## 傅立葉級數 (下)

$\displaystyle F(x)=\frac{a_0}{2}+\sum_{k=1}^{\infty}a_k\cos(kx)+\sum_{k=1}^{\infty}b_k\sin(kx)$

\displaystyle \begin{aligned} a_k&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(kx)dx,~k=0,1,2,\ldots,\\ b_k&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(kx)dx,~k=1,2,\ldots\end{aligned}

$f(x)$ 是一奇函數，則 $f(-x)\cos(-kx)=-f(x)\cos(kx)$，故 $a_k=0$$k=0,1,\ldots$。另一方面，若 $f(x)$ 是一偶函數，則 $f(-x)\sin(-kx)=-f(x)\sin(kx)$，故 $b_k=0$$k=1,2,\ldots$

$T$-週期函數的傅立葉級數

$\displaystyle F(t)=\frac{a_0}{2}+\sum_{k=1}^{\infty}a_k\cos\left(\frac{2\pi kt}{T}\right)+\sum_{k=1}^{\infty}b_k\sin\left(\frac{2\pi kt}{T}\right)$

$dx=2\pi dt/T$ 代入 $f(x)$ 的傅立葉係數的積分公式，可得

\displaystyle \begin{aligned} a_k&=\frac{2}{T}\int_{-T/2}^{T/2}f(t)\cos\left(\frac{2\pi kt}{T}\right)dt,~k=0,1,2,\ldots,\\ b_k&=\frac{2}{T}\int_{-T/2}^{T/2}f(t)\sin\left(\frac{2\pi kt}{T}\right)dt,~k=1,2,\ldots\end{aligned}

$\displaystyle\cos x=\frac{e^{ix}+e^{-ix}}{2},~\sin x=\frac{e^{ix}-e^{-ix}}{2i}$

\displaystyle\begin{aligned} a_k\cos(kx)+b_k\sin(kx)&=a_k\left(\frac{e^{ikx}+e^{-ikx}}{2}\right)+b_k\left(\frac{e^{ikx}-e^{-ikx}}{2i}\right)\\ &=\left(\frac{a_k-ib_k}{2}\right)e^{ikx}+\left(\frac{a_k+ib_k}{2}\right)e^{-ikx}\\ &=c_ke^{ikx}+c_{-k}e^{-ikx},~~k=1,2,\ldots,\end{aligned}

$\displaystyle c_k=\frac{a_k-ib_k}{2},~~c_{-k}=\frac{a_k+ib_k}{2}$

$k=0$，就有 $c_0=a_0/2$。將以上結果代回 $2\pi$-週期函數 $f(x)$ 的傅立葉級數即得指數傅立葉級數：

$\displaystyle F(x)=\sum_{k=-\infty}^{\infty}c_ke^{ikx}$

$\displaystyle c_k=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)(\cos(kx)-i\sin(kx))dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$

$\displaystyle F(t)=\sum_{k=-\infty}^{\infty}c_ke^{2\pi ikt/T}$

$\displaystyle c_k=\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-2\pi ikt/T}dt$

$\displaystyle\left\langle f,g\right\rangle\overset{\underset{\mathrm{def}}{}}{=}\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(x)}g(x)dx$

\displaystyle \begin{aligned}\Vert f\Vert&=\left\langle f,f\right\rangle^{1/2}\\ &=\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(x)}f(x)dx\right)^{1/2}\\ &=\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}\vert f(x)\vert^2dx\right)^{1/2}.\end{aligned}

\displaystyle \begin{aligned} \int_{-\pi}^{\pi}\cos(mx)\sin(nx)dx&=0,\\ \int_{-\pi}^{\pi}\cos(mx)\cos(nx)dx&=\pi\delta_{mn},\\ \int_{-\pi}^{\pi}\sin(mx)\sin(nx)dx&=\pi\delta_{mn},\end{aligned}

\displaystyle \begin{aligned} \int_{-\pi}^{\pi}e^{-imx}e^{inx}dx&=\int_{-\pi}^{\pi}(\cos(mx)-i\sin(mx))(\cos(nx)+i\sin(nx))dx\\ &=\int_{-\pi}^{\pi}\cos(mx)\cos(nx)dx+\int_{-\pi}^{\pi}\sin(mx)\sin(nx)dx\\ &~~~+i\int_{-\pi}^{\pi}\cos(mx)\sin(nx)dx-i\int_{-\pi}^{\pi}\sin(mx)\cos(nx)dx\\ &=2\pi\delta_{mn},\end{aligned}

$\displaystyle F(x)=\sum_{k=-\infty}^{\infty}\left\langle e^{ikx},f(x)\right\rangle e^{ikx}$

$\displaystyle c_k=\left\langle e^{ikx},f(x)\right\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$

Parseval 定理

$\mathcal{V}$ 為一有限維向量空間，$\boldsymbol{\beta}=\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$$\mathcal{V}$ 的一組單範正交基底，則 $\mathcal{V}$ 中任一向量 $\mathbf{x}$ 的長度與其正交分解展開式 $\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n$ 的係數 $c_1,\ldots,c_n$ 之間具有以下簡單關係：

\displaystyle\begin{aligned}\Vert\mathbf{x}\Vert^2&=\left\langle\mathbf{x},\mathbf{x}\right\rangle\\ &=\left\langle c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n,c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n\right\rangle\\ &=\vert c_1\vert^2\left\langle\mathbf{v}_1,\mathbf{v}_1\right\rangle+\cdots+\vert c_n\vert^2\left\langle\mathbf{v}_n,\mathbf{v}_n\right\rangle\\ &=\sum_{i=1}^n\vert c_i\vert^2.\end{aligned}

$\displaystyle \Vert f\Vert^2=\frac{1}{T}\int_{-T/2}^{T/2}\vert f(t)\vert^2dt=\sum_{k=-\infty}^{\infty}\vert c_k\vert^2$

$k>1$$\vert c_k\vert^2=\vert c_{-k}\vert^2=(a_k^2+b_k^2)/4$，且 $\vert c_0\vert^2=a_0^2/4$，故複傅立葉係數平方和亦可寫為

$\displaystyle\sum_{k=-\infty}^{\infty}\vert c_k\vert^2=\frac{a_0^2}{4}+\frac{1}{2}\sum_{k=1}^{\infty}(a_k^2+b_k^2)$

\displaystyle\begin{aligned} \Vert f\Vert^2&=\left\langle F,f\right\rangle=\frac{1}{T}\int_{-T/2}^{T/2}\overline{F(t)}f(t)dt\\ &=\frac{1}{T}\int_{-T/2}^{T/2}\left(\sum_{k=-\infty}^{\infty}\overline{c_k}e^{-ikx}\right)f(t)dt\\ &=\sum_{k=-\infty}^{\infty}\overline{c_k}\left(\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-2\pi ikt/T}dt\right)\\ &=\sum_{k=-\infty}^{\infty}\overline{c_k}c_k=\sum_{k=-\infty}^{\infty}\vert c_k\vert^2.\end{aligned}

$\displaystyle a_k=\frac{16}{\pi^2k^2}(-1)^k,~~k=1,2,\ldots,$

$a_0=8/3$，則 $f(t)=t^2$ 的傅立葉級數為

$\displaystyle F(t)=\frac{4}{3}+\frac{16}{\pi^2}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\cos\left(\frac{\pi kt}{2}\right)$

$\displaystyle\Vert f\Vert^2=\frac{1}{4}\int_{-2}^2t^4dt=\frac{16}{5}$

$\displaystyle\sum_{k=-\infty}^{\infty}\vert c_k\vert^2=\frac{a_0^2}{4}+\frac{1}{2}\sum_{k=1}^{\infty}a_k^2=\frac{16}{9}+\frac{16^2}{2\pi^4}\sum_{k=1}^{\infty}\frac{1}{k^4}$

$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90}$