Posted by Nicole on Sunday, December 7, 2008 at 1:01pm.
Hello,
I'm trying to find the Fourier Series of a function which is 1 from pi/2 to pi/2, and zero everywhere else
inside of pi to pi. I realize this is a square wave and I began the problem with a piecewise function
s(t) = 1 abs(t) < pi/2
0 abs(t) > pi/2
Then I had
a_k = 1/pi integral(from pi to pi) of [cos(kt)dt]
and
b_k = 1/pi integral(from pi to pi) of [sin(kt)dt]
but I don't know what to do from here.
Any help is greatly appreciated, thank you!

Reiny Can't Help Here  Reiny, Sunday, December 7, 2008 at 2:21pm
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.

Linear Algebra  Damon, Sunday, December 7, 2008 at 2:36pm
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

Linear Algebra  Damon, Sunday, December 7, 2008 at 6:19pm
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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