f(x)= -1 if -pi
f(x) has period of 2pi
Find Fourier Series
You want to create a function: \[f(x) = a_0+a_1\cos(x)+b_1\sin(x)+a_2\cos(2x)+b_2\sin(2x)+\ldots\] the formula for the a's is: \[a_n = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\cos(nx)dx\]and the formula for the b's is: \[b_n = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\sin(nx)dx\] ima do the calculations on paper, i hate typing integrals in latex >.>
Thanks Joe, I am gonna print those out and try to figure it out
first one finds a_n , second finds b_n , what does third one do?
here is what the function looks like with the first couple of terms, you can see it starting to form the function (although it would take quite a bit of terms to make it look better) http://www.wolframalpha.com/input/?i=%284%2F%28pi%29%29sin%28x%29%2B%284%2F%283pi%29%29sin%283x%29%2B%284%2F%285pi%29%29sin%285x%29%2B%284%2F%287pi%29%29sin%287x%29%2B%284%2F%289pi%29%29sin%289x%29%2B%284%2F%2811pi%29%29sin%2811x%29%2B%284%2F%2813pi%29%29sin%2813x%29%284%2F%2815pi%29%29sin%2815x%29
third is the finish of the b_ns
oh, okay
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