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) $$x(t)=1+t^2,\qquad y(t)=\frac{1}{t(1+t^2)}$$
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2$\begingroup$ Michal Bizzarri & Miroslav Lávička (2009) has a Mathematica implementation for computing rational parametrizations. $\endgroup$– Michael E2Oct 31, 2018 at 15:03
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$\begingroup$ Yes, I saw this, but the link to their code is broken, AFAICT. Unless you happen to know an alternative link.. $\endgroup$– Paul Zinn-JustinNov 1, 2018 at 1:52
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1$\begingroup$ @PaulZinn-Justin Did you try writing the (corresponding) author of that paper? See zcu.cz/en/about/people/staff.html?osoba=11521 $\endgroup$– Jules LamersNov 2, 2018 at 1:21
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1 Answer
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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.
answered Oct 31, 2018 at 10:40
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$\begingroup$ 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
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1$\begingroup$ That is not a rational parametrization (it has fractional exponents). $\endgroup$ Oct 31, 2018 at 17:28
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$\begingroup$ 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