how to decompose a polynomial

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Saeed rhmt 2018-12-14

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https://ww2.mathworks.cn/matlabcentral/answers/435708-how-to-decompose-a-polynomial

编辑： John D'Errico 2018-12-14

Dear All,

For a given polynomial of the minimum order equals to 1, as follows:

F = mu1^2*x1^6 - 2*mu1*mu2*x1^6 + 6*mu1*mu2*x1^5*x2 - 6*mu1*mu2*x1^4*x2^2 + 2*mu1*mu2*x1^3*x2^3 + mu2^2*x1^6 - 6*mu2^2*x1^5*x2 + 15*mu2^2*x1^4*x2^2 - 20*mu2^2*x1^3*x2^3 + 15*mu2^2*x1^2*x2^4 - 6*mu2^2*x1*x2^5 + mu2^2*x2^6 

where m1 and mu2 are coefficients.

Having defined X= [x1 ; x2];, how can I decompose 'F' to the following Structure:

F = X' * M * X; or F =X' * S' * S * X

Where M or S are matrices in terms of the powers of x1 and x2 or constant coefficients;

Is there any special matlac command for doing this?

many Thanks.

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Jan 2018-12-14

Do you need for convert this function only? Then doing this manually is easy. Or is there a reason to perform this automatically?

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Answer 1

John D'Errico 2018-12-14

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https://ww2.mathworks.cn/matlabcentral/answers/435708-how-to-decompose-a-polynomial#answer_352360

编辑：John D'Errico 2018-12-14

在 MATLAB Online 中打开

It is not a polynomial, a multinomial is a better description. I'm not even sure that what you ask is well-posed, in the sense that it looks like it does not even have a unique solution.

However, that does not mean you cannot write it yourself. Just extract each additive term of that multinomial sum. The function children can do that.

C = children(F)
C =
[ mu1^2*x1^6, mu2^2*x1^6, mu2^2*x2^6, -2*mu1*mu2*x1^6, -6*mu2^2*x1*x2^5, -6*mu2^2*x1^5*x2, 15*mu2^2*x1^2*x2^4, -20*mu2^2*x1^3*x2^3, 15*mu2^2*x1^4*x2^2, 6*mu1*mu2*x1^5*x2, 2*mu1*mu2*x1^3*x2^3, -6*mu1*mu2*x1^4*x2^2]
C(1)
ans =
mu1^2*x1^6

Now, can you write that single term in the desired form? Lets see, it has no factors of x2. So we can write it in a simpleform as:

[x1,x2]*[mu1^2*x1^4 ,0;0 0]*[x1;x2]
ans =
mu1^2*x1^6

You should see that you can always write any such subexpression of total degree 2 or higher in such a form, but that it is not always unique. In the above case, it was. But consider this subexpression.

C(7)
ans =
15*mu2^2*x1^2*x2^4

If we factor out x1*x1, then we could write it as:

[x1,x2]*[15*mu2^2*x2^4 ,0;0 0]*[x1;x2]
ans =
15*mu2^2*x1^2*x2^4

But we can also write it as:

[x1,x2]*[0 15*mu2^2*x1*x2^3;0 0]*[x1;x2]
ans =
15*mu2^2*x1^2*x2^4

Any bets that I cannot find infinitely many ways to write this? (Never take a sucker bet like that.)

[x1,x2]*[0 7*mu2^2*x1*x2^3;8*mu2^2*x1*x2^3 0]*[x1;x2]
ans =
15*mu2^2*x1^2*x2^4

Regardless, just extract each term of the multinomial. I show how to do that using children. Then pick any of the infintely many possible forms that I showed how to create. The result will be a factorization in the form of a sum that will look like...

[x1,x2]*S{1}*[x1;x2] + [x1,x2]*S{2}*[x1;x2] + [x1,x2]*S{3}*[x1;x2] + ...

But we know that these terms can then be recombined as:

[x1,x2]*( S{1} + S{2} + S{3} + ... )*[x1;x2]

Again, the solution is not at all unique. And yes, you will need to write it, since I seriously doubt anyone has bothered to do so for you.