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This question already has an answer here:

Is there some identity that shows a connection between $\pi$ and the golden ratio, $\phi$?

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marked as duplicate by Asaf Karagila, Pragabhava, Stefan Hamcke, Pedro Tamaroff, Ayman Hourieh Dec 28 '13 at 1:44

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$$\pi= \frac{5\sqrt{\phi+2}}{2}\sum_{i=0}^\infty \frac{(i!)^2}{\phi^{2i+1}(2i+1)!}$$

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  • $\begingroup$ Proof sketch? $\,$ $\endgroup$ – Ian Mateus Dec 28 '13 at 0:15
  • $\begingroup$ proof by "found it in Wolfram Mathworld" mathworld.wolfram.com/PiFormulas.html, formula (28). I thought it's so abstruse that it should be in the list :) $\endgroup$ – benh Dec 28 '13 at 0:17
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    $\begingroup$ I see, this comes from the identity $$\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}=\sum_{k\geqslant 0}\frac{(k!)^2(2x)^{2k+1}}{2(2k+1)!},$$ stated as $(26)$. Beautiful one! $\endgroup$ – Ian Mateus Dec 28 '13 at 0:21
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    $\begingroup$ and the sum converges incredibly fast, e.g. $\pi= \frac{5\sqrt{\phi+2}}{2}\sum_{i=0}^5 \frac{(i!)^2}{\phi^{2i+1}(2i+1)!} = 3.1415918$ instead of $3.1415926$. $\endgroup$ – benh Dec 28 '13 at 0:24
  • $\begingroup$ The Ramanujan-type formula for $1/\pi$ using $\phi$ converges faster, with already a $10^{-31}$ difference for the first five terms. $\endgroup$ – Tito Piezas III Dec 28 '13 at 1:17
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What about: $$\phi = \frac{1}{2}\csc(\pi/10)$$

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  • $\begingroup$ Bravo! This is most likely to be the perfect one. But I suspect functions are marvelous that way $\endgroup$ – Nick Dec 28 '13 at 2:26
  • $\begingroup$ Wow, that is amazing. How did you come to this formula? $\endgroup$ – Daniel Marschall Nov 6 '14 at 20:14
  • $\begingroup$ @rinntech - this is just because $\sin$ of $18^\circ$ involves $\sqrt{5}$, as does $\phi$. $\endgroup$ – nbubis Nov 6 '14 at 20:23
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Here's a curious identity connecting $\pi$ and $\phi$ via the Watson triple integral $I_1$ and Gauss' constant $G$,

$\begin{aligned} I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}\\ 2G^2 &=2\left( \frac{2}{\pi}\int_0^{\pi/2} \frac{dx}{\sqrt{1+\sin^2 x}}\right)^2\\ &= \frac{25\phi^6}{\sqrt{\phi^{24}-4}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \left(\frac{-\phi^{16}}{4(\phi^{24}-4)}\right)^{3n}\\ &= \frac{\Gamma^4(\frac{1}{4})}{4\pi^3} = 1.393203\dots \end{aligned}\tag{1}$

The second belongs to a class of formulas for $1/\pi$ found by Ramanujan.

$$\frac{1}{\pi} = \frac{10\,(5^{1/4})}{\phi^6} \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{6n+\phi^{-2}}{(2^6\phi^{24})^n}\tag{2}$$

$$\frac{1}{\pi} = \frac{1}{2\,K(k_{25})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{24})^n}}\tag{3}$$

They rely on the fact that $\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = -2^6 \phi^{24}$ where $\tau=\frac{1+\sqrt{-25}}{2}$ and $\eta(\tau)$ is the Dedekind eta function, implying,

$$e^{\pi\sqrt{25}} \approx 2^6 \phi^{24}-24.00004\dots$$

The third involves the complete elliptic integral of the first kind $K(k_{25})$ and can be calculated using this WolframAlpha command. Similar ones are,

$$\frac{1}{\pi} = \frac{1}{2\,K(k_{15})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^{12}\phi^{8})^n}}\tag{4}$$

$$\frac{1}{\pi} = \frac{1}{2\,K(k_{5})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{6})^n}}\tag{5}$$

which uses $\tau=\frac{1+\sqrt{-15}}{2},\;\frac{1+\sqrt{-5}}{2}$, respectively. The last one leads to,

$$\frac{4\sqrt{10}\,K(k_5)}{\sqrt{\Gamma(\frac{1}{20})\,\Gamma(\frac{3}{20})\,\Gamma(\frac{7}{20})\,\Gamma(\frac{9}{20})}}=\frac{\phi^{3/4}}{\sqrt{\pi}}\tag{6}$$

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One such relation is

$\phi \cos(2\pi/5)=1/2$.

This is a geometric relation: from the canonical pentagon, and in particular from the triangle with angles of $72, 72$ and $36$ degrees.

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See Guy and Matiyasevich, A new formula for $\pi$, Amer. Math. Monthly, 93 (1986) 631– 635, MR1712797 (2000i:11199), Zbl 0614.10003; if we write $f_1,f_2,\dots$ for the Fibonacci numbers, then $$\pi=\lim_{n\to\infty}\sqrt{{6\log f_1f_2\cdots f_n\over\log{\rm lcm}(f_1,f_2,\dots,f_n)}}$$

So $\phi$ isn't explicitly in here, but the Fibonaccis are, that should be good enough.

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There can be no non-trivial algebraic relation between $\pi$ and $\phi$ since $\pi$ is transcendental whereas $\phi$ is algebraic. However, other identities can be obtained like for instance: $$\phi = \dfrac{\pi+\sqrt{\pi \cdot \pi + \pi \cdot \pi + \pi \cdot \pi + \pi \cdot \pi + \pi \cdot \pi}}{\pi + \pi}$$

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    $\begingroup$ I am so not falling for this again. $\endgroup$ – Nick Dec 28 '13 at 1:59
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The answer is there isn't a truly astonishing relation. Probably because one of them is algebraic and the other is not.

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    $\begingroup$ $\phi=e^{i\pi/5}+e^{-i\pi/5}$ is a decent relation. $\endgroup$ – robjohn Dec 29 '13 at 0:06

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