sin x的无穷乘积式
MDNTCT
·
·
个人记录
观察以下两个式子:
\boxed{\begin{aligned}
&\sin{3x}=3\sin{x}-4\sin^3{x}\\
&\sin{5x}=5\sin{x}-20\sin^3{x}+16\sin^5{x}\\
&\cdots
\end{aligned}}
可以发现
\sin{(2n+1)x}=\sin{x}\cdot\mathrm{P}(\sin^2{x})
其中\mathrm{P}(x)是关于x的n次多项式.
因为\lim_{x\to0}\dfrac{\sin{(2n+1)x}}{\sin{x}}=2n+1，所以\mathrm{P}(x)的常数项为2n+1.
同时，\sin{(2n+1)x}的根为\dfrac{k\pi}{2n+1},k\in\mathbb{Z}，所以\sin^2{\dfrac{k\pi}{2n+1}},k=1,2,\cdots,n恰为\mathrm{P}(x)的n个根.
\begin{aligned}
&\mathrm{P}(x)=(2n+1)\left(1-\frac{x}{\sin^2{\dfrac{\pi}{2n+1}}}\right)\left(1-\frac{x}{\sin^2{\dfrac{2\pi}{2n+1}}}\right)\cdots\left(1-\frac{x}{\sin^2{\dfrac{n\pi}{2n+1}}}\right)\\
&\mathrm{P}(x)=(2n+1)\prod_{k=1}^{n}\left(1-\frac{x}{\sin^2{\cfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{(2n+1)x}}{\sin{x}}=(2n+1)\prod_{k=1}^{n}\left(1-\frac{\sin^2{x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{x}}{(2n+1)\sin{\cfrac{1}{2n+1}x}}=\prod_{k=1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{x}}{(2n+1)\sin{\cfrac{1}{2n+1}x}\displaystyle\prod_{k=1}^{m}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)}=\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
\end{aligned}
左边对n取极限，使n\to\infty.
\frac{\sin{x}}{x \displaystyle \prod_{k=1}^{m}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}
对于右边，有这一不等式
\boxed{\begin{aligned}
&\frac{2}{\pi}x<\sin{x}<x,x\in\left(0,\frac{\pi}{2}\right)\\
&\sin^2{\frac{1}{2n+1}x}<\frac{x^2}{(2n+1)^2}\\
&\sin^2{\frac{k\pi}{2n+1}}>\frac{4k^2}{(2n+1)^2}\\
&\frac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\cfrac{k\pi}{2n+1}}}<\frac{x^2}{4k^2}\\
\end{aligned}}
于是就有
\begin{aligned}
&1>\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)>\prod_{k=m+1}^{n}\left(1-\cfrac{x^2}{4k^2}\right)>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)\\
&1>\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)
\end{aligned}
这个式子是否成立与n的取值无关，所以代入n\to\infty的极限可得
\boxed{1>\frac{\sin{x}}{x \displaystyle \prod_{k=1}^{m}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)}
再在两边对m取极限，使m\to\infty.两边极限都是1，所以根据极限夹逼准则，可得
\boxed{\frac{\sin{x}}{x }= \prod_{k=1}^{\infty}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}
\boxed{\sin{x}=x \prod_{k=1}^{\infty}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}