Solving a non-linear system of equations in Python using Newton's Method

To make your code more readable, I would suggest reducing the number of function definitions. They obscure the relatively simple computations which are happening.

I rewrote my own version:

def iter_newton(X,function,jacobian,imax = 1e6,tol = 1e-5):
    for i in range(int(imax)):
        J = jacobian(X) # calculate jacobian J = df(X)/dY(X) 
        Y = function(X) # calculate function Y = f(X)
        dX = np.linalg.solve(J,Y) # solve for increment from JdX = Y 
        X -= dX # step X by dX 
        if np.linalg.norm(dX)<tol: # break if converged
            print('converged.')
            break
    return X

I don't find the same behavior:

>>>X_0 = np.array([1,2,3],dtype=float)
>>>iter_newton(X_0,function_exercise,jacobian_exercise)
converged.
array([9.26836542e-18, 2.00000000e+00, 1.00000000e+00])

even works for far worse guesses

>>>X_0 = np.array([13.4,-2,31],dtype=float)
>>>iter_newton(X_0,function_exercise,jacobian_exercise)
converged.
array([1.59654153e-18, 2.00000000e+00, 1.00000000e+00])

I tried to rewrite your code in a more Pythonic way. I hope it helps. Maybe the error is the sign of vector_b_f_output in x_delta_by_gauss(jacobian, vector_b_f_output)? or some missing term in the Jacobian.

import numpy as np

# Example from the video:
# from youtube https://www.youtube.com/watch?v=zPDp_ewoyhM
def jacobian_example(xy):
    x, y = xy
    return [[1, 2],
            [2*x, 8*y]]

def function_example(xy):
    x, y = xy
    return [x + 2*y - 2, x**2 + 4*y**2 - 4]

# From the exercise:
def function_exercise(xyz):
    x, y, z = xyz
    return [x + y + z - 3,
            x**2 + y**2 + z**2 - 5,
            np.exp(x) + x*y - x*z - 1]

def jacobian_exercise(xyz):
    x, y, z = xyz
    return [[1, 1, 1],
            [2*x, 2*y, 2*z],
            [np.exp(x) + y - z, x, -x]]



def iterative_newton(fun, x_init, jacobian):
    max_iter = 50
    epsilon = 1e-8

    x_last = x_init

    for k in range(max_iter):
        # Solve J(xn)*( xn+1 - xn ) = -F(xn):
        J = np.array(jacobian(x_last))
        F = np.array(fun(x_last))

        diff = np.linalg.solve( J, -F )
        x_last = x_last + diff

        # Stop condition:
        if np.linalg.norm(diff) < epsilon:
            print('convergence!, nre iter:', k )
            break

    else: # only if the for loop end 'naturally'
        print('not converged')

    return x_last

# For the exercice:
x_sol = iterative_newton(function_exercise, [2.0,1.0,2.0], jacobian_exercise)
print('solution exercice:', x_sol )
print('F(sol)', function_exercise(x_sol) )

# For the example:
x_sol = iterative_newton(function_example, [1.0,2.0], jacobian_example)
print('solution example:', x_sol )
print( function_example(x_sol) )

If you want to verify using fsolve:

# Verification using fsvole from Scipy
from scipy.optimize import fsolve

x0 = [2, 2, 2]
sol = fsolve(function_exercise, x0, fprime=jacobian_exercise, full_output=1)
print('solution exercice fsolve:', sol)

The guys that answered this question helped me. However, modifying one line of code made everything work in my implementation.

Since I am using the approach described on the YouTube video that I mentioned, I need to multiply the Vector-valued function by (-1), which modifies the value of each element of the vector.

I did this for the function_example. However, when I coded function_exercise, the one that I needed to solve for my homework without the negative sign. I missed it.

Now, it is fixed and it works fully, even with very diverse starting vectors.

import numpy as np

#### example from youtube https://www.youtube.com/watch?v=zPDp_ewoyhM

def jacobian_example(x,y):

    return [[1,2],[2*x,8*y]]

def function_example(x,y):

    return [(-1)*(x+(2*y)-2),(-1)*((x**2)+(4*(y**2))-4)]
####################################################################


### agora com os dados do exercício

def jacobian_exercise(x,y,z):

    return [[1,1,1],[2*x,2*y,2*z],[np.exp(x),x,-x]]

#print (jacobian_exercise(1,2,3))
jotinha  = (jacobian_exercise(1,2,3))

def function_exercise(x,y,z):

    return [(-1)*(x+y+z-3),(-1)*((x**2)+(y**2)+(z**2)-5),(-1)*((np.exp(x))+(x*y)-(x*z)-1)]

#print (function_exercise(1,2,3))
bezao = (function_exercise(1,2,3))

def x_delta_by_gauss(J,b):

    return np.linalg.solve(J,b)

print (x_delta_by_gauss(jotinha, bezao))
x_delta_test = x_delta_by_gauss(jotinha,bezao)

def x_plus_1(x_delta,x_previous):

    x_next = x_previous + x_delta

    return x_next

print (x_plus_1(x_delta_test,[1,2,3]))

def newton_method(x_init):

    first = x_init[0]

    second = x_init[1]

    third = x_init[2]

    jacobian = jacobian_exercise(first, second, third)

    vector_b_f_output = function_exercise(first, second, third)

    x_delta = x_delta_by_gauss(jacobian, vector_b_f_output)

    x_plus_1 = x_delta + x_init

    return x_plus_1

def iterative_newton(x_init):

    counter = 0

    x_old = x_init
   #print ("x_old", x_old)

    x_new = newton_method(x_old)
   #print ("x_new", x_new)

    diff = np.linalg.norm(x_old-x_new)
   #print (diff)

    while diff>0.0000000000001:

        counter += 1

       #print ("x_old", x_old)
        x_new = newton_method(x_old)
       #print ("x_new", x_new)

        diff = np.linalg.norm(x_old-x_new)
       #print (diff)

        x_old = x_new

    convergent_val = x_new
   #print (counter)

    return convergent_val

#print (iterative_newton([1,2]))
print (list(map(float,(iterative_newton([100,200,3])))))