Formula lui Wallis

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

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

A fost descoperit de John Wallis în 1655.

Demonstrație utilizând produsul lui Euler pentru sinus

Se utilizează produsul lui Euler:

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

Aceasta se scrie:

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

Fie x = Format:Frac:

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

Demonstrație prin metoda integralelor

Fie:

$I(n)=\underset{0}{\overset{\pi }{\int }}{\sin }^{n}xdx$

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

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

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

Acest rezultat va fi utilizat mai jos:

$\begin{array}{cc}I(0)& =\underset{0}{\overset{\pi }{\int }}dx=x{|}_0^{\pi }=\pi \\ I(1)& =\underset{0}{\overset{\pi }{\int }}\sin xdx=-\cos x{|}_0^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\ I(2n)& =\underset{0}{\overset{\pi }{\int }}{\sin }^{2n}xdx=\frac{2n-1}{2n}I(2n-2)=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I(2n-4)\end{array}$

Repetând procedeul,

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

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

Reinterând procesul,

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

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

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

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

Conform teoremei cleștelui,

$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{I(2n)}{I(2n+1)}=1$

$\underset{n\to \mathrm{\infty }}{\lim }\frac{I(2n)}{I(2n+1)}=\frac{\pi }{2}\underset{n\to \mathrm{\infty }}{\lim }\prod _{k=1}^n(\frac{2k-1}{2k}\cdot \frac{2k+1}{2k})=1$

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

Legături externe

• Format:En icon Wolfram MathWorld