SkyLadder

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傅里叶级数(指数形式)

推导过程

根据欧拉公式 \(e^{i\theta} = cos(\theta)+isin(\theta)\) 可得

\[ cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2},sin(\theta)=-\frac{i(e^{i\theta}-e^{-i\theta})}{2} \]

将上式代入 (1)

\[ f(x)=\frac{a_{0}}{2}+\sum^{+\infty}_{n=1} a_{n}\frac{e^{in\omega x}+e^{-in\omega x}}{2}+\sum^{+\infty}_{n=1}b_{n}(-\frac{i(e^{in\omega x}-e^{-in\omega x})}{2}) \]

化简

\[ f(x)=\frac{a_{0}}{2}+\sum^{+\infty}_{n=1} \frac{a_{n}-ib_{n}}{2} e^{in\omega x}+\sum^{+\infty}_{n=1} \frac{a_{n}+ib_{n}}{2} e^{-in\omega x} \]

由于 \[ \frac{a_{0}}{2}=\sum^{0}_{n=0}\frac{a_{0}}{2}e^{in\omega x},\sum^{+\infty}_{n=1} \frac{a_{n}+ib_{n}}{2} e^{-in\omega x}=\sum^{-1}_{n=-\infty} \frac{a_{-n}+ib_{-n}}{2} e^{in\omega x} \]

\[ f(x)=\sum^{0}_{n=0}\frac{a_{0}}{2}e^{in\omega x}+\sum^{+\infty}_{n=1} \frac{a_{n}-ib_{n}}{2} e^{in\omega x}+\sum^{-1}_{n=-\infty} \frac{a_{-n}+ib_{-n}}{2} e^{in\omega x} \]

显然

\[ f(x)=\sum^{\infty}_{n=-\infty}C_{n}e^{in\omega x} ,其中 C_{n}= \left\{ \begin{aligned} &\frac{a_{0}}{2},n=0 \\ &\frac{a_{n}-ib_{n}}{2},n>0 \\ &\frac{a_{-n}+ib_{-n}}{2},n<0 \end{aligned} \right. \]

\(C_{n}\) 的值

\(n=0\) 时,根据 (2)

\[ C_{n}=C_{0}=\frac{a_{0}}{2}=\frac{\int^{t_{0}+T}_{t_{0}}f(x)dx}{T}=\frac{1}{T}\int^{t_{0}+T}_{t_{0}}f(x)e^{-in\omega x}dx (n=0) \]

\(n>0\) 时,根据 (3) 和 (4)

\[ C_{n}=\frac{a_{n}-ib_{n}}{2}=\frac{1}{2}(\frac{2}{T}\int^{t_{0}+T}_{t_{0}}f(x)cos(n\omega x)dx-i\frac{2}{T}\int^{t_{0}+T}_{t_{0}}f(x)sin(n\omega x)dx) \]

化简得

\[ C_{n}=\frac{1}{T}\int^{t_{0}+T}_{t_{0}}f(x)e^{-in\omega x}dx \]

\(n<0\) 时,根据 (3) 和 (4)

\[ C_{n}=\frac{a_{-n}+ib_{-n}}{2}=\frac{1}{2}(\frac{2}{T}\int^{t_{0}+T}_{t_{0}}f(x)cos(-n\omega x)dx+i\frac{2}{T}\int^{t_{0}+T}_{t_{0}}f(x)sin(-n\omega x)dx) \]

化简得

\[ C_{n}=\frac{1}{T}\int^{t_{0}+T}_{t_{0}}f(x)e^{-in\omega x}dx \]

因此,当 \(n\in(-\infty,+\infty)\)

\[ C_{n}=\frac{1}{T}\int^{t_{0}+T}_{t_{0}}f(x)e^{-in\omega x}dx \]


结论

综上所述,任何满足狄利克雷条件的周期函数 \(f(x)\) 可写成以下指数形式

  1. \[ f(x)=\sum^{\infty}_{n=-\infty}C_{n}e^{in\omega x} \]

其中

  1. \[ C_{n}=\frac{1}{T}\int^{t_{0}+T}_{t_{0}}f(x)e^{-in\omega x}dx \]

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