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Series expansion of Coth(pi)

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data

    show coth([itex]\pi[/itex])=1/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]n=1 (1/(1+n^2)

    2. Relevant equations
    The fourier expansion of e^x is
    Sinh([itex]\pi[/itex])/[itex]\pi[/itex](1+[itex]\sum[/itex] 2(-1)^m/(1+n^2) (cos(mx)-n sin(nx)


    3. The attempt at a solution
    I subbed in
    Coth(Pi)=1+e^-pi/sinh(pi) =1+1/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]n=1 (1/(1+n^2)

    But this is wrong because of the one out the front - I know that needs to stay there, which means I somehow need to stick a -sinh(pi) into the front of my fourier series of e^pi I would say that it would be my term for n= zero, but I already took care of that one, which was sinh(pi)/pi...
    Please help - I've spent about fifteen pages on the algebra for this and I'm getting really sick of the stupid question...
     
  2. jcsd
  3. Aug 29, 2012 #2

    vela

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    You're missing closing parentheses and have two index variables, m and n, so I'm not sure what you meant to write exactly. Did you mean
    $$\frac{\sinh \pi}{\pi}\left(1+\sum_{n=1}^\infty \frac{2(-1)^n}{1+n^2}(\cos nx - n\sin nx)\right)?$$

    How'd you come up with ##\coth \pi = 1+\frac{e^{-\pi}}{\sinh \pi}##?
     
    Last edited: Aug 29, 2012
  4. Aug 29, 2012 #3
    Coth([itex]\pi[/itex]) =(e[itex]\pi[/itex]+e-[itex]\pi[/itex])/(e[itex]\pi[/itex]-e-[itex]\pi[/itex])

    which equals
    1+2e-[itex]\pi[/itex]/(e[itex]\pi[/itex]-e-[itex]\pi[/itex])
    which equals
    1+2e^pi/2sinh(pi)
     
  5. Aug 29, 2012 #4

    gabbagabbahey

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    As vela pointed out, this makes no sense. What is the actual Fourier series expansion of [itex]e^x[/itex]?
     
    Last edited: Aug 29, 2012
  6. Aug 29, 2012 #5
    It's the one that vela used - I usually use n's but the lecturer uses m's so it's a bit messed up, and they are both standing for the same thing. And the e^pi is a typo - I have it right in my working...
    I've found another page that uses this as the fourier expansion of e^x, so I'm fairly sure that it's right (after making all the indeces the same)
     
  7. Aug 29, 2012 #6

    vela

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    Hint: What does your Fourier series converge to when you set ##x=\pi##? Keep in mind there's a discontinuity there.
     
  8. Aug 29, 2012 #7
    It'll be... halfway between the limit of the fourier series at Pi and -pi?
    Which means....
    1/2(e^pi+e^-pi)=Sn(+/-pi)?
     
  9. Aug 29, 2012 #8

    vela

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    I don't know what you mean by "Sn(+/-pi)", but yeah, it'll be the average.
     
  10. Aug 29, 2012 #9
    And that lets me calculate e^pi+e^-pi from my fourier series at Pi; but how do I get e^pi-e^-pi for the denominator then?
     
  11. Aug 29, 2012 #10

    vela

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    Here's another hint: What's ##\cosh \pi## equal to?
     
  12. Aug 29, 2012 #11
    wait, no, I got it. The denominator is 2 sinh(pi) and the numerator is twice the fourier series at Pi - which is sinH(pi) multiplied by the series I want.
    Thanks!
    :)
     
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