I love the balance between being straight to the point and giving enough explanation so that the concept sticks. Very few math educators are able to walk that tight rope as good as you sir.
I have a question if you don’t mind. With constrained optimization, we look at the bordered Hessian matrix and we say that if the determinant is > 0 we have a MAX. This rule seems counterintuitive and out of line with my previous understanding of D2 > 0 translating to a minimum. Is there a way for me to reconcile these two differences in methods intuitively?
Sure! I will do my best to answer. The previous understanding of D2>0 translating to a minimum is spot on. In the Hessian determinant, we are looking at the four different partial derivatives that exist and creating products. Every statement that follows will be for a function of two variables. Imagine a local maximum... from the 2D perspective, it would be concave down from every orientation, meaning that both the partial derivatives f_xx and f_yy would be less than 0. In the Hessian determinant, they are multiplied, which produces a positive number, meaning now that D>0. The same is true for a local minimum being concave up from every orientation: both f_xx and f_yy are positive, meaning that their product is positive as well. This is why a positive D value means a critical point could be a local max OR a local min. It's not until we look at the values of the individual second partial derivatives (f_xx and f_yy) to determine which of these critical points we have.
@@davidfriday7498 thank you so much for your quick response! I’m actually studying the Fundamentals of Mathematics for Economics by Alpha C. and I hope to finally wrap my head around the concept with time. Thank you so much once again for your help!!
@@AverageAlaya I took a look at the graph using Geogebra: www.geogebra.org/3d?lang=en I see what looks like a local minimum at (0,0) and saddle points at the other two critical points, including the one you sent. I'll try the problem and see if I get the same result as you.
@@AverageAlaya I have completed the problem. I believe I see the issue. When I set the partial derivative with respect to x equal to zero, two cases arose: x=0 or y=-1. I believe you plugged in y=1 rather than y=-1 to the Hessian determinant. This would result in a D value of 0 rather than a D value of -8, which is what I got for both (sqrt(2),-1) and (-sqrt(2),-1). I hope this has helped you in your interpretation of your problem :-)
@@davidfriday7498 Hi I also encounter the same problem. May I ask for your help?, the function is: f(x,y,z)=x^4+x^2y+y^2+z^2+xz+1. There are 3 critical points, one of them is at (0,0,0) and the first principal minor(the bordered Hessian determinant) evaluated at that point is 0. Since the graph of the function is in the 4 dimensions... so I can't use any graphing utilities to help me see the behavior of the graph at that point. Maybe plotting its level surface and with inspection help?
I love the balance between being straight to the point and giving enough explanation so that the concept sticks. Very few math educators are able to walk that tight rope as good as you sir.
Amazing video. You presented the Hessian Matrix in a way that was easy to understand and visualize. Thank you so much David.
Clear and straight to the point, thank you!
short but clear and easy-understanding video, thank you.
I'm a student from Morocco
Thanks a LOT !!!
Maths student from the UK, thanks for this video
I'm glad you took something positive from this!
Hey buddy
swear i was looking through my notes for like an hour and a half before just googling it haha
I have a question if you don’t mind. With constrained optimization, we look at the bordered Hessian matrix and we say that if the determinant is > 0 we have a MAX. This rule seems counterintuitive and out of line with my previous understanding of D2 > 0 translating to a minimum. Is there a way for me to reconcile these two differences in methods intuitively?
Sure! I will do my best to answer.
The previous understanding of D2>0 translating to a minimum is spot on. In the Hessian determinant, we are looking at the four different partial derivatives that exist and creating products.
Every statement that follows will be for a function of two variables. Imagine a local maximum... from the 2D perspective, it would be concave down from every orientation, meaning that both the partial derivatives f_xx and f_yy would be less than 0. In the Hessian determinant, they are multiplied, which produces a positive number, meaning now that D>0. The same is true for a local minimum being concave up from every orientation: both f_xx and f_yy are positive, meaning that their product is positive as well. This is why a positive D value means a critical point could be a local max OR a local min. It's not until we look at the values of the individual second partial derivatives (f_xx and f_yy) to determine which of these critical points we have.
@@davidfriday7498 thank you so much for your quick response! I’m actually studying the Fundamentals of Mathematics for Economics by Alpha C. and I hope to finally wrap my head around the concept with time. Thank you so much once again for your help!!
Thank you! I couldn’t find online what D=0 means
This is also known simply as..... the D
My personal favorite way to refer to it :-)
oh ,why nobody actualy showes a solution of problem with actual numbers???
Then I believe you should probably watch this video: th-cam.com/video/r4AQw1PdDRk/w-d-xo.html
I got a case where my determinant is zero... help :')
If you provide more information, I can certainly help. What was the original function?
@@davidfriday7498 Thank you! The function is f(x,y)=x^2+y^2+x^2y; I got the roots but the root -sqrt(2),1 seems to have a determinant of 0. Thank you!
@@AverageAlaya I took a look at the graph using Geogebra: www.geogebra.org/3d?lang=en
I see what looks like a local minimum at (0,0) and saddle points at the other two critical points, including the one you sent. I'll try the problem and see if I get the same result as you.
@@AverageAlaya I have completed the problem. I believe I see the issue. When I set the partial derivative with respect to x equal to zero, two cases arose: x=0 or y=-1. I believe you plugged in y=1 rather than y=-1 to the Hessian determinant. This would result in a D value of 0 rather than a D value of -8, which is what I got for both (sqrt(2),-1) and (-sqrt(2),-1). I hope this has helped you in your interpretation of your problem :-)
@@davidfriday7498 Hi I also encounter the same problem. May I ask for your help?, the function is: f(x,y,z)=x^4+x^2y+y^2+z^2+xz+1. There are 3 critical points, one of them is at (0,0,0) and the first principal minor(the bordered Hessian determinant) evaluated at that point is 0. Since the graph of the function is in the 4 dimensions... so I can't use any graphing utilities to help me see the behavior of the graph at that point. Maybe plotting its level surface and with inspection help?
slashy slashy
Thanx for video
THANK YOU
Very good ! Thanks!
trauma inducing indeed when D=0 !
bump