Miscellaneous formulae for pi

Infinite products

The $\text{sinc}$ function is defined for $x\neq0$ as

$$\text{sinc}(x)=\frac{\sin(x)}{x}$$

It can also be expressed as an infinite product:

$$\frac{\sin(x)}{x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{\pi^2n^2}\right)$$

The normalised $\text{sinc}$ function is

$$\frac{\sin(\pi x)}{\pi x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$

See this post on the Riemann zeta function at even values for a short overview on expressing functions in this form, or look into the Weierstrass factorisation theorem for a more rigorous explanation. Letting $x=\frac{1}{2}$ gives

$$\frac{2}{\pi}=\prod_{n=1}^{\infty}\left(1-\frac{1}{4n^2}\right) = \prod_{n=1}^{\infty}\left(\frac{4n^2-1}{4n^2}\right)$$

This is Wallis’s formula for $\pi$. And taking the reciprocal:

$$\frac{\pi}{2}=\prod_{n=1}^{\infty}\left(\frac{4n^2}{4n^2-1}\right)$$

If instead we let $x=\frac{1}{4}$ we get

$$\frac{2\sqrt{2}}{\pi}=\prod_{n=1}^{\infty}\left(1-\frac{1}{16n^2}\right) =\prod_{n=1}^{\infty}\left(\frac{16n^2-1}{16n^2}\right)$$

Again, taking the reciprocal gives

$$\frac{\pi}{2\sqrt{2}}=\prod_{n=1}^{\infty}\left(\frac{16n^2}{16n^2-1}\right)$$ $$\frac{\pi\sqrt{2}}{4}=\prod_{n=1}^{\infty}\left(\frac{16n^2}{16n^2-1}\right)$$

These are both quite nice formulae for $\pi$. However, they converge VERY slowly. The first expression approximates $\pi\approx3.1415848$ after ten thousand terms.

Euler

Euler famously proved that $\zeta(2) = \frac{\pi^{2}}{6}$. He started with the function $\frac{\sin x}{x}$, i.e. the $\text{sinc}$ function, which has the Taylor series expansion

$$\text{sinc }x = \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}x^{2n} = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \ldots$$

The $\text{sinc}$ function can also be expressed as the following infinite product:

$$\text{sinc }x = \prod_{n=1}^{\infty} \left( 1 - \frac{x^{2}}{\pi^{2}n^{2}} \right) = \left( 1 - \frac{x^{2}}{\pi^{2}} \right) \left( 1 - \frac{x^{2}}{4\pi^{2}} \right) \left( 1 - \frac{x^{2}}{9\pi^{2}} \right) \ldots$$ $$= 1 - x^{2} \left( \frac{1}{\pi^{2}} + \frac{1}{4\pi^{2}} + \frac{1}{9\pi^{2}} + \ldots \right) + \ldots$$

Equating the $x^{2}$ coefficients of the above to the Taylor series expansion gives the desired result:

$$-\left( \frac{1}{\pi^{2}} + \frac{1}{4\pi^{2}} + \frac{1}{9\pi^{2}} + \ldots \right) = - \frac{1}{3!}$$ $$\zeta(2) = \sum\_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$$

A similar result can be derived for the odd squares. Instead of $\text{sinc } x$ we start with $\cos x$. The series expansion for $\cos x$ is

$$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}x^{2n} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \ldots$$

$\cos x$ can also be expressed as an infinite product:

$$\cos x = \prod_{n=1}^{\infty} \left( 1 - \frac{4x^{2}}{\pi^{2}(2n - 1)^{2}} \right) = 1 - x^{2}\left( \frac{4}{\pi^{2}} + \frac{4}{9\pi^{2}} + \frac{4}{25\pi^{2}} + \ldots \right) + \ldots$$

And again, equating the $x^{2}$ coefficients:

$$-\frac{4}{\pi^{2}}\left( 1 + \frac{1}{9} + \frac{1}{25} + \ldots \right) = - \frac{1}{2!}$$ $$\sum_{n=1}^{\infty} \frac{1}{(2n - 1)^{2}} = \frac{\pi^{2}}{8}$$

Going back to the infinite product form of $\frac{\sin x}{x}$,

$$\text{sinc }x = \prod_{n=1}^{\infty} \left( 1 - \frac{x^{2}}{\pi^{2}n^{2}} \right) = \left( 1 - \frac{x^{2}}{\pi^{2}} \right) \left( 1 - \frac{x^{2}}{4\pi^{2}} \right) \left( 1 - \frac{x^{2}}{9\pi^{2}} \right) \ldots,$$

and noting that each term is the difference of two squares, we can separate them out:

$$\sin x = x\left( 1 - \frac{x}{\pi} \right)\left( 1 + \frac{x}{\pi} \right) \left( 1 - \frac{x}{2\pi} \right)\left( 1 + \frac{x}{2\pi} \right) \left( 1 - \frac{x}{3\pi} \right)\left( 1 + \frac{x}{3\pi} \right) \ldots$$

Taking the $\log$ of both sides:

$$\log \sin x = \log x + \log \left( 1 - \frac{x}{\pi} \right) + \log \left( 1 + \frac{x}{\pi} \right) + \log \left( 1 - \frac{x}{2\pi} \right) + \ldots,$$

and then taking the derivative:

$$\frac{\cos x}{\sin x} = \frac{1}{x} - \frac{1}{\pi - x} + \frac{1}{\pi + x} - \frac{1}{2\pi - x} + \ldots$$

Letting $x = \frac{\pi}{4}$ gives Leibniz’s formula for $\pi$:

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \ldots .$$

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