Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. In case you're wondering, the dots on his shirt are the digits of pi
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  2. 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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  3. 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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  4. Stop it this is just a ClickBait title!!
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  5. 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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  6. Ty
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  7. 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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  8. 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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  9. What’s amazing is how this represents the three aether modalities. Dielectricity, magnetism and electricity.
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  10. 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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  11. 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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  12. 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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  13. 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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  14. Except for the wizards, I think- they’re the sanest of the bunch.
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  15. Another great video! I always tell the class about your channel and Ramanujan
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  16. 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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  17. I loved the video, great as always!
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  18. 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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  19. 7:52 "Kasteleyn's watching you" i'm... genuinely scared now, im sitting in a dark room
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  20. 4:27 The blooper: the characters should have been stacked up vertically according to the traditional Chinese-Japanese-Korean writing system.
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  21. I guess I'm the first to point out the "Pi mal Daumen" pun in the intro?
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  22. 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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  23. 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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  24. Thank you so much what a great video
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  25. 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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  26. 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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  27. 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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  28. 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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  29. This was heartachingly beautiful. Thank you.
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  30. 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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  31. 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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  32. 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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  33. 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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  34. I define phi as "where addition meets multiplication", BECAUSE (1/phi)+1 is phi.
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  35. 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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  36. Nobody has finished the video yet!
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  37. 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)
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  38. 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.
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  39. 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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  40. What do you do when you’re not making these videos?
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  41. @Mathologer, is it the promised crazy Galois theory video ? _
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  42. 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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  43. Hey. 3b1b Daniel Radcliffe Avatar Euler Mandelbrot 'My son' Are your patreons
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  44. 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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  45. Be still my beating heart, another Mathologer video!
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  46.  @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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  47.  @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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  48. same but in pakistan
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  49. 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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  50. 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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