Second order partial derivatives, the Taylor series and the degenerate case.

Hello!

I am wondering if someone could let me know if my understanding is right or wrong. The Taylor series gives the function in the form of a sum of an infinite series. From this an approximation of the change in the function can be derived:

$\displaystyle f_{a}$ and $\displaystyle f_{a,a}$ are the first and second partial derivatives of a, respectively.

$\displaystyle \Delta f(x,y) = f_{x}(x-x_{0}) + f_{y}(y-y_{0}) + \frac{1}{2}f_{x,x}(x-x_{0})^2 + \frac{1}{2}f_{y,y}(y-y_{0})^2 + f_{x,y}(y-y_{0})(x-x_{0}) + ...$

The first few terms in the Taylor series (those that appear above), can be used to make an approximation of the change in function. At a critical point: $\displaystyle f_{x} and f_{y} = 0$

And so we are left with:

$\displaystyle \Delta f(x,y) = \frac{1}{2}f_{x,x}(x-x_{0})^2 + \frac{1}{2}f_{y,y}(y-y_{0})^2 + f_{x,y}(y-y_{0})(x-x_{0})$

Which is a quadratic approximation of the change in the function at a critical point.

For me, its easier to understand what is happening by factoring out y:

$\displaystyle (y-y_{0})^2 (\frac{1}{2}f_{x,x}\frac{(x-x_{0})^2}{(y-y_{0})^2} + \frac{1}{2}f_{y,y} + f_{x,y}\frac{(x-x_{0})}{(y-y_{0})})$

From this, the discriminant can be used to determine local min, local max, saddle etc properties. If the discriminant, $\displaystyle f_{x,y})^2 - 4\frac{1}{2}f_{x,x}\frac{1}{2}f_{y,y}$, equals zero. Then it is not possible to tell what is happening. This is because higher order terms are then important.

Edit: realised attempt at an explanation is wrong.

Thanks in advance.