Date: Feb 24, 2013 5:56 AM
Author: Milos Milenkovic
Subject: Re: Automating an iterative procedure

Dear dpb,
yes, intermediate values of B0 and B1 have to be stored, I have to see when they converge.
Any help would be greatly appreciated!
Best,
M

dpb <none@non.net> wrote in message <kgbdd4$isu$1@speranza.aioe.org>...
> On 2/23/2013 3:01 PM, Milos Milenkovic wrote:
>
> ...[top posting repaired--please don't: hard conversation follow makes]...
>

> > dpb <none@non.net> wrote in message <kgb86j$446$1@speranza.aioe.org>...
> >> On 2/23/2013 2:04 PM, Milos Milenkovic wrote:
> >> ...

> >> > K1=[-0.2582 -18.596 0 -0.2582;
> >> > -34.737 -1300.188 0 -34.737;

> >> ...
> >> > for i=1:100
> >> > B1(1)=[0.01 0.02 0.07 0.03;
> >> > 0.02 0.05 0.06 0.02;
> >> > 0.07 0.09 0.01 0.03;
> >> > 0.01 0.03 0.02 0.04];

> >>
> >> ...
> >>
> >> No, but why would you want to duplicate a constant array anyway?
> >> Sometimes for matrix operations there's a need/desire to do so, but
> >> generally a temporary via repmat() or some other technique would be
> >> more reasonable than just holding multiple copies of the same array.
> >>
> >> B1(1) refers to a single array location whereas the rhs is an array.
> >> "Can't put 5-lbs of 'taters in a 1-lb tote" to use the E TN vernacular.
> >>
> >> Only a cell array element can hold another array.
> >>

>
> ...
>

> >> B1(1) and B0(1) are given matrices in iteration 1, then for the next 99
> >> iterations I want to calculate updates of these matrices.

>
> Oh.
>
> Move the initiation outside the loop then store new values into the
> desired location(s) in the array inside the iterative portion (the loop
> iow). To do that you don't need to subscript B1 at all other than to
> address individual elements therein if required. But, if you can
> vectorize the computation of B1 then there's no need for the indexing.
>
> Are you needing to save all iterations to the bitter end?
>
> --