**Wallisin kaavat** ovat menetelmiä, joilla voidaan laskea piin likiarvoja mielivaltaisen tarkasti. Kaavat on johtanut englantilainen matemaatikko John Wallis[1]. Wallisin kaavojen mukaan:

(1) {\displaystyle \prod \_{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}=\lim \_{n\to \infty }\left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \ldots \cdot {\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {\pi }{2}}}
(2) {\displaystyle \lim \_{n\to \infty }{\frac {(n!)^{2}\cdot 2^{2n}}{(2n!){\sqrt {n}}}}={\sqrt {\pi }}}.

Kaavojen todistus
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Wallisin kaavat pystytään todistamaan  osittaisintegroinnin avulla.

Merkitään jokaiselle {\displaystyle n\in \mathbb {N} }

{\displaystyle I\_{n}=\int \_{0}^{\frac {\pi }{2}}\sin ^{n}x\,{\text{d}}x}
{\displaystyle a\_{n}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \ldots \cdot {\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}}
{\displaystyle b\_{n}={\frac {(n!)^{2}\cdot 2^{2n}}{(2n!){\sqrt {n}}}}}
Tällöin

{\displaystyle I\_{0}={\frac {\pi }{2}}} ja {\displaystyle I\_{1}=1}
Jos {\displaystyle \scriptstyle n\geq 2} , niin osittaisintegroimalla nähdään, että

{\displaystyle I\_{n}=-\cos {\frac {\pi }{2}}\sin ^{n-1}{\frac {\pi }{2}}+\cos 0\sin ^{n-1}0-\int \_{0}^{\frac {\pi }{2}}-\cos x(n-1)\sin ^{n-2}x\cdot \cos x\,{\text{d}}x}
{\displaystyle =(n-1)\int \_{0}^{\frac {\pi }{2}}\cos ^{2}x\sin ^{n-2}x\,{\text{d}}x=(n-1)\int \_{0}^{\frac {\pi }{2}}(1-\sin ^{2}x)\sin ^{n-2}x\,{\text{d}}x}
{\displaystyle =(n-1)(I\_{n-2}-I\_{n})}
Siispä saadaan  rekursiivinen kaava {\displaystyle \scriptstyle I\_{n}}:lle:

{\displaystyle I\_{n}={\frac {n-1}{n}}\cdot I\_{n-2}}
Tämän avulla nähdään, että

{\displaystyle I\_{2n}={\frac {2n-1}{2n}}\cdot I\_{2n-2}={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot I\_{2n-4}}
{\displaystyle =\ldots ={\frac {1}{2}}\cdot {\frac {3}{4}}\cdot \ldots \cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-1}{2n}}\cdot I\_{0}} ja
{\displaystyle I\_{2n+1}={\frac {2n}{2n+1}}\cdot I\_{2n-1}={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot I\_{2n-3}}
{\displaystyle =\ldots ={\frac {2}{3}}\cdot {\frac {4}{5}}\cdot \ldots \cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n}{2n+1}}\cdot I\_{1}}
Näin ollen

{\displaystyle {\frac {I\_{2n+1}}{I\_{2n}}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \ldots \cdot {\frac {2n-2}{2n-3}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\cdot {\frac {I\_{1}}{I\_{0}}}=a\_{n}\cdot {\frac {2}{\pi }}} , eli
{\displaystyle a\_{n}={\frac {I\_{2n+1}}{I\_{2n}}}\cdot {\frac {\pi }{2}}}
Koska {\displaystyle \sin ^{2n+2}x\leq \sin ^{2n+1}x\leq \sin ^{2n}x} kaikilla {\displaystyle x\in \left[0,{\frac {\pi }{2}}\right]} , niin {\displaystyle I\_{2n+2}\leq I\_{2n+1}\leq I\_{2n}}. Siten

{\displaystyle 1\geq {\frac {I\_{2n+1}}{I\_{2n}}}\geq {\frac {I\_{2n+2}}{I\_{2n}}}={\frac {{\frac {2n+1}{2n+2}}I\_{2n}}{I\_{2n}}}={\frac {2n+1}{2n+2}}\to 1} , kun {\displaystyle n\to \infty }. Siis
{\displaystyle \lim \_{n\to \infty }a\_{n}=\lim \_{n\to \infty }{\frac {I\_{2n+1}}{I\_{2n}}}\cdot {\frac {\pi }{2}}={\frac {\pi }{2}}} , eli väite (1) on todistettu. {\displaystyle \Box }
Koska

{\displaystyle b\_{n+1}^{2}=b\_{n}^{2}\cdot {\frac {(2n+2)^{2}}{(2n+1)^{2}}}\cdot {\frac {n}{n+1}}} ja
{\displaystyle a\_{n}=a\_{n+1}\cdot {\frac {(2n+1)(2n+3)}{(2n+2)^{2}}}} , niin  induktiotodistuksella nähdään helposti, että
{\displaystyle b\_{n}^{2}={\frac {2n+1}{n}}\cdot a\_{n}} kaikilla *n*. Siten väite (2) seuraa väitteestä (1). {\displaystyle \Box }

Katso myös
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* Luettelo piin laskukaavoista

Lähteet
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1. ↑  Lehtinen, Matti: *Osittaisintegroinnin ihmeitä*:
*Wallisin ja Stirlingin kaavat*