Oddiy Nyuton Usuli bo'yicha python kodi:

           
import numpy as np

# F(x, y) funksiyalarini aniqlaymiz
def F(x):
    # Misol uchun fayldagi sistema:
    # 2*sin(x-1) - y - 1 = 0
    # 10*cos(y-1) + x - 4 = 0
    f1 = np.cos(x[0]) +2*x[1] - 1.5
    f2 = x[0]-np.sin(x[1] - 0.5) -1
    return np.array([f1, f2])


# hosilalar
def J(x):
    df1_dx = -np.sin(x[0])
    df1_dy = 2

    df2_dx = 1
    df2_dy = -np.cos(x[1] - 0.5)

    return np.array([
        [df1_dx, df1_dy],
        [df2_dx, df2_dy]
    ])


# Nyuton usuli
def nyuton(x0, eps=1e-3, max_iter=100):
    x = np.array(x0, dtype=float)

    for i in range(max_iter):
        Fx = F(x)
        Jx = J(x)

        # Δx ni topamiz
        delta = np.linalg.solve(Jx, -Fx)
        # print(delta)
        x = x + delta
        # print(x)
        # to'xtash sharti
        if np.linalg.norm(delta, ord=np.inf) < eps:
            print(f"Iteratsiya soni: {i+1}")
            return x

    print("Yechim topilmadi")
    return x


# Boshlang‘ich yaqinlashish
x0 = [1, 0.5]


solution = nyuton(x0)
print("Yechim:", solution)