Re: Psi function on the complex plane

Phred <nospam@xxxxxxxxxx> writes:

On Wed, 10 Oct 2007 05:11:19 -0500, Oscar Lanzi III wrote:

Well, Gamma(z) itself has simple poles at all nonpositive integers. So
Psi = Gamma'/Gamma will also have simple poles; moreover, the residues of
all poles of Psi will be -1.

Apart from repeating what I have already written in my question, is that
of any use?

Let me restate my needs. I don't need the Laurent expansion of psi, which
is clearly trivial.

On the contrary, I think the Laurent expansions are exactly what you need.
Namely, for |z+n| < 1,
Psi(z) = sum_{j=-1}^infty a_j(n) (z+n)^j
where a_{-1}(n) = -1, a_0(n) = -gamma + sum_{k=1}^n 1/k and
a_j(n) = (-1)^(j+1) zeta(j+1) + sum_{k=1}^n 1/k^(j+1) for j >= 1.
This gives you arbitrarily good approximations to Psi(z) for z
near -n.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

.

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