Effortlessly Solve C^{2} Functions with Quick and Simple Tips | Get Help Now!

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In summary, the problem is to show that for a C^{2} function u in R^{n} and an nxn matrix A, the gradient of the function v(x) defined as v(x):= u(Ax) is equal to A^{T} . \nabla u(Ax). This can be solved using the chain rule.
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coolalac
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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 can't 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 don't know how to put it in the same line!)

Pleaseeee help!
Thanks
 
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  • #2


coolalac said:
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 can't 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 don't 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].
 

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