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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16a21c793e9504de11ae568d40f735792501d08d)
A fost descoperit de John Wallis în 1655.
Demonstrație utilizând produsul lui Euler pentru sinus
Se utilizează produsul lui Euler:
![{\displaystyle \sin(\pi z)=\pi z\prod _{n=1}^{\infty }\left(1-{\frac {^{2}}{n^{2}}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c91503d33e4bac762260b2849881cd2a5f7b8919)
Aceasta se scrie:
![{\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f74dce42e79aff73ee96dbfb03216bd02fc23c68)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06c1c775d24d34a9394c8e7dc6b48b5d2839a5ae)
Demonstrație prin metoda integralelor
Fie:
![{\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}xdx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c85b81c0f27e787aa0a2a33ee62c98a080d790f)
(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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27c1f56fad19c2a04f7387a59ed4d4ff66504d98)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a683238e7cbe9c2a9a46d175702a2864dd1d08f)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bc85279a6b8b301b7a0a1c8e382830624ecd7d4)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd9bc4b863591db0d769e9b84733eb682f66f802)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bd30b886bf011747cf644ca3483582f6faf0eb7)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5eda3cca37e87b58e35173f4b79c6b124a66b90c)
![{\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3381e5223fdeb96426c56355e403f7049c53dbf8)
![{\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a5b396dc3b7d480bbc41c23bc2b07f661aa0b99)
, din rezultatele de mai sus.
Conform teoremei cleștelui,
![{\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3d284350606423ae6aba21e6b32f724853857aa)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6d2765e8c0d367700dda3ef1521390d44fc9841)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/414a124c282cbace8fd083b949eae278cefab585)
Legături externe