From: bousch%linotte.uucp@topo.math.u-psud.fr (Thierry Bousch)
Newsgroups: sci.math.symbolic,sci.math
Subject: Re: Laurent series for 1/(1 - x*cot(x)) at x=0
Date: 11 Apr 1998 00:10:51 +0200

In sci.math.symbolic Glenn H. Davis <gdavis@best.com> wrote:

: How would one show that all the coefficients in the above
: Laurent series are negative, except the first one ?

: Maple V3 takes "series( 1/(1 - x*cot(x)), x, 5 );" and expands it to:

: 	3*x^-2 - 1/5 -1/175*x^2 - 2/7875*x^4 + O(x^6)

: So these three coefficients are negative,
: and it stays that way up to x^100.

The classic way to estimate the coefficients of the Taylor series of f
is to find out the singularities of f with smallest modulus. In your
case f(z)=z^2/(1-z*cot(z)) has two singularities +-a of minimum modulus
(which is the convergence radius of your series), and since these are
simple poles, it is easy to compute their contribution to the Taylor
series, and bound the rest.

More precisely, write f(z) = k/(z^2-a^2) + r(z), where k is chosen so that r
has no singularity at +-a. Since r(z) has a greater convergence radius than
f(z), its coefficients will soon be much smaller than the coefficients of
k/(z^2-a^2). Use path integrals to get explicit bounds.

Thierry