HELP PLEASE Quick and simple!

  • Thread starter coolalac
  • Start date
  • #1
1
0

Main Question or Discussion Point

HELP PLEASE!!! Quick and simple!

Hey I was wondering if you'll could help. I was told that this is a simple proof but have been working on it for ages and cant get it:

Suppose u is a [tex]C^{2}[/tex] function in [tex]R^{n}[/tex]. Given an nxn matrix, A define v(x):= u(Ax). Show that ([tex]\nabla[/tex]v)(x) = [tex]A^{T}[/tex] . [tex]\nabla[/tex]u(Ax)

(Note the grad u(Ax) isn't supposed to be superscripted I just dont know how to put it in the same line!)

Pleaseeee help!!
Thanks
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955


Hey I was wondering if you'll could help. I was told that this is a simple proof but have been working on it for ages and cant get it:

Suppose u is a [tex]C^{2}[/tex] function in [tex]R^{n}[/tex]. Given an nxn matrix, A define v(x):= u(Ax). Show that ([tex]\nabla[/tex]v)(x) = [tex]A^{T}[/tex] . [tex]\nabla[/tex]u(Ax)

(Note the grad u(Ax) isn't supposed to be superscripted I just dont know how to put it in the same line!)
You put every thing on the same line by putting the entire formula inside [ tex ] and [ /tex ], not just bits and pieces:
[tex]\nabla v(x)= A^T\nambla u(Ax)[/tex]
Much simpler to write and to read!

Pleaseeee help!!
Thanks
It is, essentially, the "chain rule". v(x)= u(Ax) so [itex]\nabla v(x)= \nabla u(x) d(Ax)/dt= \nabla u(x) A = A^T \nabla u(x)[/itex].
 

Related Threads on HELP PLEASE Quick and simple!

Replies
3
Views
8K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
1K
Top