1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding singular points of a non-algebraic curve.

  1. Sep 18, 2012 #1
    Let [itex]F : \mathbb{R}^2 \rightarrow \mathbb{R}^2[/itex] be the map given by [itex]F(x, y) := (x^3 - xy, y^3 - xy)[/itex]. What are some singular points?

    Well, I know that for an algebraic curve, a point [itex]p_0 = (x_0, y_0)[/itex] is a singular point if [itex]F_x(x_0, y_0) = 0[/itex] and [itex]F_y(x_0, y_0) = 0[/itex].

    However, this curve is not algebraic, so I'm not sure if that still applies. If it does, then

    [itex]F_x(x, y) = (3x^2 - y, -y) = (0, 0)[/itex] and [itex]F_y(x, y) = (-x, 3y^2 - x) = (0, 0)[/itex] at the point [itex]p_0 = (0, 0)[/itex]

    Is that the correct way of determining the singular points? Are there any others?

    I graphed it in Mathematica.

    Untitled_zps5e89a03b.png
     
    Last edited: Sep 18, 2012
  2. jcsd
  3. Sep 18, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You don't really have a curve here, but rather something like an algebraic surface. Does your text say how singular points are defined in a surface?? There should probably be a determinant condition.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding singular points of a non-algebraic curve.
Loading...