Newton methods for solving nonlinear
Afficher commentaires plus anciens
Please help me fix this code programe
main()
function main()
X0 = [25; 20; 4; 1];
% Áp dụng phương pháp Newton
X = newtons_method(X0)
% Hiển thị kết quả
disp('Kết quả:')
disp(['Nhiệt độ nước lớn nhất là: ', num2str(X(1))]);
end
function F = equations(X)
F = [X(1) - (X(2) + 0.5 * X(3) - 0.2 * X(4));
X(2) - 20;
X(3) - 5;
X(4) - 2];
end
function J = jacobian(X)
J = [1, -0.5, -0.1, 0.2;
0, 1, 0, 0;
0, 0, 1, 0;
0, 0, 0, 1];
end
function X = newtons_method(X0)
% Phương pháp Newton
max_iterations = 100;
tolerance = 1e-6;
X = X0;
for i = 1:max_iterations
F = equations(X);
J = jacobian(X);
delta_X = -J \ F;
X = X + delta_X;
if norm(delta_X, inf) < tolerance
break;
end
end
end
8 commentaires
Torsten
le 15 Nov 2023
Ok, the first row of your Jacobian is wrong and your system of equations is a linear one so that you could have solved it easier than with Newton's method, but the iteration converges. So what exactly is your question ?
Khai
le 15 Nov 2023
Khai
le 15 Nov 2023
Torsten
le 15 Nov 2023
We have enough problems in the world - why inventing a new one ?
John D'Errico
le 15 Nov 2023
Your "solver" works. At least it does on the trivial example problem you chose, which is purely linear, so it will "converge" in one iteration. So where is the problem?
Khai
le 16 Nov 2023
Khai
le 16 Nov 2023
Réponses (0)
Catégories
En savoir plus sur Contrast Adjustment dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
