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If we have a function $f: \mathbb R^3\rightarrow \mathbb R$ of the form:

$$f(x)=(a_1x_1^2+a_2x_2^2+a_3x_3^2)e^{-x_1^2-x_2^2-x_3^2}$$

where $a_1>a_2>a_3>0$.

We can find the critical points by:

$$\frac{\partial f}{\partial x_1} =2x_1e^{-x_1^2-x_2^2-x_3^2}(a_1-a_1x_1^2-a_2x_2^2-a_3x_3^2)=0$$

And solving symmetrically for $x_2$ and $x_3$. We get critical points: $(0,0,0) , (\pm 1,0,0) , (0,\pm 1, 0) , (0,0,\pm 1)$

We could determine the nature of them by looking at the eigenvalues of the Hessian, but is there a way to just analyse it directly? Obviously $(0,0,0)$ is a strict global minimum, but I can't quite see that the other points are. Any pointers/hints would be appreciated.

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  • $\begingroup$ Per your question it is more suitable to say "find extrema without using eigenvalue's"; a critical point is by definition a point of the zero set of a gradient. $\endgroup$ – Megadeth Oct 27 '15 at 23:33
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We have $f(x,y,z) \geq 0$ and $f(0,0,0)=0$ so $(0,0,0)$ is a global minimum. The other critical points give rise to the function values

$$\left\{\frac{a_1}{e}, \frac{a_2}{e}, \frac{a_3}{e}\right\}$$

Since $f$ is continuous and bounded on $\mathbb{R}^3$ and approaches $0$ as $||x||\to \infty$ it has a maximum point(s). Since $\frac{a_1}{e} > \frac{a_2}{e} > \frac{a_3}{e}$ we must have that $(\pm 1,0,0)$ is a global maximum.


Close to $(0,\pm 1, 0)$ we have

$$f(\epsilon,\pm 1,0) = \left(\frac{a_2}{e} + \frac{a_1}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_2}{e} + \frac{a_1-a_2}{e}\epsilon^2 > \frac{a_2}{e}$$ and $$f(0,\pm 1,\epsilon) = \left(\frac{a_2}{e} + \frac{a_3}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_2}{e} + \frac{a_3-a_2}{e}\epsilon^2 < \frac{a_2}{e}$$

so $(0,\pm 1, 0)$ are saddle points.


Close to $(0,0,\pm 1)$ we have

$$f(\epsilon,0,\pm 1) = \left(\frac{a_3}{e} + \frac{a_1}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_3}{e} + \frac{a_1-a_3}{e}\epsilon^2 > \frac{a_3}{e}$$

and since $x=1$ is a maximum for $x^2e^{-x^2}$ we have

$$f(0,0,\pm 1 \pm \epsilon) = a_3(1+\epsilon)^2e^{-(1+\epsilon)^2} < \frac{a_3}{e}$$

so $(0,0,\pm 1)$ are saddle points.

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  • $\begingroup$ Thanks, that's very helpful. One thing I don't quite understand, where does the $\frac{-a_1}{e}$ come from in this step $(\frac{a_2}{e}+\frac{a_1}{e}\epsilon^2)e^{-\epsilon^2}\approx \frac{a_2}{e}+\frac {a_1-a_2}{e}\epsilon^2$? And how can we tell that the approximation is close enough that it is $>\frac{a_2}{e}$? $\endgroup$ – James Oct 28 '15 at 1:08
  • $\begingroup$ @James It's sloppy notation on my part. I'm assuming $\epsilon$ is small so we can neglect terms of order $\epsilon^3$ and larger in the Taylor expansion of $e^{-\epsilon^2} = 1 - \epsilon^2 + \frac{\epsilon^4}{2} + \ldots$. The $-a_1$ term comes from multiplying in the Taylor expansion above. If you do it more properly you will get something like $\frac{a_3}{e} + \frac{a_1-a_3}{e}\epsilon^2 + O(\epsilon^4)$. We can always choose $\epsilon$ small enough so that the $O(\epsilon^4)$ term is negligible. $\endgroup$ – Winther Oct 28 '15 at 1:34
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Other than $(0,0,0)$, which as you point out is a strict global minimum, the other critical points all lie on the unit sphere $S^2 \subset \mathbb{R}^3$. Restricted to that sphere, the exponential factor is just a constant $e^{-1}$. The critical points of the restricted function $g: S^2 \to \mathbb{R}$, $g(x_1,x_2,x_3) = \frac{1}{e}(a_1x_1^2 + a_2x_2^2 + a_3x_3^2)$ should be pretty easy to classify. Then you remember that all of those points are at a maximum with respect to the radial dimension.

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