Error in line 12. I cant use 'inline' command. How to fix it? There are 3 iterations. Equation is f(x)= e^-x-x and the derivative is -e^-x-1. Using Newton Raphson Method

%%%% MATLAB CODE of Newton Raphson Method 
%%% Method for finding the ROOT of
%%%% equation f(x)= exp(1)^-x-x
format short
clear 
clc
syms x
% Write the function here
% f= @ (x) exp(1)^-x-x;
f = @ (x) exp(1)^-x-x;
df = diff (f,x);
dfx = inline(df);   %derivative of funtion df
x0 = 0;     %initial guess
n = 4;  % number of decimal places
Variables={'Iter','x','f_x0','Error'};

iter = 1;
 err = abs (f(x0));
epsilon = 5*10^(-n-1);
itermax = 70;
HG = [];
if dfx(x0)<10^(-9)
    disp('Wrong choice of initial Guess');
else
    while (iter<=itermax && err>epsilon)
        x1 = x0-f(x0)/dfx(x0);
        err = abs (f(x0));
        HG = [iter x0 f(x0) err];
        iter = iter+1;
        x0 = x1;
    end
end

disp ('======================================')
disp ('Output Table with Iteration wise')

Result= array2table(HG);
Result.Properties.VariableNames(1:size(HG,2)) = Variables

x0 = x0-rem(x0,10^-n);

fprintf('Converged solution after %d iterations \n',iter);
fprintf('Root is %1.5f \n',x0)

[Stephen23]

I get the feeling that INLINE and symbolic is not the ... neatest approach.

Perhaps other regular users have some suggestions?

Answer 1

[Askic V]

Try this:

format short
clear 
clc
syms x
% Write the function here
% f= @ (x) exp(1)^-x-x;
f = @ (x) exp(1)^-x-x;
df = diff (f,x);
dfx = @(x) eval(df); 
%inline(df);   %derivative of funtion df
% inline will be removed in future releases
x0 = 0;     %initial guess
n = 4;  % number of decimal places
Variables={'Iter','x','f_x0','Error'};
iter = 1;
err = abs (f(x0));
epsilon = 5*10^(-n-1);
itermax = 70;
HG = [];
% Use abs function here!
if abs ( dfx(x0) ) < 10^(-9)
    disp('Wrong choice of initial Guess');
else
    while (iter<=itermax && err>epsilon)
        x1 = x0-f(x0)/dfx(x0);
        err = abs (f(x0));
        HG = [iter x0 f(x0) err];
        iter = iter+1;
        x0 = x1;
    end
end
disp ('======================================')

======================================

disp ('Output Table with Iteration wise')

Output Table with Iteration wise

Result= array2table(HG);
Result.Properties.VariableNames(1:size(HG,2)) = Variables

Result = 1×4 table
    Iter       x          f_x0         Error   
    ____    _______    __________    __________

     4      0.56714    1.9648e-07    1.9648e-07

x0 = x0-rem(x0,10^-n);
fprintf('Converged solution after %d iterations \n',iter);

Converged solution after 5 iterations 

fprintf('Root is %1.5f \n',x0)

Root is 0.56710 

[Christian]

The root is correct. I mean there should be 3 iterations in the table with 3 column.

[Askic V]

The number of iterations generally depends of initial guess and the epsilon.

In your while loop the stop criteria is maximum iteration reached or error is below predifed epsilon. If you want 3 iterations, the easiest way (for the same initial guess 0) is to make epsilon larger, such as:

epsilon = 5*10^(-n+1);

This would produce the following result:

Iter       x         f_x0         Error  
    ____    _______    _________    _________

     3      0.56631    0.0013045    0.0013045