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Hessian matrix question.

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Given a function f: R^2 -> R of class C^3 with a critical point c.

    Why CANNOT the hessian matrix of f at point c be given by:

    1 -2
    2 3


    2. Relevant equations



    3. The attempt at a solution

    So first i want to clarify this.

    When it says f: R^2 -> R, that means the function is of two variables (x and y)?

    And when it says class C^3 that means the third derivative of the function exists and is continuous. So would a function be x^3 or x^4? the third derivative would be 24x for x^4 and is continuous. The third derivative of x^3 would be 6.

    I'm not sure about the answer..
     
  2. jcsd
  3. Nov 13, 2011 #2

    Ray Vickson

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    IF your matrix A above was a Hessian, what would the number a(2,2) = -2 represent? What would the number a(2,1) = +2 represent?

    RGV
     
  4. Nov 13, 2011 #3
    Ah ok. I think i got it. In the hessian which is given by

    fxx fxy

    fyx fyy

    fxy is not equal to fyx which should be the case for mixed partials?
     
  5. Nov 13, 2011 #4

    Ray Vickson

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    Yes, exactly.

    RGV
     
  6. Feb 27, 2012 #5
    I have a question considering the applicability of Hessian matrix.

    So, Can I use Hessian to prove that x^y > y^x whenever y > x >= e.

    At first I start by multiplying by ln() => y*ln(x) > x*ln(y)

    Is it enough, if I take g(x,y) such that g(x,y) = y*ln(x) - x*ln(y) > 0 and show det(H(g)) < 0 whenever y > x >= e?

    My purpose with this is to show that there are no real local or global critical points in g(x,y) when y > x >= e, and conclude that x^y - y^x diverges. I am not sure if I can use Hessian to draw that kind of conclusion.
     
  7. Feb 27, 2012 #6

    lanedance

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    Hi Viliperi, welcome to PF, please start a new thread if you have a question as opposed to resurrecting an old one. You're more likely to get an anser that way as well - thanks
     
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