欧拉公式

Euler's formula

三角函数的指数表达

证明

根据泰勒级数有：
$e^x=1+x+{\displaystyle \frac{1}{2!}}{x}^2+{\displaystyle \frac{1}{3!}}{x}^3+{\displaystyle \frac{1}{4!}}{x}^4+\cdots$
$cos(x)=1-{\displaystyle \frac{1}{2!}}{x}^2+{\displaystyle \frac{1}{4!}}{x}^4-{\displaystyle \frac{1}{6!}}{x}^6+\cdots$
$sin(x)=x-{\displaystyle \frac{1}{3!}}{x}^3+{\displaystyle \frac{1}{5!}}{x}^5-{\displaystyle \frac{1}{7!}}{x}^7+\cdots$

$e^z=1+z+{\displaystyle \frac{z^2}{2!}}+{\displaystyle \frac{z^3}{3!}}+{\displaystyle \frac{z^4}{4!}}+\cdots +{\displaystyle \frac{z^n}{n!}}\cdots$
$e^{ix}=1+ix-{\displaystyle \frac{x^2}{2!}}-i{\displaystyle \frac{x^3}{3!}}+{\displaystyle \frac{x^4}{4!}}+i{\displaystyle \frac{x^5}{5!}}-{\displaystyle \frac{x^6}{6!}}-i{\displaystyle \frac{x^7}{7!}}+\cdots$
实部： $1-{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^4}{4!}}-{\displaystyle \frac{x^6}{6!}}+\cdots$ 即 $cos(x)$
虚部： $ix-i{\displaystyle \frac{x^3}{3!}}+i{\displaystyle \frac{x^5}{5!}}-i{\displaystyle \frac{x^7}{7!}}+\cdots$ 即 $isin(x)$