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Fourier Series for Cosh

  1. Oct 5, 2006 #1
    Hi folks,

    I found the Fourier Series for Cosh(x) in the range -1 to +1 and discovered that:

    [tex]f(x)=sinh(1)(1+\sum_{n=1}^\infty\frac{(-1)^n}{1+n^2\pi^2}cos(n\pi x)[/tex])

    Now, I have to integrate the series twice to prove that:

    [tex]\sum_{n=1}^\infty\frac{(-1)^n}{n^2\pi^2(1+n^2\pi^2)}cos(n\pi x)=\frac{1}{2}(\frac{1}{sinh(1)}-\frac{5}{6})[/tex]

    I've been looking at this for ages! I can obtain the Left hand side pretty easy, but nothing like the Right. I'm really stuck here and I've really been trying. If anyone can help I'd be really grateful.

    Thank-you in advance.
     
    Last edited: Oct 6, 2006
  2. jcsd
  3. Oct 6, 2006 #2

    HallsofIvy

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    I'm not sure what you mean by "nothing like the left". It's evident, isn't it, that if you integrate the right hand side of the Fourier series term by term, you get the quantity on the left of the second equation?
     
  4. Oct 6, 2006 #3
    swap left for right! :)

    (I've edited the main post now)

    I cannot get:

    [tex]\frac{1}{2}(\frac{1}{sinh(1)}-\frac{5}{6})[/tex]
     
  5. Oct 6, 2006 #4

    matt grime

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    so, you've got


    f(x)/sinh(1) - 1 = some sum

    and you've integrated twice... so what have you now ended up with?
     
  6. Oct 6, 2006 #5
    Hi,

    Here goes:

    [tex]\frac{f(x)}{sinh(1)}-1=\sum_{n=1}^\infty\frac{(-1)^n}{n^2\pi^2(1+n^2\pi^2)}cos(n\pi x)[/tex]

    Integrating the LHS:

    [tex]\int_{-1}^1(\frac{f(x)}{sinh(1)}-1)dx=\int_{-1}^1(\frac{cosh(x)}{sinh(1)}-1)dx=\left[\frac{sinh(x)}{sinh(1)} \right]_{-1}^{1}-2=\frac{2sinh(1)}{sinh(1)}-2=2-2=0
    [/tex]

    This is only integrating once. Obviously I can't integrate zero.

    Maybe I don't put the limits in on the first integral?

    If so:

    [tex]\int(\frac{cosh(x)}{sinh(1)}-1)dx=\frac{sinh(x)}{sinh(1)}-x[/tex]

    and now integrate again with the limits.

    [tex]\int_{-1}^1 \frac{sinh(x)}{sinh(1)}-x=\left[\frac{cosh(x)}{sinh(1)}-\frac{x^2}{2}\right]_{-1}^{1}=\frac{1}{sinh(1)}(cosh(1)-cosh(-1))-(0.5-0.5)=0[/tex]

    :(


    Help svp.
     
    Last edited: Oct 6, 2006
  7. Oct 7, 2006 #6

    matt grime

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    I see you intergrated it once. But the question says integrate twice.
     
  8. Oct 7, 2006 #7
    I got zero when integrating once with the limits included, this will give me a constant if I integrate again.

    I also got zero when I integrated twice with the limits put in on the second integral, but not the first.
     
  9. Oct 7, 2006 #8

    matt grime

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    Why are you picking the lower limit of -1? What do you aim to determine by putting in some, or any lower limit?

    if f(x)=g(x), then that means there are functions F(x) and G(x) with F'=f, and G'=g, and F-G is a constant. Find out the constant, then repeat this, find the constant again and THEN evaluate at some point to get you final answer.

    So, let's do it: integrate once and we get

    sinh(x)/sinh(1) - x + K = the integral of that series.

    for some constant K. If we put 0 in the rhs is zero thus K=0

    Now, let's do it again.

    cosh(x)/sinh(1) - x^2/2 +K = series integrated twice.

    pick a sensible x to find constant K.
     
    Last edited: Oct 7, 2006
  10. Oct 7, 2006 #9

    matt grime

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    Actually, it's probably better to start this way:


    [tex]\cosh(x)=\sinh(1)+\sinh(1) \sum_{n=1}^\infty\frac{(-1)^n}{1+n^2\pi^2}\cos(n\pi x)[/tex]

    integrate

    [tex]\sinh(x)=K + x\sinh(1)+\sinh(1)\sum_{n=1}^\infty\frac{n\pi(-1)^{n+1}}{1+n^2\pi^2}\sin(n\pi x)[/tex]

    evaulate at x=0 to find K=0

    integrate again

    [tex]\cosh(x)=K + x^2\sinh(1)/2+\sinh(1)\sum_{n=1}^\infty\frac{n^2\pi^2(-1)^{n+1}}{1+n^2\pi^2}\cos(n\pi x)[/tex]

    and of course we have the orignal identity for cosh to use.
     
    Last edited: Oct 7, 2006
  11. Oct 7, 2006 #10

    mathwonk

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    what are you doing? cosh is defined as a sum of two exponentials, and a fourier series is by definition an expresion in terms of exponentials, so its fourier series should only have 2 terms. or maybe i don't know what a fourier series is?
     
  12. Oct 7, 2006 #11
    Matt, thanks for your guidance here. Let me try and complete this with you watching:

    Using what you obtained for Cosh, and comparing with the original identity we get:

    [tex]K+x^2sinh(1)/2+sinh(1) \sum \frac{(-1)^{n+1}}{n^2 \pi^2(1+n^2\pi^2)}cos(n\pi x) = sinh(1)(1+\sum \frac{(-1)^{n+1}}{n^2 \pi^2(1+n^2\pi^2)}cos(n\pi x))[/tex]

    Now, let x=0 and we get:

    [tex]K+sinh(1)\sum \frac{(-1)^{n+1}}{n^2 \pi^2(1+n^2\pi^2)}cos(n\pi x)=sinh(1)(1+\sum \frac{(-1)^{n+1}}{n^2 \pi^2(1+n^2\pi^2)}cos(n\pi x))[/tex]

    and so K=sinh(1)...I don't know, this feels circular...I'm probably not following your instructions correctly. I need your insight. :(
     
  13. Oct 7, 2006 #12

    matt grime

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    Fourier series use complex exponentials


    The Fourier series for cos has two terms, when written in exponential form, but cosh doesn't.
     
  14. Oct 7, 2006 #13

    matt grime

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    You're using the wrong series in at least on place. The one on the right is incorrect. You have to use the series for cosh from post one.
     
  15. Oct 7, 2006 #14
    Ok, I got an n^2 pi^2 wrong. sigh! Here we go again:

    Integrating twice and rearranging for K then setting x = 0 I obtain
    [tex]K=1-sinh(1)\sum \frac{(-1)^{n}}{n^2\pi^2(1+n^2\pi^2)}cos(n\pi x))[/tex]

    Plugging this into my expression for cosh(x) I get something that ultimately tidies up to:

    [tex]\sum \frac{(-1)^{n}}{n^2\pi^2(1+n^2\pi^2)})=\frac{1}{2}\left[\frac{1}{sinh(1)}-1-\sum \frac{(-1)^{n}}{(1+n^2\pi^2)} \right] [/tex]

    I feel really close...do you know any tricks that would make that last term become 1/12, then that would be the answer!! :)
     
    Last edited: Oct 7, 2006
  16. Oct 10, 2006 #15
    its ok, I think I've got this now...(after 3 days!)

    :)

    Thanks for everyone who helped direct me.
     
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