Repeat the example using the same Fourier series but at x=π/2.

Short Answer

Expert verified

At, x=π2:

f12=14+1π11cos3x+15cos5x-17cos7x+...+1π11sinx-22sin2x+13sin3x+15sin5x+26sin6x+17sin7x+...π4=11-13+15-17+...

Step by step solution

01

Given

The given function is fx=0,-π<x<0sinx,0<x<π

The given point is x=π2.

02

Definition of Fourier series

The Fourier series for the function f(x):

f(x)=a02+n-1(ancosnx+bnsinx)a0=1π-xxf(x)dxan=1π-xxf(x)cosnxdxbn=1π-xxf(x)sinnxdx

If f(x)is an even function:

bn=0f(x)=a02+n-1ancosnx

If f(x)is an odd function:

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

03

Write the Fourier series of the function and input the value

The Fourier series of function is shown below.

fx-14+1π11cos3x+15cos5x-17cos7x+...+1π11sinx-22sin2x+13sin3x+15sin5x+26sin6x+17sin7x+...π4=11-13+15-17+...

Now, at x-π2.

role="math" localid="1659238212014" f12-14+1π11cos3x+15cos5x-17cos7x+...+1π11sinx-22sin2x+13sin3x+15sin5x+26sin6x+17sin7x+...π4=11-13+15-17+...

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