How can i implement my method ?

I know the classic newton method but i can not implement the fourier form please help me.

%Newton-Raphson method 
clear; 
clc 

%first plot the function 
plot(f) x=0:0.05:4; 
f=@(x)(x^3)+(x^2)-(x)-1; 
plot(x,f(x));
grid 
fd=@(x)3*x^2+2*x-1; 
x1=input ('x1='); 
tol=0.000001;
i = 0; 
while abs(f(x1)) > tol 
    f1=f(x1); 
    f1d=fd(x1); 
    x2=x1-(f1/f1d); 
    f2=f(x2); 
    x1=x2;
    i = i + 1; 
    fprintf('%9.6f %13.6f \n',x2,f2) 
end

[Walter Roberson]

What do you know the classic method to do? What difficulties are you encountering in implementing whatever it is you are implementing ?

[Rena Berman]

(Answers Dev) Restored edit

[Bruno Luong]

EDIT: Format the restored question

Answer 1

[Morgan]

Here is a function I've created that hopefully answers your question:

function [x,fval,fevals] = NewtonFourier(f,ab,tol)
% NEWTONFOURIER     Function that will efficiently find the local zero
%                   of an arbitrary function within a specified initial 
%                   guess and tolerance.
%
% [x,fval,fevals] = NEWTONFOURIER(f,x0,tol);
%
% INPUT ARGUMENTS
% ================
% f             Function handle to function to be searched
% ab            Interval [a,b] in which root is assumed to be
% tol           Desired relative error
%
% OUTPUT ARGUMENTS
% ================
% x         The coordinate where f(x) = 0
% fval      The evaluation of f(x)
% fevals    Number of times the func was evaluated

    % CHECK IF PROBLEM IS LIKELY UNSOLVABLE
    if f(ab(1))*f(ab(2)) > 0
        warning('f(a)f(b) > 0, there likely is no solution in [a,b].');
    end

    % DETERMINE APPROXIMATE DERIVATIVES OF func
    fp  = @(x) (f(x+tol)-f(x-tol))/(2*tol);
    
    % INITIALIZE fevals
    fevals = 2;

    % INITIALIZE xn AND zn
    xn = ab(2);
    zn = ab(1);

    % MAIN LOOP
    err = Inf;
    while err > tol
        
        % Update root estimates
        fxn = f(xn)/fp(xn);
        xn  = xn - fxn;
        fzn = f(zn)/fp(zn);
        zn  = zn - fzn;
        
        % Determine if converged
        err = abs((fxn+fzn)/2);
        
        % Update function evaluation counter
        fevals = fevals+6;

    end
    
    % RETURN OUTPUTS
    x      = xn;
    fval   = f(xn);
    fevals = fevals+1;
    
end

I've also written a demo file that solves the specific problem you've referenced in your question here:

% demo_NewtonFourier.m

% Initialize MATLAB
clear variables
close all
clc

% Define NewtonFourier.m Inputs
f   = @(x) x.^3 + x.^2 + 1;
tol = 1e-6;
ab  = [ -2 +2 ];

% Plot Function
x = ab(1) : 0.05 : ab(2);
y = f(x);
    figure(1);
    hold on
    plot(x,y,'-k','LineWidth',2);
    plot(x,0*x,'--k','LineWidth',1);
    hold off
    xlabel('$x$','Interpreter','latex');
    ylabel('$f(x)$','Interpreter','latex');
    title(['$ f(x) = ' latex(f(sym('x'))) '$'],'Interpreter','latex');
    set(gca, 'FontSize', 12, 'FontName', 'Times', ...
             'XMinorTick', 'on', 'YMinorTick', 'on', ...
             'TickLength', [0.015, 0.0015]);
    
% Call NewtonFourier.m 
[x,fval,fevals] = NewtonFourier(f,ab,tol)

The outputs of this demo file include the following figure showing the only real solution to be approximately -1.46,

And

x =

   -1.4656


fval =

   4.4409e-16


fevals =

   153
   
   

If you require explanation of the code snippets or need anything else, let me know!

• Morgan Blankenship, B.S., M.S., EIT

[Hazel Can]

Morgan Thank you so much for your answer and effort! I want to ask to you 2 questions.

First : I misspelled the sign of the question, so the question should have been x=x^3-x^2-1. Changing the sign in the function will be enough, right?

Second: Question is x=x^3-x^2-1, not F(x)=x^3-x^2-1. Have you considered this? Is there any differences?

[Morgan]

No problem Hazel, my pleasure.

First Answer: Yes changing the sign in the code will still work. Newton-Raphson method and Newton-Fourier method are generic zero-finding techniques. The only problem with these techniques is you have to provide either an initial guess or the bounds [a,b] where you think the zero is located between.

Second Answer: My apologies, I misread the question at first. The only difference with this is you have to solve

to be

The right hand side of this equation becomes your new function handle that you give to NewtonFourier.m function.

[Hazel Can]

Thank you so much Morgan! It's very useful answers for me.

Have a good day :)

[Walter Roberson]

Are you prepared to pay Morgan private consulting fees for providing a private solution to your question?