? Mathologer smiles impishly and wiggles the hook Command voice:" Lock all targets and fire at will! All ahead Warp 6! Take us in Mr Sulu!" Punches the big red button...
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  1058. Is anyone going to download all the videos to some cloud, for a backup in case the channel will be taken down again? I'm planning to do it, but i'm not sure this is technically possible for one person.
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  1059. Praise the lord, the legend lives on 🎉
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  1060. Three minutes in, and Mathologer has already far exceeded the videos this video is about.
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  1061. Exactly what I needed today EDIT: My favourite part was the no 9-s proof. It is just simply elegant.
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  1062. Animated proofs should become routine in classrooms.
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  1063. This video made me fall in love with Mathematics again. To know that people have been doing such amazing things from 1400s is absolutely humbling to me.
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  1064. For the blackboard problem, it occured to me that 10²+11²+12²+13²+14²= 5 × 12² + 2 × 1² + 2 × 2² because the double products of (12+x)² and (12-x)² cancel. Knowing 12²=144, it is straightforward that 5 × 12² = 1440/2 = 720. The remaining squares sum up to 10 and we get 730, which is 365 × 2.
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  1065. The similar triangle thing: if r1+r2+...+r5 = R (the decomposed red points as complex numbers add up to the red point in the starting pentagon) and b1+...+b5 = B (for the black points) you can just notice that the construction of any of these "ears" can be done by taking the vector from the r's to the b's (b_i - r_i) rotating and scaling it by some amount which is the same as multiplying by some complex number m, and then adding it back to the red position. So the coordinates of the new points is r_i + m(b_i - r_i), adding those up gives you r1+...+r5 + m(b1+...+b5 -r1-...-r5) = R+m(B-R) the same construction on the original pentagon.
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  1066. 13:57 how many reg triangles and hexagons? minecrafters: YES
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  1067. BUT big loans are socially* bad - so your point is moot. Big loans lead to kleptocratic Capitalism. A problem of scale - unbound credit. *?how much debt is sustainable in society?
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  1068.  @jannegrey  One must remember that complex numbers are not complex as hard to understand but complex like an apartment complex eg. just two numbers joined together.
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  1069. that is like saying 3 shouldn’t be prime because it is the only prime divisible by 3
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  1070. So it’s like a Fourier analysis of a wave?
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  1071. The determinant or closely relevant tricks are still intriguing topics nowadays. Leslie Valiant even introduced the name and opened a new subarea, “Holographic algorithms”, for these types of reductions (from seemingly irrelevant problems to linear algebraic ones).
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  1072. What is interesting is that you can also turn any magic square into a new one by adding a non-zero integer constant to every square (the summations will change by the side length times the non-zero integer constant). Additionally there is probably some way to generate a new magic square by taking the modulus of every square (need to be careful about creating repeating numbers with certain values for the modulus. Edit: This actually might not be possible. I don’t have an example that it would work without causing a duplicate value).
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  1073. 24:40 the sum of all the triangles that lower-approximate the blue areas is: 1/2*(1*(1 - 1/2) + 1*(1/2 - 1/3 )+ 1*(1/3 -... = = 1/2*(1 - 1/2 + 1/2 - 1/3 + 1/3 -... = 1/2*(1)= 1/2
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  1074. I wouldn't say it was missed, but rather everyone who noticed it never bothered to write a paper on it. It's all part of the beautiful symmetry of mathematics in nature.
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  1075. The best math comes from Side Projects. The things you think about when you should be doing something else.
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  1076. Great video. Although wasn't this fraction in particular discovered by Laplace and proved by Jacobi. Of course, Ramanujan os a genius to have rediscovered it all by himself, but I was really hoping we'd get more about similar fractions. Digging deeper I found a book by S Khrushchev which discusses a whole theory of continued fractions like these along with great and largely unknown work done by Euler in this field. I think it can be found online as a pdf.
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  1077. Yes mathologer video when i'm bored is the best of the best
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  1078. The real question is: will a rubber band stretch the way it is presented?
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  1079. The way Mathologer pronounces ramanujan makes me so happy
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  1080. Youtube didn't show me this video until now and I'm annoyed so I'm commenting to try to convince the Almighty Algorithm to show it to everyone else.
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  1081.  @mikemthify  He said "My original notes date from 1971." I don't know if that means he came up with the proof then but if he did he really would've been 19 and that just blows my mind.
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  1082.  @ckq  Doesn't that mean it should be countable? 2ⁿ should be countable.
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  1083. I didn't know that Homer Simpsons likes to teach math.
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  1084. 3:20 minor formatting bug at the 53rd decimal place - the values are 2 and 1 but they're formatted as equal.
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  1085. 20:16 This is brilliant! That's the very reason this theorem is about primes.
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  1086. I dont even know math but your personality and enthusiasm has drawn me to watch
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  1087. Thank you for all the effort you put into these videos. I very much enjoy the learning experience from watching this channel!
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  1088. Thank you Burkard. Another fine video with lots of interesting and surprising twists and turns!
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  1089. Great video! Another way I noticed of describing why that (1st power) integer sums pattern works is that the left hand side always begins with a square number n^2, and apart from that term, there are n terms on each side, with each of the terms on the right side each having a term on the left side which it is exactly n larger than. For example, with 9+10+11+12 = 13+14+15, the terms 13, 14, and 15 each have a term on the left hand side they are 3 larger than (10, 11, and 12) making the difference between those be 3 times 3, which equals the square number 9 on the left.
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  1090. Very original video. I watch a ton of math edu content.. and I’ve never heard of math walls. Well done!
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  1091. Thanks! Lovely reminder why I love elegant mathematics like this.
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  1092. Here it is... Another interesting and deep connection between two (seems) "basic" things in Math. Really, this is all about: It's about deep connections, not just like "Did you know that a cup and a torus are equal, technically?". Math is beautiful with all its concepts and etc. what really amazes me are these connections. Man, It is such an art . It IS worth to spend the entirety of life with Math, really... No matter how it can be challenging sometimes.
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  1093. I made it to the very end! Love the visual proof as always. My first guess at the second "what comes next" is 1, 3, 8, 21, 55, 144, just the Fib numbers.
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  1094. "Of course, it would have been very hard to put into action in 1400 without a computer...." That's what graduate students are for!
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  1095. It's actually the square root of 10, or about 3.162. It's the number whose logarithm to base 10 is exactly one half.
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  1096. Love your work, you're rigorous and funny! Thank you. <3
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  1097. Mandatory rewatch everyone! Do your thing, share with your friends who dont care about math!!
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  1098. Great way to handle the found error. Might not be how the algorithm of Youtube wants and promotes it, but integrity in how you present information is one of the many reasons I'm a long time subscriber.
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  1099. I wish I were a mathematician.
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  1100. Animated version of the Pythagorean windmill proof: https://youtu.be/0VYYK-3tT1E
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  1101. Lets see if I can finish this one before the end of 2021 =) Thanks Burkard
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  1102. I liked the motivation and the path it took us down, but now I want to know if there are more of these anti-shapeshifters, besides the class we explored!
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  1103. Such a lovely insight! This explains the error term in any base, but certainly easier to notice in base 10. I wonder if Madhava also considered series for Pi based on the series expansion for arctan(1/sqrt(3)) = Pi/6, which converges faster. It is hard to imagine how they discovered these results without using our present day notation.
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  1104. The most memorable for me has to be Kempner's proof, just due to how counterintuitive it is after seeing so many divergent series, but how intuitive the proof is.
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  1105. Since 2020/4 = 505 which has prime factors 5 and 101, we just have the power set of {5,101} which is {1,5,101,505}. Since both 5 and 101 are congruent to 1 mod 4, they are both good and so are all their products. You definitely gave us an easy one. So 4 good - 0 bad gives us 16 ways of writing 2020 as the product of squares. To find them, we'll first use the usual trick of dividing out the 4's. We'll find the ways to write 505 as squares and multiply each component by the square root of 4. True confession time: I'm adapting this from work by Dario Alejandro Alpern, whose fsquares program I ported to Gnu GMP. We can find the ways to write 505 by finding the way to write its factors and using the fact that (a^2+b^2) (A^2+B^2) = (aA+bB)^2 + (aB-bA)^2 We'll find the solutions in positive integers, and then convert each such solution (a,b) into 4 solutions, {(a,b)(-a,b)(a,-b)(-a,-b)}. We know that every prime congruent to 1 mod 4 is the sum of two squares. For 5 this is easy: 5 = 1^1 + 2^2 and that's all. For 101 it's not hard either: 101=1^2+10^2, and this confirms that you are pitching us your softest softball. A quick check vs {49, 64, 81} confirms that this is the only way to write 101. Again, this is just the positive/positive solutions. So we have: (1^2+2^2)(1^2+10^2) = (1+20)^2 + (10-2)^2, which gives us: 505 = 21^2 + 8^2 2020 = 2*21^2 + 2*8^2 = 4*441 + 4*64 = 1764 + 256 and consequently 3 other solutions, 2020 = (-42)^2 + (16)^2 = (42)^2+(-16)^2 = (-42)^2+(-16)^2. We also have: (1^2+(-2)^2)(1^2+10^2) = (1-20)^2 + (10+2)^2 = (-19)^2 + 12^2 = 361 + 144 Thus we find 8 ways of writing 2020: 2020 = 42^2+16^2 = -42^2+16^2 = 42^2+-16^2 = -42^2+-16^2 = 19^2+12^2 = -19^2+12^2 = 19^2+-12^2 = -19^2+-12^2 According to the formula, there must be 8 other solutions out there, but I'm not seeing the permutation of these equations that gives them.
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  1106. I learned Heron's formula when I was in middle school in the state of California (4 years ago) and I'm pretty sure it's still being taught, but I've never used it outside of math competitions. Lots of great information in the video that I didn't know before though. Thanks!
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  1107. I have a habit every time you release a video, of taking a cold bath, then I turn my air conditioner on to remain cold to make the video more fun, and then I watch the video and try to do every challenge in it, just to make the best out of this video.
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  1108. 4k+1, 4k-1
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  1109. Saturday morning fun with Mathologer! What a pleasant surprise! 😎
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  1110. Fermat invented it first, Newton's fluxions make the most physical sense. Leibnitz's notation is way the best.
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  1111. A bit late on the lower bound for gamma, but... You can take every blue piece, place it into a rectangle of dimensions 1 x 1/(2^n), and split that rectangle in half with a diagonal from the top left to the bottom right. If you were to take the upper triangle from every one of these divided rectangles, you would get an area of one half of the square. Because every piece has a convex curve, it will stick slightly outside of the upper half of its rectangle. This means that every piece has an area greater than half of the rectangle, and the sum of all the pieces is greater than one half of the square. Because the square is 1x1, the area of the blue pieces is greater than 1/2.
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  1112. Same here. Am a teacher myself, but not of this level - not by a long shot. Burkhard and his team are something else 😊
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  1113. Great stuff! I did find myself looking at the zig-zag, and thinking - ugh, that's the one that gradually de-centres the lace each time you tighten it until it's moved so far out of centre you have to take the time tease it back into position. It can also easily look wonky, as you can pull it out of shape pulling on the long diagonal. Enter what you called the French lacing, which I use on boots, Doc Marten's, that kind of thing - horizontals are always on top, diagonals always underneath; you just see the horizontal pieces neatly perpendicular to the seam since the diagonals are hidden, but at the same time solving both the de-centring problem and the skew problem. 8-)
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  1114. Archimedes' bunch of bananas 😀
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  1115. 8:58- I think it is 292 (one less the result you gave us due to the fact we exclude the possibility of using the 100 cent coin).
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  1116. This is perfect! I have a phd in applied mathematics, and still have to stop and ponder quite a bit during the last few parts. Also I would love more videos on group theory/Galois theory :)
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  1117. I love playing with differences and sums of sequences! (I had taken to calling them "discrete derivatives" but that might be sort of a backwards name.) So cool to see a proper mathematician talk about them!! You went much deeper than I did, but I did use this on 2^x and x^2 to see how it worked similar to a derivative, and on the Fibonacci sequence to note that it is exponential in nature :) That the number of rows is the same as the degree of the polynomial feels so nice to me
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  1118. Great video! I recently had a project related to hyperbolic trigonometry and it was really nice to see a completely different approach.
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  1119. The {7/2} animation looks like a rotating pentagonal prism to me.
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  1120. My goodness! These people were so smart. This is crazy. Wonder is essential to beauty, and I am feeling both wonder and beauty right now. It is a travesty that I was unaware of Madhava's existence.
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  1121. Love your stuff and how much fun you seem to have presenting it....I get lost pretty easily because I'm old but enjoy the journey....anxious to see what you have up your towel next
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  1122. That was sum homework assignment.
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  1123. I was looking forward to this whole december :). Merry Christmas from Czechia!
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  1124. 48:47 my feedback: i'm a undergrad engineer, and loved the video. The explanation was very good, and very gradual. please make more :)
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  1125. Interesting video! I ran into vortex maths videos some years ago. I drew the number circle in photoshop and played around with it awhile, trying to understand what I was doing. I'm not great at maths so in the end I didn't properly understand what you taught in this video, but I did come to the conclusion back then that the vortex diagram was just an emergent feature of our arbitrary numbering system and you could make any number of them and make them seem mystical by highlighting patterns and features. It does make pretty patterns. Was Tesla really fixated to 3,6,9? What I was left wondering then and now is what kind of number theories Tesla was really thinking about! I don't think I've seen... objective video on the topic, and he must've had better stuff in his genius brain than what the youtube videos on vortex math are showing! This is a topic for a different channel, but did Tesla really believe that 3,6 and 9 were the key to the universe, and if so, surely not because of the "vortex diagram"? He's a smart guy so he must've understood that our numbering system is just one of many. What do we actually know about his fixation with numbers? Is the whole vortex math thing even in anyway linked to Tesla, other than by coincidence through youtube conspiracy mythos? Anyone know good sources that try to understand how Tesla thought by referencing his notes and such?
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  1126. I absolutely agree with you, it is a crime that schools don't introduce proofs. Even the most simplistic proofs, like the ones you showed. The only time they are put into my curriculum is if you take the hardest math course, which only 5% of the cohort takes. 95% of students aren't even introduced to proofs.
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  1127. It's a good thing King Solomon didn't follow through with his apportioning of the contested baby itself between the two candidate mothers, and went for a conservative "winner takes all" approach 🤣
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  1128.  @Mathologer  Don't erase everything on it, or else it will become a black hole, and suck you in !
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  1129. I feel like a kid who just found a $100 bill on the roadside. He will now burn his weekend to think about all the toys he can get ;)
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  1130. The closed formula for the mortal enemy's sequence is: E_n = 0.8469 (1.5747)^n + 0.2636 (1.3802)^n cos(1.7805n + 0.9507). All the decimals are rounded so this won't be very accurate past n = 15 or so, but we can use this to get the asymptotic formula for E_n. Since 1.3802^n << 1.5747^n for large n, we can drop the second term and say E_n ~ 0.8469 (1.5747)^n. So for large enough n, each term in the mortal enemy's sequence will be 57.47% larger than the previous term.
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  1131. I remember using this "unusual" formula around 1995 in a math olympiad (16 years old at the time). I picked it up in a textbook, and managed to memorise it to use it "blindly"
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  1132. Iris-free Proof: join the six exterior points consecutively, calling them A' A'' B' B'' C' and C''. Note the numerous isosceles triangles in this diagram and mark the relevant congruent pairs of base angles. Next, observe that A'A''C'C'' is a quadrilateral with supplementary opposite angles. Thus it is a cyclic quadrilateral and A'A''C'C'' (circle 1) are concyclic. The same reasoning shows the sets of points A''B'B''C' (circle 2) and B'B''C'C'' (circle 3) are concyclic. Circle 2 and 3 have three points in common, and so A''B'B''C'C'' are all concyclic, since three non-collinear points determines a circle. This circle has three points in common with circle 1, and so A'A''B'B''C'C'' are all concyclic.
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  1133. @praveenb9048 You have it backwards. Even today the definition of the parabola is the conic section you get when you cut the cone parallel to a side. A defining property of the parabola is that all lines parallel to the axis of symmetry of the parabola cross at a certain point when they're reflected on the parabola. Ancient Greeks knew this too (as it's a purely geometric property). The fact that second degree polynomials' graph is a parabola is not a definition. You have to prove it using the defining property above.
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  1134. Torricelli's argument about the volume of the horn equaling the volume of lots of circular discs reminds me of Archimedes' argument about the area under the parabola, except his argument also used mechanics like moment of inertia.
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  1135. When he put the stack leaning over Oresme I lost it.
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  1136. There's a simple process to get the first step answer without having to count all of the boxes. The pattern is made up of 4×4 rows of boxes that equal 16 boxes in total. Divide 16 in half to get 8. Divide 8 in half to get 4. Divide 4 in half to get 2. 16+8+4+2=30
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  1137. I like how the words "unit" and "Fourier" are interchangeable. Just like "Lie" and "differentiable".
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  1138. Wonderful proof. The incentre is slowly becoming my favourite triangle centre! (overtaking the orthocentre as my previous favourite)
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  1139. ​ @royalninja2823 he says that when he was younger, he used to get upset that people only know his game of life, because it was a quick and draft invention. But I also only know his game of life.
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  1140. Pre pandemic I used coins everyday, I can't wait for the day the markets reopen and I can use them again!
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  1141. In C↑D↓, the reason it only mixes up the rows and not the columns is because both C↑ and D↓ use the same vertical sequence (i.e., top, middle, button, top middle, bottom...). Only the horizontal sequence changes.
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  1142. Sum, product, exponent, factorial... how does this pattern keep growing? Is the next step a factorial exponent? Where does product summation fall within this pattern? What do we do with the levels that haven't been named, let alone conceptualized yet? What can we do with them once we've figured that out?
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  1143. You did WHAT??? You scrapped and reuploaded your whole episode to fix an error? You got my enormous respect, sir! I wish every YouTuber was as diligent about their content accuracy as you are. Meryy Christmas and all the best wishes for the New Year! Gregorian or Julian doesn't matter – your uploads will definitely be on my calendar.
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  1144. I'd love to see this channel growing big. All the immesureasurable effort put into your videos is just awesome. Merry Christmas :)
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  1145. Glad you remembered password after 3 months
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  1146. I never learned this at school or university, but now I'm in love! It seems like a way to teach/reach much richness of calculus without the (potentially) intimidating infinite limits and infinitesimals. I mean, we want to learn those too, but I remember it was a lot to adjust to at once, and it made calculus seem magical for a while, rather than logical.
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  1147.  @Mathologer  It was a long time ago. Tempus fugit! I think learning a little Latin has a long lasting effect of helping us understand the syntax, grammar, and vocabulary of other languages, including our own.
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  1148. For chapter 6, I think we're still missing a small but important detail that the collections still have to be nonempty when we remove the overlap. Since they add to the same number and are not the same, neither can be a proper subset of the other, which proves that they will not be empty when we remove overlap.
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  1149. At 0:18:33 it should be ...N+9/(4N), and at 0:21:16 the two "4n+" should be swapped with the two "n+" (also: I would replace the "n"s with "m"s in this yellow box, since they don't need to be equal to the "n"s above). However: this video made my day; I've been waiting for the new Mathologer video for 2 months! Great work Burkard, most of your videos are in my "Favorites" list, and Mathologer is definitely my favorite channel of all. 🤗🍻
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  1150. 3:24 oof... Again with the ai art? You definitely received considerable backlash last time, and i dont really see the point if doing it again! It also just looks really wonky! im sure it cant be difficult to put clip art of faces onto clip art if vessels. Great video otherwise as usual though :)
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  1151. The GOAT has returned
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  1152. Welcome back! So happy to see everything back to normal!
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  1153. Nice π-zza shirt
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  1154. Yay! I think I need to invest in Mathologer T-Shirts!
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  1155. The tiling of 2 by grids is Fibonacci series with first term 1 and second term 2
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  1156. New subscriber sir
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  1157.  @Mathologer  A cool thing one can do on an E-6B, is "enter" winds aloft for several altitudes simultaneously (direction and speed), thus one can optimally choose the altitude of flight for time and/or fuel. In a mainly head/tailwind case, isn't worth the effort (unless the winds are very strong), but when crosswind, then there may be surprising options.... (entry is a pencil mark on the vector plate).
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  1158. It's taught very carefully in India, but i doubt how many students actually care about it.
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  1159. I think this video shows why math education channels like Mathologer are so important; a simple, easy to explain proof didn't exist, so you made it! Maybe it isn't the most rigorous, but being forced into the visual medium instead of a dry, mathematical paper makes things WAY more understandable.
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  1160. This has some fascinating connections to a type of random number generator called an MLCG, which is a special case of the commonly used LCG.
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  1161. 13:43 the numerators follow the pattern a1=1, a2=3, a(n)=2*a(n-1)+a(n-2)
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  1162. With the Magic Cube 4D app sitting in a dark, forgotten directory for years, this video gave me the tools to solve it in a single day. Thank you. Your other videos are also fantastic, you explain things very well and have great charisma. Keep it up!
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  1163. I don't remember which proof I saw first but a few weeks ago I wanted to show a proof to one of my kids and realized I forgot all the proofs I had learned in school. The second (algebraic rearrangement) was the one I worked out first.
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  1164. My background in maths: two years of an undegraduate degree in theoretical physics before dropping out due to ill-health, then haven't really done any formal maths for the ~15 years since. My experience with this video: I was following everything fine up to the Bernoulli numbers; from that point on I was following enough of what was happening to still find it interesting and enjoyable to watch, but I'd have to go back and rewatch -- probably with multiple pauses -- if I wanted to be able to say that I truly understood it all.
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  1165. Finnish engineer student here. Heron's formula is in our textbook and might have been mentioned like once during a lecture. But we mainly used laws of sine and cosine to solve triangles.
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  1166. Is it weird that as a break from studing lineal algebra im watching a mathologer's video? xD
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  1167. Most Memorable : Every seconds of this video. I couldn't choose a single thing. I am sure that this is the best video I have ever watched in my life related to anything. Thank you so much Mathologer.
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  1168. A great way to look at familiar stuff. Top 5 Mathologer video of all time.
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  1169. I used to solve towers of hanoi on fights because I had no internet. Solving the 10-peg version with 1023 moves must've been one of the proudest moments in my life :D Edit: huh, I didn't know the "smallest disk" trick explained at 12:18. I just tried to follow the recursive algorithm in my head. Way to trivialize my life's achievment, Mathologer! /s
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  1170. I find it astonishing how numbers align so well. Much better than stars. Beautiful video as always.
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  1171. A mathologer video with less than 50 comments? That must be brand new. Must watch now!!!
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  1172. Observations: If one circle has a radius of A and the other has a radius of B, and the radius-B circle revolves around the radius-A circle, then there are two equations depending on if the revolution is internal or external. The tilt of the radius-B circle per revolution is (A - B) ÷ B × 360° if the revolutions are internal. If the number of revolutions is external, then that value is (A + B) ÷ B × 360°. The easiest explanation for these equations is that the tilt of a circle is always 360° times the distance the origin of the circle travels divided by its circumference. The radius can also be considered negative for internal revolutions.
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  1173. Thanks for recognizing Indian Mathematician Madhava for his work on Calculus and infinite series relating to Pi. More than the credit, I am interested in bringing in awareness among talented Indian youth towards Maths which is currently more influenced by cash-generating Wall-street jobs and related courses.
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  1174. Is it a known fact that you can get from any domino tiling to any other tiling by rotating 2x2 squares? Seems true for non-hole boards...
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  1175. Neat proof of an amazing result! I'll have to watch that show because it sounds like the best use of math in a film / tv show ever. Also, nice pi shirt!
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  1176.  @conanedojawa4538  No because it is a root of polynomial equation r^3-r^2-2r+1=0
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  1177. Reve's table of Cheeses is an extract from a leaning Pascal's Triangle
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  1178. was that an arctic cardioid at the end? how... cold-hearted of you :P
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  1179. 2:10 I saw the sqrt(2) version that uses the hypotenuse of a triangle with the other sides equal to one. Happy holidays!!!
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  1180. My first thought with seeing this was a way of defining a Dual of a triangle (up to scaling), following up with some theorems saying "A triangle has property X iff its dual has property Y". Time to explore.
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  1181. I've just discovered there's no greater way to start a Christmas day than with a Mathologer video x
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  1182. only comments after 36 minutes are valid
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  1183. This portrait of Madhava looks suspiciously similar to a certain Mathologer...
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  1184. Greetings from Czech Republic and Thank You for educating me about the achievement of my fellow countryman I was not aware of.
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  1185. I'm going to tile my floor with randomly generated aztec square arctic circles, in shades of grey.
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  1186. I see Mathologer's new upload. I just literally drop anything else I do, and watch. Cat video after this, maybe? :)
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  1187. Would the maximum number of moves to turn all 10 cards over be 55? By flipping the digits furthest to the right on every move, you decrease the binary value by the smallest amount, and it would take 1 move for the first bit, 2 moves for the second bit, all the way until 10 moves for the 10th bit. So 1+2+3+...+10 = 55?
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  1188. THAT is a COOL STORY!! - WOW!! (damn... I love the comments on these videos!)
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  1189.  @Mathologer  Some of us know about him, but I agree, the guy was a legend and deserves much more attention. India's advancements in Maths and Sciences were severly affected in the last millenia as Indians had to deal with islamic invasions and subsequently european invasions.
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  1190. I did a bit of trigonometry to express the six angles with the coloured dots in terms of the angles of the given triangle. Here's what I figured out. (I'm sure this is known to the triangle experts.) With the usual notation, let's call A, B, C the points of the triangle, a, b, c the edges and alf, bet, gam the angles. The median through A divides alf into the angles alf_b and alf_c (to the side of the edges b and c respectively). Similarly, bet=bet_c+bet_a and gam=gam_a+gam_b. With this, one gets: cot alf_b = 2 cot alf + cot gam, cot alf_c = 2 cot alf + cot bet and two similar pairs of equations. (The proof uses the law of sines and the addition formula for cot.) Btw., it can be checked that cot(alf_b+alf_c)=cot(alf). Now the folded triangle has angles alf_F = bet_a + gam_a, bet_F = gam_b + alf_b, gam_F = alf_c + bet_c, and one obtains cot alf_F=(-cot alf + 2 cot bet + 2 cot gam)/3 and two similar expressions for cot bet_F and cot gam_F, i.e. a linear relation between the cotangents of the angles! So, if one forms a 3-vector from the cotangents of the angles, then the folding operation from the video is the multiplication of this vector with the 3×3-matrix M which has -1/3 on the diagonal and +2/3 in all other entries. This matrix satisfies M^2=1, reflecting the fact that folding twice reproduces the triangle up to size.
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  1191. Does the distribution of numbers around this wheel correspond to Benford's law about the distribution of the first digit of random values?
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  1192. Your animations were amazing, I would have never been able to visualize this without your excellent animations. I was truly impressed, I learned a lot thank you!
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  1193. Got to the very end and I'm wondering if this could be used to pay taxes more fairly. For that I would change how we see the creditors as they are now debtors and so, the one with more to gain from the budget (the richest) would be the one to pay first, then the second richest up until the poorest, then when the poorest are maxed paying, then the second poorest... And finally the richest. We would be taking the inverted picture of the areas and still be using communicating vessels, having the ones with more to gain paying more at the beginning and at the end and also having the ones less to gain in the middle, paying all and maxing out first but when all others have equalised with them already.
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  1194. After preparing for competitive math Olympiads i hated geometry, this video has shown me the true light of how beautiful it is.
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  1195. This is one of the clearest mathematical expositions I have ever heard.
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  1196. Mathologer love ❤ fav channel
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  1197. No. You must make stokes theoreom videos
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  1198. Damn, this is one thing I've know about for a while but no-one else seemed to know about, so thank you for showing this to the world.
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  1199. You left out the fact that the Hanoi constant 466/885 gives the average distance between any two points in a sirpinski triangle with a side length 1
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  1200. Combo Class has a video as well on the {2,2}, {1,2,3}, {1,1,2,4}, ... infinite family. But they're very different and complement each other nicely.
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  1201. The entire video I was wondering what the 2 in the (x+2) formula was really referring to. Sooo satisfying to see the recurrence equation at 25:00, makes so much sense!
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  1202. Undergraduate student in physics here, more into theory side of it so I know some difficult maths Finished the video and went: yep i understand everything and yet remembered nothing xD went back and looked through the formulas again, beautiful stuff oddly it doesn't feel like even 20 minutes had passed for me... ah yes, i was watching at x2 speed with a lot of pausing and backtracking :l the math used in this video is considered 'basics' for me, yet the application is mind blowing nonetheless soooo... "yes please make more videos i would pay for them even" "I am thinking of finally putting up a Patreon page. What sorts of things would you like to see there?" ... well, yes. also maths and perhaps some math story you find fascinating (as for 'is the video too hard for the general audience'... well... i am not an accurate reference) Thanks for the lesson!
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  1203. My vote definetely goes to Kempner's proof, it is extremely elegant, since the concepts used are individualy simple, such as the calculation of numbers without nine or geometric series, but when cleverly combined they form this amazing result. Besides that, great video as always. Edit: typo
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  1204. As a mid-level PowerPoint wizard, just be glad you didn’t have to use google slides. Those animations looked difficult enough as is.
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  1205. Numberphile also a Ptolemy video 4 years ago, featuring Zvezdelina Stankova. Also enthusiastic, but with very different twists.
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  1206. A short video on change
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  1207. I mean I cheated or crammed my way through high school I was too bored to try, so this is the first time I’m seeing these design elements explained from a mathematical standpoint which helps me understand them deeper: It’s amazing how intertwined art, beauty, mathematics, numbers, algorithms, patterns etc. Truly all connected. I am all that is, because all that is - is within me. 🙏🏻
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  1208. @ Everyone would divide their loans into the smallest chunks possible, until the admin costs outweigh the gains. The admin costs will determine how small the chunks will be.
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  1209.  @Mathologer  yes, I remember to have watched that one a long time ago. I just rewatched it. (It's now 2:41 am, I should be sleeping already). I am pretty sure I understood everything you did there, but to be sure I'll rewatch it a second time in a few days. But I'm really waiting for the "completely insane" (yes that was your real promise) Level 7 of this video. Galois theory. I love completely insane video's. It will, I already am sure about that now, probably be the most interesting and exciting masterpiece you've ever made as mathologer. All the best, A.
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  1210. animation autopilot is getting smoother day by day (I can see the damn hard work) Merry Christmas mathologer and wishes for another year of masterclasses.
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  1211. It's those diagrams and cool connections that make me love number theory. Please do more number theory videos.
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  1212. Marble: Most memorable Idea: No integers among the partial sums. Your excitement is always a great feature not your presentations. And why you're my favorite Mathematics YouTuber!
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  1213. I reckon the Iron Man title is more enticing than the +/- title. I had not cheched the video before, seemed such a dense, intimidating subject. Now it's like discovering an easter egg of the MCU, seems worth enduring the hard maths somehow.
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  1214. Such a scheme can be used against any creditor, which is why we have a regulated bankruptcy process: you can seek and get future protection from your creditors, but you must reveal under oath all your assets and claims against others in return. The bankcruptcy court will divide most of your assets among your creditors but grant you bankcruptcy protection from your creditors for the future. However, if you hide that scheme you and your friend have agreed upon, you commit bankruptcy fraud, which is a felony.
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  1215. If we somehow were able to get back the burnt books of Nalanda University, then I am pretty sure that more than 90% of today's "modern" discoveries in maths and science would be found in them.
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  1216. Wonderful video as always. The more videos I watch the more I'm convinced that Euler must've been a time-travelling Mathologer viewer who really wanted to look smart by appearing in every video
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  1217. reads how to make googol dollars Nobody: Bill Gates and Jeff Bezos:That’s not a dare, that’s Tuesday.
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  1218. This channel is great like its viewers. We always get excited when notification popped out.
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  1219. just wanna compliment the pacing of this video. first time i didnt have to pause/rewind to absorb, except when you prompted me to for the last chapter, which was when i was planning to take a break anyway lol
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  1220. When I heard geometry and turtles, my first thought was Logo.
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  1221. With a little bit of trigonometry on the heptagon, the set of equations for r and s can be used to show that cos(π/7) is the root of a cubic polynomial (8x³-4x²-4x+1=0). This immediately shows why the heptagon cannot be constructed (cube roots are not constructible). It's the same reason that one cannot construct a cube of volume 2. Using "origami math" (look it up, it's cool) one can construct roots of quartic polynomials. Geretschläger (1997) used this to construct a heptagon starting from just a quadratic sheet of paper (no straightedge or ruler required), but where you can make paper folds as in origami. The proof is given in the form of an origami folding instruction.
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  1222. For anyone working on the problem at about 16:20 with derivatives, do yourself a favor and use (1-x)^(-1) instead of the fraction. Power rule & Chain rule is sooooo much easier than Quotient rule when it comes to taking more and more derivatives in this case.
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  1223. So essentially if we were to rank lacings on strength and length 1 being the best and infinite being the worst we're looking for the lowest sum of the two factors to have the optimal lacing. The natural choice here would be the crisscross lacing. The crisscross lacing at best has rank 1 in both situations for a sum score of 2 and at worst it falls behind in strength on the zigzag lacing for a sum score of 3. The zigzag lacing at best will be 1st on strength and 2nd in length for a sum score of 3 which is merely equal to the crisscross in the same case. The case where they do equal at a sum score of 3, it would be a competition of preference where some situations might demand short laces by lack of material but some may prefer a higher tensile strength because they want to climb a mountain today.
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  1224. It is nice to see this geometric approach to the topic. The result is not new to me, but the path 😊
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  1225. You miss so much to talk about this more!! Do part 2 !!!!
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  1226. The third edition of Proofs from THE BOOK also contains the proof, with all the details filled in.
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  1227. I recently watched an interesting video featuring Roger Penrose in which he expressed the view, which I share, that mathematics is discovered, not invented. When I see patterns like this it further strengthens my view that maths fits together too beautifully and with too many hidden but elegant patterns to be invented. This was very interesting and no, I hadn't heard of Moessner. Thank you for another informative and well-explained video. The internet isn't just stupid cat videos.
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  1228. I think the fact Gabriel's trumpet is infinitely long, makes it feel paradoxical. Because in my opinion, something like the Koch Snowflake extended in a 3rd dimension, to make a prism of finite volume and infinite surface area, intuitively feels less paradoxical. You can easily contain it within a cylinder. Yet, you can take it apart into infinite triangular prisms. Then you can stack those atop eachother to get a roughly trumpet-like object; which has the same finite volume, and has infinite surface area.
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  1229. Wonderful! Spectaculary clear and teacherly. Richard Hamming's book "Numerical Methods for Scientists and Engineers" (a staple in engineering education 50 years ago) discusses "Difference Calculus" and "Summation Calculus" and describes "Summation by Parts" (!). For those of us who wrestle with numerical problems, all this material provides powerful tools and the basis for software algorithms that do the heavy lifting in science and engineering.
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  1230. Everyone: maths is boring :( Mathsloger : let me take care of it. ;) Btw your videos are very interesting and full of knowledge...... Love from india 🇮🇳❤❤
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  1231. I know "Numberphile" might be a complicated word around these parts.. but i will never forget watching that interview Brady did with him [Conway] when he's looking out the window and says "i wish i knew whyyy..." .. in reference to the strange universe-implying numbers in group theory - why the monster and no more?.. why any of the sporadjcs?.. especially when symmetry seems to play such a big role in fundamental physics
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  1232. The challenge at 15:33, solved in a different way: The number of rotations of the smaller coin per each time it rolls around the bigger coin is equal to the ratio between the smaller coin's radius, and the distance between its center and the center of the larger coin. In the case of equal-sized coins, that ratio is 2, which explains why the outer coin rotates twice. For a smaller coin with diameter 2/3 of that of the bigger coin, its center is at a radius of 1/3+1/2=5/6, while its own radius is 1/3. (5/6)/(1/3)=5/2, so it should make two and a half rotations around its center for each time it rolls around the bigger coin.
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  1233. You are very special. Thank you.
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  1234. This is how far I could get equilibrating the cards from ace to king of diamonds: imgur(dot)com/gallery/6wJ657Q (I didn't put the link normally because youtube moved to spam)
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  1235. ​ @josepeito Because their chief weapon is surprise 😆
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  1236. Professor Polster, The Mathologer is THE best thing that happens to me in all of the media including Internet, blogs, youtube, social media, podcasts, TV, radio, movies, lectures, courses, and print. I can not describe how much pleasure I derive from watching Mathologer. At times, it's like a detective solving a crime (the crime being the difficult math problem at hand). Thank YOU, MARTY ROSS and OTHER GOOD PEOPLE who make Mathologer, such a good unusual program, possible. May you continue to produce your programs for at least another 50 years. Behnam Ashjari, PhD EE
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  1237. Merry Cristmaths!
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  1238. I remember coming across this formula at the corner of one page in my textbook in high school, and we just glossed over it.
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  1239. 5^3 + 6^3 = 341, which is short of 7^3 by just two. I love that because you'd already mentioned trying to wrap a cube in square slices, I immediately imagined such a wrapping where two opposite corners were missing (though without checking I'm not sure that could really be done without further slicing the slices).
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  1240. I live ~250km away from the birthplace of Saṅgamagrāma madhavan.
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  1241. 13:00 Spoilers below! Using a=1, b=c=d=...=2, the highlighted sequence is 1, 2+2, 3+2+4, 4+2+4+6, ... In each term of the sequence, we can split up the first term of the sum into k 1s and spread them out over the rest of the terms of the sum, giving 1, 1+(1+2), 1+(1+2)+(1+4), 1+(1+2)+(1+4)+(1+6), ... Each term of the sequence is the sum of the first k odd numbers, which are the squares. That means that the sequence generated by the squares is 1, 2*1^2, 3*2^2*1^2, 4*3^2*2^2*1^2, ... In general, the kth term is k(k-1)!^2=k!(k-1)!
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  1242. Congratulations on hitting 100 videos! That’s quite a milestone! Love your content, Mathologer!
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  1243. Okay..Half way through the video..They do teach this to us..In highschool..But not in calculus..But in Sequences and Series under the name : Method of difference..Our teacher said.."When u find no other way(like the classic Vn method as they call it here) to find the General term of a sequence..Use this method to find it"
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  1244. The hot topic in the last month has been AI. I have neutral feelings on the topic as it is just another tool and one that can be unplugged. The reason I bring this up is because the Indian Mathematician had to express mathematics with what was available to him, which was verse and not the concise mathematical language we have today, to borrow the term used in this video. At any rate, back onto my comment, Humans speak and think in verse, and maths is purely a symbolic form of logic to express patterns. AI works by using filters and approximations in much the same way as these approximations work and learns over time in much the same way. It is the next leap as it is even more efficient and concise than symbolism and mechanizes pattern recognition without the middle man of symbolism. The reason why it is frightening is because it does not give us symbolism to understand how it arrives at the answer. With each AI solution to a problem, cracking open that black box and seeing what steps the AI took to arrive at the answer might reveal a new level of mathematics or give insights for mathematicians to chew on.
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  1245. You know what is really magical? You have made your instruction and animation line up, to almost create a gamified experience. This was wonderfully engaging and explantory, well done! Mathematical Tetris :)
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  1246. 6:20 No, as you can isolate a corner, which clearly not be tiled over
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  1247. I came back after long time ! what a beautiful voice 😗🙃
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  1248. I studied electrical engineering and at University we learned something called symmetrical components theorem, of wich we use the specific simpler case for triangles. In reality we're not taught it as a triangle, but rather, as the three components of our three phase electrical grid (be it current or voltage). You see, in the electrical grid, the 3-phase system is not always symmetrical (with 3 phasors of the same length and 120° apart) because of asymmetrical loads and asymmetrical faults and such. So it would be very complicated to model and visualize it as a 3 phase system because of the symmetries. So instead, we model it as a sum of 3 symmetrical systems (analogous to those 5 symmetrical base pentagons). And seeing them as triangles would be like one equilateral triangle, one "mirrored" equilateral triangle and one "triangle" with all 3 vertices in the same location. It's sad the in my university we didn't go into much depth with this theorem, we just saw a glimpse of it and were immediately learning the application. But it's really nice to now see a video about it explaining everything in detail in an easy to comprehend format.
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  1249. Hey I found 2019 in your π-shirt after the 244th digit after the decimal #yay
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  1250. Professor, I am looking forward to you aiming at presenting the Galois theory the Mathologer way in the future. I have always considered the inability to solve quintics by radicals mysterious, hopefully your video will shed some light on the concept to non-mathematicians like me. Thank you very much for all the stuff you create, it is unimaginable to comprehend the amount of thought, time and effort to get such things done, with the above video being an excellent example.
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  1251. Funny piece of trivia : the Arctic circle theorem is linked to the alternate sign matrices, described by no other than Lewis Carroll
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  1252. At first sight, there is something slightly unsatisfying about Aumann's game-theoretic solution, because it assumes a happiness function. However, standing a bit further back, among the uncertainties of the original lending decisions might be ignorance of where you stand in the list of creditors, ordered by size. And another might be uncertainty what your happiness function would be, at the time the dividend is being determined. From this perspective (originated, as far as I remember, by Harsanyi) there is an intuitive basis for wanting small creditors to be favoured when the dividend pool is small relative to the total liability of the estate, and for them to be subordinated to large creditors when the pool is larger, but still deficient. It could be argued that it is an even greater merit of the Talmudic solution that it responds to this intuition than that it is optimal given the Aumann utility function. It is also an attractive feature of the hydraulic model that it visually exhibits the ranking between creditors when there is a super-dividend, as well as generating the solution at every level (whereas the garment model generates the solution without showing how it works).
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  1253. I love the structure of this video. The moment when I understood how the visual proof would go (just before we moved to visual representations of it) is why I watch videos like this.
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  1254. Increbile, the amount of astractions one's mind can demarcate around a given subject. Just wow, maths is beautiful
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  1255. Most Memorable: The fact that it is possible to arrange the bricks on the table such that the last brick can be as far as the size of observable universe from the table, and yet be perfectly balanced!!!🤯🤯🤯
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  1256. "How satisfying was that?" .... Very! That was such a perfect full-circle moment!
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  1257. My interest in maths came back after my exam .
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  1258. Does anyone else smell the Silver Ratio hiding around here somewhere?
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  1259. I love the 1,1 solution 😂 technically correct is my favorite kind of correct
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  1260. Already seen Henry Segerman's video and left a comment there. :) Fantastic to see this explained in great detail and crossing my fingers for that part two. I really like the map projection. Instead of Antarctica, you can take your home location as the centre of the map and your homeland will then be the least distorted country on the map.
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  1261. 26:48 - Cavalieri works for parallel slices, but those slices you depicted aren't parallel, even in the limit. I think there's a small gap in the proof here.
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  1262. Very beautiful demo and excellent explanation!
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  1263. Can the King's Algorithm construct a Parker Square?
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  1264. 34:36. Also how I look every time I realize I forgot +C
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  1265. I love all the visual intuition for why logs and e fall out of 1/x with all the squeeze maps. This video gave me so many aha! moments.
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  1266. What about for lacings where the holes don’t have to be directly opposite each other? The zigzag lacing shown is very similar to spiral lacing, used for stays/corsets until some time in the 19th century, except for two things: the holes (except for the top and bottom pair) are offset so that the crossing angle stays constant, and the lace is simply tied off at the top and bottom, with no long lace running between them, but I suspect that’s also the case for the shoes. EDIT: after hearing about which lacing is strongest, I have to take a look at whether the typical distance between eyelet became denser over time, in which case it would make sense if the bones bodices of the renaissance were spiral laced, but the Victorian corset crisscross laced.
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  1267. In any triangle, tanA + tanB + tanC = tanA x tanB x tanC.
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  1268. Proof that γ>1/2 (24:30): Replace all the blue parts with right triangles inside of them and the resulting sum is 1/2 (visually this is obvious, writing out the sum isn’t too bad either).
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  1269. For the first half of the video, I was wondering how relevant this would be to numbering systems other than base 10. Then you addressed it. Now I feel like we’re getting somewhere… 😏👍🏻 Great presentation.
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  1270. Wouldn't the Talmud method incentivize big loans if the amount being split up is more than half of the sum of the claims?
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  1271. I wonder was he on the very cusp of something just as he was killed by that Roman solider? He was so deeply engrossed in his diagrams in the sandbox that he had tuned out the collapse of the defence of Syracuse and all thoughts of personal safety.
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  1272.  @jannegrey  how long would it take you to translate those 30 pages? Sight unseen, I bet we could come up with sponsors to encourage you to do so.
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  1273.  @jannegrey  So in order to understand it, we need to reverse polish notation?
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  1274. There are so many comments see if you can find mine
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  1275. Your infinite fractions/sum videos have been absolutely amazing. Please don't ever stop making videos, they are super clear and entertaining.
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  1276. Problems would be easier with a copy of this "Algebra Autopilot". Where can I get one ;-)
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  1277. "Step back and squint your eyes." Brilliant guide to this insight!
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  1278. •Always loving your videos. 🥰🥰🥰 Love from India 🇮🇳🇮🇳🇮🇳. •By the way, if you plug in 3 for n, you get 《946/243》(=3.893) •Please make videos on ☆☆☆ Unsolved problems in Mathematics ☆☆☆.
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  1279. Everytime I watch a long Mathologer vid it feels like I'm watching a movie. Thank you!
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  1280. You can call 1 a prime of inconvenience – most of the time you would be justified in regarding 1 as not prime except in rare instances, such as I was reminded of recently, exempli gratia, back in 1732 Euler gave a list of Mersenne exponents starting with 1…
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  1281. Yeah, what was his name? Alvin Engels?
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  1282. ​@themathhatter5290 no i think it was gilbert something
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  1283.  @tomkerruish2982  we are sorry to inform you that you failed the sarcasm test.
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  1284. holy shit i found the convergance speed ups like a month ago bruh why did an indian have to find it 600 yrs ago, i felt like such a gamer
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  1285. I was taught about Heron's formula. My high school teachers said that there is never a need to use it. I think it's always better to use some other tricks to get the area unless all sides are known
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  1286. Me when I see a mathologer video: "My excitement is immeasurable and my day is made ."
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  1287. 36:57 "Outramanujan" is now my new favourite verb!
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  1288. 25:08 "Where does he enter the picture?" Right there, on the left!
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  1289. I do not understand all of it , but these are very interesting things to study next . I find an exploration of my ignorance to be a rewarding journey . Thank you for working so hard to bring them to my attention.
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  1290. You really the best man! 🙌 Stay cool! Your contents really hits me hard and inspired me to create my own channel and be an inspiration also to others. More power everyone
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  1291. ​ @Mathologer  heptagons always seem like the "cursed" one, especially when it's surrounded by hexagons and pentagons and octagons, so huge thanks for giving it the well-deserved limelight this time
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  1292. A wonderful video! By the way: because alpha is the arclength in sin/cos and the area in sinh/cosh the inverse functions are called arcsine/arccos and ar(ea)sinh/ar(ea)cosh.
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  1293. Very interesting video - thanks for putting this together! The {7/3} star and {7/4} stars are essentially the same stars, just traced in reverse. I noticed that the triangles in the {7/3} star are rotating in the opposite directions as the squares, which I think we can take as squares rotating in the {7/4} star. So perhaps what’s going on here is that if you have a {p/q} star (with p and q coprime) we expect to see q-gons rotating one way and (p - q)-gons rotating the other way? Perhaps that would make it easier to prove why this works.
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  1294. 36:53 okay yep it''s cat videos for me
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  1295. A brilliant video illustrating how law manages complex situations. Kudos to you. I hope the water level test appears in law cases, and argument on setting it up.
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  1296. Ok...now what do I do for the next 23 hours today?
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  1297. Nosferatu? I've always thought of you as an extremely affable villain from a spy movie. Also I saw the pattern in the intervals betweens +s and -s almost immediately. Finally, a chance for my "genius level" IQ to get used for something other than detecting fake patterns in random noise.
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  1298. Had to pause so many times just to truly appreciate the beauty of this visual proof. So much to reflect on.
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  1299. 4:03 – Unfortunately, I can’t guess “what the more is” because I can read German and already got the answer at 2:18 :-/
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  1300. These videos are categorically beautiful. With such a transparent love of mathematics, and talent for education, this truly was a gift to the world. As per your request, I’ve a BS in physics here, and while everything seemed reasonably intuitive as you explained it, I’ll need to spend a few hours going over it carefully to deal with the technicalities. I’ll wait with bated breath for your next video.
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  1301. U make video about at the rate of 1 per month. But that whole 30-50 minutes video made my day, seriously. Lots of love ❤❤❤ prof from India🇮🇳. But I would love if u increase the rate..
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  1302. I think it's even simpler than that. Each bug always moves perpendicular to the motion of its "target" bug, so the target's motion doesn't affect the rate of closure. Therefore, each bug must travel exactly one unit to reach its target, at the center. Fred
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  1303. My intuition on the initial paradox was thus: even in the limit, there's a point of the initial square not on the circle. That point will always not touch the circle, so that's where rhe extra perimeter comes from. I like the conjugate explanation that having the normals of the approximating curve not getting closer to the actual curve also explains the discrepancy.
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  1304. As an armchair maths fan I sometimes wonder what the world today might look like had Ramanujan lived longer. He seems like another Euler.
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  1305. I'm going to look into a history of Indian mathematics and science.
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  1306. Great video! I have been watching these videos since I was in high school, and now I'm at the end of my degree in mathematics! One of my favorite topics is set theory, notably ZFC, which you are most likely familiar. Is there any chance for it to be covered in the future? It's a very mathy technical topic, but once we get the gist of it, it's definitely perplexing. Cheers!
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  1307. Both Mathologer and 3b1b are great and I love their visual representations!
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  1308. ”French are best” Cars and military cries**
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  1309. For me, the most memorable part was the optimal setup for 20 bricks because it made me glad I'm not an architectural engineer.
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  1310. ​ @Mathologer the crazies are the best lol
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  1311. These are incredible. The speed and shape options on the first link (Kieran Clancy) are really handy. I just spent over 15 minutes playing with that one. Marvelous!
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  1312. Feedback: Science Teacher with BS in Molecular Biology and Chemistry Highest Math - Multivariable Calc Overall, I was able to follow through to the end fairly easily. You always do a great job in explaining these higher-level concepts by stepping us up from the basics. The part I got a little bit confused on was where the factorials came from when deriving the Euler-Maclaurin Formula. Then, I went back and watched that part again and figured it out. 50 mins is nothing when you’re deeply immersed in the video!
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  1313. The reason why they teach the continuous/analytical stuff is because it is still useful, and one should understand how to derive actual, precise solutions even if one only works with discrete approximations in practice. Hopefully analytic solutions for those problems which currently have no known such solution will be derived/discovered in the future. As George Polya amusingly but somewhat truthfully said on the topic of PDEs: "In order to solve a differential equation, you look at it until a solution occurs to you." Perhaps as an engineer or someone in your particular field, analytic solutions are not useful to you — that's fair enough — but they are still useful, and the practice/art of finding them is still useful as a whole to mathematics. Just as solving the heat equation led Fourier to devise Fourier series, so too may further attempts to solve other problems yield other techniques with myriad applications.
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  1314. I'm only 7 minutes in and this is my favorite video of yours by far
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  1315. Glad you are back! I was concerned after the channel was hijacked. You are a tier above the other math youtubers in my opinion! :)
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  1316. 23:56 I guess that the formula doesn't work because the binomial coefficient "isn't defined" if the number on the top is smaller than the one on the bottom. I think that, if we define the choose function to be 0 in those cases, it works just fine. When you replaced them by algebraic expresions, you indeed set them to be 0 in those cases, so the last formula works in general.
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  1317. All I can say is, pi seconds is a nano-century. :-)
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  1318. When he flipped that triangle for the difference rule 14:07 I almost jumped out of my seat, Great work as always !
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  1319. 25:15 "This is really just high school stuff" Too bad I didn't go to school in Germany
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  1320. The most memorable part is the part about Kempner's no 9s series. Loved how intuitive it got after after seeing the square that became more and more white (gray) :)
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  1321. I learned the derivation of the Pythagorean Theorem when I took geometry. We were on the topic of similar triangles and found that a right triangle could be made up of two smaller right triangles. It was very straight forward and consistent with the topic we were on.
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  1322. I would LOVE a series of videos on measure theory and conceptions of length in fractaline structures. I've been trying to solidify my intuition of some of the bizarre units of measure come upon by manipulation of formulae, like units in m^±½. I have a feeling such a series of videos by you would help me to connect some of these abstract dots in my mind. Thank you for this video on limits and the potential for arbitrary endpoints in their calculation. This is fascinating!
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  1323. One Day I am Gonna beat you for 50 mins sake!! ( I am just a student in grade 9 btw)
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  1324. Your office is like a toy shop 😀
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  1325. Fiboghoras and Pythanacci, my favorite duo
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  1326.  @SuperJuiceman11  yes true I noticed that too lol
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  1327. Wait a minute, does that mean that if we extend the domain of x and y into the complex numbers, it works for any (real) prime? 4^2+(3i)^2=7, for example
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  1328. This channel reminds me why I love maths. The most beautiful and perfect stuff in the universe and beyond!
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  1329. I love this, I was reading some old English textbooks from the 50s and was impressed by their derivation of power sums for high powers using a recursive model. And you've generalised it!
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  1330. Ah, I did it. Here is the proof of the cube shadow theorem (with terms defined at 13:50): Without loss of generality, assume S is the lowest point of the cube. The area of the shadow is then just the area of the parallelograms formed by the projections of the faces spanned by (u, v), (v, w), and (w, u) onto the x-y plane, respectively. These areas are given by the z components of the cross products u x v, v x w, and w x u. By construction, these cross products are just the vectors w, v, and u, so the area is given by the sum of the z components of all three vectors. For the length of the line projection, assume (without loss of generality) that S is at height z=0. Since we also assumed it is the lowest point, the distance between S and the opposite vertex u + w + v (which is the highest vertex by symmetry) is the length of the z projection. Therefore, this length is also the sum of the z components of u, v, and w. This completes the proof.
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  1331. I came across Zagier's one-sentence proof a while ago. I couldn't understand it until I read a longer explanation in "Proofs from the Book". It's really neat!
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  1332. I love this guy, he shows me beauty of the mind. Have a good New Year. I hope you make many more videos like this. Thank you.
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  1333. Each angle of star corresponds to arc of circle. In sum they correspond to full circle. But each angle of star is inscribed angle so it is one half of corresponding arc, so they sum to half of circle which is pi or 180 degrees.
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  1334. Wouldn't the preferred lacing simply be the one that has the least friction throughout, and so, requiring the least "loop-pulls" to both loosen and tighten the shoe?
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  1335. Thank you very much for another excellent demonstration of the amazing beauty of mathematics! I love the 700 years old divergence proof. Also, the unbelievably slow pace of divergence is absolutely amazing.
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  1336. Amazing stuff as usual!! Thanks Mathologer :) Especially the spinning projections in 29:06 completely blew my brain up. I think it's because we're so used to interpreting overlapping lines on a 2D plane (on the page) as faux-3D objects, that interpreting them as just what they are (lines) when they move around basically short-circuits my brain.. once again, well done Burkardt :D
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  1337. Your videos do have a tremendous impact on me, making me wanna attend you at Monash and enjoy the rest being of my life in that kind of Maths you’re reciting to us in every single dope video of yours!
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  1338. Welcome back!
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  1339. Thank you for showing the brilliance and true history here! Often non-western mathematicians, scientists, inventors, artists and others are not given due credit so it is wonderful to see the opposite be the case ❤
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  1340. The real question is whether Mathologer got the reference, or just gave this comment a heart because he's giving all the early comments on his video a heart...
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  1341. 20:33 “Where did that come from?” The big “whoa” moment for me! Thanks for making it so much easier to understand why this equation is significant. I could never understand why when I was taking a course in number theory.
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  1342. got to know about him today. Its such a same, even I have studied PCM in my class 11th and 12th and in my drop year. What a legend he was. When I learn about these people I love mathematics more. My fellow Indian please don't only know about them we have to get to the top in every field to regain our legacy of Golden Bird. Let's do our best to make India great again
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  1343. Ypu can also do (12-2)^2+(12-1)+12^2+(12+1)^2+(12+2)^=5×12^2+2×1^2+2×2^2=720+10=730
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  1344. The transformation of the pentagon ratios to a golden rectangle and the heptagon ratios to a golden 3-d box seems to suggest that the ratios present in a golden n-dimensional hyperbox (probably the wrong word) are present in a (2n+1)-gon. At a quick glance, Steinbach's paper doesn't seem to address this, so is it true?
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  1345. Most memorable: Sad looking portrait of Nicole Oresme along with the Leaning Tower of Lire
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  1346. Mesmerizing. Reminds me of being a kid in early years of learning math and just drawing and playing with shapes. Wish I had this kind of instruction back then.
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  1347. I tried tiling triangles with small triangles. Every row has 2 more small triangles than the row above it, so that the total number of small triangles is 1, 4, 9, 16, 25 etc. (I.e. square numbers.) You can then slice such a tiled triangle so that it envelops another tiled triangle. For example, 9^2 + 12^2 = 15^2. The slices look pretty bad though, and not intuitive at all. 😀
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  1348. Merry Christmas! Still waiting for that fancy Galois theory video you mentioned...
    10
  1349. Every video is so damn interesting and explained incredibly well. Words can't explain how thankful I am to have found a channel like yours.
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  1350. One of the best math channels out there. Your glee is contagious!
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  1351. 8:07 If you can make a 12 pointed star with this, you can design a CLOCK with this fancy movement!
    10
  1352. To take it one (very small step further), this shape of constant width has the smallest diameter of all shapes of constant width containing all 6 points!
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  1353. Me every time a Mathologer video comes out: visible excitement
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  1354. Your explanation did work on me prof. Burkard. My background is engineering, and I played a lot with Optimization, Partial Differential Equation and Numerical Methods. The 50 mins long does not bother me. I believe 50 mins is just right to sums up millenia of math discovery. The moment you said 10 chapters really sparks my joy, and finishing the whole video is really satisfying. This is indeed a Masterclass professor. Can't wait for the next. Thank you very much!
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  1355. Your videos never disappoint, great work. This proof reminded me of that for the Pythagorean theorem that we take 3 copies, scale and bring together.
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  1356. 3,4,7 is actually really useful in music. Intervals of 3 and 4 are periodic complements in mod 12, because they also function as the smallest interval in basic trichords, intervals of 7 are the most common. I often visualize the structure of chords in a way similar to the graphic representation used in this video. The triangles and squares form a continuum that you can rotate along to find relatively valent harmonic structures in a set of closely related keys.
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  1357. Жаль, что я не нашёл этот канал раньше. Это просто потрясающе! Жаль, что у нас в универе нет таких классных учителей по матеше))
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  1358.  @Achill101  That is correct, but since taxes come out first the government doesn't have to make any allegation, investigate anything, or prove anything. They just take the money. Non-governmental creditors can avoid fleeing by this scheme only if they take actions and make investigations.
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  1359. I enjoy partitions as one of the many studies in mathematics that can get mind-numbingly complicated, but starts from a place an elementary school student can understand. Amazing.
    10
  1360. 15:50 The missing chunks look approx triangular, so I calculated the size of the missing triangles, and used that to find a correction factor for the (1+1/n)^n approximation of e. If you multiply that approximation by (2n+2)/(2n+1), you get a much more accurate approximation. For n=100, (1+1/n)^n has an error of 1.35%, but ((1+1/n)^n)*(2n+2)/(2n+1) has an error of just 0.001%
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  1361. Thanks for crafting such a lovely video. The hours spent in photoshop yielded a beautiful result, and the delightful subject matter made the video a joy to watch.
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  1362. ​@muskyoxes That established the existence of photons and was instrumental in the foundation of quantum mechanics. You know, that field of physics which says that God plays dice with the Universe.
    10
  1363.  @funkybibimbap6972  Really? Even with the irony contained in my final sentence?
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  1364. I don’t understand why, when you are explaining how to grow a random tiling at 22:41, you split up the empty 2x4 region into two 2x2 regions. Wouldn’t there be more possible tilings if you were allowed to also split it into 2x1 + 2x2 + 2x1? It seems to me that there would be. And while the tiling you get would be identical in tile positions to one you could have gotten another way, the arrow directions would be different and they would grow into different tilings at the next step.
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  1365. Another nice trick: If you make any move, you can just repeat the exact same move and you will always get to the starting position. That way you can develop your own algorithms too although they ore often quite long and tedious
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  1366. I used my maternal uncle's slide rule at school: 15" boxwood with engraved ivory scales. I believe it goes back to the early 20th or late 19th century. The case was made in Harland & Wollf's shipyard in Belfast, in "repurposed" steel, probably in the early 1920's. I still have it, but I haven't used it for ~60 years.
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  1367. Some energy absorbing crushable elements in cars, called 'crush cans', are designed to buckle like the tin can shown, even to the point of having the first band of triangles pre buckled to make sure that the buckling progresses in the correct manner.
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  1368. As you started explaining the "decomposition", I was reminded of fourier transforms. However, when you then mentioned that this in fact is a fourier transform, I was quite surprised. Not at all obvious that these two things really are the same (to me anyway). Fascinating!
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  1369. Very interesting ! Let's try this : Assuming there's really a Planck length in the Universe and you work with real world coordinates, you couldn't keep shrinking the polygons forever without them converging to a single point. Would that mean there's no such thing as a square grid in the Universe or that triangles are not a thing?
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  1370. It’s amazing how algebra and geometry can be connected by such a pretty formula. And the derivation using recurrence is simple and… simply stunning.
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  1371. pretty sure everyone who ever played around with a casio in their school life will know what the 36th triangular number is. (since when you add all numbers from 1 onward, exactly 36 will fit on most casios, except for newer ones. They allow for more)
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  1372. 2nd problem is 3^n because there are n discs, each of which has 3 'choices' of pegs - same as the number of vertices on the sierpinski triangle 3rd problem can be done by comparing coefficients of 12/59 and 18sqrt(17)/1003, upon which irrational parts drop out - not sure I want to do this.. -btw, thanks Mathologer!
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  1373. Thank you for the fantastic presentation!! 🙏
    9
  1374. That was a wonderful video. I understood most of the things except 41:01 I'm in class 12
    9
  1375. It worked brilliantly. By now you know how much this means to all of us, but thank you, yet again.
    9
  1376. You are the greatest math teacher on the internet. I am grateful and happy whenever I watch any of your videos. Thank you.
    9
  1377. 00:10 Yes, the overlords will abolish all cash and your digital money will only be in one central bank governed by the oligarchs.
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  1378. When I was taught about scientific notation as a schoolkid the slide rule never entered into the picture as to why it was useful, but one of the great things about it is that it removes a lot of the potential for error from the "moving the decimal point" part. By forcing the mantissa (the part that's not ten to the something) to be between 10 and -10 you get a problem that's easy to work out with a slide rule (or if you're fancy a log lookup table) and the significand (the part that's ten to the something) becomes simple addition of the exponents. This is also how floating point works in computer logic, that's just a clever binary form of scientific notation. Multiplying in a digital computer relies on repeated addition plus some bit shifts to double and half the numbers and the size of values has a defined limit in terms of number of bits/significant figures.
    9
  1379. My high school math teacher (this was back in the 60's; I'm 75 now) made and marketed a very nice circular slide rule. His name was Stanley Arlton. I wish I still had mine.
    9
  1380. Cool paradox this also shows why the curve area length formula has to be secant lines instead of just horizontal and vertical. Approaching something doesn’t always mean that they have the same properties
    9
  1381. Now find the ways of making change for a googol euros. (coins denominations 1, 2, 5, 10, 20, 50 cents, 1, 2 euro)
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  1382. Nice debunking as usual. Nodded my head the whole way through. Base10 isn't special beyond its human convenience. It's shocking to think so many people can be wowed by lack of connecting some dots in their own basic mathematical knowledge. But I didn't know much about these interesting diagrams, nor had I thought about decimal places of divisions by two as powers of 5. As always beautiful video.
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  1383. Yes - I like to think of it as adding a new layer of paint, and adding up (integrating) all the layers. For some curves/surfaces the tricky part is making sure the new “layer of paint” is the same thickness at all points. Measured normal to the surface, that is. It can be a challenge for some curves (surfaces) where the offset curve (…) is a different type of curve to the original, such as for a parabola.
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  1384. The 153, 104, 185 triple at 12:00 is the box 9, 4, 13, 17 and you get there by navigating right right right and left :)
    9
  1385. Finally. Amazing.
    9
  1386. Extremely cool! Made me realize there’s a very practical strategy for handling these problems that, in my opinion, beats the pure programming AND pure math approach: programming this sort of thing correctly is nontrivial, and even with a smart implementation isn’t that fast (my quick and dirty python version takes about 20 seconds for $2500 - I just remembered I only bothered with 1, 5, 10, and 25 cent coins). The pure math approach to obtain a closed form expression through generating functions is mechanical and straightforward (and beautiful) but incredibly unwieldy for generating functions of this size. Even if you use Mathematica to do the algebra, you still face the problem of convincing yourself you didn’t make a mistake in writing the code! My approach, inspired by your video, is to simply observe that since the generating function is a rational function, its coefficients will be exponentials times polynomials in the index (and I think a bit more reasoning gets you to the fact that if you only extract coefficients with the same residue mod 100, they’ll just be polynomials (as you show for residue 0, of course)). For N coins, this polynomial will be of degree N-1. So pages and pages of computer algebra, in the end, are just to get N rational numbers! A much more direct way is to write a piece of code that is just barely fast enough to deliver N values of the function, and then solve the resulting linear system for the coefficients of the polynomial! No need for super slick iterative implementations, which will top out way before a million of dollars anyway; no need for huge amounts of algebra. Fast, easy, and delivers an essentially O(1) algorithm for a huge number of problems (ie. anything with a rational generating function, so anything produced by a finite state machine)!
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  1387. Just one line is crossing my mind watching this "It's a kind of magic, MAJIK!!!" ✨
    9
  1388. 28:11 "Euler" ...
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  1389. Mathologer, I have an argument at my end. I'm saying infinity IS used in mathematics and my friend says it ISN'T, that it has no place in maths. Of course we both understand that infinity isn't a number and all the rest, and I know about square root of -1 (is that imaginary number i? Which is really useful in maths and just as silly as infinity? Isn't zero itself in a similar category?) too. Riemann's paradox seems to be just one of many examples in which infinity is used (in the practical sense?) in mathematics. Please do a video explaining how the layman's concept of infinity differs from or is the same as the mathematician's!
    9
  1390. Wow that blew my mind that Pascal's triangle mod 2 gives the Sierpinski triangle, and mod 3 makes a similar pattern. I wrote a script to play with this more and it seems that mod any prime number it gives a nice Sierpinski-like pattern, and mod composite numbers give more messy overlapping patterns that relate to the factors. It's always amazing when seemingly disparate areas of math connect together in unexpected ways!
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  1391. At 12:00 I naturally thought of that formula on my own while I was taking calculus 2.
    9
  1392. 39:16 tile an (ab)x(ab) grid with (a)x(b) rectangles and draw the top-left to bottom-right diagonal. If it intersects the bottom-right corner of a rectangle then a square is formed by (n)x(m) lots of (a)x(b) so na=bm, n!=b, m!=a so a,b cannot be prime.
    9
  1393. This is a beautiful example of SSS congruence / AA similarity
    9
  1394. I have a question regarding 32:19, the challenge at the end. You claim that the existence of integers x,y with x^2 - y^2 = n (> 0, for simplicity) leads to n being odd. As i found the counter example x = 4, y=2 and therefore n=16 - 4 = 12 being not odd , I probably misunderstood you. Any help is kindly taken. Greetings from Germany.
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  1395. Your videos are greatly awaited and they are always worth waiting for. Your videos always generate love for Mathematics. I wish I had a teacher like you in the school days. Lots of love and respect to you. Always ❤
    9
  1396. Your work is simply amazing. I like how the difficulty rises. I always have something to pause and ponder. As usual I will re-watch it a couple of times to try to understand the last part.
    9
  1397. Challenge answer: no two collections from {1, 2, 4, 8, 16, 32, 64} have the same sum.
    9
  1398. 5 minutes gang
    9
  1399. Those crazy polynomials are going to hunt me in my dreams
    9
  1400. This is the reason why I'm getting a PHD in mathematics: the infinite beauty of the numbers.
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  1401. For the second problem: the bogdanov-belski painting. I already knew it was 2. -) But yeah following the video: It's 2x(10²+11²+12²). And 100+121+144=365, goes indeed fast. (memorized) - ) 13²,14² (never memorized those for some reason): I tend to do 10x16+9 and 10x18+16, which is also still fast. (explanation: 13² = (13-3)x(13+3)+3²). So for instance if I have to square 62: 60x64+4=3844, goes decently fast. It's somewhat similar I think to what was described in the video: 7² = 6x8+1 = 5x9+4 = 4x10+9 = 3x11+16 = 2x12+25 = 1x13+36. Note that the added numbers are all squares. Each time you take a column of a squared number and you turn it into a row, you'll lose a square.
    9
  1402. I did this in class in first semester. I was a kind of dizzy, thank you for helping me understanding and appreciating this after so many years.
    9
  1403. Hearing "difference equation" just gave me flashbacks to my second year electrical engineering courses, where we worked with discrete signals and needed to use difference equations and z transforms
    9
  1404. Excellent point, and an easy way to remember this characteristic of cyclic quadrilaterals!
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  1405. So glad I stumbled on some of your videos today. I remember when we both worked at The University of Adelaide. Your enthusiasm and humour have inspired many, many people for several decades. I am thrilled to see you are still actively exciting people about Mathematics. I will have to view all your videos when I can! Best Wishes, DB
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  1406. Wow ... sincere thanks ... wasn't aware of this at all ... Great video, as usual ... Cheers
    9
  1407. I stumbled on something similar to this in my last year of high school, 31 years ago... but my analysis of the nth powers resulted in a reduction to n! with some additional terms that I couldn't figure out what to do with at the time. I put my expanded tables into a spreadsheet that I still have... and I'm going to have to have another look at my data with this video in mind.
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  1408. This wasn't discovered too soon because notation wasn't very precise and polynomials were not useful those times.
    9
  1409. Awesome! Your visual sum of Ln(2) @22:04 can be also be Ln(m) = \sum_{i=n}^{m*n}1/n for large n
    9
  1410. This is such a brilliant video. I am so happy I watched it. Initially I wanted to watch it in two sittings but I could not take my eye off it.
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  1411. Last puzzle: The area of the middle square is 0.2, since you can divide the big square of area 1 in five equal squares of the desired area. 1/5=0.2
    9
  1412.  @Mathologer  😢
    9
  1413. Bad ones? Why not naughty?
    9
  1414. ​ @Mathologer  cos(π/7) is transcendental or not
    9
  1415. As a colorblind, can you not color like you did with good and bad numbers? It may be very clear to people without vision problems, but green/red is the most common colorblind problem, using blue/red, blue/green, yellow/red is not a problem, even a strong red with green is generally okay. Great video again, nice job!
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  1416. Yeah! When we got to the sum of 5 pentagons I started thinking a lot about how it might relate to the Fourier series.
    9
  1417. Wow!! It's always a great pleasure to see your animated proofs! Challenge myself as a teacher to improve my skills and imagination for my lessons. Thank you a lot!!
    9
  1418.  @Mathologer  He must have had an amazing childhood in mathematical wonderland.
    9
  1419. "Oh , I know about this puzzle" "how tf I never thought about multiple pegs"
    9
  1420. 6.10. : Me: What if you cut out 2 black and 2 green squares? 6.20. : Mathologer: What if you cut out 2 black and 2 green squares? It's like he read my mind.
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  1421. Taylor series expansion of tan inverse x........then put x=1
    9
  1422. Laughted a lot with the t shirt
    9
  1423. Cool shirt as always doc!
    9
  1424. "YOU'RE GODDAMN RIGHT I AM"
    9
  1425.  @Mathologer  I think it is because when we saw the West with all these technological advances and compared to our own country, We just see that we have nohope for the future with ever increasing population. So nobody even bother to look at the history, because we never had any kind of industrial achievement. But we forgot that it is just the brain and knowledge which is a necessary recipe for advancement. Can you describe the mathematics and engineering required to build Ellora caves in India. We just have a feeling of inferiority based on the current condition of India in comparison to the country like U.S.
    9
  1426.  @kinshuksinghania4289  this one was one of my favorites for sure. Completely peaked my interest in the pythagorean theorem as I was already interested because it's basically the foundation of trigonometry which also makes waves and fourier series and a lot of stuff in physics ofcourse. And I have a particular interest recently in the connection between Fourier series, Taylor series, ect, and Calculus. I mean have u noticed that the Fourier transform is basically using an infinite number of curved lines to make a straight line, and an integral is using an infinite infinitesimal straight lines to make a curved line?
    9
  1427. Great ways to conceptualize and visualize properties of logarithms. One of my favorite videos.
    9
  1428. One of the best teacher of mathematics.❤️❤️❤️❤️❤️ u sirji
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  1429. I've been recently working on Tower of Hanoi animations for my channel and in the process discovered some of the stuff mentioned in the video, for example the algorithm to solve Tower of Hanoi without recursion by moving the smallest disk every second turn. I also proved why it works. Answer to challenge questions: Direction to move the smallest disk is: if (n is even) then Right else Left That's because to move (N)-tower in some direction, you first have to move (N-1)-tower in the opposite direction, so the direction to move the smallest disk changes every time and it starts with Left for N=1 disk The number of turns is (as mentioned in the video) equal to the number of reachable states, which is = 3^n, because: there are 'n' disks, each disk can be placed in one of 3 towers, so there are 3^n valid states. Each valid state is reachable with a standard recursive algorithm, so there are 3^n reachable states. Another interesting puzzle for you: given a number of disks (N) and a number of turns (T), how do you quickly find the state of all disks in the optimal solution of N disks after making T turns? I had to solve this puzzle while making my Tower of Hanoi animations, so I can instantly preview my animation at any moment in time (instead of re-playing it from the start), which becomes increasingly important with several-hour animations. Check out my channel for these animations: I've already published the shorter ones and longer once, including 10-hour 16-disk solution animation, coming soon:)
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  1430. I have a pet peeve about math videos that claim to discuss inherent properties of numbers but are really discussing representations of numbers in base 10, without generalizing. We have here a video in which the unmentioned special treatment of base 10 appears to come from the numbers themselves, not from the person describing them. Incredible! This marks the first time I decided to pay a creator using this donation system, as well as the first time I noticed this option exists. Imagine my frustration when I learned that donations are limited to values defined apparently by the prime factors of 10. When I was an engineer at Google, at least half my coworkers had some kind of degree in math. I guarantee you someone there is annoyed by this as well :)
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  1431. Just dreaming here: Mthologer creates a full course on class field theory. With historical references and proofs. That book you briefly flashed might be a backbone.
    9
  1432. The formula shown at 24:00 works for any dollar amount if you let n choose m = 0 for n<m, which is a natural definition. It only falls apart when you expand the formula.
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  1433. Good idea
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  1434. Like the short videos. Call then "mathologer quick bois"
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  1435. I have not verified this myself, but I was told that the fact that R^n has exactly one smooth structure for all n except n=4 (where it has uncountably infinitely many) is related to the fact that 2*2 = 2+2 = 4
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  1436. snooker fans are horrified
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  1437. The are of the tree @16:00 I’m assuming the angle is 45 degrees. The side length of the trunk is 1. (1x1=1) Focusing on just the first branch we know that the side length is sqrt(0.5). (sqrt0.5 ^2 + sqrt0.5 ^2 =1) So the area of one first branch is 1/2 (sqrt0.5 x sqrt0.5). But there are 2 of them so the total area of first branches is 1! By similar logic the area of all second branches is 1, and so on. Total area of tree =5.
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  1438. 31:38 For a long time I was hoping the pi^2/6 identity and its relatives could be held up as a counterpoint to "everything is easier with tau" zealots. Then, wanting to know if there was a general formula for the multiples of the powers of pi in those identities (whose numerators and denominators have OEIS sequences, btw), I learned about Euler's general form for those sums and how it contains 2*pi as a unit. 😐
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  1439. "On two occasions I have been asked, — "Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out?" In one case a member of the Upper, and in the other a member of the Lower, House put this question. I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question." Charles Babbage.
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  1440. Wow, ich hätte nie geglaubt, dass etwas das mit Analysis zu tun hat auch Spaß machen kann... :) Sehr cool!
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  1441. For the challenge near the end related to the Wallis product, if you simply eliminate the 1 in the denominator then each factor is < 1 converging to 1, meaning the overall product is finite.
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  1442. Amazing story. Makes one wonder if those old European guys ever caught wind of and (perhaps) were "inspired by" the much earlier, related mathematical work in India. ...seems unlikely but still fun to ponder
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  1443. First challenge: Π(1 + x^2^i) from 0 to n = Σ(x^i) from 0 to 2^(n+1) - 1. Nice! :D
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  1444. Yeah, infinite sets make sophistry an easy task, because standard logic dictates that there are no greater or lesser infinities. Problems like this one, among others, prove that this is not the case; you just have to add to the infinity in a different direction.
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  1445. It will NOT reduce to proportional splitting. You can only game it if you are the only one splitting your loan, if everyone else divides their loans too, the distribution of money will not change.
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  1446. Most memorable : Euler comes in all fascinating formulaes !!! And ur math relating shirts😁 Pls can u pls make a video on Euler's line???
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  1447. i can't help but smile whenever self-similarity pops up out of seemingly nowhere!
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  1448. Most memorable: Kempner's Proof. Just pipping the 100 zeros weirdness.
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  1449. I love it! What a great video! Your deep dives into maths are amazing and I can't imagine just how much editing goes into all these presentations.
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  1450. You've shown me the true beauty in math. Your videos are truly intellectually stimulating
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  1451. Have you been 4k+1 this year?
    9
  1452. It is a good day when Mathologer uploads
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  1453. Most memorable : " Convergence of no 9 series ".
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  1454. All hail lord curriculum
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  1455. 9:21 Let's find length with more than 10 identities of type 1+...+1+A+B=A*B*1*....*1 (that turns out to be relatively easy): We need just A*B-A-B = const for sufficiently many pairs of A and B. So we rewrite A*B-A-B = (A-1)*(B-1)-1. Thus we take some number with many divisors (M) and take all different pairs of numbers C and D, such that C*D = M. Then we take A=(C+1) and B=(D+1) and we get indentities of length A*B-A-B + 2 = C*D + 1 = M + 1. EDIT: should keep watching before posting, this is exactly what is explained later (around 14:45)
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  1456. Great video as always! 2:28 "1/x was hiding in a spinning cube" because a hyperboloid of two sheets is a ruled surface. 9:00 (Wish we could use A(X)=ln(X) here :P) Consider the region bounded by {x=1, x=Y, y=0, y=1/x}, after squishing and stretching it becomes the region bounded by {x=X, x=XY, y=0, y=1/x}, thus A(Y)=A(XY)-A(X). 25:40 We have the parametrization (cosh(a), sinh(a)) for x^2-y^2=1. But I never realized the parameter a is indeed the area.
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  1457. Cool! I wonder how many mathematicians/geometrists have realised this in the past but either assumed it was already widely known or thought that it was trivial, so never bothered to publish it.
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  1458. 47m20s – "What's next 3 1 4 5 9 ? :)" Did you forget something? No, apparently, π forgot something! Thanks for another enjoyable safari through some mathematical curiosities! Fred
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  1459. I made it to the end but... Der anfang ist die ende und der ende ist der anfang.
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  1460. I'm 65 years old. I graduated in mathematics with a first class in the year 1977 from Madras University! It's a shame none of my Indian professors had anything to say about these great Indian mathematicians! It's all Newton and Lebnitz and Euler!
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  1461. Ok after playing with python and some databases, I've realised that the outcome can only be determined by nested statements or recurrent algorithms. Recursively, the 666th statement is too far outside my ability to calculate. However, I do realise there is a way to work it out faster, but from what I can see up to pausing the video at that point, I could not program it using a simple equation without recursion. If there is a simpler way, pleeeeease tell me. Stuff like this sets my head on fire and I don't want to spend months thinking about this, I have another project I'm already beginning to procrastinate on because I'm thinking about this one lol
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  1462. Okay, this is actually one of the most beautiful things I've seen in math.
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  1463. Used that device at work making elliptical coffee tables with the aid of a small woodwork router, very useful device. Way better than two nails and a loop of string. Great video bye the way.
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  1464. Took me way too long to realize that I've already learned some of this (nearly all to the cyvlic quads, including proofs for them) in highschool (special, math-heavy class, it is not in the base material)
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  1465. This is an absolutely amazing video. I needed a few minutes to come back to my senses after watching this
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  1466. I have a vague memory of seeing a circular slide rule once and thinking that it would make sense to do it that way. Of course, it would be harder and more expensive. Slide rules were almost universal when I was in engineering college (1973 grad). In the last year or so a few people had HP calculators with four functions and very little else that cost about $200.
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  1467. Leibniiz would have gotten hold of many mathematical discoveries from India.Kerala school school of mathematics was very advanced in the 13th and 14th centuries in India. Taylor series was also discovered in India
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  1468. What a pleasure! So many new perspectives that seem to have been just waiting to be noticed: 1. Looking at y=1/x as a shape invariant to stretch&compress 2. Defining ln(x) and e^x using the curve 3. deriving cosh+sinh identity from their geometric definitions (as opposed to the exponential identities definitions)
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  1469. Every time I see something like this, I wish (again) that I had nothing else to do than to learn Mathematics.
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  1470. 11:50 Arrange the bricks in a two layer formation where, on the bottom layer, you place a single block with its center of mass on the cliff's edge (akin to a single-block maximum overhang position). Then, on the second layer, place both remaining blocks with their centers of mass aligned with each of the bottom block's edges. If done correctly, the block placed over the cliff should create precisely a 2 unit overhang (assuming all blocks are 2 units long), with the other brick on the same layer acting as a counterweight. We would need to assume all blocks weight exactly the same, have perfectly equal shapes and there are no external forces aside from gravity acting on the system. Below I'll try to make a small ASCII schematic to illustrate the formation ______ ______ ______ -------------| Cliff |
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  1471. Note that a proof of an equality that only works in a special case that's still sufficiently generic (in other words, where there are just as many free variables in the special case as in a generic situation) can often be generalized by analytic continuation. Inequalities need more careful handling.
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  1472. I didn't learn Heron's formula in school and I was taught in Portugal. Very good video as always, just found a typo at 25:03 (the third term is A+B+D-B)
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  1473. Mathologer just roasting introverts that go to parties
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  1474. Given the 1/x shape, I was kind of expecting the squeeze and stretch method to be used again...: If you squeeze the width and breadth (i.e. diameters) of the horn by a factor 2, and stretch its length (i.e. axis) by a factor 2, then its volume halves but the resulting horn is the same as the original with the part between x=1 and 2 (in the graph) cut off. Similarly, if you squeeze width and breadth by a factor (1+eps) and stretch the length by (1+eps), then the horn has a factor (1+eps) less volume, but equals the original with a tiny eps-thick disk missing. That disk is like a thin cylinder and has volume pi*eps. So if the whole horn has volume V, then we get V = V/(1+eps) + pi*eps which simplifies to V*eps = pi*eps*(1+eps) or V = pi*(1+eps) which in the limit for small eps becomes V = pi QED But then you would admittedly have missed Torricelli and his anagram. Maybe I may suggest my username which is an anagram of my real name...? 😋 Unfortunately, that trick doesn't work for the area for reasons that the reader may ponder (but neither does Torricelli's).
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  1475. Most memorable: The proof that the bishop came up with, beautiful simplicity
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  1476. Hi, 5 minutes into the video and first thing I thought was an exercise in an old projective geometry book of mine that uses the false position method to solve the Cramer-Castillon problem. Somehow this reminds me a lot of it. I'll see if tomorrow I can link the two somehow.
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  1477. Awesome video. From the animation to the soundtrack to the explanation. Just amazing.
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  1478. In case you are wondering, the notification for this video works for me. With crazy YouTube algorithms many creators are talking about these days we need these notifications.
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  1479. You know what the worst part is? The colonial education system that exists in India still does not acknowledge this. We were taught the very same things parroted in the West. We called it the Leibniz Series and we always acknowledge Newton and Leibniz as the inventor of Calculus. Their contributions have been innumerable but not even an acknowledgement of someone as great as Madhava back in his home country irks me to the core.
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  1480. I'm forever grateful for the visuals lol as always. Thanks so much for being such a great teacher. I'm not even into maths, but i love learning. So many answers to grown up questions can be found in these videos of yours
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  1481. Your commant proves you do not understand the value behind Judaism, because this is a moral dilemma masquerading as a purely financial one. Real old Judaism mostly occupied itself with court judgment which are problems that occur between humans, some of them are financial and god has no part in them. To illustrate the point, giving charity has rules in the talmud/old testament and that to relates to moral-financial problems in the same vaine.
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  1482. What happens is you make a circle with the cards? I guess you'd get some kind of indeterminate result (half ones and half zeros) ?? ... a probability distribution of 'n' cards after 'x' trials??
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  1483. Question for Mathologer: Why does the Fibonacci sequence start with 1 1 and not with 0 1 ? I'm guessing it has something to do with the reputation had by zero at the time Fibonacci discovered his sequence.
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  1484. Please make the video on e^ (gamma) gamma is euler's special number you were talking about ,and what is solution for continued fractions containing e^(gamma) You had already said to make one but i didn't find it Please I'm starving for it😂
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  1485. 14:40 Those glasses should work with the determinant, because you can build the matrix as given and give the 4 squares in the holes the highest numbers (23 and 24). That would give the matrix for the board without holes. For our case: Just ignore the "new" last two rows and columns, so get the sub-determinat which would be exactly the same as if we had built the matrix just as is. So why does this not work for any holes? Where could it go wrong?
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  1486. I love your channel, but the AI images and videos are really unsettling.
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  1487. For added “fun” this shows that the extra path length for the finite cases is the sum of the non-orthogonal movement introduced
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  1488. YES! The guy with the towel hat!! I've always always wondered about this image of Euler, and what he was wearing on top of his head! Lol!!
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  1489. Man I've been waiting FOREVER for someone to bring this up!! This pops everywhere in quantum mechanics, Apollonian Gaskets, Ford Circles, Fractals..You Name it! Always had the feeling that the Theory of Everything would somehow be related to this! We need to keep it alive at all costs! THANK YOU!
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  1490. SOOOOOOOO coincidental! I was just going over some of these ideas this morning. Question: if 2^n is the analog of e^x, does that mean (2^(i)*n +2^(-i)*n)/2 is the analog of cos(x)?
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  1491. "Or...make yourself famous."
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  1492. Yes. Very strange. I confirm that this video was hidden from me too. Anyway I'm glad I watched it. A lot of hard work to explain rings.
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  1493. I'm looking forward to the Galois theory video(s i hope :)) a lot! Great work, but i have to admit that i'm not Gauss - reincarnated so rewatching the last approximately third will be necessary. Thank you. I especially enjoy that you seem to have so much fun at explaining, I guess your students appreciate this very much. And the jokes are always extraordinary :D
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  1494. Also worth noting about this sequence (the first few terms are shown at 41:11) is that the odd-numbered terms produce all the Pythagorean triples in which the legs of the right triangle differ by one: 1/1: 1 = 0 + 1; 1² + 0² = 1² (trivial case to get started) 7/5: 7 = 3 + 4; 3² + 4² = 5² 41/29: 41 = 20 + 21; 20² + 21² = 29² 239/169: 239 = 119 + 120; 119² + 120² = 169² ...and so on. Every Pythagorean triple of the form x² + (x + 1)² = y² is hit.
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  1495. Triangles come in pairs that you turn into each other by folding them inside out, lovely! And thinking about the vertex angles: assigning identical angles the same color. an Isosceles triangle (RR, GB, GB) turns into a different isosceles triangle (RG, RG, BB) but only an equilateral triangle actually turns back into itself? edit. Whoops, i got too ahead of myself and wrote this, right before you explained the isosceles
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  1496. The most memorable is definitely Kempner's proof. Seeing it, I thought to myself "damn, I should have seen this coming!". PS. if I get lucky, please don't send me "The Mathematics of Juggling", I already have it :)
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  1497. x to doubt, actuaries are bad at math and worse at calculus - src am actuary (gladly left)
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  1498. p=x^2-y^2=(x+y)(x-y) => if p-prime, then x=y-1 => p=2x+1 (proof of the unique)
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  1499. Wow, your content is amazing, and even more so, because it is actually original. How are we worthy? Schöne Grüße aus Deutschland :)
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  1500. 13:36 - Numerator is equal to 2x the denominator of the previous term plus the numerator of the previous term, denominator is equal to the numerator minus the denominator of the previous term. Maybe not the simplest rule but it’s the first one I saw, by looking at the sequence of partial fractions. I appreciate the little challenges included in these videos. Not many math YouTube channels include them. Most of the time I don’t go for them, but whenever I do and find the solution, it’s rewarding 🙂
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  1501. Youtube tried to give back the IQ points I have lost watching Schiff argueing for impeachment by recommending this video.
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  1502. @mikemthify Roger Heath-Brown was 19 in 1971. Could you post some sources?
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  1503. Beautiful theorems. Elegant presentation. Bravo!
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  1504. I wonder : if one were to 3D print these pseudo3d timings into actual 3D shapes, would the « arctic circle » turn into an arctic section of a sphere of sorts? What about matching pairs? And what about higher dimensional tilings ?
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  1505. A slight alteration to the overshoot/undershoot algorithm I believe already gives you uncountable infinite ways of doing it. The alteration is that you can overshoot and undershoot by as many terms as you please, and you can choose a different number each time. This still works because the positive and negative terms sum to infinity. Because you’re choosing a natural number by which to overshoot/undershoot at each stage, the number of choices you can make is equal to the cardinality of the reals (because each choice corresponds to a sequence of natural numbers) which is uncountable.
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  1506. Would it make sense to get a mathologer video someday on the continuum hypothesis? I know that it's undecidable but I have no idea why. If there were a way of making that fact familiar and intuitive, that'd be a big deal to at least a very countably finite subset of your audience
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  1507. Fun fact: if you have a circular slide rule as an avatar, there is a lot of questions like "What a strange scale on this stopwatch?"
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  1508. I work at a bank and one flaw I can see with this method is that when providing a loan there is no transparency as to how much the bank can expect to recover upon non-payment of the loan. This is an important variable for banks and is why they can provide mortgages with relatively reasonable rates, since they receive (a portion of) the mortgaged property upon default of the home owner. If I provide a loan to a company, I can estimate from financial records an expected total recovery amount, but I would typically not know how many other creditors there are.
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  1509. To answer your question about other countries (although you might already know this): Suggesting to include something like Heron's formula into the mathematics curriculum in Germany would make the other commission members look at you as if you had told them to leave the building because a marsian space ship has landed on the roof.
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  1510. Very nice theorem! This folding process gives some sort of "sides-medians duality": -The sides of the folded triangle are each 2/3 the length of the corresponding medians of the original triangle; -The medians of the folded triangle are each 1/2 the length of the sides of the original triangle. This proves the 1-time-folded triangle is in general not similar to the original one, but the 2-times-folded one is similar to the original one, with a lengths ratio of (2/3)*(1/2)=1/3.
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  1511. When I saw the penrose tiling at the end I gasped! So interesting as always ❤
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  1512. Try to find a mathematical branch where you don't find a speck of either Euler's it Gauss' contribution. I'll wait.
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  1513. Now that is an interesting way to deal with fairness. Impressive.
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  1514. The little German that I know, I learnt from a Dick Smith Electronics print Ad from decades ago. It is how to complain about rip off retail store prices: "Gross schoppendollarpreiss, upderjumpersticken!!"
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  1515. I liked the crazy optimal overhang tower the most. Didn’t expect that at all.
    8
  1516.  @Mathologer  Oct 1973 Scientific American, reprinted in "Knotted Doughnuts and Other Mathematical Entertainments".
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  1517. I'm so impressed by the way you make difficult mathematics easy to understand. Thanks for inspiring me to be a better teacher :)
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  1518. You are too pretty
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  1519. Fantastic lecture. Btw no equilateral triangle actually follows quite quickly from picks theorem & area =(1/2)s^2sin(60)
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  1520. I get the same, n²(n²-1)/12
    8
  1521. Hello there
    8
  1522. 24:13 The location of the center of Feuerbach circle for the 3-4-5 triangle is on the same horizontal line as the incircle, therefore the outlined procedure will produce a degenerate triangle (a line).
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  1523. Mr Mathologer, do you know any visual proof of Pick's theorem? Using that, it's easy to prove that no equilateral triangle fits into a square grid: Let's suppose that one of these triangles exists. By Pick's theorem, the area of any polygon with its vertices on grid points must be n/2 for some integer n. (For the equilateral triangle, this is also easy to see by inscribing the triangle into a rectangle and then subtracting three right triangles with integer legs.) But if we name "s" the side of the equilateral triangle, then its area is (√3/4)s². As s² is integer (because it's the Euclidean distance between two grid points), and using the fact we previously saw, then (√3/4)s²=n/2, meaning that √3 is rational. As always, thanks for your amazing videos!
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  1524. So, there is no way to advance 5 steps in in this solitaire army game. And there is also no solution to a general 5-degree polynomial involving only basic arithmetic operations and radicals. Coincidence? I don't think so ....
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  1525. As always T-shirt on point
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  1526. I have loved each depth of unveiling the mathematical concepts ❤ Thank you, mathologer, for building concepts deeply always😊
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  1527. 10:43 The number of rotations equals the distance between the center points between both coins divided by the radius of the rotating coin.
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  1528. Hello early gang!
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  1529. I am very happy that Your channel continues to work. Your wonderful deeds, o Lord.🙊🙃
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  1530. I am so grateful to you sir, for giving the credit of discovering Calculus back to Medivieal Indian Mathematician from that of Newton ... Kindly search for more Mathematicians like Bhaskaracharya and you'll be amazed to discover most of the discoveries of Maths as modern world know of ... Thanks once Again sir
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  1531. Love your channel =) one of the best math channels
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  1532. you have blown my mind once again. absolutely loved this video, and the animation was amazing! amazing work!
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  1533.  @Filipnalepa  the banana has a concave shape. You might think that if you were dealing with a convex shape your intuition would have been correct. Unfortunately, it is not the case. However, any line that goes through the center of the mass of a shape, cuts it in two halves that have equal, but oppositely signed "first moments of area". You see, the area of a shape is the integral of (dA). The first moment is the integral of (x dA), where x is the location vector of the infinitesimal area element dA. You can imagine the area as the "zeroth moment" of area. Your intuition told you that the halves had equal zeroth moments, while the reality is they have equal but opposite first moments.
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  1534. I wish the video had been 10 minutes longer with just a tiny bit less appealing to the gods. Thanks for this. (High school math teacher / enthusiast)
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  1535. Fermat was where "The proof is left as an exercise" started.
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  1536. Ah yes, I gave a talk to the math club at my school demonstrating the difference calculus applied first to polygonal numbers, then I revisited the triangular case and derived the tetrahedral formula, ending with a pascal triangle surprise. The interplay between the discrete and continuous is, in my opinion, an understated mathematical motif which I felt the need to highlight in my one and only pedegological presentation.
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  1537. 13:19 That is an easy one. For any n number of disks, you just change the direction depending if n is even or odd. N == odd -> direction == clockwise. And N == even -> direction == anti-clockwise. ;)
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  1538. 14:25 I think it's no coincidence that the opposite side of 1 (half a circle) is about 3.14, which is pretty much pi.
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  1539. It's no coincidence that it is the sqrt(10). The distance is log(10)/2, (let's say natural log), and so to get the units we do exp(log(10)/2)) which is sqrt(10).
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  1540. I love the little giggles. Could be a bady in a bond movie
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  1541. Mathologer Are in Melbourne. ?? From a Melbournian
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  1542. I love it. Congratulations! Euler is a spectacular mathematician, even today.
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  1543. I recently solved that problem on codewars and came up with the 10 rule myself (looking at the results of many different triangles). Helped me solve it fast enough
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  1544. Mathologer videos are always such an inspiration to me. I'm no mathematician, but I enjoy a bit of mathematical dabbling. Most of my exploration is what might be called empirical mathematics. In short, I look for patterns without bothering too much about proofs. To test my pattern-finding ability, I paused the video at 29:01, to see whether I could identify the next few members of the family. I got the following: 9² + 40² = 41²; 11² + 60² = 61²; 13² = 84² = 85²; 15² + 112² = 113²; 17² + 144² = 145². The general pattern could be expressed as (2n + 1)² + (2(n² + n))² = (2(n² + n) + 1)², where n is a natural number. The pattern for the family shown at 30:12 was even easier to identify. The next few members were: 63² + 16² = 65²; 99² + 20² = 101²; 143² + 24² = 145²; 195² + 28² = 197². The general pattern for this could be expressed as (4n² − 1)² + (4n)² = (4n² + 1)², where n is a natural number.
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  1545. I honestly don't know any mathematicians who aren't willing to acknowledge mistakes. I mean it's always hard but it's part of what I like about math. I mean if you aren't making mistakes you aren't working on hard enough material and the nice thing about math is you can't just pretend that your mistake was correct.
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  1546. But what if we take Input = {n^k} for any k > 2 ?! If you post the answer before me, you get a cookie.
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  1547. That's true, the primes page on Wikipedia doesn't exist in any other language yet.
    8
  1548. ​ @zlodevil426 "you only say the food is disgusting because it's made out of trash"
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  1549. RIP Graham, a true GOAT 🐐🐐
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  1550. Loved the video! Definitely really easy to understand compared to other Master Class videos. I think it helped that it was very interesting so taking it one step at a time didn't cause me to lose interest
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  1551. 20:57 As a French, I was taught "ln" stands for "logarithme népérien" after John Napier, discoverer of logarithms (John Napier's name is often francizied as "Jean Neper").
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  1552. math good yes very math number
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  1553. The problem I see with the contested garment rule is that it promotes people claiming as much as possible, preferably all.
    8
  1554. To honor Madhava of Sangamagrama, there's a (rotated) indian flag in the lower left corner at 11:40. Unintentional Christmas egg?
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  1555. I'm from Kerala and I am feeling proud after seeing Malayalam.
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  1556. 2nd puzzle is easy. (2*365)/365. And yes, I also noticed those patterns and extended them to 0. Tried to look for other ones with different exponents (after I learned of Fermat last theorem, I looked for non-positive and non-integer exponents), but to no avail. I always liked that first pattern doesn't have a "hole" - so it uses all numbers. I always wondered if the unused numbers in 2nd pattern can be used for something. I guess I will sit down with my pen and paper later this evening to check if there isn't some interesting there. Even if not as beautiful.
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  1557. How do you prove ramanujan’s continued fraction for phi? Or formula for pi?
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  1558. I did not expect to ever see the trollface in a mathologer video 😂
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  1559. 15:35 very nice way to teach binomial coefficient
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  1560. It always kills me that Tesla died alone and penniless.
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  1561.  @jannikheidemann3805  I don't know. I'm from Germany. I grew up in the 90s an 00s.
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  1562. My school in Britain taught Latin. I think Latin was an essential requirement if you wanted to be accepted as a student in Oxford and Cambridge Universities. A little indicator of a well educated candidate. Not any more of course!
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  1563. 8:10 that's a big ππ formula indeed
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  1564. Hallo Burkard, Ich schreibe Dir einfach mal auf Deutsch, weil mir das ein bisschen einfacher fällt. Ich möchte Dir einfach mal Danke sagen! Dein Video zur quadratischen Reziprozität hat mich zum ersten Mal mit der modularen Arithmetik in Kontakt gebracht und, was soll ich sagen, jetzt bin ich im JugendForscht-Wettbewerb (in der Altersparte "Schüler Experimentieren", weil ich noch etwas jünger bin...) Landessieger in Berlin mit einem Projekt aus dem Fachbereich Mathematik, inspiriert durch Dich! Falls es Dich interessiert, die Preisverleihung ist auf dem Youtube-Kanal des innoCampus der TU Berlin zu finden. (Es ist ja alles digital gewesen; ich bin ab 40:24). Es ist wirklich schön und auch irgendwie beeindruckend, was für ein Projekt aus dieser Inspiration und diesem tollen Video entstanden ist. Ich weiß noch, als ich diesen ersten "Wow!"-Moment hatte und bis nach Mitternacht wach war, um die ersten Ansätze auszuarbeiten. Ich finde es überwältigend, was die Mathematik bisher alles Großartiges bereitgehalten hat, was ich dann entdecken durfte, und das alles nur Dank Dir! Ich bin mittlerweile sogar soweit, dass wieder einen Bogen zur quadratischen Reziprozität selbst bekommen habe! ;) Aber auch generell möchte ich Dir einfach mal danken für Deine tollen Videos, die mir durch diese schöne Mathematik immer wieder den Tag versüßen, wenn nicht sogar der Höhepunkt sind! Sollte Dich vielleicht interessieren, was ich so gemacht habe, kannst Du mich gerne kontaktieren! Viele Grüße und noch einen wunderschönen Tag, Tim
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  1565. Well, you’ve set me down a rabbit hole again for at least the next week. Amazing video as always!
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  1566. Thanks for the mention! I love when I stumble across a comment like this, especially on a channel as great as Mathologer :)
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  1567. 46:28 okay, now that is the sneakiest conjugation I think I've ever seen
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  1568. The thing about Euler's solution to the Basel problem is that the maths in it isn't actually very hard to follow, it's sixth-form-level stuff really. But the reason the Basel problem stumped the likes of Fermat and Bernoulli was because the solution requires thinking outside the box, and skating on thin ice with regards to mathematical rigour. Euler took a few steps that more experienced mathematicians might have disregarded as nonsense - but he got the right answer, didn't he? It was Euler's creativity that was what really unlocked the problem.
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  1569. No. The counterexample is easy, remove a2, a3, b1, c1. a1 is isolated and therefore no tiling is possible
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  1570. Video uploaded 1 minute ago and already had 5 likes. I am really impressed by the amazing mind power of mathologer's viewers where they fast forwarded a 40 minute video to watch in 1 minute and still understood it well to like the video at the end. Or may be they used some kind of algorithm i am mot sure. Great job guys!
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  1571. 31:00 I guess the Queen of Hearts is the missing card.
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  1572. Had an incredibly rough week, this was precisely the pick me up I needed!
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  1573. Hero's formula for the area of a triangle is one of those things introduced as a curiosity in a math textbook way back in middle school that I never really got a handle on. To a sixth grader, that is an impressively complex formula for an ancient to have discovered, and no proof was forthcoming. Through high school I saw it a couple more times, always in passing, a sort of neat oddity that seems compact but rarely gets used in practice. It was really neat to see an intuitive proof and motivation after all these years. That said, it doesn't exactly seem useful. Even if you somehow do know the lengths of a triangle but not its angles, this formula is still not the fastest way to find the area. Typically, if you're doing this by hand, you will either have a table of square roots (for Hero's method) or of logs and logs of sines (for the law of sines method). That method is still faster, because you skip all the multiplication steps. If you want to compute the area of the triangle with a computer, you can use Newton's method to get the square root, and I assume Heron's formula really is faster. But the thing is, you basically never know all the side lengths of a triangle (and nothing else) before trying to find its area. Rather, you probably have coordinates, in which case the shoelace formula is by far the fastest. So like, what is this formula actually good for? Is it just a novelty like the quartic formula? If it's never used, then no, I don't think it should be taught as part of a standard curriculum. The brief mentions in books for interested students are probably enough. There is so much I want to add to the math curriculum, and the curriculum is already packed as it is. It's hard to justify cramming in more random formulas to teach, prove, and memorize. (BTW, although the phrase "Heron's formula" is seen pretty often in mathematical texts, in pretty much all other contexts in English, "Hero" is far more common than "Heron." Similarly, we say "Plato" rather than "Platon." The practice of Latinizing ancient Greek names is pretty standard in English. In classical Latin, the nominative singular would be "HERO," and the genitive singular would be "HERONIS." Since the Latin stem is still Heron-, the English adjective would be "Heronic" rather than "Heroic." Again, that's like the adjective "Platonic" rather than "Platoic. Other examples include "Pluto/Plutonic" and "Apollo/Apollonic." Admittedly, there are some exceptions, like the word "gnomon.")
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  1574. It's 12:05 AM here, what a Christmas gift mathologer thanks! 😍
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  1575. fascinating, I could watch you for days!
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  1576. Can we get a round of applause for the most inefficient way to calculate factorial?
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  1577. What a delight to watch your video's! Being a math teacher myself, I cannot help but notice the similarities in how we teach. Especially the animated (sometimes hand wavy ;) ) proofs are sublime. Most educational math videos on YouTube sure lack proofs and just summarize/explain statements. Hats off to you dear Sir! Hopefully you'll keep on educating us all!
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  1578. For some reason, I'm gonna remember this video as the Pigeon Pizza Pi Hole Principle.
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  1579. The connection to physics: If you have a physical wave with frequency 10Hz, and another with frequency 4Hz, you get an interference pattern. Map this with a fourier transform onto a circle and you see a recurrence. Now if you change one of these waves to be non-constant frequency…. Which is not mysticism, but understanding the effects of repeated physical interactions.
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  1580. 28:34 Wait a moment! Take a closer look at the second row... where's the 5? xD
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  1581. If you find the “infinite tail of zeroes” hard to swallow, it might help to realize that the only thing that controls whether a particular fraction has an infinite tail of zeroes or an infinite tail of something else is your choice of numeric base (which is not a fundamental property of numbers); for example 1/3 is 0.333... in decimal, but it’s 0.1 in ternary. If you’re still unwilling to accept tails of infinitely many zeroes as equivalent, then take this: 3/4 = 0.75 = 0.75000... but also exactly = 0.749999999... a sequence of infinitely many 9’s.
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  1582. This is really interesting! Thank you 💕 You are my favorite math youtuber! <3
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  1583.  @maypiatt3766  and this channel is also inspiring mathematic thinking and curiosity in many 13 year olds (like me)
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  1584. Thanks for this video. It was a wonderful presentation on a usually very stuffy and difficult subject. I honestly cannot say that I (PhD applied maths) was able to follow everything while I was watching the video, but it showed me one very important thing: that I skipped numebr theory, Galois theory, extension fields and so on for all the wrong reasons when I was studying maths at uni. If they had explained to me then that it would give a lot of insight into the boundaries of what is possible with Euclid's geometry I would have been all over it.
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  1585. 42:05 My Memorable part: As Always It Goes to Euler. The Gamma : )
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  1586. I don't remember doing anything in primary school involving indices including ^2 for 'squared', we did areas of shapes by counting squares on squared paper, nothing about algebra either. It was all new in secondary school. The first year there (ages 11-12) was basically doing catch up for what we should have/could have all learnt in primary school, but without a 'national curriculum', each junior school had probably taught different things and missed some things out. I think things have probably improved since then with maths teaching, but some (10-12 year olds) still seem to struggle with the basics - because of not learning times tables, or because new maths concepts are not introduced in a simple way, avoiding using complicated language.
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  1587. Makes me wonder just how much of mathematics can be reduced to stuff that's easier to understand.
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  1588. and here i always wondered how mathologer decides on topics - another mystery solved
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  1589. Mathologer I think there's a mistake at 16:49 - the x at the top is missing a power of 0
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  1590. Wow astounding how they could have done all this so long ago!
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  1591. I learnt Heron's method of finding area of any triangle by myself from a mathematical formulas Book , when I was in 8th class in the year 1988. Even today I find it useful in my field works 👍
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  1592. my high school had a copy of that same book, Thompson's "Calculus Made Easy". i remember using it to teach myself calculus before i ever took a class because i wanted to help my girlfriend at the time with her homework lol
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  1593.  @cliveso  Yes, if each creditor divides their loans proportionally to the loan size, then you give disproportional advantage to the largest creditor. But if assume everyone is equally determined and thus able to divide their loans as much as everyone else, then you get a stalemate and no one gains any advantage
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  1594. A motivational video for discovering mathematics.
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  1595. The problem with the sphere seems to be a matter of curvature. The angles of triangles with flat curvature will always add up to 180 degrees no matter how small they become, and so there will always be extra area at the points.
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  1596. The way I went: there are 33*33=1089 tiles on the board. The sum of all tiles 1 to 1089 is the 1089th triangular number: 1089*1090/2=593,505. Divide that by the 33 tiles in any row, and you get 17,985.
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  1597. Such another great piece of art you were able to produce! I like the pure mathematics combined with the archeology. Isn’t is amazing that we just slowly discover how far advanced medical scions was advanced? I am so glad that we are now more open minded and allow to combine the facts of the past. Past scientist would have been happy and proud of us seeing the collaboration and joy.
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  1598. It’s interesting to consider this prospective of the logarithmic product rule in the context of primes, the prime number theorem, harmonic series, and Reimann Zeta function.
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  1599. Pick's Theorem! Very nice idea. You don't actually need the full Pick's theorem, only the fact that the area of any polygon with vertices on the grid points must have rational area, which is pretty easy to show.
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  1600. It’s been years since I’ve seen these proofs but I was taught the second one first. I love the visual explanations. Great video.
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  1601. I came up with an interesting anagram a few years ago. We've all heard that saying "Rome wasn't built in a day". If you rearrange the letters in the phrase, it explains itself: "I want years to build, man". The interesting thing is: the apostrophe in the word "wasn't" in the first phrase becomes the comma after the word "build" in the second phrase, so every symbol is accounted for perfectly.
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  1602.  @Falanwe  that suggests it might be a very valuable curve to quickly generate for certain applications, especially manufacturing and construction. Minimization is a huge part of engineering & design.
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  1603. I collect coins as a hobby so double thumbs up!
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  1604. Infinite repeating series: Example 0._14_ It is equal to sum(0.14*0.01^(n)) The series is 0.14* 1/(1-0.01) = 14/99 In general, 0._abcdef_ = abcdef/999999 (as many nines as repeating digits)
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  1605. let's go early gang
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  1606. 8:13 Going with the previous method, we can have two on the equator, and of the remaining five at least three would reside in one of the hemesphere, totaling 5 out of 7.
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  1607. Made it to the very end! Wanted to complain a bit, but after getting there I have nothing to complain about :)
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  1608. Interchangeability of nested summands (or integrands) is usually glossed over and to me this video/result emphasizes the rigor and caution to be exercised when performing such. Thank you!
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  1609. These animations were better than my whole calculu's teachers. Congrats for this awesome job
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  1610. This is one of the best mathologer videos so far, incredible. I inspires to play with math.
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  1611. 8:05 "woooooah.. not hard to see why you struggle" School in a nutshell
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  1612. 15:58 Since each new pair of squares corresponds to a right triangle with the hypotenuse of the previous square's side length, and due to Pythagoras, the sum of the areas of these squares is equal to the area of the square in the previous generation. Therefore each new generation of squares has a total area of 1. Since there are 5 total generations in this tree, the area of the tree is 5. It's always cool when fractals turn out to have infinite area.
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  1613. @8:50. There are 6 different patterns. They remaining ones are redundant due to rotational and mirror symmetry.
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  1614. thank you for always having captions :) auto-generated aint toooo terrible these days fortunately but the extra effort to have your own is noticed and appreciated
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  1615. 1:22 knowledge and hair doesn't fit to the same head? 😀
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  1616. Thank you for giving the credit that Indian mathematicians really deserve.
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  1617. The realization that you get closer and closer without actually touching any integer number other than one is really memorable. Definitely something to talk at the table with your parents.
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  1618. Thank you for the mention, Burkard! And thank you for bringing Steinbach's work back into the light. Some years ago I tried some exposition on the same topic, seen from a very different angle, in a triplet of iPython (now Jupyter) notebooks. Apparently I cannot include the link in this comment.
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  1619. Ha, echo of Fermat. "I have a link to a notebook proving this, which this comment field is incapable of accepting". 😁
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  1620. I don't know if it's going to come up here but I once spent some time looking at 4x4 Panmagic Squares. There are the usual rows, columns, and diagonals, but, for instance there are also all 2x2 sub-squares many more sub-patterns.
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  1621. The radii of the regular polygons derived with the PDN theorem can also be derived by treating the reference vertices as complex numbers, with origin at the vertex centroid, and then taking the discrete Fourier transform of the set of points. The modes derived represent the vectors of the radii of the regular polygons. Neumann uses this method to derive his proof of the theorem. In Neumann’s paper on this “Some Remarks on Polygons”, he references “symmetrical components” which is what this DFT method is called in power system theory, which is equivalent to the DFT. Neumann’s father was a power system engineer and gave him the idea to use this technique from power systems. In power systems we use the specific case of the triangle, and the radii of the Napoleon triangles provide a physical interpretation of different types of unbalance that occur in power systems. The radii of the Napoleon triangles also have a deep connection to the Steiner Ellipse which I discuss in one of the videos on my channel. The ratio of the radii of the outer and inner Napoleon triangles are equivalent to the “third flattening” value of the Steiner ellipse.
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  1622. Hi, Burkard. would the right name for a 10-gon not just be 'decagon'?? We also say pentagon, hexagon and octagon, after all.
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  1623. This video is a great introduction to the helicone numberscope. I appreciate the clear explanation of how it works. The visuals are very helpful in understanding the concepts. I'm definitely going to try making one of these myself. Thanks for sharing this fascinating project!
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  1624. This video is really something special, the level of insight, beauty, deep concepts and even mathematics history is staggering. Thank you so so much!
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  1625.  @davidmeijer1645  Very glad to here that you do. How about your colleagues? Would you say that showing these proofs to kids is something that is commonly done in Canada?
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  1626.  @Mathologer  i would say that the reality is..no. Most teachers are concerned with covering the curriculum, which doesn’t direct teachers to explore these expansive variations on a theme, say this Pythagorean Theorem. That’s why your videos are great…for us teachers to recommend (to the interested student). I went through 3B1B lighthouse explanation of the Basel Problem yesterday with a class of 6 Gr. 10s. I really think it was quite a bit above them, but I’ll see when we next meet. And I think that’s why most teachers are reluctant to delve….concerned that it’s a bit above the students. Have you been amongst high school students lately? I mean, most are at the stage of struggling with something like, how to get the angle, when a trig ratio is known….I.E. learning to use the inverse trig function on a calculator. It’s quite amazing actually how developmental stages are quite uniform across geography and time.
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  1627. My brain REALLY wants to see a cuboctahedron in the figure where the triangles and squares are connected, but the closer I look, the less it even looks like a possible polyhedron.
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  1628.  @Mathologer  Yes, good point! I guess it is so well known that we take it for granted. Poor Pythagoras!
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  1629. Brilliant! Thank you! Love your approach to explaining things! ( Merry Christmas! ... let's make 2020 the best year so far! )
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  1630. It's too bad that I had teachers who complained that I was too slow and would never be good at math.
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  1631. Chess is being popular day by day everywhere becoz of queen's gambit😂
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  1632.  @Mathologer  I stand corrected! I had first learned about star polygons via a different construction - lay out p points on the perimeter of a circle. Jump from one point to the point q steps ahead of it, and repeat until you return home. In that context {p / q} and {p / (p - q)} do end up being identical but traced backwards. It’s interesting to hear that the nested circle construction treats these differently!
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  1633. Memorable: The fact that every positive number is the sum of a sub-series, and that the convergence is very fast!
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  1634. You only need to check 27 since only the difference between the colors matter. So you can ignore the first one and use relative values to that one instead of absolute ones. This leads to 3^3=27 combinations. Greetings from Germany :)
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  1635. 8:00 Just filling in the details of this proof because i havent seen any other commenter do so yet: (unless i missed them, idk) Set R = outside radius of sphere= outside radius of cylinder. h = height of our cut plane r = radius of the circle x-section And a = inside radius of ring. Area of the circle cross section: A1 = pi*r^2 = pi(R^2 - h^2) via the Pythagoran theorem. Area of the ring: A2 = pi (R^2 - a^2) A1 will equal A2 if we set a = h, so then the subtracted cone will have straight sides with rise equal to run, and therefore the cone is actually a cone.
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  1636. I was fortunate enough to hit that sweet spot in educational timing that included manual and mental calculation, log tables, slide rules (the main focus in the 60's), mechanical calculators of various sorts, electronic calculators, and some building sized computers that could not even compete with a smart watch. A major part of side rule use was the quick mental calculation of what you would approximately expect the answer to be, then you would know where the decimal point went.
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  1637. I wasn't very impressed with Gabriel's horn having finite volume, since it's only unbounded in one direction. However, you can create infinite copies of the horn, then rotate and fuse them into each other, such that their circular bases share the same point. If you make each horn point at a rational latitude and longitude, then scale the horn based on those denominators, you'll have a 3d solid with finite volume that is unbounded in *every* direction. Any other solid, no matter how small or far away, would be pieced by infinitely many horns.
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  1638. Just my method: there are 100, and the average (linearly distributed) is the average of 1 and 100. So 100 * 50.5. That just "feels" right, more than n(n+1)/2. But frankly it's nearly instantaneous to evaluate 50 * 101 (I always "gather terms" first), whereas taking the average and multiplying is a tad more work. So which to choose: on these matters, if both attain the results with comparable elapsed time and accuracy, I think you go for whichever has more style :) Or go for the most teachable approach. Or for some of those out there, they go for the more arcane solution. Yikes! I had some of those teachers!
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  1639. 6513516734600035718300327211250928237178281758494417357560086828416863929270451437126021949850746381 = 16120430216983125661219096041413890639183535175875^2 + 79080013051462081144097259373611263341866969255266^2
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  1640. I always go back to time, distance, speed, and acceleration whenever I need to get an intuition about differentials, it's the simplest indeed, and it's so natural to all of us
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  1641. Your T-shirt looks odd to me.
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  1642. I didn't even notice that the movie and tv references were gone, nice to see a revival!
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  1643. I have a book of my father's titled "Turtle Geometry" from MIT, 1982. It is this exactly.
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  1644. Wow. Such a simple way of looking at it, that the first reaction is to see if something is missing. But it is not a paradox... thanks.
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  1645. This is really beautiful. It's even more beautiful than the theorem itself, which was hard to beat.
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  1646. I was also trying to think of the weird n = 2 case of digons. But that really is a degenerate case - all digons are already perfect because the "two sides" always exactly coincide so of course they're the same size, and both angles are also obviously the same 0 degree angle. And indeed with the "n - 2" rule of how many times you need to apply the ears, n - 2 = 0 in this case so this tells us we don't need to apply the ears at all to get a perfect polygon. Boring, but consistent, which is kind of what we would hope from a degenerate case.
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  1647.  @dsdsspp7130  Dude is your teacher Flammable Maths? I think he talked about giving his students that exact homework. I'm just asking because I don't think that is going to be a common task to give to students.
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  1648. Für deutschsprachige Interessierte: --> https://youtu.be/Vv3Rve3yXBY Sehr guter Vortrag zum selben Thema. Schaut euch auch die anderen Weihnachtsvorlesungen von Herrn Weitz an!
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  1649. Thank you for this great video. A long time ago I heard that Zagier did a one-sentence-proof without knowing what it was until two weeks ago. I did a bit of thinking on my own and want to share what I found (probably not as the first one) because it might be interesting. In his original paper Zagier states that his proof is not constructive. In itself both involutions (the trivial t:(x,y,z) --> (x,z,y) and the zagier-involution z as discribed in the video) don't give many new solutions starting from a given one. But combined they lead from the trivial solution to the critical, from the fixpoint of the zagier-involution F := (1,1,k) to the fixpoint of the trivial involution t. Proof (sry no latex here): Let n be the smallest integer with (z*t)^n(F) = F. So t*(z*t)^(n-1)(F) = F (multiply by z on both sides). And therefore (t*z)^m * t * (z*t)^m (F) = F with m = (n-1)/2. Bringing (t*z)^m to the other side proofs that (z*t)^m (F) is a (the) fixpoint of the trivial involution, ie a critical solution. Note that n is always odd, assuming n is even results in a contradiction: If n is even we have t*(z*t)^k * z * (t*z)^k * t(F) = F with k=(n-2)/2. So again we see that (t*z)^k*t(F) is a fixpoint, this time of z, and therefore equals F. Multiplying by z gives us (z*t)^(k+1)(F) = F contradicting the choice of n.
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  1650. 28:17 I came up with a proof of complex numbers: if you know two points a, b of the triangle as complex numbers, to make the third one you just take the vector from A to B (A.K.A b-a), then to get side AC you rotate and scale it (A.K.A multiply by a complex number (named z)) , and then you add it to point A to get C. overall the function you get is (b-a)z+a=(1-z)a+zb. overall a linear function, so you get this linearity, that when you add up two vertices, the sum of the third vertices is the third vertex of the sum.
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  1651. 2. I mentally wrote the numerator as 10^2 + (10+1)^2 + (10+2)^2 + (10+4)^2, and applied the squared sum rule. So I have 5 tens, that's 500 in total. Next I have 1^2 + 2^2 + 3^2 + 4^2, this adds up to 30, and finally 2×10×(1+2+3+4), which is 200. So In total I have 500 + 30 + 200 = 730 = 2×365 It is surprising because 365 just happens to be the number of days in a year.
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  1652. Proud to be born in this Legends place. He used to lie down on a rockbed inside the temple for observing the sky for long hours. The sad fact is that many of the Indians especially people from Kerala are not aware of his existence.
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  1653. I'm a maths student. Honestly my path is so much easier thanks to interesting Youtube channels like this one. Thanks for that.
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  1654. Nice work! Gives me some flashbacks to Paul Lockhart's lovely book Measurement, which is like the one textbook that I, as a soon-to-be math teacher, actually really like; of course, I can't base my teaching on it because of these soul-crushing standardized requirements and tests, though (I guess I can at least use it for inspiration). One of my favorite related proofs when it comes to understanding the logarithm based on the graph of the function defined by f(x)=1/x that could have been mentioned, though, is a lovely visual argument that the alternating harmonic series sums to the natural logarithm of two. You restrict yourself to the little region from 1 to 2 and ponder the area under the graph – it's the natural logarithm of two, of course, but you can also approximate it with rectangles. If you use a 1x1 square, you're overshooting a lot, but you can remedy that by taking a rectangle out horizontally to remove the second half, where the overcounting is most egregious. You then add the smallest rectangle you can that still covers the desired area in that region, which has a height of 1/(3/2)=2/3 and a length of 1/2, so its area is a third. I think most of you will know where this is going, right? Our estimation, which still overcounts but is much more accurate, is 1 - 1/2 + 1/3. Now, we remove the second half of the first rectangle (area 1/4) and add a better fit back in that has an area of 1/5, and we also remove the second half of the second big rectangle we have (area 1/6) to get a better approximation with an area of 1/7. And so on and so forth – it's not too hard to convince yourself that this approximation strategy does indeed amount amount to evaluating partial sums of the alternating harmonic series, and the argument shows that they approach the logarithm of 2. Isn't that beautiful, even if it's tricky to describe everything just with words? This is exactly the kind of thing I want to teach eventually.
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  1655. BEST MATH CHANNEL! MERRY XMAS TO THE MATHOLOGER AND HIS FAMILY!
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  1656. At 17:37: an A-A-A equilateral triangle clearly has area A²F and nestles into the corner of the larger A-B-? 60 degree triangle. So notably, the two have the same height. Now scale our A-A-A triangle horizontally (i.e., along the base, not the height) by B/A. This new triangle has area A²FB/A = ABF. It's not the same size as our large triangle, but it does have the same base (A*B/A = B) and the height wasn't changed by our horizontal scaling. Any two triangles with the same base and height have the same area, hence the A-B-? 60-degree triangle has area ABF, QED.
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  1657. In fact, I've seen these diagrams as divisibility test. Make the multiplier equal to 10 (or whichever base you're working in), and the modulus equal to the divisor you want to test divisibility/remainder by. Then, with the arrows as you suggest in your third wish, start by moving clockwise from 0 as many steps as the first (most significant) digit, then follow exactly one arrow in the diagram, move clockwise as many steps as the second digit, follow one arrow, so on. Do not follow an arrow after your last digit. If you follow these instructions correctly, you'll end up at the remainder (and if and only if you end up at zero, the number is divisible by the modulus).
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  1658. Thanks for giving his due credits . You are a nice teacher 👍👍👍👍👍
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  1659. Christmas came a little early this year! Thanks for the gift!
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  1660. Ofc we will!
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  1661.  @FLScrabbler  And then you discover that it's square at three feet off the floor and at no other height.
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  1662. Is there a combination, which leads to a vortex, in which (at least one) remaining number ist 42? Which multiplier and modulus does this one have? 🤔 That would be a sign of the key of the universe.
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  1663. Ramanujan was God gifted not like me, A normal students wasting time in Indian Schools to memorize everything for exam.😢😢
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  1664. Convention, mostly. The equivalence classes mod n are normally named by the smallest element in them >= 0. Though honestly it's often more intuitive to think of the classes just under n as negative numbers instead.
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  1665. Damn, this was, I think, your toughest masterclass yet. Almost all of it worked for me, but I was really stuck at the point where you used the powers of two and got the fact that the new permutation will be obtained by adding 3 to the new natural permutation. I would have benifited from a little step-by-step at that point. This was a lot of fun, though. I'm looking forward to learning more about these fields and the inversions and other properties of permutations. Thanks for this!
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  1666. 11:50 ab = 153 2cd = 104 ad + bc = 185 For my fourth equation I decided to use the radious of the in circle: ac = r I have one small issue that being i completely forgot the formula for the in circle so i made my own: I remember how to graphically find the center of the in circle so knowing that I will create 2 function which each draw one of the lines that halve one of the angles of the traingle. Where my functions overlap is the center of the in circle. I imagined the triangle orientated as show in the video intro and set the origin of my coordinate system to the most left point of the triangle. My functions draw the lines which halve the alpha and beta angles. 1. equation: f1(x) = x*tan(alfa/2), alfa = arctan(153/104) 2. equation: f2(x) = -(x-104)*tan(pi/4) = -x + 104 Combining f1(x) = f2(x), solution x = 68 The radious r = f2(68) = 36 ac = r = 36 Using the 4 equations I found the 4 variables to be: 9, 4 17, 13 ------------------------------------------------------------------------------- Would it be much easier if I also remembered: a+c=d c+d = b ? Yes. Yes, it would have been much easier.
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  1667. Thanks for the awesome video! Do you have a video on how to make such cool animations? I think a lot of people would be interested!
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  1668. Mind blowing & amazing! Thank you sir for sharing this ancient wisdom with us lay folks. Please share other insights that you are aware from the ancient sages. It might not be too far fetched for us to assume that Madhava and his fellow mathematicians at the Kerala School of Mathematics might have discovered many other concepts that we might not be aware yet.
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  1669. You know it's going to be a great video when it's posted by Mathologer.
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  1670. A quotation by David Mumford (a nobel prize recipient in Mathematics and is considered to be one of the founders of Algebraic geometry). Only a fraction of this mathematics has become generally known to mathematicians in the West. Too many people still think that mathematics was born in Greece and more or less slumbered until the Renaissance. It is right time that the full story of Indian mathematics from Vedic times through 1600 became generally known. I am not minimizing the genius of the Greeks and their wonderful invention of pure mathematics, but other people have been doing math in different ways and they have often attained the same goals independently. Rigorous mathematics in the Greek style should not be seen as the only way to gain mathematical knowledge. The muse of mathematics can be wooed in many different ways and her secrets teased out of her. In another instance he says Though Panini is usually described as the great grammarian of Sanskrit, codifying the rules of the language that was then being written down for the first time, his ideas have a much wider significance than that. Amazingly, he introduced abstract symbols to denote various subsets of letters and words that would be treated in some common way in some rules; and he produced rewrite rules that were to be applied recursively in a precise order. One could say without exaggeration that he anticipated the basic ideas of modern computer science. Chandrasekhar
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  1671. You really did keep that self contained. I understand the principles of calculus from school but I never realised how beautiful mathematics was, mostly because I had undiagnosed ADHD so something that required extreme focus and care like mathematics just became frustrating because I would always seem to make silly or seemingly careless mistakes and so I just avoided it. Thanks for your videos!
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  1672. Take a tank of water and spin it as a whole; the surface becomes a paraboloid.
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  1673. My favorite channel strikes again! I've been going through the Mathologer backlog, waiting patiently <3
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  1674.  @MizardXYT  I noticed that a few minutes after I wrote the original version of my comment, that's why I then thought about rounding up! @SmileyMPV was talking about the current version that always rounds up. Hope that helps! :)
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  1675. "Is the no 9 series finite? You life depends on this!" Me: suspicious, has to be finite! "Believe it or not, it is finite!" Me: YAY
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  1676. İn the next video: "Here is a little challenge for our astronomer friends, go to the mun on your own"
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  1677. Beautiful. I usually do not comment on channels, but i had to on this one. Amazing. Never stop making these videos.👏👏👏
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  1678. 4k+3 = x^2 + y^2 so x^2 + y^2 ≡ 3 (mod 4) Mod 4 table: x: 0 1 2 3 x^2: 0 1 4 9 x^2 (mod 4): 0 1 0 1 since x^2 and y^2 can be either 0 (mod 4) or 1 (mod 4), x^2 + y^2 (mod 4) has 4 cases: x^2 ≡ 0 (mod 4) and y^2 ≡ 0 (mod 4), where x^2 + y^2 ≡ 0 (mod 4) x^2 ≡ 1 (mod 4) and y^2 ≡ 0 (mod 4), where x^2 + y^2 ≡ 1 (mod 4) x^2 ≡ 0 (mod 4) and y^2 ≡ 1 (mod 4), where x^2 + y^2 ≡ 1 (mod 4) x^2 ≡ 1 (mod 4) and y^2 ≡ 1 (mod 4), where x^2 + y^2 ≡ 2 (mod 4) x^2 + y^2 ≡ 3 (mod 4) never happens. Q.E.D.
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  1679. "Le principe des tiroirs" en français ;)
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  1680. I always enjoy your lovely stuff. Retired physics/maths teacher.
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  1681. I covered finding the equation for a sequence in the way you did for the Fibonacci sequence in my discrete structures class, but the reason it worked wasn't well explained at all. Seeing it in context makes the process so much more sensible and natural! Thank you for this awesome video.
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  1682. Wow, so this sequence stands as another proof of infinite pythagorean triples! (As the Fibonacci sequence is, itself, infinite) Neat!
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  1683. When you do Euler's polyhedron formula, here is an interesting bit you could include. For any polyhedron*, the angular deficits at the vertices sum to 720 degrees (4 pi steradians.) This can be very quickly proved via Euler's polyhedron formula, using for a polygon sum-of-angles = 180 x number-of-vertices - 360. The appeal is that this is about a 30 second proof. For example, consider a square pyramid with regular triangles. The 'top' vertex has 4 triangles, so the deficit is (360 - 4x60)=120 degrees. The other four vertices have a square and two triangles so the deficit is (360-90-2x60)=150. The sum of the deficits is 4x150+120=720. I expect (I haven't looked into it) that this is a special case of a theorem which says integrate-curvature-over-a-topologically-spherical-surface = 4 pi, and in turn gives surface area of a unit sphere = 4 pi. And probably integrate-curvature-over-any-surface = 4 pi (1 - number of holes in surface) * Not self-intersecting, topologically equivalent to a sphere.
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  1684. Ho, Ho, Ho...
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  1685. Looking forward to the hardcore video!
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  1686. Guys, I was hoping until the end for a combinatorics' argument for the (x+2)³ formula. I'm still trying to figure it out. Isn't there such an argument?
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  1687. Please make a video about your favorite math books when you were a student, sir
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  1688.  @Mathologer  Oh that's true! I missed that at first! Thank you very much for pointing it out and explaining it to me! Now I got it. I kept thinking about this argument through the weekend in the case of the 3-d cube. And now with your answer I completely got the general case!! Thank you so much! You called this way boring 😂 but I love the combinatorics arguments in maths. Although Tristans' argument indeed is really beautiful... But I kinda place his argument in combinatorics, wouldn't it be a recurrence relation argument? Now I see two arguments in combinatorics: one using the fundamental principle of counting and the other using a recurrence argument! I'm not sure I am right, but kinda feels even more beautiful now... A peaceful symmetry in the world 😂❤️
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  1689. ngl, when i saw where the visual proof that the base of the "area function" is e was going i had the biggest grin on my face
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  1690. THANK YOU! This is a fantastic presentation you created here... with easy steps to follow the logic. I personally consider Newton the greatest scientist/mathematician of all time ... Tesla second. This presentation shows what level he thought on ... and consider he was in his late teens and early 20s when most of this was formulated.
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  1691. The primitive pythagorean triples are given by the general formula (a,b,c)=(u^2-v^2, 2uv, u^2+v^2) where u and v are positive integers, u>v, coprime, and of different parity (i.e. one is odd the other even). This shows that the 4k+1 primes are always also pythagorean triples' hypotenuses. The converse is not true though, for example (7, 24, 25) is a primitive Pythagorean triple but 25 is not prime.
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  1692. Taking a shot at the "Multiplication Partition" problem at the end... Empirically, the numbers seem to follow the pattern F(2n), where F(n) is the Fibonacci function. So the pattern is every-other Fibonacci number, henceforth called the "Skiponacci sequence". I will prove the hypothesis that the sum of products generated from partitions of the number n follows the Skiponacci sequence. Here, S(n) is the Skiponacci function, and P(n) is the partition-product-sum function. Firstly, analizing the stacks of equations, we can utilize the "recursion" mentioned early in the video. Looking at the example given for n=4, we see that the products are: 4 3 * 1 1 * 3 2 * 2 2 * 1 * 1 1 * 2 * 1 1 * 1 * 2 1 * 1 * 1 * 1 We focus on the products ending with "1", and removing the "* 1" we see: 3 2 * 1 1 * 2 1 * 1 * 1 Oh look, its the products for n=3 ! Looking at the products ending with "2" and removing it: 2 1 * 1 Its the products for n=2. The pattern is becoming clearer. Looking at the remaining products: 4 1 * 3 The "1 * 3" clearly follows the pattern, being 3 times the n=1 product. The 4 sticks out, but for now its easy to write it off as just "n". The final formula for this pattern is: P(n) = P(n-1) + 2P(n-2) + ... + (n-1)P(1) + n The reason for this formula makes sense. The "recursion" is because the partition products that are multiplied by 2 are made from partitions that are 2 less than n. Hence the "+2" in the partition list becoming a "*2", giving us the 2*P(n-2) part of the equation. Now how does this fit into the Skiponacci sequence? It becomes clearer if we write the terms out into a pyramid. For instance, for n=5, the answer is the sum of these numbers. Each row has n copies of P(n-1), except the last row, which is written as n "1"s, for the "+n" term. 21 8 8 3 3 3 1 1 1 1 1 1 1 1 1 To aid in making sense of this, here is the pyramid for n=4: 8 3 3 1 1 1 1 1 1 1 Notice the recursion? The n=5 pyramid contains the n=4, just with the extra diagonal. This makes sense, since every time n increases by 1, each P(n-k) factor's coefficient increases by 1 (and the "+n" term increases by 1, naturally). All this means that this equation holds: P(n) - P(n-1) = P(n-1) + P(n-2) + ... + P(1) + 1. This gives us a neater equation for P(n) if you add P(n-1) to both sides, but for now lets test our hypothesis and replace P(n) with S(n). S(n) - S(n-1) = S(n-1) + S(n-2) + ... + S(1) + 1 The left side is easy to simplify, because S(n) = F(2n) S(n) - S(n-1) F(2n) - F(2n-2) F(2n-1) For the right side, we can recursively replace the two right-most elements with another fibonacci number, until we are left with F(2n-1) S(n-1) + S(n-2) + ... + S(1) + 1 F(2n-2) + F(2n-4) + ... + F(4) + F(2) + F(1) F(2n-2) + F(2n-4) + ... + F(6) + F(4) + F(3) F(2n-2) + F(2n-4) + ... + F(8) + F(6) + F(5) ... F(2n-2) + F(2n-4) + F(2n-5) F(2n-2) + F(2n-3) F(2n-1) This leaves us with this equation, which is obviously true: F(2n-1) = F(2n-1) Therefore, because we were able to replace P(n) with S(n) in our equation, we showed that P(n) = S(n). QED. Also I did the math and found that the general equation for S(n) and P(n): S(n) = 2/sqrt(5) * sinh(2 * ln((1+sqrt(5))/2) * n) This is a long way of saying I think the next number is 55 :)
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  1693. Congrats on 100 videos, mate. You really made it as a maths educator and content creator on YouTube, and I'm looking forward to seeing you do even better in the future. Hope you blow our minds again with each video you make That said... CHALLENGES! 11:18 I have no army of middle school minions but I am still ready to attack Same reasoning as before. This time we start with the second paraboloid, the one carved from the cylinder. Makes the maths a little easier. The paraboloid has radius R and height H. Cutting it at height h will leave a ring with outer radius R and inner radius r. The paraboloid is modeled after a parabola y=ax^2, and so we should have H = aR^2 and h = ar^2. So it's possible to solve for r and get r = Rsqrt(h/H). The ring thus has area pi * R^2 - pi * R^2 * h/H, or piR^2(1-h/H). The first paraboloid should also have that area. Thus its radius should be Rsqrt(1-h/H). Now the inverted paraboloid can be modeled by another quadratic, but the important takeaways are H = bR^2 and H - h = br^2. Solving for r this time gives Rsqrt(1-h/H), exactly the same as what was predicted by the circles area argument. Or you can use integrals. Whatever floats your boat 17:42 The layers of the onion look almost like surface areas stuck together. That can be written as: V(R) = the integral from 0 to R of SA(r)dr By FTC1, we can also write this as V'(R) = SA(R) And so the derivative of the volume is the surface area. Even in 420 dimensions. 18:04 The base of the cylinder has area piR^2. The height is 2R, and the circumference is 2piR. In total, the surface area of the cylinder is 6piR. With the surface area of the sphere being 4piR, the ratio of the surface areas of the two shapes really is 3:2.
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  1694. I am absolutely trash at math but when I stumbled across a numericology video on 369 and they explained the theory I immediately commented that it was clearly because we use a base ten system. For instance base 14 was used in the Middle East for a long time. Based on counting the inner pads of the fingers, I am a behavioral scientist and Buddhist philosopher so I am well studied but I really am horrible at math. I can’t even do basic division if the numbers go into the hundreds without taking a long time. It was actually obvious because from a behavioral psychological point of view when you have something people can’t explain the culprit is most often a blind spot in your paradigm. Anytime we attempt to build a thought model of what we believe is true we create a type of tunnel vision because our imagination becomes constrained by your base assumptions. If you think a certain action is always bad you will be blind to anything good that comes from that action and when it’s pointed out you won’t be readily able to accept the evidence, it literally requires an individual to rewrite the most fundamental assumptions about life because our thoughts are like cards stacked upon each-other our conclusions come as a result of previous conclusions, as this happens our ability to recognize anything new or aberrant narrows and our ability to explain concepts becomes reliant upon what we have “figured out” thus creating a blindness in our personal philosophy. Most people assume that math is always a base ten system. But there are as many systems as there are numbers and historically societies that weren’t connected to the western tradition came up with novel calculation systems. Most notably the Indians who were the first to recognize the need for the number zero. Incendently this comes from Hindu philosophy which posited emptiness or void and singularity or oneness. Both of these concepts are represented by the number zero which like consciousness can be posited and observed in its function but cannot be qualified or countrified as it is immaterial like space. Anyways from a purely psychological perspective the answer was obvious I didn’t need to comprehend the math to see that this was merely a blind spot made by a false axiom. Rant about Ether vs Space: I will add that Tesla was correct about ether. When people talk about bending space they are talking about bending ether. Space is not a medium and cannot be acted upon it is a potentiality for expansion and position. If something bends in a given position that which is bent must rest within something. Without a medium the distance between all things would be zero, the fact that we can imagine folding space means that something which has qualities is being manipulated, however actual void ness, emptiness, the expanse of nothing allows for all laws of physics to be unobstructed and to function within their own constraints. If space was not completely void there would be no possibility for anything to bend, 0 the absence of all is what gives freedom to bend the ether. When you bend ether that fold is still a location in space and relies upon the void as a platform for existence. Nothing has freedom or capability to be anything without a complete void to appear within. This void exists outside of time and was present as the externality of the singularity but void or emptiness is not a thing and so it is completely logically coherent to say nothing existed outside the singularity because emptiness has no qualities and cannot be proven to exist. It can only be pointed out as being the function of unobstructed freedom of material phenomena. If space was void you could not bend it, it has qualities and so it cannot be said to be the expanse which allowed for the Big Bang. Without this freedom of nothingness and infinite indescribable void the singularity would never have been able to exspand as it’s size would be fixed. Even the diagrams of bending space require the illustration of grid planes which automatically describes a medium which has properties. True space ie the infinite void has no properties it is the absence of properties and can only be proven to exist because the bending of space happens within a free unobstructing openness. It a real shame because I suspect this distinction is likely an incredible resource for computation. Again if there was no ether than all of existence would not have any measurable distance, distance in space is a fixed property which is measurable as space/time, but something has to exist that undergirds and existed before space time in order to allow it to come into being and to exist in an exspansive way. Nothingness is that thing. Without the absence of all nothing can expand out into infinity. Without void everything would be permanently constrained. Space itself is a constraint as it is governed by the speed of light. Space what I would call ether acts on objects because it is a medium with properties subject to the qualities of other objects like planets whose gravity actually warps it. True void is inert and has no properties it cannot be affected or measured it is merely the absence of all qualities which allows for qualities to come into being without being constrained. Can you comprehend emptiness or is your paradigm constrained by preconceived notions of what your science books have told you. I have made a coherent argument that we are missing a fundamental piece of reality in our current scientific paradigm. Exists in math as zero it is already proven to exist in math and many higher calculations cannot even be done without zero proving that as a model of reality zero is acknowledged as indispensable to model our reality. ... yet we ignore emptiness. We already know zero cannot be affected by any other number zero cannot be divided subtracted or multiplied because it is an absence of qualities. It can be added to because it is inert and unobstructed and so it is a vessel which is truly and absolutely infinite in its allowance for qualities to exist within it. We already know all this yet our current models of physics refuse to acknowledge that what we call space has qualities that occupy the infinite void as a basis for all. Without void we cannot posit existence we understand that two things cannot occupy the same place which means for anything to occupy any space there must be a void which does not obstruct position, characteristics or behavior. As I said if what we call space can bend it is bending within emptiness. There must be a basis for reality which is without qualities or existence would be constrained. If what we call space was truly the basis for reality it wouldn’t have qualities. If space was not a medium there would be no time or distance to travel between objects, you could move from earth to Mars in an instant because there would be nothing to travel through. Traveling through something means it is a medium especially if it has coherent properties, it’s qualities are consistent not chaotic meaning like atoms it has a particular measurable property which can be counted on to act and behave in a uniform manner. Having inherent measurable qualities means it exists as an object. Please don’t reply if you’re just going to quote text book physics as my entire point is to point out something missing in the current assumed model of reality. We know the number zero is real functional and vital yet we deny it’s actual existence in the world despite our model of reality requiring zero to be calculated. The absence of everything came before anything this should be obvious as only an absence of things allows for an unobstructed exspansion. Without an absence of qualities qualities could not come into being without being affected by constraints. If there was no external void the singularity would have nothing to expand to. This is important for positing the big crush as there may be a limit to this ether or space time, it may have an elastic property where it can only stretch so far, space time may be capable of thinning metaphorically like a gas perhaps creating a gravitational vacuum. Just like as gas thins it creates a vacuum which can literally implode or crush objects. Once we see space as a medium this new paradigm means measuring the qualities of this ether may actually allow for incredible breakthroughs. I believe humanity is constraining it’s progress by resting on a preconceived notion that space/time is the undergirding foundation of reality when In my mind you always must have nothing before you have something. If
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  1695. Very interesting! This reminds me of a result I came across recently. Take a regular polygon positioned anywhere in a circle. Extend the sides to form two sets of segments with lengths x_1,x_2,...,x_n and y_1,y_2,...,y_n, so that each y_i is counterclockwise from x_i. Then applying power of point to each vertex, we get x_k(L+y_(k+1))=y_k(L+x_(k-1)), where L is the side length of the polygon. Adding all these equations and cancelling the x_ky_(k+1) terms, we get x_1+x_2+...+x_n=y_1+y_2+...+y_n. So the two groups of segments have equal sums.
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  1696. We can also extend the result from chapter 6: it is possible to prove that for any 9 positive integers less than 100, there is are two disjoint non-empty sets of these numbers with the same sum. Also there is a set of 8 positive integers less than 100 such that any two different sets of these numbers have different sums. The proofs are pretty hard though, so I’ll leave that for someone else. Also, I’m not sure what souped-up version of the IMO was occurring with 10 problems, because usually they only have 6!
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  1697. I suspect the Talmud's authors were informed by a history of negotiations where the parties reached agreement. This amounts to an empiric solution of the game theory problem and validates the game theory theoretical result.
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  1698. When you showed the equal area segments on the hyperbola I finally felt like I "got" why it was a rotation. I was aware of the various taylor expansions and exponential related similarities but this was a really good thing to comfort my gut about it.
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  1699. I was just thinking about how many ways there were to make change for a dollar like a week ago. Are you telepathically spying on me?
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  1700. The most memorable moment was when I earned the Mathologer’s seal (I am now wondering what I got an approval for 👀)
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  1701.  @VAXHeadroom  :)
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  1702. Ah yes, I always like to express the polynomial 0*x as the product of all real numbers times a constant. Since f(x) = 0x has all zeros.
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  1703. Iterative Turtles... Damn, so it is turtles all the way down.
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  1704. Is there a geometric reason for the integral on your shirt?
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  1705. 12:19 I was sure that this had to use integrals. e.g. the stretch factor of the rubber band when the number x is at the top of the rubber band is 1/x, so the distance from 1 to x around the circle must be the integral of 1/s ds from 1 to x, which is log(x). I guess I'm too grounded in my non-Mathologer ways to come up with this much cleaner proof!
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  1706. This will come in useful in the gym later when I am scratching my head figuring out how to load up the barbell.
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  1707. Personally I would argue that Tesla was more concerned with the behavior of the the numbers 3,6,9. For example the alternating pattern of the 3 and 6 or reflective of an alternating current pattern (the 3 being a trough and the 6 being a crest and alternating back and forth). The 9 always equates back to itself which is indicative of a cycle or the balance (or nullification) one gets when they combine the equal but opposite ends of a wave (i e crest/trough or in this case 3/6). In short, 3,6,9 just reflects the constant "movement" of the universe and everything else in nature. It's a WAVE in numeric form. Just my thoughts. Loved the video!
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  1708. Not first
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  1709. Me at 22:25 "This looks like a fun exercise." Half an hour of calculating generating functions later. "Well that's certainly not the Mathologer® way of doing things."
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  1710. "Pretty obvious this pattern will continue forever." Fascinating how infinity is so intrinsically numerical...
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  1711. These videos are so fun! Once I begin, I get so hooked that it's impossible for me to stop.
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  1712. Definitely missed out on a great joke by calling them naught and nice numbers :') Great video once again, however. Merry Christmas to you and your team! :)
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  1713. 12:58 At this point I thought "Why not just use triangles to do the argument?" so I've tried to do it, but then realised that the triangles, when you apply the rotation, actually grow in size rather than shrink, so the infinite descent argument won't work here. 21:00 I guess I'll prepare for that video more.
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  1714. Damn Euler! The greatest mathematician of all time. True superhero.
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  1715. yes please ❤️
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  1716. This is the first Mathologer video I understood at one go. Keep it up professor 👍👍
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  1717. The story of how Euler got into formulating quadratic reciprocity is fascinating. It's contained in "Primes of the form x^2+n*y^2" by David A. Cox. HIGHLY RECOMMENDED because it's an historical recount.
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  1718. ​ @jannikheidemann3805  If you mean the United States, then yes, they teach Latin in many high schools, it's gaining in popularity, many Catholics study it independently to deepen their study of the faith, and many more people study the Latin roots of English in preparation for standardized tests like the SAT and GRE.
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  1719. The most memorable part is the connection of the 'γ' and the log() function to the harmonic series! Really amazing!!
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  1720. Top notch video! I was enthralled the entire way through it, and it was all easy to understand! Very well done!
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  1721. I tried to prove all rational numbers have repeating decimal expansions before this video and found the proof down in it
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  1722. Merry Christmas mathologer,thanks for our gift 🙂
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  1723. When I was a kid, I learned the "casting out 9s" trick in some puzzle book (possibly by Jerome S. Meyer) my older brother had. I didn't learn why it worked until I read Martin Gardner's column about divisibility tests.
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  1724. Dirichlet's approximation theorem states that any irrational number x is approximated well by infinitely many rational numbers p/q, in the sense that x differs from p/q by at most 1/q^2. This is proved by the pigeonhole principle!
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  1725. Many civilizations discovered this triangle, India, China, Persia. If we have to hyphen the names of all the mathematicians who discovered it, it's gonna be a very long name. Also, In Persia, Omar Khayyam is not the one who discovered it. It's Al Karaji. Though it's true that later on, Khayyam continued Karaji's work on the triangle so successfully that it's called Khayyam's Triangle (مثلث خیام) in Iran
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  1726. Man, Post's proof really was an amazing proof. That was one of those proofs that just makes me go "Wow!"
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  1727. Despite the lockdown, the comforts I appreciate in life have really shown themselves to me and I feel grateful for what I have. These maths animations are among that!
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  1728. This is great, I wish we would go even deeper into the maths
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  1729. The thing is that, if the straight line assumption overlaps, then we always measure a bit more. Because, we know that the sum of the two sides of a triangle is always greater than the third one. If our assumption overlaps then it gradually leads to a fractal.
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  1730.  @candiman4243  dangit I was gonna make that joke
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  1731. Isn't it much simpler to see just the center of second coin if coin is outside than its center cover more distance than if it is inside 😃 we can also find the no of rotation by this 😌
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  1732. Search again. Maybe we’ll get another video tomorrow!
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  1733. I couldn't concentrate after chapter 3 because all I could think about was that amazing modified machine!
    7
  1734. Matholgger just confirmed the Antarctic ice wall and flat earth. Wake up sheeple!
    7
  1735. Matt Parker must have invented that anagram
    7
  1736. Two years later this is still just as enjoyable and informative.
    7
  1737. Most memorable: that cool proof that gamma < 1 by sliding all the blue regions into the unit square
    7
  1738. The proof sketch was brillianty displayed. That "aha" moment you experience when it finally sinks in is priceless, almost addictive.
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  1739. As a math teacher myself, I must say that this is beautifull and it's not very known even at graduate levels; I've loved this difference stuff for many years now and had very few people with whom share talks about it. Greetings from Argentina!
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  1740.  @cliveso  Its an 1800-year-old problem for a pastoral society, it's not like you should be concerned with outside companies in the Bahamas or derivative securities. And optimal solution on R2 with the usual Euclidian metric it's not necessary and the optimal solution on all Rieman manifolds in R(n). The papers optimize for a set of reasonable conditions, more general than pastoral but not all possible solutions in all possible scenarios.
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  1741. I’d love to hear aboeut Galois theory
    7
  1742. Homeworks : 1) Considering a general nxn square, it is simplest to add the main diagonal, whose points stay in that diagonal. It becomes the middlemost row post transformation. Their points are (i,n-i) (1 at (0,0)) and the values are n+(n-1)*i for i from 0 to n-1. Adding, we get n^2 + n(n-1)^2 /2 = (n^3 +n)/2. For a 33x33, we get 17985.
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  1743. I want that t-shirt in the shop!!
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  1744. Definitely the most impressive part is the animated Kempner's proof, I've expected something extremely complicated and yet the whole thing was "nice and smooth".
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  1745. I love this guy! Keep 'em coming!
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  1746. This is the best possible christmas gift 😁😁 Thank you MATHOLOGER Merry Christmas ❤️❤️
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  1747. I wonder if there are any similar fractal-like paradoxes? ("volume" slightly less than 3 dimensional finite, yet infinite "surface area" slightly larger than 2 dimensional? Any trade-offs here as the two dimensions approach each other?) Hmm, something for me to ponder!
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  1748. I won't be surprised if this video would suddenly get dislike bombed by rabid fanboys with in a week. Some people really don't like when they're told that their branch of woo isn't special.
    7
  1749. Can you do one on why the platonic solids fit so nicely inside each other?
    7
  1750. New IAU-approved prime definition: 1. must have only 2 indivisible factors (1 and itself) 2. must be odd 3. must be large enough to have cleared its orbit of all other integers.
    7
  1751. 21:30 As soon as you started talking about magic, I knew recursion would be involved. Recursion is the basis of all reality, my friends.
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  1752. Weird curve's diameter= L1+L2+L3 which is a chord of Conway's circle not containing its center < diameter of Conway's circle. BTW, (Conway's Radius)^2 = r^2 + p^2, here r is the incenter's radius and p is both the triangle's semiperimeter and the weird curve's "radius" (there's no center). You can get r = H/p, using Heron's Formula (H)
    7
  1753. The world is always a bit better with a video from the great Mathologer. I am glad you put this together, thank you Mathologer!
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  1754. The solution to the puzzle at the end is easy, you can consider scanning across 2-wide columns and 2-high rows. Each time you move the square, you know the colors in the newly covered squares have to be exactly the ones as just left. This gives us half the solution for free, since it it implies corners along an edge must be different, since the rows and columns have different parity. It's not a full solution because it doesn't exclude the case where a color is the same on one corner and the one opposite it. You can use an inductive argument to prove this. If the opposite corners really have the same color, then that forces the inner cell of the 2x2 squares at the other corners to have that color (by similar row/column parity reasoning). This means we have a (N-2)x(N-2) square with same colored opposite corners. In the 2x2 case, this is excluded by the puzzle's rules. Thus we can form an inductive argument, (1) a 2x2 square trivially cannot have the same color on opposite corners, (2) a (N+2)x(N+2) square cannot have opposite corners with the same color, as that would require the NxN square inside it to have opposite corners with the same color. Note that this only applies to grids of even sizes. Although I think that proves more than you wanted now that I rewatch that. I'm still confident in my solution and that it holds even in cases like 6x6. Here's a diagram to help visualize the tricky case for 6x6: a_?_?_ __?_a? ??a___ ___a?? ?a_?__ _?_?_a
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  1755. Yes, you are right. My intuition was wrong. I love how these videos get people talking—any ideas on getting 15-16 year olds talking like this?
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  1756. I put on the captions for my deaf mother and we both appreciate not only the effort you put into them but also the smilies :)
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  1757. Love me some mathologer masterclasses... still waiting on that Kurosawa length Galois theory video :)
    7
  1758. "Coins may become an endangered species" Or better yet, endangered specie!
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  1759. The column sum of the 33 square is 17985. It's just the number of squares (1089) into Gauss' sum formula n*(n+1)/2 over the number of columns (or equivalently rows).
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  1760. I am so glad that you are still around!
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  1761. 17:30 Just remember that the triangle can be thought of as a half of a parallelogram of sidelengths A and B.
    7
  1762. this is a great topic for an undergrad first year calculus class.
    7
  1763. What perfect timing! I just remembered these claims about Vortex Math and mentioned the explanations of their patterns in the comments of a math meme about how 9 is only special due to the decimal system. :)
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  1764. I do NOT remember it being taught in my school in England in the 1960s. I do remember coming across it in my mothers school textbooks - but not with Heron's name attached. Always thought it was fabulous - particularly like the way it gives a zero if S is any one of A, B or C which corresponds to the triangle collapsing into a line.
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  1765. Bioware put this in KotOR, and I had to memorize this thankfully simple recipe because I did multiple playthroughs. Years later, they put it in Mass Effect 1 and I still remembered it. And years later I saw this video and still remembered it. Stuff you do repeatedly as a kid really sticks with you, huh?
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  1766. I actually really enjoyed the blinking eyes. Well done!
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  1767. Ah that bug explanation is lovely! Wish I’d given it more thought myself first haha.
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  1768. first
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  1769. Great, as always! Loved the animations of Thompson and Leibnitz and the metal soundtrack of the end (though I've always been a fan of the usual wistful and nostalgic guitar theme)--it was distracting, but worth the distraction!
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  1770. When you laugh it very hard to differentiate you from a James Bond movie villain.
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  1771. it’s amazing to also know that the flavors of derivatives and integrals were already discovered by Indian mathematicians before Newton ❤
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  1772. My favourite Mathologer video thus far. Props to Archimedes et al.
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  1773. Prof C.K Raju [INDIA], also complied with previous old mathematician work, you can also see that. thank you for the amazing video.
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  1774. So he is a doctor and a plumber....and also a mathematician 😱
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  1775. Thank you for this new, great Mathologer video! I am really enjoying this long video about that absolutely interesting topic! 😄
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  1776. This was the best of your videos. I am a PhD student of math, and today I learned something new which I have to look into much more. Definitely more of this please. Complicated math is my bread and butter and you are one of the few youtubers who dares to go into the details. The length of the video was absolutely appropriate.
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  1777. I have a different perspective on "the king's method". When I went about proving the method, I looked at the form the columbs take when we unwind the magic square back to the lunch box formation. Under that perspective each columb corresponds to two diagonals with a combined leangth equal to that of the columb. Than the key to proving the method is to show that when shifting from a columb to its' adjacent, the sum over one of the corresponding diagonals increase by exactly as much as the sum over the other will decrease.
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  1778. This is a masterpiece!
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  1779. I had always thought of myself as a mathologerist, but now I see that I was always a mathologerer.
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  1780. On Ramanujan's channel this video is 20 seconds long and the explanation consists of him saying "I saw that this identity must be true"
    6
  1781. Exciting, awesome video. Absolutely amazing.
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  1782. I am an engineer, and I have to say that the video is crystal clear. I really wish I had such clear presentation back in the days!
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  1783. I often wondered why you didn't have a patreon and ads on your videos. Please, monetize NOW (and do employ someone for the editing, etc.)
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  1784. Amazing content, as always :)
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  1785. Honestly it blows my mind when I see these mathematicians from India generated some of the most popular integrations, limits and whatnot with sheer simplicity yet mysteriously and seamlessly embedded them either in a poem or a mantra or a prose so not only the willing one is able to synthesize the literature written but able to practically implement the math encrypted in it. Even more interesting is that these mathematicians always dedicated their discoveries to God and let the discovery have an open access for all irrespective of their background without claiming to be the founder of the said discovery; precisely why I am convinced to believe why several of their discoveries that we today are studying/ using don't bear their names.
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  1786. This was super enjoyable. Loved the bit of history along with a fantastically clear explanation with great graphics. Just great fun.
    6
  1787. 25:02 small typing mistake here inside the third pair of parethsesis, you wrote "A+B+D-B" instead of "A+C+D-B"
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  1788. 3:40 After you had got to the line of 1's I immediately got to my mind what Babbage was trying to do with his machine. Automate the process of calculating differences. So this is calculus because calculus is the mathematics of differences.
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  1789. Thanks for your videos.
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  1790. The Talmud is in some ways more a legalist text than a religious text. It's basically a collection of legal opinions, and it's also full of dissenting opinions.
    6
  1791. 42:49 Is that the same Markus Persson that created Minecraft?
    6
  1792. It works if the surface "moves outward at the same speed" everywhere when the parameter is increased. So the perpendicular thickness of the shell (the dV) is equal everywhere (the dr). For instance, the area of an ellipsoid is NOT the derivative of its volume (for common parametrizations). Even simpler, it doesn't work for non-regular polyhedra. Like, the volume of a pyramid with square base with side x as well as height x equals ⅓x³, but its area is (1+√5)x²...
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  1793. 0:35 "on the look-out for the mathematical soul of things". Haha - nice shout out to Shylock ("not on your sole, but on your soul...").
    6
  1794. 7:39 the coloring proof I suppose. The static picture that you can look at and check over feels more comforting somehow.
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  1795. Now we are just waiting for Mathologer to compare the Laplace transform to the Z-transform in a simple way :)
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  1796. my comment is 1st your video has made me felt proud the most ❤️💕 india ❤❤
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  1797. For many years, and from the first time I knew that I should support this wonderful math channel!, I really feel happy that you're still going forward!
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  1798. Yeah the recursion for the 3 and 9 divisibility rules was something I'm not sure was taught because it seemed so obvious to me I think they don't mention the remainder thing because they don't really care about remainders
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  1799. Here when there's less than 1000 views! Woohoo! The last time I was this early, I didn't exist
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  1800. Another great video as usual, I will definitely revisit this one after doing more studies. I just love the attitude towards math you have, and that you poke fun at those "intelligence tests". Those tests are so easy if you just know the trick, it's hardly an 'intelligence' test as it is a knowledge test at that point. Maybe very clever people might come up with the solution with no prior knowledge though.
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  1801. I think mathologer deserves millions subscribers .
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  1802. Fascinating stuff indeed! I remember spirograph as a kid, but got quickly frustrated with it - cogs kept slipping lol
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  1803. I am a math uni student and I did not know like half of these proofs... This was amazing. The sin(a+b) formula in particular is just so beautiful.
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  1804. Excellent introduction to logarithms, I liked the video! Thanks.
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  1805. 14:02 Yes. If I lay the leftmost domino of the 2xN rectangle on its side, there's 1 option to fill out the bottom and X(N-2) ways to fill out the N-2 columns to the right. If I put it upright, there's N-1 columns left to fill. So X(N) = X(N-2) + X(N-1), with 1 way to fill a space of 0 columns. Thus, we get the Fibonacci sequence.
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  1806. "That's a good time to stop" 42:35
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  1807. I'm kind of confused at the 180 degree step for the 10-gon. Shouldn't it do nothing since the triangle with 180 degree would be a line?
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  1808. depends on your school. we were thought about that. we were also given homework to come up with divisibility tests for other numbers like 7 and 11.
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  1809. Loved the long and in depth video! I'm still in high school and really had to fight through a lot of this, but absolutely loved learning about the more technical side of mathematics! More like this!!!
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  1810. Dear Prof. Polster, your videos are always so clear, so entertaining, so well done, and so energetic and with such good vibes, that I thank you to infinity ♾️. Please, keep'em coming!
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  1811. I'm from kerala and my father's caste is of "kaniya caste" basically astrologer caste. You have to learn mathematics to calculate the positions of planets in particular days and stuffs so my uncle who was interested in it as a toddler went on to study mathematics and became a mathematician. I was also fond of mathematics during highschool and in college went on to study mathematics but later within 1 month due to family pressure had to change to engineer because studying mathematics doesn't offer fancy job and salary in India. Anyways i still had to learn all these theorems in engineering mathematics for 5 semesters and i realised i was never into maths lol. I was only into highschool level maths which were simple. I had a very hard time doing engineering level especially applied mathematics but i still love the fact that i was able to solve it and graduate somehow. The only mathematic area i was really into was probability because i could use it in many real life scenario which i really liked and it was easy to learn as well for me. Anyway thanks for introducing a mathematician from where i am to the world. Long love maths.
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  1812. 11:20 "None of what I said so far is really terrifying" Are you sure?!?! You just talked about suddenly stopping and going backwards on the Germany Autobahn, that is terrifying.
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  1813. I literally just came from a video of Douglas Hofstadter talking about his INT function, and now there's 46 minutes of Mathologer right after? Better take a break before I take this in. :)
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  1814. 18:50 Euler's portrait is improved by the Santa hat.
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  1815. Turning property can be thought of intuitively if you rotate the entire 2D plane 90 degrees clockwise. Each time you do this you are essentially rotating the line counterclockwise. You can do these turns 4 times, each with 90 degrees before ending up at the original shape. This also explains why the turning property doesn’t hold in 3D, you can’t rotate the figure the same way with with the grid staying the same.
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  1816. The most memorable thing was that the sum is never an integer. On one hand it feels ok, because there are so much more rationals than integers for the sum to land on, but on the other hand, it gets closer and closer with each next integer and still manages to always avoid them!
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  1817. Yeah WHY? I don't need sleep, I need answers!
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  1818. I found this problem before 2002! http://mathcentral.uregina.ca/mp/archives/previous2000/ (Problem of the month June 2001) I was really proud back then because I managed to prove why 3^n+1 where the only solutions! (http://mathcentral.uregina.ca/mp/archives/previous2000/june01sol.html) The webpage says the problem is from "Crux Mathematicorum 27:3 (April 2001) pages 204-205 - it is problem 3 from the Ninth Annual Konhauser Problemfest (Carleton College, prepared by David Savitt and Russell Mann."
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  1819. I wish if I could give you 100 likes at once.... Lots of Love from Kerala. Thank you for making such wonderful maths videos ❤💚💙
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  1820. I think your czech is quite good ;-) I have read the original paper in the link and although I have understood each word, your explanation in english was much clearer to me.
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  1821. It is, but we didn’t consider this as a proper lacing.
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  1822. Thank you for the corrected version!
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  1823. Very nice. When one consider that the pins reflect the resources of an army, that is soldiers, transport, fuel, food, etc., than it reflects how far an army can conquer land. Without logistics, as the pin game is, it will starve out. That's why Napoleon lost its expedition to conquer Russia.
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  1824. First off, let me tell you how great these animations are. For my calculus students, I will definitely show them the part with the leaning tower of Lira. I also enjoyed the 700 year-old proof from a Bishop. I knew the integral test proof which I usually do in class. Of course, if we replace every other term by a minus sign, then the series partial sum converges to ln(2).
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  1825. Thanks for shining the light on Indian mathematics contribution. Kerala school of mathematics invented many things before the westerners did.
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  1826. A relatively easy proof: Any time you finish adding positive numbers (i.e. any time you go above the limit), it doesn't do any harm to add up to a fixed number of extra positive numbers. Let's say add an arbitrary amount of extra positive numbers that is between 0 and 9. Because the sequence of positive numbers converges to 0, the distance this gets you away from the target real number will be arbitrarily small (namely, in total 10 times the epsilon from the definition of being a zero-sequence), and so, if you do this, your rearrangement will still converge to the same real number. But this means that any real number R on [0, 1) can be turned into a new rearrangement: Take the decimal expansion of R. Then when creating your rearrangement and deciding how many extra positive numbers you want to add at the end of one consecutive sequence of positive numbers, just choose the next decimal digit of R, and add as many extra positive numbers as that digit is high. Because the set of real numbers on [0, 1) is uncountably infinite, and each number gives a distinct rearrangement, the set of all rearrangements resulting in the same sum must also be uncountable.
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  1827. My favorite math teacher is back let’s learn a new thing :D
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  1828. Every single video of yours has so many interesting mathematical connections! I always get excited while watching them!
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  1829. Actually it's not calculus that is hard: it's analysis that is hard, and it is often mistaken for calculus, since all of the results in calc. are derived via rigorous proofs in analysis.
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  1830. My favourite Pythagorean triple is 20, 21, 29 because it so close to an isosceles triangle and thus if you actually make it physically out of wood or meccano with inherent measurement errors, it will be more accurately a right angle than say 5,12,13 would be, subject to the same measurement errors.
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  1831. 11:00 - a proof by induction, my favorite. 🙂
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  1832. 💙💚🙏🤲🫂😅
    6
  1833. Nowadays parties became a thought experiment
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  1834.  @neradaensabahakasuna8560  Exactly, [-1] = [3] mod 4
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  1835. i learned to use a slide rule and vernier calipers despite going to school in the 2000's lol. my grandpa was just adamant that i learn how to use it, but more as a way to impress older engineers i think. which has come in handy! as for backup though, i have a solar scientific calculator that would survive any kind of disaster i could think of that a slide rule could also survive, short of going to the moon. radiation is one element that will always beat an electronic calculator and not a slide rule
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  1836. Mind blowing BTW at 39:25 it looks like these fibonacci-generated pythagorean triples have a pattern you missed: A alternates between 1 over and 1 under B/2, in a similar way to those other families in the tree. Makes me wonder what other families of triples can be generated from other sequences compatible with Euclids formula
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  1837. Please make a video on group theory please.....∞
    6
  1838. slight error at 18:48, there is no green circle on the 2nd row third column. As for the second method of creating the magic squares is to go up and to the left by 1 square (looping when going off the square) and if that square is already occupied going down 1 instead.
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  1839. Answer to the riddle of chapter 6: 1, 2, 4, 8, 16, 32, 64 (just the powers of 2 up to 100).
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  1840. Most memorable: Chapter 1: "Let's assume that the grey bar does not weigh anything - thought experiment - we can do this - hehe" Top notch video!
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  1841. I remember getting Q-Bert to jump around on a pyramid like that when I was younger.
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  1842. 16:26 the only squares mod 5 are 1 and 4. -1= 4(mod5) so the negative of the square 1 is also a square, and -4=1(mod5) so the negative of the square 4 is also a square, thus all the square on z/5z are squares. Q.E.D
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  1843. The real key here is that we now have a transformation in which all the CURVED longitudes in 3-space map to a single flat surface with the same length - no distortion. The curved longitude really is a flat 2-d curve if you just look at it from the right perspective but now ALL of them map to the same plane! Then the area stuff follows by INTEGRATING over closer and closer longitudes. For me, this elusive mapping is a real game changer. I mean it really is. Curved arcs on the sphere are now lying on the same flat plane with no distortion from which it follows (by integration) that curved area of sphere = flat area of circle (like adding up the arc lengths of an infinite number of undistorted semi circular longitudes to get the area). With radius of 2R to boot! No stretching or squeezing. Well done and thank you ALL so, so much! This has resolved a long time major conflict in my own mind trying to understand what curved area really means and how we can transform a curved area into an equal flat area with no distortion - which what area really is defined as all along - flat.
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  1844. personally, never seen either twisted square proof. but I think I am one of these children you speak of :)
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  1845.  @MizardXYT  This is as expected, since youre solving 3^39+...+41^39=x^39, so my claim is that this should give approximately x=41 + 0.458, which it does. In fact, I was originally expecting x-(n+2) to either converge to 0 or diverge to infinity, but experiments suggested otherwise, so I had to find the limit!
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  1846. Always quality content 💪💪 one of the best channels on YT 💕
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  1847. The most impressive thing about Ramanujan is that his problems are solvable using incredibly simple techniques. I can't imagine what would have happened if he had access to the breadth of knowledge that is available today.
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  1848. 3/4 is also .74999..., so infinitely many 9's as well as zeroes for 3/4 ; )
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  1849. 21:20 And therefore pi is a Sum of Two Square. That Excitement Nearly Killed me.
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  1850. First challenge: It can always be done when removing an even number of tiles as long as there is a round trip that always carves out green and black tiles one after the other and not for example two blacks before one green space is removed
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  1851. Great video! A few years back I wrote my bachelor's thesis about homomorphic encryption. I would love to see how you explain how it works 🤠 I'm sure the different perspective will amaze me.
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  1852. If instead of n-cubes, you look at n-tetrahedra, the number of vertices, edges, faces, etc. exactly match Pascal's triangle (with the last 1 in each row removed).
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  1853. all the time i am trying to wrap my head around on the thing that is printed on your shirt
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  1854. Not so discrete now that you blabbered it all over jnternets
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  1855. Best content on The Tube (and elsewhere)!
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  1856. You'll probably appreciate the recent Action Lab video "This material can un-crush itself" that shows a simple application of the Schwartz Lantern geometry
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  1857. Made it to the end, and I love the visual proof. Of course it is clearly fair in that no one can argue that it is unfair, but that is not quite the end of the story. That's because it treats gains and losses as equal, and psychologically, people feel losses more intensely than gains. The estimate I read was that we feel loss twice as intensely as gain. It then begs the question: How we should modify the communicating vessels model if we want it to be psychologically fair instead of economically fair?
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  1858.  @Filipnalepa  For n>3, there are different kinds of 'center of mass'. There is the vertex center of mass, the edge center of mass and the area center of mass. In this video, Mathologer refers to the VERTEX center of mass. Your intuition looks a the AREA center of mass.
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  1859. Another gem 💎 of number theory! Only Mathologer could translate it into an accessible youtube video without compromising rigor! Gauss said of the proof, 'It tortured me'! Reading that I knew I wouldn't last one lemma before it, in my first encounter with it in Burton's Elementary Number Theory. Sure, the Preliminary Gauss's lemma and the main combinational argument of the proof was tough to understand, at the introductory level I first studied it an year ago😧. Now, most of that no longer dwells in my mind 😅 (forgotten! ), so imagine my ecstacy at this mathologer video😍! I'm sure it will help me comprehend it better than before, and I'll remember it longer (once I finish watching 😁)!
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  1860. You can´t challenge me with visualisations. Am doing it all in my head! :face-purple-crying:
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  1861. This reminds me of my high school Math teacher, he used to tell us " and this is the art of mathematics"
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  1862. You're a pedagogical genius!
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  1863. One minute into the video, and I was already surprised: I had never learned/noticed that the 3-4-5 triangle has inradius 1! The fact that the sum/product identity generalizes to any triangle with inradius 1 is amazing.
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  1864. I remember being amazed at slide rules. We did only the "numbers" with it. The scale (decimal) was in our head only. I remember in a class we were proof checking the results given by the then "new calculator device" in order to see if it was calculating properly!! At the time everyone was amazed at the degree of precision of the calculator but we all knew the slide rule was faster!!! What really killed the SR was the TRIG tables... no more need to look up the tables with the calculator...
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  1865. 19:44 Potential crossover with Michael Penn?
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  1866. 20:48 Never realised, that he has an non-native-english accent, until I hear him speaking German. Grüße aus Leverkusen, der Heimat des Aspirin.
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  1867. I am happy that video was not that big (I mean not as long as 40min) because it's not easy for me to find such time in which I can see your videos 😊😊
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  1868. it is a coincidence
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  1869. 10:41 “enhance!”
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  1870. No words could explain the infinite joy I get while going on this journey with your way of telling this story. Thanks 🙏
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  1871. I like way when you get pumped we too get pumped.... And that's what I think is showing true love to THE MATHEMATICS
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  1872. @ that seems fortuitous for you. I have had many professors, both math and non-math professors, that have refused to acknowledge a mistake. In one instance, the mistake was on an exam…professor refused to acknowledge the mistake, so I wrote down the trivial counterexample, wrote down what the professor intended the problem to be, then solved the intended problem. I’m not insinuating that a majority won’t acknowledge a mistake, just that they exist and I conjecture that as one’s popularity rises, the willingness to acknowledge a mistake decreases.
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  1873. 18:41 should be {5/2}, and that would be either 3 line segments ("2-gons?"), or 2 triangles.
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  1874. When I was young this game was being sold under the name "Hi-Q" by one of the big game/toy companies (Hasbro? Ideal?). A quick search reveals several companies were in on it. This was long before haiku become a thing in North America, though Hi-Karate (after-shave, etc.) was being marketed.
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  1875. Haha 28:09 you have Euler in yours patreons. That's something!
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  1876. What if we have a Penrose tiling grid?
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  1877. It's always exciting when these come out!
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  1878. There very often is. That's why it's a clever place to hide sneaky things.
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  1879. Ooh ooh I've been working on polylogarithm functions and Dirichlet series recently and what you have shown is very exciting to me.
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  1880. I love seeing Magic Squares used in Sudoku variants. Thanks for sharing.
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  1881. It begs the question about just how much amazing maths is out there but has just been lost to time. Amazing video thanks Mathologer!
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  1882. As a straight C, occasional B, math student I’m in awe of this channel. I discovered this channel just yesterday and I’ve already devoured several hours worth of math I otherwise would not have cared about. Idk if you are a teacher or were a teacher at one point but your presentation, explanations and enthusiasm is something I wish I had when I was taking math.
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  1883. The brilliance is that they all were "oral" recitations - imagine a full paper worth of mathematical notations in 4 lines of Sanskrit stanzas! YES - it "has" to be exact 4 lines of Sanskrit stanza - otherwise they will flout the Sanskrit grammar rules. So imagine the plight!!! It has to be the mathematical the full proof; needs to be Sanskrit grammatically correct- and YES -easy to learn and record in minds as well - no paper. Very small order indeed! This exact proof it like so - it is written like this in Madhava's mother tongue Malayalam script but in Sanskrit : വ്യാസേ വാരിധിനിഹതേ രൂപ ഹൃതെ വ്യാസ സാഗരാഭിഹതേ ത്രിശരാദി വിഷമസംഖ്യാഭക്തം ഋണം സ്വം പൃഥക് ക്രമാത് കുര്യാത് Those 4 lines - summarize the entire summation series like so : വ്യാസേ വാരിധി ->4 x Diameter ത്രിശരാദി -> 3, 5 etc വിഷമസംഖ്യാഭക്തം -> Div by odd nos ഋണം സ്വം -> plus & minus Which goes like this – Do the 4 times of the diameter of a circle and then …
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  1884. Hey! I've seen your videos since last year, and I really enjoy it. I turned 16 couple days ago and I'm really used to studying for olympiads. In fact, I was one of the 4 people from Brazil to go to "Conesul", it's a south America olympiad, and I'm studying really hard to go IMO. You and 3b1b are the only foreign channels I know that make videos about the "real math", and I truly love watching your videos. And my request is, would you make a video solving the problem 6 from the 1988 IMO? It's a very famous problem and I'm sure you know the problem and the solution to it (me too btw), but I would love to see a video of yours solving this problem. Jokes aside, I would watch it every morning lol
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  1885.  @leif1075  Yes, I have :)
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  1886. I'm a simple man. I see mathologer upload hair, I click
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  1887. honestly, I kinda wish this was broken up into several half hour videos for each topic, it seemed like it was so jam packed to fit all the segments into one hour long video that you could have gone more in depth and explicit if you had broken it up.
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  1888. Why is 1 = 1 excluded? It looks like a base case of length 1. If you picture addition and multiplication as functions rather than infix operators, +(1) is just as valid as +(2, 2).
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  1889. Mr. Burkard Polster, Mathematician, I really say from the bottom of my heart, you are a very valuable analyst in explaining the theorems of the mathematical world. Every time I watch your videos, a window opens to me deep into the infinite world of mathematics. Your 19 yrs fan :')
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  1890. In the card trick, how is the direction of rotation communicated?
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  1891. Monetizing videos is fine (I block ads anyway) but please don't put sponsored content inside your videos, like many other great Youtubers seem to do for the past three years. Any video with sponsored content won't get my upvote.
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  1892. I love this! I am curious about two parts of the dream proof -- How do we know that the TSP criss cross shortening trick can be applied to exploded lacings (maybe the TSP graph interpretation only applies to real lacings)? And, when carrying out the shortening trick, how do we know there will always be a "0" in the right position to un-cross with? 🤔❤️
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  1893. In Italy, heron's formula is taught in the middle school, when pupils are about 12 years old. Many among them will forget it forever, some of them see it again in high school.
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  1894. Wow! I’m this early for a Mathologer video!
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  1895. Every other triangular number
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  1896. for the record, T. Bousch also has a strong contender in the "best title for a math paper, ever" category : Le Poisson n'a pas d'arête (Annales de l'Institut Henri Poincaré, 1999)
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  1897. I didn't here a lot of the words you said, I was too busy trying to figure out which non-regular polychoron you generated by having both the squares and triangles overlayed on the points. It's a prism formed with triangular prisms on each end
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  1898. Actually Herons method for approximating square roots is still taught here in Germany from time to time, depending on the teacher.
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  1899. Do they even have Latin class in the country most viewers here come from?
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  1900. Very interesting! This seems like a different method from what I know for odd sizes. Mine is algorithmic, and I have no clue how to prove that it is correct. It is the method of 5x5 square in the king's room.
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  1901. Q of hearts?
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  1902. Gratulation zum 100. Video! Alles ist wirklich ein Hochgenuss, einfach perfekt, vielen Dank! 👍
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  1903. „ - How it even make sence to you ?? , it's not real ! - Ah , in mathematics we have general trick for this one , called not careing about it ” (c) Numberphile
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  1904. I cannot tell you how much I enjoyed this lecture. It reminded me of several other Maths, and that in itself is one of its delights. Thank you.
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  1905. It is interesting to note that when reducing numbers to DR using this method 9 & 0 are of equal value. eg 19 & 10 Both come down to 1. This rings true regardless of frequency or location. In short, we can calculate by ignoring these values. 9 = o
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  1906. it was literally 5 marker question in IOQM (1st stage of maths olympiad in india)...😭
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  1907. Truly amazing. Finding wonderful hidden connections among the known things.
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  1908. Can we trust that a Mathologer analysis of Velcro / hook-&-loop would be short? The bunny comes out of the hole ....
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  1909. * screams in Hooke
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  1910. Most memorable part: the visualization of the "no nines sum convergence" What an awesome way to look at it.
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  1911. The best of the best channels ❤️ with high qualitt content, we love ur animation and appreciate every second u spend in preparing them
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  1912. Very nice! It seems to me that, by putting these things together, Mathologer has just added one more proof of Pythagoras' theorem to the long list: Because can use Pythagoras' theorem to prove that Cavalieri's areas cut through the sphere and the cylinder-minus-cone are the same, you should be able to use the conveyer belt alternative prove with the same result to backwards proof Pythagoras' theorem.
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  1913. 9:45 In school they taught us about the normal divisibility, but I vividly remember summing again and again till I got a single number (which helped a lot in school), but I never realised that it was the remainder until really later, when I was around 13-14 and had just learnt some modular arithmetic and was just playing around with it
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  1914. I was moved by this, almost to tears. what a great treatment of a rich subject. Thank you, thnkyu, thku, thx...
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  1915. But that shirt tho!
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  1916. First one to comment
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  1917. American math student, even though we got up to differential equations I never saw either proof of Pythagoras until much later on YouTube lol
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  1918. For n disks, the shorthand trick I remember is that odd heights start by moving the small disk to the destination peg and even starts by moving it to the intermediate peg.
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  1919. "Pascal-Khayyam* triangle." Ftfy Come on... I really love and respect your work, you are a true mathematical mythology logger, but please get this right that Khayyam used the triangle 600 years before Pascal made famous. Edit: mathematical mythology logger Edit: timeless truths are not invented, and Karaji used it before Khayam
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  1920. "Still here? Whoa! Need another coffee?" And I was just about to say "yes!". It was just before that point where I was beginning to fall asleep, LOL
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  1921. This is how I was taught the logarithm definition in school, great video
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  1922. Amazing, awesome video, thanks. I'd like to add that moving into 3D is also good with golden triangles. Yes, the pentagon is filled with phi, but also the icosahedron. In 2008 I discovered a way to use icosahedral geometry to create structures with some amazing properties which one can see via search for towerdome. The golden ratio contributes (I believe) to a unique routing of stress forces to make it perhaps the most efficient way of holding weight above ground with multiple floors.
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  1923. As a Keraliya myself, the representation of Madhavan of Irinjalakkuda fills me with pride. It is interesting how Mādhavan wrote in Malayala Bhasha and not in Sanskrit, the standard among scholars of that era.
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  1924. Great pizza golden ratio t-shirt!
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  1925. It is as simple as stated. The algorithm is always treated as a whole. There are no intermediate steps.
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  1926. As usual, one of the best videos on Youtube. Thank you, especially when showing that the locus of any point on the smaller circle of radius r rolling inside the larger circle of radius 2r is a straight line. I've taught high school mathematics for almost 20 years and didn't realize that. Beautifully explained!
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  1927. At the beginning of every Mathologer video, I always think of Kate Bushes’, Babushka. 🤔
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  1928. Isn't the double folded triangle exactly the tiling triangle? Both are similar and have 1/9 of the original area
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  1929. Oh, what could be nicer than a weekend with a mathologer video :)
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  1930. You can also use the golden ratio to drastically improve the prime number theorem. Number of primes less than (n/phi)^2=((n-n/phi)x(n/Ln(n-1))/2. For example: n=641,894 Pi(157,380,656,251) =6,361,970,514 n/Ln(n-1)=6,350,670,612 Using the golden ratio=6,360,264,553 Not sure if this shows some connection between phi, e and primes or just a neat trick that works really well.
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  1931. Hey!! This channel has helped me a lot with competition maths and i want to say keep doing what you’re doing. Not that you need the encouragement, of course, just thought I would say :)
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  1932. The audio has definitely gotten louder, and I appreciate!
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  1933. After appreciating geometry and history, I must commend the script. The coverage and sequencing is well planned and executed. Thanks for everything.
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  1934. I was thinking what a GOD Euler was right when Mathologer said the same thing near the end.
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  1935. Wow, there's so many takeaways from this video(the Fibonacci one was so subtle and satisfying). Quality content, as always! And the tshirt was so cute 😁😁 (HO)^3 😂 Happy holidays, Sir!
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  1936. What can I say, this is just a gem ! thank you !
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  1937. 23:56 It doesn't work for k<=3 because then for some nCk we get n<k, for which n choose k is not defined. However, thinking of Pascal's triangle, we can set those equal to zero (or omit these terms) and that should work.
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  1938. Man if I had a math teacher like you , I would be a rocket scientist by now ! I was in high school in an elite math class but switched to Biology going to university cause of losing interest in math, but i understood 99 % of your video and really got me excited again by math.
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  1939. 2:34, check of the calculations up to that point: For the equilateral triangle, the height is the length of one of its legs times the cosine of π/6, so the height equals 2*sqrt(3) times sqrt(3)/2 . Thus, the area is half its base, i. e. sqrt(3), times 2 times sqrt(3) times sqrt(3) times 1/2 which amounts to 3*sqrt(3). For the isosceles triangle with legs of the length 2+φ with a base of 2*φ, Pythagoras's theorem gives sqrt( (2+φ)^2 - φ^2 ) = sqrt(4 + 4φ + φ^2 - φ^2) = sqrt( 4(1+φ) ) . The golden ratio is a solution to the the equation x^2 = x+1 , so sqrt( 4(1+φ) ) = sqrt( 4φ^2 ) = 2φ . (Since we're talking about lengths, we can ignore negative results.) Finally, the area of this isosceles triangle is then half its base times its height: φ times 2φ = 2φ^2 .
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  1940. I figured out the puzzle at the end! I imagined the hexagon as a cube made from cubes. If I look at any of the three sides, I will see a solid color, and each of the three sides is the same, so there is an equal amount of each color.
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  1941. you are the mathematical superman, saving mathematical discoveries from oblivion!
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  1942. "Checking by hand would be way too taxing for people, who grew up punching buttons on the calculator". Like this line. What about people, who grew up punching keys on the computer keyboard? :)
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  1943. 19:59 I would never have recognised that pattern in my entire life ,I wonder how does Euler does this ?
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  1944. I like how for this video I sat for 23 mins straight with full attention while I can't do the same thing for maths in school
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  1945. Fantastic stuff!!! Actually I found a pattern in your videos. With each new video, the length of your channel's supporter list at the end grows enough to conclude that the length of subsequent videos approaches infinity!
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  1946. The ultimate algorithm is to solve all the problems
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  1947. That's a feature, not a bug 😂
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  1948.  @cliveso  That's another way of saying it incentivizes you to hedge your risk by sharing it among other creditors, which seems to me to itself be a societal good.
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  1949. It would make the Greeks more misogynistic. Technological progress would get delayed even more because people would think that they already reached the technological celling.
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  1950. School maths should teach more like this , love the geometric drawing conceptualiseations.
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  1951. If any Indian made this kind of video talking about Madhava he would have been trolled to death
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  1952. I really enjoyed it. I guessed no 9s has finite sum by thinking about binary numbers with no 0s. that's 1/1 + 1/11 + 1/111 + 1/1111 + .... < 1 + 1/10 + 1/100 + 1/1000 + ... = 10 (=2). I most enjoyed the Leaning Tower of Liire, because they didn't teach it in my calculus course and it's simple and beautiful. Thank you Bar
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  1953. Great video!
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  1954. Why everytime that I'm intersted in a specific branch of mathematics Mathologer does a video explaining exactly what I was searching for?
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  1955. Our turn: sum of all 7 angles is also 180°.. awesome maths magic.
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  1956. beautiful! thanks!! medians of median-triangle give back the scale-down sides -- i also re-discovered this in middle school, when i tried to compute the formula for median lengths using pythagoras and area formula, and noticed that it was a reversible formula, in the sense that you can apply the same formula to get back the sides, if the median lengths are known, with a scaling factor. now seeing the animation today looks very beautiful.
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  1957. but 2 is the oddest of all primes
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  1958. I’ve been looking for this!!!
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  1959. Solution for the cross: Just use Banach Tarski lmao
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  1960. Yay, the best suprise ever
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  1961. Thoughts on the 🧅 problem For the area of a circle(onion cross-section): finding the area is like adding up all the circumferences so Area = integral of all circumference × dx. (dx→ very small thickness) On differentiation The derivative of area = derivative of integral of circumference × dx = circumference. For the sphere (onion), Appling the similar logic on the = whole of ( \int of) Surface area× dr (dr→ small thickness of cone) On differentiation Derivative of volume = derivative of integral of surface area × dr = surface area . By Sabbir Mondal (India- studying in high school)
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  1962. Circle lovers been real quiet since this dropped 🤔🤨
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  1963. Ah, finally ! Math on YouTube. Subscribed. I was always enamoured by pentile formations. Could we have a video on that please ? 🎉❤
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  1964. I'm Ukrainian. We did learn Heron's formula and used it a couple of times. Just enough to still remember it.
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  1965. I have used such circular calculators 50 years ago for designing books, which required complex scaling computation for croping pictures or sizing type. I still have two that I use sentimentally. One is calibrated using inches, and the other in millimeters. I would often use the millimeter dial like a slide rule.
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  1966. Great video as always! 13:42 Let's focus on the label "1" on the rubber band. Initially it's at the end of the straight part of the rubber band. The original rubber band is 1 unit long. Assume the radius of the wheel is R. Denote the angle of rotation of the label "1" by x (initially x=0). For any given angle x, there is a label at the end of the straight part of the rubber band. Denote it by f(x). We have f(0) = 1 and f(2π) = 0.1. For any x, consider a tiny rotation so that x becomes x+Δx. The length of the straight part of the rubber band decreases by RΔx. The new label f(x+Δx) = f(x)(1-RΔx). Take the limit as Δx->0 and we have a differential equation df/dx = -Rf, solve and we have ln(f) = -Rx + C. From the values f(0) and f(2π) we know C=0 and 2πR=10.
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  1967. Thank you so much for continuing your amazing effort!!! You are a true gem in all this dimension.
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  1968. Some of these and others similar you can find in 1st volume of Knuth's "The Art of Computer Programming" with invitation to prove ones
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  1969. 36:34 Challenge accepted (Spoils the answer) Consider that the small right angle triangle at the top right is similar to the same triangle but including the trapezium below it. The ratio of sides is 0.5:1=1:2 (by considering the ratio of hypotenuses). Therefore, big triangle area = 4×small triangle area. White area = 16×small triangle area then. Now, add the small triangle at the bottom and you get a right triangle with sides 0.5 and 1, so area = ½×0.5×1=0.25. It also happens that's the area of 5 small triangles. 5∆=0.25 ∆=0.05 16∆=0.8 Hence, area of the square is 1-0.8=0.2 Hope my proof is pretty
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  1970. Lol fibonacchos
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  1971. Dear Mr Burkard Polster, I sincerely appreciate the fact that you give credit to the right person for the discovery of a mathematical identity/formula (in this case, the Indian mathematician, Mādhava). Earlier also, I learned through you that the fact that e^ix = cos x + i×sin x was known to Roger Cotes before Euler. Cheers and merry Christmas!
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  1972. Mathologer never ever disappoints ;) 😄❤❤❤🤗
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  1973. 37:25 this operation is the same as performing the permutation given by n 1 2 ... n-1 to the previous permutation Since this permutation clearly has n-1 inversions and signs of permutations multiply, when n is even it will change the sign and n is odd it will preserve the sign
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  1974. Finally!!! I've been waiting for it.
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  1975. I've got a proof for the last problem distinct from ilonachan's very elegant sudoku-style proof, inasfar as grid-puzzle proofs can be distinct, giving the same fully general 2n*2m result. I) Any solution must have an "Alternating Orientation" (equivalent solution under reflection over a diagonal or 90° rotation) where each row alternates between two colors, and the sequence of rows alternates between two such pairs of colors. II) a. For an even number of columns, an alternating orientation of a solution must have different colors for the first and last columns in each row. b. For an even number of rows, an alternating orientation trivially has different colors between the first and last rows. c. The colors in each of the four corners of an alternating orientation must be different. III) Reflection along the diagonal or rotation by 90° preserves both the property of being a solution, and the span of the colors in the corners. (Trivial) Proof of I: Ia) If the first row alternated over a pair, the second row must contain the complement pair; and that row must therefore alternate, as, trivially, two adjacent cells cannot share a color, and there are only two colors in the row. Inductively, each successive row must also contain the complement pair to the previous row, and must itself alternate. This satisfies the definition of an alternating orientation. Suppose there is a solution which does not have an alternating orientation. By Ia, the first row does not alternate over a pair. Without loss of generality, it can be shown that the following configurations remain: i. abcd[...] ii. abcb[...] iii. abac[...] Where [...] represents any sequence of successive terms, for grids of more than 4 columns. We shall first assume configurations i and ii, together. Both configuration i and ii start with the same three colors, and we may say they both satisfy of configuration 'abcx[...],' where 'x' is undefined. As no steps of the remaining steps do not depend on cells in columns 3 or later for this orientation, configurations will be denoted with only the first three color values. Cell (2, 2) is part of both the top-left 2x2 square, and the top-center 2x2 square. It must therefore not be the same color as the first two or second two cells in row 1. As all three colors are represented in those three cells, cell (2, 2) must be the 4th color. Cells (1, 2) and (3, 2) are now determined, as they each have 3 colored neighbors. In particular, row two is now in a configuration of 'cda.' This configuration is isomorphic to configurations i. or ii. of row one. Therefore, each row can be determined inductively. In particular, row three is also in a configuration 'abc,' which is the same as row one. If we look at column 1, we find it must alternate between colors 'a' and 'c.' This means that if we reflect over the diagonal, we must have a solution where the first row, equivalent to the original solution's first column, must alternate over the pair of colors 'a' and 'c.' This satisfies the condition of Ia), and so this reflected solution must be an alternating orientation. This means the original solution had an alternating orientation, contradicting the premise for configurations i and ii. We may use a similar argument for configuration iii, but shifting our perspective over by 1 column. We first remove column one from the solution, and observe that we now have a configuration isomorphic to configuration i or ii. As removing a column does not alter the property of being a solution, as it does not affect 2x2 grids not intersecting that column, the argumentation for configurations i/ii follow. If we take the resultant alternating orientation, and then reflect this solution vertically so that the top row is now at the bottom, we may notice that our new top row is also alternating, as given by Ia. If we try to replace an appropriately reoriented row at the bottom corresponding to the column we removed here, the solution must still satisfy Ia, as the new top row is not changed. The property of having an alternating orientation is trivially preserved under horizontal and vertical reflections. As such, this re-appended row must still be alternating, and our overall solution still has some alternating orientation. I: QED QED (Sorry for the haphazard alterations to the proof of I, but I realized I needed to generalize it to get the full 2n*2m result, but that proved non-trivial.)
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  1976. ​ @Mathologer  Yes I did, actually! But then it goes 23, 37, 60, ... Oh well . . . On one of his first two record albums*, ca 1960, Bob Newhart did a short bit on the famous monkeys-and-typewriters principle, playing the part of the guy who's walking up and down the line of primates reporting their output to the experiment's conductor, when he remarks: "Oh wait. I think we may have something here. 'To be or not to be – that is the gzornmplat ...' " * "The Button-Down Mind of Bob Newhart" and "The Button-Down Mind Strikes Back"
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  1977. This!
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  1978. 20:46
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  1979.  @Xanthe_Cat  1 is a REALLY inconvenient prime, when writing down prime factors. Also it would make itself not a prime because of 1 = 1*1*1*1*1....
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  1980. I’d say the proof showing the harmonic series diverges is the most memorable, given that I had seen it before and remembered it
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  1981. I used a slide rule as a kid. Much cheaper and more portable than calculators of the time (I was born in 1959)
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  1982. I tend to approach your videos with an open mind and willingness to learn and I am never disappointed. Your teaching methods are wonderful and refreshing as ever. Thank you for all your hard work and dedication. Your love and passion for math is unmatched. We’re kindred spirits, but your grasp of mathematics is astounding. The final problem I believe the answer is 1 if my math is correct. Though I have been wrong before. Thank you for such a wonderful video!❤❤
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  1983. 25:29 Your whimsy is boundless ^_^
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  1984. Great video. I really loved learning about Bachet's algorithm and geomagic squares. I was also reminded of orthogonal latin squares when you described Bachlet's algorithm.
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  1985. Requested info: I was able to follow the whole thing (something I have not been able to do in some other videos) . BS in Mathematics/Computer Science. Work online as a math tutor today.
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  1986. I didn't know that you were german. Only this video showed me, how perfect you pronounced the names of the mathematicians. Gutes Video wie immer und Gruss aus der Schweiz ^^
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  1987. Excellent work! And yes, you should be wary of discontinuous "logarithms" lurking in the dark :) The axiom of choice guarantees their existence (but that also explains why you can't find any of them - because they're non-constructive). Essentially, the multiplicative group of positive real numbers forms a vector space over Q (taking vector addition to be multiplication and scalar multiplication to be exponentiation). So the axiom of choice guarantees that there is a Hamel basis for the multiplicative group of positive real numbers over Q, and we can assume e is part of that basis. Similarly, the additive group of all real numbers is a vector space over Q, so the axiom of choice guarantees us a Hamel basis here, too, which we can assume contains 1. (Note that we can prove the dimension of both of these spaces over Q is |R|.) Then we can create a function which maps the Hamel basis from the multiplicative positive real numbers to the Hamel basis of the additive real numbers, choosing e to get mapped to 1, and choosing it so that some basis element d of the multiplicative positive reals is mapped to something other than ln(d). But a function defined from a basis of one vector space to elements of another vector space defines a unique linear transformation between the two vector spaces. Let's call this linear transformation L. We can then check by the definition of these vector spaces and the definition of linear transformations that it has all the algebraic properties of the natural logarithm. If a and b are two positive real numbers, then L(ab) = L(a)+L(b) because vector addition in the domain is actual multiplication but vector addition in the codomain is actual addition. For any positive real number a and any rational number c, we have L(a^c) = cL(a) since a^c is scalar multiplication of a by c in the domain but scalar multiplication in the codomain is actual multiplication. We also have L(1) = 0 since 1 and 0 are the identities respectively, and L(e) = 1 by choice. So we have the "properties" of a logarithm that you were able to pin down without continuity. But we have L(d) = something other than ln(d), so L is not the same thing as the natural log. (As an added bonus, I made sure L is bijective too!) But your argument pretty much shows that ln(x) is completely pinned down by ln(e) = 1, ln(ab) = ln(a)+ln(b) for all real positive numbers a and b, and continuity. So the only conclusion here is that this nasty L function cannot be continuous. And since it has the other properties of logarithms, it is in some sense a discontinuous "natural log".
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  1988. That card trick is very clever.
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  1989. Wow... I liken watching Mathologer to watching Bob Ross. I’m watching a man explain something far beyond my capabilities but just listening to him talk about something he loves and the intricacies therein is beautiful. I can follow the minutia but I can’t keep track of all of it. It’s just lovely watching these long strings of equations getting simplified and expanded and simplified again. Never stop, Mathologer, never stop.
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  1990. but are those subgroups normal?
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  1991. 16:42 so satisfying! I had to explain to my roommate why I audibly gasped :D
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  1992. Wow, somehow I have never completely understood why the inverse Fourier Transformation is calculated as it is .. thanks to this video now I know.
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  1993. With Mathologer you can always count that the subject will be deeply explored :)
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  1994. "calculate the 36th triangular number" HAIL SATAN
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  1995. I was thinking about that Pythagorean triple tree. If one associates going straight with a 1, going left with a 2 and going right with a 3, then one can encode every primitive Pythagorean triple (PPT) with a rational number between 0 and 1 in base 4 as follows: 0 corresponds to (3,4,5), 0.1 to (21,20,29), 0.2 to (5,8,17), 0.3 to (5,12,13), 0.12 to (77,36,85) and so forth. That way one gets a 1-1 correspondence with the finitely representable base 4 numbers between 0 (included) and 1 that do not contain the digit 0. The three families of PPTs correspond to the numbers 0.11111..., 0.22222 and 0.3333... I wonder if one can get anything geometrically meaningful with that correspondence. Can one interpret the periodic or irrational numbers as interesting infinite paths/families of PTTs in the tree? What about interpreting numerical manipulations like multiplication of 0.1111 with to to get 0.2222 in terms of the associated PTTs?
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  1996. I only remember some of this vaguely from university. They glossed over most of the calculus of differences. Although when you get into Lebesgue integration and measure theory they just sort of start assuming you know this stuff already.
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  1997. The dance algorithm is incredibly cool. I think my favorite “aha” or maybe even a forehead-slapper in this video was “but how does the powers of two accommodate the deleting of some pairs?! Oh!!!!! Because those add a degeneracy which can be resolved exactly with a multiple of 2!” That was very satisfying.
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  1998. 8:40 This is probably the more straightforward Mathologer homework question ever. Let's start by defining an essentially different triangle. Let's say two triangles are essentially the same if we can get from one to the other by relabeling colors or by flipping across the vertical axis. Some have mentioned rotational symmetry but I'm not including that. Now let's canonize the colorings: We'll relabel the colors so that the top left hex is always red and the next color as we go rightwards is always yellow. If there is no next color, we're OK because we know there's only one solid red triangle. Let's also canonize the orientation: We'll also say that if we can flip the triangle before relabeling to get more red hexes on the left half of the top row after relabeling, we do. If we had larger triangles, we would need a more complex rule using yellow hexes as a tie-breaker, but triangles of 4 are too small to ever need that. Now all we need to do is count the triangles that have red in their upper left, yellow next, and can't be flipped before relabeling to get more red hexes in the left half of the top row. There are 10.
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  1999. video suggestion: covering systems you can talk about modulo arithmetics, prime numbers (fragile primes), trees (solution for minmod 40)
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  2000. ⭐️⭐️⭐️⭐️⭐️THAT WAS FOR THE LACK OF A BETTER WORD: BRILLIANT! Thank you, Mathologer! 😀 👍
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  2001. Knowing there is some trick to guessing quickly I looked back at the examples. I noticed that the top left and right corners added to the bottom one in each. Let's see if that hypothesis is right!
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  2002. We can rewrite the correction 1/( N + 1/( 4N + 4/N) ) as 1/N * 1/( 1 + 1 / ( 4N² + 4) ) which, using 1/(1+eps) = 1-eps, approximates as 1/N * (1 - 1 / ( 4N² + 4 ) ) = 1/N - 1/(4N³ + 4N) which once more gives 1/N - 1/(4N³) * 1/(1+1/N²) which approximates as 1/N - 1/(4N³) * (1-1/N²) = 1/N - 1/(4N³) + 1/(4N⁵) So the errors when N equals a million occur near the 6th (N), 18th (N³) and 30th (N⁵) position, explaining the sizes of the series of correct digits even a bit better. (This works because the error in 1/(1+eps) = 1-eps is also of order eps³, hence why the coefficient of the 1/N⁵ term seems wrong compared to the video; a better approximation would account for that term, but I didn't want to complicate things too much...)
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  2003. if you can put up some videos for group theory,then probably i would be the happiest person in the world.
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  2004. Crazy day! I literally told my friend about the Bernoulli Numbers (and finding them using Pascals Triangle) and the Euler-Maclaurin Formula today (we found the formula for summing 66*x^6 to the n'th term) Approximations are really my thing I'm doing with mathematics lately, and I've used the Euler-Maclaurin formula for approximating the Zeta-function. I have a whole spreadsheet with approximation of special functions. (I also fell a mathematical tear when you mentioned the Stirling Approximation as it in my opinion is the most beautiful thing in mathematics even surpassing Pythagoras and Euler's Identity, hoping for a video on it one day) Education is High School, so I get most of your videos (but the videos on transcendental numbers I still struggle on), due to my previous workings with the Euler-Maclaurin Formula I understood this video really well, but for someone who's not heard about Bernoulli Numbers and the Euler-Maclaurin formula before, it might be difficult. Looking very much forward to more videos on this (and that Stirling's Approximation video!)
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  2005. I just watched it for the second time today. It's even better the second time around. The length actually adds to the clarity because it attacks the problem in multiple ways.
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  2006. I understand every point, it was very interesting.. Thank you Sir..
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  2007. Absolutely amazing! Don't worry about length, this was too beautiful not to share!!
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  2008. Mathologer videos slowly becoming true epics in mathematics
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  2009. Thank you for presenting this. Haven’t had a math class in more than 40 years but I did have formal logic which helped a bit when following this video. If I had seen this in high school I might have had a whole different career path.
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  2010. Watched 28 minutes video, but only thought: "why so short? I wanted more pi formulas!"
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  2011. Phew! Got through it but need to rewatch how those crazy calculations in a pre PC era where done. My background is in computer science.
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  2012. Great video as always! I think it’d be really cool if you did a video on Minkowski’s Lemma and you could use it to prove the Fermat’s Christmas Theorem. The proof is relatively simple, but I didn’t see this until Uni
    5
  2013. To infinity and beyond... talk about tilting at windmills.
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  2014. This is honestly the best video on magic squares on YouTube & the best comprehensible video on the topic I've ever seen. SUPER! B)
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  2015. My dad gave me his circular slide rule that he got back when he was in university. It was produced in 1957. Really cool for me to have and play with, but I love the rubber band explanation!
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  2016. ​@Mathologer I found it uncanny but still pretty cool
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  2017. how you develop 'wonder' thing? I am not wondering at anything
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  2018. After seeing this proof it looks like that this theorem could have been put in an Olympiad problem also
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  2019. Hope this is good
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  2020. "Why halfen the container size?" Simply to make the model's numbers (containers) match the Talmudic numbers. In this step, Mathologer is just exploring how water containers could be used to try model the Talmudic problem. Again, the original problem is actually to try to figure out what fairness-reasoning the Talmud is using; but the first step is just to try some model that produces the same numbers, and then to look at extending/modifying the model to see if it gives insight into the original problem's reasoning.
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  2021. In other words, yes, seven million people CAN be wrong. But I already knew that. And I really, really dislike Tesla-groupies.
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  2022. Australian
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  2023. Noice
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  2024. I am studying math in university right now, and I am not put off by such a long video. I had to split it up into chunks for when I had time, but it was intriguing, so obviously I watched it all the way through, and if you continue to produce videos like this, I would watch it even if it was 2 hours.
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  2025. To me the fact that you can get from one convergent to the next is amazing. At first it looks like putting one more coefficient means you need to calculate the fraction again from scratch. But the picture makes the quick way so obvious!
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  2026. 44:25 did everyone REALLY miss the fact that 'Markus Persson' AKA NOTCH! watches mathologer!!
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  2027. I came up with a poem of my own for the cubic formula: x^3 + n*x = m x = cbrt(m/2 + sqrt(D)) + cbrt(m/2 - sqrt(D)) D = (m/2)^2 + (n/3)^3 When x cubed and x times n, Are added and equal to m. The values of x, The goals of our quest, Here's how to calculate them. Cube roots to add, Square roots they had, Both of a term we'll call D. Square half of m, Cube third of n, Add together and see. Half of m, adds to the root, First with a sign of plus. Its little brother, Is just like the other, Except with a sign of minus. Cube rooting time, of both the brothers, Add up the roots with glee. We found our first x, But where is the next? I know there have to be three. With help from DeMoivre, Who's theorem, we love ya, There's cube roots all over the plane Yes, they're complex, But do not perplex, A new kind of numbers we gain.
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  2028. And thank you so much for getting another video out before Christmas. Especially one with things to ponder over the break!
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  2029. Now you've set yourself up to mathologerize the representation theory of Sn, and I can't wait for it.
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  2030. Funny thing, in the film clip the proof relies on the 'fact' that starting with a positive integer and reducing its value must eventually result in a negative number, but in this problem the number goes to zero in finitely many steps and never becomes negative.
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  2031. This is the reason why I mayored in math: to watch Mathologer master classes from start to finish
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  2032. yes, very special! I wondered about the Chinese (or Japanese, as it turns out) flavoured music, but I guess it was suggested by the sinuous line of the baggage carousel. I was really skeptical about the preservation of the area through the meridional lay-down, but the travelling circle argument convinced me. So much to think about in this video!
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  2033. I am very very fascinated by 1) How hardworking you are with all these presentations 2) How kind, positive and interested in math you are. It's perfect that you make these videos, it literally makes me much happier because i fall in love with math more and more. P. S. Sorry for my english, it's not my language.
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  2034. @Vihart will love this episode :)
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  2035. I love the no comment animation at the beginning. Well paced to be able to capture the sequence. Maybe do some more of this conzent. Not for shorts though as there's not enpugh time.
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  2036. I think it's also nice to mention two certain patterns cos'x = -sinx sin'x = cosx Buuuut what if it's like cos'x = sin(-x) sin'x = cos(-x) sin(-x) = -sinx cos(-x) = cos x Pretty neat. And the other is the derivative of a constant: C' = (Cx^0)' = C(x^0)' = C * 0 * x^-1 = 0 Which also shows that there is no x^n that results in x^-1, but that function must have an integral, right? And it's the only one that connects to lnx, as ln'x = 1/x = x^-1
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  2037. He caved in and gave us the clickbait we deserved 🗣️🗣️🗣️📣📣📣
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  2038. In all seriousness: yes, calculus is easy. Or at least not as hard as most people think. But you need to approach it like you did here: by discussing a real-world case you already know about. I learned calculus exactly like that. In physics class, not math class. And I suspect I am not the only one.
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  2039. I will always have a soft spot for the Wolfram spiky
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  2040. Ahhh endlich!
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  2041. Yeah it just got recommended even though I've been waiting for this to be done on a big math channel for a while now
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  2042. yes, we were actually taught the "digital sum is the remainder" thing, our school system here loves to emphasis on the little tips and tricks instead of heart of the matter, take that as a negative or positive, up to you
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  2043. For each even n (you mention n=4), you have a round with 'flat ears' (180 degrees). Instead of looking at the 'tips of the ears' for constructing a new n-gon, there is anothe way of looking at it: in each round add regular n-gons to all sides and construct the new n-gon by connecting the midpoints of these added regular n-gons. The difference in the rounds (the different degrees in the 'ear wording') is by aligning not two consecutive vertices of the regular n-gons, but such that k vertices ly on one side and n-2-k on the other side, for k=0, 1, ..., n-3. So it is like every round of constructing a new n-gon is by adding regular n-gons 'sunken one vertex more than previous round'. (And indeed the rounds can be permuted.) You can see this visualised in the Geogebra application found at Geogebra ... /classic/sdftbe2y. So there are two equivalent formulations of the Petr-Douglas-Neumann theorem: - using tips of triangles (ears) with different angles per round; or - using centers of regular n-gons 'sinking' through the edges per round. The intriguing idea is that Napoleon's theorem uses the second formulation and Van Auble's theorem uses the second formulation only for the 'first round'.
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  2044. 6:22 Remove (0, 1) and (1, 0) (=> same color) and 2 others of the opposite color anywhere except the corner (0, 0). That's 2 black + 2 green removed and you won't be able to tile the corner (0, 0) :-)
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  2045. Can you please make a vedio on the very less known, but very interesting kaprekar's constant After seeing this vedio i had a feeling that it can be proved using the same pegion hole principle
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  2046. The videos are always worth the wait! ❤❤❤
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  2047. The trick I always used for the tower of Hanoi is based on odd or even quantity of discs. If you have an odd number of discs then the first move will be to the target tower. If you have an even number, then your first move will be to the other empty tower.
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  2048. I like the connection of this pattern with Sierpinski's triangle. I was expecting it to be a direct Sierpinski triangle, not a "higher order" one with many more smaller copies of itself in the central triangle though. I also saw the symmetries of rotating the triangle and running the rules that way. I now wonder if it's possible to "run the rules in reverse" and build up the triangle that way? It doesn't really look it's possible with this rule set, but is there one that can?
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  2049. Great video! The concept of what shapes can be made from a number of small squares has interested me in terms of number theory and prime numbers. Obviously a number of small squares can form a rectangle if and only if the number is composite and conversely if a number of small squares is prime it can not be formed into a rectangle. If anyone knows of any material that explores that aspect of the square tile re-arrangement subject it would be greatly appreciated. Thanks again for the great content!!
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  2050. Vielen Dank für das tolle Video!
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  2051. 5:37 I'm going to be that annoying person and point out that each vertex can be no more than √2/2mm from each grid point.
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  2052. I must say, I love your shirt!
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  2053. Excellent explanation. I especially liked the explanation of the hyperbolic trig functions. It's always puzzled me why they're called hyperbolic. I appreciate the work you've put into this. It must have taken hours and hours to get right. Well done.
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  2054. I'm still waiting for a follow up video on rooty expressions by the way :P But this was also a very interesting topic and I was able to follow along quite well for the most part. Only during chapters 7 and 8 there were some calculation where I just believed you to be right because I could not check them in my head.
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  2055.  @FumblkruschLP  You're numberphiled!!!!
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  2056. "There should be an equation in there; let's go and find it!" For a moment, I felt that spark of adventure I used to feel when doing math in my youth. I missed that feeling.
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  2057.  @alflud  you neglect that although mathematics governs the physical universe, the physical universe does not govern mathematics. numbers (including 0) and other mathematical objects would exist even in the absence of a physical universe, and it's not only unhelpful but detrimental to attempt to restrain mathematics according to some preconceptions about how the physical universe ought to behave. ignoring the affects on mathematics, there are plenty of instances in the physical universe where something can be quantified correctly by 0: the electric charge and mass of a photon, for instance
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  2058. Nice, never heard of him, but glad to learn about him now.
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  2059. solution to puzzle from the beginning of the video: (in a reply)
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  2060. I think it's uncountable: Take any order of the terms that converge to the desired value. As long as this value is finite the signs of the terms will change infinitely many times. Let's define a chunk as an unbroken sequence of positive terms followed by an unbroken sequence of negative terms. If we rearrange the positive and the negative terms inside one of the original chunks the final sum will still be the same since the sum of any finite series ending after the chunk will stay fixed. So to construct a new ordering from an old ordering simply pick any real number between 0 and 1, express it in binary, for every chunk if the corresponding digit is 0 leave it alone, if it's 1 swap the two terms on the border between the positive and negative halfs of the chunk.
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  2061. Wow, just realized Sir Geoff Smith is my friend on Facebook, amazing person indeed!
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  2062. The problems start to settle in when your curvy curv happens to be a fractal. Where the more dots you use on your segment, the lenght continues rising without converging. So not all curvy curves have a lenght, so I conjecture that only the ones that don't contain fractals can have lenght (i.e. curves where for any two there exists a dot that admits a tangente line)
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  2063. Glad for your recognition of Madhavan Nambudiri of Sangama-grama. Way overdue! I hope one day that an important theory of math derived from his insight will be bestowed with his name, like the Boson particle in quantum physics. If he were around, he would have accepted it with humility and grace. And you have a new subscriber!
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  2064.  @bjornfeuerbacher5514  Yeah, that was long ago. Lambacher Schweizer is still there, but anything beyond "draw a rectangle around the triangle and compute half the area of it" is nowadays beyond expected student abilities regarding "proofs". (Actually, most students don't even (need to) get the difference between "proving" and "using" the formula.) Fun fact: Just last week I was asked to comment on a complaint of several principals (obviously urged by students and parents) about the recent "Abitur" in the state of Niedersachsen. One of the exam tasks could not be solved by the students, because they had to calculate the area of a triangle without being allowed to look up the formula (one part of the exam is to be solved with pen and paper only).
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  2065. I would have never related the armonic series with a leaning tower, amazing!
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  2066. Bachelors in math checking in: The video worked very well! I feel like I have a good high level grasp of the concepts now. As far as feedback, the pacing was excellent and I think you should continue omitting some of the “dryer” parts of these proofs. This keeps the pacing very good and all the content interesting. I recognize that they’re very important, but I never much cared for proofs of things such as convergence or uniqueness. I think acknowledging that some steps are omitted and providing details in the description is a perfect way to handle these necessary evils of proofs.
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  2067. Look at the 👕 Pythagoras and Einstein fighting for C^2 😂😂
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  2068. I now understand why the edges of 3d shapes of constant width look just like the animation at 27 minutes! It's a sphere being mapped between the vertices, so elated! Thank you, Mathologer, for another wonder full lesson!
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  2069. If your art is as lifeless as AI generated art, you're a bad artist
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  2070. Great video, as always! I think that, at 28:43, the AM-GM inequality for 3 numbers should really be (ABC)^1/3 =< (A+B+C)/3
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  2071. It's the same here
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  2072.  @Mathologer  that makes sense. However, what I choose to believe is that this is why clothes tended to get skimpier over time: from generation to generation, the heirs went from full garment, to quarter garment, to bathing suit, to bikini, to mini bikini, to micro bikini.
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  2073. 31:03 Is it Queen of Hearts or did I miscalculate?
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  2074. An hour ago I wondered when Mathologer would uploud a new video. I started to search answers in the whole internet an now he uploads !!! Coincidence?
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  2075. Mr.Mathologer you are incredibly wonderful thanks for everything ❤
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  2076. the connection this sequence has to the power of twos is that, if the line of 1s extending down continued forever never terminating at a row of all 1s, it would be the powers of 2. the reason it works for 1-16 and then breaks at 31 is because 31 is the first number in the sequence that is effected by the horizontal row of 1s.
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  2077. "Whatever you want comes next"... the sentence that blew my mind 🤯
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  2078. He didn't fold the right triangle twice. If he had, it would have turned back into a smaller right triangle.
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  2079. Great video, as always! Fun fact: in French, ln is usually referred to as "logorithme népérien" or "Neperian logarithm", from Scottish mathematician John Napier who created the first log tables.
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  2080. I followed a lot of math youtuber for years. Mathologer is really the only one that consistently blows my mind.
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  2081. you knoiw its going to be a great video when its mathologer
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  2082. You know it's going to be a good video when the intellectuals watching the channel make idempotent statements just to emphasise how great the video is. No judgment, @neonblack211 :)
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  2083. Most memorable part was the proof at the end showing that the no nine sum is finite.
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  2084. Mathologer There are actual integral sum parallels if you remove the dt from the integral, such as the sum for SIN Radians, except from -∞ to ∞ . The sum and integral equations are identical summing multiple complex number results which thus cancel except when only one for all 360° pluses or minuses multiples. Then only integer t exponents don't cancel. Of course reciprocal-factorial not reciprocal of factorial! Likewise other series. Maybe t all complex I'm not sure
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  2085. I want that shirt man lol
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  2086. What an elegant proof!
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  2087. Welcome back i dont know why should i watch and rewatch your videos over and over again ??????? Thanks much prof .
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  2088. Looks like n-simplexes are being used to describe the triangle numbers, tetrahedral numbers and so on
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  2089. 12:55 you mention your Mathologer T-shirt shop. It would be nice to have a direct link in the description. Love the short videos and the long videos. Something missing in the proof … it isn’t really specified what happens when you get down to just the last digit .. 00001. Of course, this can only happen if there are an ODD number of cards, and all the examples chosen have an even number of cards so will always get to zero. But and odd number of cards will always finish at 0000…001. So either the question could be reposed as “for any positive even integer number of cards”, or the special case of the rightmost card as the only face-up card needs to be formally considered / have its own rule.
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  2090. 20:20 I wonder what the set of all possible limit points looks like. Without having given much thought to that yet, I guess it would look like some kind of fractal?
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  2091. I’ve noticed the rounding pattern, too (likewise, for 7^4 + 8^4 ≈ 9^4,…; so: (2n-1)^n + (2n)^n ≈ (2n+1)^n); but, unfortunately, when I tried to prove it with my Friend (who’s a long-time Mathologer-viewer, and actually introduced me to the channel), we failed to come up with a definite answer. 🤔
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  2092. Try vertices: 23 Density 11... shape to the right, speed to the left Wow!
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  2093. This comment is no longer pinned. I think it stops being pinned when you edit it.
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  2094. Great video, as always! The online applets from the challenge are awesome. I also spent some time playing around with GeoGebra to make this animation (thanks for the mention in the description :)). Kind regards.
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  2095. You've taken me back 39 years (!) to my days at Technical College and the seemingly endless amounts of Calculus theory that we had to learn and then had to apply to physical laboratory works it was great to predict the answers mathematically and then prove them on a test rig (we were taught to think in terms of 'analogues' so we could apply the theory equally happily to mechanical or electrical systems) Today my work involves less mathematics but it still stretches my brain and memory. I'm working on a problem at the moment that involves a lot of electrical relays and, after having spent three days going through electrical wiring diagrams, it has occurred to me that I need to recreate the original logic diagram for the system in order to identify the cause of the problem. I wonder if I still have a good chart template kicking around...
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  2096. Goat
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  2097. 15:23, I arrived at the same answer as the Talmud, by the following reasoning: the contested child ("red") who has a possible claim to the 1st son's estate takes first pick, and is given half-way between what he would receive in the two scenarios: (1/2 + 1/3) / 2 = 5/12. What remains (7/12) is given to his two siblings, and is shared equally between them, so they each take 7/24. My reason for giving the "red" child precedence was more of a practical than a legal one: it's messy to modify the fractions for all the children at once, and you can easily end up with something that doesn't add up to 1. By first settling the "red" child's claim, it's then trivial to divide what remains between the "green" children. Interestingly, we can go about it the other way, settling the "green" claim first: (1/2 + 2/3) / 2 = 7/12, then 5/12 remain for the "red" child. So, the solution is the same regardless of who takes first pick, which is definitely a plus. A more modern approach would be to assign the two scenarios some probability, then divide between them in proportion to how likely they are; but given that no information is available about the paternity of the "red" child, a 50-50 divide is kind of forced.
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  2098. I did a basic irrational test for the triangle. Given a rational base 'a', the perpandicular to the base (the height) of such equalateral triangle is sqrt(3)*a/2. Given a square grid, the proportions of all lines to other lines are rational (aka, a scale of one unit per smallest line exists and all point-point lines can be described using this scale as a base and the slope between points). Therefore, I concluded there exists no equalateral triangle on a square grid as such would imply the existance of a relation between perpendiculars that is both rational and irrational on a grid containing only rational relations.
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  2099. i found that formular for the number of m-dimensional struktures in a n-dimensional cube with 16 by my self and was vary ecited about it. it was the first mathmatical problem i ask my self and solved it. this broad me into studying math in the first place^^ later i tried to finf a similar formula for n-D tetraherda, but i dont remember it any more.
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  2100. Those glasses are beastly things.
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  2101. Wow, windmills! this is one straight from what Paul Erdős calls "the book" Now we know what Peter and Hanna really did in that windmill...
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  2102. I figured out best solution for tower of hanoi when i was 6... My dad gave to me tower of hanoi puzzle with 8 peaces and so that i don't bother him said that i should find best solution and left me like that, and then was very surprised that 3 hours later i came to him and said "255", he already even forgot that he gave me the puzzle, but back in those days i figured out the recursive solution to tower of hanoi which impressed my father quite a bit... I was always good with recursive algorithms...
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  2103. Nice vid’ bro
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  2104. Wow. Just wow.
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  2105. I see an immediate application of this theory. It could be used in power system engineering. The three-phase voltages and currents for any imbalanced system could be broken down into three balanced components. The first balanced component would represent positive rotational torque, the second component would represent a negative rotational torque, and the third component would represent a non rotational component. Since these components are themselves balanced we could call this idea Symmetrical Sequence Components. You can take the transformation matrix shown in the video at 27:30 and create a 3x3 version of this to use for three-phase systems - but it would be good to change the Greek letter used for the rotational operator to the small letter alpha and then try and come up with a catchy name for this 3x3 transform. Could I suggest that we call it a Fortescue Transform and back date it to 1918.
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  2106. 23:09 Turn on subtitles here
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  2107. I was taught Heron's formula in Belgian high school. 12th grade, the most intensive math option they had (8 hours/week). And only as a side tangent, without proof.
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  2108. One thing I noticed is that for example at 24:35 you need the fact that those two segments intersect each other. I don't know if that's something you can assume or you need a proof of that.
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  2109. "just" 😅
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  2110. This reminded me of my Calculus classes back in University... I have master degree in physics, which required a lot of calculus class... The only thing that this introduced to me (or maybe i have just already forgotten) is the source of Bernoulli numbers... But approximation you showed in the end was not new to me... I wasn't required to know it for exam but we were introduced to it in passing... I just like your explanations that is why i was watching... Not for new stuff but for new insights into stuff i more or less already known so far... A different angle you take to show known mathematical results is super awesome...
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  2111. I made it to the very end. Partitions are amazing. My first introduction to them was through Ferrer diagrams and then later again with generating functions. It was nice to see yet another connection with pentagonal numbers. That one was new to me.
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  2112. concerning the one-liner: f(x) = 1/x => 1/c * f(x * c) = 1/c * c/x = 1/x Q.E.D.
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  2113. This algebra seems a little abstract... amirite?
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  2114. Very sad to say that he is not very well known ... Even if you ask maths graduate in India there is a great chance that he never heard of his name...
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  2115. Your videos are a marvel to behold (and dim memories of school learnings sometimes flit into view, briefly).
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  2116.  @Θρησκόληπτος  I'm brazilian and Bolsonaro (alias BOZOnaro) is really a nightmare: incompetent and evil. Unfortunatelly, part of the population (decreasing everyday) still supports him.
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  2117. base 10 haters will notice this works in any base :)
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  2118. Really appreciate the perspective of a determinant as a sum of permutations, I've never seen that before. I feel like an actual intuition for the determinant may not be too far behind. Trace has been similarly confusing for me my whole life - what possible simple interpretation could it have, and why is it so conserved?
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  2119.  @want-diversecontent3887  2x is cheating because you skip half the frames
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  2120. I'm not the most mathematically trained or clever person, so I don't know if it was subtly covered in the video and I didn't catch it, but I noticed that the 4k+1 primes (Pythagorean primes as it turns out?) are all hypotenuses of pythagorean triples, that is, a^2+b^2=c^2, where a and b are integers and c is the pythagorean prime, or a 4k+1 prime. So, the pythagorean primes can be written as the sum of two integer squares, and the square of the pythagorean primes can be written as the sum of two integer squares. This is I guess part of the definition of a pythagorean prime, as well as being a prime of the form 4k+1. Is this totally obvious from what the video was about?
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  2121. In hindsight, I saw my math teacher was showing us some of the math-related to famous anecdotes, like the sum 1..100 and also Collatz conjecture. Without ever mentioning it. As a planted memory, that might be a reminder. I like my math teacher very much. He was an Egyptian, and though he wasn't our main "Tutor" teacher, he invited the Leistuingskurs home and showed us a way to make a sweet (Mr. Tom) with peanuts and sugar sirup ourselves, without going to a shop in the school breaks and leave the school area, which actually was forbidden. I got the idea to group 1+100,2,+99, too. It's not that far fetched to get this idea. But powers is another story. But thanks for another reminder of a great time in school.
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  2122. I think the transformation of the hollow space within the claw's sphere-like configuration would be interesting to see in animation too. Just off the top of my head, I believe it should correspond to the divets between the lunes in the disc-like configuration, but I've no solid proof for that right now
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  2123. So where is 0? Actually I realized before 5 years that 0 has the same digital properties as 9. in general, in vortex-based mathematics, the first and the last number in any numaric system have the same propeties, So 9 is not special, it just happened to be the last digit in our number system.
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  2124. Came for a fractal that looks like an iris, stayed for the beautiful visual geometric proofs
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  2125. Dear Mathologer, Karl and Lara, thanks for this new year's gift! 🎁 Have a great year. 😀
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  2126. Perfection deserves attention...
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  2127. ​ @JackDraak  The E6B was just a specialized circular slide rule, of which there were many. About 50 years ago, you could get them as promotional material to calculate things relevant to your industry like concrete volumes or resistor sizes. Still a lot of interesting ones on ebay.
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  2128. I'm a mandala artist and I find this fascinating. I've been creating eggs before the chicken... :-) Geometry has always been a passion of mine and I thank my high school teacher Mrs. Locke for making geometry interedting and thank you to you too for expanding my understanding !
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  2129. Hahaha, i figured it out :)
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  2130. I tried to watch all the way to the end; but first I had to watch halfway to the end, and then half of what was remaining, and then half of that, and then my friend Zeno texted me and we went and got pizza. (We're still trying to divide up the bill.)
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  2131. You here use some magic to put this video together so nicely! 😍
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  2132. Es un placer ver, escuchar y entender! Muy bien logrado Mathloger! 👍 Esperamos el próximo. 😀
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  2133. 7:22 Proof Commutative Law for any Z/n: Since Z/n is just Z mod n for any whole number, we just have to add a+b and divide by n, then take the remainder, and that's our answer. Let C be the sum of a and b. But no matter what order you add whole numbers a and b, you always get the same number C by the ordinary commutative law of addition. So you will divide the same number by n no matter what order you add a and b in. Distributive Law for Z/n: The argument is basically the same, except you use the ordinary Distributive Law. You get the same number, which we will now call D, on each side of that second equation, so you will always be doing the operation D/n. This completes my proof for any modular ring Z/n being valid for both equations. QED. Let me know of there is any error in this logic.
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  2134. Most times I take him for granted but damn... Euler was an absolute genius. Imagine if he never existed. We'd probably be at least 100 years less advanced mathematically
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  2135. Ooooooo, nice thought, I'll have to verify. Well played mate.
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  2136. I actually find the optimal stack of books quite aesthetically pleasing
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  2137. @ Ohh that's fair. I guess I'm thinking of interacting with them as a colleague when I think back to being a student there were certainly profs who resisted admitting they were wrong to students.
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  2138. u bet I am
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  2139. ([π]-1)st comment
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  2140. I found the constant by multiplying the side by the midpoint number. A 33x33 magic square has 1089 squares, the midpoint is 1089 divided by 2 rounded up: 545. So 545x33 (using this method of creating magic squares leaves the midpoint number in the middle of the square also), is 17985.
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  2141. As one does
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  2142. As always sir, I appreciate the free Educational Videos. You are keeping the love for Numbers alive.
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  2143. not in a mathologer video
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  2144. Everything is coincidence. It's just your brain that feels the need to give you the illusion that it's otherwise.
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  2145. Funny, I just answered a Numberphile vid on Grandi, where they ask what's the answer .. I wrote an explanation with equations why no unique sum could exist.. the partial sums would always differ by one Now I could just refer them here
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  2146. ​ @Wecoc1 wait really?
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  2147. 2019 is in pi. AMAZING
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  2148. just recently got a C in calculus 2…
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  2149. Crazy, how unintuitive things can get
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  2150.  @Mathologer  thats the tragedy , i do not want to be political , but the British as well as the islamic invasions wiped out a lot of ancient knowledge . for example the burning of Nalanda University whose Library burnt for 40 days and several manuscripts were burnt. This happened two hundered years before Madhava
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  2151. 14:09 In this situation, the Torah is clear that it doesn't matter whose biological son the third son is, he would be legally counted as the son of the brother who died childless, regardless, so that his family line does not die out. That's the whole point of her marrying the nearest kinsman in the first place. The Talmud should not be contradicting the Torah on a matter like this, or on any matter for that matter.
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  2152. I think one problem with the derivation is that we need to know if the infinite product converge before using it and that if we write sin(x) in function of its zeros, how we know that it doesnt have complex zeros that we need to add. Great video by the way.
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  2153. I'm actually surprised and pleased with myself for being able to understand the solution from the clip alone, haha. Of course it is such a simple that anyone should be able to understand it, was is not for he using a binary number representation, and most people being unfamiliar with binary. If, for instance, he decided for no particular reason to use 0 and 9, everyone who isn't familiar with binary would immediately have an easier time understanding it. You can also describe the solution without using any numbers at all. You can just use the letters D and U, and show that the only two things that can happen to the sequence of letters is two Ds turn into Us, or a D swaps places with a U to its right. As a result you either end up with two Ds next to each other, or they are separated by some run of Us. In the latter case you can move the rightmost one until its all the way to the right, at which point the only remaining option is to move the leftmost D closer to the rightmost one until you're back to the former case, meaning you must turn both Ds into Us.
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  2154. Hello sir , Namaste, I am big fan of your teaching.
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  2155. This video was a joy to watch.
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  2156. Imagine having a dad who can always help you with your math home work🤣
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  2157. The # of steps is known. You always take those steps CLOCKWISE! The First person of the trick has a pair of cards to choose from. He determines the shortest distance between those two cards (circular measuring) and chooses to take away the card that is that many steps anti-clockwise of the other card. Example: player1 has to choose between Jack and 3. Shortest Distance = 5 (other distance =8). The 3 is the card taken away because then the player2 sees Jack + "code for 5 steps". So player2 knows he has to go 5 steps clockwise from the Jack and lands on the 3.
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  2158. It will become...1M.....very very.....soon
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  2159. You completely deserve it. A Mathologer merry chrismas to you
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  2160. h³o³
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  2161. Math is easy if you love it and understand why it happened :)
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  2162. I'm not much into history or race when it comes to math. It's nice to notate who made what discoveries, but I care more about the how they figured it out than the who. Pride gets in the way of many things this way. ESPECIALLY in math as mathematics has nothing but facts to give us. No politics, no pride, no religious...just facts. Noting that the discovery was made earlier gives us factual view of history that says we weren't just mud and straw builders and DID have the knowledge to accomplish great feats of engineering. I enjoyed your video immensely.Thank you so very much.
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  2163. I studied maths for six years after my high school degree and, still, I learned so much in this video! Thank you Mathologer for all the wanders you bring us. :)
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  2164. My God that was mindblowing, this channel has me obsessed!!! Everytime I watch a Mathologder video, I can't wait to explain it to everyone I know (though they aren't math nerds like me :D)
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  2165. First
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  2166. Dear Professor, another shining explanation about a brilliant gem. Thanks a lot. Again: this is my favourite channel EVER!
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  2167. Kudos! To think that such simple principles could go unnoticed for so long. Which is why I love this channel, come to think of it. Not only is the content absolutely amazing, I always feel like I've learned something new. (Almost as if taking a glimpse into the realm of "sacred geometry" or something!)
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  2168. Sometimes when I'm just goofing off with math or science I notice weird things and think "there's no way someone hasn't discovered this yet" and just don't think about documenting my findings. this is what happens when every mathematician on the planet did that for 5000 years straight
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  2169. 13:51 I count 54. Each 2D plane has 6: 4 small squares, 1 large square, and 1 diagonal square with corners at the midpoints of the edges. There are 3 of these planes in each orientation, and 3 × 3 × 6 = 54. Intuitively, I felt like there were also 6 more, each with two corners at the centres of opposite faces of the large cube, two corners at the midpoints of two of the edges that link those faces, and each side of the square being the long diagonal of a grid cube. It turns out these aren't actually squares, since one diagonal is √2 times longer than the other.
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  2170. So much great content packed into 45 minutes! Something I’ll always remember will be that the no nines series converges, and how simple the proof was!
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  2171. My non-mathematician intuition tells me that the Aztec diamond is some sort of parallell to how observable time is only able to move forwards, as colliding time cancels out and disappears. The four corners would be the 4 observable dimensions we live in. And; may the 4th be with you all as we travel inevitably into the year of 2021.
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  2172. 9:03 On the last row, the 3rd and 6th are the same pattern. Blue-blue-yellow-blue seems to be missing, but so does blue-blue-yellow-yellow, so there must be another duplicate somewhere.
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  2173. Something I like underpinning the two proofs is the duality of representations of triangles implicit in them. Three sides, or three angles and area. That one proof exists demands that the other one should be there as well. Nice.
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  2174. 24:42 is not only a beautiful palindrome timestamp, but in a flash it gave me a deep understanding of both the Euclidian algorithm and of continued fractions. Thank you!
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  2175. It's always beautiful how different fields of mathematics combine. Like geometry and linear algebra in this case.
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  2176. I'm a simple man, I see mathologer video, i make popcorn and watch it!
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  2177. That T-Shirt is quite harmonious 😃
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  2178.  @dr.doppeldecker3832  Getting any of the concepts addressed on this channel into less than a minute would be impressive.
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  2179. Love the shorter videos as well, keep'em coming 👍
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  2180. Ein kürzeres Video von Zeit zu Zeit würde mich sehr freuen. Aber bitte keine "You Tube Shorts" LOL
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  2181.  @jannikheidemann3805  There are Latin language classes available in some universities in the United States. But Latin Math class is not in the states!
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  2182. "Just tell our brains to shut up." one of the nicest phrases concerning the understanding of math problems.
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  2183. Welcome back! We missed you, mein Herr! Hertzlich willkommen!
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  2184. i always wanted to dig into partitions but never got around to it. Thank you for outlying it and making it so easy to follow! Euler used to be my favourite as well, that dude was amazing. Good job Mathologer, keep it up
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  2185. I have a special relationship to Ptomelys Inequality because it's the first theorem where I came up with an original and beautiful proof of it. And it was actually because of you :D Back then, I was re-watching some video about proofs of Pythagoras and one of the proofs involved scaling the original triangle by the factors a, b and c and then rearranging the parts in a clever way. You can do the same thing with Ptolemys Inquality! Take any quadrilateral with side lengths a, A, b, B, c, C like in your video but make sure that c is "in between" a and b. Take three copies of the quadrilateral and scale it by a, b and c respectively. Notice that two of your copies now have a side with length ab and diagonals ac and bc respectively. Join them on the side ab. Now notice that you can fit the third copy perfectly on the diagonals of the other two, as the third copy has two side lengths ac and bc too and the angle also matches. Put it in place and look carefully. You will have formed a triangle with side lengths aA, bB and cC. Ptolemys Inquality therefore follows. And the equality will hold exactly when the point lands exactly on the cC side. Which will happen exactly when the opposite angles of the original quadrilateral add up to 180 degrees which is exactly the case when it is circular. qed I stand by my opinion that this proof is absolutely marvelous and the best proof of Ptolemys Inequality ever. And it's perfect for an animation so if you ever want to show it, I absolutely allow you to do so :)
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  2186. This video made my day,i was waiting eagerly for a mathologer video
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  2187. Thanks. Based on the little I know about Chinese numerals, some of the ones shown seem to be being used in an Arabic way - e.g. 九八七 = 9 8 7, whereas I would expect it to be 九百八十七. Is that the point you were making?
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  2188. What you noticed is essentially equivalent to the correct thing—we know that the right column turns out to be the sequence of numbers one less than a corresponding Fibonacci number. (Specifically R(i) = F(i+2) - 1) Because Fibonacci numbers have the add the last two to get the next relationship, adding R(i) = F(i+2) -1 and R(i+1) = F(i+3) - 1 gives F(i+4) - 2 and so adding one does give exactly F(i+4)-1 = R(i+2).
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  2189. The Greedy algorithm has to be the most memorable. simple, yet brilliant
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  2190. I was never taught the proof.
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  2191. Is that the me Zachary Kaplan? Thats very exciting, if so! I really enjoyed this video; combinatorics is one of my favorite subjects, and the arguments used were very clever. Next up on my list is to really read about how that crazy monster formula for the square chessboards is derived!
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  2192. But have you seen Vi Hart’s video on edible cyclides? (This one wasn’t a collab - just a coincidence.)
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  2193. I remember learning that the volume of a sphere equals the volume of the respective cylinder minus that of the cone back in school, but it completely blew my mind to learn the equivalent is also true for a parabola. Also, it's amazing how the derivative of the volume of a shpere is its surface area. Made me realize how I've been taking things for granted withhout actually analyzing them. Thank you for providing me with all this insight! Keep making great videos like this!!
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  2194. Marvelous content! The hidden theme of balancing the numbers around the mean itched my mind, but once I saw the solution I was in awe of the elegance of Bachet's algorithm. Can't wait to implement it in python.
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  2195. As promised, I put the like also for this new version
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  2196. At 12:00, why isn't 15 listed as divisor of 15, or 19 as divisor of 19?
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  2197. 31:20 That magic moment when one needs to watch the Mathologer to finally understand where the abundance of "this is the way" memes is coming from all of a sudden...
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  2198. Very interesting, thank you!
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  2199. I think the number of tight lacings is (n!^2)/(2n). So that would be 1440 for n=5.
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  2200. Thank you for introducing us to Peter Steinbach's work; your mission is accomplished! This has to be one of the most beautiful math lectures I've experienced. My eyes welled up when we got to the powers of phi, and I had to pause the video and go for a walk when I realized there's a Pascal triangle coming. Math (for me at least) can be overwhelmingly beautiful at times.
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  2201. That was very very awesome, thank you ;^> Also, your (ten minutes in) guess at how some of us would prove the result was exactly spot on; once you told me the result, that's how I had worked out it was right.
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  2202. I recall a different discussion about the paradox with one quarter rotating around the other relating to astronomy with the rotation of the earth around the sun, and the difference between a solar day and a sidereal day. Rather than a mathematical description of the paradox, the explanation points out that from the point of view of an observer at the center of the stationary quarter, the rotating quarter APPEARS to rotate only once. At the beginning, the observer in the center of the stationary quarter sees the bottom of the rotating quarter. As the rotating quarter rotates to the bottom of the stationary quarter, the stationary observer sees the top of the rotating quarter. As the rotating quarter continues its movement back to its origin, the observer again sees the bottom of the quarter -- each part of the rotating quarter was closest to the center of the stationary quarter only once, even though from a distance, the rotating quarter made two revolutions to produce that effect. Still confuses me!
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  2203. I really enjoy watching your enthusiastic explantation of these concepts. The animations are beautiful.
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  2204. You made me do the Rubik's cube experiment twice! The first attempt made me tear the cube apart and reassemble it in its solved configuration. I lost track of the algorithm and "got lost. After 30 spent "fixing" the cube, I tried again. I repeated your algorithm, with certain configurations teasingly close to the actual solution. Eventually, the cube did cycle back to the solved state!
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  2205. Sophie Germain primes and safe prime are important in cryptography, those involving prime fields. Because if the size field is a prime p, the size of its multiplicative group is p-1. For the group to be secure, the multiplicative group must have a large subgroup of prime size so p-1 must have a large prime factor. So, having a prime q such that 2q+1 is also a prime is a good candidate. Even though it is not strictly required as long as p-1 has a big enough prime factor, this is what students are taught as a start.
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  2206. you are a genius and such a good teacher who is willing to share to the world, thankyou
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  2207. 17:31 Slight mistake, the sum is supposed to be x/1 + x^3/1.3 + x^5/1.3.5 and so on.
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  2208. As others have shared, I also really enjoy Mathologer's take on these great topics even if other math channels I follow like 3B1B cover them! Every minute spent on this production by the team was worth its weight in gold (or cheese disks) I always get this sense like it's Christmas morning when I see a new video from you guys, thanks for pumping up my weekend!
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  2209. Ohh I'm early! Yo mathologer sir ! Big fan I'm of yours! It's my physics exam in couple of days so I'll watch the video later! Excited ! ❤❤❤
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  2210. I think i missunderstood the challenge in the end at 32:18.. because this would imply, that 12 = 4² - 2² would be an odd number (which is an odd statement), what do i missing here???
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  2211. Most memorable: every positive number having its own infinite sum. It’s very obvious afterwards, but I would never believe it without your explanation. Thank you for all the interesting videos!
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  2212. Congratulations Mr. Polster (and Marty) for your 100th video! The more I watch, the more I ❤ it.
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  2213. I think I need to build that calendar. Just putting the julian dates into 2 squares. And as Simone Giertz said for her Every Day Calendar, the leap day is the day you should take off, sleep in and have ice cream with a movie.
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  2214. Recently stumbled back across mathologer videos, finding them so relaxing and interesting at the same time :) keep it up❤️
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  2215. Should ban fake accounts and bots
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  2216. I'm proud to possess a magic log wheel since I was about 17yo. Now I'm 70. Great presentation!
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  2217. Challenge 1: 1 less than with it:) Challenge 2: binary number system tells me that any from 0 to 15 with one way
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  2218. When I first watched the video about solving the standard 3x3 rubik's cube a few years back, I was blown away by the simplicity of the solution. I stumbled upon this 4D cube introduction a few days ago and decided to give it a try. It truly is about as easy as Mathologer makes it out to be. The part about seeing a 7-cell section of the hypercube as just any other 3D rubik's cube feels quite ingenious. I didn't give the 4D cube a try before watching this video though, so it might be more evident than it seems. Anyhow, this puzzle was a pretty fun way to spend a few hours, and now I get to enjoy the everlasting glory of having my name on the hall of fame.
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  2219. Imo you only really need the chain rule to rule them all. The product rule is the chain rule for multiple variables. And all others can be derived from it and might sometimes be trivially easy Like this: d/dt f(x(t), y(t)) = df/dx dx/dt + df/dy dy/dt That's the multivariate chain rule, and it's extremely close to the product rule. If you simply take f: KxK -> K to be the product on K, you get: d/dt x(t) * y(t) = y dx/dt + x dy/dt which is literally just the product rule. If you take it to be the sum instead, you get the sum rule, etc. And the quotient rule is one I never ever use: Just transform to powers and apply the product rule! (Which nets you the power rule too)
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  2220. Seeing this as a computer science person the most fun application I could think of is steganography. You could take a uniform tiling of the plane e.g. with regular hexagrams, and then slightly perturb them with small values in their other frequency components. The data would be hidden in these small values. Then you can go and tile a real floor with them and now you have a secret message in your floor. I think it also has a lot of potential as a technique for lossy compression of polygonal meshes, by quantizing different elements of the frequency vector representation differently (e.g. dedicating more bits to the regular components, which would tend to produce lossy meshes that move slightly more toward regularity, which would be visually pleasing).
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  2221. I have to vote for the Kempner's proof animation, it was simply stunning to see such a seemingly complex problem; being broken down into techniques that a school student could understand 👏
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  2222. 0:56 that puzzle becomes really hard really quick if allow to input not only number disks but sticks as well.
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  2223. Must be a more recent edition than the one on my bookshelf :) Maybe also have a look at the links in the description of this video :)
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  2224. Yes it is. This is also a more intuitive way to see the three parts are of equal area: The tiled and then median-cut original consists of 18 triangles, all of which are exactly half of one of the tiles, so they have equal area. And since each of the parts has six half-tiles they are all of the same area. It seems like you should be able to show the rest like this as well, by colouring in the six different types of triangles. (Three median directions to cut a tile, each cutting a tile in half.)
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  2225. They are never long enough my friend. I could watch these for days.
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  2226. Are you really a real mathematician if you can do arithmetic in your head?
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  2227. Very nice geometric proofs. It might be also interesting to look at it a bit more algebraically. Let's say a',b',c' are the big parts (as in 2/3) of the respective medians. From Steiner theorem (or law of cosines) we have a'^2=-a^2/6+b^2/3+c^2/3 and similar equalities hold for b', c'. So if we represent the triangle by the vector (a^2,b^2,c^2), "folding" is just a matrix multiplication (a'^2,b'^2,c'^2)=M*(a^2,b^2,c^2), where M=[[-1/6,1/3,1/3],[1/3,-1/6,1/3],[1/3,1/3,-1/6]]. Since M^2=I/4 (where I is the identity matrix), folding twice means making the squares of sides 4x smaller, i.e. scaling the triangle down by a factor of 2.
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  2228. First like from me
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  2229. This is truly some mind boggling math Also Heron's formula isn't really taught in school here in Singapore, except in Math Olympiads.
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  2230. omg domotro
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  2231. I had the same reaction
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  2232. That Hex board Reminds me of One qbert video game.
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  2233. Tenagon ? I prefer Decagon. Maybe because I'm Greek ?
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  2234. My brain took 10 minutes to get the HO^3 joke.
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  2235. Let's run the Gregory-Newton formula on the prime numbers !
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  2236. This is my favorite mathologer video in a while. Quite easy to digest, and beautiful.
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  2237. Great
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  2238. Not segmenting risk aka big loans is a bet by the lender. The water method remains fair. Big loans is a risk decision taken upfront. Just as fair if the method of distribution is know at time of betting on a big loan.
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  2239. Great. Another thing I've never heard of!
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  2240. I liked the colored line proof the best. The other one seems like something someone would show you and it would not turn out to not be true, but just looks like it. Love your channel!
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  2241.  @Mathologer  How many videos are in the pipeline ahead of Archimedes's'? I CAN'T WAIT!!!
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  2242. Do another juggling video?
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  2243. Challenge: Make a similar YT primer on the Calculus of variations. That's not gonna be that easy due to the intrinsic unintuitive nature of the subject. Anyways, great stuff.
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  2244. Learning calculus is easy, learning about calculus is hard
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  2245. Nice π-digits tee shirt there, Burkhard! 😊
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  2246. As always: beautifully and clearly presented.
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  2247. Its due to channels like yours that makes youtube videos worthwhile to watch.
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  2248. He said the numbers came to him from another place in his mind, not of this world.
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  2249. @pierrebaillargeon9531, Exactly. The Talmudic/Game Theory solution assumes that the actors are rational. But we know that, in reality, that isn’t true. People are irrationally loss averse (which has likely always been true including the times when the Talmud was written). Irrationality/human nature needs to be taken into account.
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  2250. Quite wonderfully lucid! I particularly enjoyed the 2D & 3D animations - they lend a new perspective to otherwise very dry algebra. Since you ask, I'm a physicist with advanced degrees in Cryptography and Applied Math.
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  2251. most memorable: ln(n+1) + γ and everything with missing numbers series ^^ !
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  2252. My favorite part was when you revealed that the sum of the 100 zeroes series is greater than the sum of the no 9 series. Absolutely mind-blowing. In truth, my favorite part was the entire video you just made me pick :) Thank you!!!
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  2253. yes i know how to do that puzzle. no there's no way in hell i'd get through the whole thing without screwing up at some point
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  2254. Slightly disappointed that the Pell numbers (by name) and the silver ratio went unmentioned. It makes the continued fraction even simpler since, as a metallic mean, its expansion is just an integer repeating (i.e., the golden ratio's is 1 repeating, the silver ratio's is 2 repeating, bronze 3, and so on).
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  2255. 14:53 Fun fact: The entrance to the Museum of Math in New York City is a glass cube, with this hexagon drawn.
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  2256. No wonder Knuth mentions Concrete Mathematics every so often in TAOCP.
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  2257. 7:14 Well, that's great and all but you won't be able to explain the OTHER gray number so easily. 7:46 Oh. Well, I'll see myself out.
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  2258. Anyone else used these slide rules for "O" level maths ? Shows my age,eek!
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  2259. One of my favorite channels on Youtube. Congratulations to Burkard and his team. Ausgezeichnete Arbeit! I always look forward to your new videos.
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  2260. Somehow I guessed the rule. What do I get?
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  2261. For those wondering what's wrong with the numbers - they are using decimal notation - they write 22 instead of the normal 2*10 + 2
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  2262. Yea I think I could figure it out if I was in the Doctor's shoes, I figured it out when I was, Idk, early in my teens? And it's also mindlessly easy to execute if you get the first move right, everything else just falls into place. The Doctor is far smarter than me, and more experienced than I was or even am now
    5
  2263. Being honest...I rarely make it to the end of a Mathologer video. In this case I've surprised myself by not only making it to the end, but in a seemingly too short amount of time. No video stretches like a Mathologer video.
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  2264. By Pythagoras' theorem, the square of the radius of Conway's circle is the square of the radius of the incircle plus the sqare of half of the chord.
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  2265.  @Hiltok  Here is a much trickier question: Which is larger, the area of Conway's circle, or the area of the shape of constant width?
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  2266. For the Rubik’s cube order, I got 315 repetitions when I calculated the orders of all the edge and corner cycles (keeping track of piece orientations) and took the lcm. Recently I’ve been thinking about the connection between the pigeonhole principle and the Steinitz exchange lemma, since the pigeonhole principle could be seen as a consequence of the lemma with each of the pigeons and each of the boxes both being regarded as vectors over F_2, with the boxes forming a basis. While the latter perspective is overly complicated in this case, if you could find a different vector space that some problem could be interpreted in terms of, you might be able to use the exchange lemma in a similar way, but get better results than you would by using the pigeonhole principal directly. (The way in which the exchange lemma is used is that any collection of vectors with cardinality larger than the dimension of the space they are contained in must be linearly dependent)
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  2267. 11:24 Looks like an origami Flower Tower. 😀
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  2268. This has been a very helpful video. I'd been pondering how any island has an infinite coastline if measured to infinite precision yet has a finite land area, and had become stuck. This video came at a fortuitous time. Thanks for another great video!
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  2269. Let A = Infinity 😂
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  2270. awesome video! it's nice to see these types of problems be recognised
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  2271. Mathologer it’s beautiful. Keep more of it coming!
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  2272. I know about Moessner's miracle because I have the go-to Math bible - the Book of Numbers by Conway and Guy 😁
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  2273. I would imagine it's true for all odd sized squares, but I noticed on the 5x5, that all pairs of squares that are 180 degrees opposite add to the same sum. I thus conjecture that for any 2k+1 square constructed in this way, taking k such pairs and summing them along with the center square will always sum to the magic total.
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  2274. For the list which begins with 3^2+4^2=5^2. You had a short cut (formula) at the item before the equal sign. I have a short cut (formula) to jump from the last item of the previous line to the first item of the next line. 1st line ends with 5 2nd line begins with 10. Because 5 (last of previous line)/1(number of items on the right side)*2(number of items on the left side) 2nd line ends with 14 3rd line begins with 21. Because 14 (last of previous line)/2(number of items on the right side)*2(number of items on the left side).
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  2275. I made it to the very end.. but it was NOT the same "I" that began.. i think..?
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  2276. Although I'd be exaggerating if I said I understood this without reviewing some portions of the explanation, I believe the explanation was extremely well done! These videos are, for the most part, very satisfying mathematically. As some have already suggested perhaps we could one day find another human with the skill set of Ramanujan - but I'm not holding my breath!
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  2277. This was fun. Thx. Peter PhD Theoretical Physics.
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  2278. The diagonal construction of magic squares and the geo magic squares were both superbly presented... Really interesting!
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  2279. plese make a video on master therom of RAMANUJAN
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  2280. 16:43 when the doctor is sus
    5
  2281. I was curious about what kind of wall would be generated from the prime number sequence.
    5
  2282. plugging real numbers into the Fibonacci function and using complex numbers gives a nice exponential helix: the center follows exp(x) while the radius follows exp(-x) (or really, exponents of phi).
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  2283. Amazing as always. Love it when you can start with a concept which is easy to formulate like ways of summing to an integer and end up needing e, pi, infinity and derivatives to express a general solution. Makes you appreciate how interconnected maths really are.
    5
  2284. Hey can you please make a video on Telescoping Sums? Its a great method to solve sequences and series :)
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  2285. this animation looks like a 4d shape of some kind i know it probably isn't one but it's still cool
    5
  2286. K. Ramasubramanian is a professor at Indian Institute of Technology Mumbai. He is my favourite professor. The manuscript is his production.
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  2287. Almost as excited to see the t-shirt as I am to hear the math!
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  2288. 2:12 Consider the vertex of the triangle at the bottom-left. There are two lines going through it, both with red on one side and green+blue on the other side. Thus, the angle bisector of these two lines consists of all points that are equidistant from the red endpoints, and (separately) equidistant from the green+blue endpoints. But applying the same reasoning to the other vertices, this leads us to the intersection of all three angle bisectors of the triangle, namely the incenter. We have six equalities between the distances from the incenter to the endpoints, which combine to show that they are all the same! q.e.d.
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  2289. 11:48 The box for 153^2 + 104^2 = 185^2 is 9 4 17 13. The upper two number can be calculated via: b = sqrt((z - x) / 2), a = sqrt((z + x) / 2) - b, where x^2+y^2=z^2. Then c = a+2b, d=a+b, in the form of: a b c d. The path there is: right right right left The algorithm to find the path to a given pythagorean triple in the tree: (x^2+y^2=z^2 , x,y,z ∈ N (0) Reduce triple to primitive triple by dividing by their greatest common divisor gcd(x, y, z) x = x / gcd(x, y, z) y = y / gcd(x, y, z) z = z / gcd(x, y, z)   (1) let a, b: b = sqrt((z - x) / 2), a = sqrt((z + x) / 2) - b   if a is even or either of them is not an integer or either of them is 0, repeat the calculation with y instead of x!   (2) Calculate the Category of a and b: C: N^2\{1,1} -> N   C(a,b) = 2 - floor(2 / ceil(2 * b / a)) for a != 1 or b != 1   if C(a, b) is undefined, jump to step (6)   (3) Save C(a, b) in memory   (4) Calculate a' and b' let c = C(a, b)   { a-2b for c = 0 a' = { 2b-a for c = 1 { a for c = 2   { b for c = 0 b' = { a-b for c = 1 { b-a for c = 2   (5) Repeat from (2) with a = a' and b = b' as long as a != b and a * b != 0   (6) Read the saved C(a, b) from (3) in reverse order 0 corresponds to going left, 1 corresponds to going straigt, 2 corresponds to going right!   DONE! Example code and Algorithm here: https://pastebin.com/T71NP8Z9 JS-Applet here: https://jsfiddle.net/ptu6f5cr/4/
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  2290. 6:24 Sometimes you have to think inside the box
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  2291. Herr Dr Polster: That was lovely. My chemistry teacher was like you, so I became a chemist. My maths teachers were terrible; but I like (love?) math anyway. I did have the calculus, some algebra which included Power series (which I never got), some differentials, partials, numerical methods, and so on. But I never got power Series at all until I just watched the Power Series Master Class. I got through all of it in two nights; I will have to watch it again and get out a pencil and paper to get more (Erdos: "if your fingers don't hurt, you're not working enough"). I got most of it, except the animation at the end, and I'm sure I must have missed something as I'm not that smart. But, thank you so much!
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  2292. Finely an other mathologer video
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  2293. Most excellent follow-up to the Fibonacci Flowers and the Most Irrational number videos. Loved the tree bit on the helicone! I appreciate the work and time you took to correct your small mistake. Your dedication to accuracy and truth is a worthy endeavor with manifest positive impact, and while it may be a given for you to be that way, I'll just point out that many many people don't operate like that, and in my eyes, this is a huge win for you because being open to correction and then correcting it looks better and is better than never making mistakes, because human. ❤
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  2294. ??? The sum of the series is exactly equal to pi, 355/113 is not :)
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  2295. dislikes are from casino managers
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  2296. 15:00 Also; adding the 10th multiple-angled ears (in the case of the 10-gon; 5th multiple-angled ears, in the case of the pentagon), a.k.a. 360°-ears, a.k.a. line segments, to the point, will result, in another regular 10-gon (or pentagon, or n-gon). Obvious, I know. I just thought I’d point that out, lol. 😅
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  2297. To be honest, "what comes next" questions don't even make sense. What relation is simple enough to be considered a "pattern"?
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  2298. 21:15 shows the shrinking algorithm in real life. 🙂
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  2299. The clip surely has olympiad level quality in playing on the audience's emotions with the visuals, ambience, and timing in delivery of lines.
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  2300.  @Mathologer  MIND... ABSOLUTELY... BLOWN! Slide rules are complicated enough for me since I well and truly grew up in the age of digital calculators that can surf the internet, but to think that slide rules actually exist for complex numbers is beyond me! :learning:🤯
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  2301. Awsome video! In the x^2-y^2 problem at the end, all solutions divisible by 4 are also possible (if you assume that x and y are coprime then you can get all odd numbers as well as numbers divisible by 8).
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  2302. Im letting this play when I fall asleep so I can have mathematical revelations in my dreams
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  2303. Thanks for the video! I felt compelled to try to find a proof, so here it is. The center of the big circle must be the intersection of the three lines cutting the angles of the triangle in two. This is also the center of a small circle, inscribed in the triangle. Consider a segment of length equal to the sum of the three sizes of the triangle. If we glue the middle of this segment to side of the small circle, and rotate this construction, the extremities of the segment draw the big circle (I skipped a few details that were a pain to write concisely : ) Thank you for making me feel worthy! I hope I got it right
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  2304. Always, fascinating! I followed, surprisingly, until the last example. That's when my brain went flat and would inflate no more, until patched with a second cup of tea. Why, at the end of each video, is there a smile on my face, and a chuckle in my heart? Perhaps they have much to do with your own gentle encouragement along the way with a chuckle here and a giggle there, however, I'm also quite fond of that sweet tune, pleasingly harmonized on a collection of folksy strings, that serves as a segway and ending to your videos. As always, thank you and I'll look forward to the next time. 😊
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  2305. AI Sophie Germain in the thumbnail/transitions was an OK experiment but please never again. Not what she looked like
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  2306. 1. 7 pigeons. Well, by the principle said in video, that's 5
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  2307. 5:18 is the first proof I ever saw (or that I remember). For some reason I think it was featured in the “Cosmos” book but I have it in storage and I can’t check right now; I’m probably misremembering.
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  2308. I had seen the twisted squares in a textbook before, but I didn't connect them to the proof of the theorem until I received a book specifically on the Pythagorean Theorem called Hidden Harmonies. It had a full chapter of different proofs, half of which led back to the twisted squares. I didn't quite understand the rest of the book at the time, but it really is quite fun to see the history of humans proving the same thing dozens of different ways.
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  2309. Love you mathologer. Great video. I was particularly excited by the proof of the value for the base, it was my first time seeing it.
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  2310. The pattern I found for the numbers at 9:20 was that the next term was three times the previous term minus 2. Or t(n)= (3*t(n-1)) -2. so (2*3)-2=4, 4*3-2=10, 10*3-2=28 28*3-2=82; the next term would be 82*3-2=244.
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  2311. The weird optimal overhang towers were absolutely stunning. I'm not even sure why it works, but hey, math says it does.
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  2312. I really like this guy's videos :) keep up the good work!
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  2313. 4:50 already every power of 9 plus one is special, because the triangles can be collapsed in like so: every 9th hexagon is sufficient to account for 9 generations in the future, so ignore the eight between that and the previous one. 10 is good, and a group of 82 would behave just like a group of 10 (with 8 between each one, 8*9 = 72, which is 82-10), and generation 10 of 82 hexagons would behave just like generation 1 of 10 hexagons. likewise generation 82 (assuming the first row is generation 1) is determined only by the top corners in the same way. Therefore 82 behaves like 10, and so 9^3 + 1 will behave to 82 the way 82 behaves to 10, and so on. My first expectation is that since there are 9 possible rules that this is the root behind why 9 is so special, but it may be that 4 generations also obey this property (powers of 3, plus 1 instead)
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  2314. awesome
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  2315. I applied at my school to get my money back, and now they charged a processing fee.
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  2316. This was one of the clearest mathologer videos yet. Everything worked. I loved it! <3
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  2317. Just being curious: Do we get anything interesting from using x+3, x-1, or some other number instead of the 2?
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  2318. Triangles are awesome!
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  2319. Wow the logarithmic trick was amazing!
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  2320. 4:07 ... Ron Swanson traveled back in time? This is the proof!
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  2321. 24:33 "How on Earth did they found that one?" What? That's what university librarians are for.
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  2322. I like the Pigeons at the Olympiad proof best. For the challenge at the end of chapter 3: Because every repeating decimal can be written in the form (x+y)/10^n + x/10^2n + x/10^3n ... for some integers x,n,y. We can factor out x, and we can rewrite: 1/10^n + 1/10^2n + 1/10^3n ... as 1/(10^n - 1) Noticing that when n is an integer, 10^n and (10^n)-1 are integers, we have the sum of two rationals, y/10^n + x/(10^n - 1) which is therefore rational, QED. For the challenge at the end of chapter 6: 7 integers from 1 to 100 such that no 2 different collections from that set have the same sum: {1,2,4,8,16,32,64} This is easily seen by putting the sum in binary form. Each bit in the sum is set by exactly one number from the collection, so if collection A sets a given bit, disjoint collection B lacks the only element that can set it. For the think-about-it pause at the beginning of chapter 7, I found a much harder and clunkier solution: There are 2*3*4 = 48 permutation of 4 cards. That's almost enough to pick out 1 card among 52 - if only somebody removed 4 cards from the deck. But somebody did! The mystery card can't be any of the 4 cards you passed to your assistant, which leaves 48. So we place some natural ordering on the 4 cards. Say, suit is the primary sort key and number/jack/queen/king is the secondary. We also number the permutations of 4 items - so we have a table that gives us a bijection from a permutation like C D A B to/from a unique index. And we number the remaining cards in the deck. Same idea: force an ordering, make a table. Then we look up that card in the deck table, look up that index in the permutations, and arrange the cards in that permutation. Our assistant does the reverse: Sees what permutation we handed him, looks up its index, constructs a deck table for the ordered deck minus the 4 cards he sees, looks up the permutation index in that, and announces that card as his answer. Ta daah, except that we need to build large tables on the fly. On the plus side, we can do it for any mystery card, we don't have to choose it ourselves from among 5 cards.
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  2323. My thoughts too. I'd be suprised if the pythagoreans, ancient Indians and who knows before didn't know of this.
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  2324. I enjoyed this despite it's simplicity at a glance, your explanations are always insightful and provide different and novel perspectives on approaching problems. Please continue with these interlude videos and keep up the great work!
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  2325. An application which immediately comes to mind is in creating a novel measure of ‘randomness’ for a set of points. Or at least to gain better intuition for those measures (or tests) of ‘randomness’ already existing which make use of the discrete Fourier transform. Great video, lots to consider! 🤗
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  2326. Remember, when Karl (a.k.a. Mathologer Jr.) was, like, 6 years old? Feeling old, already? 😅
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  2327. Missed an opportunity to make it 50:50.
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  2328. 50 minutes is fine, complex topics require time to explain. My advice, which I think is reflected in the Youtube output of other math content creators, is to try to keep it under one hour. I seldom comment, but have been watching for years and love your videos and beautiful animations. Looking forward to future episodes on this topic, keep up the great work!
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  2329. I've always pronounced sinh as sinch.
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  2330. 23:27: A little-known proof that blurry bitmap angles are rational multiples of crisp, vector-based angles.
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  2331. Well, if the leading 1 is the 0th term, and the second 1 is the 1st term, then the 666th partition number seems to be a 10 digit number ending in 26125... ..wait, no, gotta account for overflow... some 26 digit number ending in 29485. Why did you make me do this???
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  2332. Wow. Nobody?
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  2333. Both proofs have their merits, but I prefer the second one just a tiny bit. On the one hand, when I follow a proof I like to be sure that I haven't missed some tricky step that might undercut the whole proof and that was easier with the first proof by just following the angle dots, but on the other hand the second one is quite short which is a big advantage.
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  2334. This was a really great journey.
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  2335. I noticed that the power of fourth pattern is replicated with storage on microSD cards. I figure it has something to do with binary but I cannot quite place my finger on it.
    4
  2336. Morbius
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  2337. Please make more of such contents
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  2338. Unfortunately, the Wikipedia link in the description doesn't work, probably because of some problem with character encodings. The correct URL is https://en.wikipedia.org/wiki/Petr%E2%80%93Douglas%E2%80%93Neumann_theorem or https://en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem .
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  2339.  @bobosims1848  you've missed the point entirely sir. Twice. Accessibility isn't about intelligence, it's about meeting an audience at a carefully considered level of education. This video does a fantastic job of meeting several different levels of education in one package. The goal of educational content is to plant a seed of curiosity. The point of the theorem is that this works for all polygons. Potentially sending viewers off on a side quest to Google what a decagon is detracts from the goal of producing this content in the first place. Most importantly, however, the point of my previous comment was that you have no credibility on this platform. With no videos of your own, and now your own admission that you lack the knowledge of how to produce this type of content, or an original thought to turn into a cohesive presentation, when you assert that you would have stated things a different way it holds no merit. Thus, I kindly encouraged you to remember the old saying, "If you don't have anything nice to say, don't say anything at all."
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  2340. What an unexpected fibonacci 'aha' moment!
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  2341. I think we Indians did not appreciate the mathematical writings of our heritage, and gave more importance to philosophical literature which was unfair.
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  2342. Pilots are still examined on using an e6b circular calculator :) And I also once used a cylindrical slide rule that was more accurate (to 4 sig. digits compared to 3 for an ~11" rule) due to scale expansion. The reference to "Napiers bones" has a deeper connection because slide rules were constructed using bone or ivory (plastics were not invented) and bone does not expand/contract too much and wears well.
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  2343. The most impressive part in my opinion was the fact that the 100 zeros sequence converges to a bigger sum than the no 9s sequence. Greetings from Germany by the way and keep up that great work. It is always a pleasure diving into your mathematical discoveries!
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  2344. These mini rings are what music theorists call modifications. Basically, you think of a clock with n values and you count around the clock face until you find the remainder
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  2345.  @1stlullaby484  Sorry can't provide the link, but if you search orthogonal polynomials form Euler's point of view pdf, I think you'll find it online. If not, I'll try adding the link.
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  2346. I'm not sure what your shirt is supposed to mean this time. "To infinity and beyond"? Or what is it? Edit: well someone else already figured it out... I'm late
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  2347. Imagine uploading Couldn’t be burkard
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  2348. Oh man the timing of this video couldn't have been any better. Thank you so much!
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  2349. the greatest disservice hollywood has done to germany is its depiction of its citizen during wwII thankfully, some germans gracefully provide a reality check. thank you for your excellent vid.
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  2350. My daughter learned in 2nd grade last week to make "number rainbows" where you write the numbers from 0 to n, then connect 0 to n, 1 to n-1, 2 to n-2, etc. The idea was to visualize all the ways to combine 2 numbers to sum up to n. I showed her that if you count the rainbows, skip counting by n, you can add up all the numbers quickly. So my 7 year old correctly summed the numbers from 1 to 10 in only a few minutes with that technique.
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  2351. I've been following your channel for years and have always loved the way you've explained math. I get excited every time I see a new video from you. Congrats on the 100th video milestone!
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  2352. You know, Algebra Autopilot would be a great name for an app aimed at teaching algebra. Demonstrate all the algebra basics with these animations, maybe some short clips from our favorite geometer explaining the concept...
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  2353. The problem today, from my perspective, is that mathematics and math education is much less geometric in its way of thinking.
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  2354. One more thing to add. When I was eleven realized that the sum of the first and last term and the sum of the 2nd and 2nd last term were equal. So were the 3rd and 3rd last and so on. One just had to take the final number, add 1, divide the last term in half, and then multiply the two together to reach the sum. I suspect that Gauss did it the same way. 101 X 50 = 5050.
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  2355. 3:41 1046—256 BCE Look into Anatoly Fomenko's Empirico-Statistical Analyses regarding Isaac Newton's and other chronologists' timelines. It's more plausible that Chinese Imperialists added fifty-nine years to Li Qingyun's life, thus extending their chronology and thereby the longevity of the empire's mandate. He claimed to be 197 when he died. They said he was 256 and that he lied about his ago pretending to be younger. He was a mountain herbalist. The imperialists were autocrats, totalitarians. Fomenko's New Chronology holds that this type of invented history is more common than not, that Pharaoh ruled Egypt through the 1700s, that 2022 CE (7530 Slavic Aryan Vedic Calendar) is 869 AD, and that The Great Wall of China is still being built. Or maybe you had a typo, and it was 146-256 BCE? I didn't look it up yet, I was just reminded of all the above. Thank you for excellent reference!
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  2356. It took me thirty minutes to decipher the shirt 😂
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  2357. Thanks a lot mathologer... But I'm badly missing videos on number theory and algebra... You seem to have gone full board to geometry. Still waiting for your promised video on Abel's proof of why there can't be formulae for some higher degree polynomials
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  2358. Learned both when i was 13 by my teacher. I loved math more than i can remember! Thank you for your content once again! Hope to see an ecology "ecosystem stability" math video from you, they are quite intriguing!
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  2359. What a nice story from the 2D world. Thanks for finding time to share it with us.
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  2360. I was the 999,999th view
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  2361. Nothing better than seeing a new video when wake up.
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  2362. 13:07 No, it's too complex to be real
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  2363. 7^3 + 8^3 = 9^3 Because the pattern is 2^1, 2^2, 2^3 for each pattern
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  2364. I'm from Brazil, and I was taught Heron's formula very early on, around 3rd or 4th grade. It always puzzled me, because it seemed to have been dropped on my lap out of nowhere (as far as my education was concerned, it really was). Still, I'm personally thankful it happened, because my first memory of doing maths by myself, out of sheer curiosity and without any obligation to do so, was trying to verify it from the usual formula for the area of a triangle, equating the two and working through the algebra. I wasn't successful, of course, but it got me here eventually, so that's nice.
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  2365. @AJMansfield  ah ofc, it only needs to make one revolution, this is embarassing my bad 😅
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  2366.  @Mathologer  I have tried Google search with no avail, I would like to find out how the scales on the various types of slide rules are actually made. Are they measured with standard linear tools, or is a machine tool used that by the nature of its geometry, automatically marks or cuts the scales on the reference dies used to create the final products? Do you know?
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  2367. Wonderful video Every maths lover must watch.
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  2368. I think I got it. Think of the Laplace expansion along the first row. You get a problem, when you can't number a board with holes such that both "1s" are non-adjacent to a hole. A counter example would be a 3x3 ring, ie a 3x3 board missing it's center. It's determinant is 0 but there are 2 tilings.
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  2369. Nice content
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  2370. sir, this was a marvellous throwback to your previous video. ❤
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  2371. 30:07 Minimum number of moves for 9 disk and 4 pegs (I think): 41 (^.^). You explained how to divide the disks into 3 superdisks for 8 disks but not for 9. So, just for fun, I tried every possible division in a very inefficient c++ code. I found that you can solve optimally in 41 moves if you divide the disks into 3 superdisks of 2, 3 and 4 disks (from top to down); also dividing them into 4 superdisks of 1, 1, 3 and 4; 1, 2, 2 and 4; AND 1, 2, 3 and 3. I didn't check if these divisions provide different solutions (Challenge for anybody?), but my guess is that, at least, the 3 and 4 superdisk divisions are different optimal solutions. Calculation (hopefully) explained: Assume that you got to solve for 8 disks and 4 pegs, and you divide the 8 disks into 3 superdisks of 1, 3 and 4 disks (from top to down, like Mathologer did). Also note that f(n)=2^n-1 is the optimal number of moves to solve n disks and 3 pegs. Then, (hopefully) it's easy to see that the number of moves, following the Mathologer path, is equal to f(1)+f(3)+f(1)+f(4)+f(1)+f(3)+f(1)=4f(1)+2f(3)+f(4)=2^2 f(1)+2^1 f(3)+2^0 f(4). Note that the arguments of the f's add up to the 8 disks and also note the decreasing powers of 2. (hopefully) it's easy to see that, for n disks, 4 pegs and m superdisks of a1, a2, ..., am disks, such that a1+a2+...+am=n, the resulting number of moves is equal to 2^(m-1)f(a1)+2^(m-2)f(a2)+...+2^0 f(am). For the code, I simply apply this formula to every possible division of the 9 disks into superdisks. I cycle through the possible divisions in a very inefficient and lazy way: I go though all possible lists of between 1 and 9 elements with all possible values between 1 and 9, checking if the accumulation of the elements add up to 9. Seeing the formula, it's easy to see that it's not worth it to check superdisk divisions that are not in decreasing amounts of disks, but whatever. Here's the code for anyone interested: #include <iostream> #include <vector> #include <numeric>//accumulate algorithm using namespace std; int pow(int b,int e){//b to the power of e with e being a non negative integer int ans=1; for(int i=0;i<e;i++)ans*=b; return ans; } int h3(int n){//optimal number of moves with 3 pegs return pow(2,n)-1; } bool operator!=(vector<int>v0,vector<int>v1){//how to compare vectors if(v0.size()!=v1.size())return 1; for(size_t i=0;i<v0.size();i++) if(v0[i]!=v1[i])return 1; return 0; } int main(int argc, char *argv[]) { int N=9;//total amount of disks for(int i=1;i<=N;i++){//i is the number of superdisks used in this attempts vector<int>v(i,1),v0=v;//v holds how many disk in each superdisk, v0 is the initial attempt do{ if(accumulate(v.begin(),v.end(),0)==N){//the total amount of disks has to be N int m=0;//total number of moves in this attempt for(size_t i=0;i<v.size();i++){ m+=pow(2,i)*h3(v[i]); cout<<"2^"<<i<<" f("<<v[i]<<")+"; } cout<<'\b'<<'='<<m<<endl; } for(size_t i=0;i<v.size();i++)//cycle through all possible v's if(v[i]==9)v[i]=1; else{++v[i];break;} } while(v!=v0);//all possible attempts has been searched when v loops back to v0 } return 0; } Enjoy. (^.^)
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  2372. finally my insomnia cured!
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  2373. BSc and MSc in maths here: You just keep outdoing yourself. Always great and interesting content no matter what level your'e at; keep it up.
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  2374. yes please. make more such.
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  2375. Very cool.
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  2376. I only just saw it today.
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  2377. yep, it was totally hidden for me, too. it appeared in my feed today, and i was like, wait, it's 2 weeks old? how did i miss it?!
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  2378. I FINALLY have a greater understanding of Finite Fields.... I shunned a Finite Maths course in High School (I was not offerred AP Calculus!) because of a certain rumor about the mathematics teacher that I almost verified (brushing against female students during his perambulation of the room during tests) and that damaged my mathematical development and maturity (male gossip!!) to some degree
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  2379. It is really interesting how eulars comes into all these different formulas. Personal favourite was the very neat proof at the end that the sum was less than 80
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  2380. I can't believe for your effort in video Please never stop making such beautiful videos .
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  2381. If math is a garden, you consistently show us its prettiest flowers and sweetest fruits.
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  2382. Most memorable: Thought that I knew what was a harmonic series until watching your video! Now I know something about what it can be and what part of it can be.
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  2383. Happy new mathologer 🎥
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  2384. I was just finishing sperners lemme thinking when's the next vid coming out when ...
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  2385. 23:03 For, abc = a + b + c + N - 3 abc - a = b + (c + N - 3) a(bc-1) = b + (c + N - 3) ca(bc-1) = cb + c(c + N - 3) ca(bc-1) = cb -1 + c(c + N - 3) +1 Being lazy 🦥, Fill out the details! (cb-1)(ca-1) = c(N - 3 + c) + 2 Put c = 2 Done! 😴
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  2386. Deep thought and imagination.
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  2387. Also term 5 is 55. Every 2nd Fibonacci is the next onr
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  2388. 36:19 The x's on the right-hand side of the second row are supposed to be n's. Thanks for the wonderful video! I think this would be a wonderful approach to teaching calculus in school. Start with differences and sums, and then go over to differentials and integrals. Just like we teach probability theory, even in college: We also usually start with discrete probabilities and then go over to continuous concepts of probability.
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  2389. Thank you!
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  2390.  @Mathologer Nice coincidence; fortunately, I was only 10 when commies went away and I could enjoy my puberty in free and wild 90’s. The most difficult thing for everybody in the world including Slovakians is to correctly pronounce Ř, the “Put your finger through the neck” twister is piece of cake in comparison.
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  2391. I'm only halfway through the video but so far I've learned a new geometric fact every 30 seconds. This is absolutely insane.
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  2392. ​ @Mathologer  yeah there is a similar melody in that song!
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  2393. 1+2+3+4+5+6........ = [(infinity)^2 + infinity] / 2 Please reply and correct me if I am wrong. Please.
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  2394. You are awesome. Your explanations are always very good. 😄
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  2395. I was disappointed that the original reddit post was a computer generated animation. Ok, the premise is a virtual rubber band, which sticks to the wheel at the exact point it touches initially. Further, the rubber band is infinitely stretchable. Is there any reason to think a real rubber band could perform a similar (obviously limited) function and accurately stretch out in a logarithmic scale as shown? I kind of doubt it … and, having grown up using slide rules in high school, this topic seems to require a much lower level of intellect than your usual gems of genius. Its 1:30am here in Adelaide, insomnia and hunger make this old man very grumpy. I should point out that I think Burkard is brilliantly clever, entertaining and a thoroughly nice person in real life.
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  2396. The whole idea of "summoning" satan with pentagrams/pentacles comes from horror movies heavily misrepresentating thing. First, the pentacle/pentagram, if pointed up, is a protection symbol, and has been used by many religions, including Christianity, Pagans, and others. Second, if pointed down, the symbol is linked to Satanist, which doesn't really have the right name since we don't even believe in Satan, and most Satanist are basically atheists with a few more rules (which are mostly common sence, ex. dont hurt kids; dont hurt innocents; if someone invites you, respect them; and so on) and the few that do believe in an entity believe in either Lucifer (which is NOT the one from the Bible) or Baphomet, which represents balance as they are both female and male, light and dark, good and bad, and so on.
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  2397. In the section on returning to the solved state on a Rubik's cube we have to be careful about what we mean by an algorithm. Some rule that dictated different rotations depending on the color of a specific square, for instance, could create a non-injective (and thus non-reversible) mapping on the set of states, which can cause you to get stuck in a loop not containing your initial state.
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  2398. @Mathologer how does one partecipate in this competition?
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  2399.  @rubenverg  I second this question. Until we have an answer, I'll try the poor man's way of pasting a link into a comment... which will likely be auto-removed for containing a link
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  2400. How do we submit our code? I posted my solution in another comment to this thread, but it seems to be filtered out due to an external link...
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  2401. 35:45 Because the motion is smooth, we consider the local movement of one bug (being chased) relative to another (which I will refer to in the first person - I am the bug that chases), and we see that it's linear. Further, in this case we see that the movement is perpendicular (movement being it's change in position all happening relative to me) - this is because I am always facing it, and more importantly, the angle between me, it, and the bug it's chasing (the direction of it's movement) starts off at 90° and never changes - THIS is due to symmetry. It follows that (locally) relative to me, it's moving in a circle around me (and also I'm rotating in place to face it) - never getting either closer or farther from me. I'm the only one that has any affect on the distance between us, so the final distance covered is the initial distance I needed to cover.
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  2402. I don't understand why mathematical theories attract me. I don't even know the multiplication table well
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  2403. Very kind to put the solution in a reply to prevent accidental spoilers
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  2404.  @manavgunnia  2^countable is uncountable afaik. PS: @tom13king has a wonderful argument above.
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  2405.  @josephmathes  Definitely a great topic. We'll see .... :)
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  2406. Thank you so much for your wonderful and inspiring videos!
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  2407. Man.... i used to cry because of math.... I am now crying because of you... Amazing... and thank you
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  2408. We need a Mathematica plug-in for algebra autopilot!!
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  2409. Did you read The Archimedes Codex by Reviel Netz and William Noel? I would highly recommend, it includes both a very entertaining history of the remaining copies of his works, as well as insights into what he was working on :)
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  2410. When I was at the Gymnasium in Germany, in 9th grade, approx. 1990, the formula was included in the problem section of the book we used (Lambacher Schweizer). If I remember correctly, the problem even consisted in proving the formula (given many hints on how to proceed).
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  2411. Roger Penrose also presents the "cube-hegaxon" visual proof in his 1994 book " Shadows of the Mind: A Search for the Missing Science of Consciousness ". It is in chapter 2, section 2.4 (" How do we decide that some computations do not stop?")
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  2412. Amazing....
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  2413. I've never been more ready for a math video
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  2414. Mathloger, one of the few channels that I'm excited to see a nearly hour long video posted. Love it! Thanks for the maths!
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  2415.  @ffggddss  Wow nice argument!
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  2416.  @gammaknife167  Thanks. But I think I may have simply remembered it from the Mathematical Games column. Not sure. After digesting enough beautiful arguments, they tend to become ingrained in your thinking, without your knowledge of their origin. But whoever came up with it, it is indeed a nice insight.
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  2417. YES! Saturday morning Mathologer! 😎
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  2418. :face-red-heart-shape:
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  2419. I admit, I enjoyed this very much. The most fun with differing rates of convergence, and the distributive property, I've had in years. You have reawakened me. Thank you.
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  2420. I first encountered the coin problem studying engineering over 60 years ago. The context was designing what we referred to as epicyclic gear trains. The rule of thumb is : if the small gear [ coin ] rolls round the outer of the centre gear [ coin ] its motion relative to an outside observer is the ratio of diameters [ in gearing parlance the pitch diameters ] PLUS 1 rotation. If the smaller rolls round the inner surface of the larger then the number of rotations of the smaller is the ratio of the pitch diameters MINUS 1 rotation. If you imagine the problem of an interconnected train of epicylic stages it's essential to have simple rules of thumb to work with. It's clear that the coin problem is actually only a simple system that one is almost never going to encounter in the real world of gearing. The basic real-world system would be THREE gears interconnected and referred to as a sun, planet and ring gear combination. Any one of the three elements of this system can be held stationary with respect to an outside observer. As a final glimpse into the world of gearing, it's important to remember that if the basic function of a gear train is as a speed reducer there is a concomitant increase in the torque [ and vice versa ]. It can happen then that if the train is driven in reverse this can lead to excessive torque or tooth forces which can cause failure of the gear teeth at the former input end.
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  2421. Slide rules are *awesome*. I love them. They're like a physical manifestation of mathematics - the most abstract field of study there is. What would it take to make a slide-rule type instrument where the calculation performed is not one number multiplied by a second number, but instead, one number raised to the power of a second number? Bonus question, but maybe only for the most hardcore of mathematicians: what about a slide rule where the operation is addition?! The mind boggles.
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  2422. What I disliked about calculus was seemingly endless questions asking me to integrate or differential nonsensical mashups of sins and cosines like wtf would be the actual utility of integrating sin^2(5x)cos^3(x) aside from exam questions so annoying
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  2423. Great video. I missed the -1/12 in the end somehow 🙂
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  2424. 0:04 I found it - it's in the bottom right corner.
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  2425. For the 3D version, the fourth color d we put on top of three colors a, b and c is a⊕b⊕c where ⊕ is the nim-addition https://en.wikipedia.org/wiki/Nimber#Addition_and_multiplication_tables . We also have a⊕b⊕c⊕d = 0, which explains why rotating any such pyramid gives another pyramid of the same type.
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  2426. So this is why sudoku named one of their optional rule as magic square.
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  2427. I asked ChatGPT to flesh out your idea Title: Pentagon’s Shadow Synopsis: In the heart of a bustling city, a series of seemingly random crimes baffle the local police. Each crime scene forms a perfect pentagon, its vertices marked by the locations of heinous acts: arson, robbery, assault, kidnapping, and murder. The pattern is unmistakable, but its origin and purpose remain elusive. Main Characters: 1. Detective Laura Mathis: A seasoned detective with a sharp mind for patterns and an unconventional approach to solving crimes. She has a background in mathematics, which becomes pivotal in cracking the case. 2. Dr. Petr Novák: A reclusive mathematician known for his work on complex geometric theorems. His breakthrough, known as “Petr’s Miracle,” describes the transformation and center-shifting of pentagons, a theorem that holds the key to the mystery. 3. Lucas Grey: A brilliant but troubled mathematician turned criminal mastermind. Obsessed with proving his intellectual superiority, he uses Petr’s Miracle to orchestrate his crimes. 4. Captain James O’Neil: Laura’s superior, a pragmatic and seasoned officer who struggles to understand Laura’s mathematical approach but trusts her instincts. Plot: Act 1: The Initial Crime Scenes • Scene 1: Detective Mathis is called to a grisly murder scene, the fifth in a series of crimes that seem unrelated except for one peculiar detail: each crime scene location, when connected, forms a perfect pentagon. • Scene 2: Laura revisits each crime scene, mapping out the pentagon. She notices subtle clues suggesting the involvement of higher mathematics. • Scene 3: Captain O’Neil is skeptical but authorizes Laura to pursue her hunch. Laura reaches out to her old professor, Dr. Petr Novák, who explains his theorem and how it could apply to the crimes. Act 2: Uncovering the Pattern • Scene 4: Laura discovers that each pentagon transforms according to Petr’s Miracle, with the center point and vertices shifting in a predictable manner. • Scene 5: The team realizes that each new pentagon is formed based on the previous crime’s locations, creating a series of nested pentagons. • Scene 6: Laura and Dr. Novák work together to predict the next vertices. They identify a pattern and pinpoint the next potential crime scene. Act 3: The Chase • Scene 7: As they rush to the predicted location, they narrowly miss catching Lucas Grey, who leaves behind a taunting message: “Catch me if you can.” • Scene 8: With the stakes higher, Laura delves into Lucas’s background and discovers his motivation: a twisted game to prove his genius. • Scene 9: A cat-and-mouse game ensues, with Lucas always one step ahead, using Petr’s theorem to stay elusive. Act 4: The Final Showdown • Scene 10: Laura finally deciphers the ultimate goal of Lucas’s geometric game – the final pentagon points to a significant, symbolic location in the city. • Scene 11: A high-stakes confrontation at the final predicted crime scene. Laura uses her knowledge of the theorem to outmaneuver Lucas, predicting his moves. • Scene 12: Lucas is captured, but not before a tense standoff where he reveals his admiration for Laura’s intellect, acknowledging her as a worthy opponent. Epilogue: Resolution • Scene 13: The city breathes a sigh of relief as Lucas is put behind bars. Laura reflects on the case, realizing that her unique skills in mathematics have not only caught a criminal but also saved lives. • Scene 14: Dr. Novák publishes a paper on the practical application of his theorem, crediting Laura for her intuitive leaps and deductive prowess. Themes: • Intellectual Duel: The battle of wits between Laura and Lucas showcases the power of intellect and strategy. • Math in the Real World: The plot highlights how abstract mathematical concepts can have real-world applications, even in crime-solving. • Moral Ambiguity: Lucas’s genius is both his strength and downfall, raising questions about the ethics of using intellect for harmful purposes. Style: The narrative is a blend of crime drama and intellectual thriller, with detailed explanations of the mathematical concepts woven seamlessly into the plot. The tension escalates with each new crime, keeping readers on the edge of their seats, while the unique use of geometry adds a fresh twist to the genre. Potential for Adaptation: “Pentagon’s Shadow” is ripe for adaptation into a TV series or film, with its intricate plot, compelling characters, and the unique blend of crime-solving and mathematics. Each episode could explore a new facet of the theorem and its application, culminating in a thrilling final showdown.
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  2428. The same you would do with a half baby
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  2429. 5:00 Isn't there a Pascal Triangle also? I have just started the video and its already intriguing
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  2430. This entire process of an “adding ears algorithm,” along with the 4 “basis pentagons” for every pentagon, along with the importance of directional adding of ears, along with the points being described by complex numbers, etc etc all seems like any polygon can be described using some sort of complex polynomial where each power term refers to a different “basis polygon” that when rotated and scaled in some way then added with all the others can form any polynomial. This has some interesting implications as it would mean that an n-sided polygon can be described using a polynomial of nth degree where the roots of the polynomial correspond to the vertices of the polygon, and would also be another demonstration as to why one of the “basis pentagons/polygons” goes to 0 + 0i when the ear adding algorithm is applied as it is just taking the derivative of some complex polynomial where that power goes to zero, and also showing that when enough “derivatives” are taken of the polygon it becomes regular then to a point like going to some constant then to zero. This could also imply the existence of polygons described by power series, like a circle or any continuous loop being described as an infinite degree polynomial which could also indicate the power series forms for sine and cosine and e^(ix) = cos(x) + i sin(x) describes more than just a cool identity but also a polynomial describing a circle as a polygon in the complex plane. I wonder how this could be applied in terms of sequences to differential equations like the fibioci sequence and sequences being described by polynomials through taking “discrete derivatives” of the sequence the have the first numbers be coefficients into a formula which forms a polynomial describing the sequence, could there be a way to term a sequence into a polynomial into a complex polygon, or find complex sequences using complex polynomials and then complex polygons which shape could describe some property and maybe get into topology? Sorry I’m rambling lol Has this been studied at all or does anyone know if this is a valid way to understand polygons? I noticed it again and again throughout the video and thought it was very interesting but I don’t think I have the skill to explore it on my own quite yet
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  2431. This might be the most mindblowing piece of math I have ever seen. I had problems focusing on the rest of the video because my mind was reeling from the extreme and simplistic beauty of this structure!
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  2432. Amazing video (I am assuming) EDIT: It was indeed a good one
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  2433. I really like the design of your shirt :)
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  2434. lost at 50:00 will watch again
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  2435. Math(s)ologer!!
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  2436. Wow, I am early haha :)
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  2437. As someone who's just now getting to grips with non-integer applications of the binomial theorem (I have a weird idea of fun) I'm now wondering if there's any way to extend Moessner's Miracle to the reals, or beyond...
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  2438. And indeed if you look at the formula for rectangular boards, it always spits out an even number.
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  2439. Kudos to the discoverer of this. A very organized mind.
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  2440. Brought my jacket and a tie for the grand premiere :) Thank You for the great format of Your videos. They help me stay sharp long time after university studies.
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  2441. I made it all the way to the very end, but I'm still not sure how to split the pastrami sandwich. These are real-world problems Professor! 😁 (jk - I see the scenario where this works for that purpose too ... unless someone either does or doesn't like the bread crust and so each cut isn't valued equally.)
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  2442. I think the proof that shows that you can make a dual triangle by composing the three medians is a more rigorous and easy to understand proof. It also immediately shows why isoceles triangles duals are isoceles: because two of the medians are the same!
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  2443. Maybe 2000 years from now, we solve quantum gravity.
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  2444. At 5:20: the original marble you consider cannot be the final marble simply because it’s movement is always in steps of two. If you label all possible places it can move on the board, you see that the center is not one of those spots. Using the same argument for the other marbles, only marble ‘e’ can survive.
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  2445. This man is straight up a beast.
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  2446. I'm so sad humans are going extinct :(
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  2447. the polynomial you observe this pattern from could be manipulated in other ways, i.e. (x+3)^n instead of (x+2)^n , I wonder what those patterns correspond to.
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  2448. I loved it as always. Didn't want it to end.
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  2449. Reminds me of the matrix exponential. The power expansion of { e^(A) * e^(B) } for commutative matrices A & B can be rearranged to fit the binomial theorem. Fun!
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  2450. This is a really great presentation, but at every point I can think of an architectural example from history built and still existing today which demonstrates the points of the presentation. For example think of Bernini’s Baldacchino at St. Peter’s. The twisted ‘rope’ columns have rational and calculable areas and volumes. This was built a few years before the mathematics presented by Cavalieri. Imagine what the ‘3D printer’ used to build it was like, - 400 years ago. 😊
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  2451. In case you're wondering, the dots on his shirt are the digits of pi
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  2452. A bit late, but I wrote an implementation in Python that can run on the command-line or provide a user-interface: https://github.com/pierrebai/AztecCircle
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  2453. mathematicians are not people who find mathematics easy, but instead the people who like the challenge of it... So don't get discouraged and challenge yourself
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  2454. Stop it this is just a ClickBait title!!
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  2455. 5:00 "Fasten your mathematical seatbelts" is one of yours (and mine) favorite phrase. But how does a mathematical seatbelt look like ? Perhaps like a giant string made of your amazing T-shirts ?
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  2456. Ty
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  2457. I don't properly know any calculus in a practical sense, but just knowing that "derivative" just means "rate of change of the function" and "second derivative" just means "rate of the function's rate of change changing" helps a lot watching this.
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  2458. That table at 37:31 really shows how outstanding 355/113 is as an approximation of pi. It is astonishingly good for a three digit denominator. . Here's a question: Is there any other famous irrational number that has a rational approximation with a three digit denominator that is anywhere near as good as this one? To decide this, we need to think about making some measure of the "goodness" of a rational approximation (q) of an irrational number (r). To start with, we should calculate the proportional difference, (r-q)/r. We then realize that we must take absolute values to ignore the sign of the difference. We could then take the reciprocal so we can express this amount as being accurate to within 1 part in N, where N=|r/(r-q)|. (I use N here to indicate nearness of the approximation.) If we care only for how accurate the approximation is, this is a good measure and the higher the value of N, the better the approximation. But, that would have us just choosing bigger and bigger denominators as "better" when really some preference should be made for smaller denominators as these are generally found sooner and with less effort than bigger denominators (and perhaps smaller denominators are easier to remember). So, we could apply a penalty for larger denominators by dividing the score, N, by the denominator, d. But is this a sufficient penalty? Perhaps a better measure is to divide by the square of the denominator. I will settle on having a "rational approximation goodness score" calculated as: RAGS = | r / (r-q) | / d² Here are some scores (N and RAGS) for approximations of π (I have used the fractions given in the table at 37:31) Rational N-score RAGS 3/1 22 22.188 13/4 29 1.811 16/5 54 2.152 19/6 125 3.480 22/7 2,484 50.704 179/57 2,530 0.779 201/64 3,247 0.793 223/71 4,202 0.834 245/78 5,541 0.911 267/85 7,549 1.045 289/92 10,897 1.287 311/99 17,599 1.796 333/106 37,751 3.360 355/113 11,776,666 922.286 52,163/16,604 11,801,038 0.043 Note: The fractions given by the convergents from the continued fraction representation of π are 3/1, 22/7, 333/106, 355/113 and then the following: Rational N-score RAGS 103,993/33,102 5,436,310,128 4.961 104,348/33,215 9,473,241,406 8.587 208,341/66,317 25,675,763,649 5.838 312,689/99,532 107,797,908,602 10.881 Here are some scores for rational approximations of √2 Note: The sequence is given by a/b → (a+2b)/(a+b) Rational N-score RAGS 3/2 16 4.121 7/5 99 3.980 17/12 576 4.003 41/29 3,363 3.999 99/70 19,600 4.000 239/169 114,243 4.000 577/408 665,857 4.000 1,393/985 3,880,899 4.000 3,363/2,378 22,619,537 4.000 8,119/5,741 131,836,323 4.000 19,601/13,860 768,398,423 4.000 It is interesting that while the nearness (N-score) increases at a rate that converges towards 3+2√2, the denominator increases at a rate that converges towards 1+√2. Since (1+√2)² = 3+2√2, the adjusted score (RAGS) converges to a constant (=4) as the sequence progresses. Here are some scores (N and RAGS) for approximations of the golden ratio φ: Rational N-score RAGS-score 3/2 14 3.427 5/3 33 3.697 8/5 90 3.589 13/8 232 3.629 21/13 611 3.614 34/21 1,596 3.620 55/34 4,182 3.617 89/55 10,945 3.618 144/89 28,658 3.618 233/144 75,024 3.618 377/233 196,419 3.618 610/377 514,228 3.618 987/610 1,346,270 3.618 1,597/987 3,524,577 3.618 2,584/1,597 9,227,466 3.618 4,181/2,584 24,157,816 3.618 6,765/4,181 63,245,986 3.618 10,946/6,765 165,580,143 3.618 In the case of approximations of φ the improvement in nearness (N-score) occurs at a rate that converges towards 1+φ = φ². Then, as the denominator increases at a rate that converges to φ, the RAGS-score also converges to a constant which happens to be 2+φ = 1+ φ². The approximations of φ and √2 are generated by formulae that create convergence to fixed multiples for increases in the denominator. There is also a fixed rate of convergence towards the rational, as evidenced by the continued fraction representations of φ and √2 being 1+1/(1+1/(1+1/(1+1/… and 1+1/(2+1/(2+1/(2+1/… respectively. I think it is fair that this scoring system gives these formulaic fractions equal standing. It is also worth noting that the RAGS for φ is less than for √2, which is consistent with φ being the "more irrational" number that is harder to approximate. I have also calculated the N-scores and RAGS for √3 = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/… The continued fraction pattern is 1,2,1,2,… It shouldn’t be a surprise that the N-scores increase at a rate that converges to 2+√3 while the RAGS converges to an alternating sequence of 3 and 6, with the higher RAGS coinciding with the approximations that are slightly greater than √3. This is because the approximations that are greater than √3 have a proportionately smaller increase in denominator than those that are less that √3 – i.e. if you go from above √3 to below √3, the denominator has increased by more than the numerator to obtain a smaller fraction. The scoring for the approximations of π is certainly more interesting since the continued fractions are not in a fixed pattern and so the quality of the approximations relative to denominator as indicated by the RAGS varies considerably. And just to come back to it: How good is the 355/113 approximation for π? It is the absolute stand-out amongst those shown here. [Note - these calculations were done "quick and dirty" in a spreadsheet and so the values are likely inaccurate past 10 digits.]
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  2459. What’s amazing is how this represents the three aether modalities. Dielectricity, magnetism and electricity.
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  2460. Maybe it's just me, but I immediately identified the four corners of the diamond with the four kingdoms of Oz: Gillikins in the North, Munchkins in the East, Quadlings in the South and Winkies in the West. Which would place the Emerald City at the center of the circle. Pretty fitting given our math wizard host's country of residence, don't you think? ;-)
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  2461. 23:05 He says that you can take any set of numbers that make 15, and the corresponding objects will also make a square without corner. This doesn't hold when you take two numbers though! Clearly 9+6 and 7+8 don't fit nicely into a square. Still very impressed about this property when taking more than 3 numbers!
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  2462. Jaw-dropping proof! I love each and every one of your videos, but this one really caught me off guard. Thank you Burkard! 🙏🏻
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  2463. The total area for the figure at 15:58 is 5. At first I didn’t know how much the areas got smaller with each new color, but it’s clear that each new color diminishes by the same ratio, and at the same time the number of squares doubles. For the total area, you would then have an expression like 1+2r+4r^2+8r^3+16r^4 for the total area. To figure out r, I saw that the angle at which the squares were branching out had to be 45 degrees, since two branches made a right angle, which meant that the side lengths were decreasing by a ratio of 1/sqrt(2). This means that the area decreases by a ratio of 1/2, and plugging in r=1/2 into the expression above gives 5. Basically, each color contributes the same amount of area (1), and since there are five levels you get a total area of 5.
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  2464. Except for the wizards, I think- they’re the sanest of the bunch.
    4
  2465. Another great video! I always tell the class about your channel and Ramanujan
    4
  2466. That looks like my ratio calculator. I bought it in an art supply store. It also reminds me of a video on Binford's law. My dad had a slide rule. He was a veternary toxicologist.
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  2467. I loved the video, great as always!
    4
  2468. It makes total sense that the derivative of the area of a circle with respect to radius is the circumference of a circle, using "onion reasoning". If you took a circle of radius r and increased its radius infinitesimally, you would basically be adding a "line" of area to that circle, which would have a length of the circumference of that circle. Do this many times and you will find that the area of a circle must be equal to the antiderivative of the circumference of a circle.
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  2469. 7:52 "Kasteleyn's watching you" i'm... genuinely scared now, im sitting in a dark room
    4
  2470. 4:27 The blooper: the characters should have been stacked up vertically according to the traditional Chinese-Japanese-Korean writing system.
    4
  2471. I guess I'm the first to point out the "Pi mal Daumen" pun in the intro?
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  2472. 13:42 (here is my attempt, but I'm no math demon😊): 1) If at the beginning the rubber band as length = 1, then the length(L) of ⅟₁₀ will have a length L₀ = 1/10. Now we start rotating the wheel (w/ radius r), the rubber band stretches and L increases... 2) Say at some point in the rotation L = L₁. If we rotate the wheel by a very small amount Δθ, the new length L₂ will be L₁ scaled by the factor ≈ (1+r∙Δθ)/1 = 1+r∙Δθ (the "1" comes from the fact that wall to wheel distance is always 1) 3) Given #2 one can therefore write L₂ ≈ L₁ (1+r∙Δθ) ⇒ L₂ - L₁ ≈ L₁∙r∙Δθ ⇒ ΔL ≈ L₁∙r∙Δθ. In the limit, when Δθ → 0, we have: dL = L∙r∙dθ 4) dL = L∙r∙dθ is a separable differential equation that can be solved by separating the L's and θ's and integrating: dL = L∙r∙dθ ⇒ dL/L = r∙dθ ⇒ ∫ dL/L = ∫ r∙dθ ⇒ ln(L) = r∙θ + C (C is the const. of integration) 5) When θ = 0 then L = L₀ = 1/10. That means that C = ln(1/10) ⇒ C = -ln(10). So the final solution for L (the length of the original ⅟₁₀) as the wheel turns is: ln(L) = r∙θ - ln(10) ⇒ L = ⅟₁₀ ∙exp( r∙θ ) 6) Finally, when θ = 2π we want L = 1. Plugin this in the equation above we get: ln(1) = r∙2π - ln(10) ⇒ 0 = 2π∙r - ln(10) ⇒ 2π∙r = ln(10) Q.E.D.
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  2473. Hi Mathologer! I love your channel as well as your animations, and think you should take advantage of GameStop's new NFT marketplace to mint some of the animations as NFTs :)
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  2474. Thank you so much what a great video
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  2475. 1:50 In the Sidef programming language, we have the built-in "faulhaber(n, k)" function which efficiently computes sums of k-th powers of the first n positive integers. In this particular case, faulhaber(1000, 10) = 91409924241424243424241924242500 is computed in about 0.002414 seconds.
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  2476. You might be a nerd when your brain automatically says hey must watch a 20 min video on a math concept. You might be a teacher when your brain is automatically converting this into something the students can understand. You might be insane when you're contemplating doing this with elementary school students. Parents thought common core was insane? Wait til your kids are talking about a guy named Moessner lol
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  2477. Thank you for all the lovely videos you made this year! I wish to you and Marty all the best in the next year, and I wish myself many more of your great videos... Merry Christmas!
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  2478. Reminds me of my doodling in high school, in math class (in 2016/2017). I had discovered the fractal pattern that appear when doing the Pascal triangle mod p, with p prime, and the weird combination it creates when you do it with a non-prime integer.
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  2479. This was heartachingly beautiful. Thank you.
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  2480. In school here in the US, I was taught to say sinh as “cinch”. Saying it as “shin” works also, but I’m surprised the English speaking world hasn’t agreed on a standard, or maybe that is the standard and I learned about it before? Regardless, Excellent presentation! I would have loved to have these back in the day.
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  2481. 20:04 My first thought was 2(a+b). My next thought just seconds later was 3-(a+b) which is just the same as -(a+b) mod 3
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  2482. Great video, as always. I just thought you should mention that although those two constructions don't converge to the actual length/area, they do converge to the area/volume, respectively. I think stating this point may help some people who are still confused. Edit(s): Typo.
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  2483. One of the best, clearest, bits of number theory. Students have it so much easier than I did 40 years ago. Great video.
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  2484. I define phi as "where addition meets multiplication", BECAUSE (1/phi)+1 is phi.
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  2485. Curious. I recall being taught that children from the levirate marriage were considered children of the dead brother. But maybe only the first born and only if the late brother was childless, (or probably son-less.) Guess I need to refresh on the laws of the Jewish faith that are no longer practiced.
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  2486. Nobody has finished the video yet!
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  2487. True fastest solution; If each rod is a processor register... Just rename the processor registers so that r0 (rod A) is now r2 (rod C)
    4
  2488. The surface area of a sphere is its circumference multiplied by its diameter. The distance around times the distance across. The circumference of a circle is its diameter multiplied by pi. The surface area of the sphere is the SQUARE of the diameter multiplied by pi.
    4
  2489. 37:14 There are an equal number of tiles of each colour because if you were to think about it as a 3d tiling of cubes, looking at it from above would make it look like a perfect square of orange square tiles. This is also true for looking from the other two orthogonal directions. Of course, since the squares are all the same size, they would contain the same amount of tiles.
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  2490. What do you do when you’re not making these videos?
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  2491. @Mathologer, is it the promised crazy Galois theory video ? _
    4
  2492. You always explain things so clearly that I know exactly where I stop understanding the maths. I often don't quite understand the last section or two of your long videos, but they're still fascinating.
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  2493. Hey. 3b1b Daniel Radcliffe Avatar Euler Mandelbrot 'My son' Are your patreons
    4
  2494. Here is another cute way to derive the product formula for sin(x), starting from the series expansion of the cotangent function - Observe that if f(x)=sin(x), then f'(x)/f(x) = cos(x)/sin(x) = cot(x), hence cot(x) is the logarithmic derivative of sin(x) - Write out the series for cot(x) = sum(1/x+n) [where n ranges from minus infinity to plus infinity] - Integrate the expression from step (2) term by term - Take the exponential
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  2495. Be still my beating heart, another Mathologer video!
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  2496.  @Mathologer  I am 90% sure they won't know or even the proof for it. The original poster has an Indian last name [sarkar meaning goverment]. So he is probably preparing for jee in a coaching. The method of teaching in coachings is highly utilitarian because the exam is highly competitive and not a single second is used on something which won't be directly useful in the exams. Only how to solve questions is taught. Somewhat sad but this is what high competition for resources does. Deriving proofs yourself turns detrimental . I think this harms scientific aptitude of India. Though it does teach working under pressure and being extremely practical while avoiding perfectionism if someone does pass the exam.
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  2497.  @pravinrao3669  That actually sounds similar to Greece! It must be different in it's ways, for example we all take private tutorials during cenior high school although it's supposed to be optional. We had though one math teacher in cenior high school who taught "useless" things and made fun of tutors and shared math books. Most tutors and students hated his guts but I love him! "The student says I'm tiiired! I solved 30 exercises! And what got tired? His hand got tired. Not his mind. He solves and solves and his mind shrinks and shrinks... Open a University book, to open your mind!"
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  2498. same but in pakistan
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  2499. In the mental arithmatic painting you can use the squre rearangement trick to turn 10²+11²+12² into 13²+14² which turns the expresion into 2×(13²+14²)/365 Which is 2×365/365. There is another square rearangment trick you can use here: X²+x+(x+1)=(x+1)² So the top of the fraction can be written as 10²+11²+12²+13²+14² =10²+11²+12²+13²+13²+13+14. =10²+11²+12²+2×(12²+12+13)+13+14 And so on... Eventually we are going to get 5×10²+4×10 +(3+4)×11+(2+3)×12+(1+2)×13+1×14 Which is equal to 540+77+60+39+14 = 730 = 2×365.
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  2500. Awesome vídeo! I like the increasing difficulty through the lesson, It keeps the video accessible and still challenging for everyone. I'm a nerdy high School student from Brazil xD I followed everything until the integrals after 38:00
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  2501. Honey, wake up, new Mathologer video just dropped!!
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  2502. Why can't you start the pattern a^3+b^3=c^3+d^3 or a^3+b^3+c^3=d^3+e^3 or similar?
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  2503. Just open Youtube to take a break from Algorithms exercises and sees this(
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  2504. Finally, one of my favorite theorems! I want to leave you an exercise about this theorem that made me appreciate it even more: Let p be a prime such that p is either equal to 2 or 3 modulo 5. Prove that the sequence n!+n^p-n+5 has at most a finite number of squares
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  2505. Always love your videos, a great sunday afternoon
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  2506. The video so nice I liked it twice.
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  2507. 24:30 For a while, I struggled with figuring out, why the ”>”-sign can’t become a ”<”-sign, straight away, as the quadrilateral becomes flat. But, I’ve finally got it: The transition between 3D and 2D is continuous; and, as such, the difference (Aa+Bb)-Cc must either stay positive (corresponding to the ”Strictly greater, than” -case), or, on its way to negative (corresponding to ”Aa+Bb < Cc”), it must pass through 0 (Aa+Bb = Cc), as you indicated, in your video on the ”Turning the wobbly table” -theorem (in a slightly different context). But; because, in a continuous transition, there is only 1 instant, where the quadrilateral is flat (and, therefore, the ”strictly greater than” -inequality doesn’t necessarily hold), the difference is always either positive or 0; and therefore, for all quadrilaterals, Aa+Bb >/= Cc. Very satisfying, I must say. 😌
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  2508. Who else started thinking about how you can create a dictionary of 52C5 cards such that you can give 4 cards to someone and they can predict the last card. Of course, the last thing you would have to do is fit that dictionary in marty's head. Sorry marty!!!!!!!!!!
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  2509. First! :D
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  2510. The necklace made me think of the jewel-division problem from an earlier video. I'm now imagining a band of burglar-mathematicians trying to divide a necklace into equal-value pythagorean triples.
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  2511. Thank you sir for this explanation
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  2512. Seems like an E6B paper flight computer, to me....
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  2513. Big fan of Heron's formula here, so I may go on a bit. Here in the US, New York State, under the "Regents" math curriculum, in the late 1960's, we did learn Heron's formula. We were not required to memorize the derivation, but it was in the textbook. Decades later, I decided to test my algebraic chops, and try to derive it with the only clue I remember from the book, that it relied on factoring the difference of two squares. If you drop an altitude on one of the sides, you can solve for the altitude and one of the unknown segments on the base simultaneously, using the Pythagorean Theorem. Mathologer takes a shortcut by using the law of cosines, but my method is how you get the law of cosines as well (and you can get rid of the fraction in the Mathologer's version with some convenient cancellations). But once I was there, and I got the formula knowing what I was looking for, I thought of four justifications for "discovering" Heron's formula instead of calling it a day having a formula for the area in terms of the sides: 1) it lacks symmetry. There is nothing special about any side, other than that you chose one to drop an altitude on; 2) it's pretty nasty to calculate from. Many of us have probably calculated nastier ones, but we can do better; 3) it is badly scaled. You end up raising numbers to the fourth power, which usually results in something large, then subtracting them, leading to truncation or round-off errors if you don't keep a lot of decimal places (I realize Heron wasn't thinking about floating point calculations. Heck, he didn't even have a decimal system) 4) it's not obvious from the original formula, at least to me, that your area isn't going to turn out to be the square root of a negative number. If you expand the trinomial and collect terms, you do get something symmetrical, but all the other objections remain. That might tell you that, since expanding didn't work out very well, maybe the opposite--factoring--is the solution. In addition to factoring the difference of two squares, twice, at one point you have to collect some terms and recognize it as the square of a binomial. It's all quite pretty. But it's 100%, algebra, none of the geometric insight of this video. With the final formula, in addition to being much prettier and easier to calculate from (if the need arises which, truthfully, it rarely does) you can see at once that for a legitimate triangle, no one side being longer than the sum of the other two (or equivalently, no side being longer than the semi-perimeter) you'll never get a negative number and moreover, if a "triangle"s one side is exactly equal to the sum of the other two it has zero area, as expected.
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  2514. IT'S MATHOLOGER TIME!!! runs around the room crazily
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  2515. In truth, that is exactly the only rule that i use to solve any classic Tower of Hanoy puzzle. I only have to care about the "sub-stack" of disks where the smallest one is at the time and if it is even or odd. In between i just do the moves that are unique, avoiding any "repetitions".
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  2516. only in exactly one case, the equilateral, the triangle is the same as it's folded counterpart.
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  2517. I think the answer to the problem on the board is 1 and 364/365. I was adding 100, 121, and 144 when I realized that makes 365, so the problem is just 1 + (169 + 196)/365 Edit: Apparently it's 2. I guess I forgot to completely add a number.
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  2518. I really really love this type of video, please continue Master Class videos :D
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  2519. That panel at the end...a reminder how easily momentous things can be overlooked, by anyone not understanding what they're holding.
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  2520. Glad to hear it! Much love to you and yours!
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  2521. Coin paradox blew my mind. Great video, great visual and explanations.
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  2522. I have always wondered who could watch this and give it a thumbs down?
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  2523. 2019 is the last 4 digit on the second line of the christmas tree stem thing on your clothes
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  2524. I must have Stockholm Syndrome. I didn't mind being kidnapped at all. :)
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  2525. Loved this Video! Always wondered whether there was any research on the addition of odd numbers to get squares. I knew how to get the cubes, but was pleasantly surprised to know that the pattern continues! Made my day, (the week actually). Loved the fact that some of the approaches in the video share a big similarity to the ones in your power sum video, grids and various others (And pascal's triangle!). Also, how you can use geometry to not only visualize, but also solve what is, essentially, a number theory and algebra problem. Gently makes you appreciate how interconnected everything in mathematics is. I hope you can keep making these videos, and that they reach an even larger audience, so everyone can fully enjoy and learn advanced mathematics. PS : Is your geometric proof the first one? PPS : I think, the output sequence might be n!*(n-1)! ? if a =1 and b, c, ... =2?, Since then each term would become n + 2(n(n-1)/2) = n^2.
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  2526. I was supposed to study english for an exam tomorrow, but this is just way too fascinating
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  2527. Thank you, for your videos. "fresh" ideas and content and ton of work behind them.
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  2528. Wow, I love this mix of algebra, calculus and clever manipulations ♥️ For me the most beautiful part of the derivation was 1/1/2/3/4/...=sqrt(pi/2) Thank you for your work 👌
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  2529. fixed 👍
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  2530. you kept using the wrong infinite sum from 17:13 to the end
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  2531. Please make your vids in dark mode
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  2532. How are we able to draw equilateral triangle on a paper when its simply an infinite grid with small spacing
    4
  2533. As always, I really appreciated the history bits that one comes to expect from this awesome channel :) But I especially liked the closing bit about ideal mathematical objects vs reality. Seems to me that often, a lot of otherwise great mathematical content tend to get a bit lost in idealism, and lose sight of that difference. It's a small detail, but it makes a big difference. Thank you for all the work that goes into these videos!
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  2534. Beat me to it
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  2535. Burkard, many thanks for your very informative videos on mathematics. I am a retired professional engineer and it is a pleasure to look back on the areas of mathematics I had to learn for my engineering degree, and you make it so much more interesting and enjoyable as a subject. Kind Regards Tony
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  2536. I've always been intrigued by this paradox, calculus-involved approach or not. The visual aesthetics of your videos are really sleek as well. Thank you Mathologer for posting such interesting content in such approachable and easy-to-grasp ways :)
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  2537. 여기서 한국 드라마를 보게 될 줄은 꿈에도 몰랐네요. Never expected to see K-drama in this channel!!
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  2538. If there is 33*33 tiles The number of sum must be ((33*33) * (33*33 + 1) / 2) (gaussian sum) Suppose every row must sum to the same number. That's mean ((33*33) * (33*33 + 1) / 2) should be 33*n Each row then must be ((33*33) * (33*33 + 1) / 2) / 33 or 33 * (33*33 + 1) / 2 which is 17985 Generally (n * (n^2 + 1)) / 2
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  2539. I now have a panic attack whenever I see a du/dx written somewhere. Goddamn integrals
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  2540. My favorite channel!
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  2541. just watched mathvengers , so this is like a sequel to your post credit scene
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  2542. Galileo was one of the first to notice the "part equals the whole" weirdness with infinities. I find it interesting to contemplate how this then plays out in another puzzle he wrestled with - the shape of cables or ropes under gravity. (He got it wrong- he guessed that they were parabolic.)
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  2543. Before people knew the analytic solution to the shape of hanging cables under gravity, the architects would build models upside down, with hanging strings and fabric, and then study the shape to specify how to build arches and domes right-side-up. They recognized that since a string hangs with a purely axial load, and so should an arch be built in a shape so it also carries a purely compressive axial load.
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  2544. first time I got confused on these videos, heh warning about baby calculus wasn't strong enough to include baby differential equations as well I wasn't ready
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  2545. Yes!! I was thinking of linoleum floor patterns myself.😊👍
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  2546. 15:08 please make a complicated video about 4d stuff :] might not be for everyone but i´m sure many people will appreciate it!
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  2547. I created a Geogebra application on Petr-Douglas-Neumann myself a year ago (see at Geogebra ... /classic/sdftbe2y). It has a slider to switch from triangle (n=3) to decagons (n=10). And there is a play button to run through the 47 steps. (If you choose a smaller n, then note that some of the 47 steps are hidden because they are meant for the larger n's.) I choose a different but equivalent approach to Petr, Douglas and Neumann. I prefer using regular n-gons instead of the ears. Imho, this is (mathematically) more elegant. Instead of connecting the tips of the ears, you have to connect the centers of the added regular n-gons. And instead of using ears with different angles each round, I add the regular n-gons where each round has a different distribution of vertices that get to each side of the edge. One round has all (n-2) vertices on one side, another round has (n-3) vertices on one side and 1 vertex on the other side, etc. You have to see the example in the app to understand what I mean.
    4
  2548. A solution like this was probably inconceivable before expressions like "liquidate" and "financial liquidity" came into common use.
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  2549. That visual proof that e^a = cosh(a) + sinh(a) is really satisfying! I've always wondered how the exponential related to hyperbolas, i.e. why we call them hyperbolic trig functions. I'd seen the construction in terms of hyperbolas, and I'd seen the representation in terms of the exponential, but whenever I tried to find an explanation how these two were connected, all I got was, "That's just the definition." (Read: I don't know, and I've never thought about it.)
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  2550. Another fun fact you missed about the Letters. „Æ“ is the sound you make when you hear about the paradox for the first time and „O“ is the sound you make after someone explained it
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  2551. I think it has been agreed beforehand, because no magician chooses a random assistant
    4
  2552. A y=ln(x/m-sa)/r^2 to everyone!
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  2553. Actually india was way way ahead in science and math in the era of Gupta empire and maurya empire in compare to rest of the world , nalanda was first multinational university of the world . But most of that was destroyed by foreign invaders from middle east and central asia .
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  2554. you were right, that animated proof at the end was absolutely worth the wait! and thank you SO MUCH for explaining natlogs. I took trig in high school and in college and never understood where e came from or what a natlog was :/ this made sense! thank you!
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  2555. 🪖
    4
  2556. It has been nearly a decade since I sat in a maths classroom and it wasn't until I began to watch your videos that I realized how much I miss it.
    4
  2557. I guess that you have a background in graphic design aswell as maths
    4
  2558. Glad you enjoyed this explanation :)
    4
  2559. In 1976, I was an sophomore engineering student in college. My calculator broke and I couldn't afford a new one. I got through that year with a slide rule and a book of log tables. I still have the slide rule in it's original leather case.
    4
  2560. 13:35 : Nicole Oresme took the overhang problem too seriously and tried it with himself 😊😊
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  2561. If I had learned math this way in school, I think I would less suck at it today. Still I am learning things here.
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  2562. Now all I need is a googol dollars :)
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  2563. Here we've only looked at number walls with a finite number of rows. Are there ones with an infinite number of rows? What strange sequences would produce them?
    4
  2564. That was interesting. It's too bad I was having an allergy attack (at least, I hope it was). I'll try again when I'm not so congested.
    4
  2565. Great animation, very good explanation
    4
  2566. I watch and rewatch all your videos almost nightly, before going to bed, they’re all so beautiful. The music at the end of this one made me emotional enough to post today. Thank you for the wonderful work you do.
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  2567. All monkeys in the world are looking forward to showing us how easy Calculus actually is.
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  2568. Keep up the good work!!!
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  2569. For n is the number of color blocks, show 4^n always has 'a' top-left, 'b' top-right, 'c' bottom-left, and 'd' bottom-right. Color block, 4x4: [a b a b] [c d c d] [a b a b] [d c d c] Where n=1 for 4n x 4n blocks = 16n^2 = 16 blocks. Corners of color block, 2x2: [a b] [d c] Tiling this color block in a configuration of x-rows by y-columns will result in the same corners. This works too if x=4n and y=4n. Checking any 2w x 2w of the configuration will show 4 unique colors in the corners. Where w=1, is our normal 2 x 2 checker for 4 unique colors in the corners.
    4
  2570. This would actually have been a relevant question for Zimbabweans in 2008-9.
    4
  2571. I love that card trick! The one drawback is that it needs an assistant, but aside from that, I think it definitely is the best mathematical card trick I've seen.
    4
  2572. This is one of two reasons Futurama is the greatest animated show of all time. The other is it‘s portrayal of the future...
    4
  2573. 2:44 I think that was meant to be "except for a scaling factor"
    4
  2574. Wow, great. Thank you. Surprisingly, simple mathematical reasoning remains - "always true". Is this a law or a craft? Thanks again.
    4
  2575. I had a Post Versalog (10") linear slide rule in junior high school that my Dad gave me. I didn't see a circular slide rule until my senior year in high school when we bulk ordered some for our Physics class. It was a cheap little plastic thing but I had fun with it. It had a vinyl slipcase with "FZIX IS PHUN" stamped on it.
    4
  2576. Visual mathematics is the easiest way to learn mathematics.
    4
  2577. Some YT'ers are just old-school/long-written proofs aficionados and some are just not very good at showing, visually, these complex problems (or they don't have the imagination to do so). The former are probably too conceited and are mostly showing off and the latter are patronizing us because they believe we cannot handle seemingly difficult equations. Your Mathologer videos are a perfect hybrid of both camps without the conceit or the patronizing. Your videos are very beautiful and thought-provoking while at the same time making complex concepts easy to understand (especially for visual learners as myself). Many times, I feel like I'm in over my head at the beginning of your videos only to have that "AHA" moment towards the end when you bring it all together with your impressive animations. I cannot help but think that some of these past great mathematicians agonized over how to write down the images and animations they were seeing swirling around in their minds-eye in exactly the same way you have shown us time and time again. Also, you always give credit to these past geniuses much more often than other channels and so we also learn some history along the way. Thank you for all yo do and please keep it going!!
    4
  2578. I've only watched 45 seconds of this video, so far. And I have to say... "Wow!" A circle... Wow. Greetings from Dallas, Texas!
    4
  2579. I used to have a lovely 5" (linear) slide rule with very finely engraved scales, but misplaced it somewhere along the way. I still have my 10" slide rule that I bought from WH Smith's over 40 years ago. It hasn't seen much action recently, though!
    4
  2580. It was row 7, now row 10. We are making progress
    4
  2581. Noice
    4
  2582. "Without a computer" underrated words x
    4
  2583. I love the beautiful sequence of adding elements starting at @6:40
    4
  2584. the length of this one is perfect, and i love the geometric proofs. the logical proofs get so boring! and its always helpful to confirm what you know by coming from a different starting place and taking a different path
    4
  2585. You didn't mention the most exciting part of Lee Sallow's squares: in 21:46 we can have disjoined line segments. Magic squares using numbers thus account for no more than a small fraction of all 1-D geomagic squares!!!
    4
  2586. Great content to have breakfast with :). Would be great to see the prime numbers number wall :0
    4
  2587. 1st
    4
  2588. After reading your suggested references, I remain confused as to Madhava's original work. Were these actual proofs (as per the better known Western contributors), or were these more insightful observations and approximations? The parallel would be how Babylonians and Egyptians both had versions of Pythagorean triangles and relationships, but it was the Greeks who provided the proofs. Is that the situation here?
    4
  2589. The best shoe lacing can be also defined by: Quickest to loosen, Quickest to tighten, impervious to spontaneous loosening. Having regard to friction, a lacing that tensions segments in place without dragging the laces through the eyelets would be best.
    4
  2590. my OCD expected ❤️👸
    4
  2591. One should remember that a multiplication always is repeated additions, and a power always is repeated multiplications. Of course there must be a way to replace the one with the other, that's called an algorithm. Mathematics: Fine, now write that down as a proof ...
    4
  2592. Sir please tell me through which software you record your lectures?
    4
  2593. But what about the imaginary green Power Ranger? 🤨
    4
  2594. The best channel posted!
    4
  2595. Just hobby mathematician, but clearly understood everything. Dark magic! Really loved this one
    4
  2596. All this ending 1s really tickle my Collatz conjecture obsession
    4
  2597. I claim the entirety of the video.
    4
  2598. I swear, I thought he said Goth Shoes instead of God's Shoes for the 100 eyelet example.
    4
  2599. Michael Penn had a great video a week ago about alternating harmonic series and a proof that any pattern of positive numbers m and negative numbers n can be expressed as ln(2)+ 1/2 ln(m/n)
    4
  2600. ​ @Mathologer  why do you stay up so early edit: tomorrow is a sunday so you will probably not go to work
    4
  2601. I'm from India and study in a particular syllabus called ICSE. Heron's formula is a compulsory part of the mathematics syllabus starting from Year 8. However, we only learn and use it in the √s(s-a)(s-b)(s-c) form.As for my opinion on whether it is useful, I think it is very handy to have learnt because you wont always have the height of the triangle given as data. Thanks for the amazing clarification on how the formula was derived.
    4
  2602. Nathan's problem: Prove a sequence of "1"s for face-down cards terminates in a sequence of "0"s for face-up cards. Only valid moves are "11" becomes "00" (net of +2 face-up cards or -3 in binary) , "10" becomes "01" (net of +0 face-up cards or -1 in binary), and "1" becomes "0" (net of +1 face-up cards or -1 in binary). Each of the moves results in [0 to 2] face up cards or [-3 to -1] in binary.
    4
  2603. When I learned calculus, I always seemed to screw up the algebra and not the calculus.
    4
  2604. Amazing! As any of the Mathologer's videos. This is by far my favorite channel on YouTube. Thanks!!
    4
  2605. I think the paradox partly comes from the fact that you can't compare surface area to volume at all. If some object you measured has the same area and volume, it's the consequence of your units and not the object itself.
    4
  2606. That trick of finding a less restrictive form and finding that your proof is simpler there is a common one.
    4
  2607. "Very nice song" is a link back to this video Prof. Hardy's life appears to be increasingly anticlimactic. Always overshadowed or outdone, it seems.
    4
  2608.  @kenhaley4  Fixed the link :)
    4
  2609.  @andlabs  Fixed the link :)
    4
  2610. I'm a brazilian olympiad student, the time I traveled to where I was going to participate in Conesul (South America Olympiad) there was a training during 3 days before the test. One of the problems that my professors showed was this one, pretty fun problem, by the way. My solution at the time was, without loss of generality fix the point that's gonna be reached by the soldiers, and say it has an energy of 1. And then, the natural ideia that came in my mind was to say that any square that was K squares apart from the square we're going to has an enery of (phi)^{-K}. Like this, as well as your solution, the energy of the whole table is always, at least, preserved, and summing the energy of every square below the red line equals something less than 1 (which you can easily compute by some infinite power sum tricks). It's always great to watch your videos! thanks🇧🇷
    4
  2611. 22:02
    4
  2612. @19:04: to get the sequence of fractions, start with 1/1 = a/b. Then, given the current faction is a/b, the next fraction is (a+2b) / (a+b).
    4
  2613. Another great video! Thank you. BTW that rotating hypercube at the end is a torus whose surface rotates around the poloidal axis while remaining fixed on the toroidal axis. An interesting video would show rotations about both axes simultaneously.
    4
  2614. Whoa, getting name-checked in a Mathologer video! Achievement unlocked!
    4
  2615. Yes! You can also prove that for any n different coin denominations, the number of ways to make change out of S bills, each bill with value the lcm of the coin denominations, is a polynomial in S.
    4
  2616. "Our summation formulas are some sort of weird inverse of Pascals triangle" - Mathologer 2019
    4
  2617. Third puzzle: :o 3D: :0 30-45-60: :O This was so cool, thanks!
    4
  2618. MATH QUESTION A |\ | \ |. \ B |.___\. C ABC is right angled triangle, AB = 1728 BC = 1050 AC = ? (Write you answer in reply section) HAPPY NEW YEAR EVERYONE
    4
  2619. Beautiful. And that is just one of the all-dimentional superfigures (I just made up the term) : hypercube. There is also simplex (triangle-based), and a strange weak superfigure, which is a hexagon in 2d, dodecahedron in 3d, another figure in 4d, and then it dies. But... a question: What about the doubles? the double of the cube is an octahedron. Does it have its own formulas, extending through all dimensions? What about icosahedron (the double of the dodecahedron)? Thanks.
    4
  2620. "are there any more triangles that work" wouldn't building a width 10 out of width 10's work... you can treat the width 10 as a single unit of three then this pattern can be expanded forever...
    4
  2621. I didn't rewatch it entirely, but I enjoyed the new part and I did play it entirely nevertheless; now I also commented, and I gave it a like, which I rarely do. Lets see if we can let THE ALGORITHM bite its own tail!
    4
  2622. this blows my mind
    4
  2623. Reminds me of something i looked into once: the random harmonic series. That is, the harmonic series with the sign of each term chosen by a coin flip. The resulting series converges almost surely, and it turns out it has some neat properties as a random variable.
    4
  2624. Thank you so so much
    4
  2625. 11:58 One small correction. The golden spiral is not made up of quarter-circles. It is the logarithmic spiral with growth factor φ. The succession of quarter-circles approximates the golden spiral. A Fibonacci spiral also approximates the golden spiral, with the approximation improving as the number of squares increases.
    4
  2626. My problem with Mathologer videos: do I watch it now or do I save it until I cook up a gigantic traditional Asian meal to go with it?
    4
  2627. hello mathologer. your content brings me such incredible joy and I am so grateful for the effort you put into making this beautiful mathematics accessible. although I had to replay the level 4 section once or twice to understand the recursive field logic (I am also rather drunk), I feel I have taken a relaxing tour and come out with a strong barebones understanding of the proof. at the very least, I leave this video with a deeper appreciation for our human struggle which unites us over time and place to evolve our logical tools. what an incredible triumph for wantzel, whom I have learned died only at 33 but will live eternally in legacy. (although he had more time than galois.) let us learn not only from his proof but also from his example, and practice moderation in excitatory drugs and late night mathematics!
    4
  2628. to the window to the wall
    4
  2629. I’m glad I’m not the only one hung up on that. I was half-expecting him to go, “Oh by the way … figure out what’s wrong with this picture.” He’s done it in other videos before, so it would be on brand!
    4
  2630. Answer to the card question: Q of H?
    4
  2631.  @Mathologer hi
    4
  2632. Now we all know why atmospheric pressure is measured in TORR (a unit of pressure that is equal to 1/760 of one standard atmosphere. One torr is approximately 133.32 Pa)!
    4
  2633. 28:13 my high school teacher showed this to us once, and then he proved it to explain how proofs by induction work!!
    4
  2634. By bloopers, do you refer to the formatting inconsistency? For example, 二百五 is "two hundred five", so 205. Next to it is 九八七, being "nine eight seven", which is contexts like this would refer to 987. In these settings, you'd just list the numbers off one by one. You wouldn't give the full name. A bus numbered 64 would be read as 六四, which is "six four", rather than 六十四, which is "six ten four", so "sixty four". Basically, you wouldn't have things like 白 (hundred) in this sort of thing. Just the base characters listed like a phone number.
    4
  2635. did you know that if you square the Fourier series of f(x)=x between [-T,T] and equate its infinite coefficients with the coefficients of the Fourier series of f(x)=x^2 in the same interval, one gets all the identities of even powers of pi at once?
    4
  2636. Channel 6 Challenge: (Some space to prevent spoilers, you have been warned.) 1, 2, 4, 8, 16, 32, 64 Using binary, any given number can be uniquely made by some combination of those 7.
    4
  2637. Long waited 💪
    4
  2638. This lends itself to turtle graphics, via the The Total Turtle Trip Theorem, which states that the turtle will draw a closed figure with n sides when the sum of the angles turned is a multiple of 360. (From Turtle Geometry by by Andrea diSessa and Hal Abelson)
    4
  2639. Με αρεσκει.
    4
  2640. I would guess same diameter ;) Nice content, Nice video name, thank you for the chronological facts too.
    4
  2641. Banger video as always!
    4
  2642. There is something weirdly relaxing and also beautiful watching the animations and the number somehow forming! ❤
    4
  2643. really liked this one!
    4
  2644. I am Dutch and a huge fan of the Mathologer way of explaining mathematics. F.J.M. Barning is Fredericus Johannes Maria Barning, in daily life he was called Freek Barning. He died in 2012.
    4
  2645. SOOOO AGREE!! This guy makes math MAKE SENSE!
    4
  2646. Domotro!
    4
  2647. Now I see why it is "one sentence proof". Because most part of the proof is in visuals, not in sentences :)
    4
  2648.  @wolvenedge6214  Yeah, I realized as he went along that the user is choosing more terms in a certain direction to ENSURE arrival at predetermined sum. This still leaves my intuition feeling that if the chosen number is positive, then the sum MUST contain more positive numbers than negative ones...then again...even that can be shown to be untrue, cuz one could arbitrarily choose more (but very small) negatives, and fewer (but very large), to arrive at the same number. Finally, however, the systematic "rule" that Mr Polster used, isn't arbitrary! I still feel the infinite positive set is larger than the negative set! BAH!
    4
  2649. Well done with the video. I had to watch it a few times to understand it all! I was OK to the halfway point then started to struggle, but stayed until the end. Thank goodness for the chapters. Keep making them.
    4
  2650. Absolutely amazing! blown away! what a great presentation, Mathologer!! You made my day!
    4
  2651. I'm always looking forward to these monstrously long videos from the many Math-oriented content creators on YouTube. In particular, I feel the animated approach in these videos really helps to visualize the most intricate arguments. I should also mention that your work is very enjoyable to mathematically inclined people, mathematicians and non-mathematicians alike. And even if one has a deep background in Mathematics, there is always something new to learn.
    4
  2652. From where you got all this?
    4
  2653. Exactly.
    4
  2654. I don't understand any of this. But whatever it is, I'm glad it got solved...
    4
  2655. Thank you for this Christmas gift for us all.
    4
  2656. Given that shapes on the area-preserving transformation become more distorted the closer they are to the edge of the circle, it makes sense for that to be something boring - like a large expanse of water. What point on the Earth's surface should be chosen to minimize the amount of land near and on the perimeter of the circle? What would such a map look like?
    4
  2657. Why is 27 not on your t-shirt?
    4
  2658. I just logged in to leave exactly this comment.
    4
  2659.  @cliveso  That's a fair point. But given that law would need to be modified in order to implement such fair divide, it's not far fetched to suppose that a provision for debt consolidation would be included. Especially if that's sort of obvious. Or else, it's about supposing the incompetence of the law makers, which is a far different topic that I don't want to go into. Given the debt hierarchy presented in this video, you can then proceed to fairly divide amongst a given debt type, then move on.
    4
  2660.  @Vaaaaadim  Exactly. If the amount to be split is more than half the total claim, the largest creditors get the highest percentage of their claim. That's what he said in the video. If there's only a little to split, then the loans are all bad, but everyone gets a little. If the amount is only slightly short, then the loans are all mostly good, with only a slight loss for each creditor.
    4
  2661. Despite being a Physics graduate, I had never seen this. And the very first time you built up to the 7th tow, at the start of the video, I actually let out a little involuntary gasp of delight. I LOVE it when this happens. Thank you, Mathologer!
    4
  2662. Excellent as well as beautiful
    4
  2663. no, in australia we didn't learn the formula, but luckily i came across it somewhere and was so enthralled by it, i used it in class to my teacher's dismay.
    4
  2664. So the true genius of Leibnitz and Newton was to invent the proper symbols for expressing the ideas in their heads. So it would follow: to spread great ideas, it is not enough to 1) prove them, 2) show they work in practice but also 3) develop expressions that make it easy to parse with our brain and 4) open doors to new realms.
    4
  2665. The benefit of this solution is that, in the case of a small estate, it makes a greater number of creditors less unhappy.
    4
  2666.  @mr.johnson3844  Totally agree. High school teacher from USA.
    4
  2667. I saw the first twisted square proof first, and it was from either you or numberphile lol. Am in America for context
    4
  2668. This is awesome. Thank you!
    4
  2669. Great video as always! Personally, I would have preferred if Torricelli's construction for the exact volume also included a top-down view (looking down the axis of the horn) after 14:34. This way we could see clearly how each point on a radius of the base circle was assigned its own circle to construct the cylinder.
    4
  2670. i have never seen the bridge from conic to catenary explained this way! now i understand why so much effort went into trying to find the catenary as a conic section in early days of math
    4
  2671. This is so awesome! I'd never gotten around to learning this, because (and I quote) "if anyone ever asks me to find a formula for 1^10 + ... + n^10, then, well, I know it's a polynomial starting with n^11 / 11, so I'll just calculate the first 12 sums and find the unique polynomial that fits". But I'm so glad to see such beautiful maths behind these power sums! And yes, keep the insane stuff coming! Especially on the thing about the "asymptotic expansion" at 47:11.
    4
  2672. After a fashion, I think you have shown the nature of how slide rules work. Logarithmic scaling of two sets, and when compounded, brings to value in magnitude, similar to a pantograph.
    4
  2673. Rewatch 100%. Liked. Already subscibed. Congrats on your integrity.
    4
  2674. 2:42 “Why on earth would anybody be interested in doing that?” Well, one thing you can do is to show that a prime number actually can be the product of two integers — two complex integers, that is. The fact that p = a^2 + b^2 implies that (a + bi)(a - bi) = a^2 -abi + abi +b^2 = p So, when someone says, “A prime number can only be divided evenly by itself and one”, you can say, “Not so fast! Is it congruent to one, mod four?”
    4
  2675.  @Mathologer  Please keep it that way.
    4
  2676. Great. Both the proves ant the music.
    4
  2677. new vid pog
    4
  2678. This is one of the many things I enjoy about traditional geometric patchwork design quilt patterns , (like Flying Geese, Irish Chain, Feathered Star, Grandmother's Flower Basket, Disappearing 9-patch, Bear's Paw, Log Cabin and all it's variations, Pinwheels, etc.) with 1/2 square or equilateral triangles, rectangles, squares, etc. and shuffling them around to create different patterns. Now I have a name for them, and may design quilt patterns with the names of Pythagoras or Trithagoros, or the long ago Chinese mathematician's name! Great visual explorations of these relationships, thank you. (of course I enjoy even more, the random sewing together of odd shaped and sized pieces, called, "crumb quilting"! lol) (see also, the 'golden ratio' seen so much in nature and used from antiquity in architecture to produce unconsciously, naturally pleasing, building facades.)
    4
  2679. Thank you, Mr. Mathologer. You explain the geometric meaning of mathematical formula precisely. We are happy to see more.
    4
  2680. Guess It's time to get really interested in a topic I didn't know existed!
    4
  2681. Yeah, it took me a while to see it from the diagram, but I started doing the French lacing since finding it in a store because it looks nice. I like to imagine that it's less stressful for the lace to be getting pulled against the edges of each side, but when has a lace ever broken there?
    4
  2682. What does 2 being prime have to do with Pluto not being a planet lol
    4
  2683. The reason all even numbers are not prine is because 2 is! You're elementary brain wants to look at even vs odd but you don't understand the structure
    4
  2684. What a coincidence. Tomorow, 1 december is the national day of Romania and the flag is red yellow blue
    4
  2685. In the 'Second Proof' you showed that the folded triangles are all the same, but not that the fold of the fold is similar to the original. However this is a simple extension when you see that the black line in the center is one of the medians of the folded triangle and it is also half of a side length of the original triangle. Since all three foldings of the original are the same, we know that it's medians are all half the length of one of the sides of the original triangle. Then folding again and applying the 'Second Proof' again, we get a triangle with side lengths are the medians of the folded triangle, ie. a triangle who's sides are half the side lengths as the original. In the 'First Proof' we did not know what the ratio between the original and folded folded triangle was, but the 'Second Proof' gives us that.
    4
  2686. I love watching these videos of advanced math and I am inspired to get into it, could you guys make great tutorials that would help beginners to learn basic key math concepts, that would allow to understand bigger concepts easier?
    4
  2687. I had requested long back as comment in one of the earlier videos, an intuitive graphical explanation for why the Heron's formula for area of the triangle works, without needing to use trigonometric formulae. Thanks a ton for releasing a video that exactly answers to my request. I haven't found such a great shortcut to reach to the Heron's formula anywhere else in internet.
    4
  2688. I vote for the second one.
    4
  2689. @24:20 is a little tougher I think if you try to take it’s parent you get a square of 1,0,1,0 So I suspect the Feuerbach centre is directly on the incircle. In other words you get a 0,0,0 pythagorean triplet. Which... works? For some definition of work!
    4
  2690. How do you solve the integral in your shirt, btw?
    4
  2691. For the puzzle at 11:37, I was too tired yesterday to get it, but now that I had time today I gave it another shot. My square was defined as: ┌──┬──┐ I defined the layout of any pythag.. triple as:. │B1│B2│ X^2 + Y^2 = Z^2 . ├──┼──┤ . │B4│B3│ . └──┴──┘ . I started out pretty standard with the main three equations of: B1*B4 = X 2*B2*B3 = Y B1*B3 + B2*B4 = Z I also included these: B1 + B2 = B3 B2 + B3 = B4 To define the innercircle, I used r_i to denote its radius. The equation is then: r_i = B1*B2 I started by creating an equation to find r_i. To do this, I used the slope of the hypotenuse (Y/X) to solve for an angle, divide it by two, and convert it back to a slope. This gives a slope which bisects the angle XZ. The equation is: y=tan(1/2 * arctan(Y/X)) * (x+X) (my origin is placed at the corner XY so x+X places the lines origin at the corner XZ) Using trig identities, namley half angle identity and inverse trig identities, I could simplify the expression: tan(1/2 * arctan(Y/X)) => (sin(arctan(Y/X))) / (1 + cos(arctan(Y/X))) => (((Y/X)/sqrt[1 + (Y/X)^2]) / (1 + (1/sqrt[1 + (Y/X)^2])) => (Y/X) / (sqrt[1 + (Y/X)^2] + 1) => Y / (sqrt[X^2 + Y^2] + X) This means that the equation is: y = (Y / (sqrt[X^2 + Y^2] + X))(x+X) Which is our equation for the first bisector line. Since the angle XY is 90°, it has a bisector with a slope of -1. This results in the equations: y=-x Setting them equal to find the centre of the incircle results in a simplified equation of: x = -XY / (X + Y + sqrt[X^2 + Y^2])) Since X^2 + Y^2 = Z^2, this becomes: x = -XY / (X + Y + Z)) OR: r_i = (XY) / (X+Y+Z) Plugging in the values, we get: r_i = 36 since r_i = B1*B2 = 36, this means B1 = 36/B2. plugging into the equation: B1 + B2 = B3 yields: 36/B2 + B2 = B3 => 36 = B2*B3 - B2^2 using Y = 2*B2*B3 = 104, we know B2*B3 = 52. 36 = 52 - B2^2 => B2^2 = 52-36 => B2 = 4. So, using B2*B3 = 52 we find: B3 = 13 Since B1 + B2 = B3: B1 + 4 = 13 Or: B1 = 9 Since B2 + B3 = B4: 4 + 13 = B4 Or: B4 = 17 To Conclude, my guess as to the numbers in the square are: ┌──┬──┐ │ 9│ 4│ ├──┼──┤ │17│13│ └──┴──┘
    4
  2692. The magic square construction at 18:45 was the method I was shown many years ago. You start with one in that location, then go up and to the right one space for the next number. You pretend the grid wraps on itself, the top going to the bottom and the right to the left extremes, and if the square it is headed to one that is already occupied you go down one. This construction works for all odd grids as far as I know
    4
  2693. Here’s my answer for @12:00 9,4,13,17 And you go Right, right, right, left
    4
  2694. I found the hidden 2019 on your t-shirt in 11th row (whole row is 165271*2019*)
    4
  2695. This is one of the most beautiful things I’ve ever seen in my life.
    4
  2696. That’s more than a bit dismissive.
    4
  2697. they didn't obviously otherwise they would have given a general solution not 4 cases like they gave, the idea is to achieve contested garment type solution for more than 2 people for any type of distribution and any amount of garment, they obviously couldn't do it in general sense hence they gave ones which they were able to solve as the cases they presented.
    4
  2698. I am curious if this visualization can be attributed to @Mathologer. The limit is well-known, but I never saw the limit presented in this way before.
    4
  2699. The question in the thumbnail: Do you have a hair doppelganger? The answer: No idea. I just know that someone has.
    4
  2700. I was looking for an Intuitiy proof of this some time ago. Excited to see you explain it. Thank you :)
    4
  2701. 19:45 wait this isn't Michael Penn
    4
  2702. This is awesome! I do not remember being taught the ‘divisible test’ thing in school!
    4
  2703. If it was my birthday, This would have been the best birthday gift I would ever get
    4
  2704. @SigmaSharp, as a creator in YouTube studio, you can view how many returning Subscribers are coming to a video from notifications but not how many people are Subscribed to a channel with notifications.
    4
  2705. Master! You have to work a lot to make these awesome animations and it's a demoivre set
    4
  2706. When I was at school in the 1960's we all used slide rules and I actually had a circular one. We got very quick at using them.
    4
  2707. A 31 minute youtube video Ya , it's mathologer , we can watch it 😍
    4
  2708. Beautifully done, beautifully explained. Is it known, or can it be even remotely guessed, how Cardano/del Ferro got to their solution? Just by trial an error?
    4
  2709. This is a really nice visualization. Here’s another you may appreciate… draw a 45 [deg] line through the origin intersecting with f(x)=1/x and label these points B and B’ , with B being the intersection in the positive quadrant. Now draw a circle with diameter BB’ (i.e. centred about the origin with radius |B|). Pick a point, C, on the circle. Take the tangent to the circle at C and intersect it with the line BB’ to get D. From D, draw a perpendicular line and intersect it with the line B’C to get E. The point of intersection, E, will lie on f(x)=1/x and |CD|==|DE|. As a bonus, reflect f(x)=1/x about the circle to generate a very special shape 😉
    4
  2710. I was intrigued with the title of this video, 'Magic Log Wheel'. I have been recently diving into the Antikythera Mechanism, and was seeing if this was another mystery. I instantly recognized your devise; 'log' for logarithm; a circular slide-rule. I used two of these in the building industry and construction from 1968 until around 1987 until I relied entirely on my hp-15c. I still have both rules, a pocket version and a ten-inch diameter. Thank you for making my day.
    4
  2711. Wow I’m early
    4
  2712. This video is a great example of why Mathologer is the BEST Mathematics channel all over YouTube. Thank you Burkard, for making my day!!
    4
  2713. Your video on the many proofs of Pythagoras is one of my favorites, so I’m sure I’m gonna enjoy this one!
    4
  2714. You did an absolutely beautiful job of explaining this in a simple and beautiful manner! This is the way math should be taught.
    4
  2715. This is crazy. I was just exploring recursive sequences and the characteristic polynomials to generate a function that's the sum/product of exponentials and polynomials. I was doing it to generate coefficients to best predict a time series with noise.
    4
  2716. Always something new and cool! And in the back of my head I’m wondering about whether there might be a lot of other applications here to turning a multidimensional problem into a 2D problem? Can we use this to help us visualize multiple dimensions?
    4
  2717. counting is a miracle
    4
  2718. I reached the end without difficulties, i am in second year of prepa in France
    4
  2719. 35!
    4
  2720. 5:35 1.63 dimensional!!
    4
  2721. Oh, this reminds me of large parts of my first calculus class, in particular when my teacher stated that calculous was easy, it's the algebra you have to go through to DO calculus that's hard. Which is true to an extent. It too me a long time to realize that d/dx meant "As X changes". I was a lot happier when I figured that out. I don't know if I'd have been less intimidated if I'd known at the time that if you have a number of datapoints, you can figure a formula that fits them and then to calculus to THAT. Ah, well. Don't forget that the Cs cancel! But your teacher will be mad if you don't put them in! I wonder if calculus can be used to prove that the distrebution of numbers in pi is normal. I got lost in analytical geometry. I understood the bits but putting them together was beyond me. And ultimately, I was more interested in humanities. But I did do very well on the math portion of my GRE's. :) I wonder what the Austrian Autobahn's like.
    4
  2722. One has to be careful when dealing with infinite series as not all "infinities" are equal. By taking m positive terms from one infinity and subtracting n negative terms from the other infinity, you no longer have a one to one correspondence between the terms of these two infinities so as you pointed out the difference can be any arbitrary number you choose these to converge to.
    4
  2723. A mathologer video 😊😊😊😊😊 I keep waiting for the great videos like it. Just love watching mathologer videos.
    4
  2724. ​ @aaaaaa8410  That’s the typical objection based on uniqueness of factorisation demanded by the fundamental theorem of arithmetic. As I said, it’s a prime of inconvenience – not convenience. However if you’re teaching primary school children about prime numbers and you use the simple rule "a number is prime if it is only divisible by itself and one" then the rule seems to obviously include one. (At some point you will have to tell students the terrible inconvenience of one.)
    4
  2725. Galois theory... I cannot wait to see how you approach it. Even the chapter 7 and 8 here exceeded my expectations. These are topics where the teacher learns more each time they try to teach someone.
    4
  2726. 3:00 Hmm... It looks very sexy...
    4
  2727. Ah, approximately 0.7, the number that shows up all the time when you deal with things like logarithms and roots.
    4
  2728. Regarding the arrangement of dominoes in a 2 by “n” grid ... I see a pattern for the first five grids that’s featured in previous Mathologer videos. Perhaps one that gave rise to a certain partition / pentagonal number theorem in recent memory? ;)
    4
  2729. God save internet! I can not even imagine a life no more without having access to these wonderful, interesting, and enlightening videos like the ones you present every now and then! Herzlichen Dank, fröhliche Weihnachten, und bitte weiter so.
    4
  2730. 7:05 yes if a number has last two digits which can be divided by 4, the whole number can be because 100 can be divided by 4 so any multiple of 100 can be, like 83500 and you can check by delete all the other digits like 83516 it will be 83500+16 83500 can be divided and you have to check 16 now with the prime number it end with 81 which is 80+1 4(20)+1 :) the rest don't not matter because they can be divided by 4 any way
    4
  2731.  @EebstertheGreat  I use Pythagoras all the time in structural engineering, as every problem has to be turned into, first, their vertical and lateral components (think diagonal braces, roof members, trusses, anything not perfectly vertical or horizontal), and then every member needs to be checked for axial stresses and flexural stresses (column buckling and failing vs beam bending and sagging).
    4
  2732. 9:49 binary representation of all the digits
    4
  2733. I've read about or seen videos about this extremely-slowly converging series for Pi (or for Pi/4) many times. Why did the idea of an error correction term never occur to me? This is brilliant, and it's amazing that this mathematician figured this out in the 1400s!
    4
  2734. From india, i am big fan of Ramanujan ji
    4
  2735. That is why it's an "anagr." instead of an full anagram, I guess.
    4
  2736. This is amazing.
    4
  2737.  @Mathologer  Since I made the comment I was investigating the group structure of flipping the square across the different axes and until I saw the wiki article you provided I completely missed that the group was isomorphic to D8 and could be simplified from 4 operations (a flip across the 4 different axes) to 2 (90 degree turn clockwise and a flip across one of the axes). Thank you for sharing the wiki article.
    4
  2738. 10:46 For negative integers, A(x^n x^-n) = A(x^n) + A(x^-n) = nA(x) + A(x^-n), so A(x^-n) = -nA(x). For rationals, qA(x^(p/q)) = A(x^p) = pA(x), so A(x^(p/q)) = (p/q)A(x). For reals, I'm pretty sure we have to assume A is continuous? For any real number c, we get a sequence of rationals q_1, q_2, ... approaching it. By continuity, the limit of A(x^(q_i)) is A(x^c). To compute this, we want the limit of q_iA(x), which is cA(x). If we don't assume A is continuous, I'm not 100% certain. My real analysis class has traumatized me to the point that I'm pretty sure there's a weird discontinuous counterexample lurking around the corner, but I can't find it.
    4
  2739. Wait until Terrence Howard finds out about this.
    4
  2740. We were taught this in school (7th/8th standard Maths) in Maharashtra, India. No proof was derived sadly but I did one by myself later when learning about incircles. I was talking about it to an Aussie kid years later and he very proudly said they didn't learn any of that. Like I was the idiot for learning something extra.
    4
  2741. MY NEW YEAR'S PRESENT!!
    4
  2742. "Whether one thinks this additional argument is necessary or not is up to the examiner :)" :)
    4
  2743. To infinity and... greater than?
    4
  2744. Love the Pi Tshirt, Nice Visualisation of Pi.
    4
  2745. At 6:13 you say that we get triangles arbitrarily close to being equilateral in the square grid. This is clearly true if you are measuring closeness by differences of the angles from 60 degrees. Is it true though if we take different measures of how "close" a triangle is to being equilateral? An example of such a measure of closeness could be the distance from the triangles centriod and circumcenter (or distance between two such triangle centres).
    4
  2746. This presentation remind me a experience. At high school my math is bad, at exam my teacher give one question, how much summary number from 1 until 1000, i slove this one with draw a diagram from 1-1000 like stair case , then divide that draw to be a large triangle and many minor triangle, and calculate that's area. And next day, she call me and said to me how dumb i'm, she said i'm not even use right formula and get right answer, said that i'm cheated and give me 0 score i just laugh at thats time. But my economic teacher pass by on righ time and right place like superman, hearing her bit anger he said to me "what have you done?". She explain to him and give my paper to him, then suddenly he said "ok... i get it, leave it to me, you can back now." I saved by economy teacher at math problem, truly i cant forget that. 😂😂😂
    4
  2747. Great video. Thank you very much for revealing something that was never taught to me in my calculus classes!
    4
  2748. 3^n-1 actually (for the path length)
    4
  2749. Hi there, Mathologer. Really nice explanations -- thank you. One method that I've used to convince my calculus students that expressions are indeterminate is to give them, or have them try to construct, functions with the specified limits which show that these indeterminate forms could be "reasonably" assigned any number that they like. For example, for a given nonzero complex number w, the functions f(z)=wz and g(z)=z give rise to a quotient f(z)/g(z) which takes on the indeterminate form 0/0 as z approaches 0. But of course the limit is w, so this gives a "convincing" argument that 0/0=w. Note that f(z) and g(z) can be taken to be holomorphic functions, and one can run similar arguments for most of the other indeterminate forms. Edit 12/6 : Incorrect reasoning in the following paragraph. Please see my reply to this comment. One curious thing I've encountered recently is that I run into problems in applying this process to the indeterminate 0^0. For example, it is easy to make an argument that 0^0=1 -- simply take f(z)=z and g(z)=0. Then f(z)^g(z) clearly approaches 1 as z approaches 0. But suppose I sought to make a convincing argument that 0^0=w for a given complex w using this method. I seek two (holomorphic, say) functions f(z) and g(z) such that f(z) and g(z) both approach 0 as z approaches 0, but f(z)^g(z) approaches w. By L'Hopital's rule, I determine that a necessary condition for this to happen is that the limit as z approaches 0 of the expression [(f'(z)/g'(z))(g(z)/f(z))](-g(z)) should equal ln(w). But by L'Hopital's rule the two quotients f'(z)/g'(z) and g(z)/f(z) must be the reciprocals of each other and therefore the limit of their product must be 1, which of course forces w=1 (since g(z) approaches 0 as z approaches 0). Thus, in order to have f(z) and g(z) behave like I want them to, at least one of them must not be differentiable at 0. Can you comment to what extent the indeterminate form 0^0 is different from the others?
    3
  2750. Barely 10 minutes in and I'm already blown away. This ACTUAL 'numerology' is just so much better than the simple minded numerology that gets so many people carried away. If only they knew...
    3
  2751. I took your advice and bought the book you recommended from Amazon AU. Not cheap, so I hope it's as good as you say. :-)
    3
  2752. kako bih voleo da sam imao ovakvog profesora
    3
  2753. This is definitely going on my rewatch list
    3
  2754. Great video! Thank you. I believe that there is a mistake at 23:00 . The denominators should be 1*3*5*7 and not 1*2*3....
    3
  2755. Bold move, making the “one last thing” worthy of a video in itself! It’s a great example of a really useful general principle - many tricky problems with lots of moving parts reduce to the simplest case by holding most of the variables constant. (Foreshadowing multivariate differential calculus, for those who haven’t had this pleasure.)
    3
  2756. Nice choice of colours! Red yellow and blue are the colours of the Romanian flag, and tomorrow December 1st is the national day of Romania.
    3
  2757. On a lower level, how do you come up with interesting abd challenging problems every week, when you are a "puzzler" like Dudeney ? That's a mystery to me.
    3
  2758. 2:25 - "Why? Who knows..." and that's why there's this expression "talmudic reasoning" or "talmudic problems" - meaning obscure and convoluted (and often illogical) way of thinking using all sort of arbitrary "laws" and "precedents" - or bizarre/ weird interpretations. Why so? Who knows, or as Tevye the Milkman has said: You may ask, how did this tradition start? I'll tell you! - I don't know... ;-) 11:00 - the creditors may suffer "the same loss" in absolute numbers, but certainly not in terms of "relative loss". Someone who can lend $300 (i.e. is capable of giving such a loan) must, obviously, have MORE than that amount (say/ assume $600 of cash) - while someone who loaned $100 might be much poorer (say, $200 of ready money at most) - so the same AMOUNT of lost money (say, $50) would hurt the "$200 guy" much more than this "$600 chap". But then you never really know what is their financial situation/ status - it could be just that this "$300 sucker" loaned all he had, while this "$100 scrooge" holds $10,000 - but he's a scrooge and a penny pincher, and that's why he shelled out a mere $100. And since this part of equation is beyond any reasonable judgement of any "inheritance court" it should be left out and never taken into consideration, as this "talmudic do-gooders" approach/ reasoning can actually create worse and less fair outcome for the debtors. Which only goes to show "why this talmudic reasoning expression exists". When it's the time to light up Shabbat candle? "When one can see stars on the night sky" (i.e. they become visible) goes one (of probably gazillion other interpretations. So when a first star appears, it's only ONE star, not "stars", hence it doesn't count ("einmal ist keinmal" as Germans say - or "once/ single occurrence is no occurrence/ it doesn't count"). So, when the SECOND star appears it's still only a "single" star, as the first one "doesn't count, so only after a third star appears you can say there are TWO stars, so it is "stars" (not "a star"), so there you go. I mean, "there you can light up a Shabbat candle". Simple, no? ;--)
    3
  2759. 10:32 This is a form of chiasm . Extensively used in ancient Hebraic writing. As such, it was probably so immediately obvious to readers of the Talmud that it didn't need further explanation. https://en.wikipedia.org/wiki/Chiasmus
    3
  2760. This is so extraordinarily satisfying!
    3
  2761. Always loving your knowledgeable videos. Thanks for that.
    3
  2762. The solution to the final puzzle is 1/5. Perhaps the most visually pleasing way is to rearrange the 4 small triangles so they each fit onto one of the trapeziums to form 4 identical squares. We know that these are also identical to the middle square with unknown area because the ratio of original lengths create similar triangles. The result is the large square of area 1 is separated into 5 smaller squares of equal area, one of which is the middle square.
    3
  2763. Strange but true! Didn't know about the infinite pattern of first powers & squares, but I did stumble onto the single equation with cubes many years ago (3³ + 4³ + 5³ = 6³). Tried to find some general rule to extend it, but was never successful. As for the attempt to extend your 1st- and 2nd-power patterns to cubes, the obvious try almost works, for the first line: 5³ + 6³ = 341 = 7³ – 2 Fred
    3
  2764. Gamma > 1/2: For each of the blue pieces we can find a triangle that is smaller than the blue piece, but all the triangles add up to 1/2.
    3
  2765. Absolutely amazing. Even I have the same hair style I cannot proof Fermat's theorem.
    3
  2766.  @bazzers  I can't believe I'm back here after 4 years just to say that there is a short throw projector shining downwards on to a wall (what ever colour it may be). At 3:41 when he is pointing at the 'a' label on the triangle and you can see the sqrt(b) text on his hand. The bright light on his arm that disappears mid way down his arm is the border of the projector light. These 2 points show that the projection is coming from above. I understand there is a translucent overlay of the lecture images, it was previously mentioned in my first post "The clean slides are superimposed in post." - Itzmist 2020
    3
  2767. Coloring proof!
    3
  2768. 13:00 Answer: n! * (n-1)! Edit: I was a few hours too late apparently. We start by using what you showed at 11:56 and choose a = 1 and b = c = d = ... = 2, because from that follows that 1^2 = 1a 2^2 = 2a + 1b 3^2 = 3a + 2b + c --- and so on. The Moessner sequence is then 1^a, 2^a * 1^b, 3^a * 2^b * 1^c, which we can simplify since a is 1 and all the other variables are 2. The general formula is: n * 1^2 * 2^2 * ... * (n-1)^2 = n *((n-1)!)^2 = n! * (n-1)! 1, 2, 12, 144, 2880, 86400, ...
    3
  2769. Exactly!
    3
  2770. For once, I was able to follow it all the way through, which doesn’t often happen. Nicely explained!
    3
  2771. I understood all of it except 41:01... please set up a patreon!
    3
  2772. Mathologer I am a 7th grader who is passionate in math and knows some basic calculus. I could comprehend this perfectly (besides a couple of derivations) and am waiting for more of this amazing content!!!
    3
  2773. yes, couldn't blink an eye for 50 minutes.
    3
  2774. I'll take all the Mathologer I can get - 50 minutes is fine by me. I'm looking forward to Galois theory - maybe give us a cartoon during intermission?
    3
  2775. Omfg this video is so awesome mate keep it up
    3
  2776. Loved the video. (i have a phd in math). I was expecting also the closed form formula for the sum of k-th powers up to n: S(k) = Sum(j=1, k) Sum(i=0, j) (−1)^(j−i) i^k Binomial(j, i) Binomial(n+1, j+1) check it here (pag 8): http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.540.8640&rep=rep1&type=pdf
    3
  2777. Maths has magic indeed. Thanks for such fascinating videos. From Siberia with love
    3
  2778. thank you for the really well done video! i have studied theoretical physics for the past 5 years now, so the bernoulli numbers are not that new to me (eg. casimir effect). i could quite well follow each and every step, except that i was too lazy to proof your replacement of f(0) by f(1) in chapter 7. i have to admit that i am the kind of person who just believes identities to hold in all practical cases, so i don't really care about mathematical details, but i would have loved to see some more applications and maybe rearrangements of the euler-maclaurin formula to see its use. i am excited to see the next videos on this.
    3
  2779. Though your videos are not visually beautiful as 3Blue1Brown's and don't require appointments with geniuses as Numberphile's, your channel is no way worse (I think it is even better for me) because you explain hard things simple, and by adding a bit of humour you make this quite entertaining. Really motivates do some interesting math, though I graduated my IT-university many years ago and don't have to strive for grades :))
    3
  2780. Psssst…Thank you Isaac Newton
    3
  2781.  @Mathologer  danke schön
    3
  2782.  @Mathologer  I'm sorry! I haven't seen your letter. (Most likely it was sent to spam automatically or you did not write to the email address I use.) I generally answer everyone, and I love and respect your channel, so of course I would answer you immediately. The proof was invented by me no later than 2005. After some time, the opportunity arose to insert it into my book "Arithmetic-2". Thanks for making the video. They did it very well. Almost a million views. Thank you!
    3
  2783. Thank you very much!!! I wish you long life!❤
    3
  2784. I never understood how this could be considered a paradox (apparent or otherwise). Finite area under an infinitely long curve in the plane is completely analogous, but we take those for granted. (Or a 2-D 1/x^2 horn that could fit inside or outside one of Oresme's towers.) Fun informative video from Mathologer nevertheless, as always!! ❤
    3
  2785. Excellent video as always! Thanks for your work!
    3
  2786. @Mathologer Concerning the higher dimensions: I always assumed that, say, in 5D you can construct a regular pentagon by taking points (1,0,0,0,0) and (0,1,0,0,0) ... and (0,0,0,0,1). Arguably, these five points are very regularly distributed, forming (seemingly) something 5-fold rotationally symmetric around axis (1,1,1,1,1). But your proof suggests, that this is not a pentagon. So what is this? Do these five points not lie in a 2D plane? (Very nice video btw, I am huge fan!)
    3
  2787.  @bzduso  oh yes, that's right! thank you. and you are right about what you said too (at least i would say haha). hmm so let's see. we should construct FOUR vectors instead of 5. and that by substracting the first one from remaining 4 and then we will get a matrix 5x4 with rank 4. so you are right!
    3
  2788. Peter Steinbach took the Golden Ratio from 2 to 3 dimensions. Does anyone know if there is also something analogous in four dimensions?
    3
  2789. For me, the Dots 😮
    3
  2790. Where can I buy an Algebra Autopilot?
    3
  2791. Too late to watch tonight. Will finish in the morning.
    3
  2792. This is shockingly beautiful
    3
  2793. I came up with this just before seeing your comment : r2-1 by 1-(r2-1) makes r2-1 by 2-r2 for the small rectangle which compare to 1 by r2. So (r2-1)/1 = (2-r2)/r2 <=> r2(r2-1) = 2-r2.
    3
  2794. Tristan's Fractal was absolutely gorgeous, I was really inspired by it, it was my favourite!
    3
  2795. solution to the ending puzzle (49:52) (SPOILERS!!!) the solution as indicated by others is the odd-indexed fibonnaci numbers if you count f(0) = f(1) = 1. with f(n) being the n-th fibonacci number. here is my proof: first ill mark p(n) = the n-th term in this series, p(1) = 1. and for convenience, p(0) = 1 as well. as shown in the beginning of the video, every way to break up n into a sum of integers can be expressed as n-1 "spaces" between n boxes, when each way of closing some of the spaces is a unique integer sum of n. this means that you can express every integer sum of n with a vector with n-1 terms, all either 0 or 1 . example: take n=5 and the vector 1101. this represents 3 + 2 = 5. now i will organize the vectors in a row by their values in binary. meaning that for example for n=3, the 4 vectors will be ordered as such: 00 01 10 11 and the corresponding sums like this: 1+1+1 = 3 1+2 = 3 2+1 = 3 3 = 3 now for n, there will be 2^(n-1) rows in this kind of ordering. the first half of the rows all have 0 as their left most term, meaning all sums look like 1+ integer sum of n-1. and since the rest of the vector goes from 00...0 to 11...1 this covers every integer sum of n-1. meaning that in the first half of the rows there are exactly, all integer sums of n-1 with an added 1 on the left. that 1 does'nt change the product of the rest of the numbers, hence the sum of the product of each row is just p(n-1). now ill look at the second part, and again at its first half. this time all vectors start with a 1,0 and the rest corresponds to an integer sum of n-2. now every row looks like 2+ integer sum of n-2. meaning that the sum of the product of those rows is 2*p(n-2). continuing like this by induction we get: P(n) = sum from k=1 to n-1 of k*p(n-k)+n , p(1) = 1. first, f(2*1-1) = f(1) = 1. ill now show that p(n) = f(2n-1) satisfies this recurrence relation. meaning that i need to show that: f(2n-1) = [sum from k=1 to n-1 of k*f(2(n-k)-1)] +n proof by induction: (n=2) base: [sum from k=1 to 1 of k*f(2(2-k)-1) ]+2 = f(1) +2 = 1+2 = 3 = f(3). step : ill assume the reccurence holds up to n. now ill show it for n+1: first, f(2n) = f(2n-1) + f(2n-2) = f(2n-1) + f(2n-3) +f(2n-4) = ... = f(2n-1) + f(2n-3) +...+ f(1) +f(0). (*) now : f(2n+1) = f(2n) + f(2n-1) = = f(2n-1) + f(2n-3) + f(2n-5) +...+ f(1) + 1 by (*) + f(2n-3) + 2f(2n-5) +...+ (n-1)*f(1) + n by induction. = f(2n-1) + 2f(2n-3) + 3f(2n-5) +...+ n*f(1) +n+1 = [sum from k = 1 to n of k*f(2(n+1-k)-1)] + n+1. hence for every n, p(n) = f(2n-1).
    3
  2796. I would love to see a fluid dynamics simulation of the claw opening and closing. What happens to the fluid inside it?
    3
  2797. The geometric flower, beautiful.
    3
  2798.  @Mathologer  another point on Steiner ellipse. A Steiner ellipse can be generated from any affine transformation of a regular or star polygon, and the affine polygon can be generated from just two components of the decomposition. For example an affine pentagram can be created by applying the reverse DFT to the two regular pentagrams in your video.
    3
  2799. Burkard, the underlying maths here is identical to Symmetrical Sequence Component theory that was discovered by the Canadian mathematician Charles LeGeyt Fortescue. The mathematics of Symmetrical Sequence Components is used in electrical power systems engineering. Fortescue came up with his theory that states that any 'N' phase power system can be represented as an equivalent 'N' Symmetrical Sequence Components. Since three phase is the most common multi-phase system then 'N' is equal to three. In a three-phase system any imbalanced three-phase quantity can be represented as the sum of a positive, a negative, and a zero sequence component.
    3
  2800. 15TH! Thanks for the deep math videos Mr Mathologer!
    3
  2801. 14 😅
    3
  2802. ​ @shoam2103  No, it counts as you, looking for something to call click bait.
    3
  2803. i would like to know the names of your music choices, they are nice
    3
  2804. They are "discovering" the things.
    3
  2805. Send me link please I would like to read as well
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  2806. Hey, thanks for hearting another comment of mine. I am only in high school, but I’m thinking that I might want to be a mathematician one day. How can I go about emailing you about a mathematical phenomenon I found. While messing around, I found an infinite series that converges to 2e and I don’t know why. Thank you!
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  2807. Where can you submit solutions to the challenge?
    3
  2808. 7 millions can't be wrong right ? Me : proceeds to disclose the election results of any country with more than 50 million inhabitants xD
    3
  2809. FIRST!
    3
  2810. It is indeed!
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  2811. 3blue1brown, mathologer and standup maths are my favourite maths channels. What are yours?
    3
  2812. The idea of Navagraha (nine planets), distance between sun and earth, gravity, value of pi, calculus, archedes principle, decimal system,binary system, principles of accounting etc etc were known to Indians 1000s of years before their European counterparts appropriated these ideas. Please do a series on it. It'll be a big series and it'll greatly catapult the view count of your channel
    3
  2813. It's all about the base.
    3
  2814. .
    3
  2815. I loved this even though I have no mathematical background besides a great interest in these topics!
    3
  2816. As a sneakerhead and mathematics student I greatly appreciate this.
    3
  2817. Who is depicted in the thumbnail? Shouldn’t it be Sophie Germain?
    3
  2818.  @Tumbolisu  I know, just trying to make the point a different way. I don't think they'd treat Newton that way.
    3
  2819. The most memorable part for me was the proof by the bishop 700 years ago. And I'm looking forward to the German Mathologer video you once promiesed ;-)
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  2820. A very pretty proof! The 2D case is also a corollary of Pick's theorem.
    3
  2821. Very Great! Indian Mathematicians are getting the attention they deserve
    3
  2822. In Rabbinic Judaism, the Torah has rules about civil jurisprudence as well as ritual observance. The Talmud records discussions about all of these topics.
    3
  2823. Secant lines work inside, or tangent lines outside.
    3
  2824. I'm not much of a mathematician myself, but I find that what in the so called Dunning-Kruger graph is called "the peak of ignorance" at least to a reasonably satisfying degree describes the reasons for the WOW feeling that many get when confronted with cool tricks from the world of number play. There is always "more" ... right?
    3
  2825. Finally someone said it.
    3
  2826. Can the coin paradox solution be used to predict the number of days a planet will have? What happens if the center of rotation is between the two centers of the circles?
    3
  2827. Four minutes in and this feels like a slide rule.
    3
  2828. this guy is such a legit lecturer thanks alot
    3
  2829. I see a similarity with this: The intergal of 2x is x^2 (original theorem, aka "taking the sum of numbers skipping one gives you the squares)
    3
  2830. 17:18 I rather think this construction works for all n ≠ 3, 4, 6. As I see it, we are constructing a non-degenerate star n-gon with side length equal to that of the original polygon. Taking the convex hull of the star gives a smaller n-gon, and we are done. A non-degenerate star has Schläfli symbol {n/q}, where 2 ≤ q ≤ n-2 and gcd(n, q)=1, so the construction can work as long as such a q exists. (To do the octagon for example, we construct an {8/3} star.) If such a q does not exist, we must have (phi() being the Euler totient): phi(n) = 2 <=> n = 3, 4, 6.
    3
  2831. well... for me it is 5. 2 on the equateur then 5 is 3+2 i les alone the 2 and put the 5 others together. How do u get 6 ?
    3
  2832. After having spent years of my life watching math videos like this one, I've concluded that math only has 3 original ideas and the entire field is just their remixes.
    3
  2833. I still have my circular slide rule that comes with a periodic table and all kinds of physical constants, formulas, math identities and such.
    3
  2834. Math would be scary thing if mathaloger would not be there .😌
    3
  2835. Brahmgupta's formula for area of cyclic quad
    3
  2836. Thank you for your wonderful, clear and fascinating methods for solving polynomials. Inspiring!
    3
  2837. I'm going to paste a response I gave to someone else asking the same question earlier: There's certainly nothing special about 2 just from the perspective of saying 2 is the only even prime. You're absolutely right about that. But 2 does behave differently from a lot of other primes, probably because it's simply too small for a lot of the results for other primes to hold. For example, in this situation, all odd primes can be written in the form 4k+1 or 4k+3 and the "+1" and "+3" is the critical information to determine whether it can be written as a sum of squares. On the other hand, 2 = 4k+2, so the "remainder after division by 4" information is different from any other prime. There are plenty of theorems in mathematics that work only for odd primes and some that work only for primes greater than 3. The reason for the exclusion of 2 (and possibly 3) in these cases is often dependent on the specific details of the theorem and/or proof. And usually, the reason 2 (and 3) work differently in those cases is because 2 (and 3) are just too small to behave like every other prime does for that specific problem :)
    3
  2838. i spent way too much time doodling in class on my square grid paper to construct a 60deg angle using just the grid and a straight edge. Though fruitless, it was a fun exercise
    3
  2839. I almost didn’t make it to the very end. The credits made me tap away. But then I came back and I'm very glad I did and made it to the very end.
    3
  2840. I was taught this formula and found it to be a hidden gem for many problems. Unfortunately, I cannot remember whether we proved it or not, but we did prove most things we have learned.
    3
  2841. I thought somebody added that design to the seashell by hand. First time I've seen that.
    3
  2842. Holy cow, the pythagoras fibonacci thing works for other variations too! Look! If you start the fibonacci sequence with 3 and 4 and check the 29, 18, 11, 7 part it makes 203^2 + 396^2 = 445^2 which checks out.
    3
  2843. For math noobs like me I'd enjoy a video with any tricks about the even squares.
    3
  2844. 25:55 No, not full credit yet. You have to show that removing common numbers doesn't make any empty sums. Not difficult, but possibly important enough to cost you the full score on the IMO.
    3
  2845. He demystifies math like no one. Magnificent.
    3
  2846. I still have my old Faber-Castell slide rule I used in the 6th and 7th form at school in the late 1960s. I also have my father's engineering slide rule which is about 600mm long and has functions on it I'm not entirely sure of. I show them to young people and they are amazed. In credible to think they were used for hundreds of years and became redundant almost over night when the electronic calculator came in the late 60s, early 70s but then so did the abacus.
    3
  2847. I wonder who the hell can dislike such videos. Love your work though! Your videos are an inspiration to me sir🙏🙏🙏🙏
    3
  2848. South Indians are smart and I am one of them
    3
  2849. This is perhaps my favorite Mathologer video ever. And for me the existence and derivation of 𝛾 is the best most amazing fact. Also this video was really fun to watch while slightly inebriated. 😁
    3
  2850. Most memorable: how the sum of any number of consecutive reciprocals is never an integer.
    3
  2851. I've always been weak in Geometry, and I won't pretend to have a ready sense of the correctness of what you're presenting even with the help of your excellent graphics. But, even so, I can tell that you really do the graphics well, and I'm a bit envious of those for whom the graphic presentation is intuitive.
    3
  2852. It looks a bit like the projection of a rotating teseract.
    3
  2853. 🥇🥇🥇
    3
  2854. Great stuff as usual!
    3
  2855. I'll probably never be able to say which of your videos is my favorite but if I had to try, this one would have to be on the list.
    3
  2856. I’ve always been told sinh was pronounced like “cinch” (or “sinch”) and I’ve never heard it called shine before but honestly I like shine better
    3
  2857. Wow, this is so fascinating. To answer your challenge 9,4,13,17 and from the root you gotta go right 3 times then left once. Previous group is 1,4,5,9. Had so much fun!
    3
  2858. Shoot, the quickest I've caught a mathologer.
    3
  2859. 32:25 my calculator says: 6859356963880484413875401302176431788073214234535725264860437720157972142108894511264898366145528622543082646626140527097739556699078708088.000… (and many many zeros after) which is amazing 😁
    3
  2860. Anyway, here's Number Walls.
    3
  2861. Solution to Tristan's challenge: the new loops are just all the numbers from the previous loops, just (in the case shown) multiplied by 3. The loop obviously holds from this, given that ca mod cb = c(a mod b): if a = kb + d where d < b: ca = kcb + dc, which has a factor cb and a part dc which is less than cb, as d<b and c=c.
    3
  2862. 14:40 That's the first thing I thought of. The silhouette of a cube viewed at an angle is a hexagon. So, if you were to squash down a cube from opposite angles, you'd get a hexagon and equilateral triangles. That'd kinda be cheating but it's probably considered true in a non-euclidean way.
    3
  2863. That last geomagic square has some nice structure build in. Number one is of course just a little square, but from that point on you just build towers of three next to each other as the corresponding number increases. To me that seems so elegant - the Mother of all geomagic squares almost!
    3
  2864. Calculus is easy. It’s algebra that’s gets you while calculating.
    3
  2865. It's resemble building protein chains!
    3
  2866. Also, as an aside, I was very surprised you didn’t talk about these tricks as “conservation rules”. They provide a beautiful connection to Noether’s theorem via that interpretation.
    3
  2867. "Explodes to infinity" More like crawls to infinity.
    3
  2868. Made it to the end. You need to change your artist, whoever drew the oicture at 3:24 has no talent.
    3
  2869. First comment
    3
  2870. Euler mascheroni constant being greater than 0.5 can be seen in the figure at 25:13 by noticing that the blue convex triangles contain straight triangles whose area sum to 1/2. Great video Mathologer.😉
    3
  2871. 9:24 Ok, that got me off guard 😆
    3
  2872. The guy is so enthusiastic that you want to nod your head in understanding. But the fact of the matter is that his teaching takes so many fundamental concepts for granted that you really have to understand in your bones so much. I hope there is another video that does not assume so much in teaching calculus. If you are not fluent in speaking/thinking certain math concepts like "function" or "derivative", you will quickly become lost.
    3
  2873. 11:47 My logs aren't rusty at all, since wood doesn't rust
    3
  2874.  @Mathologer  Yep, I answered myself two minutes after you did, before I just saw yours. Thank you!
    3
  2875. Congrats on the centenary! Cant wait for the next 100.
    3
  2876. One important thing to mention is that for unintegrable functions you can use the numerical aproach to calculate the aproximate area under the curve between two points. For instance using just Excel.
    3
  2877. Every time I talked about Newton, I should've started with Madhava. Thanks for sharing this info.
    3
  2878. Did you also know that you can unfold a sphere's surface area into the area under a sine wave with important values that are directly related to the important values of a sphere? The amplitude should equal the circumference of the great circle of the sphere, and the period should be the diametre.
    3
  2879. I made it to the very end. Also, seen the video 3 times to get the tricky parts, very nice once you get the rhythm :)
    3
  2880. The pattern at 17:27 of the path following all Hanoi states once reminds me a lot of the Hilbert curve for space-filling a square grid. I'm interested in what the path for all solutions of the 4 peg version of the Hanoi puzzle is going to look like.
    3
  2881. Glad u recognised the Kerala mathamatcian. He has a great importance in the qorld of mathematics
    3
  2882. The music in this video is great, and also the video is great.
    3
  2883. This is called the Gnomon Conquest.
    3
  2884. The maximum for any number of cards n should be (n+1) * n/2 I would prove why this is true, but a proof would be too long to fit in this comment section
    3
  2885. I refuse to believe that the ‘exactly 100 zeros series’ is larger than the ‘no 9 series’ (40:51). I mean... what?!
    3
  2886. Hi
    3
  2887. I initially read what is written on his t-shirt as MÆTH...🤭😌
    3
  2888. Thanks for monitoring the comments so assiduously.
    3
  2889. Using a short java program I wrote it looks like once you get past the initial 13 terms to overshoot pi and subtract it takes between 8 and 9 additional terms to overshoot pi again and then you subtract one term. Additionally, it alternates adding 8 and 9 terms before you subtract one term (you always only need to subtract 1 term) but eventually theres 2 9 terms intervals and the pattern continues. What is interesting is the number of "intervals of intervals" aka the number of times this alternating takes place before you get 2 9's varies initially but settles on the pattern 37, 39,39,39, then back to 37. However, every now and then theres 4 39's before going back to 37. It looks like this pattern has a pattern as well. I suspect these inner patterns continue on due to the irrationality of pi and there is never a "straightening out" of the pattern. Edit: I am trying out different values for what you want the series to converge to and it's pretty interesting. First of all, for every number you only ever need to subtract one term to get the current value to go below the desired value. Every integer you try the number of terms before it overshoots generates a pattern with little deviations in it. Im thinking the pattern of the little deviations have their own patterns, and this extends infinitely. Is anyone aware of someone coming across this and looking into it more? I'm thinking it would be fascinating to generate a "3d graph" of desired values vs how the series behaves but I need to get to my computer to try this out. I'll let you guys know what I find
    3
  2890. To complete the proof you need to show that the second bug is moving orthogonal to the first (which follows from the symmetry of the setup)
    3
  2891. At 12:29 the square on the right with the "6" on it disappears
    3
  2892. Holy macaroni :0 I am simply in shock-
    3
  2893. I used to love numberphile, I still do, but you are my favorite maths professor
    3
  2894. 8:55 There's a town called Walhalla in Australia. 😮
    3
  2895. Thanks for the peek at the original Indian manuscript. It has a certain calligraphic elegance,- as if he "felt" the beauty of the mathematics embedded in it.. Any communication with an alien (as in extraterrestrial) civilization will begin with mathematics,- this certainly gives the "flavour" of what that might be like.
    3
  2896. My Favourite Profess to establish Kerala Story. He has again become famous due to his take https://www.youtube.com/live/ctrROj3Tv-E which has been cited in new mathologer video and made more famous and now after eight years of this endeavour.
    3
  2897. Unexpected symmetries are always delightful.
    3
  2898. Thank you for making math visually interesting.
    3
  2899. 18:55 - bit of a blooper by you, you forgot to hit row 2 column 3 with a dot! Also, the pattern is “If you can, move right 1 and up 1 (modulo 5), else move down 1”
    3
  2900. No, it just sat in my queue for 2 weeks. Quarantine has upped my YouTube binge game...
    3
  2901. Are you going to make Right Triangle Fermat's theorem?
    3
  2902. I came up with an even simpler visual proof. Take a cube of side length x+2. This cube has a volume (x+2)^3. Now, slice the cube six times. Each slicing plane is parallel to a face and 1 unit deeper than the face. Don't throw away any volume. What you're left with is an inner cube of side length x (volume x^3), 6 square pieces of volume x^2, 12 edge pieces of volume x, and 8 corner cubes with volume 1 each. Adding up these volumes gives you the original (x+2)^3 volume, so it's proven. This works in any dimension.
    3
  2903.  @hydra147147  I was just about asking about the clepsydra shape made of either a quadrilateral or (let me count in memory: one, two, three...), a pentagon (or a hexagon, etc.). ThnX!
    3
  2904. It heartening, hilarious, and humbling to see the various people reporting "I discovered this when I was much younger" (or a variation on it). I am fascinated by math and terrible at it. So I was 26, did everything by hand without a calculator (not because I'm hardcore, but because that was all I had), and twice ran into worsening and worsening polynomials to try to deal with, before I finally came up with a one-line formula that spat out the polynomial for a sequence of numbers. In this attempt, I started over twice and I don't know what I did differently the third time (when I say I started over twice, I mean over months of hand calculations). I certainly had no names for the sums of sums, but I did see Pascal's triangle buried in it, was multiplying the falling "n choose k" statements by coefficients derived from differences of differences. I was using horizontal rows, rather than diagonals though. Watching the video, I can't put together the relationship between what's happening here and what I did exactly; this is obviously much more straightforward and sensible, but both end with elegantly simple statements. I have a really ugly Excel spread sheet (the "function finder") that crunches a list of numbers into a polynomial. The background calculations are just as ugly as my stumbling attempts, but it gives you the correct polynomial. Just to show the prettiness, if you have a sequence of four numbers [A, B, C, D] that generates a degree 3 polynomial (variable n>3), the not-simplified formula that kicks out is (I think): [-(n)(n-1)(n-2)(n-3)]/3! * [A/(n) - 3B/(n-1) + 3C(n-2) - D/(n-3)]
    3
  2905.  @captainunicode  Out of curiosity what's the answer?
    3
  2906. 4:30 I'd say that the mistake is that chinese numerals are not directly positional, you need to add the appropiate symbol to mark the tens, hundreds, thousands and beyond, but given that the 1080 at the top of the picture is perfectly correct (it does have a 1 before the 1,000, which is not done usually but is done in places where there is a need for accuracy) and knowing that there are indeed times where they are used with no position markers in places that look to keep a "traditional" ambient (mainly in restaurant menus, at least outside of China, although I don't know how modern the practice is), they may be using a positional system as a kind of shorthand. (EDIT: 7:45 shows the tens with the marker, so the theory is quite likely, although I don't know if it's anachronistical or not.)
    3
  2907. A new wonderful video the same week we got a new wonderful president ❤
    3
  2908. Mathologer, I'm a second year college student studying applied and computational mathematics currently taking a course in group, ring, and field theory. I'm not gonna lie, I understood very little of this video 😅. But, I still definitely learned something and I hope I can pick more up on my second (and maybe third!) viewing. Your videos are incredibly educational and enjoyable. Thank you for all you do.
    3
  2909. Ok. Im am a huge fan of k-dramas. And now you gave me a double reason to watch this one.
    3
  2910. The magic square proof is brilliant! To generalize it, i turned the tiles' numbers into coordinates for a n*n square, T=x+(y-1)*n, 1<=x, y<=n, T->(x,y) is a one-to-one map therefor we have (1,1) to (n,n). 1->(1,1), 2->(2,1) etc. arrange the tiles into the big tilted square as the video shows, for every tile T, the only tiles that share a common x or y coordinate are the ones that share the same diagonal with T Thus, every tile on the same horizontal/vertical with T must have different x and y coordinate (that is the first part) Now we define horizontal and vertical distance, which is the number of horizontal and vertical steps required to move a tile to another square (empty or another tile) As the video says, when creating the magic square, all tiles outside moves exactly n steps to it's destination, which gives us a distance (horizontal or vertical) of n However, as there are only n squares on a diagonal, the maximum (horizontal or vertical) distance between T and any tiles that share the same x or y coordinate <=n-1 (that is the second part) Thus, on the final magic square, all tiles on the same horizontal (vertical) with T must have different x and y coordinates this holds for every tile T so every horizontal and vertical sum= n(n+1)/2+(n(n+1)/2-1)*n=n(n^2+1)/2 which is correct the diagonal sums are n*(1+n)/2 (same x coordinate (1+n)/2) + (n(n+1)/2-1)*n (y coordinate 1-n) and n(n+1)/2 (x coordinate 1-n) + n*((1+n)/2-1)*n) (same y coordinate (1+n)/2) this proves that it is a magic square My wording is really bad, but this is the proof in the video
    3
  2911. 11:02 "The proof is elementary." Aha! A true mathematician!
    3
  2912. 🥽
    3
  2913. 25.8069 is actually only approximately the root of evel ;-)
    3
  2914. i discovered almost all of that myself in secondary, excepted for the bernouilli numbers part.
    3
  2915. I got to believe every astronaut on the Apollo missions earned their academic degrees using slide rules.
    3
  2916. Mind Blowing information, love this explanations and love the giggles.... thanks for making it easy to understand.
    3
  2917. Among other things, your videos prove the beauty and elegance of mathematics.
    3
  2918. This episodes shows again that no one can escape Euler in math.
    3
  2919. The pink identity can also be demonstrated as follows. Looking at the position of the cards after the permutation, each card makes an inversion with all other cards that are either higher and to the right or lower and to the left of that card. Such cards form smaller rectangles of cards that go either up to the top right corner or down to the bottom left corner. Counting all cards in all these smaller rectangles, we would count every inversion twice, so we only consider say the top right rectangles. To get the total number of inversions we sum up the number of cards in all possible smaller top right rectangles, which is Sum{i=1..p-1; j=1..q-1} i.j = Sum{i=1..p-1} i.(q.(q-1)/2) = (p.(p-1)/2)(q.(q-1)/2).
    3
  2920. 35:30 Bugs start being distance 1 from their target. From point of view of one of the bugs, you always travel towards your target, but your target always tries to travels perpendicularly around you. Always momentarily traveling perpendicularly is circular motion, so your target always tries to move in a circle around you, meaning they don't actually try to get further or closer to you.
    3
  2921. I love Zahlentheorie :)
    3
  2922. Finally something good with a Peter term. It's not Petering out.
    3
  2923. He was born into a FAILED SOCIETY...and thats tragic!!
    3
  2924. Well now, the outro music was absolutely excellent. 💪🏻💪🏻 As always, the video content is outstanding too. 👌🏻👌🏻👌🏻
    3
  2925. @Mathologer: I tried with 9 colors. Each color has a number from 1 to 9. The result of the addition of 2 colors is determined by the addition of the digit until the result is a 1 digit number. For example color 3 + color 8 = color 2. Then I tried the shortcut you proved by applying the rule 9^n+1 to find the row of the triangle but it did not work. It works for length 2 for sure but not for length 10 and 82. Do I make a mistake or the shortcut just work with 2 or 3 items ?
    3
  2926. ​ @Mathologer I found the references from 2001: https://cms.math.ca/crux/v27/n3/CRUXv27n3.pdf (pages 204-205) and https://www.macalester.edu/mscs/studentopportunities/competitions/konhauserproblemfest/kp2001/ (Problem 3)
    3
  2927. Thanks! Truly mesmerizing and super engaging :)
    3
  2928. 11:54 A single glance reminded me of 17^2 - 2(12)^2 = 1, the flashbacks on the marching squares video hitting me like a truck. It reminded me of the approximated hypotenuse of isoceles triangles, and I suddenly feel like I know where the video's headed
    3
  2929. I was aware of this fact but never looked into much deeply. Thank you for addressing it.🙏🏼 Knowledge knows no boundaries.
    3
  2930. Very good you explained it in a way other people can understand. 👍
    3
  2931. So for the solitions that work, at the infinite iteration the curve is continuous everywhere and also differentiable (hence why they are smooth). But for curves that are not differentiable (hence the buckling), the approximation fails? In fact now that I think of it, they're examples of curves that are continuous everywhere but differentiable nowhere.
    3
  2932. Sir, please make a vedio on Fermat last theorem, i am from India🙏🙏🙏🙏🙏
    3
  2933. I get so many german science channels recommended to me now. I love it. I'm studying in the field pf socio-political sciences so it's fun to get more math input for laymens like me!
    3
  2934. At the moment I am writing this comment, I am only 2:26 into the video, but I would like to share an question/curiosity of mine, what happens when we workout with negative exponents? What kind of weird patterns can we find? Does that even make sense? I don’t know, I only thought it would be cool to share this question, and maybe some of you, just like me, might get a note book and try to experiment yourselves! :)
    3
  2935. 6:30 can it be 4k - 1 instead of +3?
    3
  2936. 31:54 I am still here, and it is very good. Also, this part reminds me of Shapley value in cost-sharing problems, is that related?
    3
  2937. Can a number really be "the most" irrational number? Isn't it a binary choice of irrational or rational? It is like putting 21 pregnant women in a line and declaring one of them "the most" pregnant.
    3
  2938. It is the least able to be approximated by a rational number
    3
  2939. 7:20 Expected a secret message but all I got was scripted.
    3
  2940. Beautiful 😊 what a mind Ramanujan had
    3
  2941. I have paused the video at 14:03, but I believe I can solve the programming challenge for any number at least up to 2 googol. Producing the whole sequence is of course not feasible but just the result for X for any number X < 2 * googol * googol, I think I can do. Would you be interested in that or was the point that you don't get very far if you go through the sequence from the beginning? Edit: After watching the rest of the video, I guess the general formula for k could simply be implemented for any number. But I had something different in mind, similar to fast matrix exponentiation.
    3
  2942. I think a part 2 would be great. Geomagic squares. Self tiling, where all of the pieces combine to make larger versions of a single pieces and rearrange the pieces and get any of the other pieces. And....... Geomagic squares that are fractal. With any row combining to make the target shape but also all of the pieces combning to make the same shape but bigger.
    3
  2943. Your tee-shirt shows sound wave harmonics. Will you be calculating the sound of Torricelli's horn in the next video? 🎺😆 If not you, who will?!
    3
  2944. Originally,Satan was not associated with the 5-point star; only Pythagoras. Wo sind sie Doktor Faust?
    3
  2945. For real. Great way to show so many topics in this one. The diff eq. in here was great and so much more accessible. Now I get what it can do a lot more clearly.
    3
  2946. Amazaing visualisation and explanation! You do great job!
    3
  2947. My teacher never explained why we use generating functions now i understood
    3
  2948. "that's probably enough new stuff for today", he said. I glance up at my brains dripping from ceiling.
    3
  2949.  @tridivsharma2342  There are several math youtubers with multiple integration videos: blackpenredpen, flammablemaths, Dr.Peyam and Michael Penn are some examples. Many of these techniques are pretty specific to a certain equation and thus limited in use.
    3
  2950. 😺1/2+1/4+1/8+1/16......=1😽
    3
  2951. For every tilted square draw the smallest aligned square that touches the tilted one from outside. Now take the top-left corner of the aligned square and the top corner of the tilted square and draw an aligned square whose two upper corners are these. We can see how the final, small aligned square is unique for the beginning tilted square. But from the other side, the tilted square is not unique. There are more tilted squares that will produce the same small aligned square. But when we have both small and big aligned squares drawn, then we can reconstruct the tilted one. So for unique tilted square we need a small aligned square and the difference of side lengths of the smaller and bigger aligned squares. We now organize the counting by that difference. For difference 1 we can take any aligned square of full nxn grid, for difference 2 we can take any aligned square of top-left (n-1)x(n-1) subgrid, and so on... So the number of tilted squares is the sum of aligned squares for all sizes from 1 to n. AND I DON'T WANT TO FINISH THE EXCERISE HAHA
    3
  2952. This was very educating. I found something out you maybe do not know, did you know that if two integer fractions are the same, the product is an integer square(a/b = c/d than it is always an integer square: (a*d)*(b*c)) ? Also I found out that difference of squares can be done with rectangle's instead of squares, when the factor pair difference is the same of the rectangle. For example (a*(a+c))-(b*(b+c)) = (a-b)*(a+b+c) rectangle 1. factor pair difference is c: (a*(a+c)) rectangle 2. factor pair difference is c: (b*(b+c)) Than factor 1 is (a-b) and factor 2 is (a+b+c)
    3
  2953. congrats ! :D
    3
  2954. What are the odds? I was giving a lecture today on geometry to some architecture students which included a discussion about Euler's formula. Thanks for the great content. I told my class that they are to watch your video for homework/fun! Still by far my favourite channel on youtube.
    3
  2955. Funniest ever from Mathologer I think 🤣🤣 @mathologer Great job 😂
    3
  2956. Just amazing sir. I spent three months in my collage days on this series so I just love all that what you are saying here ......
    3
  2957. To echo everyone else: amazing videos. Thanks so much for making them I think they will last down the ages, like the Feynman lectures books.
    3
  2958. 1:50 right? WRONG!! *Vsacuce music intensifies
    3
  2959. At about 26:40 you talk about treating each creditor equally vs treating each dollar equally. In a world where people can buy and sell debt, you have no choice but to treat each dollar equally. Otherwise, the bank would benefit from official having the loan come from 10 (or even a million) independent companies instead. Companies are really cheap to set up. (Or the other way 'round, debtors can cooperate to consolidate debt, if it benefits then to have fewer people behind the claims. I think that's the case for over 50% recovery with the Talmud rule.)
    3
  2960. Student: Calculas from Where does it come in this? Sir: Randomly from nothing
    3
  2961. haha, cool T-shirt 😀
    3
  2962. You really are the GOAT of mathematics youtube. Fantastic as always!
    3
  2963. a nice way of showing the row permutations are "identical" modulo cyclic ordering is to, say, subtract 5 from the second row and see it immediately with one own eyes ;D
    3
  2964. 19:00 I was starting at this plot, after you mentioned the paper on eigenpolygons, when it suddenly hit me that's a discrete Fourier transform!! of course! the degenerate pentagon point is the constant term, and each of the others is a constant (size and angle) times the phase e^2πik/5 and adding the ears is a linear operation, so it plays nicely with the DFT, and had the property that it knocks out the pentagon with phase/angle opposite to the ear!
    3
  2965. Yes sir
    3
  2966. Great effort , thank you
    3
  2967. Unbelievable :O
    3
  2968. I really felt mindblown by the new perspective offered by this video. Feel like I acquired a new superpower with what you have shown.
    3
  2969. Stopped at 18 minutes and started over. Just saying, it's interesting but complicated.
    3
  2970.  @byornski  That is not the case. At 25:00 he shows that there are must be at least two different sub-collections of the group of 1022 that must fit into the pigeon-holes. After you find two such collections you can just cancel out the values that appear in both collections, leaving you with two disjoint, but non-empty collections that sum to the same value
    3
  2971. The first dislike will get cursed by a giant circle.
    3
  2972. What’s the shirt mean/reference
    3
  2973.  @Mathologer  Hey! So was my uncle!
    3
  2974. Even in Egypt civilization they where using 25/8 as π approximation at the era of the great Pyramid
    3
  2975. My guess is it's because the star is regular: The tips on the star are equally spaced, so the distance from the first tip to the second tip is the same as from the second to the third, and from the third to the fourth. (And because it's a {7/3} star, the fourth picks up the same point as the first, and so on.) The distances between picked-up points are the same during one full rotation, so the points have to form an equilateral triangle. Not sure if that was rigorous, and it still kind of leaves open the question why the star has to be regular.
    3
  2976. Because it's valuable and worthwhile. Don't expect anything of value to be even "talked" about at a public school.
    3
  2977.  @ffggddss  I really like that argument. I originally encountered the problem when the bugs were in an equilateral triangle, so I not think that the argument applies to other regular polygons. A fun exercise is to use the method mentioned above to prove that the distance traveled by a bug starting in a regular n-gon configuration with side length 1 is 0.5*sec^2(90(n-2)/n). Another interesting thing to look into is other arbitrary starting configurations of the bugs.
    3
  2978. At 20.40 a leading factor 1 in the numerator has just been suppressed. Pop it back in and the product inside the red box becomes (1/1)(2/3)(4/5)…..Each term is less than 1 so the product is convergent.
    3
  2979. The fulfilment i draw from this is beyond comprehension!
    3
  2980.  @QuantumHistorian  I can agree with your point, my comment was more about my immediate reaction to seeing AI art. Clearly the animations in this video are not emulating anyone's style specifically, however, I explain my point more thoroughly in my reply to the other person (above this one).
    3
  2981. Reminds me some work I've done with Phi as a numerical base :D The dimensionality expansion of all this is... staggeringly amazing !
    3
  2982. Hmmm ... this seems almost too useful and simple at the same time. I wonder if it's so simple that it could be implemented on actual silicone hardware and make some operations faster.
    3
  2983. Top shelf stuff Burkard. Extending the Fib S "leftwards" we obtain: ...-8,5,-3,2,-1,1,0,1,1,2,3,5,8... and note that just as F(n)/F(n-1)->φ for n->+∞ it also->-1/φ for n->-∞.
    3
  2984. As a non-native English speaker, I sometimes find it hard to follow when your voice goes low. I hesitated to mention this, but I hope this feedback helps you reach an even wider audience. Your content is truly marvelous! Thanks for your generosity in sharing with us the results of such a hard work of yours!! ♥
    3
  2985. A slide rule for addition is easy: just slide two rulers (with the same scale, obviously) next to each other. Exponentiation is harder to explain, but my dad’s old school slide rule has scales for exponentiation on the back of the tongue, so it’s obviously possible.
    3
  2986. Dream solutions are extremely powerful and convincing until fully awake. The same brain that is totally unhinged when devising a dream solution keeps the problem intact as it is posed in the awake state which is fascinating.
    3
  2987. Thank you for this gem. You're right - I think it should be taught! At least in Greek schools!! 😃The most important formula taught in school (after Pythagora's theorem and the area of a circle) is the solution to the quadratic but I have lost count of how many times I wanted to quickly calculate the area of a triangle and in the end I stopped bothering. I wish I had learnt this sooner. Btw, can something be done about this bright white background? It's really painful for the eyes - literally, not aesthetically.
    3
  2988. When I first heard the idea of magic squares, my initial thought was "no way! that's way overconstrained!" (though I wouldn't have used that word.) I thought sure you could get all the rows, or all the columns, but not both and definitely not the diagonals too. Then you learn more about how they work and how to make them and it becomes "well, of course! how could it be any other way?"
    3
  2989. He meant that those are the only numbers less than 100, then the rule goes on.
    3
  2990. The "windscreen wiper" method can be used to prove the same result on a sphere, which by extension also proves the 2D case as well. 1)Draw a great circle around a sphere, marking two points at random on this circle. 2) Mark a third point at random, but not on the same great circle, and connect this to the first two points around two new great circles. This creates a spherical triangle. 3)Rotate the sphere so that the "north pole" lies within the triangle in such a way that the northernmost points on the three great circles all lie at the same latitude. 4)Draw the "whiskers" by extending the sides of the triangle as shown in the video, following the great circles in each case. 5)The great circles all have exactly the same radius of curvature at every point, so the windscreen wiper method can be used to rotate and tilt any one of the extended sides around each vertex to bring it into superposition over the adjacent side. Three rotation and tilts brings a side back to its original position, but pointing the opposite direction, just as in the video. This proves that the three extended sides are the same length, and the northernmost points are the exact midpoints of each arc. 6)Starting at the northernmost point of any of the three great circles and measuring the same arc in each direction takes you to two points which lie at the same latitude. Since the three arcs are all the same length, all six end points must lie at the same latitude. Ta dah. 7)Let the radius of the sphere tend to infinity. The surface of the sphere tends to a plane surface, and the original Euclidean result follows directly. Ta dah!
    3
  2991. Still waiting for a dream proof. I have one "Jog in the forest proof" (which guaranteed a top grading for my thesis) and a half completed "drive on the highway at constant speed proof". Most important issue probably is: do something where your mind does not have a predefined job.
    3
  2992. Just in time when I'm taking a break from learning set topology😄
    3
  2993. OK, so here's my feedback after the first watch (I may add some additional points after really thinking about the ideas presented and convincing myself of some details): Overall, I found this video extremely interesting and approachable. Ever since I got interested in math, the Bernoulli numbers have felt utterly mysterious to me, and it was really nice to find out about some of their origins and the reason for their appearance in the Euler-Maclaurin formula. However, one of the things that kind of irritated me was at around half of the video's runtime, where the pattern relating the Bernoulli numbers to the sums of integer powers was established very quickly – I'll definitely have to ponder that for some time, which is why I'll try to find a proof of this fact that is accessible to laypeople like me when I have the time. I'm really excited to see your announced video on the Riemann Zeta Function and the Bernoulli numbers as well! One mystery that remains for me is the appearance of the Bernoulli numbers in certain power series – I'll also try to read up on that. I personally have no mathematics background – I'm a high school student who started getting interested in the subject around one and a half years ago and largely learns by herself through textbooks and YouTube videos.
    3
  2994. 1st one to comment 😀
    3
  2995. Great as always! A question on a detail: Why is it clear, that the sliding trick is guaranteed to give you a unique new tiling?
    3
  2996. Most memorable: The beautifully simple proof for finding center of mass for 2 weights on a scale
    3
  2997. Incredible how Pascal's turtle also happens to be a fractal
    3
  2998. The constant width shape of your video has 6 points, but in your previous video about constant width shapes, they all had an odd number of sides. Is there a way to generalize this to create more even sided shapes of constant width ?
    3
  2999. I knew it, the universe is rigged.
    3
  3000. Love this. You have a pretty awesome teaching style. Even I could follow this. Take my upvote and sub 😊
    3
  3001. Fantastic! I would love to see a mathologerisation of the pattern for the continued fraction of e. [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, . . . ] It's so beautiful and so hard to wrap my head around...
    3
  3002. For the riddle at the end: instead of rearranging triangles I came up with algebra and the cartesian coordinate system. You get the corner points of the blue area with solving linear equations like -1/2x+1=2x --> x=2/5 y=4/5 and so on. You get the difference by subracting two of these points from another and then you have data to work with Pythagoras. The width of the square is then sqrt(5)/5 and its area 1/5. ^^ Spectacular video by the way. A++ P.S. after that I came to the conclusion that the areas of the smaller triangles are 1/20 and therefore fit 4 times into the blue square. So all the trapezes together make up 3/5.
    3
  3003. Art as the inspiration for math, now that is something!
    3
  3004. Oh my god, just yesterday I was wondering exactly about that: how many "elements" (vertices, edges, ...) do n-dimensional cubes have! And now you made a video on it!
    3
  3005. Yesterday, I checked the date and it was september 5 2020.. and I said to myself, "wow time flies so fast". Today, I opened my notifications on youtube and then see a new mathologer video. I thought to myself, "wow, I never even got to live sept 6 - dec 24"....... BECAUUUUSEEE ITS CHRRRRRRRISSSSSTMAAAAAAAAS!!!!!!!!!! thank you for uploading still :'( :'D
    3
  3006. I was shaking my head with disbelief half the time while watching this video. Amazing
    3
  3007. This is the most Whaaaaaaat? thing i have seen. Thank you
    3
  3008. an extremely high effort video with great payoff. Classic Mathologer!
    3
  3009. Wwow wow amazing Thank you. I am flabbergasted. Lovely😮🤩
    3
  3010. OMG wau....❤️❤️❤️
    3
  3011. Welcome to another Mathologer video.
    3
  3012. Yeah! A new Mathologer video. And a great shirt
    3
  3013. Calculus Made Easy was my go-to book when I was learning!
    3
  3014. Nice video, as always! Really interesting topic; I once thought about the sequence difference being similar to the derivative, but I didn't formalize it. Glad to see it was a thing!
    3
  3015. Thank you for your amazing dedication to teach us. I watched the whole video even though it's midnight here in Finland.
    3
  3016. I have a feeling this will be used in a few flat earth videos!😂
    3
  3017. The result of folding a right triangle gives a triangle which has two medians perpendicular to each other. As a consequence, folding a triangle with this property (and such a triangle will not be a right triangle) gives a right triangle.
    3
  3018. I found the pattern. Each of those equations start with an nth triangular number, where n is an even whole number, and 1 is the first triangular number. Each equation has 1 more term on each aide than the previous one. For example: 21 is the 6th triangular number, so if my conjecture is true, this is also true: 24 27 Σ n² = Σ n² n=21 n=25 And since 36 is the 8th triangular number, we see another one of those equations: 40 44 Σ n² = Σ n² n=36 n=41 If you don’t know what the weird symbol Σ means, you should learn it by the time you take algebra 2.
    3
  3019. Wow this actually cleared a lot of my confusion regarding the properties of e. Thank you
    3
  3020. =SO FUTURAMA IS RIGHT=!!!!!!!!!!!!! =BEST SHOW EVER,HEHEHEHE
    3
  3021. Ah, Dynamic Programming! Excellent.
    3
  3022. Love your shirts, the animation at the end just gloriously simple. Just a math hobbyist, helps to spot patterns in nature. Cosmology my real passion. You are an excellent presenter, keep it going!
    3
  3023. Thanks a lot for this video. I don't recall another Mathologer video that so baffled me in its underlying relations.
    3
  3024. So a 1^2+0^2=1^2 triple, which corresponds to the first two terms of the Fibonacci sequence 0,1
    3
  3025. By the way, the second method you demonstrated makes use of the Pell number, which can be defined as P(0) = 0, P(1) = 1, P(k) = 2P(k-1) + P(k-2), The limit of the sequence of P(n+1)/P(n) = 1+root2 which is the silver ratio. In fact this is related to the Metallic Ratio. Metallic ratio is the root of the equation x^2 - nx - 1 = 0, if n = 1, you get the golden ratio. If n = 2, you get the silver ratio. If n = 3, you get the Bronze ratio, and so on. And the sequence of number you can obtain for each n = {1, 2, 3, 4, 5, ...} can be defined by f(0) = 0, f(1) = 1, f(k) = n.f(k-1) + f(k-2). So if n=1, then you get the Fibonacci number. If n=2, you get the Pell Number, and so on. So to get the approximation of root2, you go like this: Since x^2=nx+1, then x=n+1/x, then 1+root2 ≈ 2 + P(k-1)/P(k) or root2 ≈ 1+P(k-1)/P(k) = (P(k)+P(k-1))/P(k). For instance, the Pell sequence: 1, 2, 5, 12, 29, 70, 169, 408, ... root2 ≈ (P(7)+P(8))/P(8) = (169+408)/408 = 577/408
    3
  3026. I have seen a lot of comments before saying that they needed the video before, but I didn't expect this one to be one of those. This was the topic of my research paper for my gifted class last year, specifically the smallest amount of moves to win a k-poled Hanoi Tower, and I plan to carry on this year. I haven't finished the video, but thank you for this wonderful video anyway. Cheers from Korea! Edit: Never thought about 4-peg Hanoi Towers with the concept of superdisks. Thanks again.
    3
  3027. The pairs of integers are ordered (since we are counting points on the plane. E.g. (1,4) and (4,1) are both a distance sqrt(17) from the origin.) So 4^2 + 1^2 and the three others from their negatives count as another 4.
    3
  3028. Most memorable: definetely Tristan's Fractal! I loved its beauty
    3
  3029. 26:01 "a" and "b" are variables , diagram looks like b side is the longer one , but the growing process is the same, add a square to the longer side depending on values of "a" and "b"
    3
  3030. Most memorable: Euler is everywhere
    3
  3031. Like the video!
    3
  3032. this was a revelation..what an amazing genius Madhava of sangamagrama is !! ..the palm leaf inscription is in South Indian language script..perhaps ancient malayalam ..so he can be rightfully called the father of calculus..thanks mathlogger for bringing this gem of a video..
    3
  3033. to check how many sum=prod identities are there for a given length n, you first check what is the maximal number of non-one numbers in your identity. you do this by taking a list of n ones and seeing how many ones can you replace by twos s.t. the product is less than the sum. now you manually check cases for each number of non-1 numbers below the suprimum. so for 10 you've got 2 identities: 4,4,1,1,1,... and the trivial case 2,10,1,1,1,...
    3
  3034.  @ಭಾರತೀಯ_ನಾಗರಿಕ  on the german keyboard, the "y" is right below the "a" key
    3
  3035. I was sort of waiting for this , subhanAllah !
    3
  3036. Homework: 1- 5 2- a) {5;2} or {5;3} (they are equivalent, and proving this equivalence could have avoided introducing the coin rotation paradox) b) 2 equilateral triangles (equilateral polygon of 3 vertices) and 3 lines ("equilateral polygon" (might be abbuse of notation...) of 2 vertices) 3- For the app, I am trying to make something on geogebra (I will update this comment when I have it).
    3
  3037. Best thing was gamma - it's scary how often it pops up. Damn you Euler and your ubiquitous constants!
    3
  3038. I am bothered by the rectangle boxes being cut off the wrong way somehow... The spiral was made over the surface of the boxes and I think the spiral should instead follow the diagonal of the box's interior (A^2 + B^2 +C^2 = D^2) and therefore should spiral inwards somewhere in the interior and not end up at the surface of the initial box.... hmmmm.... 🤔
    3
  3039. 14:00 For odd number of disks, move clockwise; for even, move counter-clockwise. Easy to see. if one disk (or odd), move directly to ending peg; otherwise, not.
    3
  3040. I loved this presentation! The beauty of the trithagorean triples is that 60 and 120 degree triangles are the only ones besides right triangles that can have all rational lengths. No other angles work! no other formulas like this are possible. I do have a question about your quick generalization of GM and AM to more numbers. I think you want to change the root to a cube root when multiplying C the third number on the GM side and also dividing by 3 on the AM side when you add C, and so on as you add more numbers.
    3
  3041. Off topic, but anyone else notice a weird comment from a user with a baby's stomach for their profile picture, and in their bio they have a link for a pron website... and i saw this account commenting on several videos. What is going on with youtube, man?
    3
  3042. India has been a leader in science and technology since time immemorial. We don't know much about this partly because they were interested in neither using technology to disrupt nature nor turning their advanced knowledge into weapons to conquer other civilizations. It's sad.
    3
  3043. 17:35 I had that hypothesis as well, but never got around to proving it. It was just something I felt when I was playing around with a Rubik's Cube
    3
  3044. hes obviously only referring to finite sequences of moves. thats what we refer to as algorithims. what you describe are more abstract notions that dont apply to a construction like an algorithm. ofc an infinite sequence of moves would never return to the solved state at its completion, because it doesnt have a completion.
    3
  3045. Patterns. Animation. Understanding. One follows the other. Explanation in its purest form without danger of distraction. This is educational genius.
    3
  3046. I thought "Dudeney" was pronounced the same as "Dewdney" (A.K.), i.e. /dudni/. Or is it /djudni/? The Wikipedia article of neither of them gives a pronunciation.
    3
  3047.  @Mathologer  I'm reminded of the Mathemagician from the excellent children's allegory, The Phantom Tollbooth.
    3
  3048. To unlimited plus greater than? dafuq his shirt say?
    3
  3049. do you speak german?
    3
  3050. Yes it is a Cartesian product of three Cantor sets: all points (x,y,z) where x,y and z all lie in a different Cantor set. Rather than the standard ratio of 1/3 in the cantor set construction, the ratios vary and, well, are some other ratios that probably have nothing to do with the rest of this video. 😉 If you consider the equivalent thing with the rectangle and φ, then the cut points in each segment are 1/φ and 1/φ² , so the Cantor set is full of golden ratios. This is fun to look at in base φ.
    3
  3051. you always seem to have the best t-shirts
    3
  3052. This is the ultimate introduction to Maps Projections.
    3
  3053. I always give u a LIKE before I look your video.
    3
  3054. I thought of a 3d version of your 3d polyhedron formula proof before it was mentioned. I'm slightly proud that it's an actual proof and not just something I initially thought could work.
    3
  3055.  @SmileyMPV  my man busted out every weapon the big calculus artillery, well done
    3
  3056. 35:30 maaaan! My life was depending on that!
    3
  3057. Harvard MATH245
    3
  3058. Lovely. Great work!
    3
  3059. Loved the video as usual and enjoyed playing with the animations. Found Kieran's the easiest to control and the cleanest looking. 57 vertices with the density at 31 and very low speed produced some particularly nice effects at the centre.
    3
  3060. 24:40 its because area of the blue part at certain x can be approximated by the area triangle between x and x-1 base = 1 height = 1/(x-1) - 1/x the area of triangles = 1 * [1/(x-1) - 1/x]/2 when we sum the values of area from 2 to infinity, it becomes a telescopic series 1/2 * (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...............) which equals 0.5
    3
  3061. this super cool not gonna lie, i never thought about the golden ratio in so depth
    3
  3062. For the solution of 153^2 + 104^2 = 185^2, I've found it with some algebra... If the terms of the sequence are f1, f2, f3 and f4, we know that f3 = f1 + f2, and f4 = f2 + f3. 153 = f1 x f4 104 = 2 x f2 x f3 185 = f1 x f3 + f2 x f4 The first one gives us: 153 = f1 x (f2 + f3) = f1 x (f2 + f1 + f2) = f1^2 + 2 x f1 x f2 From the third: 185 = f1 x (f1 + f2) + f2 x (f2 + f3) = f1^2 + f1 x f2 + f2 x (f2 + f1 + f2) = f1^2 + f1 x f2 + 2 f2^2 + f1 x f2 = 153 + 2 f2^2 Then f2 = sqrt(1/2 x (185 - 153)) = 4 And f1 = sqrt(153 + f2^2) - f2 = 9 f3 = 13, f4 = 17, and finally: (9 x 17)^2 + (2 x 4 x 13)^2 = (9 x 13 + 4 x 17)^2
    3
  3063. Just had math through high school calculus, but I am rather curious about nearly everything, and the visually intuitive explanations of the underlying beauty of math you and 3blue1brown present keep me mostly hooked.
    3
  3064.  @aioia3885  I get it now. The d/delta notation is used to derive/prove the elementary derivative equations, and these equations/rules are used to solve classes of calculus problems. I mistakenly thought it was a notation to be used for specific calculus problems, so I was looking for something that wasn't there. Thanks!
    3
  3065. For those who don't know: the reason you only saw a still and heard an audio is that only photos from the TV broadcast and a duplicate audio of the telecast are what BBC have in the archives of the first three episodes. Addendum: there was a proposed sequel to "The Celestial Toymaker" called "Nightmare Fair". Rumour had it that an early draft of the script included the Doctor solving the minimal path problem discussed here for the Trilogic game.
    3
  3066. One of the yardstick of Vedic Mathematics is the Vedic Measurement and its spiritual values. Take calculation of Time in Veda. It's mind boggling at least from the point of view which era it was used into.
    3
  3067. It comes down to Leibniz notation 😌💭
    3
  3068. Another master class video from mathologer. Your videos make maths so cool.
    3
  3069. I remember asking myself the exact same question (which regular polygons can be embedded into the grid) long time ago, but I didn't find a proof. Really nice to finally see a proof, and what a beutiful one!
    3
  3070. It's likely relevant that the only primes which are conjectured to give rise to windows are precisely those which can be described as the sum of two squares.
    3
  3071. 9:33- I don’t know if it is correct but I think we can make all the integers between 1 and 15 (including 1 and 15) only once. I didn’t use the product it’s just the same idea of binary notation. If you have infinitely many coins with all the powers of two it is the same idea: all the integers only once.
    3
  3072. because (n+1)(n-1) = n² - 1, the 3rd binomic formula, class 8
    3
  3073. That's a good thing. Don't get brained.
    3
  3074. Thank you for another amazing lesson!
    3
  3075. Medieval mathematical proof written as poetry! Medieval Christianity and European imperialism really did stunt humanity in so many ways... How much better could the world be now if not for the arrogance of my ancestors?
    3
  3076. Interestingly. This is only taught in lowest level maths in Australian year 12 (in Vic at least) but not in the 2 Higher level maths
    3
  3077. background: no degrees yet, saving to get into maths at uni (first year out of school - finished year 12) planning on becoming a professor and obtaining field medal in mathematics. analysis: understood everything, though not all the dots are connected (abstractly, they make sense, but if one asked me to connect all the strings together in this video, I would have to rewatch it a few times)
    3
  3078. I'm somewhat pleased with myself that I realized this video was about slide-rules about 01:50 into the video, even though I've never used a circular slide rule. However, I did have multiple really nice high-precision slide-rules in high school and the first year or two of college.
    3
  3079. Wow I'm early!
    3
  3080. I wasn't expecting that part on the last lol. I was so captivated with the proof that i forgot about the Easter egg lol
    3
  3081. Breath taking, awe inspiring and spellbinding. Thank you Euler/Mathologer.
    3
  3082. Is it weird that I was already looking for exponential proofs using factorials at eighth grade?
    3
  3083. A wall made out of numbers? :D
    3
  3084. Is there a formula for the magic number itself if you don't know the three-by-three is supposed to some to 15 can you know beforehand
    3
  3085. 30:15 phi can be written like (1+sqrt(5))/2, which is what you get from the positive solution. for the negative solution the square root is subtracted, so we end up with (1-sqrt(5))/2, and doing some algebra leads to: -(1+sqrt(5))/2 + 1 which is -phi + 1. at this point i would try find a correlation between the two solutions (or try finding a mistake in my calculations), but i've known for quite some time now that 1/phi = phi-1, and i will try prooving it now: if you have a golden rectangle with sidelengths 1 and phi, and then remove a square, the new sidelengths are going to be 1 and phi-1. but since the ratio is the same, we conclude that 1/phi-1 = phi/1, which (hopefully) proves phi-1 is the reciprocal of phi. 31:50 for puting real numbers there is already a video on youtube online, it basically just results in complex numbers because of the exponents. for the negatives, it just goes backwards and switches between positive and negative, which, now that i think about it, makes the sequence even better! when taking the ratio of two consecutive positive fibonacci numbers, their ratio approaches phi as you go bigger and bigger. if you go in the opposite direction, the ratios approach -1/phi, which is the second solution to the previous equation! i find that funny as it seems as such a clear fact, the equation we use to get the fibonacci numbers uses -1/phi afterall. 41:35 my guess is that it will produce the mixed r and s identities.
    3
  3086. Yes, much of the famous terminology in math derives from Latin or German terms.
    3
  3087. Doesn't eulers formulae work for all non-intersecting manifolds, not just convex ones?
    3
  3088. How do i join the ICOMM
    3
  3089. Baruch Hashem!
    3
  3090. Thanks for such a masterpiece. love your shirts as well.
    3
  3091. 😊 Gorgeous to see a new Mathologer video,, and thank you for introducing so much hearty content!
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  3092. I think this is the most useful Mathologer I've seen yet. TY!
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  3093. Hey Mathologer! Super-kudos for your vids. Probably other people answered along the same lines to your question at 07:05 already, but here's my two cents on how to get an efficiently approx for \pi using the McLaurin series for \sin(\pi)=0: mapping it into a fixed point problem and then solving iteratively. you truncate the series to a certain term, (say \pi^7) and rearrange it to obtain an equation for \pi = \pi^3/3! - \pi^5/5! + \pi^7/7! Choose a certain starting value \pi_0 and numerically solve the this in an iterative way i did a little worksheet for that.... aand curiously you find that for 0 < \pi_0 < \pi the iteration converges to 0, and for \pi < \pi_0 < 3\pi it converges - very fast! to 2\pi. this has to do with the periodic nature of the zeros of sin(x) I guess. Anyways, with \pi_0 = 4 truncation at n=29 (14th element of the series) and 5 iterations you get an approx of 2\pi \approx 6,28318530653, i.e. correct to the 8th digit. Thanks for making me think and spend a dreary rainy evening with some good gymnastics of the mind.
    3
  3094. Hhh iknow i have not seen the video yet but iwanna say welcome back Dr ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
    3
  3095. I was able to follow your (quite wonderful!) video. I studied theoretical physics several decades ago. I work as a handyman, doing maintenance in apartment buildings.
    3
  3096. Hi Burkhard, the 3-variable equation plays a role in integrable systems. In fact, it appeared in Sklyanin's work on quadratic Poisson algebras (around 1982) providing solutions in terms of elliptic functions, and (with an extra constant) as the equation for the monodromy manifold of the Painleve II equation.
    3
  3097. Could you please explain in detail why -*- =+ elaborately..no school teaches us properly and if you did,we will get a great foundation to work with and understanding what we do...if we don't understand, it tends to accumulate like a garbage in our mind..pls make a video😌😌
    3
  3098. Thanks a lot. This is the first time I have seen someone prove the Heron's formula in this manner. The proof of Brahmagupta's formula is also very interesting and unique.
    3
  3099. double the points
    3
  3100. Is there still a video in the works about the unsolvability of the quintic?
    3
  3101. Kind of like something I found considering a circle inscribed within a square and differences between them. AB = A-B Where: A = √2 - 1 B = 1 - 1/√2 In practice these values are scaled by the radius of the circle, without that the radius is an implied unit length (making the square that inscribes it a 2 by 2). AB = (√2 - 1)(1 - 1/√2) = √2 - √2/√2 - 1 + 1/√2 = √2 - 1 - 1 + 1/√2 A - B = (√2 -1) - (1 - 1/√2) = √2 - 1 - 1 + 1/√2 This is similar to a form of the golden ratio: (a+b)/a = a/b = Φ (a+b)/a = a/b a/a + b/a = a/b 1 + b/a = a/b 1 = a/b - b/a a = a^2/b - b ab = a^2 - b^2
    3
  3102. I've been a follower since the beginning of the channel and my hype for newcoming videos is growing as the content keeps on improving. Congratulations and please do keep that amazing project running. We need math videos if we are due to stay home !
    3
  3103. ​@daniel57345 - if you go through Fortescue's original work then you see that he made it generic for any number of phases. The very first AC grids were actually 4-phase so Fortescue came up with a generic framework which is identical to this theory.
    3
  3104. and adding ears removes one frequency until only the base frequency for a regular polygon is left (and the next step also removes that so that only the constant term remains, leading to a ... point). Ears - the ultimate polygon notch filter!
    3
  3105. Love this video
    3
  3106. Doesn't really matter, since we never evaluate that function in the variable x :)
    3
  3107. Solid, it gave life to hiperbolic trignonometric that felt allways pretty blury. Thank you!
    3
  3108. I feel like this must have some application in cpu register allocation or cache hierarchies. Especially interesting in light of the idea of grouping disks into super disks etc.
    3
  3109. I say that for every one of his videos! This one was special - I easily followed most of it 1st watching.
    3
  3110. It's so beautiful! Also, it is the first time I can clearly see why is ln(a) = integral 1/x dx from 1 to a. It somehow feels... natural. :D
    3
  3111. Jesuits were in Kerala india and brought back lots of Indian documents. Thanks for giving credit where is long overdue “Hindu Mathematics “
    3
  3112. The most memorable was the optimal towers
    3
  3113. MY EYES NOOOO PLEASE FOR THE LOVE OF SCIENCE STOP WITH BRIGHT WHITE BACKGROUNDS
    3
  3114. My head hurts. Thank you for your efforts. May you and yours stay well and prosper.
    3
  3115. What an amazing video. Also sending a salute to Andrew for the great ideas ;)
    3
  3116. Elon musk likes this video
    3
  3117. 7:20 Of course; assuming that the missing circle of negative space has equal area to the circular cross-section of the hemisphere (c); in order for the ring and the small circle (c) to have the same area, the large circle (C), that is, the ring plus the circle missing from it, must have exactly twice the area of the small circle (c). This amounts to the radius (R) of the large circle (C) being exactly √2 times the radius (r) of the small circle: R = √2r —> A(C) = (π(√2r)² / πr²) * A(c) = (√2)² * A(c) = 2A(c). Then, removing A(c) from A(C) leaves: A(C) - A(c) = 2A(c) - A(c) = (2-1)A(c) = A(c); and the ring and the (cross-section) circle have the exact same area.
    3
  3118. Little did he know that he overcomplicated his water computer
    3
  3119.  @Mathologer  He is not,i have studied till highschool and i haven't heard of him till now.
    3
  3120. I saw something very interesting about the diagonal dealing permutation. Let's say we have two distinct odd prime number p & q, what on earth does doing a diagonal dealing operation have to do with squaring numbers and taking remainders? The idea is, you draw a rectangle with sides p, q and area p*q and observe when you do the diagonal dealing operation you will be in this analogy drawing a line whose length is equal to the area of the rectangle p*q. Then since the length of this diagonal line is p*q, if it were possible to find two integers a,b such that (p*q)^2 = a^2 + b^2, we can draw a different rectangle of side lengths a & b whose diagonal is also a line with length p*q. To see where the remainder operation comes in to play in this analogy, cover the plane with fundamental rectangles of sides p,q and make the rules that any 2 points in the plane are equal if they are in the same position in the fundamental rectangle.
    3
  3121. At a rural high school in PA a few years ago, I was taught Heron's Formula in a Trigonometry class.
    3
  3122. I can only speak for myself but I am pleases. You make excellent, educational videos sir. Keep up the great work!
    3
  3123. Sorry about the slightly offtipic comment, but I couldn't find where to post this suddenly. I'd like to ask about your opinion of Geometric Algebra and whether you are planning to make a video in that topic in the future ot not.
    3
  3124. Another great video with many wonderful proofs. Though I suggest using n and k as variables instead of n and m. That makes it a lot less confusing to listen to and n and k are also easier to differentiate visually.
    3
  3125. It seems that the expression (x+1)^n corresponds to tetrahedron… if so, we want one more chapter ;)))
    3
  3126. in some way, this is also tied to one of the most famous pieces in literacy, "around the world in eighty days." remember when they thought they gonna lose the bet but didn't adjust for the fact, they went around the earth one time? which also means that one must either add or subtract one day in regard weather you're going "with" the sun or "against" it. to me there's a very similar, underlying principle.
    3
  3127. "Are you going to sit around and drink beer all afternoon?" "I'm building Schwarz Lanterns."
    3
  3128. I have a theory that we could make math more interesting for students if we taught it in chronological order of discovery so we could travel down the same paths as our intellectual forefathers. I would be ready to take it even farther and have them doing abacus math with Roman numerals so they could see the brilliance of Fibonacci's contribution to the world of math by bringing the Hindu-Arabic numeral system to Europe. Everyone knows his name because of the rabbit problem but his contribution of a numeric system with a built-in abacus was far more important.
    3
  3129. 28:45 Isn't the 7 supposed to be a 5🤔
    3
  3130. Is this the first strand type game?
    3
  3131.  @Mathologer  You named him after C.F.G? Of course you did so. My favorite mathematician after B.P. 😀 He is cute. He must have taken after his mother. 😁
    3
  3132. Great video! Get him to 1 million already! :)
    3
  3133. This was a particularly fun one. I feel like I was able to follow pretty much every part of the proof. Enough hand-holding for casual viewers but no tricky stuff shoved to the background. Even the algebra autopilot is fun to watch, as its pace and animation make it easy to understand and apreciate the cleverness involved. Many channels like to focus more on geometric proofs only (which are also nice) but algebra has some charm too.
    3
  3134. Just buy your paradoxical horns pre-filled! Saves you a lot of time thinking of things in unhelpful ways ;)
    3
  3135. I had a circular slide rule for many years in my job as an engineer back in the 60s' 70's. When it was stolen I caved in and bought a newfangled electronic calculator. I missed the wheel though.😥
    3
  3136. High School level algebra and geometry, and I see these longer in depth explanation of concepts that I have not studied as interesting. Thank you.
    3
  3137. I am shocked by the formula at 4:20 :)
    3
  3138. Conway's circle has the large diameter. Diameter of the shape of constant width is the cord formed by extending the triangle. Cord < circle diameter
    3
  3139. The peg solitaire kept me happy when I was 6-7 years old visiting Grandma's bungalow, then I inherited it and played from time to time and occasionally it worked, though I don't think I ever memorized the strategy! I wonder if the soldier version could be adapted to give the soldiers that are 'jumped over' more than one life so they only disappear after being jumped over twice or 3 times or n times? It could advance a lot further?! Or how many wars might be prevented if the leaders battle it out individually with armies of small plastic (or wooden?!) soldiers instead of real people?!
    3
  3140. Technically 18 minutes if someone watches in 2x speed
    3
  3141. The talmudic system is not resistant to some (smaller) creditors putting on a trenchcoat and acting as one with a bigger claim. For example: Suppose Huey claims $12, Dewey also claims $12 and Louie claims $24. Then if the estate is $36, the talmudic system gives $8 to Huey and Dewey, and $20 to Louie. If however Huey and Dewey put on a trenchcoat, the claims are $24 and $24. Both get half, $18. Then Huey and Dewey split their share and both get $9 > $8.
    3
  3142. I am not the only one thinking at first Madhava on the Thumbnail looked very much like someone else known here, am I?
    3
  3143. I’m pretty sure (for the three color game) that is it MULTIPLES of powers of three plus one. (m*3^n)+1. For example, 19 works, 2* 3^2 + 1. You may have said this in the video and I just missed it, but if not I think this is very interesting because it means that 2,4, 7, 10, 13, 16, 19, 28, 37, ect. all work. Lmk if I’m wrong
    3
  3144. Awesome video, as always. What is the music at 1:48 please ? Thanks
    3
  3145. This fascinating proof put me in mind of a very impressive trick that one can do (if one has the time) with a normal pack of 52 playing cards. Put the cards in suit order and then deal the pack into four hands of 13 cards. Assuming, in bridge notation, you are South. Pick the piles of cards clockwise starting with west - hence WNES. Now deal the cards a second time (D2) and pick them up in the same way. Continue until after D13. Now examine the cards and you will find that they are back in the original order. My low grade graduate maths is not up to explaining this, and, of course, 4 is not prime, even though 13 is. But it is another of the wonders of theories of ordering things.
    3
  3146. Multivariable calculus would like a word.
    3
  3147. I like how your shirt changes once you start explaining the cubic part! The video was also very interesting and good!
    3
  3148. 3:56 there's a bug in the illustration! The top number has 10582 and the bottom has 10581, yet the 2 and 1 are marked as the same 😅
    3
  3149. Last time I was this early, it was 2019's Christmas
    3
  3150. 4.7 to 5k viewers *views 87th,88th commenter
    3
  3151. 37:30 one of the pairs (a,b) and (b,a) is an inversion. Cycling the permutation reverses each pair formed with n. There are n-1 pairs so the sign changes n-1 times, no change if odd, sign changes if even.
    3
  3152. Is there a limit to genious ?
    3
  3153. It's really crazy that I just heard about "Napoleon's theorem" and looked for a deeper explanation, and here it is Mathologer posted one just 6 days ago!!! Love your work!!
    3
  3154. That one will definitely take quite me a few times get my head completely around :) P.S. Glad to hear that Galois is on your back burner. I'll be looking forward to it
    3
  3155. Challenge 3: 14:00 The corresponding matrix is n by n with the main diagonal composed of only entries of i with the diagonal above and below consisting of 1 and all other entries are 0. Performing the cofactor expansion on the first row reveals that the determinant (d_n) is i * d_(n-1) - d_(n-2). Computing the first two values and following the formula reveals that the number of ways is precisely the nth Fibonacci number! Very neat!
    3
  3156. So how does Ramanujan compare with Madhava?
    3
  3157. 37:13 That would be because in each seed hexagon as the board grows, there must be exactly one rhombus of each direction.
    3
  3158. Does anyone else see a cardioid forming as the number of sides increases? Id be interested in Mathologer revisiting this with a connection to the interior points of the Mandelbrot Set. Naturally that would also connect to the Logarithmic Map. It also feels like there is a connection between the regular polygon at the end with the Dot Product and the regular star with the Wedge Product. Then the whole thing would describe the Geometric Product.
    3
  3159. This also seems related to path counting in lattices. Specifically, in mod m, canceling occurs at a point B for a starting value A for which there are exactly m paths from A to B. I presume, and am currently too lazy to check, that there is a combinatorial way to compute the number of paths from one point to another in such a simple lattice (factorials will be everywhere probably), so these facts should be provable by algebraic manipulations of those path counts and, in fact, might tie into some combinatorics structures as well. Thoughts?
    3
  3160. I can understand all the things you say with my poor English😂😂 ❤️❤️
    3
  3161. Definitely a glitch in the matrix lol
    3
  3162. Calculating the area of a triangle is part of the Mathematics Olympiad program (IMO training), because counting things in multiple ways can reveal complicated identities. The proof used the cosine rule. Ew. I actually discovered (and proved) the formula myself, when I was about ~13, just by using the right-angle theorem (Pythagoras). More recently, I found out (thanks to Wikipedia) that the maximum area of a quadrilateral happens with a cyclic quadrilateral. This is beautiful, because we can extend this for any polygon. I used some ugly math to prove this, but I did found out about some nice things. I also found out about the correction term. I think that many people discovered this, even before 1800. Your video's encourage me to seek for simple proves myself. The method for the proof of p=a²+b², can be extended for all natural numbers. You can prove that numbers that shouldn't be written as sum of two squares, have an even number of representations. Next, you parameterise n=a²+b²=c²+d², to show that n is the product of two numbers (greater than one), as the sum of two squares. Next, you can say that we had the lowest number that works (with some extra things), to finish the prove.
    3
  3163. Stop dragging my cat around...
    3
  3164. Mathologer dropped
    3
  3165. It's 2:30AM in Tehran and I'm going to bed. Ooooh look a Mathologer video. No sleep tonight
    3
  3166. Regarding the frozen regions in the diamonds, I would have liked to also have him showing that along any of the edges only two types of tiles can be placed (e.g. blue and red along the upper right edge), which can easily be seen of the squares are colored as chess squares as in the beginning. And considering any of these edge pieces, as soon as you place a red piece on the upper right edge, then all other edge pieces all the way to the red corner must also be red. This means that each edge have to be subdivided in the two colors of the connected corners and any random subdivision should tend to divide the edge close to the middle. Then the same procedure can be repeated for the squares inside the edge and so on. The farther you get from the actual edge, the likelier it is for the subdivision to stray from the middle. Also, at each change from one corner color to the other, there is a possibility for the other colors to appear, which will cause the blending in the middle. I don't argue that the explanation given in the video is bad. It is really beautiful to connect it to the random construction of any tiling. I just would have liked to see this side as well.
    3
  3167. You're Awesome 😎😎
    3
  3168. There is a similar growing process for tetrahedrons: at each step add a single vertex that is the current side distance from all current vertices, then draw edges between all existing vertices to the new vertex. Octahedrons are a little trickier: basically, draw 2 points orthogonal to the current shape the current side distance from all current points, then edges to them. However, there are some hiccups: 1) the meaning of orthogonal for 0-d, and n-d for n > 3 is tricky, and 2) you sort of need to remove the previous shape each step.
    3
  3169. "I'll indicate the argument in the description of this video"
    3
  3170. Thank you very much for this amazing gift!! Happy new year professor, and keep up the good work!!🥳
    3
  3171. Anarchist math
    3
  3172. Great video. I actually had a project in my number theory class to verify the one sentence proof. Very fun, but this is way more enlightening.
    3
  3173. I was fully expecting a mention of benford's law tbh, still a great video!
    3
  3174. I still have my dad's Keuffel & Esser sliderule that he used from 1940 to 1975, including during development of artillery firing tables at Ft Sill. And of course an E6-B from my own student pilot days. Timeless tech.
    3
  3175. What's the best way to submit entries to these coding competitions? Whenever I post a link as a comment, youtube deletes it.
    3
  3176. yo Mike Wessler's is SLICK
    3
  3177.  @Mathologer  Thank You for sharing this fact. It is probably getting even funnier, as I live really close to the border of Bavaria, like 15 kilometers from the town Furth im Wald and there is a hill next to it - Schwarzriegel - where, as per my knowledge, some of such surveillance was being perfomed during Cold War, it would have been a nice coincidence if You had served there :-).
    3
  3178.  @Mathologer  First of all, your Czech pronunciation is pretty good, and this is one of my favorite videos on your channel. Great job! I have a similar story: I once interviewed a Norwegian guy for a job, and surprisingly, he spoke some Czech. Apparently, he was doing military service in a Norwegian interrogation unit assigned to potential Czech PoWs. I told him that there was an alternative universe meeting where he would do some interviewing/waterboarding on me, and he said no, only Americans do waterboarding. :)  Once again, thank you for your work. Great channel!
    3
  3179. I found the 2d stuff elegant and enjoyable. I felt totally lost when we jumped to 3d.
    3
  3180. Nice! It seems to me that the wheel actually integrates 1/x, which is of course the log.
    3
  3181. Numero uno della divulgazione matematica mondiale ❤ congratulation
    3
  3182. I always hate it when someone shows up at a party with an acoustic guitar and plays numberwall...
    3
  3183. I found a mistake at 3:31 10581 and 10582
    3
  3184. This seems like it would arise in a special case of an elementary proof of ceva's theorem, that seems like the most natural generalisation, although you do need to scale the triangles. But people are too concerned whether someone else made the same discovery as they did. This is definitely the case with open problems, but presenting mathematics in an accessible way is here just as important, if not more important.
    3
  3185. Gonna be another banger!
    3
  3186. I made it to the very end... ;) and thanks for always making me smile whenever I see one of your videos pop up in my subscriptions!
    3
  3187. It's sad to think of all the knowledge and insights that have been lost over the millennia.
    3
  3188. The perfect build up for MathVengers: Euler's game
    3
  3189. I like the idea of coins with powers of 2. Having coins of 1, 2, 4 and 8 allows us to make charge for all values up to 15. It is basically binary numbers. If each coin is a bit and you have 4 of those bits (1, 2, 4 and 8) you can make charge for 0 to 15, exactly 2^4=16 different amounts.
    3
  3190. Saturday night in Rockport Texas. Reasonable time here too.
    3
  3191. Love your vid
    3
  3192. Thanks again for this amazing video. I figured out the relation between x and y before you tried to do it algebraically but then stopped. I'm really interested to know how to evaluate these infinite fractions algebraically (without matholorization ;) but I don't want spoilers because I want to do it myself one day. Peace!
    3
  3193. I love that thumbnail
    3
  3194. 4:14 Reminds me of my circular slide rule. Unlike the foot long version, it fit in my back pocket. The engineer's friend.
    3
  3195. Amazing quality kept for another Mathologer video. Thank you so much for spreading glorious mathematics ideas to mathe-maniacs like us, Mathologer!
    3
  3196. I learned "finite differences" in 1975, when studying for actuarial exams. It gave me a really useful intuition for all sorts of mathematical problems, and a quick and easy tool to start productive analysis. I see they've stopped doing that now, and instead teach more useful modern things like diversity awareness.
    3
  3197. This reminded me of that video where Adam Savage (former host of Mythbusters) and Michael Stevens (of Vsauce) did the tower of lire out of accurately cut planks So tower part is my favorite
    3
  3198. First to press like :)
    3
  3199. By the way, I learned when I was a teen that if you take the rows of pascal's triangle, mod 2, not only do you have a fractal, but if you turn that into a digital sound stream, say go out 1024 elements then skip down and do the same, it plays a little tune all made up from the same note in different octaves and beats that are also at multiples of that frequency.
    3
  3200. Is anybody else more excited about Mathologer videos than Hollywood videos. I can’t explain the excitement I feel when I get the notification.
    3
  3201. Some of the resulting forms are very beautiful, independent of their mathematical origins, some fractal reminiscent, some just joyful. Thank you for a beautiful afternoon's half hour.
    3
  3202. There is a very pleasing pattern in the series of fractions at around 14 minutes into the video. Each numerator is the sum of the denominators of the current and previous terms. For example, 41/29 and 99/70 appear sequentially and as you well know, 29 + 70 = 99. This feels related to the Euclidean algorithm presented later in the video, but I can't put the pieces together. A very pleasing pattern, though! Edit: I see this was pointed out numerous times in the comments. Ah well, cool anyway. Edit 2: Oh, this is proven later in the video. Very cool!
    3
  3203. colouring proof works better for me
    3
  3204. Today is pi day 2020
    3
  3205. Thank you for your videos sir, very relaxing, interesting and informative, this channel is one of the reason i'm a math major
    3
  3206. THIS VIDEO IS FANTASTIC!!! THANK YOU
    3
  3207. The part that I found most memorable was Kempner's proof. I find that result very unintuitive and I expected a far more technical proof.
    3
  3208. ​ @Mathologer  The Malayalam subtitles are pretty neat :) A lot of the words are taken literally, but that's fine by me
    3
  3209. This is nice: r = 2 * cos(π/7), (and s = r^2 - 1). I saw that Mathologer typed something similar somewhere, but it disappeared :-) And oh, we could also write s as: s = 2 * cos(2π/7) + 1 . Loving it!
    3
  3210. Hi! the pdf file link returns a 404 page :(
    3
  3211. Thanks for your integrity and the effort you put into this great video during holidays!
    3
  3212. Hey, I have written a piece of code in C# (also in Python) that generates formulas for cosine waves of any frequencies, and I think it is pretty cool that we can do such things very easily in an automated way. Would you be interested in seeing the code? Do you have an email that I can send it to? Perhaps in the future, you might make a video related to this topic (or at least mentioning this property of cosine/sine waves).
    3
  3213. For the cube sequence, presumably the 1st line would be: 5^3 + 6^3 = 7^3 which isn't quite correct, but is very close (left hand side = 341, right hand side = 343)!
    3
  3214. 666000 subscribers what a crazy number of subscribers to have exactly on Christmas.
    3
  3215. Well i knew that digital root is remainder by division by 9... Not from school but because i have proven it myself when i was doing recreational mathematics and needed to find closest multiple of 9 for some reason
    3
  3216. About to teach pigeon-hole to a bunch of year elevens, just did groups. That Rubik's cube example is absolute magic thank you
    3
  3217. Mathologer: Explains mathematical induction in under a minute Teachers: take notes furiously
    3
  3218. True. Only lack of knowledge
    3
  3219. The rapidity of the harmonic series converging to infinity makes the tortoise in the fable seem rather rapid
    3
  3220. Props for mentioning the Celestial Toymaker... Jeez, I'm old. Remember watching this on the original transmission!
    3
  3221. A minor mathematically unrelated improvement: you should use the color pairs red-green and blue-yellow for tiling of the opposite parity.
    3
  3222. It's amazing how sensible this is while being so non-intuitive.
    3
  3223. Finally somebody recognised contribution of South Indian mathematicians into calculus,on youtube. Thankyou very much Sir. I am your regular viewer.
    3
  3224. How the introduction i.e. spinning Rubix Cube features a lot of foreshadowing; the link between rotation (i.e. circles), hyperboles (which essentially are perpendicular to circles) and 1/x (which is a hyperbole rotated by 45°).
    3
  3225. 1. I discovered a part of this as a kid when i was playing around. I showed it to my parents and they told me i'm doing numerology and its worthless, thus (doing similar things on other occasions many times over) killing my interest in playing around with math. THANK YOU parents and teachers. 2. I am actually interested in actual vortices and the math behind their fluid dynamics and these stupid vortex mythology is driving me nuts -.-
    3
  3226. Next video about 4 dimensions?
    3
  3227. Favorite part was the cat saying "μ" γ though, I liked everything about γ
    3
  3228. I learned about heron formula in Senegal. I remember we did the proof in the dorm of an exercise....
    3
  3229. Fascinating 🤭 One little mistake I found. During the explanation of the uniqueness of “9”. On the multiples of “9” chart, the last number is 2340... it should be 2304 Really, I’m terrible at math but I see patterns. Watched this twice, and did the exercises along with you the second time... I was able to work through most of it. Well done.
    3
  3230. Most memorable. That the 'no 9's sum' is finite. My intuition was way off.
    3
  3231. Some of it is easy but it soon becomes incredibly difficult when you get to differential equations or anything containing trig functions or log functions... or a bunch of other things. Integration by substitution is a nightmare. It never clicked, nor did integration by parts. Even partial differentiation which looks conceptually easy is not so.
    3
  3232. Noice. In IT science you would call this a BACKPACK algorithm, where you memoize the results in a 2d matrix
    3
  3233. Promeeth.... I bet he's working while he's " hiding".
    3
  3234. That's unbelieveable amazing! Or to say it in German: Scheiße, ist das geil! :-)
    3
  3235. "Can you remove any 2 black and any 2 green squares and still tile the chessboard with dominoes?" No; if we remove the squares at (1,2) and (2,1), for example, the corner square is now detached and not connected to any other squares, and cannot be part of a tiling.
    3
  3236. "2" should be kicked out of the primes.
    3
  3237. A very good video. Also one of those where I managed to grasp everything you explained. I studied some of these things just this year in 12th grade so learning about the history behind it was really fun.
    3
  3238.  @ChefSalad  Oh crap I remember those things. Your computer has a mouse, why not a cat as well?
    3
  3239.  @ChefSalad  There are also USB cueCats.
    3
  3240. As a CS-student, I of course studied the Tower of Hanoi and knew of the Reve's puzzle, so I thought what could he possibly teach me, that I don't know already. Well as always your explanations are not only beautiful to look at, even when I watch a video of yours where I think I know my stuff, you leave me speechless. You are very talented in educating others and I hope I will learn this skill in time when I become a father.
    3
  3241. 11:30 ach, das ist beinahe das Schönste, was ich an Erklärungen gesehen habe. Außerdem auch ist es so ein typischer "waaaas?" Moment
    3
  3242. First Pi =22/7 as discovered by Chinese Mathematician and Later by Greek and then Egyptian and then after, it was Indian Mathematician... That too Tamil Mathematician framed to reduce pi to 22/7, No other Indian Mathematician did...
    3
  3243. The reality theologically from Jewish Tradition is that a decimal in front of the 333 is totally evil once again. Think about it. :D
    3
  3244. Excellent! This video really answers to tons of questions that I've being asking myself without doing enough research to answer them! Your animations make it really simple to follow (the math behind is not that simple though...) and your explanations really help me understand! I might however have been a little bit lost at 45:35 when you replace the yellow sum of some strange patterned fractions with the sum of inverse squares... You seem to use the result as if it were true so I assume I simply can't see it... Thank you for your brilliant videos, please continue explaining hard math like this, that's terribly interesting!! (P.S.: I'm a french student in first year math "classe prépa" (university equivalent), so forgive me for my english and math inexperience 😉)
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  3245. I guess it's "rock paper scissors lizard Spock"
    3
  3246.  @Mathologer  Thank you so much sir for being so considerate, unfortunately i was unable to clear the Olympiad i was preparing for, but yet thanks to your videos my love for mathematics has not died down and hopefully never will.
    3
  3247. Coolest thing I've seen in arithmetics for many, many years. I want to see programs for this, too!
    3
  3248. @[05:05]: If I had to guess, it involves the omni-directional symmetry of a circle.
    3
  3249. Beautiful, just one word! It must have been such a joy to discover this for the first time.
    3
  3250. That video was a tour de force! I think my favorite was the crazy towers
    3
  3251. In higher dimensions (> 3), any quadrilateral is contained in a 3-dimensional subspace (the vertices are… co-3-spatial??), so the 3-dimensional version of the theorem applies directly.
    3
  3252. My dad showed me the straight and circular rules when I was in junior high and for a while I knew how to use them, but didn't at the time have any clear idea why they worked. I wish they'd bring more of this kind of thing back into education.
    3
  3253. interesting to think about the consequences if you change the rule for flipping the far right card, instead it wraps around to flip the far left card. the game would never be guaranteed to terminate right?
    3
  3254. My favourite Yt channel
    3
  3255. Yes; you can just pick a direction and it will work out, because you always have a choice of at least 2 cards to take out. Say the magician and assistant have agreed always to add the encoded number to the first card value. Say the 5 cards include the ace of spades, and there is one other spade in the group. If it's the 2, 3, 4, 5, 6 or 7, you take out the larger card and use the ordering of the others to encode a 1 ... 6 and all is well. But what if it's the 8? Well, we're using modular arithmetic, and (8+6) mod 13 is 1. So you take out the ace and encode a 6. Similarly for all cards larger than an 8, and for all other combination of cards; just choose whichever card makes the gap between 1 and 6.
    3
  3256. The Wikipedia article I have seen starts with integer pairs (2,1) and (3,1). Call these pairs (m,n). It then applies the following algorithm separately to each pair: Branch 1 = (2m-n, n); Branch 2 = (2m+n, n); Branch 3 = (m+2n, m). The algorithm in the video starts with the same seed pairs, arranged in a 2X2 array, which you can alternatively view as rational pairs (1/3,1/2) The Branch 2 algorithm reproduces the same “middle children” as those derived in the video. The other two algorithms reproduce the left and right children but swap one column in each for one column in the other. So in the video left child (3/5,1/4) and right child (1/5,2/3) become in the Wikipedia algorithm (3/5,2/3) and (1/5,1/4), starting with an initial pair (1/3,1/2) as in the video. So the two algorithms are equivalent.
    3
  3257. I believe it's just a fixed sequence of moves as you said.
    3
  3258. Question: why is Augustus de Morgan in the thumbnail?
    3
  3259. The most difficult question to answer is what part to vote for. If my life depended on it, I would pick the “no integers in the partial sums” topic ... and surrender my life for stealing gamma. Awesome video, wonderfully addictive, as usual!
    3
  3260. I liked Tristan's light at the end of the tunnel - and claim my prize as over a sufficiently long time period this late reply approximates to 'within 2 weeks' :)
    3
  3261. Those aren't pentagons; they are petr-gons! Absolutely fascinating subject and video!
    3
  3262. Perhaps take some of the funds to market those shirts you design?
    3
  3263. Besides Madhava, please upload other mathematicians like Brahmagupta, Aryabhata, Bhaskara, etc. Thank you.
    3
  3264. In Talmudic law, no claim can be made against an estate without proof. Period. The end. The principle here is the garment rule. We first determine what each party agrees belongs to the other. Then split the difference. If one claims half and the other claims the whole, the former has no claim to half so both are agreed that it belongs to the latter. Then they split the disputed part. So 1/4 3/4. In both the estate case and the garment case, all claims are of equal status; the former with proof and the latter without.
    3
  3265. Does every irrational number have interesting infinite formulas? 🐯🐯🐯🐯🐯🐯🐯🐯
    3
  3266. the magic method in 18:47 is simply drawing diagonals but using modular arithmatic or if you perfer imaging the board goes on forever and then collapsing in back to a single board so using (row,column) notation : we have (5,3) draw a diagonal thats (6,4)->(1,4) then (2,5) diagonal (3,6) ->(3,1) then (4,2) . once the first 5 diagonal has ended go down 1 and draw the second 5 diagonal thats (3,2) then (4,3) MARTY : you've made a mistake no (4,3) green dot! (5,4) , (6,5)->(1,5) (2,6) ->(2,1) go down (1,1) , (2,2) and so on this ensures every column has all (x+5*y) sequences (x being the sequence in the 1-5 drawn by the y'th diagonal)
    3
  3267. U r into programming if u read it as "2 infinity OR beyond"
    3
  3268. Euler again ! He was really a badass his work is everywhere メ!
    3
  3269. 11:37 mind blown! This must be the Laplacian matrix of the graph associated with this choice of numbering the black and green tiles, then by the spanning tree theorem all we have to do is take the determinant to get a handle on the total number of tiling's!
    3
  3270. @18:46: Answer is {7;2}. We start from one corner and count to reach the next edge of straight line. This gives us 2. Now we know this is a 5 point star or flower pedal. Using a-b=c, we can solve for a where b=2 & c=5. 2+5=7. We get {7;2}. This means small circle does 7 rotations on straight line perimeter of larger circle. The 2 counterclockwise rotations from coiling back. This gives the 5 total rotations.
    3
  3271. Congrats on 100 videos! Here's to many more!🎉
    3
  3272. @15:33 Answer is 5 rotations. 3 clockwise rotations for the 2 times the straight line perimeter length of large circle. Then coil the large circle back 2 times, which rotates small coin 2 clockwise rotations. 3+2 = 5 clockwise rotations.
    3
  3273. Nice subject and nice proof! I'm specially impressed by the generalized Moessner Miracle. I think we can Mathologerize even more the part starting around 22:00 - no need to resort to binomial coefficients (though, of course, they continue lurking in the backstory). Take the triangle starting from the powers of 3 on the diagonal: 64 48 36 27 16 12 _9 _4 _3 _1 In the starting diagonal, each number is 3x the number on the lower-left side. That property extends to every parallel diagonal, by an easy recurrence. One then easily see the number at the lower-left end of each diagonal is 1+3 times the number below it. Bingo!
    3
  3274. The most memorable part for me was the greedy algorithm that was used to prove that every positive number is the sum of a series included in the harmonic series
    3
  3275. Wow Mathologer and Standupmath videos on the same day! great work as always! many thanks and respects Mathologer!
    3
  3276. Really looking forward to the hardcore video.
    3
  3277. Great video as always! 8:24 The sum of all tiles from an order-n magic square is S = 1+2+...+n^2 = n^2(n^2+1)/2. Thus the magic number should be S/n = n(n^2+1)/2. It's 17985 when n=33. 18:50 Extend the square to the whole plane by translating horizontally and vertically (similar to the flat torus). Start from the middle tile in the top row, go towards top-right diagonally 4 times, then go down 1 tile. Repeat. P.S: There are 5 different starting places of tile 1 for this method to work (to make sure that 11 through 15 are the five numbers on the "/" diagonal). Actually, it's better to start from tile 11 - we just need it to be on the "/" diagonal. BTW, are there any sum-and-product double magic squares? I vaguely remember having read about it before.
    3
  3278. Mathologer: 1+2+3+4+...+ is not equal to -1/12 also Mathologer: A-A is not 0
    3
  3279. To prove this, consider two intersecting chords. If you establish that the product of the segments created by their intersection point is equal for both chords, then they are points on a circle. This is evident from the equation r(b+g)=r(g+b), indicating they lie on a circle. To further prove that this circle shares the same center as the inner circle, examine, for example, a red isosceles triangle. The center of the larger circle lies on the bisector of these red arms. This bisector also serves as the bisector of two sides of the circle where the center of the inner circle lies, confirming they share the same center.
    3
  3280. I wonder what the 3 dimensional equivalent. Like if we work in reverse, we would have all our linesrly independent eigen polyhedra(whatever that entails), and we can have some linear transforms that map each of the miscellaneous polyhedra to 0. I havent finished watching the video, but if im to guess, the eigenpolygons are all the possible orderings of vertices that can be attained from connecting the points of the regular polygon by skipping n of the vertices(i.e original clockwise pentagram attained by connecting without skipping, clocksise pentagram attained by skipping 1 at a time, anticlockwise pentagram skips 2 at a time, anticlockwise pentagon skips 3 at a time). This doesnt translate well to 3d, but I can guess that this specific process of forming eigenpolygons is just one of many possible basis polygons, and that there might be another basis that translates well to 3d. If we dont care much for clenliness, we can just have a non linearly independent basis by having every single permutation of our vertices, modulo rotations, which certainly spans all of our possible polygons just the same, but I dont think this will work. As in the original basis was isotropic(i think thats the right word), we could send eigenpolygon components to 0 by adding ears on, which was a process that had no preference of starting vertice, and the action was identical for each pair of vertices, but if we wish to send arbitrary permutations of vertices to 0, I'd have to guess this isnt going to turn out the same. Which makes me wonder, is there a convenient basis for polyhedra, where an isotropic process will send each eigenpolyhedron to 0, and the basis spans all of our polyehdra. I think i would be comfortable in saying it isnt possible. But now that I've taken another step back im beginning to think its possible again. Instead of thinking of a process that relied on an explicit ordering of the vertices, why dont we think of all the possible choices faces of polyhedra that are rotationally invariant, that way an action on a single surface is the same for all surfaces. I still dont think there is enough of them to form a basis here. Additionally, the process of adding ears was linear in the 2d case, but would it be in the 3d case? I suppose we can always modify our action so that it must be linear, we have plenty of degrees of freedo. there(takes 3 vertices as input, must be isotropic, must be linear, and must send a specific face to 0, feels like we have the freedom here). And now that i think about it the eigenpolyhedra dont even have to be nice whole surfaces, we only care that all the faces are identical, and its rotationally invaiant, and that it forms a basis. The more I think about it the more I feel its possible, though whether our 3d analogue of adding ears is natural in any way, thats debatable. But now im doubting myself again and thinking its impossible, and my reasoning is now impossible to explain.
    3
  3281. A mathologer video with a full section about cubes! Am I dreaming? :O
    3
  3282. Learned it in USA; knew it for the test, haven't stumbled on many applications. Heron's formula is numerically unstable in floating point. I feel like there should be a straightforward algebraic proof involving a system of two quadratics, but it's more interesting to know how the ancient Greeks convinced themselves.
    3
  3283. 19:45 "a good place to stop", I see what you did there 😉
    3
  3284. As my junior-high-school math teacher put it....... "Mathematics is to science what woman is to man. Its queen-and-servant." +-×÷<>
    3
  3285. I didn't know the sign of a permutation as the -1 to the power of the number of inversions. I knew it as -1 to the power of the swaps you need to get to the natural order. I could follow the video without a problem.
    3
  3286. :face-red-heart-shape:
    3
  3287. Why m i always first here
    3
  3288. The second differential equation in this video is a first order linear differential equation and it has a well developed theory using integrating factors and so on..
    3
  3289. Using the coin counting trick, we have the product (1+x)(1+x²)...(1+x^(2^n)) If we multiply the product by (1-x), we have (1-x)(1+x)(1+x²)...(1+x^(2^n)) = (1-x²)(1+x²)...(1+x^(2^n)) = (1-x^(2^(n+1))) So the product is (1-x^(2^(n+1)))/(1-x) Notice that this is equivalent to the geometric sum 1+x+x²+...+x^(2^(n+1)-1) So there is exactly one way to make change for values from 1 to 2^(n+1)-1 (This can be visualized in binary)
    3
  3290. Now I can use this to place farms in Age of Empires 2. Thanks.
    3
  3291. Alaways a pleasure to watch your videos, @mathologer! You are an heir to Martin Gardner IMHO.
    3
  3292. If the available estate covers less than 50% of the combined claims, it unfairly favors smaller claims. If it covers more than 50%, it unfairly favors larger claims. The simple "there is only x% of the total claim available, so you get x%" seems far easier to calculate and actually fair.
    3
  3293. my apologies for asking an unnecessary question yesterday. I was watching this wonderful lesson on continuous fractions using my cell phone and was unable to navigate to your full explanation which included the credits and titles for the background music. Thanks for helping us readers get really excited and interested in furthering our math education well beyond what we learned in high school.
    3
  3294. I don't always use tricks, but when I do I prefer 700 year old ones.
    3
  3295. Incredible as Always!
    3
  3296. It's beautiful how complex behavior emerges from simple rules. Great video!
    3
  3297. Heron's formula was one of my favorite from trig class! Good stuff! Thank you for sharing!
    3
  3298. Cool, here's my story: In school our physics teacher had such a wheel, he used it but didn't explain.. Years later I saved such a slide ruler from trash, but until Today I knew it's for multiplication and division and I knew that it works with log. Know I have a better understanding how it actually works.
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  3299. School obviously let me and my classmates dowm, as I've never seen this before.
    3
  3300. All of your content is welcome, short or long or monstrous. All good!
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  3301. Please yes
    3
  3302. Shorter videos are always welcome!
    3
  3303. More mathologer = more better
    3
  3304. Seems perfectly OK to me. 🤷‍♀️
    3
  3305. At first blush, rs = r+s appears to violate dimensionality, but with your proof, using dimensions of length, there are "hidden" unit lengths in the equation, so, if our units are meters, we have rs m^2 = r*1 m^2 + s*1 m^2. Wonder if considering the dimensionality for some of the other development might yield more interesting insights?
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  3306. I actually did the same thing as Gauss when my 5th grade teacher asked us to add the numbers from 1 to 100. She wrote "1-100" up on the board, and I just kind of mentally changed the dash to a plus sign, and then realized 2+99 is the same, and it all falls out from there. I spent a little while deciding if there were 50 or 51 copies of 101, but then I went up and asked what we should do when we were done. I didn't generalize a formula or anything, but I suppose if you'd asked me to do another similar sum at that moment I probably would have made the connection and done the same thing again. Years later, I talked to the teacher and she said she gives that problem every year, and that once every few years someone gets it. I had this as one of my proudest memories for so long, especially once my college professors spoke so glowingly about Gauss for doing the same thing, but apparently it's not that uncommon.
    3
  3307. Please make the permutation video soon . Waiting curiously to watch it.
    3
  3308. Ooh, another problem hit me based on modern finance: most claims are made by corporations. But a corporation represents many investors. So, how do you treat it: as 1 entity that gets an equal split with a single human creditor, or as many claimants who get an equal split? These are *vastly* different results.
    3
  3309. 28:30 consider your two original triangles A and B. Since they are similar, one is just a rotated/scaled version of the other, a.k.a. its points have been multiplied by a single complex number. We can write this as A=cB, where c is the complex scale factor. Now when we add the two together we can simply substitute: A+B=cB+B=(c+1)B. The new triangle can be expressed as triangle B multiplied by a single complex number, and thus it is a rotated/scaled version of triangle B. This proves that it is similar to triangle B!
    3
  3310. Fredericus Johannes Maria Barning (1924-2012)
    3
  3311. The animation at the end is a thing of beauty. It lets me intuitively understand what it means. Thank you.
    3
  3312. This reminds me of the identity tanA + tanB + tanC = tanA x tanB x tanC, but I cannot see any connection.
    3
  3313. I learned the first one of those proofs first, and the second one before I got out of high school. I had no idea that they aren't taught anymore! 😮 Geometry has always been, as you called it, pretty to me. Thanks for this!
    3
  3314. POG
    3
  3315. My favorite proof technique❤️
    3
  3316. Hey guys, I am from the future (2125 AD). Proving Riemann hypothesis is so damn easy! I don't get how it took 200+ years to prove it. If this comment section was a bit wider, I would write the marvelous proof here.
    3
  3317. 42:41 Continued fraction to be continued!! BRILLIANT!!
    3
  3318. The final claim, that an integer is a difference of two squares exactly if it is odd, is incorrect. In fact, an integer is a difference of two squares precisely if it is EITHER odd OR divisible by 4. The reason is that being a difference of two squares is equivalent to being the product of two factors that differ by an even number (and thus can be represented as x-y and x+y, which differ by 2y), and this is possible for any number that can be factored as a product of two odd numbers (1 included) or as a product of two even numbers.
    3
  3319. 10:02 Suppose that m and n are odd, therefore [m/2]=(m+1)/2 and [n/2]=(n+1)/2 ("[x]" means round up in this case); therefore, in the last factor, j=(m+1)/2 and i=(n+1)/2, therefore the cosines end up being cos(pi/2)=0; therefore, the last factor is 0. The formula is a product (a product of products), therefore, if at least 1 of the factors is 0, then the formula is equal to 0. (^.^)
    3
  3320. Soviet mathematician/programmer Alexey Pajitnov was working on something like that, on tiling pentaminoes using computer. But computers of that time weren't fast enough, so pentaminoes were too difficult to them and he decided to consider easier problem with tetraminoes. While playing with them he accidentally created computer game and called it TETRIS.
    3
  3321. When I was in high school, I tried to use this method to develop a formula for prime numbers. I didn't really understand its significance back then and I barely understood calculus. I just knew how to work the formulas. I think I gave up because I eventually reached an alternating series of 1 and -1 and I didn't really know what to do with that. Looks like it's time to revisit the problem 😃
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  3322. This tickles just the right part of my brain. Thank you.
    3
  3323. (5:44) "In this part of the mathematical world we start at zero" Me (a programmer): First time?
    3
  3324. More like zeroth time :-)
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  3325. The proof is in the end of the video 29:00.
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  3326. Haven't even watched yet, but when YT showed me a brand NEW Mathologer vid, I immediately smiled.
    3
  3327. Wow, the Doc has a huge office.
    3
  3328. This all indeed is very neat! 🎄🧡 This triple tree reminds me of a tree of Markov triples a bit, though their constructions aren’t at all analogous.
    3
  3329. chapter four was F**** EXCITING!!!!!!!!!!!
    3
  3330. great stuff. the 'evil oracle' version of what's next is an interesting logical idea. every time you guess, he says "no, that's not it" and gives you another number in the sequence that doesn't match your guess.
    3
  3331. What do mathematicans do when they compete in a race?... they step on the gauss
    3
  3332. Tesla had very poor limited skills knowledge in Advanced math and physics. His two phase AC 6 wire motors were not used . Steinmetz phaser equation s are used in Electrical engineering as i his Three wire 60hz systems for design and engineering since 1900s world wide. Sadly he became insane and homeless later in life. Many myths and legends are not accurate about him. He was a genius though in certain areas. I live near his Museum on Long Island his old experiments still on old RCA land in Rockpoint forest exist and should be a national park. He died poor and broke. I respect him.
    3
  3333. You're right. It's not that difficult to check the identity from the Triangle involving Phi. The golden ratio is the ratio such that 1 + 1/phi = phi. Then, multiplying both sides by 2×phi and writing 2 × phi + 2 as phi + 2 + phi gives us the identity we wanted to check! :) Nice and enlightening video! Here in the Philippines, I was taught Heron's formula in a math training guild, but it's not taught to all students. :) It also is a tool math enthusiasts like us can use to detect impossible triangles in math tests. I saw impossible diagrams of triangles in 2 separate BuzzFeed quizzes. One of them had an answer of 12 corresponding to ½ × base × height, but applying Heron's formula would give a decimal rounding to 14 [another answer choice!] :)
    3
  3334. Just started watching. It is amazing with the n-side-gons. Thanks for making this video!
    3
  3335. The enumeration as "九百八十七" is a relatively modern way of writing Chinese numerals, influenced by the Arabic way. In ancient Chinese texts, the digits are always listed first, followed by the magnitude. In other words, it would have been written as "九八七百", aka "nine eight seven hundred" - sort of like modern scientific notation.
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  3336. Since you asked in the video.... I think i grasped more than half of what you said and understood a fairly continuous thread of ideas from beginning to end. I think I could get it all with a few more viewings. My skill level is weak freshman, unused for 4 decades. Btw did you notice that the ideas presented have strong links to major unanswered questions in math?
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  3337. Czechia mentioned!!! 🦁🇨🇿
    3
  3338.  @Mathologer Yes it is! I'm so honored to receive a reply from you 😍😁 keep up the great math content!
    3
  3339. Hey Burkard! Would be great if you did a video on how to animate math videos in your style,this will enable other ed. Tubers to bring more visuals in their content which directly engages more audience towards the glory of math
    3
  3340. It's interesting that the symbol ∫ was long used for discrete sums as well. For instance, in Summae Potestatum, Jakob Bernoulli wrote ∫nn = ⅓n³ + ½nn + ⅙n. The interpretation is that the left side is the sum of the first n squares and the right side is a polynomial in n. It kind of annoys me that n is playing a dual role here (both as the index and the upper bound), but I guess it wasn't considered confusing at the time.
    3
  3341. speaking of freaks let me introduce myself 👅👅👅
    3
  3342. Merry Christmas Mathologer!
    3
  3343. I SWEAR I JUST DISCOVERED THE FORMULA MYSELF YESTERDAY!!!
    3
  3344. It was Douglas R. Hofstadter's GEB that introduced me (and I guess many of us) to this fascinating man from India, while the intriguing man from Germany quasi introduced himself, through these nonpareil YT videos of his.
    3
  3345. He gave us 2 Christmas gifts this year (first gift was the sphere video last month and second gift is this video)
    3
  3346. The whole video is a collection of masterpieces! There are multiple kinds of beauty in there, that makes me love mathematics. The simple, yet powerful proof by Kemptner, the unintuitive, yet real-world brick-tower, the magic of the crossing of finte to infinite leading to the approximation formula, ... My favourite part was the γ with it's amazing properties like the e-gamma-prime formula. Thank you for such interesting content!
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  3347. Always having a great time watching Mathologer !
    3
  3348. Although your work is always intensely brilliant, about four minutes in, you started a quick summary without the requisite image. I assume you didn’t notice the image wasn’t present. Cheers
    3
  3349. These visual proofs are so comforting to watch :)
    3
  3350. Hello Mr. Polster. Thank you for the videos. I really like your style! The only problem I have with your videos is probably (?) uncompressed audio. Some parts are really quiet, the next moment it's frighteningly loud. I'm not an audio engineer, so I might be wrong, but it will be more comfortable for listeners if you apply compressor to the audio. Then it will be easier to find an appropriate sound volume to watch your awesome videos.
    3
  3351. Student pilots actually use this device in training. Never in practice, but in theory classes and exams. It's a modified slide-wheel that can calculate pressure altitude, speed corrections, temperature correction, etc. We refer to it as "flight computer". Edit: I just saw the bit at 10:50
    3
  3352. actually after a few experiments is not hard to find examples satisfying a generalized form: 3^3 + 4^3 + 5^3 = 6^3 3^3 + 10^3 + 18^3 = 19^3 4^3 + 17^3 + 22^3 = 25^3 5^3 + 30^3 + 40^3 = 45^3 6^3 + 8^3 + 10^3 = 12^3 7^3 + 14^3 + 17^3 = 20^3 8^3 + 48^3 + 64^3 = 72^3 ....
    3
  3353. Great video! Lots of other YouTubers might have stopped after giving one explanation but you really went the extra mile with the animations and multiple proofs. Thank you for sharing this with us!
    3
  3354. At 23:00, you need to take at least 3 numbers that add to 15. It does not work for 8+6 or 6+9.
    3
  3355. Interestingly (and perhaps not too surprisingly), while there is an infinite solution to Conway's Soldiers in a certain sense, there is no well-founded solution. That is, every solution involves sequences of steps which are forwards-infinite (no final move) and also ones which are backwards-infinite (no initial move). You can't do it in the intuitive way of having some function f(α) on the ordinals α≤κ (where κ is some large countable ordinal), where f returns a legal position for each ordinal in its domain, each proceeding from the last by a legal move, where f(0) is the starting position and f(κ) is the winning position. If we do allow arbitrary forwards- and backwards-infinite sequences in a solution, we can actually do better. We can make a soldier appear anywhere on an empty board! That's clearly not acceptable, so an additional requirement is necessary. Specifically, there is some number N such that no space on the board flips from empty to occupied more than N times. With that extra condition, the best we can do is reach row 5, using up every soldier in the process.
    3
  3356. That was super amazing. One of the best Mathologers I've seen.
    3
  3357. Yep.
    3
  3358. This concept can be extend to tetration (and higher hyperoperations).
    3
  3359. Its really cool to learn that mathematics I am familiar with often turns out to be generalizations of even more powerful mathematics
    3
  3360. I always get sad when your videos finish 🙁
    3
  3361. Is it me, or do these look like hyper polyhedra?
    3
  3362.  @Mathologer  Yes. But I doubt there is one for this one, though; since it’s not an integer, the leading coefficient is 1, and the constant term is non-0. :)
    3
  3363. I think the method for the 7x7 was trial and error. Like for the purposes of the drama the plot was that these methods haven't been discovered yet so they made a deliberate effort to not use the same method for any two of them, they likely wrote a program to brute force it with trial and error.
    3
  3364. Aber klar doch :D Hier auch noch eine Aktivität in der Kommentarsektion
    3
  3365. As to the (once) folded triangle, there is no such ratio in general, since it is not congruent to the original one. As to to the folded folded one, the video explains tha the ratio is 1:3 for the lengths and 1:9 for the area.
    3
  3366. Cool!
    3
  3367. I felt that
    3
  3368. The pentagon star is {5/2}, which, by simple pattern matching, means that there are "2 sided polygons", ie lines, and 3 sided polygons, or triangles.
    3
  3369. Make a video on circle inversion
    3
  3370. Mathologer, I have a book titled MATHEMATICAL MASTERPIECES, published by Springer Verlag, where you could find the necessary demonstration why Archimedes is Euler's dad or grand dad, where you could discover many more of his masterpieces. I want you to find out the source of these masterpieces. I think some of the Indian mathematicians or some unknown Ptolemaic mathematician may be the culprit. However, your friends Andrew and Henry made my day along with you. Thank you.
    3
  3371. I always drink hot coffee, choco or tea in the morning to wake me up... this is far stronger than 4 cups of coffee
    3
  3372. Ah that's better
    3
  3373. This video will blow up
    3
  3374. The last proof also made my day :)
    3
  3375. Nice video before i get a nice lunch
    3
  3376. Our plan was that the extra flat-earther subscriptions will push mathologer over 2²⁰.
    3
  3377. All n-polygons in a 2D plane, can be described by a sequence of n complex points. Using a discrete Fourier transform they can be decomposed into their spatial frequencies (which are also complex). The inverse Fourier Transform of each of these n points, will result in a regular convex or star polygon.
    3
  3378. I followed the proof ALMOST to the end. I fell short with the P divides BLUE. I mean, I understood it, but it would take me a lot of time to do it myself. Unfortunately Primes are my Achilles Heel. For many of us Math fans, they are super interesting. For me they are just quite interesting. The only theorem or rather conjecture about primes that fires me up is twin-prime conjecture. But since it has been 15 years since I've dabbled into scientific Math language (and never on that level), I can't read many proofs. That's why your channel is so great - it allows me to look at the same level as real mathematician, but doesn't require knowledge of rigorous Mathematical language. And frankly there is nothing better for me than your proofs by contradiction. Even though towers are sometimes hard to follow, I adore your video on transcendence of Pi by proving that tan(x) can be written as infinite tower (I think it was by Lambert). Sometimes I have to pause and analyze, but it is such a beautiful thing. Thank you and sorry for long comment - I tried to be brief, my usual comments look more like treatises :D
    3
  3379. Christmas glasses: The squares numbered 20r and 20g must be part of their outer 2x3 "arms", or they'd leave a 5-square board that can't be tiled. These two 2x3 sections can be tiled 3 ways each. So far: 3 x 3 = 9 ways to tile the outer two rectangles. Then, the red and green squares numbered '8' and '15' can only be part of 4 different pairings (e.g. 8r and 15g must both be paired to their left or their right), otherwise they'd isolate a section with an odd number of squares. Call these 4 pairings First, Outside, Inside, Last (like FOIL). EG: 'First' = both sets of pairs made with the left neighbour. Using 'First': Working left-to-right, we get a 2x3 grid (3 ways), then 13g-9r and 14g-10r must be paired, then another 2x3 grid (3 ways), then 17r-7g and 16r-6g must be paired, then a 2x2 grid (2 ways). This multiplies to a total of 3x3x2 = 18 ways. Using 'Outside': Working left-to-right, we get the same 2x3 grid (3 ways), then 13g-9r and 14g-10r must be paired, leaving a 2x2 grid in the middle (2 ways) and a 2x3 grid on the right (3 ways). This makes a total of 3x2x3 = 18 ways. Using 'Inside': 17g-7r and 16g-6r must be paired, leaving a 2x2 grid to the left (2 ways); likewise, 17r-7g and 16r-6g must pair, leaving a 2x2 grid to their right (2 ways). This all leaves a 2x4 grid in the middle (5 ways). Total: 2x2x5 = 20 ways. Using 'Last': Same as 'First', but mirrored. Total: 18 ways. Multiplying all this together gives: 3 x 3 x (18 + 18 + 20 + 18) = 9 x 74 = 666 Truly a 'beast' of a solution.
    3
  3380. That "new discovery" at the start is completely obvious to everyone.
    3
  3381. At 17:15 and beyond the first series became 1,1x2,1x2x3 denominators rather than 1, 1x3, 1x3x5 at the beginning of the video which looks like a mistake. Or to test how well we are paying attention :)
    3
  3382. The best Mathologer video I've ever seen (and the level of Mathologer videos is always very very veru very high...)
    3
  3383. This channel deserves many millions of subscribers, it's very underrated. Unfortunate, except for the current followers! :)
    3
  3384.  @Mathologer  THAT MEANS A LOT TO ME SIR!! THANK YOUUUU! ALL THE BEST IN EVERYTHING YOU DO SIR! 👍👍👍
    3
  3385. The so called "bow-tie lacing" here is also known as the Army boot lacing and is by far the best balance between speed, convenience, strength and style there is in my opinion.
    3
  3386. Me: The max overhang can't be that complicated Mathologer: ■■■ ■■■■■■ ■■ ■■■■ ■■■■ ■■■ ■■■■■■■ ■■■ ■■■■■ ■ ■■■■ ■■■■■ ■■■■ ■■■■ ■ ■■■■■■ ■■■■■ ______________ Me: Holy S---
    3
  3387. This may be the one Mathologer Video to approach being "age appropriate for all". Thanks.
    3
  3388. This probably isn’t a very common opinion, but I think that the ugly optimal stacks were one of, if not the, most interesting things in this video. I’d really love to see the maths that went into finding them.
    3
  3389. 11:22 Working out the width of the cross section of such a paraboloid at a given height h yields two times sqrt(t - h) where t is the total height of the paraboloid. The corresponding cross section of the cylinder minus something has the width 2 times (sqrt(t) - x) , where x is again the radius of the inner circle we don't know yet. The respective areas are pi times sqrt(t - h)^2 and pi times sqrt(t)^2) minus pi times x^2 . Simplifying: pi times (t - h) equals pi times (t - x^2) , i. e. h = x^2. In other words, the sides of the part cut out of the cylinder are another paraboloid, and since it is described by the same parabola as the paraboloid with which we started, its volume is also the same.
    3
  3390. My brain is overheating!
    3
  3391. (8x1),4,4 and (8x1),2,10 are the only solutions for length 10!
    3
  3392. Another truly amazing video! Ramanujan was amazing, and so are you!
    3
  3393. 19:49 if your proof uses directional angles, then the proof is not a lie!
    3
  3394. For any two integers a and b, you can find a Pythagorean triple with (a² + 2ab; 2b² + 2ab; a² + 2ab + 2b²). These won't always be primitive triples, unless of course you pick 2 consecutive numbers out of the Fibonacci series. What is more the GCD between a and be will be a factor of the GCD of the triples. This is as far as I've gotten, but it is fascinating. I hadn't gotten to the proof section yet when I left this comment. I was playing in Excel.
    3
  3395. You remind me of the sweet genius 😂
    3
  3396. The last move is a trap. Take shelter and "Advance to move 1023."
    3
  3397. Mahalanobis-. Which house no. Does the man live in Ramanujan- all right, so we take this infinite fraction... True genius, man
    3
  3398. Beautiful stuff! I'm a first year engineering student and I have to say, you really make all these deep stuff really easy to understand. I understood almost 95% of the video. The more I watch your videos, the more I feel like I should follow a career in pure Mathematics. It's just so intriguing. Only if our teachers taught like this :(. You are really doing a great job!
    3
  3399. Mind blown. I love it. And you explain it so well I even get the feeling I understand it a bit. Great vid, thanks. Now I'm off to watch the others you mentioned.
    3
  3400.  @Trucmuch  Thanks for the detailed answer. I had forgotten about Karaji. So I'd be happy to call it a natural tetractys :) But why give Pascal all the credit!? You're right though I'm having a righteous inflammation :D
    3
  3401. I liked the way you saw it as an ordinary number rather than a binary number. It removes one layer of obscurity. Also, yes, short problems between your larger more involved ones would be great. Regarding the incomplete rule problem: There are three options: - You can't flip the last card. - If you flip the last card, you don't flip any other card. - If you flip the last card, you flip the first card. The first two options work but using the last option, it is possible that it may not terminate though I'm not sure if I can prove it except that the upturned cards can no longer accumulate to the left. Also, if you reverse the labels (0 for down-turned card, and 1 for up-turned card) then, obviously, the number will progressively increase till you reach the maximum number of 11111111 (or, in non-digital form: 11,111,111).
    3
  3402. My math teachers were generally so mind-numbingly awful and painfully mechanical that it's amazing I can add. There were a few -- but important -- stars in the darkness of that firmament, to whom I remain grateful.
    3
  3403. Thanks Burkhardt! Eve & I were super thrilled to see this episode. We like watching you & 3b1br. I had mentioned to her all of this, with appropriate references to mandarines, and as she doesn't know Andrew she was amazed that he would do that for her! Afterwards she was a bit surprised, I think, that mathematicians get paid for such fun! 😂 . Keep it up please!
    3
  3404. 46:00 the attack
    3
  3405. You pronounce Madhava perfectly
    3
  3406. first comment
    3
  3407. There's an implied 2019 hidden in the ... !
    3
  3408. The 347 animation also reminds me of a hypercube projected on a plane and rotated.
    3
  3409. aaw, you really got my hopes up for a octagon in a tesseract grid for a second 😅.
    3
  3410. Everyday you upload feels like christmas. I was always fascinated by generating functions. Thanks for the great content as always! Greetings from Germany :)
    3
  3411. Another interesting topic related to logs is the "Irish Logarithms" used to implement digit multiplication in the Ludgate mechanical calculator around 1900. I would like to see a Mathologer talk on that one!
    3
  3412. I just had a lecture on Digital Signal Processing last semester, funny how this fits
    3
  3413. loshu is never old!(just inscribed on a turtle shell🐢).Love this channel!
    3
  3414. The solution to the math problem in the painting is 2. 10²+11²+12² → 100+121+144 = 365 (denominator) Since we know 13²+14² is the same quantity, we're effectively doubling the numerator to get 2/1.
    3
  3415. Binomial, it is everywhere, gotta love it
    3
  3416. This is why we should never stop playing with all the subjects.. there's a lot of beauty still hidden! Thank you for the video :)
    3
  3417. Our math teacher in france asked us to watch your video ;)
    3
  3418. It's weird, I often worked with 4d (and 6d) polyhedra, including rotating them, but it was always just numbers for me, I didn't consider that a 4-d cube rotating would make such a mesmerising shadow!
    3
  3419. "... a journey back to some school Maths …", Me: "Yay, always my favorite subject."
    3
  3420. I am a 10 th class student from India and this high class maths are not taught to us . But amazingly I am following Mathologer from a looooong time , and his presentation is such that high class maths are not seemed to be such hard or such not understandable. Thank u Mathologer , I had started research on various topics of math just by inspirised by you ☺️☺️☺️☺️☺️☺️
    3
  3421. sliderule... יוי
    3
  3422. Fortunately, he didn’t use the 7 that replaced the 5.
    3
  3423. WOW!!! Bravo again! This is another indisputably perfect example that supports my theory, metatheory, proofs & metaproofs that show how and why nature's astrophysical geometry, geometry, numbers, maths, and logic are enabled & sustained by the natural metalogical principles of being (i.e., the "cosmos"). I will definitely cite (& link) this episode in my next draft of "Astronomy, Geometry, and Logic" (and formally request permission to use a pic or 2 from the video). Dear Burkard & Marty, thanks again for doing the best, most useful maths series on Youtube.
    3
  3424. ?
    3
  3425. 4≠1
    3
  3426. JC is king.
    3
  3427. Okay, but the golden ratio appearing everywhere in art is such a pernicious myth with no basis in reality. That's about as true as when Veritasium claimed that 37 was everywhere just because you start noticing it when you're thinking about it.
    3
  3428. Hi Mathloger! Great video. I made an arctic circle generator for web, its hosted here https://wrsp8.github.io/ArcticCircle/index.html please visit it on a desktop/laptop, for some reason the page doesn't behave well on mobile devices, I might fix it later.
    3
  3429. Didn’t Matt Parker do Puzzle One as a Matt Parker’s Maths Puzzler a few weeks ago?
    3
  3430. But I just want a seperate video on the pattern formed on your T-shirt if it really mean something 🥹😢😢
    3
  3431. There are no ways to make change for a googol dollars. A googol dollars is much larger than the entire global money supply.
    3
  3432. I think you should make videos on research done by some other ancient indian astronomers such as aryabhatta, brahmagupta, bhaskaracharya, varahmira. These are some of the ancient scientists who were actually ahead of their time.
    3
  3433. This is my favourite math video ever, for some reason. I'll certainly be trying the programming challenge. ~ earlier thoughts ~~~~~~~~~ With the triangles and squares drawn in, it looks like a hypercube gone wrong. Ohhh... The coin rotation paradox is no paradox if you count rotations from the point of view of the line; the edge of the center coin. It only seems strange when you count rotations from an external point of view. Adopting the line's point of view is much the same as unrolling it.
    3
  3434. I love the way this guy teaches! Amazing when someone loves what they do...
    3
  3435. Why is math so beautiful?
    3
  3436. I don't think this was offered as a serious suggestion, but that twisted triangle version of the logo looks REALLY good. Maybe change it for the month of october, as a "Halloween costume" for the channel :)
    3
  3437. I had a small slide rule in high school, but my prize possession is a large slide rule from my dad dating back to 1953. The common joke was that an engineer would give as answer to the multiplication 2 times 3 that it was “about 6”.
    3
  3438. At 6:16, if you substitute -2 for x, on the first line you get 0^0 = 1 :)
    3
  3439. When i was kid i played around with my dad's sliderule...
    3
  3440. I really enjoy watching your videos Burkard, thanks to you I've rediscovered the thrive and passion to take pen and paper and try to prove things. Thank you so much.
    3
  3441. I really enjoyed that. As a follow-up, could I suggest that you become the Maptologer for one video only to reveal the Math behind various map projections and the various possible compromises when moving from a sphere to a subset of the plane?
    3
  3442. First?
    3
  3443. 21:10 nice job, editor.😉
    3
  3444. 5:25 I guess the ”null” function (f(x) = 0) is also in that family of f(x) = ce^x; just setting c = 0 🤔. Because f(x) = 0 sure as hell is its own derivative, as you can very easily check.
    3
  3445. It is right when the original triangle is equilateral! Then the "funny" circle coincides with the second, outer circle.
    3
  3446. 5:00 One minor thing, he mentions the area of the harmonic series rectangles is infinite and therefore the area under the curve is also infinite, but doesn’t clarify that the corresponding surface area of the revolution of both of those around the axis is a constant times those cross-section areas. (I.e. the curved surface area of a cylinder is 2πrh, and in the example here h=1 and r is the harmonic series). He does clarify this a little more for the next part involving the volumes.
    3
  3447. In the description it reads "CHAUCHY Schwartz"
    3
  3448. Stop giving it sugar, it won’t get so hyper and complicated. (Though in seriousness, I heard that is a myth sugar makes you hyper)
    3
  3449. There was a Finnish book series published in the 1930s and 1940s called "Tietojen kirja", that is 'book of knowledge'. I have a copy of the series. It had "courses" on many subjects like Finnish language, Latin, economy, biology, chemistry etc., and also of course mathematics. Calculus was introduced saying that many people are apprehensive of this "higher mathematics" but said that many methods are more straightforward than complicated arithmetic problems. They showed it to be true. Then there were some examples of mechancial integrators etc.
    3
  3450. Wieder super! Frohe Weihnachten, lieber Burkard!
    3
  3451. Spoiler for anyone interested in the square within a square puzzle : https://youtu.be/5VaQVXECaVM
    3
  3452. I wanted to sleep but couldnt stop watching. Now I have to wake up early tomorrow :')
    3
  3453. I have known Ian's website for ages, being myself a kind of knots geek. I have used Ian's knot for my shoelaces for almost 20 years (more than 15 for sure), and it really drives me crazy when I see someone who ties their laces with a granny knot. Lastly, my 4yo niece was proud that she could tie her shoes, alas, her mother learned her, and she always ties them the wrong way. My niece was so disappointed when I told her, as gently as I could, that she learned it incorrectly and that it would be better for her to learn it again... I think I will try to teach her Ian's knot. It's really easy and fast, once you have the hang of it. By the way, am I right to suppose that the boat illustration comes from the Ashley's book of knots?
    3
  3454. Also this shows a general method for generating curves of constant width. Given a set of strait lines of general position (no two lines are parallel, no three lines intersect at the same point) you can always draw a constant width curve as complex as you want. I don't remember the algorithm exactly but I suppose you'll manage to reconstruct it.
    3
  3455. 8:05 YES! 14:07 YES WE DO!
    3
  3456. Nice t-shirt
    3
  3457. 28:30 but this theorem is NOT TRUE :) - what have i won? I started proving it, read about the normed spaces, Tichonov 3,5 topology and conformal LINEAR transformation and that helped me to disprove your theorem about similar triangles :) similar triangles in 2D euclid plane will not necessarily add up to similar triangle because of reflection - simply take a △ and a reflected one ( -△ ) its trivial but even mirrored vertically will not keep up the similarity. In the case of not reflected similar triangles the theorem is trivial and comes out from the not-reflected similarity definition and normed space definition ;)
    3
  3458. One of the cooler features of those flight computers is on the back side is a device that you can mark with a pencil and figure out wind speed and direction without doing any trig. Then you can reverse it to get the heading to hold to fly a strait path to your destination and effective ground speed. Flip it back over, then use that speed and distance on a map to figure out the time until destination.
    3
  3459. 14:18 Shouldn't that be a + in the denominator?
    3
  3460. 6:20 fun thing is, I've encountered enough people to know you can make zero cents in many more ways than one Pun intended ;)
    3
  3461. 🔥
    3
  3462. For me you could go further, im having that feeling that I need more. I’m a physics student and I really miss those beautiful relationships in math.
    3
  3463. Like that shirt! Sieht echt cool aus, danke für das tolle Video!
    3
  3464. This was awesome to watch. I felt my eyes shining.
    3
  3465. I noticed the first few seconds of Babooshka by Kate Bush as this video started. We're not ignorant, you know!
    3
  3466. Lol he started reading in German and I just shook my head quickly and said gesundheit 😂😎✌🏻
    3
  3467.  @ABruckner8  You got me all confused since I thought that your first instinct is right. But I kept thinking and was wondering if you meant, that the positive direction would grow infinitely. Then I thought some more and I realized (hopefully that's right), that when you cancel the terms with each other what is left is an infinite series that converges.
    3
  3468. Loved this video so much. Thanks for all the great content this year. Also, this was directly inspiring for my research - what a treat!
    3
  3469. Even if ur videos are long still it keeps us engaged. Good job 👍 appreciable ❤️love from india 🇮🇳 Ramanujan was great🙏
    3
  3470.  @Mathologer  Maestro!! If primary school kids in Australia play with enneacontatriagons then they're in another level , indeed. Megagon is very easy . It's " Ecatomiriogon". "Mega" is Greek too and means BIG , by the way. I would very much enjoy a visual presentation of you , of Gauss proof of Decaheptagon (Heptadecagon is wrong). OK ... 17-gon. Many respects from Greece , Maestro. We are watching you (lol).
    3
  3471. try to disprove the following: for an N sided star we have an N sided polygon P formed by the intersections of the lines of the star. the sum of angles of P is 180 * (N - 2) and for each angle a in P there is an angle b adjacent to it by the star where a + b = 180. the star itself is a polygon S , if you ignored the intersections inside, formed by 2N sides where the sum of angles of S is 180 * (2N - 2). the star has 3 different magnitude angles A,B,C where sum of angles of S = AN + 2BN + CN , CN is the sum of angles we want to find. AN = sum of angles of P = 180 * (N - 2), b = 180 - a, therefore BN = 180 * N - AN = 360. substitute we get CN = 180 * (2N-2) - 180 * (N - 2) - 2*2*180 = 180 * (N - 4).
    3
  3472. There's something beautifully simple about this, once you get to the end it just feels completely obvious that it works. Love all your content, definitely helps me to see things in different perspectives and to engage with problems better! Now and then I get into little mathematical rabbit holes of my own and end up taking between a couple weeks and months to intermittently chip away at them til I get somewhere satisfactory, I actually just got started right now on a new such thing and probably am gonna have to break into programming cause the spreadsheets won't cut it. Thanks for the inspiration <3
    3
  3473. I'm learning differential equations right now and I'm very happy to say that at 34:50 you said exactly what I expected to hear.
    3
  3474. A related problem for programmers: see Project Euler problem n°31
    3
  3475. What does the Committee of Magical Mathematics think about the Parker Square?
    3
  3476. There is actually an implicit speed limit and its the limit of reasonable driving, and going way to fast than your car can handle is considered reckless and will get your license revoked and a couple of points in flensburg!
    3
  3477. 31:30 hmmmm this suspiciously looks like a geometric sum... let's look at the formula again: Φⁿ=F(n)Φ+F(n-1). this gets us F(n)=Φⁿ⁻¹-Φ⁻¹f(n-1). plug again, and again, and again until you get to F(0), which gives you... a geometric sum! How Cool!
    3
  3478. Shame on me watching this for free😞
    3
  3479. This is very interesting! I've only watched intro and while I did find the connection between first 4 numbers and 1st pythagorean triple on my own in school (I was bored), for some reason I never extended beyond it. I thought it was coincidence. Especially since I had to double the 2nd number and din't know about the radii of the circles. Only numbers and area.
    3
  3480. Challenge time! I will write out the answers, but I don't have time to write the whole proofs, so I should conjure Fermat and only write the main points for each challenge. 1. 5. Big circle through 2 points leaves 5 points to be split, so 2+3=5. 2. Trivial, as "a"."b"_"c"_, with a, b, c length l, m, n can be turned into a fraction by multiplying 10^m and 10^(m+n) respectively and subtracting. "abc"-"ab"/10^(m+n)-10^m is the result. 3. Ramsay theory. For each point there is either 3 segments or 3 non-existing segments. Focus on the three segment/non-segments and the three end points. If there exists a same type interaction between the three, a triangle is created. If there are none of that type, the three automatically make a triangle. 4. 315? Not really sure, as I do not do Rubik's cubes, but each part seems to have 7, 5, 3, 9 cycles. Lcm is 315. 5. {1, 2, 4, 8, 16, 32, 64}. Binary represantion, trivial as 1, 2. 6. Queen of Hearts. First card shows that the suit is hearts. Dictionary order is 3rd, so 3 spaces away means queen.
    3
  3481. That "weird" french lacing at 2:09 is the one I've always used and for many years thought everyone else did, too. It is quite surreal to me that the "criss-cross" lacing may be the most common, especially with dress shoes in formal settings.
    3
  3482. 93. An hour EARLY...
    3
  3483. Those shapes reminded me of cellular automata.
    3
  3484. 28:10 - This is the exact set of shapes that is used for the pieces in the board game "Blokus," produced in the US by Mattel. (But there is a mistake in the video. The stairstep 5-square piece has an extra square, making it a 6-square piece.)
    3
  3485. Yes!
    3
  3486. Waited the whole month for your vid :D you're the best!
    3
  3487. Another gem!! Here's a little side-note –– At 28min+, where you go into approximating √2 with PRT's that have legs that differ by 1 (b = a+1), you can get "best" rational approximations by adding the legs and using the hypotenuse as denominator (this amounts to averaging the two legs): √2 ≈ (a+b)/c = (2a+1)/c = (2b–1)/c For the 3-4-5 PRT, this gives √2 ≈ 7/5 For the 20-21-29 PRT, this gives √2 ≈ 41/29 For the 119-120-169 PRT, this gives √2 ≈ 239/169 All of these are solutions of the Negative Pell's Equation for N = 2: (num)² = 2(denom)² – 1 which makes them "best" rational approximations of √2. Fred
    3
  3488. this proves flat earth
    3
  3489. Beautiful proof, beautifully explained!
    3
  3490. "Off with their heads."
    3
  3491. Unfortunately im a big Euler fan and i already knew everything in this video, but i love to see it over and over again. There is never enough Euler.
    3
  3492. 12:30 Why does one of the green squares disappear during the fade? That really confused me for a minute xD
    3
  3493. This was fascinating! I enjoyed the shorter form for a change, and I look forward to the next deep dive.
    3
  3494. Danke! Viel zu viele Videos über Tesla gehen leider ins esoterische, natürlich befeuert durch diesen ganzen "freie Energien" Quark.
    3
  3495. I solved the angle problem differently. Inside of these regular n-pointed-stars is always a regular n-polygon. N-polygons have an angle sum of 180*(n-2). So the angle sum of a pentagon is 180*(5-2) = 540. This means a single angle of the pentagon (because it's regular) is 540/5 = 108. Every angle of the inner polygon builds a triangle with two outer points. So we have a triangle with one angle of 108 degrees and two of the unknown angle. And a triangle has an angle sum of 180*(3-2) = 180. In formula this means: 2x + 108 = 180. So quick math: 180 - 108 = 72. And 72/2 = 36. So each angle is 36. The same goes with the 7-pointed-star. 180*(7-2) = 900. 900/7 = 128,571428. 180 - 128,571428 = 51,428571. And 51,428571 / 2 = 25,714285. And if you multiply these angles by n you always get 180. So in the end you just have to divide 180 by n. Quick Proof: [ 180*(n-2) ] / n + 2x = 180 [ 180n - 360 ] / n + 2x = 180 180 - 360/n + 2x = 180 | + 360/n 180 + 2x = 180 + 360/n | - 180 2x = 360/n | / 2 x = 180/n | * n x*n = 180
    3
  3496. Learned about divisibility by 9 and 3, but you're right, the remainder piece was not mentioned in my school even though why it works follows quite naturally after a few seconds of thinking about it.
    3
  3497.  @jannegrey  Where to find it?
    3
  3498.  @Mathologer  I have a Ph.D. in Mathematics, but I can't count that high. :)
    3
  3499. This video is so elementary to any math buff, I predict it will overtake all other videos on this channel… …except maybe the one about -1/12.
    3
  3500. 9:49 has the interesting conclusion that 1/(1-x)=sum[x^i,{i,0,Infinity}]=product[(1+x^(2^i)), {i,0,Infinity}] (at least if i didnt do anything wrong). very nice!
    3
  3501. I made it to the very end. :) Wonderful video, so many fascinating properties coming from something simple.
    3
  3502. Great T-Shirt! :) I love "Nightmare before Christmas"...
    3
  3503. الفكرة أنّه إذا كان لدينا قطعين زائدين يحصران بين مقاربيهما نفس الزاوية فسيكونان قطعين متشابهين، تحويل السحق والتمديد لم يغير الزاوية بين المقاربين ولذلك صنع قطعاً زائداً مشابهاً، لكن إذا كان هناك شكلين متشابهين يمتلكان نفس المساحة فسيكونان متطابقين، هذا يثبت الخاصية في بداية الفيديو
    3
  3504. The timing of this video—was it inspired by the ongoing bankruptcy case of Alex Jones in the US? Just 10 days ago, the judge overseeing the case blocked—for the second time—a proposed distribution of Jones' assets between two groups of plaintiffs. (Most people missed the news because of—well—crazier things happening in the US.) The case has all the hallmarks of a Talmudic deliberation—what's fair, what's legal, and what's the best way to get closure amidst conflicting objectives.
    3
  3505. Mathologer is reading my mind... During the last video, I had to make a presentation about tilings, now I have an exam about discrete mathematics including these infinite generating functions in 3 days. Coïncidence? I think not!
    3
  3506. Kerala was overrun by Islamic rulers in 1565? Which ones? Btw his name is Madhava. That's how he himself spelt it
    3
  3507.  @jursamaj  That's how I did it also. Very simple and straightforward 🙂
    3
  3508. 37:30 if you look at the hexagon like a space filled with cubes, you notice that grey walls appear in every rank on every row, orange walls appear in every rank in every file, and blue walls in every file on every row; meaning for a size M hexagon, you have M^2 of each colour
    3
  3509. Thanks for injecting a bit of mathematical joy into my day, and informing me of this Indian mathematician.
    3
  3510. Definitely the way to go if you are a flat Earther :)
    3
  3511. Have a look at the description of this video :)
    3
  3512. The pattern of the 1st 5 numbers of the last "What comes next?" problem is: 1, 3, 8, 21, _55_; since it is the sum of the products of all of the partitions (including non-distinct duplicates) of each positive number; and the next positive number in this problem is 5, so we have: 5=5 4=4x1 4=1x4 6=2x3 6=3x2 3=3x1x1 3=1x3x1 3=1x1x3 4=2x1x2 4=2x2x1 4=1x2x2 2=1x1x1x2 2=1x1x2x1 2=1x2x1x1 2=2x1x1x1 +1=1x1x1x1x1 ---- 55
    3
  3513. amazing!!! and a question: is there a method of constructing a magic square containing only primes? that'd be the next level, isn't it?
    3
  3514. The infinite fraction formula blows my mind. How do you discover something lie that!? How do you think to take those steps!?so damn cool
    3
  3515. "I hate to break the news to you...." LIAR! You love it! :D
    3
  3516. 4:26 If the mini-blooper is (probably spoiler below) is that there are no character for zero (零/〇), that's not a blooper. Ancient Chinese numerals do omit these zeroes when written out. The notation of "〇" as a zero digit actually comes from Hindu-Arabic numerals, and the usage of "零" for zero digit comes from its original meaning "remainder". During calculation, zero digits are represented by empty space with counting rods, or no beads moved in a position in abacus. For example, "二百五" is "two hundred (and) five" or 205; in comparison, modern Chinese write 205 as "二百零五" ("two hundred (and) remainder five"). To disambiguate the position of zero digit, the position value of the present digits will write out if it is not at 1s place: "一千八十" is "one thousand (and) eight -ty", the presence of "十" (ten) indicates the eight is in 10s place, so this number is 1080, not 1008. And actually, ancient Chinese numerals also likes to omit ones if they are not in 1s place; the position value will indicate the presence of digit, and defaults to one on that position. So 1080 may also omit the "one" and written as "千八十".
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  3517. this also motivates the jump from spherical geometry to the hyperboloid model of the hyperbolic plane
    3
  3518. Mathologer, Just a note to thank you for all your videos, this is great work, so inspiring. Just the right level of compromise between rigour and popularisation to deal with such amazing topics! Thanks! Sebastien
    3
  3519. For people in Calc, this will make sense given integration work with logarithms. The integral of 1/x is ln(x), and the integral is the area under a curve.
    3
  3520. Sooooper....... And great explanation and visualisation.... But there's a need to quench the thirst of Euler McLaren series.....pls do it as soon as possible
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  3521. Fun :) its also fun to see the difference in relation to poisson ratio of some incompressible solid that can deform in any given direction like a fluid. Its quite intuetive, if you unfold a sphere into a circle just by bringing down all the meridians you will get a larger area, but the difference between the circles of latitude on the sphere and the circles of radius r that corresponds, gets you a number for how much you should retract the distance between each circle of some radius on the disk to be area preserving. I wonder whether the angle of the meridian turned at 90 degrees onto the disk to the radius of the disk says something important ;).
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  3522. 1. . . . . . . If the cube pattern had worked, b would be equal to 6*1=6, and we get 5³+6³=7³, which is off by 2: LHS=341 and RHS=343
    3
  3523. How are your videos still free to watch ??
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  3524. I love how the 9 flips at 24:45! Lovely video!
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  3525. By 14 00 india was being strangulated by violent Islamic invasion which destroyed the universities in nothe hence many scholars ....earlier north south west east scholars in India were traveling and ideas spreading esp with pilgrimage and temple towns but all these being destroyed and destruction spreading to both many knowledge got destroyed and those got survived by people who might have lived in poverty away from political hotspot and exile without much resources and intellectual company ......we should respect who saved. Some of the knowledge of yoga, Ayurveda , music etc
    3
  3526. 30:15 The phi is equal to (sqrt(5) + 1) / 2 the other solution to the equation is (-sqrt(5) + 1) / 2. The difference between them is sqrt(5). If we rearrange the phi = (sqrt(5) + 1) / 2 we get sqrt(5) = 2 * phi - 1. So the other solution in terms of phi is: phi - (2 * phi - 1) which simplifies to: - phi + 1 = - (phi - 1) and since phi - 1 = 1 / phi, -(phi - 1) = - 1 / phi.
    3
  3527. Love it! My favourite was "The Leaning Tower of Lire" overhang by natural progression revealing the harmonic series. Nice!
    3
  3528. Now that 300K people have watched this video only 99.995% of humanity haven't heard of quadratic reciprocity
    3
  3529. This was excellent. I’d love a part 2 that goes into the other scales on a slide rule
    3
  3530. Wow!! I derived this amazing formula years ago but it never occurred it could be obtained by a generating polynomial. :O Thank you as always to the Mathologer team! Fun facts: the dimension, m of the most numerous bits of an n-cube converges to n/3 as n increases. This can be shown by setting the derivative of 2^(n-m)*(n choose m) wrt to m equal to zero, while using the derivative of f(m)^g(m) and Sterling’s approximation. For a large n, we get a bell shaped curve presumably due to the DeMoivre-Laplace Theorem.
    3
  3531. 2:10 I cant be the only one thats been tying their laces like this forever.
    3
  3532. I absolutely love your channel Burkard. I bought your book "QED: the beauty mathematical proof", I enjoy it so much. I am happy every time I read or wacht your explanations :) ans I even get insights from them for my research! Thanks for this and please keep it as much as you can :)
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  3533. I love mathologer
    3
  3534. #1: Nine disks need forty-one (41) moves. My solution is the "minimum of (two times a smaller number of disks plus the minimum of the remaining disk moves) method. #2: An observation: if d (# of disks) < p (# of pegs), then you ALWAYS only need 2*d-1 moves. Basically, having more pegs than disks means you can move every disk to its own peg, including the base disk (to the designated ending peg), then move all of the other disks back on to the next bigger disk, which you should have already moved to that ending peg: (d - 1) + 1 + (d - 1) = 2d - 1. #3: Likewise, if d = p, then m = 2d + 1, since you will have to move whichever disk is on the ending peg (or move another disk, if the ending peg is the only one available to the biggest disk), move the biggie, move the disk you had to move beforehand, and then move all the other disks onto the biggie: (d - 1) + 1(clear off the ending peg) + 1(biggie) + 1(split the double stack up with the starting peg) + (d - 1) = 2d + 1.
    3
  3535. You are very engaging to your audience, and coupled with the excellent visuals this video was made very easy and pleasant to watch. Thank you!
    3
  3536. That would be like saying 5 shouldn't be prime because all numbers ending in 5 we know are not prime. Numbers ending in 5 or 0 we know are divisible by 5 just like even numbers we know are by 2....5 and 2 are special any the only numbers this works for in base 10 number system. 5 and 2 are the only numbers that evenly divide into 10 so we can project the structure
    3
  3537. First challenge: If you cut out, for example, the two black squares adjacent to the upper left green one and other two green squares away from them, then you won't be able to tie the chessboard with domino pieces because the upper left green square is isolated. Second challenge: if m and n are odd, then m+1 and n+1 are even. If you take j to be m+1 divided by two and k to be n+1 divided by two, then you get pi over two inside of the cosines, so the two cosines evaluate to zero for those two particular values of j and k. Then you have to multiply that zero by the other factors, which will just give you zero.
    3
  3538. That reminds me of binary finger counting: so the thumb is one or zero and so on; two fingers counts to three; Three fingers counts to seven; Four fingers counts to fifteen; You can count to thirty-one on your first hand in stead of just five, and with two hands, you can count to two thousand and forty-seven! I also developed trinary finger counting, by bending a knuckle in stead of just folding the whole finger closed, so then the thumb cam be zero, one OR two, and you can count to two hundred and forty-two on the first hand!
    3
  3539. In india people would just say it's a right wing propoganda and ignore this stuff
    3
  3540. The tension distribution analogy was great. I can't believe slide rule and circle rule were dropped. They give a intuitive volumetric connection to things. Math is geometric first and only numerical when you try and pin it down
    3
  3541. Amazing theorem. It's a funny description, that Ptolemy "beats" Pythagoras... Like ellipse does to circle. The proof video at the end is amazing. Thanks for the super work. I'm wondering if there is a work for finding the sides with - smallest - integer values of the convex ones (incl. diagonals). BTW I liked also "1/s+1/r=1/q".
    3
  3542. 34:58 See the rightward facing open boxes, left ones, or upward facing, but not all three.
    3
  3543. 5:53 Madhava looks like Mathologer.
    3
  3544. 2:40 mathemagical wand*
    3
  3545.  @Mathologer  Absolutely! When I saw the film I thought, of course that's a binary number, you don't have to explain it.
    3
  3546. second
    3
  3547. Magical. How did Ramanujan think? How did he come up with crazy identities like this?
    3
  3548. At 12:32, one square magically disappeared
    3
  3549. You are by far the best math channel on YouTube
    3
  3550. Wow! 100 videos and 8 years. When I first found your channel I was a high school student learning calculus and now I’m a graduate student studying statistics and coding for public policy. Your passion for math helped me find wonder in what was being taught in class and put me on the path for success that I’m on today!
    3
  3551. Those were some very beautiful animations 😍 … where do you get your background music? It really completes the experience.
    3
  3552. I had both kinds at one time. You can also do square and cube roots by dividing the linear distance between one and your number in half or by a third. The advantage to the circular was the printing was on the wheel and it had two clear pointers you could set the distance between and unlike the stick kind you never had to worry about it falling apart on your way to and from class. Since accuracy is based on distance to print on the circular at four inches around was about as accurate as a 12 in rule. I think they also had trig functions on the wheel or with slide rules along with a linear scale and maybe one marked off from 0 to 90 or 360. Our high school math books came with trig and ln functions printed as tables in the books.
    3
  3553. 0:21 "mathematicians, artists, wizards, and a lot of crazies" ... It feels like you've said "mathematicians" four times!
    3
  3554. It took me a while to figure out what the circles on the T-shirt represent. What do the changes in light and dark colors mean? It looks like a simple repeating light-dark-dark pattern, but that's as far as I got.
    3
  3555. Hi! My name is Alex 15 years old and I'm a big fan of yours. I discovered the exact same formula about 2 years ago while... well let's say I had too much free time. I was really really excited to see that you've made a video about it. And thanks for the proof. I kinda missed that part when I showed it to my classmates by the name " I can guess your polynomial ".
    3
  3556.  @Supremebubble  this connection doesn’t surprise me - it is my favourite Pythagoras theorem proof. … and nice work on nerd-sniping Burkard.
    3
  3557. I know our ancestors weren't stupid. ...But aren't we giving the guys who wrote the Talmud too much credit, there? The first part of the video explains it all IMHO. I would be very surprised if their thinking went beyond what we see at about 10:00
    3
  3558. Mathologer is by far the greatest math "niche seeker" of all times. I mean, the pattern he seems to follow is : 1) Explore old scientific libraries, read books and memoirs. 2) Find some obscure but strange result in the works of some 19th or early 20th century number theorists mathematicians. 3) Try some examples, explore them and find astonishing patterns. 4) Find a way to show them visually, using all ressources of modern computing power. 5) Use your great teaching skills to build up a video clip, turning those extraordinary obscure things into seemingly obvious visual demonstrations 6) Watch you audience collapse in awe. Awesome work, by all means. God, I was mathologerized!
    3
  3559. I am so excited that I could come across a Chinese element here!
    3
  3560. 10:30 i guess another thing to mention is that it doesn't matter which pair of angles, since a+b+c+d = 360 in a square, meaning a+b = 360-(c+d), and so cos((a+b)/2) = cos(180 - (c+d)/2) = cos((c+d)/2)
    3
  3561. When all the lines are connected, they look like oscillating 4 dimensional shapes.
    3
  3562. i like the dot proof because i wouldn't have thought of it, and it's very beautiful.
    3
  3563. Never learned the proofs at school. However i found both on my own when I wanted to check if I was able to do it :)
    3
  3564. Great video! Quite approachable and always enjoyable. 😁 And RIP Ron Graham. ❤️
    3
  3565. I remember finding a photo on google once about this. I showed my teacher it. The proof was very simple. It’s really interesting how there are an infinite number of ways you can construct these types of numbers. I am looking forward to watching this on your Channel. I don’t usually comment before watching a video. So I might have some questions.
    3
  3566. The video was amazing. I was playing with something similar to this with the exp expansion minus one, a few week back and it is really helping me kind of tie things together. As with many your videos i will have to rewatch and do a bit of extra reading and calculatorizing before i can say i understood it all, but that is exactly why i love this channel. As for my background in math I took a calc class once and count back change on a daily basis lol. Please hurry with that next video this is truly YouTube's best math channel.
    3
  3567. You explained rather hard mathematics with the simpler language than my teachers. But English is the foreign language for me. This short 'lecture' in english I understood better, than 2 university courses of maths in my native language.
    3
  3568. Q of hearts
    3
  3569. 15:30 If the larger coin is x times the size of the smaller coin, then the smaller coin does x+1 rotations on the outside and x-1 on the inside. Did I get that irght?
    3
  3570. Truly the first strand type game. Hideo Kojima has outdone himself once again
    3
  3571. Just revisiting this one. It's so cool! I'm now wondering... do the differentials of the polynomials have any geometric meaning in the same diagrams?
    3
  3572. 11:00 Also; √2 is also a component, in the left side of Ramanujan’s identity; it’s just in the denominator. The left side is, basically, just: (√π*√e)/√2. 🤔
    3
  3573. whos the one who disliked it
    3
  3574. My background in math is sleeping through Math 31 (which included Calculus AB)... and still getting 96% equivalence on the exam.
    3
  3575.  @jaspermcjasper3672  who said it's otherwise?
    3
  3576. To find when the Pizza Cutting sequence is a power of 2, one must solve P(n)=2^m, where P(n) is a particular quartic. This is analogous to finding Figurate Fibonacci numbers, say solving n^2=F_m, where F_m involves powers of Phi. Nothing new upto f(10^10), where f = the polynomial. I suspect 256 is the last.
    3
  3577. Love this. Thank you.
    3
  3578. Avoid red/green contrasts in diagrams. 10% of people can't distinguish them.
    3
  3579. I solved this 3 years ago. Everyone just needs to go learn binary and forget all else. Most math youtubers will not tell you the complete truth because they want to monitize on their books and puzzles similar to the toy creator.
    3
  3580. Labeling the endpoints CW from top left: A, B, C, D, E, F Any circle through A and B is centered somewhere on the angle bisector of the top angle Any circle through B and C is centered somewhere on the angle bisector of the left angle (similar for C-D, D-E, E-F, F-A) QED: all the points lie on a circle centered at the intersection of the angle bisectors. (The intersection of the angle bisectors is the center of the incircle.) Not proved here, but true: the angle bisectors intersect at a single point.
    3
  3581. Wow, a video uploaded at 12:28 AM AEST by someone who lives in Australia...
    3
  3582. Watched the whole video. Crazily enough I would say that this was maybe easier to follow than some of your previous videos. I was studying maths and physics but quit btw...
    3
  3583. Потрясающе. Доказывали в школе, что на сетке нету правильного треугольника, но совершенно не так весело.
    3
  3584. It's so amazing and beautiful!
    3
  3585. World class quality! So beautiful. Impossible to not understand. Almost forgot to breath for 50min...
    3
  3586. I love how some of the transitions make the image of Ramanujan smile.
    3
  3587. Nice π T-shirt !
    3
  3588. Great video as always! There's a gear-rotation argument which is equivalent to the coil-uncoil one. Happy holidays!
    3
  3589. The magic square at 18:50 is constructed by step up and to the right (wrapping as needed) for each number in the sequence, and then shifting down one by every sequence of 5. There is also a green dot missing on the 7 (which confused me for a bit while determining the sequence).
    3
  3590. Personally I didn't mind the use of Ai art, but I'm this case the slide hurt my eyes to look at
    3
  3591. Due to the PHP, you'll always have a pair with the same suit. Put one of those two cards in the front, so you get the suit. The same does not work for the value of the card, though, so we need another plan... At this point, you have 3 cards of encoding to figure out which of the 12 remaining cards (same digit, not the card that encodes the suit). My first idea for encoding the digit of the card was to use the order of the cards. Since the first card is necessarily at the front, you only have three cards, or 3x2x1 permutations to work with, which isn't enough. My second idea was to use high-low order of the cards, which could use the first card for encoding. This gets to 2^3 or 8 different values, still not enough. In the second idea, though, an edge case popped out where the card indicating the suit is the highest or the lowest value, and can't be "higher or lower" than the next. That can be solved by giving different suits higher values (i.e. diamonds are "lower" than hearts, etc. However, the edge case does point out that we can choose which of the cards to take, which gives us another "digit" to encode with. For clarity, we'll call them the numbers 1 through 13. So how do we use those digits? If you always take the lower card, you get a worst case of 1, which still gives you 12 possibilities, which is not less than the 8 we need. What if we always take the card closer to 7, though? Well, 7 is the worst case, and give us 12 cards to dig through still. 6 would give you 1-5, 8-13, so 11 I gave up here and watched the video. Makes sense.
    3
  3592. Answer for third challenge (glasses) is 666=3^2(3*3*2+2*3*2*3+2*2*5), calculated it out of curiosity
    3
  3593. So if at y(0) it explodes why not just take value of y at some other value where it doesn't? Or are all the other values difficult to estimate?
    3
  3594. Dafuq are you on about? That's Dirichlet's drawer principle.
    3
  3595. Oh no, I've been TRICKED into learning about Calculus and e-related mathematics! 😆
    3
  3596. So many nice reminders of maths I haven't thought about in a while. Great pedagogical approach.
    3
  3597. Every comment has a heart!!!! Why not this then??? I Liked the video mathologer
    3
  3598. 15:34 - That's precisely what I did. But today we have MS Excel and other tools for number crunching and comparing if the solution is correct 😉
    3
  3599. 27:40 Oh wait, I have that book! I devoured it in 1 or 2 days because of how fascinating it was.
    3
  3600. If I had a time travel based wish, I'd go back to one of these early mathematical geniuses and teach them modern notation. See what they could do with it.
    3
  3601. I had a circular slide rule when I was at school, and I'm less than 400 years old.
    3
  3602. Burkard the troll strikes again. 😆
    3
  3603.  @nycoshouse  Yes, that is, the Fibonacci recursion rule applied to those starting values. But the point was, it eerily resembles the first several digits of π, but then after spitting out that 14 (to fill the earlier 'gap' it left?), it goes onward looking nothing like the digits of π. Like the monkeys in that Newhart routine.
    3
  3604. you can also get all those derivatives (multiplied by n!) by expanding (x + a)^n. This information is in Burkard's video here but it's cool to see all of the derivatives together in one equation! Amazing what just spills out of the binomial theorem
    3
  3605. I loved this video. I was able to follow it, and learned as well. Very interesting.
    3
  3606. As somebody who has taught (or at least, tried to teach) this level of calculus, I'd be curious to know how many people are able to complete calculus questions (e.g. differentiate this complicated function) as a result of watching this video. In my experience, a huge blocker is not understanding the pre-requisites. E.g. what a function is, algebraic fluency A couple of interesting examples: 1) A graph of a function is provided (without any algebraic counterpart). I had many students who could answer 'find x so that f(x) = 10' but not 'find f^{-1}(10)'. The reason is that many students do not know what f inverse actually is. They just memorise the symbolic manipulation of writing f(x)=y, rearranging, and swapping symbols around. 2) Most students cannot do this instinctively and rely on some kind of algebraic manipulation: "A straight line goes through the coordinate (340, 23.7) and has gradient 0.2. The straight line goes through the point (341, y). What is y?"
    3
  3607. People in the Netherlands feel so relaxed they even put their name and home address openly on public documents. However maybe Mathologer would like to hide that in his video to prevent any annoyances. Great video by the way.
    3
  3608. Including 1=1 like that also opens the door to include any n=n identity for any value of n.
    3
  3609. I had never seen the product and chain rule derivation before. Honestly, I probably would have had the professor I had for Calc 2 and 3 been my prof for calc 1. My Calc 2/3 prof didn't like just giving out formulas.
    3
  3610. For higher dimensions we just need the limites of any and all angles to go to 0 or 180 for the area to be miniminzed and non wrinkly, so you can just look at the substitutions seperately and garuantee a nice limit by overpowering the bad substitutions that increase angles by a larger substitution per step or something lile that for the other kinds of substitutions that effect that spesific angle affected by the naughty kind of substitution, then chrismas can be saved.
    3
  3611. Wow! That was awesome. Thoroughly enjoyed this new way to think about an equation I was already familiar with.
    3
  3612. Some new ones for me there. always nice to see. out of the many Combs books I have, I have to say that many of the PHP examples feel a bit contrived in that you have to have a go through a lot of hoops to spot the pigeons and holes. my favorite is "10 players in a complete Round Robin chess tournament. 1 point for a win, -1 for a loss, 0 for a draw. more than 70% of all games ended in a draw. Show that at least 2 players had the same total score." From Principles and Techniques in Combinatorics by KOH, KHEE-MENG; CHEN, CHUAN-CHONG
    3
  3613. Monsieur Polster, vous êtes le roi du "mathematical slow-mo". C'est adorable!
    3
  3614. Geometric infusion like infusing a triangle with an inner and outer circle with each side of the triangle representing a ratio. This can work for triangle mesh ratios between 2 nurb or dsub models. All part of a big growing science geometric engineering.
    3
  3615. Every one of your videos is so amazing, Professor Burkard. Best on the internets
    3
  3616. 1:24 Welcome to the wonderful world of conspiracy theory and woo (they just love Tesla's 369 and the vortex) :~D
    3
  3617. Geometric proofs are the best
    3
  3618. The most memorable for me what the no 9's series. I've just learned all about Benford's law because of the conspiracy nuts, but of course I did not think about the impact of increasing orders of magnitude before the 5 seconds was up.
    3
  3619. Probably you are that Madhava in your previous life 😊
    3
  3620. math suggestion: foias constant it's so weird to have only 1 seemingly-random number to diverge
    3
  3621. No it is not reversible. There are an infinite number of choices at each reversed step. In terms of the visualisation of the pentagon as the sum of the eigenpentagons, each step zaps one of the components to size 0. So this component could have had any size - the “before” sum can’t be deduced from the “after” sum. The maths is that of orthogonal projections in the vector space C⁵ (5-dimensional complex space). Projections are the natural example of non-invertible linear maps in linear algebra.
    3
  3622. The videos where calculus, geometry combines are awesome
    3
  3623. 12:32 for whatever reason I thought of this as a binary number and was confused about how this might be true at a glance, and distressed when I thought of 3^2 = 9 = 1001. I then realized you obviously meant decimal, and that numbers of this form are 1 less than a multiple of 3 since 100...002’s digits sum to 3.
    3
  3624. 37:15 seeing a few solutions, mine is to use an alternative definition of the sign of a permutation as the number of "swaps of 2 elements" in the set its instantly clear from the definition why the number of elements in the set you are doing the permutation on will determine if the number of swaps is even or odd the trick is you have to see that a right shift is equal to starting all the way on the right and doing swaps for each pair of adjacent elements and working from right to left one card at a time
    3
  3625. Excellent video I enjoyed very much . Please upload more videos on this topic .
    3
  3626. I got the 2:50 question right
    3
  3627. With the first part I thought he was gonna go into the "x lines divide a circle into y different pieces" thing, which is a sequence that closely resembles 2^n for the first few terms Edit: He eventually got there but definitely went on some interesting tangents along the way
    3
  3628. actually good pronounciation of "Adrien Marie Legendre", good job!
    3
  3629. https://tradeideasphilip.github.io/aztec-tiles/ Try it out live!
    3
  3630. I love your math videos ❤ Lots of Love 😊❤
    3
  3631. Very interesting, this video expanded my knowledge on how infinite sums behave, thank you!
    3
  3632. This is the sad reality of BHARAT that we are not taught about these all interesting things. In fact,we are forced to forget our own history.But,a gradual change is happening in the world.
    3
  3633. Genius dimensional harmonics.
    3
  3634. Area of the small tree is 5 as each branching square has factor 1/2 from its original square but also doubles the number of branching squares. Meaning you only need to count the unique colours as all coloured areas =1
    3
  3635. Great stuff. It's always amazing how you manage to find such intuitive explanations. Most memorable is probably the "no 9s" visualization.
    3
  3636. Thanks for another fantastic video! I really liked the first proof in the introduction because during school I only learned the second algebraic proof. It was neat how you could prove identities about sin, AM-GM, etc using these diagrams!
    3
  3637. Excellent video. Thank you. Folks should also check out: What's 1/2 Doing in Φ? - the Golden Ratio What's inside Φ? - the Golden Ratio on the Imaginary Angle YT channel.
    3
  3638. Mathologer would have sent the prince a lunchbox with the sums written, leading the prince to feel the need to at least discover more ^^
    3
  3639. What does the phrase "key to the universe" even mean?
    3
  3640. Cool
    3
  3641. Fun fact: at my school (not American nor European) we weren't taught the quartic formula per se, but a method of solving it by calling it a different variable (like x^2 = y) so you could solve it through quadratic formula then solve it again through the roots found in it.
    3
  3642. He was a member of the first Dutch Math Group to attend a course of a brand new revolutionary computer language ALGOL, in Darmstadt, Germany, 1955.
    3
  3643. It's certainly easier when you don't have to deal with trigonometric functions
    3
  3644. 28:14 but what if you optimize the chain in a way, that at every point it is just strong enough to not break under its weight. This way, it could be thin and light in the middle, but gets stronger and heavier the further out you go. Interestingly the result involves the normal cos... it has the form -ln(cos(x)) (insert some constants for a more general result)
    3
  3645. Amazing video!! one of the best youtube channels!
    3
  3646. I don't know if this means anything but congrats for finding this fact and you just got a new subscriber! Let's blow your channel up everyone!
    3
  3647. I went to school in the 80s and we had a BBC Micro with a game that had a turtle. You could type commands into it to get it to draw a line behind it. Not sure if it supported equations, I was only 5!
    3
  3648. Velcro anyone?
    3
  3649. 01:15: Me: "Why not multiply through by 4?" 01:17: "Oh."
    3
  3650. I'm a level designer, I make levels for video games, and I use the golden ratio for similar reasons as described in the Nature's Numbers section. If you have a limited number of art assets to place in a level, it's important to get as much use as possible from each one. In situations where I need to place one asset multiple times, I rotate each one by the golden ratio times 360 degrees (or 222 degrees, which is close enough). Imagine a row of bushes along a street. If each bush is the same exact model and placed with the same rotation, small details that repeat on each bush will stand out, and the human brain is really good at seeing that sort of pattern. But if each bush is rotated in the horizontal plane by 222 degrees more than the last one, then as much 'new' side of each bush will face the player as possible.
    3
  3651. OMG. I wrote my first (FORTRAN) program 55 years ago to find the solution for x = cosh(x). And yet I didn't know that cosh and the catenary were the same thing until just now! Thank you, Mathologer :)
    3
  3652. Me, seeing the visualisation: cool i should put that in geog- Me at 2:36: oh you did it awesome
    3
  3653. Still kinda stunned by 2^n being analogous to e^x. What about e^n? Is that still its own difference? Is there a concise way to account for the difference between 2 and e when going from discrete to infinite?
    3
  3654. iSeeker says, on StackExchange: “For those interested in the origin of spinors, one might take a look at a couple of papers by Jerzy Kocik, in which he argues that Euclid’s parametrisation of Pythagorean triples is (by a couple of millenia!) "the earliest appearance of the concept of spinors”
    3
  3655. Great explanation! Logarithms are fascinating
    3
  3656. A very nice video!
    3
  3657. You can find a nice discussion of the history of this sort of paradox here http://tinyurl.com/3pe2eaav
    3
  3658. Aww, this is sweet!... I knew it, that this clever way of dealing with the coin change problem must have been done already! I've played with this coin change problem recently, deliberetely not knowing about backtracking approach or dynamic programming - I just wanted to see what can I discover on my own (later on, I found that I did not do anything radically new, but I was pleased anyway). I'd loved, of course, to find some formula for it, but theoretical approach was not my goal back then, I just wanted to make my computer to calculate the damn thing. I was not particularly interested in calculating the minimum number of coins (of certain denominations) that add up to a given amount of money, but only in the number of ways such amount of money can be presented. I was also not very keen in writing computer program to do the job, just to make a table, filled with formulae that are calculating everything. The reason for all this is that for many years dealing with various problems I developed for myself a lazy approach to programming - part of the work (a huge chunk of it, if possible) can be done in table forms, another part - with much simpler program, if writing such a program is inevitable, otherwise custom functions are good enough in many cases (I mean, in Excel spreadsheets). So, practising such "programming" style helped me to learn how to compartmentalise any heuristic problem in smaller, more "edible" problems. Firstly, I made a counter that gives me the answer, just for verifying the results (for relatively small amount of money, of course; for bigger ones it has to count for hours). Easy-peasy. The analysis of the problem resulted in following table: 1. The number of columns = the number of 'n' distinct positive integer values (whole numbers), arranged in increasing order as c(1) through c(n); oddly, the results do appear in the leftmost column, associated with the smaller coin in the set, c(1). The set of coins may be whichever one wants {1; 2; 5; 10; 20; 50; 100;...}, {1; 2; 5; 10; 25; 50; 100;...}, or some exotic: {3; 7; 12; 19}, {1; 2; 3; 5; 7; 11; 13; 17; 19;...}. One even can play dumb and use a set as {6; 9; 18; 30}, but the table will give back zeroes for all odd and not divisible by 3 sums anyway. 2. The number of rows is not limited above, with increment of 1 cent per row. Naturally, you should have 100 rows if you want to calculate the function for 1 dollar, or 250 rows if the target is 2 dollars and 50 cents. 3. In every cell [x, y] there is a formula that does one simple thing (for the leftmost column is slightly different) - it calculates the ways 'x' dollars can be presented by coins with nominal value >= c(y): cell[x, y] = sum( cell[x - c(y), j] ) for j = y to n / for 1st column: cell[x, 1] = cell[x-1, 1] + sum( cell[x, j] ) for j = 2 to n. This is all, now every row will give you the corresponding number, for example, in the sequence A000008 (it is only for set [1, 2, 5, 10], doesn't include coins of 20, 50, 100 cents). If we use all coins and banknotes up to 20$ , {1; 2; 5; 10; 20; 50; 100; 200; 500; 1000; 2000}, the sum $20,21 will be presented in 30,399,653,516 ways. Clearly my counter will outlive me here ;-). Now on, the hypothetical program is easy to be written and if one looks closer a little bit on the method, will see that this program will have to memorise maximum n*c(n) numbers to do its job (or even less, if you are greedy), for arbitrary large amount of money. I was delighted that this problem has such simple practical solution, but thankfully, here I learned about the heavy, theoretical stuff, and they are also so elegant. The practical approach has some similarity with Pascal's triangle, so, no wonder that binomial coefficients do appear here and there.
    3
  3659. Most memorable moment was when you showed the e^gamma identity and there was a giant capital Pi and I felt like I was going through a 2001: A Space Odyssey montage because I thought only Sigma did that.
    3
  3660. These videos should really be called 'Mathologer Masterpieces' instead. Thanks to everyone who contributes to making them.
    3
  3661. this one was one of my favorites for sure. Completely peaked my interest in the pythagorean theorem as I was already interested because it's basically the foundation of trigonometry which also makes waves and fourier series and a lot of stuff in physics ofcourse. And I have a particular interest recently in the connection between Fourier series, Taylor series, ect, and Calculus. I mean have u noticed that the Fourier transform is basically using an infinite number of curved lines to make a straight line, and an integral is using an infinite infinitesimal straight lines to make a curved line?
    3
  3662. Beautiful stuff Mathologer!
    3
  3663. This thumbnail is absolutely unhinged lol
    3
  3664. I've always held the opinion that calculus is easy since the second time I tried to learn it.
    3
  3665. (x+1)^0=x , but how? I think it's a mistake.
    3
  3666. I found the pattern where the sum of odd numbers produces squares in high school. I actually came at it the other way, I was looking at the differences between the sequence of squares and it popped right out, clear as day. I had no idea why that pattern existed, so I took it to my math teacher and asked her to explain what was going on and why. But she didn't know either.
    3
  3667. 24:36 I assume that this way of generating random tilings does give the same probability to each possibile tiling: ie, I assume it has been proven that by this method we don't reach a certain type of tilings with a greater probability than other tilings. Thanks for the really amazing video.
    3
  3668. I independently stumbled onto the Fibonacci cascade when I was in high school… ended up using the name as my gamer tag and my account here… still find it really cool, and sometimes I wish I went down an academic path in mathematics, but had no idea how to go down that road. Still don’t. Oh well, cool videos though, keep up the good work (and I want pretty much all your t-shirts, man)
    3
  3669. Wait a second. ln(2) = nice pi? 3:02
    3
  3670. Cool!
    3
  3671. Finding minimal polynomials for r and s: r^2 = s+1 -> r^3 = rs + r = 2r + s -> r^3 - r^2 = 2r - 1 -> r^3 - r^2 - 2r + 1 = 0. Similarly, s^3 - 2s^2 - s + 1 = 0. Meanwhile, using trigonometry on the regular heptagon we get r = 2*cos(π/7) and s = 2*cos(2π/7) + 1 = 1/(2*cos(3π/7)). This means we can get exact values for cos(nπ/7) with the cubic formula. Also, for completeness, φ = 2*cos(π/5).
    3
  3672. 3b1b not only has a video on the Basel problem, but also on thus problem with this solution. It's interesting to see 2 different people explain the same proof.
    3
  3673. 16:00 is xor logic :)
    2
  3674. Newton - 🔥
    2
  3675. Good question :) Luckily @bsmith6276 has already provided a very good answer :)
    2
  3676. Posted 2h ago!! Since last week i was starting to be a bit worried. Ok time to watch.
    2
  3677. Excellent video as always :)
    2
  3678. I'm 34 and I just learned about Pascal's triangle yesterday.. Very interesting indeed.
    2
  3679. Interesting! I was actually informed of the Benjamin Franklin magic square (and magic circle) just a day before this video released! Now I'm thinking about things like geometric magic circles.
    2
  3680. I have a round slide rule like this. Never called this magic, just logarithmic. The manual calls it a "Concise" Circular Slide Rule, No. 28N
    2
  3681. This should be taught in the computer science curriculum. There is a lot of need to find closed forms for sums in algorithms design. It seems that there is a tantalizingly close way to handle discrete and continuous differences with the same formulas. ex: f[x + d_x] = (x + d_x)^3. d[f] = (f[x + d_x] - f[x]). You can take d[f]/(d_x) as a function, and plug in d_x equal to 1 for the discrete case, or zero for the continuous case. There are two particular problems that I have been trying to figure out with the d[] operator: 1) generalize to handle discrete cases for computer science applications. 2) handle non-commutative inputs, which happens with Geometric Algebra when you multiply vectors, because basis vectors anti-commute. Most of machine learning is just a "million-variable calculus" problem, where you implicitly differentiate a function of millions of variables; to make the weights fit observed data.
    2
  3682. You're as awesome as ever 💖🤝
    2
  3683. Agam bizimle dalga geçir :)
    2
  3684. Aaahhhh, I played with these a lot in highschool and college (and beyond) and it was always something I was doing on my own so it is so nice seeing something familiar in the math world @___@ I remember in a basic math class (learning to write proofs) we were given a {1, 4, 7,...} sequence and had to come up with a formula, so I made 11 come next :P I also remember in physics (which I suck at), we had ~30 data points and we were supposed to approximate a function to fit the data points (it was gravity, so supposed to be quadratic). Instead I found a formula to fit every point exactly XD [*EDIT*: there were a lot of small imperfections in the measurements and equipment setup, so it wasn't exactly quadratic, but it was supposed to be pretty close and I thought a ~30-degree polynomial was hilarious at the time)
    2
  3685. Maybe "shine" is nice, but I learned "cinch". It seems to depend on country of origin.
    2
  3686. Great video
    2
  3687. Mr. Burkard, first of all thank you so much for your Mathologer channel. It is wonderful. I enjoy each and every video and keep waiting for the next one :) I would like to share one thought I've got after watching the latest episode about the Ptolemy's theorem - suspecting it is quite trivial and seems different to me only - for which I apologize in advance. Here is what I thought: the Ptolemy's imparity can be generalized a little bit. Let say, we have a quadrilateral - any in fact. Add the diagonals, i.e. get all vertices mutually connected. We will have a number of line segments (original shape's sides and diagonals). Now split those in pairs such that the segments in each pair would not have common vertices. Then apply that naming convention you've been using in your video, where one pair would have A and a segment, the next - B and b, and C and c. Then it seems that the rule Aa+Bb>=Cc will hold despite which pair is a''s, b's or c's, that is c-pair should not necessarily be of the diagonals. And it is pretty obvious based on the proof you have presented. Then, the case of equality follows with all vertices located on a circle, and c-pair being the diagonals. Thank you so much for your time - and again sorry for being not enough educated to recognize a trivial result. Am already waiting for your next video! Best regards, Mike Faynberg
    2
  3688. I always sleep if I'm stuck on something. I always have the best ideas when I wake up.
    2
  3689. Nothing and none respectivelly. Sorry
    2
  3690. Tree with deep roots needs a Subscription 😢
    2
  3691. Math is just simply gorgeous
    2
  3692. YouTube suggests Graham and Trees, but here we have a Master Class quadrillions of powers sums. Epic I must say.
    2
  3693. Dear Mathologer, I remember doing that overhanging bricks thing for hours in the science museum in London when I was little! About 35 years ago... It looked exactly like your solution with counterweights and bridging bricks, and it set the record! I was there so long that my parents had to put a tannoy announcement out for me! Only thing is, that since bricks are three dimensional, as opposed to your two-dimensional diagram, I had turned the bricks so the overhang was based on the diagonal length of the bricks rather than the length on one side. Also, I had turned some of the counterweight bricks through 90 degrees, allowing them to be placed closer to the fulcrum, giving more stability and greater overhang potential... Like I say, hours....... Love your content. Andrew
    2
  3694. I guessed the number of rotations by the size of the circle being traced.
    2
  3695. Such puzzles and mathematical "mysteries" are like a concentrated food for always hungry young mathematical intellects - they're teaching them to mathematical beauty, elegance and aesthetics, and hone their logic, heuristic ability, pattern recognition, intuition (I saw the shortcut too, yesss!), making seemingly impossible connections between semantically distant objects or facts. For not such young minds it is pure pleasure and recreational activity to tackle such problems. So, thank you, sir, for making my day.
    2
  3696. I enjoy a short interlude!
    2
  3697. Thanks for creating this video and giving credit to the right person (genius). Regards
    2
  3698. The "collection 1: 12 , collection 2: 12" scenario is picking the same pigeon twice and pretending that it is two different pigeons. The pigeonhole principle guarantees that there are two different pigeons in the same pigeonhole, meaning we are getting two different subsets (corresponding to two different pigeons) with the same sum (in the same pigeonhole).
    2
  3699. What kind of arcane trickery is this?! Magic, I proclaim!
    2
  3700. What about complex numbers?
    2
  3701. The thousandth like 😊
    2
  3702. So as shown in the proof, we know what the (x + 2)^n formula actually represents: we start with a 0-d "cube", which is just a point, so it gets assigned a polynomial p_0 = 1*x^0. Then every time we go up a dimension, we make two copies of the previous object (multiply the previous polynomial by 2), then another copy where all the elements are lifted one dimension higher - vertices become edges, edges become faces, etc - so we multiply the previous polynomial by x. That "lifted" copy connects the two normal copies together, and you get an (n+1)-dimensional hypercube from an n-dimensional one. Or in other words, p_(n+1) = (x + 2)*p_n. Now you can think of what objects you could represent if you took a different "base" instead of x + 2. x + 3 doesn't seem to make much sense, as you're making 3 copies of the lower-dimensional objects, then connecting them together with only 1 of the higher-dimensional copies. Seems like one of the pieces will be left by itself, just floating out there. What if you took 2x + 3 as the base instead? It's easy to imagine getting three points connected by two edges after the first step, then you get a 3x3 square grid after the second step, and so on. In fact, you can take any k and get a k×k×k×... grid, then count its features with the polynomial ((k - 1)x + k)^n. These seem like the most obvious options, maybe you could create other shapes by picking different numbers, but I'm not sure if they could be expressed with such a simple formula.
    2
  3703. Maybe (k*x + k) could also work - start with a k-cycle in one dimension, then you get torus-like structures once you go into higher dimensions.
    2
  3704. Nice, I actually understood it.... I noticed the 'beautiful' map 4 color theorem on the list. Is it the same as the fact the max number of objects you can join to every other object in 2D without crossing paths (into the 3rd dimension..) is 4?... My favourite scalable computer architecture is processing cells each connected to each as higher level cells that themselves can repeat the pattern, fractral-like.... The connections are also the (shared) memory.
    2
  3705. isn't it possible for the sum to be divergent? If you add positives terms in such a way that it always increases then you can surely get your sum to diverge to +infty right?
    2
  3706.  @Mathologer  That's awesome! It's paradoxical that rearengement changes the result but it's awesome.
    2
  3707. I think you work for Aspirin and drum up business... My head hurts ;-)
    2
  3708. OMG mathloger is alive! Looks like triangles are spinning counter clockwise like in an invisible ferry with equilateral triangles 🔺 When I've seen the animation 2:40 I chuckled as I got to know something so deep in a trivially easy manner Looks like a star approximation algorithm so chilling..
    2
  3709. Amazing display of the art of mathematics, brilliant video ! Many thanks
    2
  3710. Closing mini-vid, that animation is great.
    2
  3711. I wonder if Ramanujan knew of Madhava's formulas.
    2
  3712. This was a very enjoyable and crazy journey (as most things ramanujan seem to be). However, I felt like the slight handwaving of using a non-converging expression as a bound for a differential equation at the very end was very unsatisfying. Maybe if there had been a couple of words on why that might make sense it could have been nicer. One bit I didn't understand was at 20:15 wouldn't removing the leading 1 change how that expression behaves if it's evaluated as a product 2/1 * 4/3 * ... Instead of 1/1 * 2/3 * ...? That latter one should even converge to 0, so I'm not sure divergence weirdness can explain how you'd reasonable get it to be sqrt(pi/2) Edit : fixed typo 1 - > 0. Also, looking at the partial products I funnily enough find a sqrt(pi) and a gamma function, so basically an e...
    2
  3713. I so appreciate this.
    2
  3714. Awesome Burkard
    2
  3715. 21:10 shouldn't these equal x, not 1?
    2
  3716. It was awesome, I understood every part of it except how the formula in zeroes turned into in ones. I will think about it. I feel i understood the rest. Really appreciate the effort you put in your videos. I am from India and I study in 11th standard
    2
  3717. You deserve over a million subscribers!
    2
  3718.  @Mathologer  Captions are also sometimes helpful to get the exact spelling of the name of something e.g. the mathematician Simon Jacob you mentioned
    2
  3719. Tschuus!
    2
  3720. For me, the most impressive part is that the harmonic series narrowly misses all integers by smaller and smaller amounts. I cannot think of any real event, process, or procedure that behaves similarly.
    2
  3721. Showing the 3d pictures made me think of electron orbital shapes. Any geometric mathematical intuition behind these? If so, please make this video!
    2
  3722. 11:45 do we need the overhang to be >1, or >=1? I see an =1 solution but would need to check some math to find a >1 example.
    2
  3723. ah another amazing fantastic awesome mathologer video
    2
  3724. I swear everytime you publish a video I drop what I'm doing and try my hardest to follow what's going on. :)
    2
  3725. Thought of a magic square while watching this. 1 1 1 1 1 1 1 1 1 Adds up to 3 in every direction.
    2
  3726. squares or 'cubes'? THough as that honestly brings an impressively complicated line of thinking if applied towards dimensional logics. Though as a for a gride, the number of squares is for a flat grid. is for each to points where it's height & length are the same, I am sure this is poorly worded, but assuming a grid of 6x6, there are 36 points, & a shrinking list of pairs, & of that list x&y must be the same. (though a rough concept is in (0,0) (x,y) where y is 6x, 6y and the limit of grid size 6 squared, & the number of squares would be [Grid size=>]6 - Z[<=Largest value of X or Y] = Squares of (x,y) pair [where Z the number of highest value of the number pairs for that point, that is valid] & is repeated for each valid number pair. -GS = Grid size (6x6 = 6, etc) -- that said... the list of valid points would be [GS -1]^2 = VL (valid list of pairs that'd produce atleast 1 square) y = [1, .... GS-1] (independent from) x = [1, .... GS-1] (treated as a range/array, that's lowest value is 1& no repeating numbers, sharing the same numbers - being a square) - The number of squares for that grid would be ---- initial Z value is 0 ---- for each x, do for each y, [GS - {X or Y, Highest value} = tmp] & z = tmp + z Though for a cuboid structure.. it would that on the total assumption of cubic forms (GS^3)*z = TZ (total Z) appologizes for the half programmatic thought on how it'd work. Not entirely sure how it's be written down, but that's how I thought. This is of course, assuming 'squares' equal sides & flat.
    2
  3727. I still have the circular slide rule I learned to use as a freshman in high school. It just felt easier to use than a linear one, and was compact enough I could put it into the case with my TI-30 (in case the battery ran out).
    2
  3728. Lol, I would recommend going to theoria ap… something and checking out the comments, same thing, or any other kronky channel, its always the same, ty notes and appeasement mixed with stupid uncritical questions. Thats YouTube pseudo knowledge channels for you
    2
  3729. Made it all the way to the end. There are some distinctions to be made between "utility" in game theory and "happiness" as in your video, but thank you for yet another fascinating excursion.
    2
  3730. Playing around with Wolfram Alpha I recently found out that (1+10^-x)^10^x approaches e for x to inf.. I'm trying to wrap my head around why that is.
    2
  3731. You can find anything on YT these days, no matter how useless. Some people just don't know what to do with their time.
    2
  3732. 10:00 heard that before from Zvezdelina Stankova on Numberphile 😄
    2
  3733. My brain exploded out of excitement at 14:44 when I realised your argument might be the most beautiful one in existence!😂
    2
  3734. 1:00 to cover the whole surface with paint you need exactly 0 liters of paint since mathematically surfaces are two 2 objects with no thickness and hence no volume
    2
  3735. I love mathologer video 😍
    2
  3736. My brans cells not brain celling
    2
  3737. Euler kind of reminds me of Gary Oldman. Plus, I like his hat.
    2
  3738. To sum it up: 1) Not actually a paradox. Neither in logic nor in counterintuitive result. This is essentially an apparent coincidence that is then explained away. 2) Maybe you should insist in the beginning that this coincidence only happens for n=1 million, and not for 1,000,001 or 999,999. Importance of the decimal base. I guess the presence of 4 in the correction terms leads to very funny coincidences if the numbers are written in base 4. 3) What is truly amazing is the rest of the video, the history, the formulas, the calculations. Thank you and bravo! 4) Did Madhava et alii find 355/113 through continuous fractions? The correction terms have definitely a cont.frac. look. 5) How come Powell did not (or did he?) notice the evidence of the form of the substraction term, one millionth minus one fourth of a billionth (international billion, not US) etc.?
    2
  3739. Time for golden tesseract 😎
    2
  3740. There is also Bhaskara 2, maybe can be subject of another video! https://en.m.wikipedia.org/wiki/Bh%C4%81skara_II
    2
  3741. What Mathologer means by "algorithm" in the Rubik's cube part of the video is fixing a sequence of moves.
    2
  3742. You completely lost me at 5:00, unless I am missing something obvious. You compared the plane area of a series of rectangles with the surface area of a 3D volume. I did not understand how you could make a deduction from that? Unless you are saying that when projecting the surface area of 3D shape onto a plane surface, the area of the projection must always be less than the area of the 3D surface? And if the area of the plane projection is infinite, then the area of the 3D surface must also be infinite? Maybe some additional commentary might help viewers when making that leap?
    2
  3743. Mathologer never disapoints. I'm so glad to know this channel. <3
    2
  3744. Ramanujan was any doubt something different when it comes to write down some crazy identities. The theory of infinite fractions has some very interesting results. On of my favorite is "Lochs´s-Theorem". In a nutshell and imprecise: For allmost all real numbers in (0,1) the continued fraction carries "more" information then the decimal expansion. This sounds so weird. For more precision please take a look at this theorem. It is wonerful.
    2
  3745. If r^2 = s + 1, we cannot have r = s = 0 or 2. But yet, 0 and 2 are valid solutions to the original equation!
    2
  3746. Mathologer!!! Make a shirt that has a complete proof of Pi being irrational by Lambert!!!!
    2
  3747. at 17:53 I laughed so hard that people came in to check on me. Absolutely phenomenal proof delivered wonderfully, as only Mathologer can do!
    2
  3748. Hey mathologer I ❤️ your videos ❤️❤️ Can you make a video about error function..., (Integral of e to the -(X square))
    2
  3749. I actually own that book! Totally did not register that that was you though.
    2
  3750. How delightful! I'm gonna 3D print one of these for sure.
    2
  3751. Not a mathematician, but watch a lot of your stuff, and find it fascinating. It just seems on some level, that because you don't use an equal number of terms, you're cheating. That it's not just a series, but a series accompanied by another rule that says how many terms you can use. Anyway, it was fascinating.
    2
  3752. Fascinating. Thank you for the clarity of the demonstration & the inspiration that comes with it!
    2
  3753. Can you please continue your video on the Euler-Maclaurin series? That was probably my favorite video
    2
  3754. 19:00 Funnily; I’ve been re-consuming ”Toy Story” -content, recently (including 100%-ing the PS1 -video game: ”Toy Story 2: Buzz Lightyear to the Rescue”, in 2–3 sittings), and the nostalgia is real. I’m surprised to see it pop up, in a Maths-video, of all places. 😅
    2
  3755. 2020 factors 1, 5, 101, 505 4 good - 0 bad -> 16 solutions 42^2 + 16^2 = 2020 gives 8 solutions, permuting 42, 16 and +,- 38^2 + 24^2 =2020 similarly 8 solutions Total 16 solutions
    2
  3756. 13:54 I JUST THOUGHT ABOUT THOSE SHAPES AND THE WAYS TO MAKE THEM THIS WEEK!!!!!!!
    2
  3757. I noticed that as well! That is a good insight into the gaps in the errors.
    2
  3758. There is a beauty when it is explained with visuals which you have well explained. You have seemed have done a beautiful job at that for which i can't thank you enough.Great job.
    2
  3759.  @Mathologer  I am too, surprisingly soon. 👴🏻 P.S. Thanks for the heart (and like?) 😌.
    2
  3760. use a plum bob.
    2
  3761. Thanks for another fine work of videomath! Some of my thoughts going into this: There are methods for accelerating the convergence of certain series, especially ones that are slow to begin with. Sometimes they can be applied multiple times, in order to really streamline convergence. Sometimes this will work at first, but then make matters worse after a repetition or two. For this starting series, the "alternating odd harmonic (AOH) series," a very simple acceleration method is to add half of the first omitted term, as you showed early on. This amounts to averaging each pair of consecutive partial sums. BTW, another fairly good computational series for π is the Taylor Series for arcsin of one-half: sin(⅙π) = ½ π = 6sin⁻¹(½) It's not nearly as simple as the AOH series, but it converges a lot faster, with odd powers of ½. It would be most interesting to apply the techniques in this video to that series, I expect. Of course, the Machin formula, ¼π = 4tan⁻¹(⅕) – tan⁻¹(1/239) is much faster still, but it has those pesky powers of 1/239. And then there was Ramanujan's series and its refinements... Fred
    2
  3762. Beautiful
    2
  3763. This seems like a really complicated way of showing that 3 + 4 = 7
    2
  3764. This is not an isolated issue where the orginal inventer has been forgotten or ignored. There are many many many inventions there in astronomy, mathematics, metallurgy, medicine etc which were orginated from India but later the credit claimed by europeons. The europeons have made very disappointing false claims.
    2
  3765. its amazing! i only had basic calculus and lin algebra in uni, i didnt watch the last chapter but everything up until then was amazing
    2
  3766. i would wait for part x-2=0
    2
  3767. Leonhard Euler Vs ramanujan , who will win
    2
  3768. 50 minutes for 10 minutes problem
    2
  3769. PhD in mathematical physics (actually had the pleasure of being taught by Marty!) Followed 90% of it. I was thrown by morph bit (40:40) ... I think I could piece together what's happening, but a dash of motivation would be very helpful. Love the channel, KUTGW!
    2
  3770. Actually, I hit like the second I saw it's a new mathologer video and I made sure I did the second I noticed the runtime. (:
    2
  3771. To be honest, your longest video somehow didn't felt that long at all. Felt much more easy to consume high density of content spread evenly over long time. Enjoyed your effort put into this video.
    2
  3772. go for it, if you go too far, we'll let you know!
    2
  3773. Great effort. For me length was not an issue, I just took few breaks here and there. Your style of presentation is excellent. Those funny bits keep the interest without interrupting the flow, something I'd like to copy in my lectures.
    2
  3774. Background in maths: studied it for 5 years (15 years ago) and continued as a hobby. The video was clear and very interesting. Euler-Mac Laurin foemula reminded me of my first years :) The level is fine, I wouldn't mind a longer level 8 or having a level 9 (if that's one of the things you have in mind).
    2
  3775. 50 minutes works for me - but I'll be watching it in parts. Thanks.
    2
  3776. This video will keep me going for some time. I like these as I have to rewatch them many times. That's alot of value. Thanks.
    2
  3777. This was an awesome video. I would watch any length video or video series put out by Mathologer. I have my B.S in Physics, and I just started working towards a masters in mathematics. The animations for the sum of the powers are wonderful.
    2
  3778. Hi, I felt that I understood everything at least conceptually. Whether I could reproduce it by myself is of course a different question ... I am doing a PhD in Theoretical Physics and I am also generally interested in Pure Mathematics. Some topics that I have heard about and would be interested in (and which I think would fit the Mathologer style): -Complex Analysis especially stuff like Riemann surfaces -Topology and topological invariants (Chern numbers are quite a hot topic now in Physics with topological insulators) -Asymptotic Series -Projective Geometry But I will be watching whatever comes up! Thank you for the great work!
    2
  3779. Magic Video!
    2
  3780. Velcro.
    2
  3781. I learned some of this during my senior design project while taking engineering classes, where i realized that differences were a simple stand in for derivatives for real world problems.
    2
  3782. The contended garment solution seems like what king Solomon pretended to do in the famous case of the contended child. He must have been knowledgeable in the Talmud, as expected :D
    2
  3783. In fact we were taught this in our correspondence math school in the USSR in the 1970s.
    2
  3784. comment for the algo
    2
  3785. Very intersting as always. Did you or Marty play Asturias? Nice choice ;)
    2
  3786. 36:17 1) In a 3x3 square grid, where one single grid square has side A, draw a twisted square with vertices at 1/3 of the 3A side length square 2) take a moment to realize that the twisted square drawing corresponds to the puzzle presented 3) the blue area is therefore A², but given the problem data (2A)² + A² = 1. So A² = 1/5.
    2
  3787. Great for bedtime watching
    2
  3788. About the parametrization of the unit circle using Pythagorean triples scaled down to have a hypothenuse of unit length: Take the first box, about the 3-4-5 or 4-3-5 triangle. The fractions 1/2 and 1/3 are the Y-axis intercept (the parameter t of the parametrization) of the line between the point (-1,0) and the points (3/5, 4/5) or the (4/5, 3/5) respectively that exist on the unit circle. Each box is for one triangle, representing the two possibilities of which side lies on x axis. This is a very rich subject! This rational, pure math parametrization is explained beautifully by Norman J. Wildberger on his YouTube channel. Also there's another subject worthy of Mathloger exploration& perspective perhaps one day, the Miller Protractor and the core circle, x²+y²=x as key between projective (2 variants, an eighth of a turn of rotation away of each other, basically the x=y asymptote of the hyperbola of the red projective quadratic norm x²-y² rotated on the coordinate axis to yield what Norman calls the the green geometry based on the projective quadratic norm 2xy, (x²+y²)²=(x²-y²)²+(2xy)² is the among founding identities of chromogeometry) and the Euclidean, x²+y² -based blue geometry, it is marvelous. 😁 Thanks again for the great video I have been waiting for a long time! 😁🙏🙋
    2
  3789. i think the F_n*F_(n+2)+F_(n+1)*F_(n+3)=F_(2n+3) thing would allow us to calculate a fibonacci number in a logarithmic amount of time, if multiplication and addition took a constant amount of time (but sadly, it takes longer) edit: still, the naive way of calculating F_n would take O(n^2) steps, while this way takes only O(n*log(n)^2), so i'll go into more detail. you first calculate some 5 consecutive fibonacci numbers, let's say 2,3,5,8,13 because that's the first time it skips anything. then you do 2*5+3*8=34, skipping 21, and 3*8+5*13=89. calculate the 89-34=55 between 89 and 34, and then the next two numbers 89+55=144 and 144+89=233, and then you have 5 more numbers, so you can repeat: 34*89+55*144=10946 and 55*144+89*233=28657, and then fill in the numbers around that with simple subtraction and addition. now, this on its own doesn't help calculate all fibonacci numbers, but you can adjust by shifting around the 5 numbers. so for example instead of using 34,55,89,144,233, we could calculate 144+233=377 and use 55,89,144,233,377, and get to different parts of the sequence. but you can figure out when to do that based on the index of the fibonacci number in something like binary (or maybe even exactly in binary)
    2
  3790. I started to play around with fibonacci sequence after you demonstrated how to bend it. I wanted to make a zig-zag pattern and see it if it takes me anywhere. And I got something like this: 1 1 2 3 5 8 13 21 34 55 89 144 ... If the zig-zag pattern is split down the middle you get two sequences: 1 2 5 13 34 89 ... and 1 3 8 21 55 144... First sub-sequence is sequence of odd terms of fibonacci (F1 F3 F5 ...) and second one is sequence of even terms (F2 F4 F6 ...). Now if you look closely you can see that to compute n-th term in odd terms of fibonacci sub-sequence you can double the n-1 th term and add all the previous terms: 1 + 2*2 = 5 1 + 2 + 5*2 = 13 1 + 2 + 5 + 13*2 = 34 1 + 2 + 5 + 13 + 34*2 = 89 Similarly you can do this with even terms: 1 + 3*2 = 7 1 + 3 + 8*2 = 20 1 + 3 + 8 + 21*2 = 54 1 + 3 + 8 + 21 + 55*2 = 143 and if you add 1 you get the right terms =) I am not sure if this is a known fact so I hope someone here in comment section can tell me.
    2
  3791. Nice to see you again after a longer time than usual. Thanks a lot for the video. For a change, you haven't given too much homework this time.
    2
  3792. "... due to the Indian mathematician Rama-" Me: "-nujan." "-gupta." Me: 😯
    2
  3793. George Washington had a chad neck!
    2
  3794. I used to own a circular slide rule. In fact I used a slide rule (a straight one) until about 1990. While much fun was made of me, the entire group learned that me and my slide rule were reasonably likely to be correct. I finally moved to a calculator when a Pratt Whitney engineer gave me one - he was embarrassed by my use of a slide rule.
    2
  3795. 5:00 looks like n^k sequence has a wall of (k*(k-1))^(k+1)
    2
  3796. 7:59 -- "erecting ears" sounds like something from fantasy-themed live action role playing, rather than from maths.
    2
  3797. Is this somehow related to the fact that A+B = A*B is also true for 1/sin²(x) and 1/cos²(x) for any value of x?
    2
  3798. Your videos are great.Lots of Love from India and keep it up.
    2
  3799. Yay! New video
    2
  3800. how did marijuana solve the what? o you meant that guy
    2
  3801. 2:30 I would give every point coordinates (x,y), and proof that when 2 edges of this triangle are on some points, then third edge cannot be on any point (it's because √3 is irrational)
    2
  3802. The best thing is that it also works for other bases too. Just use ln(8), ln(654) or ln(n) instead of ln(10).
    2
  3803. Do you know that you are my favourite youtuber? This is incredible.
    2
  3804. Also, I really like the falling factorial notation. I learned that combination C(n, k) = n! / (n-k)! k!, but I've always disliked that formula. First, if n is big enough, this makes it uncomputable as n! is too big. Also, it hides the fact that terms cancel out between n! and (n-k)!. If I note the falling factorial like this: n^k, then n^k = n! / (n-k)!, which seems like a much more useful notation, also easier to compute, and yields C(n, k) = n^k / k!, nice.
    2
  3805. Wow!
    2
  3806. yes, they don't lie in a plane. make a matrix of these vectors ang you'll get that it has rank 5 so the smallest space that it can lie in has dimension 5, not 2
    2
  3807. Isn't it a bit logical to not be 0? I mean even in the series depicted in 3.05 you see that you have more positives than negatives so, for every manipulation of a number you do, you present more positive numbers.... 🤔
    2
  3808. Did you edit the video? I cant find the part with different do-yourself equations. Because i searched for them and i'm disappointed not to have found them.
    2
  3809. So, the 3 chords to the circle are all the same length r+b+g. Easy to prove that the centre of the circle must be on the bisectors of the angles of the triangles. Drop perpendiculars from that centre onto each side and show that the 3 cords are all the same distance from that centre. My gap is showing that those perpendiculars fall on the mid point of the chords. Well, it's either messy trigonometry or watch the rest of the video. The video wins!
    2
  3810. I attended a flight school between 2007 and 2009 (abandoned at 80 hours), and we were taught to use this E6-B computer. As I watched this episode, I picked up that analogue computer. I also have two linear slide rules stashed away. I remember using the '100' scale, as I didn't understand the wrapping around thingy when I was learning to use it. The E6-B computer also solves triangles (on the other side) via geometric means. @Mathologer, I challenge you to make a video on it.
    2
  3811. If any of your slide rules has multiple scales, such as 1-10, 1-100 and 1-1000, you can use them to calculate squares, cubes, square and cube roots, and even combining them (useful for scaling area and volume).
    2
  3812. Very compelling to watch. Nice presentation! Thanks for making this accessible for the "rest of us" LOL
    2
  3813. Steinbach's work shows that this happens with any odd-number-of-sides regular polygon. So a regular nonagon will have side length and three different types of diagonals. This would lead to a golden "4-dimension-box" of sorts. And with a regular hendecagon (11 sides), you would get the golden "5-dimensional-box" counterpart using the side length and the four types of diagonals. etc.
    2
  3814. There is the neoplatonic 5x5 square built from the word SATOR-AREPO etc. However no one knows what the "words" mean. An incantation of some sort?
    2
  3815. This Video Sums up Most of my "Aa Ha" Moments from my Calculus Classes.
    2
  3816. Great and beautiful.....on this Christmas
    2
  3817. One question came to my mind for the 3D square grid: cubes can be found within that grid. But how about the other 3D n-gons? Tetrahedron, Octahedron, Dodecahedron and Icosahedron? Can that be answered (or is it just trivial)?
    2
  3818. Very much enjoyed this one! Thanks a lot for the nice and intuitive insight into Euler-Maclaurin Summation formulae!
    2
  3819. You can’t get any information about which proof people prefer with this polling method, because you didn’t control for the order you presented them!
    2
  3820. Are there any (chances for) generalizations in 3D?
    2
  3821. Funny thing about those two ancients; Pythagoras didnot discover "pythagoras theorem" and Pascal didnot discover "Pascal triangle". Those concepts had been known long before those guys..Because both are about "triangles".and everyone loves triangles.
    2
  3822. Nice stuff as always! :) Most memorable part for me was Tristan's fractal! Cheers
    2
  3823. I would imagine that Alpha geometry knows about this theorem :)
    2
  3824. The coefficients of the 5 pentagons are the coefficients in the DFT. Think of FT as picking out frequencies. There are only 5 possible frequencies in the pentagon (or the cyclic group of order 5) and these correspond to the 5 components, as Burkard’s visualisation shows.
    2
  3825. Burkard, I am sure you are aware of this, but I still feel compelled to point it out. You can find objects with infinite surface area contained in a finite sphere, so simply saying that the horn contains the staircase is not enough to show infinite area of the horn. I agree it is obvious from the picture that you can project the staircase onto the inside of the horn at the cost of increasing its area, so the argument does work. I am just worried some viewers might walk away with the wrong impression.
    2
  3826. 0:53 That was one psychedelic music video
    2
  3827. Fascinating Captain. Life form seems to belong to a form of social organization based on pure mathematics! I believe it is attempting to communicate with us.
    2
  3828. You got me into continued fractions Mathologer. Now I have published work on folding continued fractions.
    2
  3829. The video may have given some viewers the idea that a circular slide rule is new, rather than just the connection to rubber bands. William Oughtred invented both straight and circular slide rules about the same time, about 1622.
    2
  3830. I am new to the channel and very impressed by your talent to explain and present those « things » Thx
    2
  3831. 20:30 Well, yes I can: 2 = 1 + 1/2 + 1/4 + 1/8 + ... You never said it had to be a finite sum :-)
    2
  3832. Most memorable: missing all the integers
    2
  3833. No!! Stop!!! flat-earthers will now go crazy!!!
    2
  3834. Another wonderful video!! I love the music at the end too. What is it?
    2
  3835. Came for the T-shirt, stayed for the math.
    2
  3836. All i heard was "heroin formula". No wonder I'm adicted to this channel.
    2
  3837. As usual, very cool stuff, but I am still waiting for the Mathologer on Galois theory !!
    2
  3838. I am realy sorry for the people who think math is just learning formulars and boring math.
    2
  3839. i dont see the paradox, if the paint is allowed to be infinitely thin then it can cover an infinite surface area
    2
  3840.  @nictinkers  Talking about missing the point: Credibility doesn't come merely from making and posting videos. Knowing not to do what you don't have the talent and inspiration for, is just as important. That's something entirely different than not having the skill. It's the wisdom to not needlessly make a fool of yourself. Besides, in several instances, I've complimented Burkard on his work, and added some facts which seemed to be appreciated. I never saw you open your mouth in those instances. So maybe you yourself should follow your encouragement.
    2
  3841. The animated proof at the end was really nice
    2
  3842. 25:40 when I was in school I learnt about this formula. I realized that I can replace 1 in log by any number and it will remain asymptotically true. And by taking the difference between the consecutive partial sums, I figured out that 1/2 works best. For example, at 26:06, (ln(5.5) + gamma) is nearly equal to 2.28196375..., way better then the estimate you used.😅
    2
  3843. Such a wonderful, wonderful channel. This and your cubics video were both absolutely superb.
    2
  3844. Please, what's the song that starts at 11:00 ?
    2
  3845. I once noticed the square pattern had triangular numbers somewhat at its heart. Happy to have stumbled upon this video!! The pattern involving the squares seems like another extension of the Pythagoras Theorem just like De Gua’s Theorem.
    2
  3846. A very small criticism. With my poor colour vision I found it hard to distinguish the colours used for good and bad, esp. the dots around 12 min. More tonal contrast would have been helpful. That cavil aside ... most interesting!
    2
  3847. 15:25 - A fun thing about this strategy that you didn't mention is that you will always only ever need one negative term following each sequence of positive terms to bring you back down below pi. I'll leave proof of this as an exercise for the reader.
    2
  3848. Amazing "upgrade" of the classic relationship. I also liked the OpenAI logo that references this cone/sphere/cylinder relationship and includes the following shape-to-letter mappings: A -> cone G -> sphere I -> cylinder This sequence references AGI, an acronym for artificial general intelligence.
    2
  3849. Same here. I decided to try the nonagons and see how far I would get.
    2
  3850. what ever iam bad at maths any way
    2
  3851. Lovely Video! Most memorable: the “left over area” gamma relates to not only e but the sum of every prime.
    2
  3852. The crazy maximum overhang tower was the most memorable for me. There's something really mind-blowing about generating a complex structure from a relatively simple problem, a bit like how the Mandelbrot set is defined.
    2
  3853. slightly less satisfactory, as you have not offered an explanation of how to come up with the "irregular" geo-shapes featured at the very beginning.
    2
  3854. 0:45 "start of infinite identities" whatever the case, this is going to be an interesting video!
    2
  3855. ​ @shoam2103 it might be, but not to the point of necessarily being malicious. Sophie Germain is a woman.
    2
  3856. since you are a big youtube personality, you might not read all the comments but if you ever did . you will find out the pythagores theorem was also made by bhaskara 2. and lot of other things are in his book like ratio proporation, algebra, etc. even trigonometry .you can ask me for more proof if you want. ty for making detailed explaination of topic )
    2
  3857. what if you had a graph of all the incircles all originating from the 3,4,5 incircle? like making the triangles around the excircles, and then showing the excircles of that triangle and so on. I can't think of the best way to describe it.
    2
  3858. 14:23 I got 666. Hahaha, what's the obsession with the number of the Beast? (^.^) I'm not 100% sure I got this right however. Here are the matrices I got: Arbitrary incremental number assignment of the 2 sets of colored tiles. 1 1 2 0 0 2 3 3 0 0 4 4 5 0 0 5 6 6 7 7 8 8 9 9 0 10 10 11 11 0 12 12 13 13 14 14 0 0 0 15 15 16 0 17 16 18 17 0 18 19 19 0 0 0 0 0 0 0 0 20 20 21 0 0 21 22 22 0 0 0 0 0 Magical 22x22 matrix (rows are green and columns are red). 1 1 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 i 0 0 0 0 0 0 0 0 i 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 i 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 0 1 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 0 1 0 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 i 0 0 0 0 0 0 0 0 i 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 1 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 1 0 0 0 0 i 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 i 0 0 0 1 Determinant of last matrix: 666 (positive real) I actually calculated everything with this c++ code I wrote: #include <iostream> #include <vector> using namespace std; struct Ci{ int re,im; Ci():re(0),im(0){} Ci(const Ci&z):re(z.re),im(z.im){} Ci(int n):re(n),im(0){} Ci(int a,int b):re(a),im(b){} Ci operator+(Ci z){return Ci(re+z.re,im+z.im);} Ci operator*(Ci z){return Ci(re*z.re-im*z.im,re*z.im+im*z.re);} bool operator!=(Ci z){return re!=z.re||im!=z.im;} };/*complex integer*/ ostream&operator<<(ostream&out,Ci z){ if(z.im) if(z.re) if(z.im==1)out<<z.re<<'+'<<'i'; else out<<z.re<<'+'<<z.im<<'i'; else if(z.im==1)out<<'i'; else out<<z.im<<'i'; else out<<z.re; return out; }/*complex integer console print*/ static Ci i=Ci(0,1);/*i constant*/ int pow(int b,int e){ int ans=1; for(;e>0;--e)ans*=b; return ans; }/*integer power*/ void minor(vector<vector<Ci>>&ans,const vector<vector<Ci>>&m,int r,int c){ int h=m.size(),w=m[0].size();/*height and width of matrix m*/ ans.resize(h-1,vector<Ci>(w-1)); for(int k=0;k<h-1;k++) for(int j=0;j<w-1;j++) if(k<r) if(j<c)ans[k][j]=m[k][j]; else ans[k][j]=m[k][j+1]; else if(j<c)ans[k][j]=m[k+1][j]; else ans[k][j]=m[k+1][j+1]; }/*minor matrix*/ Ci det(vector<vector<Ci>>m){ if(m.size()==1)return m[0][0]; Ci ans=0; for(size_t k=0;k<m.size();k++){ vector<vector<Ci>>m0; minor(m0,m,0,k); if(m[0][k]!=0)ans=ans+m[0][k]*pow(-1,k)*det(m0); } return ans; }/*determinant*/ int main(int argc, char *argv[]) { vector<vector<char>>m0={ {1,1,1,0,0,1,1,1,0,0,1,1,1,0,0,1,1,1}, {1,1,1,1,1,1,0,1,1,1,1,0,1,1,1,1,1,1}, {0,0,0,1,1,1,0,1,1,1,1,0,1,1,1,0,0,0}, {0,0,0,0,0,1,1,1,0,0,1,1,1,0,0,0,0,0} };/*sunglass*/ char r=0,g=0;/*numbers to assign to tiles*/ for(int k=0;k<4;k++) for(int j=0;j<18;j++) if(m0[k][j]){ if((k+j)%2)m0[k][j]=++g; else m0[k][j]=++r; }/*number assignation to red and green tiles independently of each other.*/ vector<vector<Ci>>m1(r,vector<Ci>(g,0));/*final matrix*/ for(int k=0;k<4;k++) for(int j=0;j<17;j++) if(m0[k][j]&&m0[k][j+1]){ if((k+j)%2)m1[m0[k][j]-1][m0[k][j+1]-1]=1; else m1[m0[k][j+1]-1][m0[k][j]-1]=1; }/*1's of the final matrix*/ for(int k=0;k<3;k++) for(int j=0;j<18;j++) if(m0[k][j]&&m0[k+1][j]){ if((k+j)%2)m1[m0[k][j]-1][m0[k+1][j]-1]=i; else m1[m0[k+1][j]-1][m0[k][j]-1]=i; }/*i's of the final matrix*/ for(int k=0;k<4;k++){ for(int j=0;j<18;j++){ cout<<int(m0[k][j])<<' '; if(m0[k][j]<10)cout<<" "; } cout<<endl; }cout<<endl;/*glass print*/ for(int k=0;k<r;k++){ for(int j=0;j<g;j++) cout<<m1[k][j]<<" "; cout<<endl; }cout<<endl;/*final matrix print*/ cout<<det(m1)<<endl; return 0; } (^.^)
    2
  3859. Another very enjoyable video to learn from!
    2
  3860. Starting 4 days after the video was posted I expected to be the 2^(1+2+3+4)th programmer to finish the "dance", but scrolling through the comments I don't see any other implementations, so here is mine: https://github.com/hjp/aztec_diamond/ Quick and dirty JavaScript and it gets quite slow as the board gets bigger. I'm sure that can be optimized, though.
    2
  3861. Thanks!
    2
  3862. Can modular math be used to attack Fermat's last theorem (a^n+b^n=c^n n<3) because a^3 mod m becomes 0,1or m-1. 0+0+0 won't work because all would have a common factor. 1+(m-1)=0 has to true for all m less than c^m.
    2
  3863. Proof 2, first learned on YouTube.
    2
  3864. I think I first encountered the second of them, by deriving it at a math test in school. The task was simply to prove Pythagoras' theorem.
    2
  3865.  @davidmeijer1645  Proper understanding of formulae should always involve seeing a proof. Otherwise learning and using a formula is a rather pointless exercise.
    2
  3866. Implex+C?
    2
  3867. Great video on the golden ratio! If you’re interested in how this concept can be applied in design, check out this video about a plant pot based on the golden ratio (called phi-pot). It's amazing!: https://youtu.be/pHeTEpVirFU
    2
  3868. I knew there was a gem in herons formula ... just couldn't see it until your video.... thank you for your amazing insight!!...
    2
  3869. I'm regretting on myself, why I didn't get your channel soon
    2
  3870. 7:40: The cubes work out great: 3³+4³+5³=6³, but no such luck with the fourth powers. The sum is 143 short of 2401, and is not the fourth power of an integer.
    2
  3871. Very well explained, with those graphics, it's a question of watching this carefully. Bravo signore!
    2
  3872. This was a fun one!
    2
  3873. 37:27 looking at the tiling as the walls of a box, it is quite easy to see that each of the three orientations make up a kind of shifted wall. However, the total area of the wall doesn't change when you shift around the tiles it is made up of. If one were to look at the box from a certain angle (such that there's no "depth" from that point of view), you'd simply see one of the walls. It's impossible for that wall to have holes in it.
    2
  3874. Haven't you already done a video on conditionally convergent sums?
    2
  3875. Incredible: this reduces the problem to a double mutually constrained Sudoku.
    2
  3876. One great use of Heron's formula, which I used not long ago, was to calculate the distance from a point to the line joining two other points, all in 3d space. I'm sure there is a formula just for that, but I couldn't easily find one. But I did remember Heron's formula, so worked out that I could calculate the area of the triangle using Herons formula and then the perpendicular distance using the ordinary formula for the area. I had to calculate that for several million points, so all done very fast on a computer.
    2
  3877.  @Mathologer  Ah cool thanks. I'm struggling to find it in my copy, but that's me not knowing how to look something up in there. It seems to be in the section about women, which makes sense.
    2
  3878.  @Mathologer although it might still be a good question as its probably split between Baba Metzia the garment rule and some tractate in Kiddushin(marriages) or nashim women.
    2
  3879.  @darren.mcauliffe  and often the same type of case due to the tangent nature will be in very different places.
    2
  3880. For the non successful primes you can also use 4k-1 ;)
    2
  3881. "Since this will probably be my only ever Mathologer video on magic squares" ... I am already waiting for the next video on magic squares (or maybe magic cubes?) 🤩🤣
    2
  3882. Congratulations on #100!
    2
  3883. OK, so another video on the fact that addition when applied to infinite sequences of terms isn't commutative.
    2
  3884. Interesting 🤔 I
    2
  3885. I made it to the very end 🎉
    2
  3886. WOW .. Power of the folding Triforce .... ähm Triangle 🙂
    2
  3887. I'm so happy to have seen your video on anti-squish shapes before this one! That first proof was a real beauty, of which I would not have been able to fully appreciate otherwise
    2
  3888. Brilliant and Fantastic steps those few from 15:10.. looked like some magic done
    2
  3889. O wow! I didn’t know Euler was one of your patrons!
    2
  3890. Applause for the translators of these old texts 👏👏👏 No, no don't stop.
    2
  3891. Also, when dealing with summation, I realized something: if our f(n) is 1/(n+1), our resulting area is 1 + 1/2 + 1/3 +... which is really close to ln(n) PLUS THE EM CONSTANT. Basically the equivalent of the integral of 1/x = ln(x) We dealt with polynomials, exponentials, rationals, and logarithms. Basically the only things left to do are to rediscover the chain rule and winder about sine and cosine. We'll just call cos(πn) χ(n) for now. The function is 1 0 -1 0 repeating. The difference of this function will be -1 -1 1 1 -1 -1 1 1. It still oscillates. Just not quite in phase. Now instead of the pase shifting by π/2, we shift by π/4. Differencing again, we get 0 2 0 -2 0 2 0 -2... Alright. Sine and cosine down. I'm too lazy to do tangent and the other functions as well as the chain rule (I'm writing this at midnight) so someone fill me out here.
    2
  3892. 1st part is obviously the best!
    2
  3893. 18:09 3^n-1 ?
    2
  3894. 8:42, i was going to comment this: we did learn this at school in Italy, where the construction is known as the 'scodella di Galileo'( Galileo's bowl). I am not sure where Galileo comes into this, but i thought of adding this for the maths history afficionados. The proof we learnt is that one that uses Cavalieri's principle . :)
    2
  3895. I swore that was Stalin on the thumbnail when I saw it...
    2
  3896. Euler uses this same approach in his epochal book on differential calculus.
    2
  3897. He is bald
    2
  3898. In Bava Metzia 61b the Talmud explains the importance of accurate fluid measurement. The Romans were aware of thee water level principle in containers. Perhaps they had water computers.
    2
  3899. @Mathologer  thanks for sharing your enthusiasm and pedagogical talents!
    2
  3900. 30:38 This is really poorly written so please don't judge but I threw this javascript code together. It runs in nodejs and contains all of the steps, but only shows the completed result in the console at the moment. If I have the time I might try to make a webpage around it and make it a lot nicer, or if anyone else wants to feel free to steal any of this code if it helps function getRandom(num = 2){ return Math.floor(Math.random() * num); } class ArcticCircle { constructor(num = 1){ this.size = num; this.grid = new Array(num * num * 4); for(let i = 0; i < this.size*2; i++){ let offset = this.getLineOffset(i); let length = this.getLineLength(i); for(let j = 0; j < this.size*2; j++){ if((j >= offset) && (j < offset + length)){ this.grid[this.getIndex(j, i)] = "_"; }else{ this.grid[this.getIndex(j, i)] = " "; } } } } getIndex(x, y){ if(x >= (this.size*2) || y >= (this.size*2)) return -1; if(x < 0 || y < 0) return -1; return y * (this.size * 2) + x; } toGridCoords(line, column){ let nl = (line < this.size) ? line : (this.size*2) - line - 1; let offset = this.size - nl - 1; return [line, offset + column]; } getLineOffset(line){ let nl = (line < this.size) ? line : (this.size*2) - line - 1; return this.size - nl - 1; } getLineLength(line){ let nl = (line < this.size) ? line : (this.size*2) - line - 1; return nl*2 + 2; } print(offset){ for(let i = 0; i < this.size*2; i++){ let s = ""; let padding = ""; for(let j = 0; j < offset; j++)padding+=" "; for(let j = 0; j < this.size*2; j++){ let tile = this.grid[this.getIndex(j, i)]; let c = "\x1b[0m"; if(tile == '>') c = "\x1b[41m\x1b[36m"; if(tile == '<') c = "\x1b[43m\x1b[34m"; if(tile == '^') c = "\x1b[44m\x1b[33m"; if(tile == 'v') c = "\x1b[42m\x1b[35m"; s += c + tile + " "; } console.log(padding + s + "\x1b[0m" + padding + "."); } //console.log(this); } fill(){ for(let i = 0; i < this.size*2; i++){ for(let j = 0; j < this.size*2; j++){ if(this.grid[this.getIndex(i, j)] == "_"){ let tiles = (getRandom(2)) ? ['^', '^', 'v', 'v'] : ['<', '>', '<', '>']; this.grid[this.getIndex(i+0,j+0)] = tiles[0]; this.grid[this.getIndex(i+1,j+0)] = tiles[1]; this.grid[this.getIndex(i+0,j+1)] = tiles[2]; this.grid[this.getIndex(i+1,j+1)] = tiles[3]; } } } } static GenerateNew(){ let circle = new ArcticCircle(); circle.fill(); return circle; } static EmbiggenCircle(oldCircle = this.GenerateNew()){ let circle = new ArcticCircle(oldCircle.size + 1); for(let i = 0; i < oldCircle.size*2; i++){ for(let j = 0; j < oldCircle.size*2; j++){ let newI = i+1; let newJ = j+1; let tile = oldCircle.grid[oldCircle.getIndex(i, j)]; if(tile == "v"){ if(oldCircle.grid[oldCircle.getIndex(i, j+1)] != "^"){ circle.grid[circle.getIndex(newI, newJ+1)] = "v"; } } if(tile == "^"){ if(oldCircle.grid[oldCircle.getIndex(i, j-1)] != "v"){ circle.grid[circle.getIndex(newI, newJ-1)] = "^"; } } if(tile == ">"){ if(oldCircle.grid[oldCircle.getIndex(i+1, j)] != "<"){ circle.grid[circle.getIndex(newI+1, newJ)] = ">"; } } if(tile == "<"){ if(oldCircle.grid[oldCircle.getIndex(i-1, j)] != ">"){ circle.grid[circle.getIndex(newI-1, newJ)] = "<"; } } } } circle.fill(); return circle; } } let circle = ArcticCircle.GenerateNew(); let iterations = 32; for(let i = 0; i < iterations; i++){ circle = ArcticCircle.EmbiggenCircle(circle); console.log(); circle.print(iterations-i); }
    2
  3901. Most memorable: the bishop's proof.
    2
  3902. Can you make a video about the insolubility of the quintic? I’m dying to understand it. :((
    2
  3903. I'm the 666k view. I'm in the"V"...
    2
  3904. 20:00 aha! He has proven that math is the devil! I knew it this whole time, so torturous 😭
    2
  3905. 2+2=2*2=2^2
    2
  3906. I teach math and introduced Heron's formula as part of an exam question. There the students had to use the formula to calculate the area of a triangle with sides 2, 7 and 11 cm. The trick about this question is that this is an impossible triangle (A+B < C), which quite beautifully does not work in Heron's formula as you get the square root of a negative number. Sadly nobody figured out that the question itself was incorrect and probably assumed that the fault was either in the formula itself, or in their calculations. Once the solution became known I got some complaints about it (lol), but I think they learnt a good lesson.
    2
  3907. This is great! Another awesome video. I have one question though. I understood the formula works with the binomail coefficients you presented. But isn't the denominator of the binomial coefficient missing a (n-k)! ? Can we still call it binomial coefficients then? Or did I miss something?
    2
  3908. If we remove 2 white and 2 black squares such that white and black come alternatively along the path you said then we can tile it. If two similar coloured square come consecutively along the path then we can't tile it
    2
  3909. Could you please do a video on minimal polynomials as you said you would in your 2019 Xmas video (the rational root theorem one)?
    2
  3910. Wow! I actually got it! In the time given, and for the right reason! I couldn't justify the reason, but, still kind of impressed with myself.
    2
  3911. That ln(2) trick is really something special. Thanks for showing it.
    2
  3912. at 23:00 the 7 and 8 clearly do not combine into this shape ;-)
    2
  3913. no integers along the partial sums, the proof was very elegant and simple for what seemed to be quite a challenging problem
    2
  3914. Thank you for bringing out to laymens sight the no nine's sum concept.
    2
  3915. It's like black hole singularity.
    2
  3916. 25,807 is more accurate than 25,8069 - root of EVIL
    2
  3917. Sad to see India not being as good at math as it was before, nowadays youth is busy solving jee level ques.
    2
  3918. I loved this video and would definitely watch more. Your channel is one of my favourites, but sometimes I do kind of wander off for a bit when I don't have time for the long videos, so videos like this would make it easier to keep with it when I'm pressed for time and attention. That said, I also love the longer ones when I have the time, so I wouldn't want to see and end to those.
    2
  3919. I am from India mathloger if you read my comment I want to say you are explain maths in very eassy method so I like you and I also like maths and I want be a Mathematics my name is aditya bagyan .discuss my name in your next veido. Please 😉
    2
  3920. Very interesting topic!!
    2
  3921. 12:34 To move the entire tower on step anti-clockwise, start by moving the smallest disk clockwise and then every second move is the smallest disk clockwise. In other words, the smallest disk move opposite the direction of the entire tower. But then, at ... 26:54 ... the smallest of the orange super-disk move in the SAME direction. Is this dependent of if there are an even or odd number of disks? I really don't want to start of wrong if I meet the Toymaker!
    2
  3922. insane!
    2
  3923. You made my day! Incredibly beautiful !!! ❤
    2
  3924. 14:30 Wow... I remember during my maths degree we went through the proof that lim{n->inf} (1+x/n)^n = e^x, during some long tutorial session, but that visual argument was like a mind blown moment!
    2
  3925. Fantastic! Looking forward to another video from Mathologer.
    2
  3926. 13:38 A simple rule that relates consecutive fractions: p(n)/q(n) = {2q(n-1) + p(n-1)} / {q(n-1) + p(n-1)}
    2
  3927. 4:30 To be fair, this does use calculus… you snuck a Riemann integral in here!
    2
  3928. I would have enjoyed math if shown more of the concepts you have posted in your videos! I was in the talented and gifted program in school and loved puzzles if they had shown us more like this I would have seen math as more of a puzzle than a means to a solution!
    2
  3929. Thanks for another amazing math video! As always, you've inspired some mathematical curiosity in me. I hope you dont pay any mind to the rabid anti AI people and just keep making these videos in whatever way works best for you!
    2
  3930. pleeeeeeeeeeeeeese
    2
  3931. 10² + 11² + 12² + 13² + 14² / 365 (12-2)² + (12-1)² + 12² + (12+1)² + (12+2)² / 365 5*12² + 2( 1² + 2² ) + ( 2*1*12 + 2*2*12 - 2*1*12 - 2*2*12 ) / 365 12² + 2 / 73 146 / 73 2 вот и всё
    2
  3932. Would've been nice if you had explained up front what "make change for" actually means.
    2
  3933. Ohh, this time I did not get the T-shirt reference :(
    2
  3934. Thank you mathologer. You are a great fountain of knowledge. And genius, in your ability to provide visual explanations. You have a gift--that you know this stuff--and a great gift that you teach it to us for free.
    2
  3935. For volume we added units (in this case liters) but for area we did not. If we add units to the surface, we will soon reach the plank length in our calculations and at that point we can add no more area either.
    2
  3936. so if we do mod with a prime number p the rule works with triangles of size p^n+1. If we do mod with a non prime I don't know. For example with 4 it doesn't work with 5 values
    2
  3937. Makes me wonder what other simple discoveries are waiting out there.
    2
  3938. Great video, interesting, entertaining and well made. Your mathematical enthousiasm is contageous. But you don't really believe this was missed the whole world over for 5000 years, do you? Not that I mind. As long as it's nice effective click bait it doesn't bother anyone. 'Love your shirt by the way.
    2
  3939. 37:15 If you think them as a building in 3d, it's obvious that the number of orange tiles is always constant because they are the top of the building. And it works the same for blue and gray tiles
    2
  3940. Loved your dedication, Professor Polster. A side story: I interviewed with Ron Graham years ago for a PhD position at UCSD. While I did not get the offer, my conversation with him is one of my treasured memories.
    2
  3941. I like the shorter video. When they're short i can watch them almost immediately rather than having to wait to have time.
    2
  3942. hi
    2
  3943. First challenge: no. You can cut off 2 corners, for example. Aha!: 2xn tilings follow the fibonacci sequence. Neat! You can see this inductvely. The upper left domino can be placed one of two ways. Placing it vertically leaves a 2x(n-1) board. Placing it horizontally means another domino must be placed directly underneath, leaving a 2x(n-2) board. So the total number of tilings always winds up being the number of the (n-1) board plus the number of the (n-2) board. In other words, you're adding the previous two numbers in the sequence. Glasses challenge: 666. The 2x3 sections on each end are independent and can each be tiled 3 ways. Next you can make a C-shaped path around the left eye that leaves either a 2x3 section on the left (option A) or a 2x2 (option B). You have similar choices on the right (options C and D). Options A and C leave a 2x2 in the middle. B and D leave a 2x4. A and D or B and C leave a 3x3. Now just add up all the ways and multiply by the independent pieces. (3)(3)[AC+BD+AD+BC] = (9)[(3)(2)(3)+(2)(5)(2)+(3)(3)(2)+(2)(3)(3)] = (9)(18+20+18+18) = (9)(74) = 666
    2
  3944. I don’t know how I’m subbed or why it was pushed, but I’ll watch your content. I saw your community post to like, so I liked. Happy 2025
    2
  3945. Very clean
    2
  3946. WHAT'S NEXT? 14?
    2
  3947. Ausgezeichnet! 😊
    2
  3948. This link to Penrose tilling is amazing!
    2
  3949. very interesting, but leave some questions unanswered. how did you get to (X+2)^n? what happens with (X+1)^n or (X+3)^n? is there a formula for an n-dimension pyramids? or other platonic solids?
    2
  3950. 21:30 ok, but what if we'll evaluate it as 1/1 * 2/3 * 4/5 ... ? I don't think this way of thinking proves either because we can multiply top or bottom of the fraction by how many 1s we want to shift stuff to get even bigger or smaller partial products
    2
  3951. These days I've got an Artlist subscription and just trawl though their collection every once in a while to identify all the pieces I like :)
    2
  3952. Magic, as always! 👍🏼👌🏼👏🏼
    2
  3953. What a roller coaster ride from mathematics to statistics, a fiction of sherlock to a lockdown of world #1729 👌
    2
  3954. 5³+ 6³ = 7³- 2 (1)². 16³+ 17³ +18³ = 19³ +20³ - 2(3)². 33³ + 34³ + 35³+ 36³=37³+38³+39³ -2(6)². (56³+ ... + 60³) = (61³+ ... +64³) -2 (10)². ... ... 7⁴+8⁴=9⁴-4³. 22⁴+23⁴+ 24⁴ = 25⁴ +26⁴ - 12³. (45⁴+...+48⁴) = (49⁴+.... +51⁴) -24³. (76⁴+ ... +80⁴)= (81⁴+ ... +84⁴) - 40³. ... ...
    2
  3955. Can we make a formula for Proportional range ?
    2
  3956. 11:38 The 153-104-185 triangle is generated by the following square of numbers 9 4 17 13
    2
  3957. This is so cool.
    2
  3958. One video is missing in the "100", and I remember which and you too 😉
    2
  3959. An restriction like "2 is the only even prime so we discard it" allways baffles me One could just as easy say "13 is the only prime that is a multiple of 13" Is there another property of eveness that makes this so useful?
    2
  3960. Greta video, missed you
    2
  3961. The other nice heptagon fact is the construction by neusis, based on the same cubic. A fun problem may be to find a natural connection between this geometry and the equations for r and s. There will be a purely algebraic one, but is there a natural way to use these lengths of the heptagon’s diagonals in a neusis construction? .. or demonstrate that in one of the neusis constructions (or origami constructions) of the regular heptagon, the values r & s occur naturally in a way where rs=r+s etc are obvious.
    2
  3962.  @Mathologer  I have, thank you! This just led me down the rabbit hole of A055034 in the Online Encyclopedia of Integer Sequences (degree of minimal polynomial of cos(π/n) over the rationals). This is equal to phi(2n)/2, where phi is Euler's totient function. As a consequence of the known lower bounds on phi, n=7 and n=9 are the only cases where that degree is 3. The heptagon and nonagon are the only regular polygons where origami help with the construction.
    2
  3963. 1 : r : s = sin(π/7) : sin(2π/7) : sin(4π/7)
    2
  3964. I started working with magic squares 50 years ago. Find new stuff every year. Love your video, thank you for the deep dive. Working on constructions were the center cell is surrounded by adjacent pairs of numbers whos sum is twice the center number. Question: if a line of consecutive numbers is 2d to wrap them into a grid would they pass through the edge, where the edges are Zero Dimension. Like Pac-Man.
    2
  3965. Yes, that's really beautiful, isn't it :)
    2
  3966. I'm Italian, my teacher in probably 1st/2nd year of high school DID teach us this formula, but I'm pretty sure that it's kinda dying here too 😟 It's sad that these little interesting things stop being teached because they are considered useless, I'm a math lover and these are the things that keep me alive!
    2
  3967. Great video sir 👍☺️☺️ really enjoyed.,.. thank you
    2
  3968. I didn't know you where german until you read out the title at 2:08 xD
    2
  3969. 2:26 Sometimes the best things are free... and this is one of those times.
    2
  3970. Great video! :) Your Simple is King remark reminds me of how the divide-and-average method for calculating square roots is occasionally explained to a layman audience via incorrect reasoning in educational websites or modules. The very first “explanation” I came across in 7th Grade is if the initial guess x you pick to calculate √N is too small, N/x would be too large & vise versa. Then they erroneously infer that since √N is between x & N/x, the average must be closer to √N than both x & N/x [this doesn't immediately follow because 2 is between 1 & 1000 but I think we can all agree that (1+1000)/2 = 500.5 is a worse approximation for 2 than 1] This lead me to grow a dislike for the method especially since the first legit explanation I heard of later on involved the Newton-Raphson method which I think is rather overkill for most smart 7th Graders. So, eventually I found an explanation that I think is way better using the AM-GM inequality & basic algebra. Essentially, after applying the procedure of divide & average once, we end up with a new guess which is greater than √N which would then get closer and closer while stay greater than √N. So, it must converge to some limit L which, once we know it exists, we can solve for algebraically & get that L does in fact equal √N :) The claims in the previous paragraph's first sentence can be proven with the AM-GM inequality & algebra, while the only bit that requires calculus to rigorously justify would be the rather intuitive fact used in the next statement that a decreasing sequence with a finite lower bound must have a limit. I'm a bit saddened that the simple & valid mostly algebraic explanation is still way less popular than the simplified but blatantly flawed “explanation” that I heard in 7th Grade. Still, I'm glad people like you show simplified explanations without having to risk their validity! :)
    2
  3971. Merry Christmas to the Mathologer, and to my son, a mathematics major in his first year, who will receive a copy of "A Dingo Ate My Math Book" tomorrow morning from Santa. I'm hoping he enjoys it as much as I have. Thanks, and Happy Holidays to all your viewers too!
    2
  3972. 5:51 it's usually generalized to "for mn+1 objects over n spots at least one spot would contain m+1 or more".
    2
  3973. Well, in my opinion it is equal to -1/12 Just kidding!!
    2
  3974.  @Mathologer  yeah just kidding!! 😂😂♥️♥️♥️
    2
  3975. (HO-)3 this is Chemistry
    2
  3976. Isn't the rotation angle between the 1 and a number just 360 (or 2π) times the log₁₀ of that number? Log₁₀(3)≈0.47712, so the 3 is about 47.7% the way around the circle or 171.7°. Log₁₀(5.62341)≈.75, so this number is at 270° So the math could be seen as adding angles. Measuring angles would be easier than measuring circumferences.
    2
  3977. 13:19 I want to tie my shoes like that to be honest.
    2
  3978. Is z/1a ring? Like 0*0=0 0+0=0 not sure it s useful though ...
    2
  3979. I have a hunch, if we replace the points above with molecules in a gas, we could arrive at some fundamental laws!
    2
  3980. 32:40 It would be the area if you use step function instead.
    2
  3981. Choosing orange would have made the claw pumpkin-looking.
    2
  3982. 1:04 while the 4-color theorem produces beautiful results, the proof is super ugly since we just had a super computer brute force the answer
    2
  3983. Your shirt is Pi!
    2
  3984. The generalization of AM-GM on 28:43 is incorrect, e.g. take A=B=C=10, then LHS=sqrt(1000) and RHS=15. The correct generalization is that the cube root of ABC is less than or equal to (A+B+C)/3. Or more generally the nth root of A1*...*An is less than or equal to (A1+...+An)/n.
    2
  3985. So essentially you just ruined a whole lot of IQ tests. Though using the right strategy at the right time will still be tricky, that's a powerful tool against the hardest series.
    2
  3986. Today, I was writing a solution for the question in Michael Penn’s video by using this method and I did not know that you did this video (great video by the way), what a nice coincidence! If I were doing a youtube channel on math, this (something similar to that) could have been one of my first videos. I have been using this for more than 15 years already in action. So I love using Lagrange Interpolation/ Newton’s Divided Differences and use it in places where people don’t expect to use, as a number theorist, it’s brother Chinese Remainder Theorem is something I like too. Once, when I was in high school I wrote down first 25 primes in a big poster to see their pattern looking at the consecutive differences of primes and tryed to find a pattern for prime numbers. Later on, I figured that primes do not satisfy any polynomial. By the way, there is Gilbreath Conjecture on primes (considering consecutive absolute differences), as a number theorist, I am still interested to see if there is a connection between Newton’s Finite Differences and Gilbreath Conjecture. This is a very hard problem in professional math level… Also I once trolled a facebook post, a what comes next question like the following: The question was asking 9->90 8->72 7->56 6->42 3->? I just thought of the sequence 9,90,8,72,7,56,6,42,3,19. In my post I said that the answer is 19 because I like 19!!! And explained that the terms of the huge 9th degree formula/polynomial (I used Wolfram Alpha to get the expanded polynomial of course! ) of the sequence exactly match with 9,90 etc… 😀. Because you can literally continue any finite sequence in any way, Lagrange Interpolation Theorem guarantees unique polynomials for each value of ?
    2
  3987. Not Parker's square?
    2
  3988. When I learned about imaginary numbers and their exponentials, it became much easier to remember and derive the angle sum formulae.
    2
  3989. 21:29 On the other hand, 1*(2/3)*(4/5)*(6/7)*(8/9)*... goes to 0. This seems to be a problem of grouping.
    2
  3990. how did you know I went and got a cup of coffee?
    2
  3991. Very nice! Thank you.
    2
  3992. I finally got the courage to watch this video after several months of stalling. As I expected, it wasn't easy, but was worth it.
    2
  3993. So, I learned something today. That thing that my brain does with numbers sometimes, is called the digital root. What I figured out when I was in my teens made me more interested in maths than any of my maths teachers could do. (Note, Im only at 5 minutes in, so this might even be talked about.. who knows). This is probably really obvious to most, but if you have a string of numbers where each double adds up to a multiple of 9 (eg 728145 [7+2=9 etc]) then that number is divisible by 9. So, 7+2=9, 8+1=9, 4+5=9, giving a root of 999 which is then rooted to 9+9+9=27, 2+7=9. Obviously the example looks divisible by 9 at first sight, but if the numbers were re-arranged to something less obvious like 785124, due to the fact that we can root sum to 9 means this number is also divisible by 9. Now, you may be asking, in what situation would you need to know if a number is divisible by 9, where you wouldnt simply just divide the number by 9 to check. And youve got me there. My brain just likes to point out useless weird facts to me every so often. :) Another side note. I dont think its esoteric or anything. Its just a strange interaction between base 10 and the weirder numbers (3, 6 and 9). Im now at 8m30s... wait... you were taught this in school? >>> Ahh, ok, so the first root is maybe taught, but the rooting to a single number isnt. OK, fair :)
    2
  3994. I often use dimensional analysis to do reasonableness checks. Areas have dimension of length squared, volumes have dimension of length cubed. Comparing quantities with different dimensions is invalid. Hence there is no paradox.
    2
  3995. In my opinion Kempner's proof at the end was the most fun since it's one of those proofs where one (or at least I) would have no idea where to even start, but once you see it, one suddenly realizes how obvious it is ;). Thanks for the very interesting lecture :)
    2
  3996. Pineapple on pizza, yay or nay?
    2
  3997. Oh, how I miss the days when such elegant demonstrations would come sooth my brain like butter on hot toast...
    2
  3998. It is getting close to an uncountable infinity, the number of times I had to pick my jaw off the floor watching this magic. Love your erudite but accessible expositions. If only my math teachers were clones of you 50 years ago...
    2
  3999. I made it to the end! Great problem, thanks for explaining.
    2
  4000. That music is not funky at all!
    2
  4001. Thank you for the good holiday class. My answer the the first question is Can the mutilated chessboard with 2 green and 2 black be always tiled? No. Reason: it can be like the first mutilated board, where 2 squares of the same colours from either half of the board giving two disconnected ones
    2
  4002. At first I was tripping, but then I realize angle of incidence/reflection stuff makes it so obvious. Oh and like u mentioned the English is what makes it confusing
    2
  4003. When the differential equation was introduced, I thought: huh, we used to solve that using Laplace transforms in calculus (esp. electrical engineering), and I vaguely remember solving the discrete version of that using Z-transforms (20 years ago, never used it since unfortunately). Does this mean the Z-transform is Laplace transform's equivalent in the difference calculus?
    2
  4004. I love your content. I only hope, one day, we get more Euler-Maclaurin stuff! :D
    2
  4005. That's exactly what I learned in school. The not so easy part is when you actually have to use this rules especially in an stressful environment where you have only a limited amount of time and aren't allowed to use any tools except of pen and paper.
    2
  4006. 13:55 When the right zero to add is a fucking INFINITE SERIES
    2
  4007. 00:52 Manjaro Linux Logo hjfghf :glasses-purple-yellow-diamond::glasses-purple-yellow-diamond::glasses-purple-yellow-diamond::glasses-purple-yellow-diamond::buffering::dothefive:🖥💻
    2
  4008. Trying to get from A to B, I think if N is odd, you move the smallest disk clockwise, otherwise, move it counterclockwise. As far as the other disks, I think each move is forced by where the smallest one is.
    2
  4009. in Turkey, in the 60s, Heron's formula and the related stuff about incircles, excircles, was thought in the 1st year of high school. Not using trigs. I had used it in engineering CAD early on (80s). I always remember that high school geometry book's page since it was such a scary jumble of circles triagles, lines... But actually deriving the formulas was not too bad.
    2
  4010. I remember doing some of this stuff in school, I think in combinatorics as a first year grad student. I remember regarding it fondly, but since it doesn't really have much use outside of its own contexts, it's not something I remember much about in terms of details. I would have to look through the text again to even see what kinds of things we did, but it certainly was very similar. I think we were focused the differential equation analogs and how all this relates to counting things, which is the general goal of combinatorics.
    2
  4011. 28:44 the second row should be 1 3 5 7 not 1 3 7 9
    2
  4012. Most memorable: the partial sums are never an integer. Wow.
    2
  4013. Golden Pizza
    2
  4014. They are the only three numbers that can be added or multiplied and you get the same answer.
    2
  4015. Furthermore, there is the distributive property: log(1+2+3)=log(1)+log(2)+log(3) :-P
    2
  4016. Thanks!
    2
  4017. 18: 4(2-1)=4 ways (3)^2+(3)^2=18 (3)^2+(-3)^2=18 (-3)^2+(3)^2=18 (-3)^2+(-3)^2=18 2020:4(2-0)=8ways
    2
  4018. I mean.. 1+2+3 = 6 (2+3 = 5 and 5+1 = 5) 1*2*3 = 6 (2*3 = 6 or 1*2 = 2 and 1*6 = 6 or 2*3 = 6) Sounds very logical to me.
    2
  4019. What is the relationship with 1/7=0.142857142857..... ? Same numbers in another order and lack of 3-6-9 ?
    2
  4020. That's really neat stuff. For the answer to the triple 153 104 185 I think it would be 9 4 17 13 To figure it out, I turned it into a quadratic and then used the prime factors of the 2 solutions as the 4 numbers. x²-185x+153*104/2. I don't know if there's an easier way. That way didn't seem very difficult, though. For the area of the tree, I think the area might grow by 1 with each branching. So A=1+n. But that does not make sense to me. However, I have no alternative ideas, so I'll stick with it.
    2
  4021. Here in Illinois, we learned Heron’s formula in Geometry class, which is typically taken between 8th and 11th grade, depending on how accelerated you are in math.
    2
  4022. since today is fibonacci day, maybe look up a proof why 144 is biggest square in that sequence?
    2
  4023. This was a neat video. Learned a couple of things I really hadn't conceptualized or thought about. I may have to write some VBA to generate a table where integer squares work. Also... based upon your shirt, I see where the shorter sides of ⦜△ correlate with physics of E=mc² whereas a²+b²=E/m or √(a²+b²)=c. You only now need a multidimensional △ where it could be possible to have lengths of energy, mass or speed of light where which to extract the other integer sides. >>:=p The bug problem was a little bizarre; I wasn't expecting that, but once you mosied over into the geometric explanation, it was like "huh, ain't that weird?"
    2
  4024. got it! excellent
    2
  4025. 7:40 https://drive.google.com/file/d/10z88GUBLV_UMwi6_GibkMyCahJylXOcY/view?usp=drivesdk well, that are five book, but it works. perhaps in the perspective it looks like the first book is not completely over the edge, but it really is, when you look from upside
    2
  4026. I wish my mind was still young and flexible enough to embrace this fully
    2
  4027. Huh, it's look like rotated Crytek logo =D
    2
  4028. So what happens if you apply it to a sequence of prime numbers?
    2
  4029. Never imagined that there might be a relationship between calculus and continuous fractions🤯
    2
  4030. Here's my solution for the vortex math web app: https://bnelo12.github.io/vortex-math/
    2
  4031. I went to school with Gregory Newton. I never guessed he was that smart.
    2
  4032. Great as always
    2
  4033. Is there a geomagic square for which you can use all the shapes at once in approximately correct magic square positions to make one larger square?
    2
  4034. Wow, this is an amazing result! I'm a 59-year-old mathematician and I've never heard of it before, go figure!
    2
  4035. First
    2
  4036. How about Ramanujan's master theorem for a sequel?
    2
  4037. So you said proof for fallen m thing is easy... So i decided to do it in my head, it was a bit weird at first but then i realized that x^(falling m)=(x-m) x^(falling (m-1))
    2
  4038. Beautiful 😊
    2
  4039. This video clarifies a lot! Four years ago, I found the envelope curve of a family of curves of same shape (for a few shapes) whose area underneath is constant and bounded by the x and y axes and in the first quadrant. The envelope curve was always of the form xy=k (const area). Now I know why! To the Mathologer: you are welcome to peruse my new maths channel: "Sobermath". I would be honored by any comments.
    2
  4040. The approximate sum leading to Euler's constant gamma is my favourite. Because it has a geometric interpretation very intuitive and easily memorable!
    2
  4041. Fantastic video! I love these long excursions into deeper topics - please keep them coming. 😁 I don't have much of a formal background, but I have always enjoyed math, and use it regularly in my work as an engineer and software developer.
    2
  4042. That would be an amazing feat
    2
  4043. Any magic square can be turned around by 90degree, twice or trice so three varieties can occur or at least seem. But does that resemble some cheat with the wrong potential to abuse?
    2
  4044. If a pentagon's diagonals produce a 2-dimensional shape, and a heptagon's produce a series of concentric 3-dimensional solids, would a nonagon's diagonals produce a series of concentric 4-dimensional solids?
    2
  4045. Wow - amazing proof. Fortunately I did learn the Ptolomey's Theorem when I was in elementary school. Unfortunately this implies that I was in elementary school a long long time ago.
    2
  4046.  @Mathologer  It really is 😌. I also noticed another nice property about the lead digits of π (in base 10), in your video on Heron’s formula (wherein π shows up, as the area of the incircle (unit circle) of the 3–4–5-triangle); all having something to do with said triangle: π = 3,1415… :) 3: The length of the shortest side, of the 3–4–5-triangle. 1: The area of the unit square, in the right-angled corner. 4: The length of the medium side, of the 3–4–5-triangle. 1: The radius of the unit incircle, of the 3–4–5-triangle. 5: The length of the hypotenuse, of the 3–4–5-triangle. 🙂
    2
  4047. Your T-shirt choices are impeccable! I <3 mathologer
    2
  4048. Yet another amazing video:)
    2
  4049. 2:50 I came across twisted square proof (first version) in a 3b1b video last year. Around that time, I fell in love with math by watching videos from the 3b1b channel and your channel. 17:27 If one side of F is stretched by B, the area of the stretched triangle becomes BF. This is because the height of the triangle does not change if we consider the stretched side as the base. Likewise, if the other side of BF is stretched by A, the area of the stretched triangle becomes ABF. 35:51 At every moment, each bug moves along the shortest path to the bug it is facing. Therefore, the distance each bug traveled is equal to its initial distance from the bug it is facing. (I know this because I've seen the answer before.) 36:19 The diagram can be rearranged into 5 squares, so the area of the middle square is 1/5.
    2
  4050. If your math trick only works in base 10 counting, it is in no way fundamental to the universe.
    2
  4051. I’ve been waiting for this one! Merry Christmas!
    2
  4052. 5:12 Not that good in writing down the math here. But.... Please look at the start of the sequences. The 1, 4 and 9 start as 1^2, 2^2 and 3^2. Of course the next one is 4^2 to begin with. And of course, you can keep counting in between. Either way, the intervals are 1. 1+1=2 2+1=3 Then X^2 The 3, 10 and 21 are the sums of 2, 4 and 6. Or in other words. (X^2 + X)/2, with intervals of 2. 2+2=4 4+2=6 etc. The next start is at 6+2=8, then (8^2+8)/2=36 Thus 36^2+37^2+38^2+39^2+40^2 = 41^2+42^2+43^2+44^2 Now to look at the end of the left side of the equations. We see 2, 6 and 12. Then for the power of 2 sequences we see 4, 12 and 24. That just so happens to be the double. The 2, 6 and 12 follow the sumation again. What I mean is (X^2 + X)/2, but then times 2. Or just X^2 + X The power of 2 is 2* (X^2 + X) And you can see this at the 4th of both sequences as well. 20 and 40^2 Not sure what logic we can use for the cubic numbers though.....
    2
  4053. 5 pigeons in a hemisphere
    2
  4054. fucking awesome !!!
    2
  4055. 28:29, there's a "5" missing in the 2nd row. It should go "1 3 5 7 9...", instead of "1 3 7 9..."
    2
  4056. Chapter 5 with gamma: weird constants that show up where you least expect them is a favourite of mine!
    2
  4057. I’m not sure, if I’m missing something; but that doubling of the product (like the 1 * 2, in the first example) seems, to me, a bit forced, and ”ad hoc”; like they were thinking: ”Damn! Can’t connect 1 * 2 to the ”3-4-5” -triangle. Let’s double it, and get 4; and if that didn’t work, we could have tripled it.”. Like, there’s no clear reason for, why you should double it; other, than getting it to work. Seems very random. The same kind of goes for the adding of the products, to get the area; like, now we suddenly went from multiplication to addition? What gives? 🤔😮🤷🏼‍♂️
    2
  4058. What is the name of the formula 26:35
    2
  4059. I want to know if there's a generalization of this formula to any cyclic polygon
    2
  4060. Does this mean that all valid Sudoku puzzles—or at least some of them—are magic squares?
    2
  4061. some sort of results limited auto-generator might be - and also works as some sort of a proof... this: sum(1, for 1 to N-2) + 2 + N = N x 2 x 1's where sum(1, for 1 to N-2) + 2 = N or merging the 2 into the sum: sum(1, for 1 to N) = N thus the first formula is reduceable to: N + N = N x 2 (yes, it is just the basic equivalent of adding a number to itself to multiplying the number by 2.) Examples: 3 + 3 = 3 x 2 6 + 6 = 6 x 2 ;-)
    2
  4062. I'm coming to get that shirt, watch out now.... WHAT... IS... THIS?!?!??
    2
  4063. Thank you for this amazing video! That's absolutely beautiful and interesting! :)
    2
  4064. ٣π=٩.٤٢٨٥٧١- ٩.٤٢٩٥٧٠=٠.٠٠٠٩٩٩×١٠٠٠=٩٩٩-١٠٠٠=١ ١/٧=١٤٣×٧=١٠٠١ Each second equal 100 year's So 83/6= 13.833 M/ year's
    2
  4065. @15:33 Answer is 5 rotations. 2/3, so 3 rotations clockwise of small circle over straight line of 2x perimeter of larger circle. Then plus 2 clockwise coiling of large circle back gives 5 rotations.
    2
  4066. Wait one minute, there's an error in your t-shirt: the number 27 is missing!!!
    2
  4067. So the Horn Of Plenty only contains a finite amount of stuff, then?😁
    2
  4068. You are probably the best math teacher in the world...the way u bring out the magic in math is mesmerising....one can get hooked on for hours
    2
  4069. Thank you for what you do. I will go to sleep dreaming of discs and pegs now :)
    2
  4070. Sir, what's your name I wanna purchase your book from Amazon
    2
  4071. There goes a shiver down my spine when I saw that hexagon minimizing
    2
  4072. Not possible because if you had two overlapping sets where one of them has a single element, then the sum of the other set is necessarily larger than that one value.
    2
  4073. The 7x7 square shown at 19:01 is not magic. For example, the sum of entries in the 3rd column is 137, not 175.
    2
  4074. I prefer the dots proof. If you do the right triangle again do you get the right Triangle? Yes I think you do. Love these videos on a nice Sunday afternoon. Love the shirt btw
    2
  4075. thank you so much for this video. i haven’t really felt like i’ve truly learned something useful in a long time and this is so amazing and simple. you are classic. i am grateful
    2
  4076. the Pi=4 thing forgets that it is creating synthetic ordinates that overlap. While still infinitely small overlaps they ARE overlaps and effectively they're producing an infinite number of infinitely small overlaps. So technically that proves that (4 - pi)^2 = infinity.
    2
  4077. I found that image of Madhava guy eerily familiar, lol you based it on yourself.
    2
  4078. Conway's circle diameter > curve's diameter, one more interesting thing to observe is that, perimeter of curve = pi * perimeter of triangle.
    2
  4079. First!
    2
  4080. The formulas in this video became easier and easier. I anticipated the last one with a litte bit fantasy. Nope. 😎 - Just kidding!
    2
  4081. New sub. Didn't have time to watch the whole video but great presentation 😕 Man I miss doing math. Spend all day in a windowless building processing mail but it pays the bills.
    2
  4082. All this was great! Background = undergrad in math+physics
    2
  4083. 5³+6³=/=7³ or 125+216=/=343 (it's off by 2). To follow the method: 5³+6³ = 6³+5³ ==> 6³ exists of 6 slabs of 6² or 36; it needs to be wrapped around 5³. Place the 5³ cube on a surface, place four of the 6² squares around it, seen from above you've got 7x7 (note the top layer is just a rim going around, it still needs 25 squares to fill up the middle and get 7x7x6). The height measured from the table is still 6 so place the 5th 6² square on top of what you've got, you still need to fill in half of the edge with 7+6 = 13 squares. To fill up all the remaining gaps you still need 25+13 squares, you only have 6² squares, so there's 2 missing squares. Or put in another way building from 2D: To surround a 5x5 square: you can do that with 6x4 (you only need 4 sides): 25 + 24 =49 =7²: for one slab. Stack 5 slabs above each other: 7²x5. What you used: 6x4x5 (from the 6³cube) + 5³. (What you didn't use up: 6x2x5+6x6 = 96. And you still need 7²x2=98 to get to 7³.). It comes down to the fact that 7x7 isn't equal to 6x8. You miss one square. (twice) 7x7=6x8+1x1. To find that exact line 5³+6³=/=7³. I did... 1D: 1x2 (covers 2 'sides'); 2D: 1x4 (covers 4 sides); 3D: 1x6 (covers 6 sides), and so the number before the equal sign is 6... etc...
    2
  4084. Disclaimer : I only watched up to around 14:00 before making this comment. I had a thought for my program : This looks like the infinite series for 1/(1-x), but composed with the function x^n (for a n-cent coin), namely 1/(1-x^n). So, instead of doing this giant product, which would get extremely hard extremely quickly, I just take my "vocabulary" of coins, find the product of (1-x^k) for k in my set of n coins, then do a partial fractions decomposition which gives me easy terms like 1/(1-x) and 1/(1+x)^2, which can easily be further expressed into their power series. Finally, I add them all up to get a general formula for the number of ways to make a certain amount of money using my coin vocabulary. I'll admit, I didn't do the whole thing myself, most of the heavy lifting is done by SymPy, but I'm pretty happy about my program, especially after spending the last 4 hours on it. Now I can finally watch the rest of the video.
    2
  4085. Did you consider that the eyelets are separated into two different arrays who’s distance from each other changes as the lacing is tightened? This changes the spacing aspect dramatically.
    2
  4086. Indian religious books talk about age of universe, multiverse, inter dimensional travels, medicine and operations and they were written 5000 years back. The more we learn the more we know they knew it way back. Do drugs and read these books you will get addicted to the books and not the other way round
    2
  4087. Haven't we seen the one-sentence proof at 19:00 in another video somewhere? I seem to remember figuring it out once, but I've forgotten the details.
    2
  4088. Well, the total area is 5. Many in the comments already explained how to get to this number. Some also say that if you add more and more branches. The area increases by 1 for every new branch. With an infinite number of branches, we get an infinite area. But here is a math question for everyone. What if you indeed add an infinite more branches. The branches start overlapping at the 6th level. What is the total area with infinite branches?
    2
  4089. Thank you Burkard. I needed to learn something beautiful today. 💙
    2
  4090. Wow, a natively English speaking person who properly pronounces German names! Gratuliere!
    2
  4091. There's a third proof: tessellate the plane by the original triangle and draw all medians. Then a tessellation of the plane made of the "folded" triangle can be made by erasing some lines.
    2
  4092. This can be called discrete calculus?
    2
  4093. I absolutely love this. I'm going to attempt to program it for sure. The hexagon version intrigues me. If the random tiling is turned into a 3d pile of cubes, what shape will they "carve out" of the large cube? Spherical?
    2
  4094. Easily one of the best channels on all of youtube. Thank you once again for a very interesting, entertaining, learnful, and inspiring video, professor Burkard! 🙌
    2
  4095. 0:52 That is the Millennium Thinker AI voice xD
    2
  4096. *especially when he chuckles 😊
    2
  4097. Very nice video. Excellent work, as always. Proving that there are no equilateral triangles in 2D square grids doesn't seem hard. Basically, if there's an equilateral triangle, you can split it in half to get a 30-60-90 triangle whose sides are d, 2d and d*sqrt(3). A little bit of similar triangles shows that this leads to sqrt(3) being a rational number, which is false, hence no such equilateral triangle exists. But this proof doesn't generalize to other polygons or other dimensions.
    2
  4098.  @Mathologer  Ah, I didn't realize! Small typo: where you've written "(on e-daf Ketubot is spelled Yevamos)" it should of course be "(on e-daf Yevamot is spelled Yevamos)"
    2
  4099. My fav YT Channel. Amazing Video.
    2
  4100. Nifty AF
    2
  4101. Is it a European thing to pronounce sinh as "shine"? I've always heard it as "sinch", as in "sin-H" but you elide the "ay"sound from H out of existence. Anyone else?
    2
  4102. Weine nicht, wenn der Regen fällt 34:04 Es gibt einen, der zu dir hält 34:04 Marmor, Stein und Eisen bricht...
    2
  4103. 6:14 Please fix the 3 X 9 = 28
    2
  4104. (1) If you consider triangular, square, pentagonal, hexagonal, ... numbers, do you get an infinite pattern of patterns! (2) Note the starting numbers of each row. On the tower with the sums of consecutive integers, the starting numbers are squares. On the tower with the sums of squares, they're alternating triangular numbers: 2n(2n+1)/2
    2
  4105. How does the "exactly 100 zeros" series not diverge? After that last zero, you can have a 1 and then infinitely many ways of combining the digits 1-9, making it possible to sum infinitely many numbers between 1/googol and 1/10*googol. Wouldn't that give infinity? What am I missing?
    2
  4106. Why didn't we write it as 4k-1 ??
    2
  4107. The First Doctor! The One! The Original!
    2
  4108. 50 minutes of incredible mathematical discoveries. Thank you a lot !
    2
  4109. This is hella cool! The GameStop NFT marketplace is almost live, I look forward to seeing your animations minted there soon!
    2
  4110. cool
    2
  4111. Thanks for the citation info, Herr Polster—I'd have died of old age trying to decypher all those subscribed deltas without it;-)>
    2
  4112. log u + log v= log(u*v) when moving the two circles u AND v then= multiplication u*v
    2
  4113. nice!
    2
  4114. Another beautiful experience. Thank you
    2
  4115. Can you expand this to any dimension... 🤔???
    2
  4116. I wonder, would things look even prettier if we start at 0 rather than 1?
    2
  4117. Wow, its just incredible how you always come up with some wonderful new regions or facts in math I never heard before that are always useful or at least very interesting to watch! Also you explaing everything very understandable with the animations, love these aswell. Keep up the great work, you really help a lot :)
    2
  4118.  @dougr.2398  Some of those cowboys do not even wrap the hitch around the rail; they simply slap it against the rail, and inertia does the rest. BTW, if you know boats, what do you think of the claim that it is possible - using just ones arms - to push surprisingly large craft (even ships) away from the dockside?
    2
  4119. One of the channels for which I will always give a thumbs up, even before watching 😅
    2
  4120. Thank you Matholger for so much.
    2
  4121. What great Video and thank you so much for all the effort so the curious in us who are not professional mathematicians can follow along I create educational videos too, so I know how much effort would have went in. Thank you Sir :)
    2
  4122. Does anyone know the name of the song played at the end?
    2
  4123. 15:29 5 times it depends on the ratio of the smaller coin I think it is just: Inside: when ratio smaller coin is a/b it is a-b times When outside: then a+b times We are expecting that shortening fractions for a/b is not possible.
    2
  4124. The comments are real gold.
    2
  4125. thank you Sir!🙏
    2
  4126. I live in sangamagramam, Which is currently called Irinjalakuda.
    2
  4127. abacabadabacaba
    2
  4128. I ❤ MASTERCLASS😊
    2
  4129. 10:05 Pascal triangle ?
    2
  4130. great t-shirt
    2
  4131. I did enjoy this! As for the four by four grid of colours, it is a forced pattern. In each half step of the two by two square, you are forced to add the two colours you just left. Add in the vertical component and it forces each colour to its own corner. The pattern repeats for every four unit lengths added to both dimensions in a square grid. Make it as big as you want - in fours.
    2
  4132. Just recently rewatched 3b1b's old video on this sequence. Really cool that you did a totally different perspective on it. Grant's video was geometric with circles and lines and graphs and Euler's characteristic formula.
    2
  4133. I loved playing with the cube puzzles like those when I was a child. The manipulation of the pieces, not just in hand, but also in my mind lead me to a greater understanding of mechanics, physics, and engineering. Thank you for this great video!
    2
  4134. Very high quality Mathologer
    2
  4135. How did I go this long without finding this channel? Thanks, YT - you finally got me a great new channel to subscribe to, instead of pop-culture 💩!
    2
  4136. That actor now has acted in Vincenzo :face-red-heart-shape: I also love Mr Queen ^_^
    2
  4137. That number can be written as a sum of two integer squared. Indeed, that number minus 1 is divisible by 4, since the number formed by its last 2 digits is 80, which is divisible by 4.
    2
  4138. Wow. That's a lot to take in. I get the idea, but I feel like to truly get an intuitive grasp, I would need to take some time to think it all over. Amazingly well explained. Well done!
    2
  4139. My favorite part was the approximation of the nth harmonic number (partial sum with n terms of the harmonic series)
    2
  4140. Obvious gamma > 0.5 because for each "element", it represents slightly more than half the rectangle it fills (blue curve goes beyond the line segment connecting top left and bottom right of its bounding box).
    2
  4141. Brilliant!
    2
  4142. I keep trying to post about whether this can be related in any way to Branko Grünbaum's 5-set Venn diagram, which I played about with and found that it collapses into a geometric figure consisting only of pentagons and quadalaterals, 31 in total if you try to average out the size of the sets. I don't know whether Branko Grünbaum found this out. Unfortunately, my post gets deleted possibly because I provide a link to it, so I'm posting without the link if you want to discuss it.
    2
  4143. I really liked this, although I would like some more detail on the ch 8 stuff. Definitely looking forward to more videos on Euler Maclaurin summation, as it does seem powerful for approximation. Does the fall of in accuracy with more terms have anything to do with the radius of convergence of the taylor series for 1/x^2? :) I have an engineering degree, so I am on the higher end of background familiarity with math. I love all of these, but the chaptered ramp up to crazy videos have been my favorites 41:00 Add and subtract f(1) . The forms of f(1) you need are already on the page :)
    2
  4144. 746/365 Solved in a bit over 30 seconds... my strategy was calculating (13*13+14*14)*2/356 and since 14*14 is just 4 away from 200 it didnt take me much to figure that out edit: got that wrong, it was 2, i added insted of subtracted
    2
  4145. The coin paradox reminds me of poi in-spin and anti-spin patterns
    2
  4146. That moment when you realize one of the authors of the Arctic Circle Theorem (James Propp) was your calculus professor in college
    2
  4147. No Nine November :)
    2
  4148. Another great mathologer video. Sorry I was late watching this, this week, I was making my latest Lego
    2
  4149. Love from Bhaarat ( India) 🇮🇳
    2
  4150. I have a slide rule! It even came with a little book with instructions and practice problems to learn how to use it! While I do know that it can do much more, I have only figured out multiplication and division. Also, my slide rule has several very nice constants, in my case, π and e
    2
  4151. This is simply a brilliant explanation. Your videos are getting better and better. I am astounded at the ideas and creativity you have in the topics you cover. Thanks!
    2
  4152. 1:50 thank you and gooni—uahOooomg there’s more.
    2
  4153.  @AB-Prince  It is mentioned and displayed in the video, but it is easy to miss the details of it in his explanation. (Sorry, Mathologer, I should have noticed this could be a point of confusion!)
    2
  4154. Everytime I get a notification from Mathologer I jump in euphoria
    2
  4155. Thank you for this comment. Thanks to this I figured out why only odd numbers worked for my visualization program!
    2
  4156. Great video as usual! Your content always makes me more interested in math :)
    2
  4157. Damn I'm tired. For a moment there I thought you were multiplying by an exponent and I just didn't get it. The I realized that is X as in the letter X commonly used by mathematicians as an unknown. It's moments like this that make me apreciate the use of an asterisk for multiplication on keyboards and a dot or outright absent in most day to day mathematical operations.
    2
  4158. What’s on your shirt?
    2
  4159. Hello, I like to do the homework you leave in videos. I'm no differrential eq. connoisseur, but you can brute force the solution for y'= 1 + xy. Asuming I'm not dividing by 0 and that we apply fundamental theorem of calculus (FTC) the right way (i dont put constants while calculating because they will be 0 by the way we use the FTC): y' - xy = 1 y(y'/y - x) = 1 factor y y'/y - x = 1/y not dividing by 0 y'/y = x + 1/y ln(y) = (x^2)/2 + int_0^x (dt/y ) integrating, forming the ugly integral (from 0 since ln is defined in positive numbers) y = e^(x^2 /2) e^(ugly integral) get rid of ln and do sums into products. If we derive this last expression, by Leibnitz rule we get y' = (e^(ugly integral))' e^(x^2 /2) + x e^(x^2 /2) e^(ugly integral) , but we know y' = 1 + xy , i.e. y' = 1 + x e^(x^2 /2) e^(ugly integral) = (e^(ugly integral))' e^(x^2 /2) + x e^(x^2 /2) e^(ugly integral) implies 1= (e^(ugly integral))' e^(x^2 /2) then (e^(ugly integral))' = e^(-x^2 /2) integrating e^(ugly integral) = int_0^x (dt e^(-t^2 /2)) so the solution is y(x) = e^(x^2 /2) int_0^x (dt e^(-t^2 /2)).
    2
  4160. https://pastebin.com/qnBcJGkF answer for christmas glasses is 666
    2
  4161. Fun fact: Notch, the creator of the game called Minecraft, a.k.a. Markus Persson, funded this video, as you can see at 23:53.
    2
  4162. Please change the White background IT burns my eyes and battery
    2
  4163. This is not quite the same as sum = product, but it might be related: Consider the pair of pairs of numbers (1, 5), (2, 3). The sum of the first pair equals the product of the second, and the sum of the second equals the product of the first. Aside from the obvious (2, 2), (2, 2) and the trivial (0, 0), (0, 0), I can't find any other pairs of pairs of positive integers for which this relationship (ab = c+d and a+b = cd) holds. If we allow negative numbers, then (-1, n), (-1, -n+1) is a general solution. I have no idea what non-integer solutions exist. I have no idea if this is an already-explored topic, or whether it's of any significance. But I find it quite interesting.
    2
  4164. 7:20 answer is 2
    2
  4165. Bookss dont exist
    2
  4166. I've always done a bow tie lacing with my oxfords, because its the only practical way you can guarantee that the crosses will happen on top of the shoe at regular spacing. The regular spacing also looks nice. Would recommend, not just because it's generally the shortest (but thanks for additional excuse Burkard)
    2
  4167. I'm starting as a substitute teacher. I'll just play this in math class then ask 'any questions?'
    2
  4168. Question: Is it ok to differntiate and still keep the equal-sign?
    2
  4169. Ahhhhhhhh lovelyyyyyy
    2
  4170. You have show formula for reiman zeta function for zeta(2k) by Euler in one of your video...but you promised us that you will also tell zeta(2k+1) in another video,I am waiting for that... please do that video as well...take care...stay safe...
    2
  4171. The 4x4 square with a missing "tile" looks like the game of sliding squares with which I used to play as a kid, trying to order the numbers by sliding them. My real question, however, is whether this is related to the non-repeating pattern found in Penrose tiles.
    2
  4172. Ramanujan got his answers from Goddess Namagiri, If one does enough penance to attain that siddhi from that goddess there you go we get another ramanujan
    2
  4173. Please upload videos on sequence and series...most of us love those content too much I believe...
    2
  4174. How come Hermann Schwarz, who has a very German first name and last name, published his paper in French? Was he actually French? Was it the common mathematical language of the time?
    2
  4175. I love mathologer video 😍
    2
  4176. That make me remember that time in high school when we went from summation to integrals. We started by doing the summations using smaller and smaller intervals. Back then, pocket calculators where just starting to get popular. I also remember doing stepped fractions in fourth grade... Yes, I DID resolve this one, and a few others, by hand : 1+___________________________1____________________________ 1+_________________________1___________________________ 1+_______________________1___________________________ Going 12 levels down.
    2
  4177. Love these videos. Saying I understand half of what he says would be exaggerating :) But his love for Math is keeping me going . Well done!!
    2
  4178. Instead of logarithms, I saw it more in terms of square roots... I.e. the "midpoint" of this circular number line, or the number exactly opposite the 1, appears to be the sqrt of 10 or ~3.16... Ya you could consider it a number line on a logarithmic scale, but each number's place is based more on what you have to raise that number to in order to get 10.. if that makes sense? LoL
    2
  4179. That seals it
    2
  4180. The most memorable part for me is probably the proof that H_n - ln(n+1) converges. Such a beautiful proof.
    2
  4181. Thanks for this clear explanation. As a non-mathematician I could still follow along and appreciate the elegance of the math here. Also, your little giggle at 5:52 sent me. I love your enthusiasm!
    2
  4182. I was lost at the beginning of the video??? Lol
    2
  4183. 2:15 It fits on one page because it's a German proof regarding the natural numbers, which means there are no spaces between the words ("mathematisch-naturwissenschaftlichen") and no reason to put 2 digits after the decimal point of each number.
    2
  4184. 6:44 actually this one is very confusing to me. The circle matches the cone at the extremes, but in between the circle is always wider than the triangles Edit: thinking more about this, it is obvious that they are not the same because one is a ring and the other a circle. But still it’s not obvious that those areas match
    2
  4185. the thing is some of the famous math texts from antiquity have been lost , you can find only references to them
    2
  4186. I wonder if the optimal solutions for 5 or more disks are unprovable even in principle.
    2
  4187. Hello, Because there could be a circle, rectangular, trapezoid, oval, or other shapes that the observer should solve, no one can teach the whole *universe*S in one universe. Genuinely,
    2
  4188. 34:20 For tribonacci numbers: 2nd Difference + half of 3rd Difference. Didn't try the differential equation though.
    2
  4189. 20:29 Here it is- 1+1/2+1/4+1/8+1/10+1/40 = 2 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/15+1/230+1/57960 = 3
    2
  4190. I met Conway once. I guess he didn't seem anybody there to be worthy of attempting the proof. He was, however, very excitable about rainbows.
    2
  4191. Simply brilliant!
    2
  4192. Because the three values add up to 0mod3. Since + is commutative, you can go whichever direction you choose. :)
    2
  4193. Sequences are always fascinating .Amazed you can convert it into a formula.
    2
  4194. amazing and simple solution to use instead of usual banging your head against wall :D Thank you
    2
  4195. Under 3k views here. Never been this early
    2
  4196. Bro I swear those mathematicians were just like bro but what if I did this.... and then they find some buzzin equations and relationships no cap g
    2
  4197. That little laugh is infectious. Lols. 😂
    2
  4198. There are often convergence issues with the derivative of a Fourier transform.
    2
  4199. As a teacher, I really liked chapter 1 and 2 ! Great introduction
    2
  4200. I did get the pattern and calculate the Yellow grid, but mostly was an educated Guess
    2
  4201. The last part i liked the most. About an idea of how the determinant helps to find number of permutations. But i couldn't understand about why physicists are interested in this
    2
  4202. Awesome video! I appreciate the effort you put in to these!
    2
  4203. 22:55 2XY-(2+X+Y)+1=N-2 2XY-(X+Y)-1=N-2 2XY-X-Y+1=N 2(2XY-X-Y+1)=2N 2X2Y-2X-2Y+2-1=2N-1 2X2Y-2X-2Y+1=2N-1 factor left side ... (2X-1)(2Y-1)=2N-1
    2
  4204. The final puzzle answer is 0.2, You can see this by completely surrounding and attaching the original figure with identical figures. You can identify a cross shape pattern made of five identical squares that equals the area of the original figure and where the center square of the cross is the center square of the original figure.
    2
  4205. I made it to the very end.
    2
  4206. All these formulae would be much more simple if we used a constant whose value was 6.2831853… instead of 3.1415926…
    2
  4207. Well said, man! We should redefine ℼ = 6.28..., because 𝜏 is just so ugly.
    2
  4208. I'm too young to ever see a slide rule used, but back in elementary school our teachers used these kinds of wheels to grade tests. Specifically you could choose the denominator (how many problems were on the tests) and then it would give you the percentages for how many of those problems were correct.
    2
  4209. Just in time!
    2
  4210. Also what is that remote you use? Looks nice!
    2
  4211. It appears tesla invented VolksWagen symbol...
    2
  4212. If we look at the finished magic square, we can imagine that it has been filled using the following process: Put the average number 13 in the center. Fill the anti-diagonal using consecutive numbers with 13 in the center, that is, with 10+1, 10+2, 10+3, 10+4 and 10+5. These sum up to 65 as was explained in the video. Now we fill the main diagonal with the numbers 3+0, 3+5, 3+10, 3+15 and 3+20, again with 13 in the center, also summing up to 65. Now start at any remaining number on the main diagonal, e.g., 3+5, and move diagonally down/left counting up and diagonally up/right counting down until all numbers of the respective group, i.e., 1+5, 2+5, 3+5, 4+5, 5+5 (or 1+k.5 to 5+k.5 in general) have been filled in. Do this for each number on the main diagonal. When you reach the edge of the magic square, cycle around! Just imagine a second square adjacent to where you reached the edge and see which field of the adjacent square the next diagonal step would end up in. Filling up the square diagonally this way ensures that all numbers of a group 1+k.5 to 5+k.5 are all in different rows and columns. All numbers of the form k+0.5 to k+4.5 are also all in different rows and columns, because that is true for the main diagonal and the filling up process starts from there. This ensures that all rows and columns add up to 65. The cycling around when you reach the edge of the board is equivalent to the royal shifting method. This is not quite rigorous but you get the idea.
    2
  4213. I think this guy is mathological person.
    2
  4214. Hey man! You should have shown why the alternating sun converges to ln(2). That proof is just amazing.
    2
  4215. The power of 2 is the same as first multiplying by 5 and then dividing by 10
    2
  4216. They teach this in Turkey, but they dont use Heron's name oddly enough.
    2
  4217. That was so cool!
    2
  4218. The sum of the angles of opposite cyclic quadrilaterals is 180°, because the angle of a cyclic quadrilateral is half of the angle between the two radiuses joined to the same points as the quadrilateral, and the sum of the angles of the radiuses of the opposite quadrilaterals are obviously 360°, since the radiuses divide the circle in two parts, and 360° / 2 = 180°
    2
  4219. Sir who are u..??......u r amazing.....i'm from India 19yrs old engineering student and your videos are very insightful very very interesting very knowledgeable and very easy to understand.......are u a math professor(by proffession)....in which university ??
    2
  4220. Ramanujan was an amazing genius.:hand-orange-covering-eyes: Love from Sweden💛💙
    2
  4221. Ramanujan😎
    2
  4222. Thanks!
    2
  4223. Just want to thank you for this incredibly interesting content. I love this channel.
    2
  4224. 5:41 electromagnetic wave functions superimpose up by multiplication (Psi) (or sum of intensity) and addition (Euler's Theorem) (or multiplying of wavevectors (momentum))... I probably remember unclearly...
    2
  4225. "Mathematics is the key to the universe"... agree, but is too long to write on a t-shirt... so 3,6,9 wins :-(
    2
  4226. For the 3rd assignment at 9:40, I first looked in what ways 2020/4=505 can be written as a sum of two squares. Mod 5, the only ways that two quadratic residues (0,1,4) add up to 505=0 are 0+0, or 1+4. [Here observe that (0,0) is not a possibility, because it would imply that 25 would have to divide 505] Mod 7, the only ways that two quadratic residues (0,1,2,4) add up to 505=1 are 4+4 and 0+1 Mod 9, the only way that two quadratic residues (0,1,4,7) add up to 505=1 are 0+1 Mod 11, the only way that two quadratic residues (0,1,3,4,5,9) add up to 505=10 are 1+9 and 5+5 Listing the candidates 0 through 22, these conditions each exclude some possibilities, leaving only the squares 1^2, 8^2, 12^2, 19^2, 21^2. By multiplying everything by 2, the 16 (ordered) ways to write 2020 as a sum of two squares are easily found now: (+/-2×8)^2+(+/-2×21)^2=(+/-2×12)^2+(+/-2×19)^2=(+/-2×19)^2+(+/-2×12)^2=(+/-2×21)^2+(+/-2×8)^2=2020
    2
  4227. Fantastic video thank you for increasing my love of maths with every video you make
    2
  4228. Good Stuff Burkard! :)
    2
  4229. This made me think of this Vi Hart video! https://youtu.be/Yhlv5Aeuo_k
    2
  4230. In this paradox, a finite volume can paint an infinite surface but it need an infinite time
    2
  4231. I guess that's why it's also called Gabriel's horn...plenty of time in heaven!
    2
  4232. i like how he call this a 'mathematical creature'. how bizarre
    2
  4233. Amazing...
    2
  4234. 13:22 Every time I hear SOHCAHTOA, I shake my head anymore.
    2
  4235. Some people call them cosh, sinch, tanch.
    2
  4236. The worlds first Strand type magazine
    2
  4237. Man I love when the algebra autopilot hits you with the "tingletingletingletingle"!!
    2
  4238. Golden rectangle in the shape of a Kit-Kat chocolate wafer bar. Kit Kat -- Break the Leg! Even golden bars are in shape of golden bouillons of 24K. Except for price taste of Kit Kat bar rhymes with Kakis and it :)
    2
  4239. Ich liebe diesen deutschen Akzent. Englisch ist damit so viel einfacher zu verstehen ;-)
    2
  4240. The number wall you show at 10:28 shows getting to all zeros at row 5. I did an Excel and I get to all zeros at row 6; I show row 5 as being 1 1 0 0 1 2. Am I missing something?
    2
  4241. I retract my comment. Nevermind! My Excel logic has a very simple error; the logic checks to make sure the cell two above the one being calculated is “greater than zero”. It should be “does not equal zero”. Rookie mistake! :)
    2
  4242. Math is easy if all the wrong steps are taken out
    2
  4243. 4:58. Yes I would be happy to, though when I saw this when originally transmitted I was only 11 and it was beyond me then. Another BBC programme, Blue Peter, ran a competition in parallel with the initial transmission of this story inviting people to submit solutions and I remember that the winners of the coveted Blue Peter badge submitted their solution on a l-o-n-g strip of paper from a bus ticket roll. Years later i independently discovered the recursive algorithm. I was at Uni when I first was shown the iterative algorithm in a computer class to demonstrate that recursion is generally time and data hungry. The take away was to only use recursion if we didn't know any iterative way. And it was only because I pestered the lecturer that I was shown the validity proof. That's because I know the iterative solution. That's one that is harder to explain why it's valid but easier to do in practice. Incidentally it is a lot less machine cycles if you track the program on a computer for each algorith. If you get the first move right the rest is automatically correct.
    2
  4244. Now generalize it to all reals and complexes.
    2
  4245. The frustrating thing is that no amount of decimals can ever tell you whether the number is truly rational or irrational... His approach can tell you how close it is to fractions with denominators up to a certain scale though. The typical error of an ordinary fractional approximation m/n is of the order 1/n, whereas these best approximations have an error of the order 1/n².
    2
  4246. Whilst "is the number able to be expressed as a ratio" a question with a yes or no answer it can be interesting to consider how well any irrational number can be approximated using rational numbers, which is called Diophantine approximation, and I would suggest you look into that if you're still interested!
    2
  4247. @phyphor  It is a bit more of a language issue than a math issue, but just like math, some words have well defined and exact meanings. It is just inaccurate to say any irrational number is more irrational than another. Perhaps math just needs a different word for grading irrational numbers within the binary.
    2
  4248. The most pregnant one is the one who is going to give birth first.
    2
  4249. s a t o r a r e p o t e n e t o p e r a r o t a s
    2
  4250. Great stuff!! Thanks for such insightful videos.
    2
  4251. 12:13 proof that Burkard is a robot
    2
  4252. Lovely video on the magic square by a gentleman who is NEVER SQUARE!!! Gary in dreamland. Have a nice dream!!🙂☁️☁️☁️⛅💫🌟🌟🌟🌟🌟👉☁️⏰☁️👈
    2
  4253. Just pure extasis. Some times we have to sumup weeks to get the 50 min of Mathologer this Is a gift. (Phisycs bachelore)
    2
  4254. Beautiful!
    2
  4255. My favorite guitar pick!!
    2
  4256. 1) I was never allowed to take math beyond math (I failed algebra many times. I usually got the right answer the wrong way which made me a pariah among math teachers.) 2) In a college statistic class I asked what was the probability of probability and got ejected.) That said when I look at your moving diagrams they look like extrapolations form the real world... for example a circle in a circle and a straight line is a driving wheel on a train.... or a differential is another example..... so which came first? Real world use or mathematics?
    2
  4257. Your enthusiasm for maths is really infectious 😊
    2
  4258. I liked chapter 5 the most. Well, five is a prime number, I think that's enough for a reason... and yeah of course Gamma came in! ;)
    2
  4259. Kinda reminded me of perk system in Ghostrunner
    2
  4260. 4:26 Is it because they don't use 百 and 十? I don't see any other thing that could be wrong. For the record I only speak intermediate level Japanese, not Middle Korean.
    2
  4261. Great job! By the way, just a trivial remark: I've realised that the "Greek identity" reads in complex numbers just |z|^2 |w|^2 = |z w|^2. Cool!
    2
  4262. one symbol kills it all
    2
  4263. Gamma must be greater than 0.5 because 1/x is an convex function. You shifted the blue area to the left, but you also could have shifted the red area of the same hight (equals the white area in the marked square) to the left. ...I can't explain better, but for me its obvious. Best part for me: The crasy tower in 9:50
    2
  4264. Great video - I really learned a lot from this one!
    2
  4265. The most memorable was no 9 series.
    2
  4266. These geometry videos are always so enjoyable :) Even when the subject is something I thought I understood reasonably well in uni, the videos always help to make it more intuitive instead of just being an algebraic truth. I really hope one day I'll get to see mathologerised versions of Lie groups/algebra. Like it's one thing to sit down and work out that SU(2) and SO(3) algebras are isomorphic but there ought to be a way that that's visual right? Given how visual SU(2) and SO(3) are...
    2
  4267. dammit - good video, but now youtube thinks I love quack math and numerology
    2
  4268. These are Slide Rules and how people used to compute before digital calculators. Awesome!!!
    2
  4269. Your explanations and animations are just beautiful poetry!!! Thanks a 10^6!
    2
  4270. Lovely - and already 80,000 views
    2
  4271. This is the first strand-type puzzle
    2
  4272. Mathologer, your shirt-gamecis criminally underrated!
    2
  4273. About the coding competition: Is there a preferred coding language (example HTML5 with javascript, Processing, Unity, Scratch, Python+Pygame, C# windows application, Python Turtle, other)? keyboard interaction?, mouse interaction?, webbrowser?, windows/mac/linux application?, android/iphone?, zooming?, slow motion buildup?
    2
  4274. You forgot to mention the Parker square! ;) Great video though.
    2
  4275. Is there any analog of conditional convergence over the complexes? Or are conditionally convergent series sort of just a pathological case of real numbers?
    2
  4276. Great presentation. You made it looked easy.
    2
  4277. Let me guess. More than twenty! 😀
    2
  4278. 9 4 17 13
    2
  4279. I think the long cross rule can be used to compute the two numbers directly below a 2x2 square, but then to compute the two numbers below those 2, you still have to use the horseshoe rule.
    2
  4280. this is amazing! i wonder if there are any interesting properties about the triangles you get when folding right triangles
    2
  4281. Amazing as usual
    2
  4282. Four thumbs up to you (mine and a friend's)! This is at the bare minimum really interesting and likely more than that :) the complex ones are fascinating and leave me with awe and wonder but I understand them only superficially for lack of math training. This one and the Die Hard ones are very welcome for their accessibility while still being all maths. I also like maths applied in every day life and stuff like the game theories were also super cool 👍 Here is my answer to your puzzle: stated as "What is the maximum number of turns if one flips any 2 FACEDOWN cards over at each turn?" I would say the maximum number of turns is also 5 since there is no room for anything else than the straight path. (Since my answer is obvious it must be wrong 😂)
    2
  4283. This was a truly great video. I'm homeschooling my children, and this makes me completely impatient to get to teaching them about roots of polynomials. Too bad they are only 10. We've got a lot to cover before they can appreciate this. But they are going to love this when the time comes. You are making a difference 1 child at a time.
    2
  4284. Brilliant and very satisfying video!
    2
  4285. In David Wells's (The Penguin) Dictionary of Curious and Interesting Numbers, in the entry for triangular numbers, there is an analogous pattern: T1 + T2 + T3 = T4; T5 + T6 + T7 + T8 = T9 + T10; T11 + T12 + T13 + T14 + T15 = T16 + T17 + T18, and so on, attributed to M. N. Khatri. I don't understand everything about how this happens, but the fact that the sum of consecutive triangular numbers is a square (whose base is the larger index of the two triangular numbers), plus the fact that the difference of consecutive triangular numbers is the index of the larger triangular number, seems to be at the heart of it.
    2
  4286. Nice!
    2
  4287.  @sebastianjost  thank you :)
    2
  4288. EDIT: this is a long one Since you can choose in an order you want, and order ultimately doesn't matter within groups, choosing whether or not to include an object means that every choice you make you have 2× the number of previous choices. Since the number choices are infinite, you the ways to pick and choose is 2^∞ which is of the form k^∞ The infinity in the exponent is a countable infinity that has the same cardinality as א‎₀ If we take a detour and switch to base k for a second, we see that Cantor's diagonalization argument can be applied here. Cantor's diagonalization argument shows that the reals are uncountably infinite and specifically they share cardinality with א‎₁ [n^א‎₀ ≈ א‎₁] meaning........ The cardinality of the number of ways to approach an arbitrary real number exactly the same as the cardinality of the real numbers!!!!!
    2
  4289. ​ @jaafarmejri3361 except you can't ever reach any of the irrational reals between PI and 4, you can iterate towards them with a finite sequence but that finite sequence would be the same for all reals in a range where there are finitely many possible ranges, the number depending on the number of terms in the sequence. While there exists an infinite sequence of terms that approaches any real, you can't get there and then approach a different real because there is no "after" unless your sequence has an element number infinity, infinity+1, and so on.
    2
  4290. it's uncountable as you can have an infinite number of deviations from the "shortest" path to Pi (quickest converging) after your partial sum crosses your target number, and you can either have a deviation or not at each crossing, as long as you then correct your path (to make sure the sum doesn't actually converge to something else, if this is even necessary), so the number of paths is at least 2^(aleph_0) which is uncountable.
    2
  4291. My guess is uncountable because the plus/minus corresponding to 1/0 will give us binary numbers which is comparable to the set of real numbers.
    2
  4292. ​ @kappasphere good argument! I think you can simplify it a bit by just choosing between 0 and 1 overshoot terms each crossing, which gives you 2^(aleph_0) rearrangements, which is already more than aleph_0 (ie uncountable) 🤔
    2
  4293. I got quite confused at the end there, I’m going to have to watch it again. Anyway, I learned a lot, great video
    2
  4294. Somebody show Newton's Fluxion!! No! Not like that... The way HE did them... The method that died away during the 1700s.
    2
  4295. Very NICE.........
    2
  4296. Now I'm curious if this will continue to generalize into higher dimensions, if the nonagram will help develop hypercubes along with a hypergolden ratio.
    2
  4297. This is a favourite topic of premier engineering or research oriented colleges while conducting entrance test in the colleges in India. These really blow off the mind especially when variants of these problems are given in "one or more than one correct answer" type questions.
    2
  4298. I learnt in both these proofs plus the proof involving similar trianges (Brahmagupta's proof) back in my High school in India. I am surprised you say that most of us are not aware of the first two proofs
    2
  4299. 25:25 not sure why the relation to the area remains the same when moving to the hyperbolic functions. Additionally, using the same symbol "a" for the area before and after the animation is confusing, it seems like we're talking about the same quantity but it's not.
    2
  4300. Reason why the mentioned article was ignored? I think the title was too generic.
    2
  4301. This video was amaaaazing. Wow. I enjoyed it a lot
    2
  4302. I'm pleased to say I actually have heard of some of the things in this video before! I'm just a simple humanities person so I get excited when not everything in a maths video is completely new to me. :)
    2
  4303. this is : "Thinkful of a dollar"
    2
  4304. I like having this length video for simpler concepts
    2
  4305. I like the shirt, to infinity and beyond!
    2
  4306. I really like your videos that have a pop culture reference, Simpsons, Futurama, movies. I also really enjoy shorts closer to 10 min videos over 20. I would have liked if you went into more detail about binary.
    2
  4307. wonderful !!!!
    2
  4308. The 3d-rule corresponds to adding in the group Z_2 x Z_2. If the bottom layer is a b c d e f then the top ball will be a+d+f+2(b+c+e) = a+d+f. Then we can do the collapsing argument giving the special sizes 2^n+1. Rotational symmetries: If the hexagons in a small triangle have numbers a, b, c, then the negative mod m game says that a + b = -c (mod m). But that is equivalent to a + c = -b and b + c = -a, which are the rotations. Since all the small triangles can be rotated, the whole figure can as well. For the pascal (positive) mod 2 game, a+b = c = -c, since 1=-1 in Z_2, so the rotations work here as well. The same applies for the 3d game: a+b+c = d = -d.
    2
  4309. A little late to comment, but I actually read the paper this is from a few months back! I was browsing the Bridges Archive in my free time, and it is so satisfying to see it covered here.
    2
  4310. Hmm, I think I've seen the hints for this in Numberblocks toon series :)
    2
  4311. Maybe it was intentional and not a blooper. Maybe it was secret knowledge.
    2
  4312. My favourite part was about Gamma and its connection to ln(x) as integral of 1/x
    2
  4313. This is absolutely mind blowing. Please never stop producing content like this! I absolutely love the channel.
    2
  4314. I'm special too, a special boy. They say my mind is special.
    2
  4315. I always love your videos Mathologer, going to grab a laptop to watch this on full screen and yeah, I'm one of those who never got any proofs. I've missed the "Ready to be amazed?"
    2
  4316. We need a person cutting in to say "wait did you just say a ten block overhang using one method needs 571 blocks and the other way takes 12,307!?"
    2
  4317. The hyperbola in the spinning cube intrigued me. Suppose the unit cube rotates about its main diagonal (from vertex (0,0,0) to vertex (1,1,1)) The largest distance to the main diagonal is at an edge, the interesting edges are those not connected to the rotation axis, for example x=0, y=1, 0<=z<=1. The squared distance from a given point (0,1,z) to some point (t,t,t) on the main diagonal is (0-t)^2+(1-t)^2+(z-t)^2. The distance to the diagonal has its derivative to t equal to 0, so that z=3t-1 We find that the curve is d = sqrt(6t^2-6t+2). The interval for t that is featured by the cube is then [1/3 , 2/3]. My quadrics / cone sections are a bit rusty, who can show that this is a hyperbola?
    2
  4318. Very nice video and such pretty graphics. I have a small python program on github that shows the graphics for various modulos and multipliers and even gives you a nice animation. Search for Modulo-graphics on github. I did that after I saw your first video on the subject as a python exercise.
    2
  4319. What a ride!!
    2
  4320. Too much fun, ill try to prove the equal sign in the limit of your approximation N(r)/r^2. The difference between the approximation and the real area is in the boxes around the edge. These boxes are no more than a distance of sqrt(2) from the edge of the circle, whose circumference is 2*pi*r. The "error" e is then bounded by e<=2*pi*r*sqrt(2). I.e. the error grows linearly with r, while the approximation grows quadratically. In the limit, the proportion of the error -> 0.
    2
  4321. 🎖🎖🎖🎖🎖🎖🎖 incredible again!!! Those transformations and metamorphoseses... always open a totally new view on a 'known' issue.
    2
  4322. It took me a long time to get to what the Mathologer pulls out of his hat after the first minute (following "whoa! What just happened?") Part of the problem was I steadfastly refused to accept that the three angle bisectors of the triangle meet at the center of the inscribed circle, without proving it according to Euclid--perhaps because of it's similarity to the flawed proposition that leads to the "theorem" that all triangles are isosceles. Getting past that hurdle, though, it's pretty straightforward to derive a formula for any triangle with a unit circle inscribed in it. Each colored segment (BTW of course you wouldn't use "blue" because it's name would be the same as side B) is simply equal to the cotangent of the half angle of each vertex. For the 3-4-5 you can apply the half-angle formulas to get the cotangents in terms of the cosines (sqrt(1+cosine/1-cosine) ), which for the 3-4-5 are 4/5/ 3/5 and 0, and you miraculously get whole numbers for each side. Time to unpause the video. It follows that the sum and product of the cotangents of any 3 angles that add up to 90 degrees must be the same. I guess the only hope of proving that is expressing them in terms of imaginary exponentials, and grinding through a lot of algebra. No geometric insight in that though. For another time. BTW while Heron's formula in terms of the sides instead of the semi-perimeters might be preferable in some ways, if you're actually calculating it you annoyingly have to form 4 sum-difference combinations of the sides, whereas in terms of the semi, you can just stick it in a memory on your pocket calculator and re-use it. Did Heron have a pocket calculator? He was an engineer, wasn't he? In all seriousness, Heron being an engineer clears up a question I've had for years: were this Heron of Alexandria and the one credited with inventing the first steam engine-which consisted of a steam boiler with jets on the side that set it spinning- one and the same? Apparently so. It's inefficiency, and other logistical problems problems, probably explains why the chariots of the day weren't powered by Heron engines.
    2
  4323. Happy Christmas mathologer ❤️❤️❤️
    2
  4324. Rats only 1 minute in - now I have to get the bandsaw out and make those 9 shapes in wood.
    2
  4325. Feliz Natal para todos do grupo
    2
  4326. re: the questions at the end im in year 12 and everything clicked for me. great presentation! also, i cannot wait for more euler-mclaurin stuff :)
    2
  4327. 13:00 I believe if you use the "1a, 2a+1b, 3a+2b+1c" sequence transformation you get the answer of n! * (n + 1)! (I find it interesting that for the squares a=1 but all the other letters equal 2)
    2
  4328.  @Mathologer  Huh, I wrote a whole explanation but I guess yt didn't like it. Oh well, nothing you can't google. And yes, woo is always dangerous :)
    2
  4329. Wow, really beautiful!
    2
  4330. Awesome video! Keep going please. Most memorable: The Lira tower.
    2
  4331. What happens if you use this to try to find the next prime? Actually, I’ll give it a a go! (Does is just find a polynomial that fits the numbers provided but doesn’t actually predict the next prime?)
    2
  4332. Doesn't seem to be a popular option, but my vote goes to the "no integers" fact. It's such an obvious pattern and such a simple insight (and apparently a not-too-complicated proof), but the end result is surprisingly deep. Isn't that what we look for when we seek for "elegant" or "beautiful" mathematics? Simple insights that lead to complex and perhaps counterintuitive results?
    2
  4333. We need more)
    2
  4334. We always like it!!! 👍
    2
  4335. Awesome channel! 😌
    2
  4336. Always. Love it.
    2
  4337.  @tbttfox  riiiiight thats where i got confused. thank you so much!
    2
  4338. Just noticed: I love that shirt! :D
    2
  4339. I used the following reasoning to think that the sum of no 9s was finite: You can also say that the theorethical largest number closest to infinity must contain a 9, because if it didnt I could just replace the first digit with 9 and its larger. And thus the no 9s sum is finite, since the sum is always stopped by some 1/9xxxxx... like in the fractal that stops it from being infinite. I dont think its waterproof and probably only applies to 9 but it was just my intuition lol
    2
  4340.  @Mathologer  As was mine - a Builder of Targets! First as a SeaBee in WWII, then as an engineer with the Manhattan Project in Hanford and Oak Ridge...
    2
  4341.  @Mathologer  I was taught to find a center point and approximate axes then tile outwards from the centre.
    2
  4342. Cavalieri was about 50 years before Newton and Leibniz, if I recall. It's standing-on-the-shoulders-of-giants all the way down.
    2
  4343. 2+2=2×2=2^2
    2
  4344. Can anyone recommend some resources to be able to understand parts or follow a long to parts of videos like these? I find them interesting but overwhelming!
    2
  4345. what would happen if you skip every prime numbers? my guess is that since prime numbers are getting rarer the higher you go, the result would be close to just the sum of all integers (maybe the difference is something like the log of n)
    2
  4346. You know, reading these comments, it's funny to see how many of us did something like this when we were young and thought we were alone and were actually discovering calculus.
    2
  4347. I cannot believe we missed this. Does this extend beyond triangles? And is there some nice group theory ramification?
    2
  4348. Isn't the folded triangle the same as the triangle created by joining the midpoints of each side of the original triangle? It seems to even be the same size and orientation.
    2
  4349. name of music around 1:00 ?
    2
  4350. https://youtu.be/Yy7Q8IWNfHM is the answer to this is 666?
    2
  4351. Could you still get an infinite overhang on a planet like Earth considering that it’s spherical, and it’s gravity goes down with height?
    2
  4352.  @Psycho0Robot  That's not why artists' jobs are in jeopardy. To give a brief summary, generative AI models hoard images from the web to train themselves more and more. Recently, Twitter (one of the biggest artists hub) has changed its terms and conditions on the users' part to allow virtually unrestricted access to its database, provided of course that the company agrees to it (in exchange for money, obviously). This is causing a massive exodus and migration of artists on other websites, for example bluesky, as well as artists using countermeasures like digitally "masking" their art with special tools that render the images unreadable for current AI models. Keep in mind that generative AI can be trained to emulate the works of a specific artist fairly easily, and while not perfect at present time, it's bound to get there in a few years at most if unregulated. While this video's use of AI does not of course directly influence the data hoarding, seeing AI art still provokes some bad emotions in me.
    2
  4353. Ah, I came back to this because of the notifications and realized my other comment, the one I referred to, got deleted by Youtube, probably for mentioning "the bird website". To sum it up, I quickly mentioned the fact that "the bird" recently changed its terms and conditions to be able to give away/sell its images database basically unconditionally. Lots of artists are migrating to other platforms or starting to mask their work with AI countermeasures, because of this, to avoid stylistic forgery.
    2
  4354. "We've been familiar with 345 triple since kindergarten"... I think German kindergarten must be a little different than American kindergarten because we were learning that G is for gooey gum and A is for apple here in America.
    2
  4355. Definitely great stuff! So much math revelations in one video. Totally mind blown.
    2
  4356. I found out about this theorem myself when I calculated some Legendre symbols of the form (5/p) and noticed a pattern (I then didn't know that they were called Legendre symbols). I then was disappointed that I wasn't able to prove it, but now I understand why I wasn't able to. 😅
    2
  4357. Taking verisatium TM advice about multiple icons and how clickbait helps in getting your message across
    2
  4358. Very pretty result. It looks like with the middle column of the tree, you can just take the ratio of the two top numbers to get an approximation of root2. Also, it appears these pairs of numbers have products equal to the balancing numbers (i.e. the solutions to Pell's equation @ N=2) : 3 x 2 = 6 1 + 2 + 3 + 4 + 5 = 7 + 8 7 x 5 = 35 1 + 2 + 3 + 4 + ... + 33 + 34 = 36 + 37 + 38 + ... + 48 + 49 17 x 12 = 204 1 + 2 + 3 + 4 + ... + 202 + 203 = 205 + 206 + ... + 287 + 288 41 x 29 = 1189 1 + 2 + 3 + 4 + ... + 1187 + 1188 = 1190 + 1191 + ... + 1680 + 1681 etc...
    2
  4359. I have to defend my maths teacher... He showed us not only the Pythagorean Theorem with squares but also the "Lune of Hippocrates". I was expecting it to show up here in the video - it would have been a great addition in my opinion. Anyway, I don't quite remember whether we were told the theorem holds for all kinds of shapes in school... So thanks for the reminder! :)
    2
  4360. Great video. I had never thought of it that way even though I have probably read it and forgot its beauty. I also advocate student to do these "discrete derivation" steps and see if they get to a row where all values are similar and don't understand either why it's so forgotten. It's such a great problem solving methodology. Then why it is a indication that the points might be a polynomial of k:th power, where k is how many times they have "derived". I usually take the sum of squares as first example, close to what you did. I have on the other hand told them to set up the desired polynomial as ax³+bx²+cx+d and then replace x with the order the number have in the sequence. I usually include 0 so to be nice. This will make them get the four equation d=0 a+b+c+d=1 8a+4b+2c+d=5 and 27a+9b+3c+d=14 which they then solve with Gauss Elimination making a=1/3 b=1/2, c=1/6 and d=0. So I sort of make them do a vandermonde matrix without them even noticing. They will have to make the indexing correct though our their polynomial might get transformed on the x-axis a bit, but would still be able to predict upcoming values. Probably easier with the method you suggest here when it's an even bigger power though as the Gauss elimination or matrix inverse processes are prone to introduce accidental mistakes by students and myself :). Thank you so much for the insight! Maybe even better even for lower powered polynomials :) It all depend on the circumstances! But this is more beautiful I agree. Another nice example of this potential is make this exercise: Find the second-degree polynomial that intersects these points (2,1) (3,6) and (5,28) making them realize we can set up such a matrix even if we are "missing" a value if we know n+1 points and that we are seeking a polynomial of the n:th power. Your channel is really great as you put that little extra concise effort in the sentences and phrasing, making it unlikely for viewers to take it out of context, which seems to be a big problem on YouTube nowadays.
    2
  4361. Zagier's gonna like this geometric intuition
    2
  4362. @[06:44]: This will also involve the omni-directional symmetry of a circle, won't it... (;
    2
  4363. 5:41: The two yellow segments are congruent because they are formed by two tangent lines where they intersect the circle and each other. If you connect the two points where the tangents intersect the circle, the newly made angles are interior angles that intercept the same arc. If you use the formula { measure of angle = 1/2 measure of arc in degrees} you can set the angles equal. This makes the triangle isosceles, and the two yellow segments of the triangle congruent.
    2
  4364.  @Mathologer  thanks that's awesome, I can't wait!!
    2
  4365. Best thing I watched this week even though it was uploaded 4 mins ago
    2
  4366. Pi abacus shirt- love it!!
    2
  4367. Well, this is some crazy stuff! I'm wondering what would similar constructions in higher dimensions give?
    2
  4368. I learned/used the digit sum trick for checking addition/multiplication when I was at school. It was called "casting out nines".
    2
  4369.  @ragnkja  Yes! I had forgotten that. There's a logarithmic scale from 1 to 10 on the sliding part, and a mirror of the same scale on the frame of the ruler - this is for multiplication. And, there's another line of numbers on the frame of the ruler. Where the second scale has '1', the third has e, Euler's constant. Where the second scale has '2', the third has e^2. Where the second scale has '3', the third has e^3, and so on. So, if you slide the '1' of the slider to '2' on the second scale, the same place as e^2 on the third scale, then the slider numbers '2', '3', '4', and '5' match up with the second scale at 4, 6, 8, and 10, and match up with the third scale at e^4 (or, e^2, the starting number, squared), e^6(=(e^2)^3), e^8(=(e^2)^4), and (e^2)^5. Putting the '1' of the sliding scale at any value 'x' on the third scale means '2' will match up with x^2, '1.5' will match up with x^1.5; putting the '10' of the sliding scale at any value 'x' on the third scale means '5' will match up with x^0.5, '3' will match up with x^0.3, etc.. But the third scale only works for numbers between e and e^10. There's a fourth scale for numbers between e^0.1 and e, and depending on the slide rule, a fifth for e^0.01 and e^0.1, a sixth for e^0.001 and e^0.01, a seventh for e^10 and e^100, and an eighth for e^100 and e^1000. For multiplication, the second scale works for all powers of 10, but this method for exponentiation needs another printed scale for each power of 10. I'm curious if there's a way to make a (finite) slide rule with just two scales that will compute one number to the power of another, generally.
    2
  4370. I studied this in college, with a slight notation change. Instead of "falling powers" the text notated superscript "(m)", which my profs called "upper m". At the time, mid 70s, Finite Differences was part of the life insurance actuary's professional exams.
    2
  4371. Like many others in Mathologer-land, this video has helped me with some elementary school students I tutor. They need exercise in simple arithmetic, and need even more a window into how exciting and powerful math can be. As for me, I am now 4 episodes into Tree With Deep Roots. I got goosebumps watching the king solve the 33x33 square. Now I'm reading all this 15th century Korean political history. What a gas!
    2
  4372. if i may make a pun: "of course, we also get that one way of making zero sense"
    2
  4373. I made it interactive: https://www.desmos.com/calculator/isxalxvdcd
    2
  4374. I'm glad you mentioned that simple geometric series equation. I'd never thought about the Lorentz gamma,  𝛾(v)=cosh(atanh(v/c))=1/√(1-v²/c²)=dt/d𝜏, being the square root of a geometric series before. Yep, it's a weird tangent but I'd just been deriving functions to map from c+v√1 to cdt+dx√1 and dt+dx√c⁻², namely c exp(atanh(v/c)) d𝜏 and exp(√c⁻² atanh(v/c)) d𝜏.
    2
  4375. Very interesting
    2
  4376. YES, NEURAL NETWORKING SHOES 13:12
    2
  4377. The car analogy is great. On a related note, I have wondered how the average speed displayed in an onboard computer changes during a long trip with faster and slower roads. I did think how integrating my "speed function" and dividing by t to give me an average speed function, but never got much further with it :-)
    2
  4378. You are truly a great science communicator — a wonderful teacher. I humbly thank you
    2
  4379. We learned about this a little bit. I remember studying sequences and series and Calculus 2. However, to answer your question, it may help to look at the Curriculum from a business perspective. Higher education is no longer focused on cultivating the intellectual capacity of its students. The focus of higher education in our time is to give students information that they can apply to the benefit of some future employer. So, if something, such as sequences and series, doesn't have the potential to help an employer offer a good or service to their customers, there is little incentive to add it to the curriculum.
    2
  4380. I was taught pronouncing sinh like "cinch". Is that an American thing?
    2
  4381. You just amazing sir! Respect from Bangladesh 🇧🇩
    2
  4382. I'm not sure but i feel like Stephen Wolfram could use this for his physics project. Hypergraphs evolving into the universe based on some simple rules is the aim. They've made huge leaps in exploring what they call "rulial" space. The supra manifold of all possible rules and more recently how this connects to combinators.
    2
  4383. it's an extra extra extra special video
    2
  4384. I'm a 3rd year maths student and Im currently taking abstract algebra but have not started analysis. The only part of this video I couldn't perfectly follow was solving for the sum of (1/n^2). I know you said you glossed over a lot of technical details, which could be why. What I'm struggling with is how you justified the yellow to be an approximation of 1+1/2^2 +1/3^2+...=... and how you rearranged the equations to take advantage of the n terms. If you could please go over that in greater detail in your follow up video that would be amazing. I love your content because it feels as though I am getting a personal, relaxed lecture from a professor that has a great way of explaining and illustrating complicated ideas. That's one of my favorite ways to talk about maths and why I will happily sit here and gain knowledge for 50+ minutes. Please never stop doing what you do!
    2
  4385. 3 bricks with 1 completely overhanging: put 1 brick flat as usual extending a ways over the edge. Tip the other 2 bricks up on their ends, and balance them on opposite sides of the first brick. (inre the question from 11:53) This gets the nearest-to-the-table part of the overhanging brick much further away than the naive solution where all bricks stay in their usual flat orientation (just how much depends on the shape of the bricks).
    2
  4386. Excellent video as usual. Perhaps you forgot to mention how to arrange the terms to get positive or negative infinity.
    2
  4387. You patch it with fabric that would other wise make 3/4 of a shirt.
    2
  4388. Depends which quarter
    2
  4389. one question: it seemed that we just assumed we should look at the face down cards as ones and the face up cards as zeros. couldn't we have just imagined the opposite of that, and then the number would be ever increasing? would that be ok?
    2
  4390. I love that the answer to the clickbait title is that the original paper didn't have a clickbait title. That's some Alanis Morissette ironic.
    2
  4391. If I ever become a medieval Korean monarch, I will definitely command Mathologer to be my Court Mathemagician. 😄
    2
  4392. I’ve been studying linear algebra and have been really interested in its applications. When I got to the end and realized that he was going to use linear algebra to tie it all together I got sooo excited! Super cool video and I’m excited to delve deeper!
    2
  4393. Gold Standard for math content on youtube
    2
  4394. Anyone nonproportional rubs me the wrong way.
    2
  4395. 6:32 "How?" disappears
    2
  4396. I was taught Heron's formula in school here in Ukraine. Always treated it as some overcomplicated exploit :D
    2
  4397. This also explains why the 1st, 2nd, and 3rd roots of unity are the only roots of unity with Z+Z^(-1) a rational number. This is a fact that is used in the representation theory of dihedral groups, but I didn't really know a nice reason for it until now.
    2
  4398. Could the dimensionality of each measurement be a factor? The surface area is a two dimensional quantity, but the volume is a three dimensional quantity. How many surfaces can you fit into a three dimensional volume?
    2
  4399. how do you tie the devil and angel knots
    2
  4400. I think the Talmud system was great when creditors were actual people. It is not the case anymore, and worse than that, credits can be traded, split and merged. Now creditors are typically banks, and banks have shareholders, is every shareholder a creditor or is the bank as a whole a creditor? Using the Talmudic system would be a nightmare in the modern banking and market system. A proportional system simplify things a lot.
    2
  4401. Ahh, not theory ;-)
    2
  4402. I was always terrible at math and great at science and now since I'm 40 years old I'm trying to learn all sorts of things I love your channel you explain things nicely and calmly 😊
    2
  4403. Now I'm wondering about B. F. Sherman's "4th side of a triangle", and how long it is, and whether that can be tied in with all of this interesting
    2
  4404. Awe! Mathematics is discovered. It is in the mind of God.
    2
  4405. When you handshake someone, they also handshake you. Therefore, the total amount of handshakes having orcured, appears to be even. 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21, and 21 isn't even. So seven people can't hold one by one distinct values from the set: (0, 1, 2, 3, 4, 5, 6). So there must have been a handshake twin.
    2
  4406. This is true but will only help you prove the result in the cases where that sum is odd. For example with 5 people the set of possibilities {0,1,2,3,4} adds up to 10 and you can't rule it out by this reasoning. But the argument in the video always holds.
    2
  4407. M.C. Escher approves of the spinning 4-D cube.
    2
  4408. @21:01. Answer 180 degrees. Arrow makes 3x360 degree rotations or 6x180 degree rotations. 180- each of the angle is the rotation the arrow makes. It does it 7 times. 6x180 cancels the 6 of the 7 x 180 degree. Remaining is the angles are equal to 180 degrees.
    2
  4409. that is an awesome christmas t-shirt!!!!! happy holidays Mr. Loger!
    2
  4410. Don't you think Ramanujan solved the two simple related functions first. Then he saw that they they could be added together to give an even more fascinating result.
    2
  4411. 14:53 after that point I expected a 4D geometrical animation and explanation. Also, I like how people and even the smartest mathematicians are calling everything "irrational" when it comes to the decimal points.
    2
  4412. My take: Imagine the coin rotates around a point. How much "distance" does it travel on the point? Zero, because a point has no dimension. It simply rotates around the point without spinning at all. The circle itself is simply forcing this "point rotation illusion" accross the entire surface evenly. I think the real lesson is that the circle itself is an illusion. Or rather "curvature" is an illusion. Because if I take any physical object that is straight, and I "curve" it, it is actually creating MORE surface on the outside of the curve, and creating LESS surface on the inside of the curve. Math accomodates this illusion by calling these expansions and contractions, "infinitesimals" or "integrals".
    2
  4413. Next Video Mathematically best way to tie a knot
    2
  4414. Some one could make a maths parody of Oasis's song Wonderwall about these.
    2
  4415. Your videos make may days… everytime… thank you.
    2
  4416. 2:32 P+2+P=2(1+P)=2*P^2 given P's definition P^2-P-1=0
    2
  4417. The things that should be explained the most is that why stretching rubber band equal to natural log
    2
  4418. I suspected that it was the top two corners that had to do with the bottom colour because I noticed the pattern in the previous random triangles and I'm so ecstatic that I got it right
    2
  4419. It's a pie tshirt.....😃😃
    2
  4420. 16:00 By Pythagoras theorem the sums of the areas of the darker squares always add up to those of the brighter squares, so every colour adds an area of 1.
    2
  4421. Lol I lost it when you brought up the real world logistical nightmare of counting everyone's body hair!
    2
  4422. Game of life worse game ever 0/10
    2
  4423. It's not quite the same as the addition/multiplication talked about here, but in music theory there is a special class of tone rows that spit out the same result with a specific transposition vs. an inversion.
    2
  4424. What is the music at the end?
    2
  4425. Might I suggest that you shall find the source to be the associates of 'Mike Hockney' (Author by pseudonym) speaking for the 'Pythagorean Illuminati'. You can find that material in 32 books on amazon and others. It is the internal geometries of a static point on a circle... within Euler's Unit Circle...tracing a moving point on that dimensionless circle. The Ontogical Math of Reality insofar as it generates frequencies which then, subjected to Fourier Transform (in a complex domain), become material SpaceTime. That is my summary, so please do forgive the shortfallings in such a limited summary. 32 books, however, might appreciate your insights having reverse engineered the math, as it were. An excellent video Sir. Thank you
    2
  4426. Ooh!! You're doing Galois! So looking forward to that. <3
    2
  4427. Now I wonder what you can proof when you use this visual in 3D
    2
  4428. This may have been overlooked because of two other papers that took center stage, relatively speaking.
    2
  4429. We were taught Heron's formula in the United Kingdom. They used the alternative name "Hero's formula" and that was over 40 years ago.
    2
  4430. Another great video from prof. Burkard. I love it! Thank you and stay safe professor. Note: I'm amazed that even Euler is Mathologer's Patreon Supporter.
    2
  4431. Just solved the 3x3x3x3. Saw your video’s all these years ago, thinking this was amazing. Couple of months ago solved the 3x3x3 blindfolded and wantend a new challenge. With the help of your and a couple of other video’s I succeeded. Thanks for this and all the other facinating video’s on your channel.
    2
  4432. A slight variation on the theme in the form of a primary school magic trick... 1. Take one 3 digit number 2. Mix the digits to get a new number 3. Calculate the difference of these two 4. Then tell me either the first or the last digit of the difference 5. I will be able to instantly tell you whole difference number Eg. 654 and new number 456, difference is 654-456=198; last digit is 8, and I tell you the full difference is (9-8)&9&8=198 Ta-daaaa The magic of course again is that the digital root of the difference and the middle digit are 9.
    2
  4433. Welcome back, so great to have you back. I can relate to being busy. Am working at three location teach math at Adult education. 7 students alone in one location. So I really appreciate that you still have time for all of us. :). It not only amazes me the things you find to talk about, but to be able to find ways to explain things including things that are new proofs.
    2
  4434. The Kempner's proof animation is amazing
    2
  4435. Thank you for your videos. Happy New Year.
    2
  4436. Most memorable: finding out the various no 1's, no 2's, no 3's ... no 9's infinite sums are all finite and that you can find an approximate value for the sum of the Harmonic series with a simple formula.
    2
  4437. At school, I was out of my depth with trigonometry, and calculus just terrified me - i'd like to say it was bad teaching, but many of my peers strolled through it 😖
    2
  4438. Wrt chapter 0, there seems to be an assumption that all 0s of sin(x) are real. To me it is not immediately clear how this is assumed. Also, exp(x) can obviously be written as an infinite polynomial but there are no roots in the complex plane. So apparently this factorization does not work for all functions...?
    2
  4439. For 13:36 the sum of numerator and denominator of fraction is the denominator of the next the fraction The numerator can then be obtained by adding the new denominator with the old denominator
    2
  4440. Dear Mathologer follower, why is 5 exponentiated (by n) always ending with the cipher 5 (3125=5x5x5x5x5) like with 6 (216=6x6x6), is this a rule set to having a number system based on ten? Would the the same appear with the boarder numbers at the half of another number base, like 6 and 7 in a system based on 12, or like 8 and 9 in a system based on 16 ? The latter video here made my "sinfully" think of if there were magic cubes, not even having filled a single sudoku yet... enough for a guy like me to kinda feel dwarfed in some manner.
    2
  4441. Dear MATHOLOGER THANKS A LOT FOR YOUR QUALITY CONTENT.PLEASE ANSWER MY QUESTION IN BEST POSSIBLE WAY AS YOU CAN.[NOTE :READ FULL QUESTION AFTER ANSWERING]. Theorems have proof and formulas dont have proofs .We can understand theorems better because they make sense with help of proofs but we can't make sense of formulas beacuse they dont have proofs.SIMPLE QUESTION :HOW CAN FORMULAS EXIST WITHOUT ANY PROOF.WILL BE GRATEFUL IF YOU ANSWER MY QUESTION.
    2
  4442. There's another paradox here, I think. As shown in the video, the volume of the horn is Pi. But we also know that the integral of 1/x from 1 to infinity is infinite. That means the area between the curve and the x axis is infinite. We also know that the volume of the horn can be obtained by rotating that (infinite) section area 360º. So, an infinite area rotating 360º generates ... a finite volume.
    2
  4443. When you rotate a surface, there will be r^2 factors in the resulting volume. For most of the length, the r is extremely small.
    2
  4444. Some degree of rigor could probably be restored to the question "what comes next?" by asking for the formula with the minimum entropy.
    2
  4445. Great answers !!!! For your last question have a look at this http://math.mit.edu/~borodin/aztec.html :)
    2
  4446. The "no 9s series" part was the most memorable for me, and this is probably because of the way it was presented and because it was the last section in the video
    2
  4447. "Hey programmers..." Uh, hang on, I haven't taken Automata Theory yet. Soon.
    2
  4448. witty, intresting. 2d visualization is a sulotion.
    2
  4449. I made it to the very end and it was worth it!
    2
  4450. Wow, just WOW! Especially the final animation, The Lotus!!
    2
  4451.  @Mathologer  I'm in New York and have friends who have studied for more than half a century. If it's not in there directly, they'll analogize it to something else. But as you know, the Talmud is the compilation of the Gemara which explains the Mishnah which explains the Torah ... it's more complicated than that but that's the general idea. There are modern-day commentaries as well because it does address a body of laws. Meanwhile, my Irish-Catholic law school bankruptcy professor was very familiar with this and it is still considered a legitimate technique for legal analysis even though it's not enshrined in statutory law in the United States.
    2
  4452. Beautiful! When I was first learning college calculus, I tried making my own proofs for simple questions. When I brought my answer to my instructor, he warned me against extending a pattern into infinity without good reason, and he used the maximum divisions problem to instill in me a proper sense of paranoia! One more reason to read Newton.
    2
  4453. Why couldn't the card in the example been the 2 of spades since it's also 4 away from the 6?
    2
  4454. I thought I was too old for it.. but MIND BLOWN!!!
    2
  4455. wtf is this anti-shapeshifter propaganda????
    2
  4456. A few months ago was was playing around with phi^n and 1/(phi^n). There are some interesting recursion relations in there. Another great video from Prof Mathologer!
    2
  4457. I thought the jump from arithmetic to algebra was harder than the jump from algebra to calculus.
    2
  4458. Yes, calculus and algebra were both invented in India
    2
  4459. the cross puzzle is blowing my mind!! the area of two crosses sum up to the original cross i assume, i can't see any possible cuts wow hmm
    2
  4460. More memorable: No integers among the partial sums. How is this pattern related to prime numbers?
    2
  4461. 뭐여
    2
  4462. 4k+1 and 4k-1 is much nicer looking than 4k+1 and 4k+3.
    2
  4463. I have proven in my head... It just creates bunch of isosceles triangles, so it is easy to prove then that for every quadrilateral that you can make out of 4 consequtive points opposite angles add to 180... So all of them of them are cyclic. And as all adjecent ones share 3 points and those 3 points define a circle all six are on a same circle.
    2
  4464. In 2000 years quantum theory is old stuff and science historians, will wonder, how people could do physics like that.
    2
  4465. Onion proof = happy tears
    2
  4466. One of the first commenters! Result!
    2
  4467. Love your videos! - A mathematically inclined electrical engineering student
    2
  4468. Marvelous 😃. My Dad bought me a slide rule when I started my Engineering apprenticeship in the 1960s. A Faber Castell. He was an Aircraft designer so knew his slide rules. I used it continuously until calculators came of age in the 1970s. I still have it. Thanks for this great video.
    2
  4469. Are you german? You sound 100% like a german speaking english.
    2
  4470. Watched again
    2
  4471. Love 😍 from india 😍😍❤❤❤
    2
  4472. 26:11 - Geometry is cool !😎 ^.^
    2
  4473. I suspect that our dutiful "professor" is really the Toymaker, and the "leave your answer in the comments" is implicitly followed by "if you want to live". I always reach a point where I'm in WAY over my head in the math(s), but it's brought sufficiently close to my level where I could, with a bit more research (or Mathologer videos) I would have a clear "AH HAH!" moment! Also, "Should be fun!" really means "I hope you didn't plan to sleep tonight without solving this problem! MWAHAHAHA!"
    2
  4474. Your blue square is area 1/5.
    2
  4475. I haven't seen either of those square proofs but i prefer the second one just cause u dont have to rearrange stuff
    2
  4476. A truly enjoyable video. Did a lot of math on my time but did not know about this. Very cool!
    2
  4477. This is vaguely reminiscent of particle wave functions, Feynman's path integrals, the many-worlds interpretation of quantum mechanics, and the nail-and-board carnival game, where balls dropped into the setup bouncing randomly around will settle into a pattern reminiscent of Pascal's triangle. Could there be anything to this?.
    2
  4478. Looks like that's what I missed.
    2
  4479. Very interesting stuff; when I was about 15, I ran into these Moessner sequences when I was trying to figure out how to compute roots and produce good guesses for Newton–Rhapson methods. To this day, the fact summing the odd numbers gives you the squares is still such a fascinating thing to me; when it comes to learning about infinite sums, and sequences, I personally think it is a better introductory puzzle than Gauss' famous story of adding the integers, just because the output here is so recognisable.
    2
  4480. Does this generalize to larger (odd?) polygons giving golden numbers in higher dimensions?
    2
  4481. Why is zero to the power of zero = one?
    2
  4482. I heard a lot of the Oily-Maccaroni constant e.g. in number theory, but I didn't expect it in the harmonic series. Astonishing!
    2
  4483. many scientific discoveries including ones in mathematics are still named after European/arab translators than the original ones from India, are yet to be acknowledged
    2
  4484. Strange to wear this shirt while talking about squares ! 😉
    2
  4485. In ancient China, Qin Jiushao found another formula about the area of any triangle. That's a very complicated formula. However, this two Formulae, in fact, are the same. Because we use characters, they're not as convenient as letters, so he cannot make the original one simpler. Also, he uses the algebraic method. S=1/2*absinC And cosC can be expressed as a form of a and b. Then use: sin(α)^2 +cos(α)^2 =1 We can get a complicated form to express sinC by a and b. At last we can get the area.
    2
  4486. Amazing video mathologer as usual. I really liked the tiny bits of sneak manipulations with calculus in the video.
    2
  4487. at ~38:16, u say that 113, 106, 99, 92 are all spcial as they are all part of the 7 spirals. Yet, i see eight "7"s in yellow. What does that indicative of? 8 spirals? thanks,
    2
  4488. To me, the most amazing thing here is that you demonstrated the multiplicative property of a wheel generated this way, without even mentioning logarithms. I was anticipating some explanation involving the physics of uniform elastic stretching onto "sticky" wheels. NB. I grew up in the era of the slide rule, so I was already familiar with the circular variety. This was still a new & fascinating way to look at them! And BTW, an important disadvantage of the circular ones is the increasing loss of resolution with scales that are ever closer to the center of the device. Fred
    2
  4489. Mathemagier
    2
  4490. This is strange... I remember this theorem from early highschool where it was illustrated in a mathsy publication for kids. Admittedly, I slightly misremembered it as I thought it required equilateral rather than isosceles triangles. Hence it never appeared again in my life, but it was a fond (mis)memory that always remained, but lacking the details. But then the algo recommended me a video this week ( https://youtu.be/pUSI91Mo7LU ) that mentioned Napoleon's theorem and some other stuff, and that allowed me to search for and find precisely this theorem. I was happy to tie those loose ends. And now here is ye ole Mathologer explaining it further still! What are the odds of that happening in one week after almost 40 years!? (That is a rhetorical question.) Thanks!
    2
  4491. he died at age 33? man what a shame
    2
  4492. I guessed it right! I am taking my free subscription! 😊 I was actually looking at the 2 corners but had no idea why. Just somehow felt right and consistent with the first example.
    2
  4493. I have the 1998 revision of "Calculus Made Easy" with modern annotations by none other than Martin Gardner, my favorite math communicator. Next to Mathologer, of course. 😉
    2
  4494. I've recently gotten into euclidean constructions (compass + straightedge) while doing woodworking. I wouldn't necessarily use them to build a house, but for laying out cuts for visible jointery it's a lot easier and relaxing to grab a compass and start marking off ratios on the lumber itself and ignore the precise dimensions. The expression of the euclidean algorithm via snugged-up squares is really neat and a heck of a lot easier to do as a geometric construction than long division. Since it also uniquely expresses the simplified form of a fraction, it gives the largest squares that tile the rectangle. Not sure if I'll ever be able to use that, but I'll certainly keep it tucked in the back of my mind.
    2
  4495. Yay, I always look forward to Saturdays with Mathologer drops
    2
  4496. An n-cube is the n-fold product of (1 line + 2 vertices). In the formula, the 1 line becomes 1 x^1 = x and the 2 vertices become 2 x^0=2. The geometric shape described by this (x + 3)^n is the n-fold product of 1 line and 3 vertices. Ie. just draw a line with two end points and then another point on its own, next to that line: o-o o Now "multiply" that with itself n times. (A line times a line gives a (square) face, a vertex times a line gives a line, a vertex times a vertex gives a vertex.) The result is described by the formula (x+3)^n as in the Mathologer video. For example: (o-o o) x (o-o o) = o-o o | | | o-o o o-o o You can check that the number of vertices, lines, and faces of this are described by (x+3)^2. This formula generalises - start with V vertices, E edges, F square faces, C cubical cells, ... and represent them as a polynomial V x^0 + E x^1 + F x^2 + C x^3 + ... Powers of this polynomial will describe powers of the associated geometric shape.
    2
  4497. This one was nice
    2
  4498. Best
    2
  4499. That r x s = r + s reminds me of logarithms, is there a connection? ...could it perhaps be said that phi is the 'true' base of natural logarithm since it's just an 'identity' (of multiplication/addition) rather than a computational constant (e) ?
    2
  4500. Most memorable: the fact that Euler-Mascheroni constant is Ramanujan summation of harmonic series.
    2
  4501. To expand this proof to any odd prime look up the Dressler and Shult [DS75] paper on the Zolatev Forbenius theorem, basically you have to think of the number now not just as power of 2 or the prime generator, but as the canonical tuples you get from the Chinese remainder theorem inmersion (for example in reality n mod 15 are a 5x3 grid, where you do operations component wise, to get for example 6 in this representation you take the remainder with respect to 5 and with respect to 3 to get bot your cordinates) then you can use primitive generators in every one of the components as they exist for every prime p^k and you can see that the permutation decomposes in cycles where you can apply what was demonstrated here to every cycle. In essence every cycle is one component of the legendre symbols that form the jacobi symbol.
    2
  4502. Next challenge idea: p-adic numbers? quantum probabilities? Anyways: congrats!!!
    2
  4503. Not mentioned are shoes with an odd number of eyelets (like mine). 😳
    2
  4504. I thought Isaac Newton invented calculus???
    2
  4505. These days kids have velcro fasteners. When I was a kid, I laced all my running shoes. What I learned early on was that unusual lacings caused uneven wear. The worst was the zigzag form you showed at the beginning -- this always resulted in a worn out, then broken lace on the long diagonal lace. Today whenever I buy shoes they usually don't have criss cross lacings (no idea why) but I always re-lace them to criss-cross style so that I get the most even wear.
    2
  4506. This is definitely the random name you should pick <3
    2
  4507. 09 04 Path: R, R, R, L 17 13
    2
  4508.  @bscutajar  I think that's just a coincidence. If you follow the box construction for the numbers 0, 1, 1, 2 you get a Pythagorean triple 0²+2²=2², and if you try to follow the tree structure none of the three children are our 1, 1, 2, 3 box.
    2
  4509. The most interesting part to me was that there are no integers on the way to infinity for the harmonic series. Thanks for the video. It was excellent.
    2
  4510. I'm bald with a beard. I'm sure I have many hairy doppelgängers 😂
    2
  4511. Ptolemy's Theorem was taught in my basic geometry class, is it not taught anymore?
    2
  4512. I have always loved the unbridled beauty of mathematics and even studied it at university, but don’t get around to flexing that mental muscle much in my day to day. Your videos & clear passion for the field always make me fall in love again and I can’t thank you enough for that :) Merry (belated) Christmas!
    2
  4513. These videos always impress upon me how unimaginative and utterly unoriginal my mental workings are. Marvellous!
    2
  4514. 6:53 I am from australia and I do actually remember being taught this formula in high school. However it wasn't referred to as herons formula. But this definitely was taught where I went to school in perth, maybe its different for eastern states.
    2
  4515. 24:43 You connect the points to make triangles between the intervals, so sum of the ares of the triangle should be 1/2(1)(1).
    2
  4516. 25.8070 is the better rounding!
    2
  4517. Anyone notice the video was released on pi day (3/14)? Probably. Anyway, I missed the link to the permutation videos, is that one out yet?
    2
  4518. I don't understand what was the motivation behind performing such huge calculations in medival times.
    2
  4519. This was amazing!
    2
  4520. 9 4 13 17 16:00 I think it's 5. The area of the big square is the same as the two little squares directly above it, namely one. You can just continue it: the 4 littler squares' area is equal to the two little square's area, namely one. So step one you start with area 1 (one big square), step two is two little squares also an area of 1, step 3 is 4 squares also area of 1... Since there are 5 of such steps, each step having an area of 1; you get 5. 24:07 Just a guess here: The 3-4-5 triangle corresponds with the 1-1-2-3 sequence. It's parent triangle would correspond with 0-1-1-2 sequence, 'would' because that isn't a triangle anymore: the center point of the feuerbach circle lies on the same line as the inner circle, so you don't get a triangle but a line.
    2
  4521. Regarding the prompt at 24:40, gamma must be at least 0.5 because each blue region takes up slightly more than half of its corresponding bounding rectangle. This follows from the fact that ln(x) is convex-downwards.
    2
  4522. I made it to the very end! And now, it's bedtime... Fred
    2
  4523. I'm 166.5
    2
  4524. Legendre's name is in the Eifel Tower ... in very good company, BTW.
    2
  4525. So beautiful explanation of calculus problems 💛
    2
  4526. I looved that part about gamma and realizing that the partial sums of the harmonic series can be a finite amount close to the logarithm even just having countably many terms, but the integral to get the logarithm uncountably many. Thanks for your amazing work
    2
  4527. The fractal pattern is amazing, but it makes me wonder how does it relates to the reciprocals geometrically
    2
  4528. What according to you should students learn first, INFINITESIMAL CALCULUS or DIFFERENCE CALCULUS? and why?
    2
  4529. I loved the algebraic proof at the end the most. So much simpler than I could have possibly imagined. And as a runner up, I've never seen the worm on a rubber band paradox before. Very fascinating, but also intuitive if you imagine the worm along for the ride at the far end that is moving without accelerating.
    2
  4530. ok, but where is the explain of almost prime in the video description
    2
  4531. Well, at 13:37 , the simple rule seems to be twice the fraction (both numerator and denominator) plus the previous number (again, same side of the line) equals the next fraction. In context take the fraction 41/29: 41*2=82, +17=99, that's for the numerator. 29*2=58, +12=70, the denominator. That gives us the next fraction, 99/70. I have no idea why this works, but this is the relation in this series.
    2
  4532. Hi! I loved the lead up and visualization for "sum of cubes equals sum of integers squared". How interesting! I love how this motivated the general approach for formulas based on the binomial identities. After that I felt less smart but still quite interested. Thank you! (MS in computer science).
    2
  4533. ps ,I am a tiler, to get around non square rooms, you tile off the centre lines. one wall may be square, to another , and yet the other walls may not agree with your initial mark out. Working off centre lines, you fade out the difference towards the edges
    2
  4534. Great video.
    2
  4535. Hi Mathologer, I might have a solution to the red cross puzzle you described in this video. But before I post the solution, I need to clarify an uncertainty of the 2 small crosses. After cutting the larger red cross into 5 pieces and assemble them to make 2 smaller crosses, do the 2 smaller crosses have to be in the exact same size? Thanks.
    2
  4536. The generalization of the am-gm inequality is incorrect :(
    2
  4537. I'm not sure I'd consider 0.98 million "and change"
    2
  4538. Indra built the seamless cylinder, show me its beginning
    2
  4539. You are the best Math teacher. NOBODY does proofs better than you in my experience. The more you understand something the easier it is to explain to others, so, you really know your stuff. If you can make me understand, you are a true master. Yeah, I think you're better than 3b1b (No offense!) If you happen to read this I'd also like to state I've memorized most of your videos. Deep learning occurring here. I went to school for Math but you make it intuitive. The Pi formula video and the Summation masterclass are some of the best Math content ever produced by man.
    2
  4540. This video is a great journey 👍
    2
  4541. 1+x¹+x²+x³+…=(1-x)^(-1) Now take n-1 derivatives. Every term where the power of x is less than n-1 initially goes to zero. The rest of the terms give us factorialish coefficients. As for the right side of the equation, for each derivative, the Power rule multiplies the coefficient by successive negative integers and the Chain rule always multiplies the coefficient by -1, so we exactly build the coefficient to be (n-1)!. x⁰(n-1)! + x¹n!/1! + x²(n+1)!/2! + x³(n+2)!/3! +… = (n-1)!(1-x)^(-n) Clean both sides up by dividing through by (n-1)!. x⁰ + x¹n/1! + x²(n+1)n/2! + x³(n+2)(n+1)n/3! + … = (1-x)^(-n) Remember that the formula for b Choose k (bCk) is b!/(k!(b-k)!), so it gives us numbers of the form b(b-1)(b-2)...(b-(k+1))/(b-k)!. This is exactly the form our coefficients are in where k is always (n-1) and b is always (n + <power of corresponding x term> - 1), eg (n+5) for the x⁶ term. (n-1)C(n-1)x⁰ + nC(n-1)x + (n+1)C(n-1)x² + (n+2)C(n-1)x³ + … = 1/(1-x)ⁿ EDIT: Made a few mistakes in the bit where I was explaining how I understand the Choose function to work.
    2
  4542. Regarding the "application" question: I'm currently visiting a class on complex analysis (/calculus) and finding a "complex circle" description is essentially the alpha and the omega of what we're doing so far. Integrating and deriving off of parametrised curves in the complex plane is almost trivial, and as such being able to find a closed curve of any polynomial shape is really useful. It also becomes useful when trying to do calculations with rather complex shapes like e.g. the wing of an airplane, and being able to just transform it down into a simple shape (e.g. something with a circular cross section). Doing the maths on the simplified shape is usually the way people approach these calculations in computational mechanics, and as such it has a huge amount of "behind the scenes" real life application.
    2
  4543. If you define M x N as the cellwise multiplication of two images, M, N, then after applying the cross rule to seq (assuming no 0s) down = (center**2 - right*left) * 1/up to get an image M, then we'd have shiftUp(M) == (MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M)) M is a fixed point (i.e. f(M) = M) under the function f that takes an image M to f: M -> shiftDown((MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M)))
    2
  4544. Logarithmic Time self-defining relative-timing ratio-rates of reciprocation-recirculation Singularity-point nothing in Eternity-now Superspin Relativity of No-thing, the Conception of Existence/Everything. Merry Christmas, thank you for your fabulous teaching videos.
    2
  4545. First!
    2
  4546. e^i pi+1=0 is MUCH prettier than e^i pi=-1 !
    2
  4547. Great work as always. I hope you will show us the astonishing beauty of math for years
    2
  4548. Amazing as always. These always make my day. 😊
    2
  4549. If this doesn't deserve a like I don't know what else does. You just get intuition from this kind guy for free. His channel is amazing.
    2
  4550. Not down with putting money in Chuck Lorre's bank account with that t-shirt. He'll just use it to buy more Wayfair cabinets.
    2
  4551. In Brazil we learn about Heron's formula in high school, but it's extremely rare for it to be on college admission tests (Vestibular/ENEM). And I had never heard of Brahmagupta's formula.
    2
  4552. that 333 reminds me of the book "die stadt der träumenden bücher"
    2
  4553. The deal D works with number p q which share no common factors because in this case Z/p x Z/q is isomorphic to Z/pq
    2
  4554. Pythagoras would have said: I told you! It's divine, isn't it?
    2
  4555. My father was an Air Force cadet in WWII, and spent some of that time at Randolph Field, Texas. Classes were held at a nearby university, where there was a huge slide rule, about 12 feet long, run by hand cranks, capable of 5 or more digits of precision. And I once a double scale slipstich, so you didn't have to do the 'PacMan' thing.
    2
  4556. 26:07 The m-th falling power of x, from now denoted as x^(m_), equals x(x-1)(x-2)...(x-m). Δx^(m_) = x^(m+1_) - x^(m_) = x(x-1)...(x-m)(x-m-1) - x(x-1)...(x-m) = x(x-1)...(x-m)(x-m-2) = (x^2 - 2mx + m^2 - 2x + 2m)Δx^(m-1_) How do you turn that additional polynomial into m?
    2
  4557. Great! You go quite deep in this problem. I didn’t find the reduction argument as compelling as I would have liked and had to pause and think a fair bit.
    2
  4558. 28:10 one of those shapes is made of 6 squares
    2
  4559. Have not seen anything relevant (and nice) while surveying the literature, but who knows :)
    2
  4560. Homework @12:08 9 4 13 17 Right - Right - Right - Left
    2
  4561. Imagine if we could learn this as toddlers
    2
  4562. If we look at (1+1/2 x^a + 1/3 x^2a + … +1/(n+1) x^na) n-> ♾️ This series = -ln(1-x^a)/x^a: -1<x<1 So if we take lim (b-> 1) of ln(1-b^c)/b^c -ln(1-b)/b, computer estimation shows we get around ln(c), but I’m too lazy to verify.
    2
  4563. Does it work in 3D? I.e. can we transform irregular polyedrs into regular ones, adding tetraedrs as ears?
    2
  4564. GORGEOUS VIDEO 👏🏻👏🏻👏🏻👏🏻👏🏻
    2
  4565. You know I've always wanted to see a cone turned insideout ever since I was a kid. Impressive as always! I thought when I was a kid that the animation shown would have made some kind of other cone, but I guess its a sphere
    2
  4566. phi-phi^2=e^ipi
    2
  4567. … and your observation was where this all started with me, years ago. I thought it was just a curiosity until I found that it was much better than the usual “zigzag” trick when you want to start cutting up other objects, like the sphere.
    2
  4568. I always want more videos on just this topic. I do.
    2
  4569. 'I made it to the very end'...It is kind of an achievement for me..... xD Great stuff from Mathologer as usual... It's something really worth praising..... :)👏👏🙌🙌👌👍 Since I made it to the very end, I must say that I was most fascinated and amazed by your slightly modified proof of the Kempner's No 9 series converging to a value less than 80.....The other attractive thing in this video was how you intuitively showed by animation, the supergood approximation formula for the sum of the first 'n' terms of the Harmonic series with the addition of Gamma....It was also really clever of you to demonstrate how the Harmonic series pops up in calculating the overhang in the Leaning Tower of Lire via the centre of mass concept....All in all, I must say that this was simply a Marvellous video all math lovers should watch.....👏👏👌👍
    2
  4570. Legal pads are 14x8.5 in the usa (13.75/8.5 approximates phi) Letter pads are 11x8.5 (12/8.5 approximates root 2)
    2
  4571. one odd way of interpreting -(1/12) is that it's the last triangular number.
    2
  4572. My favourite Part was The Visualisation of Haromic Series approximation as Area under the curve + the Correction Term referred to as Gamma(Euler constant)...I thought It was just a correction term for making things work Fine but was surprised to came across that it also Features in many other Important Formulas...It made my day to learn that...❤..
    2
  4573. This is a really cool explanation of natural logarithms. I remember learning about regular logs and natural logs in algebra 2 but we never really learned what they were or how they came about so to learn this is very cool. Really tempted to get a refresh on logs now lol
    2
  4574. For a 33x33 magic square, the constant should be 17985 calculated with the formular (x^3+x)/2 = (x^2^2+x^2)/2/x what takes the sum of all numbers of the square and divides it by the number of columns or rows of the square
    2
  4575. if m,n both are odd => m+1,n+1both are event, and we will always have situation with both j/(m+1)=1/2 and k/(n+1)=1/2.
    2
  4576. 1+2+3 = 1*2*3 is nothing. 6+9+6*9 = 69 is a real miracle.
    2
  4577.  @Mathologer  hehe
    2
  4578. Nice 6 square staircase in the 5 square figures. Great Video anyways.
    2
  4579. Wiederum ein Genuss! While 3cubed + 4cubed + 5cubed = 6cubed, a consecutive pattern appears as follows: 2cubed + 3cubed + 4cubed + 5cubed = 6cubed + 8 7cubed + 8cubed + 9cubed + 10cubed + 11cubed = 12cubed + 13cubed -10 14cubed + 15cubed + 16cubed + 17cubed + 18cubed + 19cubed = 20cubed + 21cubed + 22cubed -90 23cubed + 24cubed + 25cubed + 26cubed + 27cubed + 28cubed + 29cubed = 30cubed + 31cubed + 32cubed + 33cubed -280 34cubed + 35cubed + 36cubed + 37cubed + 38cubed + 39cubed + 40cubed + 41cubed = 42cubed + 43cubed + 44cubed + 45cubed + 46cubed -640 etc. with differences -x/2(x+3)(xsquared+3x-8) for x = 1,2,... (i.e. +8, -10, -90, -280, -640, -1242, ...)
    2
  4580. Here's my proof (I did not watch beyond 2:22). Clearly the three line segments have the same length, being each composed of one segment of each colour. Then any two of them are interchanged by the reflection in the inner angle bisector of the vertex where they intersect. It is well known that the inner angle bisectors intersect at the center C of the inscribed circle of the triangle. Then the mentioned reflection implies two equalities of distance from C between a pair of end points of the reflected segments (one of them being the ends of a pair of same colour whiskers, the other the remaining pair). One gets that for each of the three inner bisectors, and by transitivity these equalities imply equal distance of each of the six end points to C. Viewing a bit more of the video, I can add that for the successive application of the three mentioned reflections in some order, there is in each case one of the segments S that is mapped successively to each of the other two segments and then back to S again, but with its end points interchanged. The combined action is the colour-reversing one in your "swiveling proof" that the midpoint of S is it point of tangency with the inscribed circle; you realised each swivel as a rotation, but a reflection does the same with less drama. Now the composition of any three reflections in lines concurrent in C is again a reflection with respect to some line passing through C (since the only isometries fixing C are rotations and such reflections, and the composition reverses orientation). Here that composition coincides with the reflection with respect to the perpendicular bisector of the segment S. The fact that this axis passes through C means that the midpoint of S (on this axis) is the point of tangency of S to the inscribed circle of the triangle. When composing reflections through three concurrent lines, the result is the reflection in the line obtained by rotation the axis of the first reflection by the angle turning from the axis of the second reflection to that of the third. Applying that general fact in this situation gives a relation between the angles formed, at C, between the angle bisectors and the perpendiculars to the sides, and interesting triangle fact. Finally the fact that any reflection is its own inverse menas that if we apply the three reflections and then all three again in the same order, the composition will be the identity. The "orbit" of any point of the plane under these successive reflections gives six points, clearly all at equal distances from C; that's your windscreen wiper theorem.
    2
  4581. these should be taught as base information in school, would give a much simpler mental map for maths.
    2
  4582. The rule to determine the partial fractions of sqrt 2 is: -start with 1/1 -the denominator of the nth partial fractions is the sum of numerator and denominator of the n-1th partial fractions - the numerator of the nth partial fraction is the sum of the denominator of the n-1th and n-2th partial fraction So, apply the first rule, and then the second rule (since it needs the denominator) 1/1 (1+1=2) => .../2 (2+1=3) => 3/2 (3+2=5) => .../5 (5+2=7) => 7/5 And so on What have I won? :D Edit: I saw that later in the video you gave out the answer (not exactly the same form but yeah)... My moment of glory has vanished :c
    2
  4583. yummy dish
    2
  4584. I kind of dislike calculus because it's more of engineering tool, rather than pure mathematics one. In other terms, - calculus is approximate, which would do for an engineering problem, but not for mathematical one.
    2
  4585. Plz make such video lectures for NUMBER THEORY & CALCULUS course 😭plz. ❤️always waiting for your videos love it
    2
  4586. I was looking at MSs a month ago. This threw new aspects into the mix. 🙏I like the music 🎶
    2
  4587. So... the row sum of 33x33 is 17985.
    2
  4588. I have find out the General Nth term of the series of 1,2,4,8,16,31,57,99,163,... that is a_n = 1/24 (n^4 - 6 n^3 + 23 n^2 - 18 n + 24) (for all terms given)...
    2
  4589. Вы заставляете меня краснеть от осознания собственной тупости... Но всё равно, крайне интересно!
    2
  4590. Happy Anniversary Mathologer! Thanks for you've done to educate us.
    2
  4591. From now on, we should call using log tables "rolling the logs".
    2
  4592. Q: What’s the only thing better than a Mathologer video? A: Another Mathologer video
    2
  4593. Harmonic series is really special
    2
  4594. Why is this man obsessed with satanism?
    2
  4595. Is that a buzz lightyear t-shirt? (to infinity and beyond)
    2
  4596. Really fascinating. I love mathematics.
    2
  4597. Great, as always!
    2
  4598. At 13:50 , does it depends on the number of disc ?
    2
  4599. I've been seeing your channel around for a while now, but without checking what it is about exactly. First time I've been watching one of your video, I instantly want to say thank you for the neatness and pace of this video. Very impressive and informative. The beauty of maths is very well transmitted. Best regards.
    2
  4600. Ramanujan is impressive....but you are not less..... at list for us normal people... lets say that you are ramanujan to us as ramanujan is to you! ..... i hope it doesen't sound offensive...haha
    2
  4601. Feedback: Probably tell your editor to make your voice louder. I have to put my laptop volume to maximum to hear you.
    2
  4602. At 10:30 let us note a corollary: 2*q is the harmonic mean of r and s. :cat-orange-whistling:
    2
  4603. My vote goes to gamma: an important, unusual and intriguing constant I didn't know existed.
    2
  4604. 4k+3 = 4k'-1, So 4k+1 and 4k-1 would be neater.
    2
  4605. Super nice video as always. I really like the note at 28:10. A few years ago, my brain completely exploded when I realized continuous interpolation of Riemann zeta function partial sums, with this method, when integrated between -1 and 0, gives you the actual value of the zeta function. (Consequence of your 1+2+3 video ^^)
    2
  4606. Did you overdub yourself several times in this? I don't recall you doing that as much before. Made me think my audio was lagging at first :D
    2
  4607. Leonard Euler comes everywhere in mathematics . What a genius 🙏🙏
    2
  4608. Ratio multipliers on which Newtonian mechanics industrial revolution based gears runs the world not change energy conservation law so always equal to unit one or 100%. Physics have philosophical or metaphysics layer of understanding - multiplication always equal to unit one. Pure mathematics and engineering is more utilitarian.
    2
  4609. 9:40 + C ;)
    2
  4610. Im always fascinated by your discussion of proofs!
    2
  4611. There's so much more number theory here! When you consider specific paths through that tree of triples, you're dancing around the solution to Pell's equation that is one of the pillars of Gauss's majestic solution of the quadratic Diophantine equation in _Disquisitiones Arithmeticæ_. Which I'm sure you're aware of - it would be great to see you dive deeper.
    2
  4612. That imbalanced amount of positive and negative terms (meaning that the entire sum doesn't actually converge to zero) reminds me of another somewhat similar (seemingly) paradoxical result: Start adding consecutive natural numbers, starting with 2, to a container. However, every time you add a square number, remove its corresponding square root from the container. The paradoxical result is that even though the number of elements in the container always grows or, sometimes, stays the same, if you keep doing this infinitely many times the end result is an empty container (because all of the numbers will have eventually been removed). Or is it? Because here, too, there's an imbalance in the amount of added and removed elements. There are way more added elements than removed ones. Does the result still work?
    2
  4613. F.J.M. Barning --> Fredericus Johannes Maria Barning
    2
  4614. Touching dedication to Ron Graham at the end of the video ❤️
    2
  4615. Had a similar topic in my book recently. Thanks for touching on the same more in depth
    2
  4616. Just wait till Marvel introduce Mr. Fantastic! I can't wait the amount of Mathologer video that can be made with Reed's equations <33
    2
  4617. Slap a ham on Omaha, pals.
    2
  4618. The most interesting part was surely the different sums that can be made by the absence of a certain digit or a group of specific digits. It is so counterintuitive that it makes it more beautiful in a way because the beauty isn’t surface level it is more deeply rooted inside of the sums!
    2
  4619. I wasn't going to watch this video because I already know what the pigeonhole principle is, until I saw it's by mathologer ;)
    2
  4620.  @Mathologer  I was not! Great video, once again!
    2
  4621. This video is such a gift ! Are there also formulas for the other Platonic solids ?
    2
  4622. So many such manuscripts are lost to time, invasions, colonization, partition etc. I wonder how much the world would have benefitted from it if all these manuscripts are available today. Or are they?
    2
  4623. i could not get the glass challenge . Can you help? Any hint would also do
    2
  4624. Really neat! Another little geometry gem discovered in "modern" times, that seems as though it should have been discovered in Euclid's time, is this. In any triangle, draw all 3 pairs of angle trisectors, extending each until it intersects another from a neighboring angle. The 3 points of intersection are always the vertices of an equilateral triangle! The surmise is that it didn't appear in ancient times because angle trisection isn't possible with the classical straightedge-and-compass technique, so no one back then even considered it. This fact was mentioned in one of the early Mathematical Games columns by Martin Gardner in Scientific American. Fred
    2
  4625. if you consider x = 3 (floor of π) u = ceil(1/(π-x)) = 7 x += 1/u (x = 22/7) u = ceil(1/(π-x)) = 113 x+= 1/u (x=553/113) and so on
    2
  4626. High School Maths teacher here, so glad I happened in this video! Researching how to bring life into functions for my Precalculus class. I think I found it! These connections will live through students and maybe future mathematicians.
    2
  4627. I was just playing around with the no 9s problem and saw a proof to show that the sum is less that 30. Here goes .. Let's define S8= 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8 Then for terms from 10 to 19 there are 9 numbers with no 9s 9(1/10)>1/10+1/11+1/12+1/13+...1/18 9(1/20)>1/20+1/21+1/22+1/23+...1/28 9(1/30)>1/30+1/31+1/32+1/33+...1/38 ................... 9(1/80)>1/80+1/81+1/82+1/83+...1/88 So sum of 1/ 10 till 1/88 is less than (9/10+9/20+9/30+...9/80) or 9/10(S8) Then for 100 to 200 there are 81 terms with no 9s 81(1/100)>1/100+1/101+1/102+1/103+..1/188 81(1/200)>1/200+1/201+1/202+1/203+...1/288 81(1/300)>1/300+1/301+1/302+1/303+...1/388 ................... 81(1/800)>1/800+1/801+1/802+1/803+...1/888 Then sum of numbers from 1/100 to 1/888 is less than (81/100+81/200+81/300....81/800) or 81/100(S8) There are 729 numbers from 1000 to 1888 with no 9s So the sum from 1/1000 to 1/8888 is (729/1000+729/2000+729/3000.. +729/8000) or 729/1000(S8) This pattern continues forever Therefore the sum is less than S8 +9/10(S8)+(9/10)^2(S8)+(9/10)^3(S8)..... Taking S8 common Sum is less than S8(1+9/10+(9/10)^2+(9/10)^3.....) (1+9/10+(9/10)^2+(9/10)^3.....) is 1/(1-9/10) =10 so the sum is less than S8(10) = 27.1785 Sum of no 9s<27.1785
    2
  4628. 24:35 "But can you also see at a glance that gamma has to be greater than 0.5? If you can, tell us why this is obvious." Draw a horizontal line between any two consecutive blue regions and a diagonal through every single resulting rectangle. All upper right rectangle halves are completely blue and all lower left rectangle halves are partially blue. Hence, gamma must be strictly greater than 0.5.
    2
  4629. I actually have been trying to do something similar with expanding viesca pisces groups (nothing crazy I swear). All formed by bubbles that are expanding. It all was going off the discovery of a boundary layer in super conductors. Whith that relating to black holes and atoms... maybe.
    2
  4630.  @PC_Simo  You have (2n-1)^n / (2n)^n -> e^(-1/2)=0.606… and (2n+1)^n / (2n)^n -> e^(1/2)=1.648… It follows that (2n-1)^n+(2n)^n and (2n+1)^n in the limit differ only by a factor of 1.648…/1.606…=1.026…
    2
  4631. Aha !!! Short but rich .
    2
  4632. Wonderful video highlighting the genius of Ramanujan and the power of continued fractions.
    2
  4633. What happens if n is a fractional or negative number?
    2
  4634. Can you discuss Riemann Hypothesis?
    2
  4635. I'm almost done with my own implementation, so don't start drawing a winner just yet! :)
    2
  4636. But did you know that 3+4 is also 7
    2
  4637. Thanks!
    2
  4638. Amazing shirt!!
    2
  4639. Is there such a thing as a magic cube?
    2
  4640. Aw man, poor number 2 has always been the black sheep in the prime family, and as a result it gets frequently shunned by prime fanatics. Don't worry number 2, some day people will accept you for who you are.
    2
  4641. Proof is not that hard. First look at the triangle with red and blue on two sides and the one with only green sides, those two triangles are similar isocoles triangles which implies the two points at the bases of each of those triangles lie on the same circle. Now that you proved 4 of the points to be on the same circle, you can do the same with 2 other sets of 4 points. I left out a few easy details that I would have included if I could draw on a comment! The second proof to show that circle is the original triangles incircle is even easier. See that all the chords drawn are the same length (red + blue + green) which implies they are all the same length from the center, which implies there is an incircle tangent to all of them! Color proof was my favorite and much more elegant than mine!
    2
  4642. What's been killing me is the question, "If 2^n is the corresponding function to e^x, what significance do the corresponding cos and sin functions have?" (Sum alternating signs of alternating terms of the exponential series). These functions even satisfy the sequence version of the wave equation.
    2
  4643. Great video, Love it. I actually found out about those theormes recently when I was trying to solve an olypmpiad problem about sums of two squares. It was pretty difficult problem and I'm really proud I could solve it. Merry Christmas 🎄🎄🎄
    2
  4644. Mind=blown, thank you !
    2
  4645. clockwise for odd n, counterclockwise for even n
    2
  4646. 6:03 Can someone help me on this first challenge? I thought, and read, that the answer is no, because you can take 2 tiles of the same color in order to cut a single corner tile, which results in an impossible to-tile board. However, this feels like a bit of a cheat. Except for those cases, I think that all of the resulting boards can always be tiled with dominoes. Is that so? Why?
    2
  4647. Me, too, either.
    2
  4648.  @Mathologer  Even the seal swimming by approves!
    2
  4649. Loved the video
    2
  4650. First!😊
    2
  4651. 11³+12³+13³+14³=(2.10)³=20³ is probably my favorite Also: 3³+4³+5³=6³ and 2⁴+2⁴+3⁴+4⁴+4⁴=5⁴
    2
  4652. Now he cooks some math.
    2
  4653. The average of the furthest apart wavelengths of purple light and red light falls on green. But I doubt this is what you had in mind.
    2
  4654. Thank you for another great video! I was wondering if you would ever do a video about the Collatz conjecture and how such a problem could ever be proven in theory.
    2
  4655.  @rasputozen  the thing is that by thinking abstractly about very small differences you can derive the table of derivatives shown in the video without having to actually deal with any numbers any practical use of derivatives just applies the rules, like (x^n)' = nx^(n-1) the whole df/dx is only used to showing where those rules come from
    2
  4656.  @rasputozen  for example if you want to find the slope of f(x) = x² at x = 3, you then find the derivative f'(x) = 2x, and then plug in 3 to get 2*3 = 6 if you use a computer you can approximate this result by choosing a small difference. for example you could say ∆x = 0.001, then ∆f = f(3 + ∆x) - f(3) = 0.006001. so the slope is approximately ∆f/∆x = 0.006001 / 0.001= 6.001 but the beautiful part about calculus is that you don't need to approximate the result, because the rules that we derived allow finding the exact value using just some transformations
    2
  4657. Very cool, been a while since I’ve broken out let alone used my old slide ruler. Also I want that shirt.
    2
  4658. True
    2
  4659. Kal ioqm hai
    2
  4660. And also Cubes???? 🧐 3³ + 4³ + 5³ = 6³ ✅️ 3³ + 4³ ... + 22³ = 40³ ✅️ 6³ +7³ ... + 30³ = 60³ ✅️ 6³ + 7³... + 69³ = 180³ ✅️ 11³ + 12³ + 13³ + 14³ = 20³ ✅️ 11³ + 12³ ... + 109³ = 330³ ✅️ 15³ + 16³ ... + 34³ = 70³ ✅️ 34³ + 35³ ... + 158³ = 540³ ✅️
    2
  4661. 2nd
    2
  4662. Awesome video that gave me insights in things that I thought I new :) So "normal" and hyperbolic functions are like dual entities, and both are real because relativity is governed by a hyperbolic version. What interests me, is there more of them? Even further, is there a generalization? 🤔
    2
  4663. And you can do the same with (x+1)^(n+1) for the n-d simplex. In the case of the simplex, the dimensions are shifted by one. For example, for a triangle: (x+1)^3 = 1x^3 + 3x^2 + 3x^1 + 1 which is 1 face, 3 edges, 3 vertices, and 1 extra "-1" dimensional object! You can also neatly arrange this on Pascal's Triangle.
    2
  4664. Love to see the mathematica
    2
  4665. puzzle at 11:50 : One solution is having the first brick's centre of mass right at the edge of the cliff and then 2 more bricks side by side on top of that first brick, so that the centres of mass of these are at the edges of the first brick. This achieves an overhang of exactly one brick through use of a counterweight.
    2
  4666. One thing about this "pi stuff" that is NEVER mentioned is that the "look" is always done using DECIMAL notation of the quantities. Unlike the value of pi, the radix of the study being 10 has nothing special about it, yet this is what everyone uses. What about 2? 7? 16? or 113? If you did this study for binary, you would need to use a power of 2 number of terms (such as 2^20 which is close to 1 million) & you should see mismatch at the 20th digit.
    2
  4667. A magic square square.
    2
  4668. 15:24 well, it does have the turning property, just not at 90 degrees. If you stay on the plane of that hexagon and move any of the segments 60 degrees, you will find another grid point.
    2
  4669. I haven't seen a comment on this yet but the intro with the "o" lacing connecting with the word Mathologer right at the beginning of the video put a smile on my face.
    2
  4670. 29:22 The Harmonic series Ramanujan-summed is γ? But why, then, does the analytic continuation of the Riemann Zeta function have a pole at 1?
    2
  4671. When you try to like the video after every proof but you can only like once 😔
    2
  4672. Love the shirt, but "coswine" with two pigs would have been even better.
    2
  4673. Im finally finding something to dive my curiosity into
    2
  4674. I find the algebraic answer doesn't provide a good intuitive answer for me, so here's a visual one that I like: We can picture n^2 as n objects arranged in a square (example 3: * * * * * * * * * ) (n-1)(n+1) is then a rectangle (example for n=3 = 2 x 4 rectangle: * * * * * * * * ) Since that rectangle has one more column than the square but one less row, we can rotate the column and make it an additional row to get closer to the shape of the square: * * * * * * * * But since we already know the height of the column is n-1, when we rotate it, we get a row that is n-1 long. So doing this transform is always one object short , so (n-1)(n+1) is 1 less than n^2.
    2
  4675. Whats also cool is thinking about this in the imaginary number plane with n equal 1, you have two vectors pointing 90 degrees apart which get reduced to a single vector pointing 45 degrees away from either beginning vector. Whose length is the square of the original vector length.
    2
  4676. Awesome I don't know calculus but I somehow still predicted that we would have to start at zero right from the beginning
    2
  4677. PUZZLES! 14:57 How many 1-S-S isosceles triangles are in the diagram? 7 * something, that's for sure I'm gonna count the ones where the tip of the triangle is at the top most corner. There are 9 of those. 7 * 9 = 63 Or alternatively you can count by base. On a 1 line, there's only 1 triangle. On an R line, there are 3 triangles positioned like a W. On an S line, there are 5 triangles positioned again in a zigzag pattern. There are 7 of each of those lines, so there are 7 * (1 + 3 + 5) = 63 triangles. 30:12 Show that -1/phi is a solution to x^2 = x + 1 just like phi itself. -1/phi + 1 = 1 - phi^-1 = (1/phi^2) * (phi^2 - phi) = (1/phi^2) * (phi + 1 - phi) = (1/phi^2) * 1 = (1/phi)^2 * (-1)^2 = (-1/phi)^2 31:51 What happens when you try other indices of fibonacci numbers? Negative numbers are easy. You can just as easily extend the fibonacci numbers backwards as forwards. 2nd: 1 1st: 1 0th: 0 -1st: 1 -2nd: -1 -3rd: 2 -4th: -3 -5th: 5 It seems to be the case that the magnitude is the same but the sign alternates. Let's see: F(-n) = (phi^-n - (-1/phi)^-n) / (phi - (-1/phi)) = ((1/phi)^n - (-phi)^n) / (phi - (-1/phi)) = (-1)^n * -F(n) For non-integers, you have a (-1/phi)^n term in there. You're gonna get a complex number. As n gets larger the complex part disappears for the most part but generally it's... a bit weird 41:20 What happens when you do the R maneuver on a 1x1x1, then an S maneuver on that? Algebraically, the R maneuver does this: As + Br + C => (A+B)s + (A+C)r + B Notice that the R term becomes the constant in the term below. In the pascalian grid, the same applies: the R term in one tile is the same as the constant term in the next one below in the R direction. So there are only TWO pascalian grids, not three. Sorry I got sidetracked let's apply the R maneuver, then the S maneuver 1, 1, 1 => 2, 2, 1 => 5, 4, 2 Because RS = SR, we should get the same result if we do the S maneuver first, then the R maneuver 1, 1, 1 => 3, 2, 1 => 5, 4, 2 43:43 PF: Everything we've been doing with all these maneuvers is basically one big linear transformation. The matrices correspond exactly to the reverse R and S maneuvers earlier. Their inverses should be the forwards R and S maneuvers, then. The basis is (s,r,1) btw Then it follows quite obviously that since we modeled these matrices exactly from the properties of the numbers themselves, then the matrix multiplication should obey those exact same rules. The bottom row corresponds to what would happen if we multiplied this component matrix to the column vector (0,0,1), interpreted as a constant 1. 44:51 The chase is on for finding rho and sigma in nature! Idk, look for anything heptagon-shaped. Pretty rare in nature because it's almost always 2's, 3's, or 5's Obviously it's not gonna show up with something as simple as ruler and compass, you can't construct a regular heptagon using ruler and compass 47:14 What's next? depends on how I'm feeling, really. feeling with the theme of the video? 14. It's a fibonacci sequence. feeling like one of those crazies you mentioned at the beginning? 2, it's the next digit of pi. close enough feeling overly rigorous? literally anything, the operation isn't given to us. it could be the square root of R for all that is holy
    2
  4678. Please start the next video with "Good news everyone!"
    2
  4679. Most memorable: Tristan's visualization where the grey squares fade to white
    2
  4680. 3Blue1Brown did a good video about the circle partition problem too
    2
  4681. Volume of a sphere ÷ Area of a sphere = 1/3 Radius
    2
  4682. Excellent graphics presentation.
    2
  4683. I effectively have the math education of a 12 year old. I cant understand even the most basic algebraic well anything. These videos entertain me. Please never stop making them.
    2
  4684. It always amazed me that a few fundamental concepts like pi, and e, ln(x) can be used to describe our universe.
    2
  4685. I love the teeshirt. To avoid spoling it for anybody else, it is RPSLS, isn't it?
    2
  4686. The gurukul system and education in mother tongue must become come back.
    2
  4687. A nice fact from combinatorics or field theory is that the middle binomial coefficients can only all be divisible by n when n is prime. So for 2 and 3, we get powers of 2 and 3 +1 as special numbers, for 4 colors there should be no special numbers (except 2), but then for 5 each power of 5 +1 is special, and so on only for every prime
    2
  4688.  @Mathologer  oh, "Victorian" as in in state Victoria for 40 seconds I believed you were bashing education cirriculum coming from era of queen Victoria xD
    2
  4689. A pleasing combination of several of my favourite themes, @mathologer: puzzles, Dudeney, Diophantine equations, and Ramanujan. The puzzle from the Strand magazine is also puzzle 168 "The House Number" in the book 536 Curious Problems and Puzzles by Dudeney, edited by Martin Gardner (another favourite topic). Dudeney was arguably THE greatest puzzle maker; several of his puzzles have solutions involving Pell's equation. Thanks for the URL of the Strand archive. I hope you can find more material for future videos. E.g. Would love to see something on geometric dissections (Dudeney is famous for these).
    2
  4690. First two challenges I saw solutions to, but the third challenge seems to be 1, 2, 3, 5, 8 so I'm guessing the fibbanoci sequence minus the first 1(or if you think about it, it's like 0! when there's no chessboard the only way to tile it is with nothing.) Edit: for the final puzzle, we can see that on the smallest hexagon board, we need exactly one of each piece to tile it. Expanding the board like with the previous dance leaves us with three smaller boards uncovered in the larger regular hexagon. all of which need exactly one of each piece to be filled. While you may be able to swap pieces, they'll all still be needed.
    2
  4691. Take the face of Jesse Douglass and Bernhard Neumann and morph them together... and you come up with Karel Petr... (Ok Poetic licence, but Take the glasses and moustache and mix and match the rest and your getting fairly close)... 6:50
    2
  4692. 'Six Eyes is another innate attribute inherited within the Gojo clan. It grants the bearer extraordinary perception, allowing them to see the flow of cursed energy. The constant influx of information can be overwhelming, leaving the bearer overstimulated and tired. Six Eyes is an extremely rare trait.'
    2
  4693. Thanks for mentioning my Quora-post at 19:46
    2
  4694. Isn't it the proof of Fermat's theorem?:)
    2
  4695. 15:55 The total area of the tree is 5. Each smaller square has half the area of the previous square; giving the formula: 1 + (2 * 1/2) + (4 * 1/4) + (8 * 1/8) + (16 * 1/16) = 1 + 1 + 1 + 1 + 1 = 5 * 1 = 5. 🟩🌳
    2
  4696. Infinite math is the first subject where you can go really wrong if you’re not careful. In arithmetic and basic algebra, you learn what you can do. Avoiding dividing by 0 is the first hint of this world, but it’s momentary. But calculus is all about avoiding the paradoxes of infinity. It’s the equivalent of moving from a steep hike to mountain scrambling. Then when you get to subjects like algebraic geometry and Lie algebras, the abstractions become so complicated they can become incomprehensible to those not experienced in the field.
    2
  4697. I actually understood what your T-Shirt says ... That the game we play rock, papers and scissors can be just treated as a Triangle and similarily your T-Shirt is about a game which which is represented by a pentagon. Thereby we can make any game with N (N is odd) tools similarily like rock, papers and scissors and the rules will be that any tools can destroy (N-1)/2 other tolls and can be destroyed by other (N-1)/2 tools.
    2
  4698. Fibonacci? more like Phibonacci!
    2
  4699. Base 10 may seem arbitrary to the pure mathematician but what if base 10 is fundamental to the nature of reality as experienced by us? The numbers 2 and 5 do seem to be very crucial for experiential existence. 2 is the number of 'Duality', and 5 is the number of platonic solids/elements/senses/fingers, etc… Using a base 10 system (combination of 2 and 5) makes counting such experiences (and their multiples) efficient. Base 12, 30, or 60 would also bring out efficiency in counting certain aspects of reality! The choice depends on what we want to model.
    2
  4700. Simply fantastic
    2
  4701. Because with the divided estate being 90% of what's owed, then only a relatively small loss is suffered by each creditor. The one who is owes $1 loses only 50 cents - not a huge tragedy, while the one who is owed $1M suffers a loss of almost $100K. As he said in the video, if the amount is small, then the loans are seen as a loss, but with a little bit of compensation; while if the amount is large, the loans are mostly good, with only a small loss. And the closer the amount is to exactly half, the closer to exactly proportional the division is. Brilliant, IMO!
    2
  4702. I noticed something interesting (try it in one of the linked apps). If you select a high number of vertices on the outer circle, the dots form a shadow image of a polygon in the centre. A {13/12} star has a shadow point rotating in the centre. A {13/11} has a shadow line segment, a {13/10} a shadow triangle, a {13/9} a square, etc. The points tend toward a formation of a polygon that has the number of sides that is the difference between the two numbers.
    2
  4703. Eighth
    2
  4704. "very deadly times" — 666 likes 😱
    2
  4705. At around 16:55 the orange and green look very similar to me.
    2
  4706.  @Mathologer  I think you mistyped those values. r and s are greater than 1. cos(pi/7) and cos(pi/7)^2-2 are less than 1. I find r = 2*cos(pi/7) And the r^2 = s+1 formula directly gives s= r^2 - 1.
    2
  4707. Well, the numbers used in the grid are the numbers 1 to 9, so the sum of the numbers in the grid must be 45. Yet you should be able to get the same total by adding the sum of each row. So, 45 total ÷ 3 rows = 15 in each row (Magic Constant) If you apply this reasoning in general for an N by N magic square [from 1 to N²] we get the following formula: Magic Constant = ½ N (N² + 1) Hope that helps answer your question! Stay curious! :)
    2
  4708. Is this related at all to the fact that Fibonacci numbers can be found by taking powers of the matrix [ 1, 1 ; 1, 0 ]? We have [ 1, 1 ; 1, 0 ]^n = [ F_(n+1), F_n ; F_n, F_(n-1) ] which is not quite the same as the matrix that leads to the Pythagorean triples here but maybe there's some connection?
    2
  4709. Still have my linear and circular slide rules somewhere. Yes, the linear was more accurate, but the circular one fitted essily into my pocket. I also have a helical slide rule that I had for extra accuracy.
    2
  4710. i remember my linear algebra professor back in the days didn't like that it was called "the pidgeon hole principle" because he said that two pidgeons in the same hole would always fight and one would leave.
    2
  4711. You guys should make a video on Euler characteristic of higher dimensions and then relay those topological spaces into algebraic structures like modules or rings.
    2
  4712.  @Mathologer  This isn’t fulled fleshed out since I haven’t grasped homological algebra yet. This is more precise but incomplete. There is a Fourier series of simple polytopes. d-dimensional simplexes are equivalent simple polytopes. See equation 3.34 for Fourier transform of general d-dimensional simplex, section 8.2 for simplexes and simple polytopes, and section 8.5 which covers the Fourier transform for a simple polytope by way of Brion’s formula at “A friendly introduction to Fourier analysis on polytopes” (2023) by Robins. The geometric realization of a simplicial set is a CW complex. This means the CW decomposition of a topological space corresponds to a simplicial set. A simplicial set is a presheaf on the simplex category. n-Simplexes are homeomorphic to n-balls. n-balls are homeomorphic to R^n Euclidean space. Maps from R^n -> X to a topological space X are that of smooth manifolds when they are local homeomorphisms. This suggests using a more general definition like sheaves since they generalize the smooth structure of manifolds. Using homological arguments which I still need to learn, a (derived) sheaf should determine a homology theory. I would like replace the simplicial set of the equivalent CW complex with its presheaf. Then, I think I can argue with more work that the the sheaf of topological space X and its corresponding homology theory H*(X) is equivalent to one given by the sheaf of its CW approximation (= simplicial set) Y and its homology theory H*(Y). Next would be to calculate the homology of its CW approximation Y using the Fourier series implicit in the Fourier transform of its simplicial complex in each dimension as its nth homology.
    2
  4713. cool trick
    2
  4714. Notch is a patron? 29:44
    2
  4715. Because I'm from Italy I most vote for the "Leaning tower of lire" :)
    2
  4716. The picture of the magic square with Chinese characters is erroneous -- one and nine are swapped, and three and seven are swapped -- the result is not a magic square.
    2
  4717. Hey mathologer ,fan from india. Just wondering the amount of work that ramanujan did on infinite sums is astonishing. Can you tell me about any other mathematician who did exhaustive work on infinity?
    2
  4718. @ I’ve learned so much from you. About 3^2 years. Always waiting for more! Gesundheit!
    2
  4719. Love the higher dimensional and geometry based videos!! Very inspiring and helpful!
    2
  4720. Has nobody yet mentioned that a common engineer slang name for slide rules was a "slipstick"? I went through high school just as electronic pocket calculators were becoming affordable. I started in high school's electronics course learning how to use a slide rule, and finished the series with a pocket calculator. I still have both, but the slide rule is the only one that still works ;)
    2
  4721. The "Crushmetric" water bottle look just like that!
    2
  4722. I did your euler machine challenge. Heres the C++ code: https://github.com/Bynar0b11001001/mathologer_challenges I wanted to use String Arithmetics instead of something like bigint, but the functions for the string arithmetics turned out very messy idk if by 666th u mean partition(665) or partition(666) so here are both 665: 11,393,868,451,739,000,294,452,939 666: 11,956,824,258,286,445,517,629,485
    2
  4723. Can we build a dome with this? 8:25 what would its mathematical shape be? No books please.
    2
  4724. Most memorable part: is that as a middle school boy I was obsessed with infinite sums, some of which are in this video, I never knew how important they were.
    2
  4725. Don’t show all the ancient Indian Inventors as all Brahmins 😂
    2
  4726. i just don´t quite get how this represents all tillings, since random tilings do not come up by this method of growing the shape etc., but by simply randomly selecting spaces of the tiles.
    2
  4727. That's a beautiful proof.
    2
  4728. You Sir, in certain way, are a great influence for me. Thanks a Lot!
    2
  4729. I love that "Fresh Fallen Snow" song used throughout the video, I hear it on various channels.
    2
  4730. Another beautiful video. These geometric proofs are really something for the soul. Thank you.
    2
  4731. You end up with all cards facing up eventually because at least 50% of your action always decreases membership to the same group.
    2
  4732. Also! So early. that number of views is Two-digits
    2
  4733. This is what I love about your videos. I listened to the problem in the film clip and could barely follow the description. You talk for a few seconds and I understand what needed to be solved. You’ve covered so many things that my teachers used to talk about and I could barely follow them but you always explain it so simply and accessibly. Thank you :)
    2
  4734. Prime Three 1x1=, 1+1=2, 2+1=3, 5=1x2=2, 1+2=3+2=5, 7 1x3=3, 1+3=4, 4+3=7 now this will also create 19 as 3x4+3+4=7+12=19, how to get to.... 11 is created from 2x3=6, 2+3=5, 5+6=11, as well 1x5=5, 1+5=6, So 23 and 15 or 32 and 51 create 11 27 and 35 will crteate 23, as 7x2=14+7+2=9+14=23, 29 is created from 14 as 14 then creates 54 then creates 20+9 as I multiply and add you can also subtract... Crazy numeric messages and the link to the geometic dots and lines. Where are the lines , HOW to do..Quadrilium paper I have 5000 videos some with sanity
    2
  4735. I hope the inventors of Bootstrap coding was inspired by shoelace maths.
    2
  4736. My favorite parts of the video were the parts where you showed how you can use basic integration to figure out how to approximate the sum of the series and the part where you showed the visualization of the no-9s variant. I like how your videos explain the topics quite well even for people who haven't fully learned everything leading up to it.
    2
  4737. Sklyanin's paper is: Some algebraic structures associated with the Yang-Baxter equation. Functional.Anal.i Prilozhen 1982 vol 16, issue 4,pp 27-34 ; look at equation (27). Furthermore, in L.O. Chekhov etc al., Painleve Monodromy Manifolds, Decorated Character Varieties and Cluster Algebras , IMRN vol 2017, pp 7639--7691 you can find in Table 1 a close variant of the 3variable equation with extra parameters. BTW it seems we met in 2000 in Adelaide.
    2
  4738. Wunderbar Unterhaltsam und Lehrreich
    2
  4739. I have a minor in mathematics and was able to follow till the end
    2
  4740. n^2(n^2+1)/2 is the sum of whole square divide by n and get n(n^2+1)/2 sum of a line 33x33 -> 33*(33^2+1)/2 = 33*545 = 17985 value of the line
    2
  4741. (33^2 + 1)*33 / 2 = 17,985, since there are 33 rows (divide by 33) and you are summing from 1 to 33^2
    2
  4742. These videos are so interesting that I use them in my math essays' citations. Great job!
    2
  4743. What about higher dimensional surfaces? You showed that for the length of a line (dim=1) you need 1 parameter (number of points). For a 2D surface you have 2 parameters (b and p). What about a n dimensional surface? Do you need n parameters? Is there a set of it that converges to the true surface area?
    2
  4744. When he says 'wait, there's more', he s so cute
    2
  4745. FYI The relevant subpage linked in the description links to the wrong subpage.
    2
  4746. Amazing video! Keep up the good work!
    2
  4747. 2:47 THAT is such a true elementar point to the moment we are living in right now. The tools are simply exceeding exponentially our capabilities to design their (the tools) own structure. Like a production vacuum that pulls all the elements needed (imagination, creativity, actually material things, their arrangement, their singularity), and then more is produced over the previous things. Life momentum.
    2
  4748. So you have found the relationship between √2 and phi! I have been thinking recently that those numbers are special in geometry, after studying about polyhedra.
    2
  4749. Outstanding animation, doctor! 👍👍👍👍 , yeah, same Timelord math trick indeed 🤓🖖🤓🖖
    2
  4750. I once heard about these great Malyalam mathematicians in medieval India in one of Dr. H.C Verma's lecture on mathematics, at that time i didn't realise that there impact on modern mathematics was so profound.
    2
  4751. In the complex plane, no less!
    2
  4752.  @NatePrawdzik  :)
    2
  4753. I had so much trouble understanding a2 +b2 =c2 when i was young. this proof woulda made it so much simple
    2
  4754. Convergence of Kempner series was just so mind-boggling to me at first... I have felt for the trap and assumed that by just getting rid of 9s we couldn't possibly make harmonic series converge.
    2
  4755.  @Mathologer  i forgot that we're in Diophantine equation s world where the closest you could hope to get is 1 , anyways excellent video as always , by excellent i mean getting frustrated trying to solve the challenges :)
    2
  4756. I feel there is some analysis which considers the total surface of the claw "fingers" as being composed of the interior of the sphere and exterior of the sphere. During the "lay down" transformation this becomes the top and bottom of the disk. The interior/exterior ratio isn't 50:50 for finite fingers but top/bottom is. The transformation being a smooth process should have all intermediate ratios. If the area of the fingers can be shown to be double the area in the limit that eliminates the issue of which region of each finger counts keeps changing as the contact contours morph during the laydown as the total is invariant.
    2
  4757. One of my favourite identities is E = e^D (or equivalently ln(E) = D) which unifies the E, Δ and D operators. E = 1+Δ (eg Ef(x) = f(x+1) j and Δ is the forward difference operator and D is the differential operator, ie dy/dx. It also highlights how operators can be separated from the functions they apply to and manipulated like variables (although care is required because the communicative law does not usually apply). Solutions to linear differential and difference equations can often be solved by separation of operators. I have a wonderful book published in the late 1800s which makes an art form of manipulating such operators, including also the sum operator (Σ = Δ^-1 = 1/Δ). BTW There is also a backward difference operator equivalent of the forward difference (an upside down Δ).
    2
  4758. Back in 2016 when I saw Beck present the proof on Numberphile, I looked it up - I remember reading it in 1990 when it came out. Since I was teaching a number theory course at the time, I considered inlcuding it in my course, and so I studied it. Zagier says that it was not constructive , but I discovered that it is easy to use Zagier's involutions to give a constructive proof. Of course, I should say I rediscovered it for the ?th time. Letting S represent the switching involution and Z represent Zagier's windmill involution, take the fixed point of Z, namely x=1, y=1, z = k=(p-1)/4 as the start, compute SZ((1,1,k)), (SZ)^2((1,1,k)), etc. That is, the orbit of the solution (1,1,k) under the mapping SZ. Then RIGHT IN THE MIDDLE OF THAT ORBIT is found a fixed point of the switch involution S, that is, the sum of squares solution. Try it!! Also, this is quite general: TWO INVOLUTION LEMMA. If S and Z are two involutions on a finite set and v is a fixed point of Z, the the orbit of v under the mappint SZ will contain another fixed point of Z or a fixed point of S RIGHT IN THE MIDDLE! WHICH OCCURS DEPENDS ON WHETHER THE LENGHT OF THE ORBIT OF v is even or odd. In Zagier's case, Z has a unique fixed point so the fixed point found must be a fixed point of S, which is the sum of squares solution. (This is the constructive version of Zagieer's proof.) As Zagier points point, many of his statements are true when p is just an integer of the form 4k+1. What happens when you start with p = a product of two primes of the form 4k+1 OR with two primes of the form 4k+3 (so their product is of the form 4k+1)? I know some answers. Try it!
    2
  4759. I don't get it, but sounds important.
    2
  4760. Nice.......
    2
  4761. I've always been very bad at math. I really like the way you explain things but youtube randomly sent me here from a Megadeth video. Instead of just ignoring this video, I'll take it as a challenge and see you in 6 months when I can hopefully understand a small percentage of wtf your taking about.
    2
  4762. Really much entertaining and informative. 🤗🤗🤗
    2
  4763. Pretty magical!!
    2
  4764. thank you, thank you, thankyou, thank you.
    2
  4765. (HO)3 😂👍👍👍
    2
  4766. The long awaited Mathologer video, finally!!!
    2
  4767. :)
    2
  4768. Speaking about different countries - Heron's formula was taught in Russian schools 20+ years ago, probably still being taught. The only point is that I haven't seen no teacher in our schools explaining what's behind it as you do 🙂 so it's just a formula with limited usage for usual schoolboy.
    2
  4769. 54th!
    2
  4770. 1. Puzzle: 5³ + 6³ = 7³. 2. Puzzle: ((10² + 11² + 12²) + (13² + 14²)) / 365 = (365 + 365) / 365 = 730 / 365 = 2. 3. Puzzle: 3³ + 4³ + 5³ = 6³; and: 3^4 + 4^4 + 5^4 + 6^4 = 6,893360391134855…^4 ≈ 7^4.
    2
  4771. Title: Why is calculus so...Easy? Every IITJEE student in india : TriGgeReD
    2
  4772. Fascinating. I am guessing that this can not be used directly to start with a regular n-gon and generate a non-regular n-gon.
    2
  4773. Woah 13th comment
    2
  4774. My brain HURTS!
    2
  4775. Kinda wondering what could be learned from setting the generator row to the digits of pi… or e… or sqrt(2)… or Liouville’s number…
    2
  4776.  @Mathologer  From your visual description, I suspected this was the case. I've derived much enjoyment from your videos, even though they are at times contain highly visual explanations. It keeps the brain matter running smoothly...
    2
  4777. I don't know why YT recommended this video and I didn't understand half of what you said (lol) but it was very interesting
    2
  4778. I have no idea why but this is one of the most interesting videos I've watched in a while.
    2
  4779. I definitely enjoyed Kempner's no 9s series proof the most. I thought your animation was surprisingly easy to follow for such an unintuitive claim!
    2
  4780. I don't see anyone who's mentioned this, so I was wondering why of all possible positive integers, you chose 666...
    2
  4781. Most Memorable: Kempner's proof at the end. Remarkable video!
    2
  4782. If you add 1 + 1/2 + 1/3 + 1/4 + ... and you note the denominator when we have a new ones digit, we get the series (1, 4, 11, 31, 83, 227) and if s is the sth term of this series then s/(s-1) approaches e.
    2
  4783. I didn't think magic existed, until I watched this video.
    2
  4784. woah im early
    2
  4785. 4:24 - radii*
    2
  4786. Most memorable moment: hard to pick one, the whole video is so thrilling. Here is one most: when the small reminding area turns out to be gamma.
    2
  4787. The g-conjecture is so cool!
    2
  4788. but can you odd?
    2
  4789. Hey Burkard, do your students talk about Mathologer much? What do they say to you about it?
    2
  4790. Proof of 1/A^2 + 1/B^2 = 1/D^2: Scale the right triangle by a factor of 1/BD so that it has hypotenuse H = 1/D and left side 1/B. Call the bottom side E. Since the left triangle is similar to this triangle we know A/H = D/E, which tells us A = 1/E. Thus by Pythagoras we're done. I also have a geometric proof of X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) that uses similar ideas to the proof of the cosine formula presented in the video, but it's hard to explain in a youtube comment.
    2
  4791. Had fun with this a few years ago. Base-1 and base+1 has special rules (and the factors of Base-1 and base+1) so base 8, you have these type of div rules for 7 and 9 (and thus 3). Base 16 has add/subtraction digit rules for the numbers: 15 (and thus 3 and 5) and 17)
    2
  4792. gnomon
    2
  4793. "Nine, nine, nine, fantastic number nine -- times any number you can find, it all comes back to nine!" Did anyone else watch Square One growing up? Many fond memories for me! I'd say Square One still has about 8% of my love 🧡
    2
  4794. Lambert is my favorite. I’m reading on of his translated books right now
    2
  4795. The moon argument is one of the most beautiful proofs I've ever seen
    2
  4796. Colored one
    2
  4797. ​ @Mathologer  Before British rule in India literacy rate used to be 60-70 % , but the British destroyed the Hindu education system of gurukul's and imposed the British education system and just after independence literacy rate was left 13%. We are still following the western education system like fools idolising European mathematicians even when Indian mathematicians were way intelligent than them and ahead of there time.
    2
  4798. Can’t wait for yet another amazing video! It’s waiting for me when I’m done with work for today :)
    2
  4799. for the challenge at 13:58 - it's very cool how the 2xN chessboards produce Fibonacci numbers - 1,2,3,5,8,13,etc. - but even cooler is how, if you separate the totals for each chessboard into subsets based on the number of vertical dominoes, you get Pascal's triangle.
    2
  4800. I think watching this video make me have some kind of 3b1b flashback 😅😅
    2
  4801. Your videos are AMAZING....and the explanation is always very brilliant!!!
    2
  4802. 2:07 dude speaks a mean bit of German 😮
    2
  4803. hey, is it true that for the simplex (triangle pyramid) there is the same formula, (x+1)^(n+1), so there may be formulas for other figures
    2
  4804. When it comes to “your favorite number”, 7 is overrated. When it comes to polygons, the heptagon is underrated.
    2
  4805. These paradoxes are always related to infinity.
    2
  4806. WOW, very inspiring. Easy to understand. TOP animations. Thank you!
    2
  4807. Perfect timing Prof. I'm taking a calculus course and would love to do it the Mathologer way :)
    2
  4808. I am uniquely placed to authoritatively answer your question, Ser Polster. It's because YouTubers use the English Wikipedia. On 2021-09-24, Jade Tan-Holmes on the YouTube channel Up And Atom covered this, with a notable "What is this? Physics?" aside in the middle of the video. In response the next day I improved the English Wikipedia article vastly, based upon several university press mathematical textbooks. You will find me in its edit history. Until that point, the English Wikipedia article had only laid out the calculus approach. YouTubers, working from it, took its approach. I added all of the things that you mentioned to the English Wikipedia article, which at that point hadn't mentioned Oresme, hadn't properly laid out Torricelli's original proof, hadn't included Torricelli's original shape with the cylinder, or even supplied Torricelli's original name. I also added the explanation of the differences between "mathematical" and "physical" paint, and the contemporary 17th century philosophical and mathematical debates that ensued. The French Wikipedia had already gained a graphic for Torricelli's shape, fortunately. Although its coverage at the time was better than the English Wikipedia's, it too didn't address Torricelli's proof in detail or go into things like Hobbes and Wallis and Barrow. I like to think that the French Wikipedia was spurred, by my efforts, into attempting to leapfrog the English one and be ahead again, because it has gained a lot more in 2022 and 2023. I wrote in 2021 in a comment on the Up And Atom video that "One day, video makers referencing Torricelli will actually show the shape that Torricelli used in his proof. It's not that day yet, though. Hint: There's a cylinder on the end." It has taken over 2 years for that day to finally come. Well done.
    2
  4809. I am ready for the tomaker now! Thankyou Mr Mathologer
    2
  4810. I have often thought about a chronological math path as well!!! In fact, that is how I taught decimals and fractions to younger students— started with ancient man being happy with whole numbers until bartering started happening, and then needing parts of a whole. The human brain responds well to story!!!
    2
  4811. Sad that they were so protective of maths they discovered but which belong to no individual, refusing to freely disseminate such material, or even at the very least allowing it to be spread. What does coveting it do?
    2
  4812. Excellent!
    2
  4813. Wow this is really fascinating. i wonder if there is a general way to construct magic squares but with an even number as side...
    2
  4814. freaky ahh identity
    2
  4815. Amazing! I was talking to my girlfriend this week, about one of the things that annoyed me at school since I was ten! They give you part of a sequence, and assume there can only be one correct successor. I have waited fourty+ years for the proof! Not that I think of myself as a mathematical wizard, but I thought 1, 2, 3... what comes next? The answer might be 4 as well as 5. Either add 1, or add the previous two together. There is no way to determine which method is desired. If both are equally likely, how can only one be correct?
    2
  4816. If you go on applying the difference to the fibonacci you end up with the fibonacci on the left with alternating signs.
    2
  4817. Woah, I thought the Mathologer only lived in the white void of pure mathematics
    2
  4818. This is beautiful
    2
  4819. This quarantine just got a bit better thanks to you. Amazing video as always
    2
  4820.  @Mathologer  Oh... the proportionality theorem isn't even mentioned on that page, is it not also known as the Thales theorem?
    2
  4821. in French it is for sure! But the way people call elementary results like this one vary a great deal depending on the country. In France at least, we needed a name for this fact since we do not teach similar triangles. I guess in places where similar triangles are taught, you don't need a name for this fact, as it is basically equivalent to the similar triangles stuff
    2
  4822. ❤ every one of your episodes! curious to know why you chose at 7:25 to horizontally stretch the squished A(3) from the x=1 line and then right shift it, instead of simply stretching it from the y-axis with no need to shift.
    2
  4823. Awesome as usual!! Still waiting for Galois, though, 😊
    2
  4824. Kirigami
    2
  4825. Faszinierend, wie immer, danke! 👏 Die beiden Varianten sind schön, jede auf ihre eigene Weise. Im allgemeinen, ich finde es nur gut, wenn es mehrere unterschiedliche Lösungen gibt.
    2
  4826. 20:10 A long time ago, I discovered some algorithms to produce Pythagorean triples and subsequently discovered that I was not the first person to discover these algorithms. I extended my research to the 60 and 120 degree cases that you have shown but I suppose now I know that it has also been discovered before. No Fields Medal for me😅
    2
  4827. Many many many years (decades) ago, I had a copy of the guinness book of records (GBoR), and it listed the largest magic square as 107x107. So I used pen and paper to make a 111 x 111 magic square. When I was done, I sent a letter to the GBoR. They replied back to me that they didn't accept entries any more, because magic squares could be created by computers. Also: at the 28:36 mark, one of the 5 block figures contains 6 blocks. (I'm sure that someone else pointed that out already, so my streak of being "just too late" will remain intact.)
    2
  4828. It might also help to notice that both the determinant and the trace are coefficients in the characteristic polynomial.
    2
  4829. You probably won't find this helpful but the trace is the linear map End(V) ≅ V* ⨂ V → K where K is the base field. The isomorphism V* ⨂ V ≅ End(V) is given by φ ⨂ v ↦ (x ↦ φ(x)∙v). The linear map V* ⨂ V → K is given by φ ⨂ v ↦ φ(v). Since both of these are base-independent, the trace is also base-independent. That's why it is conserved.
    2
  4830. Before watching the video, my intuition is that if the horn if infinitely long then it would take an infinite amount of time for the paint to reach the bottom, and thus it is impossible to ever actually paint the whole surface.
    2
  4831. Not if the paint travels infinitely fast, or if it is filled from infinitely many lined-up sources inside the horn. 😉
    2
  4832. i must comment your channel after watching this chapter, My boldness is feed from your phrase that repeat, "is this brillant,isnt?" and i must say as hardcore student of your teachings that yes it is brillant. Thanks for sharing some awesome math issues. Really make my life better. Gracias por tanto y saludos desde Bolivia.
    2
  4833.  @jkvoot  is that how it works? Oh and there’s also YT browser extensions that are even faster. 10x goes brrr…
    2
  4834.  @want-diversecontent3887  so if you skip half the frames and the corresponding sound it may actually be another video in there like when using a flipbook backwards.
    2
  4835. 72 minutes. I have to watch the complicated ones at least twice.
    2
  4836. Great video, watched with my husband! My favourite part was the visualization of the no 9s series, thank you Tristan for the idea!
    2
  4837. Wouldn't it be possible to transform a "spiral circular" map like in 21:54 into a "straight circular" map like in 29:25, if you just applied a specific "rotation of infinitely thin concentric circles" transformation that would straighten up the meridians, on the spiral map? The transform would be area preserving and result in a more realistic, less distorted map.
    2
  4838. Hey Mathologer. I'm a second year university math student, so take my opinion with a grain of salt I guess, but I feel like that was an exceptionally clear video. I love how you split your videos into sections. Not only does it provide an opportunity for everyone to get something out of your videos, but it also gives the viewer a chance to pause and ponder (as another math youtuber likes to put it). Please keep it up, I've learned so much from your channel!
    2
  4839.  @jaapsch2  Yes, clearly not all pythagorean triple hypotenuses [hypoteni? ;^)] are pythagorean primes. I saw something about the formula you gave when I looked it up, but it isn't something I'm familiar with from anything I ever studied, though it makes sense... after the fact. Very interesting though.
    2
  4840. What is the max position we can achieve if we play in n dimensions? for n=2 its 4, for n=3 its 7, etc. Is there a nice closed form for it?
    2
  4841. Really boom
    2
  4842. Excelent video Mathologer!
    2
  4843. A little late to the party, but I loved this video and accepted the challenge to write a domino-dance program, stretching my modest Python skills a little bit. It doesn't use a square array but instead a dictionary to represent the grid. Keys are grid coordinate pairs; to represent the state at each coordinate pair, values are one of five vectors (four directions, 0 for empty). Domino flow patterns occur in the values of the dictionary. Probably this is quirky, but it makes a form-fitting (diamond-shaped) grid that extends naturally and performs decently, thanks to fast dictionary lookups. I don't know anything about graphics or web coding, so I used Processing and Trinket to share it. It has a couple of niceties for the user. Here's a link: https://trinket.io/library/trinkets/5b574f6671 Version without code window: https://christopherphelps.trinket.io/sites/arctic_circle_theorem_simulator Signed, a Mathologer fan, math tutor & enthusiast, first-time commenter
    2
  4844. With a little bit of experience in algebra and number theory the first 25 minutes of this video didn't really contain anything surprising or too interesting. But then you still managed to capture my interest in the last 5 minutes. I don't think I was ever explicitly taught anywhere that the digital root of a number in base 10 is equal to the number mod 9. But I definitely intuitively knew it already. It might be because of a book about mathematical magic by Martin Gardner in which I read about modular arithmetic and divisibility as a child.
    2
  4845. what does the image on your t shirt refer to, please?
    2
  4846. Wow! Amazing video
    2
  4847. It isn’t super famous because mathologer hadn’t made a video about it yet
    2
  4848. The last few minutes of music were just fantastic
    2
  4849. I believe it's fairly easy to hang one brick beyond the cliff even with 3 bricks. WLoG let the length of each brick be 8 units. So basically we need the CoM (centre of mass) of our hanging brick to be be >= 4 units beyond the cliff and and the CoM of our entire system to be inside the cliff. Lay the first two bricks such that the right end of both of 'em sticks out by 2 units, i.e. both are perfectly aligned in the vertical direction. Then lay the last brick such that it's left end just aligns with the end of the cliff - this brick is the one which'll be beyond the cliff. Job done. You can check that the CoM lies exactly on the edge. This solution also takes care of the one-brick-per-layer condition and remains stable as we progressively add each brick. Note that this solutions is just an intuitive, special case out of the multitude of arrangements possible which satisfy the condition. After all, it ain't too tough to just confine the centre of the system...
    2
  4850. 57 minutes to 1hr late
    2
  4851. Mindblowing
    2
  4852. Always eat Chung King Chow Mein when doing magic squares.
    2
  4853. Once time again a Wonderful Video from Mathologer !!! Thank you.
    2
  4854. The proof at the end is just as pretty as it gets!
    2
  4855.  @Mathologer  , Danke für deine großartigen Videos...
    2
  4856. 👅👅👅
    2
  4857. I think your math tutorials are excellent. My anecdotal story is that while copying a maths table across while sitting in Epsom Library wearing a helmet and a red cape, a Korean mother of a little girl who was sitting nearby, looked me in the face and called me "Asshole!"
    2
  4858. What a clever mathematical exploration! I loved seeing the circle come in the limit of both the diamond and the hexagon, it was so elegant. I suspect that this area of study has applications in metallurgy, crystallography, and materials science.
    2
  4859. Wonderful video. I happen to think that it gets "more magical" (i.e. more significant or special or amazing) when introducing the mathematics behind the vortex. Math is like that! :-)
    2
  4860. The 20eth minute gave me the most satisfied feeling since a long time. ♥
    2
  4861. Most memorable: partial sums are not integers.
    2
  4862. Do the normals of the Schwarz lantern not approach the normals of the cylinder? In the case where p is fixed it makes sense that they don't, but as long as p increases without bound, I don't see how they wouldn't. Or is this like with integrating 1/x, where they not only have to approach, but do so "fast enough"?
    2
  4863. I love how you slipped in some Gray Coding, very important in Information Theory; think Hamming Distance.
    2
  4864. It seems like the caption always displays two 'f's as '?'. For example at 2:27 'differ' turned into 'di?er'. Not sure if this is only me though.
    2
  4865. I like the first proof better than second one, and vice versa! 😂
    2
  4866. And where is the proof for the good minus bad rule?
    2
  4867. Ah, nerts. Still over my head. Videos like this make me WANT to learn more advanced math, though! Thank you for sharing your insights, and thank you even more for sharing your unmistakable and infectious love of the subject! We’re so very fortunate to have people like you, OC Tutor, 3blue1brown, etc who create amazing content which inspires curiosity and imparts knowledge (for free, at that) to anyone who seeks it. Imagine what Euler, Gauss, or Newton could have done with so powerful a means of communication!
    2
  4868. Gald to see Euler is contributing to Mathologer from pateron
    2
  4869. >>> sum((i+1)**10 for i in range(1000)) 91409924241424243424241924242500 Took 2 seconds.
    2
  4870. о боже! слава тебе говорящий истину! от неправильного видео у меня случился когнитивный диссонанс, это видео вернула меня на понимание вселенной в верной парадигме
    2
  4871. Why 4k+3 though and not 4k-1? Would look better next to 4k+1
    2
  4872. I'm gonna comment cause it looks like every comment is liked by mathologer
    2
  4873. Mathologer I see you have studied this in great length. Squares and cubes seem easy. How about a math equation for hexagons? After the one center a first ring makes seven, it is all multiples of six and then adding a one after counting the rings. 7, 19, 37, 61 and 91 to get started. A hexagon with four rings around the one is 61 items, they could be apples , oranges or rocket exhaust ports. Five is 91. Can you make a simple formula for these numbers?
    2
  4874. I loved the dotted angles proof!
    2
  4875. The most memorable part for me was that the sum of the no nines series is smaller than the sum of the exactly 100 zeros series, it blew my mind a little.
    2
  4876. In reference to the problem at (8:24), the magic constant of the King's 33x33 magic square is 17,985. In general, the Magic Constant for the King's Magic Square is n*(n^2+1)/2 where n is odd. As can be seen at (11:46), the magic constant is given by a sum natural numbers, and a sum of multiples. Specifically, it is the sum of the natural numbers up to and including n [1+2+3...+n], and the sum of the first n multiples of n [0n+1n+2n+3n...+(n-1)n]. We can be sure that this will always be the case, since when labeling the tiles, we always begin by filling a diagonal (as seen at 6:50) and then the squareness of the setup guarantees that the other diagonal will always also have the same number of tiles. Important to this proof, we know that 1+2+3...+n = n(n+1)/2. So, the King's magic constant (M) is: M = (1+2+3...+n) + (0n+1n+2n+3n...+(n-1)n) M = (1+2+3...+n) + n*(1+2+3...+(n-1)) M = n(n+1)/2 + n*n(n-1)/2 M = n*(n^2+1)/2
    2
  4877. 7:57 Just like the Coriolis force in a Flat Earth's video...hahaha
    2
  4878. Do people have difficulty seeing 3+ dimensions? 1: Linear line of Points, going along One Axis. 2: Square Grid of Points, going along Two Axis. 3: Cubic Array of Points, going along Three Axis. And continue... 4: A Linear line of Cubes, where each Cube has Three Axis of Measurement. Picture Multiple "Rubik's Cubes" laid out in a line. The 4th Dimension refers to which 3 Dimensional cube we are looking at. 5: A Square Grid of Cubes, where each Cube has Three Axis of Measurement. Picture Multiple "Rubik's Cubes" placed in a Grid. The 4th and 5th Dimensions refer to which 3 Dimensional cube we are looking at. 6: A Cubic Grid of Cubes, where each "Internal Cube" has Three Axis of Measurement. Picture Multiple "Rubik's Cubes" stacked into a Cube. The 4th, 5th, and 6th Dimensions refer to which 3 Dimensional cube we are looking at. So, a 4 Dimensional scenario, such as the Location of a moving object, could have a set of 3 Dimensional Coordinates, with the 4th representing different times. A 5 Dimensional Scenario, could be the Location of a moving object through different Mediums, could be 3 Dimensional Coordinates, with the 4th representing different Times, and the 5th includes all previous information, but each "Array" would represent a different Medium. A 6 Dimensional Scenario, could be the Location of a moving object through different Mediums with different Engines.
    2
  4879. Fibogarous Theorm 😂
    2
  4880. very nice work. love the graphic content. using 0 and 1 (rather than 1 and -1) is a nice segway into information theoretical stuff tho. do snake-in-the-box! gwan gwan gwan
    2
  4881. I've always wanted to learn calculus.
    2
  4882. I feel like this will be in my next algorithms exam...
    2
  4883. iirc this paradox works because it sneaks in a hidden /0 which is an illegal move.
    2
  4884. Can you please have a store where you sell your shirts? I'll buy every single one of them!
    2
  4885. i love this shirt!
    2
  4886. Only the proportional split makes sense.
    2
  4887. So, this is like the cylinder version of honeycombs, where conjoined circles collapse i to hexagons. Neat
    2
  4888. Thanks my friend. I always watch you early-anticipated videos on a nice Sunday afternoon here in Ontario Canada, bi a big mug of coffee. :)
    2
  4889. 11:13 Ptolemy's theorem is actually completely useless for math competitions. In my three years of IMO training, I have never used Ptolemy's theorem. In order to use Ptolemy's theorem, you have to know at least four different lengths in order to use it, and that is just never the case. Most Olympiad geometry problems don't even mention lengths, let alone four.
    2
  4890. @TonyNaggs I m interested in this 3d model as well 🥰
    2
  4891. 26:10 I’m still acquiring the taste of proofs for breakfast; but I’ll get there, at some point, I promise 🙃.
    2
  4892. I am only disappointed about the Video length not being 20:19 min. long :)
    2
  4893. Wow wow wow wow wow! Mathologer is back with another gem! Made my day too ;)
    2
  4894. This is a nice CHANGE. :D :D
    2
  4895. Bringing something completely new to my attention is why I love this channel.
    2
  4896. Mathologer and 3 Blue 1 Brown are my favourite Maths channels. While I love maths but I wish these 2 channels existed 20 years ago when I was in 12th and studying calculus
    2
  4897. I was able to prove this using angle bisects. Which also proves the more general case of windscreen wiper theorem as well. With my proof there is no need to "notice" that the shrinking outside circle coincides with the inscribed circle, it follows from the proof logic. It is kind of similar to @Math_oma's proof.
    2
  4898. 19:48 using the inscribed angle theorem the sum of the opposite angles is equal to a half of the length of the arcs, but in out case the length of the arcs is equal to 360 degrees so the sum of the opposite angles is half of that which is 180 degrees
    2
  4899. 😀
    2
  4900. THANK YOU! Understanding QR has been a goal of mine. I didn't fully follow the argument on first pass, but now at least I have a Mathologized argument to study. Again, thanks for all your effort.
    2
  4901. The most memorable fact for me is the fact that computers are still way behind human ingenuity
    2
  4902. 14:05 my ahaa was that the answers are fibonacci numbers
    2
  4903. Hi Mathologer. You have a circular slide rule on many Citizen Eco-Drive watches. It’s very handy - for example - if you are abroad in a country with different currency. You set e.g. the currency ratio EUR/USD once and then you can quickly calculate prices. Also it can serve as an easy „tip calculator“. Set to 1.1 for 10% tip and read off your watch.
    2
  4904. Some very pretty maths! I wonder how this let‘s one generalize ordinary calculus, to a more general algebraic setting
    2
  4905. 0:55 anyone else saw Manjaro logo?
    2
  4906. In Bandhan Sulba Sutra, similar square and remaing portion divide by 3 was done . This pre- Pythagorean priestly ritual was used for value of root over 2 but not in continued fraction. from New Delhi.
    2
  4907. Made my day as always your videos did. 🤗😍
    2
  4908. I love the last bit about how these formulas were expressed in verse. I wish you gave some rough English approximation of how that would have sounded. I'm always curious about how complex mathematics were expressed prior to the development of our present-day concise mathematical language.
    2
  4909. Now it’s making me wonder if there is a wall for the periodic table with regard to sub atomic particles and relationship to volume. 🤯 I’m doomed 😂
    2
  4910. Since 1 + 1/2 + 1/3 + ... + 1/n ≈ ln(n+1) + γ, and since each successive number he uses there is the next higher power of 10, we get that each difference should be approximately ln(10^(k+1)+1) + γ - ln(10^k+1) - γ or, it should be approximately ln((10^(k+1)+1)/(10^k+1)). As k gets very large, this approximation gets better, and it tends to ln(10^(k+1)/10^k) = ln(10). And if you check, the first several digits of ln(10) are indeed 2.302585. Nothing particularly special here! This is an artifact of Mathologer using larger and larger powers of 10 to illustrate his point.
    2
  4911. The sad thing is, the other 369 videos will get way more views than this even though they are utter bollocks. It's the sorry state of the world sadly...
    2
  4912. Is there any book that reports the mathematical part of the Almagest in modern language?
    2
  4913. Same thing :)
    2
  4914. Integration has always been some of the most interesting math to me, the process of paring down what makes a difficult problem unique, until you're left with something familiar and simple
    2
  4915. I personally find the proof with the two colors clearer than the proof wit the swivelling. Maybe it's just me, but I find saying that the sum of some numbers is equal to the sum of some other numbers more obvious than saying that swivelling some line segments around makes them coincide.
    2
  4916. 19:00 You seem to think that the name "slide Rule" is because the straight slide rule slides and the circular one does not. As far as I can see, BOTH types slide, so they have equal right to be called slide rules. It's just that one slides round in a circle and the other slides in a straight line. QUERY The other shape which slides along a copy of itself is a HELIX .... anyone want to design a HELICAL SLIDE RULE ???????? You could even have three helices. That way, EVERY one of the three helices would touch each of the others. HELICAL SLIDE RULES RULE Every helix would be the same size. You could have a transparent cover with a line (lines) drawn on it so that you could line up the position of one helix with another. The lines would have to be drawn to be circumferential. You could even take it apart by removing one helix and inserting ANOTHER helix, as the situation demanded.
    2
  4917. I gave up with maths when they started cheating and using letters for numbers.
    2
  4918. This was fun to watch!
    2
  4919. 13:27 I would guess there is a O(N^2 * M) dynamic programming approach for that (N = the amount we want to partition, M = the number of distinct coins). I will assume we have a sufficient amount of every coin type. Let's define count(i, j) as being the number of ways to partition i dollars with the first j coins (coin values are sorted in ascending order). Then count(0, j) = 1 for all j and count(i, 0) = 0 for all i. The recursion comes out to be count(i, j) = count(i, j-1) + count(i - c[j], j-1) + count(i - 2*c[j], j-1) + ... The answer will then be stored in count(N, M). If you do some clever stuff with the memoization, the time complexity can also be reduced to O(N * M), I think.
    2
  4920. When stacking blocks, I think I could reach further by turning some counter weights 90 degrees. This may be cheating though...
    2
  4921. I see it further down.Pi
    2
  4922. Never learned either in school... Just told to believe a2+b2=c2!
    2
  4923. Sofort alles liegen und stehen gelassen und das Video geschaut... 👀✌️👌👌
    2
  4924. It’s a thing of beautiful simplicity.
    2
  4925. I liked most the "Terrible aim" part. It's a surprising insight with easy proof, I love it! Funny T-Shirt, as always! :-)
    2
  4926. (10^2+11^2+12^2+13^2+14^2)/365 = 2 Easy.
    2
  4927. That centre of mass stuff really enhanced my concepts.👍🏻
    2
  4928. 0. Selection logarithm's base - At least from 2 to 16.
    2
  4929. Can you please make a place to see all past submissions. I'm not sure where to find it. I checked the description. Also, great video.
    2
  4930. Wonderful video. Recently I have become fascinated with older methods of computation, especially nomographs. Being the cousins of the slide-rule I have often wondered how to construct a mechanical equivalent to one, a sort of generalized slide rule. I think it is such an interesting topic it would make a great mathologer video!
    2
  4931. What would we ever do without electricity and screen time. Lol.
    2
  4932. Thank you so much! I always wanted to understand this.
    2
  4933. You are a godsent angel, I've had my mouth open the whole video, I wish I could subscribe twice
    2
  4934. Chinese numbers use "+" and a few similar signs for powers of 10, while in the Arabic number system, these powers of ten not necessary because of the place-value convention. So, for example, in Chinese the number 21 would be written as 2 + 1, and 12 would be written as 1 + 2, etc.
    2
  4935. I love you π Shirt
    2
  4936. No, I do not follow at time index 15:20 I do not see that proving an odd number of solutions proves anything other than we have an odd number of solutions.
    2
  4937. Crikey. I'm history,😌 I grew up with a slide-rule at school.....
    2
  4938. ... and a book of log tables; logs at the front and antilogs at the back.
    2
  4939. For some reason, I feel like I wanna see you dance 🕺
    2
  4940. Luckily that's easily fixed :)
    2
  4941. Step 1. You Tube recommendation Step 2. What a nice title, A+B=AB ---> Golden Equation... Step 3. Liked 👍 and Saved in "Watch Later" folder. Step 4. Post this comment😅 Step 5. Enjoying the whole video...now....
    2
  4942. 15:52 I love the Michael Ende quote <3
    2
  4943. Juhuuuu, neues Material. U made my day
    2
  4944. Bitte mach mehr von deinen Mastervideos. Ich studiere Mathe im ersten Semester und liebe die langen Videos die richtig tief gehen, je tiefer desto besser ;)
    2
  4945. Your videos are just beautiful ! ❤ CONGRATULATIONS ! 🎉 🎉🎉
    2
  4946. Fun fact regarding the partial fractions from the continuous fraction of sqrt(2): If m/n is one such fraction, then m^2*n^2 is a square triangular number. You can figure out all the square triangular numbers by doing this (1, 36, 1225, ...). The reason is that when looking for square triangular numbers you get immediately to the same equation as the one this video presented in solving the Strand puzzle: 2x^2 = y^2 + y.
    2
  4947. I thoroughly enjoyed this video, but it definitely reminded me why I've done my best to avoid serious combinatorics :)
    2
  4948. Golden πzza
    2
  4949. 30:20 "...which sums to 2." Zeno: What!!!?
    2
  4950. Your "prettyfied" diagrams (and by extension Wayne Roberts' piece) remind me of work of Dutch abstractionist painter Piet Mondrian. Not famous enough IMO.
    2
  4951. I was aware of this fact but never looked into much deeply. Thank you for addressing it.🙏🏼 Knowledge knows no boundaries.
    2
  4952. Your animation of sliding a straight rule along a curve reminded me of an antique tool that predates the tape measure. I know it as a walking rule, others as a stepping rule. It is a disc with circumference 12 or 24" with a handle on the centre of the disc. It could be used to measure quite well from a wood workers POV. Craftsmen working with wheels or long beams used them extensively. I just thought you might find this interesting.
    2
  4953. Chapter 4 challenge: 2=1/1+1/2+1/3+1/6 3=1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/20+1/60+1/252+1/2520 How I came up with these? I just picked a highly composite number (6 for the 2 problem, 5040 for the 3 problem) and picked factors of the number such that the sum is 3 times that highly composite number (e.g. finding the factors of 5040 that together add up to 15120). I did this manually but I can imagine a simple computer program can perform this task for larger numbers relatively quickly.
    2
  4954. is your card the queen of hearts? (hint: it is)
    2
  4955. T-shirt: No wonder I have a hard time reducing formulas and get them into a lean solution.
    2
  4956. In the beginning of the video, Mathologer mentions the Gaussian integral as ∫[−∞,∞] e^(−t^2) dt = √π. However, at 10:30 ish, he's not using the exact same integral anymore; instead, he's using ∫[−∞,∞] e^(−t^2/2) dt. Starting with the original Gaussian integral ∫[−∞,∞] e^(−t^2) dt = √π, we can perform a change of variables u/√(2) = t to obtain 1/√(2) · ∫[−∞,∞] e^(−u^2/2) du = √π , from which we conclude ∫[−∞,∞] e^(−t^2/2) dt = √(2π).
    2
  4957. For the final puzzle: The solutions can be described as ᵏ√[3ᵏ + 4ᵏ + ... + (k+1)ᵏ + (k+2)ᵏ], which if you calculate for a few values, actually seems to be linear. As k → ∞, the solution approaches k + 2.458675145387.
    2
  4958. I found the adjacency matrix for the glasses and the determinant is 666. Maybe I did something wrong because you said the determinant doesn't give the answer straight away? Matrix that you can input to a CAS: [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,i], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,i,1], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,i,1,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,i,1,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,i,1,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,i,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,i,i,1,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,1,i,i,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,i,1,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,1,i,0,1,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,1,0,i,1,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,1,i,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,i,i,1,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,1,i,i,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,i,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,1,i,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,i,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,1,i,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [1,i,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [i,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    2
  4959.  @Mathologer  Hahaha. As for the direct proof, there's probably a cleaner way than this but here goes: First, we can remove the leftmost and rightmost 2x3 blocks and just multiply by 3*3 = 9 at the end. Now we do casework. Case 1 is a tile at r7-g17. Then the tile that covers r6 must cover g16. There are 2 ways of tiling r4-g18-r5-g19. There are 5 ways of tiling the 2x4 block "between the eyes", and tracing through, there are 2 ways of tiling g4-r18-g5-r19. There are also 3 ways of tiling between the eyes where the right eye has tiles r13-g9 and r14-g10 (instead of r14-g9), and then there are 3 ways of tiling g4-r18-g5-r19-g7-r16. Putting it all together, we have 2(5*2+3*3) = 38 tilings in Case 1. Case 2 is a tile at r8-g17 (the only other option for the tile covering g17). Then the tile that covers r6 must cover g15. By repeating the steps above, we get 3(3*2+2*3) = 36 tilings in Case 2. So the total number of tilings is 9*(38+36) = 666. Yay!
    2
  4960. I really love these videos. I hope I get to be a math teacher one day
    2
  4961. I like the mini-mathologers. They hurt my brain less.
    2
  4962. I remember some time in the early 2000s visiting the Science Museum in London with a friend, and interacting with an exhibit that challenged visitors to get the longest possible overhang out of a tower of blocks. After some experimentation, we hit upon the idea of using some of the blocks as a counterweight and managed to extend our overhang into the region the exhibit deemed "impossible". Who knew, if we'd followed up on it we might have had the opportunity to generate some truly novel mathematics :)
    2
  4963. Aqua? Magenta? For those of us who are colorblind, you may as well have used Amaranth and Atrovirens. Fortunately, you persauaded me that .999=1 ... so I keep coming back. 😢😅
    2
  4964. I've looked at 7^x + 8^x = 9^x 22^x + 23^x + 24^x = 25^x + 26^x ... and found that interestingly enough, while it isn't 4 exactly, it appears to be approaching 4 as we continue down the line, with its limit being 4.
    2
  4965. Interesting. The Pascal's Triangle extension of Moessner seems to imply quantization at the edges. Everything starts with individual units, value of 1 - quanta.
    2
  4966. I saw Pascal instantly for the Fibonacci Squares!
    2
  4967. Incredible! We should have known about this 7 years ago. There is an application: melting ice blocks in a piston! As any physicists, we said that the blocks are discs. But we missed the essential! Thank you to your video!
    2
  4968. 37:15 This follows from the fact that you can interpret it as stacked cubes. Imagine looking at it from the top- you see an n*n square tiled with orange. Now imagine looking at it from the left- you see an n*n square tiled with gray. Finally, imagine looking at it from the right- you see an n*n square tiled with blue. This must be true for any such tiling, because that’s the way the tiles are facing by definition. Let’s Go Mets!
    2
  4969. Answer to the puzzle: The circle has a larger diameter. Because the weird curve's diameter is exactly the same as the sum of the 3 sides of the triangle (can be seen because it meets the lines perpendicularly, by its construction). The Conway circle is larger as it also passes through the same two points on any of those lines, but has a center which is not on the line.
    2
  4970. Amazing videos you are making by the way. I wish I have this when I was in my teen. My brain hurts 60% into your video. I need to sleep this off :). I love your giggles :).
    2
  4971. I'm a little sad you left out the hyper triangles: they can be categorized in a similar fashion by Pascal's Triangle, which of course contains the coefficients to (x+1)^n. That is, 1 1 1 (if we truncate the final one in each row, this marks 1 vertex) 1 2 1 (again truncating the final one, 1 edge and 2 vertices) 1 3 3 1 (similarly, a triangle has 1 face, 3 edges, 3 vertices) 1 4 6 4 1 (the tetrahedron has 1 cell, 4 faces, 6 edges, and 4 vertices) &c.
    2
  4972. This video has in it the first thing that I ever saw that was interesting (to me) as mathematics. As far as I knew before then, mathematics functioned only to accomplish something else, in particular far-out physics and miraculous inventions. I therefor took algebra (and got all A's) because that was what I was told Einstein was writing on the blackboard in those nostalgic, old photographs. I was disillusioned and frustrated (as teenagers are) to find by the second semester of algebra that there was nothing yet that could honestly be claimed to be useful or practical whatsoever, and no clues to how or when that could ever happen. Then a thermonuclear device was set off in my brain, by an incidental, optional section of the textbook, just there for its own sake, unrelated to the rest of the chapter. The well-known observation that the differences between successive squares are the odd numbers, also means, they said, that the second differences are all 2. Likewise with cubes, the third differences are all 2*3. Likewise with fourth powers, the fourth differences are 2*3*4. Mind totally blown, I had to calculate this by hand (no personal calculators in 1959 and I hate arithmetic), for larger and larger numbers, and higher and higher powers, over the next few weeks, just to see it (not that I doubted it), just like watching Mr. Wizard (Don Herbert) freeze a hot dog in liquid air and break it on the floor never gets old. I thought I should be able (since I got A's) to figure out why this works from what had been in the book, so I spent the next couple of months working on it (while higher priority things needed doing) without success. (But not without finding chains of things that were not in the book.) Of course the cause of the phenomenon would be in some book somewhere, but that's not the point, anymore than some one else looking at the sunset is the same thing as you looking at it. Blessing to the textbook author, who probably had to lawyer to keep it in, for putting that incidental aside in the book. And blessings to mathologer.
    2
  4973. The t shirt is nice
    2
  4974. I like mathologer💗💗💗
    2
  4975. du sprichst aber sehr gut deutsch, du musst deutscher sein
    2
  4976. Everytime i watch old videos from you, you upload a new one, at the same day.... Could you please do like a behind the scenes of the animations and that stuff? Grüße aus Deutschland
    2
  4977. Mate be safe
    2
  4978. Next video in this series please
    2
  4979. Mathologer damn, am at unimelb and wish I was at Monash now, haha
    2
  4980.  @xDosReis  Doesn't make much of a difference at the moment, both campuses are ghost towns during lockdown :(
    2
  4981. I have a suggession for Mathologer, please do research about 1. Āryabhaţa I : writer of Āryabhatīyam, etc. 2. Bhāskara I : writer of Mahā-bhāskarīya, Laghu-bhāskarīya, etc. 3. Bhāskara II : Bījgaņta, Līlavati, etc. And many more, I'll suggest you later.
    2
  4982. Could there be a sliding tile "dance" for the triangular grid as there was with the square one?
    2
  4983. I wonder how this compares to the "Mirat" distribution in Islam.
    2
  4984. 3:02 I almost fainted when he said that the answer was pi lol.
    2
  4985. i don't get the shirt. What's its meaning ?
    2
  4986. I was losing it around part 5, but these are always entertaining.
    2
  4987. Doesn't half of a sqaure, is one of two equal lateral triangles?
    2
  4988. And I thought that I was smarter than a chimp
    2
  4989. The drawback of having it in my back pocket was that I ended up cracking it from sitting on it. Oh well.
    2
  4990. This is one of my favorites
    2
  4991. Is there some rule for interference in the shadow when spinning the higher dimensional cube? Like in the 2D shadow, the edges cross, but they don’t in the 3D shadow of the 4D cube. Do the faces or volumes interfere with each other in the 3D shadow?
    2
  4992. The remaining primes which are not in the form of 4k+1can be written as 4k-1or 4k-3. It is easier to write in the form of 4k-1.
    2
  4993. I love the geometric intuition you continue to provide in your videos. I can't wait to see what you have next with this series.
    2
  4994. For the challenge at 20:24: I found a certain sequence of reciprocals (don't get too excited, I only have finitely many terms and don't even know how to extend it without using trial and error) that 1) has ever-decreasing terms (like the true harmonic series), and 2) when you calculate its partial sums it hits every integer from 1 to 3. It'll be under the "read more" button so I don't spoil it! 1 + 1/2 + 1/3 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + 1/16 + 1/18 + 1/20 + 1/24 + 1/25 + 1/30 + 1/36 + 1/48 + 1/60 + 1/100 + 1/140 + 1/180 is my sequence (the sequence adds to 3, first four terms add to 2, and first term is 1). I'm quite worried about how I can extend this to the integer 4, since my largest possible term available is 1/181 and my largest term with many divisors is probably 1/189 or 1/192 - which means it'll take a hard lower limit of 200ish additional terms to write it and almost certainly it will also take many more.
    2
  4995. What does his T shirt mean?
    2
  4996. egyptian fractions are even more closely related
    2
  4997. Awesome, interested in knowing more about Gregory-Newton series for negative values.
    2
  4998. Best channel for discovering math beauty. J'adore.
    2
  4999. Somebody should write a book on these neat algebraic tips and tricks. It would be very handy!
    2
  5000. Here is a very simple intuition for logarithms: The length of a number is roughly 1 more than its logarithm. As an example, the number 1234 is 4 digits long. So the log (base 10) of 1234 is roughly 4 - 1 = 3. [The true result is about 3.09131516.] Changing the base of the log basically means that you are changing the base in which the number is written, followed by counting the digits of that. Using this intuition, it's almost obvious why log(a ⨯ b) is equal to log(a) + log(b). The number of digits in the product of two numbers is roughly the sum of the numbers of digits in the two numbers. 1234 * 4321 = 5332114, so two 4-digit numbers multiply to an 8-digit number. Not always exact! But close enough to build intuition with.
    2
  5001. Pure Magic!!!
    2
  5002. That t-shirt 🤔
    2
  5003. At 24:01 @Mathologer asks "what if you did 1 round of 72, 2 rounds of 144, 3 rounds of 216". Just quickly off the top of my head you will still get a perfect pentagon with an origin at the same point, but it may be rotated and scaled?! This is because once a round of 144 has done a pass it has removed its special component, and the other components get rotated and scaled. Is this correct?
    2
  5004.  @Mathologer  awesome! This feels almost like the inverse of finding eigen values by applying an operators. BTW i was a student of yours probably close to 10 years ago now. "The nature and beauty of mathematics". Im now a maths and physics teacher :) love your work
    2
  5005. Congratulatios, 100 !!! 👏
    2
  5006. Frohe Weihnachten nach Down under und vielen Dank für's Video!
    2
  5007. If the outer wheel is 1 through 100 and the inner wheel is 1 through 10, the outer 10 lands on the inner 3.1...(ok my model is too small but its darn close to pi).
    2
  5008. Second 🙂👍
    2
  5009.  @abderrahimelouafi1373  18th
    2
  5010. How can it be that I'm not subscribed to this guy already? I love his pacing and enthusiasm for the subject - looking forward to more content like this!
    2
  5011. 11:53 - [9 4; 17 13] R R R L
    2
  5012. This was great. I wish math in school was this interesting.
    2
  5013. I'm thoroughly enjoying a reentry to calc. Still catching up with the language, so it's a bit bumpy. So the first part was, as always, a challenge. The last section is an eye opener. When these solutions are presented in that way it all becomes very clear. Thank you, please continue!
    2
  5014. There is no evidence to suggest Tesla ever said this about 3, 6, and 9. Instead, the work of Marko Rodin, who seems to be the originator of this approach (mod9 and the "doubling/halving") isn't mentioned at all. Why is that?
    2
  5015. I notice you linked GoldPlatedGoof's video in the description! That was the explanation that first got the original pi = 4 meme to really click for me
    2
  5016. It's amazing to see that for a given number "A" in the series of fractions (1 + 1/2 + 1/3 ... + 1/N) exists such N that may sum or exceeds "A" At first glance, (1 + 1/2 + 1/3 ... ) looks like it should converge, and I remember myself in the school time using QuickBasic testing that this series of fractions indeed diverge. Nice! Burkard, thanks a lot 🤗
    2
  5017. Hello Mathologer, I have a small (and probably simple) question for you: what is the surface area of a ruled surface delimited by 4 points in 3D space? All (4) edges are straight lines. Can you make a quick video about that, or point me in the right direction? Thanks a lot. :-)
    2
  5018. There is a problem with the extra magic of the geomagic square on 23:06 ... 6 and 9 add to 15, but they don't form the 4x4 without the corner... There might be more such examples, but this is the first I found
    2
  5019. @6:19 I fought 4k-1, before ML said 4k+3...
    2
  5020. I have to watch Toroflux again ... and also think about the Archimedes wheel paradox. (Escher would have appreciated this, too.) Yes, this raises all sorts of inquiries.
    2
  5021. That is wonderful to see. This way of rendering mathematics is very very pleasant. Well, you make it look joyous and effortless to yourself. The animations allow the imagination and working - mathologising - brain to gen and enjoy. Thank you.
    2
  5022. SS marijuana nirvana
    2
  5023. Same! Though in retrospect it makes sense: A triangle (or n-gon) is a sequence of points in 2-space. That can be written as a sequence of complex numbers, and taking the DFT of a sequence of complex numbers is sort of an obviously-possible thing to do. Extending it to 3-space seems to involve quaternions, but a quaternion Fourier transform is apparently a thing, so.
    2
  5024. Such a great video so far🐦 I'm halfway through and it's already one of my favorites! Nth differences are close to my heart.💙
    2
  5025. My brain refuses to do math when hearing "dollars" so many times!
    2
  5026. The answer is yes when putting equal masses at the vertices only. So, it is the vertex barycenter. So it is (generally) not the barycenter when putting equal mass(distribution) on the edges or the complete n-gon.
    2
  5027. Speaking of magic squares, does anyone else remember the line: "Atop a crystal pyramid spins a glowing polyhedron."?
    2
  5028. The 4 ways to write 18 as the sum of 2 squares are the 4 ways of (+/-3)^2 + (+/-3)^2 For 2020, there are 8 ways (odd factors are 5 and 101, both are 'good'). The 8 ways are composed of the squares 16^2 and 42^2, with all combinations of signs and both orderings (ie 16^2 + 42^2 is a different solution from 42^2 + 16^2).
    2
  5029. Wow, this is beautiful!
    2
  5030. I have used a circular slide rule in the 1970's. There were two disks and a transparent "wing" with two lines on it. You place the 0 on the inner wheel against the wholesale price on the outer wheel. Then you turn the wing so that the "without VAT" line aligns with the desired margin on the inner wheel. You then read the retail price on the outer wheel at the "with VAT" line on the wing. It became obsolete when the VAT rate changed.
    2
  5031. Mostly the whole video worked for me but I feel like the Bernoulli numbers themselves were left hanging as some kind of unstated mystery. I happen to know that these are themselves non-trivial to calculate directly because I've stumbled into this before when I was investigating sums of powers using discrete calculus and falling down some wikipedia rabbit hole. My background is basically up to calc2/3 and 20 years of recreational mathematics in whatever random direction it took me.
    2
  5032. My brain crying watching this video...
    2
  5033. Sad but memorable part: the seal. Happiest memorable part: see all.
    2
  5034. Incredible job, as always! Your visual derivations make many of the familiar ideas so much more magical to me. I’d love to see you talk about partitions, modular forms, and theta functions at some point :)
    2
  5035. 0:25 Have you heard of Moessner's Miracle? No? I thought so. It's not a thing normal teacher would tell you
    2
  5036. This is simply beautiful. I plan to use this when I next come across a cubic or quadratic
    2
  5037. It's a subtle point (and easy to miss), but you always measure the path from one card to the other card in the "clockwise" direction - (going in ascending card order but looping from king to ace). So the path starting at 6 and ending at 10 has length 4, but the path starting at 10 and ending at 6 has length 9. The starting and ending cards are important here! The one you keep from this same-suit pair is not an arbitrary choice! You find the smallest distance path between the two and you keep the end card of that smallest distance path, putting the start card at the front of the cards you give your assistant. To be fair, you could choose to measure in a counterclockwise direction if both you and your assistant agreed with it beforehand, but in this video, Mathologer is using a clockwise direction.
    2
  5038.  @MuffinsAPlenty  Ahh yeah that clears things up, thanks.
    2
  5039. I gotta admit, that was an awsome proof. Not long-winded, just windmilled.
    2
  5040. Windmills.... I sense don quixcote
    2
  5041. 7:24 indeed very well pronunciation. I did not even notice an accent.
    2
  5042. There seems to be a lack of photographs of this Indian man. All the videos about Ramanujan use the same picture. 😮
    2
  5043. My eighth grade science teacher Mr Lefebvre gave me Calculus Made Easy and it was life changing. Such an amazing introduction. Very cool to hear it was your intro as well.
    2
  5044. Cant you use these designs for future tech
    2
  5045. Wooooooow Jesus Christ bless you
    2
  5046. Steve from Gamers Nexus is giving you the Thumbs Up
    2
  5047. @Mathologer  I didn't catch the error but weirdly I knew what was it about by the timestamp, memory for useless stuff paid out this time.
    2
  5048. I came up with my own mathematical microscope that started as an equation I made that generates a very good often perfect approximation for some decimal and then I modified it to graph all multiples of that approximation and show the line of best approximations my equation works by taking 1/ the fractional part of 1/ the fractional part of the original decimal and making that the numerator so 0.5 becomes 2 which becomes 0 which would become infinity but any 0 I turn into a 1 so it becomes 1 then I use this numerator to get a matching denominator by numerator/ fractional part and I can round both to get an approximation or keep both as an improper fraction for example 5/8 becomes 1.6 which becomes 0.6 which becomes 1.666666... and the matching denominator is 2.666666... you could run this recursively and multiply out common factors or you could scale both, by the way for pi it generates 15.99644067 / 112.976160847 which when rounded is 355 / 113 and it comes up with 2 / 3 for phi and most types of numbers get nice approximations but it breaks for negative numbers for 22/23 it generates 22/23 but for -22/23 it generates 1/23 -1, here is the desmos graph I made with it https://www.desmos.com/calculator/bukv9u8w2j
    2
  5049.  @landsgevaer These are the coefficients of the continued fraction for pi. The coefficients for the continued fraction for phi, on the other hand, are 1,1,1,1,1,1,1 and so on, which is what makes it “most irrational” because every next approximation is as little more accurate as it can be. IIRC this is what the “most irrational” video discusses.
    2
  5050. 🕉
    2
  5051. Wow 😮 Beautiful and very ingenious ❤
    2
  5052. Preparation for fibonacci february
    2
  5053. Holy cow, the mathologermobile must have gotten upgraded into an fzero machine at the end of the video, complete with an fzero soundtrack
    2
  5054. At 0:53 you made a false statement. The 6 little triangles are equal in size !!! All what you presented here did we all discover in the planimetry classes in the secondary school at Büren an der Aare in 1958 ! It was just a fun class of mathematics. No use at all but a lot of satisfaction.
    2
  5055. 2:40 Puts Bible and Qaran to shame!
    2
  5056. Whoa, whoa, wait a second! There's another picture of Euler out there?? 😲
    2
  5057. Super knowledge video
    2
  5058. .
    2
  5059. (took me a couple of seconds to get your T-shirt, I'm ashamed to admit)
    2
  5060. Love your video. One minute remark: your derivate quotes (as in f') are upside down...
    2
  5061. Long days back I'm seen your updated video welcome 🙏👍
    2
  5062. In the original animation, when the squares and triangles are both visible at the same time, my brain tries to interperet it as an isometric wireframe of some sort of prism.
    2
  5063. Amazing video!! Even the music at the end is beautiful! I wonder if somebody could tell me from whom it is?
    2
  5064. Regardless of which proof people prefer, your explanation is, as usual, on point.
    2
  5065. Magical !!!
    2
  5066. Man to see my all-time favorite obscure K-drama come up in a Mathologer video apropos of nothing is going to mess with my head for a while.
    2
  5067. please make a video on bahrmagupta how he created algerbra
    2
  5068. This is a great piece and exposition of calculus, differential equations and continued fractions. The only thing lacking is a reason for guessing that the great Ramanujan approached the problem this way himself. Do we have even a hint of a reason that this is even remotely his own approach?
    2
  5069. Cool ! So from Euclide's algorithm, we also get decomposition of products into squares: 38*16=2*16^2+2*6^+1*4^2+2*2^2 p*r0=d1*r0^2+d2*r1^2+d3*r2^2+...+dn*gcd^2 with gcd<rn<...r2<r1<r0
    2
  5070. Hello Mathlodger. I love your videos and I always learn a lot including how to problem solve. Can you please do a video on how the Taylor Series can be dervied using your usual creative ways?
    2
  5071. God that scratches my brain hole in a COOL WAY!! LOVED IT!! (I think I have always loved math - said the musician!)
    2
  5072. Ooh the discussion of Leibnitz's notation cries out for a video on surreal numbers and non standard analysis and why infinitesimals work to a certain extent
    2
  5073. Make one video on papers of grigori perelman on pionere conjecture
    2
  5074. Sir your laugh in between seems you read some viewers minds 😅
    2
  5075. The lesson is: if you're going to be a historical mathematician, always invent the computer first. Will make your life a lot easier! Failing that, at least invent the printing press so your writing has a better chance of surviving.
    2
  5076. This would make mathematicians burn during the middle ages 🔥
    2
  5077. As for the painting, I can't honestly say that I worked it out in my head, nor in any short time. But long-hand, this was my approach. Each square in numerator can be expressed in the form: (10 + a)^2 = 100 + 20a + a^2 for some value of a. Five such squares all together, so we have: [ 5*100 + 20*(0 + 1 + 2 + 3 + 4) + (0^2 + 1^2 + 2^2 + 3^2 + 4^2) ] / 365 = [ 500 + 20*(10) + (30) ] / 365 = [ 500 + 200 + 30 ] / 365 = [ 1530 ] / 365 = 2
    2
  5078. Don't you mean [730] / 365 = 2 ??
    2
  5079. 25 dec = 25×12 = 30 × 10 = 30 oct So 31 oct= 25 dec =30 oct😎
    2
  5080. 690k subscribers. Nice
    2
  5081. Calculus is an introductory drug to make you an analyst. Calculus, not even once.
    2
  5082. Too beautiful. Keep making such videos
    2
  5083. Hero is the Latin form of the Greek Heron. In the same way, you get Nero = Neron and a choice of 616 or 666 for 'the number of the beast' from Revelation. Both are represented in ancient manuscripts. Edit: that might not be the originally intended meaning of the number, of course.
    2
  5084. You only say it’s disgusting because it’s AI
    2
  5085. ​ @VagueHandWaving  It's an interesting statement. If you showed 100 people an set of images (that either weren't AI generated or it isn't obvious to tell) and told them to rate each on a scale from 1 to 10, but you initially told 50 of them the image was AI generated, those 50 people on average would probably rate the set of images lower than the other 50. I don't know. I didn't realize the image was AI generated until I read the comments. Not sure why he didn't use a real image of her.
    2
  5086.  @nightytime  It became obvious that AI was used once other images were shown in the video because the image from the thumbnail is completely different. On top of that, all images were obviously animated with AI. Please keep in mind that, as with any technology for creating or manipulating images, over time people will simply learn to recognize it. Personally, I have been paying close attention to AI for many years, back when everything made by an AI was incomprehensible and utterly disgusting.
    2
  5087. Thank you a lot professor Burkard, you are truly enlightening my life
    2
  5088. That bell between chapters ..... it gave me goosebumps !
    2
  5089. Awesome!!! Another great video!
    2
  5090. It honestly didn't feel like 45 minutes. I am always so glad that your videos make me not only understand but fascinated with complex math problems! Much love from Poland :D
    2
  5091. I first learned about the calculus of finite differences from reading Martin Gardner. I had no idea how rich the subject was. The Newton-Gregory formula is a real gem.
    2
  5092. Proof of the rainbow square corners: I'll be referring to the example at 13:00 a lot, as "the canonical example". All right, we need some lemmas. Lemma 1 ("alternating stripes"): In any 1xN subregion in which no run of three squares have three different colors, the region alternates between two colors. Example: the top row of the canonical example. Proof: consider the two leftmost squares (rotate WLOG an Nx1 region into 1xN). Since they are inside a 2x2 subgrid, they must have different colors; call them A and B. Consider now the third leftmost square; since no three run of squares have three different colors, it must also have color A or B. But since it is adjacent to the second leftmost square, which has color B (WLOG), it cannot also have color B; hence it must have color A. Consider now the second, third and fourth squares; they must have colors B, A, B by a similar argument, and in general every square must have the same color as the square two squares to its left. Corollary 1: if the board has no 1x3 region (in either orientation) with three different squares, no two corners have the same color. Example: swap yellow and green in the bottom row of the canonical example; it now conforms to this condition. Proof: by the same argument as above, every square must be the same color as two squares to the left, and two squares up. Hence if you divide the grid into disjoint 2x2 squares (this is possible since the side length is even), each square has the same color pattern as every other square. But then the bottom left corner of the whole square has the same color as the bottom left of the top left 2x2 square; and likewise, the corners of the top left 2x2 square has the same color as the corners of the whole square. But the corners of every 2x2 square are all different. Lemma 2 ("zig-zag bands"): if there is a vertical run of three squares of three different colors, this determines the complete coloring of the containing three rows. Likewise for horizontal runs. Example: the lower three cells in the leftmost column in the canonical example. Proof: consider three squares with coordinates (r-1, c) and (r, c) and (r+1, c), colored A, B and C respectively (example: yellow, red, green). Consider now (r, c+1). It is in a 2x2 square with (r, c) and (r+1, c), so it cannot be color B or C. It is in a 2x2 square with (r-1, c) and (r, c) so it can not be color A or B. But then it must be color D (e.g. blue). Consider now (r-1, c+1). It is in a 2x2 square with (r-1, c) and (r, c) and (r, c+1), colored A, B and C, so it must be color C. Likewise, (r+1, c+1) is in a 2x2 square with (r, c) and (r+1, c) and (r, c+1), colored B, C and D, so it must be color A. But note that (r-1, c+1) and (r, c+1) and (r+1, c+1) are themselves of three different colors. Repeat this analysis, extending to the right edge. Now mirror this analysis and also extend it to the left edge. The pattern you will see is that the middle row (row r) alternates between colors B and D, and the two other rows alternate between A and C, one row starting with A and the other row starting with C. Every intersection of this three-row band with a column contains three squares of three distinct colors. [I call them zig-zag bands, as the yellow color zig-zags in knight's moves around the alternating red-blue central row; so does the green color.] Corollary 2: there cannot both be a vertical and a horizontal zig-zag band (= run of three squares of different colors). Example: none, they don't exist ;-) Proof: the middle column of every vertical zig-zag band is an alternation between two colors. Every column intersects a horizontal zig-zag band at three squares of three different colors; this includes the middle columns if vertical zig-zag bands. Three squares cannot simultaneously have exactly two and exactly three distinct colors. Corollary 3: if there is a horizontal (vertical) zig-zag band, every row (column) is an alternation of two colors; and each row (column) uses the opposite pair of its two neighbouring rows (columns). Example: the rows of the canonical example. Proof: let a horizontal zig-zag band be given, and consider its central row, colored A and B in an alternating pattern. Every square in an adjacent row is in a 2x2 with two squares from the central row, colored A and B, hence the two adjacent rows must be colored C and D. But then by lemma 1 they must alternate. But then by a similar analysis their neighbouring rows must alternate, using colors A and B. Apply this reasoning inductively until you get to the edges of the board (in both directions). Theorem: in every rectangle with even side lengths, with each tile colored in one of four colors such that every 2x2 square of tiles has all four colors present, the corners have different colors. Example: the canonical example. Proof: if there is no horizontal (vertical) zig-zag band, this is corollary 1. If there is a zig-zag band, the top row alternates between colors A and B, and since the vertical side length is even the bottom row alternates between colors C and D. Since the horizontal side length is even, any row in which two colors alternate (that's all of them), the leftmost and rightmost squares have opposite colors. This applies in particular to the top and bottom rows, and so the four corners have colors A, B, C and D, all different. (4n-by-4n squares are all examples of 2m-by-2k rectangles, if we choose m = k = 2n.)
    2
  5093. Can the fact that the elastic band wraps around in a ‘log-like’ manner be derived from Hooke’s law or other such equation relating the applied force to the stretching/elasticity properties? Would be nice to see this physical law translated to the math of the wheel.
    2
  5094. Your excitement is absolutely contagious! What an enjoyable watch!
    2
  5095. Pretty sure that you are the first to comment on this glitch :)
    2
  5096. Saw this more than a month ago. Watching again because coronavirus. Apparently I have forgotten everything.
    2
  5097. Ein super tolles Video ( wie immer). Wie erstellst du die Animation mit der Fibonacci-Spirale am Ende. Würde gerne auch Animationen dieser Art erstellen.
    2
  5098. 2:00 our teacher showed us this proof in 5th grade that is 8 years ago
    2
  5099. That proof for width 10 triangles was absolutely beautiful!
    2
  5100. You should get yourself a :CueCat as a technological mascot to go with your website. The best part is that it's cat-shaped. It might even still work, if you can find a computer with a PS/2 keyboard port to use it with.
    2
  5101.  @Mathologer  Oh, it's much weirder than that. It's not a remote control but a barcode reader that you slid along barcodes. These things were given away in the US with magazines and at RadioShacks as part of some kind of weird advertisement venture involving the Internet. But they looked cool. At one point I had 10 of them, and probably still do, if only I could find them.
    2
  5102.  @tomkerruish2982  I use the ReaderWare programs to catalog my collections (books, CDs, DVDs) and it uses a CueCat as a free barcode reader to scan the UPC or ISBN barcodes from the item. The obfuscated input was quickly reverse engineered and code published openly to decode the typed mess and recover the actual scanned string.
    2
  5103. I think it's most easiest and simplet Explanation and Vid on youtube . Thanks a lot
    2
  5104. I've only a slight ideawhats going on but I find it fascinating
    2
  5105. Love Your Mind, My Friend,,, Keep up the Incredible Work; You have a #1 Fan Here!!!
    2
  5106. That is a great video! I had seen most of the various pieces before but this video puts them together very nicely. I used to be a bit wary of the Bernoulli number but have now started to appreciate them. I am looking forward on any upcoming videos on the Euler-Maclaurin formula and would love to see more on the connections to the Riemann zeta-function. Terence Tao has this nice paper that puts sense into "identities" like 1+2+3+4+5 = -1/12: https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ And: I have a PhD in computational group theory.
    2
  5107. It reminds me of a math problem a teacher gave us in school: At noon, the hour and minute hands of a clock both point at 12. How long will it be until the two hands are again exactly aligned? I solved the problem very easily, by thinking: "The hour hand goes around once in 12 hours, and the minute hand goes around once in 1 hour. 12-1 = 11, so the answer is 12/11 hours, or a bit more than 1 hour, 5 minutes, and 27 seconds (and 3/11 of a second, if you want to be precise.)" The teacher didn't accept my answer because she had a much more complicated way of doing it as a sum of an infinite series. 1 hour + 1/12 of an hour + 1/144 of an hour ...
    2
  5108. 18:56 The second method is the first I learned - it's applicable to the Lo shu magic square.
    2
  5109. Brilliant! Just remember the legend about Gauss that he noticed that 1+100=2+99 and so on, so (100+1)*100/2 — arithmetic progression sum. Anyway, all the staff you have shown are real mathematical pearls presented in vivid mannear as you would do. Thanks a lot!
    2
  5110. 24:30 that pattern of adding looks a lot like pascal's triangle, viz the binomial which kicked us off!
    2
  5111. Well, not everyone got their copy of Mathematics Today on time. We have to remember that there's not always one inventor of things. You have people whose work is the basis of later work and you have people who are separated in time and space and don't know and don't know of each other. I'm not trying to dilute anyone's credit, just suggesting that there's a lot to go around. Still, pretty fun.
    2
  5112. Well this was really fun!
    2
  5113. Please do "Why There's 'No' Quintic Formula (without Galois theory)" Mathologer style.
    2
  5114. There couldn't have been a better maths video than this. I was amazed how fractions and irration number can be written as infinite series. This video is a proof that "Maths is beautiful!". I completely agree with this statement
    2
  5115. Great video!! 8:28 I have created a quick and inefficient program to animate the magic square construction. (Keeping the link as a reply to this comment as it seems like links are disabled in the comment section)
    2
  5116. I imagine that this must be in some way connected with the fact that triangles, squares and hexagons are the only regular polygons that exactly tesselate the plane, but I can't see how.
    2
  5117. I really liked the Euler-mascheroni constant and the way it linked to -1/12!!!!
    2
  5118. For the last question, engineering degree and computing post-grad, and could follow it all but have just taken your word for the transformation at 40:46.
    2
  5119. I solved this problem some years ago by calculating a formula and copying down in Excel. In the first column it starts with 1. In the second column =SQRT(A1*A2/2). By copying down the formula (and the 1), on the first column you end up with the natural numbers. On the second column the resulting square root. Looking at the numbers it was immediately apparent which were the integer numbers out of the "forest" of fractional numbers. The ones smaller than 50 (1 , 6 and 35) and the one I wanted (204). On the left of the door number (when integer) is the total of houses on the road for each case.
    2
  5120. 20:06 "two vertices are linked if and only if they differ by only one coordinate". This reminds me of binary Gray code : Gray(n) and Gray(n+1) differ only by one bit. Interesting parallel, isn't it ?
    2
  5121. The circular slide rule was in regular use on US submarines as recently as the early '90s to quickly perform calculations related to target motion analysis (TMA). We'd run a string through the hole in the middle and wear them around our necks for easy access.
    2
  5122. Be careful! The "expected value" of losing your life at least once is 1/2 here.
    2
  5123. The MATH shirt with the anarchy symbol is awesome. 👊 Let's show people that math isn't just some oppressive tool used by governments & corporations. We the people can use math to solve real-world problems WITHOUT relying on corrupt bureaucrats.
    2
  5124. 7:00 This looks like a Bubble Sort algorithm
    2
  5125. I like all the tree toppers. 🌟 😊
    2
  5126. Did done been there secrets out
    2
  5127.  @richardgratton7557  right on Richard.
    2
  5128. Watching your videos always pleasure for me.....I really really like your small laughing :) !!!!!
    2
  5129. I'm fast
    2
  5130. I've paused on 5:04, because my answer is YES. I'm definitely sure, that I will do this exactly in 1023 moves. Course I read a lot about Hanoi puzzle, couple of time programmed it and taught other people to program it on the lectures as exercises on recursion, but only last year I bought my own physical version (with 8 discs), what leads me to so brave assertion. Only after couple of times "physical/manual" solution I really grocked key principles needed to solve this in circumstance similar to Dr. Who's. First one is really mentioned recursion but in not very computer form rather intention: you want to move subtower on n-1 discs on auxiliary peg to be able to move the disc just under subtower (n-th) to destination peg. Second and this is very important for us human to not solve recursion every time: if your subtower (including full tower) consists of odd number of discs you very first move should be on destination (for current subtower move) peg, if your subtower consists of even number of discs you very first move should be on auxiliary peg. Having this principles you will be able to move Hanoi tower as a calming exercise/solitaire as I do. Now I will continue to watch your video to find what is Mathlogger solution.
    2
  5131. Big fan from Greece here, thanks for the shout-out! :):) By the way, a friend once asked me "you actually use greek letters in everyday language? They are not just for math?" :D:D
    2
  5132. So a Halden Calculex huh? Remember from numberphile
    2
  5133. Fantastic! Surprise discoveries from the beginning to the end!! If there is the video about the complex solutions would be nice to put some reference in the description or link frame at the end of the video!
    2
  5134. In the Doctor Who episode, even the toymaker got bored of watching the Doctor moving disks, so he accelerated the pattern, making hundreds of moves at once and then the Doctor had to continue. What is the best way for the Doctor to proceed? He either needs to move the smallest disk or make the only other valid move.... but which one? If he knew the move number, then it would be trivial, but here's the question: Given an state within the optimal move sequence, how do you continue? I once decided to see what the Towers would sound like as music. I gave each disk a note in C major, 1=C, 2=D, 3=E and played each note for each moved disk. So CDCECDCFCDCECDCG... etc. Played as simple 8th notes, it gives a very mechanical feel with the CDC making a continuous pattern and the other notes seeming sort of random. It gives a very mechanical vibe, like something you could play over Metropolis.
    2
  5135. Someone please explain Mr Mathologer's t shirt Thank you 😊
    2
  5136.  @HansMaurer.  Omg, Sheldon, right? Thank you so much!!!
    2
  5137. Most memorable: That the no-9s series converges
    2
  5138. The Celestial Toymaker! such a weird coincidence -- i was literally just watching the one surviving episode when this popped up in my notifications
    2
  5139. Seeing the numbers all click perfectly into place over and over is giving me the warm fuzzies.
    2
  5140. The most beautiful part for me was the last proof of sum of no. 9's. The other parts are known and I have read quite a few of articles but those series with 9's... Or any other numbers were never known to me ....and you know how much a headache we get when we know something new and can't figure out about it. And in the end you give me the proof along with the beautiful square visualization with black and white colour... Really made my day... Thanks....
    2
  5141. Has anyone mentioned that houses on the same side of the street - sometimes - aren't numbered sequentially but are numbered all odd or all even?
    2
  5142. Great video! Loved it , hope I can find one of them slider ruler here in India!
    2
  5143.  @DrZip  Hmm, good luck with that :)
    2
  5144. Loved that you used Asturias played by Andres Segovia for this .... Perfect match
    2
  5145. Watch it people! ;)
    2
  5146. You keep using that word...
    2
  5147. Hello professor ❤
    2
  5148. Because you'd then have to include 2 = 2, 3 = 3, 4 = 4, and so on, and while those are all of course true, they provide no interesting insight into anything.
    2
  5149. I think you can extend the proof for the opposite angle sum by using directed angles
    2
  5150. Well, that was fun. I imagine Amazon will be selling a lot of circular slide rules now. Do you have one with a Mathologer logo?
    2
  5151. 545 x 33 = 17985 is the sum of each row
    2
  5152. Great video as always, even after chapter 1 I was amazed. Thanks for this :)
    2
  5153. Even if you discover a new theorem in geometry, it doesn't matter unless you are a mathematician who can publish it. If you are the average person, you will be ignored forever.
    2
  5154. 7:18 FJM seems to stand for Fredericus Johannes Maria, from my very quick Google search
    2
  5155. 24:13 is it already known that you can do a similar thing with the positive integers series? If you graph the partial sums formula of the positive integer series(triangle number formula), then graph y=x, the area between the two intersections of the graph has an area of 1/12.
    2
  5156.  @Ramses_II  Oh, I see. A direction is agreed upon and the end and start cards are chosen with respect to that direction.
    2
  5157. You keep the higher-valued card and give the lower-valued card to your assistant. If there are more than two cards of the same suit, give the lowest-valued one to your assistant as the first card and keep the next value in ascending order.
    2
  5158. Does anyone know off the top of their head whether the infinite sum of all the positive terms in the series converges or diverges? And if it converges, to what?
    2
  5159. 24:27 I was curious about the area of the thin green rectangle that's leftover. For a 5x8 rectangle it's about 0.721 units. Interestingly, as you scale up to higher Fibonacci rectangles, that leftover rectangle gets thinner and thinner, and the rate it gets thinner cancels out the rate that the Fibonacci rectangle grows, so that the area converges to a fixed value of 0.7236. After some fiddling around, I found that the exact value that the area of the leftover rectangle converges to is (phi^2 + 1)/5 or (1 + (1/sqrt(5))/2
    2
  5160. What exactly is your definition of an "algorithm" for a Rubik's Cube? You seem to be claiming that reversibility is a property of all algorithms, which is obviously not the case. For instance, there is an algorithm to simplify fractions, and it will simplify both 2/4 and 3/6 to 1/2. This resembles real Rubik's Cube algorithms. A typical set of algorithms will take any unsolved cube to the solved state, so they are definitely not reversible. Is the "algorithm" here just a fixed sequence of moves that does not depend on the current state of the cube?
    2
  5161. My childhood explorations of Fibonacci and Pascal brought me about 25% of the way to where you took us today. Thank you so much for introducing me to its formal discoverers and a plethora of aspects that I had never considered.
    2
  5162. great video, as always! :) my favorite part was you when you split the larger numbers into 1s in the beginning to show why it balances at 1/3, 1/4 (from the right), etc you're great with visuals!
    2
  5163. 15:58:Each level in the tree adds 1 to the total area. I believe this will be true for asymmetric trees (e.g. 3 4 5 triangles) too. Pythagoras guarantees that the sum of the areas for each level will be the same.
    2
  5164. Magic squares seem to be 2D analog of how Gauss added numbers from 1 to 100. I wonder if there could also be 3D, 4D ...ND analogs.
    2
  5165. I made it to the very end
    2
  5166.  @Mathologer  woot! Mathologer merch!
    2
  5167. ZomB1986: Complete amen brother!
    2
  5168. This channel is criminally underrated
    2
  5169. Happy 100th. Again I love you videos and always watch them Sunday afternoon. Relaxing.
    2
  5170. Wonderful video! I've always wanted a way to find primitive triples only, and here it is! The parent of the 3 4 5 triangle is quite amazing. The triangle is 0 1 1 and the corresponding square is 1 0 1 1. Its three children is one clone of itself and two copies of the 1 1 2 3 square for the 3 4 5 triangle.
    2
  5171. 15:47 Easy: 0 = 0 , the trivial case.
    2
  5172. When I saw the 33 and the 3, I was thinking aha: combine the pattern for a 3x3 with the pattern for 11x11 and you get your 33x33: Copy the 11 9 times over, and then for each copy add 11x(magic number in the 3x3 for that copy). Never imagined the fold-out rule!
    2
  5173. Is there gonna be the Christmas video this year?
    2
  5174. Op👍👍👍
    2
  5175. Thanks!
    2
  5176. Have you got a 3D model of that perfect universe thingy?
    2
  5177. this is a nice hypothesis, and you've hit an important point - the problem does assume that no creditor has proof showing they have a legal reason to receive more (for example, if the three loans were given at the same time with everyone present, and there's a signed document explaining the terms of the loans, saying that a specific creditor will get more in this scenario). it's not that proofs didn't exist back then or weren't accepted by courts (there are many talmudic discussions dealing with legal proof), rather that this specific problem assumes we're dealing with a case without any proof
    2
  5178. Material viscosity is not part of that theoretical construct. What happens when material viscosity = 0 is put in that that theory with zero tolerance? The whole discussion is based on the illusion of theory=reality and the quantum difference...in other words " tolerance of measuring failures"*!* How about that ? 0∆πΩ∞
    2
  5179. It's not entirely related. But this identity came into my mind, when I watched your video. For 2x2 matrices there is a cool trace identity tr(MN)+tr(M^{-1}N)=tr(M)tr(N). I think M needs to be an element of SL(2). But otherwise it's generic and it also has this sum product relation.
    2
  5180. You can create a torus field if you add enough points
    2
  5181. Hexagon: find the 1x1x1 cube that's made of three different colors. If you replace these shapes, you'll like push this cube so it's gone. Now you have sama amount of each color, but one cube less. By that, you can get rid of all this cubes and leave the empty box. On that, walls and floor are the same, so the amounts of colors are equal. But they didn't change the whole time, so on the start, they were equal at start.
    2
  5182. "close enough is fun enough" haha that's a nice saying
    2
  5183. Long time man
    2
  5184. This video jogged my memory about another mechanical arithmetic device. Before handheld calculators became commonplace the addiator was popular for addition and subtraction. I am not sure how you could do a video about it, perhaps by discussing how modular arithmetic allows you to subtract by adding.
    2
  5185. 1+1/2 1+2/3 1+3/4
    2
  5186. 18:41 Stretching... reverse of squishing... yes good please keep the mathematical insights coming
    2
  5187. no non-trivail special number when colour >3, i.e. can't predect far away cell colour when no. of colour >3. but i am confused may be infinith cell can divide every middle term by 4,5 and on. By the way great content loved it ,signin out Dev Agarwal from India.
    2
  5188. probably the best Mathologer video.
    2
  5189. Heron`s formula is told in Russia
    2
  5190. 37:58 We have a self-similar Fractal here with a dimension of 4•Log(3) = 1.908485... very close to 2, but strictly smaller, which immediately proves the Fractal’s area (i.e. 2-dimensional measure) is zero.
    2
  5191. Of course you weren't in Toy Story, only in Futurama.
    2
  5192. p(666)=11956824258286445517629485
    2
  5193. Euler slides in down 1/x - Oh, there's Euler. There's no escaping him. So many favourites but there's no escaping Euler!
    2
  5194. I made it to the very end.
    2
  5195. Love your work!! Beautiful & Inspiring stuff :)
    2
  5196. That's a "Pizza Phi" on your T-shirt, isn't it?
    2
  5197. Looking at the thumbnail I went like "Einstein Chacha!". Hopefully the Indian viewers will understand, hehe.
    2
  5198. Can't believe I've never seen the Oresme trick applied to the sum of 1/n^2. I guess it's much more commonly done via comparison to 1/n(n-1), which gives a telescoping series. (of course you need to treat n=1 differently so the overall bound is 2)
    2
  5199. Wow! Wonderful discovery! Thanks a lot for the video!!! I like the second proof more, due to its simplicity. But the first proof is also inspiring!
    2
  5200. There is another example for the paradox of finite volume vs. infinite area: imagine slicing a finite marble cube into thin tiles to cover a floor. If the slices are infinetly thin the floor can be infinite large
    2
  5201. Thanks Mathologer to shown the proof Brahmagupta formula because I live in India 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳. And I am watching all vedio. Your maths every vedio is mind-blowing🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰🥰
    2
  5202. Can I just say that, while your videos are always excellent (this one is no different), what I really want to know is, where I can get those cool t-shirts you always have? :)
    2
  5203. The sum of sums.. e=1😎
    2
  5204. 5:50 Even though "final formula", I can still smell some a^2-b^2.
    2
  5205. My introduction to the parity of permutations was playing with the 15-puzzle as a kid.
    2
  5206. The unconventional greater hanging ‘ugly’ towers was amazing. Who would know to even look for one!
    2
  5207. CHALLENGES! 3:20 Let y = 1/x. Squishing vertically by a factor of c corresponds to replacing y with cy. Stretching horizontally by a factor of c corresponds to replacing x with x/c. Now substitute: cy = 1/(x/c) cy = c/x y = 1/x And we recover our original shape. Everything in the interior stays in the interior for obvious reasons. 8:53 A(X) ranges from 1 to X, and has width X - 1 and height 1/X on the X end. A(Y) ranges from 1 to Y, and has width Y - 1 and height 1 on the Y end. The A(Y) shape needs to squish by a factor of X to fit. And it should stretch by that same factor of X. The total width should be (X - 1) + X(Y - 1) = X - 1 + XY - X = XY - 1, exactly the width of the shape from 1 to XY. Or you can use FTC and notice that the antiderivative of 1/x is ln(x) + C, and that ln(X)+ ln(Y) = ln(XY). Provable with the integral definition but that's for later [°1]. 10:48 Just use the subtraction formula! N = 0: log(X) - log(X) = log(X/X) 0 = log(X^0) = log(1) N = -1: log(1) - log(X) = log(1/X) = log(X^(-1)) N = -M: Nlog(X) = log(1) - Mlog(X) = log(1) - log(X^M) = log(1/(X^-M)) = log(X^N) Rationals is slightly trickier. R = P/Q: log(X^(P/Q)) = P * log(X^(1/Q)). Now multiply and divide by Q (the analytic version of shape-shifting I guess) P * (Q / Q) * log(X^1/Q) = (P/Q) * log(X^(Q/Q) = R * log(X). For reals, I'm just gonna say log is continuous. I'll leave a non-rigorous "proof" for later [°2]. [°1] Integral definition Suppose ln(x) was defined by this integral: Integral[1/t dt; from 1 to x] We want to prove ln(xy) = ln(x) + ln(y). Integral[1/t dt; from 1 to xy] can be split into two: Integral[1/t dt; from 1 to x] + Integral[1/t dt; from x to xy] The first half is ln(x) by our definition. The second requires a little work. I want to take x to 1 and xy to y. Setting u = t/x will do the trick. u = t/x du = dt/x lower bound = 1 upper bound = y Integral[x/t * dt/x; from x to xy] Integral[1/u du; from 1 to y] which is exactly equal to ln(y). Notice what we did here. We split the integral into two, and rescaled one of them to fit. That's the reverse of the shape-shifting magic! And the u-substitution is that magic shape-shifting squash and stretch in the geometric proof! [°2] Probably Continuity of the Logarithm We know the area function is defined for each rational number. Clearly the area function is also monotonic since if A(x) < A(y) if and only if x < y. I can approximate any real number using a sequence of strictly increasing or strictly decreasing rationals. For example, I can approximate sqrt(2) with the sequence 1, 4/3, 7/5, 41/29, 239/169, ... from below, or 2, 3/2, 17/12, 99/70, 577/408, ... from above. Since A(x) is monotonic, we expect the first increasing sequence to output areas that increase, and the second to output areas that decrease. The LH limits and RH limits exist, and they can't explode because LH limit <= RH limit. Also, the integral of a function is usually continuous, especially of the integrand itself is continuous. We know quite easily 1/x is continuous everywhere ln(x) is defined (the interval (0, +inf]) so we expect ln(x) to also be continuous.
    2
  5208. Definitely worth pondering ... :)
    2
  5209. I like to think of the case where the smaller circle is inside of the larger circle as a variation of the circle outside the circle. For the case where the smaller circle is just 2/3 of the larger circle in terms of diameter, we just double the diameter of the smaller circle. getting 4/3. Then we need to find the diameter of the "other" circle. we subtract that from 1, we get -1/3. This means we can say that the setup means a circle of diameter 2/3 is rolling around a circle of diameter "-1/3". Their ratio is -1/2. Using the formula we add 1, so we get 1/2. So 1 complete roll of the smaller circle is just 1/2 of the "larger" circle.
    2
  5210. Gotta get that comment engagement tsfrfromm
    2
  5211. You made my day!
    2
  5212. 24:25 Wild guess: The ”parent” of the 3-4-5-triangle has irrational side lengths; so, no matter, how much you scale it up, they will never be integers. 🤔
    2
  5213. Most memorable thing to me is no integers among the partial sums. How the infinite and integers could so closely intermingle but not connect is amazing. Great video. Thank you.
    2
  5214. #മലയാളം
    2
  5215. Another great video to share with my DP HL students. Thank you!
    2
  5216. wow!
    2
  5217. Most memorable: Visual representation of gamma and how you can see its greater than 1/2 and less than 1
    2
  5218. THIS is the Calculus video I was looking for. THANK YOU SO MUCH.
    2
  5219. Is this an intro to a NIN song or am I going more mad?
    2
  5220. ham first ham first ham first!
    2
  5221. I made it to the very end, very entertaining video
    2
  5222. "Playing around with my snakes"💀
    2
  5223. The music spooks me
    2
  5224. Maths is one of the most beautiful thing i have ever seen apart from physics, music, and my girlfriend.
    2
  5225. I am here for the aha fibonacci moment
    2
  5226. I just encountered this today, and it made my day as well! And I'm not a math guru in the slightest. This is fascinating and beautiful! Thank you for sharing this and showing it in a way that I was able to understand!
    2
  5227. I like easy videos like this. The other day I fell asleep with the difficult one (which is also acceptable for me)
    2
  5228. I just started the video, so I bet this is acknowledged, but having just seen this clip for the first time, isn't it weird that he notes it's a binary number? The argument works perfectly with a decimal number lol, feels like it was "written" to say "this is a maths movie!"
    2
  5229. Let's find out together... In five easy chapters. Such a mathologer thing to say...
    2
  5230. wow amazing, just amazing, you sir just made me go and open my rings and fields notebook and got me inspired so much
    2
  5231. Given a real shoe (with a given number of eyelets and fixed spacing) and given a particular lacing pattern (like for example the criss cross), is there a way to actually compute the overall lenght of the lacing? I always find myself struggling when I put laces on my shoes and the two loose ends of the string are either too short or too long to make a stable knot, or worse, of two different lenghts. It would be nice to know in advance how much string I'll need so that I can make sure that the one I have will fit.
    2
  5232. Sometimes I wonder what Ramanujan would have been capable of if he had grown up being taught contemporary mathematics. Truly a monster that man.
    2
  5233. That was fun, thank you! As an aesthetician, I found the first method to be intricately beautiful, as a novice mathematician I loved the simplicity and directness of the second.
    2
  5234. Educated in Ontario, Canada, here, with a Physics degree, 4 years of writing Putnam through university and a lifelong passion for recreational mathematics. I had never heard of Heron's Formula until I was over 60. So it not only isn't being taught - it seems to have not been taught for at least a half century.
    2
  5235. Interesting! Yesterday I just found a generalized version of Geometric Mean Theorem when I was trying to find an easier and more intuitive way to understand the trigonometric addition formula. I believe my finding is too trivial and probably discovered by many a long time ago but I still can’t help commenting because it’s such a coincidence. 😂
    2
  5236. Fantastic!
    2
  5237. u r simply the best!! 🤸🤸🤸💕💕💕💕🙏🙌🧘
    2
  5238. I have a Jeppesen Computer from my grandfather who was a pilot since the 1930s. It was/is used to determine airspeed for a leg of your flight to find your way long a map heading. I just pulled it out to learn more about it after watching this video. It appears to be capable of these calculations too. Thanks.
    2
  5239. I learned of Ptolemy’s Theorem, from a Numberphile-video 🙂.
    2
  5240. The scheme works if the number of colors is prime. This is due to the fact that the binomial coefficients for prime numbers are all divisible by that prime number. (proof is left to the reader). From that, the special triangles are of size p^n + 1.
    2
  5241. 4:25 Isn't it a form of the birthday problem? If so, then perhaps we can have a 99.9% certainty with only a few hundred thousand people, or something like that.
    2
  5242. 1905 was Dominated by Einstein
    2
  5243. Absolutely incredible
    2
  5244. I really enjoyed all of your videos. As for your question I studied basic mathematics at Lund University for about 1½ year thirty years ago. I watched the full video and would need to pause and think more when revising it in order to be honest with that I get all of it.
    2
  5245. omg that looks like half a semester's worth of lessons.
    2
  5246. Can you use this to arbitrarily bump up and down hyper-operations?
    2
  5247. So if you start with the 2x4 in the middle, and fully tile that which forces the outside frames to be be 2x3 and 2x2 on each side. So then it comes done to how many ways can you tile the 2x4 in the middle. Four horizontal, four vertical, 1 vertical 2 horizontal 1 vertical, 2 v 2 h, 2 h 2 v = 5 permutations possible. So that’s 3*2*5*2*3 = 180 plus the 162 Is a total of 342 But this time I’ll say at least 342 in case I find more 🤔 What happens if you tile 9/12 and 14/10... no that gives you an uneven number of green and red. 342 final answer 😇
    2
  5248. This made me wonder... Is the sum of a series invariant under a rearrangement where the maximal distance any term is translated is bounded? Too lazy to google myself...
    2
  5249. In addition to always trying to generalize from base 10, I also always try to mentally convert facts about pi to tau, which is more natural and really should be the number we all talk about. Much easier to do in realtime here then trying to figure out the base 10 generalization
    2
  5250. Oh dear. It turns out, near the end of the video, that the special treatment of base 10 came in the choice of number of leading terms, and on fact from the presenter, as usual. I was hoping to learn that a similar pattern holds in all bases for all leading term counts somehow. Still, it was interesting to see where the coincidence came from; it was certainly more esoteric and surprising than usual.
    2
  5251. But a good solution should be indifferent to the form of the debts. If you want to borrow money from me I should get paid back the same amount if I lend you $100 and become your creditor or if I say: you can borrow $100 but I want you to pay $50 each to my two kids. This system fails that.
    2
  5252. removing two squares of each colour can block a tiling if the two squares removed cut one or more of the corner squares out from both sides (say, if we remove a2, b1, g1 and h2).
    2
  5253. Thanks
    2
  5254. 12:49 I think the first time I encountered the harmonic series, was about 3 years ago. I just started to learn programming in c# and I programmed some for loop that calculates some some big number of terms probably around 200,000. The answer was 20 and something and when I increased the number of terms the sum barely increased, so I thought I found that the harmonic series equal that finite number. With late middle school/early high school understanding of math, I don't think it's obvious that the harmonic series diverges, except maybe with the incorrect argument, that because this series has an infinite number of terms, it's sum can't be a finite number.
    2
  5255. Who is the guy on the thumbnail?
    2
  5256.  @Mathologer  Thank you. I had a feeling you would respond.
    2
  5257. Stay discreet
    2
  5258. ❤❤❤❤❤ ❤❤❤❤ ❤❤❤ ❤❤ ❤ Not just a pretty face
    2
  5259. The most memorable fact for me was that the sum slowly diverges without stepping on any of the infinite integers. That's just amazing!
    2
  5260. One thing I find fascinating is the connection between the Hanoi graph and the Sierpinski triangle. I wonder if the number 466/855 fits into the Sierpinski triangle in any natural way, without regard to the Tower of Hanoi puzzle.
    2
  5261. In 32:32 he said multiplied by 2 and edited in multiplied by 1/2.
    2
  5262. Wow!
    2
  5263. There was no Easter egg at the end, but I always watch the entire (always wonderful) video(s). Thank you, Mathologer!
    2
  5264. 14:02 "The universe makes sense again"...hahaha This reminded me of the winding number and the argument principle. Happy New Year!
    2
  5265. Funnily enough, I have had to prove Fermat's Christmas theorem in an exam this year
    2
  5266. why did I read the thumbnail as “fimple”😭 math letters make me crazy
    2
  5267. At 6:42 ... that is truly amazing. I have seen Ptolomy's theorem before, but this time it really hit me.
    2
  5268. Today has been something else for me. My wife and I had a break through in our relationship. We came to terms with something incredibly taxing on us. Second, I discovered a video on the Richat structure in Africa. Stirring a curiosity in my mind I haven't felt in... maybe ever. Third, through that curiosity I found you. I have never sat through a lecture, not without fidgeting, smoking, going for a snack. Nothing would keep my interest. I barely took a breath watching your explanations of 3,6 and 9 and now this video. As someone who has felt like he was awful at math, only ever memorizing material. Even in college. I now understand why. The way I learn was never catered to, in the least it seems. I perhaps could've spent my whole life not understanding why we use Pythagoras for things, other than it just works. But this has shown me the beauty in mathematics. No wonder so many strive to this. Wow. I feel like an empty cup, eager to be filled. Thank you for making these, genuinely from the bottom of my curious little heart.
    2
  5269. Dis is pure beauty ♥️♥️😍😍 much love math mathologer ♥️♥️♥️♥️
    2
  5270. Well, here's my "ahaa" recorded: [Spoilers ahead] The numbers of tilings with identical 2×1 pieces of 2×n plains seem to match up with the Fiblnacci numbers, such that for a plain of 2×n the number of possible tillings is F(n+1). So let's see why. Before proceeding, note that any 2×n board can be covered with "vertical" tiles, whose height is 2 and fills completly the 2 unit long side of the board. Also, note that any 2 parallel tiles can be rotated over their joined centre (so that vertical tiles now become horizontal). These are the only two ways of tilling a 2×n board. For the 2×0 board the one possibility is to use 2 tiles. Duh. For the 2×1 board the one possibility is to use one vertical tile. Duh. For the 2×2, things start to grt interesting. Now, you can use 2 vertical tiles to make the all-vertical tilling, or rotate the 2 to make a tilling with 2 horizontal tiles. 2×3 allowa for the all-vertical tilling and for 2 other tillings: tiles 1 and 2 may be rotated, or tiles 2 and 3 may be rotated intead. There still isn't enough space to rotate more that one pair of tiles. Vertical: 1; 1 horizontal: 2; 2 horizontal: 0. With a 2×4 board, the results are as follow: vertical: 1; 1h: 3 (=4–1); 2h: 1. K, we need formulas. For the all-vertical tilling, it is allways possible, and thus v=1. For 1 horizontal (one pair rotated, h1), since the pair itself is 2 units whide, it is n–1 for n greater that 1 (and I guess it also works for 1, as it just spits 0), where n is one side of the board (2×n is the board). For more than one pair of horizontal tiles, this gets complicated. For 2×4, there's one possibility. For 2×5 there's 3. I will further subdivide these into groups defined by how much space there is to the left of the leftmost horizontal tiles. If they are right touching the side, then there's as many possibilities as for covering a 2×(n–2) board, due to the space they take up. If they were 1 unit away, there would be as many possibilities as for a 2×(n–3) board. Since I cannot type epsilon, I will just explain the infinite sum that is about to take place. Say t(2×n) is the amounth of ways to tile a 2×n board; then t(2×n) = t(2×n–2) + t(2×n–3) + t(2×n–4) + t(2×n–5) + t(2×n–6) + t(2×n–7) + … + t(2×3) + t(2×2) + t(2×1) + t(2×0) + 1 The last one is for the all-vertical tilling. But don't you worry, we can simplify this a bit by noticing that t(2×n–1) is the sum of all of the above except for t(2×n–2), and thus t(2×n) = t(2×n–1) + t(2×n–2) Beautifull.
    2
  5271. 16:14 i guess technically you don't need to mention exponentiation since you can always say exp(ln(base) * exponent); though in the same regard you also don't need to mention multiplication or division: exp(ln(f) + or - ln(g)) = f * or / g
    2
  5272. This is black magic!😁
    2
  5273. This reminds me of Jansen's linkage planar leg mechanism but in reverse, where the angles of the central rotary axle can be adjusted minimally for maximum dynamic leverage in the extended polygon structures at the outer limb. The animations of the polygons twisting and moving in response to the different starting polygons, make this comparison particularly striking, see Theo Jansen's 'Strandbeesten' sculptures as an example
    2
  5274. Man, this channel is awesome. Keep up the great work!
    2
  5275. Bachelor's in math here, really liked the video, I can't wait for more!
    2
  5276. 4:55 Having fun so far? Definitely Yeah!!
    2
  5277. If you work with GPUs, all matrices are real magic! 😁
    2
  5278. Very nice , keep sharing such knowledge at YouTube platform :)
    2
  5279.  @Mathologer  I was a bit embarrassed about how long it took me to catch on. But when you showed the 'base' polygons, I thought 'Oh, it's like linear combinations and Eigenvectors!"
    2
  5280. Truly amazing representation of this geometrical gem!! I love geometry so much and I'm really sad that it hasn't the place it deserves in the school curriculums. PS: Can you please tell me the song at the end? I loved it!!
    2
  5281. Euler, Gauss and Beethoven..?
    2
  5282. Merry Christmas Mathologer! It's 4:17 AM here but I'm going to finish your video before I sleep >:)
    2
  5283. The digital root is fun with 11 as well. 10 is congruent to -1 mod 11 so you just add the even digits, subtract the odd digits of a base 10 number, then you get an easy analog for "11" digital roots.
    2
  5284. thank you so much for bringing this hidden glory of india(Bharata|)
    2
  5285. Using my code I tested the series up to 100 Million Terms and I didn't hit a single power of two other than 1, 2, 4, 8, 16 and 256.
    2
  5286. Nice video, as always.
    2
  5287. How far can the analogues go? Partial Difference Equations? Jacobian? Variational calculus? Complex analysis?
    2
  5288. Mathloger THE VERY BEST MATH AND LOGIC TEACHER ON THE PLANET. I missed him already when I was a school boy without knowing him. I enjoy every clip and its so well presented and easy to understand. He*'s a NATURAL AND A PRO... go on bro
    2
  5289. Dear Mathologer, An ever nicer way to show Eulers hyperformula is by showing it as a lattice. Now you miss the top and bottem elements of the lattice. The top element is the n-polythope itself. The bottem element the empty set, with dimensionaal -1. The -1 can be explained from homogenious coordinates, which rises all dimensions with 1. The sums of all levels with alternating signs then becomes always 0 instead of alternating 2 and 0. For a remark on complementary hypertopes like hypersimplexes and hyper octahedrons see my comment elsewere in this discussion.
    2
  5290. This reminds me of the infamous 1+2+3 ... = minus 1/12. Or some of the explanations which were given for that odd one. Here comes my personal take on that one, which may provoke some smiles... or frowns... The Sum of All Natural Numbers: S = 0 + 1 + 2 + 3 ... ∞ or alternatively, S = ∞ + ∞-1 + ∞-2 + ∞-3 ... 0 therefore, 2S = S + S = 0 + 1 + 2 + 3 ... ∞ + [∞ + ∞-1 + ∞-2 + ∞-3 ... 0] ------------------------------------ 2S = ∞ + ∞ + ∞ + ∞ + ... ∞ 2S = [∞ x ∞] 2S = ∞² therefore, S = ½[∞²] We can debate whether terms like "infinity squared" have any meaning, but I would suggest it means Aleph_1.
    2
  5291. Calculus is easy...differential equations is the difficult part.
    2
  5292. I'm totally good with both the long and the shorter videos.
    2
  5293. I think it's a good idea!
    2
  5294. This particular one is extremely trivial, as opposed to simply trivial :)
    2
  5295. do it! i love your long videos, but i would also love to see you post more frequently (and so would the algorithm, i expect)
    2
  5296. That’s great! I get excited similarly for every upload, dependent only on subject matter rather than video length to change how quickly I click. I do view everything eventually, of course.
    2
  5297. Your color problem reminds me of a similar one I came up with a few years ago in which you not only want every 2×2 to have every color, but you want every possible coloring of 2×2 to fit nicely together all overlapped and such so that every 2×2 is a different one. It's similar to a Debrujn sequence, in which every permutation of n replaceable objects is included all overlapped, but generalized. 10011 is such a sequence. 00 01 10 11 So is 1000101110 (a nice symmetrical one). (Hint: 0011 works for n=2 if it's looped!) (My apologies if you've heard such a problem before. I haven't watched All your videos yet.)
    2
  5298.  @thek3nger  Bellissimo!😆
    2
  5299. Sure, the more videos the better!
    2
  5300. great!
    2
  5301.  @Mathologer  Sorry I'm not sure what happened to the other comment, I think I accidentally deleted it. But I was informing of Season 2 Episode 12. It's a math geometry problem that went like this: "In the figure on the right, cubes with side length "a" are stacked and arranged repeatedly, with an atom placed at each vertex and in the center of each individual cube as shown, in the form of a body-centered cubic crystalline lattice. Assume that the structure is composed of sodium, potassium, or some other alkali metal. Within the body-centered cubic crystalline lattice, the atom at the center is point A{0}. Inside the cube, the region that is closer to A{0} than any other atom is D{0}. Find the volume of D{0}," You're gonna need the figure but I haven't found a good clip. If you see the episode you can see how two students arrive at the answer using similar means. A channel called CyclicSquares did an explanation (Video: Weird Geometry) but I'd like to see people try a "simpler" explanation I guess... I was going to link a writeup but I guess youtube doesn't like links in the comments, google it and you'll find it. Good Luck! :D And thanks for the videos!
    2
  5302. You would probably think that the worse possible odds for guessing the right color of the final triangle would be one in three, but I actually managed to lower it to zero in three by forgetting what the valid choices were and guessing green at the last second... Hahaha
    2
  5303. Ramanujan developed formulas for sums of consecutive squares and proposed more dimensional (square quadratic and quintic powers) for the first time.
    2
  5304. Took about 20 seconds to figure out and 5 min to write out in a comment. Being able to draw diagrams and being able to refer to them would help considrably :p text is not optimal for explaining things, would probably have put more work into the presentation and clarity if this wasnt a youtube challenge :)). Neat puzzle tho.
    2
  5305. 1/5, but I do not have enough time to describe the proof ... ;-)
    2
  5306. very nice
    2
  5307. I introduce my highschool students to this, but without the proof, as an alternative way to find the area of a triangle. This video is going to my "maths plus ultra" students, who like things like this. It's super accessible. Thank you.
    2
  5308. 10:55 I once stumbled across all of this by myself but I thought this just had to do with "normal" calculus and didn't expect there was another kind of calculus behind it
    2
  5309. That was fun! For power = 1 or 2 the differences are zero, as shown in the video. For power = 3, the differences are 2* triangle(n)^2. 2 18 72 200 450 For power = 4, the differences are 64* triangle(n)^3. 64 1728 13824 64000 216000 For power = 5, the differences are 2002 162066 2592552 20002600 101258850 .... and I'm stumped! (2002 = 2*7*11*13, not that THAT is any help!) Here some R code for it. multipliers = c(2, 4, 6, 8, 10) pow = 5 results = sapply(1:5, function(n) { tri_pow = triangle(n) * multipliers[pow] a1 = sum( ( (tri_pow - n) : tri_pow)^pow ) a2 = sum( ((tri_pow + 1) : (tri_pow + n) ) ^pow ) diff=a2-a1 guess3 = 2* triangle(n)^2 guess4 = 64 * triangle(n)^3 return(c(n=n, pow=pow, tri_pow=tri_pow, a1=a1, a2=a2, diff=diff, guess3=guess3, guess4=guess4)) }) results
    2
  5310. Not sure if anyone knows the answer to this, but a question I’ve been trying to solve for about four years is what the formula for a cyclic pentagon is, or if it even exists.
    2
  5311. (HO)³ = HO HO HO... All I can say is (HA)³
    2
  5312. At the end of the video, still waiting for either pi or e to show up!?!? 😅
    2
  5313. I never realized how fortunate I was to be born in an age where children are taught the value of π with as much precision as we've enjoyed. 🤯
    2
  5314. Seeing that table of logarithms brought back memories of using the "CRC Handbook of Mathematics" ... the bible for scientists and engineers around the world. Probably still useful for some, although calculators (and Wolfram Alpha) have largely replaced it. Thumbs up if you still own one!
    2
  5315. Fredericus Johannes Maria Barning, Freek, b. Amsterdam 03.10.1924, master's degree in mathematics Amsterdam GU 1954|a|, employee Mathematical Center (1954-), deputy director Mathematical Center, later Center for Mathematics and Informatics (1972-1988) Deceased. Amstelveen 27.06.2012, begr. Amsterdam (RK Bpl. Buitenveldert) 04.07.2012.
    2
  5316. Definitely make more videos on Euler-Maclaurin. I wanna see how it ties in with tangent and the trapezoid rule, and all that stuff!
    2
  5317. Thanks from anybody! 😊
    2
  5318. Plan F == Plan Feierabend. :)
    2
  5319. Gut Ding braucht Weile, and having the Mathologer as catalyst isn't so bad either.
    2
  5320. In the last chapter, you have cos120° = -1 It should read -½
    2
  5321. just awesome. love from India.
    2
  5322. Great explanation as always! I love how you manage to derive results without using calculus that seem to require them, and this is a good case in point!
    2
  5323. Pythagoras' theorem plus Einstein's special relativity equation on one shirt should equate to cxc shared ;)
    2
  5324. Can someone decipher the t-shirt for me?
    2
  5325. I'm an idiot. Of course it's obvious what it is! :)
    2
  5326. (HO)^3 is HO cubed, isn't it? Shouldn't we be saying 3(HO) ?
    2
  5327. You can do the invasion in less steps if you allow for diagonal jumps. for example, instead of the l for two in, you need two in the first row and one down and to the side on the second row. You jump diagonally over one in the first row in front of the other in the first row then that one jumps over the one now in the first enemy row.
    2
  5328. Madhava Uchiha.
    2
  5329. 18:40 I didn't understand how you obtain that "in general, it is always true that the footprints of the twisted windmills come in congruent pairs".
    2
  5330. 2^-1 = 1 - 1 + 1 - 1 +... 2^-2 = 1 - 2 + 3 - 4 +... 2^-3 = 1 - 3 + 6 - 10 +... which is the same as expanding 1/(1+x)^-n and substituting x=1
    2
  5331. I really want to thank you Mathologer for your work. I'm a maths teacher and I run a science club in my school and I often take subjects from your videos. I really really like the long format. It's peaceful. :) And of course, the maths is always so interesting! I kwen about this principle and about the magic trick at the end thanks to the channel Up and atom. I really enjoyed the video anyway! Maybe my favourite proof is the olympiad one...
    2
  5332. So is this contemporary maths? 🤔
    2
  5333. This is a MUST VIDEO for all professors in mathematics ! ❤ Mathologer makes my retirement very colorful. Thank you very much.
    2
  5334. arctanx=x−x33+x55−x77+... The Leibniz formula is obtained for π4 by substituting x=1 into the series. This will yield the Madhava-Leibniz series which is given by: 1−13+15−17+19−...=π4 . It can be proved as thus: Proof: Let π4=arctan(1) =∫1011+x2dx =∫10(∑nk=0(−1)kx2k+(−1)n+1x2n+11+x2)dx =(∑nk=0(−1)k2k+1)+(−1)n+1(∫10x2n+21+x2dx) The integral parts yieds 0≥∫10x2n+21+x2dx≥∫10x2n+2dx=12n+3→0 as n→∞ By squeeze theorem as n→∞ π4=∑∞k=0(−1)k2k+1 The series converges very slowly exhibiting sublinear convergence. Calculating π correct to 10 decimal places using direct summation of series requires about 5 billion terms because 12k+1<10−10 for k>5×109−12 . If the series is truncated at the right time, the decimal expansion of the approximation will agree with that of π for many more digits 3.1415924535897933238464643383279...
    2
  5335. 7:57 that reminded me of integrals!
    2
  5336. I think, because of how common the Tower puzzle is, and how little I've thought about it before, this is one of the most beautiful videos you've done. Despite many years of studying more complex topics, making "simple" problems elegant often gives me the greatest appreciation of mathematics. Thank you as always!
    2
  5337. 17:08 π ≈ 22/7.
    2
  5338. The legend is back
    2
  5339. Dear Matholiger, could you provide a link to 3b1b's video? Couldn't find the link in neither the description nor the comments section.
    2
  5340. I will remember many moments of this video but perhaps the most memorable one is your balance warm up thing; it was very nice and intuitive.
    2
  5341. So anyone can claim half of your garment and get 1/4, leaving you with a useless 3/4 garment. What an ancient, backward people.
    2
  5342. I'd like to submit my entry for the animation contest: https://youtu.be/CCL77BUymSY
    2
  5343.  @jannikheidemann3805  Yep! My high school had it. Though, we were a big, well funded high school, most high schools would not have a Latin class.
    2
  5344. Just watched earlier an old matholoder video😉
    2
  5345. You chose the labels: ”RGP” (Red🟥Green🟩Purple🟪), instead of ”RGB” (Red🟥Green🟩Blue🟦), to keep the Purple-would-be-Blue (B) from getting confused with the side length B, of ABC(D).
    2
  5346. Always nice to see a new Mathologer video :D
    2
  5347. 💓💯Thanks #teacher#youfanny I joined last 5 month what ever I I learning video put part of them not understand and so I am bachelor of GEOLOGY. Also I interest maths Expert I wish you Academical role model in this future From#somalia
    2
  5348. Most memorable to me was the odd/even thing. Mainly because i noticed that on each nth term, whenever n+1 was prime the numerator was a power of n+1. Makes me wonder if this is always the case. (for example 4th term is 25/12, 4+1=5 which is prime, numerator is 25 which is 5^2)
    2
  5349. What a joy! Just amazing! Thank you!
    2
  5350. The fact that such a channel can hit 500K+ subs, and maths videos with millions of views, gives me hope in human kind.... happy to +1 on Patreon
    2
  5351. Salute to your hard work sir, I'm a viewer from India and love the way u teach. You help me to feel and visualise maths which I always wanted. This was little bit exhausting for me at the end because it was getting little confusing for me because my brain processor😅😅 is slow but really enjoyed this video and learned a lot on how to proceed in different problems...thnk u very much and u deserve a lot more than 500k subscriber.
    2
  5352. I would call an odd number 2n+1 cos I'm ODD like that
    2
  5353. This is amazing. Thank you for making these videos for the world to see. They make it a better place.
    2
  5354. What a masterpiece, this is definitely a much needed video on the channel and in general! Although I now do feel dumber than a monkey, so I'll watch it again to get the click. :)
    2
  5355. visual thinker to the point of mathematical difficulties, so these proofs scratch an elusive itch, so satisfying to watch those shapes fit
    2
  5356. Hi Laura and Kyle!!! You have an awesome dad who is one of the great explainers.
    2
  5357. I think the "best" segment should be determined by comparing the local minima in the derivative of the viewer retention curve.
    2
  5358. 23:00 I don't see how 9 and 6 can be combined in such a way. Or 7 and 8. Three or more pieces are needed.
    2
  5359. The most memorable part for me was seeing the 700 year old proof.I I didn't pay much attention to math in school, despite the fact that I learned it easily. Math class was always boring to me because once I learned and understood an how an equation worked, I had no interest in running the same equation again with different numbers dozens of times for homework. My teachers never made the subject as interesting and engaging as the many resources I have found since graduation. But now that I have finally found an interest in the topic, I realize that I am woefully undereducated on anything past basic high school algebra. This is the reason why I liked seeing the simple proof. It was so easy that I could have done it myself with my current skills! That gives me hope that, while I may have MUCH to learn, there are still intricate problems that even I can grasp. Thank you for another great video, and have a nice day!
    2
  5360. the scissors and rock NEED to switch places on your t-shirt, it would make the graph so much more beautiful
    2
  5361. 28:29 The difference of squares shows 1,3,7,9... when it should say 1,3,5,7...
    2
  5362. 31:12 Contrary to whats stated in the video, its not possible to find such a subseries for every positive number using a greedy algorithm, because the set of positive numbers is uncountably infinite, while there are only countably infintely many algorithms.
    2
  5363. 😯😯nice pattern
    2
  5364. I think the answer to the initial puzzle is off summing i**10 from 1 to 1000. Double checking with Ruby I get 90,409,924,241,424,243,424,241,924,242,500 whereas the dear professor got 91,409,924,241,424,243,424,241,924,242,500? My program was merely: def sum a=0 for i in 1...1000 a+=i**10 end return a end
    2
  5365. The best part was the animation of Kempner's proof (and the Mathologer seal of approval lol)
    2
  5366.  @Nikolas_Davis  good one, lol
    2
  5367. From the n*n grid, I found that it is (n^4-n^2)/12 squares.
    2
  5368. So happy to see the video for this month. Wish they were more times then once a month, but am sure it takes a long time to make these, and happy that you make them
    2
  5369. 15 4-triangles?
    2
  5370. My favorite insight was that there are no integers among the partial sums of the harmonic series. Kempner's proof was actually simpler than I expected but initially counterintuitive all the same. Thanks for another great video!
    2
  5371. 3:04 whoa where’d your legs go lololol
    2
  5372. 27:15 Get OskarPuzzle on the job!
    2
  5373. The world's first STRAND type problem solved
    2
  5374. edit: this was such a cool video For an n-by-n square, the sum is n(n²+1)/2 The entire square is 1+2+...+n*n = n²(n²+1)/2, then divide that up into n "slices."
    2
  5375. 11:52 why is 15 not listed as divisor of 15 and 19 not as divisor of 19, but all other numbers as divisor of themselves?
    2
  5376. https://imgur.com/JfpClXR
    2
  5377. I discovered that the square that proves the pythagorean theorem can be generalized into an octagon that proves the law of cosines! I made a video on it if you’re interested in seeing how!
    2
  5378. 2^4 = 4^2 is my favorite
    2
  5379. The most memorable part for me, as a math & comp sci student, was learning about Baillie's paper! It promises to be a fun weekend read haha
    2
  5380. Merry Christmas and happy new year! Congratulations on the success of your channel. It's well deserved.
    2
  5381. I wonder if Napoleon's theorem is related to Morley's miracle, and if Morley's miracle generalises in a similar way to other polygons.
    2
  5382. Third challenge: fibonacci sequence!!! f(n)=if(n-1)-f(n-2), n>2 where f(1)=i and f(2)=-2
    2
  5383. In Field Artillery as a Fire Direction controller, I used slide rules for calculating the firing solution (tube elevation , fuze times, etc) for our howitzers for years. Before we had transitioned to completely digital systems, I could calculate the firing solution as fast as my Soldiers could using the computer systems. We had different slide rules for each charge (type and number of bags of powder), but the could all still be used for normal math functions as well.
    2
  5384. Fun fact about falling powers: The coefficient of x^m in x(x-1)(x-2)...(x-(n-1)) is the number of permutations on n letters with m orbits. e.g., the 4th falling power is x(x-1)(x-2)(x-3) = x^4 - 6x^3 + 11x^2 - 6x, and there are 11 permutations on 4 letters with 2 orbits, namely the 3-cycles and the products of disjoint 2-cycles.
    2
  5385. I thought he was going to iterate to 3n-1, and then show the general formula, and then explain how that proves that there are finitely many lengths with only 1 identity. (No need for infinitely running computers!)
    2
  5386. 0:35 you cheated, just a little, by turning a piece over.
    2
  5387. The final challenge is a classic. If I make the legs of the right triangle sqrt5 and 2sqrt5 then the hypotenuse is 5. That hypotenuse is in three segments, the short leg of a small RT, the side of the square in question, and the long leg of a congruent small RT. As the small RT has hypotenuse sqrt5, its legs are 1 and 2. The square's side length is 5 - 1 - 2 = 2, and the area is 4. The large square's side was 2sqrt5 so its area is 20. Thus the small square is 1/5 the area of the large square. As the stated problem has a large square with side 1, the small square's area there is 1/5.
    2
  5388. "multidimensional polyhedra" "two-dimensional polyhedra" Oh you mean polytopes. Why wouldn't they work on concave shapes? You can just squish them about and their numbers of topes wouldn't change. Also, the universe is mathematical so since Miegakure is a thing I am now convinced that 4D+ and multitemporal universes exist according to the multiverse theory and different universal constants. We just can't access them since we're in a separate mathematical structure.
    2
  5389. I watch your videos because currently I am learning German, and I love math and would pursuit a career of Engineering in Germany. Currently also live in Melbourne.
    2
  5390. 2³+2³=2²×2²
    2
  5391. Yay, a pretty simple one to follow for the festivities, and finish the year with less insecurities, haha Just a bit dizzy with all those rotations and toasts this time of year ;P
    2
  5392. 38:51 on top 3blue1brown
    2
  5393. 2-nd. At school
    2
  5394. I have a question? If with-out removing any numbers the series goes to infinity (initial case in the video); and if removing the “9’s” the series does not go to infinity (special case you had us wager). Does that mean if you only sum up the series with “9’s” does it grow to infinity? A is the set with out removal (1,2,3,4,5,6,7,8,9,0*) B is the set with removal of 9’s (1,2,3,4,5,6,7,8,0*) C is the set with just 9’s (9) •The set A = set B + set C ) The set A grows to infinity Then (set A - set C); are the ones with the 9’s removed and Converges (not infinity) So then does set C which is (set A - set B) diverge grow to infinity ???? Thank you in advance for your response
    2
  5395. in 28:37 isn't it should look like this (abc)⅓=(a+b+c)/3
    2
  5396. I wait with impatience the step next episode
    2
  5397. gamma (because i got a new constant to memorize)
    2
  5398. I always red Mathologer as Math-loger and it always breaks my mind when he says Mathologer all joined.
    2
  5399. Went to High School early 70s in Texas where Chem teacher taught class to use circular slide rule. Loved it and used it through college and even into first job where others used linear rules. Even some of college profs had never seen circular type. Electronics ended it and the nerdy competitions we used to have racing each other doing slide rule calculations.
    2
  5400. the magical number of the odd-sided square (side n) will be the sum of the integers from 1 to n + the sum of the multiples of n from 0 to (n-1): sum[from i=1 to i=n] (i + n*(i-1)) for n=33, the magical number is 17985
    2
  5401. Deriving calculus rules while metal music hits 🤓🤘
    2
  5402. What is the relationship of the polygon theory in regards to the areas of each iteration vs the original area?
    2
  5403. An episode on Buckminster Fuller's geometric discoveries could also save a few ideas from oblivion.
    2
  5404. Since phi appearing in nature seems to come about as a result of nature trying to pack together most efficiently, I wonder if we might see rho and sigma in a 3D analog. E.g. are there data from 3D sphere packing that we can analyse?
    2
  5405. Great
    2
  5406. Hi I've only just discovered your math vids and the ones I have watched are amazing :) so nice to see different ways of doing the math. not doing the homework, but will go back through and take more time with the vids, and attempt them :)
    2
  5407. Torricell's proof pre-dated the invention of calculus, and was based upon the work of Bonaventura Cavalieri.
    2
  5408. I can create infinite number of magic square
    2
  5409. pi × e = pie
    2
  5410. Magnificent as always! (By the way, I managed to build a square with an area of ​​3 in a simple way, the person who built the 17th square lived "above the mind":). By the way, now I am fascinated by the theory of probability (after all, chemistry now operates with the periodic table, and there only algorithms - over mathematics). It was possible to build an algorithm "even, odd" to the theoretical 50/50 (it is clear that the "excerpt" should give more). But then it is still unclear how to arrange 5 colors in a box with 6 positions :).
    2
  5411. Man your thumbnail just bought 71k over period of 2days.. Actually it did bring non math like me too Out of the curiousity
    2
  5412. Really cool equation, thank you for the video! I feel like the explanation of when thunder-equality holds means is missing. You could explain this in a following video. 🙂
    2
  5413. I made it to the end! Small mistake in the animation at 39:14. The water on the right keeps going up even when there is no water coming from the left to push it up.
    2
  5414. Amazing video, I enjoyed a lot! I'm a theoretical chemist and the relation between sum of powers and Bernoulli numbers was unknown to me, but you explained very very well. The mathematical detail you put in your videos is perfect for me! :)
    2
  5415. ANOTHER MASTERPIECE LETS GO!
    2
  5416. When I saw your fascinating animation, it reminded me of a monograph I read in the Little Mathematics Library when I was young. Those were some very nice little math books! I am moved a lot by your work! One of my friends once said that people are moved by art and music, but why not science and math? Now I understand what he meant!
    2
  5417. Another day of mathology.
    2
  5418. For those of you wondering what his shirt means, try reading the number of dots in each row one by one :)
    2
  5419. Another amazing maths video, well paced and brilliant reveal. I had no idea that was going to connect and there are many more concepts that can be built from this one phenomenon.
    2
  5420. The most memorable: Like you said, the most perfect overhang does not look perfect or beautiful at all. It looks messy and random.
    2
  5421. The fact that the "no 9's" sum is smaller than the "at least 100 0's" sum is what I will probably remember most from this video, because it's what most defied my initial intuition. So, if I have to vote for one of the chapters then it's the last one. :)
    2
  5422. Explain to homer the Riemann sphere. Because I’m a dummy And Wikipedia has failed me
    2
  5423. A like the 🐇 🐰 🐇 🐰 hole.hop hop hop hop hop not lost not lost not lost.♾♾♾♾♾♾♾♾
    2
  5424. this is perhaps one of the most beautifully rewarding videos you have made 👏🏻👏🏻👏🏻👏🏻
    2
  5425. Why is it written as 4k+3? I think it looks a great deal better as 4k-1. Then you have the groups 4k+1 and 4k-1.
    2
  5426. 9/10 is my birthday and I consider this video a lovely gift. Thank you very much, Mr. Polster!
    2
  5427. I increasingly believe it's a mistake to ever leave the field of two elements.
    2
  5428. Interestingly, for a quadrilateral that is both inscribed and circumscribed around the circle since a+b=c+d it follows that the area is just square root of the product of all of the sides!
    2
  5429. Incredible! Why – it's a miracle!! And that it is, folks; Moessner's Miracle. Thanks for this, Prof. Pollster! I remember noticing a decade or two ago, when playing around with figurate numbers, that the hexagonal numbers are just differences of consecutive cubes. And given the formula for the former, hex(n) = 6·trian(n–1) + 1 = 6·½n(n–1) + 1 = 3n² – 3n + 1 that fact becomes obvious: = n³ – (n³ –3n² + 3n – 1) = n³ – (n–1)³ I also noticed how this could be visualized with the very same cubical shells you show here. So it's great to see this falling out of a more general rule . . . Fred
    2
  5430.  @Mathologer  The video's called "These Simple Equations Are Levels of an Infinite Pattern" - somewhat different target audience, so it's mostly a slower exploration of finding the identities themselves. He does point out that it's the "integers only" restriction that makes there only be finitely many solutions for each size.
    2
  5431. First comment
    2
  5432. re: 00:14:03 Mathologer: "For example, the distance between 1 and 2 is equal to the natural logarithm of 2.". It is not natural logarithm, it is common logarithm - a.k.a. decimal logarithm - of 2. Same applies to distances between 1 and 3 and the remaining distances.
    2
  5433. I learned Heron's formula in high school and it is so great to see a proof of it! Thank you for such a great video! :)
    2
  5434. what a coencidence book numerical analysis recently started which is related to this topic thank you sir for giving such kind of present i love your channel and love also math and your teaching way math is always be beautiful i can see nature with the help of it clearly without moving anywhere with deep connection😌😌😍😍
    2
  5435. Thanks for this reveal, simply based on modular arithmetic, and on the fact that the digital roots trick works in any base b, for checking remainders mod (b–1). "Vortex" is to mathematics, as astrology is to astronomy. Fred PS. So then, clearly, by your t-shirt, 3 is 1/222 evil, 6 is 1/111 evil, and 9 is 1/74 evil. And 18 is 1/37 evil. Etc...
    2
  5436. 28:12 there’s a hexomino lying among those pentominoes 🤨
    2
  5437. Since the wiper segments pivot from the same point when they wipe, the traced path from when they start to end wiping is on a circle with the pivot point as the midpoint of the circle. This means that the "weird curve" shape is made of overlapping circles!
    2
  5438. For me the most memorable part is the fact that the 100 zeros series converges to a larger number than the no 9s series. Thank you Mathologer :) Great video as always!
    2
  5439. 24:40, not only does it pass through the center, it cuts the square into two congruent halves. Now I need to look at the shapes when all three bisectors are constructed.
    2
  5440. I think this improperly encourages debt. In the extreme, if I'm in poor health, I borrow $100M. My family gets a nice inheritance, at the bank's expense.
    2
  5441.  @andrewkepert923  I have seen the latest Crescent Rolls one a couple of days ago, yes. :) Not sure if that's the edible cyclides one you mean though...?
    2
  5442. 1. No integers is amazing 2. Gamma has my second place 3. All other your work also marvelous, about infinite sum and stack I've know from childhood, but all your analysis and methological mathology just the best. Not only in this video) Thanks a lot!!! It's a real pleasure.
    2
  5443.  @Mathologer  I think he meant elegant, just misspelled
    2
  5444. When you got to the 4k+1 part I thought on the next set you were going to say 4k-1 but went for 4k+3 instead. I like 4k-1 better just for the mirrored simplicity.
    2
  5445. Thank you for bringing this up in a video 👏 😃'Madhava Series'
    2
  5446. What about magic cubes?
    2
  5447. I guess it’s RGP and not RGB, because B is reserved for the full side of the ABC triangle.
    2
  5448. There are so many hidden formulaes in the Shlokas of Indian text . They closely associate it while praying the god . Basically chanting a Mathematical formula to a God . Interesting isn't it .
    2
  5449. this was hanoice
    2
  5450. 4:20 "The Argument is really weird [...] Although we can be absolutely certain that Australia has hair-doppelgängers, the argument gives us absolutely no clue about how to find such hairy twins." I actually think you are doing your proof injustice here ;) The argument you gave, by itself, gives a perfectly fine algorithmic description of how one would go about finding hair-doppelgängers. More abstractly it's an easily implemented algorithm for finding a pair of elements which both have the same value for a (decidable) property and the side condition "there are more elements then values for the property" guarantees termination of that algorithm. So we actually know exactly how to find those twins! The weirdness then rather comes down to the infeasibility of actually doing this in real life. I would compare this to some situations in complexity theory, where there are problems which are known to have solutions; Solutions however that are well beyond practical.
    2
  5451. This is the best thing ever. On our last two COCI programming competitions we had triangulations of polygons (with 3 possible colors of edges) where each triangle had all sides differently colored.
    2
  5452. The King's construction seems to be similar to Ben Franklin's. That looks like the Chinese poem already discussed, except there is never a space already occupied -- it's just diagonal line, wrapping at the edges. Anyway, it would be interesting to understand how the cells can be systematically relabeled to give all the different solutions having the same used numbers. You can swap rows (or columns) for example and that preserves the row and col sums but in general does not preserve the diagonal. You need to swap symmetric rows, equally spaced from opposite edges.
    2
  5453. Most memorable: the bishop's proof was outstanding
    2
  5454. sir you are outstanding
    2
  5455.  @Mebasically  wow nice
    2
  5456. Calculus poetry. that's something!
    2
  5457. Vielen Dank für das tolle Video, es ist in höchstem Maße faszinierend. Es löst Glücksgefühle aus. Unglaublich...
    2
  5458. I almost closed the video at 8:00
    2
  5459. Quite amazing D=D
    2
  5460. The first time I learned about Toricelli's horn was in an elementary calculus class, and until now, I was unaware of the existence of its algebraic form. I'm not sure how typical my experience is or how it compares to those of YouTube mathematics influencers, though. It did seem miraculous and mind-blowing to me as a beginning calculus student—as did so many other aspects of the calculus—whereas I'm not sure I'd have appreciated the paradox as much without Newton & Liebniz.
    2
  5461. What I don't understand is how pi on your t-shirt is related to the magic squares
    2
  5462. Watching this, I intuit that magic squares have something to say about Shroedinger Wave Equation. Maybe because Mathologer kept using the term collapse
    2
  5463. "solitaire" just means a solo/1-person game
    2
  5464. For all tight lacings the best thing is x=1,814,400. Edit: this is a joke
    2
  5465. I know this is irrelevant. Usually, Prof Polster's t-shirt has some connection with the video. But I don't see any connection of magic squares with pi...
    2
  5466. Unfolding the Earth: Those are green bananas.
    2
  5467. The most memorable for me was the "Terrible Aim" section.
    2
  5468. Laughed hard on the music change for the hardcore parts))))
    2
  5469. 18:45 I suspect it's because this shifting construction doesn't actually shrink the polygon for small numbers of sides; it produces a new one that's as large or larger than the original.
    2
  5470. That was interesting squared.
    2
  5471. Loved the whole thing The most memorable part was the Bishop's proof.
    2
  5472. does this have any real world applications?
    2
  5473. Repeating the step of growing the "square tree" of the puzzle infinite times, it's area grows to infinity. Yet it will not fill the entire 2d space it lives in.
    2
  5474. I think Madhava of Saṅgamagrāma, many of his followers and the people who went before him were brilliant and definitely require more recognition than they have today. I do hesitate to call it 'Calculus' though. They had no FTC (Fundamental Theorem of Calculus) as they never made the link between derivatives and integrals (or their equivalents in their framework). They made some brilliant discoveries when it came to derivatives, infinite series and other related things but to say they discovered 'Calculus' as some claim isn't quite true. We can acknowledge people's great contributions without needing to say they discovered everything.
    2
  5475. ≈ 5½ min – Lucas (pron: LOO-kah) numbers, are sums of pairs of Fibonacci numbers that are next-but-one: 0 + 1 = 1 1 + 2 = 3 1 + 3 = 4 2 + 5 = 7 3 + 8 = 11 . . . Lᵢ = Fᵢ₊₁ + Fᵢ₋₁   from which their adherence to the same recursion rule as the Fibonacci numbers, follows. But I wonder – is there a relation of these to the Suzanne Vega song, "Luka"? It's pronounced the same! I know we've touched on this in previous comments, that you too are a Martin Gardner fan; so I'm curious whether you got the idea for this from his early MG column on Fibonacci numbers? In it he mentions extending the Fibonacci recursion rule to k'th order, still with all 1's as coefficients, and that as k → ∞, the limiting ratio of consecutive terms → 2⁻⁻. That limiting ratio, R(k), is always the root of the characteristic polynomial, that's between 1 and 2. R(2) = φ;   φ² – φ – 1 = 0                 (Fibonacci) R(3) = x;    x³ – x² – x – 1 = 0           (Tribonacci) R(4) = y;    y⁴ – y³ – y² – y – 1 = 0   (Tetranacci) . . . etc. The "Infininacci" numbers (k=∞) are just the powers of 2. Another curiosity – the discarded solution, "phi-red", –1/φ, is also the consecutive-term limit of the Fibonacci sequence –– but in the negative direction, where the signs alternate! ..., 13, –8, 5, –3, 2, –1, 1, 0, 1, 1, 2, 3, 5, 8, 13, ... So the ratio F(n)/F(n-1) → –1/φ as n → –∞; and it → φ as n → ∞. Question: Is there an analogous rule when the recursion order, k > 2? Danke schön, Herr Polster, for a splendid presentation!
    2
  5476. Either the area has to be infinite or the perimeter I think. And in the perimeter case it honestly has to be something fractal. The demo at 15:00 is pretty clever. You are right, I didn't finish it, know what you are gonna do, loved it. Very clever.
    2
  5477. I just saw the Celestial Toymaker in the thumbnail, that was most unexpected. It's a rubbish story. This puzzle is superior to its quality.
    2
  5478. Very neat video. That little twist at the end resembled somewhat the rotary wheel of a Wankel motor. Yes, I'm that old...
    2
  5479. Loved the video it taught me a lot
    2
  5480. focus on one intersection of two supposed chords and notice that it has the same power wrt one pair of endpoints as the other (the colors show it) and this means the ends of those two chords lie on the same circle, now if we show that an endpoint of horizontally oriented chord lies on the smame circle as the circle formed by endpoints of other two chords, we are done. Label P,Q;R,S;U,V the endpoints in the positive angular direction starting with horizontal red endpoint and A the red vertex of triangle, B blue vertex and C green vertex. Since PBV is isosceles, angle <BPV is pi/2-beta/2, <APQ is pi/2-alpha/2, so <VPQ is pi-(alpha+beta)/2 and because <VUQ is equal to <VUC which is pi/2-gamma/2, it follows that pi-<VPQ=<VUQ which means VPQU is a cyclic quadrilateral, which finishes the proof
    2
  5481. Most memorable (& disconcerting) was no 9s converging. Thanks!
    2
  5482. Thanks for making my visualise strong🙏
    2
  5483. Avoid positivity f(x)=-|x| 😝😝😝
    2
  5484. Your visual demonstrations are great and make very complex ideas simple to grasp.
    2
  5485. I've been waiting for it. Thank you <3
    2
  5486. If they had taught things like this when i was in school, i might have a totally different job now.
    2
  5487. The Gregory-newton theorem can be thought of as a formula for the smallest degreed polynomial to pass through the points- (n,f(n)). Is there a formula which finds you the minimal polynomial to pass through any set of points?(if the points don't share the same x)
    2
  5488. Jaaa, ein Mathologer-Video auf Deutsch!!!
    2
  5489. You do great work....Huge inspiration to see the amount of work you put in to make such detailed videos...Please carry on doing this ...
    2
  5490. I made it to the very end. And this is really it for today until next time. Thank you for explaining the pentagonal number rule. Using the infinite product form of the generating function P(z) = Product_n (1 - z^n)^{-1} had previously shown me how it works, but not why those particular values. (I'm a sucker for generating functions.) Speaking of which, using them it's easy (for a particular value of 'easy') to show that, for all n, the number of partitions of n into distinct numbers equals the number of partitions of n into odd numbers. Perhaps you have some clever method using Ferrers diagrams to show this next time?
    2
  5491. Hello sir you videos has been a delight, recently i have studying perron Frobenius Theorem ,Its proof has beautiful concepts behind it and ton of applications
    2
  5492. 27:44 3blue1brown, your son. 28:34 Euler I hope that your son will be good at math.
    2
  5493. I mean... Gauss was in the thumbnail.
    2
  5494. I really really like that visual explanation. Thank you!
    2
  5495. this is the only if not one of channels that i watch all its content. amazing as always!!
    2
  5496. To find an O(1) solution of a Project Euler problem (problem 28) I found the formula of one of these sequences. It took me one entire afternoon back then when my mathematical knowledge was not that developed.
    2
  5497. I don't know if this video is beautifully awesome or awesomly beautiful...
    2
  5498. Wow an authentic Mathologer's Seal of Approval!!!
    2
  5499. Ha! The dots on your shirt correspond to the digits of pi.
    2
  5500. The 7 from 100: 1, 2, 4, 8, 16, 32 and 64
    2
  5501. I saw a variant of that tiling 2 by n board problem during a competitive programming contest, where you were allowed to use not just 2 by 1 tiles, but also 1 by 1 and L pieces. Really awesome stuff.
    2
  5502. First challange: no Second challange: it m and n are odd then there will be such l and k between 1 and m and n respectively such that l and k will be 1/2 of m and therefore we would have cos^2 of π/2 which is 0 and multiplying with anything we will just get 0 Third challange: I found good ol' fibonnaci sequence in there:)
    2
  5503. Très rigolo, ce truc-là !
    2
  5504. Also noticed that in the case of the squared sequence the last number from the previous term and the first number from the current term, are exactly 3, 5, 7, 9, ... apart from each other, which follows the difference of the squares! 0² = 0 3² + 4² =5 ², ----------------------------------->3 - 0 = 3 10² + 11² + 12² = 13² + 14², ---------> 10 - 5 = 5 21² + 22² + 23² + 24² = 25² + 26² + 27², ---------> 21 - 14 = 7 Pretty cool
    2
  5505. 9:04 the background is looking at you
    2
  5506. The order is red, black, orange, blue, green. Getting snooker enthusiasts traumatised.
    2
  5507. "Anyway, ignoring the French..", subbed
    2
  5508. Hi Mathologer. I am a big fan of your channel and maths in general. I am a 2nd year pure maths PhD student. As for feedback, I don't really have much critisism (I did just spent 50 minutes watching this video), but I particularly like when proofs are fully nailed down. Most of the time your videos do this, and even when they don't, you often point it out yourself which is something I particularly appreciate. Keep up the good videos :)
    2
  5509. 15:07 had me open my mouth in shock 😲
    2
  5510. The most memorable part for me was to show that the error between ln(n+1) and the n-th partial sum of the harmonic series was smaller than one. Using it you don't only get a good approximation but also show that the harmonic series doesn't converge.
    2
  5511. Most memorable: The no 9's series.
    2
  5512. Is 4k-1 the same as 4k+3 ?
    2
  5513. foxabilo They’re the same set of numbers, but the k in the first equation is one more than the k in the second. The composition of the set is what matters, the value of k is trivial so, basically, the answer to your question is yes, it’s the same. You can combine 8k + 1 and 8k + 5 to generate the same set. Or 12k + 1 with 12k + 5 with 12k + 9 .... GO CRAZY!
    2
  5514. great!
    2
  5515. This system has the huge flaw that the distribution changes if credits are merged or split. If the recovered money is less than half then it is convenient to split a credit in smaller credits, if it is more than half then it is convenient to merge credits together. The proportional system does not have this flaw.
    2
  5516. Too in a WolksWagen logo? okno is bad joke!
    2
  5517. So many thanks to you mathologer for your tireless work. we all salute you.
    2
  5518. Thank u.
    2
  5519. Yup! I guess it right! 0:14 Another great video as always! You always provide us with AMAZING vids, and I LOVE them
    2
  5520. I think so too. It's what I got, except that I forgot the factor of 2 in the denominator.
    2
  5521.  @tomkerruish2982  yeah I forgot that at first too!
    2
  5522. Burkard, your videos fascinate me. Every. Single. One. Of. Them.
    2
  5523. My favourite part is the really large Lira Tower. The fact that it has an overhang that is only slightly larger than that of the 20-high tower was interesting, but at that size, I am thinking that material strengths become a significant consideration as well as where the ISS is orbiting in comparison to your position so it does not bump into to your tower. ;) I also liked the appearance of the quadratic tower as well.
    2
  5524. Ausgezeichnet! 👍😃💖
    2
  5525. The rubik cube thing is group theory, each element of a finite group has a finite order, which divides the total number of elements in the group (Lagrange's theorem).
    2
  5526. Did it feature the magic square? I didn't know it at all...
    2
  5527. So this Number is gonna be based off The Knuths Up Arrow Notation as it is: - Level 0: Counting (+1) - Level 1: Addition (+) - Level 2: Multiplication (×) - Level 3: Exponential (^) - Level 4: Tetration (^^) - Level 5: Pentation (^^^) - Level 6: Hexation (^^^^) So let's start off with 2^5 = 2 × 2 × 2 × 2 × 2 = 32, 2^^5 = 2^(2^2^2^2) = 2^(65536), 2^^^5 = 2^^(2^^(2^^2^^2)) = 2^^(2^^(65536)) 2^^^^5 = 2^^^(2^^^(2^^^2^^^2)) = 2^^^(2^^^(2^^(65536))) HBUN(1) = 2^^^^5 HBUN(2) = 2(HBUN(1)^)5 HBUN(3) = 2(HBUN(2)^)5 Tip: Keep going on and on until you reach HBUN(100) = 2(HBUN(99)^)5 Note: HBUN means Hector's Binary Upper Number Function: HBUN(C) = 2(HBUN(C - 1)^)5 Size: This number is way bigger than G(64)
    2
  5528. For those interested in applications or related subject, look up Crystallography, it is an awesome physics subject related to Solid State Physics... Motivation:: You can explain why diamond is extremely tough solid.
    2
  5529. The fact that gamma is greater than 1/2 can be proved as follows: Take one individual blue bit and join the points where it intersects the graph of 1/x. Then, as 1/x is concave, the curve will lie slightly below the chord formed. So, the right triangle formed will have an area that is slightly less than the area of the blue bit. Now, on the interval (n, n+1), the area of the triangle is (1/2)×{(1/n) - (1/n+1)}. So, the sum of all the blue bits must be greater than the sum of all the triangles that is the sum of (1/2)×{(1/n) - (1/n+1)} as n goes from 1 to infinity. This sum telescopes to leave only 1/2. Therefore, the sum of the blue bits is greater than 1/2, that is gamma is greater than 1/2. Q.E.D.
    2
  5530. 5:57 Oh this one! I just recently watched a Vsauce2 video on it. It still feels surprising to me even after knowing the reasoning.
    2
  5531. Archimedes Claw should be called "Archimedes tangerine" :) . You can do the proof during breakfast as long as you're not too hungry!
    2
  5532. Hi , your proof was nicer :) and please make more short video. I really enjoy all your videos. Shorter videos are also nice. Thanks a lot Dr. Burkard Polster.
    2
  5533. While playing with Excel one rainy afternoon, I discovered on my own how any two random numbers can seed a Fibonacci-ish sequence in which the ratio of adjacent elements converges to phi. I thought at first I'd done something wrong, then much later saw a hand-wavy Numberphile video that explained it. Amazing stuff. There's beauty to be found in all these dusty little corners...
    2
  5534. 17:00 You are in dangerous territory to suggest to innocent people that the surface area of a volume is its derivative. This is not true. It may work for Platonoc solids but it is not mathematically sound in general. 21:04 Most Flat Earther's won't have a clue as to what you are talking about but they are gonna love yourproof that The Earth is Flat when this video goes viral :-) Already folk in the threads below are believing the calculus coincidence as if iot was real. You have fallen short here.
    2
  5535.  @Mathologer  Good save 🙂
    2
  5536. Ironically, the circle in 5:00 is presented to us as a collection of square pixels =)
    2
  5537. I think, the more general version of Pythagoras is the law of cosines (not cowsines as shown on the T-shirt).
    2
  5538. 27:07 what does (1/2) ^ _2 mean? It it defined in terms of gamma?
    2
  5539. These kinds lf problems are my favorite, solvable in a flash with no thinking at all because i can see the limit being formed for all kinds of cases in all kinds of dimensions. And it takes but a second to see the angle and area correlation.
    2
  5540. Thanks!
    2
  5541.  @Mathologer  Actually he does it by converting the series into a limit of partial sums. Then after rearranging, you can get partial sums that can be expressed as a harmonic number. Adding and subtracting some specific logarithms to each partial sum, you can construct the Euler-Mascheroni constant times some coefficient. These coefficients times the Euler-Mascheroni constant for each partial sum cancel each other out, leaving him with only the logarithms he added in, which in their limit become the formula mentioned in my previous comment.
    2
  5542. Always on the look out for the mathematical "sole" of things. :P
    2
  5543. I think this illustrates the relationship between different orders of arithmetic operations, no?
    2
  5544. “Hi, we are Shrinking Regular Polygons and we’re here to play your prom!!“
    2
  5545. beautiful
    2
  5546. Did you have an error in your graphics @8:15? "All the numbers on the left are exactly double those on the right. 200 x 2 is 400, 50 x 2 is 100, and so on." I can't figure out what you're talking about here. I don't even see any 2 on the screen. And the only 400 is on the top left of the screen. ??
    2
  5547.  @Mathologer  Thank you for the explanation, I see what you meant now. And I also realized an interesting effect in my brain. I think your wording (left, double, right) and our expectation of reading from left to right caused me to look from the left to the right and expect the double on the right (which would've been 100 -> 200, 150 -> 300, 150 -> 300), which didn't exist either. Funny how our ingrained expectations so strongly override the words.
    2
  5548. Nice to see the Sign of the Beast in there
    2
  5549. Also, as is often the case with pseudoscience - the real stuff seems far more interesting and certainly deeper than the junk.
    2
  5550. Simply Beautiful, agreed. Made my day also, first thing in the morning.
    2
  5551. I love your content, hope you have a nice day!
    2
  5552. This is the most satisfying video of 2021 so far ^^
    2
  5553. I am always amazed by your videos. Thank you.
    2
  5554. Hi, Mathologer! I have discovered a nice relation between partitions and prime numbers too! I have shown that the way the prime zeta function combines with itself to produce the Zeta Function forms a partition-like structure! Search for Preâmbulos Aritméticos, chapter 5, section 2, example 12!
    2
  5555. Ooh, I have a PDF of "Calculus Made Easy"; it's fantastic!
    2
  5556. My mathemaical self thinks it would be much cleaner to just assume the first n-gon is of smallest possible area. But my artistic self enjoys infinite shrinking much more :)
    2
  5557. This is peak Mathologer
    2
  5558. Was wrong to the coin problem:) should have known better. The coin Problem is also from intrest, when some counts how many Times a year the earth makes a full turn. Mabye one could say its 365 rotations because a year has 365 day. But there is one more or one less rotation because of the rotation around the sun. Should have known the right answer to the coin Problem, cause i know the sun earth Problem. but didnt recognize the similarity. Definitley its because i had not drunk my morning coffe:)
    2
  5559. Ohh... What's the ratio of the scaled down triangle compared to the original? :)
    2
  5560. outstanding, thank you
    2
  5561. Best way to laces your shoes 30 minuet video Me: y e s
    2
  5562. The title itself is the greatest clickbait 🤣🤣
    2
  5563. First from the Czech Republic!
    2
  5564. Dear Professor Polster, I wish Your videos became a part of mandatory study materials at tech. universities. Sincerely, I learned more and many mathematical concepts "clicked into the right place" within my mind while watching Your videos. I have a PhD. in electrical engineering where I worked with Markov chains and complex calculus on a daily basis. Had I had the chance to have such a quality study materials 15 years ago, I feel could have learned 10 times more in the limited timespan. Your visual proofs and way of calculation helped me to perform quick calculations by just manipulating symbols in my imagination rather than writing everything down. One big thank You, since You have the talent not only to understand the topic, but also explain it as real παιδαγωγός.
    2
  5565. and after aaaaaaaaaall, you're my number waaaaaaaaaall
    2
  5566. Man i would like to know his process of thoughts from beginning to finding such an identity
    2
  5567. The area down the curve of 1/x is the integral of this inverse function. And the integral of 1/x = Ln (x). This is why you use the 1/x function to explain logarithms. Nice.
    2
  5568. @mathologer Following the same logic, a number N with just one product=sum identity also has the property that either 3N+1 is prime or (3X-1)x(3Y-1)=3N+1 has no integer solutions different from N (besides the properties of N-1 and 2N-1 being primes). Does this make sense?
    2
  5569. 23:00 I was wondering this as it seemed like the next logical step!
    2
  5570. 256 is the last power of two it seems...
    2
  5571. Is there proof of that?
    2
  5572. 5:27 Iron Man proves the beard man formula at the bottom of the left page! Let E(m,n) be the number of m-cubes in an n-cube. An n-cube is made up of two (n-1)-cubes whose corresponding parts are all connected. So, for example, E(2,4) (the number of 2d faces in the 4-cube) equals 2*E(2,3) (the number of 2d faces in the two 3-cubes) + E(1,3) (since connecting a pair of corresponding 1d edges creates a 2d face). In general we get E(m,n) = 2 E(m,n-1) + E(m-1,n-1), which is the recurrence relation on the page there. From here you can prove the beard man formula by induction. Or, if you want to be fancy, you can use generating functions to prove the (x+2)^n expansion thing directly :)
    2
  5573. Always LOVE Your stuff..!!!
    2
  5574. The Galois video is going to be epic!
    2
  5575. Hi :)
    2
  5576. The fact that the partial sums of the harmonic series are always odd/even and thus never add to an integer was surprising. Definitely an interesting and cool fact that you brought to light.
    2
  5577. Most memorable is that 700 year old proof, I'd say
    2
  5578. Please consider making a video explaining how planineters work in the venerable, inimitable Mathologer style!
    2
  5579. Firstly: brilliant video, explanation. Consider my mind blown. 1. There are many examples of where processes and decisions expressed in flow can be used to solve problems in an ideal way. Traffic jams are one example, where most people would assume it would help to have higher throughput on the main roads by having longer traffic light phases of go, when in actuality, a "proportionally" bigger share of green light should be given to side roads leading to and from main arteries. I would love to see if communicating vessels could be used in a similar way here. Possibly these issues are equivalent? In the traffic jam example, some solutions are found by simulating the traffic as a directed graph and "water" is pressed into the "pipelines" and you check where the water leaks :) 2. In a lot of evolution simulations there is a significant issue with predation. The mere existence of predation in a simulation affects survivors, survival strategies and life expectancy much more than energy density, regeneration and most other factors, one would assume to have a significant impact. I am wondering if the idea of using a solution like this, that prefers a happy outcome, could be used to in simulations to "curb" predation. Without dragging this into a political discussion, these patterns can conversely be seen in the real world. It would be fascinating to see if such an approach on a fundamental level could lead to an overall "happier" outcome for everyone.
    2
  5580. ​ @Mathologer - Loved this vid! I am still not seeing many people tackle the quartic. What struck me about the last puzzle is x^4 = x^2 * x^2, so with 7^4 being off by 143, or 144-1, it was an instant realization that what we have in the quartic is slightly different squares multiplied together, with the trivial 11 times 13 = 143.
    2
  5581. Best way to end your day
    2
  5582. Thank you Mathologer for another amazing video! I have always been fascinated by dynamical systems (in this video discrete) with beautiful charts. Would love to see more in the future!
    2
  5583. For the first challenge I will make the path of the ball with two square width
    2
  5584. Thanks again for anther great video. As I said i watch these on a nice Sunday afternoon to relax. :)
    2
  5585. This was a great video! Even I could follow along without much problem. I'll always click on a new Mathologer video!
    2
  5586. Amazing explanation of logs. I've never thought of them that way, even though it seems quite obvious when explained
    2
  5587. You come back after 3 months and expect me to just jump right back in? You think you can play with my emotions? Well fine, I'll watch the video, but I won't like it!
    2
  5588. Yes you will
    2
  5589.  @disoriented3971  Fine! I'll like it, but you can't make me share it!
    2
  5590.  @Mathologer  OK, I shared with just my math class, but that's it, I'm done!
    2
  5591. 38:32 TORUS
    2
  5592. I find it odd that when he showed the thumbnails for the previous 99 videos, he chose to do it with a 10 X 10 array, forcing him to use a placeholder (the "100") in the first cell. Why not use a 9 X 11 array for your 99 objects?
    2
  5593. 7:30 Hmm, at first I was wondering if this could be described as Pascal's triangle mod 3, but it quickly became apparent that it couldn't, and also that this algebra isn't even associative. (R+B)+Y = Y+Y = Y but R+(B+Y) = R+R = R All that idempotency must be getting in the way. Maybe there's still a mod 3 isomorphism going from the tip (like Pascal's actual triangle), but using colour shifts for different bands of hexagons or something. With the rotational symmetry of this "algebra", any tip would do. 18:22 Well, I guess it's going to be clearer soon! 20:50 ! Guess I wasn't too far off with that thought about colour shifts, in the end. Flip red for blue every other row (but not starting from the tip). 28:30 If C = -A-B, then -A-C = -A-(-A-B) = -A+A+B = B. Hmm, yep, that works out.
    2
  5594. I thought I would never ever see more involutions after Projective Geometry. Scary!!
    2
  5595. Yay! New video!
    2
  5596. First
    2
  5597. for the sq root of wallis product which is the ramanujan's identity. Its clear that a subsequence converges. If you treat the product as (2/1)*(4/3)*(6/5)... then it might explod as each term is greater than 1. But Now treat it like the Wallis product that it is. 1*(2/3)*(4/5)*(6/7)... Its a product of fractions less than 1. Thus it too must be less than 1. The convergence of the product is subsequence dependent. Now you should argue which subsequence makes more sense from the integral expression it is being equated to.
    2
  5598. The geometrical videos are awesome
    2
  5599. Amazing video. I loved it!
    2
  5600. Interesting approach on this problem. Having already seen Numberphile's two videos on Conway Checkers I don't think I would have thought to use a monovariant that doesn't involve exponentiating a constant.
    2
  5601. Epic shirt
    2
  5602. 18:41: "... and this is something you always have to be wary of when manipulating infinite series ..." (Cough, cough, Numberphile, cough, cough)
    2
  5603. Wow what a video i m very happy to watch this beautiful video
    2
  5604. I found the hidden 2019, it is here: 20:19
    2
  5605. Glad that this video worked so well for you :)
    2
  5606. Great... And now, I feel the urge to code something to draw number walls given a sequence...
    2
  5607. I would like to encourage everyone planning to give on patreon to do so with an appropriately mathematical pledge. $3.14 is a favourite pledge, of course, but there's also $2.72, $1.41, $1.68, $6.28 and so on and so forth.
    2
  5608. Very much Martin Gardner feeling on this episode.
    2
  5609. Thanks again. I love to watch these on Sunday afternoon with my coffee.
    2
  5610. If y(x) = x + ce^x, then y'(x) = 1 + ce^x. In order for this to be a solution, we need to be able to substitute both of these into the equation and get a true statement. Substituting into y'(x) = 1 + y(x), we get 1 + ce^x = 1 + x + ce^x Subtracting 1 and ce^x from both sides, we get 0 = x which is a false statement (since the 0 function is not the same as the function x), so it's not a solution.
    2
  5611.  @p2beauchene  Everyone makes mistakes, even mistakes they think are very silly, so as long as all confusion is cleared up, that's all good! Feel free to do whatever you want with the post :)
    2
  5612. Some people can take 112 and run. Thankyou
    2
  5613. I've always liked the tower of hanoi because it is a nice binary system and there is an intuitive algorithm to follow
    2
  5614. Your laugh absolutely makes my day! The sound of the joy of learning and knowledge.
    2
  5615. Pretty sure that I could solve one of the Math exam questions if I used this method. It was so much hassle with counting of right triangles inside the triangle but if I were to turn it inside out it would turn into a one big right triangle.
    2
  5616. Something that stood out for me in the "highlight every third" pattern was that the highlighted values on the second row looked to form very pleasing ratios with oneanother. Some quick math shows that ratio of each to the one following it is a ratio of consecutive perfect squares. 3 and 12 form 1/4 or 1^2/2^2, 12 and 27 reduce to 4/9 or 2^2/3^2, 27/48 reduces to 9/16 or 3^2/4^2, and the next which isn't shown here ought to be 48/75, which equals 16/25 or 4^2/5^2. I wonder whether this pattern (or a similar one) also emerges when the spacing between highlighted numbers is higher. With the "highlight every fifth" pattern, the second row shows the same ratio of consecutive highlighted numbers, though they all are a factor of 10 higher than those squares. I wonder where that 10 comes from.
    2
  5617. Nice
    2
  5618. Love yor work! Perhaps you could try mathologerising the Lévy-Steinitz rearrangement theorem.
    2
  5619. The even/odd sum-of-product-of-coordinates is the first time I have understood how to calculate the determinants of matrices larger than 2×2!
    2
  5620. Thank you so much for sharing such an amazing and beautiful result. I hope that you all have a merry Christmas and get plenty of rest.
    2
  5621. 19:37 The real latin name of Torricelli should be Tærricellius.
    2
  5622. Careful the rain over there hope everyone is staying safe
    2
  5623. The spinning 4D cube was beautiful, but was also slightly painful to watch.
    2
  5624. Tristan’s proof is beautiful!
    2
  5625. They thought e^(i*pi)=-1 was the most beautiful theorem? That's so dumb! All that is, is a specific case of e^(i*x)=cos(x)+i*sin(x)! It doesn't even count as a theorem IMO since it's just a special case of another more general and actually useful thing!
    2
  5626. "every other*turn*" 11:30 - hmm.. having struggled a bit with writing a nice essay on discrete (and then continuous and finite, and games, by which point even I thought it had got a bit overly messy!) - but on second, briefer! thoughts.. maybe "move" would have put a bit more distance than"turn" between the rotation of the 1 tower!
    2
  5627. 17:31, oooh, it is related to harmonics And this process essentially cancels out (destructive interference) the other harmonics
    2
  5628. Andrew just confirmed that the earth is indeed flat 😂
    2
  5629. Windmill summary is 404ing
    2
  5630. Prove that every odd number can be written as the difference of two squares: 2n + 1 = (n+1)^2 - n^2
    2
  5631. THAT WAS AMAZING!!! Guys, I don't even know how to describe my emotion at 26:08. Congratulations! THANK YOU VERY MUCH!!!!
    2
  5632. 3:59 "The new blue spike has the same area as the old blue spike." It's not obvious to me. The equality of the area of the whole shape under squashing and stretching doesn't imply that of it's parts. And the visual proof used to demonstrate the result for the whole shape shows that the new shape is missing the green bit. So, at least to my understanding, you can't make that statement here.
    2
  5633. More.
    2
  5634. Another master class video from mathologer❤. Sir, recently I came across magic squares. They were fascinating. I watched several videos on YouTube on them. All of them were just about tricks to make magic squares. There was no explanation behind them about how and why these tricks work. I also checked the internet but without much success. I request you to make a detailed video on magic squares and math behind them.
    2
  5635.  @Mathologer  first to comment on your first comment
    2
  5636. I haven't thought that I would be able to survive those 50 minutes but you are the one who made it possible.
    2
  5637. There are 10 essentially different triangles of width 4: BBBB, BBBY, BBYB, BBYY, BBYR, BYBY, BYYB, BYYR, BYRY, BYRR. Now, I have no idea why...
    2
  5638. This is exactly the type of video I love! I didn’t even go through the step of wondering at myself for enjoying a shoelace video! I ponder this type of stuff too! Perhaps there a mathematician struggling to come out!😁 Thanks for the great video!
    2
  5639. We can use the remainder's cycle to find the prime factors of numbers of the form power towers... for example i know that ("by hand") the gigantic number 2^3^5^7 + 7^5^3^2 is dividible by the prime 144883729.
    2
  5640. I found a beautiful theorem and I want to share it with you via email. What adress should I mail?
    2
  5641.  @Mathologer  American schools are focused on the "utility." They teach you what works but not how and it's really sad. I didn't understand how integrals literally functioned until I was in college despite taking 3 years of calculus in high school
    2
  5642. 32:00 this reminds me of the Bresenham's line algorithm
    2
  5643. Been following your videos for 4 years and I have never been disappointed. Great stuff mate 👍!
    2
  5644. I was not expecting Sejong and k drama from this channel
    2
  5645. About the chalkboard problem: I have no idea what the nice way of solving it is in my head, but 10^2 is 100, so let's approximate all 5 of the squares with 100. Now we know that the sum is more than 500, but 15^2 is only 225 so the upper estimate is 1000. Now assume the answer is a whole number and only 730 is a notable candidate for the sum. (995 seems too much). So the answer is 2. Or something. Not at all rigorous, but this is the best I could come up with in 30 seconds.
    2
  5646. 9:58 Even better, if you want the total number of vertices, edges and faces, you only have to count the edges, double it, and add two. Thanks, algebra !
    2
  5647. Second puzzle is just the Fibonacci, you can get any new tiling by adding a line vertically (|) or two horizontally (=). This makes f(x) to be f(x-1)+f(x-2), when f(1) = 1, f(2) = 2 The 5 tilings of 4 (for example): == ||= |=| =|| ||||
    2
  5648. 2020 happy ending !! Best gift ever
    2
  5649. Very, very cool stuff. I suspected that old ppt generator would be the mechanism behind this dense and elegant tree. The equivalence between adding two numbers and drawing a triangle remains fascinating thousands of years later. And the circle stuff is pure magic.
    2
  5650. This is insane! Thank you and wait for more interesting math!
    2
  5651. It's amazing how much more intuitive this proof is over Conway's even though at heart the two are very similar.
    2
  5652. Good pronunciation , Thank you for the Video . I get it also as a grown up I refuse to say that Gabriel's horn is paradoxical , ...... If I can paint it , it's finite .
    2
  5653. 30:42 music by Ardie Son "Counterparts"
    2
  5654. The most memorable thing for me was the introduction of Gamma and the strange fact that as the numbers get larger, the approximations get closer and closer!
    2
  5655. This is the best channel on Mathematics
    2
  5656. My first thought was: Can this be introduced to higher dimensions? Eg. can you have 4D cube extended similiar way to get some ratioconvergence etc.
    2
  5657. Thanks for your videos, I just love watching them and trying not to wreck my brain. I wasn't a fan of Math but you are making a difference.
    2
  5658. nice
    2
  5659. And a happy 2022 to you as well.
    2
  5660. Man... i literally thought this was a Gothic horror video... the "curse of the lantern". On that note, I'm out. BYE!
    2
  5661. Can't wait for an hour long video at 600k subs
    2
  5662. Everytime I see a math video ..it makes me jealous about Euler and today ahh I'm jealous of ramanujan though ..no hate I love them both they are my favourite ❣️😊
    2
  5663. Anyone know the problem with the Chinese character tiles
    2
  5664. Its looks like overcomplification because we don't know right way to deal with such numbers and just brutforsing neat looking sings about them. Same story with sums of powers: geometric 4d rectangle packing was confusing but then algebra fixed it with pascal triangle for ALL powers. Maybe in future we are going to see ratio math like graph teory, matrices, negative nubers and imaginary numbers
    2
  5665. best is the ugly maximal overhangs. how ugly is the proof?
    2
  5666. Made it to the end. Truly impressed and the interplay of mathematics and justice from 2000 years ago.
    2
  5667. Merry Christmas and Feliz Navidad!
    2
  5668. When I was in my mid-twenties, I had to retake algebra two when I got to my university. Then I breezed through calculus 1, 2 and 3. I love calculus so much.
    2
  5669. Imagine if Riemann, Ramanujan, Galois, Poincare, S.Lie were all contemporaries of 1800s
    2
  5670. I can't help but point out that that thumbnail has a lot of subtle similarities to the Enneagram theory xD
    2
  5671. only 2, 4, 10, 28, 82 between 1-100.... What I need a paper to think about this...
    2
  5672. Oooh 4:17 i thought it's astronomia :)
    2
  5673. 16:00 it is 5
    2
  5674. 13:50 honestly my mind blown
    2
  5675. I like your Pi t-shirt. It's really cool.
    2
  5676. I love your approach to math. You take such complicated topics and make them so intuitive and easy to understand conceptually. I love you :D
    2
  5677. Dear Sir, I am a big fan pf yours. I have proposed a new lemma to easily understand why the harmonic series is divergent. Here is my paper https://doi.org/10.20431/2347-3142.0810004 if you get free time please have a look on my very short paper. Regards, Salman
    2
  5678. What do these look like as colours?
    2
  5679. Glad you liked it!
    2
  5680. Prior to completing a calculus course taught via uncannily mundane and l laborious methods, I loved math. Watching this video felt like reuniting with a friend I hadn't seen since I was 15. Thank you!!
    2
  5681. You may not want to do any more on this subject, but would "divided differences" be worth showing? The parallels to calculus extend there.
    2
  5682. What about opening a new discord server? It would be cool if people could talk to each other about these topics
    2
  5683. This is batshit crazy. Thank you so much.
    2
  5684. 1# puzzle: I guess there is a turnaround point at n=3 and the numbers go up from there? Was that the pattern or did I miss something? 2. Well I didn't have a specific tactic to handle this, I just did 100+121+144 in my mind and got 365. Then I've done 169+196 and got 365 again. So, result is 2. 3. Well, you got me there. Thought there was a pattern because 3³+4³+5³=6³, so I thought the next one would be 7⁴ but no it isn't. Big Mathologer always reminding us of the law of small numbers, great! Keep up the wonderful work, and happy holidays!
    2
  5685.  @Mathologer  yes I did. Well not just this one, the other ones were very good too!
    2
  5686. I don't like the visual proof. It doesn't help to re-concentualize the theorem in ways other than "if you construct such squares and rearrange them like this, you'll get another square, neat huh?"
    2
  5687. 6:15 I should think that representation theory is a good place to go digging for those sorts of things
    2
  5688. where can i get shirts like you wear
    2
  5689. The legend is back!
    2
  5690. Just watched an old Video and now there is a new Video! Da steckt wirklich großartige Arbeit in den Videos, vielen Dank <3.
    2
  5691. Once you've explained how the geometric magic squares could be constructed, they actually seem much less magical — although I would consider that ability to demystify the magical to be a form of magic in and of itself :D
    2
  5692. The most memorable part was the greedy algorithm. I tryed to use it before, to cut big tubes into different length smaller tubes, the best way possible with minimum waste... and man, it does not work for that (not always anyway haha)... If only I "knew" math better at that time...
    2
  5693. Can't seem to find one video on 5th degree polynomials. Care to help a mathematician out? @mathologer
    2
  5694. The Hydraulic Press channel shows many examples of thin walled tall "cylinders" folding into triangles when under extreme vertical pressure.
    2
  5695. the triagonals are taunting me.
    2
  5696. 4-1-k is just a reminder to prime yourself with a good 401k
    2
  5697.  @Mathologer  Thanks, and back to your video, your ability to display math/geometry with CG is so good, that I wish that students could have curricula that use these techniques and content. For a long time I've been distressed that so few people seem to know that the Pythagorean Theorem is "just" a special case of Ptolemy's Theorem. The latter seems so powerful that I always wondered how it could be used, and your video with the inscribed pentagon and heptagon and connections to phi ratio and the "golden R/S" is truly mind-blowing.
    2
  5698. heres an interesting one for you... if you think that the way children learn mathematics is correct... then you might wanna think twice about it... a simple question would be: how much is 4 - 1. easy right? thats 3 ofcourse. well, thats only half true. because 4-1 has actually 2 answers. 3 and 5. youd think "but you said 4 - 1... not 4 + 1." Well. thats what i said yes, but 4-1 CAN also be 5. Take a piece of paper with 4 corners, cut off 1 corner. which is equal to 4 - 1. you will notice that instead of your expected answer of 3, you will end up with an extra corner, which means 5 corners. ofcourse it is not a 90 degree corner, but ive never said that the corners had to be 90 degrees... a 110 or even a 125 degree corner is still a corner. conclusion: the results in mathematics are not always what you would expect and depend on the situation before you, and some mathematical situations can require from you to open your mind to other possibilities and answers.
    2
  5699. 5:36 1.nice
    2
  5700. I watch your videos at night please make the background black it's easier on the eyes
    2
  5701. Just wow
    2
  5702. People using manjaro raise hands.
    2
  5703. The leaning tower approaches the graph of ln(X) + gamma right? This is most memorable!
    2
  5704. Fredericus Johannes Maria Barning // Born 3 October 1924 Amsterdam (NL) // Passed away 27 June 2012 Amstelveen (NL)//
    2
  5705. first
    2
  5706. You can just barely see the script for the previous few seconds right at 7:20
    2
  5707. isn't it like the limits the result of f(a) can be different than limf(x) x>a so
    2
  5708. I 'discovered' the squares for myself during an exam in high school. I always was fascinated by exponents since
    2
  5709. 1:00 - How did Burkhard get a picture of me…?
    2
  5710. So it takes foreigners to tell us about our own glorious past. Such pathetic beings we are.
    2
  5711. Love your videos man. Wish I had mathematician and programmer friends 😪
    2
  5712. There is Python code here, http://rosettacode.org/wiki/Partition_Function_P#Python, together with P(6666). P(666) = 11956824258286445517629485 (Note this is three sixes, the link shows P of four sixes).
    2
  5713. The true gift in the gift to your wife would be the ability to create a right angle any time she pleased. Very thoughtful of you.
    2
  5714. I'm proud of myself for having come up with -a-b mod 3 rule after I paused the video at 3:10. I just knew you had to take the linear combination of the first row but just didn't bother working out what the coefficients are.
    2
  5715. 8:24 Me, a microtonalist: 3/4 is an inverted perfect fourth, 5/3 is a major sixth
    2
  5716. p(666) = 11956824258286445517629485 glad I could help. If this gets in any way a like I can post my dirty python code.
    2
  5717. Actually I got the trick from the 4 examples u mentioned in the beginning of the video.....where there were 10 hexagons in 1st row
    2
  5718. Sum[i^10, {i, 1, 1000}] 91409924241424243424241924242500 Good job, Mathematica =P
    2
  5719. Respected sir you are outstanding
    2
  5720. Please someone tell me, what software is used for making these animations
    2
  5721. Your adoring public is dying to know -- what does the T-shirt represent??
    2
  5722. I paused around 40:00 and took a shower break. Managed to figure out the finiteness proof and I was a little bit excited that I figured out even better approximation than Kempner (< 80), but alas, it seems that there's actually error in the video when the result is first mentioned and when I watched till the end, my proof was exactly the same as Kempner's...
    2
  5723. Great lecture prof 👌 👏 thank you sir
    2
  5724. 7:20 I see.
    2
  5725. the puzzle at 2:30 with the 1 big cross into 2 small ones can be solved with 4 pieces. Since the fabric was cut too big and none should be wasted and I figured it can be folded. The puzzle says to 'form' the two smaller ones. The solution I found was to cut the big one straight down the middle horizontally and then vertically so you end up with 2 L shaped pieces. Then by folding the ends in you will have an L shaped piece with a 1 thick corner and 2 two thick ends. this would form 4 small L's which can then be sown at the middle point. Here is a link to my thought process. https://imgur.com/a/OwD2Ndu
    2
  5726. absolutly genius
    2
  5727. Why didnt they teach this in school? It has nothing to do with common core is why.
    2
  5728. 2:30makes sense even if you take the og it would be 1-1/2+1/3-1/4... which will become 1/2+1/12... so it will be over half but under .75
    2
  5729.  @Mathologer  You... probotic! Why I never.
    2
  5730. I love your videos sir
    2
  5731. Great vid! Are equilateral triangles the only ones that fold into scaled (and/or reflected) copies of themselves? I would guess so, but am struggling to prove it...
    2
  5732. 7:00 So this is a plot of OEIS-A033178.
    2
  5733. Thanx sir for something easy and interesting than that all heavy alzebra 🖒🖒
    2
  5734. You've just asked (24:27) what happens if you apply some ears repeatedly, in a combination of ears. Looking at the "eigenpolygons", I notice they're not exactly axis-aligned, although three of them are close and the other is almost diagonal-aligned for one of its points, but I'm guessing that's a distraction and what we're doing is, for an original p-gon with corners represented by complex numbers z(n) for n in p = {0, ..., p-1}, a decomposition of form z(n) = sum(j in p: q(j)*cycle(j*n/p)) where cycle(t) = exp(2*pi*i*t) is the unit-period traversal of the unit circle; q(0) is the centroid (cycle(0) = 1 so j = 0 just contributes q(0) to the sum), the phase and scale of other q(j) configure the orientation of the p-gon traversed j steps at a time. The real fun must be in proving that, for any z, there is a q that makes this work; but, give or take normalisation (my guess is 1/p), I'm guessing q(j) = sum(j in p: z(n)*cycle(j*n/p))/p will do the job. Then application of each of the "ears" is linear in the two end-points of the edge to which it's applying the ears, so maps onto doing the same to each cycle(j*n/p) polygon contributing to the sum. The effect on these regular polygons is a simple scaling, scaling each q(j) by a j-dependent (complex) factor that's zero when the ear is either j/p or 1 -j/p turns (depending on the direction of our traversal, I suspect). If you apply k/p turn ears for each k from 1 through p-1 you'll thus annul all q(j) except q(0) and be left with the centroid. Repeated application of a given "ear" angle thus just repeats that scaling of each q(j) by the same j-dependent factor, so doesn't annihilate any q(j) that the first application didn't. So if you apply k/p turn ears for all but one of the k from 1 through p-1, you'll get a regular p-gon; and repeating some of those ears will only change the scaling. Now to resume watching and see if I'm wrong ...
    2
  5735. You answered a question I had a long time ago in undergrad when the professor used a shift operator on the recurrence relation that leads to the golden ratio. Great video as always! I'm in Greece rn on holidays so a little rho and sigma would go nice :)
    2
  5736. 13:49 it was not immediately obvious to me that orange and blue portions would cover the white square completely. I was initially expecting that there will be a white leaf shape leftover. It would be good to at least mention that problem.
    2
  5737. Where do you get your shirts from?
    2
  5738. 20:55 Uh-Oh! @DaveMcKeegan and @SciManDan are going to have to work much harder in the near future! 😁
    2
  5739. Incidentally, the mod 2 Pascal triangle is the same as the mod 2 color game since -1 (or 2-1, if you prefer) is odd. So in some sense, the same starting point generalizes in two different interesting ways.
    2
  5740. I count 644 ways to tile the glasses, by decomposing them into several separate blocks. Is there something special about this number or am I missing something?
    2
  5741. I'm from Italy, I went to school in the 90s, and I learned Heron's formula both in elementary school and in middle school, which have the same curriculum for everybody. So in theory everybody in my country should know it exist, unless they forgot. In high school the curricula split. I chose the one called a scientific lyceum (a popular one, almost a quarter of my fellow countrymen and women choose it) which again is fixed but has a focus on math and sciences, including learning to prove theorems using Euclid's Elements in the first 2 years (ages 14-15) and calculus in the latter 3 (ages 16-18). All in all I'm quite satisfied with my education. Whatever I didn't learn was not due to lack of resources, but due to excessive dumbery on my part. By the way, ϕ + 2 + ϕ = 2 + 2ϕ = 2(ϕ + 1) = 2ϕ² = ϕ × 2 × ϕ using the characteristic property of the Golden ratio: ϕ + 1 = ϕ²
    2
  5742. That was fascinating. By the way, bottom row is all ones 1, next row is natural numbers n, row after that is 1 more than triangular numbers or central polygonal numbers n(n+1)/2+1 OEIS A000124, row after that are cake numbers (n^3+5n+6)/6 OEIS A000125, finally last is Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24 OEIS A000127. Great video as it's far more satisfying to derive them than cheat with OEIS though it contains a wealth of interesting facts, recursive formula, etc
    2
  5743. Great video as always (I know this comment comes a bit late)! What is even more mind-blowing (and would require another Mathologer video!) is that even the set of numbers that we can describe with a finite number of words (and symbols) is countable. This implies that the vast majority of all real numbers remains for us a complete mystery beyond our grasp. To be more precise, let us decide a set of symbols, for example S=English Alphabet U Digits U parentheses U logical connectives (and, or, not, imply) U {symbol for belonging} U quantifiers. We may want to add some more special symbols (for example +, /, etc.) so in the end we may agree that S has at most - say - 200 symbols (or in general N symbols). You can list all the possible finite "sentences" made with the symbols from S starting from length 1 (each individual symbol), then length 2 etc. So, all the set of finite sequences of symbols from S is countable. Some of these sequences will not make much sense (e.g. "2+"). Some others will represent a number (for example "pi" will be pi, "square root of 2" will be the square root of 2 and so on). Well, the numbers represented by these sequences will constitute a countable set, thus a negligible part of all the real numbers. Notice that the same applies if instead of a finite set S we use a countably infinite set S. The resulting set of sequences is the union of S, SxS, SxSxS (Cartesian Products) etc. which is a countable infinity of countably infinite sets, thus it is countable as well and as a consequence it covers 0% of all real numbers! I bet that this will beat the 0.999... = 1 incredulity...
    2
  5744. 13:36, isn’t that a 2x2?
    2
  5745. Because of abstract polytopes and their hasse diagrams, and recurrence with different types of multiplication?
    2
  5746. Brilliant and enjoyable, and all through simple marking and counting. "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy" indeed. Somewhere in that labyrinth of marking and counting is, I bet, a key to the primes. Watch out, Riemann!
    2
  5747. 9:45 Who doesn't have 988,816 as change 😅
    2
  5748. Sorry mate, don’t see the point. What’s wrong with ai images. We can discuss the estetics, but I don’t agree with a general critisism.
    2
  5749. My favorite definition of gamma is the limit as n->infinity of: (1 + 1/2 + 1/3 +1/4 ...+1/n - 1/(n+1) - 1/(n+2)... - 1/n^2 - 1/(n^2+1) ... - 1/(n^2+n) )
    2
  5750. This should be named the Aussie Hair Principle
    2
  5751. Burkard wears two rings because he is married to his wife and to math
    2
  5752. After you drew the incircle for heron's formula my mind immediately went to the Tan identity i.e, Tan x +Tan y + Tan z= (Tan x)(Tan y)(Tan z) if x+y+z=nπ Using it, it was pretty straightforward to see how heron's formula worked.The visual proof was really elegant to see though.
    2
  5753. This video gave me an early Christmas gift. I don't want any gift for the rest of the year.
    2
  5754.  @Mathologer  I have my total faith if this thing ever built with superconductors in a matter-antimatter fluid, it will surely open something like a wormhole portal of multiverse , maybe we should seek for Dr. Stephen Strange on this ?🧐🙃🤨
    2
  5755. Watched the whole thing. Math background: undergrad in electrical engineering. Please go even futher in future videos
    2
  5756. I liked the 700 yr proof. Math was so advanced even back then
    2
  5757. I've managed to solve the 3^4 and the 4^4, I'm going to try the 5^4 next. This video really helped me to understand the puzzle. Thanks!
    2
  5758. awesome video :) thank you Burkard!
    2
  5759. 16:57 Is he going to say it? derivative??? 17:00 yaaaay!!! Is the volume of a sphere the (3d) surface area of a hypersphere? Would an integral work for calculating the hypercubic volume of a hypersphere?
    2
  5760. As an Indian nowadays I wonder what history have we been taught after the colonisation ended.
    2
  5761. For me the most memorable part was the fact that Gamma is actually a thing. Because when I watched the video i thought: "Well okay, this is going to be one of those approximations where nobody cares about how "wrong" it actually is." Turns out I am wrong and this is massively important.
    2
  5762. As always, the presentation was great and the closing music was awesome!!!! Thank you so much!
    2
  5763. At age 65, I had never seen either of the proofs even though math was one of my strongest subjects.
    2
  5764. y'know, the warm up puzzle i actually once thought of in like 8th grade in geography class cause i was bored, but i had no knowledge of calculus, so lets just say it stumped me for a while (until i aproximated and than guessed e-1 cause you know, e is pretty famous), so seeing it as a warm up puzzle in this video made me feel a bit nostalgic, so thanks i guess
    2
  5765. WOW.
    2
  5766. That was great (as always.) and for once I followed all of it. Thanks.
    2
  5767. #74 LIKE!🦋 8:34AM 02/04/2023.
    2
  5768. perfect timing for a long video! <3
    2
  5769. 3:29 - And you're... Wright!
    2
  5770. In this case, your proof was too complicated.
    2
  5771. Professor, at 10:08 I believe this proof fails. I don't know this, but I'd like to show you what I have come up with and get your opinion. May I make a response video and tag you so I can show you my work?
    2
  5772. Yes
    2
  5773. It might be helpful to choose colors less similar to watch other for those of us that are colorblind.
    2
  5774. Your videos are always interesting and fun to watch, thank you for making them!
    2
  5775. Great video!
    2
  5776. Very nice. I never heard of this before (IANAM) but as someone familiar with image and signal processing this is not strange to me - it is just a curvy analogue of the Dirac pulse. Consider a signal plotted along X as a rectangle with unit width along X and unit height on Y so it has a finite area (area = 1, finite). Then simply shrink the width of the rectangle while preserving the area: the top goes up and up and up. As the width tends to 0 so the top tends to infinity in Y. The area, however, is constant - finite area with infinite height. If we started with an area of 2 (units squared) then the height would become infinitely higher than the first case. A very nice and easy way to explain even to kids that 'infinity' is not some constant number and that there can be infinitely many infinities, all infinitely bigger or smaller than some other one. Also a good way to teach that if you know where an infinity came from - you can tame it (in the above examples, first infinity over second infinity = 1/2, not 'undefined'). Thanks for the video - excellent as always.
    2
  5777. If only economies were treated this way. Wonderful episode!
    2
  5778. Wow! This is awesome!
    2
  5779. I used to use a circular slide rule in the 1970's at school. Brings back memories.
    2
  5780. Nice Pi T-shirt!
    2
  5781. Wait, you have Paul Dirac as Patreon !?
    2
  5782. IIRC, a round slide rule is shown in Hunt for Red October when the navigator of the Red October is calculating the timing for the turns through the undersea canyons.
    2
  5783. I believe that the 6 and 9 pices at 23:38 cannot be arranged into making the dented square. But maybe I missed it !
    2
  5784. Visual animated proof at the end + good music = piece of art !
    2
  5785. 13:49 I got Fibonacci numbers. I think it's because you can get all the 2xn tiling combinations by adding 2 horizontal dominoes at the right of all 2x(n-2) tiling combinations and by adding 1 vertical dominoe at the right of all 2x(n-1) tiling combination. Thinking about those 2 sets of tilings combinations, they do seem mutually exclusive and exhaustive. (^.^)
    2
  5786. Those coins remind me of the difference between sidereal day and solar day. In one year, one day is contributed by the earth travelling around the sun (if the earth wasn’t rotating on its axis at all a day would be one year long but there would still be a day). Depending on you rotation direction you have to add the day or subtract it.
    2
  5787. I made it to the very end
    2
  5788. You are one of the Great knowledge provider.... Thanks from India...
    2
  5789. Mind-blowing mathematics channel ❤...
    2
  5790. Using property of chords that product of 2 segments is constant So use it ...yeh
    2
  5791. Very nice as always!
    2
  5792. Shalom Mathologer. I have said this several times before, and it is very interesting to hear what is your opinion about it. There is a small problem with the GOOGOL number And I will explain: Only numbers with an amount of zeros that is multiple of 3 can have a "name". The number 1,000 has a name. Thousand. The number 1,000,000 has a name. one million. now, this million, with one more zero will be called ten million. 10,000,000 (7 zeros) Another zero and it will be one hundred million. 100,000,000 (11 zeros) Only after we add another zero (12 zeros in total) will we get a number with a new name. Billion. Therefore the number GOOGOL, 1 with 100 zeros, can only be ten "something" ... Assuming there is a number like 1 with 999 zeros, suppose its name is Boomboom So only if we multiply it by ten we get one and a thousand zeros. And it will be called Ten Boomboom. GOOGOL can't have a name. It can only be 10 "something"
    2
  5793. For a change? Wow...
    2
  5794. Good day, thank you for a great presentation. Would it be possible to generate all combinations and variations of bits in a source code? We could be running whatever we could imagine. But only with your help. Bye.
    2
  5795. When we're considering a googol, even your 50+ minute videos are pretty smol
    2
  5796. Totally unrelated, but @3:10 I wondered whether there was any “sculpture to real” AI...then watched in great amusement at your use of Great Nostalgia! Nice touch!
    2
  5797. I use the term "Arctic Circle Method" to mean a method I came up with for finding the declination and right ascension of any star or object in the sky in some past or future year at a given place. The idea is to view the object on the Arctic Circle at sidereal time 18h. At that time and place the entire ecliptic and zodiac are along the horizon. Because of precession, where the zodiac constellations are is dependent on the year. So what I do is compute (other year-present year)/(26000 years), then rotate like a kitchen timer the sky by that amount, so for 15,020 AD, I would get (15020-2020)/26000 = 1/2, telling me to rotate the sky 180 degrees so that Sagittarius is the summer constellation (N. Hemisphere) and Gemini/Taurus is the winter constellation the Sun is in. After that, I take the new altitude/azimuth at the Arctic Circle and translate it back to the original place's right ascension and declination.
    2
  5798. I made it to the very end. Also, where can I find the solution for the final puzzle? Edit: The next numbers are 55 and 144 Edit2: I think that the pattern is X_{n-1} + X_{n+1} = 3 X_n
    2
  5799. I excelled fairly well in math and often watch and study math related things for fun.... yet my mind was blown from the beginning just with the added odd numbers to get the squared integers lmao.
    2
  5800. About the kidnaping services provided by the mathologer I only have one thing to say: beam me up Scotty
    2
  5801. Hey, I hope to meet you in person sometime. Live nearby. This video is amazing.
    2
  5802. ​ @Mathologer  Thanks for the interesting point 2) above. 👍 So yeah, it doesn't matter that it's specifically a constant multiple of the projection, just that it's larger.
    2
  5803. Hello
    2
  5804. For the people who don't get this one: https://www.youtube.com/watch?v=pPkWZdluoUg :)
    2
  5805. 11:40 this reminds me when I was in 7th grade and I ended up with a fraction in a fraction and my female math teacher gave me 0 points for that and said "there is no fraction in a fraction" insane how stupid she was when I look back at it
    2
  5806. SATOR AREPO TENET OPERA ROTAS
    2
  5807. That infinite product for sin x, just brilliant.
    2
  5808. Why doesn’t your shirt say 25.8070?
    2
  5809. You should not use ai but cool video
    2
  5810. What’s the Easter egg between the numbers on the slide at 23:08? I mean, I’ve noticed that one is pandigital, but I still don’t get it.
    2
  5811. Mathologer Oh, that’s cool!
    2
  5812. Great video!
    2
  5813. Final problem: Tile the whole plane and move the 2x2 cover to the position that covers the four corners. You only need to demonstrate that you can "cross the boundary", which is easy to show as we have alternating pairs when we move the 2x2 cover horizontally or vertically and we have an even by even setup.
    2
  5814. Looking forward for that next video, hopefully you can touch on the subject of imanants generalized determinants, I find the subject fascinating, but very obscure and confusing
    2
  5815. Is there a number that, if we put instead of 1, makes the harmonic series dont miss every integer?
    2
  5816. A Cow Orker once told me that “the juice of one lemon, if spread thinly enough, can cover the entire state of Texas.”
    2
  5817. 14:43, this wicked sense of humor had me in stitches.
    2
  5818. I introduced my high school math class to Heron's formula - we used it, along with 2-variable calculus, to prove that the equilateral triangle has the largest area given a fixed circumference.
    2
  5819. At one point it looked like you were going to solve the Y = 2K problem. ;-) But seriously now, if you intend to take on proof #4 (there are 5 regular polyhedra), please don't forget that there are other cases to consider. IE when allowing self-intersections (and why not?), there are 4 regular star Kepler-Poincet polyhedra. And when you allow polyhedra living in repeating finite spaces (EG the Asteroids game), there are also the 3 Coxeter "skew polyhedrons". I argue that together, these 7 are absolutely just as deserving of being called 'regular' and Plato's original 5. And when we keep all the vertex and face requirements but allow for more than one kind of edge, there admits a beautiful zoo of repeating polyhedra that I call "infinite regular polyhedra". See my treatment and collection of exceptionally beautiful examples at http://superliminal.com/geometry/infinite/infinite.htm
    2
  5820. Ramanujan, gone too soon. Anyway, why is everyone talking about Ramanujan lately?
    2
  5821. I really want that pilot watch with the integrated slide rule now.
    2
  5822. I still have the E6B that I used in 1966 when I got my pilot's license.
    2
  5823. Although the proof by Nicole is graceful, the most graceful proof is the sum of arithmetic progression by GAUSS, reversing and adding back. A proof that makes you think " huh, I could have done that too. " , but inside you know that it might not have struck to you. 😂
    2
  5824. 2^n being parallel to e^x blew my brain, the whole video is just a testament to elegance of mathematics.
    2
  5825. What is the best way to lace your shoes? Me: Well...just tuck it inside your shoe! Edit: sorry guys, don’t try that, it really hurts my foot now! 😅😅😅😅
    2
  5826. I don't get how @ 24:35 area of sector is equal to the angle.
    2
  5827. It's definitely hard to decide which is better: the block stacking or the thinner and thinner no-9s illustration. I would probably go with the block stacking. Thank you Mathologer! As always, everything is clear, with very easy to understand illustrations, and bright colors for dark math magic ;-)
    2
  5828. This is all quite fascinating, as I implied in my earlier comment. As for the relation of circle (disk), to surface of sphere, to volume of sphere, I discovered a little gem in '65, just after graduating HS, which makes me wonder whether it has been known for longer. It relates, using Pappus' Theorem, the capacity of an N-ball to the surface capacity of an (N+1)-sphere, which then gives the capacity of an (N+2)-ball using the " surface=d(volume)/dr " relation [the (N+1)-sphere being the surface of the (N+2)-ball.] Then, starting with the formulae for N=1 and N=2, the formulae for all greater N's can be obtained. And all of them fit neatly into the formula V(r;N) = (πr²)^(N/2) / (N/2)! where the factorial definition is extended, using the gamma function: x! == Γ(x+1); more formally, replace x! with ∏(x) == Γ(x+1). If you think this could be a good theme for a Mathologer video, contact me for details; I don't have the expertise to make the wonderful animations you & your team make. Fred YT seems to prevent posts that contain emails, so if there's a way of contacting me through my channel...?
    2
  5829. Good to hear from you again :) Yes, all this is very beautiful mathematics. Have a look at this wiki article https://en.wikipedia.org/wiki/Volume_of_an_n-ball (skip to "Two-dimension recurrence relation")
    2
  5830. Thanks, this was not shown in my schooling
    2
  5831. That mask is not going to help that bird.
    2
  5832. This is beautiful and elegant. Thank you!
    2
  5833. But isn't the hairy example rather a normal distribution with an average number of hairs being h? It would be odd to have a uniform distribution of the number of hairs. So the argument is not the best one? It's the average of the normal which is more than then pigeon hole principle predicts?
    2
  5834. Funnily enough, if u take the multiple paths by taking the top peg and putting it in the middle peg, you’ll still have 13 moves, while taking the right path straight down will be 8.....
    2
  5835. From a reaction on another comment I found your update in the description with a translation of his info on this website, love your dedication! Want to add also: Thanks for bringing wonderful mathematical facts into my life :)
    2
  5836. 11:15 You went for RGP because we already used the Term/variable "B" as a corner of quadrilateral
    2
  5837. @3blue1brown is a patreon sponsor of Mathologer!!!
    2
  5838. I wonder, if there is a Pentagorean Theorem...
    2
  5839. Considering squares found in a 3d-cube grid: I could only find squares that lay within a plane formed by the grid itself; that is to say, any two vertices on any squares I formed always shared at least one coordinate value. Is this true of the greater infinite euclidean 3d-space? Or is there some large scale in which a square exists that is "tilted twice?"
    2
  5840. 7:32 Because there was no YouTube back then? ;)
    2
  5841. Best chapter was : Credits ;)
    2
  5842. I loved this video and all recent videos. I'd love to add a couple of lessons about this to my students at the end of this year. I have a question: are you in contact with an Italian professor called Mario Barra? I studied under him some ten years ago and he was researching many similar topics - Pascal's triangle and n-dimensional geometry, gnomons, falling and rising exponentials, "discrete" calculus, so many things in common. Of course it might just be a coincidence but it that case there could be so many things that could be integrated and complemented to your work. Thank you for the video and the excellent presentation as always!
    2
  5843. I wonder if the 27,110 number at 29:50 has anything to do with fifteen sliding puzzles, or if that's just a visual coincidence =p
    2
  5844. The second set (4k+3) can also be expressed as 4k-1.
    2
  5845. misread the title as Conway's IRS. i was very curious to learn about John Horton Conway's tax collection system.
    2
  5846. Proud of having Indogenic DNA in me.
    2
  5847. This is a really cool video and reminded me of when I first came across generating functions when figuring out the probability in regard to the sums of the outcome of dice, and this actually deepened my understanding of them a bit as they always felt a bit like magic to me.
    2
  5848. I vote for the partial sums not being integer. It's the only old result I didn't know exactly how to prove.
    2
  5849. I vote the no 9s sum proof. It's very simple and beautiful, and the way you animated it is a biy more different than all your other proofs. Coming in second is that crazy optimal tower. You may not like it, but this is what peak performance looks like.
    2
  5850. I have done a so much work on Z=X²-Y² it nearly drove me insane! Spent like 2 years on it with the hopes of making progress in solving the prime factorization challenge. Anyhow...
    2
  5851. it looks like it is related to amicable numbers. can it be extended to more than 2 numbers like A+B+C = ABC or A+B+C+...Z = ABC...Z? If yes, would it make amicable numbers easier to calculate?
    2
  5852. Just yesterday I was memorizing the formulas for the volumes of 3D objects for a quiz on centroids for my Statics subject and I thought about how fascinating it was that the volume of a hemisphere, paraboloid, and cone of similar dimensions turn out to be 2/3, 1/2, and 1/3, respectively, of the volume of the cylinder that you can exactly fit all of them in, though I never really understood why until now, just a day after the start of my query. Thanks so much! This has got to be one of my most favorite YouTube videos of all time now for explaining simply such a seemingly complicated topic.
    2
  5853. I love ur videos, sir
    2
  5854. that's not enough to prove it's the log.. At least, for rational numbers, the function "sum of power of the product of primes" also allow f(a*b) = f(a) + f(b) for instance 8 = 2^3 so f(8)= 3 and 6 = 2^1 * 3^1 so f(6) = 2 f(8*6) = 2+3 And for that function all the other properties of log follow.. The only problem is I don't know how to define F for irrational numbers.. I think it could be generalized : any function that transform a number as a product of powers (of fixed values) into the sum of powers (with fixed coefficients) have that property (ok maybe I went too far) In fact if you take the decomposition of a number in primes q and consider the linear combination of powers p[q].. any function in the form sum ( p[q] * x[q]) where x[q] are different from 0 also have this property.. I believe Log is the continuous function.
    2
  5855. i like the music at the end 😌
    2
  5856. Greatly enjoyable ❤ as always. Please add links to your other related videos to the description section, and consider summarizing other contributions of Madhava and of other forgetten geniuses like him from India and other parts of the world.
    2
  5857. Who is Ptolemy? I know just Ptolemaios.
    2
  5858. I found the video really interesting. I could follow most of it (mainly not 100% following some of the later derivations and substitutions, but understanding the overall gist of what was happening). Background: I studied mathematics at A-level and am familiar with the maths involved here (sigma/summation, matrices, calculus, pascal's triangle, etc.), but was not familiar with the Bernoulli numbers or the Euler-Maclaurin formula. I noted the similarity to the Taylor series for calculating sin/cos/e/...
    2
  5859. I was just searching about it and suddenly your video came in notification .what a coincidence
    2
  5860. Happy Holidays
    2
  5861. If we taught math like this and this learning method to 3rd and 4th graders, our population might be more math literate and knowledgable... now with the spelling lesson ; )
    2
  5862. This class is sooooooo wonderful, I have gone through a lot of ‘AHA!’ moments. It's so beautiful to see every field of mathematics to be connected— geometry, algebra, Fibonacci series, Pascal's triangle, matrices! For the formula about n-th Fibonacci number, if we put -1 in it, we get 1.
    2
  5863. Great video. I'm new to this channel I gotta say I am blown away how you're able to talk about the intricacies of mathematics for a half hour and at the end I was disappointed that it was over.
    2
  5864. So hard. Got lost quite soon😟
    2
  5865. But we indians learn in our school books that the west invented everything and we indians were barbarians before the british. Lol. Still we follow the same education system thanks to congress being loyal to the west.
    2
  5866. Beobachtungen über das Gefühl des Schönen und Erhabenen..Immanuel Kant ...1764....42^2
    2
  5867. 39:10 here is my wild geometric proof for the diagonal dealing problem. step 1 - cover the plane with an infinite grid of fundamental rectangles of area p*q where p & q are relatively prime. dont do the wrap around thing and instead draw diagonal lines and cover the plane with these rectangles and make a rule that any 2 points in the plane are "equal" if they are in the same part of the fundamental rectangle (if the diagonal line you are about to draw passes through one of the grid squares in this analogy you place a card in that grid square while diagonal dealing) step 2 - sit at the origin of the plane at coordinates 0,0 and find out how many "grid line intersection points" you can "see" (the rule is you can't see through a grid line intersection point, its an obstruction that will block your view; you can only see things if you draw lines connecting them to the origin without hitting one of the grid line intersection points, the theorem is you can only see the grid line intersection points of relatively prime coordinates when you are sitting at the origin, since otherwise your view will be blocked, this is an easy consequence of a result in a previous video about rational angles which is not hard to prove) step 3 - realize that if p & q were not relatively prime, your view would be blocked at least once by one of the grid points before you finished drawing the whole line. if you tried to draw the diagonal line just starting from the origin and going along the line with the required length you would have to overlap cards on each other corresponding to the number of times your view is blocked in this analogy
    2
  5868. This is the video I needed around 1965. It hits points from different careers spread over a lifetime. Background: bachelor’s in physics, master’s in computer science. Never finished PhD, busy applying math. Even had to bet my life on math a few times. (Don’t tell me your orals were brutal. I can top that story.) Once upon a time a teacher showed me that sums of powers formulas could be derived using Legendre polynomials. I used a system called FORMAC to actually print out more formulas than anyone would really want. This was done with punch cards, which were only slightly better than chiseling bits out of granite. To illustrate the level of perversity, ask an old timer about punching cards with the scientific character set using an O26 keypunch. Recent reference, check out proof using Hermit polynomials on zeta and xi function. https://www.pnas.org/content/116/23/11103
    2
  5869. 30:44 "i give SUM links in the desciption" - Mathologer idk if that pun was intentional xD
    2
  5870. Are these stories of Ramanujan’s instant insights really true, or are they perhaps apocryphal? I wonder if Ramanujan had synesthesia, to get these insights so quickly all in his head.
    2
  5871. You make me proud, as an Indian!
    2
  5872. here in india, we are taught heron's formula... but just the formula no proof, no reasoning just handed the formula to apply to a bunch of problems its quite sad that we just plug and chug numbers in this formula in school : (
    2
  5873. a
    2
  5874. What of a sequence for a circle or any form of loop sequence. if it's possible to list simply. Will probably think it out later
    2
  5875. Each piece moves according to the "abacaba" pattern, with each successive letter representing the next largest piece: A ABA ABACABA ABACABADABACABA ABACABADABACABAEABACABADABACABA ABACABADABACABAEABACABADABACABAFABACABADABACABAEABACABADABACABA
    2
  5876. Beautiful as always 😍
    2
  5877. 23:40 Nice ”Countryballs” -reference/ -parody 😅.
    2
  5878. Dang, 33x33 requires sets of numbers adding to 17985... the wrong permutations you'd work through would be soooo maddening. For anyone who doesn't feel like coming up with the formula for magic square sum S with side length L I'll add that a couple of scrolls down. It arrives from a simple combination of mean value, area of a square, and the triangular sum formulae- (L^3+L)/2 Let L=8 and you arrive at rumanajin's 260 Let L=3 and you have 15 I'm very fond of patterns like this.
    2
  5879. This video calmed me down with all its magic glory
    2
  5880. I wonder what Tesla was ACTUALLY referring to. He was a super genius, so he obviously was talking about something very profound. There is so much misinformation about Tesla, and this vortex diagram is a great example of Tesla mysticism. Formula: 1. Take a Tesla fact or a quote about Tesla. 2. Take a new age religion concept. 3. Squish #1 and #2 together into an awkward, nonsensical theory. 4. Claim Tesla believed this theory. 5. Make a youtube video about it. 6. Profit.
    2
  5881. For the programming challange, I think dynamic programming is an efficient solution for you.
    2
  5882. holy moly. i was watching a numberphile video a couple years ago i think during a family christmas party, where they display 10^5, 20^5, and so on until 90^5 and i was very bored and like, hm, what if we subtract these numbers but remove the farthest right non-zero digit? i went off to calculate and WTF. i was amazed to see the "12" appear on the 5th iteration of the 5th power list that i subtracted i thought this was just a "math trick" and was only trivial, didn't know i ended up in a mathematically parralel universe xD if ur confused, basically 1^5 = 1 >> 1 2^5 = 32 >> 3 3^5 = 243 >> 24 4^5 = 1024 >> 102 5^5 = 3125 >> 312 6^5 = 7776 >> 777 7^5 = 16807 >> 1680 8^5= 32768 >> 3276 9^5 = 59049 >> 5904 then, arrange em in a triangle just like the first minutes of the video and BOOM YOU GET A DIFFERENCE OF 12 IN THE 5TH ITERATION!1!1!11!
    2
  5883. 4:33
    2
  5884. Math-o-logger
    2
  5885. The answer to the puzzle at the end is that the small square is one fifth the total square. To see this extend the grid which is in the orientation of the smaller square. The big square re arranged fills 5 squares on the grid. Thank you @Mathologer that was a nice ah ha moment. :)
    2
  5886. Dear Mathologer, thanks so much for your great videos! For this one, I admit I am a bit puzzled (ha) as I cannot square (haha) the last demonstration. The demonstration states (around 28:40) that "since all the spaces between the dot shrinks to zero, they cannot possibly be all integers." If I follow this demonstration, what prevents me to (incorrectly) "prove" that cosinus of 60 degrees (1/3*Pi) is also an irrational? I take all the demonstration, I change "octagon" by "hexagon" and 3/8*Pi by 1/3*Pi, and I (wrongly) conclude that cos(1/3*Pi) is irrational. Where is the flaw here? Thanks for your reply and/or replies by other viewers.
    2
  5887. Shortest solution for 9 discs: Superdiscs: 2+3+4 Order: 2, 3, 2, 4, 2, 3, 2 Sum: 3+7+3+15+3+7+3 = 41
    2
  5888. I keep thinking, "No! Don't stretch it further! You'll break the rubber band!"
    2
  5889. Drat! Looks like I lost my "heart" by editing in some extra explanation. :–(
    2
  5890. One more tidbit – that relation involving 4 consecutive Fibonacci numbers (≈40min) can be boiled down as follows. Decrement the indices [i.e., replace n with n–1 throughout], then put all four of them in terms of the middle two: F[n-1] F[n+1] + F[n] F[n+2] = F[2n+1] (F[n+1] – F[n]) F[n+1] + F[n] (F[n] + F[n+1]) = F[2n+1] F[n]² + F[n+1]² = F[2n+1] Which may be a bit more familiar to Fibonacci aficianados.
    2
  5891. Fun fact : the number pi miraculously appears in random tilings of large rectangles. Looking at the middle of the lower edge of the rectangle, there are three possible tiles that cover a given square: two horizontal and one vertical. We may naively expect a probability 1/3 for each possibility, but instead there is a probability 1/pi for the horizontal tiles and 1-2/pi for the vertical one (Stephenson & Fisher 1963). Hence you can try on your computer to evaluate pi using random tilings, in (arguably) the most inefficient way known to manking !
    2
  5892. Nice! I could have used this recently when working out hole coordinates in a metal place knowing nothing but the distances between them
    2
  5893. This is no doubt one of the very best Mathloger videos. Thanks for creating this, I thoroughly enjoyed watching it.
    2
  5894. 15:40 please do tell that story some day! Sierpinski triangle is my favorite fractal!
    2
  5895. E P I C 💯
    2
  5896. Still waiting for that video on this channel which i won't get on first watch...
    2
  5897. Maybe Point Nemo: https://en.wikipedia.org/wiki/Extremes_on_Earth#Oceanic
    2
  5898. So beautiful! This is why I LOVE Euclidean geometry and ratios more than numbers. My eyes are moist.....
    2
  5899. he was the one to beat the first strand type game
    2
  5900. The slide rule is an addalog computer
    2
  5901. The most incredible part is definitely in chapter 6. Amazing to think that there's a subseries that sums to any positive number.
    2
  5902. This video was quite a change of pace, with a lot of novelties! Creepy deepfakes, metal music, "common" subject like calculus... This video is different, I got a bit surprised! But one thing remains constant: the quality of the explanations. Well done!
    2
  5903. You red the Hilbert book 40 years ago? What on earth: I guessed you around 45. Merry Xmas from germany.
    2
  5904. I was utterly destroyed by the fact that I can get to the end of a whole ass (extremely good) video about the maths behind the Tower of Hanoi puzzle and be told that, if I want more, I can go read a 270 page book someone has written.
    2
  5905. You know that when Mathologer posts, it will be a good day
    2
  5906. I think I need to watch this one again
    2
  5907. Square root of 9th!
    2
  5908. 2:10 ah.... Tasmania??
    2
  5909. after a long time I am this early
    2
  5910.  @SupGaillac  Yes, the premise of the video presupposes a certain naivete.
    2
  5911.  @jaimeduncan6167  I'm just questioning the supposed "natural fairness" of the system. It probably did work well back then, in a primitive and financially inefficient society. You would think that all the modern genius financial engineers would have opted for something like it rather than the proportional method. But they haven't, and there's got to be a good reason for that.
    2
  5912. I agree completely. People would conspire to split debt into smaller chunks. Even if the court decides consolidation. Consolidation is very subjective and we want to come to an objective mathematical solution here. One that is agnostic of any conspiracy.
    2
  5913.  @kent_hdd  The word "claim" has more than one meaning. One is an assertion, regardless of its merit. The other is an assertion backed by legal evidence. Without evidence, you have no legal claim.
    2
  5914. 1. 33*(33²+1)/2
    2
  5915. I love the animated algebra :)
    2
  5916. That was pretty fun. I didn't get the answers right. I had an understanding of the 1/1 coin that solved the question but didn't solve the question for the others, so I learned something cool. :)
    2
  5917. I think C to power N + 1 give all the combination where C is color simple mathematical trick
    2
  5918. Now if only the bugs remained the same size yet their paths became smaller. (:
    2
  5919. As a Phi fan, I loved this video especially. Great work. Thanks.
    2
  5920. I learn mathematics here more than our school and colleges.
    2
  5921. I already knew most of the stuff in this video. But the thing I found the most interesting was the method to create towers that grow off the edge faster than logarithmically because I had never seen or considered that before. The last two chapters were also interesting.
    2
  5922. Most memorable: everything.
    2
  5923. Continued fractions, and their connection to the Euler-Maclaurin formula would make a nice addendum. Bravo!
    2
  5924. For people who like to see the algebra version of these patterns, if we say that the largest term to the left of the equals sign is n², then the pattern with squares looks like [ ∑ (n-i)² from k=0 to k ] = [ ∑ (n+i)² from i=1 to k ], where k is the number of terms on the right (or 1 less than the number of terms on the left). In fact, solving that equation gives n = 2k(k+1), or n = 4(1+2+⋯+k), so the only way to make that pattern with is with 4(1+2+⋯+k), as revealed at 1:21. The pattern without squares is [ ∑ (m-i) from k=0 to k ] = [ ∑ (m+i) from i=1 to k ] with m being the last term on the left, and this is exactly equivalent to m = k(k+1) = 2(1+2+⋯+k).
    2
  5925. You're the man. I just loved your video.
    2
  5926. The closed formula for the mortal enemy's sequence is: E_n = 0.8469 (1.5747)^n + 0.2636 (1.3802)^n cos(1.7805n + 0.9507). All the decimals are rounded so this won't be very accurate past n = 15 or so. If you want the exact numbers, 1.5747 and 1.3802*exp(+/-1.7805i) are meant to be the roots of the characteristic equation x^3 = x^2-x+3, which will involve square roots nested in cube roots. The other coefficients are found by matching the formula to the initial terms E_1 = 1, E_2 = 2, E_3 = 4.
    2
  5927. Woo!
    2
  5928. I am going to go out on a limb here and say that this is your most amazing presentation to date.
    2
  5929. For the first challenge no because if you took out the two squares next to the corner the corner will be isolated and could not be filed
    2
  5930. Couldn’t be happier to see this video and try the challenges!
    2
  5931. Hello Burkhard. Thank you for yet another truly excellent video. I was playing around with some sequences trying to find formulas using “The Master Formula” and was having a few problems as my results were out of alignment with the correct positions. Of course, this could be corrected by substituting (n-1) in place of n in the formula, but this was a bit of extra algebra work after already going through the algebra to tidy up the formula in the first place. To avoid having to rehash the algebra, it is simpler to use (n-1)C(r-1) instead of nC(r-1) when setting up the G-N formula. Was the choice to use nC(r-1) for the video to keep it visually tidy? I don’t think it would have been any worse using (n-1)C(r-1) with the benefit of arriving at the correct formula without adjustment. As part of my checking, I thought I would look at the sequence used at 16:26 Sequence 1 8 17 36 75 146 263 "Δ1" 7 9 19 39 71 117 "Δ2" 2 10 20 32 46 "Δ3" 8 10 12 14 "Δ4" 2 2 2 The formula for our sequence is then given by: 1 × (n-1)C0 + 7 × (n-1)C1 + 2 × (n-1)C2 + 8 × (n-1)C3 + 2 × (n-1)C4 = 1×1/0! + 7×(n-1)/1! + 2×(n-1)(n-2)/2! + 8×(n-1)(n-2)(n-3)/3! + 2×(n-1)(n-2)(n-3)(n-4)/4! = 1 + 7(n-1) + 2(n-1)(n-2)/2 + 8(n-1)(n-2)(n-3)/6 + 2(n-1)(n-2)(n-3)(n-4)/24 = 1 + 7(n-1) + 2(n^2-3n+2)/2 + 8(n^3-6n^2+11n-6)/6 + 2(n^4-10n^3+35n^2-50n+24)/24 = 1 + 7(n-1) + (n^2-3n+2) + 4(n^3-6n^2+11n-6)/3 + 1(n^4-10n^3+35n^2-50n+24)/12 = (1/12)( 12 + 84(n-1) + 12(n^2-3n+2) + 16(n^3-6n^2+11n-6) + 1(n^4-10n^3+35n^2-50n+24) ) = (1/12)( 12 + 84n-84 + 12n^2-36n+24 + 16n^3-96n^2+176n-96 + n^4-10n^3+35n^2-50n+24 ) = (1/12)( 12-84+24-96+24 + 84n-36n+176n-50n + 12n^2-96n^2+35n^2 + 16n^3-10n^3 + n^4 ) = (1/12)( n^4 + 6n^3 -49n^2 +174n -120 ) So the formula for the n-th term in the sequence is: f(n) = (n^4 + 6n^3 -49n^2 +174n -120) / 12 Checking the first few values: f(1) = (1×1+6×1-49×1+174×1-120)/12 = (1+6-49+174-120)/12 = 12/12 = 1 ✔ f(2) = (1×16+6×8-49×4+174×2-120)/12 = (16+48-196+348-120)/12 = 96/12 = 8 ✔ f(3) = (1×81+6×27-49×9+174×3-120)/12 = (81+162-441+522-120)/12 = 204/12 = 17 ✔ f(4) = (1×256+6×64-49×16+174×4-120)/12 = (256+384-784+696-120)/12 = 432/12 = 36 ✔ … The formula generated by using nC(r-1) in the video at 16:38 is f(n+1) = (n^4 + 10n^3 -25n^2 +98n +12) / 12 This takes essentially the same working as above but then requires the following working to get f(n): f(n) = ((n-1)^4 + 10(n-1)^3 -25(n-1)^2 +98(n-1) +12) / 12 = ((n^4-4n^3+6n^2-4n+1) + 10(n^3-3n^2+3n-1) -25(n^2-2n+1) +98(n-1) +12) / 12 = ( n^4 -4n^3 +6n^2 -4n +1 +10n^3 -30n^2 +30n -10 -25n^2 +50n -25 +98n -98 +12) / 12 = ( n^4 +(-4+10)n^3 + (6-30-25)n^2 +(-4+30+50+98)n + (1-10-25-98+12) ) / 12 = ( n^4 + 6n^3 -49n^2 +174n -120 ) / 12 as above
    2
  5932. the hidden 2019 is in the last 4 numbers of the 11th row on his shirt
    2
  5933. Nice! Using whoosh and megawhoosh I was able to solve infinite peg solitaire. I've made a website with a solution, but YT is not allowing post the url here.
    2
  5934. Steinbach's ratios are extremely pretty, thanks for showcasing them! I'm going to have to look at the ratios α, β, and γ for the enneagon. I'm surprised you didn't mention that A, B and C, the Fibonacci-like recurrent functions are actually only a single function. In particular, taking their definition r^m s^n = A(m,n)s+B(m,n)r+C(m,n) and multiplying through by r and applying the relations for r and s give recurrences for A(m+1,n), B(m+1,n) and C(m+1,n), and similarly multiplying through by s gives recurrences for A(m,n+1), B(m,n+1), and C(m,n+1). And one obtains that C(m+1,n) = B(m,n) and C(m,n+1) = A(m,n). Therefore not only is C a shifted copy of A, but so is B, as B(m,n) = A(m+1,n-1). So r^m s^n = A(m,n)s+A(m+1,n-1)r+A(m,n-1). That can make building the values up by recurrence slightly awkward, but you need to build them from a set of initial values anyway, so it's not any less practical. And this means you only really need a Binet type formula for A(m,n), and the formula for B(m,n) is just a re-indexing of that.
    2
  5935.  @Mathologer  Most things work out similarly nicely for the enneagon, with one exception which is particularly curious. There's still a little I haven't quite figured out, so I'll explain what I do know. In the paper Steinbach uses α,β and γ for the ratios in increasing size, but I'll be using a, b and c. Applying Ptolemy's theorem we obtain the following 6 relations, after simplifying a few things: a^2 = b+1 ab = a+c ac = b+c b^2 = b+c+1 bc = a+b+c c^2 = a+b+c+1 Just as in the case with the pentagon and heptagon, we can use these relations to reduce any non-negative integer powers of a, b and c to an integer linear combination of 1, a, b and c. However, unlike the pentagon and heptagon, we don't get all integer powers in general. The relations alone can be used to obtain 1/a = a+b-c-1 and 1/c = c-b, however 1/b cannot be obtained from the relations alone, for reasons we shall see later. We can then define maps P, Q, R, S: ZxNxZ->Z, where Z and N are the integers and the non-negative integers respectively, such that: a^i b^j c^k = P(i,j,k)c+Q(i,j,k)b+R(i,j,k)a+S(i,j,k) for all i, k in Z and j in N. Multiplying through by a, b and c gives forward recurrences for P,Q,R and R in the i, j and k indices, and dividing by a and c gives reverse recurrences for P, Q, R and S in the i and k indices. Like with the pentagon and heptagon, we see that Q, R and S are just shifted copies of P. In particular: Q(i,j,k) = P(i,j+1,k-1), R(i,j,k) = P(i+1,j,k-1) and S(i,j,k) = P(i,j,k-1) for all i, k in Z, j in N. I do wonder if it's perhaps better to write P, Q and R in terms of S, as then P, Q and R are positive shifted versions of S in a fairly symmetric way. To understand why we don't get 1/b from the relations, consider that there are multiple solutions to the relations. Any triple (x,y,z) satisfying the relations of (a,b,c), must also have: x^i y^j z^k = P(i,j,k)z+Q(i,j,k)y+R(i,j,k)x+S(i,j,k) for all i, k in Z and j in N. What are these solutions? Well we can use the relations to show that if (x,y,z) is a solution, x, y and z must be roots of the polynomials x^4-x^3-3x^2+2x+1, y^4-3y^3+3y and z^4-2z^3-3z^2+z+1 respectively. Each has four real roots which match up to form 4 distinct solutions and can be expressed as (a,b,c), (1,0,-1), (-c/a,b/a,-1/a) and (-1/c,-b/c,a/c). We can now see the problem with deriving 1/y from the relations; the solution (1,0,-1). If 1/y could be derived from the relations, we'd be able to express negative powers of y as a linear combination of 1, x, y and z. But negative powers of y in the solution (1,0,-1) correspond to dividing by 0. We shall see this isn't all to the story though. Now that we have our solutions to the relations, we can form a matrix equation A = MB, where: A = {a^i b^j c^k, (-1)^k δ_j, (-c/a)^i (b/a)^j (-1/a)^k, (-1/c)^i (-b/c)^j (a/c)^k} B = {P(i,j,k), Q(i,j,k), R(i,j,k), S(i,j,k)} M = {{c, b, a, 1}, {-1, 0, 1, 1}, {-1/a, b/a, -c/a, 1}, {a/c, -b/c, -1/c, 1}} Note that 0^j for j in N is just δ_j the Kronecker Delta at j, where we're using the combinatorial convention that 0^0 = 1. We can then invert M and obtain B = M^(-1)A, where: M^(-1) = 1/9 {{c-b+1, -3, -a+1, b}, {-c+2b-1, 0, c-b+a, -a-b+1}, {a-1, 3, -b, -c+b-1}, {-b+3, 3, -a+b+1, c+1}} So we do indeed get Binet-like formulae for P,Q,R and S, for i, k in Z and j in N. But the story doesn't quite end there. What would happen if you try to use these formulae for negative values of j? Do we just get nonsense? We no longer get integer outputs for P, Q, R and S, we get 3-adic rationals in general, but they do indeed allow you to express negative powers of y as a linear combination of 1, x, y and z for three of the four solutions (x,y,z). And the formula actually works correctly! I don't currently have a complete explanation as to why, but I did observe one thing. If y =/= 0, we can obtain a new relation z = x+1. We see this relation is indeed satisfied by three of the four solutions, but is not satisfied by the solution (1,0,-1), so perhaps this additional relation is enough to show that the Binet formulae indeed works for three of the four solutions for negative values of j.
    2
  5936. During Summers, I want to follow your Number Series videos along with Elementary Number Theory by Burton but I don't know if you have a playlist or a series in which I should watch your videos and take notes in order. Thank you so much. Suggest me some videos to start with Number Theory.
    2
  5937. i think there are 9 unique configurations for 4-triangle (we can divide initial 81 configurations by 3 assuming colors symmetry and also divide by 3 assuming rotation symmetry)
    2
  5938. While swimming I discovered the difference in consecutive natural number cubes is 6 x triangle number + 1. 2^3 - 1^3 = 8 - 1 = 6*1 + 1 3^3 - 2^3 = 27 - 8 = 6*3 + 1 4^3 - 3^3 = 64 - 27 = 6*6 + 1 Proof is that (x+1)^3 - x^3 = 3x^2 + 3x + 1 Factor 6x out of the first two terms. 6(x)(x+1)/2 + 1. x(x+1)/2 is a triangle number for natural number x. I thought it was cool. Embedded in triangle numbers are square numbers. Embedded in cube numbers are triangle numbers. 6 triangles to cover each face + 1 more to complete the cube.
    2
  5939. On balance (hahaha) I think it is Bishop Oresme's proof that will linger longest in my memory. It has that Mathologer-like quality of crystal clarity... at least the way that Mathologer presents it. I'm guessing that the Bishop wrote it in medieval latin and in a hand that I couldn't start to decipher!
    2
  5940. Where is the 2 3 5 8 Fibonacci box in the tree?
    2
  5941. as for me its such a charm to see the Mathologer seal of approval, I really dont know, but it really made my day!! thanks for the great video!
    2
  5942. Wow..
    2
  5943. Are there "magic cubes" even as a concept? Not just messin around with 3D objects in a 2D matrix but a proper "magic cube". By the way (23:00), if you take 6 and 9 (they fit though) or 8 and 7 (that you can arange them together, that is true) they don't make any kind of square... Isn't it funny? P.S; Nice video by the way, I enjoied it.
    2
  5944. 22:10 Q: Where does Calpis soda come from. A: From Cals
    2
  5945. What is the name of those polynomials? I would like to read more about them
    2
  5946. That grin at 20:24 following the "Let me know what happens" needs to become a meme
    2
  5947. After playing around a bit with the bias for filling a block in one of those two orientations with a higher probability I think it approxximates an ellipse with the ratio of the axis of the elippse (the axis parallel to the brick orientation) being the square root of the ratio of the probabilities of the bias. With the chances for filling a block for the given orientation p and q and the axis-lengths X and Y -> X/Y = sqrt(a/b). I am not really sure if this is correct but after some experimentation it looks like it. I'd be very interested in a video about biased aztec-diamonds.
    2
  5948. Topic Idea: Isoperimetric Inequality & exploring its application to Archimedes' non-circle polygonal approximation of pi. Hint: Archimedes' method suffers this problem because his pi was/is always calculated using polygons (thus it necessarily follows to suffer it.) Use Pythagorean theorem on right triangle sides (π/4), (π/4)² and hypotenuse 1. Therefore (π/4)² + ((π/4)²)² = 1² for π = 3.1446... > 3.14159...
    2
  5949. Can you tell something special about my username's no. 6174?
    2
  5950. Only a mathologer video to cheer me up ^^
    2
  5951. This actually seemed a lot easier to me than your other masterclass videos
    2
  5952. like 2 years ago when i first heard of the fact that the golden ratio was the diagonal of the pentagon i thought "hmm, that's interesting, what about other shapes", and since i already knew the diagonals of the hexagon i jumped to the heptagon. i independently figured out pretty much everything in this video, and because of that 1.801 and 2.247 became my favorite numbers, and i cannot express the euphoria i got when i saw the letters r and s at the beginning and realizing where it was going. thank you so much i am so giggity right now
    2
  5953. Put this on while scrolling in bed, feel asleep and woke up 20 minutes later without a clue what he talked about. That's the real miracle.
    2
  5954. Of course i know how to find general formula for Fibonacci formula, in fact I don't really remember formula, but i know it correspond to golden ratio quadratic and i just need to find coefficients based on first 2 terms
    2
  5955. Once again you @Mathologer nailed it...gotta love mathematics more than before
    2
  5956. I was hoping to see something related to parabolas, since my first thoughts reminded of Matt Parker's video about the one and only parabola, which i suppose it's a very correlated topic. Very nice video nonetheless.
    2
  5957. Very nice video, thank you. Und fröhliche Weihnachten!
    2
  5958. Most memorable part for me is the grader giving your proof a zero. I got a few A's in math at IU generating "original" proofs for homework or on tests. Not that I ever found a proof that wasn't already known somewhere else, but just not the one given in class. At least the professors or students grading my work went to the trouble of checking if my proofs were valid. It reminded me of my method of getting A's in all my math classes in College: In the summer before the next school year, I simply spent my time studying all the old textbooks in the school's Math library that covered my upcoming classes for the Fall and Spring semesters. I mean, they don't make any secret about what the upcoming classes will cover, and, for undergrad classes, most of the Math is not exactly new, cutting-edge stuff. And as a full-fledged Math nerd, I didn't have anything else to do all summer that was half as fun. So, by the time I took the classes, I already knew the material through and through. (I also discovered that some lazy professor's hand-out tests contained problems that were culled from other older textbook's worked examples. Tsk, tsk.) Strangely enough, in all the hours I spent in the Math library getting ahead of my studies, I never met even one other Math Geek doing the same preparatory research.
    2
  5959. Very interesting. I wonder if there'd be some wonderful things to say about the different trees of pythagorean triples. Those have the 3,4,5 at the root, every node has exactly three children, and they allow for all primitive pythagorean triples to appear somewhere along the tree, without exception and without repeats.
    2
  5960. Loved the presentation so much. First timer here. :) Cheers,
    2
  5961. Bro triggered me and my whole class with the title of this video
    2
  5962. I don't know about the other countries but in India it's been taught in high school.👍😃😄
    2
  5963. Happy Christmas day to all my friends.
    2
  5964. What I read out of this video is that for this type of series the terms somehow cancel each other out with a delay and you are left with a series that converges to the target number. Can you say that?
    2
  5965. You can paint an infinitely long fence as far as you like with just one (finite) tin of paint. Assume the fence consists of an infinite number of identical finite sections. Use half the tin to paint the first section, then half of the remaining paint to paint the second section, then half of what remains to paint the third section, and so on for as many sections as you like. The thickness of paint on any section will be half the thickness of that on the previous section, so every section you painted will be covered in paint, yet there will always be some paint left in the tin at the end. It is similar to painting the inside of Gabriel's horn by introducing paint into it. The thickness of the paint at any point on the inner surface is equal to the radius of the horn at that point, being ⅟ₓ at point x on the axis. Starting with a pot of π volume units of paint, fill the section from x=1 to x=2. This will use half the pot, the paint thickness varying from 1 to ½. Then fill the section from x=2 to x=4. This will take half of the remaining paint at a thickness from ½ to ¼. Continue with the next section from x=4 to x=8, and so on. Each section is twice as long as the previous one, but only requires half the quantity of paint, so you can fill/paint as far as you like but there will always be some paint left in the pot. Since the paint thickness varies inversely with length from the "mouth" there will always be some paint on the inner surface of the horn that you have painted so far, and you can carry on "painting" the horn like this indefinitely from the same pot of paint.
    2
  5966. YOU SAID NO PIGEONS WERE HARMED! :(
    2
  5967. 18:48 Because in these cases his construction only increases or doesn't change the polygon's size
    2
  5968. 20:10 was the big aha moment for me: it explains why the proof fails for composite numbers like 21, which can't be written as the sum of two squares even though it's 1 mod 4. 21 has a second straight cross (3 3 1), which makes the number of solutions even, so the pairing argument no longer works.
    2
  5969. Observation: Given we have a traingle ABC and n discs are on vertex A then - If n is even and our first smallest disc move is to B and then we followed the algorithm: then all discs move to C If n is even and our first smallest disc move is to C and then we followed the algorithm: then all discs move to B If n is odd and our first smallest disc move is to B and then we followed the algorithm: then all discs move to B If n is odd and our first smallest disc move is to C and then we followed the algorithm: then all discs move to C
    2
  5970. Совпадение? Не думаю.. 18:04 Thanks for the new video, and greetings from Russia.
    2
  5971. I liked the swivel proof best, but that’s probably because when I tried to prove it I found the color proof
    2
  5972. Yet another great video. My favourite of the channel is the one about the cubic equation ("what do they think you cant handle? ") and every time I see an upload I'm eager to know if this would be the one about the quartic equation... Or about Galois theory
    2
  5973. ur vids are better than any netflix web series
    2
  5974. Yeah, but there are problems. Like, the hyperbolic angles go from -∞ to ∞ which only covers one quadrant of the plane. You can cover another quadrant by allowing negative distances which flips the hyperbola over its asymptotes, but you're still missing half the plane at that point. As for what kinds of things this is used for, my first thought is Special Relativity. In SR changes in speed are hyperbolic rotations. So if there is some event happening at your location in your future, then when you change speed that event will rotate along a hyperbola in space-time to a new time and place in the future (relative to you). In other words, now the event happens at some place behind you. The angle that it rotates is called the "rapidity" and is an analog of speed in SR. So every point along that hyperbola are all the events you could witness after a fixed amount of your time, given the right "rapidity". This forms a kind of 2D grid of future events you could reach if you could move fast enough.
    2
  5975. Absolutely deserves a follow! This was amazing. There are too many amazing channels on Youtube. It really is a golden age ot maths learning! How will I ever have time to watch them all!
    2
  5976. 10:27 Do I hear excited liberal noises, in the distance? 🤔
    2
  5977. This is why sun is the only star, it broke through the only black hole , we call it space and stars are just sun in past present future, kline bottle inside a kline bottle inside a kline bottle full of infinite kline bottles
    2
  5978. When I studied electrical engineering in the 1990s, my maths class focused on algebra, trigonometry, and calculus. Now I understand calculus. Yay me!!
    2
  5979. OMG, your Pytagoras+Einstein T-Shirt! This one is so fun!
    2
  5980. That was great. One of the few videos of yours where I was ahead of the curve. I KNEW it was a slide rule. My Dad showed me his as a kid. One of my favourite scenes it the movie "Apollo 13" is when Lovelle asks them to check his math, and three guys in turn use slide rules to figure it out. Great scene! So, slide rules rule!
    2
  5981. There is a minor mistake in the yellowing of non-matching digits. This is at the 10581/10582 pair
    2
  5982. 13:56 Very nice duality property.
    2
  5983. You saved my day with this ingenious math stuff! Thank you! :)
    2
  5984. ......a matter of simple matrix multiplication.... Yeah right, I'll just break out the calculator. One of your best ones yet. 'Loger! I just wish I could keep up. 🏃‍♂
    2
  5985. Great video! You make the shoelace method so easy to understand. Thank you!
    2
  5986. "2020" is a magic number: Who can proof: 2020^n+2020^m is never a square number with n and m are both positive integers
    2
  5987. The formula for the first problem is (n^4-n^3+3n^2+27n+14)/44
    2
  5988. My vote goes to the "no 9's series". Definitely surprising result and very insightful.
    2
  5989. at 22:24 the green area is equal to the integration of l/x from 2 to 4 which is ln (4 )-ln(2) but in the vedio you divide this by 2 why !!?
    2
  5990. I prefer Stay Positive f(x) = |x|
    2
  5991. I first came across this in Concrete Mathematics by Ronald L Graham, Donald E Knuth and Oren Patashnik. A wonderful tome full of splendid ideas.
    2
  5992. 28:26 I don't feel "rooted," more like "stumped."
    2
  5993. I made it to the very end, too.. Ans(52)- last question.. Thank you for these info
    2
  5994. Excellent video! Logarithms are fascinating
    2
  5995. What a great video (as usual)! I can't wait for the fifth video of an increasingly inaccurately named trilogy of four.
    2
  5996. Dearest Mathologer Guy, You are the best instructionals presenter on the entire interwebs, and you definitely have the best laugh (giggle). If you take all the alephs, but them all to the power of themselves, then take their power sets, and do that forever, over and over, your Mathologer videos are still infinitely better by a much larger factor than even that than any other Youtube videos. Fantastisch! BTW: Any videos involving Cantor and/or Gödel and/or Cohen and unintuitive set mappings involving infinite sets most welcome.
    2
  5997. 1:58 The average of the numbers 1 to 9 is 5. Multiply that by 3, you get 15, which just so happens to be the magic constant. The average of the numbers 1 to 64 is 32.5. Multiply that by 8, that's 260, which is the magic constant. Is this always the case? If so, that's very neat. 8:24 (1+33)/2*33 = 561 18:48 You forgot to put a green dot over 7.
    2
  5998. I'm no mathematician, but I did finish my schooling before calculators were mandatory, and went into Surveying and Cartography. In primary school I recall our class doing "mental arithmetic" each morning, where we had to do as many maths problems as we could within 2mins. Our teacher always exhorted us to try and improve on our total from the previous day. At the time I was too young to understand the point, but now I'm grateful for those exercises that helped to develop my brain power. (Thank you Mr Moffat!!!!) But I do love your quip about those "way to taxing for those who grew up punching buttons on a calculator"!! (On that sentiment, I whole-heartedly agree with you Sir!! 🤣🤣🤣👍)
    2
  5999. Thank you for your efforts. You are making math very likable. Ive got a question. Can we say that any positive integer larger than 2 is made of 2^n + some number other than 1 ? Or say, lots of 2s+ some number other than 1? Except 3 as special case offcourse.The reason why i am asking this because thinking about collatz conjecture and goldbach conjecture.
    2
  6000. at 10:25 the answer is: Plug in i= m/2+1/2 (the upper bound and j = n/2+1/2. Then the factor is 4cos(π/2)^2+4cos(π/2)^2=0. Therefore the product is zero.
    2
  6001. 1+2+3=6 1×2×3=6 Thats Suspicius
    2
  6002. 23:23 the two shapes belonging to the numbers 6 and 9 (which equal 15 when added up) do not produce the 4x4 square with a missing piece!
    2
  6003. If in sin x=x(1-(x/π))(1+(x/π))(1-(x/2π)) If we take x=π We get sin x=0(no limits)? If x=0 also sin x= 0(...... We get sin x=0 ? So is 0=π?
    2
  6004. The standard narrative is that Indian contribution to the mathematics in 0 (pun intended), as though one fine morning some ancient Indian just got up and said well I guess I will give the world the 0 today. And as Indians we are so brain washed that we even stopped asking basic question why, why? Why India needed a 0 (where as Europe did not even know 0 until 13th century). Doesn't it mean Indians knew negative numbers, decimal places🤔Well well... the truth is Indian contribution to mathematics, astronomy, calendar (time keeping in general) is enormous starting from the Vedic era (with Yajur Veda we find lots of basic to semi advanced math) and these all had been hidden, stolen by the colonizers. I am sure these all come as a shock (just like I got when I first learned about this few years back) to many, but don't take my word for it, just search wiki... Now for serious folks who are really interested to dig deep here are some resources to get warmed up: https://www.youtube.com/watch?v=Rm6d-bUmmGg https://www.youtube.com/watch?v=Ldy8M-aRFtY https://www.youtube.com/watch?v=A6WpdZyase8&t=1017s
    2
  6005. ​ @Mathologer  Cause they are too good to be free! You're doing an amazing job man.
    2
  6006. Pardon thus insane thought, but what do you find when you use the method on prime numbers? Off the cuff, I think you may find a proof for Goldbach's Conjecture.
    2
  6007. 13:55 polynomial sequence 15:30 Lagrangian interpolation Builds poly formula in 1 easy step (me: nothing's easy about it) 15:38 Gregory-Newton formula (me: n choose k method hehehe) A permutation -see combinatorics 20:35 what's next formula solution concept is about 1. inferring (distributions) of a naturally occurring formula 2.guess the natural general rule of the outcomes 3. Justify why Guess is correct (& not other random number) 20:58 have to prove formula "Correctly Captures" the (right) mathematical Context I. E. Lines connecting Paris of dots Cutting cicrcle into regions; Number regions depends on Location(dots) N dots = n regions 24:08 Del(exp(x))= pow(2,n)= sum(n choose k)
    2
  6008. If you carve out 2 black and 2 green square then you will not always be able to tile it because the 2 same colored squares adjacent to the corners can be remove isolating an individual square adding 1 to an odd number makes it even and, cos (pi / 2n) = 0 where n is a positive integer, so if both numbers are odd the entire crazy formula will multiply by 0 evaluating to 0.
    2
  6009. 14:39 Ok so can I name the correction term T for the Trithagoras theorem the PARKER TRIANGLE?
    2
  6010. Maybe it's still not enough words: Whenever one talks about the incircle of a triangle, the excircles are not far away. Presumably you know that the tangent poins of the excircles also have distances a, b and c from the vertices, in particular on the edges just with the sums reversed: r=b+a etc. Therefore I guess one can obtain some more (larger) irises around the excircles with chords having their end points on the lines of the triangle at other distances made up from r, g, u.
    2
  6011. If it's on Mathologer, its survival and dissemination is assured.
    2
  6012. I am not enough of a mathematician to answer most of these questions but the first is a no! You can isolate a 1x1 box in the corner.
    2
  6013. Imagine it Ramanujan and Euler were somehow brought back to life, I wonder could they solve one of the million dollar math questions? Assuming that they are working together and are at the peak mental health.
    2
  6014. Always nice to see how simple transformations can create complex patterns. Btw, Your funny laughter made me think of a sketch by Loriot from German television: https://www.youtube.com/watch?v=5eV8XkRb8Fo :)
    2
  6015. 32:35 so the best strategy is to dicide the stack into two new stacks as equal in number as possible. Hmmm, again binary Arithmetic raises its pretty head.
    2
  6016. "Simply magic." How true.
    2
  6017. Another grand video brimming with Mathologer's characteristic good cheer and containing several playful challenges for the audience. The content is quite accessible, despite coming from relatively recent mathematical literature. "This is not the Discovery Channel" was my favorite quote. Kudos to Bjarne Fich for rising to Burkard's challenge and creating such a professional Arctic Circle animation. Here is my response to Mathologer's request for feedback. Challenge 1: No, removing 2 black & 2 green squares will not always yield a tile-able board. For example, removing (2,1) and (1,2) leaves (1,1) isolated. Challenge 2: If m & n are odd, then ( j, k ) = ( (m+1)/2, (n+1)/2 ) yields a zero term in the product. Challenge 3: 2xn squares yield (1,2,3,5,8,...) tilings for n=(1,2,3,4,5). This sequence follows the Fibonacci rule. To see why, observe that T(n+1), the number of tilings for n+1, can be computed by adding the number of tilings with the last domino vertical, which is T(n), and the number of tilings with the last two dominoes horizontal, which is T(n-1). Aha! Challenge 4: The determinant for Tristan's glasses gives 666. By drawing the 4 ways of tiling around the holes I get the same result, but have yet to see why the determinant formula should work when the holes themselves can be tiled. Challenge 5: Previous comments allowed me to see why a hexagon tiling must be split evenly into the 3 tile orientations. Unfortunately, I did not see this for myself. Most of the video was clear. However, I am mystified as to why the dance yields all possible tilings, and why, for example, a pair of adjacent 2x2 squares don't count as a 2x4 rectangle. Mathologer dropped some clues, but I guess I need to consult the references. Looking forward to more great content in 2021!
    2
  6018. I was waiting for your video from a long time thank you sir
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  6019. Most memorable part: Sum 100 zeros series is greater than the sum of no nines series.
    2
  6020. I adore your content my guy keep it up
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  6021. I first learned about the Pigeon Hole Principle while studying Computer Science. It is of fundamental importance in a variety of areas, ranging from databases to sensors. Recognizing situations where it applies always results in problem simplification, leading to savings in space and/or run-time.
    2
  6022. No, with four cut out squares it will not always be possible, because three of those squares could isolate a single square in a corner.
    2
  6023. Most memorable: Why it has been recommended and why I watched it all the way till the end and subscribed for M?
    2
  6024. A few minutes in and I am wondering if the paradox arises in say binary or base 7. Writing this comment I am wondering if Mathaloger has covered this in the remaing video.
    2
  6025. When I was a child, my dad bought a bunch of stuff from an auction, and among it was a large-diameter (maybe three or four inches; I don't know my artillery history) shell casing, such as from an artillery gun. But not an intact shell casing, quite -- nope, this one was half crushed, into precisely the type of lantern pattern you show here, but crushed vertically, parallel to the straight sides of the cylinder, to the very extreme limit of the strength of the material. It was made of. Steel? Bronze? Brass? I don't know my artillery composition, either. Someone once expressed the opinion that this casing had been crushed by getting caught in the ejection mechanism when fired, which I suppose is possible (though I don't know my artillery mechanisms), but It seems to me that the crush pattern was much more uniform than I would think an accident would produce. Surely wouldn't an accidental crushing produce some bit of bias, a tilt to one side or some other non-uniformity in the bucklings? Or does the math prohibit that and force an all-or-nothing symmetry? So someone else suggested it was a deliberate artwork, such as you have begun to produce in your final footage of that soft drink can which you score with the piece of wood. (Thank you for that, incidentally, I've been looking for a start on that technique most of my life. Score the can as you have done, then press it from above, and you can make quite an attractive little vase for some pretty flowers, exactly the way my mother used that shell casing in my youth.) I'm also surprised you can't compensate for the buckling inward of the triangles by simply calculating and applying a correction factor in determining their contributions to the surface area of the cylinder. You know the normal vector along which the surface area must point, and you should be able to calculate the normal vectors to each triangle in the lantern pattern, and from the supply a (sine theta?) correction factor to each off-normal triangle's surface area contribution. Is it simply that that technique was outside the scope of this video? I find it hard to believe you wouldn't at least mention it in passing. Or is there some other mathematical dragon that arises to defeat the valiant knight who attempts this correction? One could even envision cutting the surface of the cylinder as though to unwrap it, then examining the zigzag/seesaw pattern formed by the cross-sections of the lantern pattern at each point in its back and forth cycle, then figure out some way to correct for that, as a whole, and apply that correction to the summation over all the triangles. Now, damn it, either you need to make a video discussing this, or I'm going to have to try it myself.
    2
  6026. The volume in this video is a little low.
    2
  6027. I’ve been using Ian's shoelace site for years, and now I find our he's friends with Burkhard 😂
    2
  6028. I love this super simple explanations of logarithms. It’s just so beautiful. These videos always bring me beck to freshman year, solving problems that have already been solved and making new discoveries from that. Fun times.
    2
  6029. What a myth, not books but toys (books are of course stored elsewhere)
    2
  6030. 3:36 not enoug digits highlighted, and 18:28 more highlighted than should be... is this intentional???
    2
  6031. You are trying to pass a lot of message through your video , no way Islam is the lightning road 💫🕋
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  6032. Most memorable: The harmonic series missing all integers, specially since it will increase in extremely small steps and still reach infinity!
    2
  6033. The physical pendulum:simple pendulum is also = 3:2. Consider the centroid of the equilateral triangle. Fun! ❤️
    2
  6034. ​ @Mathologer Everything you do out here is genius! Thank you.
    2
  6035. Kerala is a gem of a place. India is lucky to have them in their country.
    2
  6036. It just clicked for me while watching this video: At 14:38, that's where all our universe's math comes from, via a simple Langton's Ant type action. Everything is ultimately just a stacked series of ones.
    2
  6037. Wonderful Video, I think it's cool to binge-watch your videos rather than NETFLIX :P
    2
  6038. I made it to the very end Jk
    2
  6039. I'm not into all the math equations and such, but you really do make it interesting. I wish I had you as a math teacher.
    2
  6040. I love your t-shirts, man!!🤘🤘
    2
  6041. Monash uni profs are the gifts that keep on giving. Rob J Hyndman, Mathologer, etc. Their students are really lucky.
    2
  6042. Watching this I had a few thoughts/questions: 1. Looking at diagonals, there was a "jump" from the pentagon to the heptagon - two shapes with odd numbers of sides. Are there any similar ratios/relationships for the hexagon or octagon? What about the square - too trivial? 2. If not, is there some reason why the even sided shapes do not exhibit this behaviour? 3. If the ratios of segments in a pentagon bear a connection to the golden rectangle, and those in a heptagon a connection to the "golden box", can we hypothesise that the ratios of line segments in a nonagon have a connection to a 4-dimensional golden object? Can we generalise to higher dimensions? Something about those diagonals is hinting at graphs and vertices but I can't make the connection. These are just thoughts - I don't think I have the maths to effectively investigate, maybe other viewers have some ideas 🙏🏼
    2
  6043. That's neat! I always prefer working with the formulas myself, visualising things is hard for me, so I always enjoy seeing your visual proofs. As for the puzzles: 1) 5**3 + 6**3~=7**3 with a difference of 2. Edit: Is it because with 6 slices we are covering only 6 out of 8 vertices, leaving 2 vertices uncounted? 2) Oh, it's 2 exactly. Before this it would probably took me way longer, but now I solved it in about 10 seconds. Summing first 3 squares is trivial since there are no carry overs, it's 365, and the total sum is double that, as we now know. 3) For cubes we get 3**3 + 4**3 + 5**3 = 6**3, but it breaks down for fourth degrees and after that. I'm not sure I can come up with an explanation on the spot.
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  6044.  @Mathologer  Mr. Polster, I made a small study. I chose 5 different type of Pythagorean triangles. These are: 1) 3-4-5 ; 2) 5-12-13 ; 3) 8-15-17 ; 4) 7-24-25 ; 5) 20-21-29 I did what you explained. There are 10 combinations (couples). I used a CAD software and the 4th diagonals are in all cases integer (for me very surprising). I thank you one more time for this very interesting information. Combinations: 1st Diagonal 2nd Diagonal C1) 1&2 65 56 C2) 2&3 221 220 C3) 1&3 85 84 C4) 1&4 125 117 C5) 2&4 325 323 C6) 3&4 425 416 C7) 1&5 145 144 C8) 2&5 377 352 C9) 3&5 493 475 C10) 4&5 725 644
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  6045. Absolutely charming video, it's amazing how products and sums are linked and how one is able to derive the known formulae so effortlessly.
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  6046. 1:phi is everywhere you just have to look.
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  6047. i can see why this was shoved onto a hidden shelf in the dusty section of the maths library. 😐🤔
    2
  6048. I wonder if you can somehow mix primes into this and get something that you can't as easily get with normal calculus aswell.
    2
  6049. I learned this stuff in grade school by obsessively filling sheets of paper with digital roots and divisions. I would have loved the vortex diagram if I had ever seen that. Another anecdote is we were supposed to invent a new culture in middle school social studies, complete with a counting system. I guess we were supposed to use tallies or symbols, but I used base 12. My teacher tried to explain that even in another culture math problems would work out the same way so I couldn’t write 6x6 = 30.
    2
  6050. The 5x5 magic square is used in The Book of The Law along with an additional "aleph" which isn't on the square itself to map out the English gematria. A=1 L=2 W=3 Alwhsdozkvgrcnyjufqbmxitep If one puts the regularly ordered alphabet in, a to right y bottom left. The kabbalah order I indicate above emerges as angled pattern. Z = 8 sideways infinity is not on the board. It's everywhere else.
    2
  6051. yee-HAW!
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  6052. Ive heard of moessner’s magic…that leads to combinatorics…factorials in that case…..
    2
  6053. Nice video. Are you able to play around like that when adding another dimension? What can fit?
    2
  6054. 23:55 If you know how long the geometric sequence is in the green row you're trying to calculate, and you know the numbers in columns past the outer edge of the square of zeros in that row, isn't the common factor of the geometric series already completely determined? What am I missing?
    2
  6055. I may have lost my eyesight to bright white background but I do not regret it one bit. HOW HARD IS IT TO CHROMA KEY OUT THE RETINA DESTROYER DAMNED BACKGROUND TO SOME ACCEPTABLE EASY ON THE EYES COLOR.
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  6056. 0:00
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  6057. on a curious note why don't we have a mathologer subreddit?
    2
  6058. Nice dominoes 🍕 mathematics. Very very very....... Tasty 😋
    2
  6059. 2^(-1/12) ?
    2
  6060. So PRETTY, good LUCK. 😎
    2
  6061. Most memorable: the 700 year old proof, so simple it really conveys the beauty of mathematics
    2
  6062. I guessed that it was the top corners somewhat arbitrarily.
    2
  6063. You get a 1^2+0^2=1^2 triple.
    2
  6064. 28:54 I'm familiar with another form where you put e^ja j^2=1 with split complex numbers
    2
  6065. The fact he was majorly self taught makes it more mond boggling
    2
  6066. 28:22 hmmm one of those fives is a six
    2
  6067. To further popularise this, get it into the program "Visions of Chaos".
    2
  6068. if you want natural numbers that irrespective of digit system pop out as weird you gotta look at the numbers from the Monstrous Moonshine conjecture, because the Monster Group and other sporadic groups are truly unusual and unexpectedly strange facts that pop out of math and Monstrous Moonshine is an even more unexpected connection that links the Monster Group family to a completely unrelated area of mathematics. And There is absolutely no logic or reason anyone can come up with for why this should be true. If there's a hidden truth about reality in pure mathematics, that's where it is imo
    2
  6069. First
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  6070. Es bewahrheitet sich immer wieder: Die wirkliche Kunst ist es, Schwieriges so einfach zu vermitteln, dass es aussieht als wäre es selbstverständlich. Top gemacht!! Vielen Dank 🙂
    2
  6071. Your videos are so fascinating! Looking forward to watching your next video! <3
    2
  6072. Why can areas not be negative?
    2
  6073. We love these long videos.
    2
  6074. I remember seeing somewhere that the Tower of Hanoi can be solved by simple binary counting. I was stunned to say the least.
    2
  6075. When you have your list at 13:43, instead of throwing away every other combination, a solution that uses all of the pairs: house number is top times bottom. To get the last house number: min of [(top^2), (bottom^2)*2], which will be two numbers in the form of (x) * (x+1), with x being the last house number. So, for 7|5 - 7*5 is the house number 35. min(7*7,5*5*2) = min(49,50), so x is 49. For the answer to the puzzle, 17*12 = 204. min(17*17=289, 12*12*2=288) = 288 for the number of the last house.
    2
  6076. Ender’s card flipping game
    2
  6077. My solution for 13:42. If we take dx to be the movement of the rubber band at the wheel, and dy to be the movement of a point y on the rubber band, then dy/dx is y. That is a point y=0.1 is moving at 0.1x of the rate that the rubber band at the wheel is. Then the total rubber band that rolls on to the wheel is ∫dx from y=0.1 to y=1. But dy/dx = y therefore dx=dy/y so ∫dx = ∫dy/y (y=0.1 to 1) = ln( 1) - ln (0.1) = ln(10)
    2
  6078. I did this on my own in high school when playing with math, but I didn't end my work at the Gregory-Newton formula. Instead, I had it written down as what i called the "sigma substitution" and devised the same scheme to find the successive sums of each row based on pascal's triangle. It's nice to know there's a better way to write it now. Also, on this same line of thought, I derived the Euler-MacLaurin formula not too long after discovering all the stuff you mentioned in this video by looking for other summations. I'm hoping you'll go over that someday so I can see if I missed anything important there.
    2
  6079. Mi inglés no es muy fluido, sin embargo veo sus videos de principio a fin, ayudándome con el traductor. En tanto que soy parte de su audiencia, me atrevo a solicitar un tema: El teorema de Villarreal. Es un teorema bello y potente que permite calcular las potencias de los polinomios. No digo más, me despido desde Perú, con toda mi atención y gracias anticipadas.
    2
  6080. 3:00 Bottom row: a_0 = 1, a_n = 1 4th row: b_0 = 1, b_n = b_0 + \sum_(k=0)^(n-1) a_k = 1 + n 3rd row: c_0 = 1, c_n = c_0 + \sum_(k=0)^(n-1) b_k = 1 + n + 1/2 n(n-1) = 1 + 1/2 n + 1/2 n^2 2nd row: d_0 = 1, d_n = d_0 + \sum_(k=0)^(n-1) c_k = 1 + n + 1/4 n(n-1) + 1/12 (n-1)n(2n-1) = 1 + 5/6 n + 1/6 n^3 Top row: e_0 = 1, e_n = e_0 + \sum_(k=0)^(n-1) d_k = 1 + n + 5/12 n(n-1) + 1/24 (n-1)^2 (n^2) = 1 + 7/12 n + 11/24 n^2 - 1/12 n^3 + 1/24 n^4 We already see a few powers of 2 when n = 0, 1, 2, 3, 4, 9. But I'm not sure if there are others. May I ask which pattern are you expecting us to find out? Thank you! P.S: After watching the video, I realized I was doing quasi-sequence integral using {\sum x^n} formulas. Now I know G-N formula is the way! 26:04 ^(_m) denotes " to the falling power m". x^(_m) - (x-1)^(_m) = 1/(m!) [x(x-1)...(x-m+1) - (x-1)(x-2)...(x-m)] = 1/(m!) (x-1)(x-2)...(x-m+1) [x - (x-m)] = 1/[(m-1)!] (x-1)(x-2)...(x-m+1) = (x-1)^(_m) 36:30 Isn't this true given the definition of "difference"? :P 44:15 A quote from Mathologer:"The infinite sum is just a finite sum in disguise." :) Great video as always! Among all these aha moments, it's the rotation into Pascal triangle that impressed me most.
    2
  6081. The most memorable is your voice... the giggle you make when telling us wonderful facts. Have a wonderful life. Stay safe... ✌️
    2
  6082. I wonder why it's not displaying parasara telugu correctly?
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  6083. Ramanujan is incomparable ... He was somehow connected with the divine source of intelligence ... ♥️
    2
  6084. Extremely nice video, like always. Just few things, we never "prove" every triangle could be tessellated or that the small medians of every small triangle actually form a line on being combined, off course its obvious visually, moreover its is the median that is split in 2:1 ratio not the altitude so you can't directly infer the small triangle's height being 1/3rd of the original. The main point is bit of lack of rigor. Personally I watch your awesome videos to get an intuition but you should still at least mention the places where you lack rigor. Keep up the Excellent Work!!!!
    2
  6085. We should not use 360° here as the angle of a full circle. We should use 1 instead. Then we get ¹⁄₅ , ²⁄₅ , ³⁄₅ instead of 72°, 144°, 216°…. https://en.wikipedia.org/wiki/Turn_(angle)
    2
  6086. I would love to see a video on the Road Coloring Problem! (Great work with this one, by the way)
    2
  6087. 10:10 Funny story. Once upon a time, I was a bored Algebra student. I decided to do this just for fun and noticed this pattern. It bugged the heck out of me because while I recognized the pattern really quickly, I couldn't prove it or explain why. Finally, I came up with an algebraic proof of this, which satisfied me. But then I tried to do this with higher powers (i. e., x^3, and so on to x^n) and ran into difficulty there. I realized, however, that do this enough times, and you will always get !n. But I couldn't prove that. I eventually realized why I was having such a hard time with this is because I was running into calculus. Then it all began to finally make sense. Thanks for this video.
    2
  6088. Gamma is obviously less than 1/2, because each of the "slices" occupies more area than a right-angled triangle with the same bounding box
    2
  6089. Not sure, but I think you just described mathematically a "cootie catcher".
    2
  6090. great video as usual! such an inspiration :)
    2
  6091. 4:37 Is there a reason you don't want to place the "pen" point directly on the circumference of the circle? Would that not remove the curves altogether? Why did the original animation cheat at all? Idle thought, but suppose you could make a usable physical compass via a disc with a pencil attached to its circumference and some idealized iris-like mechanism that let you alter the radius at will. So if you had two or more iris-compasses at hand you can construct, in the ancient geometer sense, spirography things like hypotrochoids. Is there an interesting scope of constructive geometry made accessible by pairing straight edges with iris-compasses?
    2
  6092. I see your Schwarz is as big as mine.
    2
  6093. I have discovered the "expressing angle as area inside the curve" idea myself when I first encountered hyperbolic sin and cos, but this video makes an incredible job of it and another AMAZING way to find e inside 1/x. Thank you!
    2
  6094. I am Rainer, and yes the video was exactly what inspired me :D It was a literal heureka moment where something suddenly struck me and all dots connected.
    2
  6095. I mean no? That’s like saying “this is what the problem is about, couldn’t be anything else and there’s no symbolism beyond the meat and potatoes” — like, sure on the surface it’s a technical issue that seems cut n dry, but there’s no literalistic-solo readings in Talmud so it’s not quite fair to say they couldn’t do the right answer and instead just gave a couple lesser ones when they’re really trying to showcase an expanded perspective tetralemmically as concerns its application to implicit theological issues raised by the framing device of the problem in itself
    2
  6096. The tower at the Burgplatz/Rathausufer in Düsseldorf has seven sides.
    2
  6097. I've heard of the leaning tower of lire before, but I didn't know that there was a better way to stack them than that. The most efficient solution amazed me!
    2
  6098. Always cool T-shirts
    2
  6099. 24:13 proof: let us form a sequence of integers that are used to generate the sequence 1, 2, 5, 7, 12, ..... (let this be sequence 1) using the formula f(n) = n(3n-1)/2 in same specific order. which is: 1, -1, 2, -2, 3, -3, ....n, -n, (n+1), -(n+1), .....(let this be sequence 2). now take a note of the n, -n, (n+1), -(n-1) (let this be genSeq1) part cause that gives us the generalization of the sequence we need for every 'n'. we can deduct the result for the difference between 2 consecutive numbers from sequence 1 by breaking the problem into 2 cases. case 1: difference between 2 numbers from sequence 1, where those 2 numbers are obtained by a +ve number followed by a -ve number respectively. now, the difference can be expressed as f(-n) - f(n) (n and -n are taken according to genSeq1) . which gives us: diff(+ve followed by - ve) = (-n(3(-n) - 1)/2 - n(3n-1)/2) => diff(+ve followed by - ve) = ((3n^2 +n) /2 - (3n^2-n)/2) => diff(+ve followed by - ve) = (3n^2 + n - 3n^2 + n))/2 => diff(+ve followed by - ve) = n which means that the difference between the two numbers that are obtained by plugging +n and -n into f(n) is 'n' itself. case 2: difference between 2 numbers from sequence 1, where those 2 numbers are obtained by a -ve number followed by a +ve number respectively. now, the difference can be expressed as f(n+1) - f(-n) (-n and n+1 are taken according to genSeq1) . which gives us: diff(-ve followed by +ve) = ((n+1)(3(n+1) - 1)/2 - (-n)(3(-n)-1)/2) => diff(-ve followed by +ve) = ((3(n+1)^2 - (n+1)) /2 - (3n^2+n)/2) => diff(-ve followed by +ve) = (3n^2 + 6n + 3 -n -1 - 3n^2 - n))/2 => diff(-ve followed by +ve) = (4n +2)/2 => diff(-ve followed by +ve) = 2n + 1 which means that the difference between the two numbers that are obtained by plugging - n and (n+1) into f(n) is '2n +1'.
    2
  6100. The first one with the dots — the congruence of the colors making ringlets, almost evoked a dna strand.
    2
  6101. I'm always looking for geometric interpretations for the hyperbolic trig functions, and this surely delivered 👍
    2
  6102. Looooong one, gonna save this one for later ;)
    2
  6103. E=mc^2 is my fav equations
    2
  6104. I dig this video maestro mathologer There's what I said, working on a Hilbert vector space and groups, we have here a O(1) I think, the linear transformation to 2 degrees of freedom? only 2 configurations? U(1) to work in general with two rotations only, SU(1)? EDIT Just in case groups are just a way of doing PERMUTATIONS in a easier way, like when you have to program a Rubik Cube solver. By the way, the group is S48
    2
  6105. Somebody has been playing too much Tetris in the late 80s...
    2
  6106. I've been testing my understadning of the mousatche method using the 33x33 square. It's been a journey so far. Drat you for making me do something interesting! ;)
    2
  6107. Great topic. Too great perhaps even. I think this one could have been spread over three videos or so. (For example, the differential eqn. chapter left me with lots of loose ends, I feel.) But thanks for planting new math seeds! (Meant constructively of course!)
    2
  6108. First time watching a video with a challenge in it soon enough to actually participate. This feels like something I want to be able to zoom infinitely and get full detail, so I'm going to make an interactive SVG.
    2
  6109. That's pretty cool stuff.
    2
  6110. oh my god
    2
  6111. In physics e number is unique sclalarity non-dimensional constant. For example x*y = x + y solution of equality expressed from independent t variable is: x = t + 1 y= t+1/t=1+1/t where (1+t - square hyperbola) if t=1 we get most famous math idiom easy like 2+2=4 or 2*2=4 or 2^2=4. It makes only sense if and only if x*y = x + y are non-dimensional unit, because x*y is [Area unit], x+y is [Length unit], thus definition or identity of exponents multiplication is e^x*e^y ≡ e^(x+y). And x and y must be non-dimensional units not to distort physicals reality. e - number is hidden scalarity unit or shapeshifting constant of other units, we so preoccupied with Euclid straightforward proportionality and straight linear symmetry but our World is complex and non-linear.
    2
  6112. Whaaaaat that's insane! I never realised that!
    2
  6113. Wonderful visual!
    2
  6114. comment for the algorithm
    2
  6115. That was amazing!! Definitely seems related to discrete differentials, I feel this might also be related to the Ackerman fucntion. Or more correctly the Ackerman function maybe an instance of this general idea!!
    2
  6116. The repeating tail giving a fraction can be shown by expressing said tail as a geometric series, which then has a fractional limit. Adding that to the fractional head of the number gives a fraction. QED
    2
  6117. 7:20 Too bad Ferris Bueller's dad wasn't a Mathologer lol.
    2
  6118. thanks to the mathologer videos I found myself in a wonderful math world. I really like math and I like mathologer 
    2
  6119. ⁠ @Mathologer . Hi Burkard - no problem. The real problem is that solving it this way is doing the exact opposite of what your video title says, solving the distribution problem to simulate the water levels. But setting up a robot to build the reservoirs, valves to control water volume to the amount on the on-screen slider, and a video feed to live-stream the result would have been a bit harder. I’m no Steve Mould. :-)
    2
  6120. My youtube journey tonight started with a yugioh video where some monsters "count" their attack points up to "above infinity." Now I'm here. The youtube algorithm is a mysterious thing.
    2
  6121. It blows my mind that nobody has pointed out that your shirt is actually even funnier when you realize that Einstein's equation E =mc2 is actually a very specific application of pythagorean's theorem. Because the full form of it is E^2 = (mc^2)^2 + (pc)^2 The version we normally see is for a stationary object with Rest Mass. So p is equal to 0, canceling the second term.
    2
  6122. 11:34 - easy, it's (9, 4, 13, 17)
    2
  6123. 11:51 - (right, right, right, left) going right the pattern is (?,1,?,3), (?,2,?,5), (?,3,?,7), next we get our (?,4,?,9) and go left to take the 4 and 9 as our seed to create (9, 4, 13, 17)
    2
  6124. 24:05 - the line between the incircle and Feuerbach circles' centers is parallel to the gridlines, and thus the length of one of the parent triangle's sides we're trying to create from dropping a horizontal and vertical line would be 0
    2
  6125. There's actually a nice formula for the partial sums that can be derived as follows: We want to find 1/1 + 1/2 + 1/3 + ... + 1/n. Replacing the numerators with x^(denominator) gives us the function g(x) = x/1 + (x^2)/2 + (x^3)/3 + ... + (x^n)/n. So g(1) is our desired value and g(0)=0. Then, we may take the derivative of this polynomial giving g'(x) = 1 + x + x^2 + ... + x^(n-1). We're going to exploit that this has a nice closed form expression. To see this, multiply each side by (x-1): (x - 1)g'(x) = (x - 1)(1 + x + x^2 + ... + x^n) = (x + x^2 + x^3 + ... + x^n) - (1 + x + x^2 + ... + x^(n-1)) = x^n - 1, so g'(x) = (x^n - 1)/(x - 1). Now, by the fundamental theorem of calculus, we may integrate this function to get our result: ∫[0,1] (x^n - 1)/(x - 1) dx = ∫[0,1] g'(x) dx = g(1) - g(0) = g(1). Thus this integral is our final formula. I think Euler was the first one to find this, though I only know it because I happened to rediscover it in high school. This method can also be extended to the partial sums of this series for higher powers. For instance, 1/1 + 1/4 + 1/9 + ... + 1/n^2 = ∫[0,1] (1/y) * ∫[0,y] (x^n - 1)/(x - 1) dxdy. The interesting thing about these higher power examples is that they actually converge, and that the integral forms reflect this: 1/1 + 1/4 + 1/9 + ... = ∫[0,1] (1/y) * ∫[0,y] -1/(x - 1) dxdy = ∫[0,1] (1/y) * (-ln(1-y)) dy = pi^2/6. Obviously proving that last integral equality isn't trivial, but I think this comment is long enough so I'll leave it there.
    2
  6126. 10:45 I’m confused. We’re trying to prove that p can be written as the sum of two squares, but at the beginning of the proof he says “let’s imagine (assume) p can be written as the sum of two squares.” Isn’t he assuming the conclusion before proving it?
    2
  6127. Merry Christmas, this is the best gift ever!!!
    2
  6128. This video is AMAZING!!! Thank you!
    2
  6129. I yasssss... Another genius video
    2
  6130. There are some YT math channels that regurgitate proofs they learned elsewhere or are hand waiving because they dont understand effective teaching methods. Original content and methods and visual techniques are what make great creators who share their brilliance, logic and hard work and make genuine contribution. Which is what we know we get from your channel. YouTube is saturated with a lot of borderline content now as well. But the paradox is resolved with the subscribe button and like button and the hope of the YouTube algorithm properly doing its job.
    2
  6131. This reminds me of r+s=rs
    2
  6132. I didn't know about the traveling paths in Pascal's triangle. Nice nice nice!
    2
  6133. That was amazing. Wow! My mind is blown! The feelings & emotions I am experiencing is indescribable.
    2
  6134. Neat!
    2
  6135. I love your channel. A simple argument against numerology: symply by change from base 10 to another base number, many numerology theories would fall apart.
    2
  6136. I assume that this can be extended to an infinite number of higher dimensional golden hyper-cubes.
    2
  6137. You are really the best educator I have ever encountered and make supreme use of this medium. I have an advanced degree in chemical engineering and have worked with r&d at three Fortune 500 companies, so I have encountered some fantastic people until now, yet you are the best.
    2
  6138. For the exercise at 8:20, I got y = k*e^[(1/2)*(x^2)] :) hope I was close! it's been a while since I've done DEs before.
    2
  6139. I read that article! I just hadn't made the connection that it was by you.
    2
  6140. Ah, the infamous "Moebius C. Escher" Monopoly Edition! Love it!
    2
  6141. 3 minutes in and already fascinated!
    2
  6142. This is Newton? I discovered this by accident, assumed it was a natural property that had been known about forever, and used it to teach pre-calculus because it made sense. Also, if you're at Monash, did this friend Marty also teach there in the early 2000's? Because if so, he was my favourite maths lecturer.
    2
  6143. mathologer without specs looks like johny sins 😂🤣
    2
  6144. ❤❤❤ I REALLY LIKE YOUR VIDEOS ! ❤❤❤
    2
  6145. Ima huge science nerd but I’m very bad at math
    2
  6146. You're definitely late to the party, and yet instantly my favorite video on the subject. Somehow, I'd never seen the harmonic sum approach and I loved how it fit in your 1/x trilogy. Bravo
    2
  6147. Thanks for making math accessible and fascinating 💛
    2
  6148. Merry Christmas to you burkard sir :) ☺😇
    2
  6149. Crap, so if I had published my findings in ninth grade (2001/2002) that would have been something new?! ...
    2
  6150. Which colour should the horn be painted?
    2
  6151. this describes the motion of 3d objects in 4d space no?
    2
  6152. This might be the longest video, but it's as cool as the other ones. Or even cooler. Sir, you are a huge talent of showing the beauty behind mathematics. Keep going!
    2
  6153. What I nice plate of bananas. Can we call it banana proof? Yes?! Great!!
    2
  6154. A NON mathmatician's observation , so dont laugh given any of your fraction in the pelle set i.e. 99/70 -> d1 + n1 = d2 d1 + d1 = n2 <- n1 - d1 = d0 d1 - d0 = n0 Probably has nothing to do with anything
    2
  6155. Math can be so beautiful. :)
    2
  6156. Mental Math is to Arithmetic as Metal Math is to Calculus :) Great video!
    2
  6157. Fantastic video!!! I'm wondering which software do you use to create those brilliant animations?
    2
  6158. great shirt!
    2
  6159. I used a slide rule in college (after I got out of the Navy) in all of my engineering classes from 1978 - 1981. Calculators were much too expensive then. My wife spent a whole month of her paycheck in 1982 and lovingly bought me an RPN HP-11C. I still have it and it works!
    2
  6160. /Sudoku players foaming at the mouth {Expert mode streak:10 years}
    2
  6161. I'm making this in Unreal Engine right now
    2
  6162. It's just crazy that all of this can be done without any calculus. Thanks!
    2
  6163. Phi is the solution to: x^2 - x -1 = 0, with some rearranging you can get: x - 1 = x / (x + 1), where x is phi If that is what you're thinking about Edit: In addition, phi^2 = phi + 1, so substituting that to the equation before yields: phi - 1 = 1 / phi Which is probably the thing you were thinking of
    2
  6164. You know it's a Mathologer video when he says "to finish off" and you check the timestamp and realise you're only half way through. Nice to see you back; I've missed you. Hope you've all been well.
    2
  6165. This video actually made me think a lot
    2
  6166. At work, I use an adaptation of this, called a Proportion Wheel. See the photo - https://photos.app.goo.gl/n5GwBvFSar753gg17 In the printing and copying industry it is handy to quickly determine how much of an enlargement or reduction is necessary to size an original to a specific size copy. Also can tell at a glance if one dimension will size too much. For example, if someone wants to enlarge a 4" x 6" photo to 8" x 10", you could do the division and find that 8:4 is 200% then you have to multiply that by 6 to find that 6 enlarges to 12, which is too big. So divide 10 by 6 and get 167%, then you have to multiply by 4 to find out how much that enlarges. I know with the metric system it'll be a little easier, but the proportion wheel should be much faster than typing the numbers into a calculator (for those who cannot do in their head). In the photo, line the 6 on the inner wheel with 10 on the outer, then read 167% in the window. Then look at the 4 on inner wheel to see how much it enlarges on the outer wheel.
    2
  6167. I now wonder how many regenerations it would take the Doctor to move a 64-disc tower...
    2
  6168. Does this only work with a base ten system? I know that during the French revolution there was a plan to make the number system base twelve. Would relationships like this still exist?
    2
  6169. How would this apply to taxes? In other words, one debt with multiple debtors.
    2
  6170. I would love seeing a Mathologerized proof of the Feuerbach theorem using geometric inversion
    2
  6171. Perfect, my friend! Thanks!
    2
  6172. This video gave me flashbacks to Q*bert
    2
  6173. That's cool. It would be interesting to do a series on different types of proofs.
    2
  6174. New speedrun category: how long it takes before Mathologer mentions Euler in a given video.
    2
  6175. Don't worry. We know YOU are super famous, at least...
    2
  6176. Proof by contradiction makes me feel powerful like I can clearly predict the future and thought 20 steps ahead even though I know nothing
    2
  6177. 45:34 It's Euler again... can't escape him, right?
    2
  6178. So, Ramanujan might have been a crack in solving sudokus, I understand.
    2
  6179. Excellent video as always. I'm just a little frustrated that you keep covering the topics I'm planning to do, but better than I could.
    2
  6180. OMG, I legitimately never saw that connection coming!
    2
  6181. 2:32 Because phi is defined as the number that satisfies this equation: x=(1/x)+1., therefore 2*x*[(1/x)+1]=2*(x+1)=2x+2. So this means that the left side of the equation is the same as the right side of the equation.
    2
  6182. Intuitively I'd guess that the overhang problem is impossible with only 3 bricks, since 1 + 1/2 + 1/3 is less than 2 (the original length of the brick/table setup)
    2
  6183. I love ramanujan
    2
  6184. I've always had a similar problem stuck in my head, about what would happen if you had any polygon, marked the midpoints of the sides, connected those in the same circular order as the original edges, then repeated that process on the new polygon created. Essentially put 180 degree ears on all the edges. It would shrink to a point, but what would that point be called?
    2
  6185.  @Mathologer  Ah, the centroid! Cool!
    2
  6186. Aussie is a smarter place for having the Mathologer as a resident
    2
  6187. I can tell what is magic, it's the Mathologer videos.
    2
  6188. "I decided that such a proof was most readily found on a (previously) blank page of my notebook" deserves to go down in history as the perfect counterpoint to "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain."
    2
  6189. Fantastic video, thank you so much!
    2
  6190. I learned it in school. From Pakistan.
    2
  6191. The way this makes use of addition to create higher order powers reminds me strongly of integration of polynomials.
    2
  6192. Hail Satan 😈
    2
  6193. I want one ring to rule them all.
    2
  6194. "For the really hardcore of you, can you find another power of 2 in this sequence?" checks up to 10mil with python no, no I can not sir
    2
  6195. Beautiful
    2
  6196. Master please make a video about geometric probability. Thanks for your incredible work :).
    2
  6197. You are a mathemagician.
    2
  6198. But... If we can go back from an irregular polygon that need n-2 passages to become regulare to another polygon that we can transform into the starting one with the angle of 360/n thing... It means that this new polygon need n-1 passages... But this is impossible, because also this one must need n-2 passages... But if this is true means that the starting one need only n-3 passages... And if we make this argument for all angle it will need n-4 passages, then n-5 until 0... So the starting irregular polygon... Is regular... I need a solution to this paradox please...
    2
  6199. Maybe this argument is the proof that there's not an "inverse function" of that explained in the video... But this seems strange to me...
    2
  6200. I would love a make easy for tensor calculus. 😁
    2
  6201. I also play domino means I am mathematician 😎😎😎
    2
  6202. Your shirt reminds me in that some time in 2023, three evils and the root of evil will manifest (2023 ~= 666 + 666 + 666 + sqrt(666)) I found this result using a program I wrote to look for interesting numbers
    2
  6203. I liked the last fact most: that the sum of the terms containing exactly 100 zeroes is larger than that of the sum of terms not containing a 9.
    2
  6204. I typed out, "But how does the compound curvature of the hemisphere become the simple curvature of the cone?" and Mathologer immediately started to address it. :)
    2
  6205. Exactly ten distinct perfectly sufficient variations for all possible four-tile color combinations out of three colors for the side of the large equilateral hexagonic triangle. It 's been such a long time I 've been searching for a deeply satisfying and remarkably clean formula for the combination of the most likely most complicated of all four basic types... The one of the type - With repetition, with No ordinance - it 's just... Always looks logically so similar to the classical ( n over k ) - but definitely not the same! Always somewhat tricky to determine just how many combos go uniform because of the same-color-tile cases - they always vary, one should always subtract different numbers of combos... Anyone figured out a nice formula for this one already? Or are we still searching? :) Really interested in this topic for sure! Another one awesome video and thanks alot really! :)
    2
  6206. Wait, you mentioned the related problem where you have one type of coin for each integer So, that's (1+x^1+x^2+x^3+....) *(1+x^2+x^4+....) *(1+x^3+x^6+...) : . ok, so overall we get Π1/(1-x^n) when n goes from 1 to infinity 1/(1-x) * 1/(1-x^2) * 1/(1-x^3) * 1/(1-x^4) * ... I am supper suspicious that you can do [something something prime something] with this to get something interesting not sure what exactly...
    2
  6207. I got 55 as the next number. I think it’s every other Fibonacci number, but I have no clue why it would be that
    2
  6208. You're right, I put 3×3 as 6 instead of 9, doi. It does seem to be a like double fibonacci sequence thing. Which makes sense since the partitions naturally pair up so much.
    2
  6209. Awesome video!!! Beautiful animations. Got me thinking why not use rectangles they look much tamer. : )
    2
  6210. My vote: one-glance balancing
    2
  6211. Professor, I'm starting to understand your video, great job. However, another approach to the irrational is closer to me. 2 * √3 * (√5-√3) + (√5-√3) ^ 2 = (√5-√3) * (√5 + √3) = (√2) ^ 2 = 2 = 1 + 1 Sorry for the trivial formulas. 😊
    2
  6212. this popped in my feed and watching it made me happy :) thank you Mathologer
    2
  6213. this looks like standard algorithms in computer science, and the table of shortest solutions looks a lot like dynamic programming
    2
  6214. Hey mathologer , love your videos , I think I have a very nice topic for you a new field of calculus "discrete calculus" , I also made a video on my channel, I want to see "discrete calculus" in mathologer style . Thanks for reading
    2
  6215. Ö so a 4k+1 prime can't be written as the sum of 2 squares in two different ways
    2
  6216. matholger please make a video on functional equations
    2
  6217. 01:56 Good visual explanation I was taught: If (leftWeight x leftDistance) = (rightWeight x rightDistance) then they will balance
    2
  6218. Madhava looks a bit like you. Is that on purpose?
    2
  6219. p(666)=11,956,824,258,286,445,517,629,485 Python solution: (Runtime is a few ms) import itertools _p = [1, 1] def p(n: int) -> int: if n < len(_p): return _p[n] # answer is known; return it (avoids recalculation; otherwise this'll take forever) value = sum( ( sign * p(_n) for sign, _n in reversed( # reversed to avoid recursion limit list( zip( itertools.cycle([1, 1, -1, -1]), itertools.takewhile( lambda e: 0 <= e, map( lambda e: n - e, itertools.accumulate((e for _e in zip(range(1, n, 2), range(1, n)) for e in _e)) ) ) ) ) ) ) ) _p.append(value) # save value for reuse later return value print(p(666))
    2
  6220. 11:30 Well, I'd say it's at least π²/10
    2
  6221. @39:00 I guess you mean you didn't speak about Runge's phenomenon amongst other things? It plagued me during some internships :( To summarize for those who don't know what that is, is that when one takes the limit of finer and finer discretization, the interpolating polynomial one gets using this approach may start to violently oscilate between two consecutive data points instead of approaching the real curve.
    2
  6222. The houses the man could be living in are just the natural numbers n such that n^2 is a triangular number.
    2
  6223. Dear Mathologer, Very nice, as usual ! I have a challenge for you, something that I tried to prove for a long time without ever succeeding. Maybe you know (or will find) a way ! The goal is to give a visual presentation of Fermat-Wiles theorem for threes. As in today's video, one goes quickly to cubic shells (of any width). Then one needs to prove that the border of the shell (some kind of "shell of the shell", made by 6 wedges) cannot be a cube filling the remaining shell. A proof it cannot be a cube would of course be sufficient, but it may unfortunately be a possibility. This is where I always failed... Of course, as this would seem to be naturally generalisable to higher dimensions, I guess there is a good reason for this to be a little bit difficult. Cheers !
    2
  6224. Beautiful! Thank you. I did not know about the rubber band analogy (with no slip very important as rubber probably does not have constant elastic constant). I wrote the math section of handbook of mechanical engineers commissioned by the society of mechanical engineers in Turkey some 50+ years ago. I Included explanation of design of linear slide rules using recursive log functions which is the standard explanation. By the way, I was getting paid by the page, so I decided to generate the log tables using an IBM 360. But the society staff did fall for it. Unfortunately they copied the log tables from a book. I would had received 3x pay if it worked.
    2
  6225. I wish I could like this video an infinite number of times.
    2
  6226. Legends are those who are watching mathaloger's video before sleeping
    2
  6227. Every odd number can be written as a difference of two squares in at least one way. Trivial proof: n² equals the sum of the first n odd numbers 1, 3, 5, 7, …, (2n-1), which is easily shown by induction and very easy to visualize too. Just try squared paper and see. 😎 So it is easy to see 2n-1 equals n² - (n-1)². For any odd prime number p = 2n-1 this is the only way to equal a difference of two squares. Proof by contradiction: Let's assume p is a difference of two squares n² - (n-m)² with m > 1 and m < n Then p = n² - (n-m)² p = (n + (n - m)) * (n - (n - m)) p = (2n - m) * m p is prime and p | m means m = p or m = 1, and the last one contradicting m > 1. But if m = p, then 2n - m = 1 and therefore m = 2n - 1, contradicting m < n. So there is exactly one way to express any odd prime number p as a difference of two squares. This was so much easier than Fermat's theorem. Even possible to me. 😉 (Excuse my English, but it is better than most people's German.)
    2
  6228. In case anyone needed to know (although it's obvious to anyone who's seen it), the matrix at 27:45 is the discrete fourier transform. In general it could be any size, and the roots of unity need not be complex roots of unity - they could come from any field, or similar structure with the right nice properties. Also, any results about adding ears conserving the sum are a simple result of treating any directed edges as vectors. Everything is a linear combination, so it's no suprise that everything has the linearity property.
    2
  6229. 10th viewer...
    2
  6230. (2n+1)/(2n+1)=1
    2
  6231.  @learn123  oops, typo. Fixed.
    2
  6232. What a insight, it reignite my spirit to study the quadratic reciprocity again which I did not get it in my number theory course in my university study in 15 years ago.
    2
  6233. Can you do a video for the people who don't even know how to find the papers, let alone navigate thru them and find what we're looking for. It's scary to think about highly intelligent papers and ideas that anyone can read . But i dont know if im ready for the adventure . Imagine just sitting home all day reading mathematical books written by old men decades ago. It would be very spiritual.
    2
  6234. With the help of Hideo Kojima!
    2
  6235. hey mathologer i got a question ive heard abour fractional dimention but never negative dimention
    2
  6236. The real first strand type game.
    2
  6237. Another banger of a vid, thanks ! Have you ever done a vid on self-locating strings in pi digits ?
    2
  6238. Seems that everything in math can be represented in a clever geometrical way, as long as someone clever is able to discover how.
    2
  6239. I like the swiveling proof better. Once you've observed that the two lengths from a corner to the points of intersection are equal, the rest follows with very little explanation. The colouring proof has its own appeal, but it feels a little clunkier.
    2
  6240. 13:39 I didn't check rigorously, but I think the rule is the new numerator and denominator is double the previous 1 plus the one before that, so 17/12 = (7*2+3)/(5*2+2)
    2
  6241. I'll certainly remember the convergence of the no nines sum in case I'm ever held at gunpoint.
    2
  6242. By the way, Dr. Who was a great show! Great choice for introducing the Tower of Hanoi puzzle!
    2
  6243. The image at the end of Madhava's actual writing really showed how little he had to work with. I mean, he's inventing corrections to pi while writing around holes in the piece of wood(?) he has to write on.
    2
  6244. It's amazing! Thank you so much!
    2
  6245. The three Hypotenuse lengths of the inner circle of an RGB triangle are of equal lengths too.
    2
  6246. I immediately noticed the partition numbers :)
    2
  6247. I am so excited to digest this video. I Love how you invite us to try using our Intuition and grasp these Concepts.
    2
  6248. Breaking the video into chapters was a great idea. Each chapter was a gem on its own, complementing the whole video.
    2
  6249. 10:10 Well, we also have to pick our starting point; since, on the infinity-symbol, the result will vary; depending on, where we start running; not just on the traversing-direction, as you just demonstrated, a moment ago.
    2
  6250.  @Mathologer  Yeah; having watched the video farther, I think all the discrepencies, in the results, will cancel each other out, in the subsequent rounds.
    2
  6251. EDIT: here is a p5.js sketch where you can generate your own random artctic circle. https://editor.p5js.org/kaschatz/sketches/GHCkS-FyN I just coded the visualization at 30:57 in Processing. I can try to translate it to p5.js if anyone would like. Here is a demo: https://youtu.be/q2Ksbj11WYc
    2
  6252. Visual Proof of the no 9's series is awesome and (very)*(1+1/2+1/3+...) Nice!!!
    2
  6253. The most memorable is WHY TF IS THIS SUM FINITE ARE YOU KIDDONG ME?
    2
  6254. Such a great video, hats off to you, you never made me satisfied with what I study. You bring us such a great content Thanks man 🙌👀💓
    2
  6255. The problem with the 7x7 square is that it isn't a magic square. Apparently the creators of the movie didn't bother to make sure it was correct. The moral is don't believe everything you see at the movies.
    2
  6256. Is the geometric figure at the start a visualization of projective 3-space over Z_2?
    2
  6257. I went to check out those other videos to see if the 7 million people really were right and I had to stop after like 30 seconds because I didn't want to lose brain cells hahaha
    2
  6258. Woah, that is so amazing!!! Thanks a lot sir:) PS do you remember me from the orange logo? that is "이호석"
    2
  6259. For the programming challenge for dollar amounts, this is how I might try to calculate it: if a_n is the number of ways to make change for a value n coin using 1, 5, 10, 25, 50, and 100 value coins, I have the recurrence a_n = a_(n-1) + a_(n-5) + a_(n-10) + a_(n-25) + a_(n-50) + a_(n-100). (a_k for k < 0 is 0 and 1 for k=0). You could then just use the recurrence relation to calculate the first however many terms in the sequence. If you wanted a closed formula for a_n, Because each a_n depends linearly on the last 100, you can find a (sparse) 100x100 matrix of 0s and 1s that when applied to the vector [a_n, a_(n-1), a_(n-2),..., a_(n-99)] results in the vector [a_(n+1), a_n,...,a_(n-98)]. Then you could convert this to Jordan canonical form and find the change of basis matrix. In principle, you could figure out the formula for raising each of the Jordan block matrices to the n-th power, then use this to find a closed form expression for a_n. I have a feeling that the latter method might be very inefficient, and susceptible to round off error or something like that.
    2
  6260. what sr sr² mean ? what operation we are doing ?
    2
  6261. sr = s times r, and sr² = s times r² = s times r times r
    2
  6262.  @ffggddss  Fibonacci sequence with initial values a(0) = 3 and a(1) = 1
    2
  6263. ​ @Mathologer  Bob Newhart was an American comedian, now passed on, who got his start as a stand-up act, ca 1960, with the release that year of two hit LPs, The Button-Down Mind of Bob Newhart, and The Button-Down Mind Strikes Back, then got starring roles in 2 or 3 TV sitcoms in the following decades. His brand of humor was what could fairly be called "deadpan." In his penultimate sitcom he played a psychiatrist (in Seattle?), and Suzanne Pleshette co-starred as his wife. In his final sitcom, titled simply, Newhart, he and his wife Emily, played by Mary Frann, ran a small hotel (the Stratford Inn) in Vermont, and it co-starred Tom Poston, among others. The final episode got a lot of buzz with its closing scene which started in darkness, then a light switched on showing Bob and Suzanne Pleshette in their roles in the earlier sitcom, with Bob describing his "strange dream" in which he ran a hotel in Vermont. Suzanne tells him to go back to sleep, the light is quenched, and the credits roll. IMDB will undoubtedly have more info on him. Wikipedia also has a good bio. He was quite famous in the US.
    2
  6264. His proof didn't rely on the 10 numbers at all (besides that they were between 10 and 99, inclusive), so it works for any collection of 10 numbers in the range. The way this works is that he way overcounts the "possible sums". This is essentially like overcounting the number of pigeonholes. This is not a problem, though! Even overcounting the number of pigeonholes, we still have fewer pigeonholes than pigeons, so we know that at least one of the pigeonholes has to have at least two pigeons in it. How does this work? He considers the lowest possible sum you could have. If your set of 10 numbers includes the number 10, then the lowest possible sum you could have is 10 itself. If your set of 10 numbers does not include the number 10, then the smallest possible sum will be larger than 10, since 10 is the smallest number possible. So 10 is the lowest possible sum you could have. With most sets of 10 numbers, the true smallest sum will be larger than 10. Similarly, he considers the largest possible sum you could have. If your set of 10 numbers includes the numbers 91-99, then the largest possible sum you could have is 91+92+...+99 = 855. If your set of 10 numbers does not include every number from 91 to 99, inclusive, then the largest possible sum you could have will be smaller than 855, since 91-99 are the 9 largest numbers possible. With most sets of 10 numbers, the true largest sum will be smaller than 855. So, we know that, no matter what 10 numbers you get, there is no way to have a sum smaller than 10 and no way to have a sum larger than 855. So, no matter what 10 numbers you get, any possible sum is between 10 and 855, inclusive. So there are 846 possible values of a sum we are considering here. With a fixed set of 10 numbers, there are way fewer options, but here, these 846 options account for any possible set of 10 numbers you could have. Now, when you have 10 different numbers, there are exactly 2^10 = 1024 subsets of these 10 different numbers, including having none of them and having all of them. Since we only want to consider nonempty (and hence also non-full) subsets, there are 1022 possible subsets of a fixed collection of 10 different numbers. Counting the number of nonempty non-full subsets does not at all depend on what the numbers are - just how many of them there are. So his count is accurate. So, no matter what 10 numbers you get, there are only 1022 possible subsets of them. Also, no matter what 10 numbers you get, there are only 846 possible values you could have as a sum. So at least one of those sum values must have two subsets corresponding to it. There are, therefore, at least two subsets with the same sum.
    2
  6265.  @MuffinsAPlenty  thank you very much that was really helpful :)
    2
  6266. Didn't we say the volume of the horn was less than 2, but also equal to pi?
    2
  6267. interesting
    2
  6268. The bugs meet after distance 1 because they start at a distance 1 apart. The direction of each bug is perpendicular to the path of the bug chasing it, so the total distance to be covered never changes.
    2
  6269. So, does this mean that you will get an infinite amount of light out of the lantern?
    2
  6270. I love your videos but i'm waiting for a meme inspired on your laughter 🤭
    2
  6271. when you say "wiki" do you mean wikipedia or some other secret wiki I wasn't told about?
    2
  6272. I remember this analog gasoline economy calculator my father had in his car. It consisted of a rotating disk on top of another disk where you had to align the number of kilometers driven with the number of liters that went into the tank and it would show you the number of liters per 100 kilometers. As a teenager, I was very fascinated by it.
    2
  6273. 14:20 The "unpopular" guy (with no hands): ...
    2
  6274. I first came across it when I was in middle school(3:00). It was a great mathematical leap for me.
    2
  6275. I wish we all could be taught math like this. I love this geometrical explanations!
    2
  6276. The elegance of simplicity shines once again. Thank you for this delightful video.
    2
  6277.  @Xanthe_Cat  That example isn't really a rare instance, to be fair, because the idea that 1 should be excluded as a prime number, while still the most common view among mathematicians, wasn't nearly as definitive until the early 20th century. It was Gauss's work into unique factorization domains and analogizing the concept of primes beyond the integers that caused things to start shifting from "eh 1's not really a prime but it doesn't really matter" to "yeah 1 absolutely isn't a prime". There was a fascinating survey on the history of 1's position among the prime numbers published in the Journal of Integer Sequences in 2012: "The History of the Primality of One: A Selection of Sources" from Caldwell, Reddick, and Xiong.
    2
  6278.  @will-vi9pk  Nope, it's divisible by everything except itself. In one sense, it's the most composite number.
    2
  6279. Exceptional storytelling!
    2
  6280. The peg solitaire article is so cool! An unexpected real world application of the Klein four group :)
    2
  6281. Thanks for the video!! Greetings from Brazil :)))
    2
  6282. I love how you accidentally (or sneakily?) Illustrate the general formula for the 3x3 determinant in 44:00 since the numbers just happen to match the two indices of a component in the matrix
    2
  6283. Great to see! A few avenues not explored: The four triangles and their trigonometry. Including THE heptagonal triangle (with sides 1:r:s or angles pi/7, 2pi/7, 4pi/7) r satisfies r^3-r^2-2r+1=0, and s is a root of the reciprocal polynomial. Recurrence relations, similar to those relating to the Tribonacci constant. Rational approximations to r (and s), starting with 1,2,9/5,182/101 (and 2,9/4,182/101)
    2
  6284. I am disappointed in your T-shirt. The root of all evil is 25.80697580…. Your shirt should’ve read 25.807!
    2
  6285. My approach to solving coin paradox cases right away is to trace just a quarter of the rotation in my head. It seems to help a lot and kinda does the same as the unwinding if the path.
    2
  6286. I wish I was smarter
    2
  6287. When I read the term "shrink" in the title I connected with "psychotherapist" at first. Don't know why, must be the summer heat...
    2
  6288. Lee Sallows also discovered alphamagic squares! (magic squares where the number of letters in each number when spelled out also forms a magic square)
    2
  6289. How can we get the generalized antiderivative F(x) for any arbitrary function f(x), such that F(x)' = f(x), using the sum of slices for which the width approaches zero? Are there better methods?
    2
  6290. 12:30 What happened to the green 6 there? It does not work when you remove it.
    2
  6291. You need more views. Your videos are really detailed, structured and interesting. Merry Christmas!!!
    2
  6292. Can you do a video on Penrose tilings?
    2
  6293.  @Mathologer  Can't wait!
    2
  6294. 1+1/first , next subject continuous fraction transform/moebius transform ? 🤤
    2
  6295. Beutiful video
    2
  6296. The answer to the 4 domino question: No! Imagine removing 3 tiles such that any of the 4 corner tiles is on an “island” (has no adjacent tiles). This fits within the guidelines of removing at most 2 of each color, and therefore we can create a mutilated board which we can guarantee cannot be tiled. Good puzzle!
    2
  6297. I think the calculus comes in handy when you want a rigorous proof for the area, which you hand wave slightly. It's not entirely obvious that the area of the rectangles is smaller than the surface area of the enclosing horn sections. You need to know that the surface area of something is greater than the area of its shadow or something to that effect. Proving this probably requires some calculus.
    2
  6298. For the lanterns there are two substitutions that effect the angles, we need all the angles to go to 0 or 180 depending on how you measure angles to get the minimum surface area, just like with the line segment, but now in higher dimension. If you have the verticle band substitution increase all the angles or some of the angles, but the substitution around the circumferance always reducing the angles, to do both and get a limit that is identical to the object approximated, the substitutions that always reduce the angles between triangles to reduce the angles the vertical substitutions increase for each step, such that all the angles reduce towards the limit, then we can end up with a smooth surface and not an infinitesimally wrinkly one 🍻
    2
  6299. Ooooh, that is *pretty*. More combinatorics, please!
    2
  6300. You know what would i love? A series of videos where you take the greatest mathematicians in the history and show us some of their contributions in a brief but in that marhologer level of explainability. I love to hear about the mathematical advances that the great minds performed and How these impacted maths and the humanity in general.
    2
  6301. 30:20 Homework: x² = x+1 x²-x-1 = 0 (x-1/2)²-(1/4)-1 = 0 (x-1/2)² = 5/4 x-1/2 = +/-√5/2 x = (+/-√5/2)+(1/2) x = (1+/-√5)/2 (1+√5)/2 = 1,618033989… = φ (1-√5)/2 = -0,618033989… = -(φ-1) = -(1/φ) Q.E.D. 𑀩
    2
  6302. If go in desiring to complain, and yet find nothing to complain about, you can legit complain about that, and so you will always, if you desire to complain, find something to complain about. I call it the Karen rule.
    2
  6303. def p(x): return x**10 sum(map(p, range(1, 1001)))
    2
  6304. Both proofs are very beautiful each in their own way and using color and animation is so much more elegant and easy to understand than filling the screen with Greek letters. I wonder does the original theorem generalize to 3 dimensions (and greater) where the "folding" takes place with three pyramids instead of two triangles? And the lines dropped from each apex would cut each other in the ratio 1 to 3.
    2
  6305. I’m thinking that the arbiter’s fees can be apportioned by the thickness of the connection pipes. 😳👍
    2
  6306. Without ancient Indian Maths and geometry modern world and modern technology would not be possible this is the to power of Hinduism
    2
  6307. For the card trick I dis-carded (no pun intended) the color/suit of the cards and just imagined the cards as numbers 1 to 52 in a circle with "52" right next to "1". That you can encode 24 possibilities with 4 cards is quite straight forward with (4*3*2*1) but it took me quite a while to figure out how to reduce the 48 remaining cards to 24 possibilities. The answer had to lie inside of the chosen card. Not every card of the 5 can be chosen. Since I imagined the cards in a circle you can be sure that if you choose the "right" 4 cards that the fifth is at most at a distance of +-12 cards from the edge of chosen cards if you connect the 4 chosen cards as points on the circle with a minimum distance (For example "52","1" get a connection of 1 distance and not 51. Same principle for 4 cards). If that were not the case you would have to have 5 cards where every card is at least 13 cards away from the edge. 5*13 is 65 and bigger than the amount of cards (52). Of course in principle its the same trick. The suits of the cards don't give any more information than reducing the card to a number so the trick has to be possible without looking at the suit but I have to confess its way more intuitive the way Matholger explained it.
    2
  6308. Thanks for the book recommendation. On a similar note, do you know the book "Im Zaubergarten der Mathematik"?
    2
  6309. This comes up a lot in the study of recursive algorithms. I never knew Newton had anything to do with it.
    2
  6310. As a reply to the mention at the end of the video, I’m an undergraduate in first year Maths at Nottingham, this video was perfect for my level of understanding, although I’m not entirely sure the use of matrices would have escaped most other viewers. Either way, brilliant video, and very beautiful animated proofs! Great job.
    2
  6311. excellent video but I can't help noticing your shirt, the positions of lizard and spock should be flipped so that the graph has 5-fold symmetry :(
    2
  6312. You need to sell your shirts. Your shirts are highly desirable.
    2
  6313. wow i was just reviewing a Mathologer video from 2 years ago and this came up. very nice
    2
  6314. Favorite part: Chapter 5, and the proof that the total excess over the logarithm is less than 1. That visual proof practically hands us the answer to why gamma is greater than 0.5: Each of the blue regions is larger than a triangle. It has the area of a triangle plus an extra belly. Perfect triangles would cover half the 1x1 area; area=base*height/2. I found the odd/even thing in chapter 4 perhaps less surprising. The thing is, all your denominators are going as some factor of n!. It's as if there is some formula where the "raw" denominator is exactly n! while the "raw" numerator has no direct factorial pattern. Thus putting the fraction in lowest terms cancels a few prime factors between the raw numerator and denominator. But since the raw denominators are n!, they have plenty of factors of 2, always more than the few 2's in the relatively rough raw numerators. Thus always odd/even. The same pattern emerges more weakly when you look at it mod 3: After a while (everything past 63/40, as far as I looked) all the numerators are 0 mod 3, for similar reasons. For instance, for lurkers: 7433/3960, 673/360, 8719/4680, 63373/32760, 65557/32760 This also tends to explain why the series misses all the integers, or at least make it less surprising: there are too many factors in the denominator to expect them to all be cancelled by chance, since no simple factorial pattern arises in the numerator.
    2
  6315. Lovely video as always, well presented and many fun insights! The 2+2=2•2 is also 2^2. This also reminded me of a different equality: x^y = y^x, which has the obvious solutions x=y, but also another curve of solutions with only 1 integer solution: 2^4 = 4^2. Just a fun curiosity I once thought about while trying to go to sleep. 😂
    2
  6316. Most memorable part: Mathologer seal of approval :)
    2
  6317. You almost said 666/885ths. You can´t help bringing up the Beast´s number in each of your videos!
    2
  6318. Homework from 11:56 Part 1. For the triangle given (153, 104, 185). The four numbers A B C D are as follows. A=9, B=4, C=17, D=19. To answer the bonus question. The sequence from triangle (3,4,5) to triangle (153,104,185) is (right, right, right, left).
    2
  6319. Feel like I saw the purely geometric Pythagoras proof in a math magazine under a feature called "proof without words" or something similar (Edit: around time of high school, in the US). The algebraic version always felt indistinguishable so I can't say if I ever saw it stated separately. Don't recall seeing either in public school, though that may be because by a fluke I never took high school school geometry
    2
  6320. Is the degenerated case comparable to a quadrilater in a ellipse? Is Ptolemy's theorem valid also for elliptic quadrilateral?
    2
  6321. If you play with the first 20 integers for √a+√b=√c you get that a+b can be 32.00%, 37.50%, 44.44%*, 48.00%, 48.98%* or 50.00% of c but never more. The best, ie 50% is where a=b, so a+b=½c. [Within the arbitrary limit I set] The squares 1, 4, 9 & 16 have the most integer solutions, 4 each, and the odd primes have the fewest, one only where a=b≡p because the square root of any prime is irrational. * not exactly, the others are precise.
    2
  6322. Wait the second series.. 1 - 2 + 3 - 4 + 5 - 6 + 7 - ... Can be rewritten as 1 + (3 - 2) + (5 - 4) + (7 - 6) + ... Which is 1 + 1 + 1 + 1 + 1 + ... Which is +♾️ Mathology debunked 😂
    2
  6323. A real delight to watch, thank you. I suppose the triangle is also a 1D cellular automaton with the generations running down. Similarly the pyramid could be a 2D cellular automaton with each horizontal slice being a generation. I used to love playing Conway's Game of Life. With n neighbours and c colours, I think there might be c^(c^n) possible generation rules so these particularly symmetric ones are special cases in a sense.
    2
  6324. 20:44 I see an induction here
    2
  6325. Can we just take a moment to appreciate his adorable giggling?
    2
  6326. nice pie shirt
    2
  6327. 6:20 - No
    2
  6328. The argument still works even if you take the last 10 numbers. The number of possible sums you can get is still smaller than the number of subcollections, even if you all the sum of all 10 numbers to count as a "possible" sum. But, you can actually say more. What you want to do here is find two nonempty subsets of the 10 fixed numbers which have the same sum. If all 10 numbers were used in one of the sums, there couldn't be a second collection to compare that sum to! The maximum number of numbers which can possibly be used in one of these subcollections and still end up with two nonempty disjoint subcollections with the same sum is 9. Using 10 numbers, you're overcounting the amount of possible sums, but it doesn't matter because even with overcounting, you still get the result. It's okay to overcount the pigeon holes provided that you still end up with more pigeons than pigeon holes! Actually, with some more reasoning, you can show that you can't have one of the two subcollections contain 9 of the fixed numbers either, if you're looking for two disjoint nonempty subcollections with the same sum. Here's a challenge for you: what's the maximal amount of numbers you can have in one of these subcollections if you're looking for two disjoint nonempty subcollections of the ten numbers with the same sum?
    2
  6329. Great video as always, sadly the morons who believe all the numerology BS won't understand any of it. Even words like "divisibility" or "base 10" are far beyond their comprehension.
    2
  6330. Thank you for the gifts.
    2
  6331. The puzzle at 16:35 is the one the Celestial Toymaker should use if he wants to hanoi you.
    2
  6332. @8.12 This makes me think of an animation of a rotating tesseract.
    2
  6333. I am always so surprised about how you find these topics and incredible visual and pretty proofs, even of facts I already know and know how to prove (not in a pretty way but rather more technical proofs).
    2
  6334. EARLY
    2
  6335. hi
    2
  6336. This is very cool. I do a lot of tessellations and this is related. I've thought for a long time that there are many unknown simple mathematical theorems and proofs yet to be discovered. Math is a science with infinite possibilities so it seems intuitive to think this.
    2
  6337. I remember the first time I learned that you can treat dy/dx as algebra rather than a symbol. Revolutionary.
    2
  6338. Fantastic video - loved it! 😍 I learned a method of constructing odd order magic squares when I was in Jr High School. It was much later that I became intrigued with even order magic squares. I finally managed to crack that by dividing the problem into even-odd (2, 6, 10, 14, etc) and even-even (4, 8, 12, 16, etc) cases with a different technique for each.
    2
  6339. Nothing unusual, just a typical awesome video from Mathologer
    2
  6340. He reads all comments man!!😀
    2
  6341. This sh!t here is insane, it's insane I tell you!
    2
  6342. The lazy physicist answer for puzzle 2 in a few seconds: 10^2 = 100, and the other 4 terms are slightly bigger than 100, so the five of them should be a bit bigger than 500, and 365*2 is 730, which is also a bit bigger than 500, and a puzzle like this probably has a nice round answer, so I estimate the answer to be 2 (which turns out correct!)
    2
  6343. that looks nothing like a vrotex bro
    2
  6344. Googles quantum computer could do this in about 3 minutes, but it would never come close to rendering crisis in a stable 60 frames per second.
    2
  6345. Small correction to the info: The Celestial Toymaker is more properly referred to as a Doctor Who *serial*, consisting of 4 episodes, 3 of which are still missing. Only the last episode has been recovered - interestingly enough, I see it was recovered from Australia, in 1984, and that it is the only missing episode recovered from a broadcaster (ABC) in Australia (other episodes were recovered in Australia by private collectors; some clips were also recovered by ABC, ironically mostly censored scenes).
    2
  6346. Who fan!!!
    2
  6347. Formula for finding anything: 1 - Write a python script which goes through every possibility. 2 - Wait the age of the universe (or more). 100% success rate
    2
  6348. very pleasing
    2
  6349. seen the binomical coevitionts and my first guess is : ok where are you pascal XD
    2
  6350. Thanks! This is the kind of videos where, when you want to click LIKE at the end, you realize you already did it earlier because it was so good.
    2
  6351. Most memorable: Convergence of no 9s series.
    2
  6352. 3:00 I suspect the answer has to do with the fact that rearranging the terms can change the sum of a convergent infinite series 🤔.
    2
  6353. Genius is making the impossible look ordinary. R.D. Moses (2024)
    2
  6354. The Enemy Sequence is Brazilian.
    2
  6355. The most memorable thing for me is somehow the most expected thing - the fact that Euler had to show up somewhere.
    2
  6356. I love Mathologer's videos. His videos are very informative and great for anyone who has an affinity for math. That being said, I also LOVE his shirts. I like to collect witty t-shirts, and his make me jealous that I don't have them!
    2
  6357. Should we start calling the "Dirac Delta Function" the "Oresme Delta Function"?
    2
  6358. Hmm - needs a physical model. Add it to my backlog of woodwork things to make.
    2
  6359. Did anyone else start to see cubes instead of hexagons after awhile in Pascals Triangle? And now I want to play Q-bert
    2
  6360. It does not really make it fair that the claims above the total value is clamped to the total value. If we have an estate with a value of 100 and two claims for 100 and 100000000, then both claimants will receive half the value. This, in spite of one claimant only having a claim of just 0.000001th of the other claimant.
    2
  6361. 19:47 I'd tell my friends who'd prove me wrong.
    2
  6362. 26:16 you lied here, you assumed that these three triangles can make a pyramid. lucky for you, I've got you covered :). for this patch, we only need to say that there is a pyramid with a vertex that has the three angles in the triangles. and I wonder where can this pyramid be... oh wait, the original pyramid has these triangles! problem solved! We can even get rid of this hand-waving by taking the original pyramid, flipping all of the sides near the top vertex (reflecting it across the angle bisector, such that the sides flip), and then scaling it by the correct amount!
    2
  6363. And I also really liked the T-shirt😋
    2
  6364. Good thing is, I never finished the first video, so I need to watch this one anyways!
    2
  6365. Thanks sir :)
    2
  6366. Loved the video! Great way to introduce and motivate a lot of facts that seem mysterious at first blush. Regarding the last proof, was there some reason given that the area of the hyperbolic wedge should be alpha? Visually it seems to have slipped in when you shifted from the circular to hyperbolic perspective at 25:27, but I don't see why those areas have to be equal.
    2
  6367. I would argue that in the original 3-color game a triangle of side length 1 also has the same "special" property, since the same hexagon is in the upper left, upper right and bottom corners, and if this single hexagon in the triangle has color X, then the sum of X and X, which is X according to the rules, matches that hexagon.
    2
  6368. The legend is back!!!!!
    2
  6369. dude! at the end of the video, the green part's formula for area was given, but it was the area under the curve 1/x from 2 to some mystery number near 4! what was that number near 4‽
    2
  6370. My favourite topic is covered , thanks ...
    2
  6371. "Anschauliche Geometrie" ist auch Standard sounding title that "geometry and the imagination" sounds like the book equivalent of clickbait.
    2
  6372. Awesome video! Your excitement is truly contagious through the video :D
    2
  6373. It always comes back to pascal's triangle...
    2
  6374. . All boils down to the DECIMAL system. If you can duplicate all these "anomalies" with BASE 16 or BASE 9, then YES! It is wondrous!!! . Always remember : BASE 10 was an ARBITRARY NUMBER chosen most probably because we have TEN DIGITS in our hands/feet! .
    2
  6375. Nice Pascal's triangle algorithmically - by one edge.
    2
  6376. It seems Ptolomy's theorem is just a theorem about the side lengths of a tetrahedron adapted for degenerate cases.
    2
  6377. dimensionless pythagoran triples. Its connections like this that feel like foundations rather than amusements. Stunning.
    2
  6378. Is there a way to generate magic squares with even numbered dimensions?
    2
  6379. Background: Did undergrad in math ten years ago and did so poorly in it that I didn't do any maths for a decade and just got back into it (now I'm teaching Calculus, including infinite series, maclaurin series etc) I love these because the animations are so wonderful and using the colour coding makes things easier to follow. There are so many deeper mathematical ideas that are so beyond me that reading just a little about them makes me want to give up immediately but I find your videos entrancing. What program do you use to make the animations?
    2
  6380. Just 3 mins into the video, and I'm blown by that pretty little proof. Archimedes principle yada yada yada sure, but putting that graphically in such an immaculate manner, amazing! Also turns out to be the video that's introducing me to your channel, I'm totally subscribing!
    2
  6381. Lesson learnt. There's always more to magic squares. Was not expecting that!
    2
  6382. Simply stated: math is awesome.
    2
  6383. 16:35 -- so how about you actually also enumerate the actual AREAS of the exhibited cylinders ?? >=(
    2
  6384. Very cool! When you gave your formula for generating each triple from the previous triple, that technique has a clear higher-dimensional analogue. If you generate n-tuples using this rule, I would imagine that would correspond to a golden n-dimensional hypercube, which would have a corresponding golden spiral. I’d be curious to know if these higher dimensional analogies had any interesting properties, applications, or divergences from their lower dimensional counterparts.
    2
  6385. I am looking at the super-pattern for the squares... Doesn't look so nice, but 5*(2/1)=10, 14*(3/2)=21, 27*(4/3)=36... Next would be 44*(5/4), so 55²+56²+57²+58²+59²+60²=61²+62²+63²+64²+65². Next one would start at 65*(6/5), or 78.
    2
  6386. Simon Plouffe's writeup :) Redeeming features: 7417 is a prime number and 240 has lots of factors.
    2
  6387. I'm About to plug Bitcoin Prices into the Master Formula Let's see If I become a Mathematician or a Millionaire or maybe a Millionaire Mathematician.
    2
  6388. Archimedes was indeed a genius. Seeing his way of indirectly using calculis through geometry, it would be interesting to see how he would have invented calculus. Except a monk erased it to write his prayers.. rip
    2
  6389. Total nostalgia! I discovered all of this back in 1992. Not the geometrical visualizations, but ALL of the algebra ... simply by virtually swimming in the beauty of the number sequences. In the desire to find the algebraic expressions of r and s, I ended up with x³-7x-7=0, where x=rs-1, for which there are three real solution, from none of which I was able to eliminate the imaginary part of their algebraic expressions. Major disappointment! 😅😅.. And when I finally gave up, and completely emptied my mind of all attempts at algebraic manipulation toward a non-complex expression, and instead demanded my mind to directly visualize the solution, it in fact came up with an approximation that was so close to the true value of rs-1 that my calculator fooled me to believe I actually hit jackpot! But I ran to my computer to compare with the real value of rs-1, only to find a discrepancy just a few points down the line of decimals 🙄😢😢..
    2
  6390. Ja! Mach eins auf Deutsch, bitte! :-)
    2
  6391. 14:06 Holy maths! That's the Fibonacci sequence.
    2
  6392. Thanks!
    2
  6393. Hello Mathologer! You again created an amazing video about an interesting geometric theorem and made the proof visible for us. Magic! - Thanks, Greetings and best wishes!
    2
  6394. ​ @Mathologer  what would be great is if his followers would come watch this instead. How do we reach those people?
    2
  6395. That's amazing
    2
  6396. Cool video! Really had to watch it immediately when I saw it!❤
    2
  6397. I actually do a crude Schwarz lantern to my energy drink cans when I crush them to save space. So I immediately knew where you were going with it, although I had never connected it to differentiation before.
    2
  6398. It screams "BEHOLD!"
    2
  6399. This seems relevant for sudoku
    2
  6400.  @MonstraG55  So, I once heard someone describe common ways to solve a hard mathematics problem. Sometimes we can add structure to a problem in order to make it easier. For example, if you had an 8x8 grid and you removed two opposite corners and wanted to tile the remaining grid with 2x1 or 1x2 pieces, could you do this? One way to solve the problem is to add structure to it - imagine the 8x8 grid is a chess board, where squares alternate colors. Then the two opposite corners are the same color, so it's not possible because 1x2 or 2x1 pieces always take up 1 of each color (this is essentially the argument used at the beginning of this Mathologer video: watch?v=Yy7Q8IWNfHM). So adding structure (color) to the grid helped us solve the problem. But sometimes removing structure from a problem can help us solve the problem. The example I heard for this possible way of solving a problem is to imagine you have a log which is 1 meter long. And you drop a bunch of ants onto this log. The ants are all on a single line on this log (lined up with the log, so 1 meter long) and they always move along this line. All ants move at a constant speed of 1 meter per minute. Suppose that if two ants run into each other (going opposite directions), then both ants turn around and move in the opposite direction. What is the maximum amount of time it will take until all of the ants walk off the log? It's a hard problem, but there's a nice insight. If two ants are walking toward each other, collide, and both turn around, this is the same situation as if the ants are able to pass through each other. (The ant moving left switches roles with the ant that was originally moving right but turned around to move left, and vice versa.) So we can remove this whole requirement about ants turning around when they collide and can solve a new problem where ants pass through each other. Then the answer becomes easy. The longest it can possibly take is 1 minute, since that's the longest it can take an ant to get from one end of the log to the other. That's what's happening in this scenario with chapter 6 in this video. We can remove structure from the problem. We want to find two non-overlapping sets with the same sum. That seems hard to do, so let's remove part of the problem. Remove the requirement that they're non-overlapping. If we are able to find any two different sets that have the same sum, then we can just remove their overlap (removing the same amount preserves the equalities of the sum). So really, all we have to do is fine two different subsets of the same sum. We don't have to do everything at once. We don't have to use the pigeonhole principle to guarantee that there are two non-overlapping sets with the same sum - we just have to use the pigeonhole principle to guarantee two sets with the same sum. The existence of two sets with the same sum guarantees the existence of two non-overlapping sets with the same sum. So we're done once we put these two facts together.
    2
  6401. Brilliant explanation and kudos for revealing to the world Madhava's contribution to this domain of mathematics 🙏
    2
  6402. I'd hate to say this but at my age (I'm 66) I couldn't understand much of your speech because you tend to mumble a lot (or whisper even) and your video audio is so low I have my phone speaker on full and yet I still have to stick my ears close to it. I wish you would correct these flaws for this video especially since it seems to be of significant value to everyone. Thanks.
    2
  6403. 19:54 heresy!
    2
  6404. 19:21 You can't fool me, that's Andy Kaufman! He didn't fake his death to go into hiding: he did it to time-travel to the 1800's and invent the two-square theorem under an assumed identity! IT'S THE ULTIMATE PRANK!
    2
  6405. Beautiful and brilliant
    2
  6406. did you too watched to the end for the autopilot algebra music? ;)
    2
  6407. For the T-shirt, each row contains a number of dots equal to the next digit of pi. And every other digit goes to the right. And every third digit is colored white. Am I missing anything?
    2
  6408. "Superkey to the univers" Gotta love that insulting not insulting way of speaking. Good to see a demythifying video, too many people fall for those seemingly amazing things that turn out to be very cool but very normal results.
    2
  6409. Ah, another master class by Mathologer. I recreated my own Geogebra project of this "screwed-up boring pentagon" on my phone. I found out that even if all vertices of the boring pentagon are collinear, the end result is still the same as the one you explained. And, one of the sides of the regular pentagon produced is parallel to the line of the collinear vetrices. I haven't tested the regular pentagram one. Once again, thank you, and have a good night sleep.
    2
  6410. Congratulations on 100 episodes! May you make many more ...
    2
  6411. Remember my math teacher show her "favorite" prrof of Pythagoras. Took full month explanation it. Something on Riemann geometry and some insane analysis. Parents complain lots because nobody follow her. 35 years later now I know wy she did it. She was no math teacher but physiscs teacher and later physics professor. Pythagoras with four dimensions is needed for relativity and she believed everybody must know something on relativity. But for kids science and math is boring. She trid hide everything in the explanation but failed. It become to much. Most teacher today lack spirit and enthusiasmus. They can teach the kids but it not stay in there heads. Good teachers can inspire and kids think on there self. As a kid math and physics (which is only different math) were very boring, not like chemistry. As adult now i have a good collection by math and physics books, even the Feynman lessons (German translation). To my 50th birhtday coworker gave me a modern print of Euklid, german version. More as few people asked what it is, what is sad.
    2
  6412. Put some videos on integrals
    2
  6413. I actually realized aspects of this several years ago based on visualizing the squares and cubes. This is the first I've seen of it being an official hypothesis.
    2
  6414. Watching this at 2am I think my brain’s dead now
    2
  6415. lets say q=the house that the guy is in. and c is the number of people on the street. q=+or-((3-2*sqrt(2))^n-(3+2*sqrt(2))^n)/4sqrt(2), n belongs to positive integers. c=+or-1/4 (-2 + (3 - 2 sqrt(2))^n + (3 + 2 sqrt(2)))^n, n belongs to positive integers this is found with algebraic manipulation of this equation :) also psst the 13:42 challenge for finding a relation: i figured somehow that the denominator of f(n+1) or the fraction that is after the current is 2 times the denominator of the current and you add f(n-1)'s denominator, or the fraction before the current. The same rule is for the numerator.
    2
  6416. Chapter 6 is the best one hands down
    2
  6417. Nice!
    2
  6418. I've still got a little log table booklet. Looking it's age now. I was always amazed at how these tables were calculated by hand so long ago in the past.
    2
  6419. What a fantastastic video,I MADE IT TO ALL OVER THE END but i have to do reach ALL OVER THE END AGAIN to fully grasp it completely.
    2
  6420. yay new video!!
    2
  6421.  @Mathologer  it's a wonderful gift either way. Your choice and visualization of proofs is always beautiful.
    2
  6422. Aeusserst interessant! Eine ganz schoene Erklaerung!
    2
  6423. With the "over 200" part in the number of proofs, you reminded me of Catalan Numbers and the amazing number of structures that follow them.
    2
  6424. I'm going to post my solutions to your homework assignments here: 1) The difference between the grid square approximation and the actual area of the circle is proportional to the radius of the circle. To be precise, an upper bound on the error can be found by considering the circular ring between the two concentric circles of radius r - sqrt(2) and r + sqrt(2), given that a distance of sqrt(2) or larger between the circle and the final grid square of our approximation in a given direction would enable us to place another grid square. Expanding out gives an upper bound of 4*pi*sqrt(2)*r on the error. Dividing out by r^2 gives a maximum error of 4*pi*sqrt(2)/r, which, by the Archimedian axiom within the real numbers, can be made arbitrarily small. This means that our estimate for pi approaches zero.
    2
  6425. I was proud of myself for decoding your shirt 🧐 Not everyone can be a genius
    2
  6426. the rotations work because for each triangle of 3 you either have 3 different colours or 3 of the same. However you rotate those, that always stays the same. Since the rule is followed for each individual piece, it must hold for the whole triangle.
    2
  6427. 23:05 7 and 8🤔
    2
  6428. first 1,000,000 terms vs 355/113... i'll go with the fraction. EDIT : 9:50 damn... you covered it. 😃
    2
  6429. a tower of 1cm thick, 1m long plates that hangs out 100m of the table would have the height of...... 3, 7 billion! 3,7 billion what? 3,7 billion times the distance from the earth to proxima centauri b!! s0.. i guess practically that could be hard to build
    2
  6430. HI
    2
  6431. What a wonderful teacher he is! And the visual proofs which they never taught us at school are so very intuitive and wonderful!! I have a question or hypothesis: are visual proofs always simpler and more intuitive than other algrebraic ones!? Perhaps one could count the number of geometric nodes or elements including joining lines compared with say the lines of code of an algebraic proof!
    2
  6432. Has someone already pointed out that swapping the positions of Spock and lizard on his shirt would make for a superior diagram? I can’t be the only one who is bothered by this🙃
    2
  6433. So I’m not the only one bothered by this 😆 Swapping scissors with rock would also work.
    2
  6434. "Take any sequence of numbers what-so-ever..!" Okay! The sequence of natural counting numbers from 1 to infinity: What comes next..?
    2
  6435. I always thought "e" to be greedy bankers creation. Irrational number and a rational view as to have a geometric reason for it via shape shifter functions. Your videos bring joy! And as always, " You are right!"
    2
  6436. In the maths olympiad question, there must be an additional argument, since if two equal collections overlap and you remove the common numbers, one of the collection could end-up being empty! But to prove this cannot happens is easy: since there are no more numbers in that collection, the sum is zero and the other collection must also be empty, which means the collections were the same, which contradict the proviso that they were different. Whether one thinks this additional argument is necessary or not is up to the examiner :).
    2
  6437. Woah!
    2
  6438. Finally, it works!
    2
  6439. Does Powell’s paradox occur in other numerical bases (e.g. in binary)?
    2
  6440. If you use the number of terms to add which is a power of the base.
    2
  6441. I just wanted to mention that I always enjoy your different T-shirts. And today was no exception. The only thing is…. I was so focused on the content (which is fascinating) that I was 8 or 9 minutes into the video before I thought to look at what you had on today. Wonderful video, as always! 😊❤ Best regards from Québec 📐
    2
  6442. 15:13 if there are two single solutions, isn't the sentence false?
    2
  6443. Hey mathologer a humble request from India please make a series on calculus please please
    2
  6444. 2. Since every to squares of the same size add up to the square 1 size bigger, we can get 16 dark red to be 8 red for the total of 16 red, which is 8 dark orange for a total of 12 dark orange, which is 6 light orange for a total of 8 light orange, which is 4 yellow for the total of 5 yellow, which is just 5.
    2
  6445. Funnily enough; the addition table in Z/2 corresponds to the XOR-gate, in logic; and the multiplication table in Z/2 corresponds to the AND-gate, in logic 😌. That’s even more appropriate, once you remember that, in probabilistics; in the case of P OR Q, you ADD the probabilities of P and Q; whereas, in the case of P AND Q, you MULTIPLY the probabilities of P and Q 😃.
    2
  6446. This is awesome! :) I always love Mathologer videos. Do one on symmetric polynomials! :D And connections to the infinite product for sine!
    2
  6447. I enoyed this video so much, thanks Burkard - the dream proof is beautiful! I might try picking up the book soon. Is there any obvious connection to other areas of maths you know, like juggling for example? Does adding an orientaion to each crossing affect things in any significant way?
    2
  6448. … and they are cicadas. Where I shot this is a bit north of Sydney, New South Wales. Currently mid-summer, and we have about 5 or 6 species of cicada making a racket during the day. Each one has its own “song”.
    2
  6449. Your brain is hard-wired for calculus. You can follow a thrown ball and catch it; your brain knows where it will be, and tells you where to be. Baseball pitchers fool us by making the ball defy the normal ballistic trajectory.
    2
  6450. The digits from 1 to 9 and their relation to each other through simple equations is quite stunning. It seems like there are 9 categories of numbers throughout the infity. 2 of them seems to "be alone" and work together 3-6. The 9 seems to be "alone"/the limit/"the note before the octave".. and 124875 is related to the sequence of 1 doubbled to infinity and also related to the scaling of bits in the binary code where the digits take birth in the first place. Please if you ever understand what i mean, id like to have your thoughts about it. A vid about this gets you back to roots of where we have discovered maths.
    2
  6451. Magic constant for 33x33 I got 17985. I noticed the diagonals are fixed in position before any rearrangement, so the sum of the negative diagonal is simply the sum of the middle numbers in each "row". To get these we must do 17 + 33r (where r is row number starting at 0 and ending at 32). The sum of these forms a triangle with base 33 and height 33×32=1056. The area of this triangle is then added to 17*33 to give 17985. I am too tired to give clearer explanation or simplify. haha
    2
  6452. I've loved calculus since i was 12; if a 12 year old can do it I'm sure it must be easy, it's just demonised :p
    2
  6453. As in the coin rotation paradox... What is the number of times the earth rotates on it's axis during one revolution around the sun? The answer is dependent on the relative position of the observer. If viewed from above (like the coins) it is number of apparent day/night periods + 1. Or 366. When we measure from the surface of earth we believe we see 365. Also: George Washington's nose (on the rotating coin) only sees one rotation of the coin.
    2
  6454. I assume it would also make sense to label those five components 0, 1, 2, -2, and -1. And those numbers correspond to... the winding number of the figure around the origin, I think, or something similar to it. And when n is even... ah, you didn't ever discuss what "180° ears" really entail. An "isosceles triangle" with a 180° interior angle is just three points on a line. So it's just connecting the midpoints of the sides. And the figure that's neutralized by that is... a degenerate "flat" polygon that just bounces back and forth between two points on opposite sides of the origin, I think.
    2
  6455. The transformations of your cube are glorious 🥰🥰🥰🥰👏👏👏
    2
  6456. So basically this is a particular case of cellular automata with 3 states and neighboor of size 2 ? Those draw very nice fractal-like patterns even with only 2 values (search "cellular automata" on google!)
    2
  6457. love the outro ❤
    2
  6458. I was showing the first part with the 3x3 grid to my 5th graders yesterday to teach them a technique to create their own 3x3 magic square. Their mind was blown. Thanks! Greetings from Leipzig 😄
    2
  6459. this is a teseract
    2
  6460. HOT PIZZA PUFF THIS WAS SOME GOOD STUFF
    2
  6461. I bought several attractive slide rules on closeout at a stationery store around 1980. The magnesium one is particularly nice… I think I might still have it somewhere.
    2
  6462. Chapter 5 is my favourite!
    2
  6463. This is a great introduction to generating functions.
    2
  6464. Hummm, wouldn't want to ruin the party here, but as a child I did discover that doing math, and my teacher simply told me it is not the way the tests want me to answer, thus couldn't get any point for that and needed to stick to the book... I'm most definitely sure I must not have been the only one, but expectation in class kill the creativity needed to make people rediscover this, and by the time people have enough credibility to make real reasearches, they all forgotten that little twist wich is always told has "creative, but definitely not what is expected", just seing an odd quirk that nobody is interested in
    2
  6465. Funny and awesome, as always! I've enjoyed it. I'll think about the puzzle with for and five pegs!
    2
  6466. You can remove two black and two green squares so that a single corner square is stranded, and then the board is impossible to tile with dominoes.
    2
  6467. It's a subtle point (and easy to miss), but you always measure the path from one card to the other card in the "clockwise" direction - (going in ascending card order but looping from king to ace). So the path starting at 6 and ending at 10 has length 4, but the path starting at 10 and ending at 6 has length 9. The starting and ending cards are important here! The one you keep from this same-suit pair is not an arbitrary choice! You find the smallest distance path between the two and you keep the end card of that smallest distance path, putting the start card at the front of the cards you give your assistant.
    2
  6468. Homework code: MAGIC CONSTANT Notes: Rotate the position of a square 180° around the magic square. Adding the 2 squares will yield D (dimensions squared) + 1. 4-5-6, 4 + 6 = 10, 9 + 1 = 10. For a d-sided magic square, as long as it's odd, there will be d/2 - 0.5 pairs. The middle square will always be D/2 + 0.5. 33x33 magic square, 16 pairs, adding to 1090. 1090 x 16 + 544 = 17985 Homework code: NEW METHOD We can see that there are many diagonal lines. This corresponds to the next tile being placed in the north-eastern direction. The parts where tiles loop can be seen as just the entire board looping, like an infinite repeating board. (exeption for most north-eastern tile) For the remaining tiles, they move under their previous tile.
    2
  6469. 7:23 anyone see that half red domino on the bottom???
    2
  6470. I have a feeling that this tree will be connected to prime numbers somehow in the future...
    2
  6471. Christmas has come early. What a mathematical gem!
    2
  6472. This one video is.... absolutely amazing.... Great nice job !
    2
  6473. 36911🤗😎
    2
  6474. Loved finding your channel this year sir. Looking forward to many more years enjoying the content. Happy holidays!
    2
  6475. Same cicular slide rule that Dr Stangelove used in movie "Dr Strangelove"
    2
  6476. Best point - the ant (worm) CAN reach the end of the rubber band
    2
  6477. I love a Christmas pi :)
    2
  6478. On puzzle 3, you can show with induction pretty easily that the n+1 case is equal to the n case + the n-1 case (for n>1): consider the top left-most tile: either that's tiled by a vertical domino, in which 1case all the spaces above can be tiled in the fashion of the n case, or it's tiled via a horizontal domino, meaning the two tiles below must be tiled similarly, and the tiles to the right can be tiled as per the n - 1 case. That is, 2xn board can be tiled as many ways as the n+1th fibonacci number!
    2
  6479. The only math verse I know Roses are red Violets are blue Math is hard Schnitzel!
    2
  6480. I usually never comment on videos but this video is a masterpiece. I never knew this could be conceptualized so easily. You are a legend.
    2
  6481. 7:56 “Pretty obvious that this pattern will continue forever” I don’t see it; I think it might only be obvious because you’ve been working with this proof. If I assume it does I can follow the rest, but I don’t see why I should assume that.
    2
  6482. I am the proud owner of one of the many gazillion circular slide rules that were produced.
    2
  6483. Always great to watch and be amazed.
    2
  6484. Never thought I’d see lies, degeneracy and eggs in a Mathologer video, but here we are!
    2
  6485. BUT DID THE KING COMMIT THE THING? THE SELF-UNALIVING?
    2
  6486. Hey guys, I have a question. I noticed that besides having the property of being an integer of the form 4k + 1, the primes that work are also all the primitive pythagorean triples' hypotenuses, in order. Why is that?
    2
  6487. you had me at welcome to a new mathologer video
    2
  6488. Wonderful!
    2
  6489. Wonderful!
    2
  6490. I always enjoy your videos and look forward to them coming out, so let's hope you get some respite at work ;)
    2
  6491. Phi has so many intriguing ways of popping up, it may be irrational but has a way of generating rational numbers i.e. integers, that surely makes it the most rational irrational number, but no isn't it the most irrational? Anyway when I first came across it I found it hard to remember which number was phi - was it the fraction 0.618.... or ratio 1.618... but seeing a rectangle with 1 as the smaller side helps me remember phi is the longer side so is >1. I wish they had taught us more about phi at school. It may have been mentioned briefly but nothing memorable!
    2
  6492. 1st challenge: it is obvious that the smallest piece on a n-stack and n+1 stack is always opposite. (Moving n+1-stack from A to B requires moving n-stack to C first, which is opposite of moving n-stack from A to B) So clockwise/counter-clockwise movement depends on the parity. If n is odd, we can pretend it is a 1-stack and move the top piece accordingly. If n is even, pretend it is a 1-stack and move it in opposite direction.
    2
  6493. I remember doing finite differences to solve polynomials based on some roots given in competition math
    2
  6494. I love this one. I will teach it to my kid ❤
    2
  6495. Well, now I will have this numerical value memorized for the rest of my life 😂
    2
  6496. It's Zolotaryòv)
    2
  6497. In my country INDIA , kids from 11th and 12th class are eating up entire calculus books just to get selected in IITs.
    2
  6498. Love the QED cat...
    2
  6499. Ramanujan was magical genius.
    2
  6500. Amazing released of videos this year, gracias por tanto saludos from Bo
    2
  6501. ~100 pi gramm of cereals, yum
    1
  6502. Almost at 150,000! He deserves it! Happy New Year BTW
    1
  6503. oh boy...there's something im starting to figure out...there's still a long run, but I believe one day we all will get to know about reality...thanks for all your wonderful videos...happy new year, btw...:)
    1
  6504. really appreciate this video
    1
  6505. finally some one not scared to do tangible examples of higher maths
    1
  6506.  @galoomba5559  The feature I most want from 3D puzzles is a reset button
    1
  6507. there are many infinite dimensional spaces in topology, bearing some nice properties. Hilbert's cube, for example, is compact
    1
  6508. Love this! Thank you 🙂 Happy New Year
    1
  6509. Thank you!
    1
  6510. i want an anti-cat! hahaha
    1
  6511. I liked this video and it inspired me to make a physical version with lasers, servos and a Arduino. A video of it can be found here https://www.youtube.com/watch?v=XHqZNfPvEVY and more information can be found here https://blog.abluestar.com/projects/2017-laser-kaleidoscope . Thank you for the inspiration.
    1
  6512. I usually really like your videos but I think in this one you assumed people knew way too much about calc. I also think using limit notation would have made more sense for viewers who are just now being introduced to limits and derivatives. I took calc one this year but girlfriend, who I watched the video with, took her last calc class a few years ago. She was completely lost and I had to keep pausing the video to explain what was going on. Don't get me wrong, I like your content and sub to your channel, but I think this video could have been better. I think you tried to cover too much in 12 minutes, and had to cut too much out to fit the time slot.
    1
  6513. When you as the cashier receive a thousand dollar note from a customer who buys 15 dollars of stuff But you are a mathematician: here are the ways to make your change, which way would you like your change to be?
    1
  6514. Read Dr CK Raju if you want to know history of mathematics without colonial biases
    1
  6515. Good evening, I'm from Ukraine. Studied Heron's formula at school
    1
  6516. I freaking need that t shirt!
    1
  6517. Many trigonometric ratios are approximated using something like Newton's Method or McLaurin Series. These methods can produce a decimal number version of the ratio, but these methods often do not reveal the exact result (ratio of roots and integers). I wonder if the content in this video could be used to zero in on the exact trigonometric ratios of various angles in a manner somewhat similar to how the Golden Ratio was coaxed out from the dimensions of a regular pentagon. The exact ratios might include nested roots in either the numerator or denominator. I also wonder if there would be a way to iteratively attach ears to a pentagon until a regular hexagon appeared. Ideas for future research.
    1
  6518. It's all Pascal triangle
    1
  6519. The volume level seems low on this video -- any lower and I would have stopped watching. Have you changed something?
    1
  6520. Hello
    1
  6521. Great video as always. But are you aware that the grinning face you use as a graphic is a meme associated with racist and anti-Semitic groups?
    1
  6522. The reason why it has so many views, is because we are mathematically driven. Targeted people purposely have been programmed to be disconnected but only to sound and sight! Its the music, and something everyone can identify with... a moment of FEAR! What do you do? Do you Face Everything And Recover. Movie has no essence other than to subtract from your EMOTIONS. I thank you Mathologer for reconnecting to REALITY!
    1
  6523. Very nice video! Thanks!
    1
  6524. Hmm I wonder what’s this “hardcore” video he was working on when he got sidetracked by this…
    1
  6525. I just realized the shirt matches with the threes and zeros pattern
    1
  6526. Lol😂😂 "You've never heard of shrink proof, well hehe" anime villan vibes
    1
  6527. Some of the things I like about this video - Nice music at end - Beautiful mind-blown visual animations (as always) - You made this video! ** Mathologerisation (> <)
    1
  6528. Would I be correct in thinking that, after the windmill pairings leave one straight cross solution as the odd one out, there could be only an ODD number of ways that a 4k+1 prime p could be written as the sum of two squares, and the rest of the video (with the a,b, c, and d business) is still necessary to show uniqueness?
    1
  6529. 32:20 but you never showed why the derivative of x is 1...
    1
  6530. Thanks for the physics nod at 14:47
    1
  6531. Hmm, I didn't get why is the sum of 1 - 1 + 1/2 - 1/2 + 1/3 - 1/3 + ... equal to 0 given that in one of the previous videos you showed that 1 - 1 + 1 - 1 + 1 - 1 + ... doesn't equal to 0. As far as I understood, in order a number to be declared as the sum of an infinite series, partial sums should approach this number but never overshoot but here partial sums overshoot infinitely many times.
    1
  6532. I asked Claude to calculate the radius of the circle given sides and diagonals A,a,B,b,C,c and this is the answer: 1) First, recall Ptolemy's theorem for cyclic quadrilaterals: AC * BD = AB * CD + BC * AD Where AC and BD are the diagonals, and the other terms are products of opposite sides. 2) In our notation: C * c = A * a + B * b 3) Now, let's recall the formula for the area of a cyclic quadrilateral: Area = √[(s-A)(s-a)(s-B)(s-b)] Where s is the semi-perimeter: s = (A + a + B + b) / 2 4) We also know that for a cyclic quadrilateral, the area can be expressed as: Area = (A * a * B * b) / (4R) Where R is the radius of the circumscribed circle. 5) Equating these two area formulas: √[(s-A)(s-a)(s-B)(s-b)] = (A * a * B * b) / (4R) 6) Squaring both sides: (s-A)(s-a)(s-B)(s-b) = (A * a * B * b)^2 / (16R^2) 7) Rearranging to isolate R: R^2 = (A * a * B * b)^2 / (16(s-A)(s-a)(s-B)(s-b)) 8) Taking the square root: R = (A * a * B * b) / (4√[(s-A)(s-a)(s-B)(s-b)])
    1
  6533. 33:48 There are two ways to simplify (V²+U²)²-(V²-U²)² that seem easy to me. One of them is by expanding it
    1
  6534. The sequence you gave showed approximations of sqrt(2) for 577/408. 14:49 Edited: I just realize you addressed that in the video lol.
    1
  6535. 36:01 Final Puzzle … The area of the square in the middle is 80% (4/5th) of the area of the large triangle. if a=1/2, b=1 0.5x1=0.5 0.5÷2=0.25 = area of the large triangle. Side b of the small triangles (st) is the same length of the sides of the small square. 5 (st) fit in the large triangles. 4 (st) fit in the small square. 0.25÷5=0.05 = area of (st) 0.05x4=0.2 0.2÷0.25 = 0.8 [4/5th the area of the large triangle] Which works out 1/5th the area of the large square… The large triangles overlap 1/5th of each other, leaving 4 quadrilaterals, creating the 4 small triangles, which all 4 small triangles fit in the small square, 5 total equal area piece … small square is 1/5th the area of the large square. The area of the large square is 1, the area of the small square is 0.2
    1
  6536. 10:17 The last term will be 4*cos²(pi/2)+4*cos²(pi/2), which is 4*0²+4*0²=0, so the whole product gets multiplied with 0 at the end
    1
  6537. I discovered the Eisenstein triples a while back. I love that you can use the same chord method as you do for the unit circle for Pythagorean triples, but use x^2-xy+y^2=1 as your starting conic.
    1
  6538. The second I learned at 11 from my grandfather
    1
  6539. 11:50 do the three blocks setuo and switch brick 2 and 3 as shown before? I think that should work.
    1
  6540. 3:25 I remember playing with a triangle snake toy like this
    1
  6541. You made my day... Again ❤
    1
  6542. I made it all the way to the end.
    1
  6543. Where's the wishlist for the coding challenge?
    1
  6544. Felix Wankel used this shape in his engine in the 1920s. The maths must predate it.
    1
  6545. at 6:20ish you matched the 1 up with 3.18 instead of 3.14. Surprisingly, things still work out when the wheel starts spinning... :)
    1
  6546. So you can paint the inside of the horn, but not the outside?
    1
  6547. no. you can fill it with a finite volume of water but you can't paint it with finite amount of paint
    1
  6548. Actually you can also paint the outside. Just use 1 litre to paint anything between x=1 and 2, then 1/2 litre to paint everything between x=2 and 3, then 1/4 litre to paint everything between x=3 and 4, etc. Then you paint the horn with 2 litres of paint. As with filling the horn you simply have to apply the paint thinner and thinner as you progress. Still the paint will always have non-zero thickness.
    1
  6549. How many too tight savings are there?
    1
  6550. I dont understand the strongest lacing changeover point.. At a given spacing and and number of eyelet pairs, how do I know which one is stronger?
    1
  6551. Does the relationship at 16:10 generalize to higher dimensions. Like, is the hypervolume of a 4d ball equal to 1/4 it's surface volume times R?
    1
  6552. 0,1 x (1/1-0,906) = 1,06 To make it equal to 1 instead of 1,06 we have a x [1/1-sqrt(a^2+b^2)] = 1 <=> Parabola a + sqrt(a^2+b^2) = 1 with positive integer solution a = 0 and b = 1
    1
  6553. Best videos I love it big
    1
  6554. The secret of something something!? Tell me more.
    1
  6555. 14:03 Wow. At school I remember that I had to learn a proof of this for my exams. I've completely forgotten it over the last 50 years (and have mislaid my school notebook) but this proof is so much clearer.
    1
  6556. Elementary, if you already know calculus.
    1
  6557. Loving the video. Hating the kerning nightmare of A VOID NEGA TIVITY. Aaargh!
    1
  6558. Nice as every Mathologer video.
    1
  6559. would it be possible make some stuff about Galois work that proved there cannot be formulae solving all fifth powers polynomials?
    1
  6560. I feel a little proud for pausing right before Brahmagupta's formula and just saying out loud to nobody, "cyclic?"
    1
  6561. 6:33 why not 4K-1???
    1
  6562. But you fundamentally changed the sum. It's not REALLY the same infinite sum if you do all of those operations. If you alternate the sings conveniently you can, in fact, get any number you like if your series doesn't converge in absolute manner.
    1
  6563. Would it actually be a paradox though? Something similar was shown to me as an example of how some infinities are actually larger than others. If you say that two series of numbers are both infinitely long while proving that one of those sets has more members in the set at infinity you can't subtract the two infinitely numerous members then state the number of remaining members is zero. Your first example is an infinite series, there are an infinite number of negative numbers but at infinity that series of consecutive positive fractions would be itself infinitely long wouldn't it? How would you then justify removing a smaller series of numbers from a larger series of numbers while stating the result is zero?
    1
  6564. Great Presentation; Great Music!!
    1
  6565. Where can I get the pizza t-shirt?
    1
  6566. That was a really fun lesson. Just made me think "Math DNA"
    1
  6567. Hello how are you ? I am happy to see you
    1
  6568. I have often wondered why it's always thought about as a means to find the length of a side, when a far more useful application is the construction of a 90° corner. Also its rare to see mention of the fact that 3,4,5 satisfies the equation, just ask any builder laying out a foundation
    1
  6569. I'm so sad the video is over
    1
  6570. yes
    1
  6571. The visual intuition of gamma being less than 1 was surprisingly beautiful in its simplicity :D
    1
  6572. ❤️
    1
  6573. one of the better 40min in life.
    1
  6574. I hadn't heard of the no 9s series or the non harmonic maximal overhang towers, so those
    1
  6575. Being completely new to this.. So what is special about the Tesla numbers, and what is it’s relation to nature?
    1
  6576. First thing I would think of is the number fields Q[r, s] (and Q[phi]). Q[phi] is the splitting field of x^2-x-1, in particular Q[phi] is isomorphic to Q[x]/(x^2-x-1). Q[r, s] is isomorphic to Q[x, y]/(xy-x-y, x^2-y-1, y^2-x-y-1). We can look at the ring of algebraic integers inside these number fields. For Q[phi], that would be Z[phi]. If I recall correctly, this is a unique factorisation domain. Perhaps it would be interesting to check if the ring of algebraic integers in Q[r, s] is a unique factorisation domain (else we may consider computing class numbers, it might be interesting too). Since we get them from polygons, I think these lie in cyclotomic extensions, so Q[phi] is inside Q[z] where z^5=1. It should be clear that Q[r, s] is inside Q[z] where z^7=1, and for the infinite family of polygons I would expect something similar. (Proof: identify the vertices of the polygon with the complex numbers 1 to z^6. Then r^2, s^2 are clearly |1-z^2|^2/|1-z|^2=|1+z|^2 and |1-z^3|^2/|1-z|^2=|1+z+z^2|^2 respectively. Since s^2=s+r+1 and r^2=s+1, we can solve for r and s from there.) I believe it should be clear that [Q[r, s]:Q]=3 since all powers of r and s can be written as Ar + Bs + C.
    1
  6577. I think if I watched all the Mathologer videos in succession and put all the clues together, I might just solve one or two of the Millennium problems. 😉
    1
  6578. I'm always interested in counter intuitive facts, like the optimal stacking patterns or the convergence of the no 9s series. But the great marvel has to be the formula for the n-th harmonic number and the role of the Euler-Mascheroni constant in it. As always, a really nice video. I still hope you're going to do the video in German you announced some time ago ;)
    1
  6579. Enjoyed thoroughly
    1
  6580. Beautiful...
    1
  6581. I'm just add, that in my native Polish language, we call a slide rule "suwak logarytmiczny" - literally "logarithmic slider" :)
    1
  6582. 5^3 + 6^3 = 7^3 - 6 (covered corner points) + 8 (actual corner points) 10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 10^2 * 5 + 21 * 4 + 23 * 3 + 25 * 2 + 27 = 730 730 / 365 = 2
    1
  6583. A wee bit too hard for my 16 year old brain, but I could understand most of it. Thanks Mathologer!
    1
  6584. Nifty. And you're forgiven for the "lie" 🙂- notice that MATHOLOGER anagr MOTHER GOAL
    1
  6585. The ancient Greeks must have had a slide rule considering they had the antikythera mechanism.
    1
  6586. i like how calling sinh "shine" suggests "shine and coshine", which gets us back to "cosh"
    1
  6587. Watching this made me wonder whether you could calculate these specific coefficients by integrating your generating functions over a small closed loop around 0 in the complex plane and using the residue theorem. So the amount of ways to make n cents, using coins of amounts 1, 5, 10, 25, 50 and 100, would be 1/(2i pi) times the integral of 1/(x^(n+1)(1-x)(1-x^5)(1-x^10)(1-x^25)(1-x^50)(1-x^100)) dx over a complex circle around 0 with sufficiently small radius (I think <1 might be small enough). Am I missing something here? If not, then that is amazing: being able to calculate a discrete amount of ways by evaluating a complex integral and dividing by an irrational, imaginary constant :O
    1
  6588. Bravo, Bravo!!
    1
  6589. no slide rules went the moon, space is fiction.
    1
  6590. Ha! Love it. I've been folding beer cans like that (by hand, though) for ages. They do crush great then!
    1
  6591. Funny how I wrote an article some 35 years ago, similar to the first part of this video, with the triangles ordered inside or as a square, to show visually that a² + b² = c². I also wrote that the little square in the middle of some of the figures has the area of (b-a)², and that from all of this, we can immediately derive that a³ × b³ = c³.
    1
  6592. I am an undergrad student and I find this video really interesting. While I have seen other combinatorial proofs of the positive integer power sums of integers, this geometric visualization is top notch. In a mere 50 minute video, I think this is the best that could have been done.
    1
  6593. Both a maths and ethics insight in one video. Excellent :)
    1
  6594. Merry Christmas from Spain!
    1
  6595. Ah! The Pythagorean shuriken at 23:32!
    1
  6596. stunning :-) I am glad that (x+2)^n is justified as more than just a fluke :-)
    1
  6597. Does anything change for negative cases? For example in a world without bankruptcy where the next of kin people inherit debt, how does the problem change if the thing being fairly distributed is of negative value instead of assets? Which is fairer to distribute that debt irrespective of the future debtors assets or with them included in the calculation to account for how burdensome the debt is? This could be an interesting way to determine our debts to society - like taxes and fees and other such things.
    1
  6598. Beautiful! Welcome back, Burkard!!
    1
  6599. This was an absolute banger. I can see so many uses for this. Thanks!
    1
  6600. I really enjoyed this video. I teach Calc 2 on occasion and use this example, but I’ve never had a good gut-level explanation to give to students to help resolve the paradox. (I’ll argue based on the mathematical properties and note the importance of units /dimensions and the issues with comparison that arise here.) However, the 2-d —> 1-d analogue is perfect, and I will use it next time I teach Calc 2. I even explained it to my wife (not a math person) while we were on a walk (so without drawing it) and she understood it and enjoyed the analogy.
    1
  6601. WHY ARE YOU EVE- wakes up after half hour lesson in shoe tying
    1
  6602. That is the most fantastic Mathologer I've ever seen 🤩. My mind is spinning ! I will surely dream of this - hoping it won't transform in a nightmare😵‍💫
    1
  6603. First love your videos and the Super Pi short rocks. For the solution, I will say that the proof is as it is 4(n), the (n)s cancel and that leaves 4 and therefore the 4 colours. Yes I like the shorter videos. But keep up the videos either way. I watch these with my kids so they can have a better understand of math and solutions. It's kind of a Sunday tradition.
    1
  6604. The shrinking hexagon proof is magical and I love it
    1
  6605. I think to get full credit in the task with 2 digit numbers we should also consider a possibility that 2 overlapping collection consists only from overlapping numbers. In that case removing overlaps leads to empty collections and 0 sums which I suppose is not considered as a solution. As an example: collection 1: 10 11 12 collection 2: 21 12 we can remove 12, but in the case collection 1: 12 collection 2: 12 12 can't be removed
    1
  6606. Don't be surprised to see really new stuff! Watch the thanks! Euler is a patreon of this channel.
    1
  6607. This channel always amazes me, even though I know it's good stuff. I'm a maths teacher and I learn so much here! Thanks!
    1
  6608. 16:04 what's that special geometric object we are looking for ...
    1
  6609. I think the rubber band will snap after 2 revolutions.
    1
  6610. It seems to me that the final soda can demo showed the same limit problem in reverse, I'm sure that the more coarse folds you put in the shorter the can will get, more bands and less points shrink height. Did you measure it?
    1
  6611. prof I love your laughter pls never stop smiling
    1
  6612. Versprochen ist versprochen!
    1
  6613. Jo, mach mal es ein mal auf Deutsch! Danke für die immer sehr interessanten Videos, die einen mitreißen. :)
    1
  6614. amazing
    1
  6615. Yay new video
    1
  6616. I'm just watching your shirt.
    1
  6617. Cool video! Blue, orange and gray are allways the same amount, since respectively merged they have to result in the respective surface square areas of the cube. And they are allways composed by the same number of minisquares:) Proof of my first sentence: The upper right and lower left corner has always to be gray (because it is the only fitting one). Same holds upper left and lower right: blue, upper and lower corner: orange. Switching to the 3D interpretation this means, that we allways see the six corners of the cube viewed from one diagonal. Cuting out minicubes (which results in the picture we see) does not change the number of blue, orange and grey squares respectively.
    1
  6618. 2020 can be written as a sum of integer squares in twelves ways, right?
    1
  6619. most memorable: no 9s < hundred 0s
    1
  6620. 24:39 gamma has to be greater than 0.5 because if you were to cut out each rectangle which contains the top left corner and the bottom right corner of each blue area, and then slice it along that diagonal line, no matter what the area of the square would be, it would be exactly half the area of that square because all those rectangles add up to the entire area of the square, no matter how the rectangles add up to it. since the area of the square is 1*1=1, half of that would be 0.5. you can see in the image that each blue area slightly bulges below that diagonal we cut earlier, gamma must be greater than 0.5.
    1
  6621.  @tiarkrezar  the obvious one to try is for what gets a line, a triangle then a tetrahedron, presumably etc. Unfortunately, AFAIK the generator for that is T(n) = (1+1/x)T(n-1) + 1, which doesn't factor quite so nicely (and you have to bootstrap it past n=1 where it's 0x+1) - but maybe I'm missing a trick to get that +1 out cleanly.
    1
  6622.  @SimonBuchanNz  good thinking, yeah it seems like a recursive formula is the way to go there. I think it should be p_(n+1) = p_n * (x + 1) + 1 though. Every step you're just adding one more vertex and a "lifted" copy to connect it to the previous shape.
    1
  6623. Brilliant. Has Andrew's work been published somewhere? Reference? Well done on 100th video.
    1
  6624. Mein erster Gedanke: das ist mathematisch unglaublich befriedigend, weil sie den unendlich vielen Zahlentrippeln eine wunderbare Ordnung geben. Der zweite Gedanke ist der: für die höheren Exponenten gibt es ja keine Lösung (Fermats letzter Satz). Und damit auch keine Zahlenfolge geben, die so etwas ordnen könnte.
    1
  6625. brilliant!
    1
  6626. For chapter 6, is 7 numbers really the maximum number of numbers one can choose?
    1
  6627. Kempner all the way for me :0
    1
  6628. The two things I found the most astounding are a) the sum of a connected subsequence 1/(m)+1/(m+1) +... + 1/(m+n) is NEVER an integer b) the sum of reciprocals of integers with exactly 100 zeroes starting at 10^-100 is bigger than the no 9s series.
    1
  6629. Despite being woefully lacking in math skills, I am nevertheless enthralled by mathematical concepts. The Mathologer never fails to mesmerize and entertain me. 👍😊
    1
  6630. question on the card trick: how do you signify whether the assistant needs to count backwards or forwards? or do you always keep the higher card and use the lower one to signify the suit?
    1
  6631.  @Reddles37  Thanks, but I said "some have mentioned rotational symmetry but I'm not including that".
    1
  6632. 28:55 minor vocabulary slip: you say a^2 = b^2 but mean a^2 = c^2
    1
  6633. 3:20 if we consider f(x) = 1/x, then squishing by a factor of "n" is like dividing f(x) by n yielding f(x)/n, stretching on the other hand is dividing x by n, leaving us with f(x/n)/n. Filling this in gives us f(x/n)/n = 1/n * 1/(1/n *x) = n/n(n*x) = 1/x
    1
  6634. 0:57 This man does not make videos. He hands out magic mushrooms. #MathsIsShamanism
    1
  6635. I came up with the coloring proof myself, but I find the swiveling proof prettier.
    1
  6636. I think it's possible to argue for the area of the sphere by noticing that, in the limit, exactly half the area of each individual moon lies outside the sphere.
    1
  6637.  @Mathologer  Oh. That's all very true, if only a bit surprising.
    1
  6638. Thanks much for another wonderful, excellent video. Equality and inequality are powerful subjects, and can have quantitative (applied math?) as well as qualitative (pure math?) aspects. For an arbitrary set of 4 planar vertexes, choose any subset of 3 which then specify a circle. The 4th vertex is either on the circle, if Ptolemy's formula gives zero, or non-cyclic if non-zero. My conjecture (wish I had time to work on it) is that the non-zero result gives the "degree of non-cyclicness" and maybe inside/outside measure of distance to the circle. This is related to what may be the simplest geometric equality, the standard form of the line equation. I think that early students don't get the more exciting introduction; the focus tends to be on the slope-intercept form and graphing of functions, helpful but rather boring. Strangely, they short-shrift the general form, and even get students to derive it from the slope-intercept form! But, the general form is much more useful, and is derived more simply by a cross-product type formula of the endpoint coordinates. The resulting Ax + By + C = 0 is then a formula -- if you plug in a point it gives zero if it's on the line, but if >0 then the point is on the right side of the line, on the left if <0, and the magnitude of the result gives the distance from the line, if "normalized" by the length of the line segment. This is extremely useful but students don't hear about it.
    1
  6639. Love the sprinkling of sardonic German humour!
    1
  6640. The most memorable part is the brick towers one. One may stop at the leaning tower and be quite confidente about their result while actually being far from optimum... It does require some ingenuity and inventivity to come up with the other solutions
    1
  6641. Why does the proof for surface area works? It is not true in general that ir a body is contained in another one the its surface area is smaller 🤔 I mean, maybe you can make it work because each "slice" is convex and the result about surface area and inclusion is true if the subset is convex, but still a little bit trickier that it seems. I don't know ir I'm missing something.
    1
  6642.  @Mathologer  Oh! Now I see it! Thanks!
    1
  6643. 28:12 when you attach the ears like that, the point between A and B in complex numbers can be evaluated as (A+B)(1/2 + i * lambda) for some fixed lambda (depending on the current angle and whether we're going clockwise or counterclockwise), which obviously distributes if A and B are expressed as a sum of A_i's and B_i's. easier than similar triangle chasing imo.
    1
  6644. ⛬ The ➕ sign might refer to such Primes? ☆ Eg, multiplied by itself and unity. 💚 🧙‍♂️
    1
  6645. its more akin to how the royal family works according to cgp grey ie every party is assumed alive until informed otherwise. the sons inherit from the grandfather equally and then being dead the grandchildren inherit from the fathers which predeceased the grandfather. So each son gets the same split of the father and then being dead their estate is split among their sons.
    1
  6646. I thought of analytic continuation also. This gives comfort when proving the compound angle formulae for sine and cosine, for example, given that the diagrams used in any proof only apply in very limited cases. Of course, I don't mention this to my students! On the other hand, when proving the sine and cosine rules, I do separate diagrams and proofs for the the cases of acute and obtuse angled triangles.
    1
  6647. I wonder if this is related to the weave of a window screen? Can you weave a equilateral triangular window screen?
    1
  6648. Our lead programmer showed me this in 2002. Ever since I correct people on pi. :)
    1
  6649. It's a circular slide rule.
    1
  6650. Man, infinities are interesting.
    1
  6651. In India, herons formula is taught with proof, but it doesn't extend to brahmagupta's formula, which is ironic.
    1
  6652. i wanted you as my maths teacher
    1
  6653. I had a quantum field theory assignment just a few weeks ago where we calculated the magnitude of the Casimir force. The method the professor had us use actually involved setting up an integral and using the Euler MacLaurin formula to compute it; since all derivatives of the function higher than the third vanished, the formula was exact. It and the Bernoulli numbers were just introduced out of thin air as a computational tool, and it made me curious about where it came from and how they were used, so this video kind of came at an opportune time. Long but worth it as always. Thanks so much!
    1
  6654. What's the title of song whose guitar streaming was done towards the end ....???
    1
  6655. Lattice Lacing , I so want to do that one :P
    1
  6656. Dear Prof Polster I have studied science (including Maths) upto the 12th standard and subsequently graduated as a Bachelor in Engineering. Currently working as a senior software developer in a multinational firm in the UK. Never studied abstract algebra as part of curriculum. I learnt solving cubic and quartic equations out of personal interest when I was in school, at the same time perhaps I was taught quadratic equations. At that time I did not know quintics are not solvable in radicals. I watch your videos with great interest. I learnt the basics of abstract algebra watching various channels on youtube. These channels offer excellent introduction to group theory and insolvability of quintic equations with examples - but in the end I am not able to perceive the exact connection between solvability in radicals and the descending tower of subgroups or why the quotient groups need to be cyclic. I eventually found an excellent article FranzSenEx2010.pdf on the internet which seems to answer my questions - but I am unable to fully appreciate due to lack of any formal training in abstract algebra. (e.g it cites various theorems on isomorphism but the examples are not detailed enough for me to visualize). Your step by step approach with images is of great help to someone like myself whose enthusiam does not match his skills. Accordingly can I request you to consider making videos on the Insolvability of Quintic equation by Galois theory explaining the precise link between solvalibity of an equation and the descending tower of the subgroups and why these need to have certain properties. (Is it possible to demonstrate by group theory the simpler cases e.g. a quadritic equation is not solvable unless radicals are introduced, or first degree equations ax + b = 0 not solvable unless rational numbers are introduced). And also on Abel's proof (I have read Peter Pesic's book - but there is not enough commentry on the proof) Regards
    1
  6657. 8.13, the answer for 7 pigeons is 5 In general for N pigeons we can always find a hemisphere with : 2+ceiling((N-2)/2)
    1
  6658. most memorable part- I changed my answer to convergent for no 9 series at the last moment, and I got it right, I know just intuition but still.
    1
  6659. He the best
    1
  6660. I know in physics there is the equipartition theorem which stating that the probability of each microstate should be equal... is there any way to prove that nature obeys this rule? As the way "random tiling" is defined relies on this.
    1
  6661. Make it sum to i.
    1
  6662. There is a lot going on here.
    1
  6663. Circular sliderule. Its only magical to the kids, who have not grown up with sliderules, and those of us, who had to use it. Works with the more common linear ones, too. I still have a couple. Fun Fact: In electronics, this logarithmic relationship is used to establish component values. If you have ever wondered why resistor & capacitor values are multiples of: 1.0, 1.5, 2.2, 3.3, 4.7, 6.8, and so on. Its that the successive value is a fixed multiple, of the prior one, within a given tolerance.
    1
  6664. Gamma is greater than 0.5 because in each rectangle the blue part is greater than the triangle which divides it into to same parts. Hard to explain but obvious :)
    1
  6665. I have a possible improvement to suggest for the card trick. It would be somewhat more impressive, would it not, to have the 'mark' not only randomly select the 5 cards, but then to select one of them, when they are spread face down on the table, as the 'selected' card for your assistant to guess. You are then free to arrange the remaining 4 cards to hand to your assistant, passing info using just their ordering in the same manner as you've described. The twist is that you are going form a number from 1 to 48 using the card ordering to point to the 1 of 48 position of the 'selected' card. I say 1 of 48, not 1 of 52 because obviously the selected card cannot be one of the 4 that are passed. Now this calls for not a strict mapping of cards to numbers, but one that maps a bit dynamically to the 48 available positions, taking the 4 unavailable cards into account. So a bit of mental gymnastics for you and your assistant to perform. But, I hear you object, there are only 4! = 24 ways to arrange the 4 cards, so insufficient pigeons here. Ah ha, but you can pass them to either the assistant's left or right hand, yes, to make 48 pigeons!
    1
  6666. Another great mathologer video)
    1
  6667. Regarding the box at 11:50, the answer is [9,4,13,17] and it's a grandchild of [1,3,4,7], via [1,4,5,9].
    1
  6668. 24:40 By the way, gamma is greater than 0.5 because all those curved blue bits are (slightly) larger than the corresponding right triangle with a hypotenuse drawn between the endpoints. The triangles together take up half of the unit square, because each takes up half of the rectangle it is inscribed in.
    1
  6669. Today I was searching for a contious form of the harmonic series and now this! Is it my birthday? :D
    1
  6670. Very fascinating video!!! My initial guess as to a fair split was to take normalized percentages of debt of each creditor w.r.t. the amount of TOTAL debt then multiple the estate by those percentages and divide the money up that way I.E. if creditor 1 holds 25% of debt and if creditor 2 holds 75% of the debt split the estate that way But i guess I was wrong
    1
  6671. WOW !
    1
  6672. This remainder test, does it work for other bases. Is it generally true?
    1
  6673. My guess is simply: Laziness and reuse. They found the solution somewhere and simply parrots it to have a video.
    1
  6674. .
    1
  6675. Is there a similar generating function for tetrahedrons? I tried (x+1)^(d+1). Seems to work, but I didn't prove it for all dimensions, and not quite as elegant.
    1
  6676. Now do the reciprocals of the partial sums: 1/1 + 1/(1+1/2) + 1/(1+1/2+1/3) + 1/(1+1/2+1/3+1/4) etc.
    1
  6677. Very carefully
    1
  6678. I loved the casual visual demonstration of that Euler-Mascheroni<1. I know it’s not a full chapter, but I’d still like to submit it as my answer. By the way, it’s greater than one half because 1/x is concave up and therefore the area under it is less than if we’d just used straight lines connecting the endpoints. That means that area subtracted from the rectangles using 1/x is greater than what could be achieved using triangles. Triangles have half the area of the rectangle=0.5, so the areas of the curved regions is greater than 0.5.
    1
  6679. Is that Pi hanging out on the wheel exactly opposite 1?
    1
  6680. So Pascal's triangle is just a 2d version of the Fibonacci sequence, is there some sort of analytic continuation of this?
    1
  6681. My vote is for the gamma section
    1
  6682. You keep calling them "3D moons" but I keep seeing "bananas" :)
    1
  6683. Piet Kasteleyn's first name is pronounced like the english first name Pete.
    1
  6684. Well, give me googol dollars then I'll give you the answer
    1
  6685. math is beautiful and so is your channel <3
    1
  6686. Correct me if I'm wrong, but does his shirt even make sense? Isn't Ho^3 just a singular Ho, since Ho x Ho x ho would be equivalent to 1x1x1?
    1
  6687.  @Mathologer  Oh, yeah I suppose you could see it that way
    1
  6688. Great subject!
    1
  6689. Love the video and your T-Shirt. Thank you.
    1
  6690. At 20:53, look at the ratio of each number to the number below it :)
    1
  6691. The colouring proof was definitely better just because it was quite clear. The swivelling proof was some sort of magic trick, like the one with which you prove that 63 = 64 cutting the well-known rectangle. P. S. and the white curve of constant length has smaller diameter than the blue circle, as for the curve the R+G+B (sum of the lengths of the sides of the small initial triangle) is the diameter, while for the circle it's only a chord, not containing its center, that is, it's inevitably smaller than the circle's diameter.
    1
  6692. Brilliant!! Absolutely brilliant!!
    1
  6693. How come you did a video on this today I was just searching about it 1 day before
    1
  6694. I'm guessing the colours at the two corners of the equilateral triangle decide the colour at the very bottom? It's such a nice triangle, it's got to have some kind of- Wow, I literally just guessed that by looking back at the other examples.
    1
  6695. 7
    1
  6696. For me the most memorable thing is the greedy algorithm that shows that all positive real numbers can be expressed as a subseries of the harmonic series. It wasn't obvious for me that, for a given real number, the sequence generated by the algorithm would spit out a series that actually converged to that real number. I had to give it a little bit of thought before it was clearer that it wouldn't converge to another number. I'm still wrapping my head around it, but at least it's something to occupy my mind while I wait for school to return (I'm on vacation right now). Great video as usual!
    1
  6697. @7:45 I would propose to adjust the connected dots slightly to emphasize the proof. For the first chain of each new shell (2, 5, 8 etc), it should go from the lower right corner of the hexagon, via the lower middle and the lower left corners, ending at the upper left corner, but excluding the start and end corners. For the second chain (4, 7, 10 etc), it should also go from the bottom right corner of the hexagon, but instead via the upper right, the upper middle and end at the upper left corner, and this time including the start and end corners. This is slightly different from the visualisation in the video, where the chains don't always start and end exactly like this pattern. By following this pattern, it is obvious that each chain will extend it's length by three for each new shell, since it always runs the entire length of exactly three sides (excluding corners, which are always the same for each shell - two for the first chain and four for the second chain) and each side extends by one for each new shell. The extension by three is of course directly related to the skipping of each third number in the number line. The chains in the video does not give this visual proof of why each new shell is a combination of the two new numbers, but with my proposed change, this will be more obvious.
    1
  6698. What music do you have starting at 34:43? I want to listen to it on repeat.
    1
  6699.  @Mathologer  I’ve put a bug report/feature request in to Geogebra. This was on r/geogebra, as there didn’t seem to be anywhere else to put it. This doesn’t give me a lot of confidence that anyone will see it.
    1
  6700. Another visual proof, which reminds me of the centimeter cubes/rods we used to play with in grade school, is to break x+2 down as 1+x+1: this way, a cube x+2 on a side breaks down into a central x^3 cube, six 1*x^2 slabs on the faces, twelve 1*1*x rods on the edges, and eight 1*1*1 corner cubies, basically just giving the lower-dimensional bits a "thickness" of 1.
    1
  6701. C'est cool
    1
  6702. There is a notion of density of a subset S of the natural numbers which is almost a probability: It's the limit as N goes to infinity of #{s in S : s <= N}/N. Let's call this d(S). With this definition, if you take a positive term from a harmonic series whenever a number is in S and a negative term from a harmonic series whenever it is not, the sum will be log(d(S)/(1-d(S)). So in order for our sum to be pi, we need to pick a subset S with density exp(pi)/(1+exp(pi)) = .9585761678336371732... There are many ways to do this, but one of them is to feed n=1,2,3... into floor(n * (1+exp(pi))/exp(pi)) and make those values be in S. The following Perl one-liner uses this method and computes the sum of the first 10,000,000 terms: > perl -e 'use Math::Trig; $pi=4*atan(1); $alpha = (exp($pi)+1)/exp($pi); $i = 1; $j = 1; $next = int($alpha); for $n (1..10000000) {if ($n==$next) {$sum += 1/$i; $i++; $next = int($alpha*$i);} else { $sum -= 1/$j; $j++; }} print "$sum\n"' 3.14159230879059 Another way to look at this is that we are just taking 23 or 24 positive terms and then one negative one, in such a way that in the average we take exp(pi) positive terms. Any way you can do that will give you pi as the sum.
    1
  6703. This only shows that the three dots ... are easily written, but the the meaning is nowhere defined. Shame on Math!
    1
  6704. How did Euler know how many correct digits he had?
    1
  6705. I'm starting to feel like I can express any proof in maths using Pascal's Triangle.
    1
  6706. Great video
    1
  6707. 16:00 Assuming height of the first black isosceles triangle is half of the base… ((((1/2)^(2) * 2)^(1/2))^2 * 2) = 1 which is the area of gen 2 squares. Since the area of the gen 1 square is also 1, we can say it remains the same area overall for each gen. 5 gens means an area of 5.
    1
  6708. The most memorable part was that free book I won.
    1
  6709. No integers anywhere in the partial sums of harmonic series is definitely the most memorable
    1
  6710. There's many examples of later proofs of a something being much simpler (or elementary in the mathematical sense of the word). Look at the Fermat two square problem which was first proved by Euler with considerable effort by a method of infinite descent, and now can be expressed in Zagier's "one sentence" proof. Or Bertrand's postulate: Chebyshev's proof is lengthy and that of Erdos is elementary (although not simple). Sometimes it's easier to just wheel out the big guns than looking for a clever trick.
    1
  6711. Loved this!!
    1
  6712. 6:22 You can make it untilable, but my solution is quite trivial, it cuts two non-opposite corners from the rest of the board by removing both contiguous cells, making two opposite colored islands than cannot be tiled respectively. As an extension, this also works for any number of removed tiles per color >2, just remove random squares other than the islads you just formed. However, I wonder if there are more interensting solutions in which you cannot tile a full connected board - 4 tiles. 10:21 The cosines in the formula are always <1/2 for even n and m, but with odd numbers you get that for j=m, i=n both cosines are cos(pi/2)=0, therefore the product would be 0. But I definitely can't see this at a glance XD.
    1
  6713. cool
    1
  6714. Awesome video. I have loved math but stopped doing it for some time. Now these videos make me think and am loving them. Thank you.
    1
  6715. Love the buzz lightyear tee shirt.
    1
  6716. My husband has this on his Pilot’s Watch… he’s supposed to be using it for converting units (miles/kilometers, mph/kph) and fuel estimates… pilot stuff But his ipad flight computer never fails. So he uses it for calculating Horse Race Payouts, percentages of bills to tip wait staff, and helping me adjust recipe ratios to serve however many people.
    1
  6717. After watching your wonderful video, the youtube algorithm suggested this one: https://www.youtube.com/watch?v=ZK08Z5A9xH4 which has a similar visual proof that it attributes to Derrick and Hirstein 2012.
    1
  6718. It's easy to see that there are infinitely many ways to sum reciprocals to get the number you want, e.g. in the example of the greedy algorithm for γ we could at any time choose a larger denominator than the 'best' and then go greedy again. My question is, are there countable or uncountable number of ways to get the desired sum? To me it feels like there should be uncountable ways, but I could be completely wrong. As for favourite part I really liked the way you showed that γ < 1. Since 1/x is a convex function every rectangle in that illustration is more blue than white and therefore γ > 0.5.
    1
  6719. Napoleon was quite the patron of the arts and sciences. Highly recommend Andrew Roberts' book.
    1
  6720. OY VEY THEY'RE LACING ANOTHER SHOE-AH!
    1
  6721. Wow, this video really made me want to put some effort into learning abstract algebra again. All those connections just made facts I learnt in algebra pop like old memories and make so much sense, almost as if all of this was totally obvious (which it obviously isn't beforehand). This just reminded me how beautiful algebra is (I'm an engineering student who sometimes regret not having chosen to pursue more foundational fields such as physics or mathematics)
    1
  6722. 24:00 This has something to do with singularities and black holes.
    1
  6723. Nice video. If we need to choose one thing, it would be the partial sums (>1) being odd/even. That was a new cool fact :-)
    1
  6724. Knuth, guys! See Graham, Knuth and Patashnik, "Concrete Mathematics" (1988).
    1
  6725. so why is gamma bigger than 1/2?
    1
  6726. Just use integration to find volume !!😂
    1
  6727. awesome!
    1
  6728. I want the T-shirt!
    1
  6729. Beautiful theorem.
    1
  6730. The square root of 666 to four decimal places is 25.8070 not 25.8069.
    1
  6731. Love it! 😍 Greetings from Pisa!
    1
  6732. final year in high school(from india), got stuck after the xs1+c2s2........., well maybe because it was my first viewing, and maybe a little bit more info about bernouli number when it was introduced.what about the video on Galois theory? any plans on making videos about differential equations(and maybe geometry also) and complex analysis(in mathologer style off course). and once again thanks, for making such amazing videos.
    1
  6733. This figures.
    1
  6734. Very nice video
    1
  6735. For the exploded laces, you only described laces with any horizontal movement by the amount of vertical movement. What about laces whose movement is entirely in the vertical axis?
    1
  6736. Am I overthinking/under thinking looking for a link here with Fibonacci sequence?
    1
  6737. For the square puzzle, all squares can be categorised by the direction (i,j) of their pair of sides which have positive slope. This will have height/ width i+j in the direction of the gridlines, so we must have i+j<=n where each row has n squares. You showed roughly there are (n-k+1)^2 (k,0) squares in the grid. So I'd go for an expression \sum_{i=1}^{n} \sum_{j=1}^{n-i} (n-i-j+1)^2.
    1
  6738. that proof in the end was really elegant
    1
  6739. 16:20 thats afreaking tall tower to over hang only 100m XD
    1
  6740. That is so genius.... why do I found out about it only now?
    1
  6741. About all the details that were skipped over in the visual sum = ln(2) proof: First of all, to show that the sum up to the n-th positive number is equal to the sum of rectangle n up to rectangle 2n-1, a simple inductive argument: The sum up to the first positive number is just 1 (so rectangle 1 up to 1). Then, to get from the n-th positive term to the (n+1)-th positive term, you're adding (1/(2n) - 1/n) + 1/(2n+1). Subtracting -1/n removes rectangle n, so the sum now starts at rectangle n+1, and adding 1/(2n)+1/(2n+1) adds rectangles 2n and 2n+1, so the sum now goes up to 2(n+1)-1. Now let's say the left-most edge of a given rectangle starts at x. It will therefore have a height of y=1/x. If you squish the rectangle by a factor of n and stretch it horizontally by that same factor, and instead of chosing the corner of the rectangle as the fix-point of the scaling, you choose the origin x=0, then the rectangle will move to x'=x/n, and the new height will be y'=n y. With this, the new rectangle also follows the law y' = n 1/x = 1/(x/n) = 1/x'. This transformation will also move all the rectangles to horizontally fill the interval [1,2], as the first rectangle with x=n will be moved to x'=n/n=1, and the last rectangle with x=2n-1 will be moved to x'=(2n-1)/n=2-1/n, so its right-most edge will be at 2-1/n+1/n=2 because its original width of 1 was squished to 1/n. Therefore, the new rectangles fill the area under the curve 1/x on the interval from 1 to 2, and because all we did was move them around a bit, and squish them by the same factor as we stretched them, the total area didn't change.
    1
  6742. Does this trick work in 3 dimensions? My gut says "no" but it would probably work in 4 dimensions.
    1
  6743. Don't think so. At least I did not see anything nice while working my way through the relevant literature :) But who knows .... :)
    1
  6744. pls do a vid on generating functions that would be awesome
    1
  6745. It is very much evident that an insane amount of hard-work you are putting to make the video, to be honest your all the videos. Hats off for the great work. Enjoying all of your videos and learning so many new and crazy things of mathematics. My background is from physics but I have a very keen interest in knowing the original works in mathematics which are very very helpful in understanding nature in the way physics uses those very often. I am a huge fan of mathologer . Wish to get enriched by so many new stuffs in coming videos.🙏🏻
    1
  6746. Engagement
    1
  6747. where's #27?
    1
  6748. so interesting
    1
  6749. at 29 minutes your second line missed the value 5, but thats ok
    1
  6750. do bees naturally know this mathematica’s 🐝
    1
  6751.  @Mathologer i see honey comb 🖖
    1
  6752. Best bit for me was the white out of the no nine series. All in all an engrossing iteration of the history of the harmonic series. The graphics were exemplary. But then black ‘’holes’ are always engrossing. Just a thought- in the super clear thought experiment at the beginning (aka the ‘silent stack’), you forgot to mention that as the series exploded to infinity, so did the total mass. This is really the definition of a singularity. So within the structure of the harmonic series is a metaphor of the dark energy problem. This is the sort of thing that happens when you turn a thought experiment on its head! Now just suppose the event horizon of each galactic centre was not finite. How would that affect the math of the cosmic constant??!!
    1
  6753. Something that I remember being taught in France was that if the digital root of a number is 3, 6 or 9 then that number is divisible by 3
    1
  6754. Noticed the shirt, I'm assuming it denotes frequency? As a fun aside since it seems the scales match in both dimension, I believe each layer of described curves should have equal length (something I found whilst playing with scaled ellipses)
    1
  6755. You did not put the index at 3.14 but at 3.18 How did you get to 8.54?
    1
  6756. In the case of the four colour problem: Consider the grid as a set of nodes with coordinates A1 through B2 for the 2x2 case, A1 through D8 for the 4x4 case, et cetera. Choose an arbitrary colour for A1, and a different one for A2 Once A1 and A2 have colours, B1 and B2 must have the two other colours to satisfy the criterium for the A1 through B2 square. Since no distinction has been made between these two colours yet, choose one arbitrarily for B1 and the other for B2. Once B1 and B2 have their colours, C1 and C2 must have the other two colours to satisfy the criterium for the B1 through C2 square. This means that C1 and C2 are now confined to the same colours as A1 and A2, though not necessarily in that order. This pattern of altering pairs continues, with every subsequent pair of squares in a column along the 1- and 2- rows being confined to one of two pairs: the same two colours as A1 and A2 for "odd" letters, and the same two colours as B1 and B2 for "even" letters. Equivalently, once A1 and B1 have their colours, A2 and B2 must have the other two colours to satisfy the criterium for the A1 through B2 square. This logic continues equivalently along the rows exactly equivalent to the case along the rows, and thus ever subsequent pair of squares in a row along the A- and B- columns is confined to one of two pairs: the same two colours as A1 and B1 for odd numbers, and the same two colours as A2 and B2 for even numbers. Therefore, for any given corner alpha in an x by x grid where x is an even number, the corner squares beta and gamma which it shares sides of the grid with must be differently coloured to it, as beta and gamma will each differ from alpha in the evenness of either the row number or of the column number. Thus if we can prove that alpha also differs in colour from its opposite corner delta, then it must differ from all other corners; and if it is true for any given corner that it differs in colour from all other corners, then each corner is uniquely coloured. That part of the proof, however, keeps eluding my 11pm mind and should probably wait until morning
    1
  6757. The P is silent! IPA /ˈtɒləmi/ … it had to be said. 😬 You're going to tear a ligament if you keep saying it wrong. 😉
    1
  6758. This becomes a lot less paradoxical when you realize you can paint the whole plane with paint of thickness exp(-r^2) (for example), getting thinner as you go out further.
    1
  6759. Right, it's basically just saying some infinite series have finite sums. Which I suppose was pretty crazy when people first realized it but now not so much :)
    1
  6760. Ahh bue... que buena data. Messirve
    1
  6761. I love your videos so much because they help me a lot to learn new stuffs and invites me to delve deeper into mathematics. Thanks a lot. As a student and a regular viewer of your videos, I request you to create a video on Matrix, their determinants for higher order like 4 by 4 or 5 by 5, ..., their properties, decomposition of matrices into factors and related stuffs. These are very common topics, but students really need these.
    1
  6762. I used to have a circular slide rule like this. I wish I had it back.
    1
  6763. +1
    1
  6764. https://www.trade-ideas.com/home/phil/Triangles/Circles.html An interactive version. Click on a circle on the top row to change its color, and the rows below will automatically update.
    1
  6765. So basically the Talmudic approach could be written as: Until everyone gets half of their money, every additional cent goes to the person who has received the least amount of money out of those who have not yet gotten half of their money. After everyone gets half of their money, every additional cent goes to the person who is still missing the largest amount of their money. Or: Every dollar is evenly shared between those who have not yet gotten half of their money. Once everyone has gotten half of their money, every dollar is evenly shared between everyone who has the same amount of repayments left to receive. I think this approach is fair for many reasons. Not the least of which being that if the value of the estate is unknown (random), no creditor comes out on top relative to each other on average, neither in absolute nor proportional terms. It is also impossible for any creditor to receive more than half of their debt without every other creditor also getting at least half of theirs, meaning you have a certain guarantee to at least half of your money, as the only situation where you might not get it is if the estate is worth less than half of the person's total debts.
    1
  6766. That’s great.
    1
  6767. 🙏🙏🙏
    1
  6768. Wonderful stuff, thanks!
    1
  6769. thank you very much, Mathologer, for this aspiring challenge ! - I successfully implemented my own Excel-Spreadsheet-VBA-Macro-Programming-Solution for creating those arctic circle pictures - very, very nice ! thanks a lot for all your videos and Merry Christmas from Vienna !
    1
  6770. Very nice! I‘ve been toying with the idea of teaching difference calculus for a long time. Now more than ever. I think there is a visual way of proving Newton-Gregory: count the number of ways any leftmost entry contributes, by adding, to the n-th term. It is n choose k, because n steps must be taken, and k of them „up“.
    1
  6771. Can you please solve 1/a +1/b=3/2018 For all ordered pairs of a and b Where a and b are positive integers
    1
  6772. I love his made up words ! "Curvy curve"
    1
  6773. Does Prof Pavel Grinfeld watch this channel? I wonder if there is a vector based proof for these theorems.
    1
  6774. Another video on Euler-Maclaurin formula please!
    1
  6775. The even over odd section is my favorite :)
    1
  6776. I feel like this could be carried further to prove the Law of Sines or Cosines. Or both.
    1
  6777. Thank you for delivering so nice content, which helps more people experience the beauty of mathematics.
    1
  6778. I'm most interested in gamma. Though I'm always impressed with seeing a 1 with 100 zeros after it, which you showed once or twice in this video. I taught my 4-year-old how to subtract 1 from a power of 10: the number of 9s in the answer equals the number of 0s in the minuend. We wrote this out for one googol on his chalkboard.
    1
  6779. The nontrig hardcore challenge can be done using a compass and a traight edge?
    1
  6780. Very helpful for me from 19:10 .
    1
  6781. Mathologer: "Last chapter, I promise!" Me: "I just wonna die, plz... why are you doing this to me, why???"
    1
  6782. Kempner's proof is my most memorable segment
    1
  6783. Third challenge: That's without matrix' but i hope still counts. For n=1 it's 1, for n=2 it's 2. Assume that we counted it for every n<k For n=k: look at the left of board. If it s one dominoe, the other is k-1 dominoes and for it we can say the number. If it's connected to the next two squares, the other is n-2, and we can say the number. So f(n)=f(n-1)+f(n-2) and that's n-th Fibonacci number if we count the first is one and second is two. That was not easy but mostly because of i was thinking of matrix, and i don't now anything of that;)
    1
  6784. Great shirt! 😄
    1
  6785. Why don't "they" teach Newton's geometrically created fluxions, etc? Show us a fluxion!!!
    1
  6786. Ln(2)
    1
  6787.  @thedoctordowho2022  THAT don't look like no fluxion
    1
  6788.  @russchadwell  ü = 4
    1
  6789.  @thedoctordowho2022  lol. Still no. You are obviously a Leibniz/Descartes fan
    1
  6790.  @russchadwell  interesting how did you figure it out?
    1
  6791.  @thedoctordowho2022  you're continued use of Leibniz and Descartes symbols. Rather than supplying the requested Newtonian Fluxion diagram... albeit mighty difficult using a keyboard
    1
  6792.  @russchadwell  well ,"Ln" is for natural logarithm and "ü" was the dot notation that newton used.
    1
  6793. Leibniz: dy/dx = f(x)'
    1
  6794.  @thedoctordowho2022  true. But the actual "flexion" was more of some sort of geometric diagram. It would be a bit like trying to use a keyboard to directly illustrate a Feynman diagram
    1
  6795. This is a good one. I was going to comment that all you need so ab = a + b is to (1) pick n and (2) your other number is n/(n-1). But I thought better of it.... ;)
    1
  6796. 16:25 There's a simpler way to see that: just multiply all the decimal numbers by 100000000...000 (enough 0s to get integers) and then look at the sequence in reverse: it's the 2* sequence! So if you get the 2* sequence looking in reverse it's the reverse of the 2* sequence looking in the normal order
    1
  6797. Seen one of Mr Parker video lately.. he mention infinity case where you add all natural number to the box, and remove a number whenever you add its square. This box is supposed to be empty, as you will remove all numbers eventually.. but that looks exactly as ln2 harmonic distribution case. I get the argumentation here for ln2, but why Matt's case gives us 0?
    1
  6798. the question becomes, was newton familiar with madhava's work?
    1
  6799. I deeply. Love Ian's shoelace webpage. I even emailed him a new kind of doubled shoelace style. 😍
    1
  6800. Ooh I got a tiny taste of generating functions in one of my classes, they're super neat
    1
  6801. Why aren’t the comments talking about the logo? I loved the new one!
    1
  6802. No, the argument is fine as is. We only use the pigeonhole principle to find two subsets with the same sum. We don't have to use the pigeonhole principle a second time when we remove elements. We already know the two subsets have the same sum!
    1
  6803.  @MuffinsAPlenty   @MuffinsAPlenty  what I'm trying to say is that by taking away the overlapping elements you don't create a new 1023rd set but get one you already accounted for. That's why I'm saying you should prove that even with this there's still more than 846 sets to fill the 846 possible sums. Example: 1 2 3 4 5 1 and 5 give the same sum as 2 and 4 and we leave out 3. This is one possible set let's call it X. 1 2 and 3 give the same sum as 2 3 and 4. This is another possible set let's call it Y. Let's say these 2 sets fall in those 1022 sets ( I know they dont but let's say they do). Now I tell you I don't want sets with overlap. Since Y has overlap you decide to make a new set out of it by removi g the three but by doing that you get the set X. So you didn't get a mew set from Y, because X has been already accountef for. So now you have to throw the set Y out because it has overlap and you no longer have 1022 sets that could work but 1021. This number could keep lowering if there are too many overlapping sets and you would need to prove that the overlap issue would still give you more pigeons than holes(more than 846).
    1
  6804.  @nikolamarijanovic6261  "1 and 5 give the same sum as 2 and 4 and we leave out 3. This is one possible set let's call it X." What? What is X? Is X supposed to be {1, 2, 4, 5}? Or is X the pair of sets {1, 5}, {4, 2}? The 1022 number isn't counting pairs of sets. It's counting individual sets.
    1
  6805.  @nikolamarijanovic6261  By the way, you're not the only one who seems to be thinking this way, but I just don't quite understand what your thought process is. I'm curious to know why you think removing overlapping elements is a problem that undoes the previous use of the pigeonhole principle (if that is what you're saying).
    1
  6806. Curious, why a rubber band and not a string or a very thin aluminum rod having fixed modulus of elasticity?
    1
  6807. So after watching this video I wanted to try to program something to find those squares and I realised that you could find all the windmill combinations starting form the cross by alternating between the same shadow transformation and the inversing y and z transformation, only thing that was left was finding that "same shadow" transformation, I started with a random windmill and found how the second one was calculated, these new values weren't always positive so I made another test that would go wrong with that one and found another equation and so on. I ended up with three equations that filled up all the range of values except two points that weren't possible on primes anyway. Programming that was quite easy and not only it gives you the sum of two squares when you input a prime but it also tells you if it isn't a prime if either it's one of those special cases or if no sum can be found. After having all the fun testing I went to wikipedia to check the one sentence rule and the equations found there are exactly the same I found by trial and error. I'm still amazed by the fact that something so visual like the windmill and its shadow actually represents that proof that didn't make any sense for me at first glance.
    1
  6808. Einst soll ein französischer Philosoph gemeint haben "am Grunde eines jeden Problems sitzt ein Deutscher", doch mit Burkart Polster "haben wir" anscheinend eine art anti-shapeshifter, der auch die Lösung eines jeden (echten) Problems von einem Deutschen auf den Tisch, die Strasse oder einfach nur zu den Interessierten trägt oder bringt. Oder wäre Lob hier zuviel des Guten, das er sich nicht auf Lorbeeren ausruhte, so er nicht ein Rom-o-philer sein wird? Solch "mathematisierte Unterhaltung" ist stichhaltig, nicht nur weil man das so nicht mit "infotainment" vergleichen kann, meinte ich. Euclid wurde vor einigen Tagen am Lagrange Punkt L2 , weitgehend hinter dem Mond "parkiert", doch die "probe" teilt sich den "Punkt" mit dem JWST und GAIA, wie "kriegen" die das hin, in Dynamik mit geometrischen und gravitativen Aspekten?
    1
  6809. Heron's formula was not discovered by Heron but by somebody with the same name.
    1
  6810.  @serious-goober  The Rubik's cube one? I didn't try it.
    1
  6811. What you call French lacing is called Straight European on Ian's site. If you wear Oxfords, you really must have straight bars on top. You also want the pattern to be as symmetric as possible, or else the lace will shift over time. Under these constraints, the Straight European lacing seems to be the "tightest" pattern, and hence the best.
    1
  6812. Most Interesting: No 9's sum is finite
    1
  6813. How do I contact you semi-privately? I tried emailing you once but never heard anything back. I have a couple of mathematical things I've noticed but have never seen anybody else discuss. (Oh, and I anticipated most of what you said in the first 22 minutes of this video, so I guess I might sometimes know what I'm talking about. :-) )
    1
  6814. Is the so-called paradox intended to be obviously bogus? Not a mathematician, but I understand geometry. After the first corner cut, the corners' inner angle will either not be congruent with the circle, or will not be squares; therefore, the process will not converge to a circle.
    1
  6815. Burkard, you should make a t-shirt with the picture at 3:08. That would put Heron your chest.
    1
  6816. Audio question: can whoever's doing the audio cut out less of the "not even really high" frequency? It just makes things sound super muddled and much harder to listen to than if everything up to 15KHz is kept. Plenty of ways to clean up audio without a straight "cut everything above 8KHz" approach. I couldn't watch this video, it just sounds like mumbling =(
    1
  6817. Mathologer just cursed himself with that can!!!!
    1
  6818. I liked the after-credits scene
    1
  6819. Is this just a series of cool facts, or is there insight to be gained from these patterns and this way of thinking into other branches of mathematics or science? For example, the way the Fibonacci sequence folds and then produces Pythagorean Triples reminds me a little of some of the processes of DNA and RNA interaction, conversion, and protein manufacturing. Is there any way in which this pattern is interesting for continuous/non-integer geometry? Are there any interesting parameters for non-Euclidean geometry that can reproduce similar interesting connections?
    1
  6820. Colouring proof!
    1
  6821. great timing. i have to spent some time in the train right now
    1
  6822. alright this was great
    1
  6823. The fact that the absurd number we get from reimend fonction has a use in calculus
    1
  6824. Aaand! The views is Five-digits!
    1
  6825. I cant believe something that basic is discovered only few years ago. Conway was really a genuis
    1
  6826. When you asked what they'd have in common, I didn't guess, I thought: (i) each of these three extended lines has the same length; (ii) each pair, crossing at a vertex of our triangle, is split in the same lengths, so has the same products of its parts' lengths, matching the intersecting chords theorem; each pair of the lines thus puts four point on a circle; so I concluded you were about to tell us it was a circle. That isn't quite a proof, though, and I can't immediately turn it into one, but it smells like it's relevant: we have three cyclic quadrilaterals (albeit we don't know they share a single circle), each of which shares each of its diagonals with one of the others; and those three diagonals are all of the same length. So Ptolemy's theorem's product of diagonals is the same for all three, implying their sums of products of opposite edges are the same. But after scratching my head for a bit about that surely being a clue, I think I'll watch your video and find out what JHC came up with ;^>
    1
  6827. OK, so you show the circle shrink and it looks suspiciously like its centre is also the incentre, where the angle bisectors meet. And, of course, each of our three extended lines is, by construction, the mirror image of the one it meets at a corner of the triangle in the bisector of that corner's angle. So it suffices to show one of these has its two ends equidistant from the incentre, so that there is a circle centred on it through those two ends and reflections in lines through the incentre will ensure the other two lines also have ends on the same circle. Let the length of each edge be denoted r, g, b by the first letter of its colour, with the semi-perimeter s being half their sum; each of our extended lines has length 2.s. The perpendicular from the incentre to (say) the blue edge splits it into an s-r on the end we extend by r and s-g on the end we extend by g so this is the mid-point of the extended edge and a radius is its perpendicular bisector; we are done ;^>
    1
  6828. Your later "swivel vs colouring" choice: given that the colouring answer is how I know the split of the blue-edge is at s-r into the corner opposite red and s-g into that opposite green, I'll vote for the colouring one. But the swivel one is nice, too, although I think it'd be nicer with a reflection in angle-bisector rather than each rotation about corner.
    1
  6829. Of course it was Euler again.
    1
  6830. The reason every number with a repeating tail is a fraction is because you can express it as integers i and j where (i+j/999999.....)/100000.... is exactly that fraction. Just make j the repeating part and i the non repeating part. So 0.416666.... is 0.41+0.00666666... which is (41+(6/9))/100. That's definitely a fraction because (i+j/9999....)/10000.... can be expressed as i/10000....+j/9999.....00000... or (i*9999.....+j)/9999....00000..... the latter is a fraction of 2 integers. That makes (41+6/9)/100 -> 41/100+6/900 -> (369+6)/900 -> 375/900 which simplifies to 5/12 BTW.
    1
  6831. Madhava was born in Kerala
    1
  6832. The "Rubik's cube and drill" was beautiful. And the background was worth a careful look!
    1
  6833. My favorite factoid was definitely that there are no integers in the partial sums of the harmonic series
    1
  6834. From where did you get your t-shirt ?
    1
  6835. Frohe Weihnachten 🎄
    1
  6836. I saw where you were going with this quite quickly because I've been using (straight) slide rules for 50 years and always wanted to own a circular one. They are a great way of comparing proportions such as unit pricing of goods in supermarkets where they have different sizes of packs.
    1
  6837. The real question is where is the northpole on the triangular lattice?
    1
  6838. For me it's a close call between the 700 year old proof of the divergent sum, and the greedy algorithm. I'm a big fan of mathematical proofs that seem oh so obvious when they are explained, but aren't when the question is posed (at least for me).
    1
  6839. Great sir 👌👌👌
    1
  6840. @6:22 yes coz tiling a mutilated board (removed 2n squares) is possible implies that the extracted out 2n squares could be put back sequentially 2 blocks n times. each time u do a black and red square. the construction of the path is as earlier. After every step, it still remains a solution till the unmutilated board tiling is possible. Which is always possible as we know.
    1
  6841. I had never seen any proof of the Pythagorean theorem as a student but I noticed a proof off my own when I was taught similar triangles the next year . Anyways now as a Math teacher I teach both proofs , the first when I am teaching the Pythagorean theorem and the second to students that already know the Pythagorean theorem but are now learning the (a+b)^2 identity.
    1
  6842. 40:24 You say "at least ten 9s", but did you mean "at most ten 9s"? Because the former diverges.
    1
  6843. I was caught unaware by the "get rid of the eyes"
    1
  6844. Our math teachers were mostly not interested in history.
    1
  6845. I was breathing, but this caught me by surprise!
    1
  6846. Partway through the matrix explanation of the monster formula a tile disappears from the board displayed. I was losing my mind trying to figure out why I couldn't tile it once. (12:29)
    1
  6847. Very interesting. But who figured a space probe plus a negative alien baby bash equaled half a world of planet of the apes and I still couldn’t get an unripened conjunction papaya in Papp New Guinea. Throw in a couple roust cannabiles which are fine like lady Gaga and your got a theorum nobody but nobody could ever sneeze at.
    1
  6848. Why does everything have to make absolute sense.
    1
  6849. Peg number formula: if we relabel pegs (A,B,C) as (0,1,2), then for peg, p; move m, and disk, d (where the smallest disk, d=1), then I get that the peg number is p=2^(d-1)*floor((m+2^(d-1))/2^d) mod 3, for any number of disks. At move zero, all disks start on peg zero.
    1
  6850. I deeply deeply enjoyed the videos on this channel. Watched three in a row and cannot stop. Thank you so very much :)
    1
  6851. Brilliant
    1
  6852. Short Prove that ln'(x)=1/x (=> integral of 1/x=ln(x)+c) : e^ln(x)=x | f(x)=e^x f(ln(x))=x | ()' ln'(x)•f'(ln(x))=1 | f'(x)=(e^x)'=e^x ln'(x)•e^ln(x)=1 ln'(x)•x=1 ln'(x)=1/x
    1
  6853. In the mid 1970s when I was in High School, pocket calculators were just becoming available, but they were very expensive. My science teacher forbade them and made us all learn to use slide rules.
    1
  6854. If you look at Galois's perspective, is squareroot 5 structurally different than squareroot 7?
    1
  6855. When doing the visual slicing into triangles and then pruning, why did you have to bother splitting the faces into triangles at all? It seems you would arrive at the same answer, with the same justification simply by setting by pruning away edges (each time eliminating either one face or one vertex along with the edge), except until maybe the very end, when the simple irreducibility of the triangle is necessary?
    1
  6856. thank you sir!
    1
  6857. thank you
    1
  6858. I have seen both versions in my coaching classes. The coaching class taught people for Regional Math Olympiads, kinda like qualifiers for Indian teams in IMO. The setup was in their entrance test material, and we had to prove PT using triangle congruence. They had also mentioned the visual proof.
    1
  6859. re 21:20 You neglected to add the cost of the ink jet cartridges being used to do that printing....Hewlett Packard would just love this to happen!
    1
  6860. And as a suggestion for a different video altogether, how about logistic maps, bifurcations and the Feigenbaum constants?
    1
  6861. Full comment has challenge answers (attempts really) down below OOOH I UNDERSTAND THE DETERMINANT TRICK The determinant can be described as taking the product of all terms which don't share a row or a column. If you decide that one particular square is paired to another square (oppositely colored) using a domino, then you can't tile it again, which by the determinant process, disallows you to pair that square/row with another term in your product. Two i's represent two vertical dominoes. Twisting them, involves turning them into horizontal dominoes (two 1s). The product turns from -1 to 1, and the minus sign is then compensated for by the change in the sign of the permutation. Now, to understanding why twisting gives all possible tilings 🤔. 1) (removal of 2 black + 2 green squares) Not generally, because the two black squares connected to a corner green square can be removed, isolating it and making it un-tile-able 2) (how the crazy formula spits out a 0 for oddXodd boards) the denominators of the cos input are (n+1) and (m+1), so for odd m and n, they are even. Since the products upper bounds round up, inputting the upper bounds gives the input of both the cos functions as (π/2), making them 0, and making the entire product 0. (The last puzzle of why the number of blue, orange and grey tilings) Rotational symmetry!
    1
  6862. Fascinating. Congratulations ❤
    1
  6863. I thought the man had his brains out
    1
  6864. When I asked you a couple of months ago whether any sum of rational numbers could equal an irrational number, you said: "Sure! 3+1/10+4/100+1/1000..." That one converges really fast!
    1
  6865. Excellent ! : )
    1
  6866. I am about 25% through the video, but I have to ask this question. Wouldn't 3, 6, and 9 being absent from the pattern just be a byproduct of our base 10 number system? If we used other bases, would similar patterns appear?
    1
  6867. Algebra mistake: I forgot to divide by T-shirt, answer is 15.
    1
  6868. This was amazing! Results which seemed completely inaccessible at first glance made simple to see. My most memorable part was that crazy formula you flashed on screen involving e, gamma and the prime numbers.
    1
  6869. I vote for gamma... would have liked to see more of its connections to other parts of math
    1
  6870. Merry holidays, Buckard, Andrew, and everybody else who reads this! :)
    1
  6871. The "no integers" is most amazing.
    1
  6872. An hour long Mathologer video?! It's Christmas alright.
    1
  6873. Great visualization of the proof!!
    1
  6874. I'd go for the no integer partial sums
    1
  6875. Hello
    1
  6876. Is this a valid solution for the last problem? Triangle Area is 0.25 Large Square Areal is 1 Overlapping suggests each triangle consists of 5 tiny perfectly equal triangles. Thus each tiny triangle has an area of 0.05 Counting all the non-overlapping tiny triangles in the large square makes for a total area of 0.8 Thus the small middle Square has an area of 0.2. Oh, or you just add the overlapping areas and end up at the same :D
    1
  6877. Choosing this particular t-shirt for this particular topic is a stroke of satirical genius. My highest compliments! (Not just for the t-shirt ;-) )
    1
  6878. What would this math be used for?
    1
  6879. To me, it seems like this argument about area of the lantern going to infinity due to the increase of points and planes is similar to there being an infinite amount between the numbers of 0 and 1. An finite infinity, I guess, which sounds reasonable to me. The lantern doesn't increase area infinitely, it just increases the measurable area to an infinite precision within a finite amount. (edit) Huh. I've seen some vids of crushing tubes of various materials and wondered about why the material crushed the way it did. Thanks for adding that part at the end of this video.
    1
  6880. I don't unbderstand the part that starts around 20:00. It shows that the isolated solution, that is the cross solution, requires X to be equal to 1. But in the case of 13, the solution is actually 2 squared and 3 squared. So for 13, x is NOT equal to 1. So when factored as p = x(x + 4z), x is not one, so... p is not prime? So, I don't see how crosses, that are the unique solutions, which should always have x = 1 can correspond to the 13 case. Maybe there is soemthing else going on. If so, then choosing a prime other than 37, that is a prime whose decomposition into two squares does not contains 1 squared, would have been a better illustration.
    1
  6881. 4:00 OK, we have a circular slide rule.
    1
  6882. When I saw 355/113 I knew what I have to do, sadly there aren't so many yet so I reversed the order and typed 113355, wasn't disappointed Men of culture will understand
    1
  6883. Made it to the end!
    1
  6884. Ich dachte am Anfang der Beweisführung, dass nur nur ein Quandrant untersucht werden würde, da die dann angenomme Fläche zum vierten Teil von Pi führt. In der Mathematik führen aber viele Wege nach Rom :)
    1
  6885. I use the 2nd proof on day 1 of school in the fall. That way the kids have the iron clad info and can use it whenever they need. LHS, Leesville LA
    1
  6886. The cannot's also appear to be 4k-1, which to me seems prettier than 4k+3, given its similarity to 4k+1 for the can's; but of course, I'm no mathematician.
    1
  6887. So if I want to escape the infinite orchard that numberphole trapped me in all I have to do is walk straight from one tree to another, turn 60 degrees, then repeat.
    1
  6888. That was The video! Just awesome! I must confess I got lost in some parts but the whole explanation was really great!
    1
  6889. There’s only one way to make zero cents, but there are tons of ways to make zero sense.
    1
  6890. Hey so I’m a sophomore and I have not had calculus yet but you explained everything pretty well I understood it for the most part.
    1
  6891. In binary, guess the value of the no ones series... ... and the no zeroes series
    1
  6892. First, and thank you very much for this lesson.
    1
  6893. I also wanted to say thanks, but I would have lost the place ^^
    1
  6894. (Regarding your request for feedback at the end) It did all work for me, really enjoyable video. I'm trying hard to come up with something that would need improving but I can't right now. I would love to see more of these longer videos. As to my background, I'm a computer science master student (with a theoretical CS focus).
    1
  6895. There was a common exercise in low-level CS classes to search for Pythagorean triples. Now we can enumerate them without searching? That's going to leave a mark.
    1
  6896. Very easy proof for Similar Triangle thing: for two coefficients p and q, Brown = p*Red + q*Black. Done!
    1
  6897. Well, if when you position Red at 0 and Black at 1 you get Brown = p, then Brown is always at Red+p*(Black-Red).
    1
  6898. Woooowwwww
    1
  6899. are you secretly spying on me, how did you knew i needed this..
    1
  6900. Mathologer Could you say volume of a 4D body
    1
  6901. The "deep part of the deep end" lost me mainly in the sense that I was wondering where you were going with this. I think had I been interested in following it more than superficially, I probably would've got it. As for my background, I majored in computer science.
    1
  6902. Speed 2x . So 25 min
    1
  6903. About the right pitch and duration for someone with a chem eng degree older than at least some of your viewers' dads. I suppose I encountered inverse matrices and suchlike at an age when the concepts could take root. Not so the more 'modern' stuff like group theory. They kept all that sort of thing a big secret from us, and now I fear that I shall never be able to make sense of it, and manifolds will simply remain lumps of metal one bolts onto combustion engines.
    1
  6904. Terribly sorry,but it didn't work for me. :-( I'd have preferred three videos of half the length each.
    1
  6905. Just do it! :D Cant wait :)
    1
  6906. This is a fascinating video. I need to watch it again from start to finish. Made me think there should also be a recurrence relation between the inverses of the Pascal triangles of increasing sizes.
    1
  6907. Wow, that took some digesting. I studied engineering at uni some 20 years ago. I think I understood it all the way through, except at 16:40, where for some reason (x+1)^0 = x. The only real problem is that by the time I got to the end, I forgot how we got there, so maybe some kind of diagram or animation that showed the path would be useful. Possibly a progress bar in the chapter cards :)
    1
  6908. Great video! Bachelors degree in math here. Followed it all but the last hand-wavy bit regarding the Euler-MacClaren formula. Looking forward to that video to shake off the cobwebs of my 30+ yeal old degree. Keep up the great work!!
    1
  6909. I vote for the bizarre maximal overhang towers!
    1
  6910. I'm confused about one thing with the cart trick, how are you supposed to know to count the signaled distance up or down from the value indicated by the first cart? In the given practice hand challenge, I think it could be either 9 - [signaled distance] or 9 + [signaled distance]. Or did I miss something?
    1
  6911. In the sequence 729 , 71289 , 7112889 ,711128889 , 71111288889 ... ... ... Why are those numbers always perfect square and is there a way to prove that every number continued in this above sequence will always be a perfect square??
    1
  6912. Sevengon? Wasn't it called heptagon?
    1
  6913. 24:24 It becomes a 0,1,1 triangle.
    1
  6914. Definitely like the color one better. It keeps in mind the lengths of the original, arbitrary triangle.
    1
  6915. what a beautiful result
    1
  6916.  @Mike_Greene  yes
    1
  6917. The Great Archimedes
    1
  6918. 23:59 Is it necessary, to obtain the "Arctic Circle", to use this 2x2 rule? What if one decided to fill that 4x4 square with a 2x2 square in the middle and a 4x4 frame around it?
    1
  6919.  @Mathologer  I see. Thanks a lot for the quick reply, and have a Merry Christmas! Cheers from Italy
    1
  6920. What happens if you try to brute force it by starting at L5 and working your way back behind the line?
    1
  6921. A form of the Logos manifested. One wonders about shapes in n dimensions and how they would manifest such orderly sequences, and what that might say about dimensionality itself.
    1
  6922. I might be late but I made it to the very end !, the 666th partition number is 11956824258286445517629485, I wondered about the formula of the circle regions in 2:14, it didn't work, it's because it doesn't count the edge regions
    1
  6923. i still have several slide rules
    1
  6924. GREAT !!!!!!! Already shared!
    1
  6925. Master Teacher.
    1
  6926. Great video!
    1
  6927. In high school physics we had the Book of cos sine and logs etc. until the Ti 20 came along
    1
  6928. Most memorable: the proof that the harmonic series blows up to infinity was proven 700 years ago
    1
  6929. Yeah I got it... Question Ur first example had a repeating pattern. If you turned it it'll be the same break down from any edge to the point. It's why I guessed yellow as that's what the other corners are... I spent the 5 seconds just checking the next row
    1
  6930. I first saw the geometric version of foiling (and it being used to prove a^2+b^2=c^2) in freshman year of college, majored in the history of math. Was furious nobody had ever told me, that I had to go to college to learn that XD
    1
  6931. I was taught the divisibility by 9 (and 3) test in primary school in the UK, and was taught the mod 9 (and mod 3) extension when modular arithmetic was introduced in secondary school. This was in the 90s.
    1
  6932. Which of your sources contains the formula for mutilated boards with holes?
    1
  6933. Are there names for these types of polynomials for other regular polytopes? For tetrahedra type polytopes I have a recursion P(0,x) = 1, P(n,x) = (x+1)P(n-1,x)+1 and for octahedra I have P(1,x) = x+2 P(n,x)=(2x+1)P(n-1,x) + 2 - x^n
    1
  6934. 15:36 - Umm.. no, you can't! That round surface area can't be made flat without completely redistributing it into a completely different shape. (Which is why we can't have linearly scaled 2D maps of the world.)
    1
  6935. My vote is definitely to the Robert Baillie's crazy subseries sums at the end
    1
  6936. another magic video. great stuff. for puzzles, finding primes using the sieve of Eratosthenes is a staple. mathing up sieves is fun. how quickly can you count the number of primes up to 210?
    1
  6937. I once told my friend, don't forget the plus c before a test.... apparently he thought I said don't forget the pussy🙀...31:31
    1
  6938. They don't teach that because it's counterproductive making workforce smarter, which might clue in and throw them off the throne.
    1
  6939. You are a peerless producer of mathematics videos. The medium demands more than unrehearsed monologue and artless scribbling. Thank you for setting a good example.
    1
  6940. Anyone have an resources on a related subject called sum calculus?
    1
  6941. I don't think I've ever seen these geometric proofs. My favourite remains a line from origin to (a,b) length c. (a,b).(a,b)=a^2 + b^2 =c^2. "." scalar product...The closest I ever got to traditional geometry was algerbraic topology....
    1
  6942. 7th grade and 10th grade
    1
  6943. I wonder if some people go to watch X+Y as a horror movie experience?
    1
  6944. 8:30 y' = xy You could probably do this using an integrating factor, but I'll jsut cheat and pretend like y' = dy/dx and the differentials can be multiplied dy/(y * dx) = x dy/y = xdx ln(y) = x^2/2 + c y = Ae^(x^2/2) , A = e^c 21:40 I never learnt about infinite products, so I don't have any precise mathematical weaponry to fight this product, but I assume that because each successive term that you multiply by in the sqrt(pi/2) is greater than 1, the whole product explodes in the limit. Likely it would act like an exponential because the terms visually seem to be very similar, although maybe that's wrong and it acts more like a polynomial, I aren't sure. The Wallis product converges because successive terms alternate between being greater than and less than 1, and I cannot remember the justification why this works as such because the 3b1b video about it came out a while ago.
    1
  6945. ...professor mathlogger...another beautiful video ....amazed me...happy Christmas and a great new year ahead.....ho ho ho.....u.r the math Santa gifting us joy of math ..
    1
  6946. Thank you for making me feel smart. I like that. But I know it's illusory on my part, because not in a million years would I have thought to unroll the circumference of the bigger coin into a straight line and then count the rotations plus (or minus) the act of rolling the line back into the circle.
    1
  6947. Did you get my email(s) ?
    1
  6948. A sphere is a parabola with constant slope change, is that right? That's why the cone with its straight lines equals the sphere!? It only makes sense when I see the parabola and the "cylinder minus parabola" graphics. Please tell me if I'm wrong.
    1
  6949. 39:15 for every A^2 + B^2 = C^2 in that list, sqrt(C-B) is a Fibonacci number
    1
  6950. One of the best channels! Just good quality content that requires focus and thinking. 🙏
    1
  6951. AH yes brain food thank you!
    1
  6952. Loved this video
    1
  6953. I jumped to the fact that the perpendicular of the midpoint of any chord passes through the center of the circle to which it is a chord. Given that the three lengths are equal, the points must intersect at the center of the circle. However, I didn't proved the chord, midpoint, center point so voted "none of the above" for your question about preference. You, however, didn't rely on any other theorems so I guess I get points off.
    1
  6954. Aswome
    1
  6955. What is the music playing during those proof without words -parts?
    1
  6956. Finally, just what we needed. AW, I wish I had this in highschool, it really updates your mindset and philosophy.
    1
  6957. total area of the tree is 5, because there are 5 levels, and the area of each level is 1
    1
  6958. spin to win
    1
  6959. Most amazing part is that gamma. Infact I have been working on other summation formulas , that do gets a difference of 0.01 at 50000000. I don't know if it converges or not but , I to go in other ways to discover.😀
    1
  6960. Please prove 1+2+3+4+5+...=-1/12 based on calculus not based on algebra trick
    1
  6961. I did it in the other way, by connecting the corners with lines, and it still made crappy pentagrams!
    1
  6962. I like your shirt. I wonder what 999 would be?
    1
  6963. 28:50: Bonus fact: since F(n)×phi ≈ F(n+1), it follows that phi^n ≈ F(n+1)+F(n-1).
    1
  6964. F(n+1)+F(n-1) is L(n) which is approximately phi^n for big n. In fact L(n) = phi^n ± phi^(-n).
    1
  6965. I like all your lines best! ❤🐓🦅🪿🦃🐣🦆🦜🐦🦢🦉🐔
    1
  6966. Wonderful video as always! 23:38 Challenge 1: Because all the "attaching ears" operations preserve centroids. And centroids are additive, i.e. C(polygon #1 + polygon #2) = C(polygon #1)+C(polygon #2). Challenge 2: The "attaching ears" operations are commutative. After one round of each, we already have a regular pentagon. So the result will still be a regular pentagon after another round of 144 and two rounds of 216. 27:34 Is this a special case of Vandermonde matrix? 28:36 Proof using complex numbers: Since the two original triangles are similar, there exists a complex number z=(X(2)-Y(2))/((X(1)-Y(1)), where X and Y are any two colors among {Black, Brown, Red}. As a result, ([X(2)+X(1)]-[Y(2)+Y(1)])/((X(1)-Y(1))=z+1, thus the sum is still similar. One minute into this video, I was already thinking about complex numbers since I know the Napoleon's theorem. :P Then I tried to use the same method on the Petr-D-N theorem: Let the five vertices of the original pentagon be complex numbers u, v, w, x, y. First consider the operation of attaching 72° ears, the first new vertex becomes u' = u+ {e^(3πi/10)/[2cos(3π/10)]}(v-u) = (e^(-3πi/10)u+e^(3πi/10)v)/[2cos(3π/10)], a linear combination of u and v. Then comes the 144° ears, and the first new vertex becomes u'' = (e^(-πi/10)u'+e^(πi/10)v')/[2cos(π/10)] = some much more complicated linear combination of u, v and w. That's why I really appreciate the discrete Fourier transform point of view. Thank you!
    1
  6967. Full disclosure - the referenced work is mine, but this is not a mindless plug. Intimate Fibonacci/binomial coefficients/partitions connections with generalizations over 2 integer parameters are detailed with code in Sloane's OEIS A347187 and detailed in videos on the Feral Mathematician channel. This one is a summary from earlier videos https://youtu.be/zdhdiO4uUl0 The Fibonacci numbers are generalized to F(a,b,n) =F(a,b,n-a)+F(a,b,n-b) and shown to be the first 2 terms of the recurrence for the partitions function, similarly generalized and expressible in terms of binomial coefficients, which carries over to the sigma sum of divisors function. I have no doubt the work detailed in this video fits right into this emerging framework, just needs to be sussed out.
    1
  6968. I´m glad I only discovered your channel after retiring. Otherwise it would have severely affected my professional career
    1
  6969. It always amazes me when differential equations show up in something discrete (number theory?). Pythagoreans would be horrified.
    1
  6970. Best T-shirt Ever!
    1
  6971. Both proofs made sense but I liked the binary solution just a little more
    1
  6972. Is it allowed to fold the journal's page to cut the red stripes ?
    1
  6973. 34:44 What is the name of this particular definite integral?
    1
  6974. I like the decimal explanation. To show the number always decresing helps a lot.
    1
  6975. Is there a summed sequence of three consecutive cubed integers that equal the sum of two larger consecutive cubed integers?
    1
  6976. The TShirt 😁👍🏻 Some visual math
    1
  6977. Extremely useful feature of Breitling Navitimers
    1
  6978. Even without baby calculus, I have watched you enough, @Mathologer, to be able to keep up with how this works -- Danke sehr für das Video!
    1
  6979. The en-US subtitles are misaligned - they all come at the very start of the video.
    1
  6980. Can't take it in all at once?
    1
  6981. Am I the only one annoyed by the fact his shirt is missing a 27?
    1
  6982. 7:36 I've been looking into a video for Congrua for Months!! I'm looking for 3 arithmetic set with the same Congrua, that also hare offset from one another by a consistent amount
    1
  6983.  @Mathologer  finding the right set of numbers will allow one to make a 3x3 magic square with only square numbers.
    1
  6984. So 0 isn't a ring? It satisfies all the rules you stated. 0 + 0 = 0 0 * 0 = 0 0 * ( 0 + 0 ) = 0 * 0 + 0 * 0 or did I miss something?
    1
  6985. Why is it that we were not taught this in school? THE "ORIGIN" OF THE PYTHAGOREAN THEOREM - During Pythagoras’ trip to Kemit, c. 520 BCE, he noticed that the Kemites had a very interesting strategy to improve the stability of temple walls. They used a rope, with 12 knots tied evenly spaced, which resulted in the famous 3-4-5 triangle, forming a 90°angle...
    1
  6986. DDDDDAAAANNNGGIIIITTTT I wanted to make a video about this! This is something that I had explored myself like three years ago. I came up with many of the insights you presented here. My original motivation was trying to figure out how I can come up with a formula for the sum of squares (up to some point), which had a known formula and you could obviously prove it inductively, but how do get the formula to begin with right? Eventually I thought of the idea to take differences on top of differences... then do linear combinations of the difference sequences you see for 1, x, x^2, x^3, etc to get a formula for an arbitrary sequence (such as the sum of squares up to some term), which only works perfectly if it fizzles out to a constant row of course(or at least, if there is a finite number of interesting rows, you don't have to do any further analysis). There were problems I've come across where I knew the end result should be a polynomial and I needed to do polynomial interpolation, so this was a handy tool for that. At some point I also discovered you can use starting values/diffs of 0,0,0,0,...,1 as your basis instead which turned out to be (n choose h) for "height" h. This also gives you a handy way to do polynomial evaluation without using multiplication, and given this tool you can figure out how to come up with formulas for programs that involve loops where variables just keep adding to other variables in a non-cyclic manner. I will say though, that I probably wouldn't have made this video for quite a while longer(procrastinating...). Some things that I had thought to explore but didn't(because I got distracted by other things)... what if we have infinitely many rows of these difference sequences (which you touched upon). What about trying to interpolate in between the terms, between the rows? What about extending these difference sequences into the real and imaginary axes, can this lead to a reasonable interpretation of summations where the indices are complex numbers, at least for polynomials? What if our input sequence isn't a linear function? I gotta start making videos on my ideas before I get scooped as a result of my own laziness...
    1
  6987. Haha - Paranormal distribution.
    1
  6988. Please make a simmilar video about the discovery of the exponential and logarithmic functions. It would really make my day :)
    1
  6989. That's deeply fascinating ! As a software developer, I'm used to represent geometry as memory "objects" (objects that are reserved memory storing data and references to each others).. (for instance : meshs are used in every 3D application : list of vertices, edges and triangles to draw) so you would have to represent "edges", "volumes" and "vertices" as those different type of objects storing their relationship.. An edge has references to 2 points, a face is represented by a list of edges connected to each other.. etc.. So I think this formula comes down to the "rules", that in turns are given by topology.. for instance a face is given by a list of edges but they must be following each other AB->BC->CD etc.. DA (it's also a list of points).. a volume is given by a network of faces, but they must all share edges (there must be no holes to make a closed volume).. the general rule is that higher dimension object must be formed of lower dimension objects without holes in the structure. the fact that this formula encodes all of this is amazing..
    1
  6990. Your solution is more appealing to me,and was essentially how,once I'd figured out the question being asked,how I explained it to myself. The problem with Nathan's proof...and this occurred to me on first watching...was that he seemed to arbitrarily assign "1" to the back of a card and "0" to the front. This seemed counter-intuitive to me,until I realised that it was the only way his proof would work. Am I missing something here?
    1
  6991. This demonstrates, to my mind at least, that math is just another language that arises from reality. Math is a more rigorous language with very specific rules (grammar) but it is very much like any natural language in that it can express the truth just as well as it can express untruth and still obey the grammar. Indeed, nature can be expressed in any language but there is only one truth. The moral: do not take for granted the mathematical proof of nature's secrets because the math says so, even when observations are supported.
    1
  6992. Truly a tough decision, but I think the "no integers in the sequence of partial sums" result is most interesting
    1
  6993. All those 252s around 8:23 makes me think... what is the longest sequence of identical change numbers for a given set of coins?
    1
  6994. 13:30 if you speak of the fast fourier transform, i have tried that before, however its accuracy is limited to the precision of floating point numbers (24 binary digits, always starting with 1, multiplied by some power of 2)
    1
  6995. 14:52 ah yes this is practically a cheat! For any dollar amount product you can start with the product of many (1-x^amount) terms, and then determine the product of the bases (1-x^coin), and perform polynomial long division, which has a time complexity of N (N is the biggest exponent)
    1
  6996. Would you ever consider doing a video about the boustrophedon transform?
    1
  6997. niceeeeee hahahaaaaa
    1
  6998. I love math but I am always lost when it comes to trigonometry... Sin, Cos, and Tan just don't make any sense to me. I guess I've never seen it explained well, just "put this into a calculator and there's your answer."
    1
  6999. 13:18 What sound does a drowning mathematician make? Loglogloglogloglog.
    1
  7000. Nostalgia! The towers appeared as an example for a basic (or should I say BASIC) backtracking algorithm in one of the early books on AI that I read in the eighties...
    1
  7001. Cool. I will fire up Excel and feed in some sequences.
    1
  7002. it's like a bitwise logic gate but for polynomials
    1
  7003. QUESTION: Why does a hexagon disappear? Because..................... First it's there, and then it's .... hexa ..... GONE!
    1
  7004. 40 years ago, I studied math up to, but not including, gradients, divergences, and curls. I also was introduced to Pascal's triangle but not to the extend that you developed it here. I'm very intrigued by the theories of math that you present, but can't follow everything - especially on first exposure. Nevertheless, I enjoy your videos and get what I can out of them. I especially like the geometric presentations of algebraic functions. That really makes the algebraic formulas come alive and much, much more understandable. It is truly amazing how brilliant the old mathematicians were when the rest of the world was in the dark ages.
    1
  7005. Im glad to come across your channel. You have very informative videos. Good job!!! 👍i know we are now shofting to online classes, i myself as a teacher felt what you feel. I just launched my you tube channel for the purpose of making this online teaching easier. I have been making videos for my classes lately. We have 2 weeks left of school. Good luck to all of us. Stay safe
    1
  7006. Nice vid! I only caught the lie regarding the inscribed angle theorem you confessed to in 20:04
    1
  7007. 12:00 RRRL: {{1,1},{3,2}} > {{1,2},{5,3}} > {{1,3},{7,4}} > {{1,4},{5,9}} > {{9,4},{17,13}}
    1
  7008. mhm... why is my intuition telling me that we should have a function getting rational at 15°, 75° [...]? are we missing a function? or is there a reason for a difference in... uhm... gap size?
    1
  7009. 946/243
    1
  7010. Awesome 🤯🤯🤯
    1
  7011. Is either y or z in the 1+4yz instance always going to be a square itself? Because for 37 you can rearrange the 9-long arms into a 3x3 square to get the solution of interest.
    1
  7012. Most people are strong visual learners, but most educators teach in an auditory style. The next most popular learning style is kinesthetic. So it seams that we should be teaching geometry before we teach traditional math. The Greeks solved engineering problems using geometry and not a slide rule, calculator or computer. Hmmm, so why do young children that learn best by seeing and interacting not taught using their strongest learning style. Well it could be that educators have been so programmed to be auditory learners they have lost the habit of visual and kinesthetic learning and teaching. Oh and I was a high school teacher and was amazed at how we had destroyed the creativity that students are born with and enter school with.
    1
  7013. 4:30 I think because the two removed squares are not together the domino shaped tiles won't fit the whole board.
    1
  7014. So, it's impossible to draw a equilateral triangle on a 2d x/y monitor screen. Right?
    1
  7015. I had a dream proof for an assignment once, never felt as productive as when working in my sleep, plus I had gone to bed the night before being stuck and within 10 minutes of waking up everything worked perfectly
    1
  7016. The reason powers of 5 give the same remainders as powers of 2 mod 9, is because 5 is the inverse of 2 mod 9
    1
  7017. Fan of the swiveling proof. My immediate question after watching is “can this be extended to 3d with a tetrahedron?” So each face of the tetrahedron creates its own Conway circle. Do the resulting 24 points lie on a sphere?
    1
  7018. Now I wonder what’s your favorite number greater than 1 million. Will you please make a video about it?
    1
  7019. The fact that no 9s series is finite really surprised me at first. But the visual proof and the final algebraic proof made it seem so intuitive.
    1
  7020. What happens if you start with negative terms first, will they literally just go to the negated results of the other way around? I can see how with the under/overshooting you obviously can only get to -pi (unless ofc you cheat and just start with 0 negative terms) but I wonder if that will have a similar result as to the positive-first series goimg to pi
    1
  7021. I never made it past basic algebra in school largely because I hated my teachers. I'm now getting a total hang of Calculus and am trying to figure out matrices. Wish me luck.
    1
  7022. casting out the nines, works beautifully in fictional science fictions concerning Time travel well spent. thanks again, Sir.
    1
  7023. The most memorable part was the clever proof of the fact that the harmonic series is divergent.
    1
  7024. Hah, nice. I made the pixar connection early in this. This was basically a 2d example of interpolating the vertices of 3d shapes, which is how Pixar got such nice smoothed shapes, and how they manage to make organic/natural-looking shapes out of 3d primitives.
    1
  7025. Typo in comments: 4² + ... + 38² = 39² + ... + 38² should be 4² + ... + 38² = 39² + ... + 48².
    1
  7026. 8:00 wow you've hit upon two colours that are completely indistinguishable for me. If you did actually choose two colours... 🤔
    1
  7027. for how it fails trying to find the parent of the 3 4 5 triangle, I'm guessing the in-circle's midpoint is the same as the feuerbach circle's midpoint
    1
  7028. @Mathologer I thoroughly enjoyed this "shorter format" but I have to admit that I prefer the longer ones. Though it is unfortunately not possible for me to know if my "increased enjoyment" is proportional to the increased amount of work the longer videos take to make... So to make it simple, regardless of length. if You make them I will watch and enjoy :) Best regards.
    1
  7029.  @donaastor  I think your argument assumes that the hyperplane passes through the origin. One should take into account affine transformations to not enforce that. For example, all points definitely lie in a 4d hyperplane x1 + x2 +...+ x5 = 1 (where the "x_i" are the vector components). Perhaps this is the lowest-dimensional hyperplane that contains all the 5 points(?).
    1
  7030. Five-tastic! ;)
    1
  7031. γ, definitely
    1
  7032. The best videos are made by people who are professionals in the subject, not in making videos
    1
  7033. Your videos are great review and sometimes brand new stuff. I had seen this at some other point in time in history of mathematics, but my teacher must have skipped a few steps or I was half asleep.
    1
  7034. Nice job. My first thought after watching was stargates.
    1
  7035. Math gewd. I laik math.
    1
  7036. How to prove Heron's formula using sines and cosines is written in Wikipedia.
    1
  7037. for people who are too tired of figuring out the way to solve those questions start with the second part of the question and solve it from there. Solution Spoiler Alert!!! the first part was really easy because of the second part: z^2-x^2 = y this formula can also be written as : (z+x)(z-x) = y as a prime can only be divided by 1 and itself the only solution possible is z-x = 1 or z = x+1.
    1
  7038. Ha ha!
    1
  7039. I was listening to something occult called the stanzas of dyzan or something like that.. it melted me out because it talked about this same dance between 3 4 and 7...
    1
  7040. 24:18 If we take p(n) = n(3n-1)/2 The difference can be either p(-n)-p(n) or p(n+1)-p(-n) (where here n belongs to only positive integers) by substuting and further simplyfing using p(n) we get p(-n)-p(n)=n and p(n+1)-p(n)=2n+1 which prove that their will allternate between Natural numbers and odd Natural numbers
    1
  7041. So equalatterall triangles can exist on an infinite plane, with all of the arms being infinitely long?
    1
  7042. No more Dan Brown inspired math animations... PLEASE!
    1
  7043. In Italy we're taught Heron's formula in middle and high school and we're supposed to know it's demonstration, though it's used rarely, it's like a last resort formula
    1
  7044. everytime Gregory Newton introduces himself people say: are you family of Isaac?
    1
  7045. I looooove the visual proof of Heron! I remember setting my brain on fire with cosines when I tried proving it in school, and it felt so unsatisfying when I finally did it, the proof felt so ugly. This though, this is the definition of beautiful. Twelve year old me just reached catharsis ;-) As for why it isn’t usually taught, I guess, it is because it is impractical if it is not the last formula of the problem. Its huge size and that pesky square root make it a royal pain in the ass to solve equations involving it.
    1
  7046. Would it not be easier to say, 4k+1.. and.. 4k-1
    1
  7047. in the future can you try to cover some new research papers, to keep us updated with the latest research in mathematics?
    1
  7048. Would really love if your video would be more rigoros.. For example, I see a lot of videos about the intuition of fourier transformation, but none give me the real meaning and the real connection to fourier series.. (Pretty please)
    1
  7049. It is completely appropriate that the anagram is bogus specifically on Galileo's name.
    1
  7050. Not quite on topic here, but I've been wondering: why does the infinity symbol have a sign and not refer to all possible numbers? i.e. If an infinite sum f(x) to infinity is added to an infinite sum f(x) to negative infinity it causes the causes f(0) to be included in the set twice. Is there a different symbol to use to refer to all possible positive or negative values?
    1
  7051. How would someone solve this puzzle without knowing the solutions to the Pell equation? I solved the first quadratic and generated a table.
    1
  7052. Wow
    1
  7053. The most interesting thing was the no integers in the partial sums that was fascinating
    1
  7054. If you are still not curious after watching this amazing video, you are not a human being in any way.
    1
  7055. Can we please make the angel lacings illegal? I don't accept lacings which pass directly over/ under an eyelet but not through it.
    1
  7056. Most memorable: the crazy optimal tower.
    1
  7057. Are you not allowed to flip over 2 face up cards?
    1
  7058. my pure intuition guess is that the parent construction on 3-4-5 either produces 3-4-5 as well, or the center's coincide
    1
  7059. seeing it as a stack of cubes its easy to see that if you looked straight down you would excpect there to be a cube everywhere and no missing cubes bellow ones that you see, same for the other directions
    1
  7060. You only need 10 "essentially different" combinations of 4 for a simple test. AAAA = [RRRR, BBBB, YYYY] AAAB = [RRRB, RRRY, RBBB, RYYY, BRRR, BBBR, BBBY, BYYY, YRRR, YBBB, YYYR, YYYB] AABA = [RRBR, RRYR, RBRR, RYRR, BRBB, BBRB, BBYB, BYBB, YRYY, YBYY, YYRY, YYBY] AABB = [RRBB, RRYY, BBRR, BBYY, YYRR, YYBB] ABAB = [RBRB, RYRY, BRBR, BYBY, YRYR, YBYB] ABBA = [RBBR, RYYR, BRRB, BYYB, YRRY, YBBY] AABC = [RRBY, RRYB, RBYY, RYBB, BRYY, BBRY, BBYR, BYRR, YRBB, YBRR, YYRB, YYBR] ABAC = [RBRY, RBYB, RYRB, RYBY, BRBY, BRYR, BYBR, BYRY, YRYB, YRBR, YBYR, YBRB] ABCA = [RBYR, RYBR, BRYB, BYRB, YRBY, YBRY] ABBC = [RBBY, RYYB, BRRY, BYYR, YRRB, YBBR] 3+12+12+6+6+6+12+12+6+6=81, so every RBY combination is used.
    1
  7061.  @Mathologer   Alright! After checking everything, I've noticed that there are four pairs of triangles where two sides make up one pattern and the other side is a different pattern... and two triangles where all three sides have the same pattern. A B A B A B B A A A B B A B B C C C C C B C A C B C B A C C A A B A A C C A C B A A A A A B C A A A A C A B A A B C A A So, we end up with ABAB = AAAB ABBA = AABA AABB = AABC ABBC = ABAC Then AAAA and ABCA end up as their own triangles... making 6 essentially different triangle colourations. And then, 3[AAAA] + 18[AAAB] + 18[ABBA] + 18[AABB] + 18[ABBC] + 6[ABCA] = 81
    1
  7062. Great delivery of semi-complex information. Thanks
    1
  7063. My favourite is the sum of the reciprocals of primes being infinite. It is in The Book. The proof by contradiction, partitioning the primes into "small" and "large", and concludung that the number of natural numbers below X is eventually smaller than X as X grows is really beautiful.
    1
  7064. Thank you
    1
  7065. Great!!
    1
  7066. mathologer: here's the gregory-newton formula me: ooh ooh i know this one! maclaurin series!!!
    1
  7067. Shoutout to the one that named himself n/a n/a on Patreon :D Also, I'm curious about what order you named the Patreon supporters? I noticed they go alphabetically and start over a few times. Do you simply add all the new ones in their respective order?
    1
  7068. I thought it was the root of 6000/9 😁
    1
  7069. Somehow youtube doesn't allow me to edit my comments, so I'd like to add thismway that in Poland the word "professor" is written like "profesor", that why I wrote it like that. Well, after all that's already the first difference.... Greetings, A.
    1
  7070. Great video. The most amazing thing is that the root of pi divided by 2 can be represented as the sum of two numbers. An amazing result considering it is obtained from a normal distribution. Thank you, something to think about. Thanks again for the video :).
    1
  7071. Super stuff again, thank you for making this! 41:58 Is that a promise? Looking forward to it. Edit Yay it is!! 47:05
    1
  7072. This episode reminded me of a neat rotation that can be applied to the Fibonacci squares: https://www.youtube.com/shorts/3hhonbIAKUU - do you know if anything like that exists for the 3D boxes?
    1
  7073.  @Mathologer  Indeed :) De let us know if you stumble upon its 3D counterpart!
    1
  7074. Ich freue mich schon auf das deutsche Video
    1
  7075. What an amazing video💥💥!! You should make a video on Ramanujan pi series cause you simplify nicely all this complexity
    1
  7076. Nice day
    1
  7077. PhD Statistics, all worked well. Looking forward to see more of such videos!
    1
  7078. At 28:00 why should the i1,i2,i3 still be integers when iterating from one octagon to the next? Because each octagon should be again on the grid (by the shift property) with integer coordinates (at least for the x-coordinate)??
    1
  7079. Encore merci pour prendre le temps de sous titrer en français ❤️
    1
  7080. And 4D and higher!
    1
  7081. at 6:10​ you said 3.14 yet you set 3.18 on the slide rule! oops!
    1
  7082. 17:33 I wouldn't say I see it "at a glance", but here's my reasoning: we have a triangle of base B, side A and a 60 degree angle between sides A and B. Its height is: H = A * sin60 So, its area is (H*B)/2 = AB * sin60 / 2 The area of the equilateral triangle is: F = sin60 / 2 (just set A=B=1) So, the area of the AB-triangle is A*B*F.
    1
  7083. 16:58 A = b p sin(π/p) sqrt(1/b^2 + sin^4(π/(2 p))): if 1<= b <= |p| , A = π
    1
  7084. Engagement
    1
  7085. 21:41 Still looks dodgy. If you shift the denomination’s digits to the left, you get a product of fractions, all of which are less than one. If you shift it to the right, all fractions are more than one. Leaving as is, some are smaller, and some are bigger than one. (But if you multiply each pair, they will be smaller than one, except 2 at the beginning, I guess that it is here that it’s pi/2). Same trick works on the square root identity though. Shift digits to the left (or delete the useless 1) and each fraction is less than one. I wouldn’t be surprised if it turned out not to be square root of pi/2 there, though.
    1
  7086. Interesting. Not being a grammar nazi but but missed 2 and 1 highlight near the 10582 string.
    1
  7087. hell ya .
    1
  7088. Is Ptolemy’s inequality yet another Cauchy-Schwarz inequality in disguise? The fact it works in 3D makes me very suspicious…
    1
  7089. Another neat fact that was left out is that the difference between the largest number and the 2nd number is ALWAYS equal to the square of the first number in the box. Might as well include all the families in this family I found :) eg 505 - 456 = 49 (first number in the box is 7), 305 - 136 = 169 (first number in the box is 13)
    1
  7090. One thing I didn't see pointed out (sorry if I missed it) - if you look at the highlighted diagonals at, e.g. 14:25, but read them across, there's a pattern. The third row, for example, shows 10, 40, 90... That is, 10, 10*2^2, 10*3^2,... And in general every row r whose first highlighted digit (generated by Pascal's Triangle) is a_r (where the 0th row is all 1s) generates the sequence n^r * a_r for the nth highlighted number. Since the last row in any such diagonal will always have a 1 as the first highlighted number, naturally the sequence of the bottom row collapses to n^r. I don't think that counts as a proof (at least not without a lot of additional work) but thought it was an interesting angle. So to speak.
    1
  7091. لماذا لم تتحدث أبداً عن القطع المكافئ رغم أنه شكل مضاد للتحول
    1
  7092. Thinking that it "evolves" seems to imply some temporally limited reasoning. Simply grasp the entirety of its existence as a whole.
    1
  7093. Great video, but I did not understand why choosing x=-1 gives a series of expressions valid for all polyhedron. What is special about - 1, or any other value of x for that matter?
    1
  7094. A Koch snowflake is when an infinite one-dimensional space surrounds a finite two-dimensional space. In your video - an infinite two-dimensional space surrounds a finite three-dimensional one. It's like a dimensional rule. But onward. Then infinite three-dimensional space should surround finite four-dimensional space? Even more interesting is that it should be possible to draw a finite one-dimensional ring around the point. A point could have a perimeter!
    1
  7095. So this video was connecting the the topics of number theory, algebra, real analysis, geometry, infinite series... am I missing something else?
    1
  7096. ALKSnfakls nlaksngflkangf BEST VIDEO EVER. :D
    1
  7097. Oh boy do I like these videos. I could listen to this stuff all day long. Maybe what stroke me the most was the fact that there is no integer among the partial sums eventhough I also enjoyed pretty much everything else. :) Thank you Mathologer for this quality!
    1
  7098. When I saw a row and column removed from the 10x10 matrix, my intuition was that the diagram would get sparser for larger numbers, so Tristan's "fractal proof" made perfect sense.
    1
  7099. It's my party & I cry at if want to... At the joke. The proof gets me warm & fuzzy. Also I'm seeing a bit of Hopf in this video.
    1
  7100. All through this video I was reminded of a geometric proof of sin2(x)+cos2(x)=1. At 8:10 I started to realise why.
    1
  7101. I think this is one of your best videos! It was quite a drama filled with lots of "aha!" moments, but also making sure to watch out for any sneaky moves (knowing that the Wallis Product is "conditionally convergent" primed me for the big reveal that things weren't quite what they seemed with taking the square root of it). I would say that the fantastic fractions segment was my favorite part. :)
    1
  7102. Pi is a virus!!
    1
  7103. That one lie was infinitely more pleasing than the infinite lies of my ex...
    1
  7104. 2:07 from this point on you can clearly hear he's native German
    1
  7105. 🤯
    1
  7106. Is there a way for me to send u a math problem so u can try to solve it?
    1
  7107. make a video about the ant on a rope paradox!
    1
  7108. 6:52 When I first saw the formulas for the series I thought the mega pattern would have the top value (excluding the one with the zero’s) being something along the lines of (2^n(1)-1)^n+(2^n(1))^n=(2^n(1)+1)^n, with the following being (2^n(1+2)-1)^n+(2^n(1+2))^n=(2^n(1+2)+1)^n, (2^n(1+2+3)-1)^n+(2^n(1+2+3))^n=(2^n(1+2+3)+1)^n, and so on, but with a bit of thought I realised it didn’t work (though the video did get to it first). By that rule the first line of the cubes (excluding the zeros) should be (2^3(1)-1)^3+(2^3(1)^3=(2^3(1)+3)^3 in simpler formatting 7^3+8^3=9^3 which comes to 343+512=729 aka 855=729. Obviously not true, with LHS being 126 larger than RHS.
    1
  7109. Your presentation in this video is like you're giving a toddler a sweet, and I love it
    1
  7110. I got a degree in mechanical engineering in 1987 and this is the first time I’ve seen either proof. Why do I feel embarrassed?
    1
  7111. 55
    1
  7112. cycles and circles: if we look at the T-shirt and take the root of evil and compare it to a platonic year (earth biggest cycle) we get to the very prominent number 1131: 25 920 (platonic year) - 25 806.9 = 113.1 1131 ≈ 360*Pi or 377*3 leading to the approximation of Pi/3 ≈ 377/360 ≈ 1.0472 :-)
    1
  7113. Delightful proof!
    1
  7114. See 29:33: Here is proof of why the new rectangle is similar to the original. Sorry for the clumsy notation. I'm unclear as to how to use LaTex on youtube. Proof: It is a given that the dimensions of the original are sqrt(2) : 1 The new rectangle (the white rectangle) has dimensions of: 1-(sqrt(2)-1) : sqrt(2) -1 which we can simplify to: 2 - sqrt(2): sqrt(2) -1 If the two rectangles are similar then: org. length / org. height = new length / new height or sqrt(2)/1 = 2 - sqrt(2) / sqrt(2) -1 This is easy to prove. Multiply the left side by 1 -- BUT let's represent 1 as: sqrt(2) -1 / sqrt(2) -1. This gives both sides the same denominator! (sqrt(2) * (sqrt(2) -1)/ 1 * sqrt(2) -1 = 2 - sqrt(2) / sqrt(2) -1

2 - sqrt(2) / sqrt(2) -1 = 2 - sqrt(2) / sqrt(2) -1
 We have identity. So they are the same. We can repeat this proof for any new square down the chain.
    1
  7115. I'm at second year of mathematics in university. Everything in the video was crystal clear, except for the last chapter which was a little fast for me to follow.
    1
  7116. I electronics we use "j" instead of "I" for imaginary numbers to because "I" is current. You have to remember whether you are in a math class or electronics.
    1
  7117. Wow you're actually awesome dude... I envy anybody who had got to have you as a teacher. nobody has the right to expect the awesomeness that you provide as a teacher.
    1
  7118. Towers are so cool
    1
  7119. Great video! I'd like to say "As always" but somehow you made it even better than your usually very good videos. My background: I'm studying physics (fifth semester) in Germany, so I had three compulsory lectures on Analysis (I: real analysis including sequences, series, limits etc.; II: multi-variable analysis; III: differential equations and some functional analysis), one on linear algebra (with an introduction to group theory) and I heard an optional lecture on complex analysis.
    1
  7120. Digital rooting! 😂
    1
  7121. I guessed yellow but I guessed it because it was the most common colour. Kind of a win
    1
  7122. No offense, for whatever reason I really did not want to watch this video. It pissed me off every time I saw it in my recommended feed. I've ignored it purposely and avoided it, but god d*mn it, it won't stop recommending it. To get the f**** algorithm off my back I'm going to go ahead and watch this video so I don't have to see this stupid thumbnail on this topic that I have no interest in. That being said it was actually a good video and I learned a lot so I'm glad I watched it.
    1
  7123. Aw man, I used to love peg solitaire! [wanders off to find a set] …
    1
  7124. i have dyscalculia , the more numbers, the more they move around. i can really see them jump too. i still don't know why i subscribed, but i guess it is the theories and history... the accent and charisma you have must have a lot to do with it too. your attitude is great, i like accents.
    1
  7125. Nice T-shirt, but shouldn't it be 25.8070?
    1
  7126.  @Mathologer  I was curious as to your response. Indeed any number with "69" in it is surely better (funnier) than one without!
    1
  7127. is your intro music "so long, Frank Lloyd Wright"?
    1
  7128. I'm here and learning the correct way too. 😂🎉❤
    1
  7129. Where is 27 in i can't even
    1
  7130. Mathologer videos always hit the spot
    1
  7131. My favorite part was when you explained why the partial sums of the harmonic series are never integers unless you count the first term by itself as a partial sum. That was a great video. Thank you Mathologer.
    1
  7132. Omg i just realized,,,uve been posting in the exact same style for 8 yrs👏👏
    1
  7133. Fermat did briefly describe his proof of his sum of 2 squares theorem by infinite descent in a letter
    1
  7134. Je vais m’accrocher et repasser car fin est complexe Autant j’ai compris et fais la démonstration (proof in the beginning) mais =(s) don’t understood Yu and Marty (where is Marty?) yu are genius and very sympathique
    1
  7135. Me as Oeis distributor, I can say that this video is well made.
    1
  7136. (Most humbly) At 25:04, there is a typing error in the third term of the Bharhmagupta's formula.
    1
  7137. I took a Towers of Hanoi puzzle to school and showed it to each of my (high school math and physics) classes throughout the day. The most interesting question I asked was "Where is Hanoi"? I got only one correct answer in only one of my classes. I get the Vietnam war is ancient history!
    1
  7138. Rubik's cube question : after 45 times I had all the corners in place, and after 70 times I had all the edges in place. So the answer is the least common multiple of 45 and 70 i.e. 630.
    1
  7139. Hey Mathologer! Is it possible to do a follow-up on Conway's balance sheets, esp. how Conway perceived them (if possible) and how they are used to arrive at various impossibility arguments? Many thanks!
    1
  7140. Huh..? 64-4 isn't odd
    1
  7141. :face-red-heart-shape:
    1
  7142. the directions of the arrows on your shirt are killing me ;p
    1
  7143. To solve y'(x)=xy(x), I saw that this would have to be e to some power, and that the derivative of the exponent would be x. The integral of x is x^2/2, so the solution was e^((x^2)/2). I had to shift it down by 1 as is needed in the solution, so the solution is actually e^((x^2)/2)-1
    1
  7144. Everything in the video is wonderful. However, what I liked most was that every positve number can be the sum of a subsequence of the harmonic series.
    1
  7145. Oh very nice! I'm so glad this channel is a thing.
    1
  7146. For the second puzzle, I did not recognize it as some people did so I did the calculation in my head. The sum is of the form: (x-2)^2+(x-1)^x+x^2+(x+1)^2+(x+2)^2 it can be seen that if you expand the squares, all the terms linear in x cancel out, so you're left with 5x^2+2(1^2)+2(2^2) which simplifies to 5x^2+10, which, for x=12, evaluates to 730 which is twice 365.
    1
  7147. 20:26 Would you rather be a screwed up pentagon or a degenerate pentagon? What's the point?
    1
  7148. Nice!
    1
  7149. My favorite insight was the non-integeriness of the harmonic series :)
    1
  7150. I'll leave that to sink in over the next few days then watch it again!! It was like being back at Uni .. but far more interesting! The animations make a huge difference to the quality of the presentation ... like visual proofs. I stopped a few times and did some work with pen and paper but later on that seemed like a lot of work hence the repeat needed (and some more courage). Thank you so much .. clearly a huge amount of time and thought went into this. I doubt there is any better way to cover this material anywhere on this planet .. oh, and that last video was truly awesome - I have long pondered how to connect S3 with S1 and this is just beautiful. Wish I had that when I was a teacher. Finally I am, as ever, truly in awe of those "renaissance" mathematicians (Bernoulli, Euler, Gauss etc.) and their super powers as arithmetic calculators alongside godlike mathematical intuition.
    1
  7151. I made it to the very end!
    1
  7152. I just saw the secrets of Soduko square & my favorite game of Tetris shapes cool...nice vid
    1
  7153. That visualisation of the Euclidean algorithm was beautiful!
    1
  7154. (10² + 11² + 12² + 13² + 14²)/365 = 2*(10² + 11² + 12²)/365 = 2*(100 + 121 + 144)/365 = 2*365/365 = 2
    1
  7155. لقد رأيت مقالة lee sallows سابقاً ومن خلاله استلهمت بعض الأشياء عن الطي والقص التي لا أعرف إذا كانت معروفة مسبقاً أم لا، على سبيل المثال يمكن قص الرباعي الدائري من مركز الدائرة المارة برؤوسه إلى منتصفات أضلاعه للحصول على أربع رباعيات دائرية يمكن طيها بطريقة مختلفة للحصول على رباعي دائري جديد مختلف عن الأول ولكن كلاهما يمتلكان دوائر متطابقة، أيضاً يوجد طي وقص مشابه من حيث المبدأ من أجل الرباعي الذي أضلاعه تمس دائرة؛ هذا يجعل الرباعي الثنائي المركز مثير للاهتمام بشكل خاص
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  7156. 2
    1
  7157. Another month, another banger.
    1
  7158. I came up with something similar but opposite when I was 12. Start with the nth powers, take their differences n times, and you end up with the factorials. Amazing it was being discovered and proven in a more general form so close to the time I was discovering a similar pattern.
    1
  7159. Three bricks version can be solved with a bit of cheating: place first block so that a bit less than half of it hangs out of the cliff and place remaining two vertically on both ends of it. One of them will technically be off the cliff and the other works as counterweight.
    1
  7160. Such a pretty content omg !!!! Seriously the best sir love frkm india i'm nimisha 😊😘🙏🙏🙏🙏🙏🙏
    1
  7161. Could Moessners miracle contribute to more efficient algorithms for multiplication and exponentiation given the properties here?
    1
  7162. It’s calculus but dx=1
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  7163. much thanks for all your hard work 😊 I love math (latest class aced precalc and have been self study since then) .. even now I still love learning stretching my mind and vision ... being a math nerd/artist my focus is in patterns and symmetry with a bit of chaos with fractals .. did I understand all you presented? not quite but I did love seeing the inter connectivity ... the connections/flow between the formulas .... thank you again 😊😎🦉
    1
  7164. Mathologer has a theme of "a smart 8 year old could have noticed most of this" which makes me feel lazy
    1
  7165. I love the bishop's proof, though the no integers in the partial sums is pretty good too.
    1
  7166. In my high school slide rules were still used in the early 80s because scientific calculators cost over 100 USD (about $350 in 2022). They were out of reach for many families so they weren't allowed, but slide rules were only a few dollars. I got pretty good using one too.
    1
  7167. What about proving that any p of the form 4k+3 is not a sum of two squares?
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  7168. Thanks🙏 man for such amazing stuff for students like us
    1
  7169. Wringing beauty from the mundane. Simple, grade school math, that can even be done on a napkin, expanded and elevated.
    1
  7170. Amazing video. All chapters were great, but the most mind blowing for me was the no 9s fractal and the other crazy converging sums. Makes me wonder, if there were multiple universes with different laws of physics, would pure mathematics like this be the same in every universe? Perhaps someday would could prove/disprove that.
    1
  7171. Excellent! Many thanks for the content. :) I wonder if this miracle could be used for finding aperiodic tiling. Start with a hexagon tile and move the vertices in a repeating pattern, then try other polygons using Penrose for inspiration.
    1
  7172. Whoa! At 9:44 the Mathologer says, "Yes, twelve million and change." In my world, "and change" means any amount less than 25% of the next larger number. This is because in the United States, for the last 75 years, the largest circulating coin has been the 25¢ coin, so "and change" means less than this coin. Remember that ¢ = %. I would call that number "Almost thirteen million." (Note, I said "circulating". There is a 50¢ coin, and a dollar coin. But in my entire life I have never ever seen one of either in circulation.)
    1
  7173. I am afraid I have missed a part of the proof where you show that for each equivalence class defined by its "windmill silhouette" there is an exactly even amount of members. Because I can see that it can be possible for a windmill to somehow contain 3 different square-and-rectangles that correspond to it, yet it should not be true for a proof to work.
    1
  7174. In the early 1970s, The Ascent of Man, a thirteen-part BBC series written and hosted by Dr. Jacob Bronowski, had a similar proof. Not as elegant as yours.
    1
  7175. Master of Mathematics degree in 2013. I appreciate the the video a lot, because of its "deep and crazy" content. Other channels and pages out there only stay at more basic topics. So specific higher topics do never get covered. However, if you would publish more often this deep and crazy level it would change the character of Mathologer more to a lecture, I guess (which would not be good). Hence, a mixture of higher and lower difficulty videos is the perfect way, I think. I had to look the video twice and pause it here and there to understand all the Bernoulli and Pascal interactions completely and to better recognize the switching between finite and infinite in chapter 7 for example. The argument at 45:03 by which the aqua part is vanished seems very rough to me, right?
    1
  7176. At 14:05, we see that the circle’s diameter is longer ;-)
    1
  7177. my vote goes to : Terrible Aim and for other harmonic paradoxes (15:08), check Tree Gaps and Orchard Problems on Numberphile .
    1
  7178. Finally! We learn this formula in 7th or 8 grade, by heart, but we are never presented with a prof! I was complaining with my engineer Friends, and no one had any idea!
    1
  7179. That visual proof of d/dx sinx = cos x 🫰
    1
  7180. Thx
    1
  7181. Languages made of words cannot describe the beauty of the physical world. It is only through the language of mathematics that God can bring holiness into form. This made my day, thanks.
    1
  7182. 🂽
    1
  7183. the row at the end, the numbers are just every other number in the fibonnaci sequence yes?
    1
  7184. Why not Euro cents for the coin? As Brexit traumitsed Brit I am deeply saddened.
    1
  7185. Loved this episode! the music was a bit loud for me :-)
    1
  7186.  @Mathologer  right, no kidding, but can't wait for your insights
    1
  7187. wow that's such a weird proof
    1
  7188. 3:09 Proof that the shape defined as "the curve y=1/x stays the same curve when you both stretch vertically and squish horizontally by a factor s": A curve y=f(x) stretched vertically by a factor v is just y=v f(x), and squished horizontally by a factor h is just y=f(h x). With that, we can show that in our case, y=s f(s x) still lies on the curve defined by y=f(x), because with f(x)=1/x, y=s 1/(s x) = 1/x = f(x). So the transformed curve is the same as the original curve, making it resistant to shape-shifting, and any section of the curve will turn into another section still on the curve when applying the transformation. 9:00 So you have A(X) and A(Y), and want to stretch A(Y) so that it fits on the right edge of A(X). You know that A(Y) has a height of 1 on the left and A(X) has a height of 1/x on the right, so for them to fit together, you need to vertically stretch A(Y) by a factor of 1/X (aka vertically squish by a factor of X). The bottom edge of A(Y) originally had the length Y-1 (because the edge goes from 1 to Y), but because we want the new shape to fit the 1/x curve, we also need to stretch it horizontally by X after having squished it by that amount. So the bottom length of the new A(Y) will be X(Y-1) = XY - X. The resulting total shape after adding A(X) and the transformed A(Y) together will fill the curve until the point where A(X) ends plus the length of the transformed A(Y), so in total, X + XY - X = XY. Per definition of A, this area is A(XY). But we also know that this area is just A(X) + A(Y), because the transformed A(Y) has the same area as the original. 10:47 Proof that it works for all integers: If log(XY)=log(X)+log(Y) for all (X, Y), then 0=log(1)=log(X/X)=log(X) + log(1/X), and therefore, log(1/X)=-log(X), and by extension, log(X^-N)=-log(X^N)= -N log(X). For the exponent 0, you can just use log(X^0)=log(1)=0=0 log(X). For rational powers of X, so for the rule p/q log(X)=log(X^(p/q)), we can take log(X^p)=(q log(X^p))/q =(pq log(X))/q=p/q log(X^q). Now in that equation, substitute Y=X^q (which implies X=Y^1/q), and you get log(Y^p/q) = p/q log(Y). This equation must hold for all Y, because any Y can be expressed in terms of X using the earlier substitution rule. For irrational exponents, you can argue that the logarithm must be continuous on any compact interval, as 1/x is continuous as well, and because the rationals are dense in the reals, that shows that the rule must also be true for any real number.
    1
  7189. vote for chapter 6, especially sum of reciprocals of at least 100 0s > sum of reciprocals of no 9s
    1
  7190. Baskara was an ancient mathematician that I believe is credited with the proof (or at least a demonstration) of "Pythagorean theorem" using the four squares inside the larger square.
    1
  7191. Very nice
    1
  7192. Reminds me of my python-homework😁
    1
  7193. I didn't see this proof until college number theory class for Maths teachers! LOL
    1
  7194. Can you illustrate how the long division of square roots works? Why are some digits doubled?
    1
  7195. Gotta love it. CC starts off with “welcome to mythology”
    1
  7196. You have proved that prime = x^2 + 4yz has odd number of solutions when y =x, so the windmill is a cross. Sometimes the odd solution has a single windmill which is not a cross, for example 97 = 81 + 16 or 53 = 49 + 4.
    1
  7197. You should do a video on Dirichlet's prime number theorem! Recall that Euler's proof of the infinitude of the primes uses the pole of the Riemann zeta function to show that the sum of the reciprocals of the primes is infinite! Indeed, the sum of primes less than x is about log(log x) (growing much slower than the Harmonic series, and almost implies PNT! You need to refine the error term to get the true PNT). Using Dirichlet characters, the sum of primes less than x congruent to a mod m is about 1/vi(m) log(log x) which in turn almost implies Dirichlet's PNT. A lot of this ties into the stuff here on the harmonic series.
    1
  7198. brb gonna find 15 feet of shoelace and some knee high boots and lace them with the angel lacing
    1
  7199. My mind was blown so hard, I can see a mushroom cloud. Great stuff!!
    1
  7200. The most amazin part was that the no nine series is not infinite, but is also less that the hundred zeros sum, WOW! That made my day!
    1
  7201. Ironically, when the name Weierstrass was pronounced the right way, I had to listen twice to understand it.
    1
  7202. Sensational!
    1
  7203. I disagree on a fundamental level. You claim you can rearrange the series to reach any arbitrary sum. I claim you can't rearrange infinite series like that. By that I mean, it should be a well understood law of mathematic that rearranging infinite some is problematic and must be done properly if at all. In our exemple of +2 | -1 . IF you define the positive sum as being the sum of all finite numbers, then the negative sum will only contain half of the finite number. You cannot "rearrange" the second half of all the numbers. You will argue that "half of all finite number" is still an infinite amount of it, and you would be right about it. But here lies the difficult part to understand about infinities. They are a little mind bending like this... For exemple of a way you CAN rearrange the term properly, you can start at the end. for the negative turn at 1/(∞/2) + 1/(∞/2-1) + 1/(∞/2-2) and so on... until you eventually add non infinitesimal terms near the end such as ... 1/(∞/2-(∞/2-3)), 1/(∞/2-(∞/2-2)), and finally 1/(∞/2-(∞/2-1)) also known as 1/3,1/2 and 1/1 . If you do THAT rearrangement you should still reach ln(2).
    1
  7204. 5 seconds in I NEED THIS SHIRT
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  7205. I understood most of it my math background is teaching my self and watching YouTube videos I’m in 8th grade in the USA was homed school for 5th - 7th grade
    1
  7206. Congratulations on 100 videos!
    1
  7207. This will also work for solids that start as convex solids that are deformed in such a way that they become concave without changing their topology.
    1
  7208. I had an interesting train of thought resulting from this video: First: Man these people are really crazy saying this is the "key to the universe." Second: ...the math is beautiful, though. Maybe they're really just discovering that math is beautiful. Third: But is the fact that math is beautiful not exactly why it's so useful in physics, which IS the key to the universe? Fourth: Math is the key to the universe. This diagram is just a very, very small slice of that. Huh. Maybe they're not completely crazy, just mostly.
    1
  7209. You can write any odd number or multiple of 4 by difference of two squares. for example 9 - 1 = 8. Very recently in russia was task in olympiad for informatics about difference of squares. This olympiad was for russian school students. Regarding to windmill proof, it would be nice if you mention that we don't care about many possible y=z with different x. Because among odd number of solution there must be at least one without pair - which is what we need. There is probably easier way to prove uniqueness by windmill but may be not :(
    1
  7210. I've missed two things and they bug me, if someone could help please: -Why does every tiling can be achieved by the square dance (and if so, is it reversible)? -How do we know, that there's no way two different square dances result the same pattern?
    1
  7211. The part about the one-to-one outside that always warped space for me was the notion that it does not slip. That should mean a one-to-one map between points on the two bounding circles. Then somebody pointed out to me that the center of the outer circle has to travel around a circumference of a circle twice as big. But this still seems to suggest two points are getting mapped to one point at the interface. Is curvature making this possible? It all seems a bit “axiom of choice”-y.
    1
  7212. 11:10 Weird enough, the inner gaps of that proggression (7, 13, 19) are also a proggression of d=6. 36:35 The little triangles made by the intersection of big triangles are similar to the big triangles; that means that all of their sides are scaled by the same factor. The side of the inner square is equal to the hypotenuse of the big square minus both sides of the little triangles. The scale factor of the sides can be found from the hypotenuses: it's sqrt(1+1/4) for the big triangles and 1/2 for the little triangles, so sqrt(1+1/4)*s = 1/2 -> s=1/(2*sqrt(1+1/4))=1/(2*sqrt(5/4)); the side of the inner square would thus be sqrt(5/4) - (a+b)*(1/(2*sqrt(5/4))) = sqrt(5)/sqrt(4) - (1+1/2)sqrt(4)/2(sqrt(5)) = sqrt(5)/2 - (1+1/2)2/2(sqrt(5)) = 5/2(sqrt(5)) - 2(3/2)/2(sqrt(5)) =(5-3)/2(sqrt(5)) = 2/2(sqrt(5)) = 1/sqrt(5); squaring it for the area gives A = (1/sqrt(5))^2 = 1/(sqrt(5))^2 = 1/5 (= 0.2)
    1
  7213. I did what you did, but in retrospect, of course it's 1/5; there are four sets of triangles and four sets of the funky quadrilaterals that are missing a triangle. Together, that makes four squares, so the one in the middle makes five! I mean, I guess it's smart to not always trust your eyes, but in this case... it would have been just fine.
    1
  7214. Nice knowledge teacher.
    1
  7215. First difference of 4^4, typo
    1
  7216. By dropping terms you can get a finite sum, Is there a sequnce of patterns where sequence n is finite and sequence n+1 is infinite? IE, dropping even is infinte, keeping only every 10th is finite.
    1
  7217. High School Geometry 30+ years ago.
    1
  7218. neither
    1
  7219. Is there a connection to the Gömböc?
    1
  7220. oops, forgot to add :). calculations are great! I have two TVs, it happened, on one it's just a digital signal, on the other the antenna converts the digital to analog. and now I sometimes see the future :) . By the way, the simplest spiral can be obtained from the rule of the golden section, it is especially interesting to consider rabbits as the hypotenuse of the square, although the spiral turns out to be double :). marvelous are thy works, O Lord.
    1
  7221. In reality, wouldn't Gabriel's horn reach point where the fraction equals the Planck length, thus making any further fractions impossible?
    1
  7222. https://youtu.be/fw1kRz83Fj0?t=1234 you made a mistake you wrote (x-1)^0 = x
    1
  7223. 4!th
    1
  7224. I guessed the right answer with the wrong rule. Oops
    1
  7225. I went to high school in the 1960s and we had to use slide rules for physics problems. Everyone else had straight slide rules, but I had a circular slide rule — my father (an engineer) had always favored the circular ones. Two advantages: (1) you didn't have to figure out which direction to slide it, and (2) my 8" circular model had a circumference of over 25", compared to the 8" or 10" straight rules of my classmates, so it was more accurate. In addition (no pun intended) it had a log-log scale, so I could work with exponents, eg, raising pi to the power of e, as well as trig functions to work with sin, cos, tan, etc. I used that thing almost daily for a long time. I kept it for many years, then finally threw it away, which I now sorely regret.
    1
  7226. At 11:27 aren't you missing dots on the lower row corresponding to 15,15 and 19,19?
    1
  7227. woo
    1
  7228. Really calculus has changed for me forever
    1
  7229. Sure, calculus is easy - if you happen to be a genius with a PhD in mathematics.
    1
  7230. It's not easy for math professors either; nobody thinks that partial differential equations are particularly easy, and most of those equations are actually impossible to solve exactly.
    1
  7231.  @alvingao1131 yes! I really struggle with drawing the shapes free hand so I always developed techniques to construct them without using ruler or protractor
    1
  7232. Amazing
    1
  7233. So we find these patterns in two and three dimensions... What of higher dimensions? Are some dimensions special? Hmmm....
    1
  7234. if we use 2^n and one coin each of all the smaller power of two denominations then there are 2^(n+1) amounts we can make
    1
  7235. Solution to finding the square for the pythagorean triple at 11:45 (153^2+104^2=185^2): Take prime factors of 153 and 104/2 3, 3, 17 and 2, 2, 13 Test possible combinations of two factor for each until one fits the rules of the square. 9, 17 and 4, 13 result in 9, 4, 13, 17 9+4=13 13+4=17 9*13+4*17=117+68=185 Therefore the square is composed of 9, 4, 13, 17. Some extra facts I noticed while thinking about the problem: The bottom parts of the square must each be greater than the sum of the top parts, and therefore the factorizations of the two pythagorean triples being summed must be in increasing order respectively within the square; this means that there is only 1 order that must be tested for any combination of factorizations. Every triple must contain both an odd and an even number, where the odd corresponds to the left of the square, and the even corresponds to the right. The top numbers in the squares as you follow the tree can never decrease. The position in the tree for the solution to this triple, following the pattern for expanding the tree as shown in the video by this point, is: Start, right, right, right, left. EDIT: I realized that I accidentally answered the bonus question for the problem when resuming the video after posting this comment; hooray for my curiosity!
    1
  7236. Most memorable: the amazing fractal-like visualisation showing that the portion of no-9-numbers converge at zero. I didn't expect that! Despite I'm very familiar with harmonic series subsums and gamma
    1
  7237. Blue being twice as long as green is 1/2 which is part of the Fibonacci sequence! So beautiful
    1
  7238. Chapter 6 Mind blown
    1
  7239. Fantastic. Your obvious love of this material is infectious.
    1
  7240. oh so dividing by zero ALWAYS breaks maths even if its not in a division or multiplicative form
    1
  7241. The no 9s series is still a decmial-centric series, so I wonder if there's a general proof about the convergence of series of integers 1/X where X is constructed by constraining digits from base-N representations. This video makes it sound like for N=10, all series converge? Of course for N=2 it's rather trivial... I also wonder if there's anything special about the fact that all the converged values for the no-Xs series for base 10 hover around 20 or 21...
    1
  7242. video is good
    1
  7243. Absolutely adorable and charming video about Ptolemy's theorem by Zvezdelina Stankova on Numberphile. I especially watched this video because that one taught me about Ptolemy's theorem. So the teaching is happening :)
    1
  7244. If you have an n-dimensional cube of side x, made up of C = x * x * x * ... * x unit cubes, add an additional cube on every outside "face", an additional cube on every outside edge, etc. The remaining shape is an (x+2) * ... * (x+2) cube, with (x+2)^n unit cubes; at the same time, it consists of C + H + .... + F + E + V cubes. Or, using coordinates and combinatorics: you can define an n-dimensional cube with vertices at (±1, ..., ±1). Every edge corresponds to a pair of vertices that differs in only one coordinate, (±1, ..., ±1, ?, ±1, ..., ±1). Every face corresponds to four vertices that differ only in two coordinates, (..., ?, ..., ?, ...), and so on. For an m-dimensional "bit", we have nCm choices for placing the question marks, and 2^(n–m) choices of sign, showing immediately that there are 2^(n–m) nCm of them.
    1
  7245. It seems to me to arbitrarily punish the largest creditors, which would discourage bigger loans. I can't think of any reason given in the video to justify not sticking with the ratio of debt to estate. If my debt is 1/2 of the estate and another creditor is 1/3 of the estate, you should simply keep that same ratio. All creditors should remain equal in the absence of some extenuating mechanism to distinguish between them. If I am owed $300, my brother $200, and my sister $100, but my uncle's estate is only $300, the ratios should be maintained. I should get half, my brother one third, and my sister one sixth, regardless of the diminished size of the estate. Everyone should be equally made whole to the extent that they can be and share the loss equally to the extent that they must.
    1
  7246. So I guess May the Schwarz be with you?
    1
  7247. 11:46. Actually you also need a rule to determine whether the smallest disk moves clockwise or anticlockwise. OK it goes the same way each time, but is it always clockwise or always widdershins? The extra thing to memorise is that if there is an odd number of disks you move it to the target on the very first move. If there is an even number you move it to the other empty one. To remember which case is which just solve for any odd number: I tend to use 1. That rule is
    1
  7248. What a gem. Thanks!
    1
  7249. Wow it's generating functions
    1
  7250. Tristan fractal was great!
    1
  7251. Great, clear and interesting video, blew my mind! 50 minutes is too long? I say go longer :P
    1
  7252. Heart curve, yeah sure, that looks like a ... heart.
    1
  7253. Oooh it's that greedy algorithm! Not because I'm an IT-guy, but because it's just soooo interesting a connection...
    1
  7254. Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate. Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter. Complex numbers = vertical asymptote. Split-complex numbers = vertical tangent. Dual numbers = vertical line. These algebras can be simply thought of as tensors. Delanges sectrices can be thought of as opposites of vertical asymptotes. Ceva sectrices as opposites of vertical tangents, and Maclaurin sectrices as opposites of vertical lines. The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards. Ramanujan Infinite Sum (of the natural numbers): during a Big Crunch, the smaller, central black holes, not the dominating black holes, are about a twelfth of the total mass involved. Dark matter has its singularities pressed into existence, while baryonic matter is formed by its singularities. This also relates to 12 stacked surrounding universes that are similar to our own "observable universe" - an infinite number of stacked universes that bleed into each other and maintain an equilibrium of Big Bounce events. i to the i power: the "Big Bang mass", somewhat reminiscent of Swiss cheese, has dark matter flaking off, exerting a spin that mostly cancels out, leaving potential energy, and necessarily in a tangential fashion. This is closely related to what the natural logarithm of the imaginary number represents. Mediants are important to understanding the Big Crunch side of a Big Bounce event. Black holes have locked up, with these "particles" surrounding and pressuring each other. Black holes get flattened into unstable conformations that can be considered fractions, to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with dark matter being broken up, not added like the implosive, flattened "black hole shrapnel" of mediants. Ford circles relate to mediants. Tangential circles, tethered to a line. Sectrices: the families of curves deal with black holes and dark matter. (The Fibonacci spiral deals with how dark matter is degenerated/broken up, with supernovae, and forming black holes. The Golden spiral deals with black holes being flattened into dark matter during a Big Bounce event.) The Archimedean spiral deals with black holes and their spins before and after a reshuffling from cubic to the most dense arrangement, during a Big Crunch. The Dinostratus quadratrix deals with the dark matter being broken up by ripples of energy imparted by outer (of the central mass) black holes, allowing the dark matter to unstack, and the laminar flow of dark matter (the Inflationary Epoch) and dark matter itself being broken up by lingering black holes. Delanges sectrices (family of curves): dark matter has its "bubbles" force a rapid flaking off - the main driving force of the Big Bang. Ceva sectrices (family of curves): spun up dark matter breaks into primordial black holes and smaller, galactic-sized dark matter and other, typically thought of matter. Maclaurin sectrices (family of curves): dark matter gets slowed down, unstable, and broken up by black holes. Jimi Hendrix's "Little Wing". Little wing = Maclaurin sectrix. Butterflies = Ceva sectrix. Zebras = Dinostratus quadratrix. Moonbeams = Delanges sectrix. Jimi was experienced and "tricky". Jimi was commenting on dark matter. How it could be destabilized by being slowed down, spun up, broken up by lingering black holes, or flaked off. (The Delanges trisectrix also corresponds to stable atomic nuclei.) Dark matter, on the stellar scale, are broken up by supernovae. Our solar system was seeded with the heavier elements from a supernova. I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex (algebras) and I don't know if I have all the curves available to use in analyzing them. I have made some progress, but have more to discern. I can see Fibonacci spirals relating to the trisectrices. The Clausen function of order 2: black holes and rarified singularities are becoming more and more commonplace. Doyle's constant for the potential energy of a Big Bounce event: 21.892876 Also known as e to the (e + 1/e) power. At the eth root of e, the black holes are stacked as densely as possible. I suspect Ramanujan's Infinite Sum connects a reshuffling from the solution to the Basel problem and a transfer of mass to centralized black holes. Other than the relatively small amount of kinetic energy of black holes being flattened into dark matter, the only energy is potential energy, then: 1 (squared)/(e to the e power), dark matter singularities have formed and thus with the help of Ramanujan, again, create "bubbles", leading to the Big Bang part of the Big Bounce event. My constant is the chronological ratio of these events. This ratio applies to potential energy over kinetic energy just before a Big Bang event. Methods of arbitrary angle trisection: Neusis construction relates to how dark matter has its corner/cusp singularities create "bubbles", driving a Big Bang event. Repetitious bisection relates to dark matter spinning so violently that it breaks, leaving smaller dark matter, primordial black holes, and other more familiar matter, and to how black holes can orbit other black holes and then merge. It also relates to how dark matter can be slowed down. Belows method (similar to Sylvester's Link Fan) relates to black holes being locked up in a cubic arrangement just before a positional jostling fitting with Ramanujan's Infinite Sum. General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity? Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Corner/cusp and ring/cylinder types of singularities. Connection to Big Bounce theory? "Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?
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  7255. waaaaouhhhhh! I love you !!!
    1
  7256. Wait a sec, Marcus Persson donates to you? xD Nice! 28:45
    1
  7257. what if you take the other series that follow the fib rules but start with different integers? i know all sequences of that type produce the golden ratio at least
    1
  7258. I'm a huge nerd that lives in the hyper rigorous textbooks usually, but there's still something fun about just seeing things that are "just" kind of cool! There's always a gap between intuition and rigor, and here too. Knowing that you can find relations from ratios and divergence and all is cool on paper, but seeing the visual patterns that it creates is fun too. Thank you so much for all of the incredible content :)
    1
  7259. Watching this video I am convinced that it doesn’t worth for everything to have a pattern. Sometimes happy coincidences scattered around is more elegant
    1
  7260. As everyone knows, the universe is ruled by the numbers e, pi and 42.
    1
  7261. What if you alternate the running directions every cycle? 72* to the left, 144* to the right, etc.
    1
  7262. For odd n (say 7, a heptagon), this means that you use the permutation (1, -2, 3, -4, 5, -6). But that is just (1, 5, 3, 3, 5, 1). And that doesn't give you a regular heptagon yet. For even n (say 6, a hexagon), this means that you use the permutation (1, -2, 3, -4, 5), which is (1, 4, 3, 2, 5) and is just another permutation giving you the same regular n-gon after the 4th cycle (and giving you the single vertex center after the last cycle).
    1
  7263. An interesting factoid about volumes of n-dimensional spheres (of some fixed radius) is that their volume tends to... zero with the increasing n. It goes to zero so fast, in fact, that the sum of those volumes is a finite number! It's a good lesson in the meaning of volume in n-dimensional spaces.
    1
  7264. I think I managed to follow everything, despite not remembering what eigenvalues/eigenvectors or determinants were. I got confused by odd and even permutation terms in the determinant - not really sure what makes them odd or even, though I understand how it relates to the board we are trying to find the number of tilings for and twisting the 2x2 squares.
    1
  7265. Amigos. Aquí está el video subtitulado en español. Todo vuestro!! https://www.youtube.com/watch?v=nMi8tbXqiBk
    1
  7266. Glad to come across this podcasts thank you x I love maths
    1
  7267. I liked the bit about gamma the most
    1
  7268. I made it to the very end. And enjoyed it.
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  7269. Thank you so very much. This is already becoming one of my favourite mathematics videos on YouTube!
    1
  7270. The Seven Samurai? In grainy black and white with a CGI Toshiro Mifune 😊
    1
  7271. If you wanted to do this to a random quadrilateral to make a square, would that mean adding 90 degree ears, then 190 degree ears? What would 180 degree ears be - just take the midpoint of each side?
    1
  7272. Most memorable is that the exactly 100 zeros sum is larger than the no 9s. Too bad you didn't get to show how we know that.
    1
  7273. This channel is the jewel of youtube
    1
  7274. Gamma must be greater than 0.5 because the concavity of 1/x is positive. If you approximated the blue area of gamma with straight lines, the approximation would be strictly less than the actual area. The approximation with straight lines must equal 0.5, because it is made out of rectangles which are only divided in half.
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  7275. 10/19/22 @ 22:03 GMT +8, right here. Thank you.
    1
  7276. most memorable is gamma, and that e to the gamma formula with all the primes
    1
  7277. From the world of mechanics before stepper and servo motors : the limited continued fraction is the best combination of gears to achieve a given reduction (in the interpretation of "smallest diameter/number of teeth"), take PI or sqrt(2).
    1
  7278. INFINITELY-WIGGLY LINES?!!! Almost discovered fractals, eh? And was this the inspiration for Peano curves?
    1
  7279. clearly you are a free-lace mathematician
    1
  7280. holy crap 1 view and 8 comments
    1
  7281. As some one familiar in computer science, the 1, 2, 4, 8 question comes easy answer as all the binary representation of numbers from 1 to 15. Using the (1+x), (1+x^2), (1+x^4), (1+x^8), it can be seen by multiplying all with (1-x), we will get (1-x^16), and divide that of (1-x), we will get (1+x+x^2+x^3+...+x^15), aka the same answer. This is such a very interesting connection between polynomials and binary numbers!
    1
  7282. coloring
    1
  7283. Anyone know the music for Kempner's proof at the end?
    1
  7284. Exquisit selection for the intro song. 😎Added it to my Spotify liked songs.
    1
  7285. They did teach this in my high school....
    1
  7286. 37:32 Picture the whole hexagon as a cube. Then, we can easily see that the total area of the orange part = the blue and the white, by project them into their corresponding "faces" of the cube.
    1
  7287. Thank you so much for this video
    1
  7288. Hi, I really like your videos and, as a math noob, I would like to try to challenge your gift of explaining such stuff :-D Could you explain the n-body problem, why calculus fails and whats the nature of the exceptions where you actually can do it? (Last thing is what interessts me most. I understand the problem but cannot imagine how such a system has to be configured so calculus works out and can still do predictions)
    1
  7289. The best one IMO is Kempner's proof, but Tristan's is a close second :)
    1
  7290. Lol, you need to tell us where you find those t-shirts.
    1
  7291. But what was the internal volume of the can after crinkling? Did it increase, decrease, or stay the same?
    1
  7292. What? That can't be true. I played with this 40 years ago in school - you know, paper and scissors - and noone told me that this was something new. Shame on my math teachers ;-)
    1
  7293. If we are to measure our schools by the quality of our education, then U.S. taxpayers are due a massive refund--unless we apply Barnum's theorem: "there's a sucker born every minute", in which case we deserve exactly what we got. Regarding the origin, we attribute it to Pythagoras because that's our heritage. Other cultures may attribute it to someone else.
    1
  7294. It looks like at 17:14 your algebra auto-pilot mis-substituted the sum-term.
    1
  7295. 5
    1
  7296. I have a book on my desk, "Foundations of Multidimensional and Metric Data Structures." It makes heavy use of the correspondence between dimensions in geometry and data fields in a database. When you said the turning property and the translation property hold for 2 dimensions but only the translation property holds for 3 dimensions and all higher dimensions, I nearly fell out of my chair: there MUST be a correspondence between this fact and the analogy of database records with only 2 fields versus 3 or more fields, yet I'm not aware of a theory in computer science which makes any particular distinction between the properties of a schema with records having 2..to..N fields versus records having 3..to..N fields.
    1
  7297. I love math
    1
  7298.  @Mathologer  ikr
    1
  7299.  @Mathologer  I agree
    1
  7300. 18 is 3^2 + 3^2 with either, both, or neither of those threes changing to negative. 2020's odd factors are 1, 5, 101, and 505 and all those are good, so this indicates 16 ways. Let's see... 16^2 + 42^2, 24^2 + 38^2... Ah, wait! I can switch the order, so either of those give eight ways, making 16 total. I was worried for a while!
    1
  7301. Thanks Burkard: I see now. Appreciate you taking the time. 😊
    1
  7302. But it always will be good
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  7303. I remember the Numberphile video and I'm amazed that such a simpler proof is available now! Thanks for sharing it.
    1
  7304. Stopping the video and going downloading a peg solitaire app to get in touch... Be back soon!!
    1
  7305. Awesome video! I've taken 2 semesters of calculus, but unfortunately have no background in any of the matrix stuff. So I wish you went into a little bit of detail on how the inverse matrix computation is done, but other than that I was able to understand everything.
    1
  7306. The coin "paradox" was fun to think about. Since the coins work like gears (no slipping), the circular path around the stationary coin forces the moving coin to rotate once around the axis of the stationary coin, adding 1 rotation (clockwise for the inside coin and counter-clockwise for the oustide coin) to however many rotations the moving coin makes around itself.
    1
  7307. I got it, but only b/c Nathan told me how to do it.
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  7308. The most memorable part for me is... *drum roll*... the approximation algorithm ! I already stumbled upon the harmonic-approximation-where-you-can-choose-the-signs-in-front-of-the-reciprocals-algorithm but never this lovely one :-) Just two words to finish : thank you <3
    1
  7309. Just subscribed not far away from Louvain
    1
  7310. The monkey in the intro is so stylish in his red suit! However, in the name of gender equality we also demand monkey in dress.
    1
  7311. I always suspected that thare is a simple "geometrical" explanation of hyperbolic functions, now one of them is here. I need to watch more times, to internalize.
    1
  7312. gamma :)
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  7313. Paradox literally means something went wrong on human's understanding side. And I'm not surprised to see it's about 0 or infinity or imaginary numbers. All paradoxes are. These are not numbers but ideas we should work carefully with. Because infinity = 2 infinities. It's not infinity if it doesn't contain everything including all possible infinities. But if you take infinite rays and count them, you actually get real numbers. Like 5 infinite rays. It only works when you understand what you're doing and why. But you can never grab numbers from infinity to count with because obviously infinity = infinity + x. So you always get (x=0)&(x>0). That's the infinite source of all your paradoxes.
    1
  7314. I was familiar with the second representation, since it also is nice for (a+b) squared.
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  7315. Blessed...we r blessed
    1
  7316. Honestly; I don't see any "series" paradox. It has always been clear to me that the mathematical operations themselves are not precise, and that the so-called infinity does not exist, but is an error that has been introduced since the introduction of the decimal or any other number system. I maintain that there is only "1" ie. "whole".
    1
  7317. what about starting with a^3+b^3+c^3=d^3+e^3 in the cube one.
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  7318. Nice one 👍👋
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  7319. Don't pick me as a winner im from Europe! Great video
    1
  7320. I like gamma popping up here.
    1
  7321. 4:53 - Why should the base10 digital root be so interesting? I never like it when people use the digits of numbers... yet it seems to be so popular in all sorts of stuff... (mainly for fun stuff)
    1
  7322. for P!=NP, can we just prove the characteristic number wall has infinite rows?
    1
  7323. How to lace shoes🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
    1
  7324. Why did this never come up in any of my calculus or differential equations classes? I recognized pieces of this when you moved onto difference equations, as I dealt with those in 2nd Linear Algebra section. However this underlying mathematics would've connected so many dots for me so quickly. Thank you! Unfortunately, I've seemed to have forgotten more than I remember. At least I still can be amazed by such things!
    1
  7325. I don't think either of the two proofs at the start were taught to me in school, but I do remember stumbling across the second one while playing around with sketches in my notebook in maths class.
    1
  7326. The real question is: can you buckle a soda can in such a way that its surface area approaches infinity and then get rich by selling the scrap metal?
    1
  7327. I saw that 8 13 21 34 in the Fibonacci sequence and thought "wait, those U and V are both odd" giving Pythagorean triple 272-546-610. However, I realized this simplifies to 273-136-305, which comes from U=4 V=17, and that certainly follows the even-odd rule.
    1
  7328. Ok, check your normals ... gotcha. But let us stick with the intro example (you know the pi == 4 thingy) is there an intuitive explanation why the circumference does not converge while the geometric shape is?
    1
  7329. Charlie Brown feels might alone in his house.
    1
  7330. Have you heard of the “Armenian eternity sign”? It looks like this ֍
    1
  7331. As someone who dabbles in the hobby of making music, I owe a lot to the invention of logarithms. So much nuance in both the controls and the audio of a synthesizer can be brought out by making use of logarithmic and exponential functions rather than merely using boring linear ones.
    1
  7332. yes, that mod 9 stuff was taught at school, it was part of the proof that the test works.
    1
  7333. The natural logarithm ln(n) of natural numbers n ≠ 0, 1 are transcendental by Lindemann-Weierstrass theorem. Ln(2) is η(1) of Dirichlet eta function which is a zeta function. Maybe zeta functions evaluated at most natural numbers output transcendental numbers?
    1
  7334. Pretty obvious that you can add as many 1s as you want
    1
  7335. 2:40 i want that animation playing all day on my wall
    1
  7336. Thanks sir you are doing great and I always enjoy every of your video that really Interesting...
    1
  7337. I’m colorblind so guess which one I liked best… :-)
    1
  7338. Great shirt
    1
  7339. Saw a big nothing grinder with a router making a beautiful eclipse table a few months ago.
    1
  7340. Yeah there has been one as already mentioned in the video (the infinity - infinity = pi video).
    1
  7341. 7:26 (13°+14°)*2/365 ... = (169+196)*2/365 = 365*2/365 ...
    1
  7342. 3:18 it may be less spectacular, but it's also secretly somewhat closer to proving the more general law of cosines. The 2AB part kinda smells cosine-y already. Of course, those math books aren't usually gonna point that out either.
    1
  7343. Where in the Talmud is this?
    1
  7344. One interesting thing seems to come out... If the sum of harmonic series is infinite and sum of "no 9s" subset is finite then sum of 'at least one 9' subset is infinite :-)
    1
  7345. I also much prefer the coloring proof due to its more intuitive nature.
    1
  7346. For me, "crazy maximal overhang stacks" reinforces the need to not get trapped into just considering "patterned" solutions
    1
  7347. Nice 😎
    1
  7348. Most memorable part is that there are integers among the partial sums. The second place goes to the 100 zero sum being larger than the no 9 sum. Thanks for the video, amazing content!
    1
  7349. My favorite part was seeing the euler-mascheroni constant pop up.
    1
  7350. Most memorable for me was Tristan’s Fractal. Love seeing math in action.
    1
  7351. Can calculus be used to calculate risk on leveraged money?
    1
  7352. Merry Christmas to you!
    1
  7353. ❤❤❤
    1
  7354. Chapter 5 is my best Thank U Mathologer for these excellent videos!
    1
  7355.  @Mathologer  I understand it isn't necessary; I'm just very intrigued! :)
    1
  7356.  @Mathologer  Sweet.. thats what i had in mind. It will do the Job ..zumindest bei den Fremdwörtern und Fachausdrücken. Schönen Tag ans andere Ende der Welt !
    1
  7357. Is the formula for the number of moves that visits each possible state 3^(n-1) - 1? My reasoning is also by induction. Starting with the base case we have 1 disk and that is it, 3^0 - 1 = 0. If we have 2 disks we make the triangle diagram. Each state is now a vertex of the triangle. We need two moves to pass through all states. It is a little triangle. Now, see what happens if we consider 3 disks. We will have 3 of those little triangles in the new diagram. We can consider each little triangle as a "island", and to connect these "islands" we need 3 connections. But we are interested in direct connections. We pass through all the 3 states of the first triangle (2 moves), pass through a connection (+1 move) going to another little triangle, we pass through this little triangle (2 moves), pass through another connection (+1 move) and finally, the last little triangle (2 moves) We got 3*2+2 = 8 moves. If we add another disk it work the same way: we triplicate our 3 disk triangle (making 3 islands) and add two more connections to connect the "islands". For 4 disks it should be 3*8+2=26 moves. For five disks we triplicate the 4 disk triangle and add 2 connections to connect the "islands". In general we should have M(n)=3*M(n-1)+2 By induction, if 3^(n-1)-1 is correct we should expect: 3*[3^(n-1) - 1] + 2 = 3^n - 3 + 2 = 3^n - 1 Showing that our reasoning was correct
    1
  7358. While it's a pretty fact, it's not that new. Maybe the folding aspect is new, but the fact that from segments equal and parallel to medians of a triangle another triangle can be constructed is very well known Also cool fact about the construction. You will get the similar triangle if the centroid coincides with one of Humpty points of the triangle (a projection of orthocenter onto one of medians)
    1
  7359. I think I know an even prettier proof of Euler's polyhedra formula. Take an arbitrary convex polyhedra and imagine that it is a planet floating in space and put a small dimple at the center of each face. Now introduce some water by crashing a few comets into the planet, turning those dimples into lakes. The planet now has F lakes (one for each face) on a single (1) land mass. The next step is to continue smashing comets into the planet, gradually increasing the water level. The water level will rise as if it were a sphere growing out from the center of the planet. The lake closest to the center will grow first right up until it encroaches on an edge. At that point, the water will spill over and join with the lake on the adjoining edge. Two lakes become one. This process of fusion will repeat multiple times at each edge and at some point a lake will grow so large that it wraps around the planet, separated only by a single edge. When the water level rises above that edge, the lake fuses with ITSELF, forming an ocean that splits the landmass in two. Lake fusions and ocean formations continue to happen until every edge has been partially covered by water. Now the planet has a single (1) ocean and V islands, one for each vertex. There must have been (F-1) lake mergers to take the planet from F lakes to 1 ocean. And there must have been (V-1) ocean / island formations to take the planet from a 1 landmass to V islands. And every one of these events happened at a unique edge. Therefore (F-1) + (V-1) = E, or F-E+V=2.
    1
  7360.  @Mathologer  I didn't say that. :-) I didn't downvote it or anything. I've just seen many videos, some from you, about how rearranging terms in an infinite series change the value of the total. I was wondering when I started what the "A" would be, only to be mildly disappointed that I already knew that trick. :-)
    1
  7361. When I saw the lenght of the video I thought I would watch a bit of it now and the rest later... I could not stop watching😃 It was that good! It’s hard to choose a favorite part because I really enjoyed the whole thing. But, the number of terms needed to reach a partial sum of 100 just blew me away! Having said that, compared to infinity...it is almost zero😂 I love your channel, always excellent content and high quality animation that makes things so much easier to understand.👌👍
    1
  7362. You finally had me on the 37 explaintion. Got it.
    1
  7363. Thanks for the video. A Christmas present for you. In binomial decomposition two in power 1 and 2 kernel between units-conditionally 1 (Prime number) and 2 (Prime number). And then the core (sorry slang) - the numbers are not simple. Happy new year and merry Christmas.
    1
  7364. What I took away from this was that lasers were around 150 years ago.
    1
  7365. 40:25 The final Fibonacci identity has a really nice visual proof when you interpret Fibonacci numbers as counting the number of ways to tile a 1xn rectangle with 1x1 or 1x2 blocks. (Theres some indexing differences, so F_(2n+3) would count ways to til a (2n+2) rectangle, and the subcases on the left are (n-1) & (n+1) rectangles / n & (n+ 2) rectangles.)
    1
  7366. 25,8069 is not the root of evil. 25,8069^2 equals 665,99608761. These days the root of evil has a different name: Putin. Let's see you compute that
    1
  7367. 2:01 Here you align a stretched out series of negative fractions with a more compact series of positive fractions. What could possibly go wrong? ;)
    1
  7368. well explaiend thankyou
    1
  7369. So what if the last card on the right (ones place) is face down, therefore there is no card to the right
    1
  7370. Ich bin in der zehnten und liebe Mathe. Ich finde ihre Videos großartig und wirklich sehr interessant. Ich bin Ihnen unglaublich dankbar für alles was sie mir beigebracht haben. Danke!
    1
  7371. oui !!!!! Thales et les mesures algébriques conjugables !!!!!!!
    1
  7372. Hmm so for the circle example the reason u get \pi=4 is because u are using a L1 metric to measure distances. So can u show that these normals, spikes as u call them, are just a characterization of different metrics in 3D (hope fully in N-dim space as well). The reason its important to look at it from the lens of metrics is cause metrics are needed to define integrals and derivatives. You cannot have limits without a metric. So a metric is the most fundamental quantity. However the criterion for alignment of spikes of the approximation and the true curve requires u to define the normals to both of them. Since u use straight line approximations, its easy to define its spikes using geometry alone, but its not so easy to define it (compute it) for the true curve if u don't use derivative (which depends on defining derivatives). My point is that using spikes is sort of circular reasoning. Using metrics is however not. If you can define a metric and stick to that metric, then at least you have established consistency. Metric spaces are fundamental after all.
    1
  7373. The L1 metric (taxicab metric) is an interesting thing to bring up - I’ve thought about it too, as part of preparation of this video. Since (R³,||.||₁) is not an inner product space, there is no natural way to define “normal”. Also, the set of isometries (or rigid motions) of this space is a lot less interesting than for (R³,||.||₂) - basically only translations, reflections and rotations that preserve the “grid” - no arbitrary rotations. Due to both of these it is a lot harder to get a sense of what the area of an arbitrary parallelogram is - you can’t compare it to a length x width rectangle in the same plane because right angles rely on inner products, and you can’t compare it to a rectangle in the coordinate plane because there is no guarantee of an isometry that carries the parallelogram into a congruent parallelogram in the xy-plane. Without this fundamental idea settled, measuring surface area of a curved surface (using some sort of integral of areas of infinitesimal parallelograms) has a hard time getting started. I think the only natural way to measure surface area in this space is via the “how much paint” method discussed in some recent mathologer videos in normal Euclidean space. We can define volume in (R³,||.||₁) geometrically, and for a surface S, the set T of points within ε of S is a volume we can measure. So consider the limit of T/(2ε) as ε -> 0. I haven’t investigated this myself beyond this, and I don’t know if anyone has.
    1
  7374. That last bit tho. There are only two ways, and those ways are the same. XD I love math.
    1
  7375. Excelent !! Congratulations. Amazing explanation. Best regards 👏👏👏
    1
  7376. The real answer is zero. Well before there is a googol dollar in circulation, even on paper, the very concept of money would have become so dilute as to be useless. The only alternative would have been. At some point long before that, to devalue the currency to be more in line with the actual global GDP. The race would either never end, or would end well before the finish line is reached.
    1
  7377. No integers is a fun thought on our way to infinity!
    1
  7378. Great vid! Question at the end: Everything worked. Education is MSc in a Computational Science and Engineering
    1
  7379. we studied that heron formula existed in school didn't use much but we used "p" instead of "S" S = sqrt(p(p-a)(p-b)(p-c))
    1
  7380. I am in the similar situation when I heard for the first time for that Tesla quote . I am very , very familiar with Tesla life and work but never heard or read he ever said that. I am ready to believe it is another fake myth about Tesla invented recently.
    1
  7381. For 1 moment I thought he was Sandeep Maheshwari
    1
  7382. I actually found this myself a few years ago (based on Napoleon), really interesting theorem!
    1
  7383. Beautiful visualizations!
    1
  7384. I am from Cambodia , I will try to learn more about this formula .
    1
  7385. harder than knowing what a derivative is, is calculating a derivative
    1
  7386. I am super proud to see my name at the end of the video
    1
  7387. Should anyone know where to get this shirt - please tell us!
    1
  7388. In the 3D grid the number of squares is:( (4 squares with side of 1) + (1 square with side of 2) + (1 square with side line of root 2))*(the number 2D (3by3) grid(9)) = 6*9 = 54 that was easy but there is more... There is 8 more squares with side of root 5 and there is 6 more squares with side of root 3 so... 9 * 6 + 8 + 6 = 68 squares👏🏽🤗
    1
  7389. Third challenge: Each 2xN board can fit a board of size 2x(N-1) in two different ways; additionally, if the board's edges are tiled with horizontal dominoes, then the remaining squares in the board form a 2x(N-4) rectangle. From what I can tell, these make up all the possible ways to tile the board (first situation where one of the edges has a horizontal domino and second situation where neither do), so if P(N) is the number of ways to tile a 2xN board with dominoes, P(N)=2P(N-1)+P(N-4) I think? An alternative approach is to consider that the board becomes impossible to tile if you place 2 horizontal dominos in a Z or S shape, meaning you can only either place them in an O shape or place vertical dominos (this is fairly easy to see); it probably wouldn't be too difficult to calculate the exact number of possibilities from this.
    1
  7390. Cool
    1
  7391. This really struck a chord with me. I tried solving a very similar problem a few years ago about reimbursing partners in a partnership business who put in varying amounts of work. I felt there should be a better way than just splitting it proportionally (based on the available income from the business) but I couldn't quite get my head around it. I even asked on the math stack exchange. I'm curious now what would happen if the water wasn't filling rectangular spaces but triangular (or conical in 3D) ones, and what difference would the orientation of the triangles make? Point up or point down? If all triangles were the same height but the base was proportional to the claim on the money, would this be interesting or even possible? Beyond my skills I'm afraid.
    1
  7392. Despite the astonishing leaning tower of lire, my favorite is still Nicole Oresme's proof of the infinity of the harmonic series.
    1
  7393. hello. The Bogdanov-Belsky puzzle is known in our country, many people know the answer)
    1
  7394. This reminds me of the fact that the Babylonians used Pythagorean triples extensively. Could we take advantage of this way of generating triples in some sort of engineering calculations based on their mathematics? I need to think about this…
    1
  7395. My favorite part was the gamma fitting into the unit square. The blue parts are approximately triangles, which divide the area in half. That's why gamma is larger than 0.5.
    1
  7396. most memorable to me is that none of the partial sums is an integer. they have to get infinitely close to integer since the next term in the next partial sum gets infinitely small...
    1
  7397. This is insine
    1
  7398. Infinite sum of 1/n² = you're slow. Infinite sum of 1/n = atleast I'm big way BIG... Best part is that no 9s....how someone first got that in mind?😂
    1
  7399. "Why did they not teach you any of this in school?" It's not on the standardized tests.
    1
  7400. I hope if we ever meet aliens, we're better at maths and can show them things like this that will blow their minds :D
    1
  7401. Which polygon hides 2*2=2+2? And why 2*2=2+1 is not true? I feel cheated by the only example I knew.
    1
  7402. In my case thr rubber band snapped.
    1
  7403. I wondered about 3^n or e^n. I know I can just write them as 2^nlog3 or 2^nloge. Then I just plug in. The problem? Our n is no longer at integer intervals so we can't use our GN formula.
    1
  7404. 4:00 "Also there is more, much more! And now that I've announced that there is more, can you guess what the more is? Can you see where we are heading?" Bro I don't even know where we're at; I was lost the moment you said: 'Welcome to another Mathologer video.'
    1
  7405. Всё остальное европейцы тоже украли у индийцев, китайцев и арабов.
    1
  7406. 1) Challenge: It is not possible because you can isolate a square at a corner. 2) Challenge: When you evaluate the product at the highest possible values for j and k you get half of the value of m or n plus 0,5 in the numerator in the cosine and m or n plus 1 in the denomenator. Since for example m is twice m/2 and 1 is twice 0,5 you get cos(pi/2) = 0 for both summands. Squaring the cosine and multplying by four does not matter here and this 0, when m and n evaluated at their highest possible values, eliminates all other factors. 3) Challenge: You get the Fibonacci-Sequence :) So for a two by n rectangle you get f_(n+1) possible ways t arrange the dominos. f_(m) is the m'th Fibonacci-Number. I start at n+1 because the first two numbers of the Fibonacci-Sequence are 1 and 1. 4) Challenge: After typing a 22×22 matrix in my calculator and taking the absolute value of the determinate I get the number 306, but I do not know what to do with this number 😂 Maybe subtract some multiple of two because of the holes. :)
    1
  7407. About 20 sec , top of head: (12-2)^2+..+(12+2)^2=5×12^2+2×(1^2+2^2)=720+10=730, so answer = 2
    1
  7408. I don't know why Ptolemy's theorem (and other theorems in geometry) isn't widely taught in schools, but it's definitely taught at university and Pädagogische Hochschule to teachers in training. So, who knows, maybe some younger teacher might feel like teaching Ptolemy's theorem in middle school or high school. The issue is that geometry has fallen out of fashion. These days there is a lot more of statistic and probability theory in schools, but less geometry and analytical geometry. On the other hand, The concepts of Calculus are taught more thoroughly, while back in my days we only learned to calculate derivatives and integrals of real functions.
    1
  7409. Nice!
    1
  7410. Why do you call those shapes moons, and not bananas?
    1
  7411. big brain
    1
  7412. No integers among the partial sums!
    1
  7413. Best video ever
    1
  7414. Burkard made some comment on this in the “glass half full” section, which is maybe a bit about the psychology of the outcome, which is rarely linear. It can also be seen as a wager on the part of a lender - betting that the borrower will be able to pay some proportion back before they die. If the terms of a loan (including how an estate deals with it) are known at the point of making the loan, then it is up to the lender and their actuary to determine a fair price on the debt. But as Rob mentioned, this isn’t what the video is about. If there is the weird nonlinear way of resolving such problems, how does it extend for n>2?
    1
  7415. Thank you for being a voice of rationality in this world.
    1
  7416. I haven't been for a while on Mathologger, my bad, the content never ceaze to amaze me, this episode was awesome, thanks you 🤠
    1
  7417. In those circumstances... close enough is certainly fun enough. Brilliant video. Thank you.
    1
  7418. Sir, make a video how to find the value of sin20°, cos20°, cos10°, sin10° in infinite surds and by cubic equation
    1
  7419. I think the biggest issue was that when writing these proofs Indian mathematicians weren't using numbers the way we use them. The were writing in words where alphabets represented thise numbers and words encoded the numerical patterns. So the biggest thing to note was that this was not a formal way to express mathematical proofs as showcased by Western mathematicians, and since it wasn't formally learnt by Englishmen who went to conquer the world and colonize them, this way of doing math got lost in time.
    1
  7420. Absolutely magical 25 mins !! Worth investing !!! :)
    1
  7421. I have completed the first two semesters of calculus not to mention taking a specific course in trig and one in geometry, and this is my first time seeing this. Perhaps maybe I was sleeping when it was shown to me....either way I don't remember ever seeing this before.
    1
  7422. slightly related to Euler's Polygon formula is this little bit of 2D trickery I derived for the angles of regular polygons [minus for interior, plus for exterior] ( 1 ± 2 / E ) Pi = Angles at vertices ( E ± 2) Pi = Total Angle from vertices And the obvious (E * 2Pi ) = total of all vertices you can get some interesting values and swapping below E = 2 if you chart it out
    1
  7423. Oil and Macaroni constant
    1
  7424. And A=0 B=0 C=0 A=1 to 9 B=1 to 9 C=1 to 9 If A<>B and B<>C and C<>A then A+B+C=X Next Next Next ....... The triple 6 of revelations is X.
    1
  7425. Most memorable has got to be the no 9s sum and Kempners proof. Mind blown.
    1
  7426. This is so cool, holy heck!
    1
  7427. Great video, as always! To me, the best part was chapter 5: I think I understood why the Euler-Mascheroni constant is bigger than 1/2: You can approximate it's value (lower bound) as the sum of the areas of the upper triangles: 1/2 ( 1/n - 1/(n+1) ). This is a telescopic sum, so we end up with 1/2. (Thought about using a exclamation mark here but, in this comment section, it might be misunderstood as 𝛤(3/2))
    1
  7428. Time for maths!
    1
  7429. Given an n-gon, you can build an nk-gon in several ways, such as by stuttering AABBCC or by repeating ABCABC. These must interact nicely with the nk-gon version of the theorem.
    1
  7430. I may not be no mathmatolager but logs sounds lot like the Vernier scale to me when put in circular motion. idk.
    1
  7431. I'm still figuring out why e and pi are transcendental. This is my next video in the list.
    1
  7432. Wow
    1
  7433. The power of space time itself. ^.^
    1
  7434. = ban notation [2,2,n] for all n
    1
  7435. 2↑2
    1
  7436. 7:38 the colored one is exactly the proof that i was able to get when paused at 2:12. I like this proof more as it shows, that for triangle with sides a, b and c, if we split a in two segments by the touching point of inscribed circle, then difference of these segments is equal to difference of remaining sides of the triangle.
    1
  7437. If it approaches ln(2) then shouldn’t it be shown as ~ln(2)? v =ln(2)?
    1
  7438. Now do it in base 12...
    1
  7439. The optimal 20 book stacking reminds me of a sniper gun pointing to the right.
    1
  7440. I'm a bit of a math nerd, but the best part of the video is still the music! It's so good!
    1
  7441. Why the doubling on the rightmost product and on none of the others? That seems a bit handwaivy, and unjustified
    1
  7442.  @Mathologer  thank you
    1
  7443.  @Mathologer  great video by the way
    1
  7444. The song at the end is an absolute BANGA!!! 🤘 I need your opinion though. Do you think anti-derivative should be called "integrative"? Are you okay with the name "anti-derivative"?
    1
  7445. 14 is next lol, hard to ignore the elephant in the room
    1
  7446. I would simply invoke the second law of thermodynamics and the heat death of the universe to argue that all sequences in all contexts terminate. Q.E.D.
    1
  7447. Since Fourrier transforms can be use to compress sound data (as in MP3), can this be used to compress mesh (set of points and triangles in 3D used to render CGI and video games) data ?
    1
  7448. Magic of mathematics!! Never ceases to amaze!! 👏👏👏
    1
  7449. This feels like all those math lectures that really made me realize how amazing Mathematics is. Thoroughly enjoyed this, and such a cool result!!
    1
  7450. Pretty sure I was not taught either proof.
    1
  7451. OK, time for a stupid question. How do we know the "correct" digits of Pi? What if other formulas have wrong digits peppered all over the place like that?
    1
  7452. I can't get money back from my school 😞
    1
  7453. Not familiar with either proof.
    1
  7454. My favourite part is every bit of this video! Except for the -1/12
    1
  7455. 10:40 Am I getting Ackermann function vibes here?
    1
  7456. 3rd year math undergrad at ETH Zürich (focussing mostly on topology, so pretty much not related to this) - could follow everything clearly, except that I found the example at the end with the Basel problem a bit hard to follow. Too much shuffling around too quickly. I think I would’ve liked another easier example first to get used to the formula.
    1
  7457. 7:00 - It would be 5³ + 6³ (341) ~= 7³ (343). The difference is 2 as each 6² slice of 6-cube covers 1 face, 2 sides and 1 corner of the 5-cube. Total 6 faces, 12 sides and 6 corners are covered so it falls short of making a 7-cube by 2 corners (2 * 1³). Next equation would be 16³ + 17³ + 18³ ~= 19³ + 20³. This time the difference is 18 (2 * (1³ + 2³)). 7:25 - It's obvious now. Another way for someone who has not seen this video to solve is to substitute 12 with x. Then the expression becomes (x-2)²+(x-1)²+x²+(x+1)²+(x+2)² / 365. (a+b)²+(a-b)² = 2(a²+b²). So the expression becomes 2(x²+2²)+2(x²+1²)+x² / 365 = 5x² + 10 / 365. Replacing x by 12, it's 5(144)+10 / 365 = 730/365 = 2.
    1
  7458. (IMO) your vids are top quality the Maths channel is the best channel to choose for a idea, Remember: Its just My own Opinion on the suggestion, Advice; "Try getting used to making a opinion on a topic youre interested in works for *me works for anybody".
    1
  7459. Is the value of PI exactly half way around the circle? A circle is 2PI radians.
    1
  7460. 29:50 I do not understand why the blue part can not be zero? Why is the argument now different from the red part earlier?
    1
  7461. Amazing
    1
  7462. Noice
    1
  7463. Colouring is for kids, and my partner told me she loves swivelling 😉
    1
  7464. <3 <3 <3. You did a great job on this, well done! By the way, that book you mentioned is *highly recommended*! My kind of math books, profuse with history. (And not typeset with Computer Modern :-)) )
    1
  7465. Chapter 4: Terrible aim was my favourite.... it was the most surprising given that it is an infinite series and it tied to stuff you've done before about different levels of infinity (integers, reals etc).
    1
  7466. What is a .logarithm
    1
  7467. sad
    1
  7468. What is the music at 18:38? It's so relaxing :)
    1
  7469. I liked that 700 years old proof
    1
  7470. This is an amazing proof
    1
  7471. I really enjoyed this one, John Conway did so much cool stuff. RIP
    1
  7472. I will use glue to amaze you with my leaning tower!
    1
  7473. The most memorable for me was the proof by Nicole Oresme. It's greater than infinity, duh.
    1
  7474. The most memorable was the part with the optimal towers.
    1
  7475. I'm pretty sure the best part was Tristan's fractal. But I mean, I'm a bit biased since fractals are my favorite thing in maths.
    1
  7476. When we look instantaneously inside-outside holographic positioning presence Singularity-point and see e-Pi-i 1-0-infinity sync-duration, the z-axis is a line-of-sight in/of unique certification of WYSIWYG and transverse trancendental i-reflection containment of reciprocation-recirculation relative-timing, everywhere-when all-ways all-at-once here-now-forever universally, in the Pi picture of e-Inflation No-thing definable, textured in the probabilistic range rates of 1-0-infinity temporal superposition identification substantiation. Welcome Home.
    1
  7477. Frohe Weihnachten und einen guten Rutsch ins neues Jahr!
    1
  7478. That last animation makes me wonder if the golden ratio is also present
    1
  7479. Isn't there multiple ways to attach ears to each line? You can rotate it and move it around, sort of like with a right triangle
    1
  7480. Does Petr's Miracle generalize from the 2d-plane to 3d-space? From polygons to polyhedrons?
    1
  7481. I have this weird and vague feeling it might work in 3d on tetrahedra (special case). I will go play with that when I get time.
    1
  7482. A NEW VIDEO AGAIN ;DD
    1
  7483. Really great video!! For me the most counterintuitive and therefore memorable result was that the no 9s series is less than the 100 0s series. From a density point of view it makes sense - very long numbers (a million digits, say) have a tiny probability of not containing a single 9 but high probability of having 100 0s. Yet it is mind boggling that those tiny fractions after 10^-100 add up to something bigger than the no 9s series. I am also curious about the no N series in general. Their sums were different. I wonder whether there are any nice patterns there...
    1
  7484. Wow, this was so satisfying, relaxing and just generally wonderful and elegant bit of math for a sunday evening. Thank you!
    1
  7485. Is the app supposed to be a web app? Or can you also do a python program that you run from terminal?
    1
  7486. I feel humbled by this proof 😅
    1
  7487. I liked the chapter on gamma.
    1
  7488. I couldn't even blink during the video. It was amazing. I'm really looking forward to watching the next Euler-Maclaurin chapter.
    1
  7489. Triangle lover here.. This is marvellous. Thank you so much I am very surprised this was not 'known'. It seems that it must have been known at some level or other by people making things - in cloth or clay - in tiles, especially in China, Japan, Korea, India and especially Hindu Buddhism or Islam where geometric repetition in huge number and scale are central to sacred geometric patterns, architectural relationships and texture. Cut and fold with paper cloth wood, clay or fired ceramic.. glass tiles/windows ? Embroidery, patchwork, quilting String and stick tools used as templates and jigs. Musical instrument makers workshop ? All the natural patterns that derive from simple folding, repeats or recursive actions, or string straight edge and compass. Angle dividers... My feeling is It would be hard not to know or have met this this in many ways But might never have been presented, or kept and translated across the slik roads, libraries Alexandria Istanbul Italy Germany France and England when renaissance printing took off. Hope we get a superfast Visual-Math AI tool to search across imagery via time culture place pattern/ and perhaps even in number system counting ? scaling tricks for people who make things ? Thank you for this channel and your videos inspiring! I've seen on TikTok neat triangular vertex-hinged shutters transform from door to an open gateway. This pattern is very scifi. Now I want to try to use the geometry you present here to make a human sized set of experiments here at my studio
    1
  7490. this method can be tweaked to give the correct area of any shape if you make the surface of each iteration convex. how? well, if you look at just two triangles which share a common edge, they make a tetrahedron. now if you take the two triangles which are more excentric and use them instead of the two original triangles, you've turned the original convex part into a concave part. now do it a bunch of times until you get a fully convex shape. now, the spikes method is actually checking if the shape is convex, so the shape is convex and therefore it give the correct area.
    1
  7491. Yes this works if the surface is itself known to be convex. (Strictly speaking, the volume contained will be convex, and not the surface itself, but I know what you meant.). However there are plenty of surfaces we want to measure, or otherwise approximate with a grid, that aren’t convex or concave even in an infinitesimal neighbourhood. Eg the “Pringle” saddle.
    1
  7492. There is a complex slide rule. Would love to see a video on that.
    1
  7493. 23:00 I think you lied here! By your logic of "what's true of a series is also true of the limit", π=4. How? Well, take a circle with diameter 1. Embed around it a square (imagine that it's made of string, not paper) with side length 1. What's its perimeter? 4, of course! Fold the corners of the square so that the very tips touched the circle and so that the new sides were parallel to at least one of the existing sides of the square. What is its perimeter? 4 as well! And keep folding like this until you get the circle again. What's the circle's perimeter? Still 4. But it's also π. This means that π=4.
    1
  7494. pure genius
    1
  7495. This is how it ought to be taught!
    1
  7496. Big fan of your videos! I think that gamma > 0.5, because of the following. The area enclosed by the linear interpolation of 1/x sampled at integers is larger than the area of 1/x. Furthermore, the area of each of the terms in the harmonic series is larger than the interpolation area by an amount equivalent to the area of the triangle enclosed: area_n = 1/2 height*width = 1/2 * (1/(n-1) - 1/n) * (n - (n-1) )
    1
  7497. but then if we sum together all the area of such triangles, all terms cancel out leaving that the area enclosed between the harmonic sum and the interpolated area is exactly 1/2. But we said at the beginning that the interpolated area is larger than 1/x, therefore gamma must be greater than 1/2
    1
  7498. I'm a layman and know almost nothing about math and haven't studied math in decades. But my answer for 36:17 is that the area of the square in the middle is 0.2 (or 1/5). I came to that conclusion by taking that 1/2 length of side "a", but then because those lines are parallel, using that 1/2 as the hypotenuse of a smaller triangle inside that middle square. If that's the case, then side "b" of that inner triangle would be the side of that inner square. And if we multiply that scaling factor (I don't know what to call it) to get the side "b", and then square that (to get the area), it comes to 0.2 (or 1/5). The actual calculations are as follows. We have (1/2)*2 + 1^2 = c*2. So 1/4 + 1 = 1 1/4 (= 1.25). That's c^2. The square root of 1.25 is 1.11803398... So that's "c". Now, we make 0.5 the hypotenuse, so what is "b"? 1/1.11803398... = 0.89442719... . So it's that times the side that's 0.5 units in length. 0.89442719... * 0.5 = 0.44721359. That gives the length of one side of the center square. And when you square that, the answer is 0.2 (or 1/5) for the area of the center square. Okay, well, that's the answer I came up with, at least!
    1
  7499. messing around the link between 17/12 and root 2 leads to interesting things like (17/12)*17 = approx 24
    1
  7500. A slide rule came with more than just the two 1...10 scales, and they were labelled on the left hand end. I think there must have been some sort of standard, or at least a convention, because IIRC all my slide rules had the same labels. I've just fetched out mine now (Hemmi model No 130W), and the two 1 ... 10 scales are labelled C and D. There are also A and B which run logarithmically 1 ... 100 and BI and CI which are reversals of B and C respectively; a single K scale 1 ... 1000; and some trigonometry scales that I've forgotten how to use! The centre slide is double-sided and reversible, engraved on its other side with loglog scales and other voodoo that are similarly obscure to me now, some fifty years after I bought it. Edit: the scales and labelling seem to be called "Enhanced Darmstadt" after German University that developed them.
    1
  7501. Wonder if there are other ways to do transformations with physical geometrical devices. Analog fft! What happens if you change the elasticity of the rubber band as a function of log to a function of something else? E(Log base 10) -> transformation of mult to add. Guess this works with any base?
    1
  7502. Hey dude awesome post. I graduated Engineering school 30 years ago, and this was a blast from the past. What is the soundtrack playing in the last 2:00 ? German prog metal and Calculus just go together so well.
    1
  7503. If circular slide rulers were easier to produce, maybe we would call them roll rules.
    1
  7504. Euler-Mascheroni Please pronounce Mascheroni as maskeroni. :) :) :)
    1
  7505. I particularly liked the terrible aim portion of the video! Quite memorable. Thank you for the video
    1
  7506. Wow. Well done. Engaging to the end.
    1
  7507.  @Mathologer  Yitang Zhang uploaded his priliminary to net, and indeed Terrance Tao has pointed some suggestions too.
    1
  7508.  @Mathologer  Terrance has certainly mentioned some short-comings.... But what i am more interested is, why would Yitang's proof be seen as a major breakthrough..... (it seems to point towards a special condition, and Siegel's conjecture itself is again another special condition). I always have problem with number theories.... (it is, to me, like something one either have it or don't have at all).....i am more applied math person. But i am curious about Riemann, though i struggle in my Uni third year maths..
    1
  7509. Chapter 6 was the best, broke all my intuition :D
    1
  7510. I remember the slide rule when I was very young. My dad is an accountant. By the time I entered High school the calculator was invented. We still had to learn to use log books.
    1
  7511. 9:40 that's used to encoding positions in Factorio LTN
    1
  7512. How many more kids do you think would be into mathy things if schools taught stuff like anti shapeshifters, even the name sounds epic.
    1
  7513. As usual, excellent video!
    1
  7514. 21:11: The fraction is 946/243. Very curious that the bottom number is a power of 3, looks like all of the other numbers cancel out in the denominator. The numerator is 43*11*2, which doesn't seem at all related to the numbers present in the equation. Very mysterious.
    1
  7515. Is... 169 + 196 = 365? Aha! (Answer to the second puzzle.)
    1
  7516. commenting before watching the whole thing, but when we found the approximation for finite sums with different exponents i damn near wanted to smack my head for not seing it, it´s the good old integral power rule. When you said "if you knowe some calculus, it should" i got frantic, but given how well you speak German you might be familiar with the expression "Den Wald vor lauter Bäumen nicht sehen." for non German speakers it usually means, missing something entirely obvious in plain sight. Literal translation "to not see the forest because of the trees." I damn missed the whole Amazonas here^^ Calculus is the only part of maths i´m remotely competent (excluding basic "clean up level" algebra and arithmetics of course.) but i don´t see the most basic rule for integrals there is^^. (fun challenge. try to derive that formula from scratch. up to now i failed to do that. I still need to prove the power rule for differentiation and then that integration and differentiation are inverses of each other. Our approximation for finite sums already is the closest to deriving that formula i ever got without going for the rule of differentiation. Which is just a nuisance to proof. Because, well pascal´s triangles in general form are no fun at all)
    1
  7517. can we aknowledge his tshirt i love it
    1
  7518. Great video! Please make a video on the relation between the level of Nile river and fractal dimension, the Hurst parameter. It also has a story based on water
    1
  7519. If walls have windows, they should also have doors. I also heard that walls have ears.
    1
  7520. Having not watched any further than you setting the problem up for the 1, 2, 4, and 8 cent coins, the ways to make change is just the binomial expansion, calling it now, locking in my final answer.
    1
  7521. Yaaaaaay
    1
  7522. Seven point star angle sum: Imagine walking on the inner heptagon from one edge of a star point to the other. You will turn through 3 external angles of angle 2pi/7, for a total of 6pi/7. This implies the external angle of the star point is 6pi/7. The internal angle must be pi - (6pi/7), or pi/7. So the sum of the internal angles is pi. EDIT: For irregular stars: Note that you make 3 full rotations when walking the path of the star. (Any line from the center out crosses the star 3 times, so your winding number is 3). So 6pi is equal to the sum of the external angles. So, 6pi = (pi - A1) + (pi - A2) ... + (pi - A7) = 7pi - (A1 + A2 ... + A7). Rearranging you get (A1 + A2 ... + A7) = pi.
    1
  7523. i love maths, but I totally suck at it. Thanks to the mathologer, i feel like a maths genius (and I suck a bit less)
    1
  7524. mindblown already at 9:57 and it just kept getting better after that :D
    1
  7525. The return of the squish-and-stretch! Love it!
    1
  7526. i need to learn more about this
    1
  7527. For this video I played around with generating some AI animated photos of Sophie Germain using https://www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz https://www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :)
    1
  7528. That was a good one. Again I always watch this on a Sunday afternoon. Not a mathematician but my theory is that they are the same.
    1
  7529. I love this stuff, I had bad education when I was young.
    1
  7530. I feel so stupid since those guys did this magic in 1600 when Galileo just discovered gravity and force laws and was put into prison for this
    1
  7531. The "What's next" question can be formalized with Kolmogorov Complexity and bayesian inference if you believe into Occam's Razor!
    1
  7532. Fabulous
    1
  7533. My vote goes to Tristans awesome fractal
    1
  7534. 1+2=3 and 3^2+4^2=5^2. So obviously 5^3+6^3=7^3. Conjecture that the pattern for the first term is generally (n)^n+(n+1)^n=(n+2)^n. 7^4+ 8^4=9^4-2. It seems we are always missing by 2. Though perhaps not. With the next sequence we are missing by 2002 The answer for the second is 2. Who would have guessed. hmm 3^3+4^3+5^3 = 6^3 . 4^3 is the internal cube and the remaining 3^3+5^3 is the perimeter shell. 3^3+5^3 = 3*3^2+5*5^2 = 6*4^2+8. 6 4by4 squares and 8 1by1 corners
    1
  7535. 10² + 11² + 12² + 13² + 14² = (12 - 2)² + (12 - 1)² + (12)² + (12 + 1)² + (12 + 2)² Expanding the brackets we get 5*12² each also get 2* the product of the first and last in the bracket (for example 2*(12*-2) in the first) but the first two cancel out with the last two, and we also get the second part of each bracket squared which totals to 10. so we now have: 10² + 11² + 12² + 13² + 14² = 5*12² + 10 10² + 11² + 12² + 13² + 14² = 10*(12*6 + 1) 10² + 11² + 12² + 13² + 14² = 10*(73) 10² + 11² + 12² + 13² + 14² = 730 and 730=365*2 so the answer is 2 I loved this little video, happy holidays! :)
    1
  7536. I've missed these videos on my sub feed. Glad to check the channel today :)
    1
  7537. BTW Future Me knows he says pretty much this about a minute or so later. Past Me did not, and I tend to cut him a little slack as Past Me isn't so smart.
    1
  7538. I always enjoy your videos, but, to quote the Professor from The Fifth Element, "This is really amazing." 🤓
    1
  7539. Would you please do a follow up version where your rubber band is a guitar string and explain the spacing of frets.
    1
  7540. There's an error in describing the generalisations of the arithmetic mean and the geometric mean at about 28:45. The arithmetic mean of N numbers, is their sum divided by N; meanwhile, the geometric mean of N (positive) numbers, is the product of all N numbers, and then the Nth root of that product.
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  7541. all of these are so crazy
    1
  7542. To what extent is the following definition viable? Transformer: an automorphism group equipped with type closed operators that accept its self as a parameter.
    1
  7543. Looks like symmetrical components class for EE. I'm just barely smart enough to pass my engineering classes. Definitely can't do this with triangles, and more than 3=n .... I'm done! Cool video!
    1
  7544. Math grad student. Great video. Didn't know lots of it.
    1
  7545. There is a potentially serious issue with this method of distributing claims, and that is the concept of splitting or merging claims. If the total amount of available funds is less than half of the total amount claimed, there is an incentive for a claimant to sell off partial claims in order to have a larger number of smaller claims filed, thus paying out a proportionally larger award to each of the smaller claimants than the one large claimant could receive if they filed individually. And if the total funding is more than half the total amount claimed, there is an incentive for smaller claimants to collectively file a single claim for their combined amount, then redistribute the amount awarded proportionally. This was actually alluded to in the 'three sons' example, as if each of the three sons filed a claim independently, the 'red son' would be awarded 4/9ths of the estate (more than 5/12ths), while each of the 'green sons' would be awarded 5/18ths of the estate (less than 7/24ths). The primary ways out of this situation is to either have very complex rules on how claims can be spread, thereby refusing the ability to split or merge claims, or switch to a proportionality-based claiming system.
    1
  7546. This might be a stupid question but is it possible for the det(matrix) to result in something like : 12+33i if yes what would be the number of tilings ?
    1
  7547. Shit man!Awesome
    1
  7548. thank you for the excellent explanation. would you consider making a similar video on ito calculus in general or wiener process? i have never been able to understand it, despite it obvious importance in the physical world, i.e. brownian motion.
    1
  7549. I think I got it, now to restart the discussion ...
    1
  7550. Comforting part: how Nicole Oresme proved the divergence of the harmonic series, because it feels nice to find in your video something, I am already familiar with. The best chapter: the fifth, about the gamma constant and its usefulness in defining how much terms you need in the harmonic series for it to be equal to some particular number, very nice stuff. Answer to your question: why gamma is greater than 0,5? The gamma constant as you say equals the totality of the blue area over the curve f(x)=1/x . This area (gamma) is definitely greater than the sum of quarters of areas of blue squares, because looking from the bottom the curve is convex. So, gamma > 1/4 + 1/12 + 1/24 + 1/40 + … gamma > (1/2 *1/2) + (1/2 * 1/6) + (1/2 * 1/12) + (1/2 * 1/20) + … gamma > 1/2 * [ (1 -1/2) + (1/2 – 1/3) + (1/3 - 1/4) + (1/4 – 1/5) + …] Therefore, gamma > 1/2, because everything cancels out except for the first product (1/2 *1) You can also see 1/2 coming from the formula for the area of a triangle. Is there a simpler way to do it? Oh, and it’s fun to replay the video and see in the introduction the parabolic tower on top of the letter L. Your content is great!
    1
  7551. I'd never seen either of the two proofs you showed in the beginning of the video. I remember a proof that was way more complicated and hard to understand. These are beautifully simple.
    1
  7552. The most memorable for me was the table of sums of no zeros, no 1s etc - i wonder what pattern ther may be in these and on changing base to say base 16 what changes would occur
    1
  7553. You really have to go out of your way to connect this. The reason it is remanded to oblivion is because that is where it belongs.
    1
  7554. This channel is such a joy! Loved the leaning tower of Lira :-)
    1
  7555. Let's think of the grid of points as the integer-coordinate points in the cartesian plane R^2, let's think of triangles with vertices on those latex points. Any equilater triangle would have irrational Area: sqrt(3)/2 l^2 is irrational because l^2 is an integer (Pythagoras). But we can inscribe any triangle in a rectangle with orizontal and vertical sides which also has its vertices on those latex points: it's now clear that the area of the rectangle equals the area of the rectangle (an integer) minus the area of three rectangles (clearly rational), which can't be irrational.
    1
  7556. This might be hard to apply to the other regual polygons though! Also, Beautyfull visual proof.
    1
  7557. Fractal shapes can enclose a finite volume but have infinite length (you can't see this on a screen due to features smaller than a pixel) so this isn't completely surprising ..... Is it something to do with the self-similarity property of the starting curve?
    1
  7558. Quarantine made infinitely better!
    1
  7559. 30:29 that was pretty cool :D
    1
  7560. 21.19 if you try with a fruit you ll notice the line has to be turned 90 degree since one is 3d and 2d or one does not geometrically exist in real life
    1
  7561. Amazing. The reason why 5 makes the cycle backwards is because it's the inverse of 2 mod 9. I want to share that if you put the numbers 1 to 9 in a calculator pattern and cycle through the powers of 2, 3, 7 and 8 and keep the last digit (same as mod 10), you make a square each time. I did a video a while ago if you have time to look at it please. https://youtu.be/fAHAisb_nlk I also made a 5x5 grid 6x6 grid and so on and I got beautiful patterns.
    1
  7562. At 12:31 one of the squares in the shape disappears. This threw me off when I tried to tile it by eye...
    1
  7563. theres also discussions on whether renouncing property can have conditions attached hillel says no shammai says yes.
    1
  7564. Hmm. That is challenging to make in wood. I'm assuming that if I put a pin in each layer with a stops 137.5 degrees apart mod 360 I should get a tree?
    1
  7565. A comment for the algorithm.🍉
    1
  7566. I think the multivariate calculus is a different story tho. But 1d calculus is kinda easy indeed.
    1
  7567. It’s like the outward movement creates a vacuum in the center area which is total entropy. Space abhors a vacuum and so fills the space but with random quantum flux
    1
  7568. Convergence proof of the no-9s was the most remarkable for me. what a crescendo
    1
  7569. Animating 18:18 and rotating all those arrows around must've been tedious lmao Amazing video!
    1
  7570. What bother me most in math is naming of variables: a, b, c, d, etc. Programmers will understand. :3
    1
  7571. R+1 external and R-1 internal rotation
    1
  7572. On another aspect: looking at the large rendering at 2:20 of https://www.youtube.com/watch?v=CCL77BUymSY I noticed that the areas of the artic circle close to the frozen regions have a higher concentration of the frozen area color. The pattern suggests a projected sphere illuminated by 4 different lights.
    1
  7573. That shirt is so awesome
    1
  7574. In 1957 John Conway showed me a nice proof why any nondegenerate triangle with sides in rational proportion and angles in rational proportion has to be equilateral. The angles have to be multiples of, say 1/N, of 360 degrees. The only N for which the cyclotomic extension of degree N of the rationals has just 2 automorphisms are 3 or 6. Ruler and compass construction shows there can only be two triangles with prescribed sides once one edge is fixed. Bingo! Now what I want to know is where the Mathologer got his lovely T-shirt.
    1
  7575. Conceptually, calculus is easy - but remembering the theorems and rules, and when to use them, is hard.
    1
  7576. At https://youtu.be/Yy7Q8IWNfHM?t=1280 (chapter 4) is closely related to entropy : having a solid colored corner leaves many many ways to tile the inside, but you need the inside to be increasingly simple to get tiles of a different color into that corner.
    1
  7577. Taxicab numbers t-shirts
    1
  7578. def root2(num, den, prev): newDen = num + den newNum = den + newDen if (newNum / newDen) != prev: root2(newNum, newDen, newNum / newDen) else: print(newNum/newDen) root2(1, 1, 1)
    1
  7579. The last section of the video ("Intuition for multiplier") reminds me of how spin is used in physics.
    1
  7580. Nice way to oragami down to one third. Now apply this to fractals in graphics generators.
    1
  7581. 24:52 γ seems to be bigger than 1/2 because the blue leftover bit from the 1 square is bigger than a quarter, and the rest of the blue leftover area seems to fill in more than half of the lower half of the square into which all the blue leftover area fits, i. e. it appears to be bigger than a half of one half, or one quarter. So in total the blue remainder area is greater than two quarters, or one half.
    1
  7582. que buen video
    1
  7583. 3:40 i started by trying to out the distance between the centre and 2 of the whiskers growing out of a single vertex, and that was easy because the centre is given by halving the triangle's angles - i just watched a 3B1B video proving that so that part of the proof isn't mine - the, we can deduce that the line that is obtained by splitting the angle can be extended in the other direction, giving us 2 angles, both equal to the other 2 angles we've gotten inside the triangle, therefore the points, which are the same distance from the vertex by definition, are also the same distance from the centre. then just noticed all the chords are the same length, and quickly proved that if 2 points on the ends of a pair of lines of equal length (correct me if thats not the right term) are the same distance to a point, then the other ends of the lines are the same distance from that point as well, using the same angle argument. this is the proof, i probably didnt get all the math terminology right so correct me please! thanks for making this! also, i think they have the same diameter. the corners of the wierd conway's curve are more pointed and bulge outwards, while the edges bulge inwards, although its not a proof or anything edit 2: i was wrong
    1
  7584. wouldn't it be two answers to which the card is?? I mean, you say that the card is a spade and it is 4 positions appart from 6, but that could be 10 and 2, right??
    1
  7585. Epic choice of music. "Can you see the super pattern?" "This summer. One pattern. To rule them all. ksh Luke, I am your fater!"
    1
  7586. 🐢
    1
  7587. I made it to the very end 😉
    1
  7588. I think youve been shadowbanned by YouTube! I had to forcibly find your channel
    1
  7589. I don't find a reference saying that the Eisenstein triple is named after Gotthold Eisenstein. Maybe some could me out? Thanks.
    1
  7590. Heron isn' t a greek he is egyptian from Alexandria ....like Euclide , Erathostene and a lot of other mathemacians , he wasn' t greek 😅😅😅😅
    1
  7591. Most memorable: Kempner's proof
    1
  7592. Maths in general is key to understanding any scientific model of the universe or any part of it - which might help in understanding the real universe. Sometimes science gets to points where the scientific models don't match the real world, so we need to keep adjusting them and they get more and more complicated (like fractals?!) and mind-boggling... Like learning the times tables is key to so much more in maths - it's a struggle for those who haven't memorized them at an early age!
    1
  7593. It is truly tragic we never meet these proofs in school. They just say take it on our word, it’s true. The first proof I ever knew.
    1
  7594. wow
    1
  7595. The 700 year old proof was very inspiring to me. I'm a maths student right now, can't wait to explore that kind of stuff more as I go on.
    1
  7596. maybe time was a factor; large debts most commonly to be paid in longer timeframes
    1
  7597. To help boost engagement after the reupload, I opened this video in 50 tabs and let it play in the background muted, while I went to sort out food. They all autoplayed for about an hour after the video ended, and when I got back, most of them were playing https://www.youtube.com/watch?v=5PcpBw5Hbwo - either straight away or after other videos. Just thought that was interesting.
    1
  7598. I feel extremely proud of myself for guessing kempner’s series converges based on density of reciprocals that would have 9’s. I rarely get the answers you ask for in time!
    1
  7599. Swivel proof for elegance
    1
  7600. 666K, the Number of the Subscribers ^_^
    1
  7601. Your shirt is awesome! Where can I get one?
    1
  7602. I cannot believe this content is free.. I feel obligated to pay for it lol
    1
  7603. Prove: There is not a mathematical bone in my body; Proof: Response to proof by contradiction - me: "Meh". I'll keep trying the impossible - one of them might work, after all, there is an infinite choice!
    1
  7604. A LOT of time spent on lacing up shoes. But what about tying them? We all know how to tie a basic bow: Make a loop in one free end, wrap the other around the base of the loop, and push it through. But as a young child, I was also shown an alternate method: make loops in both ends, and tie them together like a simple overhand knot. I still don't know if the bows made by two methods are topologically equivalent. Are they?
    1
  7605. sorry but this is not Newton calculus. The true one is a generalization of Hudde's rule. Fluxion-the calculus of Newton, is considered from motion of point. Newton's Method of Fluxion (with commentary of Dr. Halley) explicitly explained his own perspective of true calculus. For more insight, please consult Maclaurin"s treatise of fluxion in 2 volumes. Both books are in public domain and scanned version can be found online.
    1
  7606. Breaking the laws of PHYSICS for MATH'S - as a rubber band with an such an uneven stretch is weird - what a funny solution
    1
  7607. Amazing
    1
  7608. @7:20 we can see the script :o
    1
  7609. Best ever video, I have ever watched!!!
    1
  7610. Hah, cool I had this on my number theory homework last semester
    1
  7611. I was wondering if there would be any proofs along Symmetry groups. If we pose the problem as minimizing the Sum { abs(weight'*(A^n*start - J*A^(n + 1)*start)) } where n = 0..#eyelets -1. (We are summing the lengths of individual segments.) [A] is the rotation matrix that defines which eyelets are linked where A(i,j) = 1 if there is a link connecting ith eyelet to the jth eyelet, otherwise A(i,j) = 0. J is a (#eyelets)x(#eyelets) square matrix whose antidiagonal (northeast-southwest) elements are unity and all other elements are zero (the exchange matrix.) 'weight' is a vector where weight(i) = i-1 (that is made of the numbers 0,1,2...#eyelets.) 'start' is a vector whose first element is unity other elements are zero. By defining the problem this way we see that the problem has a rotational symmetry: Shoelace length is invariant regardless where we start measuring. We need to find [A] (the permutation group) that results in the minimum length shoelace while making sure that each link crosses to the other side by the use of the 'weight' vector. Refactoring the cost function as minimize Sum { abs(weight'*(1 - J*A)*A^n*start) } we see that (1 - J*A)*A^n*start selects columns of (1 - J*A) successively.This suggests that matrix [A] should be symmetric about northeast-southwest diagonal and non-zero elements should be as close to this diagonal as possible. This approach could be used to find the minimum shoelace length using integer programming, but it certainly is not a proof. I wonder if this idea can be worked towards a proof.
    1
  7612. He's right! I didn't see it.
    1
  7613. The answer is 1/5. Take the square from the final puzzle and copy it on all 4 sides. You will then notice that each of the triangles pair up with the trapezoids to make a copy of the blue square! Hence it is 1/5 of the area of the big square, which is 1x1.
    1
  7614. Are you Jewish, Burkard?
    1
  7615. Where does he get those lovely shirts?
    1
  7616. Brightens my day, every time I see a new upload from your channel. :))
    1
  7617. Where's that infinity hiding in the surface area calculation? The thickness of the film. An infinitely thin coat of paint is the same as no paint at all. Oh. I jumped the gun there. I see you covered that issue near the end. Good job.
    1
  7618. Great video, thanks. After my recent geometric explorations of polyhedra, I reached the conclusion that somehow sqrt(2) and phi are part of a special kind of irrational numbers regarding geometric ratios.
    1
  7619. I do agree with folks on the comments that this visual explanations completes the gaps for people like me who are learning the beauty of math after certain age. For me, the explanation is simple to follow as I already have notions of calculus and log before. I wonder if you know if there is a way of explaining it using even less mathematical knowledge so I could show my kid and hopefully get her to understand these concepts from the right age and not as old as I am now.
    1
  7620. Regarding the first problem you postulated, the answer I got is that, for N cards, then N*(N+1)/2 , (the triangular numbers!) is the maximum number of steps to finish the game. I'm trying to construct a rigurous proof, but I have failed so far. But, at least, I want to present an intuitive argument of why I think this should be the case. First, for N=1, it's easy, just one step, and the formula holds. :) Now, let us assume that for N cards, this result is valid. What happens when we have N+1 cards?. An intuitive idea would be that we leave the first card (the one at the extreme left) alone for the moment, and we take as much steps as possible before we are forced to flip that card. So, with the rest of the N cards, we take the most possible number of steps to flip those, which is, by inductive hypothesis N*(N+1)/2. So, we end with the following scenario: 1 0 ... 0 0 So, in this case, it is straight forward what happens next: we just have one card, and the rest of the moves are just 1 0 ... 0 0 0 1 ... 0 0 ... 0 0 ... 1 0 0 0 ... 0 1 0 0 ... 0 0 So, we have N+1 steps to finish the game from this state. And, of course (N*(N+1)/2) + (N+1) = (N+1)(N+2)/2, which yields the desired result. Of course, this proof is NOT complete, because the obvious question would be: "Well, what happens if you flip the first card earlier?", and I've noticed that indeed, taking other strategies, we can also obtain (N+1)(N+2)/2 steps. The argument I'm currently working on is proving that, if you make a flip in any other point of the maximal sequence, you would end up in some corresponding step of the first maximal sequence of steps at best, or in, the worst case scenario, in a faster sequence of moves. For example, let us consider the following flip: 1 1 ... 1 1 0 0 ... 1 1 In this case, the first two card would be, effectively, out of the game, and you couldn't optimize the number of moves for N cards, because then you would only have N-1 "active" cards as maximum. So, I think my answer is correct, but I'm currently working in the rigurous proof of why this formula should work for all possible strategies. Thanks for the video, BTW!. Great format, is always a pleasure to watch your content.
    1
  7621. To prove the triangle simularity its enough to show that 3 adjacent sides remain scaled by a factor Chosing one color vertex from both triangles, say red. Then redbrown1 = redbrown2 * k for some k, and redblack1 = redblack2 * k. Adding the vectors is the same as adding redbrown1 + redbrown2, and redblack2 + redblack2 We then have for each leg a vector redbrown1(k+1) and redblack1(k+1). Brownblack1 =redbrown1 - redblack1, and brownblack2 = redbrown2 - redblack2 = redbrown1 * k - redblack1 * k = (redbrown1 - redblack1) k Then Brownblack1 + brownblack2 = redbrown1 - redblack1 + k*redbrown1 - k*redbrown1 = k+1(redbrown1) - k+1(redblack1) = (k+1)*(redbrown1 - redblack1)
    1
  7622. InTeReStInG 18:10
    1
  7623. When The Professor says MAGIC and AMAZING , he means it. FANTASTIC!
    1
  7624. 30:16 I plugged it in and yeah, it worked
    1
  7625. Ramanujan is the most underrated mathematician of all time..
    1
  7626. Great video. Thank you
    1
  7627. Can you add and subtract with a slide rule? Suggestions welcome.
    1
  7628.  @Mathologer  Brilliant solution with these piles! My solution using logarithmic scales goes like this: Put the 3 on the lower scale under the 5. Then look where the lower 1 is and look at the value above it on the upper scale. In our example, that would be 1.67. Now shift the lower 1 to the value 2.67. (I.e. shift the lower scale to the right by one unit of the upper scale.) Then look at what is written above the 3 on the upper scale. It's the 8. Thanks for the interesting video und viele Grüße aus Deutschland!
    1
  7629. It's nice when my skepticism is borne out, but I must admit I would love a key to the universe!
    1
  7630. 10:30 - finally, part of a mathologer video I can understand!
    1
  7631. this is an amazing video! i love it! but geometry is also a key interest of mine. 😊
    1
  7632. 2:00 and it's ALREADY genius :) any two consecutive are different colors, so when a color is missing there could not be any solution. I was searching for a hard demonstration as proof but suddenly when you said "hardly" I felt myself stupid as it was obvious and as a chess player I knew opposite where same color very well. But what is even more amazing is that if we started with a colorless problem, then just adding the color for demonstration would be a crazy smart and clever idea. I wonder how many problem do have very complicated demonstrations that would be very easy to demonstrate with the same kind of smart ideas, but we failed to see? Maybe there are a lot. I remember when I used to code, that sometimes when you try hard to reduce the length of your code, you can find the same kind of unexpected ideas to success. This kind of success makes one coder very happy when he finds it. And very ashamed if someone else tells him he missed one. And again I wonder how many terawatts we would save wordly on computers if we knew and used all the possible smart shortcuts of this kind. Maybe we miss alot. How to know? Maybe we missed a few that are real gale changers for humanity. And anyway, there are some people who have found some very hard ones, and these people did a great service but most are unknown. I'dd love a video about this one day. Maybe some mathematicians have found some mathematics/probabilities or such that give an idea of how much we miss and how to help find these?
    1
  7633. why is it that eventually we found a point where the sequence adds up to 100 while the premise is that the partial sum ratio is always ODD/EVEN?
    1
  7634. Euler-Mascheroni Constant was my favourite. Thanks for the video
    1
  7635. I bought a slide rule after watching the Internet.
    1
  7636. absolutely beautiful! I wonder if someone (after learning basic arithmetic) would built up all math knowledge that is possible from this.. Instead of the other way around like all of us.. How much of a different view/understanding one would have.. Very interesting and powerful concept. Thanks for sharing!
    1
  7637. I was waiting if for you to pull out the tesseract :{
    1
  7638. I am third daddy
    1
  7639. This is a codewars problem that I kinda cheated on to solve a while back, but even then I only got it to O(n) (where n is the length of the original row), I believe with this I can get it to O(logn)
    1
  7640. What I like most, dear Burkard, is your enthusiasm while explaining these (for me) astonishing and beautiful things.Thank you.
    1
  7641. Speaking of Phi, id like to request an analysis video of animation vs maths. Pretty please? Other than that, awesome video as always!
    1
  7642.  @Mathologer  Surely, but it would be definitely interesting hearing your thoughts about it. Some of the analysis videos lack some or just state facts. In your videos you show us the facts and then connect the dots, which creates the AHA!-Effect thats often missing. But you are definitely right, it wouldn't be the most unique video!
    1
  7643.  @Mathologer  Even better! I like that idea. I recon you already have some background there, since your presentation often have these neat little moving parts?
    1
  7644. This video is beautifully done! I'm seriously in love with the format ❤️
    1
  7645. The American bankruptcy law visualization is absolutely brilliant
    1
  7646. ​ @Mathologer  as I said in my other comment, thank you so much for making this video. As an Orthodox Jew, the respect you as an independent outsider (and someone I've subscribed to for nearly a decade) have shown to the system of logic in our Talmud is very touching.
    1
  7647. Long awaited your new and surprising video first math lovers and what a lovely surprise it was. Keep on surprising us! 😜
    1
  7648. Have loved this puzzle since Rise of the Planet of the Apes.
    1
  7649. What a nice maths video!
    1
  7650. 37:20 Of course it's the same number, it's just a matter of perspective on the 3D interpretation. If you look from the direction of any axis, all you see is one colour, occupying a constant total area.
    1
  7651. 36:17 Consider three similar triangles [l]arge, [m]edium, [s]mall. Each with sides a,b,c where a &b are legs of the right triangle and a<b. In the large triange a=1/2,b=1, so c= sqrt((1/2)^2+1^2)=sqrt(5/4). The side of our inner square is b[m]-b[s]. Since the triangles are similar b[l]/c[l]=b[m]/c[m]=b[s]/c[s]. Based on the illustration a[l]=c[s],b[l]=c[m]. This gives us b[m]=b[l]^2/c[l],b[s]=a[l]b[l]/c[l]. So the side of the inner square (b[l]^2-a[l]*b[l])/c[l]=(1/2)/sqrt(5/4). The area is the square of the side (1/4)/(5/4)=1/5.
    1
  7652. As everyone knows, the volume of a pizza of radius z and height a is pi*z*z*a.
    1
  7653. What a nice presentation!
    1
  7654. I found the proof of γ being between 1/2 and 1 to be quite enlightening
    1
  7655. Ummm… if it can be resolved it’s no linger a paradox
    1
  7656. I'm ashamed that it took me 7 minutes into the video about think what molecule your shirt was before realizing the joke.
    1
  7657. gamma
    1
  7658. Most memoralbe: The partial sum 1 + 1/2 + ... + 1/n and the value ln(n+1) are never more than γ apart. Different ideas, similar growth. What a beast of a constant this γ-boy is ...
    1
  7659. dang ! you're a mighty force of bringing Enlightenment in this Age of Confusion !
    1
  7660. from reading paul erdos's biography i think ron graham also helped erdos and sorta hosted him every once in a while. such a generous man!
    1
  7661. Another example that shows we shouldn't underestimate the engenuity of ancient people.
    1
  7662. Prof. Thorsten Altenkirch storms in and yells: Recursion
    1
  7663. well, I usually don't do puzzles from your vids, but I was too interested to know how "parenting" pattern breaks on 3-4-5 triangle, so I've spent a fair 10 minutes in geogebra and figured out that the centers of the incircle and the Feuerbach circle both lie on a line parallel to horizontal axis (the bottom side; the 3 side) so we can't construct a right triangle by connecting two points and adding two lines parallel to the cathets. hopefully this was useful to someone p.s. does anyone use the word cathet? if not, what would be a better option?
    1
  7664. The `manim` and `sympy` python packages might make such an autopilot a bit easier to build!
    1
  7665. At last one I could follow and understand! 😀
    1
  7666. I preferred the colour proof to the swivel proof.
    1
  7667. ...Tough choice but I think the Euler constant was the most interesting.
    1
  7668. The number of tilings of a "2 by m" chess board recreates the Fibonacci sequence! I feel like whoever programmed our simulation really, really liked this sequence...
    1
  7669. He is so cute and adorable! I learned so much from this presentation. Very enjoyable for me. 😊
    1
  7670. A bit of a convoluted way of proving fib(n + 2) - Σ^n_{i = 1} fib(1) = 1. Let S(n) = fib(n + 2) - Σ^n_{i = 1} fib(i). Notice that S(n) = fib(n + 2) - Σ^n_{i = 1} fib(i) = 2f(n) + f(n-1) - Σ^n_{i = 1} fib(i); by definition of f(n) = f(n) - Σ^{n-2}_{i = 1} fib(i) = S(n - 2) Hence S(n) has to be constant. Which constant? S(0) = fib(2) = 1. So S(n) = 1 for all n.
    1
  7671. Minor correction: The cylinder on the left at 16:20 has 20 bands and 20 points, not 10 and 10 as stated.
    1
  7672. So, human brains, with all those curves and cliffs... is a Schwarz lantern?
    1
  7673. ectoplasm in your shirt, wipe it off
    1
  7674. 12:13 💀💀💀
    1
  7675. Umm, I think 7th grade in the early 70's. Am familiar with several proof's.
    1
  7676. It's hard enough to get them to teach kids the times tables, let alone anything else. !!!
    1
  7677. You know the Painter's Paradox does not require the full force of calculus to prove, because Torricelli didn't have it, and he proved the paradox! Much as he didn't want to. But the line between algebra and calculus is a lot blurrier than people think. We use derivatives and integrals with Leibniz notation and a definition of them from Cauchy and Riemann, because it is easy. But Fermat and Wallis and Torricelli struggled through proving things slowly as best they could, and they discovered remarkable things. The easy way would never have been known if people didn't stumble into the unknown first. And we don't even know exactly how Indian mathematicians were doing it for 200 years at that point.
    1
  7678. Got video after a long interval, but didn't get any notification for video uploaded.
    1
  7679. I like this video already i will check other video's after this video
    1
  7680. But it isn't infinite.
    1
  7681. Amazing. You always save the best for last. That less than proof at the end is in a superposition of simple and insane.
    1
  7682. Solved the problem on the board in my head. 10^2+11^2+12^+13^2+14^2=2(13^2+14^2)=2(169+196)=2×365 2×365/365=2
    1
  7683. Indeed!
    1
  7684. Simply amazing
    1
  7685. Nice soundtrack
    1
  7686. The first numbers of each row in the linear pattern are the square numbers (1, 4, 9, 16...), while the first numbers of each row in the square pattern are every other triangular number (3, 10, 21, 36...) and the last numbers in each row are one less than the other triangular numbers (6, 15, 28, 45, ...).
    1
  7687. Finally, he's back!
    1
  7688. neat that Notch supports a math channel
    1
  7689. since I've played the professor layton games and one of them has a pancake stacking puzzle that functions like Hanoi, I already knew the answer including where to put the smallest disc in the first move. All the odd numbered discs end up fully stacked on the same peg, and all the even numbered discs end up fully stacked on the other peg. Thus if you have a stack of 10, you put the first disc on the peg you don't want to end up at.
    1
  7690. I failed to understand anything in this video because I am distracted by that T-shirt. Specifically, by the awful kerning of type on it. It looks like it says "A VOID NEGA TIVITY", obviously the glyph A needs adjusting. I'll watch it again, because it's a fascinating subject. :D
    1
  7691. Let's see... Four of the same colour, three and one on a corner, three and one on the inside, two pair contiguous, two pair alternate, two pair nested, and three colored with the pair at one side, at the center, alternating, or at the corners. 10 essentially different triangles of order 4.
    1
  7692. does the Stern Brocot tree also produce all of the unique fractions, and if so how does this relate?
    1
  7693. Why Ln?
    1
  7694. Wonderful stuff - the elephant in the room for me is not commenting on the fact the radii of the in- ad out-circles are always integral for Pythagorean triples (not too hard to show it turns out).
    1
  7695. Harmonic series a slow boat to infinity.
    1
  7696. import numpy as np from functools import reduce def change(n, coins): # Initialise all our polynomial coefficients to 0 coeffs = [np.zeros([n+1]) for _ in range(len(coins))] # Place a 1 at the positions corresponding to our coin values for (coeff, coin) in zip(coeffs, coins): coeff[0:n+1:coin] = 1 # Convolve the coefficients to multiply the polynomials ... maybe there's a fast trick here ;) coeffs_conv = reduce(np.convolve, coeffs) # Return the coefficient at the desired position return coeffs_conv[n] # Example from the video change(100, [1, 5, 10, 25, 50, 100])
    1
  7697. Dear Mathologer it is about CHIRAL time you did this ... let me know when you are ready to investigate the associations between the proof the fake Pitagoras used for the right triangle theorem, the 3x3 Lo Shu magic square, the 5x5 Rotas Sator palindrome (Knights Templar magic square) the 12,000+ year old 2D CHIRAL swastika AND the hologram of a photon, that anticipated them all ... "let there be light" ... OK?
    1
  7698. Given N random numbers, looks like dirt, nobody cares Here comes Newton-Gregory interpolation formula, turns the dirt into Gold
    1
  7699. That rectangle looks a lot like a slide ruler!
    1
  7700. Mathologer are the most educational, the most fun and the most beautiful posts on YouTube. ❤❤❤❤
    1
  7701. I think there are no t many better examples of isomorphisms than the old school log books. Usually logs based 10 and then, if you are really really lucky Napier logs (natural logs indeed - the indignation). To a young inquiring mind what finer examples are there?
    1
  7702. Wow that was fun who knew 🤯
    1
  7703. I think it is very surprising, that it will never hit any integer. btw, I was really thinking, "is it proofable, that this building pattern is the best?" and a few minutes later you stated, it's actually not the best :D
    1
  7704. I had my own personal "aha"-moment when I realised that there are hiden Levi-civita symbols in Cramer´s determinant rule. Merry Christmas!
    1
  7705. ur soo cool mannn
    1
  7706. How come a computer gives me the answer of 0 when i ask for an evaluation of the Sum ?
    1
  7707. I always thought Santa's catchphrase is defined by a summation, and not a cubic. But that may reflect being an engineer and not a mathematician.
    1
  7708. Commenting early on. You probably get to that later: So, one more stage of the same process, and all the ears meet in one point. Edit: yes, of course you get to it!
    1
  7709. The question about the no 9s series part was the best, since my intuition told me that cutting a large part of the sum would make it converge but second guessing made me answer infinty anyway. You got me this time mathologer!
    1
  7710. Has the US National Debt gotten that big?
    1
  7711. There is also the trigonometric line: cos(π/5)=φ/2
    1
  7712. 3:00 That's funny humor!
    1
  7713. Important topic as well as the presentation
    1
  7714. Same set up as the slide rule?
    1
  7715. Phenominal, and you explained it all before mentioning logarithms (until the end)! Funnily enough, I've been thinking of getting a slide rule to have around. Maybe now I'll get two, one straight, one circular!
    1
  7716. My first time here, why I didn't hear about this channel before? awesome channel, subbed instantly, love from Jordan.
    1
  7717. How do you choose what 'soundtrack' to use for your videos?
    1
  7718. I suspect the Chinese found this already thousands of years ago while playing with ravioli's or banana's. ;-)
    1
  7719. I'm surprised you didn't mention Volkswagen
    1
  7720. 4k+3, gauss prime. Nice
    1
  7721. The most memorable part was kempners no 9s proof, still seems weird to me
    1
  7722. The proof diagrams (11:45) look like magic Runes. I think Bukard might be using some Norse magic here! the exploded lacing dream proof is fantastic. I think the criss-crossing of lacings is just inherently confusing to the human mind, and eliminating that factor by exploding them really does relieve the mind and make it less intimidating to tackle the problem. great stuff!
    1
  7723. https://github.com/michaelschuff/Square-Dance 
Here’s a link to a unity project by me. Currently has an error where some tiles aren’t calculated correctly but the pattern still emerges.
    1
  7724. I much prefer an alternative painters paradox body: Koch Snowflake base prism. The volume is trivially finite - the shape can be enclosed in a finite circle. And showing the recursive construction of the snowflake trivially shows how each level of recursion adds 1/3 the base length. Extrude to a prism to move from 'finite area, infinite circumference' to 'finite volume infinite surface' and you're done.
    1
  7725. 15:18 but what if there are a bunch of pairs with different yz AND two (or four or any even number) of solutions with the same yz? Then it WOULD give an even nr of solutions?
    1
  7726. The equilateral trianggle case can also be proven with some simple algebra. The ratio of the height to base is sqrt(3)/2. But in a grid of equidistant dots, there is no construction where an edge and it's perpendicular can have such a ratio. Ratios are always rational numbers.
    1
  7727. Never before
    1
  7728. wo ist deine Maske?
    1
  7729. So the next 4 solutions to the strand puzzle are {1189,1681}, {6930, 9800}, {40391, 57121} and {235416, 332928} in the form {house number, street length}. .
    1
  7730. I made it to the very end
    1
  7731. This reminds me of the physics problem where you can show that a square wheel will roll on a uniform matched undulating surface.
    1
  7732. 13:39 "we are changing the network but each step along the way but with each step V-E+F does not change" First you added 5 edges, then subtracted 14 edges and 5 vertices. That definitely changes the V-E+F.
    1
  7733.  @Mathologer  ok so more accurately you added 5 edges and 5 faces (5-5=0) then you subtracted 5 vertices, 14 edges, and 9 faces (5-14+9=0). Sure enough.
    1
  7734. Bailed out at 11:04. No discipline. Cool to watch. See you next time.
    1
  7735. The triangular Mathologer logo kind of looks like a warning sign. Warning: maths ahead!
    1
  7736. Your cardioids, late in this video, are epicycloids, which are what you would get if you played with Spirograph but rolled the little wheel around the outside of the big one, rather than the more typical hypocycloids, which are what you get when you roll the little wheel around the inside of the big one. Therefore, any mathematical articles and theorems regarding Spirograph patterns -- how to know how many cusps you're going to get, where they're going to lie, etc. -- ought to be equally applicable to your cardioids here -- if that's of any help to you. (What continues to defy me, though, despite your having just now explained it, is how cardioids arise here, in the Vortex patterns, in the first place, as a consequence of the cumulative straight lines of the Vortex happening to collectively lie tangent to it. It's the same thing as getting a quarter-circle by joining with a straight line the points (K, 0) and (0, K), for K = 0...N, about which I am also still waiting, after forty years, to hear a reasonable explanation. This also applies to the question of "what is the shape of the curve you see along the edge of a cube, when you spin it at high RPM around an axis that passes through opposing vertices?", another one I've been waiting since the 1980s to even find out *how to approach*....)
    1
  7737. Time to stretch our brains boys and girls.
    1
  7738. Huh, so that's why my commander told me to lace my boots like that. We did zigzag but with tied off ends.
    1
  7739. depends on odd even tower height
    1
  7740. Did you just run me through a semester of high school math in 30 minutes? I believe I studied this back in 1983, though not to this detail. This was great!
    1
  7741. The pi=4 lemma actually works if you live in a world consisting of rectangular pixels, like a screen, or presumably in Minecraft.
    1
  7742. Yes - in mathematical terms, this is the famous “taxicab geometry” where any measurement has to be taken parallel to the axes. In mathematics we say that we’re using the L1 metric or L1 norm. (Often with a lowercase curly L.) Choosing different metrics to do different things is a favourite trick of many mathematicians. In this case, choosing different metrics gives you different values of π. There are metrics for any value of π from 3 to 4. This is a fun problem to investigate with a nice theorem to prove.
    1
  7743. At a beginning I was... "What? 50 minutes? this guy don't know how YouTube works..." at the end "It's already finished? These was very well spent 50 minutes". My background: I'm a Graphics programmer and have a degree in Engineering computer science, all those shiny pixels in video games come from a lot of integrals.
    1
  7744. Wake up Babe! Mathologer just posted!!!!!
    1
  7745. I love your videos, thank you!!!
    1
  7746. My fav part is the convergence of the no 9 series
    1
  7747. 2:40 0 69 Pi
    1
  7748. Thanks for the video. I smile. great.😀
    1
  7749. Thanks for the very interesting and innovative video (I just discovered your channel). Question at 25:25 : I assume the area alpha is not the same before and after the transformation. The video makes it look the same (same color and notation) so I was confused. In the end, it seems like you never go back to the trigonometric circle so that did not matter.
    1
  7750. ​ @Mathologer no, not that one, but now i see this one too. I was still pondering about the rotations, that's why I missed it. The cut I had on mind happens shortly after 22:01.
    1
  7751. Can you start with A^3 + B^3 + C^3 = D^3 + E^3 ?
    1
  7752. Most memorable must be the fact that 100 0s series is greater than no 9s. One over a google is like 10^-100 , very small :-)
    1
  7753. Just one subtlety with the end 3D quadrilateral section: Saying there is only one case due to the symmetry of the triangular pyramid assumes that, given a set of 3 points in 3D (Euclidian) space, a 2D plane can always be found that all 3 are on
    1
  7754.  @Mathologer  Yes, it does look easy to show that from the degenerate line quadrilateral exercise
    1
  7755. I was taught this in middle school here in Poland, but the middle school I went to had a special math program so I'm not sure if it's taught in other schools.
    1
  7756. Shame Petr didn't use click-baity titles (you put that beautifully). If his paper had been called "Miracle Polygons that ABSOLUTELY CRUSH Euclid", even in Czech, it wouldn't have slept so peacefully for a century.
    1
  7757. Sounds kinda like the problem of points?
    1
  7758. very fun video! I was able to follow along for most of the video, I have taken up to Calc 2 in college (USA), and they teach us the taylor series, binomial expansion, pascal triangle.... I didn't really understand the Bernoulli numbers concept and have never played around with them anyway. Matrix notation was totally lost on me lol
    1
  7759. different rotations different walls ... same shape ... if symmetric ... if not symetric ... change walls rotation and symetry appears ... so the based on 2d axiom ... the notion is generalized!
    1
  7760. Terry Smithwick's answer is a more concise explanation than my answer and was posted a little bit before mine! Terry Smithwick also makes the very good point that 2 is handled separately, but it isn't hard to do so :)
    1
  7761. time field theory stuff. can this hep explain the 5th dimension/ weak force unification thing. See with time field theory....each point...inst necessarily a point...but rather a parameter..be it distance, or energy etc...the laws of the universe behave in this way relative to these aspects etc...its a way easier way of thinking about things.. ths kinda thing allows you to do things...weird new properties of physics appear....I already know this...there are experiments and applications proving this...I cannot be alone in doing this. if anyone else is reading...feel free to reach out!
    1
  7762. Seeing the equilateral triangle in a square grid, I thought of imaginary numbers to calculate the coordinates. Then I realised, you could set the first point into the origin, the second on 0,1 and calculate the height / the thrird point with pythagoras: (1/2)^2 + x^2 = 1. The solution contains a square root of 3. So you never hit a third point this way. Der Vergleich mit dem Dosenwerfen auf der Kirmes war witzig 😉
    1
  7763. Cocoa Proofs, part of a complete breakfash
    1
  7764. The most memorable thing Johnson making fun of the leaning tower of pisa
    1
  7765. Excellent as always!
    1
  7766. Wonderful to realise how geniuses our Indian mathematicians were. Pathetic to see current tiktok and insta kids.
    1
  7767. I love your longer format videos. I'll also be very happy to watch this smaller interludes. This was really fun.
    1
  7768. You probably have heard of these weird religious people who say math (mostly related to fractals) is infinitely beautiful therefore it has to be created by god. I would like to know what they would say when you create a pentagram from any pentagon probably something like math is evil
    1
  7769. @6:35 it's not, but it made me think of the hypercube :)
    1
  7770. i made it to the very end
    1
  7771. There's something extra special going on with those approximations. With pi, for example, Wolfram lists the following approximations. 3/1, 22/7, 333/106, 355/113, 103993/33102, 104348/33215,... First: if you check the spirals corresponding to these fractions, you'll notice they alternate between spinning left and spinning right. (I'm too lazy to say CW or CCW/ACW/whatever) This is why 333/106 is listed even though 355/113 gets even closer for pretty much the same bang for your buck: the spirals must alternate in direction. The reason for this will be clear by the end of this post. Next: these fractions also correspond to evaluating the continued fractions. Related to that point is how they appear on the microscope: 3/1 is the first good approximation, and at the beginning the spiral just looks like one arm. Then as you zoom in, the approximation loops over itself until a better approximation appears: 22/7. The seven arms. Then the arms continue to loop over and twist among themselves until the next one over lines up nicely. This becomes 333/106 and 355/113. Notice what this means on the algebraic side: 22/7 = (7 * 3 + 1) / (7 * 1 + 0) 333/106 = (15 * 22 + 3) / (15 * 7 + 1) 355/113 = (1 * 333 + 22) / (1 * 106 + 7) 103993/33102 = (292 * 355 + 333) / (292 * 113 + 106) 104348/33215 = (1 * 103993 + 355) / (1 * 33102 + 113) For starters, this comes directly from the continued fraction. But now the connection between the microscope and the continued fraction is more obvious. Given the sequence A1/B1, A2/B2, the next term A3/B3 would be: (M * A2 + A1) / (M * B2 + B1) The logic for the denominator is this: B2 represents how many spirals there were on the last iteration, and B1 how many spirals before that. When the spirals loop over, the spiral that approaches is labelled B1, and it comes over in tiny steps of B2. This also explains the alternating nature of where the spirals curve. B1 must be opposite to B2's spin. Idk it's midnight I'm yapping and idk if this is correct
    1
  7772. Yeah, but what about answer to the rhetorical question in title?:)
    1
  7773. Guarda o meu joinha mano. E obrigado pelo vídeo.
    1
  7774. Every video is beautiful miracle. Thank you The Kind Mathgician ))
    1
  7775. Fascinating, but hardly of cosmic significance.
    1
  7776. With Velcro.
    1
  7777. My immediate response to the "can you see where this is going" at the start was "it folds to the same triangle? No, that's dumb, of course it doesn't" And thus, it remained undiscovered for thousands of years 😅
    1
  7778. 5
    1
  7779. Integral of x^x from-0 to-1 without knowing how it's done. If you can do this, fine calculus is easy. Differential calculus is easy but non numerical "simple" integrals are not trivial and differential equations are very easily non elementary...(if not approximated)
    1
  7780. 700 year old proof hands down. Elegant!
    1
  7781. What's the meaning/translation of the hand-sign pentagram on your t-shirt?
    1
  7782. What would very much interest me is the link between i the imaginary unit and the probability in general. Kheyly motechakerram. ( thank you in persian ) A lord of iron :-) 2020
    1
  7783. I once designed a digital circuit to compute the remainder of large number to determine into how many 53-byte ATM cells a network packet would fragment. That was an interesting problem. Also addressed the problem of soft errors in an arithmetic logic unit where data buses were parity-protected. By using modulo-3 "parity" rather than modulo-2 parity, we could check the arithmetic of an adder or multiplier by precomputing the correct value for the parity of the sum or product.
    1
  7784. Ramanajun was great.But this does not stop.we are soon getting new Ramanajun in this century
    1
  7785. At 4:32 an expression for [( sin x ) / x ] is found, making me wonder if an integral expression can be found from the right hand side, bypassing both Feynman’s technique for finding that integral and using contour integration over the complex plane to find the result.
    1
  7786. Layered with a protractor, we can divide angles to get nth root of x.
    1
  7787. The "worm on a rubber band" I had never heard of, it reminds me of the Alcubierre drive, (expanding space behind and contracting space in front) to travel "faster than light". Obvious here you have expansion on both sides though which makes it less intuitive, and that's more fun. I am curious about what happens if the ant isn't treated as point like though, as at some expansion rate it would be ripped apart or suffering extreme carpet burn.
    1
  7788. This is extremely cathartic
    1
  7789. Hi Mathologer My most memorable part was the visual proof that the Euler–Mascheroni constant is less than 1 and greater than 0.5. Great stuff. Thank you. 😀
    1
  7790. Wonderful!😍
    1
  7791. Thx, love the way you explain. Best from Germany 🫶
    1
  7792. I'm not a mathematician by any stretch (pardon the pun) so I could not easily investigate too far into my particular question - which is: what applications might the stretching concept have in assisting in the transformation of linear equations from a standard graph into those dealing with polar coordinates, or what other sorts of circular calculation methods could come into play outside of the literal numbers on the wheel itself? The whole notion may be completely unapplicable because the stretch itself is a thought experiment, but since the result is so concrete it seems to my untrained mind that there might be a link... perhaps something like stereographic projection, or some kind of use of the Riemann zeta function, since pi is used in the definition of a certain point that is moved toward (but more like away from, in this case)
    1
  7793. This man wears a unique shirt for every video and every video I wait to see the shirt. I appreciate that.
    1
  7794. I liked the coloring proof better
    1
  7795. The glasses: 666 but i counted it by trying the different ways of how dominoes lays and counting the number of ways in them (the resulting formula was like 3•3•(2•2•(2•2•2+2•2+1)+2+2•(2•2+1))) UPDATE: please tell me the 'aha' moment that I should've found
    1
  7796. Does the pigeon hole principle apply to the birthday paradox, and if so how?
    1
  7797. Incredible, I did A Level Maths at a private school and was never taught this!
    1
  7798. Vihart taught me the first one with origami
    1
  7799. That's a rich slice of π!
    1
  7800. “Nice smooth curvy-curve”. This was the xmas gift I needed :-D
    1
  7801. When me & my 2 friends were in class 2,we somehow learned 3's divisibility test and used it to leave each other messages and feel smarter than others[maybe bc mental addition is all we were good at],it was truly pitiful but kept our bubbles from busting for 2,3 years
    1
  7802. I keep thinking about the +2 business. Interestingly, for +2 we get cubes, but if we change it to +1 we get simplexes. What then would be +3? +4? +5? ... and so on?
    1
  7803. My favorite factoid from this video is that the overhang increases like the cube root of the number of blocks, as easily seen with the parabola-shaped stacks. (And, by the way, all parabolas are similar, which is one of my favorite factoids about parabolas)
    1
  7804. I've never heard sinh pronounced "shine." At least in the US all I've ever heard is "cinch." Excellent video, very intuitive!
    1
  7805. looks like me that if you take 1 ^ x solutions are e ^ ( 2kPi ) * x I don't see what is special if x = kind of golden ration , or fraction of pi ... if x is integer or a fraction 2k x will be 1/n ---> spiral ---> if x is real 2kx wil be in neigberhood of a fraction ...1/n ---> spiral .( ie with strate arms ) .. because for that given k de difference between 2k Pi x and (2k+1) Pi x are close ie mod (2pi)
    1
  7806. thanks, my head only exploded twice.
    1
  7807. Goat of youtube math channels.
    1
  7808. Sir will you please make a vedio on the question you asked us in the strand puzzle vedio
    1
  7809. I did not learn about it at school but I did know the proofs. Sir, you would make an excellent Bond-villan, they should cast you for it. English with a heavy German accent yet understandable, the laugh and the crazy scientist vibe who can come up with an evil plan to conquer the world. 😆
    1
  7810. Amazing I just loved this!
    1
  7811. 18:17 since the solution passes through the vertices of the graph exactly once, and each graph has 3^n vertices, isn’t it just (3^n)-1?
    1
  7812. Hurrrray!
    1
  7813. I love the t-shirt you are wearing it contains many useful series
    1
  7814. I like this new format, but the problem was too easy to be not interesting
    1
  7815. I most enjoyed the Bishop's succinct proof.
    1
  7816. 23:58 Choose function is undefined for negative values, but we can easily generalise known formula using generalisation of factorial - gamma function. Then everything works out nicely
    1
  7817. is there a 3D (or higher) version of the WSW theorem? Spheres instead of circles.
    1
  7818. I made it to the very end. Not sur how, but I made it.
    1
  7819. I saw the video was half an hour an thought "well i really dont have the time right now and maybe not enough concentration, but it couldnt hurt to just watch a liiiitle". It felt like 6 minutes and now i watched the full video, so good job again on a captivating video.
    1
  7820. The traditional proof of the Pythagorean theorem uses twisted squares as shown in this video, which coincides with the 30-60-90 degrees triangle. I think it is much simpler and more elegant to use the 45-45-90 degrees triangle to show the proof of this theorem, as that requires only a single diagonal be drawn between two corners of the square, dividing it neatly into two triangles of equal area, and the rest then follows without having to twist or chase any bugs at all!
    1
  7821. Can someone explain gravity to me!!!! Read more!!!!!!????? I wanna know how atoms are not physical objects!!?? I wanna know why physical objects are not physical!!! Is the www and universe just waves? What is entropy? What is time? What is gravity? They are all the same anomaly. Light energy and objects can only travel at the speed of light because they cannot be stopped using its own entropy; energy. Entropy gravity and time. Gravitating orbits slow time entropy. Perpetual machines are orbiting around entropy and energy, creating time. An orbit is just entropy with given mass/energy flowing thru time. Gravity exists because of entropy Time is entropy! (True) In the universe; time energy flows, entropy and gravity uses the energy to creates objects in space, mass and light which all retracts back to entropy, gravity and time. A universe only has a given amount of time and entropy, therefor that universe will dictate the mass and energy .
    1
  7822. These thought experiments work best if we assume horses to be spherical.
    1
  7823. Now whole universe🌌 following💻👌 🕉️hindu🕉️ dharma☸️ because it's following the🐎 natural universe🌌 Karma ki gati Ultimately universal language
    1
  7824. 16:14 Awesome minimal cheat sheet!
    1
  7825. expected 4(3+2+1)
    1
  7826. It reminds me as well of tangram puzzles and the sort (and also fractal geometry), so maybe this property may have discovered before said date, but maybe not proven as such, proofs of which I like the dot one the best , turning this more in a Yayoi Kusama video than a Kurosawa one😊, but anyway interesting as always, thanks for sharing.
    1
  7827. Wonderful!
    1
  7828. Challenge: It is trivial that a=1 and the rest is 2, but let's actually prove it. If we let k be coefficient for a, then k*1+2(k-1+k-2...+1)=k+k^2-k=k^2 Thus, the output sequence would look like k*{(k-1)*(k-2)*...1}^2, which can be expressed as k!(k-1)!
    1
  7829. This could be used to approximate pi then. (?)
    1
  7830. Thank you for this, this answered and sufficiently explained questions about extra dimensions I didn’t quite know to ask yet but had visualized in my head all this time. Pretty beautiful
    1
  7831. Thanks again. I love to watch your videos on a Sunday afternoon with my coffee. Oh and cool T shirt
    1
  7832. 2:35 that's just the algebraic repetition of the geometric "moving triangles" directly preceding it
    1
  7833. But, what is the integral of the number e to the power of minus x squared? How to find it?
    1
  7834. SHINY
    1
  7835. I believe there's a rounding error on your T shirt.
    1
  7836. @13:00 a^2 = Σ(odd) so applying moessner gives Π(odd) although I don't know if you can simplify it
    1
  7837. I made ot to the end.
    1
  7838. Wow! I believe this was exactly what I was looking for. Thank you!
    1
  7839. 9:40 I wasn't taught the remainder extension in school, but I independently worked it out for myself years later (albeit in slightly different terms; my understanding of it until now has been that the digital root corresponds to the ones column when the number is rendered in base 9, which I now realize is an equivalent statement)
    1
  7840. I also noticed that for the 119,120,169 when you rearrange you get the factorization 119^2=49*289, which when taking the root of both sides gives the non-trivial factorization 119=7*17. That seems like a real world application to me ;-). Thanks!
    1
  7841. Am I the only one who keeps his hand on the space bar while watching Mathologer videos ?
    1
  7842. 1, 2, 4, 8, 16, 32, 64 No two non overlapping groups have the same sum
    1
  7843. 42:30 you could use an extra hand
    1
  7844. honestly I really wish you'd write a book instead of make videos. it could go into more detail, give expository views into fields and topics. it be like a collection of unknown gems in mathematics. it could end chapters with with hidden gems you know of that you hint readers to finding themselves. great videos! but think about it
    1
  7845. Allways intrested what will be the new video
    1
  7846. Thanks for another very interesting selection of related results, beautifully presented as always. I think the irregular "leaning tower" was my favourite, with the unexpected simplicity of the "finite no 9s" proof a close second.
    1
  7847. Your opening spiel I accidentally discovered that idea in high school. The next day I went up to the math teacher and described what I noticed but he seemed oblivious to the idea. I was puzzled and thought either he totally didn't get the idea or was jaded to such things. I left out some details of what I saw at home during my experiments.
    1
  7848. Nice T-shirt, but not properly rounded...
    1
  7849. This is the first proof I saw, when was in middle school. It was mesmerizing
    1
  7850. Your t-shirts are really cool. You should start making some merch.
    1
  7851. The most memorable part was Kempner's proof. Very simple and elegant.
    1
  7852. 👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏
    1
  7853. What happens when we use something other than base 10?
    1
  7854. 💛💛💛 Mind boggling content. Information provided surely will help for aspirants. I #BrainBlitzAudios recommend it to others.👍👍👍👍👍👍👍👍👍👍😍😍😍😍😍
    1
  7855. What is the ratio thou. 1 to 3?
    1
  7856. Thanks Sir, you are expanding every human’s brain and consciousness. Cheers.
    1
  7857. how does this work for an n-gon where n doesn't divide 360?
    1
  7858. My grand-dads rotating sliderule (well, as well as sliding sliderule, for that matter) are always on the shelf above my computer. It's a pity YT has no image comments...
    1
  7859. I do not like the upper dimensions, they break my brain, and they are so "empty", for the lack of better word... like a 100 dimensional hyperball in 100 dimensional hypercube basically does not take any space at all...it is so weird...
    1
  7860. I did not get why , in part 6 , p could not be replaced by n , say 325 and still output that only one way is possible, when it is provably false. I'm probably just in way over my head here.
    1
  7861. Awesome thanks
    1
  7862. I seem to recall Numberphile doing a video on Zagier's proof.
    1
  7863. That's so amazing , I love pigeon hole principle
    1
  7864. Is it true that every factorization of every natural number has associated with it a primitive Pythagorean triple, just as 6=2x3 is associated with 3,4,5 via out circles?
    1
  7865. A little insight gained into d theta from this
    1
  7866. The folded, and the folded-folded, triangles makes me think of a derivative function, and the derivative of a derivative - just in a geometric form... 0.o
    1
  7867. quartic formula derivation when? :>
    1
  7868. I was familiar with neither… despite gaining an A grade in every Maths exam I took… sadly( or possibly happily) I didn’t pay for my schooling so can’t get my money back!! 😬
    1
  7869. I was given this problem and came up with practically same solution in the mid 1990s, without the benefit of your excellent animations of course. My professor didn't "Like it", which is as infuriating now as it was then.
    1
  7870. I miss Marty 😔😔😔😔
    1
  7871. Happy New Year, Burkard! I think this was one of your best ever videos. So much fun! I got very slightly lost in the masterclass at the end, but I think I got the general idea of the argument. I did most of the challenges too, and got a lot of enjoyment from them. Are there similar interesting domino tiling problems in 3 dimensions? I'd love to see a video about that too!
    1
  7872. 24:36 gamma is greater than 0.5 because if you link the 3 corners of every blue bits, it makes a triangle with a smaller area. Their area is half their height times a width of 1 always. Summing all the heights you get one. Therefore summing all areas you get 0.5.
    1
  7873. The conchoid of Nicomedes relates to newly formed dark matter. The mysterious eighth conformation of matter as prescribed by General Relativity is when that dark matter is being impacted by surrounding black holes and the center of the central bulge becomes so energetic that it spaghettifies in an "elliptical"/eye pattern - without breaking! Delanges trisectrix is similar, but at the convergence areas of atomic nuclei. The hyperbola of eccentricity 2. Here, there is no brief stability like with dark matter. There is no thought of half-lives of isotopes. Just matter being compressed overall, but still stretched a little as it's clumped together before all being flattened at the core of the Big Crunch mass.
    1
  7874. I believe the answer to the programming challenge is: 427707562988712128 ways to make change for $2000. We can multiply polynomials in O(nlg(n)) time using an FFT. where n is the amount we are trying to find the number of ways to make change for. A fairly straightforward efficient algorithm for computing the number of ways too make change for $n is to: 1. compute the coefficient arrays of the 6 polynomials. 2. Multiply all the coefficient arrays together using the FFT. For each multiplication between two of the coefficiant arrays a and b we should do: ifft(fft(a)*fft(b)) where * is point wise multiplication. 3. Get the nth index of the resulting array. This will calculate the answer for $2000 in a couple of seconds on my 2015 MacBook Pro.
    1
  7875. New yorkers be like hey that's my phone number
    1
  7876. I followed the swivel proof in real time, so I liked that best. I have issues with concentration, so I was not focussed on the colour proof. For me to be sure which proof I really preferred, you would need to reverse the order in which you give these proofs. Maybe give me a chance to forget everything first.
    1
  7877. Is CCC in base 13 double evil?
    1
  7878. People who call Math Boring do not exist 😌
    1
  7879. Handsome man
    1
  7880. me. I am the handsome man  @Mathologer 
    1
  7881. Lots of marvelous stuff here, but the most memorable for me was the crazy convergent subseries at the end, and the fact that somebody actually figured out the value of these sums.
    1
  7882. My favourite part was the Q.E.D. 🐈
    1
  7883. try n*log(primes) | mod 9, ie, multiples of log2() of primes mapped to a ring of 9, you get the probability distribution function estimation of the primes log distribution, ie, the mod instead or same as the angle, mod is the angle from 0 to 360 of mod 0 to max
    1
  7884. histogram of the log2 primes mod 9 should be interesting
    1
  7885. swivel
    1
  7886. Burkard thank you for you wonderfully accessible, stimulating, and enlightening videos. My favorite aspect was how the parabolic and customer staged tiling provide better solutions then the greed algorithm. I also loved your "Are We There Yet" t-shirt. I don't see it on your store. Do you have a pointer to where it can be purchased?
    1
  7887. I'm teaching my kids this one day
    1
  7888. Nice
    1
  7889. I prefer your explanation because actually it is entirely equivalent to the binary explanation, but without adding in an extra step of turning things into binary.
    1
  7890. The whole "what's next scheme is fundamentally silly!" YES 🤣
    1
  7891. Nice shirt
    1
  7892. John Baez ? Great name 😅
    1
  7893. The rubber band was amazing. I have used a slide rule since 1983 (high school) and still have about half a dozen straight and circular. When in the military (1986), I was taking a test prior to a school (to determine my math skills). They stated "No calculators". I produced my slide rule and they laughed and said "If you can use that, you don't need the test." (They still made me take the test, and I didn't need it anyway.) I use the circular rule in my watch as a quick conversion between miles and KM. I had a keychain one I used for calculating MPG quickly and tank range in my car. On regular (linear) slide rules, I prefer the AB scale, as I don't have to flip ends if the answer is at the other end of the scale. Where the CD scale may have to be end swapped.
    1
  7894. I love that it took 2300 years to spot this and another century before Burkhard comes up with an elegant geometric proof. Its not hard, guys. What took you so long?
    1
  7895. Thank you - when everything else is depressing and pernicious there is always Mathloger to plunge us into a better reality!
    1
  7896. I now, wanna do a math degree
    1
  7897. They don't even teach cause and effect anymore, "Proofs" wouldn't make any sense to them. Male+Female=Baby.
    1
  7898. Thank you sir for sharing Indian Mathematicians like Madhava's contributions.
    1
  7899. 14:05 is it 56 squares
    1
  7900. I love that that the thumbnail says in under ten minutes, but the video is 50 minutes.
    1
  7901.  @Mathologer  yep, that is exactly how I did the typesetting for my honors thesis hahaha
    1
  7902. Unusually short, but as good as usual.
    1
  7903. Why is there a translation for all the languages ​​of the world except for the Arabic language? This is a strange thing
    1
  7904. Oh the coin thing hurts my brains. It’s a lovely animation and in the animation it makes sense, but the red line still is exactly the same distance, rolled or not. So why does the coiling add a rotation, the fact you just say it is with the animation doesn’t make it clearer… to me of course. ;)
    1
  7905. 15:00 I get 666 as determinant
    1
  7906. To me it looked like @mathologer used 3.18 for pi in his example
    1
  7907. Thanks and wishing all the best for 2020
    1
  7908. The triangle method is nice, and it gives a formula, but for the first sum, I had a faster method I think. 1+2+3+4+5...+100, I immediately figured that there are 50 pairs of numbers that make a sum of 101. 1+100, 2+99, 3+98 and so on, the last being 50+51. So 1+2+3+4+5...+100 is equal to 50x101. 5050
    1
  7909. One vote for Gama from Belgrade.
    1
  7910. Woww...an amazing insight into how Euler's mind worked. I sooo much enjoyed this video. It involves taking notes and a bit of thought but it is worth all the effort in the end.
    1
  7911. 3. We get a 1 0 1 triangle
    1
  7912. Great video! I went to the program PROMYS EUROPE 2018 (which I'm sure you know) and they expected me to prove this. I did not manage to do it but I still got many of the pieces before I was given the whole spoiler of the proof. It's super cool that you dare to make a video of such hard stuff! Thanks from Spain.
    1
  7913. Gama Rays
    1
  7914. Andrew Poelstra mentioned in the Patreon. If I am right he worked on Adapter Signatures and is one hell of a cryptographer. I met him once. Was Lucky. Am I right Andrew?
    1
  7915. Dat maths gnome @ :11 ‘tho 😍
    1
  7916. omg hyperbolic trig, I'm gonna love this!
    1
  7917. 13:30 I am French and I think we don't have to pronounce the "s" in his name. It is like "Nicole (or Nicolas) Oreme".
    1
  7918. Wait, the "at least ten 9s" series diverges, no? That's my intuition at least. It's easy to see that a subseries of "at least 10 zeros" diverges...
    1
  7919. The 5 tower pattern in that shirt was an excellent choice!
    1
  7920. I love that little cubes within larger cubes are now called 'cubies'
    1
  7921. 1/12 n^2 (n^2 - 1)
    1
  7922. Huh. I wonder what else we can know about this. Und eine feohiche Weinacht!
    1
  7923.  @Mathologer  ❤
    1
  7924. I think you made a typo at 28:38. Shouldn't there be a 5 in the second row there?
    1
  7925. I find it hard to believe some of these comments about coming up with the modular approach and applying it within 5 seconds. I just cheated with a meta analysis. 'If Mathologer wants me to guess the rule, then the rule is guessable.'
    1
  7926. Too complex the end i do not all understood
    1
  7927. Answer: because a 38 minute YouTube video would take 6 weeks to get through in public school.
    1
  7928. Before I press play again, nice Doctor Who miniature of yours! You look cool! Ahem. Sorry. I already know this trick. After watching, I will tell you if that's so. So make it so so ENGAGE! 🤓🖖
    1
  7929. 11:40 No no no ! Euler is the number one.
    1
  7930. This is a bit bizarre. I was just reading about this in your QED book, a few days ago.😮 Congratulations on 100 videos!
    1
  7931. 1:57 G.H.Hardy
    1
  7932. Was already familiar with the 2nd proof. Can't recall where I found it, but I don't think it was at school.
    1
  7933. I... never knew why the calculus of derivatives is called "differential calculus"... thank you!
    1
  7934. You can also draw the spirals by connecting a dot to next closest dot (by euclidean distance), then another spiral is to second closest dot, etc.
    1
  7935. Thank you. Absolutely beautiful.
    1
  7936. Wow! - mind blown! - I love it . . Mathologer is going on my maths list . .
    1
  7937. Nice!
    1
  7938. I THINK I follow the proof. I'd worked out the two-square theorem existed before I knew what it was. But what I've never found out (and I searched the net lots) is - given a larger prime, how do you work out what the two unique squares that sum to it are, other than by trial and error?
    1
  7939. Wonderful video! I enjoyed your neat animation at about 9:22 - what tools did you use to create it? When I was a kid, my Dad gave me a 6-inch (!) slide rule he had used in his engineering work. Yeah, bit of a gimmick, only 6 inches, so not terribly accurate, but it was a promotional item from one of his suppliers. He, of course, preferred using the 5-figure log tables that his father gave him ...! Must look in the op-shops to see if I can find a proper 14-inch Faber-Castell sly drool (as we also used to say in Strine - the official language of Oz).
    1
  7940. In the UK there is a maximum amount of change under £1 that is legal tender. Keeping to the legal tender limits that rather reduces the counts.
    1
  7941. So so good
    1
  7942. Solution spoiler ⚠️ . . . . . . . . . 3=1+½+⅓+¼+⅕+⅙+⅛+1/9+1/10+1/12+1/18+1/20+1/40. I'm proud to have solved it alone. If you have a better solution please show it.
    1
  7943. R G P was to do with color theory? Quadratic colors I believe? Perhaps Marty was vying for a subtractive version? Coffee not soaked in yet.
    1
  7944. I had to watch it twice to understand why you panned into the eyes. Clever stuff
    1
  7945. visualizing combinations in ths way is amazing
    1
  7946. Can you make a video on differential geometry?? Please
    1
  7947. Well can a high school student understand the book?
    1
  7948. I remember seeing a different Chinese manuscript showing graphic proof of the theorem. But never saw twisted square diagram.
    1
  7949. A sPIEral shirt? Nice :)
    1
  7950. Mathologer, could you please tell me what softwares you use to create your videos? They're awesome. My friend is fascinated by your videos and we wish to know ☺️
    1
  7951. BTW Anyone who tried to crack Collatz Conjecture knows that 4k+3 are BAD numbers. 4k and 4k+2 transform into lower number in 1 step, 4k+1 take 3 steps. 4k+3 make all the trouble - can't prove they ever do.
    1
  7952. I thought this was going to be the "Turtle Method" for multiplying with the standard algorithm and I was about to get on my soapbox and be very upset. This is a little more complex than multiplying 🤣
    1
  7953. 10:39 Is it 33 (34-1)? Like 20 = 21 - 1?
    1
  7954. No quite evil! You’re off by ~.004.
    1
  7955. circular slide rule? numberphile has a short video about a circular one from 1901
    1
  7956. transcribed problem: The inhabitants of little Wurzelfold have been considerably roused out of their lethargy by the great war. Like the rest of the world, they talk of little else, so it is natural that the particular conversation which we have to relate should have a distinct material flavor. The little group of villagers gathered together in the red lion Inn where all respectable neighbor whose ages may safely be said to range from 50 to 70 years. All fir young men of the village had enlisted, and the opinion was lead by everyone that if the English Navy and Army did not in consequence "knock the Germans into a cocked hat", it was not the fault of little Wurzelfold. There is no necessity to introduce the reader individually to these estimable men, drawn together by bonds of mutual sympathy and desire to forget their anxieties in pleasant intercourse. We are entirely concerned with the curious riddles and proses that, without any premeditation, happened to be forthcoming on the particular occasion of which we write. "I was talking the other day" said William Rogers to the other villagers gathered round the Inn fire, "to a gentleman about that place called Louvain, what well– used to visit a Belgian friend there. He said the house of his friend was in a long street numbered on his side one, two, three, and so on, and that all the numbers of him added up exactly the same as all the numbers on the other side of him. Funny thing that! He said he knew there was more than fifty houses on that side of the street, but not so many as five hundred. I made mention of the matter to our parson, and he took a pencil and worked out the number of the house where the Belgian lived. I don't know how he done it." Perhaps the reader may like to discover the number of that house. "That reminds me," said James Woodhouse " of Miss Wilkinson, down Bradford way. She promised to make a lot of red crosses, and to stitch 'em on to white bands for hospital nurses at the war. She bought a lot o material and started cuttin' out all the crosses. When she had done it she found that by mistake she'd made em just twice as big as they ought for to be. She is very economical, as you may say, is so she went to the schoolmaster and asked him how she was to cut each cross into two crosses of the same shape. He showed her an artful way of doing it in five pieces so that both crosses should be of one size, and no waste. He's a sharp man, I reckon." It is true that the red cross can be cut into five pieces that will form the two crosses shown in our first diagram.
    1
  7957. Mathologer suffering from a mild form of body hair dysmorphia.
    1
  7958. I am most annoyed with myself for mislaying my circular slide rule when we moved house 😞 some years ago. I do still have a bamboo straight slide rule inherited from my late father who acquired it when studying industrial chemistry at the Gordon, Geelong.
    1
  7959. Can you do a video that intuitively explains the TREE(3) function? I have yet to see it explained in a way that can be properly understood by a layman.
    1
  7960. The most memorable part: that one time I stacked 1.5*10^43 blocks perfectly
    1
  7961. Is it normal being French and don't understanding this page ? 5:21
    1
  7962.  @Mathologer  Thanks for pointing it out!
    1
  7963. Wow, you never stop blowing me away with your videos! I'd offer you a googol dollars for letting me watch it, but I'm afraid I don't have the exact change!
    1
  7964. Calculus recap in 40 min.
    1
  7965. 2 of my favorite channels are German, this and WWII with Indy Neidell :-) ❤
    1
  7966. Coin rotations 4 7 pointed star 440 degrees
    1
  7967. 22:07: How can the maximum cycle length be 2260? What if your algorithm has more than 2260 steps? And OBTW, what if it's prime? How can (for example) a non-trivial (do nothing) 2267* step algorithm possibly repeat in fewer than 2260 steps? And OBTW, how can an algorithm of 43,252,003,274,489,856,001 steps (clearly mutually prime with 43,252,003,274,489,856,000) repeat in a factor of 43,252,003,274,489,856,000 steps? * 2267 is prime.
    1
  7968. One step is one application of the algorithm. So if eg. your algorithm is to move the top face clockwise three times, it will cycle in twelve individual moves, but four total steps
    1
  7969. Challenge 1: 292(there is only 1 way for 100cents to make 1 dollar) Or in the expansion of the series, multiplying the product before with (1+x^100) will only add 1 new term of x^100 For the 1,2,4,8 the expansion of (1+x)(1+x2)(1+x4)(1+x8) is 1+x+x2+....x15 (simply multiply and divide by term(1-x)) In other point of view, as all coins are in power of 2, write the values in binary, which is 1,10,100,1000. It is clear that the digits won't interfere each other no matter how we add them. Their combination is simply 2^4-1=15 (excluding 0) Conclusion: 15 values can be made with 1 way each, if 0 is not considered as a value
    1
  7970. Chapter 2:Too bad I'm beginner in programming
    1
  7971. The proof for the second formula in chapter 3: Combinatorics method: As said in the video, raise both sides to power of n, we now have n copies of the original series that need to be multiplied :( Coefficient of x0 term is clearly 1 Coefficient of x is nC1 (choose 1 'x' from 1 of the series and the rest is 1, there will be nC1 choices) Coefficient of x2 is nC1+nC2 (x2 is 1 x2 or x*x) //nC1+nC2=(n+1)C2 Coefficient of x3 is nC1+2*nC2+nC3 which can be reduced to n+2C3 in similar way. From this point onwards the coefficient of xm will be nC1+(m-1C1)nC2+(m-1C2)nC3+...+nCm. We can see the result by putting the sequence into a pascal triangle and it will reduce into (n+m-1)Cm in a inverted pascal triangle matter. Also (n+m-1)Cm=(n+m-1)C(n-1) which is shown in video Therefore coefficient of xm is (n+m-1)C(n-1) The formula is proved
    1
  7972. In calculus method: d/dx(1/(1-x))=1/(1-x)² by taking derivatives n-1 times, RHS=(n-1)!(1/(1-x)ⁿ) For LHS: After taking derivatives n-1 times, x^(n-1) will become (n-1)!, x^n will become 2*3*...*n (n-1 terms) Subsequently, x^a (a>=n-1) will become a!/(a-(n-1))! x^(a-(n-1)) dividing both sides by (n-1)! and the coefficient of x^a is perfectly aCn-1 Now as a>=n-1, put a as n-1,n,n+1.... into each term and LHS will become the second formula
    1
  7973. Because A.C.A.P.? 🤔
    1
  7974. @28'50 did you see the kaprekar constant !!!
    1
  7975. Yeah Mathologer!!! I teach calculus to my students almost like you just showed.
    1
  7976. yaaay seal of approval! my favourite part of this video was the optimal block-stackings being much better than the harmonic sum-style stacking.
    1
  7977. the 'wiz-wheel' of aviation renown!
    1
  7978. Since we can say that one series is greater than the other by comparing two series - term by term - such that corresponding numbers of one term is greater than the other, why can't we disprove Ramanujan's sum of integers (=-1/12) by comparing the integer series to this harmonic series? Every term in the integer series is greater than the corresponding term in the harmonic series and hence must have a greater sum than it and since the sum of this series is infinity, so should be the sum of the integer series. Btw my favorite part was the chapter on "Terrible Aim".
    1
  7979. Another stab at animating the tiling algorithm for square boards: https://arctic-circle.netlify.app/ Maybe the animation is still a bit slow, or it needs a switch to jump forward a bit more quickly, but this was fun :)
    1
  7980. Why is it 4k+1 and 4k+3 instead of 4k+1 and 4k-1?
    1
  7981. Didn't we find an upper bound for the spacing of prime numbers? In 2013 Yitang Zhang showed that lim inf (p_(n+1) - p_n) <7*10^7, so we can use a subseries 1/(n*7*10^7). Because the denominator is linear, then it diverges.
    1
  7982. i don't get the finite path part... how is it not an infinitely long path?
    1
  7983. I have to cast my vote to the Euler-Mascheroni constant, it's my favorite number! ^^
    1
  7984. At the fist stage p=4 still. From the second stage it becomes not symetrical related to the curve of the circle ! Illusion.
    1
  7985. Hahaha, I'm just wonderng at who downvotes this. "It got too complicated! F this!"
    1
  7986. Wonderful.
    1
  7987. Cowsine
    1
  7988. Beautiful video :D
    1
  7989. At 12:30 a green square disappears, so the region is no longer the same
    1
  7990. hello! Mathologer! excuse me impetuousness , but i just wanna ask if giving a proof to Fermat's last equation using simple rules would still be a thingy or not ? thanks!
    1
  7991.  @Mathologer  so i think what i arrived at is just hallucination , thanks for your quick response
    1
  7992. This is the best t-shirt I've ever seen! I knew both proofs!
    1
  7993. I often have shoes with longer strings than I need, so I think looking for the longest shoe lacing is also useful.
    1
  7994. 5:40 C programmers be like
    1
  7995. RIP Tasmania
    1
  7996. 25.807 is closer
    1
  7997. I literally watched this 10 minutes before the lesson where we learned about this.
    1
  7998. The background noise sounds like cicadas, or maybe crickets. Definitely insects of some sort.
    1
  7999. I really enjoyed this video! It's one of your best!
    1
  8000. I was a fan of this channel before taking MTH2321 this year and after I realised you ran it I was pleasantly surprised!
    1
  8001. You are a mathematical genius with an amazing memory. I enjoyed this so much that I saved a copy of it in case it disappears from YouTube. However, my appetite is not satisfied. Does this 9 modulus system connect in any way with the golden ratio? There are 10 fingers on your hand and 10 toes on your foot. I am just curious. Maybe the amazing thing is not the mathematics of Tesla's 3-6-9 ( Vortex Math), but rather the importance of it in the design and functioning of the laws of the universe. Please, post another video on this, that answers my question. Thanks!
    1
  8002. As someone who study computing science, that One-Sentence-Proof was intuitive enough for me to understand, I'm very well used to such symbolic manipulations by intuition from being a computer programmer. But sure the visual representation is beautiful.
    1
  8003. swiveling seems simpler, thus better to me
    1
  8004. Hey! Where do you get these T-shirts from? I'd really love to buy them :) Or maybe you're creating them on your own? Anyone knows?
    1
  8005. 3-6-9 Damn, she fine...
    1
  8006. I did not click this video expecting to learn a new cursed map projection
    1
  8007. 2 is the sum of the reciprocals of [1, 2, 3, 6] 3 is the sum of the reciprocals of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 230, 57960] Had to bust out python, but the algorithm is quite simple.
    1
  8008. I think the swivel proof, it is a bit more convincing
    1
  8009. 21:38 cicadas
    1
  8010. Seems very familiar, except for the folding. Having been a working class physicist most of my life, shooting from the hip, I seem to recall that the intersection point is the center of mass for a triangular plate of any thickness. If one puts a needle through that point, the plate should be perfectly balanced regardless of the type of triangle, Also, I believe that sometime in my career, it was necessary or fun to prove that the three areas involving your pairs of triangles at the vertices are equal. Trig suffices but it's not as elegant as the folding trick.
    1
  8011. I liked the way the series skipped the whole numbers
    1
  8012. Have been watching your vids for a very long time and I finally realized today I haven't subscribed to you! Shame on me.
    1
  8013. ❤ I prefer the second proof as it explains why, which they angle counting didn't (but could have, right?)
    1
  8014. You used purple instead of blue because you like Role Playing Games?
    1
  8015. In Quebec, students learn this formula in grade 10 university stream math. We also learn the formula given side-angle-side: Area=1/2ab*sin(C)
    1
  8016. I don’t know how to come up with the first proof. The second one was the one I came up with on my own when I tried to prove it. I like it better because I understand it. If you explain the first one step by step, then I might like it better.
    1
  8017. I actually calculated the (signed) area of the triangle with side lengths aA, bB, and cC, which I call the Ptolemian, in an attempt to prove the inscribed square theorem and because I thought it would be an efficient way to check if four points lie on a common circles. It turns out to be a constant time the determinant of the 4×4 matrix, with row entries of 1, x, y and x²+y², for every point (x,y) of the original quadrilateral. This is the algebraic way of saying that the points satisfy a common equation of the form a+bx+cy+d(x²+y²), which is obvious in hindsight. It also satifies some interesting identities using area, and given a quadrilateral ABCD, the Ptolemian of this quadrilateral is equal to a constant times the area of ABC times the power of D to ⊙ABC.
    1
  8018. I'm wondering about the relation to z-transform, the formalisms seem to be the same
    1
  8019. Why we can find all the possible sums like this 23:31 ???
    1
  8020. The satisfaction while watching your videos does not compare to anything
    1
  8021. Just WOW!
    1
  8022. his shirt doesn't know how to round.
    1
  8023. The most surprising fact in your video is to find round about infinite interessant things in infinity.
    1
  8024. Good morning posted southern california
    1
  8025. Visually, the space between the digits on the first turn of the rubber and around the wheel looks to be in the proportion of Benford’s Law. Is this in fact true? And if so, is there a good explanation of how these are related?
    1
  8026. It's the same scale: log10.
    1
  8027. It would be extremely instructive to know how people come up with these kinds of proofs.
    1
  8028. Who made the beats in this video? 🔥🔥🔥 Got diatracted at the end and thought autoplay put on a heat freestyle mix.
    1
  8029. WAOWWWWW💓💓
    1
  8030. Finally! The Glabrezu demon summoning ritual!
    1
  8031. Great video! Thanks for all the content!
    1
  8032. Somehow hints always tell you only the things you saw in the first 10 seconds xD
    1
  8033. 50!!!!!!!!!!!!!!
    1
  8034. 𝓯𝓻𝓮𝓪𝓴𝔂 identity??
    1
  8035. Haha i once made a pi approximator based on the harmonic series in javascript Still took like almost 3 days to get the 11th factorial right, calulculated a billion additions and subtractions and 2 billion divisioms per loop rather than just one to speed it up significantly (+1 so you would see the value flip, otherwise you onlt see either the low or the high side, it has to be odd) Let's just say ramanujan's approximation resolves a lot quicker lol
    1
  8036. Of course. This is exactly why I like maths. I wanna become crazy :P
    1
  8037. Favorite youtube channel.
    1
  8038. The other rows of the matrix corresponds to multiplying by r and s respectively. Probably explains a lot.
    1
  8039. Hello, awesome video!! I stumbled upon your channel somehow and I loved your breakdown and visualization! I just had one question. I think you mention this earlier in the video but it seems like the arctic circle phenomena and this emergent region at each of the vertices of constant colour is somewhat a result of the "shuffling" method where you assign each domino a direction. A specific tile colour "building up" at each vertex and creating this effect makes sense, but what if the constant shuffling wasn't used? Is it truly a valid method to create a random distribution? And if you used a random number generator for each square rather than using the tiling method, would the arctic circle still occur? Thanks so much for the awesome video!
    1
  8040. 😓😓❤😅💙💓💓💓💓😏😆😘💕💕💕💔💕😘😘😘
    1
  8041. Mathologer, if you see this and want to know my background, it's as an actuary. Like almost everything in corporate employment, though, learning the science (or earning any degree) just gets you a ticket into the door. Unfortunately in the real world, theoretical fun rides like this are nothing but theoretical fun rides that I do on my own time. Lots of actuaries enjoy puzzle solving, telescoping equations and relationships, tricks for quickly doing amazing demos to others ("What's the 4th root of 4096?"), and elegant proofs. But it's, in my observation, strictly for theoretical enjoyable musings. It doesn't pay the bills. Or the Woos. Mathematical skills (memorizations, tricks, quick mental reductions, etc.) however do aid tremendously in programming. And very much for zen worthy programming "with style".
    1
  8042.  @Mathologer  Thank you! I found my analysis dead in the middle of page 6 there :) Eerily identical :)
    1
  8043.  @Mathologer  And I cannot conceive of Gauss doing anything other than than (# of elements) * (average element) for his first intuitive foray! The rest is much less instantaneous unless you already know the collapsing trick :)
    1
  8044. brilliant
    1
  8045. This is reminiscent of the 100-door puzzle. Here's one link to its description (and solution): https://www.geeksforgeeks.org/puzzle-16-100-doors/
    1
  8046. ❤
    1
  8047. The problem with the Talmud split is that it is unfair in respect with the ability to sell/buy debt and thus increasing/reducing the number of creditors. If you have insight information that the available money is less than half of what is owed you're encourage to split your share among multiple creditors, while if you on the other hand have knowledge that there is more than half of the total owed you will get more money if you can consolidate multiple creditors into one. The proportional method that we use today is the only one that is kind of fair on a market where debt can be traded.
    1
  8048. 7:40 I preferred then 2nd proof (colors)
    1
  8049. "2. no. tilings of Arctic diamond: 2^(-1/12)" You need to post trigger warning before you hit us with infinite summations of discrete formulas that dodge around the complex plane like that. Also holy hell how would you even prove that? An analytic continuation for a summation in the exponent? That's just madness.
    1
  8050. 34:50 I never thought I would headbang to a math video 🤣🤘
    1
  8051. This is a great video. I specialize in digital signal processing and this is the math that underpins DSP.
    1
  8052. I won a subscription by guessing the strategy! I don't know why it works, (yet) but it felt like the thing that'd make sense.
    1
  8053. Giga great
    1
  8054. im not good with the math (as i was back on 2d paper). Today my math is simple (or maybe not) 1+1 , 1+0 , 0+1 , 0+0 , 0-0 , 1-0 , 0-1 . In labouration (of all sorts of rules, hack's and slash's) at the end i add or take decimals in to spoil, the about.
    1
  8055. I went to public school in Mississippi, so we were never taught either proof. Just the diagram of three squares forming a triangle.
    1
  8056. Hey guys, I have a question, the leaning tower make some sort of a graph and it look like the logarithmic graph, am I wrong ?
    1
  8057. I made it to the end
    1
  8058. man really put a googol on one line
    1
  8059. Wonderful video. There is a typo at 28:30, you skipped the 5 in the second row.
    1
  8060. Excellent video but the chance that the American president, within the next 6 years, comes up with a new proof of the Pythagorean theorem is less than pi squared over six per cent.
    1
  8061. Vielen Dank, auch an Reiner!
    1
  8062. 32:20 How about 3^2-1^2=8? Or even 2^2-0^2=4? Or even 0^2-0^2=0? Seems like you were confused with differences of consecutive squares. We have 2k+1=(k+1)^2-k^2. We have 4k=(k+1)^2-(k-1)^2. However, simple case analysis shows that a number can not be written as the difference of squares if it is of the form 4k+2. In more technical terms, x^2-y^2=2 has no solution modulo 4. So a number is the difference of two squares exactly if it is odd, or a multiple of 4. Finally, indeed any odd prime is the difference of two squares in exactly one way. We have p=x^2-y^2=(x-y)(x+y), so x-y needs to be equal to 1, leaving only one option.
    1
  8063. This has so many striking implications it's making my head hurt. I'm seeing applications that travel from circuit board layouts to wormholes, to the nature of reality. I'm a carpenter, so I've used the Pythagorean Theorem on 2d surfaces on a daily basis, but I just now realized the importance of it throughout many more dimensions. Wasn't familiar with the Eastern understanding of it till now, either.
    1
  8064. Let me see if I got the One-Sentence Proof. Working backwards, if (x,y,z)->(x,z,y) on S has a fixed point, then y=z, so x^2 + 4yz = p can be written x^2 + (2y)^2 = p, and we have a sum of two squares. If |S| is odd, then any involution must have at least one fixed point (more generally an odd number of them), because the involution (being its own inverse) swaps all the non-fixed points with each application, so they must come in pairs. The three-line involution can only have a fixed point if either x+2z=x, so z=0 (meaning x^2=p, which isn't possible if p is prime); 2y-x=x, so y=x=1 (otherwise, p has a factor other than itself and 1); or x-2y = x, so y=0 (not possible for the same reason z!=0). So x=y=1 is the only possibility for x and y. Substituting into x^2+4yz=p, we get 1+4z=p, which has one solution for z and also requires that p is congruent to 1 mod 4. So there is only one fixed point to this involution, so |S| is odd, so any other involution over S has a fixed point, so in particular (x,y,z)->(x,z,y) has one, so y=z for some element of S, so x^2+(2y)^2 = p has a solution. How did I do?
    1
  8065. I needed to laught out in excitement when I begin to see where that was going at 9:30 😄. That proof.. so cool
    1
  8066. A question that immediately comes to mind, and I'm only 1/3 into the video so maybe the answer is coming: If I choose n points on a closed smooth curve, does the final n-gon approach a specific circle as n goes to infinity? To discard cleverly weird constructions, let's fix a smooth parameterization F(t) of the curve for 0 ≤ t < 1 and choose points F(k/n), 0 ≤ k < n. Does the limit circle exist? Does it depend on the parameterization, or on the curve only? EDIT: Oh. Fourier. The circle would just be the constant term + base frequency term?
    1
  8067. such a lovely video
    1
  8068. I was looking up Napier and the invention of logarithms last week! I can't believe they were only invented 408 years ago!
    1
  8069. I've never been this motivated to watch a video more than twice, until I started watching your channel👍😊
    1
  8070. 29:10 I could watch this all day
    1
  8071. I can also see extra bands in the multiplication table. They are "lighter shades of grey" when they consistently have a 7, e.g. 1700-1800.
    1
  8072. COOL! so nice channel in general! TNX
    1
  8073. i liked the no-9’s with fractal example the best.
    1
  8074. isn't the correct way to write the unsuccessful number is 4k -1
    1
  8075. In fact, I always don't understand why people love asking 'What comes next?' which is extremely silly
    1
  8076. I first ran into this in 1st year university. Needless to say it has provoked endless conversations and disagreements, some of which get somewhat heated (all in good fun though). So when I asked my Calculus prof about this (back in the 70's) he gave me an answer which made me seriously re-think about what we really mean by area when applied to a curved surface. He simply said the we may not understand area on a curved surface as well as we think we do. Because, as we all know, area is defined - square meter- on a FLAT surface like the rectangle. But when we apply that unit to a curved surface we have to distort it (stretch it) to make it lie on the curved surface - and then all heck breaks loose. And, as you correctly point out, we are dealing with an "ideal" paint so squeezing some molecule into the horn just doesn't apply here. On the other hand, the volume of the horn is based on the flat definition of volume (3 dimensional - but still "flat" space). And that means the volume concept does NOT have to be distorted or stretched when measuring the space inside of the horn. The inside of the horn simple lives in flat space. Whereas the surface area lives in curved space. And, yes, I know that as you make a small area on the curved surface shrink and shrink the more and more it starts to look and behave like a flat surface - but it really never does become flat - certainly not in the macro world. BTW, I know Mathematics Teachers who are still convinced that: .9999...... = 1 is a "paradox". But it really isn't. Infinity is not a number nor a place, it's just weird. Thanks so much for all your work. I really do like your approach.
    1
  8077. Most Memorable : Euler...
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  8078. Slow but Steady goes to INFINITY 😉 Just got the video while going to sleep and this made my day, Though I'm dead cause I lost the bet on "No 9 series"😂
    1
  8079. Your video debunking 1+2+3+.....=-1/12 is my favourite one in your channel..
    1
  8080. Please tell us what your T-shirt means. Is there anything important at the bottom below the screen cut-off?
    1
  8081. Most Memorable: The approximation formula just misses the exact value by a constant less than 1 and in the limit the constant comes out to be Euler's constant gamma... which relates to primes with Euler's constant e😳😳
    1
  8082. The number of integer squares for 2020 by the heavenly formula is: 16 The possible factors are: -42 & -16, -42 & 16, -38 & -24, -38 & 24, -24 & -38, -24 & 38, -16 & -42, -16 & 42, 16 & -42, 16 & 42, 24 & -38, 24 & 38, 38 & -24, 38 & 24, 42 & -16, 42 & 16.
    1
  8083. The first thing I did this Christmas was watch this whole video! Fantastic as usual thank you!
    1
  8084. The way I was looking at it still uses the cards as cards, but because you can't turn a face up card face down, either you flip two face up cards or you effectively move one face up card 1 space to the left. This take a bit of explanation but is the same effect as "the number keeps decreasing" so thinking in terms of binary, those few words hold a lot of information. Elegant indeed
    1
  8085. So I went and found the general case for the solution, keeping it under my other comment for spoilers. It takes x moves to move the card starting with the right most move until it reaches the left most position which makes the formula x + x-1 ... until x = 0
    1
  8086. Hey so... is it just me or does the formula for 4 consecutive fibonacci numbers at 40:00 (F_n0.F_n2 + F_n1.F_n3 = F_2n+3) look like the collatz conjecture? Specifically the 2n+3 part? Intuitively, it feels like there might be a parallel proof to prove all numbers reduce to 1 in a similar vein to how all the right angle triangles are offspring of 3,4,5.
    1
  8087. Loved all the topics, artistically enlightening being the restriction we impose on the digits which produces a subset of harmonic series which converges and the fractal interpretation of the same. Would be a fancy idea to find a generalized restriction on digits implying convergence.
    1
  8088. Kaleidoscope constructor's take on this would be: "... but of course!". This looks like crystallographer's wet dream. I enjoyed it.
    1
  8089. Definitely the 9s series!
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  8090. that is just crazy!
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  8091. I had a great Geometry teacher who taught proofs. I also went with math for four years after Calculus. My schools were good with math. I use the logic of Geometry every day in non math ways.
    1
  8092. Mathematicians really are lazy folks... They keep trying to find the easiest solution for every problem... ... ... 😅😅😅
    1
  8093. Used a circular slide rule sixty years back. But owned a straight one.
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  8094. Amazing video!! A few remarks though. At 17:22 you do not have the correct infinite sum. Furthermore, the end is unsatisfactory. We are told something is unclear and that it can be put right...but we are not told how. I hope there can be a follow-up to mend the end of this video. Does it have to do with Ramanujan Summation? I will try to find out.
    1
  8095. 6:16 that was actually not that bad for a non-czech-speaker 😅
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  8096. What I found most memorable was indeed the very precise and well planned lesson. Nice Ans simple illustrations.
    1
  8097. It's the car example this First equation is distance differentiate once we have speed Differentiate twice we have acceleration If we have one equation we can either differentiate or integrate to getvthe corresponding equation
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  8098. Not sure why I subscribed here. I don’t understand anything lol
    1
  8099. The two different coloured cirkels (yellow and white) are split up .. but they clearly have a different radius ,, than later you show two equally long red lines representing a rubber band that would match up accordingly..but that cant be true if the two cirkels arent also the exact same size.....i guess we have to assume the yellow cirkle gets enlarged to be the same size as the white one to make everything valid again? ... i wanted to also put a time stamp here but not really feeling like watching the whole thing again...anyway thanks for the very interesting videos made
    1
  8100. Masters in math. Loved the video, haven't dealt with this since school so hardly remember any of it. I understood each step easily enough, but I lost the overview. I miss something like having the blackboards in a lecture with the build up so you can look back and go "oh yeah". The video was great, made me insanely curious about so much other stuff so I've been on my whiteboard for 2 hours now :) Any video that can get you annoyed about it stopping has to have a gold star, gj, looking forward to the next one.
    1
  8101. I liked the swivel proof better, but I may have been biased by hearing it first. The color-splitting proof is also quite neat.
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  8102. Gamma was the higlight of this vid :) amazing video
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  8103. I guessed yellow not by posibilities and i was correct lol hahahah
    1
  8104. Well, many people have made circular slide rules. Hasn't been missed.
    1
  8105. I thought the slide rule was base on the common log(base 10) not the natural log(base e).
    1
  8106. He should grow several inches taller at the end of every video. 😛 (Did the camera slip?) ... awesome video, tho.
    1
  8107. Love your videos sir! Would you ever do a video on P-Adic numbers? Would love to see you break them down and explain them
    1
  8108. can someone do an edit that only has his giggles and not anything else?
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  8109. Keep it up, thank you for sharing :)
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  8110. i hate math....but i just subscribed...where you been all my life?
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  8111. Are the primes somewhere in there with a predictable pattern as well?
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  8112. I know where to find the mathematical soul of a shoe. It's underneath on the bottom. Sorry, dad joke.
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  8113. Sir, you rightfully gave credit to ancient indian scholar, but so far most westerners disregard all indian ancient talent. Another example is western flat earth when even laymen in india knew about planets orbiting the sun even in 500BC . Even now they say Galilio figured it out first
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  8114. It was truly inspired i loved it😍😍
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  8115. Hello Mathologer! Here is the link to the 2D QEDcat origami model designed by me 😊❤: https://m.youtube.com/@RibbleMaths_YifanDu/community It is posted in my math channel. I learned how to design duo-color origami models in the fantastic books 'Duo-color Origami' and 'Multi-color Origami' by the great Chinese origami artist Mi wu (Chinese name:郭嵩). I also designed a few other color models.
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  8116. I am optimizing my design recently. Almost finished with 'QEDcat origami 2.0'!😊❤
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  8117. Fake news. If this is all nonsense how come the US government is using it for free electricity (which they're hiding to protect the oil companies), faster than light travel, time travel, antimatter bombs, mind control, teleportation, telekinesis and to make a rabbit appear in a hat that previously did not contain a rabbit? Whilst you sheep think about this I'm off to the shop as they have 9 for the price of 6 on a 3 pack of tinfoil.
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  8118. Mathologer always gives AHA moments
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  8119. 1/4? Just took note of the 1/2 sides leaving 1/2 length (long side) little triangles and mentally shifted them to fill the center square area. Giving it an area of (1/2)^2 I think? Lol
    1
  8120. what the heck...
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  8121. Just read the description. Are you trying to tell me you HAVEN'T been getting buckets of cash this whole time?! I wouldn't notice because I have YouTube Premium, but I was blown away by how professional and informative this video was. Without a doubt you deserve all the patrons and ad-revenue you want
    1
  8122. You just put two words in a title I would have never thought to see in a single sentence... Why would you put "why" and "calculus" in the same sentence!
    1
  8123. Absolutely amazing video, thank you. You seem a bit under the weather, I hope you're doing ok
    1
  8124. Hopefully the Kurosawa Galois video isn't scrapped...
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  8125. I'm German and study Maths but i still have no fricking clue what Goniometrie is xD
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  8126. Do you have a List of the TShirts you are wearing? They are always so great :D
    1
  8127. That lantern number of points 7, 15, 26, 42, et cetera (at minute 14), is that some significant identifiable number sequence, also approachable somewhere else?
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  8128. I was reading about the proof of the law of the lever by Archimedes last week (not really a proof) wondering why I've not yet seen a YT video about it in all these years, and then you showed at the beginning.....Apart that, thank you so much for this great video
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  8129.  @Mathologer  OK. Thanks, for the information, and the link, Prof. Polster 😌👍🏻. Seems interesting. It’s also a nice surprise that ψ³ = 3,14… ≈ π 😃.
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  8130.  @Mathologer  Yes, true that. Well; both the area and the SSSSSemiperimeter, of the 3–4–5-triangle, are 6; and, if we find a 7, we can divide other things, with it; and we’ll get the rest of the digits covered, except 9 and 0. But 2, 7, and 8 will be taken care of :). Plus; the fact that it’s precisely the LEAD digits of π that are so prevalent, in the 3–4–5-triangle, makes it all EXTRA special, to me; because they’re the ones that usually get represented, when π is displayed, in decimal form; and because it’s already an unreasonably good approximation (to 4 decimal places). 🙂
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  8131. I feel very amused with M's videos. Trying to imagine 4d objects was one of my favorites mental hobbies. Finding the generating function was very cool too.
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  8132. Notification Squad! :)
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  8133. Edited: The result is essentially not new. The beautiful presentation is. It is well known that the medians of a triangle can be translated so as to form the sides of a new triangle. The half-turn rotation proof shown here is essentially the same thing. It is also a classical result that the lengths of the medians of the new triangle thus constructed are equal to 3/4 of the lenghts of the respective sides of the original triangle. This contains in essence the involutive process shown in the video. See for instance Altshiller-Court's College Geometry, 2nd edition (1952), pages 65-66.
    1
  8134. Kudos to Rainer for a lovely proof. The opposite angles of a cyclic quadrilateral adding up to a half turn is easily obtained, even when the quadrilateral all lies on one side of the centre, by remembering to draw the diagonal connecting the two corners we're not looking at; this is a chord, that subtends at each of the corners not on it half the angle it subtends at the centre, on the relevant side; and they're on opposite sides, so those two angles at the centre add up to a whole turn, so the opposite-corner-angles add up to half that, so half a turn.
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  8135. Formula: sum(2n^2+k)^2 LHS k=n=>2n. RHS: k=2n+1=>3n
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  8136. 21:41 An azimuthal projection is too good
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  8137. I honestly just found the remainder extention on my own. Like the rule just tells you that adding digits together preserves divisibility by 3 (and 9) so that implies that you can add the digits together until you only have one and the divisibility is preserved. Pretty neat
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  8138. Thanks for the reupload!
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  8139. The optimal overhanging tower has my vote. The other parts are neat and interesting, but none of them are too surprising (and I already knew most of them except chapter 6). However, I really expected the optimal tower to be much more regular than that awful looking tower. I wonder if the optimal solution becomes more regular as the number of blocks goes to infty.
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  8140. I love the shirt - I'd totally wear that to work.
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  8141. Mel Brooks spoke of the Schwartz. Did they merchandise a lantern for Spaceballs? Is it permitted to use the Schwartz? What about as a Shield? The Light Saber is already covered.
    1
  8142. The euler macaroni constant popping up out of nowhere
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  8143. Spirograph: Ahh it's harmonics!
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  8144. The second half of this video reminds me of a 3blue1brown video about spirals and straight arms that appear when plotting primes in polar coordinates.
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  8145. Definitely Kempner's Proof was the most memorable
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  8146. RGP - RamaGuPta... Logical!
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  8147. I love the infinite fracture of the no 9 series as a visual. I believe it shows that the sum of reciprocals of all unique integers containing at least one 9 is infinite which is also surprising. Is that true?
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  8148. I love this channel =)
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  8149. I love the little chuckles you have, your joy is contagious :)
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  8150. Is it weird that I understand this and I’m only 15?
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  8151. Right.... Some logic to the cubic stuff. If we look at the number at the left side of the = . For the ^1 sequences, we have 1* (X^2+X) For the ^2 sequences, we have 2* (X^2+X) For the ^3 sequqnces, we could use... 3* (X^2+X) Thus we get 6^3, 18^3 and 36^3 We get 5^3 + 6^3 + ??? = 7^3 16^3 + 17^3 + 18^3 + ??? = 19^3 + 20^3 33^3 + 34^3 + 35^3 + 36^3 + ??? = 37^3 + 38^3 + 39^3 The ??? seem to follow a paterns, what is missing is 2, 18 and 72. This could be 2* ( (X^2 + X)/2 )^2 When looking at the 4th answer for the kubic sequence. We get 3* (4^2+4)=60 And 2* ( (4^2 + 4)/2 )^2 = 200 Thus, 56^3 + 57^3 + 58^3 + 59^3 + 60^3 + 200 = 61^3 + 62^3 + 63^3 + 64^3 So, the reason that kubic isn't possible has something to do with this... 2* ( (X^2 + X)/2 )^2 But why?
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  8152. The root of all evil is 25,8070, if you round correctly.
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  8153. most memorable: finite or infinite animation of nicole oresme's proof edit: and the weird 20 block tower @ 10:45
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  8154. Hey Burkard! I'm a chemist with a modest undergraduate-level understanding of mathematics, but with a passion for its beauty. I've been a subscriber since the very early days, and I just love your videos! Even so, when I saw how long this one was, I genuinely thought my mind would wander away in the middle. However, it was so captivating and satisfying that the whole 50 minutes just flew by. I really hope that someday I can be as good a presenter and teacher as you are. The care you take in preparing these videos really shines through. Long live Mathologer!
    1
  8155. I do not think that a slide rule (or spin rule?) is derived from just natural logs but works with any log rhythm base. Super explanation. A pocket slide rule was normally carried just behind my pocket protector.
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  8156. 24:01 Because then, the 1st binomial distribution (k+1 5) would be (4 5) if k=3, therefore impossible For k=3, then : 325(5 5) + 985 (6 5) + 287(7 5) + (8 5) To generalize for all possible values of k here : (k+5 5) + 287(k+4 5) + 985(k+3 5) + 325(k+2 5) + 2(k+1 5)
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  8157. Congratulations. I have really liked end enjoyed all your videos
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  8158. This is so cool. Thanks for making these videos<3
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  8159. The difference of two squares at the end is wrong. It's not exactly when the integer is odd; you can have 8 = 9-1.
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  8160. Perfect topic, thank you for the amazing video!
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  8161. I'm Indian, I think some version of this with trigonometry was taught in junior high!
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  8162. Is this some gift for viewers from india? One of many now. Keep em coming.
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  8163. The formula of 4k+3 could also be written as 4k-1 as 3=4(1)-1. 7=4(2)-1 etc. Could it be written like this?
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  8164. it look like a 4d object
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  8165. Last chapter was so cool
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  8166. It makes sense that 2+2=2x2 is such an important identity, because it's the result of the two main defining traits of regular numbers: the additive identity and the multiplicative identity. And when restricted to integers, it really has a lot of weight.
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  8167. where did he get, find all of these calculations, and applied them?
    1
  8168. I recalled my memory from another math video that it said almost all the integers contained "3", which is same why of saying almost every integer contained "9"
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  8169. I saw it first on mathologer. hexagorean therom proof by special case of the trithagorean therom🎉🎉 I am truly glad to have been witness. Now cat videos😂
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  8170. I'm not sure if I would consider this as a proof or a postulate. The Pythagorean Theorem is Embedded within both the addition (+) and assignment, equality, identity operator (=). I'll start with the later. We can take the simplest expression y = x. Depending on context this expression or statement is stating that either two quantities are equivalent, or that we are assigning x to y. I also call it identity since nothing within the expression has changed. Here's a list of our pairs {..., (-1,-1), (0,0), (1,1), ...} We know that y = x is also linear from the slope-intercept y = mx+b where b is the y-intercept and m is the slope defined by (y2-y1)/(x2-x1) = dy/dx. The graph of the line bisects the XY plane in the 1st and 3rd quadrants. Since the x and y-axes are perpendicular or orthogonal to each other making a right angle of 90 degrees or PI/2 radians at the origin (0,0) we know that this line has a 45 degree or PI/4 radian angle that is between the line y = x and the +x-axis giving us an angle in standard form with respect to the origin. We also know that the slope of this line is 1. If we set the length of this line to 1 going from the origin (0,0) to some point (x,y) and draw a vertical line down from (x,y) to the +x-axis we end up with the coordinate ( sqrt(2)/2, sqrt(2)/2 ) and the lengths of x and y are sqrt(2)/2. We can take the slope formula dy/dx and simply use (x,y) since (x1,y1) = (0,0). We can see that (sqrt(2)/sqrt(2)) = 1 since sin(45)/cos(45) = tan(45) = 1. Here the angle theta (t) is the angle with respect to the line y = mx+b, the origin and the +x-axis. With this we can substitute the slope m = dy/dx with sin(t)/cos(t) = tan(t). This then would give us y = sin(t)/cos(t)x + b and y = tan(t)x + b. Why does this derived substitution from y = x work? This is where the addition operator comes into context. Here we will use the very first and simplest of all mathematical operations 1+1 = 2. With this I'll demonstrate how this expression, equation indirectly satisfies and solves the Pythagorean Theorem and I will also show how it relates to the Equation of the Circle. We know that the Pythagorean Theorem is A^2 + B^2 = C^2 with respect to Right Triangles. We know that the General Equation to the Circle is (x-h)^2 + (y-k)^2 = r^2 where (x,y) is any point on the circle, (h,k) is the center, and r is the circle's radius. Now let's see this in action. We are going to start from the left and work our way to the right. We are going to define our starting point as the origin (0,0) we will then move to the right 1 unit of arbitrary length to give us a Unit Vector at the coordinate (1,0). Now we are going to horizontally translate this vector in a linear progression towards the right by the magnitude of the original unit vector. This gives us a second unit vector that goes from the points (1,0) to (2,0). Here we have v1 = P1-P0 = (1,0) = (1,0) - (0,0) and v2 = P2-P1 = (2,0) - (1,0) where v1 = v2 = 1 and their combined length = 2. What does this have to do with a circle and the Pythagorean Theorem? It's quite simple. Here we the basis for a unit circle with its center (h,k) located at the point (1,0). We can now use the equation (x-h)^2 + (y-k)^2 = r^2 and use the center location. (x-1)^2 + (y-0)^2 = 1^2. Rearrange and simplify to y^2 = 1 - (x-1)^2. Expand, y^2 = 1 - (x^2 + 2x) + 1 and simplify then gives y^2 = x^2 - 2x. Solving for y gives us y = sqrt(x^2 - 2x). Here is a sequence of some of the first few values of (x,y): { (0,0), (1,i), (2,0), (3, sqrt(3)), (4, 2*sqrt(2)), (5, sqrt(3)*sqrt(5)), ... } okay this doesn't seem to justify anything. Yet if you noticed, when we plug 1 into the expression we end up with i the sqrt(-1). Why is this happening? This is because the unit circle is located at (1,0) as it is translated to the right by 1 unit. Let's move it to the left by 1 unit placing its center at the origin (0,0). Once we do this the Equation of the Unit Circle then becomes (x-0)^2 + (y-k)^2 = 1^2 which then becomes x^2 + y^2 = 1. Here this is a well known function and this is why the trigonometric functions have a Pythagorean Identity. Let's solve for y: y^2 = 1-x^2 then y = sqrt(1-x^2). We can now generate an equivalent sequence as before: { (0,1), (1,0), (2, i), (3, 2i), (4, 3i), ... (N, (N-1)i) } This is also gives all Real and Complex Numbers and this is why we have the expression e^(PI*i) + 1 = 0 as well as the Taylor series and so on. Also if you look closely at the equation of the circle (x-h)^2 + (y-k)^2 = r^2 this is in fact The Pythagorean Theorem. We know that a circle with a radius of 1 has a diameter of 2 hence 1+1 = 2 and we know that the angle between the point (1,0) to (-1,0) is 180 degrees which equals PI radians. And there you have, basic arithmetic{addition, multiplication, exponentiation, logarithms, etc.}, basic polynomial algebra, linear algebra with vectors & matrices, geometry and trigonometry all expanding from y = x and 1+1 = 2. And with all of these being related, we can extrapolate from them and their various functions and properties that gives us dy/dx the Derivative and with its inverse Integration. So the next time you add 1 and 1 to get 2 or the next time you set a and b equal to each other, remember that there is always a unit circle embedded within it waiting to be expanded and the limits of that expansion reaches +/-infinity! But why Triangles, right angles to be specific? For this we need to jump over to a little bit of Physics. We typically use sine and cosine functions to map or graph wave motion, rotations, oscillations, cycles, etc... We know that both Sound and Light are forms of Energy. In essence the sine and cosine functions are basically Sound and Light waves in motion. We also use them to map or graph rotational energy and or angular momentum. So everytime you speak and sound emits from your mouth, your tongue, and your lips, the energy or vibration moving through the air (medium) is a Sine or Cosine Wave. And both of these functions are defined from both the Right Triangle and the Unit Circle again 1+1 = 2. Numbers don't exist in nature as they are abstract concepts. They are a product of the mind. Yet the equations, formulas, functions, and laws of the sciences require and depend on them and nature adheres to the laws of physics and chemistry. The laws of nature obey Sound and Light, the properties of Circles and Right Triangles and things that wave, rotate and or oscillate. A clock goes tick tock with even intervals and can be represented as 1,0,1,0,1,0... and now we have a binary system. Log 2 mathematics. And it has been proven that Log 2 mathematics with an infinite amount of binary digits (bits) is infact Turing Complete. Math is Grand! Math is a Product of the Mind and the Laws of Nature adhere to the properties of Math. Food for thought! So as you can see, The Pythagorean Theorem has always been long before any human thought that they have discovered it, long before any human existed. Since the properties of such things existed long before Humans, and considering that numbers are conceptual, abstract ideas, products of the mind, then the real question becomes: Whose mind did they originally come from? I can give a hit and it succeeds 1+1 = 2 which pertains to and rhymes with 3. The Divine Trinity! "Let There Be Light!"
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  8171. Good
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  8172. This video contains so many interesting insights that just leads to more questions. 🎉 thank you for your fabulous work.
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  8173. Ich will eine andere Serie nur auf Deutsch!!
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  8174. Wow, impressing and exquisitely interesting for math addicts as I am ! To me, it is your best video so far.
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  8175. That was a very nice, and simple, proof that the Kempner series is finite. It occurred to me that the first part of the harmonic series might provide a better bound for the first part of the Kempner series. Ie, if Harm(a,b) means the harmonic series from a to b and Kemp(a,b) is similar for the Kempner series then: Kemp(1,∞) = Kemp(1,10^k-1) + Kemp(10^k,∞) < Harm(1,10^k-1) + Kemp(10^k,∞) < ln(10^k) + 1 + 80(9/10)^k = k*ln(10) + 1 + 80(9/10)^k for all k. The RHS has a minimum when k = 12, giving a bound of about 51.2 on the Kempner series. .
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  8176. The most memorable part was no integers among the partial sums.
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  8177. You blew my mind for the manieth time. 😮
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  8178. More math videos
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  8179. Please correct me if I am wrong, but the pattern does not depend on the base we are writing the numbers in. That would just change the way we label the points on the circle. In fact the two diagrams 2|7 at 2:22 and 21:40 are the same.
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  8180. ... der is doch klaar deutsch der kerl
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  8181. What kind of monster would change channels in the middle of a Mathologer video?!..
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  8182. enthralling, much obliged
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  8183. THE BIG QUESTION IS... What is the best knot to use. not just lacing.
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  8184. Thats why i hate math teacher if they using 123 as an example for the task
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  8185. Code in p5js https://editor.p5js.org/hugomosh/sketches/1Sg1NxqI7 For the math/code enthusiasts: Make it better (improve the algorithm, pimp the look, more controls) and send me a comment! 🤓 Which (n,m) you like a lot!?
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  8186. What about the triangle inequality? e.g. A+B > C, a+C > b In fact for the Quad in general there are 4 triangles, each with three sides => 12 inequalities in total. I am not going to list them all here, but the conclusion of the algebra must lead to Ptolemy' Theorem? Maybe someone would like to have a bash at that.
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  8187. The most memorable is Kempners proof!
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  8188. Thank you for your explanation. Really enjoyed it.
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  8189. 謝謝！I lke your teaching.
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  8190. Ah... #2. Cool. Cos (n×pi/2), where n is odd, results in zero. Both cos functions produce zero at 1,1. Product of values containing zero is zero...
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  8191.  @Mathologer  thank you!! ...and while you're here I can't help it... ...could you focus on triple integrals for a spat? I'll already be looking into the previous videos if you aren't able to but either way awesome shirt! ...awesome video! Once again thank you
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  8192. nice to see that a photo of my towers of hanoi game that I made, and took a photo of for Wikipedia, now turns up in your video...
    1
  8193. I really really loved the "no integers" bit. I knew many of the facts mentioned in this video, but the way you present and explain them is simply marvelous. I am an avid watcher of your videos for a long time but never thanked you. You really helped develop my interest in mathematics. Thanks a lot. :)
    1
  8194. This theorem has over 371 different proofs that were known around the 1940s. I wonder how many new proofs has been born since then. Twisted hypercube anyone?
    1
  8195. 1/5
    1
  8196. 32:20 This pattern reminds me of a fence I've seen on a photo; I think that fence is around a school
    1
  8197. Just inversing matrices to solve system and find B_n looool so easy operation
    1
  8198. it will be found that this is how light energy becomes matter
    1
  8199. Gotta be gamma. You found the compression error correction for one conscious node on the biological computer of Earth. And not a coincidence that gamma is the term for 1 GeV of radiation. Those self locating strings and loops are matter in "gravitating" form from radiation into EM. In binary, space is secured but nothing is built upward on the Reimann plane. Lots of battle chess visuals though.
    1
  8200. I am so in love with the Ramanujan's background sound effect, it always feels like they is something wrong with the proof. Whenever that sound appears, I always associate with the pink elephant looking figure, which is what they called Ganesha.
    1
  8201. colour proof made much more sense to me
    1
  8202. Thanks so much for the video!
    1
  8203. i totally meant that this game master is Leonard Nimoy aka Spock. but he was younger in the time of the 1st Doctor.
    1
  8204. Well, I bet Picard is pissed that he's wasted all that spare time on Fermat's 2 Square theorem now that it's proof has been found. =D
    1
  8205. The path that passes through each possible position, I thought that was the solution to the Tower of Hanoi back when I was a youngling. Since I thought you cannot pass a larger disc above a smaller one.
    1
  8206. I never thought about this matter, but was surprised that at the first example the solution presented was about what I already had thought to be fair.
    1
  8207. No 9 series has a finite sum?! Shock for me!
    1
  8208. This just blew my mind ! Thanks a lot. I had seen many type sequence like the one in which it had e and π and all and I was unable to get how has this come but you just solved the problem. Please make some more videos on calculus and it's real life examples.😊😊😊😊😊😊😊😊😊😊
    1
  8209. I don't need a Kurosawa-length Mathologer-Video, I'm quite comfortable with this Polster-length video. Love, peace and happiness! Cheers Célestin
    1
  8210. I was trying to make this problem continuous, where there are an infinite number of pegs and you start with an infinite number of blocks on the starting peg, and all of your strategies have to be recursive and done by taking limits.
    1
  8211. High quality math video. The ending score makes it perfect.
    1
  8212. this video is just amazing , thank you !!! (and I think you can be prod of your job when Notch is supporting you for doing this amazing stuff !)
    1
  8213. how do you discover these topics? where do you search for them? I find that for most of your videos, your the only one who made a video on the topic or you the only one who showed it the way you did
    1
  8214. i want that tshirt pleaseee😭
    1
  8215. for those who like a little binary recursion: HO(3), with Procedure HO(n) If n > 0 print(“HO “) HO(n-1) endif or if you’d prefer something with a little less recursion ... Procedure HO(n) if n > 0 print(“HO “) m = int((n-1)/2) HO(m) HO(n-m-1) endif
    1
  8216. Calculus is itself witness & speaks that Newton was (is) really really Great, and he was 1 out of Billions...
    1
  8217.  @Mathologer  I read the other comments, but it sounds like a good excuse 🤣😜
    1
  8218. @Mathologer You could have mentioned the story of Phileas Fogg.
    1
  8219. yellow, not a guess
    1
  8220. In order, for the things I'm supposed to put in the comments: * 15 * 8 * By "this plus that" mathematics do you mean modular arithmetic, or something else? Either way probably yes.
    1
  8221. I know why those videos are so popular. It's because they've been promoted by the true rulers of the Internet: cats! Proof? Cat Power's song "3,6,9": https://www.youtube.com/watch?v=LDknkiUcmxA Illumicati confirmed.
    1
  8222. If you are trying to make a video too long for me to watch, you better keep trying.
    1
  8223. visual fractal argument for sum of no nines series convergence was most memorable to me
    1
  8224. I really like the Euler-Mascheroni constant. Some while ago I stumbled I to it by chance, and discovered a deep part of mathematics that I didn't know
    1
  8225. I like the second version of the proof that the iris tangents are midpoints.
    1
  8226. The Vortex has the VW logo embedded!
    1
  8227. Here should be a picture of my books stack if it would be possible to add pictures in youtube comments.
    1
  8228. Now I want a t-shirt with the 'hexagrian' theorem drawing! lol Great video as always.
    1
  8229. trulyyyy magic, brilliant!!
    1
  8230. There is a typo in the formula at 25:02 - instead of (A+B+D-B) it should be (A+C+D-B)
    1
  8231. Its wrong to say that it's a circle after corner chopping. It's still a rectangular surface of infinate number. No paradox for me. It's just not 4*pi
    1
  8232. 👍🇧🇷🌎
    1
  8233. Hi
    1
  8234. Most surprising part for me is that the series consisting of the reciprocals of primes diverges.
    1
  8235. new video :D
    1
  8236. Gamma number No 9 fractal Your style :)
    1
  8237. What if one of the overlapping collections has only one element?
    1
  8238. I would have loved to see the Euler-Maclaurin summation formula get a mention during the integral approximation section. One of my favorite formulas, and very useful.
    1
  8239. Have you tried Game Of Seven? It revolves around the number 7.
    1
  8240. The most memorable part was guessing that Kepner's (spelling?) series had a finite sum even before the counting down started, and for much the same reason explained later. I recalled having encountered the idea that for any given digit, "almost all" natural numbers contain that digit, with the proof involving that as the number of digits increase, the proportion of numbers containing that digit tends to one. I think that idea may have been my first exposure, way back when, of the term "almost all" in a discussion of an infinite set. Anyway, in this case the complement is considered, and the fraction of numbers not containing the digit tends to zero. Not a proof, at least not without a whole lot more work, but definitely a way to see that finite summation is at least reasonable on its face.
    1
  8241. I like the coloring proof better
    1
  8242. 4k+3 triggered me. I wanted it to be 4k-1... it's more symmetrical. D:
    1
  8243. So I found the answer on Google, it seems that p = 1 is the limit. Wow, so could we we say that 1 + 1/2 + 1/3 ... diverges infinitely slowly since any power infinitesimally greater than 1 converges? o_0
    1
  8244. I love the humor you bring into these videos.
    1
  8245. "let me know what happens" lol
    1
  8246. I made it to the end. \o/ I thoroughly enjoyed the video. It's as if you created it specifically for my level of mathematical sophistication. I did well in math olympiad, I went to grad school for math (studying algebraic geometry) but I never finished my PhD. Oh, the animation at the end was a real treat.
    1
  8247. Great video !
    1
  8248. no doubt there...terrible aim is the star:))
    1
  8249. Someone send this video to Brady of Numberphile. I would not be surprised if he rearranges the positive and negative terms of the alternative harmonic series to get -1/12 + -1/12 and yet rearrange them to get his one and only favorite -1/12.. That is why I love Matloger. Not only you know what you are talking about but explain it in a way that is both mathematically rigorous and easy to understand.
    1
  8250. Can someone recommend a good function visualizer?
    1
  8251. Well it's at least the key to fractions with 7 as the denominator.... which never have 0, 3, 6 or 9 in them. No we weren't taught digital roots in school - only the quick way of seeing if a number is divisible by 3 - then obviously you can do it for 9 too if it can be divided by 3 twice! Playing with spirograph patterns certainly helped with understanding maths, like recognising highest common factors and dealing with equations in polar co-ordinates. It was not just to enjoy the beautiful side of mathematics! It also gives an appreciation for these circular ways of organising numbers!
    1
  8252. rock paper scissors lizard spock
    1
  8253. I was thinking that this would be a fun programming activity and then you brought up the challenge, so here's my implementation: https://www.thecodingfox.com/interactive/arctic-circle/
    1
  8254. Does this mean Mel Brookes was right?
    1
  8255. Hm... if you are going to cover the 4 color theorem, I would show how it is equivalent to the fact that every "snark" must be non-planar. That is also a very nice visual demonstration which is not so well known (on the other hand, both problems are equally hard so it does not really help). And also, the 4 color theorem can be "proven" by one of my favorite fake proofs -- I have even put that one into a video: https://www.youtube.com/watch?v=Ryoipkksktg
    1
  8256. Youthful Giggle + Authoritative Tone + Good Editor + Good Writer + Great Accent = Subscribe
    1
  8257. Ptolemy s theorem : what happens with a ellips instead of circle.
    1
  8258. 👋🏼👍🏼
    1
  8259. 24:50 Looking at the shape of the individual slices, it is clear that when, instead of taking the "bow" shaped path form the bottom right to the top left corners, one closes the shape with a straight line, the resulting triangle has a smaller area (rigorously, this due the the fact that 1/x is convex, therefore a connecting line between ( n, 1/n ) and ( n+1, 1/(n+1) ) does not cross below the function). So the total blue area has to be smaller than the total area of such triangles. Looking at the right side of the blue shapes, the total length of these sides add up to 1 (since 1/x -> 0 and the lengths are the differences between consecutive values of 1/n, starting at 1). And since the area of each triangle is one half times the length of it's right side, this means that the total area of these triangles is one half. This proves the desired fact that the sum total of the areas of the blue slices is less than one half.
    1
  8260. Amazing
    1
  8261. Nice :)
    1
  8262. Thanks Mathologer.. what a mind Ramanujan had..
    1
  8263. What if you take into account the curvature of the Earth and the differences in gravity as you go up, how far can you overhang?
    1
  8264. I'd say "Is the no 9 series finite" was the most memorable but my life depended on it and am now dead. Maybe a book in the afterlife will be comforting
    1
  8265. Most Memorable: I liked theBishop's proof best.
    1
  8266. area: yes
    1
  8267. You are an outstanding teacher, Mr. Mathologer
    1
  8268. Whoa proof not immediately obvious to me for that
    1
  8269. Epic.
    1
  8270. No they never taught this division remainder, pretty cool.
    1
  8271. 2:12 I don’t get it at all. Why is the volume taken out from the cylinder a cone? Shouldn’t the cross-section of the non-base edge of the negative volume be curved inwards, not straight?
    1
  8272. Burkard's proof (or sketch thereof) is enough to show that it is a cone. Or there's a more detailed geometric explanation in the supplementary playlist, where I look at the intersection of sphere, cylinder and cone. https://youtu.be/-RB9Yxh7NnM
    1
  8273. No, we're taking out slices of the semisphere and basically orient them in a way that they look straight from the side.
    1
  8274. ... And that's why pie day is August the 54th. ;)
    1
  8275. Yet another great video... But still I don't know about the quartic equation l
    1
  8276. All of the video is beautiful. What I liked most is the symmetric arrangement of bricks that give the cubic root growth for the overhang.
    1
  8277. 23:52 is it because like 3(or smaller)+1 is 4(or smaller) and is less than 5, to make it work just like sabotage the first one(few) part where the k+1(or 2,3) of it is smaller than 5. (I watched the algebra autopilot for like 3, 4 times before finally understand lol) From a high school dude (comes from China but now live in Germany) who really likes math and enjoys your video:). Danke schön!
    1
  8278. BUT (3×3×3)+(4×4×4)+(5×5×5) = (6×6×6) 4th power doesn't work
    1
  8279. Beautiful, could you show which rearrangements are valid in case another one with infinite series' comes up :)
    1
  8280. Cool t-shirt :)
    1
  8281. The complexity of math ended with 6th grade algebra for me, broke my soul enough from never learning anything more that I simply went self taught homeschool until I was free from shackles. Biggest waste of time ever, and it is straight up criminal like the majority of government intervention nowadays.
    1
  8282. The crazy overhang the with counterweights was my favorite tidbit and insight here.
    1
  8283. how can I recreated something like that pattern on your shirt so I can make a texture to put on a sphere flying at a player in VR? it would look awesome. Any tips?
    1
  8284. Thanks! And I can see in that image how to reproduce it. Very cool  @Mathologer 
    1
  8285. Now my xmas will be awesome!
    1
  8286. Delicious mathematics
    1
  8287. 10:06, is that pascal's triangle?! Where did that come from?
    1
  8288. Owww! My brain!
    1
  8289. Another coincidence is a leap year: 8² + 9² + 10² + 11² = 366 🥸
    1
  8290. This is incredible
    1
  8291. This is just mind-blowing! I love it, thank you so much (from a professional). How is this - in mean the existence and the insights of the medieval Indian mathematicians - not more common knowledge? Far from taking away from Newton's and Leibniz's achievements, who surely were unaware of their distant predecessors, it's an astonishing testament to the wonderful unity and universality of mathematics; one of the few human exploits in which all cultures have contributed equally. Historians of mathematics, we want to hear more from you, and louder!
    1
  8292. My brain is melting. This is 🔥 🔥 🔥
    1
  8293. Genius!!!
    1
  8294. I don't know if anyone asked this... but you showed the [10/3] which resulted in 7 identical rotating circles and therefor in 3 heptagon. this means there are 3 [7/3] rotating too. Would you animate at least one [7/3] in a [10/3] and if possible a single version combined both?
    1
  8295. Never heard of the Heron's formula, neither at school nor in the university (where I've got two full years of Further Mathematics, but not much geometry there if any at all). Ukraine.
    1
  8296. Yeah, I also thought of Fourier when he talked about the sum of polygons. I thought this as an analogy, but apparently, this is really the same maths.
    1
  8297. MIND BLOWN !!!
    1
  8298. I found that 700 year old proof incredibly satisfying to watch, even if it was rather simplistic - Especially in comparison to the other things on show in this video.
    1
  8299. My vote certainly goes for the chapter on the Leaning tower of Lire.
    1
  8300. Amazing video, as always. Happy whatever you have in your country 😅👍🏼 Just a very tiny "but", those orange and green color choices are not so colorblindness friendly. In the big artic diamond I hardly notice any diference between those two. Maybe a darker green and a yellow would have worked better for me 😅
    1
  8301. I love the innocent way he says "towel head" I guess he doesn't know it's a slur Protect this man, he is precious (Even YouTube is warning me about community standards before posting this)
    1
  8302. The no 9-s proof is my favourite.
    1
  8303. what is that background music around 19:00?
    1
  8304. 24:40 all the blue segments are more than half of their respective rectangles (draw a diagonal)
    1
  8305. here is an attempt on proving the circumference of the magic wheel is ln(10): If we label our rubber band starting with 0 at the wheel and ending with one at the stick fixing it, our rubber band after being stretched starts at 0, wraps around the wheel and stops at 9/10. now let's choose an arbitrary point a distance x from 0, x is in [0,9/10], when the wheel rolls and x touches the top of the wheel the rest of the rubber band has stretched to and has become length 1, so it stretched by a factor of 1/(1-x), so the neighborhood around x is stretched by a factor of 1/(1-x). So finally the circumference of the wheel is integral from 0 to 9/10 of 1/(1-x) dx = ln(10)
    1
  8306. «Clearly some posture problems due to excessive obsessing with mathematics»...hahaha. I also like the fact that no integers can be obtained from partial sums.
    1
  8307. Made it to the 🔚 😛
    1
  8308. hi
    1
  8309. You and your way of illustrating this matter re just great! Never get enough of it. Are you sure you are not breaking any law about dependency? :)))
    1
  8310. This was one of the best videos yet!
    1
  8311. 666 k subs
    1
  8312. Hi every one ! There's something more ! If you take 4 consécutive numbre in the séquence of the Fibonacci séquence and muliply them you get the area of all the triangles obtain with this method ! J³
    1
  8313.  @Mathologer  something else ! At 11 min 56 sec, you can see in the Pythagorian identity that you show ! 153 is a narcissic number ! 153=1^3+5^3+3^3 something interesting about these number ! If you take the third order residu of a number that can be divided by 3, it converge to 153 😉
    1
  8314. We learned it somewhere in the 8th grade in India
    1
  8315. 700 year old proof
    1
  8316. I was just reading about the happy ending problem which also relates to Ramsey Theory.
    1
  8317. Yeah, little seems to be known... They also say it has to be between 5 and 18. Who knows, someone might just put the right answer here. Or at least say why it is so difficult to solve. I am going to wait and see. :)
    1
  8318. Finally NEW video! Thanks for sharing!
    1
  8319. Natural logs are made of wood
    1
  8320. I love following along and getting that giddy feeling when I can tell where I proof is going. Merry Christmas! Love the Channel since day 1!
    1
  8321. Wow. Thank you!!!
    1
  8322. I love getting into the metaphysics of this.
    1
  8323. Sorry, i wrote this bevor i started watching the video😅
    1
  8324. I'm not a fan of the Fibonacci sequence or the "Golden Ratio", either. I blame the media for their overuse.
    1
  8325. Funny story: my mother a bit of a maths whiz, was still keeping relic textbooks from her school days on the shelf at home. This in combination with my curiosity and desire to read —anything in those days: hey, look at me, I'm reading! — led to me getting such a head start on numerophilia(?) that I tested out of all maths until college when I was finally forced to face that dreaded beast, the calculus. So, I don't know whether or not Heron was part of the curriculum when I went through secondary, ca mid-70s... I wasn't even aware that "analytical geometry" had a different name until then, either :D
    1
  8326. Boy did I ever c get lost on YouTube🥺
    1
  8327. If we treat the hexagonal tilling as a stack of 3D blocks, it's looks like the construction will converge to an eighth of a sphere's surface. Is it the case?
    1
  8328. 6:16 Nooo, they are all 4k-1. I mean, you could probably argue that 4k-1 and 4k+3 are same, but if 5 is 4k+1, then 3 is clearly 4k-1 :)
    1
  8329. The sequence of partial sums never being an integer is my favorite in this video.
    1
  8330. Great video, really interesting stuff!!
    1
  8331. Where did yoi get that shirt? I genuinely love it because im the know it all in my family. Lmao.
    1
  8332. Best X-mas video ever.
    1
  8333. riemann hypothesis. please please please please please please please please please please please
    1
  8334. I really like this shorter format. I often struggle to stay focused for the whole video when it's of regular length, but this was perfect for me.
    1
  8335. Sounds like English with a slight German accent
    1
  8336. My maths IQ has just gone up
    1
  8337. That's the ruler on When the wind rises!! Thanks, I was looking for it
    1
  8338. Awesome 🎉
    1
  8339. I don't get it. 3 over 2 is 1.5. So how did it turn into 2?
    1
  8340. Proud to be an indian 😄😄😄
    1
  8341. no nine vizualization is cool
    1
  8342. Excellent, as always <3
    1
  8343. If we look at the marching bug problem and change it so that our coordinate system is locked to a single bug, we have the following situation: this bug is stationary, while all other bugs move towards it with constant orientation, thanks to the symmetry of the system. Since the angle between the bugs (their center lines) never changes, and there is no lateral movement, the path becomes a straight line, and the total distance traveled will be obviously equal to the starting distance between them.
    1
  8344. Congratulations for your 100th video. Please go through the works of Aryabhata I 5th century CE.
    1
  8345.  @Mathologer  Very thanks to you to consider my request.
    1
  8346. They are all amazing facts - but the greedy algorithm is so simple and yet so powerful that I'll vote it my favourite bit.
    1
  8347. Great presentation! I remember sequences and series coming fast and thick in pre-Calculus (way back in the day), but it was never given this kind of relevance. (I also took the original sequence and went constructed "up" and "left" instead of down and right the pyramid. Each column would go from seemingly random to resolving to the next power of 2. Like a limit. Fascinating.)
    1
  8348. Love these long videos without even one advertisement in the middle. You are making a really good job, even my university recommends your channel 👍👍👍 And for the question: All of the highlights were really amazing and interesting but I really prefer the animation of no 9 series ❤️ Greetings from Germany 😊
    1
  8349. It's a 3-4 duoprism!!
    1
  8350. He looks and sounds like an Austen Powers villain. "ONE MILLLLLION DOLLARZZ!"
    1
  8351. Love your work
    1
  8352. Very Nice
    1
  8353. Sie sind verdammt gut 😂😂😂
    1
  8354. The power of 2 coins challenge is super easy for someone who understands binary very well
    1
  8355. wow
    1
  8356. 45 minutes? My body is not ready.
    1
  8357. One of your best videos, IMHO. I absolutely loved it.
    1
  8358. Best part sum of the various series.
    1
  8359. now we can cheat in IQ tests
    1
  8360. 'They' don't teach 'they' train and brainwash.
    1
  8361. It worked well
    1
  8362. I already complied some zeta function proofs.
    1
  8363. I really liked Tristan's fractal; I'm just a sucker for fractals, and visual math in general!
    1
  8364. Just found this channel. Now I have a back log..
    1
  8365. 2 infinity and beyond?
    1
  8366. Professor Polster is an amazing and enjoyable teacher, except when he chuckles 😀
    1
  8367.  @hziebicki  I suffer from ASD (Attention Superavit Disorder), so when he chuckles, I get distracted for hours.
    1
  8368. Oh, sorry to hear...
    1
  8369. Thank you. This garment rule video shows the meaning of the most appropriate allocation with the correct water storage model. Because each water tank is a metaphor for a different creditor, small creditors can bear less risk, and large creditors can bear more risk, so paying off small debts first. The most appropriate distribution and equal distribution are different. It's about Priority of payment, just like Bankruptcy Law and General Average principle of Maritime Law. God blesses you, thank you again.
    1
  8370. But eventually at the infinitely small end of the trumpet there would be an infinitely small hole through which the infinitely thin paint would eventually leak. Therefore, this trumpet could not hold 8 L of paint. Problem solved.
    1
  8371. I'm here for the Conway.
    1
  8372. Nice! Nice! Nice!!!!!!!!
    1
  8373. Really loved the proof that shows that the No 9 Series converges! The graph really helps with familiarizing and understanding such a strange concept!
    1
  8374. What was the name of that conjecture? Pekora's conjecture or something?
    1
  8375. I beg to report sir, that, you are inspirational. I wish I could meet you in person. Best regards a soon to be influent individual.
    1
  8376. My favourite was Kempner's proof because it brings back so many memories from high school. This was one of the first problems that really pulled me into mathematics.
    1
  8377. So you're saying 1/3 x 1/2 x 2/1 = 1/3 ?? Or are you saying, Area / 3 = 1/3rd Area ?? Or are you saying, 180° / 3 / 2 x 6 = 180° ?? Or are you saying, 180° = a°+b°+c° = ((a°/2)₁ + (a°/2)₂) + ((b°/2)₁ + (b°/2)₂) + ((c°/2)₁ + (c°/2)₂) ?? Or are you saying, area of triangle = (h x w) /2 = 3 x ((1/3h) x w) /2 ?? smfh
    1
  8378. Student who remembers the sum rules for sine and cosine: 🤓 Student who derives them from the Ptolemy's theorem: 🗿
    1
  8379. I think it would be 7³ + 8³ = 9³ because in the first pattern it starts like 2*(1), 2*(1+2) ... and in the second pattern it starts with 4*(1), 2*(1+2) ... so it always starts with 2^1 then 2^2 so it would be 2³*(1+2+... )
    1
  8380. Mathologer Seal of Approval! Woo-hoo! I wanted to thank you for your great videos. I teach 4th Grade students in America, and I routinely share your videos with my students. Your clear discussions and visualizations help them to understand concepts far more advanced than we would normally cover in 4th Grade content. We all have wonderful discussions about the maths that you share with us. I sincerely appreciate all you and your team do to keep this content coming out for us. Your videos are what the internet should be. Thank you!
    1
  8381. Final puzzle answer: 55 ? :) 5 = 5 4 = 4 1 4 = 1 4 6 = 3 2 6 = 2 3 3 = 3 1 1 3 = 1 3 1 3 = 1 1 3 4 = 2 2 1 4 = 2 1 2 4 = 1 2 2 2 = 2 1 1 1 2 = 1 2 1 1 2 = 1 1 2 1 2 = 1 1 1 2 1 = 1 1 1 1 1 total = 55
    1
  8382. How would 3d shadows of 4d regular triangular-faced polyhedrons look like?
    1
  8383. Mathematics is always confusing.
    1
  8384. Now why are the non-shrinking polygons the only ones that tessellate?
    1
  8385. This is just a circular slide rule - common knowledge.
    1
  8386. It will only go down if there’s gravity.
    1
  8387. HO-LY CRAP! I'm stunned. Probably it will make things worse if I listen to this a few more times. A lot of thoughts came to mind during it. What are the closed form expressions of r and s? (or did he cover that and it flew by too fast?) What is the relation of r and s to phi? What happens when we look at nonagons? Octagons, hexagons, and squares? Generalize to n-gons? You don't suppose there is a connection to prime numbers lurking somewhere in this, do you? Or to pi? Is it possible to use a quarter sphere somehow within the r/s boxes to make a 3D spiral shell?? Is phi somehow related to r and s, and if you go to nonagons do you get t, u, and v and somehow r and s are deriveable from those? What is the infinite generalization of all this? Does it somehow include all of mathematics? 😂
    1
  8388. AKA, a circular slide rule. Aviators have been using these things for as long as there have been things that go into the sky. This is the same “technology” that we used to put men on the moon and since we stopped using these, the higher functions of our brains have been turned off (computers don’t require the use of imagination) and our exploration of space has suffered ever since. I wrote this while at the beginning of the video…I didn’t know that you (he) was going to say, with more flair, the same things. I’ve always loved slide rules and I can’t explain exactly why.
    1
  8389. When you put up that line of ones along the top and suddenly all the numbers clicked into place for me and I had to pause for a minute to go "holy shit!".
    1
  8390.  @Mathologer  oh my god you replied I legitimately... π ed myself. I'm unreasonably proud of this joke
    1
  8391. IOW: Most "what comes next" math problems have a correct answer of "you didn't give me enough information to say for sure." Fun Fact: I know the answer to an infinite number of questions. The answer is one. ... "huh?" ... I hear you ask. There are literally an infinite number of mathematic equations which can be devised which will give 1 as the answer. e.g. 2-1=1, 3-2=1, 4-3=1 and so on, adding 1 to each number to the left of the equals sign each time. Since you can go on adding 1 to each number infinitely and still have 1 as the correct answer, that proves that I know the answer to an infinite number of questions. (Even More Fun Fact: So do you) :D Editing to add, more properly I could say that I know an infinite number of answers to an infinite number of questions, since it's possible to create an infinite number of mathematic equations to derive any one of an infinite number of mathematic answers.
    1
  8392. Please prove Cauchy's Mean Value Theorem without the help of Rolle's Theorem.
    1
  8393. Hate to point out an error, but... 28:30
    1
  8394. I learned about the second proof when I was 14. I wanted to know why the theorem was true. My teacher was kind enought to tell me the second one; rather than doing the teacher classic of saying "the area of the square of the hyp is the sum of the others" which is a visualization at best, not a proof
    1
  8395. Dudeney was fond of drawing tessellations on tracing paper or strips of tracing paper and sliding one over the other til interesting congruences suggested novel dissections. The solution to this puzzle can be found by this method as can his famous square to triangle dissection. Google "536 Puzzles and Curious Problems" for the solution.
    1
  8396. It is similar to the Coastline Paradox, but in this case the problem is simpler to visualize as it comes down to trying to approximate a smooth surface with a combination of convex and concave shapes, which will always have a larger area than a smooth surface.
    1
  8397. Great nuclear apocalypse prep, very timely!
    1
  8398. Most memorable = Chapter 5 because of the e^{\gamma} formula :)
    1
  8399. The power of 2 coins is just binary numbers. In example 1, 2, 4, 8 you have a nibble (half a byte) so you can "encode" every positive integer up to 15 (1111 in binary). With all of the powers of 2, you can create change for every positive integer.
    1
  8400. How beautiful!!
    1
  8401. I guess the Leibniz method is more precise
    1
  8402. I liked the proof at the end, especially the fact that the upper limit was still so far away from the actual limit.
    1
  8403. The most memorable was the fact that gamma converges despite being the difference between 2 series that explode 🤯
    1
  8404. How to like this video pi squared times? :D
    1
  8405. How on earth Price and Bernhart discovered this just blows my mind.
    1
  8406. Two and a half rotations
    1
  8407. Nevermind, I was misunderstanding how the sequence was generated
    1
  8408. it really seems by the observation of the bricks' structure that the series may have infinite terms but as a sum its value has for sure a limit... so our senses are lying to us! Actually, the most part of the bricks in the structure are used to build more and more counterweight on the table, so at least what you are building is a solid extension of the table itself. According to my calculations, in certain point our planet Earth should start gravitating around the brick structure, with the table included.
    1
  8409. Jokes on you, i hadn't actually seen the other version so i'll peaceful watch this whole for the first time
    1
  8410. The practical historical use of this is making an elliptical dining table with a hand router. The modern way is CAD/CAM which is an inferior result as the mathematical segments can be detected. It's pretty much a lot art of the trade. ⚠
    1
  8411. Wow. Fantastic🎉
    1
  8412. Burkard! What a lovely video. Thank you very much! Have you ever heard of "Project MATHEMATICS!" by Jim Blinn? It's a collection of videos that teach principles of math using lots of wonderful animations. It was produced around 1990, so the production is not nearly as slick as what we expect today, yet the results are remarkably good. I recommend it to your attention. https://www.youtube.com/watch?v=vpxWyJg4_1A&list=PL8_xPU5epJdchhIkbjCPJM7m2anGUq9JT
    1
  8413. I love your content and am proud to support your Patreon at the π level!
    1
  8414. Incredibly nice well done....
    1
  8415. Obsessed with Euler's e-Pi-i Fusion-Fission Function, which is an instantaneous way of labelling the universe as ONE-INFINITY, limit 1-0-infinity superposition logic "connecting" the Superspin pivot of nothing everywhere-when No-thing by one identifying pure-math calculation, the entirety of (Aether/eternal absence) Black-body @ potential Eternity-now Interval, and the point-focused density-intensity real-numberness sync-duration of log-antilog Singularity reciprocation-recirculation potential positioning spacing, ie relative-timing ratio-rates of 1/0/infinitesimal=> Entanglement/Spacetime..
    1
  8416. Mr. Mathologer! G'day mate! Please do a video in this vane on Elementary Differential Equstions (or Diffy Que as they say in Chicago Uni) It's my birthday in 5 months and a new video from you would Bea perfect replacement for a cake. I'm finally at that age where these sort of events aren't really celebrated. Help keep my spirit young!
    1
  8417. The swivel proof was more "ah-ha", but the other one feels more "classical". Tie :-) .
    1
  8418. wow i haven't thought about actual maths for years!! cheers!!!
    1
  8419. Plot twist: this was an April fool's videos and nothing really works
    1
  8420. What is his shirt?
    1
  8421. 14:10 the jaded calculus student would like to say that it is probably e.
    1
  8422. wow.... counterintuitive
    1
  8423. I need to upgrade my mathematical seatbelt to a full blown mathematical three-point harness at this point.
    1
  8424. And the 6666th number is 190665405206533521449176627943131346656689354710161988378090397495716755843325094208487 :-)
    1
  8425. I don`t understand a single word, because i suck in math. But i like this man`s accent and voice, so i keep watching
    1
  8426. All this depends on the assumption that Hooke's law of elasticity holds exactly. This is, however, not true; it is only a first order approximation. So I really don't believe that a rubber band is suitable for making a true logarithmic scale.
    1
  8427. Actually, on second thought, Hooke's law is not needed here. The only thing that is needed is that the rubber band sticks to the wheel with a force firmly, so the elastic forces are not strong enough to move to parts of the rubber band already on the wheel. So there is no dependence on Hooke's law.
    1
  8428. I almost screamed when you mentioned "Calculus made easy" in the video! It was my first exposure to calculus as well and it is a fantastic book
    1
  8429. I saw a fraction that looked like it simplified to two, got too clever trying to quickly add up the squares and got a wrong answer that was less than two.
    1
  8430. Oh? You love Math? Then name every number
    1
  8431. 10:48 "Let's also quickly go on a little bit of a tangent here" Was that an intentional trigonometry joke or was it just a happy little accident?
    1
  8432. I think it's 2,9551...
    1
  8433. Ron Graham seems to have been a man of many talents. Juggling, out of all things... reminds me of your video on the mathematics of juggling.
    1
  8434. 4:40 - It looks like you're building a round slide rule!
    1
  8435. The last animation was just the best.
    1
  8436. I learned about casting out 9s on my own reading math books as a kid, they never taught it in school. (Along with how to calculate square roots by hand and other interesting things.) A few quick comments - When making these sorts of diagrams, I think it's nice to, instead of using the numbers 0 through 8, use the numbers -4 through 4. (Or if you're working in something other than modulo 9, it goes from roughly -n/2 through n/2, give or take the parity). The reason is that if you number the diagram that way, with 0 at the top as usual, then you immediately see that 8 = -1 mod 9, 7 = -2 mod 9, 6 = -3 mod 9 and 5 = -4 mod 9, so the left side of the circle is just the negatives of the right side. But since ab mod 9 = (a mod 9) (b mod 9), you get exactly the same diagram as if you used 0 through 8. However, now with those negatives, it's instantly obvious why the diagram is symmetrical about the axis - once you hit a negative number then multiplying it by 2 (or whatever the multiplier is) is going to give you exactly the same result as the corresponding positive number but with the sign flipped. So if you start at 1, make a couple of connections, then hit -1, the following connections are always just the mirrored versions of the ones you just drew. That's why all these diagrams, no matter what multiplier and modulus you use, are symmetrical about the y axis. - Regarding the proof that "casting out nines" works, an alternative quick proof is to notice that 10 = 1 mod 9. So 2578 = 2*10³ + 5*10² + 7*10 + 8 , which modulo 9 is just 2*1 + 5*1 + 7*1 + 8 = 2+5+7+8.
    1
  8437. Good job, you successfully teased me into grabbing a calculator for 3^4+4^4+5^4+6^4. Also, I will now remember 7^4 until I die.
    1
  8438. Your Video is awesome sir❤️🙏
    1
  8439. I cannot express how much I like this channel... the equilateral triangle proof was amazing and reveals some really important aspects of general problem solving. Genius!
    1
  8440.  @Mathologer  So, building on the pattern, am I right a 9gon would give the diagonals and equations and ratios for 4-dimensinal boxes...? What do the even - gons (like hexagon) represent?... Intrigued 😄 Hope you make more videos on this topic...
    1
  8441. the most memorable part was definetly the introduction of gamma, it even made it obvious how it had to converge, but also be greater than 0.5
    1
  8442. At 8 minutes I predict that 16 will also be special.
    1
  8443. It feels like this is mathematically related to the "The best A – A ≠ 0 paradox" math and the Riemann rearrangement theorem - you are rearranging an infinite sequence that would converge to the desired value into a different sequence that still has elements that converge, but which in toto converges to another value. Is it related?
    1
  8444. thank you, gifted me some insight
    1
  8445. @21:45 Wallace product not exploding: Notice product of 2nd and 3rd term < 1, product of 4th and 5th term <1 while RHS all terms >1
    1
  8446. In Poland one ought to be taught this formula in a high school.
    1
  8447. Burkard, take a bow, another amazing video. I cannot thank you enough. Cheers, KCruz.
    1
  8448. This is truly gorgeous.
    1
  8449. kudos to John W. Wrench jr
    1
  8450. I made it to the very end.
    1
  8451. Book book book book book!
    1
  8452. i wrote the sums as triangular numbers and noticed some beautiful symmetry! the sum of the houses must be a square triangular number. if there are j houses before the man's house, and there are k houses total then we find (j+1)^2 = k(k+1)/2 neato
    1
  8453. nice. too bad i hadn't learnt about this is school before jumping on straight to regular calculus. and i like the change of music too, it's refreshing and bright.
    1
  8454. You say cutting out a region with a loop ensures there are no holes... What if the loop nearly doubles back on itself? Like ┌─┐┌──┐ │┼││┼┼│ │┼└┘┼┼│ │┼┼┼┼┼│ └─────┘
    1
  8455.  @HHXOXHH  Well obviously; but I can remember, as a kid, some hack SF writer (it might have been Patrick Moore) trying to be clever by picturing someone being crushed, although the mass was only modest. Campbell the SF editor, who backed the loony Dean antigravity Drive, once claimed that docking with a space-station would be impossible (even a slow incoming vessel would plow through it, he said) ... but that was mainly because he did not comprehend the physics of collisions. BTW the Shaolin monks regularly use age-old physics tricks to fool the public. One trick combines both the mass/weight and the collision misconceptions. They have 6 suckers heft a large section of telegraph pole which they then use to ram a monk who - supposedly - stops it dead. Except that he doesn't: he decelerates over a couple of feet, and the 6 guys cannot get up any real speed precisely because of the weight of the ram ... which they can barely lift. Meanwhile, the monk has to contend only with the mass [sic] of the ram and the modest required deceleration.
    1
  8456. Thanks for the informative and joyful videos! Your enthusiasm is contagious. I believe the asymmetry---in the final decagon size---demonstrated with your random decagon example could be used to communicate the concepts of 'left' and 'right' to an alien civilization, without relying on physics' favorite weak interaction. Assuming Petr's Theorem only works when 'ears' are drawn to the left for angles <180° and to the right for angles >180°, could we use this to establish a shared understanding of direction? -Pick a random decagon. -Choose a direction of motion. -Choose an orientation ("Alien Right" or "Alien Left") so Petr's Algortihm produces a regular decagon. -Repeat with the same Alien orientation but reversed direction. -The "Alien orientation" with respect to the direction resulting in the larger regular decagon is equivalent to our "left." Is this correct?
    1
  8457. The main point I take away from these proofs is that their appeal relies on GEOMETRY and not ALGEBRA. The Greeks worked geometrically and for me this has a greater appeal - is more elegant, more beautiful - than often long and sometimes tedious algebraic manipulation, IOW the geometry leaps out at one,, whereas the algebra is a slower discovery of the truth. As usual the Mathologer videos are examples of what can be little known - even abstruse - theorems presented in an elegant way. mit bestem Gruess.
    1
  8458. Very important video. thank you.
    1
  8459. Liked the swiwelproof best
    1
  8460. Did anyone notice the animation at https://youtu.be/Y5wiWCR9Axc?t=1337 is a bit wonky? The conclusion is correct, of course, but the process should show two red segments from the larger square pairing up with two from the smaller in order to form the final two red rectangles. But instead only one large segment and one small one cross "sides", requiring one small segment then to grow into a large segment, and one large segment to shrink down to the smaller size. Otherwise, lovely animation.
    1
  8461. Oh my GOD my brain hurts!! Numbers and letters do not mix!!
    1
  8462. God bless you! In India, People never say such info, at free of cost. Wish you are already in a stable and secured job! Above all, YouTube recommends of my interest, among many videos
    1
  8463. I could be wrong, but isn't Calculus Made Easy the same book that Richard Feynman used to learn Calculus? Too lazy to use Google right now lol
    1
  8464. What happens if you plug in a sequence like the individual digits of pi?
    1
  8465. Most memorable: irregularity of maximum leaning tower (but I loved gamma too ^^)
    1
  8466. Quadratic Reciprocity is in "Proofs from the BOOK", which is from what I've read of it a fantastic resource for beautiful proofs. There's also a very boring and unenlightening proof by J.S. Frame that was given in my lecture notes, which could be done as algebra autopilot.
    1
  8467. 2:32 I verified both triangles with the shoelace method. Please remember (phi)^2 = phi + 1. The coords for the equilateral triangle I used, (0 , 0) (2*sqrt(3) , 0) (sqrt(3) , 3) The coords for the golden triangle I used, (0 , 0) (2*phi , 0) (phi , 2*phi) Very easy with the shoelace method!
    1
  8468. 2k + 1 = k^2 + 2k + 1 - k^2 = (k + 1)^2 - k^2
    1
  8469. A practical use for this is in optometry or any field where you're trying to create a corrective lens that fixes a lens with an irregular or non-symmetrical shape.
    1
  8470. I should have fucking watch the cat video. My background is computer science. right now i studying data science, a lot of maths is there. Und du hast mich mit dem video getötet. Danke für die schönen animationen, diese helfen mir wirklich.
    1
  8471. Definitely the part with the graph of 1/x vs the harmonic series and gamma. Maybe I'm just biased because I love that constant though its amazing the places it shows up
    1
  8472. I have a slight... bug with the shirt from the video: why have you floored the "Root of all evil"? Am I missing a math trick from some of the previous videos? Because by general conventions it should be 25.807, if we are rounding it up.
    1
  8473. Optimal towers: Largest overhang with fewest blocks! Very cool.
    1
  8474. in 9:30 the pattern I saw was An+1 = 3*An -2
    1
  8475. This induction hypothesis is correct only if you define your A_1 term. Otherwise, it can be proven incorrect by assuming A_n=5. Also, if do some rearrangement you get A_(n+1)=3(A_n -1)+1 which shows to be 3^n+1
    1
  8476. yeah I figured
    1
  8477. The last proof was so beautiful
    1
  8478. Memorable moment: Fractal visualization of Kempner Series.
    1
  8479. Each line extending from the edges of the triangle is the same length, since its length is the sum of the lengths of all three edges. The lines are also the same distance from the incenter. If you construct a line from the incenter to the closest point, and a line from the incenter to the ends of the lines, it will form a right triangle. Then you can calculate the distance from the ends of the lines to the incenter. If the closest point on the line exactly divides it in half, then the distance from the incenter to every point is the same.
    1
  8480. The CGI manipulation of equations and geometry really helps me a lot. I know there is a connection between algebra and geometry but few every actually show this relationship. I am not a master. Does anyone know of similar CGI used to illustrate algebraic expressions and geometry while building up from the basics? My thought is to mentally cover as much ground as possible. I'm only really interested in mentally being able to follow the mathematical patterns. Repetition only diminishes my rate of success. Once the concept is grasped I just want to move along. Thank you Mathologers.
    1
  8481. Di Vine HI 5
    1
  8482. “There is some debate over whether Heron really discovered this formula” Was it Euler?
    1
  8483. Mr. Mathologer, what is that sound at 11:20? Is it a bird?
    1
  8484. Thankyou
    1
  8485. Made in through just fine but did so in two sittings on two consecutive days. The "hardness" level was just about perfect all the way through. One reason I stopped on day one was that I was inspired/shamed into getting back to some of my pet hobby mathematics for the rest of the evening. I have a "minor" in math from way back...it was a brutally rigorous proof-based program, and since then I've had little interest in math without proof.
    1
  8486. I had the hardest time understanding logarithms in school. I wish this video was around then as seeing everything explained geometrically makes it click for me. Wonderful video!
    1
  8487. Не смотрите это видео за праздничным новогодним столом, не повторяйте мою ошибку.
    1
  8488. The vortex diagrams also remind me of the "Pentagram of Venus": https://en.m.wikipedia.org/wiki/Venus#Pentagram_of_Venus
    1
  8489. So will I 😌.
    1
  8490. I'm watching it 6 years down the road and i'm it still serves me quite the laugh, to see the Mathologer amongst the greats. But what's so laughable about that... is it implying i question that status?
    1
  8491. This is beautiful marvellous and magnificent .... thank you
    1
  8492. Lots of marvelous generating functions-- moment generating function, Laplace transform, Fourier transform, and my personal fave, probability generating function. A bunch more that I don't know, I'll bet.
    1
  8493. Most memorable/interesting fact was that the Ramanujan sum of harmonic series is gamma
    1
  8494. 1+(1/(n-1)) and n
    1
  8495. 4:30 Wouldn't the music be called a Luney Tune? Or, the display be called a Luney Toon?
    1
  8496. Thanks to this equality, we have log(1+2+3) = log(1) + log(2) + log(3) 🤣
    1
  8497. beautiful proofs! I wanted to mention that I am colorblind and had great difficulty distinguishing the orange circles.
    1
  8498. -1/12 is back. You gotta slay that dragon again
    1
  8499. At around 14:38, why is the correction term in the blue square 1/(n-1/4n)?
    1
  8500. I always watch your shows as a nitecap, always very interesting stuff.
    1
  8501. I don't like the symbol on your shirt. It is a satanic symbol.
    1
  8502. I can't help but wonder if this relates, in some way we are not aware yet, to particle spins...
    1
  8503. Neat!
    1
  8504. I really enjoyed this super easy visual proof, that γ<1. And in General the connection between the harmonic series and the natural logarithm
    1
  8505. I wonder if nicely filling 3d volume by some kind of spiral is achievable by using similar algorithm. All the spirals here are 2d, the helicone is basically 2d surface too. Or maybe it's impossible but perhaps you can combine 2 spirals in 4d as there are two independent rotation axes?
    1
  8506. My favorite, fairly succinct proof of Fermat’s two-squares theorem goes a little like this: (1): A whole number c can be written as the sum of two squares c=a^2+b^2 if and only if it can be written as (a+bi)*(a-bi), for integers a and b. Complex numbers a+bi with a and b both integers are called Gaussian integers. (2): A number is called a Gaussian prime if its only Gaussian integer factors are itself and 1. Just as every integer can be written uniquely as a product of primes, a Gaussian integer can be written uniquely as a product of Gaussian primes. (I’ll reply to my own comment to prove this later this evening) (3): If a prime number is not a Gaussian prime, it must factor into some (a+bi)*(a-bi), and so it must be a sum of two squares. (4): For a prime p, if we can find some number m such that m^2+1=kp for some integer k, then we have (m+i)(m-i)=kp. (5): But if we multiply p*(x+yi), we always get px+pyi. That means that p is a factor of a Gaussian integer c+di if and only if it is a factor of both c and d. But in (4), p is not a factor m, 1, OR -1, so p is not a factor of m+i or m-i (6) This implies that, if we find m such that m^2+1=kp for any integer k, that means that p is not a Gaussian prime, and therefore can be written as the sum of two squares. (7) Under what conditions can we find such an m? Well, we know that m^2 divided by p must give a remainder of (p-1), so we can answer this question by asking “What remainders can a square number give when divided by p?” (8) Let’s assume we have some whole number r<p such that m^2/p never gives remainder r. For every number s<p, there must be some other number t<p such that st gives remainder r when divided by p. This follows from the fact that p is prime and the pigeonhole principle, since there are (p-1) whole numbers less than p to multiply s by, and (p-1) possible remainders, and if there were some t1 and t2 such that st1 had the same remainder as st2, then if we considered st1-st2, the remainders would cancel and we’d be left with a multiple of p, even though neither s nor t1-t2 is a multiple of p. Furthermore, t=/=s, since we know s^2 does not give a remainder of r. (9)By Wilson’s theorem (which was proved on numberphile) (p-1)!+1 is divisible by p. But by (8), we can pair up the factors of (p-1)! Into (p-1)/2 pairs such that the product of each pair has remainder r. So r^((p-1)/2) gives the same remainder as (p-1)! when divided by p, that is, a remainder of p-1. (10) We now consider p=4n+1 and r=p-1. In this case, r^((p-1)/2)= (p-1)^2n=((p-1)^2)^n. Since (p-1)^2=p^2-2p+1, the whole expression will give a remainder of 1 when divided by p. (11) But by (9), if there were no m such that m^2 gave remainder p-1 when divided by p, the remainder of the whole expression in (10) would be p-1. So we conclude that there is such an m. But by (6), this means p is a sum of two squares. QED.
    1
  8507. Now the proof that we really do get unique factorization in the Gaussian integers. It’s easiest to see the logic side-by-side with the logic that shows we get unique factorization in the integers. (1): First, we show that there is a concept of “division with remainder” in the Gaussian integers. That is, for any two Gaussian integers z=a+bi and w=c+di, we have two more Gaussian integers q (the “quotient”) and r (the “remainder.) such that z=wq+r and such that the |r|<|w|. (2): To show that this is true for the integers is easy. For two integers a and b, we can say q=a/b rounded to the nearest integer, and r=a-qb. Dividing both sides by b gives r/b=a/b-q. It’s easy to see that |r|<|b| if and only if |r/b|<1. And what is r/b? The distance from a/b to the nearest integer. But on the number line, the furthest you can be from an integer is ½. So r/b is certainly less than 1. (3) Now we carry out a similar proof for the Gaussian integers. z/w is a point on the complex plane, so we can declare q to be the nearest (in terms of absolute value) Gaussian integer to z/w, and say r=z-wq. To show |r|<|w|, we need only show |r/w|<1. But |r/w| is just the distance from z/w to the nearest Gaussian integer. Since the Gaussian integers form the corners of a grid of 1-by-1 squares on the complex plane, and the furthest a point inside a 1-by-1 square can be from a corner is sqrt(2)/2<1, we can surely say |r/w|<1. (4) We can now use the Euclidean algorithm,which is described on wikipedia, to find the least common divisor of w and z. Since this algorithm constructs the least common divisor through a series of steps which each involve multiplying the inputs w and z, as well as the outputs of the previous steps, by (Gaussian) integers and adding them together. If follows that the least common divisor of w and z can be expressed in the form LCD(w,z)=jz+kw, for (Gaussian) integers j and k. (5) Using this fact, we can prove the following: if a (Gaussian) prime p divides a product of (Gaussian) integers a*b, it must divide either a or b. (6) Assume p divides a*b, that is, ab=pq for some q, but p does not divide a. Since p is prime, this means LCD(p,a)=1. It follows that 1 can be expressed by 1=jp+ka for some Gaussian integers j and k. Multiplying both sides by b, we get b=jpb+kab=jpb+kpq=p(jb+kq). So p must divide b. (7) Now let’s take c to be an arbitrary Gaussian integer. If it cannot be written as the product of two Gaussian integers of strictly smaller absolute value, then by definition it is a Gaussian prime. If it can, say, c=ab, then either one or both of them are Gaussian primes, or we can break down one or both of them further until we have expressed c as a product of Gaussian primes. (8)But we can use (6) to show that we could not get a factorization that ends up having different primes. So factorization in the (Gaussian) integers is unique
    1
  8508. A koch snowflake extended to the 3rd dimension has a finite volume and infinite surface.
    1
  8509. 5
    1
  8510. Just a brilliantly done video, thank you for the amazing explanation and the hidden gems! ❤
    1
  8511. My vote goes to the swivel proof.
    1
  8512. 1/1=1 2/1=2 3/2=1.5 5/3=1.666... 8/5=1.6 13/8=1.625 21/13=1.615... 34/21=1.619... 55/34=1.617... 89/55=1.618... 144/89=1.6179... ... lim n→∞ F(ₙ+₁)/Fₙ = (Fₙ)+(Fₙ_₁)/Fₙ =1+ (Fₙ_₁)/(Fₙ) lim n→∞ F(ₙ+₁)/Fₙ = lim n→∞(1+ Fₙ_₁/Fₙ) x = lim n→∞ F(ₙ+₁)/Fₙ x=lim n→∞ (1+ Fₙ_₁/Fₙ) x=lim n→∞ (1) + lim n→∞ (Fₙ_₁/Fₙ) x=1+ lim n→∞ (Fₙ_₁/Fₙ) x_1= lim n→∞ (Fₙ_₁/Fₙ) lim n→∞ (Fₙ_₁/Fₙ)= 1/x x_1=1/x ,x²_x_1=0 =>x=1±√5/2 x=1+√5/2 lim n→∞ F(ₙ+₁)/Fₙ =1.61803398875...
    1
  8513. 17:21 The sum should use odd numbers but uses the other one with all natural numbers. Same at 22:51
    1
  8514. my vote is for the logarithmic approximation of the partial sums, and in particular that (before adding in gamma) it is never off by more than one.
    1
  8515. [{(K^2)•((k^2)+1))}/2] will be the result after summing squares, k is the indexing.
    1
  8516. Of course I made it till the end of this video, two times. But I'm still very much in favour of proportional distribution. The more and better a person has worked (and therefore the more money he is owed), the more "important" his level of happiness is, compared to other workers. 🍻
    1
  8517. Best part had to be the no nines and 100 zeroes part. To me its insane how slowley they converge
    1
  8518. What’s the little motif beginning every chapter from? Bach’s inventions?
    1
  8519. I have seen the first visual proof before but not in school. I saw it sometime after I started watching YouTube regularly which was about 20 years after I graduated college in 1996.
    1
  8520. oh lord that joke….
    1
  8521. The object on Euler's head isn't a towel.
    1
  8522. A-A=0. Zero.
    1
  8523. Who else was staring at his shirt the whole time
    1
  8524. The Helicone Numberscope: Mathematical Superpowers Hidden in a Simple Toy 🙂
    1
  8525. Does this proves the fundamental theorem of calculus?
    1
  8526. Regarding 3 family trees (left(L), mid(M), right(R)), all properties holds no matter what we choose as our initial step. For example if you go R and then M to infinity, the difference will be 7 all the time (at least the absolute value).
    1
  8527. This video has allowed me to finally pin down the annoyance I feel when I see one of those 'complete the sequence' puzzles: "There are infinitely many functions containing this sequence, and you expect me to read your mind and come up with the one that YOU thought of?"
    1
  8528. By complete chance I was feeling sufficiently paranoid and I was able to figure out the pattern (and what the question would be) when you went through all the examples. The corners are the most distinctive part, if you just keep looking at them it's not implausible for anyone to notice.
    1
  8529. I guess this is kind of meta cheating, figuring out what clues the clue-giver wants to give.
    1
  8530.  @Naeddyr  :)
    1
  8531. I just very recently ran into the Pell Equation for a problem, thanks for helping to make the conection to more maths. I think it's worth noting that since the recursion is linear, it can be calculated with powers of a matrix, which leaves a very short expression to calculate them in the same time order as an explicit formula but without roots (avoiding having to use a rounding function in a computer due to the numerical errors). If you wanted, you can diagonalize it to get the formula as well!
    1
  8532. In fact a lot of conventions are never questioned they're taken for granted
    1
  8533. So I take it (adding a few bits about fractals that I know) that when using rectangular corner-cutting to approximate a circle, you approximate the area of the circle, but the perimeter of your approximation is a fractal curve that has infinite right-angled discontinuities which are not present in the smooth circular curve. Fractal curves have "fractal dimension" of at least 1 - they can e.g. be area-filling curves that have an area rather than a length, or space-filling curves that have a volume. Fractal surfaces have fractal dimension of at least 2 and can be space-filling surfaces. Either way (and including all the cases with non-integer dimension), arguments about measures being preserved/increased/whatever as the approximation is improved break down when the measure (length or area) ceases to apply, so you can't then take those measures from fractal approximations and apply them to the non-fractal thing being approximated. On the other hand when subdivision forms smaller and smaller in-between angles, in the limit you have infinitely many zero degree angles - the discontinuities have disappeared so that the infinite pieces (lines/triangles/whatever) form a smooth, continuous curve/surface. There's a different non-convergence problem (Runge's phenomenon) which I find interesting when approximating smooth curves using high-order polynomial curves, or when approximating smooth surfaces using high-order polynomial surfaces. The normal solution is using piecewise curves/surfaces using low degree polynomials for each piece, though usually based on a single specification for the full curve (e.g. B-spline) rather than explicitly working out the polynomials for each piece. Converting into a connected sequence of simple polynomial curves (e.g. Bezier curves) is easy enough but unnecessary. Subdividing into linear B-splines into lines is the linear-polynomial-pieces special case that (like linear B-splines themselves) no-one uses. I mention this because I'm now wondering if the Runge's phenomenon convergence failure indicates another kind of fractal, with infinite extreme "oscillations" rather than infinite discontinuities.
    1
  8534. Most memorable: the "irregular towers"/"unexpected" maximum overhangs
    1
  8535. The physicist and inventor Charles Wheatstone wrote a single mathematical paper ('On the formation of powers from arithmetical progressions', Proc. Roy. Soc. London, 1854), with some nice results. For example, Any cube n^3 is the sum of an arithmetical progression of n terms, the first term being n and the difference 2n. Say, if n = 4, so that 2n = 8, then n^3 = 4 + 12 + 20 +28 = 64. Also a neat proof of the standard formula for the sum of the first n cubes. The paper is probably trivial to an expert mathematician, but might be worth a look.
    1
  8536. Excellent video!! A good (simple) book about this is "A Primer on Number Sequences" by Shailesh Shirali.
    1
  8537. Solution to the task given at 11:55: place 2 planks on top of 3-rd plank and move 3-rd plank at the edge until it is balanced.
    1
  8538. The Bernoulli numbers have the important distinction (outside of proper math) of being the subject of the first computer program ever, appearing in Ada Lovelace's note on Babbage's Analytical Engine in 1842. But hardly anyone knows about it ;-)
    1
  8539. This would also prove that a triangular mesh or honey comb would not contain any squares
    1
  8540. This is the only channel which gets an instant watch as soon as the notification goes off.
    1
  8541. Amazing stuff - had no clue!
    1
  8542. yes.
    1
  8543. There are cyclotomic polynomials lurking in there.
    1
  8544. Mathematics class at Wynstones Steiner school in Gloucestershire, England. I was lucky enough see these proofs
    1
  8545. Dutch math teacher here. Wonderful video, no real problem in following what you were doing. It was somewhat long though, so my mind started wandering a bit after 35+ minutes (and having a beer while watching probably also didn't help :-). Keep up the good work -- you're doing an amazing job.
    1
  8546. So...what happens in the real world, when we divide up until we get to the Planck Length?
    1
  8547. I really love γ so I can't say otherwise, γ and the sum relationship is my favourite!
    1
  8548. Also, chapter 5 was my favorite if I HAD to choose... although chapter 4-6 are all equally delightful. The most memorable thing for me will be from chapter 4, that no partial sum of the harmonic series will ever hit an integer.... All the way to infinity.... That's so profound.... Like.... Woah!
    1
  8549. Anytime I feel I know enough (like just after a scolding session to my daughter), I come see a mathologer video and become a child again, go back to my daughter and say sorry.
    1
  8550. There are more scandals in 'science' to be educated to the 'sleeping beauties'. But hey, the whole idea of 'science' is now claimed by the UN when Under-Secretary-General for Global Communications Melissa Fleming told the world "We own the science".
    1
  8551. I like the tower of lire
    1
  8552. I was this close to posting the joke "Heron has no egrets," but fortunately I was able to stop myself.
    1
  8553. If you make a follow-up video: what lacing produces the least wear?
    1
  8554. Strange..reminds me of a physical process..cant remember which one
    1
  8555. First
    1
  8556. Considering what the principle taught in part 3 means for other base systems, primarily binary, was a fun throught experiment!
    1
  8557. Ramanujan taught himself but if he would get an education he would be worse
    1
  8558. Great sir
    1
  8559. “12 million and change” is a lot closer to 13 million.
    1
  8560. Brilliant as usual
    1
  8561. Great video once again! Great mathematical content for number lovers and of course incredible geometric illustrations! PS: Would it be possible, to refer me to the link that describes an algorithm(with code or math) that explains Petr's miracle? Because as a programmer I am unable to hide my enthusiasm to visualize it :) Thanks in advance!
    1
  8562. Most memorable: the Odd/Even non-integer fact
    1
  8563. The most surprising fact in the video certainly had been that the "no nines" sum is finite. My gut feeling was that the thinning out itself would be somewhat logarithmic, which would not be enough to make the sum finite. And the proof at the end was very pretty!
    1
  8564. Part of the sequence appearing around @3:24 is reported to be the same even if last digit is off (10582 vs 10581).
    1
  8565. This is just a circular version of a slide rule. And I once had a watch with a slide rule bezel. Casio DW-403.
    1
  8566. low key just verified the flat earth as reality. and im sure in a hour , because after the quick 5 min math proof, the oth er 55 min would be coming to grips with it, is that the earth cant possibly be a globe, using his fav thing, called math. even disk to sphere contraption when made smaller and smaller slivers, uncovers the universal yin and yang, the polarity, which would map onto a taurus
    1
  8567. I got 730/365, but my brain started to overheat when I tried to reduce that to its simplest form. Then I realized it's just 2. Do'h!
    1
  8568. Hexagons are really the bestagon
    1
  8569. Changing the lacing pattern on your shoes voids the warranty.
    1
  8570. The rearrangement theorem is a proof by contradiction of the non-existence of infinity.
    1
  8571. Indeed, pretty cool.
    1
  8572. His satisfaction with the animation is so delicious.
    1
  8573. Must be bigger than 0.5, since if you imagine that unit square divided into rectangles, then if you divided each rectangle diagonally, and picked the top half of each, you would get an overall fraction of half. But the curvature means that everywhere, the lines from the curve lie below a linear diagonal, so the area of the shaded area of each sub rectangle must be more than half of that rectangle. So the total fraction of the shaded top region of each rectangle is more than half or the corresponding rectangle so the overall fraction must be more than half.
    1
  8574. Only two other people noticed the error at 6:11
    1
  8575. Excellent video! Very cool identity and - as always - cool and useful animations (I'm a German studying Mathematics and Physics)
    1
  8576. Some people knows infinity ❤
    1
  8577. Okay, next video is Fermat's Last theorem proof, right?
    1
  8578. Am I wrong in thinking that "simple" relational mechanics like this is similar to how abacus math is performed?
    1
  8579. My final thought on how to prove it's a circle was to use the intersecting chords theorem. The ICT obviously holds at each vertex of the triangle, so we have a circle. That seems a bit glib: can I run the logic in that direction? And now to watch the rest of the video to discover that I'm totally wrong.
    1
  8580. super cool! please keep showing lesser-known mathematicians, i had no idea about these beautiful parts of mathematics :)
    1
  8581. By the way, nice video
    1
  8582. It's amazingly awesome
    1
  8583. @Mathologer  No. Please tell me more. And yes it is!
    1
  8584. I watch the Mathologer at 75% speed because I'm not that smart.
    1
  8585. What happens if you took ovals instead of circles?
    1
  8586. Wonderful video as always! 9:20 Length 10: Case 1: 8 copies of 1, a and b (2≤a≤b), 8+a+b=ab, (a-1)(b-1)=9, (a,b)=(2,10) or (4,4). Case 2: 7 copies of 1, a, b and c (2≤a≤b≤c), 7+a+b+c=abc. c is not an integer when a=b=2. Thus ab≥6 and c≥3, then 7+3c≥7+a+b+c=abc≥6c, contradiction! Case 3: 6 copies of 1, a, b, c and d (2≤a≤b≤c≤d), 6+4d≥6+a+b+c+d=abcd≥8d, contradiction! By the same argument as in case 3, we know there are no solutions with fewer than 6 copies of 1. More than 10 identities: Consider the equation (a-1)(b-1)=2³·3²·5 i.e. 359+a+b=ab where 1≤a≤b, there are (3+1)(2+1)(1+1)/2=12 integer solutions. Length 361 is one solution.
    1
  8587. I really liked the optimal leaning tower of lire. I've never seen anything like that and I had to try it out with 3 books to see how far I could go.
    1
  8588. Sum of “no 9s series” is square of fractal, which is tends to 0... awesome:) Intuition of prove in secs. Sum of harmonic is not an integer also made my day
    1
  8589. Dear Mathologer. I decided to share with you another view of Euler's famous topological formula: E-V+F=2. Here is my video about it, feel free to make it on a Mathologer way if it's has any value for our community. https://youtu.be/i0QoEtrhjlc https://youtu.be/M2ssnIv37jk Thanks for the beautiful works! Youcef Ammar-Khodja
    1
  8590. The numerical proof has a pair of orthogonal Latin squares. Have you ever considered discussing pairwise orthogonal Latin squares and finite affine geometries?
    1
  8591. waiting for the galois theory video
    1
  8592. @Mathologer--Thank you for another great video, I look forward to these all the time! To answer your question at the end, I followed this fairly well. It was excellent taking a deep dive here into these sums--my number theory course in undergrad glossed over these, not going much beyond the sum of cubes. I have a BSci. in math and have kept up with it since college. The "master class" videos you've done have been excellent, and I would love to see those boundaries pushed a bit further. Even going beyond the lovely animations and usual style, I'd happily buy a ticket for some more traditional lecture style videos, maybe covering some cool geometry stuff. Regardless, A+ content.
    1
  8593. Where can I get a T-shirt like that.
    1
  8594. This video is so hard to watch on a phone in the dark 😟
    1
  8595. Nowadays, reading a book is a small revolution. Reading creates smart people, and a smart person is always a
    1
  8596. What beautiful and magical transformations-Moessner to Pascal to the binomial theorem.
    1
  8597. I love the shirt.
    1
  8598. Odd/even <>integer is so obvious I never realised which is why I was never a mathematician Good video Well explained + never heard of the gamma constant before
    1
  8599. with the card trick, how does the assistant know if they should count clockwise or counterclockwise?
    1
  8600. It's a subtle point (and easy to miss), but you always choose "clockwise" - going in ascending card order (but looping from king to ace). You know at least two cards have the same suit, so you pick two of them. But the one you keep from this same-suit pair is not an arbitrary choice! You find the smallest distance between the two and you keep the end card of that smallest distance path, putting the start card at the front of the cards you give your assistant.
    1
  8601.  @MuffinsAPlenty  ah, so you always choose the card that makes it clockwise. I feel as though that was something he didn't mention
    1
  8602. About 7 minutes in, my mind was blown when I realized the integral of 1/x dx is ln(x). That is a hell of a way to visualize it.
    1
  8603. Pie slicing to number of fractions to equal measures for pizza or round food taste experiences to refine recipes... thanks for sharing!
    1
  8604. That is Mathology for you 🤝
    1
  8605. enjoy your "part of the first thousands views" ticket
    1
  8606. "warning complex maths ahead" -- oh boy, time for another coffee!! :)
    1
  8607. Pi really is equal to 4 for sufficiently small values of 4
    1
  8608. What a beautiful video, love it <3.
    1
  8609. Weird: suggesting something super natural, uncanny.
    1
  8610. 18:38 isn't it going on with squares of natural numbers with "something related with 4" somewhere? EDIT: 21:15 LOL
    1
  8611. For 2 there's an easy one 2 = 1/1+1/2+1/3+1/6. Sadly, it doesn't seem that easy for 3
    1
  8612. 3Blue1Brown is his patron.
    1
  8613. 42:27 now i see why you're getting that necklace to your wife--because it tastes delicious!
    1
  8614. A nice way to mesure happiness, but humans being what they are, everyone would probably bitch about the amount they got anyway unless they got the whole amount…oh, and made it to the be end…now it’s really the end of my comment….
    1
  8615. Why do you call them special cases? They work exactly the same way, don't they? As far as you count a 0º, 0º, 180º triangle as a triangle, of course.
    1
  8616. ​ @saityavuz76 I expect that you run into the same 'problem' with odd composite values of n. But you may not have run into that, because they start with n=9.
    1
  8617.  @RonVandenBurg  In my case, I tried to find the intersection between two parallel lines. I was confused as to why it kept glitching till I remembered euclid. Thanks to @hydra147147 I just take the middle point of the two points for when the lines are parallel.
    1
  8618. I made it to the very end - took me 2 years, though
    1
  8619. I hope you can read this and it is not upside down, I use a projector and flip the output when watching video's from the southern hemisphere. Anyway, after seeing the formula that what was used in the paper, I, like many others in Math(s)(?) often do, thought about extending it to the next 'level' operation higher. The results are golden at best. Given the implicit curve (xy)^2+x^2+y^2+xy+x+y-1=0 y=0@ -phi, 1/phi and x=0@1/phi and -phi The curve is symmetric oval. Keep the truth coming and fight back against Big Math a.k.a. -1/12 #theresistance is actually just #ohms
    1
  8620. The summer I was 13 years old, which is to say in the year 1964, I was anachronistically attached to a slide rule my father got me. It was linear rather than circular. I carried it around with me everywhere, learning how to use it for multiplication and division, along with a number of other tricks that it could do, such as exp and log, on special scales. In retrospect, I was a weird kid!
    1
  8621. I never thought about applying DFTs to polygons, but its truly amazing :D However this method only works in 2D. Is anybody aware of 3D extensions to this method?
    1
  8622. How do we get to our destination , if we are always halfway there?
    1
  8623. I made it to the very end! :)
    1
  8624. Wow. I should not watch these videos for self protection. You always make me want to jump full in ...
    1
  8625. The question from 11:44 is rather easy once the relationships are written out, I'm convinced the question is just warm-up or to drive engagement. Getting started, note that C=A+B and D=A+ 2B. This turns the square's relationships into: 1) 153 = A²+2AB 2) 104 = 2AB+2B² 3) 185 = A²+2AB+2B² From here it's a short step to the square [9, 4, 13, 17]
    1
  8626. That's a pretty neat way to count functions between sets. ;P
    1
  8627. What a fascinating video! One of the members of our associated LinkedIn group recommended to us, so we are now promoting it on our channel.
    1
  8628. 1:44 The last number of one series and the first number of the next are in type: ab, a(b+1)
    1
  8629. That's the best vedio i ever seen about my love......euler and mathematics 😍😍
    1
  8630. 17:24 It follows easily from the formula T = ABSin(60)=ABF
    1
  8631. The area of tree is 5 which is equal to the area of the first square multiply by the iteration number. Because the number of leaves grows as power of two and the area of leaves divides by power of two at each iteration so this two process cancel out each other.
    1
  8632. Serious Jedi mind tricks at 12:35 :O
    1
  8633. Soon 1000000. Merry Christmas!
    1
  8634. I'm 72 yrs old, took Maths A level, never seen the twisted square before!!
    1
  8635. I was familiar with the second version, because as an undergraduate in philosophy, I was attending a seminar on the Platonic dialog, Theaetetus, where there is a description of the figure with the one square nested in the other. The proof itself wasn't explained in the class, and wasn't really important to whatever it was we were talking about, but upon the mere mention that this was related to the Pythagorean Theorem (which I couldn't remember ever having seen proved, but maybe I didn't pay attention in high school math) I dropped everything and and quickly drew the diagram in my notebook, and started in on the algebra, proving to myself the very same relation in the video here. The flash of insight and revelation when I finished the calculation was like a religious experience. That was three and a half decades ago.
    1
  8636. My answer to the quizzes: 1. The theoretical pattern for higher powers would have the highest root of the left hand sum as 2^m * n(1+n)/2 since the sequence for the integers is 2 * n(1+n)/2 while the sequnce for squares is 4 * n(1+n)/2. So you would get for the cubes the left hand sums 7³ + 8³, 22³ + 23³ + 24³, etc., for the tesseracts the sums 15⁴ and 16⁴, 46⁴ + 47⁴ + 48⁴, etc. up to higher dimensions. 2. Although I didn't manage to calculate it in 30 seconds but using the equation you showed, 10² + 11² + 12² + 13³ + 14² can be reduced to either 2(10³ + 11² + 12²) or 2(13³ + 14²), both of which can be calculated in the head with the binomial formula (I used the right sum but left sum might have been easier). The sum you get for the sum itself is 365 which also is the numerator. The fraction thus equals to 2. 3. 3³ + 4³ + 5³ = 6³ which I kind of have guessed. However, 3⁴ + 4⁴ + 5⁴ + 6⁴ doesn't have this kind of pattern since the sum isn't divisible by 7 but rather by the primes 2 and 1129.
    1
  8637. I appreciate this video, but I have to admit your comment at 9:50 bothers me a bit. If I didn't know any better, I'd think you were assuming that most of your viewers were straight males. Given the equity issues in STEM, I think educators like you and me have a responsibility to use more inclusive language. Easy fix in this case: just use "they" instead of "she."
    1
  8638. So the first folded triangle is in a sense a complement of the original.
    1
  8639. Not that it matters but the OCD in me forces me to point out that at around 4:08 the Toymaker asks for the disks to be moved to point C and not B, as it is repeatedly quoted elsewhere. ;-)
    1
  8640. Aesthetically pleasing way to look at calculus. Wish they taught this in Calc 1, it's absolutely beautiful.
    1
  8641. Top get the number of rotations per revolution you can divide the distance covered by the center of the rotating circle by the circumference of that circle.
    1
  8642. It was really a mind opener. Thank you very much for wonderful insights
    1
  8643. <3
    1
  8644. This idea at the start of the video is referred to in Tetris as "parity," in this case referring to the single- or double-evenness of the light squares. Just like in this puzzle, you overlay a checkerboard pattern on the Tetris board and count the number of light squares covered. The parity (or double parity, rather) only changes when you clear a line or drop a "T" piece (the piece literally shaped like a T). Knowing this helps players of modern Tetris games set up boards they know can be perfect-cleared, or to assist them in resolving parts of boards. For instance, if you are trying to clear garbage on top of a clean well, considering the parity of that garbage can help in deciding the best way to clear it, including in deciding whether to burn a line. Since you usually know if a T is coming up with the multiple next boxes in these games, you can make a decision a few moves in advance.
    1
  8645. Finally, im not late
    1
  8646. At 12:30 a green square with a six on it is removed
    1
  8647. The animation during 'autopilot' mode is amazing! If I may: What software/automation stuff do you use?
    1
  8648. 14,500 have watched this video. 1,500 likes. I'll never understand why 90% of viewers don't click like. Something wrong with most of them.
    1
  8649. LOL! I just either got distracted or suddenly paid attention to your shirt! That's beautiful!
    1
  8650. For me, the bizarre maximum overhang was the most memorable because it reminded me that I could use my computer to find the ridiculous solutions to simple sounding problems. By the way, this is how to set up 3 bricks with 1 overhanging brick: you make an upside down triangle like this: |__________||__________| |__________| 000000000
    1
  8651. order emerges within the randomness
    1
  8652. What you call the "optic equation" is what I call the "parallel sum".
    1
  8653. Most memorable:Log formula for the partial sums and gamma
    1
  8654. I love ab=a+b. It's also closely related to my favorite pair of rational functions, y=1-1/x and y=1/(1-x).
    1
  8655. G
    1
  8656. Isn't what your describing, a cicurlar slide rule?
    1
  8657. Golden angle relationship:We have a circle with an angle of 360 degrees and we want to divide this circle into two angles so that the ratio of these two angles is equal to the golden number. We assume that these two angles are a and b and b>a. a+b=2π , b/a=φ & φ=1+√5/2 b=a.1+√5/2 , b+a=2π => a(1+√5/2 +1)=2π a(3+√5/2)=2π a=720/(3+√5) => a=720/5.236 =137.50° 360°_137.50°=222.5° 222.5°÷137.50°=1.618... =φ
    1
  8658. This was a really neat video.
    1
  8659. The shirt is epic as always :-D
    1
  8660. Love your videos, but why do you have DCLXVI & 1010011010 on the intro black board? 666 in Roman numerals and in Binary. Can't be a coincidence. Those are the only 2 I've noticed so far.
    1
  8661. Absolutely shocking!!!!!
    1
  8662. wow
    1
  8663. An amateur's way to generate the recurrence relation is to start with an approximation m/n, divide that into 2 giving 2n/m and then take the mediant of m/n and 2n/m. If you start with m=n=1 you get the desired sequence.
    1
  8664. You wonder what came first. The chicken 🐓. Or the. 🥚
    1
  8665. I found that making a T shape gets you to 3 as well
    1
  8666. Finally getting a chance to watch this video
    1
  8667. 18:51 He meant “chaotic”.
    1
  8668. the problem is nowhere in my entire school curriculum did we EVER approach the concept of proofs. we avoided the foundations of mathematics ENTIRELY for the sake of simply learning computation. and it drained us all and everyone hated math. so not only did we not learn math, we were taught to actively hate it. math education needs such an enormous rework in the USA
    1
  8669. Such a beautiful result! I am wondering now: would this be extensible to 3 dimensions?
    1
  8670. Amazing
    1
  8671. I've been adding together all the positive integers for the past few years, to see what their sum is, and so far it's around -1/13, so things are looking up.
    1
  8672. 45mins let me get my popcorn and drink ready
    1
  8673. Airline flight engineers still use this for load & fuel calculations, even today.....
    1
  8674. Did (100 + 121 + 144) * 2 or 300*2 + 65*2 Managed to do that in my head but failed to realize 730/365 is 2, lol.
    1
  8675. I just didn't think Nathan's explanation needed to be further explained.
    1
  8676. 256
    1
  8677. Here in Italy a water bottling company used to construct bottles with that shape, I think to ease the flattening and recycling
    1
  8678. The number of squares in an nxn grid is (n-1)(n^2)(n+1)/12
    1
  8679. The fractal for the no nines is exactly the same as the cantor's carped for the no twos, and the proof resembles the proof that the area of the cantor set is zero
    1
  8680. I'm wondering about what happens when the number of sides is not a prime number. When getting the set of basis polygons it isn't really clear what to do for the k'th basis where k divides the number of polygon sides. For instance the second basis for hexagons would be... two overlapping equilateral triangles? Just one equilateral triangle? One hexagon with cyclic overlapping points that looks like a triangle so traversing the points goes around the origin twice? The third basis would be a straight line? Or maybe three? Or a straight line that you traverse back and forth three times? Having written this I think the last answer for each case is correct. Which choices of ears you choose to apply to the original hexagon might produce a line, an equilateral triangle, or a regular hexagon. You also need to recognize that 180-degree "ears" amount to taking the midpoints of the existing sides.
    1
  8681. You'd fail the IQ test (of course you would not really fail it, because you know the following very well). The "correct" answer to a series in an IQ test is not a number which you derive from a arbitrarily complicated formula – it's sort of trivial, at least for a mathematician, that any number can be the next one. The implication in an IQ test is: find the next number with the most simple algorithm. So in the next number in 2 4 8 16 ? is 32. And once you begin arguing that you haven't understood the question and don't get a point in the test. Conclusion: to be overly smart gives a low IQ :-)
    1
  8682. we didn't even learn the basic divisibility test in primary school!!! :( damn... watching mathologer as an adult is realising how much you missed out on and coming to terms with why you now sick at math because there's gaping holes in your understanding :/
    1
  8683. I think I would stretch the shapes with respect to x=0 rather than x=1. It's simpler and in this case, the additional shifting part isn't needed
    1
  8684. The proof at the end was really nice. Ive seen the result before but never the proof. I also really liked the proof that the partial sums of the harmonic series is never an integer.
    1
  8685. Since the sum of the prime reciprocals also diverges incredibly slowly and primes are arbitrarily large, the same process can be applied to get an alternating sum of prime reciprocals to converge to any real number. Fun stuff!
    1
  8686. What are the music of the transition between chapters and of the credits ?
    1
  8687. I found the most memorable proof the proof of the divergence of the series, it's so simple and intuitive
    1
  8688. it's amazing how all these formula are somehow "encoded" inside spatial manipulation, meaning they come from properties of "2d space" and just moving figures.. I wonder if somehow we could just infer them from mathematical definition of what 2D space is (symmetries etc)
    1
  8689. hurricanes, galaxies, sea snail shells, pine cones, dna, even dolphins are a giant golden ratio :)
    1
  8690. A lot of crazies, is that a reference to Mario Livio?
    1
  8691. I don't know if someone already told you, but F.J.M. stands for "Fredericus Johannes Maria"
    1
  8692. Decimal numbers are also possible too, 1.4+3.5=3.5×1.4 1.5+3=3×1.5 1+1+1+1+1+2.5+5=5×2.5×1×1×1×1×1 but some decimal numbers are not possible, 1+1+2+3.5≠3.5×2×1×1 1.5+3.2≠3.2×1.5
    1
  8693. Thanks!
    1
  8694. On the 4k+1 and 4k+3 part. Why doesn't 4k-1 work for the right side (3,7,11,19,23,31, and so on)?
    1
  8695. My favourite is the fact that the harmonic numbers never equal an integer (or even that a sum of consecutive reciprocals of natural numbers is not an integer) because it is so extremely unintuitive at first, but once you see the odd/even pattern it is basically trivial and totally sensible.
    1
  8696. 25:24. I have made a decision that ANY channel that uses anti-free-speech Patreon will not get my support.
    1
  8697. The hair problem was the best..😍😍
    1
  8698. Radius = singular. Radii = plural. What is “radiuses”?
    1
  8699. So, does root(pi) /2 also come from analytic continuation of some L function?
    1
  8700. Can you discuss the mathematics of numbers on a waiter’s check pad? I believe the branch of mathematics is known as bistromathics.
    1
  8701. My favorite fractional approximation of Pi is: 355 ÷ 113
    1
  8702.  @Mathologer  Mine is, "A day less and 2 primes." (One less than a year, followed by two primes.) I would also like to extend my sincere thanks for these videos that you create. I discovered the beauty of math late in life (post school), so I'm trying to catch up, and remember. You help me to see and often understand the joy and beauty that is the language of the universe. And for that, I am very grateful.
    1
  8703. I like the quickness of this video, I don't always have time to watch the full videos that you normally do. I also imagine they are a bit easier to produce.
    1
  8704. I like the Devil lacing. I wouldn't use it, but it looks really neat.
    1
  8705. I minored in both Religious Studies and Mathematics in college so it's awesome to see the two subjects overlapping. If you could find some way to collab with someone like Dr. Mark Henry (aka ReligionForBreakfast) I would be in heaven :)
    1
  8706.  @Mathologer  The math minor basically came free with my computer science major since my school made it very easy to earn minors, and I've always liked history and social sciences alongside "harder" sciences and math.
    1
  8707. Halfway through the video, I was wondering if an aluminum can would crush into a Schwarz lantern if the material strength and thickness and the forces on it were perfect.
    1
  8708. 0:57 I spotted another relationship. 2*34 = 88 - 20 (20 is the 3rd number behind 88); 2*21 = 54 - 12 (12 is the 3rd number behind 54); 2*13 = 33 - 7 (7 is the 3rd number behind 33). I like this proof better Zvezdalina Stankova's proof on Numberphile. It's shorter and more elegant in my opinion.
    1
  8709. It seems to me that the intention of the Talmud is that smaller creditors. who usually are poorer people, get a bigger share up to a certain point. After that point it would make sense that all parties get a proportional share. That would mean that the columns in the upper half all have the same height, but widths proportional to the amount owed. The split that you suggest would mean that smaller creditors are disadvantaged if more than 50% of the owed money has to be split. If for example 75 Dollars are missing, the smallest creditor in your example would lose 25%, while the biggest creditor would only lose 8.33%. I have some issues with the contested garment rule. Firstly it rewards claiming as much as possible, because claiming more means that you will get more. Secondly it does not always make sense so simply split the contested parts into equal parts. You could construct a situation that makes it much more likely for example that one brother was the father and not the other. The fairness of a split rule also very much depends on what share of the owed money usually is available. Does it really happen in half of all cases that the debtor can only pay back half or less of what he owes? If most debtors for example cab pay 80% or more, the system favours the big creditors. I also need to mention that it is very unfair that taxes always have to be paid before a split between creditors. That practically means that the creditors have to pay taxes on money someone else owns them. In many cases no money at all might be left for creditors once taxes were paid.
    1
  8710. 2:28 simplicity*
    1
  8711. 29:43 One of your prettiest animations yet.
    1
  8712. The most memorable is Euler’s constant Gamma
    1
  8713. Beautiful !!
    1
  8714. I'm coming to the view that watching a piece such as this is character building. Let me try to explain : my background is engineering, so it's natural for me to admire what mathematicians do, but the pragmatic side of me rebels against doing it myself ! In other words, this material has a huge wow-factor - I'm content to view it from afar with admiration for the powerful intellects who have participated. To coin a phrase - c'est magnifique mais ce n'est pas moi ! So it's character building because it affirms what I am and gives be permission not to pretend to myself that I am something else, while at the same time be made aware that there are beautiful abstractions beyond my view. I thank you for this work of great beauty.
    1
  8715. "333 Only half evil" haha That joke is so bad it's good. At least in my opinion.
    1
  8716. Jeez! Lovely video! In addition to the gold rush in finding all the great stuff about r and s (rho and sigma), those ratios from all the higher odd-number sided regular polygons seem like another gold rush waiting to happen.
    1
  8717. I like the greedy algorithm.
    1
  8718. e to the (e + 1/e) power: is this the way to understand the speed of light? Consider e, the base, as the ring/cylinder/horn singularity. Raising it to the e power acknowledges that gravity is being released. The extra 1/e acknowledges that photons are filamentous, and can travel faster than, say, an electron, because of being so much less massive.
    1
  8719. I like the colour proof a lot more! The swivels matching points is not very intuitive - it looks good, but I have a hard time imagining why that would be the case. The colours make the case immediately that it would be true of any triangle.
    1
  8720. I once knew this? I'm 65. The set of things I once knew is quite large.
    1
  8721. That's what a 3 dimensional shape looks like in 2-D. This is a 2 dimensional dodecahedron.
    1
  8722. Just amazing! I like the part about Germany XD because i am from germany
    1
  8723. Does this video have 4 versions of the thumbnail?
    1
  8724. as wel as the solution where very other move you move the smallest piece there are many others such as always make a move between a and b then a and c then b and c and repeat there will only every be one possible move that you can make and you will always succseed.
    1
  8725. Your work is always between interesting and amazing along excellence.
    1
  8726. Yeah, the first time I ran into 3^3+4^3+5^3, I looked at the fourth powers and was annoyed. And while rearrangement of cubes doesn't work out, it doesn't work out by a rather precise amount so a pattern of sorts can be salvaged from it, but it's nowhere near as nice as it is for the first two patterns.
    1
  8727. I can't help myself but the T-shirt drives me nuts... Why 25,8069???? Why not 25,8070???? Did rounding go wild during pandemic?
    1
  8728. Wacht CTC. This might be a new set of rules😉
    1
  8729. You knew iron man was a real mathematician when he calculated the "eigenvalue of an inverted mobius strip factoring in the spectral decomp"
    1
  8730. Thanks
    1
  8731. Most memorable: the partial sums never hit an integer, even though the differences become almost infinitely small and the sums themselves tend towards infinity. That blows my mind.
    1
  8732. Do the buckling approximations "accordion" in some precise sense? If so, I think we can spot a contradiction: the same buckling could accordion to approximate cylinders of the same radius but different heights, but it could correctly approximate at most one of those cylinders, so the buckling can't in general be correct.
    1
  8733. What a fascinating result. Thanks for gifting us with this masterpiece of a video!
    1
  8734. In 3D cube there are 204 squares
    1
  8735. The mathematician in me wants to say that the square-dance is not yet maximally elegant! You can define the A(0) Aztec diamond as the 0x0 square, and it, obviously, has 1 possible tiling: the empty tiling! Then you can march the nonexistent tiles of A(0) out to A(1), and fill in the empty 2x2 square that they left behind! This way, the square-marching algorithm always starts with the unique empty seed, and out of this one seed grows a whole universe of Aztec diamonds. So elegant.
    1
  8736. 12:04: 153^2+104^2=185^2 is represented in the tree as 9 4 13 17 (read clockwise from top left.) From 1 1 2 3, you can get there by going right, right, right, left. I got this by factorizing the first two numbers: 153=9*17, 104=2*4*13, testing that 9 4 13 17 fulfils the fibonacci formula and then working my way backwards by trying to complete 9 ? 4 ?, ? ? 4 9 or ? 4 ? 9, of which only the last one works (as 1 4 5 9), which corresponds to going left at the last step, and so on until I reached 1 1 2 3.
    1
  8737. 1, 2, 4, 8, 16, 32, 64
    1
  8738. I love you shirt
    1
  8739. 20:02 The Second Fundamental Theorem of Engineering: near enough is good enough.
    1
  8740. Does anyone know where to get this i can't even tshirt .. i need that !
    1
  8741. ❤
    1
  8742. Excellent video! I wish this stuff was available when I was a child learning math for the first time. My daughter is 12 and it is so sad how slow and boring her math class is. I can see how children interested in math will soon lose all interest due to this. How many potential mathematicians do we lose per year?
    1
  8743. School problem: 1
    1
  8744. Cool
    1
  8745. 37:13 Since you can think of it as a 3D object, you just have to rotate it. If you look at it from the side, you will see a grid with a single color. This applies to all sides except the back.
    1
  8746. I made it all the way to the end, Professor. Dissatisfaction tends to zero as t[sub]vid → 39:57.
    1
  8747. Excellent video! I understood more than usual. I think this was due to having seen checker board parity tricks before and having done quite a bit of matrix/eigenvalue/determinant related things in the past (numerics). Having some background helps. Not really surprising of course. So I guess the only thing you could improve in the video imho is to put some extra emphasis maybe on what background would really help and maybe give some hints on how to obtain that, should one care! I also like to just take the back seat in your videos and just enjoy the nice graphics and your enthusiasm, so on the whole: fantastic. Thank you!
    1
  8748. 16:00 The area of the tree is the same as the number of plys, and so asymptotically infinite. Notice that the first "child" squares, being Pythagorean squares, have the same total area as the parent square, and that accounts for that entire ply of the tree. Then notice that for successive plys, the pattern continues so we can always use that relation to show that the total area is the same as the area of the previous ply. So each ply adds 1 unit of area. And since the "height" dimension of the tree is clearly growing sublinearly, the area of the tree is eventually going to exceed the available area that it can grow into and so the tree must eventually intersect itself.
    1
  8749. Never have seen this. Loved it!
    1
  8750. hi, thank you so much for adding closed captions. youtube can't automatically pick up on math notation and wording so it's suuuper helpful.
    1
  8751. i know the second part about matrices and coefficients and pascal's triangle as related to fibonacci series is mostly dry algebra, but i wished it was not brushed past so fast. there should be some salient observations along the way, but nothing is stopping me from researching or working through it. i am not sure why the video ends with 314159 what's next. is there a 314159 sequence in the golden ratio, not encapsulating pi's infinite non-repeating sequence, but just the six digits? not in phi's first 20000 digits.
    1
  8752. oops it really said 31459. what is next? i googled it and the answer was 31459 Barrel Road. that is probably an orthogonal answer. :)>
    1
  8753.  @Mathologer  this could be useful on pi day.
    1
  8754. Hi What's the name of this tiling?
    1
  8755. Lol. Tesla was a charlatan.
    1
  8756. gamma
    1
  8757. I want to do two things after watching this video. First, try it in Base 6, and then in binary, so see the patterns emerge. Then, write some software to show the 4th power proof using computer animation. Do I have time to do this? No.
    1
  8758. Three-brick solution, if we allow an overhang of precisely one unit: place one brick precisely halfway over the edge, and two bricks on top of it symmetrically so that they both balance on the edges of the bottom brick, such that they keep the overall center of mass the same as with just one brick. The right brick on the second layer is completely off the table.
    1
  8759. From Italy, the most memorable part of the video to me was Kempners proof definitley
    1
  8760. Что будет, если мы будем наливать краску в этот сосуд? Сосуд наполнится краской полностью или краска будет бесконечно стекать по внутренним стенкам сосуда?
    1
  8761. First puzzle: just going off the pattern that the 1D starts with 1, 2 -> 3, and 2D goes 3,4 -> 5, I'd say 5^3 + 6^3 = 7^3. Obviously it doesn't (5^3 + 6^3 = 341, 7^3 = 343), it's off by 2. But that jives with what would happen if you took the 6 slices of the 6x6x6 cube, and tried to fit them around the 5x5x5 cube - you'd be missing two corner pieces! Second puzzle: 10^2 + 11^2 + 12^2 = 100 + 121 + 144 = 365 (this one I know simply because I have those memorized). From the square pattern though, this also equals 13^2 + 14^2, so that's 2 * 365 in the numerator, the result is 2. Don't know of a quicker trick for this one though. Final puzzle: I couldn't find any integer solutions for the extension past 3rd powers (3^3 + 4^3 + 5^3 = 6^3), so probably similar to how 5^3 + 6^3 ~= 7^3, I'll guess ~7 for the 4th powers, ~8 for the 5th powers, and so on.
    1
  8762. I made it to the very end
    1
  8763. i liked the old proof most
    1
  8764. I wonder if this is related to ellipses.
    1
  8765. to the end...cheers :)
    1
  8766. The "log" wheel was done more than 10,000 years ago. Lol.
    1
  8767. 3D moons? Surely, they are bananas!
    1
  8768. Second problem numerator is, using the gnomons between adjacent squares: 2 * (11^2 + (11+12) - (10+11)) = 2 * (121 + 23 - 21) = 2 * (363 + 2) = 2 * 365.
    1
  8769.  @Mathologer  Agreed; but the first time I saw this painting I didn't yet know that.
    1
  8770. I think 1/(# of creditors not fully paid the fraction) is the limit to any individual creditor's recovery.
    1
  8771. It's like a discrete calculus
    1
  8772.  @Mathologer  Oh I see! Thanks for clearing that up just a coincidence with my uni's acronym haha!
    1
  8773. Grt to know that you have an Indian wife. Much love as always ❤
    1
  8774. Very high precision slide rules were made by essentially making them verrrry long but cutting them into pieces and arranging the upper scale sections in parallel along the outside of a cylinder somewhat like a cage. The lower scale was etched onto a solid cylinder that perfectly nested inside the cage. The equivalent motion of sliding the lower scale a great distance could be accomplished by spinning and sliding the inner cylinder. A quick combination of slide and spin motions could quickly set up the solution. Because this slide rule was the equivalent of several feet long, very precise calculations could be made. I have seen examples displayed in the Arizona Historical Society museum in Tucson and the California Railroad Museum in Sacramento. I shudder to think that these were probably made from elephant ivory in that age. The invention is credited to Edwin Thacher, and fine examples made by Keuffel & Esser in the 18th century can be found online. Some are the equivalent of 66 feet long! Imagine a circular slide rule of similar circumference!!
    1
  8775. Chicago-series math books taught this trick to get the order of a series in either Functions-Statistics-and-Trigonometry of Precalculus-and-Discrete-Math. So I did learn this, but it wasn't in Calc. And of course I learned it with no background. The only time you get background is if you take algebra and chemistry at the same time or calc and physics at the same time.
    1
  8776. I wonder how different the number turns out if we add in standard bill denominations (e.g. USD has 1, 2, 5, 10, 20, 100 dollar bills) and what the lowest value of k becomes (if it changes at all) with the adjusted formula.
    1
  8777. Always excited for a new Mathologer video! I especially enjoyed the final animation (not counting the closing credits!) of the spinning 4d cube. It helped to conceptualise the otherwise imperceptible nature of the hypercube. Hope you and family are well and flourishing in the current lockdown. Otherwise come back to South Australia where the total # of covid cases have jiggled between 2 and 4, over the past few weeks! 😎
    1
  8778. Two Squares tile well in 2D, two cubes not so much in 3D, two tesseracts less so in 4D, two hypercubes no go in nD>2D?
    1
  8779. I'm a math Phd specifically trained in knot theory, but after viewing this video, I have no clue as to how to tie my shoelaces. ;)
    1
  8780. You could've called it a Doppelgänghair in the thumbnail
    1
  8781. (IMO) your vids are just high quality the determination is the most to pick for a idea, Remember: Its just My own Opinion on the suggestion, Advice; "Try getting used to making a opinion on a topic youre interested in works for *me works for anybody".
    1
  8782. You forgot about lace orientation. Do the laces go from bottom to bottom and top to top, or do they go top to bottom, only? Any "real" lacing has the ends coming up through the topmost pair, otherwise it is too hard to tie into a tight bow.
    1
  8783. The power of the roots of unity
    1
  8784. Do you know a video system more suitable than youtube where for Ramanujan-themed posts you could apply fractional downvotes? -1/12 for instance?
    1
  8785. Would you consider using a black background? I love to watch videos like this while going to sleep but the bright white background makes watching your videos impossible :(
    1
  8786. Guess yellow. I am guessing that only the top ends matter?
    1
  8787. YAY!!! Got it!!! Bragging rights FTW!!
    1
  8788. I liked the greedy algorithm.
    1
  8789. I remember that when I was young I was wondering if there is anything like 2x2 = 2+2, and I found 1+2+3=1x2x3
    1
  8790. I think, but I am not sure, the lengths y r b of the yellow red blue lines can't just have any value. I mean, perhaps one can calculate one given two others (and also given some values of the original triangle)
    1
  8791. Your videos are always an inspiration. Thank you. ♥️
    1
  8792. yep, really enjoyed that, thx.
    1
  8793. 9:54 Ok, that much I obtained in the same manner. But then I got stuck.
    1
  8794. Just curious, given that these elegant relationships work with n = 1 and 2, but not n >= 3, couldn’t this be used as part of another way to prove Fermat’s last theorem? For example, the fact that the geometry doesn’t work out for n>= 3 could be used as part of a much simpler proof of Fermat’s last theorem.
    1
  8795. Euler-Maclaurin is my favorite tool and I use it regularly ^~^ I'm so happy to see a video about it! My attachment is because I got obsessed with summing and rediscovered the general-purpose algorithm in high school and thought it was a novel technique :P
    1
  8796. If you begin your understanding of the universe with straight lines and right angles, circles and spheres are going to be a challenge.
    1
  8797. 11:58 9,4,13,17; from 1,3,4,7, go right, then left
    1
  8798. It was a fantastic video,really!! Please make more videos to make it a better year.
    1
  8799. Thank you for reminding me how beautiful math is. Subscribed!
    1
  8800. Very nice proof. I used to play solitaire with pegs but never knew the math behind it. Very illumunating.
    1
  8801. In computer science there is this algorithm called, "The Ham Sandwich Algorithm" (HSA); which is shown to be true via Arrangements. HSA could also be show to be true with pigeons, lovely.
    1
  8802. Does all this remain true if we were using a hexadecimal numbering system? In other words, does vortex math depend on a 10 base numbering system?
    1
  8803. Lots of mindblowing stuff but I'll go with the absence of integers in the partial sums.
    1
  8804. Gamma being greater than 0.5 was the most fun bit for me in this video ..
    1
  8805. Awesome lecture! Thank you
    1
  8806. I'm just having trouble calculating the area of the sector rescaled. Wouldn't it be πr² • (θ/2π) = r² • (θ/2) Let θ=a√[2]; r=√[2] (√2)² • (a√[2]/2) = 2 • (1/2) • a√[2] = a√[2]? area = a√[2] ≠ a
    1
  8807. This is a beautiful clarification. Thank you and Happy Christmas
    1
  8808. You need neural networks for that 🙄
    1
  8809. Cantor would not be pleased. Or would he?
    1
  8810. infinity-infinity is an indeterminate form so im very curious whats gonna be in this video 🤨
    1
  8811. I found the most amazing part to be the lack of integers among the partial sums, despite the partial sums getting closer and closer to integers...
    1
  8812. 6:12 you say you set the wheel to 3.14, but actually it shows to be at 3.18. Those lines between the thicker 3.1 and 3.2 lines have an interval of 0.02 (only 4 thin lines instead of 9). You are doing it right at 6:41. (Before people ask "How can it be wrong if the result is correct?"; that's because it is an animation made to show to end on correct value, not an real physical wheel.)
    1
  8813. I used a straight slide rule in Physics class in 1970. Sliding the center revealed the metal holding the top and bottom. I had scratched the formulas on the metal with a straight pin. 'never got caught.
    1
  8814. The T-shirt: »To infinity and beyond«? ;)
    1
  8815. liked the missing whole numbers :-)
    1
  8816. Apparently i am not even on the same level with a Monkey :) Wonderful Video. <3
    1
  8817. I love the fact that it's a stupidly large number and one.
    1
  8818. Good video 16:49 This method is faster than the method I know I found the other method a while back and was very exciting
    1
  8819. 16:00 The area is 5. Every layer of the tree has an area of 1.
    1
  8820. Ramanujan summation interpretation of gamma is certainly going to stick with me. When you started showing the squares from 1 to 2, I was thinking it is deliberate choice to not show the square on top of 0->1.
    1
  8821. For me.... Whole video is beautiful and not just any single part or chapter...... But if I have to choose then it will be chapter 1...... I knew it but the proof here was completely new to me.....
    1
  8822. let me go into the other direction of your pyramid of equations... above the already shown top you could add these ones: -1 -1 -1 + 2 - 1 = -1 x 2 x 1 x 1 x 1 -1 -1 + 2 + 0 = 0 x 2 x 1 x 1 -1 + 2 + 1 = 1 x 2 x 1
    1
  8823. Most memorable was the approximation of the sum. With the Euler–Mascheroni constant😁
    1
  8824. The classical algorithm is a recursive one and pretty simple. However, there also is a non-recursive solution. Figuring that one out (+ implementing it) makes for a fun afternoon
    1
  8825. There is a rounding error on your tee shirt.
    1
  8826. As someone dabbling in writing fantasy, this gives me magic system ideas.
    1
  8827. I maybe more than 2 weeks late but it's always worth it
    1
  8828. Best part was the first understandable definition of euler-macheroni constant I've heard. Thanks for the video!
    1
  8829. 10651st view :P
    1
  8830. It's clear to me that all the comments here are from serious math heads... Mine, however, is just that of a cosmic minded psychedelic ingested with a lifelong fixation on the numbers 3 6 and 9... They proved pretty significant, however, in my life, as you may see in my more concise comment below...
    1
  8831. next to the big spinning wheel,. roulette is the worst game to play odds wise in the entire casino.
    1
  8832. I first encounter the second twisted square on the brilliant app. I never saw it in school.
    1
  8833. First proof but not until a late teenager reading The Ascent of Man
    1
  8834. It really makes me cringe that you're saying 4k+3, rather than 4k-1. So inelegant.
    1
  8835. I already knew some of the things shown in this video, so for me the most memorable would be Kempner's proof. Although crazy maximal overhang stacks are a close second.
    1
  8836. I once executed the Rubik's cube sequence FRBL (turn front turn right turn back turn left) until it got back to the solved state. I believe it took 1200 turns. I didn't count to 1200. Instead, along the way, it alternately solved the corners and the edges over and over, and I believe 1200 is the LCM of the number of turns to solve edges and to solve corners. I think it was 60 for the corners. (my memory may be off and the numbers are slightly different.)
    1
  8837. You sir is worthy of subscription.
    1
  8838. Early!
    1
  8839. The method that first occurred to me when summing 1 to 100, was to recognize that there were 49 pairs of numbers that add up to 100 (99 +1, 98 + 2, etc.), plus 50 and 100 that would be left over. No surprise that the described method is much better though.
    1
  8840. Pretty sure it's closer to 25.8070
    1
  8841. My favorites part was euler's constant
    1
  8842. [Oh, by the way, I found an error (technical but minor) in the book “Eye Twisters”. Dan Brown’s main character in “Angels & Demons” is ROBERT Langdon, not JOHN Langdon!! ]****[ACTUALLY, this was MY ERROR as a result of reading in haste and failing to review & fact check!!! YOU HAVE MY APOLOGY!!]. Should I delete what is below or just leave it?
    1
  8843. FJM: Fibonacci Juicing Mathematician (as well as Fredericus Johannes Maria) 😆
    1
  8844. I liked the coloring proof better. Swivel...I thought there was something we were assuming that wasn't proven/self-evident. Maybe it's just me.
    1
  8845. I actually... have very little background in maths. I utterly failed calculus in highschool, but then passed trigonometry in college with flying colors when I /finally/ understood the unit circle. I just really enjoy watching advanced maths videos. PLEASE do a Galois theory video, though. I did a report on Galois in high school freshman year. I'm trying to teach myself Calculus and advanced maths even though I'm a lit/English major.
    1
  8846. The numbers are 1, 2, 4, 8, 16, 32, and 64. How can you talk about hairy balls, without talking about the hairy ball theorem? Not to be confused with the no-hair theorem, which is melanopology, not mathematics.
    1
  8847. Therapist: "Smiling Pythagoras isn't real, he can't hurt you" Smiling Pythagoras in my room at 3AM: 1:37
    1
  8848. I cast my vote to the fact that no partial sum of consecutive terms ever adds up to an integer. Beautiful maths and beatiful explanations. Love your videos ❤
    1
  8849. Can you explain Indian mathematician and thier work...please
    1
  8850. May you please design a shirt that has a proof of the irrationality of pi or e?
    1
  8851. To proof the irrationality of e is not a big deal and could conceivably fit on a t-shirt :)
    1
  8852. As usual, just too good!!!.....and I'm gonna ask for my school money back but I fear I'm a little late as it was 60+ yrs ago...lol
    1
  8853. I enjoyed this video very much, just as with all of your content. I'm just a tinkerer; lover of science & maths but some of this was a bit hard for me to follow. BS Biology. Much appreciation for all of your hard work, and incisive explanations. Cheers!
    1
  8854. I was watching the whole video waiting for you to shoutout 3b1b's Basel and Wallis videos. Not dissapointed. Cool video btw, even though I didn't like you glossing over the zooms out to 1/3 and so on...
    1
  8855. Is it that why the derivative of log(x) is 1/x? O.o
    1
  8856. You are on the right track there. I'd say check out the previous video. That one was all about visual logarithms ;)
    1
  8857. this is weird, because this is the only proof i knew about. it may not be exact proof, it was scaling sides, and then matching the two copies. i recall the "scale and match" part, since when someone showed me this proof, i had a small debate, how can you scale a side with another side? you could scale a side with a scalar, but how could you do with a measurement? the resolution, i set at that time in mind, is that you could either divide it by Unit length; or start measuring the length of the scaled version in Unit^2, where Unit was the original metric. math is built on the shoulders of the giants. while this proof is beautiful visually, but i think there were shoulders of scale-n-match type proofs. of course, thanks for pointing out many of its applications and making a very beautiful educational video.
    1
  8858. ​ @Mathologer  on a separate note, we know all integral right triangles. do we know 1. All cyclic quadrilaterals with integer sides? 2. All cyclic quadrilaterals with integer sides and integer diagonals? these 1. and 2. are separate questions.
    1
  8859. Merci !
    1
  8860. Hello Burkard, Today swiping across Youtube, I saw that article where you asked why almost no one watched this video about logarithms the first weeks. Earlier, I estimate some months ago, I watched several of the Mathologer videos, for example that one about the "do nothing grinder". And Youtube gave me recommendations for the other videos. Then some time no more of the recommendations appeared. As the information below told, all the the videos seemed to be released at least several years ago. I had already assumed, that you didn't produce new videos anymore. In the meantime I watched lots of other math channels, as there are many Youtube members also producing content. Sometimes I am not ready calculating an answer to a math puzzle, then there are already three new recommendations. I don't know how the algorithm decides, which recommendations the users get, but perhaps channels you didn't choose recently, slide down in the list and are replaced by other similar contents, perhaps to avoid staying in a "bubble". But I didn't forget you and Mathologer. The videos are great 👍, even though I have to admit, that sometimes I'm not able to understand everything at once. English to me (German) is a foreign language and the videos proceed very quickly. And school didn't show some of the things, and school is in my case long time ago (I am 55). And I have not always as much time as I want, busy with work and so on. But I don't want to miss Mathologer. Greetings and best wishes!
    1
  8861. I think I need to buy some gears. For the first problem with equal circles, when Mr. Mathloger asked how many rotations, I thought " I dunno, maybe π rotations???"
    1
  8862. The animation at the end was marvelous.
    1
  8863. This was one of the more enlightening 17 minutes I have spent staring at my laptop screen.
    1
  8864. Here's my slightly scuffed proof Note that all the final lines are the same length because they're made up of red, blue, and green. Since they're all from the triangle, they all touch the inner circle of that triangle (inscribed? It's been a while since I've done Geometry), and are tangent to it. Take one of these lines and imagine it is a chord of a circle. The circle that it is a chord of must have a center on the line that intersects at the midpoint, perpendicular to the chord. If you can show that the midpoint of the line makes a radius of the inscribed circle (I'm pretty sure that's true but I forgot how to prove it), then the center of the circle where that line is a chord could be the center of the inscribed circle. Since each line is not parallel to the others, if they share a circle, it must have the same center as the inscribed circle. Finally, note that you can calculate the radius by the inner circle and half the chord, since they're at right angles, Pythagoras can be used, and so you get that all the circles where the lines are chords and the center is the center of the smaller circle all have the same radius. Since the radius and centers are the same, the circles are the same
    1
  8865. Pythagorus solved math and we've been splitting angles every since. Prove me wrong... : )
    1
  8866. I learned the theorem at school but not the proof. At uni it was assumed that everyone knew the proof so it wasn’t proven to me there either. I commented to a fellow student that I’d never seen the proof, and he explained your second proof to me. Today at 56 is the first time I saw the first proof!
    1
  8867. Historians attempting to reconstruct the Claw of Archimedes have long debated how the weapon actually worked. The sources seem to have trouble describing exactly what it did, and now we know why. Turns out it was a giant disc that slid beneath the waters of a Roman ship, then raised countless eldritch crescents which inexplicably twisted into a sphere, entrapping the vessel before dragging it under the waves, all while NEVER LEAVING ANY GAPS in the entire process. No escape, no survivors, fucking terrifying. No wonder that Roman soldier killed Archimedes in the end, against the Consul's orders. Gods know what other WMDs this man would unleash on the battlefield if he were allowed to draw even one more circle in the sand. The Roman marines probably had enough PTSD from circles.
    1
  8868. I really liked the gamma part. The blue bits are more than 0.5 because each blue slice covers more than half of their respective rectangular addition to the 1 by 1 square
    1
  8869. Most memorable: Leaning Tower of Lire, leaning over Nicole Oresme head
    1
  8870. Very very interesting ! Very beautuful too. Just a present for my birthday ! Thank you!
    1
  8871. For Christmas fun, the un-tilable checkerboards (say with 2 green corners missing) becomes a 2-player game. Take turns with a friend placing a domino. If you can't, you lose. And for a second game, same rules, but if you can't place a domino, you win! (One more rule: one player for each direction of domino.)
    1
  8872. Gamma (and the gamma function) have been fascinating me for some time now, so my vote goes to gamma!
    1
  8873. Amazing content. Explanations are really well thought out and splitting the video up into sections is great. The sequencing ensures the maths are covered and flows like a story, hooking you to see the next part. Thank you so much for your efforts! I have really learned a lot! I am currently teaching mathematics in China, and I studied mathematics for my undergrad. I will definitely find time to introduce your videos to our students!
    1
  8874. How you managed to do these animations is beyond me... very very good
    1
  8875. I got kind of blase about seeing sin, cos, i and exp in discrete problems after I first read about the partition count formula. (Not that I actually understand any of it)
    1
  8876. Very good video sir. 99% understood. 50 minutes completed. Second year bachelor of mathematics student. Thank you so much sir. One more thing to say that was told by my sir, Let s2 = 1x2 + 2x3 + 3x4 + ... + n.(n+1) s2 = 1x2 + 2x3 + 3x4 + ... (n-1).(n)+ n.(n+1) (Subtract below equation from above) 0 = 1.2 + (1-3).2 + (4-2).3 + .....(n+1-n+1).n - n(n+1) n(n+1) = 2 (1 + 2 + 3 +...+ n) which implies formula for S1. Let s3 = 1.2.3 + 2.3.4 + 4.5.6 + ... + n(n+1)(n+2) and do the same process again we get 0 = 3(1.2 + 2.3 +3.4 +...+ n(n+1)) - n(n+1)(n+2) 1.2 + 2.3 + 3.4 +...+ n (n+1) = (n(n+1)(n+2))/3 (sigma)(n(n+1)) = (n(n+1)(n+2))/3 (sigma)(n2 + n) = (n(n+1)(n+2))/3 (sigma)(n2) + (sigma)(n) = S2 + S1 = (n(n+1)(n+2))/3 Solving for S2 we get the formula. One more, Let s4 = 1.2.3.4 + 2.3.4.5 + ... + n(n+1)(n+2)(n+3) Do the same stuff, 1.2.3 + 2.3.4 + ... + n(n+1)(n+2) = (n(n+1)(n+2)(n+3))/4 (sigma)(n(n+1)(n+2)) = (n(n+1)(n+2)(n+3))/4 (sigma)(n3 + 3n2 + 2n) = (n(n+1)(n+2)(n+3))/4 (sigma)(n3) + 3(sigma)(n2) + 2(sigma)(n) = (n(n+1)(n+2)(n+3))/4 S3 + 3S2 + 2S1 = (n(n+1)(n+2)(n+3))/4 simplify to get S3. Again thank You for such awesome videos sir.
    1
  8877. This is one of the best (if not the best of all) Mathologer's videos. It shows not only cool stuff, but points out that Mathematics is not just finding stuff about numbers. Mathematics is also about how to apply those findings, in order to make our lives easier. Most of our programmable machinery and equipment works, because somebody centuries ago discovered how to perform quick calculations with important case specific numbers. Congratulations to Mathologer - once again you presented super cool content.
    1
  8878. 665 thumbs up. So this being mathologer, I had to give mine. :)
    1
  8879.  @Mathologer  Youtube won't let me. :P
    1
  8880. 16:00 Assuming the triangle to be isosceles, I think I have an idea. If it isn't, I'm a little sad 1st box=1 2nd box=1/2+1/2 3rd box= 1/4+1/4+1/4+1/4 4th box=1/8+1/8+1/8+1/8+1/8+1/8+1/8+1/8 So I think it just adds one each time And I can show it. If you have b^2+b^2=a^2, or 2b^2=a^2, then you have your answer 2 of the smaller squares is always going to be equal to the bigger square, holding always. So it equals 5. Should have been obvious to me, but it was not.
    1
  8881.  @Mathologer  yeah! You get to these comments so quickly!
    1
  8882. The Moessner Miracle is the Golden Mean/Fibonacci Sequence in 3D with the potential to prove the existence of interdimensionality. 0 + 1 = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55....INFINITY. Furthermore, it proves the existence of a FIRST CAUSE....THE LAW OF ONE within the VOID. The place of 'matter' in the void is that of an exquisitely beautiful precipitate established by the First Cause-Law of One through the unified field interface as determined by a geometrical necessity. (Restatement of John G. MacVicar) In other words, the universe is a holographic projection from an unquantifiable (in mechanistic terms) source, inadequately expressed (mechanistically) as ONE (1) IN CONTRADICTION TO NONE (0).
    1
  8883. Windirstat displays the relative size of folders and files using the graphical example illustrated in this video. This windows application was written for windows 95 and is still working fine under windows 10.
    1
  8884. Gamma
    1
  8885. complex numbers are a great crutch :) there are too many zeros in the definition :) but we have obviously become more stupid and use too many templates, the values ​​of which we no longer know :(.
    1
  8886. I don’t know how to uses cell to answer. I just see it as something like 11117511532. Reflections of 1113
    1
  8887. Interesting how things are rediscovered all the time. The fact that the formula of a surface of the sphere is the derivative of its volume gets us back to some of the core attributes of derivatives and integrals. Since the integral of a curve is also the surface under the curve and similarly the integral of a 3d curve is the volume thereunder, it would make sense that the reverse of the integral would provide for the reverse effect. volume -> surface in 3d and surface to curve length in 2d... Thanks !
    1
  8888. It took me longer than 30 seconds, but I think the intendent way to solve the problem on the painting is to notice that the numerator = (12-2)^2 + (12-1)^2 + 12^2 + (12+1)^2 + (12+2)^2 = 5*12^2 + 10, and then it is not too difficult to simplify the fraction.
    1
  8889. Most memorable: Kempner’s proof
    1
  8890. If the center is a black hole.
    1
  8891. How many times can one click the like button?
    1
  8892. i once found a formula that divides a cube similarly and got a cubic formula for the sum of squares...
    1
  8893. Good morning 🌞
    1
  8894. Thank you ... could do a video for the most beautiful equation. You know e^opie is 1.
    1
  8895. First with magenta is better! Also, continue one more turn and you will get the Feynman belt trick.
    1
  8896. What was most memorable was that if the scenario about making a decision whether the series is diverging to infinity or not was real , it would have cost me my life :) . Damn i need to study inf series more just to be sure i survive the test !
    1
  8897. I thought that there was Mathologer on the opening screen looking like Indian.
    1
  8898. The Beauty of Math? It is simple.... Math is Never Complete.
    1
  8899. The background noise is a cicada rubbing itself in hopes of convincing another cicada to join in, because Andrew evidently lives far enough south that the wildlife has not realized it is winter.
    1
  8900. I think my favourite bit is Kempner's proof at the very end, simple and elegant, reminds me of the harmonic series divergence proof in the beginning.
    1
  8901. 13:33 People on our imaginary party are shaking hands. (Feelings start)
    1
  8902. Two minutes silence for those who don't know trapezium rule, bisection method,Newton Raphson method used for computing programs.(*Easy for computer).
    1
  8903. The problem with algebraic Math sequences is that they never represent the 'real' world which is random; at best these can be used for theoretical simulations where too the application is limited to a few sequences, such as Fibonacci.
    1
  8904. Wondering if we're getting any closer to that Super Infinite fraction formula relating pi, e, and phi :o
    1
  8905. How did they calculate this such a long time ago?
    1
  8906. I didn't make it to the very end because I was cheating
    1
  8907. I attempted my own proof that the no-9s series was finite before finishing the video - it turns out to use a similar geometric series to reach a smaller upper bound. The sum from 1/10 to 1/18 is less than 9/10 (9 terms, the largest is 1/10). Similarly, 1/20 + ... + 1/28 < 9/20, up to 1/80 + ... + 1/88 < 9/80. These sums are just 9/10 times the terms from 1 to 1/8 - so the sum of all the 2-digit reciprocals is less than 9/10 of the sum of the 1-digit reciprocals. By the same logic (1/100 + ... + 1/108 < 9/100 = 9/10 * 1/10, 1/110 + ... + 1/118 < 9/10 * 1/11, etc), the sum of the 3-digit reciprocals is less than 9/10 the sum of the 2-digit reciprocals, etc. That gives us an upper bound of the sum of the 1-digit reciprocals (which happens to be a little under 2.718) times the same geometric series of powers of 9/10, for a total of 27.18.
    1
  8908. "Preparations A through G have all failed"
    1
  8909. Somehow this video feels like a prelude to a visual, mathologerized proof of Fermats Last Theorem.
    1
  8910. Here in Poland we are thought it
    1
  8911. I like the first shown proof of the length more!
    1
  8912. Absolutely superbe like always! When do you do a video about the Godel incompleteness theorem?????? All other video are so bad on this point!!!!!! Pleaaaaaaaaase :)
    1
  8913. great solution, cringe scene
    1
  8914. Very very nice video, great method of explaining the differences & principles. Kudos to Maharshi Madhav. ❤
    1
  8915. good lord im dumb.great stuff!
    1
  8916. What a way to start off my day while I was having my breakfast! Admittedly, chewing food and swallowing it down your throat does not help to to get some 'homeworks' that you quickly skimmed through at around 40m+. But times sure flies and I have surely had a pleasant morning! Cheers for the hardwork! It was a joy to both learn something new and relive some of my fondness memories. And yeah, my background is a automation engineer bachelor with a bit of intensive math courses during pre-uni days.
    1
  8917. I was teaching myself some relativity when this came out
    1
  8918. What are your tools to make those fantastic animations in every video ? Is that your own software ?
    1
  8919. Well done!
    1
  8920. I loved it.. and all your videos. I am an Electrical Engineer (math minor). You lost me about 32.21, but Plan to go back and see if I can figure it out.
    1
  8921. I’ve never seen the words calculus and easy used in the same sentence before
    1
  8922. I'm certain my idea isn't new, but if the derivetive of sine is cosine, and any function can be expressed as a sum of sine waves, does that not mean that the derivetive of any arbitrary formula be extrapolated from a fourrier transform.
    1
  8923. hi, always nice to see your videos.
    1
  8924. Really made my day!!!
    1
  8925. as always: EXCELLENT
    1
  8926. 3:22 Did he really bob down? 🤣
    1
  8927. I wish they did deriving in school... all just BS memorization and I don't have that kind of memory, I learn via understanding... not memory...
    1
  8928. Indeed, this is cool
    1
  8929. Remember I was young and there was Conway's fan club on the next street: they listen punk rock, drink beer and has Conway's iris emblem on the jackets.
    1
  8930. 2^n has a lot of nice mathematical properties
    1
  8931. The most interesting was gamma :-)
    1
  8932. So... is the sum of any "n zeroes" series where n is finite always greater than the "no nines" series? It kinda feels like it shouldn't be but I'm kinda guessing it is. If not, how many zeros do you have to require for it to be less than the "no nines" sum?
    1
  8933. The proof in CH6 of no9 series is just like fractal! Love it so much<3
    1
  8934. I remember sniffing at this, inspired by the 4 digit case - which appeared in a Norwegian math olympiad some years ago. I didn't find as much as you did. One little detail I thought was a bit funny, was that you can replace any 4, 6 or 8 with 2*2 = 2+2, 1*2*3=1+2+3 or 1*1*2*4=1+1+2+4=1*1*2*2*2=1+1+2+2+2 to generate new numbers that work . Like, 1*1*1*1*2*6=1+1+1+1+1+2+6, replace 6 with 1*2*3=1+2+3 and shuffle: 1*1*1*1*1*2*2*3=1+1+1+1+1+2+2+3
    1
  8935. A fantastic video as always.
    1
  8936. The paradox where you add numbered balls to a bucket and at every square you remove its square root ending up with 0 balls in the bucket, doesn't that also have an imbalance?
    1
  8937. What's the area of your pizza?
    1
  8938. in the 90'thies i used a slide ruler just for fun. My mother had one i a drover, i picked it up and borrowed a book from the library about it, that was fun. ;O) I got nothing out of it, just used a calculator ;OD
    1
  8939. I MADE IT TO THE VERY END. (thanks for that lovely music, would you please tell me what it was?)
    1
  8940. 12:51 Ahhrg! How do you even come up with such things??!
    1
  8941. actually the correct formulation of the pigeonhole problem is: if there are more holes than there are pigeons then at least one pigeon must have at least two holes :^)
    1
  8942. 5:17 I just love that little bit of editing that made Mathologer dodge out of the way of the arrow 😆
    1
  8943. Thank god you didn't upload this yesterday, I would have thought it was an April Fools joke. This is truly "magical" xD
    1
  8944. Ive just noticed that at 20:52 there are to pegs and all disks missing from the bottom two corner arangements. very good video though, I love the tie-in with dr who.
    1
  8945. Fantastich! Now I'll have to make one of these.
    1
  8946. Every one press like 👍
    1
  8947. It was visually apparent that no-9s wouldn't be infinite as you had 'cordoned off' sections of series much like a crossword puzzle. Tristan's opacity / resolution presentation is much more intuitive and helpful, for all us guessers out there.
    1
  8948. I'm a french high school student and i don't understand a lot, but what a pleasure to listen to !
    1
  8949. 7:06 if proving that 0^0 = 1 isnt interesting then i dont know what is
    1
  8950. I had watched the Numberphile videos with Matthias Kreck about Zagier's "One Sentence Proof" already a while ago and I thought that I would never ever get close to understand this. But now, after having seen this "windmill incarnation" of the proof, I think I could have a chance. Regarding the difference of two squares and the challenges raised at 32:20 - an entertaining introduction to the topic is this: https://youtu.be/LkIK8f4yvOU
    1
  8951. I loved these proofs - its a nice idea to have in mind! I wonder if there are other fun ways to prove this - maybe by using the fact that the square grid is bipartite or by looking at homomorphisms of Coxeter Groups! Edit: runways to fun ways
    1
  8952. Numero Uno LIKE! 7:54AM South Texas 12/03/22. 12/03/22 = 1 2 3 4 I know! I know!🦋
    1
  8953. I would give ten likes from my account if that was allowed. Oh wait! It's allowed but it results in just one like in the like counter.
    1
  8954. I liked the second proof better (non-swivel one). Also, wasn't there an earlier Mathologer video with exactly the same swiveling logic? I'll see if I can find it later.
    1
  8955. Oh that last proof was wonderful. Thanks for sharing it with us!
    1
  8956. I've been watching your videos for years and respect you as a creator, so I did not expect you to post a thirst trap near the end of this video.
    1
  8957. Here is more about the circular slide rule https://youtu.be/xRpR1rmPbJE
    1
  8958. do you have references to who found this?
    1
  8959. Shirt is on point again
    1
  8960. What struck me most is that the sum of the "no 1 series" < the sum of the "no 2 series" < ... < the sum of the "no 0 series" (at 39:30) --- I wonder if that is caused by Benford's law -- removing everything with a 1 affects more numbers than removing everything with a 2 and so on.
    1
  8961. Infinity exists only in imagination because infinity of time would prevent anything from moving if it existed. pi also is a figment of mans imagination it does not exist either in a world or universe made up of quanta.
    1
  8962. Huge Respect for your sir for giving an Genius Mathematician Sir Madhava, a credit and fame he deserves !!. Being Indian, I felt so proud about how brilliant our minds are, specially in Mathematics.
    1
  8963. From the idea this is related to measuring the length of a coast line. The finer grained you measure it the longer it gets. Or the surface area of the intestine, If you take the length and the circumference and multiply you don't get the 200-300 m² scientists approximate it to.
    1
  8964. Great video as usual! A minor nitpick: At 23:41 , when you say that this proves that the partial sum is also dragged to infinity, you are not actually using the <1 inequality being shown, but the implicit inequality that the difference is >0.
    1
  8965. 2 x 2 in the real world ? It's complicsted since everything is 3d ,
    1
  8966. I think the most memorable one is the Leaning Tower of Lire, because it's the easiest to recreate and one of the coolest to show off.
    1
  8967. As ever with Mathologer videos, I start to follow...then it gets a bit tricky, then i get lost... but still watch to the end. Then rinse, and repeat. Hoping that the Hermeneutic circle will kick in. But Merry Christmas Burkard, and to all out there who love being dazzled by these superb videos!
    1
  8968. Awesome! That's a fun conundrum.
    1
  8969. I thought the same thing.
    1
  8970. Wouldn't one have to prove that the center of the incircle is the same as the center of the "whisker" circle? Only for the first proof. For the second, it seems pretty obvious.
    1
  8971. I liked the colored version best because it's a purely geomentric proof doesn't rely on imperfect human vision.
    1
  8972. For a digon you have 2-1=0 steps to change it to a regular digon so that works too 😂
    1
  8973. That is a nasty math pun: Expressed in verse, or is that inverse?
    1
  8974. It's always how the \Gamma function can be made to show up everywhere ... somehow: that impresses me the most.
    1
  8975. You could (via Euler's continued fraction formula) turn the sum monster in the intro into another continued fraction!
    1
  8976. Regarding the question at 24:05, the center of the incircle of a 345 triangle and it's Feuerbach circle lie on a horizontal line, so you can't form a right angle triangle with those two points, and the parent construction fails!
    1
  8977. Why in 12:30 to 12:32 the board loses a square?
    1
  8978. Can it be useful for Fermat Last Theorem?
    1
  8979. Can you explain this, 12X12 = 144 + 13X2-1 = 169 ( 13X13 = 169 ) 20X20 =400 + 21X2-1 = 441 (21X21 =441) 90X90 = 8100 + 91X2-1+92X2-1+93X2-1+94X2-1 = 8836 ( 94X94 = 8836)
    1
  8980. 25:37 how about shinus and chosinus? Not my idea, suggested by Russian math youtube.
    1
  8981. Mathematics is a Universal and Dimensional language.
    1
  8982. 22:43 Not sure if this is an intentional joke given what you say right after - but this animation is wrong. The leftmost (and rightmost) triangle only gets rotated 180°, indicating that the sidelength of the large blue triangle equals the sidelength of the small blue triangle - of course, this is not true, so there is a scaling involved in the animation, making it not suitable for an area argument.
    1
  8983. The no 9's grid was my favourite part
    1
  8984. happy 2s day! 2/22/2022
    1
  8985. I made it to the end. To think I was planning on sleeping, when behold, I saw a new Mathologer video.
    1
  8986. A bad use of the Schwartz?
    1
  8987. Spoilers ahead: My solution to challenge 1 was not elegant or clever, but it was indeed the solution. From the parameters available, I chose a and b as my free variables and the resulting system of equations was a pain. 1.)a^2+ 2ba= 153 2.)b^2+ba= 52 3.)a^2+2ab+2b^2=185 3.)- 2( 2.) ) yields that a^2= 81 which in turn solves for b, c, and d. With the second challenge I found through construction that the line connecting the incenter of the 3-4-5 to the other center was parallel to the segment with length 3 and perpendicular to the other leg, so no parent triangle could be formed.
    1
  8988. Thank you for everything you do. What a great teacher!!
    1
  8989. Lesson 5:
    1
  8990. since we didn't have calculators, i had to get a slide rule for school. then i got my hands on a circular slide rule. i ended up using the circular one more for most things.
    1
  8991. Mathologer - love your videos. Will this work for prime powers to get the special numbers?
    1
  8992.  @alre9766  try -(3 +8) mod 9 = 7. Does it work then?
    1
  8993. Hey Mathologer, have you heard of the following paper: https://arxiv.org/abs/1908.03795 ? It's a breakthrough in linear algebra in quite some time.
    1
  8994. Where can I get that shirt?
    1
  8995. If the largest angle between triangles approaches 0 or 180 depending on what angle you count the different planes or the angle at the intersection, and the surface approximated is differentiable and smooth and nice everywhere then it works. If the angles do not change, the area does not changes if the angle grows the area grows, if it shrinks then the area shrinks and approches the area of the shape. Talking about the angle between the planes the neighbour triangles are in, two triangles in the samenplane having 0 degrees between them, while the intersection spans 180 degrees, you cannspit the different definitions and how to convert the statement i am sure, this has to be true, because you can show it is so, take the initiam clue example meme, there you preserve the angle at 90 degrees, so the estimate will not change, of you clipped away edges such that angles keep getting sharper then the lenght will increase, and so on, andnthis generalizes to any dimension. :)
    1
  8996. Amazing video as always, and although I am little late and I used Thales to solve it, I still got the solution. The square is formed of various triangles, but we only need to pay attention to four of them, the one that gave us the units. we know that the hypotenuse of the biggest triangle is sqrt(5)/2, and that this hypotenuse is formed by segments, lets label them AB, BC and BD, and lets label the bottom of the triangle (a full side of the original square) FG and GD. AB is a leg of ABF, which is a similar triangle to AFD, so we can extract that AB : 1/2 = 1/2 : sqrt(5)/2, and so AB = sqrt(5)/10. Then, we can extract that the BFD is similar to AFD and to GCD, so GCD is similar to AFD by transitivity, from this we can extract that sqrt(5)/2 : 1/2 = 1 : CD, and so CD = sqrt(5)/5. Now, we know that AB + BC + CD = AD, and so, sqrt(5)/10 + BC + sqrt(5)/5 = sqrt(5)/2. Turning on the proverbial algebra autopilot, we get that BC = sqrt(5)/5. Returning to the original problem, we need to get AB squared, and so (sqrt(5)/5)² = 5/25 = 1/5. Although solved via a different method than the one expected, we get the same result, but well, that is one of the many beauties of math I guess :).
    1
  8997. The leaning tower of lira is by far the best chapter and I can't help wondering if I can find a better (longer) overhang solution... thumbs up for your channel. We all love you! (at least, I do).
    1
  8998. Hi! I like cars! In other words, this flew right over my head! I won't say I'm dumb, as I've tested my IQ to 130 on a couple of occations. But DAMN! Mathematicians are on a totally different level...
    1
  8999. Mind blown.
    1
  9000. I think knowledge has been gained, then lost, then gained, then lost, much like an infinite series. I bet there were people who discovered these things before even the indian mathematicians, including Madhava. We just forget the knowledge because 1. poor communication to future people and 2. periods of time where we lose interest or are too busy trying to kill each other to worry about these mathematical things.
    1
  9001. 11:09 at this point you can already show that this is not a square grid the way it was shown in the beginning or am i wrong? What's the point it's already clear here.
    1
  9002. So, you covered differential calculus and integral calculus. Are you also going to cover dental calculus?
    1
  9003. 432 Hz
    1
  9004. The no 9s grid visualization made my day
    1
  9005. Sometimes a longer lacing is more desirable because the ends of the laces where you tie them are too long
    1
  9006. if tesla was obsessed with 3 the way the introduction suggests, it sounds much more like OCD than an actual divine insight. and knowing the man's genius, i doubt he considered a mathematical curiosity the root of all knowledge.
    1
  9007. A very erudite effort.
    1
  9008. Where can I get the tshirt you're wearing now?
    1
  9009.  @Mathologer  English is not my native tongue, and haven't touched math since school. Cut me some slack 😅
    1
  9010. the circle is an 16 cylinder radial engine.
    1
  9011. Hello sir, I am touching your feet. I am biggest fan of you, please sir make a vedio best book of mathematics because I want to become a Mathematician.
    1
  9012. If my brain isn't completely tricking me, it should be 5 moves minimum and 55 moves maximum to flip over all ten cards. Unless, of course, we assume the rule for flipping the far right card would be to also flip the far left card, in which case the sequence never ends.
    1
  9013. I actually came up with this 2nd proof on my own. I knew that others had thought of it before me, but until now, I hadn't seen it from anyone else.
    1
  9014. Is there maybe something like 3^3+4^3+5^3=6^3? With the third power. With maybe more items?
    1
  9015. Nice video. Ramanujan is my favourite mathematician.
    1
  9016. Could 4k+3 also be found as 4k - 1?
    1
  9017. I thought I was so clever seeing the connection to primitive roots, and you go ahead and mention it anyways. 😖
    1
  9018. Should call it the Zing Square
    1
  9019. The continued series of fractions are not the only best approximations of root 2, right? Like 4/3 and 24/17 are the best approximation up to X=3 and X=17 respectively, but they are not included in the list 1/1, 3/2, 7/5, 17/12, 41/29, etc. I calculated: | 17/12 - sqrt(2) | = 0.002453 | 24/17 - sqrt(2) | = 0.002448 So although just a little smaller, 24/17 is a better approximation of root 2 than 17/12 is. But it is not included in this series. So these are not the only best approximations of root 2 out there. Or am I missing something? Also, I visualized the best approximation for every X in Desmos. Looks very beautiful: https://www.desmos.com/calculator/bynul01qnr It shows for every X what the difference between the best Y/X approximation and sqrt(2) is.
    1
  9020. I love math constants so gamma is my favorite
    1
  9021. someone once told me that the median, incentre and circumcentre (but not the orthocentre) lie on a straight line. i managed to prove it, but also discovered something else. if you join up the three sides' midpoints to form an interior triangle, the three centres remain except two of them switch roles.
    1
  9022. I hope you had the orthocentre in your proof and not the incentre.
    1
  9023.  @WK-5775  TBH i can't remember now. one of the centres didn't seem to relate to the other three.
    1
  9024.  @MusicalRaichu  It's the incentre which doesn't (directly) relate to the others. Together with the three centers of the outcircles, it has other nice properties instead. Btw, the line you mentioned is called the Euler line. As you noticed, the triangle formed by the midpoints of the edges has the same Euler line as the original triangle. What's more interesting, the nine point circle and Feuerbach's theorem do establish a connection between the stuff related to the Euler line and the incircle. Have a look at the Wikipedia articles on these topics.
    1
  9025. Sigma Rule : You are a Super Mathematician
    1
  9026. Congrats on your centenary! Interesting you rank Archi ahead of Euler at the top mathematician of all time ... is it a stretch to say he invented calculus thru his use of leverage as you alluded in his many proofs including sphere volume. Perhaps you could cover the Palimpsest in future videos.
    1
  9027. Here is another piece in the jigsaw. The link to Pascals triangle. It only works for the Fermat series of triples (ie the set of "middle children"). Choose any row in Pascals triangle. Multiply the odd entries by 1, 2, 4, 8, ..... and add to get the top left entry in a Fibonacci box. Do the same with the even entries to get the top right entry. For example taking the 1 5 10 10 5 1 row, we have top left number = 1*1 + 2*10 + 4*5 = 21. Top right number = 1*5 + 2*10 + 4*1 =29. For example taking the 1 6 15 20 15 6 1 row we have top left = 1*1 + 2*15 + 4*15 + 8*1 = 99. Top right = 1*6 + 2*20 + 4*6 =70. Not obviously useful, but it seems to make things more complete.
    1
  9028. Wow I'm early lol
    1
  9029. OK but no-one has noticed that Madhava looks like Mathologer? At first I had thought his face was photoshopped!
    1
  9030. My grandfather had a mechanical gadget with such a wheel, and I played with it when I was a kid. He said it was a "round logarithmic ruler."
    1
  9031. Awesome shirt!
    1
  9032. The lack of intergers in the harmonic series feels quite unlikely until the fractions become visible.
    1
  9033. Superb .... as always.
    1
  9034. I love Maths. I wish I were smarter. I reached my level of incompetence in Maths at the advanced undergraduate level, say 19 + 29 years ago, near enough to fifty years (I made a career in engineering), successful enough to be able to retire in Germany. Now I'm learning German. My head hurts.
    1
  9035. Nice, I intuitively knew that you were doing an IFT but seeing all the parallels and the path you took to get there was really informative
    1
  9036.  @Mathologer  As always :)
    1
  9037. Me too! Every time I do an IFT my brain goes KUL! It has to be the Bose's effect on TGM...:body-blue-raised-arms:
    1
  9038. To infinity and beyond? Your shirt?
    1
  9039. It was quite brief but I liked Euler's gamma's crazy identity at 25:07. HOW CRAZY IS THAT! really.
    1
  9040. I've seen 2d projection of a 3-d cube. I've seen a 3d projection of a 4d cube. Beautiful. In general, an n-1 projection of an n-d. Is it possible do do an n-2 projection?
    1
  9041. The reason all the "by 9" math works is that our "decimal" system is really a base 9 system, unreduced. Decimal numbers can span multiple columns, but their inner workings are base 9.
    1
  9042. Wut
    1
  9043. The favorite pastime of mathematicians is to mislead both themselves and people.
    1
  9044. It is a paradox how math in general seems very intuitive, yet it contains many hidden patterns that are not intuitive.
    1
  9045. Great video Professor! Great algebra as always. Merry Christmas and Happy New Year - let it be a home year (it's better not to wear red, it's better to wear green). I have a question - is there a geometric way to check that a number is square (let's skip modular arithmetic :). All health and Happy New Year. 🖖
    1
  9046. You stressed how slow the logarithmic growth of the sum is, but it boggles my mind much more that the sum of the prime reciprocals grows as the log of the log.
    1
  9047. Reminds me of the CS trick for writing out the list of powers
    1
  9048. 21:09 I try to avoid commenting before the end of the video, but (since I'm watching on a cycle of: watch 5 minutes, rewind 6 minutes) it might take a while to get to the end, here it is. I noticed something that I found interesting and that hadn't been pointed out in the next 5 minutes (yes 'next': see rewind strategy above). But is probably explained later. The coefficients of the top right diagonal's terms are 1, 2, 3, 4, 5 ... (adding one each time) which is 1x(1,2,3...) Of the next diagonal down are: Minus 1, 3, 6, 10 (adding 2, 3, 4....) (haven't figured out how the even rows work, yet) Next: Plus 1, 4, 10 (Adding 3, 6, which is 3x(1, 2...) The next section is Pasqual's Triangle, which has to do with adding things, so I may find my answer later. I could cook up a "pattern", differing on odd and even row, but I'd need to calculate more data. Probably I'm looking for something that either isn't there or is explained in the video. Sorry for the ramble: so far, so great! Keep up the good work
    1
  9049. 25:03 Turns out the patterns are column-wise. Neatly illustrating why I should hold my fire or keep my powder dry, or whatever the analogy is. Why are they all martial analogies? Because shooting people never solved anything? Sorry: off-topic.
    1
  9050. 25:03 Turns out the patterns are column-wise. Neatly illustrating why I should hold my fire or keep my powder dry, or whatever the analogy is. Why are they all martial analogies? Because shooting people never solved anything? Sorry: off-topic.
    1
  9051. 1 of each 2^x coins is basically binary right? So only one way of making any number up (its binary representation!)
    1
  9052. Wow, very nice!
    1
  9053. I thought Kepler and Leibniz were geniuses way ahead of their time, but Bernoulli just stunned me (Pascal was famous for inventing computer, but now I appreciate and his mathematical genius)
    1
  9054. Isn't there a contradiction in the counterexample here? When increasing P by 1 and B by >1, you are necessarily creating isosceles triangles. These surely can't trend to all three sides being equal to zero? I'm not aware of any special rules at the limit that states that isosceles, equilateral, scalene, right etc triangles are interchangeable - otherwise we could use Pythagoras on scalene triangles and then just drop extend it out to infinity. What am I missing please?
    1
  9055.  @Mathologer  thank you for replying so quickly # especially at Christmas! I guess that as we approach the limit, everything becomes everything else anyway, so the type of triangle (or even quadrilateral, pentagon etc) is meaningless once area = 0. Many thanks for another wonderful year of learning. Merry Christmas to you, and a Happy New Year!
    1
  9056. You are a genius, I am glad you are back' you have made my brain hurt again :} ;>
    1
  9057. Wow! A 3D-Moebius-Monopoly-T-Shirt. I gotta have one, Burkhard!
    1
  9058. simply too good .....u r indeed a great mathematician....
    1
  9059. Liked the clip. I don't think I've heard of this approach before. How clever.
    1
  9060. One of the prettiest ideas I remember from long ago and never forgot. Phi is everywhere. To contribute a tiny gem, in N dimensions you can get at most 3N-2 positions into enemy terrain.
    1
  9061. The weird optimal overhang towers were absolutely stunning. I'm not even sure why it works, but hey, math says it does.
    1
  9062. undergrad physics: H A R M O N I C O S C I L L A T O R also sin theta = theta :)
    1
  9063. For yuo data base, I learned casting 9's in elementary school in the 1950's in Maryland. My wife did not get it in the New York Schools. What I wrote about was an observation: The Tesla sequence is 124578. The reciprocal of 7 is 0.142857142857,...the set of integers. Is this interesting?
    1
  9064. I've just completed the 20.000 thumbs up. 😃
    1
  9065. Hello, Great video ... as always. I am not a mathematician, so I ask the expert: Is this way of generating Pascal's triangle correct? watch?v=CL8XIWtcx0c Thanks.
    1
  9066. I learned this stuff in school. I didn't learn that specific sequence, but those techniques are all familiar to me. Edit: Okay, not all of them! I wasn't aware of the Gregory-Newton formula.
    1
  9067. I'm early!
    1
  9068. the claim made at 13:10 that the ratio of the surface area of the cylinder and sphere is also 3:2 is incorrect, they are both equal to 4*pi*R^2
    1
  9069. if you include the two caps then yes, I agree, the ratio of the areas is 3:2; just saying "the area of a cylinder" is confusing, it can mean the side area or the total area; I think that the equality of the area of the sphere to the side area of the circumscribed cylinder is even more striking and I assume this is what Archimedes alegedly had engraved on his tomb, not the 3:2 fact
    1
  9070. My life is a slide rule calculation 😁
    1
  9071. Respect!!!
    1
  9072. More comments mean more engagement! Go algorithm go! :D
    1
  9073. The main question is: why is the regular heptagon(5th regular polygon) the chosen one?
    1
  9074. Indeed 🙂. Mathologer rarely drops something without introduction, justification or explanation. So whence these specific r and s values, out of all the infinite rs = r + s possibilities? Fantastic story, nevertheless,. Thanks 1000x
    1
  9075. 11:47 The box is 9 - 4 -13 -17. To reach it from the bottom (1-1-2-3), go right (1-2-3-5), then right (1-3-4-7), then right again (1-4-5-9) then left (9-4-13-17). In summary, R-R-R-L. I'm in awe! Thank you for sharing all this wonderful math.
    1
  9076. Magic!
    1
  9077. For bonus points, prove it in a way Euclid would understand.
    1
  9078.  @Mathologer  Thinking on this further, consider the 18-gon. To even do the construction, you need to create an angle of 360°/18=20°, which is provably (via group theory) impossible with straight edge and compass. So using Euclidean methods, you're already restricted to the n-agons which are constructable with straight edge and compass. You couldn't possibly prove the general case, so you'd need a separate proof for each n. I'd bet Napoleon's Theorem wouldn't be too hard, but proofs would get increasingly tedious for higher n. Incidentally, I love how the concept of a basis pops up in so many unexpected places.
    1
  9079. So so good!
    1
  9080. I've heard that the alien 'b' in mathematics/computer modulo operations is base 12, vice our decimal-weighted numerical system; and, for this reason, functionality of our Earth-paradigm of a true quantum computer's ALU (and all mathematical/scientific/engineering calculations) will always 'crunch' for our paradigm, but be in dissonance with the universal/cosmic mathematical structure of things.
    1
  9081. i think there is an error in animation at 12:29, you removed the green #6 square
    1
  9082. Draw the lines between neighboring points. Notice that each one forms an isosceles triangle with (alternating) the inner and the outer angles of the original triangle. Choose any four consecutive points to form a quadrilateral. This is a cyclic quadrilateral as two opposing angles will sum to 360 degrees minus the sum of the angles of the original triangle, i.e. 180 degrees. Since I can repeat the argument for any group of four neighboring points, all these possible cyclic quadrilaterals must have the same circumscribing circle.
    1
  9083. 4K+1=a²+b² For K=5 4•5+1=21 21=25-4 21=5²+(2i)² a=5, b=2i Lol, just kidding 😆
    1
  9084. Thanks so much for the great content sir
    1
  9085. Lmfao I love that 1/x squish and stretch part
    1
  9086. As always, thank you for your videos !
    1
  9087. 0:37 If the points of the plane are each colored Red, Yellow, and Blue....then What? Or is the answer just David Wells?
    1
  9088. 3:00 On one of the diagonals there are the powers of 4 but beside that I didn't find any powers of 2 in interesting places. Please tell me, Burkard, are there any more patterns?
    1
  9089. Top row: 1, 2, 3, 4, 5, 10th place, ... done? Row below: 1, 2, 3, 4, 8, ... 24th place, thats it?Middle row: Powers of two on 1, 2, 3, 6, ... 91st place, done? No more columns in open office ;-) The only pattern i found was the doubling (5-10, 4-8, 3-6), but no idea, why that is so.
    1
  9090. Nice job here. Future video idea: how Riemann integral fails to be "continuous" and then introduce the Lebesgue counterpart to solve the inconvinience, concluding with a mathologer-proof of the DCT.
    1
  9091. 12:22 "... similar to what Fermat came up with a couple of hundred years ago." 9:41 Fermat image has his death as 1665. Proof by Contradiction, Pierre is not dead.
    1
  9092. Here's my (relatively non-rigorous) go at proving that log-exponent property beyond the positive integers. Log-exponent property for rationals: ---------------------------------------------------------- For any positive integer q, (1/q)A(x) added to itself q times is (q/q)A(x) = A(x). Likewise, A(x^(1/q) * x^(1/q) * x^(1/q) * x^(1/q) ... ), where we multiply q times before applying A, is the same as A(x^(q/q)) = A(x). Therefore, (1/q)A(x) = A(x^(1/q)) for all positive integers q. It's already known from the video that for any positive integer p, p*A(x) = A(x^p). Combining with our above proposition, we can conclude (p/q)A(x) = A(x^(p/q)) for any positive integers p and q, and thus this log-exponent property holds for the positive rationals. For (positive) real numbers: ---------------------------------------------------------- By definition, any real number can be written as an infinite convergent sum of some arbitrary sequence of rational numbers (for our real number in question, r, let's call it q1, q2, q3...). We know that our log-exponent property holds for all the rationals, and so q_i * A(x) = A(x^q_i) at all indices i >= 1. Consider A(x) * r = A(x) * lim(sum(q)). The log-exponent property holds at every finitely indexed step of this summation, so as our number of steps approaches infinity, the whole expression's limit must inevitably approach A(x^sum(q)) = A(x^r), i.e. A(x) * r = A(x^r) for any positive real number. Probably not 100% adequate but I think it makes sense?
    1
  9093. Funnily calculus is the only part of maths that ever came easy to me. I struggled with integral calculus because I failed to see the fundamental theorem. I knew it. But it didn't click until much later. But once it did, it just turned easy.
    1
  9094. So touching... :)
    1
  9095. Got the yellow right..! Love this episode, very interesting and something that i actually do understand..!:)
    1
  9096. Watched the whole thing. Probably the best at explanations on YouTube. I would really like to see you take on TREE(3) and the fast growing hierarchy. The most recent Numberphile video was very confusing. I’m doing my own research, but not much is out there.
    1
  9097. There is no doubt that Mathologer videos are worth waiting for.
    1
  9098. Great video! I watched it al the way through plus went back afterwards on some points I have missed and the quality of it is just ... exceptional! Much better than some math class I took during college. I have a PhD degree in physics but I've always been most interested in math so I would say I'm at a graduate level in math. I think the pace was quite slow at the begining and maybe a little bit fast at the end although I mostly follow all the way through but I guess the pace was to fast from the begining for certain people. But I think it's good that way for two reasons. First, if the pace had been slower, this video might have been way longer and it is quite long although it didn't feel like it to me. Second, if one misses something, as it's a video, one can go back a few minutes and watch it all over again as much as one needs/desires without slowing anyone down. So I'd say it's an excellent video all the way through! @47:34 : for the reason why the correction first gets better but in the end worse, I guess it's due to the fact that the power series representation of 1/x^2 only converges in a certain radius of convergence. So maybe the correction term diverges at fixed n (meaning adding more and more correction terms) but at fixed number of correction terms it converges as n gets bigger...
    1
  9099. It’s a primary tool that I used in my flight school. The battery never runs out, and you have to think what you calculate. It’s really easy to make a simple mistake of placing the decimal point in a wrong place when just using a pocket calculator. Also 6/60 can be used for hour-minute conversion.
    1
  9100. Why does your Christmas tree with only green and red appear to have 2 spirals? One spiral to the right and one spiral to the left. Is it similar to what takes place in a sunflower seed pattern shown latet in the video?
    1
  9101. if it squishes, it can die!
    1
  9102. I remember learning about Dijkstra's Algorithm and the Traveling Salesman Problem in highschool (in VCE 3+4 Further Maths for my fellow Melbournians/Victorians) but wow I'd never have thought that would translate to shoelaces
    1
  9103. Beautiful!
    1
  9104. Finally enough I tutor second semester calculus and kinda intuited something similar to this about two months ago when I helped someone do an integral similar to the cone cylinder one.
    1
  9105. I really thought, that just a U lacing would be the shortest
    1
  9106. I'm one of those know-it-alls that was going "yeah, yeah, get on with it" And then my jaw dropped. Amazing video! I think I was following everything up the point where the Bernoulli numbers appeared. I had never heard of them, that I can remember. And my problem with that part is that they simply appeared. Or rather, "these fractions are the Bernoulli numbers". Sorry, what? I think I followed most of this. I would have to revisit everything from chapter 5 or 6 onwards, but I think I followed the argument. Physics major, PhD in CS.
    1
  9107. The 52 nd mersenne prime found?
    1
  9108. I'm intrigued by why the Euler formula is said to only work with convex polyhedra. I understand why it might not work for self intersecting polyhedra or non-closed surfaces etc., but what changes as a polyhedra goes from convex to concave (e.g. by morphing the position of vertices) that suddenly violates Euler's formula??
    1
  9109. It seems like a variance in the tension of the rubber band would dictate a variance in the placement of the numbers on the wheel - and if not, that there must also be a correspondence between some aspect of the change in tension as the wheel turns, and the consistency of equalities around the wheel. No?
    1
  9110. I learned it!
    1
  9111. This is great if you want to understand what calculus is about, and you are mostly interrested in application. But I would say that if you are interrested in understanding the justifications for some of this, you need to dig a bit deeper. At least the chain rule and (sin(x))' needs some further justification to be actual proofs, but is sold as such. I was surprised about the amount of stuff swept under the carpet given that fact that you have thrown others under the bus for shady maths in the past. Great video none the less, but I feel like it was missing a small disclaimer of some sort.
    1
  9112. Generally, how many moves does the convoluted solution take?
    1
  9113. please do an explanation/breakdown video for the Animation vs Math video from Alan Becker!
    1
  9114. The first diagram screamed 12" x 12 " self-adhesive vinyl flooring tiles.
    1
  9115. Most memorable part: kempners proof :)
    1
  9116. Hmm, I love the simplicity of the no 9s proof, so I'm going with that. Figured it'd be connected to a geometric series. Very nice!
    1
  9117. All first order linear differential equations such as y'=xy+1 may be solved by the "integrating factor" method, leading to the integral formula solutions the Mathologer pulls out of his hat.
    1
  9118. Comment for engagement
    1
  9119. Funny. I had everything that was established till 5:45, and thought it wouldn't lead anywhere.
    1
  9120. i got the coin paradox right the first time on all 3 examples :D
    1
  9121. Another great video, I always watch on Sunday. Of course today its Christmas Eve. HO HO HO Merry Christmas
    1
  9122. this is the most clickbaity video yet from mathologer. I obliged with no regrets.
    1
  9123. The reason why the formula around 23:00 doesn't work for k < 4 is that we have some extra options we should not have due to the way the formula was derived. For the k={1, 2, 3} cases one must not include the terms higher than k*100. The general formula works beacause there is only limited [and fully utilized] choices for 1st and 2nd part of the product only one valid chice raimains for the infinite 3rd part. While the 3rd part is infinite it behaves well. Only 5 of the terms work at the time. when including quite simple corretion factor. For k = 1 we have: (1)* (1)* (6*x^100) + (1) * (287*x^100) * (1) = (6 + 287) * x^100 = 293 *x^100 For k = 2 we have (I'll skip all the ones): (21 + (287 * 6) + 985 ) * x^200 = 2728 *x^200 For k = 3 we have: (56 + (287 * 21) + 6* 985 + 325 ) * x^300 = 12318 *x^300 For k > 3 we have (C representing the combinations operator and done before multiplications: ( 1*(k +5)C5 +287*(k + 4)C5 +985*(k+3)C5 +325*(k+2)C5 +2*(k+1)C5 )*x^(k*100) Now we can see that for k < 4 we use only k + 1 first terms of the general solution.
    1
  9124. 26:19: Should we also tell those kids Garfield was assasinated shortly after taking office?
    1
  9125. Something about the properties of regular polyhedra if I had to guess at 40 seconds in...
    1
  9126. Time for maths!
    1
  9127. Hmm.. generating reduced fractions this way is a cool method to do a breadth-first search of important rational numbers. Like if you had an algorithm that needed a few thousand fractions (p/q's) as an input, only wanted one of each, you could generate them with a breadth first traversal of the tree :D
    1
  9128. Years ago, I concluded that this is a lot of work to find out something that I never needed to know. Eg How long it takes for my coffee to get cold.
    1
  9129. Swiveling proof is my pref
    1
  9130. Voting for the optimal irregular overhang at 10:00. I would love to see the optimal sequence for different numbers of blocks.
    1
  9131. Another reason that 7 is considered a lucky number. 🎄
    1
  9132. ahaha the -1/12 in the thumbnail
    1
  9133. What a wonderful Christmas present, you just saved my life. My mind is buzzing with mathematical activity, I really needed that. :) Me during chapter 3: "I wonder how this plays out on hexagonal grids" You: "Chapter 6: Hexagon" <3
    1
  9134. Most memorable (but not my favourite) was seeing the cat saying mu, and then gamma; leading me of in a totally wrong direction thinking about Schrödinger's cat dying if a gamma ray came... definitely don't want that :(
    1
  9135. Colouring one would be easier to follow for a younger age group.
    1
  9136. Great video as always. And once again, you've managed to make exactly the same video I was working on, but much better than I could. I'd even bought a new slide rule for this one.
    1
  9137. so in the final equation with the triangles, pi represents the amount of radians the triangle set rotates with an infinite number of triangles! that's neat!
    1
  9138. ♡BEAUTIFUL♡
    1
  9139. I remember being asked the leaning tower problem in my doctoral qualifying exam. I remember being shocked in the exam by the result that the leaning tower can extend to infinity. Thanks for bringing back my good memories 😄 As for the most impressive chapter, I vote for the linkage between fractals and no 9's sum.
    1
  9140. For the last one... 3^4+4^4+5^4+6^4=?^4. Any number raised to the 4th power ends in 0,1,5 or 6 by Fermat's little theorem. This series ends in an 8 and cannot be a perfect 4th.
    1
  9141. That porcupine one looks a bit like some of the stuff found in some fetish catalogs. I like this guy Schwarz already! It stops at the Planck length.
    1
  9142. It is amazing that no partial sum of the harmonic series is an integer, yet every integer can be formed from a subset of the harmonic series.
    1
  9143. the secend prouf i made it up when i was 8 without knowing it was pythagoras therom
    1
  9144. yeah, it can. but we like to use non negatives.
    1
  9145. The graphics in this video reminded me of that N64 game Tetrisphere.
    1
  9146. We need to stop thinking of Pi as an irrational number with fixed value (with infinite digits beyond the decimal point) but instead as a vibrating number much like an undulating string. In reality, Pi will change its value to match the fluctuating nature of reality at the Quantum-scale.
    1
  9147. A very nice theorem and observation. Thanks for sharing this with us. The second you suggested folding the two neighboring triangles I paused the video and both got the result and the (second) proof, with the conclusion of the result of a second fold being similar to the original taking only a few extra seconds. Obviously, I prefer the 2nd proof 😊
    1
  9148. The 700 year old proof....that, for me, was hands down genius. I got to remember this name...Nicole Oresme
    1
  9149. Watched it until the end and found every explanation easy to follow. Especially the coloured moving objects in the "technical algebra stuff" was really helpful as I often struggle with these "easy and lengthy" claculations. My maths background is a finished master's degree in mathematics. So for people like me you can get away with way more difficult topics, however I cannot be the average viewer you are trying to teach mathematics to ;) Anyways great video, love the great work! :)
    1
  9150. A portal toward a new world has opened for me today!
    1
  9151. The proofs of the rules are very nice, but there’s always a bit of uneasiness with the claim that df, dx, and dg are the result of the limiting process such that they become almost zero but not actually zero, such that df/dx is nonzero while (df/dx)(dg) is zero in the product rule case.
    1
  9152. Dammmmmmmmm what a great content Thanks for the vídeo
    1
  9153. Woosh. I don't feel so smart right now.
    1
  9154. 547 slides, does he use powerpoint
    1
  9155. New trapper keeper art
    1
  9156. Impressive stuff. You've convinced me that it's worth learning more than my very basic understanding of mathematics.
    1
  9157. Thanks!
    1
  9158. Just came here from https://www.youtube.com/watch?v=VYQVlVoWoPY and I didn't really get how to identify that the pi=4 was wrong, the expanding cylinder was great! It also explains the infinite area/finite volume funnel "paradox" too, very nice!
    1
  9159. 1:31 looks at German election in 1932...
    1
  9160. I sense a part 2 coming where he shows the (4/7) star with 3 squares inscribed in circles 4/7 the size of the larger circle, as well as showing a septagon traced out by a 1/7 size circle.
    1
  9161. Fascinating video. I had no idea about eigenvalue decomposition of polygons. For the question, what would happen if you used 1 round of 72 degrees, 2 rounds of 144 etc. I expect that the result would be the same up to scaling and rotation, because further earing does nothing to the degenerate pentagon and keeps all the other eigenpentagons the same up to scaling and rotation and thus within their original class. Keeping them in their original class was essential to the proof in the first place.
    1
  9162. my first thought was rs=r+s => rs-r=s = >r(s-1)=s => r=s/(s-1) isn't this also valid?
    1
  9163. Geometry was never my specialty unfortunately, so I don't have anything like the intuition Conway had for coming up with his six-point theorem or the windscreen wiper theorem, but I'm still quite awed by both proofs! I'm also awed by the connection with curves of constant width, which I already knew about and strangely I had a feeling would be mentioned later in the video.
    1
  9164. Most memorable was the fact that no partial sum, or cut-off partial sum, hits any integers. Blew my mind :-)
    1
  9165. Thank you.
    1
  9166. How do you capture a unique toymaker? Unique up on them.
    1
  9167. At 4.00 I was about to go the other way round, and use the harmonic series (starting at 1/2) to prove the blue area was infinite. Are there any nontrivial antishapeshifters that are not derived from the reciprocal function?
    1
  9168. This is one of the weirdest math things I've ever seen
    1
  9169. I remember the proof that .99999... =1 and in the proof they shifted the terms or moved a decimal or something and then subtracted the shifted terms and I'm saying, but it's an infinite series. Prove to me that operation is valid for an infinite series. They were not able to do that.
    1
  9170. Omg thats awesome!!!! So simple
    1
  9171. It'd be cool to have some type of compilation of homework problems given at the end or in the description (for example in the form of timecodes of where they are in the video). I watched the whole video and wanted to go back to solve some problems but I don't want to rewatch the whole 45 minute video just to find them.
    1
  9172. Brilliant, brilliant, brilliant!
    1
  9173. Here's a problem. How can you prove that pi is a member of R, the set of all real numbers. There's no algebraic solution that will pop out the number. How can you prove that an infinite set of non-repeating digits is there "somwhere" between (probably) 3 and 4. (HOWEVER, I can derive pi in Q, the rationals using a Taylor series. Ergo....)
    1
  9174. Lets label the block ((a,b),(d,c)) where (a,b) is the top row and (d,c) the bottom row. For the triplet 153, 104, 185, we need ad=153, 2bc=104, ac+bd=185. By definition of the Fib block, a+b=c and b+c=d. Then, our block is ((a,b),(a+2b,a+b)). We eventually find a=9, b=4, c=17, d=13. For the path, notice that the only possible parent is ((a,4),(9,c)). Then c=5 and a=1. The grandparent is ((1,b),(d,4)). So b=3 and d=7. Its parent is ((1,b),(d,3)), with b=2 and d=5. Finally, we reach the bottom, ((1,1),(3,2)). We then start there and go right three times, then left.
    1
  9175. The inverse powers also have nice sums. I once found them but never investigated their strange existance.
    1
  9176. Yes, Tesla was a great electrical engineer, but he was pathetically worthless for any other study of physics. I have always been astounded by the people who think Tesla was some type of god.
    1
  9177. 😂😂😂😂 wow!❤
    1
  9178. Genius content, genius video. Worked for me well until Chapter 7... from there onwards it became WTF. Level: PhD in engineering.
    1
  9179. 7:04 I've studied Heron's formula in middle school, but I've never really used it
    1
  9180. That claim is not the only BS claim associated with Tesla. The nutjobs believe he invented "free electricity" and that the power companies are in cohoots in trying to keep this out of the market.
    1
  9181. Christmas came early!
    1
  9182. Wouldn’t the bugs maintain their speed? Why should they slow down? If for every turn they traveled the same distance, wouldn’t they meet a lot sooner?
    1
  9183. Thank you! 😎👍
    1
  9184. Unless my understanding is completely wrong, real world use came first. Mathematics is just the language we developed to describe and communicate the way these patterns work. But the pattern, aka real world use, was always there before we realized and then described it.
    1
  9185. Super nice video! The no 9 series really surprised me
    1
  9186. nice
    1
  9187. Kepners proof is best
    1
  9188. That thing works in base 10, but I'm not sure this 3-6-9 vortex would show up in another base... hence its esoteric nature.
    1
  9189. When I was in high school, I had the idea that mathematics is the Western version of enlightenment. Anyone agree or disagree?
    1
  9190. First time in first thousand!
    1
  9191. 3 minutes in and you already lost me
    1
  9192. GREAT work sir .... Thank you....
    1
  9193. You guys are weird... I love it!
    1
  9194. Great video , thanks. Small error at 28:30 in the second row, there is 1 3 7 9 and should be 1 3 5 7 . Paying attention everybody ?
    1
  9195. That was really cool. My only complaint was that it was not longer! xD Oh, and I wish that this video had existed, like, ten years ago. It would have saved me a lot of trouble in trying to understand some of these things. (Undergrad degree in theoretical mathematics).
    1
  9196. 🥺
    1
  9197. BTW One formula (that I found myself) for generating infinitely many unique Pythagorean triplets is: For any integer n > 0: A = 2n+1 B = ((A^2)-1)/2 C = B+1 It outputs: [3,4,5], [5,12,13], [7,24,25], [9,40,41], and so on...
    1
  9198. not even a paradox. at some point down the line you stop adding but you don't stop subtracting. If you really want to use such a "infinite" pattern to do math with it and you need to stop th e WHOLE pattern at one single point and take all the positives TO THIS POINT and all the negatives TO THE SAME POINT and not all the positives to one point but all the negatives to a completely different point of the pattern. You wouldn't even need something that complex to say what you are saying. You could just say: 1 - 1/2 + 1 - 1/2 + 1 - 1/2 should be 0 because by taking the first 10 1s and the first 20 1/2s, they all cancel out, but the whole pattern never drops below 1/2 so it's a paradox.
    1
  9199. I took "long street" to simply mean that a single house (0 neighbors on each side, 0 = 0) was not allowed. Once you require 50<n<500 it's clear, though. Btw, http://oeis.org/A001108 gives all possible street lengths.
    1
  9200. Great video again. Real throw in was, I had just pondered a question related to the first puzzle. What numbers can/ can't be expressed with (sum 1 to k) / (sum k+1 to n), with any integers k and n?
    1
  9201. This is a good video, but I really hope you go into the generalizations of this like the one published by Dexter Kozen and Alexandra Silva in 2013. There is a lot to unpack here.
    1
  9202. I would really like to thank you as your previous videos on fibonacci sequence, fibonacci-like sequences and the strand puzzle made me follow a deep rabbit hole and led me to discover a super generalisation of the fibonacci numbers with similar properties, I generalised the Binet formula and many fibonacci identities and used them to find general integer solutions to infinitely many pell-like equations all super fun Oh and I'm particularly waiting for the video you promised at the of the strand puzzle video
    1
  9203. 🎉
    1
  9204. Thanks!
    1
  9205. Well we can get the same Series by using the Taylor Series expansion of arctan(x) and if we put x=1 then we get the same Series But I think at that point of time concept of Taylor Series was not developed.
    1
  9206. I feel stupid AM i the only one?
    1
  9207. No. I still watch these videos even though I don't understand them and always end up feeling stupid and inadequate.
    1
  9208. …from non-mathematician: this made me wonder if the transformation process involved could be in any way hitched to Serpinski’s bringing order out of chaos process? - just musing..I wouldn’t understand any answer!
    1
  9209. 7:50 DNA
    1
  9210. Another beautiful video, Thanks so much.
    1
  9211. All of the points made in this video are amazing and new to me. But if I have to pick one the most memorable to me is the insight that there is no integer in the partial sum of harmonic series due to the odd/even pattern.
    1
  9212. Incredible explanation!
    1
  9213. Wonderful video. Thank you for making these!
    1
  9214. I like the last proof with the comparison of the length that kills the magic of this discovery. The angle thing doesn't give me a clear cut intuition of what is happening.
    1
  9215. the cardioid genuinely scared me
    1
  9216. Me: I finally counted the squares and it's right. Mathologer after 1 second:Wrong
    1
  9217. Not accepting merry christmast
    1
  9218. I am Wowed i never relieved matematiks kod be fun
    1
  9219. Superficially the volume calculation looks a lot like the calculus method without invoking calculus terminology. I'm curious if pre-calculus students grasp the argument. For I all know it's comfortable to me because I'm used to those methods
    1
  9220. No partial sums being an integer is a real surprise.
    1
  9221. Math by "Dr. Evil" (reformed) ?! 😆 Thanks for wishing everyone a "Merry Christmas" ! btw, I would tend to understate Santa's signature polemic as 3 x ( ho ) . . . rather than his 'hyperbolic' graphic t-shirt version.
    1
  9222. Nice ending music! The part I liked the most were the weird optimal towers, because I was talking recently about the tower of Lire.
    1
  9223. I was very interested in discovering the fact that the partial sums of the harmonic series is always a franction odd/even. thank you for your very interesting work.
    1
  9224. My brain hurts. In a good way. LOL
    1
  9225. bro that creepy smile edit at 9:13 😵‍💫
    1
  9226. 8:37 lol
    1
  9227. Thank you for your video :) My favourite part is how those brick stacking problems are still being researched while every person can understand the problem :D
    1
  9228. Kempner's proof was so cool!
    1
  9229. Amazing stuff, as always..
    1
  9230. I try to find different of 1^n , 2^n , 3^n , 4^n ... m^n in 3 years ago, I found constant at n row, and I found factorial formula, I use delta symbol like in this video, but I don’t know the origin of formula (I guess, factorial formula is from derivative of a^n) , now today I know it more and clear haha
    1
  9231. in that case: the rotating one surely already is in THE BOOK 😉
    1
  9232. One glance balancing was my favourite 😂 I liked all of the proofs especially the last one, but the first proof was really simple and unexpected definitely my favourite!
    1
  9233. For the algorithm
    1
  9234. Is a perfect timing between spectator found the idea and you explain that THX!!
    1
  9235. Overhang of 1 with three blocks would be an inverted pyramid.
    1
  9236. Oida sehr geil!
    1
  9237. Now can't wait for RS machine, like Fibonacci machine you've used to generate pythagorean triples.
    1
  9238. Awesome. The visualization of gamma was stunning. What a precise formula for a sum... ln plus gamma
    1
  9239. On the power of 2 coin question at the start of chapter 2: Hopefully I won't be spoiling this for anyone I have done so much messing around with binary counting I just intuitively knew this answer without actually trying the trick we just learn unfortunately
    1
  9240. Is there any particular reason why turning the paraboloid inside out transforms it into a cylinder minus the same paraboloid but in the case of the hemisphere, it is transformed into a cylinder minus cone and not a cylinder minus the same hemisphere (as I intuitively thought it would)?
    1
  9241. Woo!
    1
  9242. Ty for covering the base-1. Which is, imo the key here.
    1
  9243. Amazing!
    1
  9244. Sad news about Ron Graham, most famous for the "Graham's number", a number a bit larger than a googol but smaller than TREE(3). He also studied at Berkeley, where I visited as a student.
    1
  9245. Poor Mathologer just tried to add a bit of fullness to the video by using AI and now hes gonna get beat up in the comments😭😭😭😭.
    1
  9246. Great video. We did Herons formula in highschool and it was one of the most usefull ones when it comes to the problems we solved
    1
  9247. The rules didn't state that you have to use each peg, therefore for higher peg solutions you can ignore all but 2 and the starting pegs and it will be optimal.
    1
  9248.  @Mathologer  Well then, my question would be, what is the optimal number of pegs? Or does it only get more efficient the more pegs there are? My guess is that there is an optimal number of pegs for any number of disks, but that optimal number changes depending on number of disks.
    1
  9249. I've done the third one on my high top chucks.
    1
  9250. 333. Only half evil. Margarine of evil. Diet Coke of evil.
    1
  9251. Speechless, especially about the last proof featuring the QED cat.
    1
  9252. Where you got "2+2=2x2" you forgot 2^2 =4.
    1
  9253. I still remember doing the equivalent of pi = 4 when first meeting arc length in calculus. Fortunately, way too much homework disabused me of that notion. (The only time in my life that I thought that that many drill problems was worth it.)
    1
  9254. the best channel😁☠
    1
  9255. without even watcthing your latest quanta of mathematical thought, 1+2+3=11*2*3. I know that 1+2+3+4=10, I also know that 4 factorial =24. Which also turns out to be the differential between metric and imperial measurements of distance. Did you know that? 1024. Ring a bell? 2.4% every 2 to the ten, then do it again. Genius
    1
  9256. I wonder if we'll get enough rankings to know there must be repeated ones 🤔 Mine is 7134652, by the way
    1
  9257. 4k - 1 is more elegant
    1
  9258. How did you do that.....Respect!
    1
  9259. i'm thinking about how, if you split the containers in 3 instead of 2 you get a different distribution, and if you take the limit of this splitting, you'd get the proportional splitting at the limit
    1
  9260. 9:46 There is one way to make each value. Basically, it's showing that there is only one way to make a number in binary.
    1
  9261. Not 4k + 3 but also 4k - 1 ( +3 = - 1 in. Cyclic 4)
    1
  9262. For the hexagonal three color equal problem, I understand the 3D projection proof but how to show every configuration must resemble 3D cubes? Maybe some configuration might look like some strange irregular (for example, non-manifold or having holes or floating cube or other strange whatever) shape. I know that won't happen but how to prove it?
    1
  9263. Great video! By the way, it was quite nice to see that derivation for the formula for the n-th Fibonacci number
    1
  9264. I've only seen proofs 3 & 4 before...
    1
  9265.  @Mathologer  Mumbo Jumbo is a minecraft youtuber who makes videos in which he explains Redstone contruptions, and there is a joke that He sometimes says something is simple, when it is really complex
    1
  9266. Pretty basic, but very beautiful and powerful mathematical wonders were shown here. If anyone wants to see more weird stuff - look at the Ramanujan works on the infinite series... I still cannot comprehend what mathematical intellect one should have, what intuition and inner vision indeed , to see all this beauty in progress.
    1
  9267. This solution is art, not math
    1
  9268. Look t the 2 ends
    1
  9269. My attempt: in one of the vertices (call it A) of the triangle, draw the bissecting line. Let A₁ and A₂ be the extensions. Every point X on the bissecting line makes an isoceles triangle XA₁A₂. Do the same on the other vertices B and C. The intersecction between the bissecting lines shows us that A₁, A₂, B₁, B₂, C₁ and C₂ are all in the same circunference. :)
    1
  9270. 7:17 oh, man. You broke the symmetry: left is fine, top and right are wrong😂. While walking on the line it should be always triplets: yellow, red, green.😢
    1
  9271. 36:20, I guess it's 0.25, because one pair of parallel lines splits the square in half, so the small square is half of half the big square, so 0.5².
    1
  9272. No integers among the partial sums
    1
  9273. Good stuff. I'd never heard of Heron. He seems like a fun guy. Makes me wonder how we ever learned anything before animation.
    1
  9274. Truly amazing! Thank you for sharing. What are the initial setup sizes of the line and circle at the start? 0:47
    1
  9275. My favorite part was the 20-block chaotic maximal overhang configuration, mostly because I was shocked that there was anything better than the orderly arrangement I had previously known.
    1
  9276. I check everyday for a new Mathologer video. Today is a blessed day. 🙌
    1
  9277. 3 is a pretty nice tho its the first odd prime and i like odd people, but these vortex people are a bit too odd for their own good
    1
  9278. The most memorable part is to discover that visual fractal. Because I voted that the no-9-sum would be infinite. And I found not only I was wrong but also a nice "proof" :D
    1
  9279. Does the dream proof hold up with the eyelets at any width from each other? My brain is not quite up to dreaming up proofs but it would seem to me that if eyelets were further away from each other than the total height of the eyelet structure then having more vertical segments would be better than having horizontal segments.... Not sure if i have explained myself well but i am about to get paper, pencil and tape measure out..
    1
  9280. Ok after 5 mins with pen, paper and ruler, It still works...ignore all previous posts from me. Perhaps If i had coffee before watching mathologer i would not have needed the pen and paper.
    1
  9281. the answer to the challenge is : ready to be amazed? ;) if you take the little triange and move it on top of the Trapezoid you get a square (the same as the blue one) . thus the area of the big square is the area of 5 little squares hence the blue square is 0.2
    1
  9282. The interesting thing to me, computationally, is that this is technically a scientific-notation multiplication calculator. All it's lacking is the ability to take the exponent into account
    1
  9283. makes seal noices
    1
  9284. Ein Mathologer-Ausgabe!
    1
  9285. I need to buy your 333 shirt.
    1
  9286. 9:06 I have that exact kind of slide rule, except pocket sized.
    1
  9287. You know what's crazy? I actually had this hunch that I can use Ptolemy to get a nice proof of the trigonometric equalities but I never really figured out how this could work out nicely. Finally I know that my hunch was right!
    1
  9288. The gift we don't deserve, but we needed
    1
  9289. is it 15?
    1
  9290. Magic, love it
    1
  9291.  @Mathologer  awesome, thanks, I'll check it out! Also, love your videos every time. Always a good morning when a new one drops 🙂
    1
  9292. The optimal stacking solution not being the most beautiful one was the most remarkable thing in this video imo.
    1
  9293. 36:01 I thought about this one for a long time… And I came to a similar “sparseness” conclusion, though I wasn’t able to definitively prove that it converges One thing I came up with is that for large N, the expected distance between the integers used for this series should be N^0.0458, which is somewhere between the 21st and 22nd root of N (To be more precise, the exponent is exactly Log_10(10/9) = 0.04575749… Where the inverse of 0.9 shows up here, precisely because we are removing another 10% of the integers every time we approach a new power of 10 The way I’ve worded that might make it sound like I’m over counting by removing things that would already be removed… But it’s easier to think of it in terms of a weighted average of what has already happened. For instance, every 10 integers before 90 keeps exactly 9 integers, which is 90%. But then the 90s keep 0%, so we have 0.9(90%) + 0.1(0%) = 81% of all integers kept up to 100 Speaking of which… This whole pattern of removals reminds me very strongly of “negative space” fractals such as the Sierpinski Triangle, or the Cantor Set. I bet the fact that this series forms a fractal ultimately proves to be a huge factor in the reason why it converges, despite the fact that the “generation 0” series diverges Edit: oh, you actually drew a representation of said fractal! That’s awesome. Thanks Tristan ❤) My inspiration for this idea came from the prime number theorem, where we know the expected distance between primes is about log(N). As you mentioned earlier, the inverse prime series actually diverges, so I knew the integers for this Kempner series needed to be more sparse than the primes (if we want it to converge) And of course, that turns out to be true! N^a grows faster than log(N) for all a > 0, even if it leads to something “very slow” like the 22nd root of N 😅 But unfortunately, I haven’t been able to see how I would prove that it actually converges. (I even went back and reviewed all the series convergence tests that I might have forgotten about, but nothing came to mind…) I guess I’ll see how it works at the end of the video 🙂 Edit: Man, I totally could have come up with that proof! Kinda disappointed I didn’t figure it out, but oh well… That’s what I get for trying to do most of this in my head at the computer, without writing much down lol
    1
  9294. I was puzzled at some point, with an impression that something had happened to your face, like you'd had a stroke and your left upper lip was drooping or something. But then later in the video, it went back to normal. Then I found out you had flipped a whole segment left-to-right when explaining the race car thing (probably because your gesture in the camera was the other way around wrt your commentary). Mystery solved, and no need to see a neurologist, phew.
    1
  9295. If you plug in a negative integer, the formula gives what you might call the Fibonacci prefix, and gives it correctly. It's like the Fibonacci numbers but with alternating sign: 1, -1, 2, -3, 5, etc. If you plug in non-integers, it stops generating integers and gives complex results - in both senses; they have a complicated nest of square roots that seem to include sqrt(-1) aka imaginary parts.
    1
  9296. I have no preference between the usual Mathologer video and the short clip. I like them both. I'd love to see a Mathologer take on Geometric Algebra and/or Clifford Algebra.
    1
  9297. 6:48 took me this long to realize that his shirt is a zombie math joke.
    1
  9298. This Christmas Tree structure resembles solenoid of chromatin.
    1
  9299. Great video as always! My favourite part was that about gamma, the areas and integrals! When you talk at the end of the "at least ten 9s" series, I think that one has to diverge as when the number of digit grows, almost all integers have at least ten 9s.
    1
  9300. I made it to the very end, fantastic!!!
    1
  9301. "we just found the best visual proof in 2000 years" -- what took you so long?
    1
  9302. I wondering how different splitting rules would manifest graphically on the simplex triangle (or might be we even need come back from special 2D projection (on a plane of constant sum) of 3D which is simplex triangle to original 3D).
    1
  9303. Recently I've been trying to wrap my head around 4D geometry. Now I'm wondering, is it possible to create a 4D equivalent of Petr's miracle with the help of quaternions?
    1
  9304. Thank you!
    1
  9305. This method of differences is similar to what I learnt the chapter of numerical methods etc in my mathematics 4
    1
  9306. Mathologer and 3b1b make vids
    1
  9307. Since when we dropped the remote. Best thing that my daughter noticed this😅
    1
  9308. Had some fun with Python. For the first puzzle, I used the formula 6*(1+2+3+...+n). The first difference was 2. The second was 18. Third was 72... Turns out the difference is always 2*(1+2+3+...+n)^2. I wonder where does that comes from? Tried to do the same for the fourth power, and the difference is always 64*(1+2+3+...+n)^3. I don't know enough to figure out where do these come from. I sadly couldn't find a pattern for the fifth power. (2002, 162066, 2592552, 20002600, 101258850, .... Do these ring any bells?). For the last puzzle, it seems that it goes 5,6,7,8,9... when you round the number, but the pattern breaks at power 39 where the sum rounds to the same number as power 38.
    1
  9309. Flawless Video!
    1
  9310. Wonderful convergence and divergence towards the regular polygons with isoceles triangles. Sometimes I remembed origami when geogebra was used to distort the irregular pentagon, relaized all irregular pentagons are deformed form of regular pentagons and some massaging with isoceles triangles are required 😂 to make them regular. Yeah kidding 😂 eigen values and eigen vectors preserves orthogonality and ensures maximum separation of vectors or reduced overlapping I hope. Long time with such a wonderful video, congrats 🎉🎉🎉
    1
  9311. Velcro
    1
  9312. Same, I hadn't fount out this deeply but when I was in 10th, figured out that squared terms have a constant after 'taking the difference' 2 times, cubes have a constant after taking the 'difference three times', and also that the the constant term came out to be the factorial of the power, not anything beyond that though
    1
  9313. Gr8 Mathologer!! 🐺 🌙 🌙 🌙
    1
  9314. This is also how Charles Babbage's Difference Engine was supposed to work - the method of finite differences. I mean his first, limited design, not the more ambitious Analytical Engine, which was a general-purpose, Turing complete computer (or would have been, if it had ever actually been constructed). Given the first few values of any polynomial of degree < 7*, it could calculate the rest - and therefore tabulate the polynomial's values as densely as one would care to. In practice, of course, it would not be used to make tables of polynomials, but more likely tables of transcendental functions that could be approximated by polynomials. * 7 was just the "depth" it could go into, due to its mechanical design.
    1
  9315. Fantastic
    1
  9316. ,,, gotta luv' this man ...
    1
  9317. Looks like a Christmas present: wunderbar!
    1
  9318. When I was a junior in high school I convinced a group of friends to help me build a tower of Lire while we waited for the bell to ring using copies of Frankenstein. Out of the 60-ish copies we had to our disposal, we only managed to balance 15. Chapter 2 really did remind me of the "good ol' days" xD
    1
  9319. I like your T-shirt🤣🤣🤣
    1
  9320. Color Proof
    1
  9321. 54?
    1
  9322. 2:06 You lost me there while the math hasn't even started yet .
    1
  9323. Masstheorie war damals ganz schön anstrengend, zur Erholung gab es dann im vierten Semester Funktionentheorie… :-)
    1
  9324. Pythagoras family can be easily constructed by adding 2 to the first component, and the next multiple of 4 to the second. So starting at (3,4,5) add 2 to 3 and 8 to 4 which results in (5,12,13). The next time you add 2 again to 5 and 12 to 12 => (7,24,25) and so on.
    1
  9325. Solve for: (10^2 + 11^2 + 12^ + 13^2 + 14^2)/365 Because we know that 10^2 + 11^2 + 12^ = 13^2 + 14^2 We can calculate the value of the first 3 squares and multiply by 2 The method i used is by transforming the squares into a sum squared like this (10+n)^2 And because there are 3 terms each one with 10^2 we add 100 three times and we get 300 Now we calculate the values of all 20n + n^2 and sum it all together (20*0 + 20*1 + 20*2 + 0^2 + 1^2 + 2^2 ) = 60 + 5 Now we add the 3 values we got 365 and as we said before we multiply by 2 and get 730 Now we know that the top part of the expression is equal to 730 Now we calculate 730/365 whicj is equal to 2
    1
  9326. Ive realized that you can also add the 2 squares of the other side of the expression and get the same result, which i guess makes it a little easier.
    1
  9327. Does anyone know where i can find the music that plays during the algebra autopilot (19:35 for example)? Amazing video, i am looking forward to try these challenges!
    1
  9328. Total number of squares for a 5 x 5 grid is 50. And the general formula for an N x N grid is (n^2)*(n^2 - 1)/12
    1
  9329. How come infinite area? Here the area is part of volume of Gabriel's horn/Torricelli's trumpet, meaning less than π. If math says that a flat area included in a volume is bigger than the volume, then the math is somehow broken. Back in the Torricelli's days they had not yet abandoned mereological coherence...
    1
  9330. The real question is 2+2=2×2 What the previous ?
    1
  9331. Wow beautifully explained.
    1
  9332. in reality, the rubber band will break
    1
  9333. math is good for describing .... but tells us nothing
    1
  9334. yes it is taught here in Nepal in 9th graded in the mensuration chapter right at the beginning although it is not given much important... love from Nepal ❤❤❤ Your videos are so awesome
    1
  9335. whats the music used in the animations? nice video!
    1
  9336. 21:50 windmills==zagier
    1
  9337. I was thinking more like "in any situation when from an eyelet there is more than one step the lace takes to reach another eyelet, you can rearrange the lacing in a way to put the lace just one or zero steps apart from the original eyelet. That closest eyelet now has three lace parts, rearrange one of them to be to that second eyelet we just took the lace from. Now the local lacing is the shortest for that eyelet. Take a random lacing. Let's rearrange the lacing to make it locally the shortest across the whole lacing. Let's start at the bottom, rearrange the lacing to make it so two bottom ones are connected, every other change happens above the part we have already, because there is no below, making it impossible to screw up the optimized part. Now rearrange the lacing to make to make it connect just one step above the lowest row, it naturally creates the cross lacing. Second row, if we try to connect them with step 0, we cannot connect all the other eyelets, thus another rearrangement to create cross lacing occurs naturally, and since no eyelet underneath is screwed up, the change only happens with the eyelets above, this will be true to every step. Thus no matter the amount of steps the cross lacing will naturally occur to locally optimize every eyelet. And since there is no way to create any line that isn't 1 or 0 step lines, but being shorter than 1 or 0 step lines, at all and specifically by creating longer lines elsewhere, which comes from the fact that eyelet columns are at an exact distance from each other (which doesn't necessarily mean that the solution would not be cross lacing for the inconsistent distance, but it means we don't have to think about those cases), if every eyelet is optimized, the whole system is optimized. The only way we can try optimizing the system to trouble individual eyelets, is to create more 0 step connections, which geometrically can be proven to only make the total length worse" Or something like that
    1
  9338. I remember when I first began studying calculus I had a hard time understanding the proofs. I kept at it and worked with a tutor until the light went on. When I really understood the concept, it was a marvelous feeling. After that the rest was easy.
    1
  9339. 7:39 Colouring proof over swivel proof for me.
    1
  9340. actually, bugs will never behave like that
    1
  9341. hold up.... Do each of these form a solid in a higher dimension?
    1
  9342. RS = R + S is symmetrical with respect to interchanging R and S. So R squared = S + 1 and S sqared = R + S + 1 are both true only when S is zero
    1
  9343. ThankU, I waited 50years for a reasonable explanation. Namaste
    1
  9344. That was refreshing. Very good job. Subscribed :)
    1
  9345. If you call a five-sided polygon a pentagon, you should call a ten-sided polygon a decagon.
    1
  9346. I’ve been playing with the differential operator D=d/dx lately, specifically in the case of exp(D). Right now I’m focusing on the effect it has on polynomials as they have finitely many derivatives. I’m taking a linear algebra approach and trying to construct matrices for exp(D) exp(D) is really interesting in regard to the trig and exponential functions exp(D) e^x = e^(x+1) exp(D) sinx = cos(1)sinx + sin(1)cosx exp(D) cosx = cos(1)cosx + sin(1)sinx
    1
  9347. So cool!
    1
  9348. 16:30 ish, as he reveals the constant e, the video's progress bar lines up perfectly with the arrow in the graph :)
    1
  9349. You have shown that knowledge has various facets, some of which are quite stunning and amazing! Thank you.
    1
  9350. Maybe a naïve way of looking at the problem: assume that instead of reverting to the negative terms right after reaching pi, we choose to overshoot until some x between pi and 4 is reached, then we revert back to the original algorithm, we eventually converge to pi in this case too. But as there are uncountably many reals between pi and 4, there must be uncountably many rearrangments.
    1
  9351. ​ @Your2ndPlanB I think is countable from you argument, if I have an infinite sequence of 1 and 0 I can assign it to a unique integer, counting in binary, the n_i has to be bounded to converge so if n_i<b I can count the sequence in a base b number system
    1
  9352. So, the natural question is how many ways there are, but the real question is whether they are countable or not? Hm... it sounds complex, and I'm not sure it's rational at all. I'll see myself out.
    1
  9353. They has to be countable, every sequence that sum to pi swap a finite number of terms, let's say the first n, so I have a total of n! permutations up to the first n number, so, given a permutation of the first n elements I can assign in a unique way a natural number up to n!, so they are countable! I remember something similar in game theory where you have sequence of 1 and 0 and they are similar only if they differ by a finite number of terms.
    1
  9354. ​ @gdclemo ah! Yes you are right!
    1
  9355.  @rielco8442  "every sequence that sum to pi swap a finite number of terms" Are you sure? Because if so, pi would not be transcendental...
    1
  9356.  @dlevi67  No...? Maybe I was not clear with my explenation, if I say that pi=a_0+a_1+..., is still true that pi=a_2+a_3+a_0+a_1+(all the other term UNpermutated, starting from a_4), the important thing is how the sum asimtotically procede, you can't tuch the "terms at infinity"
    1
  9357. 1) le numbers : 9 4 13 17, observable by factorization. the sequence is RRRL 2) 3 4 5 would have the parent as 1 0 1, where the box is 1 0 1 1. It's siblings are itself and 1 0 1 1 again, which would give us an infinite repeated nesting of the tree. 1 0 1 1's parent is itself, and itself flipped over but one of the diagonals in negative. Perhaps all possible fibonacci seeds are navigable using the modified traversal moves of RML and R',M',L', where RR' = MM' = LL' = not moving.
    1
  9358. Just my intuitive observation about why gamma > 1/2. From the diagram at 24:50 I notice that 1/x is convex. Therefore the blue remainder sections each have area greater than one half of their width times height. This is clear to me if I imagine a diagonal drawn from the upper left to lower right of each blue remainder and observe that the upper right half of the bounding rectangle is always filled with blue.
    1
  9359. ians shoelace sight is my go-to reading when i'm bored
    1
  9360. 6:30 Too easy.. Here's an extra stipulation for the puzzle: The remaining tiles must be orthogonally connected. Not so easy now :)
    1
  9361. Hi, 38:39 : and that's a good place to stop (and nice music)
    1
  9362. I have never heard about this, and it's really super interesting and surprising. Thank you!
    1
  9363. Gamma.
    1
  9364. This "magic" was actually common throughout XIX. and XX. Centuries in mechanical computers, like those used for heavy artillery, naval or rockets. You should really reinvent these very very soon, for sure no electronics devices will survive first EMI effects in upcoming nuclear war.
    1
  9365. At the balance warm up, I always end up thinking of moving the 2 kg to 1/2 of the distance from the center of the beam to the end. Or to put it mathematically, when balancing 2 objects n and m where m is heavier, we must move m to to n/m of the distance from the center to the end. Is this true?
    1
  9366. Supreme awesome and amazing. I love math and you teach to love on math.
    1
  9367. You relied on an extremely similar thumbnail style to Numberphile to try syphoning off some of their viewers. Miserable.
    1
  9368. loss and gain is the same
    1
  9369. Bla math is easy, just have to break the head open and arrange all the information so that you can easily binary search it.
    1
  9370. Honestly, it was all great, but my favorite part is that 👕!
    1
  9371. My background is that i took calculus up to multivariable/vector calc and i took a class on ODEs, and I tried to take linear algebra but failed it. I was able to understand every part that you expected me to understand with only a couple pauses. Anything that you intentionally glossed over I probably didn't understand that well. Also I commented a long time ago I asked for a video about bernoulli numbers and you finally delivered haha 😊. It was very satisfying, basically everything I wanted to know. I actually knew about the "bernoulli numbers being the matrix inverse of the pascal numbers" thing from some other video I watched (dont remember where). Anyway I really liked this video and all these longer ones. Your presentation style is inspiring especially for someone that loves pattern spotting and tends to think very algebraically. Definitely my favorite math channel if nothing else just for the great topics you choose to cover! Thanks for that and I can't wait for the followup! ❤️ Also back in May I started messing with zeta(3) stuff. I accidentally rediscovered a constant which I found on OEIS that has to do with the eta function, and some of the stuff I was doing involved gamma(1/3). My intuition tells me a closed form for zeta(3n) would not include pi, but probably some other constant, maybe even one that isn't very well studied. I also discovered a family of things which I call partite functions, and I started a secondary math youtube channel to start documenting some of it. I hope to work on that more this year and next year but I havent had much free time. I think my passion for math this year was mostly from watching videos on your channel. Thanks for all the inspiration~
    1
  9372. Problem 1 . . . . . For n=3, the last term on the left is 6 times the triangular numbers, for six faces of a cube. So 5^3 + 6^3 = 7^3. You can find the difference without calculating. If we have a 5 length cube and put 6 6x6 faces on it, we will have 6 of the new faces, 12 of the new edges, but only 6 of the new corners, so we’re missing only 2 units. 5^3 + 6^3 + 2 = 7^3.
    1
  9373. best video ever.
    1
  9374. weird counterweighted optimal tower O:
    1
  9375. The final proof of the sum of no 9 sum is the most impressive, although all the video is excellent, as allways. Thank you.
    1
  9376. Thank You Very Much for Share and Teach !
    1
  9377. Most fun thing to me was the longer than simply harmonic overhang with one block per layer at 11:35 But wow ... It was all amazing
    1
  9378.  @Mathologer  sorry for my dry sens of humor 😂
    1
  9379. Because he was not a normie
    1
  9380. Please explain Euclid book 5 definition 5 🙌💯💯
    1
  9381. Das ist der Weg ! 😂😂😂
    1
  9382. i just started watching this video at 1:03, and without spoilers, I can already tell that 36^2+37^2+38^2+39^2+40^2=41^2+42^2+43^2+44^2.
    1
  9383. Interestingly, when choosing U and V, the constraints on choosing the numbers have very little meaning. If you choose V<U, you still get a Pythagorean triple, you just reflect the triangle about the y axis. If you choose U and V such that they're both odd or both even, you get a primitive triple that is scaled by a factor of at least two. there seems to be a pattern for determining the scaling factor, it would seem to involve powers of 2 and powers of the greatest common factor, although I only tested with small numbers so it could be more complicated than that. Still, interesting to know its possible to get EVERY triple, even those that aren't primitive, if you allow yourself to choose any U,V such that V>U. I wonder if that still only makes each triple show up exactly once?
    1
  9384. Nice one
    1
  9385. Slide rules….yes! Amazing invention and instrument! I have a collection of those, and they keep on amazing me. The inventors of the different scales definitely knew what they were talking about. Thise channel is gold if you are interested in how to use the slide rules and all their aspects: https://youtube.com/c/ProfessorHerning
    1
  9386. Write-in vote for the mu cat.
    1
  9387. Am überraschendsten fand ich mit Abstand die Konvergenz von Kempners Reihe, was zunächst völlig unintuitiv erschien. Der toll animierte Beweis war dann ein „Aha“-Moment. Danke für die tollen Videos - weiter so!
    1
  9388. 0^3 =
    1
  9389. Does 2+2 = 2x2 also being "2 to the power of two" bring anything three dimensional, like on the axxis x, y and z? like in some "identity tree" or something?
    1
  9390. This is awesome, definitely up for more of this style of video Adam, Final (MSci) year Natural Sciences
    1
  9391. Sehr cool 🙃
    1
  9392. I nice samller video. I always enjoy watching your videos on Sunday with a cup of coffee. Relaxing.
    1
  9393.  @Mathologer  And just in time for Christmas
    1
  9394. The so-called, and poorly named "vortex diagram" is in fact the "enneagram", which has it's roots a long way back into esoteric traditions. It may be fair to say that throwing those traditions, as well as Nikola Tesla himself, into your "criticism" about the online phenomenon of "vortex math", has nothing to do with a possible analysis of the subject itself. Your very well made and enjoyable analysis of the math itself, like it is with all "math", is a great deep dive into the "how", yet doesn't say anything about the "why" of course.
    1
  9395. it looked scary but then i just started clicking on clickbait calculus videos and now it is easier than arithmetic
    1
  9396. I find it disturbing that some series will resolve exactly to a particular sum within a finite length, while others will only approach an arbitrarily close figure that can only be considered exact if the series continues to infinity. This seems to also be a way to determine whether a given number is irrational or not.
    1
  9397. This method also works well in computer science and programming.
    1
  9398. I have a high school background and I loved every second even though I understood only a few basic concepts 👍 The warnings and relaxed feel makes it all fun
    1
  9399. i see you changed the title following veritasium's suggestion
    1
  9400. I love you shirt which resembles the golden spiral.
    1
  9401. Didn't learn the second proof until I found it on the internet my first year of college!
    1
  9402. 14:03 Welcome to Euler's geometric object! (It's not a number, you can eat two or three apples, but not e apples).
    1
  9403. As soon as you pointed out that the number of vertices, edges, faces, etc. of an n-cube sum to 3^n, I had a total "aha moment", where I thought, "of course they do, just as a Rubik's cube has 1 mini-cube for each vertex, edge, face, and (hypothetically) an interior mini-cube for the 1 cell to make 27". And then I paused the video and worked out a visual proof sketch (of the (x+2)^n coefficients counting the elements of an n-cube). I haven't kept records, but maybe once in 4 videos or so, something early will give me an "aha, I see where he's going" moment and that's always fun. And usually my aha moment pans out later in the video. But this time you never went in my direction, so I'll present my proof concept (EDIT: Well, Tristan's proof is the same concept, but it treats the vertices and edges directly as algebraic objects instead of using n-volumes like my concept below. I thought my idea was missed because of the Euler tangent.) If you have a 1+x+1 line segment partitioned into segments of length 1, x, and 1, you can raise it the nth power and get an n-cube with n-volume (1+x+1)^n that is partitioned into 3^n n-cuboids. Each vertex is adjacent to exactly one n-cuboid with n-volume 1^n*x^0=1. Each edge is adjacent to two vertex-cuboids, but also exactly one n-cuboid with n-volume 1^(n-1)*x^1=x. Each face is adjacent to vertex and edge n-cuboids, but also exactly one additional n-cuboid with n-volume x^2. And so forth. (And finally there is one with n-volume x^n in the center.) It is clearer to start by illustrating for n=1,2,3: https://imgur.com/a/VIajEwu (1+x+1)^2 creates a square with 9 pieces, 4 of area 1 at each vertex, 4 of area x along each edge, and 1 of area x^2 in the center. (1+x+1)^3 creates a cube with 27 pieces, 8 of volume 1 at each vertex, 12 of volume x along each edge, 6 of volume x^2 in the middle of each face, and 1 of volume x^3 in the center. Now, I'm not an expert at turning visual proof sketches into proofs, but I think it's very pretty.
    1
  9404. Of course, when you went in a different direction than I expected from my "aha", and produced a generalized Euler's formula from x=-1, that's fun too!
    1
  9405. Why do physicists play dominoes? Because they can't shoot dice?
    1
  9406. why is 0+0=0x0 not a solution?
    1
  9407. It is, but 0+0+...+0 = 0×0×...×0 just isn't interesting at all.
    1
  9408. Loved this!
    1
  9409. I found the bizzare overhang tower the most memorable!
    1
  9410. I made it to the end Masters Mechanical Engineering
    1
  9411. There are N-1 distinct squares whose corners are on the 4 edges of an N*N lattice. That lattice can be shifted to (M+1-N)^2 positions on an M*M lattice (M≧N). So the total squares in an M*M lattice is ⅀(N-1)(M+1-N)^2 for N from 2 to M. For the 5*5 in the video, that's 50 squares.
    1
  9412. Does your shirt read, "To infinity and beyond"?
    1
  9413. I figured out what your shirt says :-D
    1
  9414. I made an animation of this: https://www.youtube.com/watch?v=Ijgc61XRL1Y Source code is linked under my video.
    1
  9415. w00t
    1
  9416. 140 years ago we had US presidents thatwere doing mathematical proofs. Now we have moronic loser business men like Trump that want to inject themselves with bleach to fight COVID... What a regression!
    1
  9417. oh purée c génial !!!!! merci merci !!!!!
    1
  9418. I like the Swivelling Proof best.
    1
  9419. My answer for 11:42 is the following 9, 4 17, 13 My logic? We have a function that takes four whole numbers as input a,b,c,d and outputs three whole numbers ac, 2bd, and ad+bc. In this case ac = 153, bd = 104/2 = 52, and ad+bc = 185. Next, we can determine the factors of ac & bd to guess & check possible solutions. ac = 153 = 3 x 51 or 9 x 17. Setting a = 9 and c = 17, my remaining equations become bd = 52 and 9d + 17b = 185. You can either do algebraic substituion or continue guessing & checking using the factors of bd; I did the latter. bd = 52 = 13 x 4 or 26 x 2. I set b = 4 and d = 13, And all I had to do was check my final equation was true, 9*13+17*4 = 185, which it is! ---- Now I wish I could solve the next part but I ran out of steam.
    1
  9420.  @Mathologer  I'm sure the colorful pictures were in the appendix. Right?
    1
  9421.  @Mathologer yes but GeoGebra didn’t exist in 1904 ;-)
    1
  9422. My favorite mathematician.
    1
  9423. Dann, your pronunciation of the german was near to perfect. Did u study german or are u just a genius?
    1
  9424. One minute in and HOW HAVE I NEVER NOTICED THIS?!
    1
  9425. 2:07 this was perfectly pronounced
    1
  9426. 25.807 is actually a more accurate approximation.. 😀
    1
  9427. new video will see later
    1
  9428. The dots are cooler than the lines.
    1
  9429. People who want to have fun with this, remember that the definite integral and all of the derivatives of exp(x) are exp(x). Sure, you can sum it as a geometric series, but plugging e^1+e^2+e^3+... in to the Euler-Macluarin formula is sooo satisfying. You can do the same for sine, and for a challenge, use these to find the sum of all of the positive B_n/n! and also the sum of the negative B_n/n! . If you do a great job simplifying, you end up with cotangent in the expression!
    1
  9430. After watching the whole video, I'm tempted to make a program that can do this for any set of coins to any dollar amount
    1
  9431. 9, 4, 13, 17
    1
  9432. You really have a gift at explaining math in an accessible way. I love math and am grateful for the time and dedication you put into each of your videos. Thank you for taking the time you put into reaching out with math to us! -- Rick
    1
  9433. It’s all very cool and some of it is above my regular math practice. My favorite part was no integers in the sum. That’s a piece I could explain to my jr high students.
    1
  9434. Einstein's published papers
    1
  9435. I KNOW that something in there has to do with Ptolemy's theorem. I learned in a Numberphile video that pythagorea is a specific application of Ptolemy's theorem.
    1
  9436. I don’t see any paradox.. Strange to think any mathematician sow it as a paradox either. If you have perfect paint that can fill out such a horn, it would obviously have 0 thickness. Then let say you need 2kg of that paint to fill out the whole horn. Exactly 0kg from that would be needed to paint the whole area of the horn.
    1
  9437. I didn't grasp everything, but I didn't have that bad of a time. I have only Calculus 1 by education. Otherwise, I am self taught.
    1
  9438. Couldn't this also be an interesting way to project the map from a globe to a flat plane? With lines of longitude becoming semicircles arranged around the north pole.
    1
  9439. Pretty cool stuff. The first thing that came to mind when I saw this was a vernier scale which also uses some sort of sliding ruler to make measurements and read between two graduations markings more accurately. I can imagine there are similarities in the principle but one important difference is that the markings on a vernier scale are spaced evenly. I'd be curious if anyone know whether the two are related at all.
    1
  9440. In the approximation ln(n+1)+γ, the leftover blue bits get closer and closer to triangles, so should add up to approximately 1/2n, so an even better approximation would be ln(n+1)+γ-1/2n.
    1
  9441. I used to see circular slide rules. I seen to remember circular slide rules optimized for solar navigation.
    1
  9442. @35:10 I always ask my students to cold prove theorems discovered by 20th century math geniuses !
    1
  9443. Mathologer Love from india Wanna contact you
    1
  9444. 18:37 golden spiral, is that you ?
    1
  9445. I was having a really hard time trying to logically pick my way through the claim that e^a = cosh(a) + sinh(a) right up until I got to the proof of such at the very end. Bravo, maestro. Bravo. :)
    1
  9446. I wonder about the number of ways for generating Sierpinski triangle. Recently, I did it using chaos game. :)
    1
  9447. Commenting!
    1
  9448. Those bizzar brick stackings kind of blew my mind - How could one even prove an optimality theorem for these where shapes??
    1
  9449. I was almost waiting for a Mathologer end credit scene
    1
  9450. 12:31 a square is removed, making the board untileable
    1
  9451. No degree whatsoever, but a genuine interest in everything mathematical. This was great stuff, and quite easy to follow. Looking forward to other videos on the subject. Hope the mysterious Euler-Mascheroni constant will get some attention as well.
    1
  9452. Definitely the bizarre maximum overhang is the most remarkable part
    1
  9453. Can you please make a video on the Gaussian integral ??
    1
  9454. I got some really weird looks using a slide rule in freshman college courses (2013). They are particularly handy for the types of calculations you tend to find in physics and chemistry... The ancient physics professor I became good friends after that. Quickest way to make them useful is solve for the variable you want without calculating anything, just write the numbers out as a large fraction, then you do all the calculations at once, sliding through incrementally each number. You keep track of the exponents in your head as you do so (remembering the +1/-1 for falling off the end of the scale). At the end, you get your answer from the scale x 10 ^ (exponent)... Scientific notation was MADE for these devices and it rocks.
    1
  9455. Wir hatten das 'Kunundrum' in Form einer Stufenfunktion die gegen eine einfache Diagonale konvergierte (Länge der Diag. = sqrt(2), Länge der Stufenfunktion = 2) Die Verfeinerung der Stufenfunktion konvergiert gleichmäßig gegen die Diagonale aber das Wegintegral bleibt 2. Wir sind dann einfach zu Prof Hofmann (TU-Darmstadt) gegangen und der hat das dann innerhalb von 10min an seiner Tafel rausgepuzzelt: (Gleichmäßige) Konvergenz der Funktionenfolge garantiert noch keine Konvergenz der Ableitung. Weiß ich noch nach 20 Jahren ohne jemals wieder mit Analysis zu tun gehabt zu haben. So ist das mit echten Aha-Erlebnissen. (So, und das ganze hab ich auf Dt geschrieben, weil mir das englische Vokabular fehlt. Das einzige was Ich übersetzen kann ist Aha-Erlebnis: epiphany :)
    1
  9456. But, wait! Why (ho)^3 ? shouldn't that be? 3(ho)
    1
  9457. Thanks. Another great video on a Sunday afternoon with Coffee. Lovely.
    1
  9458. I made it to the end but the proof they're made things murkier for me. If you define happiness by comparing to the Talmudic solution then it's no surprise that that solution makes everyone happiest overall. It seemed more natural before I watched your explanation at the end
    1
  9459. 0+0=0x0! :D
    1
  9460. 9:23 i hate this "paradox" and this is NOT a full rotation!!! From the rotating coin's perspective, it is currently upside down!!! Consider if you were a ball the size of earth, and you rolled around the equator of earth from 0 degrees to 0 degrees. You are inside the ball, and you start with your feet pointing at the ground, at 0 degrees on the equator. You start rolling around, and when you reach 90 degrees, your HEAD IS POINTING TO THE GROUND! You would never call this the "first rotation"!!!!!!!!!! Only when you reach 0 degrees on the equator again would you return to your starting position of feet down, making your SINGLE FULL ROTATION. The coin is exactly the same! From the coins frame of reference, it only makes one rotation! It is only when you place both coins in a different frame of reference and observe them arbitrarily from the side than it appears to be two rotations. Really what you are asking with the coin "paradox" is if you rotate a coin about a point that is 2 times its radius away from its center such that the rate of rotation of the coin about its own center is twice that of the rotation rate about the point 2 times the radius, how many times will the coin rotate about its center per each rotation about the further point? This is obviously a stupid and trivial question because you are literally asking "if a coin rotates twice around its center for each time it rotates around an arbitrary point, how many times does the coin rotate for each time it rotates about its arbitrary point?" This is exactly the same as saying "if 2x = y, then what is y?" In math we often toss out the trivial solution, or at least set it aside, but it often TELLS US NOTHING. The coin paradox is a case of this. The "it rotates twice" answer is not only wrong, its the trivial solution to an even more trivial question.
    1
  9461. (loved it all)^{ᴵ ˡᵒᵛᵉᵈ ⁱᵗ ᵃˡˡ }
    1
  9462. Most memorable: no 9s sum!
    1
  9463. I never knew why the h... are they called hyperbolic - now i know thank you! - I better like coming out from the imaginary numbers and general exponentiation and coming into the trigonometric (real) and hyperbolic functions - i feel more conscious of the way it came out , visual proofs dont talk to me because teachers neglect them (but in middle school the good old teacher Mrs Kubala was happy when i answered the question of how do we know where does the limit of (2x+log(x))/x comes to - i answered thats because the higher terms calculated on a calculator reach 2 and nowhere higher (i didnt want to say that the terms reach it but that the calculator showed the result in case it was wrong and it would be my fault :)
    1
  9464. I just spotted that A^2 + B^2 = C^2 + AB is equivalent to (A^2 - AB + B^2) = C^2, and the LHS is the norm of a+bω, where ω is a primitive third root of unity in the ring Z[ω]. It's unbelievable that I've never seen this very elegant connection of abstract algebra to elementary geometry before.
    1
  9465. i can't count how many times i said "HOLY SHIT THATS SO COOL" i have been reading so much maths lately and i'm actually able to follow everything and go ahead even! EXCUSE MY SWEARING BUT THIS IS SO FUCKING COOL
    1
  9466. I WAS LOOKING AT THE MATHOLOGER CHALLENGE AT 8:30 and i kinda just... saw y=e^((1/2)x^2) as a solution and the connection with the gaussian integral shows up from there! wow! just woooow!!!! SO COOL
    1
  9467. Thank you. This video was very nice and the tessalation at the end was very interesting!
    1
  9468. Rearrangement of infinite series breaks my brain. Associativity and commutativity seem like fundamental qualities of addition, just from what addition means...it can't not have those properties. And it does have those properties for any series with any number of elements. But as soon as your series is infinite, addition just *breaks*. It almost feels like the opposite of things like complex numbers, where you introduce a counterintuitive element (inventing a number that squares to -1) but when you work it out everything still works consistently. Or geometric algebra, where allowing scalars to be added to vectors and defining multiplication in a pretty intuitive way Just Works.
    1
  9469. I admit when you folded the Fibonacci line near the beginning I thought of matrix algebra right away.
    1
  9470. Another great video. Love these. Sorry I was late, I usually watch them on Sunday but I was making a Lego set. :)
    1
  9471. Enjoyed this presentation. Oughtred's Scale used for navigation was a serious technical breakthrough.
    1
  9472. Pretty awesome. I think you would be fun to hang out with.
    1
  9473. Think its right if i didnt bungle the comment or lose my thread during the whole thing.
    1
  9474. Just a small correction to the description: For anyone looking for the music/songs in this, it appears they're both by Nate Blaze, not Nick Blaze.
    1
  9475. Amazing. I love all of your videos. School maths generally depressed me for most of my life. Although the maths involved is great, the presentation of it was awful. For me, I felt they didn't spend long enough trying to be like Einstein: "I never teach my pupils, I only provide the conditions in which they can learn." It feels to me, schools just never answer children satisfactorily when they ask: "What's the point of learning this stuff? It looks boring." I think a perfectly natural response to a poor answer to this question is the very saddening things we hear: "I'm just not a maths person", "my brain doesn't work that way" and so many children get a bad relationship with maths from the beginning sadly. Now Einstein's quote clearly went to an extreme and suggests he spent all his time only trying to give a strong answer to that oh so common question above. In reality, it would likely be split between teaching and inspiring but I'd say right now schools are 5% inspiration and 95% teaching (perspiration), when we really need to get to the exact reverse. If we develop interest and show people where maths can take them and why it's fun, so much would be better in this world I feel, as a purely natural consequence. I think many teachers in other fields are closer to the extreme of Einstein's quote and we need to get there with Maths too. Thanks so much for your efforts, always love the little touches like the shirt also :)
    1
  9476. The animations would look prettier if you rendered them at higher than like, 4 fps. Any reason you picked such a low fps?
    1
  9477. I got math exam in 10 days im gonna dieeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
    1
  9478. I love this channel
    1
  9479. All you did was using the reverse integral test to estimate the integrals with series (i.e. limits) rather than computing them exactly, so you did use calculus. However you are right in that the calculus is hidden and the explanation is much easier than the usual. Great video as always!
    1
  9480. So many "ohs" and "wows" in 25 min, need to watch again
    1
  9481. Ramanujan is the smartest person in the world on my opinion and that's why he is my idol in mathematics
    1
  9482. Made it all the way to the end
    1
  9483. Coloring proof is best by a long shot. The generalized version is trivially a fourth color.
    1
  9484. David Schwartz
    1
  9485. A couple minutes into the video, I know this reminded my of Reuleaux triangle.
    1
  9486. the method described starting at 22:22 cannot generate every possible tiling. it seems to be missing tilings for subregions that are greater than 2x2 (like the 4x4 region that appears somewhat later)
    1
  9487. That is great video that includes history, story and applyful context. This is the content youtube has to promote.
    1
  9488. Pizza Phi
    1
  9489. Something worth noting is that in the original puzzle at one point you have to put a smaller piece on a way bigger piece (e.g. piece 1 on piece 5) but in the puzzle at the very end of the video that is not the case. This seems like something worth exploring. Is it a property of the number of pieces or poles or both and if it is how many pieces and/or poles are needed for this to be avoided?
    1
  9490. There is an optical illusion in Fermat's tiling; it looks like the lines are not horizontal or vertical but slightly skewed.
    1
  9491. Only two minutes in. It’s weird that the hexagons obey the rule regardless of which side you’re looking from. So if you were to rotate the triangle 120 or 240 degrees it’s still a “valid” triangle
    1
  9492. It should also be noted that a man of the name b bergren discovered this tree first
    1
  9493. Actually, we have to project the 3d-plane again to a 2d-plane to be able to see the result on a screen.
    1
  9494. THEN TO IMPROVE
    1
  9495. 10:23 "If he was a hair man,I would be jealous"..lol
    1
  9496. I knew of the first proof of the Pythagoras theorem from 'Shulbosutra' perhaps of Katyana, which had many theorems copied in Euclid's Elements.
    1
  9497. @Mathologer  Well, we have situations like that. It's frowned upon to buy and sell claims to children, but people still benefit from negotiating agreements for child custody in case of eg a divorce. More mundane, even at work people sometimes don't like to put a monetary value on things, nor trade them around freely. Eg if your team has on-call duties and you are trying to assign shifts in a way that's fair and reasonably efficient. See also https://en.wikipedia.org/wiki/Repugnant_market
    1
  9498. 🤌🤌
    1
  9499. Inbelievable cool! Thanks!
    1
  9500. Astounding video. Thank you so much for the service to humanity.
    1
  9501. Very cool. Also, it looks like at 39:06 if you add together the vertically aligned numbers on the left side of the equations it fills in the gaps in the Fibonacci sequence; 3 + 5 = 8, 5 + 16 = 21, and 16 + 39 = 55.
    1
  9502. This video right befor my my discrete math exam <3
    1
  9503. I was thinking "what if we go up a dimension and consider quadrilaterals in 4d?" But I guess you can always find a 2d plane for 3 points, and then find a 3d slice of 4d space containing the point and the plane. Anyway, great video!
    1
  9504. I liked both proofs. But i liked it better when you completed your proof with “looking good. still looking good. oh, definitely looking good. beautiful.” 😂❤8:00
    1
  9505. Can you make a video for really nice prime formula it is called Willans Formula
    1
  9506. I have first heard about that proof at a math and physics summer camp :-) We did not talk about it in school.
    1
  9507. That's why the anagram guy wrote only "Anagr." instead of "Anagram", cause anagram was not a complete anagram. 🤣
    1
  9508. Absolutely the sum of no 9s! As soon as I saw the rectangle my gut feeling told me that it must converge and I was right! Amazing!
    1
  9509. My father taught me. But you are more fun...
    1
  9510. Coolest was "Chapter 6: Kempner's no 9s series" for Tristan Tillij's squares going to gray and than fading away toward infinity. :-)
    1
  9511. Oh Burkard, that was great! On the level of Concrete Mathematics (Graham, Knuth, Patashnik), one of my very favourite books. I love your videos.
    1
  9512. Wait, did you know that there's a direct correlation between the decline of Spirograph and the rise in gang activity? Think about it. - Dr. S.
    1
  9513. When pithagoras doesn't work, cosine law does for any geometry 🤓🤓🤓🤓🤓 Get curvature with a negative sign Yeaaaaaah this video of yours is very good with your animations 😳 Hey, don't start with Lorentz covariant regime, I'm a quantum eigendude (UCLA)
    1
  9514. I’d rather see something covering the real unideal characteristics of shoelaces. Like the lacing technique with the highest strength to friction ratio, or the lacing technique that applies tension the most uniformly. When you pull on your laces the top tightens up but the bottom basically doesn’t change whatsoever because of the cumulative friction, so a lacing method that changes as a function of distance from where it’s being tightened from would be preferable. If you assume a constant percentage loss due to friction at each eyelet, this could still be a relatively ideal mathematical question. Though maybe it’s more correct to assume a friction percentage that’s a (linear?) function of the angle at which the links bend through he eyelet. A straight line through an eyelet will have no appreciable friction, after all. Such a model would actively discriminate against the narrow angles favoured by all tight lacings, so I’d guess that the tightness of the lacing technique will vary over the length of the shoe. Still, it’s a neat rabbit hole.
    1
  9515. Your T-shirt contains an error. Should say 25.8069758. 🤓 Also, what is the spring constant of the rubber band?🤔
    1
  9516. Maybe silly question, but what happens if you take the primes as a series, and try to calculate the polinomial?
    1
  9517. Somehow at 18:00 this reminds me of buddhist mandalas https://bit.ly/381M4zW
    1
  9518. Nathan Is more visual to me
    1
  9519. Initial guess: wouldnt it just be a googol choose (the number of ways to make change for $1)?
    1
  9520. I guessed correctly yellow but for a different reason. Thank you for the exercise.
    1
  9521. Does point has no dimensions or infinitely small dimensions?
    1
  9522. I love discrete mathematics! Super relevant in computer sciences.
    1
  9523. For me the "no nines" proof
    1
  9524. I like the leaning tower of lyra because of its apparent (till I test it) symmetry to the Brachistochrone curve, which is my favourite geometric shape. :D
    1
  9525. Could one prove this with first-year calculus, or is first year calculus the result of this?
    1
  9526. This video takes me back... I first learned about infinite continued fractions in 2012 through David Burton's "Elementary Number Theory," and the methods used in this video make a lot clearer what's going on under the hood in that text. I let out a squeal of joy when I heard Pell's equation would feature here today!
    1
  9527. Hi Edit: 00:30 : commented “Hi” 01:01 : WAIT WOW THAT’S AMAZING
    1
  9528. apparently the egyptians used a string with 12 marked segments to make sure the foundations of the pyramids had perfectly orthogonal sides
    1
  9529. Gamma was the most exciting
    1
  9530. For the regular poligons with 6 or fewer vertices the distance from the points to the center is shorter or equal to the lines between the points, making the shape grow or stay the same shape rather than shrink
    1
  9531. So much for my Merry Christmas🤔🤔
    1
  9532. The strange looking optimum tower of lire is not the most elegant but certainly the most memorable.
    1
  9533. i much prefer descriptive titles because i don't like Marvel stuff but i watch anyway. might be hard to find again later without knowing what the video is really about though
    1
  9534. They did teach us a lot of this in school; but 90% of my classmates would today tell you "they never taught us that." Just like "stop drop & roll" was taught every year for 9 years, K through 8, by the fire department coming out to our school, and that was one of the high points of everyone's year. Yet in high school a lot of them didn't know what to do if someone's clothes caught on fire in chem class. You can lead a horse to water...
    1
  9535. 7:25 I remember seeing this as an exercise in Spivak's Calculus in the chapter on integrals. Sometimes analytical mathods brush away all the beauty there is to a mathematical fact. Take this example as an instance: claiming ln(a)+ln(b)=ln(ab) is much less impressive than claiming the sum of those two areas equals the area enclosed up to the product of a and b.
    1
  9536. Because they don't want us to understand, just obey
    1
  9537. Only watched to about 6:30 so far and noticed the ‘cants’ coulda also be 4K - 1 ‘ Idk :P just seems simpler to me and was an interesting observation
    1
  9538. How is the theorem with the good and bad factors called? I'd like to see a proof
    1
  9539. At MIT the joke was... How does a mathematician start every discussion? "Imagine a circle..."
    1
  9540. I always liked math but not math teachers.
    1
  9541. Your opening statement about heads or tails being a thing of the past got me wondering how accurate flipping a coin is as a way of simulating a random choice between 2 options. You know, because coins technically have 3 sides.
    1
  9542. Im wondering if this manipulation of the sum of an infinite series containing both infinitely many positive and negative numbers can be applied in order to solve the 3x+1 problem
    1
  9543. 👍
    1
  9544. I love your videos
    1
  9545. What about the symmetry of a lace, is there a lacing between the crisscross and zigzag that is "better" than crisscross, but symmetrical (unlike zigzag)? Zigzag pulls one side more up than the other and I don't wanna use crisscross if there's something more efficient... Wouldn't wanna have our lace anything less than optimal!
    1
  9546. Beautiful video! I really cannot think of anything unclear, and the highly visual approach is always incredible. I'm a freshman in college taking functional analysis; it'd be neat to see you cover something from our course perhaps?
    1
  9547. EEExxxxxcelllentttt t-shirt.
    1
  9548. I liked the first section. As a physics teacher I would use moment arms to have students determine various distances and masses. Fun stuff and can be demonstrated on a playground. Nice video. Thanks.
    1
  9549. Publish this ASAP, you have a novel research here.
    1
  9550. one glance balancing, simple and yet teaches to think.
    1
  9551. Professor Polster, I always enjoy your videos and the ways you have visual animations to represent your topic of discussion. I try to do the same with my students, but your depictions always blow me away. This was a very helpful and informative video, and I especially enjoyed Part 2. I tried the leaning tower with Jenga blocks, as they are the same mass (for the most part), and I was able to get up to a stack 8 high. Well done Sir, and thank you very much for your thougthful and enjoyable content!
    1
  9552. ... and those cubes towards the end Nice - very nice.
    1
  9553. Actually impressive, that there are multiple other simple proofs right here in the comments. :o PS: Yes my spider senses were tingling right after the first iteration and said, „ah that‘s the proof“
    1
  9554. Try the digital roots of prime numbers.....very surprising ;)
    1
  9555. Amazing! Thanks for this gem
    1
  9556. It's a great,but what's about that's calculated by Fourier series
    1
  9557. Hello is this another mathologer video?
    1
  9558. Brilliant! Really enjoyed this video!
    1
  9559. With high a enough order, you can fit a cow to a polynomial. That's what my numerical calculus teacher used to say
    1
  9560. Your videos are really amazing. Thank you, Dr. Polster.
    1
  9561. <3
    1
  9562. Write it on a TORUS because this is the topology of the magic square. (2D surface wrapped round inside 3-space) Next interesting challenge: make a magic CUBE :) ...which will have the topology of... a 3D surface wrapped round in 4-space.
    1
  9563. 27:00 “Aumann and Maschler explain the Talmudic approach this way: they suggest that half of the total claim, representing half a container, serves as a psychological threshold between two perspectives. If a creditor receives less than half of their claim, they focus on the loss—seeing the situation as a complete loss with only a small portion salvaged. If a creditor receives more than half, they focus on what they recovered, perceiving it as a full repayment with a minor shortfall.” It’s almost the reverse of counterfactual thinking in Olympic medal winners. The bronze medal winners are, supposedly, happier than the silver medal winners. If you win the bronze medal, you engage in downward counterfactual thinking, looking at all the competitors who won nothing at all, i.e., focusing on your win; if you get the silver, so the theory goes, you engage in upward counterfactual thinking, i.e., looking up at the gold medalist and thinking “That could have been me!” focusing on the loss. The algorithm given by Presh Talwalkar in his 2008 blog post, “How Game Theory Solved a Religious Mystery,” (which is how I first learned of this distribution problem and its Talmudic solution), makes the logic clear: (1) Order the creditors from lowest to highest claims. (2) Divide the estate equally among all parties until the lowest creditor receives one half of the claim. (3) Divide the estate equally among all parties except the lowest creditor until the next lowest creditor receives one half of the claim. (4) Proceed until each creditor has reached one-half of the original claim. (5) Now, work in reverse. Start giving the highest-claim money from the estate until the loss, the difference between the claim and the award, equals the loss for the next highest creditor. (6) Then divide the estate equally among the highest creditors until the loss of the highest creditors equals the loss of the next highest. (7) Continue until all money has been awarded. That algorithm ensures that, to the extent possible, each creditor, from lowest to highest, gets at least half of the claim, and only after that is satisfied, the losses are equalized in absolute terms among the remaining creditors.
    1
  9564. 25.8069 is the root of the evil??? Ever heard of rounding?
    1
  9565.  @Mathologer  :-P
    1
  9566. I made it to the very end.
    1
  9567. I liked the color proof better
    1
  9568. Am I the only one who thinks the Toymaker looks a bit like Bill Nye?
    1
  9569. >see thumbnail >read video title Thanks, Terrence Howard!
    1
  9570. I knew I wasn't the only one! A+A = AA suck it haters!
    1
  9571.  @Mathologer  Terence Howard is a mentally ill man who came uo with "Terryology" in which he "proves" that 1 x 1 = 2, and a host of other crazy claims and quantum woowoo. There's plenty of material about him. Be prepared to lose some neurons though.
    1
  9572.  @VikingTeddy  Heh. Wait until Electric/Plasma Universe becomes mainstream. The neuronal loss will be inevitable. Fun times ahead, baby!
    1
  9573. The optimal overhang was the best! Also your videos rock
    1
  9574. Turning a sum into an x-power series and/or a definite integral, then calculating the derivative, is also very reminiscent of the Feynman integration technique.
    1
  9575. I like the idea of a bishop 700 years ago solving this in an environment with little more than oil lamps and quill pens.
    1
  9576. I really enjoy the constant masterclass uploads
    1
  9577. 29:50 another neat property of this family of triangles is that the square of the shortest side will equal the sum of the other two (discarding units). Eg, 5^2=12+13
    1
  9578. This works because the highest two numbers always differ by 1. With A being the smallest of the 3, (1) A^2+B^2=C^2 [Given] (2) B+1=C [Given] (3) A^2+B^2=(B+1)^2 [Substitute (2) into (1)] (4) A^2+B^2=B^2+2B+1 [Foil (3)] (5) A^2=2B+1 [Subtract B^2 from both sides of (4)] (6) A^2=B+C [Substitute (2) into (5)]
    1
  9579. natürlich
    1
  9580. crazy long one with big greek Ds. Hahah
    1
  9581. I never liked short math videos, so these hour long videos are super awesome and at least for me the work you put into this was very worth it.
    1
  9582. I can't wait for all that knowledge
    1
  9583. I stumbled on this in HS trying to solve 'what comes next' problems. Once I got to the line of constants, I'd take the integral to find the original equation. After a while, I found the Pascal's triangle and the shortcut equation. One thing I realized is that if the numbers are not exact, you get a wildly oscillating solution, similar to a sine wave, since an nth degree polynomial can change directions n -1 times.
    1
  9584. Definitely we like it!
    1
  9585. As always your videos are fantastic. Please consider making videos on probability and statistics and topics like Markov chains
    1
  9586. There's the necklace, generally generating with a generous generator. Lol no.. It's my other fractal program, de moivre set 🤓🖖
    1
  9587. 00:07:15 - Pythagoras' theorem is just a special case of Ptolemy's theorem—math magic! 00:13:43 - Discover the hidden beauty of geometry: ratios, identities, and perfect fits! 00:20:17 - Infinite ways to create golden spirals, not just the classic one. 00:26:40 - Start with any two numbers and grow a Fibonacci-like sequence; the ratio between consecutive elements will always converge to the golden ratio. 00:39:45 - Pascal's triangle isn't just a math trick—it's a gateway to infinite patterns and identities!
    1
  9588. with the question of the collection i didn't understand why it is ok to remove numbers from each collection. wouldn't that make each collection less than ten numbers? at 26:00
    1
  9589. If I understand your confusion correctly: He's talking about the sub-collections that add to the same number. Pause at 22:58. You see 3 blue and 4 green. If we had a case where 25 had been included in both the blue and green collections, you could just remove it from both and they would still sum to equal numbers.
    1
  9590.  @tbttfox  The problem is that he included collections with only the same numbers in his total count. If you remove all the shared numbers then both collections would have zero elements which isn't correct.
    1
  9591. Very elegant and interesting . But the question is : has this result not been discovered long time ago , but not published ?
    1
  9592.  @Mathologer  thank you. I would really like another video explanation of collapsing numbers as used in Numerology (78 = 7+8 = 15 = 6, etc.) that I saw you use in at least one of your presentations. Would you point me in the right direction for that?
    1
  9593. Thank you,  @Mathologer  .
    1
  9594. Like I have "always said", Math is Life and Life is Math! We use Math everyday, whether we realize it or not... General Practice allows us to "retain" the basic principles with everyday basic principals!
    1
  9595. This video should've been posted in 2016
    1
  9596. The one I always use is the lacing at 2:09. I just like the way it looks. I first learned it at a Summer camp decades ago.
    1
  9597. to infinity and beyond
    1
  9598. This is great! Inspiring!
    1
  9599. My geometry teacher showed us the algebraic proof in class and I saw the visual proof in AP calc AB
    1
  9600. the history of mathematics is the history of updating rules from "this always works" to "this always works, except for annoying contrived stuff like Frank came up with"
    1
  9601.  @Mathologer  nobody in particular, it just seemed like a funny arbitrary name :)
    1
  9602. Very enjoyable and well explained. This is quality 👌
    1
  9603. Chapter 5 (liked best). Enjoyed "discovering" gamma and its tie-ins to superstars Euler & Ramanujan and related videos I've also enjoyed, thanks to this gifted mathematician and communicator.
    1
  9604. This is the first I've heard of Cavalieri's Principle, but it sure does sound a lot like an integral.
    1
  9605. AMAZING - this clip was just pure joy. thx for the video. alos yesterday i watched "die vermessung der welt." was great too.
    1
  9606. now i wonder what's the behavior of (sum n zeros series) as n tends to infinity
    1
  9607. Queen of hearts
    1
  9608. What was the deadline for the coding challenge to be included in the draw? I might have some free time in a few days to give it a go.
    1
  9609. Why are they equilateral triangles? Why does "picking up points at the tip of the star" necessarily form equilateral triangles? (I believe it and it's probably a simple proof, I just hope someone in the comments says it)
    1
  9610. Thanks!
    1
  9611. An easy (and general) way for the second exercise is to note that the numerator is of the form: (n-2)² + (n-1)² + n² + (n+1)² + (n+2)² if we were to expand the binomial squares, all the terms in the first power of n would cancel as they have opposite signs in pairs. So what we are left with is: n² + 2² + n² + 1² + n² + n² + 1² + n² + 2² = 5n² + 2*(2² + 1²) for n = 12, this is obviously 5*144 + 2*5 = =1440/2 + 10 = 730, which is twice the numerator of 365. That's what I did in my head in less than 30 seconds. Alternatively, one could do 169 + 196 or 100 + 121 + 144 in less than 30 seconds of mental calculation, and know from the video that it's half the numerator.
    1
  9612. Wow is the only thing i can say.... it is indeed a miracle!!
    1
  9613. This sequence is also the key to proving the 3n+1 problem.
    1
  9614. 44:35 Did you say "other things worth talking about" or "other things worse talking about"? Both fit.
    1
  9615. One can find Integers among the sums of the diagonals. It is not a magic square, but close. One diagonal sums to ~16 and the other ~14. Sums of ~7 and ~8 appear among pairs of elements.
    1
  9616. MathLoger using grade school math to prove that Hippees are stoopid.
    1
  9617. Many many years ago I read in a book by Ian Stewart ("Another Fine Math You've Got Me Into. . ") about the well known puzzle of crossing the river with a goat, a lion and a cabbage. If I remember correctly - or pause to think - this is isomorphic to Hanoi with three discs. Ian Stewart discusses Hanoi in that chapter and quotes Dr Andreas Hinz's wonderful article about the average path length of 3-peg n-disc Hanoi. I could not resist the temptation to try to prove the sqrt(17) infested formula and thought I discovered a mistake. I seem to remember there was actually a printing error in Stewart's book. Anyway, I spent many an evening with our 9-year old daughter on the balcony of our rented Swiss holiday chalet, enjoying the balmy evening weather and doing maths with a little girl, trying things out, sketching out attempts, thinking out loud, in short, something like real research mathematics. I wrote out my results and sent it to Dr Hinz who kindly sent me a reprint of his article with a friendly letter discussing my "paper" and gently pointing out one error of reasoning, in the form of a question, in the best Socratic fashion. I will never forget the joy of doing real maths - never mind the fatal error. I learned so much. Best of all was the wonderful interaction with Dr Hinz and his generosity. I cannot help but feeling sorry for anyone who has not had the experience of immersing oneself in the beautiful world of mathematical ideas. In my Covid-induced isolation I spend one hour each week with a high school student who wants to study mathematics. My aim is to cure her from high school maths and, after a fashion, show what maths is really about. After introducing her to the rationals and the notion of countable infinity, I showed her Cantor's diagonal proof of the uncountability of the reals. She exclaimed "But there are no holes between those fractions! Where do all those decimal numbers go????" Next, I will have to introduce her to Dedekind cuts and similar ideas. If I can convince her that the real line is very mysterious - and she needs at this stage little convincing - I believe she will be hooked on maths forever. Like me. High school teachers who never showed the young minds in their class something like Euclid's proof of the infinity of the primes, should be lined up against the wall and shot 😎. How many people end up mathematophobes without ever having seen real maths? ​ @Mathologer , you are one of my heroes - with Ian Stewart, 1B3B and the other soldiers fighting the good fight.
    1
  9618.  @Mathologer  Stewart hat wahnsinnig viel für den guten Zweck getan. Ich habe gerade ein paar goldene Oldies von ihm bestellt, die ich irgendwie vermisst habe, als sie veröffentlicht wurden. Mathematik auf YouTube ist großartig, eine Erleichterung unter all dem Unsinn. Lieber Mathologer, immer weiter so!!!
    1
  9619. This is an absolutely wonderful video. Thank you for your effort.
    1
  9620. Awesome !!! 🔥🔥🔥
    1
  9621. A Menger sponge is also infinite area but finite volume.
    1
  9622. 2 minutes in I remembered that I actually independently came to a formula for calculating the number of n-cubes in a m-cube. I then ran the binomial formula through my head and found that this trick works for cubes in all dimensions. Cool.
    1
  9623. Ever considered making black/dark background versions of your vids? would be helpful as a relaxing watch at night
    1
  9624. 8:29 The way you said "cell" made me think of Dragon Ball Z. 😄
    1
  9625.  @Mathologer  Thanks! Will definitely check them out!
    1
  9626. 2:33 HES GONNA SAY IT 2:37 :( 2:39 :)
    1
  9627. Was not taught this in school (UK) but discovered it when doing a final year project for my math degree. The context was computational, and I needed to find the area of some surface made up of triangles (approximation). I could access the point mesh easily so I used the coordinates of the vertices of each triangle to find the length of sides. To my surprise, I had no idea how to find the area of a triangle using only the lengths of sides without first using the side lengths to get an angle and then use one of the formulae for that. But I knew there had to be a more elegant way. I managed to derive the formula without much trouble, then make a function which could do the calculations at each iteration in the simulation. After asking around it seemed only about half the math profs/students at the university knew about this formula already. Definitely should be taught more in schools, it can be extremely useful.
    1
  9628. Booster comment
    1
  9629. I don't understand why Ptolemy theorem isn't taught more widely not only in western countries but in Asia countries as well. It's a much deeper theorem than the Pythoragas theorem.
    1
  9630. There are YouTube math videos and then are these Mathologer videos. 😮
    1
  9631. I wanted to like this so hard, I clicked thumbs up over and over - but stopped on an odd number so I actually liked it, but then I realized I had done it nine times so did it two more times so it was prime then had to do two more to get a number in fibonacci sequence :)
    1
  9632. Thankyou
    1
  9633. Thanks! Was the concept: sometimes while surface area is bound, length can approach infinity?
    1
  9634. The Bishop's proof. We've seen it often, but I am in respectful awe of the pioneers in the field upon whose blocks we now stack ours, striving towards an "infinte overhang" of knowledge and understanding. Thank you for the seal.
    1
  9635. There is a really big problem with that. It sounds far too much like the pronunciation for the sinc() = sin(x)/x function. As an electrical engineer, I see sinc() far more than I see sinh(). The difference between the pronunciation of the words sink and cinch can be subtle -- especially when heard from somebody with a non-English accent. So I have heard sinc() pronounced pronounced very similarly to cinch.
    1
  9636. Yes; no; [left the ah-ha for later]; treat the glasses as three boards?;
    1
  9637. The Gibson slide rule company built several circular slide rules in the 40's. One was a special version for submarines. The regular version gave same functionality as the sliderule style but the outer scale was close to 2 feet in circumference and you never had to redo like sliderules do for some calculations because being circular you couldn't "run off the end". It had a log scale on the outer ring such as he describes and two "hands" of clear plastic with a black line down the center. Moving the big had moved the small hand. Moving the small hand did not disturb the position of the big hand. I'm blanking but somehow that let you add logs to multiply just like a sliderule. I used one all through college as an engineer. Never saw another. Never saw any advantage to changing to a sliderule. I duckduckgo'd : "Gibson slide rule" and got a nice write up. I lived just a few miles from the "factory" but never went there - I regret it now.
    1
  9638. 9:56 3n+1, where have I seen that before? 😜 BTW nice video!
    1
  9639. My vote is the Animation of Kempner's proof
    1
  9640. To me the most memorable was the blue regions sliding to the one unit square to the left (and the fact that it’s more than half of it since the blue areas are convex). Great stuff!
    1
  9641. Thats the best Christmas video I ever watch!
    1
  9642. cool
    1
  9643. the most memorable part was the sum of any consecutive part of harmonic series can’t be an integer. and i still have some troubles to proof odd/even_1 + 1/even_2 is not an integer.
    1
  9644. Euler strikes again
    1
  9645. Made my day!
    1
  9646. I am not surprised by this news or revelation that it was not the Europeans that had this knowledge of calculus first. And I can even speculate beyond the facts we know of today anout the Indians and ask just how far back does knowledge of calculus really go ? And who did it originate from ? Were they human ? Before you scoff, let me remind everyone that speculation and the human imagination are the most powerful elements before starting down the scientific road to knowledge.
    1
  9647. 8:18 - I would like to test this against the Brachistochrone or half of a cycloid ... hm ...
    1
  9648. You think the Mathologer seal of approval will help me get into grad school?
    1
  9649. First time, I saw the proof of divisibility by 9 by the sum of its digits! Quite easy, but I have never seen it!
    1
  9650. 22:59 Why not take x^2 from the first 1 from the second x^200 from the third? Or x^4 from first, 1 from second and x^100 from third etc.? IMHO there are more terms giving x^400 than the ones you showed. Am I missing something?
    1
  9651. Have you considered which side the lace enters the eyelet (from top or bottom), esp under a constraint that both laces need to come from the bottom at the final eyelet?
    1
  9652. E-6B flight computer says "hello"
    1
  9653. (IMO) Mathloger is 1 of the best channels to exist for a idea, Note: Its just My own Opinion on the suggestion, Advice; "Feel free to exchange eachothers own Opinion even mine* to eachother".
    1
  9654. I followed what we call in France "Math sup Math spé", which is the equivalent of a math and physics license I think. All this stuff is amazing but it kind of bother me a bit that we manipulate infinity so easily without taking precaution :p
    1
  9655. How much each person needs the money should be a factor.
    1
  9656. Dang, for a sec I thought the title was lost NumberWang.
    1
  9657. cool!
    1
  9658. Geometry is so weird. So is algebra. Putting them together is, wow. Sorry. Exhausted. I'll have to watch again.
    1
  9659.  @Mathologer  I did. Still have to. Not a lot more rested. Dratted migraine.
    1
  9660. Did you know the uniform width of a plotted π annulus is equal to the inverse of the so-called golden ratio? For major x² + y² = 5/4 = R² and minor x² + y² = 1/4 = r², find the uniform width w = (R - r) = (√5/2 - 1/2) ≈ 0.618... for annulus π(R² - r²) = π. There is indeed in fact a geometric relationship between these two universal constants.
    1
  9661. Greetings from the 5th dimension
    1
  9662. The vortex cycle 1-2-4-8-7-5 is an anagram of 1-4-2-8-5-7, which is the repetend of 1/7 in decimal.
    1
  9663. Is 20C bigger or smaller than 4hours? That question doesn’t make sense because temperature and time are not comparable units. Neither are volume and surface area. So this “paradox” is really an attempt of comparing two incompatible units.
    1
  9664. So a new type of relativity. Unifying doesn't really mean correct. Things have connections that point to things that really are just effects, not causes. Is the point the effect or the line. That is the next step.. can we take the points and make them fluid yet keep the same area, or can we take lines and move them around. It seems it's the points.
    1
  9665. I am a victim of that crime. Neither proofs have been presented to me.
    1
  9666. I just stumbled across the sierpinski visualization of the solution a few months back after knowing about this puzzle for decades! Great video as always. I’m curious about nice visualizations off the table of minimal solutions leading to later minimal solutions—each is an integer partition weighed by the table values so I wonder how to elegantly display that.
    1
  9667. Now Burkard's German has a slight English accent ...
    1
  9668. Would have whished your videos were available when I studied math 20 years ago. Things become very simple when you explain them in the right way.
    1
  9669. Great video and fun narrative! At first I was daunted by the 50 min length, but the visualization makes everything easy to follow! I'm doing a Physics PhD.
    1
  9670. Colouring proof was much more intuitive for me! Transforms around a circle might be simple enough to show in a graphic but to do it rigorously seems more complex.
    1
  9671. In the beginning of the list of these primes I found a chain: 2, 2*2+1=5, 5*2+1=11, 11*2+1=23, 23*2+1=47. I wonder if there are more and longer chains among the Sophie Germain's primes.
    1
  9672. As someone with only a tentative grasp of algebra.. trying and failing to understand the maths here, I cannot help think that God must have had one hell of an alien brain! XD
    1
  9673. I prefer the second solution because it is more direct and is in agreement with the principle of Occam's Razor.
    1
  9674. You guys are amazing! Thank you for the amazing videos!
    1
  9675. Where's the paradox ?
    1
  9676. Great Video as always. I think people lean on the supernatural when they cannot explain interesting findings. Arthur C Clarke once said “Any sufficiently advanced technology is indistinguishable from magic”. Your explanation has the benefit of almost 100 years of math GIANTS behind it. It is also very necessary because superstition has no place in Math and Science until proven otherwise. .
    1
  9677. Since you asked, another example where the sum of the harmonic series comes into play: Because the sum of the harmonic series is finite, it follows that there is no limit to the sum of the factors of an integer n divided by n. For perfect numbers, this ratio is exactly 2. For so-called triperfect numbers, the ratio is 3. To get an integer where this ratio is greater than any given value, find the n where the harmonic series exceeds this value. A number that has all the integers up to and including n as part of its factors will exceed this ratio.
    1
  9678. Yey hello mathologer
    1
  9679. I really enjoyed this, though, i know you lost me about half way. Thinkni got stuck on a triangle tripling a square times plus or minus not enough or too much sleep. Will watch a few more times for sure.
    1
  9680. Loved the counterpoint music at the end.
    1
  9681. There are, of course, many lattices one might use, and other curves. Area of an ellipse is Pi*a*b, as Archimedes knew. I can hardly wait for the Mathologer video on Minkowski’s Geometry of Numbers. There are so many directions one might go from here.
    1
  9682. Is the number of generalized 4k+1=p solutions also prime as well as odd? If it is prime, are the number of solutions of the form 4k+1 or 4k+3 or mixed?
    1
  9683. Everything explained in this video is just Mega Beautiful
    1
  9684. круто! Интересно посмотреть на лектора! Однозначно максимальную оценку! Но не осилил....
    1
  9685. The most memorable was the optimal towers.
    1
  9686. Theorem: Every Mathologer video is definitely a piece of art. The proof is trivial and left as an exercise for the viewer :-) By the way, my favorite part was that we can assign gamma as the sum of the harmonic series. But I wonder isn’t the harmonic series the only one who can’t be assigned to finite value according to Riemann (Zeta(1))? 🤔🤔🤔
    1
  9687. I am starting to get posture problems like Nicole Oresme due to excessive obsessing with these videos. I died at 14:50 when you put the overhead over his head and deadpan stared at us!
    1
  9688. $ python -c 'print sum([ n**10 for n in range(1,1001)])' 91 409 924 241 424 243 424 241 924 242 500
    1
  9689. I certainly don't remember the twisted square as proof for pythagoras.
    1
  9690. I hope the thumbnail wasn't clickbait. I did some quick calculation before watching and have determined (by expanding square binomials in n) that it is no coincidence. (n-2)^2 + (n-1)^2 + n^2 - ((n+1)^2 + (n+2)^2)) = 0? n^2 -4n -2n - (4n + 2n) + [square and constant terms that are opposite and equal.] = 0? n^2 -12n = 0? This works whenever n=12.
    1
  9691.  @Mathologer  Not at all, it was excellent. Also, apropos de nothing, you're sponsored by Euler. Here I thought Erdös Numbers were nothing to sneeze at. Euler number = 1??
    1
  9692. Tristan’s Fractal
    1
  9693. What struck me as I saw the ears being added and the final regular polygon emerging is it looks like a type of self-healing. I don’t know how it would be applied in the mechanical or other realm, but it would be nice if it could.
    1
  9694. I made it to the very end
    1
  9695. 9:33 5-8+5=2 broke my mind for some reason...
    1
  9696. Why did I never learn hyperbolic functions in highschool nor my undergrad math education.
    1
  9697. second challenge. all we need to know is cos(pi/2)=0 and of course if we multiply a number sequence and one number is 0, the result is also 0. for odd numbers, say m=7 and n=9, you will get pi/2 in the fractions when j=4 and k=5. Or in general, you half them and round up. Since this term is guaranteed to exist if both numbers are odd, the result will be 0.
    1
  9698. Yeah, as soon as you had two positive then one negative term, I was troubled. It's like, "You have infinite positive and infinite negative terms, but my 'gut' says you have "more infinite positive terms" even though that's not exactly how 'infinity' works. Then the 'rearrangement of infinite terms' thing. I'd say, "Addition and subtraction are associative for any finite number of operations, but it is NOT associative for an inifinite number of operations." (even though that probably doesn't really stand up to scrutiny, it's how I look at it.)
    1
  9699. I made it to the very end
    1
  9700. Maybe itʼs too easy for IMO?))
    1
  9701. 2+2=4 and 2x2=4 is still mind blowing to me
    1
  9702. 14:07 the diameter of the circle is larger unless you're playing an illusion trick on us
    1
  9703. I like in depth mathematics topics. Not having formally done anything more complicated than differential equations, I often wonder about analysis, functionals, topology and a bunch of things I lack the adequate background to understand. I still remember with amazement a linear algebra teacher generalizing linear span from vectors into polynomials to show how you could use this to represent a function. I often wonder (or maybe he explained and I don't remember) if you can generalize this, and have a set of infinite "vectors" (that represents polynomial coefficients) and with those generate "any" function. In any case, I guess I miss the math that I didn't get the chance to learn in undergrad or gradschool. What worked and what not worked for me: I guess you did some jumps there in the end and there it required trust (he did the algebra, it probably works out); otherwise great work! If you are wondering what to show next, I guess I personally would like to know a bit about non-linear spaces and why it was hard for Einstein to come up with the general part of relativity theory; I am not particularly familiar with the difficulty of it and why, for instance it would be hard to adapt anything to a non-linear space, for instance, why is it hard to adapt quantum field theory to a non-euclidean space (in my head there is just some sort of jacobian somewhere and then you're done). tl;dr. great video. Got to the end, didn't even notice it was 50 minutes long. Cheers
    1
  9704. This hints at the holographic nature of our universe. The smallest piece contains the whole.
    1
  9705. Tristan is the best)
    1
  9706. Most memorable: the thinning out of the no 9s series
    1
  9707. Nice Czech. No worries :)
    1
  9708. The right boundary could also be interpreted as a forbidden move, since there is no card right to it. Interestingly, when the number of cards is even, then you still end up with all cards face up, since the total number of cards facing up would always change by two or zero, so it must be even.
    1
  9709. Only half an hour? You are slipping. :)
    1
  9710. I like the monster videos! But a short snack now and then is nice too.
    1
  9711. This is amazing in its simplicity and apparent obviousness... how it went so long unspotted (before the dot proof, of course... see what I did there?) is a testament to how the human brain works (or, often, doesn't work). My undergraduate degree is in cognitive science, and I never went to graduate school due to a combination of the effects of late-diagnosed neurodivergence (autism diagnosed in 2005, five years after graduating; ADHD diagnosed earlier this month) and shifting gears from academia to tech work the year after I graduated (which, again, was probably the ADHD talking). Mathematics, especially geometry and number theory, have always been fascinating to me, and your videos and those of 3Blue1Brown have done more to maintain that fascination than anything else. Thank you.
    1
  9712. Since volume and area are two different dimension, it is possible, a simple way is to take a portion of putty with a finite volume, now you can roll it into infinitely thin sheet covering a infinite surface area, simplest explanation without any maths involved
    1
  9713. \sum_{n=1}^{1000}n^{10} did it in seconds ( couldnt do it in my head though :/ )
    1
  9714. Listening to Morning Mandolin is a reason to leave the video rolling to the end 😁
    1
  9715. Noice video as always, and very clear explanations too 😃. Btw could be the Patreon list be sorted? In what order, it does not matter 😄😃.
    1
  9716. 2
    1
  9717. Always very good content of maths, thanks!
    1
  9718. We met years ago in Kathleen Lumley, Adelaide. Thanks God, I can watch your channel here. All the best.
    1
  9719.  @Mathologer  I did hear that just recently. I read from the internet that it closed in January 2022, as a result of the impact of the COVID-19 pandemic. I am sorry to hear that.
    1
  9720. I see why you chose to wear that shirt for this video! Good choice.
    1
  9721. Please make a video on abstract algebra
    1
  9722. love the t-shirt.
    1
  9723. knowing math makes you have super powers...literally
    1
  9724. Sigma
    1
  9725. Choosing any single fact out of that amazing world of couriosity would put me in disagree with myself constantly. It is like finding these constants in the infinities. I have to pass on this book oportunity, sry!
    1
  9726. It's (multi?) scale (independent?) (free?).
    1
  9727. Imagine that a 4x4 grid has the same color in the two corners along one side. That means that central 2x2 grid along that side will not contain the color in the corners. By extension for larger 4nx4n grids as you move inward starting with 2x2 grids from the corners when you get to the middle eventually the same problem will occur and the centermost 2x2 grid along the side will not contain all four colors. The same problem occurs in the central 2x2 grid when two diagonally opposite corners of the 4x4 grid have the same color. Again, by extension for larger 4nx4n grids as you move inward starting with 2x2 grids from the diagonally opposite corners when you get to the middle eventually the centermost 2x2 grid will not contain all four colors. Thus all four corners must have different colors to start with.
    1
  9728. Ofcourse you are an expert
    1
  9729. 11:09 “Oh damn no lacing, but we can fix that”😂
    1
  9730. At this point I'm starting to wonder if Fermat ever actually recorded ANY of his proofs lmao
    1
  9731. 13:57 I plugged the sequence 7, 15, 26, 42 into OEIS, and only got two results: "[ n(n-1)(n-2)/8 ]" (sequence A011890) and "Row sums of triangular array T: T(j,1) = 1 for ((j-1) mod 8) < 4, else 0; T(j,k) = T(j-1,k-1) + T(j,k-1) for 2 <= k <= j" (sequence A131076) No mention of Schwarz or infinite area in either one, which makes me wonder if you've found a new sequence.
    1
  9732. 31:48 "What does this formula say if you plug in a negative number?" If I take the recurrence relation F(n) = F(n-1) + F(n-2), run it thru a change of variables n -> n+2, and rearrange terms to re-isolate F(n) on the left, I get F(n) = F(n+2) - F(n+1). Together with the base cases F(0) = 0, F(1) = 1, in this way I can iterate the sequence in the other direction. That is, F(-1) = F(1) - F(0) = 1, F(-2) = F(0) - F(-1) = -1, F(-3) = F(-1) - F(-2) = 2, F(-4) = -3, F(-5) = 5, and so on; where in general, F(-n) = (-1)ⁿ⁺¹ F(n). Substituting a negative integer for n in Binet's formula gives me the same result as iterating manually in this way. 31:53 "...or a number that's not an integer, like π, or φ (hehe)?" Whoa, hold on there cowboy! Before I go trying to plug in craaazy stuff like F(π) or F(φ), I'm gonna first try plugging in an "easy" non-integer. Like maybe n = 1/2. I'll try F(1/2) at first; then maybe from that, I'll learn what I need to know to plug in F(π) or F(φ). I'm gonna use the form of Binet's formula you alluded to at 32:11, by noting that -1/φ = 1-φ & plugging that in to simplify the denominator to √5. I'm also gonna change notation by n -> x, just because n is often an integer by convention & I wanna remember that now we're generalizing to real arguments of F. That gives me this formula: F(x) = [φˣ - (1-φ)ˣ] / √5. Plugging in x = 1/2, and noting that x^(1/2) = √x, I get F(1/2) = [√φ - √(1-φ)] / √5. Wait a sec... 1-φ ≅ -0.618 < 0... now you got me taking the square roots of negative numbers!! If I take the principal square root √(-1) = i and factor it out from the √(1-φ), I can rearrange to F(1/2) = [√φ / √5] - [√(φ-1) / √5] i = √(1+√5)/√10 - √(√5-1)/√10 i ≅ 0.569 + 0.352i. Backing up a bit now, what if instead of factoring out a √(-1) from the substituted formula, I factor out a (-1)ⁿ from the original formula, i.e. F(x) = φˣ/√5 - (-1)ˣ (φ-1)ˣ / √5? Now I can handle all the "complex" stuff in that (-1)ˣ bit, as everything else is real-valued. If x = 0, I have (-1)ˣ = 1; when x = 1/2, I had (-1)ˣ = i; and when x = 1, of course (-1)ˣ = -1. This is like a unit vector rotating thru the complex unit circle... so to interpolate, I'm gonna use Euler's identity, e^(iπ) = -1. Then I get F(x) = φˣ/√5 - e^(iπx) (φ-1)ˣ / √5. If I wanna split this into its real & imaginary parts, I can use e^(ix) = cos(x) + i sin(x) ==> Re[F(x)] = [φˣ - (φ-1)ˣ cos(πx)] / √5, and Im[F(x)] = -[(φ-1)ˣ sin(πx)] / √5. So to answer your question for x = π: Re[F(π)] = [φ^π - (φ-1)^π cos(π²)] / √5 ≅ 2.117, and Im[F(π)] = -[(φ-1)^π sin(π²)] / √5 ≅ 0.042. Thus F(π) ≅ 2.117 + 0.042i. And for x = φ: Re[F(φ)] = [φ^φ - (φ-1)^φ cos(πφ)] / √5 ≅ 0.900, and Im[F(φ)] = -[(φ-1)^φ sin(πφ)] / √5 ≅ 0.191. Thus F(φ) ≅ 0.900 + 0.191i. In addition to F(π) ≅ 2.117 + 0.042i, I also get F(π+1) ≅ 3.226 - 0.026i, and F(π+2) ≅ 5.343 + 0.016i. If I add F(π) + F(π+1), that gives me (2.117 + 3.226) + (0.042 - 0.026)i = 5.343 + 0.016i = F(π+2). Would you look at that -- the original Fibonacci recurrence relation continues to hold! :)
    1
  9733. You like Gandhi
    1
  9734. Ah, it's mathologer time!
    1
  9735. Instead, if each block used simply increases in length by 1 unit, when infinity is “reached” all the blocks will suddenly disappear leaving only one block that is inset away from the edge by 1/12 of the block length.
    1
  9736. thank you
    1
  9737. great teaser/intro for hyperbolic rotations in special relativity
    1
  9738. Hello there. I'm just a stupid high school student and a bit of math video lover. I watched this video on the transit and it's really really good. I would say that sometimes I would get lost but would quickly jump back to pace. Good stuff.
    1
  9739. One of your best videos imho, thanks
    1
  9740. As mentioned, the venerable e6b is still used in aviation today, as a backup if all power or electronics dies.
    1
  9741. About a week ago I watched Another Roof's video about phi and now your video about phi
    1
  9742. ' I know! Let a computer do it. > Why not? We use calculators to do math & trig.. And then we can shop for groceries & do our quarterly taxes with ease..
    1
  9743. 6:15 no, remove the two bottom right black squares and the two green ones on top of them, that leaves a single green square disconnected and it can't be tiled. but wait, maybe thats against the rules, maybe they have to be connected. i think as long as the squares are connected by a side they can be tiled, if however you only leave a square connected to the rest of the board by a corner it will not be able to be tiled but that breaks the rules. just my guess
    1
  9744. nice videos
    1
  9745.  @vik24oct1991  While I understand the point you're making, your mathematics is incorrect here. In the scenario where there is one person owed $200 million, and ten people each owed $1 million, and there is an account of $20 million to be divvied up, then the Talmudic solution is the following: Each person owed $1 million will receive $500,000, and the person owed $200 million will receive $15 million.
    1
  9746. 3-4-7 is far more superior to 10-3-7
    1
  9747. Do you know that twisting inside your skull when you watch something that just rearranged your brain? That. Thankyou.
    1
  9748. nice
    1
  9749. The nice diagrams with volume splits reminded me of atomic orbital population dynamics, particularly when 'pumped' optically, as in Einstein occupation numbers. I wonder if Einstein was mindful of the contested garment and talmudic wisdom for bankrupt estate settlement .. without playing dice ;-}
    1
  9750. 1:18 That third-degree difference formula, for some reason, is used in the MMORPG Tibia for the formula of the required experience for a given level.
    1
  9751. what does it mean a stacked pair? i thought the number is repeated like (3,3) but why does (1,7) and (4,1) is a stacked pair?
    1
  9752. I do not get, why the squaring of the circle is supposed to show, that the circumference is 4. That is not a lim of the outline of a circle. The area removed from outside of the circle should equal the area removed inside of the circle. But you are only reducing outside of the circle. So starting with one square inside sin(45°)² and one outside (√(2)-sin(45°))², which divides the diagonal by the ratio of 0,7071, which is √(0,5) or √(2)/2.
    1
  9753. @Mathloger - there's still a space for a significant improvement there with this somewhat discreet distrubution change presented now - wouldn't it be better if the vessels had a continuously curved walls instead of such a nasty steps ? What kind of curve would it be and how weighted values of its parameters to guarantee a continuous transition in the split with the overall part of the original debt change from 0 to 100% ?
    1
  9754. Really good video, thank you (: 29:35 - is it just me or is that 7 supposed to be a 5, and that 9 is supposed to be a 7?
    1
  9755. 19:44 good place to stop?
    1
  9756. Watch video. Few hours later. "Wait. Could this be used to trisect an angle since there is a 1/3 value baked into the geometry."
    1
  9757. Math is endlessly amazing. Chess without checkmate.
    1
  9758. Surely anything that gets back to where it was in the beginning has gone through angles that add up to a multiple of 360 degrees? OK I thought straight away the points angles add up to 180 degrees because the more of them there are the smaller the angles are so probably stay adding up to the same amount. Also they are all points on the edge of a circle, so the angle to those 2 next points is half what it would be at the centre of the circle and those add up to 360 degrees. If some other number (besides 360) had been chosen for how many degrees there are in a full rotation it might have been one that has more prime factors including 7. (It's awkward dividing 360 by 7, 11, 13 etc.) Then more angles would have been integer numbers of degrees when dividing the whole circle by different numbers. None of the [small] angles in the exact integer right angled triangles (all 3 sides integers) turn out to be an exact number of degrees - they are irrational.
    1
  9759. You might keep the new Logo for a second channel. ;-)
    1
  9760. Anyway, ignoring the French
    1
  9761. In Bulgarian this device is called "Logarithmic Line" Thanks for the video, great as always!
    1
  9762. Saw the title and thought "hmm... that will be terrible...." Saw it was by mathologer. Nuff said.
    1
  9763. I had a similar type of problem in my high school research project last year and got this long recursive solution. https://math.stackexchange.com/questions/3551099/choosing-numbers-with-the-sum-s/3817369#3817369 I'll try to simplify it using things i've learned in this video.
    1
  9764. I do truly have to comment. The way you explained this l, was a) amazing b) very enlightened. Thank you please keep up the amazing work. I've learned much for one single video. .
    1
  9765. To make such overhang with only three blocks place the upper two at the lowest level and place a third one at some angle so it contacts the previous two.
    1
  9766. Great! As ever you do...
    1
  9767. Doesn't it bother you that all the eights in the video are flipped horizontally?
    1
  9768. It could also be said that n^(_k) = n!/(n-k)! ? With n^(_k) I have indicated n to the falling power of k
    1
  9769. It does have to do with dimension. For example a simple rectangle with finite area is made up of infinitely many line segments. Finite in 2d, infinite in 1d. So we can paint it but not color it in with a pencil. And these paradox problems are similar. Enclosing the volume with tiny 3d volume cubes is always possible, and a limit will exist as the cubes keep decreasing of the tiny size times the number needed. Covering the surface with with tiny 2d (flexible) rectangles is never possible.
    1
  9770. One thing I didn't quite get at several points, most notably at the very end of chapter 3: were you saying that 2, 4, 10, 28, 82 were the ONLY special numbers for the 3-color game — that is, that 244 (the next one in 3^n+1) isn't actually special?
    1
  9771. The chapter 6 "greedy formula" is pretty much what i spent a lot of time doing at school with my calculator, eccept with decimal places in order to calculate pi or the log of a number, sometimes the square root, by comparing it to the calculator's value i got closer and closer. That is untill the Calculator ran out of significant figures.
    1
  9772. Hi mathologer!
    1
  9773. A book that uses visual math: Visual Complex Analysis Tristan Needham From the book: Imagine a society in which the citizens are encouraged, indeed compelled up to a certain age, to read (and sometimes write) musical scores. Al quite admirable. However, this society also has a very curious-few remember how ti al started— and disturbing law: Music must never be listened to or performed!
    1
  9774. You should write a follow-up book about wobby tables theorems.
    1
  9775. Thanks!
    1
  9776. Triangles are serious fun. Thank you!
    1
  9777. is there a maximum depth we can advance if there are infinitely many circles below the line?
    1
  9778. I can't get over his posture. Never seen someone hold his hands up like that
    1
  9779. I taught 7th and 8th grade math, so my students and I had such a fun time discovering arithmetic ways of solving rich problems or, more often, answering a student’s question of why , that were traditionally relegated to algebra or calculus. Thus I think: It’s more about how we were taught to solve problems and whether we were encouraged to verify in other ways.
    1
  9780. Oh holy crap. Holy CRAP. Less than three hours ago I was poking around YouTube thinking "why do i have these three 'magic trick' videos in my 'maffs' playlist, and why does the happy little kitty say 'QED' on him and oh wow REALLY now oh man HOW long has it been oh jeez a whole month i hope things are okay…" and now this video!
    1
  9781. I remember taking a problem-solving seminar, which included prompts for proving different divisibility tests, including the general case of the divisibility test for k-1 in a base k system, so when you got to that section, I was screaming it excitedly! This was a delightful video. I'll try out those questions.
    1
  9782. i like your MOTH t-shirt
    1
  9783. Hey, you're coming up to one million subscribers! Rightly so, a brilliant channel.
    1
  9784. Many years ago I built a slide rule using octal numbers. Still working.
    1
  9785. that voice dub at 30:10 really caught me off guard lol
    1
  9786. Language of gods
    1
  9787. My maths course (so long ago) went: Evaluation of finite sequences Evaluation of infinite sequences, including testing for convergence Evaluations of infinite sequences "in the limit" Calculus!
    1
  9788. One word - AMAZING!!!
    1
  9789. Your shirt is bothering me a lot! The arrows are quite chaotic. Can't you just swap rock and scissors, so we all could enjoy some nice regular arrows? Please?!?
    1
  9790. WHat works for me is the geometric intersection of trig. Too bad I don't know Trig...
    1
  9791. i went to sleep watching this for some reason i think i ate something to much potasium and i fell asleep dreaming about this.
    1
  9792. ​ @Mathologer i actuly had the most riddiculus nightmare ive ever had. But other than that no i did not sadly. i had a pretty severe headache after that.
    1
  9793. Amazing video, resparked my interest in higher dimensions and got me researching again!
    1
  9794. Must the chapter transitions be so ominous?
    1
  9795. So cool! Thank you Mathologer!
    1
  9796. Yes, the math is beautiful. But, please for the love of music, what is the name of the background music?
    1
  9797. I can never turn down a number theory video. Truly the Queen of the sciences!
    1
  9798. To solving Euler project I go
    1
  9799. Phenomenal! What a surprising thing!
    1
  9800. His shirt is advertising constant sin.
    1
  9801. Sir your videos are just so amazing,I just am aspiring to make animations as such,please please tell me which software u use,it would mean a lot to me,
    1
  9802. Wow, I don't usually manage to make it to the end of masterclass vids without getting completely lost, no idea why I did it this time. Fascinating subject, loved it! (Background if curious: High school education up to pre-calc, technically calculus if you count learning the barest rudiments of polynomial derivatives, but I don't since it didn't even cover integrals. I'm very autodidactic, everything since then has been informal and self-directed.)
    1
  9803. The scaling by a factor of 4 is the thing that confuses me
    1
  9804. I really appreciated this treatment of a favorite subject of mine. Thank you. It is a very different geometrical take on the subject than my own. I love particularly the whiff of the historical logarithm approach in it. *sigh*. I have GOT to find time to get back to my videos. You would appreciate my approach, I think. My (later, still-unpublished) proofs of the same concepts are quite different, highly geometric, and (I think) at least as mathologer-like as the ones you gave here.
    1
  9805. I discovered this several years ago while sitting in a restaurant bathroom looking at the black and white checkered floor, that squares are the sums of odd numbers.
    1
  9806. Just a quick comment as i'm going through the video. For a colorblind person like me (deuteranopia) the orange and green colors you're using around 16:50 look very nearly exactly the same.
    1
  9807. 5:29 Well, actually no, the approximation will not yield the exact answer. It will near the answer but never be the answer. The answer would be really really close, and yet for each distance >0, the curve would yield straight line from one point to another, and although, when fine enough, approximate the answer even by whichever number of digits, sooner or later the digits would fail to represent the answer completely. That's why it's approximation.
    1
  9808. I love this channel.
    1
  9809. Most memorable: The bizarre overhead towers, and in particular the parabolic tower with cube root growth.
    1
  9810. The equation you show in the end isn't the same as the initial one. There the denominators of the first part, only had odd integers. Now all integers are shown. Small thing to correct, I'm afraid.
    1
  9811. 1/x is convex for x>0, so γ>1/2
    1
  9812. My Uncle gave me a circular slide rule 55 years ago. I still have it..
    1
  9813. 24:00 Here's another proof: connect the centre to the vertices; you'll get an inscribed quadrilateral because its opposite angles are right; the sides that are the halves of the diagonals of the square are equal, so they subtend equal arcs, therefore, the angles based on them are equal, i.e., the line is the angle bisector, and so is the line we constructed originally, implying that they are the same
    1
  9814. Sensational presentation. Just 3 points. 1) The sine notation, which makes such neat formulas, was invented by Indian mathematicians. Prolemey's trigonometry was much more cumbersome as he used the chord instead of the sinus. I am sure you know this, but maybe it is important to emphasize this to clarify what a great mathematician Ptolemy was. 2) I would recommend also the 2 books by Aaboe that were my introduction to the subject: Aaboe - Early episodes in the history of mathematics Aaboe - Early episodes in the history of astronomy 3) What tool do you use for drawing production?
    1
  9815. @Mathologer what if we put non integer values for m and n in the moster formula?
    1
  9816. Tight lacing formula = (x!)^2 Where x is equal to number of pairs?
    1
  9817. Now the sides and three different diagonals of a nonagon; I was kind of expecting the video going towards all "odd-gons". Works in a highly similar way... (Actually, the triangle is somewhat of a degenerate case.)
    1
  9818. Exciting proof.)) Especial longest. 😅 I even try it! 😄 In end of video I had remember the scene of the Beautiful Mind movie, when Nash say "There could a mathematical explanation for how bad your tie is". 😊
    1
  9819. I would never have called an integer optimisation problem a "dream"!
    1
  9820. The most memorable thing for me was the leaning tower, because it reminded me of a physical puzzle in the museum at Norwich, which my son solved when he was about ten, in which one had to build a bridge over a river with a limited number of "logs", high enough for a boat to pass underneath. He's now in the process of choosing a university at which to study engineering, and I'm so proud of him! :D
    1
  9821. I really liked the part where you showed why 0.5<γ<1. So simple.
    1
  9822. Most interesting- it was a pleasure to watch to the end
    1
  9823. Somebody needs to make a ytp of this video
    1
  9824. As you pointed out, we now need to denumerate all tilted squares, and one way to do that is iterate over the rise and the run of the top line. We have already completed some of them! I'll use the notation (rise,run): how many. Note that run can be taken to be positive WLOG, but run must be negative and positive. Also, the sum of the absolute values of rise and run must be <=4, else it wont fit! We will also find, for reasons that become obvious when you try it, that the number that are present is only a function of |rise|+|run|. Also (-x,y) turns out to be the same square as (y,x), except the coordinates are now the right side of the square not the top. (0,4):1 (0,3):4 (0,2):9 (0,1):16 (1,3):1 (-1,3=3,1):1 (1,2):4 (-1,2=2,1):4 (-2,2=2,2):1 (1,1):9 Total: 50 Now for a formula -- sum from (run=1) to n: sum from (rise=0) to n-run: (n+1-rise-run)^2=n^2+2n-2*n*rise-2*n*run-2*rise-2*run+1+rise^2+2*rise*run+run^2 Now completing the sum offline (totally didn't use wolfram alpha): (1/12)*n*(n+1)^2(n+2) And we can confirm that this gives 50 for n=4
    1
  9825. The spacing of laces in exploded lacings seems rather arbitrary
    1
  9826.  @Mathologer  Very extensively detailed! Change and adapt some things to make it look like your own idea and you can submit it to a Hollywood producer.😎
    1
  9827. The beauty of this proof is that, back in Pythagoras time, all math was based on geometry.
    1
  9828. 9:08 it was also working before, but cute cheat Prof 😏
    1
  9829. The first method made pi=4 is correct for a higher dimension: fractals have higher dimensions.
    1
  9830. One thing that doesn't get taught much is how logarithms were used in practice by engineers in the past before calculators were a thing, to more easily make calculations that would have otherwise been much more laborious. Could this be a good idea for a video? Showing such practical real-life examples where logarithms were being actually used? This would make logarithms much less esoteric and abstract, as it would show their practical uses.
    1
  9831. What if the garment is plus-fours?
    1
  9832. Use it as a washcloth?
    1
  9833. Back in those days patches of cloth was a thing.
    1
  9834.  @user-iu1xg6jv6e  bury it in the woods before the authorities find out?
    1
  9835. Was definitely impressed with how elegant the no 9s proof visualisation was
    1
  9836. Beautiful! This is exactly what I was trying to show in my comment to your Moessner Miracle video, that summing a sequence is roughly equivalent to integrating the sequence.
    1
  9837. It's funny how common dreams about proofs actually are, most of mine were nonsense, but multiple times I've woken up to having the right idea to a proof I struggled with.
    1
  9838. Using a 3-4-5 triangle to check right angles has been around since when the ancient Egyptians used it to re-mark out their post-flood fields. Yet still, I have not yet realized that I should use it to check squareness of room corners during home remodeling! Our usual remodeling assumption is "Nothing is square, nothing is plumb, nothing is level, not in this world." Personally, trained in the long-ago to reify Physics, I blame General Relativity when non-Euclidean geometry becomes involved in construction.
    1
  9839. Seen a billboard just now in Philly. You wouldn't happen to be the man on it would you?
    1
  9840. https://www.youtube.com/channel/UCt8XLqRKsLSyr1gkPIaNXng This is for the lovers of science
    1
  9841. yo, square root of 2 be craycray!
    1
  9842. Puzzle for really advanced people: Find the equivalent formula for a Penrose tiled plane OR prove it doesn't exist (Start by considering what you need as dominoes)
    1
  9843. It's always a joy to see these videos come out -- I remember reading your articles from the days of QEDcat on The Age.
    1
  9844. Interesting!
    1
  9845. Please Turkish subtitle!
    1
  9846. Hello Mathologer! Wish you a very happy new year! <3 from India
    1
  9847. I wonder if it is just a coincidence that only the regular polygons that can fill the plane are the ones that can appear on the square grid in 3 or more dimensions.
    1
  9848. The video was amazing as usual. I actually learnt the Pigeonhole Principle from a book, but your representation was very clear and fun to observe. Though I didn't get what was special about 113355...
    1
  9849. Fun Fact: the shape formed by the simultaneous squares-and-triangles may, or may not represent a 2-D version hyper cube rotating in 3-D space.
    1
  9850. (HA)3 = LOL
    1
  9851. Sembra la scala Logaritmica
    1
  9852. Frohes neues Jahr und vielen Dank für all die schönen Mathe-Videos!
    1
  9853.  @Mathologer  that's the comment that inspired mine. That obviously means they never watched the full video before commenting. You could maybe even say that their comment was clickbaity😁
    1
  9854. बहुत सुंदर❤️💐👌
    1
  9855. Excellent Thanks sir
    1
  9856. My favorite chapter was chapter 1.
    1
  9857. Can't stop looking at his shirt!
    1
  9858. I want to major in maths in college and boy is this scaring me out of it and pulling me in more at the same time.
    1
  9859. @2:20 I'm missing something on the explanation here. You say to look at the products of pairs, but 3 is the only number you obtain that way. You get 4 by taking 1x2 and doubling it, for no reason I can see other than because that is what it takes to get 4. Then you seemingly change the rules again to get 5, now adding 1x3 and 1x2.
    1
  9860. I like the first proof better, it is more elegant
    1
  9861. 7:05 The 100-digit prime ends in 81 so it's of the form 4k+1 so can be written as the sum of two squares?
    1
  9862. My wife gave me a slide rule - purchased online from Eastern Europe - for Christmas, but I wasn't sure how it worked. It is in great condition and has a few extra marks made on it (i.e. pi and e) from the previous user. Thanks to your video, I now know at least some of the first principles associated with using it.
    1
  9863. How many ways to make change for Graham's number of dollars?!?! Might take a few more tricks to solve that one!
    1
  9864. I've seen the crazy maximum overhang towers before but still love them. I honestly don't know what was best here. It was all great. So I'ma just go with the towers :)
    1
  9865. Now I have to code it myself !
    1
  9866. 11:07 - priceless. You sir, are a legend!
    1
  9867. Before 17:27 I was thinking we could identify equivalence classes of polygons. For example maybe some set of pentagons that are one step of 72deg ears away from becoming a regular pentagon, another set of pentagons that just need the 72deg step followed by the 144deg step. I was imagining these steps would be the basis of some group actions on these equivalence classes. Of course, given the explanation that what these operations do is actually zero out components of a linear combination, the operations don't generate a group. I guess since these operations are associative, not invertible, and can't make a neutral operation, it is a semigroup. Since these operations are linear transformations, it should be straightforward to see what these equivalence classes look like. Polygons in the same equivalence class in the sense I described would basically match in being zero or non-zero in the eigen-polygon components. It might be interesting to take a look at the polygons 1 operation away from a regular n-gon. And it should be relatively easy to visualize them... we just need a 2D "slider" for each component, set all but one of the components to 0.
    1
  9868. These "corner cuts" are so wonderful and promising for explaining physio-biological fundamentals!
    1
  9869. A physicist, a biologist and a mathematician are standing outside a pub. They see 2 people enter and then 3 people leave. The biologist: The 2 people must have reproduced. The physicist: There was clearly a hidden variable we hadnt accounted for. The mathematician: There are now -1 people in the pub.
    1
  9870. Nice.
    1
  9871. 23:25 and you just made my day with your awesome explanation, walk-through and visualisation. Such a privilege to watch you, thanks for sharing
    1
  9872. leaning tower of lire is my fav. and your t-shirt.
    1
  9873. I made it to the very end (in... 2-3 sittings 😅) Great video, thanks for all the dedicated explanations.
    1
  9874. Dance, dance, little dots.
    1
  9875. LIKED. AND. SUBSCRIBED. 💪💞🌟
    1
  9876. As a proof for the similarity of triangles at 28:30, start with T1 and T2 similar triangles (with size ratio r), overlap the red points at the origin and put the black points on the positive x-axis. bc all T2 edges are T1*r we can add the vertices pairwise to form T3 of side-lengths T1*(1+r).
    1
  9877. 10:02 thus ln(0) = negative infinity
    1
  9878. ❤
    1
  9879. 4 min, 437 clicks.
    1
  9880. My favorite is the visual explanation of the Euler-Mascheroni constant and how the approximation becomes better and better. So neat.
    1
  9881. I only learned the basic divisibility by 9 test not the extension back in the school. I am from Palestine BTW. Really nice 👌🏻 to know it after those years.
    1
  9882. Congrats on 100!
    1
  9883. the 3-4-5 triple is a quadruple. A=3 B=4 C=5 AND AREA=6
    1
  9884. Kempner's proof was a masterpiece, so easy to follow.
    1
  9885. what about monopoles? lol
    1
  9886. Is there a general formula OR procedure for calculating for any estate, any number of creditors and any number owed?
    1
  9887. Hey there! As the year comes to a close, I wanted to take a moment to send my warmest wishes to you. May the upcoming holiday season bring you joy, relaxation, and cherished moments with loved ones. May you find inspiration and creativity in the coming year, and may it be filled with success, growth, and exciting new opportunities. Your content has been a source of knowledge and entertainment for many, and I want to express my gratitude for the valuable content you've shared. Thank you for being a part of this online community and for your dedication to your craft. Wishing you a peaceful and joyful holiday season, a fantastic New Year, and continued success in all your endeavors. Stay safe and keep up the fantastic work!
    1
  9888. I made it to the very end, but on third take ;)
    1
  9889. 8:14 wait, i never got taught that in school!
    1
  9890. is my intuition that a regular nonagon ratios would follow a 4D cutting pattern and yield a similar set of identities? And is there a generalized way to write an equation that somehow covers all regular polygons?
    1
  9891. This is a great question, and I also wonder if there are any fractal-like paradoxes with horns with volume slightly less than 3 dimensions, and surface area slightly larger than 2 (is that even possible?)
    1
  9892. Finding optimal overhang stack is my favorite part :)
    1
  9893. I looked at this from an angle view for your triangle puzzle and geometry. There would have been 120 degrees or 30 degrees from a 90 degree right angle. No matter how many dots or how large of a grid you can never touch 3 dots to make the triangle. No combination of bisected will work. 90 degree to 45 degree to 180 degrees.
    1
  9894. You are the best math teacher I had in my 40 years of life :-)
    1
  9895. 1:46 if you remove the circle and the color, you get a triangle with whiskers, which is also a union of three lines. Adding the colors back in, each of the lines is made out of one red, one blue, and one green. Something something rotate the lines
    1
  9896. Fascinating. Thanks for the video. If I understand hypercubes, no matter what the dimension, a face is always enclosed by 4 vertices, never three or 5 or any other number. And a cell is always enlclosed by 8 vertices and 6 faces, never any other numbers. What about other Platonic solids, like tetrahedra? I imagine there are n-dimensional hyper-tetrahedra. yes? What about non-Platonic solids, like a Buckyball, in which the faces can have either 5 or 6 sides? Are there n-dimensional hyper-Buckyballs?
    1
  9897. Great as usual! There is stuff for lots of videos, not just one.
    1
  9898. I really wish that you were one of my high school maths teachers. These videos are always great.
    1
  9899. Plot twist: Mathologer is the original creator and is now gloating. 😹
    1
  9900. One-glance, no question
    1
  9901. 26:09 all number can be expressed as binary in a unique way, so just pick powers of 2 which are 1, 2, 4, 8, 16, 32, 64.
    1
  9902. I vote for kempers proof. I didn’t know that before and it’s Great.
    1
  9903. I don't understand the coin thing still. What justifies the extra +1 or -1 rotation from coiling? The large circle is already a circle, it's not like the smaller circle is being wound around a line segment as it becomes a circle. To me, "no slipping" means a rolling object covers the same arc length on itself and the surface it's rolling on, which would make the 0.5 radius circle go around the 1 radius circle once
    1
  9904. high school geometry
    1
  9905. This is amazing!
    1
  9906. It actually is easy. One of the most fun subjects I've ever studied.
    1
  9907. I had heard of euler constant before but never knew what it was exactly... until this video. Great video with chapter 5 as my favorite !
    1
  9908. In today's episode, the Cult of Musk discovers Numerology
    1
  9909. I'm from Bangladesh and I learned Heron's formula when I was in 9th Grade
    1
  9910. After a lot of fideling I managed to stack four boxes of playing cards such that the top one just about cleared the edge. Very fun challenge I must say.
    1
  9911. I wish I could hit like on this video like four times (minus one clockwise).
    1
  9912. I preferred the coloring proof.
    1
  9913. Every one press like 👍
    1
  9914. 17:49, not obvious at all to me :(
    1
  9915. Random circular mazes https://cp4space.hatsya.com/2014/10/29/random-circular-mazes/ Permanent determinant method https://cp4space.hatsya.com/2013/03/09/permanent-determinant-method/
    1
  9916. Very nice. But over 10% of us are colourblind.
    1
  9917.  @Mathologer  I was thinking more of the epicycles than the geocentric belief. As you suggest, the centre is almost arbitrary, but Kepler's ellipses revealed Ptolemy's epicycles for the folly they were. Of course, epicycles have made a recent comeback with those elegant outlines drawn by cycles on cycles on cycles...
    1
  9918. Yeah yeah yeah, but how do I cut up the big cross to get the little ones?
    1
  9919. Scale of 1 to 10? I think it's really deeply beautiful, but I'm afraid I'm compelled to give it a 4...
    1
  9920. I think the best system is to pay the lowest claim in full, then the next lowest, etc. until the money runs out. That way the largest creditor takes the greatest risk - which means the more you loan the more careful you should be in assessing risk. Prudent lenders surely do just this? If I lend you 10 coins and you default, I don't care. If I lend you 10,000 , not so much ! PS Reached the end.
    1
  9921. I love how these first chapters are thing I'd seen before but thought unrelated, and the later you blow my mind sliding all those blue wedges over to make Gamma. I did however think of why it is at a glance greater than 0.5, since every wedge can be expanded to a rectangle and together cover the 1 area square, and each wedge fully includes the triangle that would be half the area of that rectangular section.
    1
  9922. A great way to start Christmas morning - thank you. Also i really like many of the teashirt designs. Could you not put the digital design files for all your t-shirts on sale in your online store so that people could buy and then send to an online t-shirt producer to get both their prefered t-shirt and design. It would expand your market. Best wishes for the new year.
    1
  9923. Índice
    1
  9924. Brilliant !! This was precisely the video I had been waiting for all my life !!!
    1
  9925. I don’t know why, but i could also see the Newton’s method to approximately calculate the sqrt(2).
    1
  9926. That's what I got it as 🥰
    1
  9927. why the queen?
    1
  9928. I've used these tools my whole life, so the error for 3.14 made me laugh. The animation stopped on 3.18.
    1
  9929. I am a maths grad student and I love your videos. I understand that you simplify a lot in your videos to cater to audiences of all backgrounds, However it would be much appreciated if you could designate a time post which you explain higher mathematics.
    1
  9930. I said I would only watch the first 4 Ch but hung on to the end easily. Given me a great outline for about 3 hours of algebra exercise. M Stat.
    1
  9931. 24:52, gamma must be greater than .5, because the sum of the difference of the first six terms exceeds .5, in fact its equal to 0.5041
    1
  9932. Man... you must have been tired making this video. 5:49 the "1" in 3×2×1 is visibly too high. And the two "1"s in 4×2×1×1 are visibly different distances from their preceding "×"s.
    1
  9933.  @Mathologer  Don't worry for me. I notice such things but don't care about them.
    1
  9934. Where can I buy one of these? Its accurate enough and a lot faster than using a calculator
    1
  9935. I have a masters in physics, and was worried that the later chapters would be like a steamroller of maths that would require both more knowledge than I have and time to ponder. However, I found it easy enough to follow in a video format like this. In fact, this way of making equations animated is probably better at quickly illustrating ideas like this than a lecture on a blackboard. You might be able to simplify things by introducing even more new notation, but I've learned that such a way of conveying ideas is very quickly the opposite of illuminating.
    1
  9936. 11:57 “Let’s switch to Genius Mode…”. — love that quote!! 😅😅
    1
  9937. 25:00 i can tell you why euler-masc constant is obviously greater than 0.5 When you take the first "blue wave", the unit square intersects the curve at x=1 and x=2. It intersecets 1 at 1 and 2 at 0.5. That means you have a rectangle of units 1 and 0.5 that contains the first blue wave. But 1/x is monotone decreasing, that means that inside that rectangle, the curve lays under the diagonal, giving the blue part a bit more surface than half the rectangle surface. This pattern of rectangles of side 1 and 1/2, 1/3, 1/4 repeats, meaning the sum of all blue parts has to be greater than 0.5x the sum of all rectangles's area, which is the area of the unit square 1. Hence the sum of the constant is over 0.5, less than 1
    1
  9938. Made it to the very end.
    1
  9939. Thank you for this kind of video, it really helps to drive home some of the concepts that are often explained in such convoluted terms
    1
  9940. Runmi
    1
  9941. 1 * 2 < 1 + 2 1 * 2 * 3 = 1 + 2 + 3 1 * 2 * 3 * 4 > 1 + 2 + 3 + 4 The balance point happens to happen on an integer. Obviously the video is going to talk about far more interesting things than the answer to the title :P
    1
  9942. Ive been wondering this for years lol even made a spreadsheet to add the nexr odd to get the next square thinking ir could be a way to figrue out square roots if somehow the odd number in the sequence was related to the sequence of the square somehow
    1
  9943. It's VERY beautiful
    1
  9944. Square root of 666 is 25.8070, according to my calculator.
    1
  9945. Using a harmonic series bricks to optimise a thrust vector.
    1
  9946. AxA+BxB=CxC
    1
  9947. Could it be working because of 24303030303…
    1
  9948. “And after all… You’re my Number Walllllll”
    1
  9949. why 4k + 3 and not just 4k - 1?
    1
  9950. Modulo Arithmetic , useful for checking results of computation. Base (10=1) Tesla had some strange ideas, such as perpetual daylight. He did not invent the Alternator, but did use it to good effect.
    1
  9951. “Where does he get those wonderful toys?” 🙂
    1
  9952. I am ignorant, but why is (2y)^2 = 4y^2? (@ time marker 11:23) Until I understand that the rest is lost to me..
    1
  9953. It works for me. How many times to someone come right up to the point of inventing calculus and simply fail to do it? It's amazing.
    1
  9954. Aspirin please!! Can you made a deal with some company, you promote their headache remedy and we get a promotion code for 20% off? Would help both sides!
    1
  9955. Insane video, loved it and understood everything quite well. I'm currently preparing exams (physics, mathematics, English, &c.) to enter an engineering school (I'm French and that means I'm two years older than someone who ended high school). Voilà. I'm looking forward for more videos on Euler-Maclaurin and all topics surrounding calculus.
    1
  9956. Kenyon taught me multivariable calculus!
    1
  9957. El pi(π) es solo la mitad de la sumatoria de los lados de un polígono regular que se logra al calcular los lados de polígonos anteriores de menor perímetro sepuede empezar con un pentágono, cuadro, hexágono y se usan algoritmos como √2-(√4-l²) Pero la idea de tomar a π como la relación perímetro diámetro de la circunferencia está equivocada pues una circunferencia es una curva cerrada no un polígono regular de infinitos lados Es acaso que una línea recta y una línea curva son lo mismo creo que ni en la concesión mental No entiendo porque los matemáticos y científicos creen que todo lo deben analizar con un teorema para hallar el valor de líneas rectas el teorema de Pitágoras 3,1415... Es un error
    1
  9958. If you have the guts, make a video on Ito's stochastic integral... If you dare...
    1
  9959. You have today 666 k subscribers
    1
  9960. Hey, mathologer's bank! Awesome, right?
    1
  9961. Simple answer that i just found right away: 1+2+3=6 and 1×2×3=6 Try doing it from the left first
    1
  9962. The volume sucks (pun intended).
    1
  9963. I feel like I am missing a lot at discovering maths because I don't know how to write little programs to calculate these kind of things. I really am curious, and I love Mathologer's videos, but I also lack some of the knowledge needed to understand everything at 100%. Any tips, exercises, tutorials or anything like that I could follow to help me? I really am ready to invest some time into this, I really like playing with numbers, thanks in advance! :)
    1
  9964. Today is when I saw this proof and the squares
    1
  9965. Really enjoy your videos. Some are a little too technical to me - Im a family Dr (sorry re Dr in my your tune username- its linked to my work email!) who loves mathematics - not particularly gifted but enjoy it all the same. Ive always thought that people who really understand something can explain it to dummies in a simple manner and you are the proof of the pudding!! Thank you
    1
  9966. In Sweden we have 1, 2, 5, 10, 20 ,50, 100, so if I really anted to know, I'd have to do a different formula. Nice.
    1
  9967. OMG! You just answered a 40 year old question from calculus class. We were studying the techniques for deriving the area under a curve and the volume of surfaces of revolution. In the former the shape of the partition needed to be a trapezoid and in the latter the partition needed to be a frustrum of a cone. My question at the time was, since the explanation given of why this works was because when integrating the width of the trapezoid or the frustrum approaches zero and the number of partitions approaches infinity, why could one not use a simpler shape such as a rectangle or cylinder as the partition? (The motivation being that it made the math simpler.) I tried that and remember demonstrably getting the wrong answer but never knew why what the difference is. If I understood your presentation correctly, one answer is that the in the case of the rectangle or cylnder, the edge or surface normals do not tend toward the normals of the shape being calculated. Thank you!
    1
  9968. Lol. Now I must thank you again for refreshing my memory. You are exactly right. It was the length of curves and the area of surfaces. I really enjoy your channel. Thanks again.
    1
  9969. Oh just differential rings. I know someone who researches these generally in model theory.
    1
  9970. R u a mathematician ??? I am only 11th std
    1
  9971. Wow. I am blown away. I have never heard of Moessner's Miracle before. It is truly mind-blowing what kind of system is hidden behind just a adding integers in a consistent manner.
    1
  9972. Havent heard about that drama - but will be happy to hear about it :) Thank you! 1:30 - ok now i know why i didnt bother taking notice of that problem - its not a problem for a phycist - the diagonal of a square is still SQRT(2) - they could prove the perimeter is less than pi that way and that would be a contradiction.
    1
  9973. I hate infinite series.
    1
  9974. So if the area of the hypotenuse square is 1, the summed area of the "next generation" of two squares on the tree must also be 1 because Pythagoras. If we continue to generate the tree in this fashion, 1/1 + 2/2 + 4/4 + 8/8 etc. is equal to 1 + 1 + 1 + 1 + 1... forever, which seems counterintuitive to the picture but there you have it. So the total area of the tree must be infinite. I originally said -1/12 but that joke doesn't even work because I got the wrong infinite sum! Dang.
    1
  9975.  @Mathologer  Which leads to a good question: what's the total area of the tree?
    1
  9976.  @Mathologer  Thanks very much for your reply! I am no mathematician so "what is the area of the plane occupied by the non-overlapping regions of the graph" is so far beyond my abilities I have absolutely no hope. I did conclude later that day in the shower that because of this, any node of the graph, when combined with its descendants, has infinite (overlapping) area. I thought that this was the interesting point you were trying to make.
    1
  9977. 👍
    1
  9978. 25:02 proof that the blue area in the 1x1 square is > .5: For each blue excess draw the diagonal line from the upper left to the lower right. The upper right of that is all blue, and some blue extends across the line. So the sum of all the blues must be > .5
    1
  9979. I think the most interesting part is that the sum 1/p is infinite. It raises so many questions, obviously the sum of every alternate prime is infinite, but can things be pruned out more drastically? What if we take every pth prime? I have a suspicion that it might be, and diverges as ln (ln (ln (n))). If this is true it raises another question. Let Si be a monotonic increasing integer series whose harmonic sum is divergent. Take the Si th term from that series and form the harmonic sum from that (which must be smaller than that formed from Si ) when does this diverge? Is there something unique about the primes making it the sparsest sequence with this property? Clearly the integers are the opposite because forming such a sub-series just results in the same series. It's as if they are the densest.
    1
  9980. I totally understand that. What exactly are you trying to say?
    1
  9981. Fantastic Video. I wonder if that Roman solder hadn't interrupted Archimedes musings, could we have had calculus 1500 years earlier? For that matter, if he had some involvement in designs that later became the Antikythera device, could we have had mechanical calculators? A bit of a stretch, but an intriguing possibility.
    1
  9982.  @Mathologer Fair point. I didn't know that.
    1
  9983. Most memorable part to me is how simple the argument for the "no 9s" sum was in the proof at the end. Pure beauty!
    1
  9984. My graduation present from high school was a slide rule - with a leather holster. Circular slide rules were also available. Then in a couple of years, calculators replaced them. And the last slide rule manufactured ended up in the Smithsonian.
    1
  9985. You had me going there. I tried to solve this using real numbers instead of integers. Had that y**2 - 2*x**2 = 0 and was off into space. Doing math with only integers is like eating without moving your jaws. What you came up here though is the best I've seen in years. Guten Appetit.
    1
  9986. My favorite was how Kempner's proof echoed Oresme's, by grouping terms to form a tidy inequality. Maybe that's a common technique? But it made for a nice presentation.
    1
  9987. For #2, I ignored the answer from the video (too easy and a cop-out!) and noticed it splits into three easy groupings: (10 + 0)**2 + (10 + 1)**2 + ... + (10 + 4)**2 = 500 + (0 + 1 + 2**2 + 3**2 + 4**2) + 2*10*(0 + 1 + 2 + 3 + 4) which is doable in the head in short order to get 730 = 2*365.
    1
  9988. Great stuff. Thanks.
    1
  9989. Good videos, good insights. But you dont really expect that by the end we already understood implicit differentiation
    1
  9990. At the age of 7 i used a slide rule as a guillotine for chocolate bars
    1
  9991. Least memorable: connection between this sum of all fractions and logarithm (for me intuitively obvious). Most memorable: Ramanujan summation is not "true" sum (adding all members), just playing with parts of the infinity (and "borrowing" from the rest of the series without giving it "back") => so we can this way easily get that part is bigger then the whole. Cesàro summations <> summations and should be as such explicitly told (feel cheated there). BTW, do you know An Alternative set theory (https://en.wikipedia.org/wiki/Alternative_set_theory) from Petr Vopěnka (https://en.wikipedia.org/wiki/Petr_Vop%C4%9Bnka)? His work is mostly just about those games which are possible to see "on horizont" (inferface between what mathematican can (barely) see) - between finite and infinite...
    1
  9992. Your german pronounciation is perfect .. some german branches in your family tree?
    1
  9993. Great video!!!!
    1
  9994. Lagrange? Oh! Please!
    1
  9995. PhD student in fluid numerical modeling. As always a fascinating video. I like the part where you took the chance again to disproof the nonsense of 1+2+3+...=-1/12
    1
  9996. This might be the ultimate algorithm for the Tower of Hanoi. But "the" ultimate algorithm is surely Marcus Hutter's algorithm that solves every well-defined problem. If the fastest possible algorithm to solve an arbitrary problem P(n) takes f(n) steps, then Hutter's search solves it in 5*f(n) + d time(n) + c steps, where c and d are constants depending on P but not n and time(n) is the time required to calculate f(n). Since time(n) << f(n) for almost all interesting problems, this is about 5*f(n) + c. (However, c is usually so large, this is considered a galactic algorithm for almost all interesting problems.)
    1
  9997. Came here to learn how to make money , stayed here to learn how change money btw for that 1 coin , 2 coin , 4 coin puzzle we can uniquely represent each number upto 2 * highest denomination -1 method 1 : The corresponding product in this case will be (1+x)(1+x²)...(1+x^{2ⁿ}) = (1-x^{2^{n+1} })/(1-x) [By multiplying numerator and denominator by 1 - x and using difference of squares identities ] now the obtained expression can be interpreted as a sum of the GP 1+x+x²+...+x^{2^{n+1} - 1} hence we are done as the coefficient of each power is "1" method 2: express the number in base 2
    1
  9998. Amazing video!
    1
  9999. Willans' Formula for primes: 2 to the n part = vertical asymptote and p-adic numbers. 1/n part = vertical tangent. Factorial part = vertical line. These tensors from differential calculus determine singularities in stable matter as represented as primes.
    1
  10000. About your request at the end. I’m in 11th grade, in AP calculus BC. I figured everything up to the end of chapter 3 out independently last year. I understood all of the steps you did on the second viewing with a little bit of research in between viewings. Very cool video!
    1
  10001. yo math teacher got the half evil?
    1
  10002. What is the ratio of the triangle after cutting and folding to the original? From a glance it looks 1:3 but I need more than the minute I have to confirm or deny this. Edit: I am reffering to the length of the sides, not area.
    1
  10003. I went to middleschool in Helsinki, Finland in 70's. The first slanted square proof was how my mathteacher and our textbook proved the theorem. My favorite proof of larger Pythagoras that says for each similar blob attached to sides of a right angle tringle holds that the area of the blobs attached to the short sides combined is the area of the blob on the hypothenusa is the fact that cutting the original triangle by the height against the hypothonusa gives you to similiar triangles whose areas are sized by the smaller sides.
    1
  10004. Fantastic video!
    1
  10005. Dear Mathologer: First of all, thanks! What a beautiful & elegant tour de force to end a crazy year! Now, I wish to tell you a little story, & I hope you can answer a question for me. This has been bugging me for 35 years! I can't seem to find anything in the literature about it. Here goes... In the early to mid 1980s, I was playing around with an idea, & I developed a formula for it. I'm sure it's some standard thing, but it was neat for me to be able to find it. It's the formula for how many lines it takes to connect a given # of points - every one to every other one. 2 points takes 1 line; 3 points takes 3 lines; 4 points takes 6 lines, etc. The formula (of course) is: x= {n*(n-1)}/2, where n is the # of points, and x is the # of lines. Now, I wouldn't be asking you anything if that were the end of the story! Almost 20 years to the day later, I was playing an old video game called Qbert. You can easily look it up online & find out what it was like, but here's a quick description. The playing field is a triangular grid of solid 3D blocks, as though they're stacked up higher like bleachers, but toward a middle point (apex). I wondered if I could find a formula that would tell me how many blocks would occupy the face, based on the number of blocks along one side (all 3 sides being equal, of course). I found it, & to my utter astonishment, it's this: x={n*(n+1}/2), where n is the # of blocks along any one side, and x is the number of total blocks on the face. How can this be? I've never been able to explain how these are related! How can they be identical except for a +/- sign?!? I hope you can help me kick off the new year by finally solving this weird conundrum! Thanks again. tavi.
    1
  10006.  @Mathologer  Thanks so much! :D Now it makes sense. I hate it when I can see there's a pattern, but can't figure it out! tavi.
    1
  10007. 6:05 How do you get the water in if they're sealed?
    1
  10008. Degree in Maths (and computing science) Owens 1980. I didn't understand Euler - Maclaurin at the time, Mathologer filling gaps!
    1
  10009. Very elegant proof!
    1
  10010. Slide rules give you an arithmetical alertness: you must always estimate the answer first, to avoid getting confused by decimal places. This is a good thing. Have you ever come across the QAMA calculator, by the way? It takes that concept to a new level.
    1
  10011. Why can we not say that for an infinite 2d square grid, an equilateral triangle can only exist on the boundary with vertices at infinity? or are we only including the natural numbers as part of our grid set?
    1
  10012. 2:00 and already reordering conditionally convergent series. Old news ^^ ofc everyone sees this a first time some time.
    1
  10013. The way of the Mathologer is long and difficult. But I think I'll take a nap instead.
    1
  10014. My favorite part of this video was learning about the logarithmic growth of the harmonic series. The proof was quite neat, and I was also very pleased with myself, since logarithmic growth was my first intuition upon seeing the leaning tower of lira (and was further strengthened by seeing the partial sums). The whole video was amazing, though, as always. Every time you come out with a video, I make sure to find time to set aside, because I know I will end up pausing the video to solve some of the great puzzles you pose and to think about the elegant proofs you show.
    1
  10015. If you take the earth sphere and spread it down the other way with Antarctica around the far edges of the circle, you get the map which those masquerading as "flat earth" theorists use. If you sail a ship from anywhere toward the edge, you reach Antarctica.
    1
  10016. Amazing.
    1
  10017. What about the starting configuration OOX - it doesn't have any valid moves, we will thus never turn over the X.
    1
  10018. Saturn + Jupiter + Venus makes sense to me
    1
  10019. For future videos it might be a good idea to use some other colours than green and orange like at about 15 mins. Particularly those shades you chose are really difficult to distinguish for a colorblind person. Just as a tip, if you want to make something colorblind friendly then instead of trying to come up with colorblind friendly colors try using something else to distinguish things like shapes.
    1
  10020. Thank you for this educational video!
    1
  10021. Oh there there! You put the matrices and the mapping!!! With that graph I can easily program this one 🤓🖖
    1
  10022. 20:37 makes this officially a 'Cat Video"!
    1
  10023. Hallo Ich bin abonniert in deinem Kanal Ich verfolge deine Videos und bewundere deine Art die Mathematik zu erklären. Ich habe vor 4 Monaten angefangen Vodeos über Mathematik zu machen, bis jetzt habe ich 11 Vodeos über verschiedene Themen, eins auch über den großen Ramanujan Vielleicht kannst du mir Tipps geben Vielen Dank Gruß Elayachi Ghellam
    1
  10024.  @Mathologer  Danke schön für deine Antwort Ich bin Marokkaner, habe hier in Deutschland studiert und english ist halt international, ich bin aber noch nicht in der Lage auf englisch zu erklären Vielleicht erstmal auf deutsch oder auf französisch 😀
    1
  10025. Out of two proofs in the beginning i saw the visual one in polytechnic museum
    1
  10026. Chapter 5 was the most memorable to me.
    1
  10027. So when he started the video and said that this probably screams Pythagoras, I was like well not exactly. Still I was fortunate enough to be taught both proofs in High School. Still it was only briefly considering the amount of time put into making sure we knew the formula for the Pythagorean Theorem. The second proof was a lot more familiar, but that might be me because I was more algebraic minded. It is hard for me to say which proof I prefer, the second I like because in the end we want a formula so it clicks better for me, but the first is more beautiful to me so I guess it is a tie but for different reasons.
    1
  10028. Speaking of the "pi = 4" paradox, I found a non-video submission to 3blue1brown's SoME3, the "Blunder of Millennia" which used the isoperimetric inequality (with equality if and only if the containing length is a perfect circle) which Archimedes was unfamiliar with (hence the "blunder"). The document established an algebraic pi, specifically a root of X^4 + 16 X^2 - 256 = 0, evaluated to 4 sqrt((sqrt5)/2 - 1/2) (approx. 3.144605). When I saw this, I was like, what about e^ (j pi) + 1 = 0 where j^2 = -1? The proof that e is transcendental has nothing to do with pi.
    1
  10029. mathologer trolling the amateur programmers with an np-hard problem XD
    1
  10030. But wait there's more...
    1
  10031. 9:55 It is beautiful! По-русски имею возможность сказать несколько больше: эта штука прекрасна своей сложностью и контрастом этой сложности с простотой вопроса, который её порождает.)
    1
  10032. Nice proof of the p-test 😀
    1
  10033. I liked the swivel proof slighlty better than the coloring proof.
    1
  10034. The overhanging blocks for the bridge got me wondering if it would change if the block were glued together.
    1
  10035. Just because no one has described it doesn't mean that no one has noticed it before. It is even questionable whether all geometry books have been reviewed. It is a simple relationship.
    1
  10036. Three "extras" because it is cubed. If scientists can't figure it out they cube it, sometimes square it like that lame brain Einstein
    1
  10037. I believe you have your numbers of vertices and faces backwards for the dodecahedron at about 9:30. Love your stuff!
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  10038. 15th
    1
  10039. For some reason, I have this weird feeling that the determinant trick can be interpreted as det(H +iV) where H is the horizontal pairing matrix and V is the vertical pairing matrix with only ones and zeros, and since H and V are interchangeable by rotating/flipping the board, each tiling has the same coefficient. It would be really interesting if the rotating by 90 degrees added an i factor to the determinant and every flip added a -1 factor.
    1
  10040. Thanks!
    1
  10041. I really enjoyed you. I'm so glad you're here.
    1
  10042. I wish I had a teacher like you when I was younger… My brain refuses to understand maths of any kind beyond 1+1=2 😄 But I’m curious, so please bear with my stupidity. This is cute and I’m sure very interesting for many, but for me it’s a bit boring because it’s unidimensional (I guess it is two- dimensional as in length and height, but it is unidimensional, really). You cannot squish and stretch in 2 dimensions (you see? I have the mind of a preschooler who plays with plasticine, so the thingie would expand towards me and on the opposite side, creating the volume, as in tri-dimensionality). Is there such a thing as multi- dimensional maths? To use your prop as an example, I imagine a cube made up of many identical cubes, each having its properties and functions. I imagine each little cube having a specific colour. The cubes interact with each other, so the colours would blend (would they necessarily? If yes, how?). How does the individual velocity of one or more of them spinning, or the spinning of the whole assembly, affect them and the whole (do they change shape or colour?). Do they remain cubes and if not, why not? If they turn into something else, does the big cube remain a cube? Can a cube be made of smaller non- cubical items? If the big cube insisted on remaining a cube (could it?), what would be the space left amongst the other non- cubical sub- structures? What if we don’t have a name for that space? Would that space have any colours, physical properties and mathematical functions? If yes, are those determined/ assigned by the other non- cubical sub- structures and/ or the big cube that still lingers in your mind because I mentioned it as a possibility? I am probably not just stupid, but also crazy 🤪, but it’s fun to do thought experiments even though I don’t have the mental maths apparatus to describe and analyse it; it’s a very rewarding aesthetic experience. More seriously now, is there anything such as multi- dimensional maths? There must be because I heard of multi- dimensional chess. Why do you have algebra, trigonometry, geometry, and physics as distinct? Does one necessitates the other? If so, why? It’s a bit of a series of circular arguments. And, last but not least, how does all this knowledge help in the ‘real world’, anyway? PS. Do these cubes smell? In school I was as good (not) at chemistry as I was at maths 😄 but I remember something about molecules as in smell molecules. What shape is the smell of chocolate and why is it so delicious (especially if it’s hot chocolate with a few drops of rose oil)? Why is human nose of a shape different from a dog’s nose; it cannot be explained by the evolutionary theory. Why do snakes smell with their tongues? What is the function of smell? Why is the function of smell not a mathematical function? What are the disadvantages of defining people in terms of the function they (must) fulfil in society? Who says that the society is really a cube? Sorry, I am really not trying to pretend I’m smart, but I’m curious if anyone has thought about these things. Thanks. 😊
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  10043. Great, great video. Time really flew by and it was interesting to watch despite the fact, that I haven't got invested in every detail (I just trusted the gods, as you say). I'm a mechanical engineer and with my work I rarely use anything more hardcore than a derivative, but these kinds of mathemathics really relax me and make me more curious about the world.
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  10044. Mathologer is back, cool
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  10045. Merry Christmas, Burkard and company! :)
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  10046. the most awesome part was the block arrangement to achieve highest overhang, it just looks so random,yet makes full sense ! I'm pretty sure someone can train an AI to find the arrangement for any number of blocks, cause that reminded me of how a child would try to do it ,just randomly placing blocks and then feeling if they will topple over :)
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  10047. AHA! beauty makes me cry 🥲 wonderful stuff. also my favorite autopilot music was also there too ^___^
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  10048. The music pieces you pick though🌺
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  10049. 17:34 The area of a triangle is proportional¹ to both its base and its height. You've scaled the base by B and, by similar triangles, the height by A so of course it's ABF. I think. Bit rusty, tbh. ¹ as we all have known "since kindergarten": it's exactly half the product, but we don't need to know that.
    1
  10050. Great video! But at 22:55 it says cos(120) = -1. I believe that is not correct, it should be -1/2.
    1
  10051. 0:38 i am immediately suspecting 4D-rotation shenanigans, looking at that edit: ah, plenty of others did, too; regardless of whether it's correct, that's reassuring :D
    1
  10052. Get a haircut, kid.
    1
  10053. My guess is if no one will be getting paid fully you want the little guy to be the least screwed over proportionally.
    1
  10054. So would you say this method of teaching is Carculus?
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  10055. 25:31 I think people often pronounce 'sinh' as 'cinch'?
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  10056. It only works on 10 digit base system. It cant be universe key if is not universal. Won't work for aliens with 4 fingers in their 3 hands, for instance Edit: nevermind, I commented before minute 20 of the video.
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  10057. Hallelujah! Partitioning problem
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  10058. 8 is a boundary volume we know will "paint" the inner area. How much is actually needed?
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  10059.  @Mathologer  I was thinking of the paint as a bounded 3d topological blob and wetting a surface is finding a point in the blob, such that there is an arbitrarily small sphere that includes points on the surface. And to paint the surface, all surface points must have at least one such paring in the blob. I think that is how it is done. How you use a small radius to define contact or whatever. And since a finite sized blob has an infinite number of points, the paint can be arbitrarily small, but non-zero in volume.
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  10060. I know I'm being picky but at 9:25 you list the number of faces of the dodecahedron as 'F=20' and vertices as 'V=12'. I play DnD and I know a 12-sided (12-faced) die when I see it. Fun to think about the platonic solids and their inverses though :)
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  10061. My son and I enjoyed the once glance balancing as it was the most useful for everyday knowledge. thanks!
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  10062. Oh wow, I was thinking for many many years that we are done with triangle! But new things were discovered!
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  10063. 36:29 As I've seen in other comments, blue area = 1/5. If instead of 1/2 and 1 we had side lengths a and b (with a < b), then I think the blue area would be (a-b)²b²/(a²+b²). Not as pretty as "1/5", of course, but I was curious what it would be, so I went ahead and calculated it.
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  10064. Really awesome videos sir as always. The best part is Mathologerize version of convergence of no 9's Harmonic series. Thank you so much sir.
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  10065. The ending where you showed a different way to sum up the terms left me befuddled. You didn't rearrange the terms but the result is still different. So is the first method leading to ln(2) even valid?
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  10066. But he did combine them in a different order than the original. Hence the different answer. So the ln(2) method is valid.
    1
  10067. Not gonna lie, the ones after 1+2+3 = 1 x 2 x 3 are a bit boring.
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  10068. I slept in my 1-hour lectures but I didnt even get tired watching this haha
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  10069. If F(n) = φ^n - (-1/φ)^n) / φ-(-1/φ)], then φ^n - (-1/φ)^n) / φ-(-1/φ)] at thee limit must be a rational number.
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  10070. Since the area of the trapezium plus the small triangle form another equal square A, we have five squares A in total, so 5A=1 --> A=1/5
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  10071. Awesome video, thank you so much for all the work You put in it. I've been watching Mathologer video since start of channel, but I've never study math. I just have passion for all those satisfying math rules in nature and abstract universe. And You are really the One who can show me what I want to know even I don't know that itś exist. And its really nice to see 3blue1brown in your patreons. I'll try to become your next patreon, in 2k21
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  10072. I'm so disappointed that the Mathologer seal of approval is... not an actual seal (like, with flippers and whiskers).
    1
  10073. What if the card on the right is connected to first card so we can turn last card and card to the right (first card)
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  10074. Calculus is easy if you only learn steps to take. Where it can get challenging for some is when you have to actually prove certain theorems and formulas. For instance, many students probably know that a p-series diverges when p is less than or equal to one, but they might not know how to actually provide a proof for that.
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  10075. To conjecture a brief lemma from this video: every vertex of an N-D cube is incident to N orthogonal edges. I think.
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  10076. Absolutely wonderful! I would like to ask 2 things. First, why is it called 'harmonic'? Second, if we change the base, how do the value change for 'the sum with no Xs'? I'm mean, at the extrem, what happens in base 2 for the sum with no 0s for instance?? I have a little guess but last time, for the no 9s, I should be dead.
    1
  10077. That is so interesting.
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  10078. From 17:12 on you used the warm up series instead of the Ramanujan series. I misprint I guess.
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  10079. Ha! Windmills. Love it!!👍
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  10080. I like the color proof best. Seems like something I might have been able to come up with myself. Whereas I never would for the swivel proof.
    1
  10081. What would a number wall of a random sequense look like? Is there a way to prove a number wall is from a random sequence?
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  10082. https://www.khanacademy.org/computer-programming/primitive-root-confirmation-program/5981942307373056 A program to tell if p is the primitive root of num.
    1
  10083. I don't think I ever saw either twisted square proofs in school. I must've discovered it while doing some reading online. Maybe I don't remember since we learned Pythagoras's Theorem really early and I didn't pay attention much. After that we basically just assumed it was true without proof. Maybe we proved a higher-dimension version in Analysis class, but that was more rigorous.
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  10084. Always great to wake up to a new Mathologer video!
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  10085. Okay, I need that shirt. That's hilarious.
    1
  10086. Just cut from the 4 concave corners and assemble new cross.
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  10087. I believe there is always a choice of "all of the above are interesting!".
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  10088. Interesting he was able to make the connection with a problem like that, though, haha...even if he had studied continued fractions...to take a word problem that might not seem related to continued fractions and instantly see the connection, haha...but maybe I just don't know much about all that stuff...
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  10089. Did I miss something in Burkhard's statement of the challenge (at the end), or did he? Because clearly, any integer that isn't congruent to 2 mod 4, can be written as a difference of two squares, simply by writing it as a product of two integers of the same parity (congruent mod 2), then using the difference-of-squares formula. Wanting n = x² – y² Write n = ab Then x = ½(a + b), y = ½(a – b) If n is odd, a & b will both be odd, and x and y will both be integers. If n is a multiple of 4, n = 4k, then take a = 2k, b = 2, and again, x and y will both be integers. Only when n has just a single factor of 2; i.e., is congruent to 2 mod 4; will it be impossible to write it as a difference of two squares. The second part of the challenge works, however, because if n is (an odd) prime, then a=p, b=1 (or vice versa) must be true, and x = ½(p + 1), y = ½(p – 1) is the only possible solution. [Swapping a & b only negates y, so the squares being 'differenced' are the same.] The even examples for n: 0 = 0² – 0² = 1² – (-1)² 4 = 2² – 0² 8 = 3² – 1² 12 = 4² – 2² 16 = 5² – 3² . . . 4k = (k+1)² – (k–1)² Fred
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  10090.  @Mathologer  Thanks for the reply. I, and I hope many others, find you easy to maintain one's attention to. Fred
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  10091. Don't remember what I was ever shown as a kid when they taught us this (might've just presented it with no proof at all, honestly) but I saw the second proof on a numberphile video with Ben Sparks recently
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  10092. ... nice animation. equally nice music. thanks for 'gem'
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  10093. Commenting to remember to watch this later. I could always use some extra calculus review hah
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  10094. He mentions this at 49:00
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  10095. Smartest joke in Rick & Morty: Fart Dumbest joke in Futurama: New Theorem
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  10096. why did I stop watching these videos so many years ago?
    1
  10097. They dont want you to know the real reasoning ,thats why they make long videos like these, convincing you THIS is the truth. So, did it work?
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  10098. 6:05 First challenge: Remove the horizontal and vertical neighbors of a corner square leaving it disconnected from the rest of the board (and remove any other two squares except for the disconnected corner). Since no domino tiling can cover this disconnected corner square, there is no domino tiling of this board. 10:18 Second challenge: The formula spits out 0 for odd m,n, because in this case the upper index boundaries j=⌈m/2⌉=(m+1)/2 and k=⌈n/2⌉=(n+1)/2 yield jπ/(m+1)=π/2 and kπ/(n+1)=π/2, so both cos terms in the sum vanish simultaneously in one factor of the long product. Hence, the entire product vanishes. 13:50 Third challenge: The number of tilings for the (2,n)-board is the n-th fibonacci number Fn starting with F1=1 and F2=2. Proof: If we number each square with its columns number (or row number, if you consider the vertical board) the resulting adjacency matrix An has i's on its main diagonal and 1's on the two neighboring diagonals. Using Laplace expansion we see that its determinant Dn:=det(An) satisfies D(n+2)=i·D(n+1)-Dn for all n≥1. It follows inductively that Dn=i^n·Fn. Hence, after removing signs and i's from Dn the number of tilings for the (2,n)-board is |Dn|=Fn. 14:34 Unnumbered challenge: To count the number of tilings for the pair of glasses, we can break it up into smaller, simply connected pieces (i.e. pieces without holes) on which we can apply the determinant-method. Firstly, denote by H(row,col) or V(row,col) a horizontal or vertical domino with its left/upper square at coordinate (row,col) with coordinates starting with (1,1) at the top/left-most square (like a matrix of squares). Now to the fun part. The square (1,8) can only be tiled in two different ways, either H(1,7) or V(1,8). Similarly, the square (4,8) can only be tiled with H(4,7) or V(3,8). Together those four possibilities give us four groups of valid tilings. Call those four groups HH,HV,VH,VV (first letter representing the choice for the upper square, second character representing the choice for the lower square). Notice that there are no tilings for HV and VH, because trying to extend HV or VH to a valid tiling will always leave one square untiled. Therefore, all tilings belong to HH and VV. The same holds for the squares (1,11) and (4,11). Call their corresponding groups HH' and VV'. In summary, the set of all valid tilings is a disjoint union of tilings HH∩HH', HH∩VV', VV∩HH', VV∩VV'. By the board's symmetry HH∩VV' and VV∩HH' contain the same number of tilings, so we only need to count the number of tilings in HH∩HH', HH∩VV' and VV∩VV'. To count the number of tilings in any of those groups, we only need to remove the squares under the dominos on the squares (1,8),(4,8),(1,11),(4,11), apply the determinant-method to all resulting connected components and multiply them together. Here are the results: HH∩HH' has 6*5*6=180 valid tilings, HH∩VV' has 6*3*9=162 valid tilings, and VV∩VV' has 9*2*9=162 valid tilings. Summing everything up (remembering to count HH∩VV' twice) we get 666 valid tilings for the pair of glasses. 36:42 QUESTION: To me (and I am surely not the only one to notice this) it looks like the hexagon tilings (interpreted as 3d stacks of cubes) build a sphere at the origin together with three coordinate planes. Are there any theorems/papers on this as well?
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  10099. The maths is cool but your genuine enthusiasm is the hook that keeps me watching. Thanks
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  10100. Yessssss
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  10101. cavalieri?
    1
  10102. Math!
    1
  10103. I saw a CD, not an iris.
    1
  10104. Yo!! Good to see your doing well!
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  10105. nice h3 merch
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  10106. I thought the "thinning the monster" segment was the best, as the question that always comes to my mind when looking at this kind of series is, what is the line between a series that diverges vs converges. It's similar to the Reimann zeta function and the Mandelbrot set in that it explores the limits between the infinite and the finite.
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  10107. That was great. I clearly haven't spent enough time thinking about geometry or transformations.
    1
  10108. Help, after watching this, the YouTube algorithm keeps recommending a lot of BS Tesla vortex videos …
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  10109. The digital root/remainder thing is also only an artifact of the use of base 10. Any supposedly deep mathematical truth that vanishes when the numerical base changes is automatically suspect.
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  10110. Mathematics is Art
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  10111. π*Daumen ;-)
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  10112. I know a lot of Pythagorean triples and I know how to calculate them and I notice that at infinity the series of triangles converge to a straight line. I have figured out how to calculate two kinds of Pythagorean triples; but THIS is very interesting! There are new Pythagorean triples I have never heard of before featured in this video
    1
  10113. the integration of 1/x leads us to logarithms
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  10114. You sometimes have the same laugh as Comissioner Dreyfus from Blake Edwards Pink Panther movies. And that's hilarious. 😂
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  10115. in a set of 5 different cards there is always a set of at least 2 cards with the same suit & in that set, one is always bigger than the other, you always pick the bigger one, so the assistant always count up if mathologer has 2 of spades instead of 10 of spades in that example, then he would pick 6 of spades edit: sorry, not the bigger one, but in two cards there is always a card you can count up to the other card the fewest after factoring in the 13 cycle so if you have 2 & 10 in the same suit, then you should pick 2 because 10 count up 5 will get you to 2 but 2 needs 8 to count up to 10
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  10116. Over last summer, I programmed this in processing (KA was faster than Repl): https://www.khanacademy.org/computer-programming/triangles-1d-cellular-automata/6678150943440896. It seems to correspond to rule [2,1,1,0,2] in the program. Hmmm... EDIT: Found that this was one of the "notable rules", named "WHAT THE" in the comments. The other rule outlined later, with modular arithmetic, was actually what inspired me to make this program in the first place. The rule was [1,2,2,0,1], called "The original pascal" in the comments. Other rules can be used as well, but when I saw that thumbnail, I IMMEDIATELY opened up that program to see what would be discussed.
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  10117. "a sine of madness" would be a great name for a book
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  10118. Been Watching You Forever, as a Mathematician; You Are My Favorite One On The "Tube" ~
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  10119. I think that its interesting to think about it in the following way: I) Case where the two coins have the same radius: the center of the outher coin will trace a circunference wich radius is equal to the sum of the radius of the two coins, so it has a perimeter equal to 2π(r1+r2), that is the lenght of the path that te outher coin will pass through. so when r1=r2, the coin will have to pass through a path that is the double of the coin perimeter, therefore, it will rotate 2 times if r2=k•r1, it will rotate k+1 times, even if k isn't a integer
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  10120. Thanks for your great work
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  10121. No integers!
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  10122. 06:06 - looks like right circle doesn't touch one line.
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  10123. Thanks for the education. i really need this way of understand calculus, Gracias por tanto desde Bolivia . Saludos
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  10124. How do Jacob check his answer
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  10125. Its interesting how trivial it is to find the solutions to these problems: 1. rs=r+s ==> 1/r+1/s=1. Pick r, s automatically fixed by this conjugacy relationship. 2. r^2=s+1. ==> pick r real, and s is automatically fixed. 3. s^2=r+s+1 ==> pick s and r is automatically determined.
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  10126. Most likely, somebody has already posted this, but I thought a little bit about the challenges raised at 32:20: I start at with the fact (do I need to prove this?) that summing up the odd numbers delivers the sequence of square numbers. Since I cannot typeset formulas here, I try to write it this way: S(n) = sum(2*i-1, i=1..n) = n^2. And S(n-1) = sum(2*i-1, i=1..(n-1)) = (n-1)^2. The difference between these two sums is just the last summand of S(n): S(n) - S(n-1) = (2*n-1) = n^2 - (n-1)^2 This works also for a subset of even numbers, because S(n) - S(m) is even if (n-m) is even, e.g.: S(n) - S(n-2) = ... = 4*(n-1) = n^2 - (n-2)^2 Obviously, not all even numbers can be constructed this way, they always jump in increments of 4. We can also get alternative representations (n^2 - m^2) for some odd numbers if (n-m) is odd and greater than 1. Regarding the uniqueness for prime numbers: p = n^2 - m^2 = (n+m)*(n-m). If (n-m) was anything else than 1, p would be composite and not prime. So there must be: (n-m) = 1 and (n+m) = p. n and m must differ by exactly one, so only the difference of two consecutive sums S(n) - S(n-1) can deliver a prime. Finally we get: n = (p+1)/2 and m = (p-1)/2
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  10127. I've used, well rarely now, a slide rule for 60 years. My dad taught me how to use one and how it worked when I was in 3rd or 4th grade. Upon seeing the answer to 13 times 13 my automatic reflexive action was to say "THAT'S WRONG!" However, upon closer inspection the 13 mark was on 169. So what hit me so hard as to evoke such a knee jerk reaction? I grabbed the last slide rule I ever bought for academic work (In 1968 when I was 15 years old) and stared at it intensely. The scale setup on it is identical to that shown @9:21. I think the P scale (Pythagorean) was its main attraction for me at the time. Now I understand what caught my instinctive eye. There is no where on any slide rule in my collection (I have more than 10 now), in any place, on any scale, that has an intermediate graduation (a short line between two long lines) that doesn't represent half as 0.5, i.e. having long line 168, short line 169, and long line 170 would NEVER be used in any 'real' slide rule. Yeah, it's half way, but it is not 5, 0.5, 0.05, etc.. The 'rule for the rule', without exception, seems to be that all graduations between graduations represent 0.5 (for 1 short graduation between 2 long ones), 1 (for 9 short* graduations between 2 annunciated marks), and 2 (for 4 short graduations between 2 long ones). *Here too 0.5 or the 5th mark is lengthened. Between annunciated marks the '5 position' is always longer, e.g. the mark between "6" & "7" or 6.5 is lengthened on the C & D scales. This was all just instinct until I analyzed it and wrote it down. Thank you! There's a secondary lesson that's also 'taught' by familiarity with the slide rule's layout. I find that virtually none of the engineers I've worked with over the years realize that a constant distance on a logarithmic plot represents a constant percentage or factor. This simple, easily understood concept, always seems to be a surprise. In recent years I've never heard someone say, "Yeah, I knew that."
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  10128. He is laugthing like Julius Hibbert
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  10129. I wonder if we can invert Moessner miracle. I mean, we can calculate square root of an integer just by substracting odd numbers from this integer until it becomes very close to zero. And a number of times we made the substraction is the root. Can we use Moessner miracle to count cubic roots, n-power roots?
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  10130. Brain exploded at 24:30
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  10131. It makes me happy that there are at least two people on the planet who have thought deeply about laces
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  10132. I guess this deity must have 10 fingers as all this changes when using a different base.
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  10133. What a neat way to explain the basics of calculus. I like the usual mathologer music with it too - but not the noisy bit near the end :( Also a new word 'odometer' I thought it was called a mileometer or mileage counter. It must have changed its name for countries that use km. Who would want a 'kilometremeter'? I forget what the other new word was. Maybe it was 'run' in a new context, as in 'rise and run'. Not heard the x co-ordinate being called that before. It gets complicated when you get a function to integrate that doesn't have a simple 'elementary' solution. I like using a Mathematical handbook that has pages and pages of definite and indefinite integrals to look up. It also has infinite series for functions - that might help with integration too! Also very useful are its geometric formulae and diagrams. It's got lots more in that I don't use or don't remember - or at a level way beyond where we got to at university. I just remember we had some differential equations to solve - probably for equations of motion, pendulums, orbits, vibrations or waves. It got so complicated with del or div or curl - partial differential equations in 3D (or 4D including time). We could have done with a mathologer style animated explanation of the Del operator and the different kinds of derivatives it makles! It just kept getting more and more complicated with fluid dynamics and atmospheric physics! I can see why supercomputers are needed to solve them numerically for climatology and weather forecasting. PS I started to wonder how the odometer measures distance when you go at a constant speed in a circle because the wheels on one side are covering more distance than those on the other side, but I presume they still all rotate at the same rate. Part of the tyre must be sliding on the road surface and not gripping it as well.
    1
  10134. I feel like there was a standardized test (the SAT maybe) that had a question with the coin roll paradox, but none of the multiple choice answers was correct because the question writer didn’t understand that there should be an “extra” rotation. The question had to be invalidated after the fact.
    1
  10135. Love the rock-paper-scissors-lizard-spock tee
    1
  10136. Good stuff!
    1
  10137. For me, this is one of your best!
    1
  10138. Take a Roots of a transcendental number and you finally bring it back down into this fucking dimension so we can work with it
    1
  10139. neurons
    1
  10140. No removing 4 tiles cannot always be tile. Example remove 1,2 and 1,3 and 2,1 and 2,2
    1
  10141. I’m pretty sure the answer is more than -1/12
    1
  10142. I have some questions. Since in the wheel there is a notion of equivalence between numbers whose ratio is a power of ten, isn't it be possible to define a set of equivalence classes between them? What is that set called? Does it have any interesting unique properties or something? Also, it would be a field, right?
    1
  10143. Being more of a computer scientist than a mathematician, the first thing I thought of when I saw A - A != 0 is that A is Infinity or NaN (not a number), which in IEEE-754 computer floating-point arithmetic cause A - A to evaluate to NaN, which is by definition not equal to any number (or even itself!).
    1
  10144. Is that on your shirt "To infinity and beyond"?
    1
  10145. This result can ofcourse be extended to the complex number (you have the same notion of convergence and absolute convergence). However it is not possible to create i by shifting terms of the mentioned sum over (-1)^n/n. For a given (non-absolutely convergent) series, is it known what its range of possible limits is? For example, is it always a line (like the real-axis in this example), or is it possible to get the entire complex plane?
    1
  10146. I cast my vote for the bizarre leaning towers. Gamma has to be greater than 0.5 because each excess blue bit takes up slightly more than half of the part of the rectangle they fill. Each bit starts right below the previous one, so the 1x1 square can be divided in rectangles with a height equal to each blue bit's height; you can divide each rectangle by half by drawing its diagonal, and you obtain two equal right triangles of half the area of the original rectangle. If you trace a line that joins the non-straight vertices of the blue bits, you get the same right triangle from its related rectangle plus an extra bit, which means that each blue bit is more than half of its rectangle. Therefore, the sum of all of the blue bits, gamma, should be less than the total area of the rectangle (1), but more than half of it (0.5).
    1
  10147. He made it symmetrical bc he is adding length of side b to where angle B intersects (etc)
    1
  10148. 36:22 A FIFTH! Bad way to prove: coord-bash(set a point as 0,0 and use formulae for lines, perpendicular lines, shoelace theorem for area, etc etc) Good way: look at one of the right triangles with hypotenuse 1. This must be similar to the 1, 1/2 right triangle (shares an angle). Thus it has area 1/5 after a bit of math. Thus 4 of them have area 4/5 and the square has area.. 1/5!
    1
  10149. That moment when you have to make a presentation abt polyominos and mathologer comes to give you ideas
    1
  10150. Just when you think you know everything there is to know about Pythagoras!
    1
  10151. 17?
    1
  10152. This reminds me of a topology problem i found on YouTube: How to invert a sphere.
    1
  10153. Nice T-shirt, Mathologer! The rock paper scissor lizard spock digraph can also be written with rotational symmetry.
    1
  10154. by switching rock and scissors on the T-shirt.
    1
  10155. 2
    1
  10156. heron's formula is the most used formula in architecture, surveying and civil engineering. It is used to find area of literally anything. any curved area is simplified into a polygon, polygon is divided into triangles, area of triangles are calculated using heron's formula, areas of all the triangles are added to get the area of the whole polygon. That is how you calculate area of everything on a plane. But this is the first time i have heard that this formula is called heron's formula. Thanks for the information.
    1
  10157. most memorable: gamma
    1
  10158. This is an example of a geometric combinatorial problem (counting edges, faces, &c) having a "nice" generating function (x+2)^n. Does it work in the reverse? Is there some way to identify a "geometric counting problem" underlying other ordinary generating functions such as x/(1-x-x^2) for the Fibonacci numbers?
    1
  10159. Mindblowing. Thank you for these videos its amazing.
    1
  10160. The Oily-Macaroni constant!
    1
  10161. Coolest Christmas present I ever Had .THANK YOU
    1
  10162. Really liked the visualisation of the no9. Simple and effective. I love it when things become less abstract and visible like this. Im not competing for the book. You can make other people much happier with that. And i really wanted to say something else. This was YT advised to me and i watched all of it. Paused a few times to process it. But, a German mathematician, speaking english is in itself, interesting. Then you would probably expect someone in a suit (why?) and you are not. You would expect someone less masculine (why?), but you seem pretty masculine. And then you have this laugh/giggle and that caught me most i think. I am wondering now how we get our laugh. Our natural laugh. When i was an adolescent i remember i was told at some point that i had this particular laugh. At that time i copied it from others i think. It had changed and wasnt my natural manner of laughing. Since then it changed back to natural and on several occasions i was mocked in some mild manner bc i had a somewhat similar giggle as you have. I didnt really understand why. Also, it didnt matter to me really. Then some 10 years ago in a House MD episode there was some chap that had an odd laugh like that (followed by a snore) and it became clear throug the dialogue and such that this was not appoved of. It wasnt considered masculine or even mature. That was the message. But it makes no sense. Why would a laugh say anything about a persons masculinity or even adulthood? I couldnt place it and it seemed silly to me. Like how women nowadays disapprove of white socks. Or if a man crosses his legs or something. To me it seems a natural reaction to if you are hot or cold. Not to if you are a man. So seeing you now with this same giggle i have, i must say; Sir, i approve of your giggle! It is real, it is honest and i love it ;] (ok, has nothing to do with mathematics, or maybe it does, but i wouldnt know how). As you can see, i am less into math and more into psych. I will now be pondering for a week why we laugh like we do. I will also make sure to write 'Euler' correct in the future. I had quite some of your questions right! But i realise there was a 50% chance in most cases to get it right, without overhang. So it was more luck than wisdom, i guess ;]
    1
  10163. man i just want some of your shirts
    1
  10164. Superb. The visuals really help.
    1
  10165. really astounding geometrical facts (trees, spirals,etc) in usual numerical objects like Fibonacci sequence (now bent like inception movie)!!!
    1
  10166. Loved the weird optimal overhang the most. It dispells the illusion that 'beauty' has some universal 'power'.
    1
  10167. This did make my day. What a beautiful proof.
    1
  10168. is there a similar book in German to Thompson's "Calculus Made Easy" ?
    1
  10169. I did not solve the one on the board in under 30 seconds. I get that it's 2(13²+14²) but I didn't see an easy way to get 13² + 14² because I only know the squares up to 13 and I'm slow at doing arithmetic in my head. I may have missed something. For the next two I got 6³ And (∜2258)⁴? I'm not sure if I'm supposed to see something there. Have a MERRY (y-b)AS
    1
  10170. Great Video as always <3 You can simply explain the formula [Area of Triangle] = AB * F by taking the equilateral triangle with side lengths 1 and scaling it in x and y direction so that its new side lengths are A and B
    1
  10171. I was immediately disappointed that your anti shapeshifting shape wasn't a shape. I.E. a closed region with finite area. My intuition that such a thing would be impossible is preserved, and my hope to be astounded, shattered. Anyways, time to see the actual point of the video haha.
    1
  10172. When you go to 'infinitely thin' and turn the lines from curves to straight lines makes it more confusing...
    1
  10173. Absolutely LOVED this video! As a University Student, my background terminates at only basic linear algebra + calculus, set theory, and the basic maths before it. However, I always greatly enjoy your videos regardless of how difficult is sometimes is to make sense of them!
    1
  10174. I'm going to have to watch this at least two more times: once to note all the places you lost me and once more to figure them all out. Thanks for the vid.
    1
  10175. “Mathologerization” has just become my new favourite analytic approach; and “cubies” my new favourite geometric technical jargon.
    1
  10176. 2:15 you accidentally put 00.9 instead of 0.09
    1
  10177. Most of the problems are solved from 0 to infinity, this is obviously a good approach, but it would be nice to think in the opposite way, infinity to 0, maybe we'lll find something new.
    1
  10178. 2nd proof was taught at school for 6th graders in Albania, before the 90's. No idea what is going on now.
    1
  10179. Angles and cyclic quads.
    1
  10180. For the hexagon tiling interpreted as blocks, do you get an eighth sphere in the limit?
    1
  10181. Trying to solve the problem at 9:08 without actually doing maths, I would say instinctively there is exactly 1 way to make any amount between 0 and 15. I haven't calculated this, but I'm fairly sure I'm right this time since this looks very very similar to counting in binary. Using more power of 2 coins you get exactly 1 way to make any amount between 0 and (2^x)-1, where x is the number of coins used. This is very interesting way to describe binary counting, so thanks for that.
    1
  10182. the ln(x+1) + gamma bit was my favourite. I don't get why gamma is obviously more than 0.5 but it being obviously less than 1 was really neatly visualised.
    1
  10183. Just a few weeks ago I was playing around with Fibonacci numbers. I discovered that F_(k+n+1) = F_n * F_k + F_(n+1)*F_(k+1). This generalises the Formula at 40:18 (simply set k to n+2). The proof isn't that hard. Can you manage to prove it yourself?
    1
  10184. 😳 ..which Pi 🥧 ..is fresh?
    1
  10185. Did the school in Romania. 5th grade, 11 years. I've seen the first demonstration in school. Also my grandfather was a math teacher, showed me both of them during the summer.
    1
  10186. I would put e=mc^2 is number two. :-)
    1
  10187. test
    1
  10188. There's no hidden 2019... or is there?
    1
  10189. 3:21: 1/2, 1/3, 1/4, 1/5, ..., I believe I see a pattern there. I don't want to be too hasty, though.
    1
  10190. Three cubed plus four cubed plus five cubed equals six cubed got my fascination from long time ago.
    1
  10191. I prefer the swivel proof because it's closer to the proof I came up with while I had the video paused. I'll admit I did use the thumbnail as a hint that the inscribed circle would be relevant, then proved each two adjacent points were the same distance from the incentre. Basically the same logic but without actually mentioning midpoints. I realised after the video that I'd actually seen the windscreen wiper generation of a curve of constant width before - it's in Matt Parker's Things to Make and Do in the Fourth Dimension as a way to make curves of constant width with an even number of arcs, while regular Reuleaux polygons are always odd.
    1
  10192. Wie immer super!
    1
  10193. You are a math angel sent by God 😂
    1
  10194. ok so dont roast me too bad im not a mathematician but how does 1x2 + 4x2 =17 i thought it was 18
    1
  10195. I just realized that you have 666k subs . Math devil be thee ?!
    1
  10196. Should I be surprised or not surprised that the solution to the puzzle is somehow connected to the square root of 2?
    1
  10197. We were taught this also here in Romania!
    1
  10198. So the sum of the squares of the integers between 10 and 14 inclusive is twice that of the number of days in a year? That's quite a odd concurrence but I don't see any reason other than coincidence but that seems unlikely.
    1
  10199.  @Mathologer it may be coincidence but it is conspicuous
    1
  10200. Thanks. Great math, great music.
    1
  10201. Why don't they teach all math visually so we can figure out ways to apply it to real life? Sitting in classrooms and making numbers fuck on paper with nothing to tie to the real world is just creating useless forgettable fluff.
    1
  10202. Incredible!
    1
  10203. Pope of Hearts
    1
  10204. The hexagone that vertices are the six cocyclic points defines two alternated inscribed triangles such that the angle between two sides is the angle between two bissectors of the original triangle
    1
  10205. Hi, Very nice video.This suggested the attached figure to me. https://drive.google.com/file/d/1yjp6IZvwB5kPihXdFXRB_ZFd6FgSwpAN/view?usp=sharing
    1
  10206. It equals a necker cube
    1
  10207. Like a chocolate orange.
    1
  10208. If we have such constraints in Euclidean Universe, we can escape to Hyperbolic Universe :)
    1
  10209. I see a Parker Christmas tree in Aztec Gold ....
    1
  10210. Familiar with the first proof (geometric) and some others from the book "God Created the Integers" I think it's Mario Livizio.
    1
  10211. 7:06 Someone could make a clip and meme this "Nice!!"
    1
  10212. I think the calculus solution to the paradox is prevalent on youtube simply because people are lazy to find new topics and settle with retelling the same story, though sometimes with different visuals/storytelling.
    1
  10213. This is great! Side note - there's a similar 'shrinking' argument called deflation (with a 'growing' argument called inflation) that you use to prove that Penrose tilings are aperiodic... You show that if these tilings can all be inflated/deflated infinitely often then any periodic shift of such a tiling must be preserved, but that would mean that the shift is then a zero vector! This was generalised by Shen in 2010 for any 'self-similar' tiling. :)
    1
  10214. 🥺🥺... 😖😖....... 🥲🥲 Thanks sir
    1
  10215. Nice shirt ;)
    1
  10216. I have my doubts on the “7M can’t be wrong” quote. Take for instance all those millions that drive on the wrong side of the road….
    1
  10217. I absolutely love your videos. and there are times you have me laughing so hard like this one. I really have learned a lot from your videos. "Whoever has problems with this pattern should change channels now..", lmao brilliant
    1
  10218. Interesting
    1
  10219. If you take the next set of ears, in each case, I believe you get the center point of the final, regular n-gon. How does that relate to the initial n-gon? And perhaps all the intermediaries?
    1
  10220. Looks like I jumped the gun... 😁
    1
  10221. The operation of adding ears and constructing a new n-gon maintains the vertex center of mass.
    1
  10222. The shrinking polies gives another meaning to proof by infinite descent.
    1
  10223. For fun, python code to calculate sum at 2:11: sum([x**10 for x in range(1,1001)])
    1
  10224. I really enjoyed the video. Ever since i heard the story of little Gauß solving the summation exercise with ease, i started wondering about different power sums. I came up with formulas i could prove with the concept of induction myself when i started studying electrical engineering at university. Thank you so much for the amount of effort, that went into the production of this video!
    1
  10225. Most memorable for me was the proof on how the difference between the harmonic series and the logarithm actually converges.
    1
  10226. First Fifty-third
    1
  10227. I often say that math is the pursuit of highly structured, abstract beauty. 😀 Your posts show that I am right!
    1
  10228. The memorable part was: the unusual optimal brick stacking pattern. Great video, Thank you!
    1
  10229. Why did the title change? I wanted to learn why calculus was so fast 😢
    1
  10230. "Let's roll!" I laughed.
    1
  10231. the most memorable for me ist the fractal representation; you can see early on that the sum must stay below 81/100 if you count only the big white stripes (right and down)
    1
  10232. I have to wonder, can this somehow be applied to decimal numbers? Could you somehow for example take every pi-numbers, then do some incremental reduction of of pi all the way down to get the powers of pi? Maybe there's more work to be done here :)
    1
  10233. Proud to be a mathimacian from the same county as ramanajun
    1
  10234. 5:16 this moment would have benefited if you had some transparency, so that we can still see the triangles and square after you overlay A^2 and B^2
    1
  10235. For R=S=2 R**2=S+1 not satisfied
    1
  10236. I believe I learned about digital roots and remainders from the Martin Gardiner math books I checked out from the library as a pre-teen. So many good things from those books!
    1
  10237. Amazing!
    1
  10238. I'm sure someone noticed, but I could find no reference to some interesting "not so random" aztec Diamonds: https://drive.google.com/file/d/1i_l8yLflquTz-TJYspiY-Db3N_bt0DgO/view?usp=sharing
    1
  10239. Still in high school in Australia. Don't remember doing it as part of school but as part of another assignment I got curious about being able to represent triangle area in terms of its side lengths and learnt about this theorem
    1
  10240. In a self intersecting diagram/curve, going through the dots is not always in clockwise/counter-clockwise. Add a point to the intersection point (this will not change the result calculated) and the shoelace formula now goes clockwise for one part and counter-clockwise for another. The result is exactly the difference of the area of two parts. Generally, for a self intersecting loop, point out all the points including the intersections. Draw out the loop from any point counter-clockwise. The areas which you passes the dots clockwise will be negative, otherwise it's positive. The result of the formula will be the sum of all those positive and negative numbers, not the actual area and expected to be smaller than actual.
    1
  10241. I am amazed by the hard work you've don and equally, with the nice visualisations. Thank you for sharing this golden idea with a huge numer of audience! Keep up the good work😊 Here's an idea: What if we try to describe the ratios "s" and "r" (or even matrices "S" and "R" in terms of infinite fractions similar to phi=1+1/(1+1/(1+1/...))? Imagine how much additional tools we already have for playing around the infinite series... Any thoughts?
    1
  10242. The colouring proof is nicer.
    1
  10243. The human brain is pretty much incapable of visualizing 4 dimensional space and objects. Trying to visualize even a simple 4D cube is pretty much impossible. However, I once realized a method by which visualizing, or at least understanding it, becomes slightly easier. The method is like this: Let's start with the "one-dimensional unit cube", ie. a line segment of length 1. (This method works even if starting with the "zero-dimensional cube", ie. a point, but let's start easier with the one-dimensional cube.) Now, consider this 1-dimensional cube to be on the X axis. In order to get from here to a 2-dimensional unit cube (ie. a unit square), duplicate the 1-dimensional cube and translate it by 1 unit in the second dimension, ie. the Y axis. While doing this, keep the corresponding endpoints connected with straight lines. When you do this, you get a 2-dimensional unit cube. In order to get from the 2-dimensional unit cube to the 3-dimensional one, we do the same trick: We duplicate the 2-dimensional cube and translate it 1 unit in the third dimension (ie. the Z axis), while keeping the corresponding corner points connected with straight lines. This way we get a 3-dimensional unit cube. Obviously, the same trick works to get a 4-dimensional cube. And it is precisely this trick that helps one understand it a bit better: Take the 3-dimensional unit cube, duplicate it, and translate it by one unit in the fourth dimension, while keeping the corresponding corner points connected with straight lines. You get a 4-dimensional unit cube. You don't need to precisely visualize this operation. Just think of it as moving the copy of the 3-dimensional cube in a direction that's not any of the three dimensions. This trick helps you, among other things, deduce the number of corners, vertices, faces and 3D cubes in the 4D cube, and helps you understand where the multiple 3D cubes come from.
    1
  10244. 40 mil wow!
    1
  10245. That Mathologer seal of approval was the highlight of the video for me! (and the fact that the optimal leaning tower was not beautiful)
    1
  10246. I seem to have missed something: if folding a right triangle twice doesn't give a right triangle then the end result isn't the same triangle we started with?🤔
    1
  10247. So the 5 (x,y) vertex coordinates for any pentagon can be transformed back and forth between a single (x,y) and 4 (size, rotation) values of regular pentagons/pentagrams? Cool but hmm how can this be used somehow? Makes sense I guess.. 10 dimensions <-> 10 dimensions
    1
  10248. 2020 = (±38)² + (±24)^2 = (±42)^ + (±18)^2 Eight variant
    1
  10249. A working greedy algorithm always sticks out to me as a programmer. They so often fail.
    1
  10250. 18:08 That's easy: it's the number of vertices of the respective iteration of the Sierpinski triangle since we are visiting each vertex exactly once. So n disks require 3^n steps.
    1
  10251. Gonna make some Vermicelli horns
    1
  10252. As a total jerk in math I especially don’t understand when something is sufficient as proof and when not. Nice juggling around with triangles, sure.
    1
  10253. most memorable is gamma
    1
  10254. i would like a piece of that golden pie
    1
  10255. The intersection may be in these vids: Why can't you multiply vectors? (or how Algebraic Geometry told me to stop worrying) https://www.youtube.com/watch?v=htYh-Tq7ZBI&ab_channel=FreyaHolm%C3%A9r Smooth interpolation in one dimension https://www.youtube.com/watch?v=vD5g8aVscUI&t=614s&ab_channel=EpsilonDelta Be awesome, math people
    1
  10256. Lacings are tights ! 😃 Yes sir they are …
    1
  10257. In before he says 216 is 6*6*6
    1
  10258. 4:18 i was wondering, before substituting the original infinte sum into the derivative, shouldn't you prove if/when the original sum converges?
    1
  10259. I think what's really most memorable has got to be, not thatno-9s converges, but that there are infinitely many convergent subseries based on either excluding some digit or chain of digits or demanding some chain of digits. I wonder if the "always 9s" converges? It surely can't, because the infinite harmonic can be subdivided between having and lacking 9, and the lacking series converges, therefore the having must diverge. Yet requiring 100 zeroes converges, so the series which always has less than 100 zeroes must diverge
    1
  10260. My Australian Rabbi would remind that you glossed over the "Rational Solution" it's based upon fair distribution, fair relations, & fair judgements are EQUAL TREATMENT! I love your approach!
    1
  10261. Cicadas
    1
  10262. I WANT that T-shirt!!!
    1
  10263. Thank you for existing
    1
  10264. I got a paradoxon. The sum of no-9s is less than 80. The proof at the end is valid for every digit 0 to 9, so every sum of no-digit is less than 80. If I apply that to each digit 0..9, every fraction in the harmonic series is in at least one of the ten sums of no-digit. Therefore the sum of the harmonic series is less than 10x80=800. Please find my mistake :)
    1
  10265. Many terms of the harmonic series are missing when you add up all of the "no digit n" series. For example, 1/1234567890 is a term that is not present in the sum of the sums, since this term appears in none of the individual sums! Indeed, terms which don't contain all ten digits tend to get spread out very far, so you will be missing out on almost all of the terms. ;)
    1
  10266. ah! I see.. yes. the remaining fractions are still infinite. thank you :)
    1
  10267. Swivel proof
    1
  10268. Indeed, I have been somehow thought at school that geometrical proofs are not as rigorous as algebraic ones. But I think there is no difference, and also proving things by geometric means would help people learn more and better. Doesn't it?
    1
  10269. more of Ramanujan's autopilot please
    1
  10270. What a gift. Thanks santa
    1
  10271. 0:30 - was afraid to click it earlier because of that socialist / atanist entagram... :( how much time is left for coders - not sure if i watch the video till the coding problem arises :) ?
    1
  10272. Isn't 4k+3 = 4k-1?
    1
  10273. The cat at the very end is so sweet 🥰🥰🥰🥰.
    1
  10274. Kind of sad that m1 wasn't reduced to 2*m0+1 with an explanation that a tower of size 0 is already in its finished state, hence m0 = 0. Would have been more fun.
    1
  10275. how do you know a 0-tower is finished?
    1
  10276. The perfect evening entertainment. Thanks!
    1
  10277. well... variable A could be seen as a placeholder (or a reference) to a number... so placeholder A - A, is the same exact number, no matter how you arrange it on one side, it will reference the same layout. if you change one side the other side will automatically be arranged to the same. So it must be zero.
    1
  10278. But, an African or an European pigeon?
    1
  10279. How is the square handled? Can't "slap on" 180degree ears easily.
    1
  10280. "the maximal minimal length", you manimal!
    1
  10281. I like the dot one.
    1
  10282. what i find interesting, though, is the oddity of the second term, which is always half the last term. sum of integers, need n/2. squares, n^2/2, so on.
    1
  10283. What is the title, and who are the performers, of the cool outro music?
    1
  10284.  @Mathologer  My apologies. I didn't look close enough.
    1
  10285. :face-fuchsia-tongue-out:
    1
  10286. do one about 2+2=2*2=2^2=2[tetration]2. thank you
    1
  10287. I'm not on board with the numerology key if the universe guys but just cause you understand the property that led to 9,6-3 being unique doesn't make them any less unique, I don't think you addressed why they think it's special properly and that may just lead that group to think you're making fun of them if they held onto that idea with any strength
    1
  10288.  @Mathologer  Thanks for replying. So then, There's no significance to the observation that there are 8 yellowed "difference" numbers. Further, can I conclude that 57 is on the same spiral as 113? Thanks! (Your vids are AWESOME. PS: I think 3blue1brown does a looksy into this too. And relates his findings to Mandelbrot). All very very cool!
    1
  10289.  @Mathologer  Thank you!
    1
  10290. For many years I have wondered why the properties of sums of finite sequences were not the starting point for considering the sums of infinite sequences, both taught as the starting point of calculus. My memory was that was the way I was taught, but when I went back and acquired a copy of the text we had used (Apostol's "Calculus" volume 1) I found that sequence theory came after the derivation of calculus from a point-set, modern analysis perspective.
    1
  10291. Thank you Mathologer for a nice Sunday
    1
  10292. Can you use this in 3d graphics processing? Is this an optimization for polygon based rendering engines?
    1
  10293.  @Mathologer  Yes indeed, but I remember it generalized to n-gons by repeating the operation multiple times. So I actually puzzled for a bit long ago, before the YT era, attaching perfect squares to the outsides of a general tetragon, and such. Never got it exactly working. But now I can rest! 😜 The link with Fourier and seeing each extension as a band-stop filter is really elegant btw! 💚
    1
  10294. How to find sin20⁰ = 0.5×√(2-√(2+√(2-√(2+√(2+√(2-√(2+√(2+√(2- .................∞
    1
  10295. Naive question: is the hexagonal artic circle an extension from 2D to 3D of the 'regular' arctic circle? (And projected in the [1,1,1] plane)
    1
  10296. Nice vedio. Learned a lot from the video. Can you tell me the music used in the bramhagupta formula deriving and thank you period?? Nice one.
    1
  10297. Question for Mathologer: does the (x+2)^n formula works for any convex polyhedra and not ‘merely’ for the ‘cubic’, very regular ones?
    1
  10298. Awesome!
    1
  10299. Danke! Thanks! This practical peace strategy !
    1
  10300. wait you didn't pronounce 'reciprocals' right! @20:35
    1
  10301. Most memorable: Almost no numbers do not contain a nine (when looking at the big picture). Made a lot of sense to me when you said it and I actually thought about it. But before that would have never thought about it.
    1
  10302. im doing my part too MATHOLOGY FOREVER
    1
  10303. 8/10
    1
  10304. One of the very few channels I have subscribed to! This video saved to favorites. Keep up the fantastic work!
    1
  10305. hi
    1
  10306. "Also for the programmers among you…": Done. https://chridd.nfshost.com/tilings/diamond
    1
  10307. Second challenge: If both m and n are odd, then m/2 and nwill round up to the half of m+1 and n/2 will round up to the half of n+1 . Therefore, one of the factors in the product becomes 4*cos^2((m+1)/2*π/(m+1) + 4*cos^2((n+1)/2*π/(n+1)) = 4*(cos^2(π/2) + cos^2(π/2)) = 4*(0^2 + 0^2) = 0 . Any product with 0 among its factors is equal to 0.
    1
  10308. Dear Sir, wish to send U lots of thanks for this nice, simple explanation of logs!!
    1
  10309. Does a Schwarz lantern emit black light? Is the sequence of the number of bands in the OEIS?
    1
  10310. My python imlementation for working out the number of ways to make change: *this works in cents coins = [1,5,10,25,50,100] maximum = 1000 ways = [] for i in range(0,len(coins)): for x in range(0,maximum): way = 0 if(x + 1 == coins[i]): # We can +1 if our index is our coin denomination. Only happens 1 time so probably better to not repeat this. way += 1 if(0 <= x - coins[i] < len(ways)): # Now see if we can make the remaining coins with lower denominations of coins. way += ways[x-coins[i]] if(i > 0): # Ways already populated ways[x] += way else: ways.append(way) print(ways[-1]) For reference: # 1 / 1 # 1 1, 2 / 2 # 1 1 1, 2 1 / 2 # 1 1 1 1, 2 1 1, 2 2 / 3 # 1 1 1 1 1, 2 1 1 1, 2 2 1 / 3 # 1 1 1 1 1 1, 2 1 1 1 1, 2 2 1 1, 2 2 2 / 4 # 1 1 1 1 1 1 1, 2 1 1 1 1 1, 2 2 1 1 1, 2 2 2 1 / 4 # 1 1 1 1 1 1 1 1, 2 1 1 1 1 1 1, 2 2 1 1 1 1, 2 2 2 1 1, 2 2 2 2 2 / 5 # 1 / 1 [1,0,0] # 1 1, 2 / 2 [1,1,0] # 1 1 1, 2 1, 3 / 3 [1,1,1] # 1 1 1 1, 2 1 1, 2 2, 3 1 / 4 [1,2,1] # 1 1 1 1 1, 2 1 1 1, 2 2 1, 3 1 1, 3 2 / 5 [1,2,2] I looked at the patterns with these small cases with coins = [1,2] and coins = [1,2,3] respectively and remembered a similar problem to do with counting number of ways to climb steps if you can make 1 or 2 steps at a time and using dynamic programming. I realised I could introduce the coins in ascending order and then use previous answers to see if we can make the remainder when introducing a new coin. P.s. im tired so I haven't written my explanation up very well but hopefully someone can understand the code :) Runtime: O(n) if number of coins is constant. I haven't really looked for more efficient ways as this was what popped to mind first.
    1
  10311. thank you very much for the lesson and the precious links
    1
  10312. In one journey around the stationary coin, the moving coin makes s/r rotations, where: s = the straight line distance between the centres of the two coins, r = the radius of the moving coin. This is because it is the path taken by the centre of the coin that matters - it is the one part of the coin that is invariant under rotation and so it moves linearly by a distance proportional to the angle turned and therefore conversely the angle turned can be deduced from that distance. The path here is a circle of radius s and the distance travelled is 2πs. For the moving centre to travel that distance the coin must rotate through an angle of 2πs/r radians i.e. s/r full turns. The circumference of the stationary coin is not significant here and all that is required is for there to be no slippage at the points of contact between the two coins.
    1
  10313. Most memorable: guessing (incorrectly) that the no 9 sum was infinite. It was worth a try :) I hope you can find that proof you wrote -- would like to see it!
    1
  10314. 55. Gauss's sum trick seems to work here.
    1
  10315. That was beautiful. Thanks
    1
  10316. The puzzle on the board, I solved by seeing that 12 squared was the average of the 5 numbers on top of the equation, so it had to be 144x5. Or 144 x 10 / 2 (easier to do in my head), so 1440/2 is 720. 720 over 360 is just 2.
    1
  10317. Good question… i wonder too. I could not find it for (x+3)^n, but (X+1)^n correspond to tetrahedron…
    1
  10318. Thanks for video , Please Mr can you make video about Euler-Compretz constant and the continued fraction for it 🙏
    1
  10319. Can someone put my jaw back in place??? Amazing stuff. When math and physics meet, magic happens.
    1
  10320. bachelor’s degree in computer science. Everything is ok except formula at 40:45.
    1
  10321. I liked "no partial sum (or difference between two partial sums) is an integer". Are the partial sums mod 1 dense in (0,1)? Are they uniformly distributed? Is there a number x s.t. the partial sums mod x are not uniformly distributed on (0,x)? Loved the video, got the brain ticking over.
    1
  10322. The way the proof at the end kept falling into place was so cool. Why can't all proofs be like that?
    1
  10323. Beautiful maths
    1
  10324. U earned many subs
    1
  10325. Todos tus videos son espectaculares. Gracias por tomarte el trabajo de hacerlos, son realmente inspiradores.
    1
  10326. implex + C
    1
  10327. The circumfrence of a circle is the sum of all points where the line through them is perpendicular to the duameter. The shrinking square always has 4-3.14159 excess where its line is not perpendicular to the diameter
    1
  10328. Calculus is easy. Algebra is hard.
    1
  10329. The ‘inverted’ sphere is the void space of a cylinder of heigh r, where the increasing radii of the sphere from pole to equator are the cross sections of the cone/void cut from the cylinder. It’s like an anti-integral. Remove from a solid cylinder whose radius is r, the shape of changing radii (0 to r) integrated across the height equal to the radius. I could go one further and fully derive the equations of geometric cosmology (A=pi(r)2 is to e=mc2 as V=(4/3) pi(r)3 is to … 😉), but you’re on the right track, and I enjoy watching these and don’t want to ruin the surprise. If this blew your mind, just wait.
    1
  10330. I certainly wasnt taught the remainder/modulus function in school. I learned it on my own through learning programming, and as you mentioned, it is EVERYWHERE. It's SO useful. The way i learned about it was actually not as a division thing, but rather as a "subtract the second number from the first one until it 's smaller than the second number". for example, 12 % 5 = (12-5) % 5 = 7 % 5 = (7-5) % 5 = 2 % 5 = ... well, just 2. cause 2 < 5. I don't even remember how i came to this explanation (which is essentially just a redefinition of division, the same way multiplication is just adding over and over) but it's probably similar to how i came to learn that sine and cosine are the vertical and horizontal (respectively) components of a vector because i kept seeing them used as a way to "move" an object in cartesian coordinates based on an angle.
    1
  10331. Oh no! Flat earthers will use this to make a map of flat earth where the area of Australia is now correct!
    1
  10332. Captivating
    1
  10333. 21:00 wird endlich mal Zeit :)
    1
  10334. Someone help me here. The horizontal area of the sphere and its inverse are meant to be the same at any given height. However, as you move the cross section down, the area of the sphere does not increase linearly but the area of a cone does, therefore the shape on the right cannot be the inverse of the shape on the left.
    1
  10335. I'm definitely going to chose the Tower of Lire. I like the idea of reaching infinity with small fractions. :)
    1
  10336. Thats the stuff! Übrigens, ich freue mich auf das kommende deutschsprachige Video;)
    1
  10337. My first proof was: Take 4 points and show that the angles opposite each other of that quadrilateral add up to 180 degrees. And that this is the case is actually not very difficult, you can find a lot of equal angles in the picture. Repeat the argument to cover all 6 points.
    1
  10338. 9:38 there is one way to make any number of coins, this somehow looks like writing numbers in binary form
    1
  10339. I didn't even think of any particular connection with complex numbers in the linear algebra step at the beginning, I just noticed that you can interpret a pentagon (and we can do an analogous process for any polygon with n sides) as a 5-dimensional vector with coordinate pairs as the elements. Then it's clear that any set of 5 random pentagons can be added together with appropriate weights to form any other, provided they are linearily independent. This is to say, 5 5-dimensional vectors form a basis for a 5-dimensional space, and a polygon can be interpreted as a vector in a 5-dimensional space if we allow the elements to be coordinates in R^2 or C or whatever 2 dimensional space you deem appropriate.
    1
  10340. Thanks for this beautiful and smart exemple from real life of ancient people. I'd love to know what would be the role of the water vessels constant level means in mathematics language and iits equivalent in economics? Youcef AMMAR KHODJA
    1
  10341. To infinity and beyond. Lol
    1
  10342. A small typo in the chart [6:16] Excircle drawn with r=20 (correctly it is 3*8=24) I can't begin to describe how many fascinating themes and connections are hidden in this video
    1
  10343. "Difference calculus"
    1
  10344. But supb.....
    1
  10345. you make math really imnteresting, not the painfull experience like in school...
    1
  10346. Another 6 related combination is 3^3+4^3+5^3=6^3 = 6*6*6=216 iliuminati confirmed :))))))))))))))))))
    1
  10347. 25:01 I was just thinking... "Wait, isn't it 1/sqrt(3)?" but it isn't.
    1
  10348. Computer science degree here. In CS, my math continued away from advanced calculus and more toward combinatorics, analytical topology, automata theory, etc. But I think I understood everything here. I might not have remembered enough calculus to do all the stuff you did, but I understood it, and could see how it worked out in the end. That's the nice thing about mathematics, which I tell my own students: Numbers behave themselves. If you ignore one part of an equation and come back to it later, it will sit there patiently waiting for you. It's not going to change or do anything weird. And if you can see how a part of your function is going to quickly get infinitesimally smaller, then you can ignore it as being close enough to zero, that you don't need to worry about it for a good-enough approximation.
    1
  10349. I got the mental arithmetic problem, but I didn't see a trick to make it easy, so maybe I missed the point.
    1
  10350. Could there be a cubic patter if we were to start at a three or four term sum instead of starting with a two term sum on the left hand side?
    1
  10351. How was this missed for 400 years?...majic?
    1
  10352. what about circle?
    1
  10353. Is this a sort-of 'inverse factorial' ? 🤔🤷🏽
    1
  10354. Okay, but how do I make a space/time machine out of this... 🧐
    1
  10355. Now I want to build my own physical helicone !
    1
  10356. Such a clever proof and simple enough for me to understand!
    1
  10357. Why only 30 years ago finding the family of phi cousins? Do mathematicians have something better to do than play with #'s? This didn't take computers to discover, it could've been seen even in the 60's on a chalkboard.
    1
  10358. I love math and all but I can't make it to the end of some of Mathologers more complicated videos. I loved this one because I was able to follow it to the end and, as a bonus, I was able to figure out the last puzzle in my head.
    1
  10359. I don't know about 292 not being a square, all I know is that he never shows up to my parties because he always "has to study", soooo. I'm not gonna say it.
    1
  10360. Doctor Polster, Please consider changing the background color from white to something a little bit easier on the eyes (possibly grey?). I watch your videos frequently, and no matter how much I enjoy squinting my mind through the amazing journeys you guide us on, the strain from squinting the eyes takes some of the fun out of it.
    1
  10361.  @Mathologer  As you guessed, most of the lights I use indoors are warm ( < 4000 Kelvin ). I also use inverted colors in most of my PC use (night mode in browsers, darker background colors in text editors while I code..etc). Since I spend most of my waking hours in front of a screen, avoiding bright colors has decreased my eye fatigue/headaches by a lot. Perhaps other viewers who complained about the same issue share my screen usage protocols or possibly have photophobia (discomfort or pain to the eyes due to light exposure or by presence of actual physical sensitivity of the eyes). *This study suggests that these concerns might be universal : "The Effect of Blue Light on Visual Fatigue When Reading on LED-backlit Tablet LCDs"
    1
  10362. Real magic, for real
    1
  10363. Simply delightful !
    1
  10364. Clanator.waiting.your.videos
    1
  10365. Perron Perron, que grande sos
    1
  10366. im 13. i learnt calculus is just a few days. at school we were doing ratios. its safe to say im a little bit ahead
    1
  10367. Love the Chris Haugen music ❤
    1
  10368. Warning: do not try this with actual ears.
    1
  10369. I mean I assume it's not how it is since you didn't mention it but it would be real nice if a 150 deg triangle had a -2T :D
    1
  10370.  @Mathologer  I mean that's not as perfect as one would hope but still real cool :D
    1
  10371. This is some great stuff, one of the best Mathologer videos!
    1
  10372. wut i'm really interested in is how much u can thin out the harmonic series while still keeping it infinite.
    1
  10373. The b=1 p=5 lantern shape @10:32 is called a pentagonal antiprisim.
    1
  10374. Here before 3 million views
    1
  10375. I just watched a 50 min video..... and I couldn't be happier.
    1
  10376. My favourite is the 100 zeroes. It's mind blowing because it seems to me that the exclusion of 9s is less of a penalty than relying on 100 zeroes, yet 9s are smaller. It's whack!
    1
  10377. Finally soneone agrees with my pythagoras conspiracy :D
    1
  10378. Thanks for this very well done video. Also, thanks to it, i was able to raise my IQ to arbitrary levels. 😊
    1
  10379. >So, he had a string of numbers. >He applied a pattern of his choosing >That pattern gave him pattern. How is this helpful at all?
    1
  10380. PLEASE PLEASE PLEASE Make a TikTok! Black pen red pen made one and a lot of gen z prefer TikTok over YouTube, and probably haven’t found your content yet. I’m sure a lot would love to see it! :) This is amazing!
    1
  10381.  @Mathologer  See, that’s where you’re gonna have to be creative haha!
    1
  10382. love depth and homework
    1
  10383. I wasn't taught this, but I did something just like it in school
    1
  10384. I like ur laughting :D
    1
  10385. i see you uploaded after a long time
    1
  10386. 13:52 The number of possible tilings of 2xn = n!² Examples: 2x1 = 1!² = 1 2x2 = 2!² = 4 2x3 = 3!² = 36 2x4 = 4!² = 576 2x5 = 5!² = 14,400
    1
  10387. 09:25 The four ways to write 18 as the sum of four (integers squared) appear to be: 3^2 + 3^2 3^2 + (-3)^2 (-3)^2 + 3^2 (-3)^2 + (-3)^2 But for 17, mentioned at 17:00, I can't work out the second block (of four) to get a total of eight ways. The first block are, 1^2 + 4^2 1^2 + (-4)^2 (-1)^2 + 4^2 (-1)^2 + (-4)^2
    1
  10388. Ever wonder, if such things are quite consistent with us "living" in a simulation, and we are realizing granularity and trick aspects of hardware acceleration in the simulator hardware? I'll show myself out.
    1
  10389. I do ponder. You think you can correlate the degrees of the contact points on the inner circle to the angles of the triangle? Or is that just something a clueless mind is distracted by? I paused the video at 6:17 to pose this question.
    1
  10390. This visual proof was described in a MathOverflow post by Moritz Firsching: https://mathoverflow.net/a/299696/88133. See that for more information, including a reference to some 2007 notes of A. Spivak.
    1
  10391. Coloring proof. In my head I started with the rotations proof but I don't like them as "proofs" since it just leads to statements like "these two points are the same" which I can't visually differentiate. I could write code to simulate it, but in the end, that is just solving it algebraically and presenting the output visually. The coloring proof actually shows the detail that the rotational proof needs in order to justify doing its rotations. For me it really came down to realizing that the center of the chord must be the tangent to the inscribed circle. Then its obvious that the distance of any point depends only on the radius of the inscribed circle and the length of the triangle perimeter. Which, of course justifies the rotational symmetry that I started with.
    1
  10392. Dissapointed, that 42 never appeared 😊
    1
  10393. i have come around to the idea that the use of " + ... " is too ambiguous. Recursion seems to be the only situation where you can avoid doing anything suspect. You can't just say "+ ... " and chop the tail off; because when you say A=B, they need the same information to be equal. Without specifying a recursive tail, the two sides might not have the exact same information. But it's really clear when the tail is there. S = -1 -S = 1 = (1-2)S = S - 2S = 1 S = 1 + 2S = -1 = 1 + 2(-1) -1 = 1 + 2(1 + 2(-1)) -1 = 1 + 2 + 4 + 8 + ... + 2^n + 2^{n+1}(-1) // it's obvious why S=-1, and there's no use of infinity // it's just that recursion expands to patterns. S = (1 + 2 + 4 + 8 + ... + 2^n) + 2^{n+1}S S = Sum(S) + Tail(S) S - Tail(S) = Sum(S) = Sum(S) = S - 2^{n+1}S = (-1)(1 - 2^{n+1}) = 2^{n+1}-1 ie: 1 + 2 = 3 1 + 2 + 4 = 7 1 + 2 + 4 + 8 = 15 ... ie: using recursion, we solved for the closed form for the sum of n terms. The S=-1 doesn't mean anything like what people assume it means. The S=-1 was crucial to correctly caclulating the sum. It makes sense, because when Sum(S) is large, Tail(S) must be large and negative to cancel it out. You have a convergent series when Tail(S) goes to zero as n gets large. ie: (S = 1 + (1/2)S). You can reproduce this with -(1/12) = T = 1 + 13T = 13^0 + 13^1 + 13^2 + ... + 13^n + 13^{n+1}T I'm pretty sure that there's a way to implement these series in code, where the paradoxes clear up. " + ... " doesn't really seem to have enough information to do clear and machine-checkable calculations.
    1
  10394. 36:20 Look at the small triangle in each corner of the figure, and call it A. You can tile the trapezoids with 3 copies each of A. That makes 16 total white copies of A. You can tile the blue square with 4 copies of A. So the blue square is 4A/(16A+4A) of the whole figure, or .2.
    1
  10395. Fantastic 😍💋 💝💖♥️❤️
    1
  10396. For the case of writing 2 as the sum of the repricols of positive integers, there are infinitely many ways as attributed to the infinitely many perfect numbers. Also the sum of the factors of a positive number can never exceed 3 times the number itself E.g 2*28 = 1+2+4+7+14+28 2= 1/28+1/14+1/7+1/4+1/2+1
    1
  10397. We always called sinh 'sinch', but I like yours better.
    1
  10398. I thought one should not refer to headdress as a towel.
    1
  10399.  @Mathologer  Well, right. I know. But, for some today that are expected to wear certain items in order to be in keeping with customs and rules to be followed, it could be that they already feel out of place in certain instances. And it could be that the words "towel" and "head" used together, even if referring to some European at the time could bring unnecessary tensions into play.
    1
  10400. Q-h
    1
  10401. The inherent flaw at the core of mathematics allows it to act as the foundation for all manner or irrational thought. It's great fun as the Mathologer regularly demonstrates in all his lectures but it also accounts for the failure of mathematics to account for 'realities' at the quantum and cosmic level.
    1
  10402. I learned this in late 3rd grade ,early 4th grade but not in the USA ,in a 3rt world country.
    1
  10403.  @Mathologer  Most circle geometry has been removed/reduced here but that one is still included.
    1
  10404.  @Mathologer  Utah, USA I teach 7-12 (12-18 yr olds) math/physics in a school out in the desert with 40 students. Geometry is not a specific class but has been broken up and incorporated into all the math classes.
    1
  10405. You know why the don't teach this in school. Because they think you're stupid. And maybe it's true. Smart people always have to find ways to learning new things outside of school.
    1
  10406. I graduate high school this year - I found this video extremely fascinating, but many aspects a little hard to grasp. Does anyone have any recommendations for college courses I can take next year to help me better understand complex mathematical proofs?
    1
  10407. Question: From the periodicity and the physics I am led to believe that the rubber band should be under some tension when this whole process starts. However, if it's not under tension then any arbitrarily small angle of rotation will cause at least some tension. Does this mean that no initial tension is required after all?
    1
  10408. The windmills of our minds :) Nice elegant proofs!
    1
  10409. The second proof is surely simpler for showing that the 3 folded triangles must be congruent, but doesn't it require an extra step (e.g. based on angles) to show that those triangles are also equivalent to the original triangle (thus making it a little less simpler)?
    1
  10410. With regard to turning the rightmost card, if the move was not allowed could we not just say that the problem must terminate at either 0 or 1? Or, if it wraps around turning the leftmost card face down again, could we not prove that there is an order to turning the cards that will terminate and that we will eventually hit upon that order?
    1
  10411. I just wrote a cryptography module in Java that uses some of the maths in the video to encode and decode messages. No idea how secure it is, and it certainly isn't very efficient... but it was fun! :) Thanks for the inspiration.
    1
  10412. This is so cool! I immediately paused the video and programmed it to see for myself. Thx for making my day
    1
  10413. ---------❤❤ -------❤ ❤ -----------❤
    1
  10414. How many ways to make a change for 30 dollars and 40 cents?
    1
  10415. Recreational mathematics..Indeed
    1
  10416. I knew most of the stuff. I would vote for Kempner's proof as it was something new for me. Thank you for teaching me all these new things.
    1
  10417. How Gauss would have done it? Too simple to me. No formulas I am using. Only elementary insight. Take the sum of first and last term. It's 101. Then take the sum of second and second last term. Again 101. Keep doing this 50 times. You get 101x50. A kid, if genius and have no background knowledge, can do only by this method at his first sight. I guess same applies to Gauss childhood:)
    1
  10418. I would be interested in seeing a difference version of Laplace transforms. I guess that it would be an interesting case study.
    1
  10419. What I am more intrigued by is the pronunciation of halving with F as hal-F-ing instead of the usual pronunciation h-AL-wing 😂
    1
  10420. I worked in NYC in the graphic arts industry. Back then most calculations for enlargements or reductions in that business were done on a circular slide rule. I've always been pretty good at math. One day my manager was getting frustrated using her circular slide rule to find a reduction percentage, so she said "Instead of fiddling with this thing, I should just ask you for the answer". She told me she wanted the percentage reduction for a size of 32 down to 17. I replied ".53125", the answer to five decimal points without use of a calculator. The boss asked her if that was correct, as she checked it on the calculator. She silenty nodded "yes", but of course no one gave me any credit for knowing the answer. In that business everyone made the same hourly pay, with no consideration for skill, education, or even if you produced 400% of the work output of your colleagues. It was a totally different world from nowadays where millions of people "work from home" and actually produce nearly zero actual work product in a day or week's time.
    1
  10421. Wow. Mostly men to give paypal and patreon. Women do not want maths on youtube? Are there famous female mathematicians out there in history? I remember just one from mathologer videos.
    1
  10422. funny that because our device screens are square grids, none of the triangles you showed could have been equilateral
    1
  10423. Where can I get that shirt!?!?!?
    1
  10424. fun illusion: at 35:19 or so, if You look at the tree in the right way, it looks like a lattice of square tubes
    1
  10425. Thanks
    1
  10426. My proof at 2:48 is this: All three lines are the same length (one green, one red, one blue in various permutations). Each pair of lines is connected at the same distance from an end and the second equals the first only turned around that connection point. So the four endpoints of each such pair of lines must lie on a circle (out of symmetry reasons). Only the question remains why all three pairs of lines form the same circle. I guess because it can be shown that all three lines have the same distance from the center of this circle. Let’s go on with the video.
    1
  10427. Squeeze out some more information from your "symmetry reasons", and you have a very nice proof: the pair of lines are the diagonals of a symmetric trapezoid. Its symmetry axis is an angular bisector of the diagonals and contains the centre of its circumcircle. So the centre of the circumcircle lies on an (interior) angular bisector of the given triangle. As this is true for all three pairs of lines, the six points lie on a common circle concentric to the incircle of the triangle.
    1
  10428. The end Animation made it worth it
    1
  10429. Mathematics in real life!
    1
  10430. Does this have a meaning for origami lovers?
    1
  10431. just like that learned like 5 new things
    1
  10432. was taught, but having a root and weird terms without knowing the proof made it difficult to use. I couldn't recall it until the very moment it showed up on my screen, smh.
    1
  10433. If the derivative is the slope of the tangent line then it is also the tangent of the angle opposite of the small change in x dx and the difference of this angle dx with the angle dy ie the angle opposite of the small change in y dy will be analogous to a |dy/dx| where a number 0 < dy/dx < 1 is increasing/decreasing slowly or angle dx > dy and dy/dx > 1 is increasing/decreasing quickly or angle dx < dy how you’d find these angles dy/dx I have no idea maybe convert it to polar coordinates
    1
  10434. I like how this guy laughs
    1
  10435. Today is gonna be the day that I'm gonna watch another Mathologer video.
    1
  10436. "there is one way to make zero sense" 😅 I thought there were more. Seems like writing something that makes sense is easier than we thought.
    1
  10437. Cristopher Columbus nickname came from word Columbarium - pigeonhole (box) in 17 century ? Masonic Columbarium graves of 17 century but Washington Columbia district is not a hole LoL
    1
  10438. Stigler’s Law of Eponymy - a discovery is never named after the first person who actually discovered it.
    1
  10439. first time i saw this ... and ... damn, really easy ways... come on, why dont they teach this....
    1
  10440. I don't really get what benefit a decimal representation of Pi has. It may be more convenient for some reallife applications, but mathematically speaking, they may as well had used a binary or whatever representation system. On a side note I'd had prefered a symbolic system anyway, where new potential digit values are introduced on the go by adding the set of all arithmetic combinations of prior elements as new valid digits, thus completing the set of constructable values. Currently our number system is only based of increments of '1', which may be convenient, but mathematically uninteresting.
    1
  10441. it is dec. but what about hex or oct?
    1
  10442. This must be the reason behind my deep hatred of shoelaces
    1
  10443. Thanks! I always wondered why that formula worked, now I understand the underlying mechanics of it. PS Merry Christmas and Happy New Year.
    1
  10444. The take away is that the order of computation of an infinite series matters? Schöne Grüße aus Nordrhein/Westfalen.
    1
  10445. Mathologer, I have been doing number theory in my own time for 6 years now. I have been fascinated by Bernoulli numbers, it's connections to taylor series expansions of trigonometrics functions, and particular values of the zeta function for a long time now. From my own work, I was able to show many interesting results, but several identities I found online that I knew were true were impossible for me to prove. After watching this video, everything makes sense!! I was using partial summation for a long time to prove some results, but now I have a more powerful tool, and I know why it works! Thank you so much! A lot of my future work will be partly due to you. Don't forget to make the follow up video!!!
    1
  10446. I have been taught this at school.
    1
  10447. The diameter of the weird curve is just a chord of the circle, so it's smaller than the diameter of the circle.
    1
  10448. 25:02 shouldn’t that third factor be A+C+D-B, not A+B+D-B?
    1
  10449. A really math firework 😁
    1
  10450. Dang it, now I got a hankering for nachos.
    1
  10451. The leaning bricks over the edge were definitely the most interesting part
    1
  10452. Not me watching Mathologer video instead of watching my professors lessons so I dont fail
    1
  10453. As for as the code challenge:I got the change of base (5), easy, will implement hex notation and base decimals if you wish. Got it so you can choose what number is on top on the outer ring and inner ring, so wish list (1). Wish list (2), zoom for "infinite precision" has been the greatest challenge. I do have the zoom feature implemented. But getting into javascript precision cruft, getting 2.01000000000003 for 2.01. Considering the logs and trig functions involved as well as multiplying to zoom in, that is asking for that. I will work on removing it from the finished product. I opted to skip the labels and tick marks that fall off screen so as to save memory and processing, I should also apply that to the circle that falls off screen. I implemented the squaring. Presently you tell it what number you would like squared and it places that number on the top for the inner ring and on the out right top the mod(inner_top^2,base). By change of base did you intent it to happen on both inner and outer circle, which is what I did, or individually;example: outer ring is base ten, inner ring is base 2. I could also work in hexadecimal notation if you would like.
    1
  10454. Great video, but I'll watch it a few more times :) The area of ​​any rectangle (square) is equal to the sum of the areas of two equal triangles (which means either the root of 2, or the root of 5). I probably did not understand the condition, 7 + 8 = 15 = 1 + 5 = 6 and 6 + 7 = 13 = 1 + 3 = 4. Thanks so much for your video, but have you really found the connection between pi and fi. I will wait for the continuation.
    1
  10455. Was this a development of Napier's Bones?
    1
  10456. So a flow on the surface in all directions is ????? How do you go a level up rather than a level down? That would be the total flow of a sphere rotating?
    1
  10457. the fact is, if the series is infinite then there will never not be twice as many positive actions than negatives which means they can't perfectly cancel out.
    1
  10458. I made it to the end; I am a high school junior who just really loves math and calculus. Everything was good up till the part where you changed the 0s to 1s. After that I just took your word for it lol
    1
  10459. Hyperinflation solves it all.
    1
  10460. While I really enjoyed all of this video, my vote would have been for the bizarre and interesting overhang solutions, as even though they were ugly, they were completely non-intuitive, and hilarious at the same time.
    1
  10461. Is it just me or I feel like this is reflected in particle math re quantum spin. Maybe it’s just the matrix math but ....
    1
  10462. The video title doesn’t English.
    1
  10463. Go one by one.
    1
  10464. Finally, I've waited for 50 years for a proof that I could understand. I did an under-graduate number theory course and it wasn't covered. Thank you.
    1
  10465. Welp, here is another video that will get me half a year to go through!)
    1
  10466. Definitely "no integers in the series" (apart from 1)
    1
  10467. First time I see a Mathologer video where I have already seen everything on the university. Great stuff by the way. ...Probably last time as well.
    1
  10468. horn with a mute --- is that a mime with antlers
    1
  10469. Nice shirt
    1
  10470. I like the shorter video every now and then
    1
  10471. Gamma, followed by 700y proof
    1
  10472. Why not 4k-1 instead of 4k+3 ?
    1
  10473. Best weekend ever! Ha ha
    1
  10474.  @Mathologer  Just that it is nice that two great YouTube channels released videos on the same weekend.
    1
  10475. ​ @ಭಾರತೀಯ_ನಾಗರಿಕ Now I'm trying to figure out how the MythBusters could get involved....
    1
  10476. The unsuccessful primes could also be written as 4k - 1. Not sure if the whole thing would still work though...
    1
  10477. The first time i really understand Quadratic Reciprocity law !!!! Thank you a lot for this.
    1
  10478. Precision is acceptable if you multiply or divide the last figures in your head. Lovely Mathemagics. The "Pacman Entanglement Universe" is also the Modulo of concentric log scale orbital-orbits of e-Pi-i sync-duration connectivity in the axial-tangentially Fluxion-Integral Temporal Calculus structural-coordination of ONE-INFINITY => United Pi*i zero-infinity axis orthogonal-normal alignment in the Universal Circle of vertices in the demonstrated exponentiation-ness spiralling vortices (of hyper-hypo transverse trancendental complexity, Hydrogen-Neutron formation). WYSIWYG, a sum-of-all-histories in a conic-cyclonic modulo-geometrical example of QM-TIME In-form-ation, continuously created probability objective-aspect of Quantum-fields pulse-evolution in logarithmic Time, visual representation of AM-FM Duality. The Rubber Sheet demonstration of Gravity should be reviewed in this context too, it's the axial-tangential orthogonal-normal tangency surface envelope, line-of-sight projection-drawing of e-Pi-i logarithmic resonance floating point coordination functionality. More Fluxion Calculus at Singularity and Integral Calculus at numberness dominance symmetry of the Unit Circle. Wow realization! Superbly presented micro documentary, thank you.
    1
  10479. This is an amazing video :) I mostly enjoyed the no digit sums
    1
  10480. Interesting split of your viewers. Over halfway through and the credits roll, then several more minutes of actual teaching to prove how simple the communicating vessels properly distribute the various estates.
    1
  10481. Doesn’t matter how long it is. I’m going to watch it.
    1
  10482. The Newton Gregory formula is mind blowing
    1
  10483. Your German was really impressive (saying this as someone born there), did you train those words or can you actually speak fluently?
    1
  10484. Okay, NOT the same answer as given at 39:37. I wonder what this means. EDIT - Okay, turns out I simply skipped a step intuitively. I'm relieved.
    1
  10485. Maths got the wildest true conspiracies
    1
  10486. If the way the digit representing one was more round the curve would have looked more smooth.
    1
  10487. I have no idea what is going on but I can feel this guy's passion in teaching so I had to subscribe
    1
  10488. Terrible aim 😂😂😂
    1
  10489. 17:25 multiple of 2 not power of 2 Hyperventilating intensifies
    1
  10490. At 12:30 the green 6 got lost
    1
  10491. Archimedes’ claw makes me want some schupfnudels. Don’t know why
    1
  10492. Another great one! My first thought was that you'd arrive at the fixed point of some mapping through a symmetry group. The later use of roots of unity, or the cyclic group in disguise, reminds me of the trick for finding a function invariant: for "any" function f, take the formal sum of f + g(f) + g^2(f) + g^3(f) + ... Here it would be a finite sum of course. Finally, you're on to something with the smoothing of a segment chain - notice how adding trivial points make the ears approximate the centers of curvature...
    1
  10493.  @Mathologer  Would you be up for discussing how this could lead to curve-shortening flow?
    1
  10494. I vote for (maximal) overhang. Mike and Uri's paper really is beautiful.
    1
  10495. dans un réseau nxn hier g supposé qu'on ne peut le couvrir en dupliquant la figure je sais bien que sur un échiquier on parle d'un cheval capable de couvrir toutes les cas en partant de la sienne
    1
  10496. The rubber band analogy is new to me. But it's genius. I'm having a 'Aristo' slide rule in my drawer. And as a amateur astronomer I have a circular calculator for siderial time called "planisphere". Even drawed one myself about 40 years ago as I was 15. The only drawback of slide rules is that you cannot use it on accounting where you have to chase the penny. @Burkard, circular scales reminds me on calculation with modulo. Maybe a topic for the future, like encryption or FFT. Or did I missed it.
    1
  10497. There is no problem with using 1's. The problem is your broke the pattern of using unique integer (so using each integer only once per side). If we follow that rule there is there a four integer line after the 1,2, 3 line?
    1
  10498. I see what you did there; "Pirodox" @ 0:38
    1
  10499. Absolutely love this!
    1
  10500. I can relate to the unfounded sense you've stumbled onto some fundamental set of rules for a finite system only to realize that it's baked in as part of the numerical base and really not that remarkable. When I studied music I got really into Hugo Riemann's "tonnetz" and figured out a way to truncate the pitch classes into a toroid when I was looking at a square bifrustum with a bore through it's square facets. I was convinced it was some blueprint to harmony... until someone who studied topology showed me that, while 12 is interesting, changing bases elicits a multitude of toroidal figures. Some that are frankly more useful musically speaking.
    1
  10501. Since it is often desired to start k at 0, or 1 if necessary, why not define the primes which cannot be expressed as the sum of two integers as 4k-1 (instead of 4k+3)?
    1
  10502. Sorry for injecting political content into this excellent lesson, but, as an American, I can't help but identify Burkhardt as a Democrat and the Tesla key-to-the-universe crowd as Republicans!
    1
  10503. If we cross Alfred Moessner with Charles Babbage, we can design a very strange difference-engine to do these calculations.
    1
  10504. I am not highly educated.  This was wonderful.
    1
  10505. if all of calculus can be done with these nice simple sequences of friendly numbers, why are we bothering with infinitesimals and limits of secant lines etc etc
    1
  10506. I solved the final puzzle by method of re-arranging shapes. The entire area is exactly 1. Now by focussing on the 4 smallest right triangles (each with the hypotenuse of 1/2). Rotate each one CW by 180 degrees about its larger acute angle, and each completes a square with the neighboring quadrilateral. The result is large tilted "plus sign" made of 5 congruent squares. The center square is one of 5 and therefore has one-fifth the total area or 0.2.
    1
  10507. When we discussed divisibility for 9, the question came up about repeating the process and everyone agreed it was pretty much obvious that it would boil down to 9 given enough iterations. I guess I lived in an era where we didn't have to be taught it explicitly.
    1
  10508. How did he know what ? :)
    1
  10509.  @Mathologer  I'm only partially kidding. I've been a long time viewer of your channel and I enjoy your videos. As a matter of fact your video on e^i*pi cleared up a lot of things for me. I don't dislike the video. It's Clickbait but it's the good type of Clickbait which actually satisfies. I guess I should've put '/s' to indicate sarcasm. Great video and keep up the amazing work.
    1
  10510.  @Mathologer  Also not much of a point for me to leave a dislike anymore since youtube removed that shit. But also because I like your videos.
    1
  10511.  @Mathologer  Although differential equations are entirely different beast lol. Love the visual proofs and animations in the video.
    1
  10512. I wouldn't try this rubber band trick with hamstrings, no sir.
    1
  10513. Well done again Mathologer. I think that 22/7 is practical - most people with basic arithmetical skills can do this by hand or even in their heard. For example - if I have a circle radius 10cm - we have 22/7 * 100 = 2200 / 7 = 314 cm squared - good enough if you are icing a Christmas cake.
    1
  10514. this is the magic
    1
  10515. Try integrating tan(x)^(1/3), for instance. That's not very easy :-)
    1
  10516. Great video
    1
  10517. But the cat had definative boundaries in the x and y directions. 1/X is effectively infinitely large in both x and y. So it doesn't seem like a fair comparison. The squash and stretch rule is true of all defined 2d shapes but clearly cannot be applied to non definef (I. E infinity large) shapes. You can divide the length of the cat by 2 and can multiply the height of the cat by 2 but you can't divide infinity by 2 or multiply infinity by 2.
    1
  10518. Rapaz, conheço matemática teológica de cabo a rabo por isso. Quando Newtão não dá pq a sociedade enlouqueceu... olavar é preciso Amo músicos e artistas inclusive Caraio, hein, sociedade?
    1
  10519. WOW
    1
  10520. I never knew this was not known to mathematics. I never knew the math behind it nor the specifics as presented here, but as a kid I loved playing with geometries and jotted down mandalas and stuff in school when I was bored, which was quite often. I did notice symmetries and relationships, but always thought that I only "discovered" things that were already old knowledge. Now it feels the math I learned in school was mostly a waste.
    1
  10521. why it seems to me that at the beginning of the puzzle i should move the minor piece to the stick that i don't want the pile to go to, but in the straight schematic is the opposite?
    1
  10522. 👏👏👏 So excellent lecture. Thank you, Doctor, your scientists and your colleaques. and With luck and more power to you. hoping for more videos
    1
  10523. When senpai finally drops the long video you always wanted
    1
  10524. I would be super about more esoteric (hermetic, occult, etc.) math videos. Some fun stuff.
    1
  10525. I wonder what kind of patterns one would get is alternating between inward and outward facing isosceles triangles for each sequential line segment.
    1
  10526. Great video. I think you meant the first B to be a C in the third factor under the radical.
    1
  10527. Most memorable part, wondering whether the number of tricks that allow these questions to be solved converges ;-)
    1
  10528. The greedy one was quite good for me. That may be because I was too lazy to figure out the 700 year old proof. Greed and sloth are among the classic 7 deadly sins. If you remove these two habits from your routine, will your total number of sins be finite, or still be infinite?
    1
  10529. This got an instant subscribe, awesome video, super fun to watch, perfect pacing and i love the tshirt
    1
  10530. You weren't fibbing about the 14:10 challenge. I found it easiest to see it this way: When going from the set of 2xn configurations to the set of 2x(n+1) configurations, first partition the set into configurations that have one or more vertical tiles on the right and those that don't. And we'll extend them always by adding onto the right side. And let's denote the cardinality of the first set A(n) and the second as B(n) There are two ways to extend those with at least one vertical tile on the right: add another vertical tile, or remove the vertical tile and add a 2x2 block of horizontal tiles. For those with no vertical tile on the right, there's only the first way. Notice that there are two ways of ending up with at least one vertical tile on the right: Adding a vertical to an A(n) configuration, or adding one to a B(n) configuration. A(n+1) = A(n) + B(n) And there is only one way of getting a configuration with no vertical tile on the right: Starting with an A(n) configuration and transforming the right to a 2x2 block. So: B(n+1) = A(n) Let's also notice that if we remove the 2x2 block we just added, the number of configurations of what remains is the same as for 2x(n-1). B(n+1) = A(n-1) + B(n-1). So with a little algebra, fib(n+1) = A(n+1) + B(n+1) = A(n) + B(n) + A(n-1) + B(n-1) = fib(n) + fib(n-1) Which is a formula that any son of Bonacci would recognize.
    1
  10531. At 14:25 it might have been clearer if you (and Torricelli) had used an arbitrary shape instead of a disc. I tried to imagine where the disc came from until I realised that any base area for the prism would have worked. But this is still the problem part of the proof for me. To a pro mathematician it may be obvious why the infinite shells are exactly as "many" as the infinite discs, but when I tried to use similar reasonings at my exams, I tended to end up with the wrong result, zero points and a stern remark that "You can't do that!". 3b1b has a video about when beautiful proofs are not proofs at all: https://www.youtube.com/watch?v=VYQVlVoWoPY It resonates with me and probably lots of other students that got that remark on exams.
    1
  10532. Linternas hechas de anti-prismas.
    1
  10533. This is also related to monopole searches in physics.
    1
  10534. What about on the molecular level? Wouldnt there eventually be 1?
    1
  10535. The incircle is a concept wholly new to me!!!!! Why have I not heard of this?!???
    1
  10536. 1+1/2-1/3+1/4-... is ln 2, but what is 1-1/2+1/3-1/4+1/5-...? I guess it diverges to infinity.
    1
  10537. My friend introduced me to this channel.
    1
  10538. in taiwan we teach Heron's formula after we learnt the trigonometry, but teachers tends to tell us the formula in junior high when we didn't even know the trigonometry. this is a great proof for students in junior high to understand where it comes from ouo
    1
  10539. I like your pumpkin pi... especially at Halloween.
    1
  10540. I write so much code based on your videos.
    1
  10541. ok ok I knew I wasn t the only one... you said: just checking you heeh- hee - hee. ;-)) but most of your classroom just got asleep by this time.... anyway your graph skills are awesome :-))
    1
  10542. I never get what you deliver but considering the positive comments, your videos must be very interesting
    1
  10543. I really liked this one.
    1
  10544. Love it! Is there also an iterated way to generate random hexagonal boards?
    1
  10545. So if we truncate a highly regular infinite series for PI, we can correct the finite series by a highly regular infinite fraction! Somehow the decimal representation of PI completely obfuscates its highly regular structure.
    1
  10546. as a simple analogy, two steps forward, and one step back will always move you forward, even if, at infinity, you can match every step forward with every step back. Saying it this way does not make intuitive sense. but when rearranging numbers, it can seem somewhat logical.
    1
  10547. Requested Feedback: As always your animations make things generally quite easy to follow, though a personal pause every so often doesn't go astray. Similarly the finer points only come when working through the full calculations, but the overview is generally perfectly sufficient. I teach high end high school math and physics at a low SES school in Victoria, Australia. I have a BSc maj Math & Physics, plus the original BComm maj Eco, Fin, Intl Trade. I watch all your stuff, and periodically find a way I can use it in class, but mostly I just highlight you and others exist to the properly interested so they can follow the rabbit hole themselves. Commonly in class they require a few more gaps filled in a bit more explicitly, but that can be for all sorts of reasons, none of which is associated with the delivery, just their (mis)thinking. Continue your great work, it has been greatly appreciated for some years now.
    1
  10548. can you work the number 2520 into all this some how? I'm sure you know the significance of 2520, but incase you don't LOL , 2520 is the first number divisible by all numbers 1-10.
    1
  10549. Thought the formula was just a coincidence... never knew there is such a simple geometric transformation from one to the other...
    1
  10550. Once you know either the formula for the surface area or volume, you can derive the other by thinking of a sphere as being made of an infinite number of infinitesimal-base pyramids with height r where all the vertices meet in the center.
    1
  10551. Dürer folium: corresponds to the two ring/cylinder singularities in protons and neutrons. Its inverse is the Delanges trisectrix that concerns the elliptical bonding in atomic nuclei, where there are only (other than cases such as hydrogen without any neutrons) protons and neutrons. Dürer folium at the hyperbolic singularities.
    1
  10552. Enjoyed it! 👍
    1
  10553. I managed to understand the factorial part using similar method: the first time when we move to the "left block" we have (1,1,1,1,1) to (1,2,4,8 16) which is the same as the power proof, But when we move to the left block the second time, since the height decreases by one so we are left with (2,4,8,16). Writing it as 2x(1,2,4,8) and it becomes 2x(1,3,9,27). Now the third time, we start with 2x(3,9,27) which is 2x3x(1,3,9) therefore it becomes 2x3x(1,4,16). The last time we start with 2x3x(4,16)=2x3x4x(1,4) and end with 2x3x4x(1,5). This directly ends with 2x3x4x5 which is the factorial number.
    1
  10554. If there were two ways of writing prime as two integer squared, wouldn’t the total number of solutions for x^2+4yz be even?
    1
  10555. I saw the first proof on TV some time in the 1970's on one episode of Jacob Bronowski's series The Ascent of Man. Never seen the second one.
    1
  10556. M - I - I = V
    1
  10557. ...like: the "simplest" 3D-object should be a tetrahedron. If you take any irregular tetrahedron and add "ears"(*) to every plane of it, the endpoint of these "ears" should form another tetrahedron. After a certain times of iterations you should end up with a regular tetrahedron... (*) wich should be also be tetrahedrons, made out of 3 isosceles triangles of a certain angle and a plane of the original tetrahedron.
    1
  10558. Already smashed the like button multiple times before Intro was finished. Such was the enthusiasm to learn.
    1
  10559. Darn, I never learned a lot of this at school
    1
  10560. Just love the crazy optimal overhang solutions. How does one approach such a challange?
    1
  10561. I love this guy. Everytime you upload a video my heart jumps in joy, and I cant wait to watch it. Its the same excitement when I watch baby elephant videos.
    1
  10562. 2 1/2 minutes in and my eyes are crossing. I am not meant to be a mathologist 🤦‍♀️
    1
  10563. Ten-gon? Decagon!
    1
  10564. You mentioned relativity. A lot of ppl get mixed up thinking you can move relative to space…moving is just a shapeshift….stretching time and squishing length for some other observer. But as far as you, you moving persons thinks: you just still a regular 1/x hyperbola. This is why you can never reach the speed of light: no matter what you do , it’s alway “c” away in all directions.
    1
  10565. your video is precious man, thnx
    1
  10566. Hi Burkard can we have access to the presentations?
    1
  10567. I loved the way we go from equations to matrices
    1
  10568. THANKS!!!
    1
  10569. Greetings from his grandson, Martin Sperner 😅
    1
  10570. Messing with infinities and zeros can get you in trouble 😱 But then again, zero is always zero, but infinite, do they come in many different sizes? It's mathematically proven that continuum hypothesis cannot be proven true nor false. So there is a possible universe where continuum hypothesis is false. Do we live in a universe/multiverse/hyperverse where continuum hypothesis is false? Are there many different sizes of infinite multiverses? 🤔
    1
  10571. Very cool! Also I didn't know Ptolemy's theorem before - that's wonderful!
    1
  10572. Mathologer never disappoints.... Infact I should be disappointed in myself that sometimes I don't get stuff lol. I am still learning cool stuff at least XD
    1
  10573. Can't say I can quite follow but I enjoy these videos very much nonetheless. Heads-up: his name is mis-spelled in the first line of the description...
    1
  10574. Very nice! This method is great for sequence prediction but it gets tricky if the sequence is noisy or it has sharp edges like this sequence: 123412341234 but in general it's a very intuitive way to get into the differential equations. We used a similar method to this one few years ago for sequence prediction and it works nicely as long as the sequence is smooth and not very noisy.
    1
  10575. I read the same book at school! I got differentiation but struggled with integration. Was not dedicated enough to finish the book - probably why I ended up not being a mathematician
    1
  10576. i particularly enjoyed the cantoresque visualization of the no-9s series
    1
  10577. great stuff as always, thumbs up first time patreon, seeing my name felt real good bit disappointed it was only 30min long rather than full hour
    1
  10578. A thought i had about your program is you cna make this a fun form of encryption for maps, or locations and even messages. The process is you make a line and pick a number like 18 points along that line. Then the distances of the line and how many numbers are chosen would determine the pattern and shape and size, this could in turn create a map. Further increase its complexity with how you right it down which could force you to alternate or counter or clockwise simply by the number orientation, laying a few roles that its always in sequential order, but how they are listed determines the other shape. TO further make a complex encryption you could take it further and put a curve into the system, ,plus you can denote the terms of measurements also. Interesting how distance effects location... so its a interesting way you could encrypt a very topographical map with recognizable features. A test could be attempting to encode a location at the school in such away. It may require two orthogonal lines with perpendicular intersection, and one letters and the other numbers.
    1
  10579. 8:26 y'(x) = xy(x) y'(x)/y(x) = x ʃ y'(x)/y(x) dx = ʃ x dx ln|y(x)| = x²/2 + c |y(x)| = e^(x²/2 + c) y(x) = ± e^c * e^(x²/2) Since ± e^c is a constant, let a = ± e^c y(x) = a * e^(x²/2)
    1
  10580. More details about Madhava, their desciples (who were also great mathematicians in their own right) can be seen in this lecture series. https://www.youtube.com/watch?v=yWZ15EKElH0
    1
  10581. Nice extension of the classic! If only Martin Gardner were still alive to enjoy this!! Fred
    1
  10582. so, a few naϊve observations in 12:08, if you take the determinant of these 2x2 matrices: 1) det( left_child ) == -det( right_child ) 2) det( parent ) == -det( center_child ) 3) abs( det(...) ) is a prime number (except for det == 1, if one excludes it from the list of prime numbers to begin with) Not sure if the pattern breaks the more you go down the branches though...
    1
  10583. Hello in german
    1
  10584. Just commenting to say the shirt should say 25.807 instead of 25.8069. The square root of 666 to seven decimal places is 24.8069758, which means the digit following 25.8069 is 7 and should be rounded up to 25.807. Not only is this more accurate, it also increases legibility when plugging in into a calculator, as 25.807^2 = 666.001 while 25.8069^2 = 665.996.
    1
  10585. Very nice! You've especially taught me that I shouldn't have skimmed so much of Conway and Guy's The Book of Numbers so I could get to the part about the surreals. I should not be so mathematically provincial!
    1
  10586. 3^4 + 4^4 + 5^4 + 6^4 = nothing great. Just 2 times a prime number. Waste of time.
    1
  10587. 3:12 No, not at all.
    1
  10588. Singing: Don’t know what a slide rule is for… (Wonderful world- Otis Redding )
    1
  10589. Evil genius villain voice, but probably the nicest guy in his town.
    1
  10590. OMG! Lets "imagine" we know the Taylor Series for both cos(x) and sin(x) respectively; cos(x) + sin(x) = [1 - (x^2)/2! + (x^4)/4!...] + [x - (x^3)/3! + (x^5)/5!...] making x -> ix get, cos(ix) + sin(ix) = [1 +(x^2)/2! + (x^4)/4!...] + [ix +i(x^3)/3! + i(x^5)/5!...] cos(ix) - isin(ix) = 1+x +(x^2)/2! + (x^3)/3!+(x^4)/4! +(x^5)/5!... or e^x ≡ cos(ix) - isin(ix) notice that both cos(ix) and isin(ix) are real functions, i.e. have real ranges. so if we define cosh(x) = cos(ix) and sinh(x) = -isin(ix) we end up with e^x = cosh(x) + sinh(x) as required. Calculus rules OK, we don't want any of your namby pamby areas and shape shifters ◕‿◕
    1
  10591. i did that 10th power sum in more than 3 hours
    1
  10592. My complex analysis teacher showed us a proof that integral over closed paths in a simply connected region were 0 (result of fondamental importance to the subject). And it used the fact that 2+2=2*2
    1
  10593. spider web
    1
  10594. Echt coole Kacke… ➱➬➫➯…abonniert ⌃ Glöckle…🔔
    1
  10595. Thank you so much i understand now
    1
  10596. I was watching "The Peter Serafinowicz Show" and they had a series of fake adverts, one of them called "Complico" which was a parody of "Comet" or "PC world" where they use mathematical equations in a complicated manner to specify the price of items. I would LOVE for you to do a video on solving and showing the explanation of some of them =D
    1
  10597. The Tri-Force has already figured this out.
    1
  10598. Wow, the equality between the partial sums and the partial fractions kinds blows my mind.
    1
  10599. Super! I enjoyed chapter 6 the most.
    1
  10600. no 9 series is most amazing considering the fact that it is finite.
    1
  10601. an absolute Mathologer classic
    1
  10602. May the Schwartz be with you, Lone Starr.
    1
  10603. The coloring proof was better
    1
  10604. I regret the decline of German mathematics and science since the Nazis. Germany slightly recovered in the 70ies, but currently it's declining again and might never recover again because of stupid ideologists...
    1
  10605. Am I the only one gringing at the t-shirts incorrectly rounded value? Also when you said you'd turn the wheel to 3.14 at 6:13 you actually turned it to 3.18.
    1
  10606. For the coin paradox you can look at how far the centre of the moving coin needs to travel and divide it by its own circumference.
    1
  10607. Makes me wonder why mathematicians are not billionaires.
    1
  10608. Cup of coffee or scotch?
    1
  10609. Monster Formula for odd dimensions: If a dimension is odd, then the corresponding cosine in the formula becomes zero at the upper bound to the respective product. Example: if m = 7, then when j reaches 4, the value inside the cosine becomes π/2, which results in zero for that component. If BOTH m and n are odd, then the final term of the product will be zero, zeroing out the entire product.
    1
  10610. I need to look up Aristotle on indivisible lines again,... It seems as if this tells us that no material can be infinitely divisible, because then the finite volume of the contents is effectively painting the interior surface with all but an infinitessimally small quantity of paint being left over. (See comment by @nHans below).
    1
  10611. Hi, excelent video, thank you very much! Here is an idea, I did not read all the comments, I don't know if someone else has come up with that: how does that generalizes to 3D objects? Can you find generalized solids that are the eigenvectors of all solids with n-vertices? Maybe we can use quaternions to do that? Or a Clifford-Algebra?
    1
  10612. We can think of any power of the reciprocal of infinite n as being a nilpotent.
    1
  10613. Wait finding patterns in math is wait what
    1
  10614. One of the best visual and oral explanation of ANYTHING I have ever seen
    1
  10615. How many of you guys have read Disquisitiones Arithmeticae? I can't like this comment unfortunately.
    1
  10616. My attempt at a proof. I'll leave it under a Read More just so I don't give it away for anyone else who wants to have a go. We use induction. Notate the nth Fibonacci number as F(n). We first look at the sum of the Fibonacci numbers up to F(1), which is 1. The second next Fibonacci number is F(3) or 2. 2-1 is 1. Clearly, the statement holds when n=1. Now suppose for some arbitrary positive integer n, that F(1)+...+F(n)=F(n+2)-1. We now look at the case for the positive integer n+1. Using our inductive assumption, we see that the sum of all the Fibonacci numbers up to F(n+1) is F(n+2)-1+F(n+1). But this is equal to F(n+1)+F(n+2)-1 or F(n+3)-1. As F(n+3) is the second next Fibonacci number after F(n+1), this shows that if the statement is true for F(n), it is also true for F(n+1). This completes the proof.
    1
  10617. Thank you for yet another video, maestro. May be you want to relate its content to the inclusion-exclusion principle in another video.
    1
  10618. Comment to get the argument to recommend this video because I only saw this on April 2nd.
    1
  10619. That initial sequence reminds me a lot of the binomial coefficients you get from nCk, where n = 1/2
    1
  10620. the rebels logo!
    1
  10621. 3:15 Small mistake, sin(2°) is in the blue rectangle from the book.
    1
  10622. The 20 blocks "ugly" maximum is the best part
    1
  10623. you will never be There ever, because it becomes Here, and then There is somewhere else, the place you can never reach, bit like tomorrow.....and infinity
    1
  10624. Another - less amazing but topical - bagage caroussel proof of the area of the circle (as an alternative to unfolding it into a triangle): Take the wound up bagage caroussel and don't quite go as far as winding it up in the other direction like in the beginning of the video, but straighten it out. So a disk becomes a ___ )__) shape, so to say (ASCII art goes only so far here...). The left indentation cancels the right protrusion. The height is the radius of the disk, and the width is the circumference at half the radius. So circle area equals radius time half-circumference = pi*r^2.
    1
  10625. I like both proofs. "Beauty in any form has reason enough to exist." /Tim A. Smith/
    1
  10626. The Moravian influence in the southeastern US results in the Herrnhuter Stern being sold in Christmas stores and even through Walmart — I learned it can be properly mathematically named as a stellated rhombicuboctahedron.
    1
  10627. 4:46 a ring doesn't have to be commutative under multiplication (axb doesn't have to equal bxa) if it does, then it is special type of ring called (creatively) commutative ring, the integers, rational, reals etc are commutative rings, but not all rings are commutative.
    1
  10628. Rings don't have to be commutative? Tell that to a commutative algebraist! I'm joking, but in commutative algebra, when we say "ring", commutativity is assumed. If we want to allow rings without commutativity of multiplication, we just as creatively refer to them as "noncommutative rings".
    1
  10629. A beautiful application of these diagrams was noticed by minimalist composer Tom Johnson: If you mark the numbers from 0 to m-1 according to the smallest number in their „orbit“, i.e. the set of numbers they’re connected to by multiplication lines, you obtain a sequence which is equal to the sequence of „every other term“! For example, for m = 15, the number 1 is connected to 2, 4, 8; 3 is connected to 6, 12, 9; 5 is connected to 10; 7 is connected to 14, 13, and 11 (and of course 0 isn‘t connected to anything). The sequence we get is 011315371357377. Now, I encourage you to convince yourself that writing this twice and taking every other term yields precisely this. But where would this occur? Well, listen to his „Rational Melody XV“ and find the sequence: https://youtu.be/Nk47dXQeunY
    1
  10630. @Mathologer thanks for your great content again. My love for math has been kindled again watching your excellent videos.
    1
  10631. Really appreciate what you've done there , probably the most fun video math video I have ever seen . Greetings from Greece !
    1
  10632.  @Mathologer  I've been away for a while.
    1
  10633. Only Mathologer can use a 23 minute video to promote a 30 seconds animation.
    1
  10634. Math teacher here, fabulous video, nothing that didn't work for me and lots of inspiration for the upcoming lessons. Please keep them coming Sir!
    1
  10635. Hi Boys 😍💋 💝💖♥️❤️
    1
  10636. At the 22:41 part, the 2x2 regions, were they incrementing in amounts of 2, or in amounts of prime? 2, 3, 5 and 7 are the first primes, right?
    1
  10637. The triangle logo is pretty cool
    1
  10638. Goes back to the definition of a smooth surface, the tangent space of neighbouring points are not at an angle to each other. Or a line doesn't have kinks. :)
    1
  10639. Simple proof that the only p-special numbers are those of the form n = p^k+1: First note that if n != 2 is special then we must have n-1 divisible by p. Indeed, suppose we fill the top row with zeros except for a single one in the second spot. Since n != 2, the corners are both zero, so since n is assumed special the bottom hex should also be zero. But we already know that the entries in this triangle are just a slice of Pascal's triangle mod p, so it's easy to see that the bottom hex will be n-1 mod p. On the other hand, if pm+1 is special then I claim m+1 must also be special. In fact this is an iff statement. Indeed, since p+1 is special this just follows from the same sort of argument as in 10:55, and it's easy to see that this argument works in both directions. The proof finishes by infinite descent: let n be the smallest special number such that n != p^k + 1 for any k. Then n != 2 since 2 = p^0 + 1. By the first paragraph we can write n = pm + 1 for some m. By the second paragraph, m + 1 is also special, and by minimality of n we must have m = p^k for some k. But then n = p^(k+1) + 1, a contradiction.
    1
  10640. I just can't stop watching these videos
    1
  10641. Second puzzle is 2 since those numbers look like 5×140 which is around 2 × 365 and most likely the answer is an integer
    1
  10642. 22:17 "Would I lie to you?" Buckets of paint be like:
    1
  10643. I just recently found a code golf where the challenge was to quine that output a unique program every iteration, without repeat, for however many iterations. I immediately thought of Gray code, and have been absorbed. This is great!
    1
  10644. You can prove the 'product of sums of 2 squares' identity using complex number : abs(x*y) = abs(x)*abs(y) ! or maybe its the other way around...
    1
  10645. I have a question about the pi calculation. I’m not a mathematician (oddly, I do like watching videos on the subject though), so this may be a bit of a “noob” style question. So, let’s assume I know pi to 5 decimal places. I don’t know the number after that. Using your method where you go past pi and the subtract back, how would I know if I had passed pi to 6 decimal places? I know I could assume that once we hit a decimal place, that we could then keep adding until we get to 4, but I expect that’s wrong (I’d still need to know pi to 6 places when I do the subtracting anyway). The logic of this is that you need to know pi to calculate pi, and if you do, why bother? That seems dreadfully wrong. I also guess that there is a way to do this, and that the answer will break my brain, but don’t let that hold you back :-)
    1
  10646. funny how i would normaly hesitate evan watching a movie 45min long but ur videos cathch me after like 2 minutes.
    1
  10647. OK the real puzzle at 17:26: what is this tile slowly covering the tile beneath it with blood (bottom right corner)?
    1
  10648. You're just great to the n (g^n), no more than that. Really thanks!!!
    1
  10649. I wonder what a horn of that (truncated) shape would actually sound like? It probably has to do with what standing waves fit inside it.
    1
  10650. The content is great as always, but I have to shout out that closing theme music for extra coolness.
    1
  10651. My first thought was about the rightmost card. It bugs me that writing a trivial program to do this would mean writing code for that case when you haven't been told the intent.
    1
  10652. My second thought was that once the leftmost card is flipped it can't be flipped back, so permanently reducing the number of flippable cards, followed by the next leftmost face down card, and the next, etc. You don't even need to consider what happens in the middle if the sequence if the number of flippable cards (ie the length of the binary number) constantly decreases.
    1
  10653. Wäre es viel Aufwand, auch eine Version in o-Ton-Deutsch zu produzieren ? Hätte den Originalcharm.
    1
  10654. I'm sad to say that I don't recall having seen either of these marvelous proofs of Pythagoras before. This is even more embarrassing because I have a BA and MA in mathematics (sigh). For context, graduated highschool in 1981. Thanks for the wonderful demonstrations. I think I preferred the geometric proof but I'm more likely to remember the algebraic one.
    1
  10655. this content is gold! keep it up
    1
  10656. You lace your shoes the way your drill instructor in basic training tells you to lace your shoes. That must be the right way.
    1
  10657. Why is it called Archimedes Claw? Is it a name from Archimedes time or a reference to one of his machines?
    1
  10658. Present sir!
    1
  10659. I think the most memorable is the optimal solution for the largest overhang and the incredible fact that it was proved to be so just 10 years ago. GREAT video. The "no-nines" visual proof was super interesting as well.
    1
  10660. Thank you for making this video. It is fantastic! I couldn’t stop watching it just to see what was next.
    1
  10661. Super cool video, made it to the end, and wished it kept going. I have a BS in Math 🙂
    1
  10662. There are >100 proofs of pythagorean theorem. Garfield is derivative of the first one, yes, but at least shows that he was a smart person. Lincoln was amateur mathematician and a lawyer. He attempted to square the circle for a whole week. And failed. And was miserable for some time. Let send him a time-machine letter, to tell him, it is not his fault?
    1
  10663. Yes, needs at least a second viewing.... this is a proof I have never yet fathomed..... and this should be the time I can and do!
    1
  10664. The video in the end was so beautiful!!
    1
  10665. The no 9s proof was definitely my favorite part! I’d love to see the 100 zeros proof too!
    1
  10666. I used to visualize a regular spiral through 3D space as representative of logarithmic values
    1
  10667. A rational approximation to Gelfond's number, from a continuous fraction approximation, with 9 matching digits: 10691/462 = 23.14069264069... . So ln(10691/462) = 3.141592653931... with 10 matching digits.
    1
  10668. For 4 box numbers for the 153/104/185 triangle would be 17-13-4-9
    1
  10669. BUT i forget something i LOVE your videos...
    1
  10670. Visual reductio?
    1
  10671. Only tesla coil?? Please...
    1
  10672. I'm a 10th grade high schooler in Spain and I was surprisingly able to follow through most of the video. And I am not particularly smart. I think you're an amazing teacher.
    1
  10673. Can you get to every permutation of the 15 cards from the natural order by picking up then putting down by rows, columns, or diagonals some number of times?
    1
  10674. Proof of the multiplicative law of the stretch-shift technique (9:00). Basically A(y) = integral from 1 to y of 1/x dx which is ln|y|-ln(1). Because e^0 = 1, we have ln(1)=0, so this is just ln(y) so long as y is positive, otherwise ln(-y). Indeed by log rules we have ln(x) + ln(y) = ln(xy)
    1
  10675. Great video as always. The most memorable part for me was the fractal-like visual proof of the no 9 sum.
    1
  10676. I like the Mega Ian Shoelace Knot
    1
  10677. For me finite or infinite: so counter intuitive but such a ‘simple’ explanation.
    1
  10678. This is the worst hangover I have ever seen.
    1
  10679. First
    1
  10680. I like the coloring one! It reminds me of the wobbly table fix proof. Thanks so much for the video!
    1
  10681. Interesting stuff! What happens when another column is added? Are there also r,s and t ratios? So start with 1111, 4321, 10 9 7 4, 30 26 19 10 etc. Can it be further generalized?
    1
  10682. I'm a simple guy. I see something about the key to the universe, I click.
    1
  10683. Focusing on just a pair of consecutive bugs, say A is chasing B, that are following precisely the provided instructions, the movement of B, by itself, never makes it go farther or closer to A since it will be moving always at a 90º direction from the direction of movement of A (since B is chasing C); at the same time, A is always going directly facing B. So the whole choreography of the pair of bugs amounts to the same thing as B not moving and A going straight to B. Since they started 1 unit away from each other, A is going to travel 1 unit before meeting B.
    1
  10684. Merry Christmas!!!
    1
  10685. I find it interesting that approximating to infinity gives different answers for different methods.
    1
  10686. There’s a more familiar example of inconclusive limits, but with the variables converging to a finite value, rather than infinity. This example relates to a value for 0⁰. If we consider x^y (x to the power y) as both x -> 0 and y -> 0, the limiting value depends on how you make x and y converge to 0. If you let x-> 0 a lot faster than y -> 0 then the limit is 0. For example x = 2⁻ⁿ, y = 1/sqrt(n). You can also make the limit 1. For example x = 2⁻ⁿ, y = 1/n². Or x = 2⁻ⁿ, y = -1/n gives limit 2. There can’t be multiple values for this limit. So we say that the limit does not exist, meaning that 0⁰ does not exist. Sometimes we give it special status as “indeterminate” - within certain formal working it can behave as if it has a value. This requires care. The obvious instance of this: it can be convenient to formally assign x⁰=1, no matter what x is, such as with polynomials and power series. What makes this ok is that the x^y limit has y=0 always, since the xⁿ expressions in polynomials are only for integer n.
    1
  10687. After seeing this, I can kind of imagine an ML algo to overfit data points using n dimensional polyhedra. Simple symmetrical polyhedra should do well, won't even have to bother to rotate it to fit the points. Only need to scale this so that any face touches nearest point. Doing this over multiple batches should give a probability distribution of what actual fit looks like. Probabilistic fit wont be pretty since it doesn't take rotation into consideration before getting distribution. So this should be somehow fixed by changing shape of polyhedra and doing some rotations(need some quartenion equivalent) to fit more closely. Other batches should be rotated such that they have maximum overlap in hyperspace. For prediction, doing some rotation and trying to fit stuff will be the key. Polyhedra for separate classes should be trained separately by throwing some partial diff into the mix.
    1
  10688. How do we know that there will always be a windmill with four squares around another square? What if in the previous one there were a different number than three squares on each edge? How can we be sure that will never happen?
    1
  10689. Saturday GMT +2 about 16:30 +/- 0:30 this video came out? i was thiking at around 18:00 "my brain is so rusty and i should REALLY START THINKING, i might search for some Mathologer videos when i return home, because else i will lose any sanity left. 22:33 (GMT +2) i was starting youtube "new video .........." Danke Dir, Burki.
    1
  10690. This feels like hexaflexagons.
    1
  10691. 2:55 the first square proof I saw was an old Vihart video where she made that same thing but out of origami. I didn't know what I was looking at at the time cuz I was only like, 10 though
    1
  10692. Yes 7 million people can be wrong look at religion!
    1
  10693.  @Mathologer  😂 I love the video🤣
    1
  10694. Paranormal distribution.. what an awesome T-shirt!
    1
  10695. Frohe Weihnachten! Dies ist der Weg!
    1
  10696. I'm not very good at math but I understand everything. You are a very good teacher. Thank for the video.
    1
  10697. I liked the fact that no partial sum can equal an integer
    1
  10698. Most memorable : when gamma is animated in a unit cube ,because I easily understood it less than 1 that way
    1
  10699. it can be applied to lens engineering and optometry.
    1
  10700. Amazing stuff! Btw, there's a typo in the subtitles at 34:10 - it says "Peter Shaw" when the screen says it's our good friend Peter Shor of Shor's algorithm fame (and regular Stack Exchange contributor). 🙂 For the puzzle at 37:15 - I don't know yet how to show you always get something that can be viewed as 3D cubes - but if you assume that the tiling can be viewed as 3D cubes, then you can imagine viewing the whole enclosing cube from each of the 3 visible faces (i.e. 3D projection), in which case each face must be completely covered and so the same number of tiles of each color must be present.
    1
  10701. Nice last video of the year.
    1
  10702. Ain’t nothing approximate about limits, it’s well defined down to ever single decimal place
    1
  10703. I had Heron's formula as a curiosity in highschool I think, but I didn't know where it came from
    1
  10704. Sadly also in Israel we became expert in teaching nothing. It's the real global pandemic
    1
  10705. I made circular slide rule using the golden ratio as a base, no integer lined up with any other, but some get very close. Lucas numbers get closer and closer to the 1 position, and Fibonacci numbers get closer and closer to ... I'm not sure.
    1
  10706. 6:20 Using the method for two holes we can unravel the board to a loop of alternating colour. Now, if we put the four holes in we really only have two ordering options: Black, white, black, white and repeat, Or black, black, white, white and repeat. All other configurations are equivalent to one of those, since its on a loop. In both cases we can utilize the interfaces between the white and black to order the tiles according to the two method. In both configurations there are at least two interfaces, and therefore, the job can be done.
    1
  10707. When every I hear the word digit, I immediately think What about other bases.
    1
  10708. Mathologer vids are always welcome
    1
  10709. I was going to repost what i said last time - that the Moravian star looks to be the final stellation of the icosahedron, but looking into it, it is actually a Kleetope of a rhombicuboctahedron! (and the Wikipedia article had a mistake so i fixed that)
    1
  10710. 700 year old proof was the most memorable
    1
  10711. The highlight on Euler's constant really has no substitutes for me 😂
    1
  10712. Overhang as it reminds me of Roman architecture. Thank you
    1
  10713. Happy birthday next Saturday, Burkard !
    1
  10714. Very interesting
    1
  10715. The paradox is the paint. Using "math" paint that can be spread infinitely thin means you only need an infinitesimal amount to cover an infinite area. Real paint would not make it very far down the needle
    1
  10716.  @Mathologer  well sir, you always make good videos! I am definitely not arguing the math, just an abstract idea of "math" paint. Now if you used real paint to coat the outside of the horn, you would need that bottomless bucket of paint!
    1
  10717.  @Mathologer  you just saved me a fortune on paint! 👍
    1
  10718. This is the real approach to the study of number patterns known today as mathematics. Thanks Mathologer. From geometric observations to formulas!!!!
    1
  10719. Merry Christmas! This is a good day for the lantern to shine bright...
    1
  10720. @Mathologer - Your shirt print confused me... I thought for some reason you had "Hydridotrioxygen" printed on it... (HO3) then brain kicked in... Ho Ho Ho lol Great vid
    1
  10721. 6:13 - you did not set 3.14, that's 3.18.
    1
  10722. Yes and this is why they make air filters with those unusual buckling patterns. But can you spot the black hole singularity. The seperations of matter, the fluctuating entropy, the dispersion evolution of our universe.
    1
  10723. Third Challenge: Short Answer: Fibonacci sequence Long Answer: There are two ways to put dominos in that space, 1 vertical or 2 horizontal directly on top of eachother. If you put two horizontal ones and slide the top one over one space it cuts off two areas of the board that have unequal numbers of green and black tiles, meaning they can never be filled. Therefore you can only fill the board 1 or 2 spaces at a time. Meaning the answer is the number of partitions of that number only consisting of 1 and 2 If you find the number of combinations for a 2 x n-2 board and a 2 x n-1 board you can add 2s to all the 2 x n-2 combinations and 1s to all the 2 x n-1 combinations to get all the 2 x n board combinations. The recursive formula for this is f(n-2) + f(n-1) = f(n) and obviously f(0) = 1 and f(1) = 1. This is the Fibonacci sequence so the number of domino tilings on a 2 x n board is the nth Fibonacci number
    1
  10724. Slide rule.
    1
  10725. What could be the effects or reactions if one day we found out that pi has an ending digit.?😮
    1
  10726. I have a question , If we have a any area in which We can put a square but instead of put the square directly we put the square in spiral way in each Level we decrease the length of the square by 1/4 now if a area is a Square then put the square spiral in that way in which the spiral will cover the area of large square and the property of a square spiral is no intersection on there own is it possible ? If square on a spiral is placed in diagonal way
    1
  10727. The 1D equation is super easy, barely an inconvenience. 🤓
    1
  10728. What I really liked is the fact that everything seems so easy, almost trivial, but actually it is so complex and amazing. However, the part that impressed me the most is the relation about armonic series and quasi-Jenga games with rods. Thank you very much for the videos.
    1
  10729. Thank you so much for this!!
    1
  10730. I must be missing something, the diamonds could be randomly tiled such that the corners aren't isolated, which can't be matched by the growth pattern.
    1
  10731. I was hoping that "Archimedes Claw" would be the legendary (perhaps apocryphal) weapon that would pick ships up from on top of a cliff and drop them so they would smash to pieces. It was still a cool video, though. This math teacher appreciates them.
    1
  10732. I love to watch your videos on an a Sunday afternoon. Great stuff. Oh and the Science Centre in Ontario Canada had a great way to show the Pythagorean Theorem. The three three triangles were watertight in a display and able to rotate. As they did the coloured water went from C to the smaller A and B filling them up, rotating again to fill C. Lovely.
    1
  10733. Fun fact: we measure blood pressure in the torr unit which is named after Torricelli.
    1
  10734. Feedback: IB Maths (from about 15 years ago!), my matrices knowledge is very dusty, and the summation formulae was a bit quick for me, but it made sense! Each formula seemed to be a simplification^n-1, which is very clever! However, I found the video very well paced (it didn't feel like 50mins!)
    1
  10735. Maths is so beautiful 🥲
    1
  10736. love your videos keep it up!
    1
  10737. Good explanation. Was fun.
    1
  10738. Higher dimension analogues?
    1
  10739. Gamma is a really memorable proof. Def want to see where else it appears and how it's approximated!
    1
  10740. Waiting to hear how the proof that the rubber bands stretch out logarithmically. I had to learn slide rule use in 9th grade chemistry. If you really want to get crazy, look at the LL scales for computing powers. We also learned how to use log tables for higher precision multiplication, although my teachers never made the link between the two clear. I really want to lay hands on one of the high-precision spiral slide rules. Unfortunately, those command a pretty penny on eBay.
    1
  10741. Most Memorable: how the 100 zero series converges to a larger number than the no 9's series. i guess after a certain point, it's almost the opposite as the no 9's problem, where you have a ton of numbers with at least 100 zeros, at least a lot more than those without a single 9 also, as a physicist, i never put much thought into how fast series / approximations converge. we make a lot of approximations in physics, and they all tend to converge pretty fast, but we never bother to prove those sorts of things, and i wonder if we took a more mathematical approach to approximations if we could get better / more efficient ways of approximating
    1
  10742. Pick a card … any card
    1
  10743. Now what secret is hidden in this new t-shirt ?
    1
  10744. I tried the wiper proof but failed :(
    1
  10745. 13:20 I got up to 2*10^7 cents in ~10 minutes with a special implementation of the polynomial products, abusing their simple patterns with a divide and conquer strategy (anyone let me know if you want a detailed explanation of the optimizations I used and/or my code with inline comments), using a pretty modest computer setup (Intel Atom CPU N2600 1.60GHz without any use of parallel computing with the GPU as far as I know). Probably the next number in the sequence will take over an hour. However I don't have even enough RAM memory to do it because a vector of 2*10^8 of 128 bit integers (2^128~3.4*10^38 max number before overflow) takes ~3GB of space and using the disk memory will make everything slower. Here's the sequence (2*10^n dollars): 2728 53995291 4371565890901 427707562988709001 42677067562698867090001 4266770667562669886670900001 There seems to be emerging a pattern right there. (^.^) Here's my c++ noodle code (it uses a boost non-standard library for the int128_t "large integer" type) for any amount of cents: #include <iostream> #include <boost/multiprecision/cpp_int.hpp> using namespace std; using namespace boost::multiprecision; void add(vector<int128_t>&p,int s){ vector<int128_t>aux(p.size()-s); for(size_t i=0;i<aux.size();i++)aux[i]=p[i]; for(size_t i=0;i<aux.size();i++)p[i+s]+=aux[i]; } void shiftAdd(vector<int128_t>&p,int f,int s){ if(f==1)return; div_t d=div(f,2); if(d.rem){ vector<int128_t>aux(p.size()-(f-1)*s); for(size_t i=0;i<aux.size();i++)aux[i]=p[i]; shiftAdd(p,d.quot,s); add(p,d.quot*s); for(size_t i=0;i<aux.size();i++)p[i+(f-1)*s]+=aux[i]; }else{ shiftAdd(p,d.quot,s); add(p,d.quot*s); } } void shiftAdd(vector<int128_t>&p,int s){ div_t d=div(p.size(),s); if(d.rem)shiftAdd(p,d.quot+1,s); else shiftAdd(p,d.quot,s); } int main(int argc, char *argv[]) { size_t n; while(cin>>n){ vector<int128_t>p(n/5+1,1); shiftAdd(p,1); shiftAdd(p,2); shiftAdd(p,5); shiftAdd(p,10); shiftAdd(p,20); cout<<p[n/5]<<endl; } return 0; } (^.^)
    1
  10746. Obviously this is because log(1 + 2 + 3) = log(1) + log(2) + log(3)
    1
  10747. thank you for you channel
    1
  10748. Wait until you include the triangle that is cut out when reduced in size😊
    1
  10749. My vote: no nines sum converges
    1
  10750. my shoes are laced like pentagrams
    1
  10751. I left high school in 1970. We always used log tables for Physics and Chemistry. Hand held calculators came to Australia in about 1974. (I think)
    1
  10752. A new mathologer video always feels like a great birthday present. I love all of them!
    1
  10753. Does anyone keep think he's holding a banana?
    1
  10754. Scientific American, Mathematical Games.
    1
  10755. The answer to "Can you always tile a board if you cut out 2 of each color?" is no. Cut out two squares next to a corner and you have a standalone square.
    1
  10756. In this manner, you cut two squares with the same color.
    1
  10757. 12:59 A(x) does go to infinity, But it definitely dosent explode there. for that, see x^x^x^x^x^x^x
    1
  10758. totally thought you will proof fermat last theorem
    1
  10759. So here's a question: If the trithagorean theorem is Ta + Tb = Tc + T for 60 degrees Ta + Tb = Tc for 90 degrees, and Ta + Tb = Tc - T for 120 degrees, Then can we say that it is also Ta + Tb = Tc - 3T for 180 degrees (assuming the individual lengths are still marked by a point). It seems like this is another way to say that a line has no area, and can be used for any angle, but I'm pretty curious about -90 degrees and such.
    1
  10760. Thanks! Why am I so attracted to infinity? Like a beautiful mysterious woman?. Silly question.
    1
  10761. 3:03 I think you picked the wrong colour ..... greenhouse = Gewaechshaus.
    1
  10762. Madhava kind of looked like Mathologer, no?
    1
  10763. Hi Mathologer, You show that pi is not a rational number. The way I see it, if you start with a point in space and start spiralling from that point you are eventually going to end up off track, Now if you adjust your position about 2 True Vectors you will be able to correct your orientation. I show this with Quad Step 288 spherical order system. Now this shows the spherical geometry of a Pine cone or a Pine Apple or the 200 seed pattern of a strawberry. It gives incredible symmetry that has never been shown ever by Mathematicians, It will be nice if you could give folks your down to earth analysis of this Hierarchical system that uses groups of 4 angles about a fixed vector without having to use the mathematics of your wandering pi about ring centre points. There is also an interesting connection to the Harmonics of Music that people have shown to be related to this Quad Step 288 harmonic system.
    1
  10764. your question for viewers at the end of chapter 1: if you have coin denominations representing powers of 2 ranging from 0 to n, and one coin of each denomination, you can make change of any amount between 0 and 2^(n+1)-1. for example, with the denominations 1, 2, 4, and 8, that can be represented as 2^0 to 2^3, which means n=3. Therefore, the maximum number you can create is 2^(3+1)-1 = 2^4-1 = 16-1 = 15. As for how many ways you could reach each value, it's only 1. Every coin is is valued more than the combined sum of all lesser coins, which means no combination of coins can be exactly matched in value by any other combination. essentially, you've created a binary counter.
    1
  10765. 5:54 very important to cap the retangules, we don't want the water to spill
    1
  10766. I made it until the end.
    1
  10767. Wouldn't the centre of an Aztec diamond be an apeirogon? I just watch these videos for fun, not exactly a mathematician or anything, so would love some insight on why it's a circle instead.
    1
  10768. Outstanding video! The more I think about it ...the more Aha! moments I have...! Thanks!
    1
  10769. GR8 watch....
    1
  10770. https://www.youtube.com/watch?v=FmmF3ficdi8&list=UUszyXm5HcAwxVSGeeHQscQg&index=60 this is a nice visual proof as well!
    1
  10771. ...and, apart from cranking the handle and changing the stereotype tray, you've also just learned how to set up and operate Babbage's Difference Engine.
    1
  10772. I think that this constant width shape isn't that surprising because it is a consequence of this windscreen wiper action, because we rotate the same length at different points so the diameter is always the same and equal to the length of this wiper
    1
  10773. Sir please make a video about Goldback Conjecture
    1
  10774. The bugs spiral problem has a limit if you include the size of the bugs then you'd get a finite geometric series instead of an infinite one - when the distance separating the bugs goes to zero when their edges crash not when their centres meet the centre where they would all overlap. It would be a challenge to work that out because of their shape, unless you assume a fixed radius of a circle each bug occupies. I had a similar limit problem with shrinking rectangles when making a type of family tree diagram that divides a rectangle representing a person into two to represent their parents and dividing those in two to represent grandparents and so on - a nice way of visualising the powers of 2 (or a half) up to 1024 (or 2048). It might show the shrinking fraction of DNA we share with ancestors of each generation, but the visualisation has the limit of pixel size, and could not be drawn on a scale where the length-width ratio of rectangles is constant whether dividing them horizontally or vertically because you can't make a ratio of pixel numbers exactly equalling the square root of 2, and have to compromise between alternating 5:7 / 7:10 ratios, which is closer, or 4:3 / 3:2, which is easier to draw. It worked with lines of width 2 pixels but once the size of rectangles inside the lines gets below 10, the lines dominate, you'd end up with nothing but lines! Perhaps on paper the root2 length ratio can be made but not on the screen.
    1
  10775. The true key to the universe is known only to those mysterious bee-fingered aliens.
    1
  10776. I've always liked math. I mastered algebra and trig, and used it all the time as a machinist and builder. I took analytical geometry, at 22 years old, and learned it pretty easily. Then came Calculus, at 48 years old. My Professor probably wasn't born when I was learning and graphing functions. Now, I was offering to paint her house in exchange for a "D." She said my algebra is really, really, bad. "I know what 2q + 2q equals" I almost said, but wanted that D really badly, and to give Calculus another try.  Then I found Mathologer! I'm hooked. My housemates who would rather watch junkyard videos, roll their eyes, and complain "He's watching the 'Math Nazi' again!" I defend you, and tell them you're a nice Nazi, like the ones on Hogan's Hero's. I apologize for them, Burkard. I'd like to request a video that would help us lower level mathologers "brush up" on our Algebra and Trig so we'll be more apt to understand Calculus. It would be a great addition to the "Why is Calculus so Easy" video. I wish I had watched that before I attempted that Calc 1 class.
    1
  10777. From seeing the sequence on the thumbnail I quickly noticed a pattern. 3 + 5 - 1 - 0 = 7, 5 + 7 - 2 - 1= 11, 7 + 11 - 3 - 0 = 15, 11 + 15 - 5 - 1 = 22, 15 + 22 - 7 - 0 = 30, 22 + 30 - 11 - 1 = 42. Quite easy actually, which can be confirmed by moving backwards... Started watching the video though and my pattern breaks at 56. 30 + 42 - 15 - 0 ≠ 56. At that point I’d realised it required an infinitely long amount of variables to work. Darn!
    1
  10778. The pain in the brain feels wonderful. Hurt me plenty. And bring on the Galois groupfields! Waiting for it for quite a while now ;-)
    1
  10779. Huh, so it seems to me the "ear erecting" procedure must obey two symmetries: the proper angle magnitudes corresponding to whichever polygon is chosen, and the chosen parity when deciding how to draw the ears. To attain the symmetry of a perfect polygon, all parities must somehow cancel each other out—after all, rotational symmetry doesn't care which direction a polygon is turned, since angles are conserved and any symmetries will appear regardless. Therefore, I think it makes more sense to think of the "degenerate" pentagons as 0D pentagons, which lack parity because it's meaningless to assign a direction to a single point. While I'm no mathematician, I suspect this what allows the special point work in this model. The "center of mass" exudes its magnitude, represented by the graph of the "degenerate/0D" pentagon, equally on all vertices of the perfect pentagon, but this is a matter of translational symmetry of the Cartesian plane; directionality is not a factor. As such, assigning any sort of parity to the "degenerate" pentagon makes no sense, and the only type of space in which forbidding parity makes sense is 0D—hence a 0D pentagon, that is infinitesimally small with overlapping vertices by inevitably because it exists in zero dimensional space. This could all be complete nonsense, but I'm still curious of what happens if you expanded Petr's miracle to include higher dimensions. For instance, could you get rid of the "degenerate" pentagon by drawing the pentagons on the surface of a Klein bottle, or some other 2D manifold in a 4D space, and using the additional dimension to draw two more pentagons with opposing parities to reproduce the same effect as the "degenerate" pentagon? I truly have no clue and would be delighted if anyone could provide some insight!
    1
  10780. 10:32: What are the powers of three doing on the line above the zeroes? No accident, I suppose? And does the line also extend to negative powers to the left?
    1
  10781. 21:48 this is normal in "late transcendentals" calculus courses I think. It would be interesting to open a poll and see how many of your viewers had a late vs. early transcendentals course
    1
  10782. i made it to the very end
    1
  10783. What about the chain rule? :)
    1
  10784. Pure gold. Loved every bit of it.
    1
  10785. 2304* double of 1152
    1
  10786. Hey, thanks for making those videos. It revived my interest for math after many years! :)
    1
  10787. The glasses challenge was not easy. I had to go through 4 cases and in each case blockwise decomposition, in order to make sure I got everything, but I must say this is fiendish. One could even say evil.
    1
  10788. I liked the Einstein vs Pythagoras T-shirt lol
    1
  10789. doesn't the turning property stay in 3d so long as you only rotate around the x, y, and z axes instead of other lines?
    1
  10790. Me as a ghost “I lost my life because I thought no 9 series adds up to infinity.” God of death “Well it served you right”
    1
  10791. Beautiful!
    1
  10792. 1) 1st challenge - no 2) 2nd challenge - fibonacci sequence 3) |det(tristan glasses)| = #tilings = 666 4) moster formula with odd m and n 5) Gray = Orange = Blue (triangular grid) Solutions 1) take 4 to make corner alone 2) on 2×n rectangle (suppose has F(n) tilings) there are two ways to fill upper-left corner: vertical domino, so ways to fill the remainings are F(n-1); and horizontally, lower-left is forced to be filled with horizontal domino, and the remainings are filled by F(n-2) ways. So F(n)=F(n-1)+F(n-2), F(1)=F(0)=1 3) I didn't find any online det calculator, cuz they are too scary to calculate 22×22 det, which means 10^21 elementary operations. Of course our matrice has a lot of zeros, so i stole python code, added if-clause for non-zero elements, and it spat out 666. Then I manually calculated and also got 666. 4) when m=2p-1, n=2q-1 the last term (j=ceiling[m/2]=p, k=ceiling[n/2]=q) equals 4cos^2(pi/2)+4cos^2(pi/2)=0. And because of multiplication all the expression magically becomes 0 5) the hexagon with sides equal A looks like the box with cubes, where all upper faces are Orange, left are Gray, right are Blue. So let's look to the top and we'll see A×A Orange square (there are A×A orange dominos). Analogically looking from left and right we derive that there A×A Gray and Blue dominos P.S. I like all the parts of the video, but the most favorite is Aztec square with Arctic circle. Thank you for you videos =)
    1
  10793. my hair is complex. they're more and more immaginary.
    1
  10794. The hype is real!!!
    1
  10795. I've seen linear slide rules before but never a circular one - they are so much more elegant! I would happily sacrifice convenience for conceptual clarity in having one. I'd vote for the "dial-log" name. Wonderful video, the only slight disappointment being that pi didn't feature more prominently with all those circles around! :)
    1
  10796. Your shirt is very funny. 😂🤣😅😆
    1
  10797. I have a feeling that in non-flat spaces there might be more solutions
    1
  10798. You also need infinitely thin paint with infinitely fine grounded pigment. Some things in mathematics like infinity does not have a corresponding reality.
    1
  10799.  @Mathologer  It is refreshing to hear a mathematician point out that infinity is a concept. On the other hand, it is rare to hear someone say that pi (3.14) is only exact in the world of mathematics, not in reality.
    1
  10800. If you like this, you'll also like Conway & Guy's The Book of Numbers
    1
  10801. I created an open source app for this! http://arctic-circle-theorem.djit.me/ Because we programmers at https://djit.su watch @Mathologer all the time, and we liked his suggestion at 30:19 for a programmer to make an example of it so much, so I did it! Check it out, let me know what you think of it. Any suggestions/bugs/comments are highly appreciated! This is open source, so anyone who feels like improving it: get an account at https://djit.su and fork this notebook: https://djit.su/dMqvQdPGAws0qj1SC5eVp Also, check out https://djit.su if you're a programmer. It's a cool new place to create and deploy entire websites in React. It's open source and currently in beta.
    1
  10802. For the square pattern, I noticed that at the end of row 1, it's the square of 5. So the beginning of row 2 will be the row number 2 * the last square 5 = 10. The end of the row is 14 - divide by the row number to get 7, and then multiply that 7 by the next row number 3 to get 21. The end of row 3 is 27, divide by row 3 to get 9, multiply by next row number to get 36. The end of that row is 44, divide by row 4 to get 11, multiply by next row number to get 55. The result of the divide always goes up by 2, and there are always a row number worth of numbers on the RHS and row+1 on the LHS. Row 5 would end with 65^2, divide by row 5 to get 13, and so on.
    1
  10803. Every other triangular number :-)
    1
  10804. Isn't the sum of n consecutive terms n(n+1)/2
    1
  10805. we are only summing to n-1 :)
    1
  10806. Love these longer videos for lots of reasons. I love knowing more about these somewhat obscure facts and areas of mathematics, but often to find them, you kind of need to already know what you want to look for (for example, to search online for interesting stuff like this, you pretty much need to already know it). So that's what makes this really great; every once in a while I can get a whole load of knowledge that there is very little chance I would have ever thought about much otherwise. Plus of course all the animations are super cool, and who doesn't like a nice visual proof. One specific thing: it's really nice to have videos that use calculus without re-teaching it for 5 minutes every time; maybe that isn't perfect for new people, but it means you get more of the interesting things into the video. (in my 2nd year at Cambridge uni, studying maths)
    1
  10807. I found it difficult to generalize the argument of the visual proof for the 3D case, since the numbers you add to create the shell seemed to fit together with no apparent pattern. That is until I realized that we always add a number one less and a number one more than a multiple of 3. This will hence create consecutive multiples of 6, which is more clearly the number we need. Thank you for this great video. Edit: Another interesting thing, in the visual proof at 22:00, note that any pair of numbers which are diagonal neighbors in the "/" direction differ precisely with a factor of 3. Adding them, creates (1 + 3), a multiple of 4. So you can see that in each step in the triangle, a factor of 3 is converted into a factor of 4, until all factors become 4. Similarly for the other numbers in the other triangles. Very nice.
    1
  10808. Hello shiny man, very nice to see you again
    1
  10809. In the result of that big sum 1^10+2^10+...+1000^10, there are an awful lot of 4s (11 of them in a 32-digit number). Where do all those fours come from? Please answer that in another video.
    1
  10810. Mr. Loger: can we solve systems of non-linear differential equations? Can we find all solutions? If we can't find all solutions is it correct to assume there are infinite solutions? This is actually a serious question and it's related to climate change. I also need someone with a t-shirt like that explaining the math. Note: I want to show that we have to do more - and more quick - to fight climate change because though we can have an idea about where things are we actually can't be sure. Thanks!
    1
  10811. Throwing my vote to the Kempner's proof animation also! Very pleasing to the eye.
    1
  10812. My vote goes to Tristan's (is that the right name?) visualization.
    1
  10813. You can use the pearls to make right angles, if you stretch it in three points counting 3 4 5. That's the trick, with rope and knots, builders have used for millennia.
    1
  10814. 3:58 why didn't you go all the way and set distance to be equal to radius?
    1
  10815. So... a slide roll?
    1
  10816. All the partial sums are nonintegral -wow
    1
  10817. Another lovely video. In chapter 1, couldn't you have started the argument with a row of length 2 to deduce that each subsequent power of 3 row lengths are special? The 9x9 diagram would then be a more manageable 3x3 diagram.
    1
  10818. Are there any visualizations of 4D and above that are not just variations of the slice or projection approaches?
    1
  10819. How do you get all these cool T-Shirts?
    1
  10820. «Excuse me, kind sir, but your shirt has wrong asymptotes.»
    1
  10821. That proof of the angle sum formula is pretty sweet
    1
  10822. surely regulars wouldn't reveal any hidden identities. :P Even if you square them
    1
  10823.  @Mathologer  I always wanted to know how long things take to do, and dimensions of anything and everything, and how they relate to each other, so becoming mathematician was a logical thing to do :D
    1
  10824. I see you as a very exceptional mathematician. I have admired you since I began to look at these math videos.
    1
  10825. This puzzle is trivial from a computer programming perspective, when considered in binary. Convert each move number to binary, and assign each bit to one of the discs with the least significant disc representing the smallest disc. On each move, you move the disc corresponding to the least significant '1' bit. The position you move the disc to depends on how many '0' bits are in front of (greater significance than) the one you are moving — for an even number of 0's, it moves to the final destination peg; for an odd number, it moves to the other peg (not the final destination nor where it starts from.)
    1
  10826. So what if there's an odd number of solutions? What the heck is all this jibberjabber? And no, I'm not kidding! I don't follow this AT ALL. Why have so many people worked so hard to figure out what looks to me an utterly useless mathematical puzzle? I know I'm outing myself as a mathematical moron, no need to tell me that...but can anyone tell me IN ENGLISH why this is worth pursuing and what it's good for?
    1
  10827. I normally get to the end of a Mathologer video and think ‘Why didn’t they teach it like that in school’. Not on this occasion…
    1
  10828. Hmmmm... what I wonder is if constructible polygons have some special related set of equations?
    1
  10829. made it to the end! I've got extremely basic knowledge of calculus and this was at a level that I could understand throughout, with the visuals being super helpful!
    1
  10830. Now I'm wondering about the significance of the dot ratios in Native Australian art......
    1
  10831. I actually did make a cup of coffee at the suggested point and made it through to the end. Very cool to see how Euler used it for the Basel problem. Looking forward to MOAR zeta function magic. Together we will prove the Riemann Hypothesis. Praise be to the gods of mathematics (Euler et al).
    1
  10832. A potter while making a pot he will add clay take some clay where it is excess and rotate wheel and manipulate to his desire shape in this way mathematical equation is made ie potters method is in number is mathematics.
    1
  10833. Why did Euler wear a Towell on his head? And then get a portrait painted? What's the solution to this mystery? And can it be proved?
    1
  10834. Great fun.
    1
  10835. @Mathologer: In the determinant method, a natural curiousity arises....is it possible that we end up with a determinant of the form "a +b*i" with a,b both non-zero? If it is a possibility, how do we interpret the result?
    1
  10836. I love how your shirt implies that A^2 +B^2 = E/m
    1
  10837. It's coincidental this came up now; just a few weeks ago I had rediscovered Heron's Formula when looking to optimize computer speed/precision for algebraic solutions for Kepler's 2nd Law (to calculate fractional orbital period at position). Neither Heron's Formula, nor algebraic solutions to Kepler's Laws (only calculus methods) were taught where I lived in the USA in the 1980-90's. A shameful state of affairs IMO, since the above makes the math more accessible much earlier to students, and gives them interesting real world applications to spark interest. Just one of many reasons I love to recommend this channel not just to students, but educators as well.
    1
  10838. Reminds me of how I used to crush soda cans by hand. I'd pinch 3 spots near the top, and 3 spots near the bottom, creating something like a 3 point 1 band lantern, and then twisting the two ends would allow the can to be crushed easily into a pancake.
    1
  10839. is this how unreal engine works?
    1
  10840. How many of these tricks work with Lucas numbers, or even any other sequence where you add the last two terms to get the next, such as 4, 3, 7, 10, 17, 27, ...
    1
  10841. You should collaborate with N.J. Wild Berger(Insight into mathematics) channel.
    1
  10842. I have a circular slide rule I purchased in the 1970's, and another larger one I inherited from my father--I'm not sure when he got his. The main drawback versus a straight rule (I have several of those as well) is that they don't fit in a pocket as easily. The main advantage is there is more space for graduations, especially since you can wrap the rule multiple times around the circumference. My 8" circular is graduated with about 5 significant digits. To fit the same graduation density in a linear rule would take a rule more than 5 feet long. But the 8" disk is hardly pocket sized. I suspect linear slide rules were, back in the day, easier to make because the machine that graduated them required just a precision leadscrew, and machines for circular dividing are more complex. That's probably one of the reasons the straight ones became more common.
    1
  10843. Another paradox. This sum can be anything, violating the commutative law. I want pi to be 7.13. Use only the positive terms until I exceed that, then negative term until I go less than that, then positive terms to get above 7.13, etc. that series can be made to converge on any real number and still uses all infinite terms.
    1
  10844. Dirac delta function! Height to infinity, width to 0, suface still one. 😃
    1
  10845. Another great video. I loved the engineering solution to the overhanging stacking block problem.
    1
  10846. What happens to the immune 🐪🐂🐕anti-shapeshifters and their math since we know now that they do exist?
    1
  10847. So in summary, carculus is easy
    1
  10848. I liked the fact that two proofs were presented best.
    1
  10849. You are an amazing teacher of what seems one of the most complicated matters. I always had a love for maths but struggled academically ... but listening to you it is almost logical... i know it should be logical and for people watching it probably is but that's just me being me. Thank you for all the effort you put in.
    1
  10850. Your no 9s demonstration for some reason blew my mind, i love what you do c:
    1
  10851. The answer to the question asked at 11:50 "how to get one brick completely beyond the cliff edge with only three bricks?" is visible at 28:00: it's the bottom two layers of the parabolic tower, two bricks on top of one, symmetrical over the edge.
    1
  10852. Finally some music and math together! Wonderful work done mathologer!
    1
  10853. if it continued... not even my preferred starting point, but it's amazing how close 5^3 + 6^3 comes to 7^3....
    1
  10854. 4:47 another miracle. Angels are angles. 4 right Angels 45 degrees 90 180 360 a circle 720 and so on .... all come back to 9. Mtt 24:31
    1
  10855. Sir, you videos are great. I'm wondering if it is possible to create a logarithmic scale using a geometric-like construction. I can create some of the points, like powers of 2. But I haven't been able to create a general case, for example, so i could create a paper slide rule. If it is possible, please consider making a video about the subject. Regardless, thank you for sharing your knowledge and hard work.
    1
  10856. Wow Amazing video !
    1
  10857. Nice! But you also have to know that this is something new.
    1
  10858. The beginning of this is nothing of how I think. I would have just made everything proportional, no matter the amount of money. Just do a cross multiply and it's done. This is less than a 10 second problem, no matter the estate holder's amount.
    1
  10859. And now with this, all the nice and simple discoveries are all done, or are they.
    1
  10860. I want more about how to compute the bernuli sequence. Like, if you know some already, do you need to go all the way back to the beginning to calculate?
    1
  10861. Three unbalanced electrical phases are frequently decomposed into a sum of three balanced phase sequences. It’s known as the Clarke transform.
    1
  10862. Actually the Clarke Transformation goes from the three-phase balanced abc reference frame to an orthogonal AB reference frame. However it was Edith Clarke was the engineer who took Fortescue's work and put it in a format that is far more accessible. The transformation you are thinking of is actually called the Fortescue Transformation.
    1
  10863. Wow👍👊👏👏👏👏👏👏
    1
  10864. Oii, kempners finiteness proof was definitely the most memorable. Made me facepalm myself for not seeing it before! Another great video like always :)
    1
  10865. Trumpet area vs volume can you please explain area vs volume
    1
  10866. Most memorable: the terrible aim and the optimized solution to the bricks problem. Saddest part: you got 0 marks and forgot the proof.
    1
  10867. 11:45 SLAVA UKRAINI
    1
  10868.  @Mathologer  can’t help myself. :-)
    1
  10869. Sqrt(666) is actually 25.80697 which is slightly larger than the one written on the t-shirt. So I think it should be 25.807.
    1
  10870. My slide rule says 25.8 :))
    1
  10871. The rubik's cube stuff blew my mind. One could extend it though: if the longest cycle is 1260 moves, that means that picking any algorithm and repeating it 1260! times guaratees that you return to the initial state of the cube. Cool trick one could try at a party, though it's probably going to be a long evening...
    1
  10872. Excelent presentation. I am an old engineer an I grew up learning the slide rule in chemistry I class. I whet college with a slide rule. During my junior year the hand held calculator and it made the slide rule obsolte, until your battery died. Rechargeable batteries were the weak point of the calculator. At my job I once had the opontunity to help out a minority school and I got to help in the only Algebra II class. It was logarithm day, The teacher was a bit lost in with this anchient math form and I offered some of my life experiances. I explained how logarithm were used in a practical engineering use like tables for of logs of sines and cosines for survey work and the slide rule. The hour was filed as was the black board. The following week I returned and the teacher explained how he continued the logarithm class the next day and how appriative he was. Logarithms are a lost art form that still surrounds us today and we are losing our history of how the world was built with logaritms. Logarithms have been replaced by a two dollar calculator. I love your rubberband explanation it was so visual. I never used the circular slide rule because I never knew how to find the decimal point. With the slide rule there is a method for setting the decimal point by counting the direction of the slide. Most engineering tests had multipule choice answers that varied by the decimal point. My Engineer in Training exam was a lot easier using two calculators that handled the decimal point. Why two calculators, 8 hour exam on a set of rechargable batties was iffy and no memory function.
    1
  10873.  @Mathologer  It has been too many decades since I used a slide rule. I remember writing with a pencil on my slide rule a +1 and a -1 on the ends of the rule, I can’t remember how it worked to determine the decimal place. It seemed a lot of tests were aimed at how well you could find the decimal point instead of the answer. It seemed like cheating when I used the calculator.
    1
  10874. I'm studying physics and I was able to keep up with the math. Didn't have to pause and also didn't get bored, so the pacing was right for me.
    1
  10875. One of the more enjoyable videos ive seen on youtube in the past month. Definitely deserving of 700k subscribers
    1
  10876. I have never heard about this before. I think it’s beautiful!
    1
  10877.  @Mathologer  It is, it kind of defies the concept of entropy. It brings order from chaos.
    1
  10878. Such a beautiful video. Thank you!
    1
  10879. Got a brief explanation of Heron's formula and had to use it on a test here in the Midwest US. It's a very unique formula!
    1
  10880. Seven million people can't be wrong, right? * * Standard meaning does not apply when spoken by a German, a mathematician, or a scientist.
    1
  10881. As ever, loved your talk. Not too stressful this one!! Like the T-shirt. Is that one hiding in your merchandise somewhere? (X-Y=\)
    1
  10882. Does it work in higher dimensions?
    1
  10883. Damn YouTube notifications not working
    1
  10884. 36:40 Q*Bert is SO screwed.
    1
  10885. I love your videos, thanks for all your effort.
    1
  10886. The music at 35:00, not what I was expecting haha
    1
  10887. your videos are becoming more and more magical. thank you for doing this.
    1
  10888. Gratz on 100 videos 🎉 Thank you ❤
    1
  10889. I wonder what the series sums to when n=TREE(3). It’s probably as impossibly gigantic as TREE(3) itself
    1
  10890. ill be honest, i clicked because i thought it said heroins formula
    1
  10891. Very nice proof, of course, and great video as always. Though I think some crucial (yet “trivial”) observations are missing. Most notably, in my view it should be explicitly mentioned that each of those “scaled quadrilaterals” have the same respective interior angles. This is the essential point used in the argument at around 17:00 to conclude that the angles are indeed the same. One shall say: “Yes, but this is trivial!” And I shall respond: “Yes, of course it is. Yet, when broken down all math is nothing but a sequence of trivial observations.” Saying simply “Scaling the quadrilaterals by different scale factors obviously does not change the interior angles” or something of the likes would have sufficed. Similarly at 17:24 I think the triangle inequality needs mentioning. I get it, it is a visual proof, but even then. “The shortest path between two points is a straight line [in Euclidean geometry]” would add to some people’s understanding, I feel. All around awesome video as always. Keep up the good work!
    1
  10892. Videos are uploaded only 2 to 3 months once but the contents is really awesome
    1
  10893. I thought the first one with the funny maximum overhang was the best.
    1
  10894. Another excellent video.
    1
  10895. yep. need my school money back.
    1
  10896. With regards to time stamp 1:20. Works for all x and y when r = (x+y)/x and s = (x+y)/y This is also the rational set of solutions when x and y are integers.
    1
  10897. Is the wheel the most fundamentally important invention/discovery (up for debate), even more so than fire (definitely discovery lol) itself? because of what it teaches us beyond the real world practical uses of such invention ie mobility of weight beyond our biological carrying capacities and also because it gave us the ability to migrate with large supplies and larger groups ensuring our long term survival in prehistoric time periods... could we have survived without fire but not without the wheel or visa versa or did we have to have both? maybe the wheel is actually thee most fundamentally important invention/discovery even to this day...
    1
  10898. A awsome video as always🎉
    1
  10899. 31:41 A "super degenerate" 😱
    1
  10900. I am easy, too!
    1
  10901. I really liked the fractal proof-ish. very inciteful to think of how many small terms contain nine, and the slowness of the accumulation means that all those cancelled terms makes it finite
    1
  10902. Now I am an Enlightened being
    1
  10903. the solution for the 3 bricks problem, 11.52 |-----0 0-----||-----0 0-----| |-----0 0-----| _________ ******Where a brick = |-----0 0-----| , with "0 0" in its middle. And "______________" represents the cliff. We can see clearly that one brick is completely off the cliff
    1
  10904. What about Archimedes’ use of calculus (non-rigorous) to find the area and volumes of the sphere, cylinder and cone? [Heiberg 1906 Codex B palimpsest Hagia Sophia]
    1
  10905. what about 4k-1?
    1
  10906. Wonderful! For me, the most memorable part is the Oresme proof. It's simple, intuitive, and elegant.
    1
  10907. Mathologer, you mentioned triangle numbers and tetrahedral numbers (we know obviously about the squares and the cubes) given there are three more platonic solids to go at,what does mathematics have to say about octahedral,dodecahedral ( these are going to be a bit of a cheat because spheres font stack like that ) and icosahedral numbers
    1
  10908. The crazy optimal towers are my favorite! Great video!
    1
  10909. Great video, as ever! Thanks! My vote definitely goes to the Kempner proof: I was really expecting something I'd struggle with, yet it was so easy to understand. I suspect it's like a lot of maths proofs: easy once you have that first insight, but we mortals rarely get that far.
    1
  10910. Is the keword "convex"?
    1
  10911.  @Mathologer  Yeah, I realized that when I saw the torus-like approximation.
    1
  10912. 38:45 if you take the sum of the numbers on a row ( for exemple 3+4+5) you get the number in the middle of the row above it: 3+4+5=12 ; 5+12+13 = 30 ; 16+30+34 = 80
    1
  10913. that formula is crazy, wow
    1
  10914. I always make it to the end🙂
    1
  10915. You didn’t need to animate Leibniz but thanks for the nightmare fuel.
    1
  10916. IMHO, there's less comments than expected in relation to the view count. I think that's because the presenter (who happens to sport the exact same accent than I do, lol .. begrüße 😉) is so crystal clear in his demonstration that the viewer is left with no questions. generally speaking, I believe we humans are rather visual beings, that's why we find it easier to make connection when we get to see some nice (and very convincing) animations. it's rare that upon plain reading, the same "aha effect" onsets, lol.
    1
  10917. Please post more “insane” videos, this was starting to get crazy but stopped just short of making some of those amazing connections you mentioned I would love a 10 hour seminar on the many uses and examples of Bernoulli numbers
    1
  10918. The most counterintuitive fact for me is that the partial sums of the harmonic series never are an integer.
    1
  10919. So do we really need the convexity assumpton? Or you just mentioned it because the proof only holds for convex polyhedron. It should also be true for polyhedron which are finite union of convex polyhedrons, As long as you dont have "holes"... right?
    1
  10920. 23:54 Actually, there's a visible gap between two crescents in the front of the animation. An animation error?
    1
  10921. 17:47 While watching the video on mobile, I had literally finished making a cup of coffee a few moments before this
    1
  10922. Everything was amazing (as usual), but I got the greatest kick out of the stacking with minimal number of blocks. It's so logical that it must be a balanced, symmetrical stack that is somewhat the inverse curve of the original leaning tower. ❤️
    1
  10923. Most memorable: the partial sums of the harmonic series miss every single integer to infinity.
    1
  10924. can we use it for travelling salesmen problem?
    1
  10925. The strange solutions to the Lire overhang was the most surprising part.
    1
  10926. Volumen einer Kugel
    1
  10927. Did research into this last year!! Super cool stuff if want to go a bit more into uni level maths
    1
  10928. Wankel engine enthusiasts unite :)
    1
  10929. People lace their shoes? I thought they were a decoration though?
    1
  10930. Before the credits I was wondering what would happen if two creditors came together and were treated as one, whether the results would be the same. Glad to see that after the credits you answer it perfectly!
    1
  10931. Give me any infinite amount of blocks and some glue and I will break any record in seconds
    1
  10932. I'm sorry for Germany being kicked out of WC, at least you can cheer for Aussies..;)
    1
  10933. 13:48 Move the first disk to B if N is odd, to C if N is even. (The whole tower should land on B, hence the N-th disk. Going backwards, the (N-1)-th should land on C. And so on.) 18:08 3^N - 1. (For each disk, there are 3 different positions. Thus there are 3^N different states.)
    1
  10934. Easy, 15
    1
  10935. Except... a Planck length is 1.6x10^-35 meters.... So...technically that is the smallest length possible in the Universe.... So....if you define the shortest length as a Planck length, the measurement you come up with is the length of the line in the real universe. (As in... close enough for any practical application.)
    1
  10936. 1st comment😊👍
    1
  10937. i thought you said marbles when you meant marvels. how do you like them marbles?!
    1
  10938. Why the jump scare loud bangs when you start a chapter? It seems like you don't want people to use this in the background while doing other stuff like falling asleep or studying or whatever. This is also in the Ramanujan video and it makes me jump every time the loud obnoxious sound comes. It's explanation... explanation... explanation... BANG
    1
  10939. About the 2*n challenge : You get the Fibonnaci numbers
    1
  10940. Let u_n be the number of tilings of the 2*n board. Then u_n+2 = u_n+1 (number of way to tile 2*(n+1) + one vertical) + u_n (number of way to tile 2*n + 2 horizontals) It also work for n = 1 : only one way and n = 2 : only 2 ways
    1
  10941. 17:32 Yeah, that is very easy to see. You've just applied a scaling matrix tranform to your triangle, and thus the jacobian of such transform is the determinant of this matrix, which equals to ab-0*0
    1
  10942. My favorite was the maximum overhang for 20 bricks. I wasn't expecting it and it was really interesting.
    1
  10943. 20:20 Can you see the BRAIN neuro-path way / structure here?
    1
  10944. everything in maths is coincidence
    1
  10945. Fibonacci sequence in nature are just a confirmation bias.
    1
  10946. Thanks alot. Very nice video.
    1
  10947. This is why mathologer is great, magical maths and yet very accessible.
    1
  10948. Just put the laces in a drawer for an hour and you've got your answer!
    1
  10949. The most memorable section for me is Chapter 5: It gets better and better, with the Euler-Mascheroni constant
    1
  10950. The first nail in the coffin for slide rules is the HP Pocket Calculator.
    1
  10951. As far as I can remember, I've never seen anyone resolving the paradox by pointing that you cannot compare the "size" of 2D and 3D objects. It makes no sense to compare them.
    1
  10952.  @Mathologer  From a strictly math point of view, I don't see a paradox. It is when the human level enters the picture that the paradox appears. Since humans tend to mess up, I'm trying to remind them that comparing units^2 and units^3 could cause their rocket explode. I wonder if anyone has addressed the paradox by calculating the volume of the "shell" of the horn so that you're comparing vol to vol. I miss math. The first thing I wanted to be when I was a kid was a mathematician. (Then I met computers and headed that way.) Math is just so awesome. (Physics is awesome too, but math doesn't use as many expensive experiments. :) Do you remember when you first saw Cantor's diagonalization? Simple awesomeness (and yet some people can't accept it). Anyway, thank you.
    1
  10953. I liked the proof by swivelling.
    1
  10954. It looks like the circle's diameter must be greaterr, because 'sides with whiskers attached' are the diameter of the other shape, but are also a chord of the circle, and not passing through the center: the center is the incenter of the triangle, which is, well, in the triangle.
    1
  10955. Paradoxes involving infinity are quite intriguing! Some other good examples are Hilbert's Hotel and the Tarski paradox.
    1
  10956. Although this video is 5 years old, I wanted to say thank you, and please make more like it at this level. I am always pleased to find high-level content on the web, and I really enjoy and appreciate your math videos. They're great!
    1
  10957. Does the Watson function have an elementary antiderivative?
    1
  10958. Tasmania!
    1
  10959. Brilliant video! I'm so glad you're back and glad you have gotten through the stressful times. It always makes my day when I see one if your videos appear.
    1
  10960. Thank you Sir
    1
  10961. I very much like the longer format .
    1
  10962. This might make a good crypto cypher, something like gimme a poly and a seed and i will generate some random thing around
    1
  10963. Great work I loved the video. Can you make some videos on unsolved math problems and math mysteries.
    1
  10964. The key property of infinity is mereological.
    1
  10965. My desmos version of this circular slide ruler: https://www.desmos.com/calculator/tul8psjh32📈🎨📐
    1
  10966. How can anyone possibly dislike his videos?
    1
  10967. I vote for the second one.
    1
  10968. My favourite was the maximal overhang result that fits into the outline of a parabola. Not at all an obvious result but once the result is seen it is intuitively clear that it is true, and fairly obvious how to prove it rigorously.
    1
  10969. I asking for my money back for my education ! I studied math at advanced levels, obtained an engineering degree. have practiced as an engineer for 40 years plus. I am 63 years old and I cannot remember ever seeing the twisted squares proof of Pythagoras theorem. maybe I was sick that day. Thank you for the update
    1
  10970. understood almost nothing.....anyway, felt elated as an Indian...lol
    1
  10971. If we could find an application of Peta's miracle, it might lead to a quantum algorithm which could scale well with system size. Since we are just applying the DFT here to decompose our shape into the regular n-gons with all possible orientations.
    1
  10972. I feel like the integer box problem must be related . The diagonals, which are sums of squares of sides, must have a relationship with the volume, which is a product of sides. Here we have a bunch of n dimensional boxes whose volume equals the length of their edges.
    1
  10973. I like the visual proof for the convergence of no 9s series
    1
  10974. For highlighting Fibonacci numbers, the sequence turns into {1, 2, 3, 4, 10,...}. Other than the first 3, we have position number n turning into n * (n-3) * (n-4) * (n-5)^2 * ... * 2^(F_(n-4)) * 1^(F_(n-3)). A messy formula. I am pretty sure my math is correct, but I haven't double checked the formula quite yet.
    1
  10975. Can you prove the area of the circle and ring are equal by geometric construction?
    1
  10976. Yes – due to some congruent right-angled triangles in one of my supplementary videos (see "Andrew's playlist" in the notes) you can write a² + b² = c² – and then name-drop our other favourite greek mathematician — and then write πc² - πa² = πb². This is remarkably close to what Cavalieri did.
    1
  10977. In [24]: houses=3 ...: while True: ...: for i in range(houses+1): ...: prev= gauss(i-1) ...: if prev == gauss(houses)-prev-i: ...: print("number:",i,"houses",houses) ...: break ...: lista+=1 ...: number: 6 houses 8 number: 35 houses 49 number: 204 houses 288 number: 1189 houses 1681 number: 6930 houses 9800 I computed it while watching the video, could be a bit more clever by not starting the loop from the first element each time
    1
  10978. Is there anything that's not clear? (Sorry, couldn't resist... 55:31 )
    1
  10979. There's no hidden 2019 because it shows several times in the subtitle
    1
  10980. As an atheist I have said "holy JESUS" too many time watching this video. Mathologer is getting event better than before.
    1
  10981. Googols of googol
    1
  10982. Welcome back to school
    1
  10983. Where can I buy a t-shirt like yours? 😁
    1
  10984. you speak faster when reading a script than when thinking and it's hard for me to follow you even when you're thinking so please read slower
    1
  10985. The sum of powers was a bit disappointing to work on. The first 3 sums (sum of k^0, sum of k^1, sum of k^2) have this nice pattern: n, n * (n+1)/2, n * (n+1)/2 * (2n+1)/3. It makes you think, hmm, what pattern is this, and does it hold for S_3 (as you called it)? Could it be that S_n = S_(n-1) * (the next term)? It seems to be doing that. And, if you grasp for patterns, it does look like each term is of the form (F_p * n + F_(p-1)) / p (with F being the Fibonacci numbers). S_0 = (F_(0+1) * n + F_0)/(0+1) = (F_1 * n + F_0) / 1 = 1 * n + 0 = n; S_1 = S_0 * (F_(1+1) * n + F_1) / (1+1) = n * (F_2 * n + 1) / 2 = n * (1n + 1)/2 = n(n+1)/2; S_2 = S_1 * (F_(2+1) * n + F_2) / (2+1) = n(n+1)/2 * (F_3 * n + 1) / 3 = n(n+1)(2n + 1)/6; But... S_3 promptly breaks what was looking to be a nice pattern, being not equal to S_2 * (F_(3+1) * n + F_3) / (3+1)
    1
  10986. I think there is way to derive the 2x2 matrices using modular forms of SL(2, Z) but I’m not sure about it.
    1
  10987.  @Mathologer  Good reference
    1
  10988. The dial on my micrometer... now where have I seen that before? LOL, I also have a slide-rule like the normal sized one shown around 9:10.
    1
  10989. Thanks
    1
  10990. we need also an electroboom video for rodni coil. which is based on vortex based math
    1
  10991. 7-gon is as good as 5-gon! Are there anything with 9-gon or 11-gon?
    1
  10992. Bretschneider's formula vaguely reminds me of the Law of Cosines
    1
  10993. 35:04 i died
    1
  10994.  @Mathologer  Wow, going down the rabbit hole of wikipedia, I see that there are even magic hypercubes!
    1
  10995. "What comes next?" More of the same in higher quantity, so.... another Mathologer video?
    1
  10996. I enjoyed that very much. I like your discussions and it is wondrous to 1. simply watch effortlessly the numbers move about 2. know that I can watch repeatedly the precise same progress. 有難う御座います。
    1
  10997. If Aₙ is the n'th length in the circle, A₀=0, A₁=1, ..., then the product of two of these can be calculated as Aₙ₊₁Aₘ₊₁=Aₘ₋ₙ₊₁+Aₘ₋ₙ₊₃+...+Aₘ₊ₙ₊₁ Note that this sequence has to become negative for the identity to hold. This makes all the diagonals look really nice in the multiplication table, since two diagonally adjacent multiplications only differ by a single factor (which also follows from Ptolemy's theorem, and this implies the general result because it holds if n=1). In a pentagon: φ=A₂=A₃, A₂A₂=A₁+A₃, so φ²=φ+1. The only other identities involve linear dependence, like A₂+1=A₄ in a nonagon, although you can see this because the expansions of A₃A₄=A₂+A₃+A₄ is equal to the sum of the expansions of A₃A₁=A₃ and A₃A₂=A₂+A₄.
    1
  10998. feel like taylor approximation in a way!
    1
  10999. Answer to if you remove two black and two green tiles, can one always domino-tile the board: Obviously NO: if you remove two black tiles from a corner diagonally next to each to each other (leaving an isolated green tile) and two green tiles diagonally next to each other (from an adjacent corner) (leaving an isolated black tile), then there is a corner black tile with no green tile next to it which cannot be domino-tiled and also a corner green tile which also cannot be domino-tiled.
    1
  11000. Thanks for another great video, Sir! I also like your t-shirt, even though some would claim that 24.8193 is the root of evil. ;-)
    1
  11001. I don't really get it, why would anyone think sum will be zero? "Every item on the top has equal negative" - no it's not, unless you skip one-third of the input.
    1
  11002. 12,5,13 and 8,6,10 are the only two equable right triangle (integer) length
    1
  11003. The total area in the tree area exercise at 16:00 is 5.
    1
  11004. Really like that t-shirt, so creative.
    1
  11005. Of course you can't "always be able to tile the resulting mutilated board" with dominoes if you remove two squares of each color, because you can isolate one square or either color or one square of each color that way.
    1
  11006. phew. I followed all the way.
    1
  11007. Loved the relation of this diagrams with the cyclic multiplicative group of the field Zp
    1
  11008. Next time a video with unsolved math problems plz😊
    1
  11009. Thanks for your videos! Please crank up your volume though, I always have to turn you up waaaaaaaay high.
    1
  11010. Ramanujan was a bit like an idiot savant. The typical idiot savant performs an arithmetical operation enormously difficult (time wise) and cannot explain how he did it. We also do that all the time when we open our eyes and recognise 1000s and 1000s of objects in a 1/10 of a second. (Try to program that on computers: difficult ask AI experts!!!!) The difference was that Ramanujan dealt with abstract algebraic equations rather than trivialities like the typical idiot savant. But like them he often could not explain his thought processes : we cannot explain our visual pattern recognition ability. We just have it. I wounder if it was a mistake by Hardy to force him into pure mathematics......he perhaps should have left him produce books and books of non-trivial relations..... What do you guys think? Amazing video Mathologer. Very nice.
    1
  11011. For the final problem: Are there really 7 times 4! ways to color this board? Looks like a small proof by exhaustion, but too big to fit in a comment. Or maybe I'm too lazy.
    1
  11012. Imagining if such machines exist for primes or other seemingly incalculable sequences is making my mind go wild. It seems the hardest part is finding the machine that works.
    1
  11013. No 9s sum was pretty interesting
    1
  11014. If we're allowed to consider unbounded shapes, the entire plane was my first thought for an "anti-shapeshifter"
    1
  11015. You deleted the one with a link so hear it again. And I make copies...This is a reasonable representation of the explanation given in the Timaeus and not credited hear hear it the sound is deafening...
    1
  11016. In Germany we learn this first prooves in "Mathematikdidaktik 1". But the schoolbooks are a strong force to make bad lessons. All seen in school. But there are a few teachers with a other way of thinking how to work.
    1
  11017. He's back! It's an early Christmas
    1
  11018. As someone who loves both supernatural stories and math, I'd DEFINITELY buy that T-shirt if I lived in a country that didn't tax imports so much...
    1
  11019. I made it to the very end.
    1
  11020.  @Mathologer Excellent point, communication is an essential skill. My (short lived) disappointment had to do with not looking for subtext. Now that I have, great video.
    1
  11021. Ironic that the exciting thing about this is that it is the most boring fractal ever.
    1
  11022. This not mathematics but magicmathics!
    1
  11023. 32:00 Matt Parker did a whole video on non-integer and even complex inputs on this formula https://www.youtube.com/watch?v=ghxQA3vvhsk
    1
  11024. Wondering if the approach in this video can be extended to give an explanation of the solution to the Basel problem. Basically, start with the volume of a 4-dimensional ball (possibly without explaining why) and then count the number of representations of an integer (i.e a point) as a sum of four squares and tie them to get pi^2/6. I looked around and found proof 14 on the link below that seems to approach it this way, but I need to spend some time to understand what it really says. http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf .
    1
  11025. The 2xN boards were the last place I would have expected the Fibonacci numbers to turn up
    1
  11026. This method is dependant on the decision of how to split the "parties". Consider, for example, the case of the babies. There, we made an arbitrary decision to split the babies into two parties, one with the two babies and one with the red baby. This gives one result. On the other hand, one could argue that, since they are separated individuals, each of the two green babies should be considered as one "party", thus totalling three parties claiming parts of the heritage. However, if we split the parties this way, the result changes. That is concerning, as it allows ways for the results to be manipulated by a partisan judge, for example. Note this doesn't happen with proportional sharing.
    1
  11027. who is elso happy about the new video? :)
    1
  11028. seeing the simple beauty made mine! thank you
    1
  11029. Amazing... Beautiful...
    1
  11030. Yes. You are always RIGHT
    1
  11031. Tough choice, but the final proof was so elegant , it get's my vote.
    1
  11032. Thought: the next row after 1331 is 14641 which has GCD 2, but then 1 5 10 5 1 which I suppose means if you have a rule (C = A+B mod 5) then a size 6 (or is it 7?) triangle has its corners also following this rule because the 5 10 5 will mod out. Proof/disproof?
    1
  11033. @Mathologer what a coincidence, my plan was to watch these videos until I'm 100 years old
    1
  11034. Thanks so much once again! Let me get my popcorn.🍿
    1
  11035. Well, looking at this problem, I know the circling triangles and circles squares are just based on the fact that vector addition is commutative. I'm not too sure about the 7-pointed star, but it must have something to do with the sum of the two vectors.
    1
  11036. The circle and curve have the same diameter. As soon as I saw them together I was sure their circumferences were the same, so that would imply the diameters are too. Or at least that's my hunch. The windshield wiper thing reminds me of a neat problem which is whether a bicyclist in the rain will collect the least water by going fast or slow. It's not difficult, but the solution becomes obvious when you find the right visualization.
    1
  11037. The most interesting for me was definitely Kempner's series being convergent, and the grid-visualization was the whoa moment that made me giggle.
    1
  11038. What's the formula for the length of rubber band to start with, relative to the size of the wheel you're wrapping it around? I see how fixing the start point of the band and letting it stretch as the wheel turns means that the part from 9 to 10 uses about 1/10 of the band's initial length, from 8 to 9 uses about 1/9 of the band's initial length, and so forth, until you've ultimately used up about 1.92 times the original band length to cover the whole circumference... but that's inexact, because with a real rubber band you'd have some calculus in there -- once you've wrapped 9.9 to 10 onto the wheel, for example, the bit from 9.8 to 9.9 on the rubber band is a bit longer than the 9.9 to 10 part, and so on with a bunch of tiny errors that nevertheless might add up enough to be relevant.
    1
  11039. 👍🇧🇷🌎
    1
  11040. The authorship of "EUCLIDE of Megara" indicated on the cover of the book is a medieval delusion of European book publishers. Euclide of Megara is an ancient Greek philosopher. And Euclid, who is a mathematician, who is the author of this book, lived 100 years later. 🥸
    1
  11041. If you abuse math to infinity you can get any number you want...
    1
  11042. 2 mins into this video and already mindblowed me
    1
  11043. I dont want to exaggerate so Im just going to say that he is an unmatched international treasure !!
    1
  11044. 9:08 1st "aha!": having exactly one of each coin you can make change for any number 0-15 2nd "aha!": There is only one way to make them each. 3rd "aha!": Using a 16 cent coin, 32 cent coin and so on would...WAIT THIS IS JUST BINARY! :o
    1
  11045. Thanks. Another great video for a sunday afternoon
    1
  11046. I'm going for the proof with Euler's γ and the terrible aim proof ;) Especially those nice-to-know properties found along the way were quite interesting :D
    1
  11047. Thank you for the animations. Not sure why the inverted hemisphere has constant slope ? My expectation would lead to something like infinity at bottom of. cone ??? I did these integrations point to circle to sphere as an undergrad by hand. I was pleased with myself 🎉
    1
  11048. That 4 coloured one at the end is impossible to decipher for the colour blind. It's (probably, :) ) green and yellow but they look the same to me.
    1
  11049. I did make it to the end. 50 min video is ok for me. Anything below 1 hour works. My background in math... Well, I have a masters in theoretical physics and participated in math olympiads when I was at school, so math was my first passion and I learned a fair bit of extra stuff. Sometimes I feel sad for not being able to share things I learned and it wouldn't make much sense anyway with the current state of technology - too slow... I can only share a tiny fraction of what I know, yet I know only a tiny fraction of what I want to know. Sometimes I feel sad when I forget something I once have known, and when that something isn't easily available for me to refresh. I go to wiki and can't find derivation of Webber functions, can't find methods for solving quasi periodic differential equations, can't find some non-linear mesoscopic physics topics. Hardly anyone makes a good video on quantum physics and general relativity, which are much better understood these days than some think. Quantum field theory and string theory sometimes seen as non-existent... Riemann zeta pops up in QFT all the time as well as those "divergent" sums. Renormalization groups topic didn't exist on the wiki when I needed it, at least it is there now... Some baby level math isn't always taught at school and gets missed at universities later. Like Ceva's theorem, continued fractions(and their connection to hypergeometric function and Pade approximants), game theory(for cases when the next move of each player is restricted by the move history), all kinds of coordinate systems(and useful tricks to solve physics problems, like orths derivatives), (least) quadratic non-residues.
    1
  11050. It was a parker anagram!
    1
  11051. It's not just 3 6 9 numbers all of them are awesome
    1
  11052. The binary proof makes more sense to me
    1
  11053. This is a brilliant and well made video. Wish you'd post stuff more often
    1
  11054. Has to be the bishop Probably caught up with angels dancing on the head of a pin.
    1
  11055. My most memorable was the no-9s series. I could not fathom how it would be convergent.
    1
  11056. This is the first time I ever grokked e
    1
  11057. Some day I want to go through all your videos with a big notebook and just learn my butt off! Thanks for always producing such great content :)
    1
  11058. I'm a 3rd year student in engineering school in France, 2 years of Prépa of that makes sense to you and just started first year out of three of applied math(s). I understood pretty much everything, I could have replayed some parts for a 100% where calculations were a bit fast but as we can pause if we really want to check the arithmetics it's not a problem! I remember having a harder time understanding (or not actually) other videos of yours, like the one involving the axiom of choice. I am really looking forward for your next videos! Also I was wondering if it was possible to make an even better approximation of the sum of n^-2 by Euler-Maclaurin-seriesing the first Euler-Maclauring we had.
    1
  11059. Can we switch to the powers of 2 coins? They are much nicer mathematically...
    1
  11060. That pumpkin "pie" T-shirt
    1
  11061. 🌀
    1
  11062. FJM stands for Fibonacci Juice Master.
    1
  11063. It's amazing!
    1
  11064. What are the odds of all my favorite German vlogers being bald?
    1
  11065. Yes it's math explanation that has nothing to do with physics. So in quantum mechanics, if you measure close to infinity you will not get infinitely many points, you will get a black hole because measuring device always interferes with measurement. So yes, math OK, physics not so much. I know, you try to tell there is infinity angles in the universe, which also is still to be proven. Math works out, irl not so much. Measuring so small you know you will reach planck length (yes, you conveniently ignore this) and planck radius, angle etc... So, until you do these I'm very much unimpressed. You can calculate lim towards these but still nope, sorry. You can claim in physics there are so many dimensions there's no limit to directions, but you need to prove that. Math is very good but it's math, not reality. You did not even calculate amount of possible measurements on radius of r ball, come on, you didn't even try with this video. IK, gets more comments and such but c'mon.
    1
  11066. Most memorable part: @14:50 You placed the picture of Oresme under the leaning tower, and if his head would have hit the tower if not for his posture problems 😅😅😅 Jokes aside, I really enjoyed this video, one of the best!
    1
  11067. You can't change a square into a circle by cutting off the corners in this fashion. No matter how fine the steps get from cutting off the corners, the original square drawn around the circle will never, ever become a curve. It will remain a polygon; a series of minute steps made up of straight lines that are so small you cannot perceive them by conventional means.
    1
  11068. Fortunately Peter Steinbach isn’t a chemistry teacher…. (is he???)
    1
  11069. I've still got my British Thornton slide rule in my desk draw. I must be old!
    1
  11070. To read more on the work of Indian mathematicians, see KV Sarma and Dr CK Raju. Not only this but Europe's longitude's problem in the 15-16th C was solved using the results of these scholars.
    1
  11071. I tried it before watching and got this solution in about 5 minutes (use GeoGebra for visualization): x^(2)=(y)/(2) (y+1)
    1
  11072. Swivel proof is cooler for my taste, but the second one is more rigorous
    1
  11073. They don't, but I do :-) This approach to logarithms is used in chapter 1 of Analysis by Its History by Hairer and Wanner.
    1
  11074. As a short answer to the easy challenge of putting out squares: As long as you carve out alternating (black and green) squares along any path through the labyrinth tiling the residue with dominoes is always possible, otherwise not.
    1
  11075. I liked the coloring proof better. It was easier to visualize. Edit - Also, you didn't ask which proof was "better"... just which one we liked more. :P
    1
  11076. @Mathologer 35:45 I feel your pain, the same happened to me when my Math professor gave us a difficult double integral and I saw an easier way to calculate it geometrically.. zero marks also for me grrr, but talking of video generic most memorable for me is the convergence of the sum of the 100 zeros series vs no 9s series, let's you feel the power of the infinite!
    1
  11077. Wow great video on a new way of thinking about what I thought were old and familiar formulas and identities!
    1
  11078. Than you very much. Exceptional as always. About the IQ tests... it would be nice to talk about Solomonoff's Algorithmic Information theory. ;)
    1
  11079. Most Memorable : The thinning out of the 9 series by the square diagram. When you first showed the square for the first 100 terms of the harmonic series and asked whether it converges or not, I too tried to create a similar larger box in my mind for 10,000 but apparently 5 seconds were not enough for my brain to do that!
    1
  11080. delta : a normal segment (e.g. of length 1, 0.5, 2 etc...). d : an infinitesimal segment (very small almost a point but not a point length = 0.00...)
    1
  11081.  @Amine-gz7gq  Right I just don't see how, in practice, I'm supposed to efficiently use these concepts to determine the slope at that point. Do I pick a random small number for d and then from the difference df/dx - f/x I'm supposed to use that to help guess the slope at x? And if that doesn't work I pick a smaller number for d and guess again?
    1
  11082. Great video as always! I'm wondering is there a formula for the sum of 1/(n^3)?
    1
  11083. The mu joke was great, almost woke up my parents!
    1
  11084. Mathologer never runs out of golden ratio stuff; that's why I like him.
    1
  11085. Burkhard's giggle. The best.
    1
  11086. My pick is Tristan's fractal!
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  11087. Dear Burkard, great mind-bending stuff, the animations really help get at the underlying patterns. It was reassuring to have our mathematical seat-belts on, but I don't think we needed the crash helmet because you're a reliable and safe driver! I studied maths up to Masters level, though mainly applied maths and theoretical physics. I was never aware of the Bernoulli numbers before, so it is wonderful to see how they shape the patterns of the Riemann zeta and other sums. Thank you so much for all your time and creativity. I'm looking forward to more Euler-Maclaurin magic. PS A request (again) that you one day do something on chaos, Feigenbaum's constant etc... I never felt I got to grips with all that.
    1
  11088. Loved this one! Thanks :)
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  11089. 7:35 I liked the colouring proof more, because it can easily be done on paper. In fact, that is exactly how I proved it earlier in the video when you asked for proof (the inscribed circle helped as a clue), except I didn't colour it, but labelled it with letters (same thing, but colours are prettier). EDIT: yes, the swivel proof is superior. I'm convinced. EDIT2: the circle has a greater diameter. It's clear from 14:00, but can be easily proved: the chords are the "diameter" of the strange curve, and since they do not pass through the centre of the circle, they must be shorter than the diameter of the circle.
    1
  11090. Most memorable: Nicole Bishop's proof
    1
  11091. I'm not sure I get the first explanation of why it works, without using logarithms. Of course if one wraps two scales, one being double the other, around concentric circles, they will retain their relationship and the numbers will retain their "double by two" relationship after wrapping. But wrapping it plainly, not logarithmically, that would remain true. And it wouldn't allow for sliding the two scales against each other and have their keep the multiplying property...
    1
  11092. I think that this must relate to the polar cordinate equation r=cos(theta) describing a circle? I was quite surprised when that one fell out.
    1
  11093. Related to Sierpinski's and Pascal's triangles. Awesome
    1
  11094. I think it would be neat to make an animation of a reflective lantern getting more and more bands and converging to occupying the space of a cylinder while reflecting the surroundings very differently from a cylinder.
    1
  11095. 1:32 oh interesting. I noticed it was also taking the 2nd 1 in fibonacci was x2 + (3 numbers below on the other side) and it repeats up.
    1
  11096. No link provided for Henry's video.
    1
  11097. a to b: small disc moves clockwiase for an odd number of discs, anticlockwise for even, this because the intermediate position for the n-1 stack is the other peg, so the solution for n-1 is the opposite direction.
    1
  11098. 3³+4³+5³=6³. Are there other examples with powers higher than 2?
    1
  11099. Ich bin sehr erstaunt, ich hätte die irrationalität von sin oder cosinus genutzt, um zu zeigen, dass bestimmte n-gons nicht drin sind, denn rotieren und verschieben ist nur ein Basiswechsel und mindestens eines bleibt irrational dabei. Wenn also diese irrationale Zahl nicht im grid darstellbar ist, dann können wir diese Form nicht darstellen. Das lässt sich über die Vollständigkeit (bzw hier gerade eben nicht) zeigen. Es gibt nicht immer Folgen, die gegen diese irrationale oder algebraische Zahl in einem abzählbaren grid konvergieren. Das wäre aber genauso algebraisch
    1
  11100. just happy to learn about gamma from this vid, great stuff
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  11101. Fascinating example of modeling something from the real world into something formal :D. For me this is teaching a certain way of thinking. Thank you for your work! Cheers!
    1
  11102. awesome video
    1
  11103. Is it bad that the stretching rubber band made me nervous?
    1
  11104. Fifth
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  11105. I'm old with stiff hands so years ago when I started velcros  I had to ask in this modern world why do we still fasten shoes with a string.
    1
  11106. This is only true for any finite numer of steps. After an infinite number of steps, the result will be 0.
    1
  11107. Is it applicable for primes
    1
  11108. How lucky you are to have met Ron Graham! But I am sure the great enjoyment of meeting was mutual... 😊
    1
  11109. Hmmm. It has been a while since I learned Pythagorean Theorem, but I don’t remember learning either of the twisted square proofs. I just remembered learning the equation. I also don’t remember learning about the unit circle in dealing with pi. 🥸
    1
  11110. If someone can imagine from the outset that ANY finite amount of paint, no matter how small, is enough to paint ANY finite surface, no matter how large, then I wonder why for the imagination that powerful it is a problem to vision an infinite surface covered with a finite amount of paint.
    1
  11111. After many trials at shortening the list : 3=1+1/2+(1/3+1/6)+(1/4+1/8+1/9+1/72)+(1/5+1/10+1/12+1/15+1/20) !! Each sum between parentheses is worth 1/2.
    1
  11112. I wanted to write a program to do figure out all the different change you make out of a amount on an AS400 after seeing a Saturday Night Live skit. My teacher was not happy with me.
    1
  11113. Sum of no 9s series is my favorite piece, at least for today in all of Mathematics.
    1
  11114. The herons formula is taught in India at 9th or 10th standard (maybe earlier I don't remember)
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  11115. I’m still on this video ( a week later ) Been chipping away at it bit by bit. It’s like a little candy I can enjoy at the end of each day after I’m done with all my school and work related math.
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  11116. I got it =)
    1
  11117. Why does repeatedly dividing by two give the opposite sequence of digital roots? It's actually the same sequence if we rearrange it a bit... ... 0.0625 -> 4 0.125 -> 8 0.25 -> 7 0.5 -> 5 1 -> 1 2 -> 2 4 -> 4 8 -> 8 16 -> 7 32 -> 5 ...
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  11118. If the number of disks is odd you have to begin with the smallest disk on the pin you want to end up on. If you have an even number of disks you have to start at the pin you don’t want to end up on.
    1
  11119. Thank you for your explanation and visuals! In no way I'm a matematician, but when I saw you summing up the coordinates of the points on circles with different radii and staring angles ("phases" so to say) I was like... Heeey! That looks really familiar!
    1
  11120. Most memorable part was the ugly maximal tower that can not be built layer by layer.
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  11121. What a great explanation!
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  11122. 28:40 Was I tricked into leaving a comment? If so, it worked. 7 is wrong!!111!! ;-)
    1
  11123. How crazy would it be if some variation of this riddle at 4:15 solves hodge conjecture.
    1
  11124. I noticed a couple of little patterns not directly mentioned. In the first sequence, the first terms follow the sequence 1**2, 2**2, 3**2, ... In the second sequence, the last term on the left is consecutive multiples of four squared. These observations make it trivial to figure out the Nth row of the patterns in your head.
    1
  11125. First seen on jacob Bronowski's Civilisation/Ascent of Man video. Elegantly done with square tiles on a hillside where Pythagoras lived. Stunningly, brutal aimplicity.
    1
  11126. @11:49 9 4 17 13 Since 104 is even, it'll have to be the double of the right two numbers. So their product is 52. 52 factors to 2*2*13, so the possible pairs are (1,52), (2,26), and (4,13). 153 is the product of the left two numbers. 153 factors to 3*3*17, so the possible pairs are (1,153), (3,51), and (9,17). Since the numbers progress in a Fibonacci manner, you'll need the top two numbers to sum to the bottom right and you'll need the right two numbers ti sum to the bottom left. The only way is for the top two numbers to be 9 and 4 and the bottom two numbers to be 17 and 13.
    1
  11127. Interesting... I was wondering where you were going with this....then my heart started beating faster
    1
  11128. 16:14 x*2 = x*10/5 which mean we move backwards from x*5/10. No DR, no long explanations
    1
  11129. Never seems this before
    1
  11130. Interestingly Prof John Robison of Edinburgh, writing in the mid-1700s, developed the epistemological basis for what we now call the Method of Successive Approximations. An equation need not perfectly describe a natural system, to be useful. One simply keeps improving the model with equations that fit the data more closely. A Swiss contemporary named Euler grabbed Robison's concept and ran with it, proving that in the case of power series that can be proven to converge, one can work out how many terms of the series to calculate, until the difference between further terms is less than the part of the number having practical utility. By Euler's advancement of Robison's method, power series to calculate pi, sines, cosines etc. have been found. We can know the circumference and diameter of a round object, to greater accuracy than we can measure them, and this gives us the edge we need to build things with wheels in them. So yes...the whole of mathematics is a tool for understanding the Universe.
    1
  11131. The slides need to be rearranged so that they are entirely visible. They are truncated to the right in some instances.
    1
  11132. Wow! Best part was that the sum skips every integer! You’d think it hit at least one, SOMEWHERE
    1
  11133. That Pascal's Turtle part at the end was beautiful and the music chosen to go with it was great. I almost cried (in a good way) actually. Thanks for the video and for showing us Lill's method.
    1
  11134. Love to watch your videos! Though this time I must admit I could not really follow, probably have to watch it again and stop at many steps to think about. : )
    1
  11135. Most memorable: bizarre maximal overhang towers. Nice music at the end.
    1
  11136. I wish you had been my maths teacher at school. 👍
    1
  11137. One can also exactly get to pi (or any other value) by greedily choosing signs of each term in a single harmonic series (as opposed to mixing two series). I.e., the n-th term equals ±1/n where the sign is + if the partial sum of n-1 terms is short of pi or - if it overshoots pi. That is also guaranteed to converge. Interestingly, then the ratio of the number of positive and negative terms approaches 1. In other words, there are "relatively equal amounts" of positive and negative terms. It can also be done non-greedily, but then that ratio of 1 is no longer guaranteed. Proof in the margin.
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  11138. Always a good day when one wakes up to a Mathologer video
    1
  11139. I have a pocket slide rule set up like this
    1
  11140. Oh you said it. Spoke too soon lol
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  11141. Not sure if anyone will see this, but I wanted to ask for a video about the power of a circle, which includes the intersecting secants, and intersecting chords. Obviously, the point would be to see a nice, coherent discussion that ties all these circle theorems. I don't know if the right lens is projective geometry or something else, but I have seen beautiful solutions, yet they still feel disconnected. Thoughts? Anyone else?
    1
  11142.  @Mathologer  I was referring to videos like Numberphile's Epic Circles and several of 3blue1brown, like "Why do colliding blocks compute pi", or many here! But of course, I don't know if any of these analytical approaches have any relevance here. I simply have the intuition that there is an image lurking behind the power of circle. For instance, I knew the intersecting chords and intersecting secants theorem separately for math competitions, but never saw them as part of the same concept. Ever since I learned that they are related through the concept of power of a circle, I've been trying to find any discussion that goes beyond simply stating the theorems algebraically and using them in a sort of obvious example. I'm praying to the math gods that there is some deeper meaning, and something worthy of one of your videos!
    1
  11143. Who gets credit for supporting the portion of this video’s content which appears after the credits? 🤔
    1
  11144. If we imagine the hexagonal tilings as stacks of cubes, we can extrude each color to fill in the 3 visible faces. These faces are each squares of equal size, and thus each color must be represented with an exactly equal number for all 3 faces. The thing I haven't fully figured out: how to prove that each hexagonal tiling must match a 3d stack of cubes.
    1
  11145. Never explained only asked to memorize
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  11146. Description for infinity and process paradox definement: I). If there is one or more mathematical process is defined among limited elements, you ain't norm-ally have any result as a paradox and/or infinity sort. But, if at least one of those elements is not well and concretly defined, then you may receive conclusion at pardixial and/or infinity sort. For example: element c is the number (excluding zero) one of Reel leauge. Elemet z is equivalent to zero. The process is division and orders element z to divide c, means c/z=? C/0=infinite or defineless. Let's say element c=z then division process among them is 0/0=N/A defineless. Infinity-inf., inf./inf., inf-inf, inf.-c, etc are all defineless or infinity again. We have to make a total concrete definement among our elements, process and processing order rule. Our description mentioned above says/asserts: if there is one or more process among limited elements, and if the elements concretly and really defined, and processing syntax is definitly defined (exchange, disribution, replacement rules must be applied linearly not rondom, (somtime like that sine time like this during processing is forbidden!) Then you may not have any paradox or hardly any result as infinity nor N/A leauge. Infinity is not and cannot definable precisely and concretly, so that even if the elements limited under processing you may not have any other result except 0, infinity, N/A leauge or paradox. For Processing 0 with 0 or with infinity, it gives simillar results. C is element if reel leauge, infinity-c is equivalent to infinity. BUT if c is equivalent to zero, infinity-0 is not equivalent to infinity the result is N/A. Ahh why? How? It is because of 0 is not a well and concretly defined number, even as if it is not a really reel number 0 is simething like infinity it has double face like blade edges and binary quantal appearence in pricessings. In limit pricessing it has somethime -epsilon maner and simetimes +epsilon maner and breaking the pricessing synrax rule!!! We in mathematical pricessings to have a reel result and avoiding fall into infinity result trap, breaking processing rule. So, uncertainity defunement if zero, and zero's hermaphroditial +/- binary epsilonial janus-face nature causes the problem The definement for zero must be checked up and recorrection it needs. Our acceptances for zero are: If the process is +/- we may put 0 to left/right side of equivalency. This is correct you put -/+ a to the left and/or right. But not correct durectly puting zero to the wings, because 0=a-a and zero occurance after process aming reel numbere is nit equivalent to any zero absolute. I mean you may not put any absolute zero in your processes because the zero at the process +A-A=0 is not the zero absolutly exists in your assertation. Zero effectless at -/+ process among reel numbers leauge, Like this Zero is terminator (singular blackhole) at process multiply among R leauge 0/0 is N/A C/0=infinity 0/c=0 As you see here there is something sounds paradoxial already and there is definitly a great problem about definition of zero as a number!
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  11147. Swivel
    1
  11148. I had the Tower of Hanoi on my graphing calculator way back in middle school, and used to 'speedrun' 10 disc solutions when I was bored in math class, so I can actually say with relatively high confidence I could still pull it off without a mistake. There's a nice rhythm to it once you get the hang of it, almost meditative. Interestingly, because it was presented as a straight line of pegs instead of a triangle of them, I never stumbled on the clockwise/counterclockwise solution to figuring out where the top disc goes, another example of how how you frame a problem affects how you solve it. Instead, my method was to think in terms of a start peg, a storage peg, a goal peg, and partial pyramids. Move the top part of the pyramid to one peg (the target peg if the pyramid has odd number of discs, the storage peg if it has an even number), move the top disc of the bottom part of the pyramid to the other peg, move the top part of the pyramid on top of it, and now you have a bigger top part and a smaller top part. Repeat until there's just one disc on the start peg, move it to the other peg, and then you have a new whole pyramid, and you recurse back to the start, just with the start and the storage peg switched places, and a pyramid one disc shorter.
    1
  11149. What's the scale
    1
  11150. Been starving for mathologer content
    1
  11151. 15:15 it's 5. if you continue you'd get +1 for each step, provided the squares don't get on each other (they do)
    1
  11152. It is unclear why face ups are 0 and face downs are 1 when calling downs 0 and ups 1 is far more intuitive. That way, total number gets bigger and there is no limit of 0 as lowest integer because number will increase, so Nathans proof messes everything up and makes thing obscure to occasional viewer who may ask a pretty legit question wtf is going on and isn't it just made up. I mean, i understand that they will be stuck at 111...11 that way, but it also says that explanation about number always decreasing and something about negative looks kinda stretched on common human terms to be easier to understand but it isn't - it just raises more questions like why all that was said should be exactly that? Basically adding some sort of mystery from nothing. That's why i don't like math much and therefore bad at it - such "problems" are so abstract and look completely detached from life by that mystery.
    1
  11153. A request if it's not too much: Could you link to the shirts you wear in the video descriptions in the future? So many shirts I'd love to buy!
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  11154. Is it too late to do the 11:49 puzzle yet? 153²+104²=185² Since the pythagorean triple's coefficients are products of a pair of numbers in the box(forgive me my math is rusty), I factorized them. 153 = 17 x 9; 51 x 3; 153 x 1 104 = 52 x 2; 26 x 4; 13 x 8 185 = 37 x 5 Working with the numbers in the prime factorization of these numbers within the defined rules, I found this grid to work: 9 4 17 13 9x17 = 153 2(4x13) = 104 (9x13) + (4x17) = 117 + 68 = 185
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  11155. 22:52 it must be 1.3 / 1.3.5 in the summation
    1
  11156. Instead of asking where the circle is hidden, wouldn't it be better to ask for the hidden triangle?
    1
  11157. After just a little bit of checking, I'm finding that 4 colors has no special numbers, 5 colors has them at 2, 6, ..., 5^n +1, 6 has none, 7 has at 7^n+1. Do only primes have special numbers??
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  11158. Great! Coincidentally, I stumbled on the first property of this video just today, while reviewing errors of a student of mine on matrix product (how did she get this 5 here? Wait... What about... This kind of "stumbling"). About the area of the 1-1-2 pythagorean tree, it seems to be 1 for every new step of the tree, so I'd say the one you display has an area of 5.
    1
  11159. Years ago I stumbled on how to solve these after seeing someone show analogies between finite difference equations and differential equations: Let's say you're trying to solve the recursive definition for the sequence S(n): S(n) = S(n-1) + 5. Start with the homogeneous equation by only keeping terms involving S(n): S(n) = S(n-1). Clearly S(n) = C for some constant C; we'll call this H(n). Now guess that S(n) is the sum of H(n) and some other function of n based on the sum of the remaining terms, in this case 5. The trick is to always guess a polynomial one power above the inhomogeneous term, in this case a linear function: I(n) = A + B*n. By checking the recursive definition, S(n) = S(n-1) + 5 <==> C + A + B*n = C + A + B*n - B + 5 <==> B = 5. So if S(n) = C + A + 5*n, or by removing the redundant constant term, simply S(n) = C + 5*n, then the recursive relation is satisfied and the variable C is determined by setting the initial value of S(n), e.g. S(0) = 0. This approach works for summation as follows: If S(n) is the sum of the integers from 0 to n, then there is a recursive definition of S(n): S(n) = S(n-1) + n. Applying the same technique: S(n) = A + B*n + C*n^2 ==> A + B*n + C*n^2 = A + B*n - B + C*n^2 - 2*C*n - C + n ==> (1 - 2*C)*n - (B + C) = 0. Since this relation has to be true for all n, each coefficient in front of a power of n must be zero: 1 - 2*C = 0, and B + C = 0 ==> C = 1/2, B = -1/2. This means that S(n) = A - n/2 + n^2/2 = A + (n * (n - 1))/2, just like in the video but with an additional constant which can be set by setting some initial value of S(n). It's actually possible to continue this method in a general form for any sum of powers, which results in a convenient definition of coefficients which is again related to the definitions in this video. Just thought I'd share a parallel method for generating these sums.
    1
  11160. Great video as always! It seems sqrt(2) is not special. Instead, sqrt(2) +/- 1 is. (Why 2 of them? Similar to 1.618... and 0.618..., one greater than 1 and the other less than 1.) And I remember Mathologer introduced one of them as the silver ratio in an earlier video. The golden ratio: cutting off 1 square from the original rectangle. The silver ratio: cutting off 2 squares from the original rectangle.
    1
  11161. Fantastic Video! Thank you.
    1
  11162. Mathflix sounds uber cool :)
    1
  11163. Merry Christmas from my simulation to yours. Gotta keep it real on the Q-tip(what does Q say your IQ is if you are Q? StarTrek the next Generation does an amazing mocking of Q clearance & what it means in governments you know like John G. Trump or Charlton Heston or Snowden all who had-have Q clearance this isn't top secret problem is most people are things here they act as computer player easy red dress people the walls or why W-HILL-S is used in Star Wars remembering that HILL is Q or where Snowden said he went for secret training when he was on the Joe Rogan podcast the fist time about 1 hr 23 min into it) for all those out there with the 2005 IQ try adding the LeveR & break on through to the 2112 other side. What a RUSH
    1
  11164. First time I jhave seen face nirmals being called 'spikes'. Nice analog.
    1
  11165. When are we gonna get into z-transform
    1
  11166. 6:41 - I don't get this... In the 15/6 example, couldn't I just fill 5 of the holes with 1 pigeon and put the remaining 11 in the last hole...?
    1
  11167. The statement is that there is at least one hole with at least 2.5 pigeons. 11 is at least 2.5. No problem.
    1
  11168. 🤯
    1
  11169. To answer your first question, I learned about these two visual proofs only recently from YouTube that's about two years ago although I consider myself a math fan and I am 49 years old and by the way I am from Iraq, maybe this why I didn't learn math properly from school!
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  11170. not so great with higher mathematics, but I feel like this is related to how planets and perhaps even electrons fall into standardized orbitals.
    1
  11171. That why using therm by therm approach on infinite operation is flawed. Formula that depend on therm by therm also exist and should be avoided in this context. That the -1/12 “discovery” all over again. And it's just the proof of the limit of the formula we use.
    1
  11172. Excellent
    1
  11173. This is analogous to decomposing a three phase voltage system into thr positive, negative and the zero system...
    1
  11174. The first chessboard thing. I don't know in which one, but I had that in a lecture just a week or two ago
    1
  11175. i liked this shorter video
    1
  11176. Our 10 fingers makes 9 special and so the specious arguments of vortex videos are specist.
    1
  11177. Neat! Yes I made it to the end 😀
    1
  11178. Did this video answer the question?
    1
  11179. I think i have an idea for the 2*10^x series. Which amounts are we including? 1, 2, 5, 10, 25, 50 cents. 1, 2 euros, 5, 10, 20, 50, 100, 200, 500 euros? All of them? Just the coins?
    1
  11180. Merry Christmas!!! What a beautiful proof. Amazing.
    1
  11181. Great insight. My thanks!
    1
  11182. Music comment and math question: At one point I thought, okay I like the music now, but as soon as someone starts singing, I'm going to hate it. But she started singing and I liked it even better! Also why are the two triangles in the proof that the derivative of sin x = cos supposed to be similar?
    1
  11183. The most memorable thing was the gamma cat
    1
  11184. hello from 1035
    1
  11185. Beautiful!!!
    1
  11186. colouring proof
    1
  11187. Continuous fractions, finally got it..!
    1
  11188. I don't know if I feel smarter or dumber for watching that
    1
  11189. Wow, ich habe erst jetzt bei der Aussprache des Titels auf deutsch aufgehorcht und feststellen müssen, dass der Mathologer aus Würzburg ist. Unglaublich! Hätte ich nicht gedacht!!!
    1
  11190. Thanks! Clearly I am very basic with my math understanding. 😂
    1
  11191. Asking halfway through. Does it work for four points? 180 degrees goes to infinity!
    1
  11192. Before I watch, thank you. We see a computer beat us at go and chess and admit defeat as a species. Nah, bro. I like you. I looked at elliptical curve encryption; no bueno. I can retoroduct the code in the patent that is supposed to maintain cybersecurity for the future. NO CHANCE it will withstand quantum power boosts and my dynamical arithmetic processing. Stochastic processes can rebuild the system perfectly. Randomness and time are compressed out in dimensional transform. And a plot wraparound function a lot like your mutilated chessboard is how limits are set in ECE. It is child's play to steal the world now. Good thing I believe in something above myself.
    1
  11193. Anybody else noticed "the nudge" at 15:20 like me?
    1
  11194. Ops, looks like somebody else already beat me to it.
    1
  11195. It looks to me that the number of degrees of freedom is not conserved. Of course it has to be conserved, so could someone explain this to me? I start with a general polygon, which has 2N degrees of freedom (denoting with N the number of vertices, we have 2 d.o.f for every vertex). The transformation, also looking at the papers in the description, are matrices that does not depend on the polygon, so they have no free parameters. I end up with a regular polygon, that can be rotated or scale (2 free parameters), and translated (other 2 degrees of freedom). So it looks to me that I start with 2N free parameters, but I end up with only 4, and I do not understand why. What is wrong in what I said? Thank you, amazing video!!
    1
  11196. (IMO) this is a amazing video to watch for a idea, Disclaimer: Its just My own Opinion on the suggestion, Advice; "Be very proud of having a Opinion on something that counts for eachother including *me.”
    1
  11197. :face-red-heart-shape:
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  11198. Most memorable to me is the definition of gamma, because I’d heard of it but had no idea where it came from.
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  11199. Normally I don't stay for videos this long, but this time was an exception, loved the visual proofs the most, thank you for this, I'm looking forward for more
    1
  11200. BONUS: Try it with other polygons! 3: Nothing interesting happens when you try it on an equilateral triangle, we don't have enough points to use ptolemy's. Unless we consider one point as a duplicate... L^2 + 0L = L^2 yeah there's nothing here, next polygon 4: Pythagoras shows up. 1^2 + 1^2 = [diag]^2, diag = sqrt(2). You still have golden rectangle magic but this time it's in this form: take a rectangle with aspect ratio 1 : sqrt(2) and cut it so the long edges are split in half, you'll now get two identical rectangles each with ratio 1 : sqrt(2). This is the basis for A4 paper. 6: Two kinds of diagonals, call them C for the short one and D for the long one. C^2 = 1 + D D^2 = 1 + C^2 = 2 + D CD = C + C Very quickly you can figure out what the exact values of D and C are. (Spoiler alert: D = 2 and C = sqrt(3)) If you draw out the diagonals here the symmetries do check out (if it wasn't obvious by the star of david and rule of thirds shenanigans going on) 8: We still have the same diagonal before. In order of length let's call the diagonals X, Y, and Z. X^2 = 1 + Y Y^2 = Y + X^2 = 1 + 2Y Z^2 = X^2 + X^2 = 2 + 2Y XY = X + Z XZ = 2Y YZ = 2X + Z (That last identity is redundant btw but it's useful to have around) What would a reverse X, Y, and Z maneuver look like? Trying to imagine, rescale, and rotate a 4d box is not the best idea for a 3d brain so let's reason through this algebraically. But I assume it involves the same "removing squares trick" (unless of course, something weird happens) X(AX+BY+CZ+D) = (B+D)X + (A+2C)Y + BZ + A Y(AX+BY+CZ+D) = (A+2C)X + (2B+D)Y + (A+C)Z + B Z(AX+BY+CZ+D) = 2BX + (2A+2C)Y + (B+D)Z + 2C There's a somewhat nice pattern here, the As only ever get added to the Cs, and the Bs only ever get added to the Ds 9: Call the diagonals in order of length from smallest to largest F, G, H. Can you tell I'm running out of letters? F^2 = G + 1 G^2 = G + H + 1 H^2 = F + G + H + 1 FG = F + H FH = G + H GH = F + G + H Notice that the 2s only ever seem to crop up when the number of sides is even. They're gone here. The magic bars? 1¦F¦G¦H F¦G+1¦F+H¦G+H G¦F+H¦G+H+1¦F+G+H H¦G+H¦F+G+H¦F+G+H+1 This feels like a weird magic multiplication table where a bunch of combined areas are all suddenly the same as each other For example, H^2 = F^2 + FG = F + G^2 = GH + 1 = FH + F + 1 Anything larger: We can use some logic for something: what happens when we repeatedly apply Ptolemy when one of the diagonals is the shortest one? Let's go in alphabetical order. A^2 = 1 + B AB = A + C AC = B + D AD = C + E The magic happens when we loop back around somewhere, but what if that never happened? Well then we get the natural numbers. A = 2, B = 3, C = 4, etc... The repeated identity is 2N = (N-1) + (N+1) But then Ptolemy under the reals is this (remember, Ptolemy applies to any circular quadrilateral, hence reals): Assuming A > B > C > D, (A-B)(C-D) + (A-D)(B-C) = (A-C)(B-D) You can multiply this out yourself to confirm this BONUS: What exactly are R and S, anyway? Two ways to go about this: one using trig and the other using polynomials TRIG: Let's solve for R first. The interior angles of a heptagon are 5pi/7, so the isosceles triangle formed by 2 edges and the R diagonal must have base angles of pi/7. R = 2cos(pi/7) We can use the formula R^2 = S + 1 to solve for S. S = 4cos^2(pi/7) - 1 The equivalent for phi is that it's 2cos(pi/5). POLYNOMIAL: The two solutions of phi came about from x^2 = x + 1. Can we do something similar here? RS = R + S RS - S = R S = R/(R - 1) R^2 = 1 + S = 1 + R/(R-1) R isn't 1, obviously. Otherwise S = 1+S which is just nonsense R^3 - R^2 = R - 1 + R R^3 - R^2 - 2R + 1 = 0 Similarly, R = S/(S - 1) S^2 = S + R + 1 S^2 = S + S/(S - 1) + 1 S^3 - S^2 = S^2 - S + S + S - 1 S^3 - 2S^2 - S + 1 = 0 Feel free to use cubic formula but I'll stop here
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  11201. I really like the logarithm approximation of the harmonic series. And of course the gamma constant representing all that remaining area which is less than one :)
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  11202. what is the piano song used in the intro?
    1
  11203. 32:34 I hope there aren't any viewers who rely on lip reading to understand the video.
    1
  11204. 19:53 Numberphile on blast
    1
  11205. Double helix lacing for the win! It's a variant of criss-cross, because In practical application you also have to decide for each island if you go from top to down or from down to top. Here the double helix makes less resistance than the criss cross. BTW: I love Ian's side, years ago after changing the laces of my dance shoes, I just pulled out the old ones, and then was stumped... how does one actually do this correctly? After some searching I found Ians site and it was a great resource to a) learn that there is no "one correct" way, b) helped me to decide for double helix.
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  11206. THE calculus.
    1
  11207. I first saw the twisted square visual proof of the Pythagorean Theorem online when watching maths videos online out of interest in year 11
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  11208. Maths Background : French student, did olympiad mathematics before starting cgpe this year (2 year school before Grande Ecole entrance exams) Suggestions: I really liked the step by step introduction of more sofisticated ideas (rather than having something like the Euler McLauren formula be presented then proved without much explaination). Maybe some more abstract algebra or number theory would be cool!
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  11209. I have dual degrees - comp sci and electrical engineering. I was okay with most of it except the substitution of the lower ntegration limit from 0 to 1.
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  11210. The problem with proof without words is that its not always rigorous. For the case of the sum of consecutive integers, you would still have to prove that that for each n, we will get an nx(n+1) rectangle. I am interested in knowing if there exists a proof without words which could be considered as rigorous. My guess is that there isn't, but I am a bit of a fan of proof without words and will always keep up my hope.
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  11211. Was I the only one who watched half of the video before noticing the T-Shirt and laughing too much, that they had to pause the video to be able to watch and listen to the rest of the video properly?
    1
  11212. The circle must have a larger diameter. The areas of the blob that stick out more stick out less than the areas of the blob that stick out less stick out less.
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  11213. Enjoyed the video, and watched all the way to the VERY end. Thanks!
    1
  11214. We used to use these at school even in the 70s - and called them slipsticks.
    1
  11215. I definitely did not learn the 3, 4, 5 rule in kindergarten. American education really, really failed me
    1
  11216. I posted some hours ago the URL where anybody can freely and lawfully get access to the article by Karel A. Post, but this comment was removed because I put a direct link to the paper (I didn't know that Youtube forbids that). You can still go to the "Elemente der Mathematik" page "ems . press / journals / em / read" (remove the spaces). There, it is indicated that everything older than 5 years is "freely available on the Swiss Electronic Academic Library Service E-periodica". Following that E-periodica link to "Heft 2" of "Band 45" (Issue #2 of Volume 45), you can download the article "Moessnerian theorems : how to prove them by simple graph theoretical inspection", which is the third of that issue.
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  11217. Kempner is defintitely the best part. Harmonic Christmas!
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  11218. Bravo Mathologer! This was amazing.
    1
  11219. Woah! I just got done animating a classic visual proof of one of these identities. I’ll tag you when it eventually posts and link to this one. Thanks!
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  11220. This is absolutely incredible, and very much on my subject. Actually, I am now working on "1.5" sequences and sequence triangles, which are formed by adding 1 to the previous number, then 2, the 1 again, 2 again, etc. (This is how integer partitions are formed.) Question 1: I know that adding all numbers up to X leads to triangular numbers (1+2=3, 1+2+3=6, etc.), which, when replaced with circles, form triangles. I also know that adding triangular numbers leads to tetrahedeal numbers, which correspond to tetrahedrons. And then to higher dimensions, etc. All these geometric shapes formed on the basis of triangle (triangles, tetrahedrons, and beyond) have a general name of "simplex". Here we have a different beast. By adding every SECOND number we get square numbers, then cubic, then hypercibic, etc. As well as the cube's doppelganger, the octahedron. The overall name for these shapes is not "simplex," but ... WHAT? Question 2: What about he dodecahedrons and icosahedrons? Are they too formed of some basic numerical principles? They are doppelgangers of each other, so I'm pretty sure they too come from the same numerical basis. However, so far I've been unable to figure out the gnomon formula for either of them. (I figured out the ones for tetrahedron, cube and octahedron without trouble). Do you know one? Or can you recommend a book on the subject? I'd be very grateful.
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  11221. Do you have a recommendation for a visualization software for math enthusiasts to play with?
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  11222. I don't think that Tesla is involved with any of the 3-6-9 mysteries at all.
    1
  11223. Yellow! I got it right!
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  11224. 👑
    1
  11225. Numbers are weird again! :D
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  11226. its obvious because the blue leftovers are bigger than a right triangle and therefore use more than half the space of every block they are in
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  11227. Is there anything special about the fact that the "special" triangles also work when rotated? Or is that just propogating from the symmetry of taking the negative of the sum mod n? (2 minutes later he points it out. D'oh!)
    1
  11228. Expanding my thought about the rotational symmetry of the negation of sum mod n operation: a ⊕ b = -(a+b) rotate counter-clockwise: b ⊕ -(a+b) = -(b + -(a+b)) = a rotate again: -(a+b) ⊕ a = -(-(a+b) + a) = b Since the algebra works out with only the abelian group rules, it'll work in any abelian group, which includes the reals, complex numbers, rationals, integers, any modular addition subgroup of the above (including transcendental divisors like radians), even polynomials and analytic functions if you're feeling frisky.
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  11229. I hate that I saw the medians snapping into place on the side angles while you were describing the fractals lining up bottom to top.
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  11230. For 6 disks and 4 pegs, you made 3 super disks; but for 7 disks and 4 pegs, you only made 2 super disks. I don't understand why this is the case.
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  11231. "Welcome to another Mathologer video" Me: clicks like
    1
  11232. in 1150 AD , indian mathematician bhaskara 2 wrote a book to teach mathematics and astronomy to his daughter . the book is named after his daughter lilavati . book has 13 chapters , covers topics like plane geometry , solid geometry , arithmetic and geometric progression , interest computation etc . book contains poems to explain these topics . for example: Whilst making love a necklace broke. A row of pearls mislaid. One sixth fell to the floor. One fifth upon the bed. The young woman saved one third of them. One tenth were caught by her lover. If six pearls remained upon the string How many pearls were there altogether? please make one video on this .
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  11233. Your office looks like a toy store. So cool.
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  11234. typo at the start of the "Integral = sum" portion of the video...
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  11235. How does one go about sending you a result concerning triangles that may be of sufficient interest to you for making another video? I am quite impressed with the presentation quality of your "Simple yet 5000 years missed?" triangle video!!!!
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  11236. Kempner's proof was incredible
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  11237. Finally I'm early in a video
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  11238. Took me more like a minute but 9^2+10^2+11^2=13^2+14^2 so that makes the top 2(13^2+14^2) and I know 13^2 is 169 and so 14^2=(13^2+2*13+1)=170+26 then add 196+169 to see the result on top is 2*365 and the final answer is 2
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  11239. You really saved the best for last! I’m shocked at how straightforward the proof that the no 9’s series is less than 80 is, I was expecting some wicked analysis approximations. That’s my pick for the highlight of this video, with the crazy block pattern beating out the Tower of Lira as an honorable mention.
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  11240. How can we explain that a larger-term by term-sum of 1+1/2+1/3....explodes to infinity so much slower than the one in the proof?
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  11241. This is one of the reasons why i love matgs !!! Gabriel horn paradox .
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  11242. Perfect pedagogical presentantion of mathematics! Showing it a the wonder it is, showing the magic, lighting that spark of fascination - thats how schools should do it. i enjoy your videos massively even if i have a pretty solid maths background from physics; but many of your approaches via geometrical concepts give me a fresh perspective. Fun and enlightening!
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  11243. I did the harmonic stack with bricks and broke my foot.
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  11244. I think the physical answer is finite if the minimum distance you can move a block is a Planck length.
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  11245. Amazing video, keep producing this great content !!
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  11246. Now that's some cool math history!
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  11247. The biggest repository of Indian science and arts was burnt at Nalanda University by Islamic Invader Bakhtiyar Khailji in 1200 CE. The manuscripts burnt for weeks. Imagine the loss done to the Indian civilization and to the world by these barbarians. Work of thousands of scientists over thousands of years was burnt to ashes. Imagine where we would be if we hadn't lost this knowledge. It pushed humanity back by several hundred years. Just imagine the brilliant work that could have been done by our western counterparts if they already knew all that we had already done. They could have started from that point and made progress ahead instead of using precious time on existing knowledge. The sad thing is these Islamic invaders didn't spare any effort in destroying India for 800 years. They saw anything of value they burnt. We will never know what we lost. I hope someone made copies and hid it somewhere and it's discovered soon. But chances are slim. Nalanda alone had 10,000 students and 1,500 faculty members. It was active for 800 years. The number of documents would be in hundreds of thousands.
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  11248. the most memorable part of this video was the adding any combination of the harmonic series can add up to any number. tho I think you can't do negative numbers.
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  11249. I prefer Nathan's solution because it is truly mathematical; he takes a problem that appears graphic and transforms it so that he can exercise a mathematical solution. This short clip will likely help you gain audience from the non-mathematical types which will definitely help them gain more insights into maths. However, I really enjoy the deep dive that you take because as a physicist my 'toolbox' is definitely filled with 'maths'.
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  11250. m=5 is the fifth, but m=4 I don't know where that is yet. Greater than 20 at least. Anyone have a link to that Balak Ram paper?
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  11251. That's definitely not just an Australian thing. In Ohio, USA, I read about Hero's formula in my textbooks a few times in middle school when first learning about geometry, but it wasn't even in any of the chapters of the book we covered in high school geometry. It's one of those "by the way" moments, or a "well how about that?" kind of thing. It's funny, because the various triangle center properties were considered essential parts of the curriculum, but those feel pretty much the same way. Especially the coincidence of angle bisectors at the incenter. Like seriously, so what?
    1
  11252. why 6(2x3) wouldn’t act as good as 16(2x8), and form spiral lines?
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  11253. Now I wonder if the Pythagorean Theorem is also named after Pythagoras in China... 🤔
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  11254. 666 doesn't mean evil, it's the number for man, or rather the symbol. It's less a number than a symbol.
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  11255. In answering the first coin rotation question. the An hour in and no comments containing the word 'sidereal.'
    1
  11256. Paint has to have a thickness to have a volume. Mathematically perfect paint would not fill anything (and, conveniently, can be carried home from the shop without a pot) 😂
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  11257. In the second problem I doubled 10 to 12 squared, instead of adding 10 to 14 squared, mainly because I forgot what 13 and 14 squared was. 2(12^2+11^2+10^2) 2(144+121+100) 288+242+200 730 730/365 =2 When writing this I realized I could have calculated a different route and canceled 365 out earlier but that didn’t click. (Writing on mobile so no proper math writing) Solved it in way less than 30minutes…
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  11258.  @Mathologer  I thought we would divide their sides by how many degrees they have... Ph.D, DSc, ScD, DEng, DESc, DES, EdD etc...
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  11259. A nice lesson in mathematical elegance — not using a cannon when a fly swatter will do.
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  11260. At 9:00 - For adding A(X) and A(Y), we widened the shape with width (Y-1) by a factor X to have a new width of X(Y-1). We then moved the shape's left side to X units on the horizontal axis. The right side hence touched X + X(Y-1) or "XY" units on the horizontal axis. Hence A(X) + A(Y) gave A(XY).
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  11261. Your video description is very thorough. An area often overlooked. Thank you for delivering top quality.
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  11262. I taught my 8yo nephew his 7-times-table using the 63 move.
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  11263. Thank you for bringing this to light. Indian civilization has been under-credited historically
    1
  11264. The 314 in ln(2)'s decimal expansion is actually followed by 718, the first 3 digits after the dot in the decimal expansion of e...
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  11265. So would this also work in 3 dimensions?
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  11266. Most memorable: That it is easier to figure out that the harmonic series adds up towards infinity than it is to know precisely how many terms you need to add to get past a "low" number like 100. I don't really know why but the fact that it has been known since the 1300s that it diverges yet only relatively recently has the number of terms needed to get past 100 been calculated struck me a bit. That said, the were many other things that I found memorable, but I think this one was the most memorable.
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  11267. Superb !!!
    1
  11268. Another month, another banger.
    1
  11269. ugly maximal overhang solutions would be delightful to learn more about
    1
  11270. Thank you :)
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  11271. the logic of measuring something with line segments that go all the way to zero is escaping me. How do you find a finite distance from a number of segments of length zero? This really sounds like the explanation of fractals, and how you have a surface bounded in a finite area with infinite area, and another one in the same area with zero area.
    1
  11272. Every math video is better with metal in the background of algebra auto-pilot. 🤘
    1
  11273. paused at @4:57 to note that the overlapped transformations of the 10-gon appear to approach an approximation of a fractal cardioid shape? Will now watch 'til end of your video to see if the marvels of Mandelbrot set can indeed explain this mathematical miracle, chuckle.
    1
  11274. also anthropologically speaking, one wonders if awareness of this reproducible rule lies behind the ancient skill of flint knapping (in German knoppen ) where a hard cobble stone is slammed against a different conchoidal-fracturing chert pebble at particular percussive angles to sharpen the resulting facets via lithic reduction into a suitable cutting edge, such as a primitive axe?
    1
  11275. I can cut the cross into four pieces and build the smaller ones. How would five pieces be done?
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  11276.  @Mathologer  ok, that's a surprising one.
    1
  11277.  @Mathologer  Yes, you are absolutely right. For me - a non native speaker - the "is an irrational..." sounds like "isn't irrational". I heard the "an" after like 10 tries :)
    1
  11278.  @Mathologer  Anyway, I love your videos, that's why I'm your patreon
    1
  11279. Fantastic visual proof and amazing new insights unit the building of powers from sums
    1
  11280. Primes are sure bizarre. Don’t know if there is any pattern though cause nature needs to be able to produce transcendental numbers like the golden ratio or pie or even A-periodic tilings.
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  11281. Excelsior
    1
  11282. The area of the square in the final puzzle is 1/5.
    1
  11283. 🤪
    1
  11284. Some time back, you indicated in the comments section that you wished to do a video on Conway's surreal numbers. Is that making its way through your deep, deep stack of projects?
    1
  11285. My sister broke her elbow in a car accident in 1971. She was given one of these things to exercise her elbow to try to prevent it from stiffening. It didn't help that much, by the way. Thanks for this mathmeticsl analysis!
    1
  11286. The argument at 15m isn't even oversold. Beautiful stuff. Might have been nice to have the 1 = x^2 - y^2 form of the hyperbola explicitly introduced at some point.
    1
  11287. The shirt is making me smarter.
    1
  11288. Does Mr Mathologer teach in some university/college?
    1
  11289. The line segment and point cases are still cyclic. For the point, it's clearly cyclic, since everything coincides. And indeed, 0*0 + 0*0 = 0*0. For the line segment, the points lie on a "circle" of infinite radius. So the converse still holds true when properly interpreted.
    1
  11290. make more sense if you rotated the polygon while you added the ears.
    1
  11291. The maths are cool….yeah maths! What no one can explain though, what is meant by “the key the the universe”. Is this like the answer to life, the universe and everything else is…42 (Hitchhiker’s Guide to the Galaxy). 42…digital root = 6 AHHHHH!
    1
  11292. 28:17 Now I want a tshirt with axes and a "cosh(x)" label for when I'm wearing my shiny dukey ropes
    1
  11293. 333
    1
  11294. For me it was the greater overhang stacks. Their complexity was surprising. It reminded me of trying to get the largest overhangs possible while playing with Jenga blocks.
    1
  11295. Is your 333 t-shirt a Primeval reference ?
    1
  11296. I'm thinking why not give the value of reimann zeta function of -1 equal to gamma
    1
  11297. Late Xmas gift?:)
    1
  11298. Most memorable to me was that the sum of 100 zeros > no 9s sum. Just goes against every intuition I would have. Thanks for the video!
    1
  11299. Teacher-student nothing can travel faster than light Me- then how was I able to click this video at 5 times the speed of light????
    1
  11300. Why is it still a theorem instead of a law? Is there any doubt as to its validity?
    1
  11301. I'm guessing rattlesnake .
    1
  11302. Most memorable part was looking at the blue curves inside the 1x1 block and thinking "well sure it's less than one, but is it really negligible? It's still more than a half" and then you asked us to state why it's obvious that it's more than one half.
    1
  11303. Swiveling proof
    1
  11304. I was surprised😊 to find 1x1x1x2x5 in Prime Newtons a month ago. https://youtu.be/qhj4KXPqXQE?si=p0IM-GE-ApaHspUZ
    1
  11305. MATHS
    1
  11306. Marin Mersenne 25 December 1640
    1
  11307. The visual proof at the end was by far the most memorable part
    1
  11308. The challenge at 18:00 seems straightforward: increasing the volume by increasing r infinitesimally is equivalent to adding another infinitesimally thin onion layer around the sphere, and that layer has the volume of sphereArea(r) times the infinitesimal depth of the layer, so the derivative of sphereVolume(r) by r is sphereArea(r) dr (QED). You can do the same for disk area adding an infinitesimally thin ring.
    1
  11309. Yeah, I bought a circular slide rule from Engineer's Bookstore during their fire sale.
    1
  11310. In the midnight's shroud, a tale is spun, Of a lantern dark, its curse begun. The Schwarz lantern, a flickering light, A mysterious glow in the hush of night. In shadows deep, where secrets dwell, The lantern's curse, a spectral spell. Its light, a dance of eerie shades, In the labyrinth of time, where fear cascades. A whispering wind through ancient trees, Carries the legend on spectral breeze. The lantern's glow, a spectral might, A haunting glow, in the ghostly night. Its keeper, a soul in shadows draped, A pact with darkness, secrets escaped. A bargain struck, in the twilight's haze, The curse of the lantern, a spectral phase. Guiding lost souls with flickers dim, A spectral lantern's spectral hymn. Through realms unknown, where shadows play, The curse unfolds in the lantern's sway. Beware the light that draws you near, The Schwarz lantern, a phantom seer. Its glow, a beacon to the spectral kin, A dance of shadows, a dance within. Cursed to wander in perpetual gloom, The lantern's keeper, a ghostly loom. In echoes of the past, the curse is sung, A haunting melody in a spectral tongue. Yet, amidst the curse, a glimmer of grace, A chance for redemption, a spectral chase. Break the chains of the lantern's plight, In the dance of shadows, find the light. So, in the moonlit dark, the story's spun, Of the cursed Schwarz lantern, its course is run. A tale of shadows, a spectral trance, In the haunted dance of the lantern's advance.
    1
  11311. Being the only house on that street is also a valid answer
    1
  11312. 26:05 dont worry, i nutted already
    1
  11313. I'm sorry youtube kinda screwed you over for this one, I dont know if reuploading would help?
    1
  11314. I had a circular slide rule in high school back in the early '70s
    1
  11315. I came across the first proof of (a^2 + b^2 = c^2) like 10 years ago - many years after I graduated from university. Never saw that before. I was quite upset, so easy a proof, yet it's not mentioned in school.
    1
  11316. @11:03 I think of the same size quarters like a 1:1 gear box, each shaft makes 1 rotation, both rotating. For 1 gear to orbit the other saft is 2 rotations.
    1
  11317. I think I have an excel solution for minimum moves with N pegs: All of Row 1 =1 All of Column A =ROW() B2 and below needs a very large number, since it's impossible to move more than one layer with only two pegs. I used =POWER(10,299), because 1e999 didn't roll over to floating-point infinity as I'd hoped. Finally, C2 gets the somewhat awkward Array Formula (ctrl-shift-enter after pasting) =MIN(2*C$1:C1 + INDEX(B$1:B1,N(IF(1,ROW()-$A$1:$A1)))) and then that cell gets copy-pasted to fill everything below and to its right, so that the mix of relative and absolute row numbers automagically change as they should. Much of that formula is devoted to pairing one column bottom-to-top with another top-to-bottom, working around the way INDEX() is hesitant to accept an array of indices and spit out an array of values (thanks, google). Ultimately, it's calculating MIN(2*(best for D-x disks) + (best for x disks and P-1 pegs)) the long way, but it's a 2D table so all the intermediate values have already been computed elsewhere, which makes for a "nice" spreadsheet formula, before excel quirks are factored in. Edit: Oh. that exact algorithm appears to have been mentioned just after I paused to write this. Well, wrangling excel to do the work is still nice, I hope.
    1
  11318. Yes, it doesn't work for this case: If I'm owed 100k and Elon Musk is owed 1 billion but the estate is 999950000 then it's fairer if I get 100k and Elon Musk gets 999850000. Not if I get 50k and Elon Musk gets 999900000. Or proportional would be MUCH better than the Talmudic one.
    1
  11319. No hexagon number continuation of this pattern?
    1
  11320. german :3
    1
  11321. I would bet that Vector Analysis facilitates some if not all of these proofs and there might be an identity that contains the quadrilateral formulae.
    1
  11322. The solution for sqt (6+2*sqt (7+3*sqt(8+4*sqt(9+... would seem to be 4.
    1
  11323. Why the actual fuck there is a dislike?
    1
  11324. Good one as per usual. Strange that such a cool channel hasn't reached 1M subs after so long. Not sure it helps but I always at least hit 👍.
    1
  11325. Stunning!
    1
  11326. My ranking: 6 - pidgeons at the olympiad 5 - repeating rubik 7 - the best mathematical trick ever 3 - repeating decimals 4 - partying pidgeons 2 - five pidgeons on a sphere 1 - hairy twins
    1
  11327. Yes! Another huge video! Good to see you back :)
    1
  11328. Love your work! Thank you!
    1
  11329. I made it to the very end, and it was worth it hehe ♡
    1
  11330. Why is your t-shirt (ho)^3 and not 3(ho)?
    1
  11331. Fibonacci cats are the most gracious!
    1
  11332. Wonderful and very well produced video as always. I liked the gamma - moving the blue sections into one box to show they were less than one was a real eureka moment for a Sunday evening!! Thanks all, great stuff.
    1
  11333. Hi. I have a question about one thing that you mention about the sum of 1+2+3 is a perimeter and 1x2x3 is a volume. And the question is there a way to relate a volume of a square in to an area? Like you take a cube and by slicing it and put all the slices side by side and you have a surface. This can be done? It will be a infinite surface? I start to imagine this thing and I went for many ways like the construct of a cube from a square moving in the speed of light and so on. :) It was a nice trip! I'm grateful for your time and detection.
    1
  11334. Very nice presentation. I learned some thrilling mathematics today.
    1
  11335. The 13th century proof is still my favorite.
    1
  11336. I amazed to see that the surface of the sphere and the curved surface of the cylinder that is circumscribing that sphere have the same area!! That is 4πr² Mr. Polster has proved that the surface of the sphere is of area 4πr². So, I am not going to that. Rather I observed the cylinder. If the cylinder is circumscribing the sphere, then its floor and roof surface must touch the south pole and north pole of the sphere respectively. And its curved surface must touch the equator of the sphere. Hence, the cylinder has the the height= the diameter of the sphere= 2r, and its base surfaces have the radius= the radius of the sphere= r. Now, let A(x)=Surface area function. And P(x)=Perimeter function. Then, A(cylinder)=A(roof surface) +A(floor surface) +A(curved surface) =πr²+πr²+(P(floor surface)×height) =2πr²+(2πr×2r) =2πr²+4πr² =6πr² Hence, the surface area of circumscribed cylinder : the surface area of inscribed sphere =6πr² : 4πr² =3 : 2 Q.E.D
    1
  11337. 21:07 and the Earth is flat. Or at least it's planar projection is :D
    1
  11338. 9,4,13 and 17 are the numbers in the Fibonacci box for the Pythagorean triples 153, 104 and 185.
    1
  11339. I like the 700 years old proof. Just because it is 700 years old :-)
    1
  11340. Can we put any arbitrary pair of whole numbers in the 2x2 box, continue them in a Fibonacci way, and get Pythagorean triples?
    1
  11341. 3b1b as a patreon why am I not surprised <3
    1
  11342. All chords have the same length and all chords with the same length will always be tangent to a given circle.
    1
  11343. EXCELENTE VIDEO! ES SUPREMO! Maestro por favor agrega subtítulos en español, entiendo parcialmente el ingles y con los esquemas y numero lo comprendo, pero me gustaría saber el 100% de lo que dices
    1
  11344. What a joy to see those color-coded rectangles and fractions side by side! It just feels "ahh". This makes me curious about a correspondence in general between finite lists of integers (as they appear in the fractions) and the rationals (as represented by rectangles). Do all lists "work" as rectangles? Is the correspondence 1:1? To noodle...
    1
  11345. Hi, A couple of years ago, I founded a series that, like your's, "automatiquely converges towards any number": 1 - add 1/n if you do not exceed the number 2 - otherwise add nothing 3 n=n+1 , goto 1
    1
  11346. LOL ❤️ You killed me with this shirt😃😃😃😃👍 X - Y and the answer is just a little diagonal line, which is what remains from the X after subtracting Y... Great!!
    1
  11347. i'm curious of what the math would look like for the second tower with the counterweights
    1
  11348. The part of the video I liked the best was the one that goes from 0:00 until 51:49
    1
  11349. Mind blowing
    1
  11350. the secret is VolksWagen
    1
  11351. I'm early!
    1
  11352. Did you have an appointment?
    1
  11353.  @godfreypigott  xd
    1
  11354. So if we draw a line perpendicular to those mid point midsection lines. Those will surround the original triangle with its "complement"? Each triangle has a complement triangle. Equilateral triangle is its own complement.
    1
  11355. No 9 sum is finite! You really surprised me there.
    1
  11356. Finished the video in 2 sittings. It is pretty clear and easily understood. Especially glad that I tried out the 40:40 part and succeeded.
    1
  11357. Don’t know why I stayed cuz I suck at math (cuz I haven’t used hard equations for over 9 years). Glad I did though.
    1
  11358. in last puzzle sum of 3^4 + 4^4 + 5^4 + 6^4 will result in a number whose last digit is 8, but no digits to the power of 4 will yield 8 as it's last digit, so there is no solution. and it does not seem to work for 5, 6, 7, 8 and 9. maybe no pattern.
    1
  11359. 33:10 Notch donated :D
    1
  11360. Whole of first quadrant is also anti-shapeshifer.
    1
  11361. All these squares make a circle!
    1
  11362. What made this video interesting for me is: "what makes a math video go viral?"
    1
  11363. I think this Fourier series of polygons can be extended polytopes in n-dimensions. These polytopes are simplicial complexes which make up CW-complexes. Simplicial complexes have their own homology and cohomology. Any topological space is weakly homotopy equivalent to its CW approximation with an induced isomorphism on their homology and cohomology. Assuming this can extended this way, it should be possible to compute the homology and cohomology of any topological space by doing Fourier analysis on its CW approximation. As an aside, this also has applications to the homotopy hypothesis since CW complexes have easy to derive fundamental groups.
    1
  11364. This song sounds like white stripes.
    1
  11365. You also need the paint to be able to travel arbitrary distance down the funnel in a finite amount of time. That's another "mathematicness" of the needed paint.
    1
  11366. If a spiral is an extrusion of a 0-dimension point, along a 1-dimension line with curvature on a 2-dimension plane, and vertices of golden rectangles are used to generate a golden spiral, then shouldn't the golden rectangular prism be used to generate not a spiral, but rather an object of the next higher dimension, possibly a logarithmic spiral surface? A 1-dimension shape, extruded along a 2-dimension planar surface with curvature in 3-dimension? Whereas, a golden spiral passes through vertices of squares within golden rectangles, the golden logarithmic surface would pass through edges of square cubiods within golden rectangular cuboids. I am likely incorrect, as I am not a mathematician. Is there a golden (snail) shell?
    1
  11367. Brillance, is an understatement
    1
  11368. I one day called an old friend from my work phone. He asked the number and I gave it to him. The last 4 digits were 7854. He said, "oh, pi divided by 4" I asked, "Bill how on earth did you know that?". He said as a young engineer I found the area of a lot of circles using a slide rule., He would have loved the infinite series.
    1
  11369. Empezó mi novela 😊
    1
  11370. Mind blown. I've always liked all your videos, but for this one, with each new step, I kept exclaiming to myself: "Oh my goodness", "Oh my goodness", "Oh my goodness".
    1
  11371. 14:20 Why isn't math man shrinking further as the logs get more
    1
  11372. Nice video!
    1
  11373. I used 2 and then generalized to 10 so basically I got the answer right although through wrong means. So Im here to brag aboat my annual membership 😬
    1
  11374. Thank you very much for the wonderful presentation of the convergence techiniques for these infinite series. The followers of Madhava, like Neelakantha Somayaji, Jyeshthadeva, Shankara Variyar, all of them called this convergence technique as "antya samskara" and its translation is end correction or end refinement.
    1
  11375. I'm currently working on a formula to find how many ways there are to make change for n cents using those coin values (1, 5, 10, 25, 50, 100)... I'll update when (or if) I find one! EDIT: Looks like OEIS already lists a formula for the number of ways to make change for n cents using coin values 1, 5, 10, and 25, not including 50 and 100 cent coins - now all I have to do is incorporate those two coins into the formula.
    1
  11376. 1: (2 + A^3 + B^3 = C^3) if B has factors 2 and 3, and A,B,C are consecutive integers (need 6 slices to wrap, but they miss 2 corners) 2: Don't know why I didn't reuse the earlier identity, since I know 100, 121, 144 from rote memory but I timed out trying to be clever 3: stumped me for an overarching pattern, 6 and 6.89... are the answers, however... ...a funny thing about the first 2 is if you subtract the exponent (eg 3- exp) from each term instead, the equation still balances, and first term + exponent equals final term without the exponent
    1
  11377. Mathologer Guy doesn't age.
    1
  11378. Also, thanks for bringing joy to the masses with your laugh.
    1
  11379. In line with the dedication, perhaps this video should have been about making change from Graham's number of dollars. (Or to make it very slightly more manageable we could go for Graham's number of cents). Here is some extra homework. The invitation to try these goes not only to fellow viewers but also to Mathologer himself Can we can predict that the digit 3 will dominate? Only if the last digits of Graham's number are zeroes. How many zeroes would we need? Is there a way to predict how many trailing zeroes the number has, either as a count or in proportion to the total number of digits?
    1
  11380. Wonderful 😊
    1
  11381. Chapter 3.
    1
  11382. Whenever a problem involves infinity, I know a petty crime scheme is going on, quite finite yet nasty.
    1
  11383. ​ @Mathologer  math is not fun, it's sad. Why is 2 x 2 not 27 on Thursdays and 13 on Saturdays? Now that would be fun... otherwise we just get what we get and pretend happy...
    1
  11384.  @Mathologer  If 2 by 2 was 13, wouldn't you be just as happy? I guess! You've just got a lucky good natured nature. Any day you can chuck math and have a career of a phychoterapist! I remember in school when I was like 10, I was bugging out teacher about Euclid's theorems, like why do you pretend to give us this triviality under the sauce of science? Now I realise how important they are, those theorems, even from the practical standpoint, and yet I still have a feeling we get tricked somewhere. Would love to see you video on those theoretical issues. For example, does math stands BEFORE physical things, sort of pre-exists reality? I think you are beleiving something like this, which is a total non-sense of course. hehe.
    1
  11385. This is probably one of your best videos ever! Every single part was totally mind-blowing! The no-nines bit and the proof at the end were super fun surprises :)
    1
  11386. At 6:35... that animation kinda reminds me of a 2 dimensional projection of a rotating 4 dimensional object. I bet there's a connection.
    1
  11387. I waited Shoelace formula of Gauss area formula
    1
  11388. The first term in every row of the one dimensional sequence is a perfect square too!
    1
  11389. coloring proof is more proofy
    1
  11390. Anomaly positions: 6, 19, 20, 31,32,33, 34, 41,42,43,44,45,46,47, 48, 54 to 62 Difference: 13,1,11,1,1,1,7,1,1,1,1,1,1,1,6,1,1,1,1,1,1,1 Results: Their different, seems like a 3D map 🗾 of sorts?
    1
  11391. 🤔 fractal picture..someone😊
    1
  11392. Everything was very easy to understand
    1
  11393. 9:03 :|
    1
  11394.  @Mathologer  The approach was: Determine for each number n from 1 to e.g. 20,000, how many ways there are • to write n as a sum of 1s • to write n as a sum of 1s and 5s, using at least one 5 • to write n as a sum of 1s, 5s and 10s, using at least one 10 • to write n as a sum of 1s, 5s, 10s and 25s, using at least one 25 • to write n as a sum of 1s, 5s, 10s, 25s and 50s, using at least one 50 • to write n as a sum of 1s, 5s, 10s, 25s, 50s and 100s, using at least one 100 Now, if you already have computed the corresponding 6 values for n-100, n-99, n-98, etc. up to n-1, these 6 value for n can be calculated just using those precomputed numbers e.g. the number of ways to write n as a sum of 1s, 5s and 10s, using at least one 10 (n > 10) is the same as the number of ways to write n-10 as a sum of 1s, 5s and 10s i.e. the sum of: • the number of ways to write n-10 as a sum of 1s • the number of ways to write n-10 as a sum of 1s and 5s, using at least one 5 • the number of ways to write n-10 as a sum of 1s, 5s and 10s, using at least one 10 The program then operates by keeping the 6 values for the last 100 numbers around, while working its way up to the target, e.g. 20,000.
    1
  11395. For the huge prime challenge, the answer is yes. A general rule to determine if a number is multiple of 4 says that you just need to see if the last 2 digits of a number is a multiple of 4, and since 81 falls into the 4k+1 category, the huge prime can be written as sum of 2 integer squares
    1
  11396. But then how do you explain a 9 being a rotated 6? and the little half-circles in 6 and 9 make a 3 (kind of)?!!! All kidding aside, I wonder if the symbols that have naturally evolved for numbers (though arbitrary) do reflect a deeper human feeling towards them.
    1
  11397. 3:28 What if we bend this mutilated chessboard like a tube, connecting squares of the same colour?
    1
  11398. The bizarre maximal overhang towers.
    1
  11399. All this reminded me of the nine-point circle and the Euler line, though in general the incenter doesn't lie on this line.
    1
  11400. that book for sure made making maps 100 times easier
    1
  11401. 🇮🇳➕❤️
    1
  11402. It's Fabolous and amazing man what a way of developing such a deep intuition of these misunderstood concepts of mathematics.
    1
  11403. 26:43 I don't think the central argument here should be glossed over like that. What do you mean these are both roughly the diameter? The blue line will always be curved, no matter how many circle cross-sections you make. I mean, I can see that this would probably work if you straighten it, but I though this channel would be a bit more rigorous when putting out 30' videos
    1
  11404. Yes there are details left hanging - a strategic decision on how the video wrapped up. I have a few more details in my playlist of supplementary material, including a bit of a discussion on the extent to which this is a proof - have a look at that if you're interested. How formal you go is a choice - before we can talk about surface area being preserved, we need to define what we actually mean by surface area. If we have to go as far as integrating |∂u/∂s × ∂u/∂t| (or similar) then we may as well just keep using calculus. However, we're saved from all of this, as the video already contains a proof. We know V=4πr³/3 and also V=Ar/3, so nothing else is required. You can view this section as an enlightening visualisation (hopefully) of how a sphere's surface area is equal to an area in the plane. It's an attempt at what Grant Sanderson (3B1B) was trying - an explanation that was more geometrically direct than Archimedes brilliant trick of distorting everything to fit on the surface of a cylinder. Thus the "meridian fold" map appears as an area preserving map as the main tool. This could be shown to be area-preseerving using a few lines of calculus. But that's against the spirit of the rest of the video: prove areas/volumes are the same by moving shapes around. It can be turned into a formal proof - the way I have uses the observation that cyclides are locally as close to cylinders as you like (in relative terms, as n→∞, due to conformal mapping) so that rotating them keeps area as close as you like to the original area. The process of equating blue and pink lunes is a version of this. ... and I'm really happy that we had an excuse to share the cyclide glider.
    1
  11405. Done! ❤
    1
  11406.  @WarmongerGandhi  is that (x+1)^n ? i think so, hey look i see how the binomial expansion correlates to the pascals triangle and NOW even the higher dimensional triangles.
    1
  11407. You meant Fe-man
    1
  11408. Anyone here from India 🙋‍♀
    1
  11409. Silly mathematician. Pi doesn't equal 4. It equals 1, because it is less than 10
    1
  11410. I liked most the approximation of the harmonic series with the ln-function, and especially the easy geometric argument, that gamma is between 0.5 and 1.
    1
  11411. BTW, is there a similar connection between Pythagorean Quadruples and Fibonacci (or other) numbers?
    1
  11412. dr.watson: who has mortal/arch enemy nowadays ?
    1
  11413. Yes, he's always right. He once thought he was wrong, but he was mistaken... 🥸
    1
  11414. Very well explained, I found this a very interesting video!
    1
  11415. Cool! A 5/3 star is just a "backwards" 5/2 star and vice versa.
    1
  11416. Coloring proof was best for me,, easier to visualize
    1
  11417. The most memorable part for me was that the proof that the no 9 series converges, and it does so to a sum less than 80
    1
  11418. There are 675 comments as I write this at 10:39 into the video, so I expect someone else might have mentioned the following: The choice of sides creates tetrahedrons, not just triangles and so you are mixing apples and oranges and should not expect to get a useful answer. This tiling the surface with polyhedra instead of triangles is hidden by ignoring some of the surfaces of the polyhedral tiles. Different connection rules between the points yields a single fully convex polyhederon and that will converge to the surface of the cylinder. The same sleight of hand occurred with the circle, where you tile the perimeter with right triangles then ignore the hypotenuses to make them appear to be 1 dimensional.
    1
  11419. Looking for the t-shirt in the shop :)
    1
  11420. Which movie is it I want to watch it?
    1
  11421. So was anyone able to solve Ramanujan's problem, or did he have to present the solution himself?
    1
  11422. I would call this mathic. Like magic, but with math.
    1
  11423. -1 + 0 + 1 = 0 ........ -1 * 0 * 1 = 0.
    1
  11424. I had to unsub. Years of exposure to searing white mathologer background finally took its toll and I am legally blind now. Thanks mathologer.
    1
  11425. Problem at min. 14:10 is Fibonacci starting from 1,2. The proof is simple: you start at the two options for placing the first tile at the beginning; they lead to either the previous rectangle (vertical tile at the lhs) for the remainder of the rectangle, or the one before that (two horizontal tiles at the lhs) so you get the Fibonacci recurrence.
    1
  11426. I am not a smart man
    1
  11427. math is magic, just with better spells
    1
  11428. I persevered by skipping back and reading the transcrpt. There is hope!
    1
  11429. No turtles were harmed in the making of this video.
    1
  11430. If you look to the expression (2•4•6•8•)/(1•3•5•7) Multiplying top to 1 It would be 1•2/3•(4/5)•6/7•8/9.... And seem to converge
    1
  11431. Haha that school picture had me laughing, despair, deep thinking and one kid visper the answer
    1
  11432. I don't see why you can't continue the tree backwards from the 3²+4²=5² tripple. Depending on which route you go, you either end up with the parent: ┏━━┳━━┓ ┏━━┳━━┓ ┃ -1 ┃ 1 ┃ ┃ 1 ┃ 0 ┃ ┣━━╋━━┫ ┣━━╋━━┫ ┃ 1 ┃ 0 ┃ or ┃ 1 ┃ 1 ┃ ┗━━┻━━┛ ┗━━┻━━┛ It is a triangle where one of the lengths is zero (i.e. a line of length 1). The incircles and excircles are the same and circumscribe that line. The fraction pairs give 1 & 0 (or possibly -1 depending on the route). This is like the seed of the tree. The interesting thing is if you keep going backwards, then the line becomes triangles again but now there are negative sides lengths. This is like you are going down into ground through the tree roots, now forming all of the negative fraction pairs. Here is one such possibility, the negative version of the 3/4/5 tripple: ┏━━┳━━┓ ┃ -3 ┃ 1 ┃ ┣━━╋━━┫ ┃ -1 ┃ -2 ┃ ┗━━┻━━┛ (3)² + (-2*2)² = (6-1)²
    1
  11433. I never saw either of the 2 twisted square proofs. And I totally agree that that is just mathematical criminal negligence.
    1
  11434. I like how a triangle is a special case of a cyclic quadrelateral, which is a special case of all quadrelaterals... Are there an infinite levels above this and everything is a special case of everything else?
    1
  11435. So if use linear interpolations to determine any value on arbitrary line in the grid not in those angles. The distribution of weight is close to uniform distribution.
    1
  11436. Animations inspired by Think Twice😄😄( just commented that) ...few seconds later you said it
    1
  11437. This seems pretty obvious. The medians and area is pretty well known. The folding is something I haven't thought of but should've been done before
    1
  11438. Amazing. Thumb up! As for your request (I made it to the end!), I have a master degree in engineering. Only thing I couldn't catch is the relation with the Riemann Z (no knowledge about that).Thanks for these wonderful videos.
    1
  11439. Everything in this video is some great maths but i can never get away from how great the leaning tower of Lire is!!
    1
  11440. And for those children of the 1950s, that is why we were taught to cast out nines in order to check manual calculations in the 1960s, everyone else: you are spoilt children :~D The next thing you need to do Burkard is look at the patterns when you join the numbers of a magic square with straight lines... mystic patterns (acutally used for ventilation grilles according to Martin Gardner).
    1
  11441. My answer to the question at 13:02 is: n*((n-1)!)^2
    1
  11442. The kerning on your shirt is killing me. Where is Knuth when you need him?
    1
  11443. In another 7 days, will there be a π video?
    1
  11444. Another examples of this sort would be given by the area u der the Gausian curve. The area is finite, while the length of the curve is infinite.
    1
  11445. lol is that acoustic guitar music when shifting out from the equilateral triangles an homage to the game polybridge?
    1
  11446. I think the tiling proof of Pythagoras can be made simpler by making more use of the fact that it's a tiling. We have two unit cells: one is the two small squares and the other is the single large square. They both tile the plane when translated by the same distances along the same axes, therefore they have the same area. No need to chop them up further into puzzle pieces.
    1
  11447. Is it really true that this fact was first discovered only 10 years ago? I'm not 100% sure, but I think in my high school (>30 years ago), we were given a problem to prove that the medians of a triangle form a new triangle, and medians of that triangle form another triangle that is similar to the original triangle. This is not exactly the fact that medians of a triangle divide it into six triangles, which can be reassembled into three congruent triangles. But this fact looks to be the most natural way to solve that high school problem. So the claim that no one thought of it is very bold. And extraordinary claims require extraordinary evidence.
    1
  11448. My first thought on seeing the sequence at 1:49 is that it works in reverse, too: (-1)^2 + 0^2 = 1^2...
    1
  11449. 9:48 joke's on you - I'm going to get a diamond with concavities!
    1
  11450. It's sad when the easy ones aren't taught. From an early age, you should be able to prove: - pythagoras - infinite primes - root2 is irrational What else is super easy? Edit: Just have to say what was new to me. Had no idea you could prove am-gm with this.
    1
  11451. This is AMAZING! this actually describes the structure of th universe and movement ;)
    1
  11452. This was (half) evil. My Tesla dream is crushed so I will probably go with some Toyota hybrid.
    1
  11453. ONLY for you, I click on LIKE before watching the video. I am always sure of the HIGH QUALITY of what you do
    1
  11454. nth root of x = e^((ln x)/n) = 10^((lg x)/n) Formula i wrote when i was fiddling with logarithms. I was wondering how a calculator can calculate roots and i came up with that.
    1
  11455. No costumes for this one? Hardly anybody nose about it? KnowsFeratu knows with his nose. If you scale this harmony from ONE Leibniz holomorphic ring of compression in notation to Global Coordinates, all of physics becomes coherent mathematically. I showed you how, 360 x360 degrees for the Global sheet. Take time out for simplexity in energy-matter, and Boltzmann is unhung. Thanks for doing some heavy lifting. I will watch tomorrow.
    1
  11456. Wow. With so many clean ratios between different volumes, I guess I can see how the ancient Greeks were so hesitant to accept the existence of irrational numbers.
    1
  11457. A wonderfull aha Erlebnis once again. Thank you for the beautifull mathematics.
    1
  11458. I learned the second one in, I believe, middle school.
    1
  11459. Very good video as usual. Thank you mathologer!
    1
  11460. Well, I'm a programmer and I happen to be bored, so here's the square dance in TS + React with a bunch of buttons. https://codesandbox.io/s/inspiring-browser-6mq10?file=/src/CpArcticCircle.tsx Here it is running https://6mq10.csb.app/ It starts to chug a bit when the square gets to 150, but there's a mountain of optimization headroom for anyone that wants to experiment. The entire thing could be done in WASM quite easily, and in fact it should be possible to do both the calculation and the rendering in webgl, which would allow this to scale to into the thousands.
    1
  11461. 16:40 You mean “...which, remember, also works in *2D* so .... pentagons also don’t exist in the *2D* grid.”
    1
  11462. It seems like you made a few 'vocal typos' when you said "natural logarithm" rather than "common logarithm" after time-stamp 00:14:03 "For example, the distance between 1 and 2 is equal to the natural logarithm of 2."
    1
  11463. For the +/-1 alternating approximation: If an approximation is of the form (L/S), the next approximation is (2S+L)/(S+L), and let's assume (L^2 - 2*S^2) = 1 Then using the same formula: (numerator of approximation)^2 - 2*(denominator of approximation)^2 we get: (2*S+L)^2 - 2*(S+L)^2 = (4*S^2 + L^2 + 4*S*L) - 2*(S^2 + L^2 +2*S*L) = 4*S^2 + L^2 + 4*S*L - 2*S^2 - 2*L^2 - 4*S*L. here a few terms cancel out, and so = 2*S^2 - L^2 = -(L^2 - 2*S^2). However our assumption was that the expression in the parenthesis is 1, and so: = -(1) = -1. If we assumed the original expression equaled -1, we would have got -(-1) = +1, so the successive approximations indeed alternate between +1 and -1, so long as the first one evaluates to either +1 or -1. The first expression we used was S = 1 and L=1, and so L^2 - 2*S^2 = -1, and so our series in fact starts with a -1.
    1
  11464. Where can I get this T-shirt? It's amazing!
    1
  11465. The most memorable part was definitely that meme 20-block tower, since I have seen most of the rest before. It's funny how the best known (is this a proven result?) tower is a huge mess. The other thing that was new to me was just how slowly the no-9s series converges, to the point that the sum is hard to calculate. The convergence of the no-9s series is not clear at all. Almost all large numbers contain a 9, but that by itself is insufficient because the primes also have density 0 but diverges. Interestingly by PNT we have p_n = n ln(n) approximately, so the partial sums of Σ(1/p_n) are about ln(ln(n)) which is just an amazingly slow growth rate. Necessary joke insertion: Q: What noise does a drowning analytic number theorist make? A: log log log log...
    1
  11466. Most memorable: bringing in -1/12.
    1
  11467. Why do you think mathematics so closely replicates reality? Is it something we made up to explain what is happening, or is it a fundamental element of reality?
    1
  11468. classical reduced mass of two body system A & B (muAB)=(A+B)/AB
    1
  11469. aha
    1
  11470. Hi! DOES anybody know if there some BOOK that explains this log stuff with the same approach the video took? Thanks.
    1
  11471. I should invest all my money into Pascal triangles. It's completely obvious there are only Pascal triangles in the future.
    1
  11472. Seems like a method for storing energy in crystals with "eary" lasers. In the end, you will always have the zero point in predicted space and time.
    1
  11473. In terms of squishing and stretching any geometric shape with one side going to infinity will be antyshape shifter
    1
  11474. I have a question: 30:19 what if the card is 2 which also has a distance of 4 away, how can we distinguish whether to increase or decrease?
    1
  11475. This is easy to miss, but the algorithm Mathologer established for the card trick always deals with the card value increasing (going clockwise in the animation he has). The "start" and "end" cards are important. You keep the "end" card and use the "start" card as the first one your assistant sees.
    1
  11476. If you close the vessels, the water levels will not automatically even out.
    1
  11477. You asked, about if it was possible to create a Toro flux with two extra rings when collapsed for an example with two odd numbers. 13 and 15, i do not think this is possible. My reasoning for this is due to the explanation of where the extra ring comes from when collapsed. if we say the number of coils when expended is a variable of N. The extra ring when collapsed is 1 / n * n. this is Due to traversing consecutively from one coil to the next. We would have to skip one coil each time, meaning we would have to be able to go from 1 though all odd number coils and then cycle though a second time for alll even number coils. The could be achieved by, eg taking an existing Toro flux cutting it once then extending one end to continue to making coils between all existing coils and then re connecting the new end to the origin. original toy coils, 1 2 3 4 5 6 7 extended version. 1 8 2 9 3 10 4 11 5 12 6 13 7 14 (back to one) im also not sure if this version would be very stable or not, would be interesting if sombody made one. this would give us total extra rings when flat as 2 / n * n. and suggests that a Toro flux with and even number of extra rings compared to coils must have an even number of coils also. Forgive me if my suggestion to cut end extend in the manner i suggested stops the Toro flux form being a Toro flux. I am not a mathematician. just presenting some intuitive thinking. (as a note you could have an odd number of coils but this would result in slightly less or slightly more then 2 extra rings.
    1
  11478. If you're interested in seeing more buckling of cylinders, The Hydraulic Press Channel made one about steel & stainless steel tubes buckling 2 months ago.
    1
  11479. great video prof!
    1
  11480. Most dramatic music ever for a calculus video! :D - Nice explanations. I will recommend it to my daughter when they cover that at school...
    1
  11481. I’m seeing a bunch of bananas. If I knew how to search the comments I would be liking every comment that mentions bananas!
    1
  11482. This should've been discovered in 2014 BC.😂
    1
  11483. It's interesting that the sum of the number of parts is 2 in odd dimensions and 0 in even dimensions. This reminds me of the form of the solution for the wave equation being different for an odd and even number of dimensions (see Hadamard's Method of Descent). It seems far fetched but I wonder if these two things are somehow related.
    1
  11484. Every number is a long long way to infinity! 😀
    1
  11485. Yeah everybody's gangsta of calculus, until they suddenly run into integro-differential equations.
    1
  11486. Its sad That South asians Don't know Anout great people in the past Like Madhava.
    1
  11487. You good looking, bad mother lover, you just earned a subscriber! Love your video! I’m a construction worker btw. I wish your formulas were on blueprints! Lol. You’re awesome! Keep the videos coming. You’ve fanned the flame.
    1
  11488. guys this is the difference between math professor and math enthusiast... thank you professor Burkard
    1
  11489. Whatever these mathematicians smokes, count me in
    1
  11490. Least favorite part: Including a reference to -1/12. At least you said it is not equal.
    1
  11491. Final challenge task: 1) p is even 6 is counter-example, because otherwise 6 = a^2 - b^2 = (a - b) (a + b), but there are no such a and b. Just look at all possibilities of a and b between 0 and 6. 2) p is odd p = a^2 - b^2 = (a - b) (a + b) Let a = b + 1, and you get p = 2b + 1 But any odd number can be written as 2b + 1, so this is the case 3) p is normal prime (that is, bigger than 2) p = a^2 - b^2 = (a - b) (a + b) = a + b, because a - b = 1 since p is prime p = c^2 - d^2 = (c - d) (c + d) = c + d, the same reason Since c - d = a - b, and  a + b = c + d, we conclude that a = c and b = d. So p has unique a^2 - b^2
    1
  11492. Reminds me of "Pi hiding in Prime numbers" from 3Blue1brown. That was the video that inspired me to do maths. Lovely job, Mathologer
    1
  11493. I'm in undergrad, first year. This was a wild ride. The geometry interpretations really helped alot.
    1
  11494. I knew they knew calculus to some level but I didn't know they were this advanced!? Maybe they actually had a full working theory of calculus but it was lost? Otherwise it must take some godly genius to figure them out
    1
  11495. This is great, but for some reason the video was extremely quiet for me. Even at full volume. Just a bit of feedback :)
    1
  11496. I remember the first and last time watching you like 10 or 11 years ago, avoiding u because it was so hard for me to watch and understand math videos in english (spanish is my first language), now i'm very happy because i can watch, understand and learn from u
    1
  11497. Here in India we learn Heron's formula in 8th grade. Most people...including me dislike it because multiplying large numbers and taking the square root of them is tiresome. But now I know the beauty behind it and I think it's pretty cool!
    1
  11498. My vote is for Oresme's proof. So clever, like much great math so obvious once you've seen it (but not until then), it contains the germ of an important theorem of Cauchy (weren't they all?!) and to my mind it manifests the same spirit and methodology as Kempner's proof, which perhaps it inspired, at least in part: divide the series into chunks of unequal length that you can compare to another series that you can easily sum.
    1
  11499. Can you do a video on non standard algorithms and why they don't teach them at school? I mean doing multiplication, addition and subtraction left to right, or division using the partial quotients (scaffold) method. I really wish I knew about these back then.
    1
  11500. most memorable moment - Euler making an cameo to find Gamma
    1
  11501. Calculator drawer: calculators, slide rules, abacus. I was told that my great grandfather had a spiral circular slide rule a foot across, with a spiral of logs on its face. From it's description, I figured he got six digit accuracy.
    1
  11502. Nice video! Math PhD candidate here. I made it through the video and had a pretty decent understanding of what all you explained. I will probably watch it again, though, since I don't have much experience with this topic, and would like to dig deeper with a second watching of it (and actually working out the exercises). Thanks for all your work on this! :)
    1
  11503. Whoever writes the captioned transcript is rocking it!!!
    1
  11504. I liked the divergence proof of the harmonic series the most, I knew it before but I think it is pretty neat and intuitive.
    1
  11505. I like the color proof much better!
    1
  11506. Lovely globe trick, enough to drive Mercator mad.
    1
  11507. I grew up with slide rules. Computers, especially hand held computers, didn't really become popular until after I graduated from college. With that in mind I have an interesting, humorous story to share. First you should know that years ago nearly all engineering and math classes had a large slide rule mounted above the chalk board. The professor would often use the slide rule to demonstrate or to do involved calculations in from of the class of students. Before I tell the story, I should let you know there was an expression in those days called "slide rule accuracy". It is actually a pretty useful concept as it says that only the first two or three digits of a number are really important in a calculation. In fact, I taught this idea when I myself was a professor many years ago. The idea is if you have a number like 314,265,473 you might as well call it 314,000,000 because that is accurate enough in most cases. The additional digits may add accuracy, but are often not of any significant value to the overall result. Now for the story. There was a professor very involved in a long and complication calculation for a class of students. I don't know what the subject was, but let's say it was thermodynamics. The problems in thermodynamics could get very involved with lots an formulas and involved calculations, so it was a pretty messy and complicated problem he was solving. He was sliding the slide and the cursor (the little glass window which enables you to read the numbers off the slide rule easily) back and forth furiously and shouting numbers as he hurried through the calculation. At one point he could be heard to shout, "Two times two - three point nine eight - call it four".
    1
  11508. Three Six Nine.. Tesla's Fine Hope he can sock it to Edison just one time.. Get low.. get low..
    1
  11509. when studying physics euller came up, naturally. this partition stuff is kinda interesting. very nice presentation
    1
  11510. Fascinating, I would have expected the devil and angel lacings to be the strongest.
    1
  11511. pi = 8 liters
    1
  11512. i cannot remember either of these proofs. Indeed I can't remember what proof we were shown! I don't think it was either of these.
    1
  11513.  @Mathologer  it was around forty years ago... we were certainly expected to learn the result, if we were given a proof it was unexciting.
    1
  11514. Most memorable: Bizarre maximum overhang tower with 20 blocks! 👨🏻‍🎓
    1
  11515. Piet is pronounced "peet", not "pee-yet".
    1
  11516. thanks!
    1
  11517. 11:28 I'll be honest, it never occurred to me that the b in logb(a) stood for base
    1
  11518. Why would one run the alternating sum instead of combining pairs of terms? Sum of pairs of terms has reciprocal quadratic convergence, instead of just reciprocal.
    1
  11519. Interesting curiosity: around 14:15 is mentioned that the number of solutions of prime=x^2+4yz are even, but also this number is "mostly" prime! (for 37=7, but also for 5=1, 13=3, 17=5, 29=5, 41=11, 53=7, 61=11, and finally a counter example 73=21, 89=21, 97=27, 101=11, 109=17, 113=23...)
    1
  11520. The image at the left at 16:20 actually shows p=20 and b=20 instead of p=10 and b=10. The case p=10 and b=10 is shown at 20:24.
    1
  11521. This is yet another wonderful lesson from Mathologer. It could have been included why this sequence is called Harmonic: any term is the harmonic mean of the neighbouring terms. I don't know why this type of averaging is called harmonic, that could also be explained. The part I liked most was the static balancing at the introduction, and for the history of the h. series, the demonstration that it tends to infinity by Bishop Oresmes.
    1
  11522. When I watch mathologer videos my mouth just remains open ...because i get stunned alot of times by the marvelous proofs which I will never think of
    1
  11523. Very nice video, I've learnt a lot of new stuff! Just go crazy with all your videos, as a physics student I had no problems understanding what was happening :)
    1
  11524. @goodplacetostop2973 19:43
    1
  11525. Where can I sign up for "Integral Baby Calculus 101"?
    1
  11526. There's another approach: https://www.math.auckland.ac.nz/~butcher/miniature/miniature2.ps I think this is probably THE continued fraction which Ramanujan was thinking of .
    1
  11527. 10582 is different from 10581
    1
  11528. Magnifique démonstration et plus rigoureuse magenta et aqua par rotation est magnifique extraordinaire jamais vu félicitations
    1
  11529. Well, technically the book wasn't in print because it wasn't printed; it was hand-copied for over a thousand years.
    1
  11530. 24:45 If you build a rectangle on each blue part and then you draw a straight line between corners in which blue and white parts connect you can see why it is really obvious that gamma>0.5 Sorry for bad English, it's not my native language
    1
  11531. @24:12 Isn't this uniqueness problem analogousx to the solution the necklace splitting problem or the ham sandwich problem?
    1
  11532. Made it to the end 😁 Now asking how long it took you guys to produce this marvel?!?!
    1
  11533. I'm curious as to the history of the 355/113 approximation. How did they know it was close? Did they have precise enough circles to compare to?
    1
  11534. 25.8069 is bad rounding, 25.807 is actually closer to the square root of 666, unless bad rounding is part of the joke.
    1
  11535. I think a little lie was that the Almagest was in print for over a thousand years. I don't think it was in print until the fifteenth century, after the invention of the printing press, and it hasn't been a thousand years since that!
    1
  11536. 32:32 multiplied by twONEHALF :D
    1
  11537. Electrical engineers like to use j for sq. root of -1.
    1
  11538. I was going to find a movie to watch while eating dinner. I found this. I am happy😊
    1
  11539. We could also simplify the last argument by saying... This implies the existence of a regular octagon in the square grid, which we had previously proved to be impossible, hence the cosine of (3/8)*360 is irrational.
    1
  11540. Thank you so much for this video. Its awesome!
    1
  11541. Most memorable: the optimal leaning tower. Makes me want to see all of the other types of optimal towers
    1
  11542. General formula for externally rotating coin going around the circumference of the static coin once: rotating coin, (ratio of the static coin) = (n/m) static coin = (n/n) formula: (n/m)^-1 + (n/n) e.g. (1) The rotating coin has half the diameter of the static coin: (1/2)^-1 + (1/1) is the same as 2/1 + 1/1 = 3 e.g. (2) The rotating coin has two thirds the diameter of the static coin: (2/3)^-1 + (2/2) is the same as 3/2 + 2/2 = 5/2 N.B. If this coin goes around twice: 5/2 x 2 = 5
    1
  11543. Didn't get it all :) but cool video :) greetings from Poland :)
    1
  11544. Yes, exactly what is required to replace amateur associations with a clear picture from the Real Mathematicians. But I will have to reiterate the video a few times. Adding to Favourites. Because a real Professor of Mathemagical Thought processes is an actual Intuitionist, observing in resonance, coherence-cohesion chemistry-bonding with a purely functional concept in Conception Completeness.., the definition of a philosophical absolute, abstract process of such component that is self-defining quantization cause-effect materialisation of conceptual elements is the Camera Obscura setup, discovering the Universal Singularity-point positioning of the Centre of Eternity-now Interval Time, macro-micro reciprocation-recirculation magic mode of mathematics.., or some thing, holography-quantization from no thing in the general Conception of real-time logarithmic> Singularity-point <relative-timing ratio-rates Perspective of log-base numberness. (It's only the beginning/Origin orientation to Actuality) *** Superspin Superposition-point location logic is inherent in e-Pi-i axial-tangential, point-line-circle-> horizon as infinite Black-body surface of orthogonality, ..so transparent that it's not a clear idea of Actuality, until you get the Trancendental Meditation of point location nothing in No-thing definable Relativity, ie what every Mathematical Disproof Methodology Philosophy begins and ends with in perspectives vanishing-into-no-thing Singularity-Duality Conception. Relative-timing turning points are made-of-making elemental e-Pi-i Calculus cause-effect, no wonder we think the universe is consciousness, but it is simply the Function that is Absolute Zero-infinity reference-framing containment of harmonic mathematics in Truth, not the other way around, which, by default, is what explains observable quantization properties.
    1
  11545. I was most surprised to see the leaning tower is not the optimal solution for overhang.
    1
  11546. I loved this video. As a university educated physicist it is no problem to follow this video. I loved it! In my daily work I almost never see math like in this channel. Most of it I encountered in the math classes at the university (and far deeper differential equations) . This is a nice refresh in my fundamental math knowledge.
    1
  11547. I really liked this video, it introduced some things that were new to me and became like a journey of discovery. A big thumbs up.
    1
  11548. Great video again, thanks :) However, I couldn‘t get my eyes of that shirt, wondering which local anarchist group Mr. Mathologer belongs to 😂 Kein Gott, kein Staat, M🅰️the für alle 🚩🚩🏴 😅
    1
  11549. One observation on the “curvy curve”: yes, the measured length gets longer and closer to the actual length as the intervals get closer to zero - but if they actually get to zero, you’re just summing an essentially infinite number of zeros. So the length gets closer and closer and then drops off a cliff. 😊
    1
  11550. I first encountered the skgrbraic Pythagorean Theirem proof in the appendix of Simon Singh's book Fermet's Last Theorem.
    1
  11551. oh my, This takes me back to my Fourier analysis classes.
    1
  11552. So if I have a large matrix of this shape, can I always use a number wall to find the determinant of that matrix? That's pretty crazy
    1
  11553. As soon as you started turning the wheel around I realized that you were creating a circular slide rule. We used the regular slide rule when I was in high school. However there is a great scene in Apollo 13 where they whip out a circular slide rule to do emergency calculations, much faster than the computers of the day could do.
    1
  11554. What a beautiful T shirt you have :)
    1
  11555. How does this continued fraction 'prove' that root 2 is irrational? I can see that you can prove that root 2 is not a ratio of 2 integers or natural numbers. Or, there is no rational number that equals to 2 when squared. To prove that something is irrational you first have to establish what an irrational number actually is.
    1
  11556. At 18:04 odd and even numbers are marked with almost exactly the same colors as my university uses to mark odd and even weeks. Is that a coincidence or are those some special even/odd colors? :)
    1
  11557. 3 7 11 13 ...... can also be written as 4K-1 so it would be symmetrical to the other half series of the prime like one series is 4K+1 and 2nd one is 4K-1
    1
  11558. Feuerbach pronunciation sehr gut. I bet the Dutch guy is called Henk.
    1
  11559. At 12:30 you seem to have accidentally deleted a square off of the grid you had (green square 6), rendering it untileable and making the following statements not apply to the grid being shown. Whoops?
    1
  11560. the run towards a straight line of pythagorus triples is easy to see in either direction, 345 is the clue. A balanced rise over run would suggest an integer relationship of 1/root2 ,which is clearly untrue
    1
  11561. Excellent video! Thank you!
    1
  11562. Do I need to program an app as in an iPhone app or am I able to make any program? Does the program have to be sharable?
    1
  11563. Are there any connections to a mirror view, something looks familiar?
    1
  11564. I only recently became aware of how crazily useful for proving stuff the DFT is. I was aware of results, but not what was behind the scenes. E.g., Green's work on arithmetic progs in primes. I watched a vvideo lecture he did from about 15 years ago (cant remember which, IAS or Princeton, one of the usuals). basically if you want to decide how ""pseudorandom" a set of stuff is (like coin flips) wrap it round a circle. if the "centre of mass" isnt the centre, then it isnt (very). but then, you can also discount APs (as not being very "random"). but that, basically, just describes all the possibilities.
    1
  11565. ​​​ @Mathologer  When I saw the degenerate point I thought this looks like the zero frequency term. Btw, I think it was confusing when you said the special pentagons run in different directions, they don't. Do you plan to elaborate on that later?
    1
  11566.  @michael-j-harrison The first electrical grids were 4-WIRE, not 4-phase. They were 2-phase, at a 90 degree phase shift. Edit l: to clarify, 2-phase systems are not (notice the present tense - 2-phase systems are still in use in a handful of places in the US, perhaps abroad) necessarily a 4-wire configuration. The two phases can share a neutral conductor, but it must be oversized as compared to the rest of the conductors. I believe the requirement is a 1.5x multiplier, which leaves a bit of margin compared to the 1.41x multiplier from sqrt(2)
    1
  11567. Well dang. I was gonna say the set of complex numbers representing each n-gon that are added just feel like the DFT but you already did. Can we get a Homer Simpson?
    1
  11568. The root of evil: 25.806975801127880315188420605149140896082606671872206858524136923712280803989051038349992672096886193188550075761727345910961596375588434332985885744038259074792756060915875182845943860310128816167264637737512822723494310385564832857611197868935774695335629895358928362192068099642020110905005852090502683790063981842505985987241524826932153872975456324829652269176946115344600813527862313389470583024390256057330197818944600803447317135881850966133199032957881377748270387460966461547310364719699718316661155854414949...
    1
  11569. If we use the simple formula for sum of first k integers we arrive at: x = sqrt(k*(k+1)/2), where k is the total number of houses and x is house number which balances the sum on left and right. So, for all k, as long as k*(k+1)/2 is a perfect square, we have a solution. It happens for k=8: as 8*9/2 = 36. x is 6. I wonder if we started with this - the video would be much shorter. Another thought I had was that infinite fractions are often solved using quadratic equations (equating part of the infinite fraction to the unknown value). I will work on it later and post another comment. Thanks for making this interesting puzzle accessible and available, sharing the interesting story and deeper mathematical context around the general topic.
    1
  11570. Absolutely beautiful.
    1
  11571. I remember seeing this episode of Dr Who on the ABC when I was 8 or 9 years old and being so stunningly impressed that I was converted to seeing maths as patterns making everything explicable. I'd have been a great acolyte for Pythagoras - or the Kaballah if I'd ever heard of them in Warragul - in those years!
    1
  11572. I am 11 and I found another technique to chapter 1 (n÷2)×(n+1) Edit: Wait, it's same with (n(n+1))÷2
    1
  11573. Nice of you to wear a shirt celebrating the great state of idaho :) I live there
    1
  11574. Ahh yes. The "Parker anagram". Not exactly an anagram but close enough 😂 @standupmaths
    1
  11575. When you change the exponent in the Mandelbrot, you get this extra "petalization" effect as well - of one less than the exponent. Call me weird, but I always found it "buttlike" ever since I was a kid (like the odd reflection in the bottom of a cup).
    1
  11576. Pigeon party!
    1
  11577. I think in civilized countries schools are free and available to all kids, actually also mandatory, up to high school, somewhere.
    1
  11578. Nice video! Even better this time 😊
    1
  11579. Next Up: find the link to log (xy)=log (x)+log (y)
    1
  11580. 15:38: So, how do we know x meets the condition of -1<x<1?
    1
  11581. I will call the layered sphere volume proof the Shrek proof
    1
  11582. Amazing!
    1
  11583. 700 year old proof gets my vote
    1
  11584. how do you keep track of the assets on the screen? is it projected?
    1
  11585. Sorry. I lost you on the 1,000,000th term.
    1
  11586. Hi , Mathologer. Your videos are quite interesting and I love your videos. Your tone and way of making things visualise is amazing. But I face difficulty to visualise 4D... Please help me out with it. I love to discover science and mathematics, and after this video, I am excited to search more about it. If you have good sources for more information related to the topic please do share. Thank you for uploading such informative video.
    1
  11587. The only thing that's bugging me is: why we can't do better than one using rational numbers?
    1
  11588. Im.looking for the link to your heavy metal videos? That's very interesting! I love it! It helps me concentrate on math!
    1
  11589. Noticed another "super-pattern" in the natural numbers. Each sum is divisible by 3. When you divide each sum by 3, you get 0, 1, 5, 14, 30.... It just so happens that the difference between each term in that sequence is the perfect squares! 0+1^2 = 1, 1+2^2 = 5, 5+3^2 = 14, 14 + 4^2 = 30, and so on and so forth.
    1
  11590. Does 0+0+0 = 0*0*0 count?
    1
  11591. Which are the best lacing to pull up/tighten by the top?
    1
  11592. Genius. Yes, you are always right! Geometry is what stays in our head, because humans are visualising creatures.
    1
  11593. Love the shirt.. and the video, 😜
    1
  11594. Perhaps this should be called the Inverse Arctic Circle Theorem - the permafrost in the Arctic is within the circle, while here it's outside.
    1
  11595. Calculate = countings, +, memories.🙁
    1
  11596. I would love more if thee upload more videos. Thy videos are maths blowing in my mind, I mean mind blowing in maths.
    1
  11597. An old buddhist koan: "If you see the Buddha on the road, kill him!" Mathologer isn't the Buddha but he is a true disciple of one way of seeing reality truly. Keep up the good work, truly!
    1
  11598. My school did'nt explain it, but a few month ago I though, bro why does A²+B² equal C². so I played a bit around on a piece of paper and came to the secound proof you showed, the first one is even more elegant though! very cool!
    1
  11599. The root of evil to 4 significant figures is 25.8070, not 25.8069.
    1
  11600. (I made it to the end.) There was one bit that was hard to follow. When you're considering pairs of partitions, I kept waiting for the explanation of how we can derive a partition from the pair, or what the correspondence is between pairs and partitions. I was a little confused when you got to the end and just kind of ignored that, and I had to think about it for a bit to realize that the pairs don't actually matter; the formula for p(13) comes from working with the pairs indirectly. I guess it's a case of preconceived notions, since I was expecting a different kind of recursion. Really cool video, though! I learned a lot!
    1
  11601. There seem to be a prime in the bottom row.
    1
  11602. The reason that the shrinking argument doesn't work for hexagons (and be extension, triangles) in the cubic grid is because their radius is identical to the length of each side, so if you shift all the edges to instead be spokes, you have just defined a new hexagon with the same dimensions as the first one. Likewise using the shifting argument on the triangle in 3d along the same lines as the other regular polygons quickly show that triangles actually grow, rather than shrink, when the methodology of the "shrinking" shift argument is applied to them. This is the case for all regular polygons with less than six sides. There's a deeper connection with the fact that the only regular polygons present in any nxnxn grid are are the polygons that tile the plane, and that all regular grids can be defined a cross section of the points in a cubic grid, but I don't have nearly enough knowledge of mathematics to explore this in depth on my own.
    1
  11603. Mja aagya mkc
    1
  11604. I actually got it because I did secretarial work for a robotics company and you pick a little up over time.
    1
  11605. I really enjoy your videos. They are a staple of a relaxing Sunday afternoon.
    1
  11606. The version of the t-shirt in the zazzle link you gave in the description is bad. it ruins the joke completely by not using the custom font Y. So where did you REALLY get the t-shirt? Please let us support artists who actually understand what they are putting on the shirts.
    1
  11607. I swear you've geeked out over sqrt(2) in every video ive watched
    1
  11608. Lieber Professor Polster, ich habe noch nicht das Video bis zum Hälfte gesehen, aber schon hat das Video ein LIKE von mir verdient. Herzlichen Glückwünschen.
    1
  11609. I watched this video on my phone while lying in bed. I was able to solve the first two puzzles while pausing the video but not using any other aid. 5,4,9,13 produces the required triplet. We can reach it by going RRRL. It was also easy to calculate the total area of five by counting the five branches. I didn't even try to tackle the other questions asked. I don't know if I can do it even if I try because I have limited mathematical knowledge, just some interest.
    1
  11610. Thank you for sharing so much!!! Hyperbolic function and the saddle shape of the universe in terms of our spiral galaxy having a cosh/sinh; if the unit circle/galaxy touches a similar hyperbolic slope…how does the negative infinity approach the speed of light? Reimann's  infinitesimal primes?
    1
  11611. I can say ther is a value of x= M so that the area from 0 to M of 1/x is equal to the area from M to infinity of 1/x :^)
    1
  11612. The function y = 1/x is symmetric over the line y = x, so this means that the "upward tail" has the same area as the "rightward tail". In other words, infinite! And this will actually be true no matter what M you take. The area from x = 0 to x = M will be infinite for any positive M, just like the area to the right of x = M will be infinite for any positive M. So, I guess you could say that it works for any positive real number M.
    1
  11613. This was so cool to watch, thank you so much for this and I will definitely stick around for another.
    1
  11614. why does the coin have to travel a distance of more than one circumference to cover a distance of one circumference?
    1
  11615. 13:55 "pretty much any number" shouldn't this be "almost no numbers" :)
    1
  11616. What happens if the lines connecting the points are not flat? Does the inequality hold when quadrilaterals are drawn on more exotic surfaces?
    1
  11617. It would also seem the proof requires that the distance of A to B is equal to the distance of B to A?
    1
  11618. Ten cards flipped by max 2^10 (1024) steps as it would be basically counting binary from 0000000000 (0) to 1111111111 (1024)
    1
  11619. Haha the 36th triangle number. Mathologer you clever mathematician
    1
  11620. Isn't a "3D polygon" generally called a "tetrahedron"?
    1
  11621. Most memorable: the fact that gamma is the Ramanujan summation of the harmonic series.
    1
  11622. Early Christmas gift, thank you!
    1
  11623. * Russel has entered the chat * * Godel has entered the chat * I jest, but I wonder if fundamentally paradoxical sets fit the model. Let's watch, shall we!
    1
  11624. This answered so many questions from high school that I didn't remember I had
    1
  11625. The horn's volume may be finite, but it would take forever to fill it.
    1
  11626. You really should check out this potential manifold. Neutron decay cosmology. Geometry demands. Oh..btw..I'm sure any sensible person thinks I'm nuts but...the manifold... Surface(cos(u/2)cos(v/2),cos(u/2)sin(v/2),sin(u)/2) 0<u<4pi 0<v<2pi Notice that Tau only gets you half of the way around the surface? Electron half spin is artifact of this topology. I believe it to be minimal single sided compact surface. Is this Klein bottle? What is it? If not Klein and unclaimed "Shirley's Surface" for my mom. :)
    1
  11627. This is exactly right up my alley and exactly what I've been studying. This all relates to music, my friends. Plus other things 👍 😉 We Will Win.
    1
  11628. Most enjoyable Mathologer video yet :) However, one must be a "Flat-Earther" to take the magic blocks (cantilever) demonstration as complete. It would take an infinitely large spherical planet to support it without gravity variance corrections over distance of "overhang" also being accounted for in the gamma formula. Thank you and keep them coming!
    1
  11629. Is this not related to a time series re integration to invert diferencing? It feels like a connected idea
    1
  11630. I still have yet to think/math through the the fact that the traditional spring force equation F=kX is linear and not logarithmic. It’s still early, I didn’t get much sleep, and I haven’t had coffee yet.
    1
  11631. I was definitely most surprised by the fact that the sum of the "Exactly 100 zeroes" series is greater than the sum of the "No nines" series. At first glance that didnt seem very intuitive to me
    1
  11632. Yeah I've come across a similar diagram (in the complex plane) following the positive and negative square roots of 1, and of those roots, and of those roots and so on.
    1
  11633. Does anyone else find perfect triangle very mysterious? I mean pie is irrational which makes sense cause it should be hard to find perfect ratios. irrational ratios should be the norm. So how in the world does a 3 4 5 triangle exist and why are there so mamy perfect triangles? I mean I can see the odd perfect triangle existing with much larger numbers but 3 4 5 is so small.
    1
  11634. The question I keep asking now is does the pattern N-1, 2N-1 continue onto either 3N-1 or 4N-1, I suspect either it continues by integer multiples or powers of 2. Because that would place great restrictions on the numbers of each frequency which would be basically have to be unsatisfiable after a certain point on the number line.
    1
  11635. I really appreciate the work you are doing. I wouldn't find(look for) this nice proof on my own and if you didn't post the video I would spent this limited time I had today on something useless... Your videos boost my inspiration and thus make me feel better. Keep going!
    1
  11636. Wait a second. I tried sketching this up in Alibre, and with 10° slices, they don't fit together like that. There are small gaps in the inner portion and interferences in the outer portion of the cylinder. Looking at the pattern, I'm not convinced the error goes to zero with more segments, since it exhibits stair step patterns, at least for the surface area.
    1
  11637. Great that you’re having a go yourself! Have a look at my playlist for the rainbow-coloured “area of circle” animation. There are a few tricks, like which area to include in the lune, or which radius to choose for the arcs. In the limit, the adjustments to area can be made as small as desired, so don’t cause a problem.
    1
  11638. to prove at a glance, note that all the rotations are by 180º, so doing an AFFINE TRANSFORMATION reduces to the equilateral case.
    1
  11639. Wanna see a simulation of 100 super disks for self-hypnosis--
    1
  11640. Einstein
    1
  11641.  @Mathologer  And throughout the ages. It is human nature to focus on “the bright, shiny things” and often miss the splendor in the background.
    1
  11642. Well you do get a parent triangle for 3,4,5 if you consider a shape with sides 0,1,1 a triangle
    1
  11643. For the third square: Since a is 1/2 and b is 1, the base of the parallelogram that's created inside each triangle by the overlap must be 1 - 1/2 = 1/2. Since the hypothenuses of the triangles are parallel this means, the sides of the inner square must also be of length 1/2, so the square has an area of 1/4. PS: great videos! Lot's of things I didn't know before in this one as well. Always fun to learn, especially when the person teaching is as excited about the information as you are!
    1
  11644. Oooh, i see my error now. Wow my head made a lot of sense of this... After reading other comments and looking at the diagram again, I now also think that the area of the small square is 1/5. (For anyone whose head had the same idea: the "base" of the parallelograms isn't parallel to the sides of the inner square, so they can't be of equal length)
    1
  11645. The no 9 sum can actually be changed to no arbitrary string of digits in any base, so for example in base 8 if we ban the string 0123456 then the sum will still converge.
    1
  11646. Pilots still learn this as E6B
    1
  11647. This is a nice Christmas present. Thank you. ❤️
    1
  11648. Rounding error on your t-shirt!😀
    1
  11649. Is it just me who wasn’t focus on what he was saying at the beginning of the video because of his shirt?
    1
  11650. Euler was a true genius
    1
  11651. It was not missed for 400 years. I hat one of those at school (Gymnasium Bern, Schweiz 1965-1969)
    1
  11652. Around 30:00: But isn't there another tiling added every time two 2x2 squares are adjacent (if they are, say, on top of one another, you can cover the top of the top square, the bottom of the bottom square, and two vertical rectangles in between)?
    1
  11653. I prefer the swivel proof
    1
  11654. PAZA M C
    1
  11655. Kempner's proof is the most brilliant part of this video 👏
    1
  11656. Claudius Ptolemaeus (Ptolemy) was definetly underated ... its actually strege this guy isent more known by ppl
    1
  11657.  @Mathologer  a thousant, hum, and ive barely been underated for 40 .. perspective perspective ;D
    1
  11658. My favourite part was the proof that the "no 9 series" at the end. I felt stumped on how to prove it when you mentioned it in the video, and seeing the actual proof was illuminating, specially since it was so simple.
    1
  11659. Long waited 😊
    1
  11660. What about regular n-gons in m-dimensional grids? Edit: I asked too soon. I see you answered it
    1
  11661. German mathematician makes "windmill" drawings.
    1
  11662. gem work . .!
    1
  11663. Mindblowing :D Things like that should be taught in school!
    1
  11664.  @Mathologer  lol
    1
  11665. When I went to school in the '60s and '70s we were forbidden to use calculators, even though they were around from the mid-'70s. I used a circular slide rule, and a really cool linear one, I think it was made by Faber-Castell. I kind of wish I still had them.
    1
  11666. Common cycle: measure -> derive mathmatics -> predict measurements not done -> measure to prove or disprove predictions
    1
  11667. Hey, Mathologer! Some your old videos are missing. I think I remember two videos named major league word problem video and negative times a negative is a positive. I don't remember anything else missing. Maybe somehow they are unlisted ? Please upload them again if you can.
    1
  11668. Does that shirt say to infinity and beyond?
    1
  11669. It's close for me. I'd have to do some serious searching, but I just might have more slide rules (including 3 or 4 circular slide rules up to 10" in diameter) than calculators. It would depend somewhat on whether to include a couple of "toy" slide rules (accurate to maybe 1-1/2 decimal places), and whether to count every calculating app on my computers, tablets, and phones in addition to any actual calculators. (Do I include my TI-2500 which no longer works?) Here's a few of them: bit(dot)ly(slash)slide_rules As a closing note: can you consider yourself a true slide ruler if you can't say "log-log-deci-trig" really fast?
    1
  11670. Being almost always right is naturally almost as good as being always right can be, and it dodges the consequence of the widespread notion that isn't right to always be right.
    1
  11671. Great video. Looking forward with bated breath for the Galois theory video.
    1
  11672. This video skyrocketed to my top 10 favorite YouTube videos!
    1
  11673. Pure magic!!!!
    1
  11674. Definitely the Kempner series. My original thought while watching was that's a base-9 sum of everything similar to a base-10, how could it converge? But the proof is such clean and nice, math is just beautiful!
    1
  11675. It is a real shame that so many of the really smart folks were stuck living in places where they could not pass on their genes for intelligence. It's too bad for us, anyway. :/
    1
  11676. I'm terrible at maths but I love maths history and maths documentaries :D.
    1
  11677. "The proof is elementary." The sign of a good and lazy mathematician 👍
    1
  11678. 23:00 is anyone else curious about what happens with (mx-1)(my-1)=mn-1 for other m than 1 and 2?
    1
  11679. That was neat. I liked your phrase "it looks like X and quacks like X, ....". Since mathematcians are all demons who can construct counter-examples: Can you make a video where X actually reveals to be 'only a wooden duck'? Lookin' forward to your next video :)
    1
  11680. I'm first
    1
  11681. This blew my mind! <3
    1
  11682. Finally getting around to this one. I would LOVE some more hyperbolic trig.
    1
  11683. Very elegant use of the lever law of Archimedes.
    1
  11684. Magnifique tjs magnifique cette chaîne 👍
    1
  11685. Guten Rutsch.
    1
  11686. Brilliant explanation! Science stands on the shoulders of giants. It does not see - white, black, brown or yellow. Lets acknowledge where credit is due!
    1
  11687. Seconding the Euler-Mascheroni constant being the Most memorable (I recall the interesting beast when I was doing my undergraduate... I barely scraped a pass in real analysis) We still don't know if gamma is irrational!
    1
  11688. I’m a second year math major and found the video to be very enjoyable overall. If anything, I would have appreciated more justification for some of the results you showed. It’s very impressive that formulas for infinite sums can be obtained using pascals triangle and the Bernoulli numbers, but I still don’t feel I can intuitively understand why, other than by checking and seeing that it works for a given sum. Hope to see many more of these videos from you in the future!
    1
  11689. I feel like mathematicians think more “logarithmically” than “logically” sometimes… get it? :)
    1
  11690. 37:38 it's because if you tilt the "cube" to a certain perspective it will look like one solid side of a color
    1
  11691. 2:08 you can't do that because the series is not absolute convergent
    1
  11692. Can I please be excused? ...my brain is full!
    1
  11693. 6:50 would've been 5^3 + 6^3 = 7^3 in my opinion that gives you 125 + 216 = 343 but in reality left side is 341, meaning that there's a 2 unit difference between both sides
    1
  11694. Swivel!
    1
  11695. Actually, you always know.
    1
  11696. There are special properties about N-1 and 2N-1, but how about 3N-1. Any properties related to 3, and if not, why not?
    1
  11697. Mathematische im deutsche? 9! (Ich spreche keine deutsch, except this bit I memorised)
    1
  11698. Fantastic! I have had so much fun over the last couple years telling my students, "maths is broken". I often show your videos to help create that bigger picture feeling about math.
    1
  11699. For me, it was definitely the gamma part, although it was all really great. Just the thought that Euler had one more important number that I had yet to hear about was very interesting, and it had some great properties
    1
  11700. "It gets better & better" - Starting to add calculus to the mix is always a winner, IMHO :)
    1
  11701. I going to think about coloring rules for cubes where 9 cubes define color of an one beneath.
    1
  11702. 28:44 take that, Numberphile 32:34 your mouth says two though :^) Great video as always!
    1
  11703. What about 3.3^(2) + 4.4^(2) = 5.5^(2) when 3 4 5 is a pythagorean triple.
    1
  11704. My try for a proof: First, if those point are indeed on a single circle the center of this circle should be the center of the inscribed circle of our starting triangle (as it is the center of all bissectrices). Let's call O this point. Next, let's pair the 6 points we are trying to proove are on a circle according to wether they are linked by a straight line on the figure at 2:17. Each of these pairs is distant by the same amount (the sum of the length of our starting triangle's sides). Some not-so-simple trigonometry lets us know for all these pairs adding O makes an isocele triangle, and because O is the center of the inscribed circle all those triangles are similar (they have the same height). Thus we can conclude.
    1
  11705. These triangleceptions made my day
    1
  11706. At 32:28 you say non-zero n can be a difference of squares exactly when n is odd, but this is not true, it is iff n = 0,1,3 mod 4.
    1
  11707. In the MCU, if you look closely at the tesseract when it's not quite so glowy and bright, you can see the inner cube inside it. They did in fact render it as a 3d projection of a real tesseract.
    1
  11708. 👍🏻 Another great T-shirt. Of course the lecture too. 😉
    1
  11709. it reminds me of solid state physics. our teacher used to ask why there are only twofold, threefold, fourfold and sixfold symmetrical axes in crystal lattice
    1
  11710. Thanks!
    1
  11711. If you expand the Fibonacci sequence to a depth plane (4th squared(?) integer) what is then explained? And a 5th plane would then be (maybe rate of expansion of a curve of a line within the cone)? 6th? And would tracking the intersect values of one of these planes possibly line up with Pi, or another number?
    1
  11712. This video is amazing. "The Book of Numbers" by Conway and Guy details Moessner's miracle as well. A few years ago, I found a proof of this result using Horner's method.
    1
  11713. 3^3+4^3+5^3=6^3... May it be a start of another infinite pattern? 3^4+4^4+5^4+6^4... Not equal to N^4 for natural N.
    1
  11714. I've missed your videos for so long, good to see the king talk about another king of math..
    1
  11715. 24:20 didnt try it but are the 2 centers on the same point?
    1
  11716. @2:42 I somehow feel I have grown up every day as a kid hearing that sound :)
    1
  11717. Wouldn't e -1 = 1/1! + 1/2! ... etc. Just be the same (and look more beautiful) as: e = 1/0! + 1/1! + 1/2!....etc. (Since 0 factorial is defined as one, we can get rid of that -1 on the left side and just show e by itself.
    1
  11718. Beautiful video! I liked the IMO proof the most. It was amazing! P.S. Mathologer, why are some of your videos link only? Like the hidden cube in Simpsons one. Also, Are you still working on the gallois theory video? I am waiting for that amazing video :) And also, I want to learn problem solving like that in IMO. Can you suggest some nice books for the same? Thanks for making such wonderful content on YouTube!
    1
  11719. There are 16 ways of writing 2020
    1
  11720. It's difficult to find harmony in harmonic series
    1
  11721. Imagine Arabs and India did not invent modern math (decimal point and Zero),algebra) numbers and we still used silly roman numerals.. China would advanced before thw west. שלום
    1
  11722. thank you so much for subtitles! really helps!
    1
  11723. This video was so good!
    1
  11724. I really like calling sinh "shine", since it sounds similar to sine, along with cosh sounding like cos. In my calculus classes decades ago we called sinh "sinch" /sɪnt͡ʃ/, which sounds quite different.
    1
  11725. Hereby another interesting geometric triangle where the horizontal fractions can be factorized from the outer to the inner of the triangle. The outcome of that factor is the location and the number in the Pascal's triangle. 1/1 2/2 2/2 3/3 x 2/1 x 3/3 4/4 x 3/1 3/1 x 4/4 5/5 x 4/1 x 3/2 x 4/1 x 5/5 6/6 x 5/1 x 4/2 4/2 x 5/1 x 6/6 7/7 x 6/1 x 5/2 x 4/3 x 5/2 x 6/1 x 7/7 8/8 x 7/1 x 6/2 x 5/3 5/3 x 6/2 x 7/1 x 8/8 Where 1 is the unit of length: 1/1, 2/2, 3/3, 4/4, .. The length is going up 1 unit: 2/1, 3/1 , 4/1 for each row.. The "fibonacci triangle" variant of the Pascal's triangle (the number is the sum of the 2 diagonal numbers above that number) is : 1 1 2 1 1 3 2 1 1 5 3 2 1 1 8 5 3 2 1 1 can also be seen as answers to the fractions, factorized from right to left (in this excample below): 1/1 1 2/1 1/1 1 3/2 2/1 1/1 1 5/3 3/2 2/1 1/1 1 8/5 5/3 3/2 2/1 1/1 1 So ,1,2,3, Cheers to the catalan numbered sencorship networks.
    1
  11726. Really love your channel. The math is great, amazing.  But the mathologer is even better!Congratulations on hitting 500 000 subs.  Merry Christmas and Happy New year to you. :)
    1
  11727. 6:23 yes because there is a equal number of green and black square so answer is yes u can put the Domino in the board even when 4 square 2 black 2 green square is missing
    1
  11728. Why not 4k+1 and 4k-1 ?
    1
  11729. Happy new year mr mathologer!
    1
  11730. The mathematicians among us see you touching on algebraic number theory here. How about a video about Evariste Galois and his amazing discovery on polynomial roots?
    1
  11731. I believe I was taught Heron's formula in school geometry in the U.S.
    1
  11732. Mathologer, please stop putting important formulas or text at the bottom of the video where the captions appear when we turn Youtube's captions on. Captions help me to learn.
    1
  11733. please continue your quest for "ever more insane videos" :)
    1
  11734. That's a very satisfying final animation
    1
  11735. ..if a Oriental Mathematician discovers something and no Western Mathematician is there to know about it..
    1
  11736. I worked until late last night to complete, only to now the next afternoon realize that Youtube blocked my comment. I uploaded a video of it and added links there: https://youtu.be/MnZwz2kOuhc
    1
  11737. Doesn't seem too different from the coastline paradox
    1
  11738. Hm, das hätte ich vor dreißig Jahren gut gebrauchen können...😄 Sowohl die Infos, als auch die Präsentation, toll gemacht! 👍🏻 Ich schaue mir den Kanal mal genauer an
    1
  11739. Tell me about graph theory without telling me about graph theory
    1
  11740. How does our view of the "limits" of this proof change when we view each component as vectors and apply Clifford/Geometric Algebra? That shows that the complex number "i" is simply the product of two vectors. So x^2+y^2=z^2 -> |\vec{x}|+|\vec{y}|=|\vec{z}|. This would seem to imply that the parent vectors x and y create the child vector z. Would that also imply that there are two more parent sets before 3,4,5 that are only 2 and 1 dimensional instead of 3 dimensional?
    1
  11741. "Please let me know what worked and didn't work" - i didn't get any sense of how area gymnastics can lead to the derivative of e^x being itself
    1
  11742. I think the problem that can arise with transforming a closed curve(with posibly self intersections) to a circle is exactly those self intersection points. Even with a finite polygon if the self intersection is a vertex the procedure break ends as a sum of multiple points.
    1
  11743. The most memorable part was the leaning tower of lira for two reasons: first, it was a fantastic exercise to have high school students actually perform; second, because it was "coined" by Paul Johnson. Third time I showed this video I realized what was said and started laughing, then had to explain myself.
    1
  11744. Do a video with Archimedes proofs!
    1
  11745. Bizarre overhang
    1
  11746. Lol, I did neither. I did the algebra proof. x+y=a y+z=b z+x=c From which you can get x=(1/2)(a+c-b) y=(1/2)(b+a-c) z=(1/2)(c+b-a) You can then prove the midpoint stuff, (example, using a and the whiskers that extend it, we have x+b=y+c) Not as aesthetically pleasing though.
    1
  11747. Well...if it comes from china and is ancient, it's probably a scam
    1
  11748. 😂
    1
  11749. What was that word? Nomond?
    1
  11750. 1/x is divergent so no mystery there :P
    1
  11751. What music did you use in this video? I love it
    1
  11752. Moessner and Fibonacci are related. Please make video on it
    1
  11753. I'm Turkish, in high school we're given Heron's formula but without any proof whatsoever.
    1
  11754. Entirely dependent on using base 10 ~~> Key to the universe -__- It's like they weren't even trying. So much of math to choose from to spin your mysticism around, and they pick something that boring. At least the Fibonacci crowd picked something at least somewhat mathematical interesting. I rate this conspiracy 1/10. At least there was a somewhat pretty diagram.
    1
  11755. Alas, no analysis in the 3D grid of any lack of deltahedrons or 3D regular polygons. Maybe next time?
    1
  11756. He'd buy two Talmuds and cut one of them in half to find out how it works.
    1
  11757. nice video Take care of yourself sir you are continuously making the videos for thank you so much for support and kind gesture Lots of love ❤️❤️❤️
    1
  11758. I saw a version of this that was attached to a rolling ruler. You could draw lines with the length of the results on a paper with it. Cool af
    1
  11759. Daaaaaamn
    1
  11760. In terms of differential geometry, a cylinder is a flat surface. That's because the curvature at a given point is the product of the major and minor curvatures, which for a cylinder equals 0 because the minor curvature is 0. This also explains why you can eat a slice of pizza by curving it, and it will not bend down. Because it started as a flat surface, the curvature of 0 is preserved. Also, a torus can be modelled as a flat surface, although its embeddings in 3-space may not appear to be. Maybe you could do a video about that. Anyways, good job with this one.
    1
  11761. 2:33 why did you double 1x2? it just feels completely arbitrary. You didn't double 1 x 3
    1
  11762. Could you also explain why it only works for Euclidean geometry?
    1
  11763. About the T-shirt. It's a myth that the left brain hemisphere is the logical one and the right is the emotional one. These qualities are not associated with hemispheres.
    1
  11764. Wow very good!! How long does it take to make a video like this?
    1
  11765. I was lucky enough to be tought divisibility tests in school and now I'm doing my best to pass it onto my students.
    1
  11766. Too bad you can't demonstrate this in terms of sound would be nice.
    1
  11767. Hi Burkard, I take this opportunity to thank you for your great videos! There are many memorable aspects of harmonic sums that you highlighted, but I particularly like the fact that all the integers manage to escape the ever finer net represented by the partial sums. Keep up the great work! PS: Is there a way you prefer for people to buy your books, so that the most of the money ends up in your pocket? Cheers!
    1
  11768. In my 3rd to last semester of CS undergrad and taking a class on probability, where we are dealing with sums. I have yet to take linear algebra, though I have some knowledge there. Thanks for doing these!
    1
  11769. Danke ➕🟰✖️
    1
  11770. By the way, it seems that the 38 million number 4:00 seems to be missing a zero at the end, it should be 380 million from applying the given formula?
    1
  11771. I didn't understand the last argument in the proof in Chapter 4. In the example A(3) there were 2 patterns with colliding. I didn't get why these 2 patterns would always appear in pairs.
    1
  11772. I definitely learned it in American High school but it was not as interesting or as in depth as your videos!
    1
  11773. I really like the infinite series type videos you make. Those are my favorite. 💙
    1
  11774. Does it work for scalene triangles?
    1
  11775. Why does this remind me of x-ray crystallography and other studies of the shapes of molecules?
    1
  11776. 2:48 - My head exploded! 🤯🤕😁
    1
  11777. Does this equation have positive integers (with integers different form one another) solution for all n and m and z? a^z+b^z+c^z+....+x^z+y^z (n terms) = ā^z+ă^z+ą^z+....+ć^z+ĉ^z+ċ^z+č^z+ď^z (m terms)? We know it doesn´t for n=2 and m=1 with z=!2, but wondering if there is one general result on this. Thank you!
    1
  11778. Fav part was that the harmonic series is the sum of sums! Apparently, even a "son" of a sum is in it!
    1
  11779. I loved this video, although I still haven't completely digested the last one I saw about the cone and the hemisphere. (I specifically can't get over that if you bend the graph of sine from max to min in a half circle it will lie on the surface of both a hemisphere and a cone of twice the diameter.) I only missed two things here but if the proof is just ten years old, maybe this hasn't been done yet. Anyway: 1: How can this be applied to some practical problems? I have a new hobby with 3D-printing, and 3D has triangles everywhere, but not necessarily in a plane. 2: Does it generalise to something more abstract? Could it be reformulated in for example geometric algebra?
    1
  11780. Seems like the circumference and surface area approximations require chains of line segments that minimize the sum distance of the line segments. Approximating with zig zag line segments doesn't seem to minimize distance over the line segments.
    1
  11781. "show me the way to go home" my grandfather used to mention like this: show me the way to go home, ive had my drink and want to go home. That's how I've understood the signs from the movie, he's had his drink of protective solitude and the obsession with your kind wanting to protect a pure but bastardized meaning; its time to go home.
    1
  11782. Yay for me: I determined the 10-row-triangle's bottom-cell's rule correctly without pausing or guessing :-) though not the full (n^3)+1. :-( Too much time with Mirek's MCell and Wolfram's NKS? :-D
    1
  11783. Thanks Mathologer, great work!
    1
  11784. Thank you for the video!!! Merry Christmas!!!
    1
  11785. At about 15:00 I began singing “Oh Tangent Baum” 🎄
    1
  11786. Feedback: great video, as usual. Some calculations are a bit too fast for me, but pausing here and there is not a problem. Math background: grad student of Computer Science (Mathematical Optimization).
    1
  11787. Oh! Domotro from combo class is your patron :) that's great :)! - regarding the closed curve approximation with... circles? I think its of no use! Its unstable as it degenerates, and Compter Science already got polynomial approximation mechanisms like engineering splines, Chebyshew ortogonal polynomials and that french guy polynomials... Legrande? Legendre! ;)
    1
  11788. 🌷🌷🌷
    1
  11789. Cool t-shirt.
    1
  11790. Why is it (x+2)? What would happen if 2 was replaced by a different number? Are the expressions not as meaningful?
    1
  11791. 3B1B covered this exact topic some time ago (https://youtu.be/NaL_Cb42WyY), including more details on how the "good" / "bad" numbers contribute to the Jacobi formula. For the question re. limit of (whole number part of (r^2 / k)) / r^2, if you split the numerator into r^2 / k - its fractional part, then the first term simplifies to 1/k and the second vanishes as r -> inf because the fractional part of any real number is bounded above by 1.
    1
  11792. What an odd shirt...
    1
  11793. I really enjoyed this! I already used stuff from your Cardano's formula video for a little maths academy and this one will surely provide material for the next one. PhD degree in Maths.
    1
  11794. Engagement
    1
  11795. Surely paranormal distribution is carried out by the Ghostal Office?
    1
  11796. YAY! thank you for best maths
    1
  11797. Prime P; even X = 2n; odd Y= 2m+1; , If (P == X^2+Y^2) then P = (2n)^2 + (2m+1)^2, which follows that P=4n^2 + 4m^2 + 4m +1 , simplifying more we get P =. 4(n^2 + m^2 +m) + 1. Let k =. n^2+m^2+m. We get P = 4k+1 <> Seems to me a very simple proof, did I miss something, is the proof wrong?
    1
  11798. For the last exercise: n= x^2 - y^2 = (x+y)(x-y) If n is prime if can only have factors of itself and 1. This means that x+y = n and x-y= 1 . So the x and y must always sum to the prime and only differ by 1 e.g. 7= 4^2 - 3^2
    1
  11799. 1st part The best in my opinion was the glace proof for balancing (Though that is probably because I have seen the others before :) ) 2nd part- Not integer part, again because I had seen others (And also rediscover one (gamma) once while playing around with graphs) :) Final: Woah! Such cool stuff! Difficult to choose but I vote for- Tristan's fractal Because its so intuitive :) Great video
    1
  11800. Kempner's proof
    1
  11801. Fifth time I'm rewriting this, it should be okay now (well that's what I said the other four times). Let A, B, C be the points of the circle, and let's say D and E are the two points that lay on the line AB and AC (respectively, when extended) such that AD = AE = BC. Similarly, F, G are in BC and AB such that BF = BG = AC, and H, I are in BC and AC with CI = CH = AB (this is just the setup). Draw the angle bisector of the angle EAD and let its intersection with ED be J. The triangle ADJ is equal to the triangle AEJ by the side (AD = AE) angle (half of the angle EAD) and side (common side AJ) theorem. Then the side EJ is equal to the side DJ and the angle EJA is equal to the angle DJA and since they lay on a straight line they must be right angles. Then, the angle bisector of EAD is also the perpendicular bisector of ED, and because opposite angles are equal, it's also the angle bisector of BAC. Let K be the point where GH intersects the perpendicular bisector of ED. AH = AC + CH = AC + AB and AG = AB + BG = AB + AC and so AH = AG. Then the triangle AGK is equal to the triangle AHK by the side (AK) angle (half of the angle BAC, see previous paragraph) and side (AG = AH as we have just seen) theorem, therefore the angle AKG is equal to the angle AKH but together they lay on a straight line so both are right angles, and since GK = HK (by the triangles) then the aforementioned perpendicular bisector of ED is also the perpendicular bisector of GH. By symmetry, the perpendicular bisector of FG must be the same as the perpendicular bisector of DI and the same as the angle bisector of ABC, and the perpendicular bisector of HI must be the same as the perpendicular bisector of EF and the angle bisector of ACB. The mentioned perpendicular bisectors must intersect in the incenter of the triangle ABC since they are also the angle bisectors of the inner angles of the triangle ABC as we have just seen. Recall that any point on a perpendicular bisector of two points is equidistant to said two points. Let's call the incenter O. Then, OE = OD, OG = OH, OF = OG, OD = OI, OH = OI, OE = OF (because O is the intersection of the perpendicular bisectors). Combining all that we have OD = OE = OF = OG = OH = OI and so all those points lay on a circle with center O and radius OD. Definitely a very messy proof. It's hard to explain things without being able to point at a picture but I hope this all makes sense and I didn't make any (more) mistakes. Now, time to watch the video!
    1
  11802. It kind of reminds me of the do nothing machine. Should have waited a few moments before this post --- sorry
    1
  11803. best part was the proof og no 9 series convergence! and seal of approval
    1
  11804. Am I right in thinking that the paradox arises because we're attempting to sum two infinite series with different numbers of terms? When the alternating-sign harmonic series is first modified by splitting the negative terms, we create a series with twice as many negative terms as positive terms. When we then separate the negative terms from the positive terms, we create two infinite series with different numbers of terms.
    1
  11805. I do not recall ever seeing this diagram, only forced to memorize the formula.
    1
  11806. The swivel approach resonated better with me.
    1
  11807. 30:24 Hey, wait a minute. The “7” should be a “5” !
    1
  11808. When I took Calculus the hyperbolic sine was always called sinch .
    1
  11809. Woah!! I love that shirt... where can I get one.. and if so, where can I get a onesies with that design!!
    1
  11810. The sum on no 9‘s was really counterintuitive but after your explanation very clear.
    1
  11811. Mind blowing!'😮thanks'
    1
  11812. Here's another thought. Imagine a long thin cylinder (which is what the horn basically turns into waaaay out there). Section off a chunk of length H. V= (pi)(R^2) H Lateral Surface Area = 2(pi)RH Take the ratio LSA/V and simplify to 2/R As R ---> 0 the ratio ---> inf. The volume decreases proportionally faster than the lateral surface area. It's like the reverse of the square/cube growth thing (limiting the size of animals) because the dimension (R) is ---> 0
    1
  11813. Excellent! We should have done this at school.
    1
  11814. I also want to participate in the fantastic prize lottery. My favorite part is the No 9's series. I would go about and tell everyone about it since that shocked me really hard specially the fact that it is less than sum of one over googols.
    1
  11815. This video just jumped to the #1 spot of my favorite Mathologer videos! ...which also means is vying for #1 of ALL YouTube math videos. (But please don't make me pick the-best-of between different math-YouTubers, as there are many gems on YouTube.)
    1
  11816. What year? Hello - Oh! Re. Carl. 💖😎
    1
  11817. C’est très fort
    1
  11818. Merry Christmas and happy new year mathologer. That was a great video! Greetings and wishes from Greece.
    1
  11819. Has anyone thought of this convergence being a possible result of a comparison between a Pi(infinite pixel circle or perfect circle) vs how small the pixel used to create the circle is? (like in MS paint). The bigger the circle the closer to Pi, because smaller pixels mean closer to real Pi
    1
  11820. Thank you for great explanation.. I like your t shirt
    1
  11821. Unless you somehow get an infinite horizontal plane with an infinite mass below distributed in just the right way, no real world gravity field would remain constant for the height required for the "tower" with 10^100 blocks. In fact, this "tower" would have more blocks than particles in the observable universe, and most surely would have it's own gravitational field rendering the calculation of the centres of mass quite useless.
    1
  11822. this is not super famous because nobody cares about "maths" (aka math)
    1
  11823. Most memorable: I never linked the harmonic series and the natural log before, it's like two parts of my limited understanding are now for the first time talking to each other. Ps. Like always: thanks for making Math visual! It truly shows its beauty...
    1
  11824. I love the equation on the t-shirt 👕. The video is amazing as usual
    1
  11825. So is there a nice connection to the golden ratio in relation to right angled triangles?
    1
  11826. Sort of. If you look at the Fermat series of PPTs you can see that both numbers in the top row of the squares form a series where each entry is equal to 2* the previous entry plus the entry before that (look at the column of boxes at around 9.40 - for example 7 = 2*3 +1, the next box would have 17 in the top left corner, and 17 = 2*7 +3 and so on). So you have a slight variant on the Fibonacci series. You can express the nth term of the Fibonacci series in the form a*phi^n + b/phi^n. Look up "characteristic equations" to see how this works. With the Fermat series of PPTs, solving the characteristic equation (x^2 - 2x - 1 = 0) leads to the general term being of the form a*(1+sqrt2)^n + b*(1-sqrt2)^n. You need to set a and b so that the first two terms of each series correspond to the first to boxes in the Fermat series. So there is a congruence between the way the boxes driving the Fermat seris of PPTs evovles and the way the Fibonacci series evolves. You can decide whether you think its a nice connection!
    1
  11827. THIS IS AMAZING!! :D
    1
  11828. Hi
    1
  11829. I enjoyed the part with the integral and gamma the most. I already knew about the Leaning Tower of Lire because of Vsauce (he also made a great video about that) and it's still so unbelievably cool. Greetings from Germany :)
    1
  11830. The only incorrect assertion in this presentation is that the outro music is funky.
    1
  11831. I actually just did the chess board domino problem as a homework in my graph theory course. My proof assumed a bipartite graph (such as a chessboard) and looked to find a perfect matching in that graph. First I found an arbitrary Hamiltonian cycle in the bigraph. It was easy to show that when two vertices (tiles) were removed, they must come from differing partitions for the resulting graph to still have a perfect matching. I then used the fact that such a graph could express a board which must then also contain a Hamiltonian Cycle. This makes the shown method of tiling along the path that touches all vertices (tiles) the obvious solution. In addition it helps with the reasoning behind the matrix for the formula for the ways to tile a given "board". (adjacency matrix of sorts) great stuff! As to the question of whether the removed of two additional (differently colored) tiles would still be able to be covered with dominos, I think the answer is yes. There are two cases to consider. First, the two removed tiles were next to each other in the initial circuit and thus the rest of the board is now a single path. If two more tiles were removed from this board, it would always be able to be tiled by dominos using the reasoning above. Second case is if the two initially removed tiles were not next to each other in the initial cycle. Their removal would then divide the cycle into two paths. If both of the 2 additional tiles to be removed are from the same path, then the board can be tiled. If they are from different paths however, then there is no guarantee under what we have established so far. However, if an alternative cycle in the initial graph can be found such that the two newly removed tiles would fall into the same path, then a domino cover can be constructed. I hypothesize that such an alternative cycle can always be found, at least for an 8x8 graph and removing 4 tiles.
    1
  11832. Hi ! I had been playing with the shoelace formula derivation and geometrical meaning for 3 days.....did grasp only a few aspects but most of them still seemed evasive.... THIS LESSON is so satisfaying !! THANKYOU VERY MUCH .....the only thing I just feel a bit beaten by not being aneble to come throough these topics on my own.
    1
  11833. Felix Wankel would like this.
    1
  11834. From my knowledge the overhang paradox is used a lot in architecture
    1
  11835. 1- 5^3+6^3=7^3. The difference is only 2. 2- 2 in a minute. 3- the first equation is 6^3, and the second is… I didn’t get it 😅 it may be 7^4 in a difference of 143… and… Merry Christmas 🎄
    1
  11836. Answer to the second challenge puzzle: When we try to get the "parent" triangle for the 3 4 5 triangle we discover that the 2 centers of the circles are on a line parallel to the 3 side, so instead of a triangle we get just a short line.
    1
  11837. Could follow it all! Only couldnt figure out how you did 41:06, but that doesnt matter for following the video. Studied math at university level :)
    1
  11838. mh is there no cool way to get rid of theses nasty ones ?
    1
  11839. I'm not at all sure of the details but I think the harmonic series of primes diverges because the nth prime, pn, is approximately n*log(n) (Prime Number Theorem) 1/pn is approximately equal to 1/n*log(n) and that's a sort of multiple of the harmonic series.
    1
  11840. Can we say for trithagorian theorem... We can write T(A) +T(B) =T(C) +2Tcos(theta), where theta is the angle of the red triangle which is mentioned in this video... For 60°, T(A) +T(B) =T(C) +T For 120°, T(A) +T(B) =T(C) -T For 90°, T(A) +T(B) =T(C) It's absolutely amazing how pythagoras theorem works for any shape... 🙃 Amazing stuff after amazing stuff... How Gp-Ap inequality and sin addition formula came... Absolutely fabulous✨ The area of the square will be 1/4 unit. And, for the beetle puzzle, if we consider any one of the beetle, Since for symmetry, they will meet at last at the center of the square, so, average displacement of a beetle is equal to 1/2 of the diagonal of the square. <x>=a/√2 And also, we know average displacement is equal to average velocity times time. <x>=<v>t And <v>=vcos(45°) =v/√2 So, elapsed time to meet = t= a/v Where a=side of 🔲 and v= constant speed of each beetle
    1
  11841. That shirt tells you're a true scientist
    1
  11842. The area of the square is ⅕ Lets call the smaller triangle A, with sides a,b and hypotenuse c and the bigger triangle B, with sides d,e and hypotenuse f We can see that A is from the intersection of the 2 bigger triangles, B, and is congruent to them We solve for the sides of A We find: 1/a = ½ • 1/b = ((√5)/2) / ½ 1/a = 1/(2b) = √5 So: a = 1/√5 and b = 1/(2√5) We can see that the hypotenuse of B equals a + b + some number x And its also equal to √(1+½²) = √5/2 Therefore: 1/√5 + 1/2√5 + x = √5/2 We find that x = 1/√5 Therefore the area of the square is: (1/√5)² = ⅕
    1
  11843. Hi mom!
    1
  11844. Hi! It's been a while!
    1
  11845. At school, this ( functions) was presented as a slight of hand, wherein the student must guess how it is done. Now I know, 55 years later
    1
  11846. 24:40 Each pair blue and white areas is bounded by some rectangle. If the areas of the blue and white parts were the same, then they would each take up half of the rectanges and thus half of the square. However, just by looking, we can see that the blue takes up more space than the white in each rectangle. So in total, there must be more blue in the square, so it must be greater than 0.5.
    1
  11847.  @Mathologer  Odd, Youtube didn't tell me you replied. I'm aware I have something backwards, I'm always doing something backwards, just didn't know what exactly. Thanks.
    1
  11848. what annoys me about most socalled proofs of Pythagoras theorem is that you only do it for a given right angle triangle. Which doesnt automatically mean it applies to all, without also doing the step to prove that. Like the initial visual proof you show here for example. Yes it's great for showing that Pythagoras applies to that particular right angle triangle, but why should that automatically mean it applies to other right angle triangles with different a and b lengths; after all, it is not the case that all right angle triangles are self-similar
    1
  11849. Did not enjoy the music in this video, though otherwise it’s a great little episode :(
    1
  11850. The Cauchy Schwarz inequality - not to be confused with the Swoosie Kurtz inequality which says that no matter how much you put into your character, she will play hers up even more.
    1
  11851. Loved the video! I noticed that the last step of the 2d animation appears to draw the edges of a cube in the final step, and a bipyramid in the concave case. Is this pure coincidence, or is there more 3d shenanigans?
    1
  11852. The most mem. thing is how slow they add to inf.
    1
  11853. My school is opening soon! I needed that man!🥺
    1
  11854. This is dangerous to our democracy.
    1
  11855. 面白い😄
    1
  11856. swivel proof.
    1
  11857. While there are many amazing things over here, it was actually astonishing for me to find that there are better solutions to overhang problem than Tower of Lire. Of course it makes sense once you've seen it and I feel a bit stupid ;) So chapter 2 I guess, would be my favorite. Runner up would be Chapter 6 - because I of course thought that no 9's sequence would still blow up to infinity. Sorry my math days are 15 years behind me - but doesn't it mean that "only containing at least one 9" sequence would blow up to infinity? Because in this black square in white square the white would thin out the black eventually so it would be "infinity - small number (~22.92068) = infinity", right? And while you have sequences that blow up to infinity and what is left also blows up to infinity, when 1 sequence converges, the remainder MUST blow up to infinity. I think?
    1
  11858. 3³ + 4³ + 5³ =6³ , but 3⁴+4⁴+5⁴+6⁴ is not 7⁴
    1
  11859. By the way, there's a video by Michael Penn(https://youtu.be/5lR3y1bTFZ8?si=4HPor9sInY8UFfiX) about the sums that you mention in the description!
    1
  11860. I want to say the coloring proof because it’s the one I found, but the swiveling proof is way cooler so it gets my vote
    1
  11861. If a prime number is the sum of two squares, it has to be of the form 4 k + 1. Well, obviously. As you showed, one square must be odd, and the other even, and their roots too. Let's write the odd one as (2 x + 1)², and the even one as (2 y)². Then p = (2 x + 1)² + (2 y)² p = 4 x² + 4 x + 1 + 4 y² p = 4 (x² + x + y²) + 1 What did I miss?
    1
  11862. Your videos, and your shirts, never fail to deliver.
    1
  11863. ah yes, of course, it has an area of 2∞+1 🤔
    1
  11864. I like many shorts here & there in between the long. 🤓 Of course this all depends on the problems you are solving. But Simplified complexities are often anonymously gratifying. For me…. That is. 💫 Thank you for your brilliant Mind! ✨
    1
  11865. My vote for most impressive thing is the sum of exactly 100 zero numbers. When it was larger than no 9’s I gasped.
    1
  11866. Something tells me this is related to the problem of map projection. That is, it is impossible to accurately draw the earth on a map, because maps are flat, and the Earth is curved. Is this another example of the same buckling problem?
    1
  11867. It got interesting when rational approximations came in to play. I'll have to rewatch. As always, thank you for bringing out the beauty in math.
    1
  11868. crazy interesting subject. i could easily watch another 3 hours of such "exploration" of sequences!!
    1
  11869. wow
    1
  11870. I'm thrilled that you're still making these incredibly well-produced, expertly-explained videos.
    1
  11871. Pi is beautiful. This could have been a 'Pi-day' video! Very well explained, illustrated and yes, it's full of 'gems', as mathologer videos usually are! Incredible how ancient mathematicians managed to do such complex maths (in a very complicated looking language!) without calculators - just pen and parchment, which could easily be destroyed. They must have been well off to have time to pursue mathematics. I'm glad it was valued as something worth working on. I also wonder about those correction terms if there might be a way to see a pattern better (or get pi approximation to converge faster) if we look at how the alternate values (every other consective approximation) decrease (or increase) as they both approach the same limit from different directions. PS Sometimes I wonder how many digits of pi-related numbers have been calculated like pi-squared, the square root of pi or something to the power of pi, which might be just as useful. Or how many accurate digits of pi you need to get pi-squared or some other pi-derived number to a given number of decimal places. PPS pi-trivia: I recently had a look around the 'million digits of pi' Website for alternating sequences of numbers and the longest I found was a sequence of 4s and 2s. ...1324242424201... or there is ...6398989893...
    1
  11872. @15:30 - the rotation is not strict to the normal against the plane, a rotation against a line in 3d gives you a circle with many points, not just 1.....this means that the object can be pushed outward, and you find a similar sized hexagon against the same grid, just shifted. This means the rotation method is allowed, you just have to include all of its endpoints.
    1
  11873. Want this book
    1
  11874. honestly i didn't remember any proof of pythagoras' Th, however seeing the diagram in the first few seconds of the video i decided to figure one out. seeing that the whole square had sidelength a+b, the algebraic proof of (a+b)^2=c^2 + 4*1/2*a*b seemed much more intuitive than shuffling the triangles around to arrive at the simpler visual proof. but i think that's just because the binomial formula & the area of a triangle formula have been so deeply ingrained into my brain, otherwise i'd probably find the visual proof much more intuitive, tho not as easily derivable
    1
  11875. The area of the small square is 2\5 square units. Am I right? I am a long time fan! Thank you for what you do!
    1
  11876. Nice vid, and, coincidentally, Lee Sallows lives in my neighborhood...
    1
  11877. i made it to the very end
    1
  11878. 1:34 no idea if its interesting or not, but you can also get to the corresponding sum on the right, if you double the nth Fibonacci number and add n-1th and minus 1, i.e. 88=34*2+21-1
    1
  11879. I wanted to prove (x + 2)^n having those n-square coefficients, but kinda got tired of doing so. However, as an intuitive argument, take an n-square, and extrude it out to an (n+1)-square, so that it's obvious that all k-edges will double, because you have two copies of the n-square, plus all k-edges of the n-square will extrude out to new (k+1)-edges. You can see that's what multiplying a polynomial (in x) by (x + 2) does: double each term, and add the coefficient beneath. Having said this, I think it probably would be easy to formalize, but I'm done working the problem, because I want to actually finish the video :)
    1
  11880. Video starts, gets a thumbs up, because it's always good
    1
  11881. The most memorable is the mentioning of the series 1/p is infinite because that blew my mind the first time I heard that. I have since loved the proofs of it so much I have a tattoo that features this. :)
    1
  11882. Hey Mathloger! This is a really nice video, just wanted to point out they're not called sevengons but heptagons.
    1
  11883. if Karl brings all this toys as dowry, I would like to marry him 😂😂😂😂😂
    1
  11884. Extraordinary! Merry Christmas to you!
    1
  11885. I'm a CS Grad student, did several years of high school math and some undergrad level. Couldn't follow some stuff in second half but still blown away. Awesome video!
    1
  11886. Can we talk about the limit set of points in the pi = 4 proof? How does this set compare with the set of points in the actual circle?
    1
  11887. To infinity and beyond!
    1
  11888. nice. thx. video graphics are so useful for this. god, what did we used to do before.
    1
  11889. AUGH!! Where are the cute cat videos??
    1
  11890. Absolutely loved the video.
    1
  11891.  @Mathologer  An office full of great toys! After that teaser I'd be interested to see videos about them
    1
  11892. Since this scaling anti-shapeshifter is so intrinsically related to e, and since we know that integrals and derivatives (and thus the rates of change) of 'e^x' are also invariant... and since both sine and cosine are intrinsically related to e... and in general 'e' seems to show up whenever you have some kind of invariance, that leads me to this question: Is there any non-trivial system that is invariant under certain transformations that has no relation to e?
    1
  11893. The best and most accurate math wheel is on the iconic Navitimer watch by Breitling.
    1
  11894. Could you discuss how angles work in ellipses, in terms of arc length, area, the inscribed circle, ..., please?
    1
  11895. The best part was everything. The animations, the clarity of arguments, the enthusiasm and, of course, the t-shirt. Clearly, this is hours and hours of work just for our enjoyment.
    1
  11896. French swiss guy here, we had classes of algebra and geometry at school, and had to do a lot of demonstrations, many of which involved the angle at the center, Thales,... I loved that stuff, and it really framed my mind. I enjoyed that video the same as I did then. Nowadays my children do no more see any proof, not even Pythagore... it's a pity! Btw I still have the book, Géométrie plane, by Delessert
    1
  11897. Only 6 comments?
    1
  11898. Wow didn't think you would reply
    1
  11899. I love see a new Mathologer video pop up in my subscription box! For me the proof that the harmonic series never produces an integer in its partial sums. I just feel like somewhere on their journey the sums should bump into an integer, but no Math always throws some curve balls.
    1
  11900. My intuition suggests that if plain figure have a center of mass, then each line passing through the center of mass divites the figure on two halves with same area. That's an intuition, not watertight definition, simple counterexample is e. g. a banana.
    1
  11901. Dear Mathologer, thanks a lot for this amazing video. I noticed that many people selected "no integer as partial sum" as most memorable subject, although you did not provide any proof about that. So I began to look for such a proof and I found a nice one, that I hope you will appreciate. It runs like this: Among the first n integers, there is one (say p) including a higher power of 2 than any other. p is the highest power of 2 not higher than n (eg n=23, p=16=2 power 4). This is because the first multiple of 2 power x is 2 power (x+1). When summing 1/n, the denominator gets the product of numbers from 1 to n, and the numerator gets the sum of the products of all combinations of (n-1) integers. Among these products, the one where p is missing has a lower power of 2 than any other, so after simplifying by 2 as many times as possible, the numerator is the sum of n-1 even numbers and one odd number, and as a consequence is an odd number. Example: n=5 p=4 1+1/2+1/3+1/4+1/5=(2*3*4*5+1*3*4*5+1*2*4*5+1*2*3*5+1*2*3*4)/(1*2*3*4*5) =(1*3*4*5+1*3*2*5+1*1*4*5+1*1*3*5+1*1*3*4)/(1*1*3*4*5)=(60+30+20+15+12)/60= odd/even I suppose the same proof applies for the sum of 1/n to 1/m but this is a bit more difficult in case there is no power of 2 between n and m. Although I am sure there is in all cases one number with a higher power of 2 than any other, I did not find an elegant way to prove it.
    1
  11902. Really enjoyed the video. I followed almost all of it. Not bad for a Fedex delivery driver who took a few calculus classes over 30 years ago.  I self-educated myself with the help of YouTube. :)
    1
  11903. Incredible
    1
  11904. I actually figured out that 2nd proof of Pythagoras on my own.
    1
  11905. can you imagine sitting down and creating this...so far beyond my ability since I have problems getting through Cal II
    1
  11906. I showed my maths teacher the diagram of a square and another square with width being the diagonal of the first. Therefore, half the area of the first square was a quarter of the other. This meant the diagonal is sqrt(2)xside of the square. I was ten and it was in France, and because it was my fourth country, I was always learning the language of my new country. However I spent my younger years loving maths and walking round the playground (clockwise). And am proud of that proof with a stick in the Southern French dust.| However I was told I would be taught it next year. So glad to watch this video where no-one delays the learning
    1
  11907. can you please suggest a book to learn these tiling and parity/coloring problems in combinatorics
    1
  11908. The glib animation makes no sense to me: Near the center, the tangent is near horizontal. Near the rim of the sphere, the tangent is near vertical. Shouldn't the tangent near the new outside be likewise near horizontal of its former location? And shouldn't the tangent near the new center be the near vertical of its former outside position? Yet your inside-out thing shows straight radii with constant tangent, top to bottom.
    1
  11909. I liked the gamma part
    1
  11910. your shirt is mathematical poetry
    1
  11911. is the power of 2 coins deal just representing the change in binary?
    1
  11912. My maths background is undergrad electrical engineering. By and large I found this video approachable. I'd say the only thing that bugged me was not getting at least a hint at the end as to when the correction factors stop improving the value and start to diverge - it's the main thing that made me go "it would be difficult to start using this as a tool without that piece of information". I get that you're planning to make another video, but a reference along the lines of a) there's a neat cutoff expression, b) there's a complex cutoff expression, or c) it's not consistent but there are tools for working out when it's started to diverge towards the end, even as a teaser for the next video, would have helped immensely. Otherwise, I'm all for more videos along these lines.
    1
  11913. Beautiful. Reminds me of the golden period of screensavers. Math is trying to tell us something. Now apply this to the wave viscosity of the boson field to illuminate fundamental particles.
    1
  11914. Isn't thales theorem the proportionality theorem and not the angle in a semicircle as he said?
    1
  11915.  @Aerendil97  ah i see, even in my country (India) it is called the thales theorem despite being taught similar triangles... i guess it just varies region to region. The people who wrote the article on thales theorem on Wikipedia are probably from a region where proportionality isn't a separate topic from similar triangles.
    1
  11916. Challenge (when can n be written as a diference of two squares of integers?) We have: x^2-y^2=(x-y)(x+y)=n, If n is odd, we always can: take x=(n+1)/2 and y=(n-1)/2, and x^2-y^2=(x+y)(x-y)=(2n/2)(1)=n. The only factorization for a prime n is 1×n, so for odd primes the only solution is the one given above. Even numbers n must either have x and y both even or both odd, so that the product n must be a multiple of 4. However it is not true that this is exactly possible for odd numbers (in the sense that even n are never writeable as a difference of squares): 8^2-6^2=28. Edit: I noticed you added this to the notes below the video.
    1
  11917. The hidden 2019 is in the second row after the widest row of numbers on his shirt
    1
  11918. OK, with my basic secondary school geometry and lifetime of setting out drives and houses on building sites, I got completely lost by the phrase "length of edges reduces to 0". If it does, and you are trying to work out an area of a cylinder, surly you end up with 1/2 base times height times the number of triangles "Y" which is 0/2 x 0 x Y= 0 So what am I missing?
    1
  11919. Fun coincidence! Michael Penn did a similar video just a few days ago. https://youtu.be/5lR3y1bTFZ8?si=HhPraedbWZTG5AzR
    1
  11920. It’s really hard to pick just one part of the video that is my favorite, but the most memorable definitely is the original demonstration of the one-glance balancing because by itself inspired me to make a Facebook post and is the most easy to be conveyed to a mixed audience of mathematicians and non-mathematicians alike.
    1
  11921. When I was a kid, I observed that in the original puzzle, if we label the discs as 1,2,3 etc from smallest to the largest, odd numbered discs only touch even numbered discs. That’s enough to tell you where the number 1 disc go and therefore the whole solution. Also, for the 10 disc version, you can imagine at the destination peg (B), there is a number 11 disc at the bottom. Therefore number 1 disc should not go there because 1 cannot touch 11. It should go to the other peg (C).
    1
  11922. thanks for this nice video. please make a video about the concepts behind Infinite. for example.. why Rational set is bigger than Natural set if both are infinite... i know that this is not that complicated, but it will be nice expand the classical concept of Infinite (geometric infinite) to others that already exist. it also can be very attractive to the public.
    1
  11923. How does this generalize to fractional dimension, for example on a sierpinsky triangle?
    1
  11924. I'm 67 and this is the first time in my life to learn that adding the odd numbers gives the squares. Amazing! Why didn't I learn this in school? Why is school so boring? And another question is: Why did it take thousands of years for someone to notice the next simple step (the Moessner miracle), especially since people have been in love with the proof for about the same amount of time?? That is the most interesting question surrounding this whole topic.
    1
  11925. I love this video. The most amazing part to me was that no integer could be derived from any segment of the harmonic series, with the rate of growth being so small, that is almost unbelievable. Thank you!
    1
  11926. Using python I found that the partition for 666 is 11956824258286445517629485
    1
  11927. The number of partitions of 665 (666th starting from 0) is 11393868451739000294452939
    1
  11928. Mindbending ❤
    1
  11929. 16:21 E
    1
  11930. i have been seeing all programme any time
    1
  11931. I am going to sell my Tesla and buy a Plouffe.
    1
  11932. Mission completed.
    1
  11933. I would point out that the Chinese diagram is referring to specific numbers and so it is not a general proof more like Egyptians mathematics. The Pythagoreans proved the theorem but the credit of any discoveries was always credited to Pythagoras because of the cult.
    1
  11934. Mathologer - champion of infinite recursive convergent fractional series. Bravo!
    1
  11935. I've always wondered if there was a way to create an "algebra of operations" (like a counterpart to algebra of variables, where we're given the operations and have to find the numeric value of variables). So for example 5 # 7 = 12, obviously # is the familiar + operation, 5 # 7 = 35, # '=' *.... But I realize that this would involve some comparison operations roughly equivalent to comparing numeric values with <,>, etc. AND some sets where we classify mappings between the Reals...probably relying HEAVILY on the information presented in this video. Does anyone know what I'm driving at?
    1
  11936. Wouldn't the proof of Sum of no 9s, work for any other Sum of no other digit?
    1
  11937. It would have been nice to have had this resource when I was a child. My mind could have handled it. But now is all gibberish. Thanks for trying to make this simple and accessible for people.
    1
  11938. Another wonderful aesthetic video, with amazing background music, came to rekindle our mathematical passion! And of course, as a Greek and a lover of geometry, I love both proofs, but I prefer the final "at a glance" proof, because beauty, lies in simplicity :) And as always, thank you for the video!
    1
  11939. Early???
    1
  11940. One of the coolest things . It's revolutionary at that time of analog computer techs.
    1
  11941. Anyone else shopping for a spirograph now?
    1
  11942. I liked the coloring proof better 😅
    1
  11943. Number of moves required for N number of dishes are 2^N-1. So for 10 disks 2^10-1=1023
    1
  11944. In top view if the pegs are arranged in an equalateral triangle. All odd number disks move is a clockwise direction and even number disks move in counter clockwise. With every disk moving double the number the disk just below it.
    1
  11945. The background music reminds me of "To Live Is To Die" by Metallica.
    1
  11946. 39:34 Benford's law strikes again! And I see I'm not the first one who noticed. It's pretty easy to see the reason for this pattern if you just take the visualization at 37:05, imagine removing some different column (and row) instead, and consider where that puts the mass of numbers and what it does to their sum.
    1
  11947. Your channel is like taking LSD
    1
  11948. That tshirt 👏🏻
    1
  11949.  @Mathologer  Yes!
    1
  11950. I thought the no integers in the partial sum was unexpected. I knew that the sum would go off to infinity so it passes by each integer. It’s interesting how the partial sum “knows” how to miss all the integers. ( I think this also implies that each integer is missed by a unique amount). Not unimaginable, but a nugget of info that I enjoyed.
    1
  11951. 23:36 Looks like an orange. I think you've made a proof for playing with your food.
    1
  11952. I love both of the (perpendicular) bisection proofs SO MUCH.
    1
  11953. Diameter of the circle is larger. If you superimpose the circle over the shape, you can form a triangle of the shape's width and two radii of the circle. Due to triangle inequality, the two radii, and therefore the circle's diameter, must be longer.
    1
  11954. Can it have application in String Theory? ;-)
    1
  11955. Your feed is always worth waiting for. Well visualized and explained.
    1
  11956. Thank you very much, Sir.
    1
  11957. Chapter 6 answer: 1,2,4,8,16,32,64 :-)
    1
  11958. You’ve totally lost me and I’m falling asleep. I took Algebra one semester with a teacher who obviously did not understand it. Pescadero High School, I want my money back.
    1
  11959. Wow!
    1
  11960. I saw the entire video, and i'm confident on saying that i understood all the video, but i'm almost graduate (on december) in mathematics.
    1
  11961. neither and non. Wow 😲:D
    1
  11962. skips
    1
  11963. 1st valid commment. (Not really, I just started watching)
    1
  11964. Hi
    1
  11965. He had the shoe lace dream!
    1
  11966. Brilliant! Very well explained. Now I'm ready for a night on the tiles.
    1
  11967. I got the number of ways in which those glasses can be tiled to be 306. Can't guess what is special in it.
    1
  11968. I was just thinking would this be good for a heat exchanger with this design be good for a extractor thank you very much I’m thinking
    1
  11969. I think they don't teach that, because school adhere so strictly to the one idea they have (usually some form of method from a new study on "how to teach kids things easier"). I witnessed this with my kid who has a very high IQ (whatever that means of course). He simply used his own methods, which were also valid, but called "tricks" by the teacher. Super frustrating, but the key to what the problem is in schools I think.
    1
  11970. The way I see it, it's a philosophical view rather than a purely mathematical one. The human brain exists to recognize patterns There are mathematical patterns all over the place we can see, things like the Fibonacci sequence, the pareto distribution. I don't see why it would be impossible to say there are patterns that are not as easy to just "see" right infront of our faces like these. Math a tool what we use to categorize the impossibly that is the infinite pattern of reality. So maybe this is a some mathematical representation of a sect of reality we don't fully comprehend. A synchronicity of math and reality, something a little too weird to be a "mathematical coincidence". At any rate it does have some pretty serious symbology, beauty, and historical context around it dating back to some of the oldest texts in human history (afaik).
    1
  11971. Portuguese subs for my friends, please. I need to share this content
    1
  11972. I am yet to see a video going deeper than just the slight surface layer scratch as to what concepts can be formed, representative of numerous elements of the number theory field, just from exploring relationships between numbers, functions & operations of number interaction - from the numbers used in this symbol. For me its just a symbol for entry to one entry into a beautiful realm of relearning how I consider numbers & geometry. The slight edge of ego towards those who see something in this symbol as being a 'key to understanding the universe' rather threw me to be honest. I don't really go with the vibe or flow of whether or not there is a key to the universe from this symbol, that seems like unnecessary noise around something not really needing a label as such. If some/most want to stick with that layer, that's up to them. Sacred geometry tattooists are glad for it as too are those who get the cool designs not always having the layers of awareness as to the mathematical beauty in such shapes and symbols which also doesn't matter :) There is without doubt from my side, interesting and mathematical observable elements beyond just this symbol and special characteristics which could be declared for each whole integer or single digit, even decimals... grouping around characteristics, whether results of functions or influence in functions, pure number play from just 1-9... personally this opens up one of the most stimulating parts of my brain... to figure things out without googling them, to then search for the true math behind the sandboxed math which leads me to concepts to further my knowledge now based on that self exploration seed. That truly personally, is something of a key to the universe, or my universe... to not see things only as we are shown but to explore seeing things from different perspectives to realize some quite fundamental truths that our truths are our truths but that some things just seem to be true whoever observes them... At best, even if people play with digital root for numerologist intentions, I am still glad that numbers are explored for different utilization beyond the mundane counting of what we earn or spend or watch on a clock :)
    1
  11973. Love these connections between geometry and algebra (former becomes so hard to navigate beyond triangles and quadrilaterals)
    1
  11974. Thanks a bunch.
    1
  11975. Its nice how the famous Varignon parallelograms can be viewed as a special case of this construction: it corresponds to putting 180° ears on each side of a quadrilateral. Now this of course is nothing else than midpoint of this side. The next step, adding 90° ears, then creates a square.
    1
  11976. Can this be used to solve NS equation?
    1
  11977. The graph proof is absolutely fantastic!
    1
  11978. Tesla was a snakes oil salesman in physics, and it seems he tried the same in math. Taking something relatively simple and sell it as deep secrets isn't a sign of science, but of religion.. Key to the universe? Nonsense...
    1
  11979. WOW!!!!!!
    1
  11980. When I saw the start of the video, I was expecting you to Fourier Series out some periodic polynomials to get the Basel problem series. If you take the Fourier series for x^2 x in (-pi, +pi) and equate the values at x=0, then you can get a series for pi^2. I think x^2n type functions do the trick for higher powers.
    1
  11981. You get an alternating series if you take x^2 in [-pi, pi) and same sign series if you take x^2 in [0, 2pi) I guess.
    1
  11982. We had been studying number sequences in algebra (I was 14, I think) and one day on the way to school, I was playing with numbers in my head and "discovered" that odd number sequence and how it gave perfect squares at each new addition. I excitedly told my teacher about it, and she gave me a very skeptical look and told me to go look up Gauss. As I recall he made the same "discovery" when he was eight years old. Now I find that it goes all the way back to Pythagoras. I don't care. I figured it out all by myself.
    1
  11983. second is better
    1
  11984. Thanks so much for sharing my proof! My new book, Proof and the Art of Mathematics https://www.amazon.com/Proof-Mathematics-Joel-David-Hamkins/dp/0262539799 devotes a whole chapter to this theorem and its generalizations. For example, there is a similar proof by shrinking that works in an arbitrary lattice: the only regular polygons to be found are triangles, squares and hexagons. The cover design of my book is based on this proof.
    1
  11985. 9/17, 4/13
    1
  11986. Huh, I've never heard sinh pronounced as shine. In my region people call in sinch (pronounced the same as the english word cinch). Cosh is the same though.
    1
  11987. In which "dimension" would the operation in question turn in to tetration?
    1
  11988. 34:43 Things got unexpectedly metal :D Not that I'm complaining.
    1
  11989. 36:20 arithmetic solution: blue = square - 4*big triangle + 4*small triangle. big triangle = 1*(1/2)/2 = 1/4. Scale factor between triangles = hypotenuse litte triangle/hypotenuse big triangle = (1/2)/(1^2+(1/2)^2)/(1/2) = (1/2)/(5/4) = 2/5. little triangle = (2/5)^2*(1/4) = 1/25. blue = 1^2 - 4*(1/4) + 4*(1/25) = 4/25.
    1
  11990. Can you prove that the octagons (and such) become arbitrarily small?
    1
  11991. so, the schoolmaster was a german spy, right?
    1
  11992. Very interesting and well-presented!
    1
  11993. I have to give the trophy for most interesting to the old proof by the monk. It was the proof that first brought home infinite sums and their proofs for me. I didnt like infinite sums until I met that proof about 15 years ago.
    1
  11994. I vote for the swivel version of the proof. It's dynamic and fun.
    1
  11995. The most interesting and informative channel !! So good mathematics in this channel
    1
  11996. Great video. For sure, most interesting to me was the sun ;-) of the no-9 series.
    1
  11997. The hardest part of calculus is the algebra, and the hardest part of algebra is the arithmetic. To make calculus easiest, master arithmetic.
    1
  11998. Yes - use the formula at 27:00 (p^2 = (a*c+b*d)^2 + (a*d-b*c)^2) and make c = b, d = a (if a > b). This shows that p is the hypotenuse of an integer right-angled triangle. It is equivalent to the formulas Jaap Scherphuis mentioned.
    1
  11999. ​ @juttagut3695  Right! I thought of that, but then it gives p^2 = (a*b - b*a)^2 + (a^2 + b^2)^2, or p^2 = (a^2 + b^2)^2, which then just gives p = a^2 + b^2 back when you simplify. So then I thought I was mistaken, because I was over simplifying and not paying attention. Nice!
    1
  12000. 11:42 This number box must be | 9 4| |17 13| because 9·17=153 2·(4·13)=104 (9·13)+(4·17)=185
    1
  12001. Aha! Found a way to wake my brain up first thing in the morning. Just as I suspected, infinity always begins and ends with 1. Interesting very interesting and the next one I run into claiming nothing ever changes... I will simply refer them to this very 'simple' video😃
    1
  12002. Just knowing that there is a chance TheOneThreeSeven may never see this saddens me
    1
  12003. The most outstanding fact is the no integer in any arbitrary large consecuent sum of terms excluding 1 from the harmonic series. Just out of words. Who can can say that math isn't beatiful???
    1
  12004. In Dutch we call this kind of proof "bewijs uit het ongerijmde" or "proof from the absurd". One of the most poetic logical terms I know. It is one you should always try when trying to prove anything.
    1
  12005. How the heck could be the 4-color "theorem" is in the first 10 without having a proof???
    1
  12006. Hi
    1
  12007. Cara, a abertura desse canal parece o início do Babooshka da Kate Bush https://youtu.be/6xckBwPdo1c
    1
  12008. A guy does binary substraction and gets 40 mil. views. This is the world we live in people
    1
  12009. Congratulations on the one hundredth vid!
    1
  12010. i am sorry but no. Modern math has this wrong. The area and volume are both infinite. the domain of 1/x is all reals such that x>0, which means that there is not way to have anything other infinity for both area and volume. the horn will forever extend in the positive direction.
    1
  12011.  @Mathologer  I have calculus 3 in college. I am a computer programmer. I am over 50 and seen the things and lengths people have gone to to seem smart. No, the use of infinity is wrong there. The limits says it goes to 2 but we are only ever infinitesimal away forever and ever. It is a paradox in that the same function is infinite in one sense and finite in the other, which is contradictory. We know that logic cannot be true and false at the same time but the law of non-contradiction,. We therefore have something wrong.
    1
  12012. So theres infinit negatives, but theres a bigger infinit of positives. Its not a maths problem its a philosophie one
    1
  12013. Not sure how it works but I’m all 4 it
    1
  12014. One can only understand and entity as a whole if they can look at it as a whole…since the universe is infinite, and we do not have a complete knowledge of infinity, nothing within this system can completely explain how this system works…the key to the universe is not mathematics, but devotion…maybe the goal of life is not understanding, but devotion…devotion is the most important way to achieve ones goal…no amount if explain the universe will help someone become devoted to their purpose in life…you sir, are a briliant man who has devoted his life to a purpose, and have therefore discovered the actual key to the universe…your devotion to your duty to discover and share the glory of mathematics is very inspiring, and increases my devotion to my own duty. Thank you
    1
  12015. The petals appear based on the multiplier…p=m-1…just like the numeric key is one less than the base system Nk=Bs-1
    1
  12016. I can reduce the sum to 1 + 1/3 + 1/3*5 + ..... but then I get stuck
    1
  12017. I have a do nothing grinder. It's a dog.
    1
  12018. Thanks
    1
  12019. cool
    1
  12020. The image shading at 25:40 implies the area was unchanged after the triangle was scaled. Visually this seems impossible.
    1
  12021. excellent work
    1
  12022. Via the inscribed angle theorem, in a 7/3 heptagram, the angle at any star point is exactly half the angle that the opposite corners would make with the center of the star. Since all angles are the same, their sum is half of what the sum for those angles at the center would be. Since those angles naturally add up to tau, since they describe one rotation, the sum of angles at the tips of the 7/3 star is pi. This holds for the regular 7/3 star, and for all irregular 7/3 stars, due to continuity.
    1
  12023. Nothing wrong with this shorter video. My brain often starts to have trouble keeping up with the longer ones. I always envied the sharp guys who could follow the long arguments, "sure, sure, got it", while I'm thinking "wait, how did you get step 6 from step 5?"
    1
  12024. We learned the digiral root trick for 3 and 9 Similarly divisibility by 4 is equivalent tk the last 2 digits being divisible by 4, similar to chdcking the last digit for divisibility by 2
    1
  12025. Hello @mathologer what is the software used to make the animations?
    1
  12026. Chapter 5: It gets better and better was my favorite part. Seeing how gamma and ln(x) make the calculation of the partial sum “child’s play.”
    1
  12027. Combining several different types of polygons into three dimensional objects and treating that similarly results in crystal-like structures?
    1
  12028.  @Mathologer  I am so shamefully lazy that I am hoping for a follow up video from you.. :)
    1
  12029. In the conventional pascal's triangle, the sides are 1s, you fill in the rest by adding the two numbers above, and in the resulting triangle every row sums to the powers of 2. If you alter this to: the sides are 1s and you fill in the rest by adding the two numbers above PLUS the sum of the entire row above, you get a triangle where every row sums to the factorials. I bet there's a way to interpret or prove this with the Moessner framing
    1
  12030. 25.8069 as the root of all evil just killed me!!! I love your work man!! much ♥ .. um this is basically a slide rule right? I mean understanding how we get to a theory or system is important, but this is looking alot like my slide rule.....
    1
  12031. "The first to nail it all down was ..." And i guessed Euler and then you said it. Nice pause!..
    1
  12032. To intuitively understand the fact that the percentage of no 9 numbers is small, one can ask the following: if you write 1000 random digits, what’s the probability that none of them is 9? One can easily see that’s small. That’s the percentage of no-9 numbers within 10^1000
    1
  12033. I always called sinh sinch, which seemed more intuitive. But I had no idea where they came from. Quite impressive and confusing on first pass. Must persevere.
    1
  12034. Great, now I'll have this making extra ticks in the back of my head ;) Great videos !
    1
  12035. I think divergence makes sense if integers are halted. But if objectmotion has complementarity like spacetime, and integers are nonhalting, does divergence make sense? I don't think so. Thanks for the vid! I'm referring to logs as unavoidable.
    1
  12036. 15:51 the area of the tree, as shown, is 5.
    1
  12037. Beautiful!
    1
  12038. Your content never ceases to impress me 🤯🤯🤯 makes regret changing from pure math to applied math...
    1
  12039. Swivel proof
    1
  12040. Because were in a Computer....
    1
  12041. Regarding the final question: the square of the diameter of the curve of constant width PLUS the square of the diameter of the incircle EQUALS the square of the diameter of Conway's circle. If Conway had lived in another era, he could have been the Little Euler.
    1
  12042. For the limit of floor(r^2/3)/r^2 ; We have the identity : x-1<[x]<=x. Therefore ; r^2/3 -1<[r^2/3]<= r^2/3. Then , 1/3-1/r^2<[r^2/3]\r^2<=1/3 By the squeeze theorem , we get Lim [r^2/3]/r^2 = 1/3.
    1
  12043. Of course cosines make sense when dealing with integer numbers! 🤣
    1
  12044. Wow what an elegant proof. So fascinating and a lot of respect to the ancient Talmud solution - the people who created it must had a very good sense of what fairness is in the distribution of assets to creditors... Thanks so much for your very clear illustration.
    1
  12045. CS student, I really liked this video. I personally don't like when some rigor is overlooked for the sake of brevity, but I also understand that the video can't be ten hours long :)
    1
  12046. 9:25 - Taboo part :)
    1
  12047. this is really neat.
    1
  12048. I love your shirt!
    1
  12049. At 26 minutes in, we learn that an American President presented a mathematical proof of the Pythagoras' theorem. I don't think it was Joe Biden.
    1
  12050. Great video, thank you, and thank you for the uploaded correction!
    1
  12051. Mind before this video: 🧠 Mind after this video: 💥
    1
  12052. Eventually became a competent engineer, but lost interest in pure math before ending high school. Calculus and matrices simply seemed irrelevant and a waste of time. Got most of the engineering done by mimicking calculus using Excel iteration function and looking for con or divergence. All good, but then had to get to grips with PID controllers both early hydro-mechanical versions and digital. Got there in the end but OMG wish I had listened more carefully in school or had seen a video like this. Nice work - you're a natural as a great teacher
    1
  12053. Yes I very much indeed love algebra autopilot, the music, the pace, the ending...
    1
  12054. I'll take the 5 pigeons hemisphere.
    1
  12055. I made it to the very end But I m cheating a little bit because I watch your videos whilst ironing. Great stuff btw
    1
  12056. Wow! This so cool.
    1
  12057. We were taught both the proofs back in school. My memory is a bit hazy, but if I remember correctly, the main textbook that we had also had both the proofs in it
    1
  12058. In "Geometric Algebra": A = 1/2 (a^b + b^c + c^a ) or even A = 1/2 (a^b + b^c + c^d + d^a) if you triangulate a rectangle. With "A" the triangle area, "^" the wedge product of two vectors and a,b,c,d perpendicular vectors
    1
  12059. beautiful thought process- real world usages please=or are these just complex puzzles? I mean I can create a problem ( I know the answer) and you are challenge to develop the answer, but how does this work in real life; when no one created a problem while knowing the answer..??? anyway thank you...
    1
  12060. He's not going to pick a random name. He will pick a psuedo-random name.
    1
  12061. Dream proofs are a thing. Good, I thought I was just crazy
    1
  12062. I see a Schwartz lantern as a concertina accordion.
    1
  12063. But what is the sum of the yes 9 series that was kicked out while making the no 9 series?
    1
  12064. Odd number of disks first move to target, even first move to other one.
    1
  12065. But the square dance proof isn't complete, right? It has to be proven that this scenario at 30:00 always works, so that if two "yellow pieces" appear, the board will look the same, and that in the other cases each board is different
    1
  12066. Excellent video
    1
  12067. The biggest surprise for me was the absence of a scene from The Simpsons ... 😎
    1
  12068. For me the most amazing was the crazy maximal overhang stack, because I did not expect it at all. Never thought of having multiple block in the same layer. The animation of Kempner's proof was excellent as well :D Something I have not been confortable with is using the divergence of the logarithm as another proof of the divergence of the harmonic series ([23:30]). If I recall correctly, the proof of the divergence of the logarithm comes from the divergence of the harmonic series in the first place (easy proof: simply take the picture at [22:00] and use the rectangle below the curve instead of above. Their sum is clearly less than ln(n), but it's 1/2+1/3+1/4+1/5... + 1/n ie the harmonic series with its leading 1 missing. Thus the harmonic series diverging makes the logarithm diverge). Is there another easy way to prove the logarithm diverges so the proof is not circular?
    1
  12069. 1:09 when you realize 3 of the top five most beautiful thermos are Euler’s and (4) infinitely many prime was proved by Euler but after Euclid which is likely to what they are referring.
    1
  12070. My mind is extremely blown. Can I please reincarnate into the Phythagorean triple tree?
    1
  12071. i got 9, 4, 13, 17. i factorized 153 into 9 and 17 and then through a not so elegant process of elimantion i simply compared if the smaller addends of 17 plus 9 equals to the second. 1+9 ≠ 16, 2+9 ≠ 15, 3+9 ≠ 14, 4+9 = 13 ✓
    1
  12072. That is a great book. I have a Polish edition of Concrete Mathematics. I love it.
    1
  12073. Awesome video!
    1
  12074. Second
    1
  12075. Actually the twisted squares diagram was added to the Chinese manuscript in the 3rd century CE, almost 800 years after the death of Pythagoras (see here https://en.wikipedia.org/wiki/Zhoubi_Suanjing)
    1
  12076. I just found this channel keep up the great work
    1
  12077. Yes, there is a super important application for this in electrical engineering. It is called symmetrical components. In a 3 phase system there are three wires called phases, and one called neutral. All phases have a sinusoidal current and these currents collect in the neutral. If the system is balanced 2 things happen: 1) The 3 phases have sinusoidal currents of same amplitude but 120 degrees out of phase with each other. 2) When these currents add the cancel out, thus having no need for returning wires, allowing for copper economy In real life the system is never balanced, and that is what the neutral is for (it returns the what is left from an imperfect cancellation) Every sinusoidal current (and voltage) is seen as the horizontal component of a spinning arrow. These arrows form a triangle. It is too hard to study how the system behaves when it is not balanced, because of electromagnetic fields interfere between wires. But these effects cancel out in a symmetrical system. So engineers break the assymmetric triangle into three symmetric ones, called symmetric components. Different kinds of defects affects each component differently, so we study each one separately, as a kind of sygnature. You showed my a different way to filter out each component. Thank you. Worth noting: in Russia the system is 5 phase, but the same ideas apply.
    1
  12078. By the way, I recall a few years ago solving an Euler problem involving computing the number of ways of making change for 2 (British) pounds using British currency, but I did it the hard way, by brute force! I got the right answer, but my solution involved computer code with about 10 nested loops! I wish I'd known at the time this much simpler method!
    1
  12079. 🙂 👍🏻
    1
  12080. Kempner's proof animation very cool
    1
  12081. Most memorable: one glance balancing.
    1
  12082. In the proof that you can reach pi exactly we didn't provide an exact sequence of how many positive and then negative components etc. we take. I wonder if it's possible to derive an exact formula for that like we did for any number of the form ln(n/m) by always taking m positive and then n negative numbers.
    1
  12083. This was a great video, I hope to show it to a class one day when explaining calc 2 to them
    1
  12084. For the first million entry at least, I only found these: Number: 1 Position : 0 Number: 2 Position : 1 Number: 4 Position : 2 Number: 8 Position : 3 Number: 16 Position : 4 Number: 256 Position : 9 :(
    1
  12085. If you wanted to have a horn with volume 8, you could extend the "bell" towards the y-axis, to (if my crazy calculations are correct) pi/8, you will have a horn with volume 8. BTW, love the harmonics on your shirt!
    1
  12086. It's so beautiful.
    1
  12087. This video really blew my mind! By the way, my vote goes for gamma!
    1
  12088. 17:56 But Octogon *DOES* exist on the square grid! Isn't it? No no sorry 😁The model i had in my mind was an Octogon with 4 of segment lengths being 1 and other 4 being √2
    1
  12089. Frequently unexpected… Always interesting…
    1
  12090. You forgot dooper
    1
  12091. vortex math, more like bollocks math
    1
  12092. Thanks!
    1
  12093. Might it be possible if you used hyperreal spacing?
    1
  12094. Video Request (is there a place for vid requests?): Counting parallelograms in a triangle with each side cut into n equal parts, and all lines connecting these points and parallel to each side drawn. https://www.youtube.com/watch?v=UiJFl072kdw She writes a formula at 4:20. I know the formula, but I don't understand it. Where does it come from / why does it work? How is it related to Gauss's Sum(n)|[n=1 to N] = N(N+1)/2? How is it related to similar problems (say #of rectangles in an orthogonal grid of known dimensions) Seems like it's esoteric and complicated enough to warrant your treatment :D I hope it's related to some crazy physics, like how cells work or something, i don't know, but I would love to have your take. Apologies if it's already been done, I couldn't find an explanation for the formula. This problem reminded me of your Christmas water bucket video , because, ...the diagrams are similar. :) Much love, thank you for all your hard work, it's a joy
    1
  12095. Left field ( apologies). Given the necessity of triangulation for calculating a given position in 2/3 dimensions. Could this possibly be useful in calculating positions including time/space? Also , within acoustics, man made oscillations can and will produce constant dissonance, whereas natural oscillation will and does resolve into harmony. I'm thinking Petr's miracle may well explain why?
    1
  12096. I suppose information is being lost in each generation as pairs of cells are combined into one and it's irreversible in that we don't know which colour was on the left. With the level 10 triangle, only 50% of the leftmost cell and 50% of the rightmost cell makes it to the last generation with all information about the eight intervening cells disappearing. Reversible rules with the same number of square cells in each generation might be interesting. Presumably the generations would cycle eventually.
    1
  12097. Yes, any row determines everything below, but not everything above :)
    1
  12098. 😈666 thousand subscribers😈
    1
  12099. Faster way to make the next fraction: add numerator + denominator, then add both previous denominators. 1+1=2, 2+1=3, 3+2=5, 5+2=7 ... 239+169=408, 408+169=577...
    1
  12100. That 666...if there's a god joke was fantastic!!
    1
  12101. Why is it that p divides only the blue or the red? Isn’t it supposed to divide everything on the right side?
    1
  12102. Most memorable are the optimal towers. Never thought an optimal solution can look this ugly
    1
  12103. Those sums are related to a pattern in series expansions for log (ln) and certain zeta/eta functions. Imagine the sum of all Reimann zeta(n)/n .. Write it out in a matrix, with columns being the zeta series. Look at the rows of the matrix. These are sums for log (x-1)-log(x) with x being the row number (so your first row does have log(0)... disregard this for now)). Now go ahead and flip the signs of the zetas. make them (-1)^(n+1) zeta(n) / n... This makes the rows log(x+1)-log(x). We get rid of that log(0). Now do zeta to Dirichlet eta conversion on all the columns. Just switch every row's sign to make the etas. Look for the Wallis product for Pi in the log rows. Convert the etas to zetas, using zeta(s) = eta(s) / (1-2^(1-s)). Now make them etas again by flipping row signs. You now have log(pi/4). Repeat the past 2 steps.... you'll find that... well... (1-2^(1-s)) applied repeatedly is, well, essentially a binomial expansion, and certain patterns arise in the logarithms that each row sums up to.
    1
  12104. 3:35, also known as 3blue1blue iris.
    1
  12105. Something I figured a while ago is that the cubes (1, 8, 27, 64, 125) go up by a series of numbers (7, 19, 37, 61) that go up by a series of numbers (12, 18, 24) that go up by 6, and there’s something like that for squares and powers of 4 too
    1
  12106. Natural log of ln(google +1) + gamma = 230 If by the time this function gets to 1000 the universe has ended Is it really infinite?
    1
  12107. What in the world is that a cardioid I see forming in the negative space of the two hexagons with a shared vertex?!
    1
  12108. 18:36 is it my or does this second solution looks one move longer?
    1
  12109. One day, buried within plant-growth genetics, we will probably discover that Nature has been using this Triangles Property for millions of years.
    1
  12110. God... what a brainstorm!
    1
  12111. Math hack!
    1
  12112. Great ! I was very suspicious of the list of best fractional approximations of pi given in the first video, but I did not expect them to behave so nicely ! Thanks for the reupload :) By the way, do you know how to determine all best fractional approximations of a given number using its continued fraction expansion ? It looks feasible, and perhaps then one could prove that the patterns we observe for pi are true for arbitrary irrational numbers ! (ie, the list of the denominators of these fractions is a concatenation of finite arithmetic sequences (with possibly only one term, as for the golden ratio) whose ending terms are given by the odd order truncations of the continued fraction, and whose reason (or common difference ? that's the way we say it in french) is the last term of the previous arithmetic sequence) ! Plus it's been quite some time since your last video about continued fractions !
    1
  12113. Ia this related to the fact that taking any random polygon and following the step to create new vertices from the existing vertices by using their mid points you eventually get to a polygon approximating an ellipse (I think at the limit all vertices will be on an ellipse)? This page shows a proof of that by...you guessed it... Fourier Analysis: https://almostcomplexstructure.wordpress.com/2020/04/16/from-polygon-to-ellipse-a-proof/
    1
  12114. Pet cubic :-)
    1
  12115. The no9er series (and relatives) look like fractals, at least the graphical proof does. I wonder, what the fractal dimensions would be and if there is a relation to the finite sums.
    1
  12116. I enjoyed the 700 year old proof. I'm continually amazed by all the sheer mathematical genius that has come before us.
    1
  12117. Dat intro do
    1
  12118. Also, gamma > 0.5 because in every "rectangle" blue area > white area
    1
  12119. Might be a silly question, but we see that 2^n is the discrete analogue of e^x, in terms of differencing/differentiation. Is there a continuum between those? So, some sort of sense in which you can move from discrete differencing to continuous differentiation? In which case, the analogue of e would go from 2 to e in a (what I assume would be) continuous manner.
    1
  12120. Am I mistaken, or can’t you just label all the adjacencies 1 (instead of some being i) and take the permanent instead of the determinant?
    1
  12121. 15:50 Every level adds 1 to the total sum area. so if their are 5 levels so the area is 5
    1
  12122. Loveit
    1
  12123. there is a few things I never knew. very nice and informative presentation 🙏 thank you
    1
  12124. Yeah, you rule.
    1
  12125. The thing that surprised me most was the sum of 100 zeros > sum of no 9s. I knew no 9s did not converge. Numberphile did a video that the % of numbers with the digit 3 grows to 100, which is weird. But it generalized, so I knew most numbers had a 9 in them. So my question is: If The percent of numbers with a nine grows to 100, how can the sun of the series be anything but zero? All numbers are excluded! Yes, infinities are weird. But it is a fair question.
    1
  12126. Stunningly beautiful!
    1
  12127. I liked best the optimal leaning tower.That is so cool. Also, I am so impressed: Euler supports you. Even though the greatest maths-man is dead, he likes your work enough to support you through Patreon.
    1
  12128. He said first that the sum is going to explode in infinity , and pooof he says later thatbit converges to γ just like the ramanujan serie of all positive integers
    1
  12129. I like your shirt but it bothers me.
    1
  12130. Solution to why gamma > 0.5: It's made up of a bunch of those individual blue blocks of area. Surround each of those blue blocks with the smallest rectangle that encloses it. Each block takes up just over half of its corresponding rectangle. The sum of all those rectangles makes up the 1x1 rectangle. So the blue shaded region of the 1x1 rectangle takes up more than half of it.
    1
  12131. 👏👏👏 So excellent lecture. Thank you, Doctor, your scientists and your colleagues. and With luck and more power to you. hoping for more videos.
    1
  12132. For a general nxn square (in terms of side length, so we have an n+1 by n+1 grid), the total number of squares is given by the following equation: sum{k=1}{n} [(n+1-k)^2 * (k)]= sum{k=1}{n} [(#of bounding squares of side-length k) * (number of squares where all four corners "kiss" a box of side length k)] Note that every rotated square can be "bounded" by an unrotated square. Consider the largest squares which can be fit into a box of side-length k (i.e. all corners kissing the boundary). For example, a box of size 1 has one unrotated square inside of it. A box of size 2 has 2 largest squares (one unrotated one, and one rotated 45 degrees). A box of size 3 has 3 squares it bounds, so on so forth. Continuing this pattern, a k x k square has exactly k rotated squares where all four corners kiss the boundary. Going in the opposite direction, note that every square we find in our grid IMPLIES the existence of a larger bounding box in our grid (so we only need to find these bounding boxes). (i.e. unique bounding box inside of our grid for every rotated square.) Thus, we count the number of bounding squares of size k, which is (n+1-k)^2 (the incomplete formula that we had earlier in the video) Each bounding square yields k unique rotated squares Putting these all together for bounding squares of all sizes yields us the formula sum{k=1}{n} [(n+1-k)^2 * (k)]
    1
  12133. Your channel usual white background is getting genuinely painful. I can't watch at late hours as it lights-up the all room like it was midday. For a few episodes i've been watching with the `filter: invert();` css rule applied on the player. Literally. May i humbly suggest a tiny try at inverting the colors ?
    1
  12134. Très 😎
    1
  12135. You know you are a maths Mafia boss when you make a video to correct numberphile 😁
    1
  12136. why did they scalp Nicole Oresme?
    1
  12137. fibonazi pizza.
    1
  12138. Always love your videos and this one was particularly engaging. Have a degree from Caltech, so following most of what you did was no problem. Didn't see the transform from f(0) -> f(1) at around the 40 minute mark, but I just took your word that it was correct. Such is the trust that I have in your abilities. :-) I have come across the Bernoulli numbers several times but never really knew what they were or where they came from, so this was a real education.
    1
  12139. hey you all, comment what you want, but you cannot beat 3:23. just the driest joke, dry like e.g. the paper of the Sitzungsberichte ........ der Bayrischen Akad. d. Wiss. zu München
    1
  12140. I actually guessed right wth
    1
  12141. Here is one way to show that the Euler Mascheroni constant is greater than 1/2. Let's see how much does each blue piece at 24:20 contribute to the sum in proportion to the area of the rectangle which contains it. To be clear, the top horizontal segments of each piece split the 1x1 square in rectangles of length 1 and height 1/n(n+1). Now notice that each blue piece is defined by a convex curve joining two corners of its rectangle and thus occupies more than half the area. Since this is true for every rectangle, the total sum must be greater than 1/2. : )
    1
  12142. it's Pascal triangle from different prospective
    1
  12143. If you had all the powers of two, you coud make all the numbers. It is a lot like a binary system.
    1
  12144. They never taught this proof, just expected us to accept that formula! This proof is so elegant! I suspect the schools make things much harder than they need to be.
    1
  12145. I got so excited about the final proof ! Thank you very much Mathologer.
    1
  12146. Most memorable: Visualization of the Euler-Mascheroni constant in the 1x1 square that unambiguously demonstrates why the value is between 0.5 and 1.
    1
  12147. It's 1 45 am rn but mathologer uploaded
    1
  12148. I was familiar with the 2nd of the 2 "twisted square" proofs. Don't know where I first came across it, because it was so long ago. The first one (the simpler of the two), I actually hadn't seen before! Beautiful!
    1
  12149. i don't think that the molecule on your shirt is stable
    1
  12150.  @Mathologer  Thanks! It immediately reminded me of this one, of course. The one in this Heron’s formula video was just perfectly timed for some reason. And might I say, just that extra tad more magical.
    1
  12151. When I saw the inverted spere I instantly thought singularity. Anyone else? And is there any merit to that?
    1
  12152. I hated coding algorithms for the TSP, but that's because I don't like coding. My computer science degree was a mistake
    1
  12153. Today I made my own proof of this
    1
  12154. 2020 will be over soon i hope
    1
  12155. My Dad received one of these as a business gift before any electronic calculators had come into being. He gave it to me just before we were required to buy a slide rule for Physics class in high school (1971/72). To the skepticism of those around me, I learned how to use my free circular slide rule while everyone else took instruction on using a straight one, and became quite adept with it. No answer was ever "off the scale!"
    1
  12156. Put the point on the edge of the small circles. Now, your star gets really pointy.
    1
  12157. Can we get a video all about how math would change when we change the base? Base 12 is very popular.
    1
  12158. Most math would not change in the slightest. The only thing which could possibly change are things which depend on using the digits of the numbers you're working with. So, like in this video, divisibility tests change. But anything which doesn't rely on the digits of a number (such as which numbers are prime, which are composite, calculus, abstract algebra, geometry, etc.) wouldn't change in the slightest, since they don't depend at all on the digits.
    1
  12159. Ok video paused. If i draw a circle connecting the three points of the triangle, then add some lines equivalent around it, im just doing a sort of scaling up. Sorry it makes easy sense to look at it but i dont write maths.
    1
  12160. Ooo. Ya, each chord is a combo of all three sides, thus the same. Neat!
    1
  12161. Now im curious about the offset of the circle on each point vs conways circles...
    1
  12162. I knew it, there was something fishy about that vortex myth
    1
  12163. I was surprised at the no 9 series, I hadn't seen this before but it makes complete sense. Always happy when there's a new mathologer video!
    1
  12164. "Mathematicians are weird, get used to it"
    1
  12165. "the first to nail it all down with a proof was Leonard Euler". Well that was a surprise...
    1
  12166. Leaning tower was very cool, I never thought about the most block efficient way to form the longest overhang. Very nice! Btw I'm student from Monash... I love your videos.
    1
  12167. I couldn't resist and calculated the possible change for 1€: it's 4563. I didn't expect the difference to 1$ (293) this big. Apropos big, here's the polynomial as provided by GiNaC, enjoy: 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    1
  12168. I pronounce sinh like "cinch" (sɪntʃ in IPA or maybe sɪnʃ depending accent).
    1
  12169. 11:15 I'm now entitled to say "I know calculus!"
    1
  12170. Fascinating. Btw.: I think your shirt has a rounding error... 😁
    1
  12171. I love his shirt this month. Anyone know where I can get one
    1
  12172. I suppose the most remarkable thing to me is that it's unknown whether the Euler-Mascheroni constant is rational or irrational, let alone whether it is transcendental. (I'd imagine most mathematicians would wager that it is transcendental, and there are a lot of results that come close to proving this, but it is an open question.)
    1
  12173. Not magic. But a very convenient aspect of nature! And a fascinating one. It further supports an idea I call the Ultra-Anthropic Principle: Not only do we live in a universe which allows us to exist, we live in the best of all possible universes in which we could exist.
    1
  12174. This is the most compelling proof I've ever seen. It's truly miraculous as you say.
    1
  12175. Every shape that has this property is unbounded: suppose a shape is bounded, then there is a unique rectangle aligned with the transformation axes that snuggly fits it (proof below). Note that this rectangle has nonzero area as it contains the shape. After transformation, that is a different rectangle, but it still fits snuggly. So it can't be the same shape, as that rectangle was supposedly unique. Existence of the rectangle: bounded shapes fit in a ball, and this ball fits into a rectangle. Take the intersection of all encompassing rectangles, it's still a rectangle and there can be no smaller, as it takes part in this intersection. Uniqueness of the rectangle: if there would be two, the shape would fit in their intersection, which would be smaller.
    1
  12176. Im taking pre calculus in highschool right now, I'm 15. These videos are extremely interesting 😸😃
    1
  12177. Are there equivalents for this formula for the other regular n-polytopes?
    1
  12178. Beautiful! The last animation makes me think of a rotary engine! Constant diameter shape seems like another way to design a rotary engine where the outer part could be simpler than the one actually used in rotary engines.
    1
  12179. You can find sum of harmonic series with any common difference using this formula :- integrate (e^(-ax) - e^-(a+nk))/(1-e^(-kx)) from 0 to infinity and you will get your sum. Here 'a' is first term and 'n' is number of terms upto which you want to find the sum and 'k' represents common difference.
    1
  12180. And now what is the use of this paradox and its solution - none. In terms of practical value, it corresponds to spitting at a distance. Accordingly, the benefit of this video is zero.
    1
  12181. Astounding and beautiful. Bravo!
    1
  12182. Based-10 counting system :)))))) 6*9=42. M. Douglas English person will never make joke in base-13. I disagree 12 hour of day is 2 modules of 12or 60 seconds and 60 minutes based system of degrees if pi=360 was 334 45degree 42 degree is much more reliable 336=pi. 3+3+6=12 (12 hour of north) 336/2=168 1+6+8=15=6 (6 hour) 84=8+4=12=1+3= 3 hour east 252=2+5+2=9 west or 9 hour why we use 360 ? because one year is about 364 days. Maybe Marko Rodin who popularized or Tesla who discovered have something. 360 is just convenience and people lazy and like round numbers 10 better than 9 because that is round, same with 2000 years celebration was not millennium, 2001 was new millennium but 2000 was nice round number and all people jump on advertainment train of new millennium.
    1
  12183. On a scale of 1-10 the argument at 10:14 broke the scale.
    1
  12184. Reminds me of integrating powers of x.
    1
  12185. Uh oh. I'm not a monkey.
    1
  12186. At first, I thought this wasn't so interesting as any integer not of the form 4^n(8m + 7) can be written as a sum of three square but then I noticed that they're consecutive...
    1
  12187. 19:11 Damn... couldn't you use rainbow order for colors instead?
    1
  12188. Brillant !!! Thanks for this beautifull démo !!! It seems not only Indian mathematicians but also precolombian ones knew a bit of maths nobody knows about... https://youtu.be/qszK66xJZp4
    1
  12189. What if every point is directly next to each other, spacing is 0, then while its also still a point you also have an infinite number of points and an infinit number of 0 length side triangles on the grid. (Its a trick to use a singularity of course, I know, but why do we never mention it?)
    1
  12190. Counterexample to the idea that simply connected, equal-color grids can be tiled at least one way: 3x4 "H" shape.
    1
  12191. If you go to OEIS web page, and the A217290 sequence, you will have an URL in order to view the Willy Scherer paper. German written text.
    1
  12192. Twenty eight plus greater than.
    1
  12193. 2 3 5 7 11
    1
  12194. with the given rules, max turns before termination is 55 with a set of 10 cards
    1
  12195. I made it to around 30 or 40 minutes mark but watched the rest anyway. Im a highschool student but I'd love for you to do more complicated stuff just so we'd know all the different beauties in mathematics!
    1
  12196. At 29:10, did the third row from the top get an entry left out? Shouldn't it be "1 3 5 7 9"? The row above it seems to treat the 7 as a 5 in any event.
    1
  12197. Sorry, I didn't see the correction at the bottom of the description and now YT does not want to let me delete this comment.
    1
  12198. a= lim as x approaches 0 of function ln(x).
    1
  12199. You sir, are a (YouTube) treasure. Thank you!
    1
  12200. Of course I made it to the end! What an amazing topic!
    1
  12201.  @Mathologer  🤓
    1
  12202. Nais one.😢
    1
  12203. @Mathologer Thanks as always for amazing effort. Seems like the distribution that maximizes the minimum happiness across all coalitions is the average of the "extreme" distributions, where an extreme distribution is one that makes someone maximally happy, a next one the next best happy, a next next one the next next most happy, etc. (for the case of 200 level of estate, the 4 extreme distributions to the 3 creditors of (100, 200, 300) are: (100,100,0), (100,0,100), (0,200,0), (0,0,200), and whose average is (50,75,75)).
    1
  12204. There is a more beautiful non-euclidian geometric solution, have a think.
    1
  12205. Well, you can scale up a side of the triangle by copying it n times, slanting them, and putting them together
    1
  12206. It took me way too long to connect the Gregory-Newton formula as the discrete version of a Taylor series and I am deeply disappointed in myself. Edit: at least I got there before you spoiled it near the end of the video, so I still have a few shreds of dignity.
    1
  12207. Is there a way of defining what you can minimally remove from the harmonic series to make it finite. Or, related, what you can maximally remove and it still be infinite.
    1
  12208. best parts when he says pchkh
    1
  12209. Thank you for this :)
    1
  12210. 28:26 can I also add that as a colorblind person I can't see the difference --- it's black and two reds
    1
  12211. Lieber Mathologer, vielleicht wäre es sowohl für Dich als auch für Deine wissbegierige Gemeinde interessant, etwas über die Kaprekar-Konstanten zu erfahren. Mir ist bisher nicht gelungen, einen Beweis zu finden. Vielen Dank im Voraus!
    1
  12212. There is any video that explains to the layman how Von Newman conceived non integer-dimensional lattices in his "continuous geometry"? Did he used fractals?
    1
  12213. If all the holes can be tiled by dominoes, then the determinant formula works perfectly.
    1
  12214. Mathologer Rules OK. Best wishes infinitely
    1
  12215. Voting for the lining tower of lire.. that's the best real life example of the harmonic series
    1
  12216. outstanding! MIND B L O W N
    1
  12217. The answer is 42!
    1
  12218. It really relates to math <3
    1
  12219. Swivel preferred...
    1
  12220. 24:35 WTF? Wasn't the arc length and the area the same number in radians when the radius is 1?
    1
  12221. Thanks for putting out such great content. This is way better than my video on generating functions. Do you have any tips for someone with a new math channel looking to improve?
    1
  12222. Could you please do an animation of this if you were to use the harmonic series instead?!?! As in... X × 1, ×2, ×3, ×4 etc?!?!? I'd be VERY curious how the harmonic series would translate into this. I feel as though you'd see a very interesting pattern.
    1
  12223. In my opinion the most memorable part is the first, the one glance balance. I can't evaluate how much I spent thinking about balancing a system; I'm genuinely amazed.
    1
  12224. Fantastic video!!!
    1
  12225. 19:49
    1
  12226. I have a truly marvelous proof of this theorem, which this Youtube comment is too narrow to contain.
    1
  12227. It's merely a passing fad. My rotary abacus can do all the maths I'll ever need. ;)
    1
  12228. I achieved the same result by setting the answer equal to a+b*sqrt(2) and equating rational and irrational parts on both sides.
    1
  12229. There is an infinity that occurs in the real universe: the redshift at the event horizon of a black hole is infinite. This infinity is causing a lot of controversy in physics.
    1
  12230. It seems that there is a simpler proof, merely by inspection, that there is no possible equilateral triangle on the square grid:; dividing an equilateral with side length 2 into two right triangles, you would have to have the perpendicular sides of lengths 1 and root 3. But root 3 is irrational, so, even on an infinite grid, no matter how much you scale up your equilateral, it can never fit perfectly. This seems so simple, I fear I must be missing something, but cannot tell what?
    1
  12231. 20:29 2=1+1/2+1/3+1/6, 3=1+1/2+1/3+1/6+1/4+1/12+1/7+1/42+1/5+1/20+1/13+1/156+1/8+1/56+1/43+1/1806 Not really pretty, but since I couldn't see anyone commenting about it, I guess it's fine.
    1
  12232. please learn how to use the audio compressor filter because youtube's "stable volume" is a joke
    1
  12233. The fact that the no 9s sum converges is memorable, though it also begs the question of what happens in other bases, it makes me wonder if there is a base for which removing a digit the sum still diverges, there probably is one as the primes also thin out but still diverge
    1
  12234. Programmer here: Can you elaborate what you want for the Aztec diamond thing? Like do you want something that just shows the whole process of random generation of a single aztec tile of any size? Also its a little tricky to think not a good internal representation of an Aztec board tiling off the top of my head. Could you link a paper or introductory material that constructs the set of aztec diamonds without using visuals? I can probably solve it without doing that but it would likely speed things up and improve the quality of code it I could start from a known-useful representation
    1
  12235. If you take a look at SVG files coding, you will recognize that the arc formula makes more sense now. 😁 It has the same behavior in angles and swapping when an angle is greater than 180°
    1
  12236. Why is a green tile disappearing at 12:30?
    1
  12237. It’s really hard to choose the most memorable part of this master class . The whole video is very interesting. If I have to choose I would say that the fact that no consecutive terms of the harmonic series could add up to an integer is the most surprising and memorable
    1
  12238. Awesome video!
    1
  12239. So the generalized formula is that if there are n squared terms on the left side and (n - 1) squared terms on the left side, the smallest number in the equation will be (n - 1)(2n - 1). In other words, (n - 1) times the nth odd number. Pretty cool!
    1
  12240. Just one word: ASTONISHING
    1
  12241. Application for Petr's miracle...? How about the "cocked hat" problem in celestial navigation? Typically, when trying to achieve a celestial fix using a sextant on a number of celestial objects (stars, moon, sun, planets) you achieve a number of "lines of position" on a chart. If the sighting were perfect, all the position lines would meet at one point, your actual location. But in fact there are experimental errors in the sightings so that for three sightings on three celestial objects, you get a triangle formed by the position lines, the "cocked hat". For sightings on five celestial objects, you would get a pentagon. There has been a long running debate in the navigation community on how to recover your actual position from the random polygon of your sightings. Petr's miracle looks like a possible solution...?
    1
  12242. I think the swivel proof was more visually appealing but the coloring proof was more clearly a proof that that holds for any configuration and not just this one. It was more obvious with the way it was colored that this is always the case.
    1
  12243. using these processes is this how mathematicians turn a sphere inside out without ripping or bending it
    1
  12244. For hexagon case, number of different colored tiles are the same - that is obvious when you start accepting the picture as 3D. Every of the 3 sides has the same area, 1/3 of the total number of tiles, because no of these sides has gaps or holes. So when looking from perspective from top/left/right side, you'll see it completely covered with the appropriate color. That's really nice!
    1
  12245. THE POWER OF MODULAR INVERSES!
    1
  12246. 24:24 But, do they all have the same chance of showing up? Maybe I will test in program.
    1
  12247. LLLR 9 4 17 13
    1
  12248. Start by factoring 153, it has factors 1x153, 3x51, 9x17. possible values of b then are 76, 24, and 4 the smaller factor of each pair=a the larger factor=a+2b factor (104/2)=52, it has factors 1x52,2x26,4x13 the smaller factor=b and the larger factor=a+b the only b that matches both is 4, and 9+4=13. On the chart because 4 is in the upper left corner, either the previous step was x 4 9 y or x y 9 4 the first option is the only one that avoids negatives and results in 1 4 9 5 the parent of this is is then 1 3 7 4 and the parent of this is 1 2 5 3 reading forwards is LLLR
    1
  12249. Analog computers: they are library functions of the universe's API !
    1
  12250. Lol
    1
  12251. Elon Musk probably won't be pleased to hear that the Tesla vortex is actually nothing special... 😉
    1
  12252. the equations at the beginning and at the end are different, at first the denominator has odd numbers.
    1
  12253. I'm about to cry, thank you for the proof at the end
    1
  12254. long time no see!
    1
  12255. Already having a decent grasp of binary I preferred Nathan's approach. As soon as he wrote them out as binary I could see where he was going with it.
    1
  12256. Is it complicated to prove that (1+1/x)^x=e , rather than just trying a large value? This part seemed a bit too "fast" maybe, otherwise great video
    1
  12257. I made it to the very end.
    1
  12258. Can you make a video of real world uses of this type of math? For us dummies please.
    1
  12259. not much use for math in australia. the till at the bottle-o does all the calculations that are required
    1
  12260. There is another shape with similar properties to the circle and hyperbola. It too has it's own form of Euler's formula and its own trig identities, though perhaps the most shocking part is how simple they all are; too simple, until you realize that it really is the trivial case. That, along with the hyperbola, and the circle form the foundation of the ℂomplex numbers and their 2 cousins.
    1
  12261. 11:46 Hmmm.. that is a simplex (i.e. a pyramid with a three-sided base), so you could inscribe it in a sphere. Is there no end to this? 😁
    1
  12262. The most memorable part is the leaning tower, a true beauty that anybody can appreciate in my opinion.
    1
  12263. I can't pick a favorite. My observation regarding "favorite" is this is a high favorite Maths discussions because it conveys some modern Maths results approachable by non-"real" Mathematicians, like me.
    1
  12264. I check everyday for a new Mathologer video. I might not understand it, but it's the only thing I look forward to.
    1
  12265. At 6:30, are you sure that is not 3.18? Turning the wheel the other way, you could also unwrap the numbers >10.
    1
  12266. more like the key to base 10
    1
  12267. Easily the best channel on YouTube. Yes, I said it.
    1
  12268. The best advise I can give to people is...be polite,be on time,learn to read a tape measure...I enjoyed the lesson
    1
  12269. Why was it not 4k+1 and 4k-1?
    1
  12270. Tristan's fractal was gorgeous and brilliant!!!
    1
  12271. Perfect interpretation for per fact of visual al-go-rithms🌹
    1
  12272. These animations are neat as a pin!
    1
  12273. Another thing to point out: the inverted inverted triangle has dimensions that are exactly a third of the original.
    1
  12274. Anyone see spiderman in the 27 loop?
    1
  12275. I liked the last proof the most.
    1
  12276. I was using power sums recently to evaluate how much operations my LDLT-based symmetric matrix inversion method takes (needed to implement it on obscure embedded system). Now in your video I've seen triangular matrix and first thought: "Nice! Triangular matrices are so easy to work with!". Then shouted to the screen: "DONT INVERT IT, JUST USE FORVARD SUBSTITUTION FOR PARTICULAR VECTOR OF N POWERS!!!". Yup, computational math does something with people :)
    1
  12277. Ptolemy was geocentrist, that's why he fell out of favor. Also, he was favored by the papal church, and that was a red flag for the thinkers in Renascence.
    1
  12278. At 4:50, I'm not sure this is that impressive. I could do the same with a D6.
    1
  12279. Mathematics major and I loved all of it
    1
  12280. Amazing video, as usual!!
    1
  12281. I love your videos professor thank you Love from India
    1
  12282. Before you even start, where can I get that T-shirt please?!
    1
  12283. Brain blast at 14:33
    1
  12284. if zeta(1) is infinite and zeta(2) is finite is there a value of s that is the inflection point from finite to infine?
    1
  12285. I always roll my eyes when people talk about galaxies and nautilus shells in relation to phi, it's not true, but also just observing patterns is far less interesting than understanding them. This meant I had always had a "resistant" feeling towards phi. Mathologer has well and truly turned me into a phi appreciator, first with the continued fraction representation explanation, and now this, I'd somehow never seen Binet's formula, or at least if I had, I had forgotten it, but the derivation and wider connections was absolutely wonderful. I do want to note that as a colour blind viewer, I do struggle with some of the coloured lines in some videos (at the start), it's not an easy thing to solve, I know I tend to prefer dashed, dotted, double, ect. lines instead of colour indicators but that can get messy fast, it's just something maybe worth considering.
    1
  12286. I liked the Euler-Mascheroni constant (γ)
    1
  12287. What is phi?
    1
  12288. Thanks for your amazing content!
    1
  12289. Collapse fold, twist reapply yzA
    1
  12290. I didn't understand any of this but you still get a thumbs up from me my guy
    1
  12291. There were a year or so ago, probably still available, Russian made circular slide rules which were pocketable, much handier that the 12-inch log straight ones. But they were all a bit more than I wanted to pay - So I'll stick with my British Thornton one for now.
    1
  12292. 1+1 ='s 11
    1
  12293. omg, i want this t shirt
    1
  12294. I tried my hand at programming the Aztec Diamond tiling in Python. It isn't the prettiest, but it works! https://github.com/ginop/AztecDiamonds
    1
  12295. I really liked the whole video, but somehow the fact that there are no integers among the partial sums is what impressed me the most! So it gets my vote :) Thank you for this amazing video, as a math student I enjoyed it a lot and discovered very interesting facts!!
    1
  12296.  @Mathologer  wow, math really is unintuitive sometimes
    1
  12297. Since you asked, I'm a recently retired computer scientist (MS/BS in CS). Math background is 9 credits in calculus, 6 in statistics, 3 in theory of computation, 3 in matrix algebra, 3 in electrical engineering where we used matrix algebra to build circuits (I'd forgotten how wonderful matrix algebra is). I'd never before seen Pascal's triangle or Bernoulli numbers. I started watching your videos after viewing Numberphile's disastrous -1/12 infinite series. Many thanks for that!
    1
  12298. Do they have to be actual hexagons, or can I just use cubes as viewed from one corner?
    1
  12299. I don't agree with 4:56 I can enclose an infinite area in a finite area. For example put a Menger sponge in a sphere.
    1
  12300. 8:34 Ah, simple binary math.
    1
  12301. 29:30 : Lambert avenging Antarctica :D
    1
  12302. No matter which operation on those 15 cards, 1,8 and 15 stay put, no matter how many times you permute with R↑C↓ or C↑D↓ or R↑D↓ That's the real magic here, the math is just a bonus.
    1
  12303. Merry Christmas 🎅 and a marvelous video.
    1
  12304. his t-shirt burnt my brain
    1
  12305. Embrace measure theory and your sum is no longer ordered
    1
  12306. The length of each strut of an iris is equal to the perimeter of the triangle. Thus, the midpoint of each strut is at the semi-perimeter! There must be some connection with the various formula involving the semi-perimeter, such as Heron's Formula!
    1
  12307. 4, 9, 13, 17
    1
  12308. When a mathematician tells you a video game character is always greater than you...
    1
  12309. 6:04 but now the air is going to get trapped in the top and prevent them from filling all the way :)
    1
  12310. cool. now do multivariate :p
    1
  12311. wut
    1
  12312. The YouTube math G.O.A.T.
    1
  12313. @Mathologer  great observation skills! 🏆 I hosted an event about 15 years ago where someone was showing off an Enigma machine, and we were allowed to touch and handle it. 😀
    1
  12314.  @Mathologer  Not Simon Singh, I have never had the privilege of meeting him. Nor James Grime, who I have met. I am in a pub, and don't have the details to hand - wouldn't be too helpful as I only knew their handle/nick. If you have a 3d printer I commend my friend Craig Heath's "Replica Enigma Machine Rotor" model on Printables.
    1
  12315. Are you a French native speaker? Because the way you said "récréation" witch such a perfect accent is impressive.
    1
  12316. I was introduced to the concept at some point in grade 6 math, but we did not work through the algebraic proof until 9th grade geometry. It was part of our introduction to proofs.
    1
  12317. I like the three color proof best
    1
  12318. I learned of Pythagoras, in Freshman Algebra, Fall, 1962, and how to solve his famous theorem, in Mrs Ferguson's 10th Grade Geometry class, Fall, 1963, where I also heard the news of the murder of JFK, a few moments before noon, Nov 22, 1963. I have used the Pythagorean theory much of my professional life, for a multitude of calculations, along with various other proofs that apply to area, volume and related derivatives. I cannot imagine not understanding this simple, infallible tool, although I am not surprised by young people who haven't a clue.
    1
  12319. Google E6-B flight computer. Circular slide rules have been around for years.
    1
  12320. Spot the chap who comments before watching all the way through...
    1
  12321.  @jonathanrichards593  Guilty as charged
    1
  12322. Thanks!
    1
  12323. Great video as always - thank you. I was very confused in chapter 2 when the formula yielded 88 tiling solutions when there are an unequal number of green and black squares. Rewatching the video I can see that the green "5" tile vanishes somewhere around 12:35 which I guess was a mistake. Anyway, great video, thanks
    1
  12324. So ... use crisscross or zigzag and the rest is waist of time
    1
  12325.  @Mathologer  That's odd. Boot camp taught me "crisscross right-over-left as worn and no bridges" as the only acceptable way of lacing. ("no bridges" means the horizontal segment closest to the toe goes on the inside (not outside) of the shoe.)
    1
  12326. Really enjoy this kind of videos. I wasted another day to rewind again and again to really learn what you write and try it myself. Just like in the old times at university calculus class. A wonderful way to waste my holidays. :)
    1
  12327. So simple, but blows my mind, which is almost impossible.
    1
  12328. The number for ways to tile the glasses is 3*3*(2*5*2+2(2*3*3)+3*2*3), which equals 666. You seem to really like this number!
    1
  12329. The happiness number must be normalized by the stake of each coalition. Either I misunderstood the end or I think the happiness score assessment at the end is fundamentally not accurate. The most Fair solution is it still the proportion one. Maybe Talmud's should be called the minimizing average dissatisfaction solution. But a very thought-provoking presentation as a whole. Thank you
    1
  12330. Your videos are mind-ticklers! Thank you
    1
  12331. All those triangles must have been smiling to themselves for ages....
    1
  12332. Is there a real reason slide rules are still used today as backups? You could add 100 spare calculators with batteries for a negligible cost, and all of them will be easier to use by a slide rule, especially in an emergency by someone who may not have ever used the emergency slide rule under pressure in his life. And the calculators will also be faster and more accurate. And anyone could use them even without training. EDIT: I looked into it, and it mostly looks like they're used in training or as backups to backups and in practice don't get used by pilots much or at all. Pilots and copilots do bring backup batteries and devices, so these are pretty much obsolete.
    1
  12333. This video is one of my absolute favourite videos of all YouTube. And I've watched it so many times that I know exactly the 2 places where Burkard says "sed zeven" instead of ℤ/7. 🤗 By the way, one thing that I always miss in this video is a simple verbose interpretation of the quadratic reciprocity formula. I think about it like this: as we all know, every odd integer is either of the form 4k+1 or 4k+3; that ugly looking formula (with "(-1)ⁿ") simply tells that the "quadraticness" of p in ℤ/q is the same as the quadraticness of q in ℤ/p, if at least one of p and q is of the form 4k+1; and they are the opposite only if both of them are 4k+3. 🤠
    1
  12334. I always disliked the notation (-1)ⁿ, abused for example in the alternating sums. We could (once and for all) define a function "even(n)", which gives +1 if n is even, and -1 if n is odd. Easy. In computer programming, it's enough to check the least significant bit of the number, in a single clock cycle. 😇 But I'm utterly disgusted by the idea that some "programmers" would make the dear CPU use floating-point exponentiation then multiplication to achieve the same result. ☠️
    1
  12335. In x86 machine language (the only programming language directly understood by all modern computers), the entire Quadratic Reciprocity formula is: mov eax,p and eax,q test eax,2 jnz .opposite ; (otherwise: equal) This above is written in the legacy Assembly syntax; in my own Rix syntax (of the assembler Rix.exe I wrote in year 2004, to be published soon I hope - I'm completing the documentation these days), the next single line generates exactly the same 4 instructions as above: eax=p&q, if eax&2: .opposite
    1
  12336. I don't understand but I wish I could remember! We all knew this ! We don't remember!
    1
  12337. This reminds me of how a Planet, when orbiting the Sun, sweeps out equal areas in equal time. The equal areas are all (almost) triangles, with some being really long and skinny and others being squat and chubby.
    1
  12338. Nice. Remember some of this from Conway's writing on frieze patterns.
    1
  12339. Great video
    1
  12340. The last mathematical tidbit I've found is the existence of just 17 wallpapers
    1
  12341. (6:50) yay! (n+1 choose 2) and the howManyHandshakes problem for n+1 people!
    1
  12342. Beautiful result. Really liked it. Small comment: Complementary angles usually means that they add to 90, not 360 :) Since the name for angles adding to 360 in English is hardly ever used, I guess it is OK to use it in this context as well.
    1
  12343. Because school maths education only has the purpose of making you hate math
    1
  12344. Pretty cool.
    1
  12345. Great video as always. I’m actually a student at Monash that unfortunately took MTH1030 in semester 2 (so I had Norm)... I’m in 2nd year with a decent maths background. and planning to major in pure maths. This video was really eye-opening and helpful in approaches to problem solving! My favourite part was the insights into how linear algebra can help with general formulas... definitely really unexpected and I love how you brought seemingly disconnected ideas together in a easy-to-understand manner.
    1
  12346. Love the t-shirt. Makes you look like you're holding a banana in your right hand every once in a while.
    1
  12347. "LOL, this just proves you cannot prove heliocentrism is 100% fact, because it also simultaneously proves Earth is in the center, and everything is revolving around us! Huh! Bet they didn't teach that in class. Nor why we went with heliocentrism. Because Einstein and the like understood, out of these 2 options, one demands God, and one just leaves the door open for God, not proving it outright, so they went with option B, there is no God, and cannot possibly be a God, so we have to see it from the heliocentric model way. Period. Case closed. P.S. Man, so many videos just recently, indirectly are proving the Earth is not a sphere, nor are we in some solar system. That the Theory of Relativity is laughably bunk, and that the constant speed of light isn't a constant at all, and gravity and time-space are also not real. Hilarious that those inferences can be made from the info provided, yet, because of the depth of the education, and the utter impossibility of God making us the center, like I mentioned about Einstein, that these scientists literally cannot even see what they are showing. It's like the story of the Indians not seeing the ships because they had no mental framework for it, so they physically could not see them. TL; DR. The Earth is factually and provably not a globe. Basic logic and first principles reasoning proves this. Most science claiming globe and gravity is all based on conjecture and theory, not provable in court level of certainty. If I had a gun to my head, and was told to prove one, I'd be forced to prove we cannot prove at all the Earth is a globe. It's the biggest lie the managers of this world ever foisted on to us slaves. REALLY TLDR - you aren't an amalgam of space dust and particles from the Big Bang, just happened to form due to billions of years of evolution, and our existence is all happenstance, and has no meaning. Quite the contrary. A creator of some kind made Earth, made it hospitable, created us, and everything we see. We are the center of creation. We must be, using logic and reason alone."
    1
  12348. You cannot prove the existence of a sky daddy just as I cannot disprove his existence, but I can calculate the distance to the horizon based on the height of the observer. You cannot.
    1
  12349. Marvellous!!! U r an excellent teacher. U know the nuances of voice modulation while teaching. Excellent write up.
    1
  12350. I learnt about heron's formula in 9th grade but our textbooks didn't give a proof of it. Luckily our math teacher didn't like that and he showed us a proof of it by deriving it with just algebra. The geometric proof in the video was something I've never seen before, it is really nice.
    1
  12351. amazing video as usual, Thank you!
    1
  12352.  @Mathologer  I explained this that way cause I might have got this proof from the teacher
    1
  12353. Ok, I need a chocolate now
    1
  12354. I guess I am of the second twisted square. I discovered this on my own. I let the outer edge be C length, then use an isosolese triangle with hypotenuse C and sides A and B, and rotate that within the larger square. Equating, C^2 = 4 (half AB) + (B-A)^2.
    1
  12355. Maths is beautiful. Thank you.
    1
  12356. Great video! I would love to make a similar video for a class of mine. What software do you use/recommend for mathy animations?
    1
  12357. No integers, That really really blew my mind!
    1
  12358. 所以法国电影＜第五元素＞的底层逻辑就是：公元前45年古罗马历法的5季度10月份=春季，march april; 夏季， marius, juno, 仲夏， quintility &sexitility , 秋季， sept &oct , 冬季= nov&dec. Ten gon 这个发音和含义与 【夏小正】历法的【天罡，天干】完全吻合。 这说明：古罗马历bc45 ----＞中欧捷克----＞东亚 的历法有跨时空的2联系。 谢谢这么精彩深刻的天文几何视频。🎉🎉
    1
  12359. Nice shirt!
    1
  12360. Oh, dang. I predicted 28 and 82, but forgot 2.
    1
  12361. oh that inside out trick is nice
    1
  12362. Very nice. That video clearly explains how to spot patterns using visualizations. However, in practice creating nice drawings and explaining them is more time consuming then doing raw algebraic proofs. In my works I usually use algebraic proofs, even if they are heavily motivated by visualizations of ideas. And if time is short, due to project deadlines, the required graphics may never be created.
    1
  12363. Excellent presentation
    1
  12364. Swiv
    1
  12365. It's infuriating how everyone talks about the Fibbonaci numbers, but not about 1 3 4 7 …, which is the sequence to which the powers of φ actually converge
    1
  12366. Why was the preprint never published?
    1
  12367. i cant remember what it was previously, but the new logo to suit is really cool too, from a branding perspective.
    1
  12368. can you talk faster?
    1
  12369. I was hoping this video might have a trick I could use on the standardized test I’m taking in a bit, no calculator. It probably won’t help much but this was still really cool.
    1
  12370. just from the first look of it, it looks like a projection of the 4-cube down to the plane.
    1
  12371. As a CS Major, I'm wondering why I watched and hooked up to this video :D
    1
  12372. I don't know, but perhaps Amita Eppes nee Ramanujan knows…
    1
  12373. The palmscript at the end is written in Thamizh!
    1
  12374. 8:08 Using the same method and we get 5. (Is there a simple example of 7 points where no 6 are on the same hemisphere?) 10:10 An alternate proof using PHP(the Pigeon Hole Principle). Consider the sequence 9, 99, 999, ... Claim: 113 must divides one of them. Proof of the claim: Divide each of them by 113. And we get a new sequence of the remainders 9, 99, 95, ... PHP tells us that the new sequence must repeat, say the remainders of (10^m - 1) and (10^n - 1) are the same where n>m. That means 113 | (10^n - 10^m), thus 113 | 10^(n-m) - 1. Then 10^(n-m) * 355/113 - 355/113 = an integer which is the repeating tail(or multiple copies of it). 16:24 Define two relations: S for "shake hands with each other", and N for "not shake hands with each other". Then person A either S with 3 or more people, or N 3 or more people. WLOG say A S with B, C and D. Senario 1: There exists some S relation among B, C and D. Then those two, together with A, are the 3 people who shake hands with each other. Senario 2: No S among B, C, and D. Then they are the 3 none of whom shake hands with each other. 22:20 I tried (RULD)^3 and found a 5-edge-cycle, a 7-edge-cycle with corner-twists and edge-flips. So I guess it's 3*5*7*3*2 = 630. 26:10 1, 2, 4, 8, 16, 32, 64 31:02 DCJ stands for 3, thus Queen of hearts. I like the proof of 355/113 one most because it's new to me.
    1
  12375. My high school calculus teacher brought his playstation into class one day and had us play a racing game to show the exact same differential calculus example. It's such a fun and clever way to introduce the concept to teenagers.
    1
  12376. I probably don't even need to mention this, but if Euler's formula for polyhedra is covered in a future video, it would be nice to have discussion of more general result of Euler Characteristic, including for higher dimensions, and connection to topological charges. There is a definitely enough material there for a good video.
    1
  12377. Incredible video on an amazing problem!!
    1
  12378. I like math. This is a very well known and important problem. The question is when a sequence can be summed. It has taken decades, if not centuries, to answer this question. A sequence is summable if the sum does not depend on the order of the sequence and if the sum is finite. For positive components, the answer is obvious. But if the sign changes constantly and the components are sufficiently large, any result can be achieved.
    1
  12379. how about 4^2+5^2+...+38^2 = 39^2+40^2+...+48^2 ? or 16^2 + 17^2 + ... + 141^2 + 142^2 = 143^2+144^2 + ... + 178^2 + 179^2
    1
  12380. > Your challenge is to show that a non-zero integer can be written as the difference of two integer squares exactly if the integer is odd. This has me very confused. 12=4^2-2^2=16-4, and 12 is even.
    1
  12381. re: intro, coins gone. Here, in Ukraine coins for the smallest denominations were recently introduced instead of the paper currencies of those denominations. So, no, it's not looking like that at all.
    1
  12382. I usually think of Knuth as a computer science person, but of course his work would stand as valid maths even if computers had not been invented.
    1
  12383. Great video and explanation of the material. Which I would of had you as my Maths teacher in high school might of amounted to something. Thanks.
    1
  12384. great video, and ur giggle makes the cherry on the top
    1
  12385. It's complicated.
    1
  12386. Heron's formula is useful for finding the area of a triangle whose vertex coordinates lie in R^3 (or higher dimensions). Fun fact: the square of the area of such a triangle is the sum of the squares of the areas of each of its projections in the orthogonal two-dimensional basis planes.
    1
  12387. Never knew the sum of no 9s converge... ramanujan silently subscribes to mathologer yt channel
    1
  12388. Wish you had taught me maths at school.
    1
  12389. I am a Year 12 high school student. Your videos are the best. They are hard and requires more than one tackle, but I am proud to say I watched to the end and took notes until 35 minutes. I think I get the Bernoulli numbers, and most of what's after... My calculus is dodgy so I just went along with the Euler-Maclaurin formula instead of engaging with it.
    1
  12390. When I first learned about adding fractions in elementary school (so many years ago ...), I was wondering that searching for the common denominator (to be able to add them) should result in a pattern to easily tell the sum of the first "n" elements of the "1/1 + 1/2 + 1/3 + ..." series. However I could not find anything, and of course I had no knowledge back to then about "harmonic series" or "limits" or anything, so it was a bit stupid to try, but anyway :) When I asked my teacher, he said: "oh boy, don't be stupid, it won't work". Not so much helpful either :-/
    1
  12391. Dear Marhologer, I am the one who challenged you yesterday to make a vid about the Galois Theory. But I have, only out of curioisty and only that, one other question: I see you are a profesor (I knew that for years already of course, but now I saw it fot real), and now I wonder what does the word "Profesor" mean in Australia. Of course, you teach students at university, this is NOT meant as a stuoid question, but is there also something else? In different countries around the world, the job may have different meanigs and a profeosr might have other tasks than in another country. For instace: do they call high school teachers "professor" as well, or may be the piano teacher. You are in a different country, on a different island and even on a different hemisfere, so yeah, this is a reasonable question I think, All the best from Agnieszka
    1
  12392. Coloring proof ftw
    1
  12393. Common sense: 2+2=2x2 1+2+3=3x2x1 2x(2+2)=2x(2x2) 2x(1+2+3)=2x(3+2+1) ……
    1
  12394. Now, what is quantum or complex about!
    1
  12395.  @Mathologer  No offense at all. Sometimes paths may lead to lead same results, but it may depend on conditions, costs, time and choices….
    1
  12396. While Nathan's proof is super cute and may be better for Olympiads, I strongly prefer the Mathologer proof: Introducing an abstraction to a problem like the mapping the state to a single number always requires a proof that no essential dynamics of the problem were lost. (e.g. do the faces of the cards matter?) Mathematicians often defer that into "without loss of generality" but there's a required mini-proof, the burden of which gets shifted onto the reader. Requiring the proof-reader to be at least as smart as the proof-writer is a sign of a bad proof even if it is technically correct. Useful for Olympiads but makes other peoples lives difficult in industry.
    1
  12397. Sir, CAN YOU PLEASE🙏🙏MAKE A VIDEO ON NUMBER SYSTEM (in a visualizing and historical manner, as u always do)... It is very interesting... Plz sir plz🙏🙏
    1
  12398. Ok, forgive my small brain, but how does infinity tie into this? Assuming there are infinite integers, wouldn't the octagon just keep shrinking indefinitely and cause the unit of the grid to keep subdividing itself into smaller and smaller numbers? I fail to see how you'd end up with a unit of 0 if you keep going. When I try to do the opposite and start with a unit of 1 and keep expanding the octagon into larger octagons, you would just end up with an infinite set of subsequently larger octagons, who's vertices would be on the grid I assume. Starting from unit 0 would make no sense to me. I feel like I'm overlooking something fundamental here, perhaps my assumption is wrong? Mind, I'm not a mathematician, just a hobbyist.
    1
  12399. I just ordered your new book! 🙂
    1
  12400. This is basically what I discovered on my own when before I knew what calculus was. It started with trying to find a pattern in n^2, which lo and behold the difference was the odd numbers. I never got beyond trying it with n^m functions, but I explored those to around the point shown at 12:30.
    1
  12401. Oh wow I've been doing this in algorithms class.. the tree is showing the NEIGHBOURHOODS of each state!
    1
  12402. Also holy shit where do I get that shirt?
    1
  12403. In my ten years or so of watching YouTube, this is the video that made me feel like I have to comment. My favorite was learning the 100 zeros series was finite. I thought in my mind, that series is more than the harmonic series multiplied by 1/Googol, and if the harmonic series is infinite, the 100 zeroes series should be too. I know I’m probably wrong, would love to figure out why! Anyway, more importantly - Thank you for your amazing videos, your energy, and your excitement for maths. It’s been a pleasure seeing these videos over the years and really getting to understand some fairly complex math without any academic background in it. :)
    1
  12404. You are a brilliant mind. Leaving me smiling from what you teach. I'm literally talking to myself in full conversation when I watch your videos. Thank you for the intelligent video and teaching.
    1
  12405. The transition music being so close to my old iPhone alarm was disturbing me
    1
  12406. Most memorable: the proof that gamma is less than one by moving every little surface to the left up to the vertical axis Inside a square of edge 1! So beautiful.
    1
  12407. Hey ppl, something really wierd I found while fiddling with my calculator. 1^2 + 2^2 + 3^2... + 24^2 = 70^2 I dont think this has any extreme hidden meaning, but yeah.
    1
  12408. Pure math is nearly as lazy as handwaving. Conveying something intuitively is hard and takes work. Source: My engineering degree. Lots of math, little intuition, lazy teaching.
    1
  12409. Vote->Bizarre maximal overhang towers!
    1
  12410. this was kinda mind blowing, I loved it!
    1
  12411. Great video again! Would be awesome if you can do one about Bertrand's Paradox.
    1
  12412. One cute fact left out of this video: the number of domino tilings of a 2xn board replicates the Fibonacci sequence. (Specifically, the number of ways a 2xn board can be ruled with dominoes is the (n-1)th Fibonacci number.) I was puzzled for a moment on why this might be, but it's pretty easy to see why, based on splitting the tilings into two cases depending on whether or not the rightmost pair of squares is tiled horizontally or vertically. I leave it as a bonus challenge to others.
    1
  12413.  @Mathologer  Oh I missed that! Merry Christmas, thanks for all the hard work on this channel.
    1
  12414. WOW! This Mathologer video made me better understand this monstrous-turned-Mathologery madness especially in auto-pilot mode. This reminds me discovering the sneaky Bernoulli numbers as a kid alone in the dark without the infernal fancy maths in mind. (I even didn't know who Bernoulli was, and what the heck calculus was.) So, do I have the spirit of the Gauss? (Pun intended, but my story is legit.) P.S. Kudos to the hard-working men (or man) for preparing and presenting this video to the mathematically-needy people like me.
    1
  12415. At 15:45 the difference converges to 2.302585. Is there anything special about this number?
    1
  12416. anniversary of his death on 11th. man i hate the fact that despite this being 21st there is si little of his presence saved online. wish he was on lex podcast :(
    1
  12417. The real question is, if we operate with a number system in non-"base 10", is there a base we can use which doesn't repeat? I bet a base system of 0.9999~ or a base system on complex numbers might yield unexpected results, right?
    1
  12418. I like your shirt. I didn't think I was going to watch a calculus video, but I love it.
    1
  12419. I prefer the swiveling proof.
    1
  12420. I liked the disguisting optimal configuration of 20 bricks. I like seeing math shatter my expectations
    1
  12421. Great video again! Interesting historical fact: Torricelli’s and Cavalieri’s work was seen as highly critical by the Jesuits, who found the notion of the infinite paradoxical and heretic. See Amir Alexander (2014). Infinitesimal: How a Dangerous Mathematical Theory Shaped the Modern World.
    1
  12422. Most memorable from this video since I seen a few of these before is the idea behind the number γ.
    1
  12423. A bit cognitively loading but enjoyable for the sheer beauty of it. Thank You. Kudos.
    1
  12424. I'm in the 8th grade and love math but this one hit me hard
    1
  12425. Your german pronouncing is perfect.
    1
  12426. Wow, ausome vedio ♥️ ,mathaloger, and I want to say one thing that at the power sum vedio is one of my best favorite vedio ever seen I like that vedio lot and I made an general formula for summation((1/n^m),1,n) .and I like all your vedios and what a animation my special thanks to all of your team and you.....!
    1
  12427. First
    1
  12428. This is the most impractical container ever.
    1
  12429. Thank you for the excellent video as always Burkard :) Just one note, as someone who is colour-blind, I sometimes struggle to differentiate between the colours you choose. Namely, the orange and green in chapter 2. Think of the problem like this, some hues with a given brightness and saturation can look the same or similar to other hues with a differing brightness and saturation. Interestingly, I think there is probably some nice maths behind which colours are conflated and a "colour-blind coefficient" could be prescribed ...anyways. If I could offer a suggestion, maybe you could try using colours with matching brightness and saturation. Also, if hue range is a rotation from 0 (red=rgb[255,0,0]) to 1 (blue=rgb[0,0,255]) try using hue values equidistant along the range. Graduations of purple can be tricky. As red/green colour blindness is the most common maybe a range 0 (green=rgb[0,255,0]) to 1 (blue=rgb[0,0,255]) would be even better, with red=rgb[255,0,0] and purple=rgb[255,0,255] as discreet additional colour options. Just some food for thought if you see this :) As a side, I am red/green colour blind and I personally find it impossible to differentiate between green[0,255,0] and yellow[255,255,0] without conscious deduction. Cyan[0,255,255] appears smoke white and I see colours in grey-scale.
    1
  12430. At 18:12 all the denominator are not only even. But they are all either whole number multiples or divisors of all adjacent denominators . so for example 140 is evenly divisible by 20 on its left and divides evenly into 280 on its right.
    1
  12431. log(1+2+3)=log(1)+log(2)+log(3)
    1
  12432. There is spirit
    1
  12433. I totaly forgot about partion numbers
    1
  12434. why strip the 1's from the right and not the left?
    1
  12435. A result of how I set up the equations. I can also arrange so that we have to strip away The ones from the other side.
    1
  12436. I think I’d like an Escher’s Solid star!
    1
  12437. my first impression was that both conway's iris and the windshield wiper effect constructions are similar to modeling the moire effect. on that note, i expect breadcrumbs from the mathematical traditions of cultures where netting made from fine threads or webs evinced moire pattern mysteries in antiquity, like vietnam. could be conway rediscovered the wheel...
    1
  12438. Your shirt is great!(as usual)
    1
  12439. My most prized possession that I kept from attending Brooklyn Technical H.S. (back in the days of the Summer of Love, Vietnam War, and the Civil Rights Movement when it was 1 of the 4 entrance exam only schools in NY, and going COED in my junior year!) was my Pickett's 12-inch slide rule (steel with a yellow finish). It had a belt holder/sleeve, and I would wear it like a sword like the nerd I always was. The school store also had circular disc rules, and a cylindrical model (way too expensive IMHO). I think I spent like $30 or so for it.
    1
  12440. I think you spawned a cottage industry. Expect the Mid Wit imitators to make it a booming business. https://www.youtube.com/watch?v=CXDxNCzUspM
    1
  12441. I do not see why the Aztec boards require brick work tiling in the corners. Can anyone help here? I found another solution with all dominos either vertical or horizontal. Is it that other solutions are not random enough to qulify?
    1
  12442. Sorry, I was to fast. It's well addressed here 20:42.
    1
  12443. Thank you for the Christmas present. Another great video.
    1
  12444. X ^2-NY^2=1, solved by Brahmgupta in solving indeterminate equation of second degree just 1000 yeas ago Before Pell. Pl see his other form of equation. X^2-Z= Y^2in Chkravala. Misnomer that Algebra started by al khwarijmi.
    1
  12445. what on earth is eknok?
    1
  12446. OUTSTANDING ......
    1
  12447. Thanks!
    1
  12448. There are lots of gr8 things to check out. Like the CAL of a gyro. Yes. Your new view of algebra awaits and the Gestalt of limits. I mean 3 months on limits. Like it's some kind of religion. Man.....
    1
  12449. Most memorable part: Proof of no 9's convergence.
    1
  12450. Pure math can be experimental! Thanks for pointing out that aspect of the joy of discovery.
    1
  12451. I was pausing and pondering, trying to figure out what's going on throughout the video, then Mathologer said "Complex numbers," my brain immediately said, "Oh, it's just DFTs," and the rest was painfully obvious. I wasn't thinking about "ears" so much as similar triangles as points on perpendicular bisector that are a proportional distance away from the edge, and both of these properties carry across to the DFT decomposition. Saying "ears are just filters," would be a stretch, but it's not far off. Now I'm wondering if I can do this with polyhedra in quaternion space, or if the 2:1 mapping of SU(2) to SO(3) is going to break things. And whether this would still look like some sort of a discrete QFT or if I'm going to only pick up some of the DQFT terms.
    1
  12452. Hm, well, the decomposition certainly works, but most of the neat DFT properties that aren't just inherent in general linear decomposition aren't there. If you start with points that can be decomposed into regular polyhedra, some analogies remain, including the "ears" (which are now constructed from centroids along the normals scaled by face area - there might be a solid angle connection here), but it's not as exciting. A general polyhedron can be decomposed into a set of polyhedra with matching topology, but the choice is almost arbitrary. They just have to be linearly independent under the operation. In practice, there could be interesting applications where the basis is not complete, using pseudoinverses, resulting in some sort of mesh compression or applications in 3D reconstruction, but at that point, it's just projection onto a basis and any real connection to the elegance of DFTs for planar polygons is gone.
    1
  12453. It seems somehow sad that Pascal might not have know about his lovely turtle.
    1
  12454. A consequence of that 1,260 limit on the order of elements of the Rubiks group is that there isn't any operation (no matter how complicated) that you can perform over and over from any starting position and be guaranteed that it will eventually result in a solved cube. Instead, your operation will bring you back where you started within 1,260 repetitions, and for 99.999999999999997% of starting positions, your orbit won't have included the solved cube :)
    1
  12455. 4:10 5x2=10, 7x3=21, 9x4=36
    1
  12456. Waouw ! Merry Christmas and happy new years ! This is a wonderful explanation ans so intuitive ! But now I ask myself, how do you do the same in higher dimensions ?
    1
  12457. I always love watching Math videos and looking for ways to incorporate it into my music composition... It's interesting to see a "harmonic series" that, at face value, isn't connected to sound!
    1
  12458. The area is the derivative of the volume with respect to r, because as you increase r, you grow the volume by adding on layers to the onion. In other words, the rate of change in the volume of the onion as r changes is the surface area of each onion layer (at the limit, of course)
    1
  12459. As an Italian that's "differently young" and has lived more time with liras than with euros... I find the proof of divergence pretty amazing! Wonderful stuff as always!
    1
  12460. I preferred the color proof in general but it seemed like the color proof still relied on the first part of the swivel proof to establish that the two segments in the corners were equal which would make the swivel proof win in the end. Just a guess but I have a sneaking suspicion that the two areas are equal.
    1
  12461. 28:42 hold on, markus persson? notch??
    1
  12462. "Making sure the diagram isn't fooling us" reminded me of "near-miss Johnson solids". Multiple solids have been defined which appear to be arrangements of regular polygons, but actually are not. One or more of the polygons is actually irregular (has different angles and/or side lengths), often by a small amount.
    1
  12463. Is there any significance to the fact that left hand side of these equations always contain one Prime Number?
    1
  12464. LOL cool shirt
    1
  12465. nice to see you, as an engineer I knew that the key was the pulley, and I have to tell you that my best way of shoe tie is the one that looses the shoe easier and faster, and I came to a solution not considered by you and is the one of MANY that includes MORE PULLEYS and is like this, when you start a criss cross just go back to the same side making a pulley in the middle between laces, this way, intuitively you have a large number of pulleys multiplying each time force times 2 ,mathematically, with 0 diameter lace it is also as short as criss cross, it is very easy to loosen also, only drawback is is hard to keep middle pulleys centered but as one side straight it comes to multiply force a lot more, now you have another equation to dream off.... HEEE heee heee ( friendly lol)
    1
  12466.  @Utesfan100  combo class had that right?
    1
  12467. ​ @Mathologer  pun intended?
    1
  12468. ​ @briancooke4259  All equations of the form axy = bx+cy are hyperbolas. This can be derived with some algebraic manipulation (we are essentially solving for y in terms of x). axy = bx + cy can be rearranged as axy - cy = bx. Factoring y on the left hand side gives y(ax-c) = bx and isolating y gives y = bx/(ax-c). Note that we may write bx as bx - bc/a + bc/a, which becomes (b/a)(ax-c) + bc/a. Substituting this back into our equation gives y = b/a + bc/(a(ax-c)). This is the classic y = 1/x scaled vertically by bc/a, shifted right by c/a, and shifted up by b/a, all of which preserve the hyperbolic shape of y = 1/x. To answer your actual question, the desired hyperbola is y = φ + φ^2/(x-φ) according to the general formula or more simply y = φx/(x-φ), where φ is the golden ratio (sqrt(5)+1)/2.
    1
  12469. Classic stuff! I'm an undergraduate mathematics (and music composition) student, and so I followed the video without much trouble at all. There were points at which I paused to see if I could work things out on my own, namely 23:30, and 40:40, to name a few. I was really impressed by just how naturally you were able to motivate this discussion. For that reason, I'd like to see a video with your take on "free objects" culminating in a description of free functors. Moreover, I think a collaboration with Grant (3b1b), in which he does an accompanying video on universal properties would be beyond cool.
    1
  12470. Those overhang towers provide a nice visual explanation of why arches work in engineering.
    1
  12471. Section concerning gamma gets my vote
    1
  12472. like just for the intro - calming right away!
    1
  12473. Absolutely bonkers video
    1
  12474. you are crutial for my mental health
    1
  12475. Notice that the arguments around ~4:30 doesn't prove that the series converges. Had you known it converges(absolutely), your argument shows it converges to 2. You cannot simply exchange numbers between the series elements in negative signs. E.g., your misleading argument can show that 1-1+1-1+1-1+1-1+... converges to 0, but it's not. I think that the easiest way to show your series converges for newcomers is to show it's absolutely bounded by a geometric series.
    1
  12476. Mathologer’s videos are always artworks and often masterpieces.
    1
  12477. Wow !!! Always the best of best. I am mind blown.
    1
  12478. Breathtaking.. astonishing.. great presentation. Thank you.
    1
  12479. I think symmetry is an important criterion for a good lacing too. With the zigzag lacing I find it hard to tell what to pull where to get it tightened up evenly. With the crisscross, it's easy -- just start at the bottom and pull on both sides at each eyelet. Non-shortest lacings could have uses if you have some extra length to use up so your laces don't drag on the ground. Also there are other things affecting the aesthetics of a lacing that this model doesn't capture, such as which segments pass over or under other segments. Seems like some knot theory or braid group theory could be applicable.
    1
  12480. aren't all the extra dimensions(beyond 4, if you count time) so small humans can't observe them?
    1
  12481. I very much appreciate and like the mathematical contributions of this group; no doubt instructive and inspiring. Be that as it may, it always remains recognisable that the numbers assumed as an approach mostly coming out of the blue, and have not been calculated. But that does not diminish the contributions in any way!
    1
  12482. The sound is a cicada.
    1
  12483. The gamma and prime connection is fascinating
    1
  12484. Sum = product can help explain this weird equation: log(1 + 2 + 3) = log(1) + log(2) + log(3) Matt Parker has a good video explaining this: https://www.youtube.com/watch?v=OcTMBrUutfk
    1
  12485. This is pretty amazing. Is there a relation between 1/x and the split-complex imaginary unit j, where j²=1, j≠1?
    1
  12486. Let's use it to guess the next prime number :D
    1
  12487. For years I had a slide rule hung above our driveway entrance. It was a very fine Deci-log-log slide rule that was just over 6 foot long. It was originally manufactured to be used in a classroom to teach/demonstrate the proper use of the slide rule. I really loved that. But the harsh conditions at 8,500 ft elevation in the Colorado mountains and the hard UV radiation took their toll and it finally fell apart. I still have my pocket slide rule (6 inches) I used until I purchased my first calculator (HP-67). I can't seem to bear parting with the slide rule. Even after I got rid of the HP-67 and bought an HP-41CV with all the extras.
    1
  12488. Thank you for such a nice and informative video. I liked The colour proof more than the swivel proof
    1
  12489. Another absolutely amazing video. Thank you Mathologer for all your work and research you put into these awesome videos!
    1
  12490. another way of showing the 2nd fold triangle is the same as the original is by seeing that when taking a first third of the original triangle, and separating it in three thirds, the final sets of two triangles are either already a scaled down version of the original as per the triangulation, or one that is part of it, and the other that is vertically opposed to the other half (it's hard to explain through text)
    1
  12491. 19:45 is that a Michael Penn reference?
    1
  12492. Thank-you. Very neatly described and demonstrated. 🙂👍 My slide-rule is still working properly, despite never being plugged into an electricity supply, over the past 50 years!.   I was with you until you said 'Can you code...' at which point my answer to almost any following question is 'No.'. Coding has never quite jelled, despite trying to learn it for decades. I have degree qualifications in an engineering field, speak some Russian and can even read Cyrillic script easily, but coding is much worse than a foreign language, because it's so illogical.  Yes, I know you'll intake breath sharply with that word! But I have no idea about where, or how, to start to answer your question. My answer would be to point you to a slide-rule, and to say 'This is all you need to calculate figures to any scale, with an accuracy of 2 to 3 decimal places. It gives even more precision if how to use it properly.' Every one of us has our strengths and weaknesses, of course, but coders seem to assume that their favourite subject is automatically easy for anybody. I'm living proof that they are wrong with that assumption.
    1
  12493. Thanks so much for this video! This felt like a damn good lecture, with the pacing nailed down. I'm a 2nd year undergrad, and I found all of this very clear (although I do want to check some of the sums in the 7th chapter...) - and all you've done is excited me more for further videos on this topic!
    1
  12494. It seems we can "stretch" any of the 3d quadrilaterals and the non convex ones into a convex quadrilateral by just rotating two sides around one of the diagonals. When this is done the only side that changes size is the other diagonal, and it seems to always be "streched" (i.e. yielding a c' for the new quadrilateral that is greater than c). By having a proof for the convex case one could then argue that Aa+Bb>=Cc' (for the new streched quatrilateral) and Cc'>Cc as c'>c
    1
  12495. Most memorable: the greedy algorithm
    1
  12496. As ever you pitch this perfectly for me. I like to think this is your skill and not simply my good luck but it seems whenever I feel that twinge from rusty memory of maths unused too long, you swoop in to clarify. We know mindfulness is valuable, well also brainfullness so thank you!
    1
  12497. is anyone ,outside of the creattors, archiving these videos? just curious
    1
  12498. Often come along with a bunch of trig identities and geometries and boundary conditions. Derive Snelll's Law from Maxwell's Equations ... ok but it's quite a lot of work.
    1
  12499. Ah taxicab numbers. I have to admit, I had to google that one.
    1
  12500. Yay I have lifetime subscription just by rewinding the video once and finding out the pattern and the answer in half a minute lol
    1
  12501. 11:53 I imagine you can solve the puzzle with three bricks if you stack the 2nd brick in the vertical position right on top of the inner most edge of the 1st brick. This would apply the most leverage so that it would be possible to push the 1st brick on the other end farther away and hopefully clear the cliff completely.
    1
  12502. Visual mathematics like this isn't taught because it's more confusing that formulae on paper/blackboard. It would also be much easier to introduce a couple fallacies into a reasoning based on visuals.
    1
  12503.  @Mathologer  I probably listened to your video only with half of my attention. Not trying to put your video down, it was interesting. Just saying that if you are suggesting your video is how the maths ought to be taught then I give this argument. It sounded to me like you were saying "why don't we teach using visual maths instead?" but I may not have been paying proper attention.
    1
  12504. 27:05 the area cannot be created or destroyed only divided by tessellating polygons
    1
  12505. The visual proof of pyth. seems to originate from the Dutch writer Multatuli who claimed in his notes "Ideeën 529 tot en met 532 ( II )" to have found this proof. Ref. online.
    1
  12506. One thing that I’ve been curious about with that Chinese diagram - was it just a proof of the 3-4-5 triangle or was it a general proof. Looking at it from today’s perspective, it’s easy to see that this diagram can be used to prove the theorem in general but did the Chinese know the general theorem or just specific examples? The grid lines make me think they were just talking about the 3-4-5 triangle but I am not able to read the text to see if they talk about it more generally.
    1
  12507. I can't remember discovering this
    1
  12508. My dad had a circular slide rule like this. :)
    1
  12509. Hahaha I found thr t shirt funny
    1
  12510. Great video! :) Here the solution with three rectangles overhanging: ◻︎◻︎◻︎◻︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎ ◻︎◻︎◻︎◻︎◻︎◻︎◻︎◻︎◻︎◻︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎ ◻︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎ ◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎◼︎ (Empty boxes are air, the fourth line represents the table.) And the solution for the gamma proof: gamma>0.5, because the blue parts are all a bit larger than a triangle with the same corners (since 1/x is a convex function), and those would fill exactly half of the unit square. And to decide for the favourite part is very difficult, but I vote for Chapter 2, since I used to live in Pisa. 😉
    1
  12511.  @Mathologer  No... UK. I dread to think how long it would have taken me to learn to use it upside -down...
    1
  12512. Those 99 videos, does that include the rogue upload by your disgruntled former coworker? 😘
    1
  12513.  @Mathologer  well congratulations, can't believe it's been so many videos already! 🥳
    1
  12514. I laughed out loud at "rectangle snake charmer". Love this channel.
    1
  12515. The Italian guy is pronounced Torikeli, not Toritseli...
    1
  12516. Entertaining and informative! As always.
    1
  12517. cool af. very useful in real life applications. i wonder it can be applied to polygons greater than 3 sides.
    1
  12518. Most memorable: the proof that 1+1/2+1/3+... is infinite.
    1
  12519. 8 houses can be a very long street if the lots are very large.
    1
  12520. Great video, but I did find one relatively large mistake. You said, "here's another kitten" but it was clearly the same kitten as before.
    1
  12521. What about the area? They look like they have the same oriented area at the begining and the end.
    1
  12522. 54:02 oh heck, please do that again, anytime !!! 🤩😍🤩
    1
  12523. The recursive nature of that proof is beautiful.
    1
  12524. Amazing class!!!!!!!! Unforgetable! (and the final music is chilling:)
    1
  12525. I am not sure how the referenced Reddit post is credited as an invention unless it is the rubber band methodology in the demonstration. I have a circular slide rule that I have owned since I was in high school a half century ago. This was just before the advent of affordable pocket calculators.
    1
  12526. It is indeed the rubber band methodology for inscribing the logarithmic scale that's the novelty. It's just lovely how the stretching and sticking of a length L band around a disc of circumference ln(L) will come out with the logarithmic scale. It really seems to justify the title "natural logarithms".
    1
  12527. Each "and so on" in mathematics says a lot in few words and leaves much still unsaid. 😏
    1
  12528. I am an indian and Ramanujan said that his goddess namagiri showed him everything
    1
  12529. Did a 100c(1$/100) coin(s) really circulated(like massive) or this is just for assuming?
    1
  12530. During my last or second-to-last year in school (2005 or 2006), we visited a nearby university in Germany for a "trial lesson". I visited the maths department, and among the things they showed us were those domino tilings of rectangular boards. We did, with a little help, derive the general formula for the 2-times-n rectangle (Fibonacci); then they asked us for a guess about the general m-times-n board. Nobody came up with an answer of course, but I remember they showed us the "monster formula" in the end. I think today was the first time since then that I saw that formula again. :-)
    1
  12531. Now I really hope that the next video will be about the connection between gamma and -1/12.
    1
  12532. According to Saint john the number 666 is a number of a man, not the devil.
    1
  12533. I just spit half a coffee cup over my desk at 1:37.
    1
  12534. Interesting. Two questions: 1. To what extent (if at all) does the 'degree of irrationality' depend on the base of the number system used? 2. What happens to this 'degree of irrationality' if an irrational base number system is used (for example, let's say we count in base Pi - as we do for the radian angle system - or base Phi)? Clearly when using base Pi, Pi is no longer irrational, likewise, when using base Phi, Phi is no longer irrational. But when using base Pi, is Phi still the 'most irrational' number of all? Thanks (PS - I am not a mathematician so forgive me if these are dumb / obvious questions to you).
    1
  12535. Great vid as always... but what's the track playing over the exponential functions animation (34.54+)? 🤔
    1
  12536. Excellent video. Well done!
    1
  12537. 16:00 5
    1
  12538.  @Mathologer  Of course I have. I am a retired tech teacher, so only needed enough maths to teach electronics (degree in engineering...). But we had training days and I had this huge argument with a primary school maths 'specialist' who insisted that all pythagorean triples were just multiples of 3,4,5. She would not believe the 5,12,13 was a right angle. Where to start...
    1
  12539. your shirt is wrong! the square root of evil rounded to 4 decimal places is 25.8070.
    1
  12540. So many alien tricks: keeps me wondering! Fascinating and mindblowing stuff.
    1
  12541. The strange maximal overhang shapes!
    1
  12542. I like the second proof better, because Sallows' theorem is really a theorem in affine geometry, where lengths, angles and rotations don't make sense in general, but translations and scaling about a point do make sense, hence length ratios exist for parallel line segments, as do rotations by \pi (scaling by -1). In particular midpoints of lines make sense, hence so do medians of triangles. Now for each midpoint, apply a -1 scaling to the two smaller triangles that meet it. The result is a hexagon whose sides are parallel to the medians. The latter extend to diagonals of the hexagon which bisect each other at the centroid (this gives another proof of the 2:1 property of the centroid) and hence cut the hexagon up into 6 triangles. Sallows' theorem follows from the fact that these 6 triangles are all translates of each other, perhaps after scaling by -1 about the centroid. Furthermore, if the sides of the hexagon are extended to form a star of two triangles (related by -1 scaling again), then the additional triangles in the star are also translates of the Sallows' triangles, so the two big triangles are rescales by \pm 3 of translates of the Sallows' triangles. The medians of these new triangle are parallel to the sides of the original triangle, so repeating the process with either of the new triangles gives back a translate and rescale of the original.
    1
  12543. Hum, I'm amazed that I couldn't find an actual latern based on this on Amazon.
    1
  12544.  @Mathologer  I like the pens.
    1
  12545. Burkard: The harmonic series is the sum of sums! Me: 🤯
    1
  12546. Omg, this gave a real insight into a part of my work. You should be teaching mathematics to molecular biologists.
    1
  12547. Most memorable - Nice application of the greedy algorithm !
    1
  12548. oh yeah you can easily grow at more than o(sqrt(n)) overhang by just building an inverted triangle, it's even buildable 1 brick at a time. It's crazy how Humans can focus on one thing, find it amazing, and not consider relatively simple and down-to-earth alternatives like a simple brick pattern
    1
  12549. "Here's the problem: a man d*es."
    1
  12550. wow , to da się rozwinąć na ułamkowe potęgi a^(x/n) ale będą wtedy szeregi nieskończone, dla n=1/2 będą kule (zależne od wymiaru ) , dla n=2 okrąg, dla n=3 kula ect...
    1
  12551. Thanks I really enjoyed that. Some of it seems vaguely familiar from my schooldays back in the days of using 6 figure mathematical tables and hand crank calculators when more accuracy was required than could be found with a slide rule.
    1
  12552. The demoed crease pattern is for the flapping bird, not the crane, it seems
    1
  12553. $1 pledge level? No brainer. I made a comment on a recent video (maybe the most recent) about wanting to see more about how theorems have been used to solve other theorems. I'm not sure if you meant to do that here, but I really like learning about putting together these "looks suspiciously related" theorems to build another result. Thank you very much!
    1
  12554. The piano at the end was nice. Real piano too; I could hear the pedals moving.
    1
  12555. At 20:15, you removed a 1 in the numerator (presumably because 1 is the multiplicative identity and so does not affect the multiplication at the top.) But you DIDN'T remove the 1 in the denominator, which you could do for the same reason. This has the effect of changing the fractions you use to generate your infinite multiplicative series at 21:30. Options are: (2/1)*(4/3)*(6/5)... your choice, with inconsistent 1 removal, which results in all fractions > 1 and a divergent series, or (1/1)*(2/3)*(4/5)*(6/7)... consistently not removing 1s, which results in all fractions <= 1 and a convergent series, or (2/3)*(4/5)*(6/7)... consistently removing both 1s, which results in all fractions <1 and a convergent series (the same convergent series as before * 1/1 = the same.) Edit: 4th option is inconsistently removing the 1 in the denominator, resulting in (1/3)*(2/5)*(4/7)*(6/9), again convergent. I am genuinely curious as to why the inconsistent treatment of the 1s in the numerator and denominator.
    1
  12556. Legend
    1
  12557. Favourite part was how counter-intuitive the performance of the leaning tower was vs those manic designs :D
    1
  12558. 20:28 2 = 1/1+1/2+1/3+1/6 3 = 1/1+1/2+1/3+1/4+1/5+1/6+1/8+1/10+1/12+1/15+1/20+1/24+1/30+1/40+1/60+1/120 (I started out knowing that 1/3+1/6=1/2 and then used that to generate more instances of 1/2, e.g. dividing 1/3+1/6=1/2 by 4 on both sides gives 1/12+1/24=1/8 => 1/8+1/12+1/24=1/4 => 1/4+1/8+1/12+1/24=1/2, and then the other 1/2 comes from dividing the 1/3+1/6=1/2 by 5 to make a 1/10 and by 20 to make a 1/40, with the goal of making a 1/5 without directly using 1/5 and 1/10, since 1/5+1/5+1/10=1/2.)
    1
  12559. I need your T-shirt
    1
  12560. I made a few modifications to my submission. here is the new link: https://www.geogebra.org/m/cufneprj
    1
  12561. hi
    1
  12562. GA = 4/3π Diameter Density - Cosθ (latitude) (Velocity^2/Diameter)
    1
  12563. If you cut out 4 tiles, 2 of each color, from the chess board, some configurations are tile-able, but if you were to, say, cut out A2 and B1 as your two green squares, you isolate A1 as a single square on its lonesome that can never be tilted by the dominoes
    1
  12564. 7:25: The answer is 2. It took me a second. But that only since I first double-checked that indeed 10²+11²+12²=13²+14² at the beginning of the video, so I already knew the sum is 365.
    1
  12565. masterclass
    1
  12566. Wow this is so fascinating! Loved learning about all these amazing properties!
    1
  12567. Again, WOW!!!
    1
  12568. My understanding of engineering principle suggests the method is to make two measurements. (a) construct a measurement where each segment is smaller than the object. (b) construct a measurement where each is bigger. The truth is in between. Drive (a) and (b) to infinitissimal sizes. Now to check the video.
    1
  12569. Native Americans are not Asian Indians, the other paradox you forgot to mention. You made the same mistake as the early explores.
    1
  12570. What happens in other base systems, ie, Fermat four square theorem?
    1
  12571. But of instead a line, there cards were done in a circle this may never terminate
    1
  12572. Concrete mathematics... Good old times :D
    1
  12573. My mother done Ptolemy.🤪
    1
  12574. Excellent. Fabulous. We take any square. The vertices of a square can be built on any circle. You just need a compass :). We take two opposite vertices of the square. We measure conditionally 1, for example, clockwise. We connect the resulting lines with a segment. Exercise - we got the Pythagorean theorem. And then there are many numbers and spirals. But chance, how to understand it. Can an algorithm create an algorithm? I dont know. Great video. Bravo. Yes, I forgot, the Fibonacci spiral is double (it is enough to build a hypotension :). And the last question :). Can "time" be a topological characteristic (for example, how time passes on the Möbius surface, but there is "width" in mathematics :). И пусть удача будет с нами🖖.
    1
  12575. Really love this video, and Archimedes is my man as well!!
    1
  12576. The second puzzle at 7:25, the answer is 2. I did this mentally using the sun of square formula to figure out the sun of the first 14 squares, and the sum of the first 9 squares, and then did a subtraction. It’s (14)(15)(29)/6 - (9)(10)(19)/6, which simplifies to (7)(5)(29) - (3)(5)(19). This becomes (5)(29*7) - (5)(3*19), which is (5)(203 - 57), which is (5)(146). We are dividing the entire thing by 365, which is (5)(73), and 146/73 is 2. So with all of the cancellations, the answer is 2. Not the easiest thing to do in my head, but the answer does reveal itself.
    1
  12577. Goodness I love this video. I'm going to watch it at least five more times to let it sink in :D
    1
  12578. So, Bernhard Neumann was another famous German-Australian mathematician?
    1
  12579. Is there a mathematical equation for the speed of smell?
    1
  12580. I love you and I also love palindromes ❤
    1
  12581. The most memorable part for me was finding out about gamma for the first time!
    1
  12582. ...I have a couple of questions though... must the cards be adjacent? does the 10-card problem operate in the same manner as the previous problem? In other words, can face up cards be turned face down as in the previous problem?
    1
  12583. Nice, now I'm going to want to use this to solve for my Pascal's simplex point order sequence formula... And I'm still not done playing with dividing higher dimensional cubes to explore higher dimensional spheres! I also want to compare the behavior of higher dimensional spheres inscribed to higher dimensional cubes with higher dimensional spheres inscribed on equal dimensional simplexes. Btw, I have all kinds of knowledge on how to use Pascal's Triangle to generate the terms for Pascal's simplex in multiple ways and would love to chat about it with someone who is likely to understand me ☺️.
    1
  12584. i got that too
    1
  12585. I decided to look at other paths through the tree and was surprised to find some unusual sequences appearing. There are infinitely many paths which produce a hypotenuse to smaller side ratio approaching sqrt(2). However, the paths where the two smaller sides differ by a constant are limited in a surprising way. I'll notate the paths with L, M and R for left, middle and right child. We saw in the video that M^n for n >= 0 produces triples where the smaller two sides differ by 1. Are there paths where the two smaller sides differ by 2, or 3? The answer is no. It turns out that if (a,b,c) is a Pythagorean triple with |a-b| = k, then the triple can only be primitive (and appear in the tree) if the prime factors of k are all congruent to 1 or 7 mod 8 (A058529 on OEIS). The first few values are 1, 7, 17, 23, 31 and 41. We get k = 7 with paths LM^n and RM^n, k = 17 has RLM^n and RRM^n, k = 23 has LLM^n and LRM^n, k = 31 has RRLM^n and RRRM^n and k = 41 has MLM^n and MRM^n for n >= 0. Note that they all begin with travelling to some node, then repeatedly choosing the middle child to maintain the constant difference between the two smaller sides. Now consider paths where the ratio of hypotenuse to the smallest side is sqrt(5). This can be achieved by having the larger of the two smaller sides differing by a constant from twice the smallest side. We get a similar restriction as before, though this time more complicated. If (a,b,c) is a Pythagorean triple with |a-2b| = k, then the triple can only be primitive if k is squarefree and the only prime factors of k are 2 and primes congruent to 0, 1 or 4 mod 5. I couldn't find an OEIS entry for this sequence. The first few values are 1, 2, 5, 10, 11, 19. We get k = 1 with path L(ML)^n, k = 2 has (MR)^n and R(MR)^n, k = 5 has ML(ML)^n, k = 10 has LR(MR)^n and RR(MR)^n, k = 11 has LL(ML)^n and RL(ML)^n and k = 19 has M(ML)^n and RML(ML)^n for n >= 0. Here we see that the repeated unit to travel along the paths involves two steps, to maintain the constant difference between the middle side and twice the smallest side. Another curiosity is which sequences have the property that the ratio of the smaller two sides approaches the golden ratio. I can't find a pattern to them, but I did observe that down to the 11th level of the tree the two paths that gave the closest approximation to the golden ratio were LLRLLLLLMRL and RRLRRRRMLR. And instead of prefixes of these sequences being the best approximations for earlier levels, instead suffixes are better. Which means these aren't really describing two paths, but that the best path is moving further away from the middle of the tree at each level after the first few, with the occasional move towards the centre. It is quite curious.
    1
  12586. Is this all because they hadn't discovered fractions?
    1
  12587. The arguments that the video brings are, of course, correct. There is no uniqueness in the mathematics of 3-6-9; this is indeed a private case. The argument of Tesla proponents is not in the mathematical uniqueness of the vortex but in the fact that every living creature replicates cells to the power of two. The sum of the digits of powers 2, as stated, does not include the numbers 3-6-9. Therefore, it belongs to the other dimension perceived as divine and sublime!
    1
  12588. Is (ho)³ the same as (ho )x 3?
    1
  12589. Hi, at 6:20 , that looks' like 3.18 not 3.14.. I was taught to use a slide rule in high school here in Australia, early 1970's, one of the best things I was taught.
    1
  12590. 6:22 Not always possible. I can remove the two black tiles in the top left corner to leave a single green tile that cannot have a domino placed on it.
    1
  12591. And the winner is... Leaning tower of lire! (Obvious, I’m from Italy 😄). Anyway, the whole video is amazing and beautifully realized. Please note that the pronunciation of “Mascheroni” is not Masceroni but Maskeroni 😁 (for the sake of italian language)
    1
  12592. He sounds like the third reich professor from a movie.
    1
  12593. I think the two programs would have been more impressive with some loops. Then they'd really look like eachother.
    1
  12594. I studied applied physics, master level. I can well follow it. Also able to intuitively follow the part in the 40th minute. The matrix formulae were great, need some more experimenting (e.g. with Cramer's rule instead of inverses, if that works). I can't reproduce the B_n expansions off the top of my head, but if reviewing the video a few more times, that would also likely sink in. Not gonna do that now though. Perhaps a summary cheat sheet with the entire argument in formulae might be helpful?
    1
  12595. ASTIG!!!🍻
    1
  12596. Don't recall where I first saw it, perhaps in high school 50 yrs ago, but when I taught algebra, the moment my students knew how to square a binomial I would show them this theorem.
    1
  12597. Background: Bachelors in pure maths, masters in physics. This is one of my favourite subjects in mathematics and you're the first person I found that is interested in in answering the questions I'm asking in the simplest way possible. Thank you, I love it. Your explanations are generally clear. I could go with or without the visual proofs. I'm more just really impressed that you made/ animated visual proofs than actually follow them, but I follow the algebraic proofs just fine. I love all things divergent and look forward to more videos on the subject. I'm also greatly anticipating your Galois theory lectures, as I've never studied the topic before. It would also be cool to see your thoughts on the following identity: Integral(from -1 to 0) S_n(x)dx = Zeta(-n) Where S_n(x) is the sum of the nth powers of the first x integers as defined in the video. I know that, in order to perform the integral, you need to reinterpret x from an integer to a real number, but it's interesting to see yet another way that the sum of these integers is related to Riemann-Zeta function.
    1
  12598. I made it to 7
    1
  12599. I got (n^4 - n^2)/12 as the number of squares in an n x n grid.
    1
  12600. Amazing! Danke! Love your channel
    1
  12601. So I just deleted a huge comment, and this is me giving up on life. 4 years of work ( not continuous, but anyway ) for a puzzle my math teacher gave me in highschool... and all I got was a shifted version of the Gregory-Newton formula ( shifted one position to the right by using (n-1) instead of n... I am literally having a crisis right now, and I'm dying inside, but please continue. At least I'm learning how dumb I am, reinventing the wheel... Anyhow, one note I feel needs to be more emphasized: if the abscissae are not evenly distanced, this is useless to hell and back. I know that oh so very well because that's what I did for about half a year, every night. Now, if the resulting polynomial is 4th degree or lower, no problemo, you can apply Gregory-Newton on the abscissae, and then get the inverse of that function, but for 5th degree polynomials and above, you're fucked cause you can't get the roots. That being said, I was going to try and use fourier series to try and solve that issue somehow, but now I lost all motivation, someone resurrect those two dudes and let them have at it. It's their formula, not mine. Yes. I am really not happy right now. The general Newton interpolation uses abscissae as well in those differences, which account for unevenly spaced abscissae and remove shifting. That being said, for the same case scenario of evenly spaced abscissae, the general formula yields a bigger error than the Gregory Newton formula... Now, there's spline interpolation and shit out there, also the Newton Raphson approximation formula ( which I hate, cause some dick put us through doing that by hand... and it took 30 pages each time, for 4-link mechanisms and also for crank-piston ones, even tho crank-piston mechanisms have a nice analytic formula to them ). ANYHOW, thanks for the video! I mean it. I'm salty cause I realized how much time I wasted doing something someone else did 400-ish years before my time... and if someone told me this like... idk, earlier, I'd not have wasted 4-5 kg of paper and about 1 year worth of free time on this (cause of course I did other things with my free time in 4 years )... oh, god, I feel so bad. Also, I already know the proof :)) I had to write it for myself... but I bet yours is prettier, let's see
    1
  12602. Following the pattern on your t-shirt makes me dizzy.
    1
  12603. Great job illustrating and presenting. Sadly, first time I recall seeing any of this. I recall being told to just memorize a2+b2=c2.
    1
  12604. That IS the best math card trick I've ever seen
    1
  12605. Not that it matters but from upper left clockwise 9 4 13 17
    1
  12606. Holy moly... What kind of thinking does it require to realize something like this? I was always fascinated by math, but that may be because I was always unable to grasp it. But this is amazing.
    1
  12607. As I continued watching the video, I thought the leading tower was the most interesting. That is, until you showed Tristan's fractal, which ended up being my favourite because of how intuitive and clever it was.
    1
  12608. The English letter ‘r’ is so much more confusing than the Greek ‘ρ’ because of your personal non-rhoticity. When you say the name of the letter ‘r’ it sounds like “aaah” but the name of the letter ‘ρ’ sounds something like “row” - ironic really 😅
    1
  12609.  @Mathologer  I’m so embarrassed that this was the comment that this was the comment you engaged with 😂 I love your content and your style; you remind me of Mr Morris, my all time favourite maths teacher 😊 I just feel that for mundane reasons of practical reality the actual sound of the name of the letter ‘r’ (and to a lesser extent ‘s’) as realised in non-rhotic dialects of English is problematic. Using fancy Greek letters solves this by putting consonant sounds at the start of syllables. For context, I am a native speaker of a non-rhotic variety of English, however, I am dyslexic and I struggle with slow auditory processing; so I think about this a lot, and since you mentioned the decision to use English letters 🤓
    1
  12610. Please make a video on real analysis
    1
  12611. 3³+4³+5³=6³
    1
  12612. In these animations the smaller polygons are constructed on the edge of the larger ones, but it can be seen easier. If you can shift just one corner of a regular n-gon so it ends up inside it, then by symmetry you can do it to all of them and get a smaller regular n-gon. This only fails when you can't shift a corner to end up inside (triangle and square) or when the only point you can get is the center (hexagon).
    1
  12613. I liked the colour proof more than the swivel proof. I think it is easier to show for less experienced people, because it is more intuitive. You're just comparing the lengths of segments and their sums being equal.
    1
  12614. It may be a "lantern" in a cylindrical shape, but with all those internal folds it’s simply not a cylinder. I don’t get how this was a confusing math question.
    1
  12615. Most memorable was Kempner's proof. It's deceptively simple once you see it.
    1
  12616. Amazing video
    1
  12617. Nice video presentation. This is just an "old fashioned" slide rule. I have 3 of the linear slide rules, and I once had a circular one, but I forgot where it was long ago. Of course, the whole idea is to implement a logarithmic scale for convenient multiplication. The innovations on some of my slide rules was to add additional scales for squares, cubes, logs, and trig functions. This whole technology of slide rules was made obsolete by hand calculators that became common place in 1976 or so.
    1
  12618. By the way, I’ve always really enjoyed your degenerate hairstyle.
    1
  12619. Clearly! Matt Parker gives it a go, but can only manage "Parker bald" most days. Poor fellow.
    1
  12620. Nice Pi Christmas tree shirt
    1
  12621. (IMO) great vid mathologer Roger Apéry is the best mathematicians to exist for a idea, Note: Its just My own Opinion on the suggestion, Advice; "Feel free to exchange eachothers own Opinion even mine to eachother".
    1
  12622. at last! generating functions on Mathologer!
    1
  12623. theodolites have that. it has nice precision and it is very intuitive
    1
  12624. These sequences of 1, w, w^2, etc immediately reminded me of FFT, though matrix was arranged in unusual way. Most logic way is make element (n,k) equal to w^(n*k), where n,k = 0..N-1. So for 'zeroth' row we have all ones. Then we have 1, w, w^2, w^3 etc. Next row: 1, w^2, w^4, w^6 etc, but of course as w is Nth root of unity, we can subtract N each time from exponent. Some time ago I developed balanced ternary fast Fourier transform. Not that big achievement, but it was very convenient to work with in practical applications. In usual binary FFT there are several ways to perform transform 'in-place' but all samples get 'binary inversion' that is they are rearranged as if their indexes expressed in binary form are now most significant bit became least significant and so on. There was 000, 001, 010, 011, 100, 101, 110, 111 (0,1,2,3,4,5,6,7), but now we read them 000, 100, 010, 110, 001, 101, 011, 111 (0,4,2,6,1,5,3,7). Well, with balanced ternary FFT it's the same but with balanced ternary number system! And that's very well, as we have same number of positive and negative 'frequences' and no nagging Nyquist frequency which is neither negative or positive. Doing diffraction pattern on some symmetric aperture works as a charm!
    1
  12625. The cubes on the t-shirt are really distracting.
    1
  12626. Can someone please explain the joke on his shirt EDIT: never mind, saw the comment a few comments down and it says that the joke is “two infinity and beyond” (2 [infinity sign] + >)
    1
  12627. Simply because of the excellent choice of music, I will watch every mathologer video.
    1
  12628. First one i saw was the one at 1:08. 6th grade or something.
    1
  12629. The most memorable for me was definitely Kempner's proof.
    1
  12630. No capital pees, but capital pies. But I guess that's because pi is pronounced pee in german.
    1
  12631.  @Mathologer  and in Dutch! But in English P is also pronounced pee so using the correct greek pronunciation for pi can lead to confusion while it doesn't in German or Dutch.
    1
  12632. Coloring proof
    1
  12633. ;)
    1
  12634. I like the color proof better than the swiper proof. Simplicity wins.
    1
  12635. I suggest to you to treat at this point the finite differences calculus and the parallel with differentiation/integration theory
    1
  12636. Thanks!
    1
  12637. Ugh! I wish I had my 17 year old brain back 😢
    1
  12638. I just realized that this square (09:02) is made up of 4 squares + 1 square + one square that is 1×1. So the area of the squares could also be summarized as ab⁴/2×1.. I think. I've just become interested in math so I'm new to this. But I would like to try 🤭
    1
  12639. I had a different take on the product of evens divided by product of odds: the 1 on the denominator contributes nothing (just as the 1 that was in the numerator was removed without issue), and so it could be described as the product of the ratio of each consecutive even-odd pair -- that is, as the product of (2/3)(3/4)(4/5)(5/6)(6/7)... . This shows that something fishy is indeed going on -- that these methods of evaluation are not appropriate here. And yes, the Wallis product doesn't have this problem of infinite expansion because the fractions alternate between above 1 and below 1, and each pair with like denominator combines to less than 1 -- that is, the 2/3 and 4/3 combine to 8/9, the 4/5 and 6/5 combine to 24/25, the 6/7 and 8/7 combine to 48/49, in general taking the form of a square in the denominator and the number preceding it in the numerator. This value gets closer and closer to 1 for each pair of combined terms, and so it will converge.
    1
  12640. General solution at 36:03. Define O = (0,0) A = (0,a) B = (b,0) D = (d,0) a = OA = short side = ½ (= BD) b = OB = long side = 1 c = AB = hypotenuse (c² = a²+b²) d = OD = b-a θ = OBA sin θ = a/c cos θ = b/c A-small = A-large - 4A-triangle = b² - 4(½)(b sin θ)(b cos θ) = b² (1 - 2 ab/c²) = b²/c² (a²+b²-2ab) = b²d²/c² In this case b² = 1² = 1 d² = (1-½)² = ¼ c² = 1² + ½² = 5/4 A-small = 1/5
    1
  12641. @Mathologer how do you do your animations? they’re great!
    1
  12642. I hope the next video will explain the beautiful logic of a^2 + b^2 = E/m :D
    1
  12643. If we take an infinite number of points and connect each point to 2 other points with the idea that they represent an infinite sided degenerate polygon, then by applying this principle would we not recreate an infinite sided polygon e.g. a circle. Following this line of thinking would it be correct to assume that the entire complex number plane considered as a degenerate polygon [where every number is connected to 2 other numbers ] could be recreated as a circle under this transformation? As an example if we look at any number and connect it to its 2 square roots then this entire structure could be considered from the perspective of a degenerative infinite polygon to simply be a degenerative circle? Such a circle would be a circle of alternating + and - values both real and imagined ?
    1
  12644. Tristan's fractal visualization is the most impressive for me, I like such graphical insights into the infinity.
    1
  12645. My mum heard you on the radio talking about you and your team offering to help confirm correct answers for a test(I think it was a Victorian school test thingy), I showed her a bit where you laughed and she just burst out laughing. Laugh: 11:19. Great video
    1
  12646. @16:40 Shouldn't that be (x+1)^0 = 1?
    1
  12647. Absolutely Brilliant. Thanks for wising all of us up!
    1
  12648. Crazy intelligence out there!
    1
  12649. Look at the screen at 22:26 and hopefully you'll see why this stuff is complicated to an Art Major like myself. I'm trying to improve my math skills and I thought this whole turtle business was going to help this make sense... If I don't have something REAL to relate this to, like how this is used to make a thing or understand a phenomenon, it's all just numbers and symbols floating around. What does this math do? Is it used to figure out something we need to know?
    1
  12650. Another trick for solving a rubik cube is to have up your sleeve an already-solved cube.
    1
  12651. Fredericus Johannes M(unnik?) Barning
    1
  12652. The guy's face in the thumbnail is hyperbolic
    1
  12653. This video is the pinnacle of "blink and youll miss it" It's hal an hour but its too short
    1
  12654. My vote for Kempner's Proof because It's less than a geometric series? Always has been
    1
  12655. int(1/a da, a goes from 1 to x)= ln(x)-ln(1)=ln(x) int(1/b db, b goes from 1 to y)= ln(y)-ln(1)=ln(y) int(1/a da, a goes from 1 to x)+int(1/b db, b goes from 1 to y)=ln(x)+ln(y)=ln(xy)
    1
  12656. My jaw literally dropped when you derived the quadratic formula from completing the square. Like, I was sitting in my room with my mouth open. I'm amazed.
    1
  12657. >Why did they not teach you any of this in school? Lol have you seen all of the behavior management that USA schools are legally obligated to do instead of cool stuff because they have shitty kids? No time for cool stuff because the children (and parents, and teachers tbh) suck.
    1
  12658. The colouring version of the proof appears much more intuitive to me.
    1
  12659. absolutely loved the discussion on baillie's series!
    1
  12660. I have a circular slide rule my father gave to me from his days as a production engineer. Irs one of my favorite things, my only regret being that these days I don't get much excuse to use it :)
    1
  12661. Good old times indeed.
    1
  12662. JFYI. in ancient India, Calculus was used from centuries ago. believe it or not, Vedic maths is really awesome.I have read that Madhavan an ancient Mathematician lived in Kerala, Southern India was the first pioneer who introduced Calculs or Kalana Sasthra(Sanskrit) ever first in the history of math....
    1
  12663. problem archimedes?
    1
  12664. The ending izz soo elegant, the music, the animation, and the proof……
    1
  12665. What an amazing video! I wish I was given enough time to use this to get my students started with Calculus!
    1
  12666. 15:26 Uncoil twice, three rotations timerwise (that's what I call the opposite of clockwise), recoil two more rotations timerwise, so total rotations are five rotations timerwise.
    1
  12667. Every time you said it's pretty natural, my mind went "there's nothing natural about this"😂
    1
  12668. Where is this not taught? Basic stuff for senior engineering students in my place. One must understand Finite Differences, Interpolation, etc. before delving into numerical solution of PDEs.
    1
  12669. Zagier's proof is not that it is 'complicated'. He just used pure math language to describe it concisely, and a professional mathematician can understand it. For us lesser humans, if you show the proof like kindergartners with pretty colored blocks, then we can get it. :-)
    1
  12670. In high school, at the time the TI30 was a new thing, we were taught how to use a slide rule, and these techniques were also taught. But only in the "college prep" (today it would be called "AP") math classes. The "required to graduate" math classes didn't teach a lot of these advanced concepts, as it was explained to me because it took too much time and people wouldn't do it because they couldn't understand it enough to trust it, so was discarded as a waste of time.
    1
  12671. Why isn't it called Antarctic Circle ?
    1
  12672. My favourite part was when you made it look like Oresme was craning his neck under the leaning tower.
    1
  12673. Wow, this video really blew my mind! Some very amazing combinatorics! Now I think I know how to solve one of the projecteuler problems that stumped me for several years, namely figuring out the number of domino tilings of a 3 x 2n board.
    1
  12674. Number 9 is most amazing
    1
  12675. professor Beautiful?!!!!!?????
    1
  12676. Napoleon's hat
    1
  12677. 👍👍👌
    1
  12678. What if we add a joker to the pack of cards, to make the trick more challenging?
    1
  12679. 1+1/2+1/3+1/6=2 1+1/2+1/3+1/4+1/5+1/6+1/8+1/10+1/12+1/15+1/20+1/24+1/30+1/40+1/60+1/120=3
    1
  12680. Ist er.
    1
  12681. So many memorable things here. Tristan's fractal visualization of the no-nine series and the progressive whitening of the structure was at the top for me.
    1
  12682. Holy wow, that was beautiful!!
    1
  12683. Genius
    1
  12684. I was confused by your statement at the 13:13 mark "we can think of this extended horn as being made of thinner cylinders". It wasn't clear that you were referring to the volume of the extended horn being made of the thinner cylinder *surfaces*, not the volumes of the smaller cylinders --- which would have meant you were overfilling the volume of the extended horn. It was when you mentioned the method of shells from calculus that, after some thought, I understood what you meant: you're filling the volume more or less like an onion --- a mathematical onion, made up of an infinite number of infinitely thin layers. It wasn't immediately obvious what you meant, as it is a little bit ambiguous in the way you phrased it: I had to rewatch a couple of times to confirm I understood what you said. Of course, this is a wonderful proof (which I realized once I understood fully what you were doing), and I thank you for showing it to us. Loved your video, as always.
    1
  12685.  @Mathologer  thank you for your reply. Yes, what I said probably would have been confusing in a different way. In any case, it only took a little bit of thinking, and frankly that made it sink in better, and made me appreciate the proof even more, thinking in retrospect. Very enjoyable, definitely. After all, if one has to work for it, it feels more like one has earned it. Once again, thank you very much for the video 🙂
    1
  12686. 1/1 = 1 + 1/2 + 1/3 + 1/6 = 2 + 1/4 + 1/5 + 1/20 = 2.5 + (1/8 + 1/12 + 1/24) + (1/10 + 1/15 + 1/30) + (1/40 + 1/60 + 1/120) = 3
    1
  12687. Gamma, divergence of harmonic series and "regular" tower as well as famous series you mentioned before were known to me, all the other staff are just a miracle. Sometimes I had to deal with "long arithmetic" and I wonder how the large numbers have been obtained with such a precision.
    1
  12688. Awesome Video, Professor Polster. Stay safe. Waiting for the next video in the series :-)
    1
  12689. Am enjoying your clear and exciting lessons in mathematics (you say Maths). Please tell us about the musical piece at the end of your video. It is beautiful. Perhaps who is the performer and what is the name of the piece?
    1
  12690.  @proth1951  From a @Mathologer response further down: The music in this video is by Chris Haugen, Fresh Fallen Snow (playing in the video) and Morning Mandolin (for the credits) and Nate Blaze 'Tis the season (chapters), all from the free YouTube music library.
    1
  12691. Very enjoyable
    1
  12692. Mathe macht Spaß
    1
  12693. amazing
    1
  12694. Rising powers? Riemannian integrals bound the area under a curve by rectangles both below and above the curve, so I am surprised this wasn't mentioned.
    1
  12695. Wow!!! Ich bin außer Atmung. (....und das ist nicht wegen Corona.)
    1
  12696. gamma is larger than 1/2 because each "blue bit" is slightly larger than a right triangle, each if which is half the area of rectangles that (in the limit) fill up the square
    1
  12697. Im quite early so no dislike (2^7)+1th like
    1
  12698. euler: original neckbeard.
    1
  12699. 35:17 Damnit. Damnit damnit damnit. Damnit !
    1
  12700.  @Mathologer  your video fits and probably extends work i've done on Pseudo-Random Numbers Generators based on Fibonacci/Pisano. So I now have a theoretical basis to extend my system from 2 to 3 variables and... potentially more, since the "boxes in boxes" could be extended to any number of dimensions, right ?
    1
  12701. I wish I could get my money back from school, except it was public school, and thus free. I suppose you get what you pay for :)
    1
  12702. The 2nd, which I found in a book from general study. Not taught at school
    1
  12703. 13:55: But doesn't the Torah say that the first son born of a Levirate marriage shall be accounted as the child of the deceased brother? (The first deceased in this case). Thus regardless of the biological father the youngest should be accounted as the child of the brother that died first. I feel like I'm missing something here. Though if the wife was pregnant and the time of the first death there is an argument the Levirate marriage would not be required and so would be invalid, though I vaguely recall this being a point of contention. I'm not a Talmud scholar so I may have gotten some details wrong here.
    1
  12704. This makes me happy.
    1
  12705. I did not understand why 3-6 is marked a "i" at 12:14. What is stacked pair? :(
    1
  12706. @Mathologer I think this would be another excellent video. Just like the procedure and instructions in Torah and Talmud (Judaism), there is a more complex procedure and algorithm in Quran and Hadith (Islam) for "Almirath الميراث - - > Inheritance".
    1
  12707. Question, how is the proverbial singularity at the bottom of a black hole any different? Isn't it just another Infiniti in a finite space just four dimensions?
    1
  12708. PhD econometrics. Needed more time on the last chapter with the infinite sums. I couldn't quickly see how the polynomials were related to the infinite sums. You could have explained that more slowly for me. I think the polynomials when summed over all values of x turn into the expressions with S(i) that you show? Is that right?
    1
  12709. Bravo! This is an actual physical proof of Pythagoras’ theorem. This is what Plato demonstrated in his Meno dialogue, where a “mere” slave boy could recreate the most advanced geometrical knowledge. This is why Socrates was such a threat to the Athenian oligarchy.
    1
  12710. With all due respect, the step right after 'Engage' in which you describe scaling all interior quantities by the incircle radius r, pointing out that the sum of three quantities is 1-D, area 2-D, and the product of 3 quantities 3-D, and then jumping right to a slide that shows the scaled Area = Sr and the scaled product of 3 quantities RGP to be Sr^2, to be completely unclear as well as rushed through (after spending significantly more time on much easier to follow steps). You aren't showing how you are scaling the sum by r at all, and you aren't explaining why scaling the area amounts to multiplying by r (why not r^2, or would it be r^(1/2)?), or why scaling the 3-D RGP involves multiplying by r^2 (why not r^3, or would it be r^(1/3)?). The inference would appear to be that scaling the sum of 3 quantities amounts to multiplying by r^0. Clear as mud why the scale factor drops by 1 in the exponent, but the way you rush through it as if one is just supposed to grasp it immediately, makes me not want to watch any more.
    1
  12711. count the dots. 3 1 4 1 5 9 ... digits of Pi
    1
  12712. I made it the very end :)
    1
  12713. Tristan’s visualization!
    1
  12714. How can you hate trig? That was the first class I took in the advanced math curriculum (over half a century ago).
    1
  12715. Jai Sarda maa🙏🏻🙏🏻🙏🏻!!
    1
  12716. "this number is actually another number. " what? I wonder what is going on here. It reminds me a bit of the p-adics thing (where you have an infinite integer which is actually a fraction ), but I am sure something deep is going on here with these divergent sequences. I wonder if I could understand the explanation
    1
  12717. It is easy to make a mistake. Recently, in the clouds, I had been writing E/m=0/∞, when it is E/m;E=0;m=∞. I had to go back and correct them all. I really like how you taught us to remember 355/113, and yes mathematical heaven actually exists.
    1
  12718. This was the first pythagoras theorem proof, where I understood it's essence
    1
  12719. Solution to the question posed: With 3 blocks in hand, one block can be completely outside with the following arrangement. First block only 25% out and the second block completely out and the 3rd block only 25% over the second block
    1
  12720. The dodecahedron has diagonals That are sqr(2)Phi, sqr(3)phi and (phi)^2 the geometry demonstration is beautiful by showing the square, cube, and large pentagon that lays on the surface of the dodecahedron - love your videos makes me happy to see other math enthusiasts even though I am not a mathematician
    1
  12721. I just realized why proportional system might not be all that great. It isn't based on the money that was given, but money that was promised back. So they can continue taking loans under higher and higher percentage. So for example if someone is worth 1m and they owe 1m, whats stopping them from still taking another 800k loan and promise 9m back? If they "burn" those 800k and go bankrupt, wouldn't the last lender be entitled to 900k? Surely there is something that will prevent that, right?
    1
  12722. Thanks new insightful
    1
  12723. Please do a video on something graph theory related:)
    1
  12724. Great ❤️❤️❤️
    1
  12725. No, an "equilateral" triangle specifically means all three sides are equal. When just two are, as in your example, that's "isosceles."
    1
  12726.  @TJStellmach  I see thanks. So isosceles are irrational.
    1
  12727. Would be interesting to see if making quadrats 80 degree, as the matrice. Distorted, and from x,y point in opposite direction. so -n8 to +n8. How many squares and triangles then?
    1
  12728. Most intriguing part is that teaser at 25:00 about relation of γ to Merten's third theorem involving primes. I think digging deeper here will not require solution to non trivial zeros of Zeta function for a pattern in primes. This is it, the shortcut to heavens!
    1
  12729. I am sure Martin Gardiner touched on this in a column once. Always neat how everything old is new again and how things come up in different forms. Not everything is published either, often they were just fun distractions.
    1
  12730. "If you've got access to JSTOR" How can I get access to JSTOR?
    1
  12731. Isn't the "trapezium" a trapezoid?
    1
  12732. At 12.31 we can notice that 1+1/2+....+1/8=2.7178 quite a nice and unexpected approximation of e !
    1
  12733. This must be related to the Moleeds theory. All things are Moleeds
    1
  12734. I didnt made it to the very end - asked for a comment.
    1
  12735. this blows my freaking mind! why haven't i seen this before? astounding. this shou;d be taught with the diffy -q and linear algabre. you math guys hide some of the coolest things.
    1
  12736. 17:30. Why the area of the red triangle is ABF: If we draw a height from side B to the top vertex, this height will be the same length as in an equilateral triangle of side A. This is because the little triangles to the left of the height have two equal angles (60 and 90 degrees) and an equal side (A). So the area of the equilateral triangle is AH/2, and the area of the red triangle is BH/2. The ratio of their areas is B/A. The area of an equilateral triangle is proportional to the square of its side. Thus the ratio of the area of an equilateral triangle of side A to an equilateral triangle of side 1 is A*A. Overall, the ratio of the area of the red triangle to the area of an equilateral triangle of side 1 is A*A*B/A = A*B.
    1
  12737. Sir, I really like your content, like I said in previous videos. I also said that I should have had better teachers or paid more attention to math classes... :D Still, this are not that complicated, you chop the info in really tiny bits. However 45 minutes (or 22,5 because I watch YT videos at 2x speed) are still very long videos. Anyway, I've seen it all. :D Congrats on the video and regular lessons! Cheers from Portugal! P.s.- Your giggles break me every time! :D
    1
  12738. I think the real question here is: Why are we putting numbers in a circle in the first place?
    1
  12739. I love the short form videos :)
    1
  12740. Are there any convergants of the n^n infinite series less than 0?
    1
  12741. You can see a lot equal triangles. This square is 4 of this triangles, the others =16. S=1=20*x=>x=1/20=> the area 4x=1/5
    1
  12742. When Mathologer casually drops a fact like the exact number of eyelets in God's shoes, I believe him.
    1
  12743. The shadows don't really interfere with each other. Imagine two lines that don't intersect in 3D can still have shadows that appear to intersect, but nothing special happens at that point. It's just an artifact of the way you've chosen to do the projection.
    1
  12744. I am half-way into and I know what I am going to do on Sunday. Thanks. Mind blowing (Cliched term, I know - but true)
    1
  12745. I find it hilarious to claim something is universal when it is specific to the base 10 system.
    1
  12746. former Yugoslavia in 1990s, using an old Communist-era curriculum from 1960s, we learned this in 7th grade primary school
    1
  12747. My vote goes to the fact that the partial sums are never integers, quite amazing! Tied for second are the bishop's proof (which I did know, but not that it is 700 years old), and Kemper's proof.
    1
  12748. I used my slide rule to figure out what your shirt meant.
    1
  12749. @3:20, there's a 2 in pi and a 1 in the 1,000,000 calculation, but they aren't highlighted as being incorrect.
    1
  12750. 7:46 You already did that with that initial cardioid, if that wasn't the ultimate mathematical 'but' (I resist the second t on principle). OK I'll stop now (I blame our host for bringing out the inner schoolboy!)
    1
  12751. Proposed answer to the riddle in the end: SPOILER! Denote the number sequence a(n) where we sum all a(n) is the sum of products of non-unique (not up to permutation) partitions of the number n. For partitions of n, look at all the ones that start with 1, then the sum of products of such partitions excluding the factor 1 in each is a(n-1). For the ones that start with 2, then the sum of products of such partitions excluding the factor 2 is a(n-2). Hence a(n) should include the terms 1*a(n-1)+2*a(n-2)+3*a(n-3)+...+(n-1)*a(1) which covers all partitions starting with 1 to n-1. The only partition we are forgetting is the trivial partition, i.e. partitioning n as n. Adding this term n to the sum we get a(n) = n + 1*a(n-1)+2*a(n-2)+3*a(n-3)+...+(n-1)*a(1) = n + sum_{i=1}^{n-1} (n - i)*a(i) as the final formula. Trying this formula we get the beginning of the sequence as 1 3 8 21 55 144 377 987 One might also want to express the term n as n*a(0) or something like that, but it did not make much sense to me. Also, one might want to simplify this expression, but I only come up with adding a(n-1) to a(n), which gives a(n) = 1 + 2*a(n-1) + a(n-2) + a(n-3) + a(n-4) + ... but this for me is losing all its intuition and is not that much nicer.
    1
  12752. How a finite volume can have an infinite area can easily be demonstrated by the following thought experiment: Take a cube with sides equal to 1. Then the volume is 1 and will remain 1 even if we cut up the cube. The area is 6 x (1 x 1) = 6. Cut the cube in half. We get 2 cuboids with dimensions (1, 1, 1/2). The area of each cuboid is equal to 2 x (1 x 1) + 4 x (1 * 1/2) = 4, so 8 for both cuboids. What we've done is adding two 1 x 1 sides at the cut. If you repeat the process we'll add two 1 x 1 sides for each of the new cuts, so the total area will become 12. After another iteration the area will be 20, and so on ad infinitum, where the area will be infinite, while the volume still equals 1. (This is also the reason coffee gets ground)
    1
  12753. The Red Cross puzzle looks like it is a 45 deg rotation. So, the edge length is 2*root(2).
    1
  12754. I was looking for an explanation of how to build a gear driven clock to travel at variable speeds while rotating. Oh well, . My Dad gave me a circular Slide Rule that was a given to him by a salesman as a present that only engineers or math mavens could enjoy. The one thing that the slide rule assures is a lesson on significant numbers. The modern day calculator implies accuracy that is not justified.
    1
  12755. I seem to recall a bit in "Connections" TV series that during the French revolution surveyors were surveying the countryside and they used a series of triangles. Now the last time I saw this was in the 1980s and that's a long time ago but it may be the start of unwinding why it's called 'Napoleon theorem'
    1
  12756. What did I miss? Area is base times height over 2, but the height you used is not perpendicular to the base.
    1
  12757. This time you have(got) me, with the hot Schokolade ;)
    1
  12758. I enjoyed the no 9's series and the fractal animation explanation
    1
  12759. Favourite part is the balanced warm-up. I never thought about why it works.... but that visual makes it so clear and at the same time still give the feeling of wonder.
    1
  12760. 14:33 The O figures are pretty restrictive, as they cut off a block on the outer left resp. outer right, which must have an equal number of red and green squares in order to remain tileable. This leaves 4 options for those O figures, giving a sum of 4 products for the total ways to tile these glasses. Likewise, the two parts of each outer block cannot be bridged, and so they can be viewed separately for tiling: 3*3*2*3*3 + 3*2*3*3*3 + 3*3*3*2*3 + 3*2*5*2*3 = 3 * (3*18+20) * 3 bruh
    1
  12761. I just like to say the divisibility test was never taught in school I was never taught to add digits together and then see if they're divisible by something to figure that out that was never taught to me ever not even in college in remedial math which I only took as a refresher course not cuz I had to
    1
  12762. 11:00 Fractals =D
    1
  12763. I still don't think "Pi is the sum of two squares." as claimed in the video. A PR disaster, worth correcting.
    1
  12764. When you first mentioned the no 9's sum, I misunderstood it to be "no leading 9's" sum (ie 1/9[n])... It makes sense that no 9's sum would converge, and therefore no [n]'s sums would all converge, as the table you showed illustrates. However, the "no LEADING 9's" sum I mentioned above, surely that would diverge? Has anyone worked on this? It seems obvious to me it should diverge, though I'm not sure my intuition is correct.
    1
  12765. Feeding the algorithm to give it the "boost" you asked for. Haven't watched your stuff in a while... Your mistake post with the 🤦picture is the first one to pop up in my feed for a while. Happy new year mathologer!!
    1
  12766. The color question has to do with the color wheel and contrast. Two primary colors have an opposite color on the wheel. Triangles and Circles relationship.
    1
  12767. seems everyone is so obsessed with analog computer recently🤔
    1
  12768. This video (and some of the referenced ones) always remind me that I really should get around to truly learning about continued fractions... one of these days!
    1
  12769. Okie Dokie 🤔 I don't consider this a good approach for homo sapiens Sapiens except people who code lil programs, iterative method very numerical. Now finally around 16:30 the Euler, convergence. 😳 Oh, professor, now you will get the value of e? 😅 I see at almost 20:00 you are basically using not much scratch, since you know the way to proceed. There's it, complex variable and applications, Churchill Brown. Finally, fundamental theorem! Often we have to start from contour integral and connected domains multi or single, multi is not much, just we work with any region like that but we put them inside a surface with boundaries, which serves to approximate the maybe odd figure
    1
  12770. A stunner as always! I’m still waiting for the Abel Ruffini proof to get the mathologer treatment someday :)
    1
  12771. Very hard, but amazing!
    1
  12772. Definitely intrigued by the equation at 25:05. My first thought after seeing that was to take the natural log of both sides to get an explicit formula for gamma, leaving us with a sum of logs whose first term is ln(ln(pn))... but what's with the double natural log here?
    1
  12773. the most memorable part is the "terrible aim" part of the video! its really got me thinking about the four parities of rationals, as well as the no integers in the sums part :)
    1
  12774. Similar to all programming languages , you start with multiplication and division first then you do the addition and subtraction .. Easy , right ? .. Now complications occur when you don't follow the rules . Conclusion : FOLLOW THE RULES .. Lol.
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  12775. OMG! It is brilliant!
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  12776. , if you counted points in space, how many would there be? so, we can show a square meter has 6 square meters of surface area, its going to need a little doing, but we will get there. To get a circle by points, we say; 2x2, 4x4, 5x5x5, by representing a number of bits to turn on a light bulb, it takes a multiplex of the number of bits of the signal type, such as liquid crystal display, but an array of liquid crystal displays. Due to this complexity, a knob or a slider of any rotation excedes any surface area as shown to achieve transform mechanical heterodyne work. Si an example is the ultimate micro chip. A modest 256 megabyte square array, for mathologers on a tight budget, but you gotta win right? So, theres all sorts of ideas, except, why didm't they make this? Only the garbage videa card and CPU? Because of the number of possible busses makes the chip infinitely high stacked, far exceding the chips surface area.
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  12777. I teach Art of Problem Solving Geometry and I've taught the proof of Herons formula and even the Area=(perimeter • inradius)/2 for years. I never even thought that perimeter/2 is S from Herons formula. I had no idea there was a connection and my mind was blown at every step along this beautiful journey.
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  12778. 21:50 is interesting because it shows a flat earth. Usually the flat earthers' map puts the north pole at the center. Maybe you can make a version of that for clickbait :D
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  12779. The swivelling proof is better because it opens the idea to so many other uses in geometry.
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  12780. Well, that was exiting. Thank you for providing new videos. This one reminded me at another aspect of Eulers formula: Graph duality demonstrated e.g. by "Euler's Formula and Graph Duality" (3Blue1Brown). We'd love you to talk more (as you indicated) about 'meta-cubes'...
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  12781. This video is quite the Magnum Opus! It makes a result from deep in the heart (bowels?) of mathematics accessible to mere mortals, and introduces a bunch of mathematical constructs (squares in Zn, rings, sign of a permutation) along the way. For me, watching this video felt like being a tourist on an eloquent expertly-guided tour of a hidden room inside a massive museum. I am left in awe of the inventiveness of the mathematical minds of yore and supremely appreciative of Mathologer's efforts to spread his enthusiasm for mathematics. In response to Mathologer's query of what worked for me, with the first viewing I felt I was on top of the material until maybe the last 10 minutes. I expect that a second viewing will fix that, so... Congratulations Mathologer! I think you achieved your goal.
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  12782. 👌👌👌👌👌👌👌
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  12783. An idle thought/unanswered question about rearrangements of conditionally convergent series I've harbored for a a long time If you think about these rearrangements as permutations on N applied to the series 1 + -1 + 1/2 + -1/2 + 1/3 + -1/3 + ... the greedy process of selecting positive and negative terms is a way to link any real number to a permutation on N. It can be proved that, among permutations on N, the "bounded" ones do not result in a different sum: if p: N --> N is the permutation and |p(n) - n| is bounded over all n, then the sum will remain zero. This is true in general: for any conditionally convergent series, bounded permutations of the terms do not alter the sum [I've seen this proved in an article in a Math magazine!] Within the domain of permutations we can also establish the notion of two permutations being bounded from one another: p and q are bounded/close if |p(n) - q(n)| has a finite upper bound. My question is, is it true that whenever p and q are permutations that correspond to the same real number, are they bounded from each other? ------------------------------ The rest of this is fleshing out the details of what I mean, and why the answer either way seems very unintuitive (to me) Firstly, let's pin down that there is more than one permutation associated with any real number. For instance, let r be the permutation on N resulting from greedy method from the video for making pi using the above conditionally convergent series. Now construct a new permutation, s, by repeating this process, except we pretend we mess up with an early insertion of -1/3 to get 1 + 1/2 + 1/3 + [-1/3] + 1/4 + 1/5 + ... + 1/13 + -1 + -1/2 + -1/4 + ... (note lack of of -1/3 in the subtracting section) and from there continue without error -- adding terms till you exceed pi, subtracting terms until you're less than pi, and so on. Though it doesn't strictly adhere to the method outlined for making a permutation that sums to a desired number, that doesn't mean s does not correspond to a series that validly sums to pi. After all, it's equivalent using the same algorithm to construct a sum to pi + k, for some rational number k, when starting with a conditionally convergent series made by dropping several terms from the original conditionally convergent series, then reinserting the dropped terms. On their face, r and s would be very different permutations. By close examination you wouldn't be able to tell there's anything that links them unless you backtrack through the processes that defined them. It would be surprising if r and s were bounded from each other. Yet they both correspond to the same sum
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  12784. Did not get a notification. Had to manually find it. Information
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  12785. The gap between sample points can only approach zero, never reaching zero.
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  12786. Definitely chapter 5, the visual proof that the blue square bits sum to less than 1 is so simple. And then to find out it sums to gamma, wonderful!
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  12787. You should do a video with Sabine Hossenfelder ....
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  12788. At 3:00, what sticks out to me immediately is that the negatives are cancelling out earlier positive terms, and there are always going to be positive terms that aren't cancelled-out. It's a bit like the dodgy guys from Numberphile trying to tell us a divergent sum totals to -1/12.
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  12789. Pretty much as I say at this point, there is a conflict between the mathematically sanctioned definition to calculating the sum of an infinite series (=limit of the partial finite sums), vs. what we'd expect the sum to be based on the pairs of terms cancelling out. The main message from this video, apart from the nice visual proofs at the beginning, is that the sums of infinite series don't necessarily commute. But that does not mean that these sums are not useful or don't make sense :) I guess the next question is why don't we define the sum to be zero when terms cancel, right? Well, that is because for most infinite series terms don't cancel in pairs and so our definition would not apply to most infinite series and so wouldn't be very useful.
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  12790. - This is a lovely video. The heavy lifting is done by all the properties of the limit (in a mathematical sense), which are disguised in the clever notation. - 35:10 e = lim (1 + x)^(1/x) as x goes to 0. Swap the usage of lim and write e = (1 + dx)^(1/dx) instead (abusing/exploiting the notation). Raise both terms to the power of dx, then subtract 1: dx = e^dx -1. Finally, divide by dx to get 1 = (e^dx-1)/dx - By using the concept of the car as a vehicle for the explanation (pun intended), some hard issues are avoided, such as "can I differentiate any function at any point?" or "can I integrate any function on any interval?" - We Spanish speakers have access to a very neat opening interrogation symbol, "¿". This is very handy, given that you know that the phrase is an interrogation from the very beginning. - The rule for the derivative of the inverse function is graphically very intuitive after noticing that the plot of the inverse function is the same as the original function but mirrored about the line y=x. Hence, the slope of the inverse function is the (algebraic) inverse of the slope of the original. I'm sure Mathologer is aware of this fact, and probably decided to leave this property out for the sake of the total length of the video. - 27:22 That math joke is brilliant. Here's a similar one I found elsewhere: take the equation x^2 = 25. Of course (?) you can cancel each number 2 appearing on both sides, which leaves x = 5, the correct solution for the original equation. I wonder if there are more of these... - 32:15 Fun fact: the quotient rule for h = f/g can also be obtained by applying the product rule to h = f * (1/g) first and then using the chain rule with the functions 1/x and g. - 33:17 The rule (x^n)' = n x^(n-1) is the same for ANY value of n, not only positive integers, but for any real number. But the proof is more involved and I would start by writing x = e^(ln(x)). I liked the music of the final part!
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  12791. OK, but what are r and s, algebraically? They must be of higher degree than quadratic, because if they were quadratic, then a regular heptagon would be constructible with compass and straightedge, and we know that it isn't. These lengths could be expressed as ratios of absolute differences between some pairs of 7th roots of unity, but similarly phi could be expressed as ratios of absolute differences between some pairs of 5th roots of unity, but phi is really only quadratic and not quintic. So what degree are r and s? Maxima just says things (cos(4π/7)-1)/(cos(2π/7)-1), but that's the question, not the answer.
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  12792. Man personally thanks for picking this subject. Make one about intuition of logarithms
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  12793.  @Tumbolisu  Nice way of putting it :)
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  12794. I cannot help myself watching until the end of the video. the gamma is less than 1 really shock me.
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  12795. Only because you asked for notes: The bit around 26:00 felt very much like it fell out of nowhere. And then saying that I can plug a few values in to see that it is true makes me feel even more distant from the proof-y reasoning I’m hungry for. Additionally, the matrix inversion derivation for the Bernoulli numbers feels so labor intensive and mechanical, compared to a closed form for the nth Bernoulli number (although it did connect very naturally to the ideas being presented). The rest of it all clicked, although I have experience with all the prerequisite number theory, linear algebra, and calculus, and was very interested in the sums of powers at a young age, so it’s not much of a surprise that I followed along with and ate up this video. It goes without saying, I loved this, and eagerly await more Riemann-Zeta, Bernoulli, or, uh, any content from this channel, really.
    1
  12796. Can someone explain in more detail why and how stretching the rubber band leads to a logarithmic scale? I have a vague idea for it with differential equations.
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  12797. My favourite is no 9's series and Robert Baillie's paper :)
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  12798. I loved this video as I do all your videos. Long ones are fine because you keep them interesting and don't bog down into stuff that is too technical. When you say 'are you ready to have your mind blown?', I say "yes!", and I'm never disappointed. I've studied math formally through differential equations and linear algebra and informally through videos and reading. I'm surprised you've gotten this far without hiring people to help w editing etc. I'd be very willing to contribute via patreon, since you are my favorite channel. I already give to 3 blue 1 brown and Matt Parker. The only part of this video I did find awkward was the 'morph' thing in chapter 7... I can see that if the '+f(1)' form is the standard way of writing it, you might have wanted to show that for people who had seen that form before... but for me it was not really very 'mathologerlike' because it didn't seem like it had any purpose or elegance... it seemed just kinda wedged in there with little explanation. Looking forward to your next video!
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  12799. Taking the derivative of 4/3*pi*r^3 = 4*pi*r^2 or the surface area of a sphere. But in ancient times, without modern calculus, how did they express or derive the latter from the former?
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  12800. Very beautiful and simple proofs of Pythagoras. Thank you. Now three easy challenges for you, dear mathologer: a) prove the horizontal line of sight of an observer looking across the sea from a beach makes a right angle with the earth's radius. b) prove that a 2,000 foot mountain 60 miles across the sea cannot be seen by the observer if earth is 24,901 miles in circumference. c) find a large body of water with nearby mountains and see them from a beach on the other side and prove by contradiction earth is not a ball 24,901 miles round. Good luck!
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  12801. Sorry did a mistake at first... it was Vsauce not Matt parker... Matt parker did rhombic dodecahedron
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  12802. Lost a heart it seems, but misinformation is never good
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  12803. Some absolutely beautiful maths in this one. I rather love the concept that all n-sided shapes are the interferance pattern of n perfect shapes.
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  12804. This is beautiful.
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  12805. Hey, you're right. Carculus is easy!
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  12806. Not quite an aside, but if all "calculation" is reiterative reintegration of here-now-forever, then division by zero-infinity is eternal sync-duration and division by one, the reciprocal, unchanged "solidity" of state-ment, equal to a frozen phase-locked substantion. From observation of the e-Pi-i sync-duration dimensionality coordination analysis, ie the inherent mathematical intuition, under-stood circum-stances of ultimate Superspin Modulation positioning, the first principles reasoning about why this "works" is the i-reflection containment connection between e-scaling and Pi-bifurcation multiples proceeding from unchanged Unity, to twoness symmetrical probability distribution in Spheroidal Toroidal proportioning, sync-duration connectivity as described/demonstrated in the video. The point-in-perspective being .., how Mathematicians develop intuition about numberness by holding a variable, either e or Pi and picturing numbered gradients in their minds. A natural identification, pictographic symbolism from learning by doing experience. (Genius everyone has the mechanism for, if not the motivation) More beautiful Mathemagical orientation.
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  12807. Coloring
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  12808. 25.8069 is not the root of evil, unless somehow you are not using the normal rules of rounding. It should be 25.8070 when written to 4 decimal places. Am surprised that a major maths channel would allow that to slip through the net!
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  12809. I'm terrible at math, so I am all the more amazed by this explanation I can understand. almost to the end. This is remarkable instruction.
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  12810. my answer for the powers of 2 question is "the binary chapter of my intro to computer theory notes from the late 90s", because oof. I used to know this answer.
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  12811.  @MathAdam  f(🙂)=😐
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  12812. Video still on before start ads and i already love it. That image next to 'skip ad' is 👌
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  12813. My favorite part was gamma, finally I got an idea what is it :)
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  12814. Very good video! Professor sorry, I got carried away with the "Golden section" again - maybe it's better to consider the sum of squares like this https://www.geogebra.org/geometry/daxcs75f ? Thank you again for your excellent work in this video...
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  12815. 1: Assume x = y+1 Then x^2-y^2 = (y+1)^2 - y^2 = (y+1) + y (Or alternatively y^2 + 2y + 1 - y^2) = 2y+1 for all y in Z Thus, all odd numbers can be represented by x^2 - y^2 where x and y are in N. 2: (Assuming you meant unique for all positive integers x and y.)* Let p = x^2 - y^2 be an odd prime number where x, y in N. Since all primes p must be positive, x > y. We know that x^2 - y^2 = (x+y)(x-y). let a = y - x so that y = x + a. Since x > y, a > 0. Then p = (x+y)(x-y) = (2y+a)(a) Since p is prime, then one of the factors must be 1. But 2y+a > 1 when a > 1, thus, a must be equal to 1. Since (x+w)^2 - (y+w)^2 > x^2 - y^2 where w in N, x = y+1 and y must be unique in x^2 - y^2 when it is set to evaluate to a prime. *Assumption made because (-x)^2-y^2 = x^2 - y^2 and so on.
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  12816. ended up changing email addresses a few months ago as part of a name change. I am so glad I remembered to re-sub and ring the bell. Your work is amazing for any time I have the brain power.
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  12817. i think the best part is the gamma.
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  12818. Most memorable: Tristan's fractal blew my mind!
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  12819. 15:55 This explains why court jesters often wore patchwork garments. It was obviously the result of contorted inheritance arrangements.
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  12820. Lovely! and, the music score through the end is terrific.
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  12821. You should be aware of where the captions get placed in your videos and not put content there. How are deaf people supposed to keep up? Hell, I can hear fine but still have better comprehension when reading the captions while listening- and if those captions are covering what is being discussed than it is very frustrating.
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  12822. t= 21:04.... is that a flat-earth proof (LOL... Nevermind, I know Exit)
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  12823. 14:40 - How do you know this: "Well, because every square pops up exactly twice , exactly half of the non-zero numbers must be squares"? For Z/7 it's true. In the multiplication table there are 18 non-zero "squares" and 18 non-zero "non-squares".
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  12824. He's not talking about the entries in the multiplication table. He's talking about perfect squares. In other words, numbers of the form x^2 where x is a number in the mini-ring. (Though I guess the same will also be true if you look at the full multiplication table, but that's not what he means.) The perfect squares are precisely the diagonal entries of the mini-ring's multiplication table. But yes, exactly half of the nonzero numbers in the mini-field will be squares (meaning they will appear on the diagonal of the multiplication table) if we're going modulo an odd prime. Why? Well, if we're in Z/p for an odd prime p, then there are exactly p numbers total, an odd amount. So if we remove 0 from consideration, there are an even amount left: p-1. Next, notice that because of how weird the addition is on Z/p, every number can be seen as both a positive value and a negative value. For example, in Z/7, 5 is both +5 and -2. Adding 5 has the exact same effect as subtracting 2. (Try it out for a couple of numbers!) This can be seen because adding/subtracting 7 is the same as adding/subtracting 0 in Z/7 arithmetic and because -2 and +5 are 7 units apart. Now, because of how multiplication works, opposites square to the same thing. So in Z/7, 2 and 5 (= -2) must square to the same thing (in this case, 4). So now, we're approaching the result. Since there are an even amount of nonzero numbers in Z/p, if we show that each square comes from squaring precisely two different numbers in Z/p, we will have then proven the claim. To show that each square comes from at least 2 numbers, we have to show that no number is its own negative. Suppose n is a number in Z/p so that n = -n in Z/p. Then adding n to both sides, we get 2n = 0 in Z/p. But the only things in Z/p which are 0 are multiples of p. Since p is an odd prime, p doesn't divide 2, so p must divide n (since p divides 2n). This means that n is a multiple of p, so n = 0 in Z/p. This shows that there is no nonzero number in Z/p which is its own opposite. Since opposites square to the same number, this shows that every nonzero perfect square in Z/p comes from at least two different numbers in Z/p. All that's left to do is show that you can't have 3 or more numbers square to the same thing. We will do this by showing that if two numbers square to the same thing in Z/p, then they must be opposites. Suppose n and m are two unequal nonzero numbers in Z/p so that n^2 = m^2. Then n^2 - m^2 = 0 in Z/p. But we can factor this. n^2 - m^2 = (n + m)(n - m). So we have (n + m)(n - m) = 0 in Z/p. Since p is a prime, this implies that n + m = 0 in Z/p or n - m = 0 in Z/p. (If a prime divides a product, it must divide at least one of the factors.) The first option gives n = -m in Z/p, meaning they are opposites. The second option gives n = m in Z/p, which cannot happen because we assumed they were different numbers in Z/p. So the only things that can square to the same number are opposites in Z/p. So what we have shown is that in the mini-field Z/p for an odd prime p, there are an even number of nonzero numbers. Every number squares to a perfect square, and numbers which square to the same thing come in pairs. So there are precisely two nonzero numbers in Z/p for every nonzero perfect square. This means that exactly half of the nonzero numbers in Z/p are perfect squares. This leaves that the other half cannot be perfect squares!
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  12825. If you want to avoid talking about "negatives", I've come up with another way to show that two different numbers square to the same number. Fundamentally, it's the same thing. But you don't have to think about negatives. Suppose m and n are numbers in Z/p so that m+n = p in normal arithmetic. Then m^2 = n^2 in Z/p. Why? Multiply both sides by of m+n = p by (m-n). m^2 - n^2 = p(m-n). But since p(m-n) is a multiple of p, this tells us that in Z/p, we have m^2 - n^2 = 0 (in Z/p). Adding n^2 to both sides gives m^2 = n^2 in Z/p. So if you have two numbers that add together to give p normally, then they square to the same thing in Z/p. And since p is an odd prime, n and m have to be different. If m = n, then m+n = 2m. And, like I said in my original reply, if p is an odd prime, 2m cannot equal p. And these numbers definitely come in pairs, since if you take m to be a number in the range 0 < m < p, then you can take n = p-m.
    1
  12826. Hi! I just finished the video. Undergraduate Engineer here, I found this video easy to follow even with the hard concepts. Definitely enjoyed the humor that went along with it as well because it kept me captivated :)
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  12827. Math is beautiful
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  12828. So the 3d example with 4 colors has rule where the vertices are (1,1,1,1), (2,2,0,0), or (4,0,0,0). Which feels like quaternion algebra can be integrated into this in some way. Though, I thought the same with the pascal's triangle with complex numbers so I might just be crazy.
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  12829. Does it work on any sequence? If so why aren't prime number sequence yet answered?
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  12830. My vote for the unexplained expression of gamma by primes ...
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  12831. To me this does not seem to be a paradox. The problem occurs because of treating the outer surface of the horn differently to the inner surface. I solve this by applying the same trick outside like inside as following: Let's build a second horn using the function g(x):=sqrt(2)/x. This function g(x) is completely outside the function f(x):=1/x. Each point of the function g(x) defines a radius being sqrt(2) times the corresponding point of f(x). As the radius contributes with the square to the volume, I can immediately calculate the voulume of the outer horn being equal to 2 pi. Now let's focus on the gap between the inner horn and the outer horn. This gap has a volume being the difference between both volumes being equal to 2 pi - pi = pi. Now I fill this gap the paint and I will need the amount of pi paint for this. This amount will totally cover the outer surface of the inner horn. Therefore the ammount of color to paint the outer surface of the horn is equal to the paint for the inner surface. There is no paradox at all.
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  12832. Most amazing part for me was the addition of constant gamma.... I get that the extra area would be greater than 0.5 but involving gamma in the approximation, dramatically changes the situation ! It is so cool. I love it :)
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  12833. Great video and so helpfull for experimental physisist🤔. Color is great 😉! Is there a link to higher dimensional spheres, should build one ..... (physisist goes away mumbling int his beard)
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  12834. The first mutilated chessboard was one of the first real math puzzles i solved, perhaps 50 years ago
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  12835. Makes me think of electron orbitals...
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  12836. happy holidays
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  12837. Something that doesn't 100% make sense to me. At least with how we describe the operations we do to the polygon at the start, rotating the points doesn't make it a different polygon. By which I mean, if we describe a polygon with an ordered list of points, and we rotate that list (relabel the points), the polygon doesn't actually change, and nor does it affect what the operations do. So actually, we don't need all 10 degrees of freedom that the basis eigen-pentagons give us for example. Using the basis shown at 17:28, let's say the components are E1,E2,...,E5 a regular pentagon could be described as c1E1 + c2E2 or c1E1 + c5E5. Actually, wouldn't even a linear combination like c1E1 + c2E2 + c5E5 work? So wouldn't doing the steps of 144deg and 216deg ears suffice for getting a regular pentagon?
    1
  12838. Ok, having made something which lets me play with a linear combination of eigen-pentagons... having both E2 and E5 does not make a regular pentagon. But moving around c2 or c5 makes it look like I'm rotating a pentagon in 3D space.
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  12839. Thanks!
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  12840. Going 'viral' is so 2000s ...I think in 1910 they would have said the book 'spread like the plague'.
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  12841. In India, Heron's formula is taught at school level.
    1
  12842. This was an amazing video!
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  12843. Omg I proved it exactly as Tristan
    1
  12844. The Colouring proof.
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  12845. Love your videos!!!!!!!!!
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  12846. 2020 = 2^2*5*101 Good: 1 5 101 505 Bad: none Number of ways = 4(good-bad) = 4(4-0) = 4(4) = 16 2020 = 1444 + 576 = 38^2 + 24^2 = (-38)^2 + 24^2 = 38^2 + (-24)^2 = (-38)^2 + (-24)^2 2020 = 1764 + 256 = 42^2 + 16^2 = (-42)^2 + 16^2 = 42^2 + (-16)^2 = (-42)^2 + (-16)^2 Then just swap the order of the numbers to get the other 8 solutions.
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  12847. Wonderful, wonderful! Even a mathematical baby like me could follow it (mostly). But will someone indulge this baby by answering the following question: at 6:22, we see that e = the infinite sum + 1. But everything on the right is a rational number. Can a sum of rational numbers, even infinite, ever equal a transcendental number?
    1
  12848. That Kempners no 9s series konverges.
    1
  12849. Interesante
    1
  12850. Great thing to watch while reading JoJo Steel Ball Run 😂
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  12851.  @Mathologer  its a manga were the golden ratio is very prominent and important for the story :)
    1
  12852. Never thought you can take a DFT of a triangle
    1
  12853. 10:07. It appears to actually tile somewhat.. So, it is kinda beautiful.
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  12854. i keep seeing 3d rocks in the artic circle
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  12855. Most memorable part: Thinner and thinner, intuition that no 9 sum is actually converges.
    1
  12856. Interesting... hmmmm....
    1
  12857. I don’t understand the role of the sea pickles in all this
    1
  12858. what if it was two spheres on inside the other. or one sphere in a bowl?
    1
  12859. this would be fun to make in Desmos
    1
  12860. I want that ti-shirt
    1
  12861. Arithmetic class 1973. 1st year Scottish secondary. Can’t remember teachers name. Possibly Kenneth Head. Used the areas as described in the first 4 minutes of the video.
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  12862. Beautifully mind bending
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  12863. 12:14 what was that 😂
    1
  12864. 2:03 Mind Blown Visual proof that 1+2=3
    1
  12865. 28:00 I'm a little confused. We seem to be assuming that all of the points on the octagon are a part of a square grid (since we're using our shifting property with those points)... but why though? I get that their x coordinate-points are all guaranteed to be part of a square grid, but I don't understand why the other points would be (and hence why we're allowed to use them in our shifting).
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  12866. The shifting property is being applied to the x coordinate only. It’s as if the grid of points has been replaced by a ruled page of vertical equally spaced lines.
    1
  12867. I don't think I get the challenge at the end. 8=3^2-1^2, but 8 is even. Could someone tell me what I'm missing?
    1
  12868. In the first part you prove the volume is less than 2π but then a bit later claim it's equal to 8. A check to see if we are paying attention?
    1
  12869. Most memorable was The Euler-Mascheroni constant (GAMMA). I've even started doing some research on/about it😁
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  12870. at 7 minutes, checking 81 is sufficient; because, subtract 81 and you have a number which is a multiple of 100, therefore divisible by 4. 81 is if course 20*4+1, so yes, it satisfies the requirement
    1
  12871. I am en engineering student...and the great video which inspires me learn more mathematics. Thanx for your great work.
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  12872. I've done some STEM tutoring/mentoring, and the reasoning behind the "Balanced warm-up" of Chapter 1 got me thinking about how to better illustrate (if not animate) a variety of other problems. Thinking about this may mean I may not have been paying very good attention to the subsequent chapters...
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  12873. Idk. I learned abt hyperbolic trig in school
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  12874. 13:15 is the point we end up after N-1 iterations (when N is the number of sides) the barycenter of the polygon?
    1
  12875. Love the humor 😉
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  12876. 33 = 15 + 14 + 4
    1
  12877. We want a video on graph theory. I hope you try it.
    1
  12878. For a different way to prove the thing with power of two at the beginning (without knowing that the binary system identify any number exactly once), it's easy to prove by induction. We have our n factors of the form (1+x^(2^k)), let's prove that the product of those is indeed the sum of x^j with j form 0 to 2^n-1. First (1+x)(1+x^2) is indeed (1+x+x^2+x^3). Then, the induction gives us that the n first factors give us (1+x+...+x^(2^n-1)). Let's multiply it with the next factor (1+x^(2^n)). By distributing, we have the sum of (1+x+...+x^(2^n-1)) + (x^(2^n)+x^(2^n+1)+...+x^(2^n-1+2^n)). That last exponent is equal to 2*2^n - 1, which is equal to 2^(n+1)-1). Induction done. It's pretty easy to visualise this proof, but a bit hard to do in a comment, so I'll let you the exercice to do it.
    1
  12879. He nailed it once more.
    1
  12880. the shirts alignment annoyed me...so i figured out how to make the arrows flow perfectly. im happy now.
    1
  12881. So cool
    1
  12882. Hap-Pi Christmas
    1
  12883. Tbh the balanced warmup was surprisingly satisfying (yes, I watched the whole video lol)
    1
  12884. The most memorabla part for me was the part where we find that the partial sums do not hit an natural number larger then 1
    1
  12885. If the two coins were rolling at the same time they would each complete one full rotation, with one rotating clockwise and the other counter, and it would be a 1:1 progression. But since it is not the outer coin must work twice as hard and cover twice as much area to return to start. Now - imagine if one was a role of tape with exactly one full piece, sticky on the outside. The other an empty roll exactly the same size. If you roll them together to transfer the tape is the principle the same? Because that is hurting my brain atm. Thanks 😊
    1
  12886. The most memorable thing for me are the bizzare maximal overhang towers. I knew about the leaning tower of Lire but I don't remember seeing the maximal towers. It reminds me of the sphere packing problem and how people had to consider all of the irregular packings to show that the trivial packing is the most optimal. Only here the irregular way is the most optimal.
    1
  12887. Okay, so I'm a high school student. I know a fair bit of calculus and have watched a shitload of Maths videos online. Didn't have a problem wrapping my head around this, but probably cuz I didn't try filling in the details myself. More of these please.
    1
  12888. tight lacings for n eyelet pairs = (n!*(n-1)!)/2.. Hope this is correct.. below is my long explanation (I am sure there is a short and better explanation for this). For tight lacings with 5 eyelet pairs, we have a choice of 10 eyelets to select at start. next we have 5 eyelets to select (which are on other side of first selected eyelet), next we have 4 eyelets (which are on same side of first selected eyelets).. If we continue to select this way.. we get 10*5*4*4*3*3*2*2*1*1 (i.e. 5!*5!*2). but in these combinations there are lot of lacings that will repeat.. To know how many are repeating, arrange eyelets in a circle for a given lacing. And we are calculating same lacing 20 times in our above calculation.. how? we are selecting the starting point 10 ways and for each way you can go clockwise or anti clockwise.. these are calculated separately as well. So total tight lacings for 5 eyelet pairs are (5!*5!*2)/(5*2*2) = (5!*4!)/2 = 1440 in general the solution is (n!*(n-1)!)/2
    1
  12889. 21:11 🙊
    1
  12890. For me chapter 2 was very nice, especially the "ugly" solution that I would never have thought about.
    1
  12891. Shipping would cost too much for an infinitely long horn, that's why you can't find them on Ebay.
    1
  12892. Whole thing was great, I especially liked the outro proof. 👍😀
    1
  12893. Great video!
    1
  12894. Are you from Germany? Or do you speak German?:o
    1
  12895. Too too much many good !!! thx
    1
  12896. I hate to say it but among the meme numbers 69420 > 69314. Also just in general
    1
  12897. How to paint this infinite horn with only one drop of ideal paint : 1. Use 1/2 of the drop to paint the first cm² of the horn 2. Use 1/4 of the drop to paint the second cm² of the horn 3. Use 1/8 of the drop to paint the third cm² of the horn ... n. Use 1/(2^n) of the drop to paint the n-th cm² of the horn ... In total, you'll use 1/2 + 1/4 + 1/8 + ... drop of paint, so exactly one drop for the whole area.
    1
  12898. Yeah, I figured it would be about one one-and-a-sixthth.
    1
  12899. Here to see the corrected edition!
    1
  12900. mathologer auf deutsch, ich freu mich jetzt schon :)
    1
  12901.  @nHans  I don't think I can learn Galois theory to save my life. 😂
    1
  12902.  @Mathologer  It will probably make me have PTSD flashbacks. I'm not kidding, two years after I conclude my graduation, I still used to awake in the middle of the night sweating after a nightmare in which I didn't got grades to pass Galois and was expelled. But I think with your talent for teaching complex math I will definitely learn something. I will wait cautiously for the video.
    1
  12903. Most memorable: You coming up with a novel proof and the professor rejecting it. I had a similar situation in organic chemistry when I came up with an industrial scale pathway rather than a lab scale one and I got zero marks on it. All the other parts (except one) were equally mind-blowing.
    1
  12904. Thank you so much for your easy to understand videos. The visual proofs are a joy to watch.
    1
  12905. Once you get there, it´s so obvious where all those threes come from! (but you have to get there first; thanks for taking us along in this trip)
    1
  12906. The most memorable was that any integer can be expressed as an infinite sum of a subset of the harmonic series! It’s probably just me not looking hard enough, but are there any videos around on the series 1+1/2-1/3+1/4-1/5+1/6•••?
    1
  12907. I like the parts about gamma. I'd never heard about gamma before in this context so it was very interesting to learn about. I also have a fondness for small constants.
    1
  12908. Fascinating. And the presenter is very good. :)
    1
  12909. 11:30, "leave in the comments if you like being kidnapped". OMG, it scalated quickly!!
    1
  12910. This was characteristically great, as usual, but if I may be so bold, please at some point bring back the number theory! While the visual geometric proofs are marvelous, I never quite “get” them the way I do with some of your other topics (nothing new, it’s been that way for me since grade school). I’ll admit it’s not like I have any leverage here. Whatever you put out I’ll keep watching anyway!
    1
  12911. Any relationship to convolutions and Fourier transforms perhaps? Dunno why I think that but feels similar to the way multiplication and convolution are equivalent after Fourier transform...
    1
  12912. Happy new year!
    1
  12913. Great dedication..! 💙
    1
  12914. Are there any triangle shuffle diagrams for pentagons, heptagons, octagons, or any other regular polygons? or does the fact that those shapes can't be tiled on a plane prevent there from being any diagrams for those shapes?
    1
  12915. For potential, the definition of gamma. For surprise, Kempner.
    1
  12916. :face-blue-smiling:
    1
  12917. Your videos are amazing. Thank you! RE: Feedback: It all worked - the pacing was good, the visuals were great. Had a little bit of "fraction overload" with some of the expansions but they helped. Background is lots of practical math oriented towards things like signal processing , control law, RF, filters,phasors, lots of linear algebra. I live half of my professional life in the Z-Domain . Masters is in EE. Write lots of code. Keep 'em coming!
    1
  12918. I really made it to the end
    1
  12919. 15:30 oi! clever!
    1
  12920. Coloring
    1
  12921. Do we know if the transformation is angle preserving (conformal)? My gut is telling me that it is and that it has to do with the spheres that define the moon shape, but I don't have a proof.
    1
  12922. Hey ... You have 666k subs .... You are satanic ... 🤣🤣🤣🤣😂😂😂
    1
  12923. I'm guessing yellow, because I think the corners must either match or all be different like every other three near each other.
    1
  12924. 👏👏🔝
    1
  12925. Not sure 2021 went any less crazy than 2020 (AH)², but thanks anyway for the video :)
    1
  12926. 35:25 Since the bugs are always moving at a right angle to each other, they never actually move away from each other. Therefore, to reach each other, it makes sense that they only have to travel the original distance that they were separated by.
    1
  12927. I like your shirt
    1
  12928. A conjecture... if there exists an area, it should be the minimum of all possible triangle arrangements when the max triangle size goes to zero. Makes sense?
    1
  12929. 20:57 Why can’t the windmills come in triplets? Why must they come in pairs?
    1
  12930. Hlo sir ,Is there any relationship between ln(m+n) to[ ln(m) or ln(n)].?
    1
  12931. The k=<3 doesn't work because the 1st term in the last brackets is (5 5) and (k+1 5) is impossible
    1
  12932. The solution is to only keep terms where the top is not less than bottom
    1
  12933. My version: https://youtu.be/mRu_i1vyIl8 3rd attempt to post this. I guess youtube doesn't like external links, so here's a video. There's a link to the web page in the video description.
    1
  12934. Something's not clicking for me with the last trick. You use the shortest distance to the next card so it's never more than 6 away, but there's always 2 cards that share each shortest distance. If you indicate that it's 3 cards away from a 9 of hearts, it could either be a 6 or a queen. How do you distinguish between these?
    1
  12935. Too bad vultures don't do math. The fastest, greediest, most vicious bird takes as much as he can gulp down while fighting off the others...
    1
  12936. Most memorable: what gamma is and how it relates to -1/12.
    1
  12937. This is a video I need! thank you.
    1
  12938. why use a decimal point to mean multiply?
    1
  12939. That spinning tesseract (?) at the end just broke my mind! God damn.
    1
  12940. 30:41 Given that the video describes essentially discrete Fourier transform, the limiting case of a smooth shape will become the continuous Fourier transform (on the complex plane). Considering that the smooth shape would describe a periodic signal, the composing polygones would be an infinite set of discrete 'simple' shapes. My feeling this is just an infinite set of circles with different radius and centers, but I'm not sure if one could associate these to spherical harmonics. Amazing!
    1
  12941. Wouldn’t this be easier to do with Lego?
    1
  12942. MOTH
    1
  12943. mmm but how I code it?
    1
  12944. I had never seen this before now. this is super cool
    1
  12945. The easiest language to program the coding challenge in is Mathematica. It's a one-liner: PartitionsP[666]. XD I did program my own using recursion just to check that Wolfram did his correctly. He did. indexes[n_] := Accumulate@PrependTo[Riffle[ Table[x, {x, 1, n}], Table[2 x + 1, {x, 1, n}] ], 1][[1 ;; n]] signs[n_] := Table[If[Mod[x - 1, 4] < 2, 1, -1], {x, 1, n}] part[0] = 0 part[1] = 1 part[n_] := part[n] = Plus @@ Times @@@ MapThread[ List, {part /@ ((n - indexes[n - 1] + Abs[n - indexes[n - 1]])/ 2), signs[n - 1]}]
    1
  12946. Pascal had no idea!!!
    1
  12947. 11:40: 1 1 2 3 --right--> 1 2 3 5 --right--> 1 3 4 7 --right--> 1 4 5 9 --left--> 9 4 13 17 gets you 153^2 + 104^4 = 185^2. 15:50: The area would be 5 as tree is grown 4 times, and every time the area increases by 1.
    1
  12948. I was taught both of the simple visual proofs of pythagoras in high school.
    1
  12949. Colour proof was better
    1
  12950. p(666) = 11956824258286445517629485 Amazing video !
    1
  12951. Triangles are the bestagons.
    1
  12952. I just want to know what is the reason of the 2 in (x+2)^n if we put a +3 would it give the formula in some different space can anyone please help and give some material to reference thank you great video
    1
  12953. For the 2x(integer) boards I got the Fibonacci sequence!
    1
  12954. Ignoring the math, I haven't seen this concept of splitting based only on the contested claims before. I can see some possible ways to abuse it, by claiming more, but it's an interesting idea.
    1
  12955. 12:50 the left bottom green tile disappears
    1
  12956. Pumpkin Pi!
    1
  12957. My IQ literally went down after watching this video. Jokes apart, great video as always!
    1
  12958. Answer 1: (x-1)^f+1 Where f is the power and x is the variable
    1
  12959. Sir, off topic, but what do you think of this: https://www.nytimes.com/2020/02/05/science/quadratic-equations-algebra.html
    1
  12960. Because x2 = also :5 x10 (instead of x5 :10) :) !
    1
  12961. After few years using zig zag lacing, I conjectured a behavior that this one had : unstability. aka when you expand and tight and expand and tight your lacing over the time, your lace usually move, aka you pull from one side more when expanding and you pull to the other side when tightening. So at the end when you want to do our node you have one very long part and one very short.
    1
  12962. How do you define those surface normals?
    1
  12963. This is amazing. But I am too stupid to truly understand it
    1
  12964. Burkard is back ❣️
    1
  12965. I feel so angry that I wasn't taught this in calc 2 but I was expected to make proofs about infinite series. It is no wonder I got wrong answers all the time! I could make most series converge on any number!
    1
  12966. White background is killing me dude
    1
  12967. Great music for the perplexed.
    1
  12968. it tooks me less than a minute to calculate Gauss's sum 1-100 if we imagine the sequence 0-100 as a line and we took 50 as a pivot, then actually we realize that all the numbers that are above 50, let it be x, are actually x=50+y if we took the y and added to the symmetricaly counterpart according to the pivot which has such value z=50-y then we get 101 times number 50 which means 101 * 50 = 5050
    1
  12969. 15:15 Hahaha...My kind of joke! Congrats on the 100th video...and thanks for the amazing content.
    1
  12970. 4:06 Mathologer: In the next two sections I will be lying a tiny little bit here and there try to spot the lies before I come clean Me the whole two sections: this wise man speaks absolute facts!
    1
  12971. Well done, as always. Thank you!
    1
  12972. AB= A+B, A=2, B=2
    1
  12973. Why the hell all the "no-X" series, as well as the "100-zeros", have so similar sums? Always around 20. Damn, there are unlimited possibilities. The sums could be anything, but... they equal 22. There must be some deeper truth here.
    1
  12974. This is GOLD
    1
  12975. hmmm "i" squared is -1 right.... is an imaginary number an integer? Would the unsolveable ones be solved by using imaginary numbers squared to add a negative value?
    1
  12976. Beautiful work, Burkard! Norman Wildberger has also done great lectures on what he claims is the more fundamental (than Heron's) Archimedes formula - see https://www.youtube.com/watch?v=iMWEiPuFhBQ
    1
  12977. So where's that ln10 proof at?
    1
  12978. 28:16 - That shape made my hanging cables is called a catanary .
    1
  12979. If the shape contains just two points who aren't vertically or horizontally aligned (on one ofbthe transformation axes), that shape must contain an entire hyperbola (arm) resulting from all the images of these two points under the described transformations. (If they are aligned that hyperbola degenerates into a line)
    1
  12980. Amazing highlights. Need to check some papers after this. Starting with that bizarre optimal stack.
    1
  12981. The algebraic one - YouTube
    1
  12982. thank you Sir!
    1
  12983. The Chinese names for the famous approximate ratios are delightful.
    1
  12984. Thank you for this corrected video.
    1
  12985. I like the coloring proof more than the swivel proof. Recording here as ordered by great master Burkard.
    1
  12986. I love it when a demonstrating a solution makes you giddy.
    1
  12987. I wrote an implementation in C#. On th Github page I not only published the code but a rendered 2000x2000 image too. https://github.com/Sickthon/Arctic-Circle
    1
  12988. Gamma segment wins! But how can a segment, being just a segment, win? When is a segment a faux segment, hiding infinity???
    1
  12989. This reminds me of another problem. A number of friends, (4 say), share a taxi to get home. However, their homes are at different distances along the taxi journey. (Assume that the taxi travels in a straight line dropping off each passenger along the way, so it doesn’t take a detour). Assume that the taxi driver simply charges for the total distance travelled, (the same as if there was only one passenger travelling the total distance). How should the total taxi fare be split? We would expect no one to pay more than the passenger who travelled the furthest, and no one to pay less than the passenger that travelled the shortest distance. But is there an indesputible way of splitting the total taxi fare fairly? Mike.
    1
  12990. I don't want to generate a formal solution, but this idea came to mind. Assume a 4 mile total distance and it's $10 per mile, so a fare if $40 in total. The passengers are spaced out by 1 mile. The total distance traveled by the passengers is 1+2+3+4, or 10 miles. Passenger 1 pays 1/10 or 10%, passenger 2 pays 20%, etc... This seems to be fair, as passenger 4 traveled 4 times as far as passenger 1 and paid 4 times as much.
    1
  12991. That was fascinating, thank you. I wish I wasn't under the weather.
    1
  12992. My circular slide rule in collage had all the usual scales, including the square scales, the trig scales, and the log-log scale. There were separate inner wheels on the front and back, and the index wrapped around allowing you to transfer values between the front and rear scales.
    1
  12993. Coloring proof is cleaner.
    1
  12994. shall have to experiment, but for long hikes (30+ miles a day) I need the lace near the ankle to be tight, but near the toes looser. criss cross makes everything evenly tight, so i use zigzag. but after seeing this I wonder if i can combine bow-tie (toes) and crisscross (ankle).
    1
  12995. I'm doing a master's in physics. Could follow it all conceptually. maybe some steps in the final two chapters were a bit fast to follow the rearranging. I worked through a blog post of Terrance Tao a couple of years ago on exactly this subject.
    1
  12996. Just... THANK YOU!
    1
  12997. how bout tetrahedron in 3d grid
    1
  12998. What a card trick... And in the next video Mathologer will fool Penn & Teller :)
    1
  12999. 21:43 yes but there's no need to put the 1 in the denominator, in which case the infinite expansion is then 2/3 * 4/5 * 6/7 * ... each term being <1 means it cannot explode to infinity
    1
  13000. The Mathematica standardform i symbol at 26:57 brings back memories!
    1
  13001. 2 wrongs dont make a right... but 3 lefts do.
    1
  13002. I doubt you remember, but I apologize for getting so angry at you and avoiding your content.
    1
  13003. 6:09 That's 3.18, not 3.14.
    1
  13004. Thank you for another outstanding episode! I'd not heard of this mystical conjuring!
    1
  13005. Everything in this video worked for me. What great stuff this is. I myself am a math teacher, but this is stuff I have never needed/learned/used in my life before. My eyes pop out and my jaw drops just from being astonished about this! This is something one hardly learns in university.
    1
  13006. This comes up whenever the pention for the retired people is raised. They want something other than by a flat X percent.
    1
  13007. Indian people like here
    1
  13008. The existence of Euler's Number means that the Harmonic Series is close to the Log Function. I have found that if a(n) = 1 + 1/2 + 1/3 + . . . . + 1/n -- log(n) then Each a(n) > 0, and it is a decreasing sequence. Therefore a limit does exist, which is the Euler Number.
    1
  13009. This is weird. I was recently working on some problem and dealing with some approximations, and somehow found myself having to learn about the asymptotic expansions for the digamma function. I had never heard about series that gave better and better approximations when adding terms, but after a point started to diverge. I also found that the terms in that expansion were written in terms of the Bernoulli numbers. And, two weeks later, The Mathologer releases this video. I'm looking very much forward to learn more about, specially the Euler-Mascheroni constant! My exact problem was on how good was to approximate the digamma function using the natural logarithm, to try to approximate sums of the form 1/(x+k) for k=1..N and x>1.
    1
  13010. Pi-uper Man!
    1
  13011. Loved the dotted one.... really gave a good overview after the flipping. Thanks for sharing.
    1
  13012. 9:35 Just see it as binary numbers and there you go
    1
  13013. Hey Mathologer could you do me a favor and do a video on homology/cohomology and de-rahm cohomology? I never 100% groked that back in school and if you are gonna do galois theory it doesn't feel like a big ask. btw are you a mind reader and just picking all the subjects I didnt fully master to make videos about so I can finally get it? Thank you so much! You are the best =) P.S. I also have some unresolved questions about how to apply lie groups to the solution of differential equations as well as applications of lie groups to invariant theory and algebraic geometry in general, you probably knew that already but if you are looking for more video ideas that would be fantastic too
    1
  13014.  @Mathologer  YES!!!!!!! Thats gonna be an awesome video. I can't wait to find out why even though I think I know all about Jones polynomials that actually this entire time I had been missing another valuable perspective
    1
  13015. Euler Mascheroni constant was the best (it doesn't get mentioned enough). I hadn't heard of Kempner's series before though so that was a close second.
    1
  13016. Rock paper scissors lizard spock!
    1
  13017. Best mathematics channel so far I came across
    1
  13018. I agree.
    1
  13019. If only we had $.25 coins in Australia… Then it would make sense
    1
  13020. The part that I liked the most was that Finding the value of Gamma.... And that absolutely amazing visual proof that Gamma < 1 through the graph of the harmonic series And also how the approximate partial sums of the harmonic series gets better and better as n gets larger.......... Hat's off to the 🐈.....
    1
  13021. Is there a Vernier Scale to go with this?
    1
  13022. Es ist derzeit schon etwas schwarz bzw. dunkel in der nördlichen Hemisphäre, aber die Tage werden nun länger. Bei Ihnen in Australien, Herr Polster, werden sie aber nun kürzer. Liebe Grüsse, schöne Weihnachten und ein gutes neues Jahr!
    1
  13023. Your videos are so interesting and inspiring
    1
  13024. For you c# programmers out there: using System; namespace Program { class Program { static void Main(string[] args) { int x = 1; int y = 3; int z = 1; int w = 1; int v = 1; int a = 1; int b = 0; int c = 1; bool isx = true; bool isy = false; int[] array; while(c < 3) { b = 0; if(c == 1) { Console.WriteLine(1); array = new int[0]; array[0] = 0; } for(int d = 0; d <= c; d++) { if(a == v & isx == true) { if(z == 1) { while(w > 0) { w = w - 4; } if(w == -3 || w == -2) { b = b + array[d]; } else if(b == -1 || w == 0) { b = b - array[d]; } else{ Console.Write(w + " "); Console.WriteLine("Houston we have a problem"); } a++; w++; } isx = false; v = z + x + y; z = z + x; while(w > 0) { w = w - 4; } if(w == -3 || w == -2) { b = b + array[d]; } else if(b == -1 || w == 0) { b = b - array[d]; } else{ Console.Write(w + " "); Console.WriteLine("Houston we have a problem"); } x++; w++; isy = true; } else if(a == v & isy == true) { isy = false; v = z + x + y; z = z + y; while(w > 0) { w = w - 4; } if(w == -3 || w == -2) { b = b + array[d]; } else if(b == -1 || w == 0) { b = b - array[d]; } else{ Console.Write(w + " "); Console.WriteLine("Houston we have a problem"); } y = y + 2; w++; isx = true; } else { } a++; } array = new int[c]; array[c] = b; Console.Write(b); c++; } Console.ReadKey(); } } }
    1
  13025. 07:20 -> seems to be Fredericus Johannes Maria Barning (1924-2012)
    1
  13026. 7:36 Back in the day, they really made GEOMETRY papers with NO DIAGRAMS!!! How could anyone make heads or tails of it?? That's why it slipped thru the cracks.
    1
  13027. I loved that no 9's fact, and when you expanded it to no-any-set-of-digits that was also cool. Thanks for being my favourite maths YouTuber, your videos always have good vibes.
    1
  13028. That 700 year old proof
    1
  13029. You know, typefaces of the 17th century (usin a pretty broad deffinition) have a similar problem to the Fermat's Christmas Theorom thing. Often, someone would design a face and their successor gets to be the name of it. And because they've stolen their predicessor's thnder, their successor does the same. I don't know how they stopped it, but be aware that no all typefaces are named after (or by) their creator. That every odd number is the difference between two squres is trivial. I did it walking down the street. What I didn't know is that you could do it more than one way for any number, let along non-primes.
    1
  13030. “The hardest what comes next problem”. I figured out the next number by the thumbnail and have concluded 3 possibilities; 1 - I am extremely smart 2 - I was click-baited 3 - the sequence is actually pretty easy to figure out I would like to say it’s number 1 but I think it’s probable the other 2.
    1
  13031.  @authenticallysuperficial9874  I just so happened to reinvent Cunningham chains while awake in bed. I did not know about them 2 weeks ago, but read up on them when I discovered I was pipped by a few centuries.
    1
  13032. Excellent. I should have taught my students this years ago
    1
  13033. Toll erklärtes Youtube Video. Ein echtes Highlight. Danke.
    1
  13034. I was taught by teachers who didn't get math. Out of 12 years pre-college only two of my math teachers could I consider competent teachers. And I'm kind of old now, so this was America's "golden age" of public education. So I never got this stuff. Studied languages instead. This is brilliant. And I want my money back too.
    1
  13035. Tardis is not satisfied with no proof that element contained in outer one has less volume;)
    1
  13036. Great Xmas gift 🤗🤗🤩🤩🤩
    1
  13037. Bonus comment question: . . What if you add another class of irrational coin, equal to 1/e^n where n is the "level" of coin?
    1
  13038. Sounds a bit like a saddle point?
    1
  13039. I love that little jingle introducing the chapters.
    1
  13040. Yay new upload
    1
  13041. Ramanujan Is the first strand type game
    1
  13042. Slide rules are for real numbers. Genaille-Lucas rulers are digital. So if the rounding error bugs you, go for the Genaille rulers. Unfortunately they're bulkier and take a bit longer.
    1
  13043. Youtube took 2 minutes for the video to load at all, that was weird
    1
  13044. 13:33 Hazards of obsession with mathematics!!
    1
  13045. Last scene was too scary to remember, especially Newton. What's wrong with his nose and eyes? His eyes make his eyesbones more distinctive.
    1
  13046. What a wonderful Christmas present! I'm a physicist working on something called "valence-bond solids" which is basically the quantum version of these domino tilings that realizes in actual materials. I knew about the Arctic circle from a textbook called "The Nature of Computation" (great book, pretty sure many Mathologers would love it!), but never really worked through it so it's really nice to see a visually stunning video about it. What a treat! XD One thing I noticed from my background is that the inner region of the Arctic circle must have a "power-law correlation" meaning that the circular "messy pattern" in the middle has no characteristic length scale, and is in a sense fluctuating in the maximal possible way. This property is called "critical" in theoretical physics, and is essentially the same to materials going through a universal type of phase transition (like water-to-vapor etc). Critical states are also super important in particle physics, since that's where quantum field theory gains relevance to empirical macroscopic observation. I'm not an expert in particle physics, but sure think it's very cool that the Arctic circle we can directly see is connected to the most fundamental physical theories!
    1
  13047. That was an incredible proof. Well done 👍
    1
  13048. So to generalize, if we have m,n€N, gcd(m,n)=1, then we can take the radius of smaller circle to be m cm, and that of bigger circle to be (m+n) cm, then we can produce (m+n) pointed star, m regular n-gons and n regular m-gons. I was wondering if something interesting would happen if gcd(m,n)>1, but turns out nothing better would happen, if m=xd,n=yd, such that gcd(x,y)=1 then the new diagram would just look like a (x+y) pointed star. That's a little disappointing. ☹
    1
  13049. Can this be expanded to higher dimensions?
    1
  13050. Is the next number 55?
    1
  13051. egglose
    1
  13052. I don't think tengon is a thing, more likely decagon ;)
    1
  13053. (x-1)^2 + (x+1)^2 = 2(x^2) + 2 (x-2)^2 + (x+2)^2 = 2(x^2) + 8 so 10^2+11^2+12^2+13^2+14^2 = 5(12^2) + 10 = 5 * 144 + 10 = 144/2 * 10 + 10 = 730
    1
  13054. Splendid, but is this useful somehow? Sure might be beautiful, but can this be used to simplify or speed up computation? I don't see a feasible way.
    1
  13055. The last curve is making nodes and antinodes with the circle.
    1
  13056. nice
    1
  13057. this is what I got for the 666th iteration: 11393868451739000294452939
    1
  13058. here's my python code: user_number = int(input("What is the position of the sequence you wish to find the number for? ")) user_number -= 1 plus_minus_sequence = [1,2] even = 2 odd = 3 for i in range(1, user_number): if i % 2 == 1: plus_minus_sequence.append(plus_minus_sequence[i]+odd) odd += 2 else: plus_minus_sequence.append(plus_minus_sequence[i]+even) even += 1 actual_sequence = [1] for k in range(user_number): # print(f" round {k+1}") next_val = 0 w = 0 for p in range(len(actual_sequence)): for i in range(len(actual_sequence)): if plus_minus_sequence[p] == i+1: if w % 4 == 0 or w % 4 == 1: # print(f" - adding {actual_sequence[-i-1]} to {next_val} which is {next_val + actual_sequence[-i-1]} ") next_val += actual_sequence[-i-1] elif w % 4 == 2 or w % 4 == 3: # print(f" - subtr. {actual_sequence[-i-1]} to {next_val} which is {next_val - actual_sequence[-i-1]} ") next_val -= actual_sequence[-i-1] w += 1 actual_sequence.append(next_val) if user_number+1 == 1: print(f"The value for the {user_number+1}st position is: {actual_sequence[user_number]}") elif user_number+1 == 2: print(f"The value for the {user_number+1}nd position is: {actual_sequence[user_number]}") elif user_number+1 == 3: print(f"The value for the {user_number+1}rd position is: {actual_sequence[user_number]}") else: print(f"The value for the {user_number+1}th position is: {actual_sequence[user_number]}") # for b in range(len(actual_sequence)): # print(f"{actual_sequence[b]}, ",end= "") # print("") # for o in plus_minus_sequence: # print(f"{o}, ",end= "") # print(len(actual_sequence))
    1
  13059. Another delightful video illuminating the connections between various infinite sums (and an infinite product) and powers of pi by using the product formula for sine. Glorious. In response to your query about what is missing from the Nike argument, two things stand out for me. If you start with the product formula for sine as a postulate, you need to show that a) the product converges and that b) the leading coefficient needs to be one. I expect there is a theorem that allows you to then conclude that this infinite product which has the same zeros and scale factor as the sine function is indeed the sine function. If you are looking for more potential material, I think the Gamma function, its various identities (the reflection formula, the Legendre duplication formula) and its use in evaluating infinite products is pretty cool.
    1
  13060. Man those videos are treasures! Thanks
    1
  13061. I really enjoyed this.
    1
  13062. 3:22 The 16th decimals from the left aren't highlighted even though they differ from eachother
    1
  13063. Like the mandelbrot set has as an infinite perimeter but finite area
    1
  13064. I wonder how this can be used to prove cubes?....
    1
  13065. i wonder what happens if its closed curve if we imagine it as infinte sides polygon
    1
  13066.  @Mathologer  well if "ears" arent infinitely long its kinda feels its gonna trace a curve that paralel to original curve
    1
  13067. The natural thing to think of is the NDFT = non discrete Fourier transform. … aka Fourier series. There are now infinitely many steps (projections) that need to be done to knock out all but 2 of the frequencies, but after you’ve done that, yes you’re left with a circle. The trouble is that the geometric realisation of this is similar, but not quite the same as Burkard’s “ears on a polygon”.
    1
  13068. Wait, they don't teach this? I remember having to do these kinds of problems in middle school, at least for integer sequences.
    1
  13069. Love the shirt! That's hilarious! :D
    1
  13070. I had a circular slide rule in 1975
    1
  13071. Total number of squares for n*n grid of points = summation(i in range(2, n)) (n - i + 1)^2 * (i - 1)
    1
  13072. CHALLENGES! 10:27 Pretty much all the details about the shape-shifting rectangles are explained in your previous video about the shape-shifting invariant. All that's left to explain is why ln(m/n) shows up. Notice something strange about how the Mathologer groups these. The header group is the only group that doesn't have m+n terms, and all the rest do, but it's in a way such that the resulting shapes fit neatly into boxes after squishing and stretching. Well then let's try and force this into a box starting at n and ending at m. Let's rearrange the terms a little bit but still preserve this balance group (technically it's a set not a group but shush) structure, in that every m+n terms we encounter m positives and n negatives. I will do so like this: (first m-1 positives) (first n-1 negatives) (last positive) (last negative) Our header group will consist of only the first m-1 positives and n-1 negatives. Each subsequent group will contain in this order: (one positive) (one negative) (m-1 positives) (n-1 negatives) Side note: we can throw these terms in any order within the groups, but each group must always contain it's respective elements. Then we can safely squash and stretch our way into a nice integral. For example, if I wanted 3 positives and 2 negatives I would do: HEADER: 1 + 1/2 - 1 GROUP 1: + 1/3 - 1/2 + 1/4 + 1/5 - 1/3 GROUP 2: + 1/6 - 1/4 + 1/7 + 1/8 - 1/3 GROUP 3: + 1/9 - 1/6 + 1/10 + 1/11 - 1/7 Do you see where I'm going with this? The header spans the range 2 to 3. Group 1 will span the range 4 to 6. Group 2 will span the range 6 to 9. Group 3 will span the range 8 to 12. By squish and stretch we can target the range 2 to 3. Once we go infinitely far into the series we will have essentially just computed the integral of 1/x dx along 2 to 3, which is ln(3/2). You can already see how this applies to m and n.
    1
  13073. just wow ... i am a maths teacher to high schoolers but let me tell u ... i feel like i am a happy little kid with his toy when i watch your videos . you are amazing . thank you
    1
  13074. nice video really nice stuff : so no one is spoilered: is the card the ace of hearts or am I mistaken?
    1
  13075. Lovely explanation and illustrations.Really a nice proof.
    1
  13076. Gamma must be > 0.5 because the area of each blue bit of gamma in the stack of rectangles is larger than half its box's area (i.e. the white part). My favorite part is the irregular maximal hanging blocks. Now my question for you: What fractional series lies exactly on the border between convergent and divergent sums?
    1
  13077.  @Mathologer  Just made a down-payment on the mold! Assuming it works, production could be as soon as a couple months. If not, I will cry and then try to fix it.
    1
  13078. nice
    1
  13079. I vote for the maximal overhang towers.
    1
  13080. Hi
    1
  13081. The link to the summary that you mention at 19:06 doesn't seem to work.
    1
  13082. First EDIT: third =(
    1
  13083. Grats on 100 ^_^
    1
  13084. I learned Heron's formula at some point, don't remember if it was in school or in a Numberphile vid or something
    1
  13085. what about polygons on the other grids
    1
  13086. I have a PhD in Electrical Engineering and occasionally used mathematics (calculus, matrix algebra and signal processing) in my professional life. I made it all the way through the video, but I hesitate to say I understand everything. The idea that a divergent sequence can yield good approximations when truncated appropriately is bizarre, but other than that I expect that doing a bit of homework is all I need to solidify my understanding. Thanks for another magnificent instalment to your growing body of entertaining and informative explanations. I'd be glad to sign up as a patron should you ever decide to use Patreon.
    1
  13087. I think asking "Why was XY not discovered earlier?" is a pretty stupid question.
    1
  13088. Mathologer , plz make a video to explain this statement "Mathmatics is unreasonably effective "
    1
  13089. i liked the second one the most
    1
  13090. ILOVE HOW YOU SIMLIFY MATH, I WHICH I HAD YOU AS A TEACHER WHEN I WAS YOUNG I AM YOUR LITTLE BROTHER FROM WASHINGTON D C 10 4
    1
  13091. 5:41 "Maybe not" Me: That's so simple I could maybe come up with it!
    1
  13092. Tristan’s graphical argument
    1
  13093. In France . As any kid born aroubnd 1955 , I was told about 'la preuve par 9' which is this technique of digital root.
    1
  13094. Really frickin crazy is that these numbers are even found in the great pyramid, they knew this 5 thousand years ago
    1
  13095. Love the pattern in trig to explain this! Worth trying them out for sure!
    1
  13096. For me the most surprising part was the missing of all the integers.
    1
  13097. My solution for the final puzzle is 0.2 the small triangle is similar to the big triangle because their 3 angles are equal. The scaling factor of them = ratio of their hypotenuse = (0.5 / (√5/2)) = (1/√5) side-length of the small square = hypotenuse of big triangle - (sum of the 2 shorter sides of the small triangle) = (√5/2) - (1.5)/(√5) = (1/√5) so the area of small square = 1/5
    1
  13098. To assist your engagement; I liked the coloring better, the idea of the semi-sides each composing the half the chords is pleasing.
    1
  13099. Oh, you didn’t use Blue because that would be represented by B. And B is usually assigned to “B” the length of one side of any triangle
    1
  13100. Das Mathologer Video auf Deutsch wird bestimmt nice :) Freue mich drauf. Grüße aus Hannover :D
    1
  13101. I have a string of plastic camping lights that look like that! 😂
    1
  13102. This is so cool!
    1
  13103. I'm wondering if one can deal with all these various cases by using "directed angles" instead.
    1
  13104. Dr. Strangelove (1964) shows him using a circular slide rule. https://youtu.be/zZct-itCwPE?t=30
    1
  13105. Wow! This definitely felt like magic!!!
    1
  13106. Only remotely related, but I have found the "long division" algorithm to be hilariously and frustratingly circular, to the point of being almost useless. How come it's circular? Well, what is the division of some number A by another number B? The result is what number must B be multiplied by in order to get A. In other words, essentially, how many times B fits into A. So, suppose you have a number A with quite many digits (let's say, 10 digits), and a number B with about as many digits (like 8 or 9, or even 10 digits), and your task is to divide A by B, using long division. How do you do this? By finding out how many times B fits into A. Well... duh. That's what we are trying to find out in the first place! The long division algorithm is essentially saying "to find out how many times B fits into A, we start by first finding out how many times B fits into A". Which is a bit hard to do if A is 10 digits long and B is, say 8 or 9 digits long. What a wonderfully circular, redundant and useless algorithm.
    1
  13107. Please, generally how to construct a 4X4, 6X6, 8X8 .... even numbers magic square? Danke!
    1
  13108. Neither. Right now. lol
    1
  13109. Powerful Mathologer, your video is (as usual) amazing! I just don't like the usage of the center of mass to show that the scale is balanced. The position of the center of mass proves nothing (unless a few other conditions apply). For example, if you glue the two masses in the same position but you move the tip of the "thought" tray so that it applies the force at a different point, the scale is no longer in equilibrium!
    1
  13110. Can you maybe make a video about the new aperiodic monotiles?
    1
  13111. The real Mathologer badge should go to who stood until the QED at the end.
    1
  13112. Would this be true of spheres of dimensions greater than the second?
    1
  13113. most memorable: your "are we there yet" t-shirt!!! and of course 0.57721566490153286060651209008240243104215933593992 -1/12 and the no-nine sum
    1
  13114. woa
    1
  13115. A beautifully done video & presentation! :washhands: Thank you for the mention too! 😉
    1
  13116. Danke! :D
    1
  13117. Wouldn't be easier to say that the height of the triangle is 1/2 the width times the square root of 3 Since the square root of three has an infinite number of digits....you'll need an infinite number of dots on your grid before you could have three dots line up equilaterally.
    1
  13118. I am definitely going to remember the bizarre overhanging tower
    1
  13119. I WANT TO BECOME A SRINIVAS RAMANUHAN
    1
  13120. Is there anything in math in the last 100 years that was not discovered by Conway ????
    1
  13121. Super cool! Thanks! Love it. Something tells me there are significant applications of these relationships.
    1
  13122. Why is the product rule expression not symmetric with respect to interchanging f and g?
    1
  13123. What worked for me: Youtube What didn't: My brain after watching the video :') Either way, I got a long way without losing you this time and actually learned a few new tricks with prime numbers so I'm making some progress here nonetheless. Also, judging by the fact it was about primes and modular maths I guessed it was Fermat. But I guess Gauss has caused me plenty of headaches too when it came to complex analysis so I could've known it was that pesky man :P
    1
  13124. 25:00 Okay, but for this "pseudo-cyclic" quadrilateral, if you draw the diagonals Cc, you've just completed a proper cyclic quadrilateral with sides BCbc and diagonals Aa. So for this quadrilateral, Bb + Cc = Aa, right? This should extend to any quadrilateral which is inscribed in a circle (provided you properly label the sides as you noted.) So a more general rule should be to draw all four sides and two diagonals. Then the sums of the products of the lengths of the two opposing pairs of non-intersecting segments should equal the product of the two intersecting segments.
    1
  13125. I prefer the coloring proof.
    1
  13126. Introducing the chain fraction. Me: "Yawwwwnnn." Introducing the infinite sum of odd denominators. Me: "Yawnnn. Mnmnmnmnmnm". Showing the result with pi an e. Me: "WWWHHHAAAAT THE F......" Another great video of Burkard. You enlightened my day :) Thanks!
    1
  13127. I'm pretty sure 2 + 2 = 2 x 2 is required for Euclid's formula to work.
    1
  13128. Pythagorean Triple: a^2 + b^2 = c^2. We rearrange: b^2 = c^2 - a^2 Substitution for Euclidean algorithm: a= m^2 - n^2 b = 2mn c = m^2 + n^2 Substituting: (2mn)^2 = (m^2 + n^2)^2 – (m^2 – n^2)^2; (2 x 2)(mn)^2 = m^4 + n^4 + 2(mn)^2 - m^4 – n^4 + 2(mn)^2; (2 x 2)(mn)^2 = (2 + 2)(mn)^2; This will not work for: a^k + b^k = c^k; k <> 2. So: k x k = k + k is required for the Euclidean algorithm to work, so it only works for k = 2. Not really sure about the formalities of proof. Sorry.  @Mathologer 
    1
  13129. From the moment when I came across the name of this argument, then always stressing out that this is charging argument, and always saying that it's important in math and informatics olympiads.
    1
  13130. My Favorite part was Chapter 6
    1
  13131. ASTIG😃 I have a great breakfast😄
    1
  13132. I've found out how to generate clean examples of cubics, to give as examples that simplify nicely to integers at every step. Given the standard form of a monic depressed cubic, x^3 + p*x + q = 0, your clean examples will have the following properties: 1. The p-term needs to be a multiple of 3 2. The q-term needs to be a multiple of 2 3. It doesn't matter if you pick positive or negative for q, you'll simply negate the root when you do (given an example with just one real root). Given a single real solution, a negative q cubic will have a positive real root, and a positive q cubic will have a negative root. 4. When you find a p-value that makes a nice cubic, there will be a related p-value that is its negative, which also makes a nice clean cubic. I've generated several examples by setting up a spreadsheet. I have a column for p, a column for q, and a column for the discriminant, D = p^3/27 + q^2/4. Then I calculate sqrt(D) and -q/2 + sqrt(D), and then -q/2 - sqrt(D) in the next group of columns. Finally, the two cube roots, and the sum of the two cube roots. Hold your trial values of p as a constant, and increment even values of q. Then duplicate formula and look for rows with all integer terms across the board. I came up with only two related examples, that A) are non--trivial, B) have a positive discriminant for one real/distinct solution, and C) are within reason to expect students to solve without a calculator. x^3 + 9*x - 26, and x^3 - 9*x - 28. The discriminants will explode to much larger square numbers that very few people will memorize, even if you do get nice numbers to cube root later on (e.g. cbrt(1000)). A nice clean example with a repeated root is x^3 - 12*x - 16. The real cube roots will add up to the distinct root of this cubic (+4), and the complex cube roots will form two redundant paths of finding its repeated real root (-2). For cubics with three real solutions, any clean example for Cardano's formula that avoids irrational numbers at every step along the way, can only have one rational root. The other two roots will involve a conjugate pair of square roots, like -1+sqrt(3) and -1-sqrt(3).
    1
  13133. It is math. Not maths.
    1
  13134. Aha-moment: 15:18
    1
  13135. In my 9th grade honors geometry class in 1996 ( Geometry for Enjoyment and Challenge , McDougal Littel/Houghton Mifflin), we learned Heron's formula and Brahmagupta's formula, but we learned the latter under the name Hero's formula .
    1
  13136. A cool related thing to this I've seen is the power expansions of (x+1)^n It also gives you combinatorial numbers (which appear in Pascal's Triangle), and tells you the properties of n-dimensional triangles/tetrahedra.
    1
  13137. Wow. Thanks. You get me every time.
    1
  13138. coloring proof
    1
  13139. When the Moors left Spain, they left behind 400,000 books on science, mathematics and medicine. They were translated by the Europeans and helped them out of the dark ages and into Enlightenment. Kepler, Copernicus, Descartes, Fermat, Leibniz, Newton, Euler etc., got copies of the books on mathematics, enabling to discover these beautiful mathematics, including calculus, first discovered by the Indians, then taken to Europe by Arabs. Madhava gave the mathematics of infinite series and calculus around 1200 ad.
    1
  13140.  @Mathologer  Archimedes introduced the geometric concept of limit, Madhava gave the arithmetic concept sum of infinite series and calculated the value of pi, his idea let Bhashkara and perhaps Aryabhata to give the concept of calculus, Eudoxus gave models of the motion of the planets with concentric spheres. However, Pascal's triangle was obtained by a linguist. Ptolemaic mathematicians were great indeed. The great advances made by Arab mathematicians, kept in the library at Alhambra are frantically searched by scholars from Oxford and Cambridge, feared lost and never to be found anywhere. Your series on Newton is enlightening and very absorbing.
    1
  13141. At 15 minutes, my brain hurts. Mathologer, thank you for your efforts. May you and yours stay well and prosper.
    1
  13142. Please, can someone explain his shirt to me?!!
    1
  13143. Gamma was easily the most interesting part of the video, because up until now I'd heard about the Euler-Mascheroni constant but had never really developed an intuition or understanding for what it meant.
    1
  13144. Why does Edward Lucas look like Minecraft Steve?
    1
  13145. 7:22 2
    1
  13146. Lol your art isn't good enough to compete with an algorithm
    1
  13147.  @Tumbolisu  Fair. I remember using GPT-2 back in 2020 and it wasn't very good at text-based prompts. Never thought of AI art until after the AI boom. I do think at some point, AI art will become too dissimilar to preexisting art for the average person to tell a difference.
    1
  13148.  @nightytime  Yes, the standard for images of a historical human being should be how advertising customers rate her attractiveness on a scale of 1 to 10. GTFO
    1
  13149. ​ @Mathologer No, not disgusting or offending at all, imo. It's just that some people like it and others don't. Anyway, thanks for answering so frankly.
    1
  13150. Very enjoyable. Thanks m8.
    1
  13151. This just reminds me of the duality between lengths of medians (p, q, r) and length of sides (a, b, c): p*p = 1/4 (- a*a + 2 b*b + 2 c*c) On solving the three equation gives the proportional inverses for sides and medians: (9/16) a*a = 1/4 (- p*p + 2 q*q + 2 r*r) This would mean medians of triangle formed by medians of triangle would be 3/4 of the corresponding sides. Which is also how the last triangle turns to be 1/4 of the original triangle. Amazing!
    1
  13152. Stuff with gamma is so cool
    1
  13153. Sounds and acts just like Astrid Dienard who works with Indy Nidell and Sparticus Olsen on the World War video series! 😂😂😂
    1
  13154. Yes that number can be written in form 4k+1, because the given number - 1 is divisible by 4
    1
  13155. We just learned Pythagorus theorem as a fact to be memorized, no proof nor explanation, England 1980s.
    1
  13156. the proof of the pudding is in the eating - therefore the last proof gets my vote : the sum of the reciprocals without nines < 80 !
    1
  13157. What miģht happen if they taught that all energy is in constant distribution and transformation everything even social energy and how we think
    1
  13158. the book went viral in 1910? im mindblown, its well ahead of its time😜👍
    1
  13159. brag brag brag brag
    1
  13160. 23:25 me too
    1
  13161. I really liked the visual proof of the finiteness of the "no 9 series" with the square getting lighter and lighter. It brought me back to the Numberphile video that claims all numbers contain a 3.
    1
  13162. My God! If you were smart enough to write the textbook in verse, the mathematics would have been a breeze! When I first saw the thumbnail for this episode, I thought: Burkhard's dressing up again! Where does this image come from?
    1
  13163. As someone who doesn't know anything about higher dimensional geometry, I have a question. What is your definition for the higher dimensional analogues of the cube? It semed like the 0 through 3 dimensional were defined well, but generally speaking what is an n dimensional cube?
    1
  13164. Number 91 = 9 + 1 = 10 ? .But 10 = 1 + or - 0 = ? What next ?
    1
  13165. A circular slide rule and scientific notation is all you need then! Cool
    1
  13166. this is the first one of your videos i've been able to follow in a while
    1
  13167. For a guy who is unable to straighten his elbows, Mathologer has really contributed a ton of good content to youTube.
    1
  13168. Can't we just say that the height of an equilateral triangle is irrational
    1
  13169. Truly amazing, as always. Thank you so much for making these videos, I always get so happy when I see a new one in my subscription feed!
    1
  13170. I just came to this video just for comments from the same community.
    1
  13171. I love this channel so much, thanks Mathologer
    1
  13172. I don’t get the bit about odd and even permutations; I just view the matrix as a cylinder and the sign is determined by the direction of the diagonal.
    1
  13173. For 7 pigeon , there are two hemispheres , 7/2 = 3.5 rounded to 4.... so 4 .... final answer is 4+1
    1
  13174. 19:02 actually...I guessed the pattern kind of ...when you asked before...do you see any pattern...I guessed that everything -1 might be prime numbers...😅
    1
  13175. Video auf deutsch wär klasse!
    1
  13176. Next step: connect this with knot theory? ;)
    1
  13177. Because my teachers were idiots driven by ideology.
    1
  13178. does anyone find the proof that “1 + 1/2 + 1/3 + 1/4 + 1/5 … = infinity” to be extremely disturbing?
    1
  13179. May da Schwartz be vit you!
    1
  13180. I learned math from a tired old woman that should have stopped teaching years earlier. I thought I hated math. Turns out I hated to be hated by her.
    1
  13181. 1:52 hey Vsauce' kevin here. but what is a square?
    1
  13182. interesting . 14:52 you end up with a pyramid of line .Let apply this to giza pyramid missing the cap . Do we end up with one half of earth and its magnetic pole region ?
    1
  13183. =<2.
    1
  13184. At 12:30 you can see that one of the green tiles goes missing. 14:56 That determinant is 90 but I don't why that should be special. And to say something positive, merry Christmas to you, Burkard, Marty and to your families as well.
    1
  13185. I liked the first proof better! but both are nice!!
    1
  13186. This is one of my favourite videos of yours, and that's a very high bar! Looking forward to more on this
    1
  13187. I did the 10 tower while drunk. It took several hours of attempts but I finally did it.
    1
  13188. UK Senior school, back in the day, for ages 11 to 16 y.o., week one in maths class was to use log graph paper to make a cardboard linear slide rule... 8 years later I then got my first Casio 'green' numbered calculator... 4 years later both a new slide rule and a Casio "BASIC" programmable calculator... the via CPM into DOS into Windoze via main frames, skipping most of the HP reverse Polish, etc., etc. But if you make one, and get the real basics of 'log' tables (6 digit ones, not those with 4 digit training wheels) engineering is just so much easier. : ))))) Many thanks for this content, excellent new 'stuff' and memory revivals.
    1
  13189. I am fammiliar with both, and I came to be aware of it a few minutes ago...
    1
  13190. Marvelous
    1
  13191. Area being the derivative of volume is obvious. The calculus way to calculate the volume is by integrating a whole lot of infinitesimally small area’s, so how could it not be?
    1
  13192. So Cavalieri got a smell of calculus before it was discovered.
    1
  13193. 9:49 Next time you buy an elaborately cut diamond for your beloved and she won't be happy unless you also tell her how many vertices edges and faces it has: remember that you only have to count two of these numbers. The third follows from euler's formula. rolling on the floor, laughing!
    1
  13194. Most memorable: the beautiful visual connection of fractal thinning and the convergence of the geometric series in Kempner's elegant proof.
    1
  13195. 3:10 Agree! BTW, more videos about "why we learn math". The explanation to your question is, in most cases, lack of training for teachers and the idea that the most important teachers are the ones tutoring post-docs and the less important are the kindergarten ones - an idea that is completely upside down!!!! The most important teachers are the ones in kindergarten and the least important one tutor post-docs. The only teachers more important than kindergarten teachers are the ones that teach even younger children. OMG! Our parents are the most important teachers!!!
    1
  13196. When I was an undergraduate, I was originally interested in going into mathematics education. (I eventually ended up just doing mathematics, in the end.) Anyway, my education advisor had an EdD and who specialized in mathematics education. I always remember him mentioning to me an article which showed that there was a correlation between an educator's fear of mathematics and what level they taught. The higher their fear of mathematics, the lower grade levels they taught (with those most scared of mathematics teaching things like kindergarten and first grade.) This, no doubt, affects how students view mathematics.
    1
  13197. The 3D moons comprising the claw can be called as Lunoids (extended from the 2D version of Lunes). You are welcome! Please acknowledge me in the papers later :))
    1
  13198. The formal name is “cyclide”
    1
  13199. but why? why does the stretching rubber band get stuck at exactly the log distances around the circle?????
    1
  13200. Astonishing. Thank you a thousand times over.
    1
  13201. This was a really nice proof!
    1
  13202. The pattern to spot for the 4 color problem is that any rectangle with sides a multiple of 2 will also have all 4 colors in the corners.
    1
  13203. The longest sequence would be one where a "smile" is introduced as far right as possible, and carried as far left as possible before a second smile is introduced and that process is repeated. That would mean the number of moves would be in the form of an "additive factorial" if you get what I mean. N + (N-1) ... + 1, or for 30 cards, 30 + 29 ... + 1. In reflecting on this, I'm reminded heavily of the sorting algorithm channels.
    1
  13204. fermat's room is not on Netfllix :(
    1
  13205. 27:44 Ah, such a nice son.
    1
  13206. Reposting and slight editing of recent mathematical ideas into one post: Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate. The so-called triplex numbers deal with how energy is transferred between particles and bodies and how an increase in energy also increases the apparent mass. Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter. The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards. Mediants are important to understanding the Big Crunch side of a Big Bounce event. Matter has locked up, with particles surrounding and pressuring each other. The matter gets broken up into fractions of what it was and then gets added together to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with all stable conformations of matter being broken up, not added like the implosive "shrapnel" of mediants. Ford circles relate to mediants. Tangential circles, tethered to a line. Sectrices: the families of curves deal with impossible arrangements. (The Fibonacci spiral deals with how dark matter is degenerated/broken up and with supernovae. The Golden spiral deals with how the normal matter, that we usually think of, degenerates, forming black holes.) The Archimedean spiral deals with matter spiraling in upon itself, degenerating in a Big Crunch. The Dinostratus quadratrix deals with the laminar flow of dark matter being broken up by lingering black holes. I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex and I don't know if I have all the curves available to use in analyzing them. But, I can see Fibonacci and Golden spirals relating to the trisectrices. General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity? Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Connection to Big Bounce theory? "Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?
    1
  13207. Why would he assume that anybody who doesn’t know calculus would watch this video? Are there actually casual math appreciators?
    1
  13208. No integers among the partial sums
    1
  13209. I just know that math is just a big scam! As soon as i learn how to count, I'll prove it to you all!!!
    1
  13210. the equal of blue, orange and gray diamonds in the hexagonal boards is just beutiful. you can see it, by thinking in 3D, and looking at projection of each face of the 3D cube it forms. You will only see one color, and it will cover entire cube face with same area.
    1
  13211. This shift property can be applied to simplifying the search for monohedral convex pentagonal tilings. Not an obvious insight until you show your visual transformation. Delightful and elegant. Thank you.
    1
  13212. Pi looks like a exercise equipment to me.
    1
  13213. A cute hyperbolic solid!
    1
  13214.  @Mathologer  yep. That too!
    1
  13215. watching the starting section of this video which contained Animated maths...was like watching the trailer of "Avatar-The Way of Water"
    1
  13216. For the 1-2-4-8 challenge: The product is (1+x)*(1+x^2)*(1+x^4)*(1+x^8), which expands to 1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10+x^11+x^12+x^13+x^14+x^15 - so including 0, you can make 16 different amounts of change, with one way for each amount. For the further entries in the series, you can always make all amounts up to the sum of the values of the coins (or twice the value of the largest coin minus one), with one way to make each amount - which becomes obvious once you realise that this is simply binary numbers packaged as change.
    1
  13217. 23:55 Who else among you can see the gap in the bottom left quadrant?
    1
  13218. No one has said it isn’t! These however are esoteric relationships between Pi and infinite series of powers of integers. The relationship between one of these expressions and parts of a circle is shown in a video by a different mathematician
    1
  13219. TIL there are people who don't watch Mathologer videos right to the end.
    1
  13220.  @Mathologer  Very
    1
  13221. The 100 zeros series having a sum greater than the no 9s series is the most mind blowing result. One lacks any small numbers and the other lacks large numbers, so I would definitely had guess the opposite, which just shows that intuition doesn't exist for these guys.
    1
  13222. Amazing stuff! I liked all of it, but the ugliness of that tower will continue to haunt me ... with its beauty :)
    1
  13223. α β γ δ ε ζ η θ ι κ λ μ ν ξ π ρ ς σ τ υ φ χ ψ ω ϐ ϑ ϒ ϕ ϖ Ϛ Ϟ Ϡ ϰ ϱ ϲ There is some greek letters for you. You can copy and paste them to tidy up the text if you want. :)
    1
  13224. Do 3- or more-dimensional grids contain other higher-dimensional polytopes other than hypercubes though?
    1
  13225. Thanks!
    1
  13226. I like color proof better I think.
    1
  13227. I find very interesting that the cos and sin of the multiple of a angle conserve their rationality..is it a unique singular property of "cos" and "sin" ?
    1
  13228. A doppelgang-hair? ( ͡° ͜ʖ ͡°)
    1
  13229. Seeing is believing indeed. I tried triangle in Fusion 360 and got perfect triangle :) Thank you!
    1
  13230. I feel like what is difficult is applying processes. here it's that we should have used iteration on the original idea. I can't help but to think about the importance our social, and political environments have on what we might tend to look at in the first place as well. is this discovery due in part to Mandelbrots focus on self similarity? we may never know. the realization of the significance of these factors does however highlight the fraud and oppression of systems of private property considering it is always shared human experience that brings us forward. NOT the fences so common in our backward modern societies of capital today. I'm happy to elaborate. either way its just a thought... and I always love your work and videos. thanks!
    1
  13231. This reminds me of symetrical components from power engineering, so I expected complex numbers to show up.
    1
  13232. How this shrinking graph will look like for the good angles? The ones that do not contradict the assumption. I suppose, points there shoud start to overlap after a while, right? But how exactly will it look like?
    1
  13233. What? Coins becoming an endangered specie?
    1
  13234. The most memorable part to me was seeing all those bizarre towers, with half their length over the cliff, yet not falling down.
    1
  13235. I love recreational maths problems they're wonderful, thanks for sharing your little passion! Consequently, I learned that I had actually given up tying the starting shoelace knot entirely! (Straight to the bow, no slipknot)
    1
  13236. 16:14 probably more advanced answer but 2 and 5 are inverses of modulo 9 (2*5=10==1 mod 9) So basically 2^n == 5^-n mod 9 for all n, making the remainder (digital root) reverse each other.
    1
  13237. 24:01 the loop 1,2,4,8,7,5 times 3 is 3,6,12,24,21,15 This has to do with Euler's theorem a^phi(n) == 1 (mod n) where gcd(a,n) = 1, in this case a = 2 (second element), n = 9, meaning the cycle has phi(n) elements. When we times k = 3 we have 3*a^phi(n) == 3 (mod n), same thing occurs where the cycle has 6 elements. If gcd(a,n) = d > 1 we can always reduce it to gcd(a/d,n/d) = 1 and we can observe a cycle on case 3,6, this case d = gcd(a,n) = gcd(6,9) = 3, a' = a/d = 2, n' = n/d = 3. The cycle from 1,2 in n' = 3 is from same Euler's theorem a'^phi(n') == 1 (mod n'), where a' = 2, n' = 3, phi(n') = 2. Basically, all cycles either contains 1 or doesn't contain 1. If it contains 1, the cycle will always have phi(n) elements on graph of n elements, and we can always multiply by some constant k and still maintain the cycle and the shape of that cycle. Else, the cycle will contain phi(n/d) where d is the smallest element of the cycle, which we can reduce the n elements to n/d elements and maintain the structure of the cycle.
    1
  13238. Whoa! These sequence things act sort of like numbers, or should we say mathematical entities that form ... a group? ... an algebra? ... What exactly?
    1
  13239. 主播会吹唢呐 给预约送走
    1
  13240. Claims: Kempner no 9s sum is less than 90 at min 39:00. Proofs: no 9s sum < 80 at min 44:10. Was this his original proof or a typo? Liked this part the most because it was new to me (beside the optimal overhanging bricks) 😊
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  13241. Thank you. Your work is reaching into the future. My students LOVE watching your videos even if they just grasp the very edges of what you're talking about. They continue to think about your videos long after they have watched them. Thank you to you and all your team. Please keep up these great works!
    1
  13242. This is wonderful I would like to understand this better. Would the seven sided star be used in something Mechanical or biological or ?
    1
  13243. Why obvious? Because those shapes have greater areas than triangles.
    1
  13244. We US engineers pronounced "cosh" as written but "sinh" as "cinch".
    1
  13245. Loooved this interlude. It was simple but make turn on my interior math spark. Hope this really catches on and have more small videos in between the cool ones!
    1
  13246. 2:45 there is one house on the street. the man lives in it update: mathologer is the self-proclaimed mean school
    1
  13247. The reverse journeys also remind me of the 2048 game.
    1
  13248. What is the difference between Iron Man and Aluminum man? One stops the bad guys and the other just foils their plans.
    1
  13249. Prime Spirals at various levels of zoom look suspiciously like the 'Microscope' Fibonacci spirals. If this coincidence becomes important in proving the Riemann hypothesis, I wish to be credited in the paper :)
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  13250. ​ @Mathologer  If you change the number of radians in a full circle, you can get plots with very few spokes rather than spirals, that neatly demonstrates the point made by this video. That is, a spoke with all primes ending in 1, 3, 5, 7, 9 etc. The interesting point there is the angle at which these spokes radiate from the core. which you think would be the same, but instead often resemble Y shapes I moved on to plotting twin primes, but didn't find a solution to any open clay institute problems, then I had to go back to my day job :)
    1
  13251. FIrst laugh at 0:02 X-Y = \ Now I continue to watch ;-)
    1
  13252. For the sum of 10th powers, for fun I derived the general formula by first taking the generating function of the sequence F(x) = (xD)^10 (1/(1-x)) / (1-x), taking the partial fraction decomposition of F to get the polynomial as a sum of coefficients times terms (x + k choose k). Expanding out those terms and factoring yields the general solution (1/11)n(n + 1/2)(n + 1)(n^2 + n - 1)(n^6 + 3*n^5 + 2/3*n^4 - 11/3*n^3 + n^2 + 10/3*n - 5/3). Substituting n=1000 gives the result. For more detail on these ideas I learned it from Wilf's generatingfunctionology.
    1
  13253. Thanks!
    1
  13254. Did not know about this „proof“ until half a minute ago! Pure beauty… 😅
    1
  13255. The most satisfying fact for me was the cube in intro.
    1
  13256. I liked chapter 5 and the visual representation of the sum and the introduction of the Euler-Mascheroni constant. Of course Euler comes along, because, as James Grimes once said in a video "that's what Euler does"...
    1
  13257. One last observation from me. Most primitive triples only have one pair of legs associated with the hypotenuse. But there are examples where there are more. Without the algorithm shown in this video it is hard to find them. Its fairly well known that the smallest hypotenuse that shares two different pairs of legs is equal to 65: (56, 33, 65) and (16,63,65). Hypotenuses of length 85 and 145 also have this property. Note that these numbers are equal to the length of the hypotenuse of the first parent triple (5) multiplied by the lengths of the hypotenuses of the first three children (13,17 and 29). This property ( ie having more than one possible pair of legs) extends to the products of the hypotenuses of any three children with the hypotenuse of their parent. And if that is not a big enough surprise, then the product 5*13*17*29 = 32045 is the length of the hypotenuse of the smallest primitive Pythagorean triple that has eight different pairs of legs. 31964 2277 30956 8283 27004 17253 24124 21093 22244 23067 15916 27813 6764 31323 716 32037 Easy exercise for the reader. Confirm the above, using the Fibonacci boxes. Its not difficult.
    1
  13258. The video was on point! You've not lost your edge. Let's face it, the video was excellent.
    1
  13259. Tristans fractal
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  13260. I did learn the "twisted square" proof at some point in high school, I am sure. So I won't be asking for a refund of the school taxes my parents paid in the 1960's. The "official proof", though, that we were required to memorize, involved drawing an altitude on the hypotenuse, which creates 3 similar triangles. An elegant proof itself, though it does require a fair amount of algebra. The twisted square proof involves no more algebra than the square of a binomial, which itself can be easily visualized by geometry. Areas that add up is very intuitive--though the Mathologer's complex re-arrangements are beyond my capacity to visualize. There's just one thing about the twisted square proof that, to my mind, takes it down a notch in elegance: you actually need four identical copies of the triangle. I've tried to think of a way to dispense with the 4 copies. But President Garfield's proof does dispense with half of them, a good thing. The idea of any of our recent presidents coming up with a geometric proof is laughable, with the possible exception of Bill Clinton (a voracious reader, I'll bet he reads STEM too).
    1
  13261. Podrías.algo.sobre.geometria.rropical
    1
  13262. I learned the second proof in school, maybe at 14-15
    1
  13263. I wonder how Euler knew how many decimals of his sum estimations were correct. Did he just calculate a few more terms and guess which decimals probably wouldn't change anymore or did he have some way to estimate the remainder? Maybe I'm just missing something though.
    1
  13264.  @Mathologer  Thanks for the reply!
    1
  13265. 4:26 did you just build a calculation circle for multiplication ?
    1
  13266. dear Burkard today we saw this video for the first time ... we were 'compelled' to add your great video to this blog of ours ... WINDMILLS are the common denominator ... https://at37.wordpress.com/2012/02/23/windmills-swastikas-seeds/ Einstein would comment on why our work collides sir. "the coincidence is god's way of remaining anonymous" ... well it appears 'god' is speaking to us both, and we use whatever skills we 'got' to receive 'gott' arriving as CHIRAL and ACHIRAL messages from 'command central' ... https://at37.wordpress.com/?s=137 Is it a coincidence Burkard that you mention in your video, the persona theonethreeseven and you use the prime number 37 as an example when both of those prime numbers 37 and 137 are vital/central and highlighted on our blog/diary with the address at37 ? Burkard whether you like it or not, your fine work, your brain fARTS and heART work are associated with our heART work and brain fARTS.
    1
  13267. 9:08 With one of each of the first n powers of 2 coins, you can make any amount up to 2^(n+1)-1 exactly one way. This is like the base 2 representation of natural numbers in (I believe it's called) positional systems, which we use with arabic numerals. There's a general theorem for this, so, for example, you can make any amount up to 3^(n+1)-1 with TWO of each of the first n powers of 3 coins (1, 3, 9, 27, etc...) exactly one way; the obvious example is NINE of each of the first n powers of 10 coins (1, 10, 100, 1000, etc...). ^.^
    1
  13268. You're awesome!
    1
  13269. On those glasses, a rank 2 topology. I bet the ECE rank/hole connection is obvious. Did "we" solve climate change? I assume you know how the planet works. Lorenz was right, and Sabine's best work was on the strong claim he made that she pulled out of the legend of the Butterfly Effect and Chaos Theory. A dynamical friction generator, and from the poles cold and from the equator heat. Yikes. Those bubbles of methane in the ice. The Arctic Circle Theorem is the essential difference between binary 2D recursion (square space) and dynamical processing, implying a spherical space. Did Boltzmann hang himself too soon? Isn't that entropy in 2D-3D dynamic transform?
    1
  13270. In the 2D proof the end diagram appears as a 3D square. This means there is a projection angle, which is possibly the amount needed to “fix” the inequality.
    1
  13271. in the last couple of years watching people like yourself - I was at school hundreds of years ago - and yes, beautiful
    1
  13272. Algebra and arithmetic is much harder than calculus.
    1
  13273. 34:57 Aren’t you assuming that the integral of f at zero is zero, which it isn’t always? So the constant is F(0), where F’(x)=f(x)
    1
  13274. Honestly that was your best video yet. Beautiful.
    1
  13275. cool
    1
  13276. Pascal's Principle.
    1
  13277. Down to translations a unique square on the grid can be characterized by the vector (l,m) where l is a positive integer, m is a non-zero integer and l+m are less than or equal to n. There will be x^2 copies of such a square where x = n - l - m. So every time we increase n by one we add a number of squares equal to the sum of the squares on the natural numbers up to n. This means that the number of squares should be the 4D pyramidal numbers: N = n^2*(n^2-1)/12 In particular for n=4, N=50.
    1
  13278. The video helped me understand what's behind this, I just saw this formula in a mathematical physics class (undergrad in physics), but it was a bit cloudy yet
    1
  13279. It doesn't really look like a paradox since it requires the horn to be infinitely long (for the harmonic series).
    1
  13280. Let N be the positive natural numbers. Let N* := N + N^2 + N^3 + ... Let O(n) := {a=(a_i) in N* | sum a_i = n} ordered partitions of n Let p(a) := product of a_i Let q(n) := sum a in O(n) p(a) For a in O(n) define: a^ := (a_1, ..., a_k, 1) in O(n+1) a* := (a_1, ..., a_k +1) in O(n+1) Here a_k shall always denote that last component of a In the video it was shown that: O(n+1) = O(n)^ + O(n)* := {a^|a in O(n)} + {a*|a in O(n)} Now q(n) = sum a in O(n) p(a) = sum a in O(n-1) ( p(a^) + p(a*) ) = sum a in O(n-1) ( p(a) * 1 + p(a) * (a_k +1)/a_k) ) = sum a in O(n-1) p(a) * (3 - (a_k -1)/a_k) = 3 * q(n-1) - ( sum a in O(n-2)^ + sum a in O(n-2)* ) ( p(a) * (a_k -1)/a_k ) 1st sum = 0 because a=b^ => a_k=1 2nd sum = q(n-2) because a=b* => b_k = a_k - 1 => p(b) = p(a) * (a_k -1)/a_k => q(n) = 3q(n-1) - q(n-2) This is the Fibonacci sequence skipping each second number: F(n) = F(n-1) + F(n-2) = (F(n-2) + F(n-3)) + F(n-2) = 3 * F(n-2) - (F(n-2) - F(n-3)) = 3 * F(n-2) - F(n-4) Same recursion relation, same initial values => same sequence q(n) = F(2*n) In particular, next number is q(5) = F(10) = 55
    1
  13281. 29:49 A lot of what we discover about psychology or politics is modern rediscoveries with our current pedigree of knowledge. It's super interesting going back and reading, say, the discussions about such things from Babylon or India that just died out for one reason or other.
    1
  13282. I used to basically speedrun Tower of Hanoi as a kid 😂 once I did all 3 pegs in 3 minutes. All because I didn't have a gameboy 😭😂
    1
  13283. having recently watched the mathologer animations of ptolemy's theorem, i wondered if a quatrilateral inscribed in a circle can pivot like a gimbal, and generalize the pythagorean theorem, then maybe a 5 sided polygon, pentagon or pentagram inscribed in a circle could underlie another powerful theorem. the "poster" for this youtube video looked like Mathologer was going to answer just that. alas, Petr's Miracle is fascinating, almost like a mesmerizing whirlpool in action. still, i bet there is some kind of theorem when a simple pentahedral figure is inscribed in the circle, and it will of course, have some kind of complex but visualizable gimbal action.
    1
  13284. Now all we have to do is show this to dota 2 developers so that they can fix their shapeshifting bug.
    1
  13285. It's Amazing
    1
  13286. Only matholger will send his viewers to think of a proof in the beginning of the video! Huge respect for this 🙏🏾. Also- the shape of constant diameter at the end appears to be smooth… is it really?( guessing it is only C^1 smooth)
    1
  13287. What's the partial sum at Graham's Number of terms? Let's get crackin!
    1
  13288. I pity those who learned about matrix multiplication without learning linear algebra.
    1
  13289. Phizza shirt
    1
  13290. I think it's more complicated than simply which holes to connect, because there are two ways to connect each pair of holes: With the lace going in the same side as it came out, or going in the opposite side it came out. Generally you want the laces to come from the inside to the outside at the two topmost holes, as that's the most convenient. But all the other holes have two options.
    1
  13291. I made to the very end!
    1
  13292. I'm struggling with why a super complex Rubik's cube algorithm can't reach way more cube configurations. I mean an algorithm with 4.3e19 different moves would reach every configuration. I guess I don't know the definition of algorithm.
    1
  13293. For me favourite thing about this video was realizing how the armonic series never lands on an integer. I remember trying some finite sums of the reciprocals with my calculator, when they told us that the complete series was divergent. And now watching this I was amazed by the simple of the argument on how the partial sums always fall outside of the natural numbers.
    1
  13294. Hope I win the book!
    1
  13295. Well, extra pegs would still allow you to use an identical strategy, so that’s the upper bound for the minimum. I’m not sure how helpful it would be though.
    1
  13296. @Hoppy Spadge Should be 16!/2^8? 16 moves, in 8 pairs each of which has only one valid order
    1
  13297. Jesus was not afraid of the dark side man. Atheism will get you money, but it can't ransom your soul from Hell. Good luck, Rubber Duxky.
    1
  13298. I want my money back. Have never seen any of this!
    1
  13299. Merry Christmas mathologer
    1
  13300. I like the swivel proof because it was nice, and the coloring proof seems like the more obvious sort of approach to take.
    1
  13301. So did I, for the same reason.
    1
  13302. So they had computers already! 😁
    1
  13303. And all of the "specialness" of 3,6,9 goes out the window in any other base than 10. To me that is proof they are not special.
    1
  13304. 9:08 I've managed to acquire both circular and straight from a local antique shop.
    1
  13305. I made it to the very end!
    1
  13306. How could I miss a new mathloger video??? I need to see it rn lol Edit: Such lectures never fail to amaze me ^^
    1
  13307. Very informative and humorous videos! Love them.
    1
  13308. what happens if you use a non 10 digit count? Like using 87 points around the circle, could the 3 6 9 thing is just a product of how we quantize numbers or sets of them
    1
  13309. Surely that depends on the elasticity of the rubber band?
    1
  13310. Thank you ❤️💚💜
    1
  13311. Thats how quarks might work!
    1
  13312. Love the title and the video. Remembering another humble human being. Being 4 years on april the 11th... Thank you Following all kinds of math related channels, this, really gets to my heart. Sometimes the game of life just isnt fair.
    1
  13313. Great stuff
    1
  13314. Amazing explained... so simple counting can give so Pi-eautiful results!
    1
  13315. my favourite Chapter is Chapter 6 which had taught me that every positive number can be generated from the marvelous harmonic series!
    1
  13316. I'm a bit confused by who is Moessner or what Moessner name refers to. I've found about "Moessner's theorem", "Moessner's triangles", "Moessner's sieve"... but none of the article mentions Moessner full name. Edit: Finally found, after more research that his name is actually Alfred Moessner and discovered his theorem in 1951. Only that page mentioned it : https://www.futilitycloset.com/2016/08/06/moessners-theorem/
    1
  13317. 50 seconds in you describe Tesla as the inventor of the Tesla Coil. It might be better to point out that Tesla was the inventor of the AC electrical system which still dominates the world today.
    1
  13318. Pausing for necklace, as a hobby clockwork maker, it seemed obvious that it is clock face markings. meaning "Always" or Any/All times"... Hah! And I was wrong because it's also the sides of a 3,45 triangle which I should have guessed from the video theme I also have to say that all the correspondences between Fibonacci, Pythagoras, and even extending to rational fractions took me by surprise... shouldn't have done, but did
    1
  13319. definitely caught off guard by the optimal stack of 20 blocks -- definitely a fun surprise that breaking the "rules" can give you a better result than you expect. but i think my favorite insight from this video was "no integer besides 1 appears in the partial sums of the harmonic series."
    1
  13320. Are you trying to make me smarter? It won't work, but I will give you a sub and let you continue to try.
    1
  13321. Internet meme space has concluded that a way to jokingly spell disgusting is "discostan". Your shirt makes me think of a potential religious mathematic pun- sin is dis costan.
    1
  13322. Surely some philosopher must have taken issue with the existence of the bell of a horn jutting out from infinite. Even if only in thought form, there is something about observing the finite portion of an infinite object that should stir imaginations. Beautiful math!
    1
  13323. Anyone knows a link of Zagier's amazing proof?
    1
  13324. I'm fascinated by the problems around 7 in plane geometry, especially construction of a regular septagon. You gave a solution using a spirographer (sp.?), a partial radius of a circle rolling inside a larger circle, but I did not see how to set the length of that partial radius to get a regular 7-figure. Now you seem to have a simpler solution using this vortex to form multi-petal figures, although you did not show a 7-petal figure. Is it true that a regular 7-petal figure & therefore a regular septagon can be made this way?
    1
  13325. My guess onto what happens when trying to find the parent of the 3-4-5 triangle is that both points coincide, so they don't form a triangle.
    1
  13326. i dont quite have proofs for breakfast, but I really enjoyed watching this with my eggs and toast :)
    1
  13327. The maximal number of moves has a point if one move cannot repeat twice and we can also add that the we always have to turn two cards and the number of 1s is event. This problem is more elegant. Transformations under an invariance rule appear a lot in math. For example, the rotation group SO(n) never changes orientation.
    1
  13328. Love folding sciences.
    1
  13329. This is so fascinating! Really enjoyed watching this video :)
    1
  13330. Great presentation ...your enthusiasm reminds me of how much I really love math ! Love it! 🤗
    1
  13331. 7:48 - BUT, 42 has digital root 6, and we all know that's the answer to everything!
    1
  13332. 7:26 since 10^2 + 11^2 + 12^2 = 13^2 + 14^2 we can reduce the equation to 2(13^2 + 14^2) we can then calculate what 13^2 + 14^2 is : ( 169 + 196 ) which gives you... 365 ! Answer ends up being 2(365)/365, cancel out the 365 and you get 2 as the final answer. ( trick to do 169 + 196, just do 169 + 200 and then subtract 4 )
    1
  13333. Every time I see another video that illustrates a miraculous pattern that was discovered in numbers, I wonder if Fermat had spotted it and if it had any relation to his "last theorem" comment that it was very easy to prove...
    1
  13334. We used circular slide rules in our Print Shop to resize pictures.
    1
  13335. Insane work, congrats with that huge video ! Super captivating honestly, and fairly followable for a bachelor's degree math student. Cheers from France ;)
    1
  13336. The number of states in the transition graph is given by: S(1) = 3 S(n) = 3S(n-1) S(n) = 3^n S = 3, 9, 27, ... The length of the path that visits each state exactly once is 1 less than the number of states. L(n) = 3^n - 1 Interestingly, the number of states is equivalent to the perimeter of the Sierpinski triangle. In fact, you can create a 2-graph (borrowing terminology from category theory) where: The vertices are the vertices of the Sierpinski triangle. The 1-edges are the edges of the Sierpinski triangle and the states of hanoi. The 2-edges are the edges of the transition graph. I drew these 2-edges in a different color on top of a Sierpinski triangle, and it looks kinda cool. You get these sawtooth patterns because of how the smallest triangles are rotated. To understand how these two structures are related, consider the construction process. In the case where n=1, both graphs are a simple triangle. The number of nodes is equal to the number of edges, so both are 3. Each combines 3 graphs of the next smallest size to build itself. They differ in how the smaller versions are combined. The Sierpinski triangle overlaps 3 vertices, while the transition graph doesn't. No new edges are created with the Sierpinski triangle, while 3 new edges are created for the transition graph. This gives 3 different recurrence relations: V(n) = 3V(n-1) - 3 S(n) = 3S(n-1) E(n) = 3E(n-1) + 3 Each could be solved using the formula for geometric series, but there's an alternate way of determining the number of edges. L(n) = (2/3)E(n) The length of the path that visits each state can also be constructed recursively. L(1) = 2 L(n) = 3L(n-1) + 2 Once you leave a subtriangle, you never need to return, so you ignore an edge at each level of iteration. This ends up multiplying each term of the series by a constant. E(n) = 3 + 3^2 + ... + 3^n L(n) = (2/3)3 + (2/3)3^2 + ... + (2/3)3^n And this constant can then be factored out.
    1
  13337. I liked that the no 9 series converges.
    1
  13338. I vote for the parabolic solution for the overhang. I will leave a trail of that solution in every science museum I visit.
    1
  13339. (ho)^3 is incorrect; 3(ho) is correct ...
    1
  13340. ???? this is the weirdest paradox I have seen
    1
  13341. I definitely liked the Log and Gamma approximation!
    1
  13342. Very beautiful video. I really saw this proof first https://www.geogebra.org/geometry/qnyq85cf. One of the proofs of the Pythagorean theorem (4 + 1 = 5) and the square of the sum. I wonder if there is a connection between such evidence and the way presented in your laser turtle video. Thanks again for the great video.
    1
  13343. hmm I wonder if there are squares in a triangular grid...
    1
  13344. Great video! Peace out
    1
  13345. Takk!
    1
  13346. Halfway through the video & my mind keeps blowing up again & again.
    1
  13347. I take what he says at face value, brain not quick enough to decipher as he presents it.
    1
  13348. There's a song in my first language whose lyrics have a line meaning "If I subtract myself from myself, it doesn't make zero" lol.
    1
  13349. Im india we are taught this but not about heron lol
    1
  13350. What didn't work for me: trying to watch this at 1.5 speed to save one-third of fifty minutes, lol. I wasn't a math major, but I'm an electrical engineer, so I'm familiar enough with sums, calculus, matrices, differential equations, etc. to follow your videos on more pure mathematics. Other than that, I just like watching channels like yours and 3blue1brown. I was able to follow your presentation, though I needed some time to think about some of it. That's fine though, because it's a video, and I can rewind and pause as much as I want.
    1
  13351. As a conséquence the médianes satisfy the triangular inequality 😅
    1
  13352. Gamma and that weird tower at the beginning were my favourites, personally. I love when unexpectedly weird, complex and irregular structures like these emerge from what looks like a perfectly regular construction.
    1
  13353. Nobody remembers the teachings of Pythagoras' brother Preposterous. He favoured the 3 x 5 x 6 triangle, which turned out to suggest a P time solution to the maximum clique problem.
    1
  13354. A simple program using dynamic programming took just under 2 minutes to compute change for $2,000: 427707562988709001. By the end of the video I just programmed in the polynomial formula, and then I could compute change for 2*10^(2^20) in less than a second.
    1
  13355. long time no see, glad to have you back
    1
  13356. I do not know what is situation in Croatia now, but when I was in elementary school (some 18 years ago), we learnt that formula for area of triangle in Cartesian coordinate system. I just do not remember teacher mentioning it by this name.
    1
  13357. If I saw either of them in school I didn't understand it and therefore forgot it. And yes, I should get my money back - my geometry teacher absolutely destroyed math for me.
    1
  13358. If we would rise x+2 to the power of negative numbers would this still have some corresponding uses? Or would that have no meaning
    1
  13359. 🙏 Thankyou sir. Lots of respect from an Indian student
    1
  13360. I was most surprised by the weird optimal balancing block configuration, but the thing that will probably stick with me is the visual intuition why the Euler-Mascheroni is less 1 and greater than 1/2.
    1
  13361. Honestly, your videos are F**KING AWESOME!
    1
  13362. No. I have not heard of "the mysterious Schwarz lanterns and the existential crisis in maths that was triggered by them?" Yet lacking this knowledge in no manner prevents me from recognizing Obsessive-compulsive disorder (OCD) when encountered. Pass... Or possibly, En Passant? ♾
    1
  13363. My favorite super-hero, Towel man 😌
    1
  13364. Mathologer is the very best channel ever
    1
  13365. BUT: what if you can't flip a face down card if there is no card to the right to also flip - a perfectly reasonable assumption since is wasn't specified? Then if one can arrange for a situation where only the rightmost card is face down, one ends up stuck. And this is easy to arrange if there are an odd number of cards by just starting at the rightmost card move two to the left, flip them, then move two to the left, etc. This only applies to trying to get them all face up, not that the sequence terminates...
    1
  13366. ♥️♥️♥️♥️
    1
  13367. I learned nothing of this in 55 years. I want my time back from my school.
    1
  13368. 0:42 current caption: "It's some really nice material starting with the super famous tower of an oil puzzle". Sounds messy.
    1
  13369. I haven't seen all your videos but this is the best one I've seen. I love the geometric explainations and proof, I could watch an endless amount of anything connecting higher mathematics to geometry. I'm a mature student in my 3rd year of a physics degree, and one of the things that got me back in school is yes, prepare to groan, numberphile's -1/12 videos, as annoying as mathematicians seem to find it haha. So anything relating to infinite series' gets my attention. I had heard of Bernoulli numbers but dind't know what they were until this videos and now I'm SUPER interested to learn more. I'm also learning a lot of linear algebra at the moment so the matrix expressions you used were really eye opening. As for the length, i often see your videos and say to myself, "ah don't have time" But honestly anything over roughly 12 minutes I begin to think that, however, whenever I do watch one of your long videos I leave it thinking "wow I wish I watched that sooner" I never mind spending the time. Maybe you could do something like have a companion video for the longer ones, you could get away with putting a lot more clickbaity stuff in it, tease some of the more fascinating questions answered in the video, and if people watch both, you're getting double the views. I mean the videos are almost as long as movies now, and movies have trailers, so why not!?
    1
  13370. My favourite proof is the chord version. Even after playing around with this in Geogebra, I spent extra time with that. Conway discovered so many things in math, he truly was a genius.
    1
  13371. The 😀 proof was easier for me to get.
    1
  13372. Hi! With non Fibonacci numbers i would argue that the spirals are still there although its beads are more apart.
    1
  13373. 👍✌👏👏👏
    1
  13374. I did not learn this in school, and when I did learn it, it fell out from the fact that cot a+cot b+cot c = cot a cot b cot c when a + b+ c = 90.
    1
  13375. Another random phi fact I found recently while messing around doing multivariable calculus: the global minima/maxima of x*e^y with the constraint that x^2 + y^2 = 1 are located at (+/- 1/sqrt(phi), 1/phi).
    1
  13376. ​ @Mathologer  I've been in love with the 2 x 2 = 2 + 2 equation since childhood. As a more adult mathematicians I have been looking for 'categorizations' of this equation, i.e 2-dimensional vectorspaces V where V \tensor V = V \oplus V in some natural, or meaningful way. So far I came up emptyhanded. The two natural candidates: the general equation V \tensor V = S^2 V + \bigwedge^2 V applied to the dim V = 2 case and the decomposition into irreducible representations of V \tensor V in case V is the defining rep of the group SL(2) both reduce to 2 + 2 = 3 + 1. (One can argue that they are in fact the same example, just viewed as special cases of different general phenomena). I find it quite annoying how 2 + 2 = 3 + 1 seems to beat 2 x 2 = 2 + 2 to the punch every time, but maybe you or one of the other readers knows an example. I have one example that sort of works, but in an artificial way. View C as a two-dimensional R-algebra. Then, if I am not mistaken, C \tensor_R C is isomorphic (as an R-algebra) to Mat(2, R), so simple and hence we get 2 x 2 = 4 rather than 2 + 2. However. If we interpret C \tensor_R C as a C-algebra in the same way that we can interpret C \tensor_R A as a C-algebra for any R-algebra A, then we get, as a C-algebra that C \tensor_R C = C \oplus C. So in a way this is an example of 2 x 2 = 2 + 2, although here it is perhaps fairer to say that we have an example of 2 x 2 = 1 + 1 for some very fat notion of 1
    1
  13377. Very impressive way to understand how to construct a higher dimensional cube from lower dimensional cubes and their relation! Can this idea help us to understand hyperspheres?
    1
  13378. love this. rep-tiles (would make a good t-shirt> 'my favourite reptile') ad frac-tiles are also videos I'd be interested in. rep-tiles with 'holes' is something that also makes for some nice t-shirty patterns
    1
  13379. oh and the hexagons too me back. reminded me of the first 3-D arcade game that used perspective: https://en.wikipedia.org/wiki/Q*bert
    1
  13380. Very entertainment, thank you.
    1
  13381. In next vedio explain the pi formula of ramanujan,chudnoviski,newton,c.f gause,wisetres,and more 😊😊
    1
  13382. I love math
    1
  13383. Very nice animation at the end
    1
  13384. At 45:33, why not just compute the value of the yellow sum? Surely that would converge faster than anything else, and Bernoulli numbers seem like the kinds of things that would come in prepared tables.
    1
  13385. Mario D Meza. You made math a dynamic and beautiful piece of arts! Thanks you for This
    1
  13386. I earned a PhD in mathematics in the U.S. but was unaware of Heron's formula until reading William Dunham's great "Journey through Genius." And then you say there's a formula. for. the. area. of. ANY. quadrilateral. (convex or not) Thank you, Mathologer!!! 💥💥💥 ❤❤❤
    1
  13387. the 700 year old proof really struck me as intuitive.
    1
  13388. Second Challenge: If 'm' and 'n' are both odd, then for the last 'trig monster', 'j' and 'k' will be equal to 'm/2 +1/2' and 'n/2 +1/2' respectively. This is the same as '(m+1)/2' and '(n+1)/2'. The numerators of these values of 'j' and 'k' will be canceled by "m+1' and "n+1' in the denominators of the arguments of cos's leaving just pi/2. Since cos(pi/2) is zero, both terms will be zero rendering the entire product zero.
    1
  13389. I started the video, looked at your shirt, and started laughing.
    1
  13390. I wasn't taught heron's formula in school but in competition math during high school in Texas.
    1
  13391. 19...? Σ (Fibonacci)_k
    1
  13392. 4:57 the coach never said anything about terminating on all face up cards
    1
  13393. I'm wondering whether the Mathologer has alopecia universalis.
    1
  13394. Can you do a video to show a more general solution? For instance, can you show us how to know how many solutions to a^2 + b^2 = n exist for ANY instance of n (whether or not n is prime)? n = 1105 is one of my favorite examples. Another thing I like is figuring out solutions to a^2 + b^2 = 10^n You can show that there are 4(n+1) solutions including trivial ones ... BUT ... when you stipulate that a>b>0 AND that a,b cannot divide ten there is always one and only one solution for any n>0. I happen to think that the solutions themselves (3,1);(8,6);(26,18);(96,28);(316,12)... are beautiful and that learning to derive them is fun.
    1
  13395. Mathologer I guess you mean the Fermat Christmas Theorem. I will look at it. I hope I made it clear what my question was.
    1
  13396. Caution Mathologer, there's a huge green arrow! Oh, he dodged it. All's fine. 😃
    1
  13397. 10:23 This happens because cos^2(pi/2) = 0. Say m and n are both 7. One of the terms will be 4cos^2(4pi/8) + 4cos^2(4pi/8). 4pi/8 simplifies to pi/2. cos^2(pi/2) is 0. This term comes out to 0. Multiplying 0 by all of everything else gives...0. In general, the last term of any case where m and n are both odd will be 4cos^2(m*pi/2m) + 4cos^2(n*pi/2n).
    1
  13398. at 30:23 why the big triangle has 2 equal sides with length x?
    1
  13399. Whenever people tell me something about digital roots, i tell them that if there was any sense in the universe we would have six or twelve fingers (how about four opposable thumbs for 3d grabbing+holding?) instead of the stupid 10 we got. They usually stop talking to me then.
    1
  13400. Brilliant &/- nice
    1
  13401. My (informal) reasoning as to why the initial degenerate polygon is the centroid of the final polygon. I'm late to this video, but I'm writing this comment mainly to solidify the idea in my head. Each point on the polygons can be considered a vector from the origin. Consequently, without the offset degenerate polygon, the final polygon would also be centered at the origin. The degenerate polygon adds the exact same vector to all of the points of the final polygon, which has the effect of shifting it (and its centroid) by that vector.
    1
  13402. Squiiiiiiiiiiiiiiiiiiiiiiiiiiiiiish
    1
  13403. To Herr Mathologer: I like the unique design on your shirt 25.8069=x=666^1/2 Maybe it should be 18.2482 for being only half evil 333. Ha!
    1
  13404. 'sustainable a. f.' on a single-use tin. with a liquid that also comes from the tap instead of being transported by lorry.
    1
  13405. Harry Maguire
    1
  13406. Minor point; they are not folded they are rotated around a point.
    1
  13407. This polynomial looks like the Ehrhart polynomial of some 5-dimensional polytope. I wonder if it is... Edit: I got it! Each configuration of coins that together forms one dollar can be thought of as a point in Z^6 where x_1 + 5 x_2 + 10 x_3 + 25 x_4 + 50 x_5 + 100 x_6 = 100 and where each coordinate is positive. This gives us a polytope with vertices (100, 0, 0, 0, 0 0) (0, 20, 0 , 0, 0, 0) (0, 0, 10, 0, 0, 0) (0, 0, 0, 4, 0, 0) (0, 0, 0, 0, 2, 0) (0, 0, 0, 0, 0, 1). The Ehrhart polynomial of this polytope will then be a formula E(k) = #ways to form k$ in coins. From the fact that these vertices form a simplex (they are all linearly independent) and there are 6 of them, we can see that the polytope is 5-dimensional, which implies that the Ehrhart polynomial has degree 5. A polynomial of degree 5 has six coefficients, so it's uniquely determined by six values. These can be computed explicitly like in the video: E(0) = 1 E(1) = 293 E(2) = 2728 E(3) = 12318 E(4) = 38835 E(5) = 98411 From this, we can see that E(k) is given by E(k) = 40/3*k^5 + 65*k^4 + 112*k^3 + 161/2*k^2 + 127/6*k + 1 which is exactly the polynomial in the video!!!
    1
  13408. I'd be careful about installing things that can run Java applets in a browser. There's a very good reason browsers stopped supporting this - it is VERY dangerous. It would be far better if the owners of this website would simply redo the work in Javascript.
    1
  13409. The most memorable part for me is the visual proof for gamma being less than 1. I always thought that the difference between two diverging series being a finite number in its own right was interesting but the quick and easy proof of it being less than one really makes it more interesting.
    1
  13410. If calculus is easy, then why do I hear boss music ? 34:46
    1
  13411. Thank you very much for your work and greetings from Würzburg!
    1
  13412.  @Mathologer  No, I did not know that, thanks for the info. Yesterday I learned that you are from Germany and refering to the german Wikipedia from Würzburg. "Die Welt ist ein Dorf"
    1
  13413.  @Mathologer  I'm a civil engineer who likes to "look beyond one's nose". For example, about 10 years ago I worked as a surveyor for two years to better understand their work. I also worked as a scientist at the university for six years. In my opinion, the main problem is always the interfaces.
    1
  13414. Can i ask why do you double the fourth number in the matrixes, when you dont do that to the rest?
    1
  13415. Finally someone explained logarithms in a very clear way.👍❤️
    1
  13416. yes very good
    1
  13417. cannot resist to 😍
    1
  13418. Easy relative to what? I took honors classes in humanities, Latin, etc. but then took remedial maths with all the dropouts on the extreme other end of campus. To this day I have trouble adding numbers together, and I've completely forgotten how division works. Calling calculus easy is like saying that learning English is easy to someone that was born and grew up in Dayton, Ohio.
    1
  13419. Hey! You again. Nice!
    1
  13420. does the odd/even pattern imply that 1 is both odd and even?
    1
  13421. So...what is the solution to the one Big Cross that divides (via 5 pieces) into two Smaller Crosses? I've been trying all evening but I can't sleep untill I see the solution :-D.
    1
  13422. I made it to the “true ending” of this game 🎉
    1
  13423. The most memorable is that every time I see your videos I like maths more and more.
    1
  13424. scissors cuts paper paper covers rock rock crushes lizard lizard poisons spock spock smashes scissors scissors decapitates lizard lizard eats paper paper disproves spock spock vaporizes rock rock crushes scissors (as it always has).
    1
  13425. What happens if you get negative integer differences when subtracting in the sequences???
    1
  13426. Dream proof because it came from a dream or because it's so dreamy?
    1
  13427. beautiful
    1
  13428. Best part : Chapter 4 - Terrible aim I love the fact that it never hit a integer even if it goes to infinity. The argument is so easy to understant (but not to come up with)
    1
  13429. The same chess board color counting is used for proving that it's impossible to tile a rectangle with the 35 hexominoes.
    1
  13430. What happens if not base 10?
    1
  13431. Each side of triangle ABC can be partitioned into two sub-segments by the tangent points of its incircle. Let the length of the segments meeting at A be called a, at B be called b, etc. Then it's easy to see that each one of the 3 extended line segments comprising the whiskered triangle ABC is of length 2(a+b+c) with the incircle tangent to it at its midpoint. So if the incircle has center O and radius r, then their endpoints are a distance of √(r²+(a+b+c)²) from O. In other words, the whisker tips all lie on a circle centered at O. ∎ p.s. now I'm wondering if the "windscreen wiper" phrase in the video title refers to the fact that since these secants are the same length, you can just slide one around into another, and in the process sweep out an annulus. ... or i could just keep watching and find out 😜
    1
  13432. Please change the background to black one
    1
  13433. PBS SpaceTime expanded when they dropped the white-screen background, and hired an animation design team. ;)
    1
  13434. 100xpi comments! hahaha
    1
  13435. Fascinating, but wanted to say, given the topic... you should get a shirt that says "Hyperspace Invaders"
    1
  13436.  @Mathologer  Thanks! It is always great to see one's work appreciated by someone who understands your fascination with Maths. Inspired by your video (and your appreciation), I created a second version (search ztvrbdyy on Geogebra) that contains: 1. The extended version resulting in a degenerated n-gon. 2. A selection box with - the standard sequence - or a shuffle of the standard sequence - or a manual sequence. The last option allows you to test the {1, 2, 2, 3, 3, 3} homework practice from your video. Perhaps an idea to add this ggb file to your qedcat / petr repository?
    1
  13437. ⭐⭐⭐⭐⭐
    1
  13438. I need my money back. I came across the algebraic proof when I tried teaching my kids. The schooling I had never taught us the proof. I came up with my own crazy algebraic proof in desperation but promptly stopped using that once I learnt the two squared proof.
    1
  13439. okay, maybe one can do that for stuff like triangle numbers and such hmm. let's say a function tri(n) gives a number that is triangular number #n. tri(n) = sum(n+n-1+...+2+1) = n*(n+1)/2 = tri(n-1) + n 1 3 6 10 15 21 28 36 triangle numbers can't have a trivial example of tri(x-1)+tri(x)=tri(x+1) because tri(x+1) = tri(x) + x+1 which would mean tri(x) = x+1 which would mean the difference between tri(x) and tri(x-1) is just 1, and there is no such triangle number pair. But what if they don't have to be consecutive. tri(n) outpaces n, so for example tri(3)+tri(5)=tri(6), tri(4)+tri(9)=tri(10) by definition, and there is an infinite amount of these triples; n+tri(n-1)=tri(n) by definition for every n = tri(x), x>2 But not easy to analytically get patterns for more numbers. square numbers proven to work, as well as linear next is pentagon 1, 5, 12, 22, 35, 51, 70, 92 12+22 is barely lower than 35, 22+35 barely higher than 51, no consecutive pent(A)+pent(B)+pent(C) either. I guess cause its n value is actually barely higher than 2 for A^n+B^n=C^n hexagons? 1, 6, 15, 28, 45, 66, 91, 120, 153 15+28 barely under 45, 28+45 way above 66. same problem Now question, how do we calculate the n value for A^n+B^n=C^n in these cases.. it's related to some surface dimension shit isn't it, the sequences we have have some bs inside.
    1
  13440.  @Mathologer  oh that's indeed super nice
    1
  13441. My vote is for Tristan's idea of the fractal animation. Very intuitive and pretty cool!
    1
  13442. Hi Mathologer. Biologist, 1 semester calculus, but I've continued studying calc a very little bit. Your video was understandable and enlightening. Your hint that you will continue with something on the Riemann Zeta function is just so cool. I look forward to it. Oh, and BTW, thank you for doing these wonderful videos.
    1
  13443. If you imagine first experiment with point moving around bigger diameter, you can easily solve coin paradox
    1
  13444. My prefered part is that the partial sum is never an integer !
    1
  13445. Me when ln(2) shows up: "Oh, you would be involved in this kind of nonsense, wouldn't you?"
    1
  13446. @23:45 I can't help but see a bunch of bananas... Curse your packing efficiency nature.
    1
  13447.  @Mathologer  Your video drives me bananas; this makes me want to apply similar logic to find a well behaving ellipse perimeter equation. :)
    1
  13448. 39:00 this triples also have the 2 first numbers going closer to a 1:2 ratio
    1
  13449. Pi/2 does not explode, because you are constantly multiplying a fraction greater than one, with a fraction less than one. The SQR equation is infinity, because you are always multiplying fraction greater than one, although you can also get ~0 if you remove the denominator's 1, ie. 2/3 * 4/5 * 6/7... Therefore, 0*infinity = pi/2. Whooo, I have solved it! Where's my Nobel?
    1
  13450. nice!!
    1
  13451. Yes, reminds me of my Mono Rietz from VEB Mantissa. No kidding :-)
    1
  13452. Second ;)
    1
  13453. A slide rule but round
    1
  13454. I did say that before you did LOL
    1
  13455. Most memorable: maximal overhang!
    1
  13456. 18:00 what you just showed us was a subtle STEMerch advertisement! Maybe I’ve been watching too much Flammable Maths...
    1
  13457. I know the second proof by heart, it has sentimental value for me. The video was absolutely beautiful!
    1
  13458. Not for all 4-mutilated boards we can fill with dominoes. For example, you can take out the 2 black top-left corner squares, and any other 2 green squares. Then the top-left green square can never be filled.😁
    1
  13459. Don't remember seeing either in school
    1
  13460. Could you check the colors? "good" and "bad" look different, but the numbers look the same color until I put on my 3D glasses.
    1
  13461. inconceivable!
    1
  13462. Mathloger, make New Calculus famous. Let canceled dude be uncanceled. Brutal cuckademia.
    1
  13463. 1024
    1
  13464. First!
    1
  13465. And this is how we do math in Germany Dr. Jones...
    1
  13466. Why does this video has so few likes? I loved it ;)
    1
  13467. Professeur ! I don't know if you remember, but it would be neat if you did something fun on Metronome-styled pendulums. Just a thought for ideas on moment of intertia and torque in maths. ¡Saludos desde México!
    1
  13468. Okay, this video was just awesome. I watched it entirely (in one take ^^) and I found it particularly interesting. I've just been taught formulas for sums of k up to the third power. Your video really adds a new perspective to how to find new formulas and really meets my curiosity. Plus the video being an hour long, I just love what you do. Your channel is superb. Keep on going ! I love it !
    1
  13469. The remarkable fact that a 3-4-5 triangle has an incircle of radius 1 was new to me. It just goes to show that Truth can be sitting there, just 1 step away, and you may not see it-- for decades. I am both humbled and joyous whenever Mathologer opens my eyes in this way.
    1
  13470. 2098765413 = 17 * 123456789 11322 = 17 * 666
    1
  13471. Is there an English translation of thierry bousch's paper anywhere? the one in the description is french :(
    1
  13472. So you've gone over square grids and their turning property of 90 degrees... Would a triangular grid have a turning property of 120 degrees? What about hexagon grids? Can these grids even exist? If so, can any n-gon grid of 5, 7, 8, 9, 10... exist?
    1
  13473. Another interessting fact about Torricelli's horn is when you slice the horn in a flat 45 degree way and you then look at the cut shape area you now get the perfect egg shape of nature.
    1
  13474.  @Mathologer  As far as I know there's this German pdf document on Google search result: "Die Ei-Kurve als Schnitt des Hyperbolischen Kegels".
    1
  13475. Another great video, Burkard. In retrospect, it's weird that I had never considered that your name wasn't actually Mathologer.
    1
  13476. So, what did I miss playing with LEGOs? I could have been a child prodigy just by counting colored blocks!
    1
  13477. coins are an endangered specie 🥁
    1
  13478. My vote is for Gamma!
    1
  13479. Very happy 😍💋 💝💖♥️❤️
    1
  13480. I agree with JF Tremblay, Euler Mascheroni const, increasingly BETTER approximation., and the no nines less tha 80 animation. All chapters fantastic though BTW all those blue areas are concave up, so each is larger than a triangle with the same 3 corners, the sum of those being 0.5, hence sum of blues is greater than 0.5
    1
  13481. How do you prove the convergence of series of [r2/(2k+1)]*(-1)^k/r2 to series of (-1)^k/(2k+1) as r goes to infinity? Uniform convergence is mandatory since you don't even have absolute convergence.
    1
  13482. Request show 4/1 instead of just 9:29 4 for nice display of series .
    1
  13483. Hi Mathologer, I am a second year university studying physics and applied math. I love your videos, and often start my day by watching them as I'm eating my breakfast, so I can give my brain something to chew on as well so to speak! Your videos are an absolute joy to watch, and please don't be afraid to keep making long and not particularly easy videos such as this one. While I don't think I quite managed to get every detail of this video on my first viewing, I am looking forward to giving it another shot very soon. As a fellow math enthusiast, I implore you to continue making quality videos such as this. Thank you.
    1
  13484. The most memorable is gamma, I just love where everywhere it pops up :)
    1
  13485. Hahaha "Slide rule apps", wie witzig ist das denn? :D Wie wäre es denn wir programmieren einen Rechner auf der Basis von Slide-Rules und richten auf dem dann wieder eine Slide-Rule-App ein. Das wärs doch xD
    1
  13486. I dont get why you would get to any sum you want. Cause in the pi example it would be like leaving out many negative term
    1
  13487. Fantastic!
    1
  13488. Hey Mathologer! To answer your questions near the end, I'm currently finishing my last year of high school, and am planning to get a degree in mathematics/computer science in university. I don't really have a special background in maths, just standard maths courses in school and internet videos like yours. In fact, internet resources like this are what really got me into maths. Being able to see higher level mathematics like this, instead of glossing over concepts in a textbook, really helps to make things interesting! I can't think of any major thing which would make these types of videos better. Besides, I think it's unrealistic to expect to understand everything in the first viewing. The beauty of these videos is that you understand more and more each time you watch them! And I gotta say, seeing the Bernoulli numbers defined in such a simple matrix equation was beautiful. Thank you for taking the time to teach us mere mortals!
    1
  13489. Thank you very much for this inspiring video!
    1
  13490. I think you do pretty damn good. Curious what you use to display / make your math video display on screen. And yes Euler sum formula is usually not explained enough so nice to see it again in new light.
    1
  13491. The Cat's Meow !
    1
  13492. the reason is because a surface doesn't have a volume but real paint has so it's combining weird assumptions if you think about it
    1
  13493. The three diametral chords of the weird curve are off-center chords of the Conway's circle, which must therefore have a larger diameter.
    1
  13494. Thank you
    1
  13495. The horn is about as paradoxical as the story of Achill and the turtle. In other words, infinitely many infinitesimally small things are not exactly intuitive.
    1
  13496. great video
    1
  13497.  @Mathologer  I'll still be around then waiting :) love your videos
    1
  13498. I think the direction is counterclockwise if the number of discs is even and Clockwise if the number of discs is uneven. ((-1)^n for n being the number of discs and -1 being clockwise and 1 being counterclockwise) So in this case counterclockwise i think. (sorry for my bad english)
    1
  13499. Where can we get a copy of Madhava’s calculations work in Sanskrit?
    1
  13500. I have no idea why this channel doesn't have as many followers as 3B1B. I'm a huge fan or both and they both seem to be of the same quality. Don't get me wrong, Mathologer is huge, but 3B1B is like 6 times bigger. I feel like 3B1B's essence of calculus and linear algebra really set him off. Because a lot of people were going to him to actually learn a subject, and not just to learn about a fan fact. Maybe Mathologer should consider doing something like that. Maybe a essence of probability (although he should probably call it something different). Since 3B1B seems to be taking his sweet time with it lol.
    1
  13501. 5^3 + 6^3 = 341 = 7^3 = 343. The difference between cube numbers is 2. 7^4 + 8^4 = 6,497 = 9^4 = 6,561. The difference between quad numbers is 64. 9^5 + 10^5 = 159,049 = 11^5 = 161,051. The difference between pentagonal numbers is 2,002. The difference between 6 is 69,264.
    1
  13502. In the Wikipedia article on coprime integers there is an algorithm for generating all coprime integer pairs exactly once, and it also has the structure of a ternary tree, but it generates all ordered pairs, i.e. all reduced proper fractions AND their reciprocals. How does that algorithm correspond to the one in this video, which only generaters proper fractions?
    1
  13503. Mmmm, forty plus minutes of discrete derivatives. Love it 😁
    1
  13504. Would love to see Euler's face watching this video.
    1
  13505. This should be made into paper, a superb math picture book!
    1
  13506. havent watch it yet, but its a gift to watch your videos.
    1
  13507. Amazing video, Mathologer! You call out that the s and constant grids are the same. Is there no shifting that would make the r pascal grid the same as the other two, even if not as simple as down-one?
    1
  13508. "Andrew also found an American patent dating back to 2025" - the description
    1
  13509. I love when he smileys. It's a mix of nervous and sarcastic smile.
    1
  13510. The path in the tree to the pythagorean triple [153, 104, 185] goes like this: [1,1,2,3] -> right -> [1,2,3,5] -> right -> [1,3,4,7] -> right -> [1,4,5,9] -> left [9,4,13,17]
    1
  13511. It's a little silly, but the most memorable proof for me is the odd/even no-integers proof. That's just such a clever property to notice and utilize! (Also, even/even fractions are never in reduced form, since you can always cancel factors of two!)
    1
  13512. Honestly, for the insanity in this video, I, a Calc 3 and Linear Algebra student with very rudimentary background knowledge in complex functions, understood it FAR more than I expected. Absolutely incredible work with this one. I'm looking forward to more insanity.
    1
  13513. Isn't that desk too low?
    1
  13514. Simple explanation on steroids
    1
  13515. I made it to the very end. Very interesting as always. :)
    1
  13516. I have a bachelor's in math and started to get lost around 26:28 at the "magical merge" and didn't really know what was going on when f(x) kept getting referenced around 36:00 but I didn't go back to rewatch and felt I got something out of it anyway. Great video!
    1
  13517. To solve the problem of building 2 and 3 as sums of reciprocals, you can leverage perfect numbers. The sums of reciprocals of the proper factors of a perfect number always equals 1. There are some subtleties involving factors which repeat across several different perfect numbers, which ends up being the powers of a half. Those are rather annoying to work around, but it's perfectly doable.
    1
  13518. so, you are saying the universe is finite?
    1
  13519. I don't know whether r & s occur in nature, but there are similar bi-ratios occuring in architecture, like the Dom Hans van der Laan who built a structure with the bi-ratios of the plastic ratio which also arises from a similar simple linear three-variable sequence defined by a similar system of three recurrence relations.
    1
  13520. First
    1
  13521. TheOneThreeSeven is releasing a mathematics visualisation tool so you can create plots like 3blue1brown's - https://youtu.be/IQAhv8JM0Jo
    1
  13522. Interesting
    1
  13523. 9:11 yes i remember that. Imre Lakatos used this to try to expand the finding of truth after Popper used it and later other scientists (philosophers, mathematicians, physicists, ....) enhanced it until Lakatos tried to prove the development of systematic of scienctific research programmes
    1
  13524. Watched the whole video - in spite of losing it completely after about two minutes :)
    1
  13525. I think I like your proof, it's a little simpler to understand.
    1
  13526. The "no 9's" series was the most interesting. The whole video built up the mantra "it is infinite", only to be broken with this revelation. I loved how the fractal formed. Cool T-shirt.
    1
  13527. Most memorable is easily the fact that no partial sum can be ever be an integer! The harmonic series is brutal. How the reciprocal primes still manage to be infinite seems truly brutal-- this series never quits.
    1
  13528. can't believe you haven't mentioned universal hyperbolic geometry ;(
    1
  13529. No pie inside is merely a pie plate.
    1
  13530. Fascinating. I'm glad you did this one that is so much like one or two I saw about square orbits by "All things physics", only with more maths details on Fourier series. I bet they saw your video first! I can't do animations but just formulae on Excel - and have one set up to make spirograph patterns and can just change some figures at the top of some columns to change the shape, number of points, etc. It's based on the toy version with its limited cog number ratios - but on Excel you can input any numbers. The problem with trying to make the square orbit was if you make the 'hole distance' on the small wheel further out towards the edge of the wheel the orbit corners are too pointy and sides too curved and if you put the 'hole' closer to the centre the square sides are straighter but the corners more rounded, and at some point there is a compromise between square corners and straight sides. Anyway that other video inspired me to get the Spirograph out and draw loads more patterns, and my Mum's cousin appreciated a card covered in them for his birthday! I also printed out some computer generated patterns and left them at church for children to colour in when they are bored or have nothing better to do...! At least that is kind of educational for them if it gets them thinking about such mathematical patterns. It would be nice to see pictures of Fourier transforms of the sounds that various musical instruments make. Wouldn't they just be spiky plots of the distruibution of frequencies / harmonics. Some sounds look fascinating on audio files (I use Wavepad aduio editing) and sometimes I look at the Fourier transform time plot which shows how the proportions of each frequency change over time. (Maybe because I want to find out the frequency of an annoying background whistle or hum!)
    1
  13531. 6:50 In russian school we learn this formula, but without proofs.
    1
  13532. Wow. Simple maths, but mind completely blown. Easily one of my favourite episodes. So, so beautiful.
    1
  13533. He smiled like Sheldon Cooper
    1
  13534. Proving the Riemann hypothesis: Moments after a Big Bang, dual numbers to be considered at the origin, repulsive and gravitational forces are balanced, but dark matter is breaking up. "Dual numbers?", you might ask. Isn't that the square root of zero? (Actually, it's a great way to start a proof that ends with division by zero - a vertical line, and it corresponds to what I call Doyle's constant that deals with a brief window in time starting with taking the eth root of e and ending with a vertical line. Delanges sectrices for the unflaking of dark matter, Archimedean spiral for spinning up dark matter, Ceva sectrices for that dark matter breaking into smaller, primordial dark matter and black holes, Dinostratus quadratrix for impacting surrounding black holes, and Maclaurin sectrices for dark matter being slowed down over time. Split-complex at one, noting the more gravitational nature of diagonal ring/cylinder singularities of baryonic matter from broken dark matter. Using repetitious bisection (from arbitrary angle trisection), one gets added (SSA) for when dark matter dominates a universe and one gets added for when black holes and rarified singularities make things more gravitational, leading to a Big Crunch. At 2 = a Big Crunch, the solution to the Basel problem, and the "bellows method" drives things. A Big Crunch event needs the net gravity from the localized area AND gravitational effects from surrounding universes. The 2 represents how locked up black holes are making a stacking, planar contact. Everything else are harmonics - effects on surrounding universes. I see this proof as like trying to climb a steep cliff with the process of dark matter breaking up, getting to the top of the hill (like stable atomic nuclei) and then skiing down the "hill of many black holes" past an inflection point where black hole formation and spaghettification happens, leading to the final vertical line tensor of the Maclaurin trisectrix. This is much like I have posted about considering the recursive form of Euler's Identity as being like an old incandescent light bulb that you can see having first, second, and third degree critical points. Ramanujan's Infinite Sum of negative one-twelfth: each positional jostling, mass, and gravitational/attractive exchange between universes (defined by Big Bounce events, considering Ramanujan during both Big Crunch and Big Bang sides) ADDS up to evenness (or a partner in marriage - like half of the whole, I guess). Net gravity increases over time from the breaking up of dark matter, much like a vertical asymptote. It continues to increase, but hits a critical point, much like a vertical tangent, when enough dark matter has weakened and been broken up into the more net gravitational baryonic matter (split-complex numbers). Then, the effects of gravity reach their upper limit - much like a vertical line - at e to the negative e power (where black holes have been flattened). Over a decade ago, I said that the potential energy of a Big Bang was 1/6 times (the speed of light, cubed). If I had said Big Bounce instead, I would have been correct, in a sense. Now, I see how the speed of light would have to be the fourth power.
    1
  13535.  @cmuller1441  "… reminds me why I love maths." Right?
    1
  13536.  @hamiltonianpathondodecahed5236  Now, to find 2020 … :-D
    1
  13537. I suspect (and obviously hope) your channel will grow much quicker now due to the way Youtube recommendations prefers larger channels and view counts. I reckon you'll hit a million.
    1
  13538. ❤❤👌👌👌🎖🎖🎖🎖🎖
    1
  13539. 4 9 13 17
    1
  13540. For me Nathan’s explanation, explained by you, is the best.
    1
  13541. Great video as always. And I noticed Euler is one of your Patreons! 😊
    1
  13542. oo wow!!
    1
  13543. mathematica tech tips :)
    1
  13544. Amazing
    1
  13545. Have to say, the gamma proof was great. I love how you make challenging mathematical concepts approachable.
    1
  13546. 1) What's the 2 pieces of music for the "Algebra Autopilots" in Chapter 3? 2) MARKUS PERSSON IS ONE OF YOUR PAYPALS!?!?!?
    1
  13547. Ignore question number 1.
    1
  13548. My favorite Hamonic Marvel has to be the Euler-Mascheroni constant. I remember "discovering" it myself in high school when first learning about the integral of 1/x. I did exactly the proof that it's between 0.5 and 1 you mentioned in this video.
    1
  13549. @7:20 F.J.M. stands for Fredericus Johannes Maria. Just his name :)
    1
  13550. So you are saying mathematics is not the key to the universe.
    1
  13551. Another wonderful video.
    1
  13552. The lesson being, of course, be vewy vewy careful with infinities. And wabbits. I remember when I discovered Gelfond's constant. It has something to do with my app, probabilities and and getting imaginary numbers for answers. I can't even remember what I figured out. I think it was something to do with being able to get that constant but only when doing so would get you a negative 1 if a was raised to the power of bi. Man, I'm a German major, why do I find this interesting? 'Cause it is.
    1
  13553. Great loss to the world of mathematics as Ramanujan passed away too early.
    1
  13554. For the card trick, when it is decided what is the suit, we can use an other method to tell what is the lable on that card by flipping some specific cards out of 4 cards that we give to assistant in such a way that when we consider a flipped card as 1 and a non-flipped as 0, we will have a binary no. of 4 digits. Which can easily specify what is the lable on the card that we retained.(by assigning A to 0001,2 to 0010,...upto K to 1101)
    1
  13555. I was watching and I was like, "wait, can you do that?" ?
    1
  13556. 36:34: +C…😂🤣
    1
  13557. Thought I had to do a task at the end of this video.
    1
  13558. The "Convergence of no 9 series" was incredible
    1
  13559. I would really like to see what is going on when one of the angle is 180 degree.
    1
  13560. Interesting stuff, this video and the one about Moessner miracle are some of the most insightful videos I've encountered in YouTube
    1
  13561. Mathologer, your videos prove something you never mention in your videos, that mathematics is the language of physical reality and proves divine design. Do you agree? Or don't you, please explain.
    1
  13562. Help -- does the fact that the 'arctic circle' emerges result from the method used to generate a random tiling, or is it truly the case that random tilings of a large aztec diamond predominately feature the circle, regardless of method of tiling? If you used a different method -- like start in one corner and expand outward -- would a random tiling still show up?
    1
  13563. I'm surprised the genius Blaise Pascal didn't discover this centuries ago.
    1
  13564. What is the music you use on the channel? Is it a custom piece?
    1
  13565. Where can I buy that shirt?
    1
  13566.  @Mathologer  Thanks!!!
    1
  13567. I miss the explaination, why the showed construction method leads to all possible tilings. Without this a prove of the formula of number of tilings is incomplete/useless.
    1
  13568. hats off. You hit it out of the ball park.
    1
  13569. the animation at 22:20 changes the size of the triangles as they move. The small triangles all need to come from the smaller square, and the large all from the large square, but the animation moves them and resizes ! Whoops !
    1
  13570. This video was amazing! As an uneducated math enthusiast it was very enjoyable. Thank you for all you do! I would never have the opportunity to have these sorts of lectures without people like you.
    1
  13571. Made to the end. Understood what was explained as it was explained. I will need to rewatch a few times in order to wrap my head around everything that was presented in this video. (Some Uni, non graduate)
    1
  13572. What are we listening to at the end of the video?
    1
  13573. cool shirt!
    1
  13574. Absolutely outstanding! Great presentation as always.
    1
  13575. I, like the previous commenter GreenMeansGo, have a bachelor's in math and followed everything pretty clearly aside from the sleight of hand around minute 41. Great video!
    1
  13576. I think this also nicely illustrates the relationship between 1/x and ln(x).
    1
  13577. Who else just had to check what the square root of 666 was?
    1
  13578. Is there a typo at 28:30 where it goes 1 3 7 9 ?
    1
  13579. I feel special having voted for the title
    1
  13580. I wish people I hear babbling continuously endlessly about God knows what would talk about stuff like this
    1
  13581. I bought your book about juggling, it's fantastic
    1
  13582. The exact scenario mentioned in the Talmud is as follows (tractate Ketubot page 93a): "In the case of one who was married to three women and died and the marriage contract of this wife was for one hundred dinars and the marriage contract of this second wife was for two hundred dinars, and the marriage contract of this third wife was for three hundred" Sefaria adds in the line "and all three contracts were issued on the same date so that none of the wives has precedence over any of the others." This is not in the original; I'm guessing the case with precedence was discussed somewhere else in the Talmud. The Talmud goes on to say: "Similarly, three individuals who deposited money into a purse [i.e., invested different amounts of money into a joint business venture]: If they incurred a loss or earned a profit, and now choose to dissolve the partnership, they divide the assets in this manner [i.e., based upon the amount that each of them initially invested in the partnership]." This is actually all a quote from the Mishna which was reprinted in the Talmud, rather than being originally from the Talmud itself. In fact, most of the Talmud is essentially just a commentary on the Mishna. After the Talmud prints this quote, it records various discussions about the potential logic behind it.
    1
  13583. So much cool stuff, but I think my favorite was the infinite span with stacked blocks.
    1
  13584. I'm watching every SECOND to find the fixes!! Good decision.🎉
    1
  13585. 3rd grade
    1
  13586. i would go with the gamma approximation. having a self-correcting algorithm that is so "simple" (comparatively) is amazing.
    1
  13587. Love from Bangladesh. Good video
    1
  13588. very reminiscent of a 4d cube/tesseract rotating animation, intriguing indeed
    1
  13589. Your videos are amazing! This is the first one I've watched in a year or so, and I'm just as amazed by this one as by your earlier ones. I learned two important things from this video. First, I learned a very intuitive visual proof that the alternating harmonic series converges to ln(2). Second, I learned another very intuitive proof of Riemann's rearrangement theorem, which I never even knew how to prove before! As always, excellent job!
    1
  13590. Great work ❤️🙏
    1
  13591. The third one, when you can rearrange all the small triangles. Six times three identical triangles give the three new triangles. Beautiful picture though. 🙂
    1
  13592. My vote: the greedy algorithm for finding any sum within the harmonic series. Just a small jump from there to see how any number has an infinite number of infinite partial sums yielding it hiding in there!!
    1
  13593. In Poland we have heron formula somewhere in the middle school but it's more of a curiosity than real tool
    1
  13594. Yeh, The giggle, but also Content.. I won a free lunch in 1965 by helping solve Archimedes Cattle problem, but this and stuff on primes of form 4*k+1, Also computed Euler's constant gamma, on 1620 2 years before for Dr. Ralph Gordon Stanton. But the giggle...
    1
  13595. No, cut the corner.. any corner by removing three tiles plus one more tile which isn't the mentioned corner.
    1
  13596. The brief discussion of higher-dimensional lattices makes me wonder if anyone has considered, for any purpose (not just jumping pegs), an infinite number of dimensions
    1
  13597. In topology I’ve seen infinite dimensional spaces be used to describe spaces of functions, though this isn’t something I know much about
    1
  13598. What did Tesla, himself, say about those numbers and why they were some key to understanding the universe?
    1
  13599. Please give a pause so I don't have to run up and press pause to spot the pattern. Some of us are in the middle of winter sitting wrapped in blankets, so it can take a few seconds to get up. But I got it before continuing in a minute or so. (difference between the two columns is one less than two up.
    1
  13600. I knew the algebraic proof in high school. I never knew about the geometric proof. I love math.
    1
  13601. Frieze Patterns https://youtu.be/0mXz-NP-raY
    1
  13602. 3^3+4^3+5^3=6^3. I hate how much this is a coincidence. I tried brute forcing things on desmos by using sum functions to write out sequential integer power polynomials defined by left and right side counts of terms and fighting to find a root that was an integer- but alas, to no avail. Just wish there was another 3, 4, 5, 6 out there.
    1
  13603. Just more stuff out there to find then? I'm sure we'll get more Mathologer videos then! 😆
    1
  13604. I am (-) less smarter than a monkey 😭
    1
  13605. Wow. Your best video until now! Definitively. I will need at least a couple of days to decide what part I liked most. But I think like the most the proof at the end - yes, the part about the no 9s sum. It is amazing how much structure one can find in this infinity sum. I think the Riemann zeta function has also a vast wealth of structures that we have no hints about.
    1
  13606. Another cool property of Fibonacci numbers: Take any 3 consecutive Fibonacci numbers: 55,89,144. The difference of squares of the larger 2, divided by the smallest, is the next Fibonacci. (144^2 - 89^2)/55 = 233. 233 is the next Fibonacci after 144. 14930352, 24157817, 39088169. (39088169^2 - 24157817^2)/14930352 = 63245986. 63245986 is the next Fibonacci after 39088169. Of course, excepting 0 as the first Fibonacci, as dividing by it is undefined, use 1 as the first. Thus, in interesting ways the Fibonacci numbers are intertwined with the squares. Wonder what other connections are yet to be discovered.
    1
  13607.  @Mathologer  So in fact the tower was computed by computer?!
    1
  13608. Me encantaría ver la torsión de esas paletas. Gran video.
    1
  13609. Is it possable to generate the PI'th Tribonacci, Fibonacci and Luca numbers and if so how? it would be a good video if these series can be generalised that much
    1
  13610. 13:16 - =U wait... 42-13=29... hmmm...
    1
  13611. Easily one of the top three math shirts of the channel right here.
    1
  13612. as a fan of mathematics, I am so proud my mother gave me that name! :)
    1
  13613. 2:28 People who have somebody's hair? Well that's next level thievery
    1
  13614. I guessed it would be the most common color in the top row, so I'm exercising my accidentally guessing the right answer bragging rights.
    1
  13615. Imagine a triangle of 270 degrees. That would be a bit... Off
    1
  13616. I love Ian's Shoelace Site. At one point I checked it whenever I got new shoes to find a new way to lace my shoes, nowadays I have my fave picked so I don't really need to anymore, but it's still nice
    1
  13617. I once advised Dr. Amy Mainzer of how to cross-reference Euler's Identity with conic sections to learn about stable particles, although I did not specifically quote or understand Euler's Identity at the time back in 2009 (or about). Bubbles. I always use an electron, an incandescent light bulb, or similar shape to understand this connection: newly formed dark matter has a bubble at its center bulge. Circles and the conchoid of Nicomedes correspond. Compressed atomic nuclei (or neutron stars) - more bubbles, at their contact points. Elliptical part. Delanges trisectrix. The inflection point is parabolic and corresponds to the birth of a black hole and the regular trifolium/the rose with three petals. The ring/cylinder singularity is hyperbolic and corresponds, in this viewing of gravitational increases over time until a Big Crunch - with respect to conic sections, to many, many black holes and the Maclaurin trisectrix. It is no coincidence that advanced mathematical proofs often talk about elliptical and hyperbolic curves. The movie "Dark Matter" had a similar concept. The part where the student is boiling water and exclaims, "bubbles".
    1
  13618. Brilliant!! I am hooked on your math.
    1
  13619. Wonderful video as always, Merry Xmas !
    1
  13620. The visual proof for removing the 9s was my favourite part. It highlighted the difference of a finite series from an infinite series best for me.
    1
  13621. In America never taught however I knew it and it has made a few school math problems easier back in the day
    1
  13622. Why are you including this woke "Lebnitz and Newton and ... quote INVENTED quote calculus". Yes, they discover it because they didn't know anything about Madhava.
    1
  13623. OMG, I was just reading about writing numbers as a sum of two squares when I open YT and I see that :DD
    1
  13624. If you will notice, at this point, you have 3.3 million views, 68K likes and 900K subscribers
    1
  13625. In Galice, anh is aŋ, so sinh is "sing"
    1
  13626. I knew only the second proof, because i was born long after arithmetic equations were invented, but i am still working to understand the proof documented in EUCLID's Elements who only used logic, a ruler and a circle to draw in sand; Partly hard because the old greek language used was obviously much subtler structured than contemporary languages, whose translations give us the impression that the sentenses handed down were somehow corrupted during its way into our timeframe.
    1
  13627. So, each triangle has the ability to oscillate between two different similar triangles, then, if you continue this process indefinitely. Interesting. I suppose this also results in a kind of fractal pattern if you don't remove the previous triangles.
    1
  13628. I sat and worked this on paper and now it is clear
    1
  13629. Gamma must be greater than 0.5 because if we look at each blue piece individually, it's like a right triangle with the hypotenuse pushed out. If we consider the rectangle with sides matching the non-hypotenuse sides of the triangle, the blue piece is more than half of the rectangle. This is true for all rectangles. And the rectangles areas sum to 1. So the blue pieces areas sum to more than half of 1.
    1
  13630. Amazing as always. Keep your good work!
    1
  13631. This is very cool stuff! I'm surprised, however, that you got all the way through this without once mentioning Charles Babbage's Difference Engine.
    1
  13632. Thank you for the beautiful Christmas gift. This video should be under my Christmas tree! Congratulations and continue with your work!
    1
  13633. Made it 🎉
    1
  13634. I saw nothing like that before
    1
  13635. 好！
    1
  13636. It's look like ----> (Friends : friend = s) Kenape jd nyambung ke Clark Kent sih??? Hadeuuuuhhh...
    1
  13637. 11:00 isn't it cheating saying that sometimes x=y or x=z?
    1
  13638. a
    1
  13639. Thank you Burkard!
    1
  13640. I can feel a Galois theory video coming...
    1
  13641. Semi-private question out of curiosity: where did you study? Nice video, by the way.
    1
  13642. Imagine someone watching this 20-30 years later who've never seen coins
    1
  13643. I learned a lot of this from the book, "The Number of Things" by Evans Valens. It is the most salient book on the nature of existence I have ever read.
    1
  13644. Totally awesome. It is no surprise that when the Golden Ratio is involved that the Fibonacci numbers and Continued Fractions are lurking in the background. The fractional part of divergence angle, Δ, can be thought of as Δ mod 1. So each leaf is placed on a circle of circumference 1 (radius 1/(2*pi)) after a rotation of Δ mod 1. From Ergodic theory, the rotations will be periodic iff Δ mod 1 is rational. In the aperiodic case, when Δ mod 1 is irrational, the leaves will become dense on the circle as number of leaves go to ininity. It certainly makes sense that nature (biologic) would want to use "irrational" Δs, but what about Interstellar Space? Can the arms of spiral nebulae, be modeled with the helicone and if so, why?
    1
  13645. Did not know these 2 proofs
    1
  13646. For some reason I'm craving a cheese log.
    1
  13647. You got my attention back! Great video!
    1
  13648. I still think we dont exactly know what happens in the infinity. I mean in between the dots here for example: 1/2+1/4+1/16+...
    1
  13649. Me watching a mathologer video for the 5th time trying to understand it: mmmm yes makes sense
    1
  13650. Thanks!
    1
  13651. thought this was a polybridge thing. not dissapointed when i saw mr.math
    1
  13652. I like Sophie Germain more than Twin primes. They are actually very similar conceptually to twin primes as sort of a natural pairing. Instead of +2 it's kinda x2. Of course, you cannot just multiply by a factor, you have to add a bit to get to a prime (same way with twins, they're not just the next integer, just n+1, you have to make another +1 to get to a possible prime). SG primes are most beautiful in binary. Just shift everything left and add another one in the end (and primes in binary always end in 1 anyway). Then there are SG primes of second kind, 2p-1 instead of 2p+1... Same thing in binary, but while adding another one in the end, change the one that was there to zero.
    1
  13653. Amazing!
    1
  13654. For the magazine puzzle, I quickly noticed that it could be expressed as x^2 = y(y+1)/2, or equivalently, that any triangular square number could provide a solution. e.g. 6^2 = 1+2+...+8 -> 1+2+3+4+5 = 7+8
    1
  13655. Are you going to tell us later about all the continued fractions that were hiding in this tree?
    1
  13656. this channel is a gem
    1
  13657. Thanks for the video. You've given me lots of ideas for coding programs! Hours and hours of fun!
    1
  13658. 14:00 I couldn't find the pattern in the number of possible tilings. The results I found for 2x1, 2x2, 2x3 and 2x4 were i, -2, -3i and 2, respectively. I assume I made some mistake I can't spot on the last, and the pattern should be 1, 2, 3, 4...
    1
  13659. At 16:32 the first line says (x+1) to the power of zero equals to x, obviously incorrect, it has to be (x+1) to the power of zero equals 1.Then the typo continues at x-1 to the power of zero...
    1
  13660. You make me a better person.
    1
  13661. The visual "proof" of the sum of no 9s... Magnificent :-)
    1
  13662. A part of the schwarz-lanturn as origami-model can be crafted into a simple frameless paper-kite fold the paper into 8-10 bands for the lantern-structure with valley folds and add the diagonal mountain folda in a way, that they start at the corners and center of the top Edge and divide the valleyfolds with their own cross-points alternating in two halves and ome half between two quaters. This makes 1/4 of a 8-8 lantern or 1/5 of a 10-10 lantern Attach a kite-string to the seccond cross-point along the middle axis from one curved edge and some paper-strips as a tail in the middle of the other curved edge.
    1
  13663. 6:50 Two other nice patterns in the left-hand terms (ignoring the powers): the top one is the square numbers, and the second one is alternate triangular numbers :)
    1
  13664. Woah! That was faster than I expected!
    1
  13665. Did you also know that Tesla wanted to marry a pigeon? People also forget that he finished in an institution as he was getting mad with age.
    1
  13666. Your shirt is brilliant
    1
  13667. thank you for another wonderful explication of a proof that even my block-headed self can at least taste -- and BTW: great T-Shirt.
    1
  13668. Amazing video! Electrical engineer here, the explanations are intuitive and the visuals (when possible) also great. I do think that sometimes the train of thought is broken by fast conclusions, so I find myself going back and pausing to figure out exactly what happened.
    1
  13669. I want some fibonacchoes
    1
  13670. Hey, I got a nice idea. We have learnt from this video that in order to measure the length of a curve, we have to put several points on the curve and then to join every consecutive points by segments. But, I think if we draw tangent segments from every point, the measurement will be more accurate. And it is the way by which Archimedes measured π
    1
  13671. after watching this i'm getting recommended those misleading vortex math videos :-(
    1
  13672. In an earlier video you mentioned the fact that the integral of n(n+1)/2 from -1 to 0 is equal to -1/12 = zeta(-1). I'm sure many of us would love to know more about that. Please do a video explaining more about it. Whiy does it turn out that way, and what does it tell us about the very idea of divergent sums ?
    1
  13673. So ... it's a war where you kill your own men to advance into vacant territory ...
    1
  13674. On the basis that a circle encloses the greatest area within a minimum length circumference, the circle will be the largest area because any distortion can only reduce the enclosed area while increasing the length of the circumference.
    1
  13675. But which is the easiest to pull tight (lowest friction)? Most comfortable (equal tension after pulling tight)?
    1
  13676. To answer your question @5:03, I made it this far in the comments ➡🔴
    1
  13677. Very good. And with some sound level compression could be even better!
    1
  13678. The identity $(u_nu_{n+3})^2 + (2u_{n+1}u_{n+2})^2 = (u_{2n+3})^2$ is an exercise in Burton's number theory book. I thought it was neat and moved on!!!
    1
  13679. The only A-ha moment I got was when I realised why he was wearing that tee shirt.
    1
  13680. All a+b+c have same length, and all have same distance to center of the circle within the triangle. So rotating them around this center gives the same outer points on the circle(, its actually three times the same full line, just with different angles).
    1
  13681. Ok i forgot the center of the line part😢
    1
  13682. Much learned mind blown thank you
    1
  13683. The most memorable thing for me is how γ showed up unexpectedly in the infinite sum
    1
  13684. That's a wax burner.
    1
  13685. Gracias
    1
  13686. This is actually amazing. Any time you can visualize math and then use imagination to do calculations I think it expands our understanding and ability to calculate in our heads exponentially.
    1
  13687. 🎉
    1
  13688. Second proof is better
    1
  13689. Wow it is very interesting)
    1
  13690. wow!
    1
  13691. These paradox are called "paradox" and doesn't happen in real life because of Planks constant.. Phew!
    1
  13692. OMG I get goosebumps when you talk German. You and Sabine Hossenfelder. Best. Math. Channel. Ever.
    1
  13693. Did he say that no regular 3D polygon fits inside of a 3D square grid? A tetrahedron does. It is more likely that I did not understand him (very likely), but I just wanted to point that out.
    1
  13694. I made it to the very end. Thanks Mathologer, awesome material as always!
    1
  13695. Would you consider using a dark background?
    1
  13696. yay some fun from Matholiger thank you
    1
  13697. Nathan, distracted by glances from his two girlfriends, starts by assigning 0 to face down cards and 1 to face up cards. He soon realizes that the binary number is increasing, not decreasing. He has second thoughts about his, let's face it, knee jerk decision that the cards had to be replaced with numbers. But he can't change it. This isn't just a math problem; it's a math problem in a move. Disaster?
    1
  13698. I made it to the very end! It was so interesting
    1
  13699. 35:30 this is not formal by any means, but intuitively I would say it is because the bugs are travelling perpendicularily at every instant. I will try to formalize it using calculus and vectors: let b1=(x1,y1) be the first bug, while b2 be the second bug. Let R be a clockwise rotation of 45 deg about the center of the square. WLOG we say that point is the origin, meaning that R can be expressed as a matrix. It is clear that b2=RR*b1=(y1,-x1)=(x2,y2) let d(t) de the difference between b2 and b1: d(t):=b2(t)-b1(t)= (RR-I)b1(t) Computing the matrix of RR-I=[[-1 1],[-1 -1]] we can see that this is a cw rotation by 135 degs, and an enlargment by sqrt(2) => RR-I=sqrt(2)*RRR let D(t) be the size of d(t): D(t):=|d(t)|=|sqrt(2) RRR b'|=sqrt(2)* |b1| because rotations don't affect length size. We also have D(t)=sqrt(d.d). We will prove D(t)'=1: D(t)'=sqrt(d.d)'=1/2*1/sqrt(d.d)* (d.d)'=(d.d')/2*D(t) let dT be the transpose of d (d.d)'=(dT * d)'= dT'*d+dT*d'=(d'.d)+(d.d')=2(d.d')=2 * Transpose(sqrt(2)*RRRb1') * (sqrt(2)*RRRb1) = 4 * Transpose(RRRb1') * (RRRb1) = 4 *transpose(b1') * R^(-3)*R^3*b1= 4 (b1.b1') Since b1 moves in the direction of d, then the angle(b1,b1')=angle(b1,d)=135 degrees because d=sqrt(2)*RRR*b1. Since the speed is 1 we have |b1'|=1 Solving for (d.d)' we get (d.d)'=2*d.d'=4 (b1.b1')=4 |b1| cos(135) Pugging back in we have D'(t) = 4*|b1| cos(135)/2*D(t) =4*|b1| cos(135)/2/sqrt(2)/ |b1| = 1 D(0) = sqrt(2)*|b1| => T = D(0)/v=sqrt(2)*|b1| which is equal to the time it takes to reach on point of the square. This was much harder than I expected. Probably you can get better proofs using drawings and geometry. Not only is my proof long but it requires a lor of familiarity with linear algebra and some non trivial properties
    1
  13700. Nothing like a little Cauchy Condensation in the morning :)
    1
  13701. I love it! It's beautiful ❤️!
    1
  13702. I would like to send multiple likes if possible.
    1
  13703. 34:34 What I see: Vcube 6
    1
  13704. :-)
    1
  13705. 03:00 ... I first came across this proof in the 5th grade (age 10) in my elementary school math class (yes, in the USA when we actually taught the fundamentals)
    1
  13706. In my opinion the simplest way of looking at bugs at the corners of a square--or for that matter, a pentagon, hexagon, etc.--is to draw, at some instant of time, a circle circumscribing the square. If the bugs were chasing each other along the circumference of the circle instead of directly toward each other, the circle's radius would remain constant. Instead, there is a component of their motion in the tangential direction and a component in the radial direction. Since the figures are similar at any point in time, the angle of the velocity vector between the tangential and radial directions is always the same. Hence, the radial and tangential velocities are both constant, so the radius shrinks to zero at a constant rate. The constant tangential velocity means the angular velocity becomes larger and larger near the center, reaching infinity at zero radius. The centripetal acceleration does too--impossible for bugs of non-zero mass. However, the path is simply a spiral. The time it takes to reach the center of the square is simply the half-diagonal of the square divided by the radial velocity, and the total path length is that time multiplied by the total speed. Since, for a square, the tangential and radial components are equal, it's pretty easy to see why the path length should be as the Mathologer claims. I'll have to think a bit more about an equation for the spiral. It won't be a constant pitch.
    1
  13707. I made it to the very end.
    1
  13708. Mosner/Matholedger miracle can be applied to PRIMES
    1
  13709. Did anyone else think of a hyperpyramid at 12:05?
    1
  13710.  @Mathologer  I posted this comment before, but when I tried to edit it to thank you for the heart the heart disappeared. So I reposted it. Thanks for the heart and the reply.
    1
  13711. Matter of definition... a 1x2 board can be covered with a single 1x2 tile in 2 ways. Is tile orientation to be included in the sum?
    1
  13712. You should prove (using you visual of the hexagon) that r and s do fit the equation rs=r+s. R and S are supper special in that they are defined by the hexagon. A bit off your topic of the video, but to see the relation for axb=a+b, run the following python script. # ab = a + b calculation for integer values for a in range(2,20): b = 1/(1-(1/a)) print(f"For a = {a} the value of b is {b:.5f}") x = a*b y = a+b print(f"a*b=a+b ==> {x:.5f} = {y:.5f}")
    1
  13713. He proves it using Ptolemy's Theorem at about 7:33.
    1
  13714. I loved the near misses (terrible aim).
    1
  13715. I get that his shirt is supposed to be a math joke "Ho Ho Ho" but I would assumn but if "Ho Ho Ho" is 3, then "Ho" is just 1 and 1³ is just 1 so (HO)³ would just be "Ho". Right? I mean (HO) * 3 would make more sence.
    1
  13716. 7:30 ... wait shouldn't they be called "sames" then ?
    1
  13717. Interesting takeaway from this gives Babbage’s motivation for constructing his Difference Engine, arguably the first general-purpose computer.
    1
  13718. Visuals are wonderful.
    1
  13719. Anyone training as a pilot will be familiar with these in the flight computers we use to do our theoretical exams. We need them to calculate all sorts such as speeds, fuel calculations, decent rates etc. Fascinating to actually see the mathematical principles behind it!
    1
  13720. Why couldn't an inverted pyramid be used in the overhang problem?
    1
  13721. Amazing video
    1
  13722. 5:58 how is G+P=A? The green + purple parts are clearly bigger than the single length of A.
    1
  13723. It’s really interesting that the 0 #’s series (and even more, the at most 1 or 2 #’s series all have a limit of around 20-ish
    1
  13724. Yes, I could make those exact moves. To move a tower of height 1 to point B, the first move is the smallest disc to B. For a tower of size 2 the first move is the smallest to C. This alternates as the tower gets bigger; odd-height towers have the first move to where you want the tower to finish & even-height towers have the first move to the other empty spot. After determining the first move, they all fall naturally because you never move the same disc twice in a row (that would be inefficient), and you never move the biggest available disc (it has no valid moves).
    1
  13725. Andrew Wiles proved Fermat's Last Theorem using "a special case of the modularity theorem for elliptical curves". That was step one! He got us all to the "top of the hill". What an amazing view there is to be had! I cannot get into that math, dealing with stability, very easily, as I have to mind how it relates to sound silencing technology - just the same as I did back in the '90s when I talked about this on the POP3 server at UW-Parkside. But, I CAN get into how this relates to Ramanujan's Infinite Sum and some of its 12 parts. Step two is using hyperbolic curves, along with dual, split-complex, and complex numbers as tensors (and triangular positioning and angle trisection methods), to help prove the Riemann hypothesis and its close connection to Big Bounce events. I may need to have some Target Motion Analysis mathematics declassified. I cannot say for sure on this, but would be a fool to not anticipate a potential clash or partnership with the Navy over this "Ancient Bubblehead Knowledge".
    1
  13726. Like is for the shirt.
    1
  13727. Burkard, any video with Euler in it is awesome, thanks! And I just read that you are planning to do one on Galois theory? That would be amazing!!! Please, PLEASE, teach us how Abel proved the generic quintic equation has no solution in radicals, and how Galois found when such an equation can be solved.
    1
  13728. Neat!
    1
  13729. 11:51 firstly done it with computer help, next by hand(help only for simplifications/expansions and less ink), solution 9,4,13,17 in clockwise order and x(=9) square and y(=4) square, z(=xy=36) popped up as a byproduct of the formulas you stated, as a random observation, now i am not so impressed since you can pick any 2 numbers as x,y and you get a "triangle" with said properties no matter the domain.
    1
  13730. This episode reminded me of a years older video. 👌🏻
    1
  13731. The optimal stacking pattern of the books was the most impressive. I had seen most of the other stuff before, but that surprised me.
    1
  13732. Is it possible to simplify that as the length of the triangles' sides approaches zero, at each step, all sides should be reduced by approximately the same factor, or is there a counterexample for that too? Or is that simply not possible to do in complex shapes, where the triangles are by necessity very differently sized?
    1
  13733. Not sure if I got it right, but if the input sequence is the square numbers, then this could be encoded by the sequence (a,b,c,d,...)=(1,2,2,2,...). That works, because 1^2=1, 2^2=4=2*1+1*2, 3^2=9=3*1+2*2+1*2, 4^2=16=4*1+3*1+2*2+1*1. In general: n+2*(1+2+3+...+n-1)=n+(n-1)*n=n^2. So feeding (1,2,2,2,...) into the Moessner machine yields: n!*(n-1)!=n*[(n-1)!]^2.
    1
  13734. How did I unexpectedly "Bump" into this very Video? Well, a very odd cause! A few months ago, I was treading this hall of Clifton Communicare Center on Probasco Street in Cincinnati, Ohio! This is where I have lived in the last 4 years. This very hall has my Room 319B on the side facing the rear view through the windows, giving on some evenings a beautiful scene of the orange setting Sun over the Ohio River and Downtown Cincinnati way below the hill. Well, on a certain day, I was about to turn to the drinking fountain at the turn of the hall to the front entrance/exit out of the building. This is where the cold-water fountain is. It probably was in August on a very humid day. But then my eyes were looking at my tennis shoes, when I suddenly noticed the floor area near me. This suddenly prompted me to close in on this particular floor spot: it was not only near to that cold water to fill in my drinking cup, but also quite near to the nurse's stand, from whence to get a straw to sip the water in my cup. Well, at that particular moment, with no one near around me to notice what I was seeing, I found this specific aged paper wrapper on that very Spot that once came from a now long-ago discarded straw inside it! What was strange was that it was twisted out of regular shape somewhat for being on that spot and trampled on by passersby for many hours. When my eyes focused on it at close range, being no longer a straight line, to originally enclose a brandnew straw, the very shape perfectly resembled THE SQUARE ROOT OF 3 lying on the floor! UNUSUAL! I wish I had a cell phone to save the very image of this to show others or save in my album. That was months ago. I suddenly recalled seeing that square root of 3, while viewing U tube videos, hence, to look up anything unusual about that number. And that's how I bumped into your video!
    1
  13735. Inductive proof sketch: Suppose Fn+1 = 1 + F_0 + ... + F_n-1 Then prove that F_n+2 = 1 + F_0 + ... + F_n-1 + F_n F_n+1 + F_n = 1 + F_0 + ... + F_n-1 + F_n F_n+1 = 1 + F_0 + ... + F_n-1 which we know from the induction hypothesis and of course 2 = 1 + 1. So it works for any n>0
    1
  13736. I liked Both Versions of the Proof, the coloured one is a Little Bit more intuitive. Thanks again for a Great Video ❤
    1
  13737. Minute 21:48.. please explain why.. my brain Is fooled
    1
  13738. There aren't many videos so when I get a notification I get super excited
    1
  13739. Looking at how the triangles... it looks like there's also a 3D cube in it... technically 2 of them.
    1
  13740. Followed it all and want more. BS in math and PhD in computer science (many, many years ago). I really like that you're not dumbing it down.
    1
  13741. Thanks for showing the optimal 20-block stack. I was indeed surprised by that. Will have to verify it experimentally... and then find the pattern for N blocks?
    1
  13742. easy challenge: cut out the two black squares in the top left, disconnecting the green corner. Arbitrarily pick two unrelated green squares to remove as well. The lone top left corner now makes it impossible to tile.
    1
  13743. Most interesting part: -1/12 is to sum(n) as gamma is to sum(1/n) It suggests there's a whole family of integral-sum differences that converge at infinity, which really helped my understanding of the Riemann-zeta function
    1
  13744. Question: i have subed to the Channel and all notofications on, but the first notification or community post from your Channel since 2 years was the Post because of this reupload and it was shown today on 25.01….very wierd. Any one an idea why ?
    1
  13745. It was not clear to me… both sequences converge to PI or not?
    1
  13746. I’m kind of confused. Given an odd prime p, and that we know p = a² + b². As you noted, one must be odd, the other even, assume (as you did) a is the odd one. Then let a = 2x+1, and b = 2y. So p = (2x+1)² + (2y)² = (2x)²+2(2x)+1² + (2y)² = 4x² + 4x + 4y² + 1 = 4(x²+x+y²) + 1, thus p mod 4 must be 1. In fact, this should hold true even if p is not prime, as long as p is odd and can be written as the sum of two squares… which if we allow for the case of y=0 would imply that all odd perfect squares mod 4 is 1. Which, given f(n) = n², then for n=1 we get f(1) = 1, which mod 4 is 1. Given a base case f(n) that is odd, then the next odd perfect square is at n+2, and f(n+2) can be shown to be larger by f(n) by the sum of two sequential odd numbers, let those be 2x-1 and 2x+1, which when added together give: 2(2x)+1-1 = 4x, which means that for all for all natural numbers, f(n) mod 4 = f(n+2) mod 4. (This later identity holds for all natural numbers n, not just odd ones.) And since f(1) mod 4 = 1, then all f(2n+1) mod 4 must also be 1. So, yes, all odd perfect squares mod 4 are 1.
    1
  13747. I guess this is proving a much more general, although related: “all odd numbers that can be written as the sum of two integer squares can also be written 4k+1” which doesn’t actually prove that every odd prime p mod 4 = 1 can be written as the sum of two squares.
    1
  13748. Just finding the center of mass seems like a practical application. Like where to put a well for a village to minimize walking for each family (nature permitting). But that doesn’t take hills into consideration. Hmm.
    1
  13749. thank you. i actually wondered the same question many years ago and ended up using similar techniques to figure out how many m-dim 'objects' in a n-dim hypercube. and then i proceeded onto simplexes as well as cross-polytopes. re-inventing the wheels, i know. but the feeling of figuring out all of those things by myself is still one of my most precious moments.
    1
  13750. Probably a naive question, but at what point does the sum of the straight sides on the "pi = 4" circle start diverging?
    1
  13751. Background in math: basic undergrad calculus with some linear algebra. Otherwise I'm just a neurosurgeon. The animations are very useful. I didn't follow but took for granted the change in limit of integration.
    1
  13752. When gamma appeared, I completely lost my mind. Great video!
    1
  13753. Merry christmas, mathologer!
    1
  13754. :)
    1
  13755. can you please explain the star trek games like tongo..I hear its pretty complicated..a mix of cards and roulette
    1
  13756. the no integers was interesting. I knew that beforehand, but having it explained and shown put it in a new light
    1
  13757. Pretty neat!
    1
  13758. The number of possible collections of 10 numbers you could have chosen out of the 90 you were given was (takes a deep breath) 648,721,197,647,800,200
    1
  13759. I enjoyed the explanation in Chapter 2. I’ve done that experiment with my kids, and they loved how the tower seems to defy gravity.
    1
  13760. List of Patreon=1+1/2+1/3+....
    1
  13761. Operations Inverse: 4/2 = 4-2 (quotient/difference)?
    1
  13762. 6/3/2 = 6-3-2
    1
  13763. Top of the head kind of stuff. But probably a lot of good stuff out there. It’s worth the effort to find symmetry in the paired operations.
    1
  13764. I'm sorry to say that there is no new insight from this: 2×2 = 2+2 = 4 → 4÷2 = 4−2 = 2 1×2×3 = 1+2+3 = 6 → 6÷3÷2 = 6−3−2 = 1 1×1×2×4 = 1+1+2+4 = 8 → 8÷4÷2÷1 = 8−4−2−1 = 1 1×1×1×2×5 = 1+1+1+2+5 = 10 → 10÷5÷2÷1÷1 = 10−5−2−1−1 = 1
    1
  13765.  @Tumbolisu  good to know. But it gets challenging with powers and roots and factorials. Would be nice to find more than two operations connected.
    1
  13766. I was also looking for a path that only uses the number 1 ONCE, as would be the rule for any and all numbers that make an appearance in the sequence.
    1
  13767. Anyway, I should probably wake up first before doing any math. But input appreciated. Thanks.
    1
  13768. ​ @jay_13875  Who knows the citation ist a man of culture
    1
  13769. interesting. i would more like to know how they actually did these calculations, what expressions, notations and symbols they used.
    1
  13770. In the argument at 24:24 on reducing an exploded lacing, I missed an explanation why there always is a zero segment that intersects the segment larger than one.
    1
  13771. 16:43 I think all the topologists in the comments died inside. 😂 Is that legal?? We’ve compressed the positively curved onion layers into flat circles??
    1
  13772. Oh superb stuff! One surprising thing for me is the fact that, if the no 9s series converges, the rest, the 9s series diverges: 1/9+1/19+1/29+...+1/89+1/90+1/91+1/92+...+1/99+... +1/109+1/119+...
    1
  13773. One of my first dream proofs was solving P=NP through the clique problem. Only later did I figure out the number of nodes in my proof was constant.
    1
  13774. The weirdest thing you showed I think is the unusual optimal brick stacking pattern. Mainly because most of how well you explained the other topics, which took away some of their mystery. While the question of the best way to stack a given number of bricks feels like it might have some surprises still hidden in there.
    1
  13775. I cant imagine how much time you put into this video, all those animations, not to speak about the derivations! Thank you so much for your effort!
    1
  13776. great sir
    1
  13777. Students are on a need to know basis and so they do not need to know the proofs just that the proofs work.
    1
  13778. 6:13 isn't what you composed on the top wheel actually 3.18?
    1
  13779. 2:34 did you just... sing giant steps?
    1
  13780. Magic
    1
  13781. 6:13 that's showing 3.18 instead of 3.14
    1
  13782. 21:38 so it`s 42 then?
    1
  13783. The dots were more interesting to me. However, the animations for the entire video really showed how much geometry is contained in just one triangle.
    1
  13784. Thank you, very cool :) I most liked chapter 5, although honestly this entire video is great!
    1
  13785. Wonderful video! I was talking about those summation formulas with my colleague a couple of weeks ago!
    1
  13786. My favorite part was the 700 year old proof, simply because of how simple yet powerful it is.
    1
  13787. What a great video, it is one of my favorites. Since your last master class, I have sort of explored this path of math by my own, so i wasnt overwhelmed by the rate of info, it was a nice linear progression of complexity. what to do better? dont really know.
    1
  13788. Nice
    1
  13789. e^(iπ) < e^(iτ) e^(iτ) = 1 is the most beautiful thing in mathematics.
    1
  13790. I can relate to professor wildberger. He teaches this type of math. Sorry I tied you up so much. It was silly for me to do.
    1
  13791. Someone please tell the name of the background music played at last section... Thank you...😇
    1
  13792. I just love this channel and the way things are shown, and I also really like the shirts, this one from Space Invaders is really cool, especially because I'm from the oldies and I love this game!!! congratulations for this beautiful educational channel!!!
    1
  13793. Is geometry the basis of math? Hm. What about other metalic ratios? It took me a bit, but I did figure out why you only used the fractional part. I wonder if I should work the gold angle into my app. That is a lot of fun.
    1
  13794. Any time. I don't think I'll add it. It feels a bit specialized. Not that I don't have a lot of speciealized units. But do I need more?
    1
  13795. Wish I had watched this earlier :) That way my whole day would have been better :)
    1
  13796. Loved this!
    1
  13797. The first proof I leaned was the Chinese one. I first saw this proof on the TV series The Ascent of Man by Jacob Bronowski (and in the book too).
    1
  13798. Around the 28 minute mark -- took me a few viewings to see how the formula for calculating Bernoulli numbers emerges from Pascal's triangle. But it's still a fair bit more helpful than, say, the Wikipedia article on them. The animations showing how the summation formulae emerge from blocks/squares were really neat. I enjoyed those. No official math background---took three semesters of calculus in college.
    1
  13799. One day I'm thinking about this infinite series, next day mathologer upload about it
    1
  13800. Tristan's fractal
    1
  13801. This general solution by Moessner and Post is quite amazing. When I could not sleep, I thought about the Cubic Version of the Squares (being solved visually). This involved thinking of unit cubes as the building blocks for a Cube . One needs to add a three-sided wall (each wall being rectangular) to get the next Cube. Each rectangle is a sum of 2 Triangular Numbers (+ one). That is, 6 Triangular Numbers + 1. For example: 3^3 - 2^3 = 6 T(2) + 1 = 19 {where T(2) is 2nd Triangular Number} In general: (n + 1)^3 - n^3 = 6 T(n) + 1 (these 'pebbles' can be arranged into an hexagonal form) Leibniz, when deriving his notation, first used it for discrete mathematics (and later for infinitesimals): dx 1 1 1 1 1 ... x 1 2 3 4 5 6 ... Int 1 3 6 10 15 21 ... (the Triangular Numbers) Int2 1 4 10 20 35 56 ... (sum of Triangular Numbers) AND dx 2 2 2 2 2 2 X 1 3 5 7 9 11 ... Int. 1 4 9 16 25 36 Int2 Sum of Squares
    1
  13802. So, looks like rotating the triangle by 120 or 240 degrees will still have it work.
    1
  13803. I searched and I searched, and found nothing.
    1
  13804. charming
    1
  13805. This number walls are very similar to frieze patterns.
    1
  13806. I'd love to get that book but for some reason it's $45. Education for the wealthy only? LOO
    1
  13807. While it's clear to see how the number of results in the divide+slide method is 2^(1+2+3+...), I somehow missed the proof for the fact that the divide+slide way gives all possible tilings of the Aztec diamond. I guess I'll need to read the actual papers for this.
    1
  13808. Oh, I like me some Dr. Who. :) This is a puzzle I like and am glad to see it here.
    1
  13809. How do I chose a single mind blowing fact in a 46minutes mind blowing video. Probably the 100 0s > no 9s, which is quite unintuitive. Or the crazy towers.
    1
  13810. Very nice example of divergent series!! -1/2 ln(pi/2) = ln1 - ln2 + ln3 - ln4 +... = zeta'(0) - 2ln2 zeta(0).
    1
  13811. Nice!
    1
  13812. dear mathologer, would you be willing to consider covering the german tank problem(*) on your channel? this one has been interesting me the most for a long time, but i have a hard time wrapping my head around it on my own.. (*) which is really a coined term for estimating the maximum of a discrete uniform distribution from sampling without replacement, which wiki explains in the following way: "In simple terms, suppose we have an unknown number of items which are sequentially numbered from 1 to N. We take a random sample of these items and observe their sequence numbers; the problem is to estimate N from these observed numbers."
    1
  13813. Regarding the change of the sign after cycling:We have p cards. The card to be cycled has (p-1) cards to it's left, of which a certain number, call it N, will be bigger and the rest, (p-1)-N, will be smaller. | By cycling it our card has no more cards to it's left, so the number of inversions decreases by N (since it had N cards larger than it before), but the rest of the cards now have our card as a new card to it's left, so all the cards which were smaller, (p-1)-N, now have a card to their left which is bigger, so for each such card the number of inversions increases by one. | This results in a total change of the number of inversions of -N + ((p-1) - N) = (p-1) - 2*N. This means that our sign will be multiplied by (-1)^((p-1)-2*N) = (-1)^(p-1) * (-1)^(-2*N) = (-1)^(p-1). | So for an odd number of cards, we get (-1) to an even number, so the sign stays the same, for an even number of cards we get (-1) to an odd power, so the sign flips.
    1
  13814. What is the math of a rubber band stretching? Formula?
    1
  13815. I think we can name the proof with Archimedes claws garlic proof.
    1
  13816. Gaussian's gravitation constant (with help of Isaac Newton) is also being obsolete now because of big numbers?
    1
  13817. I kept seeing the straightedge and compass method for finding the incircle of a triangle, just the intersection of two angle bisectors in the triangle -- and I kept seeing angle bisectors of the vertical angles.
    1
  13818. If the exponent is 1 or 2, a^x + b^x = c^x has solutions. But not if x = 3, and not if x = 0 either (as 1+1 != 1). In fact, among the integers, a^x+b^x=c^x has the same solution set as x^2-3x+2=0.
    1
  13819. 12:09 - here's one crazy fact mathematicians don't know want you to know!
    1
  13820. I wish I learned these shortcuts at school
    1
  13821. They look more like bananas than moons...
    1
  13822. This video litterally begins with "Welcome to an X-X-X special Mathologer video" 😮 (X being common shorthand for extra)
    1
  13823. What's that "smallest perfect universe"? It looks like four Fano-Planes stuck together plus an extra point in the center?
    1
  13824. This video reminds me what if A+B+C=Pi then ctg(A/2) + ctg(B/2) +ctg(C/2) = ctg(A/2) * ctg(B/2) * ctg(C/2)
    1
  13825.  @Mathologer  This video is fine for me too.
    1
  13826. How about the maximum overhang with N rows that can be built row by row without toppling?
    1
  13827. In addition to great math and visuals, your music is nicely chosen
    1
  13828. no-9 series is a fractal
    1
  13829. This buckling with the periodic diamond pattern reminds me of mach diamonds in rocket thrust
    1
  13830. So beautiful! It's like the curve version of infinite fractions summing to previous fraction. 1/2+1/4+1/8.....1 1/3+1/9+1/27....1/2 etc
    1
  13831. Sir something on magic square
    1
  13832. This is basically how harmonic reducer (harmonic gearing) work
    1
  13833. 2 n d proof. 1983
    1
  13834. Nice t-shirt. I finally understand how a2 + b2 = E/m.
    1
  13835. I finally had a chance to share this mathematical curiosity with my eight year old niece today. She thought it was really cool and gave me a hug. She's the best!
    1
  13836. I used to use a z-type lacing. This was because shoes usually came with that one. I moved over to the criss-cross lacing as it was easier to ensure that both ends were of an equal length. I suppose the bowtie and devil's lacings would be useful if you had to replace the laces with ones that weren't of the ideal length for the criss-cross pattern. I've watched several videos about optimum lacing strategies. This one is by far the most rigorous. How about a video about the optimum knot? :) With particular emphasis on knots that won't spontaneously come undone!
    1
  13837. 9:00 A(X)+A(Y) =A(XY) ∴ ∫1/x with lower bounds 1 upper bounds A, + ∫1/x lower b as 1 upper b as B is = ∫1/x lower bound 1 upper bound AB ∫1/x = ln(x), so following definite integral calculations, we get ln(A)-ln(1)+ln(B)-ln(1) = ln(AB) ln of 1 is 0, so ln(A)+ln(B)=ln(AB), his follows laws of logarithms! First time I could solve a puzzle presented! :D
    1
  13838. I wonder if the animation could have a little higher frame rate to be easier on the eyes. Otherwise great video.
    1
  13839. any number that ends in an infinitely repeating string can be split up into two parts - the non-repeating part at the start, and the repeating part at the end. the non-repeating part will just be some integer divided by a power of 10, so clearly rational. the repeating part can be expressed as a fraction using the same kind of argument in the 0.999... = 1 argument, and then "sliding" it into place by dividing by some power of 10. each of these numbers is rational, and the sum of any two rational numbers is also rational.
    1
  13840. This is obviously witchcraft!
    1
  13841. 21:51 I can't wrap my head around that (pun intended).
    1
  13842. A (translated) quote of Galileo is still "boggly" (mind boggling) to me: "Wer die Geometrie versteht, mag in der Welt alles zu begreifen" (german) . Who understands "trigonometry" is enabled to mentally fathom everything in the world. And i got a video proposal for (matrices?) mathematricians https://www.youtube.com/watch?v=A9WY_HZUK8Q (nine minutes) about a swiss' book or table that was ahead of its period or chaptre. If there were a female (called) Mathematica she'd be or even stay too beautiful for most of us (with or withour retri-beauty-on), not compareable with a Medusa (wearing tiny snakes on her head) we're not supposed to observe with our watching organs. And everybody would be allowed to know too moch, if not invited.
    1
  13843. Awesome video! I've been studying summation formulae for about 2 weeks now (for whatever reason) and this answers pretty much all the fascinating questions I came up with! Some other things I found you might enjoy: 1. Summation formulae 'transcend' the ring over which you define them, for example because sum(x^(n-1)) necessarily contains the multiplicative inverse of n in R, then in any ring Z/nZ there must exist polynomials whose summation formulas are not expressible as polynomials because they don't contain a notion of n^-1 2. The iterated inverse of 'taking a summation formula' on a constant function is always an alternating binomial function in any ring 3. The process of 'taking a summation formula' is always an automorphism of the space of functions over which it is defined
    1
  13844. I'd really like to hear your take on any mathematical relationships between brain wave states and sound phenomenon, solfeggio frequencies etc.
    1
  13845. An interesting observation about "the easy angles" specifically 45 degrees. Here in Oregon they have been planting Hazel nut orchards, often also called Filberts. The young saplings ares planted with a white tube to protect it from mice. They are planted using GPS and they are VERY exact. You can stand at the edge of the field and pick a direction where they all line up and call it the zero line. Without moving, turn your head and you will find the 45 degree line. If the field is big enough, and some of them are, you can also see 22.5 degree line as well as its multiples. Then you can begin to pick out 11.25 degree line and ITS multiples. I am certain if the field was larger you could pick out the 5.625 degree line as well. Just thought you might find this an interesting note.
    1
  13846. I like that you dont have to wear suits anymore...
    1
  13847. The sound at the end of the animation is the sound of cicadas, or the "17 year locust".
    1
  13848. I was never taught either of these a^2+b^2 = c^2 proofs (usually my math texts would just show an image of Euclid’s proof with no explanation), but I derived a variation of the second one: (a+b)^2 = c^2 + 4ab/2, on my own when I was 18.
    1
  13849. pls search krishna's butterball stone and make a video on it
    1
  13850. For all you vintage HP pocket calculator aficionados, XYZU should actually be XYZT.
    1
  13851. All miracles must be confirmed by the Vatican.
    1
  13852. I think that the thing about no-9s serie is astonishingly amazing... wow I do not have watched the end proof that it converges (I'm trying to prove it by myself), but it is so counterintuitive that it is mind-blowing
    1
  13853. In Poland we are taught this formula
    1
  13854. Oh, and btw, P=NP.
    1
  13855. those prettified diagrams are Mondrian inspired for sure!
    1
  13856. Outstanding sir
    1
  13857. Truth in Labelling.., ONE-INFINITY Pi-bifurcation is not the same as Pi-e unity on the same mechanism of i-reflection containment, but everyone is entitled to their opinion and this (word) form of truthyness is subjective in the context of the Singularity-point Observer's POV. Angles not Angels.
    1
  13858. Super, amazing!
    1
  13859. Dear Mathologer, like (x+2)^n in this video works for the n-cube, the formula (x+1)^n seems to work for the n-simplex: https://en.wikipedia.org/wiki/Simplex#Elements But what about the n-orthoplex? Its numbers also seem to be asking for such a formula: https://en.wikipedia.org/wiki/Cross-polytope#Higher_dimensions
    1
  13860. The best Mathologer video in my opinion. This does not deserve 500k subscribers, it deserves ((500000*(500001))/2)^2 subscribers.
    1
  13861. 2 is the inverse of 5 mod 9 :)
    1
  13862. oh boy burkard doesn't kid himself when he says ,"fasten your mathematical seatbelts! " does he?
    1
  13863. Ah! That is such an excellent proof for <80. The visual of the no nine series getting whited out was my favorite part
    1
  13864. So why do you think it’s not usually taught?
    1
  13865. Did I miss something with the card trick? It doesn't look like you have enough information to be able to discern whether the fifth card is four more or four less than the six of spades.
    1
  13866. Ah, you forgot to disclose that you and your assistant would mutually understand that the mystery card is either higher or lower than the first ordered card, before the trick. You could even memorize a higher-lower sequence if performing the trick multiple times for the same scrutinizing audience.
    1
  13867. Ah! Tricking the audience again with the 'no calculus' talk...😁
    1
  13868. No problem with the length. I just paused at the chaper breaks which really work well for me. My background is degree level engineering so happy with appled maths. I'm not so stong on pure maths so your videos are very interesting. P.S. I bought a book on Galois Theory about 30 years ago. Still got it. Still don't understand it.
    1
  13869. After 5 minutes watching this video, I am like : "this stuff can not become more interesting". ...50 minutes after... still listening...
    1
  13870. Ah, "The Schwartz is with you!" I can't believe that in 3 days, no one has made a Spaceballs reference (at least that I could find).
    1
  13871. NO MA'AM
    1
  13872. Love your T-Shirt! What about a cat bending playfully like that inside the profile of a chambered nautilus?! Wouldn't that be cool?
    1
  13873. I wasn't taught the remainder rule at school, but I discovered it for myself. Starting with any multiple of 9, add any number from 1 to 8 -- ie stsrt from the remainder you want By the rules of arithmetic you must increase the sum by that number, after you have cast out just one more nine. Therefore the eventual sum (digital root) is the number mod 9.
    1
  13874. One of the most beautiful videos i have ever seen
    1
  13875. The real beauty of Fermat's Last Theorem is that it is so intuitively true. But even Andrew Wiles' proof is so long and ugly, I feel there must be a simpler proof waiting to be discovered. Although his use of apparently unconnected fields of math was brilliant.
    1
  13876. 9:27 I thought that this was the actual definition of the golden ratio. At least that is how you define it at school. A/B=(A+B)/A.
    1
  13877. I remember trying out the areas of equilateral triangles and semicircles to see if Pythagoras' Theorem still held true. I became fascinated with mathematics and was so proud to tell my teacher what I had found out! It gave me so much confidence in the subject, it became my firm favourite.
    1
  13878. We kind of learn, here in Brazil, part of this in school and then we 'finish' it in college. Pretty boring for those not following a mathematical career
    1
  13879. My favorite was still the "chaotic" optimal stacking. I love seeing chaos emerging from "neat at a glance" problems, especially if they don't require delving into "naturallly chaotic" systems like infinite complex number series.
    1
  13880. This deserves an applause
    1
  13881. I am really interested in bizarre maximal overhang towers which don't have a pretty pattern. wanted to know if there exists any hidden algorithm that produces them for any number of bricks.
    1
  13882. I was wondering the same thing, but my maths skills are too weak. Things to consider: Any polyhedron can be tesselated to have triagular faces. Gluing tetrahedrons to the faces seems the obvious choice. But that turns faces into corners for each step - I don't know what that means. In 2D the results are nice regular polygons, but in 3D is the result always platonic solids?
    1
  13883. Start of the video: the closest I can get to a proof for Conway's circle is that the three extended segments are the exact same length, and for some reason, three intersecting segments of the same length can always be contained inside a circle that touches both extremes of all of them, so the proof is lacking the most important part.
    1
  13884. I like how symmetrical algebra is....🤗
    1
  13885. Gamma was the coolest part to me. I always love seeing fundamental math constants pop up. The more optimal but 'uglier' solution to the tower of lire was a close second.
    1
  13886. Wonderful. Love seeing how Pascal's triangle and the binomial formula can help us understand the sequence of integer powers.
    1
  13887. Ooh if you're doing that theorem next you can do the 4 square theorem while you're at it! I think the proofs are basically the same.
    1
  13888. 9:14 That is: 2 = 3^0 + 1 4 = 3^1 +1 10 = 3^2 +1 28 = 3^3 +1 82 = 3^4 +1 My guess: it works for natural numbers of the form 3^n + 1, and it would because the 3^n rows would cancel amongst themselves, leaving us with the resulting row of the cancelation and the following one.
    1
  13889. Honestly, I thought calculus was a cakewalk after advanced trig.
    1
  13890. I like the swivel proof and the color proof equally. The Action Lab made a good video about shapes of constant width, he also constructs one. Fun fact: if you make Conway's iris with a regular triangle, the radius of the outer circle is roughly 5.29 times (twice root 7 times) bigger than the radius of the inner circle (which was a fun algebraic challenge to compute).
    1
  13891. 4:05 Why don't you just put the orange point on the perimeter of the smaller circle so that it makes a perfectly pointed star? Would that mess with the squares and triangles later?
    1
  13892. I’m assuming 3-4-5 has no parent, because its parent if self referential. I’m not sure how to prove that though.
    1
  13893. Re: end of video question. I consider myself a recreational mathematician. I started with undergraduate degrees in chemistry and physics, followed by a PhD in physical chemistry, and ultimately an MD. I’ve been a practicing physician for years but mathematics has always been my favorite. I never tire of your videos and much appreciate the effort!
    1
  13894. Very nice :)
    1
  13895. I am obliged to tell you how grateful I am watching your wonderful videos. Regards, Mike
    1
  13896. we can live in a hyper universe
    1
  13897. Your t shirt is wrong, if you are rounding to 4d.p. the answer is 25.8070
    1
  13898. I really liked the fractal animation in the "no nines" series. Should have seen that pattern a mile away. This channel continues to enrich my knowledge of mathematics. I'm a master's student that had to leave my program twice because of money and GPA issues, and videos like this keep me going.
    1
  13899. I literally said "a-ha" out loud right before you said "did you already have an a-ha moment?" 😀
    1
  13900. I had only been exposed to the 2nd pythagorean proof. I think it was from numberphile or something. Wasn't from school (in the US).
    1
  13901. du brauchst echt nen nachtmodus für deine videos :P
    1
  13902. Amazing video, as allways. Thank you so much for sharing those incredible proofs. The first time I stumbled with that equilateral triangle problem was in my childhood: basicaly, I wanted to draw a pixelated equilateral triangle in paint, for some reason. When I got a little older and had some contact with mathematics, I managed to formulate the problem in a more rigorous way (some way how its formulated in the video, basicaly), and, eventualy, managed to prove the impossibility using sines and cossines. Pretty cool to see such a beautiful alternative proof for that. Needless to say, the proof shown in the second part of the video was also amazing.
    1
  13903. 21:43 all the properties of the natural log under control? Well so long as you stick to the ln of positive reals ...
    1
  13904. The most memorable part is definately the bizare and caotic optimal stacking that you showed us. It's such a "what...???"-moment.
    1
  13905. What if you start from the "straight-cross" windmill, and then alternate finding the y<-->z conjugate (rotate the blades) and finding the "same-footprint" conjugate? This gives you a "walk" through the space of windmill solutions. At least in your one example, this walk visits every solution, it ends at the Fermat square solution, and the "eccentricity" of the blades decreases monotonically along the way (starting with the longest/skinniest rectangles, and eventually becoming squares). Do these properties hold for all 4k+1 primes? It would give an alternative proof of uniqueness for Fermat square solutions.
    1
  13906. I've been working with maps and maybe this would be a nice way to make really compact representations of the polygons I use for drawing places.
    1
  13907. Witchcraft!!!
    1
  13908. Huh, only a few minutes in but I just remembered how the method for solving difference equations is similar to differential equations. It'll be nice to find out why
    1
  13909. I'm a bit late to the party, but excited as always for a Mathologer video!!
    1
  13910. Thank you!!!❤
    1
  13911. your giggling is soo spooky....
    1
  13912. It’s sinch
    1
  13913. Just in time for dinner! Lets go!
    1
  13914. The "combine F_n through F_{n+3} to get F_{2n+3}" identity reminds of a similar elegant identity: if you square the matrix {{F_{n-1}, F_{n}}, {F_{n}, F_{n+1}} then you get the matrix {{F_{2n-1}, F_{2n}}, {F_{2n}, F_{2n+1}}. These kinds of identities are useful because they allow you to skip ahead in the Fibonacci sequence without computing all intermediate positions, while using exact integer arithmetic. In fact the identity I mentioned comes directly from the algorithm for computing large Fibonacci terms by repeated squaring of the matrix {{1,1},{1,0}}.
    1
  13915. Thank you, this made my day
    1
  13916. 333= half evil 222= one third evil 111= one sixth evil ___________________ =666 = pure evil
    1
  13917. I made it to the very end, lol.
    1
  13918. Best channel!!!
    1
  13919. I wouldn't be surprised to find out that this had been discovered numerous times in history. In fact, I wouldn't be shocked if this is discovered every single generation, it's just that there has never been a time until now where access to mathematical literature was so readily available and searchable. The same is true for the cost of writing things down, as it has never been cheaper in the past. The longer back, the less likely they'd be inclined to spend the resources to write this down. The more recent, the less likely they would've felt that it was a discovery at all and so they also would've not been inclined to write it down. This is something that I could imagine a caveman showing his caveman buddy late at night on the cave wall.
    1
  13920. Did this thing have anything to due with the creation of the metric measurement system?
    1
  13921. Well I made it through, phew! I loved it a lot, it all worked great! One complaint that I have is during the video you mention a couple of things would be in the description but are not. I'm majoring in Astrophysics right now in university. I would be interested in seeing your Creator Statistics for this video: like how long people watched for on average and where they dropped off
    1
  13922. They are related problems, but they're not the same. The birthday problem is about when you have less than 365 people in a room, that is, less pigeons than holes, so nothing assures you that there will be shared birthdays.
    1
  13923.  @inigo8740  I'm aware of that. I'm simply expanding the question about when you get absolute certainty to a question about when you get a certain amount of confidence, and pointing out that it leads right to the birthday problem.
    1
  13924. implex + c
    1
  13925. You are a great mathematician. I love Mathologer a lot!
    1
  13926. Once you color all the hexagons to the bottom, then you have...Q*BERT?!
    1
  13927. Thank you!
    1
  13928. Hello! yes, I'm getting to the end of your videos and really enjoying them thanks. I've taken the bait, and I must say that I was also inspired by Martin Gardener and other similar writers while I was in my teens and I went to the University of Leeds and got a degree in Mathematics (1994). I was one of the few you mentioned in another video who took a third year course in Galois theory (partly because the other option was something to do with statistics and by my third year I'd had enough of that). Unfortunately I was really, really disappointed by the lecturing style and lack of engagement at university. Added to that was Godel's incompleteness theorem and Maurice Kline's book 'Mathematics, the loss of certainty'. The final straw was my discovery of a world of literature and philosophy and before I finished my degree I had turned thoroughly against everything "scientific" and "mathematical" (meaning things mired in mainstream scientific methodology). A few years later I spent a semester lecturing Cultural Studies and Art students about postmodernism, psychoanalytic approaches to art criticism, and other similar things, and some time writing about ethics and the crisis of rationality before throwing myself into setting up small charities and working with people mainly discarded by society. Now I'm living in a remote part of Eastern Poland building a little homestead and helping my daughter with her schoolwork and rediscovering the beauty I found in maths all those years ago. I think that your videos and our mathematical achievements are amazing but I probably won't watch too many more. It seems very likely that man made climate change is going to destroy our civilisation. Maybe we got so enchanted by the abstract beauties we've constructed, that we were blinded to the destructive nature of the lives we've been living. Very easily done, and I don't blame anyone, but I need to get back to my garden, my bees and my goats. Thank you Mathologer and all he best for the future
    1
  13929. My mind is blown
    1
  13930. gnomon technic in 3d
    1
  13931. ho ho ho ! it's cough !!!
    1
  13932. For the ugly optimal tower I wonder how those towers look as the number of blocks gets larger. I could imagine there might some interesting structure there that's hard to see with only 200 blocks.
    1
  13933. So that means, A squared plus b squared plus c squared equals Deez squared?
    1
  13934. The Bernoulli numbers also appears in the asymptotic expansion on the Euler-Heisenberg Lagrangian in quantum electrodynamics. Between this and Cardano's formula of last video you just has shown two thing that appeared in my PhD thesis. If your next video is on asymptotic series I would be speechless.
    1
  13935. This video was amazing! I am on the first year of computer engeneering (in brasil) and i chose my final calculus project to be about power series, this video could not have been better to inspire me! Cheers from brasil!
    1
  13936. I don't watch movies. I'm pretty happy with just this stuff. 09:30
    1
  13937. This is really a great video! Which reminds me Hilbert polynomial in algebraic geometry:)
    1
  13938. most memorable: sequence of no nines infinite or finite? me: infinite (because I recently prove that number of numbers up to 10^n without nines is approximately 9^n) it's finite! oh, 9^n/10^n -> 0 oh, well... indeed.
    1
  13939. We like this turtle math archaeology!
    1
  13940. Hey dude I guess you're my new math teacher and I thank you keep up the great work
    1
  13941. Amazing! Thank you maestro <3
    1
  13942. I was told about Pythagors' theorem "it just works" and was surprised much later when I worked out a proof for myself. I used the first diagram.
    1
  13943. Beautiful as always! Thank you for an amazing year. Merry Christmas and Best Wishes for the New Year.
    1
  13944. Sharing another nice video on related topic. https://youtu.be/n-e9C8g5x68
    1
  13945. 10:19 The last element of the double product will be a multiple of cos²(π/2) i.e. zero, making the hole product zero.
    1
  13946. 8:56 293 - 1 = 292 9:12 1 through 15 and all unique because that's how binary works. Using the polynomial product, (1+x)(1+x^2)(1+x^4)(1+x^8) = (1+x^16)/(1-x) = 1+x+x^2+...+x^15. 24:00 because we don't need all the 5 highlighted terms from the 405-degree polynomial to make the product x^(k * 100). And to make the formula work, the answer is at 24:12 right?
    1
  13947. That was great
    1
  13948. Astonishing proof
    1
  13949. Vote: Coloring proof best. Great video!
    1
  13950. Les deux démonstrations sont magnifiques j’arrive pas distinguer mais j’ai un faible avec la rotation magenta et aqua
    1
  13951. Well. It’s nifty. Thankyou
    1
  13952. I liked the no-9 series, the seal of approval, the prime numbers lurking around. And all the rest of course. Thank as always. Thumbs up!
    1
  13953. Incredible, so many answers to questions I aked myself when I was at high scool
    1
  13954. Intuitively, it seems at first that in five dimensions there should be pentagons available; although on reflection I can see that this would entail the assumption that you could intersect the grid at these points with a two-dimensional plane, which of course is an unwarranted assumption. However, this does appear to say something interesting about higher dimensions, although it's melting my head to try to work out what that something might be.
    1
  13955. Merry Christmas!! 🎄 For the second challenge (👓) at about 14:35 There are three ways to tile the first 2x3 (2 rows x 3 columns), namely three vertical tiles; one vertical and two horizontal; and two horizontal and one vertical. Likewise there are three ways to tile the second 2x3 grid. For the first “eye hole” on the left—because the two squares that connect it will always be taken (as those two are already accounted for in the previous 2x3 matrix)—there is only one way to tile the first eye. This means the first vertical of the 2x4 connection section will always be taken, meaning the 2x4 is really a 2x3 as the first column is taken. So there three ways to tile a 2x3 as before. By the same logic, on the right eyehole the 2x3’s first column will always be tiled, as there is no “degrees of freedom” there, so it’s essentially a 2x2 (which can be tiled with two horizontals or two vertical only). Finally the last, right-most 2x3 follows the usual logic, three ways to tile. So, the grand total is a permutation bc the grids are treated as independent and discrete (by which I mean I can how I tile the each 2x3 doesn’t affect the others, hence multiplication) 3 x 3 x 3 x 2 x 3 Or 2 * 3^4 = 162 Right?? 😄 I hope!! 🙈 Edit: 342 final answer! See my replies below!
    1
  13956. I just saw two other way sets of ways! But these are forcing and so these ways can be added to the 162 total instead of multiplying. Namely, you can start with he 2x4 in the middle
    1
  13957. And another comment, actually don’t know if I commented in the first one
    1
  13958. Well, the favourite part's gotta be the teaser to you own proof of the no 9's sum, even though we probably won't ever get to see it. Apart from that I really enjoyed the greedy algortihm to rewrite every number as a subseries of the harmonic series, since I have never thought about that at all, it just never crossed my mind. Doesn't seem complicated, but sometimes the initial question is the most ingenious and also complicated part of a problem. Thanks a lot!
    1
  13959. I remember I had a proof that P=NP in a dream. That was the most disappointing thing ever when it didn't work.
    1
  13960. In india it is still taught in 9th grade i suppose
    1
  13961. Or for a more practical example, the large coin has a radius ~365.2425 times the smaller coin. How many times does the small coin rotate to go around the big coin? How many times does the small coin point towards the centre of the big coin?
    1
  13962. Hello Mathologer, Best Season’s and Holiday’s Greetings to you, the Mathology Team, and to every viewer!! Please can you help me by telling me something - very simple - but it’s just REALLY bugging me, like an itch I can’t reach!!!! Ok: at about 8:20, you mention that we can “ignore the ‘strange brackets around (in this case) the ‘4s’ (the fours).’ I know what they mean, and what they do, how they can be “upper” or “lower” (with the squiggly line at the top or bottom of the bracket); BUT.......WHAT ARE THEY CALLED? WHAT’S THEIR CORRECT NAME?? Aaaaargh it’s driving me nuts - PLEASE can someone help me out!!!! Thanks in advance, and all have a good holiday period, and goodbye to one of the craziest, weirdest years EVER!!!!
    1
  13963. iM 10 hOuRs L@Tln(e^e)
    1
  13964. you go zig-zag, you place the first piece with an overhang of 1 then the second one with the same overhang and then the third on top goes back to overlap the first position.
    1
  13965. I have lost my eyesight due to being exposed to years of bright white Mathologer backgrounds.
    1
  13966. Leaning tower was the most approachable id say, thanks!
    1
  13967. One aspect of these tiling problems that always struck me as amazing is that coloring the checkerboard is seemingly irrelevant to what is in the end a purely geometric problem. The fact that we play games on alternately colored checkerboards is really a historic accident; the tilings would remain the same on perfectly white checkerboards. Colors are just handy labels for our logistics. Try to imagine, for a moment, how a race of blind mathematicians ('bathematicians'?) might approach the same problem. They'd have no intuition for color, and no natural reason to suppose that a domino covers two squares of different color. Could they, then, effortlessly invent some analogous label, e.g. "hot"/"cold" squares, and proceed with the proof? Or would that step be so counter-intuitive to them that it would be considered a touch of genius? To what extent does our biology and history affect the progress of our Platonic math?
    1
  13968.  @Mathologer  Indeed, I think I first came upon the mutilated chessboard problem in John Haugeland's book "Artificial Intelligence", where it is presented as an example of how formulating a problem correctly can make all the difference in solving it. Raymond Smullyan also mentions it as a brilliant proof. If I'm not mistaken, mathematicians call such quantities as the total number of black/white squares covered by dominoes an "invariant" of the tilings. Finding an invariant greatly simplifies a problem, but of course there's no systematic way to go looking for one. However, it's certainly possible to train a computer through machine learning to learn how to "sniff out" possible invariants after trying out a large number of examples, just like a human can gain intuition through practice.
    1
  13969. I followed along until the no 9's animation at the end, I got lost immediately. I liked the part with gamma, what I inferred was that gamma is the identity of the sum. Not sure if that makes sense.
    1
  13970. Love the t shirt
    1
  13971. Say, maybe this is a stupid question but instead of a number wall could you have a number cube for example? or any other type geometric "3D" shape?
    1
  13972. Can I just throw a related problem into the ring [ since I'm not wearing a hat, I cannot throw it in ! ]. What is the area of the largest square than can be inscribed in a right-angle triangle ? [ I encountered this question in solving the classic ladder and box problem ].
    1
  13973. I went to college and got an A.A.S. in Electronics. I have never seen either proof, that I can recall. Although, I do remember lots of work involving Pythagoras. I do believe I still have all my math and physics books, as well as, my electricity/electronics books/materials in a box of reference materials, somewhere. Now, I'll have to find them and look through them. 🤦‍♂️
    1
  13974. notice that r and s satisfying rs=r+s implies a log troll. take the ln of each side to get: ln(r+s)=ln(r)+ln(s) of course, not true in general, but you could trick someone with these numbers
    1
  13975. I am having a little bit of trouble with Couchy's proof. Maybe someone can help me :) The proof hinges on the fact that the shadow has non-intersecting lines. But is this always possible? It seems to be, if there is one face sticking up from the rest of the surface, since in that case one can simply bring the light closer to the object until the shadow of each line surrounding that face does not cross the rest of the object. For cubes and all the shapes shown in the video, this works fine. But what happens if I take the cube and subdivide each square that is a face of this cube into four smaller squares separated by new edges? There does not appear to be any place I could put the light above the shape, such that the shadow lines are non-intersecting. My intuition tells me that one could deform this shape in a way that the numbers of bits do not change, but I do not know how to formulate this so it is obvious that this works even in higher dimensions. Thanks in advance for any insights :)
    1
  13976. I did learn both methods in school, but in different years. But not the "ah-ha" diagram as far as I can remember.
    1
  13977. Not a dumb question. It's a natural question to ask, and related to the pinned question about how many rearrangements are there that sum to the same number
    1
  13978. 14:55 It is not a "posture problem", it's just to avoid hiting the bricks! :P
    1
  13979. This reminded me of a video by ViHart, this one to be precise https://youtu.be/Yhlv5Aeuo_k
    1
  13980. As a musician, all I see is two happy elephants converting to each other.
    1
  13981. I really enjoyed the visual fractal "proof" of the no 9 series. I love your videos mate and as a structural biologist at the Monash Institute of Pharmaceutical Science it's great to see someone at Monash making such amazing outreach videos. Thanks!
    1
  13982. That's so wonderful. Is it possible to find a circle everywhere such as Gaussian integral?
    1
  13983. I liked the parabolic overhang. Statics and dynamics were some of my favorite problems in freshman physics.
    1
  13984. I wish you could zoom out the animation in the end a bit more.
    1
  13985. This video was full of surprising facts that made sense once you thought about them for a bit, but the irregular stacks and how on earth one would arrive at them still boggles my mind.
    1
  13986. This blew my mind. People just do math and crazy correlations happen.
    1
  13987. Epic video😊😊😊
    1
  13988. I was really wowed at the end where you showed a derivation for the formula for cosh
    1
  13989. Most memorable is certainly the question about the no 9s series. Never would have though it converges although it makes perfect sense now.
    1
  13990. Wonderful to get to see your office, really 🥰
    1
  13991. Numerical analysis what's one of my favorite topics.
    1
  13992. Namaskar ji I'm from India I request you to give the Hindi version of your videos like 'VERITASIUM'. Your videos are very knowledgeable.
    1
  13993. The coloring proof was the one I came up with. It was obvious that if the points were all on a circle then it had to be concentric with the inscribed circle of the triangle. Since the six pieces obtained when cutting the sides at the tangent points to the inscribed circle would be pairwise identical I could show that the cords were indeed bisected.
    1
  13994. Great video as always! 8:26 It's a separable equation if 1 wasn't there. dy/y=xdx and we have y=C[e^(x^2/2)-1]. 21:42 The Wallis product W = (4/3)(16/15)(36/35)...>1, at the same time W = 2(8/9)(24/25)(48/49)...<2. Thus 1<W<2 (not sure if the argument is rigorous).
    1
  13995. the thinner and thinner really got me i didnt expect that, definitely my favorite part! love these vidoes
    1
  13996. 18:16 for 1 it's 2, then for 2 it's 2 moves for smallest disk, then 1 for big, then 2 for small, then 1 for big and 2 for small again. In general, m(n)=3m(n-1)+2. If m(n-1)=3^(n-1)-1, m(n)=3^n-3+2=3^n-1. m(1)=2=3¹-1, so this is right
    1
  13997. My college girlfriend's dad had one of those big teaching slide rules on the wall in his computer supply shop. He had it set to calculate sales tax (which was something obnoxious like 7.85 percent) so he could impress customers by figuring the tax on hand-written bills without a calculator.
    1
  13998. I have a round slide rule I purchased in Hong Kong in the late 60's and used it in college. Works like this. Never could understand the straight ones,
    1
  13999. My highlight has to be that the harmonic series misses every single integer (by narrower and narrower margins!!) on its way to infinity
    1
  14000. Love your videos. Accept my thanks sir.
    1
  14001. Love it!
    1
  14002. I found this to be an informative , well presented an complete addition to my own line of studies an though process. Thank you.
    1
  14003. When you went into the hyperbolic functions, you should have started with the basics. Show how Sinh and Cosh relate to the x and y values off the hyperbola as you did with sin and cos at time 23:29
    1
  14004. Very entertaining .tq
    1
  14005. A -A is logicaly same - same you can not use different values for the same variable letter
    1
  14006. To answer the title question "Why do physicists play dominoes?", I believe it's because they can not finesse a tiddly wink as yet.....
    1
  14007. I am a ph.d. candidate in algebraic geometry and I went through the whole video. I'm leaving feedback as requested. It was difficult for me to keep attentive from 33:39 to 37:45. I could follow the steps okay, but I wasn't sure where you were going, so it felt like you were just fiddling around without an aim. Once I saw that you were relating a finite sum to a power series I was okay again. I think if you say what role f(x) will play from the outset that would help keep this section motivated.
    1
  14008. The part around 15:20 and later is a bit troublesome - we need some extra knowledge about the behaviour of [r2/n]/r2, i.e. the rate of convergence.
    1
  14009. Yooo!! 🍹🏖
    1
  14010. Fold the cross down the middle horizontally then vertically. Then you have an L shape with even legs, cut down the center of each leg the long way and you should be left with a cross and 4 L's that make a cross.
    1
  14011. The number of Choronzon.
    1
  14012. But I would try a^3+b^3+c^3＝d^3 where a,b,c,and d are integers.
    1
  14013. Learning about gamma has made my week. Additionally, the way you showed it visual in the blue bits in the square was very help. Would love to see videos exploring where else gamma shows itself. I love your videos.
    1
  14014. r2-1 by 1-(r2-1) makes r2-1 by 2-r2 for the small rectangle which compare to 1 by r2. So (r2-1)/1 = (2-r2)/r2 <=> r2(r2-1) = 2-r2. That’s the very purpose of A paper norm as said by the Professor in the video :-D
    1
  14015. a^2 + b^2 = E/m
    1
  14016. Madhav is not living in a peaceful country , India is under continuous attack from invaders who are destroying our universities and our kings maybe not able to properly fund Madhav and his school because he has to increase spending on Millitary.
    1
  14017. Cool music, like in fps Doom from 1993. Love it!!!
    1
  14018. 14:54 alright now do it with a hexagon and you'll have dont it with all the basic simply tessellating polygons lol
    1
  14019. Towel Man and Beard Man: name a more iconic duo
    1
  14020. My mom says I know more than her. So I try to extend my knowledge and videos like these help. Thank you.
    1
  14021. Wow! I found that to be one of your best! Very nice!!
    1
  14022. I really enjoyed the video. Very well visualised and explained :) If you take some time to think about what's on the screen it's very accessible!
    1
  14023. 100th video, first collab, and an obscure topic?! We're eating good today! Great video, man.
    1
  14024. If only we’d reached out to Vi Hart we could have timed it to match her recent croissant video. Too much cyclide-shaped food is never enough.
    1
  14025. Any pilot has used the circular slide rule. One can be seen in episode of tv series UFO. We had competition at work, we could do calculations quicker than pocket calculator. Also visually you can see many alternative answers without readjusting.
    1
  14026. God these videos are still great. You're still the best mathematician on youtube, in my opinion!
    1
  14027. Can you also make videos on Number Theory and more on Geometry please.
    1
  14028. 27:18 markus persson? Notch is watching mathologer?
    1
  14029. Who doesn't love giving their wife a pearl necklace?
    1
  14030. That was just beautiful, I didn’t understand all the little detail but just the over arching idea opened my eyes to just how beautiful the series and sequences are considering I really hated them in my calc class
    1
  14031. Praise Euler for his book, foundations of differential calculus. thank you for the video
    1
  14032. I love math as eastern eggs on series
    1
  14033. Gamma is greater than 0.5 because the blue objects have areas bigger than triangles joining the endpoints. The area of these triangles is 0.5. Favorite part: The visual proof that gamma is less than 1. Second-favorite part: The proof that the no-9s sum is finite.
    1
  14034. when you said that these binary numbers can be thought of as decimal if it's easier- does this same rule about flipping cards this way always leading to a string of zeroes also apply to e.g. ten sided dice (or any numbered dice) and does the same proof work?
    1
  14035. I hope next time is sooner than 3 months — that was rough.
    1
  14036. Awesome video as always
    1
  14037. 16:11 2 and 5 are multiplicative inverses mod 9
    1
  14038. It's appropriate for Einstein to be here. My understanding is that Pythagorus is the basis of his time and space equasions. 16:53 Man, I was planning on my comment noting that you can do triginomety to this. I'm glad to see the people have. :) I've just realized that this is the literal basis of triganometry. It's too late for it to be this late. 36:19 I think it's 25/16 or 1 9/16.
    1
  14039. x = -(((3 - 2 Sqrt[2])^i - (3 + 2 Sqrt[2])^i)/(4 Sqrt[2])) y= 1/2 (-1 + 1/2 ((3 - 2 Sqrt[2])^i + (3 + 2 Sqrt[2])^i)) from Mathematica. The first ten solutions (x,y) (1,1)(6,8)(35,49)(204,288)(1189,1681)(6930,9800)(40391,57121)(235416,332928)(1372105,1940449)(7997214,11309768)
    1
  14040. The sum of the integers equals -1/12. So the reciprocal of the sum of the integers, which is (obviously) equal to the sum of the reciprocals, (meaning our harmonic series) is equal to -12/1 = -12. QED maths is easy
    1
  14041. To me the best part was the realization that the ln(n+1)-gamma approximation becomes better for large n and not only relatively better as its the case usually. Great video :)
    1
  14042. There is a "classical" proof of the two square theorem using some basic arithmetics on Gaussian integers (one of the "textbook proofs" I guess). It is my favourite one because first of all it shows the existence and uniqueness at the some time, and also the proof itself shows you "why the theorem is true". In other words no black magic is involved. In fact the way you showed uniqueness is kind of related because the identity that you introduced at 26:27 can be obtained by writing up the norm of the products of the complex numbers a+bi and c+di in two different ways.
    1
  14043. Wow, Gregory-Newton, fantastic, first time to see this construction :)
    1
  14044. Obviously I am going to download a slide rule app... why wouldn't I emulate a centuries old low performance calculator on a modern high performance calculator....
    1
  14045. The birthday paradox is about likelihood. The pigeonhole principle is about guarantees.
    1
  14046. 34:45 - er, if you're integrating over 0 to x dt, shouldn't every x on the right be a t?
    1
  14047. Silly me - must be thinking backwards.
    1
  14048. Is he reading script?
    1
  14049. When I was in school the nerdiest of all the nerds carried circular slide rules on their belts.
    1
  14050. I love this !
    1
  14051. I saw the first of the two proofs when it was given to me by Martin Gardner -- or rather when Martin Gardner was passed on to me by the librarian in my school. I was probably a little too young for the more 'algebraic' second proof, but of course, one proof is as good as two proofs, so long as the one proof is correct. This is probably also the time when I realized that librarians were the best teachers in any school, if they're allowed to teach at all.
    1
  14052. I used lego for a better overhang Much further , less math
    1
  14053. sweet
    1
  14054. Sir i am an indian and i would like to add that Ramanajun is a different approach but it seems as though you are a better mathematican as well, sir
    1
  14055. Pi becomes 4 because of the nooks and crannies. The same thing happens when you zoom into a coastline. The length is almost infinite depending on the scale. When we say that Pi = 3.14, then we take for granted that a circle is a distinct object that can be described more precisely than Pi = 4 but this is not always the case :0 If you want to have a maximum precision ? then Pi = 4 is the only value. If you want to have a practical approximation ? then Pi = 3.14 is the correct value but it's not a real value. Sit down and have some water.
    1
  14056. Chapter 5
    1
  14057. Wow, this is really a gold mine, and very inspiring! Thank you so much! 😀
    1
  14058. Is it important to get a general formula for polynomial higher than degree 4 just like quadratic formula
    1
  14059. I have several slide rules that I actually used back in the day. My high school participated in state wide competitions. I think I was 12 or 13 when the first 4 function calculators were being developed. There were many variations of circular rules developed for engineers. OHMITE was an OHM's law slide rule; SYLVANIA produced a circular rule for calculating doppler shifts; the GREENLEE tool company produced a circular rule for drill sizes for screws and nails that included characteristics of woods - a must have for carpenters. I have many of these pieces of history and still use them from time to time.
    1
  14060. To be honest, doesn't the fundemental theorem of algebra has as many proof of itself?
    1
  14061. I guessed that the no nines series was convergent because I've dabbled a bit in that area. It turns out that the minimal sequence of natural numbers that has no arithmetic triples are all the numbers that in base three contain no twos. It's clear how sparse the sequence is when you mess around with it.
    1
  14062. I actually remember learning this x / 9 concept in 4th or 5th grade, and then never touched on it again. There's a certain type of learner that would benefit to such a huge extent if this stuff was taught in school... I actually graduated high school in 5 years, and barely graduated at all, but once I got to college and was able to study what I wanted (Physics) I graduated with a 3.8. The whole obsession with passing tests regardless of comprehension is why we're falling behind the rest of the world in the US.
    1
  14063. what about 0.75=0.74999999999999999... Because it is true
    1
  14064. another one!! i m lucky
    1
  14065. Wow. thank you so much! I'm a neuroscience guy with a background in astronomy. Whenever I see a sequence I represent it in my mind as a time series (can't help!), and I was wondering if I can use what you wonderfully showed with fMRI signals. Will try! For sure!
    1
  14066. Too bad this didn't come out on Fibonacci day, 11/23. But 12/3 is not too bad. Anyway, great video, thanks for making it!
    1
  14067. Great episode. And I loved the outro music!
    1
  14068. Just to note: The movie has a big big big stupid error inside: In the scene when both boys travel on the train, looking out of the window they see a rainbow. That rainbow has: 1) arc not in the form of a semi-circle 2) the sun not at the exact back of the observer (look at shadows) 3) the alexander zone not darker as the inner first arc 4) both arcs has the same red outside (non inverted) it's a very miserable prove of ignorance. :)
    1
  14069. That was an Amazing video. Love it!
    1
  14070. You’re adorable, Burkard. Each time I find you here, I see an excited little Boy, speaking of his passion, smiling whilst saying numbers; you’re absolutely beautiful, my dear Friend. Don’t you ever change. I so love your Soul’s most brilliant colours. I spend a lot of time here with you, thank you for the amazing and top calibre Company.
    1
  14071. Brilliant videos
    1
  14072. Beautiful nature
    1
  14073. There are 10 kinds of people, those who understand binary and those who don't.
    1
  14074. 2020 is 2*2*5*101. leaving out the twos because the theorem hates them that leaves 5, 101, and their multiplier 505. all three of these are 1 higher than 4 which means they are all good. which means there is 12 ways. But I did the typical lazy thing and started a spreadsheet to find these 12 ways. 16 and 42 pair up so that's 4 ways, 24 and 38 pair up that's another 4 ways for 8 total. I couldn't find the last pair of integers which must mean that I misunderstood the theory or didn't set up the spreadsheet correctly. Edit: I forgot 1, which must mean there are... 16 ways? ??? I am confused. Edit 2: Okay I'm not confused anymore. this is just like the 5 case. 16^2 + 42^2 is a "different way" from 42^2 + 16^2.
    1
  14075. Magic! 👍
    1
  14076. Most most memorable: bizarre maximum overhanging towers :)
    1
  14077. Can’t sleep- lets learn math! Was totally hooked. Still up. 👍🏼
    1
  14078. Before watching: I assume you talk about super interesting property of phi that phi^2 = phi+1 or rather related one that might use small phi, because AB=A+B doesn't fit exactly. Perhaps there is some generalization to other numbers? Or at least some extension for phi that is relatively unknown? Now I will watch it and learn.
    1
  14079.  @Mathologer  It seems so. I didn't know there was this whole family of related stuff. I knew of Silver and Bronze ratios, but that's all.
    1
  14080. Indeed, the gradual whitening of the graph is only an hint, not a proof, because if you produce the same graph with 1/prime series sum, you get a white graph as well.
    1
  14081. why 4k+3 and not 4k-1?
    1
  14082.  @Mathologer  Best solution for BANKRUPTCY (rather than DEATH) is for the candidate to sell everthing and pay who he wants to and then go bust with ZERO assets, Note- this is probably illegal !
    1
  14083. Mathologer. This will have great use for me and my Synthesizer, Re: Neo Rheimanian Theory now I just have to program it ;)
    1
  14084. A very intuitive reading of calculus! Great!
    1
  14085. Great video as always! I bet some Indian mathematician must have written a poem about this proof 700 years ago 😂
    1
  14086. I guessed that the "without 9 sum" would converge because I thought "well, the sum of the numbers didn't count diverge to infinity, and the harmonic series is basically one of the slowest to tend to infinity, so for sure if we substract something that foes to infinity the harmonic series will have a hard time diverging" lol
    1
  14087. I already knew most of it, but Ramanujan summation was new for me. I read the article on Wikipedia, but I still don't quite understand it.
    1
  14088. Made my day!😘
    1
  14089. This is much more interesting and easier to follow without having to bring out the calculus, which actually obscures the mathematics. Brilliant as usual. Thank you. Logarithms are fun, any chance of a tetralogy!
    1
  14090. The even denominators of the harmonic series are collectively just 1/2 times the whole series, which means the odd denominators are also one half. But these are subtracted from one another term for term in the expansion of ln 2. This way perhaps the paradox is the more impressive. But indeed (rather like the partial fraction separation of even denominators in this video) the odd terms like 1/3 = 1/2-1/6 which leaves 1/6=1/(2*3) when cancelling the 1/2; 1/5 = 1/4-1/20 laving 1/(4*5). So then ln 2 is really 1-{1/(2*3)+1/(4*5)+1/(6*7)+... which converges fairly quickly ; i.e. ln(e/2) = 1/3!+3!/5!+5!/7!+... (though one would use the first form computationally).
    1
  14091. 18:45 PIXAR I remember you as a good deal older, playing chess with yourself. :-)
    1
  14092. This shape is well known for carpenters when you drill rectangle hole. John Conway died of covid.
    1
  14093. Wow this is the first and definetly not the last video i saw from you! I wonder if my classmates in school would be amazed at the same level as I was. Carry on with that passion until every other kid in my age stops hating math and starts seeing it as interesting tool to explore and explain many many parts of the universe! (We will pretty sure never be able to proof and explain everything!) I really like the design and structure of the video. (and your voice) Personally I found the video easy/easier to follow on x0.75 speed. I wish you and all the people in the comments a blessed day ... and I also want to ask if I could use the graphics and formulas for a presentation in class. <3 Stefan
    1
  14094. I don't really know about this one. It doesn't seem very expedient to get to any power by using this method unless there is a closed form for B_n. EDIT: Since you asked for feedback, here I go. Everything here was pretty accessible to me, but I'm also in the process of writing a Master's Thesis regarding a 1994 PhD thesis about how three-dimensional flat tori are spetrally defined, i.e. if two flat tori are isospectral then they are isometric, so perhaps I don't really reflect on the average viewer of this channel. For my tastes, you could make a video that's like, 50 times more abstract than this and I'd still have a grand old time. For one, I'd like you to tackle something where numbers don't even help. Something about topology or fundamental groups, or something measure theoretic. I dunno, might be too complex to digest in a video.
    1
  14095. Part i learned an interesting trick is with your balancing at a glance part. Really a nice way to show it
    1
  14096. Just when you thought infinite series were safe.
    1
  14097. I like the new outro music.
    1
  14098. (very minor bug in minute 12:00: on the 2nd line 396 should be 369)
    1
  14099. Yes, the perimeter of a circle is 4d, but the perimeter is also a fractal :)
    1
  14100. A note on English: "first" does not mean "item 1", it means "the thing with nothing before it", likewise, "second" is just the thing immediately following the first. I propose we write "first" as "0st" instead of "1st", "second" as "1nd", and then figure out how to say "twoth" or something.
    1
  14101.  @Mathologer  Oh, yeah. I'm all for unironically adopting 0-based numbering (and half-open ranges, which I found recently that Dijkstra wrote a nice short article on). I am a programmer, so I'm used to it.
    1
  14102. You lost me after the rubik cube and the drill.
    1
  14103. I figured out the second proof of Pythagoras' theorem by myself, after high school. In high school they taught us something else.
    1
  14104. If both Mozart and Ramanujan lived to their old age, the world would be a better place today!!
    1
  14105. For some reason, a gift shop at an exhibit of films that my sister, father and I attended when I was a child had a "shadow of the 4D analogue to the cube" sculpture. I cannot remember for the life of me why, though. I am also somehow reminded of how the Museum of Science and Industry in Kensington had a Klein bottle (sort of!) for sale. I regret not buying that, in a way - but I would have had to ship it to Canada without breaking, alas.
    1
  14106. The rotating shapes at the end were very nice, but it would have been cool to show them rotating to other shadows, like the necker cube and the tesseract that looks like an octogon.
    1
  14107. Question: if the pythagorean triple tree contains every reduced fraction, does it also contain every prime as part of each fraction?
    1
  14108. W. Paulsen has established an online tetration calculator with complex heights (!!) . Do you understand complex-heighted tetration ? 😅
    1
  14109. Help! So each cake slice of a hemispheric cake corresponds to a crescent-curved cake slice of the cylinder-minus-cone cake, right? The base of the slice is a "curved triangle", where two edges are half-circles of radius 1/2 (assuming the hemisphere and cylinder-minus-cone have radius 1) and thus length pi/2, and the third edge is an arc of radius 1, and a length of 2pi/n (n is the number of slices). Looking at the cylinder-minus-cone, its cross section is of course two right triangles with a vertical height of 1 ( the height of the hemisphere and the rim of the cyl-min-cone), and horisontal side 1, the same radius. But that is not the shape you get from flattening the infinitely thin but crescent-become-half-circle curved shape of the cake slice limit. Its height is still 1, but its base is pi/2. These two edges are straight (one having been flattened) and at a right angle, but what is the length - and shape - of the third edge? It's probably not a straight edge, as it has to have a horizontal tangent at the top, and also at the bottom, but is it simply a cosine graph from 0 to pi scaled to height 1?
    1
  14110. Solution of quadratic equation was given by Indian mathematician sridharacharaya
    1
  14111. The shape shown at 1:00 was the proof we were shown. The A and B squares were filled with a blue liquid, and when you rotated the shape, C would fill up perfectly. That was honestly such a good way to show it.
    1
  14112. Freaking circles, they pop up everywhere
    1
  14113. great to see you sir! a wonderful video you share again!
    1
  14114. The simple answer is that the woke left cancelled STEM.
    1
  14115. the outro music fried my brain
    1
  14116. Thank you for the French and German subtitles .
    1
  14117. 46:34 min. All of them were awesome. I mostly liked your way of animating No 9 sum.
    1
  14118. Whats really funny about this is how we still subconsciously use the crisscross lacing across history like a tried and tested method even a shoe as old as 5000 yrs old has bigger eyeholes with a rather thick width and after tying firm the crisscross. This was highly education and just a fun exploration in general.
    1
  14119. 3 brick solution: if you can lean the top brick, the extruding brick between two bricks over the cliff would extend past the cliff
    1
  14120. Attempt to prove: Let an odd prime p=x^2+y^2=4k+i. Since p is odd, i is odd, and due to the 4k, we can limit i to be either 1 or -1. so 4k = x^2+y^2-i. As i is odd and 4k is even, either x^2 is even or y^2 is even. Assume y^2 is. Then 2 divides y^2, and since 2 is prime, 2 divides y. So 4 divides y^2. So 4k = (y^2)+(x^2-i). So x^2-i = 4j for some j, where x is odd. If i = 1, then x^2-i=x^2-1 = (x-1)(x+1), which 4 divides because x is odd, and thus both x-1 and x+1 are even. if i = -1, then x^2-i = x^2+1. Note if x=1, 2 isn't divisible by 4. Assume x^2+1 isn't divisible by 4 and x is odd. (x+2)^2+1 = (x^2+1) + 4(x+1). Since x^2+1 isn't divisible by 4, (x+2)^2+1 isn't. So by induction x^2+1 isn't divisible by 4 for any odd x (well even x too, but that's easy to see). so i can't be -1.
    1
  14121. This video was superb! I really appreciate your effort! 🙌
    1
  14122. I thought 42 was the key?
    1
  14123. "May de Schwarz be witcha!"
    1
  14124. Strongly suspect theonethreeseven is a physicist referring to the reciprocal of the fine structure constant. Alternative, he is referring to the Kabbalah.
    1
  14125. I think we should start calling it "Wiles's Theorem" or "Fermat's Last Troll Comment."
    1
  14126. I predicted the base e based on the exponential shape. When X reach e witch is infinity the base, the area will be 1 which is the area of the initial square.
    1
  14127. Thanks sharing valuable information and numbers after decimals….
    1
  14128. Ramanujan , most underrated genius.
    1
  14129. absolutely amazing thumbnail
    1
  14130. There is an easy way to make a new math, here. Love love love love love
    1
  14131. Sis is uan of se mohst german acczentz i heeve efer hörd xD
    1
  14132. Thank you. This is the way. I have spoken.
    1
  14133. 100squared+101squared+102squared=103squared+104squared isn't true, so yes it does seem quite random to me, you just piceked an arbitrary place to start and made it fit
    1
  14134. coloring proof
    1
  14135. Undergrad, Math Major. Enjoyed the bit where you derive the summation formula for powers from binomial expansion With your quality, long videos aren't an issue..I can watch about 3hrs of educational content on end.
    1
  14136. Queen of hearts, right? And I love the shirt. Does anyone know where to get one. I don't see it on Cafe Press or Zazzle.
    1
  14137. I watched one of those a while back. I was thinking, ok, cool patterns. Give me some cool math solutions to something then... they never had any.
    1
  14138. so basically adding these "ears" acts like a notch filter on the signal that is the polygon's coordinates
    1
  14139. My highest in Maths was from my electrical engineering course in my polytechnic days. I didn't think too much on Maths then, didn't really put two and two together, just thought it was cool that some people managed to find correlations between all these funky terms. Then, the Maths videos populated and popularised. And now I like Maths a lot, whether be it basic primary-school level logic-based or advanced calculus Maths. For the 1-100 sum, I broke it down like so: 1+2+3+...+10 = 55, 11+12+13+...+20 = 155, 21+22+23+...+30 = 255, 31+32+33+...+40 = 355 and so on. ... 91+92+...+100 = 955. So we have (1+2+3+...+9) x 100, and ten lots of 55. Basically, I compartmentalised the sums and broke it into easier bits to add. We'll get 4,500 + 550 giving us the iconic 5,050. Still pretty cool in my book, even if the triangle formula was not invoked.
    1
  14140. Great work! Thanks a lot!
    1
  14141. Regarding the rolling coin paradox, can you also roll the coin on an equilateral triangle? I am trying to reason about the paradox by looking at the circle as an infinite-sided polygon With this approach, I am not sure how one would roll a coin on a triangle I assume the coin would have to do a separate pivot step between switching sides but not sure if it breaks any rules
    1
  14142. The proof at the end was breath-taking. Also, an idea for the t-shirt. Achilles shouting not so fast to a turtle while running in half leaps.
    1
  14143. One question that came to my mind at 9:30 (it definitely wasn't "who cares?" !) -- consider the number of non-1 terms in the identities. For each identity-length N, consider all identities, and find the largest in terms of "how many non-1 entries does it have". Wonder if there's anything there, but couldn't find anything in OEIS... maybe i'll look into this later.
    1
  14144. The imbalance may stem from starting with a positive term, perhaps the same would occur inversely if starting with a negative. So, for all practical purposes, 1 - 1 = 0. But then, how does all of this conjecture over a useless math paradox help to solve the food crisis around the world?
    1
  14145. The Euler constant is greater than 1/2 because each blue piece has an area greater than 1/2 the area of the rectangle that inscribes it. That's because the 1/x curve is always below the diagonal of that rectangle, because it's concave up everywhere.
    1
  14146. It's hard to be more rigorous than that... why should a curve with upward concavity always fall below a line segment connecting any two of its points?
    1
  14147. We did not learn a prove of pythagoras in school but my math olympiad teacher has a cool book which shows you the picture of a prove and you have to figure out why it proves a theorem. I saw both proves in there but I like the second one better as it requires less geometrical juggling. Dont remember the name of the book though.
    1
  14148. Madhava of Sangamgrama and his legendary disciples aren't known by 99.99% indians who was the Originator of Calculus..
    1
  14149. 133 comment! 133 = 7*19 7 is the 4th and 19 the 8th prime! 4=2² 8=2³ :-D
    1
  14150. 35:32 Drama of my life... :D
    1
  14151. if a r is traced , and a crux to 90 join the circle , the end of the crux always turns on the circle and two others crux too .
    1
  14152. Does this work in 3 or 4 dimensions? If not, why?
    1
  14153.  @Mathologer  I agree. Especially since in 2D, the ears are segments of a regular n-gon. For higher dimensions, the higher dimensional ears do not have a regular 'base' anymore (for 2D, the base is a regular digon).
    1
  14154.  @Mathologer  I bet if you do this in 3D you would end up with an irregular polygon that can be projected into some 2D plane as a regular polygon, but the actual vertices of the 3D polygon will be at various distances above/below that plane due to having a 3rd degree of freedom that has not been eliminated. Exactly how to find the correct plane that you'd get the regular polygon as a projection, I could not say.
    1
  14155. That's an impressively mathematical maths lesson for Y9s! You must have a clever class. You could also show that the longest path covers all combinations, at which point 3^n drops out easily.
    1
  14156.  @DoReMeDesign  it was my top set, the local University came down for enrichment. I think the proof by induction went over their heads but the most capable ones could then do up 7 discs before they got bored 😂
    1
  14157. I have earned the mathologer seal of approval ❤️
    1
  14158. I have heard about the Euler–Mascheroni constant from other YouTube math channel but never knew what it was. So this was my favorite part. Thank you for your seal of approval :)
    1
  14159. I feel cheated by education system / life / luck. Its 2350 IST, I can't sleep now.
    1
  14160. Nice!
    1
  14161. 18:40 I don’t get it. The “average of all these numbers”? The length of the shortest path for two given points is just one number, and I just calculated it. Does he mean the average over all pairs of states?
    1
  14162. Sir, Could you please make a video on Fourier transforms.
    1
  14163. Can the formula be extended to other polygons with more than 4 sides?
    1
  14164. Presumably, given the proof of the convergence of the no-9s series, or the 100-zeroes series, or any of those, does it follow that you can do it with any other digit sequence? E.g. the harmonic series, minus any terms that contain my phone number in their digits - does that converge?
    1
  14165. One major weakness in this bankruptcy law is that it favours decisions to make small investments over large ones, whenever there is an element of risk. Someone with a fixed sum to invest, or lend, will prefer to distribute it among multiple small loans rather than a smaller number of large ones, to minimize risk in the event of bankruptcy or death of thecdebtor. This makes big investmenrt projects far more complex to finance than they need be. In a primitive society this may not be a practical objection, but with any degree of sophistication in a capital market, it would be unworkable.
    1
  14166. 19:10 I noticed that within the Arctic circle, there are a higher concentration of red tiles near the red corner, and less red tiles close to the orange corner. It's like a smooth gradient. Same is true for the other colours. Will this be true most of the time as it approaches infinity? Also, 14:10 I got 1,2,3,5,8 which is Fibonacci. I came up with a simple proof: for a board of size 2 by n let the number of tilings = t(n). The end pair of tiles on the right can be filled with either a single vertical tile, or two horizontal tiles. If it is filled with one vertical tile, the remaining board is just the board of size n-1. If the end two tiles are filled with two horizontal tiles, the remaining board is of size n-2, and so the total number of tilings is t(n) = t(n-2) + t(n-1) which is Fibonacci. (As t(1) = 1 this is proof by induction, and it is assumed that t(0) = 1)
    1
  14167. thank you
    1
  14168. I have to say I was a bit lost as to how exactly the cards relate to modular arithmetic, but then pac man came crashing in, packing both of those important mods.... Have to say, I love that feeling when two seemingly very different things suddenly transform into one another through a thing as neat as that :D
    1
  14169. Suddenly I see it: M likes weed. Noice.
    1
  14170. amazing
    1
  14171. I use geometry. I am a retired mechanical engineer. The relationship of lines and points and surfaces is necessary to make stuff. These special relationships are interesting but until I can use them as I conceptualize a final shape of a “thing” I don’t get excited like a mathematician. There is room and a need for both of us.
    1
  14172. One problem with the Nike method is that it leads to the -1/12 garbage...
    1
  14173. I watched this late, and I should have been sleeping. I might have missed something. In the card trick, what if you have (say) the 2, 6, and 10 of clubs, and two hearts? It seems there's lots of varieties of a full house where the trick doesn't work. This doesn't make it not cool, but I'm wondering if I understand. LOVE this channel, but now I'm tired, and a little distracted.
    1
  14174. Let's say the cards you have are the 2, 6, and 10 of clubs and also the ace and 5 of hearts. You have quite a few options for the two cards you're going to base things off of (you only need them to be of the same suit). For the sake of demonstrating the trick, let's say you use the 2 and 6 of clubs for the two cards you're going to base everything off of. The distance starting from the 2 and ending at the 6 is 4. The distance starting from the 6 and ending at the 2 is 9. The smaller of the two distances is 4. So you keep the end card, the 6 of clubs. The top card on the stack you give to your assistant is the start card, the 2 of clubs. Now, the other three cards are the 10 of clubs, the ace of hearts, and the 5 of hearts. The natural ordering of these cards has suit alphabetically first, then card value. So B = 10 of clubs, M = ace of hearts, and T = 5 of hearts. You want to communicate "4" to your assistant, so you consider the fourth permutation of BMT, which is MTB. So you give your assistant the cards in this order: 2 of clubs, ace of hearts, 5 of hearts, 10 of clubs. (You keep the 6 of clubs.) Your assistant receives the cards in the that order: 2 of clubs, ace of hearts, 5 of hearts, 10 of clubs. Since the top card is a club, your assistant knows that the secret card you kept is a club. Next, your assistant notes the following three cards: ace of hearts, 5 of hearts, 10 of clubs (in that order). The assistant notes that the standard order gives B = 10 of clubs, M = ace of hearts, T = 5 of hearts. So these three cards are in MTB order, which is the 4th permutation. This means that the secret card is 4 away from the start card. The start card is the 2 of clubs. 4 away from the 2 of clubs is the 6 of clubs! The assistant says, "6 of clubs!" You reveal your hidden card, the 6 of clubs.
    1
  14175.  @MuffinsAPlenty  - Thank you! I had missed the part where you can order cards within individual suits in the "T-M-B" series. I thought all hearts were the same. I was up way past my bedtime. Thanks for clearing that up for me.
    1
  14176. I couldn't help but notice honey containing breadcrumbs stretch as I wrapped it around a tubular milk bottle, those points lining up. This needs some work, though a liquid slide rule might be novel..
    1
  14177. Brilliantly made video! Thanks
    1
  14178. Tie between the "Worm on a rubber band paradox" and “exactly 100 zeros series” is greater than the “no 9s series”. The later really makes explicit how we as humans are terrible at estimating density and big numbers
    1
  14179. Great job. This video lacks the hardcore final part where nobody can follow but at least I understood the whole thing
    1
  14180. Can I vote γ?
    1
  14181. in india its still taught in cbse board in like 9th standard a full chapter dedicated to it
    1
  14182. For me, "Gamma" is the most amazing part of this. Another universal constant that is the exact difference between the logarithmic function and the sum of the Harmonic Series.. I know that mathematicians have know about it for ages, but I was blown away by the fact that it even existed.
    1
  14183. Just seeing the thumbnail is exiting!
    1
  14184. As the triangles in the problematic lanterns get infinitely small, don't their bases get infinitely long compared to their height? I would not be surprised that if I add up the area of a lot of triangles shaped like line segments, I get a strange result. Having the triangles be parallel to the surface (spikes be perpendicular) solves the problem, if I understood correctly, butt wouldn't it be easier to require the triangles to have a finite aspect ratio, and would that not solve the problem equally well?
    1
  14185. You made my day.
    1
  14186. amazing stuff, you made me love mathematics even more
    1
  14187. Mathologer super
    1
  14188. It's just a Primary school Maths... Let a+b+c+d+e=540 . prove that: a+b+c+d+e-72-144-216=(a+b+c+d+e)/5
    1
  14189. can i have my money back, school?
    1
  14190. Stuck with it to the end, only got a bit confused when we appealed to the gods toward the end. Have an associates in Mathematics, and the more insane stuff the better!
    1
  14191. Wouldn't mind these shorter videos coming more often between masterclasses. Merry Christmas professor.
    1
  14192. Nice t-shirt though!
    1
  14193. I still have the Martin Gardner issue! I had to prove it in engineering math without the bugs. Dammit.
    1
  14194. When you start counting in different number base systems, perfect squares take on a whole new meaning
    1
  14195. There may be a small "rounding error" on that T-shirt... :-)
    1
  14196. Can I suggest an experiment? What if the cards have 3 sides? (0,1,2) When 0 turn into 1 1 turn into 2 2 turn into 0 And we only aloud to choose 1's or 2's and turn them upside down together with the next card to the right. Will it still work?
    1
  14197. Schön und im Unterricht anwendbar. Fände es gut, immer Mal wieder so kleine Zwischenspiele unterzubringen :-)
    1
  14198. I prefer watching 15-20 min long vids.
    1
  14199. It's good
    1
  14200. Also: your Simpson or Futurama references and citations are very rare the last months.
    1
  14201. I personally prefer the long in-depth videos, but one of those short ones once in a while is a good change of pace :)
    1
  14202. I like it. Sometimes I am stuck "saving" the longer videos for a weekend, but shorter clips I can easily digest after a long day at work.
    1
  14203. While I love your longer videos, sometimes it is great to have shorter videos like this. It makes watching them less of a commitment and more of a fun adventure. Would love to see you do both types.
    1
  14204. Some shorter videos would give more options to use them in a class room.
    1
  14205.  @RandomAmbles  But such a grid is not possible, right? The set of all possible configurations as in the video has a simple structure and for instance if there is AB CD in the first square (with A,B,C,D the four colors), then it is not possible to obtain CA DB in another 2×2 square for instance. PS: @Mathologer Your videos are great but I wouldn't call this clip "very good"... In my opinion it's again another one of these Good Will Hunting unwatchable scenes, honestly. The genius and the jaded unsuspecting teacher blahblah.
    1
  14206. Having a video thats still interesting but shorter and - presumably easier/faster to make - is awesome. Hopefully a fun and less draining task for you, and a good break from the constant hard-core thinking we (or perhaps just most of us) have to do for your normal videos. Good stuff all around!
    1
  14207. Shorter vids are the go!
    1
  14208. Definitely do it bro, it would appeal to the masses and get you more views
    1
  14209. This is great! I enjoy your deep-dive videos too but something a little lighter is more than welcome
    1
  14210. I'm a huge fan of the longer videos...when I get a chance/have the time to watch them. Which is significantly less often than I would like so they just end up getting queued up 'watch later' list. 'Later' being, apparently, a point at infinity. This was fun and interesting and, importantly, I actually had the time to watch it! And, as a bonus, it let me know about a movie I had no idea existed. So, mark me down in favour of the occasional shorter videos!
    1
  14211. I’d watch that 👍🏼
    1
  14212. long or short like them alll
    1
  14213. I like these shorter ones as well, but I would have liked to see an exploration of a wraparound version. Select the in the lsb position and it will flip the msb as well. What happens then?
    1
  14214. 👌
    1
  14215. Good idea.
    1
  14216. Given the choice of shorter or longer videos, I'd prefer the longer ones. Given the choice of shorter and longer videos, both sound good. I'll agree with Dr. Doppeldecker though: I don't think I'd like Mathologer Shorts. It might be good as an experiment, just to see if you can fit something interesting in the 1 minute, vertical format limits.
    1
  14217. Thank You and good evening from Slovakia, this video made a nice closing of the day. Nice refreshment of mind before getting to sleep. Reminds me of the fast 10 minute exams we had to solve at the beginning of lessons at high school. Experiment highly appreciated and approved, QED (ČBTD in Slovak) :)
    1
  14218. I love the idea but please don't give up on the long ones either!
    1
  14219. You'd do great making short videos, but I can get short math videos plenty of other places. I like that this channel makes longer videos.
    1
  14220. it is easy. i create new face at the right and move it to the left, trying not to create extra faces, since the value with extra faces will be smaller. So i have n moves up to the left and then i put new face, move n-1 times, etc. So all steps will be n + n-1 + n-2 + ... = n(n+1)/2
    1
  14221. YES. There definitely needs to be quicker, "easier" videos slipped in that can be explained quickly and not as deeply.
    1
  14222. Any video length is good
    1
  14223. The function that maps my enjoyment of Mathologer is increasing in the amount of content. So keep them coming!
    1
  14224. YES please, Sir!
    1
  14225. Yes!
    1
  14226. Man you should cover the math problems in the anime "Assassination Classroom"
    1
  14227. I think the idea is fine. Although I don't mind the length of longer videos. What I definitely would appreciate a lot is the diction. Sometimes it is challenging to understand the speech.
    1
  14228. Auf jeden fall gerne öfters. Denke das ist auch ein guter Einstieg für Neulinge, die sonst eher abgeschreckt sind von der Mathematik und sich auf zu lange Videos nicht einlassen werden. Der Spaß an der Mathematik entsteht ja auch nur, wenn man das Erfolgserlebnis hat, sich der Nebel lichtet und diese Dinge auch versteht und nachvollziehen kann.
    1
  14229. I prefer longer videos, more context, more details
    1
  14230. I'm definitely a fan of the shorter videos. Plus I think a short video every now and again will help others gain interest in solving math problems
    1
  14231. I love it! :D
    1
  14232. Another nice video, and also a busy-beaver card game :) Maybe you could consider doing a video (using again basic properties of integers and how under some circumstances they tie to any symbolic set) on Godel's stylish idea?
    1
  14233. Please do! More Mathologer is always good!
    1
  14234. Personally I prefer the long videos and would tend to skip ones like this. But that doesn't mean I disagree with you making them!
    1
  14235. Yes, interludes are interesting.
    1
  14236. Long or short, I love Mathologer.
    1
  14237. 👍
    1
  14238. I like it!
    1
  14239. For the quickie we've got Presh, for the longer in-depth satisfaction we have Burkard and Matt. _____ At least this is our expectation.
    1
  14240. I like it. In fact I unsubscribed a few months ago, because the topics were to complicated for me to follow all the way through. I like the "short" (your words) video, because even if a topic is more complicated I can power through and still follow you to the conclusion at the end.
    1
  14241.  @monika.alt197  Yes, and Matt is O'Dowd
    1
  14242. Too much of a good thing does not apply to Mathologer!! :)
    1
  14243. Definitely like it!
    1
  14244. Finally my 6th grader can enjoy mathology too. Please make more such accessible videos.
    1
  14245. Call then heartbeat videos
    1
  14246. Plus one 👍 on the short format. I like to watch math videos while doing dishes and it takes me about fifteen minutes to wash a sink full of dishes. This short format fits perfectly with that.
    1
  14247. Yes, please! There exist math gems that do not require a 40-minute work-up, but they are still amazing!
    1
  14248. As long as it does not replace the original scope of videos, this new concept is fine. Variety is the spice of life...!
    1
  14249. I like both long and short clips. I’d say, you may spend about the same total time with each kind.
    1
  14250. Accessible is the door that everybody must come through to enter the world of math. It is only a matter of when they go through this door, and if they do so alone or if they are led.
    1
  14251. I get mathdrawals if it’s been over 2 months, so punctuated shorter videos would be very nice.
    1
  14252. I like the concept of simpler fun videos in between the monsters
    1
  14253. I really enjoy some shorter videos, too!
    1
  14254. I really enjoyed this style. I can see both explanations for the face down card problem, but I know binary, so I see the termination more from that side. As you know, there are 10 kinds of people in the world - those who understand binary, and those who don't.
    1
  14255. Yess! Please do release shorter and accessible videos in between the monsters.
    1
  14256. I enjoy getting a break from the Kurasawa length videos. It's nice.
    1
  14257. For the proof of the 4n by 4n colored grid: Any 4n grid in which n > 1 simplifies into an eventual quadrant which contains a 4 by 4 square at n = 2. As a callback to the video, consider the original problem from the film clip. Instead, express this row of cards as a one dimensional grid, with all possible configurations of the domain to color its space with one of the two available combinations: black and white. Extending this model into the complex plane results in a two element pairing of black/white options for each element in a two-dimensional grid, giving four options for a possible cell configuration: (0, 0) (0, 1) (1, 0) (1, 1) Visually, this works best as distinct colors instead of pairs of complex binary numbers. As for the base case of n = 1, again the video of the original problem demonstrates that flipping adjacent values "swaps" the least significant bit with its neighbor, creating the illusion of the smiley face moving to the left. In reality, this operation is the temporary conversion of each overlapping pair of points into a complex value, swapping the real and imaginary bits (if applicable), and returning them back into their original form. In complex numbers from the two dimensional grid, these instead form temporary values containing a real number for the real part on the left hand element, a complex value for the imaginary part of the left hand element, and for the right hand element, a complex number for the real part, and a real number for the imaginary part: (1, 1) <-> (0, 0) (1, (1, 0)) ((1, 0), 0) This extra information that is duplicated follows the same rules for the original one dimensional space, and when returned, is "lossed" when considering the real value which replaces the complex element. With a cube, 6n is used, three elements are required per cell, so I'm pretty sure there isn't a solution (I'm not a mathematician). 4n-numbered dimensions only, as further diving into 4, 8, 12, etc. dimensional grid is essentially expansion of each real part with a complex value, etc. I suspect that the net affect of this sequence of operations maps nicely to some definition of a space filling curve, but I don't think I'd go as far as say that's definitely the case.
    1
  14258. Yes please
    1
  14259. I prefer shorter videos, At about 15 minutes I really start weighing my interest level in the topic against my current state of boredom. At 20+ minutes it has to really capture my attention for me to choose to watch it... At 30+, I basically need to have already been looking for info in the specific topic, have nothing else to watch, AND I'll probably be skipping thru it for the juicy bits... 30+ minutes is a lot of time to invest into a video focused on a single topic that I may or may not find interesting (and multiple topics should obviously be broken into separate videos). But I'll waste 5 minutes on 6 different videos I have very little interest in without blinking an eye.
    1
  14260. While I absolutely love all the content you make (you're one of the few channels where I watch every single video you upload), for me, this video's main focal point was a bit too simple? I didn't feel that usual "Aha! Fascinating!" moment that I get from your longer form content.
    1
  14261. I think this is a very good idea.
    1
  14262. Your videos are fun and engaging. There is one thing I would very much like to see you add into your videos and that would be to mention the actual branch of study that applies to each problem. If there are any students watching you that would really like to study and understand the problems, it would be very helpful to give them a starting point of where to look.
    1
  14263. even though I am very interested in Mathologer videos I am often exhausted by the intellectual effort because they are quite long and starting from the initial idea they go very far in developing the subject. For me it would be better if these videos are shorter and concise to help the fruition.
    1
  14264. our very thinking works in a log manner
    1
  14265. The Monopoly on the Penrose Triangle is just brilliant.
    1
  14266. 🫡
    1
  14267. Using a calculator took me 30 seconds. Spoilers below. Just used sigma notation. 91409924241424243424241924242500
    1
  14268. why is this making me feel like continuity and quantization are two sides of the same coin?
    1
  14269. Какая большая проблема :) Аппроксимирует оказывается только касательная к "объекту аппроксимации" а не произвольная "гиперпрямая" пересекающая "оный объект" :)
    1
  14270. seven numbers <100 with no collections with repeated sums: 1, 2, 4, 8, 16, 32, 64. Every sum of distinct powers of two is unique. :)
    1
  14271. Great video
    1
  14272. $ ghci ghci> sum (map (^10) [1..1000])
    1
  14273. Thank you for your work! ^.^
    1
  14274. മലയാളി ഉണ്ടോ എന്ന് അറിയാൻ വന്നതാ
    1
  14275. 29:45 The original paper has a long/short ratio of sqrt(2). The white rectangle has long side 1-(sqrt(2)-1) and short side sqrt(2) -1, that means we need to show (2-sqrt(2))/(sqrt(2)-1) = sqrt(2). If we times both sides by sqrt(2)-1 we get 2-sqrt(2) = sqrt(2)*(sqrt(2)-1) = 2-sqrt(2). Hence the white rectangle indeed has the same proportions as the original.
    1
  14276. "3, 6, 9. Whoa Tesla. . . not quite able to follow the reasoning here.". LoL.
    1
  14277. My favorite bit was the optimal solution for 20 blocks - its just... what??? It raises many interesting questions: Is there any good strategy for working out the optimal overhang for n blocks? About how quickly does the optimal overhang grow with respect to n? Closely followed by the sum of the reciprocals of the primes diverges. This is VERY counterintuitive.
    1
  14278. 1^3 + 2^3 = 3^2 Just dodged Fermat by one digit.
    1
  14279. 🤯🤯🤯
    1
  14280. Saw this video with the non marvel thumbnail a week ago and did a double take now, i love it!
    1
  14281. Good explanation. And an excellent Xmas shirt too!
    1
  14282. In gamma each of the blue ‘segments’ has a white counterpart to its left. You can see clearly that blue is greater than white in each case (and must continue to be, due to the curve). Therefore gamma must be greater than 0.5, as if it were 0.5 white and blue would be equal. Another great visualisation that I don’t think I’d have been able to comprehend without the mathologerisation!
    1
  14283. I think Geogebra could have been a little better for those.
    1
  14284. That's a hilarious tshirt
    1
  14285. When will you run out of interesting stuff to show us?!? 😮
    1
  14286. what about acosh and asinh ?
    1
  14287. The only visually confusing part on 20:07 is that if you add all the numbers in the steps, you end up with 36 going up and left, and 33 going left and up, but I know it's just the "road there" :) It is not an exact reversible from 18:02 but the result remains identical (( similarly like internet traffic. Going from A to Z results in going through one set of countries and submarine cables, while going from Z to A can go through another set ))
    1
  14288.  @Mathologer  Holy cow, there really IS an Encyclopedia of Integer Sequences! And I thought you were pulling my leg. But yup, there it is, and nope, sequence not there. It's Darin Brown's sequence now.
    1
  14289. Herr Mathologer, I found you only recently, and your videos are exceptional. Thank you for sharing your knowledge!
    1
  14290. Actually, √5 isn't that bad. Like all real quadratic irrationals, it has a periodic continued fraction, and its largest continued fraction term is 4. It's not like π, which has big numbers like 292 and 161 in its continued fraction expansion.
    1
  14291. Another interesting way to look at this is that the complex exponential function is analytic and periodic with a period of 2 pi on the imaginary axis. exp[i (log (ab) + 2 pi n)] with n an integer, a and b real exp(i log (ab)] exp[i (log a + log b)] So if we go around the circle a distance of natural log a, and then more natural log b, we get to the same position as log(ab) modulo 2 pi radians.
    1
  14292. There is an entire chapter on Heron's Formula in grade 9 in India.
    1
  14293. Instead of rotating along the outer angles (@ 20:27), you could have rotated along the inner angles and see the 180° directly (an arrow of the length of the lines might help to do that).
    1
  14294. Maybe some already know Numberphile's golden ratio song: https://www.youtube.com/watch?v=mqK63v2Jzks&t=0s This is all incredible fun!
    1
  14295. That is why engineers are happy but mathematicians always try to get to the happiness.
    1
  14296. The equation in the intro is wrong! MATHS-MATH is not S but MATH(S-1).
    1
  14297. 0:03 Pi mal Daumen XD sehr gut
    1
  14298. As a person relatively unskilled in math, I am in love with stuff like this. I am subscribed and angry that I only just found your channel! Cannot wait to binge your content!
    1
  14299. Seemed like a fun little challenge :) https://github.com/mjasperse/aztec_diamond
    1
  14300. Is there any chance you've thought of doing a video on the great Gregori Perelman? You are the man for the job
    1
  14301. Was the music of the introduction from Kate Bush??
    1
  14302. I enjoyed the Euler-Mascheroni constant, gamma, visualization a lot. It was a really cool way to see that the constant had to be finite (and less than 1). It was also cool seeing (on my own) that the constant was more than a half.
    1
  14303. Maybe it's worth adding why the shrinking converge to zero, otherwise the proofs are not complete. Hypothetically, it could have been the case that while the polygons keep shrinking, the shrinking factor changes over iterations and approaches 1 fast enough so that the polygons do not shrink to a zero diameter. It's not the case here of course, but I think it should be mentioned. Other than that - GREEEEAAAATTTT !!!!
    1
  14304. Is it just me or is Charles Babbage skulking in the background muttering about finite differences?
    1
  14305. Mathematics are a devine devilment, blessing man while at the same time corrupting him with its beauty.
    1
  14306. I love your T shirt
    1
  14307. At about 14 minutes: Isn't the core problem that in the original method each point is connected to that point's nearest adjacent neighbors and in the Schwarz method each point is connected to its nearest two adjacent neighbors?
    1
  14308. I guess not. It looks like maybe that was a visual artifact of layering the bands?
    1
  14309. summary of the different cases for the windmill link is not working.
    1
  14310. sounds like the minimum surface problem with the soapfilms
    1
  14311. Solution to the 3 bricks problem: Join 2 bricks together with a weightless nail or so. Place the one brick exactly at the edge of the table and the other one with its center of mase above the end of the second one. And voila here is your solution.(Please accept my weightless nail because it’s only a tough experiment :)
    1
  14312. 14:09 Hello, Fibonacci, my old friend.
    1
  14313. What is the slowest growing non-converging series that we know of?
    1
  14314. It's amazing the kind of things that can be proved with such a simple idea: you cannot have each pigeon in its own private pigeonhole if there are more pigeons than pigeonholes.
    1
  14315. Shush - don't tell the Chinese - they might figure out how to use shorter lengths of lace in their manufacturing process and make a bigger profit!
    1
  14316. This is so boring in binary
    1
  14317. Modern paintings have a finite area and an infinite stupidity.
    1
  14318. Well that was a another truly magical journey in to the beauty of Math
    1
  14319. Amazing video as usual. By the way, there is a small mistake, (x-1)^0=1 not x.
    1
  14320. Is it just me or a half of all math youtubers speak German? (flameable maths, Dr.peyam...)
    1
  14321. Hey Mathologer? Try these methods on daily Covid numbers. Find out what the other "derivatives" are doing. I did a little looking around and found it illuminating.
    1
  14322. Thank you for your videos!
    1
  14323. Loved this video...! And the little challange was... Kinda easy to be honest! Stay awsome!
    1
  14324. Good stuff.
    1
  14325. If your'e baked enough you can see the corner of a cube in the inscribed triangle.
    1
  14326. I made it for the very end, really interesting Also, this time I could understand everything
    1
  14327. The most I liked is the determinant proof at the end... It's amazing how determinants appear in estrange places all time...
    1
  14328. MAZDAのrotary engineじゃないですか
    1
  14329. I dont know why, but i love the leaning tower of lire
    1
  14330. The max number of moves has to be the sum of n, n being the number of cards, because, with the rule that if you choose the card on the right it only change it self, you can pick everything card from it to the last card on the left, and end up just with 1 card facing up on the very left, and you can repeat this process for as many cards as you have. So the movements for each iteration are n+(n-1)+(n-2)+...+1=(n)(n+1)/2
    1
  14331. I think I understood everything, I have a math level of an engineer that just have graduated
    1
  14332. Chapter 4 was my favourite. Awesome video by the way - it was a lots of fun. Many thanks for making them.
    1
  14333. great vid, I liked the square problem a lot so I did the cubes as well, it turned out to be (1!^6 * 2!^6...(k-2)!^6 * (k-1)!^3 * k!^3)/k^2 where k is the number cubed on top. I was only able to check the first 3 answers by hand tho, k = 4 is like 4-6 digits so not doing that any time soon
    1
  14334. Oh er spricht deutsch, er kann kein schlechter Mensch sein
    1
  14335. Nice video. The puzzle arises naturally when stacking pots and pan, you have two hands, one full of pans and an existing stack of nested pans. The Tower of Hanoi is closely linked with the "Chinese rings" puzzle which also counts in Gray code. In the puzzle community these puzzles are classed as N-ary puzzles, as in binary, tertiary, etc. Some I can make headway with, but the Panex puzzle defeats me I can't even get my head around the recursion. See Jaap's Puzzle Page for the details and an online playable version.
    1
  14336. Ramanujan is my favorite. I simply cannot begin to comprehend how his mind worked.
    1
  14337. 22:07 -- hyperbolic sectors +
    1
  14338. The video did not end up in my subscription box. Maybe you should send out a community message.
    1
  14339. Where can I buy rubber bands (we call them GUM BANDS in Pittsburgh) that can stretch an infinite amount without breaking! :-)
    1
  14340. If m and n are odd then, m+1 and n+1 are even. So for any tiling say m=2x+1 and k=2y+1 .Then at some point j=x+1 and k=y+1. So as cos 90 =0 .The product becomes 0
    1
  14341. Thanks!
    1
  14342. So the parent of the 3-4-5 triangle is 0-1-1 lol
    1
  14343. Thank you! Aztec diamond model is flat earth, and our cells, since they’re fractals of the same😊. Between this and your multiplication model video showing the shadows of the lunar phases, I’m loving it! 💙 & ✌️
    1
  14344. Answer to the final puzzle I believe is 1/4. The area of the inner square. The sides are all 1/2 in length. 1/2*1/2=1/4. You can rearrange the triangles on the sides to make 4 squares around the center square that all have sides of length 1/2.
    1
  14345. As always Euler to the rescue
    1
  14346. Lovely video, as always. I did notice that you added an 'h' to Toricelli's name in the video description. In Italian, that would alter the pronunciation to a 'k'-sound.
    1
  14347. Fantastic video! So incredibly clear how this all is derived, step by step, from some simpler warm up expressions. Great job!!!!
    1
  14348. Somehow when i watched the clip a few days ago, i didnt understand a thing. But when you just showed the clip i understood everything crystal clear. Your channel has something to it XD
    1
  14349. beautiful
    1
  14350. I like all of your videos, but for some reason I really loved this particular video. Thank you 😊
    1
  14351. 53:33 I made it to the very end.
    1
  14352. No Integers definitely the most beautiful thing
    1
  14353. i came up (for the second differential equation) with e^(x²/2)-1
    1
  14354. Utterly fascinating.
    1
  14355. I made it to the very end.
    1
  14356. Never saw it, I was only taught the algebra
    1
  14357. I think that the most memorable part is about the missing digit sums. I simply can't handle the idea that the 100 zeros sum is larger than the no 9s sum. And that fractal proof for the missing 9 was pleasant.
    1
  14358. 6:00 this indeed was beautiful
    1
  14359. I started to proof the theorem in the moment I watched that video since we did not learn a proof for it at school or at least I cannot remember we did. I came out with the algebraic version (2).
    1
  14360. I'm using C++ and without installing special math libraries, I am unable to easily compute anything beyond the 405th partition number. The 405th partition number is 9 147 679 068 859 117 602.
    1
  14361. I listen but still have questions and I guess will always.
    1
  14362. Personally, I liked the Harmonic series not being integer part. I saw the proof of that while reading a number theory book, but it was much more cumbersome than this one shown here. Nice videos as always, thank you.
    1
  14363. That animation of the rotating 3D and 4D cubes was very illuminating. Thank you for doing these videos.
    1
  14364. Fascinating! Tnks!
    1
  14365. my vote: I found the coloring example easier to understand, while the swivel one felt more like a magic trick. I liked both! how's that for a vote? 😀
    1
  14366. For gamma > 0,5 You divide your square in two parts and cut each blue part to get a triangle You cut your triangles in two part to get a rectangle The sum of all of the rectangle would be equal to 0,5 Which means gamma > 0,5
    1
  14367. Fabulous! I hope that one day (soon) all Math will be taught "Mathologerly". Thx.
    1
  14368. Und..... Kann man damit lineare pdes in polygonen lösen? Das Problem in regulären polygonen zu lösen sollte einfacher sein
    1
  14369. That was fun. I wish I could have given it the attention it diserved. By the way, was there ever a solution to that mirror question?
    1
  14370. Where was this when I learned it at awesomemath last year
    1
  14371. That's a pretty chilly cardioid there at the end. I think it's trying to tell us what to do with 2020. Namely, that we should let it go.
    1
  14372. hi can you make a video on achievements of indian mathematician please
    1
  14373. I thought you made the picture of Sri Madhava by editing your own photo. Then when I googled I realised that is not the case. I see similarities in your looks. By the by we Indians believe in reincarnation.😑
    1
  14374. Thanks for the sharing!
    1
  14375. I loved that :-)
    1
  14376. Pearl anniversary representing 30 years? I see 12 large pearls and 4 small pearls for each large pearl being months and weeks respectively? Congratulations, any advice for newlyweds?
    1
  14377. Maybe it's just the math book scenario here, but I don't see why you would need to give it to your wife to have that much fun. 😂😂😂
    1
  14378. The largest known primes are Mersenne primes. I.e. the largest known primes cannot be represented as sums of two squares. :-( Die größten bekannten Primzahlen sind ja Mersenne-Primzahlen. D.h. die größten bekannten Primzahlen kann man gerade nicht in Paare von Quadratzahlen zerlegen.
    1
  14379. Burkard, weißt Du noch wie toll es war, wenn man ein 5 Mark-Stück bekommen hat? :-)
    1
  14380. I'm just glad people understand this now to the point where they have stopped claiming that 1+2+3+... "adds up" to -1/12.
    1
  14381. I'd bet a fiver without a second thought that it was Torricelli(us) himself who came up with the anagram... Etymologically, Italian ell derives from Vulgar Latin ull from Classical diminutive suffix ul (it lost the diminutive function in the Vulgar, common people, street Latin). Torriculus would sound more Latin. René Des Cartes did much better with his Latin name: Des Cartes = of/from Cartes (pl.) = Cartesianus. And the first name, Renatus is also etymologically correct.
    1
  14382. You are the best.... I believe you know that already.
    1
  14383. Very nice explanation how?how?
    1
  14384. Gamma was easily my favorite!
    1
  14385. Greetings from Thailand. I love the idea of being amazed by things I do not understand. It is a bit like magic. It awes me and it makes me realize all at once the apparent paradox of the infinite and the infinitesimal. Ahhh...my ignorance in all of its splendor and my capacity to come back and be hit over and over again by things I cannot ever hope to understand given my age. And please keep to yourselves the comment of "you are as old as you feel", which is another piece of crap propagandized by some new suckers. Thank you Burkard for another wonderful video.
    1
  14386. the dephinition is realy satisphiing
    1
  14387. Awesome video once again. I love the way you visualise everything to make the math understandable.
    1
  14388. Fantastic!! Thank you 🙂
    1
  14389. "Infinity" is shorthand for a properly defined process, and invoking it as if it existed outside mathematical language necessarily leads to fatal contradictions.
    1
  14390. Mathomagic. Mathomagician.
    1
  14391. when trying to prove it by myself I ended up with something close to the coloring proof so I'd say it's more intuitive but the swivel one is also very interesting
    1
  14392. Funny story. I used to play a video game roughly 20 years ago that made an allusion to this. That ignited a curiosity for numbers, eventually leading me to John Conway's work, which, in turn, led me to numberphile, then eventually, here. To see this come up on an actual mathematics channel hits a weird sort of nostalgia for me. Thanks for all the great videos sir.
    1
  14393. Answering what you asked in the end of the video, I recently started college for aeronautical engineering, and I know some things about number theory since I'm very interested by it. I understood preety much everything in the video except for that part where the 0s became 1s and the - became +. No cat videos needed
    1
  14394. Watched the whole video and after second attempt understood all ~17 years old child currently not in any particular school
    1
  14395. Parallel threads on a multicore (256) computer would also take some 100 years .. by that time quantum computers would do that while the user is making an omlette
    1
  14396. 👍👍👌😊❤
    1
  14397. Gamma gang!
    1
  14398. Mind=Blown
    1
  14399. @22:48, that is TOTALLY a drawing of a cat.
    1
  14400. 15:57 puzzle : area is 5 how crazy is that .
    1
  14401. I like this new music
    1
  14402. Please could we have a 8 hour loop of the Mathologer music to sleep to? It's a knockout.
    1
  14403. Wow.. I understood this. Great fun
    1
  14404. Great video! To answer your question, I remember in my last year of primary school (age 10 or so) I was involved in an extension maths where we did divisibility rule for 9, and tried to make ways for all the other integers up to 10. Looking at other comments I think I was fortunate to have a passionate maths teacher! It instilled in me an enjoyment of all things mathematical
    1
  14405. Has to be the refined Kempner proof for me!
    1
  14406. Voting for Kempner's proof. such a simple proof for a very counterintuitive fact.
    1
  14407. Burkard - having struggled with math all my life, you bring clarity and fresh air to an otherwise somewhat rarefied endeavor. Your efforts are most appreciated. Thank you kind sir!
    1
  14408. THIS WAS GREAT!!! MERRY CHRISTMAS
    1
  14409. thank you very much for your time and effort to make this amazing video, deeply appreciated
    1
  14410. Those crafty pigeons also inform us that our trusty zip file compressor is on average actually a deceitful file ELONGATOR!
    1
  14411. an infinite sum, an infinite fraction and an irrational number walked into a bar....
    1
  14412. All I can manage to see are 5 of the 8 c-cubes inside the 4-cube. What have I missed?
    1
  14413.  @Mathologer  frack, my reply disappeared. TY I know where they all are & think I know why I cannot see them all. Sending love to Melbourne!
    1
  14414. Great video. Cheers. One minor thing though, the title made me think that Torricelli was a 14th century monk and I got quite confused when you started mentioning Gallileo at the end. Of course it was Oresme that was the 14th century monk, but he doesn't get mentioned so much in your vid. Still a great video though. Nice one, keep 'em coming.
    1
  14415. Holy shit. How long have you been back? I thought you were gone. Thanks for another awesome video.
    1
  14416. The video was interesting as usual. And then you conjured Euler's formula out of thin air! Wow!!
    1
  14417. I was interested in this video, because i used Hurwitz zeta function to find indefinite integral of floor(x)^n dx a few days ago and i hoped to understand this function better.
    1
  14418. Ok here is a dilemma for you It's said that a man died, he had 17 camels, and he had three sons In his will, he gave the older son one half, the second son one third, and youngest son one ninth Now how to do the division without slaughtering any camels
    1
  14419. @Mathologer  So you know your tulmod, you know your Arabs, you know your math You are a knowledgable man sir! Salute!
    1
  14420. Ok, to generalize, n²+(n²+1)+...(n²+n) = (n²+n+1)+(n²+n+2)+...(n²+2n) = (n(n+1)(2n+1))/2 (2n²+n)²+(2n²+n+1)²...+(2n²+2n)² = (2n²+2n+1)²+...+(2n²+3n)² = (n(n+1)(2n+1)(12n²+12n+1))/6 (10²+11²+12²+13²+14²)/365 = 2 3⁴+4⁴+5⁴+6⁴ = 2258, not 7⁴
    1
  14421. 2² = 2 • 2 = 2 + 2 and 2⁴ = 4²
    1
  14422. I've got a bachelor's degree in computer science. I followed pretty much everything, however I was disappointed by there not being a nice closed form for calculating the Bernoulli numbers. Spending a tad more time on that section was needed for me. The transformations around 40:45 I still haven't completely figured out but I'll get there eventually I'm sure. Really interesting video though and I'm sure I'll be discussing it at work tomorrow.
    1
  14423. It works for me. I am a master of computer science and mathematics, but I have never used any advanced math since leaving university more than 25 years ago. I just watch this for fun, and you may say nostalgia. Mathologer is the best of the math channels I have found.
    1
  14424. i remember during my school days, we were given an assignment of making a report on pythagoras theorem with whatever material we find in the internet, if only i saw this video 4 years earlier.
    1
  14425. Atotal = 1 Amiddlesquare = 1-16*Asmalltriangle 5*Ast = 1/4 => Ast = 1/20 => Ams = Atot-16*Ast = 1-16/20 = 1/5
    1
  14426. without involving logs, but using integrals, we now that A(c)=Int((1/x)dx from 1 to c) Int((1/x)dx from 1 to a)+Int((1/x)dx from 1 to b)= in the second integral make change of variable y=ax; x=y/a; dx=dy/a Int((1/x)dx from 1 to a)+Int((a/y)dy/a from a to ab)= In the second integral we can make a simplification, and since the variable is dummy, we can change y with x, to pair up with the first integral Int((1/x)dx from 1 to a)+Int((1/x)dx from a to ab)=Int((1/x)dx from 1 to ab) Or A(a)+A(b)=A(ab)
    1
  14427. The current 8th grade math textbook in china is featuring the Zhoubi Suanjing diagram. The logo of International Congress of Mathematics 2002 is this exact diagram too, and appears in the textbook. Other diagrams are shown in the textbook as extra readings, such as 2:24 original Pythagoras squares, and 25:38 Garfield's trapezium
    1
  14428. Classifying fractions into 4 categories according to parity of numerator and denominator is weird. If you do not insist on reduced fractions (the video does not mention this), then every fraction can be written as even/even, so you do not have a classification at all. If on the other hand you do insist on reduced fractions (and at least in the examples this is what you do) then the even/even category is empty. So it would seem the only reasonable classification of this kind involves only 3 classes.
    1
  14429. I finished year 12 maths which included matrices but one thing I didn't understand was how to get the inverse of the square matrix at 21:47
    1
  14430. Proof of Circle : Given any two lines, they lie on a circle because the power of the intersection point is the same. This is the product of the lengths of the parts of any chord on which the point lines. If the lengths are a,b,c then the power of one of the points, evaluated through one line is a*(b+c), and evaluated through the other line is a*(c+b). This makes the 4 points cyclic because given 3 of the points, we can define a circle, 2 of the points are on a single line and that gives the power of the point. We can then determine what point gives the correct power for the other line, and since the point at which the circle and the other line intersect gives that same power (and power is linear with point position along the line), this intersection point is the same as the 4-th point. Now we find the center of the circle formed. Note that the two lines, lets name them ABC and ACB, are mirror reflections along the angle bisector since the lengths of the sections formed by the point are the same, a and (b+c). Thus the center lies along the angle bisector. Note that the center also lies along the perpendicular bisector of the two chords ABC and ACB. The distance along the perpendicular bisector is a) identical, can be found bu evaluating position of perpendicular along the lines then noticing the equal angle, hence congruence. Evaluating another pair of lines and checking the lengths, we see that the center point is the incircle center, and the distance along the perpendiculars is all the same (by congruence of triangles), and is the incircle radius, which are both known to be unique. Also, the diameter of the circle has to be larger because two of the opposite points form the chords aka whiskers of the triangle, which is smaller than the circle's diameter.
    1
  14431. You ask "Did you get it?" and the answer is yes... YES I DID! RIGHT AWAY!
    1
  14432. I've still got my 'slip stick'
    1
  14433. My vote: The leaning tower of maths; the crazy maximum overhang gave me a WHAAAAAT?!?! nerdgasm ;-)
    1
  14434. nice pumpkin pie shirt! (Not so) coincidentally, made some today :)
    1
  14435. please nn dis%continue the SeriOUS,, keep it serial,,,ifpossible,,.. ....
    1
  14436. Very cool video!
    1
  14437. By the way, thanks again for your proof from a little over a year ago that e and pi are transcendental. That was some Mathologerization!
    1
  14438. Challenge 2: There exists j and k such that they are half of m+1 and n+1, so that term of the product is zero. QED
    1
  14439. coloring proof sits better with me than swivel proof.
    1
  14440. 700 year old proof was best. Simple but elegant.
    1
  14441. Is everything OK? You've uploaded more than once in a month :))
    1
  14442. I wish i would have kept up on math in school, i had no idea that 1 wasn't a prime.
    1
  14443. It’s more of a convention thing, there are many statements about primes (fundamental theorem of arithmetic) that if going off “primes are numbers that are only divisible by 1 and itself” would have to exclude 1. Really, so long as you get the ideas going on this is just dotting your i and crossing your t.
    1
  14444. I saw this proof for first time today at 52
    1
  14445. Okay yes but Dozenel. If Someone makes a vortex visualizer based on this info please submit the diagram for dozenel
    1
  14446. In Russia we are taught to Heron's formula in school but I don't remember now how it was explained or proven.
    1
  14447. It's nice how sinh(x) and cosh(x) are the even and odd part of e^x. In general every function f(x) is equal to the sum of an even part which is (f(x)+f(-x))/2 and an odd part which is (f(x)-f(-x))/2.
    1
  14448. i like this
    1
  14449. Coins will NOT die.
    1
  14450. This is my derivation for the derivative of the inverse of F(x): F(F^-1(x)) = x dF/dx(F(F^-1(x)) = dF/dx(x) f^-1(x)f(F^-1(x)) = 1 f(F^-1(x)) = 1/(f^-1(x)) f^-1(x) = 1/f(F^-1(x))
    1
  14451. Can the harmonic series be subdivided into infinitely many (disjoint) subseries, such that each of these subseries is divergent? Or is that too much to ask? :-D (The obligatory vote on the most memorable proof: The homework assignment that Euler-Mascheroni is larger than 0.5.)
    1
  14452. Ok, I'm not a mathmatician... but to me it looks like the rules are not being applied accross the whole series. On the first negative term, why didn't he do (1/2 - 2/4) = (1/2 - 1/2) = (0)? Why is it (1/2 - 1) which yeilds (-0.5) which does reflect the original (1)-(1/2) = (0.5). Also, does BODMAS apply in infinite series... so should I be getting the yeild of (1/2)+(1/3) before i subtract it from 1?
    1
  14453. I multiplied 25.8069x25.8069 and it resulted in Trump, so that shirt is really accurate
    1
  14454. Most memorable: the link between e, gamma and all prime numbers.
    1
  14455. Please notice me matholiger!
    1
  14456. Bloody brilliant m8.
    1
  14457. A rotating cuboctahedron.
    1
  14458. The Aristotle Wheel has been around for about 400 years and the computer works on the same principle. It makes one wonder if this wasn't given to people 400 years from an alien from a different planet to see how far earth people could go with this wheel and it only took us 400 years. If these same people would use the Aristotle wheel in reverse they might even be able to figure out to reverse gravity and go further into building an engine that can go the speed of light so we can travel to other planets. It could be this simple and maybe they are just overthinking this.
    1
  14459. I know why it never caught on initially -- it's because it's about triangles and pentagons but the people making the decisions were totally square.
    1
  14460. TRISTAN'S FRACTAL IS AMAZING I DIDN'T EXPECT A WHITE SCREEN AT INFINITY,SECOND IS 700 OLD BISHOP
    1
  14461. How awesome! I wish I could give a Super-like after your trick at 25:34 ;) Wish you all the best!
    1
  14462. Stringtheorie is based on 10 dimensions
    1
  14463. If only the Key to the Universe were so simple! Then we wouldn't have to waste our time with all this studying
    1
  14464. Nicest Mathologer I've seen yet!
    1
  14465. The geometrical visualization of the "no-nine" series reminds me of the construction of the Cantor set. In fact, it has to be somehow isomorphic to a 2D version of a Cantor set, as the process of eliminating 9s in base-10 is practically the same as eliminating 1s in base-3 expansions, which is one way of constructing the points belonging to the Cantor set.
    1
  14466. Ach Ta LiebeN~!....(FroM La CANADAs'...) I was asKeD By a maTh iNsTRucTor in EaRLY~NiNeTieS iF I finD a sLiDe RuLe...I haPPeneD aCRoss one at a FLeA MaRKeT, & so, GRAbbeD iT for him! Now I'd wisheD I'd GRabbeD MyseLF oNe Too, by noW! NoTwiThsTANDinG YouR exceLLenT iNTRoduCTioNs HeReiN'..!! hahahahahaha...ThanKy-You! PPS Ich BiN EiN BRunsWicKeR~BeRLiNeR Too....by a aRTy LieuTeNANT HusieN' in 1770's QUebeC & MichiGAN!...AchTunG!...Eh???....OYE!!! LeT us MeASuRe The ToNes oF SeMi-TRANsPoRT AiRhoRNs...eh!!?? I AM CANADIAeN! I am-NoT a BieeRe'..! Eh...
    1
  14467. what is interesting is that this curse is also a blessing: for battery as well as for chemical catalyst, you want the biggest surface per volume. in this case you have to think the opposite
    1
  14468. To be honest when I saw the title of the video, I thought you were going to talk about Newton’s approach to calculus in principia, the geometric approach.
    1
  14469. Only learned this in college as part of modular arithmetic.
    1
  14470. Please make a video on vector space
    1
  14471. Oh 333 only half evil the Morley’s Miracle
    1
  14472. in the visual for gamma you can tell the area is greater than a half because each piece fills in slightly more than half of their respective rectangles, thanks to the slight curve.
    1
  14473. The most memorable part was the appearance of Euler's gamma: beautifully explained! As always, awesome video! ❤️👍
    1
  14474. I liked the coloring proof better, because that was a proof with quantities, while the swiveling "look there" could be a card trick deceiving our eyes. 😊 Of course if swiveling used the missing notion of equal length from the outside point to the touching/common points, then it would be ok. That step was missing. (Sorry for the lack of lingo, English is my 2nd language)
    1
  14475. If it's so easy, what's the solution to the Navier-Stokes equations? Me and a couple of other students would really like to know
    1
  14476. Thank you for the video - it is great!
    1
  14477. Post Easter Mathologer video! Cool!
    1
  14478. That took me a minute (dead fast, for me). Nice illustration of the Harmonic Series.
    1
  14479. Mathologer, I love you!
    1
  14480. I have discovered a truly remarkable solution to the puzzle, but this margin is too small to contain the URL...
    1
  14481. I was taught to use zig zag lacing on my safety boots in case of accidents... running a sharp blade straight down the top instantly frees the foot, or so I was told, I never had need to call upon it! The steel toe cap trainers I wear for jobs around the house these days will only lace up nicely when criss crossed and it grates on me, I am so used to zig zag that it is the one I use in all other shoes.
    1
  14482. Give the "French" lacing a try: diagonals below, straights above. It's like the zig-zag but more stable vertically.
    1
  14483.  @SimonBuchanNz  I must have misunderstood what the zig zag was in the video because what you describe is what I was taught (and use generally), straights along the top and diagonals beneath. The idea was, I assume, that it would be easy to run a sharp blade down a line of straights on top without risking further injury if there was need to release the foot easily - I am sure it was this advice and the somewhat gory videos on the dangers of finger rings that led me to software engineering as a career! The trainers I have make anything other than zig zag criss cross uncomfortable by having cloth loops at the top that are oriented vertically, when I have worn them out I shall be seeking a different brand that doesn't do this :D
    1
  14484. I love seeing those 'old' faces brought to life!
    1
  14485. 😮
    1
  14486. So what about which lacing will most evenly distribute the force applied when pulling the loose ends? Criss-cross lacing tends to pull more effectively up top and very little at the bottom thanks to friction. Will this always be the same as the shortest lacing?
    1
  14487. it’s the same diameter. you even said it yourself two sentences before the quiz in the end
    1
  14488. Wow it even works with Fibonacci going in negative
    1
  14489. F3rm4t is overrated
    1
  14490. 23:00 is my favourite part! That geometric property that makes us visuallize that the sum of harmonic series minus ln(n+1) is less than 1. That's GORGEOUS!
    1
  14491. This is a teachers teacher! Absolute best presentations!
    1
  14492. I see the slide rule is being rediscovered. How retro!!
    1
  14493. Awesome video as usual! No they never actually taught us about the nines trick. But when I was a kid I watched a great math educational show on Public Television called “Square One” where this was mentioned. It did blow my little kid mind! I had an inkling it was related to using base 10 but it is great to see exactly why. Square One was a great show and I hope you and the other great math channels on YouTube are filling that void for kids today!
    1
  14494. Most memorable: no integers among the partial sums.
    1
  14495. F. stands for Freek (Barning). I guess J.M. is Johannes Maria or so. Catholic tradition.
    1
  14496. I like "Trithagorean" .... It just means that I think of it as π-thagoras now though 😂
    1
  14497. We've got formulas for all triangles and all quadrilaterals (except maybe ones with crossing lines). Can these formulas be extended to straight-edged shapes with more sides - pentagons, hexagons, etc. - is there a generalized formula for n-gons?
    1
  14498. In England we don't touch on this till secondary school and even then its very little
    1
  14499. Can you please do a video of how Madhava derived the series for trig functions 🙏🙏🙏
    1
  14500.  @Mathologer  So cool. THANKS!!!
    1
  14501. Perfect Videography
    1
  14502. Did I hear Fun maths problems?
    1
  14503. Yeah should be impossible to do a choice of a surface area less than pi. And also if you choose pi there should be no buckling. I just think about what happens when you take something inscribing. If you'd try to buckle that it should be obvious that you cannot stretch that to make it look like it would have same length than the original approximation. That way ... couldn't you find a minimum possible value that you can choose to still approach the right value? Is that then by logic conclusion the right limit. This now obviously more refers to the 2D circumference approximation, where it can be easier be followed.... Yet if you'd follow this procedure you'd have to combine buckling and stretching to an compensating amount .... something which is not that easy to do, anyway. Dunno... (And similarly, what would be the equivalent on buckling but in the opposite way, is it stretching to be shorter/less surface area then the to approximate surface?) Seems so, that buckling or also stretching have to always be reduced respective compensatingly kept to make this work out (for buckling outside you'd have regions that then stretch to a compensating amount, then you could kind of start with a "right initial guess" and then iterate that to show it will converge.... not sure though at all if that is all possible to do)
    1
  14504. My English teachers always told me that HOMING IN ON a target was like heading home to it/zeroing in on it -- the way a homing pigeon does. They also told me that HONING a blade or a point or an argument meant sharpening it so it will cut better. I'm very confused by your emphasis on wanting to HONE IN ON an irrational number. What are we sharpening it for? Are we out to cut something more effectively?
    1
  14505. Thanks!
    1
  14506. It's fascinating
    1
  14507. Made it to the very end, but to be honest, I only did so because I started reading the comments about half-way through and saw other people talking about reaching the end. That got me curious, and when the credits came up, I waited until the end to see what else was there. 🤓
    1
  14508. 🇨🇿🇨🇿
    1
  14509. The squares question is quite interesting. I used combinatorics, but I have yet to prove my assumption is true, and it always works, but it is trivial I will leave as an exercise. The upright squares are as is. That's simple. As for the tilted squares, it's easier to reduce the case. That is how many squares can be in 2x2, 3x3, 4x4, 5x5 etc. Note that each tilted square is embedded within a slightly larger upright square necessarily since all upright squares have their sides pass through one or more of the points. 2x2 has no tilted, because there are only 4 points. For all other cases, tilted squares cannot have one of its vertices in the "corner" of a grid (proof is trivial :P). For all not corner pieces, a square can be formed among them. In 3x3, there is only one such square. In 4x4, ther are 4 3x3 squares, yield 4 tilted squares. But if we do the edge analysis again, there are now 2 non-corner pieces, hence two more squares. Total 6. We may conclude there are no more than 6 tilted squares in 4x4, because any other tilted square will be either embedded in 4x4 or 3x3---both accounted for. 5x5 same thing. How many embedded in 3x3 (9x1), how many embedded in 4x4 (4x2) and how many 5x5 (3 non corner each side, 3 = 1x3) So for the puzzle question: 1+4+9+16 upright + 9x1 + 4x2 + 1x3 tilted. Total 50.
    1
  14510. We can extend the method if we notice a pattern. Given n x n points 1+4+9+16 +... +(n-1)^2 perfect square number for upright For each upright, there is also going to be (n-2) tilted triangles embedded in the n x n large square...along with all the other embedded in smaller squares. Each only have one less tilted square Tilted 1x(n-2) + 4 x (n-3) + 9 x (n-4) + ... + (n-3)^2 x 1 (n-3, because 3x3 minimum size required) If we group them by the embedded square Number of squares formed within n x n grid = 1(n-1) + 4(n-2) + 9(n-3) + ... + (n-3)^2 (2) + (n-2)^2 + (n-1)^2 There is probably a way to simplify it but I am too lazy to figure it out.
    1
  14511. 15:50 Yeah, I had a real fun time doing the animations for my Trigonometry video.
    1
  14512. I love this. 😊
    1
  14513. I am a subscriber, and I have not a clue what is being said, reason being that there was three things I could not understand at school, that is maffs and Engrish 😊
    1
  14514. Amaze! Thank you!
    1
  14515. I think that Tristan's fractal and the "only 100 zeros" series are the most interesting facts. One would expect the no-nines sum to be much greater, but as n approaches infinity, there are less and less terms to be added compared to the harmonic series. Meanwhile, in the 100-zeroes series, numbers with 100 zeros in them will become more and more prevalent, which is enough to make it greater than the no-nines sum! This really demonstrates why you cant discount unimaginably small numbers when infinity is involved. Great video as always!
    1
  14516. The leaning tower of maths is my favourite. The unexpected longer overhang configurations that are still to be found interests me Ty mathologer. another enjoyable video in these abhorrent times.
    1
  14517. The palm manuscript seems to be in Malayalam script. It is strange none knew about this
    1
  14518. I'm going to point it out again: recent archeological developments indicate that 616 was the original number of the beast in revelations not 666.
    1
  14519.  @Mathologer  Oh, I misheard you the first time. Somehow I heard "now tilt the equator slightly and you have four", or something to that effect -- meaning perturb the great circle very slightly so it now includes the two pigeons on it within the appropriate hemisphere. Incidentally, this does means that for the generic case you can guarantee four pigeons strictly inside a hemisphere (excluding the great circle).
    1
  14520. Great video!!!
    1
  14521. Most memorable: the proof that 1+1/2+1/3+... is infinite.
    1
  14522. Another nice thing about fibo fractions a/b, so that a>b. For both a/b and b/a, let's multiply the denominator by 2. When both a=1 and b=1, both a+2b and b+2 = 3. After that, a+2b -> Lucas numbers and b+2a -> Fibonacci numbers.
    1
  14523. As a musician I work with harmonic series every day. And I'm more than happy to watch videos about true music theory made by Burkard!
    1
  14524. Fascinating!
    1
  14525. 7:40 I prefered the swivel proof. Probably because I've a fondness for math animations.
    1
  14526. 12.99999 “This is clearly 12 and change” -Mathologer
    1
  14527. I really enjoyed the Tower of Lire section, specially since it made me wonder if one could build it and other similar objects with weird centers of gravity using 3D printing.
    1
  14528. r*s = r+s for logarithms of the same base!!
    1
  14529. My favorite is definitely the 700 year old proof
    1
  14530. Seriously miraculous 👍👌👌
    1
  14531. ❤️🙏🏽
    1
  14532. Most memorable: the crazy formulation for the actual largest overhang. It's so beautifully ugly.
    1
  14533. 24:45 That's obvious because the convex curve from corner to corner means that each rectangle marked out by the pattern contains more highlighted space than empty space. This is all very second-nature to me thanks to countless hours spent doing audio synthesis-related things.. the average value of a curve like that becomes relevant in modular synthesis when using curved waveforms for various purposes, and it quickly becomes intuition that a convex curve like that has an average value higher than half its maximum value, while a convex curve has one lower than half its maximum value, and a flat line has an average value exactly half its maximum value. Sidenote: this is probably the least relevant part of this video to audio synthesis by the way, lol.
    1
  14534. Amazing explanation about Rubik's cubes!
    1
  14535. Amazing! Thank you!!!
    1
  14536. While I definitely didn't know for sure, I did assume the top two corners would determine the bottom. So I won bragging rights! Now to watch the rest of the video to learn why.
    1
  14537. I keep revisiting this video because I keep forgetting that I don't know what interest is
    1
  14538. i didn't learn about any of this in school. i of course learned algebra and trigonometry, but never this. thank you for uploading this.
    1
  14539. 23:52 this inequality doesn't imply that harmonic series divergence to infinity. It can still be finite.
    1
  14540. Math
    1
  14541. Lovely video. I actually encountered the Kempner series in university too (not as homework though). My favorites were the fractal and the Kempner-like series at the end.
    1
  14542. Nope. Doesn't work. After the second derivative, the terms start getting bigger and there's no discernible pattern.
    1
  14543. Yes a new video from mathologer ! you never fail to satisfy my mathematical palate
    1
  14544. Probably the best ratio likes/dislikes out there. Congratulation!!!
    1
  14545. Made my day !
    1
  14546. Awesome!! Thank you
    1
  14547. Thank you for your amazing video's! I divided this one over 3 sittings. But your enthusiasm is contagious! I studied mathematics back in the nineties to become a math high school teacher here in the Netherlands but never got to play that role. Instead I landed in a IT job -it had something to do with a rising bubble back then. Nowadays I'm a tour guide in my hometown.
    1
  14548. I have an old circular slide rule, it looks just like that.
    1
  14549. Takes me back to my Numerical Methods class
    1
  14550. wonderful !
    1
  14551. Very nice video really appreciate your hard work and love your way of explanation. Thank you for this video. I am feeling very lucky to learn Euler maclaurin formula being high school student. Thanks again.
    1
  14552. It's eactly the same amount of the three tiles colors because if you think of the hexagon as a 3D box, and you look at it from the top then it will be all one color, then if you look at it from the right side it is also only one color, and from the left side it's also only one color. This means all 3 colors are in exactly the same proportions !
    1
  14553. Is there any relationship between the SOCW and a cycloid?
    1
  14554. It is also funny how infinite is hidden not only in infinite objects. Snowflake fractal with finite area is bounded by an infinite line. Another thought. In this kind of objects with one finite magnitude and anothe infinite. Is it always the dimension of the finite magnitude greater than the dimension of the infinite magnitude? Horn. Infinite = 2d, finite = 3d Snow. Infinite = 1d?, finite = 2d ? Because i remember that line to have another dimension greater than 1 and smaller than 2.
    1
  14555. Thank you, it's simple and interesting, people will enjoy in math more.
    1
  14556. "A book in print, and used, for over a thousand years." Eh, try two thousand years, then you've actually got something rare. (I can think of two such books off the top of my head. One of them is a math text. If for "used" we allow non-technical uses, e.g., in literature classrooms, then the list gets significantly longer.)
    1
  14557. This is a great video, thank you!
    1
  14558. A googol dollars? That is three novemvigintillion dollars for every person in the US!
    1
  14559. ♥️
    1
  14560. Tough call... The swivelling proof is pleasing, but relies on the fact that the number of sides is odd, meaning it's less likely to generalize to other shapes. So the colouring proof seems more likely to be more useful.
    1
  14561. Maybe if all of us pronounce these names incorrectly somewhere in infinite time maybe they will approach the state of correctness. And all languages will become one universal word.
    1
  14562. Quick proof of the phi thing. By definition phi^2 = phi+1 so 2*phi*phi =2*phi^2 = 2+2phi = 2 + phi + phi just like it said.
    1
  14563. Amazing video line ever. Can you please give a list of „m“ and „n“ to exactly reach pi. Maybe there is a pattern. Thanks.
    1
  14564. Amazing
    1
  14565. Hard to pick a favourite moment but the identity for gamma was nice (and close second the QED cat saying "µ").
    1
  14566. Shouldn't you first prove that two circles have centers in one point? Doesn't look obvious to me
    1
  14567. On the first example, it feels like some very critical rules of why it can work as such were not pointed out enough on the presentation... particularly how both the amount of Australians and the amount of hair follicles are both facts ( as in, observed in reality ). Being able to say, "Australians are more in number than the average hair follicles in a particular people goup", "There is no House with number 0" and "Everyone Must enter a house" are the important rules of the underlying measurements that need to be taken into account in order to avoid fallacies ( pun intented ). In the end I do though understand that this is why the whole example is funny :) it tickles the scenario of every arbitary numbers also work the same way while ingiting the curiosity to people to learn more about it.
    1
  14568. COS 120 = -1 ?????????????????????????????????????????????????????????????????????? AT 23:00 errare graphicum est
    1
  14569. I have no background in math beyond some simple applied math which is part of engineering programs. I understand and enjoy every second of your videos. I think your levelled approach toward the topics Is very helpful in particular. I couldn’t stop watching the whole 50 minutes of this video and was thirsty for more when it ended. Finally learnt what Bernoulli numbers are and looking forward to know more.
    1
  14570. The simple statement is true, if any angle is not flat in the limit, the area is always greater than the smooth object, even if it is by an infinitesimal amount, but this is only possible for discontinuous surfaces anyway, because the angle between just two infinitesimal triangles cannot be non flat while all the other are in a sense, this is a broken point if we ever saw one. ^^
    1
  14571. Incredible video, loves the longer format! the more the better :) For those who can't wait here is a link to a paper by Terrence Tao on the connection between Euler MacLaurin, Zeta functiona nd Bernouilli numbers. I found this paper a while ago after watching your excellent vido on zeta(-1). Fasten your mathematics seatbelts for this link. https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ I look forward to watch your next video, thanks a lot!
    1
  14572. Awsome shirt
    1
  14573. What about CNN
    1
  14574. 😳🤯
    1
  14575. I used to have a circular slide rule (1970's) which I used in high school until I was able to save up for one of those fancy new calculators which actually had trig functions! Sadly both were lost in moves. My litle round slide rule worked great, though it was a bit harder to read my less portable slide rule. I've never seen another one.
    1
  14576. not only that I enjoy your vids but your t-shirt logo in this episode is funny AF 😄
    1
  14577. no paradox, just unintuitive to our brains unfamiliar with a world full of spherical cows living in a frictionless voice
    1
  14578. For me, the best lacing is one where you can just pull the top and the whole set of islands tighten together.
    1
  14579. Even i could understand this 👍
    1
  14580. Final puzzle is quite easy. Interpret as a three dimensional stack of boxes, which has the property that there's no hidden boxes. In this situation we can built up from an 'empty room' by adding new boxes to the corner. This always removes 1 orange 1 blue 1 grey and addes 1 orange 1 blue 1 grey because flipping a box from corner to convex is just a 180 degree rotation of that hexagon
    1
  14581. Волшебство математики и ловкость ума.
    1
  14582. I'm in love with maths
    1
  14583. i didn't understand anything... but i will build one of those grinders because it seems quite simple and like a fun thing to have :D
    1
  14584. I wish I could give you more than one likes
    1
  14585. I made it to the very end
    1
  14586. I don't follow "just get rid of the i" 26:25 to turn Euler's formula into the cosh+sinh formula
    1
  14587.  @yinq5384  yep - thanx. i commented while watching.
    1
  14588. That was my math project for graduation 4 years ago !
    1
  14589. Bishop Oresme had the top of his skull removed for better cooling.
    1
  14590. Nice 😎👍🏻
    1
  14591. The funny part is the meme on the shirt
    1
  14592. Penrose Monopoly! Brilliant idea!
    1
  14593. Best part was the representation of Euler's constant
    1
  14594. Yessssssss!
    1
  14595. Most memorable fact: amazing approximation formula.
    1
  14596. Very awesome video and insights. Thank you.
    1
  14597. Let G be the group of linear operators on V which preserve angles (so like, O(V) \times \br^\times ) Ah, actually, let G be the set of all affine transformations that preserve angles. If you have two labeled sets of points from V which are related by an element of G (sending each point from one set to the point in the other set with the same label) then, well, ((A(y) + y) - (A(x) + x)) = (A(y)-A(x)) + (y-x) Err... hold on.. Let A’(v) := A(v) - A(0) A’(y-x) \cdot A’(z-x) = r^2 (y-x)\cdot (z-x) ok, but what about the cross terms? (y-x)\cdot A’(z-x) and A’(y-x) \cdot (z-x) well, the adjoint of A’ will also be a multiple of its inverse.. hm, not sure that helps. Is this specific to 2D somehow? It doesn’t seem like it should be.. ... then again, maybe it is specific to 2D, Because in 2D, and sum of two rotation-and-scaling will be a rotation-and-scaling unless it is zero (because they can be identified with complex numbers) And I’m not sure if that holds true in a greater number of dimensions.. Like, the only division algebras over the reals are like, R, C, H, right? So... hmm
    1
  14598. But isn't the square root of 666 more accurately described as 25.8070?
    1
  14599. First of all: your content is awesome. But today I was tired after work and couldn't help but falling sleep at the middle :( ... The good (funny) thing is I woke up just to receive The Mathologer seal of approval :)
    1
  14600. I was so disappointed when the video ended! I was following you the whole time and was ready to have my mind blown by some more crazy proofs!
    1
  14601.  @Mathologer  I know I can't complain 😂 Your videos are just so engaging, they leave me wanting to learn more. So... mission accomplished!
    1
  14602. Now to solve for twist on round shoelaces (e.g. the bottom lace is pinched between shoe and top lace, so when it's pulled tight a twist is induced). I have this happen on my work boots all the time unless I think about it first when re-lacing.
    1
  14603. haven’t watched the video yet- but i did a bit of math out of curiosity. there are two solutions for x^2 + (x+1)^2 + (x+2)^2 = (x+3)^2 + (x+4)^2 they are x=-2, due to symmetry, and x=10, the thumbnail solution. (the quadratic equation gives x=4+-6)
    1
  14604. A part of the program your looking for has been done by the french youtuber micmath (for the record, his video is named "la puissance organisatrice du hasard") : http://micmaths.com/applis/lgn/diamantazteque.html
    1
  14605. bring back your old masterclass videos, I miss them!
    1
  14606. Im math illiterate but i like these vids, keep am up man.
    1
  14607. Final Puzzle at 36:01, Area of the inside square is a²(1-(ab/(a²+b²))) , where "a" is side of the bigger square, & "b" is the segment cut. You can solve it using similarity of triangles, and ratios of sides.
    1
  14608. This has the same beautiful "there is only one shape here" as the parabola ❤
    1
  14609. what if instead of starting at A + B = C (all cubed) we started at A+B+C=D+E (all cubed)? why and how would this not work? thanks.
    1
  14610. Bizarre maximal overhang towers blew my mind. Amazing video!
    1
  14611. You make maths understandable
    1
  14612. 9:27 It should be V=20, E=30 and F = 12
    1
  14613. It's just like mathematically proving that a round ball is best for playing soccer/football. The conclusion; a match will take very long to finish when you use a cubic ball (which isn't a ball per definition ofcourse) 😉😁
    1
  14614. My vote: Bizarre maximum overhang towers. I was asking myself if you were going to show the optimality and instead this crazy unintuitive solution pops up. How would you even come up with this stuff, let alone show optimality?
    1
  14615. lol my school sucked my gr7 text book didn't even inform us that pythogoras only worked on right angle triangles
    1
  14616. Well boys and girls - Here's a new one for ya - "PapaSans" formula: 2 + 2 = 2 x 2, and it only uses 2 elements instead of three ( and the 2 elements are both the same )
    1
  14617. Amazing video
    1
  14618. Wow I thought you are gone forever! So glad to see your video!
    1
  14619. For me the most memorable is that crazy-looking tower at 9:48.. It is ugly, but I kinda like it :D
    1
  14620. the "omg that's just some linear algebra" moment (when you show it's a sum of 5 different pentagones )
    1
  14621.  @Mathologer  indeed !
    1
  14622. Great 👍
    1
  14623. Amazing... 😇
    1
  14624. As usual, watched the whole video. understood most of it, for sure, I'm gonna watch it few more times. I have no math background. waiting for the next video.
    1
  14625. The diameter of the circle is larger, because the other lines are non-diameter chords of the same circle.
    1
  14626. I just learned this from you .
    1
  14627. 20:42 I was hoping that you'd show a "smooth" golden spiral, one that goes inside the 3D shape, and each segment of the spiral would connect the 2 furthest vertices of the rectangular prisms. 34:52 I wonder how such quadruples would look like for 4D boxes. Then a general case for n dimensions.
    1
  14628.  @Mathologer  Ah cool I didn't see that before :) I also see it "only" animates the equality part for circular quadrilaterals so I guess my version is pretty much a generalized version that also covers the inequality part. And the animation is based on work from 2012? It's always so interesting how we constantly find new beautiful proofs of things that has been known for centuries
    1
  14629.  @Mathologer  I sent you one :)
    1
  14630. i got all i needed in this video that took several days in school.... tell me again why our education system is fine the way it is?
    1
  14631. Great videos thanks. How do you make your animations :)
    1
  14632. Every one get heart tab😍.
    1
  14633. 13:06 ℝeal mathematical magic 😅
    1
  14634. So cool!
    1
  14635. Handsome man. Marty also handsome man
    1
  14636.  @Mathologer  I might have watched some time ago. I had previously seen it on Cut the Knot.
    1
  14637. I made it to the very end =).
    1
  14638. Gamma is the most memorable! Can't beat getting familiar with a named constant
    1
  14639. 3:30 you forgot to mark as different the 2 and 1 in the 53rd decimal position.
    1
  14640. But did you prove that when you fold, you actually get a triangle?
    1
  14641. No, but it's not hard - the corners of the two halves meet at the same point since they're both half the same side of the original triangle away from the hinge, and the three corners of the "triangle" you end up with are made up of the same four angles as the triangle a third of the original triangle that they were folded from, so they still sum to 180 degrees, so form a triangle.
    1
  14642. I’m still trying to work out what he’s doing with his hands. He sounds like Bruno.
    1
  14643. Anyone got the STL for the Mathologer seal of approval that I earned?
    1
  14644. challenge 4: 9*66=594? This is a hard one, I tried seperating into cases
    1
  14645. I retried, wow its 666😂😂 This number for Christmas, wow😈😈
    1
  14646. First we noticed that 20-4 is will not fit a domino in any case, so the 6 piece at the edge can be seperated. That gives us 3*3=9 tilings at the edge. Then at the 'hoop' at the middle, the tilings can only be 17-7,16-6,8-13,15-10 leaving 9-14 to the middle OR 13-9,14-10,17-8,6-15 leaving 7-16 to the edge. Now by seperating case of the 'C shape' to towards left or towards right, we have: 2*2*5+2*(2*3*3)+2*3*3=74 Total tilings=9*74=666 !! Thanks Mathologer for the numberings😉😉
    1
  14647. My favorite part was probably the optimal stacking visual at the beginning (hard choice, though). 🙂
    1
  14648. As a student of PCM and Statistics, I knew both Interpolation (which deals with difference) and Differentiation. 😎😎 I guess there won't be many students of Statistics here.
    1
  14649. The best way to lace shoes is to actually not get lace shoes, but to get slip-ons. PROOF: Time to Put on Tie Shoes: >10 seconds Time to Put on Slip-On Shoes: <10 seconds
    1
  14650. Surely the sum of the no 1’s harmonic series should be zero owing to the fact that each term features a 1 in the numerator ...
    1
  14651. I tried using a direct proof with complex numbers, however I am grateful there exists such an elegant proof with the decomposition. The fact that all the intermediate steps can be completed in any order makes me think a direct proof would not be nice now.
    1
  14652. Incredible work like always 👍👍
    1
  14653. Why is it impossible to stack 1,2,3,4...n,n-1 bricks, just like in a brick laying pattern ? wouldn't it be more optimal than the cubic pattern ?
    1
  14654. Kudos to the people who answered all questions in this video
    1
  14655. As requested: I made it all the way to the end! I'm not clever, only completionist :-).
    1
  14656. The solution to the second problem is two. We know that 10²+11²+12²=13²+14². Hence we can say the numerator is simply 2(13²+14²). Their squares add up to 365 and leaves us with the solution of two since the denominator has 365
    1
  14657. Is the T-shirt an extended rock paper scissors?
    1
  14658. Firsttt
    1
  14659. seems like a fun puzzle game
    1
  14660. The answer that I see coming next is 2: that's none-repeated digits of pi😜
    1
  14661. ​ @Mathologer Well, that's: 3+1=4 1+4=5 4+5=9 5+9=14🙃
    1
  14662. I will take care of this after I finished my diy volcano in the backyard.😁
    1
  14663. Undergrad. I Followed it all. Not sure how deeply though.
    1
  14664. Ohh, wait, I don't need a series with negatve terms. All I have to do is skip all the terms that, when added, go over pi. Oh, better yet, I can do that with the series sum(nsubi × 10^-m), where m € (0,N) and nsubi is the value of the ith digit of pi.
    1
  14665. ​ @Mathologer  😁 Just poking a little bit of fun at you. Goryus, in another thread pointed out the issue that I was alluding to, and I loved your reply. By the way, I do appreciate your videos - they have been most insightful at times. Edited to add: My favourite elements of your videos are the visualisations. The make the concepts so clear. I recommend them to all students.
    1
  14666. Really wonderful video! I am only an electrical engineer with a physics background so have studied math for three years in college, but have never come across these higher power sums and the formulae for them. I always love the visual approach you take to proving these concepts and find them very educational and inspiring. Please keep doing this for the rest of us non-mathematicians who still love math.
    1
  14667. My approach to prove the 3rd challenge more mathematically: If we consider to enumerate every tile, black and green from left to right, with 1 to n, so that every number is stacked over itself, our corresponding matrix will consist of i's in the main diagonal and 1's on the secondary diagonals. Every renumbering is just a permutation, which affects only the sign of the determinant and thus is irrelevant. If we use the Laplace's expansion wisely to calculate the determinant, we will come to the following by expanding along the first column: (The second summand has to be expanded one more time to get to A_(n-2), but if you write it down it gets obvious.) det(A_n) = i*det(A_(n-1)) - det(A_(n-2)) Above formula can also be written as (factor i out with i^2 = -1 in mind): det(A_n) = i*(det(A_(n-1)) + i*det(A_(n-2))) Since we're only interested in the volume of each matrix itself, which is a real positive number, we can get rid of the i's and therefore we get: det(A_n) = det(A_(n-1)) + det(A_(n-2)) Tadaa :-)
    1
  14668. I agree with Philosophy of Horror. Challenge answer: 153^2+104^2=185^2 => top row: 9,4 bottom row: 17,13
    1
  14669. The reason because Euler-Mascheroni constant is clearly greater than 1/2 is that each of the blue slice (see 24:40 for drawing) can be seen as sitting inside of a rectangle of sides 1 and 1/n, and the blue part fills more than half the rectangle because convexity. Since the rectangles cover the square, the blue slices cover more than half the square!
    1
  14670. 😶 hi guys
    1
  14671. the amazonians did it first
    1
  14672. You prettified the red-blue-yellow diagrams, inspired by the paintings of Piet Mondrian. Give credit where credit is due.
    1
  14673. I loved ALL OF THEM !!! :)
    1
  14674. this is a favourite subject of mine a few months ago i figured out how show via hyperbolic rotations & trigonometry that the area under 1/x was logarithmic, which felt quite gratifying i didn't write any of it down, though, because my mental health is quite bad lol
    1
  14675. Can this work in higher dimensions? O_o
    1
  14676. 24:34 The 1/x curve is concave, so if we draw the diagonals of the blue bits, the halves above each diagonal will be completely blue, while the halves below each diagonal will not be completely white
    1
  14677. hello from greece what a nice and honest mathematical experience love
    1
  14678. New year just got a little happier with a Mathologer video
    1
  14679. My head hurts.
    1
  14680. Don't understand any of it but I do find it fascinating. Enjoy your videos.
    1
  14681. Love you mathologer Please make more videos like this
    1
  14682. Eres el mejor! felicitaciones
    1
  14683. no, two of each type of square removed is not always tileable.
    1
  14684. Those first 3 and a half minutes were like 2007 YouTube lol
    1
  14685. Nice t-shirt
    1
  14686. Cool that i had such idea before it was in math. A space where you can actually move by expanding matter around your and at the same time staying with integrity of linear space. Its really beatifull.
    1
  14687. Nothing to do with heliconia, Mount Helicon or Helliconia Winter, presumably - though I could not help thinking of the rotating prison described in the latter!
    1
  14688. I wasnt familiar with any of them, but I saw something similar in a math test I took some weeks ago
    1
  14689. Aww, my other, imo, much more interesting comment didn't show up in your comments collection. (That said, it was only an aside to the video, more of a phi-fact than anything) Apparently, all such Möbius transforms of φ, (p φ+ q)/(r φ+ s) such that | p s - r q | = 1 and p, q, r, s are integers, will share the property with φ, that they cover an interval as evenly as possible. Now, (p φ+ q)/(r φ+ s) can be simplified: (Throughout this I'm using the fact that 1/φ= φ-1 and φ² = φ+ 1 because phi is the larger root of x² - x - 1 = 0) 1/(r φ + s) = (r / φ - s) / ((r φ + s) (r / φ - s)) = (r (φ - 1) - s) / (r² + r s / φ - r s φ - s²) = (r φ - r - s) / (r² - s² + r s ( 1 / φ - φ)) = (r φ - r - s) / (r² - s² + r s (φ - 1 - φ)) = (r φ - r - s) / (r² - r s - s²) So any inversion involving phi can be simplified in this way. And if you multiply two such values (a φ + b)(c φ + d) = (a c φ² + a d φ + b c φ + b d) = (a c (φ + 1) + (a d + b c) φ + b d) = (a c + a d + b c) φ + a c + b d So in particular, (p φ + q) (r φ + (- r - s)) = (p r - p (r + s) + q r) φ + p r - q (r + s) And therefore, any such Möbius Transform can be reduced to ((p s - q r) φ - p r + q (r + s))/(s² + r s - r²) but since we have the condition | p s - r q | = 1, this can be simplified further to (p φ + q)/(r φ + s) = (p r + q (r + s) ± φ) / (r² - r s - s²) and since all of these are integers, I think this condition really says (p φ + q)/(r φ + s) = (a ± φ) / b where a, b are once again integers. I'm not sure whether a and b can just be any integers, rather than just certain ones, but currently I believe so. If that were the case, even the ± can be dropped, as negative a and b are gonna cover those cases, so we just arrive at (p φ + q)/(r φ + s) = (φ + a) / b where p, q, r, s ∈ ℤ and | p s - q r | = 1 which happens to be precisely the sorts of numbers you encounter in a Helicone with a variety of leaf counts.
    1
  14690. ​ @Mathologer  I don't have a final reference, it was a statement hidden in the Extreme Learning blog post "The Unreasonable Effectiveness of Quasirandom Sequences" by Martin Roberts "It is interesting to note that there are an infinite number of other values of that also achieve optimal discrepancy, and they are all related via the Moebius transformation" (I can't link it because YouTube, so just look it up, should be the first result in search)
    1
  14691. I always called it sinch
    1
  14692. Isn't this similar to Euler's Approximations idea...
    1
  14693. I feel like, some day, Matholer will stumble onto some new revolutionary branch of mathematics and he will not even realize it
    1
  14694. Swivel proof is more “Aha”. I like the coloring proof better.
    1
  14695. 11:32 RRRL gives 9,4,13,17. 1 1 -> 1 2 -> 1 3 -> 1 4 -> 9 4 3 2 -> 5 3 -> 7 4 -> 9 5 -> 17 13 9*17 = 153, 2*4*13 = 104, 9*13+4*17 = 185
    1
  14696. Awesome. I always love your explanations. It took me a few minutes to get your T-shirt but very inventive.
    1
  14697. From the harmonic series, add two, subtract one, add three, subtract one, add five, subtract one, add seven, subtract one, add eleven,.. etc (the primes!). Thus: 1 + 1/2 - 1/3 + 1/4 + 1/5 + 1/6 - 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 - 1/13 + .... Does this converge and if so, to what?
    1
  14698.  @Mathologer  Yippie you reacted <3. And yeah, it's probably obvious that it does. I was hoping to squeeze in the primes and get a non-diverging sequence. Perhaps alternate adding p_i terms and then subtracting p_{i+1} etc, alternating adding and subtracting the next prime number terms? I'll give that a try, was writing a program that generates a lot of prime numbers first :/.
    1
  14699. Long ago people would read books to entertain
    1
  14700. I am confused how this works for a 2n-gon; your visualization with a decagon was too busy for me to make out the details. In particular, the 180° ears should just be bisecting the lines, right? What does that look like in decomposed form, five points all at one location (x, y) connected to five points all at (-x, -y), with each edge connecting one of the first set of points to one of the second? Actually I think that makes perfect sense now that I write it out, if I’m correct there. Skipping those degenerate ears would, if I’m correct, result in a decagon where alternating points tend to zig-zag back and forth in a way that’s completely smoothed out by bisecting the edges, which makes sense.
    1
  14701. Could you talk about prime number
    1
  14702. most memorable: the no 9s black and white visualization.
    1
  14703. what am i doing here? i have my physics exam tomorrow and its just for minors
    1
  14704. OMG this is math for math sake. I just love pure math. Thank you Mathologer
    1
  14705. I'll give my attempt at the similar triangles proof. It's not as concise as ones others have posted, but it may help in visualising the geometry. The two given similar triangles are A₁B₁C₁ and A₂B₂C₂ labelled so that vertices with the same letter have equal angles. We translate ΔA₁B₁C₁ by adding a complex number z₁ to each vertex, such that A₁ moves to the Origin. Then we do the same to ΔA₂B₂C₂ adding z₂ to each vertex and so moving A₂ to the Origin. Let θ be the angle subtended by A₁B₁ and A₂B₂. Since ∠B₁A₁C₁ = ∠B₂A₂C₂ then A₁C₁ and A₂C₂ also subtend θ. Let A₃ = A₁ + A₂, B₃ = B₁ + B₂, C₃ = C₁ + C₂. Clearly A₃ is coincident with A₁ and A₂ at the Origin. Also B₁B₃ makes an angle of θ with A₁B₁ produced. ∴ ∠A₁B₁B₃ = 180° - θ Similarly ∠A₁C₁C₃ = 180° - θ By the given similarity of ΔA₁B₁C₁ to ΔA₂B₂C₂ we have |A₁B₁|/|A₁C₁| = |A₂B₂|/|A₂C₂| = k (say) But |B₁B₃| = |A₂B₂| and |C₁C₃| = |A₂C₂| by simple vector geometry, ∴ |A₁B₁|/|A₁C₁| = |B₁B₃|/|C₁C₃| = k Therefore since ∠A₁B₁B₃ = ∠A₁C₁C₃ it follows that ΔA₁B₁B₃ is similar to ΔA₁C₁C₃. ∴ |A₁B₃|/|A₁C₃| = k ⇒ |A₃B₃|/|A₃C₃| = k ... ① (since A₃ = A₁) and ∠B₁A₁B₃ = ∠C₁A₁C₃ ∴ ∠B₃A₃C₃ = ∠B₁A₁C₁ ... ② From ① and ② it follows that ΔA₃B₃C₃ is similar to ΔA₁B₁C₁ (and therefore also to ΔA₂B₂C₂). Finally, translating ΔA₃B₃C₃ "back" by subtracting (z₁ + z₂) from its vertices, it becomes the triangle obtained by pairwise adding the vertices of ΔA₁B₁C₁ and ΔA₂B₂C₂ in their original positions. ∎
    1
  14706. For those who speak German, there's also Professor Weitz' "almost Christmas lecture" on the topic from 2017: https://www.youtube.com/watch?v=Vv3Rve3yXBY
    1
  14707. this ultimately disproves the 4th dimension!
    1
  14708. Really enjoyed that one, thankyou.
    1
  14709. This is great. You can feel the love for mathematics. I dream about a video like this on monstrous moonshine.
    1
  14710. For the 4 bugs, Imagine you are where one of the bugs are, the bug in front of you always moves at 90° to your sight line, so its movement in a way doesn't change it's distance to you. The only thing that changes the distance b/n you and the bug in front is your movement. So from your perspective it looks as if the bug was stationary and you moved all the way to it at a steady speed.
    1
  14711. I left an svg as a proposed means of showing animation and speed variations to cover the loops, lengths and directions instead of using colors and arrows. I don't see my svg any more, maybe a bot took it down since it looked like html. Anyways I implmented the 240 7417 using that principle. I am using excel to generate the svg code then pasting it to a notepad to save it as svg format. Anyways I had to slow time time down from 2s to 60s for the animation on that one as my computer was not able to keep up with the number of calculations needed to implement that animation. It is a really neat design to see animated though.
    1
  14712. notice he is staying away from exception , gas , dot , 1d , 2d
    1
  14713.  @Mathologer  as i wrote . 1 cloud plus one cloud etc
    1
  14714. 1+2+3 = 6 The 1 contributes to the answer. 1x2x3=6. So does 2x3=6 Here the 1 does not contribute to the answer.
    1
  14715. wow very brave
    1
  14716. Most memorable: That ɣ has to be smaller than 1, because it is not very intuitive.
    1
  14717. I always thought it seems arbitrary to start the Fibonacci sequence with two 1's. It would be more natural to start with 01, which leads to the same sequence.
    1
  14718. 0 is the 0th term in the Fibonacci sequence, while 1 and 1 and the 1st and 2nd
    1
  14719. Wow😍😍😍so intresting🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵
    1
  14720. I find myself revisiting your hugely helpful references again. Kudos to Team Mathologer!
    1
  14721. colouring
    1
  14722. Most memorable was showing the value of the no-nines sum. As soon as you said it converged, I wanted to know what it converged to.
    1
  14723. So, basically, if you try machining a cylindrical surface by grinding smaller and smaller triangles, you will need an infinite amount of paint to cover it... In my humble opinion, approximation of "curvy curves" works because it happens on a flat plane. If you curve it into a convex or concave surface, seemingly "absurd" things may happen.
    1
  14724. Thanks for bringing the works of Madhava to the world, greatly appreciate. Would like to see more of Pingal and Aryabhatta too. Thanks 🙏🏻
    1
  14725. Why gamma is "obviously" larger than 0.5, each blue piece filling the square is like a fattened triangle, if they were all triangles then gamma would be exactly 0.5, since they are all curving outward a little bit (because 1/x is concave up), then gamma has to be larger than 0.5
    1
  14726. That is exactly what I thought. We see Fibonacci numbers, and therefore φ, when things are packed in a 2D circle like a sunflower. Also the 2D surface of a pine cone or pineapple, which are not so easy to describe precisely. So we might expect to see ρ and σ when things are packed in a 3D ball. A 3D analogue to sunflower seeds might be cells in an embryo, specifically in animals with little yolk such as therian mammals, where the fertilised egg is more or less symmetrical. But I don't think that does occur in practice. In the early stages, cells divide simultaneously so the total is a power of 2, and later embryos have differentiated into cells of different shapes and sizes. Crystals are the most familiar form of sphere packing in nature, but if they use ratios of 1.80 and 2.25 I've never heard of it. The next obvious place to look is quasicrystals, but I see Burkard thought of that already. I also thought about crystals inside geodes, but they aren't usually close to spherical. And they form from the outside in, so the process happens on a 2D surface, not throughut a volume.
    1
  14727. Most memorable thing for me is the optimal but not "beautiful" towers. I like optimizations, and the complexity of it looks interesting to me.
    1
  14728. Linear slide rules only became popular in the US after Sputnik kickstarted the US space program (probably due to all the European scientists that came to the US after WWII). Before that, scientists and engineers in the US almost always used circular slide rules. Of course, slide rules only give 2-4 significant digits of precision; if you needed more than that, you would use Napier's Bones.
    1
  14729. A PROOF! AAAH! Oh, wait, that's why I came here. So basically, the weight of target squares goes up differently from triangle tip squares the further you get from the line. Those dratted duplicated ones. It has to, otherwise, 1-4 wouldn't work. I would sometimes pay the Triangle version you sometimes get in the States. I have solved it, but it's one of those situations where you have to think strangely and I can never hold onto it.
    1
  14730. I’m leaving my comment here to help out Mathologer :)
    1
  14731. This is amazing
    1
  14732. Excellent video, I love the visuals & extending the tower of hanoi puzzle beyond what's typically talked about
    1
  14733. no. of tiling in 2xm is in Fibonacci series.
    1
  14734. since we are to explore this incredible rich gold mine of mathematical miracles, we may need to find the mathematical version of the one piece. all you need to do is to sail this sea of knowledge and find it.
    1
  14735.  @Mathologer  somebody had to be the first. after all... math is also a unique treasure. although I wasn't sure about making the reference since math isn't one piece. it's many many many pieces that nobody will ever gather. that being said, the goal isn't finding the treasure. it is enjoying the search with your nakama.
    1
  14736. They are not at all "weird". π was invented to make formulas for the circumference and area of a circle. Is it then any surprise that it also is in the formulas for the area and volume of a sphere?! None at all!
    1
  14737. It should be clarified that perimeter is a shortest path, specifically for such math theoretical evaluations. Same for area and so on.
    1
  14738. It’s obvious that the area of the blue region is greater than 0.5, since the blue regions are touching all of the right hand side. Thus, If we replaced the curves with triangles, we would have triangles with an area which sums to 0.5. Since the curves are convex, their area is greater than the area of the triangles. Transitively, the blue region has an area greater than 0.5.
    1
  14739. all these paradoxes present in math lead me to believe that someone (maybe more than one. hell, that's likely) that a lot of people think was really smart really wasn't and overthunk their whole schtick
    1
  14740. Pretty cool how something this crazy is understandable with just a little bit of calculus and a couple of prior results.
    1
  14741. 7:33 "Why do you think nobody took notice?" People have a limited cognitive budget to spend on exploration. Therefore heuristics are used to assess the difficulty of analysis weighed against the personal benefit to budget the blood sugar and pure time input required. The cognitive cost versus benefit is estimated within a very short period. We all have to judge a book by its cover to some limited extent or the infinite noise produced by mankind would strip us of all our personal time and trap us on a path towards an unguided or random (unthinking) allocation of time and effort. This is also why orthodoxy always begins to dominate because all things outside of orthodoxy look to be unworthy of the time input in the eyes of that necessary heuristic we must apply to be reasonable humans. The only answer is greater risk of assessing noise (listening to nonsense and word salad occasionally), making mistakes (acting upon knowledge with lower probability of truth), and wasting some time.(entertaining unlikely ideas long enough to adequately test them)
    1
  14742. Another Mathologer Masterclass!!! 👏👏👏
    1
  14743. 48:24 Who was saying 2 4 3 1 there? It did not sound like Mathologer. :)
    1
  14744. Next 'super-pattern' is Prime number for N. 😅
    1
  14745. Just beautiful ! Thanks
    1
  14746. Y so smart 🧠
    1
  14747. ​ @Mathologer  lots of tiny LEDs in the modern ones
    1
  14748. I doubt that, increased complexity of the PCB would probably negate any cost savings. And brightness isn't really a limiting factor anymore. The LED max brightness cheaply achievable today is beyond blinding.
    1
  14749. Aha moment is strong here 😮
    1
  14750. Made it to the very end...
    1
  14751. Many people know that any signal can be decomposed into a sum of sinusoids, but I found it surprising that you can use almost any arbitrary function instead of sinusoids. I did some animations based on this idea 2 years ago: https://youtu.be/nwAKu01ESlY?si=_6At2AotIxQ9HTNY&t=118 It would be interesting to know what functions are not valid as a basis. In general, the characteristic matrix for the linear system must be invertible, but I'm curious if there is a more intuitive rule. For example: - One criteria was that the basis function must have its center of mass equal to zero. - Another is that its first fourier frequency must be non zero. It also can't be equal to its negative counterpart.
    1
  14752. is this grothendieck
    1
  14753. Such a fulfilling video!
    1
  14754.  @hedgechasing  I was wondering if there was a connection there! Thanks for sharing
    1
  14755. The most memorable part for me is how the area under the logarithm curve approximates the sum of the reciprocals, and that area undershoots the actual sum by exactly gamma, the Euler-Mascheroni constant.
    1
  14756. It's immediately apparent when you just think for 1 moment about that horn. It soon becomes a very, very, very narrow indeed. So you'd have to spread out the paint in an unimaginably thin layer. No paradox at all. Yes, you can indeed paint an infinite area if you can paint on the paint infinitely thin.
    1
  14757. Hello, freshly out-of-college and eager-to-pass -my-knowledge-further teacher here. This is the first time in my 25 years of math-loving that I've found logarithms beautiful. Thanks so much! :)
    1
  14758. I thought, I thought I watched all of these, I must have missed it. Wow, not many views. Oh! Sweet!
    1
  14759. Where can we get your book? Love the video's!
    1
  14760. Whole time I was thinking of friends regrouping, from random direction in a jungle with out gps.😂
    1
  14761. It is SO gratifying to see this topic tackled here. I did some work on it a few years ago, so I can answer a few questions and add something else. You asked if anyone had written some code to make a diagram, and I have something rudimentary that only calculates values for me to calculate by hand. Here is my spreadsheet I used to make a pentagram version: https://docs.google.com/spreadsheets/d/1VpLl383Y46w9LQkqwtimhXY5OoftPDnyQ0ipfadTvh4/edit?usp=sharing You used a digital root property to explain why 1-2-5-8-7-5-2-1 repeats, but I found something else: you already know that the highest digit in base b, (b-1), is key to the vortex. Specifically, it must be an odd square*. But it will also turn around and loop symmetrically if the *second-highest digit is the digital root of a power of two (if you're doubling). Base 10 has this property, as 9 is an odd square and 8 is a power of 2; the next base with this property is 26, since 25 is an odd square and 24 is the digital root of a power of two (in base 26). There is a video on my channel in which I show all my work on screen, if you'd like to check it out. https://youtu.be/jWHsJNm-gUM I didn't encounter these multiplication/divisibility rules until I was a math teacher, looking for shortcuts in the times tables to help my students. I'm very happy to see a channel that inspired me come around to a topic I never would have investigated without your inspiration. Cheers!
    1
  14762. I'm not sure it's the most memorable part, but the one I find most interesting is the greedy algorithm to obtain any real (calculable?) number. I think it looks a bit like Riemann's rearranging theorem, and I wonder how close the two are to each other.
    1
  14763. Looks like you can fill any aztec diamond with all horizontal or verical dominoes and the method for deriving how many tilings there are doesnt account for that
    1
  14764. First
    1
  14765. Clever
    1
  14766. That was really cool! I was talking to my dad today about slide rules. My grandfather used one (he was an engineer). I think I wanna try out some of those apps and have a play once the academic year is over. I'm a maths student and I've known the log laws for years, but I've never seen a proof of any of them.
    1
  14767. Math is cool
    1
  14768. I liked the proof of no integer partial sums! Although that seems intuitive, I appreciate seeing the reasoning behind why that fact is true.
    1
  14769. What was the music at 18:11??
    1
  14770. Industrial Engineering. Wow, amazing video. The effort put on it is just outstanding The pace for explaining is just spot on! I just missed that you didn't show what the actual Bernoulli numbers are (as a formula) and an actual final formula for S(n) = ... ! I would have loved to see how the binomial coefficients of pascal's triangle affect that formula. That would have been the cherry on top of the video! The Euler Mclaurin formula followed very nice but it looked so far from the main topic which was the sum of powers. Also, the end was not about the sum of powers but pi squared over 6., which feels weird That is my constructive opinion! Just words of admiration for your channel and I can't wait for the next ones in the series! (no pun intended, well maybe a little) PS: No more homeworkkkkk
    1
  14771. I loved the hexagonal tilings. This is yet another great video. Well done and thank you.
    1
  14772. oh wow, you can literally add, subtract, multiply and divide, all with a set of triangles!
    1
  14773. In Québec, we teach Heron's formula in secondary school
    1
  14774. pi now equals 3.18
    1
  14775. 2:30 where is Tasmania :(
    1
  14776. I barely got started with a Mickey Mouse slide rule for addition and multiplication in about 2nd grade, then we moved on to the ti-1000 which the school locked in a closet except for calculator training a few times a month.
    1
  14777. Excellent 👌 top Quality , smart guy
    1
  14778. I just watched a math professor talking about shoelaces... not a bad binge watch.
    1
  14779. I don't understand why only odd numbers of solutions qualify.. Of course I see that if y and z are different, they can be switched and form a pair. But what if there are 2 different solutions in which y=z (with x being different in both cases of course)? In that case you would still end up with an even number of solutions?
    1
  14780. Impressive collection of Rubik's everything.
    1
  14781. Nice video and explanation. Just a very small nitpick: at 19:00 it would be better to say that the condition for the rearrangement theorem is that both the positive and negative sums are infinity rather than infinite. Being infinite can just mean that the sum never ends.
    1
  14782. This really shows how line art inspires mathematics
    1
  14783. I don't trust the volume calculation technique. You must prove that the volume of the smaller cylinders is evenly distributed or you'll get a wrong result.
    1
  14784. GREAT ! i LIKE IT. NOW PLEASE ANSWER ME THIS QUESTION. HOW DO YOU FIND THE CIRCUMFERENCE OF A CIRCLE OR A SPHERE WITHOUT FIRST KNOWING EITHER ITS RADIUS OR ITS DIAMETER? i WILL LEAVE IT AT THAT FOR NOW.
    1
  14785. Bad aim astonished me
    1
  14786. Do you have any mega favourite number? #megaFavNumber
    1
  14787.  @Mathologer  Actually pretty good for someone from Australia! Amazing explanation as always, btw
    1
  14788. Simply beautiful!!! 😊
    1
  14789. Those correction terms look like they might involve squares
    1
  14790.  @Mathologer  hah 😁 I just assumed I had to be wrong 🤣
    1
  14791. At 35:36. Isn't that a Pascals triangle? And the summation before that, could that be a Pascals pyramid with a triangle base?
    1
  14792. This is what UAP’s probably are 😂 phenomenal channel btw thank you! 🙏🏽
    1
  14793. Does "almost prime" mean that they're either prime powers or square-free?
    1
  14794. Here is trick using a slide rule for a harmonic analysis. Back in the day we would measure the freqency of rotation using a strobe lamp. When teh flashing lamp made ot object stop take down the number. There will be a set of these numbers corresponding t harmonics of the base frequency. Mark them oin a scale with a removable marker, a pencit or erasable market. Now slide the corresponding scale into the marks line up with integers. If you got a good set, the fundamental with align with one.
    1
  14795. Very fascinating indeed.
    1
  14796. 17:34 Area of triangles are 1/2b*h. Without considering the Pythagorean theorem to get sqrt(3)/2 scale lets just call the height of triangle F, h[F] and of triangle A, h[A] (letting triangle A be a scalar of magnitude A of triangle F). This means h[A]=A*h[F]. The area of F = 1/2*1*h[F]. The area of the triangle in 17:34 is 1/2*B*h[A}-> 1/2*B*A*h[F].Since 1/2*h[F]=F. The area is ABF.
    1
  14797. The most suprising to me is fact that -1/12 is gamma of whole number series.
    1
  14798. ​ @Mathologer  I edited my comment and added a famous (edited) quote. Sorry.
    1
  14799. Brilliant explanation, so simple and effective. The first one is more awesome.
    1
  14800. All parts were great, but the visual intuition in Tristan’s fractal was so striking! Could you do this with other infinite series, I wonder?
    1
  14801. 3:36 for a split second i got a "prime number" alert
    1
  14802. Why am I watching this I suck at math? 🤣
    1
  14803. I think what I'm really going to take away from this video is that mathematicians have extremely loose standards of what can be called a "paradox".
    1
  14804. 314 😝
    1
  14805. 3:05 i learned this in math class in 7th or 8th grade, and it was an aha moment, but the real shock was that one of my classmates figured it out by himself, and that made me wanna dive into math!
    1
  14806. When I was at college, out of a class of 30 only two people had electronic calculators.That's because they were extremely expensive at the time, the rest of us used slide rules.
    1
  14807. I watch a chapter per day.
    1
  14808. Awesome
    1
  14809. Santa is early this year
    1
  14810. You can do the same trick with the hypotenuse if a right angle triangle.
    1
  14811. I made it all the way to the very end. Moreover, I happens to know Tamás Fleiner from our university years in Hungary. :)
    1
  14812. Has anyone attempted playing around with using Moessner's method in conjunction with primes: I did a google sheet mock up, but i'm not clever enough to notice anything other than primes being their mostly unpredictable selves.
    1
  14813. My mind is blown! Love your work!
    1
  14814. The maximum overhang is the most amazing part for me. Thank you a lot for the excellent work you do.
    1
  14815. Take the midpoint of each line - connect those for a… hey, you stole my idea! -.-
    1
  14816. Wonderful video as always! The 3D-to-2D continuity argument is brilliant! 10:42 Let θ=pi/7, the identity ⟺ 1/sin3θ+1/sin2θ=1/sinθ ⟺ sinθsin2θ+sinθsin3θ=sin2θsin3θ ⟺ (cosθ-cos3θ)+(cos2θ-cos4θ)=cosθ-cos5θ ⟺ cos2θ+cos5θ=cos3θ+cos4θ ⟺ 0=0. P.S: Not long after I wrote down the proof above, I saw the relation between the trig identities and Ptolemy's thm in the video. :P 13:30 We can move the top vertex to be antipodal to the bottom-left vertex, and the angle remains α+β. Thus the chord's length is sin(α+β). 19:50 We can consider the two smallest equal angles to be negative, then the red angle is still the "sum" of two angles and the argument works as before. 28:12 They fit together because the three angles at the vertex "inside" Aa, Bb and Cc are exactly the same as those at the pink vertex (from 27:52).
    1
  14817. I instantly thought of that solution for the sums of positive integers. Very similar to the power set formula and concept.
    1
  14818. You have the best shirts.
    1
  14819. Anarchy
    1
  14820. This time I stayed until the end! And it was great thanks
    1
  14821. Make my day: power-day!
    1
  14822. Noseferatu, the Jewish Vampire! Nice axioms. What does this mean? 4D is 2D^2. By iterating in a circular pattern, the embedder is NOT binary. A coherent 5-6D dimensionality, bruh. 5 for a point on the sphere NOT the pole, in matter monopole. A God's eye view is 6. Is that the delta from Navier-Stokes?
    1
  14823. I quite liked the gamma insight
    1
  14824. that dimer sampling trick is neat!
    1
  14825. The swivel proof is more visually striking and easier to follow, but I have no idea how I'd even start formalizing this so I don't have to wonder whether it's just a little bit off and I just don't see it. The color section proof in comparison seems pretty easy to formalize without worrying about my visual acuity. After you said your bit about the swivel proof: Okay, thinking about it a bit more it's pretty obvious that the swivel proof is also pretty easy to formalize.
    1
  14826. What a great video! I'm third-year theoretical physics student, never faced Bernoulli numbers, so this video partially worked for me as an introduction for Bernoulli's numbers and stuff. It would be great to see more magic behind those numbers as now I've got some questions about them :-)
    1
  14827. #666: 11393868451739000294452939
    1
  14828. I made it to the very end
    1
  14829. This video deserves not three, but four "extras", because it contains a hidden hint to flatearthers at 20:53 ;)
    1
  14830. Meanwhile, as the rubber band is being constantly wound onto the wheel, an ant is desperately trying to escape by walking at a constant velocity along the band…
    1
  14831. Is there a way to do adding and subtracting on a slide rule? Maybe add a+b by multiplying a(1+b/a)... So you would need to add log(a)+log(1+b/a). Hmmm....
    1
  14832. I learned both proofs at the beginning of this video decades ago. I don't remember from whom or from what book. When I taught geometry, I taught the first proof by actually giving the students construction paper, rulers, and scissors, and having them move the triangles around themselves.
    1
  14833. Experience wise, I've only gone as far as some college calculus. I followed fairly well until the last 7 minutes or so. I think you expect an intuitive leap to (n(n+1))/2 when you say "that some will recognize" where a simple illustration (ie the "folded number line") might be more approachable to aspirant mathologers. I think it might also have been worth noting, when introducing the bernoulli numbers, that the denominator appeared to be the LCD of the preceding terms of each expression. The real strength in these master classes is that, in each section, I see new ways of looking at things I've come to see as arbitrarily true. This, thanks to the clear linear progression of ideas from surface to depth.
    1
  14834. Noticed the 9 rule at school, pointed out to others as a way to cheat to see if the answers for any Times table X9 maths question was right. Didn’t know it above 100, as wasn’t taught about it.
    1
  14835. I'm not home right now to do it, but there's a quick trick to programming it for powers of ten by just appending all those 3's
    1
  14836. It doesn't get more Mathologorean than this. 😉😄
    1
  14837. Do all those diagrams especially the monster one show some mathematics of event horizon, the boundary marking the limits of a black hole? Imagine you could send 240 lasers in a specific angle from a circle a specific distance from the black hole, will it follow such lines into the event horizon?
    1
  14838. Not really. Close orbit around a black hole do produce some nice Spirograph patterns due to the precession of the orbit, but that only really holds true for a strictly equatorial orbit. Things get chaotic if the orbit is inclined. Feel free to google “black hole orbits” if you’re still curious.
    1
  14839. 24:10 Parent construction of 3-4-5 triangle returns the 3-4-5 triangle
    1
  14840. Any hint on 40:40?
    1
  14841. Merry Christmas! & Thank you for all the fascinating videos! Can't get enough of 'em!
    1
  14842. Hurray! A mathologer video! I need to study mathematics again! Arrgh! Im not keen enough :(
    1
  14843. I thought this was going to lead into another great video on the perfect numbers
    1
  14844. Didnt know christoph waltz made math vids
    1
  14845. Before you even asked how beautiful I thought the proof was, I thought to myself, "Wow! That's beautiful!" I would rate it a 10 on the scale from 1 to 10 :) Also, for the rotation property... Let's see. My guess for an explanation is that adding a, b, and -a-b always results in 0. So no matter which two of a, b, and -a-b you pick, the third will be the opposite of the sum of the other two. Picking two of these numbers corresponds to a rotation where those two numbers/colors are on "top". The reason why the rotation should not work for Pascal mod n for n > 2 because taking a, b, and a+b, the sum of a+b with either a or b is not the other one unless n = 2. (But also, for n = 2, addition and subtraction are the same thing, so the Pascal mod 2 is the same rules as the 2 coloring game.) I may have missed something in my reasoning, but I have to get going. I will come back to this later. Thanks for the great video!
    1
  14846. Nice! What can we say about the scaling factor between the original and the folded-folded one?
    1
  14847.  @Mathologer  do you have a nice visual proof of it?
    1
  14848. I was waiting when you will upload a video. Love Mathologer ❤
    1
  14849. I do love your videos!
    1
  14850. I live the shirts
    1
  14851. Fantastic. This is the first time I have watched this video and I understand it! ! ! Congratulations Mathologer, I look forward to the two hour extension you promised? ? ?
    1
  14852. There's only one way to make change for any particular value given a single coin for each power of two. It corresponds to 1 coin for each 1 in the binary representation of the number. Also, I lost my keys in July and found them in January, in the change pocket of my wallet where I had put them for safe keeping. I had not used coins for six months because of covid.
    1
  14853. Dear Mathologer. Who was - for you - the Bach of mathematics? Euler?
    1
  14854. An important thing to ask is, whether that center point correlates with any other possible center points of a polygon, for practical application, does it specifically correlate with the center of mass of a polygon slab of uniform material. Another thing this to note is that it's not only a degeneration, but a transformation too, and if applicable in 3d that can lead to interesting automatic morph animations
    1
  14855. You had me at..... Pretty pictures
    1
  14856. If you have qm positive terms and qn < qm negative, you obtain 1/(qn+1) + ... + 1/(qm) (the first qn reciprocals cancelled ). Squish by q to make them 1/q wide each. Stretch (multiply) by q to obtain the heights 1/(n+1/q),...,1/m. The reciprocals n+1/q, ..., m are evenly spaced between n and m, giving you the area under 1/x as q approaches infinity.
    1
  14857. Can something similar be done with 3 dimensional shapes/objects?
    1
  14858. Shouldn't root 666 be 25.8070 to 4 decimal places?
    1
  14859. I'm REALLY interested in the solution to what the parent of the 3-4-5 triangle is. Not sure I have the chops to do it. I'm guessing either something doesn't work OR, my intuition points to an infinite regress of triangles with sides with fraction lengths. Or maybe it breaks down that there are no (whole) number pairs "before" the first two ONES in the Fibonacci sequence. although I've seen 0, 1, 1, 2, 3, 5...., what would precede 0? Of course, it is interesting that the ratio of any two numbers in the sequence gets ever closer to the Golden Ratio, always going slightly above it after going slightly below it, when you deal with the first few numbers it is as crude as it possibly can be. If the "goal" is 0.618... then working BACKWARDS we get 3/5 = 0.6 (too small), 2/3 = 0.667 (too big), 1/2 = 0.5 (too small, again), 1/1 = 1 (too big, and in fact THE BIGGEST you can have a fraction be between 0 and 1), and 0/1 = 0 (too small, and the SMALLEST you can have a fraction be between 0 and 1). There are no fractions less than zero that add up to zero, so unless one starts considering NEGATIVE numbers..... But instead I get this weird thing where the LEFT side of 0 is not simply the negative version of the right, but ALTERNATES positive and negative. Any two integers add to the number to their RIGHT, and subtract to their number on the LEFT: ....5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5... The only thing that makes "sense" in having the values LEFT of zero alternate +/- is that by dividing any two neighbors guarantees a negative result, i.e. exact NEGATIVES of the ever-conversing Golden Mean value. So what does this mean? Can you continue creating these triangles with these sides? But what does it mean to have a triangle with a side length of 0, or a negative number? Is this connected to the "how" it breaks down?
    1
  14860. Hey! Aren't we just re-discovering algorithms for scientific calculators? I wonder if you can turn the tables and use this technique to find an approximation to convert an irrational number to a ratio of rational ones. i.e. Work out m and n from the result.
    1
  14861. why not take the points on the smaller circle at its perimeter? Wouldn't they trace straight lines then?
    1
  14862. Dear Mathologer, Happy late pi day
    1
  14863. I’ll tell you why it was missed for hundreds of years: The rubber band wasn’t invented until St Patrick’s Day, 1845. The stretchy metaphor would not have been that understandable.
    1
  14864. OK! Let me get this straight. All curves are 0 units in length. Keep approximating the segments of your curve until they reach Zero. Add all your zeros together and you get the true length of your curve. 0+0+0... add infinitum. I think I can reasonably guess the answer without showing all my working.
    1
  14865. Ah yes, my favorite algorithm U L’
    1
  14866. Jk my favorite algorithm is J Perm.
    1
  14867. Do you have any recommendations for someone wanting to learn more about continued fractions?
    1
  14868. 7:17 Answer is 2. 365 = 13^2 + 14^2, and 10^2+11^2+12^2=13^2+14^2.
    1
  14869. Gracias por tu consejo sobre "La Habitación de Fermat''. ¡Es una muy buena película!
    1
  14870. Could someone please help me understand the point at 12:12 - why we can conclude 3k+2 != x^2? How does the sequence 3k+2 relate to squares?
    1
  14871. In Z/3, the numbers we're working with are remainders after division by 3. So the "0" in Z/3 represents all integers of the form 3k, and "1" in Z/3 represents all integers of the form 3k+1, and "2" in Z/3 represents all integers of the form 3k+2. Said another way, integers are put into equivalence classes based on their remainder after division by 3. So the numbers in the multiplication table are the remainders! Everything with remainder 0 squares to something with remainder 0, everything with remainder 1 squares to something with remainder 1, and everything with remainder 2 squares to something with remainder 1. So nothing can square to anything with remainder 2. Which means that nothing of the form 3k+2 can be a perfect square. Does that help?
    1
  14872.  @MuffinsAPlenty  Brilliant that clarifies things thank you for your reply!
    1
  14873. I bailed out @19 mins. Not bad.
    1
  14874. I made it to the very end.
    1
  14875. Doesn't planck length put an end to all of these "paradoxes" involving physical manifestations of infinity?
    1
  14876. Another jam
    1
  14877. Pitagoras y Fibbonaci eran primos lejanos. Los 2 tenían 3 hijos. Y los hijos eran muy racionales. Tenían 3 nietos legítimos pero en realidad eran 3 más. Los 3 no conocidos son los más importantes pues también son de la familia. Eran los gemelos olvidados.
    1
  14878. always great to have your mind blown.
    1
  14879. anyone noticed how less he blinks his eyes?
    1
  14880. I want this t-shirt
    1
  14881. I made it through the whole video in one sitting!
    1
  14882. The coin paradox outlines the compression existing on the inside vs the outside of the line.... So when coiled a compression event occurs at a set ratio for the circle to exist mathematically within the universal system. It should also be noted that a stretching occurs to the exterior side in equal and opposite. That creates the difference. These differences can be seen with the internal and external diameters having a tiny difference when measured and compared. Those differences weren't accounted for in your model nor was the line stretching and compression required to create the circle. Meaning the distance traveled on each is calculated by the stretch and compression rates.
    1
  14883. So what does that shrinking animation look liked if you start with an almost hexagon in an actual square grid? How does it stop shrinking?
    1
  14884. True, but any straight line through the center of a square will cut the square into two congruent halves.
    1
  14885. That little titbit at the end explains why we sometimes use Torr as a unit of pressure . . .
    1
  14886. Just watched the first minute or so up to now, but wondering if one could set some conceptual radius that has infinite decimal digits (or that has repeating pattern of such?).
    1
  14887. @Mathologer I don't know if you'll see this but I believe there is an error at 23:18. The sum has factorial denominators, not products of consecutive odd numbers :/
    1
  14888.  @rogerbye6984  you are very correct. I obviously didn't think it through!
    1
  14889. Should the T-shirt be rounded (25.80697580112 rounds closer to 25.807)?
    1
  14890. Love you mathologer
    1
  14891. Girl: gets naked Me: 35:52
    1
  14892. I was with you until you started throwing apparently random fractions in the series at about 9:00. I did not follow where those came from. Sometimes shifting left, shifting right, hell, I can throw anything into a series, I didn't catch why those were OK vs tossing in 1/e, or some other random number. Numberphile did one of these, so I've seen it before, and I think I followed their explanation better. Anyway, on to the next video.
    1
  14893. American president being good at math is obviously not a general rule, my proof contains only two words " Made in America "
    1
  14894. My proof: label the 3 triangle lengths a,b,c, and their opposite angles A, B, C. connect 4 of the outer points, two from each of two pairs of the extensions (say two of extensions b, and two of extensions c, to form a quadrilateral. Draw from where the b's meet to the broad edge of the new quadrilateral a line with angle A. do the same from where the c's meet, again with angle A. Since each of those origination points have an angle C and A, (or B and A) and lie on a line, the interior angles are B and C respectively. There fore B, a, C must give you a congruent triangle to the original triangle. There fore you now have a couple more isoceles triangles. Now go to opposite corners. You can quickly find that, you'll get, for instance, angle 90 - C/2 opposite angles 90 -A/2 and 90 - B/2 on the other. These add up to 180, and so are supplementary, and so the new quadrilateral is an inscribed quadrilateral. Since there was nothing special about which of the three quadrilateral we drew, all three are inscribed quadrilaterals. You can draw more quadrilaterals in the same way, to find a bunch more of these half angles, and since 3 points define a circle, you can group them so that eventually you find that all 6 points are parts of a whole bunch of inscribed quadrilaterals that have to be on the same circle.
    1
  14895. لقد استخدمت مبرهنة lee sallows في إثبات صيغة مساحة المثلث بدلالة أطوال متوسطاته
    1
  14896. Who knew hydraulic press channels could provide math related material? Here : pipe vs press... https://youtu.be/z9IH4vb72Gs
    1
  14897. 19:48 your hair was longer back then
    1
  14898. There's something I don't understand. Couldn't the bowtie lacing be even shorter if you only crossed it once in the middle instead of going back and forth?
    1
  14899. ... 30*30+33*33+36*36=39*39+42*42 ... Regeln der Quadrate Regeln der Kubike (2+1) 🤗
    1
  14900. Okay, we have one more requirement on the method of approximation. The normals of the known areas should approach the original surface normals. Now is it all or can we expect a new bearded guy coming up with a new counterexample that would fulfill that the areas approach 0, the normals approach that of the surface being approximated and still diverging from the real area of the surface?
    1
  14901. Lovely video and a beautiful methodology. After producing all the nth powers in this manner, is there some way of extending, or reversing, the methodology, to produce a proof of Fermat's Last Theorem?
    1
  14902. Excellent 👍
    1
  14903. Beautiful.
    1
  14904. Working in the location TV Pro Sound industry I am familiar with and own many Hypercardioid shotgun mics and their patterns, As a "Sailor" I want to understand the maths behind why I can see a lighthouse with only a 50 watt bulb all the way out to the horizon and many nautical miles out. Who was this guy who crunched the numbers and invented the "Fresnel lens" way back in history without computers and fancy graphics machines etc. You would make my world complete.... if as a old psych-trance-hack like me, you could put together a vid that explains this and of course has a danceable backing track with fantastic graphics. Next up as a budding Ham radio opp , understanding propagation patterns of yagi antennas in the real world and using modeling apps as they all seem related after watching your channel . Cheers again... Kris from the ass-end of the pacific ocean, in Auckland, New Zealand.
    1
  14905. Amazing, continue with this!
    1
  14906. Valeu!
    1
  14907. Because we're clever. ;)
    1
  14908. Instead of 4k+3 the Term 4k-1 would look much nicer and fits also nicely to 4k+1 of the "first column". What do you think about that?
    1
  14909. Great to see you again sir
    1
  14910. OMG!
    1
  14911. I’m probably the dumbest person who watches these videos . I know no trr r ig
    1
  14912. Technically, there is only 1 way to make change for a googol....you get the government to mint a googol coin and sell it to the fed, the fed then gives you "change" in 1 way, and that's given basically the majority to foreign nations and interest groups, the next majority to native americans and corporate interest.....finally to congress and lastly give a few pennies to the citizens. I mean, you have to put it into perspective on real world patterns for others to really understand how it works. Or not, it doesn't matter, I'm laughing all the way to the bank either way.
    1
  14913. I remember being mesmerized by a software engineer at Nokia or probably some another big company proving pythagoras theorem with same scaling as your guy with more hair 😂
    1
  14914. @Mathologer, thanks. Yes, @ 6:10 And the engineer was from Motorola I think
    1
  14915. The audio on this seems quite quiet, for me at least. (Great video anyway!)
    1
  14916. School never charged me any money... but they did cost me a lot of time. Any ideas on how to get that back? I'll pay you handsomely in time if you can figure that out.
    1
  14917.  @Mathologer  https://www.mtai.org.in/wp-content/uploads/2023/09/IOQM_Sep_2023_Question-paper-with-answer-key.pdf ... 29th question
    1
  14918. Chapter 2: 5
    1
  14919. I was stumbled after watching the video and knowing that no 9series is finite
    1
  14920. I agree with you in general but here you are wrong. Matholiger isn't in the habit of doing this, and this title isn't at all clickblaity.
    1
  14921. I love the asian soundtrack at end. Of course the explanations are great as always. 👍
    1
  14922. Absolutely fantastic!
    1
  14923. If you are ever stranded on a desert island for an extended period of time, math might be the thing that keeps you sane.
    1
  14924. I often wonder why we do not default to beginning math education of children with similar fun proofs. Instead of teaching them memorization and rules, why don't we instead have them derived their math understanding from fun proofs?
    1
  14925. Am I to understand your T-shirt as "Pi Man" ?
    1
  14926. Very funny ! Thanks.
    1
  14927. At some point somewhere it ought to nail down that he is actually in front of a house and not between them. It matters.
    1
  14928. Science has many wonders, Math has the beauty. Good job!
    1
  14929. I just can’t get over how much Èdoard Lucas looks like Rob Swanson.
    1
  14930. I love how the octagon proof becomes a fractal
    1
  14931. What a pretty Christmas present. Vielen dank!
    1
  14932. Well, I think my job isn't the best in some sense, because my dreams about it are always nightmares!
    1
  14933. "Powell's Pi Paradox", hm, never heard of it.
    1
  14934. My favorite part was the clues leading to 0.5 < Gamma < 1. While you didn't demonstrate the 0.5 bound directly, your illustrations made it straightforward, which made the proof appear obvious and made me feel clever. Top notch content, thank you!
    1
  14935. That french shoe shop lacing is how i always used to lace up my converse 10 years ago!
    1
  14936. Only do calculus when you are sober! You should never drink and derive!
    1
  14937. for the first gauss sum, 1-100 I like my formula more. it's not in terms of n, but it works for any consecutive sequence, or even any sequence numbers that increase by a constant not just starting at 1 and incrementing 1. I add the fist and final number and then multiply by half of the length of the sequence. so for example if you want to add all the odd numbers 5-100 I would go 105*(48/2) or 2520.
    1
  14938. Thank you for the video.
    1
  14939. Another channel to add to my collection of subscribed channels. Heeheehee
    1
  14940. This is fascinating. You can well order the set of rationals between 1 and 0 and a way that seems to be a measure of the complexity of the rational expression.
    1
  14941. The Universe understanding is Math
    1
  14942. Here in Massachusetts, I remember it as a random aside in either elementary or early middle school. I don't think there was any kind of proof, and I'm not sure how much of the curriculum was really motivated by anything.
    1
  14943. 1+1+2+4 = 1×1×2×4 simply got in seconds 🙂
    1
  14944. excellent and no formula, incredible but nice
    1
  14945. This was my bachelors thesis! Nice way to approach it!
    1
  14946. A paradox related to the coin paradox: a rotation of planet earth takes 86164,09 seconds. A translation around the sun takes 31557600 seconds. If you divide 31557600/86164.09 you get the numbers of rotations of earth in a year. The result is 366.25 rotations. But a year have only 365.25 days...¿where is the mising day? (sorry, my english is not as good as i wish)
    1
  14947. The sum of the harmonic series without any 1s in binary = 0 :) Without any 0s is the of sum (1/(2^n - 1)) for n = 1 to infinity which converges to ~1.6067
    1
  14948. Thanks for a fun math video to finish off the week. #my_favorite_fact was that there are no integers among partial sums (excluding 1). This and Euler's gamma number can really mess with people's OCD.
    1
  14949. What kind of monster would change channels in the middle of a Mathologer video?!..
    1
  14950. I liked the part in which the series is approximated to ln(n+1)+gamma. And the fact that the more you go on, the better this approximation is valid. Which, as you said, quite interesting! Usually, the further you go, the worse it gets but not in this case. Very interesting! EDIT: the demonstration of the no 9s sum is super cool!
    1
  14951. Triangulating a smooth surface allows combinatorics to apply since it makes the surface into a simplicial complex. As mentioned, similar geometries in the form different triangulation can give different invariants. For example, as explain in this video the surface area which is invariant the under Euclidean motions is different while another invariant the genus remains 0. From algebraic geometry different sheaves define different coverings of topological spaces with different geometry and different invariants. Depending on how the sheaf of the smooth space is defined changes the invariants.
    1
  14952. It took me some time to realize that the involution constructed by Zagier is the same as the windmill. Btw if you are in Australia, I hope everything is well there.
    1
  14953. in russia heron's formula is taught since 5th or 6th grade and used all the way until 11th
    1
  14954. I love your videos, they are so well thought out and masterfully presented with charm and wit. I liked the 700 year old proof the best.
    1
  14955. I check everyday for a new mathologer video. These videos are how I keep track of time during the pandemic :D
    1
  14956. So remarkable!! Thank you thank you thank you!
    1
  14957. I didn't understand any of that stuff @30:00 onward
    1
  14958.  @Mathologer  not a bit, great stuff, thanks for your work!!!
    1
  14959. Most memorable for me, the 700 y.o proof of the torticolis monk
    1
  14960. I'm definitely looking forward to a proper proof that all primes equal to 1 mod 4 can be written as a sum of 2 squares. 3blue1brown did this same topic (it's probably my favorite video he ever did) and skipped that part as well, and it's been floating in the back of my mind as an unsolved problem ever since.
    1
  14961. " THE ROOT of LOVE " Do you have the number ?
    1
  14962. I feel a bit uneasy about the step in the proof where you shorten the exploded lacings. For instance, when an exploded lacing contains a zero and a two, you suppose such a sequence corresponds to a lacing and show that the lacing can be shortened, resulting in an explosion with two ones. But you seem to have implicitly assumed that the 0 and 2 cross one another in the lacing. If they do not, how do you shorten the lacing?
    1
  14963. Everything breaks down into the same triangle: outside triangles having area A, trapezes are area 3A (this relies on triangle smaller side being 1/2), center is area 4A (again), we have 4 * (outside triangle+trapeze)+ 4A =1, A=1/20. and the center is 1/5.
    1
  14964. I did some abstruct algebra from youtube so it was reletivly simple Had sone parts when i scruched my head
    1
  14965. Music at 18:09? Anyone?
    1
  14966. I love the fact that you're smiling again
    1
  14967. I actually thought the proof of the bishop is the most memorable, since it's probably the only thing i will be able to remember.
    1
  14968. Made it all the way to the end and I love your videos.
    1
  14969. where does the -1 come from in y' at 13:55?
    1
  14970. ​​​ @Mathologer ok thanks. i'll ponder the formula for the derivative of what comes after the minus sign in y. ok i see where the (e^stuff e^-stuff) comes from now thanks again
    1
  14971. If I had to chose I would go to the 700 years old proof, it's smart, easy to understand, relies on simple fact and doesn't require some crazy function you stumbled up in a dream to work, ie is not very analyisis firendly that is a very good think not to be
    1
  14972. (3^3)+(4^3)+(5^3) = 27+64+125 = 216 = 6^3 (3^4)+(4^4)+(5^4)+(6^4) is an even number, so it won't be equal to 7^4.
    1
  14973. 22:40 For me, the best conceptual realization from this video came from considering the fundamental difference between the cylinder and the sphere: to change the cylinder's relationship to the projection, you increased the height and decreased the area of the cross-section. So why not just do that for the sphere and break the whole argument? Because the sphere's height and cross-section are linked - they are both dependent upon the radius. So any change in the sphere will be expressed the same way along all dimensions. From this I realized that circles, spheres, and their n-dimensional analogues can be considered as one-dimensional objects with n derivative (or, perhaps more accurately in calculus terminology, integrative) properties. That is, they all have one dimension: r, and the circle, for example, has circumference and area. Sphere adds volume, etc. You can even extend this idea to regular polygons/hedra/N* by adding a second dimension: the number of sides.
    1
  14974. In the case of the 3-4-5 triangle, the connecting line between the centers of the incircle and Feuerbach circle is horizontal, so it cannot make a proper right triangle with another horizontal and vertical side. That's why the parent construction doesn't work for the 3-4-5 triangle.
    1
  14975. In the beginning, there was darkness and then there was - 1/12 ... and then there was light.
    1
  14976.  @Mathologer  Wow! I didn't really expect a reply. :) That is a really bizarre connection. Well, I guess it is true to say that "any two topics in XXX have some sort of relationship" occurs with nonzero probability.
    1
  14977. The teacher looks like a discovery channel Neanderthal face reconstruction
    1
  14978. Hmmm... why does Arctic Gold change into Mayan gold at 15:14? A Mistake? A Christmas Easter egg? 15:14 is Pi if you use american time stamp ......
    1
  14979. any mathematical argument involving splitting an object into basis objects apeals to me. i just love linear algebra
    1
  14980. Your videos are oxygen for the brain.
    1
  14981. 55 comes next.
    1
  14982. Most memorable for me was probably the fractal proof for the no 9s series. As a side note, the thing about the 100 zeroes series summing to more than the no 9s was shocking.
    1
  14983. When I saw this was a 46 min video I thought I would just skip it and wait for a shorter one, but I am glad I didn’t. It’s 46 mins of pure goodness!
    1
  14984. This is the first video that made me actually understand how to solve the final layer myself. Thank you so much, you seem to have a deep understanding of the Rubik's cube if you can explaine it properly. Everyone else is just parroting algorithms, but you taught us how to move the piece we want and then undo the damage in the other layers. One thing you didn't mention explicitly is that (if I understood correctly) these moves only work in pairs of pieces. That is, when you do your B move by rotating the upper side, you are going to affect another piece on the top layer too. I don't know how the exact mechanics of the puzzle work, but it seems this is ok for solving it because some combinations are impossible. Also this method requires pen and paper so that you can write down what you did and reverse it. Here are my notes from the three moves shown in the video: Flip front and right edges: A = L' R F2 L R' D L' R F' L R' B = U A-1 = L' R F L R' D' R L' F2 R' L B-1 = U' Twist back right corner CCW, front right corner CW: A = R' D R D' R' D R B = U A-1 = R' D' R D R' D' R B-1 = U' Move front, right and back edges CW: A = R L' F2 R' L U R L' F2 R' L U' R L' F2 R' L B = U A-1 = R L' F2 R' L U R L' F2 R' L U' R L' F2 R' L B-1 = U'
    1
  14985. I like this video
    1
  14986. Please master class about matrix
    1
  14987. 22:15 Why can you ignore the gaps in the green part of the sum? AFAIK, calculating the integral with only half the amount of rectangles is not a valid approximation for the true value of the integral.
    1
  14988.  @Mathologer  Aha makes sense. IMO it's easier to understand if you imagine every single rectangle being doubled up, obviously leading to a correct approximation.
    1
  14989. Pentagon (5/2) so line and triangle. What should happen if we put the gray circle on the outsite ...?
    1
  14990. Representing hypercubes as (x+2)^n reminds me of Flajolet and Sedgewick's symbolic method for counting objects. For example, to count all objects (vertices, edges, faces, cells) in a cube, you would pick one of {start vertex, stop vertex, edge} from each of the three dimensions and this would correspond to one of the objects in the cube. That would be represented as (x+1+1)* (x+1+1)* (x+1+1) = (x+2)^3
    1
  14991. Das ist ja mal der Hammer! Super! :D
    1
  14992. Is the bugs path a brachitrone cause it looks like one.
    1
  14993. My dad had a K&E 21 scale. I found a Post 22 scale in an office supply and asked how much. The guy said no one had asked about it for over 15 years. I bought it for $10 and thought I got a steal. I carried both to high school while all the other kids carried TI calculators. I could read out answers faster than them punching in all the numbers. I knew how to use all of the scales. There was a 6 foot wooden slide rule that hung above the chalk board. My algebra teacher let me teach the class one day on the use when we were learning about logarithms. I got about 1/2 way through and had to post. The L scale, common log, is equally spaced allowing you to easily see that the sum of the log of 2 numbers is the product of those 2 numbers. I later found a circular slide rule and added it to my collection. I still have all 3. I love old school and use an abacus/soroban as well.
    1
  14994. I think I would have enjoyed math for 12 school years with this guy but I’m older and on a lot of pain meds so all I can think is how could this have made me more money
    1
  14995. I remember I came up with a version of this problem about 10 or so years ago and it comes to my mind every now and then since. I was wondering why the diagonal of lets say a 1x1 square does not converge to sqrt(2) if approximated by the manhattan distance with an infinitely small cell side. I have to admit I do not understand the reason explained in this video as well as I would like. Is the reason that in order for the approximation to be correct the gradient needs to converge as well? And if so, why? The perimeter/area of the cell still tends to zero. At least now I have a name for the problem so I can look further into it.
    1
  14996. Wooow 😍 I can say that it is the most important & interesting video I have ever seen on the internet Many many thanks sir Mathologer
    1
  14997. I needdddd that shirt! Where is it from?
    1
  14998. School in my country is free so the chances of me getting my money back are actually pretty decent.
    1
  14999. Yesterday I was scribbling some stuff on paper and I remember thinking “hmm a+b = ab is an interesting equation, I wonder what’s up with it” and then I briefly checked to see if mathologer had any videos on the topic and went on with my life. I check back today, and I’m treated to this uncannily timely video 😂
    1
  15000. 25:40 my dad works in constructions and he uses this a lot
    1
  15001. I had to pause the video at 16:07 to write a program to find the 666th partition number... 11393868451739000294452939... I hope I got the right answer.
    1
  15002. 2
    1
  15003. Came for the barber's pole
    1
  15004. If you live in the country you could have 8 houses on a long street, much longer then 49 houses in the city lol.
    1
  15005. Beautiful explanation of this proof of Fermat's theorem but I think a small error or omission slipped into (or out of) the statement of the final challenge at 32:20. Shouldn't the statement be that a non-zero integer can be written as the difference of two integer squares exactly if the integer is a power of 4 (possibly 4^0) times an odd number? (For example, if n = x^2-y^2, then 4n = (2x)^2-(2y)^2.)
    1
  15006. amazing as ever... mathologer the best channel on maths... greetings from colombia. Love u
    1
  15007. Why are they all converging to nearly 23? Help!
    1
  15008. At first, I looked at a preview to this video and I thought about how does "10² + 11² + 12² = 13² + 14²". And my math intuition was like: _0² + _1² + _2² = _3² + _4² _1 + _4 = _9 + _6 The operation with dividing numbers in groups and forming a border around is a wonderful discovery!
    1
  15009. 5:14 the single block is at equilibrium when the overhang is 1/2, but it is not stable. It is an example of what in mechanics is known as an unstable equilibrium. If the block is very slightly tilted the wrong way it will topple. For it to be stable the overhang must be (1/2)-delta where delta depends on both the thickness of the block and the angle of tilt for which we want stability In a world with an atmosphere, Brownian motion would otherwise topple it at some unpredictable time. In a works with negligible atmosphere quantum effects would do the same after an appreciably longer expected time. In a similar way the whole pile of twenty blocks (or of infinitely many) will topple. A more interesting set of questions consider what happens after a topple? If the toppling is gentle enough the remaining blocks will be stable. Will the toppling be gentle enough? If so, what is the probability that exactly n blocks will remain? Is the answer the same or different for Brownian motion as compared with quantum-triggered toppling?
    1
  15010. Having fun so far? Yes!
    1
  15011. Okay so after finishing my shower and continuing the video, your proof is pretty much the same as mine (I thought up the 8/9 improvement on the limit but I felt it was kinda pointless to include), I'll just leave this comment here for posterity The no 9s series is interesting. Proof of convergence I devised in the shower: between 0 and (10^n-)1 (aka 999...999) there are 9^n numbers without a 9 in them (number of possible digit combinations that do not include a 9). Therefore the amount of appropriate numbers between 10^(n-1) and 10^n is lesser than 9^n. Each of the terms in the series between them is lesser than 1/(10^(n-1))=10*10^(-n), so the sum between those numbers is lesser than 10*(9/10)^n. To get a limit for the sum of the entire series we just need to sum over n (from one to infinity) then, and the sum converges to 9*1/(1-9/10)=9
    1
  15012. Quadratic reciprocity?! My old professor, way back, wrote a book on number theory, in which he and his coauthor proved the law of quadratic reciprocity several different ways. Then he included a question at the end of the chapter asking the reader to find their own proof! At the time he was teaching analysis but he often slipped in number theory themes in his lectures. I never picked up the book because I wasn't interested in number theory at the time and frankly I was not any good, but I always thought he was special, a genius even, who loved his subject. You remind me of him. Happy New Year!
    1
  15013. Yes, yes. Um, hmm. Easy peasy. Just like the paragraph that explains the "following proof" as trival and straight forward in my textbook... Gulp.
    1
  15014. Queen of hearts, right?
    1
  15015. Wow, a very very long video there, save first :) I'm an accounting student, which on the other side of your whole theme right there. But I just really really love your videos, especially about infinite something something. :D I subscribed about 2 years ago, and I hope I will pretty pretty enjoy your next videos 👍👍
    1
  15016. This begs a question to me: what does it mean to "converge"? A number that goes on infinitely could still be considered "convergent" if it has a label, like pi, yes?
    1
  15017. I'm afraid that if I say 666 comes next on my next IQ test, those Mensa folks just won't get it....
    1
  15018. amazing, greetings from Brazil!
    1
  15019. I study mathematics in Cambridge, but I still like this video more than my online lectures :)
    1
  15020. Very nice. I like the argument pushing the 2d vertex slightly into 3d and then considering how it moves back. I think Archimedes would like that.
    1
  15021. This channel should be renamed to "The illuminati". Too many coincidences, just.. too many.. coincidences..
    1
  15022. Reaaally nice video, since the years I'm watching I never get disapointed ! But for the first time ever I'm wondering if I haven't found an even more mathologerized "proof" of Euler's formula ! (or maybe I'm totaly wrong ^^). The basic idea is that every convex polyhedra is isomorphic (homomorphic ?) to a tetrahedron, in the sense that you can smoothly deform any convex polyhedron into a tetrahedron and vice versa if you admit that one line segment can be split into two parralel ones, and one face can be split into any number of coplanar faces (because we are not interrested into positions of the vertices, and so the lengths of the edges and areas of faces). So you choose four not colpanar vertices in a polyghedra (there will always be) that will become the final tetrahedron, you let them in place, and you move all the other vertices so that they get colpanar with the three closest fixed vertices (three among the four we have chosen). You also have to align some the vertices with te fixed ones (in order to creat new edges), but I struggle to rigorously describe where and when ^^'. In the process, you will align some pairs of vertices that have a common vertex and some faces that have common edges. Then you remove the extra vertices ; those which cut a newly created edge (created by the alignement of two edges) in two parts. Then you remove the extra edges ; those which cut a newly created face (created by the alignement of two faces) in two or more parts. You end up with a tetrahedron !! In the process of removing the extra vertices, you created one edge from two edges, so you loose a vertex and an edge, so the initial (V - E + F) becomes (V-1 - (E-1) + F) = (V - E + F). In the process of removing the extra edges, you created one face from two faces, so you loose an edge and a face, so the initial (V - E + F) becomes (V - (E-1) + (F-1)) = (V - E + F). So during the whole process (V - E + F) staid constant ; we just have to calculate it for a tetrahedron which is simple !!! Of couse, you can do the other way around (which is maybe easier to see) an construct any polyhedra from tetrahedron by splitting edges in half, which creates each time a vertex and an edge, and then by moving this new vertex where you want, which creates each time an edge and a face, so (V - E + F) stays each time constant. I know it's not a rigorous proof (althought I think we could do a solid demontration from the last two lines above), it is really "intuition based" bur I'm quite proud of having found a very visual and simple "proof" (would be simpler if I had animation rather than text but maybe someone can ;) ). Please master, tell me if my idea is good or if I made a mistake. Thanks a lot for all your content for all this years !!
    1
  15023. Awesome Mathologer video is awesome
    1
  15024. Maths is just a magic Cool video
    1
  15025. Those beautiful patterns remind me of Yantras in Hinduism.
    1
  15026. The most memorable part was learning about gamma. I didn't know about that constant. Also I was watching the video with my mom. When you said "flip" she was flipping an omelette and when you said "very" she was typing very, and she pointed it out lol.
    1
  15027. Fractal.
    1
  15028. towel man
    1
  15029. Thank you! Nice video.
    1
  15030. What happens to 1/x squish and Stretch, if the basis vectors are not normal to each other? The plane can be described as any to non-parallel pair of basis vectors. So if i define my coordinate system on those, and then take 1/x line, and do the same operation, can i maintain it's shape invariance, regardless of my choice of bases? #
    1
  15031. Requested feedback: you had me scared hyping up one of the chapters as being super super difficult, but you explained it very clearly. I appreciate how you give us a chance to pause the video and experiment with some ideas before continuing. I’m a software developer (20 years) with a degree in engineering and a love of all things math since my childhood. Thanks for keeping the faith. 👍
    1
  15032. I was taught Heron's formula in elementary school in Italy - but that was 40 years ago. Don't know about now...
    1
  15033. I was able to add up to 730 within 30 seconds even though I didn't look at the watch.
    1
  15034. the video's great!
    1
  15035. I need a supercut of this gentleman giggling. probably my favourite thing on youtube. :3
    1
  15036. There's no paradox because any layer of paint has a thickness. For any amount of surface area that is not infinite, the 8 litres of paint will spread out to form a layer of paint with non-zero thickness, but as the surface area grows towards infinity, the thickness of the layer of paint goes to zero.
    1
  15037. The first time I encountered generating functions was when I was having fun filling up paper charting how many vertices, edges, faces, etc. there are in a line segment, square, cube, 4-cube, etc. and finding the patterns. I showed this to a math teacher, and he was like, “or you could find the coefficients of (2 + x)^n.” (Where n is the dimension of the cube, and the power of x is the dimension of the face.) I was blown away and mystified and had no idea how it worked. To find the k-faces of an n-cube, you doubled the k-faces of the (n-1)-cube and added the (k-1)-faces of the (n-1) cube. Which shows that (2+x)*f_(n-1)(x)=f_n(x). And f_1(x)=2+x, because you start with a line segment, with 2 vertices and 1 edge. So you have proof by induction. You could even start with f_0=1, and start with a single point (Pointland). But the proof by induction isn’t that satisfying, and doesn’t quite bring a greater understanding. I’ll look into it tomorrow morning.
    1
  15038. I was the one who suggested this in the comment section of the last video - but I am still impressed by the number of connections you've made that I'd never thought of.
    1
  15039. 2:52 Proper pronunciation of "forte". Very nice.
    1
  15040. The most brilliant math channel
    1
  15041. Hay another video on proof of e^-x^2 has no elementary antiderivatuve.
    1
  15042. This is a joke, I can barely do 1+1, but thanks for reading 😂
    1
  15043. Should we tell him about 2² or will that break the universe?
    1
  15044. @johnrichardson7629  you mean 2 with extra 2s on top? Ya, that is some crazy stuff
    1
  15045. I can’t be the only one who discovered most of this by experimenting.
    1
  15046. Definitely another interesting video. Not only maths part was interesting, even the music at the end was very soothing too.
    1
  15047. simply magnificent!
    1
  15048. what is the best way to saw a button?
    1
  15049. I'm reminded of the vastness of our lungs area, or the immense length of our intestines, when they are unfolded, compared to the size of the human body. This folding is what creates fractal objects, and is also the cause of the normal vectors ending up so wildly overlapping. An interesting observation: although the area of the Schwarz lanterns and the perimeter of the rectangular curve around the circle do not converge to the area and length of the objects they approximate, respectively, their volume and area DO converge to the right values (if I'm not mistaken). I'm not sure if that's a general property, and if so, how it is proven.
    1
  15050.  @MatthewGilliard  🤣You won the Internet for today.
    1
  15051. 7:00 Subtract 1. The last digits are 80. By the rules of divisibility by 4, 80 is 4*20. So that huge number is a sum of 2 squares. QED
    1
  15052. I love you, Heron. Thank you for the steam engine.
    1
  15053. 'Does anything ever make sense in War?' - No it does not my friend.
    1
  15054. Wow. You're constantly outdoing yourself with the animations. Heron himself would be mind blown!!!
    1
  15055. Here's a little program that simulates the tiling of a diamond as shown in the video. To run: copy the code below and save it in a file with type .html. Then load the file with the browser. Attention! Tested only with Safari. begin of code------------ <!DOCTYPE html> <html> <head> <title>Arctic circle</title> <link rel="stylesheet" href="styles.css"> <script type="text/javascript" > const covered = 3; function draw() { new diamondPlotter( parseInt(document.getElementById("grade").value), 800, document.getElementById('showArea').getContext('2d')).plot(); } class squareCoverage { constructor(matrix) { this.matrix = matrix; } get() { this.coveredSquares = []; this.copySquaresToCovered(); for (var tile of this.matrix.tiles) { this.setCoverageOfTile(tile); }; return this.coveredSquares; } copySquaresToCovered() { for (var row of this.matrix.squares) this.coveredSquares.push([...row]); } setCoverageOfTile(tile) { this.coveredSquares[tile.y][tile.x] = covered; if (tile.vertical) this.coveredSquares[tile.y + 1][tile.x] = covered; else this.coveredSquares[tile.y][tile.x + 1] = covered; } } class tileGenerator { constructor(coveredSquares,tiles) { this.tiles = tiles; this.coveredSquares = coveredSquares; } get(){ this.fillEmptySquares(); return this.tiles; } fillEmptySquares() { for (var y = 0; y < this.coveredSquares.length; y++) { var row = this.coveredSquares[y]; for (var x = 0; x < row.length; x++) { if (row[x] > 0 && row[x] < covered) { if (this.isFreeSquare(x, y)) this.fill2x2WithTiles(x, y); } } } } isFreeSquare(x, y) { return Boolean( this.coveredSquares[y][x] < covered && this.coveredSquares[y + 1][x] < covered && this.coveredSquares[y][x + 1] < covered && this.coveredSquares[y + 1][x + 1] < covered); } fill2x2WithTiles(x, y) { if (randomTrueOrFalse()) this.fill2x2WithVerticalTiles(x, y) else this.fill2x2WithHorizontalTiles(x, y); this.set2x2Covered(x, y); } set2x2Covered(x, y) { this.coveredSquares[y][x] = 3; this.coveredSquares[y + 1][x] = 3; this.coveredSquares[y][x + 1] = 3; this.coveredSquares[y + 1][x + 1] = 3; } fill2x2WithVerticalTiles(x, y) { console.log(x + ',' + y + ':vertical'); this.addLeftTile(x, y); this.addRightTile(x, y); } addRightTile(x, y) { this.addTile(x + 1, y, true, 1); } addLeftTile(x, y) { this.addTile(x, y, true, -1); } addLowerTile(x, y) { this.addTile(x, y + 1, false, 1); } addUpperTile(x, y) { this.addTile(x, y, false, -1); } addTile(x, y, verticalFlag, pushDirection) { this.tiles.push( { x: x, y: y, vertical: verticalFlag, pushDirection: pushDirection, delete: false } ); } fill2x2WithHorizontalTiles(x, y) { console.log(x + ',' + y + ':horizontal'); this.addUpperTile(x, y); this.addLowerTile(x, y); } } class squaresExtend { constructor(oldSquares){ this.squares = []; this.oldSquares = oldSquares; } get() { this.extendExistingRows(); this.extendWithNewRows(); return this.squares; } getOldLastRow(matrix) { return [...this.oldSquares[ this.oldSquares.length - 1]]; } getOldFirstRow() { return [...this.oldSquares[0]]; } extendWithNewRows() { this.addNewLastRow(); this.addNewFirstRow(); } addNewFirstRow() { this.squares.unshift(this.widenRow(this.getOldLastRow())); } addNewLastRow() { this.squares.push(this.widenRow(this.getOldFirstRow())); } extendExistingRows() { for (var row of this.oldSquares) { this.squares.push( [...this.extendRow(row)]); } } extendRow(row) { return (this.isFullyFilled(row) ? this.addSquaresToRow(row) : this.widenRow(this.extendRow(this.narrowRow(row)))); } addSquaresToRow(row) { return [row[row.length - 1]].concat(row).concat(row[0]); } isFullyFilled(row) { return row[0] !== 0; } narrowRow(row) { return row.slice(1, row.length - 1); } widenRow(line) { var result = [...line]; result.unshift(0); result.push(0); return result; } } class tilesExtend { constructor(matrix,oldTiles){ this.oldTiles = oldTiles; this.matrix = matrix; this.coveredSquares = []; } get() { this.matrix.tiles = [...this.oldTiles]; this.matrix.tiles.forEach(this.moveOldTile); this.eliminateColliding(); this.shiftTiles(); this.coveredSquares = new squareCoverage(this.matrix).get(); return new tileGenerator(this.coveredSquares,this.matrix.tiles).get(); } eliminateColliding() { var index = 0; for (var tile of this.matrix.tiles) { this.checkCollision(tile, index); index++; }; var tiles = []; for (var i = 0; i < this.matrix.tiles.length; i++) { tile = this.matrix.tiles[i]; if (!tile.delete) tiles.push(tile); } this.matrix.tiles = tiles; } checkCollision(tile, index) { let compareIndex = this.searchCompareTileIndex(tile); let array = this.matrix.tiles; if (compareIndex > 0) { tile.delete = true; } } searchCompareTileIndex(tile) { var checkX = tile.x; var checkY = tile.y; if (tile.vertical) checkX += tile.pushDirection; else checkY += tile.pushDirection; for (var compare of this.matrix.tiles) { if (compare.x === checkX && compare.y === checkY && compare.vertical == tile.vertical && compare.pushDirection != tile.pushDirection) return this.matrix.tiles.indexOf(compare); } } shiftTiles() { for (var tile of this.matrix.tiles) { tile.x = (tile.vertical ? tile.x + tile.pushDirection : tile.x); tile.y = (tile.vertical ? tile.y : tile.y + tile.pushDirection); tile.type = (tile.type === 1 ? 1 : 2); } } moveOldTile(value) { value.x++; value.y++; } } class diamondMatrix { constructor(grade) { this.grade = grade; this.matrix = { squares: [], tiles: [] }; }; generate(oldMatrix) { if (this.grade === 1) this.getInitialMatrix(); else { this.lowerMatrix = new diamondMatrix(this.grade - 1); this.lowerMatrix.generate(); this.extend(); } } getInitialMatrix() { this.matrix.squares = [[1, 2], [2, 1]]; this.coveredSquares = new squareCoverage(this.matrix).get(); this.matrix.tiles = new tileGenerator(this.coveredSquares,this.matrix.tiles).get(); } createCoveredSquares() { } extend() { this.matrix.squares = new squaresExtend(this.lowerMatrix.matrix.squares).get(); new tilesExtend(this.matrix,this.lowerMatrix.matrix.tiles).get(); } } class diamondPlotter { constructor(grade, size, ctx) { this.size = size; this.graphic = ctx; this.grade = grade; this.squareSize = size / grade / 2; if (this.squareSize > 20) this.squareSize = 20; } plot() { this.graphic.clearRect(0, 0, this.size, this.size); this.matrix = new diamondMatrix(this.grade); this.matrix.generate(); this.startingPoint = this.size / 2 - this.grade * this.squareSize; for (var tile of this.matrix.matrix.tiles) { this.plotTile(tile); } } plotTile(tile) { var { x, y, yTo, xTo } = this.getTileCorners(tile); var type = this.matrix.matrix.squares[tile.y][tile.x]; var fillStyle = (tile.vertical) ? (type === 1) ? "red" : "blue" : (type === 1) ? "yellow" : "green"; this.graphic.fillStyle = fillStyle; this.drawTileFrame(x, y, yTo, xTo); } drawTileFrame(x, y, yTo, xTo) { this.graphic.beginPath(); this.graphic.moveTo(x, y); this.graphic.lineTo(x, yTo); this.graphic.lineTo(xTo, yTo); this.graphic.lineTo(xTo, y); this.graphic.lineTo(x, y); this.graphic.stroke(); this.graphic.fill(); } getTileCorners(tile) { var x = this.startingPoint + tile.x * this.squareSize; var y = this.startingPoint + tile.y * this.squareSize; var xTo = (tile.vertical ? x + this.squareSize : x + 2 * this.squareSize); var yTo = (tile.vertical ? y + this.squareSize * 2 : y + this.squareSize); return { x, y, yTo, xTo }; } } function randomTrueOrFalse() { return Math.round(Math.random()) === 0; } function arrayRemove(arr, value) { return arr.filter(function (ele) { return ele != value; }); } </script> </head> <body> Grade: <input id="grade" type="number" value="1"> <button id="go" onclick="draw()">Go</button><br> <canvas id="showArea" width="800" height="800"></canvas> </body> </html>
    1
  15056. I noticed when you showed your patrons, there's no PayPal. I guess you broke up with PayPal after they wanted to charge their customers $2,500 if they spread "misinformation". If so, good choice. And about 11:38: those numbers are 9 4 17 13
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  15057. 18:54 Any value for pi you can actually calculate with is an approximation of pi, it's a ratio. Pi itself is irrational, you cannot calculate by it, and it is not a number.
    1
  15058. Love your videos!!
    1
  15059. i already have ians laces app. i liked how the shoelaces dangle when you shake the phone.
    1
  15060. I love your shows but your animations are obscured by the subtitles. Might I politely request you reserve that lower part of the screen for those who use subtitles? Many thanks.
    1
  15061. That moment when the triangle normals aren't perpendicular to the parent shape (with any non-intersecting projection of a line to the parent shape)
    1
  15062. 13:00 Just being able to find the center of any polygon may be worth the price of admission. Been puzzled since Jr High school how to determine the center of a triangle, and if it makes sense that the center may lay outside the triangle: if you use the method of the center of the circle laying on all of the triangle's vertices.
    1
  15063. You need a "Well..." tee-shirt.
    1
  15064. Is it possible to Mathologerize the A+B=C conjecture? I am trying to understand what it really means and why are high quality triples so difficult to find.
    1
  15065. excelent video, but you could definitely change the xy graphic design, it somehow annoyed my eyes. Maybe the dots are too big, or the colors too vibrant, for instance, i didn't get why the xy lines are red
    1
  15066. Where were you 55 years ago. Thanks for this!
    1
  15067. Canadian Cube (1997) has gained notoriety and a cult following, for its surreal atmosphere and Kafkaesque setting and concept of industrial, cube-shaped rooms. works based on primes
    1
  15068. I remember proving the 9 divisibility test in high school.
    1
  15069. NOTE This was done before he revealed the answer on screen: There technically IS a regular hexagon hiding in a square grid. Though it won't be flat. If you turn a cube so the opposite diagonal corners are aligned, the perimiter of the cube then forms a regular hexagon (at least in orthographic projection) Moving out of that perspective though revleals that it's a zig-zag pattern in 3d space. But that doesn't really matter if viewing things two-dimensionally. Since a cube has 8 such corners, we can deduce that there has to be four such regular hexagons in it (viewing the cube from the oppsite side just shows the same hexagon) Scale this up by however many tiled cubes you have a 2x2x2 cube would have 4 × 8 + 4 hexagons
    1
  15070. This makes no sense. I prefer crowns, guineas, and pence. ... oh, and shillings.
    1
  15071. how about alternative? (0)²=0² (-1)²+(0)²=1² (-2)²+(-1)²+(0)²=1²+2² (-3)²+(-2)²+(-1)²+(0)²=1²+2²+3²
    1
  15072. There's obviously some differences in the methodology between India and the West. In the next 100 years after Newton and Leibnitz, there was Euler and Gauss and Riemann; Lagrangian mechanics, complex analysis, Fourier series and on and on. Continuous nowhere differentiable funtions, functions continuous at irrationals, discontinuous at the rationals... If India was ahead by 2 or 3 hundred years. What happened? Did they advance past calc I?
    1
  15073. great job great voice great content and great approach obviously lots of research well dome
    1
  15074. Calculus is the beautiful topic i have ever heard i am in 10th standard but felt in love with calculus..
    1
  15075.  @Mathologer  Got it! I'm listening through my phone's mono speaker. It's around 6 years old, maybe that's the problem. I'll try it with the headphones.
    1
  15076. Question: is it possible to trisect an arbitrary 2d angle by making use of 4d (with circles and straight lines)? Thoughts on why it might be possible: there's an excellent algorithm for cutting a line (call it AB) into x equal parts, by attaching a straight line BC, at an angle, composed of x equal parts. Draw the straight line from C to A, draw the parallels. Trisecting a line (1d) takes a walk through 2d. Angles are 2d creatures, so maybe there's a 4d walkaround to trisection? If not, proof?
    1
  15077. These are also very good at reducing fractions
    1
  15078. I didn't know he taught at Monash Uni, I would've thought harder about joining them in that case...
    1
  15079. I like the first proof better
    1
  15080. That's a nice hexagon @6:00 (wink, wink 😉)
    1
  15081. What a journey!
    1
  15082. The hexagonal tilings always have the same number of each orientation because if you think of it as 3-dimensional and look orthogonally down any other the three apparent axes, you would see only and all of one orientation and it would be a square with a side length that is the order of the hexagonal tiling. This will be the same square along all three axes which is required intuitively by its 3-dimensional structure. I'm sure there is a more straightforward way of thinking of this in 2 dimensions but I can't get the 3-dimensional image out of my head. I've always been fascinated with the relationship between the regular hexagon and the orthogonal cube. ^_^
    1
  15083. +
    1
  15084. 24:39 an n-tube is an n-cube minus(or pruned from it) n-1 times the (n-1)-cube????
    1
  15085. 9:40 that's a great aha moment maker :)
    1
  15086. :trophy-yellow-smiling:
    1
  15087. Area 5 as drawn... area -1/12ths if continued to infinity counting overlap. : ) just kidding, but thought that might get a rise
    1
  15088. Mind blowing!
    1
  15089. Each time I open a mathologer video,its maginficence hits me.
    1
  15090. The animations are very choppy, not so pleasant to look at. Rest of the video very well done.
    1
  15091. I am looking forward to the all-german video.
    1
  15092. I watched until the end. Not sure what here would or wouldn't work for an average person, but I have a solid background in lots of Calculus and such, so all of it worked for me. I'm currently an active (but unofficial) researcher mathematician, researching dimensionality and methods of solving "impossible" integrals.
    1
  15093. Most memorable: Kempner's proof
    1
  15094. For me the most memorable part was all the Kempner-like series stuff (how they all converge and what-not)
    1
  15095. Very interesting, but I have to watch the video again. It's ok for me till the appearance of the Bernouilli's numbers. Then, I see the general way but not more. I received mathematic formation in secondary school (till 18). Note that I'm not an english-speaker, so it's more difficult.
    1
  15096. RIP my life and mathematical career for guessing the 'no 9 digit' series wrong
    1
  15097. 20:30 This(w/ the fact that the circle is the right size) implies that if I join the center of the Feuerbach circle and an excircle with a line, the tangent of the excircle at the 'far' intersection with that line forms the right triangle derived from that excircle from the original right angled lines.
    1
  15098. 👍 for the shirt
    1
  15099. Could someone explain the t-shirt?
    1
  15100. I confirm the capital Delta made me very happy. :-)
    1
  15101. Thank you, this made my day! ❤
    1
  15102. Hello the no 9 series is my favorite. Even with the proofs I can't beleive it.😃
    1
  15103. I remember seeing the first proof of the theorem. In school we didn't prove it by rearranging but using the original Euclid's proof which I always took as the ultimate proof.
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  15104. Paul Levrie's article is available for download in the link below. It is the only article of that volume that is available for free. link (dot) springer (dot) com (slash) journal (slash) 283 (slash) 34 (slash) 4 As YouTube does not publish my comment if I include the link, I have put it in disguise, sorry. As a matter of fact, if you have enjoyed this video, there is a book by William Dunham called 'Euler, the master of us all', which shows important contributions of Euler in different areas of mathematics, along with insights of his developments. The book includes commented out tour for proofs of the series expansion of the exponential and logarithm functions, as well as how and why he actually named a particular base 'e'. And, of course, the Basel's problem. It really makes you feel Euler's genius.
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  15105. Can’t tell if these are people or BOTs, but the grammar suggests BOTs
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  15106. 9:39 mfw mathologer forgets the +c
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  15107. So much beauty
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  15108. So zeta(1) is a singularity and zeta(2) is not, and the rest is medieval poetry. I like poetry.
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  15109.  @Mathologer  Sorry, I didn't intend to disturb you in person, I was just hoping someone could confirm I got it right.
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  15110. Is it possible to find similar patterns with different parameters? Is 7 the only one provides trace similar to straight segment?
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  15111. Thing is - the paint is assumed to be covering the horn with an infinitely thin layer - so if you need to cover an infinite area with an infinitely thin layer of paint - then the volume of paint that you need is infinity area times zero thickness - which is undefined - so (evidently) 8 liters is plenty. If you think that paint has to have a non-zero thickness - then it won't be able to fit into the entire length of the horn - so you STILL won't need an infinite amount. So why is it a paradox?
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  15112.  @Mathologer  Thank you for the response. That is a tidy proof. I really should be getting to bed though, so gute Nacht.
    1
  15113. Has anyone worked with the following configuration of 4 pegs: Start with 4 pegs connected in a sort of Y configuration, 3 edge pegs connected to a center peg. If on an edge peg, you can only move to a center. If on the center peg, you can only move to an edge. Seems to me, with 1 disk starting on an edge you can get to another edge in 2 moves, with 2 disks starting on an edge you can get to another edge in 6 moves, with 3 disks on an edge you can get to an edge in 12 moves, with 4 disks you can get there in 24 moves (I think, this is where you start second guessing yourself), and at 5 disks, you guessed it... 38? No seriously, this is what I was able to actually repeat if I counted correctly a number of times! Very weird if I'm not messing up. So, then I decided to try an X shape, 4 edge pegs with one on the center. 1 disk you can get to another edge in 2 moves, 2 disks you can get to an edge in 6 moves, 3 disks you can get to in 10 movies this time, 4 disks seems to be 16 oddly, and 5 disks seems to be 24. You could do a Hanoi-ish puzzle presumably with any sort of graph structure defining the connectivity and thus possible moves you can make at each peg/node. I wonder what happens in more complicated graphs... can you get any possible configuration in almost any graph (maybe save some particularly restrictive exceptions), or does it become more and more challenging to get to certain states if you keep going? Certainly something I might puzzle over and play with a bit myself. Edit: I *think*, again easy to get lost in these that actually you can do 4 disks on the Y configuration in 22 moves, not 24.
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  15114.  @Mathologer  Thank you very much, I'll have to give it a peak ^ ^
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  15115. I had never seen either proof, but the algebraic proof was very helpful for me as I'm currently reading an algebra book and trying to get back to where my education left off.
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  15116. That is amazing and proved that everything is connected. First time to realize that 1/x is hyperbola !
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  15117. The "greying out" problem has always been a problem in 3d graphics, for the very reason that you describe about the white line forming a value change in the 3d perspective distance, as things begin to "white out" as they recede. TRON the Disney production worked on this problem extensively, and found a pseudo solution for scaling lines as they recede into the distance, going into black rather than white. This created a much more believable scale of digital environment for the completely 3d landscapes they produced, such as the grid for the light cycle battles. A lot of these algorithms are still used in graphics today, though are much more refined and less regarded of as "problems" which had very technically derived solutions mathematically speaking. Perhaps you could graph how light appears to "dim" as it recedes to infinity? This begs the question of how scale of a line (width) affects reflected light off an object at specific distances along a path to infinity... food for thought.
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  15118. That selection of music packed some powerful emotion at the end. Amazing stuff, as usual!
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  15119. The animation at circa 18:15 is the best thing I've ever encountered on this channel. Please, give us more of your original research, beautiful dreamer.
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  15120. Great!
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  15121. 🦇I am really learning the mathematics, in your choice of vocabulary and explanations. 🧑‍💻 I see as you are explaining things, that you are learning, as you are speaking. 🗝 Your teaching skills; in knowing when to present a graphic, and moving around the screen, is sovereign strength. 👾 If you only knew how much information and tuition, I am scoring by your pedagogy, you would attack me in a covetous rage, ...just one breeze of resentment... Good job. 🤩
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  15122. His shirt is pretty cool, moon signaling. I wonder if the center of the moon could physically maintain a fusion reactor. I never heard of this puzzle before. It reminded me of the Triangle Peg Board Game, with golf tee's as nodes. Mathologer should be given a massive budget to expand his video content.
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  15123. Swivel , just because I didn't originally catch the insight into the incircle... and the swivel has reflective properties that feels like decoding a secret message through a mirror - very cool. Colouring is ok. It's nice.
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  15124. 333 is not half bad.
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  15125. The memorable is the amazing approximate formula and the Euler's constant gamma . Without your animation these things won't be memorable . Thank you so much . I admire you so much . Keep doing more videos on various topics . And one thing advance happy Christmas.
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  15126. Calculus is easy, but this video is not for beginners.
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  15127. i was guessing the inverted hemisphere would be paraboloid inside. i am happy to see that it is included later on and the original shape was also a paraboloid. i am completed
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  15128. most memorable part: the smooth appearance of the super villain gamma.
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  15129. I seem to remember that what Napier actually invented was something like the importance of 10^7(1-log n). I wonder if the hardware version of the similar idea promoted the change.
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  15130. Hi Mathologer. I am a keen amateur with a few maths units in my B.Sc and a very poor maths start in year 1. I love your videos but they normally go too fast for me, especially in the algabraic sections. Unfortunately, I am usually in bed when I watch them so I don't get out pen and paper and stop and start the video to give me an opportunity to internalise the concepts. I am getting too old to remember them, even if I did. I still get a lot of satisfaction from seeing the process displayed in such an interesting way.
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  15131. With logarithm tables and slide-rule (slipstick), returned to UK schooling from Africa in 1958. German slide-rule; it felt like boasting a gold Rolex at the school Admissions test.
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  15132. Instead use this concept in reverse to "prove" the expansion of (n+2)^3. Picture a n×n×n cube (made of n^3 unit cubes). How do you turn it into a (n+2)×(n+2)×(n+2) cube? Add 6 faces (left, right, front, back, top, bottom) each of volume n×n×1 = n^2. Fill in the 12 missing edges each of volume n×1×1 = n Fill in the 8 missing vertices each of volume 1×1×1 = 1 So (n+2)^3 = n^3 + 6(n^2) + 12(n) + 8(1)
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  15133. "Later on I also present a proof that there is only one way to write 4k+1 primes as the sum of two positive integers." Are you sure? 😜
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  15134. Coloring proof was easier to see
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  15135. Visual logarithms. Never thought I'd see that in my life
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  15136. I love your maths expositions. But your German pronunciations are really second to none.
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  15137.  @Mathologer  I know...
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  15138. Looking forward to the mathologerized proof! You probably already know, but 3blue1brown has a video on this exact topic, its really good although he also glosses over the fact that every prime congruent to 1 mod 4 can be written as the sum of two squares.
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  15139. So, when thinking of the coin paradox, I realized that distance of the 2 centers of the coins divided by de radius straigh up gave you the answer: for a coin spining over a coin of the same same size, the distance of the centers are 2r, divided by r we have 2; Coin spining over a coin twice it's size, the distance of the centers are 3r, divided by r we have 3; For the coin spining inside the coin twice it's size, the distance is r, so 1 spin; Lastely, for the coin spining a coin 1,5 times it's size [or the spinning coin is 2/3 of the bigger coin], first I "rationalize" the coin sizes by multiplying both by 3, so they became 3r [bigger coin] adn 2r [inside coin], and again, the distance from the centers is r, so 1 spin again. Can some one back me up with the necessary math? 😅
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  15140. Thank you for releasing a corrected version despite the obvious drawbacks!
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  15141. It reminds me of Benford’s law
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  15142. I thought you were going to make a video about Galois theory. I would be interested to see your treatment of that subject. Maybe you haven't done one yet, or I missed it.
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  15143. Ptolemy is erased because geocentric thinking. Educators think students are too stupid to understand a person can have good and bad ideas.
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  15144.  @Mathologer  egocentric earth.😆
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  15145. I haven't given it a full go, but I'm willing to guess that the number of tilings for the glasses (or any other board with holes) is equal to the MAGNITUDE of the determinant of the same kind of matrix. Removing the i and negative sign from relevant outputs of the standard algorithm shouts "magnitude" to me.
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  15146. 39:00 smallerer :)
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  15147. There were three things mentioned that I very much enjoyed in this video, I cannot choose one of them: 1) That no integer other than 1 can be summed with consecutive partial sums. 2) The Logarithm formula for calculating partial sums. 3) Kempner's series without 9s, especially the fractal animation.
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  15148. 32:30 wait what? what about 6^2 - 4^2? The result is an even number. What am I missing? You said we can write x^2 - y^2 for exactly when an integer is odd
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  15149. Beautiful!
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  15150. This video is even better than the last one, because you fixed that error! Also, cool spiral thingies.
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  15151. Yes it was a joy to follow, thanks. BA mathematics, MA statistics and PhD in epidemiology
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  15152. Great video! Does anyone have the piano sheet of the ending music ? It’s beautiful !!!
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  15153. Merry newyears, conrgats on the 666k subscribers!
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  15154. Sequencing dimensions with a natural count of Pi divisions at the primary coordinate intersection, as the secondary coordinate intersection drifts away, is a more neutral camera perspective. It also points out when two vertices intersect.
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  15155. Stray thought after just the intro (up to where Schwarz has been introduced): consider a simple straight line of length 1. Approximate it, in the first instance, by the other two sides of an equilateral triangle on it. This approximation's length is 2. It lies within sqrt(3)/2 of the line, attaining that bound at one point. Repeatedly: construct a line parallel to the original at half the distance from it of the furthest point(s) of the path from it; fold the parts of the path beyond that down so those parts previously furthest from it are now on the line. This does not change the length of the path since, in each case, we flip an equilateral triangle about the edge of it that isn't part of the path. At each step we halve the distance from the original line of the most distant part of the path. So the path "converges" to the line, but has twice its length. So you don't need circles or pi to get the original pi = 4 paradox: you can do something exactly equivalent to show 2 = 1.
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  15156. as a graphic designer, i’m gonna need to learn keynote animation next..
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  15157. I prefered the gamma explanation cause i heard about this constant a lot and now I can understand a bit what it is
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  15158. Wear a jacket and tie for next presentation. Oh, yes, and a sleeved shirt too. 😀
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  15159. Gauss did it? It wasn't Euler? 🤦🏻‍♂️
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  15160. 13:28 looks like the CC playback on YouTube doesn't understand the ligature glyph for "ff" that whatever program you wrote the transcript in snuck in there.
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  15161. 15 answer
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  15162. These animations leave me hypnotized....and cross-eyed!
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  15163. Has anyone made lanterns out of a cone? How about a dish or sphere? Many thanks for the content!
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  15164. Newton backward interpolation will work faster
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  15165. Kudos to your animator! 14, though I thought it was a pi problem first!
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  15166. I showed V-E+F=2 and she had the epiphany of THAT's why they always say we will communicate with aliens using math. thank you for that. I crap at explaining this stuff to anyone and I never could get that one through to her.
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  15167. I loved the crazy overhang best. It shows how maths can have surprisingly creative alternative solutions
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  15168. My background in mathematics is more than this, so may you make even higher level videos? Also, can you make a video on Mahler's measures?
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  15169. I paused at 27:18 and using that equation for 1^10 + 2^10 + ... + 1000^10 was pretty simple to solve. Using n=1000 makes it really easy because all the n^x just become 10^(3x) 1/11 n^11 + 1/2 n^10 + 5/6 n^9 - 1 n^7 + 1 n^5 - 1/2 n^3 + 5/66 n 1/11 10^33 + 1/2 10^30 + 5/6 10^27 + -1 10^21 + 1 10^15 + -1/2 10^9 + 5/66 10^3 1/11 = 0.0909090.. 1/2 = 0.500000... 5/6 = 0.833333... -1 = -1.000000... 1 = 1.000000... -1/2 = -0.500000... 5/66 = 0.075757... 090909090909090909090909090909090.90909... 500000000000000000000000000000.00000... 833333333333333333333333333.33333... -1000000000000000000000.00000... 1000000000000000.00000... -500000000.00000... + 075.75757... ------------------------------------------ --------------------------------------- 91409924241424243424241924242500.00000...
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  15170. Most memorable part: that the sum always misses the integers.
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  15171. Thank you so much for providing the "missing link" at long last! Many years ago I had to present some information about Fourier Transforms and signal analysis, so, obviously, sine waves were super important. With a friend of mine we pieced together everything we could have been taught about trig functions but never were. To our surprise, they had nothing to do with triangles but everything to do with circles. You can relate the meaning of the words sine, tangent and secant to lines on a diagram based on a circle. The worrying thing was that in a previous existence my colleague had been a maths teacher and I was an engineer so both of us had been familiar with trig functions for years at that point but we never saw the "big picture." Subsequently, I decided there ought to be a link to the hyperbolic functions. I discovered the area link but could never see a reason for it - until today! The "turning circle" did it for me. Thanks again.
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  15172. Pappa smurf heavens with ice cream and real frogs be nice at Satans' Alley. I go outside, outside now time.
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  15173. I think the first version is cooler animation-wise, but I might prefer the second one actually, since it is more intuitive. (Okay and I might be biased since that is the proof I came up with when you said I should pause and see if I could prove it myself)
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  15174. Honestly I loved every part of this video and cannot chose which one was my favourite.
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  15175. I love the nice hint on your t-shirt that phi = 1 + 1 / (1 + 1 / (1 + 1 / ( ... )))
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  15176. Are we sure this "shadow trick" always works? 🤔 What If a face degenerates to a line?
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  15177. I think I prefer the longer proof, ‘cause I feel the side paths and additional demos give me the sense of how many neat facts the proven fact actually supports. The short proof is beautiful, but I think my appreciation of it is much richer with the first proof in my head already.
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  15178. Wonderful!
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  15179. I'm sorry, algebra, but I don't know your x or y she left.
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  15180. Thank you for another excellent video.
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  15181. Arrange the puzzle in a loop surrounding a 3-4-5 triangle
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  15182. Excellent! I love it.
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  15183. I used the 4-bug puzzle in my senior physics class as an illustration of vector components. Bug A steps toward bug B. If bug B stepped a little bit away from bug A, bug A would have to walk more than the length of the square. If bug B stepped a little toward bug A, bug A would reach it in less than the original distance. But since bug B steps exactly perpendicular to bug A's step, it neither increases nor decreases its distance away. Hence A reaches B in a distance equal to the side of the square. In a nutshell, this problem demonstrates that mutually perpendicular vector components are independent. Since bug B's movement is perpendicular to bug A's movement at all times, it can be neglected. The answer is the same as if bug B never moved! Each bug travels the length of the original square. Although I took a lot of words to explain it, I think the vector component (i.e. the physics answer) is the simplest.
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  15184. Yes. I am disgusted. Thanks for that.
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  15185. hello from 2015
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  15186.  @Mathologer  mth2015
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  15187. That's another thing I really should have noticed but didn't. On the plus side, there's no sign of these things running out.
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  15188. Tring to connect it with Meruprastara...
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  15189. this is just finite differences
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  15190.  @Mathologer  This is the method Charles Babbage used to create polynomial functions when calculating navigation tables. An N-layer finite difference can generate any polynomial of nth degree. If a person sets the constants correctly on the left hand side of the layers, a one can generate pure powers. You have a trivial example, hard to do here without monotype, but you will get it. 2 2 2 ... 1 3 5 7 .... 0 1 4 9 16 ... Second finite difference = 2. Setting the third finite difference to a constant can get third order polynomials. Etc. You are just pulling the numbers off the layers. Set the third finite difference to 6. 6 6 6 .... 6 12 18 ... 1 7 19 37 ... 0 1 8 27 64 .. Ah .. so not so obvious to me. Always love it when a mathematician of your insight says something is obvious .... perhaps you should add 'to a mere mortal' or something to the end of that. ;-) Perhaps some credit to Charles Babbage, though I have the impression that this method was well known in his era.
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  15191. This is probably incorrect, but I think the first time I was introduced to this concept was from the Scarecrow in The Wizard of Oz.
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  15192. A truly scrumptious proof. Is there any generalizing to more than two sides with eyelets? -for example lattice towers with three or four vertical segments linked by diagonal cross members like this image: https://5.imimg.com/data5/AL/WH/MY-3337989/lattice-type-towers-500x500.jpg Of course the cross members are in compression rather than tension so the definition of strongest might need to be modified slightly. Or alternatively tensegrity structures with cross members in tension - but that's probably too big a jump as typically the equivalent of eyelets won't be in rows anymore.
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  15193. Shame you are missing 0.0000758
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  15194. Wait a minute! The area under the 2-d graph you have proven to be infinite IS NOT the surface of the horn. This is still missing
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  15195. Remembering from calculus that the cylinder with the least surface area for the most volume has a height equal to the diameter.
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  15196. Hm! I wonder why I didn't discover this two thousand years ago. I'd be rich!
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  15197. To solve (153, 104, 185), we need to factorize 104 to 2uv - that gives (u,v) = {(4,13) or (2, 26)} 153 should be (v-u)(v+u) which fits with (4,13) which gives the matrix as 9 4 17 13 To get to the tree path, if 9,4 is picked up from the bottom 2 of -1 5 9 4 That's invalid. So we pick up from another. It turns out the parent is 14 95 Now this can iteratively be built
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  15198. Wowwww sir thnx for such hard work ... ... I never thought maths is so beautiful...🙏🔥❤️
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  15199. A killer sudoku solver program I wrote previously had a math PhD student helping me discover all these polynomial tricks. A killer cage in sudoku has a sum that must be satisfied by some combinations of typically single digits. Another application of the concept :)
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  15200. I like your video, but could you include a little history of it, too. Who were the mathematicians in the 1600s who studied it? What did they find?
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  15201. Is there a continuous version of the stack of books? I'm guessing it's hyperbolic.
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  15202.  @Mathologer  Thanks for the reply! I'm reading your paper right now.
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  15203. my favorite part was definitely the 700 year old proof that the harmonic series diverges
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  15204. Thank you :)
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  15205. sinh is pronounced sinch.
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  15206. 24:37 If you place the corners of the red squares at the centers of the blue squares, you don't need such a complicated puzzle to prove Pythagoras' theorem.
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  15207. This is a hall-of-fame level video. Really great. But I did wonder for a moment if he was secretly American because measuring a 3, 4, 5 METER triangle in an ordinary room seems unlikely. It’s more plausible in feet. Just sayin.
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  15208. Great video. Most memorable: any integer is a subseries of the harmonic serie!
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  15209. 17:55 wtf?!?!? Dude has that many people throwing him scratch?
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  15210. I like the swiveling proof more, because it uses less colors at the same time. It also seems more directed. In the color proof, the result seems to just fall out, but in the swivel proof It takes less work to understand why what we're doing will result in the proof. Unlike some of the other commenters, I felt like I could trust both of the proofs equally. Maybe that's because I imagined the swivels as reflections?
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  15211. These Petals diagram reminds me of Multibrot Set.
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  15212. 3:21 His shirt🙃🙂
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  15213. I studied math once, 1957 as I recall...
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  15214. So cause there is no repeating in the Pi digits it can not be written as a fraction. cause every fraction must have an infinite repeating pattern of digits(or 0). beautiful :)
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  15215. A(n) Aztech diamond has 4 times the nth triangle number cells. Relevant? The number of possible tilings of A(n) is also 2 to the power of the nth triangle number.
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  15216. My favorite was Kempner's proof, I found it very counterintuitive and had to think about it quite a bit.
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  15217. 22:22 cos 0° = 1 sin 30° = 1/2 tan 45° = 1 cos 60° = 1/2 sin 90° = 1 cos 120 = -1 Hmm. One of these is not like the others.
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  15218. It is this visual approach which I believe Fermat used to deduce his last theorem; the cubic nomand (not sure if I heard that Pythagorean word right) which is described here as a hexagon, is not a member of the cubes.
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  15219. CHALLENGES! The following two questions were written before I saw any proceeding chapters. So any of the following may be redundant, but I came up with it myself. 23:39 Show that the degenerate point is the centroid. A: If you know how Fourier Transform works (or at least have the intuition of it from 3B1B's video), this is trivial. But in case you haven't seen it, here's my take: The centroid of all the other pentagons is the origin. By adding together all the pentagon points the centroids do also add together. The centroid is the average of all the other points, and taking the average is a linear function, which is why this works. This also gives a way to figure out which pentagons to use. To get the center of mass you add the red point, the black point, the orange point, the blue point, and the green point, and then divide by the number of points. What if each time you sampled a point you rotated by 72 degrees clockwise first? In the composite diagram with all the regular pentagons, all of our pentagons begin to shift. The degenerate becomes a regular pentagon, one of our directional 72 degree pentagons (the one going counterclockwise) reduces to the degenerate! By repeating this with the other angles you can figure out what the composite pentagons are that make up our shape. Congratulations, you have just successfully performed a DFT... kinda. This also shows that it doesn't matter how many sides we start with, the theorem still holds. 24:03 What happens when you go for multiple rounds? I'm gonna be switching up my terminology here and there so here: The 72 pentagon is constructed by going 72 degrees. I also call it the anti-288 because applying 288 degree ears will reduce it to a degenerate. The shape is gonna be the same regular shape, that's for sure. What's not sure is the orientation and size. The anti-72, anti-144, and anti-216 all reduce to the degenerate all the same, so let's see what happens to the anti-288. 72 will affect the anti-288 all the same. 144 and 216 complement each other (in the 360 degree sense of complement) so they'd undo each other minus a rotation of some sort. Any set of ears rotates the 72 pentagon 36 or 216 degrees, the 144 pentagon 72 or 252 degrees, the 216 pentagon 108 or 288 degrees, and the 288 pentagon 144 or 324 degrees. The way to choose which one is, is by seeing if the ears "hook" around the origin, that is, if it's closer to the opposite side of the pentagon. If it does, pick the closer to 180 degrees. Heuristics. We'll fix this definition later. Also the variants to choose from are 180 degrees apart, so if there happens to be a 180 polygon in your decomposition (which will never have this "hook" property), it'll rotate 90 degrees if your angle of ears is below 180, 270 degrees if it's above 180, and it will reduce to degenerate on 180. We have an additional 144 and two additional 216s. This will rotate our final anti-288 polygon 72 + 108 + 108 + some multiple of 180 degrees = 108 + some multiple of 180 degrees. Also the ears do scale the pentagon but the gist is since the 216 ears show up on the inside of the anti-288, it's gonna be smaller. So our anti-288 is also known as the 72-polygon. On an angle of 72 and an outer radius of R, the inner radius or apothem is R * cos(36). The side length is 2 * R * sin(36). Now to get the short side we need to look at our isosceles triangle. It has base length of 2 * R * sin(36) and it has an angle of 216 degrees. If we slice and look at a right triangle of 72 degrees INSIDE and an opposite leg of R * sin(36), we have the height = R * sin(36) * cot(72). So we get that our resulting angle is inner radius +- extra (plus if the ears < 180, minus otherwise). More precisely, if R is the outer radius of the polygon , A is the angle of the polygon and E is the angle of the ears, the new outer radius is R * cos(A/2) + R * sin(A/2) * cot(E/2) (the sign of cotangent makes things simpler), and the scale factor is cos(A/2) + sin(A/2) * cot(E/2) We can now more directly define the whole conundrum about the hook thing: a negative scale factor is equivalent to a 180 degree rotation. And even more: the scale factor is negative on a theta-polygon for angles larger than 360 subtract theta if theta is less than 180 degrees, and for angles smaller than 360 subtract theta otherwise. Bonus: on a theta-polygon, the angle ears theta/2 and 180 subtract theta/2 will keep the size the same (you can easily check this by inscribing a polygon with double the number of sides and using inscribed angle theorem, or you can use the fancy formula derived earlier) We can come back to our problem: what happens to our anti-288 polygon? It gets an extra 144 and two extra 216s. Crunch the numbers and we get a 288 rotation and a scale factor of 1 * 1/phi * 1/phi = 1/phi^2 compared to the original. As a final point: talking about scale factors and rotations brings me to the final point: the ears operation is commutative because 2d rotation and constant scaling are commutative as well. 28:36 The sum of the coordinates of two similar triangles is another similar triangle. The second similar triangle is a complex number times the first one. Factor out the similar triangle, and you're done. Note that chirality matters: a triangle plus its mirror or conjugate is a straight line.
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  15220. As always, an excellent video. After watching one of your videos, I'm always left with an unformed thought, like an itch you cannot scratch. I feel we are seeing glimpses of some universal truth that we still cannot see completely or understand. It is a frustratingly delicious feeling. Thank you.
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  15221. Has Pythagoras ever realized how many proofs there are? Just in primary school we learnt at least 10.
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  15222. Several libranduus disliked due to lack of appreciation and education
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  15223. Hello Mathologer, thanks for another interesting and inspirational video. I got to the end; for me none of it "didn't work" and I love your clear explanations of some really quite complex stuff. As long as you start off from a reasonably simple starting point I think you could disappear down the deepest mathematical rabbit holes and hold my attention to the end. I got a degree in engineering, work as a software engineer and enjoy recreational maths; I should probably "do" more than "watch" but I'm no good at coming up with problems I can solve myself. I think I understood everything in this video but might have to replay some of it a few times before trying to explain it to someone else. Now going to see if I can cross 3blue1brown's lighthouse derivation of pi^2/6 with an inverse cube law to see if I can find f(pi)=1.202056903.
    1
  15224. God he is good isn't he: great stuff! Euclid would be proud: this is geometry at its finest!
    1
  15225. You overlooked a more sinister aspect of the paradox: somebody named Leslie is trapped inside the horn! Proof: GABRIEL'S HORN anag. HARBORING LES.
    1
  15226. Where can I get me some of that algebra and substitution auto-pilot? I need to fix my addiction. LOL/<Not>
    1
  15227. …still loving the start mostly ❤
    1
  15228. Nice video. My proof of the first part: it is enough to show that the quadrilateral formed by the a and c whiskers is cyclic. For this, we use the 3 isosceles triangles (two whisker triangles and one triangle with side length a+c). Then opposite angles are complementary.
    1
  15229. I just love your work. And one word that I always wait for you to speak is " Ooohkhaay ". More health to you. Looking forward for more such videos.
    1
  15230. Gamma difference less than 1 fitting in first square
    1
  15231. I did not see any mention of the PRIME Number Wall (aka PRIMEVIL WALL), of course there is no linear Fibonacci sequence in that case, so there will be no termination of the wall. So I predict there will be no pattern in the PRIMEVIL WALL, because if there was it would necessarily terminate. I suppose we could have pseudo prime number walls that never terminate, but not necessarily formed from the primes? Its very difficult to scale these walls, I found that out as a small child, so gave up then! 👾👾👾👾💣🤷‍♂
    1
  15232. I know the shape of the object that casts a 4-D Shadow out to 3-Dimensions.. And I solved The Riemann Hype in a very special way… Unknown axioms and Higher Dimensions….. My Math Skills are Lacking But I’m friends with Neutrinos and I know the Places where they hangout on the seamy -side .. Where if yo want to be Pheonixed? Come and get Parked. Ill show yo how Space Folds and makes Infinities ! And How it is simply connected to the metric circle . .. Well I have A lot of Big Talk….. !
    1
  15233. Simple,when A is a negative number (-1,-2,-3,-4,-5,etc.),A+A is not 0,the sum of negative numbers just make it more negative
    1
  15234. The date of upload of this video = 9 ! 1+9+2+2+2+2=9
    1
  15235. ​@Mathologer, oh, you bet that, Prof. Polster, you bet that 👍🏻
    1
  15236. Thank you for the video, I like it. Do you know in Nepal we follow a lunar calendar. After 3 years there is an extra month added to balance the solar and lunar year - the "mala maas". At that time there were no any mathematics to show such things but they calculated the same idea accurately.
    1
  15237. at first i was thinking you could for example move the red/blue corner up and not increase the size of the green side. but that would force the other end to move and it would fall out of the circle shape.
    1
  15238. RGP not RGB because sides of the triangle are ABC so using B would be confusing.
    1
  15239. The reason that the circle with cutting corners of a square doesn't translate to pi is because you still have a *Lot* of right angles that are all adding up their lengths. Those right angles do not smooth out, no matter how infinitely down you go. As you get smaller and smaller, the brain wants to smooth them out into a nice solid curve around the circle and ignore that they are still squares repeating around a circle. The reason Pythagoras' method works better is because the angles are chopped finer and finer too, making the more gradual angle around the circle smoother.
    1
  15240. As always: WHAT A VIDEO!!
    1
  15241. At 3:36, there is an unshaded non-match starting 10582 on top and 10581 on the bottom
    1
  15242. I like to make little interactive math demos in JavaScript for my website, and this little miracle would make a great addition! I think it could look quite pretty if the polygons were colored sequentially according to some color mapping function. If anything comes of that, I'll edit the comment again.
    1
  15243.  @Mathologer  I finally have some time to work on it...I'm probably not going to implement the eigenpolygon stuff, just the theorem itself.
    1
  15244. I have a degree in maths and physics and I am working in the field so math is all I do. Yet you manage to keep finding this amazing beautiful things in such simple ideas like Pythagorean triplets and circles which I thought for a long time have nothing to them. Absolutely amazing and beautiful thank you
    1
  15245. what the positions in the tree of triplets generated by the typical fibonacci sequence??
    1
  15246. Weird This is trivial - anti-irrational construction.
    1
  15247. Marvelous! I pretty sure that using the strategy of approximate a number like min. 16:30 we create a base numeration system! In that pi = 13, 21, 58 ...
    1
  15248. Great video as always. Beautiful proof as always.
    1
  15249. When you first showed the lantern I began to think of buckling and of the so-called zero energy buckling modes that bedevil finite element analysis of shells.
    1
  15250. 11:10 It is a beautiful thing. 10/10 for '10 is special' But it uses '4 is special'. It would be nice to prove '4 is special' without 'trial by exhaustion' If line 1 was A, x, y, B, line 2 is A+x, x+y, B+y, line 3 is A+2x+y, B+x+2y, line 4 is A+B+3(x+y)=(A+B) mod 3 If the 3 colours are numbered 0,1,2 we want 0~0=0, 1~1=1, 2~2=2, 0~1=2, 0~2=1, 1~2=0 Compare with the function 2A+2B mod 3 0+0=0, 2+2=1, 4+4=2, 0+2=2, 0+4=1, 2+4=0 all mod 3 So the single cell in line 4 is 2A+2B mod 3; x and y are irrelevant, lines 2 and 3 are irrelevant, so compacting can begin.
    1
  15251. I've known about the terrible aim property and it's proof for quite some time, but when you showed it I kind of remembered how intuitive and amazing it is! Definitely my most memorable.
    1
  15252. Programming challenge: Oh ye of little faith, it is possible to calculate answers for values like $2 googol quickly with a fairly "computery" technique. It boils down to raising a 191x191 matrix (of big integers) to the 200-googolth power (or whatever). This is a tractable number of multiplications because you can exponentiate by repeated squaring (square a matrix 340 times and you've already exceed the 200-googolth power). (None of this would be tractable if the big integers became too large, but we know they don't if the final answer doesn't.) My program calculates: $2: 2728 ways $20: 53995291 ways $200: 4371565890901 ways $2000: 427707562988709001 wyas $20000: 42677067562698867090001 ways $200000: 4266770667562669886670900001 ways $2 million: 426667706667562666988666709000001 ways $2 billion: 426666667706666667562666666988666666709000000001 ways $2 trillion: 426666666667706666666667562666666666988666666666709000000000001 ways $2 googol: same but more 6's and 0's Two trillion takes about 1.4 seconds, two googol takes about 10 seconds, 2*10^1002 cents takes about 107 seconds, 2*10^2002 cents takes about 257 seconds, 2*10^4002 cents takes about 13 minutes, and that's the highest that I let finish. The approach is to start with the standard dynamic programming approach to number of ways to make change. (Standard in that this problem is a textbook example or exercise of DP.) This approach takes time proportional roughly to (target number of cents)*(number of types of coins) so it's obviously not tractable for a googol. However, in the DP approach, when you progress from N cents to N+1 cents, there are only 191=1+5+10+25+50+100 values in the state that will ever be used again. It is possible to use 191 variables (in the mathematical sense, not the programming sense) for the state, and the transition from N to N+1 cents is purely a linear recombination of these variables, so the transition can be represented as a 191x191 matrix and exponentiated, as explained. Obviously this is slower and less elegant than the generating function simplification in the video, but it is really quite fast until the actual answer has 10000+ digits, and exponentiating big matrices can solve a variety of problems on a computer. You just have to pick the right matrix. :)
    1
  15253. at the 1:18 mark, you say minus one sixth when you shouldn't have.
    1
  15254.  @shyrealist  well, what you don't know is that Burkhard and I, we share the same accent (and fate that comes with it) 😂👍
    1
  15255. OK, we have 2D - Fibonachi, 3D-Steinbach. Any 4D? Some prolongated tesseractes? 😁
    1
  15256.  @Mathologer  same table, 4 numbers, 10th itteration. k, s, r r^2=s+1, s^2=s+r+2=r^2+r+1, k^2=sr+s+1. weird. r in nonagon same as table approximation.
    1
  15257. As with nearly everything: It is magic as long as you don't understand it and it loses it magic when you understand the concept/s behind it. Then the concept/s can become magic and so on (forever?).
    1
  15258. Amazing video SIR!!!🙏🙏🙏 ...as always inspirational ...
    1
  15259. Was first familiar with the 2nd proof since I found it myself many years ago. I also discovered a cool way to find an infinite number of pythagorean triples. The method doesn't find every triple, but an infinite number of them nonetheless with a surprisingly simple method.
    1
  15260. Ricardo, I'd be interested in seeing your simple method for generating an infinite subset of the Pythagorean Triples if you'd care to disclose it in this public forum!
    1
  15261. Anyone else think Edouard Lucas looks like Stalin?
    1
  15262. Quelle est la musique à la fin? What is the title of the music at the end?
    1
  15263. 17:34 hmm why ? I mean how can we do that ? Nevermind
    1
  15264. At around 18:00 you make a correspondence between all the possible moves and a graph. Does this mean that the statement "there is no proof that these moves are the optimal" corerspond to the fact that the graph theoretical approach hits a brick wall at some point? Because basically all one needs to do is proove that in a specific case the provided solution corresponds to the shortes distance between two points in that graph.
    1
  15265. Careful! That light at the end of the tunnel could be an approaching train!
    1
  15266. The optimal overhang with finite numbers was a big surprise for me. There is always more to learn. The guitar music at the end makes the choice hard, though.
    1
  15267. So the pattern to the fractions is that numerator N_i = 2N_(i-1) + N_(i-2), same for denominator
    1
  15268. Massive shirt
    1
  15269. Fantastic!!
    1
  15270. no way i was blown away
    1
  15271. Magic of digits.
    1
  15272. Small typo at about 17:16 . The wrong infinite sum is there (all integers instead of odds, all powers of x)
    1
  15273. Obviously 0.5 < gamma < 1. The unit square can be seen as split in rectangles of height 1/n and width 1. If every rectangle is cut diagonally in half from top-left to bottom-right corners, then the sum of all triangle above each cut would be 1/2. Every blue slice has the same corner points as every triangle, but each slice has a slight belly that protrudes from the triangle. Therefore, the sum of the slice's area must be larger than 0.5.
    1
  15274. Bonne prononciation !
    1
  15275. The most memorable part for me was the optimal towers, but as usual the entire video is amazing
    1
  15276. instead of looking for a key, try finding lock-picking tools. either way, we enter. of course, let's not talk thermite and other explosive ways of entering. :oP
    1
  15277. With any two numbers a and b in the top of the box we have a b a+2b a+b It is easy to check that this always gives a pythagorean triple: (a^2+2ab)^2 + (2ab+2b^2)^2 = (a^2+2ab+2b^2)^2 So this will work for e.g. Lucas numbers too
    1
  15278. XD problem set of MTH3170
    1
  15279. 👏 👏 👏
    1
  15280. I'm from India,i really love your videos they are interesting
    1
  15281. Oh Yes, a new mathologer video, two doses please
    1
  15282. When do you mention the term “slide rule”? [(Ah, above) and at 9:10]
    1
  15283. Actually, the anagram is perfect because Galilæus' real name was Galilous. Except it was not. 😄
    1
  15284. At 39:00 or so of your video, you mention that in the course of building out the diagonals, none of the lines intersect. That reminds me of Chris Webb's Gresham College video on math and art. At about the 13:00 mark of his video, he talks about the number of strings required to create certain patterns of Celtic knot, and if the grid consists of a pair of co-prime numbers, then only one thread is needed. It seems that these are related ideas... https://youtu.be/bOF5iCs3n2k?t=782
    1
  15285. How about some video on taking a derivative of integrals with integraton boundaires?
    1
  15286. How about using a negative base (say, -8). What circle would you then use?
    1
  15287. amazing how beautiful this is. I marvel at what God has done, we only discover what has already been created
    1
  15288. Every Mathologer video is like a YouTube version of math paper (be it any paper, or a master's thesis or doctoral dissertation).
    1
  15289. Can you use a black background next time?
    1
  15290. Happy for you to call it Archimedes tangerine. The version in the thumbnail is Archimedes pumpkin :)
    1
  15291. We missed getting it out by Halloween.
    1
  15292. I found this very enjoyable and thought provoking. Speaking of provoked thoughts - the original problem posits a cylinder with diameter 1 and height 1 giving a surface area of pi. However as soon as one starts creasing the cylinder to create rings (even or especially an abstract mathematical cylinder) perforce the indents will shorten the height of the cylinder implying that as the number of rings increases infinitely, the height shrinks to zero leaving just a circle. Now I am not sure if I just disproved the lemma by showing that it can not exist or just found an other solution.
    1
  15293. I'm rewatching this video after a while, and it really clicked for me. One viewing is not enough, but don't change anything! It was great!
    1
  15294. [9,4] [17,13] is the square that creates the Pythagorean triple in the puzzle.
    1
  15295. I wish you were my teacher for algebra...I definitely would have done better!
    1
  15296. I love that Tshirt!
    1
  15297. Team coloring proof here
    1
  15298. 22:48 2xy - (2+x+y) +1 = N-2 <=> 2xy - 2 - x - y +1 = N-2 <=> 2xy - x - y +1 = N <=> 4xy - 2x - 2y +2 = 2N <=> 2x(2y - 1) - 2y +2 = 2N <=> 2x(2y - 1) - (2y - 1) = 2N-1 <=> (2x - 1)(2y - 1) = 2N-1
    1
  15299. Brilliant!
    1
  15300. I love your stuff! Thank you! I’ve always been interested in the doszen system and its possible benefits, like when you divide the integers you end up with whole numbers instead of decimals. Could you please do a video about counting to 12 instead of 10?
    1
  15301. You're great, keep up the good work. And please don't turn into another Sabine Hossenfelder telling us about the virtues of men in women's sports and the immense value of child mutilation for society.
    1
  15302. Lawks! It's all so good that picking a best is very difficult, but I think I'll go for Baillie's work (no 9's etc.) And of all the maths channels I watch, I'll vote Mathologer as best too :) !
    1
  15303. Brilliant video as always! Thanks a lot, professor!
    1
  15304. 25.807 is actually closer to the root of all evil. Evil rounding error!
    1
  15305. Favourite graphical proof of Pythagoras: hmmmm. I'm not a great fan of "cut them out, move them around and it's obvious it's still the same area" proofs. It's really easy to make it look right while it's actually not (see "chocolate" that I refer to below). Related but not the same: Penrose, in Road to Reality, tiles the infinite plane with two squares and makes a larger grid by joining equivalent points so the two starting squares add up to the bigger square that makes the grid. That works if you are happy to believe the squares can tile the plane and that the two squares do, in fact, equal the bigger square. That's either circular reasoning or you need a really good argument for why it's true. I suspect the latter is the case, but I am a Bear Of Little Brain and don't quite see it. Perhaps counting the vertices of the grid and showing they map to just one pair of squares does it. Oo. Maybe it does. Anyway: above my pay grade, as they say. I quite like the first proof with the four triangles at the corners of the square, but I think you have to be careful to just "slide" two of the triangles to the triangle opposite. I think that can be justified rigorously. Spinning them round and shuffling them about really isn't convincing: we've all seen the "cut a bar of chocolate at an angle and make the same bar plus a bit extra" video that (I think) Vsauce did. To be honest, my favourite proof of the "Pythagoras" theorem is Euclid's: it's not visually stimulating; it's terrible at breaking the ice at a party (trust me) but it's long and convoluted and I can actually remember it, which puts it in a small set of proofs.
    1
  15306. see this other object, finite value at 2 dimensions paradox an infinite at 1 dimension: circle have finite area to one given radius R . One swirl line starting at center until that R have 1D Length infinite, as the line diameter is infinitesimal. Is the polar coordinates . The Integral from 0 to R of 2πr, gives exactly πR^2, but trying to see the integration process as that increasing swirl see the 1D line Length going to infinite. At Gabriel Horn is the same. 3D finite paradox a 2D infinite, one dimension below.
    1
  15307. I'm terrible at math but still love it and greatly appreciate your channel! Would love to buy a heather grey color Mathologer T-shirt (hint hint).
    1
  15308. Regarding the sum of the first 'n' terms of the harmonic series, those 'extra blue bits' trailing off approximate triangles beyond the first few elements, so you can approximate the sum much better with ln(n+1)+gamma-1/(2(n+1))
    1
  15309. beautiful
    1
  15310. Perhaps you are immature, but Pythagoras was a Master Astrologer, and we are part of an imperialistic force, that has and IS, crushing the truth of His culture and everything dualistic. That’s why he was cancelled. “We’r” stupid.
    1
  15311. Does this count as a solution for 3 bricks? Link in reply
    1
  15312. How great is it to open YouTube and see a new Mathologer video!?
    1
  15313. Who's Scarlet Johansen?
    1
  15314. Great! One suggestion: dark background, nicer for the eyes
    1
  15315. I have a masters degree in statistics and it was really smooth to follow along. I see the similarity (and relationship) of the last formula with McLaurin series which a lot of approximations in stats are based on. That was definitely my favorite part!
    1
  15316. The most memorable thing for me was the not integers in the partial sum, really interesting and cool stuff!!
    1
  15317. bro thats a whole ritual, although ngl his videos are just a treat
    1
  15318. Your “every number” demonstration provides a mapping, akin to the Minkowski Question Mark Function, between the reals and binary fractions. Begin with 0. and, with your chosen irrational value, write as many 1s as there are positive fractions before overshooting. Then write 0s for the negative fractions before undershooting. So, for example, ln(2) would map to 0.101010101…. In base 10, this is 2/3.
    1
  15319. Most memorable: missing all the integers
    1
  15320. My knowledge of Maths is fairly poor. My highest taught Maths course was Open University introductory Maths. I am quite a visual thinker so I like the video demonstrations, where squares and cubes and formulae move round the screen. Quite complex maths appears to me to be about pattern recognition. I like patterns and algorithms, because they sometimes have applications in other fields, therefore the more patterns, you are familiar with, the greater your chance of understanding other fields of study. Thank you for this video.
    1
  15321. No integers among partial sums for sure. Partially, because that's the only one I've never even thought about, but is so obvious in retrospect.
    1
  15322. a coin video? now that's going to pay off!
    1
  15323. I noticed one day that the y=1/x curve had this strange property. I thought, huh, this curve can also be written as 1 = xy, if you multiply x by k and y by 1/k nothing changes. It felt very unintuitive. If you think about it, the curve x-y = 0 has a translational symmetry (adding k to x is like substracting k to y). This result can be generalized to all lines, so I'm happy the multiplicative one can be generalized to all hyperbolas
    1
  15324. Circular slide rule. (C and D scales?) I used the straight kind and the circular variety in high school and college.
    1
  15325. I vote for gamma.
    1
  15326. Since my students keep complaining that they cannot understand english videos (after having had english as a school subject for years....), Mathologer-Videos in German would be great.
    1
  15327. how school couting to ∞ :1 2 3 4 5 6 7 8 9 10... me : ⅖ + ⅔ + ⅗ + ⅚ + ¾ + ⅘ + ⅞ + ⅜ + ⅝ + ⅟₁ + ½ + ⅓ + ¼ + ⅕ + ⅙ + ⅐ + ⅛ + ⅑ + ⅒...
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  15328. I think if the 1st disc turns counterclockwise also the third, fifth, seventh, and ninth disc (so the uneven numbers) will turn the same direction while the even numbers (so the second, fourth, sixth, eight and tenth disc) will turn the other direction (so in this case clockwise). (again sorry for my english :))
    1
  15329. Yep🥰
    1
  15330. German accent should be used by all maths teachers 😀
    1
  15331. You are a true master of mathematical choreography! Thank you so much for this video! It was a great way to spend 50 minutes of my life. The greatest surprise for me was the no-9-series. I can't help thinking all the fun ways of deleting terms in the harmonic series and wonder at what point things start to converge. Are you going to talk about the crazy formula connecting e, gamma, and prime numbers in more detail at some point?
    1
  15332. I think I liked the visual proof at the end the best, but then again, I already know most of the other stuff. The overhanging parabolic arrangement of blocks was also very cool.
    1
  15333. what is the best most fashionable lacing? possibly with most uniform parallel patterns
    1
  15334. I was taught Heron's formula in high school in Hungary, but without proof.
    1
  15335. lol
    1
  15336. Those flipping hexagons. ;)
    1
  15337. @Tshirt: Isn't the sqrt of 666 approx. 25,8069758011? So when we round it should be 25,8070?
    1
  15338. I had a thought about that e^x and 2^n once. because if you start with e^x and it's derivative as f(x)=f'(x), start at x=0 and then walk in x with a step of 1 you'll find: f(0) = e^0 = 1 = f'(0) f(1) = f(0) + f'(0) = 1+1 = 2 = f'(1) f(2) = f(1) + f'(1) = 2+2 = 4 = f'(2) ... and this will create: 1, 2, 4, 8, 16, 32... = 2^n now for every function A.e^x will have f'(x) = A.f(x) so if I have a continuous exponential and I wanna sample it at a discrete points can I find the other exponential that following the algorithm above will draw? Ex.: on the first case I wanted to follow the continuous curve 2^x at steps of 1, so the continuous function with the derivative to follow is e^x. I also wondered if you could do the opposite way, if you found the exponential A^x but you've sampled as intervals t, what is the real B^x? and sorry for the long comment; I guess I got inspired by your video. 😂
    1
  15339. I clicked because I thought that was Uri Tuchman in the thumbnail. Not only are you painting (coating) the infinite interior surface with a finite volume of paint, you have paint left over.
    1
  15340. ​ @Mathologer I was just being a bit silly. He's a craftsman, with a YouTube channel.
    1
  15341. Cool 🙂
    1
  15342. So much more interesting than the dry way I was taught about hyperbolic functions at college nearly 60 yrs ago ! :)
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  15343. I am great fan of Alexander Bogomolny - Author of Cut the knot... He is no more now , but his applet work speaks his name always. Thanks for the APPLET CHROME EXTENSION suggestion. I can run back all applets now.
    1
  15344. 37:25 Why is there always exactly equal of each color tile? Spoilers: Because, as the finished board is able to be perfectly represented by stacks of cubes (with one final layer on each of the 3 faces), you could always 'fill in' the cubes to complete a cube of size n (the length of a single edge) and that will never change the number of each color of tiles (this think about adding any single cube, you just shift the 3 colors around a slot), but with a fully filled in cube, you have an equal size of sides and thus an equal number of each color of tile.
    1
  15345. I like your videos even before watching them, just like when I tell my wife that the biriyani she cooked is superb even before tasting it 🤣
    1
  15346. what is the integral of ln(x)?
    1
  15347. Re: I'm mechanical enginner and I like discrete mathematics (also modular and cryptography). I took 2 days to watch video and the last part was tricky, but i got it (only understanding, not by proof on paper!)
    1
  15348. Ok but wheres my -1d cube
    1
  15349. That was brilliant - thanks!
    1
  15350. The answer to the chalkboard problem is 2. I know all the squares up to 15 by memory, so just a little quick mental addition.
    1
  15351. Here’s a link for Ian’s shoelace site: https://www.fieggen.com/shoelace/ianknot.htm
    1
  15352. I regret only watching this video 10 hrs after the release, the solution of the final problem takes, like, ten minutes, and is pretty straightforward (though I am sure there is a more clever way of dealing with this problem which involves some discrete analogue of the Stawke's theorem, or the use of group theory): 1) 4x4 board Consider the central square filled with arbitrary different colors a, b, c and d. ab cd There are essentiaaly two cases: bottom right square of the whole board can be colored in any way except d, so it can either be colored a or c/b (these two cases are indistinguishable because of the symmetry of the problem). Let's try a "diagonal" coloring (say, b) first. 0000 0ab0 0cd0 000b The choice of color for that bottom right corner defines colors of squares immediately to the left and to the top uniquely. 0000 0ab0 0cdc 00ab (*) It also defines uniquely colors of the next cell to the top and the next cell to the left to be equal to the color of the bottom right corner. 0000 0abb 0cdc 0bab We've come to a contradiction: one of squares has two b's in it. That means that once we've chosen the tiling of the central 2x2 square, the remaining coloring is defined uniquely, and each corner has the color of the 'opposite' corner of the central square, Q.E.D. 2) By induction: once the coloring of n central square layers is defined, the coloring of the remaining n+1 th layer is defined uniquely, since on every side by the same logic as in (*) and by induction we have two interchainging colors. abababa dcdcdcd and we again have 4 differntly colored corners.
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  15353. 36:22 Qbert anyone ?
    1
  15354. In the 1970s, I had a laminated paper circular slide rule that was a giveaway from some company (and my dad, an engineer, had a very nice real linear Tri-log-log slide rule). My class in high school (class of 1980) was the first that had electronic pocket calculators. My older brother's class used slide rules.
    1
  15355. A solution to the "crazy glasses" domino tiling: First, let T(n) denote the number of tilings of the 2×n rectangle. You can partition them into two disjoint classes, depending on whether the rightmost column is a vertical domino (so remaining rectangle is (n-1) wide) or rightmost consists of stacked horizontal dominoes (so remaining rectangle is (n-2) wide). This is the recurrence relation of the Fibonacci numbers, and the base cases of the induction work too, so T(n) = F(n) on the nose. Now... for the glasses, there are four disjoint classes of tilings, depending on whether the largest center rectangle is tiled as a 4×2, 3×2 on the left, 3×2 on the right, or just 2×2 in the center. Then, the dominoes around the "lenses" of the glasses are determined, as well as the remaining rectangles on the arms, which must be tiled as rectangles, as if they're disjoint. On either side, independently you get a 2×3 followed by a 3×3 rectangle or a 3×3 followed by a 3×3. Counts of all four possibilities (note the center number): T(3)⋅T(2)⋅T(4)⋅T(2)⋅T(3) = 3⋅2⋅5⋅2⋅3 T(3)⋅T(2)⋅T(3)⋅T(3)⋅T(3) = 3⋅2⋅3⋅3⋅3 T(3)⋅T(3)⋅T(3)⋅T(2)⋅T(3) = 3⋅3⋅3⋅2⋅3 T(3)⋅T(3)⋅T(2)⋅T(3)⋅T(3) = 3⋅3⋅2⋅3⋅3 Thus, the total number of tilings is T = 9⋅(20+18+18+18) = 9⋅74 = 666 😈
    1
  15356. Isn’t this this the premise of the rotary engine?
    1
  15357. Honest question: why isn't Math taught like this at school? It is fascinating.
    1
  15358. Once I bought a National Geographic wristwatch with a rotating outer wheel functioning as a circular vernier. I was able to do multiplications, divisions and extracting square roots with it. Just out of curiosity: When you wrap the numbers around the circle at the beginning, what is the tension force of the rubber band? Does it follows the Hook law, i.e. the displacement is analogous to the force? Obviously not. In fact, its not exactly a rubber band but a variable density fluid; the more you pull the less material is left, so it follows the y'=-y law. But I may be wrong.
    1
  15359. The curve of constant width at the end was neat. Also, Conway's game of life is a simplification of a game invented by John Von Neumann. Von Neumann was studying the idea because he was thinking about self-replicating machines.
    1
  15360. I paused the video to do the 2xN puzzle. I uttered a-ha after doing 2x5 before you said there will be an a-ha moment. :)
    1
  15361. The colour proof is the one I would most likely find on my own. The swivel proof I think is more elegant and more appealing
    1
  15362. 5 rotations
    1
  15363. Can i give 2 likes?
    1
  15364. 45:23 Rewriting my original puzzle answer for the reupload: In short, 1 1 0 1 1 0 (repeating) is the Fibonacci sequence mod 2. Why does this work? Each pair (or n-tuple) of natural numbers, both labeled k, is offset by a certain angle from the pair of 0’s which could be said wlog to not move. This angle is (0.618k)/2, or (0.618k)/n in the general case. Which branch of the spiral each integer will appear to correspond to is determined by which nth part of the circle its angle is closest to. (0.618k)/n ≈ m/n in the reals mod 1. This can simplify to 0.618k ≈ m mod n for some nonnegative integer m < n We can look at the first two values: 0.618 ≈ 1 1.236 ≈ 1 This is the start of the Fibonacci sequence. It’s easy to see that if two natural numbers a and b add to c, then the corresponding angles A and B add to the corresponding C. So we can ignore the precise angles and just use the rounded results to get the next one! With two branches, 1 + 1 = 0 (opposite + opposite = same), 1 + 0 = 1 (opposite + same = opposite), 0 + 1 = 1 (same + opposite = opposite) And the cycle repeats. (Not rigorous part) Now since these are Fibonacci numbers, as the sequence continues, approximations using these numbers as denominators will become better and better, so the angles will approach whole multiples of 1/n. There is no chance of this pattern being disrupted by accumulating errors. With three branches the pattern will be 1 1 2 0 2 2 1 0 repeating, where zero is a spiral of all the same alignment, and twos and ones are spirals of cycling alignment in different orders. With 4, the pattern is 1 1 2 3 1 0 repeating, with zero being same-alignment spirals, 1 and 3 cycling in different directions between the four, and 2 cycling between one alignment and its exact opposite.
    1
  15365. I also recall ads back in the 50's & 60's in Scientific American for a cylindrical, telescoping slide rule, that I guess was helical, and whose main scales were 55 inches long. The name of the thing escapes me right now, but it did come up in some other YT video a while back.... Fred
    1
  15366. 13:33 That picture of Nicole Oresme looks like he has his scalp pulled back.
    1
  15367. I vote for the "no partial sum is an integer". Nice surprisingly simple proof.
    1
  15368. The best lace is the absent one. I loath laces. Why aren't all shoes lace free these days?! :(
    1
  15369.  @Mathologer  Nein nein, ich habe Ihren Akzent erkannt und eine kurze Suche ergab, dass Sie aus Würzburg stammen. Daher hab ich das einfach auf Deutsch geschrieben. Studiere momentan Informatik in Aachen und bin ziemlich an so kleinen Einblicken in die (reine) Mathematik interessiert.
    1
  15370. 🦄
    1
  15371. for 7 pigeons there will always be a hemisphere that contains 5
    1
  15372. Great video thx!
    1
  15373. the sneaky audio edit at 7:46 >v<
    1
  15374. Master, in italian you should write ToRRiCelli
    1
  15375. The maximum number of moves for n cards is n(n+1)/2, or the nth triangular number.
    1
  15376. I liked the swiffle proof because it looks like some hidden symmetry.
    1
  15377. Odd×odd boards have zero tilings according to the formula because at some point m+1 = 2j and n+1 = 2k, so both cosines are 0 and thus the corresponding factor is 0. Both can be simultaneously 0 only if both m and n are odd, so this never happens with even×even or even×odd boards. Edit: I was beaten to this by the (better-written) response from Ben Spitz.
    1
  15378. No 9 sum proof is the most memorable part for me. Tristan's visualization reminds me of the classic Cantor set.
    1
  15379. what about 2 cent coins? many currency like the Euro uses 2 cent
    1
  15380. Mind blowing stuff man. How great is to invent all of that from nothing ....
    1
  15381. My answer to the question at 36:32: there are 3 different sizes triangle in the image: call them Small, Medium, Large (Large having sides 1/2 and 1). Medium = 4 * Small (similar triangles, hypotenuse twice as long) Large = Medium + Small = 5 * Small Therefore, every Small triangle takes up one fifth of a corresponding Large triangle. This is also the part that overlaps with one other Large triangle. The 4 Large triangles together have area 1 and would fill the entire large square if it weren't for these overlapping parts. So the remaining (blue) area is equal to the four Small triangles, and to 1/5.
    1
  15382. There's an old mill near me that makes/made picture frames, and they have machines used to turn the profiles of both circular and elliptical frames. It baffled me how that could work, but now I suspect that mathematically at least it's like a nothing grinder where the handle is locked in place and the base is rotated.
    1
  15383.  @Mathologer  incidentally, I like the 2nd proof best: it is closer to the way I proved it for myself, which was to write everything in terms of vectors for the sides and medians of the triangle.
    1
  15384. There’s a Knuth book that goes over the 2 by n tiling as an example in a chapter. We worked through it as the first example in an Analytic Combinatorics class, which besides that introduction with Knuth’s book, mostly covered the book Analytic Combinatorics.
    1
  15385. What, if any, relationship does this have with the following facts about the quadratic equation a*x^2+b*x+c=0? If X and Y are the roots, then Sum of the roots (X+Y)= -b/a Product of the roots (XY)= c/a Then XY - (X+Y)= c/a - (-b/a)= (c+b)/a 🤔
    1
  15386. nice
    1
  15387. Math is as math does.
    1
  15388. Well, let's see! Actually it is very easy. No need for so much complexity. In Arabic the letter JIM is written as JIM , YA , MIM. The gematrical values of these three letters is 3, 10, 40. So, if you skip the 0's you get 3,1,4 = 3,14......... And look at this beauty of Arabic. The letter JIM is ج Do you see the circle with a point at the centre. That's π !!!!!!! Easyyyyyyyyyy 🤦‍♂️😁
    1
  15389. Amazing video, as always. Keep up the great work!
    1
  15390. Good video 👍
    1
  15391. In Real world it can be only infinite to the right to the zero
    1
  15392. 6:22 That's not Pi, thats 3.18. Someone fudged the scale rotation after that.
    1
  15393. joined from bangladesh
    1
  15394. I love your video's. You lose me pretty fast on the mathematics stuff, because I'm not that good in maths, but I just love the magic of maths and you make it look so easy and nice with your animations. And I learn stuff of it. Keep up the fantastic work. See you in your next video👌
    1
  15395. omg it is so remarkable that sum of numbers with 100 zeros is greater than sum of no-9 numbers! My favourite part of this video!
    1
  15396. 18 1 Ru Ra
    1
  15397. Visual proofs are all that speak to me :). If youre interested the electronics multiply/divide functions simply count the logarithms of numbers, then add or subtract them and then use the inverse log (of the same order) - to count e^pi its easier - you just take inv log of Pi :). For large multiplications (order of 1024bytes long numbers) there is even a nicer algorithm based on Parseval theorem :) but you need to implement fourier transform
    1
  15398. As always your videos are amazing, keep it up. I can’t wait for the Galois theory one
    1
  15399. Any time a new Mathologer video comes out, it’s a chance to get the tea and biscuits and relax on the sofa!
    1
  15400. Why don't they teach simple visual logarithms (and hyperbolic trig)? so that cryptography lasts for longer???? (*me secures tin-foil hat*)
    1
  15401. 👍👍
    1
  15402. Euler formula, Taylor series, and of course the mysterious pi! Thanks for the video, very inspiring. once I can not follow the auto formula, I can easily go to meditation mode, thanks for the music.
    1
  15403. Didn't our host promise to reveal how to properly resolve the π=4 conundrum?
    1
  15404. I am so very sorry. I have accidentally figured out the formula for making a change for Graham number dollars. Earth is going to turn into a black hole shortly. I apologise for the inconvenience.
    1
  15405. I could watch this, or any, video ten times and appreciate something different each time. You are an International Treasure.
    1
  15406. I find it weird how slide rules instantly know the most significant digits of a multiplication without having a clue about the least significant, whereas any multiplication algorithm has to figure out the least significant before it knows the most significant.
    1
  15407. Recursive differences show it as: 1 2 4 8 16 31 57 99 163 256 386 ... _1 2 4 8 15 26 42 64 93 130 __1 2 4 7 11 16 22 29 37 ___1 2 3 4 5 6 7 8
    1
  15408. I saw this process in high school from "Mathematics for the Million" by Lancelot Hogben, use it all the time for polynomial sequences. He calls it "Vanishing differences" and mentions Gregory's formula on p. 281.
    1
  15409. Wonderful video
    1
  15410. Learned the first one from the internet, never seen the second one In school we were presented with only the formula, no proof :/
    1
  15411. hey Congratulations on your 100th video!~ given your number of subscribers, it just reflects on the quality per video. thank you ❤
    1
  15412. I watched the video and still don't know what the curse is?
    1
  15413. what is the angle of (or tan of) A_infinit_°? take a paper pencil and (for easy drawing) "Geo-Dreieck (equilateral_rectangular_ruler)", and choose a unit_ONE. Draw horizontal One + draw a vertical "ONE_line" up. Then draw the hypothenuse. Rectangular to that hypothenuse draw the next line of length ONE, downward; connect the origin with the end_point, now. (Length = Sqrt(3)). Draw rectangular to this last connection a ONE upwards. Connect the origin to the endpoint. The Length now is square_root of (4). Continue down and up until infinity; You get all the Sqrts of all integers. What is the lim of the angle between the last "hypotenuse_infinite" and the baseline? If You find the solution, could You make a video of it?
    1
  15414. The most memorable is the maximal tower
    1
  15415. In the 3-4-5-triangle the midpoint of the incircle and the midpoint of the Feuerbach circle lie on the same horizontal line. So there's not really a triangle to form in this case and thus 3-4-5 has no parent.
    1
  15416. I vote swivel. The coloring proof is definitely the one I would have come up with, so I found the swivel proof more interesting.
    1
  15417. I really do like your videos! Thanks for your kindly and hard work. By the way, what software you use for your animations? It does look nice :)
    1
  15418. 2+2 = 2x2 = 2²
    1
  15419. The only thing that didn't work for me in this video was your distorted face at 22:00😂.
    1
  15420. Ah yes, another 40 minute of mathologer video
    1
  15421. Most memorable was intuiting that the no-9-series might converge based on a similar idea: For each finite n, arrange all numbers in the harmonic series less than 10^n in a 10x10x...x10 n-cube. The corresponding partial no-9s is now the 9^n sub-cube. Using the intuition that for large n; "most of the volume in an n-cube is near the boundary", one gets the idea of just how many of the terms actually are taken away. Then 5 seconds later be 1-up:ed by someone else visualizing the same concept with a FRACTAL!
    1
  15422. 3 diagonal one way and two diagonal the other way. Rearrange and two small crosses
    1
  15423. I've checked your shirt, it has a rounding error, better use 25.807 😛
    1
  15424. Bravo!
    1
  15425. Yess
    1
  15426. Most memorable: existence of gamma (yet another Euler constant lol)
    1
  15427. It drives me a little crazy that he didn't make it a challenge problem for the viewers to derive s^2 = 1 + r + s from the other two identities. It's actually really nice (at least if you do it right). SPOILERS BELOW: s = s+1-1 = r^2-1 = (r-1)(r+1) s^2 = s(r-1)(r+1) = (rs-s)(r+1) = (r+s-s)(r+1) = r(r+1) = r^2+r = 1+r+s
    1
  15428. Why do we necessarily have to choose face down as 1 and face up as 0 why cant it be other way round
    1
  15429. First to press like button :)
    1
  15430. congrats
    1
  15431. Wow. I came across this vortex thing 5 years ago and was blown away by it at the time. I really did consider that there was some deeper "key to the universe" or something in there. But I eventually forgot about it until now. I think the 3-6-9 thing is a manifestation of this pseudo-intellectualism we all start out with in the early stages of learning. We think there is a magic shortcut to understanding something, a kind of special snowflake mentality. Eventually you move past this and are humbled by what you did not know. I like how you disprove the 3-6-9 key to universe idea but offer a more profound view of this digital root thing. The truth is way more interesting. Mostly cause the "key to universe" theory hits a dead end pretty quickly.
    1
  15432. ​​ @Mathologer ❤ Thanks for taking the time to explain.
    1
  15433. I followed every last bit of the ending autopilot throughly enough to be able to predict where it was going at times, besides understanding where the “/2” came from in the expression “(ln(4)-ln(2))/2”. Because “ln(4)-ln(2)” already defines the area in green (at least I’m 99.9% sure it does), so why do we need to divide by 2? What does that represent? At 22:24
    1
  15434.  @Mathologer  Ah yes! Of course. The space being completely filled in the animation just threw me off. I noticed that as well and paused to think about it for a second, but guessed it had something to do with how limits work that I don’t quite understand and moved on — wasn’t able to connect it to my confusion though. Thank you for the reply!
    1
  15435. @murderousmaths taught be this first part
    1
  15436. this proof is so funny, I love it
    1
  15437. Stand-up Maths on F(z): https://www.youtube.com/watch?v=ghxQA3vvhsk
    1
  15438. 22 minutes in, and I see the inverse matrix of the binomial coefficients and the Bernoulli numbers AHHHHH GENIUSSSSS. Ok I finished the video. The formula was very amazing and exciting, but the niggling red question mark and convergence problems for those wonky functions are eating me alive. :( I'm a second year university student, so I have some real analysis under my belt. I've watched your previous videos on Riemann Zeta, and the one on algebraic numbers and the cubic formula, and I've understood them all (but not well enough to give the arguments from memory yet). The animations were great -- if only all math classes were taught like this!
    1
  15439. Ah, you DID mention it!
    1
  15440. Probably been said, but you should say this is US focused. Makes no cents otherwise.
    1
  15441. Construct the angle bisectors of the triangle. They meet at the incenter of the triangle. Draw a circle around the incenter through one of the endpoints, that are supposed to lie on a circle. Choose an angle bisector and do a reflection operation across it. The reflection maps the circle onto itself and two pairs of endpoints to each other. All endpoints can be mapped to each other by a sequence of reflections across angle bisectors. In conclusion, all these points lie on a circle. QED
    1
  15442. mathologer can you say hi to me
    1
  15443. Is your nickname MathoLner? :)
    1
  15444. my problem with calculus is all the epsilon delta xi and neighborhoods. Put multiple of those in a definition and I can‘t get through that labyrinth to the answer until you show me how…
    1
  15445. Beautiful!
    1
  15446. same
    1
  15447. "Can you write the numbers 2 and 3 as sums of distinct positive integers?" It really depends on what you mean by 'write'... For 3, [edit] 1/1+ 1/2+ 1/3+ 1/4+ 1/5+ 1/6+ 1/7+ 1/8+ 1/9+ 1/10+ 1/15+ 1/230+ 1/57960 The sum of the reciprocal of all powers of 2 of positive integers sums to 2.... 1/1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + 1/2048 + 1/4096 + 1/8192 + 1/16384 + 1/32768 + 1/65536 + 1/131072 + 1/262144 + 1/524288 + 1/1048576 + 1/2097152 + 1/4194304 + 1/8388608 + 1/16777216 + 1/33554432 + 1/67108864 + 1/134217728 + 1/268435456 + 1/536870912 + 1/1073741824 + 1/2147483648 + 1/4294967296 + 1/8589934592 + 1/17179869184 + 1/34359738368 + 1/68719476736 + 1/137438953472 + 1/274877906944 + 1/549755813888 + 1/1099511627776 + 1/2199023255552 + 1/4398046511104 + 1/8796093022208 + 1/17592186044416 + 1/35184372088832 + 1/70368744177664 + 1/140737488355328 + 1/281474976710656 + 1/562949953421312 + 1/1125899906842624 + 1/2251799813685248 + 1/4503599627370496 + 1/9007199254740992 + 1/18014398509481984 + 1/36028797018963968 + 1/72057594037927936 + 1/144115188075855872 + 1/288230376151711744 + 1/576460752303423488 + 1/1152921504606846976 + 1/2305843009213693952 + 1/4611686018427387904 + 1/9223372036854775808 + 1/18446744073709551616 + 1/36893488147419103232 + 1/73786976294838206464 + 1/147573952589676412928 + 1/295147905179352825856 + 1/590295810358705651712 + 1/1180591620717411303424 + 1/2361183241434822606848 + 1/4722366482869645213696 + 1/9444732965739290427392 + 1/18889465931478580854784 + 1/37778931862957161709568 + 1/75557863725914323419136 + 1/151115727451828646838272 + 1/302231454903657293676544 + 1/604462909807314587353088 + 1/1208925819614629174706176 + 1/2417851639229258349412352 + 1/4835703278458516698824704 + 1/9671406556917033397649408 + 1/19342813113834066795298816 + 1/38685626227668133590597632 + 1/77371252455336267181195264 + 1/154742504910672534362390528 + 1/309485009821345068724781056 + 1/618970019642690137449562112 + um... I'm going to run out of space, aren't I? now @ 31:52 ... So it looks like I wasn't cheating. and trivial way to represent so here's 4: 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20 + 1/21 + 1/22 + 1/23 + 1/24 + 1/25 + 1/26 + 1/27 + 1/28 + 1/29 + 1/30 + 1/200 + 1/77706 + 1/16532869712 + 1/3230579689970657935732 + 1/36802906522516375115639735990520502954652700 + 1/1!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 (Replace each ! in that with 1000 zeros, and you get an answer which is accurate to 112120 digits. (Today is 11/21/20, which is why i used that number.) the 'bc' code to get that answer: ms=112120; scale=ms; s=0; for(i = 1; s < 4-(1/10^(scale )); ) { r=1/i; i=i+1; if(s+r < 4) { s=s+r; print "1/",i-1 ," + "; if(s < 4 && s > 4- (1/i) ) { scale=0; i = 1/(4-s); scale=ms; }; }; }; print "\n";
    1
  15448. I got 50 squares for the first puzzle.
    1
  15449. Finally, a long video :D :D popcorn, and coffee is brought for watching it uninterrupted :)
    1
  15450. Tricky proofs are the most exciting.
    1
  15451. I love your videos!!! I have a suggest, projective geometry. thanks
    1
  15452. After I watched this video, the only prove I can remember is the visualization prove of gamma < 1. Nice and beautiful.
    1
  15453. This is the sort of math I love. Thanks for saving this result from obscurity!
    1
  15454. Really good video!!
    1
  15455.  @Mathologer  This video is going onto my pile of "important thoughts". I have the feeling this may be the key to a number of problems I have been thinking about. First I have to work through P vs NP and hopefully get it done before everyone on earth is dead :D
    1
  15456. This was very interesting to watch. I was thinking about making some video animation about the golden ratio and pentagons, but this video also gave me some new food for thoughts. It's like the discoveries around this ratio is endless.
    1
  15457.  @Mathologer  One step ahead of you, already watched those videos :) But I'll probably need to watch one of those again in the future.
    1
  15458. Couldn't Napoleon's theorem be named after the tricorn hat he wore?
    1
  15459. End screen: Well, I think that 3^3 + 4^3 + 5^3 = 6^3, but I can't find a solution to the quartic, so I guess I made a mistake? By my figures, the left hand side is 2258, and the obvious candidate for the right hand side is 7^4, but 7^4 is too big (2401). Given 2401 and 2258 differ by 143, I was hopeful that the left hand side might be equal to something nice like 7^4 - 7^3, but 7^3 = 343, not 143. Best I can do is say that the question mark is the 4th root of 2258 (only real solutions between 6 and 7, or between -6 and -7 if you want to be fancy). Being fancier still, we have, for some z, z^4 = 2258 and so, apply De Moivre's theorem to find all complex solutions, I guess it would be +/- i (2258)^(1/4) , + / - (2258)^(1/4) and I'm honest enough to admit, I thought De Moivre would be more helpful!
    1
  15460.  @YodaWhat  Yes, I also looked up the prime factorisation of 143. As you say, it's 7^4 - (12-1)*(12+1) and so we very nearly had 49^2 - 12^2. But not quite!
    1
  15461. i was working along with e^itheta formula, and i just had gotten a crazy idea, replacing theta with i so it becomes e^i^2 = cos i + isin i . 1/e = cos i + isin i and i found out complex trigonometric functions exist. when i derived the value of cos i, I found it is similar to cosh 1. I have not learned about hyperbolic functions. Could anyone say how this happened
    1
  15462. 15:54 the area is 5
    1
  15463. That Euler dude must have been tripping 25/8
    1
  15464. The end rofl
    1
  15465. @0:56 I hope you take Det(*) soon Awesome as usual! Review: 1 minute in, already thumbs up material
    1
  15466. Amazing video: I really enjoy it. Thank you Mr. Mathologer.
    1
  15467. Thats some crazy animation skills.
    1
  15468. This is the kind of thing that I'd possibly discover but probably wouldn't publish for fear of being ridiculed, because it looks so simple when you see it that it must already be common knowledge, right? Moral of the story, follow your convictions because you never know!
    1
  15469. Debe.de.ser.el.clima.en.australia.un.amigo.champion
    1
  15470. 4:25 "Okay. Today's mission [...]" My brain was like "should you choose to accept it" 😂
    1
  15471. "If I was a hair man, I'd be jealous" :-D :-D :-D
    1
  15472. What if you make distances of the - terms the absolute value of an l-function?
    1
  15473. 28:35 the 2nd row should be 1 3 5 7
    1
  15474. the bugs' path length converging to 1 means their average separation for an infinite period of time (the infinite number of steps) is equal to 1 -- the bugs, travelling at a constant velocity, never meet.
    1
  15475. Imagine you go all the way to infinity just to discover that, all this time, the way you were walking was silly. Would you then become the minister of silly walks?
    1
  15476. Dangerous thought here, but what happens when you do this with the sequence of prime numbers?
    1
  15477. Merry Christmas Burkard! Thanks for making this hard year a more bearable one.
    1
  15478. Typischer Fall von Der Prophet im eigenen Land ist nichts wert. 😶
    1
  15479. A teacher showed us a cartoon in elementary school, featuring Donald duck playing pool. That is how I was "taught" these rules.
    1
  15480. at 3:48 he squishes/stretches the blue part, and then claims, that the blue area stays the same, while it visually seems that the blue area gets smaller, while the orange parts gets bigger. If you know beforehand, that the blue part is infinite, than it makes sense, but then it would result in a circular argument. If feel like there is a step missing
    1
  15481. Number of ways to make changes for $2000: 427707562988709001. Python program: https://onecompiler.com/python/3wkxjyf2n Number of ways for larger numbers: $20,000: 42677067562698867090001 $200,000: timed out. Not sure what the pattern is from small sequence. Looks to be 42 (666...7) 70 (666...7) 562988 (666...7) 09 (000...0) 1.
    1
  15482. I’m taking an online (pre) calculus course right now, I feel like I shouldn’t be here 😬
    1
  15483. Since I am an engineer, I choose the tower balancing
    1
  15484. You are the best math teacher I never had.
    1
  15485. I absolutely LOVE that shirt!
    1
  15486. Algo bump.
    1
  15487. No more than 5 steps ? hmm this sure does remind me something :)
    1
  15488. Most memorable for me was definitely the convergence of the 'no 9 series'
    1
  15489. I discovered the second proof on my own a couple years ago while playing with four identical right triangles.
    1
  15490. Damn, you should've really used the connections with iron man to make a more clickbaity title. Now this masterpiece won't be shown to as many people
    1
  15491. Loved it! ♥️
    1
  15492. It's awasome
    1
  15493. Ohh yeah calculus will never be the same from now 😂 great stuff though !
    1
  15494. Because our computer screens are 2d, at 26:45, for example, we're seeing the shadow of the shadow of the cube together with the shadow of the cube, but somehow our brains can make perfect sense of it.
    1
  15495. Those square root of 17s coming up are really weird - reminds me of Ramanujan. Also, I've always wondered - why Hanoi? Sometimes one sees it referred as the "Towers of Brahma". which I am not sure would fit with the Vietnamese reference.
    1
  15496. Since the arc length of the circular angle is proportional to the corresponding sector area in Euclidean 2d geometry, is it true that the arc length of the hyperbolic angle is proportional to the corresponding sector area in Lorentzian 2d geometry?
    1
  15497. Also, do these circular and hyperbolic trig formulas carry over in some shape or form to fields of positive characteristic (discrete or extended over a single free variable)? Or to p-adic ultrametric fields (of characteristic zero)?
    1
  15498. Most memorable was the appearance of the Euler-Mascheroni constant, which felt like kind of an understable „definition“ of this number... I studied a little quantum field theory and in that you have the problem that you encounter infinite integrals, when they are four dimensional (which spacetime is). When evaluating these integrals in dimension „slightly smaller then four“, they become finite and you can „absorb“ the infinite part in a redefinition of (now infinite) constants, a questionable dirty trick every particle physicist has to use. In this procedure, the Euler-Mascheroni constant pops up and always bothered me... not that anything changed about this confusion, but it was nice to see it appear in a way I can understand.
    1
  15499. I'm going to use this, if I get investors in my business. For whatever income, minus expenses, each investor will receive a fair return, as long as they know this from the beginning, and do not expect too much in the way of leverage.
    1
  15500. I've never heard of this movie before. After seeing this clip, it's made me now want to watch it...
    1
  15501. 2 Observations I have: a) how would it look if we would pattern upwards (i.e. 2+1 =? 1x2)? Clearly the equation is not correct, but that's because we can't remove any further one from the left side. But what if we could? Would that correspond tot he -1? b) Also all equations seem to give a special meaning to 2 (which also visible in your p sequence, which reduces the 8 solutions down to 7). Which leads me to the question to whether there is a relation between the 7 and the 49? Are there then 7^3 solutions for 3 (special sum equals product identities)? Lovely video... now I have even more questions than I had before watching it :-) Well done!
    1
  15502. 32:30 I saw that little edit there haha
    1
  15503. I love you! Always smiling, wonderful and clean animation and explanation. Such composition inspires a person to work on mathematics immediately. Much of the work and love you put there. And My new co-worker looks so much like you. Also, balanced warm-up shocked Me, I am so excited to think about it and to continue the video.
    1
  15504. Kempner's proof shows in elegant, transparent steps that I just fell for another sucker bet. ☹️
    1
  15505. 20:10 "...mathematicians can be really rare some times..." excellent joke
    1
  15506. What a nice surprise! I've been hoping you would release another video soon!
    1
  15507. Beautiful, Beautiful, Beatiful !!!!!
    1
  15508. It is a bit confusing that the red '3' is the same to the green '3' in the last part. I feel that if choosing 3 as the primitive root, it may be clearer since that makes the red and green numbers different (as '3' and '1').
    1
  15509. Checkout my git repository for an example of a python implementation of the "shuffling" algorithm ( https://github.com/BaptisteLafoux/aztec_tiling ) ! It's a work in progress, feel free to create a pull request 👩‍💻
    1
  15510. Ramanujan is a very interesting mathematician. Probably stories about Gauss would be interesting, although he was not a nice chap ;) Gauss is said to had the intuition skill of Ramanujan, the imagination skill of Einstein and the calculation skill of von Neumann all combined in his person.
    1
  15511. Maximum would be 10+9+8 etc which is 55 I believe.
    1
  15512. Are you gonna make a vid on my math discovery? (I’m on an alt, my other account’s name is Freedom, and I couldn’t actually post on it, so I made a new one.)
    1
  15513. Woah, the amount the work that went into making this video... Awesome ...
    1
  15514. 3^3+4^3+5^3=6^3
    1
  15515. Does not work for ^4
    1
  15516. Fantastic stuff as always!
    1
  15517. Glad to see you again. Schools around the globe should see this. I learnt both proofs in my school days but yes algebraic one is easier to do on paper. Although while mentioning scholars around the world I hoped you would mention Baudhayana and Sulabh sutras ( sanskrit texts ) dating around 1000BC explaining the modern Pythagoras theorem and many others. Nevertheless great video.
    1
  15518. Thank you for showing the beauty in math.
    1
  15519. Я действительно понял как, спасибо
    1
  15520. 2x2=3+1 in the context of Lie groups representations. This can be linked to the categorical concept of colimits in a monoidal category. (I start to sound like a category theorist 😂)
    1
  15521. Your videos are perfect, when you think you can raise an objection you address the issue immediately and even when I think know where you are going to 2 or 3 steps in advance you frequently surprise me. I like that.
    1
  15522. Magnifique vidéo j’ai essayé Marty and Al to follow you but at the end its mathematical fellow ahah thanks a lots
    1
  15523. I'm the 460th person to solve the 4d cube thx <3
    1
  15524.  @Mathologer  Also I just wanted to thank you for your videos, they entertain and "captivate" me the best of them all, happy new year and keep it going, your channel is great
    1
  15525. I decided to answer all of your questions. Most questions are copied straight from the transcript (CC). 0:57 Easy, right? For now. 1:12 Okay, what if we have to add a color to itself? Let's just keep it the same. 1:19 That's the most natural rule. Agreed? Of course I agree; that's what I suggested. 2:29 What are you going to do? I'll just guess. 2:49 Ready? No. 2:55 What's the color of the bottom hexagon. Red? Okay, I got it wrong. Let me work this out... 3:07 Did you get it without pausing the video or just guessing? Yeah, I just guessed. 3:11 Sounds impossible? No; if it was impossible, you wouldn't make a video about it. 3:40 Pretty amazing, isn't it? Yeah, it sure is. Now prove that it always works. 3:53 Weird, hmm? No; I've watched enough of your videos to that it's not weird. 3:56 What if instead of ten hexagons at the top we started with nine hexagons It wouldn't work. I did the math. 4:41 Okay so why is 10 special? It's one more than a power of three. 4:45 And are there any other special numbers. Yes: 2,4,10,28,82,244,730,2188,6562,19684,59050,177148,... (https://oeis.org/A034472) 4:54 Notice all the smaller solid color triangles here and there? Yes, I do notice them. It's pretty hard not to. 5:05 Hmm, can you think of a famous mathematical supermodel dressed in little triangles? :) Sierpinski! 5:10 No? I just answered it. Why did you ask "No?" 5:40 Beautiful pattern isn't it. Yes; it is beautiful. 5:50 Butcher shop? Whatever you say. 6:47 Well, let's find out together, shall we? In five easy chapters. I assume that's what the next 23 minutes of my life will-- Did you say five chapters? CHAPTER 1 (6:54) - How special is 10? In the context of this puzzle, it's not too special. There are countably infinite other numbers that do the same thing. However, 10 is the base of our number system, so it's very special outside of this context. 7:01 Why does the shortcut work for width ten triangles. "I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain." ~Fermat Luckily, Mathologer can make a video about it. 7:05 And, are there any other special numbers? We've been over this... https://oeis.org/A034472 7:17 And that's where we stopped. Hmm, I wonder why? You stopped there because 4 is another one of the special numbers. 7:27 Probably also just a fluke. Right? No. 7:46 In this state can we still show that 4 is special? Yes. 8:35 How many essentially different width four triangles are there? There are 10 (reflections listed on second row): AAAA, AAAB, AABA, AABB, ABAB, ABBA, AABC, ABAC, ABBC, ABCA AAAA, ABBB, ABAA, AABB, ABAB, ABBA, ABCC, ABCB, ABBC, ABCA I think I did that right. 9:19 Did you spot the pattern? I discovered the pattern on my own before I watched this far. 10:40 I probably don't even have to say it, right? No, you don't. 10:42 Can you see what's happening. Yes. 11:24 So on a scale from one to ten how beautiful an argument is this? It's as beautiful as it can be, but it's not as beautiful as some other proofs I've seen, so... I give it a solid 2π. 12:16 Can you think of a second way to argue? Hint: switch the roles played by four and ten. Well, you just switch the roles played by four and ten. I'm not quite sure what you want. 12:29 Time for a cat video? NO. 12:32 Well, that's not what we do here on Mathologer, right? Right, it's NOT what we do. 12:37 Is it really just a big coincidence that widths four is special or is there a deeper reason? There are no coincidences in mathematics, except in very rare cases where it's just a coincidence. 12:48 ..., but how can we be certain? This same argument can be iterated indefinitely. Each iteration replaces the initial width with 3*((the previous width) - 1) + 1. Therefore, the pattern continues since 3*((3*((the previous width) - 1) + 1) - 1) + 1 = 3*(3*(w - 1) + 1 - 1) + 1 = 3*3*(w - 1) + 1 = (3^2)*(w - 1) + 1. The exponent of three will increase by 1 with each iteration. I know that I did a horrible job of describing it. Sorry. 12:51 What's up with all these similar phenomena? Sierpinski and snail shells and whatnot? If there's a horizontal chain of a single color, it will form a triangle beneath it. I don't feel like figuring out the Sierpinski thing. Sorry again. The snail shells are merely related through the Sierpinski triangle. 12:56 Are there any beautiful connections? You just listed one. Sierpinski. 12:59 Ready to go deeper? Do you even need to ask? CHAPTER 2 (13:04) - Pascal's Triangle 13:07 The basic rule of two adjacent colors in one row summing to give the color immediately underneath just cries out for us to have a look at the tip of Pascal's triangle. Right? I guess, but "adding" colors seems very different from adding numbers. 13:50 You want hexagons? YES. 13:54 What else? Turn it so the tip is at the bottom. 13:55 You want the whole thing to start from a row and not from the tip? I couldn't have said it better myself. 14:21 Interesting, huh? Okay. 14:38 Now what about colors? I was thinking the same thing. 15:04 Pretty impressive, huh? No. I've seen that before. 16:44 Is 10 still special? No. 16:48 If it were, black and white on top should give black at the bottom. Right? Right. 17:05 instead of 1 plus powers of 3, this time it's 1 plus--can you guess it--... 2. 18:13 Do any of you apply similar this plus that mathematics in anything you do in your working life? Nothing pops into my head. 18:37 So let's see what happens when we use remainder on division by three. YES! Hey, I made something for that. Take a look at my spin-off of a program on Khan Academy: https://khanacademy.org/computer-programming/_/6406985938862080 20:05 Can you see what we need to do? I tried, and I couldn't figure it out. 20:38 What about the top rules? They work! 20:42 In mod 3 arithmetic - 2 is the same as -2 + 3 which equals 1. Got it? Yes, but don't tell JavaScript. It thinks `(-2) % 3` is -2. 22:42 What about the odd-even, black-white mod 2 Pascal game. What about it? 22:52 Why? (2b) mod 2 = 0 for all integers b. 24:01 Figure out the smallest non-trivial special numbers for m is equal to 4, 5, 6, etc. until you get sick of it. Ok, a hint: just focusing on the blue entries can you see another Pascal triangle? I'm going to go to bed now. 24:44 How on earth did they find that one? The internet? 26:38 All make sense, right? Sure. 27:13 What's happening here? As I've already said, if there's a horizontal chain of a single color, it will form a triangle beneath it. 27:45 This really gives a very good intuitive feel for where this self-similar pattern comes from, doesn't it? Yes. CHAPTER 5 (28:00) - Beyond Triangles 28:04 There's one very surprising feature of the original three-color game that I did not mention yet. When you rotate-- No way. 28:32 Can you think of the simple explanation for this phenomenon? I'm experiencing that brain-dead thing you talked about earlier. It's 10 PM. 29:16 Did you get them right? Yes. 29:33 Why? I'm done answering questions.
    1
  15526. Why the thing written on your t-shirt?
    1
  15527. Haha it looks like a penrose diagram
    1
  15528. Would the tree's total area be 5?
    1
  15529. Mindblowing!
    1
  15530. I went to youtube in order to take a break from learning math but found out there is a new video, I guess it count as part of my study.
    1
  15531. I liked the "partial sum does not add up to any integer" part
    1
  15532. Thank God for boots 👢!https://youtu.be/5bHGLNB8sxQ
    1
  15533. Hexagon challenge (not a proof though): assume the tiles represent sides of small cubes. Assume further the hexagon is actually a view on a mutilated cube, such that an amount of previously adjacent cubes has been removed from it - starting from the foremost top corner. This cube has walls which count as sides facing inside the cube should they not be covered by a tiny cube. Removing cubes in this manner generates a possible hexagon tiling. The total area of tiny cube sides facing in one direction does not change by removing tiny cubes. This goes for each of the three visible sides. Any set of tiny cubes removed in the above mentioned way won't therefore change the total area of sides facing in one direction. Also, any set of tiny cubes removed in the above mentioned way will correspond to a hexagon tiling. As the colors can represent sides facing in one direction, any hexagon tiling will contain the same amount of each color.
    1
  15534. I didn't understand the 'spikes' explanation and how it resolves the pi=4 paradox.
    1
  15535. Is there anything pascal's triangle can't do?
    1
  15536. @15:40 wait, that's not right, the 3d grid does have the turning property provided you are rotating around one of the axes, that's why obtaining a smaller hexagon is illegal, it's rotating in the wrong plane. Minor point, because we may not know the size of the unit, but we might know it's direction. That wasn't specified in the puzzle, when we switched to 3d. Probably this was missed since in 2d, we only have one way to rotate, while in 3d, we have infinite axes to rotate around or planes to rotate in, so we need to specify something extra to keep the rotation-rule. Or maybe it was just not to complicate things.
    1
  15537. but ... 2 is the oddest prime ... right?
    1
  15538. This is completely new to me.
    1
  15539. Could you do a video about this problem? In this village in every house there are 3 children. We are standing front of one of the house. A boy walks out. Then I see the door opens again from the same house another child about to walk out. What is the chance for that second child to be a boy?
    1
  15540. I recognized the circular slide rule once you added the points between 0 and 1. I used a slide rule in the early 1980s at NASA Ames Research Centre in Mountain View, CA. To this day, I have five slide rules in my home. When I want to multiply or divide a few numbers quickly - I grab the slide rule before my TI-Nspire CX CAS Calculator. By the time my calculator powers up and is ready, I have my answer(s) on the Slide Rule. I am teaching Slide Rule multiplication, Division, trigonometry, and also teaching LaTeX to my granddaughter now. I learned and used TeX in the early 1980s while at NASA during that same time. Yes, I am a pilot. Since I was a teenager. Also, an Air Force Veteran. I loved my flight computer. Although now they have electronic flight computers in general aviation flight training ground schools.
    1
  15541. Well done.
    1
  15542. As someone with a mechanical engineering degree, I find that type of mathematics frustrating and annoying, and thank goodness not everybody is like me.
    1
  15543. @Mathologer  I've had some math at university, and love calculus etc., and like statistics. However, I just don't conceptually understand things like number theory and the kind of mathematics the video discusses. It is as if others can see through a window that is opaque to me. Maybe someday.
    1
  15544.  @MuffinsAPlenty  OMG yes of course. I didn't think this through. Thanks for the lesson in caution. I'll probably delete my post in 24h, hoping you've read my reply by then.
    1
  15545. I just edited the dumb stuff out.
    1
  15546. how many squares in nxn squares is n squares
    1
  15547. For the question at 17:32 on forming the new triangle , the previous one has its height increased by a multiple of A( becouse the 60° angle is constant so the ratio of the sides need to be same) and the new base is also increased by a multiple of B ( becouse earlier it was 1) so the area, which is proportional to height × base, is increased by a multiple of AB. Was this the most obvious explanation?
    1
  15548. My vote is for no partial sum ever equals an integer, despite the infinitely many integers and the eventual infinitesimally small steps and all rational numbers . Simply incredible.
    1
  15549. Simple proof, made harder by words substituting for diagrams but let me give it a shot:). I want to start with only 1 color here(because its more than enough). If you dont have a set of corner squares with 4 different colors you have two or more of 1 color directly adjacent to each in the corners of the 4x4, or you have 2 or more corners diagonally adjacent with the same color, these 2 cases are important in this simple proof. First of all every 2x2 in the 4x4 must contain each of the colors, that is our constraint, which means there are going to be 4 of each color in the 4x4, and in no 2x2 can we allow any color to repeat, because this would mean this 2x2 would necessarily be missing a color. The simple case is to prove that 2 adjacent corners cannot have the same color at the same time as this condition of equal participation in each 2x2 is upheld, this very very simple visually, since these 2 corners participate in a pair of 2x2s that cover exactly half of the 4x4 box, this means that the 3rd 2x2 in this 2x4 region can now no longer have a green(example color) in it and still avoid 2 greens landing in the same 2x2. So partial qed, we now know that adjacent unicolored corners are not going to give us a solution. For the diagonallu adjacent corners its as simple as counting and remembering our non repeating rule for 2x2s, our 2 corner squares cover exactly 2 of the 9 2x2s, leaving seven to be given a green square. The maximum number of 2x2’s one square can participate in is 4, meaning 8- the overlap(which is illegal) is the maximum number of 2x2s covered by 2 squares, and we need 8, to achieve this we have 8 squares to pick from, the remaining 2 corners, the adjacent corners and the 2 remaining central squares. A corner square only ever participates in 1 2x2, so we can rule those out right away, the adjacent edge squares participate in 2 2x2s, which means that even if we pick the maximum for the other square we need to color, which would be a central square participating in 4 2x2s we would only cover 2(from the original corners)+2(from adjacent to corner square) + 4 from the central square= 8 2x2s covered, which is not 9 2x2s and so one 2x2 is without a 2. This then means we are left with the 2 central squares that are not sharing a 2x2 with a corner square of the same color, in this case we are now forced put 2 greens into the central 2x2, which means we have broken the condition of 4 colors in each 2x2, that proves that no 2 diagonally adjacent corners can be the same color, and still allow all 2x2’s to contain 4 colors, qed or something. Long winded but i find pure text pretty annoying tbh :) nice little puzzle tho.
    1
  15550. Simpler equation:  (1 + √2 )^n  = a + b√2 and the nth  fraction which approximates √2 is a/b.
    1
  15551. Managed to make it all the way through - not particular issues. My degree is in Engineering - so some familiar territory here.
    1
  15552. Final question - sounds like you're an Eagles fan, or at least of Desperado.
    1
  15553. dot and hinge is better
    1
  15554. is the formula for the volume of an Nth dimentional sphere ((1+1/x)*pi*(r^x))?
    1
  15555. oh nevermind
    1
  15556. never heard of this paradox. thank you
    1
  15557. I Watch it tomorrow In the Morning Its already 1 Am in India....But I am happy That I will get an Opportunity To Watch You in action...Tomorrow
    1
  15558. Your videos are a joy. When I was a much younger man my first exposure to the Euclidean Algorithm was the excellent little text by Ogilvy and Anderson "Excursions in Number Theory" https://books.google.ca/books/about/Excursions_in_Number_Theory.html?id=efbaDLlTXvMC&redir_esc=y You may have come across this but, if not, it has some good history. Taxi - 1729 T-shirt -- nice tip of the hat to Hardy and Ramanujan
    1
  15559. Aweseome work! Also as you were "smoothing out" your image should have less and less resolution hahaha that would be hilarious!
    1
  15560. Yep. Made it to the end. Nice combo of my math and finance background and definitely challenged my intuitive notion of fairness.
    1
  15561. My jaw is on the floor! Amazing video!
    1
  15562. I wander if this can be used as a tool to calculate means of special sets? Or degrees of means?
    1
  15563. Taught math for 24 years. ALWAYS taught divisibility tests using sum of digits. So did all of the teachers I knew.
    1
  15564. Okay, I've successfully failed programmer challenge. I firstly thought about dynamic programming, so we have functions f1(n), f5(n), f10(n) f25(n), f50(n), f100(n), where f1(n) - # of ways to make n cents only using 1 cent coins, so f1(n)=1, n>=0 f5(n) - using only 1cent and 5 cent coins, so f5(n) = f1(n) + f5(n-5) f10(n) - using 1,5,10 cent coins, so f10(n)=f5(n) + f10(n-10) et cetera, And we will calculate # of ways to make n dollars which is equal f100(100n), and this works O(n) of time But I thought, why wouldn't I optimize it? For instance f1(n)=1, f5(5n) = n+1. So I calculated all the functions up to f100 and came up with the same formula, as you derived a few minutes later: 1/6 (N+1) (80 N^4 + 310 N^3 + 362 N^2 + 121 N + 6), so I got impressed much earlier, than 24:28 =) In program calculations actually take O(log^2(n)) of time, so i think it's success, but I didn't program anything at all =)
    1
  15565. 26:58 ahh, the voiceover monster reappears 😅
    1
  15566. Well done! This felt like a 15 minute video; I was "on the edge of my seat" the whole time and did not realize I had been watching for 45 minutes until it was over!
    1
  15567. I'd be curious if there's a 3D equivalent? Talking about Pixar, they're dealing with a mess of polygons in 3D space. I'm not sure that they have any need of finding any particular "center" but it would still be fun to be able to do.
    1
  15568. gamma is always greater than 1/2 because the rectangles taken to create gamma are curves that contain a greater area than if taken with a diagonal, which represents 1/2 the area of the rectangle
    1
  15569. Anyone else thinking about cabbage?
    1
  15570. I solved the red cross puzzle in 3d applying the generalized Banach Tarski paradox.
    1
  15571. I always pronounce ϕ FEE as the Greek alphabet. It is said that mathematicians go with PHI. But what about Greek mathematicians?😆
    1
  15572. I see so many math tricks like this where only the numbers 1,2,3,4,5,6,7,8,9, are used in the making of their wheel or circle. This leaves out one entire number and space between 0 and 1. Or 10 % of our divided by 1 to 10 numbers system. I find a problem with this. It's not a true representation of 0 to 10. It assumes 0 to 1 doesn't count the same as 1 to 2 or 2 to 3 and so on. Everything is being based on only 90% of our number system. And yes I see how all these number games work out so nicely and I used to use to always a slide rule in the days before the $200.00 four function calculator hit the market. But I still wonder about this 90% as apposed to 100% symmetry that's always used. What is math missing? I'm seeing a trick as apposed to reality. Numbers are a progression of growth by always in affect adding 1 or the next number in a sequence. It's not about pitting 100% of 1 to 10 literally. Or 90% of 0.0 to 10. Our reality works on ten equal parts from 0.0 to 10. That's equal to, 10 to 20 and so on. Not 1 to 10, 10 to 20. There is a 0.99999999999999999999% difference to these two number ranges that is exploited making these tricks work. If I buy only one to ten apples it's assumed I'm getting 100% of apple number 1. On your Number wheel I'd be lucky get barely more then just the smell of apple number 1. So if I may ask? What happens if this 10% trickery isn't used? Or more exactly 0.999999999999 against 0,09999999999999 relationship. There is a left out, long distance 0,0000000000000000000001 And this other 0,999999999999999999999999 that is also not represented. Does Math over the centuries ever talk about this. Or just because these neat tricks work out and are helpful we are to ignore this number gap that makes it all possible? And what other other number realities is and has mankind been missing, that if the entire 0.0 to 10 number was used for wheel and circle type tricks? After all, this is the 21st century. Have we not improved in a "progressive" way in math, being it itself is an ever progressing number system science? This is only one of many questions/problems I'm seeing in our sciences. Sooooo, What are your thoughts about this? ENJOY
    1
  15573. First time I can fully understand your video 😭😭😭
    1
  15574. Time 23:18 seems to leave out a factor: suppose the two adjacent orange squares get tiled as |=| then it's a fifth option from 2x2=4 choices. The (apparent?) oversight is repeated at 27:32.
    1
  15575. But your not really showing the entire surface area of the horn since it would cover the page.
    1
  15576. The no nines series certainly grabbed my attention with the graphical proof, and the mu cat!
    1
  15577. Ok, mind officially blown
    1
  15578. a) Great video! b) It could be argued that the title is clickbait-y, that “Why did they not teach you any of this in school?” parrots the language ultra-conservatives deploy to get uncritical audiences fired up over invented “culture war” wedge issues, and so any videos with such titles are pushed higher up on YouTube’s homepage—and, as I don’t know your intent, I won’t argue that this is your intent … and you seem like a great fellow!! … but it sure does look like clickbait, and that’s regrettable: capitalism may coerce us into all sorts of regrettable behaviors, but the least we can do is resist—profit IS NOT everything! c) That having been written, I don’t begrudge your building an audience or gaming the algorithm to do so … but it’s still a horribly regrettable situation, having to reinforce thought-stopping language for profit because it’s all that’ll cut through the noise and we all need to scrape together a living whatever the cost. d) During the ‘70s, when I was younger than 10, I learned the visual proof for a^2+b^2=c^2 in Latin America during grade school as well as from Mom at home using magnetized geometric figures on a little magnetic slate, and my memory is fuzzy, but I’d wager I grasped this better from Mom than from school; the “new math” was all the rage, so grasping concepts like set theory and if-then flowcharts and the underpinnings of formal logic at a young age were seen as the primary goal of elementary school maths, with memorizing tables being only the secondary byproduct of that primary goal … plus Mom had taken advanced math for extra credit and for fun during her own high school years from her delightful beatnik maths teacher, so wherever my school was still stuffy and stuck in the 19th century my mom would swoop in at home with her abacus and her binary math and her dozenal math games and fill in the gaps and make learning fun again rather than dreary—yay Mom!! Anyway, thanks for the video, pretty much everything past the intro felt new to me. 🙂
    1
  15579. 15 - 20 minutes of math is fine. Anything over that I get lost.
    1
  15580. Amazing !
    1
  15581. hi
    1
  15582. Σ1/2=-1/4
    1
  15583. I"M SHOCKED he didn't mention that it's the set of indices of square numbers that are also triangle numbers 1,6,35, etc. 1^2= 1(2)/2 6^2=8(9)/2 35^2=49(50)/2 I actually worked through the algebra, found this relationship, realtised that if it works for n it also works for 4n(n+1) for the nth triangular index by simple induction Applying this on 8 because t8=36 we get t(4(8)(9))=288 as a direct solution to the Strand problem Mathologer maybe do a part on triangular numbers next time in relation to this problem? This is the boring high-school way of doing it though without asking too many questions I guess Little did I know there was so much depth to it!!
    1
  15584. Your explanations of methods are always so silly and cool! Thank you for being you and sharing this incredible method! The ending is so beautiful and geometrical 😭
    1
  15585. All the outside corners of the cube design on your t shirt lie on the surface of a sphere. ❤
    1
  15586. In Australia this stuff was taught to actuarial students in a course named "Finite Differences"
    1
  15587. Great mathematically. Looked at the crush can pattern. I feel sure that i have seen patterns like this crushing pressurised cans in "hydraulic press channel". But some just concertina. I wonder if there is a ratio of height to radius that will get the patterns to appear more readily. It reminds me of a Prof who studied diaphantine approximations who told me soviet rockets were built to be more stable using badly approximable numbers in this ratio as you could not get standing wave oscillations.
    1
  15588. cool
    1
  15589. Piradox ?
    1
  15590. The no 9s being convergent is surpriseing, but it's still not the favourite part, because I don't like facts that involve how you write down a number. My favourite part was the optimal stacking problems.
    1
  15591. I am late to the party but regarding the area of the tree, if the area of the square is 1, then it has side lengths 1. The triangle above is a 45-45-90 triangle, therefore hypotenuse/sqrt(2) = 1/sqrt(2) = side length. The area of a square in the second layer is then 1/sqrt(2)^2 = 1/2 and there are 2^1 = 2 squares. Generalized to any layer, the area of the tree is the sum from n = 0 to N of 2^n*(1/sqrt(n))^2^n = 1 where N is the layer depth of the tree. While the tree shown has 5 layers, I interpret the question in the general case and the tree has infinite area.
    1
  15592. 25.807 is closer to the sqrt of 666 than 25.8069 and therefore a better result to use. Check your tshirt. Also, you selected 3.18 on your wheel, not pi.
    1
  15593. coin rolling paradox: what if you think of the coins rolling around each other. Trace the path that both centre points take. I believe the maximum "radius" of this shape is a ratio to the answer with respect to the smaller circles diameter. Does this sound correct?
    1
  15594. 4:40 isn't the only question peg C or peg B for the first move?
    1
  15595. Seeing your zeroth power sum reminds me of something that I don't really understand why it is taught the way it is. That would of course be, 0^0 Now, in your typical Algebra class, when they go over exponent rules, they say "You take any number to the zeroth power and you get 1. Negative numbers, fractions, irrationals, all become 1." And when someone asks about 0^0, they just gloss over it and say "Just put in the answer as 1, just like you would for any other number" Now, for me, that makes me say "Wait a second, if exponentiation is repeated multiplication, than going from the first power down must be repeated division. And therein lies the problem with 0^0" x^1 = x x^0 = x/x x/x = 1 for all numbers except 0 If you treat 0 as being a solution to x/x, you get 0/0 But, you can't divide by zero. Well, algebraically anyway, you can't divide by zero. In order to divide by zero, calculus has to come into the picture, and more specifically L'Hopital's Rule. But this still doesn't help us with the 0^0 problem, because if you take the derivative of x^0, you get 0. So you're basically screwed if you try to use calculus to solve for 0^0 So then, if even with L'Hopital's rule, you can't solve the 0^0 problem, why do algebra teachers insist that 0^0 = 1? 0^0 is indeterminate, and even calculus, which can help you solve for solutions when indeterminate forms pop up, does nothing of the sort for x^0 where x=0, you just get 0 for any value of x, or in other words, a line with a slope of 0 and a discontinuity at the y intercept of 1, becomes the origin.
    1
  15596. Brilliant! I admit I didn't follow every step of all this on first viewing, but I know there's nothing there beyond my ability to understand. I will watch it again (maybe several times), because it's easy to see that it's truly beautiful!
    1
  15597. Dankje Mats :)
    1
  15598. Third
    1
  15599. You always stop so early lol ;(
    1
  15600. My solutions: If I want it to be in position B, then I will have to move the tip to C. Therefore the answer is counterclockwise. Using the minimal number of steps, you have to move the tip to C if you want it on B and if you want it on C, then you will have to move the tip to B. The formular was (2^n)-1 for n layers and (2^n+1)-1 for n+1 layers.
    1
  15601. Its a circular slide rule. I had one in the early '70s
    1
  15602. Beautiful
    1
  15603. I am not a mathologist and was so proud that I could follow for really long. Got the base 10 stuff and remembered doing different base system in university. Then in Minute 23 he started saying stuff and it became just a colorful bouquet of worde for me. Totally lost. But so nice. Hartes Zeug und spannend
    1
  15604. This video needs dark mode
    1
  15605. Nice, as always!! For me, the thiner and thiner chapter was the most memorable and new for me! I got the answer wrong! Keep up your marvelous teachings!
    1
  15606. But what is the most beautifoul lacing
    1
  15607. I love your videos. I would have loved to have you as my teacher when I was a younger student (which I'm not). I'll suggest my son, who is 17, to watch at your channel. Great job!!!
    1
  15608. The multiples of the angles gave me "harmonics" vibes, but it wasn't until the visuals at 17:31 that I was sure it was actually related
    1
  15609. Yes each step is like a “notch filter”, except that it also changes amplitude and phase of non-targeted frequencies.
    1
  15610. I think this is the worst of all your videos. To much implicit knowledge.
    1
  15611. I made it to the very end! :D
    1
  15612. Hi… After I watched this video, I made my own (unoptimized) program in Embarcadero RAD Studio Delphi 10.2 Tokyo to test whether a prime is of the form of the sum of two integer squares. Here is the list of primes from 1 to 1,000,000: 39,176 out 78,498 primes within this range are of this form. Click https://drive.google.com/open?id=1SRUjylNgJY9WidvUA8i7kkfjvAVoOqoM to download.
    1
  15613. 6:21. I though it to be written as 4k-1
    1
  15614. amazing
    1
  15615. Well, at least it looks simple, but I have enough background for this to look simple. I find it also really beautiful and refreshing. But, when I learned about logarithms and e and so on, the teacher had a blackboard and white chalk to do the teaching. Thus it would have been a waste of time and effort to teach this way, I suppose. Nowadays one definitely may have the technology but it takes more than having the tools, isn't it so? 🙂
    1
  15616. The snake visuals were so good!
    1
  15617. Short and sweet :D
    1
  15618. It’s late. I thought to myself « I’ll watch the beginning of the video tonight and the rest tomorrow »… Nope! Watched the whole thing, and now it’s really late! Good night everyone 😴
    1
  15619. Easy proof that p not equal to 4k+3: the sum of two integers sqaured mod 4 is 0 or 1 or 2 (very easy to show), since the prime is odd the number mod 4 must be 1 or 3 so prime that can be written by sum of two sqaures need to be 4k+1.
    1
  15620. Wonderful and clear video. I have a degree in mechanical engineer and could watch everything without a problem. I think yours videos are very accessible, probably just a basic course of calculus and some patience and anyone can keep up with everything. Wonderful job!
    1
  15621. Does measure theory play a role in this paradox? It feels like volume and area are fundamentally different. If you say that you have a volume of 8 liters of paint that doesn’t really convert to surface area.
    1
  15622.  @Mathologer  thank you for the clarification!
    1
  15623. In a related way, the earth spins 366 times on its axis every year. Not 365.
    1
  15624. My levitation dreams are not ridiculous! I will levitate
    1
  15625. One proof for 2:20 : By 3 point there is one and only one circle. We will divide our 6 points into two groups of 3 points. I'll add names : starting on the horizontal line on the left we name the point A. Then we turn clockwise and name D,B,E,C and F. Our two groups are A,B,C and D,E,F. Then we can show that AB = FE, BC = DF and AC = DE using the colored length. Let be C1 the only circle defined by A,B,C and C2 defined by D,E,F. The triangles ABC and DEF have the same lenghts, hence C1 and C2 have the same radius. Does C1 and C2 share the same center ? If they are, we are done. Let's focus and the green angle between D and B. Let be Δ the bisector. The axial symetry applied on [DF] gives [BC], thus C1 become C2. Another axial symetry based on another bisector will also send C1 and C2. Hence the center of both C1 and C2 is the intersection of the 3 bisectors. ㅁ
    1
  15626. The paradox may be simply due to denominating the paint by volume but the painted surface by area. One could say something similar about a humble cube of paint applied to an infinite succession of squares, all of the same side length: Successively "paint" each square, remove the volume of paint that was "used up" but since it's a box of height zero nothing changes, and repeat. Anyways, nice video. I had seen Toricelli's paradox before but I like this presentation.
    1
  15627. I prefer Nathans proof.
    1
  15628. Love it! I think what is a fantastic migraine occurs when looking at the transformation from the area of the Earth to a 2d projection. The lines that sketch the Earth, when traced, travel from Antarctica to the outskirts of infinity. But never touch back to Antarctica. Appears like a great analogy to the idea of infinity by reminding us that the limit of a Cauchy Sequence, as beautiful and simple as the equivalence appears, will only ever be a limit. Ever closer, but nothing more than approximate. As though the Transcendentalism of a solution, like knowledge of Pi, becomes clearest to clever when seen to the change of 3d, through time (watching it morph), into a 2d object. Reminds myself that no matter the dimension we change to acquire equivalences (be it time or other dimensions), does not remove the Transcendental number. So then it must be True that the Infinity we chase to acquire a limit, is then equivalent to the infinity of perspectives possible to shape one proof of a solution into another?
    1
  15629. The link to the "very nice song" is incorrect. It simply links right back to this video.
    1
  15630. Well, Challenge 1 falls out more or less automatically from Pythagoras. I proved to my self that this was true for every prime >2. But looking at it again it is pretty obvious that it is true for any odd number. There is also a unique solution for every odd number.
    1
  15631. I didn’t know about this song. I enjoy poetry in motion. Elegantly written. Do think this could be the infinity song?
    1
  15632. AlainNaigeon thats why I didn’t know about it. Thanks for the info. 🖖🏾
    1
  15633. The geometric simplification of this theorem feels just as welcoming as the theorems of double-angle trig functions; they're welcoming for anyone willing to engage in the mechanics behind them and opens it up to easy understanding that many will appreciate. :)
    1
  15634. Uniqueness when P is prime If P = a^2 - b^2 then P = (a+b)(a-b), but P is prime, therefore a+b = P and a-b = 1 ?
    1
  15635. 23:40 Look down the page, here https://web.archive.org/web/20070729135755/http://federation.g3z.com/Graphics/index.htm the Secret of Pythagoras. Previously here https://web.archive.org/web/20040116014121/http://www.uwm.edu/~whopkins/pythagoras.gif
    1
  15636. Can you pin this to the top of the comment. I was going to comment on your question at the end but I see you've corrected it anyway.
    1
  15637. There's a typo in the video description, it should say squares where it says integers.
    1
  15638.  @SimonBuchanNz  Pretty sure this one's been around for a long time. Probably predates Einstein :)
    1
  15639.  @M0rph1sm55  Good to see that some people are really going for these challenges :) Now, what's the correct statement? And how do you prove it?
    1
  15640.  @Mathologer  the dangers of half-remembered Wikipedia articles! Turns out Einstein's Pythagoras proof is even prettier (to me) - which is hardly surprising.
    1
  15641.  @michel_dutch  Thanks, fixed that :)
    1
  15642. ​ @Mathologer  ​ As you already stated, you have to add divisibility by 4. Then for any decomposition n = a*b, where a and b are both odd or both even, we have n = ((a+b)/2)^2 - ((b-a)/2)^2. Since, if n = x^2-y^2=(x-y)(x+y), to get n as any difference of two squares, the only possibilities are the ones arising from such a decomposition because the linear system x-y=a, x+y=b has a unique solution. Therefore it is only possible if n is odd (a,b are odd) or n is divisible by 4 (a,b even), otherwise (a+b)/2, (b-a)/2 cannot be integers. This also shows uniqueness for odd primes, because then, the only decomposition is p = 1*p, which yields the unique p = ((p+1)/2)^2 - ((p-1)/2)^2. PS: Thanks for your awesome videos :)
    1
  15643. What's the resolution of the paradox?
    1
  15644. Little copy/paste error at 17:14 and onwards: the numerators should be odd powers of x and the denominators the products of odd numbers only.
    1
  15645. Most memorable: Euler's gamma sum. I remember doing the rest of the proof that the harmonic series is infinite as a part of teaching Calc II, but never really connected with the students. It always seemed unintuitive to them.
    1
  15646. Great video amazing
    1
  15647. Amazing!
    1
  15648. The no integers section is my fave.
    1
  15649. Love when the t-shirt is on theme
    1
  15650. 30:22 I'm glad the tiles aren't dominoes anymore
    1
  15651. 9:48 i just realize product( 1 + x^(3^m) + x^(2*3^m) ) = product( 1 + x^(2^m) ), m from 0 to inf. also the root x=i “seems” to be completely vanished on the LHS when the above equation was restricted in finite terms. what a fun way to factorize (1+x+x^2+x^3•••) very insightful
    1
  15652. Increase level more yarr ....
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  15653. hi I really like your adventures in the world of mathematics, I'm waiting for your videos and I hope one day to make videos on the proof of the last theorem of Fermat for n=3 (euler) and for (p=2q+1) sophie germain
    1
  15654. I'm late to comment this, but I was just back of my head on and off thinking about this proof, especially one part. You mention at one point the key aspect of the proof is that p and q are co-prime, and we know this because they are prime. This reminded me of a Numberphile video where Matt Parker talks about how order in the primes, particularly IIRC how the square of every prime has one of 2 possible remainders on division by 24 or something? By memory I think this stemmed from the fact that all sufficiently prime numbers have only certain remainders on division by 6 (I guess 5 if I'm thinking about it). The point being, it wasn't just a propety of primes but rather all numbers of the form 6k+5 where k is an integer. So I started thinking about that here, we are using the propety that p and q are coprime, but really that would work for any coprime p and q it seems then, right? Like, say 9 and 5? We could have sets like (3^a1, 5^a2, 7^a3...) which are arbitrary powers of odd primes, or (3×5,7×11,13x17,...) where you multiply consequtive odd primes and then move on, in fact any set made by multiplying powers of odd primes for each entry such that no prime (to any power) is used in multiple entries. I wonder if any more naturally occuring type sets have this property of any two unequal numbers being coprime? The type you might stumble upon rather than construct as example.
    1
  15655. Mind bending in more than three dimensions. I can not even tell how many.
    1
  15656. Wow great explanation
    1
  15657. The visualisations worked great. I'd have preferred the last chapters to be its own video, but then more expressive.
    1
  15658. I love your videos sooo much. I'm writing this after just watching the first minute of this video and I'm all excited to see what is to come.
    1
  15659. very good entertainment! thank you, for the excellent visuals and calm music selection! this is one of my favorite math videos!
    1
  15660. The mad German fellow from hellboy Has finally found a human body 🤣👍
    1
  15661. When the "crisis in mathematics" was resolved by requiring that the normal vectors of the approximating triangles (or lines in the 2d case) approach the normal vectors of the surface, who was the person who actually proposed/proved this as the resolution to the crisis? (And how easy/difficult is the proof? It presumably places requirements on the surface (or curve) to be smooth and not 'fractal', as Mathologer noted.)
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  15662. Music at the end only for true math lovers 😎.
    1
  15663. beautiful, beautiful, as always. Your vids always merit a second watching (or third and fourth in my case...)
    1
  15664. mutually exclusive components of fibonacci matrix, relate to the next mutually exclusive fibonacci matrix , whose components relate exactly to next integer pythagorian results
    1
  15665. 8-10 min is best for my brain
    1
  15666. I used to think there hair on ones body was a map there life lol
    1
  15667. Swivel was more obvious for me
    1
  15668. 10:29 that shape's called an 'antiprism'.
    1
  15669. I like mathematics, but I have a very weak mathematics background. I basically flunked out of first year algebra in college. That said I'm much better at it now, but there is still much room for improvement. I find the concepts of higher mathematics fascinating. I don't understand all of what you are explaining, but you make it so much easier to understand even the most difficult concepts. Thank you!
    1
  15670. This is all clever and is more profound than the 'proof' Infinity = -1/12 however this does have a fundamental flaw. This only works if you change the rules of math; so that Order of Operations is any order you desire. Obviously a math formula will get very different results if you do addition before exponents...versus doing exponents before addition. So this DOES work; but with a new Order of Operations rule for math...to very cleverly generate numbers in a way that is neat...but also inefficient. If a computer existed that had infinite capacity AND speed? Then changing order of operations would be fine; it could calculate Pi all the way to infinity at a speed that is infinite or Instant Result. For any computer that is not infinite; to get results that are useful for some reason...you would want to use the most efficient method that you know of. Say you need 10 million similar calculations done; one method has difficulty or efficiency of 1...and another is 5000 times as much CPU calculations to get the exact same result? Well if you NEED an answer for this question for some short term and immediate priority...you would choose the fastest method.
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  15671. He is the best teacher.
    1
  15672. dangerously beautiful
    1
  15673.  @paulbird2772  😂😂😂
    1
  15674.  @Mathologer  first to reply to the first comment that didn't include the word first (coming from someone other than you)? Lol
    1
  15675. Your t-shirt is wrong. The nearest correct 4 decimal stipulation is 25,8070 :)
    1
  15676. Well, I should probably try to keep some compassion for the people who enjoy Teslaian Numerology... Hopefully they are like Z s.t. Z^4 = -1. (Imaginary roots from which something REAL can be derived?) (best I could come up with on short notice)
    1
  15677. My vote goes to Burkard's first year of Uni's proof. I don't know it, but my vote goes to it anyway.
    1
  15678. Most memorable were the tricks I learned. In the first section with the weights splitting up the heavier weights to equally spaced ones. In a later section noting that odd/even can't be an integer, which could be generalized to: numerator not being divisible by n and denominator divisible by n means the fraction can't be an integer.
    1
  15679. Heya. I got all the way through. Love your videos so much. I did maths at university (mostly calculus). I'm not a great mathematician at all, but I love how you explain mathematics. I enjoy it personally and don't really use it for anything professionally with the exceptions of some excel stuff, accounting stuff, and some programming stuff, but otherwise watching your videos is like absolute candy to me, you make really difficult stuff simple and easy to understand and if not easy outright, at least digestible enough so I can figure it out as we go on. Euler is one of my favourite mathematicians. Which is to say, yes, more outlandish, weird, and difficult stuff. I love the progression going from gauss addition which we learnt at school. But we learnt it 1 + 2 +3 + 4 = (4+1)x2 = (Highest Plus Lowest) Multiplied by Half of Highest. Because n + 1 = (n-1) + 2 etc... So if n = 100 then 1 + 100 = 99 + 2 = 98 + 3 (and there's 50 of these pairs), so n100 = 101 x 50 = 5050 Ironically I use this trick to show off quick addition of Dungeons and Dragons dice, the numbers of a d20 add up to 210, the numbers of a d10 add up to 55. I mean it's simple if you know the trick, but folks are always impressed.
    1
  15680. Neither at school. :-( thank you for sharing it!
    1
  15681. I can has cheese?
    1
  15682. Can I have cheese?*
    1
  15683. Wow you just proved the earth is flat lol
    1
  15684. I really appreciated how you saved the explanation for "4 good minus bad" theorem until the end of the video. It cut off almost 9 minutes from the main proof and made it much more manageable. Good job
    1
  15685. At 7:20 it is said that we know how to construct all pythagorean triplets easily, does anyone know the method/algorithm to have them ? And in fact I would like an way to generate them by increasing hypotenuse length ^^ Is there a way ?
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  15686. me before the video starts: so this guy used math to predict the mystery of Death Stranding? 🤔
    1
  15687. He's back!
    1
  15688. I get doubling when starting with 2, but why not tripple when starting with 3 etc? why double?
    1
  15689. Because that's what they felt like doing.
    1
  15690. Really very exciting till the last second. Taking this opportunity I want to know about the following observation about multiplication and division: We have been telling our students that multiplication is repeated addition.  For example 3×4 = 4+4+4  or  3+3+3+3 And division is repeated subtraction For example 7÷2 = ((7-2) -2) -2 and 1 remains. If some RAMANUJAN from our class asks ( as Ramanujan did with his teacher that how many chacolets each gets when 0 number of chacolets are distributed among  0 number of pupils? )  how many times 3/7 is to be added in the case of 2/9 × 3/7 ? In other words 3/7+ ... ? ( times ) Like the same how can 2/9÷3/7 be explained in terms of repeated subtraction. In other words, 2/9 ÷ 3/7 = (( 2/9  - 3/7) - 3/7) -.....? times and could there be a remainder? Are the multiplication and division in numbers other than whole numbers have seperate definitions anywhere in secondary or highschool books? Or Are these questions have answer in higher mathematics by giving clear cut definitions to multiplication and division? Please mail your thought about this to my email id, kasiraoj@gmail.com
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  15691. i thought it was Aristotle that discovered calculus...
    1
  15692. 12:05 - the block has 9 in the top left and 4 in the top right. This makes 13 the bottom right and 17 the bottom left. This works out as 9*17 = 153, 2*4*13 = 104, and 9*13 + 4*17 = 185. I solved it by setting up a square of unknown factors ABCD, and then I set the product of AC to be 153, 2BD to be 104, and AD + BC to be 185. If you add AC + BD + AD + BC you get 390, but also the quantity (A+B)(C+D). I then simply looked at the integer factors of each number and did trial and error with A and B, given that A + B had to divide 390. Bonus question: starting from our first square, we can get to this new square going right, right, right, left. We know that the top two numbers (9 and 4) are enough to determine the entire square. The three possible combos that this square could spawn from either have 9 bottom left and 4 top right (which generates 14 on top and 95 on the bottom), 9 and 4 both on the bottom (which is impossible), or 9 in the top left and 4 in the bottom right (which is also impossible). Therefore, we then recursively look at 1459, with the two seed numbers 1 and 4. Here, we find that 13 on top and 74 on the bottom is the spawning square. This is show in the video to come from 12 and 53, which then comes from our starting square of 11 and 32. 15:53 - the area is 5. The yellow square has an area of 1. Since a^2 + b^2 = c^2, the sum of the two orange squares is also 1. We can do this recursively and find out that the sum of the areas of the four squares connected to the two orange squares is also 1, the sum of the areas of the eight squares connected to the four squares is 1, and so on. This means that the area equals the number of different sizes of squares in the diagram, as every single set of squares has a total summed are of 1.
    1
  15693. Is not only beautiful...i quit my math studies 37 years ago and i can follow You explaining so clearly, correct speed ( even for me spanishspeaker). So the question with unhappy answer is why it didnt ocurrs to me before?... A Challenge for you: would you be able to explain so clearly, directly (without a new hole theory) why can not exist a formula for 4°grade pol equation?
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  15694. I appreciate your intuitive teaching style. As you pointed out, other people tend to go deep into pedantic calculations, or keep gushing about how cool the paradox is.
    1
  15695. Very interesting mathematics video
    1
  15696. Beautifully animated formulae at the end
    1
  15697. What a beautiful video. Thank you very much ! ❤
    1
  15698. I loved the video and the animation, however there is something I dont understand: what's the point of the undershoot-overshoot approximation algorithmm? Its purpose is (I suppose) to approximate the value of some number x, and internally it uses the exact value of x itself to adjust its behaviour? If we already know the value of x what's the point?
    1
  15699. 9 is one less the base, 10. That's why it is "special". The division test works for any base, for example, if the we define base 16 digits 0123456789abcdef, the divisibility by the f test is exactly like the one for 9 in base 10: just sum all the digits and test if it's divisibly by f. (And also, since f is 3*5, this test will also work for division by 3 and by 5). And only the elementary math is needed to prove this statement. (Actually we proved this routinely in school.) Oh, yes, I've paused the video, wrote the comment, continued... and now I see there is an explanation in the video :) The divisibility by 3 in our decimal world is a bit special; it doesn't extend to other bases so easily.
    1
  15700. Merry Christmas for you.For Hindus it is vaikunda Ekadasi .But for knowledge and wisdom gnana we worship vinayaka elephant god.
    1
  15701. Yeah it's Mathologer not Discovery.That's the power of math. ❤️Love from India❤️
    1
  15702. Why does Mathologer looks like a hairless Schlafi trying to regain his fame for his discovery from Euler's more popular equation xD
    1
  15703. This triangle folding exercise is just another way of saying that the triangle tiles the plane. If instead of cutting the triangle up and folding the bits around, just fold one of the six little triangles around, dragging along uncut copies of the other five. Voila, the now duplicated whole triangle fits perfectly on the original triangle (it must!, same side is just rotated 180°), with one new side parallel to the opposite side of the original triangle (again, it must, as it is just rotated 180°). Repeat with the other two sides of the original triangle and you find that the duplicated triangles now form with the original triangle a similar triangle four times as large. Continue repeating the process and you can fill the plane. Not very surprising.
    1
  15704. You the one should have included, which are the most comfortable lacings to the type of feet one has.
    1
  15705.  @Mathologer  Danke :)
    1
  15706. Thank you for making this video
    1
  15707. I don’t quite get how popular this was? I realized right away that it would terminate.
    1
  15708. I wish I wasn’t afraid of math, and therefore horrible at it.
    1
  15709. Hey Mathologer. I loved watching your videos for all these years but recently I have lost my eyesight due to exposure to bright background white light from my favorite youtuber.
    1
  15710. Nice t-shirt
    1
  15711. I have two queries..i don't have a proof if these two are correct. (1)Let "a1" & "a2" & "a3" ...."an" be integers such that all of them are non zero .Is lcm(a1,a2,a3,...,an)=lcm(|a1|,|a2|,|a3|,...,|an|)? (2)Let "a1" & "a2" & "a3" ...."an" be integers such that atleast one of them is non zero .Is gcd(a1,a2,a3,...,an)=gcd(|a1|,|a2|,|a3|,...,|an|)?
    1
  15712. Beautiful proof. Well done to Rainer!
    1
  15713. For squares in 3 dimensions, I direct you here: https://oeis.org/A334881 I have no intuition for why this is the case
    1
  15714. Oh gosh I was craving such stuffs 😂. Love ur creations sir huge love and respect from India . Thank u sir!
    1
  15715. Most memorable: closed-form approx sum of first n terms = ln(n+1) + γ. Prime-counting formula is also ln(n) ... connection? Is the sum of the reciprocals of primes is the "closest" to converging of the divergent subsets?
    1
  15716. Most memorable - at one glance balancing, ie. The very first thing you showed, which is more or less a proof of the law of moments.
    1
  15717. Great video as always, love the geometric proofs! I just wanted to say, there are shape-shifters among closed surfaces. For example the torus (as the quotient of the plane deformed under any 2×2 integer matrix of determinant 1), and perhaps more interestingly, hyperbolic surfaces (genus 2 or higher) under the so-called pseudo-Anosov maps. sinh and cosh also appear in a lot of formulas for these surfaces.
    1
  15718. I was taught a lot of this - can't remember if it was school or my mother
    1
  15719. I don't know why, but I genuinely enjoy proving reciprocity theorems, not just quadratic ones either. For some weird reason, and its almost instinctive, I think reciprocity theorems are the key to proving the Beal Conjecture.
    1
  15720. Really nice proofs!
    1
  15721. 49:46 is the answer 55🤔
    1
  15722. Another great antishape is the entire R^2. And you may not believe me, but so is the first quadrant! Or the second quadrant for that matter! Amazing math!
    1
  15723. I don’t understand anything he says in any video ... yet here I am watching ... why ?
    1
  15724. DId you notice, that Gelfonds number contains the first digits of ln(2) too? 2 3.14 069
    1
  15725. when i did math in the 70s there was a slide rule class... and we were allowed slide rules in exams. i bought in my dads round slide rule (this thing)... the teacher said that wasnt a slide rule.. i said it was. he said - if you want to use it, you have to figure it out, cos i dontk now how that works! REALLY? so i used this in the 70s :)
    1
  15726. This helps motivate me to study for my maths exams next month.
    1
  15727. Obviously one of the best channels ever created on YouTube!!
    1
  15728. I have been fascinated by Pi since i was in algebra class, and i had a bet with a fellow student that i could remember Pi to more decimal places than he could, and i beat him by a long shot, and now at age 73m i can still remember Pi to the 51st decimal, which i realize is nothing compared to the record, of one hundred thousand.
    1
  15729. Very powerful and profound!
    1
  15730. Make a circle, draw a square, then divide one of his sides in half That wouldn't be a triangle, by using the middle point with the base in the opposite direction?
    1
  15731. Love your videos!! Greetings from Brazil :)))
    1
  15732. Best Christmas present I could ask for
    1
  15733. I dont understand why it is called miracle. It is fabulous to occur, but it always occurs. A miracle is something that never occurs, only God makes it happens. Greetings
    1
  15734. for the puzzle at around 12:00, the four numbers are 9, 4, 13, 17
    1
  15735. I was about to go to sleep right now......but who cares about sleep..... I will first watch your video despite it being 45 min long right now
    1
  15736. I wish my numerical methods classes were this interesting!
    1
  15737.  @Mathologer  brute forced it lol. check all doubles and triples of numbers
    1
  15738. Wondering if this could be applicable to Knot Theory to untangle arbitrary knots to lowest crossings
    1
  15739. Not for the first time, I've paused one of Burkhard's videos within the first twenty seconds because I'm analysing his shirt. I'll be back later to watch the actual video.🧐
    1
  15740. I don't know if it's good for the sound to paint your horn on the inside, but I like the video...
    1
  15741.  @Mathologer  The general formula of f(x) = (nx + b) mod m . It bears a striking resemblance to a pseudo random number generator. In that case the modulus would be something like 2^32, and b would be the random seed. But I guess it would just be a "twister" in the general case. An actual Mersenne twister has a repeating period of a Mersenne prime.
    1
  15742. The Tschirnhausen cubic relates to any of the six corner/cusp singularities. NOTE THE CONNECTION OF E TO THE E FOR RING/CYLINDER SINGULARITIES AND 3 TO THE 3 INVOLVED IN CARTESIAN COORDINATES!!! As dark matter rotates, it carves out a parabola of sorts when viewed like one would view a rotating snowflake or a rotated doily from their lateral planes.
    1
  15743. For the whole dancing tilings phenomenon. It seems relevant to know not only that all tilings can be generated with this method, but that they are generated with equal probabilities, right? Otherwise, some tilings might be generated more often, while others will be generated more seldom, so our generation is not equivalent to "Uniformly pick one random tiling", like the "Homer Simpsons" method.
    1
  15744. What's the music at 4:25?
    1
  15745. Colouring method rulez
    1
  15746. Fantastic!
    1
  15747. Most memorable part: greedy algorithm.
    1
  15748. So brilliant~
    1
  15749. New Year's Eve '21 light-hearted mathematical fireworks! Nice.
    1
  15750. My vote goes to the colouring proof.
    1
  15751. No matter how many times I think about this, it never ceases to put a smile on my face
    1
  15752. like a 5d version of the 2048 game
    1
  15753. Funny story. I am making a game and wanted to fill in the volume of a generalized polygon to automatically generate sprites. I was originally going to to this using the ear snipping algorithm which relies on the order of the points being one of the two clockwise or counter clockwise and use one consistently across all polygons. I needed a way to know which a polygon was, so I found this video where this algorithm gives negative or positive area based on the direction. Instead of doing this complicated 3 step process, I can just use a slightly modified version of the shoelace algorithm after finding out the direction, making it a much more efficient 2 step process.
    1
  15754. Coloring proof is my favorite.
    1
  15755. and you know where these things will NEVER be taught? In India. the inferiority complex bred into our DNA by British rule means we will only trust this when western scholars tell us about it. 😞
    1
  15756. This is giving me the same vibe as the "Do you see?" scene from Red Dragon.
    1
  15757. Meine Überlegung (ohne es zu Ende gesehen haben natürlich) Das funktioniert, weil die Abstände zwischen den Zahlen den Faktor zur nächsten Zahl repräsentieren. Das müsste dann bestimmt auch zum potenzieren mit Potenzenbasierten Abständen funktionieren oder ?
    1
  15758. Where do you get all these shirts?
    1
  15759. I hate the trig proof of Heron's formula with a passion. Thank you for providing the alternative proof with the triangular dance. I KNEW something more beautiful must exist! I am going to rewatch that part after posting this comment.
    1
  15760. I own a slide rule. It's around somewhere, I think on my bookshelf back at my parent's place.
    1
  15761. Hi, may I make a request? In this video you have a few places where you highlight parts of the screen by either putting a block of red (or black) around text of the other color. Red on black (or vice versa) is impossible for colorblind people to read.
    1
  15762. Maravillosa jugada.
    1
  15763. Thanks for showing us how to use math
    1
  15764. I love the symmetry in your augmented proof :D
    1
  15765. I heard the 700 year old proof of divergence in my first lesson for preparation for mathematic olympiad and it got me hooked up. When I found out that also the sum of the reciproces of the prime numbers diverge, I almost could not believe it. So going for the latter as most memorable.
    1
  15766. 14:23 There's titles segments counting as those parameters: 1*208 = 208 2*208 = 449 3*208 = 925 4*208 = 8250
    1
  15767. The music that plays when you shift the points make me feel like I’m on hold
    1
  15768. I don't buy all that nonsense about the "key to the universe". I just think that Tesla was a huge fan of the Volkswagen logo.
    1
  15769. Wait a minute! I feel my mind is being blown. OK, continue.
    1
  15770. I mean this as a compliment: you could totally play a European spymaster in some thriller set during the Cold War.
    1
  15771. Okay now I wanna do the 4d Rubix!
    1
  15772. Of course it's a coincidence: it coincides.
    1
  15773. Is the 1/7 fraction in the Bizets formula a coincidence, or is that somehow related to the fact that these numbers are generated by the 7-gon? I would argue that the sqrt 5 that appears all over in Bizets formula for the Fibonacci sequence comes from the relation to the regular pentagon, so it seems like this would be the case here as well.
    1
  15774. Babylonians solved the quadratic eq by completing the square circa 1800 BC
    1
  15775. This is such pretty maths. The equilateral triangle one is easy to prove that they'll always be equal because a box (a set of three rombic dominoes) can only be arranged in two ways: the box poking in or the box poking out (if thinking in terms of 3D), there is no other way of arranging three of these dominoes other than 'box in' or 'box out'. So whenever you pop a box in our out, you're just shuffling the same three dominoes around, so you never lose or gain dominoes of different types.
    1
  15776. Why would some people "dislike" this video? Can we prove something like: "all those who disliked this video must be mathematicians"? Can't we even dream about it? :)
    1
  15777. Neat.
    1
  15778. The first part of the challenge is quite simple. Simply set x to equal y+1 and you get (y+1)^2-y^2 = 2y+1 which would cover all odd numbers. In fact, we can assume that x = y+a since x^2-y^2 is a positive number. In this case, x^2-y^2 = (x-y)(x+y) = (y+a-y)(y+a+y)= a(2y+a). Since prime numbers have only 2 factors, a has to be equal to 1 which means there is only one solution for prime numbers.
    1
  15779. Thank you for this sweet new headache I am about to experience.
    1
  15780. Love the shirt!
    1
  15781. You know Mr. Mathologer, apart from loving your t-shirt except it's not same c they are supposedly fighting for.... It's a good thing to have your videos, the graphics are very welcomed. Except it would have been a great help more than 25 years ago for me, when this material wasn't available. I also get this expanding with symmetries etc. Groups etc etc. To generalise. Cheers.
    1
  15782. 1:37 When Pythagoras is sus!
    1
  15783. As a math teacher and amateur programmer I tried to find solutions for n>2: For n=3, the only "real" solution is 3³+4³+5³=6³ but there are other similiar solutions, for example: 4³+...+28³=30³+31³+32³+33³+34³ 8³+...+55³=60³+...+68³ 34³+...+158³=540³ 61³+...+71³=101³+102³+103³ 213³+...+272³=556³+557³+558³+559³+560³ For 3<n<11 I couldn't find any solutions at all.
    1
  15784. You know..for those of us who really struggle with mathematics, math is the devils workshop.. You are evil beings! LOL
    1
  15785. gamma value for me
    1
  15786. Pythagorean triple tree: Representing the square as: a b c d We know that: d = a + b c = d + b ac = 153 2bd = 104 ad + bc = 185 We will use 2bd: 2bd = 104 |/2 bd = 52 | substituting d ab + b² = 52 | -b² ab = 52 - b² |/b a = (52 - b²)/b Input different integer values for b lower than 8 (8² = 64, 7² = 49) to 1 until an integer is found, test with another formula (a²+2ab=153). We find a=9,b=4 making the square: 9 4 17 13 Since 9>4 there are only two squares it could come from 1 5 9 4 5>4 so we don't use this one 1 4 9 5 From here it's the same all the way: 1 3 7 4 1 2 5 3 1 1 3 2 Representing a b -> c d c d -> e f as x a b -> c b c d -> e f as y a b -> a d c d -> e f as z We get to it using zzzx. The other tree: The area of the connected branches equals that of their predecessor (as per a² + b² = c²), so this one has an area of 5.
    1
  15787. Really enjoyed the video. I understood most of what was said right up until Chapter 7. I have a physics background, and some of those algebraic manipulations at about 41:09 took me off guard. I've got some homework I guess. There's a video that goes a bit further into the matrix representation of the Bernoulli numbers which might be of some interest, although, it's not as stylish as the one presented here. https://www.youtube.com/watch?v=qmMs6tf8qZ8&list=PLzdiPTrEWyz4rKFN541wFKvKPSg5Ea6XB&index=16 Looking forward to the next one.
    1
  15788. I didn't get why we should double the product of right-colomn numbers?
    1
  15789. The area of the tree is 5.
    1
  15790. I really love your presentation.
    1
  15791. What do I think about wiper theorem? "Objects in mirror are closer then they appear"
    1
  15792. Chapter 2 is the best! so many maths solutions have a simpler geometric explanation, shown again in Chapter 5... This incredible capacity of demonstrate/ find solutions using visualising geometric examples is beautiful to see. So sorry to just a Engineer...
    1
  15793. This was brilliant
    1
  15794. Wow. I forgot why I loved math. Thanks for this video.
    1
  15795. These approximations are how graphic designs work. Now matter how hard we try true circles and spheres are impossible based on our digital diplays.
    1
  15796. colour proof!
    1
  15797. Merry Christmas...🎄
    1
  15798. I made it to the very end and thanks for the great video! :)
    1
  15799. I am still going to say "cinch" and "tantsh."
    1
  15800. Thank you sir! 2:46 I thought you were going to introduce us to a sponsor lol
    1
  15801. The stretching of the rubber band is triggering me at 1:26. I will endure! lol
    1
  15802. Now make a change for g64 and tree3
    1
  15803. As soon as I read the Lee Price paper I thought this discovery deserves to be better known, and I could visualise those dancing circles. Thanks for sharing in such an engaging way. PS the top pairs in the Fibonacci boxes of those middle children form a series in which each pair is equal to twice the preceding pair plus the pair before that. Like the Fibonacci series the general term can be expressed in closed form. The counterpart to phi is one plus or minus the square root of 2.
    1
  15804. I haven't put the powers of 2 coins questiion on paper, but, based on the binary system and how it has only two possible states, i guess that, with 1 coin for each power of 2 from 0 to N, any number from 0 to 2^N-1 can be reached in exactly one way.
    1
  15805. I was going to guess 333.
    1
  15806. I think that you have a very intuitive way of teaching my man. By the way, what’s your personal best time on a 3x3?
    1
  15807. The key to counting the number of tilted squares is to observe the vertices are identified by 90 degree rotation
    1
  15808. I made it to the end and I actually really liked this video because there was much less of omitting the details than usual except my main problem which was 40:36. I'll have to think about that part. 48:51: Personally I am a computer science undergraduate but I competed in my country's math olympiad when I was in high school.
    1
  15809. I enjoyed the video a lot. I believe some pieces are allowed to be more specialized than others. Reaching masses at every video should not be the goal of this channel, in my opinion. Otherwise this channel would not be about maths, but on soccer, baseball and dieting etc. Looking forward to more videos Euler-Maclaurin summation method. Very interesting indeed. I also remember a reference to the sum in the movie "goodwill hunting".
    1
  15810. Thankyou.
    1
  15811. I think I understand what you're doing. Calculus is the first mathematical area that we learn *rigorously*. Yes, we learn geometric proofs earlier, but we already knew geometry by that point. Yes, we learn the basic ideas of calculus from examples like this, but not deep enough to be intuitive. Introducing calculus by first teaching the epsilon-delta definition of limits is why "math is hard."
    1
  15812. T-shirt is wrong...25.8070
    1
  15813. The ending felt like a nice trip
    1
  15814. Ive watched it... i want moar
    1
  15815. WOW. How impresive! Cuspy function wins. Tell it to AI guys - they won't be surprised.
    1
  15816. In know a math YouTube channel (MindYourDecisions) who insisted on calling it the Gougu theorem, because the Chinese have the first documented proof. I find it honorable that he tried to be historcally sensitive, but I it was more distracting than anything.
    1
  15817. Differentiation is easy, integration is not.
    1
  15818. Pascal n modulus row == 0 encodes prime numbers (but I don't think its protected from Mersennes)
    1
  15819. How about: (-1) + 0 + 1 = (-1) * 0 * 1
    1
  15820. I believe I left a similar comment on the previous video, but the book Analytic Combinatorics covers the topic of this video extensively. It was the main reference of a class I took last year, though the introduction was through Knuth’s Concrete Mathematics.
    1
  15821. very nice
    1
  15822. Seasoned engineer here. This can only be the product of evil black magic. Why do I love it? 😁
    1
  15823. Amazing, brilliant, beautiful. I made it to the very end ;-) After half a century you’ve totally reignited my love for maths. The wonderful thing is, my 8 year old son has it too. We face a wonderful Voyage of discovery. 😁😍
    1
  15824. My favorite fact definitely was how the no partial sum of the harmonic series adds to an integer. I am a big fan of these little tricks that doesn’t look very interesting at first glance, but are just absolutely delightful. Once again, than you for an amazing video!
    1
  15825. +1 for Tristan's visualisation
    1
  15826. :)
    1
  15827. I am about to give up at 15th minute... This is as crazy as the year 2020
    1
  15828. I can't believe the "100 zeros" sum is bigger than the "no 9's" sum! That is the most amazing thing from this video, for me.
    1
  15829. 2:15 talking about no hairs with a straight face? 😬
    1
  15830. Just amazone, more on ramanujan pls
    1
  15831. I paused this video to go make one and now I don't where I got to.
    1
  15832. i belive there is simple solution for this ...
    1
  15833. @Mathologer, you might be able to gauge from the comments, the kind of anguish Indians experience that even to this day their education system and the dominant narrative in/history of the world, fails to be honest and recognise and educate them of such great achievements of their ancestors. There is a lot of such knowledge in the world that still sadly bears the taint of colonialism / imperialism. Anyway, thank you so much for this gem of a video; both for the wonderful math and for the honesty. 🙂❤🙏
    1
  15834. Two comments: 1) In applying Cavalieri's Principle, it is not enough to show a 1-1 correspondance between the "slices" in the two solids or regions -- consider, for example, a scalene right triangle. Draw a segment parallel to the hypotenuse from one leg to the other -- the segments from the endpoints of this segment, running vertically from each leg to the hypotenuse are the same length. The possible positions of this horizontal segment set up a 1-1- correspondence between the equal vertical segments. But the subsidiary triangles separated by the altitude from the right angle to the base are not equal in area! 2) I think there is an easier way to see that the area of a circle or volume of a sphere equals that of a triangle (or cone) with base equal to the circumference (or area) of the circle (or sphere, respectively) and height equal to the radius. Think of the sphere as an orange! Cut it into wedges meeting in the vertical diameter. Now take each wedge and flatten the outside surface, allowing the "pulp" of the orange to separate out into many small pyramids, each with height equal to the radius of the sphere. The bases of these pyramids have area equal to the area of the "rind" of the segment. (For a circle, the the interior breaks up into triangles of height equal to the radius when the circumerence is flattened). Now just add up the volume (or area) of the pyramids or triangles!
    1
  15835. Most memorable : Kempner's proof animation
    1
  15836. I can't remember where, but I read an interesting intuition for the formula at 23:11. Count the number of ways you can take n bits and flip k of them on, which is (n choose k). If you add all those terms, you must get the number of ways n bits can be configured, 2^n.
    1
  15837. I seem to remember that the peg solitaire I played allowed diagonal jumps. Is that not typical?
    1
  15838. 🌻
    1
  15839. the Euler characteristic is great. there are 3 proofs from the last 25 years that follow one another that I just love: 1998, Shahrokhi, Sykora, Szekely and Vrto: bounds on the number of crossings in a graph (“Crossing Numbers: Bounds and Applications”) gives a minimum number of crossings for a graph with v vertices and e edges simply using probabilistics of the Expectation function and the Euler characteristic for planar graphs 1995, Szekely: bounds on the number of incidences of points and lines (uses the above, but, of course, this proof did not yet exist, but the result had been proven differently) (“Crossing Numbers and Hard Erdos Problems in Discrete Geometry”). Simply make a graph as above out of the points/lines where they cross but there isnt a point. 1997, Elekes: bounds on the number of Sums and Products ("On the number of sums and products") bit more convoluted, but still doable? uses the point/line result.
    1
  15840. Great stuff!!! love it. well done
    1
  15841. I am really not qualified to watch this...
    1
  15842. Does the elasticity of the rubber band have any effect on the layout of the numbers?
    1
  15843. Who else would like to see the helicone tree implemented with Matt Parker's addressable Christmas tree lights from a year or two ago?
    1
  15844. If there was ever a question as to your accent then I shall send them to 2:06
    1
  15845. I liked it all. Even your TShirt. Don't burn out and have a great year!🎄
    1
  15846. There IS a pattern looking like A³+B³+C³ = D³ though, Burkhard!
    1
  15847. Truly Beautiful! I think I liked the last proof the best, probably since I already knew the 2:1 ratio of intersecting medians.
    1
  15848. that was incredible
    1
  15849. Found my error I was off by 1 a 4 tri has 3 sets of 3 across the top so a 4 tri made of 4 tri is = 3*4-2 = 10
    1
  15850. This was a very good one! I had not thought about why only those angles are convenient in those terms before, and what an interesting way to prove it!
    1
  15851. 35:30: I assume this is called the Nine? Nein! sum.
    1
  15852. Just 3 minutes in. The rubber band has to stretch right? To me the size of the wheel and elasticity of the rubber band would have to match for this to work. You couldn't use a stronger or weaker rubber band with this set size wheel but you could tailor other size wheels to stronger or weaker rubber bands to give you the same effect. Back to the video. I hate math because people all to often use calculations not based in empirically measured reality to claim evidence of distances.
    1
  15853. 41:13 Where's the argument? Use g instead of f, add 0 to both sides and end it at g(n-1), that converts it to g(0)+g(1)+g(2)+...+g(n-1). g(0)+g(1)+g(2)+...+g(n-1)=integral from 0 to n-1 g(t) dt + (g(n-1)+g(0))/2 + ... Now take g(n)=f(n+1). f(1)+f(2)+f(3)+...+f(n)=integral from 1 to n f(t) dt + (f(n)+f(0))/2 + ... Feedback: Cool. Maybe make some "table of contents" with short descriptions and timestamps to the various chapters, would double up as a great recap too.
    1
  15854. As you were making more bands (but keeping the points around the perimeter fixed), I couldn't help noticing that the triangles were getting 'more horizontal'. So the areas were not decreasing as much as they 'should' for more bands. So in the limit, each band becomes more like a flat ring (of zero thickness) that you're stacking on top of each and of course a 'ring' has a fixed limit of area depending on the points you've picked around the perimeter. Then you explained how the 'normal spike' for the triangles has to behave to get a good approximation and I realized that controls how the 'points' versus 'bands' relationship has to behave. Has to be such that the 'spikes' approach being normal to the true surface. Very nice explanation and very nice graphics.
    1
  15855. I'm getting my Master's Degree in Mathematical Physics, so I guess it could be argued that I don't know the first thing about Math! Anyway, I could follow it with no problem and even got my first introduction to Euler-Maclaurin! As always, I had a lot of fun!
    1
  15856.  @Mathologer  I'm getting acquainted with String Theory and, though I can't pinpoint anything specific yet, I have the feeling there are plenty aspects that would be neat to visualize. I really enjoyed when, at the beginning of a course in Special Relativity, I was shown that you can get Lorentz transformations if you search for the most general between inertial frames without assuming absolute time. Or, at the beginning of General Relativity, that a frame at rest with a uniform gravitational field is indistingishable from a, small, free falling one. What I like about these is that they were in front of everyone's eyes, but it took a relatively (pun intended) long time to take the leap and it opened up a new (way of seeing the )world. With these subjects you could also go wild with apparently paradoxical thought experiments. There's also a pedagocial purpose here. Maybe I'm wrong, but it appears to me that stuff like Schrödinger's cat or the twin paradox is thrown in the face of the non professionals as some kind of magic show. When they were inteded to test and develop our intuition of what is actually going on and the beatiful part is how you reconcile them. It's also necessary for the theory to not fail! By the way, I'd watch your feature length on Galois Theory!
    1
  15857. I figured out the pi digits part, but can't see the significance of the colours.
    1
  15858.  @kevinmartin7760  I think the color only makes it a beautiful looking without any special meaning. You can image there is only one color and how dull it is. The 2 digits interval should also be for a better view.
    1
  15859. Ha, but we have 10 fingers because God wanted Tesla to figure it out!
    1
  15860. Thought Mathologer spoke Czech for a second there 😂
    1
  15861. Im hooked to your videos. As a finance professional with deep interest in statistics and quant, this pure mathematical inquiry into calculus, limits and infinite series is rather illuminating. Thanks Professor!
    1
  15862. I wonder what you get if you raise Gelfond's number to the power of negative one, raised to the power of one half 🙃
    1
  15863. First of all, I'm very curious to hear your solutions. I'm quite familiar with the tower of Hanoi puzzle, it's one of my favourites. My method to solve it depends on even and odd numbers, let me explain. By all means I'm not a professional solver or something, I'm just very passionate about it. The first move is the most important so one should move the first disk to the second peg in even number cases, and to the third one in odd number cases. Odd and even refer to the number of disks one wants to move at any given time in the game.With practice you can move a large amount of disks just by muscle memory. Sometimes it can be tricky not to lose track of your moves, my way to solve this problem is to rely on said muscle memory to move the first 4 disks. By doing so I just need to mind the larger disks. I hope someone might find this useful
    1
  15864. God, I love everything you create.
    1
  15865. I had a big orgasm at 21:48
    1
  15866. I used to stack my books like this as a small child
    1
  15867. Favorite was the missing int. Also: gamma >0.5 because the curvature is left so the diagonal is above the cut hence the blue part of each rectangle is bigger then 0.5 Greetings from germany
    1
  15868. This week I finished my MsC in Applied Mathematics and this channel was the original one which inspired my interest in these matters.
    1
  15869. amazing❤
    1
  15870. 11:44 Ok, I've been trying to solve this for a little bit. Had to build my own formula for solving for the radius of the incircle of a right triangle, and it was huge and messy. Using the first 4 you gave here 2:20 I build 4 equations for the square [(a b)(c d)] and the equation x^2+y^2=z^2 Just so I can write one of the equations, I will say x+y+z=s 1. ac=x 2. 2bd=y 3. ad+cb=z 4. ab=0.5sqrt{((s-2x)(s-2y)(s-2z))/(s)} Yeah if someone could show me how to simplify that it would be awesome Replace for x, y, z ac=153 bd=52 ad+cb=185 ab=36 from here, I will assume that a, b, c, d are all positive integers. Here are all possible values for a and c if so (a, c) (1, 153) (3, 51) (9, 17) (17, 9) (51, 3) (153, 1) Now, we know that a needs to also be a factor of 36, so we can eliminate the values 17, 51, and 153. So adding b into the mix will halve the options left, quickly grabbing its values (a, b, c) (1, 36, 153) (3, 12, 51) (9, 4, 17) Now, b needs to be a factor of 52, and 4 is the only number that fits that criteria (a, b, c, d) (9, 4, 17, 13) and there we go. Checking the last criteria, ad+bc=9*13+4*17=185 checks out The only problem is the excircles, because I don't know how to find those. And since that would mean I'm not 100% confident in my answer, I can't answer the bonus just yet
    1
  15871. 2:02 Yes mhm I see, of course (nods knowingly)
    1
  15872. That was way more interesting than I expected. Thnks!
    1
  15873. wish all math teachers were like you...you make learning math fun :)
    1
  15874. I'm pretty not capable to figure out all the implication/application of this, but can't stop thinking it look alike Fourrier transforms - composition/decomposition
    1
  15875. he wasn't lying when he said it'll be magical
    1
  15876. Ha! you're using the little game I built and photographed for Wikipedia as an example.
    1
  15877. yay
    1
  15878. Recreational art, performed inspirationally. Thank you, good Sir!
    1
  15879. Oh, but we are not done yet...:cat-orange-whistling:
    1
  15880. I used to work in a factory while I studied mechanical engineering. I think it was about 2008? I spent some of that time in the drafting office and the lead draftsman in there was fond of his old Hewlett-Packard Reverse Polish calculator. He explained it to me and I found it fascinating. Now I use RPN in the calculator app on my phone. On a later occasion I was speaking with the head honcho, brought up the old calculator and he excitedly went on to talk about slide rules. He was delighted that I'd heard of them and (vaguely) understood how they worked so he gifted me a little 6 inch promotional slide rule for "Opperman Gears Ltd, Newbury, Berkshire, England". I learned how to use the trig scales on the back and took it into exams as a backup in case the calculator ran out of batteries :) I still have it and I love it.
    1
  15881. Stunning, as always!
    1
  15882. Marvelous...
    1
  15883. Hello from rural Idaho USA. I have a B.S in Mathematics '97 its been so long since I have seen power series problems I'm not sure what course it's taught in calc II, calc III? Thanks for your hard work on your site. I mostly use knowledge gained from a discreet math course I took in daily life have not done an integral since about 95/96. 8) cheers
    1
  15884. You made me learn something new, but not what you expected! I wanted to know the rest of #15 (if the points of the plane are colored red, yellow, or blue ...) because each of the first 14 I was already familiar with (the theorems, not necessarily the proofs) if any of those had been cut off in the middle I would still know what theorem it was, but #15 was new to me. I searched and found that the rest of the theorem says “for a given distance d there will be two points of the same color distance d from each other” and saw the proof using Moser’s Spindle, and yeah, that deserves to be on the list, IMHO. You probably already did a video on #15, but if not you might consider making one, it would suit your style (though it might be too short, given the elegance there isn’t much to say).
    1
  15885. What do you mean: "If you don't know any calculus, ..."? 🤣
    1
  15886. I enjoyed the whole video. Fun topic. Of course I do have a PhD in math....
    1
  15887. So how to prove the two on 41:00 are equal to each other?
    1
  15888. Very good vídeo! I liked very much. But at the minute 19:16, in the 3 dimensional axis there are only 7 vertices instead of 8. Of course it doesn't affect the overall quality
    1
  15889. I think it's quite amazing that something as simple as dropping all terms with a certain digit in them can make their respective series finite. Even more interesting that a series with exactly 100 zeroes can still achieve a high number despite the 1st term being so incredibly tiny. I wonder if that has to do with the fact that the "no 9s series" and related are ultimately more restrictive on what terms can show up.
    1
  15890. I don't get the point of "pi exactly". You approximate pi under the condition you know pi!? Can someone help me out?
    1
  15891. A value can be exactly pi, it just can't be expressed as a decimal. But in a practical sense, there are other means of approximating pi that you can continuously refine to use as your target. HTH
    1
  15892. Would the argument still work if face up cards were represented by 1 and face down by 0?
    1
  15893. Play with crepe paper sometime.
    1
  15894. Really good!
    1
  15895. my strategy for the second one: rewind 5 seconds, see the third row of the squares board, go back, compute 10ˇ2 + 11ˇ2 + 12ˇ2, realise it's 365, so it must be one! oh wait, the second part, so 2
    1
  15896. My favorite was gamma as an approximation of the error in the approximation of ln(n+1) for the partial sum. Kempner's proof was cool, but I find math related to artifacts of the decimal representation of numbers ("no nines") to be unsatisfying, because it feels like it's not revealing anything "deep".
    1
  15897. Thanks!
    1
  15898.  @Mathologer  you are welcome. I really enjoy your videos and recently bought your book. Keep up the great work!
    1
  15899.  @Mathologer  yes "Putting two and two together". I'm about halfway through :-)
    1
  15900. I first thought of the title as a reference to the scene in Stand And Deliver (1988) where Escalante explains to his students that “Calculus cannot be made easy because it already is”. Speaking of Stand and Deliver: There is a scene around the 52:00 mark where Escalante shows a shortcut for integration by parts he calls Tic Tac Toe. That sounds like a good topic for a future Mathologer video. :-)
    1
  15901. great video. i would be interesting to know how to get the number of combinations of rubiks cube
    1
  15902. #highlight "Gamma is to the harmonic series what -1/12 is to the sum of all positive integers" :D
    1
  15903. 12:40 Awesome!!!
    1
  15904. As we all very well know from the 𝜋=4 argument, one should exercise great caution when applying limit transformation to the length of a path
    1
  15905. A great treat. Stay safe everyone.
    1
  15906. Being a mathematical ignoramus, I'm not seeing what it could be used for ...
    1
  15907. Could we write 4k-1 instead of 4k+3?
    1
  15908. I think the circle has the larger diameter because the weird blob has a diameter of (red + green + blue), and the circle has a larger diameter than that.
    1
  15909. What property of sphere/circle makes the derivative trick (area = derivative of volume and perimeter = derivative of area) true, since this trick is not true for ellipses for example ? And by the way, why an "onion proof" can't be used to derive a formula for the perimeter of an ellipse, since we know the area ?
    1
  15910. Not to be confused with Heliconia.
    1
  15911. Great video !! (as always) Here's an online arctic circle generator (above size 500, it's getting quite slow): http://micmaths.com/applis/lgn/diamantazteque.html It's coming from the following youtube channel (in french): https://www.youtube.com/watch?v=2Wq6H8GMVm0 At least for this particular video, the english subtitles are pretty accurate.
    1
  15912. I checked the OEIS right after I saw the sequence 😅😅
    1
  15913. 21:22 mentions that no computer can count to 10^447. /me is reminded of Deep Thought 21:46 mentions Life, the Universe and Everything. COINCIDENCE? HARDLY!
    1
  15914. So, well explained, thank you for the choice to keep it simple. I don't mind the fib at all!
    1
  15915. This is ingenious. Thanks for the video.!
    1
  15916. Hey, I am here..
    1
  15917. The thumbnail helps a bit. Each side is extended into a line with the length of the perimeter. And they are all the same distance away from the incenter of the triangle. I believe that this proves all six points are the same distance from the incenter
    1
  15918. Periodicity...
    1
  15919. Another wonderful insight into the fascinating World of Prediction. Love the T-shirts too, have several. They went over big when I was at SpaceX.
    1
  15920. 1+2+3+4=4×3×2×1?
    1
  15921. At 12:30 a tile disappears from the region, no biggie but thought I'd make you aware in case you missed it. Nice video 👍
    1
  15922. For a minute I thought I was watching "El Pastificio de Nicola" and eating hyperbolic empanadas.
    1
  15923. he lived near my home!
    1
  15924. You always have the best shirts
    1
  15925. dear, where do I get your t-shirts, excellent video !!!
    1
  15926. Beautiful math magic - as usual...
    1
  15927. Neat application of math to Justice.
    1
  15928. I appreciate your creative knowledge from INDIA
    1
  15929. Thank you for a wonderful use of my time. I learned something, I heard a mesmerizing voice, I watched science in action. I subscribed to your channel. Thanks again.
    1
  15930. Girl: I change you little by little, and make you fall in love with me. Me: good luck, by the way I am e^x.
    1
  15931. 700 year old proof
    1
  15932. for the first part of the video. I am not familiar with either proof. I am 67 years old, BUT I am pretty sure none of my teachers even hinted at such a visual proof. WOW so nice to see that the ancients could have actually figured this out without knowing any trig function values. I've often wondered how the ancients figured PI to even significant values (3.1416) before having calculus and only having simple geometry. Thanks very much! The demonstration that a person knows his field is that they can dumb it down or explain it to a high school or even middle school person.
    1
  15933. I know you mean a right isosceles triangle with integer sides, but my pedantic self had to leave this comment about it.
    1
  15934. Great video! I really enjoyed it, and finally got the answer I was looking for "What are those bernoulli numbers all about?". I personally thought that Euler got the result for the Basel problem using his multiplication formula for Sin(x). Was I wrong?  I watched it until the end, and have no thought about how to improve this, I would enjoy even if this got more complicated. [MS.c. in Mathematical Physics].
    1
  15935. So 2+2=2x2 is the start of a different infinite identity. Taking the Ackerman extension of arithmetic where exponentiation is repeated multiplication and tetration is repeated exponentiation. Using the notation [1] = + and [2] = x and [3] = ^ and so on we get the infinite identity 2[1]2=2[2]2=2[3]2=2[4]2=2[5]2... and if you use diagonalization you can extend this identity into the ordinals.
    1
  15936.  @Mathologer  Not at all. The pattern for the Ackermann operators is X [n] Y is Y Xes interspersed with [n-1] and [1] is just addition. So 2[4]2=2[3]2 or 2^2 =2[2]2 or 2x2 =2[1]2 =2+2 =4. 2^2^2 would be 2[4]3. I expect the reason that arrow and chain notation focus on 3's is this exact operator fix point meaning that using 1s and 2s are both disappointing.
    1
  15937. Second year Math undergrad here, everything was pretty clear and really interesting. The final Euler-Maclaurin formula was a bit glossed over, but that's understandable in a 50 minute video :) I'm really looking forward for more videos on the topic, this seems like a really powerful (and beautiful!) tool for numerical analysis
    1
  15938. I feel like I could make a case for the next number being 13. Gonna make a coffee and play the vid to find out if I can understand how I'm wrong. edit... I am 1:47 in and already wrong for jumping to conclusions. Was that 'the fast maths' I've been hearing about?... anyway, I'm back to see how else I was wrong... (and the coffee is delicious; made with condensed milk)
    1
  15939. Oresme's proof
    1
  15940. I feel like there is an interesting question which is: why do these things that seem so complicated and geometric like integrals and derivatives actually have formulas? I know that in some meaningful sense most integrals don't have closed forms, but the fact that so many important integrals have a closed form is pretty mind blowing.
    1
  15941. Im pretty sure (ie speculating with my conspiracy goggles on) that the reason that the geometrical proof is not shown is that geometric proofs are used heavily in economic first principles, and tptb dont want an economically literate population that would see through their stupid economic ideas easily. [you may now flame away]
    1
  15942. Here's an explanation of why the function for the number of domino tilings of an n x m board returns 0 if both n and m are odd, under the cut. . . . . . The formula for the number of tilings of an n x m board spits out 0 if both n and m are odd, because it spits out 0 if at least one of the output numbers of the "trig monster" in the parenthesis returns 0. The cosine function returns 0 if it is fed an input of π/2, which is equal to nπ/2n. As such, both of the cosine functions in the "trig monitor" have to be fed xπ/2x to receive a result of 0. In the formula (jπ/m+1) that is passed to the cosine function, if m is odd, then the denominator of the fraction is an even number. We're going to call this number p for the sake of the demonstration. Notably, because the brackets indicate rounding up, j will exist as every value from 1 to p/2, meaning that, at some point, the value (p/2)π/2(p/2) is reached. If we define p/2 to be x, then the cosine function is fed the value xπ/2x, which cancels to π/2. This returns a value of 0. Of course, this only happens when one of the values for the dimension of the board is odd. If one is odd and one is even, then the function will add this pointless value of 0 to another cosine function, squared and multiplied by 4 of course, which received a value not equivalent to π/2- or, for that matter, any value xπ/2 where x is an integer, because the denominator is odd. However, if both M and N are even, then both cosine functions return 0. Both values of 0 are squared to get 0, multiplied by 4 to get 0, and then added to each other to get 0. This one value of 0 will then completely knock out every other non-0 value, through the identity property of multiplication. As such, the function will return 0 for any board of size m x n where m and n are odd, proven in a very long and possibly difficult to parse explanation.
    1
  15943. 数学分为两种，正逻辑数学和反逻辑数学，区别在于空间维度的不同。也就是说存在两种逻辑，在表面上反逻辑数学，在低维空间并不普遍成立，所以从正常逻辑看是错误的。但如果扩展其空间维度，就会展现出一个超越三维空间，违背代数运算法则的高维空间理论。
    1
  15944. Cutting the square corners off the circle always leaves the same amount extra beyond the circle. For it to average to the circumference of the circle, the sides of the newly created squares must be bisected by the circle. Really, it's simply that pi is less than four.
    1
  15945. > 45 minutes of harmonic series... but it is worth being even longer, such an exciting stuff!
    1
  15946. 7:28 made me die laughing
    1
  15947. the old college brain has just exploded from usage ... argh ... ahh .. haha
    1
  15948. my god this is incredible lol
    1
  15949. E-B²=MA² wait what... That shirt is amazing.
    1
  15950. Thank you very much for this very interesting video and your friendly reminder that quadrilaterals are not only convex and only 2d! This was very helpful!👍
    1
  15951. I made it to the end. This video is magnificent. A real tour de force. I really enjoyed it.
    1
  15952. Italian Anagram != Latin Anagram
    1
  15953. I have a bachelor degree in mathematics. I don't mind the long videos, since the topics are great! What about some symmetric polynomials as a follow up?
    1
  15954. Stirling numbers and the partial factorial products was how I learned to sum. The analog formulas with integrals (sums) and derivatives (differences) was very interesting. I became an engineer and somehow over 40 yrs did not come across them again. Thanks for the nice video.
    1
  15955. probably going to get lost in the comments here but.... lim v=1 with the summation of i = 0 to h for V^i = V(V^h-1)/(V-1)+1 covers any and all summations of powers of V for any summation of V Had to derive that for paper regarding 10 as a solitary number for recursion of similar factors of multiples of 5 and 10.
    1
  15956. No matter how many times I look at this, all I see is the projection of the vertices of a tesseract.
    1
  15957. 7:08 if x=0, on the left side we will get natural numbers, on the right sight we will get zeros, and how can they be equal?
    1
  15958. (ho)^^3 😊
    1
  15959. ancient Chinese knew this
    1
  15960. No Nine sum, using the same visualization, a fractal, we can prove so many series as a finite series but can you please explain the approximation at 43:39
    1
  15961. Great Video!
    1
  15962. You can also test for 3 by taking your number times 3 and checking for dr=9
    1
  15963. I've taken math in college, and can't ever remember this formula!
    1
  15964. 25:59 Ghost hands
    1
  15965. many years ago i heard about the IgNobel for this article. Wow , i'm currently a math undergrad, wow, congrats.
    1
  15966. I like the little mesh of curves the nodes of the harmonic series traces out when stacked up vertically on that shirt. 🙂
    1
  15967. I slept through calculus and was looking to relearn it so I'm glad to see this video
    1
  15968. I m pretty much excited please make it
    1
  15969. coloring
    1
  15970. If you have infinity as a tool, you can do anything. You can make A ≠ A, 10 = 0, etc.
    1
  15971. Wait, did I miss you mentioning the golden ratio?
    1
  15972. Sliding the ends of a chord around the outer circle sweeps the iris (twice). The chord's center draws the inner circle (pupil).
    1
  15973. Your shirt is wrong unless HxO= +/- √3
    1
  15974. Swivel proof seemed more elegant to me.
    1
  15975. For the 666th partition I got 11393868451739000294452939.
    1
  15976. Most memorable: the optimal way to set the leaning tower. There's even spaces in between, how do you figure out an answer like that? How many "bricks" would I need to get three spaces in between? Maybe a principle-relationship with maximum distance and the amount of bricks? Crazy stuff
    1
  15977. Lysergic... 😁
    1
  15978. Mind blown in first minute lol
    1
  15979. Terribly difficult to pick a favorite, but I love the accessibility of the Matholegerized Kempner(sp?) proof
    1
  15980. Gamma!!!!!!
    1
  15981. I think there's a somewhat more intuitive way of doing the last argument without invoking the binomial theorem. Unfortunately it's hard to explain the visual intuitiveness of this argument in a comment, but hopefully it makes sense. Suppose we're in the triangle going from the powers of 3 to powers of 4, and pick any node X. The value of X will be a sum over paths ending at X, weighted by the value of the node at the start of the path. Now consider any path starting at some node Y and ending at X. We can shift this path up northeast and get a new path going from the node northeast to Y to the node northeast of X, which I'll call Z. Clearly this establishes a 1-1 equivalence between the paths ending at X and the paths ending at Z. Under this equivalence, since the starting node of a path gets shifted northeast, and we know the nodes on the right side are powers of 3, the weighted value of the path just gets multiplied by 3. Thus the value of Z is just the value of X times 3. Now suppose by induction we've shown that the value of the k-th node on the left edge is 4^k (counting from 0). The value of the (k+1)-th node is the sum of the k-th node and the node to its northeast, which is 4^k + 3×4^(k+1) = 4^(k+1). An analogous argument works for every other triangle.
    1
  15982. i definitely was stunned the most of "100 zeros" > "no nines"
    1
  15983. Wallis product equation works, converges. Wallis product with square root also works, but if we apply substituted Wallis product rule equation of even numbers in nominator and odd numbers in denominator product rule presented in this video essay, equation diverges to infinity :/? It is very unintuitive, you can not substitute equation to more neat expression. I think you can not change priority of math operations first product rule and last operation is square root of product rule.
    1
  15984. Human need the PASS-WORD for Infinity life
    1
  15985. For me, it was Kempner's proof that was the most elegant and memorable, i wasn't expecting the proof of convergence to be that simple and elegant, truly amazing
    1
  15986. 10:50 this coin-rotation thing makes a lot of sense when i consider that the thing that people are actually intuiting is how many times the point-of-contact on the turning coin goes around that coin; intuiting the actual question requires acknowledging that the point of contact is itself rotating in a direction either with or against the rotation of the coin (depending on inside vs outside), and that single rotation itself must be added or subtracted from the "solar" (so-to-speak) frequency (the intuited value) to get the "sidereal" frequency (the requested value) realizing this, however, i have also realized (through the very analogy i just made) that i can use this property to easily calculate the length of the earth's sidereal day at will 😄 it also makes perfect sense considering vector-frequencies are similarly additive and linear
    1
  15987. I would guess that if the input is squares, the output would be numbers raised to the power of themselves.
    1
  15988. This is kind of goofy. For the terms to cancel early in the series you have to 'reach' further down the series to cancel it. So you are 'chasing' the zero infinity is approaching in this case. Which is what the equation was already representing in the first place.
    1
  15989. I’ve had a logarithmic circular slide rule for over 30 years.
    1
  15990. At first I didn't understand the problem. The wording made me think "the sequence must terminate, but what does terminate mean in this context?"
    1
  15991. LIKE #9.3K+ 9:45AM Texas 10/31/2022.
    1
  15992. I wonder if the digital root of the HTZ freq accounts for the amount of peddles seen with each??
    1
  15993. Great video, and actually not as challenging as some of your other videos, e.g. Transcendence of e and pi. Mechanical engineering master. My guess about the problem with divergence at the end: perhaps the yellow sum is not as well behaved as you pretend and actually divergent?
    1
  15994. Dtaught igital roots of 9, and remainders... only at university, and barely 30 seconds spent on it
    1
  15995. Número perfeito
    1
  15996. I'm seeing a connection between those partial fractions and your video on the Japanese marching squares. Commenting before I finish this video so I can feel smart if you bring it up later. :-P
    1
  15997. A whole course in one video 😂🤣😂🤣.
    1
  15998. 9:03 My face when the beginning of a proof seems simple, but I know my mind is about to be blown.
    1
  15999. "Why is the formula for the surface area the derivative of the volume formula? Easy: V'(r) = dV/dr = A(r) dr / dr = A(r)." I'm not so sure it's quite this simple, as it doesn't hold for all shapes/formulas. For instance, the standard formulas for the volume and surface area of a cube of side length s (s^3 and 6s^2 respectively). I remember vaguely looking into this a while back. I think this was able to be rectified for cubes (or regular polyhedra in general?) when, instead of phrasing formulas in terms of a side length, it was done in terms of ... the inradius, I think? For a cube, that radius is r = s/2, giving volume (2r)^3 = 8r^3 and surface area 6(2r)^2 = 24r^2, so it holds. Dunno if it holds for other shapes. Been a while since I thought about this and I never wrote down my notes. And I'm not entirely sure why it had to be the inradius.
    1
  16000. If anyone's interested and doesn't already know, Ron Graham appeared in a couple of Numberphile videos several years ago talking about his titular number.
    1
  16001. The Fibonacci sequence 9, 4, 13, 17 gives the Pythagorean triple 153² + 104² = 185² 🙂
    1
  16002. if you cut out 4, you can isolate a corner by removing the two neigbors then you can trivially see it can't be tiled.
    1
  16003. The terrible aim did surprised me a lot and it's really interesting. Left me thinking 🤔
    1
  16004. it isn't indian, it's hindu
    1
  16005. 12:30 here a green square disappears and then suddenly the area can't be tiled anymore. Tiny mistake but it stumped me for a while what had happened!
    1
  16006. Babbage Difference Engine!
    1
  16007. are the glasses 540??
    1
  16008. What A amazing masterpiece :)
    1
  16009. Why is it written “4K+3” and not 4K-1?
    1
  16010. At @12:28 the board loses a square.
    1
  16011. It's interesting; I've never heard the "shine" pronunciation for sinh(x). I've always heard it as "cinch".
    1
  16012. Dear Mathologer, this is - so to say - "of road" regarding this video, but anyway I would like to ask you if you could, perhaps, make a video-clip in which you would prove that any homogeneous fluid body (e.g. a drop of pure water (or of mercury, or ...) in a space station, or when the drop is "in the free fall" in a gravitational field) would inevitably take the form/shape of the sphere. When I think about it, I come to that conclusion, but I cannot find the way to prove that explicitly, that is, mathematically. Or, perhaps, you could give some link(s) (if you happen to know about such link(s)) to some video(s), or site(s), where such proof is given. Also, I think (or better said: I hope) that proving the following would be interesting for you and the viewers/followers of your channel: 1) If we assume that for any change to occur/happen, it must take some time for that (in other words: there are no changes which can happen during "no time"), 2) and if during an infinitesimal time dt only an infinitesimal change of something/anything (dy) may occur, 3) and if during a finite amount of time delta_t only a finite change delta_y may occur (to me, this is the direct/inevitable consequence of 2), but ... maybe that should/could be also proven mathematically), then (y, t) is the smooth continuum (free of singularities). In such continuum, an "infinitesimally small shape/"body"" cannot exist (in other words: the smallest shape/distribution of y in t must be some finite-size shape/distribution. And the simplest possible localized (finite size) shape/distribution (of y in t) may only and exclusively be a shape/distibution which: smoothly continually begins to rise (starting from "zero"-value (y=0)), smoothly continually rises smoothly continually reaches the maximum smoothly continally begins to fall (to decrease) smoothly continually reaches the 0-value. I wish you, and to your team, and to your fans, a happy and prosperous New Year!
    1
  16013. "What comes next?" on this channel is perfect for the second season of Squid Game.
    1
  16014. Hi! Brilliant video. In middle school I was taught the divisibility criteria (2, 3, 5, 9, 11) but they were never proved to me and I hadn't the knowledge to prove them myself. Some years ago I randomly meet the Tesla's vortex: and at the beginning it was interesting because it shows some interesting things, but when it started with the "esoteric" stuff my skeptical sense wakes up. I tried to prove something was wrong and I started proving all divisibility criteria i know and then showing 9 divisibility criteria was special because we use the decimal number system. This video not only reaches the same conclusion I do, but it goes further (diagrams with different modulus and multipliers) and I love it.
    1
  16015. Why am I so interested in math rn
    1
  16016. I really like the visual representation of the harmonic sum with balancing and the Euler's constant because I am learning about infinite products and the gamma and zeta functions.
    1
  16017. Thanks for this. Always love your videos on a lazy Sunday afternoon
    1
  16018. One of your best videos, thanks!
    1
  16019. Prefere dot profe!
    1
  16020. I wonder if I can patent something using this to optimise reduction modulo 1 in modern CPUs.
    1
  16021. I took all the maths I could in secondary school, but have no higher degree. I could follow everything except for the matrix stuff, as our curriculum didn’t really cover matrices.
    1
  16022. Here's a question I found interesting from my complex analysis course that uses the pigeonhole principle for an uncountable set not having an injection into a countable set. f:C-->C is an entire function and for each z_0 in C, the power series expansion sum from n=0 to infinity of c_N * (z-z_0)^n has at least one c_n = 0. Prove that f is a polynomial Another interesting application: for m>n, if you drill m holes into n pigeons, one pigeon will have at least 2 holes in it
    1
  16023. Nice pi Xmas tree shirt!
    1
  16024. UK college student doing maths + Further Maths (naturally) ...got to the end, the last 15 minutes or so required backtracking but the video was really interesting and it felt like I understood most of it (but I will need to rewatch parts to remember some of the smaller details). Challenging towards the end but not unbearably. Looking forward to your next video!
    1
  16025. We learned Heron's formula in Denmark in the 70's. I do not remember if it was in realen (years 8 and 9 of primary school) or gymnasiet (further 3 years for those heading to university).
    1
  16026. The only thing i dont like on mathologer are the videos are so long. We are living in quick age so shorter videos will come in handy. No offense.
    1
  16027. Loved part 5, personaly. I am partial to area.
    1
  16028. I noticed something similar at around the same time while I was playing with a programmable calculator. I stopped at powers of 10 terms. I then found that applying Aitken's delta squared process corrected the incorrect digit and doubled the precision.
    1
  16029. Amazing!
    1
  16030. Fits in the margins. Nice work
    1
  16031. A beautiful visualization of a proof by induction.
    1
  16032. Nein, Nein, Nein!
    1
  16033. Another possible solution Let n be an integer, n > 0 Compute ( 1 + sqrt ( 2 ) ) ^ n = a + b * sqrt ( 2 ) where a and b are integers Let x = a * b and y = min ( a ^ 2 , 2 * b ^ 2 ) Then, y is the total number of houses and x is the house number. By the way, if n is odd then y = a ^ 2 and if n is even then y = 2 * b ^ 2
    1
  16034. That kind of conveyor is called a crescent pallet conveyor, and sometimes a "sushi conveyor" because they were originally designed for carrying sushi plates. They are limited in that the whole run has to be flat and level due to the rigid nature of the segments, and can also be hazardous to touch when running due to the numerous pinch points.
    1
  16035. Great video! Not a fan of the 8 lie though; but, I am a fan of turning 8 on its side to equal infinity!
    1
  16036. The book "Calculus made Easy" is a good introduction. What do you think of the book "Elementary Calculus. An Infinitesimal Approach" by H. Jerome Keisler?
    1
  16037. About removing 2 green and 2 black squares from the chess board: this is not always tileable. Counterexample: remove a2, a3, b1 and c1.
    1
  16038. Love the video!!!
    1
  16039. The circle that touches all 3 vertices of the 3-4-5 (etc.) triangle was omitted (possibly irrelevant?)
    1
  16040. As the Fibonacci series is hidden in Pascal’s triangle then so must all Pythagorean triples be as well
    1
  16041. Lovely ... but. Wouldn't a grown-up mathematician just write the vertices of the inscribed quaterlateral as complex exponentials and then do the algebra.
    1
  16042. My most memorable part was how much the ending music reminded me of Löwenzahn for some reason, which made me realize what a tremendous Peter Lustig replacement you'd be! (Replacement sounds bad. More like continuation.) Sorry for this not being related to the really neat maths shown here!
    1
  16043. Beautiful 😍
    1
  16044. That pizza-nacci shirt tho
    1
  16045. My vote goes to the fractal image of the no 9 series. <3
    1
  16046. 27:15 I laughed here.
    1
  16047. The sum of 7 angles: each angle is 1/2 the angle from the center to the same arc of circle (considering the circumscribed circle here). The sum of these central angles is obviously 360°. Therefore the sum of the 7 outer angles is 180°. (Edit: But, of course, I'm not the first one to propose that proof here.)
    1
  16048. This explanation video amazed me in a hopefully sustainung manner, mathologer provides the interested with a virtual "liquid" wall of sense (a variety of a nineties plasma screen? no?). Most equations aren't graspable for me, but clearly doubtless though. I contribute some cores and graphic card processing capacity (mostly at nighttime) because a friend of mine is able to build computers, but i hardly know what the results for einstein@home and asteroids@home (through the boinc client manager) would bring for the guys closer to the crazies assembly hardware bonds, pins, nods and whatever. What would the best way to contribute computer capacity might be? i doubt the incompatible-ness of some graphic card types, why won't they simply calculate instead of seemingly being obfuscated by some nvidia stranger? Many so-called incompatibility issues seem to be artificial like "man-made", so that some firm can rob others, not deeply sporty.
    1
  16049. I'd like to add on to a point that there in your previous videos that , there were 2 harmonic series S1, S2 such that ΣS1 - ΣS2 = pi
    1
  16050. My school never taught me the proof (waste of money indeed) but I came up with it myself (or I saw a clue somewhere, I forget exactly)
    1
  16051. 26:18 ...Just be sure to give your campaign staffers patronage jobs so you don't wind up like Garfield.
    1
  16052. 2^10 - 1
    1
  16053. I can hear this music and watch those videos forever
    1
  16054. I’ve seen that cartoon face before, is it a character?
    1
  16055. But if the wheel or the band only went up to 9...
    1
  16056. I thought for sure Euler Totient function would show up at some point but i was wrong D:
    1
  16057. Loved the connection w/ Fibonacci-type sequences. Awesome!
    1
  16058. Wait... Maybe the surface is infinite but the amount of paint needed will get smaller and smaller=> Zeno's paradox? Is the problem related to your Fourier T-shirt btw?
    1
  16059.  @Mathologer  aha! 😊
    1
  16060. My favorite part of the video was the proof that the Harmonic Series diverges. We have it is > 1/2 + 1/2 + 1/2 ... good stuff
    1
  16061. Mathologer Sir, Could I request you to discuss Philosophy of Mathematics itself? Maybe the crisis of mathematical foundation by Bertrand Russell, etc?
    1
  16062. 16^2 + 42^2
    1
  16063. The most memorable was the bizarre overhanging towers.
    1
  16064. What's really nice here, that maybe wasn't stressed very hard in the video, is how the geometrical process of adding a dimension (points become segments, segments become squares, squares become cells) is modeled perfectly by the algebraic process of multiplying by x (thus increasing the exponent). This is a great example of how polynomials are their own kind of object, beyond just a functional relationship between numbers. Once we see the algebraic consequence of the geometrical process, it means we can manipulate algebra symbols and expect that to tell us about geometry. The fact that mathematicians do this is not obvious and deserves to pointed out explicitly.
    1
  16065. Jokes on you, toroidal diamonds.
    1
  16066. I am from Brazil and your videos are absolutely amazing
    1
  16067. Isn't the decimal system a crutch? How would that look in a duodecimal system?
    1
  16068. Fascinating video, loved the animations!
    1
  16069. I saw many of these paradoxes in high school. At first I was amazed, then I realized that, indeed, math is an imperfect tool, not a discovery. ... but I do love watching them! Thank you!
    1
  16070. Don't really know that if this is a right place to post this but I know that I can find here people with interest on maths and animation. There is a guy on YouTube Theonethreeseven who has created a Python project named maths inspector for creating and producing maths animation and stuff so I thought I should share it here as well if it would do any good for the maths community Theonethreeseven channel: https://www.youtube.com/user/TheOneThreeSeven Math inspector open source release: https://youtu.be/rN6qRv09M70
    1
  16071. We do hear of him on school, they tell us that he was an idiot who believed the Earth was the center of the universe, when clearly it's not. I don't believe that and clearly I know about some of his contributions because I love math and history, but it's a shame.
    1
  16072. Ikr. He had to start somewhere, didn't he? Being wrong is still preferable to being a barbarian.😂
    1
  16073.  @Mathologer  I understand what your statement from a mathematical perspective but the fact remains that the sun is the center of OUR solar system. Try and exist without it!
    1
  16074. ​ @Mathologer  I understand your statement from a mathematical perspective. The fact remains that the sun is the centre of OUR solar system. Try and live without it!
    1
  16075. I love how you defending if the key's of universe are compleks, we are to much underestimating the universe if believe vorteks is the key of universe
    1
  16076. Kempners proof was my favorite since he pretty much thought of the sums as a square and probably pondered if reducing the square would effect the sum. And turns out it did!
    1
  16077. Non-mathematician here. Just six minutes into the video and I wanted to ask whay the 4K+3 primes aren't written as 4k-1? Or does it not work further down the list? If it does then "4k+1 does work and 4k-1 doesn't work" seems more elegant to me.
    1
  16078. Saw the black background in thumbnail and though this is the best christmas present I ever had. Pressed the play button with a big grin on my face only to get my retinas burned out within 5 seconds. THANKS MATHOLOGER for ruining my hopes, dreams and christmas.
    1
  16079.  @Mathologer  Love you guys regardless :)
    1
  16080. the most memorable part is the proof of gamma value
    1
  16081. Adoro este canal!
    1
  16082. Kempner's proof is very interesting, but for me the fact that H_n ~ ln(n)+gamma will forever remain my personnal favorite fact
    1
  16083. I had the solution to your pearl necklace puzzle!
    1
  16084. he is basically doing an impression of peter sellers doing a german impression in dr strangelove.....omg i cant concentrate his accent so good
    1
  16085. I'm wondering what a 3D abacus [ or soroban ] might look like, and if it would be related to the Pascal triangle.
    1
  16086. Most memorable: the terrible aim of the harmonic series
    1
  16087. "1 is an honorary prime." Love it.
    1
  16088. I was hoping you'd discuss what happens with the version of the construction that gives a star polygon when you start with a number of sides that isn't prime, since the eigenpolygons here appear to be related to the polygons with Schläfli symbols {N/k} for each k (I'm not sure what you'd represent the degenerate case as though), and for example 10/2 is not irreducible, so the {10/2} polygon is actually two disconnected pentagons. Probably this is just an incorrect generalization I've drawn, but I'd still like to see what actually happens in that situation.
    1
  16089. So.... what would be the difference scheme of primes? Will we predict the next prime using this method ¿
    1
  16090. Thanks for yet another enjoyable math insight.
    1
  16091. Simply a fantastic explanation, as usual
    1
  16092. As a visual teaching aid, I made a simple folding example. You can make a "proof" (or instance of the theorem working) that folds the two smaller squares into the larger single square. It could be made into a jewel with a wire frame too...
    1
  16093. Nice video, and song at the end. Most memorable, that cute nerdy laugh.
    1
  16094. We were given Heron's formula (only the formula) in a little infobox in high school, never used it -- nothing about Brahmagupta. Everything about this video was magical to me. Edit: India
    1
  16095. Great video.
    1
  16096. Loved this video ❣️
    1
  16097. So I just started studying maths this year, and I could follow pretty well until chapter 7, starting then, you lost me more and more, but it was still very interesting to watch. I think it was because you glossed over more and more details, which is of course understandable; this is youtube, not a master's course maths.
    1
  16098. 4sqr + 2sqr= 20 or 5x4=20 but 21 is not 4k+1?
    1
  16099. I'm 43 and this is the first time I see it in all its glory 😃
    1
  16100. fermat theorem intrigued me arround 1980. at high school i drew this on 1991 https://youtu.be/DjI1NICfjOk?t=1478 and thought that the solution must be in there. that the numbers (x,y) must linear or proportion to the enlargement of those smaller rectangles
    1
  16101. Is there any generalisation in d dimensions, with coordinates being in R*SOd or something like that, and pyramid ears instead of triangles…? That seems interesting enough… my god do I wanna check this out !
    1
  16102. i use the E6B on my watches (im a fan of pilot chronos) all the time. super handy for all kinds of things... splitting checks, calculating tips, unit conversions, etc... its a shame people dont know how to use slide rules anymore, as a good slide rule with predefined scales for different common operations is nearly always faster than using a calculator.
    1
  16103. Mathematical gems! 💎
    1
  16104. the music is a bit off putting as if we're uncovering the secrets of US constitution or something, otherwise amazing visualization
    1
  16105. @mathologer love your videos, but please check your formulas for (x+1)^0 and (x-1)^0 around 16:30
    1
  16106. Brilliant!
    1
  16107. Amazing
    1
  16108. I watched the whole video. It is really great. It gives you a lot of ideas about what to learn. I am mechanical engineer my maths is basic calculus (derivatives and integrals). Thanks.
    1
  16109. Most memorable: Chapter 4: 15:33 Terrible Aim: No Integer Partial Sums! I've never seen the implications of the parity of numerator vs denominator! Brilliant content as always, Mathologer! :)
    1
  16110. How much time do you take to make all these animations? And how do you synchronize, without errors, your talk and the animations? Clicking too fast, or too slowly. Do you have to restart many times?
    1
  16111. Yes, to the very end!
    1
  16112. Final Puzzle: Let F(n) denote the Fibonacci numbers with F(0)=F(1)=1 and S(n) denote the sequence in question. Answer: S(n)=F(2n-1). Reason: I checked S(5)=55 and S(6)=144 so it must be true. Just kidding. Proof. The (ordered) partitions with last number k contribute kS(n-k) to the sum, so we have S(n) = Σ[k=1 to n] kS(n-k) = n + Σ[k=1 to n-1] kF(2n-2k-1) (here S(0)=1) = 1 + Σ[k=1 to n-1] 1+F(1)+F(3)+...+F(2k-1) = 1+ Σ[k=1 to n-1] F(2k) = F(2n-1). Example n=6 case: S(6) = 55 + 2*21 + 3*8 + 4*3 + 5*1 + 6 = (55+21+8+3+1+1) + (21+8+3+1+1) + (8+3+1+1) + (3+1+1) + (1+1) + 1 = 89 + 34 + 13 + 5 + 2 + 1 = 144. p.s. My mind just exploded during this video. Also, I made it to the very end.
    1
  16113. I really like that the partial sum ps(n) is between ln(n+1)+gamma and ln(n+1) I also like the formula at 25:05 you could make a video on that result.
    1
  16114. Aaaand! The first pyramid of sums always starts with a square number: 1, 4, 9, 16 ...
    1
  16115. I saw it more as each new row starts with one greater than the previous row’s right most number. But I can’t explain the next tree with the squares to generate new rows.
    1
  16116. brilliant, well done again, does not disappoint
    1
  16117. 2 dream proofs that actually worked? What, you're op!
    1
  16118. Very beautiful !!
    1
  16119. it's like i watched the whole video to learn the final few minutes where's the magic hidden ;D
    1
  16120. So you need to come to the United States to teach a few courses.
    1
  16121. Let's do infinity arithmetic. What happens when you get to the end of the Universe? Fall off? Dissolve? Can anybody do 500 million calculations by hand without making a mistake? I usually made an arithmetic error when solving Calculus problems in college during tests. But I did get an "A". Never used Calculus again. 30 years a programmer.
    1
  16122. Calculus itself is easy. It's all the nonsensical algebra around it that complicates things
    1
  16123. I'm so sorry of knowing this brilliant method only now. Of course, as an italian, I have always known the Ruffini's method on which is based wich is thaught to all italiam pupils, but the geometric rapresentation has something more I'd like to discuss about. The improvements of Burkard Polster are very important: QED and its reference to Pascal Triangle. But it is possible to do a little bit more in that direction. In my opinion there's a long way to go.
    1
  16124. For me, the most surprising thing was that the "no 9's" series converges and what truly makes it the best thing was that amazing animated proof. With just one short animation I went from "What? How could this ever converge?" to "Oh, sure that converges". Brilliant!
    1
  16125. Berlusconi is a (bad) clone of Madhava of Saṅgamagrāma! 4:50 😀
    1
  16126. The most memorable was the Euler-Mascheroni gamma 🤩
    1
  16127. It shows unity in an excellent way to construct things. :O)
    1
  16128. Total[#^10 & /@ Range[1000]] (wolfram language)
    1
  16129. Listen up my niageu, to work
    1
  16130. wow I just want more content like this, exactly. probably little more graphic, though not as important at all
    1
  16131. Interesting that some of the shrinks spiral in some way.
    1
  16132. Gabriel's horn
    1
  16133. There was no "windmill interlude" planned originally, was there? One can see the 6 turn to 5 right at the end of the animation of the 6th chapter's opening. But I'm glad you included that interlude! I think this is one of your best videos so far.
    1
  16134. Most surprising: the fractal pattern generated by the no 9s series
    1
  16135. Sweet!
    1
  16136. Solution to the Rubiks cube puzzle: The RULD algorithm divides the edges into 2 cycles: one of order 5 and the other of order 7, and divides the corners into a cycle of order 9 and four cycles of order 3. Therefore, the cube will return to the initial state after MCD (5, 7, 9, 3) = 315 repetitions of the algorithm
    1
  16137. Living in Hellas and no, haven't been taught it in general education.
    1
  16138. All these facts were new to me and the proofs were amazingly clear.
    1
  16139. But why does the magic happen at 216° rather than 288°? EDIT: I watched the rest of the video, and I think I understand now.
    1
  16140. The magic also happens at 288°: it always gives you the vertex center of mass. However, stopping one round earlier gets you one of the special n-gons and for most people that is more magic than the single center.
    1
  16141. That's it, magic is important :)
    1
  16142. I have to admit that I really could not follow the animated proofs at the end...
    1
  16143. I see. So this is why we learn series, limits and approximations - and patterns - so early in math in India. I would love a video-lecture from Mathologer on Golden Ratio and Geometry in nature. As of now, I'm irritated and convinced that we would have made tremendous progress in transmitting 3-d images if 2-d planar geometry had not been invented. So much time is spent on projections on to a 2-d plane when reality is 3-d and beyond. Nature is 100% 3-d
    1
  16144.  @Mathologer  :)
    1
  16145. The hexagon tiling question, it's obvious because the tiling can be interpreted as a stack of blocks. A stack of blocs occupying the corner of a cube seen in isometric perspective. Viewed from the top you will see the orange tops of the blocks which in projection will fill that square side of that cube exactly and for a hexagon of side n there will be n^2 little orange squares.
    1
  16146. . . . the point is ?!?
    1
  16147. 大晚上的，过来点个赞
    1
  16148. The Bishop gets the early running : )))) Mind Tristan's "graph" paper is close - Bishop's proof, final vote Excellent stuff..
    1
  16149. The most memorable is the Tristan's fractal
    1
  16150. That was amazing!
    1
  16151. So what did you do today? Outramanujaning the Ramanujan with some virtual friends.
    1
  16152. 13:50 In case of 1 disc to move to the B point we need to move clockwise, 2 discs - counterclockwise. Simple logic tells that it's (-1)^n "*" counterclockwise direction :D
    1
  16153. The swivel proof did it for me. Simple and elegant. Plus I have a bit of a spiral fixation.😂
    1
  16154. Why does your t-shirt have the right angle on the left? ;-) Thanks for a very interesting video: as well as being interesting mathematically, as all of your videos, this one also has 'practical' applications that make it particularly interesting - at least for me.
    1
  16155. it looks like anti-entropy. chaos to order in closed shape. can you use this in chaotic systems?
    1
  16156. One of your PayPal’s is Markus Persson. How amazing is that!
    1
  16157. Very nice vid!
    1
  16158. At any point, the target-bug is moving at a 90° angle to the chaser-bug. So it's not moving towards or away from it and can't affect the total distance travelled. Not sure how to put it in more mathematical terms.
    1
  16159. The way I do fractional square roots is to make the first approximation, and then use the 2Xdx rule to make the next iteration. So, 3 / 2 you multiply the top and bottom by 2 times the top then subtract 1 from the top. Then you get 17 / 12. The next one multiplies top and bottom by 34 so you get 577 / 408. The thing you can say is that when you square this fraction, it is 1 over the denominator squared larger (the error). You should also be able to make some boast about accuracy with your continued fractions, maybe it is within the ration of one over the denominator or denominator squared.
    1
  16160. I had never heard that trick for numerical integration by computing the polynomial. It's so so cool. Thanks a lot sir.
    1
  16161.  @Mathologer  prooving this requires algebra autopilot stronger than you can ever imagen sadly.
    1
  16162. I love these interesting aspects of geometry. Also, Numberphile has a new -1/12 video. I would really like your thoughts on it. You always provide so much more insight into something than the other math channels do. 👍🏻👍🏻
    1
  16163. Well, let's just say I'm a math genius who grossly under achieved. I am now in the twilight of my secondary school math teaching career. I have watched every one of your Mathologer videos and also binge on 3Blue1Brown videos. Both of you should go down in history along with the likes of Gauss, Bernoulli, Euler and company, if not for new mathematical discoveries, rather for the elegant presentation online of very challenging mathematical concepts. I love what you do, always eagerly await more, and encourage you to never stop. Hats off to you and Marty!
    1
  16164.  @Mathologer  Oh I see, there's the 4-4-8 split. That gives me, um, ohhhh. Very clever.
    1
  16165. I didn't take calculus. But I did notice that in Wednesday's Power Ball drawing, the numbers 2 and 17 came up. Therefore, I expected them to come up the next night in the Illinois Lotto drawing. They did.
    1
  16166. I really appreciate that you scootched out of the way to avoid getting bopped in the head by the green arrow.
    1
  16167. Here's my submission: https://youtu.be/dI3rIzs-tQw I'm submitting it as a video again, because youtube doesn't like external links, but there's a link to the app in the video description. Interestingly, the linked lines don't actually form loops in some cases (I think it's just when the multiplier and the modulus have common factors). This makes implementing the loop highlighting a bit more complicated, because you can't just walk around the loop and expect to hit every point. Instead you have to implement a graph traversal algorithm, so it's a fun little problem.
    1
  16168. 1. You put the bricks like this. -_- . Think of small bricks on top as equal to length of the bottom one and touching in the middle. This arrangement is symmetrical around the center. The top right brick is completely outside. 2. The blue area is >1/2 simply because the sum of areas of triangles formed when you connect (n,1/n) , (n+1,1/n) , and (n+1,1/(n+1)), which is \sum_{n=1}^{\infty}1/(2n (n+1)) =1/2. (Note that the sum is trivial if you write the partial fractions, each term higher than 1/2 will just cancel).
    1
  16169. About naming the proof after Pythagoras when it was known before: Stigler's law of eponymy strikes again https://en.wikipedia.org/wiki/Stigler's_law_of_eponymy ("no scientific discovery is named after its original discoverer").
    1
  16170. I think the gamma less than one, so simple and so beautyfull
    1
  16171. I wonder if asymptotically circular shape of the arctic circle isn't artifact of the method how the tiling is generated. The same patterns could be generated multiple different ways and that may be affecting the result, perhaps? My line of thought is, if we imagine a random pattern that fills the shape, then all four walls are a staircase that can have only up to one "notch" - a place where one stair is occupied by vertical tile and neighbor stair occupied by horizontal tile. From that point on, all tiles on the staircase can be only horizontal/vertical all the way to the corner. That means, on average this staircase will be divided to two by this notch and that will be on average somewhere in the middle. That splits the staircase to two for which the same applies again. So first layer has notch at the half, second layer has it at quarter from either side, third layer has it at eighth and so on. From this, I would say the "arctic circle" should be more of hyperbolic shape rather than circular.
    1
  16172. wtf is this, not even 17
    1
  16173. My compass has an clinometer. (A device which allows you to determine the angle of elevation of an object.) I added a circular slide rule with tangent scale to the back of my compass. This enables me to find the height of an object. The tangent of an angle is equal to its opposite side divided by its adjacent side. Walking away a number of paces from an object gives me the adjacent side. The clinometer gives me the angle of elevation. I then use the slide rule to find the tangent of the angle and to multiply this by the number of paces walked off. This gives me the height of the object (in paces).
    1
  16174. 🎉🎉🎉
    1
  16175. Yellow
    1
  16176. Final puzzle: area is 1/5 Yes, the answer was surprising. Having first solved it a longer way, it now seems almost obvious: swing the four small triangles onto the trapezia to complete four squares; use a little basic geometry to find that each is equal in area to the square in the middle; five equal squares whose area sum is 1.
    1
  16177. A real brain twister on Christmas Day. Need to revisit when I’m sober again...
    1
  16178. I was keeping an eye out for this myself, he talked a couple of times about appearances in nature such as the sunflower, but only once did he mention art that I caught. That one time was in the introduction as a more or less throwaway comment that it appears in art. No exaggeration as to frequency, just that it appears. There are instances where it is true, so I was personally relieved to see that he didn't really jump on that bandwagon.
    1
  16179. Your videos are always try top quality. Keep up the good work! :D
    1
  16180. I vote for Tristan's fractal as my fave most memorable part.
    1
  16181. Thank you for your video!
    1
  16182.  @Mathologer  oh u replied to my cmnt woowowowowowo
    1
  16183. Ooooh this video scratched a REAL bad mathematical itch I had. I think I might just understand calculus in general a lot better now Also, I'm gonna be rewatching this one, hope you don't mind!
    1
  16184. No matter how many times I view this, it's still unbelievable how unbounded overhang can be created with tiles using harmonic growth.
    1
  16185. No. You can easily isolate a corner by removing two squares of the same color.
    1
  16186. recently seen a video on balancing numbers: https://youtu.be/jMfZ9jRsHSI
    1
  16187. Most memorable are the longest overhangs. I see a competition there.
    1
  16188. 🕊️
    1
  16189. 1+1+1+1+1+1+1+1+2+10=1*1*1*1*1*1*1*1*2*10 didn't watch the video so the question may be wrong
    1
  16190.  @Mathologer  the ones are a cheat . but then again, also the 2.
    1
  16191. sqrt(3)+sqrt(3)+sqrt(3)=sqrt(3)*sqrt(3)*sqrt(3)
    1
  16192.  @Mathologer  I tried to generalize the equation, kx = x^k and the result looked familiar so I question the limit of (1/x)^(1/(1 - x)) as x -> 1, and yes, the number e appeared. I don't know if it has any meaningful interpretation in the context of this video. I couldn't find the definition of e in such way but it was presented as a "find a limit" exercise in another yt video.
    1
  16193. Always Mathemagical
    1
  16194. one of the best math videos I've seen
    1
  16195. 25.8069 is the root of evil. It took me the first 20 seconds to understand it. Not bad, ah ?
    1
  16196. 11:53; have one brick on the edge, and two bricks on top of it touching eachother at the cliff edge?
    1
  16197. H
    1
  16198. Where did you get that shirt? I want one for when I teach standing waves!
    1
  16199.  @Mathologer  Bummer. Treasure it! It's a good one. =)
    1
  16200. I use symmetrical components to analyze 3-phase power (triangles), but never thought to apply them to other polygons.
    1
  16201. In Chapter 1, how do you use the Cross Rule when the left hand square is three dots? Something is missing, isn’t it? (@ 3:09)
    1
  16202. Yes this was exactly the rabbit hole I was expecting from you! I also liked how one of the problems you assigned us toys with Egyptian fractions. One cool thing I like about the divergence of the sum of reciprocals of primes is that it has its own approximation formula and constant analogous to the harmonic formula: Sum(1/n) ~ ln(N+1) + gamma, gamma = 0.577... Sum(1/p) ~ ln(ln(N+1)) + M, M = 0.261... My favorite part though was your exploration of Kempner's series. The zoom out fractal intuition for the convergence was ingenious (and also created the black dot illusion, accidentally, as a consequence :D )
    1
  16203. That’s all well and good but where can I buy a tee shirt like that? 😂 (no, seriously)
    1
  16204. There are triangles of three sizes in the diagram at 36:24, and they all share the same smallest angle and all have a right-angle, so they all have the same proportions. The middle-length side of the middle sized triangle is therefore 2 and the area of the square is therefore 1.
    1
  16205. Very nice! It looks like the curta machine!
    1
  16206. “...there...”, he whispers with satisfaction
    1
  16207. 2 times infinite plus greater than.... i that a joke im too new in this channel to understand?
    1
  16208. About the challenge at 32:20.. 16 = 5^2 - 3^2 and it's even
    1
  16209.  @Mathologer  your friend andrew is a genius
    1
  16210. The sum of the reciprocal of the cubes is called Apery’s constant, but not much is known about it except that the sum does indeed converge.
    1
  16211.  @Mathologer  I thought about inscribed and circumscribed polygons 50 years ago. Back when the computers and plotters were slow. So I didn't include the circles. However, I didn't line up the vertices to give vertical symmetry. Some kind of spiraling. And some kind of question whether they converge to an inner or outer circle.
    1
  16212. 49:54 Only after calculating the differences I could recognize the Fibonacci sequence, so I know what is next... but I don't know why!
    1
  16213. Damnit, I worked it out using sigma notation and only found the trivial solution.
    1
  16214. Liking part 6, it gets me to watch the Riemann's Zeta function episode
    1
  16215. You tell very important topics but I can't help sleeping while listening you. Your voice has ASMR effect.
    1
  16216. I was the last to carry a slide rule, Pickett of course, at university. My freshman year was a year after the HP35 came out. The calculator was too compelling (reverse polish!) so when the price came down to $295 I bought one. Now, the same power can be purchased at the dollar store.
    1
  16217. math always like a magic I love her.
    1
  16218. Hey i enjoyed this substantially more than i expected i would, and i didn't exactly expect to hate it to begin with. Especially about normal aware refinement; this reminds me that in the likely event that i have to fit or write a decimator, that it should also be normal aware and balance the normal distortion against other sources of visual loss. This is not something mentioned or considered in the classic edge collapse papers. It also explains why Catmull Clark looks like shit on triangle topologies. There's of course Stam&Loop the old and the new and they should be interesting to look at from this perspective.
    1
  16219. Ha the flat earthers are going to love you 😂
    1
  16220. Why don't you ever talk about of what we know nowadays about unsolved conjectures, as Collatz or Goldbach?
    1
  16221. I see his Schwartz is as big as mine...
    1
  16222. Wait... we're going to find the inverse of an arbitrarily large matrix, and that's the easy part?
    1
  16223. I feel as if I already knew this years before the proof was published. I'm a professional game developer and have spent countless hours optimizing triangle counts to increase framerate and performance. I definitely saw these relationships but didn't give it a second thought. I was a few steps ahead watching this video and wasn't surprised by the conclusion. I was surprised that this wasn't put into a proper proof for 5k years and learned a lot about the process of proofing from your walkthrough. Now I wonder if there are other bits of unproofed knowledge lurking in my subconscious lol. Thx!!
    1
  16224. So cool! I've never understood the usefulness of Bernoulli numbers when I tried to learn about them on my own. I have an undergraduate degree in mathematics and pretty much everything in this video worked well for me, but I'll probably have to watch it again to really bring everything together and understand it. I have less familiarity with linear algebra though, so I would have been interested to see how some of those coefficients and Bernoulli numbers can be derived (i.e. a little grittier on the matrix inversion/multiplication). Overall it was fantastically done!
    1
  16225. the music eventually suited the power of the argument 👍
    1
  16226. "Epiphany" is the English word for "an aha moment"
    1
  16227. Awesome, as usual. I have been missing you here. Welcome back! Not quite as much fun as in the Maths dept at The University of Adelaide, but VERY informative. I thought I enjoyed Mathematics in the past. Your videos open up an infinite assortment of extra dimensions of mathematical wonder. The first class video presentations and detailed explanations are unrivalled. Thank you Burkard. Best Wishes, David
    1
  16228. This is so cool!
    1
  16229. Thanks for the video! I implemented the "magic square dance" in JavaScript, send you a link via email. If you like it and want to add to video description - I would really appreciate it.
    1
  16230. cool
    1
  16231. I. Love. You.
    1
  16232. If one extends this 3:41 to a tiling of the plane, then draws the medians for each of the triangles, the triangle cuts can be just joined to a cut from a neighboring triangle in the same manner to arrive at a new tiling. Then each side of the original tilings (belonging to two triangles) results in two new triangles so we also get this 1:3 ratio in area.
    1
  16233. When I saw the cube corresponding to (x+2)^n, my first instinct was to see if there was a formula for other hyper polyhedra. I came up with a formula for the simplex (triangle, tetrahedron, 5-cell, etc...): ((x+1)^n-1)/x (It's just the nth row of Pascal's triangle without the last 1). I am, however, struggling to come up with a similar formula for the hyper-dodecahedra (although I know they don't actually exist in any dimension higher than 4, the corresponding graphs should still exist). I shall continue to mull this over.
    1
  16234. I also believe that by using a circles with pi relations and a "mosner" approach RHIEMANS HYPOTHESIS can be solved
    1
  16235. I am from Germany. Once in school I had been the only one in class who could solve for the area because this formula was not taught but I stumbled over it in the mathematical table.
    1
  16236. Well, it...looks like that 1x1 square's area is being filled in by an infinite series of increasingly slimmer half-filled in rectangles...except that each piece is a bit fatter than a triangular half...so...gamma is more than 0.5 You gave an easy one to make us feel smart. I see what you did there. IIIIII see u.......sneaky math guy over here....👏
    1
  16237. Very nicely :)
    1
  16238. Visualisation helps me so much, always struggled with new concepts if I don't have some sort of context or narrative. Hope that makes sense. Tooth ache and whisky are not conducive to maths but helps with being poetic perhaps. 😃
    1
  16239. 8:38 That solution looked quite spooky, but I found that it's surprisingly easy to verify that it does indeed solve the equation. How does one derive such a solution though, I'm still quite intimidated to consider.
    1
  16240. The specific differential equation there can be classified as a "first order linear differential equation". There is a general technique one can employ to solve any first order linear differential equation, called the "integrating factor method". So that's the general technique. There are plenty of resources which can explain all of this, if you want to learn how the technique works!
    1
  16241. whats this like in base 12?
    1
  16242. book 1 proposition 47... ? or not
    1
  16243. I really, genuinely want to thank you Mathologer! Your videos are alway delightful. I truly appreciate your work. The things I enjoyed the most were the proof about the 2^(1+2+3+...) formula and the explanation about the fact that twisting a 2x2 square is the same than going from an odd permutation to an even one or vice versa. I really liked the animations of the square dance though! Happy hollidays, well deserved. :) I did some homework along the way, here it is. 6:20 I would say no but I can't find a way (if you say it's an easy challenge, then the answer should be no ^^). If I'm allowed to leave one square in an angle and take off its three neighbors, then it's done but it feels like cheating. I wanted to take off two square of the same color which are "next" to each other (like two back ones which are left and right of a green one for instance) but I can't find why that wouldn't work. :/ 10:19 Why does the formule give 0 if n and m are odd ? If n is odd then ceiling(n/2) = (n+1)/2 and, eventually, k will equal (n+1)/2. So, k/(n+1) will equal 1/2 and cos(pi/2) = 0. So, when k = (n+1)/2 and j = (m+1)/2, the corresponding term is 0, which sets the whole expression to 0. 14:10 1x2 : 1 way 2x2 : 2 ways 3x2 : 3 ways 4x2 : instead of going through all of them, we can either set one standing domino to the left and that leaves us with a 3x2 grid which can be tiled in 3 ways, or set two lying dominoes to the left and that leaves us with a 2x2 grid which can be tiled in 2 ways. Total : 3+2 = 5 ways. With the same logic we can see the Fibonacci series appearing. The series starts from 1, 2 but I guess there is 1 way to tile the 0x2 grid, which gives us the whole series we know and love. Haha! :) 14:38 I got 630 ways, I used neither the determinant nor any clever trick : i just tried to figure it out... I wouldn't bet much on the answer... What's the real answer, though ? I'd like to know. 37:32 If you see the hexagons as cubes, then you see, for each cube, always the same 3 faces. Looking at the cubes stack from above, from the left or from the right would always gives us a square the same size. Greetings from France.
    1
  16244. my favorite part was finfing out that you can technically make an infinitely long leaning tower. I appreciate it.
    1
  16245. The most memorable part was when you mentioned about Robert Bailey and is approximations for crazy series.
    1
  16246. I don't know why I watched the whole thing, I understood about 60% but I was curious to see how much more I could grasp. Also, I didn't look at the 50 minute run time at the beginning or I might have given up and done to watching cat videos earlier. I am a computer scientist, my favorite maths are geometry and probability. I am NOT good at algebra.
    1
  16247. so i took the time ,long ago to understand a question. How could i manipulate 1, so that at 1024 it would be 1000, easy minus 2.4%. do it again, same problem, same solution. Turns out , an infinitely recurring problem ,has an infinitely recurring solution. Cool hey?
    1
  16248. " 5, 10, 15, 100." — Mathologer, 2021 Edit: At 12:34 (one two: three four) he is talking about the repeating pattern in those numbers. I remember doing experiments with that a while back; digits raised to powers like 9^n have those too ((9)8(0)1).
    1
  16249. I've always said and heard sinh pronounced "sinch" which is basically just "sin h" said really fast.
    1
  16250. My proof would be If it wouldn't terminate, there must be a cycle. The cycle must have a leftmost card that gets flipped over again and again. For the leftmost card to flip, a card to the left of it should be flipped sometimes. Contradiction -> a cycle is thus not possible.
    1
  16251. Oooh, okay, how interesting! I'm at 18:33, and I've been thinking from fairly early in the video that this just "feels" quantum-mechanical or string-theoryish in nature, and here we are with Eigen-stuff... Is this useful in QM? Maybe an early step toward understanding why the universe is how it is? (Like, my line of thought is the universe prefers setups where this theorem is true in a sort of resonance-ish way...) In fact... the degenerate pentagon on the left feels awfully like a phase operator in basic QM... 🤔
    1
  16252. I noticed Pascal as well! And then once Mathologer mentioned it, I was surprised that was actually a major detail! Congratulations to the both of us! :)
    1
  16253. In the last section u have written the infinite sum incorrectly
    1
  16254. Why am I watching this? I cant even pass my math exams
    1
  16255. most memorable: competing densities causing my doom
    1
  16256. duh of course it works, it is Iron Man.
    1
  16257. What do mathematicians have to say about the tipping points between series that diverge and those that converge?
    1
  16258. There is an error at 6:15, he says 3.14 but sets the wheel to 3.18
    1
  16259. So this slide rule took advantage of the addition property of logarithms. The arctan(x) function has a similar addition property, but it's a bit more complicated: arctan(x) + arctan(y) = arctan((x+y)/(1-x*y)). The magic log wheel has an intuitive mechanical interpretation (rubber-band on a sticky wheel), but I'm not too sure the same can be said for a hypothetical magic arctan(x) wheel. I will be on the lookout.
    1
  16260. ❤❤
    1
  16261. 1, 2, 4, 8, 16, 31... I'm too stupid to solve this.
    1
  16262. I didnt believe when people were saying youtube unsubbed you from some channels, until now. Its true guys. Go check your favourite channels again.
    1
  16263. Since they all degenerate, I feel like this could be used as a cryptographic signature in some way....very interesting.
    1
  16264. That Indian man's name at 24:45 literally means "The Child "
    1
  16265. I guess if you force equilateral triangles it converges.
    1
  16266. 9:48 Patron Saint of Slide Rules? (In addition to St. Barbara, of course.)
    1
  16267. I liked the bishop's proof the most - it's amazing to me with how little tools those "ancient" mathematicians could prove stuff like that!
    1
  16268. 37:15 If you imagine the board as a 3-dimensional pile of cubes within a larger cube, you can rotate the pile to view it from different angles. If you view it from the top, perspective means that you'll see all the orange squares and none of the other two types, forming a larger square shape. You can do the same trick with different perspectives to view only the blue or only the gray squares. From each of these perspectives the cube will look the same, but with different colors, which means that the number of squares of each color must be the same.
    1
  16269. This one really sprained my brain!
    1
  16270. Recently got my hands on proclus' commentary on book 1 of Euclids elements of geometry and can't wait to get into proofs
    1
  16271. But this must be a sine b cos I was just dividing through sum ideas, naturally log(ged) on, voilà! There exists a new post from Mathologer! Pure magic.
    1
  16272. I’m supposed to be studying for my high school calculus exam Rn but instead I’m just watching YouTube videos about abstract geometric mathematics
    1
  16273. Merry Christmas
    1
  16274. Bravo, great video. You had a wonderful dream, although I would have dreamed of an ellipse :) You have a wonderful channel, it’s a pity that now is not the Renaissance. It seems in some ways we are getting dumber. Thanks again for the great video.
    1
  16275. brilliant (as always!), I think I might be watching it several times, whilst I'm sober to try and understand it! Frohes neues Jahr!
    1
  16276. I wonder if the optimal 20 bar stack was solved for n bars. Time for research, I suppose.
    1
  16277. Does anyone know what the chord progression is in the intro?
    1
  16278. Yes but drill a hole in center of the coin and use a pencil to trace the total distace the coin travels, And forget the circumferences for a second thats why it turns more on the outside instead of the inside. Its a trick of the eye trying to convince you that the coin traveled the distance of the circumference of the stationary coin when in reality it traveled the distance of and additional radius...... In short look at the center of mass..... Kinda like slide of hand, if u put the center of the rotating coin directly on the circumference of the stationary coin is the only way it travels that distance when its unfurled.
    1
  16279. 13:54 if n is even I move counter-clockwise and if it's odd I move clockwise
    1
  16280. I discovered this fact in 1976. Unfortunately, the margin was too small to contain it. ;-)
    1
  16281. Just amazing 😍
    1
  16282. I'm not a math guy, but there's something that just seems wrong about arbitrarily choosing the terms of an infinite series. A simple pattern that is obvious it repeats forever, makes sense as an abstract way to describe a quantity. An infinite algorithm, otoh, i don't know about that... What does it even mean to make choices forever? It's a description of something uncomputable, it feels fundamentally different to me.
    1
  16283. Everyone should read Eugene Wigner's The Unreasonable Effectiveness of Mathematics in the Natural Sciences. It's a short enough paper, and not beyond the understanding of a high school grad. It's available all over the internet as a pdf.
    1
  16284. a2 + b2 = c2 is most helpful out walking and want to cut corners to get there sooner..I got straight As in geometry .
    1
  16285. area of triangle = 1/4 hypotenuse length = sqrt5 / 2 little triangle is similar with hypotenuse 1/2 so area of little triangle = area of triangle / 5 = 1/20 triangle is made of two little triangles plus trapezium so trapezium area is 3 times little triangle area = 3/20 entire square area = 1 = 4 little triangles + 4 trapezia + centre square centre square area = 1 - 4 little triangles + 4 trapezia = 1 - 4/20 - 12/20 = 1/5 therefore answer is 0.2
    1
  16286. The fact that the hundred zeros series is greater than the no nines series was the craziest part.
    1
  16287. An ellipse works as well, if you allow the end result to be rotated, and you choose the stretch factor wisely
    1
  16288. 26:00 why did you first time add on short side
    1
  16289. Loved the T-shirt: sqr(666)
    1
  16290. Great video
    1
  16291. This seems to define a bijective mapping between the real numbers and integer sequences, that is the sequence whose 2k(+1)’th element is the number of positive(negative) elements of the sum needed before the k+1’th overshoot(undershoot).
    1
  16292. I really like spinning shadow of 3D cube over 2D plane at the end. Really well made to be seen as "parallel" to 4D animation.
    1
  16293. Most Memorable: The parabolic maximal overhang towers.
    1
  16294. Whaaat!? Awesomeness, love your work mathologer!
    1
  16295. Shorter ones are better, imo. You can even break down those long ones into shorts.
    1
  16296. 41:15 why do i smell φ?
    1
  16297. For me the most interesting was the gamma constant, which I've never stumbled upon in my math adventures yet :D.
    1
  16298. Slightly different proof. Consider the number of paths from the previous triangle or the initial column to any point in the next triangles (point A). Point B is up and to the right of point A and Point C is directly above point A. We are just considering the path from the leftmost column of the previous triangle or the original column to Point A. Point B has the same number of paths from the previous triangle or column. This is because any path you can trace to point A from the N-th row of the column has an equivalent path from the (N-1)th row to point B. For instance if you go from the 3rd row of the leftmost column of the previous triangle and go left-left-up to row A you can go from the 2nd row and go left-left-up to Point B. However each path to point B starts one higher than the equivalent path to point A. This means if a point on the leftmost column is X times greater than the point directly below it then point B is X times greater than point A. Since Points C is point A plus point B it is (X+1) times greater than point A.. So going up one row multiplies in a triangle multiples the number by (X+1) provided that the numbers in the leftmost column of the triangle to the right follow the pattern of being a multiple of the number below it. Since we start out with a column all 1s. So each number is the number below it multiplied by 1. So therefore in the first triangle we would have each point be the double the point below it. This increase to 3 times for the next triangle etc. Since the lowest point of any of the triangles is always 1 (because the only path to it is straight left from the lowest point of the original column) the leftmost column is always the powers of whatever the multiplier for that triangle is. .
    1
  16299. Because we number the matrix of neighbors by numbering black and green squares "any way you want" (11:07), we know our formula for the number of tilings should be invariant under any re-numbering of black squares or of green squares, which sure enough the formula with the determinant is. Coming from a physics background, this feels very familiar... Frequently in physics problems we must make some arbitrary choices (choice of basis, choice of guage, etc.) to express the problem using a particular mathematical formalism. One choice or another may be more convenient to calculate with, but the solution must ultimately be independent of those choices. In other words, the space of possible solutions is constrained by the symmetries of the problem. Physicists love symmetries for this reason.
    1
  16300. Arctic circle danse implementation : https://webgiss.github.io/CanvasDrawing/arcticcircle.html (require keyboard)
    1
  16301. s is a root of s^3-2s^2-s+1=0. Also, that means that r ans s have 2 brothers each, as r=s/(s-1) and the aforementioned equation has 3 roots (just like phi has a brother as polynomial x^2-x-1 has 2 roots) edit: no way he brought that last part up
    1
  16302. Amazing
    1
  16303. Really interesting, a huge "thank you !" from a french math teacher.
    1
  16304. Tesla was not a physicist and therefore this numerology is unlikely to have any relevance to the real world. Also congratulation on your delivery. The German (am I correct?) phonetics comes out a little only when you speak fast.
    1
  16305.  @Mathologer  I have worked in Germany as a postdoc and still understand a lot of the language (and notice the way German scientists speak in English). I am also a physicist and this is why I knew this numerology makes no sense even before I learned from the video that it is the result of using the decimal system. Once again, thanks for the great video!
    1
  16306. A lot of kudos for the description below your videos, the links you provided are very useful.
    1
  16307. My slide rule watch is just like thi!!!
    1
  16308. Lovely 😍
    1
  16309. Second one!!
    1
  16310. Shoreline paradox coming in hot
    1
  16311. Your video previews became darker at certain point. What happened?
    1
  16312. As always such a great video! Lately I've been thinking a lot about what makes good science divulgation and I still don't have an answer, but I'm sure your videos have every needed ingredient. As for this video in itself, I think it is the best simple and direct explanation to this "paradox" I've ever seen (and I look for a lot if them when I was studying differential geometry). The video make me even laugh at 21:57 😜
    1
  16313. 10:08 444? Two thirds evil?!?!?!?
    1
  16314. I figured that factorial play when I was in class 8 and solved many infinite series
    1
  16315. I actually propose you to come up with the proof based on this figure of another theorem: 3 medians intersect in one point. No triangle similarity needed, however doubling some medians are allowed.
    1
  16316. For me the best memorable thing is no integers in the partial sums.
    1
  16317. Most memorable: The optimal leaning tower. They seem to be far off mathematics...
    1
  16318. You can make block "puzzles" with these triangles. Great toys for kids and adults too.
    1
  16319. Challenge at 8:38 if I'm not wrong, is the number of distinct triangles 27?
    1
  16320. bro my computer is so laggy
    1
  16321. Nice job. My first thought after watching was stargates.
    1
  16322. I loved this video. Might be one of my favourites. I was able to follow along easily for steps 1-7. 8 I'll have to rewatch to properly digest (think I was a bit burnt/easily distracted by this stage). I'm doing a phd in theoretical physics, and I've used the EM formula in a paper I'm writing, so my background is probably much higher than the average.
    1
  16323. The mathematical soul 0:37 of lacing shoes?
    1
  16324. I teach high school algebra 1 and want to thank you for these videos. I show them to my students.
    1
  16325. I made it to the very end <3
    1
  16326. Most memorable part for me was the approximation of the partial sums
    1
  16327. Good
    1
  16328. Nice work on dodging the green arrow at 2:50 Professor Burkard :-). Clearly, you got some moves!
    1
  16329. OMG where do you get those hilarious shirts 🤣😂
    1
  16330. Same for 7 and 8
    1
  16331. I was going to make this comment too, but you did first so I'm liking yours.
    1
  16332. lol I knew you were going to show that first diagram. At least subconsciously. Just before pressing play I got an incredible urge to draw/measure various such diagrams. Been doing a lot of playing around with right triangle geometry lately or triangles in general so it's right up my alley.
    1
  16333. @20:17 - the video goes to an analytic proof that a 'thick cross' can't be prime, when it's immediately visually obvious that all 5 parts of a thick cross are divisible by the width of the arm.
    1
  16334. If I got the determinant right those are some evil glasses
    1
  16335. Direction can be given by nMod(2), where 0->Counter clockwise and 1->Clockwise
    1
  16336. What if you start one wheel from 1 till 2, and then start another whell at 2 till 1, and then put it together, so it become one wheel
    1
  16337. made it to the very end - I claim a full mathologer seal of approval!
    1
  16338. The Euler Mascheroni constant is my favorite mathematical constant, so that’s definitely one of the highlights. I’d say the biggest one is the fact that the 100 zeroes series is bigger than the no nines series. That fact was super surprising when you said it, but after a bit of thinking it started making sense.
    1
  16339. Mathologers never give up
    1
  16340. Ohh, what a intel machine he was. He shd hve lived longer.
    1
  16341. And what, do you ask, is unusual about this Mathologer video? It's less than 35 minutes long. 😅
    1
  16342. "hardcore mathematics channel"... love it! Merry Christmas professor!
    1
  16343. Perhaps a solution to the shirt he's wearing could involve some interesting math regarding triangulating black hole mass, spin, and charge, to compute the increase in the speed of light of the ergosphere to an earthly observer. Given that the speed of light basically increases in the ergosphere, is it possible to use this new type of trig in astronomy?...
    1
  16344. Nice
    1
  16345. It looks like Oresme had "text neck" ... he was, indeed, far ahead of his time. (But now having written that, I hope he wasn't suffering from anything that will get me cancelled.)
    1
  16346. The swivel proof is elegant, but I like the colouring proof better because it's more direct
    1
  16347. 35 and feeling like I’m still a primary school kid
    1
  16348. 2xn possibilities: Let a_n be the number of possibilities You start with a_1=1 and a_2=2. Then you can calculate it inductive. For the 2x(n+2) you can split up the last 2x1, remaining with a 2x(n+1) board with a_(n+1) possibilities and one possibility for the 2x1 board. Alternatively you can split up the last 2x2 field, if you lay the 2 dominos upwards it's the same layout as one counted before by the 2x(n+1). So you lay the 2 dominos sideward, therefore also having one possibility. The total formula is therefore a_(n+2)=a_(n+1)+a_n That's a nice Fibonacci sequence, but with different starts.
    1
  16349. The transition music is familiar. Vaguely resembles: P vs. NP and the Computational Complexity Zoo, hackerdashery
    1
  16350. More reading aloud of the original German papers required in future videos. Music to my ears.
    1
  16351. With luck and With regards
    1
  16352. I have a side rule that works on this principle! It used to belong to my grandfather.
    1
  16353. Mathologer = World’s Coolest Dude 😋 😇 🥳 🎁 ♾️ 🇩🇪 🇦🇺 🇨🇦 Thanks for the early Christmas present and Happy Holidays to you from Québec!
    1
  16354. What's the name of the song that starts whenever a new chapter begins?
    1
  16355.  @Mathologer  Thank you so much, love your videos and the way you handle complicated topics to make them easy to grasp and understand.
    1
  16356. 2:37 Have you ever seen the solutions of the equation ((4 x^2 + 8 x + 3)/(4 x^2 + 8 x + 4)×4 - 2)^2 = x + 2. If you see them you will be amazed.
    1
  16357. Do it with triangles rather than squares.
    1
  16358. Another amazing math adventure. Thank you and Merry Christmas, Mathologer!
    1
  16359.  @Mathologer  Sorry to hear about the reupload. Still, it is there for completists (like myself) to stumble upon later on. Best, Floyd
    1
  16360.  @Mathologer  The only way to live. Thank you, once again!
    1
  16361. On the second problem, I calculated (in my head) that 13^2+14^2 = 365. Since 10^2+11^2+12^2 also equals 13^2+14^2, the numerator is 365*2. The 365 in the denominator cancels, leaving the answer 2.
    1
  16362. 4:05 I expected all the corners of the picture on a dot.
    1
  16363. If you rotate a coin around the inside of a coin of the same diameter, you get 0 rotations.
    1
  16364. Great vids but get to the point plz
    1
  16365. God' s beauty lies also in (hidden) symmetry and Hell is somewhat broken symmetry. Thanks for the beautiful proof.
    1
  16366. I love this channel
    1
  16367. Simply a tour de force of seamless logic, I was gripped all the way through. Bravo!
    1
  16368. Calculus is a popular torture method in Hell.
    1
  16369. You didn't do the shrinking for the equilateral triangle because the interior angles are less than 90° right?
    1
  16370. 34:44 That is some quality metal right there.
    1
  16371. Thanks a lot. A bit of justice! I suggest read the book “the crest of the peacock” of GGJoseph
    1
  16372. Welcome back! We were all worried!Let's see what universal principle i can derive today? Scaling up the number of posts becomes graph theory? Moving particulate volumes around a mesh....hmmm
    1
  16373. What was your proof of no 9s sum that wasn't accepted?
    1
  16374. At 12:30 the green six square disappears, and the board becomes unsolvable.
    1
  16375. Too complicated for a simple brain.
    1
  16376. The Kempner series wins at the end, purely because it was new to me, and yet so easy when looked at logically. I guessed correctly that it was convergent, but that was for psychological rather than mathematical reasons.
    1
  16377. My favourite fact from this video must be that no partial harmonic sum is equal to an integer. Although, everything is incredible as usual, great work! To answer your questions: 1) To get a brick to completely exceed the edge of the table with only three bricks, you can put the first one 50% out. Then add the two last ones on the second layer symmetrically, such that they both have 50% overhang(one to each side). Essentially an up side down pyramid. 2) I believe you answered your own question at some point. How to write 2 as a sum of reciprocal of distinct positive integers. That is, 2 = 1/2 + 1/4 + 1/6 + 1/8 + ... Writing 3 like this, however, is slightly harder. I did not find a way of doing this, and I suspect it might not be possible. 3) In the square, where you inserted all the pieces of area that did not fit under the curve, you said that since they all fit into an area of 1 they must be less then one. The same type of argument can be made to say that they must be lager than 1/2: None of the rest areas overlap in height, so when we put them into the square, they each have their own little rectangle they fit into. Each of those rectangles can be cut at their diagonals, and become two triangles. The rest area in each of the rectangles always cover more than one whole triangle. If you sum up all of the filled triangles, you get one half, since they cover half of a unit square. Therefore, the rest area is greater than one half!
    1
  16378. There's a so called Napoleon's bridge near here, and nobody knows where that name comes from either.
    1
  16379. What about Pythagoras Theorem?
    1
  16380. I like the coloured proof more than the swivel one, because whilst both have formal versions in algebra and linear algebra (I think) respectively, the algebraic proof is much easier to explain simply and its what I'd be more likely to come up with on my own
    1
  16381. What threw me off for the coin one is that I intuitively answered how many times will every point on the circumference touch one time instead of a rotation. My brain equated the 2 and it took me a bit to detangle the explanation. If you track any given point on the coin it will only touch the stationary coin once in the 2 rotations.
    1
  16382. I'm too lazy to write any of the proofs in the comments, but I still want to help the channel with random engagement. Here's a banana for your disturb🍌
    1
  16383. I was curious whether there was any kind of an equivalent for 108-degree angles using pentagons, and it turns out there kind of is - but, unsurprisingly, you have to get the Golden Ratio involved. For a triangle with a 108-degree angle, a pentagon with the side length equal to the longest side of the triangle is equal to the sum of the areas of the pentagons with the two smaller side lengths, plus (2 * phi - 1) times the area of the triangle. It seems that, starting from an equilateral triangle, there is a sequence of correction terms (specifically as ratios of the triangle area) for triangles with an angle equal to one of the inner angles of a regular polyhedron. For 3 it's -1, for 4 it's 0, 5 is 2phi - 1, 6 is 6. 7 will be a pain to calculate exactly, but I think 8 is 8(1 + sqrt(2)). The general rule for an N-gon appears to be that the number of triangles you need to add as a correction term is equal to twice the sums of the heights of all the vertices (measured perpendicularly to some side of the N-gon chosen as the "base") divided by the height of one of the vertices adjacent to the base, minus N. Or, equivalently, 2*Im(Nz + (N-1)(z^2) + (N-2)(z^2) + ... + z^N)/Im(z) - N, where z is the principal Nth root of unity (and Im(x) is the imaginary part of x). (I think there's a way to simplify it from there but I need to get back to work...)
    1
  16384. I have a PhD in a physical science, but am a relative lightweight here - I followed the video pretty well, mostly because I have been exposed to much of the material here (I use finite sums to test code all the time), but I admit I could not follow many of the visual proofs without pausing the video frequently, especially the purely algebraic ones. But that's my problem... The treatment was awesome - especially the derivation of the Euler-Maclaurin formula, which is, as you mention, almost never covered in the specialist literature where it frequently occurs - reminds me of transfer matrices in stat mech... ;-)
    1
  16385. Both proofs only here on your video...and I'm 65!
    1
  16386. It's slide-rule, no? (Sorry, only watched 3.5 mins of it!)
    1
  16387. Yes you did it :)
    1
  16388.  @tamirerez2547  you can any number you want however you want, in the same way that 5 pi has a decimal representation, googol can have a 10 something representation. Whatever the name is doesn't really matter as long as it's clear what number you are talking about, the names of usefull or beautiful numbers will be used by more mathematicians and become convention like what happened to googol.
    1
  16389. Second puzzle: short answer is no, since I'm not good at mental arithmetic under pressure. But armed with a few factoids and nerves of steel, I could solve the problem in about 30 using the following strategy: first, what's (n-1)^2 plus (n+1)^2 ? We see that the middle terms in the resulting trinomials cancel out, and the end terms add together, giving 2*(n^2 +1^2). Same pattern for n-2. We can write the series in the numerator as (12-2)^2 + (12-1)^2+ ...+ (12+2)^2. Using the above observation we get 5*12^2 plus 10 (=2*2^2 + 2*1^2) in the numerator. In the denominator, we know 365 is also divisible by 5. If you happen to know the other factor is 73(I didn't) = 72+1 then you're home free, otherwise 360=12*30 is pretty common knowledge, so add another 5 to it and you get 5*(6*12) +5 in the denominator. Now, 2*72=144 (gross!) is also pretty common knowledge, so we have twice( 72 + 5) divided by 72+5 =2. I freely admit to having worked this out with an electronic calculator, ahead of time, in about 5 seconds. When you do that, you get 365 for the first three terms in the numerator and another 365 for the two remaining terms, thus proving, by brute force, the original Mathologer proposition.
    1
  16390. Super good thanks mathologer! I especially enjoyed the "making of", cool to see geogebra in action.
    1
  16391. I wish there was a book in English (not French) about Sophie Germain's life. Can't find one on Amazon.
    1
  16392.  @Mathologer  Thanks. OK, I'll look into it soon.
    1
  16393. great video!
    1
  16394. You took your marching orders well. Good videos
    1
  16395. BAM education for free!
    1
  16396. This has "proven" that 0 = ln(2) among other things. This is a contradiction. I.e. it disproves the math as stated. I can only think of three ways to resolve the contradiction. Reordering addition and subtraction is not valid?!?! The + and - operators you use are not the standard ones, but something similar. The method used to do infinite sums has a fundamental error in it. I would never consider myself a mathematician, but I did complete three years of calculus at CALTECH. So I am not clueless either. After thinking about the issues with infinity, I am convinced that anything involving is a fiction. Sometimes it is extremely useful, but other times it is flat out lying. As an intellectual game it can be interesting. Any claim it means anything requires cross checking with basic logic and reality. I have no problem with intellectual games. I am watching this video after all. But it bothers me that mathematicians make no attempt to distinguish between self consistent cases and fun but inconsistent games.
    1
  16397. 20:26 Yeah, I'm stepping out of this rombus.
    1
  16398. I think I enjoyed the vanishing fractal the most
    1
  16399. Most memorable: proof of divergence of the harmonic series . It's amazing!
    1
  16400. A lovely video. Just wanted to let you know that the title "Simple Yet 5000 Years Missed?" doesn't really sound right to a native English speaker. Something more like "Simple, Yet Missed For 5000 Years?" sounds a lot better
    1
  16401. The worst thing about Euler is that when you go ask someone to help you with a problem and they tell you to use Euler's Formula... You don't know which one to use....
    1
  16402. Thank you.
    1
  16403. It’s easy to DO calculus. What’s difficult is to establish calculus, the very painful beginning of analysis.
    1
  16404. Those animations look gorgeous. I'm sure an enormous amount of work went into them. At 16:25, I wondered about combining the Tritagorian and Pythagorean theorems for the special case of a triangle with a right angle and a 60 degree angle. (1) Ta + Tb = Tc + T But we also have: (2) Tb= Ta+Tc Substituting for Tb into (1) we have: Ta + (Ta+Tc)= Tc + T Giving: 2 Ta = T. So the area of a 90, 60, 30 triangle is always twice the area of an equilateral triangle on the side joining the 90 to the 60.
    1
  16405. The segment about finding the length of the curvy curve reminded me of the Infinite Coastline Paradox (the finer the measurement units used, the closer the answer goes to Infinity as you get in to the finer fractal-like nooks and crannies of a country's coastline) and how a smooth waveform of sound is digitised in to 'samples' (eg: the 44.1KHz sampling rate of standard CD & DVD audio being basically 2x the upper bound of human hearing with some extra buffer room for flaws to have a clean listening experience).
    1
  16406. wouldnt 4k+3 also just be 4k-1?
    1
  16407. Contemplating the realities of process over procedural assumptions made me a slight bit nauseous. Sometimes numbers can be an all out assault on the limbic system.
    1
  16408. How do you know the volume of the cylindrer is \pi r^2h?
    1
  16409. (Let's say I'm satisfied that the area of the rectangles is length times heighy)
    1
  16410. 21:00 das ist ein Augenöffner. Ich habe noch nie das so gesehen, ... da muss ich noch mehrmals darüber rauchen.
    1
  16411. Hi! I would argue that the spirals are still there for even the non Fibonacci numbers in the number microscope too, just like on the tree, although the lines are not properly drawn between the dots
    1
  16412. Gamma number gang here Pls pick meeeee
    1
  16413. That sounds like another great idea.
    1
  16414. I liked the heuristic proof with the fractal the most. Although I was also impressed how easy the real proof had been. Thank's a lot for all your videos.
    1
  16415. I wonder what percentage of the viewers happen to be eating nachos? Or, if not at this very moment, on the same day some time before watching? (I had nachos for lunch)
    1
  16416. I really liked it! A bit hard to follow, even with my German Math Diploma. But nicely done.
    1
  16417. oof
    1
  16418. You might like this article: https://www.yahoo.com/news/inmate-serving-murder-sentence-made-120000540.html
    1
  16419. Beautiful vid as usual. Maybe for a while we can say we use the base 9 plus the digit sum of 9 number system to let the Tesla 369 fan club down gently 😊
    1
  16420. Not watching to the end.. you can make 3-4-5 triangle out of these pearls where big balls mark our units.
    1
  16421. Love me some Archimedes
    1
  16422. Great Tutorial. Great Video. Thank You for making this !!! Patrick from Bethesda, Maryland, USA
    1
  16423. Hey, your work is art, so you are an artist, you do art with math. And by the way I think that math is magick, at least its spells act on nature sometimes in a hard to explain way. So you do magick too. Bravo!
    1
  16424. 13:00 You can inductively get from Input={n^2} to a,b,c,d, ... = 1,2,2,2, ... I tried to find a general solution with Input = {n^k} and k >2 by hand. Nope. Nope. Nope. Bless the power of Python and Matplotlib. Only k = 3 is nice with a, b, c,... = 1, 6, 12, 18, ... and so on just adding 6. Any ideas for k = 4 with its insane a, b, c,... = 1, 14, 50, 110, 194, 302, 434, 590, 770, ... ?
    1
  16425. 😆 The subtitling is often the script. Forgot to redact?
    1
  16426. Showing gamma being derived
    1
  16427. Does this extend to surfaces in 3 dimensions ?
    1
  16428. I propose the Mathologer Conjecture: "While there a finite number of Mathologer videos (to date), there remain an infinite number yet to be made..."
    1
  16429. 'AYYYYYYY E-6B SHOUTOUT!! :D I remember cutting my aviation teeth on one of these back in college when I took a ground school course. I still have it, too!
    1
  16430. It's nice all the connections to the matrix formed from the binomial coefficients and the matrix inverse. Although I had to watch the video a couple of times to understand what was happening. The fact that the sum of the cubes is the square of the sums is striking. It was not clear to me from the video whether this generalizes in any way. It's too striking to only work for just cubes.
    1
  16431. Most memorable for me is the fact the 100 zeros series converges to a higher number than the no 9s series!
    1
  16432. When can I buy that shirt <3 ?
    1
  16433.  @sasha-2574  True. More difficult would be to prove it using Euclidean geometry.
    1
  16434. I appreciate you bringing out the beauty of mathematics. Thank you.
    1
  16435. Got to the end. I just started yr. 12 (Australia) and am doing ext. 1 maths. I followed a decent amount, all of the visual stuff was really great. I had no intuition as for what was happening with an inverse matrix however. I have heard of matrices but that's about my knowledge of them. Really liked the binomial theorem stuff, that worked really well. Towards the end the sum manipulation really started to bend the mind in weird ways. I'm not going to pretend I understood it entirely. I really liked the geometric proofs for S1, S2... that had me grinning ear to ear. Euler was very, very smart. How on earth do you come up with something like that at the end? Thanks for the video :)
    1
  16436. Concerning gamma <0.5 : one can inscribe rectangular triangles into the little blue bits in the box at 24:46 which have side lengths 1 and 1/n-1/(n+1). The area of one such triangle is 1/2*(1/n-1/(n+1)). Adding all these areas up, one can factor out 1/2 and its left with the series 1-(1-1/2)+(1/2-1/3)+(1/3-1/4)+..... Changing the brackets yields 1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+... which is a telescope sum which adds up to 1 in the limit. Thus the blue area is greater than 1/2*1=1/2.
    1
  16437. Cool video.
    1
  16438. The most memorable part to me is the fact that the no 9s series sum to a smaller number than the 100 zeros series.
    1
  16439. I don't know whether anyone has pointed this out but it would take an infinite amount of time to fill the horn as the paint has to travel an infinite distance to reach the [never]end, but it can't travel faster than light.
    1
  16440. came for the t shirts, stayed for the maths
    1
  16441. For the programers: please remind, that only js solution will be welcomed.
    1
  16442. Since they're always on a square (provided by the symmetry of the problem), each bug's velocity vector would be perpendicular to the bug it chases. therefore, the only bug that changes the distance between the two is the chasing one. So the distance is the same as it would be if the latter just stood there while the former closed in on it. In fact, I imagine it'd be quite interesting to look at a diagram where one bug stands there while the rest are obeying Gardner's defined rules
    1
  16443. Calculus isn't hard. If it was, the dentist wouldn't be able to remove it.
    1
  16444. They're called "decagons", not 10-gons!
    1
  16445. :face-red-heart-shape:
    1
  16446. WOW that moment at 22:15 was mind blowing. Amazing video as always!
    1
  16447. All down to stretch of the elasticity of the marked stretcher and the tensile strength of it.manipulation comes to mind from start.how far can your elastic stretch, and will any be the same. .?also the elastic will be gradually giving up its both Tension to the next Mark ,so yield and dissipation will vary in temperature and surface contact grip.
    1
  16448. Chapter 6
    1
  16449. Thanks!
    1
  16450. non
    1
  16451. At 15:00, the converse is not implied. Proving that all polynomials end in a constant row is different from proving that ONLY polynomial functions do so.
    1
  16452. I'd kind of like to know why 1*3*5*7*9*11 . . . is privilaged over 3*5*7*9 . . . for that denominator. I'm willing to stipulate that there IS a good reason, but I max out at baby-calculus and that was a long time ago so I can't really see it.
    1
  16453. Ready or not. Mind blowing.
    1
  16454. Imagine smoking so much pcp that you come up with solutions involving things rotating left and lasers that reflect at right angles.
    1
  16455. fermat's little thm speeds up that projecteuler question of longest decimal repetitions
    1
  16456. The simple indian solution: "Show bobs end vegana".
    1
  16457. What a fantastic video again! Thank you very much!
    1
  16458. I remember in 92 my Calculus teacher in the 1st year of engineering... he always finished a very long exercise saying: " so, what looked hard was actually..." and we all always answered loud "impossible", that was really funny.
    1
  16459. The fact that gamma is supposed to have the same connection to the harmonic series as the sum of positive integers have to -1/12 blew my mind, so I cast my vote on that. It‘s interesting because I recall that zeta(-1) is -1/12 and zeta(-1) „belongs“ to 1+2+3+... in some sense but zeta(1) is undefined and would „belong“ to the harmonic series in the same sense. So this shows that there must be more to the story :o
    1
  16460. What did attempts to generalize the mn-1 approach lead to?
    1
  16461. Okay, so math dunce here, but wouldn't it be possible and even more likely (given that the simplest possibility is usually to most likely to be correct) that if Gauss was fascinated by math that he already did the sum of 1 to 100 on his own well before the teacher asked the question? I know it's a lot less glamorous by it would make sense. I played with addition when I first learned it, but a true math genius and lover is someone I can see playing with it in a MUCH more sophisticated way. the reason I never did this is because I never thought of anything remotely like it. Which further points to Guass' genius; he was able to think of the deeper questions. As a child no less.
    1
  16462. Does the pdn theorem work in higher dimensions as well? I can't find any articles.
    1
  16463. ​ @Mathologer  thank you! I wish my math skills weren't so rusted lol
    1
  16464. I made it to post-credits, yay.
    1
  16465. what does the 7 point star look like of the traced point is on the edge of the rolling circle
    1
  16466. 2:13 nice about this one is that all the laces on top are completly horizontal.
    1
  16467. Math is beauty.
    1
  16468. What does the wheel look like in Base 12
    1
  16469. I'm still unclear why it's pi, and not pi/4, that everyone's trying to get at. pi/4 has far more interesting and useful properties, such as being the ratio of both the perimeter of a circle to the circumscribed square (P(circle) = pi/4*P(square)), and the area of the circle to the circumscribed square (A(circle) = pi/4*A(square)), being the reflection axis of the trig functions (cosine = sine reflected about pi/4, cotangent = tangent reflected about pi/4, cosecant = secant reflected about pi/4), etc. and these series all natively yield pi/4, not pi. so it takes intentional effort to turn it into the wrong value, which makes asking the question 'why the hell is everyone doing that?' really quite obvious, and yet nobody appears to have asked it.
    1
  16470. We can definitely spot where you corrected yourself :)
    1
  16471. One thing i don't get about the overshoot / undershoot algorithm is: in order to determine whether your number is bigger or smaller than the target number you need to know the target number. Which is fine for all, but irrational numbers. I don't know how large pi is exactly, so i can't also say that any number arbitrarily close to pi is also bigger or smaller than pi. There is a kind of hen and egg problem for me.
    1
  16472. 10 sides shape is a decagone.
    1
  16473.  @Mathologer  hehehe I'm Greek, so I don't have any kind of arithmophobia :P
    1
  16474. I love it man, small small things in a gorgeously animated way.
    1
  16475. I ❤ this stuff & highly appreciate the cleverness w/o numbers. Definately going to use this for tutoring, thank you for sharing.
    1
  16476. Weird overhangs
    1
  16477. Come for the accent, stay for the education.
    1
  16478. I really, really, really enjoyed watching this video. Your channel is such a great example for how intriguing and fun mathematics can (and should) be. Even though I can't really pin down my favourite chapter of the video, for me personally, the results by Robert Bailie as well as the fact that the harmonic series' Ramanujan summation is the Euler-Mascheroni constant were the most interesting. Here in Austria, to get a high school diploma, one must write a ‘Vorwissenschaftliche Arbeit’ (basically a ~25-page thesis about your topic of choice). I wrote about the connection between primes and the Riemann hypothesis and did a lot of research about the harmonic series. I already knew the proof by Oresme, but the rest of the video was mostly completely new to me and I am so thankful for this video. Liebe Grüße aus Österreich :)
    1
  16479. Thank you 👍😃 great video 👍
    1
  16480. I first saw the second of your proofs (the algebraic one) about ten years ago (I’m 57 now). While I learned the formula and application of Pythagoras’s Theorem as a child, I had never seen a proof until middle-age.
    1
  16481. nice one
    1
  16482. If there's a limit to the square grid size, why do you keep making the hexagons infinetly
    1
  16483. The excel formula for this is going to contain a lot of IF statements; the proportional way of splitting the estate is way easier in Excel, I guess the Talmud rabbis haven’t thought of that
    1
  16484. I’m still confused by the windmill demonstration. What if there is no windmill with one square at it’s centre? Also why won’t this trick work for 4k + 3 primes?
    1
  16485. What a fantastic video !
    1
  16486. Chapter 4 challenge from perfect number 1 (1/6 + 1/3 + 1/2) = 1 from perfect number 28 (1/4 + 1/7 + 1/14 + 1/28) = 1/2 from perfect number 496 (1/4 + 1/8 + 1/16 + 1/31 + 1/62 + 1/124 + 1/248 + 1/496) = 1/2 Combining these fractions will produce 2 but the factor 1/4 is repeated. So replacing 1/4 with 1/5 + 1/20 will produce a set of distinct factors which sum to 2. the reciprocals of: 2 + 3 + 4 + 5 + 6 + 7 + 8 + 14 + 16 + 20 + 28 + 31 + 62 + 124 + 248 + 496
    1
  16487. A big fan of yours Plzz check my answer..
    1
  16488. Ok, here's the unit fraction decompositions for 2 and 3 which I've found: 2 = 1+1/2+1/3+1/6 3 = 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/2520+1/252+1/15
    1
  16489. G'day, my old mate and fellow Melburnian! Fascinating video that's natually right up my alley. A couple of comments: 1. I was surprised that your proof didn't use signed integers to refer to a positive or negative slope. That way, it would be trivial to add up all the signed integers and ensure that the result should equal zero. 2. In the real world, I also have to consider the eyelets as being more than just a node in that it has a distinct entry and exit, one of which is on the inside and the other of which is outside. 3. I liked your optimizations based on the "Travelling Salesman" solution. When I work on lacings with multiple passes through eyelets, I use a similar optimization pass to simplify lacings. Thanks also for referring folks to my website for the more practical aspects of real-world shoe lacing.
    1
  16490. I'm still impressed with the great gift. The connections presented were known to me (some I invented myself). But there are people who see the soul of a stone, thank you, amazing :). And now the task is simple :). There is a plane for which the parameter is set - time (you can use the definition of Newton, you can Einstein, you can geometrically). And now imagine that the topology of the plane is a Möbius strip. How, mathematically, you can match clocks on two surfaces. Thanks:).
    1
  16491. Math guy: 2020 was a crazy year 2021: Don't get me started.
    1
  16492. So, presumably there are 4d objects which can be sliced and unfurled into 3d space...?
    1
  16493. Also, I made a 4 digit slide rule using an excel chart. Just print it out and wrap it around a PVC pipe to form a spiral scale. There are two cursors, one rides on top of the other, as with a circular slide rule. Alas, adding log, trig, loglog, and P scale (for relativity problems) is just not feasible as things get really crowded. At most one other scale. Actually, a single ring log scale at one end can do the job if you know which turn of the spiral your number is on. Lots of material for experimentation here.
    1
  16494. SUBBED!
    1
  16495. THANKS FOR THE CORRECTION
    1
  16496. pi day :))
    1
  16497. Beautiful
    1
  16498. Here's a miniature implementation of the 3-peg algorithm in Go: --- type Plate interface {} type Stack interface { Pop() Plate Push(Plate) } func Shift(src, dst, tmp Stack, n int) { if n < 1 { return } Shift(src, tmp, dst, n - 1) dst.Push(src.Pop()) Shift(tmp, dst, src, n - 1) } --- What a beautiful algorithm!
    1
  16499. I love mathematics a lot, and after seeing ur videos I adore and respect it too
    1
  16500. Thanks!
    1
  16501. I consider myself very intellegent. I do excell at certain things but when I come to the Mathologer channel, I find out how much I truly do not know. I like how you explain things yet I have trouble absorbing it. I am in awe of your intellect but also I am in awe of the people who comment on your videos. That in our dumbed down society there are still people that fully understand what is hidden to most of us mere mortals is just amazing to me. I am drawn to these videos but much like a moth is drawn to a flame. I see and then I crash and burn! I am hoping for a breakthrough moment when I shout Eureka! Keep up the good work and I will keep coming back and try to grasp it all!
    1
  16502. Vielen Dank für das super Video! 🙂 Sie sind tatsächlich mein liebster australischer Mathematik-Youtuber, auch wenn ich den mit den Zaubertricks auch sehr sympathisch finde 😊 Darf ich aus Neugier eine Frage stellen... ist das T-Shirt schwarz? Nur weil es im Video deutlich heller ist als offensichtlich schwarze Elemente wie die dunklen Partien im Kupferstich vom Fermat 🙂
    1
  16503. Started thinking about how to implement the artic circle stuff in code, might try to get it working either later on this week or next week.
    1
  16504. A question for your viewers: Why is 60 in a special black box on the inner ring of the E6B and in red on the pilot's watch? I passed my pilot written exam with a cardboard version of the E6B and a pencil to measure distances on a map! That's all the precision you need!
    1
  16505. I haven’t used school or university maths in over a year but soon will have to again since I’m going to study and this video got me so hooked on maths again that I see absolutely no problem in doing so. Really great work in visualising and explaining! This video got me really curious again about everything maths related
    1
  16506. Talking about maths, I have just posted the most difficult puzzle in the world and no computer should be able to solve it. Please check it out.
    1
  16507. Bravo! These were some of the finest magic tricks by natural numbers! Loved it!
    1
  16508. I came for Ramanujan and stayed for the music. I saw a picture of Ramanujan and suddenly all the numbers were dancing until the end of the video. I don't know what this is. Are you a magician?
    1
  16509. If anything, it was modular arithmetic that was almost totally missing from the curriculum in my education. Only when I looked into group theory after my education I discovered its quite general usefulness.
    1
  16510. When I left High School in 1980, my math teacher had finally decided to throw out a sack full a slide rules. I asked if I could have them, and as I was the only one that had bothered to learn how to use it, was allowed to takes them home. Fast forward about 6 years... My wife found the sack and decided I only needed to keep one of them.
    1
  16511. I learned Heron's formula in high school, but that's because I was on the math team.
    1
  16512. Taking the inverse of a matrix? Rookie move.
    1
  16513. "The experts among yiu are probably yawning at this point" this line literally caught me mid-yawn
    1
  16514. HO HO to you too !! :-)
    1
  16515. Calculus is easy. What it does to trig and algebra is horrendous.
    1
  16516. Your example of Pi shows 3.18 instead of 3.14. The 4 tic marks between 3.10 and 3.20 are each worth 0.2, so you would need to only move 2 tic marks past 3.10 to get to 3.14
    1
  16517. Before Mathologer gave the challenge of rotated squares, I had this: https://www.desmos.com/calculator/ae6s3deqv2
    1
  16518. Faster
    1
  16519. wait one second! but you said that the zigzag lace is also the shortest. 0000011114 it contains a 4
    1
  16520. I remember coming across a lot of that myself about 45 years ago when I first discovered the calculus of finite differences. I was struck by how many of the ideas of calculus had analogies in the world of finite differences - often involving just a simple swap of derivative for difference operator and x^n for the equivalent 'factorial function'. It's a pity this area is not more widely known.
    1
  16521. Methodology produces mathematical anarchy.
    1
  16522. in the army, we need the lace to be always on the upper side (so it will be easy to cut if you get injured) the right way to do it: first you must tie a knot in the lower side of the lace lets start from lower right hole - goes from inside - lace goes to the left hole (lace up) lace goes up to second left hole & to the second right hole (lace only seen up) lace goes up to third right hole & to the third left hole (lace is up) lace goes up to 4th left hole & to 4th right hole (lace is always up!) lace goes up to 5th right hole & to 5th left hole (lace is always up) and so on.... at the end you tie the lace on the upper horizontal bar lace and you insert the rest of the lace into the shoe. this is called: Straight Bar Lace
    1
  16523. Here at 4711 views. Awesome as always!
    1
  16524. This is the first time that "if anything, then nothing" has made sense to me.
    1
  16525. It's really easy to see that if you remove at least 2 green squares you can remove the two adjacent to a black corner, and thus you can have an isolated black square you cannot tile, even if you keep the total green and black even by removing two black tiles elsewhere.
    1
  16526. It’s not a coincidence….It’s a plot! 😂
    1
  16527. Otis is a genius, therapy and math stuff
    1
  16528. Right, Right, Right, Left. Top row = 9 4, Bottom row = 17, 13.
    1
  16529. In chapter 4, the fact that there are no integers among the partial sums, no matter how tiny the increments are, was the most shocking and amazing part of the video
    1
  16530. Darn. I thought he was going to offer a solution to the Feds dilemma.
    1
  16531. One can quickly see that the 6 points lie on a circle using "power-of-a-point" lemma. If we're at the bottom left vertex of the triangle, we see that red * (green + blue) = red * (blue + green). This shows 4 of the 6 points lie on a circle, the rest is just repeating magic :3 Still, I like the windscreen wiper proof more as it's so much prettier to animate and also shows the relation with the incircle :D
    1
  16532. 19:55 "Of course this is Mathologer, not the Discovery channel. We won't be content until we dive into the 'Why?'". I miss the old Discovery channel. It used to be pretty good back in the 1990s.
    1
  16533. wow
    1
  16534. Im trying to understand this better and my mind always sees it more like 39693969396......? Can someone help me understand? In other words would I be wrong to describe this as 396? As opposed to 369?
    1
  16535. A challenge for the square version of this! Can you find other solutions to a^2+(a+1)^2+(a+2)^2=b^2+(b+1)^2 - not necessarily consecutive integers.
    1
  16536. I love me a good half an hour of maths
    1
  16537. My favorite part was the optimized overhang of a 20 block tower, it's mind-boggling to think of how anybody worked that out. :)
    1
  16538. Just a thought, isn't using a shape with infinite area as an ''anti-shapeshifter'' a bit cheating? I can take an infinite plain and say it is an "anti- shapeshiter", isn't it?
    1
  16539. Kempner's proof animation
    1
  16540. I stopped at analytic geometry, I couldn't hack it. I'm lost at Bernouli numbers. I'm really no sure what they are. Most everything at least kind of worked for me, but I'll have to take another pass at it at see how it all fits together. Watching this while ironing is not recommended. Or maybe it is. I'll just have to come back.
    1
  16541. At about 11:52 - slight typo when showing the n^4 example - on second row the 396 should be 369 -- differences are correct though on 3rd and following rows
    1
  16542. Sir, Infinity × infinity =?
    1
  16543. Equal.
    1
  16544. I've been watching your videos for years, THIS ONE! was the proof I was asked to solve in my grade 12 academic maths course. You brought me to a new realization that I hadn't seen before. My teacher was kind enough to give me a passing grade but I missed the mark. Now I want to make a learning tool that will physically describe the geometric relationship you've described. Thank you for your continued insight.
    1
  16545. Hey, may I recommend that you put the names of your patreon supporters in alphabetical order or something? I'm guessing that seeing ones own name in these videos is like half the reason for becoming a patreon and it's pretty annoying to have to read ~9 names per second for about half a minute with no breaks. Just an idea
    1
  16546. I learned a different algorithm from (I'm almost certain) a Martin Gardner column. If we number the discs in the obvious fashion, we only ever place odd discs on even discs and vice versa. This helps one realize that, in the standard setup, the first disc is moved to our desired final pole iff we have an odd number of discs.
    1
  16547. The most memorable part for me is GAMMA, Euler's constant. Thank you
    1
  16548. Yay, Igot the coin rotation right... only the 1st one.
    1
  16549. Math/CS undergrad. I got everything up to 40:45 completely, and was able to follow along with the rest somewhat comfortably. I think you've outdone yourselves this time, honestly. Fantastic video that cuts directly in to a really deep topic and makes it incredibly accessible.
    1
  16550. I remember my Father using a slide rule, for his college classes. I never knew how to use one until this video.Ty
    1
  16551. 15:23 hmm i wonder if it would work with 4 dimensional shapes...
    1
  16552. Great video and love the T shirt, very clever!
    1
  16553. I have a solution to the red cross problem but this comment section is too small to contain it
    1
  16554. 112114 13151818919 1311826912129 God's Name
    1
  16555. 222: ONLY ⅓ EVIL 333: ONLY HALF EVIL 666: EVIL 999: EVIL UPSIDE DOWN
    1
  16556.  @Mathologer  You could also see this way of looking at Green’s as an illustration of the fundamental theorem of calculus: the value of the integral over an “interval/area/volume” is equal to the value of the derivative on the boundary points/line/surface. Makes it easier to remember stuff in multivariable/vector calculus. 😊 Now, I’m no longer an instructor at any university, but the few students I tried this on sitting in the student coffee shop ten years ago seemed to like the analogy.  😃
    1
  16557.  @Mathologer Oh, sorry, my original comment made no sense without the time stamp. I thought I put one in before. Fixing that right now. 👍🏻
    1
  16558. I must add that the gamma constant is hard to grasp, but that was a nice way to introduce it. I have computed that constant to 128 million digits through a fast convergent algorithm. There is of course a way to get a fairly good approximation by 1/sqrt(3) - 1/7429. What I like the most about gamma is the relation to the Bernoulli numbers and/or Zeta function. There is a way to compute exactly the partial sum of the harmonic series by using the digamma function. Using the first 4 Bernoulli numbers I get Example : approx of 1+1/2+1/3+...+1/21 = ln(z) -1 /2/z - sum(bernoulli(2j)/2j / z^(2j) with z=22 and j varies from 1 to 4 AND then we add the gamma constant of 0.577215664901... to reach 3.6453587047627298142 The true value is 3.6453587047627295305
    1
  16559. Most memorable part: the harmonic series partial sums misses all integers
    1
  16560. "7 million viewers can't be wrong. Right?" Oh, Mathologer, you poor naive soul...
    1
  16561. Lol he didn't mention the argument in the description from 41:10. I've worked it out and I see that it does work, but I don't understand WHY it works...
    1
  16562. Easy! Rotate the platform 180° :P
    1
  16563. I liked your own proof of finite "no 9's", I think you should resurrect that
    1
  16564. Most memorable : the fact that -1/12 is to the sum of the natural numbers what gamma is to the harmonic series.
    1
  16565. I wish you could also explain the maximum overhang theory. I liked the mathologer to physicaloger parts in chapter one
    1
  16566. BUT Say we are given the sequence of 2 3 4 5 6 9 The next two or three numbers can be ANYTHING AT ALL!! You can even PROVE that your 'any next few numbers at all' are (is) true!
    1
  16567. still not easy
    1
  16568. Awesome
    1
  16569. That was great!
    1
  16570. Wow the sum of the hundred zero series is crazy! Math is amazing!
    1
  16571. 15:37: 2+3 = 5 times?
    1
  16572. This video helped me so much, and really helped me enjoy calculus (and mathematics in general) much more. Your reasoning made understanding the notation and mathematical processes much easier. I couldn't thank you enough Mathologer!
    1
  16573. 9,4=13;4+13=17 1+1/2+1/2+4*1/4+8*1/8+16*1/16=5 3+4+5=12 pearls
    1
  16574. A proof is beyond my abilities, But shouldn't an IRIS proof be on 3Blue1Brown's channel??😁
    1
  16575. Terrence Howard says you forgot one: 1+1 = 1*1
    1
  16576. That's pretty cool
    1
  16577. He went all Pascal our asses…
    1
  16578. Can you please explain why use 4k+3 instead of a "simpler" 4k-1?
    1
  16579. Reminds me a bit of 3Blue1Brown video - though that one went in different direction, but there was also a bit about "straight spirals" that if you zoomed out were not so straight anymore. And those occurred when you did use one of those Pi approximations. I do have to say that compared to some of your usual.... tough videos that require a lot of pausing - that one was easier. And while Fibonacci numbers being the best approximations of golden ratio is kind of their definition, I was wondering if the same wouldn't be true (just it wouldn't be the "best") for any sequence of numbers, where you add 2 positive integers to themselves and then add result with the higher of those integers and repeat for couple of steps. Then the 2 consecutive numbers divided by each other (bigger/smaller) will be close to golden ratio. I think in fact I might have learned it from your channel.
    1
  16580.  @Mathologer  Yeah. This was what I was talking about. Though I thought I might have learned about it in Lucas numbers video... Still power of phi were also very, very interesting.
    1
  16581. #vote I think I'll go with these amazing cute-ugly far-reaching leaning towers. But well, the final proof was also quite neet :)
    1
  16582. Nice video! But I think that every number that is either divisible by 4 or odd can be written as a difference of two square integers, unless these integers are required to be coprime
    1
  16583. Wat?! Cowsine??? Hahahahaha.
    1
  16584. (IMO) brilliant vid again the maths subject is the best subject to exist for a idea, Disclaimer: Its just My own Opinion on the suggestion, Advice; "Be very proud of having a Opinion on something that counts for eachother including *me.
    1
  16585. most memorable: chapter 3 - finite or infinite
    1
  16586. Make a video on Dual numbers and split complex number
    1
  16587. I guess for any modulo p with p a prime number, there are special numbers p^n+1. For compound p, those who have special numbers associated (still p^n+1) are Carmichael numbers and only those. Please tell me if I'm wrong :-)
    1
  16588. In the case of a democratic election, would this way of splitting seats be more fair than d'Hondt law? Maybe equal?
    1
  16589. The no 9s square was what stuck with me the most.
    1
  16590. I remember those flight computers, they were called Daltons computers. We all had them at FTS in the early 60s.
    1
  16591. 600th like!
    1
  16592. Superb ! as usual
    1
  16593. Thankyou. I'm glad to see a proper mathematician taking an objective look at this. I had a bit of a dive down that rabbit hole years ago. There was an interesting pattern to using the square numbers as the number base if I remember - but only the odd ones for example base 16 didnt quite fit whereas base 25 did (it was so long ago I cant even remember the details).
    1
  16594. For me the most interesting part was finding the amount of terms needed to surpass 100. Lots of interesting nuanced maths needed to understand such a simple question!
    1
  16595. I was certainly able to follow your explanation. But I'm always a bit uncomfortable dealing with infinite series. Folks sometimes use them to come up with, shall we say, 'incredible' results. Like the series that supposedly adds up to -1/12 or the +1-1+1-1... and someone says, "well that clearly averages to 1/2".
    1
  16596.  @Mathologer Well I agree, I wasn't trying to accuse. Just that sometimes I have to look very carefully when someone starts using infinite series to 'prove' things. In fact anytime an 'infinity' comes into play, I slow down and become wary.
    1
  16597. The way I think about the 1 + ½ + ¼ + ⅛ .... sum is mentally to write it in binary: 1.111111.... which is clearly 2, in the same way that decimal 0.999... is equal to 1.
    1
  16598. Vote for the strange leaning towers that can grow proportionally to x^(1/3) rather than just ln(x).
    1
  16599. Cantor approves of Tristan's proof.
    1
  16600. In Ukraine, it's taught quite early actually, in 7 or 8 grade.
    1
  16601. An amazing Video I can't choose a chapter they are all amazing. I am a math lover and teacher and you inspire me throught your channel thank you.
    1
  16602. How can we know something converges to pi without knowing the last digit of pi? My reasoning is: pi + 0 = pi 1 - .9[bar] = 0 pi +1 - .9[bar] = pi
    1
  16603. You can take a single floor tile (say 1 foot square marble), half its thickness using some mason's super thin cutting blade, and end up with two square foot tiles. In fact you can repeat the process, using just 1 single marble tile, tiling the whole of St Peter's Basilica in Rome. Mind you could not stand on them as they would be wafer thin and probably crack, but yet again theoretically still possible! Now I know how they done the loafs and fishes thing in the biblical old testament parable! 🏰 🟪🟪🟪🟪..♾
    1
  16604. Most memorable part is your whole video
    1
  16605. The coloring proof seemed more obvious and no measuring involved.
    1
  16606. I frequently use the 3-6-9 math in the lottery and it's amazing how it keeps on protecting me against greediness.
    1
  16607. Where in the video is the part with the Arctic Circle Theorem? I want to know what it is. Thanks!
    1
  16608. Once again, great work, Mathologer! However, I see a tiny error there in the distribution of terms at 4:22 You would expect the empty set in the "first column" in the cross product, but you made the leading empty terms disappear (except for the very first partial product). In Mathematica, by the way, it would look like this: with Distribute[{{}, {1}, {1, 1}, {1, 1, 1}}, {{}, {3}, {3, 3}}, List, List] you get {{},{}},{3}},{{},{3, 3}},{{1},{}},{{1},{3}},{{1},{3,3}},{{1,1},{}},{{1,1},{3}},{{1,1},{3,3}},{{1,1,1},{}},{{1,1,1},{3}},{{1,1,1},{3,3}}} Anyway: a very exciting topic! As a chemist, I see great applications there: e.g. if you ask for the possible elementary compositions for a given (exact) molar mass, you can get all conceivable "sum formulas" this way (and with a few chemoinformatics tricks even the valid molecular formulas). Looking forward to future videos! Here are a few "wish titles": Polya-enumerations of isomers, automorphism groups of graphs or even solving functional equations.
    1
  16609. 1 2 3... ...204... ...288. Coincidentally, today is the 204th day since I stopped smoking (not that I'm still counting).
    1
  16610. I’m out at 1:19 of my life I’ll never get back.
    1
  16611. Of course, this is all Euclidean, but the iris is a circular region on an eyeball. But hold on. If you start with a spherical triangle formed by three great circles, the windscreen wiper proof seems to be equally valid.
    1
  16612. This was taught to me when I did ATAR (Australian equivalent to brittish A levels) maths a few years ago. I thought it would be standard to teach it as part of pre-calculus maths.
    1
  16613. This proves I should've had more attention to math in school! :D
    1
  16614. Current name of the place Sangamagrama is Irinjalakuda which is my hometown & is a municipality in Thrissur district of Kerala state in India. Even I knew him while studying in 8th standard 19 years ago.
    1
  16615. Hi, Karl! :]
    1
  16616. The most memorable part for me was the approximate formula for the harmonic series
    1
  16617. Having watched your video, I feel like starting my channel on knot theory and related mathematics and physics (representation theory, 3 dimensional topology, surgery, Reidemeister torsion, Chern-Simons, and a little bit of string theory that I can understand, e.g. Type IIB + a little bit of mirror symmetry), but it takes me a camera and some time of talking and writing. Will see how it works out. And by the way I recently found the link between knot theory and organic chemistry, since organic chemists are the best experts on 4-valent graphs (carbon has four bonds)...probably I will make videos on that. Wish me good luck!
    1
  16618. I will cast it on big screen feeling smarter while sitting on my couch. Perfect Sunday.
    1
  16619. Anyone else got emotional watching this video?
    1
  16620. Your shirt seems to have a rounding error 😀
    1
  16621. 15:32 u got me like
    1
  16622. I remember a geometric proof, maybe using the Chinese diagram (~1985) but I don't remember the details, except that it was not on the sylabus.
    1
  16623. by 11:00 i paused to do the math, and it turns out that no matter what (a,b) you choose for the top row of the square, it always works: aa+2ab, 2ab+2bb -legs aa+ab + ab+2bb -hyp aa(aa+4ab+4bb) + 4bb(4aa+4ab+bb) -sum of legs squared (a+b)^4 + b^4 + 2bb(a+b)^2 -hyp squared a^4 + 4a^3b + 8aabb + 8ab^3 + 4b^4 -both expanded become this
    1
  16624. 39:15 (knowing fibonacci is ...-8, 5, -3, 2, -1, 1, 0, 1, 1, 2...) i'm betting this extends infinitely backwards too. choose a=0, b=1 and you get 2. a=1, b=0 produces 1 (the 1 after 0). choose a=-1, b=1 and 1 is a solution (the 1 before 0). with a=2,b=-1 you get 2 (the one behind 0). choose a=-3, b=2 and 5 is a solution, which seems to indicate to me that this goes forever both ways. i think this is also trivially demonstrated by the fact that squaring numbers discards the negative status, and that the diagonal multiplication and addition happens to align as long as the signs alternate the way they do when you construct the progression this way.
    1
  16625. I love this guy's laugh 😂
    1
  16626. 4k-1 too?
    1
  16627. it is the same as the kock snowflake which encloses a finite area in an Infinite perimeter....
    1
  16628. Dr Strangelove used one of these!
    1
  16629. "we can't repeat to infinity, neither can Pi reach 4". It's not a paradox when our own premises is wrong from the beginning. Laws and rules must be applied righteously or we'd fall into hypocrisy's double standards. If we can repeat to infinity, the same rule must have to be possible to be applied to other subjects as, zoom to infinity at the same rate and prove ourselves that we never reached a perfect circle we just divided the corners into infinity. The wavy line and dancing ruler? use a string...
    1
  16630. 13:25 it's fast Fourier transform. Have no idea how is it related to Fourier transform with complex integral, but FFT is algorithm calculating product of polynomials faster, you can find video in 3b1b-style(it was made with manim), explaining the work of such algorithm. Due to many zeros in polynomial it could be advanced to better work
    1
  16631. I liked the triangle logo
    1
  16632. I watched your video and I didn't see the time pass! 45 minutes like a 10 minute video :) Thank you.
    1
  16633. not taught until now
    1
  16634. Coin paradox is not a paradox when you are a mechanical engineer and know from the dynamics of rigid bodies course that the displacement of the center of a pure rolling circle is r*theta :))
    1
  16635. Here's my crack at a proof: Let ABC be the triangle with lengths BC=a, CA=b, AB=c. Suppose AC extends to points A' and C' such that the order of the points along AC is A', A, C, C'. Extend AB to points A'' and B'' such that the order of the points along AB is A'', A, B, B''. By symmetry, it suffices to show that A', A'', C', and B'' are concyclic. But notice that A'A=a and AC'=b+c, while A''A=a and AB'' = c+b. But now (A'A)(AC') = a(b+c) = a(c+b) = (A''A)(AB''). By Power of a Point, A', A'', C', and B'' must be concyclic. If you don't know Power of a Point, notice that triangle AA'A'' is similar to triangle AC'B''. Then angle A''A'C' = angle A''A'A = angle AB''C' = angle A''B''C'. This is a criterion for A'', A', C', and B'' being cyclic. EDIT: Ah, hold on, the symmetry argument doesn't tell us that all six points are concyclic, and I can't quite resolve that little problem. Yeah, this one is trickier than expected. My other idea was to do try doing some sort of homothety to the incircle, but it's not obvious to me how to show such a homothety maps points on the outer circle to something fruitful on the incircle. Interesting question.
    1
  16636. Awesome video ❤
    1
  16637. It seems anything can be proven if there is an infinity involved. And this is easy to grasp since anything can happen in an infinite universe.
    1
  16638. Mathologer, you are awesome, and I really look forward to your Mathologerising Gauss’s law of quadratic reciprocity.
    1
  16639. I think in the beginning you should say more about the behavior of the rubber band than just "the numbers spread out more and more". All clear? Great.
    1
  16640. Amazing video and visual proof. Thank you for making my day!
    1
  16641. more like the lost "wonderwall"
    1
  16642. Answer to the question: cutting out two of the same colour would allow you to isolate a corner square and thus make it impossible to place a domino, on an infinite surface you need to be able to cut out four of the same colour to make it impossible to fill out everything, in an infinite 3D field you need to be able to remove 6 cubes of the same colour to make filling everything out impossible. Now, how many cubes do we need in a 4D field? I don't know - maybe someone can help.
    1
  16643. I actually liked the proof of the convergence of the "no 9s" series best. Its marvellous in its simplicity.
    1
  16644. This is why I was never good at math. My real-life physics brain immediately started to wonder about whether certain types of rubber bands might have different material properties (aka stretching more or less per newton, breaking sooner or later, stretching in a linear or exponential/logarithmic way, etc etc)
    1
  16645. First
    1
  16646. Thank you
    1
  16647. I liked the nice little observation stuff a much about the odd over even as how these little observations can give rise to greater discoveries like drops of facts making an ocean of knowledge
    1
  16648. I calculate this(1+2+...+100) by kiddo method. 😅
    1
  16649. Marty, why do you hate the Fibonacci sequence?
    1
  16650. I had to stop before 11 minutes to say "What does trig and π have to do with tiling rectangles?!".
    1
  16651. Great video, and great background music. Does anyone know what that music is? I want to listen to more of that.
    1
  16652.  @Mathologer  Thanks, I had skimmed the comments but missed it the first time around. Again, great music choices.
    1
  16653. Love it❤
    1
  16654. Nice video, is Klaus Schwab your evil twin ?
    1
  16655. Of course, it also looks like the basis of the Rebel insignia in Star Wars. 9:13 They don't teach it in school? Don't they? I know I knew that, but I can't remember if I figured it out for myself, before they mentioned it or after and I'd forgotten it. I mean, it's pretty obvious right? Right? Someone agree with me here. 9:41 I didn't know the remainder thing. That's cool. 11:24 7! 11:43 Got it! But I took the long way around. Drat! 15:54 Ooh! Let me try! Ummmm, because we use a base 10 numbering system and that has two primes in it, 5 and 2 (O.K. 2 and 5!) and so when you take the reciprcal, you switch from one to the other. I assume the same thing would happen with a base 12 only with 2 and 3. I don't know what would happen with a base 60 or a base 7 system. 16:32 1. And I think I may have gone a bit overboard with my answer in 15:54 above.
    1
  16656. Professor mathologer. Rubik S54🤔 S48 Rubik was my project of Group theory at varsity 🤓🤓🤓🤓🤓🤓🤓🤓 I fancy in extrenis abstract mathzzzzzzz
    1
  16657. Thanks for the nice table (starting 36:30). It's a pity that nothing is said about the "monster fraction" (iirc) any more. (The old - erroneous - one is the one preceding the new one in the list if it were continued, I guess.)
    1
  16658. This sounds incredibly useful, but my brain does not has half a hour of focus, even with 2x it does not have the attention for something with abstract graphs and no colorful cartoon pictures
    1
  16659. Suggestion for a follow-up: please explain the lays of the Pequod in Moby Dick.
    1
  16660.  @Mathologer  I mean like if u make the wrong moves (starting with 3 disks), and then make additional wrong moves every time u restart by making the very first wrong move, u get 13 moves no matter which path u take.....
    1
  16661. 16:55 Assume handshakes to be edges with red colour and no handshake to be blue coloured edges . Considering the scenario , there we get a complete graph with two edge colours . Now if we consider one vertex , A , one of the coloured edges must be at least 3 in number (Say red) . Now considering those 3 vertices attached with the initial vertex we chose (B , C ,D) . If BC or BD or CD is red then we get a triangle with all same coloured vertices (ABC , ABD , ACD) . And if all of them are connected by blue lines , then we get BCD as our required triangle with same edge colour . The process can be similarly continued for the other case .
    1
  16662. Great video.! Such evidence shows the power of thinking.! Fine.! Amazing.! Let's show how to build a square whose area is 1, 2, 3, 5, 6, 7, 8, 9..☺
    1
  16663. So, my math classes always called sinh like cinch. I assume this comes from sin h = sin eych(yes American pronunciation but European wouldn’t be much different) ≈ sin ch ≈ sinch but other than it being called cinch in my classes the rest of all that is just speculation on why. I’ve never had a math teacher call it shine though.
    1
  16664. The maximum number of moves corresponds to moving faces from the right most position to the left most position one by one.
    1
  16665. gamma being sum of harmonic series I consider most interesting
    1
  16666. This feller keeps on tricking me 😤 he better not pull the ole' switcheroo at the end and pi was really there all along
    1
  16667. I boycott all products advertised by YouTube.
    1
  16668. Nifty as always !
    1
  16669. As in the Goldbach's conjecture, the larger the even number is there tends to be more ways to write it as sum of two primes. So, it is quite reasonable to assume there will be no more such cases. But just as the Goldbach's conjecture, it will hard to prove vigorously.
    1
  16670. The date of the Chinese math teacher seems to have lived 800 years??
    1
  16671. Zhou bi suan jing is not a person, but rather is a text. The text was mostly compiled during the Zhou dynasty. The dates listed are the dates of the Zhou dynasty.
    1
  16672. how did you derive the formula for the numerator and denominator sequences. I am very eager to know.
    1
  16673. Amazing and clear!
    1
  16674. My condolences and thank you for sharing this, I had been looking for this information myself and failed to find it.
    1
  16675.  @Mathologer  you probably already know this with your German accent, but just in case: "Freek" is pronounced as "Frake" and rhymes with "Lake". Please don't mispronounce the Dutch name "Freek" as "Freak", this would be very wrong.
    1
  16676. For the second puzzle we can use similar graphical proof to get 10²+12² = 2⋅11²+2 (after using 13²+14²=10²+11²+12²), simplifying whole expression to 2⋅(3⋅11²+2) = 2⋅(3⋅121+2)
    1
  16677. most memorable is definitely leaning tower or lira.
    1
  16678. Wheels within wheels. Beautiful.
    1
  16679. Fantastic journey once again Burkard . If you're worried about the length you could publish each chapter as individual videos that go in a playlist. This could have two benefits. 1. If a teacher was to use a series like this as a teaching guide they could play one of your videos and spend time covering the material of one video until students gained an understanding and then go onto the next. 2. If you're monetizing your videos 5 * 10min videos will generate more income for you. That way the little mathologers can go and visit Scienceworks more!
    1
  16680. Sums of sums is one of my favorite areas of Math because of what it reveals about exponents and exponential sums.
    1
  16681. 🙏🙏🙏
    1
  16682. 12:34 had to stop and slow down the video to read the text you just saind. i feeled foolish ;D
    1
  16683. 24:48 it's greater than 0.5 because all of the parts are basically a triangle with a curve side. Due to this they have a little more area than a triangle. All those triangles pot together have an area of 1*1/2=0.5. which means gamma >0.5
    1
  16684. Beautiful math and video.
    1
  16685. Volume \varpi, area \infty ;)
    1
  16686. Problem 1: If the pattern in 1 and 2 dimensions continued, the first 3D example would have used 6 in the bridging term. (6 is the number of faces on a cube). Thus the first identity would have been 5³ + 6³ = 7³. Unfortunately, the RHS is short by 2. Problem 2: Since 10² + 11² + 12² = 13² + 14² = 365, the calculation on the blackboard yields 2. Problem 3: By direct calculation, 3³ + 4³ + 5³ = 6³ (surprise!) and 3⁴ + 4⁴ + 5⁴ + 6⁴= (6.8934...)⁴ (not so pretty). Thanks for sharing another shiny nugget from your mine of mathematical gems.
    1
  16687. Hope you can do stuff on Brahmagupta.
    1
  16688. An odd question from a layman regarding maths: is there a kind of equivalent of Pythagoras theorem for a quadrangle with 4 different sides, forming a special kind of quadruplets in maths?
    1
  16689.  @Mathologer  I see, very interesting, I will check it out. I come from a musical background, and I hope I don't sound too pseudo-scientifically, but what caught my interest about the Pythagorean triples, is that the first one, 3/4/5, is basically the mathematical proportion of a major triad in musical harmony (in second inversion). The fact that it is related to the Fibonacci numbers (and to golden ratio?) is even more intriguing in this sense.
    1
  16690. Let me guess... It's using triangles and linear transformation to get something like rotations. Alright you started forming a.. Not so fast, a tree data structure ok there are those. You are using like, hmm en español, fibrado tangente, like phase space in Hamilton
    1
  16691. Excellent
    1
  16692. 3 1 53 52
    1
  16693. The Euler-Mascheroni constant was a nice visualization I had never seen before!
    1
  16694. The most memorable to me is the partial sum formula which is so simple looking and interesting to derive
    1
  16695. spoiler . . . . . . 9 4 17 13 r r r r l ich bin kluk K L U K
    1
  16696. Damn, had to watch the video two times because I got really confused 😂
    1
  16697. Chap 2s think-outside-the-box optimal brick construction is soo neat and memorable. Need to check out some of those papers!
    1
  16698. Breathtaking stuff. Hoping that someone codes a 3 dimensional analogue
    1
  16699. I learned the "algebraic" proof while in high school at a Caltech lecture. (I also learned the Newton Raphson square root method at a one of those lectures.)
    1
  16700. There are a few that look like they might be squares but aren't: make sure the diagonals are congruent.
    1
  16701. 17:32 mind bending video. That tweaking machine!! Easily using it to detect if a number is prime—OMG!!
    1
  16702. Explain your t-shirt
    1
  16703. I did a Miller Rabin primality test on it ...... 6513516734600035718300327211250928237178281758494417357560086828416863929270451437126021949850746381 and it is most probably prime. Note: I wrote my own large number library and the test in C++ some years ago.
    1
  16704. Yegor Zolotarev is also remembered for his extension of Chebychev approximation to two separate regions (he was a student of Chebychev). Cauer applied this to design the elliptic function filters widely used in electronics. Unfortunately, Zolotarev died in his early thirties after being run over by a train.
    1
  16705. Great video as always! Inspired by the coordinate-proof from the video, here's a proof of "an n-dim cube consists of 3^n bits and pieces": Consider any bit/piece, its vertices form a subset W of the set V = {vertices of the n-dim cube}. Now focus on the m-th (1 <= m <= n) coordinates of all the points in W. There are 3 possibilities: 1. All of them are equal to 1; 2. All of them are equal to -1; 3. Half of them are 1 and the other half are -1. Thus there are 3^n different bits and pieces in total.
    1
  16706. Man you are insane. So much work for us. Thank you so much!
    1
  16707. Great play Majestro!
    1
  16708. Welcome back! Long awaited, but it was absolutely worth it! Great video
    1
  16709. Outstanding! Brings back the joy into mathematics!
    1
  16710. I was never shown any of the proofs you displayed, kool. On your end puzzle, I came up with one fifth for the area of the inner square. I also stumbled onto: that if the inner square has an area of one then the outer square will have an area of five, in this case the smallest triangles will measure one by one half by square root of five divided by two.
    1
  16711. Aldo i did not study mathematics ( So i don't always understand everything you explain in your video's) i really love your channel and enjoy watching your video's. Thank you very much i am looking forward to the new video's this year. ^_^
    1
  16712. Harmonic series is also used when you have to split beam into n beams of equall intensity, e.g. 5: splitting mirrors should reflect: 1/5, 1/4, 1/3, 1/2 and 1 fraction od beam and you have 5 beams with intensity of 1/5. Greetings from Poland.
    1
  16713. Is there a reason why you can't write 4k-1 for the primes that can't be u²+v²?
    1
  16714. Regarding the bugs problem, each one moves to the next one which is always moving perpendicularly. So, apart from the rotation, they will just follow the size of the square path.
    1
  16715. Euler-Mascheroni is my fav constant, so I love that one. Nice variation on the ant on an elastic rope puzzle, don't think I saw it elsewhere yet: suppose a car is attached to a wall with a 1-unit-long infinitely-stretchable taut elastic band. Precisely at the moment it starts to accelerate uniformly away from the wall (say its distance increases as 1+t^2), an ant starts to walk along the band away from the wall. How fast should it be able to walk then in order to be able to reach the car? Hint: famous greek letter constant in there, but not euler-gamma...
    1
  16716. My favorite part was probably the "no-something" sums, or how ugly the optimal leaning tower looks
    1
  16717. I'm at 13:55 "leave your answer in the comments"... Well - let me see if I am confused here, but I think it's rather easy to answer the "clockwise / counter-clockwise" question for n=1... So from that, it's also straight-forward to generalize which way it then goes for even, vs for odd, n. Right?
    1
  16718. Be careful you don't hurt your head on that triangle.
    1
  16719. In minute 8, Dmitriev loses credibility already. There he says the series 1/2 3/4 7/8... "clearly converges to 1." Wrong. Cauchy always uses the word 'toward' (French vers) in connection with convergence, never the word 'to'. It should be obvious that one cannot go 'to' an asymptotic limit, but post-Cauchy, everyone became lazy and started saying 'to' (one easy syllable) instead of 'toward' (two syllables). The series in question (the one that is introduced in minute 8) converges TOWARD 1; it is nonsensical to say it converges TO 1. Statements like the latter are used by mathematicians so that they can speak of a certain series as possessing a 'sum', and thus move forward with their ivory-tower parlor games. This Mathologer 'debunking' is just as bad as the Numberphile presentation that it debunks -- and inadvertently, it reveals the cesspool of general sloppiness in all math thinking which gives rise to presentations such as the one that Dmitriev so loftily and sarcastically critiques. Just like the Numberphile approach, his approach too is 'Not even wrong' (except that it gets one a 'good mark' on an exam).
    1
  16720. I like the visual proof for no 9's. Very elegant!
    1
  16721. I like that this suggests that ln(0) = infinity
    1
  16722. I have to vote for the leaning towers, if only because I'm intrigued that the optimal version (with counterbalances) isn't yet known.
    1
  16723. I have seen some of these very same folds within a Kaleidoscope.
    1
  16724. That's some bona-fide handwaving at 7:23
    1
  16725. Brilliant. Where were you when I was learning math? In later life I came to use, understand and rely on trig in my work. Truly music of the spheres. As to your challenge about a circular slide rule with infinite precision - I really have no idea, but somehow my gut feeling says look towards fractals to inspire a solution. I think it will be found there.
    1
  16726. I also discover this my secondary school, but I didn’t realise can be combination formulae, I realised is power of polynomial when there is constant. This video recall my secondary school memory
    1
  16727. I find the color proof is the best
    1
  16728. minimal number of move to go through every possible state with n disk : 3^n-1
    1
  16729. Most memorable part was knowing we can actually make the towers grow as the cube root of n, was really expecting its best to be logarithmic
    1
  16730. neither proof was actually familiar as english schooling was maliciously lobotomized in favour of repetition rather than actually teaching any understanding. this system has been taking the piss for along time. damn cheap skates
    1
  16731. Love your enthousiasm and your honest giggling!
    1
  16732. How did i not realize you are german? I feel so dumb
    1
  16733. I also know the amount of elements in a 6 unique element superpermutation. It’s 2(6!)-(5!+4!+3!+2+1), I believe.
    1
  16734. Next chapter: are there regular n-gons in a triangular grid?
    1
  16735. ok.. what if we replace 1 with 0.9999999... ?
    1
  16736. One little gem I missed today was: phi - 1 = 1 / phi.
    1
  16737.  @Mathologer  Well, true, and I didn't expect you to spend a lot of time on it. I mean, you spent a moment to mention the relationship between phi and Fibonacci, which I'm sure you must have gone into before as well. But anyway, you shared a lot of surprising, interesting stuff in this video, for which I want to express my appreciation.
    1
  16738. Please talk about this. "I say that the Term of a Prime Number is , if prime is equal 3+4K, then 4+K+1, else is K+4". João Oliveira, Portugal.
    1
  16739. I didn’t see sin(1°) as 0.01745 in that table. Especially within that highlighted region. Lost interest in the rest of the video.
    1
  16740.  @Mathologer  Thank you so much for the explanation Sexagesimal as a base explains a lot - the number base of so many degrees in a circle; fractions of degrees (even used today), how time is portioned. I should have known or remembered that. Your patience is appreciated. I need to strive for more of that!
    1
  16741. Love the way he laughs.
    1
  16742. i derived this from babbage's difference engine
    1
  16743. To the end.
    1
  16744. 0:55 wait, what? mind blown and we're less than a minute in
    1
  16745. Are those solutions found -- ALL of the solutions, and the only solutions possible? Is it proven?
    1
  16746. 10:55 Pretty dance of discs, or dance of pretty discs, or pretty dance of pretty discs?
    1
  16747. That’s cool that all 4k+1 primes have this property, but is it possible to check whether an integer 4k+1 is prime in the first place? For k=2,5,6,8 the integer is not prime. Is there a pattern there I’m not seeing?
    1
  16748. Is there a specific software you are using for the animations, and is it freely available? I could use some suggestions for preparing my online lectures in these COVID days :-)
    1
  16749. Oh! I just LOVE that shirt!
    1
  16750. Since you asked so nicely 😊 I did make it to the end, I study math as a hobby and like you explain it, it seems very clear to me, I would welcome more video's of this level and quality.
    1
  16751. 205, 206, 207, ... 208
    1
  16752. Hi I have a question about a previous video, the anti-shapeshifter one The question being is cedar tree (like the one on the Lebanese flag) anti-shapeshifter in there own way, i.e they grow but they keep the same proportions
    1
  16753. ​ @Mathologer  Thank you for the response Another question if i may, is this self similarity can be expressed with a logarithmic function?
    1
  16754. ​​ @Mathologer  Okay You seem open to some discussion! There is a linguistic theory of the word logarithm to stem from the word log, the log of the tree, meaning "to stay static, to stay the same, to maintane the same shape" i.e "anti shape shifting" Somthing like the curse of io, the curse of agelessness, to stay forever the same So, that's that
    1
  16755. My brain is exploding rn
    1
  16756. Your Rubiks/Escher illusion shirt has a sideways Star of David in the Centre: Any idea what that might symbolise?
    1
  16757. When I was 15 old, I used 2*2=2+2 to teach myself about induction by trying to proof a*a = a+a. This method was usually teached by using successful proof examples but I wanted to get a contradiction.
    1
  16758. The explanation in the movie was very clear to me. But yours was equally well explained.
    1
  16759. That pideonhole principle is the bee's knees.
    1
  16760. Who is this person that dares to use the name Elim Garak?
    1
  16761. Is it applicable for primes
    1
  16762. 13:29 When the disc number is odd, move the smallest disc to B. When the disc number is even, move the smallest disc to C.
    1
  16763. 7:04 2 Specifically, for any natural number n, let t(n) be the nth triangle number (i.e. t(1)=1, t(2)=3, t(3)=6, etc.) So, the linear pattern is sum(i from (2*t(n)-n) to (2*t(n))) of i = sum(j from (2*t(n)+1) to (2*t(n)+n)) of j Squares: sum(i from (4*t(n)-n) to (4*t(n))) of (i^2) = sum(j from (4*t(n)+1) to (4*t(n)+n)) of (j^2) Cubes: [sum(i from (6*t(n)-n) to (6*t(n))) of (i^3)] +2*[sum(k from 1 to n) of k^3] = sum(j from (6*t(n)+1) to (6*t(n)+n)) of (j^3) The two is because a cube has eight vertices and only six faces. Each slice from the largest initial cube covers one face and one vertex, leaving two missing vertices (it also covers two edges, which works out perfectly with the 12 edges of a cube). Each missing vertex needs a smaller cube, with a side length of 1 for the first new cube, 2 for the second new cube, etc.
    1
  16764. The chance of the tip domino not being in persistent form = 0.5^n
    1
  16765. I vote for gamma.
    1
  16766. Is something known about the following Hanoi variant: you have n disks and m pegs, but you cannot put a tower of height more than k on all but the starting and goal pegs
    1
  16767. I made it to the very end.
    1
  16768. i learned gregory newton formula as a short cut to solve series problems now i know where it came form
    1
  16769. Still waiting for the Galois theory video...
    1
  16770. the most interesting fact that i love and i'm surprised about, is that there is no integer among the partial sums. i doubt is that memorable but it definitely is the most unexpected
    1
  16771. I'm addicted to your informative videos . 💌 from India .
    1
  16772. Damn, that was interesting. I had never even heard of the pi = 4 construction.
    1
  16773. halfway through at 16:16 an arrow is introduced under 2.71828 which is centered on our view the number line, so the video ticker and the arrow at 2.71828 are vertically aligned. awesome coincidence, no? yes!
    1
  16774. Missing part in proof about bugs is: why the limit is equal to the length of their path? Why it's not the case as with length of staircase always equal to one in diagonal of square. Also two notes: 1) if you use fixed step length instead of proportion then you end up with a mess. 2) fixed proportion of steps make bugs either walk infinitely long if they walk each part within fixed unit of time, either make bugs switch direction faster and faster, ending up with infinite number of switching directions within finite time.
    1
  16775. I guess that somebody already said it in the comments, but still: Rock, Paper, Scissors, Lizard, Spock :)
    1
  16776. Expressing any positive number as a sum of reciprocals of positive integers is actually a generalization of Egyptian fractions: https://en.wikipedia.org/wiki/Egyptian_fraction
    1
  16777. ok, it's against the rules but still very pretty 😅 (1 - i) + (1 + i) = 1 - i + 1 + i = 1 + 1 -i + i = 2 + 0 = 2 = 1 + 1 = 1 - (-1) = 1 - i^2 = (1 - i)(1 + i)
    1
  16778. 👍🏻 Great stuff. What I like the most: the Ho-Ho-Ho-T-Shirt. Very creative 😁, Merry X-mas. 🌲
    1
  16779. Mechanic and machinist for most of a half century, brain permanently broken.........
    1
  16780. I have the cheap plastic match to that giant yellow Pickett. I've had it since the 70s, and I've never needed to change the batteries even once. Ofc, that could be because I pretty much stopped using it once I got my TRS-80. ;)
    1
  16781.  @Mathologer  tyvm. will do.
    1
  16782. I find it quite peculiar that 2 is often such an important and exceptional prime. Obviously it’s the only even prime, but 3 is the only prime which is a multiple of three 😉 Does anyone have an intuition for why 2 is such a peculiar case?
    1
  16783. I like the coloring proof.
    1
  16784. Thanks. I always watch these on Sunday afternoons with a coffee. T shirts are always fun
    1
  16785. Brilliant!
    1
  16786. Nice bit of high school nostalgia here... Would you consider looking at Newton's divided difference interpolation formula as well?
    1
  16787. Definitely the nice logarithmic approximation.
    1
  16788. would love to see a video on the "wisdom of crowds" phenomenon
    1
  16789. still here... a weird 8-minute gap at the start of the credits had me intrigued me.
    1
  16790. My life bet on SUM of NO NINE SERIES is most memorable. It was jaw dropping : O I lost myself in the Counterintuitive World of Harmonic Monsters.
    1
  16791. 11:42 spoilers 9 4 17 13 9*17=153 4*13=104 9*13 + 4*17=117+68=185 and 9+4=13 4+13=17 The star/asterisk (*) is the multiplication operator in many computer programming languages From 3,4,5: r, r, r, l r: right; l: left first 2 are on screen/in video right from there gives: 1 4 9 5 left from here gives: 9 4 17 13
    1
  16792. Sydney based?
    1
  16793. I want someone to start making coins with that last shape. I remember when the UK introducted its 50p piece that being able to roll in vending machine was a consideration so they chose a "equilateral-curve heptagon"
    1
  16794. Wonderful video, thank you!
    1
  16795. Best bit was obvious: The story of the Hare (no nines) and the Tortoise (100 zeroes). But to be fair, whole video was very good. Loved the vanishing squares "proof" (of what I had guessed fwiw). Just wish I had been better at (more into) maths when I was younger. Watching your videos, 3blue 1brown, redpenblackpen etc I wish number theory, linear algebra, calculus and analysis had been taught that clearly in my Uni , and I wish you had all been around 30+ years ago! It's almost enough to make me want to re-run a maths degree when i retire! Please keep making maths accessible and fun. What you guys do here is so helpful to so many.
    1
  16796. Such an interesting and well produced video and topic. Your animation efforts are wonderful. Thank you! I hope someone can make a nice 3D animation of the Fibonacci tower and soldiers advancing uphill.
    1
  16797. I have a masters degree in computer science. This hurts my brain but nothing unsolvable here. The previous video was harder in my mind
    1
  16798. Great video!! Would love to see a video on fractal dimensions (i.e. Hausdorff and Packing dimension) with some mathologer animations. The tubes arguments I see in papers are a bit dull on the page at times.
    1
  16799. ok now do it with a circle
    1
  16800. Mallu😁
    1
  16801. ((x mod69)^23) mod69= x mod69!! why is that? I tried to use drawn Powerball numbers from wednesday to generate numbers for saturday's 1.6 Billion jackpot but this happened to return the same number as the input. 2, 11, 22, 35, 60 with powerball at 23. 5 numbers were choose 69 powerball is choose 26. Am I missing the reason this is not giving me more "random" numbers?
    1
  16802. The sum of digits trick for nine, of course, also works for divisibility by 3.
    1
  16803. no wonder when x^2 + y^2 = z^2 is circle radius formula, but the radius term is z^2 or radius is z
    1
  16804. That shirt is awesome!!
    1
  16805. Great video, I still have question though: towards the end, you gave us the condition with the normal vectors, which guarantees that our approximation for the the cylinder, has a surface area which aprroaches the correct one. I find it somewhat analogous to a uniform convergence condition, but I think that uniform converge is usually a sufficient but not a nesscery condition to exchange limits. So contining that reasoning, is there aome sort of a lantren such that the normals don't behave nicely, but the surface areas still aproaches the correct result? I don't mind changing the surface we are trying to approximate from a cylinder, to something else.
    1
  16806. Nice connection. A lot of the ways we handle swapping limits, infinite sums or integrals is to insist that our maths is well-behaved. An example, answering your question: a cylindrical lantern where the patch over which normals “misbehave” get smaller and smaller. For instance if there were b-1 “nice” bands and 1 “naughty” band (too skinny for its own good). Then the nice bands would have triangle normals converging to cylinder normals, the naughty band wouldn’t so you wouldn’t have uniform convergence of normals. However the nice bands cover most of the surface, so the area still converges. It’s seasonally appropriate for nice to outweigh naughty. Santa approves.
    1
  16807. Just spent some time with the larger algebras. The product rules for the diagonals are really pretty! Remarkable how easy it is to write out the product table for a given number of sides, though it took me working out the table for nonagon via quadrilaterals, then 11-gon via some algebra, then is clicked as I was writing out the 13-gon table.
    1
  16808. My head hurts 🥳👍👍
    1
  16809. This is insane. Mathematicians are insane. 🖤
    1
  16810. I was never that great at Math. But I love learning what ever I can about it. A constant fascination for me.
    1
  16811. I was Eagerly Waiting for Videos..And Yayyyy Hopefully I Have learned a new Intuition ❤❤
    1
  16812. You are "capo"😁... thank for bring it (too hard work)so beautiful and enjoyable chores ...👍🇦🇷🌎
    1
  16813. I stopped understanding at 15;00
    1
  16814. Where can I get that T-shirt 👕
    1
  16815. Can we use number walls to produce factorials?
    1
  16816. 369 - Unbalanced evil
    1
  16817. That's a really neat "dream proof" and great video as usual. I may go off to ponder what, if any, difference it makes when you have to miss one eyelet. Bunions, before you ask.
    1
  16818. 00:47 Yeah, Euler's formula would be even more beautiful if we rightfully redefine ℼ = 6.28... 😠😡
    1
  16819. I find it amazing that you actually made this video. This is certainly the best explanation of quadratic reciprocity I've seen. I have some suggestions, though. You introduced the concept of a primitive root, but you actually did not prove that the group of units mod p has one. The usual proof constructing a polynomial over the field is accessible IMO. Also, it would be nice if you mentioned that swapping any two cards changes the sign of the permutation, as well as the decomposition into cycles, which makes it a lot easier to follow the lemmas you use.
    1
  16820. You cannot always tile a board missing 2 black and 2 green squares. Cutting out two squares touching a corner could leave it stranded.
    1
  16821. Nice one. I understood ALL of it, for once.... Ps. RIP John Conway.
    1
  16822. is there an invarient regarding the colors?
    1
  16823.  @Mathologer  thank you <3 Frohe Weihnachten and in case you take a break till the first : Guten Rutsch
    1
  16824. Thank you for making this video...Mādhava is truly under-rated...the true father of Calculus ❤
    1
  16825. What if the 1/x shape is squished and stretched at a 135 degree angle instead of the 90 degree you used?
    1
  16826. Sometimes, math seems like witchcraft!
    1
  16827. This feels like a geometric differentiation.
    1
  16828. I really liked the "no 9s" fractal. That was really surprising but super clear after seeing the image. A great combination.
    1
  16829. Amazing
    1
  16830. I believe that I can create truly random numbers, not pseudo-random numbers, but truly random numbers. Do you think I'm wrong? Do you think it can be done?
    1
  16831. getting warmer and warmer to nature, which is numberless.
    1
  16832. Well, I'll say this --- maths would be a whole lot easier to understand if it were taught visually! Also interesting that an infinite area can simultaneously be itself, contain itself, be less than itself and yet be more than itself.
    1
  16833. Tree puzzle at 11:45 solution: 9 4 17 13 Used prime factorization. Path to solution from the start: right, right, left, left. rrll.
    1
  16834. I was just thinking about this the other day.
    1
  16835. time to play 2048 again!
    1
  16836. The biggest most important question I have it, where did you get that shirt?
    1
  16837. 6:35 almost looks like a hypercube
    1
  16838. Well that was a ride. I wonder when I'll be able to understand past a 20 minute mark in any master class mythology video... After college? Masters? Lol (After watching end): I'm an 11th grader in CBSE (Indian board). I made it around 15-20 minutes in this video haha. Also the animations really do help, and maybe a small note or too for some things you slightly gloss over (like the hw you give) might be helpful. Maybe a hint! Anyway keep up the great work!
    1
  16839. For $n$ cards, I suppose the maximum is the $n^{th}$ triangular number, by the natural interp. of flipping only the rightmost card. I’m all about the monsters, especially near Halloween.
    1
  16840. Fun, thanks
    1
  16841. It's all so elegant. A delight to watch and contemplate.
    1
  16842. I didn’t do homework in school, I'm certainly not doing it on the toilet I'm here to learn and poop.
    1
  16843. Wow! I wish I'd learned this stuff before I learned calculus! It's pretty amazing that you don't need calculus to explain why the area under the hyperbola y = 1/x is a logarithm, and you only need to evaluate a simple limit to define e. I was equally amazed by your definitions of the hyperbolic functions in terms of areas invariant to shapeshifting, which also doesn't require calculus. Pretty amazing stuff!
    1
  16844. i figured out this formula about a decade ago when i was still in school. this is the only way i can be sure pi has infinite digits.
    1
  16845. 1965 at CM Russell High School. I should have paid better attention.
    1
  16846. An opinion with no pi in sight is an onion.
    1
  16847. Most memorable: finding a sum for all positive number by using the greedy algorithm
    1
  16848. Algebra, now that scares me. Topology, now that gives me nightmares. Calc? I will willing take that any day I can get away with it.
    1
  16849. In all my years of math, I never knew what a hyperbolic trig function actually was (outside the “definition” of cosh and sinh in terms of e^x and e^(-x)) I guess all they wanted us to do was shut up and calculate… Thank you for enlightening me, that these functions actually mean something!
    1
  16850. The bible says the world was made with numbers, letters and words = gematria - universe (one word)
    1
  16851. Cool video
    1
  16852. Euler's constant: gamma = 0.5772156649... Its limit definition describes the conchoid of Nicomedes, with the natural logarithm part corresponding to black holes and the harmonic series to impacted dark matter during the most dense stage of a Big Bounce event. Its definite integral definition shows black holes that later on slow down a chunk of dark matter, with the dark matter signified by the floor function to coincide with the rotated, 6 alternating corner/cusp singularities. e to the gamma: this is one of the Ramanujan Infinite Sum processes, of which there are twelve. Not twelve days of Christmas! Just twelve. This part deals with mass exchanges.
    1
  16853. Most memorable fact is that there are more numbers with 9 than without. Still can't get my head around it.
    1
  16854. how is it not a curved cylinder?
    1
  16855. This video was absolutely amazing. I know calculus but matrix inversion but i think this video was pretty accessible. The animations and algebra autopilot is amazing. Look forward to the next video! Amazing!
    1
  16856. I read about Kempner series a few months ago and was truly fascinated by them. I'm glad more people will hear about them thanks to you!
    1
  16857. Triangle area = inradius times semiperimeter (just consider the 6 small triangles defined by the incircle). The 6 Conway points are thus simply the points where the circle of squared radius = squared inradius + squared semiperimeter about the incenter intersects a (produced) side of the triangle.
    1
  16858. This one is gem❤
    1
  16859. Which components can be extended to ellipses?
    1
  16860. So I agree that Euler's Identity absolutely NEEDS to be #1, why do SO many people write it in that form? Consider "e to the pi*i + 1 = 0". Here we have all the major constants (0, 1, e, pi, & i) plus addition, multiplcation, exponetiation and identity all tied together!
    1
  16861. 35:20 That's interesting. In my real analysis university notes there was a question regarding the same series but each term is raised to the power of s (1/1^s + 1/2^s +... + 1/8^s + 1/10^s + ...). The question was to prove that this series converges if and only if s> log10(9). Obviously this was way too difficult for me to prove :) . The most memorable proof is definitely the proof that this series converges when s=1.
    1
  16862. 2 ∞ + > I see what you did there.
    1
  16863. 18:09 M(n+1)=3*M(n)+2 and M(n)=3^n-1 Very pretty indeed =)
    1
  16864. e pi c
    1
  16865. Challenge: If n is odd, say n = 2k + 1, then (k + 1)^2 - k^2 = k^2 + 2k + 1 - k^2 = n. More generally, a nonzero difference of squares has to look like x^2 - y^2, where x > y, say x = y + d. Then x^2 - y^2 = (y + d)^2 - y^2 = 2d + d^2. If d is odd then this gives an odd number, and if d is even then this gives a multiple of 4. Using the above, if p is prime and p = 2d + d^2 it's clear that d must equal 1, which means the only way to express p is (y + 1)^2 - y^2.
    1
  16866. Swivel proof is best.
    1
  16867. Thanks for making numbers so approachable
    1
  16868. I watched all 50 mins, but I certainly did not understand it all (maybe the first 20 mins ish). I do not have any mathematical background, I barely scraped through A-Level Maths. But I do find maths fascinating, particularly the surprising results and beautiful proofs. The animation right at the end was nice
    1
  16869. Gamma > 1/2 because each blue bit takes up more than 1/2 of its bounding rectangle.
    1
  16870. Epic!.........As usual.
    1
  16871. So +m/-n gives ln(m/n), but what about +a/-b/+c/-d? Say, 3/-1/3/-2?
    1
  16872. A ray of 5 star is 144, not 180 degrees?! What are you guys calculating??
    1
  16873. [3rd root]e * (pi)^3 / ((3(3+1)/2)^2 is "really close" to the sum of cubes (1,202020694), apparently though, it shouldn't be... But how nice would it be if e actually appeared in that equation too.
    1
  16874. I saw the π²/6 and thought 'well 6 is 2(2+1) I've seen that before - let me plug in the value for the formula one higher', and got out a factor, cubed it and saw 2.718... and was like HOLD UP - apparently just a coincidence. (But so close? Man..)
    1
  16875. The swiveling proof is more elegant, but the colour proof is simpler.
    1
  16876. Some interesting generalizations. I like it.
    1
  16877. Great video! While I was watching this video, I was thinking about Dijkstra's Algorithm.
    1
  16878. Oh Grrr... I spoiled this. About your question, if you are saying that we will get a primitive with a zero, trivial set lol, well, I already commented epsilon, generator. And Oh a necklace. It's maybe similar to a program I wrote a while ago. Animated and all, aproximación a circle from a rectangle and cutting it into two sets of a, b, c elements and linear transformations.
    1
  16879. Most memorable: No integers in the partial sums of the harmonic series!
    1
  16880. Why does Fibonacci not start with "0,1"?
    1
  16881. Wow, this video had some particularly beautiful connections. Very very impressive results. The continued fractions and rectangles was new for me and much appreciated. I think I didn’t catch the measure of better approximation, but it seems like you might do it in more detail in a different video so I’ll try to check.
    1
  16882. Ho Ho Ho for you too.
    1
  16883. Nice! 2019 pops up 15 times in the first 1E6 digits of π, so read the fine print on the t-shirt ;-)
    1
  16884. That was excellent. I'm not always a visual learner, but this was a great way to get into ln and e.
    1
  16885. 10:47, ive already proven that identity without ptolemy by stacking 3 heptagons together and finding similar triangles. I thought I found the best proof but I guess not anymore.
    1
  16886. please red cross puzzle answer please
    1
  16887. Make the 3D version of the mandelbrot set.
    1
  16888. Does this have anything to do with roots of unity and permutations? Galois Theory? It seems like the directions of your pentagon permutations are very similar to Galois Roots of unity… It makes sense that not all permutation of roots would still satisfy some initial condition. But what would one of the permutations that does not preserve the starting condition look like? Much of the advanced math here is beyond my level of comfort. I was a single class short of a math minor, but that was 20 years ago, and my profession uses at most basic Algebra. Sometimes I have to set up ratios and solve for X. I do find the animations and pictures to be very helpful with moving step by step through the more complicated math behind them. Even when it might be something I know or am familiar with, I enjoy watching the videos as the simplicity and explanation are amazing even when discussing very complex ideas. I always learn something new when I watch! Thank you! Amazing job as always. This still smells like roots of unity to me, but I never quite got to Galois Theory…but I like how you answer without even having to get to in depth with advanced math concepts.
    1
  16889. i read this as heroin formula and clicked
    1
  16890. One of my favourite math class pranks is to go up to someone pretending to be confused, ask them "Hey, can you help me find the sum here?", and then hand them something that's really just the harmonic series in disguise.
    1
  16891. He is back!!!
    1
  16892. Queen of hearts?
    1
  16893. Your cylinder surface is not smooth, the direction of the normal has a discontinuity at the top and bottom.
    1
  16894. calculus is easier than linear algebra
    1
  16895. 30 minutes of Mathologer? Count me in!
    1
  16896. It seems that 2 plays the role of e in discrete sequences. Then, how do we get from 2 to e? If you define 1/2-discrete sequences as those that take value of multiples of 1/2 (and thus discrete sequences are just 1-sequences) then you can see that now 9/4 plays the role of e. The road to e seems clear at this point. or isn't it?
    1
  16897. I most enjoyed Kempner's proof! So inexorably convergent.
    1
  16898. the procedure at around 24:00 with the arrows reminds me of plate tectonics
    1
  16899. If you add together the connected numbers in the vortex 1+2+4+8+7+5 you get 27 and 2+7=9 Interesting?
    1
  16900. OH MY GOD. now you're saying Hexagoras.
    1
  16901. The most memorable part is definetely the last one, that mentioned the fact that “exactly 100 zeros series” is greater than the “no 9s series”... who whould have thought?! crazy!
    1
  16902. Ottimo video. Per un attimo mi è sembrato di vedere Heaviside all'opera
    1
  16903. Wow, can it be how crystal nucleation works in nature?
    1
  16904. "Wait what does 1/x have to do with the logarithm... oh."
    1
  16905. I think the coloring proof is more straightforward and understandable, but the rotating proof is prettier
    1
  16906. wow
    1
  16907. I'm going to start a Youtube channel called DumberEveryDay
    1
  16908. Strangely -1-2-3=-1×-2×-3 🤔
    1
  16909. As I have always lived my life, "Math is Life, Life is Math". Everything we do revolves around Math, like it or not!
    1
  16910. Great video! Using the fact that there are an odd number of solutions to p = x^2 + 4yz shows that there cannot be 2 ways of writing a p as the sum of two integer squares. There may be an odd number of ways though - which is why the proof in the final section is necessary?
    1
  16911. The pie lover in me wants happiness too 😟
    1
  16912. Great video...but the Mathologer called a three sided polygon a triangle but a seven sided polygon a "sevengon"! Why not a heptagon (or threegon earlier)?
    1
  16913.  @Mathologer  less serious more casual in the spectrum of comments. It struck me as odd. One more reason I love math more than the English, Latin or whatever language used to talk about it.
    1
  16914. You need to speak louder
    1
  16915. I've made a start on implementing that dancing algoritm: https://bitbucket.org/dj-djl/arctic-circle-generator/src/master/ I'll tidy it up, shove an interface on it and host it in the coming days
    1
  16916. Would love to see if this any similar applications in three dimensions.
    1
  16917. 11
    1
  16918. Please do make the continuation of this video
    1
  16919. Yeah! Why DIDN'T they teach this in my college Calculus? Neither Calculus for non - engineers nor Calculus for engineers had this!
    1
  16920. Mandelbrot: length of the boundary of a country depends on the size and scale of the measuring device.
    1
  16921. This video is really amazing. I give it a 10/10 grade to all parts of the video. Actually, I give a 10/10 grade to all of the mathologger videos. You make it quite obvious to see that knowledge behaves as a divergent infinite series. I only hope my brain doesn’t and up exploding some day! 🤯
    1
  16922. That is very nice! My favourite sequence is the Catalan Numbers, and really, that would give for a whole series of videos...
    1
  16923. If they chante 0 and 1 it will hit all 1s. It is also a finite max number
    1
  16924. This is so freakin’ cool.
    1
  16925. Also the king's method of finding the magic permutation is very cool.
    1
  16926. Aaaand the decomposition of the numbers into modS times there divisor plus the remainder is a very neat way to show how the final permutation along the diagonals is exceptionally special
    1
  16927. Thank u. I think this also can be built upon.
    1
  16928. (n+1)^2-n^2=2n+1 which can be any odd number. And for primes it's unique as x^2-y^2=(x+y)(x-y). x-y<x+y so x-y=1 and x+y=p, which gives us a single unique solution (which is the one presented higher).
    1
  16929. The most memorable to me is the greedy algorithm because I'm an algorithm junky ;-)
    1
  16930. Well down. Very nice explanation. 👏
    1
  16931. The chords at the beginning remind me of Kate Bush's Babooshka
    1
  16932. Hi Yes Peter Mondren Thanks, Fantastic, As Won Of My First Art Projects, We Must Look At Nature First ? Nature Is Oor GOD Of Invenchion Our History Box Is Mill Of Years? Thanks Sir.
    1
  16933. After watching the video I was tricked a further 10 times
    1
  16934. 9:40 wont stand up if coyote's on top, chasing road runner 9:50 the next building to go up in Dubai ?
    1
  16935. Interesting that the three lines going out also look like hydrogen in an atom smasher??
    1
  16936. Merry Christmas...🎄
    1
  16937. @36:32 that visual proof was truly mind-blowing edit: 'demonstration', not 'proof'!
    1
  16938. For the last question: the next two terms are 55 and 144, so it seems to be the Fibonnaci numbers but skipping every other term. I have no explanation (yet) for why that would be the case, and it is possible this is just a coincidence and the pattern changes but that is my guess.
    1
  16939. Hi Mathologer project, would it be possible to activate the automatic youtube audio transcription to spanish or another language, the whole world will thank you for it.
    1
  16940. Now I want to see an actual shoe, on a foot or foot mannequin, with the devil lacing.
    1
  16941. The 2xN chessboards follow the fibonacci sequence I think
    1
  16942. I will also become I patreon once I graduate and get some money.
    1
  16943. Excellent!
    1
  16944. why does the devils lacing remind me of neural networks?
    1
  16945. Ah, the Ptolomey I.
    1
  16946.  @Mathologer  I love the I one, but maybe your channel will do me good 🤣 Tnx for the explanation 🙂
    1
  16947. One of my teachers had a stack of uniform thin flat sticks in the leaning tower configuration on top of a shelf in his office. People would assume they had to be attached to something, since most of them were entirely over the edge, and they'd knock them over trying to figure out the trick.
    1
  16948. isn't the derivative of the function for the area equal to the circumference because the surface area is the integral of the circumference? i'm confused
    1
  16949. pajeet's mathology
    1
  16950. Endangered specie? XD
    1
  16951. 3-6-9, damn she fine
    1
  16952. Amazing explanation!
    1
  16953. So well explained, thank you. In a previous video that featured two coins and one rolled around the other. You said you made the animation in keynote. Did you also use keynote for the revolving wheels and stretching rubber band?
    1
  16954.  @Mathologer  Amazing. Your keynote and math skills are on par!
    1
  16955. The "crossed over" cases from about 23:00 onwards arise only if one names diagonals and sides arbitrarily and thus abandons their original meaning. Take one of the side pairs - let's say b and B - of a convex quadrilateral and call them now c and C, and rename the original diagonal accordingly - thus here b an B. Then you get your "crossed case". The caveat at the end about cyclic quadrilaterals is an artifact of this arbitrariness. Take the proper equality for the cyclic case aA+bB=cC. Now just switch the names of b and B with c and C. The old equality now reads aA+cC=bB (and holds true, we just changed the names). Now write the inequality for the crossed case as aA+bB>=cC (new names), then we can substitute into this the old (renamed) equality as cC=bB-aA and get aA+bB>=bB-aA or aA>=-aA, and with a and A being positive we get 1>=-1. Equality is obviously not possible, but the greater case is true. Therefore, in the crossed case aA+bB>cC. Thus, if one calls one of the side pairs of a cyclic quadrilateral "diagonals", and those diagonals "sides", creating a crossed case out of a convex one, then one gets the result at about 24:39. There may be reasons for doing this, but here it just confuses in my opinion...
    1
  16956. My vote: refined Kempners proof only because i knew Oresmes'
    1
  16957. sinh XD
    1
  16958. I am unsubscribed to this channel. I don't remember unsubscribing. Am I alone? May be I mis clicked it.
    1
  16959. For a good 3 mins there I was making equilateral triangles using the 3d grid printed on your shirt (Amazing how well all of those cubes were lined up, kudos to the artist for representing like 5 different things in one brilliant shirt) anyways you were talking about how there is no equilateral triangle while im creating them and wondering WTF, then you went on to the 3d grid and explained to me exactly what my mind was doing and why it works. Awesome vid.
    1
  16960. Now tell us the significance of 2 + 2 = 2 × 2 = 2 squared.
    1
  16961. 37 = 7² - 4(3ꔷ1)
    1
  16962. Best parts:- 1)odd/even fraction is never an integer. 2)Gamma
    1
  16963. Very nice stuff love the visual proofs!
    1
  16964. At 2:33 you double the 1x2 but you did not double the 1x3 and you did not double the 1x2 + 1x3 = 5. Where does that mysterious doubling of only one term come from? I still have to watch the rest of the video to see if it's brought up, but on this first go I had to stop and comment so I wouldn't forget.
    1
  16965. How can I get a to infinity and beyond shirt I love it great fun!
    1
  16966. Height to base of equal sided triangle is h=√3×a. So, there is no rational numbers for h and a. Isn't it enough to proof there is no equal sided triangle on square grid?
    1
  16967. It seems that Mathologer has a lot of time on his hands lately! Maybe not so fun for you, but great for all of us :)
    1
  16968. Thanks for a great video.
    1
  16969. I love problems involving Φ
    1
  16970. Great video! As I'm wrapping up Calc II, this was a really great way to extend what I've seen of infinite series.
    1
  16971. Burkard you are an amazing and wonderful person. Thank you for the great gift of your youtube channel :-)
    1
  16972. Another great video. :) As for feedback, I was able to get to the end quite comfortably and felt that the derivative of difficulty wasn't too high. Still scratching my head over more error correction terms giving worse answers after a point, though. :P
    1
  16973. Conflating units is still not a paradox. Volume is not convertable to area. The whole proposition is nonsense from the start. A two dimensional plane has infinite surface and zero volume, there is clearly no paradox in that.
    1
  16974. Hello! I am no expert. What i have learned about "e" (from other sources too) is that it behaves like a unit of logarithm
    1
  16975. Im not much at mathematics but the square in the diagram always shows one point higher in the right and the other lowe on the left. My brain just says NO.
    1
  16976. Greedy algorithm.
    1
  16977. That is pretty interesting. I'm going to share it with a couple of lawyer friends of mine, one of who is a country-person of yours. Oh, and I also made it all to the way to the very end.
    1
  16978.  @Mathologer  Wouldn't it just?
    1
  16979. why do people who love SI units because everything's decimal not know what to call a decagon?
    1
  16980. Awesome video, I never seen so detailed explanation of this formula, which I did use before, but I didn't really fully understood it fully. Also, are there some interesting matrix identities between the two matrices we got?
    1
  16981. About 5:00, it's not clear why the harmonic series being infinite means that the surface area is infinite. Is the idea that if you looked down on the horn from above then you would see more than H(x) surface area and thus A > H = \infty?
    1
  16982. Ok, random maths question that buggs me: with the hexagonal tiling we've shown that we can always interpret the tiling as a pile of cubes. What I'm wondering is, does the average volume of the cubes converge to something? Is it something meaningful? Is the "average" 3D shape produced by a random tiling of infinite size any interesting shape? I'm just wondering what it looks like
    1
  16983. Wiki taught a neat equality (A-B)(C-D)+(A-D)(B-C)=(A-C)(B-D) as we treat a cyclic quadrilateral ABCD as 4 complex numbers. Then here's how it relates to the cross-ratios. Let z1=AB*CD, z2=AD*BC, and z3=AC*BD. Through some experiment I find somehow the ratio (z1-z3)/(z2-z3)=λ is always a real number, i.e. they're colinear. I was so perplexed until I realized there exist a Moebius transformation to map the circle into a line. And λ as one of the cross-ratios is preserved and obviously real. It's just like the last animation that turn it into the degenerate case with 4 colinear points. Now, of what use is this random fact? It's quite useful. Say we're given the points A, B, C, then their circumcircle (being the loci of D) can be given by a parametric equation of λ, a real number. According to said relation of the z's.
    1
  16984.  @Mathologer  Just a line of thought (winks).
    1
  16985. For the coin paradox it's also interesting to consider a coin rotating inside a coin of the same size. The amount of rotations in this case would be zero. This falls in line with the idea of 1 rotation for the circumference minus 1 rotation for rotating on the inside.
    1
  16986. 👍👍👍
    1
  16987. Yes, but an approximation with triangles on a sphere, I don't know, because the sum of angles of a triangle on a sphere doesn't equal 180 degrees. And neither is the parallel postulate true anymore,so I can imagine this is somehow related to this lantern problem? Anyway, happy holidays.
    1
  16988. I couldn't easily find it; may somebody please explain the 50% of people have their shoes wrong or give a link to where Ian explains it?
    1
  16989. I liked the block stacking the most, it was a neat visualization. That said, there are two things I kind of wish you’d brought up. 1. the “ant on a rubber rope” paradox: an ant is walking across a rubber rope at a speed of 1 cm/s. The end of the rope he is walking towards is being stretched at a rate of 1 m/s. Will the ant ever reach the end of the rope? Turns out, the answer is “yes”, and the harmonic series is part of that proof. 2. With Euler-Mascheroni constant (gamma), to this day we still do not know if that number is rational or irrational. It just hasn’t been proven either way yet.
    1
  16990. I understand that Lucas was also the first person to give the Fibonacci numbers that name. Is that correct? Or did someone explicitly call them that before Lucas?
    1
  16991. I've always thought the Tower of Hanoi was a somewhat... well, BORING puzzle. How wrong I was! Thank you for this interesting deep dive into a world I'd not noticed before.
    1
  16992. Note that for odd m, an even root of half-unity exp(i k \pi/(m+1)) must partition the upper half plane into two equal regions. This means that one of these roots, say p_0, must lie precisely on the imaginary axis so that its real part vanishes, whence by Euler's formula cos (p_0 \pi/(m+1)) = [cos (p_0 \pi/(m+1)) ]^2 = 0. If both m and n are odd then we get a factor [cos (p_0 \pi/(m+1)) ]^2 + [cos (q_0 \pi/(m+1)) ]^2 = 0 in the formula hence the entire product vanishes. Kastelyne's proof is based on statistical models of a half-filled spin dimer (as a conformal Z/2-gauge theory), in which the formula can be interpreted as a certain prefactor in the entanglement entropy of the system near criticality. This is how physicists can compute sophisticated arithmetic and enumerative invariants in geometry without even knowing what a variety is. In essence, Gromov-Witten theory is based on this observation.
    1
  16993. I’d love to hear about all the great mathematical discoveries made by sub-saharan african cultures. How come we never hear about all the sub-saharan african proofs for pi, etc? I’m talking about centuries, millennia even, before white people ever arrived in Africa. And Egypt is not sub-saharan africa. Since I’m transracial and I identify as sub-saharan african, I would either like to know why my people are the only ones to make no mathematical discoveries, or hear some clever excuses for our failure that I can tell people to justify it. I’ve heard it’s merely that we avoided math since math is racist. But that can’t be the reason, because only white people can be racist, and there were no white people there, so "racism" wasn’t a thing until 1619, as all good virtuous people Know.
    1
  16994. so it's like a fractal with infinite perimeter and finite area.
    1
  16995. Mathologer videos are pure entertainment through learning.
    1
  16996. If continued fractions are no longer taught in schools shouldn't they be called discontinued fractions?
    1
  16997. But how do we prove that the centre of the incircle (i.e., incenter of the triangle) is also the same as the centre of the bigger circle? I do not see an entirely obvious reason why the incenter should be the centre of the circle... Isn't that step necessary to proceed with the first proof? It felt like you skipped it...Could someone please explain? I know I am missing something obvious here...but I can't figure it out...I know that the incentre is formed by the intersction of the internal angle bisectors of the triangle and that the perpendicular of a chord passes through the circle's center. I'm stuck after this. @Mathologer
    1
  16998. At 12:30, one of the green squares of the board (the green 6) disappears from the image for some reason.
    1
  16999. That's a really beautiful and intuitive proof. Congratulations for that!
    1
  17000. This is SO COOL!!!
    1
  17001. Well I can't do infinite, but 256 bit IEEE754 will be good enough, right? :P I'm terrible at UX and graphics, so I think I'll pass at this one. The fun part is implementing all the neat or discipline specific scales anyway. Until at least some very recent graphing calculator models, but realistically not till smartphones, slide rules were still the best way if you had to do a specific thing regularly
    1
  17002. First there’s no such thing as a paradox, second, you’ve taught us well that you can get any sum you want depending on how you slice and dice the original series and if the those series aren’t convergent then the results are essentially worthless.
    1
  17003. Thanks!
    1
  17004. One thing I love about methologer is that Burkard always provide some accessible paper for those who willing to investigate further for the topic(s)!!!
    1
  17005. Excelente !!!, muy bueno !!!, fantástico!!! 👍👍👍
    1
  17006. I really liked the approximation. I’d seen gamma before but never knew what it was.
    1
  17007. 18:52 when pairing the windmills, why can we always pair them?
    1
  17008. Would making a Shwartz lantern from a beer can and telling everyone about misterious increase in area without blowing it up, like a party trick with surface calculation, get a good student laid?
    1
  17009. This is the first time I’ve encountered Mossner’s Miracle, and it made my day too! Thanks for presenting a heady maths topic in such a way that a layperson can understand! That was really cool!!
    1
  17010. This is a really nice proof!
    1
  17011. 25:40 so wait. 3333.333333… + 6666.666666… + 1 = 10001. Oh… adding from the left messed me up again!~
    1
  17012. Sorry for sidetracking you from your monster-video with my heptagram question, the resulting heptagram videos were fascinating to say the least. :) Healthy new year wishes to you and the company you keep, Burkard!
    1
  17013. I proved this myself, quite nice, I particularly like the formulas for the green red and yellow lengths at 7:22 : calling the sides of the original triangle a, b, c, those lengths are (-a + b + c)/2, (a - b + c)/2, (a + b - c)/2
    1
  17014. Thanks for this Christmas present! 🎁
    1
  17015. At 6:49 you show a list a numbers said to be primes, but many of them are even. How is this possible?
    1
  17016.  @Mathologer  O.K., thanks, and I see now the last digit is 1 :)
    1
  17017. The arbitrary sequence of doubling the two, s has it's decimal value with no.'s Matching to decimal expansion of 22/7, 🧐🧐
    1
  17018. Instructor: it's obvious. Easy. Meme kids : I don't think that word means what you think it means.
    1
  17019. i knew both proofs of the beginning, i first encounterd the second proof in highschool at age 16 i think and the second proof i came across a few years later in a book about fraktals i read in my first year uf studying math in university.
    1
  17020.  @Mathologer  I´m from Germany and in regular schoolcontent I think we just saw one proof in total, it was a proof, the sqrt(2) is irrational. That proof for the Pythegorean theorem was showen to me by my math teacher, because i asked for a proof and he knew my interest in math. So I don´t think it´s so common to show that proof in german schools (or at least it isn´t in NRW).
    1
  17021. From a fed up locked down Melbournian, thanks for all your videos.
    1
  17022. Who else joins the club of watching to the end for reward?... that was cool! - who else but Mathologer to give me that split-second little insight/kenning of 4D shapes. Felt like I could aaallllllmost picture it there for just a sec' before it slipped away. Is that what it's like to be a genius?.. to be able to hold that kenning in your head for 4D for extended periods of time?... man I wish I could do that.
    1
  17023. How can I get that shirt?
    1
  17024. I thank the Mathologer team to put out this video and give overly due credits to the Indian Mathematicians they were ahead of their times.
    1
  17025. need my maths fix for the weekend
    1
  17026. If 666 is a symbol of evil, then 333 is half evil? 😁
    1
  17027. 50:50 not 50:00
    1
  17028. For the challenge at the end of Chapter 6: the first 7 powers of 2 work: 1, 2, 4, ..., 64 because binary representations are unique
    1
  17029. I came to know it now. But the question is why now?
    1
  17030. I had a slide rule in the mid '60s, but wasn't very good with it. This video makes a lot more sense to explain it and I just might get better with it 'IF' I still used one.
    1
  17031. 15:08 circles are like onions i really thought he was going to make a shrek reference
    1
  17032. So love this channel, never thought I would giggle and grin so much to mathematics.
    1
  17033. The pigeonhole principle is an "at least" statement. If you have 4 pigeonholes and 5 pigeons, then you're guaranteed that some pigeonhole contains at least 2 pigeons. It tells you what is guaranteed to be true. Because worst case scenario, all 4 pigeonholes contain a pigeon, right? But then there's 1 pigeon leftover, so it has to go in an already-occupied pigeonhole. But there are "better" scenarios. Maybe all 5 pigeons are in the same hole. Sure. But the most spread out they could be, there is a hole with 2.
    1
  17034. Greetings from your Home Town of Würzburg from the Julius-Maximilians-Universität of Würzburg. I am an engineer graduated and I love your videos. Keep them up.
    1
  17035. you could also start with 100 at 1/0.01
    1
  17036. Was about to skip this video... What a MISTAKE that would have been. Amazing stuff, congrats!
    1
  17037. Wow Amazing 😀 Mind blowing visuals!
    1
  17038. Je réfléchis aussi à des ombres et des vides portés .
    1
  17039. I made it to the end 😏 was looking for some paypal to donate but couldn`t find any.
    1
  17040. 0:55 i got 50
    1
  17041. I like Kempner’s proof animation a lot.... actually I like the whole video a lot !!
    1
  17042. Yeah, Math is not boring
    1
  17043. I guess Powell did not use floating point when calculating 10**-5 in 13.5 hours on the school computer of "Dr Challoner's Grammar School, Amersham, Bucks" 🙂
    1
  17044. Fascinating !
    1
  17045. Maths can be as beautiful as a sunset, just in a different way.
    1
  17046. After some bit long time ❤ waiting jusst for vedio
    1
  17047. Second proof for me- that's basically how I was able to more or less prove the ends of the struts were on a circle!
    1
  17048. Every chapter is more mind blowing than the previous one 🤯 (except the first one... because it's the first one)
    1
  17049. What a clever Earthling you are...
    1
  17050. “Terrible Aim”, for me, is most remarkable. That the ever more densely spaced partial sums of the harmonic series are integer-phobic. Reminiscent of the Dirichlet function.
    1
  17051. Do you have onions in Australia with roughly spherical shells? The ones I know have inner shells which are smaller along two axes, but they still reach from the bottom to the top of the onion. This reminds me of the math joke "take a spherical cow"... 😉
    1
  17052. I liked the video a lot, and the t-shirt too! Any chances of a link for buying it?
    1
  17053. The snowflake curve is an easier-to-grasp example of infinite perimeter but finite area....one can surely rotate it...The vuvuzela remarks are great!
    1
  17054.  @Mathologer  I think so, but I'm just throwing that out there...Great video, btw!
    1
  17055. 15:57 It adds an area of 1 for every step. Meaning here the area is 5.
    1
  17056. Thanks for the nice video Christmas present! Looking forward to more good things.
    1
  17057. 1. No; you can isolate a corner 2. The product contains a zero factor when j/(m+1) = k/(n+1) = 1/2, namely, the very last entry. 3. 2x1:1 2x2: 2 2x3: 3 2x4: 5; I suspect these are Fibonacci, for similar reasons as say Fibonacci cubes (a very lovely sum of binomial coefficients first found by Lucas that comes from counting veery precisely how many new combinations are added by adding one column)
    1
  17058. I vore for Tristan fractal
    1
  17059. Calculus by geometry : imagine a cube of side x, with one corner at the origin and the edges aligned with the axes. Now, increase the length of the sides by dx, how much does the volume change by ? Easy, you've just added 3 slabs to the cube, each of area x^2 and dx deep. So the change in volume dV = 3x^2 . dx (ignoring higher order differentials !)
    1
  17060. Wonder if it could be smashed with twin primes conjecture.
    1
  17061. Thanks Mathologer. Amazing and high quality presentation as usual. BTW, for 9-gons the relations are t = r+1, r+t = r*s, s+t = r*t, for 11-gons are r^2 = s+1, r+t = r*s, s+u = r*t, t+u = r*u.
    1
  17062. Finally, an explanation for coshnand sinh
    1
  17063. 50:09 Burkard be like https://youtu.be/f3kL16kZJrg?t=11s
    1
  17064. First Like
    1
  17065. No integers in the partial sums must be the winner here.
    1
  17066. I used Heron’s formula for final exam today. My brain fried after the exam. Ty :)
    1
  17067.  @carly09et  Let's keep this apolitical. I'm just suggesting that no one would want to give out big loans when the Talmud system is in place. The Talmud system may agree with game theory as far as dividing the estate is concerned, and may be natural in that sense, but no one would let themselves get into that situation in the first place. The proportional system does not have this problem and is what the Talmud system would degrade into in practice.
    1
  17068.  @Vaaaaadim  Good point. It'll come down to the purpose of the loan I guess.
    1
  17069. Thanks
    1
  17070. My favourite part of the video is definitely that unusual brick stacking. Great animations as usual!
    1
  17071. .. and in 3D please
    1
  17072. To me it's obvious that gamma > 1/2 because the 1-square is made up of rectangles that are split into a white and blue part, and the blue part is always convex in shape (because the 1/x curve is concave), so the blue area must be at least half of the rectangle.
    1
  17073. I have an old circular slide rule (slightly battered by a life in my school bag at the time) which was gifted to me by my uncle. Unfortunately (perhaps) I never really had much need to use it, as electronic digital calculators were finally permitted in class and exams in our school literally the year that I reached the point in our curriculum - we all got a Casio FX-82, the original model. But a circular slide rule can do a lot more than just multiply and divide - all the other scales on the slide rule serve a purpose as well.
    1
  17074. Reminds me of groups and their Cayley diagrams with multiples cycles. They have a name but, I don’t remember what it was.
    1
  17075. Could this approach be applied to numerically solving differential equations? Here the contested part would not be split in half, but split according to some function of derivative order.
    1
  17076. 🤭
    1
  17077. Your lecture gave me an idea. Is it possible to combine the blackmailer paradox, which is also a theory of Aumann, and the Talmud's approach to reach a further conclusion?
    1
  17078. When I was 11 years old, I had a new teacher who told us a story about being given the same problem on his first day of school. He went home crying, worried he would fail to answer the question. He told us his father then showed him a formula for doing so and he used it the next day to impress the teacher. Well I went home that night worried that I might be asked to do the same. That night lying in bed, I figured out the formula myself and went to school waiting for my teacher to pop the question. He never did and I lost my chance to achieve Gaussian immortality.
    1
  17079. With a round hangar with 2 hands in the centre, that necklace would make a great clock face.
    1
  17080. What I didn't get is the need to increase both p and b to infinity in order to get a "nice" convergence in the case of the cylinder. Doesn't it converge with b=1 just letting p go to infinity? Merry Christmas dear Mathologer!
    1
  17081. Oh my GOD. Mathematics WITHOUT NUMBERS!!!
    1
  17082. 22:47 - pic on the right looks like a very angry cat 😃
    1
  17083. Lovely video. There's a lot of math magic here, but it's positively pleasant to watch. Some feedback, as requested: I'm a Physics PhD student, currently teaching first year university calculus classes. I've never really encountered the Euler-Maclaurin formula before, but its consequences I've seen pop up a number of times. It's very interesting seeing where it all comes from, both mathematically and historically. I liked the tempo of this video, and together with the animations it makes it a breeze to follow along. That is - to sort of see how the steps go. To see exactly how they go I'd need to grab a notebook and write/derive along :D. However, somewhere around the 26 minutes mark I sort of lost track of why we were rewriting all these sums in the first place. It's easier to keep track of where you're going with a derivation when you're the one doing it than if you're just watching it. Maybe when you're multiple steps deep into a derivation it'd be nice to sort of recap the high-level kind of steps you took to get there, it'd make it easier for viewers to remember "oh right, that's how we got here". But I'm just nitpicking at this point---the video is already very good. I look forward to seeing the next one, even (or perhaps especially) if it's again an hour long! :)
    1
  17084. I had a circular slide rule in the 1970's
    1
  17085. Horn has finite volume and infinite area, no paradox on that. The paradox ONLY comes when you claim you can cover it with paint, because paint is not a math concept, paint is not well defined in math. The paint in real life has a minimal size, its not a collection of sizeless points. Even assume the paint is made up by points, you still can't conver the horn, because the horn is infinite long and nothing travels faster than light.
    1
  17086.  @Mathologer  Maybe I am just not normal, but I never find a pure math concept, like the horn, with finite volume and infinite area a paradox at all. It's just what it's. Also, if you define paint as some math concept like in your above reply, then it has lost all the real life connection. Because the paradox mandates that you POUR the paint into the horn in order to cover its surface completely. That POUR action truly matters, without that, you can't link to the volume, you can't fabricate any paradox.
    1
  17087. Going to predict the formula for the coin scenarios. R is the radius of the stationary coin and r is the radius of the rolling coin. If the rolling coin rolls on the outside, the number of rotations is R/r + 1, and if the rolling coin rolls on the outside, the number of rotations is R/r – 1
    1
  17088. With seven pigeons, by the same argument as before we can find an equator that sorts five pigeons into one hemisphere (including the equator).
    1
  17089. For any finite amount of terms, adding up both summations isnt the same as adding up 2 terms of one summation and 1 term of another. Therefore you cant expect the combined summation to cancel out to 0 for an arbitrary number of sums.
    1
  17090. Magic!
    1
  17091. should be HO*3 not HO^3. It's "hohoho" not "ho to the ho to the ho"
    1
  17092.  @Mathologer  oh^3 ... I failed, you're right, somehow my argument made sense to me then, I was thinking in coding, not math.
    1
  17093. 6:11 No, for example, it becomes impossible if you remove the two black squares touching the green top left. And the top left square has not been removed.
    1
  17094. incredibly simple and beautiful proof using contradiction and geometry. I love proofs that use symmetry.
    1
  17095. El hombre que conocía el infinito
    1
  17096. Good morning 🌞 Merry Christmas 🎄
    1
  17097. Nice Ramanujan shirt! 414^3 + 255^3 = 423^3 + 228^3 = 87539319
    1
  17098. Well first I see this and thought I understand this and it is based on a paper in nature so I must be a good mathematician. Then i went and checked the paper. The result is that I am completely illiterate in math and mathologer is just a good teacher
    1
  17099. Differencing the terms in a sequence to try to derive a rule of propagation is a strategy I came up with when I was about 14 ummmnmnnmmnn, about 59 years ago. =:-O
    1
  17100. I thought the lack of the viral videos' explanation of digital roots made them quite DR(a×b) -> DRab -> drab relative to this one. Keep up the good work on math-related videos with clear explanations!
    1
  17101. Gorgeous! These keep getting better and better. SO hard to pick one...Gamma or the No-nines sum?? Gamma it is. BTW, my handwavy demonstration of Gamma being greater than .5 is that in the rectangle brought over from each integer reciprocal bar, the portion corresponding to the partial sum of terms adding to gamma is always convex and thus larger than the leftover bits and thus must represent more than half the sum for every term. More analytically, if we can show that the area of the trapezoid under the straight line joining 1/n with 1/n+1 is always bigger than ln(n+1)-ln(n) then it follows that gamma is greater than half of the bounding rectangle. For n=1, ln[(n+1)/n] is .69 and (n+.5)/n(n+1) is .75, and it stays larger for all greater n. Any help generalizing?
    1
  17102. i want math to gift me stuff
    1
  17103. AHH! Scary thumbnail!
    1
  17104. i suggest using a dark background, it would make your videos nicer to watch at night
    1
  17105. @6:00 looks like (4k+1) and (4k-1) to me. Which immediately moves my mind to the thought that some primes can be written as sum of two squares, and others -- as a difference of two squares. Here are some: 4-1, 16-9, 36-25, 100-81, 144-121 (no, signor Fibonacci, not this time), 256-225, 361-324 etc.
    1
  17106. Well, yea, I could figure out it wasn't zero just by looking at it. Don't ask me what it would factor to, I don't know calculus. But, the series will always have a positive answer, and not be zero, because you're infinitely subtracting from a larger fraction.
    1
  17107. What was interesting to me was there is one mistake at first, then 2 then 4 then 8 but unfortunately then it digresses from this nice order.
    1
  17108. 4k+1 and 4k-1 (or as you say 4k+3)
    1
  17109. Where at Monash is your office? Would be cool to pop in and say hi sometime!
    1
  17110. Some of the video reminds me of Mondrian. But a more on topic question: does the Chinese manuscript offer a *deductive argument*? The sources I've seen discussing it suggests the presentation functions more as a way to induce from examples. (And I suspect that's why some might oppose some of the contemporary visual proofs.)
    1
  17111. 10:23 Cos(pi/2)=0 When both j and k are at last step That is when j is (m+1)/2 and k is (n+1)/2 When we put both in brackets, m+1 and n+1 cancel leaving pi/2 Cos(pi/2)=0 and 0 squared is 0 and we get everything in brackets to reduce to 0 And as all are being multiplied, any real number we were getting gets multiplied by 0 and becomes 0 Therefore answer is 0 when m and n both are odd QED
    1
  17112. This might be off the wall, but I've used this animation for years as a metaphor in my work as a psychotherapist. In the same way we can look at these floating points zooming around and tell competing, yet coherent stories about them (triangle, square, etc.), so too do we tell stories about ourselves. Sometimes we point to our stories and to the memories that accompany them and say "Look! They're true! I AM a red triangle. I AM a failure/unlovable/etc." Our story-telling mind does a wonderful job of drawing shapes to fit the narrative, and one of the things we can do in therapy is begin to take different perspectives on stories. Looking at our life of floating points, we can tell new stories: "I'm also a blue square. I'm also a yellow star." And when we are able to transcend the limiting powers of narrative, we can begin to see that we are all of those things--and also NONE of those things. WE are not the stories we tell--there is a self there, behind the competing stories, that we can connect to when it's helpful. Which is all to say I'm so grateful for this video and to see the math behind it. Also grateful to the contributors who made new implementations of the dance--they'll make it so much easier to use this idea in my work!
    1
  17113. I missed you
    1
  17114. 3:15 - I was told in high school (50 years ago). that the Pythagorian Theorem was called a 'theorem' because it couldn't be proven!
    1
  17115. "coin calculus", also known as... "coinculus"
    1
  17116. Maybe pretty oldschool and simple....but i still think the 700 years old proof ist the most amazing thing.
    1
  17117. Good thing they discover new stuff about triangles. Maybe they will also figure out how to gracefully get out of love triangles.
    1
  17118. Most memorable: The leaning tower of Lire, though I really like the no-9s series also, it's one I hadn't heard of.
    1
  17119. You can proof there are no equilateral triangles with complex numbers too :)
    1
  17120. I'm subscribed with my high school email so I'm surprised I'm not subscribed yet with my main account. Just subscribed!
    1
  17121. Fastest click known to man
    1
  17122. Made it to the end!
    1
  17123. Therefore E=m(a²+b²)
    1
  17124. Hey kids , pay attention in school . Don't spend your time smoking pot in the parking lot
    1
  17125. Mery Christmas Sir
    1
  17126. It's the giggling...In another context, it might be a really good tic for an Evil Scientist...."Now we will vivisect...giggle...I like this part..."
    1
  17127. In the last videos, it seems to me you prove less and less things and start showing things without really proving them could you please try to focalise more on the proofs them selves than on what we are trying to prove? Nevertheless, I'm still a fan and love your content.
    1
  17128. Well if you cut out 2 black and 2 green... You can isolate a single square in the corner so it won't be tiled ever... So it is not always possible to tile
    1
  17129. can't I go as far away from the edge as I want with an ever growing reverse triangle?
    1
  17130. I have only seen the first 1m and 5 seconds of the video. I can´t believe it when I see pi and e appearing as the tab total. Come on! Got to be kidding!
    1
  17131. The squares one gives you n! (n-1)!, which is a lovely thing.
    1
  17132. The shading in that painting makes that lady's balcony look pointy - kinda like early Lara Croft's.
    1
  17133. Your channel is amazing but here's something to think of when designing the animations. For us needing the subtitles due to hearing loss the text lines occupy the lower parts of the screen, meaning that parts of the info/animations cannot be seen. Try it out and maybe this will be a useful piece of advice - reserve space at the bottom of the video so that important information is not obstructed when using subtitles. The content and quality is of utmost quality as always.
    1
  17134. You're the best!
    1
  17135. At 11:25, what is in the red box vs the green box. How is it possible f(n) = 2 and f(x)=2? Isn't the green box the slope? Wouldn't the slope be 0?
    1
  17136. Certainly the 'no integers contained in the partial sums' was a surprise. But I think the most enjoyable was the algebraic proof you showed at the end that the no-9s sum must be less than 80. The idea that all possible terms can be found by permutations of digits and that they can then be collapsed to the series of (9/10)^n is elegant.
    1
  17137. 14:04 Ahhhhhhaaaaaaaaaaaaaaaaaaaaaaaaaaa !!!!!!!!!!!!!!!!!!!!!!!!!! Fibonacci is here !!!!!!!!!!!!!!!!!!
    1
  17138. Easier to grasp, still enjoyable. If you split your usual format into a series of semi-standalone videos of this format.. you could become a youtube GIANT
    1
  17139. He is an excellent teacher!
    1
  17140. Thanks!
    1
  17141. Swivel proof
    1
  17142. I still don't understand how people turn cubes like that with one finger, and have them move accurately. Every time I've tried this, the cube ends up stuck because one or more pieces move too far out of a plane. Worst case scenario is that I then go to turn it rather hard and it explodes! And yes, I have tried different cubes, including so-called "racing cubes". It's probably too late for me now as I'm older and have some arthritis, but it used to really annoy me when I was young! I wasn't a terrible cubist: my best time was around 40sec - it's just that I could never get that physical technique :(
    1
  17143. Absolutely elegant
    1
  17144. Missed opportunity to call it “The Golden Phi-dentity”
    1
  17145. i had to break out the pen and paper to go through this video just out of the fun of computing sums of infinite series. have to say, the best part was revealing one of the identities of the Euler-Mascheroni constant
    1
  17146. Hello from Russia))) Thank you for this video This is the best on the topic from what I have watched before)
    1
  17147. Dear Mathologer, I love the history and evolution of this video. Please keep bringing us more mathematics.
    1
  17148. The 100 zeros being larger than the no 9s is something I cannot fathom so it's the most memorable. Something I found funny about the no 9s sum proof at the end is that your refinement was really obvious. The mathematician who made the first proof had great insight to make the geometric series but forgot the first digit couldn't be zero, we all make mistakes I guess.
    1
  17149. because school is not for you to evolve is made to make stupid consumers
    1
  17150. 20:40 I'm sure you would have thought about this argument, not really sure why you didn't use it. I feel it fits the picture-proof-style even more :D If you look at the first triangle, we need to understand why these powers of 2 turn into powers of 3. And to do this, notice that along the diagonals, we multiply by 2s. This is obvious for the first diagonal, as we started with powers of 2. But for others, let's talk about the cells 6 and 12, how did we get them? 12 = 4 + 8 = 2(2 + 4) = 2 * 6. This generalizes very easily. Now look around at the rows, what exactly do we do when we go left? Say we're at cell x, we look at the lower left diagonal entry which should be x/2 and add it. So the result is x * (1 + 1/2) = x * 3/2. That is, we multiply by 3/2 as we go left, which isn't surprising as we had to prove these powers of 2 will eventually become powers of 3. Equivalently, we can phrase this argument by looking at the columns. When you go up, say from the cell x. You add the thing to the top right, which should be 2x. So going up amounts to starting with x and getting x + 2x = 3x.
    1
  17151. Is there any connection between curves of constant width and Bezier curves?
    1
  17152. I love your visual proof.
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  17153. Your videos are so well made and so informative! Keep up the good work, sir. Big fan.
    1
  17154. In all branches of mathematics, CALCULUS is the simplest and straightforward. Integral culculus however, is kinda tricky.
    1
  17155. 2:00 if the six points were to lie on a circle, then the 3 lines have to be equidistant to the center. Then I got frustrated because I hate geometry and watched the solution ;)
    1
  17156. Both proofs are nice, but the direct one is nicer for showing the three "turned" triangles result in the same triangle.
    1
  17157. Sorry. Some of these "paradoxes" are just so incredibly stupid that talking about them boggles the mind. It's not impressive that you can arbitrarily take groups of smaller & smaller fractions to approach any number, rational or irrational, when you know the number! Show me a function that predicts how many from each group to take without knowing PI, or e or whatever beforehand, and then I'll be impressed.
    1
  17158. Also, I want to see a video proof of why the artic circle is so circular, instead of square
    1
  17159. Imagine it in 3D and it's mesmerising...
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  17160. Is there one? I'd say it's because 1 x anything is an identity. 1 x n = n for all n. Thus it's irrelevant. What we're looking at is 1 + 2 + 3 = 2 x 3, which is saying that (1+2) + 3 = 3 + 3 = 3 x 2 is a statement defining multiplication: For n x m, this can be rewritten as n + n +... m times OR as m + m +... n times. In this case, 3 x 2 is = 3 + 3... 2 times (or 3 + 3), and that's what 1 + 2 + 3 is equal to just pre-adding the 1 + 2. You could also pull one from the three, 1 + 2 + (2 + 1), collect the 1s, [1 + 1] + 2 + (2) = [2] + 2 + (2) = 2 + 2 + 2 which gives you the other form of 2 x 3, which is 2 + 2... 3 times, or 2 + 2 + 2. The 1 x, being an identity, is irrelevant to all of that. You could also write it 1 + 2 + 3 = 1 x 1 x 1 x 1 x 2 x 3 and it would be the same thing, but sounds and looks less profound. : ) You could also use the additive identity (0 + 0 +...) on the left side. For example, is there anything profound about 0 + 0 + 0 + 1 + 2 + 3 = 1 x 1 x 1 x 2 x 3 ? Probably not. Yet they both are still equal. Adding 0 to anything leaves its value unchanged (hence an additive identity, as it returns the non-zero part of the equation unchanged), and likewise, multiplying 1 by anything leaves its value unchanged (multiplicative identity) The only weird part is when you get to things that are uundefined, since normal operations on those things don't work. For example, 1 x n/0 is undefined because n/0 is undefined.
    1
  17161.  @Mathologer  Oh but of course. :) I always try to think through problems and consider what my answer is to them (based on the thumbnails) before watching anything so I can exercise my mind a bit. Once I start watching the video, my further thoughts are going to be at least somewhat influenced by it, so I want to get down how I think through the problems before seeing anything, so I can compare my answer/analysis to those presented and revisit it with the new information/additional argument and see how that changes things. It's a bit odd sometimes, since some thumbnails don't entirely represent discussions and things like that, but I also just like thinking about problems. :) I also love seeing what I would call "complete" (though I think the mathematician word is "general") solutions to things. For example, the more complete Pythagorean theorem for non-right triangles (I won't remember this right, but I believe it has a - AB Cos th term) or the n-dimensional version (which is easy to understand/think through by considering a three dimensional vector can be mapped out as light sources casting shadows onto lower dimensional planes that merely form triangles with respect to the planes' origins) My first degree was physics (and second economics), so I'm only math-adjacent. :D I've always loved math puzzles, but I hate proofs, because they tend to require knowledge of specific identities. I like thinking through solutions, not finding them based on having memorized a remote trivia piece. So despite friends over the years suggesting I get a math degree and noting my interest in things like topography and logic mappings, I can never QUITE cross that bridge. But I still love seeing the puzzles and solutions or work on them. And I absolutely LOVE your videos since most of the time you break things down (though I did have to puzzle over how S - A = R for a moment...I'm not sure why THAT tripped me up but it wasn't that hard to work out, haha!) I will keep watching your videos because I really like them. Though this one I've paused so I can think through it a bit more. Think I will watch the rest of it. :)
    1
  17162. Can you make a circle out of any closed shape?
    1
  17163. There's a formula with e, gamma, AND all the prime numbers?!?
    1
  17164. Never seen either of these proofs :(
    1
  17165.  @Mathologer  oh dear in yiddish
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  17166. 3:23 equation is xy=1. Therefore when you stretch and squish by a factor k (really just stretching one of these by a factor k and the other by 1/k) the product remains the same.
    1
  17167. Im a class 12 student, and the kind who wants to jump over anything he finds interesting enough to learn(and in the process leave the actual syllabus pending, coz i find most of it to boring to ponder over, more so because of the competitional method in which we are taught these things, just juicing away the beauty and creativity of maths)! And maths is just one such domain, where a bit of imagination and observation and , lots of patience , can go a really long way, and anything like this, beautiful mathematics, is sure to attract moths like me to its light!. This was such a nice video! Although i had doubts about if i should watch the whole of it, after watching it, i have even more!! Almost understood all of it, thanks to the visualisations , but now im craving for more! P.S when you talked about the sum deviating from its actual value after a certain number of terms, my brain was just screaming "TAYLOR, ITS TAYLOR SERIES IM TELLING YOU, ITS GOT SOMETHING TO DO WITH IT!."
    1
  17168. I always knew Tasmania wasn't a part of Australia
    1
  17169. Woaw
    1
  17170. 2 = 1/1 + 1/2 + 1/3 + 1/6 3 = 1 + 1/2 + 1/3 + 1/5 + 1/6 + 1/7 + 1/10 + 1/11 + 1/13 + 1/14 + 1/15 + 1/21 + 1/22 + 1/26 + 1/30 + 1/33 + 1/35 + 1/39 + 1/546 The second one wasn't easy -_-'
    1
  17171. Every time you release a new video it just happens to be exactly what I needed to see! Thank you!
    1
  17172. Totally fascinating and awesome. Does anyone have a nice expression for either of the two components of the Ramanujan identity? I calculated an approximation to the continued fraction as 1.4331274267223117583171834557759918204315127679059805234344286364 (from 50 terms).
    1
  17173. Shoulda used 0 to 11 for the same reason good amplifiers "go to 11" (it's not just a meme or a marketing gag, contrary to common assumption), because you have 12 numbers involved, and that makes dealing with deciBells, a logarithmic measure, significantly more intuitive to deal with
    1
  17174. Really good video (my background is in electrical and mechanical engineering).
    1
  17175. The name for slide rule In Polish language is «suwak logarytmiczny» which literally means «logarithmic slider».
    1
  17176. Thanks
    1
  17177. Something always bothered me about the logic of Archimedes' approximation. It seems reasonable that the inscribed polygons have circumference less than that of the circle, but it is not at all clear that the circumscribed polygons overestimate. One can say (by fiat?) that the shortest distance between two points is measured along the straight line segment. But it is simply not true that if one curve circumscribes another then it overestimates the length. Of course this happens basically when the "contained" curve is very "wiggly". This phenomenon does not happen with (flat) area because there is literal containment, and so doing Archimedes' polygonal approximations to determine the area of a circle does give the right inequality for pi. But the circumference is not so constrained. By similarity arguments Euclid had already established that the circumference is proportional to the diameter, and that the area is proportional to the square of the diameter. Archimedes established the precise link between the constants of proportionality. But for circumference it is not even clear that the circumference is really finite. How does one resolve this from first principles? I suppose nowadays we use the fact that the straight line segment approximations always give the same limiting value, and hence we can say that it is the length (by fiat). Hence in Archimedes's work the circumscribed polygons are irrelevant for the determination of the circumference.
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  17178. I like the coloring proof because swiveling is hard to imagine
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  17179. Would be cool to do a video on proofs that were disproven later because of a faulty assumption.
    1
  17180. I'd never seen the visual proof from this video. It's great! I also noticed that when you pronounce Ptolemy, "p" isn't completely silent as in normal English pronunciation, but a "p" without release. I wonder if this way of pronouncing Ptolemy is from German.
    1
  17181. The last puzzle the middle square is the total area - 4 * 4 * area of the small triangle.
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  17182. Which is 1 ÷ 5 of the total area.
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  17183. I took Calc II twenty years ago and that connection between 1/x and ln(x) never made sense to me. And now it does.
    1
  17184. Is it ra man nah jan or ra ma knew jan? I've heard both now. I feel like I have heard the latter a lot more then what is said in the video.
    1
  17185. The thumbnail was great. Love the video.
    1
  17186. I'll vote for the no 9s proof as most memorable
    1
  17187. Can anybody tell me what the name of this background music is.
    1
  17188. Taiyo by Ian Post (listed in the video description).
    1
  17189. If you are programming a random generator of these Aztec diamonds, it would be interesting to add a slider with a floating value from 0.0 to 1.0 The slider would represent the chance that a tile pair was Horizontal or Vertical. Obviously, .5 should result in a better Arctic circle. But what would a value like .1 or .9 generate?
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  17190. How much times has the thumbnail changed?
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  17191. How is it possible that (blue area) > 1/2 ? Because: 1/x is concave, since second derivative always > 0 Hence any triangle whose vertices have coordinates (a, 1/a),(a+1, 1/a),(a+1,1/(a+1)) fits inside its related blue area. So for any blue area (now inside the unity square on the left) there is a 90^ triangle. It's easy to visualize that all the triangles covers 1/2 of area.
    1
  17192. "And of course we also get that one way of making zero sense". I chuckled a few times throughout the video :)
    1
  17193. That seems like a Jespsen E6-B flight computer
    1
  17194. Best maths videos in 2020 and no doubt in 2021 and next years! Thank you!
    1
  17195. Getting what looks like a smolth curve by making its zig zags infinitesimal is not well defined to be the same as shrinking a straight line, if you take a line and kink it at close to 180 degrees every so often in alternating direction you can increase its pathlenght by any arbitrary number you want if you shrink it down to infinitesimal, if you where to also stretch out the angle of the kinks as you shrunk the line segment down you would then end up with a straight line with normal pathlenght. This is about well defined limits mostly, and if you don't have a gold grip on what your limit tends towards both in shape and in scale, you will derrive seeming contradictions, but it was true all along that the limiting object was like the zig zag line and not the straight line ^^
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  17196. I vote for the "no integer" property 🤯
    1
  17197. no,not that easy.
    1
  17198. Nice ! What is the diameter of the wheel ? And for other logarithm -> diameter = f(base# ) ?
    1
  17199. Why not 4k-1 instead of 4k+3? It's more beautiful that way. 4k+1 and 4k-1
    1
  17200. Yeah sure, easy to understand lol
    1
  17201. My congrats to the wardrobe department for the T-shirt selection
    1
  17202. Loved the Lesson🌈
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  17203. 14:10 the graphic is wrong. ln(2) would have to be 1/2 for the visual proof to work like that.
    1
  17204. "Close enough is fun enough" ... the anagrammer applied the calculus limit...
    1
  17205. "trithagorean" theorm No. >=O
    1
  17206. Cool stuff! But i don't understand the t shirt. Anyone?
    1
  17207. that's cool
    1
  17208. How does this relate to sums of squares? I first thought that you cannot construct all sqareroots. Yet that is only for 2D and only would proof some lengths are not possible. For example the triangle. one length is at least related to sqrt(3) and there is no pair of squares that give you 3. which you would need to get sqrt(3) from the half triangle (pythagoras) and since sqrt(3) is irrational scaling doesnt help. In 3D ist easy because 1^2+1^2+1^2=3 and only one type of sqrt(3) direction other one is reflected. Anyway maybe often you need more than one irrational number simultaneously . Pentagon : 1^2+2^2+0^2=5 but that is not enough in 3D as it seems. So sqrt(5) yes pentagon no. Similar for the pentagon in 2D 1^2+2^2=5 Scherrers proof that sounds almost like a reason that you can use to show that there are only certain type of crystals or point groups (Scherrer and his x-Ray stuff.. looks almost like it). Maybe even space groups? Obviously if you have all other angles (monoklin) you dont have any rotational symmetry except for 360° or 0° How does it go with higher dimensional polygons aka platonic solids or archimedian solids. Is there any surprising things that do not work out? Any pentagon obviously that wouldnt work, but all the other polygon surfaces?
    1
  17209. Great topic too
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  17210. I have that book! It has a plethora of examples which will REALLY make you think...
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  17211. അത് ഏതൊ പഴയ ലിപിയാണ്! വായിക്കാൻ പറ്റുമെന്ന് തോന്നുന്നില😅
    1
  17212. MadhavaLooksLikeMathlogger?
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  17213. What happens if you start with a row of 4 or more ones and play the rinse and repeat game? Does that represent higher dimensional rectangles? Do the ratios between the larger number of coefficients that will be created as rows are derived also converge like phi, r & s to constant values?
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  17214.  @Mathologer A quick check on a spreadsheet for rows 1 1 1 1 and 1 1 1 1 1 ‘rinsed and repeated’ shows that within just a few newly iterated rows the ratios of leftmost row-entry to each successive rightward row-entry converge very quickly. Whether these represent the side-ratios of higher dimensional rectangles would need a bit more serious effort to elucidate. 😊
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  17215.  @Mathologer  Here’s something else, but is it trivial? Starting with a row of n ones, the next row is simply n, n-1 etc. for the second row, the ratio of each row-entry to its rightward neighbour is just n/(n-1) Play the rinse and repeat game to derive the ‘golden’ ratios after a few rows. From left to right call these rn, sn, tn, un… to extend your r and s terminology to an initial row of n ones. Take the ratio of adjacent ratios, rn/sn, sn/tn etc. Note that each of these ratios of ratios is smaller than the n/(n-1) that sits in Row 2 for each n. But with increasing n it looks like these ratios of ratios increases. I wonder whether, these ratios of ratios converge on the n/(n-1) that sits at the top of each of these columns and, if true, whether that is interesting or trivial. This probably needs more work than the half-hour I’ve spent playing a spreadsheet!
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  17216. For the coin paradox and as a Physicist I realized a nice solution. Be A the radius of the still inside coin, and B being the radius of the outside rotating coin, because there is pure rolling the travelled distance of the coin's center is proportional to the angle of rotation of the outiside coin distance = angle × B At the same time the center's coin make a circular movement with radius A+B, so the travelled distance of the center after one lap is 2π (A+B) = angle × B But the angle is 2π × laps, so A+B = laps × B so laps = 1 + A/B laps = 1 + radius outside / inside With this formula you can see the for the first 2 examples are satisfied 1) laps with same radius is 2 2) laps with 2B=A is 3
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  17217. You can use turning squares on a dot grid to illustrate the rotating coins paradox. Make one square move round another fixed square of equal size, making one quarter turn at a time. How many full turns will it make to return to its starting position? Now make both squares move in the same direction - clockwise or anticlockwise - round each other. How many full turns to get back to the start? Make them turn round each other in the opposite direction, how many then? How big does the grid need to be in each of these three cases to accommodate these movements?
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  17218. You're the best <3
    1
  17219. The Terrance Howard operator
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  17220. at the end when you started rotating the shapes, the projected 3D shape looks 3d in the shadow because we are viewing it on a 2d screen, that really made the 4D projection click for me. Great work! <3
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  17221. This "leaning tower of lire" problem translates into the difference between possibility and necessity - this is how Nicolai Hartmann illustrates it: Imagine a rock hanging over the edge of a cliff for centuries. It seems always possible that it might come down in the very next moment. But when it finally does come down, it is no longer possible, but necessary
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  17222. Thanks for this great video. I have taught Ptolemy’s Theorem but was unaware of his inequality - nice! Also (to my astonishment) I was introduced to kinds of quadrilateral that I didn't even know existed, particularly the truly 3D kind. Nice work! A couple of notes: I wonder how many generalisations of Pythagoras' theorem there are? For now, I can see two: the cosine rule and Ptolemy’s Theorem. I was slightly disappointed you didn't use the word "continuity" in your explanations of how 2D quadrilateral cases can be thought of as the limit of 3D quadrilateral cases.
    1
  17223.  @Mathologer  I hadn't heard of de Gua's theorem (https://en.m.wikipedia.org/wiki/De_Gua%27s_theorem ) - a nice generalisation indeed. I'll check out your two videos on Pythagoras' theorem. Thanks!
    1
  17224. I thought he would use geometry but a more convenient proof with zeroes and ones wow 🧠
    1
  17225. What's the ratio of the radii of the most inner and the most outer circles on your t-shirt? And congrats on the 100th video.
    1
  17226.  @Mathologer  So, you are saying that the infinite product of cos(pi/n) for n starting at 3 is of no interest for you? Are you a real Mathologer or just trying to sell t-shirts?
    1
  17227. use 99/70 as it's really close to being a nice round number, and easy to know it resolves a little higher than ~1.4
    1
  17228. My favorite highlight was the connection to Euler-Mascheroni constant.
    1
  17229. awesome
    1
  17230. Challenge at 11:34 If the Fibonacci box has numbers P, Q, R, S where P+Q=R and Q+R=S, then: P*S=153 and Q*R*2=104 and (P*R)+(Q*S)=185. So, either Q=1 and R=52 or Q=4 and R=13. If Q=1 and R=52, then P=R-Q=52-1=51 and S=Q+R=1+52=53 and these give P*S=51*53=2703 which is incorrect. Therefore, Q=4 and R=13 which give P=R-Q=13-4=9 and S=Q+R=4+13=17 and P*S=9*17=153 and (P*R)+(Q*S)=(9*13)+(4*17)=185. Again, the corresponding Fibonacci box has numbers 9, 4, 17, 13. Bonus at 11:51 Since the top 2 numbers are P=9 and Q=4, then, if the parent box has numbers A, B, C, D where A+B=C and B+C=D, we have either (1) B=4 and D=9, or (2) D=9 and C=4, or (3) A=9 and C=4. Since A+B=C and B+C=D, we cannot have option (2) where D=9 and C=4 because this would make B=5 and A=-1 (a negative number) which is invalid. Also, we cannot have option (3) where A=9 and C=4 because this would make B=-5 (a negative number) which is invalid. So, B=4 and D=9 which makes A=1 and C=5 and the parent box has numbers 1, 4, 5, 9. In turn, it has a parent box with 1, 3, 7, 4 which has a parent box with 1, 2, 5, 3 which goes back to the initial box with 1, 1, 2, 3. Therefore, navigate starting with parent box for triplet 3-4-5, go to the right child 5-12-13, then go to the right child 7-24-25, then go to the right child 9-40-41, and finally go to the left child 153-104-185.
    1
  17231. When you convert a 3 0 into a 2 1, is the 2 1 always shorter in any possible spacing between laces?
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  17232. No one's doing something like this, Thanks.
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  17233. Great video! East to follow
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  17234. This guy is great
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  17235. Well, I'm no Baley but I was at least able to calculate the sum of the no-zeros series in binary: 2.60669515241529176...
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  17236. The standout bit for me was the idea that the partial sums always miss the integers. The summands become so tiny that it is amazing they always miss forever.
    1
  17237. Piet is a Dutch name and is derived from Peter, it is pronounced like the English name Pete.
    1
  17238. planar spectral decomposition :) In one of the odder uses I've ever had for mathematical theorems, I used a similar concept to reprogram a sensor bank. The sensors shared memory with each other, and that memory was both their operating memory and their (immediate) storage as the data went through pre-processing per-bank and eventually sent it back to the computer. Instead of plugging into . . . a lot . . . of sensor banks, which, even if the data transfer was instant (it was not), would still take a long time, I wanted to plug into one sensor and use the fact that each sensor bank shared memory with its neighbours to reprogram them all. The direct approach, ("copy your memory"), didn't work because the sensors can only access some memory by actively mutating it. The quine approach probably held water, but I couldn't come up with one. But using the fact that they actively mutate the memory, I could define a chain of sensors where I directly modified the memory in the first, and the last one in the chain was the one being programmed correctly. Then I remove the last one from the chain . . .and so on recursively until I can just program the sensor I plugged into directly. In order to figure out how the bits were flipped across everything was a spectral decomposition, and I pictured it as decomposition of a polygon, and it ended up being an 8-cycle IIRC. So I only needed to transmit 8 different core files (plus a few for as I moved up the chain and started telling each member in the chain to also modify the memory it shared with members already in the chain), in order to reprogram every sensor from just one :) The files would mutate as they passed along the chain until they ended up in their original form. Everytime we needed to reprogram that device, it easily saved 3-4 hours of labour, (instead turning it into about 8 hours of plug it in and walk away).
    1
  17239. Speaking of that outro music, what's the surface area of a banjo?
    1
  17240. I still have the slide rule I used in school and college way before the advent of the calculator. When I took my HND in electronics in 1972, I was at last allowed to use an electronic (not mechanical) calculator. I still used the slide rule, though. It was quicker, and I was more confident of the answers. This is because the slide rule had many more functions than the basic calculators of the day. Powers, trig functions, exponentials etc. I just used the calculator to double check my results. The rule is a little sticky to use now. The plastic has aged a bit. I notice that you did not mention the transparent lined cursor that I have on mine to read off the required result.
    1
  17241. For me, the explanation is the pattern... Although every term EVENTUALLY cancels out, until they do, a sum exists. I have noticed this on a huge number of infinite series... It's like filling and draining a sink... It never goes dry.
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  17242. I solved the ending problem the hard way before figuring out what the easy way is. 1/5
    1
  17243. Coin paradox is easily solved if one first imagines a coin rolling down a flat surface, let's say it rolls a full 360 degrees, and then imagines the flat surface (with the coin stuck to it) being bent into a circle(/cylinder) such that the coin's end position intersects with it's start position. It takes 360 extra degrees to deform a flat surface into a circle(/cylinder), so that's where the extra rotation comes from.
    1
  17244. What in the what? How's that possible with the 9's??? I lost my life right there when it didn't turned out infinite. Definitely the most memorable part
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  17245. You look like Madhava
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  17246. This is one of the rare theorems that should have three blue lines and one brown line.
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  17247. I got 17688 ;-;
    1
  17248. Can we expect a sequel of this video exploring the topic of umbral calculus? :)
    1
  17249. Euclid Book 7 probably has some stuff related to this, ...
    1
  17250. What is the TAXI T-shirt? I didn't get it
    1
  17251. Awesome video loved it
    1
  17252. inside circle, outside circle used to do my head in. Till I learned to root 2 it
    1
  17253. If anyone is curious to learn more about the Weierstrass product formula which Mathologer mentions at the end of this video, you will probably want to check out the product formula for the Gamma function, which is related to the Sine product formula by the expression: Gamma(x) * Gamma(1 -x) = pi / sin(pi*x). You can learn more about the product formula for the Gamma function from the brand-new video I just posted on my channel
    1
  17254. 3:05 pause... ahhh... this reminds me of my childhood... I had a spirograph like this that I can draw neat patterns on paper... I was so amazed I kept drawing and drawing until the whole paper is blotted with ink.. LOL!
    1
  17255. >buy a paradox >look inside >infinities
    1
  17256. Is this used in programming? Seems like it turning multiplications into sums and such, which seems like an efficiency improvement for code. It also looks like you can start anywhere on your number line arbitrarily and the math stays simple. Just as I watched this it seemed like this is exactly the sort of "programming gem" that is used to save time in graphics generation and such.
    1
  17257. Really appreciated the non-calculus explanation!
    1
  17258. 1908 was the year when Einstein's annus mirabilis papers started reverberating through the scientific community, and nobody was paying attention to some triangles thst looked like a board game for children. That's why i think it was overlooked.
    1
  17259. pure magic.
    1
  17260. 36:19 a slight typo - the difference equation should be expressed with "n" on the left hand side, or "x" on the right. Love this video!
    1
  17261. When I could make time for it, it worked. I would say a longer format was good and necessary.
    1
  17262. Beautiful
    1
  17263. Gorgeous🙏🏻
    1
  17264. I was taught Heron's formula in school in France.
    1
  17265. I believe his name was Madhavan and the British ate the "n"
    1
  17266. The flat earthers would probably want you to do that globe projection with the north pole in the middle. ;P
    1
  17267. Danke und frohe Weihnachten!!!
    1
  17268. I think that the leaning tower of lire is the most beautiful of the proofs
    1
  17269. Animation of rearranging of squares to show that 1+2=3 just blows my mind
    1
  17270. I've heard stories that circular slide rules were banned on many campuses, for several reasons. 1. The people teaching intro math classes did not know how to use circular slide rules. Sad. 2. It was thought using circular slide rules gave students an "unfair" advantage over the strait slide rules, because they didn't have to deal with the left-index right-index business and adjust the power of 10 by ±1.
    1
  17271. I liked the reveal of the finite no 9s series. Another example of my maths intuition being wrong (happens quite often)
    1
  17272. Great one!
    1
  17273. Looking at the shirt and I'm so hungry now.
    1
  17274. "Since we've all been familiar with 3-4-5 triangles since kindergarten..." Hah. Haha. Ha. Not in 'Murica we haven't! Try seventh grade.
    1
  17275. In my sixties, and never saw either proof. Definitely a proof of a different kind, as well: you can still learn something new (to you) each day.
    1
  17276. 5^3+6^3=7^3
    1
  17277. Ian's site is fascinating! Can you do or have you done a Metronome pendulum video. I would be interested. :D
    1
  17278. All very beautiful ! 🥰
    1
  17279. I have a masters degree in math, but I graduated 25 years ago. Did you define the Bernoulli series?
    1
  17280. The why is why we love mathematics It's just logic, quantity based logic of all sorts. Any "triangle" has to behave according to rather simple logic, being only 3 points where each is relative to the other two. All the things that fall out make sense in a simple way, but the complexity that is built on top can be very misleading. This is why we have the problems of why does the universe work the way it does.. it's extremely simple but people get caught up in the "physics" that math easily creates. Physics is just constraint, yeah we're fully immersed in this game, but it's still important to remember how simple things are underneath. Computed.. future outputs.. after we make choices. We the network.. Great video.
    1
  17281. The one @9:08 is exactly the one I used during my school many years ago 😉 I should have known this video back in 1987, I made me a circular slide rule from plastic book covers for logarithmic calculations of shutter speed for my flash light (incredible strong and heavy ELGApress), depending on focus length, film speed and distance (these values are all connected on a log scale!). That did great, even with distances up to 50m and 500/1:5.6 lenses. That old days there was no TTL flash metering or Thyristor controlled flashes around. At least not in my country. And fotographing was still on analog chemical 35mm films... Good old days, calculated before shooting, think be4 you do...
    1
  17282. 37:12 I agree that this is place where I struggled a bit. This autopilot was a bit fast. But I didn't pause because even if I struggle I was able to keep going. I'm very good in mathematics, so this is only spot. One of reason why I didn't have any troubles with this part just because I myself for fun was deriving formula for n! approximation via integral using similar trick, so it wasn't something new to me. Also, about row vs columns is also common trick to me to derive expectation of binomial distribution.
    1
  17283. Brilliant lecture, thank you!
    1
  17284. 16:00 This is a cheat. You're supposed to be refining the grid by increasing both numbers at the same time. Here, Schwartz increase the P by 1, then increased the B repeatedly until he got an area he wanted. We should either be adding a fixed amount to each, or multiplying both P & B by N. 21:20 The noise is cicadas, a kind of large bug. We had them in Lubbock, Texas, and they're very loud and irritating.
    1
  17285. Not really a cheat - it’s a standard part of defining what we mean by a limit where we want both variables to grow to infinity. We have to consider all possible ways for this to happen, not just lock them together with one particular rule. If we don’t do this it seriously messes up how we define continuity and differentiability of multivariate functions.
    1
  17286. I had to watch this in two halves on two different days to get through it.... whee... that was long! (my background: MSEE from last century)
    1
  17287. I won! Hits subscribe button for free subscription.
    1
  17288. 333 Only half evil Roflmfao!
    1
  17289. Wonderful as usual.
    1
  17290. YOU ARE THE WORLDS BEST MATH TEACHER AND MATHOLOGER.
    1
  17291. I loved the crazy towers. I love complicateed answers t easy problems.
    1
  17292. 17:40 COLOMBIA 🇨🇴🇨🇴🇨🇴🇨🇴🇨🇴
    1
  17293. Finally Everyone is Knowing 👏👏👏
    1
  17294. as i saw there is a new video', i rushed to see it. now my day is better:)
    1
  17295. 16:16 I had t stand up and clap. THIS is the math class I needed it back in HS. Me: "Teacher, why is it called "natural" logarith?" Teacher: proceeds to ignore me and any other question. Me: (I bet he does not know a shit, just parrot.)
    1
  17296. Hahaha, amazing! In hindsight, it seems so obvious, but without this video, I could've gone my whole life without knowing.
    1
  17297. In my opinion, this otherwise extremely important and well explained tutorial was ruined by the horrible, distracting noise towards the end. What a shame.
    1
  17298. This video made me feel joy!🥰 thank you Mathologer😊
    1
  17299. But they did teach that in my high school, at least to the very upper track students. I think the reason most American High school students don't get taught this is that they are allowed to drop out of the science and math classes before they learn enough background to actually tackle such "advanced" topics. But there was only one science class that got as far as Nuclear Physics, with about 25 students, in a class of 535 students. Most of my classmates didn't even know that our advanced physics class even happened.
    1
  17300. Newton: I invented calculus! Leibniz: No, I invented calculus! Madhava: Nuh-uh, I invented calculus! Archimedes: Hold my κύλιξ 😎
    1
  17301. Yay, I paused at 2:00 and solved it. First I noticed that the three chords of the big circle are all the same length and must therefore be the same distance h from the center of the circle. That meant that the center of the big circle must also be the incenter of the triangle. Then (after trying a wrong direction in my thinking), I realized that I needed to show that if I split the side a at the incircle and added side b to one side and c to the other, the two sides would necessarily be equal. (And likewise for the other sides.) To show this - apologies for the lack of a diagram - I split the side of length c at the incircle into lengths x and (c-x), where x is the distance to the vertex connecting sides b and c. Then the side of length b is split into lengths x and (b-x). Finally, this means the side of length a is split into lenghs (c-x) and (b-x), so adding b and c to these will make b+c-x = c+b-x so that the extended segment is a chord of the big circle. And after watching your proof, it looks like just the last step is different: showing the aqua and magenta segments are equal. I think I did something more algebraic, but basically the same thing.
    1
  17302. Magicologer level! Beautiful and hypnotic…
    1
  17303. My vote is the no 9s sums being convergent. My gut feeling was way off the mark!
    1
  17304. Calculus is one of my favorite topics on math, sir!
    1
  17305. Engineers: meh. Surrrrre. Wind? Earth tremors? Water shedding? Did you even read the building code? Let's add some tension in there... and ...
    1
  17306. Thank you
    1
  17307. I WANTED TO MAKE EXACTLY THIS FOR THE VERITASIUM THING
    1
  17308. Can you make a video about non-whole dimensions? Is it even a thing? Does a 1.5 dimensional space have only 3 directions? (Because every line adds 2 directions so directions per dimension equals dimension divided by 2) These patterns also apply to none-whole dimensions? I really want to know more about non-whole dimensions Help
    1
  17309. This very much looks like a start to a solution to a 3d version of the solution to the pythagoras theorem that was solved by them kids.
    1
  17310. You lost me when you claimed the shorter spike area was equal to the longer spike area. I expected the green square came from the squish operation applied to the upward spike.
    1
  17311. 35:45 That's a really scummy move by the marker. Unfortunately it's not uncommon.
    1
  17312. Only one way.
    1
  17313. the math whisperer, "gold gold from the heptagonal american river or yellow paved road."
    1
  17314. Lol, today's t-shirt is awesome.
    1
  17315. Great video. I don't want to say something negative about Tesla, but he was surely no Gauss.
    1
  17316. Loved it
    1
  17317. First time I have ever seen this proof and I am 71yrs old and educated in England.
    1
  17318. 👍
    1
  17319. No one else impressed that he only pauses instead of cutting in edits? Just me? Ok.
    1
  17320. I knew that. HA HA
    1
  17321. i menntally turned all those the cylinder would have a weird shape ..how weird ? Like believing in sacred geometry weird
    1
  17322. 34:27 - I lost the plot for Pagoda wall. It rotates.... I don't see a pattern or repeated sequence. Something is implied in the red diamond. Poor animation 'cause I don't get it.
    1
  17323. Honestly I can't see any reason to complicate such an easy thing as proportional share As of the heritage or contested garment split this algorithm is extreme and totally unfair oversimplification It highly promotes parties to make maximum claim. If court blindly accept such claim the justice collapses immediately. Basically the more you claim the more you get If the court have to consider the lawfulness of claims then it can establish fair shares directly, without splitting the difference The logic behind contested garment instantly makes me mad, it's the reason of my comment here
    1
  17324. Interesting how emotions get in the way of understanding math, called autism-like reading letters in words backward, or seeing letters in incorrect syntax of said words?
    1
  17325. @14:34 the claim seems incorrect. The sum of the positive terms alone approaches "e - 1".
    1
  17326. Is this gonna be about generating functions?
    1
  17327. Bravo! This was a great video. Looking forward to your video on the applications of the Euler-Maclaurin formula, especially Ramanujan’s application of it to the Riemann zeta function and the infinite sum of integers! BTW, why is it called the “Euler-Maclaurin formula”? PhD, Engineering Mechanics
    1
  17328. r×s=r+s If r=0, s=0. If r != 0 and s != 0: let r=ns ns×s=ns+s ns^2=s(n+1) | ÷s ns=n+1 If n=0, s has no solutions. If n != 0: s=(n+1)/n r=n+1 For any n != -1 or 0: (n+1)×((n+1)/n)=(n+1)+((n+1)/n)
    1
  17329. UKMT. I took part in some of those when I was in highschool, I got a bronze and silver. my math teacher also gave me a postcard with a really cool math puzzle from UKMT.
    1
  17330. This reminded me of putting ketchup on french fries. You put 1 squirt on, then it's not enough so you add 1/2 a squirt, but its still not right so it gets another 1/3 squirt followed by.... the guy mopping the floor .... the crew cleaning the kitchen .... lights going out .... the girl is locking the doors.... almost done.....
    1
  17331. When I heard about it, I did not even understand why it is a paradox. It would only be a paradox, if you could paint the inner surface with a paint of constant thickness, as you would indeed need an infinite amount of paint then. However at some point the horn gets so thin that it is impossible to apply paint with that thickness of paint. So there is no paradox there. However painting the outside of the horn will really require an infinite amount of paint. Still no paradox there, because outside the horn there is an infinite amount of space. Actually even in many interior spaces with a finite surface - for example a sphere - you will already need less paint than the area multiplied with the thickness of the paint, because if the thickness of the paint is greater than zero, the surface will shrink after you painted it. As far as I know, there are no real paradoxes in maths, unless you based your maths on axioms that already contradicted each other. If there really was a paradox in maths, that would always lead to 1=0.
    1
  17332. First Proof is 90% stuff that is already known with the last twist to get the new fact (IE already known: the 1:2 ratio, etc) Both proofs need a quick (very easy) proof that after the fold you get a line. It can be seen in the first proof easily at 7:17, after you show that all 6 colors is 180 (or Pi).
    1
  17333. I heard "islet" several times, until I figured out the word is "eyelet".
    1
  17334. Thank you for blowing my mind again! Is it only regular polygons with odd number of sides that generate these special ratios? Are people working on the nonagon already for the 4D stuff? Or is the next one gonna be 5D, as in 1 1 2 3 5... :D
    1
  17335. You can figure out the reason that there are the same numbers of each type of domino in the hexagonal tilings by thinking about it three-dimensionally. That is, view the drawing as a collection of stacked cubes in the obvious way. Clearly a massive solid cube has the same number of each color domino. You can get from the single massive cube to any other tiling by just carving out the extra cubes. It's easy to check that every such "carving out" operation preserves the numbers of each domino type.
    1
  17336. ❤
    1
  17337. Loving your videos 🥰🥰
    1
  17338. I don't know which one I love more. The visual proofs or the algebraic ones.
    1
  17339. I like the shirt.
    1
  17340. Sir you are a mathematician I am high school student I find maths very interesting subject so please suggest some books for my level.
    1
  17341. I think it would be interesting to explore this idea with exponents For example 1+1+1+1+2+3=3^2^1^1^1^1
    1
  17342. The animations made it instantly understandable . Good job to the crew&you.
    1
  17343. You didn't show why the arrow direction is equal to the property of whether the left/top cell is green or black.
    1
  17344. You're back!
    1
  17345. Burkard Polster is just adorable)
    1
  17346. just amazing!
    1
  17347. You could technically argue that black holes are probably the only real world example of infinity in action, so to speak.
    1
  17348. As for feedback, it really worked well. Had to rewatch the part at around 45 min but otherwise everything was clear. As for background I'll soon finish my masters in complex systems which is a bit of a mishmash of programming, maths and physics. Real good way to procrastinate studying for my stochastic calculus exam haha.
    1
  17349. Nice work, thanks a lot for the pleasure you give to us.
    1
  17350. 42 seconds into the video... Here are some meaningless letters with exponents, add them... Me: NOPE, I'm done, see ya. I like this guy because he gives visual examples. But I still cannot give a damn about math unless I am given REAL WORLD examples of why I might need it. I work in finance and can calculate amortization tables in my head. I can work out aggressive payment schedules for a given rate, term, and balance without a calculator. But as soon as you add variables and exponents and fractions "just because" I stop giving a damn and my mind goes blank. The biggest FLAW in math education in America is a stunning and ridiculous LACK OF CONTEXT. Homework will literally be pages and pages of meaningless equations to solve and I equate this to torture of children.
    1
  17351. Variations of Tower of Hanoi with more than one stack of disks are described in U.S. pat. no. 7,566,057. The simplest Multistack T of H (2X4) involves four pegs on the corners of a square with two different colored stacks arranged diagonally to each other. The object is to reverse the positions of the two stacks. The restriction is that each stack can only be moved among its starting and ending pegs and one of the pegs on the other diagonal. Also covered by the patent are a 2X5 variant of the Reve's version of T of H. Both the 2X4 and the 2X5 are the first iterations of their respective series. P.S.: the last puzzles mentioned are in a puzzle museum on one of the U of Indiana campuses.
    1
  17352. Recently I got interested in Domino games. I'v found 142 variants for a standard 6-6 set. And you post a domino tiling video. Coincidence?
    1
  17353. Make a video on bhaskaracharya ll
    1
  17354. Nice programming challenge - I'm going to give tbhat a go :)
    1
  17355. The base finding was really cool, e creeps everywhere though. Only if you could add sin(ix) = sinh(x) And the others, it would have really been complete, without the Taylor series approx. of course.
    1
  17356. I made it to the very end. :)
    1
  17357. It's worth adding, I think, that the Bernoulli numbers were the subject of the first ever computer program. Ada Lovelace wrote example code for Babbage's never-completed Analytical Engine that would have calculated them mechanically.
    1
  17358. I lost you at the cards
    1
  17359. ❤❤🙏🙏🙏
    1
  17360. I had fried eggs for breakfast, now I have got a scrambled brain for lunch! ;0
    1
  17361. It would be pretty cool if all the check marks at 1:12 could be clicked, leading viewers to the appropriate video covering that theorem. If only YouTube had a feature where creators could do such a thing! They could even let creators create boxes with text in them, like *annotations*. That would be such a neat idea! I wonder why YouTube hasn't done that yet... oh wait...
    1
  17362. His laugh is amazing 😍
    1
  17363. Ich freue mich schon😁
    1
  17364. those are definitely bananas
    1
  17365. Wonderful. How can you work on this? Every time I approach such stuff I feel a vertigo and nausea as if being in front of the infinite Truth of Nature and being unable to sustain the vision. My PhD in physics does not help either.
    1
  17366. The visual representation of the Euclidean algorithm at 24:27 is really cool because it shows that consecutive Fibonacci numbers are always relatively prime. Try working it out yourself, but if not I have tried to write out the reasoning although a youtube comment isn't a good format for a visual proof: Numbers are relatively prime when the GCD is 1, which in terms of the visual euclidean algorithm means the last "remainder" rectangle you consider ends up being a 1 x N rectangle where 1 is any integer. Consider the rectangle 1 X 1, each side represents the first 2 Fibonacci numbers. As 1x1 is a rectangle in the form 1 x N we know they are relatively prime from the visual euclidean algorithm. Likewise with the next pair of consecutive terms 2 and 1, which are represented by a 2 X 1 rectangle. However the next term is more interesting. We can construct a 2 X 3 rectangle by attaching a 2 X 2 square onto our 2 X 2 square onto the side of the 1 * 2 rectangle. The new larger rectangle is a 2 + 1 X 2 rectangle, or a 3 X 2 rectangle. This represents the next two terms of the Fibonacci sequence. (This is the same visual construction of the fibonacci sequence as is used to construct a golden spiral.) We can also see that this new diagram shows the visual euclidean algorithm showing the GDC of 2 and 3 is 1. We can also easily see that for any two fibonacci numbers a and b, by constructing a rectangle of side lengths and b using the golden spiral method, we also end up constructing proof that the Euclid formula shows that A and B are relatively prime. It's pretty cool that the Euclid algorithm is so closely linked to sqrt(2) and the golden ratio
    1
  17367. I was taught Heron’s formula around 14-15 in New Jersey (US). I remember it being annoying to learn because we were too young to appreciate the proof and applications, and it is harder to evaluate than other formulas. Probably would appreciate it more later on when we didn’t need to work with messy numbers by hand.
    1
  17368. Sorry, but I missed what it was that was missed, and presumably reinvented. Was this rubber-band-wheel invented long ago? And then forgotten? And reinvented? At least I don't recall much time being spent on explaining the title.
    1
  17369. Well, the video is *wrong*. There are many mistakes along the way... The assumptions are not clear / correct. Starting at the leaning tower: 1.) Going infinitely far with the tower would quite soon eat up all the matter in the universe. 2.) the lowest brick gets pressed down by all the bricks above him. so the bricks have to be thicker... infinitely thick to bear all the bricks above them. So even the construction of the first brick has to be stopped, because he would eat up all the matter in the universe. 3.) Even when we put all these aside: earlier bricks would add only a miniscule amount to the total distance... and we have at least to throw away all the bricks that add fewer to the distance than planck length. So we are not able to build the first block, that would add only a miniscule amount as a distance... So the real sum would be 0 . More problems arise with point 1 and 2: You just can't say, that we assume massless bricks and/or zero gravity... Because if you do, you wouldn't have to care to balance them out 😀 To put a long story short: Completely wrong video. Just kidding: Great video - thanks!
    1
  17370. From now on when I see a dachshund I will simply think that this dog has the same area as just all the others. 😅
    1
  17371. a "shining" presentation...WOW
    1
  17372. You sir, got me hooked on morning mandolin.
    1
  17373. Rosetta code users might like to note that the method shown in this video makes a simple and efficient solution to the Rosetta Code Task "Pythagorean Triples". I have added a solution in the language Quackery, along with some notes to aid translation to more conventional languages.
    1
  17374. The coloring proof is great, but with even mild color blindness, I have a very hard time distinguishing the segments. Perhaps, to the benefit of ~10% of males, more strongly different colors, and/or line textures would be a good idea...
    1
  17375. O_____ O
    1
  17376. This is a marvelous topic.. Also, conjecture should only be a noun. Bring back "conject"..
    1
  17377. formula for the squares: 1/12*(n**4-n**2), so for n=5 we have 50 squares.
    1
  17378. Fantastic video. These videos you make have given me a whole new appreciation to the beauty and mystery of mathematics. Thank you so much for sharing your incredible knowledge and amazing teaching skills with the world!
    1
  17379. 9:40 It's a Christmas Miracle!
    1
  17380. Thinking about the proof, the step where the arrows point to each other can be improved. You're basically taking two cases where adjacent bricks point towards each other (one horizontal one vertical) and replacing them with cases where the pieces are swapped -- that is, pointing apart, still one vertical case and one horizontal. First, seems that saying "just exchange the two pieces" is better because you don't have to create another orange square which will generate two new cases and show that two cases merged together to form the same orange square (which generates two new cases). But... Consider that you only have a "problem" if you think about moving all the blocks one at a time in some sequence. You move into a spot that is either newly created or vacated by a previous move. But really, you are moving them all at once. You copy each brick into the pointed-to position in a new initially blank diagram. So, the order does not matter and conceptually you just do them all at once. This means that there is nothing special about the facing arrows ! Just do what the arrow says, and they end up swapping positions. The proof works without this complexity at all and is in a purer form.
    1
  17381. your shirt truncated instead of properly rounding... it should be 25.8070
    1
  17382. But why x+2 instead of x + 7 for exemple... It works only with x+2 objects or what
    1
  17383. Need a new video on -1/12, as numberphile just released a new one on this!
    1
  17384. I've coded that algorithm up. https://github.com/DonaldHobson/Mathomoger_artic_circle
    1
  17385. What’s up with that image of Legendre?😁
    1
  17386. x*imple + c
    1
  17387. This is the only place I can think of where my favorite formula in math can be appreciated. PT = P0 + (V * T) The reason why it's my favorite is that it can be used to plot: 1D lines (Algebra and number lined) 2D lines (Algebra) 3D lines (Calculus) And maybe more. What the equation means is The digit you are looking for PT is equal to the first digit in the sequence P0 plus the difference of digits multiplied by how many digits away it is. So let's say: P0 is 5 P1 is 6 6 - 5 = 1 6 = 5 + (1 * T) 1 = 1 * T 1 = T But if you notice there is a pattern not shown, one that will show that it tracks through multiple dimensions. It works in 2D as well P0 is now (5,4) P1 is now (6,6) Does it still follow V = (6,6) - (5,4) = (1,2) (6,6) = (5,4) + ((1,2) * T) T = 1, And if you still want/need to find B, (Y when X is 0) it's just a matter of simplifying the equation back down to X to find T, then plugging T to find Y 0 = 5 + (T * 1) -5 = T Y = 4 + (-5 * 2) Y = 4 - 10 Y = -6 (0,-6) Now for the fun bit: X,Y,Z P0 is (5,4,3) P1 is (6,6,6) ... Yes it still works. You are welcome to do it yourself but I know it for a few reasons, in Calculus it's called the Vector Form of the Line, and I had to learn it to A) program changing R,G,B values in a program and, in Unreal how to have 3D moving environments If I had known sooner what a simple powerful equation it was. But the secret each lower dimension is hiding is that all the other ones are there, it's just that their Vs are 0.
    1
  17388. Just a hobby but I love the depth. I might not grasp it all but it's fun to be exposed to concepts I wouldn't come across elsewhere.
    1
  17389. That Card Trick Mind blowing....
    1
  17390. R.I.P Tesla...
    1
  17391. There is an easier solution: the street is a court, and so is circular. Thus the houses to the left are the same as the houses on the right, and so the person lives in any (or all) of the houses.
    1
  17392. Made me wonder, is there an equivalent coordinate system for hyperbolic trigonometry like radial system for circular trigonometry, and what kinds of things would it make plotting easier
    1
  17393.  @APaleDot  thanks for the extensive answer!
    1
  17394. It would make sense because you can plot every triangle on a circle.
    1
  17395. He says 4K+3. Also looks like 4K-1
    1
  17396. 4:20 I just came from the RLM episode... too soon, mathologer, too soon
    1
  17397. Well... the golden cut (or golde ratio as you said) reminds me that there's always a real number between two natural numbers which fractional part is exactly the same as the reciprocal of the lower natural number (i.e. Phi-1/Phi=1). As formula: x-1/x=n, where x is a real number and n is a natural number. So with x=(sqrt(4+n²)+n)/2, you'll get the next real number which has the same fractional part as it's reciprocal (the golden cut for n=1 as I said and the silver cut for n=2). Does all those numbers also have properties like these two ratios (cuts)?
    1
  17398. That is a very loooooong kiiiittttyyyyyyy
    1
  17399. I think that'd be a cool way of darkening colours. Each channel is owed some intensity, but there's some deficit.
    1
  17400. 17:07 you know if one was so incline one could use that diagram to tile a floor would make for a very interesting design
    1
  17401. I think that the 1 over a number with 100 0's is not limited becas if It just has 100 0's at the end than it is just the harmonic series x 1over googol plus some more and that is infinity I think that works
    1
  17402. Is there a triple that maximizes the inner square so it kisses the middle square?
    1
  17403. oof,anagram made me wondering how many math between those permutation movements.
    1
  17404.  @Mathologer  < REAL MOTH, GO!
    1
  17405. how many squares in an n x n grid of points? • consider an arbitrarily rotated sized positioned square among the points • a square's rotation is determined by the slope of one of its lines, a/b • if a=b=0, the square does not exist. for now we ignore the position • let's find all the squares which are the same (one can be moved to coincide with the other) • the squares a/b, -a/-b, -b/a and b/-a are all the same, so to avoid duplicates let a,b>=0 • if b=0, a/b is the same square as b/a, so let b>0 • with our restrictions, a+b is the a/b square's width (distance from leftmost to rightmost point) • it is necessary to let 0<=a and 0<b<n-a. have we excluded all duplicates? • yes: two squares can only be the same if some of their slopes are parallel or perpendicular, which we have handled in its entirety • with our (now sufficient) restrictions, there are (n-a-b)^2 valid ways to position an a/b square so that the answer is sum(0<=a and 0<b<n-a) of (n-a-b)^2: now for algebra autopilot = sum(a=0...n-2) of sum(b=1...n-1-a) of (n-a-b)^2 = sum(a=0...n-2) of sum(b=1...n-1-a) of b^2 = sum(a=0...n-2) of (2(n-a-1)^3+3(n-a-1)^2+n-a-1)/6 = 1/6*sum(a=1...n-1) of 2a^3+3a^2+a = n^2(n-1)^2/4/3+n(n-1)(2n-1)/6/2+n(n-1)/2/6 = (n-1)(n+1)n^2/12
    1
  17406. Don't Breitling make a navigator logarithmic (slide rule ) watch? I used to have an old circular slide rule in the late 1960s.
    1
  17407. Here in Italy its not used, but i knew It on my own and i found a use for It to calculate the area of a triangle only knowing the position of the verteces in the cartesian plane, without using the annoying height or trigonometry. Congratulation for the video btw, very interesting even for a High School student like me!
    1
  17408. here (also in Italy) it was taught when we were like 8yo, most kids know it
    1
  17409. Anything can be equal to or unequal to anything else when you start playing around with infinity. If you have a REAL amount of something ( none of this infinite nonsense) and you subtract that same amount you are always left with zero. That's not math so much as common sense. . Rather than me going on and on about all the ways playing around with numbers like this is pointless, instead I'd like to know what, if any,are the real world practical applications of this mumbo jumbo.
    1
  17410. This is very nice demonstration of Heron's formula.
    1
  17411. I think the most memorable proof was the No integers among the partial sums. It's non-intuitive, yet easily provable.
    1
  17412. There are long streets with only 8 houses. Have you ever been in Australia?
    1
  17413. If the hexagon tiling is a projection of cubes inside a box then we can think of paths we need to take to reach from one side to another along a single axis, say X. This means we need to travel box size length path along Y and box size length path along Z. Just imaging a figure that you move from one tile to another one by one. You will be traveling box size tiles to the side and box size tiles up:) Repeat for the Y-axis and that's it.
    1
  17414. 21:00 the math looks increasingly like Piet Mondrian
    1
  17415. The missing numbers in the tree of sums of squares have a nice pattern to: 1*1 , 2*1 ; 2*3 ... 3*3 ; 5*3 ... 5*4 ; 7*4 ... 7*5 ; 9*5 ... 9*6 etc.
    1
  17416. I went to school in Bristol (UK) in the 1970s, and my year was the last at the school to use slide rules (the next intake were allowed calculators instead). I remember buying mine in W H Smith, and it's in a green plastic case. I still have it, and I never want to part with it. I think the hands-on experience of holding in your hand a physical object that embodies the entire idea of logarithms is an incredibly useful educational moment. Although we don't need these devices any more, I do think everyone should have the chance to use one at least once so that they can understand how indispensable they were before the advent of digital computing.
    1
  17417. It may also be worth considering when it was taught in any particular country. School curricula tend to get thinned out (dare I say dumbed down) over time? That’s the case in the UK anyway, though ‘school boards’ can be different.
    1
  17418. Here's half a sphere. That's why it takes 4pi to go around it. Surface(cos(u/2)cos(v/2),cos(u/2)sin(v/2),sin(u)/2),u,0,2pi,v,0,4pi A minimal single sided closed surface. A waveform over its own phase inversion.
    1
  17419. I vote for the infinite series of reciprocals with restricted digits! Especially the fact that 100 zeros is somehow greater than no 9's. Cheers!
    1
  17420. getting a lot of graph theory vibes from this :o
    1
  17421. WOW! "What's the difference" - I remember when I discovered this amazing factorial-based formula when I was in high school. I was so proud of myself ;)))
    1
  17422. Only now did I get the hang of this theorem.
    1
  17423. Great video! About 10 years ago I developed an efficient algorithm to compute the number of domino tilings of a n*m rectangular board. It is based on computing the determinant of the adjacency matrix of a graph in Pfaffian orientation. The final complexity of the algorithm is O(m^3 + m^2 log n). This seems to be much better than any other algorithm that computes the same thing. The thing is, I was unable to prove (or disprove) its correctness, but the provided implementation successfully passes the test suite for the problem 1594 - Aztec Treasure on acm.timus.ru. There is a lot of "magic" in that algorithm and I would still be interested if someone could help with providing the actual proof of relations 4, 11, 17 (appearance of Delannoy numbers!), and the last bit where the square root of the determinant turns out to be the same as multiplying every second diagonal element! Here is the PDF paper and C++ source code: https://drive.google.com/drive/folders/1drGS7TsvilB3Q6xpbjlRAzwqbyrdHDRO?usp=sharing
    1
  17424. The Oral Torah, the Talmud, was actually written down in 2 stages to prevent the Oral transmission from Sinai from being lost forever. The Mishnayot (Mishnah) was written down by Rabbi Yehudah HaNassi (Judah the Prince) who completed his shorthand outline called the Mishnah in 220 ce. This came after the Destruction of the Bar Kokhba Revolt by Rome's Hadrian in 135 ce (when he renamed Judea as Palestine 800 years after the original Philistine Sea people had gone extinct). The blow from this destruction of the 2nd Judean Revolt was population wise more destructive than when Titus under Vespasian destroyed the 2nd Temple in 70 ce. Because of the intensity of Hadrian's destruction amongst the Rabbinic population, the Oral Torah would have been lost had it not been written down within the next 80 years. Judah the Prince completed this work in Tiberius (Tiveriya) on the shore of the Sea of Galilee. The 2nd stage, the Gemorrah, of the Talmud took place in 2 separate geographical locations: 1. In Judea (now renamed Palestine after Hadrian) by the end of the 4th Century, the Rabbis now headquartered in Tiberius had to complete the text of the Gemorrah in 375 ce within about 50 years of Constantine's Roman Catholicizing the Roman Empire in 325 ce once again because of severe persecution. 2. The Babylonian Gemorrah in 2 cities in ancient Babylon (Sura and Pumbeditha) in 500 ce. also because of severe persecution by the Zoastrian Persians. I think Pumbeditha was renamed Fallujah. But it is true that Judah the Prince's 220 ce writing of the Mishnah, the Oral Torah in outline form, is at the Core of both Talmuds (the Tiberian aka the Yerushalayim and the Babylonian) Talmuds. The Mishnah and the Gemorrah together are in that Center section of the Talmud.
    1
  17425. Observation: C^2 + AB = C^2 + 2AB cos 60º C^2 = C^2 + 2AB cos 90º C^2 - AB = C^2 + 2AB cos 120º Which makes me wonder -- does this pattern hold for arbitrary angles?
    1
  17426. How did I not know about how to find the nth power by adding up the rest of the numbers?
    1
  17427. I take it you chose that shirt very intentionally 😂
    1
  17428. Do you simply get two negative values that cancel each other if m and n are both odd? In the 4cos term?
    1
  17429. 11393868451739000294452939
    1
  17430. How can I be sure that Fermat's last theorem is really true? Sometimes errors are found in proofs years after the proofs have been accepted as correct. Even if I could understand the proof, how could I be sure that I hadn't overlooked something?
    1
  17431. Which algorithm did you use for generating the random tiling of the square? (for example at 16:16)
    1
  17432. 6:57 given that pi is transcendental, does it mean that the volume is finite but it would take infinite time to fill?
    1
  17433. Professor X teaches humans
    1
  17434. 18:13, the formula is m(n) = 3ⁿ-1, where n is the number of disks
    1
  17435. I like the shorter videos
    1
  17436. Thank you, Dr. Burkard, you have inspired me to go poking about on eBay again for a slide rule to collect. :D I have an E-6B, but I want a slide rule made for maths and engineering to randomly use when I don't feel like whipping out the calculator or invoking Python.
    1
  17437. y' = xy dy/dx = xy dy/y = xdx Integrating both sides, ln(y) = x^2/2 + c' y = ce^(x^2/2) Unfortunately, don't remember how to solve y' = 1 + xy😐. But have got enough motivation to review and solve😇
    1
  17438. Question: while 10 is numerator, for which both denominators X and Y, that resulting division for each gives other number (denominator) at the zero-symetry? I] 10/X= must be 0.Y0Y0Y0Y0Y0Y..... 10/Y= must be 0.X0X0X0X0X0X..... II] X<Y, X is number as ab, Y is a number as (a+1)b for example: X=57 then Y=67. Y>X and Y-X=10. Y/X=[(a+1)/X + b/X] x (Y-X)= must be 1.Y0Y0Y0Y.... X/Y= is gonna be at the 9 symetry opposite reverse of X (ba) that is 0.ba9ba9ba9ba9ba9.... In Reel and Irreal numbers universe there is only two unique numbers reach this rule, those are 27 and 37. 10/27= 0.37037037037..... 10/27 gives the other one (37) at 0 symethry... 10/37= 0.27027027027..... 10/37 gives the other one (27) at right-left symethry of Zero. 37>27 and 37-27=10 37/27= (27+10)/27=1+0.37037037... =1.37037037... And 27/37= 0.729729729729... 27/37 gives opposite reverse of 27 at symethry of 9! See the symetrycity for 0 and 9... Secret correlation!: Remember for 27 2+7= 9 And for 37 3+7= 10 |27-37|= 10 37{27+10}>27 then 27/27+10/27= 1.37037037... symethry of 0, of 10's 0 (1,0). Numerator 37 comes at 0 symetry. [I] 27 {27+0 or 37+(-10)} < 37 then No 0 symethry but 9 opposite reverse of 27 after 0 decimal = 0.72 9 72 9 72 9 ...... Numerator 27 comes at its opposite reverse at 9-symetry [II] On the other hand we may assume while numerator < denominator after 0 decimal denominator comes with its opposite reverse mutiplied by 10 and 1 substracted That is the say: 37 Opp.Rev. 73 73x(37-27)=73×10=730 730-1=729 {here 1 comes from |27-37|/(37-27)=1 and/or {37} 3+7=10 {27} 2+7= 9 10-9=1 Remember 37 is {2+1}7 And 27 is {3-1}7 {2+0}7 Then 27/37= must be 0.729 729 729....
    1
  17439. First time here, but I subscribed as soon as I saw the t-shirt.
    1
  17440. I dream of a world where we use base 2, base 7, and base 12 primarily eliminating the human-egocentric base 10.
    1
  17441. chapter5
    1
  17442. I created a visual of the Euclidean Algorithm (25:15) being used to derive a continued fraction for Pi, check it out here https://editor.p5js.org/taylor8294/sketches/aL9FEoKx0
    1
  17443. 6:17 I didn't see 4k+3 as the answer for unsuccessful primes, but instead I saw 4k-1. Is there a reason to add 3 instead of subtract 1?
    1
  17444. Conways circle has the larger diameter, since the diameter of the other constant width curve is a chord of conways circle which doesn't pass through the center. Thus that chord must be shorter
    1
  17445. Beautiful!
    1
  17446. your shirt have ramanujan number .... nice
    1
  17447. Thanks for the beautiful explanations!
    1
  17448. It may not be Newtonian Calculus but it's still a form of general calculus on its own. Still love the geometric intuition. Geometric intuition is what calculus (and many other things) should be about.
    1
  17449. 3:03 And what if you didnt have the luxury of infinity outside the screen like an actual shape?
    1
  17450. Anyone else binge watching mathologer
    1
  17451. Whats the circumference formula of a ellipse?
    1
  17452. There's a superpattern for the squares too. It's a bit trickier, but 2*5/1 = 10, 3*14/2=21, 4*27/3=36, (n+1)*k_last[n]/n = k_first[n+1]. I haven't proven this, but will leave that as an exercise.
    1
  17453. Funny how I searched this up because I was thinking of exponents. I was like if a number to the exponent of 1 to 2 and to 3 makes its dimensions increase. Even the number one just that we right it as one but probably should use some type of notation that it's now describing a 3 D one. Then I thought what's a number to the power of zero. I thought at first zero . Then I thought if the first dimension I can use to define a length. Then if I take it away it's unitless , but not zero more like a unitless 1. If I multiply a first dimensional number , second , or third it could become it instantly. That's why I thought that zero can encompass everything. Because it can bring down a dimension with some number defined in that dimension. That's why I started thinking what about the unitless one zero can bring down the the unitless one can become a dimension instantly. Could it instantly get to infinity or at least get there faster than something approching infinity . My logic is probably flawed , but it was bothering so much I had to look this up 😅
    1
  17454. You are too kind--I went back to check the video to check what I thought I'd seen, only to find you put it right at the head of the comments. It did take me more than the 30 seconds to mentally add 169 and 195. The answer, of course, is 2
    1
  17455. Jaa mach mal ein Video auf deutsch :D
    1
  17456. The Kelly formula and number are big deal too.
    1
  17457. x² = x + 1 ; 1 = 1/x + 1/x² ; 1/x² = -1/x + 1 ; (-1/x)² = (-1/x) + 1 ; z = -1/x ==> z² = z + 1
    1
  17458. Yet another fantastic video. Given my background, MSc in engineering physics specializing in applied mathematics, none of the parts were hard to understand. The matrix parts heads me off thinking about using the Schurcomplement for fast inverting then as the desired power grows higher and higher. Really great video! ❤️❤️❤️
    1
  17459. That has to be the silliest integration method, and I LOVE it.
    1
  17460. Nice background music. What is it?
    1
  17461. Ed Catmull, another founder of Pixar, is another expert of math and complex analysis. He even offered a job to Tristan Needham, the author of Visual Complex Analysis, after reading his book !!
    1
  17462. I ran a full marathon on October 31. On November 1, I planned to go out for a short run but ended up running a half marathon. Then on November 2, I ran 1/3 or a marathon. I had started thinking about this kind of series. You can verify my distances from my Strava profile. (username finlip) I did feel that the series is infinite. Thank you for telepathically making a video on something that was occupying my mind this month.
    1
  17463. gamma was the most memorable one
    1
  17464. 23:57 The cases k=0,1,2,3 are problematic because then 4/3/2/1 of the binomial coefficients have an upper number smaller than 5. But doesn't this actually compute nicely without changing the formula since the binomial coefficient x choose y is zero for all x<y? Or do you mean a different flaw of the formula?
    1
  17465. Far from using coins, I haven't used cash in something like 5y.
    1
  17466. ..
    1
  17467. 3:30 so cool!
    1
  17468. TIL...
    1
  17469. when you showed perrons proof I got intrigued and read it with single frame forward real quick (suprised how easy it was, lots of algebra sure, but nothing more complicated than first semester stuff) - I kinda regret doing it before watching the rest of the video since it spoiled the reveal of the binomial coefficients and pascals triangle though :(
    1
  17470. At 6:13 the losing numbers are also 4k-1.
    1
  17471. "An onion-powered calculus explanation" is a new one
    1
  17472. We need a video on "How to derive algorithms for RUBIK 3×3 CUBE". I have seen many, but didn't know how they get to that algorithm. Also, you please mention the best algorithm preferred by you . Waiting... Like this comment or copy my comment. He must notice it. 😅
    1
  17473. Mind boggling and mind opening. Great fun teasing the grey matter! Thank you. Gamma was my favorite.
    1
  17474. I love that shirt...
    1
  17475. OMG. Algebra autopilot took me for a roller coaster ride at 9Gs and i passed out along the way and only woke up after the infinite product was finite. P.S: i wish i had algebra autopilot in school. I'm a big fan of roller coasters.
    1
  17476. Also, i like the algebra autopilot music. Very catchy.
    1
  17477. In the USA, I did not learn Herons formula but my math education was weird, with some grades skipped and so on
    1
  17478. yayyyyyy! Mr Bald Superhero is back :v
    1
  17479. id say the swivel proof is nicer
    1
  17480. 6:56 The dots remind me of square dancing or triangle one at that. The pattern could be traced and make a nice choreography.
    1
  17481. 8:06 When you subtracted the 1 on both sides of that top equation, I think you deleted information. Trivial information, but if you leave it, I'm thinking it would say that a point has 1 vertex, which is the point itself. EDIT: I'm also thinking I'm wrong and that a point is not its own vertex, and of course there is that alternating 0, 2, 0 pattern.
    1
  17482. Does the Pythagorean theorem even work for shapes with fractional dimension?
    1
  17483. The integral of 1/x from 1 to n is equal to ln(n). Why did you use the ln(n+1) to approximate the harmonic series? Shouldn't it be: ln(n)+γ=sum of harmonic series?
    1
  17484. Most memorable part was the visual demonstration of the no nines harmonic series. It's just so simple for what is a tricky thing to get your head around.
    1
  17485. Great video
    1
  17486. "I do it all the time."
    1
  17487. I quite enjoy Up And Atom's approach to this. In her video, she states it's almost pointless to compare the two without referencing their units. Like of course something can have infinite area but finite volume, they're different units!
    1
  17488. If you consider the thickness of the paint, you get infinite volume, no matter how thin you spread your paint.
    1
  17489. How does unit matters here? You can fill up the horn but can't paint it. If you can fill it up then you have also painted it, that's the paradox.
    1
  17490. You are looking for precision not available in this instance by the laws of physics. Funny how there is art history but not math history. Math theory of course requires all previous work to be absolutely correct.
    1
  17491. 0^2 + 1^2 = 1^2 Yes, it's not possible to draw that by hand but you'll get the primitive degenerate case of a line segment of length 1 posing as a triangle.
    1
  17492. I cannot remember ever seeing a proof of a^2+b^2=c^2. I have a degree in engineering (electrical so LOTS of maths), but do not recall ever seeing either proof.
    1
  17493. 6:18 instead of 4k+3 can we use 4k-1?
    1
  17494. this video is so good
    1
  17495. Excellent contribution
    1
  17496. Thanks!
    1
  17497. Neither ... this would have been wonderful for me as a student given I am a visual learner!
    1
  17498. discreet or discrete ?
    1
  17499. Eight minutes in and Mathologer is creating four-sided triangles .... I may need a second cup of coffee to get through this one!
    1
  17500. I think there's an error in the way you describe the matrix at 12:20 or thereabouts. I think stacked horizontals need to be -i, else the labeling of 2x2 which is ones on the first row, and and twos on the second creates a matrix with determinant 0. I could easily be wrong of course.
    1
  17501.  @Mathologer  Yeah, I had misunderstood the procedure (thinking that stacked pairs were about pairs of the same number). Really fascinating stuff. Great vid. Liked the fib aha especially.
    1
  17502. 0 = 1 - 1 = 0 ; 1 = JESUS CHRIST CHÚA THÁNH THẦN
    1
  17503. I am hypnotized by that t-shit...my brain is locked in a processing loop.
    1
  17504. Wild. Thanks for this one!
    1
  17505. Have you covered Z-transform in this channel before? I remember after learning about it, we used kt to find an expression of the form F(n) to find the nth term of Fibonacci sequence. It could be a very interesting video.
    1
  17506. 3:39 "what ill show you next is this new, spectacular way of visualizing... that it really approaches ln(2)" Equals sign gets bigger 😂
    1
  17507. What?? Another video? So soon? Can't complain, I'll take it.
    1
  17508. 😲🧐🤔 Burkard, please, if you read this - can you let me know why you chose to take the error corrections to the 121st correction?
    1
  17509. Very Nice.
    1
  17510. Is there a formula to solve 5th degree equation . I found a formula which can be used to solve polynomial of any degree and I checked it for multiple degree . Is there any importance for this fomula because if it is not i am not thinking of trying to publish it . It is a general formula . I would like to get your reply
    1
  17511. I first learned about this in "Concrete Mathematics" :)
    1
  17512. I've been using A LOT an Aristo rule first and later a Nestler rule when I was in a technical highschool (chemistry) in the end of the sixties. I used it not only to make multiplications but also to calculate logs and exponents and sines, cosines.... And, on the two main strips were also marked the special numbers like PI, e, .... The second 'magic trick' was the LL strip on the rule - the LogLog strip. It was so easy to calculate X to the power Y, or log or ln of X Of course, the accuracy doen't go further than the 2nd or 3rd digit, but it was so fast compared to the hard way (using paper log tables, sine tables ...) What a wonderfull thema to make a second part to that very interesting video :-) Standing ovation for all your videos !
    1
  17513. 10:31 because cos 90 is zero ie in radians pi by 2 for odd numbers( if we take common in denominator)
    1
  17514. What happens if you do the same with a plane of Hexagons?
    1
  17515. 6:20 This is just a guess based on what you've said so far, but I believe the answer is..."usually". It's definitely possible to remove two black and two green tiles in such a way that dominoes cannot tile the board. As an example, remove tiles [1, 2], [2, 1], [2, 2], and [3, 1] (where [1, 1] is the lower left corner). This leaves tile [1, 1] isolated from the rest of the remaining board. However, this almost feels like cheating. My gut instinct says that, as long as you mutilate the board such that both A) You remove an equal number of black and green tiles and B) No part of the board is completely cut off from any other part, you will still always be able to tile it. Also, as a bonus, it's still sometimes possible to tile a mutilated board with an isolated area. For example, remove tiles [1, 2], [2, 2], [3, 2], and [3, 1]. The isolated area is now two connected tiles, which you can put a domino on.
    1
  17516. 17:30 one of my math teachers did the trick of the telescoping sums, but more formally He chose P a polynomial such that P(X) - P(X-1) = n^k (where k is the power of each integer in S_k = 1^k + 2^k + ... n^k) I found it brilliant and later I tried to derive a general formula for S_k, without much success, but really loved every bit of it !
    1
  17517. I need a hint to pursue the question of rational tan values. It's true that tan(na) is a rational function of tan(a) for n = 2, 3, 4, ..., so that if tan(a) is rational, so are tan(2a), tan(3a), etc. But this only tells us that slopes of lines from the origin to the vertices of an n-gon are rational if tan(m*360/n) is rational. That doesn't imply that the vertices themselves have rational coordinates, no matter how cleverly I scale, so I'm not ready to apply the shrinking argument to them. For instance, if one of the vertices is (2/sqrt(13), 3/sqrt(13)), with rational slope 2/3, and another is (3/5, 4/5), with rational slope 3/4, there's no way to scale so that both have rational coordinates.
    1
  17518. Amazing
    1
  17519. That arithmétic géométry does not exist yet ! :p
    1
  17520. Made it to the very end. Many thanks for many years of making complex stuff almost understandable for me! 😃
    1
  17521. Have I understood your challenge correctly? e.g. Your challenge as 9:39 the answer is the sum of powers of x, i.e. for 1,2,4,8 coins you would get a solution of 1 + x + x^2 + ... x^15 because you would start with (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) ? Or have I missed something? So there is only one way to get each number.
    1
  17522. Good explanation.I think those are cicadas. And that animation depicts the deployment of a solar sail in orbit. (kinda---but yeah, origami spacecraft is a thing)
    1
  17523. Great vudei!
    1
  17524. Re: question at the end. I'm a physics student at his 1st year of university in Italy. Everything clicked and, most importantly, at the end I also had a chance to get a taste of what's coming in the next years of uni in a really intuitive way. Great job :)
    1
  17525. 10:17 before you asked this question I tried to plugin 1x1 in the formula and spitted out 0
    1
  17526. Can I know what is the especiality of whatever written in your t shirt?
    1
  17527. Ah, 2¹×2¹=2¹+2¹ 3½×3½×3½=3½+3½+3½ 4⅓×4⅓×4⅓×4⅓=4⅓+4⅓+4⅓+4⅓ 5¼×5¼×5¼×5¼×5¼=5¼+5¼+5¼+5¼+5¼ And so on ...
    1
  17528. Finally a mathematical proof that we really do live on a flat Earth.
    1
  17529. I wonder how (and if) this would translate to understanding 3D Modeling software calculating object data. I mean there are vertices, surfaces and normals, how the norlmals work (and why is it important), why a surface with flipped normal mess up the shading on a curved object (and what exactly does that mean) and why that does not happen with a flat surface. Why is it better to model in quads instead of triads, yet the render engine breaks all quads into triads during the final render. This video really tickles so many question, thanks for this one.
    1
  17530. I loved the part of the French weird head guy proof how elegant it was
    1
  17531. 9 million books in Nalanda University were burned in 1193 by Muslim invaders. We will never know what treasures it had.
    1
  17532. Wer sich nicht in's Bockshorn jagen lässt, der lässt sich eben darauf ein was Andere zu sagen (oder mit Fanfaren Trompeten oder Posaunen, dem Hirtenhorn, dem Englischhorn, dem Posthorn oder sonstigem Instrument zu tuten) haben. (german language)
    1
  17533. The essence of the euclidian algorithm is this lemma (it helped me a lot to understand it, so I am posting it here) Lemma: If a=bq + r, then GCD(a,b) = GCD(b,r) Proof. We will show that if a=bq+r, then an integer d is a common divisor of a and b if, and only if, d is a common divisor of b and r. Let d be a common divisor of a and b. Then d|a and d|b. Thus d|(a - bq), which means d|r, since r=a - bq. Thus d is a common divisor of b and r. Now suppose d is common divisor of b and r. Then d|b and d|r. Thus d|(bq+r), so d|a. Therefore, d must be a common divisor of a and b. Thus, the set of common divisors of a and b are the same as the set of common divisors of b and r. It follows that d is the greatest common divisor of a and b if and only if d is the greatest common divisor of b and r. qed
    1
  17534. This puzzle is somehow similar to this puzzle --> https://mathgarage.wordpress.com/2019/01/08/finding-a-triangular-number-that-is-twice-another-triangular-number/
    1
  17535. This video seems derivative.
    1
  17536. Nein. Neun. and associated new to me "no X & must have Y", though disappointed you didn't go for the pun (or Keine nein "kinda nine"), especially since the response to your homework was Nein! :)
    1
  17537. Great stuff as usual... 13:46 Without using generating functions I found 0:00:00 [ 2, 2728 ] 0:00:00 [ 20, 53995291 ] 0:00:00 [ 200, 4371565890901 ] 0:00:00 [ 2000, 427707562988709001 ] 0:00:01 [ 20000, 42677067562698867090001 ] 0:00:11 [ 200000, 4266770667562669886670900001 ] 0:02:08 [ 2000000, 426667706667562666988666709000001 ] 0:23:49 [ 20000000, 42666677066667562666698866667090000001 ] The running time is roughly linear in the amount, so that's what I can do in an hour. Enough to see a pattern emerging. The algorithm works for fractional dollar amounts, too. Written in GAP since I didn't want to deal with GnuMP. Grüße aus Dresden und frohe Ostern.
    1
  17538. the chirality of this is fascinating to me
    1
  17539. Oh, so that's what the 30 on my Jump To Conclusions Mat is for.
    1
  17540. What a great episode of Mathologer. I love that the end explains why it's a "2" in (x+2)^n. Could the "2" also be understood as "painting" the n-1 dim shape out to points -1 and 1 (a distance of 2) in the nth dim? This is what the animation suggested by Tristan seems to indicate.
    1
  17541. ​ @Mathologer  okay, I figured it out. I was tired and trying to calculate it in my head so I came to a faulty conclusion. 3 * 3 * (5 * 2 * 2 + 3 * 3 * 3 * 2) = 666 The two outer 2x3 have 3 tilings, the rest in the middle is just a combination of the following tilings {2x2, 2x4, 2x2}, {2x3, 2x3, 2x2}, {2x2, 2x3, 2x3}, or {2x3, 2x2, 2x3}
    1
  17542. Shame that there's no table content with time code in description.
    1
  17543. The original is just a Parker Star.
    1
  17544. Can you do that with any equal-sided polygon or just those with prime numbers of sides?
    1
  17545.  @Mathologer  Hm. Someone ought to. Thank you.
    1
  17546.  @Mathologer  Ok... My next answer may not satisfy you (I still don't see it), but evaluating the first 6 terms of the sum and subtracting log(6+1) gives me... 1+1/2+1/3+1/4+1/5+1/6-log(7)=0,50408985... EDIT: I did not look up the comment section. So above is pure calculator result. So... I will look up now. Thank you, Burkard! EDIT 2: Sigh. Ok.. I got it... Summing up triangles being inscribed in the blue parts.
    1
  17547. This formula brought some nostalgia
    1
  17548. I don't know the specific algorithm you would use, but a potentially infinitely zooming slide rule should be possible with one dimensional fractals.
    1
  17549. Just finished watching your quadratic reciprocity video......what a treat. I am a little amateur in mathematics..... would look forward to your video on permutations as you had mentioned there ...it would definitely help in appreciating the complete beauty of the proof (not sure if it's published already)...also sorry for posting unrelated topics to this video.... just wanted to post on an active thread.
    1
  17550. Took me six minutes to look at his shirt 😂
    1
  17551. I have the formula of the squares on the grid of nxn dots but it's gonna be very complicated to write here as a comment so please reach out to me if you want to now how I found the formula. Btw the formula is n^2(n^2-1)/12
    1
  17552. Start with a triangle with sides r, g, u. Draw the incircle of the triangle. This divides the triangle such that r=a+b, g=a+c, u=b+c where the lengths a are tangent to the incircle and meet at point U, the corner opposite u; the lengths b are tangent to the incircle and meet at point G, the corner opposite g; and the lengths c are tangent to the incircle and meet at point R, the corner opposite R. Extending by length r from point R gives two line segments tangent to the incircle that are both length r+c=a+b+c. Similarly extending by g from G gives line segments that are g+b=a+b+c, and extending by U from U gives line segments u+a=a+b+c. Because each pair of line segments tangent to the same point on the incircle are the same length, they form the secant line of a larger circle. Further because the three secant lines are the same length and the same distance from a common center (be a use they are all tangent to the same circle), they belong to the same larger circle. If q is the inradius (radius of the incircle) and s is a+b+c, the semi-perimiter of the triangle, the larger circle shares the center with the incircle, and has radius Q where s^2+q^2=Q^2. I think that's enough words, and working it out I worked backwards.
    1
  17553. You are amazing! Love your videos.
    1
  17554. The nice thing about this is that it doesn't change if you're using decimal, hexadecimal or which other number system you choose. It is always true. The magic with circles and triangles is very intriguing.
    1
  17555. Made a start on your coding competition - https://vortex-rho.vercel.app/ You'll probably need to turn animation off for large numbers (or have a supercomputer)
    1
  17556. Haha - nice one - kid had perfect numbers in 4th grade of elementary school :) i was like hey, i had to study this, but never until her classes didnt know there is a one to one relationship between perfect and Marsenne primes :), triangular numbers - here i hear about it for the second time :) nice one, but mathlogger already had some pythagorean triangles :)
    1
  17557. the key is in his t-shirt
    1
  17558. My brain shut down on the proof section
    1
  17559. Great ! I was so looking forward to the new video! I just checked yesterday your channel craving for some fun math to chew on...🥳
    1
  17560. That was fantastic. Thanks
    1
  17561. I came in thinking I found this as kid back then and it’s exciting to know that there’s more practical use to it hooray!
    1
  17562. Isn't this just a mechanical version of Benford's law?
    1
  17563. Does the base of the log function used here matter? E.g. if log base 10 was used versus log base 2 versus log base e (natural log), etc, would those have any practical difference on how the circular slide rule would be set up?
    1
  17564. For the sake of the fact that you're doing base 10 math you need to have the 10 land in the same place on the circle as the one. However every single logarithmic scale can be changed into any other logarithmic scale by linearly scaling it by a constant factor. That's the change of base formula, to change a logarithmic scale on base e to a logarithmic scale on base 10 you scale the scale according to the constant factor log base 10 of e, which is about 0.43429.
    1
  17565. In fact, the key of understanding calculus is pedagogy.
    1
  17566. 2=1/1+1/2+1/3+1/6 and 3=1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/15+1/230+1/57960. What can I say- being greedy works sometimes. :)
    1
  17567. That not quite circle looks like a candidate for elliptic curve inspired encryption algorithm
    1
  17568. Color proof ftw
    1
  17569. The color proof, seems easier on paper than the swivel. But I imagine liner algebra guys disagree
    1
  17570. The iterated arithmatic geometric mean (AGM) acutally rules them all in terms of elegance, efficiency, and not needing an inversion. I have to look into which one of these (if any) it relates to.
    1
  17571. @ @Mathologer There is a generalization of a mean, which I'm sure you're aware of, where you combine two different means by iterating. I'll struggle a bit to write this up because it's difficult in plain text: Let a0 = 1, g0 = 1/sqrt(2), and d0 = 1/4 Where a is a value of an "arithmatic" mean, g is a geometric mean, and d is a running sum denominator, and "0" is the iteration. Then: a1 = (a0 + g0)/2 g1 = sqrt(a0 g0) d1 = d0 - 2^0 (a0 - a1)^2 a2 = (a1 + g1)/2 g2 = sqrt(a1 g1) d2 = d1 - 2^1 (a1 - a2)^2 and so on ... "^" means raise to the power of, so the 2 is getting raised to successively higher powers each iteration. a^2/d converges to pi quadratically. Can actually get lower and upper bounds by using a_n and a_n+1
    1
  17572. Some say mathematics is racist. Let that sink in.
    1
  17573. Ok not simple!
    1
  17574. Why is it 4k+3 for the ones that can't be done and not 4k-1?
    1
  17575. This is video made me open my mouth in awe more than once. It made me pause and run to my notes, It made me think, It was so beautiful. Thank you
    1
  17576. Reminds me of when about age 12, for fun I programmed a cube calculator without multiplying. In algebra (x+1)^3 = x^3 + 3x^2 + 3x + 1, and (x+1)^2 = x^2 + 2x+1, so x=1, sq=1, cu = 1 loop: cu = cu + sq + sq + sq + x + x+ x + 1; sq = sq + x + x + 1; x = x + 1; write(cu); only I had flipped the loop around and ended up calculating cu(x+1) in terms of sq(x+1) and x+1, so my formulas varied to get the cubes right. (exercise for the reader.)
    1
  17577. I tried to search for F. J. M. Barning's first name in the Dutch language, but no result. :( Also tried Dutch common names like Frank, Frits, Fred, Ferdinand and see if I hit upon something, but still nothing... sorry.
    1
  17578. 2:26 just add another 1??!! ( I have made it till there only for now) its cheating!!!!!!!!!!
    1
  17579. The T-shirt ! 🤣🤣🤣 Actually 25.807 is a better root of evil...😈😈
    1
  17580. Excelent stuff - congratulation on a fun video
    1
  17581. I accidentally discovered this when I was 10 years old! I was looking at a 5-day weather forecast and wondered why we couldn't predict more than 5 days, so I took the differences of temperatures between days and then took those differences to see if there was any pattern. I then gave a forecast for the 6th day.
    1
  17582. Great work 👍
    1
  17583. Yet another intriguing Mathologer video. There's more on Riemann rearrangement theorem and related results by his contemporary Dirichlet in the book Achieving Infinite Resolution, very reader friendly about all things infinity 👍 costs about the same price as sweater on channel
    1
  17584. ❤❤❤
    1
  17585. The beginning of the video gave me the inspiration to calcolate some values. I've found a strange connection with numero 187 and this Is not the first time. Thanks for this and other contenents of previous videos
    1
  17586. I'm always glad to see your videos. I've noticed an underlying geometry to many complex mathematical proofs and even physics itself. This illustrated that idea very nicely. It's good to see your videos again.
    1
  17587. Amazing! One of the best mathologer videos ever
    1
  17588. Clearly you could remove three squares that isolates a corner of the chess board, plus remove any other square, and there's no way to tile it.
    1
  17589. Couldn't this sequence calculus be used to have an intuition about the convergence of the map in the Collatz conjecture?
    1
  17590. He is the man I love the most.
    1
  17591. As much as I am not able to parse most of the content, at least I felt good about being a regular when I recognized the cardioid shape =)
    1
  17592. I liked the greedy algorithm the most, and the fact that all other numbers can be expressed as a partial sum of the harmonic series. It's not really a sparkling, new idea, but it makes me happy.
    1
  17593. The most memorable for me was the optimal 20-blocks overhang. It looks ugly at first glance, but it's such a great design. Thank you for the video, it was really enjoyable!
    1
  17594. I don’t understand anything but the animations are pretty
    1
  17595. Heh, I think I even commented on the Moessner's miracle video about the method of differences :) A remarkably simple introduction to calculus that I picked before I even knew how to multiply.
    1
  17596. Approximating anything via a series which is basically harmonic seems like it should be a bad time – logsrithmically slow convergence.
    1
  17597. 4 squares: no, choose the three flanking the top left green and 1 more from anywhere else, the lone green proves there are boards which cannot be tiled
    1
  17598. clap clap.😂
    1
  17599. Love this video! No math education after high school, but I was good at it at the time.
    1
  17600. I am not a lie detector
    1
  17601. I explored how 2uv/(v^2 - u^2) converges for some regular patterns of Left, Right and Middle moves. In the video we see MMMM.... goes to 1, approximating an isosceles triangle. MLMLML.... goes alternately to 1/2 and 2/sqrt(5); same for MRMRMR. LRLRLRL..... goes to 1/sqrt(3). MLLMLLMLL.... makes a cycle of three of which one 'arm' goes to 1/3 (I cannot deduce the other two). Someone has pointed out already in the comments that the same convergences occur regardless of the starting point in the tree. It is as if the regular sequence of moves operates upon any P-triple and gradually pulls it towards determined proportions. I wonder whether there is, for any non-Pythagorean right-triangle, an infinite number of P-triples ever closer to its proportions. I wonder whether or how it is possible to predict convergence without using my brute-force look-and-see approach. In fact, the whole thing has left me wondering and hugely impressed.
    1
  17602. Phi is recursion outside the box. Root 2 inside.
    1
  17603. Great!
    1
  17604. But there are zero ways to make change for a googol dollars. Even having a googol dollars would cause a collapse of our money system, because it would make the value of coins and dollars in general practically valueless. A googol goes a long way to devaluing any such currency.
    1
  17605. Never ceases to amaze and amuse. Thank you.
    1
  17606. Shouldn't your shirt say 3(ho) rather than (ho)^3?
    1
  17607. This is Mathologer, not the Discovery Channel 😂💝💖💕✨💓 love
    1
  17608. I blame my 7-8th class math teacher for my aversion towards algebra based on math... Until then I used to be a top math class student but after that, due to my lack of understanding of the content and his unwillingness to explain it further, my scores dropped pretty hard and I lost all interest in the class as I was basically just getting dragged along instead of participating in it - Kinda a shame.
    1
  17609. typo: at 28:29, the second row reads 1,3,7,9 whereas it should be 1,3,5,7
    1
  17610. starting with blocking out the square numbers my guess is you end up with x^x numbers
    1
  17611. My favorite from this video is definitely the section on the Euler-Mascheroni constant; the reason I like it the most is because the argument can very easily be adapted to prove the existence of the Euler-Mascheroni constant (that is, the fact that the difference H_n - ln n converges) quite beautifully. Note that this is essentially a visualization of the standard proof comparing the harmonic series with the integral of 1/x, which I like because I never even considered this aspect before. Also, regarding your challenge about the value of the Euler-Mascheroni constant: Note that, since the function 1/x is convex amongst the strictly positive real numbers, the function defined piecewise on the interval [k, k + 1] (with k being a strictly positive integer) by f(x) := 1/k + (x - k)(1/{k+1} - 1/k) is always greater than or equal to the function 1/x (and trivially always less than or equal to the piecewise constant function 1/k on the interval [k, k + 1]). Since all these functions are integrable, it follows by monotonicity that the definite integrals from _1 to some greater integer upper bound n are also less than or equal to one another. More precisely, a simple estimation shows that the integrals of f and the aforementioned piecewise constant function (whose value, remember, is just the n-th partial sum of the harmonic series) differ by exactly 1/2 (1 - 1/n)_. This immediately implies that, as _n dashes off to infinity, the integral of their difference (which, by linearity, is the same thing as the difference of their integrals) converges to 1/2_. From this, it follows that the integral of the difference of the piecewise constant function and the function _1/x_, which converges to the Euler-Mascheroni constant as _n goes to infinity, is greater than or equal to 1/2 (1 - 1/n) (again, by monotonicity)– and, by the limit comparison theorem, this shows the Euler Mascheroni constant is greater than or equal to 1/2. Anyone who hasn’t drawn a picture at this point should do so now, because this proof is way more obvious and straightforward than it looks. Essentially, I am just cutting out right triangles that make up exactly half of the area covered by a rectangle of length 1 and width 1/k - 1/{k+1}. Translating them horizontally so that they line up in the unit square then leads to the desired result. Also, the careful reader will have noticed that I compared the n-1-th harmonic number with the natural logarithm of n_, rather than comparing the _n-th harmonic number with the natural logarithm of n + 1 (which is what was done in the video). Most commonly, you will see the Euler-Mascheroni constant defined as the limit of the difference of the n-th harmonic number and the natural logarithm of n – clearly, these definitions are all equivalent, given that they simply boil down to some routine re-indexing and a single term of the harmonic series (which gets arbitrarily small for sufficiently large _n_).
    1
  17612. Beautiful and amazing
    1
  17613. I have a bachelor's in computer science and I managed to follow everything fine I think. I only had to go back once in a while to refresh my memory on some steps but it was a great watch for sure. I love these longer more complex videos, it feels like we are starting to see the matrix for the first time.
    1
  17614. Very inspiring, thanks!
    1
  17615. Great video! The reason i's are needed in the matrix is because the determinant formula has a (-1)^(i+j) factor in very term. If we define a new formula of matrix which is similar to determinant but losing the (-1)^(i+j) factor, then all we will need are 1's and 0's and the result will always be a natural number. This way is much cleaner and easier to explain.
    1
  17616. Last chapter of course
    1
  17617. I am taking analysis course and the kempner series is given in assignment but I got it correct. Thanks to Internet.
    1
  17618. Holy mother of God, you just blew my mind with those hairs. I might be stoopid, but I had to watch twice up to 4min to get it.
    1
  17619. As seen time showed how it speeds up as it is slowed down,images are stretched but out side of cylinder two times are same rate movement as quickly quicker quick. A longer moment.math how? In invisible energy slips between that is magnified here.
    1
  17620. I want the infinite number of your identical books to build up the hanging tower.
    1
  17621. Best part is brick stack really appreciable and ofcourse more fascinating stuff is definitely seen in sir's video 😀 🙏.
    1
  17622. I'm sad :( I can't hear what you're saying. I turned my laptop volume all the way up, then the YT player volume all the way up, and still had to run on captioning because it was just too quiet and I kept missing words. Then trying to read the captioning while looking at the equations and diagrams was just too much, so I'll have to save this for some other time when I can plug in some amplified, noise-cancelling headphones.
    1
  17623. This is beautiful
    1
  17624. You kept my attention for 50 minutes, so you must be doing something right :)
    1
  17625. Most memorable is the astounding "sum 100 zeros > sum no 9s" given the first term is 1/googol but ends up as ~23 !!!
    1
  17626. ill accept calling 1 an honorary prime.
    1
  17627. Thanks for the degree in master of hyperspace. I directly tried it out by passing through the wall via the fourth dimension to shorten the way to the fridge, however turns out i still need a bit more practise. So for the moment i am back to use cut outs.
    1
  17628. Wow, I really want to know how this equation can be applied
    1
  17629. When the bugs are constantly rotating towards each other rather than incrementally, you might say the bug they walk towards isn't moving away from them - always perpendicular, with larger steps this isn't always true, they are also turning but never turning away, always turning inwards. They started out a distance of 1 away and that's the minimum distance you'd need to cover to meet one who is never walking towards you, and the maximum distance you'd need to walk to one who is never walking away from you, at any instant along their walk they'd always meet those requirements and that's sort of what limits are meant to describe.
    1
  17630. why can't we put a double like?
    1
  17631. Great video! This would be much harder to follow watching a tenured professor draw on a blackboard. You could add a sentence about what it means to invert a matrix. I hadn't looked at that kind of algebra for 30 years. People don't often calculate inverse matrices in the field, because the number of calculations blows up with even modest dimensionality. I have a MS in statistics, but I've mostly worked in construction.
    1
  17632. Can you help me please??? You take (x+2)^3 , you replace x= -1 , you have Euler's formula V+F-S=2, that's fine. But I do not understand why only x=-1 works for all polyhedra.
    1
  17633. 😏 Thank you for surprising every time!
    1
  17634. This is marvelous, I understand all of the words in the video.... Most of the math too.
    1
  17635. Could the thing with the cube slices have been Fermat's proof?!??!!?! [X-Files Theme plays while putting on a tinfoil hat]
    1
  17636. I had no idea tengon was a word.
    1
  17637. 15:19... way too close for comfort. Amazing that you do your own stunts though.
    1
  17638. 35:45 I totally feel you. But now, having felt that in your skin and being in a position of teaching; how do you handle it with your students?
    1
  17639. WOW 😍
    1
  17640. I learned these division tricks when I was young, but I can't remember where
    1
  17641. Very nicely done. I like the dot proof since it is "Aboriginalish." I like the shorter vids too. TYVM
    1
  17642. At 6:29 can't the red prime be written as '4k-1 where k is natural number
    1
  17643. Ah yes this is some difference AP thing in jee syllabus
    1
  17644. I made it to the very end... after reading the comments ;)
    1
  17645. Way...way.....w a y......over my head. I was always given the excuse that i did not inherit the "math gene". After watching this video I think it is more than clear that I did not have proper math instruction. Brilliant vid filled with wisdom which encourages me to start at the beginning and take it one step at a time. I never hated math - I just did not understand it. Thank you Mathologer for visually showing me, and helping me realize a map to understanding "aha".
    1
  17646. Coiling is a complex number equation and needs square root of negative one for the rotation?
    1
  17647. And he did it again! <impressed> In answer to the Q: I liked all parts. The intro with the missing corners from the chessboard was a bit well-known, but hey it's the intro; and the explanation of the number 2^(1+2+...) I found a bit tedious since quite easy to see once the domino-dance-construction was shown, but better a bit too clear than too obscure. Kudos again for such a video treasure! Now after the square and the hex/triangle tilings I wonder about whether similar ideas could hold on a Penrose tiling where the tiles cannot be paired 1:1 into dominoes but still occur in a nice golden ratio. Oh well, food for a Xmas worth of thought...
    1
  17648. 15:54 5
    1
  17649. Dear @Mathologer, great video as always but my tangential comment concerns your T-shirt! Yes, yes, "25.8069 IS THE ROOT OF EVIL" was funny and I got a chuckle out of it until I realised it's not correct and there is something very unmath-like. The square root of 666 is 25.80697580... Now, for practicality it's OK to choose an arbitrary level of accuracy, say 4 decimals BUT according to the rules of rounding, if there are 4 decimal digits are shown, the last digit is not 9 because there is a 7 after it which forces rounding up!! Or, to keep with the evil theme, was this done intentionally?! No matter what, the OCD community is upset! =:-o
    1
  17650. Ahh, it's now obvious the paradox occurs from taking a number of terms that's a power of 10. I can sleep peacefully tonight.
    1
  17651. i learned it in 11th grade (almost at the end of high school, i’m 17 years old) in America this year!
    1
  17652. most memorable part for me was "5 seconds to answer for my life" and learning I was wrong. Still I am alive
    1
  17653. Hi. I'm in Australia spiritually, Dreamtime is all the time. Thanks for ripping apart the 4D metric crap from a position of some authority. Banach-Tarski is based on the binary scaling of a DYNAMIC hypersphere, with thermodynamic uncertainty at the polar, equatorial and orthogonal planes.
    1
  17654. This is a little bit INTERESTINGB!!!
    1
  17655. Weird thought, but - The jingle between chapters like 21:17 kind of feels like an oddity. It made me think of, what if you take the square root of 2 and simply applied it to piano keys?.. Nice video. I can't really keep up with the finer details of the mathematics, but I do feel like I grasp the overall concept, so that's nice
    1
  17656. I think the coloring proof, probably because it seems more explainable to a grade-schooler.
    1
  17657. is it kΣn=0 [(n+1)²(k-n)] ?
    1
  17658. Dear Marty & Crew & fans - My gratitude & delight @ finding this one (my new #1 5.5-Star episode) prove a natural double-infinity (squared?)! I've been thinking about another, more organic example, but this seems to be a clincher, re: my work-in-progress on the metalogical "roots" that connect RH, GC, and the TPC (etc.). For example, can use the circular log slider (and/or the dynamic {infinitely self-optimizing} version) to prove the natural principles enabling circles, the wheel, log(n), n = pi, the unit circle (& trig), C (the complex plane), Phi, Fs (the Fibonacci series, Platonic solids, and Bucky balls (etc.)? I think I can. What do y'all think about that? ~ Thanks again! Sincerely ~ + Oh, BTW, if you want to check on my progress, my 2 (interrelated) preprint drafts (A: "Riemann, Metatheory, and Proof" & B: "Theory and Metatheory of Atemporal Primacy" and a brief bio) are accessible via RearchGate and > ORCID(.org)/0000-0001-5029-7074 < FYI: Paper B is a bit like my Special Metatheory of metalogical relativity that clears up the confusion about time, space, energy, causality (and other primal principles), QM, ontology, "materialist cosmology" (& other logical fallacies), etc. (all without reliance on insufficient authorities, logical fallacies, etc.). + I think you may enjoy them (and hope you do). Thanks again & keep up the great work! ~ M
    1
  17659. 2 = 1/1 + 1/2 + 1/3 + 1/6 3 = 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/10 + 1/12 + 1/14 + 1/15 + 1/20 + 1/28 Don't know if latter is optimal representation for 3 by largest denominator
    1
  17660. You won my mind🎉.
    1
  17661. Two steps of this process gives some sort of discrete second derivative. Can this be related to mean curvature flows for polygons?
    1
  17662. I had no idea what this video was about when I clicked it, funny part is that I am designing my own calculus wheel right now for a game, set value on outside ring for time of day & police heat, windows in the wheel reveals values for chances of getting caught doing different crimes.
    1
  17663. The Golden spiral pizza 🍕 at shirt looks tasty Yummy :) Einstein = man of (century)^2 E = m c^2
    1
  17664. In case If anyone need the book of bernoulli mentioned in the video Visit the website and click on GET http://93.174.95.29/_ads/2A824BCDB31B45A94882ACE89EAAA35E
    1
  17665. You blow my old mind! Takes weeks to recover. Keep it up!
    1
  17666. But what about A^3 + B^3 + C^3 = D^3 + E^3 ??
    1
  17667. 2:19 “Come up with a proof of your own” not even 5 seconds later I realised that all three lines are the same length, so if they’re all tangent to an inner circle connected at the midpoint they will always make an outer circle! …but how to actually make sure they’re all connected at the midpoint or show that it’s true whether it touches at the midpoint or not?… Nope, I’ll just enjoy the video (And, silly Mathologer, I didn’t click for the enticing Iris, I just trust that your videos will be interesting)
    1
  17668. Ah yes, my maths teacher in year 7 was so stunned by how much I already knew that he tested me on the linear version of this. As in: "Given the first few, what is the 100th sum?" I made a whole formula using the triangle function, he was impressed.
    1
  17669. You're right. Let's also exclude 3 as this is the only number divisible by 3. All others are not.
    1
  17670. Pluto not being classified as a planet has literally zero impact on anyone's life. If you want to teach your children that Pluto is part of the solar system, just go ahead. You can also teach them that Ganymede (third moon of Jupiter) and Titan (sixth moon of Saturn) are both bigger than Mercury, meaning they are both bigger than actual planets. Meanwhile, our Moon is bigger than Pluto itself. Also worth mentioning is the dwarf planet Ceres, which orbits the sun somewhere between Mars and Jupiter. If Pluto is a planet, then why not the much closer Ceres?
    1
  17671. Great video with great content. I'm a math newbie. Didn't really do super well in high school maths. But always love watching your videos. Lots of highlights, but for me what really gets my vote is your visualizations. Superb!
    1
  17672. You add one new baby square per A(N-1) -> A(N), thus the combinations grow in the A^1+2+..+N
    1
  17673. What software do you use to create your animations?
    1
  17674. Great video, appreciate you leaving the gaps in just the right places for us to work on! Bachelors in Math and Physics doing a career in data science now, so seeing your videos is always a treat to remind me of what I enjoyed most in my studies!
    1
  17675. Using this I was able to find a formula for the numerator and the denominator of the fraction used to find the sqrt(3) in a previous video. Not surprisingly the numerator contains a 3^floor(n/2) and denominator contains 3^floor((n-1)/2). Everything else is same
    1
  17676. Happy newyear2022
    1
  17677. Okay so, I actually found a similar pattern looking at exponents at school. What I found is that if you take the difference of exponents of a certain power, they all meet up. Squares: 0 1-0=1 3-1=2 1 4-1=3 5-3=2 4 9-4=5 7-5=2 9 16-9=7 16 This works on further powers too; it also works if you count negatives in the equations. (It does not work for negative powers) Cubes -1 0-(-1)=1 1-1=0 6-0=6 0 1-0=1 7-1=6 12-6=6 1 8-1=7 19-7=12 8 27-8=19 27 And as you continue you may notice the number they meet at increases, well I found the number they meet at is the power you chose factorial. So squares (power of 2) meet at 2!, or 2. Cubes (power of 3) meet at 3!, or 6; and this continues through infinity* (not 100% certain, but I'm pretty confident). So that's my brief little thing I found, not sure if it correlates to this in anyway but I found it pretty interesting and it loosely ties to the subject. There's a lot more I could add but it's already pretty long and confusing, sorry if you couldn't understand.
    1
  17678. I confess I clicked because I thought it read 'Heroin's formula,' but I stayed out of amazement!
    1
  17679. Interesting (for me) to see that 3,4, 5 is the only primitive pythagorean triple (I think) where the arithmetic mean of the sum of the first and last number is a part of the triple.
    1
  17680. The most memorable proof is probably the proof that the sum of reciprocals of numbers without a certain digit is less than 80. I also enjoyed the math paper with the Mathematica program in the description.
    1
  17681. Nothing can beat chapter five
    1
  17682. Lex Luthor is one o my fav teachers
    1
  17683. Most memorable: the optimal 20 block tower. I think it's beautiful!
    1
  17684. This method was not talk in 1960's-to present New York State USA public schools
    1
  17685. Challenge for the keen among you: My proof is that seeing is believing! :P
    1
  17686. I think mathematicians need to mention to non-mathematicians more often that polynomials with a dummy variable are a convenient way of saying that you're going to do place-value arithmetic where the places don't need to carry into other places. Instead of saying digits are 0 through 9 and you split groups of 10 into the next column, "digits" can be any number, and you never split anything into a new column. It looks more advanced, but it's really basic arithmetic without the trickiest step, and with really strange punctuation for separating columns.
    1
  17687. Is this really a paradox? Maybe it's clarified later but so far, 7:30 in, this "proof" that the sum = ln(2) is hardly rigorous. Just because the sum tends towards ln(2) as n tends towards infinity, does mean it is ln(2) at n = infinity
    1
  17688. Now imagine the light we suck up from the supposed end of the universe. But in reality it is like this rubber band. Some weirdos think that the universe is expanding because they don't want to understand other effects. - just for fun ;-)
    1
  17689. Awesome as always... Thank you so much. Can you make a video on how dose john napier create the first logarithmic table without any kind of computer and only by hand?
    1
  17690. Yep, seen it! It occurred to me one day that I could make a circular slide rule, and my favorite spreadsheet did it very nicely. BUT ... I didn't know your technique with the stretched rubber band. That is good. BUT ...what is the ratio of straight-length-of-rubber-band to circle radius?
    1
  17691. The correction terms 1 ,4 , 9 sound like n^2 the square numbers. Could that be that the correction numbers go up in " 1/(n^2) ? ,1,4,9,16,25 etc. Are there infinite correction terms?
    1
  17692. Awesome video, already waiting for the next one! 😄
    1
  17693. greedy algorithm
    1
  17694. That formula looks VERY similar to Taylor polynomials. I assume they're related somehow.
    1
  17695. The 54132 Row is even so the sign changes
    1
  17696. Herons formula is taught in 8th SSC Boards India
    1
  17697. Awesome video!
    1
  17698. No you can't always. Example is b1 b2 a2 and a fourth that is not a1
    1
  17699. Awesome! I wish these dropped more often
    1
  17700. I made it to the very end ;)
    1
  17701. We landed on the moon in 1969. I was a sophomore in high school in 1972. I used a slide rule for physics and chemistry. A 4-function calculator in 1972 was about $125, but that was a fair bit of funds then. The next year, they were $25 and a basic scientific calculator was $125. The next year the 4-function calc was being given away for opening a bank account somewhere, the scientific calculator was $25, and I bought an HP25 programmable scientific calculator for $125 for college math and science classes. Now the scientific calc is an app on your device for free, and you can eliminate the ads for a buck or two. Progress.
    1
  17702. Yay! Someone who knows how to pronounce Ramanujan’s name properly!!
    1
  17703. Here's a simpler proof that crisscross is shortest. Consider a (vertical) row of eyelets as like a mirror facing the other side of the shoe, and the path of the lace as like a beam of light reflecting between them. It must reflect because your lacing is tight. If the angle of incidence equals the angle of reflection, you get the crisscross. If the two angles differ, you don't get the shortest path by an easy geometric observation involving congruent triangles, reflection, and straight lines. This still leaves out the top and bottom, but you handled that bit masterfully at the end.
    1
  17704. Most memorable - the tower of Lyra. Honestly, that put the harmonic series' divergence into perspective.
    1
  17705. aight heres a quick whack at the first puzzle: in addition to the n^x count of squares (1+4+9+16 or 1^2 + 2^2 + 3^2 + 4^2, summarised as the sum 1->n-1: n^2 where n is the dimension of the point matrix), each of these squares has additional "tilted squares" that can be drawn equal to the amount of "unnecesary vertices" in a given side of said square, that is, points which a given "level square"'s side passes through before reaching the main vertice on each end of the side. so the 4*4 square (the largest one that passes through all the perimeter points) has 3 tilted squares (the first of which is given as an example with vertices {[0,1] [1,5] [5,4] [4,0]}) from their you can imagine the other two, one being a mirrored version of the example, and the other passing through the centre points on each side of the matrix. Since theirs only one of these 4*4 squares, the amount of squares is (1^2)*(1+3). Following this to the 3*3 squares, they only have 2 "unnecesary vertices" so each will yield 2 "tilted squares", as there is four of such "level squares" that gives us (2^2)*(1+2). Following this pattern we can say the reamining smaller squares amount to (3^2)*(1+1) and (4^2)*(1+0). (1^2)*(4) (2^2)*(3) (3^2)*(2) (4^2)*(1) -------------------------- 4+12+18+16 = 50 This formula can be summarised entirely by: For a square n points wide and n points tall, the amount of uniquely positioned squares that can be drawn with vertices within this square of points; summation of x = 0 --> n : (x^2)*(n-x) --------------------------------------------------- this essentially gives us our previous calculation for a points square of 5*5: x = 0 --> 5 x = 0 : (0^2)*(5-0) = 0 x = 1 : (1^2)*(5-1) = 4 x = 2 : (2^2)*(5-2) = 12 x = 3 : (3^2)*(5-3) = 18 x = 4 : (4^2)*(5-4) = 16 x = 5 : (5^2)*(5-5) = 0 Total sum = 50 Edit: x = 0 --> n : nx^2 - x^3 is also valid if you don't like brackets I'll do the cubicle grid another day
    1
  17706. I know calculus now
    1
  17707. And the black squares are always at diagonal lines
    1
  17708. You know when I realised the cube had 6 faces a long time ago it struck me as strange, as if there was something important in that number. Turns out there was.
    1
  17709. I'd have to vote for the proof of the convergence of the "no 9s" series at the end, it was definitely worth the wait. Though, the bizarre maximal overhang towers would probably be a close second for me. Also, regarding 37:00, I wonder if every convergent sub-series of the harmonic series must have this property. That is, must every convergent sub-series of the harmonic series have the property that it includes proportionally fewer and fewer terms of the harmonic series the further out you sum?
    1
  17710. A peculiar bit of bad luck related to Conway's construction: the area of the large hexagon formed by the 6 yellow points on Conway's circle is at least 13 times the area of the original triangle. Try to prove.
    1
  17711. Her we go: http://rednebula.com/html/arcticcircle.html The program for the simulation. Enjoy!
    1
  17712. Colouring
    1
  17713. The ark of the covenant is the 60° ark of a circle ⭕ 🌐??????
    1
  17714. 'calculus-free' should be said between air quotes: when you were adding the thin cylinders together (and then the equivalent disks), I was mentally giving them an infinitesimal thickness for the trick to make sense in my head) I think one could obtain the exact volume calculus-free by squeezing it between two infinite series: one corresponding to cylindrical slices with radii 1/n and the other one rwith adii 1/(n+1). Then split each interval [n n+1] in two and repeat. Obviously this still requires invoking a limit but I feel like it's a kind of limit that's a bit more accessible, as it's more similar to what we've been doing with infinite series from the start.
    1
  17715. I liked the coloring proof best.
    1
  17716. youtube never recommends mathologer.... but, it's a part of youtube we mathologists love.
    1
  17717. Wondered about this. Seen other proofs but they were not as satisfying.
    1
  17718. I am a structural engineer and have a collection of over 100 slide rules. I am fairly proficient with them, and a couple of years ago, I deliberately engineered a (small) structure using ONLY a slide rule. No computer, no calculator. Just for grins, but also kind of in homage to the engineers before me. They went to the moon on them, so surely I could engineer a dinky little building with one (Versalog 1460 is my "drug" of choice). They are fascinating analog computers and really give you a sense of proportion and an appreciation for the genius that went into devising them.
    1
  17719. Great video Burkard! Thank you Where can I get that awesome Mathflix tee shirt?
    1
  17720. Wow your voice was great when you said "the valley of feaaar". You should do voice over work lololol
    1
  17721. I answered "no" because I figured out how to do it in minimum moves, but I always get distracted and mess up much shorter towers. Maybe the evil toymaker would help me focus.
    1
  17722. I had seen the Euler–Mascheroni constant visualization before, but it seemed so arbitrary. This explanation will get me to remember it. Very well done, and my vote for favorite part of the video.
    1
  17723. 22:57 A glitch in the mathrix
    1
  17724. I wonder if there's a way to formalize the notion of and find the "slowest" diverging series.
    1
  17725. hmmmmmmm........I dunno..... the title of Petr's article does sound a little bit like clickbait for this channel's subscriber base ;-)
    1
  17726. I don't exactly have the math skills for this, but maybe someone will. I'm trying to think about the solution presented here as heuristics or tooling for the travelling salesman problem So instead of listing the one sequence of traversing the n-gon, instead consider a transformation of n points into an n-gon by connecting them, and then reducing it to the regular n-gon. What are the properties of that n-gon, especially its edge length (which will be linearly attached to the radius of the circle it is drawn into). Then, consider if there is a link between a connection of a different sequence of the n points and the resulting shape and the resulting regular n-gon. Can it lead to a better solution of the traveling salesman problem? If there is a relation between starting shape across the n points and the resulting regular n-gon, can it be used to estimate the results of different wiring? Maybe at least it can be used as a data structure holding the possible ways to traverse the points? Can we interpret the "regular" looking partition n-gons as "efficient" turns and the inward going as "sharp turns" and in this way infer which steps in the sequence to flip, instead of another geometrical analysis? Am I making any sense?
    1
  17727. You are very well acquainted with your snakes. I'm still in the process of learning how to tame and manipulate them. I hope I can become a great snake charmer as you are someday.❤
    1
  17728. 1a. Let the numbers in the box be (from top left clockwise) a, b, a+b and a+2b. Then we have a(a+2b)=153, a(a+b)+b(a+2b)=185, and subtracting we get 2b^2=32, b=4. We also have 2b(a+b)=104, and substituting b=4 we get a=9. This means the numbers are 9, 4, 13 and 17.
    1
  17729. 1b. RRRL
    1
  17730. i paused this and had a sit down early on, and proved the triangle bit to myself in almost the inverse way, scaling up (I'm paused again, just after the nested hexagons so i might be about to say something stupid that gets overidden later but f it); with a unit grid, an equilateral triangle with a base bridging 3 points on a line (can't have a single bridge, theres nothing above to connect to) would require a multiple of rt 3 for an integer height, so disproves itself, so we move to configurations with rotated bases. The length of the base scales to n/sin(atan(n)) where n is the number of steps of deviation in a direction orthogonal to the original base, but this configuration essentially creates a new grid of sides equal to the scaled base length. For the cases of odd n (where grid points exist above the centre of the base), it ends up requiring a multiple of rt 2 (+ 1/rt 2 offset for the offset,) to for integer height (still impossible) and even n (such as 0) the rt 3 case. So each new configuration basically reverts itself back to the original problem. I don't think i've made any sense and should stop drinking while i try and maths
    1
  17731. I was actually hoping this would be about lacing angles and resistance to pulling when tightening the laces. Can we get a proof on that? I suspect the bowtie lace will be on the easier to tighten side given the nearly sinusoidal curve of the laces reducing the amount of pressure on each eyelet as it gets pulled... Another thing would be the lacing with the most resistance to change, for people who want to lace their shoes up to a specific tightness and have them stay that way. I suspect something like a weave would be best there, since pulling against other laces which have texture and flex and like to bind up, would cause a lot more drag than pulling against smooth eyelets.
    1
  17732. thank you .. i needed that video
    1
  17733. A look into landau ramanujan constant...how is the idea?? Looking related
    1
  17734. The sequence of the rationals that are partial sums of the harmonic series, as we saw can never be integers. Also, this sequence is always increasing. Is there any other class of rationals that are skipped completely?
    1
  17735. 11:05 Thank you for proving very conclusively, logarithms are not numbers, but geometric objects, proportions (logoi, in Greek), even if they shade the arithmetic objects called numbers (hence the other part of the name).
    1
  17736. My solution to calculating the number of squares on an (n,n) grid would be: S=sum((n-k)^2*k from k=1 to k=n-1) where S is the number of squares. This is derived by projecting the squares on the horizontal/vertical axis of the grid. For example, on a 6x6 grid a tilted square could be the one, where you move from corner A to corner B (2,3) (=2 on one axis, 3 on the other). This square obviously projects as a size 5x5 square on the grid (you need horizontally and vertically 6 grid points to draw in the tilted square). It fits once inside the 6x6 grid, as you can neither translate it vertically nor horizontally. The table of possible squares on a 6x6 grid would therefor be: (side length): (vertical distance, horizontal distance),(...) 5: (0,5),(1,4),(2,3),(3,2),(4,1) where (0,5) is the obvious big enveloping square. These all fit once. 4: (0,4),(1,3),(2,2),(3,1) 3: (0,3),(1,2),(2,1) 2: (0,2),(1,1) 1: (0,1) Notice how some combinations appear again (but mirrored). Technically all should appear twice, only that there's no visible distinction between (0,j) and (j,0) as well as (j,j) and (j,j), which is why they aren't counted twice. The number of times a square occurs on the 6x6 grid is then easy to calculate by summing up how often its projection side length fits into the grid. For example all with projected side length 3 fit in a 6x6 grid 9 times. The number of times side length k fits into a NxN grid is (N-k)^2. Therefor we get the total number of squares as S= sum((n-k)^2*k from k=1 to k=n-1) (The sum could also start at 0 and end at n, but both terms are 0)
    1
  17737. If the center of the bricks is a black hole, and the outer reaches of the bricks the ends of the galaxy. The weight down would be represented in one form as the jets out of the black hole. While the upwards weight of the table would be the other jet. So when a star gets ejected out of a galaxy, it would be like a brick getting flung out of its half bridge (Brahmaguptaian half moon). I can literally feel myself holding a galaxy in my hands, while the sands shuffling under my feet was displacing the weight as another galaxy at a harmonic distance. Like elongating ones feet in a muddy brook, the kind at the end of a canal. Soft sand, no, more like organic matter.
    1
  17738. relation with gamma and analytic continuation :)
    1
  17739. i m an interested german year 12 student(who admittedly usually cannot be bothered to research beyond what the video explains) and while i had little trouble following along and i too must agree that convergence proofs or the like would hardly be appropriate in this format, it did bother me just how many proofs were reserved for a later date. while this structure is interesting for getting a high level overview while keeping the video relatively self contained, potentially you could attempt to make videos in such an order, that one isnt left with 5 cliffhangers :P
    1
  17740. Liebe aus Deutschland 》》》》
    1
  17741. the ":)"s in the captions are the most wholesome thing
    1
  17742. The third peg is shorter.
    1
  17743. 6:14 Why not 4k-1?
    1
  17744. I like the first solution because it is more dramatic to explain and stretches the story 😁.
    1
  17745. I love phyllotaxis, that aloe in my garden knows what's up.
    1
  17746. You only have to rotate those 8 liters by 90 degrees.
    1
  17747. This video is the most beautiful christmas gift. Thanks. - a Slovak pensioner :-)
    1
  17748. That “Tristan’s fractal” really mesmerized me when I first saw it. I genuinely thought that the sum of the no 9’s series would diverge just like any other harmonic series, but that visual proof really got me dumbfounded. What a nice way to start my day indeed!
    1
  17749. I wonder can we "expect" the need for these repetitive correction terms in some way from the simple Leibnitz formula? Is this typical for alternating series?
    1
  17750. The summary of different cases for the windmill's link doesn't work. Can you please fix it in the description.
    1
  17751. TY!!
    1
  17752. At 8:21, my guess is that 3^n + 1 is the required triangle length for the shortcut to work. That would mean 28 is another shortcut number..... Let's find out! (The suspense is kill me!!) Update: 9:24, the suspense didn't last long. I was right! Woooohooooo!!
    1
  17753. The euler constant was the most memorable. Amazing video again. Learned a lot👨‍🎓
    1
  17754. the most memorable moment not yet mentioned for me was the animation from the 4 block high leaning tower of lire to the 20 block high one. i find it to somehow be both soothing and jarring.
    1
  17755. Why? Well, I was taught that.. but not in America…. In America they are busy BRAIN-WASHING people with LEFTIST AGENDAS, and the dumber you are the more LEFT you’ll go… Good video
    1
  17756. When we start our actual working lives and get out of uni we never meet people with the same teacher soul like mathologger. Its very depressing having to live like this. I thank my lucky stars for this channel
    1
  17757. Yeah, let me know how that thing with the corners of a room works out for you. I'm not convinced there are any buildings on the planet perfect enough for that to work out. In my experience, once a building is a few years old, you're lucky if the walls are *straight*, let alone at right angles to one another. And there is absolutely no such thing as a perfectly flat floor, of that I am completely certain.
    1
  17758. The fact that, besides the very first one, no partial sum of the harmonic series is an integer really blew my mind! Really cool video, everything in it was super interesting. And explained super well :) (Shoutout to Tristan, who I actually went to Uni with 👋)
    1
  17759. Great video. Small gap where the 582 vs 581 term wasn’t highlighted in yellow. Now I’m going to have to calculate which one is right… time to start doing fractions 15:21
    1
  17760. learned the second proof in school, the first in an earlier Mathologer video. Since then I watch all your videos and bought several of your books. I would recommend your videos to everyone who has the slightest interest in maths.
    1
  17761. Such a nice video. Again. :) A theorem that "should have" been discovered 2000 years ago and A new(?) proof by a YouTuber*. Amazing. But why is the tree on his shirt up-side-down?? * ) okay okay not just a YouTuber -- but still a proof for YouTube.
    1
  17762. Dk what the title means, we were taught this method in my school
    1
  17763. 3+1/(1*2*3)-1/(2*3*5)+1/(3*4*7)-1/(4*5*9)+...=pi
    1
  17764. Why would the graph whose vertices are tilings which are connected by edges if they can be obtained through the procedure at 47:41 be a connected graph? Doesn't seem obvious.
    1
  17765. For me the its the "never an integer" part thats the most interesting
    1
  17766. I found myself wondering how Ptolemy reconned the cords in the days before Arabic numerals and decimal fractions. It turns out the ancient Greeks had a rudimentary decimal system and then he used fractions of 60th and 3600th parts. He even included an interpolation factor for smaller divisions. Details in Wikipedia for anyone interested in the details. https://en.wikipedia.org/wiki/Ptolemy's_table_of_chords
    1
  17767. Marvelous & lovely!
    1
  17768. This is a gorgeous Mathologerized proof <3
    1
  17769. What is bigger: the number Googolplex or the Number of elements in the harmonic series to sum to > googol?
    1
  17770. If you used that four green plus to red equation but swapped green for light and red for dark, or green for magnetic and red for magnetically attracted, then you would be explaining the propagation of laser light, not just a silly cylinder. I really got a lot out of this video. Thank you very much and merry Christmas!
    1
  17771. Most memorable: bizarre overhang towers
    1
  17772. Has no one done the glasses challenge? It's 666. The two 2x3 blocks on the ends need to be tiled within themselves; each has three tilings, tripling the total number of tilings of the entire figure. Removing those two pieces, we get three 2xn rectangles connected by "bridges"; thhe bridges can each be tiled in one of two ways, and the choice shaves one column off either the rectangle to its left or the one to its right. The original rectangle widths are 3, 4, and 3, and the bridges break either left/left for widths of 2/3/3; left/right for 2/4/2; right/left for 3/2/3; or right/right for 3/3/2. A 2x2 tiles 2 ways, a 2x3 tiles 3 ways, and a 2x4 tiles 5 ways, so the total number of tilings without the two removed pieces is (2*3*3 + 2*5*2 + 3*2*3 + 3*3*2) = 74. Times 3 for the three tilings of the left removed rectangle and times 3 for the three tilings of the right rectangle gives us 666. A very fitting number for what appear to be 2020 New Year's glasses
    1
  17773. Of course we start at f(0), say computer science students.
    1
  17774. I immediately subscribed this channel when I saw 3blue1brown response here.... 2 brilliant channels
    1
  17775. Thank you mathologer! It’s so amazing! This E-M formula! Yes I can follow. Not easy but I am just just able to concentrate and follow it. I watched part by part spreading several days so that I can let it sink in. Guess it helps to pause at the right times as needed . I also benefited from some “math is fun “ revisiting for the matrix part. I’m from Hong Kong and aced in Maths decades ago but they didn’t teach matrix at that time. Been reading maths tubes and books recreationally in the past few years after a 30 years hiatus. What reignited my interest in maths is stumbling upon the square root of minus one… which is such a special thing. A big thank you for your effort!
    1
  17776. 10:36 Another pattern is with the bottom numbers the next one is a multiple of 3, 5, 7, 3, and 11 respectively with the previous one
    1
  17777. Circular slide rules have been around a much more than you imply. I remember one such device for physicians to calculate the correct dose of antibiotics. However, what I am not sure I understood is the stretching argument. The rubber ban stretches less and less the more numbers have been wrapped around the wheel. But how is it obvious that it will be a logarithmic scale? The band will stretch ^L (delta length) =c * L2 where c is a constant and L2 is the total length L minus what has been wrapped,L1 (R* angle) - R is the radius of the circle. So the total stretch should be the integral of ^L * R ^Angle (from o to 2 Pie) = R 2Pie - L. Can you help me see where the Logs come from?
    1
  17778. Cubes rolling. Dimensional travel.
    1
  17779. They did not teach it because the teachers didn’t know it. Why? Many lack intellectual curiosity. It drives me mad having taught them.
    1
  17780. The teachers should have been taught this. We lost our passion.
    1
  17781. Great video as always! We sent men to the moon using slide rules. When I was a young engineer (circa 1988), I was chatting with one of my mentors that was retiring that day. As he was packing, he pulled out a slide rule. I marveled at it, but I didn't know how to use one. He gave me a quick lesson and then gave it to me with the promise I would learn how to use it and hand it over and teach a worthy new engineer how to use it when I retire. As I approach full retirement, I need to find a young mind that will honor this tradition. I am thinking about teaching STEM in the summer from home. Maybe I can pass it down to one of my students. From that day on, whenever I leave a company for newer horizons, I always give an engineering a trinket of sort to a young engineer (usually right out of school). Since I work with mostly electrical engineers, it is usually a brand new high quality Digital Multi Meter (DMM) - if I know they don't have one. It is a nice tradition. I also dispense some sage advice - I make a little ceremony of it. One made me teary eye as he thanked me for being his mentor.
    1
  17782. P.S. The name of Lorenzo Mascheroni (of Euler-Mascheroni fame), is pronounced Maskeroni. 'che' in Italian sounds like 'ke' ;-)
    1
  17783. ‘This isn’t the Discovery Channel’ Nice dis. Mathologer > MSM
    1
  17784. I haven't found it easy, but I am mostly self taught, and watching your videos, I review the sections on a site called Paul's Online Math Notes, the trick is I go back and forth between the Algebra Section and Calculus section, I just care about lines, I know that X,Y Algebra is just Calculus with Z variables being constant. I have learned 1 line/equation in this method. PT = P0 + VT The vector form of a line. I'm trying quadratic line now to get an arc but life has been getting in the way, so I can't focus
    1
  17785. Explain the negative domain of x^x
    1
  17786. Slide rule: only one child at a time.
    1
  17787. Can quantum computers dream?
    1
  17788. 😍 Awwww, did you see that (x +2)³ how cute!
    1
  17789. That is really neat. I feel like there might be more work to be done, but of course I would.
    1
  17790. Is this why Fibonacci came up with his numbers?
    1
  17791. Complex too complex i need to ré-Load vidéo
    1
  17792. 19:45 - it makes intuitive sense that the area of circle is pi x diameter of circle, too
    1
  17793. As always great video :) . Small typo at 21:10 though
    1
  17794. The most memorable were all the little tidbits about Euler-Mascheroni constant
    1
  17795. I think the most relevant and related Numberphile video is this one featuring Ben Sparks: https://www.youtube.com/watch?v=sj8Sg8qnjOg
    1
  17796. We've removed 2 squares from the chess board,now it is completely and utterly mutilated.
    1
  17797. This reminds me of a 4d object.
    1
  17798. I want to understand all this in terms of what I can only reference as "the unconventional perspective which I've seen from Wildberger's rational trigonometry approach". Because I expect that when using his approach to think about algebra, functions, and equations, I'll understand trigonometry in terms of the way we use to write down numbers without relying in any specific/concrete scheme to write down numbers. In other words, I think that there's something about the decimal point that is quite comparable to the parenthesis: parenthesis always show up in pairs; similarly, the decimal point always groups together two sides (>0 and <0). But to say more I need to know more. But to know more I need to write boring bullshit to demonstrate to a panel of experts I've understood things about first order logic which I need to do so I can finish this master's degree in computer science, but fuck me if I will put myself in a similar position... i.e. I don't want to pursue a PhD cuz academia is european colonialism which is american imperialism; yet, how else can I learn this stuff?
    1
  17799. No nines is crazy! So Unintuitive! that's my vote!
    1
  17800. What a great video, my favourite part was the algebra right at the end!
    1
  17801. as always the mathologer videos are amazing.
    1
  17802. Sophie Germain's sequence even has a pretty neat chain there. 2×2+1 = 5 2×5+1 = 11 2×11+1 = 23 sadly it ends here 2×23+1 = 67 isn't on the list, it's an ordinary prime haha Next time we get any chain action is 41→83 and 89→179, and it seems like a fluke more than anything interesting
    1
  17803. all parts were indeed very interesting and memorable particularly "the leaning tower" and the "no 9s series"
    1
  17804. I love this, amazing work !
    1
  17805. They should give this guy subtitles. Hard to understand him.
    1
  17806. Well, to me at least, being featured on a mathologer video as brilliantly crafted is more than enough compensation for the fame it lacks and it deserves!
    1
  17807. I was taught to use digital root in a method called ‘casting out nines’. We used it as an error detection method for arithmetic operations. i.e. The digital root of the output of an operation should match the operation acting of the digital roots of the inputs.
    1
  17808. This is beautiful, thank you. I remember learning this rules and asking my teacher about where they come from / proofs. And she refused to tell me. Probably she did not know. :(
    1
  17809. This is weird - if I use Shazam to find the song manually during the end credits, but have it playing back >1x speed (e.g. 1.25x or 1.5x), it finds different songs! Including at different time stamps. So far I found "T-Time (Taiko)", "Count It (Matt Lange)", and "Relief (Jincheng Zhang)" at regular speed. No idea what parts of those songs it latched onto. I'd love to pick at Shazam's song profiling algorithm to see how likely clashes are at normal speed, and how likely collisions at wrong speeds are 🤔. Also, @Mathologer, I think the song might actually be Jincheng's "Relief" as the last animations are switching over to the credits/thanks.
    1
  17810. Interesting. Are these numbers combinatins or permutations of the tiles" Or, the number of ways that the board can be tiled in visually unique ways?
    1
  17811. This is the most mind blowing thing.... can all those triangles form a tiling? Because primitives won't there always be a side which matches a different hypotenuse? L system is my fidget toy
    1
  17812. I vote for the Euler-Mascheroni constant. First, is amazing fact that it pops out everywhere even if it is hard to understand its meaning. I mean, pi has to do with circles, e with derivatives, but gamma ?!? Secondly, ... well it's our buddy Euler :)
    1
  17813. (14:47) "Patterns within patterns within patterns". Muad'dib, is that you?
    1
  17814. Thanks!
    1
  17815. Umbrella Corp!!
    1
  17816. If you think pi = 4 you cut too much corners.
    1
  17817. I found a nice relationship between Fibonacci sequences and Fermat's last theorem, but sadly it doesn't fit in this comment.^.^
    1
  17818. Merry Christmas to you too ;p
    1
  17819. Love the content. I'm by no means a math whiz but I really love these videos. Question though: Am I losing my mind or didn't there used to be a video explaining how to use a compass and ruler to make right angles, and pretty much every regular polygon? I cannot for the life of me find it. I was pretty sure it was a Mathologer video... but I can't find it on this one or Mathologer2. I tried to account for misremembering my math and science favorites by checking Numberphile, StandupMaths, 3Blue1Brown and Steve Mould with no success either. Anyone know what I'm talking about?
    1
  17820.  @Mathologer  perfect! Exactly the one I was looking for. For some reason I thought it was in the title. Thanks for the response!
    1
  17821. Main takeaway: stars are regular pentagons
    1
  17822. It is just a circular slide rule. Nothing mysterious. I used one for years.
    1
  17823. 37:24 this should be a parkour map
    1
  17824. The summing of the odd harmonics creates a square wave. The summing of the evens creates a sawtooth wave which sideways relates to the sphere+cone=cylinder in that sphere is sine function, sawtooth/cone square/cylinder.
    1
  17825. This might come in handy when splitting the taxi fare between 3 passangers 😉
    1
  17826. @Mathologer  🤣
    1
  17827. Slick proof at the end. 👏
    1
  17828. Not yet
    1
  17829. In some specific circumstances, you can say that one area is less than another by x square meters, so that would be used as a negative value like negative money is money owed. But in the real world, you can't buy or observe an area that is less than zero. This is similar to the fact that a vector can have a negative length (thus be pointing the other way) but a distance can be only positive since it is the absolute value of length.
    1
  17830. Amazing! Thank you for the insight.
    1
  17831. I really love your T-shirts!
    1
  17832. I entered uni using the slide rule and left using the Texas instruments calculator. I'm 65
    1
  17833. i knew he was german by his accent all along
    1
  17834. I very much enjoyed the whole video - the tempo and contents are fine - my background is a phd in mathematics and I always love to see methods and ideas. What I liked here was the fact that the sequences seem at the end seemed to be asymptotic sequences and I wonder if convention to pace form might give a larger set of convergence.
    1
  17835. The Manhattan perimeter of a circle shows up for real in computer graphics. Instead of a smooth curve, we march along one cardinal direction and every so often step in the orthogonal direction. The paradoxical effect you noted occurs with straight lines as well. How do you make a diagonal line? Infinite refinement of x and y steps, with the total length always x+y. Again, this is actually true in computer graphics, where the refinement is finite .
    1
  17836. Definetly the no 9 series animation!
    1
  17837. This might be your best yet!
    1
  17838. amazing
    1
  17839. I enjoyed that. Thank U my love. I feel like doing another IQ test. See if I score more than 68 this time.
    1
  17840. 24:15 An arc angle may pay you a visit if you sin too much 😜
    1
  17841. Nice! Considering that the estimated number of electrons in the universe is in the region of 10^80 though... 🙂
    1
  17842. 29:00 In the Pythagoras family, of the two smaller numbers: * the smaller of the two increases by 2 every step (3, 5, 7, ...) which is obvious; * a cute observation is that the larger of the two also has a rule: each is the product of the previous by a some rational, and the rationals form a sequence: 3/1, 4/2, 5/3, 6/4, 7/5, and so on. I'm assuming that the fact that b_i = b_(i-1) * (2+i)/i (that's the sequence of rationals) has to do with the path taken in the tree, plus the fact that b is twice the product of the right pair of numbers in the "square". I have a beautiful proof for that, but the space in comment in running out. :) If you run the sequence backwards, btw, you get the 0th triple to be (1,0,1) (a is 1, and c has to be 1 larger than b). Finally, the difference between the c_i's forms the sequence 4, 8, 12, 16, 20, ...
    1
  17843. handsome man
    1
  17844. donald knuth’s “Discrete Mathematics” covers this subject in great detail, would definitely recommend to those interested in more
    1
  17845. Beautiful!
    1
  17846. The fraction for n=3 at 21:08 is 946/243
    1
  17847. Nice ! As always
    1
  17848. Disappointed the can at the end wasn't a tin of Fosters
    1
  17849. 14:15 I found the trig derived formula to be equivalent to Heron's, but I'm sure there must be an easier way. I started with the trig formula. I wanted the denominator inside the square gone, so made (1/2)BA > (1/4) (2AB). Happily, this left (1/4)sqrt(...), so I could ignore that for the rest of the verification. (2AB) became 4(A^2B^2) when multiplied by 1 and went back to (2AB) inside the square term. The denominator cancelled, and I then had, (4*A^2B^2) - (A^2 + B^2 - C^2)^2 Is there an easier way after that? I did the expansion and simplification trip, I mixed up one multiplication and ended up with 8 mystery terms that didn't cancel, I had to go back to the start of expanding Heron's formula to find my error. In the end I found, -(A^4+B^4+C^4)+2(A^2B^2+A^2C^2+B^2C^2)
    1
  17850. Ramanujan is great
    1
  17851. I asked ChatGPT. Hilarious.
    1
  17852. It stated that 2+2+2=2*2*2
    1
  17853.  @Mathologer  "Give me three examples of identities in which the sum of two or more natural numbers is equal to the product of these same numbers, excluding the number 1."
    1
  17854.  @Mathologer  Today I tried the experiment again and it at least could not find any answer. Better than trying to sell me falsehoods!
    1
  17855. i thought this is a dynamic programming problem for a bit, the block stacking thing. don't think i can write easy algo for even do math to figure out that series.
    1
  17856. what lies beyond?
    1
  17857. 'Either believe me, or convince yourself that this is true' made me laugh out louder than I want to admit. Also, I'm using it for every argument in my life from now on.
    1
  17858. Most memorable: Chapter 3: Proof of divergence
    1
  17859. Lol, by people who hardly knows how to add. Sad stuff
    1
  17860. @mortahta
    1
  17861. I don't understand the ending of the proof. There is only one way in 4k+1 primes to be written as p=x^2+y^2, but 325 is in the form of 4k+1, with k=81 and can be written in 3 different ways, as shown in 25:52. It contradicts the proof, or I'm not getting something here.
    1
  17862. I made it till the very end!
    1
  17863. Gladly rewatching!!😃
    1
  17864. This is the best video about lacings ever. I love the production, the editing, and the segments. Thanks for the great content!
    1
  17865. In the The idéntical coins case, when George's Washington head Is pointing up, is only half a rotation not one complete Rotation.
    1
  17866. Perhaps you could do some videos on statistics? Like Student's t-test, or Bayesian inference.
    1
  17867. In a hypercube/tesseract dot grid.
    1
  17868. Mathologer also uses Wolfram Mathematica? I'm fluttered
    1
  17869. what the heck, this was crazy
    1
  17870. Blue Square area = 1/5 Each 1/2 x 1 triangle can be subdivided into 5 smaller equal area triangles. This gives a ratio of 5 triangles to 1/4 area unit. This may also be expressed as 20 triangles to 1 area unit. The center blue square divides into four of the small equal area triangles which is 1/5 of the 20 triangles that give 1 area unit, so the area of the blue square must be 1/5.
    1
  17871. At 6:16 you set pi to 3.18, not 3.14.
    1
  17872. dude...
    1
  17873. Your intro sounds like a Chord progression from Nine Inch Nails lol
    1
  17874. There are infinitely Pi infinite series.
    1
  17875. 32:00 I think stand up maths made a video about this, https://youtu.be/ghxQA3vvhsk
    1
  17876. Génial
    1
  17877. Pretty sure we were taught to use circular slide rules when I was getting my pilot's license, but that was a long time ago.
    1
  17878. Just remember, pi on a computer is 4. Pixels are square, so the stair step pattern they make cause pi to be longer than in the real world. ;)
    1
  17879. Great video! FYI Édouard Lucas is the name! He was a french mathematician... There are two other famous men with a homonymic name Edward L. and Eduard L. both non mathematics.
    1
  17880. This video is great. I started working on this problem in high school and watching this video is like deja vu.
    1
  17881. Nice. Most memorable... The ugly overhang. Yuck!
    1
  17882. Thank you for these excellent explanations 👍
    1
  17883. At 20:20 2(a+b) instead of -a-b also works.
    1
  17884. 6:19. No. for example cut off (0,1) and (1,0) squares and you will never fill that black one
    1
  17885. 19:00 I did it! Took a minute to get the idea, and then another hour to actually prove the function is a well defined involution. But I just finished an undergrad degree in math, so I guess I'm exactly where you expect I should be 🤷‍♂️
    1
  17886. I got to the end. Please do not send The Seal of Approval. The last one ate all of my fish and trashed the bathtub. I released it into the Pacific Ocean, but it kept swimming the wrong way because I am in the Northern Hemisphere. Eventually he did figure it out using polar coordinates. So please do not put another seal nor me to this amount of stress. Thanks.
    1
  17887. I feel like you some day make a video where special occasion is shown as a rule that would always work. Something like an April fools joke.
    1
  17888. 21:39 & 21:41, please don’t read numbers like 1.25 as “one point twenty-five”. It leads to confusion in the following way: if twenty-five is greater than three then one point twenty-five should be greater than one point three!” Of course it’s not. Please read any digits after the decimal point one by one (like you did at 21:05). The exception to this is prices in some languages. In the United States, where I’m from, we say things like “a dollar forty-nine” meaning $1.49 or a dollar and 49 cents. Awesome video otherwise!!!
    1
  17889. Wooo I realized this is a new Mathologer video after watching it! 🙂 Great names of the theorems and this is the best recreational math channel!
    1
  17890. Am I the only one who expected 7 proofs of the actuall pigeon hole principle? Still, amazing video as always!
    1
  17891. anyone have a link to the past video Mathologer has at 19:40?
    1
  17892. Oh no! At 12:34 we can briefly see your script.. I'm really curious why that happened.
    1
  17893. ​ @Mathologer  I don't know, but I came up with this, by using the fibonacci sum rows in the triangle and knowing that the numbers are related to (A+B)! / A!B! .
    1
  17894.  @Mathologer  Nice, I see the pascal triangle as a matrix of L(eft down)<->R(ight down) points where the number in the triangle is the number of possible R<->L step path ways from the top to that number down in the triangle. That number can be formulated as (L+R)!/(L!xR!) or in the context of your brilliant post about A and B: (A+B)!/(A!xB!). Do you know that this matrix triangle can also be seen as a circle? If the Left-down right-down points in the triangle are connected with constant distance shackles, you can compress the lowest row of points and you'll get a circle volume of possible L<->R all equal length path ways. In my comment previous to this comment you can do that too in the same way by combining the 1. 1. 's, but then you'll get a donut.
    1
  17895. Calculate the total resistance of a ladder of resistors, each rung being a 1 Ω resistor, and each rung being connected to the next rung by two 1 Ω resistors, one on each side.
    1
  17896. I've just begun to realize that I was horribly under educated.
    1
  17897. The lovely part is that "My Son ..." is also one of your supporter. You must be a proud dad. I wish a happy and healthy life for you and your family. Thank you, as always, for the great video prof. Burkard. It is also close to my birthday so I'd consider this as my birthday gift
    1
  17898. when you said “subscribe” the button started glowing rainbow
    1
  17899. High school algebra I came to the logarithm portion of our book and the teacher skipped it, "You will have logarithms in Algebra II." About midway in Algebra II the teacher claimed, "You should have studied Logarithms in Algebra I, we won't have time for this." And so High school went by without going over logarithms. However, I did have a slide rule, quite a full featured one at that. I understood it worked because log(a)+log(b)=log(a*b). The math placement exam for college had 2 questions on logs, which I aced. However, the moderating prof found my proof curious and asked me why I went about it the way I did. I explained that I was self-taught by acquaintance with the slide rule.
    1
  17900. I loved the video.
    1
  17901. 17:28 The 7 wall contains diagrams of radio telescopes. ;) Look for nearby dots making parabolas with 2 diagonal lines and 1 partial vertical or horizontal. I see one facing up on the left side and one facing to the right somewhat left of the middle. Beautiful video! :)
    1
  17902. That was interesting, I made it all the way to the end. Many people think that ancient cultures were stupid because they weren't technologically advanced. Far from it.
    1
  17903. Fantastic. I think these are also related to fuss catalan numbers too, certainly the "numbers of journeys" is same logic
    1
  17904. why the HECK is the number of tilings of a 2xn board friggin' Fibonacci??
    1
  17905. I'm early!!
    1
  17906. What most of us are going to takeaway from this video.. Cubies.
    1
  17907. I just want to say that I'm listening to this with my new Galaxy Buds ....and they're awesome
    1
  17908. Can I just say that your T-shirts are awesome
    1
  17909. My house number is 6321 and for over 30 years I've told people how to remember it by 6 = 3 * 2 * 1 and 6 = 3 + 2 + 1 and it being the only 4-digit number with the digits in decreasing order that works that way.
    1
  17910. Commenting for algorithmic purposes 😊
    1
  17911. Yayyyyyyyyy new video
    1
  17912. Does this solve Basel Problem for free ?
    1
  17913. I never learned the "digital root is remainder" trick when I was young. It was only when studying congruences/cyclic groups that I learned that fact, and why.
    1
  17914. Coloring proof. Very cool. I liked the swivel proof, but it doesn't have the same aha moment.
    1
  17915. Cardioid, shmardioid...it's stylized Romulan Bird of Prey. And why you did not mention that Tesla was heavily in love with a pigeon? This obsession of 3-6-9 was not his biggest problem.
    1
  17916. Hello
    1
  17917. I love your T-shirt! "Pizza Phi", right?
    1
  17918. Pure luck, spied at once!
    1
  17919. Awesome analogue computer.
    1
  17920. I think it should also be appreciated how amazing your t-shirts are all the time!
    1
  17921. My gut kept wondering, “Where will the 8 come in?” So it made me laugh when you explained; thus, it was a joke and not a lie at all!! And a hook for us to keep watching. Clever you!
    1
  17922. So basically the French were wrong
    1
  17923. This is a fantastic example of why Analog computers are so fantastic for calculations. Future of tech involves fusion between analog and digital computers.
    1
  17924. Is the centre of the final regular x-agon not the centre of mass of the all of the preceding ones?
    1
  17925.  @Mathologer  Thank you for responding. I admit to have jumped the gun and asking the question before reaching that point in the video, and was happy to have my assumption validated at that point.
    1
  17926. I vote for euler's gamma constant
    1
  17927. Very well done and had me amazed at every chapter.
    1
  17928. Welcome back. Nice to 'see' you
    1
  17929. 9:40 I think you can make all the numbers up to 2^(n+1) - 1, where n is the largest power of n you have. Furthermore, you will get each number only once. It's easier to think of this in binary. Finally, this works for any sequence of increasing powers Edited for the typo
    1
  17930. Are you opening your videos with the first three chords of Kate Bush’s song, Babushka?
    1
  17931. Easy to see that gamma is >0.5 due to the fact that 1/x is concave and if you break up the rectangle at 1 in smaller rectangles then you get that the blue parts sink below the diagonal and so for each of the rectangles the blue part is more than half the area, but the area of all the rectanges is 1.1=1, so gamma>1/2*1=1/2.
    1
  17932. The spiral drawn at 41:05 is mostly an irrational one - except for the first two squares and their sides, and for the square after the first square with an area of two. It starts with a rational equal sided triangle with a hypotenuses of root-of-two. then its a non-equal sided triangle with 1 and root-of-two making the 3 = 1 + 2 or as a side length notation to be a=root-of-two, b=root-of-three and c=root-of-five. resolved to c the formula is c = root(a² + b²) As shown the side length are going to be totally irrational starting with this third triangle. little rationale: ist not resembling the previously referenced 3-4-5 scheme anymore but using its totally own space. still its nice. a 3-4-5 would have a set of squares with an area of 9+16=25 whilst in above the most closest set of tagged area values is 8+13=21 as can be seen. And to make the numeric design "difference" even more clear none of these numbers at close to the 3-4-5 area is the square of any rational number. the 8 is the cubic-of-two, the 13 is a prime number and the 21 is the product of 3 and 7 which both are primes. Just for curiosity - walking a bit more along the Fibonacci sequence: a little further, the number 34 pops up which can be factorized as 17 and 2 which both are primes again. then 55 is 5 times 11 - two primes. 89 is a true prime number - and i feel surprised about it. 144 is 12 times 12 and suddenly works as a perfect square with a perfect rational side length. or further factorized: 3*4 and 3*4, or 3*3 and 4*4 233 again goes for being a true prime number. 377 is made up by 13 and 29 - both are prime numbers 610 is made up of factors 2, 5 and 61 - where the last is a noticeable high prime number. 987 is made up of factors 3 * 7 * 47 - thus also built up from a relatively large prime number. lets pause here and let it sink a bit. It's not at all clear or anyways sure to expect that two really rational numbers for side length will ever appear in direct sequence - but also its neither totally impossible that it can be when only doing it with that sort of approach by factorization. The Fibonacci sequence results in a relatively noticeable chance to find higher oder prime numbers, with also intermediate values having relatively big prime factors. That's indeed not a mathematical proof - just an observation that can be drawn for a relatively small lot from that series. Those all might be known by the mathematicians dealing with the subject for already some while - and sort of still unresolved up to now (unless someone already got it recently) pattern assumption.
    1
  17933. This is sooo cool! I'd like to see more about sum, power and factorial relations
    1
  17934. I would love to see mathologer video on why quintic is not solvable by radicals.
    1
  17935. oh wow, that's beautiful! when you showed that you can get 2 different regular polygons from the original depending on the orientation you choose, i wondered what happens if you take the ratio of their areas (or perimeters)? can this be used for something, like as a measure of the regularity of the original shape? i guess this isn't so natural once you start to think of it from the DFT perspective, since you have n-1 choose 2 components to compare, not just a single ratio. but this reminds me of how the DFT is used in music theory, where chords can be analyzed as polygons with vertices on the unit circle (which represents the circular space of pitch chroma) and then the magnitudes of the fourier components are interpreted as measures of various harmonic qualities that a chord can possess. this is really the most beautiful & intuitive demonstration of the DFT that i've seen!
    1
  17936. If you do ALL n-1 angles, you end up with a single point: the vertex centroid. So if you do all but one of them (so a total of n-2), you must end up with a regular figure. For angle 1x 2pi/n and (n-1)x 2pi/n you get the regular n-gons. For angle m x 2pi/n, you get a pointed star if gcd(m,n)=1. But you get a figure that has less vertices if gcd(m,n)<>1. More precisely: you get a figure with n/gcd(m,n) vertices. All these figures are exactly the eigenpolygons. The eigenpolygons can be characterised by Schläfli symbol {n/i} for i=1, ..., n-1.
    1
  17937.  @Mathologer  It's neat! Most discussions of it are written more for music scholars more than mathematicians, but Emmanuel Amiot's book Music Through Fourier Space is probably the best introduction.
    1
  17938. Most memorable: besides that Euler and Ramanujan appear again? That the sum goes to inifinity. After that moment I was dead.
    1
  17939. What you are saying seems to involve kaluza Klein, a funny mix of dynamics and kinetics, but I thought what was really interesting was core bosons, dianetics lol
    1
  17940. This must be assuming that the paint is infinitely thin. A bad assumption.
    1
  17941.  @Mathologer  Had a tour there in the late 1990s and the guide said so so I guess it is 7 sided?
    1
  17942. Such an awesome video. In the Euler-Maclaurin series, the cyan terms have 1/n in the reminder so naturally diverges and may not give a good approximation for larger n. Also, I learned to pronounce 'Jacob' properly now :-)
    1
  17943. Here is a nice way of viewing the numbers in each row of Pascal's triangle in terms of # of subcubes containing a fixed edge in the hypercube of a fixed dimension: https://www.researchgate.net/publication/316169431_A_double_counting_argument_on_the_hypercube_graph
    1
  17944. Actually, 3B1B also has a video on exactly the same topic. And he dives into the proof. Check it out! https://youtu.be/NaL_Cb42WyY
    1
  17945. Absolutely stunning! It is a big joy for me to see such a simple proof. Going to 3D is perhaps not too unnatural if you know Desargue. But still ...
    1
  17946. there's 2020 ways of tiling the glasses checkerboard. No idea; it's just 4am.
    1
  17947. you're the man
    1
  17948. Whenever I see that Mathologer has put up a new video, I know it's going to be a fine night.
    1
  17949. I'm going to be a person asking for the full video on why the crazy formula works
    1
  17950. 47:15 What's next? 3a+b+4c=5; a+4b+5c=9; 4a+5b+9c=x; x=14.
    1
  17951. Now I want this for 4D hyperspheres 😂
    1
  17952. lettsssssssssssss goooooooooooooooo
    1
  17953. 23:25 - ..although the scull-and-crossbones belong in the center! :)
    1
  17954. As we're trying to find the base of the logarithm, of course I'm already looking suspiciously at a certain point on the x-axis that is approximately centered in the screen. But that visual argument that pins it between 2 and 3 is beautiful. I love how it works because the slope at x=2 is the same as the 180 rotation of the slope at x=2.
    1
  17955. As a total non-mathmatician it seems to me that your shrink proof contains an act of faith. Sure your hexagons get smaller but then there's the jump to deciding that, that means one has proved none ever coincide with your grid points. You seem ok with the infinite and an infinite grid, but not with an infinitesimal grid spacing. And unfortunately no explanation why not. Take your infinite sheet of paper, apply two coats of black emulsion paint over the surface and let it dry, in order to create the grid. Then count up how many regular hexagons or equilateral triangle you find, and let us know how many :-)
    1
  17956. So Tower of Hanoi solution but +1.
    1
  17957. I was not taught the extension.
    1
  17958. If this dude can get a nature publication on shoe laces, then I can procrastinate on my online lectures by watching other (basically) online lectures
    1
  17959. I will never be able to un-see that the Bernoulli numbers are the matrix inverse of pascal's triangle. I'm not the kind of person who memorizes a lot of facts. I learn how the machine of mathematics works and when I need something I just crank the lever until it comes out. This is one of those facts that is part of the machine. Like 30+ papers I have read over the past 10 years just made SO MUCH MORE SENSE!
    1
  17960. Even i cant believe i watched it till the end, had to replay a few times on the euler-maclaurin part. Maybe need to write it as well to understand it better.
    1
  17961. Never saw this at school.
    1
  17962. Kempner's Simplicity wins for me : )
    1
  17963. Nice vid
    1
  17964. has the row beginning with four ones ever been studied
    1
  17965. I've definitely seen the pi = 4 thing before. Thankfully having taken enough maths I know the reason - the sequence of curves doesn't uniformly converge to the circle, so there's no reason you should believe that ancillary properties of the sequence converges to the same properties of the convergence target. Definitely NOT intuitive though, until you've been exposed to it!
    1
  17966. I had an exam where we basically had to prove this from scratch. Was hard af.
    1
  17967. For me the most interesting part was gamma and how it plays such an important role in approximating harmonic series
    1
  17968. Kempner's Proof
    1
  17969. Good Day Mathologer. I tried to find your contact info on QEDCAT website. But the site seems to be offline. I am writing this comment on the latest video, with the hope that you will read it. I am a longtime viewer I saw your video : https://www.youtube.com/watch?v=qS4H6PEcCCA I have been playing with the ideas, and two things have been keeping me awake at night. 1. Under what conditions is such an orbit free from intersections? For 2 circles, this is relatively easy to show. But I am a bit lost, with more than 2 circles. 2. It seems, that they can also generate fractal like sets - well the boundary of the set, really. But : 2a. if the number of circles is finite, can the set be truely fractal in nature? 2b. under which conditions are they fractals? 2c. can such a fractal boundary also be free from any intersections? under which conditions I certainly do not expect you to solve these things with your time - but i would really be grateful - if you could point me to some relevant literature, and perhaps some keywords I can search, to construct a solution to my queries. vielen lieben Dank im Voraus.
    1
  17970. Line has infinite length yet the area of zero. It's basically the same though each dimension is one less.
    1
  17971. i want to help you but Youtube does not allow me to
    1
  17972. Nice video. I understand calculus I, and it was very easy to follow. Not simple, but if I make the effort, I definitely can replicate your steps. Thanks for your efforts.
    1
  17973. I love the way you explain math. Very calming. if your interested in a new line I've discovered, Please reply in these comments please.
    1
  17974. Hi, first of all I really enjoy watching your videos, always great insight. In this case I'm wondering if the proposed procedure generates all possible combinations. In particular, if two "yellow 2x2 blocks" end up side by side (e.g. minute 27:30), the algorithmically generated tiles would be only these 4 as you show in the video (sorry for the simple diagrams): 1) || || 2) = || 3) || = 4) = = there is another tiling, not generated by the algorithm, which considers the 2x2 yellow blocks independently: 5) | = | And so the total number of combinations might actually be 2^1 * (2^2+1) for A(2). In this particular case, the extra case actually ends up being exactly equal to another case generated by the other branch (where the yellow 2x2 blocks end up stacked on top of each other). So the number of tilings 2^(1+2) is correct. This gets more complex in higher order A(n), since you have more and more yellow 2x2 blocks side by side (sometimes many more). The paper probably clears that this always happens, but is a bit hard to swallow. Best regards!
    1
  17975. Jo do schau her; 😄🤘
    1
  17976. not nice < niais < nescius := not-skilled but you are: -> well; length -> span; little R -> r; little bit: pick one; weird -> uncanny; purple -> violet: purple is a sea snail dye, the hue of grapeskins between pink and red; hard -> touh; fast -> swift; will -> shall; thick -> coarse; thin -> fine
    1
  17977. A mathematician: Brilliant! An engineer: How can I use this?
    1
  17978. 4:40 don't you mean permutations? 😋
    1
  17979. Wait, on second thought they aren't! The order of the coins doesn't matter. Only the presence of the coins matters.
    1
  17980. This is so elegant. Really nice! It explains everything. Great video.
    1
  17981. Merry Christmas. Your channel is a treasure.
    1
  17982. 9:14 pretty obvious that it is a representation of a binary number. You can also generalize to be n-1 coins each of powers of n. Take n=10 you will be familiar of base 10 number system. 23:57 Not sure if it "doesn't work" because we haven't defined C(n,r) = 0 when n < r? The formula should work for k <= 3 too. Ps: I absolutely love this topic especially on generating functions since I learned this for math Olympiad. It is very useful if you can somehow use it correctly, but it takes experiences to have such marvelous thought of doing so.
    1
  17983. I love this video
    1
  17984. Mr Programmer standing to attention :) I've made the diagrams from this video programmatically animated, available on my channel (Linking it seems to hide the comment?) You can view the source code in the description of the video and generate your own videos with it if you wish :D PS: I had you for MTH1030 back in 2017. Thank you for being a great motivator!
    1
  17985. I have a question.   We start with the well know approximation 355/113 = approx(pi) If we change in sequence, 11335577....., each odd number to the next odd up,   We are left with the fraction 577/335.   It's meaningless to us.   If we add 1 however, We are left with a approximation of e.   Out of all the possible results, why does that occur?
    1
  17986. I'll go with the "no nines", because that's the one that tripped up my intuition. It's a lot easier if you generalise it to "no n in base n+1" and try base 2, but that's probably cheating.
    1
  17987. I enjoyed chapter 3 - Oresme proof that the sum diverge Elegant - simple Ps: I am not in the US so sending books is not needed
    1
  17988. Just wonderful!
    1
  17989. You know a video is great when you watch 50 min and still want more!
    1
  17990. “1, 2, 3, 4, 5. Whoever has trouble with this pattern should change channels now!” I’m re-watching this video for the third or fourth time now and this line cracks me up every time 😂🤣
    1
  17991. Can you explain your shirt please
    1
  17992. Weed and Math are not working together :o My brain hurts
    1
  17993. The whole coin paradox thing seems needlessly complex with all the unrolling and coiling. Just think about the location of distance between the coin centers in terms of the radius of the smaller coin (r). If they're both the same size, distance is 2r = 2 rotations. If the big one is twice the size, centers are 3r apart = 3 rotations. Put the small one on the inside and there's your 1r = 1 rotation. Maybe the unrolling/coiling is the proof, but for actually figuring it out I think the radius thing works.
    1
  17994.  @Mathologer  That's probably because you're a mathematician... and I'm just an engineer who likes math. :)
    1
  17995. Both proofs are beautiful!
    1
  17996. Any mathematical proof using the word "Obvious" is ... Obviously suspicious.
    1
  17997. 23:04 English subtitles said that there is an Easter egg in that slide, but I couldn't find it :(
    1
  17998. Just looking at the thumbnail, must have something to do with benfords law.
    1
  17999. great video!! I'm really grateful that as a high school student wanting to become a mathematician I have access to such inspiring channels as mathologer and 3b1b, giving beautiful yet still accessible mathematics to sink your teeth in to!
    1
  18000. I wish we were exposed to such an awesome way of teaching math when I was a kid. Even by turning out into an engineer later on, I've never felt that math could be that interesting and beautiful
    1
  18001. I'll need to watch this n times, where n is definitely greater than 3. But I'll enjoy it. (Yes, I watched the whole thing, 'tho my comprehension waned around chapter 3.) (I studied maths about 4 decades ago, liked it, but was terrible at it.)
    1
  18002. I feel math is cheating here, because 1/1 has no one-to-one correspondence with decimals. It's both 1.(0) and 0.(9), and it breaks everything apart.
    1
  18003. Technically, all terminating decimals have two different representations: for example, 3/4 has 0.750000... and 0.7499999... At this point, all you have to do decide which one of these you'll pick. The only exception to this is 0, which can only be 0.0000... but, as Mathologer has pointed out before, 0 can often be an annoying exception to otherwise hard and fast rules
    1
  18004.  @TheOneTrueIgnus  It is the additive identity, and 1 is the multiplicative identity, so 1 is also probably an annoying exception to a certain class of rules.
    1
  18005. Most memorable: 100 0s is greater than no 9s ☝️☝️☝️
    1
  18006. Ewhat is the ratio of area change, between each set to the final? It would be a formula to create a point relation to skip points.. hmm
    1
  18007. If only we had paper that would do autopilot algebra with a flick of a wand.
    1
  18008. Purple Circle because you've direct registered your GameStop shares lol
    1
  18009. Imaginary Numbers are linked to Cosine and Sine. What are the hypbolic variants?
    1
  18010. μ
    1
  18011. If all the rational numbers between zero and one each appear exactly once in the tree and if the tree grows by three at every level starting from the single row containing ½ & ⅓ then does that mean that the cardinality of the rationals is equal to 2 (mod 3)?
    1
  18012. MInute 11:13. Why not odd=2x+1? From this follows that P=4x^2+4x+1+4y^2 => P = 4z+1.
    1
  18013. Froeliche Weinachten mein herr.
    1
  18014. Proud to see this as a Keralite
    1
  18015. why are they weird ? I never considered them weird :)
    1
  18016. Pxoelèph
    1
  18017. I like the part with the gamma the best. I think its amazing, that there are such numbers every where in math.
    1
  18018. I enjoy these 4Blue0Brown videos.
    1
  18019. Oh no, I corrected my capitalization and lost my Mathologer heart. 😭
    1
  18020. Super well explained. Thanks for the hard work.
    1
  18021. I really enjoyed this video. I'm a PhD student in applied mathematics (network science), and I always learn something new from you. I also appreciate this long format. It allows for a more thorough discussion of a subject. Fingers crossed for a lengthy video on Galois theory! ❤
    1
  18022. math got freaky
    1
  18023. yeah, the nonagon is fun the way it unexpectedly collapses. Just when it's looking neat it suddenly gets even simpler. Makes a very satisfying doodle.
    1
  18024. Archimedes figure stack, Sphere has minimum surface are compare to cylinder or cone. That is not something amazing. If you apply Fermat optimization to volume V to design cylinder with minimum total surface area you get spherical volume of cylinder result V=2piR^3 height is h=2R and diameter d=2R. Cylinder with minimum area must match fit into sphere. Same with minimum perimeter for cross-section area of duct or vent to reduce shear friction for fluids flow A=ab we get square a=b is best duct shape with minimum perimeter, it must fit inside circle. Circle the only figure that has minimum perimeter and all non-concentric figures abstractly must follow circle shape. Cube has also minimum V=a^3 surface area that fits sphere. You can prove that soap bubble minimum surface tension without physics but only based on geometrical properties of sphere of minimum surface area.
    1
  18025. 13:25 fast fourier transform?
    1
  18026. I have a bachelor's degree in mechanical engineering and I have understood your video. You explain math in such a manner that it is easy to understand. Well done!
    1
  18027. After a long time i have seen a good movie 😉🙏👌👌👌
    1
  18028. I have an antique Soviet era rotary pocket watch style slide rule based on this principle...
    1
  18029. Can you make a video talking about 2D and 3D Generalizations of Viviani's Theorem (which states that the sum of altitudes from any point inside an equilateral triangle equal the height of the triangle) and deeper mathematical facts related to it?
    1
  18030. Well yes but how do you blow in the horn after you’ve painted it??? Such a waste…
    1
  18031. Anti-Shakespeares? What have I done at thou?
    1
  18032. You have such a sweet laugh 😊
    1
  18033. This is profound.
    1
  18034. Finally a math video which mentions the Feuerbach circle! :) (As far as I know - no relationship. ;) )
    1
  18035. Happy Christmas Eve https://youtu.be/LGwK-_MpgY8
    1
  18036. Another great video! Crystal clear.
    1
  18037. Any number can be anything you want, if you're allowed to adjust by a factor of desired/actual at the end 😂🤦🏻‍♂️
    1
  18038. Thanks to this method of generating phytagorean triples, I was able to list the triples based on the incircle radius. We just have to feed the pair of factors in the 1st and second number, and complete the box ala fibonacci sequence. :) btw, 1,12 produces different triple vs 12 & 1. So I guess (although I shouldn't), the number of Phytagorean triples of a certain incircle radius, is just the number of factors that number has. :) I'm surprised ALL the numbers showed up as a member of a triple at least once except 1 and 2. At least for the numbers 1-100. Is it true for all positive numbers >2? For incircle radiuses/radii of 1 to 100, the number 96 has the highest occurence being a member of 9 triples.
    1
  18039. Thank you for a great video! This subject is clearly related to crystallography. Therefore, these two questions: 1) Do you know if the said Willy Scherrer is related to Paul Scherrer who contributed a lot to crystallography (for example the Scherrer formula and the Debye-Scherrer camera)? 2) Was the author of the "Hauptwinkel-paper" (who's name I was not able to pick up) focusing on the applications to crystallography (as the word "Goniometrie" may suggest)?
    1
  18040. This Channel Is so good I'd watch a 2 days Mathologer video without eating nor sleeping
    1
  18041. What a gift! I never expected another video so soon.
    1
  18042. the balancing scales is the same mathematicaly as resiator divider, Ohms
    1
  18043. Amazing. I know this is super profound in some way, the whole field of summation that is Pascal’s triangle and that it connects to Riemann paradox re infinite sums. I let it simmer for a while. Am I wrong (higher math newbie) but when you calculate areas under a curve and there is the little triangle left empty….if one makes a rectangle of the difference between two adjacent columns, then cut that in half you get the triangle that fills that slope area.
    1
  18044. An extraordinary lecture on geometry. Thanks a lot for it. Your T shirt made me laugh a lot... loved it.
    1
  18045. This is how Charles Babbage's Difference Engine worked--it was a machine for generating function tables using calculus of differences.
    1
  18046. Another wonderful video
    1
  18047. YESSSSSS, MORE VIDEOS
    1
  18048. You can addchapters to video by adding timestamps in the description, starting from 0:00.
    1
  18049. Truly windmills of peace
    1
  18050. Sorry to interrupt with this question..... There is this branch of mathematics that deals with large number. (when we write cube of 2, we wrote a superscript-3 on the right hand side of 2) For these numbers, the superscript-3 is written BEFORE on the left hand side of 2. What is this branch of mathematics (i watch some youtubes on it, and subsequently forgotten about the terminology)
    1
  18051. But why is it not provable that the algorithm is the most efficient?
    1
  18052. You make me fall back in love with Mathematics. Sincerely thank you!
    1
  18053. a lot to think about in that. thx.
    1
  18054. Like always, Marty you are the greatest math youtuber
    1
  18055. I upvoted in advance for the T-stirt
    1
  18056. Truly spectacular and elegant.
    1
  18057. I made it to the very end, although I did stop and start many times over the week or two it took me to do so.
    1
  18058. No integers
    1
  18059. Thank you! An excellent Christmas present.
    1
  18060. Where is the 27 on his shirt?
    1
  18061. Fascinating as always.
    1
  18062. I made a video about a unique geometric pattern that only exists in base twelve. I wonder if Mathologer has ever explored base twelve geometry? It seems like no one is exploring it right now but me. I've discovered a geometric pattern for calculating the base twelve value for Pi. It has different qualities than the base ten version. How could one test to see which version of Pi was better?
    1
  18063. Is it completely obvious why the greedy algorithm won't converge to a value less than gamma? I still had to do a little proof by contradiction. Anyway, the most memorable bit to me was the leaning tower bit. I feel stupid for never wondering about the most optimal configuration with finitely many bricks...
    1
  18064. 24:15 it's pretty interesting, as there will be a zero height and .25 wide "triangle" - which I think neatly coincides with the "true" start of the Fibonacci sequence: 0,1,1,2 ... :D Jokes aside, that means, that is a (1,0,1) Pythagorean triplet (1*1 + 0*0 = 1*1).
    1
  18065. For chapter 6 I would assume a hexagon can only have 1 of each rotation as they are locked in places which correlates to the same amount of different colors in a larger scale because of how it reacts on the first piece of hexagon itself.
    1
  18066. found another blooper: 29:11 - fourth row should by 1 3 5 7 😃
    1
  18067. I wonder how many dimensions this works for? It works in 2D and 3D...
    1
  18068. I'm halfway through and sooo excited for it to all come together!!
    1
  18069. WOW
    1
  18070. The key with this problem is the requirement to only pick cards that are face down. If you could select any card you could also reverse the pattern and count back up. So if you get down to it the secondary card being flipped over is only determining direction of the count at best. If you changed the endian that also would change it. The real answer is that there is a finite set of values those sets could represent by making such changes you would wind up at one end or the other depending on the assumptions of what endian you are using or if you are counting up or down. Doesn't really matter.
    1
  18071. Thank you for another amazing video, my mind has been expanded again. backgroundvisualisations?
    1
  18072. 20:34 Ich bin tatsächlich in einem Hausaufgaben Wettbewerb auf dieses Problem gestoßen. Ich hatte auch den Ansatz des vervielfältigen des Winkels, habe allerdings nicht den geometrischen Schritt gesehen, der dann zu einem Widerspruch führt. Bis heute hatte ich keinen wirklichen Beweis. Ich habe nur diesen Beweis gefunden, welcher meiner Meinung nach recht kompliziert ist: https://proofwiki.org/wiki/Niven's_Theorem . Ich finde es super, dass ich jetzt endlich einen verständlichen Beweis kenne. Danke! PS: Ich freu mich schon auf das deutsche Video ;D
    1
  18073. I remember your Simpson's video ... isn't that limit the growth over one year of a bacteria colony with plenty of food and space? Isn't that why it's called "natural"? Thanks!
    1
  18074. Nice... Pascal's triangle is everywhere!
    1
  18075. The music is so epic.
    1
  18076. When Professor Polster from down under takes us to the Arctic Circle on Christmas Eve, you can be pretty sure he's got some serious goodies on his sleigh.
    1
  18077. Quero a rede gamer! 😅
    1
  18078. Can you actually sleep? Fascinating...
    1
  18079. Absolutely worth watching the 50 minutes!The animations of the algebraic and geometric proofs are very elaborate and easy to follow.Other feedback: for the "master classes", the duration just fine.Background: Computer science university grade with lecture on Analysis I and II (calculus for the English guys) and Linear Algebra I.Wish: a video on the group theory would be intriguing
    1
  18080. This stuff feels like mathematicians just handwave whatever solution fits the nicest. By the same logic that says that when counting to infinity there are as much even numbers as there are even and odd numbers there are as many negative as positive terms in the infinite series. So the answer is zero.
    1
  18081. :dothefive:This video was taken down for practicing black magicks.:dothefive:
    1
  18082. I feel like plugging different numbers for x into the expression wasn't motivated. Why should plugging anything in for x tell us anything in the first place? And additionally, why should plugging in -1 give us particularly the relationship between the "bits" of the shape, as opposed to any other piece of information?
    1
  18083. Brilliant!
    1
  18084. The Eisenstein triangle seems to be what another video called the anti-Pythagorean theorem. Check out: https://www.youtube.com/watch?v=TMR1oYkCKEs (part 1) and https://www.youtube.com/watch?v=Sxq2hA7qZYM (part 2).
    1
  18085. I like the solution with numbers, but I would prefer face down cards as zero and face up cards as one. This is in my eyes more natural. Then numbers increase and you end up with the latest number you can represent with the given number of digits.
    1
  18086. I'm at the point where you ask a proof for the falling powers difference formula being analogous to the continuous calculus one. Probably already in the comments, but I'm going to guess here that the easiest way uses induction.
    1
  18087. When the best keeps getting better... 🙏♥️
    1
  18088. 27 different triangles —-81/3
    1
  18089. Since this is a number-theory proof, and the usual proof (seen on Wikipedia https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 ) uses calculus, this is a taste of Analytical Number Theory!
    1
  18090. Quite flabbergasting that this was overlooked for so long; I can easily imagine a diagram showing this alongside the Pythagorean Theorem in ancient Chinese or Babylonian texts. I wonder how many people over the millenniums have noticed this relationship and assumed it was already well known? Wonderful that it was finally brought to light by the legendary Lee Sallows, though.
    1
  18091. From the 1x1 square you drew it is visible that gamma is > 0.5 because each extra blue part is similar to a triangle with a little bit extra curvy part, all the edge is filled with triangle bases, all have the same height 1, if they were all triangles than the area would be 0.5, since they have extra curvy parts it must be > 0.5
    1
  18092. All the normal sums are divisible by 3=2*1+1 ( 3*1, 3*5, 3*14, ... [see OEIS: A000330] ) with the first sum being 3, but all the square sums are divisible by 5=2*2+1 ( 5*5, 5*73, 5*406 ) with the first sum being 5², so sums of cubes are divisible by 7=2*3+1 with the first sum being 7³, well at least it almost matches, so there is an almost matching pattern for cubes and higher potencies as well, it just doesn't match exactly as hypercubes are spiky, so there will probably almost be an offset but the offset is small compared to the respective powers: 5³+6³≈7³ (-2), 16³+17³+18³≈19³+20³ (-18), 33³+34³+35³+36³≈37³+38³+39³ (-72), ... and 7⁴+8⁴≈9⁴ (-64), ... and 9⁵+10⁵≈11⁵ (-2002), ... and so on.. There is even an OEIS sequence for those offsets of higher dimensions which starts with powers of zero: https://oeis.org/A127691
    1
  18093. As our Mathematics Professor said, "You just look at it and..." Euler's e-Pi-i 1-0-infinity instantaneous trancendental self-defining cross-sectional wave-particle coordination-identification positioning, a cause-effect of convergent sequences summed as the all-histories Reciproction-recirculation Singularity Unit Circle.., at the inside-outside holographic relative-timing numberness ratio-rates fractal boundary, of 0-1-2-ness projection-drawing probability, of AM-FM all-ways all-at-once all scale resonance, in/of QM-TIME ONE-INFINITY i-reflection Containment-Completeness.., after a bit of practice in Symbology.., and this is just the Flash beginning, Eternity-now Origin-zero temporal vortex-vertex Interval Event Forever because e-Pi-i is 1-0-infinity Entanglement, the probability range-spectrum inside-outside holographic instantaneous superposition=> interference Singularity-Lensing of Logarithmic positioning. You can say that this is the Observer's POV of Reciproction-recirculation Unity of 2-ness in 3D-T Perspective, self-defining substantiation experience of self-Self tuning-probability in the 1-0-infinity instantaneous flash recognition, universal positioning system in which Thinking Fast and Slow orientation can retrieve AM-FM memory associations in tune with our individual state-of-mind, according to empirical i-reflection e-Pi-i inside-outside holographic presence = resonance holistically. Flash recognition takes Time Duration Timing to think about when you use the Math Professor approach to look, listen, hear and see what you're probably knowing about who, what, how and why you're inside-outside this presence of conscious awareness Actuality. (No-thing definable = no labelling)
    1
  18094. Just loved it! 😍😍 Hey Mathologer, 🤔is there a way to hand calculate cuberoots like we calculate square roots
    1
  18095. Queen of Hearts
    1
  18096. Third challenger, either place one vertical domino on the left side with all possibilites for n-1 on the right side or place two horizontal dominos on the left side and all possibilities for n-2 on the right... Aha! :-)
    1
  18097. Ramblings of a village fool... 2 and 5 can be seen as each other's inverse (considering that the digital root of 10 is 1, and 10 is the product of 2 and 5). As for prime numbers. I consider 1 to be prime because if fulfills the basic rule, being divisible only by one or itself. If there was no 1 there would be no other whole numbers; all whole numbers can be seen as a sum of a given amount of 1's. But then, 1 is the "vain" prime, because dividing anything with 1 is a fool's errand. Fits me. Now, I get to 2 and 5. They are considered prime. I see them as "weak" primes, but in a sense more powerful than others just as well, because dividing any whole number by either of them gives a result with at most one digit on the right of the decimal point, and dividing any prime by them gives exactly one digit on the right of the point. The closest thing you can get to dividing primes with others and have a non-endless run of digits to the right of the decimal point.
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  18098. Best part is when your love for math shines the brightest.
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  18099. For reference, I've also heard referred to as "Napoleon's theorem" the result about a satellite going boom (with parts leaving at equal speed relative to the rest-frame just before the boom), that says they'll all (unless they hit something else on the way, such as the planet they were orbiting) meet up again one orbit later.
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  18100. sooo, can you ever finish eating that pizza?
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  18101. You have conclusive proof that I am dumb and just awaiting the emergence of the singularity. Seriously though mind blown!
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  18102. Merry Christmas!
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  18103. We Love You! <3
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  18104. I've sent you a candidate solution in GeoGebra. If it does what you want, feel free to note it here.
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  18105. let debts = [30, 20, 50]; // initial debt amounts let estateSlider; let debtInput; let updateButton; let totalDebt; let maxDebt; let scaleFactor; // pixels per debt–unit // drawing parameters const containerBottom = 350; // y–coordinate for vessel bottoms const vesselWidth = 60; const vesselSpacing = 20; const leftMargin = 50; function setup() { createCanvas(800, 400); background(240); // Create an input field for comma-separated debt amounts. debtInput = createInput("30,20,50"); debtInput.position(20, 20); // Create a button to update the vessels based on input. updateButton = createButton("Update Vessels"); updateButton.position(debtInput.x + debtInput.width + 10, 20); updateButton.mousePressed(updateDebts); // Initialize slider and drawing parameters. updateDebts(); } function updateDebts() { // Parse the input into an array of positive numbers. let inputStr = debtInput.value(); let parts = inputStr.split(','); debts = parts.map(s => parseFloat(s.trim())).filter(n => !isNaN(n) && n > 0); if (debts.length === 0) { // Fallback if input is invalid. debts = [30, 20, 50]; } // Compute total debt (estate max) and maximum debt. totalDebt = debts.reduce((a, b) => a + b, 0); maxDebt = Math.max(...debts); // (Re)create the slider: range is 0 to totalDebt. if (estateSlider) { estateSlider.remove(); } estateSlider = createSlider(0, totalDebt, 0, 0.1); estateSlider.position(20, 50); // Set a scale factor so that the tallest vessel fits nicely. // We leave a 50–pixel margin at the top. scaleFactor = (containerBottom - 50) / maxDebt; } function draw() { background(240); // Display current estate amount. fill(0); noStroke(); textSize(14); text("Estate (water) amount: " + nf(estateSlider.value(), 1, 2), 20, 90); let estate = estateSlider.value(); // Compute the common water–level h (in debt–units) so that // sum_i min(debt[i], h) = estate. let h = computeWaterLevel(debts, estate); // For each vessel, the allocated amount is min(debt, h) let allocations = debts.map(d => Math.min(d, h)); // Draw a horizontal line at the common water level (if h>0). if (h > 0) { stroke(0, 0, 255, 150); strokeWeight(2); let commonY = containerBottom - h * scaleFactor; line(0, commonY, width, commonY); } // Draw each vessel. for (let i = 0; i < debts.length; i++) { let x = leftMargin + i * (vesselWidth + vesselSpacing); let d = debts[i]; let allocation = allocations[i]; let vesselHeight = d * scaleFactor; // Draw the vessel outline. noFill(); stroke(0); rect(x, containerBottom - vesselHeight, vesselWidth, vesselHeight); // Draw the water fill (blue rectangle). noStroke(); fill(0, 150, 255, 200); let waterHeight = allocation * scaleFactor; rect(x, containerBottom - waterHeight, vesselWidth, waterHeight); // Label the vessel: show Debt and Payment (allocation). fill(0); noStroke(); textSize(12); textAlign(CENTER); text("Debt: " + d, x + vesselWidth / 2, containerBottom - vesselHeight - 10); text("Pay: " + nf(allocation, 1, 2), x + vesselWidth / 2, containerBottom + 15); } } // Given an array of debts and an estate amount, compute h so that // sum_i min(debt[i], h) = estate. // (This is the “hydraulic” interpretation of the constrained equal awards rule.) function computeWaterLevel(debts, estate) { let totalClaims = debts.reduce((a, b) => a + b, 0); // If the estate is enough to pay all debts, return maxDebt. if (estate >= totalClaims) { return maxDebt; } // Binary search for h in the interval [0, maxDebt]. let low = 0; let high = maxDebt; let h; for (let i = 0; i < 50; i++) { h = (low + high) / 2; let sum = 0; for (let d of debts) { sum += Math.min(d, h); } if (sum < estate) { low = h; } else { high = h; } } return (low + high) / 2; }
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  18106. Would an A4 page cover the pizza on your shirt perfectly?
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  18107. I made it to the very end
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  18108. It's hard to tell which is the nicer proof, but the first is harder to follow therefore more mind stimulating, especially if you've not done graphing and geometry for years.
    1
  18109. I loved the optimal brick stacking! Seemed so random!
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  18110. I was very pleaded by the Leaning Tower of Lire chapter and maximal overhang stacks. Thank you.
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  18111. Mathematicians in a disco ( Crazy Dance XOR Iterated Shuffling )
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  18112. This channel keeps getting better and better
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  18113. Hey mathologer I am a new subscriber I have a request that can you teach quadrilaterals of class 9th
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  18114. My answer to the general form for the problem mentioned around 4:25 (apologies if it's not very well-written...) Let n be the number of pairs of eyelets. To construct a lacing, start at the top left vertex. Since our lacing must be tight and use all eyelets, the next vertex in our lacing must be an unvisited one on the right, for which there are n choices. Then, our next vertex must be an unvisited one on the left, for which we have (n-1) choices. Then our next vertex must be unvisited on the right, then an unvisited on the left, and so on, until our only possible next move is to go back to the vertex where we started. This gives the pattern (n) * (n - 1) * (n - 1) * (n - 2) * (n - 2) * ... * 2 * 2 * 1 * 1 = (n)!(n-1)!. However, consider the fact that this forms a closed loop that starts and ends on the top left corner. Consider the sequence of vertices we visited to arrive at this lacing. We could have arrived at the lacing in exactly 2 ways: either by following the sequence in order, or by following it in reverse order. Hence the form (n)!(n-1)! counts each possible tight lacing exactly twice, so we divide by 2 to get the answer (1/2)(n)!(n-1)!.
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  18115. Most Memorable: no integers in the harmonic series
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  18116. Luckily I can speak German nD
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  18117. The area of the small square for the problem at the end of the video is 1/5 of the area of the larger square
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  18118. For five it’s 2(5!)-(4!+3!+2+1) Or that minus 1.
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  18119. I had that feeling and was right. Without a puase. Did not have the time to find a proof though. Had to just go with it!
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  18120. Real-world harmonic series example - a bit 'off the beaten path.' Consider that you're holding a 1-liter jug of water. You're standing next to an Olympic-size swimming pool attempting to fill it with water. In the first setup, you are allowed to start by filling the pool with 1 full jug, then 1/2 of one jug, then 1/4 of one jug, then 1/8 of one jug, then 1/16 of one jug, then... so on forever. In the second setup, you are allowed to start by filling the pool with 1 full jug, then 1/2 of one jug, then 1/3 of one jug, then 1/4 of one jug, then... so on forever. To a layman, both of these situations seem nearly identical. However, the brilliant thing is, all the effort in the first setup and you might as well have just thrown in two full jugs of water. You'll hardly make a dent in the entire pool. But, with the second setup, you will eventually fill the entire pool. Moreover, you can fill many, many swimming pools; as many as you desire, in finite time (ignoring potential issues arising when you've reached some extremely small portion of a liter, to the point that it's a few mere molecules of water and hardly discernible as liquid H20... however I imagine the sun might explode before this occurs?). It's remarkable how such a (realistically) small change in the process so tremendously changes the solution to the problem. Question for anyone reading - I am curious how long it would take (at say, one pour per second?) to fill an olympic-size pool. They have 2.5 million liters, by the way. Considering the billionth harmonic number (so after a billion iterations) is only just over 20, I think we might be in "heat death of the universe would come first" territory.
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  18121. 20:57 das wird echt cool; bitte!
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  18122. 😊
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  18123. 666K subs for the moment :D. Every time i seen your videos i want to apploud. Very cool content!
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  18124. Tested some python ide on my phone, starts to take time at 500. Didnt debug performance, but it works... Sorry for hacky code was lazy in bed coins =[1,5,10,25,50,100] mem = [] for i in coins: mem.append({}); def solve(target,l=0,ans=[0]): if target == 0 : ans[0] += 1; return; if l > 1 and l < len(coins): if mem[l].get(target): ans[0] += mem[l][target] return prev = ans[0]; if l<len(coins): if(target>=coins[l]): for i in range(0,target+1,coins[l]): solve(target-i,l+1,ans) if l > 1 and l < len(coins): mem[l][target] = ans[0] - prev; if l == 0: return ans[0] print(solve(100))
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  18125. Reversing the order of the iterated coins makes it much faster, probably because memorization is much more efficient in that case ( not sure hehe ) coins =[1,5,10,25,50,100] cons = coins.reverse() mem = [] for i in coins: mem.append({}); def solve(target,l=0,ans=[0]): if target == 0 : ans[0] += 1; return; if l > 1 and l < len(coins): if mem[l].get(target): ans[0] += mem[l][target] return prev = ans[0]; if l<len(coins): for i in range(0,target+1,coins[l]): solve(target-i,l+1,ans) if l > 1 and l < len(coins): mem[l][target] = ans[0] - prev; if l == 0: return ans[0] print(solve(10000))
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  18126. Calculus is a wonderful mathematical theory that is taken way too far and forced into obtuse places. Calculus fits in all the places it has been hammered.
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  18127. Magic all around, not the least in the graphical animations. Wonderful - as always. Viele Grüße!
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  18128. 39:28 I notice the difference between the two righthand numbers (4 and 5, 12 and 13, 30 and 34, 80 and 89) are also the square of the Fibonacci sequence. I don’t quite understand how yet…
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  18129. So what is being said: That it took until after the Spanish Conquest of the Americas, & discovery of rubber, to invent the rubber band to make use of the logarithmic scale on a slide rule! Very astute. ::: Mathologer got me with the “Root of Evil” shirt, subscribed❣️. :: We were still using and taught how to use slide rules thru late 80s in college/university, back in Seattle. calculators still weren’t very advanced so logarithm were the better/fastest way to get physical results... was never sure what the other lines on a slide rule were for though⁉️
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  18130. hummm, never heard of it in high-school nor college.
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  18131. I want a math puzzle that can only be solved if you construct napoleon's theorem to produce an equilateral triangle over some arbitrary triangle to begin with, and that in turn gives you a crazy easy solve.
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  18132. Coincidence : Video is posted on the Ashtami (8th night of the popular Indian (Bengali) festival Durga Puja (aka 8th night of Navratri) ).
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  18133.  @Mathologer  because in the start of this video you related 8 to ∞ (a symbol of spirituality and God in many religions) and indeed Ashtami (8th day of Durga Puja ) is one of the biggest day for Hindus (specially Bengali Hindus).
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  18134. I enjoy your videos on infinite fractions. They give me hours of fun coding software! Thank you so much!
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  18135. great video!
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  18136. Your channel is my front page of youtube. Did you generalize AGM correctly?
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  18137. 🤯
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  18138. Most memorable: that the sum of no nines series is finite
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  18139. The +/- sum always equals 1 if you include the number (1) of n-dimensional bits on the n-dimensional object. For a cube: 8-12+6-1=1.
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  18140. Splendid !!!
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  18141. Wow! Like always! And thankyou❤❤❤ All of my Mathematical insights and jokes comes for you. Again sin(q)/cos(q) ❤❤❤!!!
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  18142. This video is love. It is amazing how this guy takes up the most beautiful and deep of mathematics and breaks it down to super understandable and super cool videos. And along the way a lot of super interesting history and context makes for a real treat. I can’t wait for the video sequel of this one exploring Euler Maclaurin. Really appreciate the work you do, Sir.
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  18143. If for convex surfaces we keep all angles of the folds folding to the inside and not to the outside don't we avoid the outside/inside folding that creates this infinite explosion of the surface in the first place? In the one dimensional case is like approximating a curve by triangles around the curve, much like that function that is continuous in all points but has no derivative at any point. Or in other way the curvature seem to not converge to the curvature of the circle, or whatever surface, it gets going positive and negative more and more. OK, I can't imagine a way to extend that to non convex surfaces though.
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  18144. I take Maths Literacy so this shit is like watching someone trying to teach another language to me.
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  18145. About mathematician FJM Barning: A page in Dutch related to CWI, the Amsterdam institute where he worked, says, "Fredericus Johannes Maria Barning, Freek, geb. Amsterdam 03.10.1924, doctoraalexamen wiskunde Amsterdam GU 1954|a|, medewerker Mathematisch Centrum (1954-), adjunct-directeur Mathematisch Centrum, later Centrum voor Wiskunde en Informatica (1972-1988), OON, overl. Amstelveen 27.06.2012, begr. Amsterdam (RK Bpl. Buitenveldert) 04.07.2012." Source: https://www.hjmwijers.nl/CWI/Barning-FJM-kwst.htm
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  18146. "Y'know, I've often thought I'd like to write a mathematics textbook someday because I have a title that I know will sell a million copies; I'm gonna call it 'Tropic of Calculus.'" — Tom Lehrer
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  18147. Am I Homer!?! I didn't understand until now, thx!
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  18148. ✅
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  18149. 2nd challenge: We only need one term to be = 0. So, jπ/m+1 = π/2 and kπ/n+1 = π/2 for one of the terms. Rearrange to get j = 0.5(m+1) and k = 0.5(n+1). Since j and k have to be integers, m and n have to both be odd. 0.5(a+1) is the same as doing [a/2] for odd values of a, so the largest values of j and k that are iterated through will always result in a term equal to zero if m and n are odd. Did I miss anything? Merry (late) Christmas everyone 😁
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  18150. No - not always possible... example: remove the two squares adjacent to any corner.
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  18151. Another random YT algorithm: subscribed - though I suspect it will make my brain hurt 😂
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  18152. PHP has exponential sized resolution proofs, but by extended resolution there is a super neat polynomial sized proof
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  18153. I have no words. Fantastic stuff! Love seeing all the connections. Ah, I guess I did have some words after all :P
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  18154. Amazing video as always!! Love it
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  18155. Comparing the way we learned math 40 years ago, to today's way... makes me wonder: how will we learn math 40 years from now? Maybe we will not really learn. They will install a program in baby's brain at age 1, and BOOM! the baby knows math! Another program, and this toddler start speaking japanese... Don't say I didn't tell you.
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  18156. Most memorable was the visual proof that the partial sums can be approximated as the integral of 1/x
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  18157. This guy's slightly mad laugh reminded me of someone, and I finally remembered who: Chief Inspector Dreyfus from the Pink Panther movies, played by the great Herbert Lom: https://www.youtube.com/watch?v=LEcsgbwBFRs
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  18158. 9:05 :|
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  18159. I'm from India and we have Heron's Formula in our schools! I'd be sure to share this video onwards!
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  18160. My mind is blown. How have I never been taught this cool bit of math?
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  18161. swivel proof
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  18162. i knew the firest one because i saw it in an episode of "the ascent of man by dr j brownoski) i think it was the number in nature or some such tho he did it with a olive branch some tiles and on the island of samos... good tv series highly recomend it :)
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  18163. I have a master degree in mathematics ( but with an emphasis in algebraic geometry, so not really related to the topic of this video ), and I had a lot of fun with the video! It reminded me of my young math days where I myself tried to come up with those sum formulas after I have learned about the 1+2+3+...+n formula, with a lot less success than the results in your video, but I had a lot of fun trying to work things out! Your videos always remind me why I love maths so much :)
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  18164. Question: if you delete four squares such that the remaining squares are all connected, can it be tiled? This seems like it might reduce to a bipartite matching problem. Not sure if that makes it any easier to analyze though. Extensions to any 2k (with the remainder connected) would also be interesting, if 4 worked.
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  18165. Here a suggestion: make a vídeo about the relationship between the brownian movement and the riemman hypothesis...I know the theme also belongs to physics' universe, but it's an great oportunity to do a collaborative video...anyway, I think it would be awesome!
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  18166. It also looks like the VW logo Which has been changed by the Mandela Effect
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  18167. The end of the video 22:36 reminds me of "hyperbolic rotation" involved in Lorentz Transform, the transforms that preserve some properties in relativity.
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  18168. Is mathematics something to discover or something to learn.
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  18169. Made it to the very end of the video. Final part is definitely worth watching. Might watch twice when I have a few friends over. Side note, whenever I watch a math video, I try to find some sort of quality, inequality, ∴, or something similar that 8 I can learn from. I saw none of those here, which means this is one of those super rare mathologer videos that has little to no actual numbers included. Usually, there are at least one or 2 algebraic autopilot sections, and some working out of variables, but this one is especially good for non mathy friends as it is all intuition with the rigor hidden behind the curtain. Loved the video, and as long as you keep sharing, I'll keep watching. ❤️
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  18170. Quick question, when I saw there'd be a Christmas egg I started taking a detailed look at the intro image, and now I'm wondering: why are there two representations of the number 666 in it?
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  18171. I saw the whole thing and I really really enjoyed this video. I appreciate how you glossed over the inverse matrix, because that is a whole can of worms on its own. I'm having a hard time critiquing this one, because I felt like I understood it better than I usually do with your videos! My experience is a mechanical engineering degree in the united states, so a good amount of calculus, and some linear algebra.
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  18172. Unfolding the Earth globe finally brings us the proof that the earth ist flat! ;-)
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  18173. Does this cycle of numbers have a connection to the base 10 (decimals) in our numbering system? Ice use base 8 or base 12 or base 16 do we get a different cycle of numbers?
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  18174. Yes, he explains that towards the end of the video.
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  18175. hello from 1030/5
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  18176. Great content
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  18177. How about to prove the equation x²+y²+z² = w² have no integer solutions (prime not necessary)? :)
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  18178. Your t-shirt is confusing to look at lol
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  18179. Easy? No, certainly not! If there is anything that has a steep learning curve then it is calculus. But at least you will be able to tell just how steep ;-)
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  18180. They can differ by just 1, the smallest example being 9^3 + 10^3 -1 = 12^3 (thank ramanujan). For base numbers in sequence, not sure.
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  18181. Thanks for talking about this topic
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  18182. When is the video about ruffly flower petal shapes, gee I mean hyperbolic surfaces … thanks for enjoyable presentation
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  18183. Omg. I wish more ppl were interested in math to appreciate things like this, and your vid itself. Great edit job too, congrats the team. Perfect job man. +1 sub for sure.
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  18184. We were taught this for o-level I loved these ❤️
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  18185. I’ve been playing with these shapes in my art for a while now to very interesting effects. I’ve also started playing with the Golden Section and the related triangle of numbers. I hope by persisting with your videos, I might discover something new to explore. Thank you.
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  18186. I didn't finish watching the first time! So now I get to finish the corrected version. I am not disappointed! :-)
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  18187. I hate to ask this question but, what functional application does this provide to the human race other than a mathematical thought exercise?
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  18188. Spoiler alert... - - - - - - A) To move the 10-disk stack from position A to B one has to start in the counterclockwise direction (I just checked it for the 1, 2 and 3 disk stacks and lazily generalized that for an even number of disks you start counterclockwise and clockwise around for an odd number of disks). B) The exact value for the average minimal distance is 946/243 (I have a pocket calculator with enough CAS smartness to work with fractions 🙃).
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  18189. E ceva fenomenal , unii oameni se nasc geniali , a fost ceva deosebit acest film extraordinar -Multumesc mult Maestre -Romania!
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  18190. This is true with 3x3 as well if a similar approach is done.
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  18191. This is basically a more general version of the fact that the middle row, the middle column and the main diagonals sum up to the magic total
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  18192. AAAAAH SOOO MUCH BEAUTY!!!!!!!! :) :) :)
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  18193. It won't always work when you cut out two green and two black squares. You can cut out the three tiles surrounding the corner. Besides, if you have that loop thing and begin by cutting out two green in a row instead of switching between green and black, it won't work. You might be able to do it with a different loop though?
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  18194. Re-like
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  18195.  @Mathologer  likely as long as you like me (us) to re-like your (magnificent) videos. PS: I loved the office!
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  18196. About reversability. Surely it is possible to reverse this process by making arbitrary choices at each stage. After the appropriate number of steps you get A POLYGON which will map to the starting regular one. How is this NOT reversing the process. No-one said the answer had to be unique.
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  18197.  @Mathologer  Fair Enough. However I did once have to solve an engineering problem to find a function to map two arbitrary planar points to two other given points. I did this by placing the output points on adjacent vertices of a regular pentagon and working backwards to find a transfer function. So, it's been used at least once. PS I may be a pedant but imho reversing the process means following the steps in the reverse order, whatever result this gives. Reversing the result may need a different process. PPS I watch your channel because its fun. Please keep on.
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  18198. For the transformation matrix shown at 27:30 it is possible to calculate an inverse transformation. This theory is identical to Symmetrical Sequence Component theory that is used in power system engineering. The Fortescue Transformation (what is shown at 27:30) converts the imbalanced 'N'-gon into a system of 'N' balanced sequenced components. The inverse Fortescue Transformation is used to convert from a system of balanced sequence components to the corresponding imbalanced polygon.
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  18199. okay I shouldn't have been amazed that the sequence of natural numbers was generated by showing how this is just pascals triangle at 14:15 but I was. then what really blew my mind is when I realized that also shows a visual representation for the definition that 1^0 =1 and n^0=1. you can generate the outer sides of pascals triangle by actually writing an outer shell of unwritten 0's. I bet you could also show that/why 0!=1 also with this.
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  18200. Area=5
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  18201. I am here for tha aha binary moment
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  18202. ok now just need to bring a turtle and laser in exam
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  18203. Sir, can you said why divergent series is not intuitive.
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  18204. What if we have a point, then a line segment, then a triangle, then a tetrahedron and so on.... Can we have any formula for n dimensional tetrahedrons to count face, cell, vertex etc ? What about n-dimensional other polyhedrons?
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  18205. Great video, love how you have linked the ideas and concepts. Not sure you should go around calling people 't***el heads' though.
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  18206. Can you do the same experiments with other types of tesselations! Such as this grid! https://en.m.wikipedia.org/wiki/File:Capital_I_tiling-4color.svg
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  18207. Mathologer, do you watch Flammable Maths? He's done quite a bit with the infinite product for sine
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  18208. For the problem of maximum number of moves to turn all faced-down cards to face-up card, it goes like this: We start from the far right card and turn the card up. Then the card to the left of it, then the card to the left of that and so on, until we reach the far left card which then will be faced up but all the rest of the cards will be faced down which are to the right. So all the cards will be faced down except the far left card. Like this: Example (1) 111 -> 110 101 011 By moving like this, we only turn up one card per move and no more, where also no less is possible either. But on every move that the right card goes from up to down-faced, the number of faced-up cards will remain the same, and that helps to extend the number of moves without losing cards. So in the example (1) above, for 3 cards (or representing it, a 3 digit number), we could extend the turning up of card to 3 moves so that only 1 card has turned up in the end. Then after that, we can't to anything to the leftmost card, so we are left with two cards, which similarly will take 2 moves, and then we are left with 1 card, which takes 1 move at most. Then we are left with no cards to turn up, and all cards are already turned up. So it took 3+2+1 moves at most to turn all cards up. Here is the representation with 1s and 0s where 1 is faced-down and 0 is faced-up: 111 (starting state) 110 101 011 (3 moves to get here) 010 001 (2 more moves to get here) 000 (1 more move to get here) result for 3 cards: 3+2+1 => 6 moves at most Similarly for 4 cards would be 4+3+2+1 moves at most which by the formula n(n+1)/2 where n=4 we could easily calculate the sum to be 4(4+1)/2 => 10 moves at most. So That's it. For n cards, we have n(n+1)/2 moves at most.
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  18209. 153^2 + 104^2 = 185^2 4 9 13 17
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  18210. This reminded me of the fact that the sequence of differences of squares is just the odd numbers.
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  18211. Maybe I'm late but happy b-day!
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  18212. The most memorable part of this video is how the so-called "abstract math" discussed can relate to "real-life" situations (e.g. the harmonic series visualized by the leaning tower of lire). As an aspiring "applied math" person myself (i.e. I'm in bioinformatics), it's amazing to see the weird and wonderful things that pure math folks discover about the world
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  18213. finally, a mathloger video I can understand : )
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  18214. This video sort of serves to illustrate how responsibly used math can be used to define the world, but irresponsibly used math can be used to invent things out of whole cloth that you'll have to pretend are real in college, and never again.
    1
  18215. 3:39 Me, remembering the ”Why is Calculus so… EASY?!” -video: ”The area of the shape is ln(x), at the limit (so, evaluated at ∞).”.
    1
  18216. Bin c'est une règle à calcul, quoi.
    1
  18217. Any time I post a link, the comment is deleted. Just wanted to mention that I made a very early prototype this afternoon (Still has issues, but seems to work!). It's on my Github if someone wants to test.
    1
  18218. Yes, comments with links often get quarantined. However, I do then un-quarantine them if there are no problems. Did not see your comment though. I also did not see anything relevant on your Github page when I checked just now :(
    1
  18219. I've improved the page a bit, and I found an interesting strange outcome. With n=4 the angles are 90, 180 and 270. If your first two steps include the 180, everything looks good (you get a square as expected). In the other case (either 90, 270 or 270,90) you get a line! Why is this?
    1
  18220. Wait, I think I know why. The "eigenpolygons" of the square are: 90=square clockwise, 270= square counterclockwise, 180=...infinite line? There is no polygon with 4 sides whose 4 angles are 180...unless those lines are infinite and all the points are at the center
    1
  18221. Mind = Blown! 🔥 What an amazing video for the topic!!!
    1
  18222. Hi, there is typo in your last image … the first series should be with only odd numbers in the denominators. Still, congratulations … I like very much your posts and style.
    1
  18223. To show sum 1/n^2 converges, write, starting at n=2, 1/n^2<1/(n(n-1))=1/(n-1)-1/n and you get a telescoping sum bounded by 1 so the total sum is less than 2.
    1
  18224. There shouldn't be any taxes on it, since it was likely already taxed the very first time it was made. There's absolutely no good reason to keep on taxing the same damn money multiple times. It stinks of "value added" vibes.
    1
  18225. Every time you use "honed" I think you mean "homed". Homing refers to approaching a place, while honing refers to sharpening (of a knife).
    1
  18226. I felt like a real rocket genius for a brief moment when I discovered that the square of 2x plus one was equal to x times (x plus one) times four plus one. Then I thought about how long ago somebody else must have already figured that out. Not sure exactly what this could have to do with the "consecutivity of squares" illustrated in this video but coming up with proofs factoring for four will keep me busy for a bit.
    1
  18227. Always good to see a new Mathologer video! Nice shoutout to my old friend John Baez.
    1
  18228. ☮️ ❤⬅️🔼♾️💟🗽
    1
  18229. 142857 X 2 = 285714 X 2 = 571428 X 3 = 1714284 And so on, and so on...
    1
  18230. It seems that Maths runs in our veins...
    1
  18231. 26:22 In general, if j^2 = 1, then e^(ja) = cosh a + j sinh a. This is useful when j is not 1. Consider, for example, j = -1, or the matrix [ 0 1 ; 1 0 ].
    1
  18232. There is an equivalent formula for ε² = 0, which is exp(εa) = 1 + εa. Much like when j² = 1 but j ≠ 1, this form is useful even when ε is not 0.
    1
  18233.  @angeldude101  Yes! In general, if ε² = 0, then f(x + εy) = f(x) + ε f'(y) by Taylor expansion.
    1
  18234. Reporting: compute all the tangent lengths from the six points to the incircle of the triangle and see that they all equal the half-perimeter.
    1
  18235. love this!
    1
  18236. No human was hurt by flying playing cards during the filming of this video.
    1
  18237. A lengthier way for y' = xy + 1, y(0) = 0. We can solve z' = xz as z = c e^½x², c is some arbitrary constant. Let y = z f(x) (1, c≠0 since y=0 isn't a solution) be the solution of y' = yx + 1. [As y(0)=0 => cf(0)=0 => f(0) = 0] Therefore, y' = z'f(x) + zf'(x) and thus y' = cxe^½x² f(x) + ce^½x² f'(x) = xy + ce^½x²f'(x). Thus, ce^½x²f'(x) = 1 which can be solved as f(x) = 1/c ∫_0^x e^-½t² (2) [f'(x) = 1/c e^½x² => f(x) - f(0) = 1/c ∫_0^x e^-½t² => f(x) = 1/c ∫_0^x e^-½t²] Plugging (2) in (1) gives y(x) = c e^½x² (1/c ∫_0^x e^-½t²]) => y(x) = e^½x² ∫_0^x e^-½t²
    1
  18238. I hate Hanoi towers because I always end up on the middle tower first :/
    1
  18239. The video totally worked for me up till the end. Just go on like that!
    1
  18240. I saw this video has 50 minutes... Well, I'll be watching it on christmas day, because that day, I'm going to do a whole movie day. And this will be the first video to open my mind before dawn.
    1
  18241. I preferred the tri-colored segments argument.
    1
  18242. Love this presentation—I’ll go to the archives for the other “beautiful equations.”
    1
  18243. Mathologer You are the best presenter, teacher or lecturer I have ever encountered on youtube, and possibly in the brick-and-brain world as well. If you were on faculty at a nearby university, I would go out of my way to register for your classes. But fascinated as I am with mathematics, I wish you also taught evolutionary biology!
    1
  18244. Where do you get your t-shirts from?
    1
  18245. I also want that Fibonacci pizza shirt :(
    1
  18246. Nice video, as always.
    1
  18247. I'm amazed by how useful are the addition (0) and multiplication (1) identities, two of the five members of Euler's identity, btw... Nice video! Thanks a lot for this great channel.
    1
  18248. Not gonna lie I love this video, but I sat up and screamed in joy when I learned about the Chrome Extension for cut the knot
    1
  18249. Is there anything interesting to say about the ratio of b to p as related to, say, how fast the area explodes or especially for the case of targeting a fixed area? I'm assuming there is not since it was not brought up, but I thought I would ask.
    1
  18250. Just wanted to say this is one of the most interesting and entertaining channel I'm following. Each video brings out my curiosity and a smile on my face. Thank you! And as for this specific video I happen to have a copy of the book "Euler's gem" on my nightstand, a real spoiler 😆
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  18251. very wonderfull❤❣💓
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  18252. I have a bachelors in applied mathematics. The first half was very familiar to me. The Bernoulli numbers were cool and that was new to me. The rest was for the most part easy to follow. The part at 40:45 I didn't see immediately, but I feel like if I got a pen and paper and worked it out I could figure it out. I was surprised when you said the corrective factor diverges. Didn't see that one. I look forward to the next video to hear your explanation!
    1
  18253. Wow, 547 slides is not a joke ;v
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  18254. Well conceived and presented topic, congratulations, and thank you indeed! (It reminded me of the logarithmic spiral, which is an invariant shape for "scale and rotate" operations.) But why did you call it "anti-shapeshifters"? Maybe I understand it wrong, but it sounds a bit "anti" to me, what about "shapeshift-proof" or "-invariant" ;-) Anyway, I like geometric approaches to things algebraic, which you did nicely here with the log function. Other examples I saw approached eg the numbers e and pi, Fourier transforms, ... with nice geometrical methods. Another feature I'd like to see developed with all the functions you mentioned here (including the Lambert W ;), is their interesting geometrical(!) properties as complex functions, ie as 4D surfaces. An example: Circle and hyperbola equations represent the same complex 4D surface, and both curves belong to it; it's only that each surface has another orientation in 4D, and a different curve in the "real plane". Ditto for namesake goniometric and hyperbolic functions. Cos and Sin also show up nicely as combinations of the exponential and its inverse. Unfortunately complex function renderings are usually confined to 2D or 3D extractions (mappings, Re and Im, ...). "True 4D" rendering is possible though, I dedicated a webpage and channel to it. For those interested, welcome for a start at https://www.geogebra.org/m/bmuqbufn (with further links). Thanks again.
    1
  18255. For the 666th partition number I got: 11393868451739000294452939 How'd I do?
    1
  18256. So if I can approach any given number with the sum of the two series 1+½+...+1/m and -1-½-1/n, what is the the value m/n going to be then? Does it mean anything? Since m and n can be arbitrarily large for irrational numbers, will m/n converge to anything interesting? I am just curious. I don't know if it even makes sense. 😅
    1
  18257. The whole video was incredible! No 9s is my favorite
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  18258. you are the best math youtuber keep going pls
    1
  18259. Isnt 4×k-1 just as good as 4×k+3? Don't know how to prove it, but all the primes that aren't able to be factored also conform to p=4×k-1, where p=prime: 3=4×1-1=4×0+3 7=4×2-1=4×1+3 11=4×3-1=4×2+3 15 isn't prime (4×4-1=4×3+3) 19=4×5-1=4×4+3 23=4×6-1=4×5+3 27 isn't prime (4×7-1=4×6+3) 31=4×8-1=4×7+3 ... Edit: Or is this just another way of "drawing the windmill"? Or trivial?
    1
  18260. Hi :) Just finished to the end and as per your request: I am just finishing my degree in theoretical physics, so my maths is approximately LinA 1-2, Ana 1-3, function theory, classical field theory, some group theory and a bit of tensor calc (wow saying it like this sounds way more impressive than the last couple years felt) plus some stuff I read for leisure like non standard calc and some constructivist maths. The first couple of minutes are usually not the most engaging for me outside of the amazing animations, but I usually watch them still to be able to follow the narrative later on. The last chapter was probably the best, it left me unsatisfied in exactly the right way and have already started playing around with Euler-Maclaurin trying to get an intuition what happens with non-analytic functions/functions outside the convergence radius of the Taylor series. However rewriting the E-M formula in terms of f^(n) (1) felt poorly motivated, but I guess you'll come back to that once it is time for more :) Personally, I could listen to full lectures in this style, going deeper and deeper, but I understand that that would probably be almost unjustifiable in terms of the effort required on your end. Both this and your video touching on Galois theory especially gave me that feeling of "I wish I could come back next week to get some more in lecture 2" as they almost felt like introductory lectures, sparking interest and hinting at a more powerful, deeper subject that I am too lazy to explore on my own.
    1
  18261. Tight lacing thinking and solution below . . . . . Hope this is far enough. First, consider a shoe with two eyelets on each side. There's only one way to make a tight lacing. This is our base case. Now, imagine you know how many tight lacings there are for n eyelets (per side). Call this E(n). Now, imagine a shoe with n+1 eyelets. Start with the top left eyelet (the choice is arbitrary), and pick an eyelet on the right side to link to (You have n+1 choices). Then, pick one on the right side besides the top left to lace (you have n choices here). Now, you have n free eyelets on both sides, and this means you can lace those up in E(n) ways, and replace the lace going back to the one you started the n hole problem with to one to the top left eyelet. Hence, E(n+1)=(n+1)×n×E(n). In closed form, this becomes (remember that E(2)=1, not 2, hence the half being applied) E(n)=n!×(n-1)!/2
    1
  18262. It's a great video! One may ask: when does the similarities between these to Calculus broke up? And more important, what is a Calculus (in a given set)? It's easy to see in the first question the g.l.b. property, e.g. the intermediate value theorem, (topology property?); I think that the second one is answer in algebraic geometry.
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  18263. Vielen Dank. Ihre Videos sind immer ein Vergnügen. Auf einen deutschen Beitrag freue ich mich schon jetzt. Sie sprechen sehr gut deutsch. Stammen Sie bzw. Ihre Vorfahren aus Deutschland?
    1
  18264. Seems fair unless you're a little guy waiting forever for the final payout while some rich guy is getting paid more before you.
    1
  18265. The Ultamate "ugly" stack kind of reminds me of a bird in flight
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  18266. Solved the first puzzle! Answer below: 9, 4 17, 13
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  18267. The fact you mentioned near the end where we end up with fibonacci numbers as our hypotenuse with starting sequence 1,1,2,3,5,8... is really cool, but doesn't it apply to any starting sequence? If you begin: F(1),F(2),F(3) Then, your sequence of side lengths will still contain every other number in your sequence due to the final proof you presented.
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  18268. Why you (Methologer) are so happy? Talmud solution offers more money to small investor when there is small estate but more money to big investor when the estate is big! Symptomatic that there is no example how Talmud will divide estate worth 400 or 350. But there is a simple way. Percent of investment should decide. By Talmud in case of 350, first will get 50 second 100 third 200. By percentage first 58.3, second 116.6, third 175.
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  18269. Lols when I saw the thumbnail of the Top equation =6 over the bottom equation =6 all i saw was 6's but yeas good one for a thumbnail however both of them in order of the 123 order would break however the 1 on both does fit and so 1+2 and 1x2 is 3 and 2 so thus the order for the multiplying one has to then be 132 so that it reads 1+2 and 1x3 both will check out at 3 then finally adding the 3 to an already 3 is effectively doubling it so thus why the 2 has to be last in the multiplication as 2 in multiplication only doubles things. I would argue that it really should be... 1+2+3=1x3x2 just because you can follow it along on both sides on steps 2 and 3... well actually steps 1 and 2 as 1 already checks out and for it to work with 1 you would have to ADD 0 to the 1, but Multiply by 1 on the other so yes that is a weird reality so 1 instead will be initialized as the starting number and then the step rules would just be to Add 2 then add 3, and the step rules for MULIPLICATION be to Multiply by 3 then Multiply by 2 to keep the numerical equivalents the same... so the adding one you would have 10 after the 4 but the multiplying you be left with 24 and you can't square the 4 on the adding as that would leave you with the Prime 19 so then I was thinking what about Algebra so example 1+2+3+4x=1x3x2x4x but this implies that x is the same on each side thus I see no world where you can tack an extra number not even a hacky number or sequence so the buck does stop at 6=6 really. Now "If" we started at the initialized Digit was 6 and we both agreed we had to get 6 to come down then we can get as far as... 6/3 or 6-3 to get to 3 but the divide is the issue so order counts We would have to do 6/2 to get to 3 then divide by 3 to get to 1, now for subtraction because The division was 2 steps we also only do 2 steps in subtraction and that would be 6-3-2=1 so actually I can then make 2 equations that can take 6 to 1 by making "6/2/3=6-3-2" so Voila [236] is another ser of 3 numbers that also can get to a same resulting end number. I will now proceed to watch the 26 minute show but I feel there might be a chance that "6/2/3=6-3-2" fits in the scope of this haha God I love Math.
    1
  18270. Apparently some Triangle show and a Heron guy lols xD I think Heron was also like a Lake and or a Bird too. - Also I was thinking about before as in my above post because we dealing with the number 6 anyways there is something with the numbers 2 and 3 and this is why so when you get to the end result of (1+2=3= or 1x3x2=) You you tack on because the result is 6 anyways (1+2+3-)-(3)-(2)= or even (1+2+3)/(2)/(3)= would also be the same as then (1x3x2)-(3)-(2)= or even (1x3x2)/(2)/(3)= and that's effectively a boomerang so then numbers 12233 make a boomerang where you Toss out from a starting unit of 1 and it goes 6 Units out to then come back to 1, however I can make an oscillator/pendulum by instead by putting a forever line or repeat line in math over the function "/(2)/(3), *(3)*(2)" parts or "-(3)-(2), +(2) +(3)" parts and division and subtraction can swap and even addition and multiplication can swap meaning you could actually have a scenario Where I can even do like (((((((1+2+3)/2)/3)*3)*2)-3)-2)=1 lols I can even do another set like (((((((1*3*2)-3)-2)+2)+3)/2)/3)=1 where regardless the step you on on either side both sides used same numbers, and give the same resulting ending number "FINALLY" each step providing you stopped calculating after all */+- symbols were used you would noticed you went through exactly 2 of each, in all cases however I really dunno what practical reasons I would use this for other than... maybe making an enemy pace back and forth on a 6 width or a 6 height grid game that your trying to pass through lols... but yes /2=-3 or x2=+3 or /3=-2 or *3=+2 in this weird world but normally these results normally would not be on par in most use cases. On the Fall @6 because /2 gets to 3 then /3 gets to 1 is same as @6 -3 gets to 3 then -2 gets to 1 the inverse of each On the Rise @1 because +2 gets to 3 then +3 gets to 6 is same as @1 *3 gets to 3 then *2 gets to 6, now I have a Hunch cuz in chess a "Knight" moves 2 and 3 in the shape of an L I'm certain there is a way to rig up a chess board where going Left is subtraction and Right is addition where going Up is Multiplication and going Down is Division I'm almost nearly confident even if it's as small as a 5x5 board or something that we can do all of this Math on the board with just a simple "Knight".
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  18271. Ableitungen are derivatives? Then ableiten is derive? And I allways thought that "derive" is synonymous with "calculate" , hence the name of the calculator programming language "Derive". Oh the memories when a Ti-92 was more usefull than a phone of that aera....
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  18272. Amazing video again!
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  18273. The most memorable thing for me was actually none of the options that you presented. It was the thing you did at 23:12 I've heard before from Numberphile's video about gamma that you can prove that it has an upper bound, but I didn't think it would be so elegant About Kempner's proof though. I learned several years ago that almost all numbers contain a 9, or almost all numbers contain a 3, but now, of all times, I forgot about it, and I concluded that the sum with no 9s was infinite. "Damnit" is the only word I had in my mind.
    1
  18274. the 2nd proof is better
    1
  18275. This video makes me wonder how you'd find a sequence of pythagorean triples for triangles that approximate the golden ratio. (That spiral animation at the end shows triangles approximating root phi, which isn't what I'm looking for)
    1
  18276. Could there be patterns for the cubed level if you allow for some of the values to be negative?
    1
  18277. The problem with concrete examples: Does what you show us follow a general principle or is a mere coincidence? Maybe I'm too much a computer scientist that I'm worried when someone uses the principle of "proof by some example(s)". Because experience shows: It's almost never really true.
    1
  18278. I first heard of this in the early 80's in Scientific American, when Douglas R. Hofstadter (in his Metamagical Themas column) revisited Martin Gardner's treatment. Did you re-read that when preparing your script?
    1
  18279. I like the coloring proof
    1
  18280. Swivel proof for sure was nicer.
    1
  18281. Beautiful as always!
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  18282. So, the angles of the folded triangle are the angles of the original triangle's medians. I wonder if there are any trigonometric identities that could be deducted from this fact?
    1
  18283. At 17:18 looks like origami.
    1
  18284. A non-mathematical question inspired by 8:57 - Do you now live in, or would you consider moving to, Australia just to mess with people?
    1
  18285. 49:55 puzzle is the answer 55?
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  18286. Can somebody explain why in chapter 5 it is easy to show that cycles must be of equal lenghts?
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  18287. Because none of the relevant properties of the cube change when you change state. Imagine the following: You have just completed one cycle, starting from a solved cube. You have a second state you know is not part of that cycle. If you remove all the stickers and put them back on to represent the second state, it should be clear that the length of the cycle your cube will now take has not changed, but the properties of this sticker-swapped cube are identical to if you had properly navigated to that state by rotating sections of the cube.
    1
  18288. hey mathologer,1+1/2+1/3+.....=x x=1/1+1/3+1/5+...+1/2+1/4+1/6+...=pi/4+1/2(1/1+1/2+.....)=pi/4+x/2,so x=pi/2 so ,1+1/2+1/3+.....=pi/2 and 1/2+1/4+...=pi/4
    1
  18289. Is there a proof that all primes can be expressed as either 4k+1 or 4k+3?
    1
  18290. I remember PHI-1 has nice correlation with PHI, but I don't remember what it is
    1
  18291.  @cl4655  yes, totally the latter, thanks!
    1
  18292. PHI - 1 = 1/PHI
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  18293. Please do a video on why the quintic has no general solution
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  18294. Finally I understand. Thanks!
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  18295. 😮😮😮amazing
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  18296. Excellent, as always. If you are looking for practical use of the composition of non regular polygons as a combination of regular ones, look this https://en.m.wikipedia.org/wiki/Symmetrical_components A 3-phase power supply can be represented as a triangle, which is equilateral when balanced. If unbalanced, you can study the system as the sum of three "equilateral triangles". Even Napoleon's theorem appears in the Wikipedia entry on this
    1
  18297. I really like these interludes and hope you will continue to do them occasionally.
    1
  18298. Wow. Can you go on with Solfegio?
    1
  18299. 15:51 this is just genius. I never saw this formula for e appearing like that.
    1
  18300. 7 numbers from 1 to 100 you can't split up into groups that have a sum in common is 1,2,4,8,16,32,64
    1
  18301. Looks good. Not sure what title means, but for a 100+ years they taught the Principia at Cambridge and it went badly. I’ve written a paper on parts of the book; it is a hideous mess with inconsistencies, complex and unnecessary obsessions with arcane curves, and errors. It is a terrible textbook. This is not to deny Newton’s genius, just to say if you want to learn Calculus stay away from Principia. Leave it to the historians who have already spent too much time on it.
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  18302. Apparently, even going backwards from 3,4,5 gives a fibonacci box. 1 0 1 1
    1
  18303. Does anyone know what the name of the guitar music playing during the visualisation parts is? Or where it is from? :) I know I heard it somewhere but can't put a finger on it. Great video btw! Was quite easy to follow.
    1
  18304. To have an intuition on why gamma is greater than 0.5, you could take your blue curves and notice that if you encapsulate each blue curve inside a rectangle, it would cover more than half of this rectangle area, since this happens for all the rectangles no matter how small they are, in total the blue area would cover more than half of the area of the square.
    1
  18305. That formula with the cosines reminded me of finding the eigenvalues and eigenvectors of toeplitz matrices using the tridiagonal matrix algorithm! Then you brought up determinants shortly later, so I know that this must be related!
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  18306. Wow! Thank you
    1
  18307. The visualisation of gamma impressed me the most. I knew the definition of it, but I never knew the context or why it is finite
    1
  18308. Learning about gamma wins for me and the formula that relates it with all primes looked crazy. Also enjoyed how simple Kempners proof was (and all other parts of the video as well). Keep it up with these awesome videos!
    1
  18309. I'm wondering what special feature of the mind is required to be able to see the graphical interpretation of these formulas... or see the mathematical description of these graphical manipulations. I definitely don't have it... but, I love to see it!!!!
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  18310. Mathologer is requested to know about Tamil language which is in India before sanskrit lanugage arrived to this land mass. People sometimes say many things might have taken or learned from Sangam period Tamil people. During Sangam time, Tamil society was famous for their contribution to Tamil language. However, no idea about their maths skills.
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  18311. The hexagon is a 3D stack of cubes. Where you are either fill or empty. And you can only take cubes away that have their 3 sides connected to the corner close to you exposed. if you take one cube away from the full room. You expose 3 other side's or the cubes below and to its back. But always equal number of sides/colors.
    1
  18312.  @Mathologer  As this whole video is about tiling a plane (or some subsection of it) with a 2x1 rectangle it touches on a few packing problems I love. Such thinking is very visual and can therefore me demonstrated. To transform any checkerboard into a single line with the right way to cut it apart brings this down a dimension. There is a short video by CPG-Grey called "the hexagon is the bestagon" as it tiles the plane the most beautifully and is the underlaying pattern to most (non rectangular) tiling patterns. I will have a look at the link when I am home again in a few days.
    1
  18313. I bought a few packs of Rogers sugar cubes to have fun with watching this video. Happy New Year!:hand-pink-waving:
    1
  18314. I always wondered why the Gauss problem was considered difficult. Isn't it rather obvious what (1 + 99 + 2 + 98 + 3 + 97...) + 100 is? Or am I basically misunderstanding something? Love your videos.
    1
  18315. At 17:30 this diagram shows the Symmetric Sequence Component decomposition of the original pentagon and at 27:30 this is the Fortescue transformation. Perhaps there is a need to add Charles LeGeyt Fortescue as a 4th independent discoverer of this same theory?
    1
  18316.  @Mathologer  It gets interesting when you consider harmonics in a three-phase power system - the individual harmonics alternate between only existing as positive sequence (1, 4, 7, 10,...), negative sequence (2, 5, 8, 11,...), and zero sequence (3, 6, 9, 12,...) - this brings rise to some interesting effects for harmonic combinations - I.E., 6n+/-1 harmonic pairs. The 'coin rotation' paradox concept applies with these harmonic pairs - for example both the 5th harmonic component and the 7th harmonic components decompose down to hexagonal shaped symmetrical sequence components. As an interesting aside this ideas of harmonic pairs also turns up in the spin associated with elementary particles and explains the two discretely different values of angular momentum for the up Vs down spin.
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  18317. Thank you for giving me a mathematically rigorous way to fix all my polygons to be regular.
    1
  18318. I remember thinking about the expansion of binomial and seeing the connection between the 1st, 2nd, and 3rd powers in geometric form. Just seeing the first few minutes made me realize that adding up the numbers was simply like filling in the blanks of the binomial expansion (and hence the geometric forms)
    1
  18319. Most people I know can't tie their shoes.
    1
  18320. Awesome video. Thumbs up.
    1
  18321. Loosely understood most things. Just about to finish high school. Thanks Bukard.
    1
  18322. The sound is really low on my device.even with a full volume i can't hear half the words you saying.
    1
  18323. I don't recall learning about gamma before
    1
  18324. 11:39 my guess: they are both decahedrons
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  18325. so beautifully laid out, the examination and the best explanation I've ever heard.
    1
  18326. The no 9s series actually follows a fractal pattern similar to the sierpinsky triangle or menger sponge. And it is a well known fact those have an area aproaching zero so it is to be expected the sum of the no 9s series is low and not infinite.
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  18327. So a chord is a special stucture with seemingly magical properties, as greek mathematicians like pythagoras thought. Lets see how. Lets take the diameter of a circle. Its actually a two sided polygon with both sides superposed. If we bisect the polygon, and then determine the length of the bisector (0) we can then compute the length of the chord of the half angle by taking the square root of the combined square of the halfcord ans square of the radius minus the bisector. Lets see. Diameter = Chord 180° on the unit circle is 2, halfchord is 1, the bisector length 0, the radius minus the bisector is 1, the combined squares are 1 + 1 = 2 and the squareroot of 2 = chord of 90°. We can continue down 22.5, 11.25, 5.625, 2.8125, 1.40625, 0.725 So as an ancient greek the chords of 90°, 60° = 1, and 108° = magic number, could be used to derive the values of all angles divisible by three. This was like having a sine, cosine table with an error range of 0.75°. You can get rid of this error in two ways. 1. Trisect a chord on an inscribed triangle using a markov chain. 2. Derive an angle that is solely divisible by 2, for instance 128° If we take the ratio of 128 to any known angle say 120° (chord = SQRT(3)). So 128/120 = 1 and 1/15 to a greek. Now when we obtain successive halfchords what happens is the ratio of the chords ---> ratio of the length of the half chord. We can prove this by noting that an inscribe octogon has a perimeter better estimate of pi than an inscribed square. As each halving of the angle occurs the arc of the angle better approximates the chord, this is junior highschool geometry. But what is not evident that the bisector rapidly approaches 1. Once that occurs you can simply multiply the angle ratio by the known chord. So that seems offtopic, but how do we get our chord of 128° Take the small calcukated angle chord with bisector length of 1. obtain half chord. New chord = 2 chord * bisector. Determine new bisector, Repeat, Eventually you will get the chords of 1,2,4,8,16,32,64,128 degrees. This solves easily derives to fill determine all the chords on the circle (199, 198, 196, 192, 184, . . . . and each of the even number angles can be halved, etc) Many people think getting sines and cosines is a revelation. But the halfchord and bisector are the sine and cosine of the half angle. So immediately we now have a sine cosine table with an error of 0.25 degrees. But theres another magical secret in the chord. If we mount the one chord end on an axis centered about the origin , (e.g. 1,0) the halving the chord of the double angle (derived above chord * bisector) defines the y-coordinate. And subtracting from the x position 2 * bisector squared gives the position on the x - coordinate. As a result it makes it possible to plot the positions of any polygon without need of sines or cosines. What is the relevance here. As we see above chords behave in very predictable manners that made them easy to manipulate for field mathematics. Any chord of the same length on the same circle has the same bisector length and same halfchord length. Lets say you put 4 same length chords at various positions alomg a circle. Where they touch, does matter, form Ptolemy 's Quadrlateral. Lets do this with 3 chirds, they form a triangle, because they are chords, they all will form isoceles triangles with the point of tangent contact. Since the tangent point is the bisector point, then each side is the halfchord. Isnt that great, and since two intersecting halfchords form 2 sides of an isosceles triangle to the vertex. Since the halfchords are the samelength and the tangent-vertex lengths are also the same, the distance of the two distal segments of the intersecting chords is also the same.
    1
  18328.  @Mathologer  True, it's only a Sunday
    1
  18329. The fact that sum of reciprocal of numbers having googol zeros is larger than the sum of no 9 series is really mind bending. Also the music in credits is quite nice. :)
    1
  18330. Great video.
    1
  18331. Very enjoyable. It started to go crazy with the introduction of Bernoulli numbers as the answer to the guessing game for the coefficients, but the presentation of the Euler-Maclaurin formula is understandable with basic understanding of calculus. I would love to see more crazy videos on advanced topics. Master degree mathematics.
    1
  18332. Yet nowhere do you show how a slide rule 'slides'. Now isn't that odd?
    1
  18333. most memorable is Kempners proof
    1
  18334. Super-duper! Thank you!
    1
  18335. Thank you.
    1
  18336. 12:04: The corresponding fibonacci (am i spelling it right?) numbers for the square are 9-4-13-17, and from the 3-4-5 triangle you can get to it by going right, right, right, left. 15:58: Every square in the second layer adds up to the same value as the one on the square below. So at each layer the sum of the areas of the squares remains constant. The total area of the tree is therefore the area of the base square times the number of layers the tree has. Since the tree has 5 layers and its base square has area 1, the total area of the tree is 5. 24:22: The problem arises because the line connecting the in-circle and feuerbach circle is parallel to the base of the triangle. This results in a 0-1-1 non-triangle, and by using the fibonacci matrix 1-0-1-1 its easy to verify that this non-triangle is indeed the parent of the 3-4-5 triangle. 37:53: If you look at the geometric construction of the parent triangle, it becomes really obvious why there is only one unique primitive parent to each triangle. The construction locks you into only one choice. And since the in-circles get smaller each time you look for parents, and since the in-circle radii lengths are always non-negative integers (not obvious at first but if you look at the numbers this time it becomes clear why), they must stop at 1, the in-center radius of the 3-4-5 triangle (or 0 of you consider the 0-1-1 non-triangle as a triangle). In order to really prove the in-circle of the parent is smaller than the in-circle of the child, it can be helpful to look backwards. The above statement is the same as saying the ex-circle is bigger than the in-circle. Looking at the numbers (A-B-(A+B)-(A+2B)), the length of the in-circle radius is AB, and the lengths of the three ex-circle radii are A²+AB, 2B²+AB, and A²+4B²+4AB, all of which are bigger than AB since A and B are non-negative integers.
    1
  18337. There's a problem with the solution to the 2-garment problem: to apportion the garment, one would have to rend it and a rent garment is pretty worthless!!!
    1
  18338. 37:45 , a dog, in patreon?
    1
  18339. I feel like I just got issued homework. Damn you you lovely Mathologer!
    1
  18340. 15:30 - onwards. The partition counts crash the 64-bit lid past 416 and are about 85-90 bits at 666; this will require bigger numerals. The method used is: ∙ PartA(K,N) is defined as: ∙ ― 0, if K < 0 or N < 0 or (K > 0 and N ≡ 0) ∙ ― 1, if K ≡ 0 and N ≥ 0 ∙ ― Σ_{0<n≤N} PartA(K-n,n), if K > 0 and N > 0 ∙ Part(K) is defined as PartA(K,K). I think it's 11956824258286445517629485.
    1
  18341. The most memorable for me was Oresme’s proof. An effective way to look indirectly at the ever-shrinking fractions by using more familiar and regular terms, a bit like sign-posts
    1
  18342. Great video; Is there an algorithm for even sided squares?
    1
  18343. I loved the whole thing!
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  18344. Pov: After watching the videos every time Mr. Burkard, the left hemisphere of my brain:💪🏻😎 This video was so wonderful that I can only cry😭 ...<<New Year + Mathologer33... Mathematician, I also proved the golden angle and the golden number in your video, I hope it was good. P.S I hope I can come to the Mathologer's office in the future :)
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  18345. ​ @Mathologer  :))
    1
  18346. The most memorable is the proof of the divergence of the harmonic series. Because it's the only one I remember from my university maths courses.
    1
  18347. Why do physicists play dominoes? Because they lost their marbles?
    1
  18348. :o I got the first one, kind of
    1
  18349. I liked the coloring solution best
    1
  18350. Audio is almost inaudible, have to crank it just to barely hear
    1
  18351. Never saw either twisted square proof
    1
  18352. Could you please put a link to a donations page so that we can donate to you? My apologies if you have and I just missed it. You have excellent material and I'd love to donate to you to help keep future videos coming. If you have a Patreon page, let me know as well. Thank you sir!
    1
  18353. The leaning tower is very satisfactory.
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  18354. As a cube this video pleases me
    1
  18355. Book!!!!
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  18356. Ahhh, is there anything I don't understand. I suggest that a function that approaches infinity measures what I don't understand.
    1
  18357. "Luckily I can see the light at the end of the tunnel, so I'm looking forward to a lot more Mathologer action in the coming months"
    1
  18358. Schools suck. Because governments run them, and they always standardize things, and are slow to adapt, change, or promote things that work better and deprecate things that don't work as well. Thus evolutionary improvement of teaching becomes frozen and is extremely slow to change. This is what you get for wanting big government, standardizing everything to the point of stagnation, all in the name of "i don't want some people to teach religion to their kids"
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  18359. The sums converging to pi doesn't mean they are pi. One could argue they're getting arbitrarily close to pi but never reaching pi. Remember, if you're allowed to use infinite terms to reach a number and claim you're reaching it, not getting arbitrarily close to it, well then you could construct *any number that way. How is that useful then at that point?
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  18360. I have zero background in mathematics, have no idea of how mathematicians think algebraically. I work in a warehouse. I am no mathematician and don't want to be. I do however love systems. I study relationships within said systems. Your videos allow me an insight into how mathematicians think, work, and see relationships with numbers that is unnatural to me.
    1
  18361. I think the fractions is how we ended up with psi and Lagrangian stuff. f(13/4)=(((1/2 + sqrt(5)/2)^(13/4) - (1/2 - sqrt(5)/2)^(13/4))*sqrt(5))/5 =(((1/2 + sqrt(5)/2)^(3 + 1/4) - (1/2 - sqrt(5)/2)^(3 + 1/4))*sqrt(5))/5 =sqrt(5)*((1 + i)*(-2 + sqrt(5))*(-2 + 2*sqrt(5))^(1/4) + sqrt(2)*(2 + 2*sqrt(5))^(1/4)*(sqrt(5) + 2))/10 ≈2.202797555 + 0.06618966271*i We have entered the realm of more answers than we want. The four in the denominator makes the quadratic into an octic. Another thing, if that was a complex number, it would rotate and scale it. This here is what is commonly referred to as a rabbit hole. You might come out re-writing physics, or more than likely come out dumber than you went in.
    1
  18362. This is amazing. When you discussed the differences between the partial sum and pi as a stream of 0s and 9s I secretly hoped that you would plug your 0.999999999999 = 1 video 😁
    1
  18363. 1:07 yes the infinite plane
    1
  18364. I can't even tell what you're replying to rn
    1
  18365. Hi.. I liked chapter 5.. got introduced to Gama
    1
  18366. Leaning Tower of Lire!
    1
  18367. The slide rule was before my time, but isn't this the same concept as the slide rule?
    1
  18368. It's been months since I have watched your work. Great video as always.
    1
  18369. Thank you, Burkard for the wonderful journey. Please allow me to ask, where did you get that great shirt, I did not see it at your store.
    1
  18370. I'm a Spaniard and I watched that movie indeed!
    1
  18371. This proof was great fun!🤗 There is a little "speako" at 28:07. It should say: "...that red is equal to zero, ..."
    1
  18372. What is that formula that he used at 30:25 for the product rule d(fg)=(f+df)(g+dg). Anyone please help
    1
  18373.  @Mathologer  thank you very much.
    1
  18374. very, very nice
    1
  18375. This is a thing of beauty!
    1
  18376. Not sure if it's the same thing, but Douglas Hofstadter does a Feuerbach proof on the board. Nice video.
    1
  18377. Love your channel and videos. Than you so much. This one was a bit tough but extremely interesting. Was a bit puzzled by the (x+1)^0=x ?!
    1
  18378. @Mathologer: Do you ever submit a journal article with "tada, proof complete"? :)
    1
  18379. putting this out there with the knowledge it's probably a really dumb question, but curious enough to risk scorn lol. At 2:33 why do we double the product of 1×2 getting 4, but did not do the same for 1x3, which just stays 3? Thanks guys
    1
  18380. no actual tiles were mutilated for this presentation.
    1
  18381. Fröhliche Weihnachten!
    1
  18382. i find this black dot that appears in the middle of the screen at 10:25 very disturbing
    1
  18383. Did some investigating; thought I was being clever when I noticed that a^2 - b^2 allows you to get to the 4k + 3 primes. But it's pretty trivial to decompose any odd integer into the a^2 - b^2 form: a = ceil(i/2), b = floor(i/2); i = a + b, and a - b = 1, so (a + b)(a - b) = a^2 - b^2 = i. TIL...
    1
  18384. Great atitude, love you.
    1
  18385. Screen pixels worth of pegs, screen colour limit discs. Must dust off my basic interpreter. 😋
    1
  18386. I distinctly remember seeing this on our steam driven television the first time round... by then I’d graduated from watching from behind the sofa (for safety reasons...), it had a big impact on me ... I made a version of the game at school and started a craze for playing it that lasted two whole weeks! Nerd heaven.....
    1
  18387. I recently had lunch with a friend and he started talking about this kinda stuff. it just shows how poor our math education is. these are beautiful patterns, but they arent divine or anything
    1
  18388. The four colour theorem might just be false!
    1
  18389. Loved the new addition/debugging
    1
  18390. made it to the end of the video. I have a BSc in Applied Mathematics, but don't really use math in my job... :-(
    1
  18391. That's some awesome exposition!
    1
  18392. My proof: let x be the length of the sides of the blue square. The big square has area 1. The two big triangles on the right and left of the blue square have area 2(½½1)=½. The two small triangles at the top say have legs of length y and z then considering the diaganoal of the big triangle at the top we get z+x+y = √5/2. Hence z+y = √5/2 - x The two trapezoids (trapeziums) below and above the blue square have area 2x(z+y)/2 = x(√5/2-x) = (√5/2)x - x^2 putting this together we get 1 = ½ + x^2 + (√5/2)x - x^2 so x =1/√5 and finally x^2 = 1/5
    1
  18393. Fact: The typical definition of quadrilateral is 2d shaped polygon with four side. And they don't intersect itself as specified in the definition of polygons. So, using the word quadrilateral here is not good, it's actually confusing, Especially for me.
    1
  18394. Thank you sir for creating this channel.
    1
  18395. Salute!
    1
  18396. I was going to say that the square and triangle thing seems intuitively very possible. It was only when the star came in that I was a little bit surprised. Seeing that the star is fake suddenly makes this boring again.
    1
  18397. Such a pity that Euler and Ramanujan were not alive at the same time so they could have got their minds together. What wonders would that have produced? Nice video
    1
  18398. Never saw any of these. Wow.
    1
  18399. The fractal visualization was brilliant! I was honestly in disbelief that the no 9s sum was finite until that point, but the fractal makes it very intuitive.
    1
  18400. challenge 3: it‘s the fibonacci sequence
    1
  18401. Most memorable: the fact that gamma is a thing =D
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  18402. nice shirt. when i was trying to work it out. instead of squaring the number first i was trying to work out what the values associated with each letter of evil were lol. I assumed it was e multiplied by the roman numerals VIL (44). that got me 119.6. so wouldnt the root of eVIL actually be 10.9364? all joking aside though, thats probably one of my favorite math shirts i've ever seen.
    1
  18403. Very cool video! The area under the bell curve also deserves a video in this awesome channel!
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  18404. Thank you, but I'm just not into it today. Maybe later.
    1
  18405. Copying this from a comment I added to the "last call" message it's probably more appropriate here. Don't know if this one had been mentioned before but... the expression X+Y=XY reminded me of something. Rearranging it you get XY/(X+Y)=1 the expression on the left is a formula for the value of two resistors in parallel (or capacitors in series) in electronics. This results in an equivalent problem (for two numbers) of finding two natural numbers such that 1/X + 1/Y = 1. This generalises into a different problem: find a series of natural numbers such that the sum of their reciprocals is 1. Though I haven't looked into it it doesn't seem to be that difficult to characterise all solutions to this. There is always the trivial example of N values of 1/N and cases where N is composite where some can be grouped together for example (4*1/4 and 1/2+1/4+1/4). Are those the only solutions? If we allow infinite series then other possibilities appear: for example (1/2^n) for n=1,2,... When I came to write I got a nagging familiarity somewhere. Let me show you what I mean. Take the reciprocal case with three integers X, Y, Z. I/X+I/Y+I/Z. Give them a common denominator and see what happens... (YX+XZ+ZY)/XYZ. This generalises to more than three integers and each term in the numerator is the product of all but one of the integers. That's when I saw a parallel with Vieta's formulas for polynomials. I think that the orignial video and the case I mentioned above can be characterised as finding equal values for the coefficients of a certain degree of polynomial with integer roots. For example (X+2)*(X+2) = X^2+4X+4. The reciprocal version for 3*1/3 can be exemplified by (X+3)^3 = X^3+3X^2+3X+9. The symmetry of binomial coefficiencts makes it possible to find many answers where the roots are all equal. What is not at all clear is what happens if the roots are not all equal.
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  18406. I used to explain to students about how you could generate arbitrary continuations to fit a given sequences using polynomials (didn't use discrete example though). My favorite example of taking this to the "unnatural" extreme was the "Bill" function. The function would be generated by me calling up my father, Bill, and asking him what the next number should be.
    1
  18407. If I did it correctly then it's right, right, right, left to get to 153 104 185 square, with clockwise entries 9 4 13 and 17. Set up regression in Desmos to find what the entries would have to be given the known relationships then approached the tree part like a sliding puzzle since the top entries need to come from the same column in the previous steps. The regression is probably not the simplest way to find the box numbers but it seems to give the correct integer answers given the two legs of a triple.
    1
  18408. Professors Polster and Ross, I express my sincere appreciation for what you provide the world of mathematic enthusiasts at large. Your content and delivery never disappoint, thank you. While not pertinent to your specific topic, the subject matter in general is of particular interest. I would be remiss not pointing out one of my favorite aspects of the Pythagorean (et al.) Theorem. By careful geometric construction, three dissimilar non-right triangles can become the lateral faces of tetrahedra with right triangle base. Each of the three areas are correspondingly Pythagorean compliant. Thus, an oblique triangular pyramid or harmonious Pythagorean tetrahedron. Unfolded along the edges of the right triangle base, the “net” is quite counter intuitive to that of the conventional similar shape example you depict. Perhaps this of passing interest to some. Again, thanks for your amazing contributions and my enjoyment.
    1
  18409. I am not a big enough math nerd to understand the second half.
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  18410. Is there a mathologer video on logarithmic calculations?
    1
  18411. Are you sure Dudeney isn't the past self of Christopher Lee? Lee was a professional vampire on at least several occasions. It would be believable that he didn't actually pass away, rather just change identity, you know, every century or so.
    1
  18412. Best cure for winter blues.
    1
  18413. I love that shirt
    1
  18414. Happy holidays
    1
  18415. I actually discovered the Odd/Square progression, fooling around with addition sequences independently. (I'm not bragging, I'm sure lots of people have through the Centuries since the Pythagoreans.) I didn't know about Moessner's Miracle, so thanks for the video!
    1
  18416. I have no idea what you just said. But there was a part in there that made me think of PRNG's it felt like it could be somehow a useful component..
    1
  18417. Are you a native german speaker? The pronunciation was on point!
    1
  18418. It's Diwali today and I m watching Mathologer. Loved the content, hungry for more. Also, I was waiting for you to change your t-shirts.. 😁
    1
  18419. Brilliant and very insightful as usual. Thank you.
    1
  18420. Otherwise very clear, but somehow there wasn't enough visualisation of the logic of the argument for the surface area.
    1
  18421. I had been trying to solve the card puzzle for quite some time. I had figured out the part of communicating the suit but I had not worked out how to communicate the number. Finally, I found it here.
    1
  18422. newspaper-tearing
    1
  18423. Because modern society wants to keep us stoopid
    1
  18424. This is just my Atlas Circular Slide rule. I could read 3 decimals and gestimate the 4th place. Great for engineering school in the early 70s.
    1
  18425. Another happy Saturday morning.
    1
  18426. The most memorable bit for me was the proof that the no 9s sum was less than 80. Very nice!
    1
  18427. Thank you for your very interesting perspective on these subjects. I am a retired Computational Geometry professional and love expanding and rehashing knowledge in all things math. Worked a lot with Bernstein Bezier matrices which are similar to the ones you listed. Again, thank you and please keep them coming.
    1
  18428. Please make a video about Ramanujan's partition formula
    1
  18429. I've heard sinh pronounced "sinch" but now I'm thinking about how you'd say it in Portuguese. See-nyah.
    1
  18430. In the mid-80s as a kid, I set my 8 bit computer off on the 1+1/2+1/3+... series, not really knowing enough about maths to know if went to infinity. I naively thought it would converge on some number. After many hours on what was a slow computer, the sum got to about 15 before I gave up. After all these years, it's nice to see this explanation video.
    1
  18431. If Π is the product equivalent of Σ, what is the exponential equivalent of Σ? I don't know what to call it so I don't know what to look up.
    1
  18432. Did you get to meet or know Conway back in the day?
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  18433. 97 views only!
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  18434. good Ol conditional convergence
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  18435. Very nice clip! I have a little suggestion about how to read Heron's formula: Area^2 = (...). In the clip we rather looked at Area = sqrt(...). In such way, we were kind of losing the reasoning of the square in the exponent of Area. In fact, if we take linear algebra approach: instead of looking at Area of the triangle, we look at the parallelogram spanned by the two sides/vectors of the triangle and let's denote them v1, v2. We can pick any of two, which is fine. Consider a matrix B consisting with the two column vectors v1, v2. Then consider det(B^T.B) = det(B^T)*det(B) = Area^2 <-- that's why we want Area^2 not Area! On the other hand, it's not hard to work out B^T.B by inner product formula, so we obtain the "nature" Heron's formula in terms of Area^2:)
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  18436. your channel is amazing and i dont even like math
    1
  18437. The powers of five give the same digitroot loop as the powers of two but in reverse because within the modular ring of nine, five is the multiplicative inverse of two. In other words, for any integer n the digitroot of half of it, n/2, is equal to the digitroot of five times n.
    1
  18438. 6:27 No. because it becomes possible to leave a single isolated square.
    1
  18439. 6:59 In Quebec, it is part of maths in high school for many students (if not all), but I wouldn't be surprised if it is skipped often. Pretty handy to know in order to "cheat" questions where trigonometry may be expected. lol
    1
  18440. What's next in 1 3 8 21 .... from the partition products given? 55 and 135 are next.
    1
  18441. 14:00 I can see it's Fibonacci without calculating the numbers
    1
  18442. Brain fan at maximum... Not terribly lost though. I'll have to watch it again. Keep ui with the good work.
    1
  18443. The second proof because it is like the films ‘Flubber’ and ‘Son of Flubber’..
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  18444.  @Mathologer  i need math to maintain my sanity. I was getting insane! Sir I'm a big fan of yours. Thanks for all the vids you make.😃
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  18445. i count 60
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  18446. Most memorable: The visual proof that gamma is less than 1 and greater than 0.5, because each blue bit has width 1 and the heights of the blue bits all add up to 1. Each blue bit clearly has less area than a rectangle of its same width and height, and slightly more area than a triangle of its same width and height. So it is trivial to see the total area of the blue bits must be less than the total area of rectangles of the same widths and heights, i.e., less than (base * [sum of heights]) = 1 * 1 = 1; and the total area of the blue bits must be greater than the total area of triangles of the same widths and heights, i.e., (1/2 * base * [sum of heights]) = 0.5 * 1 * 1 = 0.5. Therefore, gamma must lie between 0.5 and 1.
    1
  18447. In mathematics, where you haven’t a clear definition of neither dimension nor exist, any fantasy is possible. However, not so in physics.
    1
  18448. Simple sexy math. I love it
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  18449. I love your shirts
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  18450. GREAT explanation!!!! I enjoy every video of Mathologer and can't even wait for the next one. My click for understanding calculus was the point when I thought about the d in df and dx not as kind of number or something else but as little machine (function) which evaluate the behaviour of x and f in a very, very, very close range around x and f(x). Rest is more or less algebra. Surley I know that this is from mathematics point of view not very saddle but I’m not a mathematitian but an engineer. And for engineer it has to work and not really more. ;-) - TWO THUMBS UP!!!
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  18451. Your videos work better than any lullaby ever. 5 mins in im already asleep thanks
    1
  18452. I got lost at the Bernoulli number section. I'm sheepishly admitting that I have a degree in mechanical engineering, which had about 5 years of calculus and above. And which was very interesting and very frustrating as well, since the way it is taught totally does not suit me. I love your animations.
    1
  18453. 3:00 It's the harmonic series minus the harmonic series so it is infinity - infinity which as we all know does not necessarily equal zero.
    1
  18454. I can't decide if Kempner's proof or all the partial sums after one missing the integers is more memorable. Both results were new to me and surprised me
    1
  18455. ultra-cool af!
    1
  18456. 32:00 Flashbacks from the numberphile vs. math video
    1
  18457. At least I am not the only one who guessed wrong initially, really counterintuitive. Great animation to clear it up though
    1
  18458. 19:28 Extrapolation - always a bad idea. Except for SOR (successive overrelaxation) for the Gauß Seidel method or other Numerial LES solver
    1
  18459. I learned the first twisted square sometime around 8th grade (in the US, so age 13 ish) I remember at the time seeing the picture but it not being explained really though, so I had to hunt that down myself. So I agree, at least my school did a terrible job of proving the Pythagorean Theorem. I like that the first one doesn't involve any algebra though, just geometry!
    1
  18460. Enoch Powell or Lewis Powell? Or is it perhaps Baden Powell? No ... Baden Powell is the mystery of the One Note Samba ... https://www.youtube.com/watch?v=ZYaVnunG9dw
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  18461. In Russia we were taught Heron's formula in 7th grade, but it wasn't named, just yet another formula for triangle's area. There was no proof given though.
    1
  18462. thank you, excellent as always.
    1
  18463. this is some wild stuff
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  18464. I vote for the brick overhang chapter
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  18465. I recently discovered the magic number 709,78271289338402... Why do many calculators fail at this point?
    1
  18466. Me: #thinks Fourier Also Me: #checksComments Me: #thinksToSelf Well I did actually understand something rather than just nodding along.
    1
  18467. Dr. Strangelove used a circular slide rule.
    1
  18468. I quit studying Maths at age 17, in order to switch physics classes from a terrible teacher, to a great one. A part of my brain screamed in protest and the joy of maths has been a nagging craving since. I'm too old to seriously expect to be able to learn much new maths now, but that itch continues to beg to be scratched.
    1
  18469. Those colours you use really work for me!
    1
  18470. I am a top 1% math guy based on college testing and I've never seen these two proofs.
    1
  18471. I had an epiphany today thanks to your shirt: a^2 + b^2 = E/m.
    1
  18472. 38:41 the numbers on the right differ from the numbers in the middle by the square of the n-th Fibonacci number
    1
  18473. This post is a bit long and detailed but, hey, you just went through a 50-minute Mathologer video, so long and detailed don't deter you. :) When I was preparing for math olympiad I found another way to compute the formulas for the sums of powers. If we have a sequence A[n], define the sequence of its differences as dA[n]:=A[n+1]-A[n] Similarly, the second differences are d^2A[n]:=dA[n+1]-dA[n], and generally the k-th differences are d^kA[n]:=d^{k-1}A[n+1]-d^{k-1}A[n]. We'll use the following easy-to-remember identity: A[n] = (1+d)^nA[0] If this formula doesn't sense to you, just think of it as short-hand notation for saying that, if you formally expand the (1+d)^n part using the binomial formula, the resulting identity is true. For the first few values of n, this is what you get A[0] = A[0] (duh) A[1] = (1+d)A[0] = A[0] + dA[0] (which is true because A[0] + (A[1] - A[0]) = A[1]) A[2] = (1+d)^2A[0] = A[0] + 2dA[0] + d^2A[0] A[3] = (1+d)^3A[0] = A[0] + 3dA[0] + 3d^2A[0] + d^3A[0] Now we define the sequence we are interested in. For instance, to compute the sum of the first 3 cubes, A[0]:=0, A[n]:=A[n-1]+n^3. Now let's look at the first few values of d^kA[n]: d=0 -> 0 1 9 36 100 225 ... d=1 -> 1 8 27 64 125 ... d=2 -> 7 19 37 61 ... d=3 -> 12 18 24 ... d=4 -> 6 6 6 ... d>4 -> 0 0 ... We know dA[n] is a polynomial of degree 3, and we can prove that d^kA[n] is a polynomial of degree 3+1-k. So the row for d=4 is a constant; and for greater values of d, it's all zeros. The identity A[n] = (1+d)^nA[0] now tells us that we can reconstruct all of A[n] by just looking at the values d^kA[0], but we only have finitely many of those: 0, 1, 7, 12, and 6. So putting all together we get A[n] = (1+d)^nA[0] = C(n,0)*A[0] + C(n,1)*dA[0] + C(n,2)*d^2A[0] + C(n,3)*d^3A[0] + C(n,4)*d^4A[0] = 1*0 + n*1 + (n*(n-1)/2)*7 + (n*(n-1)*(n-2)/6)*12 + (n*(n-1)*(n-2)*(n-3)/24)*6 That's a closed formula for S_3[n], which we can simplify to get the same formula we saw in the video: https://www.wolframalpha.com/input/?i=simplify+1*0+%2B+n*1+%2B+%28n*%28n-1%29%2F2%29*7+%2B+%28n*%28n-1%29*%28n-2%29%2F6%29*12+%2B+%28n*%28n-1%29*%28n-2%29*%28n-3%29%2F24%29*6
    1
  18474. I was expecting the video to be about solving Fibonacci-like sequences using eigenvalues or iterated powers. The ultimate solution at the end was much better, but I'll describe the linear algebra attack regardless. Change making satisfies the recurrence relation T(n) = T(n-1) + T(n-5) + T(n-10) + T(n-25) + T(n-100). If you imagine a state vector with 100 entries representing the results from k-1 to k-100, updating that state vector to instead have the results from k to k-99 corresponds to multiplying by a sparse matrix M = |0><1| + |0><5| + |0><10| + |0><25| + |0><100| + sum_{k=1}^{99} |k><k-1|. So one way to describe the target number is "initialize to the state vector 1,0,0,0,0,...,0 then multiply it by M a total of k-1 times then read off the first vector entry", i.e. the value <0| M^{k-1} |0>. So all you need to do is compute high powers of M. There are two common approaches. You can diagonalize M and raise the diagonal entries (its eigenvalues) to powers individually. This is mathematically elegant but computationally requires dealing with real numbers. Alternatively, you can using iterated squaring to quickly get very high powers M^(2^i) that can be combined together to form a target k. Ultimately, you can compute M^{k-1} in time proportional to the number of digits in its output. This attack will not get you to k=10^100, since the number of digits in the output would be more than 10^100 and no one has that kind of time. But it'll get you to numbers with billions of digits, whereas the polynomial equation powering algorithm will become intractable far far earlier.
    1
  18475. Mathologer, keep up the fantastic work! Since you asked, my formal background was severely truncated, but I have worked out simpler infinite summations for myself, because they interest me. I grok infinity both as journey and as destination. In moments of clarity, I can see both at once, in superposition. Today was not given to such clarity, but I watched to the end anyway, because I know my subconscious mind will go on working on this as long as needed.
    1
  18476. I see what you did there: You made the math do work for you :)
    1
  18477. If you want to be really smartsypants, you could argue that this proof is actually "rediscovering" a variation on a 1D version of Conway's Game of Life - which is a really useful approximation model in numerical simulations. Essentially what we just built is an irreversible totalistic rule (assuming we were to automate the process using some sort of updating rule): The state of a cell is dependent on the values of the neighboring cells (the ones on the left in particular) in the previous time step, and a previous configuration of cells can not be achieved again. Edit: Actually I just checked my old notes, and this would essentially be a "0-configuration" or a rule that always leads to all cells having the 0 value, regardless of your starting configuration. As someone who is doing stuff like this as a focal point of my studies right now, I find it fascinating that due to the framing I probably would not have been able to come up with this proof by myself, in spite of knowing all the building blocks you need for this proof. It probably also doesn't help that I really dislike mathematical terminology ("sequence terminates") and am studying this stuff in a different language, but still interesting that I would probably not have defaulted to a binary system. Then again, an engineer doesn't have to do many proofs in uni anyways, so that also likely adds onto the issue here.
    1
  18478. Growing hair is easy too 🤣
    1
  18479. Convergent math, from chaos comes symmetry. A very interesting vid.
    1
  18480. 🤔 A best kept secret and why they don't want proficient mathematicians and intelligent beings to (decipher ) that We live in a mathematically created universe by the numbers ( 9 ) and earth Is mathematically designed with energy / magnetic and other things I won't discuss that form earth and the universe combined it didn't just happen to creation by chance theory. 🕛🕛🕛🕛🕛🕛🕛🕛🕛🕛🕛🕛🧲🧲🧲🧲🧲👽👽👽👽👽👽👽🧭
    1
  18481. everything worked pretty nicely. Masters degree applied math
    1
  18482. These are remarkable results. Thanks for putting these together to share with us!
    1
  18483. Amazing video as always! You've got to love the geometric solutions, they once provided, when the the algebraic counterparts were invisible to these noble minds:) [Music at the end?]
    1
  18484. Now that everyone’s head is hurting explain the trig scales!
    1
  18485. Hello from Grecce ! :) you are great!! Δελτα
    1
  18486. So is there a fun thing about the sequences in between? Anything we can do with {6,7,8}, {15,16,17,18,19,20}, {28,29,30,31,32,33,34,35}, etc?
    1
  18487. e^ π Eat the pie.
    1
  18488. One answer to one puzzle, before looking at other comments: 3 ^ 4 + 4 ^ 4 + 5 ^ 4 + 6 ^ 4 = 2,258 = 7 ^ 4 - (11 x 13) Is there a better answer? Now I will look.
    1
  18489. ​ @Mathologer - Loved this vid! I am still not seeing many people tackle the quartic. What struck me about the last puzzle is x^4 = x^2 * x^2, so with 7^4 being off by 143, or 144-1, it was an instant realization that what we have in the quartic is slightly different squares multiplied together, with the trivial 11 times 13 = 143 being the difference.
    1
  18490. What is happening!1!
    1
  18491. Would definitely like to see a real life version of those devil lacings on some punk boots, just saying. But, seriously, this is some really interesting math.
    1
  18492. I came across the first proof in a documentary. I think it was by Du Santoy.
    1
  18493. Sir ,Can i obtain formula of one real root of cubic equation ax³+bx²+cx+d =0 because every cubic equation always exists one real root . And other two root can be real or compex Due to which it becomes very important for us one real root formula find out Due to which it becomes very important for us one real root formula find out
    1
  18494. sir also i want formula of nth prime number ex 2 ,3,5,7,11,13......nth nth=? vary long time to try ....but not...please help me
    1
  18495. i think deriving formula for cubics has been done in another video
    1
  18496. he already has a video on it, search about cardano or cubic formula
    1
  18497. Aren't these Helices rather than Spirals?
    1
  18498. Analog computer strikes back !
    1
  18499. Thank you so much. I love your style.
    1
  18500. This complex plane Fourier transform has important applications in Ceremonial Magic in once & for all time settling the argument for the setting up of attractive & banishing pentagram set up, as well as evocation of elemental angel/demon intelligences to effect reality in the physical universe limited by the Planck Constants & derived Planck Units. Thank You Mathologer, regards Arctuerion chaos Magician.
    1
  18501. I’ve noticed that 0 is equal to 1 in the sense that 0 needs to be equal to 1 for the sequence to perfectly match up when shifting a duplicate of the sequence over in comparison
    1
  18502. 12:00 Mind. Blown. edit: I have a degree in physics with a minor in mathematics and a masters in engineering. While I touched on some of this in calculus, my focus had been on the practical side of maths so I had never seen the sums presented in this beautiful way.
    1
  18503. I got the determinant of the glasses to be -666
    1
  18504. Len's t-shirt?
    1
  18505. How about starting with 3³+4³+5³ = 6³ instead?
    1
  18506. Damn, so all one can do is to multiply each term by a constant amount...
    1
  18507. Thank you!
    1
  18508. Very nice indeed. I prefer the second proof, nicely showing the scaling factor as well.
    1
  18509. I am the 999th thumbs up 😊
    1
  18510. Madhava = Silvio Berlusconi
    1
  18511. 369 has 2 lines it's not a triangle. People just make up crap. Is a dream world.
    1
  18512. Here's 73 https://youtu.be/zIkdWrzI92M
    1
  18513. Wow, I just noticed that if you truncate the tendon scatter diagram with a straight line, you’ve got a real simulation of an arachnid type alien insect which sure would scare the hell out of me. Something like in a sci-fi movie.
    1
  18514. No integers was the best IMO.
    1
  18515. Hammurabi is smiling. Until the House falls.
    1
  18516. 3blue1brown is an Patreon here, well this raises my love for the two chennels
    1
  18517. What is odd is that Euler's formula for 3D shapes works for the star faced polyhedra where the faces are pentagram stars (or higher) that self intersect, like the Great stellated dodecahedron and it works for the star vertex polyhedra where the faces intersect like the great dodecahedron. Except that it does not work for the small stellated dodecahedron; Schläfli simply declared that the SSD, thus did not exist. At 13:15 you forgot to divide the outside face No 6 into triangles, not that it matters.
    1
  18518. Spotted 3B1B in the patreons, so i won the book 😻😻
    1
  18519. The whole thing worked for me (in that I found it all interesting) and my background in math is mostly just that I think it’s really cool.
    1
  18520. Marvelous. Truly interesting. Thanks.
    1
  18521. 6:11 He wasn't kidding. It was easy. No. G for green, B for black, x for removed. GxGBGBGB xxBGBGBG GBGBGBGB BGBGBGBG GBGBGBGB BGBGBGBG GBGBGBGB BGBGBGBx no way you're gonna get that top left green.
    1
  18522. I already saw the old video, but now, I realized, pythagorian areas could be also any square/triangle, it has to be the same scaling in both diamensions. So also non-isosceles triangles, too: a^2 + b^2 = c^2 | /2 (a^2 + b^2)/2 = c^2/2 a^2/2 + b^2/2 = c^2/2 any triangle with: h_a=c_a=a, h_b=c_b=b, h_c=c_c=c h_a*a/2 + h_b*b/2 = h_c*c/2 | *d # scale height triangle, eg. to isosceles triangles (d*a)*a/2 + (d*b)*b/2 = (d*c)*c/2 => Any triangle with same relation d(>0) in height is possible. Btw. any square, it has to be the same quotiant: q = a_1/a_2 = b_1/b_2 = c_1/b_2 If q=1, than a_1=a_2=a So, more general, it is a 2D-scale-factor s(a) + s(b) = s(c), while s is the area-function of an shape, 2D-equal-scaled by a,b,c. Eg a square: s(x) = x*w+x*h, a triangle: s(x) = (x*w+x*h)/2, a circle = s(x) = π*(x*r)^2, while w,h,r are constants.
    1
  18523. for an n*n grid it's the sum of i^2*(n-i) for i from 1 to n
    1
  18524. So, I only got most of the way through the little challenge around 16 minutes. if we consider A = 2^n, B = 5^n, then A*B = 10^n. We can then take the digital root of this whole thing, and since 10^n will only ever have a digital root of 1, we get the following: DR(A*B) = 1, by the product rule for digital roots: DR(A) * DR(B) = 1, and therefore DR(A) and DR(B) must be multiplicative inverses. I'm not sure how that applies to them as sequences though. Does that just work as is? It seems like it should, but the math seems a bit too simple. I guess a lot of the weird modulus math is hidden within the constructing of the digital roots, which is kind of glossed over by this proof. Proof(?)
    1
  18525. Greetings from the Netherlands
    1
  18526. Woah! Mathemagic!!
    1
  18527. 11:57 those with OCD will be bothered by the use of two different “1”s
    1
  18528. That was fantastic! Thank you! 🤍
    1
  18529. The sum-of-digits test for divisibility by 9 is something I've known for virtually forever but I don't remember if I was ever taught that in school. I may have picked it up from a book of mathematical tricks that I read as a kid, along with the trick of easy multiplication by 11. On a related note, if you're reconciling your checkbook or, in accounting, crossfooting your debits and credits, if your sums disagree and the difference is divisible by nine you transposed two digits. In essence, it all boils down to casting out nines.
    1
  18530. So, tetragons would have 180° ears?
    1
  18531. tree trunk, second line?
    1
  18532. Terrance Howard needs to watch these brilliant videos
    1
  18533. 42:52 Markus Persson
    1
  18534. I'm not following the Archimedes circle problem. By removing the 4 squares you have an error ratio. By removing more squares you have less error but more of them. By having an infinity of infinitessimaly small errors you can still end up with any number. It's like multiplying 0 by infinity or deviding 0 by 0 or infinity by infinity. Infinite cumulation of approximation errors is a bad tactic so I feel. An error has value even when it is almost zero! Its like when you have a million in currency and the smallest unit is a dollar then you could have one million seporate dollars, add 0.4 dollar to each which can not be represented by those single dollars. Then you multiply by one million and get 1,4 million ! Wouldn't this be proof back in the days?
    1
  18535. 28:19 It is possible that when you construct things in geometry according to general instructions, in some specific cases you might end up being instructed to make something impossible. Where the pieces don’t fit together. This isn’t an issue here though since these tetrahedra are convex and have triangular faces (which are also necessarily convex) so all the necessary edges are always available and you aren’t required to overlap any of the volumes.
    1
  18536. Going out on a limb here ... if the the basic ABCD formula produces the area of a quadrilateral with points on a circle and if setting D to 0 creates the formula for the area of any triangle …. Does that mean that any triangle can be considered to exist as three points on a circle? Is there any particularly nice ratio in this case for the larger circle and the triangle’s smaller, bound circle?
    1
  18537. Wasn't taught this at school. Teachers themselves don't know anything. They just give you a formula and that's it. No history, no logic of where this might be used. It's all about passing a test. So many gaps in maths knowledge, why we use base 10, what is base 10, what is log, where did it come from. Nothing is taught. Then they wonder why people don't get Math.
    1
  18538. Upvoted for the T shirt alone.
    1
  18539. Rats - it needs at least a second upvote for the content.
    1
  18540. Sir we can generalize this sum under Riemann zeta function ???? . Waiting eagerly for ur answer . Love from india I'm nimisha 🙏🙏🙏🙏🙏🙏🙏😀
    1
  18541. "I could spend a couple of videos on representation on sum of squares" ME: 🙏🤞
    1
  18542. 13:26 ok, so we kind of convolve the two rotation operations and the winding numbers add up. Is that true in general? How abstract can I go with the concept of rotation (what constraints do I need to put to an operation) without losing this linearity of the winding number? Also, this is the first time I feel like I have a vague understanding of Schläfli Symbols.... Not after having to suffer through crystal physics for a year, no, after a youtube video about a funny GIF :P It's pretty cool how much useful maths stuff you can sneak into these videos without really even going into it!
    1
  18543. Google picked up I searched "large numbers list" today 😶
    1
  18544. Organic. 💓
    1
  18545. Fantastic animation! Really helps to understand.
    1
  18546. The leaning tower of lire 🙂 I wish many more teachers and professors around the could show the same engagement you do!
    1
  18547. I was today years old when I first saw these twisted square maths
    1
  18548. Welcome to Turkey
    1
  18549. The Kempler proof was the most memorable, I found it really surprising just how short, simple and elegant it is. One question that came to mind: Is there a "thinnest" series that still produces divergence? For example, if we sum the reciprocals of only every second prime, does that still diverge? Every third? What if we skip every 3rd?
    1
  18550. For that last teaser puzzle.. Are looks deceiving??? I cant even tell where it looks like its going!!! I got the next number to be 52 and I'm just completely baffled.
    1
  18551. So I get 1, 3, 8, 55. Which is 1×1, 1×3, 2×4, 3×7, 5×11.. which should make the next number 8×18 or 144!
    1
  18552. I personally liked the maximum overhang towers the most. Efficiency isn’t always pretty. The parabolic approximation was pretty cool too
    1
  18553. Come on. You pulled 6M out of your hat. The freaks with more than 6M hairs will be homeless. You also forgot about your own breed. House numbering should start from zero. (I presume we're talking about head and facial hair. Otherwise gaining entry could prove to be embarrassing.)
    1
  18554.  @Mathologer  Point taken.
    1
  18555. honestly, you're one of the most patient persons I've seen. And.... I made it to the end
    1
  18556. You can multiply two polynomials using the fast fourier transform, so in O(N log N) wrt. to the size of the exponent. 1,5,10,25,50,100 coins are simply 6 polynomials which is computationally feasible.
    1
  18557. Very clever. I'm not so sure a real elastic band would cope with that much stress! Also how [new] mathologer videos were missed the last few weeks !! You were busy making this one! Maybe it was a mistake to get rid of my Dad's slide rule. I never learned how to use it. This is starting to make sense now. A slide rule is part of the story on one of the anime films I like to watch... "The Wind Rises"...
    1
  18558. Wow im stunned let's do this
    1
  18559. While the no 9 series was cool, my favorite result was the no integer as a partial sum.
    1
  18560. Is the music based on the series? I made it to the very end.
    1
  18561. Chapter 4 was the most interesting to me
    1
  18562. Nikola Tesla : I like the number 3 everyone: omg! we using base n^2+1!
    1
  18563. The animations of the equations are so perfect to illustrate things
    1
  18564. Answer to the card question : queen of hearts
    1
  18565. Most memorable moment? For me, it had to be noting the stack of infinitely many "blue bits" would fit inside the unit square before you mentioned it! As for the proof of gamma > 1/2, each slice of that stack would equal one half if the curved part of the shape were in fact a straight diagonal; since the curve is convex with respect to the "blue bit," the shape's area is greater than half of the rectangle it occupies. It's really gratifying to be on top of a concept in a Mathologer video, considering that in a typical video (including this one!) there are usually parts that I have to go back and rewatch to make sure I understood correctly! 🤣
    1
  18566. Tiny question: is it not the numbers of the form 3k+1 and the numbers of the form 2k+1 for our mod 2?
    1
  18567. We were all right as kids, you can do math with shapes!!!!!!!!!
    1
  18568. So, I was looking at a circle the whole time. Nice. Greetings from New Mexico!
    1
  18569. Thank you for this I am an Indian but I didn't know about this
    1
  18570. I personally detest mathematics to the point of given a choice, I would rather have a car door slammed shut on a male body part vs. going to Algebra class in high school! However, even though I had not a clue about most of the equations in this video, I watched it in it's entirety! If this subject were taught by an individual that used history (My favorite subject) like the one in this presentation, both my attendance and attention level would have been 100%
    1
  18571. Dude this stuff is amazing.
    1
  18572. Seeing how the half orb turned into a cylinder minus a cone of same height and diameter, is it reasonable to expect a cone, when treated in the same manner, to turn into a cylinder minus a half orb of equal diameter?
    1
  18573. As usually great content. But stop trolling us poor Christians with those 6 repetitions! Wish you the best and keep doing these great videos
    1
  18574. The final section, the proof of bhrama gupta formula is so beautiful a soothing. Thank you.
    1
  18575. i dont get it... what's a 3 cent coin?
    1
  18576. if the 2 in 1+2=3 came from 2x1 and the 4 in 3²+4²=5² came from 4x1, maybe for the cubic series it would come from 8x1, so it would be 7³+8³=9³ but it's sum is quite off
    1
  18577. I really enjoy your content. However, I would suggest you change the background color to some eye friendly color, like dark theme. IMHO, it’s too bright.
    1
  18578. @13:17 also interesting 1 in 5 dots are green for the row corresponding to 5, the pattern is the same for all the rows, which really makes everything hit home. nice video!!
    1
  18579. typo in image formatting @3:15 {in series at the 51st decimal place) 10582 097494459 10581 556478834 however as 1 ≠ 2 they ought not be on the white background but should be included in following shaded area, like so 1058 [2097494459] 1058 [1556478834]
    1
  18580. Excellent, thanks. You put so much work and I will be ver so grateful for your explanations
    1
  18581. This is astounding, the tying together of something so prosaic as (x+2)^3 to a deep understanding of multidimensional cubes. Plus kittens.
    1
  18582. First time I get cosh and sin(c)h from geometry. Excellent.
    1
  18583. Today used Heron's formula in my maths exam :)
    1
  18584. I have learned Oresmes proof +30 years ago and forgot about it. It is really very beautiful. Amazing video Danke
    1
  18585. Every time I see infinite series and these sorts of tricks reality gets broken. Can we just accept infinity doesn't play nicely with out finite brains?
    1
  18586. Its easy if you have tooken atleast algebra 1
    1
  18587. This nifty video streched my brain logarithmically and in the process imparted a deeper appreciation of the inherent relationship between addition and subtraction on the one hand and multiplication and division on the other!
    1
  18588. The super-ugly optimal tower is certainly my favourite!
    1
  18589. Intro is from my favorite song ever 🥵
    1
  18590. 27:45 that's 9 degrees of freedom 😮
    1
  18591. What worked: the visual proof that the area from one to infinity of 1/x must be infinite is elegant, as is the proof that it is bounded below by the harmonic series, therefore also infinite. The bit about approximating e with successive areas inside rectangles was a "sinh" for me (yes, "cinch" is how we learned to pronounce it in the US. Not that it's important when you're sitting alone doing your homework). Whereas for the geometric interpretations of sinh and cosh the light bulb didn't quite "sinh". Maybe another viewing or two. But I want to share something I worked out for myself long ago with the "calculus daemons". Suppose we define a function, call it L(x), as the integral from 1 to x of (1/t)dt. Yes I know this is exactly the same as the Mathologer's A(x) but hear me out for a minute. As a shorthand, to avoid writing it out in words each time (or writing out Latex code, equally tedious and not helpful to everybody) I abbreviate integral from a to b of (1/t)dt as I[a,b]. So L(x)=[1.x]. For now, L(x) is only defined for x>0. Obviously, L(1)=0, L(x) <0 for x<1 and L. But what exactly is the relationship between L(x) for x<1 and x>1? To find out, what is L(1/a)? Make the substitution in the integral t=1/u, dt= (-1/u^2)du. With this substitution, the limits become 1 and a, and the integrand is the same as before with the sign reversed, so I[1,1/a] = -I[1,a], or L(1/a) =-L(a). This works whether you assume 0<a<1 or 1<a. What about L(a) + L(b)? For the moment, assume 1<a<b. We can't combine I[1,a] and I[1,b] directly as is, with the limits overlapping, so make the substitution t=u/a, dt=du/a. Now the integrand is unchanged, but the limits are a and ab, So I[1,a] + I[1,b]=I[1,a] + I[a,ab] =I[1,ab] so L(a) + L(b) =L(ab). If 1<b<a just do the same thing with the constants reversed and for 0<a<b<1 or 0<b<a<1 we already have our reciprocal rule. q.e.d and q.l.a.d. (for "quacks like a duck"). Of course it lacks the visual appeal of seeing it all worked out with areas, having access to graphics, "autopilot" and other tools. Is the invariance of the form of the integrand under the substitutions t=1/u and t=u/a intimately related to the "anti shape shifter" property pointed out by the Mathologer? I absolutely think so!
    1
  18592. ??
    1
  18593.  @Mathologer  factorial of 35
    1
  18594. Damn!
    1
  18595. Proof that the earth is flat at around 20 minute mark! Wow!
    1
  18596.  @Mathologer  Next video we calculate the perfect radius and surface area of our tin foil hats! Btw, I recently solved the 4d rubiks cube thanks to your video! Thanks!
    1
  18597. I got 11956824258286445517629485 My Java (11) code to print all of them up to it: import java.math.BigInteger; import java.util.ArrayList; public class Main { private static ArrayList<Integer> posSums; private static ArrayList<BigInteger> pentas; public static void main(String[] args) { posSums = new ArrayList<Integer>(); pentas = new ArrayList<BigInteger>(); posSums.add(1); pentas.add(new BigInteger("1")); for (int i = 0; i <= 667; i++) { int i0 = posSums.size() + 1; if (posSums.get(0) + (i0 % 2 == 0 ? i0 / 2 : i0) <= i) posSums.add(0, posSums.get(0) + (i0 % 2 == 0 ? i0 / 2 : i0)); BigInteger nextPenta = new BigInteger("0"); for (int j = 0; j < posSums.size(); j++) { BigInteger pentaSum = pentas.get(pentas.size() - (posSums.get(j))); nextPenta = ((posSums.size() - j - 1) % 4 > 1) ? nextPenta.subtract(pentaSum) : nextPenta.add(pentaSum); } pentas.add(nextPenta); System.out.println(i + " " + nextPenta); } }
    1
  18598. God I just love how often calculus takes a simple and intuitive problem and then twists it to a batshit conclusion. It never gets old. "You see if we take this bastardized version of math based on ignorance you can come to a contradictory result. Amazing!"
    1
  18599. It is curious how quickly electronic calculators displaced the slide ruler in engineering education. Freshman year (1972) calculators were an oddity (just three on campus). Sophomore year folks would line up to use professor’s calculator during examinations. Junior year professors would announce that you will not be able to complete test without calculator so borrow one. Senior year it was just assumed that all students would have a calculator.
    1
  18600. No nine november
    1
  18601. Whether the smallest piece should move clockwise or anticlockwise depends on whether the number of pieces you're dealing with is even or odd. If there's only one piece, you just move it clockwise directly into the target position. If there's two pieces, you need to move the small piece anticlockwise so that the large piece can move directly into the target. If there's three pieces, you need the middle one to go anticlockwise, so you move the small one clockwise first to leave room for it. Given an even number of pieces and a destination in the lower-right, every second step is going to be the small piece moving anticlockwise.
    1
  18602. Believe it or not I guessed this rule :)
    1
  18603. Watching again for the algo
    1
  18604. In 70's i had a wristwatch with this calculator.
    1
  18605. It would simpler if they had a cone .
    1
  18606. Mathologer needs to begin a curriculum to overthrow the treacherous banality we all suffered.
    1
  18607. I wonder what this would look like in 3 dimensions with platonic solids...
    1
  18608. Beautiful :-)
    1
  18609. Would you please share the animation files (Keynote?) from making this so we can look through how you did it?
    1
  18610. In America, we pronounce sinh as "sinch."
    1
  18611. Mathologer,tell us about how Indian Vedas eyplained relations...in triangle.please
    1
  18612. Ok got it! Let's speak about frequences and octaves!
    1
  18613. "Bugs chasing bugs"? As a software engineer that sounds like my daily work.
    1
  18614. nice video
    1
  18615. Congretulations! One of your best videos!
    1
  18616. I don’t know how, but I feel like somehow this set up could be used to figure out where all the primes are. It seems like if you rotate the inner ring to each of the integer positions that that should somehow reveal a pattern which would be logarithmic in nature that would show this. I’d have to build something myself, but maybe someone else already has? But basically,The primes would be the things that aren’t there as you keep adding in each integer number I think.
    1
  18617. I've recently solved https://projecteuler.net/problem=78 and ended up finding Euler's formula on Wikipedia, and using it without really understanding it. Thank you for filling out the gaps!
    1
  18618. So SO beautiful and fun!! You are a wonderful teacher! Thank you!
    1
  18619. 3:12 witchcraft!
    1
  18620. Whoaaa...
    1
  18621. I really liked the Balanced Warm-Up - it was awesome and I think that I will remember your example the most in the future.
    1
  18622. 13:50 FINALLY I remembered from decades ago Martin Gardner's column on equal-width shapes, including this as a method of generating them.
    1
  18623. "F.J.M. Barning" stands for "Fredericus Johannes Maria Barning". He probably would have been called Frederik or Fred.
    1
  18624. Jesse, jesse do some math
    1
  18625. So there’s a group of video gamers known as Sub Simmers (of which I’m one). We like to play submarine simulation video games for those not familiar with the term. One of my favorite subsims is Silent Hunter 5. I bring this up because SH5 is WW2 era, and one of the tools you use in game is the “Attack Disc” which is basically a circular slide rule. There is a separate app I use on my iPad called “Sub Buddy” so if anybody wants to check that out go ahead. I’m not sure what, if any, difference there is between a real slide rule and an attack disc.
    1
  18626. There's probably a good reason for this (and I'm just too much of a newb to notice), but at 6:20, why is 4k + 3 not written as 4k - 1 instead?
    1
  18627. I am always annoyed by the fact that true answer in mathematic need infinite precision I rather say that they are no irrational numbers if I take number amounts of digits after the zero therfore 0,00001 =0 and that's it I will call it practical mathematics and pi is 3,141592 = 3, 14159265 the definition of equal is different the classic equal it mean good enough
    1
  18628. Wonderful!
    1
  18629. Great!!!!!!!!!!!! Merry Christmas!!!! See you again in 43x47 (beginners joke... yours are way better :-))
    1
  18630. The only thing I would say about the equilateral triangles is that if you can’t find infinitely small equilateral triangles anywhere, then proportionally you should never be able to find a big equilateral triangle. Since the distance between the points is always the same, then there is no way you can shove an equilateral triangle in there.
    1
  18631. I prefer the coloring prof. It requires "less" thinking
    1
  18632. The rule is nth term=(n-1)th term x 2(both numerator and denominator) and then add the numerator and denominator of the previous fraction.ex-3rd term=6+1/4+1
    1
  18633. Another way to think about this sum is that you need to group the expanded form into groups of 3. -- The rationale for this is that the (1/2 - 1) etc. are inseparable. By shifting the terms across multiple groups you are not accounting for the adjusted denominator. I.e. 1/1 + (1/2 - 1/1) = 1/2 -- the 1/1 terms are alike and cancellable 1/3 + (1/4 - 1/2) = 4/12 + (3/12 - 6/12) = 1/12 1/5 + (1/6 - 1/3) = 6/30 + (5/30 - 10/30) = 1/30 etc. It looks like this generalizes to 1/x + (1/(x+1) - 1/((x+1)/2)) = 1/(x*(x+1)), but I'm not currently sure how to prove that. This gives the sequence 1/2 + 1/12 + 1/30 + ... which is cleary between 0.5 and 1 (the first term is 0.5 and the other terms get exponentially smaller, so the other terms cannot sum to 0.5).
    1
  18634. Heron’s formula would be considered an unnecessary speed bump in California schools.🤑
    1
  18635. holy I just watch 50min and the sky become dark
    1
  18636. Master stroke once again. A video after very long gap. But worth waiting. You made my day as well.
    1
  18637. Hey Mathologer, I just read a book that had a geometric proof (maybe?) of Fermat's last theorem. It seems good to me but I'm not really qualified to judge it's merits. I am curious though, maybe you'd be interested in taking a look? It's called "Geometric Approach to Fermat’s Enigma" by Leonello Tarabella
    1
  18638.  @Mathologer  I couldn't send you the kindle version because of Amazon region weirdness but hopefully you get a gift card in your inbox. Maybe it's interesting? I don't know. Also that video on Petr's miracle really was amazing <3
    1
  18639. Fantastic.Axis Foulding, .A great thing anyone can try out in CAD. is it converting between linear and polar
    1
  18640. And you didn't even consider whether the lacing goes up-down or down-up through each of the eyelets, resulting in another 2^2N different choices, and then I'm even forgetting about crossings that may go over-under or under-over. How many lacings are there then? ;-) A nice one for the OEIS, if it doesn't already exist, actually... BTW, I learned another bit of trivia when on a tangent: eyelet is not the same as aiglet, although they sound somewhat similar and both are lacing-related. How cool!
    1
  18641. On the 10-cards problem: (We assume that the operation of turning only the final 1 to a 0 exists. ) We now try to maximise the number of moves flipping 10 to 01. Notice that this move “moves” the 1 to the right. So, if a 1 is located on the furthest left digit, it can “consume” 9 moves before it moves eventually to the rightmost edge, and must be destroyed. (This takes a total of 10 moves. ) The same is true for 1s in other digits; the max number of moves it can consume is exactly its position (counting from the right). The maximum number of moves = 10 + 9 + … + 1 = 55. This can easily be verified as actually possible, as one can start with digit 1 from the right, then digit 2 and so on.
    1
  18642. A great pleasure during Christmas for me. Thank you!
    1
  18643. Jajajaja bro, you are amazing
    1
  18644. Why is it 4k+3 and not 4K-1?
    1
  18645. Another related comment about the pi/4 formula. In BASIC programming language, it had a math package, but no hard-coded value for pi. So you would see "PI = 3.1415926#" as a statement in programs requiring pi (the "#" denotes double-precision 32 bit math). But I discovered a much better way to "get" pi, to the full precision of the math used in the program. With this statement: "PI = 4 * ATN(1)".
    1
  18646. Why am I here. I'm barely passing highschool maths.
    1
  18647. I liked the visual proof that the no 9's series moves to grayer and grayer visual and then to white as proof that it is finite.
    1
  18648. I don’t really know imaginary numbers but isn’t e^(i*pi) =-1 and since i doesn’t exist e^pi also be a simple number idk /j
    1
  18649. 45 squares?
    1
  18650. Why did my real analysis class seem so difficult? I need the auto-algebra machine!
    1
  18651. Thank you for enlightening us. I am sorry as a Czech, that we do not learn about Karel Petr in our schools (Or very little)
    1
  18652. Very interesting! I wonder, if you were to project the paths of the vertices of one of the rotating equilateral triangles to their corresponding positions if the large circle were uncoiled to a straight line, would the paths of the vertices assume the form of 3 sine waves 120 degrees out of phase with each other? I ask this because the relative positions of the vertices of the equilateral triangle, relative to their distances from the large circle, seem to mimic the voltage values at any given moment from the positive or negative peak voltage in 3-phase electricity. I’m probably explaining this all wrong and I have no idea how to project the paths of the vertices if the circle were to be uncoiled, but their behavior as the triangles rotate seems very intuitive.
    1
  18653. Asturias <3
    1
  18654. I wonder if there are any slots left for me to make a comment...
    1
  18655. 14
    1
  18656. I rarely use the word but this is an honestly elegant proof. I do so love geometric proofs!
    1
  18657. calculating determinants by hand - nothing says Christmas like that hahaha
    1
  18658. Finishing watching this video makes me feel blessed. Into these amazing math stuffs.
    1
  18659. WTF. I wrote a pattern detection program in 9th grade that does this, and I discovered it myself. Guess I’m as smart as Mewton.
    1
  18660. Nice video as always! One of your clearest and most engaging yet. I remember long before I learned anything about any of this I tried exactly this difference process for primes numbers (I just thought it was cool and might give some insight). Of course I didn't really get anywhere with it but it was very interesting. I liked the pi seal after the credits.
    1
  18661. I liked it so much, I watched it twice! I did kinda want to see what happened if you used other irrational numbers... like the decimal part of sqrt(2).
    1
  18662. I like geometry content,so I am hoping ,that you make more of such content.
    1
  18663. 1) A triangular grid (2D) also has the shift property and its own version of the turning property, so it can be shown that it only contains triangles and hexagons. 2) As far as I can tell, a cubic closest packing grid (3D) has the shift property too. Therefore, it also only contains squares, triangles, and hexagons. 3) A grid of truncated octahedra (3D) does not have the shift property, unlike the grid of cubes. As far as I can tell, neither does hexagonal closest packing (also 3D). Someone prove / disprove the existence of pentagons, heptagons, and above in these! 4) We can go wild with this. Penrose tilings? Semi-regular tilings? Tilings made from special non-regular shapes? Rhombic dodecahedra (3D)? We haven't even talked about finding Platonic or Archimedean solids in these 3D grids! We need to know! We need a better proof!
    1
  18664. 1040+-13 = 9*81-(1,2,3) golden dragon number. 117}{711 80*2+-36 14^2 and 31*4 lychreal proof boom 196 algorithm
    1
  18665. Made it thru your entire presentation in one continuous setting, and I applaud you for your methodical technique. I’m an engineer with graduate level of maths education.
    1
  18666. How about Analysis Made Easy? That seems to be more useful conceptually than Calculus Made Easy.
    1
  18667. To me, that's Terence Tao analysis book.
    1
  18668.  @vogel2499  I agree. Best Book on analysis.
    1
  18669. About bugs path length. Starting distance is equal to one. Each bug is moving towards another bug who is always moving perpendicularly, so it's distance can't increase. So each path lenght must be one.
    1
  18670. Fröhliche Weihnachten. You can monetize these videos. You put so much work into these. Why not get something back.
    1
  18671. For the last problem in the coin paradox, it should be 5 rotations, the rotating coin is 2/3 if the stationary one, flowing the video, the stationary coin will uncoil twice, if you go along that line, it will be 3 times, he already showed that previously, it you go recoil it, it will be 2 more, because you uncoiled it twice, add them together, 3+2=5 Your eelkome
    1
  18672. I do not believe we were ever taught Heron's formula in my (American) school. However, I ran into it somewhere when I was in 6~9th grade and took the time to memorize it since it seemed really useful. I did end up using it a couple times on coursework in highschool, but that's about it.
    1
  18673. 6:45 5³ + 6³ ≈ 7³ (+2) 16³ + 17³ + 18³ ≈ 19³ + 20³ (-18)
    1
  18674. Love the video! Thank you. Not that it matters, but check out your value for pi (3.14) at 6:20. Isn't that 3.18?
    1
  18675. Whole time i look only your T-shirt logo😂😂
    1
  18676. Just discovered YouTube and loving these videos....I’ve got a university degree majoring in Mathematics but I’ve been working as an engineer for the last 15 years - bringing back good memories.
    1
  18677. I vote The leaning tower of maths for those crazy alternate stacks
    1
  18678. The valley of fear!!
    1
  18679. 13:37 : if a fraction of the list is x/y (what original, isn't it ?), et the next is a/b, so x+y=b and y+b=a=2y+b
    1
  18680. Amazing, @Mathologer, to treat PNT without showing the algebraic identity. Franklin^n for n->∞ I'd love to see you "do" the Jacobi triple product identity. Or Rogers-Ramanujan. Or some hyperbolic geometry & modular forms video... Your videos are sheer delight.
    1
  18681. I took Geometry in about 1992, in high school, in ohio, USA. We were taught Heron's formula.
    1
  18682. The most memorable part was when you showed the optimal leaning tower and then went on to explain that you often get asked about the single block per layer tower because that is exactly what I was thinking at the time!
    1
  18683. ninjago super video
    1
  18684. This guy giggling every two seconds is giving me real Palpative vibes
    1
  18685. Khayyam solved the depressed cubic in 1070 AD
    1
  18686. As soon as you showed 4 and 10 I was thinking, what about 55?
    1
  18687. What a fabulous video.The thing I liked the most in this video is how we approached to solve the problem, and that is absolutely we all need to know to sharpen our brains.
    1
  18688. wow!
    1
  18689. Thanks for a next beautiful video.
    1
  18690. Because they be?iev3d it was magic. 13:55
    1
  18691. My thought is that seems like a way around the constructability of regular polygons. Would the Greeks have thought that a regular heptagon was unconstructable if they knew advanced linear algebra?
    1
  18692. Definitely check out the Melbourne Star but please for the love of dog do not attempt to ride it 😅
    1
  18693. Most memorable is no integers in partial sums - those integers are so rare!
    1
  18694. Q: What is the ratio between the unstretched band and the circle's radio for this to work (and why)? Also, I'd love to see an explanation on how the old mechanical adidiators worked. Thank you!
    1
  18695. I respected your image, but didn't hold the flame, your good image, I will bring to you some dawn. Colours, colours, colours and fragrancies, colours, colours, stop playing around with me
    1
  18696. thabk you for your persistence real math needs it
    1
  18697. Hi mathologer, how do you come up with these concepts, how do u search or research on these topics..
    1
  18698. fun but I have too limited patience to study and understand all :) But this is very important for our civilization, so take notes and learn, if You who have brains enough.
    1
  18699. I was really hoping to also see this written in Python
    1
  18700. CHALLENGES! 9:12 Find all identities of length 10. (1 * 8) + 2 + 10 = 10 * 2 * (1 * 8) (1 * 8) + 4 + 4 = 4 * 4 * (1 ^ 8) Brute force checking 3 non-padding-1s didn't work, and it's easy to prove that checking any more than that is impossible. 9:22 Find a length N that has more than 10 identities. Just look at the graph from earlier, there's plenty of such N plotted above y = 10 ;) On a more serious note, consider the base case that all but two of the values are padding 1's. The equation is now (N-2) + A + B = AB. Rearrange a little bit... AB - A - B + 1 = N - 1 (A - 1)(B - 1) = N - 1 A and B cannot be 1 so A-1 and B-1 are both positive. We are now looking for an N-1 that has more than 10 distinct pairs of factors, or essentially more than 20 factors. N-1 = 576 has 21 factors as 11 pairs so N = 577 has at least 11 distinct identities. BONUS: 11:24 Do you enjoy being... you know... Not with a yassified AI-generated Sophie Germain staring at me, no thanks. But the journey itself is fine 22:38 The Hyper Sophie Primes It's possible to generalize this even further. If we have a bunch of 2s (T of them) and then the last 2 terms to worry about are A and B, then we have the following rule: (2^T * A - 1)(2^T * B - 1) = 2^T * (N + T - 2) + 1 For T = 0, 1 we get the familiar prime and Sophie Germain prime conditions: (A-1)(B-1) = N-1 (2A-1)(2B-1) = 2N-1 But following that come the following conditions: (4A-1)(4B-1) = 4N + 1 (8A-1)(8B-1) = 8N + 9 (16A-1)(16B-1) = 16N + 33 (32A-1)(32B-1) = 32N + 97 etc. Every single one of those up to when T = N-2 must work. All of those RHS terms must be prime.
    1
  18701. @Mathologer  noted, thanks for the info
    1
  18702. THIS VIDEO IS A MASTERPIECE thank you
    1
  18703. Great stuff as always. I thought you might get into the Euler-Maclaurin estimates for harmonic numbers. Those are pretty cool too.
    1
  18704. The most memorable part for me is with the Euler's constant -- I love the 1x1 square visualization!! Also te animation at the beginnign was awesome!
    1
  18705. The concepts of calculus are beautiful and remarkably intuitive if you one is also shown what they model. The exact calculation and algebra parts of it; yeah, that's the ugly stuff.
    1
  18706. Was waiting for the t-shirt to get swapped out for a Hyperspace Invader...
    1
  18707. obviously it is the key to solve Collatz conjecture. (Lol)
    1
  18708. Here's a thought. One sequence above produce the sum of integers up to and including a given integer. 1, 1+2, 1+2+3, 1+2+3+4, etc. Let us call this the aggregate of n. Another gives the factorials which is the same with multiplication. As the aggregate is to the factorial, so the factorial is to [function]? What sequence would produce this function and can this be used to tease out arithmetic functions beyond addition and multiplication or subtraction and division? I'm not talking about exponentiation but logical mathematical operators. Has this already been done?
    1
  18709. i understood it but at the same time i dont.
    1
  18710. Paint molecules are a finite size. You can't paint the inside of a hole smaller than the size of a paint molecule. Even is the inner surface area is infinite paint will not cover all of it.
    1
  18711. I liked Chapter 5 the most
    1
  18712. 4 sqares no: eg A2,B1 (and some others distant from there) this leaves an isolated A1
    1
  18713. The geogebra animation was great!! Thank you for it, and please use again when you can!
    1
  18714. Awesome video and intuitive visuals as always.
    1
  18715. There was a fine letter called Aitch, But people would often say "Haitch". In confusion, they'd mix, These sounds in their picks, But let us not mock their mistakes.
    1
  18716. This video was absolutely beautiful and was exactly the kind of geometrical magic that I love. The only thing that bothered be, albeit only slightly, was his saying radiuses instead of radii.
    1
  18717. Nice. I was waiting for a graph of p versus b with color coding to show convergence.
    1
  18718. Mom: What are you doing on laptop? watching utube again huh? Me: No, I'm attending my 1 hour long class. Look!
    1
  18719. Holy hell, I discovered this doodling as a 10yr old kid
    1
  18720. The swap rules being able to move between any pair of tilings sounds a lot like Markov Chain Monte Carlo. Any ideas on what's known about the natural MCMC algorithm (pick a random pair of adjacent horizontal / vertical dominos in a square and swap orientation)? Does it converge to a uniform random tiling? If so, how quickly?
    1
  18721. I was always skeptic if this expression for Pi could have originated more than 200 years before the claimed invention. Hearing from Mathologer leaves me in no doubt. Now - why on earth was this expression useful at 1300 in Kerala? and what were the foundations leading up to this magical piece? Can't be just the genius of one individual.
    1
  18722. What I do not understand: you show that the volume is less than 2 but afterwards you show that it is exactly Pi which is bigger than 2. Or did I miss something?
    1
  18723. Oh sorry. I missed that. Thank you very much for your answer. The videos are always great and you are doing an amazing job.
    1
  18724. 23:18 Bruh. 🤯
    1
  18725. Beautiful math! Thank you!
    1
  18726. Could you make a video about 'Hilberts Hotel' (i dont know the english name, i think its 'Grand Hotel')?
    1
  18727. 4K-1 you mean....
    1
  18728. 9:46 Every natural number will have only ONE assortment of coins which adds up to it
    1
  18729. I still have a slide-rule.
    1
  18730. Miracles never cease to amaze me. Danke dir!
    1
  18731. How can we observe the fractal nature of this phenomenon if we don't even have a rigorous definition for fractal? Self-similar and fractional-dimension no longer hold?
    1
  18732. Kempners proof is so unintuitional and yet beautiful.
    1
  18733. Haven’t watched he video yet. Is it because we use pi instead of tau :)
    1
  18734. I could be wrong, but it seems like if the ratio of smaller radius/larger radius = X/Y, then the smaller coin rotates about itself Y+X times if it's outside the larger coin, and Y-X times if it's inside the larger coin. This might only work if X and Y are relatively prime.
    1
  18735. So... why was it missed?
    1
  18736. Answer to end questions: I recently subscribed and would love to see more complex videos. I am currently working on my bachelors and soon to be masters in applied mathematics.
    1
  18737. It’s gotta be the fact that ln(n+1) + gamma approximates better and better for larger n
    1
  18738. Very insightful in case of quantum chromodynamics. Have to think about quarks colors.
    1
  18739. Me and my girlfriend sat on the floor for almost 20 minutes trying to figure this out, we had 4 bodies ( represented by stuff animals) I figured it out and we just needed two more so the answer is plus 2 bodies
    1
  18740. 32:34 Wow, you should be a ventriloquist! ;) It almost looks like you said two! haha
    1
  18741. Sadly, reality is discrete
    1
  18742. I would seem I jumped in too soon with my previous question. I should have known better. Cheers.
    1
  18743. How does one determine the side lengths of either the equilateral triangles or squares on order to make the shapes? Or does the size not matter (i.e. the triangles scale to meet the squares or vice versa)?
    1
  18744. hi
    1
  18745. Nice video!
    1
  18746. Only the very best nerds can pull it off to write a book about shoe lacing! Congratulations for that.
    1
  18747. Mathologer; Ask questions if you don't understand. Me (having no business watching this); I didn't understand anything. I'm sure this was an excellent explanation on something wonderful, what that was no idea.
    1
  18748. Love how the 3-d interpretation of the hexagonal aztec tiling makes the answer to the last question literally obvious.
    1
  18749. 3,1,4,5,9,14,23,37,60,97,157,254,411,665,1076,1741,2817,4558,7375,11933,19308,31241,50549,81790,132339,214129,346468,560597,907065,1467662,2374727,3842389,6217116,10059505,16276621,26336126,42612747,68948873,111561620,180510493,292072113,472582606,764654719,1237237325,2001892044,3239129369,5241021413,8480150782,13721172195,22201322977,35922495172,58123818149,94046313321,152170131470,246216444791,398386576261,644603021052,1042989597313,1687592618365,2730582215678,4418174834043,7148757049721,11566931883764,18715688933485,30282620817249,48998309750734,79280930567983,128279240318717,207560170886700,335839411205417,543399582092117,879238993297534,1422638575389651,2301877568687185,3724516144076836,6026393712764021,9750909856840856,15777303569604876,25528213426445732,41305516996050610,66833730422496340,108139247418546940,174972977841043260,283112225259590200,458085203100633500,741197428360223700,1199282631460857300,1940480059821081000,3139762691281938400,5080242751103019000,8220005442384957000,13300248193487976000,21520253635872930000,34820501829360910000,56340755465233840000,91161257294594750000,147502012759828600000,238663270054423360000,386165282814251960000,624828552868675300000
    1
  18750. I am sorry if I'm being stupid but in the math Olympiad problem you proved that there exists at least two collections which have the same sum from the total but you didn't prove that any 10 numbers chosen are capable ofgetting, through splitting, the equale sums please can some one help me understand the proof better
    1
  18751. I am in awe...
    1
  18752. I'm not a math pro, aber einmal I was playing with hyperbolas & noticed how a half imaginary factor made inverse functions.. I though of hyperbolas being like 'Eigencurves' if you will. I find it all so wonderful.
    1
  18753. the tetrahedron inside of a cube was beautiful.
    1
  18754. 35:30 If the bug path length = the length of the side of the original square, you would have 4 curved “equilateral” right triangles.
    1
  18755. is it just me ... or was hid head extra shiny today😅
    1
  18756. For the puzzle @12:08, you should go right-right-right-left to go from the fib matrix 1123 to the 9 4 13 17 one.
    1
  18757. Sir I am a Masters student. Your videos are very interesting . Can you say which software you are using it will be useful for me to present seminars in my class.
    1
  18758. i was always interested in formula of sphere. every other formula for volume of other shapes made sense but sphere was the only one odd.
    1
  18759. As noted, if you choose the middle branch every time, the resulting series of legs are "almost equal" ... therefore each iteration generates a rational upper and lower bound for the sqrt(2), The error gets better by a factor of ~6 each iteration ... nice! Now, I'm wondering what branch choices will generate rational approximations for sqrt(3), sqrt(5), sqrt(p) ... I have no life.
    1
  18760. Who can say mathematics are not beautiful? 🤩
    1
  18761. I solved the triple! The box that produces 153^2+104^2=185^2 is, reading clockwise from the upper left, 9, 4, 13, 17. To get to this box starting from the first, you go R-R-R-L. I found the key was to identify that the 1 in the upper left box is preserved when going to the right from the original box. Thanks for the puzzle! AND There is no parent triangle for the 345 because the center of the Feuerbach circle and the incircle lie at the same vertical position! Thus, the parent triangle has a height of zero!
    1
  18762. if we start calculating the base starting from x= 1, for every base of our shapeshifter A(r) is r-1 for r greater or equal to r. So to calculate the total base of A(X) + A(Y) we would use the formula (X-1) + X(Y-1) and that equals X - 1 + XY - X wich equals XY - 1. This is exactly the base of A(XY)
    1
  18763. I carry in my pen case two 15cm slide rules. Both are in leather cases. Older one made by Nestler and the other by Castell. Both have sines & tangents on back of the slide. On first year of high school we all had to buy book of tables and learn how to calculate using logarithms. Then we moved to slide rules that we had to buy. On second year all that was scratched as everyone had scientific calculator which they had bought without any mandate. Nice though that they teched how to use those things as I have few times had to do calculations the old way.
    1
  18764. I thought its not Pascal triangle but Newtons triangle :) - there were some many Pascals but only one solitude and alone Newton - its easier that way. The differencive method (differential but on discrete parts) was taught to me on discrete math - very useful in finding a way out of a strange nonsense string of numbers :)
    1
  18765. Ich liebe die Eleganz visueller Beweise in diesem Videos. Das sind meine Highlights. Sie erzeugen so ein Glücksgefühl beim Betrachten. Der No 9 Grid Beweis ist dafür ein super Beispiel. Ich kann daher die vielen Stimmen dafür völlig nachvollziehen. Dagegen, völlig verstörend wirkt diese löchrige Mauer bei der optimalen Lösung für 20 Steine. Dass das eine beste Lösung ist, finde ich irre. Es reizt mich aber, mich der Lösung mit Jengasteinen anzunähern.
    1
  18766. 6:26 i think no
    1
  18767. piano at the end is fabulous. Oh and the math is nice too (arcane arts to me, unfortunately)
    1
  18768. I tried out the coding challenge! https://github.com/weirichd/ArcticCircle
    1
  18769. 27:04 well don’t worry about 😂 I❤math. when it matters it matters. if it doesn’t matter, it doesn’t matter at all. so it is whatever i say it is because i said so moving on. 😂
    1
  18770. Mind-blowing stuff!
    1
  18771. That’s insane😱
    1
  18772. I wouldn't say it's only rearranging numbers. You effectively remove a big part of negative numbers when you try and target a specific number like pie. The calculation may continue forever and ever but there will always be a mis match between positive and negative numbers, so its obvious the answer can be whatever you make of it.
    1
  18773. Please explain what is written in your t-shirt.
    1
  18774. Thank you so much for the video, I solved the 3^4 just a couple days ago
    1
  18775. Manipulating powers is powerful.
    1
  18776. I liked Tristan fractal -- I got it right away before you even mentioned fractal.
    1
  18777. I've seen parabolic ghosts with only one parabolic mirror, a darkened room and a single fairly strong light source. I still don't quite get how it worked, but it was an amazing experience! :) My head was at the focal point, but one of the other boys got near enough that I saw a 'ghost' of his head.
    1
  18778. The most memorable part for me is how incredibly slow the harmonic series diverges. And even the fact that it diverges in the first place!
    1
  18779. Madness. I’m absolutely blown away
    1
  18780. Pure Greatness and joy of math
    1
  18781. Right now, I’m just at 14:52 in the video so you might address this later, at which time I will edit this comment, but you used the word “paradox” here, and I wonder if this really constitutes a paradox (a word casually thrown around often inappropriately) just because it’s physically unrealizable. I mean there is no contradiction created with the Leaning Tower of Lira in its theoretical limit of the (infinite) harmonic series. It’s just physically impossible to construct it: whether you consider the infinite number of elements required or the physical limit on infinitesimal distances.
    1
  18782. I concur and am now giddily delighted to have gotten a response from the actual video creator, but that’s not what we’re talking about presently. The topic of paradox did indeed inspire me to look up the term myself (yay wikipedia!) to see if I was just mistaken in my previous understanding, which also included in its list the famous tortoise “paradox” that I quickly recalled as being referred to as a paradox. Perhaps the only thing in this universe that is an actual paradox is the term paradox itself. 😜
    1
  18783. Hahaha, "Now prove that there're no prime numbers that are presented as difference of two squares". Obviously, you're kidding. In fact I was going to drop this joke here, but at the end of the video... I saw it. Very, very beautiful proofs, as usual... The (mathematical) truth is beautiful by definition, but there are really ingenious proofs, with unique approach.
    1
  18784.  @miloszforman6270  👍 Useless fact: that figure also visualises the law of sines.
    1
  18785. Pls don't heart my comment #RedcrossPuzzle: Is it legal to cut the cross by placing one piece on top of other do that still count as one cut???
    1
  18786. Great video!!! We missed you, guys 💛
    1
  18787. Great content 👍
    1
  18788. It is pretty. It looks like a flying bird.
    1
  18789. And now we are waiting for comments like: "They don't want you to know the truth" ;)
    1
  18790. Can i get a heart❤? Love u
    1
  18791. p.s. i made it to the very end
    1
  18792. This is how ancient Egyptians wrote their fractions, as sums of 1/n fractions, for various n, never repeating the one with the same n.
    1
  18793. First challenge answer is no, since you can completely isolate the corners by removing the adjacent tiles. I wonder how this would work on a 3 color board (by coloring the diagonal lines in three colors) and with 3 long peices, since each piece would still have to have one of each color under it.
    1
  18794. For the sequence of continued fractions around 14:00, I think it's something like: Numerator 1, 3, 7, 17, 41, 99, 239, ... 1, 3 - double and add 1 3, 7 - double and add 1 again, hmmm 7, 17 - double and add 3 17, 41 - double and add 7 Generally: x(n) = 2*x(n-1) + x(n-2) well apart from at the beginning. Same story for the denominator.
    1
  18795. Love your t-shirt ♥️
    1
  18796. Fascinating, but the music is a tad overwrought.
    1
  18797. Your laugh though. Its too contagious! Also I greatly appreciate this explanation!
    1
  18798. I am bother by the fact that the root of evil, namely 666, being equal to 25.8069758011..., should be approximated by 25.8070 and not by truncated as 25.8069, cause the evil is in the details.
    1
  18799. 3rd year Maths & Physics student here! This was such a beautiful video, excellent pacing and the animations are very helpful. I love videos like these since I feel like I missed out on more number theory based topics due to the physics side of my degree.. I wish my manifolds lecturer could do amazing animations like you 😄 instead we get badly drawn pictures on a chalkboard haha
    1
  18800. In general, this prove that maths are cool and that the stuff taught in high school is the boring one.
    1
  18801. Sure it beats Ramanujan's inf. fraction
    1
  18802. I'm just here for the poetry.
    1
  18803. This will come in useful when I sell my GameStop shares.
    1
  18804. But you already know that changing the addition order of any convergent alternating series (when both the negative terms and the positive terms both diverge on their own - as it does in this case) allows it converge to ANY real number you like. Yes, any real number by simply changing the order. That's why this is not a paradox. Not in the true sense of a what a paradox really is. Like "Russell's Paradox" (aka Who Shaves The Barber) - a logical contradiction which CANNOT be resolved. Things like the "Birthday Problem" or "False Positive Test Result Problem" or "The Monty Hall Problem" or the incorrectly labelled "Zeno's Paradox" are NOT paradoxes at all. They are all problems that have a correct and valid answer, just an unexpected one for some or one that seems to go against some people's intuition or common sense. Now that would make for an informative video. It would clear up a lot of misunderstanding. And mis-labelling.
    1
  18805. No, the extra pigeon will be on the floor of the loft.
    1
  18806. Such beauty
    1
  18807. This latter video comes over in such coolness i have to think of considering who to hide this from, who to avoid in the mailing list for the link of "it" (not getting them seduced to play with it in a disgraceful manner). Und dass dies mehrheitlich von der Feder eines Deutschen "verfügt" ist, ist zum liebäugeln. Fast dass einem ein dralles Weib oder eine Mathrematikerin die tollen Kugeln vorlegte (das muss nicht auf einem Billard-tisch zu liegen kommen ; ) Manche Wesen haben/hegen Zahlen im Blut? die anderen sollen die Runde bezahlen können?
    1
  18808. Most memorable: Leaning Tower of Lyra
    1
  18809. Dam reminds me of me messing around with logo programming language when I was young and wondered surely you could use logo to solve maths problem wow
    1
  18810. Thank you!
    1
  18811. I tell you again, this is a joke expression, which is true only in the context of the problem. Consider it formally. 1+2+3..=-1/12 Both parts by 12. 12 × (1 + 2 + 3 + ..) = 12 × (-1/12) We get 12 × 1 + 12 × 2 + 12 × 3 + ... = - 1 transfer -1 to another part 1+ (12 × 1 + 12 × 2 + 12 × 3 + ...) = 0 Let's notice that 1 is an odd number. A row in brackets is the amount of even numbers. 0 even number. Thus, we obtain that the amount of an even and odd number is an even number that is incorrect :). P.S. Алгебра иногда интересна как форма языка формул:).
    1
  18812. My attempt at the solution of the next to last puzzle: I think the trick is to create a 0 at the end of the chain of cards and slowly move it to the left until it reaches the last spot. As we are walking the 0, we get that the number of actions of flipping is equal to the number of spots the 0 has to walk. When it reaches the first spot (card) we just repeat the process but for a n-1 cards. Therefore, the maximum number of spots it can walk is equal to n+n-1+n-2… = n(n+1)/2 where n is the number of cards (all of it can be proven by induction). Is this correct?
    1
  18813. I will name my first born "subscribe to Mathologer" if Mathologer would exclusively make hour long videos on all of Jim Propp's and Richard Kenyon's papers relating to dominos/dimers.
    1
  18814. I loved the "terrible aim" part! Its like the series is jumping hurdles with shrinking legs.
    1
  18815. Son of a. That seems like fun. It's been forever since I've done calculus.
    1
  18816. Is there an extension of the Heron's formula for 3D solids?
    1
  18817. Just overload the subtraction operator and implement some nice side-effect... 😂
    1
  18818. Thank you for uploading such beautiful videos on mathematics sir, it really helps to understand the beauty of studing such a fascinating subject which is considered dull otherwise.(by many)
    1
  18819. is zero a prime
    1
  18820. Mmm it seems obvious in an intuitive way. Both the square method and the lantern doesn't aim to reduce the actual length or area, and is just playing a trick to maintain or even add length or area to the object. More exactly the additional point that doesnt stick to the surface to be approximated allows mischief to be done. They are like fractal, like some like to use to manipulate the lemgth of their national border. So the main thing is all points used in approximation should be on the line or surface to be approximated
    1
  18821. Title: "Dimensionless, timeless and acausal Knower (Monad) vs the Reimann Hypothesis" 1. Reframing the Riemann Hypothesis: The enhanced 0D Knower framework offers a novel perspective on the RH by reinterpreting it in terms of fundamental symmetries and information structures. The key reframing is: "All non-trivial equilibrium states of the fundamental knower network configuration occur at the symmetry condition of knower interactions in the 0D state, where entropy and negentropy are perfectly balanced." This reframing connects the mathematical statement of the RH to concepts of dimensional symmetry, entropy-negentropy balance, and fundamental knower interactions. 2. Key Concepts in the Enhanced Framework: a) Dimensional Symmetry: The framework posits that 0D is the origin point from which both positive and negative dimensions emerge symmetrically. This symmetry is crucial in understanding the critical line Re(s) = 1/2. b) Entropy-Negentropy Balance: The interplay between positive and negative dimensions creates a dynamic balance between entropy and negentropy. This balance is fundamental to the location of zeta zeros. c) Dual Nature of 0D: The framework proposes that 0D has two sides (real and imaginary) with an event horizon between them. This duality maps onto the complex plane of the zeta function. 3. Mapping Mathematical Structures: a) Zeta Function as Knower Network Configuration: ζ(s) = ∑(n=1 to ∞) 1/n^s ≡ Configuration_Density(Knower_Network(s, d)) b) Zeros as Equilibrium States: ζ(s) = 0 ≡ Equilibrium_State(Knower_Network(s, d)) c) Critical Line as Fundamental Symmetry: Re(s) = 1/2 ≡ Symmetry_Condition(Knower_Interactions, d=0) 4. Novel Insights: a) Complex Plane and 0D Duality: The framework maps the complex plane of the zeta function onto the dual nature of 0D, with the real axis corresponding to the "Singularity" side and the imaginary axis to the "Alone" side. b) Trinary States and Dimensional Symmetry: The knower states |0⟩, |1⟩, and |2⟩ are mapped to 0D, positive dimensions (entropy), and negative dimensions (negentropy) respectively. c) Prime Numbers and Knower Structures: Prime numbers are interpreted as irreducible knower configurations spanning positive and negative dimensions. 5. Potential Proof Strategies: The framework suggests several approaches to proving the RH, including: - Symmetry analysis - Entropy-negentropy dynamics - Event horizon properties - Dimensional transition analysis These strategies aim to show that the critical line Re(s) = 1/2 is a unique symmetry point where entropy and negentropy are perfectly balanced in the 0D state. 6. Challenges and Open Questions: The framework raises intriguing questions, such as: - How to mathematically formalize the transition between positive and negative dimensions? - How to rigorously map the complex plane onto the dual nature of 0D? - How to quantify the entropy-negentropy balance in terms of knower network properties? This overview sets the stage for a deeper exploration of the mathematical formalism and proof strategies in our next section. The enhanced 0D Knower framework offers a unique perspective on the RH, potentially bridging abstract mathematics with fundamental concepts of physics and information theory.
    1
  18822. This is an incredible topology of harmonic relations!!
    1
  18823. I'm interested in what the mathematical operation flipping the card/digit would be represented as
    1
  18824. With QE infinity on the way, we might very well have to make change for a googol dollars.
    1
  18825. 5
    1
  18826. The figure for 9 12 15 at 7.07 does not match the number of small squares, i.e. it should be 10 13 and 16.4, oh no ... ?
    1
  18827. The pattern with 2 seems to be the only one with no crossings in its interior. This should translate in a theorem on integers m,n,k of the form f(x)<>0 if g(x)<0, with g the cardioid. f(x) would then be something like A(n,k,m) x+B(n,k,m),as the intersection of two lines. Theorem: There exists two cool functions A,B of n,m,k in N such that, for all integers {n,m,k<n} , for all reals 0<theta<2 pi, then if 2*(1-sin(theta))<1, then A(n,k,m) *2*(1-sin(theta)*cos(theta)+B(n,k,m)<>0
    1
  18828. What a treat! Thank you Mathologer, happy holidays!
    1
  18829. basically you are dividing by 0 (infinitely thin layer to cover the surface is a division by 0 against the infinite surface itself)....that's why it works. Can't wait for humanity to finally figure out how to divide by zero, it's going to be fun.
    1
  18830. Is the algebra autopilot also done in keynotes? =D
    1
  18831. You can hear these relationships in any pieces of Bach = circle of fifths.
    1
  18832. When I was a kid, I saw this song about the divisibility test on a math show called Square One. Nine, Nine, Nine is burned into my brain.
    1
  18833. This video explains my favorite math joke: A mathematician is talking with the person next to them in a bar and says, "hey, think of a number, any number." The other person says..."ummmm...okay..." Mathematician: is it rational? Other person: yes Mathematician: how unlikely!!
    1
  18834. 40:24 The most surprising thing to me was this brief mention that the "at least ten 9s" series converges. Intuitively, for numbers with many digits, the vast majority of them should include at least ten 9s, just as the vast majority of them should include at least one 9. Not as many, but the fraction of them should still asymptotically approach 1 for large numbers of digits, shouldn't it? Any chance you misspoke, and this should have been the "at most ten 9s" series?
    1
  18835. Yes, great video tutorial, as usual. The eternal question is that it is geometrically possible to construct the root of 2, but how to calculate it :)?
    1
  18836. Can a lemniscate be asymmetric, for example by shrinking the left half about the focus? How could this be achieved geometrically, mechanically, or algebraically?
    1
  18837. I'm too dumb to understand much of this but it is strangely beautiful.
    1
  18838. Thank you so much for making videos and sharing! It is so beautiful!
    1
  18839. WOW
    1
  18840. Too genius
    1
  18841. Why doesn't the final equation work for one two or three?
    1
  18842. Why make a half hour long video over why sum of harmonic progression is ln 2 ... Just use PapaFlammmy's method . And the ans is 0.69 man What could go wrong? Sometimes all you need is a leap of faith! 🥸 - Livinhard Euler
    1
  18843. Nice!
    1
  18844. 25:15 my dad taught me this and when I brought up Pythagoras he was like who? I’m still shocked to this day that he uses it without understanding it haha
    1
  18845. I made it to the very end
    1
  18846. Most memorable for me was the crazy maximal overhang stacks, It's always interesting when complicated solutions can't be reduced to something neat and tidy. Also, how did you make the stamp for the wax seal at the end?
    1
  18847. what's funny is that the pattern continues backward, with one term on left and right too! (1 = 1)
    1
  18848. As someone who is awful at math, what would happen when you work a system backwards from a particular special pentagon to get to an irregular pentagon? What would be the factor that so skews it to result in a particular "screwy pentagon"?
    1
  18849. Very satisfying and beautiful! btw, I love binge-watching your videos:)
    1
  18850. @Mathologer  Haha Hello Mathologer :P No its just a joke 🤣 I love your videos! Do you know how to speak Greek? Or did you google it? Watching this video I remembered a funny story when I was a student and my math teacher started talking to us about imaginary numbers. We made that joke to him, that he is a "charlatan" mathematician, like those old people that used to sell fake medicine pretending to have medicine knowledge but actually had none 😛. Like sqr(-1) what is this guy saying? So the joke stuck and I remembered watching this video lol. Like a wandering mathematician, going around the country, selling math to people pretending to know but not actually knowing lol. Obviously you are NOT any of that, it was just a joke because it appeared so funny to me seeing these bizarre equations
    1
  18851. I couldn't watch the entire video in one sitting, and naturally it was kind of hard to pick up from where i left off. I would appreciate timestamps to the chapters in the description. i loved the block animations at the beginning and the end of the video. I am not a big fan of integral, derivatives and limits, so i didn't pay full attention to the latter half of the video, but i think i got it and would get even more out of it if i tried to work out some of the details myself. I like your style of leaving some steps as exercises for the viewer. I am a fourth year mathematics student in uni, with my main focus in combinatorics and graph theory.
    1
  18852. DOPE!
    1
  18853. That was the most metal introduction to calculus I've ever seen.
    1
  18854. helll yeahhhh the hero of math
    1
  18855. Start with the alternating harmonic series. Take the first a_1 positive terms, then the first b_1 negative terms, then the next a_2 positive terms, then the next b_2 negative terms, etc. The sum of the new series depends in an explicit way on the limit of the frraction (a_n) / (b_n). Sorry, I don’t recall the formula. This can be found in Pedersen’s book “From calculus to analysis”, the proof there uses the Euler constant gamma. Maybe our hero can find a more visual explanation.
    1
  18856. Hello, and a very Happy New Year (down under, at least) :-D There's a very beautiful video about spirographs, part of the Summer of Math Exposition, https://www.youtube.com/watch?v=n-e9C8g5x68
    1
  18857. A secant on the circumference will create an isosceles triangle between its intersection points and the center. As those secants approach tangent, the sum of all those tangents will become the circumference while the sum of all those triangles become the volume.
    1
  18858. Astounding video! I absolutely share your excitement - while you build on facts I've known, I definitely didn't see many topics covered coming. Thank you! :) As for my favourite, it's really hard to pick just one, but I'll go with the visual way of constructing gamma in the unit square and arguing it has to be greater than 1/2. I don't know how could I not stumble upon such clever construction before. Of course, Oresme's proof is a classic and close second, since such ingenuity really nails down what mathematics is about ;) Last but not least, say hello and thanks to Tristan from me, his fractally visualization truly captures the paradox of the solution fed by infite supply of number containing digit 9 :) Thanks again for compiling this all together, all the best!
    1
  18859. The best AHA moment for me? The "polynomial approximates powers of two" thing happened to me when solving the entrance for a mathematical institute. I thought it was simply curious and derived from linear algebra, but who knew there would be hidden knowledge!
    1
  18860. I wonder if anyone has done a keynote counterpart to the "powerpoint as a programming language"
    1
  18861. The most memorable part was to proof of the convergence of the Kempner series. I knew of the convergence already from the SMBC web comic "math-translations". And now I feel kinda stupid for assuming that it was probably too complicated for me to prove. I feel confident that I could have sat down myself and arrived at a similar looking proof. Just goes to show, that things get much more complicated if you assume that they are complicated!
    1
  18862. Thanks! My background is computer science/math. This was perfect level! Was not aware that the video was 50 minutes until after I saw it.
    1
  18863. OMG I love this identity. It is so obvious, when explained, but not intuitive until then.
    1
  18864. Just wow
    1
  18865. I have money. Other people do to. It would be great to win a book, but if you'd put in a bit of self-promotion now and then, I would have known about the book. Can't buy if I don't know about it.
    1
  18866. 2:43 you enlightened me. This is probably the first time I can truly visualise an hypercube. It has as faces ... 8 cubes in 8 parallel 3d universes, and they are crossing 2 by 2 on 3d cubes, and 3 by 3 on lines ...This formula is soo visual !!!
    1
  18867. Interesting. Where is the lookup table of log disk? Log is faster for Two number multiplication.
    1
  18868. u are right infinites do not exist in real world. that is why there is nothing infinite in the center of a black hole. we just not there yet to explain scientifically. BTW Oresme died in 1383 not in 1382
    1
  18869. Thanks for this beautiful video !!
    1
  18870. Thoroughly enjoyed the entire episode. Thank you Mathologer.Sending some ❤ and power ⚡ to keep making these videos. Just some questions 1) what would happen if we put (x-1) instead of (x+2) ? 2) does 2d squares in 4d cube, deviates from its square shape during rotation of the 4d cube?
    1
  18871. Mathematicians: only the top mathematicians may come close to solving the great Basel problem. Euler: Hold my π²/6
    1
  18872. I wondered why this popped up again when I had already watched it. Now I understand.
    1
  18873. I'm tempted to challenge a math teacher to try and find a formula for the area of any triangle given only the lengths of its sides, without using trig functions.
    1
  18874. 6:23 couldn't the non two squares be written as 4k-1 instead of 4k+3?
    1
  18875. Fancy seeing you today
    1
  18876. Wow, that's a lovely proof. Had never seen it before. I'm sure that's one that Paul Erdos would have said was "from the book" (God's book of the most elegant maths proofs). 😉
    1
  18877. I preferred the coloring proof.
    1
  18878. Number theory / modular arithmetic proof that equilateral triangles on a grid are impossible: Let's assume that one of the three points is at position (0, 0) in the grid and the other two are at (a, b) and (c, d), with a, b, c, d being natural numbers. Since all of the side lengths must be equal, we get a^2 + b^2 = c^2 + d^2 = (c-a)^2 + (d-b)^2 and some rearranging gives us a^2 + b^2 = 2(ac+bd) = c^2 + d^2. Since 2(ac+bd) is even, a^2+b^2 and c^2+d^2 must be even as well, so a and b must have the same parity (the same goes for c and d.) Since a^2+b^2=c^2+d^2, this means all of the involved integers have the same parity, since otherwise one side would be divisible by 4 and the other would not. If all of them are even (and non-zero), you can just factor and divide out 4 in every equation until you get four odd numbers. So let's assume a, b, c and d are all odd. This means a^2+b^2 is not divisible by 4, but ac+bd is even, so 2(ac+bd) is divisible by 4. Since a^2+b^2=2(ac+bd), that's a contradiction and thus this triangle cannot exist.
    1
  18879. I won a straight slide ruler about 38 years ago by using the least amount of paper in a math test. I needed two sheets of paper, the second one I needed just for half of one side, I remember I was a bit upset I didn't have the time to copy that part on some empty space on the first piece of paper. When the teacher came with the answers he had squeezed all of them on just one piece of paper, using all of the space on both sides. And now I'm searching for my straight ruler, though I already found the round one I inherited from my father.
    1
  18880. 21:14 The noise in the background are Cicadas.
    1
  18881.  @Mathologer  btw, I can't thank you enough with words for all you've learned to me
    1
  18882. Ah, the YouTube algorithm at it's best. It recommended me this splendid video just when it has 69K views :D Most memorable was the leaning tower of maths section.
    1
  18883. It remains far from clear to me why the highlighted fractions in the new table should be considered the best approximations. The CF of Pi is [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, …], if we can trust Wolfram Alpha. Changing that into zigzag path in Stern-Brocot tree, the first digits spell R3 L7 R15 L1. Ron Knott's page "Fractions in the Farey Series and the Stern-Brocot Tree" has a nice calculator for SB paths. For R3 L7 R15 L1 it gives the list: 1/1 -> R 2/1 R 3/1 R 4/1 L 7/2 L 10/3 L 13/4 L 16/5 L 19/6 L 22/7 L 25/8 R 47/15 R 69/22 R 91/29 R 113/36 R 135/43 R 157/50 R 179/57 R 201/64 R 223/71 R 245/78 R 267/85 R 289/92 R 311/99 R 333/106 R 355/113 L 688/219 The denominators correspond with OEIS A097546 . It's unclear to me why the sequence is called "Farey sequence" instead of "Stern-Brocot sequence". The "mediant factors" of Pi are 3/1 and 4/1, every convergent of of Pi is a mediant sum of n * 3/1 and m * 4/1. Denominator 7 is nice because it's the sum of 3 and 4, but the path changes direction after 25/8, not at 22/7! So we can't define "best accuracy" as the turn of direction of the SB path. Various perspectives of comparative accuracy shifts the discussion to comparison of norms. In the SB-norm, each new mediant convergent of infinite zig-zag path is more accurate, because the resolution of the SB-metric increases by computation of each new row. When we estimate accuracy only in the metric of denominator 10 and it's exponents, 10mod7=3, but 10mod8 is neither 3 nor 4, which are the whole number "mediant factors" of pi. For CF paths, the mediant factors a/b and b/a reflect each other and can be considered basically same. SB-paths compute accuracy in the metric of all coprime fractions which SB-type constructions generate in their order of magnitude, instead of limiting the computation to a single specific denominator and it's exponents. With these criteria, I would say that SB-paths are the most generally accurate pure math metric of accuracy that we have. For applied math using a ruler with decimal markings (or hexagesimal or what ever), it's natural to estimate accuracy in terms of your ruler markings, but I think we are discussing here pure math on the most general level we can find.
    1
  18884. A circle is always a circle, many squares do not a circle maketh. The sum of the sides of squares on one side will always stay 4. Pi is still 3.14.
    1
  18885. Absolutely delightful!
    1
  18886. Danke!
    1
  18887. Congratulations Burkhard! Many interesting and surprising things, my favourite one is the 7 centuries old proof of the divergence of the harmonic series: so clean and neat! Regards Samuele
    1
  18888. Is this obvious that there is the same number of orange grey and blue tiles for the same reason that supposedly pi = 4?
    1
  18889. The expressions 6n +/- 1 produce all prime numbers greater than three, and many more composite numbers. If we knew exactly where the composite numbers would appear in these sequences, we could infer the location of all of the prime numbers. Am I understanding this correctly? Of what use would this be to anyone?
    1
  18890. I feel like there's a correlation here to the decimal expansion of 1/7 etc. of .142857R.
    1
  18891. all the way to the end????
    1
  18892. WoW very Deep . I have accidentally found a recursive formula to calculate the square root of any natural number. Does it already exist.? Thanks for your videos
    1
  18893. I found the entre video quite pleasent to follow and to derive the equations, maybe the matrices part was a little bit fast, and it need of some linear algebra intro, but exept from that all the rest was very nice and understandable. I have a masters in physics, so this Is part of the every day job
    1
  18894. the denominators in the partial fraction are all multiple's of 113 - three guesses on the multiples. ?
    1
  18895. There are only two free pegs at the start, and tower n moves to one of the pegs, so the only free point for the base of tower n + 1 is the last remaining peg, which you then put tower n back on top of. Thus, the total tower moving the direction of the smallest disc reverses each time you add a layer to the tower. In the base case example n = 1, we can see that moving the smallest disc clockwise moves the whole tower clockwise, and indeed, in the n = 2 case, the whole tower moves to the third peg, confirming that the tower movement direction matches the smallest peg direction for odd n. A tower of height ten has even parity, so we must move the smallest peg opposite the direction we want to move the tower. Thus, to move the ten-tower from A to B, the smallest peg must move from A to C to B to A to . . ., which is an upward torque (ccw when examined from above the table) in the example.
    1
  18896. I made it till the very end.
    1
  18897. Nice @
    1
  18898. Finished the video. Just got a bachelors degree in math in May, and I really enjoyed this video! The animations were very nice, and leaving exercises to the reader is an excellent way of making sure that I keep on paying attention; this is even the right amount so that once I've done the exercise (good enough, with the hand waving you can do once you know how things should work) I still have the thread of the video. Its also the right length; much longer than an hour and I might have to split it in 2, and I'm not sure if I come back after such a long video. I also skip through the first bit (in 10 second chunks, just using the right arrow) since I already know most of it, but it still should absolutely be there, so that its more accessible, and other people can bail earlier. Yea, this is the kind of video that I don't know what I would change to make it better. Excellent job, and I'm excited to see you do more!
    1
  18899. A ternary computer holds the key to the universe 😏
    1
  18900. this is why you don't cut corners...
    1
  18901. I made it to the very end
    1
  18902. common Mathologer W
    1
  18903. 24:59 I'm guessing Euler's gamma must be larger than 0.5 based on that square simply because all the shapes of the extra bits above the graph are bulging outwards? If the pieces were exactly triangles, they would, by symmetry, fill exactly half of the square.
    1
  18904. They all have the same center of mass, don't they, if you assume each vertex has unit mass? ETA: Ah, ok. "Wait five minutes."
    1
  18905. Always delightful. The extension of the equation into another dimension in a way by adding that extra point…. Creating shadows of tetrahedra
    1
  18906. And of course you can’t make a 1, 2, 3, triangle as that collapses into a line. Losing a dimension.
    1
  18907. What a wonderful journey through mathematics land! :) I really didn't see that rectangular grid of equations coming at 38:10. I'm now assuming that if you start with a 9-gon, that there will be 3 constants that, together with 1, satisfy 4 equations. Super interesting video, which scratches my itch for integer sequences as well :D
    1
  18908. Circular slide rule. Used one in college, but do not remember the rubber band.
    1
  18909. Cool!
    1
  18910. How I summed 1 to 100. 49 pairs of numbers that sum to 100 (1+99, 2+98 etc.) then you're left with 100 and 50, add them all together gives 5050. Please keep doing these longer, more advanced videos. I always learn a lot from them.
    1
  18911. This feels very much like the p-adic numbers in logic. In a similar vein, isnt any finite sum actually an "infinite" series of 0's summed with your sum? like 1+2+3=0+0+0+.......+0+1+0+0+0+.......+0+0+2+3, so we can also rearrange this inifinte series to give a finite sum of 0, 1, 2, 3, 4, 5 or 6. But actually we can go even further since A-A=0, so we can substitute any number of 0's with A-A and we can now use the logic from this video to give any finite sum we want, though the infinite sum will of course reduce to 1+2+3=6.
    1
  18912. When you said "Bear with me!", I was expecting a picture of a bear...
    1
  18913. Circular slide rules are not that big a deal. I had one in high school before there were hand held calculators. I had a fancy straight one in college with extra functions such as trigonometry, very useful for my electrical engineering classes.. I also had an E6B when I was taking my flight training.
    1
  18914. The next freak identity is gonna be femcel they’re already talking about 4B Oh wait this is a math channel right
    1
  18915. I took the advice of getting an extra coffee and managed to follow with full focus till the end of the video, I'm loving these longer videos with their various degrees of complexity, and thank you so much for that final animation. I'll be waiting for more videos on the application of the Euler-Maclaurin formula!
    1
  18916. At 28'00", should be '+', not '-' ?
    1
  18917. Does this apply to removing the overround in betting markets where the probabilities add to over 100%? The overround is constant up to half of the implied probability? Compared to the alternatives of proportional (seems to take too much probability from the higher odds) or constant (sometimes zeroes out probabilities which doesn't work) The other alternatives (what I found reasonable) are to convert the probabilities to some other scale (logit, log log, arcsin √, etc) and take a constant from that. all are different ways of normalizing the odds to sum to 100% - logit corresponding to shifting the odds by a constant x factor for each probability - log log corresponds to taking each probability to a constant power - arcsin √ is about making the change in probability related to the standard deviation of the probability ( √ p(1-p))
    1
  18918. __ R Y B R R B Y Y B Y R B Y R B This operation is also found in the thinking machine: https://www.youtube.com/watch?v=27cLGuSYmxQ
    1
  18919. I actually followed that all the way through.
    1
  18920. Beautiful math for XMas; where X can be any complex holiday.
    1
  18921. The optimal overhang stacking was fascinating but the entire video was awesome!
    1
  18922. Strongly recommend checking out Matt Parker's video on the complex Fibonacci numbers! He shows what kind of cool things happen if you start plugging in non-integers to Binet's formula and there are some real left field surprises.
    1
  18923. 14:41 I'd bet the number of tilings is 666
    1
  18924.  @jespervalgreen6461  No such desires take hold of myself nor there are found among ones in my possession.
    1
  18925. Given a random polytope, if you perform this procedure to find the centroid for each face, and then connect the adjacent centroids with lines, will you arrive at some sort of standard polytope?
    1
  18926. So much in this video was super cool, but the bizarre maximal overhang towers were one of the few reveals in any math video I've seen that made me say "oh you've got to be kidding" out loud, so I vote for that.
    1
  18927. This video it is gold
    1
  18928. So what you're basically saying here is that all of this has nothing whatsoever to do with Christmas, except that you want to show off your beautiful Pi-shirt and needed a pretext to tell us something interesting about math on Christmas Eve. ;-)
    1
  18929. hmmm ... how about the pi fraction expressed in some other number base. What happens to the "coincidence" then? Is this purely a decimal thing?
    1
  18930. I missed a lot in mathmatics, and I'm not sure why. Thanks
    1
  18931. Hi, I post my answer before reading the other comments (about where to move the first disc): - if you have only one disc, you move it to the destination peg, - if you have two discs, you move the small one to the third peg, not the destination one, - so the rule is : if the number of discs is odd, you move the first disc to the destination peg, if it is even, you move it to the other peg About the rule for moving the other discs afterwards (moving the small one clock-wise, every even step, and for the odd steps : do the only move you can), I found it when I was 18, just observing what happens when you apply the recursive algorithm. I successed doing the recursive algorithm up to 4 discs, but very difficult with 5, and impossible for me with more. You don't know where you are any more. Thank you for this video. See you.
    1
  18932. Didn't have time to watch the original, so I guess I'll never know what's different :V
    1
  18933. 16:23 I haven't dug through the comments too much, but I think the answer you're looking for is that we're using negative powers of 2 when working through those fractions. The sequence on screen at the timestamp is 2^0, 2^-1, 2^-2, 2^-3, 2^-4. The fractions are ultimately part of the same sequence that produces the larger loop.
    1
  18934. For the record, I recognized it from when I learned binary math in college.
    1
  18935. 3³+4³+5³=6³ 5x5x4 wraps around 4³ to make 6x6x4+outer rim of 20 (still need to fill up the middle: needs 4x4=16 cubes) => you didn't use one layer of 5x5x1=25 cubes, so to get a 6x6x5 you're still left with 25-16=9. You still have the entire 3³ + 9 = 27+9 = 36. This is one layer of 6x6 cubes, use it to complete the 6x6x6.
    1
  18936. wow
    1
  18937. not really connectied, but I was reminded of that neat theorem that any quadrilateral can have a square shadow.
    1
  18938.  @Mathologer  and I was misciting the theorem: It states: every quadrilateral is a perspective image of a square. Its proof was the cover of “100 Great Problems of Elementary Mathematics”, which to be clear, means Elementary in the sense of Euclid.
    1
  18939. I love mathologer :)
    1
  18940. That's a very flexible rubber band!
    1
  18941. Ok, so I have implemented this https://imgur.com/gallery/NT4t7EH
    1
  18942. 24:28 The parent construction would give us a 0,1,1 "triangle"
    1
  18943. If anyone is going to figure out how to see in 5 dimensions, it'll probably be mathologer.
    1
  18944. Sorry if this was asked (too many comments to go thtough), but at 23:16, you can make x^400 in more ways, by using x^2 and x^4 from the first polynomial, or am I missing something here? Thanks for the great content!
    1
  18945. Because he didn't want anyone else to understand his writings. Marie de Chatelet translated his math into a context that could be understood and shared. Newton was a selfish, paranoid, secretive, control freak who may have stolen his conclusions from a German correspondent, Gottfried Wilhelm Leibniz, as you well know.
    1
  18946. Reclicked. 😄
    1
  18947. Goes over my head even if you do talk turtle
    1
  18948. 9 4 17 13
    1
  18949. 9:45 just from eyeballing the question, I can pretty clearly see that the answer is that you can make 15 different amounts (16 if you include 0) and they're all made in a unique way. What's more, if you have n coins, you can make every value in a unique way up to 2^n - 1 (Again, 2^n if you include 0). You can do this just like if you were counting in binary, which uses the powers of 2 to uniquely construct every whole number. Pretty cool if I do say so myself!
    1
  18950. That toymaker is very hanoi-ing.
    1
  18951. Interesting, but still, I got lost. But I really like this guy.
    1
  18952. Amazing. How about pentagon or hexagon or over else?
    1
  18953. I wonder where they would have applied this pi approximation in 14th century.
    1
  18954. in this sequence (a x c) squared +2(b x d) squared =( (a x d) + (b x c) ) squared directly relating to rotation of fibonacci sequence into new matrix, transform as in first sequence, find next unique pythagorian integer solution
    1
  18955. 9:10 I'm genuinely surprised by your statement, as I did learn this in elementary school.
    1
  18956. You went to a good school
    1
  18957. Thanks
    1
  18958. I'm one of those kids who never saw either of these proofs in school. The only math I enjoyed was Euclidian geometry because the proofs were so neat.
    1
  18959. i'm farmiliar with the second proof and I only learned it in my senior year of highschool in my multivariable calculus class. (i probably have learned it before but just don't remember it)
    1
  18960. ...23, 37, 60
    1
  18961. I am a PhD Physic student, and I am still learning from your videos.
    1
  18962. 4:40 Large square = C^2. Small square = (A - B)^2 Large square = small square + 4 (ab square triangles) C^2 = (A-B)^2 + 4(AB/2) c^2 = (A-B)^2 + 2AB Develop (A-B)^2: (A-B)^2 = A^2 - 2AB + B^2 So: C^2 = A^2 - 2AB + B^2 + 2AB ... C^2 = A^2 + B^2.
    1
  18963. There is not only "one way to make 0 "sense")
    1
  18964. 17:32 - No, I never thought that was a coincidence. Just looking at it it has to be. When you add a new thin surface of thickness dr to the outside of the sphere, you MUST increase the volume by <surface area>*dr. The surface area MUST be the derivative of the sphere's volume. It couldn't be anything else. You could make that argument without even knowing the formulae for the area or the volume.
    1
  18965. -½ is larger than -1
    1
  18966.  @Mathologer  Nothing, just funny. But in the part "π exactly", you can find any number with this method, like the square root of 196 for example. But how long will it take you to find the first value that is larger than 14? In the part "Riemann's rearrangement theorem", you say it can convert to any sum, whatsoever. 😂 Have a good one.
    1
  18967. Love the animations. Great work and artistic flair!
    1
  18968. Why is the Riemann's Zeta(1) not equal to gamma then?
    1
  18969. Rotation proof (swiveling)
    1
  18970. Musicians can get the term "harmonic series" confused with how mathematicians define it. A musical "harmonic series" is more like what mathematicians refer to as a harmonic progression - in a nutshell, the infinite set of divisions of an ideal string that vibrates when the string is driven to do so: the whole length, 1/2 of the length, 1/3, 1/4, etc. There's quite a bit of interesting math in this as well as the infinite sum harmonic series; analysis of which is pretty important in very basic rhythm, but also theories of temperaments and other pitch systems.
    1
  18971. Grüße aus Deutschland, eines der besten videos. daaannnkeee
    1
  18972. Web implementation including all the steps: https://bustercopley.github.io/aztec/
    1
  18973. I liked the color proof much better than the swivel proof.
    1
  18974. Bachelor’s degree in physics, did a lot of math, up to and including topology (mostly point set with applications to fixed-point theory and Brouwer’s fixed point theorem using simplices). Started masters in theoretical physics, quit due to poor health and I’m now doing part time teaching math to high school students. Enjoying life as it is and not thinking about returning to school, but I love the videos and maybe someday if I feel like it I’ll return to higher studies, but for math rather than physics. Straightforward to follow with some effort, thank you so much! I’d love to see some videos on more advanced abstract algebra or even topology. Keep the crazy content coming! Much love from Sweden.
    1
  18975. I really enjoyed the section where you related the harmonic series to the gamma constant. I learned about it briefly in a calculate class but this video gave it much more context.
    1
  18976. Burkard, thanks for sharing the ubiquity of this highly intuitive principle. There may be an application to distributional evenness of modular sequences. James Grime has a card trick on Numberphile implicitly relying on the pigeonhole principle (cf video "James <3 a card trick"), which pops after this video. Question on chapter 2: For the observation that we can guarantee 4 pigeons in some hemisphere by tilting the equator seems plausible given that we can assign pigeons on the great circle to either hemisphere. If we suppose all 5 pigeons are coplanar on a great circle, and evenly spaced along it, then without this criterion is it impossible to pick another tilted equator such that a hemisphere contains our 4 pigeons?
    1
  18977. Alright answer for the puzzles! EDIT: Corrected some typos and semantics First puzzle The n^1 numbers grow like 0 =; 1 + 2 = 3; 4 + 5 + 6 etc The n^2 numbers grow like 0 =; 3^2 + 4^2 = 5^2; 10^2 + 11^2 + 12^2 etc Therefore the n^3 numbers would grow like 0 =; 5^3 + 6^3 = 7^3; 16^3 + 17^3 + 18^3 = 19^3 + 20^3; 33^3 + 34^3 + 35^3 + 36^3 etc... As an extra, the third line would be 16^3 + 17^3 + 18^3 etc The logic for the third line and so on is that, the first term of the third line is equal to the sum of twice the first plus the second term (without raising them) of the second line, for example: 1 + 2 = 3 4 ... 4 = 1.2 + 2 3^2 + 4^2 = 5^2 10^2... 10 = 3.2 + 4 Therefore we can say that the first term of the cubes is 5^3 + 6^3 = 7^3 N = 5.2 + 6 therefore N = 16 For the next line, it's the sum of the first and second term 9 = 4 + 5 - For the first power 21 = 10 + 11 - For the squares So, for the cubes we'd have... N = 16 + 17 = 33 Second Puzzle Since we know from the second line of the squares that 10^2 + 11^2 + 12^2 = 13^2 + 14^2 and that both sides are equal to 365, this means the answer is two (730/365 = 2) Third Puzzle Too lazy to do that by hand!
    1
  18978. Would a ‘no 8’ series also converge following a similar logic?
    1
  18979. Made it to the end, so a comment for the algorithm. You are getting close to obtaining the Golden YouTube Button... :)
    1
  18980.  @Mathologer  probably YouTube using the carrot on a stick method on you... keep up the good work though, the Talmudic bankruptcy problem was fascinating!
    1
  18981. Here in India, it was kind of just a random (but still pretty cool) formula that appeared in high school then lost forever since trigonometry came to be. I always wondered back then how exactly you would derive it so thanks for answering my very old question!
    1
  18982. When we were learning about logs in higher school, our teacher divided us into groups with a slide rule each and our assignment for the day was to figure out A) how to use it and B) why it works. A really fun lesson that sticks with me to this day!
    1
  18983. 15:58 k1=1 2*(k_n)^2=(k_(n-1))^2 sum(n=0,5)=1+2*sqrt(1/2)+4*sqrt(1/4)... =1+sqrt(2)+sqrt(4)... A=5
    1
  18984. Hmm, beautiful relationship! I wonder if your final point about Fibonacci numbers regularly including Pythagoran triples like that, that there is a way to draw the sequence using lines of integer length in a way that results in these right angle triangles appearing...
    1
  18985. right by witch lens
    1
  18986. Can you make a video on pingala sir!! He is the father of what known as Fibonacci numbers and he is kinda invented binary numbers too by lagu guru method of chandas shastra of Sanskrit
    1
  18987. Now if only life was this easy. It is... for well behaved functions. Now try something that oscillates really fast. ;-)
    1
  18988. 5,4,3,2,1💣 You got me anxious and dead in chapter 6💀⚰️. Vote for chapter 6, no 9, 6🤯
    1
  18989. Thank you. I always wondered how the continued fraction entered into the house number problem.
    1
  18990. The most memorable was the Euler–Mascheroni constant
    1
  18991. I have a question for you if there is any proof of fermat's last theorem other than andre wiles' proof.
    1
  18992. Missed ya❤️
    1
  18993. The title made me shed a tear 😭
    1
  18994. I made it to the very end,essentially because I checked the video length once the 'credits' started rolling, and thought eight minutes remaining seemed a bit long.
    1
  18995. Has anyone else noticed the animation at the beginning of the video, where the triangles and squares and star are run together, that it looks like a hypercube?
    1
  18996. This really shows that adding and subtracting are the only operations we actually need
    1
  18997. Wonderful video!
    1
  18998. And sitting out of shot is a scowling Isaac Newton muttering about fluxions!
    1
  18999. I am not into mathematics at all but found your video so interesting, being an artist. Thank you
    1
  19000. 28:00 This reminds me conceptually of Fourier transforms, how any wave is the sum of several or infinitely many simple sine waves at different amplitudes... And the closer the wave is to a sine wave, the fewer sine waves is needed to describe it; and here the closer to a regular pentagon it is the fewer component pentagons needs to have any "amplitude". In fact come to think about it, I guess you could probably describe most closed polygons using Fourier transforms on polar coordinates somehow... I'm terrible at doing maths in practice, so I don't know how to even start proving or disproving my hunch though...
    1
  19001. 29:52 Lol.. I guess my feeling was somewhat justified
    1
  19002. Sometimes I wonder if I could have been quite good at maths if I had my ADHD diagnosis earlier* and had good maths teachers** I could probably have been decent at some kinds of maths. ... Well, technically if you consider computer programming a form of applied maths, I actually am, since I'm a professional programmer. * Btw, Ritalin is absolutely no cure for ADHD; but it somewhat increases my tolerance for boredom and massively increases my ability to hyper focus on things that are interesting without getting distracted and reduces my sloppy mistakes. That's the main indicator to me when I'm working alone on a problem if I forget my meds: without the meds (and without built in syntax checking in the editor) I average about 10-20 syntax errors and logic errors (usually inverted boolean logic) per 100 lines of moderately complex code; when I started taking Ritalin in the middle of my Computer Science degree, I realised I now would have less than one syntax error on average for 100 lines; and only only inverted logic in just relatively complex boolean constructions. I could occasionally even write a whole page of code that compiled first try! So whenever I forget my meds I usually first notice it because my code starts to have more silly mistakes and I really struggle with boolean logic. When I'm forced to attend boring meetings and talk to people I notice forgetting my meds almost immediately because I start to get painfully bored, and yawn a lot and feel like falling asleep, and zone out from the discussion. ** Not like that idiot teacher I had in primary school who yelled at me and my friend for completing the whole first year maths exercises booklet in just one evening, and then punished us with more boring exercises of the exact same difficulty... The first lesson of many of primary school that anything above the bare minimum of effort is not only not appreciated but punished... It wasn't until I was 16 started "videregående skole" (which is roughly translates to high school I think, it's the first kind of school in Norway where you can choose your specialisation and choose between several vocational subjects or further general studies meant for further University studies) when suddenly the difficulty jumped from absolutely no effort required (beyond showing up, half paying attention in class and just naturally perform well on tests); to actually requiring some effort and practice to do well, I started realising this lesson from primary school was wrong.
    1
  19003. Thank you for the video sir 🙏🙏 love from India 😊
    1
  19004. My favorite part was finding out that the Sum of no 9's, proves that the sum cannot be larger then 80, after learning that it's finite value is actually smaller then 23. I can't help but wonder that that in it of itself would make for a very interesting video, Thank you.
    1
  19005. Loving your videos 🥰🥰🥰
    1
  19006. Some years ago I used to just only pay cash in bakery, and, when I could not pay exact price, I would, like everyone else, pay more and get change back. Then, often I noticed: if I would need to pay that same exact price now, I would now be able to do so. And, I wondered, what starting conditions and what prices would realize these situation. Because, often, next day, I bought same for same price in bakery, and, be able to pay exact price. After all, one does not like to accumulate change. But one does not want to be without enough change either.
    1
  19007. I got 55 starting from far right and moving left, assuming we could
    1
  19008. Incredible proof
    1
  19009. Proffing 4(good-bad)
    1
  19010. Okay so why does gamma have to be greater than 1/2? BTW my favorite part would be the gamma part (or the Oily-Macaroni constant ;) ). Been seeing that a lot in calculations regarding factorials, never realized that it actually came from the Harmonic series :D Well, consider each of the blue partial difference and the small rectangle it occupies. Draw a diagonal along the curve and notice that the blue part completely covers one half of the rectangle made by the diagonal splitting, and then has some leftovers. (To be rigourous, we can say that the 1/x curve is concave up and so all straight lines lie above it). Since each blue part is greater than 1/2 of its rectangle, and all the rectangles add up to 1, gamma is greater than 1/2. Obviously :D
    1
  19011.  @Mathologer  Thanks for answering. 🙂
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  19012. beautiful as always ❤️
    1
  19013. I don't get the shirt 😞
    1
  19014. Rearranging terms can affect the formulation when you're not defining the amount you're using properly. 1 + (1/2) - 1 = 1/2 ( 1/3) + (1/4) - (1/2) = 1/6 (1/5) + (1/6) - (1/3) = 1/30 The top 2 terms can approach infinity faster than the bottom term. If we were to write it out, we would actually get; Sum_{1}^{infinity}(1/k) - Sum_{1}^{infinity/2} (1/k) This means the numerator would still be increasing for a half-infinite amount of times while the denominator has reached it's goal. According to my maths, a half-infinite can be expressed by (-1)!/2 = (-1)(-2)(-3)!/(2) = (-1)²(-3)! = (-3)! so I'll be using this term ahead. We can say our denominator has reached the point of -(1/(-3)!) when our numerator has hit the point of 1/(-1)!, or 1(0) If we continue to add to the denominator until we get to a full (one) infinite, we get -(1/((-3)! + 1)) - (1/((-3)! + 2)) ... = Until we get to... -(1/ 1/((-3)!) + (-3)!) = -(1/(-1)!) = -1(0) This would finally cancel out our numerator properly. Of course, this would give us a ton of expansion, but the same thing happens in the numerator and it all cancels out. If we don't add the rest of the terms into the denominator, our numerator has an additional Fraction of an Infinite amount of Zeroes to different powers. These infinitesimals combine to ln(2) over the course of the the half-infinite summation. Basically if you add 1(-3)! A half infinite amount of times, you get, (-3)! × (1/(-3)!) = 1. If the denominator slowly increases to (-1)! Along the way, it can't quite reach 1. 1 / ((-3)! + 1) = 1 / (1/2(0) + 1) = 1 / ((1 + 2(0)) / 2(0)) = 2(0) / (2(0) + 1) Divide both sides by 2 = 1(0) / (1(0) + 1/2) = 2(0) 1 / ((-3)! + 2) = 1 / ((1/(2(0))) + 2) = 1 / (((1 + 4(0)) / 2(0)) = 2(0) / (4(0) + 1) Multiply by 2 (1/2) × (4(0) / (4(0) + 1) Sacrifice blood (1/2) × ((1(0)) / (1/4)) (1/2) × 4(0) = 2(0) 1 / ((-3)! + 3) = 1 / ((1 + 6(0)) / 2(0)) = 2(0) / (1 + 6(0)) Times 3 = (1/3) × (1(0) / (1/6)) = (1/3) × 6(0) = 2(0) But eventually it'll reach halfway to (-1)! Which is ((-1)! + (-3)!) / 2 = ((1/0) + (1/2(0)) / 2 = (3/2(0)) / 2 = 3/4(0) = (3)(-1)!/4 1/0 - 3/4(0) = 1/4(0) Already at this point we can see that (((1/4(0)) × 2(0)) + (1/4(0) × (3(0)/4) / 2 < 1/2(0) × 2(0) = 0.5 + 3/32 < 1 = 0.59375 The rate it decreases also suggests ln(2) could be in range (I'm just not doing the calculations). We just need to remember that a lot of it will be close to the point where it's equal to 1(0) and our average will favor 1, increasing from 0.59375 to ln(2) as we continue to calculate.
    1
  19015. Thank you so much for this. It was truly a delight.
    1
  19016. There's something wrong with the funny picture on your T-shirt! Pythagorean theorem does not work in GR, except in the flat Minkowskian metric (SR as a degenerate case), and even so, with one sign of the four flipped: x²+y²+z²-t²=s². Einstein totally pwns it!!! 😄
    1
  19017. Is this process reversable? What happens if you start with a regular pentagon, for example, and apply the steps backwards - do you get inifinite possible results or a very specific one, or perhaps something meaningless?
    1
  19018. I completed the video, and it was interesting. I have a background of a B.S. in mathematics and B.S. in electrical engineering. Nothing was too complicated to follow - though I didn't go through the "homework". I'm a math purist, and don't really care for the question: How would this be used in real life? Instead, I'm interested in the variant: What insight does this give? The bit about how Euler estimated significant digits was very interesting. What other math proofs have used this concept as a tool?
    1
  19019. Each one of these counterparts are gems, thank. you for showing them to us! I went on a Wikipedia tangent after this, Steinbach was truly an interesting person
    1
  19020. Nearly fell down my chair at around 15 mins. Woww
    1
  19021. Haha, nice t-shirt
    1
  19022. My answer to question at 16:11 is ... After 11 months I'm still trying to figure it out... Edit: maths is like drugs with no side effects actually
    1
  19023. Calculus wouod be easy for me if i knew it where and how to use it
    1
  19024. Question for Mathologer: can you explain to us why the following set of parametric equations gives a 6-fold symmetry curve, given the fact that 6 nowhere appears in the trigonometric parameters ofthe equations, whereas 7 and 17 do? x = cos(t) + cos(7 t)/2 + cos(-17 t + Pi/2)/3 y = sin(t) + sin(7 t)/2 + sin(-17 t + Pi/2)/3 with t = 0.. 2 Pi
    1
  19025. Fantastic! Zagier and Firsching's proof shows that the simple search algorithm presented at the start of the video is guaranteed to find a solution. So does this mean that this is a constructive proof?
    1
  19026. I enjoyed the video to the end. Matrices not my thing. I'll take your word for it. BS Electrical and Computer Engineering. MS Management Science. At the end of my engineering career, I returned to school and became a high school math teacher. Having doubly retired, I now own and drive a taxi in my hometown and remain a math hobbyist.
    1
  19027. Back in college I did my senior project involving some number theory-esque shenanigans and I came across the imaginary golden ratio by complete accident. Simply phi^2=phi-1. Lot of cool stuff with that guy too.
    1
  19028. thank you very much
    1
  19029. Um vídeo sobre matemática com legenda em português eu não posso deixa de agradecer. Muito obrigado!😂
    1
  19030. 36:44 It seems completely obvious that a random cube packing should be random everywhere, yet it is not
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  19031. Marvelous!
    1
  19032. Basically infinity / 2 = infinity and inifnity * 2 = infinity so it did not change.
    1
  19033. 402th like
    1
  19034. See the cat! see the cradle??!
    1
  19035. Looks like the Crushmetric pens and water bottles I get ads for.
    1
  19036. Thanks!
    1
  19037. Gotta say, "terrible aim" was my favourite insight here. Though "thinner and thinner" came close second.
    1
  19038. He looks like Jimmy Savile 🤔
    1
  19039. Imagine if the Koreans had discovered a pen and paper. They could do a 99x99 square
    1
  19040. Nice to see Picard sharing his intergalactic knowledge on youtube in his twilight years.
    1
  19041. The No Integer among the harmonic series always surprises me, as we will get really really close to some integers, but will never hit one.
    1
  19042. 8:33 So, later on w would come ot further udnerstand that villi, the folds of an object increase its surface area, this is great becasue this allwos nutrients to be absorbed, because of cute folds in folds, and htus infntie space, about 15 years ago I made fun of solar panels, stating that most of the light could be harnessed ina similiar struture of villi method increasing your surface area expotentially increasing the output of a solar panel to rival that of a gas engine on a sunny day...., but with a more optimal storage for energy via tension mechanics of better battery system the purpsoe woudl be to make it lighter, but I don't know.. lol. plants do alot with sun energy and can sotre it to do amazing things, so battery relation could be drastically improved.
    1
  19043. The infinite precision wheel would require infinite radius.
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  19044. Just one request : " never stop making videos " love from India
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  19045. When making change for $10, are you including $2 and $5 bills? Is it better to have coins of 1¢, 2¢, 5¢, 10¢, 20¢, etc. or 1¢, 3¢, 10¢, 30¢, etc.?
    1
  19046. Is it possible that Fermat had a similar geometric proof in mind for his last theorem but has eluded mathematicians ever since?
    1
  19047. 9:03 : To get A(a) + A(b), we will have to squash the A(b) shape by 1/a, which means we stretch it out by a factor of a to retain the same area. But the width of this shape is actually b-1. So we get A(a) + A(b) = A(a + a*(b-1)) = A(a + a*b - a) = A(a*b).
    1
  19048. Logically, the volume of a cone must be the volume of a cylinder of the same size minus a hemisphere. Very inspiring video!
    1
  19049. am I a nerd for being into this stuff?
    1
  19050. "you guessed it" hmmmm...no LOL i feel like an idiot, i love math, i just survived calc 2, and i love math, just that i have to work for a living. but anyways es ist schön, danke
    1
  19051. NOICE German, das mag ich. Also really interesting stuff tbh. I always like to watch Mathologer.
    1
  19052. Difficult to choose great among equals 🐾❗. Just for voting sake I go for Gamma (1/12) thing. Always wondered how things would shape in other than decimal system. Animation is excellent. And presentation is lucid, 👍
    1
  19053. @22:52 Is it the psychotic cat staring at me?
    1
  19054. 23:49, shouldn't this be the last 10 numbers?
    1
  19055. @UCgDTCVIb7x25OrckfbOLQ8Q Ah, I see! Yes, that isn't brought up until the end of the argument!
    1
  19056.  @MuffinsAPlenty  I thought the argument would still work but I wanted to know if it should be the last ten ones or if I missed something, thanks anyway
    1
  19057. Astonishing!
    1
  19058. I like the video, but I didn't catch it all (somewhere after minute 20 it became somewhat unclear in some parts) in the first watching so I'll definitely have to rewatch. Videos being long is a good thing if you ask me. That way you can give us more info.
    1
  19059. Rubik cube right?
    1
  19060. Just WOW. Again. Last time it was odd numbers summing to square numbers. Thanks.
    1
  19061. I WAS ONE OF THE STUDENTS HE MENTIONED I FEEL SO SEEN
    1
  19062. Where is the video of all his little laughters chained and looped over a glitchy dubstep beat?
    1
  19063. There isn't really any particular advantage. It's probably because of how we often think about these as remainders modulo 4. If you were told to find the remainder of 19 after dividing by 4, for instance, you would probably give the answer "3" instead of "-1". But it is useful to recognize that -1 is congruent to 3 modulo 4. :)
    1
  19064. Found the book. Fun! But I have been lucky: I was taught calculus in about that way, and so never learned to hate it. :) Most of what I use in my daily job I learned in highschool, not college.
    1
  19065. Such a nice video. Plugging in details would be interesting. Using the Bohr-Mollerup representation of gamma function, this infinite product form of sine is basically proving that Γ(x).Γ(1-x)= π/sin(πx); 0<x<1. This can be done using the Fourier Series of cos(ax) for a non integer a; and then using the series and an uniform convergence arguement to calculate the integral form of Γ(x).Γ(1-x). A lot of effort but worth it. Can be taught to an undergrad. But the Weistrass Factorization Theorem for entire functions.....well that is a long way.
    1
  19066. Great video... thank you! Reminds me of string art I did as a kid. However, I was under the impression that Tesla's fascination with 3,6,9 was specifically regarding harmonics and resonance (not vortex math), right?
    1
  19067. this is awesome! ... but i think you missed a simpsons gag when you assumed pi was equal to 3 :)
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  19068. I watched to the end.
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  19069. I liked the tower a lot, and the parabolic tower was really cool, but what I found most memorable is the question it self. As a math theacher I will be asking it a lot from now on, because it really promps a very mathematical reflection on what are the truely interesting results in a subject.
    1
  19070. Gamma, by far Also fun to think about why it's larger than 1/2... spoilers? It's all about triangles isn't it?
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  19071. When Mathologer does YouTube #Shorts...
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  19072. That music makes me feel like I'm remembering the life of my partner and I at their funeral
    1
  19073. Mire ha mi y si digo una digo mas ha sobre de Hoyente ha distansia pertenesiendo a X y cuantas tanbien mujer los deribantes o primos (xeda halgo y es el ORDENAMIENTO ) de todas matematicas no entendibles cual primero mueve X despues cuantas el Tao es aire PI EL PRIMER suma y si es permanente entraria Reloj y cual de dos de entender donde de construye Linia Corta es igual ha local conponente ordenado de comunicasion de jenerasion continuada hasta dias de hoy
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  19074. Reply to your final question: Majored in economics with a minor in statistics (+ some funky stuff I have done in order to do some amateur research along a university professor, but nothing past ODE's and linear algebra, in general). I think I could follow ok until you start invoking the math gods all around, and I am sure you are skipping over pretty insane stuff there. Anyways, please keep on escalating on video insanity, I totally love it! It encourages me to learn more math on my own too. Thank you so much for everything!
    1
  19075. In the card trick, how do we communicate the order? Whether it is clockwise Or anticlockwise
    1
  19076. Agree before you perform the trick (let's say we always choose clockwise), and choose whichever of the two cards allows you to do this. In the example given, that's why they had to choose to keep the 10, because 6 clockwise to 10 is shorter than 10 clockwise to 6. If they wanted to do the reverse version of this trick, they would have kept the 6.
    1
  19077.  @Eagle0600  oh yes thanks Dunno why I couldn't think of it
    1
  19078. Logarithmus Naturalis? Latin? All this time I was happy. For once it was the french short designation for Logarithme Naturel...😂
    1
  19079. "Explaining the bizarre pattern in making change..." which Mathologer did, very well. "...for a googol dollars" Hope you get paid!
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  19080. Hell's Equation
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  19081. I respect this man for interacting with his fans
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  19082. Very nice
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  19083. So…my immediate thoughts are about why the (x+2) corresponds to n-cubes (squares, cubes, hypercubes, et seq) and what would be the corresponding polynomial for n-simplices 🤔 Sensible questions or just my mind burping at random?
    1
  19084. Most memorable: no integers. It is just so counter intuitive when you think about it
    1
  19085. I knew the second proof and learned it as an adult in the book "the music of pithagoras"
    1
  19086. What a fascinating explanation of logarithms! I am old enough to have used a slide rule at school. My Faber Castel 52-82 was a prized possession. It was lent to a neighbour's child who was a couple of years younger and I never saw it again - still annoyed at my mother for that! I occasionally search for one but never find one (not in Australia, anyway)!
    1
  19087. Nice. I'm particularly happy to see motivation for sinh and cosh. I've never seen these before.
    1
  19088. I like your shirt!
    1
  19089. Those opening chords are Kate Bush's "Babooshka", right?
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  19090. The photo of Madhava looks like you, is that a coincidence ?
    1
  19091. Most memorable part: the connection between the leaning tower problem and the harmonic series. Had already seen an MIT demo of that, but had never attempted to do any calculations
    1
  19092. The harmonic series (and the reciprocals of consecutive integers) hitting no integers is surprising at first, but actually kinda obvious when u think about it.
    1
  19093. Fun! I have memories of folding construction paper triangles in half and noticing SOME of this…so I like the second proof. It’s more intuitive. But the colored angle proof was gorgeous!
    1
  19094. Infinite series paradoxes usually arise because, for it to make sense, you need to use parentheses around each pair of terms.
    1
  19095. 20:54 946/243 (^.^) I got it by writing a c++ code that handles numbers of the form a+b17^(1/2) where a and b are rationals. Most of it are overloading of operators so the calculation can be expressed in a nice way. I coded close to the bare minimum of operators needed. Here's my code for anyone interested: #include <iostream> using namespace std; int gcd(int a,int b){//greatest common divisor int r; do{ r=a%b; a=b; b=r; } while(r!=0); return a; } struct Q{//rationals int p,q;//numerator and denominator Q(const Q&x):p(x.p),q(x.q){}//constructor by copy Q(int n):p(n),q(1){}//an integer is a rational: itself over 1 Q(int p,int q){//constructor by specifying the integers p and q, simplify fraction int f=gcd(p,q); this->p=p/f; this->q=q/f; } Q operator+(const Q&x)const{return Q(this->p*x.q+this->q*x.p,this->q*x.q);}//addition of rationals Q operator*(const Q&x)const{return Q(this->p*x.p,this->q*x.q);}//product of rationals }; Q pow(const Q&b,int e){//rational raised to a non negative integer power Q ans(1,1); for(int i=0;i<e;++i)ans=ans*b; return ans; } Q operator+(const Q&x,int n){return Q(x.p+n*x.q,x.q);}//addition of rational and integer Q operator+(int n,const Q&x){return x+n;}//symmetry of addition Q operator*(const Q&x,int n){return Q(x.p*n,x.q);}//product of rational and integer Q operator*(int n,const Q&x){return x*n;}//symmetry of multiplication ostream&operator<<(ostream&out,const Q&x){//how to print a rational on console out<<x.p; if(x.q!=1)out<<'/'<<x.q; return out; } struct Q17{//binomial of the form a+b*sqrt(17) where a and b are rationals Q a,b; Q17(const Q17&x):a(x.a),b(x.b){}//constructor by copy Q17(int n):a(n),b(0){}//constructor by specifying the rational a Q17(const Q&a0,const Q&b0):a(a0),b(b0){}//constructor by specifying the rationals a and b Q17 operator+(const Q17&x)const{return Q17(this->a+x.a,this->b+x.b);}//addition of Q17 numbers Q17 operator*(const Q17&x)const{return Q17(this->a*x.a+17*this->b*x.b,//product of Q17 numbers this->a*x.b+this->b*x.a);} Q17 operator*()const{return Q17(a,(-1)*b);}//conjugate of a Q17 number: *(a+b*sqrt(17))=a-b*sqrt(17) }; Q17 pow(const Q17&b,int e){//Q17 number raised to a non negative integer power Q17 ans(1); for(int i=0;i<e;++i)ans=ans*b; return ans; } Q17 operator+(const Q17&x,int n){return Q17(x.a+n,x.b);}//addition of Q17 number and integer Q17 operator+(int n,const Q17&x){return x+n;}//symmetry of addition Q17 operator*(const Q17&x,int n){return Q17(x.a*n,x.b*n);}//product of Q17 number and integer Q17 operator*(int n,const Q17&x){return x*n;}//symmetry of multiplication Q17 operator+(const Q17&x0,const Q&x1){return Q17(x0.a+x1,x0.b);}//addition of Q17 number and rational Q17 operator+(const Q&x1,const Q17&x0){return x0+x1;}//symmetry of addition Q17 operator*(const Q17&x0,const Q&x1){return Q17(x0.a*x1,x0.b*x1);}//product of Q17 number and rational Q17 operator*(const Q&x1,const Q17&x0){return x0*x1;}//symmetry of multiplication ostream&operator<<(ostream&out,const Q17&x){//how to print a Q17 number on console out<<x.a; if(x.b.p!=0)out<<'+'<<x.b<<"17^(1/2)"; return out; } int main(int argc, char *argv[]) { int n=3; Q17 t0(Q(12,59),Q(18,1003)),t1(Q(5,18),Q(1,18)); cout<<Q(466,885)*pow(Q(2),n)+(-1)*Q(1,3)+(-1)*Q(3,5)*pow(Q(1,3),n)+ t0*pow(t1,n)+(*t0)*pow(*t1,n); return 0; } Enjoy. (^.^)
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  19096. Tangentially related to 2x2 = 2+2: there is this video of Gromov https://youtu.be/bgePMb8wxv0?si=DmTf2wzbIr1F21f8 observing that 4 = 2+2 in 3 ways. That is, there are 3 ways to partition a 4-element set in two 2-element subsets. The fact that 3 is smaller than 4 is unique to 4, and produces a surjection from A_4 to A_3 - the alternating groups on 4 and 3 elements. This shows that A_4 is not simple, 4 is the only number where this happens, and this is responsible for many weird things in dimension 4. In another direction, I first encoutered Sophie Germain primes accidentally when learning group theory in my undergrad. There is an elementary proof of simplicity of the Mathieu groups M_11 and M_23, which actually gives a general simplicity criterion for subgroups of S_p, p prime https://www.jstor.org/stable/2974771?origin=crossref. When I read this I wondered if this ever works for proving simplicity of the alternating group A_p. If I remember correctly, it works precisely when p is a Sophie Germain prime!
    1
  19097. Half dollars are not standard, not anymore. The only time you ever see them are as a kid, when your grandpa shows you "something neat".
    1
  19098. 13:50 trick question.. The answer to this question depends on whether N is even or odd (I think)
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  19099. Wonderfull gamma !!!
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  19100. The curious convergent series made my day! Baillie's paper was amazing. I'm now going to look at your presentation of Kempner's proof <3
    1
  19101. F(.5) + F(1.5) = F(2.5), also F(phi) + F(phi+1) = F(phi+2), and yes negative n as well as all complex numbers, it holds everywhere
    1
  19102. Absolutely wonderful.. Such a beautiful way of picturizing n showing such an interesting mathematical concept like Cheese!!!! Just like that..!!!
    1
  19103. How does one pick a favorite?? I think mine might be the “ugly” overhanging block solution, since I immediately thought such a stack must be possible from trying the same thing with LEGO as a kid. I am halfway through reading “A Dingo Ate my Math Book” and I absolutely love it; some of the puzzles have even kept my kids’ attention.
    1
  19104. More spooky numerology: (values computed using Wolfram Mathematica) e^pi = 2 3.14 0 69 ... ln(2) = 0 . 69 314 718 ... e = 2 . 718 ...
    1
  19105. The most memorable part for me might be the visual proof that gamma has to be less than one by sliding over all the blue regions to the left and also could see immediately that it has to be greater than half as well. Amazing video!
    1
  19106. Knowing something 'in your bones' but then seeing a visualisation of a mathematical proof by contradiction just makes me so happy. Thank you for this.
    1
  19107. On the second Pythagorean tree. We are told our host "hadn't looked to closely at it". Isn't it a self-similar fractal, so it doesn't matter if you look close or from far away? ;)
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  19108. (01:05) It is beautiful. I spent a year working with a professor of mathematics at the University of Maryland, through postal correspondence, back in the 70s, learning to resolve "cos 2π/7" to an algebraic expression. I (sadly) don't remember many of the details. (It's been over 30 years. Maybe even closer to 50 years. Shhh.) I know we used something from Galois to develop a third-degree polynomial in one variable, which when equated to zero could be solved to yield the desired result. Solving the cubic was the hardest part, what with the various substitutions, but it was enriching, rewarding and really strengthened my fingers and wrist on my right arm. When I saw your video in my suggestions, I did not hesitate to click and watch. Thank you, Mathologer, for the nostalgia. By the way, I love your shirt. I really like artistic expressions of trees and to see one with square roots is just cool.
    1
  19109. Der Schwiveling proof. Edit: I knew it. Those coloring proofers are philistines.
    1
  19110. Came up to 40m and lost you there. I'm sorry for my self. I need to go back to my old books :(
    1
  19111. Back here for my yearly dose of Mathologer
    1
  19112. I have to admit this Talmudic/garment system has its merits, as a lawyer I'm still in support of the proportional system. Why? Law most be as clear, simple and concise as possible. The proportional distribution is very easy to formulate and compute, and it's still not unfair, while this garment solution is very hard to put in words, instractions, without using mathematical formulas, tables or being involved in all these complicated game theory, water level, infinite spreadsheet, level of happiness kind of shenanigans.
    1
  19113. I like the movie proof more.
    1
  19114. woah, it's not often that i upvote a 30 minute video in the first minute, but that cube thing is just too cool!
    1
  19115. Soon 1M
    1
  19116. Him and Euler both have these floppy hats.
    1
  19117. Similarly, 1+2+3+4+5... can be expressed as (2-1)+(4-2)+(6-3)+(8-4)+(10-5)... or 2+4+6+8+10... -1-2-3-4-5... Following the same logic, because every positive digit eventually gets canceled out, it cannot be larger than zero. In fact, once each even number is canceled out, only the odd negatives remain, leaving the sum to approach negative infinity. (Which is a little off from -1/12, the unequivocally correct answer 😆) It's interesting how infinite sums result in different answers depending on how they are arranged.
    1
  19118. Wow in just few min from someone i never met helped me to pull together answers to years worth of questions i thought i would never get an answer too or even understand the outcome put together that way thanx soo much....
    1
  19119. I like both this short video and the long ones.
    1
  19120. The most amazing fact that I took from the video was, that the limits of the no 9s series and the 100 zeros series are so close to each other even though, they cut out numbers from the harmonic series so differently.
    1
  19121. Missed you ♥️
    1
  19122. The "no 9s" sum was rather unsatisfying for me. Surely that is only a base 10 thing? That doesn't feel like real maths to me.
    1
  19123. couldn't you cover an infinity wall with an 8 volume paint and still have 8 vol of paint left over?
    1
  19124.  @Mathologer  True! But as a scientist and not a mathematician, my mind goes to quick estimations first, especially when challenged with a time limit :)
    1
  19125. This was perfect timing as use in an art project that illustrated (or better said, experienced) the process of filtering news and the reinforcing loops that lead to polarisation. Arctic polarisation! I manually transcribed the A(20) into Excel for a graphic (see 20omg20 IG) and it was laborious! Coders please assemble.
    1
  19126. My favorite is Tristan fractal
    1
  19127. Oh, I do love that T-Shirt!
    1
  19128. Those POV Ray wizard skills 😱
    1
  19129. None of my povray videos made it to the final version. Most of the 3d animations were from frames rendered in Asymptote.
    1
  19130.  @andrewkepert923  darn, still very skillful compared to anything I've tried to make
    1
  19131.  @NonTwinBrothers  The secret sauce is knowing the equations. Maths FTW.
    1
  19132. What's am I doing wrong. I can solve the second problem, but when I transform it in the subsequent equation (10^2 +^11^2 + 12^2 + 13^2 + 14^2) its resolve for 730 and have the right answer. But When I Ty (15^2 + 16^2 + 17^2 +18^2) it gave me another number 1094, so I cant apply the rule equalizing the two equantions.. What am I doing wrong?
    1
  19133. Amazing video❤
    1
  19134. On the card trick, I must have missed something: you can signal that it is 4 spots away from the 6 of spades, but how do you know whether it's 4 above (10 of spades) or 4 below (2 of spades)?
    1
  19135. plz do a series on very large numbers! (googology)
    1
  19136. This is similar to a few project euler problems I think.
    1
  19137. A blessing from the lord!
    1
  19138. Strictly speaking, if you cut out two black and two green such that one of the corners is orphaned, then no, you can't tile every block on an 8x8 chessboard.
    1
  19139. (x+2)^4 having an 8x^3 term lines up with the net of a tesseract consisting of 8 cubes (a common net being 4 cubes stacked on top of each other, with 1 cube on each remaining face of the cube that's second from the top, similar to the cross-shaped net of a cube).
    1
  19140. At 22:41, I'm seeing one rather annoyed cat - maybe jealous of the attention the one on your shirt is getting?
    1
  19141. Absolutely amazing sir,
    1
  19142. I don’t want a copy of one of your books. I want a book.
    1
  19143. 9 4 17 13 is the answer I got. Also fits the Fib rule.
    1
  19144. Best teacher
    1
  19145. With that thumbnail i was suddenly shocked by the Medussa's relation with calculus
    1
  19146. Still eagerly wait for a video about the quartic equation like the one you did for cubic equations, my favourite of the channel. Also, another about Gaolois and why there is no formula for grades higher than four
    1
  19147. This is math porn and I'm having a braingasm.
    1
  19148. Anyone able to prove what the highest number of chess pieces which can be on the board with no way for either player to win would be? Note that if you are able to sacrifice a piece to make an opening this does not count. The highest number I've been able to achieve is 22 but I can't prove it either way. Here's the best board I could do with what I've got: ⬜⬛⬜ ♔ ⬜⬛⬜⬛ ⬛ ♗ ⬛⬜⬛⬜⬛◻️ ⬜⬛⬜⬛⬜⬛⬜⬛ ⬛⬜⬛⬜⬛⬜⬛◻️ ♙ ⬛ ♙ ⬛ ♙ ⬛ ♙ ⬛ ♟ ♙ ♟ ♙ ♟ ♙ ♟ ♙ ⬜♟⬜♟⬜♟⬜♟ ⬛🐴 ♝ ⬜ ♚ 🐴⬛◻️
    1
  19149. Regarding the random generation of the tilings - does every tiling have an equal probability of being chosen, or are some more common than others? Like if I were to choose via the homer simpson method (using unique tilings only) and via the slide and fill, are the distributions the same? What about the more physicsy method of randomly selecting edges to become pairs until a full tiling is reached (ie throwing out all the paths that lead to incomplete tilings). It's not at all obvious to me that As a physicist, I smell that we're talking about artic circle theorems having particularly high entropy here, so that distinction seems quite relevant. Most importantly though: Frohes neues! Hoffentlich bleibt das Virus fern :)
    1
  19150. First go: miracleAnimation[b_, a_] := Module[{\[Epsilon] = 0.05, q, r, diff}, q = a/b; diff = b - a; r = q - \[Epsilon]; Manipulate[ Animate[Show[{Graphics[{Thick, AbsolutePointSize[10], Circle[], If[showInnerCircle, {Gray, Circle[(1 - q) {Cos[\[Theta]], Sin[\[Theta]]}, q], Point[(1 - q) {Cos[\[Theta]], Sin[\[Theta]]}], Line[{(1 - q) {Cos[\[Theta]], Sin[\[Theta]]}, (1 - q) {Cos[\[Theta]], Sin[\[Theta]]} + r {Cos[(1/q - 1) \[Theta]], -Sin[(1/q - 1) \[Theta]]}}]}, {}], If[showAllPoints, {Orange, Point[Flatten[ Table[(1 - q) {Cos[\[Theta] + 2 \[Pi] n/diff], Sin[\[Theta] + 2 \[Pi] n/diff]} + r {Cos[(1/q - 1) \[Theta] + 2 \[Pi] m/a], -Sin[(1/q - 1) \[Theta] + 2 \[Pi] m/a]}, {m, 0, a - 1}, {n, 0, diff - 1}], 1]]}, {Orange, Point[(1 - q) {Cos[\[Theta]], Sin[\[Theta]]} + r {Cos[(1/q - 1) \[Theta]], -Sin[(1/q - 1) \[Theta]]}]}], If[showPolygons1, {Red, Line[Table[(1 - q) {Cos[\[Theta] + 2 \[Pi] n/diff], Sin[\[Theta] + 2 \[Pi] n/diff]} + r {Cos[(1/q - 1) \[Theta] + 2 \[Pi] m/a], -Sin[(1/q - 1) \[Theta] + 2 \[Pi] m/a]}, {m, 0, a - 1}, {n, 0, diff}]]}, {}], If[showPolygons2, {Blue, Line[Table[(1 - q) {Cos[\[Theta] + 2 \[Pi] n/diff], Sin[\[Theta] + 2 \[Pi] n/diff]} + r {Cos[(1/q - 1) \[Theta] + 2 \[Pi] m/a], -Sin[(1/q - 1) \[Theta] + 2 \[Pi] m/a]}, {n, 0, diff - 1}, {m, 0, a}]]}, {}]}], If[showCurve, ParametricPlot[(1 - q) {Cos[\[Phi]], Sin[\[Phi]]} + r {Cos[(1/q - 1) \[Phi]], -Sin[(1/q - 1) \[Phi]]}, {\[Phi], 0, 2 \[Pi] a}], {}]}], {\[Theta], 0, 2 \[Pi] a}, AnimationRunning -> False, AnimationRate -> 0.1], {{showInnerCircle, True, "inner circle"}, {True, False}}, {{showAllPoints, False, "all points"}, {True, False}}, {{showPolygons1, False, "polygons 1"}, {True, False}}, {{showPolygons2, False, "polygons 2"}, {True, False}}, {{showCurve, True, "curve"}, {True, False}}]]
    1
  19151. Poor hairless homeless people in Australia :(
    1
  19152. Loved the video, 3rd year Mathematics and Physics undergraduate, you should do more of this stuff, I followed everything untill the adjustment you did to f(1) instead of f(0).
    1
  19153. Excellent: animations worked well, gave an insight into Math-investigation-behaviours. Extension a little quick i suppose. I wobbled in paces but 50mins! Time-flew.
    1
  19154. Muito bom gostei você explica bem e claro 🎓👍🏻
    1
  19155. Continued fractions are evil and I am happy I never encountered them at university (physics)
    1
  19156. Programmer answer: split the dominoes in two squares, each marked respectively Now, another thing, to remove the clashes, if a square points to something pointing back, it will delete itself Now, the cells turn into their neighbor pointing to them, is the top down? then is the right left? is the bottom up? is the left right? Now, the refill algorithm is up to you
    1
  19157. Waouh! Despite its beauty, this was a tough one.
    1
  19158. That shirt 🤣
    1
  19159. I would follow you anywhere Mr. Mathologer sir!
    1
  19160. Love this
    1
  19161. Well, this video is quite on the complex side for me, but nevertheless, your energy and humor propagates love for math in a beautiful way. Thank you :) Also, I would like to see more of the faces that names you are mentioning. Its good to have them in mind!
    1
  19162. 37:25 You are a cool guy !! Because that looks like a giant 3D hollow cube, and inside the cube the small unity cubes are stacked without "leaving holes". That means that above the floor you lay down a first layer of unity cubes. The layer has no holes. Than a second layer without holes. And so on. Now if you take the projection of that building looking at only one face, you see all the faces painted with one color, and you can do the same for other 3 faces. So the number of colored tiles if the square of the length of the hexagon and it's the same for all 3 colors.
    1
  19163. The rectangles at 18:05 and 20:30 cause an optical illusion that makes them look bulged out on the sides due to the fanning out of the triangles. It's not great for an illustration, because when you first pit them together, it made me question whether it was a perfect fit. Fortunately, you covered the maths behind it later to prove it.
    1
  19164. Wasn't FTC the fact that a contiguous function has to pass through each value between the end points?
    1
  19165. I'm a very simple man. I see new Mathologer video. I watch it.
    1
  19166. Another amazing/interesting video! Can't ask for a better Christmas present.
    1
  19167. That's definitely a miracle! Wow!
    1
  19168. Amazing 😁🤯
    1
  19169. The Fitch-Cheney five-card trick is wonderful. I had heard this trick described but had no idea how to make it work. Thanks for filling yet another hole in my understanding of the universe! I can't wait to share this trick with my friends! As for 7 2-digit numbers for which no two subsets have the same sum, how about 64+(0,1,2,4,8,16,32), i.e. (64,65,66,68,72,80,96)? If the common sum were S, then the 6 LSBs of the binary version of S dictate which of the last 6 elements are needed to yield S. Non-overlapping subsets with a common sum therefore cannot use any of these 6 elements, leaving only the first element. Since you can't form two non-empty non-overlapping subsets with only 1 element, no two non-empty non-overlapping subsets have the same sum.
    1
  19170. Enjoyed this a lot. I'm not a big fan of base-10 quirks, unless they demonstrate or hint at something more general (e.g. the recently discovered uneven distribution of the base-10 digits of pi), but agree the no 9s series is interesting conceptually. Seems if there's a formula for the sum of the no 9s, then there might also be a formula for the sum of no X's in base B with X, B as parameters. (The combinatorics of no 8s are the same as no 9s; the individual term values are different of course.)
    1
  19171. This is one of those topics that appears simple on the surface but which has fascinating depths - however deep you dive.
    1
  19172. Aren't the angel lacing just the devil lacing but with the extreme point connected vertically instead of diagonally. Also, if you can curve them around the eyelids, then you can make them as long as you want. They are fun though.
    1
  19173. I was comfortable through calculus, then differential equations made my brain implode (1988'ish). I had a poor teacher with a super horrible to understand foreign accent. (yours is fine btw!) Now I'm much older and really enjoyed this video! Once you got to the Bernoulli numbers I couldn't wrap my head around the substitutions. POOF another brain implosion. Maybe if I watch it a couple more times. I wish I had you as a math professor back in the day!!
    1
  19174. bendin my mind video by video
    1
  19175. oh that dreaded +C
    1
  19176. Look at that! Pythagoras eats a triangle and spits out Euler. Wahr-sinh-lich!
    1
  19177. I made it to the very end :-)
    1
  19178. ❤️💟❤️😊🤗
    1
  19179. Hey Mathologer, could you do a proof that shows all shapes of a planet that could possibly give rise to the sun rising due east at every latitude and longitude (except the poles) on the same day? :- )
    1
  19180. Enjoyable and looking forward to the sequel!
    1
  19181. Not sure if this has been asked. When one scales the triangle from circle to hyperbola, the angle remains alpha, but why is the new green area still alpha (at 25:25 vs 25:27)?
    1
  19182. I am sure this has been done before but on the screen I noticed that the harmonic series was actual the sum of sums, namely the sum of geometric series of all of the primes. Since all geometric series have a solvable value then it is 1+ SUM of p to infinity where p is prime (p/(1-p))
    1
  19183. I was taught divisibility criteria for 2,3,4,5,6,9,11 (also 7 but I think this one is not worth it) with proofs and the remainder connection for 9 when I was in first or second high school year
    1
  19184. Never disapointed by Mathologer.
    1
  19185. Congrats for the subs.. & Merry Christmas... as a present I have an infinitely nifty, proof of irrationality of pi. I would like to show you though.
    1
  19186. Thank you for this video <3
    1
  19187. Euler's Brick out too
    1
  19188. We need godel's incompleteness theorem video !!!
    1
  19189. Most memorable was the analogy to fractal-like behavior when visualizing the no 9s infinite series.
    1
  19190. The one with multiplier 240 and modulus 7417 reminds me of a hyperbolic projection.
    1
  19191. I don't understand what you mean by someone reinventing the circular slide rule. I'm 74 yo, and I grow up with the slide ruler (räknesticka -- counting stick in Swedish). Do you really think that we didn't understand how it easily could be made circular? I have two different such items, alongside several heavily used linear ones. The problem with the circular ones was that they didn't fit in our breast pockets alongside our pens. But we used it by our drawing boards; because it was easier to use with compound calculations when you didn't have to slide back and forth with the inner, linear one.
    1
  19192. The weird curve rotating in the square reminds me of the rotor in a Wankel combustion engine.
    1
  19193. One small mistake @5:21. In the first proof we didn't shuffle all four triangles. The triangle in upper right corner is at it was originally, see 2:06.
    1
  19194. thank you this is a great visualization of Fourier transform and has enlightened me
    1
  19195. Tout simplement genial Merci
    1
  19196. First challenge: No, we can remove 4 tiles s.t. 1 tile is isolated from the remainings Second challenge: If both are odd, the ceiling function will be equal to (m+1)/2 and (n+1)/2. When j and k equals (m+1)/2 and (n+1)/2, the cosine will be 0. Since one of the factors is 0, the product equals 0 as well
    1
  19197. Thanks!
    1
  19198. WoW
    1
  19199. Most memorable: Leaning Tower of Lire followed by Bizarre Maximal Overhang Towers.
    1
  19200. The Sagarmatha of math channels
    1
  19201. The Bernoulli numbers were interesting. I did not come across them in my studies until here. They were mentioned only in passing. Need more of this stuff please.
    1
  19202. regarding problem 37:30 identify the three colors with three directions. then starting with any position, you advance in the direction prescribed by your position. since theres finitely many positions you must end back where you started. this cycle must have every direction equally often. every position has such a corresponding cycle, so all directions in general have to appear equally often. i guess it isnt super strict just a scetch but i have a feeling that this proof might work out if properly formalized.
    1
  19203. While the mystical explanations of the "vortex diagram" certainly seem a bit silly, it seems to me that it's a bit hasty to completely toss away their creators, given the ways in which mystical-seeming patterns and numerology associated with them have stimulated the development of number theory. For instance, as best I understand it, the perfect numbers were initially seen as significant due to numerological significance (e.g. Philo's claim that the world was created in 6 days because 6 is a perfect number), and (according at least to Harold Edwards in "Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory") it was an attempt to understand perfect numbers (or really, to more effectively search for Mersenne Primes) that led Fermat to his celebrated little theorem, which is really a cornerstone of modern number theory.
    1
  19204. I vaguely remember going through that in calc 2 in high school.
    1
  19205. As a Talmud scholar and practitioner (an observant orthodox Jew), I must thank you for this satisfying explanation. The contested garment rule is one of the very known rules in Judaism. And it's extra satisfying to see the full reasoning behind it and the game applications resulting from it.
    1
  19206. mind blowing, great stuffs
    1
  19207. Ive made it to the very end :D
    1
  19208. Huh, I am positive someone else has thought of this before, but it just occurred to me there is a very easy way to prove that the harmonic series diverges. s=1/1+1/2+1/3+1/4+1/5+... s/2=1/2+1/4+1/6+1/8+1/10+... s/2+s/2=1/2+1/2+1/4+1/4+1/6+1/6+... Each term of 2×(s/2) is less than or equal to the corresponding term in s, meaning 2×(s/2)<s, but that just means s<s which is impossible. Unless I am having a brain fart and making some easy logical mistake, that might be one of the easiest proofs I have ever seen for the divergence of a sum.
    1
  19209. Omg I love your t shirt!!!!
    1
  19210. One of the best channels on YpuTube — you’re a gifted teacher.
    1
  19211. Dude! I thought I found this myself and noone else... I hate you newton! You had to discover everything yourself, right?!
    1
  19212. So if I have it right, any nonzero amount of mathematically perfect paint can cover an infinite surface.
    1
  19213. 6:47 Actually, the puzzle mentions that the street is at least fifty and at most five hundred houses long. Edit: Okay, nevermind.
    1
  19214. I was able to follow pretty much everything; I have a doctorate in physics but I had never heard of the Euler-Maclaurin formula before. Amazing stuff!
    1
  19215. Does someone have all the AHA answers in the comments? I am really lost, but would like to know the answers and how they were calculated.
    1
  19216. End of video: PhD student in experimental fluid dynamics. Background in data analysis, statistics and PDEs, mostly. Nevertheless, these problems are still absolutely beautiful when explained with well thought-out animations!! Some teachers in the past taught me to hate math. You teach me to love it =)
    1
  19217. I feel it could be ultimately easier to use a 45° turned hyperbola x² − y² = 1 when defining cosh and sinh: then we just use x and y coordinates as with the circle. It requires a little extra fuss but in the end it should be clear to the viewer that it’s the same thing, the same √2 scaling and 45° rotation but applied to different things (and stretching is now diagonal, which can be helped visually by adding diagonal lines, I guess?).
    1
  19218. 2+4=6 2*4=8 but wait 1+1+2+4=8 1*1*2*4=8 whoa
    1
  19219. Colouring proof
    1
  19220. The noise in the background sounds like cicadas to me.
    1
  19221. 1st one folding
    1
  19222. iv been following this topic,sum of powers and related journals for a while...Never thought that someone would come up with such a lovely video which demystifies all that and makes it accessible for a broad audience.Well worth the length. @mathologer Salutations from India :-)
    1
  19223. At approximately 26:00, is it "falling power" or "Forling power" or anything like this?
    1
  19224. i have not been taught this yet
    1
  19225. Tristan's illustration of the thinning out the harmonic series was really nice way to show how when numbers get bigger there is just more 9s in them. Somehow it reminded of Benford's law.
    1
  19226. It's criminal that this isn't a collab with Hydraulics Press Channel. Squashing steel pipes length-wise turns them into Schwartz Lanterns. Same surface, same circumference, lesser height, the pipes sink the extra area into the triangles.
    1
  19227. This was one of the things I suggested to Burkard as an ingredient for this video. I’ll have to try again - there must be a volume-related mathologer video where igniting weather balloons full of acetylene has some mathematical significance.
    1
  19228. Do the original polygon and the final regular one have the same area? At first glance it seems so! Fantastic video BTW (as usual)!
    1
  19229.  @Mathologer  My ADD medication hadn't kicked in.
    1
  19230.  @dinoboyoutuification  :)
    1
  19231. For the glasses: Notice that the far left and right 2x3 blocks have exactly 3 ways of tiling, independent of every other tile. Afterwards, let us do cases on the bridge of the nose. This is inspired by the choice that there are 2 ways of tiling the hole around the glasses, either removing a column from the bridge of the nose or the 'frame'. There are 3 cases: 1)the middle 4x2 rectangle is completely filled in. 2)It is a 3x2 rectangle filled in, and on either the left or the right side it 'encroaches' on the glasses 3)only the center 2x2 rectangle is filled in, and on both sides it 'encroaches' on the glasses. 1)There are 5 ways of filling in the middle 4x2, which forces leaving a 2x2 block on the left and right hand side. There are 2 ways of filling in each of these, for a total of 5*2*2=20 tilings 2)There are 2 ways to chose the side that the 3x2 rectangle goes to, and then 3 ways of filling in the middle 3x2. On one side, this leaves a 3x2 block, and on the other a 2x2 block, for a total of 2*3x3x2=36 ways of tiling it. 3)There are 2 ways of filing in the center 2x2 rectangle, which leaves a 2x3 rectangle to be filled in on both sides. That is 2*3*3=18 tiling Thus, there are a total of 9*(20+36+18)=9*74=666 tilings (nice).
    1
  19232. Sir (a³+b³+c³)=∆ then √∆ = a+b+c A,B,C=1,2,3 ; 0,2,2 ; 0,1,1
    1
  19233. @Mathologer  True. However, if we make it 100 and 200 instead, then I would say getting 1/3 and 2/3 would be fair, but with this system it would amount to 1/2 each. The larger claimant gets 66-50=16 less, making it a 8% loss.
    1
  19234. @Mathologer  It is very hard to remember something so specific over 9 years, but the earliests memories seem to revolve around your cartoon videos. Futurama and simpsons? Also a video about 9.9999... = 10. No idea which was actually first, but I do remember being surprised at how much math there was in the otherwise low-brow and often boring shows.
    1
  19235.  @Mathologer  If I may, why did you want to know? Just curiosity?
    1
  19236. @Mathologer  I remember that! Is it the one with the triangle klein bottle where three are connected together? And with a klein bottle mug? You put rubix cubes into alchemist bottles as well in that video, I think. I can not for the life of me remember how, though. Magnets?
    1
  19237. Dat crazy little German giggle in between the Dr. Strangelove monologue.
    1
  19238. I have to say I was fascinated by the leaning tower of lire, but more by the way the hanging over part could be constructed in the area above the curve of a parabola, leading me to wonder about whether it would work for any other graphical ideas r if it could be represented graphically. If the centre of mass of each block was taken as a coordinate point, then using a cartesian plane you could more easily compose or compare possible solutions to the problem.
    1
  19239. Nice Shirt!
    1
  19240. This was fabulous. Yours are among the very few long YouTube videos I'll watch. My math background is pretty high (PhD) but your rate of climb keeps folks watching as long as possible. Question about this particular video: you mention the relation s_3=s_1^2 to point out how you can bootstrap the computation of the s_k. But you only mentioned linear recurrences after that, and never returned to algebraic relations. Future video?
    1
  19241. Loved this video. I still remember learning convergent and divergent series. So many cool proofs. I like the fact that the partial sums of the harmonic series manages to miss all the positive integers > 1. Kempner is a wonderful question and result. Still there's a lot of other surprises you brought out. Thank you so much. My children both grew up in this era and the YouTube videos like this are amazing inspiration and learning tools.
    1
  19242. Shrinking shapes are indeed beautiful!
    1
  19243. The second fundamental theorem of engineering "2 = e" was finally proven, now!
    1
  19244. Very cool. It's basically a precomputed logarithm table
    1
  19245. What about other bases? Why is base 10 so special?
    1
  19246. I liked "Cubies" ;D
    1
  19247. (φ +1)^1 = φ+(1^1) Is there a similar bracket shift rule for r and s?
    1
  19248. Thank you for the video, it’s a great Christmas miracle. I am still watching, but second challenge is because if m and n are odd, the last term in the sum will be cos2(pi/2) which is 0. This of course only happens for odd sizes of the bird, since the coefficient of pi is strictly less than 1, and cos only goes to zero at pi/2 in this range
    1
  19249. Damm. I'm doing real analysis and now I have this nice gift!
    1
  19250. In programming olympiads there is following trick: if you know that you have three values in array, and you have two of them and you want to pick the one that is left, then third is sum - a - b. So, first that I thought was sum of 0,1,2 is 3, so formula for third is 3-a-b. Then, what about same colors? 3-1-1 = 1, works. but 3-2-2 = -1 so I thought ah, it's 2 mod 3, and 3-0-0 = 3 it is 0 mod 3. But had not enough time to figure out fast guess, because this is not enough to make guess during few seconds. So I guessed yellow just because it was most common :D The thing about this rule works for any array, it's easy to prove. for example array has 2 7 7. You have 2 and 7, so third is 16-2-7 = 7. Or you have 7 7, 16-7-7 = 2. This is to avoid matching which is slow.
    1
  19251. മാധവന്‍ ആരാ മോന്‍ ❤
    1
  19252. Most memorable: ramanujan sum of Hn is γ
    1
  19253. Thumbnail got me thinking it was a Numberphile video ahah
    1
  19254. Truly beautiful!
    1
  19255. (Feedback:) I discovered the sum of power recurrence formula at age 17, and then don't know what to do with them. Then I learned about the Bernoulli numbers in an Ada Lovelace biography. came to know about the Riemann zeta magic in another math book on the topic. Lastly I came to the majestic Euler–Maclaurin formula, at the age of 29. Thank Mayan gods that I found Mathologer :D
    1
  19256. FWIW: Tower of Hanoi is one of the first things CS students are introduced to in college, specifically when discussing recursion. It's usually some kind of homework assignment or something. So I'd be very surprised if there wasn't already some Tower of Hanoi code already out there somewhere. Though, they usually only work with the 3 peg version. I'm not sure how common more pegs are. So that part will probably challenge your programmer friends :D
    1
  19257. fun enough indeed
    1
  19258. I enjoyed all of it, but especially the leaning tower of maths, which I could show to my little brother.
    1
  19259. Phi also solves the Liar Paradox. In Boolean algebra, "a and not a is true" becomes x*(x+1)=1 (mod 1). So, x^2 + x + 1 = 0 (mod 1), because multiplication is AND and adding 1 (mod 1) is NOT. Z2/(x^2+x+1) = F4, and the two new elements are essentially phi and -1/phi (mod 1), so if 0 is False and 1 is True, .618... is Both true and false, the solution to the Liar. The 2nd new truth value is Neither. In #RM3 #logic, Both and Neither are isomorphic, so we identify them. It's a symmetric monoidal closed category. Conjunction is left adjoint to implication.
    1
  19260. For my money, the most remarkable fact on the harmonic series you present is that, the approximation of the sum for large n —which by all rights should be this hideous monster of an expression— is just ln(n+1) + gamma.
    1
  19261. I just watched the first minute and came up with a proof as suggested. Here it is: First notice that the proof in the other direction is very easy: Given a circle and three lines of equal length that lie on the circle you will always get this colour pattern. (1) This is because when looking at two lines at a time there is a symmetry axis between them so the line segments have equal length. For the desired direction it is to show that given three lines of equal length that overlap with the specified colour pattern their endpoints lie on a circle. First consider two of those lines. Their four endpoints lie on a circle because we know that for three points there is a circle that goes through them and the fourth is also on it because of symmetry. Now it needs to be shown that the third line is also on the circle. For any given angle of the third line we know that a line exists that is on the circle and follows the colour pattern because of (1). But there is only one line of this angle that satisfies the colour pattern so it must be the one that lies on the circle. Why is there only one? You have two degrees of freedom to move the line: Parallel to itself and orthogonally to itself. The position along the orthogonal direction is fixed by the colour condition. With this fixed the position along the parallel direction is also fixed by the colour condition. qed
    1
  19262. A minor side remark: When I first saw the identity (a^2+b^2)*(p^2+q^2)=(ap-bq)^2+(aq+bp)^2, I wondered how one can conceive such a formula. Using complex numbers it is actually quite easy to derive. The "The product of two sums of two squares is the sum of two squares"-identity (https://youtu.be/00w8gu2aL-w?t=1346) can be derived via complex numbers using the norm function: N(z)=z*c(z), where c(z) denotes the complex conjugate of a complex number. Due to commutativity in the complex field, it is easy to verify that N(z)*N(w)=N(z*w). If z=a+i*b and w=p+i*q then N(z)=a^2+b^2 and N(w)=p^2+q^2 and finally, N(z*w)=(ap-bq)^2+(aq+bp)^2. The equality obtained this way is the same as in Mathologer's video, except for the switched '-'-sign. Well, the left hand side doesn't change when b is replaced by -b and so does the right hand side. Frohe Weihnachten!
    1
  19263. Takk!
    1
  19264. Another Gem From Mathologer. Hats Off !
    1
  19265. The water visualization is fantastic! Thank you for making this video! Unfortunately, the video seems heavy on the visualization algorithm and light on the maths. That is, what is the general algebraic formula that represents this method of distribution? Also, supposing that the "volume" was non-linear, such as the diagram at 3:40, how can such a formula be expanded to account for this? Or even the diagram at 6:44 (though that could presumably be just "once all vessels at this level are full", logic).
    1
  19266.  @Mathologer  Ok, thank you.
    1
  19267. Mathologer is Dr. Evil's real son
    1
  19268. This video is amazing. Even though I'm not at all a fan of maths you make me understand it!
    1
  19269. 11 4-width triangles?
    1
  19270. I personally think that you should go more into details about Sophie Germain's life. For example, she was inspired by the story of Archimedes death where a Roman soldier speared him in rage when the geometry-obsessed man insisted, "Do not disturb the circles!" She also might have saved Gauss' life when she intervened with a French general in charge of the siege of the city where he lived during the Napoleonic wars. The story of her unveiling with Laplace is also quite interesting.
    1
  19271. 18:42 there's a mistake - the right most pair is not actually aligned, it's a mirror image
    1
  19272. Very very cool! I happy that you can explain about this hard for colleges theme so clearly! If I was a math teacher I`ll use it for studying of my students)
    1
  19273. Dear Mr. Mathologer Could make a video about Lebesgue Integral and What it really is and What are the applications of it in a very simplistic form. It will be very kind of you.
    1
  19274. Brilliant! This is the original definition of hacking.
    1
  19275. I can see why gamma is >1/2 It’s because each blue bit takes up slightly more than half of the rectangle it sits in.
    1
  19276. Fun with homomorphism.
    1
  19277. ..mathlogger rocks..and amazes us ..spellbound again..
    1
  19278. That was as joyful as playing with Lego.
    1
  19279. I think some short videos between the long ones is nice.
    1
  19280. But how does this work on the surface of a sphere? 😬
    1
  19281. I can bet know that my school teachers don't even know What is divisibility test?
    1
  19282. How do you attach 180 degree ears? They will not create any corners.
    1
  19283. You can think of them as being like whale ears: Flat and essentially invisible. :)
    1
  19284. The most memorable was that the sum of no 9s is actually convergent.
    1
  19285. 22:00 Sooo. There must be a distorted map made from the globe with its "north pole" at the Point Nemo. Am I right?
    1
  19286. Why Silvio Berlusconi is in the video cover? 😂
    1
  19287. Reminds me of when the youtube channel Flammable Maths "proved" that ∞! = √2π
    1
  19288. I'm a senior in high school, but generally have a really good mind for calculus (I came up with a formula to sum an Arithmetic sequence before we were even taught what an Arithmetic sequence was) and I really enjoy your videos, you give very good explanations that I can see visually and can almost always understand everything in most of your videos. I was with you until about halfway through where I couldn't fully understand the link between Pascal's triangle an the Bernoulli numbers. Ignoring this missing gap I still had a fairly decent grasp on how they were implemented into summing infinite series, and since I haven't seen your video on the -1/12 concept (yet) it gave a better understanding as to how those rediculous sounding answers were calculated. After watching all 50 minutes I'm realizing that I probably pay more attention to your videos than I do my actual math classes
    1
  19289. 6:43 Funny how that works. Karl Petr discovers something, and a few decades later the discovery is named after 2 chosen persons.
    1
  19290. Put the whole video on mute and it looks like he is deaf PS make that like button blue if you subbed to my channel
    1
  19291. I made it to the very end !
    1
  19292. Great one! Btw, "explodes" to infinity is the overstatement of the month. I mean, "slithers" to infinity is more accurate IMHO.
    1
  19293. The phi Ptolemy is almost the shape from prime numbers mod 10
    1
  19294. I want a copy of your book to experiment the stacking with books.
    1
  19295. Searching a sorted list only takes log_2 N steps. Yet, maintaining a sorted list can be costly in terms of space/time whilst adding and removing values. Using the concepts of the Tower of Hanoi puzzle, multiple sorted lists can be maintained while amortizing the cost of merging the list over a large time scale and using little additional space. (If one doesn't care about space then a double-linked list has constant time insert/removal.)
    1
  19296. Slide rule in a circle?
    1
  19297. Nice Weeks Manifold! ^.^
    1
  19298. This video was really one of the best so far. :D
    1
  19299. Here is how to see that gamma is greater than 0.5 at 24:35 Notice that all the little blue regions that sum to gamma are curved outward on the bottom. Straighten out the curved sides so each region becomes a triangle. Each triangle has base 1, and the total height of all triangles is 1. Hence the blue area is now 1/2. By straightening out the curved sides we decreased the area of each region so gamma > 1/2.
    1
  19300. 9 | 4 RRRL 17 | 13
    1
  19301. It makes me think of rotation around a circle in which you cut it and twist 180 % and reconnect
    1
  19302. Great and intriguing video but now I'm wondering about how equations - with x and x1 in IR - like (x+2)**x1; or even (x+2)**x can be geometrically interpreted.
    1
  19303. Thank you for this great video!
    1
  19304. The exact fraction for the Hanoi "average" formula when n = 3 is apparently 946/243. I'd be interested to examine other values of n to see if the denominator is always a power of 3... ;)
    1
  19305. What comes next is of vital importance. Let us dig deeper and see what the education system's hostility changes and what it cannot, or will not, change.
    1
  19306. So does the example at 4:35 also proves that there's infinities bigger then others too?
    1
  19307. Interesting. I haven't seen anybody use ... since my calculus prof used it to do calculations in his head which we were supposed to somehow keep up and follow as he blasted through formula after formula. I did however sense how the convolutions added surface area by adding facets without reducing the cylinders height. It was a bit intuitive if your gifts include spatial aptitude even without the math. Thanks, this was quite fun.
    1
  19308. Before watching the video, inspired by the thumbnail, I just worked out that (x+n)^2 - (x-n)^2 = (x^2 + 2nx + n^2) - (x^2 - 2nx + n^2) = 4nx. If x=12 and n=1, that's 13^2 - 11^2 = 4x = 48. With x=12 and n=2, that's 14^2 - 10^2 = 8x = 96. Put the two together and you get (14^2 + 13^2) - (11^2 + 10^2) = 4x + 8x = 12x = 12^2. And of course you don't need to know x from the start - you can evaluate ((x+2)^2 + (x+1)^2) - ((x-1)^2 + (x-2)^2) = 12x and then choose x=12 to get the 12^2. This is really just using the binomial expansion to do what the animations did at the start of the video, and it's easy enough to e.g. sum the differences for n values 1..3 giving ((x+3)^2 + (x+2)^2 + (x+1)^2) - ((x-1)^2+(x-2)^2+(x-3)^2) = 4x+8x+12x = 24x and choose x=24 to get 24^2 for the square in the middle and so on. For the linear case... well, the binomial theorem for raising to the power 1 is a bit trivial, so the difference (x+n)^1 - (x-n)^1 is simply 2n. For a sequence of 3 you need x=2 at the center, for a sequence of 5 you need x=2(1)+2(2)=6 at the center and so on. I didn't think of that until I saw the line case animations. As for cubic - well, the binomial expansion has no problem with that. (x+n)^3 - (x-n)^3 = (x^3 + 3 n x^2 + 3 n^2 x + n^3) - (x^3 - 3 n x^2 + 3 n^2 x - n^3) = 6 n x^2 + 2 n^3. The central value for any sequence we could want is x^3 = sum n from 1 to k: (6 n x^2 + 2 n^3), so for k=2 we get x^3 = 18 x^12 + 20. My calculators numerical solver gives me a single real solution x=18.06131003 approx, and two complex solutions x=-0.03 +/- 1.0518i approx. So there it is - it can be done, sort of. Nothing wrong with the Fermat proof that it can't, of course, except that only proves it's impossible for a sequence of integer cubes. The general pattern is that you derive a polynomial for the central x and then solve it, but for powers 3 and greater you won't get any integer solutions for x. I've checked the x=18.06131003 solution and I get ((x-2)^3)+((x-1)^3)+x^3 = 15001.411054ish and ((x+1)^3)+((x+2)^3)=14999.411053ish - close enough to blame the numerical polynomial solver and rounded output, but with a suspicious number of decimal places matched given there's a difference of 2 anyway, so there's a chance I messed up. *EDIT* I did mess up - x^3 = 18 x^12 + 18 (not 20) so the real solution is x=18.055521628 approx. (complex solutions -0.027 +/- 0.998i). Using the real solution ((x-2)^3)+((x-1)^3)+x^3 = 14986.2ish and ((x+1)^3)+((x+2)^3) = 14986.1 ish, with no worryingly matched decimal places, so I can confidently chalk the difference up to numerical results and rounding.
    1
  19309. Fröhliche Wheinachten!
    1
  19310. All zeros? ... It took me about 20 seconds to figure out that it was 2. ... 6^3 and about 517/75 ish
    1
  19311. I only finished highschool and liked Maths. I taught myself calculus and I could understand 80% of video on the first watching - though i had to remind myself what inverse matrices where all about - I am not good with matrices. Obviously Bernoulli numbers where tough for me, but my biggest problem (and that's me problem, I know) is binomial expansion. Higher than squares and I'm dead. But I could follow. Didn't understand that last part about E-M formula with Yellow and Aqua colored functions. I didn't get where the aqua part came from.
    1
  19312. It seems that his wording implies that Z/1 is not a valid ring. But why? Aside from being totally boring, it seems to fulfill the requirements. Wikipedia calls it the "zero ring", where the additive and multiplicative identities are just both 0.
    1
  19313. My favourite part was the fact the blue sum was equal to a constant gamma and that it is related to the Ramanujan’s famous sum -1/12
    1
  19314. Excellent video. Clear, engaging, and teacherly! I'm a practicing engineer and have always found physical/geometric analogs (using my mind's eye) to be very helpful with mathematics, so your animations feel like old friends. The derivation of the E-M formula was clear . . . and artful!
    1
  19315. I DECLARE BANKRUPTCY!
    1
  19316. They told me in school: do it one time and that is enough. But I always did it untill i get 9 or less ans tgat annoyed the teacher
    1
  19317. This reminds me of a Numberphile video which stated that in McDonalds, 43 is the highest number of McNuggets that you can't buy with the boxes they offer.
    1
  19318. Monsieur Polster, vous, vous-même êtes un miracle!
    1
  19319. I'm actually an extraterrestrial with B fingers. I was very surprised that you extended the line of reasoning for me to understand! I'm pleased to inform you that as a result of this you will be spared in the upcoming earth invasion Gronxcolictal Galactic Federation! 👽 Also haha silly humans and their numerology.
    1
  19320. Most memorable to me: the partial sums of the harmonic series do not include a single integer
    1
  19321. Where can I get shirt like your?
    1
  19322. Nice analysis of the presentation Rusty
    1
  19323. Interesting: 3.7 K likes, 37 dislikes... At 0701pm, 31 January 2021.
    1
  19324. Now I will be a Jenga master
    1
  19325. On your Patreon page, the link on "paypal.me/mathologer" is not correct (not directed to paypal.me/mathologer, but to www.paypal.com/paypalme/my/profile)
    1
  19326. Damn! How could you be so smart? What are you eating any way? Let me guess: ADD? I mean this as a complement. I can only dream of having your understanding. At my age to start my education all over again??? "Justify that your calculations are correct (If my understanding is correct)." There are a lot of mathematicians out there that need to understand that. Math can describe reality or fantasy.
    1
  19327. Turtle mathematics he tortoise
    1
  19328. Have not seen something that brilliant in a very, very long time. Thanks a lot!
    1
  19329. Thank you for the videos my children love them and I want to wish you a happy holiday
    1
  19330. 18:48 thought you were going to remove the zero so it lines up with the prior diagrams spoken about.
    1
  19331. Why 4K+3 and not 4k-1 ?
    1
  19332. i know this video seems very innocent and seemingly benign, but this features stuff that is by far the most mindblowing in my recent memory.
    1
  19333. Can't you move the point on the smaller circle further to the edge to make the star more straight?
    1
  19334. F 1+F 1= F1try that one and put it in 3D have fun.
    1
  19335. But when will calculus explain you?
    1
  19336. Its such a bumer these amazing people didnt have the computation power we have today
    1
  19337. OK I'll watch this tomorrow. Already had enough maths from Spivak today :/
    1
  19338. Earworm @ 17:26 pleeeez?
    1
  19339. Nice :D
    1
  19340. What happens if you use the inverse fibinocci sequence or the super fibinocci sequence or its inverse? What do those give you?
    1
  19341. Did you notice that you could multiply the bottom two numbers and subtract the product of the top two to get the third number in the Pythagorean triple? Only 12 min in so I don't know if you mentioned it but i didn't want to forget
    1
  19342. Why is it obvious that the sum of the blue areas on the left (equal to gamma) is larger than 0.5? Because this is a stack of rectangles. The part above the diagonal in each rectangle is equal to half the area of the rectangle itself. But the bottom line of the blue area "sags below" the diagonal in all cases. Therefore the blue area is larger than half the rectangle. If we stack them all up, until they fill the 1x1 area on the left, the total area has to be larger than the sum of all the "half rectangles". So it is larger than 0.5. QED
    1
  19343. You know things are about to get crazy when matrices are introduced only halfway into the video...
    1
  19344. That is so elegant!
    1
  19345. I freaking love the Mathologer.
    1
  19346. The swivel proof is my favorite.
    1
  19347. No circular slide rule in the mercy store?
    1
  19348. Did you change the intro music? I preferred the original :)
    1
  19349. even number of disk : smallest disk goes counterclockwise to go to B odd number of disk : smallest disk goes clockwise to go to B
    1
  19350. Video title got Me Triggered
    1
  19351. I feel like throwing the chalk was entirely unnecessary.
    1
  19352. Can you write the number 2 or 3 as the sum of the reciprocals of two distinct positive integers ? Well, is the answer "no" ? The lowest integers are 1 and 2 and 1/1+1/2 = 1.5 < 2 so how is it possible ?
    1
  19353. That is just amazing. Thank you.
    1
  19354. Okay, but do these "facts" work in other base systems?
    1
  19355. It has been a long time since I have done higher level maths. It takes me a while to to get back to that mindset but, when I do, I enjoy these videos.
    1
  19356. One of the best videos ever
    1
  19357. Thologer
    1
  19358. How did one get the intuition to use windmill geometry. Also how might this parlay to engineering efficiencies?
    1
  19359. Odd number of disc, move the first disc where you want all disc to go, even number of disc, the opposite.
    1
  19360. I wonder if there is an extension of this theorem for polygons defined in 3D, where each vertex is not a (x,y), but an (x,y,z) point, and the vertices do not describe a plane; ie side count >= 4
    1
  19361. I loved the bizarre maximizing overhang tower the most! Greetings from Germany and thank you so much for all of these amazing videos!
    1
  19362. 666 the number of the sunglasses? The glasses break up into manageable chunks when you realise that dominos can't be in certain places due to oddness, and then there are really only four cases to check.
    1
  19363. I made it to the very end!
    1
  19364. I watched this video for his little giggles.
    1
  19365.  @Mathologer  your videos are a treasure and a delight for me. Thank you for bringing your personality to your content.
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  19366. Hey Burkard! What software do you use to get LaTeX formatted equations in your editing Software? Do you take screenshots from an PDF, or do you have any special Trick? I use “Reduce Transparency” function in Adobe illustrator, but that is already pretty annoying.
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  19367. Very nice coding challenge. Answer below the fold: 11956824258286445517629485 OR 11393868451739000294452939 (depending on what is meant by 666th) Managed to get up to around 10000 terms before the system ran out of RAM. Paged memory is awful for this sort of calculation, so running out of RAM is game over. Haskell code generating this sequence: module Partitions where interleave :: [a]->[a]->[a] interleave (x:xs) (y:ys) = x:y:interleave xs ys ppmm :: [Integer] ppmm = cycle [1, 1, -1, -1] posns :: [Integer] posns = interleave [1..] [3, 5..] ixs :: [Integer] ixs = 1:go (tail ppmm) posns 1 where go pa@(p:ps) na@(n:ns) c | c == n = p:go ps ns 1 | otherwise = 0:go pa na (succ c) vals :: [Integer] vals = 1:map next [1..] where next n = let prevs = take n vals pixs = take n ixs in sum $ zipWith (*) prevs $ reverse pixs
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  19368. N A(x) = A(x) + A(x) … + A(x), n times Using the sum product rule we can say that it is equivalent to A(x*x…*x), n times A(x^n)
    1
  19369. This looks like the inverse of the Method of Differences
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  19370. Thank you very much. That was a joy to watch- I can only imagine how much effort it was to assemble. I'm a software engineer who had heard of Bernoulli numbers but didn't really know what they were so the multi-section approach was very helpful. I took your advice and took coffee fueled breaks between some sections. I loved the geometric animations (with the very trivial niggle that perspective projection might look better than orthographic). It just goes to show that the little blocks that kids play with early in their schooling are not just for counting or basic arithmetic - they can start them thinking about some amazing stuff and help them grow their brain muscles for some serious maths later.
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  19371. Perfect
    1
  19372. IF ALL DIGITAL ROOTS ARE NOT THE SAME EQUALING 1 IS NOTHING..
    1
  19373. 7:22 "... between the smallest length 2, ..." Hold on: if we look at the triangular chart (6:33) then it is clear to see that the sum equals product identity of length 1 is possible, namely 1 = 1 (and also x = x, but when we use 1 it fits the pattern nicely). Yes, it's trivial.
    1
  19374.  @Mathologer  Yes, that's the other side. Oh, decisions, decisions!
    1
  19375. I noticed that the address of Lee Sallows was visible. While I recognize that you intend no harm, this is personal information and should not be shared with a larger audience without permission.
    1
  19376. 9:08 Almost this exact question was asked in the oxford MAT (Maths admissions test) 2017 Q2 part iii d). I thought all this stuff felt familiar.
    1
  19377. Nothing magical here: IIII=IIII IIIIII=IIIIII IIIIIIII=IIIIIIII
    1
  19378. I welcome everyone to try coding the magic square dance in https://djit.su !
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  19379. I made it to the very end.
    1
  19380. Excellent! Math and math videos are a treat. Thanks Mathologer.
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  19381. The answr maybe is the bug or 4 bugs or 20 bugs , if one turns they all turn the same direction not move, so in the number line it starts with 0 not one, justt a thought, i wasnnt great at school....🤔🙄
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  19382. Hello dear Mathologer! I am a great fan of your videos. As for me, I have just begun working with logarithms and it would be very useful for me to have a general Taylor series formula for higher powers of natural logarithm, for example lnx raised to third power etc. The only formula in literature I have found so far is just taking the power series for lnx ( S ) and then expanding the product S×S×S in a term-by-term fashion . But this gets very messy in higher powers. Might there be a better general formula? Would be very greatful for advice....Antti
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  19383. By an amazing coincidence, I experienced the Einstellung effect today. I first watched Mathologer’s video on lacing shoes. Next, I watched Arvin Ash’s video on General Relativity. In Arvin’s video, at 3:33 (another coincidence!), I immediately spotted that the depicted shoe lacing is impossible! BTW, I posted this comment in both videos because I like likes. 🌎 General Relativity Explained simply & visually by Arvin Ash: https://www.youtube.com/watch?v=tzQC3uYL67U 👞 What is the best way to lace your shoes? Dream proof. by Mathologer: https://www.youtube.com/watch?v=CSw3Wqoim5M
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  19384. Thankyou.
    1
  19385. Is the Universe really just a big door with a keyhole? I had no idea.
    1
  19386. If n is even, clockwise, if n is odd counterclockwise 14:00
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  19387. After the ugly overhanging tower the most memorable for me was the beautiful parabolic one.
    1
  19388. Sir Please make some video on Laurent series And if possible on how to solve differential equations easily
    1
  19389. Beautiful.
    1
  19390. Because they don't know.
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  19391. I made it to the end! I'm graduating from my physics mphil and everything worked for me except for the final morph to the useful form. I think I can figure it out given some time though. Thanks for the wonderful video!
    1
  19392. But where did the Greatest Integer function go??
    1
  19393. I liked the seal I got at the end the best ...
    1
  19394. The example in chapter 2 is one of my all time math favorites. I've been teaching physics for high school students, and this was always the example when we were talking about moments and the center of mass.
    1
  19395. A tiny bit less elementary, but to me much more conceptually transparent, proof can be obtained by applying inversion centered at one of the vertices, and expressing distances between the images of three other vertices as A/bc, B/ac, C/ab (for the unit radius of inversion). (Works in any dimension.)
    1
  19396. Are there any points on the circle other than the 90° ones where both x and y are rational numbers?
    1
  19397. Maths gives you wings!
    1
  19398. Math was my worst in school, hated it. Going to save this and watch it til I get it......thank you for your time.
    1
  19399. That prime number limit equaling e^gamma is incredible. Definitely my favorite part of the video.
    1
  19400. I am teaching Thales to my daughter these days and Pythagore as well... Watching this video felt like, summoned from the ages, the great gods of math were watching the video right next to me
    1
  19401. Well, about the "homework" you left us... The first part is quite easy to anyone that has followed a couple of your videos. If we put x^2 = (y+1)^2 = y^2 + (2y + 1), being any odd number equal to (2y + 1), we have an easy solution for the first problem. Because, if we choose y = (ODD-1)/2 and x = y + 1, we can obtain any odd number with the formula x^2 - y^2 = ODD. Second part coming soon (I'm working on a way to clearly write it down here).
    1
  19402. Second part: we are trying to demonstrate that, if an odd number is a prime, there is only a couple of integer x, y that satisfy the equation x^2 - y^2 = (odd prime). Well we already demonstrated that it ALWAYS exists a couple of integer whose square differs by any given odd number (N). Now to demonstrate that in case odd number N is also a prime, this is the only existing solution. So, we start with the general equation: x^2 - y^2 = N. if we consider that it must be x > y we can write x = (y + a), and then rewrite the equation as: (y + a)^2 - y^2 = N y^2 + 2ay + a^2 - y^2 = N so: 2ay + a^2 = N we can write it as: (2y + a) * a = N but since N is prime, "a" can only be 1. So, for a prime odd number, the ONLY solution is : N = 2y +1 x = y + 1
    1
  19403. great video, I learn more about how the nilakatha series made, but what I know is the series 1/(1^5-4)-1/(3^5-4*3)+… in 21:53 is found by British mathematician JWL Glaisher, I just don't know whether I was wrong
    1
  19404. Piradox😂
    1
  19405. I like the coloring proof more than the swivel proof. It seems to require less mental manipulation.
    1
  19406. Bravo!
    1
  19407. The series for Pi goes like this: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220, 243, 266, 289, 312, 336, 359, 382, 405, 428, 451... and it seems that (except for the first term) you always add 23 or 24. Any ideas why this is happening?
    1
  19408. There is a very similar result to the first one in the video which explains why this is happening. It turns out that with tthe sequence 1/n (as in the video) if the ratio of the number of positive terms to the number of negative terms tends to A then the series will sum to ln(A). Now as the video stated e^pi is a little larger than 23 so a sequence with the right number of 23 and 24 gaps would add up to pi. On the other hand if a sum does converge then the ratio of positive and negative terms must also converge so we need to have something like this. The fact that there is always one negative term per group and that only the first one doesn't fit the pattern can be mostly explained by how simple the method of constructing the series was.
    1
  19409. If you've found this video interesting and would like to play with modular arithmetic and learn while playing, I can confidently recommend the number theory course on Brilliant. Even after studying up to a master's degree in engineering, I had almost zero knowledge of number theory. What I've learnt on Brilliant, I still remember now, and I was actually able to explain a lot of what was happening in this video, before Mathologer's explanations :) All this to say, I know you've probably seen a lot of ads for Brilliant, but I can personally vouch at least for the number theory course as a nice start into the field.
    1
  19410. :) I love this video! I have just been learning about q series! It's so strange how your videos are always surprisingly relevant
    1
  19411. Wurde mir noch gelehrt:-) Doch wie bereits erwähnt ohne skizziertem Beweis, jener folgte dann auf der Uni. Danke für jene Videos! Sie sind eine Bereicherung! Ein großes Komplement, und Danke für die Jahre.
    1
  19412. 19:46 Good place to stop.
    1
  19413. Loved the video. At 26:50 some people may be wondering how Wrench knew the value of gamma before plugging it into the formula. It seems circular since that's the formula that defines gamma in the first place. People should check out your 'Power Sum Masterclass' video to see how to easily calculate gamma to extreme precision.
    1
  19414. Are the intro chords from Kate Bush - Babushka?
    1
  19415. Excellent instructive video! However, why does an elastic band stretch out that way? What is the physical explanation of this physical deformation. Is it present when taffy is stretched. In wires? Would the rubber band composition affect its logarithmic expansion? Any other logarithmic variations in the natural world?
    1
  19416. I'm a high school graduate, the highest maths course i took was 1st year Calculus and 1st year Statistics, but all the wonderful videos you've posted have educated me a hell of a lot more than that. Thanks Burkard and Marty for making these possible for Numberphiles such as me to come into contact with deeper levels of mathematics :)
    1
  19417. Coloring is the best
    1
  19418. I made it to the end. I would have appreciated a detailed explanation of the notion of fairness that Barry Nalebuff describes in his book "Split the Pie". If I understand correctly, what the video is saying is that the distributions described in the Talmud match this notion precisely, which is fascinating.
    1
  19419. I, instead, am able to multiply and divide numbers with any desired precision with pen and paper: what the sliders are "useful" for is calculating logarithms on the fly.
    1
  19420. I wait for that day Mathologer find the proof of Fermat and explains it in one simple five minute video. Wiles needs hundred pages for it.
    1
  19421. This is amazing!
    1
  19422. Great idea the short videos.
    1
  19423. 6:18 That one was, to make zero sense :D
    1
  19424. OK, now my brain feels like it has been repeatedly squished and stretched. [ it will never be the same shape again . . . this might be a good thing . . . i'll know if it stops throbbing ]
    1
  19425. Great video and, as always, well explained which makes the video an easy to follow. Master's degree in physics.
    1
  19426. "Can you solve the problem on the board in your head-" Nope! "- in under thirty seconds" Well, I said nope in under thirty seconds
    1
  19427. This is the best explanation of the relationship between 1/x and log(x) I've seen anywhere. This is 3blue1brown level of intuitive understanding - the kind where to quote an MIT AI professor the equation now 'sings to you' - kudos! The only thing that bugs me slightly though is... when you were doing the calculations on the areas, the relationships that you defined could just as easily be said to be properties of the exponential function rather than the logarithmic function. I guess a consequence of the symmetry of the function 1/x is that these two are just inverses of each other and thus, the relationships found are just as valid for one or another? Or put another way, it would also be just as valid to define the intgeral of 1/x as e^x, if x and y are inverted?
    1
  19428. Exactly 13:37 I realized that it’s just the same as with the Deadly Marching Squares. Thanks for this great video.
    1
  19429. Mathologer 😁
    1
  19430. Mathologer I really appreciate your presentations so much. You’re meticulous. And the processes/workings are very elegant. I miss your little friend. I think he’s Indian.
    1
  19431. Euler strikes again 😂
    1
  19432. The most memorable part for me ,as a student, was the graphical visualization of sum and the Ɣ inequality. Mainly due to simplistic animated proof that you presented. Thanks Mathologer, I would love to get my hands on your book even if it won't contain your artistic animations :)
    1
  19433. watches intro what... WHAT????!!!
    1
  19434. Math is so damn beautiful.
    1
  19435. I'll watch it 3 more times and maybe, maybe I'll get it 😅
    1
  19436. The no 9s series really blew my mind! Amazing video as always
    1
  19437. Ok, why is square root of 5 in the function formula for the Fibonacci numbers? I can't see any particular 5-iness in the sequence.
    1
  19438. Hi Burkard a small correction: the root of all evil is 25.8070 if rounded correctly to 4 decimal places, so maybe your t-shirt needs an upgrade ;-) Thank you for all your work! Best regards Thomas
    1
  19439. My proof : playing the rightmost card is optimal. I can easily prove that by showing that eventually any sequence of card will occur. It's basically counting up in binary with a twist. Now alI I have to prove is that the system move to the left, and this easily done by doing the first 8 moves maybe to convince all the audience. Thank you. Edit : In other word, everytime you do not play the rightmost card, you either create a new fac e up card or move an existent one to the left. We will eventually run out of leftness and we can't play the rightmost card twice in a row. therefor, we run out of leftness. Edit2: One word : enthropy. thx.
    1
  19440. I think you complicated the demonstration by showing different motions and hence creating intrigue. Fundamental principle is same... we should always calculate the distance travelled by the centre of the object and then everything falls in place. Think about it.
    1
  19441. Very Nice Video !!!
    1
  19442. Infinity minus any amount less than itself is still infinity. So, proving a sum is less than infinity fails to negate infinity status.
    1
  19443. Tower of Hanoi is a super popular interview question, so much so that I bet interviewers have to tweak the question so it's new to interviewees. I imagine asking about more pegs is a frequent extension. Now I feel prepared. Great vid
    1
  19444. Those verses are actually sanskrit shlokas....during that given centuary we indians knew majorly sanskrit only One such sanskrit shloka also tells us about the distance of sun from surface earth as well as from the center of earth
    1
  19445. ​​​ @Mathologer  language is sanskrit,but script is malyalam.only "yuktibhasa" of nilkanth samyoji was written in Malayalam, "yuktibhasa" contains all the proofs of madhava and narayana works by using calculas.
    1
  19446. 258,741 x 2 = 517,482. The answer has the same digits. Do you know of any other numbers that have the same digits when doubled?
    1
  19447. Malayalam is my mother tounge and the last manuscript is just pure gibberish to in my eyes. It's like reading ajdbdixndismsosnejdidnduxnbd cidneiwowlurhr by an English speaker.
    1
  19448. Analog computers based on water have been used to solve complex differential equations.
    1
  19449.  @Mathologer  it's going to shake the very foundations of Terrology. You can't do this to him! He's hanging onto reality by a thread as it is.
    1
  19450. ​ @Mathologer haha, fair point, yikes. By "reach them" I meant in the sense that they realize how insane Terrence is and enjoy the beauty of real math taught by real mathematicians.
    1
  19451. Wonderfully intriguing…TY. A side observation…these approximation only work in flat space, like on flat paper. In curved space, like the reality of Earth’s actual gravity well, Pi increases because it is further across the circle than the diameter calculated from the circumference/Pi (A tiny bit ok…before anyone gets excited). This is because space (length…the diameter) curves in time so travelling the diameter takes longer than calculated from a know speed of travel.
    1
  19452. At high school we used books tabulated with logarims. At university level we changed over to slide rules. And then when we needed to handle complex numbers in AC current, voltage, power and impedance calculations we were told that rather than using the basic "x" scale pairs, we should instead use the '1/x" scale on half of the operations. That minimized the number of re-aligns. Worked like a charm, as long as my main "computer" was the slide rule.
    1
  19453. 2:32 "Maybe one of you can check in the comments" is the new "left as an exercise to the reader"
    1
  19454. Plz a proof of prime series diverging plz plz plz
    1
  19455. Great video!!! The first thing that came to my mind when I saw the video was 50 minutes long was "what the heck?", but it was so entertaining that I didn't realise when it got to the end. There were moments where I my jaw dropped in awe! I'm looking forward for the next video! Answering your question, I took calculus and algebra when I was in university so the concepts were not that hard to follow, but I got lost a couple of times anyway.
    1
  19456. 7:17 There is another visual way of seeing this. Call the two yellow points below the midpoint A and B, the yellow point at the top P and the midpoint M. Extend PM beyond M. Draw in the parallel lines to PA and PB through the midpoint M. Now the angle AMB is split into 4 equal parts. Through the well know angle theorems for parallel lines and the fact that PM, AM and BM are all the same length, you can quickly see how the angles APM and BPM fit in there twice each.
    1
  19457. Finally I understood the meaning of your video. Sorry, very long time. Great as usual. As usual, each imaginary cat has one less tail than the real one. So there is a cat with seven tails.
    1
  19458. Every time I see something about vortex math I feel so disappointed, because there’s so much potential to silly digital root based mysticism. Specifically, I think it’s a shame that they ended up with the powers of two cycle. As explained here, there is nothing special about decimal in this regard—the most you can say is that, if you want the iconic “butterfly” image in the vortex diagram, you need the base to be two more than a perfect power. (For example, the reason for the butterfly in base ten is that ten = 2+2³). This, of course, happens infinitely many times. In particular, it would at least be more mathematically interesting to use the enneagram sequence 1-4-2-8-5-7. It still avoids the multiples of three, so you still get the 3-6-9 triangle, but that fact is more difficult to explain. Also, its origin as the sequence in the decimal expansion of ⅐ (with seven being the ONLY number for which this works in base ten) is much more interesting. What’s more, base ten becomes part of an exclusive club; the only bases with a unique corresponding “seven” are base five, base ten, base eleven, base fourteen, base fifteen, and base twenty-one.
    1
  19459. To the great Mathologer: 1. Theorem: Σa_n<Πa_n +1 for a_n>1 Proof: Πa_n-Σa_n+1=Π(a_n-1)>0 Q.E.D. μ 2. l really like the QEDcat and designed a 2D-QEDcat origami model from duo-color paper!
    1
  19460. I really enjoyed this present! Dankeschön & Glückwunsch zu mehr als 500.000 !!! 💪 Fröhliche Weihnachten (=
    1
  19461. doesn't madhava suspiciously look a lot like you in an costume? hahahha
    1
  19462. <sarcasm>Meh. No relation to pi whatsoever.</sarcasm>
    1
  19463. I find new in maths seen my video
    1
  19464. Currently the creditor of 300 gets 300, and the rest is left high and dry, that's how bankruptcy works :(
    1
  19465. It was a brilliant exposition on the topic. I was able to understand a large portion of it (Have not studied much about Reimann zeta function). The animations are awesome. I have a masters in Electronics Engineering and hence had mathematics courses in Bachelors as well. I am out of touch with many topics in calculus but understand them if they are used in proofs. I believe if you structure your video like this one from easy to the toughest concepts, then 50 minute videos will definitely work (given that you do amazing animations and explain the material very clearly). 50 minutes of only tough mathematics may derail many viewers.
    1
  19466. Related question: how many ways can you make change for a googol U.S. dollars using any kind of legal tender, including the obsolete ½¢, 2¢, 3¢, 20¢, $2.50, $3, and $4 coins and $500, $1,000, $5,000, $10,000, $50,000, and $100,000 bills (as well as the existing 1¢, 5¢, 10¢, 25¢, 50¢, and $1 coins and $2, $5, $10, $20, $50, and $100 bills).
    1
  19467. That graphical representation of relation between harmonic sum up to 1/n and log(n+1) + gamma immediatelly yields better aproximation, by subtracting (1/2)*1/(n+1), i.e., the estimate on the blue area that represents the difference of the estimate from the correct value, by subtracting little triangles of unit length and 1/(n+1) total height. Thus the approximation would be, sum_i^n(1/i) ~= log(n+1) + gamma - 1/(2n+2). My guess is that the error will have magnitude of 1/n^2.
    1
  19468. Amazing, however around halfway in the video I forgot what we were looking for. If possible more back to the root problem (in a visual manner) would help the not very mathematical viewers.
    1
  19469. Friends, he is the Christmas gift that keeps on giving!!!
    1
  19470. Regarding sums and collections from ch. 6: Wouldn't removing common elements in effect transform collections into another ones, reducing size of collection space?
    1
  19471.  @MuffinsAPlenty  suppose you've found two groups with same sum, of 5 and 7 elements with 3 overlapping, those are two of possible 1022 collections. You remove 3 elements from both groups and end up with collections of length 2 and 4. But these two collections are still from 1022 possible ones. So original pair (5,7) is invalid, and we remove it from 1022, ending up with 1020 possibilities. If my logic above is right, we can, maybe, find enough invalid overlapping pairs to decrease original 1022 beyond 846 sums.
    1
  19472. Hmm. That was odd. Interesting, but odd. I'll get used to it. I think it's just too late at night.
    1
  19473. I was taught the multiplicity tests, but no remainder functions.
    1
  19474. As far as I know this sequence you start with is part of a counting problem: When you have a circle with n points on it, all connected with each other by straight lines the n-th number of the sequence gives you the maximum number of parts into which the circle is divided by those lines. I use this when doing recreational mathematics with undergrades, as this number guessing is fun, and when you draw the circle and again and again with more points and counting all those parts always gives the right number (unless you lose a part when more lines intersect in the same intersection point). It's fun to see how those fields, arithmetic and geometry, counting and drawing, are having a party. ;o))
    1
  19475. Bugs: well, if they are moving at the same speed as each other, of course the distance they travel before they meet each other is the same as the distance they were originally apart. Um, that seemed obvious a minute ago but I seem to have lost the insight. Something about mapping their path onto the original sides of the square and how far do they go to make the distance apart zero? I had some thoughts about their being in a uniformly-shrinking, one dimensional universe (ignoring the squares and the rotating) and how far did it have to shrink to make their separation zero. They are moving at the same speed as each other so, ignoring the scaling of the universe, they are always the same distance apart so they only get closer together because of the shrinking. The Unit Universe™ would have to shrink one unit to obliterate the separation. I'm fairly sure I don't have a good handle on this. Thank goodness I'm retired and not looking for a job (in mathematics or elsewhere).
    1
  19476. Great video--never knew there is a closed solution for exponent sums
    1
  19477. Amazing!
    1
  19478. My intuition tells me that if you take the hyperbola xy=1, x,y>0, then this squishing and stretching by the same factor will give the same shape
    1
  19479. When i was young, i hated maths, and i was a terrible math student. But when i studied the Aristotle logic books in my Philosophy career, appeared this equation (similar to that awesome young gaussian sum) about the sorites: c=p(p-1)/2, where c=number of conclusions and p=number of premises. Aristotle knew it, but he coudn't write it in math notation. I tried to search in books how to arrive to the equation, without fortune. So, in the middle of my ignorance, i "invented" (re-viscovered) my own way to arrive to the equation. It was an incredible obsession. Since that incredible fusion of Math, philosophy and logic, i love maths, i study it by my self and now i'm an Tauist. Thanks for your divulgation. From Colombia, thanks.
    1
  19480. In regards to the desire to use the 8 turning into an infinity, a Menger Sponge with side length 2 might have been a better choice? It would be a shape with finite volume of 8 and infinite surface area.
    1
  19481.  @Mathologer  Yes, but what volume of paint is required to submerge the infinite surface area? That would be 8. (I'm trying to shoehorn in a 8 to infinity.)
    1
  19482. 17
    1
  19483. Great video! Thanks!
    1
  19484. :face-red-heart-shape:
    1
  19485. South Africa, 1970's Technical High School. Algebra, Geometry and believe it or not, Trade Theory.....👍🍻
    1
  19486.  @Mathologer  technical drawing, fitting and turning, motor mechanics..... All use the twisted square in one form or another. (Gearing, Architecture and ergonomics) builders use it when laying tiles in fancy patterns, or calculating areas 🍻👍😎
    1
  19487. Most memorable: e^γ
    1
  19488. Very good video. I am impressed by what you explain, the math, but also how you explain it, didactically. Step by step, clear and concise sentences . Thumbs up!
    1
  19489. so ln(1) = 0 & ln(0) = +∞ O_O
    1
  19490. 38 minutes of me in exited state.
    1
  19491. 3:09 a corner of the wheel?
    1
  19492. 17:50 thanks, that's what I am always saying when people come up with infinity-paradoxes. Our reality doesn't like infinity much, pretty sure the existence of this universe is finite, if anything it will go in a circle
    1
  19493. I liked the parabolic brick stacking the most. My instinct when faced with the original maximal overhang question was to try to use counter-balancing so nice to see that I wasn't way off.
    1
  19494. hurray! I love you!
    1
  19495. I wish my math tests were like his shirt
    1
  19496. Glad to have you back. 3 months is way too long without new mathologer videos.
    1
  19497. How come a genius like Tesla did not know about the relevance of 42 rather than 3, 6 and 9?
    1
  19498. Thank you, @Mathologer . You inspire me a lot. Please wish my lovely daughter a success for competing in the Indonesian Math Olympiad for Elementary Schools on the provincial level. Thank you again.
    1
  19499. Anyone know how to do the chapter 2 answer? I'm pretty sure the top block won't be horizontal, but now the math gets tricky. In particular, I've gotten myself totally confused over how much torque the top block provides to the middle block vs the bottom block
    1
  19500. So, the possibilities for the infinite diamond is 1/12root(2) because 1+2+3+4...=-1/12
    1
  19501. My head hurts. I tried to apply this sequence to space time vectors. I landed up in an era without aspirin. Dang.
    1
  19502. Extra special indeed! 🟥✅
    1
  19503. Awww... my favorite proof wasn't mentioned... I think it's the simplest... squares decay at square intervals along orthogonal rays. That any odd natural number produced by multiplying an odd and an even number will not connect directly to a square by way of a line of slope -1 is neat. [x,x]+-[x]=[0,2x] spooky
    1
  19504. Dr. Spock vs Dr. Who. I never knew that the twain doth meetith.
    1
  19505. 4:55 I would, as long as I get the very first move right, which I'm pretty sure I would, after noticing a very interesting, completely evident, yet tricky property (which I hadn't noticed before since the verisons of the puzzle I had seen till the other day, for some reason, statd "to any other peg" instead of "to the last peg"). ANSWER: to C (counterclockwise). Odd discs move always in the same direction as the smallest one, while even discs move always in the opposite direction. PS. I notice that for more than 3 pegs, solutions are not unique.
    1
  19506. It's you fault I will be listening to Kate Bush for days now. Feel some kinda way about that!
    1
  19507. 19:58 The same proof still works if you allow for “negative” angles on the one isosceles triangle that overlaps with the others. There’s a sense in which that triangle’s orientation is flipped compared to the others.
    1
  19508.  @Mathologer I like it because the numbers involved continue smoothly from the familiar “all positive” case when you continuously change the quadrilateral.
    1
  19509. I used to draw diamond like this connecting points into a star
    1
  19510. 32:54 😂
    1
  19511. A very AHAmazing video
    1
  19512. To answer your call for feedback: I think everything in your videos make sense, and can be internalized, but it usually takes a few viewings to get down. Engineering, pursuing master's in control systems (somewhat mathy?...)
    1
  19513. También hubiéramos visto la Súper proporción áurea. Buen video.
    1
  19514. Most memorable: the fact that despite my mathematical intuition and the fact that I am living in a world of numbers since (1+1/2+1/3+...) years I did not manage to get the right answer to the No-9-Sum.
    1
  19515. Where do I get this T-shirt?
    1
  19516. log log log log log...
    1
  19517. Pardon me for not going through the comments. For others like myself 😁 I'll mention that my calculus teacher pronounced sinh as "cinch." Easy, no?
    1
  19518. I think some people say "sink," though.
    1
  19519. I recently came across a beautiful algorithm for generating all the primitive Pythagorean triples uniquely and completely, that was first published in 2008. To start write down two numbers at the top corners of a square (a goes in the top right, b in the top left. Now write down c = a+b in the bottom left corner and d = b+c in the bottom right. The two products 2*b*c and a*d form the legs of the triangle. The sum of the cross products a*c + b*d form the hypotenuse. Starting with a =1 and b=1 generates the 345 triangle. Substitute c for b keeping a unchanged generates 5 12 13,. Substitute d for a keeping b unchanged generates 8 15 17. And substitute c for b and d for a gives 20 21 29. Keep repeating these substitutions and you produce a ternary tree containing all the Pythagorean triples once and only once. Other information encoded in the number squares are: the area of each triangle (abcd); the radius of the in-circle (ab); the radii of the three ex-circles (bc, cd, ad); the half angle tangents at the (non-right) vertices (a/d and b/c); and still more besides. Quite remarkable. Most of this is easy to prove, although the uniqueness and completeness proofs are a challenge. The surprise is that it was not discovered until so recently.
    1
  19520. For the last problem, 3^3+4^3+5^3 = 216 = 6^3, but the nice integer pattern seems to end there. There is a subtler pattern though. I'll show the next few and their factorizations: 3^4+4^4+5^4+6^4 = 2258 = 2*1129 3^5+4^5+5^5+6^5+7^5 = 28975 = 5^2 * 19 * 61 3^6+4^6+5^6+6^6+7^6+8^6 = 446899 = 19 * 43 * 547 3^7+4^7+5^7+6^7+7^7+8^7+9^7 = 8080296 = 2^3 * 3 * 7^2 * 6871 Initially I didn't see any pattern, but look at this: 3^11+4^11+5^11+6^11+7^11+8^11+9^11+10^11+11^11+12^11+13^11 = 2962844752912 = 2^4 * 11^2 * 1530395017 So far, all the odd prime exponents have had factorizations that included that prime squared. Let's try one more: 3^13+4^13+5^13+6^13+7^13+8^13+9^13+10^13+11^13+12^13+13^13+14^13+15^13 = 3197503726481727 = 3^2 * 13^2 * 41 * 163 * 1523 * 206543 So it seems to be working for all odd primes. It doesn't work for at least some composites, eg 6 (above) or 10: 3^10+4^10+5^10+6^10+7^10+8^10+9^10+10^10+11^10+12^10 = 102769129725 = 3 * 5^2 * 17^2 * 113 * 41959 But it seems to work for odd prime powers: 3^9+4^9+5^9+6^9+7^9+8^9+9^9+10^9+11^9 = 3932252163 = 3^4 * 7 * 373 * 18593 (for exponent = 25) 982145735504627929354047360550891875 = 3 * 5^4 * 21644047 * 24201160667217252038038847 (for exponent = 49) 74949361513846128535166303063666052479479582590331165229173117402614388 = 3^3 * 7^4 * 2259460330447 * 5116904275542808306343187324898532075696262964763114757092600822327 Not sure why it works.
    1
  19521. I may have focused prematurely on the primes. Further experiment shows that it works for any odd exponent: 15: 4680823029669774000 = 2^4 * 3^2 * 5^3 * 13 * 107 * 186948759073 21: 63969345434591028564245663823 = 3^4 * 7^2 * 13 * 503779 * 2460976097647142521 33 : 1446256202065763190684242053295652107181281523397827 = 3^2 * 11^2 * 19 * 7106569577243 * 9835665292589597505635230641886979
    1
  19522. Wonderful way to describe mathematics
    1
  19523. A jaw-dropping presentation. Speaking as an honours level mathematics student, that was truly one of the best pieces of entertainment mathematics I have ever seen. I found myself unselfconsciously saying "bravo" and "superb" after watching this video. Incidentally, this presentation also made me appreciate better why the Hausdorff dimensions of fractals are ratios of logarithms. This makes a lot more sense given the self-similarity of areas under the hyperbolic functions and the immutable and beautiful relationship between that self-similarity and the much beloved properties logarithms. The sinh and cosh stuff at the end was just like a fine dessert after a tasty and filling meal. Am still a little stunned by the loveliness of this video. I have immense gratitude to you for creating it. Thank you.
    1
  19524. 🤯 Love this stuff!
    1
  19525. I would vote for the absence of integers among partial sums of the harmonic series (and their differences), even though the video presented an idea for the proof and not the proof itself in its entirety (which is nice, of course, because it gives the viewer a chance to work it out). I do so in part because I happen to quite like this result, and questions concerning integers in general.
    1
  19526. I like the background music at the end.
    1
  19527. r=s=2 and r=s=0 may be the only solutions to r+s=rs in N, the naturals, but it is an open question whether or not there are any other solutions in βN, the Stone–Čech compactification of N. If there were another solution, it would immediately solve the open Ramsey theory problem: If you colour the natural numbers with a finite set of colours, can you guarantee that two distinct numbers x+y, xy have the same colour, where x,y are natural numbers?
    1
  19528.  @Mathologer  If you ever want to do a video on βN and its connection to Ramsey theory, feel free to make any use you wish of my article in Chalkdust "Pretend numbers".
    1
  19529.  @Mathologer  Great. It is not my area at all, but I always felt that the question of solutions (in βN) to r+s= rs other than r=s=2 should be up there with the Riemann Hypothesis and Collatz Conjecture in terms of fame. The graph with natural numbers as vertices, and edges between x+y and xy is one of the simplest combinatorial structures involving both addition and multiplication, and asking if its chromatic number is finite is a very natural question. Moreover it feels like magic that you can answer this question by solving r+s=rs in βN, which feels like it belongs in a very different bit of maths.
    1
  19530. While I liked the video a lot (I still have to watch and understand the second half), claiming a connection between Pythagoras triplets and Fibonacci numbers is not quite true. It is basically true for any 4 numbers x, y, x+y, and x+2y. A sequence of any 4 Fibonacci numbers fits the bill but any Fibonacci like numbers too fit the bill.
    1
  19531. I was too soon to comment. The other half does cover this aspect and much more. So, there is a connection between the 2 after all. Great video.
    1
  19532. So the nutters are right - the earth is flat!
    1
  19533. It's funny how I wouldn't have ever believed that this was done by a 14th century guy if it wasn't mathologer video telling me thai stuff...
    1
  19534. Please keep making more of these videos, I have worked with the Euler-MacLaurin summation for complex variables and find these topics exciting.
    1
  19535. Using the specific Coin values and integer truncating with C++ I was able to come up with this Algorithm: using namespace boost::multiprecision; uint1024_t n = 2000000000; uint1024_t sum = 0; for(uint1024_t i = 0; i <= n; i += 25){ sum += ((n-i)/10 +1) * (((n-i)/5 +1)/2 +1) * (i/100 +1) * ((i/50 +1)/2 +1); } std::cout << sum << std::endl; I got 42666677066667562666698866667090000001 for 2*10^9 Cents in just 5 Minutes.
    1
  19536. There has been evidence found that the Romans had discovered calculus before the fall of Rome. I often wonder where we would be if we hadn't lost nearly 1200 years of advancement in math.
    1
  19537. If we define Gamma as the circle centered at the origin with radius 1, and z a complex point on Gamma, then it can be proved that the line passing through z and 2z is tangent to the cardioid sending z to ((z-1)^2-1)/3. This can be done by writing the parametric equation of the cardioid : x(t) = 1/3*(cos 2t - cos t) and y(t) = 1/3*(sint 2t - sin t). From there, a normal vector of the cardioid is x'(t) = -2/3*( sin 2t - sin t) and y'(t) = 2/3*( cos 2t - cos t). But if you compare this gradient with the slope of the line ]z,2[, which is delta = ( sin 2t - sin t)/(cos 2t - cos t), we can see that the slope is orthogonal with the gradient of the cardioid. Finally, since each point on the circle Gamma in the video is in the form z = e^(2*i*pi/n), we can use the statement above to explain why the cardiod shape arises amongst all these line ]z,2z[ when the modulus, and so number of considered points, increase. This explains only the situation for one petal and a multiplication by 2.
    1
  19538. I found out that we are already familiar with one of the variations of the no-9-sum! Let's do the same game in the 'Binary' world. There are two kinds of sum: no-0-sum and no-1-sum(w/ a trick*). Then the no-0-sum is (bin.) 1/1 + 1/11 + 1/111 + ... = (dec.) 1 + 1/3 + 1/7 + 1/15 ... Meanwhile, the no-1-sum is (bin.)= 1/1 + 1/1 + 1/10 + 1/100 + 1/1000 .. = (dec.) 1 + 1/2 + 1/4 + 1/8 + 1/16 ... = 2 which is our good old sum. (* In the binary, every integer contains at least one '1' in its heading digit. So those leading '1' are excluded while doing the no-1-sum.) Thinking in this way, it becomes quite obvious to me that the no-'n'-sum converges. Intuitively, it is unlikely that no-'n'-sum converges in some specific base-k number systems but not in the others because the base number---10 in our human being's case---is chosen arbitrarily in the first place while the essence of the skipping rule will stay. Lastly, I failed to calculate the (binary) no-0-sum of 1/1 + 1/3 + 1/7 + 1/15... If you come up with an answer, please share. Bests ;)
    1
  19539. Funtastic!
    1
  19540. I love your videos!
    1
  19541. Erdős–Szekeres theorem is another nice pigeon powered theorem
    1
  19542. My favourite is Kempner's no 9 sum.
    1
  19543. Some of your old videos are missing I noticed the negative time negative equals positive video missing and also the Word problem video (ab/a+b) missing Would be great if you reuploaded them or maybe they are unlisted?
    1
  19544. The first proof, the graphical one, I learned as a very young child from my grandpa and I was totally convinced. It was something I'd seen. He wasn't even educated, there was no education in the 2nd world war, the wehrmacht and russian camps for prisoners of war. Later in school nobody wanted to convince me anymore. But I learned what one can make with that old theorem. I learned it a lot…
    1
  19545. 2:06 weird flex but ok
    1
  19546. 13:00 "LOG(x) explodes to infinity..." if this way the explosions were defined there would be no wars and no emotions :), in the next 1,000,000,000,(...88 zeros),000 years :)
    1
  19547. Thanks a lot sir. I love your videos, and I have no words to express my gratitude towards you. I wish I could meet you in real and ask my doubts from you. You are really a gem of a person. Thank you for giving us such priceless knowledge ❤
    1
  19548. For some reason. It feels during rotation things can be hidden. And I don't know why. Even with credible channels like this. Some of our brains just question everything. The colors are easier to trust.
    1
  19549. It looks like that man that died was a victim of a sawing-a-man-in-half trick going wrong
    1
  19550. In diagonal recording, this is even more amazing.
    1
  19551. Just amazing
    1
  19552. Gluten tag
    1
  19553. cool, ty )
    1
  19554. You are the bold guy
    1
  19555. 2, 28, 82,... Where is 50!, and why 10 is not 8?! (joke for nuclear physicists)
    1
  19556. so how ? There is only one way .while the drop drop , the the line is turned 90 degree
    1
  19557. i vote for gamma, it is so curious
    1
  19558. This video is really good. It goes from story of gauss summing from 1 to 100 all the way up to super cool stuff with Euler maclaurin. I think it is really nice connection and very nice for undergraduate math majors. Of course, I’m bias. I have a PhD in applied math. I think this and other videos are amazing. You do an wonderful job in explaining interesting parts of mathematics and go deep without getting too complicated.
    1
  19559. Merry Christmas Prof Burkard. Thank you for the amazing years!
    1
  19560. Why didn't they teach us that Pythagoras spent 22 years in Africa and that that's where he learned everything he knew, and that he went there on Thales' advice, who also learned there everything he knew?
    1
  19561. Serious question (I don't have pen, paper, calculator handy to hash out at the moment...and my math abilities are a bit feeble): how well does this generalize (eg. Fibonacci series that start somewhere other than one, non-integers, non-positive numbers, complex numbers)?
    1
  19562. In the 1960's I bought a plastic laminated cardboard circular slide rule just to play with. After being dropped a few times it wasn't very accurate. I'd be afraid to use the rubber band device. If it broke I could lose an eye.
    1
  19563. I never seen the first proof, but I learned the second one in high school(? gymnasium, I'm from Europe). My math teacher was a crazy old guy(rip), we loved him so much. He showed us many neat proofs along with the basic ones.
    1
  19564. Yes that's right. Now do a thing on the Symmetric Monoidal Closed Category RM3 which allows the truth value Both, so it's not a paradox. First do how it arises from Z2 mod (x^2 + x + 1) plus the Bellman's rule (a^3 = 1).
    1
  19565. first proof :-)
    1
  19566. They still make high-end watches with slide rules using a rotating bezel, amazingly still somewhat common to see in watch collecting.
    1
  19567. Lovely as always. Actually, when you first presented the vortex, my first thought was that it probably didn’t work in other bases, so I was so glad when you went there.
    1
  19568. Great video as always! 12:06 Denote the 4 numbers by (b-a, a, b, a+b), then 153=b^2-a^2, 104=2ab. Solve to get a=4, b=13. The box will be (9, 4, 17, 13). Working backwards and we can find it's the left child of (1, 4, 5, 9), which is the right child of (1, 3, 7, 4), which is the right child of (1, 2, 3, 5), which is the right child of (1, 1, 2, 3). Thus right, right, right, then left. 16:00 From the Pythagorean theorem, the total area of each generation is 1. Thus the total are is 5. 24:20 The centers of the incircle and the 9-point circle are on the same horizontal line. Thus no right triangle can be formed. 25:56 The pearls are like the marks on a clock. It's the symbol of your eternal love for your wife. 41:20 It's the logarithmic spiral right?
    1
  19569.  @Mathologer  Yeah I checked the end already when I watched the video - 12 is such a wonderful number! :)
    1
  19570. Because maybe You're gonna be the one who saves me And after all You're my number wall
    1
  19571. Physicists think God is trying to tell us something through force and time. Mathematicians think God is trying to tell us something through sums. Hopefully God doesn't have a sense of humor.
    1
  19572. I just love videos like this one. Starting out with some surprising "magic", continuing with eye-opening connections and leaving me with the fuzzy warm feeling of "understanding" how some parts of what I knew already connect to each other. Thanks. (I also like the final 'how to build this in GeoGebra" time lapse.)
    1
  19573. For the pythagorean and plato triples, look at the side lengths without the squares. Take their sum. Is there a reason why their sum is the difference between the side lengths of the next child (without the squares)? Plato example: 3+4 = 15-8 and 15+8 = 35-12
    1
  19574. Нихрена не понятно, но очень интересно)
    1
  19575. Updated and error-free. Hooray!
    1
  19576. Most memorable: It was the whole video! Your animations make every non-mathematician math enthusiasts fixated 🙈 But if I have to pick one, it's going to be the partial sum can never be an integer And Thank you
    1
  19577. When nothing in my life makes sense, I come back to Mathologer and I know that some things are still alright.
    1
  19578. As usual,as soon as i see the notification, i fastly grab my headphones, a piece of paper, a pen and... here we go ! I watch the first half in X2, then i decrease to X1 with pauses to do some proofs by hand. I really love the fact that you go deep in your subject, and let some "details" on the side ^_^ I can understand all the video, but i'm almost sure i would not if i would watch it in one time ( and i encourage people to do the same to get the best of this wonderful work ! ). What else...oh yes, i am in second year of "prépa" in math (MP), google told me that it was the same as "intensive foundation degree" ( do what you want with that :) ). By the way, please excuse my english, i'm french ( and EVERY single word is underlined in red on my screen !!! )
    1
  19579. I came up with the coloring proof on my own with the video paused. I do like it better than the swivel proof, but mostly just because it feels more rigorous. EDIT: To answer your final question, the diameter of the circle is clearly larger, since one of the diameters of the other shape forms a non-diameter chord of the circle.
    1
  19580. 14:21 It's me...🙂
    1
  19581. Just amazing ! Your videos are always of high quality (choice of subjects and presentation). A suggestion : As you talked about "quasi-isocles" right triangles, why not talk about "quasi-equilateral" ones (size lenghts n, n+1 and n+2) ?
    1
  19582. I think we asked our teacher what to do if the digit sum was so large that you couldn't tell whether it was divisible by 9. He said we could just repeat taking the digit sum, until we got to a number that was obviously divisible by 9 (like 9 itself). That's not explicitly what you talked about, but it's close.
    1
  19583. Just amazing how mathologer brings less famous ideas into life and give them a whole new life altogether!
    1
  19584. https://youtu.be/zZct-itCwPE
    1
  19585. So good, so beautiful, so excellent. With regards
    1
  19586. Merry christmas !
    1
  19587. Amazing video! The solution is very simple, elegant, and surprising 😊
    1
  19588. You have promised to make a video on proving that no general formula exists for the solution of quintic polynomials and beyond. Did you make that video?
    1
  19589. In the "Towel man" segment, the polyhedra don't have to be convex. They just have to be simply connected (no holes, internal voids, etc.) For instance, Euler's formula works just as well for a (polyhedral) bucket as it does for a buckyball.
    1
  19590. 6:35 I was going to say 4k-1
    1
  19591. fourth
    1
  19592. Physics tells me that the circle is the extrem in area of the setup, but gut feeling leads me to the areas being identical. Btw. thx so much, this looks like solving an engineering problem of mine as well 😂
    1
  19593. Lovely. Merry Chistmas
    1
  19594. Love the logo on this guy's t-shirt 😂😂😂😂
    1
  19595. It is explained later on, sorry
    1
  19596. The swiveling proof is very appeasing
    1
  19597. This seems suspiciously connected to FLT.
    1
  19598. at the "heart" is all understanding. yet not void of re-flective as most prominently seen in the pairing of things. top two bottom front two back side two side. in a body "whole" is expressed a temple for understanding the "whole" beyond" all material di-mensioning. Gratitude and appreciation for all your beautiful re-flections as seen from glorious realm of math. one beautiful branch for understanding wholly.
    1
  19599. I feel so dumb.
    1
  19600. Thanks!
    1
  19601. For challenge 3: There is 1 way to tile the 2x1 board and 2 ways to tile the 2x2 board. The number of ways to tile a 2xn board is (the number of ways to tile a 2x(n-1) board) + 2*(the number of ways to tile a 2x(n-2) board). This recurrence relation is linear, so it's pretty easy to solve. I hope I got the answer right.
    1
  19602. Thanks!
    1
  19603. Eating a mandatrin will never be the same!
    1
  19604. I really enjoyed the cat pun... and the visualization of the no-9s convergence.
    1
  19605. To defend Archimedes' work, he explicitly made such assumptions (axioms) in the beginning of "On the sphere and cylinder": (1) Of all curves with the same two end points, the straight line segment is the shortest. (2), if there are two curves C1, C2, with same endpoints A and B, and both are convex (he intuitively defined convexity before stating the assumption) in the same direction, and C2 is located inside the region bounded by C1 and the line segment AB, then, the length of C2 < the length of C1. He also made similar assumptions in 3D regarding surface areas. His calculation of the surface areas of sphere and cylinder are under such assumptions. I think with under such assumptions, his calculation was valid and solid. Whether you agree with the assumptions or not, and whether the assumptions can be proved by other axioms or not, doesn't change the validity of his work. I think his assumptions about length, area, and convexity are very interesting and deep. However, it seems they are not discussed as often as Euclid's famous 5th postulate.
    1
  19606. My most memorable part was the harmonic series missing all integers up to infinity as I had thought it before you said it. I am usually trying to catch up to understand!
    1
  19607. You missed your usual statement before the first minute was up : 1 = - 1 !! Does this mean Math is Broken ? ;)
    1
  19608. If you don't find beauty in the thing you study, either you are not studying math or you are not studying it the right way.
    1
  19609.  @andrewkepert923  Oh, interesting. Yeah, they sound a little different from the cicadas we have in the Midwest, but there is definitely a similarity there too.
    1
  19610.  @andrewkepert923  We have them here in the NE US. Did you choose their song because their periodic emergence seems to be based upon prime numbers? That would be fitting. In the US, they are tracked by their emergence year, called a Brood, and most seem to be based upon a 13 or 17 year emergence cycle.
    1
  19611. 3:05
    1
  19612. Incredible just to know what is the Question , even more to find out the Answer! 🕯
    1
  19613. Reached till 50 min.... I had formal maths education only up till 12th grade, but my mother is a mathematician and I am doing my residency in urology , genitourinary and transplant surgery.....
    1
  19614. Use a string to measure the "curvey curve" or permeter
    1
  19615. Came late to the party but you need to be well rested and sharp for these videos. Amaziiing
    1
  19616. I now have to wonder why nobody here discusses methods for solving even sized magic squares.
    1
  19617. Maybe you've already seen this, but I've come across a "natural" appearance of these "heptangonal Fibonacci numbers" (e.g. 1, 1, 3, 6, 14 etc). Via the OEIS, I came across a paper called "Multinacci rijen" by Jacques Haubrich. Unfortunately, it's in Dutch, so I can't understand it. However, there's some diagram and I can sorta see what's going on. He's looking at m planes of glass and counting the number of different paths via reflections. These numbers seem to be the count of the times the last reflection is on a particular pane of glass (at least if Google Translate is to be believed). Perhaps not the most "natural" of situations compared to Fibonacci flower petals, but it's something! It also makes me wonder if the values of r and s have meaning physically. In optics, you have things like Huygens' principle, where every point on a wavefront can be considered the source of a new wave, whose mutual interference "adds up" to the wave motion as a whole. This reminds me of how r and s are limits (of ratios) - perhaps in some infinite limit of reflections, r and s pop out? But I'm not big-brained enough to work it out lol.
    1
  19618. Idk if you could make a proof out of this, but the shortest laces all seemed pretty intuitive by the idea that the hypotenuse of a right triangle is longer than either leg. If the distance between two adjacent eyelets is 1, then the shortest path will be all 1's (the block). If every eyelet has to cross to the other side at least once, you want the one that lets you go diagonal as few times as possible (bowtie). Tight one feels right intuitively, but without the TSP explanation, it's hard to intuit that the zigzag isn't shorter than the criss cross, despite having so many length 1 connections. I'm curious about this, thanks in advance for any thoughts you guys have!
    1
  19619. Conway!
    1
  19620.  @Mathologer  He was great at leading students from one point to another.
    1
  19621. Ramanujan was like "I know this is true, but you do the proof part" 😅 What a genius
    1
  19622. Great video! I did an interactive implementation for arbitrary ngons and commented with that earlier but I suspect it got caught in blocked messages because of the link. Can you check on that @Mathologer ?
    1
  19623. I love the completely relatable problems mathematicians use to introduce one to the problem at hand. Finally I have the theoretical foundation for building a staircase bridge over any distance without any supporting structures
    1
  19624. marvelous !
    1
  19625. Congratulations for the 100 video. 100 more will come. Thanks for all your insights.
    1
  19626. VW logo clearly uses this same geometry
    1
  19627. My feeling said at first that the upper center dot of the triangle must be irrational so it cannot fit. And I am not a professional mathematiker at all. ;-)
    1
  19628. Are those "spikes" or are they actually Christmas lights decorating your curve? Also, I feel like I've seen this trick a few times, where letting two sequences go to infinity depends on "path" you take in the indexing space.
    1
  19629. You have to do a 666 video now that you have 666k subs
    1
  19630. Brilliant. PITHAGOROUS is spellbinding.
    1
  19631. I loved the part about maximal overhangings! It seems to be a very nice and strange problem to explore
    1
  19632. So you've got a sum of a countably (I'm pretty sure?) infinite number of rational numbers that is exactly equal to an irrational number. Man, infinity is confusing.
    1
  19633. The fun thing about the leaning tower is that even quite good mathematicians look at the first two terms: 1, 1/2 and immediately assume that the next term is 1/4 and thus assume that the overhang if finate. I won a bet with a friend with a first in mathematics from Oxford with that.
    1
  19634. 24:06 I worked it out, and found that the inscribed and Feuerbach circle's centers are on the same axis, so if you try constructing a triangle, you can only make a line of length 1. Interestingly, one could consider this analogous to one of the forgotten Pythagorean triples, that people ignore for obviois reasons - 1^2 + 0^2 = 1^2. Try it again and you get a point, representing the true simplest Pythagorean triple, 0^2 + 0^2 = 0^2.
    1
  19635. MATHOOLͦOOGER
    1
  19636. I almost teared up towards the end.
    1
  19637. Fantastic video, as always. Love it! 👏💙👌 About the question at 24:15 : the centers of the incircle and Feuerbach circle are horizontally aligned. So constructing a right triangle would result in a degenerate triangle with vertical side zero.
    1
  19638. Merry Christmas to my favourite youtuber! You've got a new patron who donates a square of primes each month!
    1
  19639. Having watched this video and the Mathologer video with the proof that pi is irrational and some others I’d love to see a Mathologer video about the BBP formula for pi which is a spigot formula that outputs any desired digit of pi. Or even a video about spigot formulas in general with BBP as a prominent example.
    1
  19640. Dig THE xmas themed shirt. A Santa hat would add to the holiday Glee.
    1
  19641. 24:22 Oily macaroni gang rise up
    1
  19642. Nice music and ofcourse maths
    1
  19643. Personally I hate these kind of title " why they didn't teach you this at school", " your teacher hide these tricks in mathematics from you", etc. It's clickbait at best, and ignorant at worst.
    1
  19644. There is a nice way to interpret the yellow numbers in the diagonals. They represent a way to count the elements in each shell of an m-dim cube, by adding the m (m-1)-dim "faces" of the shell, which are (m-1)-dim cubes, and then alternatingly substracting the double counted elements. For example (17:05) the third shell of the 5d cube consists of 5 4d cubes with each containing 2^4=81 elements yielding 405 elements. But we have double counted the elements where these 4d cubes intersect. They intersect in 10 cubes containing 3^3=27 elements each, giving a total of 270 elements. These cubes again intersect in 10 squares 3^2= 9 elements each for 90 elements. Again intersecting in 5 lines 3 elements each for 15 elements, leaving only one element left. In total we get 405-270+90-15+1=211 elements in the third shell of the 5d cube. (This is also aparant from the adding rules the numbers in the triangle are derived of.) Furthermore this is a nice visual way of understanding the formula n^m = - /sum_{k=1}^{n}[ /sum_{l=0}^{m-1}(-1)^(m-l)*/binom{m}{l}*k^l ]. The exterior sum adds the different shells and the interior sum counts the elements of each shell in the way described above for m=5, n=3.
    1
  19645. Here's a neat little property between the first 10 fibonacci and first 10 prime numbers. I don't know if it means anything but it's kind of interesting. First I'll list them each in their own set, then I'll explain the operations to be done. fib(10) = {1, 1, 2, 3, 5, 8, 13, 21, 34, 55} P(10) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} The first step is to take each equivalent indexed value of each set and take their products generating a new list of values. {1*2, 1*3, 2*5, 3*7, 5*11, 8*13, 13*17, 21*29, 34*23, 55*29} = {2, 3, 10, 21, 55, 104, 221, 399, 782, 1595} From here there's two parts, the fist has a single step. We are going to add all of the digit values within each number of this list and reduce down to a single digit. 2+3+1+2+1+5+5+1+4+2+2+1+3+9+9+7+8+2+1+5+9+5 = 87 = 8+7 = 15 = 1+5 = 6. And this gives us a value of 6. Now, this time we will first add all of the values of this list to a total sum, then well do the same previous step as before in counting the digit values and reducing. 2+3+10+21+55+104+221+399+782+1595 = 3192 and this is our total sum then 3+1+9+2 = 15 = 1+5 = 6 Both of these gives us a value of six which also has only two factors that are not one or itself being 2 and 3 which are the first two primes. I don't know if there is anything unique or special about this, but then again, it's just the properties and relationships of numbers at work. Let me know what you think?
    1
  19646.  @Mathologer     ...and the proof is very nice 👌 : (*) AB = A + B implies AB - A - B = 0. We add the identity matrix and get: I -A - B + AB = I. This means that: (I-A)(I-B) = I . ...and now the 🎩 MAGIC happens ✨️ : (I-B)(I-A) = I We conclude that I - B - A+ BA = I , i.e. (#) BA = A + B. The conclusion AB = BA follows from (*) and (#).
    1
  19647. The most memorable part for me is the no integer miss, because it miss every integer even when it gets really small. Like the integers in the division don't like their not inverted counterparts.
    1
  19648. i thought i was weird. i used to play with numbers in middle school when i was bored in algebra and came up with something pretty close to this. teacher told me i was just wasting my time and to sit quietly.
    1
  19649. you have a 25 coin...?
    1
  19650. One of the tiles dissappear at 12:30
    1
  19651. The way of growing squares from lines, cubes from squares and so forth is quite neat. If get the centroid (average of all the vertices each time and then move off along a new dimension until you have reached a unit distance from the current vertices, you get the sequence: line, triangle, tetrahedron, 4D simplex. If you move off in opposite directions from the the centroid, you get the sequence: line, diamond (square standing on its corner), octahedron, 16 cell and so forth. So there are three infinite squences of polytopes. However the dodecahedron and its dual are a special case that only work in 3 and 4D and there's one extra special case (for regular polytopes) in 4D. I haven't fully got my head around this but it might make a good subject for the Mathologerization.
    1
  19652. I think the dot one worked better for me.
    1
  19653. Kemptners Proof was the most interesting part
    1
  19654. Loved it.
    1
  19655. Fabulously interesting and fun! Can be used as a process system for electric car motors
    1
  19656. pretty amazing......as usual
    1
  19657. I am from Slovakia and we've learned Heron's formula when I was at high school 17 years ago.
    1
  19658. Wonderful!
    1
  19659. Who came here from searching “Dream” in the search bar?
    1
  19660. This Square dance algorithm prefers outgoing arrow squares for corners. It seems that chance of getting perpendicular two dominos at each corner diminishes with each iteration (probability goes down by 2). This mean that although this method can generate any random board, but chances of getting regular one (with chaotic middle part) is much greater.
    1
  19661. This is for all the people who own a googol dollars.
    1
  19662. I think I kind of followed without knowing thus permutation stuff up to the last chapter
    1
  19663. Vieta's jump ("infinite descent") definitely a powerful tool. Never suspected that in my youth/math days.
    1
  19664. I attended high school in California and was never shown most of this. It was a real eye-opener. Great video!!
    1
  19665. Great video as always!!
    1
  19666. Its not the wheel thats magic, its the rubber band that doesnt snap!
    1
  19667. You teach math. I can watch your videos and if my basis is solid, I can follow your visual proofs and reasoning. In this way you impart your knowledge on me, and to the extent that I am capable of comprehension, my understanding upon watching the video is identical to your understanding in making it. Consider the possibility that there exists an entirely different kind of understanding, which does not follow these rules. Imagine for example that I am colour blind while you see in colours. It is my desire to understand blue, so I ask of you to teach me how to see in colour. Not only is colour vision not something you can teach, but until I learn to see colour, I would have no sense of what it means to be colour blind. I might be tempted to dismiss your endless talk of colour as gibberish and insist fervently that no such thing exists. Of course in this example, colour vision is a metaphor. We are talking about the possibility that the universe indeed has a secret and if such a secret exists, then gaining knowledge of this secret is akin to an awakening. In our example, enlightenment is possible, and so to extend the metaphor, experiencing the awakening is analogous to getting up one day to discover, miraculously, that you have gained the faculty of seeing in colour. Suddenly all the intellectual talk about colours and their different properties, which sounded like gibberish before, now makes absolute sense. There is another place where digital roots play an important role – tibetan numerology. There will never be a Mathologer video to either prove or disprove the usefulness of this most ancient of knowledge systems. There are two possibilities: Either such traditions are part of primitive superstition and those who practice it have not yet matured to understand the superiority of western thought, or those who are capable of putting this ancient knowledge to practical use are capable of seeing in a palette that is inaccessible to us. The important takeaway – neither possibility presents an absurdity. Also, there can never be a mathematical proof to either validate or otherwise debunk the Tibetans. If the universe does have a secret, then this secret is very much in a similar category as numerology – hence the “secret”. It is a secret only in the sense that it will never be accessible to anyone insisting that such a thing cannot be. And if the vortex diagram represents some kind of “key”, then it does so in the same sense as the tree of life diagram in the Kabbalistic tradition or as a koan contains a key to a practitioner of zen. It is not the koan itself (what is the sound of one hand clapping?) that contains the key, but rather that contemplation of the koan represents a potential doorway to experiencing the sudden enlightenment (colour vision in our metaphor) that the practitioner seeks. I have no deeper insight to share with respect to vortex math and I don’t begrudge you your efforts to illustrate various patterns associated with digital roots. At the same time, I can’t help but feel that your video is a little like listening to a philosopher theorize about the existence of God. Either you know God or you don’t. If you do, then God is meaningful in your life, and if you don’t, then the concept of God has no value to you. No amount of math will ever solve that particular riddle.
    1
  19668. smooth, elegant and graceful always.... thanks Mathologer
    1
  19669. This tri-color scheme works in all three directions because the seed rule is also direction agnostic.
    1
  19670. I watched till the end, I think I followed along decently, but inverting the matrix went way over my head, how a 1 turns to -1/30 i just don't know. I am in my 6th year of med school and have had no pure maths classes since secondary school, but I was then and am now a fan of these kinds of conversions from one formula to another in order to better understand what the abstract algebra represents. And you do a really spectacular job of that here on Mathologer :D
    1
  19671.  @Mathologer  After your video I looked up generating functions and watched that video also. As a side note, even though high dimensional hypercubes may or may not be "real," they have real use in communications theory. For example, see Hamming distance. Furthermore, to optimize the probability of communicating a sequence of symbols without error, one performs sphere packing in high dimensional spaces to separate encoded symbols so that error is minimally likely to confuse two symbols. This takes advantage of the fact that in high dimensional space, most of the "volume" is near the "surface" of a polytope. See "asymptotic equipartition property." Anyways, your videos are entertaining, insightful, and fun as always!
    1
  19672. seeing that last pascal polynomial brought memories back
    1
  19673. Verrrrrrrrrrrrrrrrrrrrr......rrrry cool... Paradox, parabol... paranormal: mind- blowing stuff!
    1
  19674. at 05:04- "And now one of the most iconic truths of mathematics is lurking just around the corner..." I love math humor!
    1
  19675. I love this channel! and I am a chemist
    1
  19676. Amazing video, thanks!!
    1
  19677. sweet
    1
  19678. I remember my sophomore year Geometry teacher proving it using both methods. In the math club, there was a 3 4 5 triangle using circles with a diameter of each side of the triangle. If I remember correctly, the area not covered by part of a circle inside the triangle turned out to be exactly pi.
    1
  19679. Every time I see Ramanujan turn up in a Mathologer video, I'm still afraid that goddess might return to shock me with an unexpected loud gong while I'm thinking about the equations
    1
  19680. Wow this is fantastic coverage if soo many concepts, so fast and yet followable.. I vote the leaning tower proof as the most helpful.
    1
  19681. Okay, but which lacing method tightens most uniformly when you tighten one end? For example tying boots you want all the lacing to pull tight when putting them on, and to loosen when taking them off.
    1
  19682. Never knew the reason why the divisibility test for 9 worked. Did a spot check for divisibility by F with a few hexadecimal numbers, it works 😅
    1
  19683. 2. Let's watch one guy. He shaked hands with ≥3 others or didn't shake hands with ≥3 others (otherwise there would be maximum 2+2=4 other guys, two with whom he shaked hands and two with whom he didn't). Suppose he shaked hands with 3 people. Then if one of them shaked hands with other of them, there are 3 people who shaked thair hands (1st guy and these 2) and if nobody of them did they are 3 people who didn't shake their hands. If the first guy didn't shake hands with 3 people, the proof is similar
    1
  19684. in the card trick, why is knowing that it's a spade and 4 away from 6 enough to conclude its the 10 of spades? it could be the 2 of spades as well, couldn't it?
    1
  19685. رغم الضعف في الرياضة والانجليزي الا اني فهمت 👍🏼
    1
  19686. Brilliant. Tales of the unexpected
    1
  19687. Rock Scissors Lizard Paper Spock! My fav version
    1
  19688. deeep but I finally got it....phew.
    1
  19689. Professor, I love you and I love your videos but I have to make a correction here. Anytime we're talking about area under a curve you get zero marks. You always fail to add your units which are necessary for areas or volumes. As soon as we add our units to a function like 1/x, we find that eventually the distance between the curve and the Baseline will reach a distance smaller than the Planck scale and we can no longer add any area or volume. This is due to SpaceTime becoming a quantum foam below the distance of the Planck length. At that point we can no longer add any volume or area. This is a classic problem of a mathematician trying to solve a physics problem. Please let me know what you think Professor. Thank you for everything.
    1
  19690. hi
    1
  19691. When you have 2 2x2 (orange square) side by side (let's assume horizontally) there is 5 tilings and not 2^1x2^1=4: there is a tiling with extreme left and right dominos vertical and two centre ones horizontal the centre dominos cross the orange square borders and are therefore not counted in the 2^1x2^1 formula which assume the orange squares are independent. How it is possible to discard this possibility when constructing the next level of diamond (and similar possibility when you have more orange square touching each others) ?
    1
  19692. This needs a part 2. 13:40 does the new (unwiped) curve become symetrical for equilateral triangles? Can this be done for quadrilaterals or shapes with more sides? What about tetrahedra?
    1
  19693. I was most surprised by the bizarre optimal towers
    1
  19694. For some reason, those rotating coins remind me of boson and fermion spins. More interestingly, though, those two rotating pentagons remind me of (only apparent, illusory) 3D movement in 2D space. Could it be that, in order for the 3rd (illusory, relativistic/length-contracting) dimension to arise within the context of the (black hole) holographic principle, it requires not only an expanding black hole (giving birth to "time"), but one that must also be rotating (giving birth to the 3rd dimension, in conjunction with "time")? That would certainly explain why "time" and one dimension of space (the direction in which an object is moving) are so inseparably intertwined.
    1
  19695. "Proof by animation" that's the first time I've heard that!😅
    1
  19696. 21:48 Going from 2D to 3D in this case was quite easy to understand. Are there some similar theories about four (really five) sided pyramid shapes?
    1
  19697. The leaning tower of maths or lira was my favorite
    1
  19698. Algebra "autopilot" is awesome.
    1
  19699. I was with you until "What is log_b(x)?" starting at 17:28. I saw the example but nothing went "click". Maybe that was partly due to the fact that I have never used a log to the base of e-squared! To convert log bases all I remember is that you can use the log you have and divide the result by the log of the base you want. So: LogN(x) = Log(x)/Log(N) As Log(N) is a constant then your description makes more sense - at least to me!
    1
  19700. Your animation skills are superb. Thanks so much!
    1
  19701. Not only does 2+2 = 2x2, they are also equal to 2^2, 2^^2, 2^^^2, etc.
    1
  19702. Knowing Taylor series helps a lot
    1
  19703. .... now in a more professional perspective.... Then... Shouldn't Magentics be studied in a 3D modeling perspective?
    1
  19704. First of all, it was a delicious ( fat ) menu of mathematical delicacies. The most memorable one though, was the fact that the harmonic series contains every sum ( not sun! :) ) that converges to any real number.
    1
  19705. Pumpkin Pi!!!
    1
  19706. Seems like a good tool to unravel a mess of points into a regular polygon that can then be fit back around the original points creating a closed loop. Perhaps this might have applications for the traveling salesman problem. Connect all the points randomly, complete the process to a regular polygon, shrink wrap the regular polygon back around the original points.
    1
  19707. I LOVE THE SOUNDTRACK 🥰
    1
  19708. ☝🏻 My memory of the first experience with the slide rule (1968): “We put the 1 of the tongue on the 2 of the slide rule and then slide the slider on its scale to the 2. then we slide the slider on the 2 of the tongue and we get…. 3.9999, so…. Let's say: four!!" 😄
    1
  19709. @26:40, the blue crosslines are curved, the pink are straight, so just because they both end at the circle, it doesn't mean they are the same length.
    1
  19710. Made it all the way to the very end.
    1
  19711. Still waiting for a video about p-adic numbers! ;)
    1
  19712. I genuinely love Kempner's proof, so simple, but so amazing :)
    1
  19713. I learned divisibilty rules in school. Also for x|4 <= (x mod 100)|4 x|8 <= (x mod 1000)|8 x|5 <= (x mod 10)|5 x|6 <= x|2 & x|3 x|2 <= x is even For 7 exists no simplification. We learned it because of gcd and lcm. But, this was already very useful, after school. My girlfriend didn't learned it. She heard already of this. Intersting, it also works for mutiplieng for <1.
    1
  19714. Amazing!
    1
  19715. I never knew any of that proofs when I learnt phytagoras theorem when I was a teenager.😢 Thank you for sharing this.😊😊 I'm glad I found this channel today.😊 Subbed.
    1
  19716. using matrix row reduction, at 3:00 i decided to find the polynomials that generate all the values you showed: x^4/24 - x^3/12 + 11x^2/24 + 7x/12 + 1 x^3/6 + 5x/6 + 1 x^2/2 + x/2 + 1 x+1 1 where x=0 represents the diagonal of ones on the left of your diagram. Naturally, this extends the sequence into all complex numbers, even though it's totally meaningless to do so!
    1
  19717. also i'm pretty sure whenever the leftmost column is all ones and the final row is all ones, that the initial term at every step is x^n / n!, where n is how many rows up from the bottom you've gone (the bottom row is n=0, then the rest follow), and the constant is 1
    1
  19718. A dumb guy may want to appear clever by using this approach: 10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 14*15*29/6 - 9*10*19/6 = 5 * ( 7*29 - 3*19 ) = 5 * ( 5*29 + 1 ) = 5*146 = 2*365 -> fraction = 2
    1
  19719. The most beneficial thing I got from MTH1030 was getting to fully appreciate these videos.
    1
  19720. Really enjoyed it. Can't wait until you really get down to the nitty gritty. PS Have you given up on Mathologer 2 Channel?
    1
  19721. I read it in Thomas calculus btw nice video 😐
    1
  19722. lg2 = 0,3 lg4 = 0,6 lg8 = 0,9
    1
  19723. I'm glad you used your intelligence to help others learn math instead of turning into an evil genius :) ..... or are you an evil genius part time?! >_>
    1
  19724. Just super.
    1
  19725. I find geometry compellingly beautiful in a way which triggers wonder and awe, and those emotions are also the origin of religious reverence. Taking these things to their extreme is the reason the Pythagoras cult existed as it did, and it's the reason the sacred geometry people exist. I think they're riciculous but I understand the impulse.
    1
  19726. hey ..is this a magic clas??!?!?!?
    1
  19727. i loved chapter 2 the most :)
    1
  19728. I loved all the math and can’t forget the QED cat with a gamma meow 😸
    1
  19729. This is the exact kind of stuff that made me ask for a calculator when I was a kid.
    1
  19730. Did I miss an explanation of why the ear angles are 360/n and why the number of steps is n-2?
    1
  19731. My understanding is that n-1 steps converges to a point, and subsequently, so does n steps
    1
  19732. The most memorable thing for me was the simple proof of the fact that gamma is more than a half. For those who haven't figured it out, sliver of the blue area is a subset of a rectangular portion of the square. It is easy to see that each sliver is more than half of the area of the corresponding rectangle by the concavity of 1/x.
    1
  19733. AxB = A+B seems to equate area with length. Physicists wouldn't like that!
    1
  19734. Calculus is super easy. It’s all the algebra after the Calc is done that wrecked me
    1
  19735. A particularly valuable video. If there was a big nuclear war, these videos would be a must-save.
    1
  19736. n^2(n^2-1)/12
    1
  19737. this is basically what i learned in 3 weeks analysis in uni but before that i learned convergence, which took twice as long but didnt yield any deeper understanding at all.
    1
  19738. Awesome video! I love physics, so the most memorable thing to me is your quite intuitive use of Archimedes' law of lever to build the Lire tower
    1
  19739. 12:05 Ooh, feels very familiar to my thesis in college which was about simplices and simplicial numbers. So instead of summing the squares, I summed the triangular numbers, and so you get a triangular pyramid instead of a square pyramid. And you can stick together 6 of those to make a n(n+1)(n+2) rectangular prism. So sticking together 6 square pyramids felt very natural. (The rest of my thesis was about how many ways there are to split up an n-cube into n! n-simplices, the non-steppy kind. For n=3 it’s 9 ways, modulo all the symmetries of the cube including reflections. And for n=2, it’s just cutting a square in half diagonally, and there’s only one way to do that.) So, you can stick together 3 square pyramids together to be a cube, so it would be neat if you could put together 3 “Maya pyramids” to form a rectangular prism. Since 6(sum)=(n)(n+1)(2n+1), I got 3(sum)=(n)(n)(n+1) + (n)(n+1)/2=(n)(n+1)(n+1)-(n)(n+1)/2. So if geometry follows the math, you end up getting a extra triangle sitting out (or sunken in) and that’s why you need double, so you can stick the two “almost cubes” together to make a big rectangular prism. I think that’s what the video showed. I was able to stick together 4 steppy triangular prisms with a “double” the volume regular tetrahedron, and they make a perfect cube, if I remember right. The way you put together the middle tetrahedron is like stacking together the cross-sections when you start edge-on, and go through all the rectangles with the same perimeter and end at the opposite edge perpendicular to the first. Except here, the squares making up the rectangles are exactly diagonal. So the first cross-section is n squares in a line all touching at their vertices.
    1
  19740. I was taught this in school without the remainder part, which is pretty easy discoverable. thank you for the great explanations and jokes as always. Also as a note, I remember you told something about Sharkovsky's ordering where 3 is the "root of the chaos", or maybe I'm confused.
    1
  19741. Always a pleasure to watch your videos!
    1
  19742. Lovely
    1
  19743. The Wallace Product doesn't blow up because the terms ((n-1) * (n+1) = n^2 - 1) / n^2 for n >= 3 are all < 1 (8/9, 24/25, etc) where as in the other product each (n+1)/n term is > 1. yay squares
    1
  19744. Best part is the no nines series converging
    1
  19745. My favorite was that there are no integers among the partial sums of the harmonic series because the proof is really clever.
    1
  19746. too goiod
    1
  19747. 1:00 you can apply the same method to "prove" that the length of the hypothenus of a square triangle is equal to the sum of the 2 sides ie c=a+b instead of c²=a²+b². Here the problem is obvious: Even if you cut and cut... To get tiny triangles, you are assuming that ci=ai+bi for those tiny triangles hence it's not a surprise that this is what you get for the original big triangle. I think the problem is to evaluate the error done by the approximation. It must be going to 0 faster than 1/n such that the total error for the n elements goes to zero too.
    1
  19748. Because mathematician, I assume this proof was extended beyond 2D.
    1
  19749. This is far as I got, I'm pretty sure 8 is the last power of two: The closed form of the sequence is: (n^4 + 2n^3 + 11n^2 + 34n + 48)/24 We are looking for: (n^4 + 2n^3 + 11n^2 + 34n + 48)/24 = 2^m Rearranging it, we get: n^4 + 2n^3 + 11n^2 + 34n + 48 = 24 * 2^m n^4 + 2n^3 + 11n^2 + 34n = 24 * 2^m - 48 n^4 + 2n^3 + 11n^2 + 34n = 24(2^m - 2) Plugging in some values in the LHS we get: 0, 48, 144, 336, 696, 1320, 2328, 3864, 6096, 13440... And for the RHS we get (starting from 1): 0, 48, 144, 336, 720, 1488, 3024, 6096, 12240, 24528... I can't seem to use the mod 2 trick. Both sides are always divisible by 48 and attempting to use rational root theorem causes issues: n^4 + 2n^3 + 11n^2 + 34n - 24(2^m - 2) = 0 The roots are -24(2^m - 2), while we can assume n, m > 8, this is still a lot of factors to prove exhaustively. This is as far as I got. I'm a bit tired now so I'm going to take a break. I'll edit this if I find a proof. Maybe someone with more experience in number theory can figure it out.
    1
  19750. I was going to suggest "conjugate triangle" but it seems it means something else. So I'm officially proposing "explicate triangle" from the Latin "explicare"- to unfold or extend. Don't need royalties, I'm happy with a credit!
    1
  19751. I think the reasons this is not usually presented include 1. the problem is usually introduced in elementary calculus textbooks, so that is the source material most youtubers draw from (often indirectly), 2. most youtubers think presenting shocking facts brings in more viewers than accurately explaining mathematical truths, and 3. most of these videos are nowhere near 21 minutes long. Also, while I think this is not the actual reason, there is another hypothetical reason one might choose not to take this approach: you can't compute an infinite sum without either modern calculus or some other particular set of unstated axioms. Historically, there are zillions of "proofs" of the convergence or divergence of various series, and it's not obvious to viewers why some reach correct conclusions and others don't. The proofs in the video can be made rigorous by assuming the direct comparison test (which is pretty obvious and uncontroversial), that series of positive terms are associative (which is a somewhat nuanced point), and that a convex surface with a planar bounding curve has greater area than the plane figure its curve describes (which is equivalent to an axiom due to Archimedes and is an extremely nuanced point).
    1
  19752. That was kind of unexpected and fun, thank you. What about 4D cubes? 8? 16?
    1
  19753. I wonder if the more creative cliff solution for 20 bars also works with disks like coins. Because their mass isn’t evenly distributed side to side. Hope I win a book! Keep these videos coming. Thanks!
    1
  19754. Anything can be multiplied by 1 :). Finding 1 × 1 is always interesting (considering that you can go into the irrational :). You can also make the gods laugh 1x1x1. У Вас всегда отличная математика и алгебра, завидую Вашим студентам Профессор. :).
    1
  19755. This is very useful for a problem I'm working on right now. Thanks you so much for making this!
    1
  19756. Mathematician with the wierdest hat?
    1
  19757. Beautiful! Stray thought: plus and minus form a pair, in other words, a number is positive or negative - a binary choice. But what if this choice could be taken from a continuum? No, I did NOT think this through 😂
    1
  19758. so the tangents to the circles of first four matrix of the of the fibonacci numbers, , also limit the first whole integral sequence to the the right angle triangle. Very neat mr maths
    1
  19759. This is just a complicated version of the fact that a finite volume has infinite planes, a finite plane segment has infinite lines, and a finite line segment has infinite points.
    1
  19760. Trigonometry was also invented in India sine was called jya. But "jya" travelling to Europe through middle East turned in Sine
    1
  19761. Interesting how at 7:58 you can see this is also a representation of the svastik, which means the wheel of life // turn of good fortune // the universal forces at work
    1
  19762. Well, I hope you haven't discovered that assuming the hairiest Australian has no more than 6 million hairs (20% more than average!?!) Is exactly a dangerous thing to assume.. I'm not super sure it's a safe thing!
    1
  19763. John Wallis, also presented values of ratios of continued fractions
    1
  19764. 11:45 it is 9 4 13 17... bended 4*17 = 153, 4*13*2=104 and 4*17+9*13=185
    1
  19765. Watched at 1.5x, impressive animation
    1
  19766. Awesome video! Very well done.
    1
  19767. Drop the equals sign in "0 =" and maybe even axe the zero too and leave it implied. The Eye of Providence sits atop the unfinished pyramid on the obverse side of The Great Seal. The zero representing The Eye can be left implied. :^)
    1
  19768. Very Very Nice!
    1
  19769. Yeah the craziest thing was that the no 100 zeros sum is larger than the no 9s sum. My mind is blown
    1
  19770. I'd call that the third in the row: circle, elips, trilips(??), can this be expanded ti infinity-lips?. Btw It reminds me of a Wankel engine.
    1
  19771. Most memorable part were: 1.700 year old proof by Nicole Oresme which also proves that in theory there can be infinite rods that are beyond the edge of the cliff. That's CRAZYY!!! 2. "Terrible aim" where it misses every positive integer except 1 even when the terms gets smaller and smaller. Its like partial sum's and positive integer greater than 1 are both north or south poles of magnet as they don't get along. 3.Sum of no 9's is finite and its visual proof.
    1
  19772. Hanc marginis exiguitas non caperet
    1
  19773. I remember in high-school we were shown a brief video tape of a college prof working Euclid's proof on one of those big lecture-hall chalkboard walls. The thing I remember best was how the lecturer had a yardstick to help them draw dotted lines, but everything else was freehand and darn near perfect. I remembered that the proof existed and that It made sense to me as it was being explained -- but I certainly couldn't reproduce it. I think the Mathologer channel might be the first place I ever saw animation of the visual proofs shown at the beginning of the video, although I'd certainly seen the diagrams at several odd occasions over the years.
    1
  19774. Love from Bangladesh
    1
  19775. All of the blue bits patrician the square vertically into rectangles of smaller and smaller height indefinitely. Because they are convex along the diagonal, they take up more than half of their respective rectangle. Therefore, added together these blue bits must take up more than half of the square.
    1
  19776. Hmm... the multiplication by halving and doubling looks awfully like Russian multiplication - never thought of a cross link to logarithms before. Love the graphics - just makes it so satisfying.
    1
  19777. First glance at the thumbnail: feh, they're messing with us with click bait again. 2 seconds later: oh, wait, that's actually right.
    1
  19778. Dude, it's No Nine November
    1
  19779. Was taught the visual proof sometime in the 80s, in primary school :)
    1
  19780. This is really sad. When I saw the movie I was thinking they really picked a too easy question for this demonstration. This level would never qualify for real math competition (I still hope)
    1
  19781. 35:35 Jetzt und immerdar und von Ewigkeit zu Ewigkeit
    1
  19782. that was the first "curvy origami" that i made in my persuit to make an origami Torus XD
    1
  19783. Could you please normalize the volume of video before posting, professor? It's too quiet...
    1
  19784. Theorem of Phithagoras!
    1
  19785. That was good.
    1
  19786. Looks like a common E6B flight calculator. They're used every day. SMH.
    1
  19787. You forgot to point out that the two terms in Binet's formula are the golden ratios :)
    1
  19788. If the number of disks is even you move the smallest disk in the opposite direction of the ending peg, if it is odd you move it in the direction of the ending peg.
    1
  19789. I have never seen a geometric proof like this! Very original! If any of you wants a more motivated proof of ptolemy's inequality, use inversion around any vertex of the cuadrilateral.
    1
  19790. I have this slide ruler device🤩
    1
  19791. Most memorable: the weird optimal towers. It's really strange. I feel like that could be exploited in real world physics, if it isn't already
    1
  19792. I dove into the sums of positive powers myself. I went at it like this: Let F(x) be the sum of x^4. That implies that F(x) - F(x-1) = x^4 for all x. It can easily be shown that this expression for a polynomial order n is always n-1, so F(x) must have order 5. If a general fifth order polynomial is substituted, the system of equations with Pascal’s triangle terms appears. It is very easy to solve by hand after the powers of x are grouped. I also stumbled into replacing all terms with their series. I found an elegant solution to the sum between ANY two integers. Given a sum formula for a polynomial, the sum of all integers including the boundaries is F(b)-F(a-1). It looks just like integration! I called the process of f(x)-f(x-1) “taking the difference;” it may be done term-by-term like the derivative, replacing each term with a binomial coefficient pattern. The “integral” involved replacing each term with it’s antidifference, the function that has the term as the difference between its values. These sums are like calulus with h set to 1, but way messier! This all is easy to see as spacing between points on a graph, and easy to prove with a telescoping series (replace each f(n) in a general sum with F(n)-F(n-1), which was, after all, defined to satisfy this). Thank you for making something on the Bernoulli numbers. As you said, it is definitely not accessible. I will have to watch again!
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  19793. Lieber Burkard Polster, beim Thema Mathematik und verrückt sowie viele Beweise muss ich das Buch „Fregattenkapitän Eins“ von Wladimir Ljowschin (Moskau, 1968) denken. Im vorletzten Kapitel heißt es: 32. Nullter Der Löwe in der Wüste Unsere Fahrt nähert sich dem Ende. Wir fahren durch den Golf des Humors. „Ich hoffe, daß keiner fragen wird, was Humor mit Mathematik zu tun hat“, sagte der Kapitän, als wir in den Golf einliefen. „Jedes Kleinkind weiß bekanntlich, daß Humor überall notwendig ist. Ihn braucht der Schriftsteller, der lustige Erzählungen schreibt, ebenso wieder Wissenschaftler, der schwierigste Forschungen unternimmt. Manche Leute denken, daß die Wissenschaftler alle todernst und langweilig sind. Ganz im Gegenteil! Sie lesen, hören Musik und lachen gern. Sie schätzen einen fröhlichen, geistreichen Scherz. Außerdem muß ich euch sagen: Ein Mensch, der mitunter 999 erfolglose Tests machen muß, um ein erfolgreiches Resultat zu erzielen, kann ohne Humor überhaupt nicht leben. Je ernsthafter eine Arbeit ist, desto wichtiger ist es zuweilen, sie mit einem Lachen zu unterbrechen. Deshalb denken sich die Wissenschaftler gern die unmöglichsten Aufgaben aus, stellen lustige Fragen und finden geistreiche Antworten darauf. Witz und Findigkeit haben großen Menschen nicht selten in schwierigen Situationen geholfen.“ „Sie meinen bestimmt Christoph Kolumbus“, unterbrach Steuermann Ypsilon den Kapitän. „Bekanntlich wollte Kolumbus auf dem kürzesten Weg nach Indien gelangen. Er beschloß, nicht nach Osten zu fahren, um Afrika herum, wie das früher alle gemacht hatten, sondern nach Westen, um auf diese Weise ein übriges Mal nachzuweisen, dass die Erde eine Kugel ist. Die Expedition war recht aufwendig. Doch die spanischen Würdenträger, an die er sich um Unterstützung wandte, eilten nicht, den verwegenen Seefahrer auszurüsten. Sie hielten das Unternehmen von Kolumbus für unsinnig. Sie dachten, daß, wenn man tatsächlich von Westen aus nach Indien gelangen könnte, schon längst jemand auf diesen Einfach gekommen wäre. Als Kolumbus ihre Argumente hörte, nahm er ein Hühnerei und bat jemanden der Versammelten, es mit der Spitze so auf den Tisch zu stellen, daß es nicht umfiele! Die Würdenträger versuchten es, aber vergeblich. Da stellte Kolumbus mit leichtem Druck das Ei mit der Spitze auf die Tischplatte. Die Schale platze ein, und das Ei stand kerzengerade. ‚Seht‘, sprach Kolumbus, ‚bislang ist keiner auf diese Idee gekommen, dennoch…‘ Seine Findigkeit blieb nicht ohne Wirkung, und er erhielt, worum er gebeten hatte. Seither ist das Ei des Kolumbus zu einem geflügelten Wort geworden. Übergings gelangte Kolumbus mit seiner Expedition nicht nach Indien, wie er gedacht hatte, sondern landete in Amerika. Auf diese Weise wurde ein neuer Erdteil entdeckt, und schuld daran war ein Hühnerei!“ Als der Steuermann seine Erzählung beendet hatte, fielen allen haufenweise Anekdoten über Wissenschaftler ein. Nur mir wollte absolut nichts in den Sinn kommen. Ich habe nämlich, müßt ihr wissen, unmittelbar mit wissenschaftlichem Personal wenig zu tun. Schließlich erzählte ich aber doch, wie meine Mutter an einer wissenschaftlichen Physikerkonferenz teilgenommen hatte. Man denke nur, da hatten sich bekannte Wissenschaftler eingefunden! Nach ernsthaften Disputen fingen sie an zu überlegen, wie man einen Löwen einfangen könnte, der zufällig in die Wüste geraten ist. Ein Wissenschaftler schlug vor, ein Riesensieb zu nehmen und den ganzen Wüstensand durchzusieben. Auf diese Weise würde der Löwe zweifellos ins Sieb geraten, aus dem er um nichts auf der Welt hinauskriechen könne. Ein zweiter Wissenschaftler schlug vor, die Wüste durch einen Zaun in zwei gleiche Hälften zu teilen. Klar, daß der Löwe sich in einer dieser Hälften aufhält. Diese Hälfte müsse wiederum in zwei Hälften unterteilt werden. Jetzt braucht der Löwe nur noch in einem Viertel der Wüste gesucht werden. Das ist aber schon wesentlich leichter! Das Viertel ist wiederum in die Hälfte zu unterteilen und so weiter, bis der abgeteilte Abschnitt so klein geworden sit, daß er gerade der Größe des Löwen entspricht. Dann kann man ihn sozusagen mit bloßen Händen fangen. Der dritte Wissenschaftler … Welche Methoden der dritte, vierte und alle übrigen Wissenschaftler vorgeschlagen hatten, habe ich vergessen. Aber der Kapitän meinte, daß diese zwei vollkommen ausreichend seien. Wenn euch noch irgendwelche Möglichkeiten einfallen, so schreibt mir bitte! Es macht doch großen Spaß, in der Wüste einen einsamen Löwen einzufangen!
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  19794. I liked the colour proof best because it was the way I was intuiting
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  19795. Math. The drug that becomes manifest after the fix.
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  19796. I love the dad joke Thank goodness you're a Maths channel not a comedy
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  19797. For me, the most memorable result is the "greedy sum".
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  19798. Most memorable thing is that t-shirt.
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  19799. Also a really nice task: Imagine an ant on a circle. The ant walks with 0.1 m/s on the surface of the circle. Every second the circle, at first with a radius of 1 meter, grows its circle immediately by 1 meter. How long does the ant need to make its way all around the circle? I know the answer, so let me know your ideas! :-)
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  19800. I think that circle is just a coincidence (but one that happens every time).
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  19801. i expected a proof for 4 progressions among squares
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  19802. Criss-cross lacing also has the benefit that you can make it with one hand on a high boot with open eyelets. We called this "military" (or sometimes "Nazi") lacing.
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  19803. woo
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  19804. I have a question why madava are looking so chad 🤣🤣🤣
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  19805. Wonderful. Very nice. Sometimes a question arises. There are numbers, or is it a way to perceive information. And as it is correct, the number of combinations, or the frequency of occurrence. Great performance. Thank you.
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  19806. With the tray the weight will be more than 1, 2 or 3 kg, lol
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  19807. The second T-shirt link doesn’t work.
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  19808. Yep. First time I have seen those visual proofs
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  19809. My vote is on the gamma approximate formula. PS. It ought to be possible to hollow out the parabolic tower at 27:51, oughtn't it? I wonder what the most hollowed out structure looks like. PPS. Got intrigued by the gamma vs -1/12 teaser, rewatched https://youtu.be/jcKRGpMiVTw... Ah, I can see how that may make sense.
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  19810. The sum of reciprocals of primes in arithmetic progressions is infinite
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  19811. I'm an actuary and I've never seen this proof. So simple!
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  19812. One of most memorable things is that 100 0s sum is bigger than no 9s sum though it consists of so much immensely smaller numbers!
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  19813. Ou jäa, mental massage! Thanks so much.
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  19814. 37:30 You could use the '3D' interpretation, it is easy to see that if you moved the cube to see each face front on, one would be fully grey, the other fully blue, and the top one fully orange, so we have 30*2 for each. There is no way to look at the cube up-down and see a grey or blue tile. Not sure how to make it more rigorous tho
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  19815. Sick. Also found in patreon 3b1b, Paul Dirac and Euler xD
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  19816. I wonder if in the example of 1mm spacing and 1 km triangle.... This hexagon will tend yo be a square...
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  19817. Really enjoyed that video. I wrote a math blog post a few years back on this fascinating subject : https://www.mathblog.uk/mathblog/article.php?page=EulerMacLaurin.php
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  19818. Colouring proof....
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  19819. "Wait, did you know that there's a direct correlation between the decline of Spirograph and the rise in gang activity? Think about it!"
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  19820. Donkey: Woooooooooooooow stares LET'S DO THAT AGAIN!
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  19821. 8:45 Background music plz.anyone
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  19822. May be telsa mean by saying vortex is that u fall inside and never come out if ur not a mathematician
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  19823. 4k-1 is much simpler than 4k+3😉
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  19824. In the last proof (there's an unique way of writing prime number p as sum of two squares) the fact that the prime is of the form p=4k+1 was never used right? So it is actually proving that "If can be decomposed in squares, then that decomposition is unique", right?¡
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  19825. Nvm just realized all primes are odd, and all odd numbers are either of the form 4k+1 or 4k+3.
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  19826. 666th partition number: 11393868451739000294452939
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  19827. I am a graduate student in Mathematics, and this same semester I am taking a course in Analytic Number Theory, so this video is a complete gift for me! I can appreciate even more your explanations as I am working on this stuff in this period, and I would love you to cover other such topics in this field and pushing the boundary even more (PNT, Dirichlet’s theorem on arithmetic progressions, sumsets and sieve methods, or whatever you want to pull out of your magic mathematical hat)!
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  19828. My favorite part is when you say "welcome to another mathologer video".. I always get so excited when I hear that. My favorite math part from this video was the formula for the partial sums of the harmonic series... and the no-nines proof.
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  19829. When Mathologer does YouTube Shorts:)
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  19830. Badass music at the end 😎
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  19831. Hi Mr. Polster, here is another theorem, which is familiar; about rotating squares. I would be very happy, if you would mention this theorem in one of your next videos. Many thanks. Özgür Baskaya https://youtu.be/mPbocVRGw7E
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  19832. 4/3πr³ is a great argument for the viral math problems like 8÷2(2+2)=X and X=1. 4/3 is the juxtapositional division contained within the coefficient of the variables. So 8/2(2+2)=16, with 8 halves as the coefficient of the parenthetical, but 8÷2(2+2)=1 since there is no grammatical reference to a factored coefficient.
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  19833. @Mathologer  The problem seems to stem from the issue that early computer programmers were not very good mathmaticians. So the way that computers handled the grammar was severely lacking. The Neutral Element property of multiplication is pretty clear about how coefficients work. Later generations then based their mathematics upon how computers and calculators worked. Which just happened to be incorrect. My favorite example is Distance ÷ 2πR= Distance ÷ πd= also 120"÷2π(11"+1")=1.59 revolutions 120"÷π(22"+2")=1.59 revolutions as well. Not to mention that revolutions, measured in terms of area is ridiculous.
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  19834. 19:56 just put negative angles
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  19835. Ans=implex + C
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  19836. It took me way too long to realize it, (and I still don't have a proof for it) but if you start with g[1](z) = 1/(1 - z), and let g[n](z) = g[n-1](z) + z^(n-1)/(1-z)^n, g[5](z) is the generating function for the sequence in the thumbnail. This gives an easy way to calculate a closed form for the sequence; letting C(n, k) be the binomial coefficient function ("n choose k"), then the formal power series for z^k/(1-z)^(k+1) = SUM(j=0..inf; C(j, k) z^j). Edit: and as it turns out, I don't have to, because the video contains the proof :)
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  19837. I always watch you on E=mc^2 theirtubespeed, for a deeper understanding...
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  19838. I already forgot math, engineer who turned into business... at some point of my life I wanted to tie my life to math, ended up as a businessman who identifies himself as an engineer, thank you for those videos that reminds me of where I come from, science!
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  19839. Slide rules... rule!
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  19840. Wonder what the visual implications of these identities can be considering multiplication can often be represented as an area of a picture while adding can be represented as a lineament in the same picture... Could be a worthwhile novelty artistic thesis
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  19841. Two Green/Two Black removal from Chapter 1: No, it will not always work. Example: remove the green tiles adjacent to a black corner tile. There will be no way to put a domino onto that tile and have it cover green tile as well. Same goes for removing two black tiles adjacent to a green corner tile.
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  19842. Math: "A - A not equal to zero" Computers: "I'm gonna pretend I didn't hear that"
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  19843. I read Geometry and the Imagination many years ago back in college. It's one of my favorite books. I feel bad that I don't remember this proof from the book.
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  19844. (@22:30): The 11-vortex demonstrates the superiority of dozenal (base-12) over decimal; ergo, imperial is better than metric!! ;^}
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  19845. Beautiful as always
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  19846. Is this used in engineering?
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  19847. How is 243 a power of 5?
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  19848.  @Mathologer  OK I was thinking 5^3 ;)
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  19849. 25.807
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  19850. 9:25 A dodecahedron has 20 vertices and 12 faces, yes? Nevertheless the formula still works for the numbers for icosahedron (V = 12, E = 30, F = 20). Anyways, very informative as always, enjoyed watching, ay. :)
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  19851. So how much pizza is there?
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  19852. Nice video!
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  19853. I think my favourite is the bizarre overhang maximums. I like the way it looks sort of pre-fractal.
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  19854. 1+1/2+⅓ ... doesn't exactly "explode" to infinity ... it crawls there.
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  19855. Oh my. that "orange" and "green" tiles. You nailed it! Just perfect. You find two exact shades which I have no way to distinguish. The sweet spot! I maxed out the brightness. I peer into the screen for a minute, and I could barely tell, that horizontal tiles are a tiny bit darker than vertical ones. still have no clue which is which. they both pale yellowish-green. There are about 7% of colorblind people in the world. We should differently start the "Colorblind for Better Infographics" movement.
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  19856. And this is why theorists of all sciences should always use a physical mean of testing their paper theories. Any mason or carpenter would be able to tell you that using a soft, non-stretchy string can allow you to determine the area of a circle with great precision; using a weight scale, sand and a larger container can let you determine the correlation between the weight of a sphere and its volume, later you can easily determine a value very close to pi Based on the relation between sand volume and sand weight. Also, if you place a sphere inside a large cubic container and fill it with water, then measure the water level, then remove the sphere and measure the water level again, this exercise will give you with almost absolute precision the area of the sphere.
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  19857. Thank you for this Christmas gift for us all.
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  19858. It's taught in 9th standard in Indian schools complying with Central board. I'm sure other schools compliant with various state boards also teach this.
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  19859. 11:08 Challenges! 1. Idk your choice 2. Campaign probably means politics is involved 3. You get the wrong answer out the other end. Proof: try it on a unit square but with two vertices switched. The area should be 1/2 but what comes out the other end of the equation is 0 4. No. Maybe later but it's 1 AM and I don't feel like doing it now
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  19860. Is this what Bernie Madoff did with my money?
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  19861. ok the reality will prevent paint molecules getting into a certain depth because it will become thinner than a singular paint molecule
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  19862. this is a very elastic rubber band
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  19863. 1/infinity is Zero plus 1/infinity, just not quite Zero, but infinitely close.
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  19864. Pointing UP is a relative concept, in respect to ground. Think it as standing on earth (which is a sphere), the UP direction always points outwards along the radius of the sphere. Once you understand this, there is no mystery any more.
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  19865. Kudos for the comments at the end of the section "What paradox?" As a scientist I get annoyed when things (volume and area in this case) are compared that have different units. Good to see you dealing with that.
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  19866. I have to run all this past me a few more times. But its good you made those 3d slices green and not yellow. Can you guess why?
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  19867. I was giddy like a little kid throughout!
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  19868. Fun Video! What language is on the first can? I can see some Hebraic nuances in the block script, but it's not Hebrew.
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  19869. p=4k+1 for primes that are sums of two integer primes....I get that, and the ones that aren't =4k+3. But why is the second one not written as 4k-1? I understand p mod 4 =1 versus p mod 4=3, however, I can't help but wonder if the math involved might not be easier using it as 4k-1
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  19870. I really liked the bizarre towers that maximize the lean, yet can't be constructed layer by layer. That'd be chapter 2 for me.
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  19871. Second challenge: if you take the j and k as large as possible they will be (m+1)/2 and (n+1)/2. Then the brackets will be (4cos^2 (j*pi/(2j))+4cos^2 (k*pi/(2k)))=4cos^2 (pi/2)+4cos^2 (pi/2)=4*0^2+4*0^2=0. So the one of the factors will be 0, and the whole composition will be 0. ))
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  19872. 7:50 - sobald ein Mathologer-Buch auf Deutsch herauskommt, nehme ich es in mein Sortiment auf und mache ein Schaufenster daraus. Und dort stapeln sich die Bücher dann ganz genau so! :)
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  19873. I made it to the end. I'm 76 and could get enjoyment from following most of it (& your other shows) without the irritation of having to take exams. Jeff in England
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  19874. You could be making all this up in all your videos and I would not know but I watch to the end anyway.
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  19875. Love that cyclic solution to Hanoi!
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  19876. that outro music is dope.
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  19877. At 6:50, doesn't that equality only hold if the "mental lines" are straight which is what you're trying to prove?
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  19878. I have waited long for a Math video to come out
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  19879. Second one is more immediate
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  19880. 16:08 the coding / app request seems a bit paradoxical .. you want digital software to create some analogue output with infinite precision? .. anything with infinite precision is impossible..thats not how how the world works.. also, you cant create more information than what went into the digital software..maybe a finitely precise app would be a better question....than how precise exactly do you want it to be?
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  19881. There are a couple of "arbitrary precision arithmetic" libraries for various programming languages. Of course everything that humans produce is not infinite, but at least you can scale resources to the amount you can afford. I guess that is meant here
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  19882. Huh, I usually pronounce sinh as “sinch” (pronounced like the word “cinch”)
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  19883. finally a mathologer video i was able to keep up with! Thanks
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  19884. I just saw the community post so I’ll do my part!
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  19885. Sure, a^3+b^3=c^3 cannot work, but what about a^3+b^3+c^3=d^3+e^3? Nobody says the pattern has to start with just two+one terms.
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  19886. Is the percussion noise that you play at each section marker the very same as the noise used in that famous ‘turning a sphere inside out’ video?
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  19887. What if we go down with dimensions instead of up? What does (x + 2) ^ -1 teach us about the -1 dimension?
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