Forumula for the Roots to the nth Degree Polynomial

Discussion in 'General Math' started by Jon, Nov 28, 2009.

  1. Jon

    Jon Guest

    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)

    For the Development, see:
    http://jons-math.bravehost.com/roots.html

    (c) 2009 Jon Giffen
     
    Jon, Nov 28, 2009
    #1
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