Fourier Series

1. Jan 20, 2017

Evilavatar2

1. The problem statement, all variables and given/known data
Find the Fourier series of the function
f(x) =√(x2) -pi/2<x<pi/2 , with period pi
2. Relevant equations

3. The attempt at a solution
I have tried attempting the question, but couldnt get the answer. uploaded my attempted qns with the picture attached

2. Jan 20, 2017

RUber

Your work looks solid at first glance...maybe an algebraic error, I'll go back and look at the details soon.
Remember, your final form should be something like:
$f(x) = a_0 + \sum_{n=1}^\infty a_n \cos 2n x$

3. Jan 20, 2017

RUber

You were right to use the half-period formula and notice it was an even function. You did not write it like one f(x) = |x|, not f(x) = x.
For $a_0$, it looks like you did 1/2pi, instead of 1/pi for the full period of pi. On the half-period formula, you want to double the result to get the full period, so you should end up with 2/pi as your coefficient on that integral
$a_0 = \frac2\pi \int_0^{\pi/2} f(x) \, dx .$
For your $a_n$ terms, your coefficient was incorrect as well, since it should be (L was correctly identified as pi/2).
$a_n = \frac2L \int_0^{L} f(x)\cos\left(\frac{nx}{L} \right)\, dx .$
Other than those coefficients, your integration by parts looks to be done correctly from what I can make out.
**edit** you missed a negative sign in the sine integral. Integral of sin(x) dx = - cos x. **end of edit**
Let me know if you are still having trouble.

Last edited: Jan 20, 2017
4. Jan 20, 2017

Evilavatar2

Thanks alot for helping. I have decided to redo the whole question and corrected the ao. But i can't find the mistake that you pointed out for the integral sin(x)dx= -cos x. Also my answer is still incorrect.

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5. Jan 20, 2017

RUber

I must have lost track of the negative sign somewhere--your new solutions seem to be in the right neighborhood.
Look at the coefficient on your $a_n$ integral. It is the same as the one you used on your $a_0$ integral. It should be twice as big.