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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:

[tex]f_{a}[/tex] and [tex]f_{a,a}[/tex] are the first and second partial derivatives of a, respectively.

[tex]\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}) + ...[/tex]

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: [tex]f_{x} and f_{y} = 0[/tex]

And so we are left with:

[tex]\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})[/tex]

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:

[tex](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})})[/tex]

From this, the discriminant can be used to determine local min, local max, saddle etc properties. If the discriminant, f_{x,y}^{2}- 4(1/2)f_{x,x}(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: need to rethink this.

Thanks, apologies this is long.

Nobahar.

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

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