傅里叶级数

1. 傅里叶系数与傅里叶级数

设函数 $\displaystyle{ f \left( x \right) }$ 是周期为 $2\pi$ 的周期函数，且在 $[-\pi,\pi]$ 上可积，则称

$$\begin{aligned} a_{n} & = {1 \over \pi} \int_{-\pi}^{\pi} f(x) \cos nx \mathrm{d} x & n=0,1,2,\cdots \\ b_{n} & = {1 \over \pi} \int_{-\pi}^{\pi} f(x) \sin nx \mathrm{d} x & n=1,2,3,\cdots \end{aligned}$$

为 $\displaystyle{ f \left( x \right) }$ 的傅里叶系数；称

$${a_{0} \over 2} + \sum\limits_{n=1}^{\infty}(a_{n} \cos nx + b_{n} \sin nx)$$

为 $\displaystyle{ f \left( x \right) }$ 以 $2\pi$ 为周期的形式傅里叶级数，记作

$$f(x) \sim {a_{0} \over 2} + \sum\limits_{n=1}^{\infty}(a_{n} \cos nx + b_{n} \sin nx)$$

如果函数 $\displaystyle{ f \left( x \right) }$ 是周期为 $\displaystyle{ T }$ 的函数，在周期上可积，则

$$\displaystyle{ \begin{aligned}a _{ n } & = \frac{ 2 }{ T } \int _{ T } f \left( x \right) \cos n { \frac{ 2 \pi }{ T } } x {\text{d}x} \\ b _{ n } & = \frac{ 2 }{ T } \int _{ T } f \left( x \right) \sin n { \frac{ 2 \pi }{ T } } x {\text{d}x}\end{aligned} }$$

为 $\displaystyle{ f \left( x \right) }$ 的傅里叶系数；称

$$\displaystyle{ \frac{ a _{ 0 } }{ 2 } + \sum _{ n = 1 } ^{ \infty } \left( a _{ n } \cos n { \frac{ 2 \pi }{ T } } x + b _{ n } \sin n { \frac{ 2 \pi }{ T } } x \right) }$$

为 $\displaystyle{ f \left( x \right) }$ 以 $\displaystyle{ T }$ 为周期的形式傅里叶级数

2. 傅里叶级数的收敛性（狄利克雷收敛定理）

设 $\displaystyle{ f \left( x \right) }$ 是周期为 $2\pi$ 的可积函数，且满足

1. $\displaystyle{ f \left( x \right) }$ 在 $[-\pi,\pi]$ 上连续或只有有限个第一类间断点
2. $\displaystyle{ f \left( x \right) }$ 在 $[-\pi, \pi]$ 上只有有限个单调区间

则 $\displaystyle{ f \left( x \right) }$ 的以 $2\pi$ 为周期的傅里叶级数收敛，且

$$S(x) = {a_{0} \over 2} + \sum\limits_{n=1}^{\infty}(a_{n} \cos nx + b_{n} \sin nx) = {1 \over 2} \left(f(x^{+}) + f(x^-)\right)$$

3. 周期为 $2\pi$ 的函数的展开

3.1. $[-\pi, \pi]$ 上展开

$$\begin{aligned} a_{n} & = {1 \over \pi} \int_{-\pi}^{\pi}f(x) \cos nx \mathrm{d} x & n=0,1,2,\cdots \\ b_{n} & = {1 \over \pi} \int_{-\pi}^{\pi}f(x) \sin nx \mathrm{d} x & n=1,2,3,\cdots \end{aligned}$$

3.2. $[-\pi, \pi]$ 上奇偶函数的展开

$\displaystyle{ f \left( x \right) }$ 为奇函数

$$\begin{aligned} a_{n}&=0 &n=0,1,2,\cdots \\ b_{n}&={2 \over \pi} \int_{0}^{\pi}f(x) \sin nx \mathrm{d} x &n=1,2,3,\cdots \end{aligned}$$

$\displaystyle{ f \left( x \right) }$ 为偶函数

$$\begin{aligned} a_{n}&= {2 \over \pi} \int_{0}^{\pi}f(x) \cos nx \mathrm{d} x &n=0,1,2,\cdots \\ b_{n}&= 0 &n=1,2,3,\cdots \end{aligned}$$

在 $[0,\pi]$ 上展为正弦或展为余弦

展为正弦

$$\begin{aligned} a_{n}&=0 &n=0,1,2,\cdots \\ b_{n}&={2 \over \pi} \int_{0}^{\pi}f(x) \sin nx \mathrm{d} x &n=1,2,3,\cdots \end{aligned}$$

展为余弦

$$\begin{aligned} a_{n}&= {2 \over \pi} \int_{0}^{\pi}f(x) \cos nx \mathrm{d} x &n=0,1,2,\cdots \\ b_{n}&= 0 &n=1,2,3,\cdots \end{aligned}$$

4. 周期为 $\displaystyle{ 2 l }$ 的函数的展开

4.1. $\displaystyle{ \left[ - l , l \right] }$ 上展开

$$\begin{aligned} a_{n}&= {1 \over l} \int_{-l}^{l}f(x)\cos {n \pi x \over l} \mathrm{d} x & n=0,1,2,\cdots \\ b_{n}&= {1 \over l} \int_{-l}^{l}f(x)\sin {n \pi x \over l} \mathrm{d} x & n=0,1,2,\cdots \end{aligned}$$

4.2. $\displaystyle{ \left[ - l , l \right] }$ 上奇偶函数的展开

$\displaystyle{ f \left( x \right) }$ 为奇函数

$$\begin{aligned} a_{n} &= 0 & n=0,1,2,\cdots \\ b_{n} &= {2 \over l} \int_{0}^{l} f(x) \sin {n \pi x \over l} \mathrm{d} x & n=1,2,3,\cdots \end{aligned}$$

$\displaystyle{ f \left( x \right) }$ 为偶函数

$$\begin{aligned} a_{n} &= {2 \over l} \int_{0}^{l} f(x) \cos {n \pi x \over l} \mathrm{d} x & n=0,1,2,\cdots \\ b_{n} &= 0 & n=1,2,3,\cdots \end{aligned}$$

4.3. 在 $\displaystyle{ \left[ 0 , l \right] }$ 上展为正弦或展为余弦

展为正弦

$$\begin{aligned} a_{n} &= 0 & n=0,1,2,\cdots \\ b_{n} &= {2 \over l} \int_{0}^{l} f(x) \sin {n \pi x \over l} \mathrm{d} x & n=1,2,3,\cdots \end{aligned}$$

展为余弦

$$\begin{aligned} a_{n} &= {2 \over l} \int_{0}^{l} f(x) \cos {n \pi x \over l} \mathrm{d} x & n=0,1,2,\cdots \\ b_{n} &= 0 & n=1,2,3,\cdots \end{aligned}$$