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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
</math>

where k is a constant.

To find the constant k, take the limit of both sides:

<math>

\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \Bigg( 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 \Bigg) = k.
</math>

Using the fact that

<math>

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

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

<math>\begin{align}

\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 \\
&{} = \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.
\end{align}
</math>

Put x = π/2:

<math>

\frac{2}{\pi} = \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} \left(1 - \frac{1}{4n^2}\right),
</math>

<math>\begin{align}

\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}
</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 {{2^{4k}\,(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 p_k converges to π/2 as k → ∞.

External link

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