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Homework Help: 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
    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.
     
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