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Stationary points

Petrus

Well-known member
Feb 21, 2013
739
Find all starionary points to function \(\displaystyle f(x,y,z)=x^2+xy^2+yz^2-2z\) and Also decide it character.
I find a point at \(\displaystyle (-1/2,1,1)\) now I have to check if it is semidefinit etc.. But I only know the formula for 2 variable \(\displaystyle Q(h,k)\) I try search this method in My book and cant find it.. What you call this method and what is \(\displaystyle Q(h,k,l)\) I indeed need to read about this one so feel free if you got a Link!

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Find all starionary points to function \(\displaystyle f(x,y,z)=x^2+xy^2+yz^2-2z\) and Also decide it character.
I find a point at \(\displaystyle (-1/2,1,1)\) now I have to check if it is semidefinit etc.. But I only know the formula for 2 variable \(\displaystyle Q(h,k)\) I try search this method in My book and cant find it.. What you call this method and what is \(\displaystyle Q(h,k,l)\) I indeed need to read about this one so feel free if you got a Link!

Regards,
\(\displaystyle |\pi\rangle\)
Hi Petrus!

Here's a link to the Second partial derivative test.

To apply it, you would need to determine the eigenvalues of the matrix with all second partial derivatives (the "Hessian matrix").
I kind of suspect you have not learned yet what eigenvalues are, have you?

Alternatively, you can pick a couple of points around your stationary point.
If you can find one where f(x,y,z) is positive and another where f(x,y,z) is negative, you can draw the conclusion that it's neither a minimum nor a maximum: it's a saddle point.
Can you find 2 such points?
 

Petrus

Well-known member
Feb 21, 2013
739
Hi Petrus!

Here's a link to the Second partial derivative test.

To apply it, you would need to determine the eigenvalues of the matrix with all second partial derivatives (the "Hessian matrix").
I kind of suspect you have not learned yet what eigenvalues are, have you?

Alternatively, you can pick a couple of points around your stationary point.
If you can find one where f(x,y,z) is positive and another where f(x,y,z) is negative, you can draw the conclusion that it's neither a minimum nor a maximum: it's a saddle point.
Can you find 2 such points?
Hmm.. I Dont know about the hessian matrix, this is from Taylor formula which I am searching for and it's for 2 variable that I got and I need 3.
\(\displaystyle Q(h,k)=f_{xx}(x_0,y,0)h^2+2f_{xy}(x_0,y_0)hk+fyy(x_0,y_0)k^2\) And you look for positiv,negativ,semi-definit. Or LET me post the soloution to this problem( it's an old exam and it's on swedish but hopefully it says something with the math)


does it say something?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Yes.
All those second derivatives form the so called Hessian matrix $H$ in the stationary point.
$$H = \begin{bmatrix}
f_{xx} & f_{xy} & f_{xz} \\
f_{yx} & f_{yy} & f_{yz} \\
f_{zx} & f_{zy} & f_{zz}
\end{bmatrix}$$

They are testing if it is positive-definite (global minimum), negative-definite (global maximum), can be both positive and negative (saddle point), or something else (undecided).

The matrix is positive-definite if for all h, k, l (not all equal to zero) holds:
$$\begin{bmatrix}h&k&l\end{bmatrix}
\begin{bmatrix}
f_{xx} & f_{xy} & f_{xz} \\
f_{yx} & f_{yy} & f_{yz} \\
f_{zx} & f_{zy} & f_{zz}
\end{bmatrix}
\begin{bmatrix}h\\k\\l\end{bmatrix}
> 0$$

This expression is your $Q(h,k,l)$, so:
$$Q(h,k,l) = f_{xx} h^2 + f_{yy}k^2 + f_{zz}l^2 + 2f_{xy} hk + 2f_{xz}hl + 2f_{yz}$$
If you can prove it is always positive, you have a global minimum.
If you can find values for $h,k,l$ such that $Q(h,k,l)$ is positive, and also values such that $Q(h,k,l)$ is negative, you have a saddle point.

Btw, saying a matrix is positive-definite comes out the same as saying that all eigenvalues are positive.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Vad roligt! (Sun)

Maths in Swedish... That takes me back!
 

Petrus

Well-known member
Feb 21, 2013
739
Btw, saying a matrix is positive-definite comes out the same as saying that all eigenvalues are positive.
and if all eigenvalues are negative then its negative definit? and if eigenvalues got positiv and negative its indefinit right?
I dont think we are supposed to use that trick as they want us to complete the square and they never mention anything about eigenvalues but I know how to calculate it as I am learning linear algebra 2, they never mention that matrix or I have been missing something..

But that eigenvalues will be a good way for me to control if I have calculate correct so basicly I am suposed to solve this
\(\displaystyle det (H-\lambda I)\) and find the roots of that determinant..?
right?
 

Petrus

Well-known member
Feb 21, 2013
739
I am stuck on one that I have to see what definit it is
\(\displaystyle Q(h,k,l)=(h-2k+l)^2-(k-3l)^2+10l^2\)
is it enough to check \(\displaystyle (1,0,0)\), \(\displaystyle (0,1,0)\) and \(\displaystyle (0,0,1)\) cause I get positive definit and it suposed to be indefinit :S?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
and if all eigenvalues are negative then its negative definit? and if eigenvalues got positiv and negative its indefinit right?
I dont think we are supposed to use that trick as they want us to complete the square and they never mention anything about eigenvalues but I know how to calculate it as I am learning linear algebra 2, they never mention that matrix or I have been missing something..

But that eigenvalues will be a good way for me to control if I have calculate correct so basicly I am suposed to solve this
\(\displaystyle det (H-\lambda I)\) and find the roots of that determinant..?
right?
Yep! All correct! :)

I am stuck on one that I have to see what definit it is
\(\displaystyle Q(h,k,l)=(h-2k+l)^2-(k-3l)^2+10l^2\)
is it enough to check \(\displaystyle (1,0,0)\), \(\displaystyle (0,1,0)\) and \(\displaystyle (0,0,1)\) cause I get positive definit and it suposed to be indefinit :S?

Regards,
\(\displaystyle |\pi\rangle\)
Neh. That's not enough to check.
Try to find (h,k,l) such that Q would be negative.
That is if $(k-3l) > 0$ and the others are for instance 0. Is that possible?
 

Petrus

Well-known member
Feb 21, 2013
739
Yep! All correct! :)



Neh. That's not enough to check.
Try to find (h,k,l) such that Q would be negative.
That is if $(k-3l) > 0$ and the others are for instance 0. Is that possible?
now I see \(\displaystyle (8,4,0)\).. I think I will stick with eigenvalue but how does it work if the question ask complete the square \(\displaystyle Q(h,k,l)=h^2+3k^2+2l^2-4kh+2hl+2kl\) how do I find the Hessain matrix for that?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
now I see \(\displaystyle (8,4,0)\).. I think I will stick with eigenvalue but how does it work if the question ask complete the square \(\displaystyle Q(h,k,l)=h^2+3k^2+2l^2-4kh+2hl+2kl\) how do I find the Hessain matrix for that?

Regards,
\(\displaystyle |\pi\rangle\)
Multiplying it out we found that:
$$Q(h,k,l) = f_{xx} h^2 + f_{yy}k^2 + f_{zz}l^2 + 2f_{xy} hk + 2f_{xz}hl + 2f_{yz}kl$$

From this you can read the entries of the Hessian matrix.
For instance $f_{xx} = 1$ and $f_{xy} = -2$.
 

Petrus

Well-known member
Feb 21, 2013
739
Multiplying it out we found that:
$$Q(h,k,l) = f_{xx} h^2 + f_{yy}k^2 + f_{zz}l^2 + 2f_{xy} hk + 2f_{xz}hl + 2f_{yz}kl$$

From this you can read the entries of the Hessian matrix.
For instance $f_{xx} = 1$ and $f_{xy} = -2$.
Ahh now I see!! Thanks for taking your time!:) evrything makes sense now!:)

Regards,
\(\displaystyle |\pi\rangle\)