## A quick way to classify stationary points for any $f(x,y)$.

There are several methods of going about classifying the stationary points of a function $f(x,y)$. Here I would like to show a method that I prefer which involves minimal calculation. Let us look at the example of

\begin{equation}f(x,y)=2x^3+6xy^2-3y^3-150x\end{equation}

It should be obvious from the 3D plot above that there are numerous stationary points on our surface. We can find them by partially differentiating with respect to both $x$ and $y$ and set the derivatives to 0.

\begin{align}\pdv{f}{x}&=6x^2+6y^2-150=0\\ \pdv{f}{y}&=12xy-9y^2=0\end{align}

Through some painstaking of substituting back and forth numerous times, we can find four stationary points: (-3,-4), (3,4), (-5,0), and (5,0).

It is at this point we can use our nifty little trick with the Hessian matrix ($H$), where

$H=\bmqty{\frac{\partial^2f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y}\\ \frac{\partial^2 f}{\partial y \partial y} & \frac{\partial^2f}{\partial y^2}}$

An we can classify any stationary point by taking what is called a quadratic approximation. This come from the definition of a multivariable Taylor Series, and essentially what we want to do is to evaluate the sign of the second quadratic term of the series of our function.

$x^TH\vec{x}$

The above is the coefficient of of the quadratic term and is what we are after, where $\vec{x}=\mqty[x-x_0\\y-y_0]$. But multiplying everthing else is quite laborious and leaves a large room for arthmetic mistakes. There is, indeed, a simpler way. What we want to show is the ‘positiveness’ of the function at that point, so, if our expression above yields a positive definite expression, then, the parabola must be pointing up, which corresponds to a minimum. Conversely, if the expression evaluated at a point and yields a definite negative expression, we can say that it is a maximum, and if it is neither, it is a saddle point. All of this can be found using the trace and determinant of the Hessian matrix by examining its eigenvalues. I will spare you the un-enlightening algebraic proof of this, but if all the eigenvalues are positive, then we have a minimum, and if all of them are negative, then we have a maximum, and somewhere inbetween, a saddle point. Let’s take the point (-3,-4). It is simply using the properties of the trace and determinant of a matrix where:

\begin{align}\text{Tr } M = \lambda_1 + \lambda_2 \\ \det|M| = \lambda_1 \lambda_2\end{align}

Our Hessian matrix in explicit form is

$$H=\bmqty{12x & 12y \\ 12y & 12x-18y}$$

Evaluated at point $(-3,-4)$ gives:

$$H_{(-3,-4)}=\bmqty{-36 & -48 \\ -48 & 36}$$

The trace and the determinant of H at that point can be easily evaluated:

$$\text{Tr } H_{(-3,-4)}=0$$

$$\det|H_{(-3,-4)}|=-3600$$

The negative sign of the determinant is indicative that the two eigenvalues are different in signs, and this is confirmed by the trace. So as per our definition, point $(-3,-4)$ is a saddlepoint. You can speed up the process even more by simply deducing the signs of the determinant in your head! (Easier said then done, but for this matrix H, it is rather obvious that it will be negative.