4
$\begingroup$

Say one has a plane curve defined by a polynomial equation $$P(x,y)=0$$ and one knows that the curve has genus 0. Is there an implementation in Mathematica of a (proper) rational parameterization, i.e., a function that computes a pair of rational functions $(x(t), y(t))$ generically one-to-one solving the equation above? e.g., for $$P(x,y)=1+x^2 y^2-x^3 y^2$$ the answer should be (up to Mobius transformations) $$x(t)=1+t^2,\qquad y(t)=\frac{1}{t(1+t^2)}$$

$\endgroup$
3

1 Answer 1

1
$\begingroup$

Very simple solution

Solve[1 + x^2*y^2 - x^3*y^2 == 0, y]

Out[]= {{y -> -(1/Sqrt[-x^2 + x^3])}, {y -> 1/Sqrt[-x^2 + x^3]}}

 % /. x -> 1 + t^2 // FullSimplify

Out[]= {{y -> -(1/Sqrt[(t + t^3)^2])}, {y -> 1/Sqrt[(t + t^3)^2]}}

In the general case, we solve the equation and make the substitution

Solve[P[x,y]==0,y]
%/.x->f[t]//FullSimplify

The problem of parametrization does not have a unique solution; therefore, the question of choosing a function f[t] remains open. If the equation P[x,y]==0 is not explicitly solved for x or y, then the question of parameterization remains open.

$\endgroup$
4
  • $\begingroup$ to clarify, the input is the polynomial, the output is the rational parameterization. the latter is not known a priori. $\endgroup$ Oct 31, 2018 at 12:16
  • $\begingroup$ See update of my answer. $\endgroup$ Oct 31, 2018 at 12:39
  • 1
    $\begingroup$ That is not a rational parametrization (it has fractional exponents). $\endgroup$ Oct 31, 2018 at 17:28
  • $\begingroup$ I think your answer basically ends at the start of the OPs question: Is there a function (that tries) to find a possible $f(t)$ yielding a rational parametrization (if it exists)? $\endgroup$ Nov 1, 2018 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.