Question

The function should starts as:

function [x,fs,k]=Newton1(x0)

% enter your code here

end

>> [x,f,k]=Newton1(0)  x = 1.1641, f = -1.6653e-16, k = 7

>> [x,f,k]=Newton1(0.1)  x = 0.1972, f = 1.1102e-16, k = 5

>> [x,f,k]=Newton1(1.5)  x = 1.3111, f = -1.1102e-16, k = 6

>> [x,f,k]=Newton1(2)  Warning iteration diverged, x = 2.8808, f = -0.5624, k = 1000

Let the mathematical function f(x) be defined as: f(x) = exp(-0.5x) cos(5x)-0.5 , x 〉 0 Write a Matlab function called Newton1 that would find the zero based on a passing initial guess as an input argument x0. The function returns the estimated zero location x, the function value at the zero location (f) and the number of iteration k. The iteration function converges if f(%) 10000. When it diverges, it should state so using disp() command. The function should starts as: function [x,fs,k]-Newton1(xo) % enter your code here end Test the algorithm to find the zero as follows. >> [x,f,k-Newton! (O) → x-1.1641, f=-1.6653e-16, k = 7 >> [x,f,k-Newton! (0.1) → x = 0.1972, f= 1.1102e-16, k :5 >> [x,f,k]-Newton1(1.5) → x= 1.3111, f=-1.1102e-16, k = 6 >> [x,f,k-Newton! (2) → Warning iteration diverged, x-2.8808, f=-0.5624, k = 10001

Answer 1

%%% Function %%

function [ xz fs k ]=Newton1(x0)
format long
tol=5*eps;
c(1)=x0;
syms x;
f=exp(-0.5*x)*cos(5*x)-0.5 ;
for k=1:1000
l1=subs(f,c(k));
l2=subs(diff(f),c(k));
c(k+1)=c(k)-l1/l2;
l3=(subs(f,c(k+1)));

fs=l3;
xz=c(k+1);
if (abs(l3) < tol)

break;
end
end
end

%%% Test %%%

[x fs k]=Newton1(0);
fprintf(' x= %f \n',x);
fprintf(' fx= %1.32f \n',fs);
fprintf(' k= %f \n',k);
if k>1000
fprintf('Iteration diverged');
end

OUTPUT:

x= 1.164097
fx= -0.00000000000000006861264023336157
k= 7.000000
>>

%%%Test

clc;
close all;
clear all;
format long;

[x fs k]=Newton1(1.5);
fprintf(' x= %f \n',x);
fprintf(' fx= %1.32f \n',fs);
fprintf(' k= %f \n',k);
if k>1000
fprintf('Iteration diverged');
end

OUTPUT:

x= 1.311127
fx= -0.00000000000000002916434802119459
k= 6.000000

%%%Test

clc;
close all;
clear all;
format long;

[x fs k]=Newton1(2);
fprintf(' x= %f \n',x);
fprintf(' fx= %1.32f \n',fs);
fprintf(' k= %f \n',k);
if k>1000
fprintf('Iteration diverged');
end

OUTPUT:

x= 2.399596
fx= -0.24611144691801684070142641758139
k= 1000.000000
Iteration diverged>>