Question

Apply Newton’s method and the modified method to the equation f(x) = 1−cos(x−5) to approximate the a double root 5. Compare the results and demonstrate the superiority of the modified method. Numerically identify the rates of convergence of both the methods.

Answer 1

The iterative scheme for the Newton Raphson method is given below :

provided

The code for Newton Raphson method is attached below.

%Newton Raphson Method%%%%%%%%%%%%%%%%%%
clear all;
syms x;
f=1-cos(x-5); %Enter the Function here
g=diff(f); %The Derivative of the Function
n=input('Enter the number of decimal places:');
epsilon = 5*10^-(n+1);
x0 = input('Enter the intial approximation:');
for i=1:100
f0=vpa(subs(f,x,x0)); %Calculating the value of function at x0
f0_der=vpa(subs(g,x,x0)); %Calculating the value of function derivative at x0
y=x0-f0/f0_der; % The Formula
err=abs(y-x0);
if err<epsilon %checking the amount of error at each iteration
break
end
x0=y;
end
y = y - rem(y,10^-n); %Displaying upto required decimal places
fprintf('The Root is : %f \n',y);
fprintf('No. of Iterations : %d\n',i);

Output

Enter the number of decimal places:5
Enter the intial approximation:4.5
The Root is : 4.999990
No. of Iterations : 17

The iterative scheme for Modified Newton method is

where

The code for the Modified Newton scheme is attached below.

%%Modified Newton method%%%%%%%%
clear all;
syms x;
f=1-cos(x-5); %Enter the Function here
g=diff(f); %The Derivative of the Function
u=f/g;
v=diff(u);
n=input('Enter the number of decimal places:');
epsilon = 5*10^-(n+1);
x0 = input('Enter the intial approximation:');
for i=1:100
f0=vpa(subs(f,x,x0)); %Calculating the value of function at x0
f0_der=vpa(subs(g,x,x0)); %Calculating the value of function derivative at x0
u0=vpa(subs(u,x,x0));
u0_der=vpa(subs(v,x,x0));
y=x0-u0/u0_der; % The Formula
err=abs(y-x0);
if err<epsilon %checking the amount of error at each iteration
break
end
x0=y;
end
y = y - rem(y,10^-n); %Displaying upto required decimal places
fprintf('The Root is : %f \n',y);
fprintf('No. of Iterations : %d\n',i);

Output

modifiednewton_raphson
Enter the number of decimal places:5
Enter the intial approximation:4.5
The Root is : 4.999990
No. of Iterations : 3

Comparison and Superiority

Comparing the both methods, we can see the number of iteration for Modified Newton Scheme is 3 which is less than the number of iteration for Newton Raphson scheme ,17.

Rate of Convergence

For the rate of convergence, I am hereby attaching an image file below.