From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Taylor-Laurent series
Organization: Computer Science Department,  Stanford University.
Date: Sun, 22 Nov 1992 20:57:23 GMT

In article <1541700007@gn.apc.org> antennae@gn.apc.org (Indra) writes:
>
>I see that a mathematician has taken up the cudgels on behalf of
>Prof Abian on the vexed question of whether 1/(z + 2z + 3z^2) is
>unequal to 1/(3z^2 + 2z + z) - I'm not sure that I've remembered
>this correctly - and asserts that Abian is in fact correct.

What was nice about the example (which was 1/(z^2-3z+2) =
1/((z-1)(z-2))) was that the division *process* yielded different
series valid in different regions of the plane, simply by permuting the
terms in the denominator.  I won't vouch for his actual arithmetic.

Here's what's going on behind the scenes.  (I'm a bit rusty so there
may be minor bugs.  I teach algebraic logic in a calculus-free
environment and haven't touched this stuff for nearly 30 years.)

The Laurent expansion of a complex function f(z) about 0 represents
f(z) as a linear combination of both positive and negative powers of z:

                  -2       -1                    2
	... + a  z   + a  z   + a  +  a z  +  a z  + ...
	       -2       -1       0     1       2

A Laurent expansion with no negative powers of z is called a Taylor
expansion.

Such an expansion of f(z) need not converge everywhere on the complex
plane, but where it does we say that it is a *valid* expansion of
f(z).  Its domain of convergence is a maximal pole-free annulus
centered on 0.

For example 1/(z-1) (which is its own expansion about 1) can be
expanded about 0 as -1/(1-z) = -1 - z - z^2 - ..., valid for |z| < 1.
But it can also be expanded as 1/z * 1/(1-1/z)
= 1/z + 1/z^2 + 1/z^3 + ..., valid for |z| > 1.

Note the pole at z=1, on the common boundary of the two annuli.

These two are the only expansions of 1/(z-1) for which there exists an
open neighborhood in which they are valid.  (One might offer the
one-term series -1/2 as an expansion of 1/(z-1) valid at the point -1,
which "open neighborhood" defeats.  Nicest is to define validity at a
point to mean validity in an open neighborhood of that point.)

Now where does one get the a_i's from?  I got the above two by a little
algebra and table lookup for 1/(1-z).  Somewhat more generally one can
use division as Abian does.  However the definitive method calculates
a_i as the integral of f(z)/z^{i+1}, divided by 2 pi i.  This
integration is performed once counterclockwise around the annulus.
(More precisely, around any oriented loop confined to the annulus that
winds once counterclockwise around the origin.)

Abian's example of 1/((z-1)(z-2)) has poles at 1 and 2, and hence its
expansion about 0 calls for three annuli, each with its own expansion.
What Abian has hinted at vaguely is a method of computing the expansion
for any given annulus by dividing by the polynomial with its terms
permuted appropriately for that annulus.  Since a degree n-1 polynomial
will have at most n annuli and n! permutations of terms it is clear
that there are more than enough permutations to go around.  What's not
clear is his method, which I've not seen the likes of before (maybe he
only uses the cyclic permutations).  However I don't work in this area
and for all I know it could be stock standard.

What I *have* looked for without success is a symbolic algebra system
which can perform Laurent expansion in a specified annulus.  Since the
process should be hidden to the users, we shouldn't know whether it
uses contour integration or polynomial division, though the latter may
well be faster.  How about it, algebra hackers?
-- 
Vaughan Pratt	  		A fallacy is worth a thousand steps.
Newsgroups: sci.physics