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Homework Help: Nature of multivariate points

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi,

    I have been given a set of functions for which I need to find the stationary points , and determine whether the points are saddle, or max/min.
    I think I may have solved it correctly but I end up with all the points being saddle, surely this can't be right.. I may have gone wrong with my arithmetic. Can anyone go through my working. Appreciate the replies.


    2. Relevant equations




    3. The attempt at a solution

    [itex] f(x,y) = x^4 +(2x^2)y -4x^2 +3y^2 [/itex]

    [itex]F_{x}[/itex] = [itex]4x^3 +4xy-8x [/itex]
    [itex]F_{y}[/itex] =[itex]2x^2+6y [/itex]
    [itex]F_{xx}[/itex]12x^2 +4y -8
    [itex]F_{yy}[/itex]6
    [itex]F_{yx}[/itex]4x

    The definition which I have used for delta/ determinant is : If Δ > 0 then stationary points are saddle i.e [itex]f_{xy}^2[/itex] - [itex]f_{xx}[/itex] * [itex]f_{yy}[/itex]

    The points which I get are the following:
    Re arranging (eq.1) and using eq. 2 (2x^2 = -6y)
    4x^3 +4xy-8x (eq.1)
    =>
    x(4x^2 +4y-8) = 0
    x[ (2x^2 +2x^2 ]+4y-8 = 0
    x[ (-6y-6y) +4y -8] =0
    x(-12+4y-8) = 0

    x = 0 , -8y = 8 , y=-1
    Points are:
    (0,0) , (+/√3, -1)

    Thanks!
     
  2. jcsd
  3. Jan 20, 2012 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You have this exactly backwards- the stationary points are saddles if Δ< 0.
    The point is that [itex]f_{xy}^2- f_{xx}f_{yy}[/itex] is the determinant of the "second derivative matrix"
    [tex]\begin{bmatrix}f_{xx}& f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}[/tex]
    of course that determinant is independent of the coordinate system- and since this is a symmetric matrix, there exist a coordinate system in which it is diagonal. That is there exist a coordinate system in which the matrix is
    [tex]\begin{bmatrix}f_{xx} & 0 \\ 0 & f_{yy}\end{bmatrix}[/tex]
    which means that, locally, it is like [itex]f_{xx}x^2+ f_{yy}y^2[/itex]. If the determinant is negative, one of those coefficients is positive, the other negative, a saddle point. If the determinant is positive, the coefficients are either both positive, a local minimum, or both negative, a local maximum.

     
    Last edited by a moderator: Jan 20, 2012
  4. Jan 20, 2012 #3
    Hi there,
    You see on my lecturer's note I can clearly see that he has stated that when a coordinate system has Δ >0 then it's a saddle point , he did mention that " some books use the expression the other way around" , which I have confirmed upon browsing.

    Could he be wrong? Also thanks for your in-depth answer ( as always).
     
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