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Homework Help: Help with Fourier series

  1. Dec 16, 2005 #1
    Hi, can someone help me out with the following question?

    Q. Show that the Fourier series for the function y(x) = |x| in the range -pi <= x < pi is

    [tex]
    y\left( x \right) = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\cos \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^2 }}}
    [/tex]

    By integrating term by term from 0 to x, find the function g(x) whose Fourier series is

    [tex]
    \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}
    [/tex]

    Deduce the sum S of the series: [tex]1 - \frac{1}{{3^3 }} + \frac{1}{{5^3 }} - \frac{1}{{7^3 }} + ...[/tex]

    I took the Fourier series for y(x) and I integrated it as follows.

    [tex]
    \int\limits_0^x {\left( {\frac{\pi }{2}} \right)} dt - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\left( {\int\limits_0^x {\frac{{\cos \left( {2m + 1} \right)t}}{{\left( {2m + 1} \right)^2 }}dt} } \right)}
    [/tex]

    [tex]
    = \frac{{\pi x}}{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}
    [/tex]

    I don't know what to do with it to find the function whose Fourier series is

    [tex]
    \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}
    [/tex]

    Can someone help me get started? I'm not sure what to do.

    Edit: Do I just equate the integral I evaluated to the integral of |x|(considering x positive and negative separately) and solve the equation for the sine series?
     
    Last edited: Dec 16, 2005
  2. jcsd
  3. Dec 16, 2005 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Yes, you integrate y(x) (or y(t)) over the interval and then you can find the function g(x).
     
  4. Dec 16, 2005 #3
    Thanks for the help. I'm just having some problems working out the sum.

    From the previous working I have

    [tex]
    \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}} = \frac{x}{2}\left( {\pi - x} \right)
    [/tex] for x positive or zero.

    I have another expression for x negative but it isn't needed in finding the sum so I'll leave it out.

    The sum I want to find is: [tex]S = 1 - \frac{1}{{3^3 }} + \frac{1}{{5^3 }} - \frac{1}{{7^3 }} + ...[/tex]

    If I set x = pi/2 then sin(2m+1)x = (-1)^m for all integers m >=0. So using the equation from previous working:

    [tex]
    \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)\frac{\pi }{2}}}{{\left( {2m + 1} \right)^3 }}}
    [/tex]

    [tex]
    = \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\left( { - 1} \right)^m \frac{1}{{\left( {2m + 1} \right)^3 }}}
    [/tex]

    [tex]
    = \frac{1}{2}\left( {\frac{\pi }{2}} \right)\left( {\pi - \frac{\pi }{2}} \right)
    [/tex]

    [tex]
    \Rightarrow S = \sum\limits_{m = 0}^\infty {\left( { - 1} \right)^m \frac{1}{{\left( {2m + 1} \right)^3 }}} = \frac{\pi }{2}
    [/tex]

    The answer is S = ((pi)^3)/32. I don't know what I'm leaving out. Any further help would be good thanks.
     
  5. Dec 16, 2005 #4

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Check your math in the last step. (ie, the cross multiplication)
     
  6. Dec 16, 2005 #5
    Thanks for the help. I mistakenly got rid of a factor of pi and ignored some constants. It works out now.
     
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