Question #17198

find the fourier sine series for the function f(x)=e^ax for 0<x<pi where a is a constant

Expert's answer

f(x)=a0/2+Sum(Akcos(nx)+Bksin(nx)

a0=1/Pi*Integrate(-Pi, Pi)(e^ax)dx)=2sh(A*Pi)/A

An==1/pi*Integrate(-Pi, Pi)e^ax*cosnxdx=2(ncoshA*pi*sin(n*Pi)+Acos(n*pi)sh(A*pi))/a^2+n^2

An==1/pi*Integrate(-Pi, Pi)e^ax*sinnxdx=2ncosA*pi*sh(n*Pi)+Acosh(n*pi)sin(A*pi))/a^2+n^2

Puting this constants to the series gives the answer

a0=1/Pi*Integrate(-Pi, Pi)(e^ax)dx)=2sh(A*Pi)/A

An==1/pi*Integrate(-Pi, Pi)e^ax*cosnxdx=2(ncoshA*pi*sin(n*Pi)+Acos(n*pi)sh(A*pi))/a^2+n^2

An==1/pi*Integrate(-Pi, Pi)e^ax*sinnxdx=2ncosA*pi*sh(n*Pi)+Acosh(n*pi)sin(A*pi))/a^2+n^2

Puting this constants to the series gives the answer

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