Posted by ethan on Tuesday, April 27, 2010 at 7:26pm.
Find the fourier series of
F(x) = 1, L<=x<0
= 0, 0<=<L
f(x+2L) = f(x)

calculus  Count Iblis, Tuesday, April 27, 2010 at 7:59pm
You can take the functions:
e_n(x) = exp(i pi n x/L)
to be your the basis functions. If you define the inner product as:
<f,g> = 1/(2 L) Integral from L to L
of f(x)g*(x) dx
Then the e_n are orthonormal. So, you can expand a function f(x) as:
f(x) = sum over n of <f,e_n> e_n(x)
The coefficient of e_n is thus the integral from L to 0 of
exp(pi i n x/L) dx
If n is not equal to zero, this is:
1/(2 pi i n) [1  (1)^n]
So, for nonzero n, only the coefficients for odd n are nonzero. The coefficient for n = 0 is easily found to be 1/2 . If you now combine the contribution from negative and positive n, you find that the Fourier series is given by:
1/2  2/pi sum from k = 0 to infinity of
2/(2k+1) Sin[(2k+1)x/L]
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