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If you want to solve a cubic equation

  1. May 4, 2009 #1
    There is a very handy numerical solution for cubic equations like [tex] x^3+ax^2+bx+c=0[/tex] with [tex]x_i \in R[/tex] while [itex] a^2-3b \neq 0[/itex]. Though it makes use of the method of Newton, the starting point for the algorithm gives it a great advantage to the normal algorithm. And you can use MS Excel as an ordinary calculator!

    We compute the parameter [tex]4d=\frac{-9\left(2a^3-9ab+27c\right)}{\sqrt{12\left(a^2-3b\right)^3-3\left(2a^3-9ab+27c\right)^2}}[/tex] for the composition of the equation [tex]\ f^3-4df^2-9f+4d=0[/tex].

    This equation has roots that satisfy [tex]f_j=\frac{f_i-3}{f_i+1}[/tex].

    Next thing to do is to calculate [tex]y_0=\frac{4d}{4+\sqrt{16d^2+25}}[/tex].

    We substitute [tex]y_0[/tex] in [tex] y_j=\frac{2y_i^3-4dy_i^2-4d}{3y_i^2-8dy_i-9}[/tex].

    This gives the value [tex]y_1[/tex], repeating this step for [tex]y_1[/tex] gives [tex]y_2[/tex]. There is no need to go any further than [tex]y_2[/tex] as Excel can’t handle those accuracies any more.

    Substitution of [tex]y_2=f_2[/tex] in [tex]f_j=\frac{f_i-3}{f_i+1}[/tex] gives [tex] f_i[/tex]. Then:

    [tex]x_i=-\frac{ab-9c}{2a^3-6b}+\frac{f_i}{18a^2-54b}\times \sqrt{12\left(a^2-3b\right)^3-3\left(2a^3-9ab+27c\right)^2}[/tex].

    Just an example: [tex]x^3+4x^2-18x-9=0[/tex]
    [tex]4d[/tex] = -2,655 292 734 356 2300
    [tex]y_0[/tex]= - 0,274 837 379 984 0870
    [tex]y_1[/tex] = - 0,275 027 669 928 0960
    [tex]y_2[/tex] = - 0,275 027 663 450 8140

    Substitution gives
    [tex]x_1[/tex] = 3,000 000 000 000 0000
    [tex]x_2[/tex] = -0,458 618 734 850 8900
    [tex]x_3[/tex] = - 6,541 381 265 149 1100

    Check on coefficients:
    [tex]a[/tex] = 4,000 000 000 000 0000
    [tex]b[/tex] = - 18, 000 000 000 000 0000
    [tex]c[/tex] = - 9,000 000 000 000 0000 In Excel!!

    For those who want to know: [tex]|f_2-y_0| < 0,00086[/tex] for every [tex]f_2 \in R[/tex]. So for practical use [tex]y_1[/tex] will suffice giving an accuracy of about 6 decimals.

    Can anybody please tell me what to do about this ridiculous jumping too small figures in LaTeX? \Large doesn't do any good nor itex (the way I used it). Thanks.
     
    Last edited: May 4, 2009
  2. jcsd
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