root() function gives compelx solution for cubic equation. how to get the nearby real root?

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mirewuti muhetaer 2020-6-24

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评论： Walter Roberson 2020-6-25

Dear All:

i have a cubic equation and the coefficients are set already. But it always gives me the complex roots (all of them).

As i know, root() function of the matlab (in practice and theory) gives the very approximate roots for any polynomial that has more than quadratic.

So, is there anyway i can force the root to give me the most possible real roots?.

Thanks.

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John D'Errico 2020-6-25

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As you THINK you know, but maybe you are wrong in what you think.

You cannot force roots to give you a real root where one does not exist. There is no magical flag you can set to change that behavior for roots.

For example, consider the equation:

p(x) = x^3 - x^2 - x - 1/5

This has one real root. Wanting it to have more than that is a good dream, but just a dream.

roots([1 -1 -1 -1/5])
ans =
             1.670356741159 +                     0i
         -0.335178370579499 +    0.0859672116099084i
         -0.335178370579499 -    0.0859672116099084i

Yes, Suppose we purturb this cubic by a small amount, changing the constant term from -1/5 to -5/27.

roots([1 -1 -1 -5/27])
ans =
           1.66666666666667 +                     0i
         -0.333333333333333 +  6.59783539196707e-09i
         -0.333333333333333 -  6.59783539196707e-09i

Which suggests the roots you might think are the ones you want to see, are: [5/3, -1/3, -1/3]. But for that to be true, I had to make a perhaps significant modification to the polynomial.

Perhaps a better question is what is the minimum norm perturbation to the coefficients of that polynomial that yields all real roots, and for that perturbation, what would be the roots? Of course, for that we would need to define what we mean by a minimal perturbation.

mirewuti muhetaer 2020-6-25

Dear @Walter Roberson:

Thanks for the clarification.

When i read the matlab root documents and other online documents, also based on what i learned in the past, for the polynomials (except quadratic), Newton's method is used to get the roots. That is the reason, i though matlab may also do this and there is kind of tiny difference between the actual solution and roots given by matlab root() function.

Now, it is clear to me that i was wrong.

Really appreciated.

Walter Roberson 2020-6-25

https://www.mathworks.com/help/matlab/ref/roots.html#buo5imt-5

The roots function considers p to be a vector with n+1 elements representing the nth degree characteristic polynomial of an n-by-n matrix, A. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A.

Not Newton's method.

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