Undergrad Understanding Hessian for multidimensional function

Click For Summary
The discussion centers on the visualization of the Hessian matrix for the function f(u, v) = e^{−cu} sin(u) sin(v). Participants express confusion about why the maxima and minima exhibit both positive and negative traces. Clarification is sought regarding the relationship between the Hessian's trace and the nature of critical points. The conversation highlights the importance of understanding the Hessian's properties, including its role in identifying local extrema. Overall, the dialogue emphasizes the complexity of analyzing multidimensional functions and their critical points.
SaschaSIGI
Messages
3
Reaction score
0
Hello everybody,

I have a question regarding this visualization of a multidimensional function. Given f(u, v) = e^{−cu} sin(u) sin(v). Im confused why the maximas/minimas have half positive Trace and half negative Trace. I thought because its maxima it only has to be negative. 3D vis

2D visualization
 

Attachments

  • hesse2.PNG
    hesse2.PNG
    37.7 KB · Views: 121
  • hesse1.PNG
    hesse1.PNG
    39.1 KB · Views: 131
Physics news on Phys.org
Hi,

You have me wondering what I am looking at. Is the Hessian projected as a color code on a plot of the function ?
Did it occur to you to write down the Hessian for this function ? So: what's the expression for the trace of the Hessian ? (*)

What do you mean with
SaschaSIGI said:
because its maxima it only has to be negative

Aren't there minima between the maxima ?

(by the way: single: minimum, maximum. Plural: minima, maxima)

(*) Notice the similarity with the Laplacian :smile: ?

##\ ##
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
7K
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
3K
Replies
2
Views
2K