# If you want to solve a cubic equation

There is a very handy numerical solution for cubic equations like ## x^3+ax^2+bx+c=0## with ##x_i \in R## while $a^2-3b \neq 0$. 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 ##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}}## for the composition of the equation ##f^3-4df^2-9f+4d=0##.

This equation has roots that satisfy ##f_j=\frac{f_i-3}{f_i+1}##.

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

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

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

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

##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}##.

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

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

Check on coefficients:
##a## = 4,000 000 000 000 0000
##b## = - 18, 000 000 000 000 0000
##c## = - 9,000 000 000 000 0000 In Excel!!

For those who want to know: ##|f_2-y_0| < 0,00086## for every ##f_2 \in R##. So for practical use ##y_1## 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.

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