The best A – A ≠ 0 paradox
This video is about a new stunning visual resolution of a very pretty and important paradox that I stumbled across while I was preparing the last video on logarithms.
00:00 Intro
00:56 Paradox
03:52 Visual sum = ln(2)
07:58 Pi
11:00 Gelfond's number
14:22 Pi exactly
17:35 Riemann's rearrangement theorem
22:40 Thanks!
Riemann rearrangement theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem This page features a different way to derive the sums of those nice m positive/n negative term arrangements of the alternating harmonic series by expressing H(n) the sum of the first n harmonic numbers by ln(n) and the Euler–Mascheroni constant. That could also be made into a very nice visual proof along the lines that I follow in this video https://youtu.be/vQE6-PLcGwU?si=iWTasKqo_JFn4etG&t=1321.
Gelfond's number
e^π being approximate equal to 20 + π may not be a complete coincidence after all:
@mathfromalphatoomega
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. 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.
(@kasugaryuichi9767) I don't know if it's new, but it's certainly not well known. To quote the Wolfram MathWorld article "Almost Integer": "This curious near-identity was apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe, but no satisfying explanation as to "why" e^π-π≈20 is true has yet been discovered."
Ratio of the number of positive and negative terms
It's interesting to look at the patterns of positive & negative terms when rearranging to Pi. We always only use one negative term before we switch. The first ten terms on the positive side are: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220,... If you look at the differences between terms, you get: 22, 23, 23, 23, 23, 24, 23, 23, 23, 23, 23, 23, 23, 24,...
The reason for this is that Gelfond's number is approximately equal to 23. It turns out that if an arrangement of our series has the sum pi, then the ratio of the numbers of positive to negative terms in the finite partial sums of the series converges to Gelfond's number. This is just one step up from what I said about us being able to get arbitrarily close to pi by turning truncations of the decimal expansion of Gelfond's number into fractions. Similarly for other target numbers. For example, to predict what the repeating pattern for e is, you just have to calculate e^e :)
@penguincute3564 thus ln(0) = negative infinity (referring to +0/1-)
Bug report: At the 1:18 mark, I say minus one sixth when I should have just said one sixth.
Music: Silhouettes---only-piano by Muted
T-shirt: Pi Day Left Vs Right Brain Pie Math Geek T-Shirt https://tinyurl.com/3e3p5yeb
Enjoy!
Burkard
Видео The best A – A ≠ 0 paradox канала Mathologer
00:00 Intro
00:56 Paradox
03:52 Visual sum = ln(2)
07:58 Pi
11:00 Gelfond's number
14:22 Pi exactly
17:35 Riemann's rearrangement theorem
22:40 Thanks!
Riemann rearrangement theorem.
https://en.wikipedia.org/wiki/Riemann_series_theorem This page features a different way to derive the sums of those nice m positive/n negative term arrangements of the alternating harmonic series by expressing H(n) the sum of the first n harmonic numbers by ln(n) and the Euler–Mascheroni constant. That could also be made into a very nice visual proof along the lines that I follow in this video https://youtu.be/vQE6-PLcGwU?si=iWTasKqo_JFn4etG&t=1321.
Gelfond's number
e^π being approximate equal to 20 + π may not be a complete coincidence after all:
@mathfromalphatoomega
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. 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.
(@kasugaryuichi9767) I don't know if it's new, but it's certainly not well known. To quote the Wolfram MathWorld article "Almost Integer": "This curious near-identity was apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe, but no satisfying explanation as to "why" e^π-π≈20 is true has yet been discovered."
Ratio of the number of positive and negative terms
It's interesting to look at the patterns of positive & negative terms when rearranging to Pi. We always only use one negative term before we switch. The first ten terms on the positive side are: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220,... If you look at the differences between terms, you get: 22, 23, 23, 23, 23, 24, 23, 23, 23, 23, 23, 23, 23, 24,...
The reason for this is that Gelfond's number is approximately equal to 23. It turns out that if an arrangement of our series has the sum pi, then the ratio of the numbers of positive to negative terms in the finite partial sums of the series converges to Gelfond's number. This is just one step up from what I said about us being able to get arbitrarily close to pi by turning truncations of the decimal expansion of Gelfond's number into fractions. Similarly for other target numbers. For example, to predict what the repeating pattern for e is, you just have to calculate e^e :)
@penguincute3564 thus ln(0) = negative infinity (referring to +0/1-)
Bug report: At the 1:18 mark, I say minus one sixth when I should have just said one sixth.
Music: Silhouettes---only-piano by Muted
T-shirt: Pi Day Left Vs Right Brain Pie Math Geek T-Shirt https://tinyurl.com/3e3p5yeb
Enjoy!
Burkard
Видео The best A – A ≠ 0 paradox канала Mathologer
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