Question

*1. (This problem is to be solved manually, but you can use MATLAB or any other software as a calculator only) Consider the problem of finding the minimum of the following function for x>0 0.65an 0.75 fx) 0.65- 1+x2 a) First find a bracket for the minimum. b) Using the bracket found in Part (a) above, perform two iterations of:. Golden section search method . Quadratic interpolation method

Answer 1

Matlab code for Golden section search clear all close all %function for which minimum have to find f-ê(x) 0.65-(0.75./(1+x.2))-0.65.*x.*atan (1./x) xx-linspace (0,5,100) fprintf( Function for finding minima using Golden section method.n disp(f) %plotting of the function plot (xx,yy) title( f (x) vs. x plot' xlabel() ylabel('f(x) gr-1/(1.61803398875) err-1;c 0; %bracket for Golden section x-low= 0 ; x-high-5; %error convergence cnn=0 . 005 ; fprint f ("AnBracket for the minima [%f %f].\n',x-low, x-high ) %Loop for Golden section search method i=0; fprintf 'All iterations for Golden section method.\n\n) fprintf('iter cnt \tx lowltx hightx1\tx2\tf (x1)tf (x2)\n') loop for golden section iterations while err>cnn x2-x-10壯(X-high-x-low)" gr ; In',i-1,x_low, x high, x1, x2, f(x1), f (x2)) if f (x1)<f (x2) x-high-x2 ; else end err-abs (x_high-x low);

C-c+1 %counter and error count (C)c error (c)Ferr end fprintf(InTotal number of iterations for getting minima of accuracy 0.005 is %d.\n",i) hold on plot (x1,f (xl),r*) 용욤욤응%욤욤응%号号%%号号%%号号%%응응% End of Code 응응용용응응용욤응응욤욤응응욤욤융응욤욤응응욤욤 Function for finding minima using Golden section method f(x)e(x)0.65-(0.75./(1+x. *2))-0.65.*x. *atan(1./x) Bracket for the minima [0.000000 5.000000] All iterations for Golden section method iter cnt x low x high x1 x2 f(xl) f (x2) 0 0.000000 5.000000 1.909830 3.090170 -0. 110168 0.049733 1 0.000000 3.090170 1.180340 1.909830 -0.202647 -0.110168 2 0.000000 1.909830 0.729490 1.180340 0. 285486 0.202647 3 0.000000 1.180340 0.450850 0.729490 -0. 309504 -0.285486 4 0.000000 0.729490 0.278640 0.450850 0. 281244 0.309504 5 0.278640 0.729490 0.450850 0.557281 -0. 309504-0.307102 6 0.278640 0.557281 0.385072 0.450850 0. 304314 -0.309504 7 0.385072 0.557281 0.450850 0.491503 -0. 309504 -0.309959 8 0.450850 0.557281 0.491503 0.516628 0. 309959 0.309346 9 0.450850 0.516628 0.475975 0.491503 -0. 310007 -0.309959 10 0. 450850 0.491503 0.466378 0. 475975-0.309903 -0. 310007 11 0. 466378 0.491503 0.475975 0.481906-0.310007 -0. 310020 12 0.475975 0.491503 0.481906 0. 485572 -0.310020 -0. 310008 13 0. 475975 0.485572 0.479640 0.481906-0.310020-0. 310020 14 0. 479640 0.485572 0.481906 0. 483306 -0.310020 -0.310017 Total number of iterations for getting minima of accuracy 0.005 is 15

f(x) vs. x plot 0.05 -0.1 0.15 0.2 0.25 -0.3 0.35 051152 25 3354 45 Published with MATLAB R2018a

%Matlab code for Golden section search
clear all
close all
%function for which minimum have to find

f=@(x) 0.65-(0.75./(1+x.^2))-0.65.*x.*atan(1./x);
xx=linspace(0,5,100);
yy=f(xx);

fprintf('Function for finding minima using Golden section method.\n f(x)=')
disp(f)

%plotting of the function
plot(xx,yy)
title('f(x) vs. x plot')
xlabel('x')
ylabel('f(x)')

gr=1/(1.61803398875);
err=1;c=0;
%bracket for Golden section

x_low=0; x_high=5;

%error convergence
cnn=0.005;

fprintf('\nBracket for the minima [%f %f].\n',x_low,x_high)

%Loop for Golden section search method
i=0;
fprintf('All iterations for Golden section method.\n\n')
fprintf('iter cnt \tx_low\tx_high\tx1\tx2\tf(x1)\tf(x2)\n')

%loop for golden section iterations
while err>cnn
    i=i+1;
    x1=x_high-(x_high-x_low)*gr;
    x2=x_low+(x_high-x_low)*gr;
  
  
    fprintf('%d \t%f \t%f \t%f \t%f \t%f \t%f \n',i-1,x_low,x_high,x1,x2,f(x1),f(x2))
  
    if f(x1)<f(x2)
        x_high=x2;
    else
        x_low=x1;
    end
  
    err=abs(x_high-x_low);
    c=c+1;
    %counter and error
    count(c)=c;
    error(c)=err;
    x_h(c)=x_high;
    x_l(c)=x_low;
  
end

fprintf('\nTotal number of iterations for getting minima of accuracy 0.005 is %d.\n',i)

hold on
plot(x1,f(x1),'r*')

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