Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Fourier series summation in David Griffiths' textbook

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    This isn't really a homework question, but something I've been wanting to know out of curiosity in David Griffiths' Introduction to Electrodynamics.

    On pages 131 and 132, there is a Fourier series,

    [tex]V(x,y) = \frac{4V_0}{\pi}\sum_{n=1,3,5...}\frac{1}{n}e^{\frac{-n \pi x}{a}}\sin{\frac{n \pi y}{a}}[/tex]

    The author then says that the series can be rewritten as

    [tex]V(x,y) = \frac{2V_0}{\pi} \arctan{\frac{\sin{\frac{\pi y}{a}}}{\sinh{\frac{\pi x}{a}}}}[/tex]

    The author says "the infinite series ... can be summed explicitly (try your hand at it, if you like) ..." so I got curious and decided to take a shot at it.

    I have been trying to figure out how to get this result all day, but after a few sheets of paper, I am still lost... Any help would be much appreciated.

    P.S. I have not taken a course on complex analysis.

    2. Relevant equations

    [tex]\sin{u} = \frac{e^{iu}-e^{-iu}}{2i}[/tex]
    [tex]\sinh{u} = \frac{e^{u}-e^{-u}}{2}[/tex]
    [tex]\arctan{u} = \frac{i}{2}\ln{\frac{1-iu}{1+iu}}[/tex]
    [tex]\arctan{u} = \sum_{n=0}^{\inf} \frac{(-1)^n}{2n+1} u^{2n+1}[/tex]

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 17, 2010 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Most of these summations are done using clever applications of the Taylor expansion for the logarithm:

    [tex]\ln (1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n[/tex]

    Notice that your Fourier series has a 1/n in it. However, in your case, n is odd! So you'll have to play a bit with the formulas to get a series with only odd coefficients. Notice that

    [tex]\ln (1+z^2)[/tex]

    will give a series with only even terms. Also, you can use Euler's formula

    [tex]e^{a+ib} = e^a \cos b + i e^a \sin b[/tex]

    and so in particular, you can take the imaginary part:

    [tex]\Im \left[ e^{a+ib} \right] = e^a \sin b[/tex]

    So why don't you play around with those things and see if you can sum the series.
     
  4. Jun 24, 2011 #3
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook