Posted by Nicole on Sunday, December 7, 2008 at 7:04pm.
a_k should be (for k > 0)
1/pi integral(from -pi/2 to pi/2) of [cos(kt)dt] = 2 sin(pi k/2)/(pi k)
For even k this is zero. For odd k we can put k = 2n+1:
a_{2n+1} = 2(-1)^n/[pi(2n+1)]
For k = 0 the prefactor of 1/pi is replaced by 1/(2pi). So, you find
a_0 = 1/(2pi) integral(from -pi/2 to p2) dt = 1/2
The b_k are zero because the function is even:
b_k = 1/pi integral(from -pi/2 to pi/2) of [sin(kt)dt] = 0
So, the Fourier series of s(t) is given by:
s(t) = 1/2 + 2/pi sum over n from zero to infinity of
(-1)^n/(2n+1) cos[(2n+1)t]
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