All Roots to any Power Series
From: Jon (jon8338_at_peoplepc.com)
Date: 06/30/04
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Date: Wed, 30 Jun 2004 02:53:11 GMT
DERIVING ALL ROOTS TO ANY POLYNOMIAL
Let
N=(a[1],a[2],...,a[n])
T=(t,t^2,t^3,...,t^n)
then
N*T+a[0]=0 = nth degree polynomial in t. N=(a[1],a[2],a[3],...,a[n]) is
the normal to the plane and T is a space curve in t. Define
Q=(-a[0]/|N|^2)N and let B,P be 2 points on the plane such that
(Q-B)*(P-B)/(|Q-B||P-B|) does not equal 1 or -1. (B,P are selected so
that (Q-B) and (P-B) are not parallel). The unit vector R on the plane
perpendicular to Q-B is,
"P-B minus the reflection of P-B onto the unit vector of Q-B, divided by
the magnitude of the same."
(P-B)*(Q-B)
P-B - -----------(Q-B)
|Q-B|^2
R = ---------------------------
| (P-B)*(Q-B) |
| P-B - -----------(Q-B) |
| |Q-B|^2 | also define,
Q-B
S = -----
|Q-B|
then
T = B + uR + mS where u is some ratio and m is some ratio. Taking
the dot product of S with both sides and of R with both sides, note that
R*S=0 and
m=(T-B)*S= S*T-B*S
u=(T-B)*R= R*T-B*R
T = B + (R*T-B*R)R + (S*T-B*S)S
T' = (R*T')R + (S*T')S
T'' = (R*T'')R + (S*T'')S
.
.
T^{n}=(R*T^{n})R + (S*T^{n})S where T^{n} is the nth derivative of T
with respect to t. For t=0,
T(0)=(0,0,0,..,0)
T'(0)=(1,0,0,...,0)
T''(0)=(0,2,0,...,0)
.
.
T^{n}(0)=(0,0,0,...,0,n!)
If R=(r[1],r[2],r[3],...,r[n]) and S=(s[1],s[2],s[3],...,s[n]),
T(0)=B-(B*R)R-(B*S)S
T'(0)= r[1]R + s[1]S
T''(0)=2r[2]R+2s[2]S
.
.
T^{n}(0)=n!r[n]R+n!s[2]S
n
T(t) = SUM[(1/p!)T^{p}(0)t^p]
p=0
Where T^{p}(0) is t=0 in the pth derivative of T (Maclaurin) Otherwise,
T(t) =
B-(B*R)R-(B*S)S +
(r[1]R+s[1]S)t +
(r[2]R+s[2]S)t^2 +
(r[3]R+s[3]S)t^3 +
.
.
(r[n]R+s[n]S)t^n Let B=(b[1],b[2],...,b[n])
Decoding T(t),
(r[1]r[1]+s[1]s[1]-1)t +
(r[2]r[1]+s[2]s[1] )t^2+
(r[3]r[1]+s[3]s[1] )t^3+
.
.
(r[n]r[1]+s[n]s[1] )t^n + (b[1]-(B*R)r[1]-(B*S)s[1]) = 0
==
(r[1]r[2]+s[1]s[2] )t +
(r[2]r[2]+s[2]s[2]-1)t^2+
(r[3]r[2]+s[3]s[2] )t^3+
.
.
(r[n]r[2]+s[n]s[2] )t^n + (b[2]-(B*R)r[2]-(B*S)s[2]) = 0
==
(r[1]r[3]+s[1]s[3] )t +
(r[2]r[3]+s[2]s[3] )t^2+
(r[3]r[3]+s[3]s[3]-1)t^3+
.
.
(r[n]r[3]+s[n]s[3] )t^n + (b[3]-(B*R)r[3]-(B*S)s[3]) = 0
==
This continues until we arrive at,
(r[1]r[n]+s[1]s[n] )t +
(r[2]r[n]+s[2]s[n] )t^2+
(r[3]r[n]+s[3]s[n] )t^3+
.
.
(r[n]r[n]+s[n]s[n]-1)t^n + (b[n]-(B*R)r[n]-(B*S)s[n]) = 0
Generalizing n equations into n planes,
(r[1]r[1]+s[1]s[1]-1)x[1]+
(r[2]r[1]+s[2]s[1] )x[2]+
(r[3]r[1]+s[3]s[1] )x[3]+
.
.
(r[n]r[1]+s[n]s[1] )x[n] + (b[1]-(B*R)r[1]-(B*S)s[1]) = 0
==
(r[1]r[2]+s[1]s[2] )x[1]+
(r[2]r[2]+s[2]s[2]-1)x[2]+
(r[3]r[2]+s[3]s[2] )x[3]+
.
.
(r[n]r[2]+s[n]s[2] )x[n] + (b[2]-(B*R)r[2]-(B*S)s[2]) = 0
==
(r[1]r[3]+s[1]s[3] )x[1]+
(r[2]r[3]+s[2]s[3] )x[2]+
(r[3]r[3]+s[3]s[3]-1)x[3]+
.
.
(r[n]r[3]+s[n]s[3] )x[n] + (b[3]-(B*R)r[3]-(B*S)s[3]) = 0
==
This continues until we arrive at,
(r[1]r[n]+s[1]s[n] )x[1]
(r[2]r[n]+s[2]s[n] )x[2]+
(r[3]r[n]+s[3]s[n] )x[3]+
.
.
(r[n]r[n]+s[n]s[n]-1)x[n] + (b[n]-(B*R)r[n]-(B*S)s[n]) = 0
These n planes intersect each other with various lines (they can all be
found), they have angles between all of their normals (also easily
obtained), and share a common point (the n equations can be solved for
the n unknowns using linear algebra). Supposing this common point is
calculated as C=(c[1],c[2],c[3],..,c[n]). It can be deduced that while
the point C is a solution to the intersection of all the planes, it does
not necessarily follow that c[2]=c[1]^2, c[3]=c[1]^3, etc as in
T=(t,t^2,t^3,..,t^n). Supposing the solution t=w exists. It can be
inferred that if w is a root, it must also satisfy the point of
intersection of the planes at C. Supposing, then, that C is a solution.
Selecting the last component c[n]=t^n,
t = c[n]^(1/n)[cos(2k*pi/n)+ i sin(2k*pi/n)] if c[n] is positive
t = c[n]^(1/n)[cos((2k+1)pi/n)+ i sin((2k+1)pi/n)] if c[n] is negative
where k = 0,1,2,3,...,n-1 Binomial Equation
Rendering the n roots for the nth degree polynomial N*T-a[0]=0.
END
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