Question

我正在尝试解决大学的这一练习。我已经提交了下面的代码。但是，我对此并不完全满意。

任务是构建牛顿方法的实现，以解决以下非线性方程组：

为了学习牛顿法，除了上课外，我还观看了这段YouTube视频：https://www.youtube.com/watch?v=zPDp_ewoyhM

视频中的那个人解释了牛顿方法背后的数学过程，并手动进行了两次迭代。

我为此做了一个Python实现，并且视频中的示例代码运行良好。但是，视频中的示例涉及2个变量，而我的作业涉及3个变量。因此，我对其进行了修改。

那是代码：

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 [x+y+z-3, (x**2)+(y**2)+(z**2)-5,(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>0.000000000000000000000000000000000001:

        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 (iterative_newton([0,1,2]))

我很确定此代码绝对不是完全错误。 如果我将初始值输入为向量[0,1,2]，则我的代码将作为输出[0,1,2]返回。这是一个正确的答案，它可以解决上面的三个方程。

此外，如果输入[0,2,1]（略有不同的输入），该代码也有效，并且返回的答案也是正确的。

但是，如果我将初始值更改为[1,2,3]之类的值，则会得到一个奇怪的结果：527.7482，-1.63和2.14。

此结果没有任何意义。看第一个方程，如果输入这些值，您很容易看到（527）+（-1.63）+（2.14）不等于3。这是错误的。

如果我将输入值更改为接近正确的解决方案，例如[0.1,1.1,2.1]，它也会崩溃。

好的，牛顿法不能保证正确的收敛。我知道。除其他因素外，还取决于初始值。

我的实现有任何错误吗？还是矢量[1,2,3]只是一个“坏”初始值？

谢谢。

Answer 1

为了使您的代码更具可读性，我建议减少函数定义的数量。他们掩盖了正在发生的相对简单的计算。

我重写了自己的版本：

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

我找不到相同的行为：

>>>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])

甚至可以做出更差的猜测

>>>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])

Answer 2

我试图用更Python化的方式重写您的代码。希望对您有所帮助。也许错误是vector_b_f_output中x_delta_by_gauss(jacobian, vector_b_f_output)的符号？或雅可比行列中的某些缺失术语。

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) )

如果您想使用fsolve进行验证：

# 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)

Answer 3

回答这个问题的人帮助了我。但是，修改一行代码可以使所有功能在我的实现中正常工作。

由于我使用的是我提到的YouTube视频上描述的方法，因此我需要将向量值函数乘以（-1），从而修改向量中每个元素的值。

我是为function_example做的。但是，当我编写function_exercise时，我需要在没有负号的情况下解决我的作业。我错过了。

现在，它是固定的，并且即使具有多种多样的起始向量，它也可以充分发挥作用。

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])))))

使用牛顿法在Python中求解非线性方程组

3 个答案: