# Taylor series and second derivative test: the degenerate case.

1. Oct 22, 2011

### nobahar

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:

$$f_{a}$$ and $$f_{a,a}$$ are the first and second partial derivatives of a, respectively.
$$\Delta f(x,y) = f_{x}(x-x_{0}) + f_{y}(y-y_{0}) + \frac{1}{2}f_{x,x}(x-x_{0})^2 + \frac{1}{2}f_{y,y}(y-y_{0})^2 + f_{x,y}(y-y_{0})(x-x_{0}) + ...$$

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: $$f_{x} and f_{y} = 0$$
And so we are left with:

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

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:

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

From this, the discriminant can be used to determine local min, local max, saddle etc properties. If the discriminant, fx,y2 - 4(1/2)fx,x(1/2)fy,y, equals zero. Then it is not possible to tell what is happening. This is because higher order terms are then important.

Edit: need to rethink this.

Thanks, apologies this is long.
Nobahar.

Last edited: Oct 22, 2011
2. Oct 24, 2011

### nobahar

I realise this is going to bump the thread, but I can no longer simply edit the post. Hopefully someone who could throw in some guidance may not have had the opportunity to see the thread, anyway.

Okay. So my issue with the second derivative test is: why does it not work when the discriminant is zero?

The quadratic is a ratio, x/y, and the discriminant identifies what values the ratio can take. When D<0, there are no roots to the quadratic, and the value of the coefficient on the x2 term determines whether the values are positive or negative, and therefore a local max or min. This is true for any x/y ratio, and that is why it is a local max or min, for which there are no degenerate directions (i.e. you can't move along the graph of the original function maintaining the same value; i.e., using a contour plot, there isn't a direction in which you can move to retain the same value for f(x,y)).
If D>0, then there are two roots: and the function D, can take on both positive and negative values, AND zero: since there are two roots. To me, this says that certain x/y ratios will lead to a decrease - the negative values; certain ratios will lead to an increase - the positive values; and that the roots correspond to the x/y ratios that yield a degenerate 'change' or no change. This is because the point is a saddle point, and there are directions in which you can travel in which you can retain the same value for the original function f(x,y). This is my understanding. Furthermore, there are two roots, because the ratios -x/y and x/-y are the same for x and y that yield the root values, and x/y and -x/-y are the same for the roots values.

When the D = 0, I get stuck. The quadratic is fundamentally no different from the others: instead of having no roots or two roots, it has one root. The only way I can make any progress is to argue that having one degenerate point means nothing more than that the quadratic polynomial with two variables, derived from the Taylor series, is only an approximation, and using only these terms belies what is happening as determined by higher order components of the Taylor series; which are not considered. So the function would appear to have a single degenerate direction – like a valley. This would mean neither a local maximum, minimum, nor saddle point as determined solely by the quadratic approximation of the change in the function, since a single degenerate ‘line’ means nothing. The presence of a saddle point, local max or min is sufficient to describe the behaviour of the function, and so higher terms are unnecessary, but when the quadratic term is insufficient to describe the behaviour, it appears degenerate. Additionally, I guess this would mean that the degenerate directions present at a saddle point are not necessarily degenerate, but appear that way due to it being determined by a quadratic approximation; much like the single degenerate ‘line’ yielded by the D = 0 case not actually being degenerate, and that higher order terms would reveal this.

I apologise if this is long. Any input is appreciated as this is annoying me no end! The internet has offered no solace.