From e0dcf5a960852f65442a560c3fc6cbf289a4c834 Mon Sep 17 00:00:00 2001
From: eggy <danielchen04@hotmail.ca>
Date: Tue, 21 Nov 2023 12:37:35 -0500
Subject: [PATCH] ece205: add complex fouriers

---
 docs/2a/ece205.md | 36 ++++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)

diff --git a/docs/2a/ece205.md b/docs/2a/ece205.md
index 7188490..c1a6032 100644
--- a/docs/2a/ece205.md
+++ b/docs/2a/ece205.md
@@ -453,6 +453,42 @@ Then, for any $x\in[-L,L]$:
 
 $$\int^x_{-L}f(t)dt=\int^x_{-L}\frac{a_0}{2}dt+\sum^\infty_{n=1}\int^x_{-L}(a_n\cos(\frac{n\pi t}{L})+b_n\sin(\frac{n\pi t}{L}))dt$$
 
+### Complex Fourier series
+
+By employing Euler's theorem, sine and cosine can be transformed into exponential forms.
+
+$$
+\cos(\frac{n\pi x}{L})=\frac{e^{i\frac{n\pi x}{L}} + e^{-i\frac{n\pi x}{L}}}{2} \\
+\sin(\frac{n\pi x}{L})=\frac{-ie^{i\frac{n\pi x}{L}} + ie^{-i\frac{n\pi x}{L}}}{2}
+$$
+
+Thus the **complex Fourier series** is given by:
+
+$$
+f(x)=\sum^\infty_{n=-\infty}c_ne^{i\frac{n\pi x}{L}} \\
+c_n=\frac{1}{2L}\int^L_{-L}f(x)e^{-i\frac{n\pi x}{L}}dx = \frac 1 2(a_n-ib_n)
+$$
+
+To convert it to a real Fouier series:
+
+- $a_0=2c_0$
+- $a_n=c_n+\overline{c_n}$
+- $b_n=i(c_n-\overline{c_n})$
+
+!!! example
+    The complex Fourier series for the sawtooth wave function: $f(x)=x,-1<x<1,f(x+2)=f(x)$. Thus we have a period of 2 and $L=1$.
+    
+    \begin{align*}
+    c_0&=\frac 1 2\int^1_{-1}\underbrace{xe^{0}}_\text{odd}dx \\
+    &=0 \\
+    \\
+    c_n&=\frac 1 2\int^1_{-1}xe^{-in\pi x}dx \\
+    \tag{IBP}&=\frac 1 2\left[\frac{xe^{-in\pi x}}{-in\pi}-\int\frac{1}{-in\pi}e^{-in\pi x}dx\right]^1_{-1} \\
+    &=\frac 1 2\left[\frac{xe^{-n\pi x}}{-in\pi}+\frac{1}{n^2\pi^2}e^{-in\pi x}\right]^1_{-1} \\
+    &=\frac{(-1)^ni}{n\pi} \\
+    \\
+    \therefore f(x)&=\sum^\infty_{\substack{n=-\infty \\ n\neq0}}\frac{(-1)^ni}{n\pi}e^{in\pi x}
+    \end{align*}
 
 ## Resources