SOLUTION: Find the Fourier series expansion of the periodic function f(x) of period p=2. F(x) = {{{x^2}}} 0 < x < 2

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Question 1046532: Find the Fourier series expansion of the periodic function f(x) of period p=2.
F(x) = x%5E2
0 < x < 2
Found 2 solutions by ikleyn, StatisticlessStephen:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
What do you know about it and what is the obstacle for you to do it on your own?

As a useful reading on a subject look in this Wikipedia article:

https://en.wikipedia.org/wiki/Fourier_series

https://en.wikipedia.org/wiki/Fourier_series


Answer by StatisticlessStephen(1) About Me  (Show Source):
You can put this solution on YOUR website!
So working through as much as I can on the wikipedia article and other sources...
I can see that the fourier series is defined as:

From what I understand, since X^2 is an even function... BN will always equal 0.
So, I can simplify to
SN%28x%29+=+A0%2F2+%2B+sum%28+AN+Cos%282+%2Api%2A+n+%2Ax+%2F+P%29+%29%2C+n=1%2C+N+%29+
Further reading highlights the definition of AN as...

Since I know the limits I can write it as:
AN+=+%282%2FP%29+int%28+s%28x%29+%2A+cos%282%2Api%2An%2Ax%2FP%29%2C+dx%2C+0%2C+2+%29
And with some integration by parts I can find that the definite integral of the above is:

If N = 0 then A0/2 = 0, so that means


If this is the answer, it seems incredibly complex... I see other solutions in a far more simple fashion, if anyone could shed some light on the matter it would be greatly appreciated.
P.S. Sorry for poor use of brackets, getting used to this formula writing business.
Edit: Cleared up the formatting a bit.