Wallis公式

作者: 洛玖言 | 来源:发表于2019-10-05 10:32 被阅读0次

Wallis

这个公式叫做 Wallis公式，考研的很多人都叫他 点火公式，废话少说，开搞！

$\displaystyle I_n=\int_0^{\frac\pi2}\sin^nx\text{d}x=\int_0^{\frac\pi2}\cos^nx\text{d}x=\begin{cases} \dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdots\dfrac{2}{3},&n为大于1的奇数\\ \\ \dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdots\dfrac{1}{2}\cdot\dfrac{\pi}{2},&n为正偶数 \end{cases}$

$\displaystyle I_n=\int_0^{\pi}\sin^nx\text{d}x=2\int_0^{\frac\pi2}\sin^nx\text{d}x$

$\displaystyle \int_0^{\pi}\cos^nx\text{d}x=\begin{cases}\displaystyle 2\int_0^{\frac\pi2}\cos^{n}x\text{d}x,&n为正偶数\\ 0,&n为正奇数\\ \end{cases}$

$\displaystyle\int_0^{2\pi}\sin^{n}x\text{d}x=\int_0^{2\pi}\cos^{n}x\text{d}x=\begin{cases} 4\dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdots\dfrac{1}{2}\cdot\dfrac{\pi}{2},&n为正偶数\\ 0,&n为正奇数\\ \end{cases}$

证明

$\begin{aligned} I_n=&\int_0^{\frac\pi2}\sin^nx\text{d}x\\ =&-\cos x\sin^{n-1}x\bigg|_0^{\frac\pi2}-\int_0^{\frac\pi2}-(n-1)\cos^2x\sin ^{n-2}x\text{d}x\\ =&(n-1)\int_0^{\frac\pi2}\cos^2x\sin^{n-2}x\text{d}x\\ =&(n-1)\int_0^{\frac\pi2}(1-\sin^2x)\sin^{n-2}x\text{d}x\\ =&(n-1)\int_0^{\frac\pi2}\sin^{n-2}x\text{d}x-(n-1)\int_0^{\frac\pi2}\sin^nx\text{d}x\\ =&(n-1)I_{n-2}-(n-1)I_n\\ =&\dfrac{n-1}{n}I_{n-2} \end{aligned}$

$\therefore \begin{cases}\dfrac{I_{2n+1}}{I_{2n-1}}=\dfrac{2n}{2n+1}\\\\\dfrac{I_{2n}}{I_{2n-2}}=\dfrac{2n-1}{2n}\end{cases}$

$\displaystyle\dfrac{I_3}{I_1}\cdot\dfrac{I_5}{I_3}\cdots\dfrac{I_{2n+1}}{I_{2n-1}}=\dfrac{2\times1}{2\times1+1}\cdot\dfrac{2\times2}{2\times2+1}\cdots\dfrac{2\times n}{2\times n+1}=\dfrac{(2n)!!}{(2n+1)!!}$

$\displaystyle\dfrac{I_2}{I_0}\cdot\dfrac{I_4}{I_2}\cdots\dfrac{I_{2n}}{I_{2n-2}}=\dfrac{2\times1-1}{2\times2}\cdot\dfrac{2\times2-1}{2\times2}\cdots\dfrac{2\times n-1}{2\times n}=\dfrac{(2n-1)!!}{(2n)!!}$

$\displaystyle I_n=\int_0^{\frac\pi2}\sin^nx\text{d}x=\begin{cases}\dfrac\pi2,&n=0\\1,&n=1\\\dfrac{(2k)!!}{(2k+1)!!},&n=2k+1\\\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{\pi}{2},&n=2k\end{cases}$

剩下几个根据奇偶性易得...

Wallis公式

Wallis

证明

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