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Posted by on Sunday, June 22, 2008 at 3:28pm.

I have the function f(x) = cos(x) on the interval from 0 to pi and I need to comput the Fourier sine series.

I have the integral of cos(x) multiplied by sin(nx), I can't figure out a way to integrate them! The "n" gets in the way, what do I do?

  • Fourier Sine Series Q - , Sunday, June 22, 2008 at 4:15pm

    first sketch a graph of cos x and sin x, sin 2x, sin 3x, sin 4x
    in the interval from 0 to pi
    You will notice that for n = 1 for example, the plus contribution between 0 and pi/2 cancels the negative contribution between pi/2 and pi
    in fact for cos 1x * sin n x dx, only even values of n will contribute for the integral from 0 to pi.
    The general rule for this definite integral is:
    integral from 0 to pi of
    sin ax cos bx dx
    is:
    2a/(a^2-b^2) if (a-b) is odd
    0 if (a-b) is even
    so here:
    2/(1-n^2) if n is even
    0 if n is odd

  • Fourier Sine Series Q - , Sunday, June 22, 2008 at 4:38pm

    So basically, that is the coefficient "bn" that I place in the summation from n=1 to infinity of sin(nx)?

    If that is correct, thanks, I was going into a complicated page of integration . . .

  • Fourier Sine Series Q - , Sunday, June 22, 2008 at 5:17pm

    yes :)

  • thanks Damon - , Sunday, June 22, 2008 at 4:55pm

    Okay, I got it, thanks Damon.

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