# How do I set up a particular function

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ektor on 15 May 2019
Commented: ektor on 16 May 2019
Dear all,
I have a function
f=@(x) umax(x,u, t);
The problem is that x changes at each value of t=1,...,T; that is,
LL= function umax(x,u, t);
if t==1
u=[x; u(2:T) ];
elseif t==T
u=[u(1:T-1);x ];
else
u=[u(1:t-1);x;u(t+1:T) ];
end
LL='some function of the sum of u which I skip now'
end
Is the above code correct or the code below?
f=@(x) umax(x,u, t);
LL= function umax(x,u, t);
u(t)=x
LL='some function of the sum of u which I skip now'
end
LL produces a scalar.
Thanks a lot

ektor on 15 May 2019
I modified by output. Is that now more clear?
Walter Roberson on 16 May 2019
The first one requires that a function named T has been defined that returns a scalar and that t is a nonnegative integer scalar. 0 is ok for t for the first one.
The second one is more efficient but requires that t be positive integer. 0 is not possible there. A vector or array of t is possible as long as x is scalar or the same number of entries.
ektor on 16 May 2019
Could you give me a particular code example as I am bit confused. Thanks.