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

In summary, the conversation discussed finding critical points and classifying them as relative maximums, relative minimums, or saddle points for the function f(x, y) = xye^[-(x^2 + y^2)]. The homework equations used were partial differentiation and Hessian determinants. The person asking the question understood how to compute the derivatives and Hessian determinants, but was unsure about the implication of a Hessian determinant of -1 at the point P(0,0). The expert explained that the Hessian at a point gives the quadratic terms in the Taylor expansion at that point, and a Hessian of [0 1; 1 0] at (0,0) would result in a saddle point on
  • #1
s3a
818
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Homework Statement


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.

Homework Equations


Partial differentiation and Hessian determinants.


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!
 

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  • #2
s3a said:

Homework Statement


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.

Homework Equations


Partial differentiation and Hessian determinants.


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!

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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