Formula lui Wallis

Formula lui Wallis sau produsul lui Wallis este un produs infinit:

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

A fost descoperit de John Wallis în 1655.

Se utilizează produsul lui Euler:

${\displaystyle \sin(\pi z)=\pi z\prod _{n=1}^{\infty }\left(1-{\frac {^{2}{n^{2}\right).}$

Aceasta se scrie:

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

Fie x = ^π⁄₂:

${\displaystyle {\begin{aligned}\Rightarrow {\frac {2}{\pi }&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}\right)\\\Rightarrow {\frac {\pi }{2}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}{4n^{2}-1}\right)\\&=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}\cdot {\frac {2n}{2n+1}\right)={\frac {2}{1}\cdot {\frac {2}{3}\cdot {\frac {4}{3}\cdot {\frac {4}{5}\cdot {\frac {6}{5}\cdot {\frac {6}{7}\cdots \end{aligned}$

Fie:

${\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}xdx}$

(o formă a integralei lui Wallis). Integrând prin părți:

${\displaystyle {\begin{aligned}u&=\sin ^{n-1}x\\\Rightarrow du&=(n-1)\sin ^{n-2}x\cos xdx\\dv&=\sin xdx\\\Rightarrow v&=-\cos x\end{aligned}$

${\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}xdx=\int _{0}^{\pi }udv=uv|_{x=0}^{x=\pi }-\int _{0}^{\pi }vdu\\{}&=-\sin ^{n-1}x\cos x|_{x=0}^{x=\pi }-\int _{0}^{\pi }-\cos x(n-1)\sin ^{n-2}x\cos xdx\\{}&=0-(n-1)\int _{0}^{\pi }-\cos ^{2}x\sin ^{n-2}xdx,n>1\\{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}xdx\\{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}xdx-(n-1)\int _{0}^{\pi }\sin ^{n}xdx\\{}&=(n-1)I(n-2)-(n-1)I(n)\\{}&={\frac {n-1}{n}I(n-2)\\\Rightarrow {\frac {I(n)}{I(n-2)}&={\frac {n-1}{n}\\\Rightarrow {\frac {I(2n-1)}{I(2n+1)}&={\frac {2n+1}{2n}\end{aligned}$

Acest rezultat va fi utilizat mai jos:

${\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x|_{0}^{\pi }=\pi \\I(1)&=\int _{0}^{\pi }\sin xdx=-\cos x|_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\I(2n)&=\int _{0}^{\pi }\sin ^{2n}xdx={\frac {2n-1}{2n}I(2n-2)={\frac {2n-1}{2n}\cdot {\frac {2n-3}{2n-2}I(2n-4)\end{aligned}$

Repetând procedeul,

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

${\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}xdx={\frac {2n}{2n+1}I(2n-1)={\frac {2n}{2n+1}\cdot {\frac {2n-2}{2n-1}I(2n-3)}$

Reinterând procesul,

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

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

${\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)}$

${\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}\leq {\frac {I(2n-1)}{I(2n+1)}={\frac {2n+1}{2n}$, din rezultatele de mai sus.

Conform teoremei cleștelui,

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

${\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}={\frac {\pi }{2}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}\cdot {\frac {2k+1}{2k}\right)=1}$

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

• en Wolfram MathWorld