From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Taylor-Laurent series
Organization: Computer Science Department,  Stanford University.
Date: Sun, 22 Nov 1992 00:18:47 GMT

In article <abian.722318103@pv343f.vincent.iastate.edu> abian@iastate.edu (Alexander Abian) writes:
>
>Dear Mr. Pratt:                 11-20-92
>
>You write:
>
>> Now that I look again I don't see how it
>>follows that this sum of singularities need be a singularity.  Unless
>>Abian can demonstrate this, I would say *that* was a real flaw in
>>his argument.  Mr. Abian?
>
>   I have given the demonstration (see the unabridged version that I think
>I specially posted for you). Now, you are asking for more details, e.g. that
>I have to prove that long division in descending powers gives the Laurent
>expansion.  I can give that proof (and in my proof no use of FTA is made).
>But then you may ask me for the proof of something else, and so on.

No, it's simpler than that, I just have one question so small and easy
you should have no objection to answering it directly in this forum.

Your argument mentions a series differing from

	1/z + 1/z^2 + 1/z^3 + ...

only in its choice of coefficients.  You then ask us to infer that this
is the series of a function with an essential singularity at z=0.

My question is, why should this sum have any singularity at all at z=0?

To illustrate my question, consider the function 1/(z-1), which
certainly has no singularity at z=0.  Now factor it as 1/z * 1/(1-1/z),
expand the second factor as 1 + 1/z + 1/z^2 + ..., and multiply by the
first factor to arrive at

	1/z + 1/z^2 + 1/z^3 + ...

The logic you use in your argument would appear to assign to this sum
an essential singularity at z=0.   Yet we have just seen that the sum
is in fact -1.  For additional confirmation, setting z=1/2 makes this
sum in binary ...1111110, which as any computer programmer can see at a
glance is two's-complement for -2, in agreement with 1/(1/2 - 1).  A
little algebra shows that setting z=1/p makes ...111110, interpreted in
radix p, equal to -p/(p-1), which as p tends to infinity tends to -1.
(This is a tiny part of the fascinating theory of p-adic numbers.)

>   On the other hand,  if you are accepting my wager ...

It is undignified for the myopic to bet with each other on a horse race
after it is over.

Your example of calculating the Taylor-Laurent series for "-3z+2+z^2"
was extremely nice.  Can anyone supply the names of those symbolic
algebra systems that can perform Taylor-Laurent expansions in regions
other than those connected to infinity?  If I understand my 1988
Mathematica manual correctly, the closest Mathematica can come to this
is to permit expansion about infinity, which presumably will not allow
one to use Mathematica to obtain the nice expansion Abian has given us
in the annulus 1 < |z| < 2.

Taylor series were understood by Gregory (forty years before Taylor) in
the 17th century, and perhaps by the Indians (East) in the 16th
century.  I don't know who Laurent is (can someone please supply this
information?  I couldn't find it in any of my usual sources), but
presumably the Taylor-Laurent series was well in hand by the 19th
century.

This prompts the following very interesting question.  Which of today's
symbolic algebra systems have made it into the 19th century?

Mr. Abian's evident facility with the tricky computations of Laurent
series in annuli lying between singularities would suggest the obvious
strategy for any proprietor of such a system wanting to bring it that
much more up to date.  But stock up on the Kleenex, one close encounter
with dbxtool should be the lachrymal equivalent of Tammy Baker reading
Love Story.

I understand Newton became something of a crackpot in later life.  If
TIME HAS INERTIA and Newton were alive today and battier than ever, I'm
sure Wolfram RI or Soft Warehouse would still jump at the opportunity
to number him among their consultants.
-- 
Vaughan Pratt	  		A fallacy is worth a thousand steps.
Newsgroups: sci.physics