# 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.

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Hrishikesh Das on 24 Dec 2019
Commented: Walter Roberson on 24 Dec 2019
%%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)

Walter Roberson on 24 Dec 2019
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 on 24 Dec 2019
In the modified newton raphson method we have to calculate the slope with respect to the initial value. please comment. thank you.
Walter Roberson on 24 Dec 2019
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;