method for Numerical analysis

Prashant Kumar answered . 2024-12-21 02:53:46

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,

x =

   -1.4656


fval =

   4.4409e-16


fevals =

   153
   
   

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

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