\(\displaystyle\pi\) is a beautiful number

\(\pi\) \(\text{can be expressed in many beautiful patterns and ways.}\)


Take The Nilakantha Series and The Gregory-Leibniz Series.

Nilakantha Series: \(\displaystyle \pi = 3 + \frac{4}{2 \times 3 \times 4} - \frac{4}{4 \times 5 \times 6} + \frac{4}{6 \times 7 \times 8} - \frac{4}{8 \times 9 \times 10} \cdots\)

Gregory-Leibniz Series: \(\displaystyle \frac{\pi}{4} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} \cdots \)

\(\displaystyle \textbf{But the Gregory-Leibniz Series takes MORE than 300 terms to be correct to 2 decimal places!}\) It can be transformed to

\(\displaystyle \pi = \sum_{k=1}^\infty \frac{3^k-1}{4^k} \zeta(k+1)\) where \(\displaystyle \zeta\) is the Riemann Zeta function and so that the error after \(k\) terms is \(\displaystyle \approx \left(\frac{3}{4}\right)^k\)


There is also Machin's Formula: \(\displaystyle \frac{1}{4} \pi = 4 \tan^{-1} \left(\frac{1}{5}\right) - \tan^{-1} \left(\frac{1}{239}\right)\)

Abraham Sharp gave the infinite sum series, \(\displaystyle \pi = \sum_{k=0}^\infty \frac{2 (-1)^k 3^{\frac{1}{2}-k}}{2k+1}\)


Simple series' of infinite sums:

\(\displaystyle \frac{1}{4} \pi \sqrt{2} = \sum_{k=1}^\infty \left[\frac{(-1)^{k+1}}{4k-1} + \frac{(-1)^{k+1}}{4k-3}\right] = 1 + \frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \cdots \) \(\small(\text{related to the Gregory-Leibniz series})\)

\(\displaystyle \frac{1}{4} (\pi-3) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k(2k+1)(2k+2)} = \frac{1}{2 \times 3 \times 4} - \frac{1}{4 \times 5 \times 6} + \frac{1}{6 \times 7 \times 8} - \cdots \)

\(\displaystyle \frac{1}{6} \pi^2 = \sum_{k=1}^\infty \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} +\frac{1}{16} \cdots \)

\(\displaystyle \frac{1}{8} \pi^2 = \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots \)

There is also \(\displaystyle \frac{\pi^4}{90} = \zeta(4)\)


In 1666 (Newton's miracle year) Newton used a geometric construction to derive the formula

\(\displaystyle \pi = \frac{3}{4} \sqrt{3} + 24 \int_{0}^{\frac{1}{4}} \sqrt{x-x^2} \text{dx} = \frac{3 \sqrt{3}}{4} + 24 \left(\frac{1}{12} - \frac{1}{5 \times 2^5} - \frac{1}{28 \times 2^7} - \frac{1}{72 \times 2^9} \cdots \right) \)

The coefficients can be found in

\(\displaystyle I(x) = \int \sqrt{x-x^2} \text{dx} = \frac{1}{4} (2x-1) \sqrt{x-x^2} - \frac{1}{8} \sin^{-1} (1-2x) \)

by taking the series expansion of \(I(x)-I(0)\) about 0 obtaining

\(I(x) = \dfrac{2}{3} x^{\frac{3}{2}} - \dfrac{1}{5} x^{\frac{3}{2}} - \dfrac{1}{28} x^{\frac{7}{2}} - \dfrac{1}{72} x^{\frac{9}{2}} - \dfrac{5}{704} x^{\frac{11}{2}} \cdots \)


Euler's convergence improvement transformation gives

\(\displaystyle \frac{\pi}{2} = \frac{1}{2} \sum_{n=0}^\infty \frac{(n!)^2 2^{n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!} = 1 + \frac{1}{3} + \frac{1 \times 2}{3 \times 5} + \frac{1 \times 2 \times 3}{3 \times 5 \times 7} + \cdots = 1 + \frac{1}{3}\left(1 + \frac{2}{5}\left(1 + \frac{3}{7}\left(1 + \frac{4}{9}(1 + \cdots )\right)\right)\right)\)

This corresponds to plugging in \(\displaystyle x = \frac{1}{\sqrt{2}}\) into the power series for the hypergeometric function \(\displaystyle_2 F_1 (a,b;c;x),\)

\(\displaystyle \frac{\sin^{-1} x}{\sqrt{1-x^2}} = \sum_{i=0}^\infty \frac{(2x)^{2i+1} i!^2 }{2 (2i+1)!} = _2 F_1 (1,1;\frac{3}{2};x^2)x\)

Despite the convergence improvement, series \((\displaystyle \diamond\)) converges at only one bit/term. At the cost of a square root, Gosper has noted that \(\displaystyle x = \frac{1}{2}\) gives 2 bits/term,

\(\displaystyle \frac{1}{9} \sqrt{3} \pi = \frac{1}{2} \sum_{i=0}^\infty \frac{(i!)^2}{(2i+1)!} \)

and \(\displaystyle x = \sin\left(\frac{\pi}{10}\right)\) gives almost 3.39 bits/term

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

where \(\displaystyle\phi\) is the golden ratio (not Euler's Totient function). Gosper also obtained

\(\displaystyle \pi = 3 + \frac{1}{60}\left(8 + \frac{2 \times 3}{7 \times 8 \times 3}\left(13 + \frac{3 \times 5}{10 \times 11 \times 3}\left(18 + \frac{4 \times 7}{13 \times 14 \times 3}(23 + \cdots)\right)\right)\right)\)


A spigot algorithm for \(\displaystyle \pi\) is given by Rabinowitz and Wagon.

More amazingly still, a closed form expression giving a digit-extraction algorithm which produces digits of \(\displaystyle\pi\) (or \(\displaystyle \pi^2 \) in base-16 was discovered by Bailey.

\(\displaystyle \pi = \sum_{n=0}^\infty \left( \frac{4}{8n+1} - \frac{2}{8n+4} - \frac{1}{8n+5} - \frac{1}{8n+6}\right) \left(\frac{1}{16}\right)^n \)

There is also a continued fraction form : \[\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}} \]

There is also a curious identity where \(314 \equiv 159 + 265 \) \(\displaystyle \bmod 10\) involving the first \(9\) digits of \(\pi\)


I am going to update this in the future and in the meantime, we can comment about the magical number, \(\pi\)

Watch Out! There is also Golden ratio in another note!

Note by Mohmmad Farhan
2 weeks ago

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You can add \(\dfrac{\pi^4}{90}=\zeta(4)\) also

X X - 1 week, 6 days ago

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Thank you for contributing

Mohmmad Farhan - 1 week, 5 days ago

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Will do!

Mohmmad Farhan - 1 week, 6 days ago

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Nice!

There are also the continued fractions, one of them being: \[\pi=\frac{4}{1+\frac{1^2}{2+\frac{2+\frac{3^2}{2+\frac{5^2}{2+...}}}}\]

Andrei Li - 1 week, 5 days ago

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Thank you for contributing

Mohmmad Farhan - 1 week, 5 days ago

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\(\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}\)

Mohmmad Farhan - 1 week, 5 days ago

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Strangely enough, my computer refuses to accept the continued fraction LaTeX...

Andrei Li - 1 week, 5 days ago

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@Andrei Li me too.. (so sad)

Mohmmad Farhan - 1 week, 5 days ago

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@Mohmmad Farhan \(\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cdots}}}} \)

X X - 1 week, 5 days ago

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