Re: Dividing Power Series

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 08/15/04 

Date: Sun, 15 Aug 2004 17:09:01 -0500

On 15 Aug 2004 11:12:44 -0700, contactgreg@hotmail.com (Gregory
Magarshak) wrote:

>Can I divide power series by long division as follows:

yes, at least for z in a disk where both series converge
and the divisor has no zero.

pf: the two series define analytic functions f, g in
this disk. now h = f/g is analytic in this disk, so
it has a power series. it's easy to show you can
get the series for f = g*h by multiplying the series
for g and h formally, and long division is just the
inverse of formal power-series multiplication [at
each step the criterion you use to choose a
coefficient for the quotient is exactly 'make
the product g*h come out right'.]

>(You need a fixed width font to view the following)
>
> 1/z + 1/z^2 + 2/z^3 + 2/z^4 ...
> __________________________________
>z - 1 |1 + 0 + 1/z^2 + 0 + 1/2z^4
> -1 + 1/z
> __________
> 1/z + 1/z^2 + ...
> -1/z + 1/z^2
> _________________
> 2/z^2 + 0 + ...
> -2/z^2 + 2/z^3
> ____________
> 2/z^3 + ...
>
>Here I do the long division from the other end. It seems to work;
>though is this a known result? Shouldn't be too hard to prove that it
>works, I think.
>
>Sincerely,
>Greg

************************

David C. Ullrich

sorry about the inelegant formatting - typing
one-handed for a few weeks...

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