O.k., Walter thinks we haven't answered your original question, and he may be right. As his comment on your answer shows, the loop can work - but not in the form you have displayed it. If you define a function this way,
function EqSys = makeSystem(N)
EqSys = cell(1,N);
for i=1:N
EqSys{i} = @(x)x^i;
end
and then try these commands
>> EqSys = makeSystem(2)
EqSys =
@(x)(x-i)^i @(x)(x-i)^i
the output is a cell array of function handles that works as in Walter's comment - for example,
>> EqSys{2}(1)
ans =
1
Then if you type
>> fsolve(EqSys,1)
you get an answer of 0 (preceded by a lot of diagnostic statements).
But did you really get the simultaneous solution of these equations? If your function is
function EqSys = makeSystem(N)
EqSys = cell(1,N);
for i=1:N
EqSys{i} = @(x)(x-i)^i;
end
and you enter
>> EqSys = makeSystem(2);
>> fsolve(EqSys,1)
you will get an answer of 1 (which only solves the first equation).
I don't see any way of setting this up using function handles that is worth the trouble. Instead, you could create a function like this:
function y = eqnSystem(x,N)
y = zeros(N,1);
for i=1:N
y(i) = (x(i)-i)^i;
end
which, by the way, assumes that x has as many components as there are equations - a requirement for a well-defined problem. Then
>> x0 = [2 3];
>> f = @(x) eqnSystem(x,N);
>> fsolve(f,x0)
gives the answer
1.0000 2.0075
which is surprisingly inaccurate (but you can improve this by adjusting the options for fsolve).
