Wallisin kaavat

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) ${\displaystyle \lim _{n\to \infty }{\frac {(n!)^{2}\cdot 2^{2n}}{(2n!){\sqrt {n}}}}={\sqrt {\pi }}}$.

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). ${\displaystyle \Box }$

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