The implicitize(peqs, params, d, options) function computes an implicit (polynomial) equation for the algebraic curve, surface, or hypersurface defined by the parametric equations peqs.

The output is given as a set that can be empty or have one or more elements.

The implicitize routine returns an implicit equation for the degree d if it exists.

The output will be exact or approximate, depending on the optional parameters controlled by the user.

The options argument can contain one or more of the following equations.

maxIters = positive_integer

The maxIters option controls the maximum number of iterations performed by findNullVector.

symbolic = true or false

If the symbolic option is set to true, all integrations are performed symbolically. If the symbolic option is set to false, all integrations are performed numerically. The default value is false.

tol = expression

The tol option controls the tolerance used in findNullVector. The default value is $\mathrm{10e}-\mathrm{Digits}$.

useFNV = true or false

If the useFNV option is set to true, the implicitize routine uses findNullVector. If the useFNV option is set to false, the implicitize routine uses NullSpace instead of findNullVector. The default value is true.

The implicitize routine accepts parametric equations that:

[1]  are polynomials, rational functions, trigonometric functions, functions involving square roots, or functions in polar form;

[2]  have coefficients from Q, parameters, or algebraic numbers; and

[3]  define multi-parameter families of curves, surfaces of hypersurfaces.

The implicitize routine is based on the algorithm described in the paper: Robert M. Corless, Mark W. Giesbrecht, Ilias S. Kotsireas, Stephen M. Watt. "Numerical implicitization of parametric hypersurfaces with linear algebra." AISC'2000 Proceedings, Madrid, Spain. LNAI 1930 (in print).

$\mathrm{with}\left(\mathrm{algcurves}\right)\:$

$\mathrm{peqs}≔x=\cos \left(\mathrm{\theta }\right),y=\sin \left(\mathrm{\theta }\right)$

$\mathrm{peqs}≔x=\cos \left(\mathrm{\theta }\right),y=\sin \left(\mathrm{\theta }\right)$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[\mathrm{\theta }=0..\mathrm{\pi }\right]\,2\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^2+y^2-1\right\}$

$\mathrm{peqs}≔x=\frac{1}{\mathrm{sqrt}\left(t^2+1\right)},y=\frac{t}{\mathrm{sqrt}\left(t^2+1\right)}$

$\mathrm{peqs}≔x=\frac{1}{\sqrt{t^2+1}},y=\frac{t}{\sqrt{t^2+1}}$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[t=0..1\right]\,2\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^2+y^2-1\right\}$

$\mathrm{Peqs}≔x=\frac{3t^2}{t^3+1},y=\frac{3t}{t^3+1}$

$\mathrm{Peqs}≔x=\frac{3t^2}{t^3+1},y=\frac{3t}{t^3+1}$

$\mathrm{Digits}≔25$

$\mathrm{Digits}≔25$

$\mathrm{ee}≔\mathrm{implicitize}\left(\left[\mathrm{Peqs}\right]\,\left[t=0..1\right]\,3\,\mathrm{symbolic}=\mathrm{true}\right)$

$\mathrm{ee}≔7.175755130803066010567000\times {10}^{\mathrm{-20}}-1.845883179623651348694161\times {10}^{\mathrm{-18}}y-1.620368226773049054536803\times {10}^{\mathrm{-16}}{y}^2-0.3015113445777636810112419y^3+5.208339920246452022722121\times {10}^{\mathrm{-16}}x+0.9045340337332909517727017yx+3.417979544288393196348183\times {10}^{\mathrm{-16}}x{y}^2-6.571949647625600126222411\times {10}^{\mathrm{-16}}{x}^2-3.336979304002439223779846\times {10}^{\mathrm{-16}}{x}^{2}y-0.3015113445777633127855855x^3$

$\mathrm{eq}≔\mathrm{fnormal}\left(\mathrm{ee}\,4\right)$

$\mathrm{eq}≔-0.3015y^3+0.9045yx-0.3015x^3$

$\mathrm{sm}≔\min \left(\mathrm{op}\left(\mathrm{map}\left(\mathrm{xx}\mapsto \mathrm{abs}\left(\mathrm{xx}\right)\,\left[\mathrm{coeffs}\left(\mathrm{expand}\left(\mathrm{eq}\right)\right)\right]\right)\right)\right)$

$\mathrm{sm}≔0.3015$

$\mathrm{fnormal}\left(\mathrm{simplify}\left(\frac{\mathrm{eq}}{\mathrm{sm}}\right)\,3\right)$

$-y^3+3.00yx-x^3$

$\mathrm{simplify}\left(\mathrm{eval}\left(\,\left[\mathrm{Peqs}\right]\right)\right)$

$0.$

$\mathrm{implicitize}\left(\left[x=rt\,y=r{t}^2\,z=r^2\right]\,\left[r=0..1\,t=0..1\right]\,4\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^4-y^{2}z\right\}$

$\mathrm{peqs}≔x=\mathrm{\alpha }\tan \left(\mathrm{\phi }\right),y=\mathrm{\beta }{\cos \left(\mathrm{\phi }\right)}^2$

$\mathrm{peqs}≔x=\mathrm{\alpha }\tan \left(\mathrm{\phi }\right),y=\mathrm{\beta }{\cos \left(\mathrm{\phi }\right)}^2$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[\mathrm{\phi }=\frac{\mathrm{\pi }}{3}..\frac{\mathrm{\pi }}{4}\right]\,3\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{-{\mathrm{\alpha }}^2\mathrm{\beta }+y{\mathrm{\alpha }}^2+x^{2}y\right\}$

$\mathrm{simplify}\left(\mathrm{eval}\left(\,\left[\mathrm{peqs}\right]\right)\right)$

$\left\{0\right\}$

$\mathrm{peqs}≔x=\sin \left(3\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right),y=\sin \left(3\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right)$

$\mathrm{peqs}≔x=\sin \left(3\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right),y=\sin \left(3\mathrm{\theta }\right)\sin \left(\mathrm{\theta }\right)$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[\mathrm{\theta }=0..\mathrm{\pi }\right]\,4\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^4+2x^2{y}^2+y^4-3x^{2}y+y^3\right\}$

$\mathrm{peqs}≔x=t+s+r,y=ts+sr+rt,z=tsr,w=t^3+s^3+r^3$

$\mathrm{peqs}≔x=t+s+r,y=sr+rt+ts,z=tsr,w=r^3+s^3+t^3$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[t=0..1\,s=0..1\,r=0..1\right]\,3\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^3-3yx-w+3z\right\}$

$\mathrm{simplify}\left(\mathrm{eval}\left(\,\left[\mathrm{peqs}\right]\right)\right)$

$\left\{0\right\}$

$\mathrm{peqs}≔\left[x=\frac{t\left(\mathrm{\beta }-t^2\right)}{{\left(1+t^2\right)}^2}\,y=\frac{t^2\left(\mathrm{\beta }-t^2\right)}{{\left(1+t^2\right)}^2}\right]$

$\mathrm{peqs}≔\left[x=\frac{t\left(-t^2+\mathrm{\beta }\right)}{{\left(t^2+1\right)}^2}\,y=\frac{t^2\left(-t^2+\mathrm{\beta }\right)}{{\left(t^2+1\right)}^2}\right]$

$\mathrm{implicitize}\left(\mathrm{peqs}\,\left[t=0..2\right]\,2\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\varnothing$

$\mathrm{implicitize}\left(\mathrm{eval}\left(\mathrm{peqs}\,\mathrm{\beta }=3\right)\,\left[t=0..2\right]\,4\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^4+2x^2{y}^2+y^4-3x^{2}y+y^3\right\}$

$\mathrm{implicitize}\left(\mathrm{eval}\left(\mathrm{peqs}\,\mathrm{\beta }=4\right)\,\left[t=0..2\right]\,4\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^4+2x^2{y}^2+y^4-4x^{2}y+y^3\right\}$

$p≔x^4-\mathrm{\beta }y{x}^2+2x^2{y}^2+y^3+y^4$

$p≔-\mathrm{\beta }y{x}^2+x^4+2x^2{y}^2+y^4+y^3$

$\mathrm{simplify}\left(\mathrm{eval}\left(p\,\mathrm{peqs}\right)\right)$

$0$

$\mathrm{peqs}≔\left[x=\frac{3t^2-2t}{t^k}\,y=\frac{3t-2}{t^k}\right]$

$\mathrm{peqs}≔\left[x=\frac{3t^2-2t}{t^k}\,y=\frac{3t-2}{t^k}\right]$

$\mathrm{implicitize}\left(\mathrm{eval}\left(\mathrm{peqs}\,k=4\right)\,\left[t=1..2\right]\,4\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^4-3x{y}^2+2y^3\right\}$

$\mathrm{implicitize}\left(\mathrm{eval}\left(\mathrm{peqs}\,k=5\right)\,\left[t=1..2\right]\,5\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^5-3x{y}^3+2y^4\right\}$

$\mathrm{implicitize}\left(\mathrm{eval}\left(\mathrm{peqs}\,k=6\right)\,\left[t=1..2\right]\,6\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{x^6-3x{y}^4+2y^5\right\}$

$p≔3t-2\;$$q≔t^k\;$$\mathrm{xtk}≔\frac{tp}{q}\;$$\mathrm{ytk}≔\frac{p}{q}$

$p≔3t-2$

$q≔t^k$

$\mathrm{xtk}≔\frac{t\left(3t-2\right)}{t^k}$

$\mathrm{ytk}≔\frac{3t-2}{t^k}$

$\mathrm{exk}≔x^k-3x{y}^{k-2}+2y^{k-1}$

$\mathrm{exk}≔x^k-3x{y}^{k-2}+2y^{k-1}$

$\mathrm{dd}≔\mathrm{eval}\left(\mathrm{exk}\,\left[x=\mathrm{xtk}\,y=\mathrm{ytk}\right]\right)$

$\mathrm{dd}≔{\left(\frac{t\left(3t-2\right)}{t^k}\right)}^k-\frac{3t\left(3t-2\right){\left(\frac{3t-2}{t^k}\right)}^{k-2}}{t^k}+2{\left(\frac{3t-2}{t^k}\right)}^{k-1}$

$\mathrm{simplify}\left(\mathrm{dd}\right)$

$\left(-9t+6\right)t^{1-k}{\left(\left(3t-2\right)t^{-k}\right)}^{k-2}+{\left(t^{1-k}\left(3t-2\right)\right)}^k+2{\left(\left(3t-2\right)t^{-k}\right)}^{k-1}$

$\mathrm{alias}\left(\mathrm{\beta }=\mathrm{sqrt}\left(b\right)\right)$

$\mathrm{\beta }$

$\mathrm{alias}\left(\mathrm{\gamma }=\mathrm{sqrt}\left(c\right)\right)$

$\mathrm{\beta },\mathrm{\gamma }$

$\mathrm{peqs}≔x=\mathrm{\beta }+\cos \left(2\mathrm{\theta }\right),y=\mathrm{\gamma }+\sin \left(2\mathrm{\theta }\right)$

$\mathrm{peqs}≔x=\mathrm{\beta }+\cos \left(2\mathrm{\theta }\right),y=\mathrm{\gamma }+\sin \left(2\mathrm{\theta }\right)$

$\mathrm{implicitize}\left(\left[\mathrm{peqs}\right]\,\left[\mathrm{\theta }=0..\mathrm{\pi }\right]\,2\,\mathrm{symbolic}=\mathrm{true}\,\mathrm{useFNV}=\mathrm{false}\right)$

$\left\{-2x\mathrm{\beta }-2y\mathrm{\gamma }+x^2+y^2+b+c-1\right\}$

$\mathrm{simplify}\left(\mathrm{eval}\left(\,\left[\mathrm{peqs}\right]\right)\right)$

$\left\{0\right\}$