# Series of hyperbolic function

1. Nov 14, 2005

### gdumont

Hi,

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 $a=0$ but I get stuck at the first term, namely $f(0)$, 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. Nov 15, 2005

### dextercioby

Well, here's the catch:

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

and is equal to

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

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

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

Daniel.

Last edited: Nov 15, 2005