Thread Subject: quadratic form with symbolic vectors?

I would like to differentiate x'Ax. Here my x is y-B*u and
A is the identity matrix

syms B u1 u2 y1 y2 real
u=[u1; u2]
y=[y1; y2]
I=eye(2);
pretty(diff((y-B*u)'*I*(y-B*u),B))

I don't know how to interpret the output:

-2 (y1~ - B~ u1~) u1~ - 2 (y2~ - B~ u2~) u2~

what are all the tilda's for? Am I doing this correctly?

Thanks!

David

looks like 'pretty' is messing things up (or rather, I do not know how t=
o =

interepret this..)

>> (diff((y-B*u)'*I*(y-B*u),B))

ans =3D

-2*(y1-B*u1)*u1-2*(y2-B*u2)*u2

(if you did not know, PRETTY just not change the answer, just tries to =

print the ans in a 'nice' way)


On Mon, 25 Feb 2008 14:36:02 -0500, David Doria <daviddoria@gmail.com> =

wrote:

> syms B u1 u2 y1 y2 real
> u=3D[u1; u2]
> y=3D[y1; y2]
> I=3Deye(2);
> pretty(diff((y-B*u)'*I*(y-B*u),B))

I did know that pretty tried to make it display nice, so
then I guess question 1 is why does pretty put ~ all over??

Then Question 2: If i run the following:

--------
%throughout, sigma = sigma^2
clear;clc;close all;
syms N sigma B u1 u2 y1 y2 real

u=[u1; u2]
y=[y1; y2]
I = eye(2);
%diff((y-B*u)'*I*(y-B*u),B)

lnf = -N/2*log(2*pi) - N/2 * log(sigma) + -1/(2*sigma)*(y -
B*u)'*(y-B*u);

%first derivatives
pretty(diff(lnf,B))

--------

it gives me the sum of 2 things, which are each the same
because it should just be something in terms of a vector
(the 2 components are what got broken up). Is there a way
for it to give back the result in vector form?

Thanks,

Dave

am I doing something wrong? Or is this for some reason not
supposed to work?
bump

My basic take is that the symbolic toolbox does not work with matrices as
you expect.
looks like you have to deal with seperating each row


On Tue, 26 Feb 2008 07:24:01 -0500, David Doria <daviddoria@gmail.com>
wrote:

> am I doing something wrong? Or is this for some reason not
> supposed to work?
> bump