Answer to Question #218954 in Calculus for hellena

Question #218954

using the knowledge of odd and even functions, find the Fourier series decomposition of

f(x)={2x,-π<x<π

{f(x+2π)


1
Expert's answer
2021-07-27T06:09:37-0400

For the odd function, the Fourier series is called the Fourier Sine series and is given by


fodd(x)=n=1bnsinnxf_{odd}(x)=\displaystyle\sum_{n=1}^{\infin}b_n\sin nx

where the Fourier coefficients are


bn=2π0πf(x)sinnxdx,n=1,2,3,...b_n=\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}f(x)\sin nxdx, n=1, 2, 3, ...

bn=2π0π2xsinnxdxb_n=\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}2x\sin nxdx

2xsinnxdx\int 2x\sin nx dx

udv=uvvdu\int u dv=uv-\int v du

u=2x,du=2dxu=2x, du=2dx

dv=sinnxdx,v=sinnxdx=1ncosnxdv=\sin nxdx, v=\int \sin nx dx=-\dfrac{1}{n}\cos nx

2xsinnxdx=2xncosnx+2ncosnxdx\int 2x\sin nx dx=-\dfrac{2x}{n}\cos nx+\int \dfrac{2}{n}\cos nxdx

=2xncosnx+2n2sinnx+C=-\dfrac{2x}{n}\cos nx+\dfrac{2}{n^2}\sin nx+C

bn=2π[2xncosnx+2n2sinnx]π0b_n=\dfrac{2}{\pi}\bigg[-\dfrac{2x}{n}\cos nx+\dfrac{2}{n^2}\sin nx\bigg]\begin{matrix} \pi \\ 0 \end{matrix}

=4ncosπn=(1)n+14n=-\dfrac{4}{n}\cos \pi n=\dfrac{(-1)^{n+1}4}{n}

f(x)=n=1(1)n+14nsinnxf(x)=\displaystyle\sum_{n=1}^{\infin}\dfrac{(-1)^{n+1}4}{n}\sin nx


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