% Problem 1: Approximating the root of x^2 - 3 using Newton's method


clear all
close all


% Define the function and its derivative symbolically
fun = @(x) x^2 - 3;
df = @(x) 2*x;

% Initial guess for the root
x0 = -1.5;

% Find the actual root using fzero as a reference
actualRoot = fzero(fun, x0);

% Display header for the iteration information table
fprintf('  n |       slope  |      Computed  |       Actual  | Absolute Error\n');
fprintf([repmat('-', 1, 70) '\n']);

% Iterate to refine the root approximation using Newton's method
n = 0;
while abs(fun(x0)) > 1e-4
    n = n + 1;
    
    % Calculate the slope and y-intercept of the tangent line
    slope = df(x0);
    b = fun(x0) - slope * x0;
    
    % Update the root approximation using the tangent line
    x0 = -b / slope;

    % Display the current iteration information in a table
    fprintf(' %2d | %5.10f | %5.10f | %5.10f | %5.10f\n', ...
        n, slope, x0, actualRoot, abs(fun(x0)));
end

--------------------------------------------------------------------------------------

% Problem 2: Approximating the root of x^3 - 3*x + 1 using Newton's method

clear all
close all


% Define the function and its derivative symbolically
fun = @(x) x^3 - 3*x + 1;
df = @(x)3*x^2-3;

% Initial guess for the root
x0 = -1.5;

% Find the actual root using fzero as a reference
actualRoot = fzero(fun, x0);

% Display header for the iteration information table
fprintf('  n |   slope      |   Computed     |     Actual    | Absolute Error\n');
fprintf([repmat('-', 1,67) '\n']);

% Iterate to refine the root approximation using Newton's method
n = 0;
while abs(fun(x0)) > 1e-5
    n = n + 1;
    
    % Calculate the slope and y-intercept of the tangent line
    slope = df(x0);
    b = fun(x0) - slope * x0;
    
    % Update the root approximation using the tangent line
    x0 = -b / slope;

    % Display the current iteration information in a table
    fprintf(' %2d | %7.10f | %9.10f | %11.10f | %11.10f\n', ...
        n, slope, x0, actualRoot, abs(fun(x0)));
end

---------------------------------------------------------------------------------------
% Problem 3: Approximating the root of x^3 - 2*x^2 - 5 using Newton's method

clear all
close all


% Define the function and its derivative symbolically
fun = @(x) x^3 - 2*x^2 - 5;
df = @(x) 3*x^2-4*x;

% Initial guess for the root
x0 = 2.5;

% Find the actual root using fzero as a reference
actualRoot = fzero(fun, x0);

% Display header for the iteration information table
fprintf('  n |         slope   | Computed      |   Actual    | Absolute Error\n');
fprintf([repmat('-', 1, 68) '\n']);

% Iterate to refine the root approximation using Newton's method
n = 0;
while abs(fun(x0)) > 1e-6
    n = n + 1;
    
    % Calculate the slope and y-intercept of the tangent line
    slope = df(x0);
    b = fun(x0) - slope * x0;
    
    % Update the root approximation using the tangent line
    x0 = -b / slope;

    % Display the current iteration information in a table
    fprintf(' %2d | %15.10f | %5.10f | %5.10f | %5.10f\n', ...
        n, slope, x0, actualRoot, abs(fun(x0)));
end