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NNTP-Posting-Date: Sun, 19 Aug 2007 03:40:20 -0500
From: quasi <quasi@null.set>
Newsgroups: alt.math,alt.math.recreational,fj.sci.math,sci.math
Subject: Re: Polynomial Roots
Date: Sun, 19 Aug 2007 04:40:13 -0400
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On Sun, 19 Aug 2007 03:39:22 -0400, "Jon G." <jon8338@peoplepc.com>
wrote:

>http://mypeoplepc.com/members/jon8338/polynomial/ 

It's wrong after just a few steps.

You start with the equation to be solved

(1)   a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0

You rewrite (1) as a dot product of vectors ...

(2) P_1 . T = 0 where 

         P = <a_0, a_1, a_2, ..., a_n> 

           and 

         T = <1, t, t^2, ..., t^n>

So far, no problem.

You then note that the 2-vector <a_0, a_p> is perpendicular to the
2-vector <-a_p/a_0, 1>. Fine.

You then make the false claim

(3)   <1, t^p>/(1 + t^(2p)) = <-a_p, a_0>/((a_0)^2) + (a_p)^2)

Although you neglect to use _words_ to defend your claim, I can guess
what your flawed reasoning probably was ...

Since the vectors P_1, T are perpendicular, you concluded that they
must be perpendicular when restricted to pairs of components.

There's no way to justify that.

Here's a counterexample ...

Consider the equation

   -3 + t + t^2 + t^3 = 0

which has the obvious solution t=1.

Thus, the vector equation is

   <-3, 1, 1, 1> . <1, 1, 1, 1> = 0

But there is no perpendicular pair of corresponding components, which
is what you claimed.

Thus, your "solution" is simply bogus.

Don't you think you should have checked a few examples before
launching into the monstrosity you posted?

quasi