Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. Have you been 4k+1 this year?
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  2. It is a good day when Mathologer uploads
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  3. Most memorable : " Convergence of no 9 series ".
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  4. All hail lord curriculum
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  5. 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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  6. 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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  7. 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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  8. 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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  9. I made it to the end but... Der anfang ist die ende und der ende ist der anfang.
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  10. 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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  11. 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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  12. Okay, this is actually one of the most beautiful things I've seen in math.
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  13. 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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  14. 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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  15. This is an absolutely amazing video. I needed a few minutes to come back to my senses after watching this
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  16. 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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  17. 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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  18. 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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  19. 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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  20. 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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  21. 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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  22. 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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  23. Mathologer just roasting introverts that go to parties
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  24. 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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  25. Most memorable: The proof that the bishop came up with, beautiful simplicity
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  26. 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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  27. Awesome video. From the animation to the soundtrack to the explanation. Just amazing.
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  28. 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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  29. 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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  30. 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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  31. 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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  32. 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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  33. 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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  34. 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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  35. 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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  36. I love your channel, but the AI images and videos are really unsettling.
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  37. 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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  38. 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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  39. 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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  40. 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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  41. "Or...make yourself famous."
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  42. 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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  43. 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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  44. 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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  45. 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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  46. 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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  47. x to doubt, actuaries are bad at math and worse at calculus - src am actuary (gladly left)
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  48. 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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  49. 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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  50. 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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