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Use fourier series to find sum of infinite series

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data

    wkbow.png

    Find the value of An and given that f(x) = 1 for 0 < x < L/2, find the sum of the infinite series.

    2. Relevant equations



    3. The attempt at a solution

    The basis is chosen to be ##c_n = \sqrt{\frac{2}{L}}cos (\frac{n\pi }{L}x)## for cosine, and ##s_n = \sqrt{\frac{2}{L}}sin (\frac{n\pi}{L}x)## for sine.

    [tex]|f\rangle = \sum_0^{\infty} \langle c_n|f\rangle |c_n\rangle + \sum_1^{\infty}\langle s_n|f\rangle|s_n\rangle[/tex]

    Compare this with:

    [tex]f = \sum_0^{\infty}c_n' cos(\frac{n\pi }{L}x) + \sum_1^{\infty}A_n sin(\frac{n\pi}{L}x)[/tex]

    We are given that f(x) = 1 for 0 < x < L/2.

    Therefore,
    [tex]A_n = \sqrt{\frac{2}{L}}\langle s_n|f\rangle = \frac{2}{L}\int_0^{L/2}f_{(x)} sin(\frac{2\pi n}{L}x) dx[/tex]
    [tex]A_n = \frac{1}{n\pi}\left[1 - cos(n\pi)\right][/tex]
    [tex]A_n = \frac{1}{n\pi}\left[1 - (-1)^n\right][/tex]

    For odd n, ##A_n = \frac{2}{(2n-1)\pi}##

    Either the first series or the third series works for ##A_n##:

    For 0 < x < L/2, f(x) = 1:

    [tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi}sin\left(\frac{2\pi n}{L}x\right)[/tex]

    Choosing x = L/4:
    [tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi}sin\left(\frac{n\pi}{2}\right)[/tex]
    [tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi} (-1)^{n+1}[/tex]
    [tex]\frac{\pi}{2} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)} = 1 - \frac{1}{3} + \frac{1}{5} - ...[/tex]
     
    Last edited: May 4, 2014
  2. jcsd
  3. May 4, 2014 #2

    BvU

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    The exercise asks for an expression for ##k_n##

    From your ##s_n = \sqrt{\frac{2}{L}}sin (\frac{n\pi}{L}x)## I get the impression that n = 1 is also allowed. But then the period definitely isn't L !

    Furthermore, for periodic functions the integral isn't from ##-\infty## to ##\infty## or from ##0## to ##\infty## but from ## 0 ## to ## L##
     
  4. May 4, 2014 #3
    Oops, the correct value for kn should be ##k_n = \frac{2\pi n}{L}##.

    And I've updated my answers above. Now I'm missing a factor of 1/2 on the LHS.
     
    Last edited: May 4, 2014
  5. May 4, 2014 #4

    BvU

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    Missing 1/2 on the LHS of what, exactly?
    Remind me of the relevant eqn for ##A_n##...
     
  6. May 4, 2014 #5
    [tex]f_{(x)} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)\pi}sin\left(\frac{2\pi n}{L}x\right)[/tex]

    If I choose x = L/4 or L/8, LHS should be = 1, given f(x) = 1 for 0< x < L/2. But the RHS is different.
     
    Last edited: May 4, 2014
  7. May 6, 2014 #6
    So, if ##A_n = \frac{1}{n\pi}\left[1 - (-1)^n\right]##, then for odd n: ##A_n = \frac{1}{n\pi}\left[1 - (-1)^n\right]##.

    Plugging this into the fourier series:

    [tex]1 = \sum_{n=1}^{\infty} \frac{2}{(2n-1)\pi}sin\left(\frac{n\pi}{2}\right)[/tex]

    Still the LHS isn't ##\frac{\pi}{4}##..
     
  8. May 6, 2014 #7
    What do you get when you do the following integration ##A_n=\frac{2}{L/2}\int_0^{L/2}sin(\frac{n\pi x}{L/2})dx##? You just made a silly mistake in your calculation. Show your calculations step by step.
     
  9. May 6, 2014 #8
    I have no idea about the fourier series but the problem doesn't state any restriction about using it to evaluate the given sum. I have two methods.

    Method 1:
    The given sum is:
    $$\begin{aligned}
    \\
    \sum_{r=0}^{\infty} \frac{(-1)^r}{2r+1} & =\sum_{r=0}^{\infty} (-1)^r\int_0^1 x^{2r}\,dx\\
    & =\int_0^1 \frac{dx}{1+x^2}\\ & =\left(\arctan(x)\right|_0^{1}\\ &=\frac{\pi}{4} \end{aligned} $$

    Method 2:
    Use the series of ##\ln(1+x)## at ##x=i## where ##i=\sqrt{-1}## i.e
    $$\ln(1+i)=\ln(\sqrt{2}e^{i \pi/4})=i-\frac{1}{2}-\frac{i}{3}+\frac{1}{4}+\frac{i}{5}-\cdots$$
    Compare the imaginary parts on both the sides to get the sum.
     
    Last edited: May 6, 2014
  10. May 6, 2014 #9
    I am guessing the OP is taking a class in Fourier Analysis, which is why he's posting this problem. Yes, there are other methods, but I think he needs to understand what he is doing wrong in his calculation of the Fourier series.
     
  11. May 6, 2014 #10
    Ok, I think I got it. Fe or a sine series, we extend the function to be f(x) = -1 for L/2 < x < L. So that the integrand is an odd*odd = even function.

    Thus,

    [tex]A_n = \frac{2}{L} \int_0^L f_{(x)} sin \left(\frac{2\pi n}{L}x\right) dx = 2\left( \frac{2}{L} \right) \int_0^{\frac{L}{2}} sin \left(\frac{2\pi n}{L}x\right) dx[/tex]

    I have found my missing factor of 1/2!
     
    Last edited: May 6, 2014
  12. May 6, 2014 #11
    Honestly, that is brilliant! I couldn't have thought of that.
     
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