love the way you go over every step systematically, instead of rushing the tiny calculations.

I'm over a year late but THIS IS AWESOME! THANKS FOR ENLIGHTENING ME W YOUR LINEAR ALGEBRA AS ALWAYS. Linear Algebra is my favorite and you make it even more fun and amazing for me!!!!!!!!!

Great bunny and great channel content, subbed long time ago, keep up with good work!

I would love an example of how to approach the problem if the test is inconclusive, e.g what to do next.

@drpeyam

6 жыл бұрын

I'll think about it, but unfortunately there's not much you can do! Because x^4 + y^4 (min), -x^4 - y^4 (max) and x^4 - y^4 (saddle) both have inconclusive tests at (0,0), but you have 3 very different behaviors! But I'll see what I can do about that.

@foreachepsilon

6 жыл бұрын

I imagine examining the function in the direction of the eigenvectors would show some behavior.

This is fantastic! I know a third method that I hope should be equivalent to this one. In this method, to classify the critical point of f at (x,y), you look at the second degree taylor polynomial of f at (x,y). Since the first derivatives are all zero, this polynomial is a quadratic form that approximates f around the point (x,y). Then we figure out whether the taylor polynomial is positive definite, negative definite, or neither. Each number in the signature of the quadratic form tells you which direction one variable "swings" in, I believe, and so its a min or max iff they all "swing" in the correct direction. Of course, this method also requires the secret ingredient that most Calc III textbooks don't have, Linear Algebra. If you're interested I could write it up and email you how it works!

This video has taught me how to classify critical points of a multivariable function properly. I have not seen an intelligible proof for the classification based on eigen values, and I'd appreciate if you could provide one on a video. Thank you Dr. Peyam!

If i get the opportunity to teach mvc, I'll be using this instead...beautiful

@JorgetePanete

6 жыл бұрын

Joe Potillor I'll*

The nice thing about this definition is also that you can apply it to functions with even more variables.

Just so much more beautiful than what I've learned at the University. Thanks for showing this. And btw

Very nice!

Could you do an example where you include the second derivative test into a constrained optimization problem using Lagrange? (Where you need to analyze a bordered Hessian)

@drpeyam

6 жыл бұрын

Do you mean like Lagrange multipliers?

@GonzieGr01

6 жыл бұрын

Dr. Peyam's Show Yes. Or a lagrangian

The first equation is similar to the coffin problem

Very Nice !! What can we do when the test is inconclusive ? Also, how do we know that there will be only real eigenvalues ? :)

@gaukatore

6 жыл бұрын

The matrix will always be symmetric, as d^2f/dxdy=d^f/dydx. and symmetric (real) matrecies always have real eigenvalues.

@cedricp.4941

6 жыл бұрын

Yes, of course...Thank you !

Would you mind making a video about the whole eigen-stuff, like calculating the eigenvalues, right and left eigenvectors, projection matrices, etc?

@drpeyam

6 жыл бұрын

There's one I made on blackpenredpen's channel, enjoy! kzread.info/dash/bejne/oIWKypN_hdXJhLA.html

@botondosvath2331

6 жыл бұрын

Dr. Peyam's Show Thanks! Haven't you done a video about the 3x3's nullspace too?

@drpeyam

6 жыл бұрын

Mmmmmh, I may have, but I don't quite remember! There's one on the Jordan Canonical Form on my channel, which may or may not help! :)

@botondosvath2331

6 жыл бұрын

Dr. Peyam's Show You were not showing the calculation there, because (I suppose) not that was the theme of the video, and the length of the calculation.

@drpeyam

6 жыл бұрын

Oh, my bad! I'll put it on my to-do list :)

whats the 3d equivilent of a point of inflection?

@drpeyam

2 жыл бұрын

det D^2 f = 0

@aneeshsrinivas9088

2 жыл бұрын

The proof is Taylor series right?

@aneeshsrinivas9088

2 жыл бұрын

At least In proving that having a positive or negative definite matrix results in convexity or concavity right?

is it the same det(A-lambda*I) than det(lambda*I-A)?

@drpeyam

5 жыл бұрын

Technically not quite because the second one is +/- the first one, but it doesn’t matter because either both of them are zero or not zero.

From the very beginning I learned calculus for utilitarian reasons, what is the max or min of a graph? Just take the second derivative! The derivative is the slope, and when the slope is 0 that is a max or min. The slope, you know, is a minimum when the second derivative is negative at that point and a Max if it is positive.

It actually reduces to what textbooks say, doesn't it? If a function, let's say f, has continous second partial derivatives, the hassian matrix (Hf) is a symmetric matrix (becuase D12f = D21f), so the fact that all of its eigenvalues are positive is equivalent to that fact that all of its principal minors are postive, i.e. D11f > 0 and detHf = (D11fD22f - (D12f)^2) > 0 The other cases are similar.

@drpeyam

4 жыл бұрын

It’s a sufficient condition and also it doesn’t give us any insight in the geometry of the hessian

@Edddfm

4 жыл бұрын

@@drpeyam I just wondered why you called it garbage. Anyway, nice video.

But... why is the other test garbage?

@drpeyam

4 жыл бұрын

It’s not a natural test, and a lot of the times it gives an inconclusive answer when there shouldn’t be one. Also it doesn’t generalize to higher dimensions

@fracaralho

4 жыл бұрын

@@drpeyam Thank you!