Solving Newton Raphson Method

I do not know how to add x1,x2,x3,x4 in my code!? And in command window "iteration number 4 appears with no Ea"!? , so please I appreciate any help for me and my friends in this project

%clear all
%clc
x1=input('x1 = ');  %value of x initial =1
Es=input ('Es = ');  % value of Es = 0.5%

i=1;
f=@(x) (x)^3 -9*(x)^2 +3.8197;
fd=@(x) 3*(x)^2 - 18*(x);% derivative of f(x)

while i<4
    f1=f(x1);
    f1d=fd(x1);
    x2=x1-( f1/f1d );  %Newton Raphson formula
    %the iteration proceeds till certian accccuracy is attained
    Ea=abs((x2-x1)/x2)*100;  
   
    
    f2=f(x2);
    x1=x2;
     disp('Ea= ')
    disp(Ea)
  
    i=i+1;
  %fprintf : This means you can format how the data is printed in
       %such a manner as to make it easy to read.
    fprintf('iteration number =  %d.\n', i)
    % d stand for integer
end
break

Command Window:

x1 = 1

Es = 0.005

Ea=

38.636006543619509

iteration number = 2.

Ea=

6.290737359979779

iteration number = 3.

Ea=

0.170808133804928

iteration number = 4.

>>

Answer 1

[Jan]

This is the expected behavior:

while i<4
  disp('Ea = ');
  i = i + 1;
  fprintf('iteration number =  %d.\n', i);
end

This stops after displaying the message "iteration number = 4."

What does "add x1,x2,x3,x4" mean?

[Shinji Edugai]

x1 means value of x in iteration 1 , xl that i wrote in script here means x0 or initial

[Jan]

Maybe you mean:

x = zeros(1, 4);
for k = 1:4
    x(k) = ...
end