Hi, I posted a related question (about the 2nd derivative test) a while ago but now that I've read through some relevant theory I'm in a better position to find out about the following.(adsbygoogle = window.adsbygoogle || []).push({});

In the notes I have, the behaviour of a function [itex]f:U \subset R^n \to R[/itex] near a critical point point [itex]\mathop a\limits^ \to [/itex] is considered. To do this, a 'quadratic' approximation to f near a is considered.

[tex]

f\left( {\mathop a\limits^ \to + \mathop h\limits^ \to } \right) \approx f\left( {\mathop a\limits^ \to } \right) + \frac{1}{2}\left( {D_{\mathop h\limits^ \to } } \right)^2 f\left( {\mathop a\limits^ \to } \right)

[/tex]

Where D_h is the directional derivative of f in the direction of h.

Then there's some stuff on matrix representation, eigenvalues and Hessians. The important point is when the determinant of the Hessian is equal to zero in which case the test of the nature of the critical point is inconclusive.Consequently, there is a need to look at higher derivatives.

My questions are; why is a 'quadratic' approximation considered in stead of for instance a 'linear' or a 'quartic' approximation? Is it because the 2nd derivative test is supposed to use 2nd derivatives? Also, how does looking at higher derivatives (when the 2nd derivative test is inconclusive) help?

Does the math just work out? If the 2nd derivative test is inconclusive then how does considering higher derivatives help? If I use a 3rd degree taylor polynomial to approximate f near a then that's just a very rough approximation. How can I be sure about the nature of the stationary point from the approximation? Any help would be good thanks.

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# Critical points of functions

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