Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. 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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  2.  @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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  3. I collect coins as a hobby so double thumbs up!
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  4. 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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  5. let's go early gang
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  6. 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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  7. 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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  8. 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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  9. These animations were better than my whole calculu's teachers. Congrats for this awesome job
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  10. This is one of the best mathologer videos so far, incredible. I inspires to play with math.
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  11. 8:05 "woooooah.. not hard to see why you struggle" School in a nutshell
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  12. 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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  13. @8:50. There are 6 different patterns. They remaining ones are redundant due to rotational and mirror symmetry.
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  14. 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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  15. 1:22 knowledge and hair doesn't fit to the same head? 😀
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  16. Thank you for giving the credit that Indian mathematicians really deserve.
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  17. 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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  18. 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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  19. 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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  20. 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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  21. 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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  22. 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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  23. 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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  24. 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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  25.  @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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  26.  @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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  27. 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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  28.  @Mathologer  Yes, good point! I guess it is so well known that we take it for granted. Poor Pythagoras!
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  29. Brilliant! Thank you! Love your approach to explaining things! ( Merry Christmas! ... let's make 2020 the best year so far! )
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  30. 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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  31. Chess is being popular day by day everywhere becoz of queen's gambit😂
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  32.  @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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  33. 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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  34. 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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  35. 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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  36. 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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  37. 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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  38. 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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  39. 6513516734600035718300327211250928237178281758494417357560086828416863929270451437126021949850746381 = 16120430216983125661219096041413890639183535175875^2 + 79080013051462081144097259373611263341866969255266^2
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  40. 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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  41. Your T-shirt looks odd to me.
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  42. I didn't even notice that the movie and tv references were gone, nice to see a revival!
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  43. I have a book of my father's titled "Turtle Geometry" from MIT, 1982. It is this exactly.
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  44. 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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  45. This is really beautiful. It's even more beautiful than the theorem itself, which was hard to beat.
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  46. 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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  47.  @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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  48. 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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  49. 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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  50. 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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