## 定义

e
\begin{align} e &= \lim_{x \to \infty} (1 + \dfrac{1}{x})^x \\ e &= \sum_{n=0}^\infty {1 \over n!} = {1 \over 0!} + {1 \over 1!}+ {1 \over 2!} + {1 \over 3!}+ {1 \over 4!} + \cdots \end{align}

## 推论

$$e^x = \lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

• 由于二项式定理从 0 开始，所以展开式从 0 开始
• 分母的阶乘来源于组合数的分母
• 分子的 $x^{n-1}$ 来源于 $(\dfrac {x}{n}) ^{n-k}$ 的指数

\begin{align} e^{\theta i} &=\dfrac{\theta ^0 i}{0!}+\frac{\theta ^1i}{1!}-\dfrac{\theta^{2}i}{2 !}-\dfrac{\theta^{3} i}{3 !}+\dfrac{\theta^{4}}{4 !}+\dfrac{\theta^{5} i}{5 !}+\cdots \\ \sin \theta &=\frac{\theta^1}{1!}-\dfrac{\theta^{3}}{3 !}+\dfrac{\theta^{5}}{5 !}+\cdots \\ \cos \theta &=\frac{\theta ^0}{0!}-\dfrac{\theta^{2}}{2 !}+\dfrac{\theta^{4}}{4 !}+\cdots \\ \end{align}

\begin{align} e^{\theta i}&=\cos \theta+i \sin \theta &= p\\ e^{-\theta i}&=\cos \theta-i \sin \theta &= n \end{align}

\begin{align} \cos \theta &= \dfrac{p+n}{2} &= \cos h \theta i\\ i\sin \theta &= \dfrac{p-n}{2} &= \sin h \theta i \end{align}