Question

Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$

score 432 · Answer 1 · 2026-03-25 16:03:45

432

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Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$

Published on 25 Mar 2026 - 16:03

score 130 · Answer 2 · 2016-11-26 11:14:01

130

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On the integral $\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}$ and the plastic constant

Published on 26 Nov 2016 - 11:14

score 74 · Answer 3 · 2016-12-10 20:47:55

74

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Prove that $ (\frac{1+\sqrt5}{2})^{n}+(\frac{1+\sqrt5}{2})^{n-1} = (\frac{1+\sqrt5}{2})^{n+1}$.

Published on 10 Dec 2016 - 20:47

score 151 · Answer 4 · 2016-12-15 11:42:35

151

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Another simple rule satisfied by the Fibonacci $n$-step constants?

Published on 15 Dec 2016 - 11:42

score 354 · Answer 5 · 2016-12-15 12:18:05

354

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The famous root of $x(1 - x^{12})^2(1 - x^{24})^2 = (1 - x^6)^7(1 - x^8)^4$

Published on 15 Dec 2016 - 12:18

score 385 · Answer 6 · 2016-12-26 13:33:10

385

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How to prove $_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-2^7\phi^9\big)=\large \frac{3}{5^{5/6}}\,\phi^{-1}\,$ with golden ratio $\phi$?

Published on 26 Dec 2016 - 13:33

score 213 · Answer 7 · 2016-12-26 20:00:04

Show that:$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(\cdots\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$

score 184 · Answer 8 · 2016-12-27 08:40:14

184

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Prove that:$(\phi-1)^1(2-\phi)^{1\over4}(2\phi-3)^{1\over9}(5-3\phi)^{1\over16}\cdots(F_{n+1}-\phi F_n)^{1\over n^2}=\phi^{-H_n}$

Published on 27 Dec 2016 - 8:40

score 253 · Answer 9 · 2016-12-30 18:31:49

Prove that: $\sum\limits_{n=1}^{\infty}{e^{2n\pi}-e^{4n\pi}-e^{6n\pi}+e^{8n\pi}\over n(1-e^{10n\pi})}=e^{2\pi\over 5}\ln{(\sqrt{\phi+2}-\phi)}$

score 130 · Answer 10 · 2017-01-01 18:26:29

130

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Limits - Differential Calculus

Published on 01 Jan 2017 - 18:26

score 594 · Answer 11 · 2026-03-25 07:41:59

594

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Ramanujan's Class Invariant $G_{625}$

Published on 25 Mar 2026 - 7:41

score 240 · Answer 12 · 2017-01-13 04:34:01

240

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Proof for the Fibonacci sequence: $F_n \leq \left(\frac{1+\sqrt5}{2}\right)^{n-1}$

Published on 13 Jan 2017 - 4:34

score 110 · Answer 13 · 2017-01-13 14:17:05

110

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Why isn't the golden ratio defined as the points where $f(x)=x^3-x^2-x$ are zero?

Published on 13 Jan 2017 - 14:17

score 761 · Answer 14 · 2017-01-17 21:32:39

761

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Prove that $\frac{1}{\phi}<\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx< \frac{24+\sqrt{2}}{41} $

Published on 17 Jan 2017 - 21:32

score 160 · Answer 15 · 2017-01-21 18:23:14

160

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Basel's problem: $5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)$

Published on 21 Jan 2017 - 18:23

List Question

Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$

On the integral $\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}$ and the plastic constant

Prove that $ (\frac{1+\sqrt5}{2})^{n}+(\frac{1+\sqrt5}{2})^{n-1} = (\frac{1+\sqrt5}{2})^{n+1}$.

Another simple rule satisfied by the Fibonacci $n$-step constants?

The famous root of $x(1 - x^{12})^2(1 - x^{24})^2 = (1 - x^6)^7(1 - x^8)^4$

How to prove $_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-2^7\phi^9\big)=\large \frac{3}{5^{5/6}}\,\phi^{-1}\,$ with golden ratio $\phi$?

Show that:$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(\cdots\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$

Prove that:$(\phi-1)^1(2-\phi)^{1\over4}(2\phi-3)^{1\over9}(5-3\phi)^{1\over16}\cdots(F_{n+1}-\phi F_n)^{1\over n^2}=\phi^{-H_n}$

Prove that: $\sum\limits_{n=1}^{\infty}{e^{2n\pi}-e^{4n\pi}-e^{6n\pi}+e^{8n\pi}\over n(1-e^{10n\pi})}=e^{2\pi\over 5}\ln{(\sqrt{\phi+2}-\phi)}$

Limits - Differential Calculus

Ramanujan's Class Invariant $G_{625}$

Proof for the Fibonacci sequence: $F_n \leq \left(\frac{1+\sqrt5}{2}\right)^{n-1}$

Why isn't the golden ratio defined as the points where $f(x)=x^3-x^2-x$ are zero?

Prove that $\frac{1}{\phi}<\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx< \frac{24+\sqrt{2}}{41} $

Basel's problem: $5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)$

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