From: mtrott@wri.com (Michael Trott)
Newsgroups: sci.math
Subject: Re: Polynomial Equations
Date: 18 Dec 1996 19:53:55 GMT

In article <59796g$8jj@earth.njcc.com> nahay@pluto.njcc.com (John Nahay)  
writes:

>Problem is, as far as I have studied this problem, there seem to exist
>formulae only for finding ONE of the roots of a polynomial y=P(x),

The theta function approach allows to calculate _all_ roots.
See

F. Lindemann: Ueber die Aufl\"osung algebraischer Gleichungen durch  
transcendente Funktionen II, Nachrichten der K\"onigl. Gesell. der Wiss.  
zu G\"ottingen, 292 (1892)

F. Lindemann: Ueber die Aufl\"osung algebraischer Gleichungen durch  
transcendente Funktionen, Nachrichten der K\"onigl. Gesell. der Wiss. zu  
G\"ottingen, 245 (1884)

H. Umemura: Resolution of algebraic equations by theta constants, in D.  
Mumford: Tata Lectures on Theta II, Birkh\"auser, Boston 1984

--
Michael Trott
Wolfram Research, Inc.
==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Summation of subrows
Date: 4 Nov 1996 20:40:20 GMT

[deletia --djr]
One suggestion is to consider the Jacobi theta function
	theta(t) = Sum exp( - pi k^2 t)  (for t>0)
where the sum is over all integers  k  (positive and negative). Taking
t = (1/pi).ln(1/x) gives theta(t)=1+2S  where  S  is your sum.

	For  x  close to  1  you're better off using the Jacobi identity:
theta(t) = t^(-1/2).theta(1/t). This is a version of the Poisson summation
formula. 
	A reference is Bellman, "A brief introduction to theta functions" 1961

dave