maybe, maybe not

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#1216

From: Jon
Date: Wed, 04 Aug 2004 04:16

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The attempt is made to find the roots to polynomials.

if
N=(a[1],a[2],..,a[n])
T=(t,t^2,t^3,..,t^n)
N*T+a[0]=0   is the nth degree polynomial.

dT=(1,2t,3t^2,..)dt

-a[0]=INT N*dT = INT(a[1]+2a[2]t+3a[3]t^2+...)dt

using the virtue that
a[1]t+a[2]t^2+...=-a[0],

-a[0]=INT(-a[0]/t + a[2]t+2a[3]t^2+...)dt

which results in,

-a[0]=-a[0]ln|t|+(1/2)a[2]t^2+(2/3)a[3]t^3+(3/4)a[4]t^4+...

again, using the virtue that
a[1]t+a[2]t^2+...=-a[0],

-a[0]=-a[0]ln|t|-a[0]-a[1]t-(1/2)a[2]t^2-(1/3)a[3]t^3-..

or,


             n
a[0]ln|t|=-SUM(1/p)a[p]t^p        eqn i.
            p=1

Both sides of eqn i. are expressed as two Taylor Series

f(c)+f'(c)(t-c)+.. = g(c)+g'(c)(t-c)+...

Consequently, after n derivatives,

c=((-1)^n a[0]/a[n])^(1/n)

After n+1 derivatives, the right hand side of eqn i. becomes zero, but 
the left hand side continues to produce values and is subtracted from 
the result.

If
             n
D=(-1/a[0])SUM[(1/p)a[p]((-1)^n a[0]/a[n])^(p/n)] and
            p=1

    oo
R=SUM [(a[0]/k)((-1)^n a[0]/a[n])^(-k/n) ]
   k=n+1

then

t= e^D  - R

Jon Giffen

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