Series generated by recursive formula

Suppose each value in a series depends on its immediate predecessor, say, x(n) == j*x(n-1) + k*y(n)

What is the most efficient way to express this in MATLAB?

Answer 1

[Azzi Abdelmalek]

y=0:10   % Example 
k=10 
x(1)=0
for n=1:10
 x(n+1)=j*x(n)+k*y(n+1)
end

[mutt]

Isn't there a viable matrix notation approach as opposed to a for-loop method?

[Azzi Abdelmalek]

Before calculating x(2) you need to calculate x(1), and so on. I don't see how is it possible to do it without a loop, unless you resolve the problem analytically, using z-transform

[Thorsten]

No. It would be possible only if you could transform your equation such that x(n) does NOT belong on its predecessor.

For large n the computation will be faster if you preallocate x such as

x = nan(1, Nmax);
for x = 1:Nmax
  :
end

[José-Luis]

You could always write a recursive function. Having said that, I have some gripes against those as I found them difficult to understand and generally not worth the effort.

Answer 2

[José-Luis]

You could always write a recursive function, but I would much rather use the for loop as it is much simpler to understand, IMO, and probably faster too.

Start by defining an anonymous function:

 myFun = @(x) 4.*x + sin(x); %the function you want to apply to the previous value

And placing the following function in your path:

function [your_mat] = someFunction(n,myFun,start_val)
temp_val = start_val;
your_mat = recfun(n,temp_val);
    function [temp_val] = recfun(n,temp_val) %here be recursion
        if n == 0
            temp_val = [];
        else
            temp_val = [temp_val myFun( recfun( n-1 , temp_val(end) ) ) ];
        end
    end
end

Where n is the number of evaluations and start_val is your starting value. And you could run it from the command line as follows:

 your_mat = someFunction(10,myFun,5); %10 reps, starting value of 5

your_mat will be a matrix where every value is dependent on the previous one. Note that you would still need to modify the code to make it do exactly what you want.