# Fourier Series

## Homework Statement

Find the Fourier series of the function
f(x) =√(x2) -pi/2<x<pi/2 , with period pi

## The Attempt at a Solution

I have tried attempting the question, but couldnt get the answer. uploaded my attempted qns with the picture attached

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RUber
Homework Helper
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 ##

RUber
Homework Helper
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:
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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RUber
Homework Helper
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.

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