Answer to Question #164820 in Differential Equations for fahad chachar

Question

define odd even and periodic function and obtain fourier series fx=x,-2<x<2

Answer 1

An even function is any function f such that 

 f(−x) = f(x), for all x in its domain. 

Examples: cos(x), sec(x), any constant function,

An odd function is any function f such that 

 f(−x) = −f(x), for all x in its domain. 

Examples: sin(x), tan(x), csc(x), cot(x),

A function f(x) is called a periodic function if ∃ p > 0

(called a period of f) such that f(x + p) = f(x) for all x.

SOLUTION TO THE FOURIER SERIES

Firstly, we try to get $a_0$

$a_0=\frac{1}{L}\int_{-L}^Lf(x)cos\frac{mπx}{L}dx\\ Where L=2\\ a_0=\frac{1}{2}\int_{-2}^2xdx\\ a_0=\frac{1}{2}\frac{x^2}{2}|_{-2}^2\\ a_0=\frac{1}{2}(2-2)=0\\ a_0=0\\ \text{The rest of cosine coefficient for n=1, 2, 3 are}\\ a_n=\frac{1}{L}\int_{-L}^Lf(x)cos\frac{nπx}{L}dx\\ a_n=\frac{1}{2}\int_{-2}^2xcos\frac{nπx}{2}dx\\ \text{Using integration by part}\\ a_n=\frac{1}{2}(\frac{2x}{nπ}sin\frac{nπx}{2}|_{-2}^2-\frac{2}{nπ}\int_{-2}^2sin\frac{nπx}{2}dx\\ a_n=\frac{1}{2}(\frac{2x}{nπ}sin\frac{nπx}{2}+\frac{4}{n^2π^2}cos\frac{nπx}{2}|_{-2}^2)// a_n=\frac{1}{2}((0+\frac{4}{n^2π^2}cos(nπ)-(0+\frac{4}{n^2π^2}cos(-nπ)\\ a_n=0\\ \text{Hence there is no nonzero cosine coefficient for the function}\\ \text{The sine coefficient for n=1, 2, 3...are}\\ b_n=\frac{1}{L}\int_{-L}^Lf(x)sin\frac{nπx}{L}dx\\ b_n=\frac{1}{2}\int_{-2}^2xsin\frac{nπx}{2}dx\\ \text{Using integration by part}\\ b_n=\frac{1}{2}(\frac{-2x}{nπ}cos\frac{nπx}{2}|_{-2}^2-\frac{-2}{nπ}\int_{-2}^2cos\frac{nπx}{2}dx\\ b_n=\frac{1}{2}(\frac{-2x}{nπ}cos\frac{nπx}{2}+\frac{4}{n^2π^2}sin\frac{nπx}{2}|_{-2}^2)// b_n=\frac{1}{2}((\frac{-4}{nπ}cos(nπ)-0)-(\frac{4}{nπ}cos(-nπ)-0))\\ b_n=\frac{-2}{nπ}(cos(nπ)+cos(nπ))\\ b_n=\frac{-4}{nπ}cosnπ\\ \text{When n is odd}\\ f(x)=\frac{4}{nπ}\\ \text{When n is even}\\ f(x)=\frac{-4}{nπ}\\ Therefore, \\ f(x)=\frac{4}{π}\sum_ {n=1}^\infin\frac{({-1})^{n+1}}{n}sin\frac{nπx}{2}\\.$