Sketch several periods of the corresponding periodic function of period2π. Expand the periodic function in a sine-cosine Fourier series.

f(x)={1,-π<x<0,...0,0<x<π,....

Short Answer

Expert verified

The answer of the given function is f(x)=12-2πsinx1+sin3x3+sin5x5---.

Step by step solution

01

Given

The given function is f(x)={1,-π<x<0,...0,0<x<π,.....

The sketch and the result of the given function are shown below.

f(x)=12-2πsinx1+sin3x3+sin5x5---

02

Concept of Fourier series

The Fourier series for the function f(x) :

f(x)=a02+n=1(ancosnx+bnsinnx)a0=1π-ππf(x)dxan=1π-ππf(x)cosnxdxbn=1π-ππf(x)sinnxd

If f(x) is an even function:

bn=0af(x)=a02+n=1ancosnx

If f(x) is an odd function:

a0=an=0f(x)=n=1bnsinnx

03

Calculate with the help of Fourier series

Given function is f(x)=12-2πsinx1+sin3x3+sin5x5---.

Use the solution of the example one as follows:

g(x)=f(x+π)g(x-π)=f(x)g(x-π)=12+2πk=2n+1sin(kx-kπ)k=12+2πk=2n+1sin(kx)cos(kπ)-sin(kπ)cos(kx)k

Calculate further as shown below.

=12+2πk=2n+1sin(kx)cos(kπ)k,[sin(kπ)=0 , where k=2n+1]

=12-2πk=2n+1sin(kx)k,[cos(kπ)=-1 , where k=2n+1]

=12-2πsinx1+sin3x3+sin5x5---=f(x)

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