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Q on Second partial derivative test for functions of n variables

  1. Dec 15, 2011 #1
    Hi, I would like to confirm that I have understood this correctly.

    The steps to find local maxima/minima of a function f(x1, ... , xn) are:

    1) We find all the stationary points.
    2) We form the Hessian matrix and calculate the determinants D1, D2.... Dn for a stationary point P we want to check.
    3) We have the following cases:

    i) if Di > 0 for i = 1 to n then P is definately a local minimum point
    ii) if Di*(-1)^i > 0 for i = 1 to n then P is definately a local maximum point
    iii) if Dn = 0 this test cannot help us determine whether the point is a local minimum or maximum
    iv) in ALL other cases (for example Di = 0 for i other than n or Di with sign other than what i and ii indicate) we definately have a saddle point.

    Are iii and iv correct? More specifically, I would like a clarification on what exactly happens when we have one or more zero Di.

    Thanks in advance for your time.
  2. jcsd
  3. Dec 24, 2011 #2
    what you have as the conditions is right. When the determinant is zero, the test fails. This is to say that we might have a maximum, minimum or saddle but the test will not help us identify such a critical point. You might have to plot the function or see some values to be able to identify critical points. My old calculus book had this question which you might want to do and see what happens when the test fails:
    [tex] f(x,y)=6xy^{2}-2x^{3}-3y^{4} [/tex]

    You can plot the function in Mathematica, Matlab, Maple or any programs and see the behavior. Another way to be able to see this is to see the proof for why these conditions come about and see why the test fails. I hope this was clear.
  4. Dec 24, 2011 #3
    Thanks! I wanted a confirmation that I got the conditions right. I will study the example you gave; I just plotted it with wolfram alpha and it seems interesting!
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