# Nature of multivariate points

by ibysaiyan
Tags: multivariate, nature, points
 P: 442 1. The problem statement, all variables and given/known data Hi, I have been given a set of functions for which I need to find the stationary points , and determine whether the points are saddle, or max/min. I think I may have solved it correctly but I end up with all the points being saddle, surely this can't be right.. I may have gone wrong with my arithmetic. Can anyone go through my working. Appreciate the replies. 2. Relevant equations 3. The attempt at a solution $f(x,y) = x^4 +(2x^2)y -4x^2 +3y^2$ $F_{x}$ = $4x^3 +4xy-8x$ $F_{y}$ =$2x^2+6y$ $F_{xx}$12x^2 +4y -8 $F_{yy}$6 $F_{yx}$4x The definition which I have used for delta/ determinant is : If Δ > 0 then stationary points are saddle i.e $f_{xy}^2$ - $f_{xx}$ * $f_{yy}$ The points which I get are the following: Re arranging (eq.1) and using eq. 2 (2x^2 = -6y) 4x^3 +4xy-8x (eq.1) => x(4x^2 +4y-8) = 0 x[ (2x^2 +2x^2 ]+4y-8 = 0 x[ (-6y-6y) +4y -8] =0 x(-12+4y-8) = 0 x = 0 , -8y = 8 , y=-1 Points are: (0,0) , (+/√3, -1) Thanks!
Math
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P: 39,682
 Quote by ibysaiyan 1. The problem statement, all variables and given/known data Hi, I have been given a set of functions for which I need to find the stationary points , and determine whether the points are saddle, or max/min. I think I may have solved it correctly but I end up with all the points being saddle, surely this can't be right.. I may have gone wrong with my arithmetic. Can anyone go through my working. Appreciate the replies. 2. Relevant equations 3. The attempt at a solution $f(x,y) = x^4 +(2x^2)y -4x^2 +3y^2$ $F_{x}$ = $4x^3 +4xy-8x$ $F_{y}$ =$2x^2+6y$ $F_{xx}$12x^2 +4y -8 $F_{yy}$6 $F_{yx}$4x The definition which I have used for delta/ determinant is : If Δ > 0 then stationary points are saddle i.e $f_{xy}^2$ - $f_{xx}$ * $f_{yy}$
You have this exactly backwards- the stationary points are saddles if Δ< 0.
The point is that $f_{xy}^2- f_{xx}f_{yy}$ is the determinant of the "second derivative matrix"
$$\begin{bmatrix}f_{xx}& f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}$$
of course that determinant is independent of the coordinate system- and since this is a symmetric matrix, there exist a coordinate system in which it is diagonal. That is there exist a coordinate system in which the matrix is
$$\begin{bmatrix}f_{xx} & 0 \\ 0 & f_{yy}\end{bmatrix}$$
which means that, locally, it is like $f_{xx}x^2+ f_{yy}y^2$. If the determinant is negative, one of those coefficients is positive, the other negative, a saddle point. If the determinant is positive, the coefficients are either both positive, a local minimum, or both negative, a local maximum.

 The points which I get are the following: Re arranging (eq.1) and using eq. 2 (2x^2 = -6y) 4x^3 +4xy-8x (eq.1) => x(4x^2 +4y-8) = 0 x[ (2x^2 +2x^2 ]+4y-8 = 0 x[ (-6y-6y) +4y -8] =0 x(-12+4y-8) = 0 x = 0 , -8y = 8 , y=-1 Points are: (0,0) , (+/√3, -1) Thanks!
P: 442
 Quote by HallsofIvy You have this exactly backwards- the stationary points are saddles if Δ< 0. The point is that $f_{xy}^2- f_{xx}f_{yy}$ is the determinant of the "second derivative matrix" $$\begin{bmatrix}f_{xx}& f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}$$ of course that determinant is independent of the coordinate system- and since this is a symmetric matrix, there exist a coordinate system in which it is diagonal. That is there exist a coordinate system in which the matrix is $$\begin{bmatrix}f_{xx} & 0 \\ 0 & f_{yy}\end{bmatrix}$$ which means that, locally, it is like $f_{xx}x^2+ f_{yy}y^2$. If the determinant is negative, one of those coefficients is positive, the other negative, a saddle point. If the determinant is positive, the coefficients are either both positive, a local minimum, or both negative, a local maximum.
Hi there,
You see on my lecturer's note I can clearly see that he has stated that when a coordinate system has Δ >0 then it's a saddle point , he did mention that " some books use the expression the other way around" , which I have confirmed upon browsing.

Could he be wrong? Also thanks for your in-depth answer ( as always).

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