Return vector that minimizes a function

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Hello,
I am writing an iterative algorithm that updates a vector "u" each iteration step. I also keep the previous vectors u, so I can control what happened previously, so each iteration there is a new row added to u. This row is the latest vector u if you want. So each iteration I try to update the latest u(k,:) I have by a new u(k+1,:).
The problem is not this iteration, but in each iteration step I find the new u(k+1,:) by minimizing a function "fun". However, this gives an error: "Assignment has more non-singleton rhs dimensions than non-singleton subscripts"
x0 = u(k,:);
fun = @(x)(r - (y0 + T.*x)).^2 + x-u(k,:);
u(k+1,:) = fminsearch(fun, x0);
Here is the full code: %% ILC algoritme clear all; close all; clc;
% Begin input: assistentie exoskeleton
u = zeros(1,5); %Begin met zero assistentie of je kan al een goede initiele guess maken eens je het systeem van het OpenSim model kent
max_it = 100; % Kwestie van niet te lang zoeken
lambda = 0.5; % Nu is dit nog een gewone constante, later verbeterd!
y0 = [1 0 -1 3 4];
y= [];
r = 5* ones(1,5);
e = [];
Norm = [];
T= [1 5 3 4 5];
for k = 1:1:max_it
y = [y ; y0 + T.*u(k,:)]; % Vervangen door Torque_driven model met u als input voor exoskeleton
e= [e ; r - y(k,:)];
%e(k+1,:)= r - (y0 + T.*u(k+1,:));
Norm = [ Norm ; norm(e(k,:))];
if Norm(k) < 0.1
break
end
x0 = u(k,:);
fun = @(x)(r - (y0 + T.*x)).^2 + x-u(k,:);
u(k+1,:) = fminsearch(fun, x0);
end
figure
xlabel('Time_samples')
ylabel('Norm error')
title('Norm error in time')
plot(Norm)

采纳的回答

Keerthana Chivukula
The underlying reason for this error message is that the output expected from the objective function provided to "fminsearch" is supposed to be a scalar as mentioned in the following link:
However, in the code provided above, the function "fun" returned a matrix and this resulted in incorrect indexing. Modifying the function to return a scalar would fix the issue.

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