# 傅立葉分析
:::spoiler (table of contents)
[TOC]
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## 傅立葉級數(Fourier series)
將一**週期函數**展開成一連串不同週期函數的合成
> 週期函數的分解

### periodic function(週期函數)
若 f(x) 對 x∈R 存在一正數 T，滿足 f(x) = f(x+T)
則稱 f(x) 為週期函數，其週期為 T

#### 性質
1. 若週期函數 f(x) 的周期為 T ，則 n∈N，nT 亦為 f(x) 的周期，即 f(x+nT) = f(x)，因此 T 稱為基本週期。
2. 若 f(x) 及 g(x) 周期均為 T ，則此兩函數的線性組合 $h(x) = k_1f(x) + k_2g(x)，k_1, k_2$∈常數，h 的周期亦為T
3. f(mx) 的週期為 T/m (壓縮/拉長)
4. f(x) 週期 $T_f$, g(x) 週期 $T_g$ 兩函數線性組合 $h(x) = k_1f(x) + k_2g(x)，k_1, k_2$∈常數，則 h(x) 的週期為 $T_h = LCM(k_1, k_2)$

### 奇偶函數(odd & even function)
1. for x∈R, f(x) = f(-x)，則 f(x) 為偶函數
2. for x∈R, f(-x) = -f(x)，則 f(x) 為奇函數

#### 特性
1. 奇偶性
2. 若f(x) odd，g(x) even
* $\int_{-k}^{k}f(x)dx = 0$
* $\int_{-k}^{k}f(x)dx = 2\int_{0}^{k}f(x)dx$

## 傅立葉積分(Fourier integral)
* non periodic function(非週期函數)
若週期無限大(無規律)
則其表示可由一級數趨近一積分表示式

## 三角函數
### 合角公式
1. $sin(a+b) = sina\cdot cosb + sinb\cdot cosa$
2. $cos(a+b) = cosa\cdot cosb - sina\cdot sinb$

### 積化和差

1. $sinA\cdot cosB = \dfrac{1}{2}(sin(A + B) + sin(A - B))$
2. $cosA\cdot sinB = \dfrac{1}{2}(sin(A + B) - sin(A - B))$
3. $cosA\cdot cosB = \dfrac{1}{2}(cos(A + B) + cos(A - B))$
4. $sinA\cdot sinB = \dfrac{1}{2}(cos(A + B) - cos(A - B))$

### 考慮 sinX、cosX 的積分特性
1. $n∈Z, \int_{-\pi}^{\pi}Sin(nx) = 0$
2. $n∈Z, n\not=0, \int_{-\pi}^{\pi}Cos(nx) = 0$
> 三角函數積整數週期，**其值為 0 !!**

### 積化和差積分 (結論)
$\int_{-\pi}^{\pi}sin(Ax)\cdot sin(Bx)dx$
$=\int_{-\pi}^{\pi}\dfrac{1}{2}[cos(A + B)x - cos(A - B)x]dx$
$=\dfrac12[\int_{-\pi}^{\pi}cos(A + B)xdx - \int_{-\pi}^{\pi}cos(A - B)x]dx$

\begin{cases}
\frac 12\cdot 0 = 0& \text{if $n \not= m$}\\
\frac 12\cdot 2\pi = \pi & \text{if $n = m$}
\end{cases}

$\int_{-\pi}^{\pi}Cos(Ax)\cdot Cos(Bx)dx$
$=\int_{-\pi}^{\pi}\dfrac12[ cos(A + B)x + cos(A - B)x]dx$
$=\dfrac12[\int_{-\pi}^{\pi}cos(A + B)xdx + \int_{-\pi}^{\pi}cos(A - B)x]dx$

\begin{cases}
\frac 12\cdot 0 = 0& \text{if $n \not= m$}\\
\frac 12\cdot 2\pi = \pi & \text{if $n = m$}
\end{cases}

$\int_{-\pi}^{\pi}Sin(Ax)\cdot Cos(Bx)dx$
$=\int_{-\pi}^{\pi}\dfrac12[ sin(A + B)x + sin(A - B)x]dx$
$=\dfrac12[ \int_{-\pi}^{\pi}sin(A + B)xdx + \int_{-\pi}^{\pi}sin(A - B)xdx]$
$=0$

## Fourier series :
> 若週期2π函數，於區間[- π, π]上為可積分的函數，我們可以決定a0, a1, a2, ..., an, b1, b2, ..., bn as Fourier coefficient，使 $f(x) = a_0 + \sum_{n=1}^{\infty}(a_n\cos nx + b_n\sin nx)$，此式存在 Fourier series representation.

### 決定週期 T = 2π 的 Fourier series
首先考慮下列函數集合，n∈N
1. $f(x) = b_1\sin1x + b_2\sin2x + b_3\sin3x +\cdots+ b_n\sin nx+\cdots$
* 其週期 2π, π, ... (2/n)π
* LCM = 2π
* sum = b1 + b2 + ... + bn
* 得 $f(x) = \sum_{n=1}^{\infty}b_n\cdot \sin nx$

2. $f(x) = a_0\cdot 1(scaling)+ a_1\cos1x+ a_2\cos2x+ a_3\cos3x +\cdots+ a_n\cos nx +\cdots$
* 其週期 2π, π, ... (2/n)π
* LCM = 2π
* sum = a0 + a1 + a2 + ... + an
* 得 $f(x) = a_0 + \sum_{n=1}^{\infty}a_n\cdot \cos nx$

3. $a_0\cdot 1+\\ a_1\cos1x+ a_2\cos2x+ a_3\cos3x +\cdots+ a_n\cos nx +\cdots + \\b_1\sin1x + b_2\sin2x + b_3\sin3x +\cdots+ b_n\sin nx+\cdots$
* 其週期 2π, π, ... (2/n)π
* LCM = 2π
* sum = a0 + a1 + a2 + ... + an + b1 + b2 + ... + bn
* 得 $f(x) = a_0 + \sum_{n=1}^{\infty}(a_n\cos nx + b_n\sin nx)$

>由前 3 case，可利用 sinx cosx 合成一週期 2π 的函數
反之，一個週期 T = 2π 的函數，可將其分解成一組三角函數的合成

><font color="#CD5C5C">ex
(1) $\int_{-\pi}^{\pi}$ $\cos 2x\cdot\cos 3x$ &thinsp;$dx=0 \because n\not=m$
(2) $\int_{-\pi}^{\pi}$ $\sin 3x\cdot\sin 3x$ &thinsp;$dx=\pi \because n=m$
(3) $\int_{-\pi}^{\pi}$ $\cos x\cdot\cos x$ &thinsp;$dx=\pi \because n=m$
(4) $\int_{-\pi}^{\pi}$ $\sin 2x\cdot\cos 5x$ &thinsp;$dx=0 \because n\not=m$
</font>

### 找出 $a_0, a_n, b_n$
#### 決定 a0 :
想法 : 保留$a_0$，讓其他項為0
$f(x) = a_0 + \sum_{n=1}^{\infty}(a_n\cos nx + b_n\sin nx)$
積分$\rightarrow \int_{-\pi}^{\pi}f(x)d = \int_{-\pi}^{\pi}a_0 dx+0$

整理 $\int_{-\pi}^{\pi}f(x)d = \int_{-\pi}^{\pi}a_0 dx = 2π\cdot a_0$
得 $a_0=\frac {1}{2\pi}\int_{-\pi}^{\pi}f(x)dx$

> 黑魔法：算平均值也可

#### 考慮 a1 :
$\int_{-\pi}^{\pi}f(x)\color{red}{Cos(1x)}$
$=a_0\int_{-\pi}^{\pi}\color{red}{Cos(1x)}+\sum_{n=1}^{\infty}(a_n \int_{-\pi}^{\pi}Cos(nx)\color{red}{Cos(1x)}dx + b_n \int_{-\pi}^{\pi}Sin(nx)\color{red}{Cos(1x)}dx)$
$=a_0 \cdot 0 + a_1Cos(1x)\color{red}{Cos(1x)} + \sum_{n=2}^{\infty}(a_n \int_{-\pi}^{\pi}Cos(nx)\color{red}{Cos(1x)}dx + b_n \int_{-\pi}^{\pi}Sin(nx)\color{red}{Cos(1x)}dx)$
$=0 + a_1 \cdot π + 0$

$\rightarrow a_1 = \dfrac1π \int_{-\pi}^{\pi}f(x)\color{red}{Cos(1x)}dx$

#### 推廣一般式 ak :
$(這個n和\sum_n的n不同)$
$\int_{-\pi}^{\pi}f(x)\color{red}{Cos(kx)}$
$=\int_{-\pi}^{\pi}a_0 \color{red}{Cos(kx)} + \sum_{n=1}^{\infty}(a_n \int_{-\pi}^{\pi}Cos(nx)\color{red}{Cos(kx)}dx + b_n \int_{-\pi}^{\pi}Sin(nx)\color{red}{Cos(kx)}dx)$
$=0+a_k \int_{-\pi}^{\pi}Cos(kx)\color{red}{Cos(kx)} + \sum_{n=1,n\not=k}^{\infty}( a_n \int_{-\pi}^{\pi}( Cos(nx)\color{red}{Cos(kx)}) + b_n \int_{-\pi}^{\pi}( Sin(nx)\color{red}{Cos(kx)})$
$=a_k \int_{-\pi}^{\pi}Cos(kx)\color{red}{Cos(kx)} + 0$
$=a_k \cdot π$

$\rightarrow a_k = \dfrac1π \int_{-\pi}^{\pi}f(x)Cos(\color{red}{k}x)dx$

#### bk :
$\int_{-\pi}^{\pi}f(x)\color{red}{Sin(kx)}$
$= \int_{-\pi}^{\pi}a_0 \color{red}{Sin(kx)} + \sum_{n=1}^{\infty}(a_n \int_{-\pi}^{\pi}Cos(nx)\color{red}{Sin(kx)}dx + b_n \int_{-\pi}^{\pi}Sin(nx)\color{red}{Sin(kx)}dx)$
$=0+b_k \int_{-\pi}^{\pi}Sin(kx)\color{red}{Sin(kx)} + \sum_{n=1,n\not=k}^{\infty}( a_n \int_{-\pi}^{\pi}( Cos(nx)\color{red}{Sin(kx)}) + b_n \int_{-\pi}^{\pi}( Sin(nx)\color{red}{Sin(kx)})$
$= b_k \int_{-\pi}^{\pi}Sin(kx)\color{red}{Sin(kx)} + 0$
$= b_k \cdot π$

$\rightarrow b_k = \dfrac1π \int_{-\pi}^{\pi}f(x)Sin(\color{red}{k}x)$
### 總結

><font color="#CD5C5C" size=5>$T=2\pi$
$f(x)=a_0+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)$
$a_0=\frac {1}{2\pi}\int_{-\pi}^{\pi}f(x)dx$
$a_k=\frac 1\pi\int_{-\pi}^{\pi}f(x)\cos kx\,dx$
$b_k=\frac 1\pi\int_{-\pi}^{\pi}f(x)\sin kx\,dx$</font>

#### 特殊案例
When T = 2π
1. 若f(x) is odd :
- $a_0 = a_n = 0$
- $b_n = \dfrac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx$

2. 若f(x) is even :
- $a_0 = \dfrac{1}{\pi}\int_{0}^{\pi}f(x)dx$
- $a_n = \dfrac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx$
- $b_n = 0$

### 實作
#### Square Wave (Odd)
$T = 2\pi$

$a_0 = 0$
$a_n = \dfrac{1}{π}\int_{-\pi}^{\pi} f(x)Cos(kx)dx = 0$
$b_n = \dfrac{1}{π}\int_{-π}^{π}f(x)Sin(kx)dx$
$= \dfrac{1}{π} 2\int_{0}^{π}f(x)Sin(kx)dx$
> 0 ~ π 區間 f(x) = 1

$= \dfrac{2}{π} \int_{0}^{π}Sin(kx)dx$
$= \dfrac{2}{π} \dfrac{-Cos(nx)}{n} | _{0}^{π}$
$= \dfrac{-2}{nπ} [Cos(nπ) - 1]$
$b_n = \dfrac{-2}{nπ} [(-1)^n-1]$

$f(x) = \dfrac{-2}{nπ}\sum_{n=1}^{\infty} [(-1)^n-1]sin(nx)$

#### Saw Wave (Even)
$T = 2\pi$
$f(x) = a_0 + \sum_{n=1}^{\infty}(a_nCos(nx)+b_nSin(nx))$

$a_0 = \dfrac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx$
$=\dfrac{1}{2\pi}2\int_{0}^{\pi}f(x)dx = \dfrac{\pi}{2}$

$a_n = \dfrac{1}{\pi}\int_{-\pi}^{\pi}f(x)Cos(nx)dx = \dfrac{1}{\pi}2\int_{0}^{\pi}xCos(nx)dx$
$=\dfrac{2}{\pi}[(x\dfrac{Sin(nx)}{n})-\int_{0}^{\pi}\dfrac{Sin(nx)}{n}dx]$
$=-\dfrac{2}{n\pi}\int_{0}^{\pi}Sin(nx)dx = -\dfrac{2}{n\pi}[-\dfrac{Cos(nx)}{n}]$
$\dfrac{2}{n^2\pi}(Cos(n\pi)-1) = \dfrac{2}{n^2\pi}[(-1)^n-1]$

\begin{cases}
0, & \text{if $n$ is even} \\
\dfrac{-4}{n^2\pi}, & \text{if $n$ is odd}
\end{cases}

### 題目
- $f(x) = x^2, -\pi\le x\le\pi$
- f(x) 定義於 $[-\pi, \pi]$ 之 Fourier Series
- $a_0 = \dfrac{1}{2\pi}x^2dx = \dfrac{\pi}{3}$
- $a_n = \dfrac{2}{\pi}\int_{0}^{\pi}X^2Cos(nx)dx$
- $= \dfrac{2}{\pi}[\pi^2 - \int_{0}^{\pi}\dfrac{Sin(nx)}{n}2xdx]$
- $-\dfrac{4}{n\pi}\int_{0}^{\pi}xSin(nx)dx = \dfrac{4(-1)^n}{n^2}$
- $f(x) = \dfrac{\pi^2}{3} + \sum_{n=1}^{\infty}\dfrac{4(-1)^n}{n^2}$
- $\sum_{n=1}^{\infty}\dfrac{1}{n^2}=?$
- 令 x = π
- f(π) = $\dfrac{\pi^2}{3}+\sum_{n=1}^{\infty}\dfrac{4(-1)^nCos(n\pi)}{n^2}$
> see $\cos(n\pi)$ as $(-1)^n$
- $\pi^2 = \dfrac{\pi^2}{3} + \sum_{n=1}^{\infty}\dfrac{4}{n^2}$
- $\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2 - \dfrac{\pi^2}{3}}{4}$
- $\dfrac{\pi^2}{6}$
- $\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2}=?$
- 令 x = 0
- $f(0) = \dfrac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^2}$
- $=-\dfrac{\pi^2}{12}$

\int_{-\pi}^{\pi}
\sum_{n=1}^{\infty}