Subject: Rational function parametric representations
From: reyssat
Date: Wed, 23 Sep 2009 08:01:41 +0200
Barton Willis a ?crit :
> Thanks for the book reference. For 2*x^4-3*x^2*y+y^4-2*y^3+y^2 = 0, Maple
> paramtrized
> the curve with radicals, not a rational expression. But I know little
> about Maple.
>
Of course, any quartic may be "piecewise parametrized" using radicals :
take t=y as parameter, and solve for x (by radicals). Here the equation
is biquadratic, so x is just the square root of the solution of a
quadratic equation :
(%i93) solve(2*x^4-3*x^2*y+y^4-2*y^3+y^2,x);
(%o93) [x = -sqrt(y)*sqrt(sqrt(-8*y^2+16*y+1)+3)/2,x =
sqrt(y)*sqrt(sqrt(-8*y^2+16*y+1)+3)/2,x =
-sqrt(3*y-y*sqrt(-8*y^2+16*y+1))/2,
x = sqrt(3*y-y*sqrt(-8*y^2+16*y+1))/2]
For a general cubic or quartic, the solution by radicals may involve
some complex numbers in the formula, even if the final value is real
(Casus Irreductibilis) ; this may be a trouble when plotting the curve.
Eric