If you want to solve a cubic equation

  • Thread starter edgo
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There is a very handy numerical solution for cubic equations like ## x^3+ax^2+bx+c=0## with ##x_i \in R## 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 ##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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  • #2
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Looks like an interesting approach, especially the part about calculating a starting point. One of the deficiencies of Newton's method is that it can sometimes fail to converge if you start with an unfortunate initial guess.

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.
Not sure what you mean, but I edited your post to replace all of the standalone LaTeX with inline LaTeX.
 

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