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

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!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

Thank you for contributing

Log in to reply

Will do!

Log in to reply

Nice!

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

Log in to reply

Thank you for contributing

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply