Re: Sure Polynomial Roots

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#1192

From: Brian Smith
Date: Thu, 05 Aug 2004 12:42

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Jon <jon8338@peoplepc.com> wrote in message news:<4111D4FB.8050608@peoplepc.com>...

I see that you are no longer claiming to have solved the general
polynomial equation, now its just a numeric formula.  Why should I
even consider this method for approximating roots when Newton's
iteration only needs the first derivative and only a few iterations
are needed to get the three place precesion that your method provides
in the example.  I see no need to calculate dozens of terms when it is
so much simpler to let Newton's iteration run for more cycles.

> If r is the root to the nth degree polynomial
> 
> a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0
> 
> Then r is the solution to,
> *******************************************
> a[0]*ln|r|=
> 
>    n
> -SUM(1/p)*a[p]*c^p
>   p=1
> 
>    n
> +SUM(1/p)*a[p]
>   p=1
> 
>       infinity
> -a[0]SUM(-1)^(k+1) * a[0]/(k*c^k)
>       k=n+1
> 
> where
> c={(-1)^n * a[0]/a[n] }^(1/n)
> *******************************************
> 
> This computes roots to a high degree of accuracy and precision.
> 
> For an explanation on how this was derived and examples for its use, see 
> my web site at,
> 
> http://mypeoplepc.com/members/jon8338/polynomial/
> 
> Jon Giffen

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Subject: Re: Sure Polynomial Roots
Date: 5 Aug 2004 05:42:39 -0700
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