Algebraic Curves Examples - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.

Home : Support : Online Help : Mathematics : Geometry : Examples : Algebraic Curves Examples

Algebraic Curves

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

When f is a polynomial in x and y, then f defines an algebraic curve in a plane.

 > $f:=51344{y}^{5}+53384{y}^{4}-47264{y}^{3}-415912{x}^{2}{y}^{3}-49304{y}^{2}+29070{x}^{2}{y}^{2}+247631{x}^{2}y+90164{x}^{4}y+73931{x}^{2}+40396{x}^{4}$
 ${f}{:=}{51344}{}{{y}}^{{5}}{+}{53384}{}{{y}}^{{4}}{-}{47264}{}{{y}}^{{3}}{-}{415912}{}{{x}}^{{2}}{}{{y}}^{{3}}{-}{49304}{}{{y}}^{{2}}{+}{29070}{}{{x}}^{{2}}{}{{y}}^{{2}}{+}{247631}{}{{x}}^{{2}}{}{y}{+}{90164}{}{{x}}^{{4}}{}{y}{+}{73931}{}{{x}}^{{2}}{+}{40396}{}{{x}}^{{4}}$ (1)

Puiseux Expansion

The function $f$ has two singularities and one regular point on the line $x=0$. We can obtain information (such as the tangent lines, the delta invariant, and other invariants) on singularities by computing the Puiseux expansions. One can view these Puiseux expansions as a sort of Taylor expansion (note that Puiseux expansions can also have fractional powers of x, whereas a Taylor expansion does not) of the algebraic function RootOf(f, y). Because this algebraic function is multivalued, we will get several expansions corresponding to the different branches of $f$ at $x=0$. The following command gives these expansions of $f$ at $x=0$:

 > $\mathrm{puiseux}\left(f,x=0,y,0\right)$
 $\left\{\frac{{6163}}{{6418}}{,}{x}{}{\mathrm{RootOf}}{}\left({-}{73931}{+}{49304}{}{{\mathrm{_Z}}}^{{2}}\right){,}{-}{1}{+}{x}{}{\mathrm{RootOf}}{}\left({50324}{}{{\mathrm{_Z}}}^{{2}}{-}{135641}\right)\right\}$ (1.1)

The fourth argument $0$ tells puiseux to compute a minimal number of terms. The number of terms that will be computed in this way is precisely the number of terms that are required to be able to distinguish the different Puiseux expansions from one another.

Note: It appears as though only three different Puiseux expansions were given, whereas the function $f$ has five different branches. The other two expansions are implicitly given by taking the conjugates of these expansions over the field Q((x)).

 > $\mathrm{puiseux}\left(f,x=0,y,3\right)$
 $\left\{\frac{{6163}}{{6418}}{+}\frac{{983148695281170}}{{6011940312110209}}{}{{x}}^{{2}}{,}\frac{{1089365505}}{{607721104}}{}{{x}}^{{2}}{+}{x}{}{\mathrm{RootOf}}{}\left({-}{73931}{+}{49304}{}{{\mathrm{_Z}}}^{{2}}\right){,}{-}{1}{-}\frac{{4746694785}}{{2532504976}}{}{{x}}^{{2}}{+}{x}{}{\mathrm{RootOf}}{}\left({50324}{}{{\mathrm{_Z}}}^{{2}}{-}{135641}\right)\right\}$ (1.2)

This command means the following: Give the Puiseux expansions up to accuracy 3, which means modulo ${x}^{3}$. So the coefficients of ${x}^{2}$ are given, but not the coefficients of ${x}^{3}$. To view the ${x}^{3}$ terms of the Puiseux expansions, we must compute the Puiseux expansions up to accuracy > 3.

 > $\mathrm{puiseux}\left(f,x=0,y,4\right)$
 $\left\{\frac{{6163}}{{6418}}{+}\frac{{983148695281170}}{{6011940312110209}}{}{{x}}^{{2}}{,}\frac{{6708313807195725}}{{9153721340698688}}{}{\mathrm{RootOf}}{}\left({-}{73931}{+}{49304}{}{{\mathrm{_Z}}}^{{2}}\right){}{{x}}^{{3}}{+}\frac{{1089365505}}{{607721104}}{}{{x}}^{{2}}{+}{x}{}{\mathrm{RootOf}}{}\left({-}{73931}{+}{49304}{}{{\mathrm{_Z}}}^{{2}}\right){,}{-}{1}{-}\frac{{149882029318321335}}{{285733439684206208}}{}{{x}}^{{3}}{}{\mathrm{RootOf}}{}\left({50324}{}{{\mathrm{_Z}}}^{{2}}{-}{135641}\right){-}\frac{{4746694785}}{{2532504976}}{}{{x}}^{{2}}{+}{x}{}{\mathrm{RootOf}}{}\left({50324}{}{{\mathrm{_Z}}}^{{2}}{-}{135641}\right)\right\}$ (1.3)

As one can see from the Puiseux expansions, the point $x=0$, $y=-1$ is singular, because two Puiseux expansions are going through this point:

and its conjugate. Similarly, $x=0,y=0$ is a singular point.