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 - Damon, 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
2a/(a^2-b^2) if (a-b) is odd
0 if (a-b) is even
2/(1-n^2) if n is even
0 if n is odd
Fourier Sine Series Q - Anonymous, 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 - Damon, Sunday, June 22, 2008 at 5:17pm
thanks Damon - Neutral, Sunday, June 22, 2008 at 4:55pm
Okay, I got it, thanks Damon.