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sigmund
Apr1-05, 03:12 PM
I have tried to find the Fourier series for a function u(x):


u(x)=\sin((1+3\cos(t))t)


The function is odd, hence the Fourier coefficients a_n equal zero and the b_ns are given as


b_n=\frac{4}{T}\int_0^{T/2}u(t)\sin(n\omega t)\,\text{d}t


where T=2\pi and \omega=2\pi/T=1.

Then, my problem is that u(x)*sin(n*t) is not easily integrated. I would then like to ask if there could be any way getting around this integration-problem, perhaps if I wrote u(x) as either the real og imaginary part of a complex function? I would appreciate any help.

Data
Apr1-05, 03:43 PM
Well, it is sort of close to a Bessel integral. But not really. It doesn't look promising at all. Neither Maple or Mathematica will do it, but that doesn't mean anything. I've had precisely one hour of sleep in the last day and a half, so I may be missing something~

dextercioby
Apr1-05, 03:53 PM
So your function is

u(x)= \sin \left[\left(1+3\cos t\right)t\right] ...Hmm.

U could you leave it like that,namely the coeff. "b_{n}",because you can't evaluate that integral.

Daniel.