Wallis product

In mathematics, Wallis' product for π, written down in 1655 by John Wallis, states that

$\prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}.$

Derivation

Wallis derived his product as it is done in calculus books today, by comparing $\int_0^\pi \sin^nxdx$ for even and odd values of n, and noting that for large n, increasing n by 1 results in a change that becomes ever smaller as n increases. Since infinitesimal calculus as we know it did not yet exist then, and the mathematical analysis of the time was inadequate to discuss the convergence issues, this was a hard piece of research, and tentative as well.

Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function.

Proof using Euler's infinite product for the sine function

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

Let x = π/2:

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

$\begin{align} \Rightarrow\frac{\pi}{2} &{}= \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) \\ &{}= \prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots \end{align}$

Proof using integral

Let:

$I(n)=\int_0^\pi \sin^nxdx$

$u=\sin^{n-1}x \Rightarrow du=(n-1) \sin^{n-2}x \cos x dx$

$dv=\sin x dx \Rightarrow v=-\cos x$

$\Rightarrow I(n)=\int_0^\pi \sin^nxdx=\int_0^\pi u dv =uv |_{x=0}^{x=\pi}-\int_0^\pi v du$

$=-\sin^{n-1}x\cos x |_{x=0}^{x=\pi} -\int_0^\pi -\cos x(n-1) \sin^{n-2}x \cos x dx$

$=0-(n-1) \int_0^\pi -\cos^2x \sin^{n-2}x dx, n>1$

$=(n-1) \int_0^\pi (1-\sin^2 x) \sin^{n-2}x dx$

$=(n-1) \int_0^\pi \sin^{n-2}x dx -(n-1) \int_0^\pi \sin^{n}x dx$

$\Rightarrow I(n)=(n-1) I(n-2)-(n-1) I(n)$

$\Rightarrow I(n)=\frac{n-1}{n} I(n-2)$

$\Rightarrow \frac{I(n)}{I(n-2)}=\frac{n-1}{n}$

$\Rightarrow \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n}$, this result will be used below:

$I(0)=\int_0^\pi dx=x|_0^\pi=\pi$

$I(1)=\int_0^\pi \sin xdx=-\cos x|_0^\pi=(-\cos \pi)-(-\cos 0)=-(-1)-(-1)=2$

$I(2n)=\int_0^\pi \sin^{2n}xdx=\frac{2n-1}{2n}I(2n-2)=\frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2}I(2n-4)$

Repeating the process,

$=\frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-5}{2n-4} \cdot \cdots \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} I(0)=\pi \prod_{k=1}^n \frac{2k-1}{2k}$

$I(2n+1)=\int_0^\pi \sin^{2n+1}xdx=\frac{2n}{2n+1}I(2n-1)=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1}I(2n-3)$

Repeating the process,

$=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdot \cdots \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} I(1)=2 \prod_{k=1}^n \frac{2k}{2k+1}$

$\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x, 0 \le x \le \pi$

$\Rightarrow I(2n+1) \le I(2n) \le I(2n-1)$

$\Rightarrow 1 \le \frac{I(2n)}{I(2n+1)} \le \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n}$, from above results.

By the squeeze theorem,

$\Rightarrow \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=1$

$\lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=\frac{\pi}{2} \lim_{n\rightarrow\infty} \prod_{k=1}^n \left(\frac{2k-1}{2k} \cdot \frac{2k+1}{2k}\right)=1$

$\Rightarrow \frac{\pi}{2}=\prod_{k=1}^\infty \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right)=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots$

Relation to Stirling's approximation

Stirling's approximation for n! asserts that

$n! = \sqrt {2\pi n} {\left(\frac{n}{e}\right)}^n \left( 1 + O\left(\frac{1}{n}\right) \right)$

as n → ∞. Consider now the finite approximations to the Wallis product, obtained by taking the first k terms in the product:

$p_k = \prod_{n=1}^{k} \frac{(2n)(2n)}{(2n-1)(2n+1)} \ .$

p_k can be written as

$p_k ={1\over{2k+1}}\prod_{n=1}^{k} \frac{(2n)^4 }{[(2n)(2n-1)]^2}={1\over{2k+1}}\cdot {{2^{4k}\,(k!)^4}\over {[(2k)!]^2}} \ .$

Substituting Stirling's approximation in this expression (both for k! and (2k)!) one can deduce (after a short calculation) that p_k converges to π/2 as k → ∞.

ζ'(0)

The Riemann zeta function and the Dirichlet eta function can be defined:

$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}, \Re(s)>1$

$\eta(s)=(1-2^{1-s})\zeta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}, \Re(s)>0$

Applying an Euler transform to the latter series, the following is obtained:

$\eta(s)=\frac{1}{2}+\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{1}{n^s}-\frac{1}{(n+1)^s}\right], \Re(s)>-1$

$\Rightarrow \eta'(s)=(1-2^{1-s})\zeta'(s)+2^{1-s} (\ln 2) \zeta(s)$

$=-\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{\ln n}{n^s}-\frac{\ln (n+1)}{(n+1)^s}\right], \Re(s)>-1$

$\Rightarrow \eta'(0)=-\zeta'(0)-\ln 2=-\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\ln n-\ln (n+1)\right]$

$=-\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\ln \frac{n}{n+1}$

$=-\frac{1}{2} \left(\ln \frac{1}{2} - \ln \frac{2}{3} + \ln \frac{3}{4} - \ln \frac{4}{5} + \ln \frac{5}{6} - \cdots\right)$

$=\frac{1}{2} \left(\ln \frac{2}{1} + \ln \frac{2}{3} + \ln \frac{4}{3} + \ln \frac{4}{5} + \ln \frac{6}{5} + \cdots\right)$

$=\frac{1}{2} \ln\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\cdots\right)=\frac{1}{2} \ln\frac{\pi}{2}$

$\Rightarrow \zeta'(0)=-\frac{1}{2} \ln\left(2 \pi\right)$