"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:co4a5b$49j$1@news.urz.uni-heidelberg.de...

kiki wrote:

Hi all,

I want to find the symbolic differentiation of the following term:

(w'*B*w)/(w'*A*w)

where w is a column vector, A and B are symmetrical matrices, w' denotes
transpose...

I want to find the second order differentiation of the above term
symbolically, w.r.t. the vector w... can Matlab do that?

By the way, if anybody knows smart techniques about how to find the
second differentiation of the above term by hand... please teach me ...

Let's see if I understand you correctly - you probably want to
find the Hessian matrix (the matrix containing the second derivatives wrt
the vector components)?

I don't know about Matlab, but I think the calculation is not that
hard to do by hand. For the first derivative, you obviously have
to use the quotient rule. The derivative of the numerator wrt
w_j gives two times the j-component of the product B*w,
the derivative of the denominator wrt w_j gives two times the
j-component of the product A*w. So the first derivative wrt
w_j is (B*w * w'*A*w - w'*B*w * A*w)/(w'*A*w)^2. Obtaining the
second derivative looks ugly, but is entirely feasible then.

HTH.


Bye,
Bjoern

Hi Bjoern,

thanks a lot.

Woud you mind showing me how to do taking one more derivative of that
expression:

(B*w * w'*A*w - w'*B*w * A*w)/(w'*A*w)^2 ?

A, B are symmetrical.

Specifically, the first derivative is in fact a row vector:

((w'*A*w)*(w'*B) - (w'*B*w) *(w'*A)))/(w'*A*w)^2 ?

Now how do you do derivitive of

(w'*A*w)*(w'*B) ???


I do the following:

Taking deriviative of w'*A*w = 2 w'*A

Taking derivative of w'*B = B

I should do

D{(w'*A*w)*(w'*B)}=D{(w'*A*w)}*(w'*B) + (w'*A*w)*D{(w'*B)}

but D{(w'*A*w)}*(w'*B) = 2 (w'*A)*(w'*B)

it even does not pass the syntax check of matrix multiplication...

What can I do?