I'm looking at the solution for finding maximum and minimum of

$f(x,y) = xye^{-x² -y^2}$

To do so, it first finds the critical points using the second derivative, and these points are $(0,0),({\pm {1\over\sqrt[]{2}}},{1\over\sqrt[]{2}}),({\pm {1\over\sqrt[]{2}}},{-{1\over\sqrt[]{2}}}) $

By inspection it finds that

$ f(0,0)=0 $

$ f({\ {1\over\sqrt[]{2}}},{1\over\sqrt[]{2}})=({-{1\over\sqrt[]{2}}},-{1\over\sqrt[]{2}})={e^{-1}\over 2} > 0 $

$ f(-{{1\over\sqrt[]{2}}},{1\over\sqrt[]{2}})=({{1\over\sqrt[]{2}}},-{1\over\sqrt[]{2}})={-{e^{-1}\over 2}} < 0 $

And it reasons that ${e^{-1}\over 2}, -{e^{-1}\over 2}$ are maximum and minimum.

It also uses the Hessian matrix to find out about $(0,0)$.

The determinant of $det(H(f)(0,0))=1$ but $f_{xx}=0$.

It then claims that because because the matrix is undefined but it also claims that $(0,0)$ is a saddle point.

So I have two questions:

Is it correct that if $det(H(f)(a,b)) > 0$ and $f_{xx}=0$ the second partial derivative test is inconclusive?

How can we claim that $(0,0)$ is a saddle point if the second derivative test shows that it is inconclusive?

  • 1
    $\begingroup$ I'm going to hazard a guess that, as with many test methods, when the result is inconclusive, the issue must be investigated by other means. If we write the function in polar form, we have $ \ f(r, \theta) \ = \frac{1}{2} r^2 e^{-r^2} \sin 2\theta \ , $ indicating that there is a directional dependence in the function near the origin, the value approaching zero from positive or negative values for various angles $ \ \theta \ $ . This would produce a "saddle point" at the origin. $\endgroup$ – colormegone May 29 '13 at 19:36
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    $\begingroup$ I think you can also "see" this looking directly at the function. $e^{-(x^2+y^1)}$ is always positive. Hence on the sets $\{x,y>0\}$ and $\{x,y<0\}$ we find that $f$ is positive. On the other sectors its negative. So the function has in every neighborhood of zero points that are below and above the tangent plane at $0$ $\endgroup$ – Quickbeam2k1 May 29 '13 at 20:23


Saddle point for sure. The Hessian of $f$ is: $$ H_f = \begin{pmatrix} 2x(2x^2-3) y e^{-x^2-y^2} & (2x^2-1)(2y^2-1) e^{-x^2-y^2} \\ (2x^2-1)(2y^2-1) e^{-x^2-y^2} & 2y(2y^2-3) x e^{-x^2-y^2} \end{pmatrix}, $$ thus $$ H_f(0,0) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, $$ and $\mathrm{det}(H_f(0,0)) = -1$ not $1$, which shows $(0,0)$ is a saddle point.

Now to answer your first question.

It is impossible for a smooth function that can have $\mathrm{det}(H_f(x_0,y_0)) > 0$ and $f_{xx}(x_0,y_0) = 0$ at $(x_0,y_0)$.

Let's say if $f_{xx}(x_0,y_0) = 0$, then $$ H_f(x_0,y_0) = \begin{pmatrix} 0 & f_{xy}(x_0,y_0) \\ f_{yx}(x_0,y_0) & f_{yy}(x_0,y_0) \end{pmatrix}. $$ Unless you construct some special smooth functions that bear the property that $f_{xy} \neq f_{yx}$, the determinant $\mathrm{det}(H_f(x_0,y_0)) = - (f_{xy}(x_0,y_0))^2 \leq 0$.

For the second question: when the test is inconclusive,

  • $f_{xx}$ and $f_{yy}$ have different signs, then we have a saddle point. You can look at the picture to see the geometric meaning of this (though in example they are zero): If our viewponint is somewhere on the $y$-axis and to observe the change in $x$, $f_{xx}>0$ means what we see is a concave up curve near the neighborhood of the point of interest $(x_0,y_0)$; Moving our viewponint to somewhere on the $x$-axis and to observe the change in $y$, $f_{yy}<0$ means what we see is a concave down curve. Then, clearly at $(x_0,y_0)$ we have a saddle.

  • $f_{xx}$ and $f_{yy}$ have the same signs, then we have a local maximum/minimum.

  • $\begingroup$ What a great answer! Thank you very very much. I double checked the determinant and you are right, my teacher's solution was wrong and the determinant is -1. $\endgroup$ – jjjx May 31 '13 at 8:32
  • $\begingroup$ @jjjx Hi, no problem, but I have one request from you, would you mind looking at this thread and learn how to upvote the answer you like and accept the answer you think is correct? meta.math.stackexchange.com/questions/3286/… Upvoting and accepting answers are the math.SE way of saying thanks. Also accepting answer will make your future question get good answer more likely. $\endgroup$ – Shuhao Cao Jun 4 '13 at 0:40

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