Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-10-23 10:54:04

ross
Guest

Cardano's method

Im a little stuck. Im trying to make a program that will find the eigenvalues of a 3x3 matrix which led to the need to solve a cubic. To do so I had planned to do some form of iterative search for a single root, then factorise it out and solve the remaining quadratic simply by using the quadratic formula.

Then I found reference to Cardano's method (http://en.wikipedia.org/wiki/Cubic_equation), and in light of the note about the case when p=0, I figure that by using just the positive square root in the expression for u, and taking the real cube root, a single real root to the equation (i.e. my first eigenvalue) will be the final result. This could then be factorised out to leave the quadratic.

It may just be an error with my code but I have checked it several times and cannot see anything wrong, so the question is, what is wrong with my reasoning? Should I be using the complex roots? My thought process does seem a bit suspect as Cardano's method does always throw out just two real roots whereas the cubic should generally be allowed to have three or just one.

Thanks,

Ross

Board footer

Powered by FluxBB