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Homework Help: Sum of infinite Fourier series

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that

    [tex]\sum_{r=0}^\infty\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}[/tex]


    2. Relevant equations
    The equation of the function is

    [tex]F(t)&=&\dfrac{\pi}{4}-\dfrac{2}{\pi}\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)-\left(\sin t-\dfrac{\sin2t}{2}+\dfrac{\sin3t}{3}-\cdots\right)[/tex]


    3. The attempt at a solution
    We are given the condition that t=0, so the cos terms are all 1 giving

    [tex]\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)=\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}\cdots=\frac{\pi^2}{8}[/tex]
     
    Last edited: Jul 20, 2010
  2. jcsd
  3. Jul 20, 2010 #2
    Did you write the question incorrectly? The expression that you're summing doesn't have the indexing variable in it anywhere, so it diverges toward infinity:
    [tex]\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{(2r+1)^2}\sum_{i=0}^\infty1=\frac{1}{(2r+1)^2}\infty = \infty[/tex]
     
  4. Jul 20, 2010 #3

    hunt_mat

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    I think it's just a typo.
     
  5. Jul 20, 2010 #4
    Hi

    The i should have been r.

    I know from elsewhere that the sum is [tex]\frac{\pi^{2}}{8}[/tex], but I haven't really shown that, any hints?

    Thanks

    James
     
  6. Jul 20, 2010 #5

    hunt_mat

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    You have to know the original function which the series you have written down represents. Evaluate this function at t=0 and do a little algebra.

    Mat
     
  7. Jul 20, 2010 #6
    The original function is piecewise

    [tex]f(t)=\begin{cases}
    -t & \left(-\pi<t\leq0\right)\\
    0 & \left(0<t\leq\pi\right)\end{cases}[/tex]

    With a period of [tex]2\pi[/tex]

    James
     
  8. Jul 20, 2010 #7

    hunt_mat

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    You have the value of it's Fourier series at t=0, now calculate what f(0) is and equate the two values.
     
  9. Jul 20, 2010 #8
    Hi

    Of course, duh.

    Thanks for your help

    James
     
  10. Jul 22, 2010 #9
    Hi

    I am next asked to choose an appropriate value for t and find the value of convergence of the following,

    [tex]
    \sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}
    [/tex]

    I know this converges to [tex]\frac{\pi}{4}[/tex], but I am not sure where to start, the function is still the same as at the beginning of the thread

    Any ideas where to start?

    Thanks, James
     
  11. Jul 22, 2010 #10

    Gib Z

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    Integrate both the original function and its Fourier Series term by term. Then try to substitute some value of x in to get that result.
     
  12. Jul 23, 2010 #11

    hunt_mat

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    Don't integrate but differentiate.
     
  13. Jul 23, 2010 #12

    Gib Z

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    Whoops, my mistake.
     
  14. Jul 25, 2010 #13
    I differentiate the approximation and set [tex]t=-\frac{\pi}{2}[/tex] giving

    [tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{1}{2r+1}\sin((2r+1)t)[/tex] as sin alternates sign we get

    [tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}[/tex]

    To which I find

    [tex]\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{\pi^2}{4}[/tex] and not [tex]\frac{\pi}{4}[/tex]

    Any ideas where I'm going wrong?

    Thanks
     
  15. Jul 25, 2010 #14
    [tex]
    \sum_{r = 0}^{\infty}{\frac{1}{(2 r + 1)^{2}}} = \sum_{r = 0}^{\infty}{\frac{1}{r^{2}}} - \sum_{r = 1}^{\infty}{\frac{1}{(2 r)^{2}}} = \left(1 - \frac{1}{4}\right) \, \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}
    [/tex]

    The sum:

    [tex]
    \zeta(2) \equiv \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}
    [/tex]

    can be found by expanding the function [itex]f(x) = x^{2}[/itex] in Fourier series in the region [itex][-\pi, \pi][/itex] and taking its value for [itex]x = \pi[/itex].
     
  16. Jul 25, 2010 #15
    Hi

    Sorry for my ignorrance, but I can't see how this helps with the question (see post#9)

    Jamees
     
  17. Jul 25, 2010 #16

    Mute

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    The problem is that you differentiated f(t), so your series converges to f'(t), which is either equal to -1 or 0, but you set to it f'(t=-pi/2) = pi/2 and equated the pi/2 to the series instead of the actual value of f(t=-pi/2) = -1.
     
  18. Jul 25, 2010 #17

    vela

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    Why not just set t=pi/2 in the original function?
     
  19. Jul 26, 2010 #18
    Hi

    If I put [tex]t=\pi/2[/tex] the [tex]f(\pi/2)=0[/tex] and we get

    [tex]\frac{\pi}{4}=\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}[/tex]

    Thanks everyone for your help.

    James
     
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