Optimize a function with a sum of series

Question

Dmytro Stovolos 2021-12-12

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1609365-optimize-a-function-with-a-sum-of-series

评论： Dmytro Stovolos 2021-12-13

采纳的回答： Abolfazl Chaman Motlagh

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.

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Paul 2021-12-12

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 2021-12-13

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

.

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Answer 1

Abolfazl Chaman Motlagh 2021-12-12

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https://ww2.mathworks.cn/matlabcentral/answers/1609365-optimize-a-function-with-a-sum-of-series#answer_853485

编辑：Abolfazl Chaman Motlagh 2021-12-12

在 MATLAB Online 中打开

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.