Posted by Nicole on Sunday, December 7, 2008 at 1:01pm.
The last time I dealt with that topic was over 45 years ago, and frankly I don't feel confident enough to answer.
Perhaps some of the other math experts might want to take a crack at it if they are on.
This is not a Fourier series really but a Fourier integral. The series would apply if the function were periodic.
However this function has to be zero from - oo to - pi/2
Then it jumps up one at -pi/2
Then it jumps down one at +pi/2
That is a unit step up at -pi/2
minus
A unit step up at + pi/2
I think you were doing fine now that I look at what you did.
a_k = 1/pi integral(from -pi to pi) of [cos(kt)dt]
= 1/(k pi) [sin k pi - sin(-k pi)
= (2/k pi)sin k pi
which is the answer
and
b_k = 1/pi integral(from -pi to pi) of [sin(kt)dt]
= 1/(k pi) [-cos k pi + cos -k pi}
= 0 because cosine is even
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