Path: news.mathworks.com!newsfeed-00.mathworks.com!NNTP.WPI.EDU!elk.ncren.net!newsflash.concordia.ca!canopus.cc.umanitoba.ca!not-for-mail
From: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson)
Newsgroups: comp.soft-sys.matlab
Subject: Re: Solve Equation Symbolically
Date: Mon, 12 Nov 2007 21:27:26 +0000 (UTC)
Organization: National Research Council Canada - Conseil national de rechereches Canada
Lines: 51
Message-ID: <fhagfu$rsd$1@canopus.cc.umanitoba.ca>
References: <fha99r$osc$1@fred.mathworks.com> <fhadf3$o58$1@canopus.cc.umanitoba.ca> <fhaes6$k0h$1@fred.mathworks.com>
NNTP-Posting-Host: origin.ibd.nrc.ca
X-Trace: canopus.cc.umanitoba.ca 1194902846 28557 192.70.172.160 (12 Nov 2007 21:27:26 GMT)
X-Complaints-To: abuse@cc.umanitoba.ca
NNTP-Posting-Date: Mon, 12 Nov 2007 21:27:26 +0000 (UTC)
Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson)
Xref: news.mathworks.com comp.soft-sys.matlab:437182



In article <fhaes6$k0h$1@fred.mathworks.com>,
evan klinger <eklinger1-nospam@cox.net> top-posted:

Please do not post your reply at the top of the material you
are commenting on: it makes it more difficult to hold discussions.
Your posting may be the only posting of interest to -you- today,
but I'm following literally over 100 simultaneous discussions today.

>roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote in
>message <fhadf3$o58$1@canopus.cc.umanitoba.ca>...
>> In article <fha99r$osc$1@fred.mathworks.com>,
>> evan klinger <eklinger1-nospam@cox.net> wrote:

>> >cv0, p[n], x, db and cr are all symbols

>> >cv1 = (cv0 + p1 - (x)*(db-cv0))*(1+cr)
>> >cv2 = (cv1 + p2 - (x)*(db-cv1))*(1+cr)
>> >cv3 = (cv2 + p3 - (x)*(db-cv2))*(1+cr)

>> It's a cubic in x, so there are three solutions. Two of them
>> involve imaginary numbers.

>I guess I'm confused about why there are 3 solutions since
>it can really be written as one long equation if you
>substitute the cv's in. In other words, you can write cv3 in
>terms of cv0s. Am I missing something?

Look at cv1: it has x times cv0.

Now look at cv2: it has x times cv1, so that means it is going
to have x times (x times cv0), or x^2 times cv0.

Now look at cv3: it has x times cv2, so that means it is going
to have x times (x^2 times cv0), or x^3 times cv0.

Therefor your cv3 is going to involve x^3, and is thus
a cubic polynomial. When you solve a cubic polynomial for 0,
then according to the Fundamental Theorem of Algebra, you
are going to have 3 roots of the polynomial.

If your constants -happen- to have very specific (and complicated)
relationships, then all three of the roots would be real, so
for completeness you really shouldn't ignore the two solutions
that involve imaginary numbers: those are of the form
f(x) + g(x)*I and if g(x) just happens to come out to 0 then
you have another real solution. 
-- 
  "I will speculate that [...] applications [...] could actually see a
  performance boost for most users by going dual-core [...] because it
  is running the adware and spyware that [...] are otherwise slowing
  down the single CPU that user has today"           -- Herb Sutter