Thread Subject: Vector Math vs For Loop

 is it possible to calculate X when

X(i) =X(i-1) + A * (B - X(i-1) )

using a vector script instead of a loop

**Importantly "x" is not a known vector, it is only given an initial value.
"a" is a constant
"b" is a vector (for this example lets say its rand(1000))

rand(1000) is not a vector.

Your code will error with
"In assignment A(I) = B, the number of elements in B and
 I must be the same."

If you post an working For loop, complete with numeric inputs, then you may get a better answer.

"happydude " <anonymousse@hotmail.com> wrote in message <hdv4n7$la$1@fred.mathworks.com>...
> is it possible to calculate X when
>
> X(i) =X(i-1) + A * (B - X(i-1) )
>
> using a vector script instead of a loop
>
> **Importantly "x" is not a known vector, it is only given an initial value.
> "a" is a constant
> "b" is a vector (for this example lets say its rand(1000))

You can do this in a one-liner using the FILTER function. "help filter" for details. Be extremely careful in the notation, because the documentation coincidentally uses the same variable names as you have, but with different meanings! Most importantly, I think what you are calling "x" is what the documentation calls "y", which is the output of the filter function. It is very easy to get confused, but I am 95% you can do exactly what you want with this function.

"the cyclist" <thecyclist@gmail.com> wrote in message <hdv75f$2s4$1@fred.mathworks.com>...
> "happydude " <anonymousse@hotmail.com> wrote in message <hdv4n7$la$1@fred.mathworks.com>...
> > is it possible to calculate X when
> >
> > X(i) =X(i-1) + A * (B - X(i-1) )
> >
> > using a vector script instead of a loop
> >
> > **Importantly "x" is not a known vector, it is only given an initial value.
> > "a" is a constant
> > "b" is a vector (for this example lets say its rand(1000))
>
> You can do this in a one-liner using the FILTER function. "help filter" for details. Be extremely careful in the notation, because the documentation coincidentally uses the same variable names as you have, but with different meanings! Most importantly, I think what you are calling "x" is what the documentation calls "y", which is the output of the filter function. It is very easy to get confused, but I am 95% you can do exactly what you want with this function.

Hoping I got this right Here is some code that should work for you. Again, be really careful about notation. In my code snippet, I retained YOUR definition of A and B, and I changed the notation in "help filter" according to the following scheme: (B,A,X,Y) --> (z,m,p,x).

I hope this helps rather than confuses you!

%%%
NUMBERTERMS = 10;

A = 0.3;
B = 0.5;

z = (A*B) * ones(NUMBERTERMS,1);

m = zeros(NUMBERTERMS,1);
m(1) = 1;
m(2) = -(1-A);

p = zeros(NUMBERTERMS,1);
p(1) = 1;

x = filter(z,m,p)

%%%

thank you so much for trying this problem...
i couldnt help but get a little confused sorry !

in your code you have
B=0.5
but B is actually a vector?

also in the equation the current value of "x" depends on a previous value of "x", which hasn't been asigned an initial value in the code?

"happydude " <anonymousse@hotmail.com> wrote in message <he1ajj$b7a$1@fred.mathworks.com>...
> thank you so much for trying this problem...
> i couldnt help but get a little confused sorry !
>
> in your code you have
> B=0.5
> but B is actually a vector?
>
> also in the equation the current value of "x" depends on a previous value of "x", which hasn't been asigned an initial value in the code?

happydude,

You are not being careful enough about notation. When I used the variable "B", that is because YOU called a (scalar) variable B. This is NOT the same as the B used in the MATLAB documentation, which is indeed a vector as you point out.

Also, the filter function is more general than you need, with two separate vector inputs (which are called x and y in the documentation). You have have only one vector. The starting value of your vector is encoded into the inputs to the function.

I hope that helps.

the cyclist

ah, sorry... still new, bit confused..

thanks for your help :)