I am unable to understand this. I was trying to solve a non linear eqation with newton raphson and modified newton raphson method. Both methods give different results. Below is my code. can you find the error. Both are separate files.

%%Newton raphson method
clc
clear all
close all
%%Initial condition
x0 =0.92;
maxiteration = 50;
tolX = 1e-04;
f = @(x) x^2-2*x+1; %function to be solved
%%Computation
x = x0;
xold = x0;
for i = 1:maxiteration;
    df = 2*x-2;
    x = x-f(x)/df;
    err(i) = norm(x-xold);
    xold = x;
    if (err(i)<tolX)
        break;
    end
end
msg = ['The solution converged in ', num2str(i),'th iterations.'];
disp(msg)
msg1 = ['converged numerical solution(critical values):'];
disp(msg1)
fprintf('x = %.4f\n',x);  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Modified newton Raphson method
clc
clear all
close all
%%Initial condition
x0 =0.92;
maxiteration = 50;
tolX = 1e-04;
f = @(x) x^2-2*x+1;
%%Computation
x = x0;
xold = x0;
for i = 1:maxiteration;
    df = 2*x0-2;
    x = x-f(x)/df;
    err(i) = norm(x-xold);
    xold = x;
    if (err(i)<tolX)
        break;
    end
end
msg = ['The solution converged in ', num2str(i),'th iterations.'];
disp(msg)
msg1 = ['converged numerical solution:'];
disp(msg1)
fprintf('x = %.4f\n',x)

Answer 1

[Walter Roberson]

    df = 2*x0-2;

That is incorrect. You are creating the slope based upon the initial point, not based upon the current point.

[Hrishikesh Das]

thank you for your response. could you please tell me the solution?

[Walter Roberson]

Look at the code for your other method. How do you calculate df there?

[Hrishikesh Das]

In the modified newton raphson method we have to calculate the slope with respect to the initial value. please comment. thank you.

[Walter Roberson]

Look at your existing code that works. It says

    df = 2*x-2;

Now how can you use the ideas there to modify the non-working line,

    df = 2*x0-2;