Relation between the gradient and the jacobian?
fdm
f(x,y)
has the gradient (vector):
F(x,y) = (\par x, \par y)
I have read that the jacobian is the generalization of the gradient and that
it always has the form of a matrix. I have also read that the jacobian is
the first order partial derivative of a multivariable function but that is
also the definition of the gradient.
So what is the difference between the gradient and the jacobian?
And what is the jacobian of f above?
Ray Vickson
The Jacobian is for _several_ functions, such as a vector-valued
function F(x1,x2) = [f1(x1,x2),f2(x1,x2)]. The Jacobian J of F is the
2x2 matrix with entries J(i,j) = df_i/dx_j, where the 'd's are partial
derivatives. So, I guess one answer to your question would be that
your f does not have a Jacobian at all; another answer would be that
Jacobian = gradient in the case of a single function. The second
answer is probably better, because a vector [the gradient] is just a
special case of a matrix. J does generalize the gradient in the sense
that for smooth functions F we have a linear relation between F at x =
(x1,x2) and at x+h, where h = (h1,h2): F(x+h) = F(x) + J(x)*h + o(|
h|), where 'o' stands for terms that --> 0 faster than |h| = length of
vector h, and J*h = matrix_vector product. For a single function f
this would be f(x+h) = f(x) + grad(f)(x).h + o(|h|), where grad(f).h =
scalar product. Also: remember that Google is your friend, and
searching on the word 'Jacobian' turns up numerous relevant web pages;
see http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant or
http://mathworld.wolfram.com/Jacobian.html . These sources are careful
to distinguish between the Jacobian matrix and Jacobian determinant (=
determinant of the Jacobian matrix), but not all books, notes and
papers are that careful; some of them just mean the determinant
itself.
R.G. Vickson
Gordon Stangler
The Jacobian is explained quite nicely here:
http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
There are even examples if you need them.
fdm
the i'th functin?
"Gordon Stangler" <gordon....@gmail.com> wrote in message
news:a3acf158-425e-428a...@s31g2000yqs.googlegroups.com...
Ray Vickson
> Would it be correct to say that the i'th row corresponds to the gradient of
> the i'th functin?
I don't know who you are addressing, but the answer is YES. That is
obvious, just from the definition.
R.G. Vickson
>
> "Gordon Stangler" <gordon.stang...@gmail.com> wrote in message
fdm
with the most growth while the derivative of 'f' is "how much" the growth is
actually increased in that direction for each unit of increase in the
domain. Is that correct?
"Ray Vickson" <RGVi...@shaw.ca> wrote in message
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