On Monday 12 of July 2010 23:08:09 you wrote:
> I think he wants to factor into C . (... + ... + ...).
>
> On Mon, Jul 12, 2010 at 16:11, Robert Dodier <robert.dodier at gmail.com>wrote:
> > On Mon, Jul 12, 2010 at 1:35 PM, Zbigniew Komarnicki <cblasius at gmail.com>
> >
> > wrote:
> > > When I use operator '.' I got:
> > > e(r+1, k+1) = - C.B.K2 . e(r, k+1) - C.B.K1 . z(r+1, k)
> > > - C.A . z(r+1, k) + e(r, k+1)
> >
> > I don't understand what is the problem here.
> > Can you please clarify.
> >
> > best
> >
> > Robert Dodier
Yes I want to factor but around the variables e and z, i.e.:
If we have:
e(r+1, k+1) = - C.B.K2 . e(r, k+1) - C.B.K1 . z(r+1, k)
- C.A . z(r+1, k) + e(r, k+1)
It can be simply simplified to:
e(r+1, k+1) = (1-C.B.K2) . e(r, k+1) + (-C.A-C.B.K1) . z(r+1, k)
My question is why it is not done or what I can do that? and why it is done
when I use operator "*" instead of operator "."?
The next matter is that operator "*" to not preserve the order in
multiplications between matrices and vectors, wears the operator "." save the
order but it not simplifies expressions.
Please see my derivation some simple ILC system in attached files for "*"
(derivation_with_star.mac) and for "." (derivation_with_dot.mac).
The result is quite different between "*" and "." (because the order is not
saved and no simplifications or do not collected terms is made):
Here is the finall result with the operator "*":
z(r+1, k+1) = e(r, k+1) B K2 + z(r+1, k) (B K1 + A)
e(r+1, k+1) = e(r, k+1) (1 - B C K2) + z(r+1, k) (- B C K1 - A C)
Here is the finall result with the operator ".":
z(r+1, k+1) = B.K2 . e(r, k+1) + B.K1 . z(r+1, k) + A . z(r + 1, k)
e(r+1, k+1) = - C.B.K2 . e(r, k+1) - C.B.K1 . z(r+1, k) - C.A . z(r+1, k) +
e(r, k+1)
but why it not simplifies as "*" to this form below with preserved orders
beetwen matrices and vectors ? Why it is not in this form?
z(r+1, k+1) = B.K2 . e(r, k+1) + (A + B.K1) . z(r+1, k)
e(r+1, k+1) = (1 - C.B.K2) . e(r, k+1) + (-C.A - C.B.K1) . z(r+1, k)
or what to do, what functions to use to tell maxima to obtain such result?
As I tell you this is very simple example but I have to work with about 50
different matrices (above is only 5 i.e.: A,B,C,K1,K2). If is more than 10
matrices it is very easy to make mistake if the output is not simplified and
what when we have 50 or more matrices and the order of this matrices is
important? I got wrong result if I use B C K1 instead of C B K1 in
multiplications.
I do not know that what I show and wrote it is clear to you? Do you understand
what is going on now?
I'm very sorry for my English language.
Thank you very much for your time and help.
Zbigniew
-------------- next part --------------
A non-text attachment was scrubbed...
Name: derivation_with_star.mac
Type: text/mac
Size: 6657 bytes
Desc: not available
URL: <http://www.math.utexas.edu/pipermail/maxima/attachments/20100713/8c490ead/attachment-0002.bin>;
-------------- next part --------------
A non-text attachment was scrubbed...
Name: derivation_with_dot.mac
Type: text/mac
Size: 7603 bytes
Desc: not available
URL: <http://www.math.utexas.edu/pipermail/maxima/attachments/20100713/8c490ead/attachment-0003.bin>;