Question

A cool integral: $\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$

score 247 · Answer 1 · 2026-03-26 12:58:06

247

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A cool integral: $\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$

Published on 26 Mar 2026 - 12:58

score 387 · Answer 2 · 2023-10-20 16:32:29

387

Views

$\pi$ approximation method confusion

Published on 20 Oct 2023 - 16:32

score 51 · Answer 3 · 2026-02-23 13:46:01

51

Views

How is $f_1(n)$ not computable but $f_2(n)$ is?

Published on 23 Feb 2026 - 13:46

score 185 · Answer 4 · 2023-10-27 08:12:13

185

Views

The closed-form of $1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot3}{2\cdot4}\right)^k-13\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^k+\dots$?

Published on 27 Oct 2023 - 8:12

score 70 · Answer 5 · 2023-10-27 14:22:29

70

Views

Is $\sqrt{\pi}=2\sum_{k=0}^{\infty} \frac{e^{-k^2}-e^{-(k+1)^2}}{2k+1}$?

Published on 27 Oct 2023 - 14:22

score 204 · Answer 6 · 2023-10-28 13:28:24

204

Views

Series $\sum_{n=1}^{\infty}\left(\frac{60n^{2}-43n+8}{n^{4}}\right)\binom{4n-2}{2n-1}\binom{2n}{n}^{-4}=2\zeta(2)$

Published on 28 Oct 2023 - 13:28

score 74 · Answer 7 · 2023-11-01 21:01:01

74

Views

Why $1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\cdots=\frac{\pi}{2\sqrt{2}}$?

Published on 01 Nov 2023 - 21:01

score 264 · Answer 8 · 2026-03-26 06:31:55

264

Views

Using Ramanujan-type series for $1/\pi^m$ to find formulas for $\zeta(2),\, \zeta(3),\, \zeta(4)$?

Published on 26 Mar 2026 - 6:31

score 101 · Answer 9 · 2023-11-03 00:53:06

101

Views

For what $n$ is the sequence consisting of the first $n$ digits of $\pi$ a palindrome

Published on 03 Nov 2023 - 0:53

score 158 · Answer 10 · 2026-03-28 05:21:01

158

Views

Is there a self-correcting iterative method for approximating pi without using transcendental functions?

Published on 28 Mar 2026 - 5:21

score 50 · Answer 11 · 2023-11-06 14:37:56

50

Views

Found a relation regarding the primes, is this interesting?

Published on 06 Nov 2023 - 14:37

score 87 · Answer 12 · 2026-03-27 23:14:04

87

Views

Deriving the continued fraction for Pi

Published on 27 Mar 2026 - 23:14

score 654 · Answer 13 · 2026-02-23 12:06:37

654

Views

Conjecture: In Pascal's triangle with $n$ rows, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$.

Published on 23 Feb 2026 - 12:06

score 30 · Answer 14 · 2023-11-28 19:15:19

30

Views

Identity involving the number pi

Published on 28 Nov 2023 - 19:15

score 317 · Answer 15 · 2026-03-27 04:23:55

317

Views

Euler's proof of $\frac{\pi}{6}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\cdots$

Published on 27 Mar 2026 - 4:23

List Question

A cool integral: $\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$

$\pi$ approximation method confusion

How is $f_1(n)$ not computable but $f_2(n)$ is?

The closed-form of $1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot3}{2\cdot4}\right)^k-13\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^k+\dots$?

Is $\sqrt{\pi}=2\sum_{k=0}^{\infty} \frac{e^{-k^2}-e^{-(k+1)^2}}{2k+1}$?

Series $\sum_{n=1}^{\infty}\left(\frac{60n^{2}-43n+8}{n^{4}}\right)\binom{4n-2}{2n-1}\binom{2n}{n}^{-4}=2\zeta(2)$

Why $1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\cdots=\frac{\pi}{2\sqrt{2}}$?

Using Ramanujan-type series for $1/\pi^m$ to find formulas for $\zeta(2),\, \zeta(3),\, \zeta(4)$?

For what $n$ is the sequence consisting of the first $n$ digits of $\pi$ a palindrome

Is there a self-correcting iterative method for approximating pi without using transcendental functions?

Found a relation regarding the primes, is this interesting?

Deriving the continued fraction for Pi

Conjecture: In Pascal's triangle with $n$ rows, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$.

Identity involving the number pi

Euler's proof of $\frac{\pi}{6}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\cdots$

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