Question

Use the golden Section search method to find the minimum of f(x)= x/5−sin(x) . Start with the range of 0 to 3, i.e., xl=0, xu=3 . Show two iterations of the Golden Section Search Method by populating the following Table. Again please do all calculations in MATLAB and make sure you have included it in your submission.

i xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d

Answer 1

%matlab programe

clear all
close all
f=@(x) x/5-sin(x);
xl=0;
xu=3;
tol = 0.00000001; % tolarance   
N = 10; % iterations
tau = 0.618; %golden number
k=0;
x1=xl+(1-tau)*(xu-xl);
x2=xl+tau*(xu-xl);
f1=f(x1);   
f2=f(x2);

while ((abs(xu-xl)>tol) && (k<N))
k=k+1
xl
xu
if(f1<f2)
xu=x2;
x2=x1;
x1=xl+(1-tau)*(xu-xl);
  
f1=f(x1);
f2=f(x2);
  
  
else
xl=x1;
x1=x2;
x2=xl+tau*(xu-xl);
  
f1=f(x1);
f2=f(x2);

end
  

if(f1<f2)
  
minimum_at_point_x1 = x1
minimumValu = f1
  
else
minimum_at_point_x2 = x2
minimumValue = f2
  
end

  
end

output

k =  1
xl = 0
xu =  3
minimum_at_point_x2 =  1.1460
minimumValue = -0.68192
k =  2
xl = 0
xu =  1.8540
minimum_at_point_x2 =  1.4163
minimumValue = -0.70483
k =  3
xl =  0.70823
xu =  1.8540
minimum_at_point_x1 =  1.4163
minimumValu = -0.70483
k =  4
xl =  1.1460
xu =  1.8540
minimum_at_point_x2 =  1.4163
minimumValue = -0.70483
k =  5
xl =  1.1460
xu =  1.5835
minimum_at_point_x1 =  1.4163
minimumValu = -0.70483
k =  6
xl =  1.3131
xu =  1.5835
minimum_at_point_x1 =  1.3770
minimumValu = -0.70588
k =  7
xl =  1.3131
xu =  1.4803
minimum_at_point_x2 =  1.3770
minimumValue = -0.70588
k =  8
xl =  1.3131
xu =  1.4163
minimum_at_point_x1 =  1.3770
minimumValu = -0.70588
k =  9
xl =  1.3526
xu =  1.4163
minimum_at_point_x1 =  1.3676
minimumValu = -0.70591
k =  10
xl =  1.3526
xu =  1.3920
minimum_at_point_x2 =  1.3676
minimumValue = -0.70591