# What are the extrema and saddle points of f(x, y) = xye^(-x^2-y^2)?

May 3, 2018

$\left(0 , 0\right)$ is a saddle point

$\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ and $\left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}}\right)$ are local maxima

$\left(\frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}}\right)$ and $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ are local minima

$\left(0 , \pm \frac{1}{\sqrt{2}}\right)$ and $\left(\pm \frac{1}{\sqrt{2}} , 0\right)$ are points of inflection.

#### Explanation:

For a general function $F \left(x , y\right)$ with a stationary point at $\left({x}_{0} , {y}_{0}\right)$ we have the Taylor series expansion

F(x_0+xi,y_0+eta) = F(x_0,y_0) + 1/(2!)(F_{x x} xi^2+F_{yy}eta^2+2F_{xy}xi eta ) +ldots
For the function

$f \left(x\right) = x y \setminus {e}^{- {x}^{2} - {y}^{2}}$

we have

$\frac{\partial f}{\partial x} = y {e}^{- {x}^{2} - {y}^{2}} + x y \left(- 2 x\right) {e}^{- {x}^{2} - {y}^{2}}$
$q \quad = y \left(1 - 2 {x}^{2}\right) {e}^{- {x}^{2} - {y}^{2}}$

$\frac{\partial f}{\partial y} = x {e}^{- {x}^{2} - {y}^{2}} + x y \left(- 2 y\right) {e}^{- {x}^{2} - {y}^{2}}$
$q \quad = x \left(1 - 2 {y}^{2}\right) {e}^{- {x}^{2} - {y}^{2}}$

It is easy to see that the both the first derivatives vanish at the following ponrs

• $\left(0 , 0\right)$
• $\left(0 , \pm \frac{1}{\sqrt{2}}\right)$
• $\left(\pm \frac{1}{\sqrt{2}} , 0\right)$
• $\left(\pm \frac{1}{\sqrt{2}} , \pm \frac{1}{\sqrt{2}}\right)$

To examine the nature of these stationary points, we need to look at the behavior of the second derivatives there.

Now

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = y \left(- 4 x\right) {e}^{- {x}^{2} - {y}^{2}} + y \left(1 - 2 {x}^{2}\right) \left(- 2 x\right) {e}^{- {x}^{2} - {y}^{2}}$
$q \quad = x y \left(4 {x}^{2} - 6\right) {e}^{- {x}^{2} - {y}^{2}}$

and similarly

$\frac{{\partial}^{2} f}{\partial {y}^{2}} = x y \left(4 {y}^{2} - 6\right) {e}^{- {x}^{2} - {y}^{2}}$

and

$\frac{{\partial}^{2} f}{\partial x \partial y} = \left(1 - 2 {y}^{2}\right) {e}^{- {x}^{2} - {y}^{2}} + x \left(1 - 2 {y}^{2}\right) \left(- 2 x\right) {e}^{- {x}^{2} - {y}^{2}}$
$q \quad = \left(1 - 2 {x}^{2} - 2 {y}^{2} + 4 {x}^{2} {y}^{2}\right) {e}^{- {x}^{2} - {y}^{2}}$
$q \quad = \left(1 - 2 {x}^{2}\right) \left(1 - 2 {y}^{2}\right) {e}^{- {x}^{2} - {y}^{2}}$

So for $\left(0 , 0\right)$ we have $\frac{{\partial}^{2} f}{\partial {x}^{2}} = \frac{{\partial}^{2} f}{\partial {y}^{2}} = 0$ and $\frac{{\partial}^{2} f}{\partial x \partial y} = 1$ - hence

$f \left(0 + \xi , 0 + \eta\right) = f \left(0 , 0\right) + \xi \eta = \xi \eta$

If you approach $\left(0 , 0\right)$ along the line $x = y$, this this becomes

$f \left(0 + \xi , 0 + \xi\right) = {\xi}^{2}$

and so $\left(0 , 0\right)$ is obviously a minimum if you approach from this direction. On the other hand, if you approach along the line $x = - y$ we have

$f \left(0 + \xi , 0 - \xi\right) = - {\xi}^{2}$

and so $\left(0 , 0\right)$ is a maximum along this direction,

Thus $\left(0 , 0\right)$ is a saddle point.

For $\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ it is easily seen that

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = \frac{{\partial}^{2} f}{\partial {y}^{2}} = - 2 {e}^{- \frac{1}{2}} < 0$ and $\frac{{\partial}^{2} f}{\partial x \partial y} = 0$

which means that

$f \left(\frac{1}{\sqrt{2}} + \xi , \frac{1}{\sqrt{2}} + \eta\right) = f \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right) - {e}^{- \frac{1}{2} \left({\xi}^{2} + {\eta}^{2}\right)}$

So, the function decreases whichever way you move away from $\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ and this is a local maximum. It is easily seen that the same goes for $\left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}}\right)$ (this should have been obvious, since the function stays the same under $\left(x , y\right) \to \left(- x , - y\right)$!

Again, for both $\left(\frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}}\right)$ and $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ we have

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = \frac{{\partial}^{2} f}{\partial {y}^{2}} = 2 {e}^{- \frac{1}{2}} > 0$ and $\frac{{\partial}^{2} f}{\partial x \partial y} = 0$

So, both these points are local minima.

The four points $\left(0 , \pm \frac{1}{\sqrt{2}}\right)$ and $\left(\pm \frac{1}{\sqrt{2}} , 0\right)$ are more problematic - since all second order derivatives vanish at these points. We have to now look at higher order derivatives. Fortunately, we do not really need to work very hard for this - the very next derivative yields

$\frac{{\partial}^{3} f}{\partial {x}^{3}} = - 2 y \left(3 - 12 {x}^{2} + 4 {x}^{4}\right) {e}^{- {x}^{2} - {y}^{2}}$

which is non-zero for both $\left(0 , \pm \frac{1}{\sqrt{2}}\right)$ and $\left(\pm \frac{1}{\sqrt{2}} , 0\right)$. Now, this means that, for example

$f \left(0 + \xi , \frac{1}{\sqrt{2}}\right) = f \left(0 , \frac{1}{\sqrt{2}}\right) + \frac{1}{3} {\left(\frac{{\partial}^{3} f}{\partial {x}^{3}}\right)}_{\left(0 , \frac{1}{\sqrt{2}}\right)} {\xi}^{3} + \ldots$

which shows that this will increase from $f \left(0 , \frac{1}{\sqrt{2}}\right)$ in one direction, and decrease from it in the other. Thus $\left(0 , \frac{1}{\sqrt{2}}\right)$ is a **point of inflection. The same argument works for the other three points.