Adding function handles

Question

I have in MATLAB a cell array C containing containing n function handles @(t). I also have a vector x_star containing n constants. I am now trying to make a new function handle consisting of the function on location i in the cell array times the constant on location i in the vector. I have been trying with this code but it just outputs "index exceeds matrix dimensions":

M= @(t) 0;
for i=1:length(x_star)
M = @(t) M(t)+ C{i}(t)*x_star(i);
end

Is it possible to do it like this or is there a more convenient way to do it?

And x_star and C is defined by:

w=2*pi/24; 
C = cell(n,1);       
C{1}=@(t) 1;
for i=2:n
    if (mod(i,2)==0)               
        f = @(t) sin(0.5*i*w*t);
    else
        f = @(t) cos(0.5*(i-1)*w*t);
    end
C{i}=f;
end


 A = zeros(length(t),n);     %% Initialize A-matrix
A(:,1)=C{1};
for i=2:n
        A(:,i)=C{i}(t);
end
x_star = (transpose(A)*A)\transpose(A)*y;   
r_star=y-A*x_star;

Are you sure that `length(C)` is greater than or equal to `length(x_star)`? — Andras Deak -- Слава Україні, Oct 19 '15 at 18:12
Also you are overwriting `M` every loop iteration. That means only the last instance of `M` will be available for further use. — Adriaan, Oct 19 '15 at 18:13
@Linda - Please show us the code that defines `C` and `x_star`. It's hard to believe they're the same length if you get that error. — rayryeng, Oct 19 '15 at 18:17
And yes - the point is to add all of these functions, so that they make up one function :) — some_name, Oct 19 '15 at 18:18
And are you sure this recursive definition of `M` works? `f=@(x) x; f=@(x) f(x)+1` gives me `f = @(x)f(x)+1` as output. **EDIT**: interestingly, the output of `f` will be fine, it will just make the `disp` implicit... — Andras Deak -- Слава Україні, Oct 19 '15 at 18:18
uhm, it is defined in a pretty long way but sure, I can show you :) — some_name, Oct 19 '15 at 18:21
@AndrasDeak That works because Matlab function handles have a workspace attached to them that takes precedence over all others. So the assignment `f = @(x) f(x) + 1` creates a handle that stores the definition of `f(x)` at creation time and assigns it to the newly create handle `f`. — TroyHaskin, Oct 19 '15 at 18:25
y is a vector of the same length as t. In my example they both have length 24 — some_name, Oct 19 '15 at 18:29
And are you sure that an element of `C{i}` will not make `A` longer along the first dimension? Do check `size(C),size(x_star)` right before your error. — Andras Deak -- Слава Україні, Oct 19 '15 at 18:29
Ok, wait wait wait... The input `t` is a vector? The beginning definition of your anonymous function only outputs 1 scalar. Should that be a vector of zeroes instead? — rayryeng, Oct 19 '15 at 18:31
They both have size (7 1) in my example where n = 7. It doesn't fail until i try to add them in M — some_name, Oct 19 '15 at 18:31
Are you sure that every element of `C` is a function handle? If one were an array, then `C{i}(t)` would refer to element number `t`, giving unexpected results. — Andras Deak -- Слава Україні, Oct 19 '15 at 18:33
There are some weird things. You should have `A(:,i)=C{1}(t);` with a dummy argument `t`. And `C{1}` should return a vector of length `length(t)`, like so: `C{1}=@(t) ones(length(t),1);` — Andras Deak -- Слава Україні, Oct 19 '15 at 19:01
Also, if I have this right, you can replace those loops with `A = @(t) [ones(numel(t),1),sin(w/2*t*(k - (mod(k,2)~=0)) + pi/2*ones(numel(t),1)*(mod(k,2)~=0))]; ` as long as `k` is a row vector and `t` is a column vector. — TroyHaskin, Oct 19 '15 at 19:11

Adding function handles

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