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From: fdm on 5 Sep 2009 11:14
The two variable function:

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?

From: Ray Vickson on 5 Sep 2009 13:08
On Sep 5, 8:14 am, "fdm" <f...(a)gk.com> wrote:
> The two variable function:
>
>       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?

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
From: Gordon Stangler on 5 Sep 2009 13:41
On Sep 5, 10:14 am, "fdm" <f...(a)gk.com> wrote:
> The two variable function:
>
>       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?

The Jacobian is explained quite nicely here:
http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
There are even examples if you need them.
From: fdm on 5 Sep 2009 13:50
Would it be correct to say that the i'th row corresponds to the gradient of
the i'th functin?


"Gordon Stangler" <gordon.stangler(a)gmail.com> wrote in message
news:a3acf158-425e-428a-89dd-8342e7981166(a)s31g2000yqs.googlegroups.com...
On Sep 5, 10:14 am, "fdm" <f...(a)gk.com> wrote:
> The two variable function:
>
> 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?

The Jacobian is explained quite nicely here:
http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
There are even examples if you need them.

From: Ray Vickson on 5 Sep 2009 14:07
On Sep 5, 10:50 am, "fdm" <f...(a)gk.com> wrote:
> 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...(a)gmail.com> wrote in message
>
> news:a3acf158-425e-428a-89dd-8342e7981166(a)s31g2000yqs.googlegroups.com...
> On Sep 5, 10:14 am, "fdm" <f...(a)gk.com> wrote:
>
>
>
> > The two variable function:
>
> > 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?
>
> The Jacobian is explained quite nicely here:http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
> There are even examples if you need them.

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