Jon
2004-08-04 04:16:38 UTC
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
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