Series generated by recursive formula

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mutt
mutt 2013-2-7
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?

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回答（2 个）

Azzi Abdelmalek
Azzi Abdelmalek 2013-2-7
编辑：Azzi Abdelmalek 2013-2-7
y=0:10 % Example
k=10
x(1)=0
for n=1:10
x(n+1)=j*x(n)+k*y(n+1)
end
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Thorsten
Thorsten 2013-2-7
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
José-Luis 2013-2-7
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.

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José-Luis
José-Luis 2013-2-7
编辑：José-Luis 2013-2-7
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.

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