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{{Short description|Infinite product for pi}}
{{comparison_pi_infinite_series.svg|400px|the Wallis product (purple asterisks) and}}
The '''Wallis product''' is the [[infinite product]] representation of [[Pi|{{pi}}]]:

:<math>\begin{align}
\frac{\pi}{2} & = \prod_{n=1}^{\infty} \frac{ 4n^2 }{ 4n^2 - 1 } = \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) \\[6pt]
& = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \\
\end{align}</math>

It was published in 1656 by [[John Wallis]].<ref name="WallisFormula">{{cite web|url=http://mathworld.wolfram.com/WallisFormula.html|title=Wallis Formula}}</ref>

== Proof using integration ==
Wallis derived this [[infinite product]] using interpolation, though his method is not regarded as rigorous.  A modern derivation can be found by examining <math>\int_0^\pi \sin^n x\,dx</math> for even and odd values of <math>n</math>, and noting that for large <math>n</math>, increasing <math>n</math>  by 1 results in a change that becomes ever smaller as <math>n</math> increases. Let<ref>{{cite web|title=Integrating Powers and Product of Sines and Cosines: Challenging Problems|url=http://www.sosmath.com/calculus/integration/powerproduct/problem/problem.html}}</ref>

:<math>I(n) = \int_0^\pi \sin^n x\,dx.</math>

(This is a form of [[Wallis' integrals]].) [[Integration by parts|Integrate by parts]]:

:<math>\begin{align}
               u &= \sin^{n-1}x \\
  \Rightarrow du &= (n-1) \sin^{n-2}x \cos x\,dx \\
              dv &= \sin x\,dx \\
   \Rightarrow v &= -\cos x
\end{align}</math>

:<math>\begin{align}
 \Rightarrow I(n) &=  \int_0^\pi \sin^n x\,dx \\[6pt]
               {} &= -\sin^{n-1}x\cos x \Biggl|_0^\pi - \int_0^\pi (-\cos x)(n-1) \sin^{n-2}x \cos x\,dx \\[6pt]
               {} &= 0 + (n-1) \int_0^\pi \cos^2x \sin^{n-2}x\,dx, \qquad n > 1 \\[6pt]
               {} &= (n - 1) \int_0^\pi (1-\sin^2 x) \sin^{n-2}x\,dx \\[6pt]
               {} &= (n - 1) \int_0^\pi \sin^{n-2}x\,dx - (n - 1) \int_0^\pi \sin^{n}x\,dx \\[6pt]
               {} &= (n - 1) I(n-2)-(n-1) I(n) \\[6pt]
               {} &= \frac{n-1}{n} I(n-2) \\[6pt]
 \Rightarrow \frac{I(n)}{I(n-2)}
                  &= \frac{n-1}{n} \\[6pt]
\end{align}</math>
Now, we make two variable substitutions for convenience to obtain:
:<math>I(2n) = \frac{2n-1}{2n}I(2n-2)</math>
:<math>I(2n+1) = \frac{2n}{2n+1}I(2n-1)</math>

We obtain values for <math>I(0)</math> and <math>I(1)</math> for later use.

:<math>\begin{align}
 I(0)  &= \int_0^\pi dx = x\Biggl|_0^\pi = \pi \\[6pt]
 I(1)  &= \int_0^\pi \sin x\,dx = -\cos x \Biggl|_0^\pi = (-\cos \pi)-(-\cos 0) = -(-1)-(-1) = 2 \\[6pt]
\end{align}</math>

Now, we calculate for even values <math>I(2n)</math> by repeatedly applying the [[recurrence relation]] result from the integration by parts.  Eventually, we end get down to <math>I(0)</math>, which we have calculated.

:<math>I(2n)=\int_0^\pi \sin^{2n}x\,dx = \frac{2n-1}{2n}I(2n-2) = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2}I(2n-4)</math>

:<math>=\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}</math>

Repeating the process for odd values <math>I(2n+1)</math>,

:<math>I(2n+1)=\int_0^\pi \sin^{2n+1}x\,dx=\frac{2n}{2n+1}I(2n-1)=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1}I(2n-3)</math>

:<math>=\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}</math>

We make the following observation, based on the fact that <math>\sin{x} \leq 1</math>

:<math>\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x, 0 \le x \le \pi</math>

:<math>\Rightarrow I(2n+1) \le I(2n) \le I(2n-1)</math>

Dividing by <math>I(2n+1)</math>:

:<math>\Rightarrow 1 \le \frac{I(2n)}{I(2n+1)} \le \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n}</math>, where the equality comes from our recurrence relation.

By the [[squeeze theorem]],

:<math>\Rightarrow \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=1</math>

:<math>\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</math>

:<math>\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</math>

=== Proof using [[Laplace's method]] ===
See the main page on [[Gaussian integral#By Laplace's method|Gaussian integral]].

== Proof using Euler's infinite product for the sine function  ==
While the proof above is typically featured in modern calculus textbooks, the Wallis product is, in retrospect, an easy corollary of the later [[Infinite product#Product representations of functions|Euler infinite product]] for the [[sine function]].

:<math>\frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)</math>

Let <math>x = \frac{\pi}{2}</math>:

:<math>\begin{align}
  \Rightarrow\frac{2}{\pi} &= \prod_{n=1}^\infty \left(1 - \frac{1}{4n^2}\right) \\[6pt]
  \Rightarrow\frac{\pi}{2} &= \prod_{n=1}^\infty \left(\frac{4n^2}{4n^2 - 1}\right) \\[6pt]
                           &= \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} \cdots
\end{align}
</math>&nbsp;&nbsp;&nbsp;<ref name="WallisFormula" />

== Relation to Stirling's approximation==

[[Stirling's approximation]] for the factorial function <math>n!</math> asserts that
:<math>n! = \sqrt {2\pi n} {\left(\frac{n}{e}\right)}^n \left[1 + O\left(\frac{1}{n}\right) \right].</math>

Consider now the finite approximations to the Wallis product, obtained by taking the first <math>k</math> terms in the product
:<math>p_k = \prod_{n=1}^{k} \frac{2n}{2n - 1}\frac{2n}{2n + 1},</math>

where <math>p_k</math> can be written as
:<math>\begin{align}
  p_k &= {1 \over {2k + 1}} \prod_{n=1}^{k} \frac{(2n)^4}{[(2n)(2n - 1)]^2} \\[6pt]
      &= {1 \over {2k + 1}} \cdot {{2^{4k}\,(k!)^4} \over {[(2k)!]^2}}.
\end{align}</math>

Substituting Stirling's approximation in this expression (both for <math>k!</math> and <math>(2k)!</math>) one can deduce (after a short calculation) that <math>p_k</math> converges to <math>\frac{\pi}{2}</math> as <math>k \rightarrow \infty</math>.

==Derivative of the Riemann zeta function at zero==
The [[Riemann zeta function]] and the [[Dirichlet eta function]] can be defined:<ref name="WallisFormula" />
:<math>\begin{align}
  \zeta(s) &= \sum_{n=1}^\infty \frac{1}{n^s}, \Re(s)>1 \\[6pt]
  \eta(s)  &= (1-2^{1-s})\zeta(s) \\[6pt]
           &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}, \Re(s)>0
\end{align}</math>

Applying an Euler transform to the latter series, the following is obtained:
:<math>\begin{align}

               \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 \\[6pt]
  \Rightarrow \eta'(s) &= (1-2^{1-s})\zeta'(s)+2^{1-s} (\ln 2) \zeta(s) \\[6pt]
                       &= -\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
\end{align}</math>

:<math>\begin{align}
   \Rightarrow \eta'(0) &= -\zeta'(0) - \ln 2 = -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\ln n-\ln (n+1)\right] \\[6pt]
                        &= -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\ln \frac{n}{n+1} \\[6pt]
                        &= -\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) \\[6pt]
                        &=  \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) \\[6pt]
                        &=  \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)
\end{align}</math>

==See also==
{{portal|Mathematics}}
* [[John Wallis]], English [[mathematician]] who is given partial credit for the development of [[infinitesimal calculus]] and [[pi]].
* [[Viète's formula]], a different infinite product formula for <math>\pi</math>.
* [[Leibniz formula for π|Leibniz formula for {{pi}}]], an infinite sum that can be converted into an infinite [[Euler product]] for {{pi}}.
* [[Wallis sieve]]
* The [[Nick Pippenger|Pippenger product formula]] obtains [[e (mathematical constant)|''e'']] by taking roots of terms in the Wallis product.

==Notes==
{{Reflist}}

==External links==
* {{springer|title=Wallis formula|id=p/w097040}}
* {{cite web |date=April 20, 2018 |title=Why does this product equal π/2? A new proof of the Wallis formula for π. |url=https://www.youtube.com/watch?v=8GPy_UMV-08  |archive-url=https://ghostarchive.org/varchive/youtube/20211212/8GPy_UMV-08| archive-date=2021-12-12 |url-status=live|via=[[YouTube]]|publisher=[[3Blue1Brown]]}}{{cbignore}}

[[Category:Articles containing proofs]]
[[Category:Pi algorithms]]
[[Category:Infinite products]]