Quadratic Approximation Method to find the maximum of f(x). got into an infinite loop

Question

Alper Sahin 2021-12-7

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1605555-quadratic-approximation-method-to-find-the-maximum-of-f-x-got-into-an-infinite-loop

编辑： Alagu Sankar Esakkiappan 2021-12-10

    clear all;
clc;
syms x;
fprintf('QUADRATIC APPROXIMATION METHOD\n\n');
f=log(x)*sin(x^2);
x0=1;
x1=1.8;
x2=2;
err=10;
epsilon=1e-11;
i=0;
fprintf(' i      x0      x1      x2      x3\n');
while err>epsilon
    f0=double(subs(f,x,x0));
    f1=double(subs(f,x,x1));
    f2=double(subs(f,x,x2));
    x3=(((x1^2-x2^2)*f0+(x2^2-x0^2)*f1+(x0^2-x1^2)*f2))/(2*((x1-x2)*f0+(x2-x0)*f1+(x0-x1)*f2));
    f3=double(subs(f,x,x3));
    if x0<x3 && x3<x1
        if f3<f1
            x0new=x3;
            x1new=x1;
            x2new=x2;
        elseif f3>f1
            x0new=x0;
            x1new=x3;
            x2new=x1;            
        end
    elseif x1<x3 && x3<x2
        if f3<=f1
            x0new=x0;
            x1new=x1;
            x2new=x3;
        elseif f3>f1
            x0new=x1;
            x1new=x3;
            x2new=x2;            
        end
    else
        fprintf('x0 is outside of [x0,x2]\n');
        fprintf('x0=%.30f\n',x0);
        fprintf('x1=%.30f\n',x1);
        fprintf('x2=%.30f\n',x2);
        fprintf('x3=%.30f\n',x3);
    end
    fprintf('%2d %5.5f %5.5f %5.5f %5.5f %5.5f %5.5f %5.5f %5.5f\n',i,x0,x1,x2,x3,f0,f1,f2,f3);
    err=abs(x3-x1);
    x0=x0new;
    x1=x1new;
    x2=x2new;
    i=i+1;
end
xmax=x3;
fmax=double(subs(f,x,xmax));
fprintf('\nTherefore, f(x) maximum at x=%5.5f with the function value f(x)=%5.5f.\n',xmax,fmax);clear all;

I tried to find max with quadratic approach but it got into an infinite loop what do you think is wrong

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Answer 1

Alagu Sankar Esakkiappan 2021-12-10

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1605555-quadratic-approximation-method-to-find-the-maximum-of-f-x-got-into-an-infinite-loop#answer_852150

编辑：Alagu Sankar Esakkiappan 2021-12-10

Hi Alper,

I see that you're trying to find maxiumum value of a function using Quadratic approach. There is no problem per se with the implementation. Only that epsilon is initialzed to a far more optimistic value than the original convergence point for err.

Coming to your Code, I see that the value of err converges to around 1.297e-9 even after 1000 iterations. Since the condition for your while loop is ( err > epsilon ) and err never goes below epsilon ( err converged to 1.297e-9, whereas epsilon is 1e-11), you're stuck in an infinite loop. You may allow a reasonable degree of error ( say 1e-8 ) to your epsilon (or) more fine tune your algorithm if more accuracy is needed.