Wallis product

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

<math>

\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 = \frac{\pi}{2} <math>

Proof

First of all, consider the root of sin(x)/x is ±nπ, where n = 1, 2, 3, ... Then, we can express sine as an infinite product of linear factors given by its roots:

<math>

\frac{\sin(x)}{x} = k \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) \cdots \qquad \textrm{where}~k~\textrm{is~a~constant} <math>

To find the constant k, taking limit on both sides:

<math>

\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \left( k \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) \cdots \right) = k <math>

Using the fact that:

<math>

\lim_{x \to 0} \frac{\sin(x)}{x} = 1 <math>

we get k=1. Then, we obtain the Euler-Wallis formula for sine:

<math>

\frac{\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) \cdots <math>

<math>

\frac{\sin(x)}{x} = \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) \cdots <math>

Put x=π/2,

<math>

\frac{1}{\pi / 2} = \left(1 – \frac{1}{2^2}\right)\left(1 – \frac{1}{4^2}\right)\left(1 – \frac{1}{6^2}\right) \cdots = \prod_{n=1}^{\infty} (1 – \frac{1}{4n^2}) <math>

<math>

\frac{\pi}{2} = \prod_{n=1}^{\infty} (\frac{4n^2}{4n^2 – 1}) <math>

<math>

= \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 <math>

Q.E.D.

Relation to Stirling's approximation

Stirling's approximation for n! asserts that

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

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

<math>

p_k = \prod_{n=1}^{k} \frac{(2n)(2n)}{(2n-1)(2n+1)} \ . <math> p_k can be written as

<math>

p_k ={1\over{2k+1}}\prod_{n=1}^{k} \frac{(2n)^4 }{(2n (2n-1))^2}={1\over{2k+1}}\cdot {{4^{2k}\,k!^4}\over {(2k\,!)^2}} \ . <math>

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

External link

• PlanetMath page on complex analysis, including a proof of the infinite product