Proof Of Wallis Product

In mathematics, the Wallis product for , published in 1656 by John Wallis, states that

:\begin
\frac & = \prod_^ \frac = \prod_^ \left(\frac \cdot \frac\right) \\ pt& = \Big(\frac \cdot \frac\Big) \cdot \Big(\frac \cdot \frac\Big) \cdot \Big(\frac \cdot \frac\Big) \cdot \Big(\frac \cdot \frac\Big) \cdot \; \cdots \\
\end

Proof using integration

Wallis derived this infinite product as it is done in calculus books today, by examining $\int_0^\pi \sin^n x\,dx$ 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. Let

:$I(n) = \int_0^\pi \sin^n x\,dx.$

(This is a form of Wallis' integrals.) Integrate by parts:

:$\begin u &= \sin^x \\ \Rightarrow du &= (n-1) \sin^x \cos x\,dx \\ dv &= \sin x\,dx \\ \Rightarrow v &= -\cos x \end$

:\begin
\Rightarrow I(n) &= \int_0^\pi \sin^n x\,dx \\ pt &= -\sin^x\cos x \Biggl, _0^\pi - \int_0^\pi (-\cos x)(n-1) \sin^x \cos x\,dx \\ pt &= 0 + (n-1) \int_0^\pi \cos^2x \sin^x\,dx, \qquad n > 1 \\ pt &= (n - 1) \int_0^\pi (1-\sin^2 x) \sin^x\,dx \\ pt &= (n - 1) \int_0^\pi \sin^x\,dx - (n - 1) \int_0^\pi \sin^x\,dx \\ pt &= (n - 1) I(n-2)-(n-1) I(n) \\ pt &= \frac I(n-2) \\ pt \Rightarrow \frac
&= \frac \\ pt\end
Now, we make two variable substitutions for convenience to obtain:
:$I(2n) = \fracI(2n-2)$
:$I(2n+1) = \fracI(2n-1)$

We obtain values for $I(0)$ and $I(1)$ for later use.

:\begin
I(0) &= \int_0^\pi dx = x\Biggl, _0^\pi = \pi \\ pt I(1) &= \int_0^\pi \sin x\,dx = -\cos x \Biggl, _0^\pi = (-\cos \pi)-(-\cos 0) = -(-1)-(-1) = 2 \\ pt\end

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

:$I(2n)=\int_0^\pi \sin^x\,dx = \fracI(2n-2) = \frac \cdot \fracI(2n-4)$

:$=\frac \cdot \frac \cdot \frac \cdot \cdots \cdot \frac \cdot \frac \cdot \frac I(0)=\pi \prod_^n \frac$

Repeating the process for odd values $I(2n+1)$,

:$I(2n+1)=\int_0^\pi \sin^x\,dx=\fracI(2n-1)=\frac \cdot \fracI(2n-3)$

:$=\frac \cdot \frac \cdot \frac \cdot \cdots \cdot \frac \cdot \frac \cdot \frac I(1)=2 \prod_^n \frac$

We make the following observation, based on the fact that $\sin \leq 1$

:$\sin^x \le \sin^x \le \sin^x, 0 \le x \le \pi$

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

Dividing by $I(2n+1)$:

:$\Rightarrow 1 \le \frac \le \frac=\frac$, where the equality comes from our recurrence relation.

By the squeeze theorem,

:$\Rightarrow \lim_ \frac=1$

:$\lim_ \frac=\frac \lim_ \prod_^n \left(\frac \cdot \frac\right)=1$

:$\Rightarrow \frac=\prod_^\infty \left(\frac \cdot \frac\right)=\frac \cdot \frac \cdot \frac \cdot \frac \cdot \frac \cdot \frac \cdot \cdots$

Proof using

Laplace's method

See the main page on 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 Euler infinite product for the sine function.

:$\frac = \prod_^\infty\left(1 - \frac\right)$

Let $x = \frac$:

:\begin
\Rightarrow\frac &= \prod_^\infty \left(1 - \frac\right) \\ pt \Rightarrow\frac &= \prod_^\infty \left(\frac\right) \\ pt &= \prod_^\infty \left(\frac\cdot\frac\right) = \frac \cdot \frac \cdot \frac \cdot \frac \cdot \frac \cdot \frac \cdots
\end
   

Relation to Stirling's approximation

Stirling's approximation for the factorial function $n!$ asserts that
:n! = \sqrt ^n \left + O\left(\frac\right) \right

Consider now the finite approximations to the Wallis product, obtained by taking the first $k$ terms in the product
:$p_k = \prod_^ \frac\frac,$

where $p_k$ can be written as
:\begin
p_k &= \prod_^ \frac \\ pt &= \cdot .
\end

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

Derivative of the Riemann zeta function at zero

The Riemann zeta function and the Dirichlet eta function can be defined:
:\begin
\zeta(s) &= \sum_^\infty \frac, \Re(s)>1 \\ pt \eta(s) &= (1-2^)\zeta(s) \\ pt &= \sum_^\infty \frac, \Re(s)>0
\end

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

\eta(s) &= \frac+\frac \sum_^\infty (-1)^\left frac-\frac\right \Re(s)>-1 \\ pt \Rightarrow \eta'(s) &= (1-2^)\zeta'(s)+2^ (\ln 2) \zeta(s) \\ pt &= -\frac \sum_^\infty (-1)^\left frac-\frac\right \Re(s)>-1
\end

:\begin
\Rightarrow \eta'(0) &= -\zeta'(0) - \ln 2 = -\frac \sum_^\infty (-1)^\left ln n-\ln (n+1)\right\\ pt &= -\frac \sum_^\infty (-1)^\ln \frac \\ pt &= -\frac \left(\ln \frac - \ln \frac + \ln \frac - \ln \frac + \ln \frac - \cdots\right) \\ pt &= \frac \left(\ln \frac + \ln \frac + \ln \frac + \ln \frac + \ln \frac + \cdots\right) \\ pt &= \frac \ln\left(\frac\cdot\frac\cdot\frac\cdot\frac\cdot\cdots\right) = \frac \ln\frac \\
\Rightarrow \zeta'(0) &= -\frac \ln\left(2 \pi\right)
\end

See also

* 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 $\pi$.
* Leibniz formula for , an infinite sum that can be converted into an infinite Euler product for $\pi$.
* Wallis sieve

Notes

External links

*
* {{cbignore

Articles containing proofs
Pi algorithms
Infinite products