Solve for y in function z = f(x, y)

Question

Tim Fulcher 2023-10-7

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https://ww2.mathworks.cn/matlabcentral/answers/2030319-solve-for-y-in-function-z-f-x-y

评论： Sam Chak 2023-10-9

采纳的回答： Sam Chak

在 MATLAB Online 中打开

Hi All,

I have a polynomial which is a function of x (2nd order) and y (fifth order) and I wish to solve for y. My equation is:

p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p21*x^2*y + p12*x*y^2 + p03*y^3 + p22*x^2*y^2 + p13*x*y^3 + p04*y^4 + p23*x^2*y^3 + p14*x*y^4 + p05*y^5

I have tried:

syms x y z p00 p10 p01 p20 p11 p02 p21 p12 p03 p22 p13 p04 p23 p14 p05
eqn = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p21*x^2*y + p12*x*y^2 + p03*y^3 + p22*x^2*y^2 + p13*x*y^3 + p04*y^4 + p23*x^2*y^3 + p14*x*y^4 + p05*y^5 == z
soly = solve(eqn,y)

and this has returned:

soly = root(p05*z1^5 + p14*x*z1^4 + p04*z1^4 + p23*x^2*z1^3 + p13*x*z1^3 + p03*z1^3 + p22*x^2*z1^2 + p12*x*z1^2 + p02*z1^2 + p11*x*z1 + p21*x^2*z1 + p01*z1 - z + p20*x^2 + p10*x + p00, z1, 1)

I'm afraid i don't recognize this format. Where did z1 come from and I assume the the 1 (before the closing parenthisis) implies that it is the first and only solution? Nor am I sure about the root.

I have tried something similar in the past:

syms x y z p00 p10 p01 p20 p11 p02 p30 p21 p12
eqn = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y + p12*x*y^2 == z
soly = solve(eqn,y)

and this yielded a more familiar format:

-(p01 + p11*x + p21*x^2 + (p11^2*x^2 + p21^2*x^4 - 4*p00*p02 + 4*p02*z + p01^2 + 2*p01*p21*x^2 - 4*p02*p20*x^2 - 4*p10*p12*x^2 - 4*p02*p30*x^3 + 2*p11*p21*x^3 - 4*p12*p20*x^3 - 4*p12*p30*x^4 - 4*p00*p12*x + 2*p01*p11*x - 4*p02*p10*x + 4*p12*x*z)^(1/2))/(2*(p02 + p12*x))

-(p01 + p11*x + p21*x^2 - (p11^2*x^2 + p21^2*x^4 - 4*p00*p02 + 4*p02*z + p01^2 + 2*p01*p21*x^2 - 4*p02*p20*x^2 - 4*p10*p12*x^2 - 4*p02*p30*x^3 + 2*p11*p21*x^3 - 4*p12*p20*x^3 - 4*p12*p30*x^4 - 4*p00*p12*x + 2*p01*p11*x - 4*p02*p10*x + 4*p12*x*z)^(1/2))/(2*(p02 + p12*x))

Regards and thanks in advance.

Tim

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Walter Roberson 2023-10-7

在 MATLAB Online 中打开

syms x y z p00 p10 p01 p20 p11 p02 p30 p21 p12
eqn = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y + p12*x*y^2 == z
soly = solve(eqn,y)

In eqn, the maximum power that y apears is y^2 so that eqn can be rephrased as a quadratic,

(p02 + p12*x)*y^2 + (p01 + p11*x + p21*x^2)*y + (p00 + p10*x + p20*x^2 + p30*x^3 - z) == 0

which can easily be solved.

But your original equation has y^5 and that gives you problems.

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Answer 1

Sam Chak 2023-10-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2030319-solve-for-y-in-function-z-f-x-y#answer_1328379

在 MATLAB Online 中打开

Hi @Tim Fulcher

If the values for x and z are known, then you can numerically solve for y.

syms y

x = 1;

z = 0;

p00 = 1;

p10 = 1;

p01 = 1;

p20 = 1;

p11 = 1;

p02 = 1;

p21 = 1;

p12 = 1;

p03 = 1;

p22 = 1;

p13 = 1;

p04 = 1;

p23 = 1;

p14 = 1;

p05 = 1;

eqn = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p21*x^2*y + p12*x*y^2 + p03*y^3 + p22*x^2*y^2 + p13*x*y^3 + p04*y^4 + p23*x^2*y^3 + p14*x*y^4 + p05*y^5 == z

eqn =

soly = vpasolve(eqn, y)

soly =

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Walter Roberson 2023-10-7

By the way, the Impossibility Theorem does not say that the roots of polynomials of degree 5 or higher do not generally exist: the Impossibility Theory says that you cannot generally construct those roots as being a certain kind of constructed numbers -- numbers involving addition, subtraction, multiplication, division, raising to an integer power, or taking square roots. The theorem shows that the number of polynomials of degree 5 or higher is "larger" (in some meaningful way) then the number of polynomials that can be back-constructed from numbers that are of that form, and therefor there must be polynomials of degree 5 that have roots that are not expressible as such numbers.

The situation is sort of similar to demonstrating that there are polynomials of degree 2 with real roots whose roots are not rational numbers -- that for example, x^2 - 2 has roots that cannot be rational.

And in the meantime, no-one has come up with a way to construct the roots of the polynomials of degree 5 or higher in terms of other kinds of numbers (that are not radicals.) But they have come up with some amazing results in the investigations.

Tim Fulcher 2023-10-9

移动：Sam Chak 2023-10-9

Many thanks guys. That worked really well.

Regards

Tim

Sam Chak 2023-10-9

Good to hear that, @Tim Fulcher. If you find the guidance helpful, please consider clicking 'Accept' ✔ on the answer to close the issue when the problem is solved.

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