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From: cfchiesa@bsu-cs.UUCP (Christopher Chiesa)
Newsgroups: sci.math,comp.graphics
Subject: Re: Solution of quartic equation
Message-ID: <2381@bsu-cs.UUCP>
Date: 16 Mar 88 21:41:52 GMT
References: <1656@thorin.cs.unc.edu> <7662@agate.BERKELEY.EDU>
Organization: CS Dept, Ball St U, Muncie, Indiana
Lines: 36
Summary: Cube roots of complex numbers

In article <7662@agate.BERKELEY.EDU>, greg@jiff.berkeley.edu (Greg) writes:
> In article <1656@thorin.cs.unc.edu> turner@unc.cs.unc.edu (Douglass Turner) writes:
> >Hello, does anyone know of a good, robust, method to find all roots of a 
> >quartic equation?
> 
>    [miscellaneous introductory comments]

> 
> The direct solution is prone to programmer's instability.  It's very
> messy and doesn't generalize.  The most painful step is taking the
> cube root of a complex number, which cannot be expressed in terms of
> roots of real numbers.  In any case, the numerous roots in the formula

Perhaps I misunderstand you (heaven knows I didn't follow the Sturm chains
and so on), but...  I recall taking cube and higher roots of complex numbers
quite easily by converting from "cartesian" (A+Bi) representation to "polar"
(r cis Theta), and performing a few straightforward real-number calculations.
Basically, you take the root (let's generalize and say the Nth root) of the
"r" (radius) part of the representation, and divide the "Theta" (angle) part
by N, to find the first Nth root; then you add Theta/N to the angle part re-
peatedly to get the other N-1 roots.  Works quite nicely; I've never bothered
to program it, but if it's been a bottleneck for you, this may be a way around
it.

I'd be interested to know if this has been helpful to you or anyone else...

Chris Chiesa
Senior, CS Dept
Ball State University
Muncie, IN
 
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<><><><><><><><><><><><><><><><><><><><><><><><><><><><> Chris Chiesa <><><><><>
<> {ihpn4|seismo}!{iuvax|pur-ee}!bsu-cs!cfchiesa                              <>
<> cfchiesa@bsu-cs.UUCP                                                       <>
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