Calculate the first n terms of the recursion
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Calculate the first n terms of the following recursion
function xn_plus_one = state(n,t)
if n<1 || ~isscalar(n) || fix(n)~=n
error('n must be a non-negative integer');
else
xn_plus_one = zeros(1,length(t));
for jj = 1:length(t)
if t(jj) < 0 || t(jj) > 3
error('t must contain non-negative scalars less than or equal to 3');
else
if n==1
xn_plus_one(jj)=1;
else
xn_plus_one(jj) = integral(@(s) (2.*control(n-1,s)),0,t(jj)-1);
end
end
end
end
end
function C=control(n,v)
C=zeros(1,length(v));
for ii=1:length(v)
if v(ii)>=-1 && v(ii)<0
C(ii)=0;
elseif v(ii)>=0 && v(ii)<=2
C(ii)=integral(@(s) state(n,s),0,v(ii));
elseif v(ii)>2 && v(ii)<=3
C(ii)=0;
else
error('entries in v must vary from -1 to 3')
end
end
end
Calling the state function at n=4 and t=3 takes too much time to run. How can I improve it?
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Accepted Answer
Morgan
on 7 Feb 2024
Edited: Morgan
on 7 Feb 2024
Using the built-in MATLAB profiler, the biggest computational load is in the evaluations of integral(). Which is no surprise since it is notoriously slow inside for loops, let alone inside a double for-loop inside a recursive function, when there are much more efficient ways of integration.
Below, I've rewritten your code to exclude for loops and calculate everything as efficiently as I could come up with. Also, I shifted n down by one compared to your code to be more consistent with the equations you provided.
% INITIALIZE MATLAB
clear variables
close all
clc
% DEFINE INPUT ARGUMENTS
n = 1;
t = -2 : 0.01 : 5;
% CALL RECURSIVE STATE FUNCTION
x = state(n,t);
% PLOT X
figure(1);
plot(t,x,'-b','LineWidth',2);
% STATE FUNCTION
function x = state(n,t)
% CHECK INPUT ARGUMENTS
if n < 0 || ~isscalar(n) || fix(n)~=n
error('"n" must be a non-negative integer.');
end
% INITIALIZE X
N = length(t);
x(1,:) = ones(1,N);
% BASE CASE
if n == 0
return
end
% CALCULATE VALID INDICES OF t FOR x
idx1 = t >= 0 & t <= 3;
% ALL OTHER CASES
x(2:n+1,:) = zeros(n,N);
for i = 2 : n+1
% CALCULATE u
u = control(i-1,t-1); % <-- Edit: upper bound t-1 not t
% CALCULATE STATE FUNCTION
x(i,idx1) = 2*cumtrapz(t(idx1),u(idx1));
end
% RETURN FINAL ROW
x = x(n+1,:);
function u = control(n,t)
% ASSUME u IS ZERO
u = zeros(size(t));
% FIND VALID INDICES OF t FOR u
idx2 = t >= 0 & t <= 2;
% CALCULATE CONTROL FUNCTION
u(idx2) = cumtrapz(t(idx2), x(n,idx2));
end
end
Edit: My apologies, I forgot the upper bound on the integral of . It is highlighted in the comments above.
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