Advanced Search

Question: Fourier expansion of trigonometric function

Question:Fourier expansion of trigonometric function

larsmars 8 Maple
student
December 24 2007
0
Hi guys, we've got a Fourier transform problem here. This is how we calculate Fourier coefficient: 4/pi*int(cos(3*x)*cos(n*x), x=0..Pi/2) (we try to expand cos(3*x)) so here we got the answer 12*cos((1/2)*Pi*n)/(Pi*(-9+n^2)) correcr answer, you see, is B3=1; but when we write this: 4/pi*int(cos(3*x)*cos(n*x), x=0..(1/2)*Pi)assuming n=3 the result shown is zero. we don't know how it works. because this value is evaluating correctly: 4/Pi*int(cos(3*x)*cos(3*x), x=0..(1/2)*Pi) the answer is right - 1. what could we do to get correct coefficients? please, help us!
Please Wait...