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Forumula for the Roots to the nth Degree Polynomial

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Nov 28, 2009, 3:55:37 PM11/28/09

FORMULA FOR THE ROOTS TO THE nth DEGREE POLYNOMIAL

a[0] + a[1]x + a[2]x^2 + a[3]x^3 + ... + a[n]x^n = 0

x = { (C^( (j-1)/n ))*S }^(1/(n-j)) j=1,2,3,...,n-1

ALL SUMS ARE FROM p=1 to n

C = { 1-(a[0]*SUM{a[p]})/(SUM{(a[p])^2}) }^(1/n)

D = -a[0]/SUM{a[p]*(C^(p-1))}

S = { (1/n)*SUM{(D^(2n-2p))*(C^(2p-2))} }^(1/2)

EXAMPLE

x^5+x-34=0 x=2

C = {1-(-34*(1+1)/(1^2+1^2))}^(1/5)=2.036

D = -(-34)/( 1*2.036^0 + 1*2.036^4 )=1.869

S = { (1.869^8+(2.036^2)(1.869^6)+(2.036^4)(1.869^4)+
(2.036^6)(1.869^2)+2.036^8)/5 }^(1/2) = 14.698

j=1

x = {14.698}^(1/4) = 1.958 ~ 2

EXAMPLE

x^5+x+34=0 x= -2

C = {1-(34*(1+1)/(1^2+1^2))}^(1/5)= -2.012

D = -(34)/( 1*-2.012^0 + 1*-2.012^4 )= -1.954

S = 15.500

j=2

x = (S/C)^(1/3) = -{15.500/2.012}^(1/3) = -1.974 ~ -2

x = 1.974[cos((2k+1)pi/3)+isin((2k+1)pi/3)], k=0,1,2

EXAMPLE

e^(nx)=me^x

1+nx+((nx)^2)/2+((nx)^3)/3!+...+=m+mx+m(x^2)/2+m(x^3)/3!+... i.

(m-1)+(m-n)x+(m-(n^2)/2)x^2+(m-(n^3)/3!)x^3
+(m-(n^4)/4!)x^4+(m-(n^5)/5!)x^5+..= 0

C={ 1-(a[0]*SUM{a[p]})/(SUM{(a[p])^2}) }^(1/q) ~ x

x = {1-( (m-1)*(m-n+m-(n^2)/2+m-(n^3)/3!+...+m-(n^q)/q! )/
( (m-n)^2+(m-(n^2)/2)^2+(m-(n^3)/3!)^2+...+(m-(n^q)/q!)^2 ) }^(1/q) ii.

Substitute ii. in i. and solve:

m=f(n)

also

n=g(m)

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