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Homework Help: Why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

  1. Apr 16, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    For the function f(x, y) = xye^[-(x^2 + y^2)] find all the critical points and classify them each as a relative maximum, a relative minimum, or a saddle point.

    2. Relevant equations
    Partial differentiation and Hessian determinants.


    3. The attempt at a solution
    I get how to compute the derivatives. I also get how to compute the Hessian determinants. Basically, I get all the algebraic details but what I would like to ask about (at least for now) is why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

    Any input would be greatly appreciated!
    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Apr 16, 2012 #2

    Ray Vickson

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    Homework Helper

    The Hessian at (x0,y0) gives the quadratic terms in the Taylor expansion at that point. In other words, if we write
    [tex] f(x,y) \approx f(x_0,y_0) + a_0 (x-x_0) + b_0 (y-y_0)
    + \frac{1}{2} r_0 (x-x_0)^2 + s_0 (x-x_0)(y-y_0) + \frac{1}{2} t_0 (y-y_0)^2 + \text{ higher order terms },[/tex] the constants [itex] r_0, s_0, t_0[/itex] are the elements of the Hessian at (x0,y0); that is, the Hessian is
    [tex] H(x_0,y_0) = \left[ \matrix{r_0&s_0\\s_0&t_0} \right].[/tex] So, if the Hessian is [tex] H = \left[ \matrix{0&1\\1&0}\right], [/tex] that means that for small x and y we have [itex] f(x,y) \approx xy,[/itex] and if you plot the surface z = xy around (0,0), you will see that (0,0) is, indeed, a saddle point.

    Doesn't your textbook or your course notes have this material?

    RGV
     
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