常用三角函数的无穷级数乘积公式推导详细过程与图形展示

多项式逼近理论得到常用三角函数的无穷级数乘积公式

问题: 求方程 sin⁡(x)=0\sin(x)=0sin(x)=0 的解。

1、首先, sin⁡(x)=0\sin(x)=0sin(x)=0 有解 {kπ，k=0,±1,±2,⋯.}\{kπ，k=0,\pm 1,\pm 2, \cdots .\}{kπ，k=0,±1,±2,⋯.}

2、假设 sin⁡(x)\sin(x)sin(x) 是多项式函数，由多项式有根的代数基本理论 (the fundamental theorem of algebra) 即多项式逼近理论 (polynomial approximation theorem)

sin⁡(x)=c∗∏k=1∞(kπ−x)(kπ+x)xsin⁡(x)x=c∗∏k=1∞(kπ−x)(kπ+x)\displaystyle \sin(x) = c*\prod_{k=1}^{\infty} (kπ-x)(kπ+x)x \\[1em]\frac{\sin(x)}{x} = c*\prod_{k=1}^{\infty} (kπ-x)(kπ+x)sin(x)=c∗k=1∏∞(kπ−x)(kπ+x)xxsin(x)=c∗k=1∏∞(kπ−x)(kπ+x)

令 x→0x \to 0x→0，求极限得到
c=∏k=1∞1k2π2\displaystyle c=\prod_{k=1}^{\infty} \frac {1}{k^2π^2}c=k=1∏∞k2π21

我们得到奇函数sin⁡(x)\sin(x)sin(x) 的无穷级数乘积公式（一）：

fsin(x)=sin⁡(x)=x∏k=1∞(1−xkπ)(1+xkπ)=x∏k=1∞[1−x2(kπ)2]\displaystyle \begin{aligned} \red {fsin(x)} =\sin(x) &= x {\prod_{k=1} ^{\infty} ({1- \frac{x}{kπ})} (1+ \frac{x}{kπ})} \\&= x \prod_{k=1} ^{\infty} \left[1- \frac{x^2}{(kπ)^2}\right ] \end{aligned}fsin(x)=sin(x)=xk=1∏∞(1−kπx)(1+kπx)=xk=1∏∞[1−(kπ)2x2]

三角函数sin(x) 的无穷级数乘积公式GGB展示

令 x=π2x = \frac{π}{2}x=2π, 有

π2=∏k=1∞2k2k−1⋅2k2k+1=∏k=1∞11−14k2=∏k=1∞[1+14k2−1]=21⋅23⋅43⋅45⋅65⋅67⋯=43⋅1615⋅3635⋅6463⋯\displaystyle \begin{aligned} \frac{π}{2} &= \prod_{k=1} ^{\infty} \frac{2k}{2k-1} \centerdot \frac{2k}{2k+1} \\&= \prod_{k=1} ^{\infty} \frac{1}{1- \frac{1}{4k^2}} \\&= \prod_{k=1} ^{\infty} \left[1+\frac{1}{4k^2-1}\right] \\&=\frac{2}{1} \centerdot \frac{2}{3} \centerdot \frac{4}{3} \centerdot \frac{4}{5} \centerdot \frac{6}{5} \centerdot \frac{6}{7} \cdots \\&=\frac{4}{3} \centerdot \frac{16}{15} \centerdot \frac{36}{35} \centerdot \frac{64}{63} \cdots \end{aligned}2π=k=1∏∞2k−12k⋅2k+12k=k=1∏∞1−4k211=k=1∏∞[1+4k2−11]=12⋅32⋅34⋅54⋅56⋅76⋯=34⋅1516⋅3536⋅6364⋯

参见 John Wallis’ product for πππ

因为sin⁡(π/2−x)=cos⁡(x)\sin(\pi/2-x)=\cos(x)sin(π/2−x)=cos(x), 所以有偶函数cos⁡(x)\cos(x)cos(x) 的无穷级数乘积公式（二）：
fc1(x)=cos⁡(x)=(π2−x)∏k=1∞(1−π/2−xkπ)(1+π/2−xkπ)=(π2−x)∏k=1∞(1−12k+xkπ)(1+12k−xkπ)\displaystyle \begin{aligned} \red{fc_1(x)}=\cos(x)&=(\frac{\pi}{2}-x) \prod_{k=1} ^{\infty} {\left(1 - \frac{\pi/2-x}{kπ}\right)} {\left(1 + \frac{\pi/2-x}{kπ}\right)}\\&=(\frac{\pi}{2}-x) \prod_{k=1} ^{\infty} {\left(1-\frac{1}{2k}+\frac{x}{k \pi}\right)}{\left(1+\frac{1}{2k}-\frac{x}{k \pi}\right)} \end{aligned}fc1(x)=cos(x)=(2π−x)k=1∏∞(1−kππ/2−x)(1+kππ/2−x)=(2π−x)k=1∏∞(1−2k1+kπx)(1+2k1−kπx)

同理可得

fcos(x)=cos⁡(x)=(1−2xπ)∏k=1∞(1+2x(2k−1)π)(1−2x(2k+1)π)=(1−2xπ)∏k=1∞(1+x/πk−1/2)(1−x/πk+1/2)\displaystyle \begin{aligned} \red{fcos(x)}=\cos(x)&=\left(1-\frac{2x}{\pi}\right) \prod_{k=1} ^{\infty} {\left(1 + \frac{2x}{(2k-1)π}\right)} {\left(1 - \frac{2x}{(2k+1)π}\right)}\\&=\left(1-\frac{2x}{\pi}\right) \prod_{k=1} ^{\infty} {\left(1+\frac{x/\pi}{k-1/2}\right)}{\left(1-\frac{x/\pi}{k+1/2}\right)} \end{aligned}fcos(x)=cos(x)=(1−π2x)k=1∏∞(1+(2k−1)π2x)(1−(2k+1)π2x)=(1−π2x)k=1∏∞(1+k−1/2x/π)(1−k+1/2x/π)

ftan(x)=tan⁡(x)=4xπ⋅∏k=1∞1−xkπ1−14k⋅1+xkπ1+14k\displaystyle \begin{aligned} \red{ftan(x)}= \tan(x)&=\frac{4x}{\pi} \centerdot \prod_{k=1} ^{\infty} \frac{1 - \frac{x}{kπ}}{1-\frac{1}{4k}} \centerdot \frac{1 + \frac{x}{kπ}}{1+\frac{1}{4k}}\end{aligned}ftan(x)=tan(x)=π4x⋅k=1∏∞1−4k11−kπx⋅1+4k11+kπx

三角函数cos(x) 的无穷级数乘积公式GGB展示

三角函数tan(x) 的无穷级数乘积公式GGB展示