Roots of a polynomial with variables

19 次查看（过去 30 天）

Lewis Fer 2021-6-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/857075-roots-of-a-polynomial-with-variables

评论： Lewis Fer 2021-6-16

For some problems, we have to to study some notions of stablility and zero polynomials in two variables, my que'stion how we can find the roots or zero polynomials in two variables. for example:

P(x,y)=3*xy -5y^2+7*x^2y

or a nother polynom

回答（2 个）

Sulaymon Eshkabilov 2021-6-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/857075-roots-of-a-polynomial-with-variables#answer_725560

在 MATLAB Online 中打开

One of the viable ways to solve such polynomial type equations is to setp up the solution space within which you are seeking the roots to compute and solve them using fzero(). E.g.:

x=linspace(-2,2):  % Choose the necessary solution space
for t=1:100
    EQ= @(y)(3*x(t)*y-5*y.^2+7*(x(t)^2)*y); 
    y_roots = fzero(EQ,0);
end

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Lewis Fer 2021-6-15

but this method in reality doesn't making a difference to search and find th pole of some matrices like transfer function in two dimensional case

请先登录，再进行评论。

Paul 2021-6-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/857075-roots-of-a-polynomial-with-variables#answer_725615

在 MATLAB Online 中打开

Don't know the scope of the actual problems of interest, but for the two examples in the question:

syms x y

sol = solve(3*x*y - 5*y^2 + 7*x^2*y == 0,[x y],'ReturnConditions',true);

[sol.x sol.y sol.conditions]

ans =

syms z1 z2

sol = solve(1 - z1*z2 - 1/2*z1^2 - 1/2*z2^2 + z1^2*z2^2 == 0,[z1 z2],'ReturnConditions',true);

[sol.z1 sol.z2 sol.conditions]

ans =

Apparently there are many solutions to this problem, i.e., many pairs (p,s) that make the determinant equal to zero. The pair (p,s) can be expressed as ( (z+2)/(z-4) , z) for z any number not equal to four. Check

A = [1 2;3 4];
p = @(z)((z+2)./(z-4));
s = @(z)(z);
z = 1;
det(diag([p(z) s(z)]) - A)
ans = -3.3307e-16
z = 8;
det(diag([p(z) s(z)]) - A)
ans = 0
z = 1 + 1i;
det(diag([p(z) s(z)]) - A)
ans = -1.3323e-15 + 3.3307e-16i