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Fourier Series

  1. Jan 20, 2017 #1
    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 WhatsApp Image 2017-01-20 at 9.34.39 PM.jpeg
     
  2. jcsd
  3. Jan 20, 2017 #2

    RUber

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    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 ##
     
  4. Jan 20, 2017 #3

    RUber

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    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
  5. Jan 20, 2017 #4
    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. :H
     

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  6. Jan 20, 2017 #5

    RUber

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    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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