- #1
jianxu
- 94
- 0
Homework Statement
f(x) = |cos x| if -[tex]\pi[/tex] [tex]\leq[/tex] x [tex]\leq\[/tex] [tex]\pi[/tex]
The Fourier series answer we should end up with is the following:
2/[tex]\pi[/tex] - 4/[tex]\pi[/tex][tex]\sum[/tex][tex]\frac{-1^{k}}{(2k)^{2}-1}[/tex]cos(2kx)
where for the summation, k = 1 and goes to infinity.
What I need to do is to actually go an solve for the Fourier series and show that we get the same series when we derive it given f(x)
Homework Equations
f(x) = [tex]a_{0}[/tex] + [tex]\sum[/tex][tex]a_{k}[/tex]coskx + [tex]b_{k}[/tex]sinkx where k =1 and goes to infinity
[tex]a_{0}[/tex] = 1/2[tex]\pi[/tex][tex]\int[/tex]f(x)dx
[tex]a_{k}[/tex] = 1/[tex]\pi[/tex][tex]\int[/tex]f(x)coskxdx
[tex]b_{k}[/tex] = 1/[tex]\pi[/tex][tex]\int[/tex]f(x)sinkxdx
where we integrate all the above integrals from -[tex]\pi[/tex] to [tex]\pi[/tex]
Lastly, the properties of even functions are also considered.
The Attempt at a Solution
So, to start out, since [tex]b_{k}[/tex] = 1/[tex]\pi[/tex][tex]\int[/tex]f(x)sinkxdx, cosx is even and the sinkx is odd,
[tex]b_{k}[/tex] = 0
for
[tex]a_{0}[/tex] = 1/2[tex]\pi[/tex][tex]\int[/tex]f(x)dx
due to the properties of an even function I have:
= 1/[tex]\pi[/tex][tex]\int[/tex]f(x)dx integrated from 0 to [tex]\pi[/tex]
= 1/[tex]\pi[/tex]([tex]\int[/tex]cosxdx + [tex]\int[/tex]-cosxdx)
where the first integral is from 0 to [tex]\pi[/tex]/2} and the second integral is from [tex]\pi[/tex]/2 to [tex]\pi[/tex],
solving this gives me
[tex]a_{0}[/tex] = 1/[tex]\pi[/tex](1 - (-1)) = 2/[tex]\pi[/tex]
For [tex]a_{k}[/tex]:
using even function properties again I have,
[tex]a_{k}[/tex] = 2/[tex]\pi[/tex][tex]\int[/tex]f(x)coskxdx
= 2/[tex]\pi[/tex][tex]\int[/tex]cosx*coskx dx
where integral goes from 0 to [tex]\pi[/tex].
Maybe I'm just rusty with integration, but I just want to make sure I'm heading into the right direction and how do I integrate that? ( meaning [tex]\int[/tex]cosx*coskx dx)
Thanks