🚀 go-pugleaf

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


JON, PLEASE (!!!) STOP THIS NONSENSE !!!!!

Your method will not even find the root of 

x-1=0

Please place a followup to english speaking newsgroups.

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

brianscsmith@yahoo.com (Brian Smith) wrote 
> 
> 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.


AND ... 

Jon's method calculates the solution of x - 1 = 0 as 0.7 ...
(ridiculous)

AND ...

the examples he gives on his website are NOT correctly calculated

AND

where is the notorious 'high degree of accuracy and precision' 

as I said before: PLAIN BULLSHIT !!!


Hint: Use Newton's method, refined by Householder:

                       (1/p)k             
x_(n+1)=x_n + (k+1)*------------(x_n)
                      (1/p)(k+1)
[the 'k' and 'k+1' at the (1/p)'s intends to mean k-th resp. (k+1)-th
derivative]
 
to obtain an approximation of real roots at an order of k+2 (iff you
have an appropriate starting value)