Proving extrema using taylor series and Hessian Matrix

[dispiriton]

How do I use Taylor Series to show f(P) is a local maximum at a stationary point P if the Hessian matrix is negative definite.

I understand that some of the coefficients of the terms of the taylor series expansion are the coordinates of the Hessian matrix but for the f_xy term there is no conclusion and how does it affect the Taylor expansion to show it is a maximum?

 

[HallsofIvy]

Write the second order Taylor polynomial at P. It should be of the form
$$f(P)+ f_xx(P)(x- x_0)^2+ 2f_{xy}(x- x_0)(y- y_0)+ f_{yy}(y- y_0)^2$$

And if the Hessian is negative definite, you can write that as $f(P)$ minus a square.

 

[dispiriton]

Write the second order Taylor polynomial at P. It should be of the form
$$f(P)+ f_xx(P)(x- x_0)^2+ 2f_{xy}(x- x_0)(y- y_0)+ f_{yy}(y- y_0)^2$$

And if the Hessian is negative definite, you can write that as $f(P)$ minus a square.

But when the Hessian is negative definite we can only say that f_xx and f_yy are negative. Can the same be concluded for the f_xy term?

 

[HallsofIvy]

But when the Hessian is negative definite we can only say that f_xx and f_yy are negative.k

No, that is not true- I suggest you review the definitions of "positive definite" and "negative definite".

Can the same be concluded for the f_xy term?

 

[dispiriton]

No, that is not true- I suggest you review the definitions of "positive definite" and "negative definite".

From negative definite we can conclude f_xx and f_yy are negative but the same cannot be concluded for f_xy. Thus how do we conclude that when the Hessian is negative the expansion becomes f(P) minus something. Is it possible for the terms behind f(P) to add up to a positive number if f_xy is large (and positive)