Please help me identifying how this person makes his code

i am trying to learn how this person execute his codes but i am new to the MATLAB application. please help in identifying this code

function xs = secant( funct, x0, xi, es )
  %     Numerically evaluate root of func using Secant Method
  %     Solve a nonlinear equation using the secant method
  %     func : a function handle;
  %     a,b  : x-interval between initial and enclosing value;
  %     es   : is the desired error;
  %     returns xs: which is the final numerical solution of func;
      x(1)=x0; %  initial guess 1
      x(2)=xi; %  initial guess 2
      f(1)=feval(func,x(1)); %   function for initial guess 1
      f(2)=feval(func,x(2)); %   function for initial guess 2
      
  for i=1:10
    x(3)=x(2)-( f(2)*(x(1)-x(2)) )/( f(1)-f(2) ); %   Formula for Secant Method
    x(1)=x(2);  f(1)=feval(func,x(1));
    x(2)=x(3);  f(2)=feval(func,x(2));
    xnv(i)=x(3);    %   Value of x
    fxv(i)=feval(func,xnv(i));    % Value of function x
    ea(i)=abs((x(2)-x(1))/x(2));  % Absolute Error
    if (ea(i) < es)
      break
    end %  stop iterating if error less than tolerance
    
 end 
 fprintf('iteration\t|\t\txi\t\t|\tf(xi)\t\t\t|\tea\n');
 fprintf('----------------------------------------------------------------------------------------------------\n');
 for i=2:length(xnv)
     fprintf('%5d\t\t|\t%10.5f\t\t|\t%10.5f\t\t|\t%10.5f\n',i-1,xnv(i),fxv(i),ea(i));
     
 end
 fprintf('-----------------------------------------------------------------------------------------------------\n');
 xs=xnv(length(xnv));
 fprintf('\nfinal solution: \n\tx = %-10.10f\n',xnv(length(xnv)));
end

clear all
funct = @(x) 1-(400/9.81)*(3+x)/((3*x + 0.5*x^2)^3);
secant (funct, 0.5,2.5,0.01)

[muhammad hafiz]

it suppose to generate a table but unfortunately no table presented, i am at my wits end.

Answer 1

[Walter Roberson]

funct = @(x) 1-(400/9.81)*(3+x)/((3*x + 0.5*x^2)^3);
secant (funct, 0.5,2.5,0.01)

iteration	|		xi		|	f(xi)			|	ea
----------------------------------------------------------------------------------------------------
    1		|	  -0.53791		|	  32.66419		|	   5.55617
    2		|	   2.52586		|	   0.81952		|	   1.21296
    3		|	   2.60470		|	   0.83761		|	   0.03027
    4		|	  -1.04453		|	   5.59958		|	   3.49367
    5		|	   3.24659		|	   0.92468		|	   1.32173
    6		|	   4.09536		|	   0.96725		|	   0.20725
    7		|	 -15.19096		|	   1.00146		|	   1.26959
    8		|	 549.37294		|	   1.00000		|	   1.02765
    9		|	386945.37155		|	   1.00000		|	   0.99858
-----------------------------------------------------------------------------------------------------

final solution: 
	x = 386945.3715490098

ans = 3.8695e+05

function xs = secant( func, x0, xi, es )
  %     Numerically evaluate root of func using Secant Method
  %     Solve a nonlinear equation using the secant method
  %     func : a function handle;
  %     a,b  : x-interval between initial and enclosing value;
  %     es   : is the desired error;
  %     returns xs: which is the final numerical solution of func;
      x(1)=x0; %  initial guess 1
      x(2)=xi; %  initial guess 2
      f(1)=feval(func,x(1)); %   function for initial guess 1
      f(2)=feval(func,x(2)); %   function for initial guess 2
      
  for i=1:10
    x(3)=x(2)-( f(2)*(x(1)-x(2)) )/( f(1)-f(2) ); %   Formula for Secant Method
    x(1)=x(2);  f(1)=feval(func,x(1));
    x(2)=x(3);  f(2)=feval(func,x(2));
    xnv(i)=x(3);    %   Value of x
    fxv(i)=feval(func,xnv(i));    % Value of function x
    ea(i)=abs((x(2)-x(1))/x(2));  % Absolute Error
    if (ea(i) < es)
      break
    end %  stop iterating if error less than tolerance
    
 end 
 fprintf('iteration\t|\t\txi\t\t|\tf(xi)\t\t\t|\tea\n');
 fprintf('----------------------------------------------------------------------------------------------------\n');
 for i=2:length(xnv)
     fprintf('%5d\t\t|\t%10.5f\t\t|\t%10.5f\t\t|\t%10.5f\n',i-1,xnv(i),fxv(i),ea(i));
     
 end
 fprintf('-----------------------------------------------------------------------------------------------------\n');
 xs=xnv(length(xnv));
 fprintf('\nfinal solution: \n\tx = %-10.10f\n',xnv(length(xnv)));
end

[muhammad hafiz]

can you explain how the code works?

[Walter Roberson]

Please ask questions that are more specific. I do not have time to write several textbooks to talk about all of concepts that you might perhaps not already know.

[muhammad hafiz]

i see thank you for the help anyways i'll tak my time identifying how it works myself. ty

[Walter Roberson]

We get people posting saying that they had never taken any programming course until they started 4 days before. When you do not guide us about what information you are looking for, I have to assume that you are one of those people, who has no idea how programming works and who has no idea about calculus or approximation theory. I would have to go back to basics such as fundamental principles of arithmetic. This is not a productive use of your time or my time, when you could simply describe what concepts in the program that you are having difficulty with.