Optimize a function with a sum of series

Hello. Basically, I need to optimize the following function

The minimum value for 'n' is 2. Imagine we have n = 25. I have a sum of series so I tried the following:

syms i n

x0=[2:25];

i=[1:25];

f1= @(x)(x(1)+7)^8

f2 = @(x)((0.01*x(i)-i)^2)/i

All looks good until I try to sum both parts

fun = f1+symsum(f2,i,2,25);

Check for missing argument or incorrect argument data type in call to function 'symsum'.

If the previous step was correct, I would call "fminunc(fun,x0);" to optimize the function.

Could you explain why I get this error? 'i' should be a symbolic variable because of being created by syms call. Set of 'x' is defined. I must have missed something but can't find out what.

[Paul]

For n = 25, it looks like y is function of x1 - x25, i.e., . Is that correct?

What is ? That term doesn't show up in the equation for y.

[Dmytro Stovolos]

Hello Paul, is a starting value of the corresponding x value. For example, (1) = 2; (2) = 3. Yes, for n = 25 the function represents summation from to .

Answer 1

[Abolfazl Chaman Motlagh]

Hi , you don't need necessarily symboilic functionality for this problem. here is a simple solution. (you can run it in single .m file)

clear;
n = 25;
x0 = 1:n;
optimoptions = optimoptions(@fminunc,'Algorithm','quasi-newton',...
    'OptimalityTolerance',1e-10,'MaxFunctionEvaluations',1e7,...
    ,'PlotFcn','optimplotx','Display','iter'); % this line is just for illustration of convergence process
[X,fval,exitflag,output] = fminunc(@(x)(Cost(x,n)),x0,optimoptions);
function y=Cost(x,n)
y = (x(1)+7)^8;
for i=2:n
   y = y + (1/i)*((0.1*x(i)-i)^2);
end
end

you may ask why i use optimoptions so serious. in your problem the optimal solution is obvious because every single term is squared of a linear expression. so if x(1) became -7 and for i=2:n, x(i) = i/0.01. the function became 0. with this knowlegde i use more options to see the actual convergence of function. you can test that with no options, optimization stop at very high value point.

[Dmytro Stovolos]

Hello. Thanks for the answer.

If I run with optimoptions I get "Local minimum possible." but if I run without it I get "Local minimum found."

Indeed, fval is high without optimoptions but with this variable the solution is not guaranteed. Please correct me if I am wrong.