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Math Help - Fourier series

  1. #1
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    Fourier series

    hello

    I need to find the fourier sine series for the odd function:
    defined on  [0,\pi].

     f(x) = |x-\frac{\pi}{2}| - \frac{\pi}{2}.

     f(x) is periodic of period  2\pi

    I tried to define the function different(to get rid of the absolute value)

     f(x) = -x  if  0<=x<=\frac{\pi}{2}
     x-\pi if \frac{\pi}{2}<x<=\pi<br />


    I set up the integral for  b_{n} .

    in this case  L = \pi so:
     b_{n} = \frac{2}{L}\int_{0}^{\pi}f(x)sin(\frac{{\pi}nx}{L}  )dx =<br />
 \frac{2}{\pi}\int_{0}^\frac{\pi}{2}-xsin(nx)}dx

     + \frac{2}{\pi}\int_{\frac{\pi}{2}}^{\pi}(x-\pi)sin(nx)dx

    and the result I got was incorrect.
    I went over and over this problem but couldn't find the mistake.

    help would be appreciated.
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  2. #2
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    Quote Originally Posted by parallel
    hello

    I need to find the fourier sine series for the odd function:
    defined on  [0,\pi].

     f(x) = |x-\frac{\pi}{2}| - \frac{\pi}{2}.
    Okay you have, (shown below)
    f(x)=|x-\pi/2|-\pi/2 for -\pi<x\leq \pi
    This is a strictly monotonic function whose discountinuity points are not replacable, therefore it can be represented as a fourier series.
    ---
    The constant term,
    \pi a_0=\int_{-\pi}^{\pi} f(x) dx=\int_{-\pi}^{\pi/2}f(x)dx+\int_{\pi /2}^{\pi}f(x)dx
    Now these intervals the function behaves like,
    \int_{-\pi}^{\pi/2} -x dx +\int_{\pi/2}^{\pi} x-\pi dx=-\frac{1}{2}x^2 \big|^{\pi/2}_{-\pi}+\frac{1}{2}x^2-\pi x \big|^{\pi}_0
    Thus,
    -\frac{1}{8}\pi^2+\frac{1}{2}\pi^2+\frac{1}{2}\pi^2-\pi^2=-\frac{1}{8}\pi^2
    Thus,
    a_0=-\frac{1}{8}\pi^2.

    The same idea of spliting the function on the interval is used to find the sine and cosine coefficients. I just do not want to do that because I know how long these fourier computations can take.
    Attached Thumbnails Attached Thumbnails Fourier series-picture9.gif  
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Okay you have, (shown below)
    f(x)=|x-\pi/2|-\pi/2 for -\pi<x\leq \pi
    This is a strictly monotonic function whose discountinuity points are not replacable, therefore it can be represented as a fourier series.
    The problem is to find the Fourier sin series for the odd function defined by:

    <br />
f(x) = \left\{{|x-\frac{\pi}{2}| - \frac{\pi}{2}\ \ \ \ x \in [0,\pi) \atop <br />
f(x)=-f(-x)\ \ \ \ \ x \in (-\pi, 0]\right{}<br />

    RonL
    Last edited by CaptainBlack; August 27th 2006 at 10:26 PM.
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  4. #4
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    Quote Originally Posted by parallel
    hello

    I need to find the fourier sine series for the odd function:
    defined on  [0,\pi].

     f(x) = |x-\frac{\pi}{2}| - \frac{\pi}{2}.

     f(x) is periodic of period  2\pi

    I tried to define the function different(to get rid of the absolute value)

     f(x) = -x  if  0<=x<=\frac{\pi}{2}
     x-\pi if \frac{\pi}{2}<x<=\pi<br />


    I set up the integral for  b_{n} .

    in this case  L = \pi so:
     b_{n} = \frac{2}{L}\int_{0}^{\pi}f(x)sin(\frac{{\pi}nx}{L}  )dx =<br />
 \frac{2}{\pi}\int_{0}^\frac{\pi}{2}-xsin(nx)}dx

     + \frac{2}{\pi}\int_{\frac{\pi}{2}}^{\pi}(x-\pi)sin(nx)dx

    and the result I got was incorrect.
    I went over and over this problem but couldn't find the mistake.

    help would be appreciated.
    What do you get for b_n?

    Simply evaluating what you have here (which looks OK to me) I get:

    <br />
b_n=-\ \frac{4 \sin(\pi\ n/2)}{\pi\ n^2}<br />

    That this is correct is illustrated by the attachment showing the partial
    sums of the Fourier sin series with these coefficients for 2, 8, and 50
    terms (together with the function itself).

    RonL
    Attached Thumbnails Attached Thumbnails Fourier series-gash.jpg  
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  5. #5
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    thanks alot CaptainBlack,I had the most stupid mistake ever(those are hard to find),I integrated cosine as cosine.

    after correcting this mistake I got the same as you.

    ThePerfectHacker:

    thanks for your help too.
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  6. #6
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    Quote Originally Posted by parallel
    thanks alot CaptainBlack,I had the most stupid mistake ever(those are hard to find),I integrated cosine as cosine.

    after correcting this mistake I got the same as you.

    ThePerfectHacker:

    thanks for your help too.
    Hi Parallel,

    I hope you liked the plot of the partial sums of the series, I always find
    these gob-smacking. I know that the series does converge to the give
    function but am still amazed to see it.

    RonL
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  7. #7
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    Yea I did like it.

    You know, the first time I plotted the Fourier series graph(I'm kind of new to this),it really amazed me,so I know what you mean .
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