Answer to Question #116600 in Calculus for gly

Question

Obtain the Fourier series for the function f whose definition in one period is f(x)= -x, -1≤x≤1.Sketch the graph of f

Answer 1

$f(x)=-x, x\in[-1,1]$

$f(x)=\frac{a_0}{2}+\sum\limits^{\infty}_{m=1}(a_m\cos(m\pi x)+b_m\sin(m\pi x))$

where

$a_m=\int\limits^{1}_{-1}(-x)\cos(m\pi x)dx, m=0,1,2,...\\ b_m=\int\limits^{1}_{-1}(-x)\sin(m\pi x)dx, m=1,2,...$

Let $m=0$

$a_0=\int\limits^{1}_{-1}(-x)dx=-\frac{x^2}{2}|^{1}_{-1}=-\frac{1}{2}+\frac{1}{2}=0\\ a_m=\int\limits^{1}_{-1}(-x)\cos(m\pi x)dx=-\frac{x}{\pi m}\sin( m\pi x)|^{1}_{-1}+\\ +\frac{1}{\pi m}\int\limits^{1}_{-1}\sin(m\pi x)dx=-\frac{1}{(\pi m)^2}\cos (m\pi x)|^{1}_{-1}=\\ =-\frac{1}{(\pi m)^2}(\cos m\pi- \cos m\pi)=0\\ b_m=\int\limits^{1}_{-1}(-x)\sin(m\pi x)dx=\frac{x}{\pi m}\cos (m\pi x)|^{1}_{-1}-\\ -\frac{1}{\pi m}\int\limits^{1}_{-1}\cos(m\pi x)dx=\frac{2}{\pi m}\cos m\pi -\\ -\frac{1}{(\pi m)^2}\sin (m\pi x)|^{1}_{-1}=\frac{2}{\pi m}(-1)^m\\$

$-x=\sum\limits^{\infty}_{m=1}\frac{2}{\pi m}(-1)^m\sin(m\pi x)\\$