傅里叶级数推导

物理意义：把一个比较复杂的周期运动看成是许多不同频率的简谐振动的叠加。

三角函数系

cos x, sinx, cos2x, sin2x.…, cosnx, sinnx.…

正交性

在[- $\pi$， $\pi$]上正交，即其中任意两个不同的函数之积在[- $\pi$， $\pi$]上的积分等于0.

可以证明：

• $\int_{-\pi}^{\pi} \cos n x d x=0$
• $\int_{-\pi}^{\pi} \sin n x d x=0$
• $\begin{array}{c}{\int_{-\pi}^{\pi} \cos m x \cos n x d x=0 \quad(m=1,2,3, \cdots, n=1,2,3, \cdots m \neq n )}\end{array}$
• $\begin{array}{c}{\int_{-\pi}^{\pi} \sin m x \sin n x d x=0 \quad(m=1,2,3, \cdots, n=1,2,3, \cdots m \neq n )}\end{array}$
• $\begin{aligned} \int_{-\pi}^{\pi} \sin m x \cos n x d x &=0 \quad(m=1,2,3, \cdots, n=1,2,3, \cdots ) \end{aligned}$当m=n时
  $\begin{array}{l}{\int_{-\pi}^{\pi} 1 \cdot 1 \mathrm{d} x=2 \pi} \\\\ {\int_{-\pi}^{\pi} \cos ^{2} n x d x=\pi} \\\\ {\int_{-\pi}^{\pi} \sin ^{2} n x d x=\pi}\end{array}(n=1,2, \cdots)$
  设 $f(x)$是周期为2 $\pi$的周期函数，且可逐项积分，利用三角级数得
  $f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n x+b_{n} \sin n x\right)$ 想要表达 $f(x)$得求出 $a_{0},a_{n},b_{n}$，对两边进行积分得
  $\begin{aligned} \int_{-\pi}^{\pi} f(x) \mathrm{d} x=& \int_{-\pi}^{\pi} \frac{a_{0}}{2} \mathrm{d} x+\sum_{n=1}^{\infty}\left[\int_{-\pi}^{\pi} a_{n} \cos n x \mathrm{d} x\right. +\int_{-\pi}^{\pi} b_{n} \sin n x \mathrm{d} x ] \end{aligned}$因为 $a_{0}, a_{n}, b_{n}$为常数，利用三角函数的正交性
• $\int_{-\pi}^{\pi} \cos n x d x=0$
• $\int_{-\pi}^{\pi} \sin n x d x=0$
  得到
  $\int_{-\pi}^{\pi} f(x) \mathrm{d} x=\int_{-\pi}^{\pi} \frac{a_{0}}{2} \mathrm{d} x=\pi a_{0}$
  $a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x$
  为了求 $a_{n}$，在等式两边 $\cos k x$
  $\begin{aligned} \int_{-\pi}^{\pi} f(x) \cos k x d x &=\int_{-\pi}^{\pi} \frac{a_{0}}{2} \cos k x d x \\ &+\sum_{n=1}^{\infty} I_{-\pi}^{\pi} a_{n} \cos k x \cos n x d x \\ &+\int_{-\pi}^{\pi} b_{n} \cos k x \sin n x d x ] \end{aligned}$当k=n时，由三角函数的正交性可知 $\begin{aligned} & \int_{-\pi}^{\pi} a_{n} \cos k x \cos n x d x=\int_{-\pi}^{\pi} a_{n} \cos ^{2} n x d x \\=& a_{n} \int_{-\pi}^{\pi} \frac{1+\cos 2 n x}{2} d x=a_{n} \pi \end{aligned}$其余各项均为零.因此 $a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x \quad(n=1,2,3, \cdots)$同理 $b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x \quad(n=1,2,3, \cdots)$
  整理一下得：
  $\left\{\begin{array}{ll}{a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x} & {(n=0,1,2, \cdots)} &\\\\ {b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x} & {(n=1,2,3, \cdots)}\end{array}\right.$
  $a_{n}(0开始的)，b_{n}$称为傅里叶系数。由傅里叶系数组成的三角级数称为傅里叶级数。

例
$f(x)=\left\{\begin{array}{lr}{-1,} & {-\pi \leq x<0} \\ {1,} & {0 \leq x<\pi}\end{array}\right.$

$\begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x \\ &=\frac{1}{\pi} \int_{-\pi}^{0}(-1) \cos n x d x+\frac{1}{\pi} \int_{0}^{\pi} \cos n x d x \\ &=-\frac{1}{\pi} \frac{1}{n} \sin n x ]_{-\pi}^{0}+\frac{1}{\pi} \frac{1}{n} \sin n x ]_{0}^{\pi}=0 \\ &(n=0,1,2,3 \cdots) \end{aligned}$
$\begin{aligned} b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin u x d x \\ &=\frac{1}{\pi} \int_{-\pi}^{0}(-1) \sin x d x+\frac{1}{\pi} \int_{0}^{\pi} \sin x d x \\ &=\frac{1}{\pi} \frac{1}{n} \cos n\left.x\right|_{-\pi} ^{0}-\frac{1}{\pi}\left[\frac{1}{n} \cos n x\right]_{0}^{\pi} \\ &=\frac{2}{n \pi}\left[1-(-1)^{n}\right] \\ &=\left\{\begin{array}{l}{\frac{4}{n \pi}, n=1,3,5, \cdots} \\ {0}\end{array}\right. \end{aligned}$所以 $f(x)=\sum_{n=1}^{\infty}b_{n} \sin n x\\=\frac{4}{\pi}\left[\sin x+\frac{1}{3} \sin 3 x+\cdots+\frac{1}{2 n-1} \sin (2 n-1) x+\cdots\right]$