欧拉公式的证明

欧拉公式的证明

由泰勒公式的麦克劳林公式展开可得:

\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4} + \frac{x^5}{5!} + \cdots \]

观察欧拉公式:

\[e^{i \theta} = \cos \theta + i \sin \theta \]

\(x = i \theta\) 代入 \(e^x\) 得:

\[e^{i \theta} = 1 + i \theta - \frac{\theta^2}{2!} - \frac{i \theta^3}{3!} + \frac{\theta^4}{4!} + \frac{\theta^5}{5!} - \cdots ① \]

\(x = \theta\) 代入 \(\cos x\) 得:

\[\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots② \]

\(x = i \theta\) 代入 \(\sin x\) 得:

\[\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots ③ \]

即 ① = ② + ③ 可得:

\[e^{i \theta} = \cos \theta + i \sin \theta \]

posted @ 2022-02-27 16:58  RedJACK~  阅读(636)  评论(0)    收藏  举报