准备材料
高等数学同济版 第十二章，无穷级数，第七，第八节
https://www.bilibili.com/video/BV1i341117L8?from=search&seid=6725153828974183622&spm_id_from=333.337.0.0

1.基础概念

\(任意周期函数f(x)可以转换为傅里叶级数\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n \sin nx),有点像泰勒展开\)

2.基础知识-正交性

\(离散值的正交性,內积=0\)
\(连续值的正交性,\int_D f(x)g(x) =0,即积分=0\)

2.1三角函数的正交性

\(\int_{-\pi}^{\pi}\cos nx dx=0,n=1,2,3,...\)
\(\int_{-\pi}^{\pi}\sin nx dx=0,n=1,2,3,...\)
\(\int_{-\pi}^{\pi}\sin kx\cos nx dx=0,k,n=1,2,3,...,k可以等于n\)
\(\int_{-\pi}^{\pi}\cos kx\cos nx dx=0,k,n=1,2,3,...,k\ne n\)
\(\int_{-\pi}^{\pi}\sin kx\sin nx dx=0,k,n=1,2,3,...,k\ne n\)
\(\color{red}{证明公式待补充}\)
\(结论:得到了一组正交的三角函数系,1,\cos x,\sin x,\cos 2x,\sin 2x,....\)

3.求\(a_0,a_n,b_n\)

\(令f(x)是原函数,\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n \sin nx)是傅里叶级数\)
\(求\int_{-\pi}^{\pi}f(x)dx,可得a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\)
\(求\int_{-\pi}^{\pi}f(x)\cos nxdx,可得a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nxdx\)
\(求\int_{-\pi}^{\pi}f(x)\sin nxdx,可得b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nxdx\)
\(这里当a_n的n=0时,就等于a_0,所以一般不单独写a_0,合并到a_n中\)
\(\color{red}{证明公式待补充}\)

4.傅里叶级数的收敛性

\(函数f(x)的傅里叶级数是否一定收敛?如果它收敛是否一定收敛于f(x)?\)

定理(收敛定理,狄利克雷充分条件)

\(设f(x)是周期为2\pi的周期函数,如果它满足\)
\(1.在一个周期内连续或只有有限个第一类间断点\)
\(2.在一个周期内至多只有有限个极值点\)
\(则f(x)的傅里叶级数收敛,并且\)
\(当x是f(x)的连续点时,级数收敛于f(x)\)
\(当x是f(x)的间断点时,级数收敛于\frac{1}{2}[f(x^-)+f(x^+)]\)
\(\color{red}{书上也没有证明,知道下就好,有点复杂,暂时就知道知道连续函数就能收敛就行}\)

4.通用周期形式

\(设周期为2l的周期函数f(x)满足收敛定理的条件,则它的傅里叶级数展开式为\)
\(\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos \frac{n\pi x}{l}+b_n \sin \frac{n\pi x}{l}),x\in C\)
\(a_n=\frac{1}{l}\int_{-\pi}^{\pi}f(x)\cos \frac{n\pi x}{l}dx,n=0,1,2,...\)
\(b_n=\frac{1}{l}\int_{-\pi}^{\pi}f(x)\sin \frac{n\pi x}{l}dx,n=1,2,...\)
\(C=\{x|f(x)=\frac{1}{2}[f(x^-)+f(x^+)]\}\)

5.例题

\(设f(x)=\begin{cases} 1 & -\pi \le x\le 0\\ -1 & 0\le x < \pi \end{cases},求傅里叶级数\)
\(1.先证明收敛性,略\)
\(2.求a_n,b_n\)
\(a_n=\frac{1}{\pi}\int_{-\pi}^{0}(-1)\cos nx dx + \frac{1}{\pi}\int_{0}^{\pi}(1)\cos nx dx =0\)
\(b_n=\frac{1}{\pi}\int_{-\pi}^{0}(-1)\sin nx dx + \frac{1}{\pi}\int_{0}^{\pi}(1)\sin nx dx=\begin{cases} \frac{4}{n\pi},&n=1,3,5\\ 0,& n=2,4,6\\ \end{cases}\)
\(3.代入傅里叶级数,略\)