Argir Pando Vasil Dobri Matea Karagorgovi
2009-11-24 21:30:53 UTC
P(t) = a10 t^10 + a9 t^9 + a8 t^8 + a7 t^7 + a6 t^6 + a5 t^5 + a4 t^4
+ a3 t^3 + a2 t^2 + a1 t^1 + a0 = 0
introducing the substitution:
t = (x + n1 + n2 + n3 + n4 + n5) / (x + n6 + n7 + n8 + n9 + R)
where R is random complex number and so P(t) gets the form:
a10 (x + n1 + n2 + n3 + n4 + n5)^10 +
a9 (x + n1 + n2 + n3 + n4 + n5)^9 (x + n6 + n7 + n8 + n9 + R)^1 +
a8 (x + n1 + n2 + n3 + n4 + n5)^8 (x + n6 + n7 + n8 + n9 + R)^2 +
a7 (x + n1 + n2 + n3 + n4 + n5)^7 (x + n6 + n7 + n8 + n9 + R)^3 +
a6 (x + n1 + n2 + n3 + n4 + n5)^6 (x + n6 + n7 + n8 + n9 + R)^4 +
a5 (x + n1 + n2 + n3 + n4 + n5)^5 (x + n6 + n7 + n8 + n9 + R)^5 +
a4 (x + n1 + n2 + n3 + n4 + n5)^4 (x + n6 + n7 + n8 + n9 + R)^6 +
a3 (x + n1 + n2 + n3 + n4 + n5)^3 (x + n6 + n7 + n8 + n9 + R)^7 +
a2 (x + n1 + n2 + n3 + n4 + n5)^2 (x + n6 + n7 + n8 + n9 + R)^8 +
a1 (x + n1 + n2 + n3 + n4 + n5)^1 (x + n6 + n7 + n8 + n9 + R)^9 +
a0 (x + n6 + n7 + n8 + n9 + R)^10 = 0
after expanding all the powered sums you get polynomial of type:
P(x) = A10 x^10 + A9 x^9 + A8 x^8 + A7 x^7 + A6 x^6 + A5 x^5 + A4 x^4
+ A3 x^3 + A2 x^2 + A1 x^1 + A0 = 0
where A10 is pure number, A9 is first degree expression dependent on
all n's, A8 is second degree expression dependent on all n's, A7 is
third degree expression dependent on all n's, A6 is fourth degree
expression dependent on all n's, ... A2 is 8th degree expression
dependent on all n's, A1 is 9th degree expression dependent on all n's
and finally A0 is 10th degree expression dependent on all n's. now you
break the prior form P(x) = 0 on system of ten equations:
system[ equation 1: A9 = 0; equation 2: A8 = 0; equation 3: A7 = 0;
equation 4: A6 = 0; equation 5: A5 = 0; equation 6: A4 = 0; equation
7: A3 = 0; equation 8: A2 = 0; equation 9: A1 = 0 and equation 10: x =
nroot(class 10, index from 1 to 10, - A0 over A10) ]
then sum all equations form 1 to 9 and group them by n9 in form:
P(n9) = B9 (n9)^9 + B8 (n9)^8 + B7 (n9)^7 + B6 (n9)^6 + B5 (n9)^5 + B4
(n9)^4 + B3 (n9)^3 + B2 (n9)^2 + B1 (n9)^1 + B0 = 0
where B9 is pure number, B8 is first degree expression by all other
n's, B7 is second degree expression by all those other n's, B6 is
third degree expression by all those other n's, ... B2 is 7th degree
expression by all those other n's, B1 is 8th degree expression by all
those other n's and finally B0 is 9th degree expression by all those
other n's. now you break the prior P(n9) = 0 on system of nine
equations:
system[ equation 1: B8 = 0; equation 2: B7 = 0; equation 3: B6 = 0;
equation 4: B5 = 0; equation 5: B4 = 0; equation 6: B3 = 0; equation
7: B2= 0; equation 8: B1 = 0; equation 9: n9 = nroot(class 9, index
from 1 to 9, - B0 over B9) ]
then sum all the equations from 1 to 8 and group them by n8 in form:
P(n8) = C8 (n8)^8 + C7 (n8)^7 + C6 (n8)^6 + C5 (n8)^5 + C4 (n8)^4 + C3
(n8)^3 + C2 (n8)^2 + C1 (n8)^1 + C0 = 0
where C8 is pure number, C7 is first degree expression by all the
remaining n's, C6 is second degree expression by all the remaining
n's, C5 is third degree expression by all the remaining n's, ... C2 is
6th degree expression by all the remaining n's, C1 is 7th degree
expression by all the remaining n's and finally C0 is 8th degree
expression by all the remaining n's. now break the prior P(n8) = 0 on
system of 8 equations:
system[ equation 1: C7 = 0; equation 2: C6 = 0; equation 3: C5 = 0;
equation 4: C4 = 0; equation 5: C3 = 0; equation 6: C2 = 0; equation
7: C1 = 0 and equation 8: n8 = nroot(class 8, index form 1 to 8, - C0
over C8) ]
... and so on ...
until you solve the final n (that will be n1) and then recursively
return the accumulated result up to t. there are 10 x 9 x 8 x 7 x 6 x
5 x 4 x 3 x 2 x 1 solutions out of which only 10 will be different,
try me Abel-Ruffini ! i am just demonstrating my SCHIZMATHIC
superiority.
+ a3 t^3 + a2 t^2 + a1 t^1 + a0 = 0
introducing the substitution:
t = (x + n1 + n2 + n3 + n4 + n5) / (x + n6 + n7 + n8 + n9 + R)
where R is random complex number and so P(t) gets the form:
a10 (x + n1 + n2 + n3 + n4 + n5)^10 +
a9 (x + n1 + n2 + n3 + n4 + n5)^9 (x + n6 + n7 + n8 + n9 + R)^1 +
a8 (x + n1 + n2 + n3 + n4 + n5)^8 (x + n6 + n7 + n8 + n9 + R)^2 +
a7 (x + n1 + n2 + n3 + n4 + n5)^7 (x + n6 + n7 + n8 + n9 + R)^3 +
a6 (x + n1 + n2 + n3 + n4 + n5)^6 (x + n6 + n7 + n8 + n9 + R)^4 +
a5 (x + n1 + n2 + n3 + n4 + n5)^5 (x + n6 + n7 + n8 + n9 + R)^5 +
a4 (x + n1 + n2 + n3 + n4 + n5)^4 (x + n6 + n7 + n8 + n9 + R)^6 +
a3 (x + n1 + n2 + n3 + n4 + n5)^3 (x + n6 + n7 + n8 + n9 + R)^7 +
a2 (x + n1 + n2 + n3 + n4 + n5)^2 (x + n6 + n7 + n8 + n9 + R)^8 +
a1 (x + n1 + n2 + n3 + n4 + n5)^1 (x + n6 + n7 + n8 + n9 + R)^9 +
a0 (x + n6 + n7 + n8 + n9 + R)^10 = 0
after expanding all the powered sums you get polynomial of type:
P(x) = A10 x^10 + A9 x^9 + A8 x^8 + A7 x^7 + A6 x^6 + A5 x^5 + A4 x^4
+ A3 x^3 + A2 x^2 + A1 x^1 + A0 = 0
where A10 is pure number, A9 is first degree expression dependent on
all n's, A8 is second degree expression dependent on all n's, A7 is
third degree expression dependent on all n's, A6 is fourth degree
expression dependent on all n's, ... A2 is 8th degree expression
dependent on all n's, A1 is 9th degree expression dependent on all n's
and finally A0 is 10th degree expression dependent on all n's. now you
break the prior form P(x) = 0 on system of ten equations:
system[ equation 1: A9 = 0; equation 2: A8 = 0; equation 3: A7 = 0;
equation 4: A6 = 0; equation 5: A5 = 0; equation 6: A4 = 0; equation
7: A3 = 0; equation 8: A2 = 0; equation 9: A1 = 0 and equation 10: x =
nroot(class 10, index from 1 to 10, - A0 over A10) ]
then sum all equations form 1 to 9 and group them by n9 in form:
P(n9) = B9 (n9)^9 + B8 (n9)^8 + B7 (n9)^7 + B6 (n9)^6 + B5 (n9)^5 + B4
(n9)^4 + B3 (n9)^3 + B2 (n9)^2 + B1 (n9)^1 + B0 = 0
where B9 is pure number, B8 is first degree expression by all other
n's, B7 is second degree expression by all those other n's, B6 is
third degree expression by all those other n's, ... B2 is 7th degree
expression by all those other n's, B1 is 8th degree expression by all
those other n's and finally B0 is 9th degree expression by all those
other n's. now you break the prior P(n9) = 0 on system of nine
equations:
system[ equation 1: B8 = 0; equation 2: B7 = 0; equation 3: B6 = 0;
equation 4: B5 = 0; equation 5: B4 = 0; equation 6: B3 = 0; equation
7: B2= 0; equation 8: B1 = 0; equation 9: n9 = nroot(class 9, index
from 1 to 9, - B0 over B9) ]
then sum all the equations from 1 to 8 and group them by n8 in form:
P(n8) = C8 (n8)^8 + C7 (n8)^7 + C6 (n8)^6 + C5 (n8)^5 + C4 (n8)^4 + C3
(n8)^3 + C2 (n8)^2 + C1 (n8)^1 + C0 = 0
where C8 is pure number, C7 is first degree expression by all the
remaining n's, C6 is second degree expression by all the remaining
n's, C5 is third degree expression by all the remaining n's, ... C2 is
6th degree expression by all the remaining n's, C1 is 7th degree
expression by all the remaining n's and finally C0 is 8th degree
expression by all the remaining n's. now break the prior P(n8) = 0 on
system of 8 equations:
system[ equation 1: C7 = 0; equation 2: C6 = 0; equation 3: C5 = 0;
equation 4: C4 = 0; equation 5: C3 = 0; equation 6: C2 = 0; equation
7: C1 = 0 and equation 8: n8 = nroot(class 8, index form 1 to 8, - C0
over C8) ]
... and so on ...
until you solve the final n (that will be n1) and then recursively
return the accumulated result up to t. there are 10 x 9 x 8 x 7 x 6 x
5 x 4 x 3 x 2 x 1 solutions out of which only 10 will be different,
try me Abel-Ruffini ! i am just demonstrating my SCHIZMATHIC
superiority.
