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Second Derivative Test

  1. Aug 9, 2013 #1
    Under what conditions on the constants a and b does the second derivative test guarantee that the function

    g(x,y,z)=ax^2+2axz+by^2-2byz+z^2

    has a local maximum at (0,0,0)? a local minimum at (0,0,0)?

    well, i used the Hessian matrix to compute the eigenvalues to set them above zero. but the computation is so complicated that i used Mathematica to solve them and it turned out to be some messy stuff. i was convinced that this is not the way to do it. but then how?
     
  2. jcsd
  3. Aug 9, 2013 #2
    [STRIKE]Just looking at the problem makes me want to try Lagrange multipliers.[/STRIKE]
     
    Last edited: Aug 9, 2013
  4. Aug 9, 2013 #3

    micromass

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    Won't work here.

    They ask about the second derivative test, so you'll need to calculate the Hessian. It's what they ask for.
     
  5. Aug 9, 2013 #4

    Ray Vickson

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    Why would you try to set the eigenvalues to zero? The Hessian is positive-definite if all the eigenvalues are > 0 (so the origin is a global minimum---not just local!)

    The Hessian cannot be negative-definite, so the origin cannot be a local maximum. (Note: a matrix A is negative-definite if and only if its negative B = -A is positive-definite, and since B(3,3) = -2 < 0, B cannot be positive-definite---no fancy tests needed here!)

    To test for positive-definiteness of the Hessian H, just perform Cholesky decomposition; see, eg., http://en.wikipedia.org/wiki/Cholesky_decomposition . The Cholesky algorithm is easy to carry out in this case.
     
  6. Aug 10, 2013 #5
    thanks a lot.
     
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