Sure Polynomial Roots

[Jon]

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

[Peter Pan]

Post by Jon
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

[u***@free.fr]

Please place a followup to english speaking newsgroups.

[Brian Smith]

Jon <***@peoplepc.com> wrote in message news:<***@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.

Post by Jon
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

[Peter Pan]

Post by Brian Smith
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)