Register to reply

Nature of multivariate points

by ibysaiyan
Tags: multivariate, nature, points
Share this thread:
ibysaiyan
#1
Jan20-12, 11:21 AM
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

[itex] f(x,y) = x^4 +(2x^2)y -4x^2 +3y^2 [/itex]

[itex]F_{x}[/itex] = [itex]4x^3 +4xy-8x [/itex]
[itex]F_{y}[/itex] =[itex]2x^2+6y [/itex]
[itex]F_{xx}[/itex]12x^2 +4y -8
[itex]F_{yy}[/itex]6
[itex]F_{yx}[/itex]4x

The definition which I have used for delta/ determinant is : If Δ > 0 then stationary points are saddle i.e [itex]f_{xy}^2[/itex] - [itex]f_{xx}[/itex] * [itex]f_{yy}[/itex]

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!
Phys.Org News Partner Science news on Phys.org
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
HallsofIvy
#2
Jan20-12, 12:12 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,533
Quote Quote by ibysaiyan View Post
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

[itex] f(x,y) = x^4 +(2x^2)y -4x^2 +3y^2 [/itex]

[itex]F_{x}[/itex] = [itex]4x^3 +4xy-8x [/itex]
[itex]F_{y}[/itex] =[itex]2x^2+6y [/itex]
[itex]F_{xx}[/itex]12x^2 +4y -8
[itex]F_{yy}[/itex]6
[itex]F_{yx}[/itex]4x

The definition which I have used for delta/ determinant is : If Δ > 0 then stationary points are saddle i.e [itex]f_{xy}^2[/itex] - [itex]f_{xx}[/itex] * [itex]f_{yy}[/itex]
You have this exactly backwards- the stationary points are saddles if Δ< 0.
The point is that [itex]f_{xy}^2- f_{xx}f_{yy}[/itex] is the determinant of the "second derivative matrix"
[tex]\begin{bmatrix}f_{xx}& f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}[/tex]
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
[tex]\begin{bmatrix}f_{xx} & 0 \\ 0 & f_{yy}\end{bmatrix}[/tex]
which means that, locally, it is like [itex]f_{xx}x^2+ f_{yy}y^2[/itex]. 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!
ibysaiyan
#3
Jan20-12, 02:40 PM
P: 442
Quote Quote by HallsofIvy View Post
You have this exactly backwards- the stationary points are saddles if Δ< 0.
The point is that [itex]f_{xy}^2- f_{xx}f_{yy}[/itex] is the determinant of the "second derivative matrix"
[tex]\begin{bmatrix}f_{xx}& f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}[/tex]
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
[tex]\begin{bmatrix}f_{xx} & 0 \\ 0 & f_{yy}\end{bmatrix}[/tex]
which means that, locally, it is like [itex]f_{xx}x^2+ f_{yy}y^2[/itex]. 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).


Register to reply

Related Discussions
Ordinary points, regular singular points and irregular singular points Differential Equations 3
Find the critical points of (x^2+y^2-4)(x+y) and their nature Calculus & Beyond Homework 8
Uncertainty Principle - nature or human nature? Quantum Physics 18
Objects without quantum nature, or without relativistic nature General Physics 5