Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. 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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  2. 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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  3. I really enjoy watching your enthusiastic explantation of these concepts. The animations are beautiful.
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  4. 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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  5. 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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  6. you are a genius and such a good teacher who is willing to share to the world, thankyou
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  7. 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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  8. 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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  9. 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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  10. 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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  11. 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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  12. Congratulations Mr. Polster (and Marty) for your 100th video! The more I watch, the more I ❤ it.
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  13. 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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  14. Recently stumbled back across mathologer videos, finding them so relaxing and interesting at the same time :) keep it up❤️
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  15. Should ban fake accounts and bots
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  16. I'm proud to possess a magic log wheel since I was about 17yo. Now I'm 70. Great presentation!
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  17. 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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  18. 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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  19. 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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  20. 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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  21. 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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  22. 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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  23. 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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  24. 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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  25. They are never long enough my friend. I could watch these for days.
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  26. Are you really a real mathematician if you can do arithmetic in your head?
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  27. 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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  28. First like from me
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  29. 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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  30. omg domotro
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  31. I had the same reaction
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  32. That Hex board Reminds me of One qbert video game.
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  33. Tenagon ? I prefer Decagon. Maybe because I'm Greek ?
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  34. My brain took 10 minutes to get the HO^3 joke.
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  35. Let's run the Gregory-Newton formula on the prime numbers !
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  36. This is my favorite mathologer video in a while. Quite easy to digest, and beautiful.
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  37. Great
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  38. 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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  39. Great. Another thing I've never heard of!
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  40. 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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  41.  @Mathologer  How many videos are in the pipeline ahead of Archimedes's'? I CAN'T WAIT!!!
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  42. Do another juggling video?
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  43. 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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  44. Learning calculus is easy, learning about calculus is hard
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  45. Nice π-digits tee shirt there, Burkhard! 😊
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  46. As always: beautifully and clearly presented.
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  47. Its due to channels like yours that makes youtube videos worthwhile to watch.
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  48. He said the numbers came to him from another place in his mind, not of this world.
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  49. @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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  50. 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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