Question

Optimization Technique: Consider one-variable function f(x). Use Newton Method to find minimum/maximum and plot the graph. Use Matlab.

Answer 1

clear;

close all;

clc;

f1 = @(x) exp(2*sin(x))-x; // %function //

x0 = 3;    // %initial guess //

maxIter = 1e6;    // %max iteration of the function //

epsilon = 10e-6; // %tolerance value //

[xk,i,error,errorVect] = NewtonsMethod(x0,maxIter,epsilon);    //% Calling the Newton's Method function //

fprintf('\nThe solution of the function exp(2*sin(x))-x = 0 is %4.5f in %u iteration with %e error.\n', xk,i,error)

xk; // % Printing Newton's Method Results in the console //

ymin = f1(xk);

errorVect;

//% Function for Newton's Method function //

// %Graphing the error vector //

x_axis = 1:1:length(errorVect);    // %x-axis values //

semilogy(x_axis,errorVect,'-mo'); // %graph of error vector in logarithmic y-axis //

grid on // %logarithmic grid (y-axis only) //

function[xk,i,error,errorVect] = NewtonsMethod(xk,maxIter,epsilon)

xold = xk;

for i = 1:maxIter

[f1x,df1x,ddf1x]=myfunction(xk);

xk = xk - (df1x/ddf1x);

error = abs(xk-xold);

xold = xk;

errorVect(i) = error;

if (error<epsilon)

break;

end

end

end

   //% Function for calling the value and the derivative of the function//

function [f,g,h] = myfunction(x)

f1 = @(z) exp(2*sin(z))-z;    // %function //

dx = 0.1;

ff = f1([x-dx, x, x+dx]); // % this is a 3-element vector //

f = ff(2);    // % the middle value is f1(x) //

if nargout > 1   // % (not sure why you use this) //

gg = diff(ff)./dx; // % this is a 2-element vector from diff(ff) //

g = gg(1);

h= diff(gg)./dx; // % single value from diff(gg) //

end

end