Quick question about use of the Hessain for critical points

In summary, the conversation discusses solving for the critical points of a function and evaluating them using the Hessian matrix. It is mentioned that there may be a typo in the original problem and that the Hessian matrix is already diagonal. It is also mentioned that the z coordinate may need to have a variable in order for the problem to work out correctly.
  • #1
trap101
342
0
So I am solving for the critical points of x3-3x-y2+9y+z2



I've found the critical points and I'm just evaluating them in the Hessain. Now I computed the eigenvalues for the hessian, but one of my 2nd order derivatives was a constant. So my hessian looked like this (it was diagonal, so I'm only writing the diagonal terms)

(6x - λ), (-6y-λ), (2-λ)

Now all my C.P were of the form (±1,±√3,0)

So whatever eigenvalues I solve for I am always going to get λ=2, but in the solutions they said that a maximum existed at (-1,√3,0) but if I solve for the eigenvalues I will have λ = -6, λ = -6√3, λ = 2. So how is this a local max if they are not all negative. (long winded for short explanation I know)
 
Physics news on Phys.org
  • #2
trap101 said:
So I am solving for the critical points of x3-3x-y2+9y+z2...



...Now all my C.P were of the form (±1,±√3,0)...

Is there a typo? I get y=9/2.
 
  • #3
y2 is supposed to be y3...my bad
 
  • #4
trap101 said:
So I am solving for the critical points of x3-3x-y2+9y+z2
I've found the critical points and I'm just evaluating them in the Hessain. Now I computed the eigenvalues for the hessian, but one of my 2nd order derivatives was a constant. So my hessian looked like this (it was diagonal, so I'm only writing the diagonal terms)

(6x - λ), (-6y-λ), (2-λ)

Now all my C.P were of the form (±1,±√3,0)

So whatever eigenvalues I solve for I am always going to get λ=2, but in the solutions they said that a maximum existed at (-1,√3,0) but if I solve for the eigenvalues I will have λ = -6, λ = -6√3, λ = 2. So how is this a local max if they are not all negative. (long winded for short explanation I know)

I think you're right, they may have just over looked the z contribution, could be a typo. Notice contribution from z has no local max, so I don't think there will be a local max anywhere in ℝ3.

Also, notice hessian is already diagonal, (6x,-6y,2), so it's not necessary to use eigenvalues to diagonalize it or find principal accelerations. In a sense, you're principal directions or eigenvectors are (dx,0,0), (0,dy,0), (0,0,dz). (Where we consider xTHx.) So x>0 and y<0 if and only if we have a local minimum (at a critical point).
 
Last edited:
  • #5
Ok. So that z co-ordinate should've had some sort of variable in it in order for the problem to "work out" the way it should, but for the most part I was on the right track. Thanks
 

Related to Quick question about use of the Hessain for critical points

1. What is the Hessain method for finding critical points?

The Hessain method is a mathematical technique used to identify critical points in a function. It involves calculating the second-order partial derivatives of the function and using them to determine whether a point is a maximum, minimum, or saddle point.

2. How is the Hessain matrix used in finding critical points?

The Hessain matrix, which is the matrix of second-order partial derivatives, is used to determine the nature of a critical point. If the determinant of the Hessain matrix is positive, the point is a minimum. If it is negative, the point is a maximum. If it is zero, further analysis is needed to determine the nature of the point.

3. Can the Hessain method be used for all types of functions?

Yes, the Hessain method can be used for all types of functions, including multivariable functions and functions with constraints. However, the calculations may become more complicated for higher-dimensional functions.

4. What are the advantages of using the Hessain method for finding critical points?

The Hessain method is a systematic and efficient way to identify critical points in a function. It also provides information about the nature of the critical point, allowing for further analysis of the function. Additionally, the method can be used for functions with multiple variables, making it a versatile tool for optimization problems.

5. Are there any limitations to using the Hessain method for critical points?

One limitation of the Hessain method is that it may not be applicable for functions with discontinuities or undefined points. Additionally, the method may become computationally intensive for higher-dimensional functions. In some cases, other methods such as gradient descent may be more suitable for finding critical points.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
873
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
25
Views
458
Back
Top