\(\sin x\)的无穷乘积

Theorem. \[\sin x = x \cdot \prod _{k=1}^{+\infty}[1-\frac{x^2}{(k\pi)^2}].\]
Proof.
对于\(x\in\mathbb R,n\ge 0 \),令\[I_n(x) = \int_0^{\pi/2}\cos^n \xi\cdot \cos(x\xi)d\xi.\]
首先证明：\[I_{n-2}(x) = \frac{n^2-x^2}{n(n-1)}I_n(x), \quad n\ge 2,x\ne 0.\]
由\(I_n(x) = \frac 1 x \int_0^{\pi/2}\cos^n\xi d(\sin(x\xi))\)
\(=\frac{\cos^n\xi\cdot \sin(x\xi)}{x}|^{\pi/2}_0+\frac n x \int_0^{\pi/2} \cos^{n-1}\xi\cdot \sin \xi\cdot \sin(x\xi)d\xi\)
\(=-\frac n {x^2} \int _ 0 ^{\pi/2}\cos^{n-1}\xi\cdot \sin \xi d(\cos(x\xi))\),
可得\(xI_n(x) = -\frac n x \int_0 ^{\pi/2}\cos^{n-1}\xi\cdot \sin \xi d(\cos(x\xi))\)
\(=-\frac n x \cos^{n-1}\xi\cdot \sin \xi \cdot\cos(x\xi)|^{\pi/2}_{0}+ \frac n x \int_0^{\pi /2}[-(n-1)\cos^{n-2}\xi\sin^2\xi+\cos^n\xi]\cos(x\xi)d\xi\)
\(=\frac n x \int_0^{\pi/2}[-(n-1)(1-\cos^2\xi)\cos^{n-2}\xi\cdot\cos(x\xi)+\cos(x\xi)\cdot\cos^n\xi]d\xi\)
\(=\frac n x\{ -(n-1)[I_{n-2}(x) - I_n(x)] + I_n(x)\}\)
整理得\((n^2-x^2)I_n(x) - n(n-1)I_{n-2}(x) = 0.\)

再证明：\[I_n(0) = \frac{n-1}{n}I_{n-2}(0).\]
与上一结论类似，
\(I_n(0) = \int_0^{\pi/2}\cos^n\xi d\xi\)
\(=\int_0^{\pi/2}\cos^{n-1}d(\sin\xi)\)
\(=\cos^{n-1}\xi\cdot\sin\xi|^{\pi/2}_{0}+(n-1)\int_0^{\pi/2}\sin^2\xi\cdot\cos^{n-2}\xi d\xi\)
\(=(n-1)\int_0^{\pi/2}(1-\cos^2\xi)\cos^{n-2}\xi d\xi\)
\(=(n-1)(I_{n-2}(0)-I_n(0)),\) 整理即得。

回到原题，不妨设\(x\ne 0\)，我们有
\(I_0(x) = \int_0^{\pi/2}\cos(x\xi)d\xi\)
\(=\frac 1 x \int_0^{\pi/2}\cos (x\xi)d(x\xi)\)
\(=\frac{\sin\frac{\pi}{2}x}{x}.\)

令\[P_n(x) = \prod_{k=1}^n [1-\frac{x^2}{(2k)^2}],\quad Q_n(x) =\frac{I_n(x)}{I_n(0)}.\]
则\(\sin\frac{\pi}{2}x= x\cdot I_0(x) \)
\(=\frac{\pi}{2} x\cdot\frac{I_0(x)}{I_0(0)}\)
\(=\frac{\pi}{2} x\cdot(1-\frac{x^2}{2^2})\frac{I_2(x)}{I_2(0)}\)
\(=\cdots = \frac{\pi}{2} x\cdot P_n(x)\cdot Q_{2n}(x).\)

只需证明：无穷乘积\(P_n(x)\)收敛，且\(Q_{2n}(x)\)收敛于1。然后将\(\frac \pi 2 x\)以\(x\)代入即得原结论。

无穷乘积\(P_n(x)\)的收敛性由\(\sum\limits_{k=1}^\infty \frac {x^2}{(2k)^2}\)收敛且同号给出。对于\(Q_{2n}(x)\)，我们有
\(0\le I_n(0)-I_n(x) = \int_0^{\pi/2}(1-\cos\xi)\cos^n\xi d\xi\)
\(=\int_0^{\pi/2}2\sin^2\frac{x\xi}{2}\cos^n\xi d\xi\)
\(\le\frac{x^2}{2}\int_0^{\pi/2}\xi^2\cos^n\xi d\xi\)
\(=\frac{x^2}{2}\int_0^{\pi/2}(\xi\cos\xi)\cdot(\xi\cos^{n-1}\xi) d\xi\)
\(\le \frac{x^2}{2n}\int_0^{\pi/2}n\xi\cos^{n-1}\xi\cdot\sin\xi d\xi\)(利用不等式\(\tan \xi \ge \xi \to \xi\cos \xi\le\sin\xi\))
\(=\frac{x^2}{2n}(-\xi\cos^n\xi |^{\pi/2}_0+\int_0^{\pi/2}\cos^n d\xi)\)
\(=\frac{x^2}{2n}I_n(0)\to 0(n\to+\infty).\)
两边同时除以\(I_n(0)\)，明欲所证。

    所属分类：数学分析     发表于2022-05-10