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Series of hyperbolic function

  1. Nov 14, 2005 #1

    I have the following function to evaluate in a power series:
    f(a)=\frac{\pi}{8d}\frac{1}{\left (\sinh \left ( \frac{\pi a}{2 d} \right) \right)^2}
    Maple computes then following
    f(a) = \frac{\pi}{8d} \left ( \frac{4 d^2}{\pi^2 a^2} - \frac{1}{3} + O(a^2) \right)
    When I ask Maple if this series is of Taylor or Laurent type it tells me that it is neither. I tried to compute the Taylor series around [itex]a=0[/itex] but I get stuck at the first term, namely [itex]f(0)[/itex], which is infinite. The first term of the expression obtained by Maple is precisely equal to the inverse of the first non-zero term of the Taylor series of
    g(a)= \left (\sinh \left ( \frac{\pi a}{2 d} \right) \right)^2 = \left (0 + \frac{\pi a}{2 d} +\right )^2 = \frac{\pi^2 a^2}{4 d^2}
    But I can't figure out where do the other terms come from. Anyone knows how does Maple evaluate this series?

    Any help greatly appreciated
  2. jcsd
  3. Nov 15, 2005 #2


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    Well, here's the catch:

    [tex] f(a)=\frac{\pi}{8d}\left[\frac{1}{\frac{e^{\pi a/ 2d}-e^{-\pi a/ 2d}}{2}}}\right]^{2} [/tex]

    and is equal to

    [tex] f(a)=\frac{\pi}{2d} \frac{e^{\pi a/d}}{\left(e^{\pi a/d}-1\right)^{2}} [/tex]

    and just then u can use the Taylor series for the exponentials.

    For the record, i don't get the "1/3" factor...

    Last edited: Nov 15, 2005
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