Solving a Nonlinear Equation using Newton-Raphson Method

It's required to solve that equation:

F(1)=1+(x(1)^2)-(x(2)^2)+((exp(x(1)))*cos(x(2)));

F(2)=(2*(x(1))*(x(2)))+((exp(x(1)))*sin(x(2)));

starting initials (-1,4)

5 iterations.

Please help me with the code ... I want the code to be with steps and iterations and if possible calculate the error also, please.

Answer 1

[arushi]

Hi Juan,

To solve the system of nonlinear equations using an iterative method like the Newton-Raphson method, we can write a MATLAB script that performs the iterations and computes the error in each step. Here is a step-by-step guide with the MATLAB code:

% Define the function F
F = @(x) [1 + x(1)^2 - x(2)^2 + exp(x(1))*cos(x(2));
          2*x(1)*x(2) + exp(x(1))*sin(x(2))];

% Define the Jacobian matrix of F
J = @(x) [2*x(1) + exp(x(1))*cos(x(2)), -2*x(2) - exp(x(1))*sin(x(2));
          2*x(2) + exp(x(1))*sin(x(2)), 2*x(1) + exp(x(1))*cos(x(2))];

% Initial guess
x0 = [-1; 4];

% Number of iterations
num_iterations = 5;

% Tolerance for convergence (if needed)
tolerance = 1e-6;

% Array to store errors
errors = zeros(num_iterations, 1);

% Iterative process
x = x0;
for k = 1:num_iterations
    % Calculate the function value and Jacobian at the current point
    F_val = F(x);
    J_val = J(x);
    
    % Newton-Raphson update
    delta_x = -J_val \ F_val;
    x = x + delta_x;
    
    % Calculate error as the norm of the update step
    error = norm(delta_x);
    errors(k) = error;
    
    % Display iteration details
    fprintf('Iteration %d:\n', k);
    fprintf('x = [%f, %f]\n', x(1), x(2));
    fprintf('F(x) = [%f, %f]\n', F_val(1), F_val(2));
    fprintf('Error = %f\n\n', error);
    
    % Check for convergence (optional)
    if error < tolerance
        fprintf('Convergence achieved.\n');
        break;
    end
end

Iteration 1:

x = [-0.504703, 2.012047]

F(x) = [-14.240462, -8.278412]

Error = 2.048726

Iteration 2:

x = [-0.374867, 1.308330]

F(x) = [-3.051423, -1.485110]

Error = 0.715594

Iteration 3:

x = [-0.301307, 1.175764]

F(x) = [-0.392853, -0.317060]

Error = 0.151608

Iteration 4:

x = [-0.293178, 1.172634]

F(x) = [-0.006912, -0.025662]

Error = 0.008711

Iteration 5:

x = [-0.293163, 1.172660]

F(x) = [0.000082, -0.000039]

Error = 0.000030

% Display final results
fprintf('Final solution after %d iterations:\n', k);

Final solution after 5 iterations:

fprintf('x = [%f, %f]\n', x(1), x(2));

x = [-0.293163, 1.172660]

fprintf('Errors in each iteration:\n');

Errors in each iteration:

disp(errors(1:k));

    2.0487
    0.7156
    0.1516
    0.0087
    0.0000

Hope this helps.