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
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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
 

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

1. Why is the Hessian determinant negative at a saddle point?

The Hessian determinant is a measure of the curvature of a function at a specific point. At a saddle point, the function has both positive and negative curvature in different directions, resulting in a negative Hessian determinant.

2. How does the Hessian determinant affect the behavior of a function at a saddle point?

The Hessian determinant determines the type of critical point a function has. A negative Hessian determinant at a critical point indicates a saddle point, where the function neither reaches a maximum nor a minimum and instead changes direction.

3. What does the Hessian determinant tell us about the second-order derivatives of a function at a saddle point?

The Hessian determinant is calculated using the second-order derivatives of a function at a specific point. A negative Hessian determinant at a saddle point indicates that the second-order derivatives have opposite signs, further supporting the idea of changing direction at that point.

4. Why is P(0, 0) considered a saddle point if the Hessian determinant equals -1?

At a saddle point, the Hessian determinant must be negative. If P(0, 0) is a point on the function where the Hessian determinant is -1, it meets the criteria for a saddle point.

5. Can the Hessian determinant be used to determine the stability of a critical point?

Yes, the Hessian determinant can be used to determine the stability of a critical point. A negative Hessian determinant at a critical point indicates an unstable point, where small perturbations can result in large changes in the function value.

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