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Fourier series question, either i'm wrong or the answers are wrong

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    p2YPJ.png


    2. Relevant equations
    KIjJQ.png


    3. The attempt at a solution
    i'm pretty much learning fourier series from scratch today after only looking at it in lectures (and my exam is tomorrow lol @ me) and i'm sort of stuck.
    The solutions have something different to what i have, i think it might be an error in their integration by parts (which i also learnt today, despite how simple it seems).

    this is my working out: http://i.imgur.com/6OXlr.jpg
    and this is the worked solutions: http://i.imgur.com/1RAJL.png

    doesnt the integral of (1/n)*sin(nx) equal to (-1/n^2)*cos(nx)? like i have in my working?
    because the solutions are saying it integrates into (-1/n)*cos(nx).
    or is mine set out incorrectly after i've integrated by parts?

    edit: i've only shown both workings from where i had to find a subscript n, before that everything was fine and correct.
     
  2. jcsd
  3. Jun 19, 2011 #2

    lanedance

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    little hard to follow but i think maybe you're just missing the limits in you int by parts...
    [tex] u = x [/tex]
    [tex] du = 1 [/tex]
    [tex] dv = cos(nx) [/tex]
    [tex] v = \frac{1}{n}.sin(nx) [/tex]

    [tex] \int_{a}^{b} u.dv = (u.v)|_{a}^{b} - \int_{a}^{b}du.v[/tex]
    [tex]\int_0^{\pi} x.cos(nx).dx [/tex]
    [tex]= (x.\frac{1}{n}.sin(nx))|_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{n}.sin(nx).dx [/tex]
    [tex]= 0 - \frac{1}{n} \int_{0}^{\pi} sin(nx).dx [/tex]
    [tex]= \frac{1}{n^2} (cos(nx).dx)|_{0}^{\pi}[/tex]
     
    Last edited: Jun 20, 2011
  4. Jun 20, 2011 #3

    lanedance

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    it is true that
    [tex] \frac{1}{n}\int sin(nx).dx = -\frac{1}{n^2}cos(nx)
    [/tex]

    just differentiate to check
     
  5. Jun 20, 2011 #4



    i dont understand what you're doing?

    what have u used as u and v?
    shouldnt (x.n.sin(nx) be (x/n).sin(nx) after the first = sign?
     
  6. Jun 20, 2011 #5

    lanedance

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    sorry, did that a too quick, corrected and lines up with yours now

    notice the sine terms cancel as [itex]sin(0) = sin(\pi) = 0[/itex]
     
    Last edited: Jun 20, 2011
  7. Jun 20, 2011 #6
    ok thanks.

    also where do the cos's disappear to?
    i think something about if that part of the function is odd then you make it -1 or something? and/or if it's even make it 1, i dont know :S
     
  8. Jun 20, 2011 #7

    lanedance

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    which cos's? in the first step the sin appear instead of cos's due to the integration by parts

    now you should be considering the whole integral from -pi to pi, braking it up into two pieces note that
    cos(-x) = cos(x)
    and
    sin(-x) = -sin(x)

    you should be able to show for any function
    - if the function is odd, it only has sin components (sin is an odd function)
    - if the function is even, it only has cos components (cos is an even function)
     
    Last edited: Jun 20, 2011
  9. Jun 20, 2011 #8
    yeah i underrstand, thanks a lot!

    also regarding the values for a0, an, and bn (http://i.imgur.com/KIjJQ.png) they are all integrals from -L -> L, but in all the examples i've done or seen (about 3), they've taken the integral from 0 -> L and then multiplied the entire thing by 2.

    Now i understand they're halving the region of integration and then doubling the overall answer, but can this always be done when a question asks for a fourier series of a piecemeal function, whether the fn is even or odd? (with a question of the same style as the one in the OP)
    And if so, any reason why my formula sheet gives us a0 etc as integrals from -L to L when we can just make it twice the integral from 0-L?
     
  10. Jun 20, 2011 #9

    lanedance

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    write out the whole integral, break it into 2 pieces at 0, then use the properties of sin or cos as described last time and it should become obvious. this will only work for odd or even functions
     
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