🚀 go-pugleaf

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0408040446.2c407887@posting.google.com>...
> Jon <jon8338@peoplepc.com> wrote in message news:<J2EPc.23857$iK.8050@newsread2.news.atl.earthlink.net>...
> > Polynomial Roots to any Degree
> > Vector Calculus Method
> > 
> > The projection of the curve T onto the normal N results in a logarithmic 
> > solution. However, the Taylor coefficients for the logarithmic 
> > contribution out-differentiate the polynomial, so it is subtracted as 
> > error from the final solution.
> > 
> 
> Jon, there is a circular argument in your method.
> 
> When you construct the equation '-a_0 = Integ (N*dT) dt', the equation
> reduces back to the original polynomial.  That circular argument means
> that your method fails.

This part of his "method" isn't incorrect, just not stated as is normally
done. He's saying that if (his first formula) holds, then (his second
formula, with the integral) also holds.

> Also, I would like your to see your application of your method to the
> equation x^3 - ax^2 + X - a = 0, with 'a' as a variable.  If you
> cannot extract 'a' as a root, then your method is useless to get any
> exact values.

He can't even solve a cubic (which is what his example reduces to). He
notes that 1 is a root of t^4 + t^3 + t^2 + t - 4, and thus wants to
find the roots of t^3 + 2 t^2 + 3 t + 4, which he can evidently only
approximate. A message to Jon: If you claim your procedure works, at
least find a polynomial where the procedure actually gives you the
right answers, and approximations don't count. (You've been told this
numerous times!)
     -- Christopher Heckman