Question

在以下代码中，我已经在Python中实现了Newtons方法。

import math
def Newton(f, dfdx, x, eps):
    f_value = f(x)
    iteration_counter = 0
    while abs(f_value) > eps and iteration_counter < 100:
        try:
            x = x - float(f_value)/dfdx(x)
        except ZeroDivisionError:
            print ("Error! - derivative zero for x = ", x)
            sys.exit(1)     # Abort with error
        f_value = f(x)
        iteration_counter += 1

    # Here, either a solution is found, or too many iterations
    if abs(f_value) > eps:
        iteration_counter = -1
    return x, iteration_counter
def f(x):
    return (math.cos(x)-math.sin(x))
def dfdx(x):
    return (-math.sin(x)-math.cos(x))
solution, no_iterations = Newton(f, dfdx, x=1, eps=1.0e-14)
if no_iterations > 0:    # Solution found
    print ("Number of function calls: %d" % (1 + 2*no_iterations))
    print ("A solution is: %f" % (solution))
else:
    print ("Solution not found!")

但是，现在我希望在同一时间间隔上绘制收敛图。这将是绝对误差，它是间隔[0,1]上迭代次数的函数。表示在x轴上的迭代次数，在y轴上具有相应的绝对误差

我试图使每次迭代产生一个带有绝对误差和迭代的2元组。这是我下面的代码，以及输出和图形。我的输出正确吗？该图应该看起来像这样吗？非常感谢所有帮助！我的代码中的迭代次数为3

import math
def Newton(f, dfdx, x, eps):
    f_value = f(x)
    iteration_counter = 0
    while abs(f_value) > eps and iteration_counter < 100:
        try:
            x = x - float(f_value)/dfdx(x)
            yield iteration_counter, abs(f(x))
        except ZeroDivisionError:
            print ("Error! - derivative zero for x = ", x)
            sys.exit(1)     # Abort with error
        f_value = f(x)
        iteration_counter += 1
    # Here, either a solution is found, or too many iterations
    if abs(f_value) > eps:
        iteration_counter = -1
    return x, iteration_counter
def f(x):
    return (math.cos(x)-math.sin(x))
def dfdx(x):
    return (-math.sin(x)-math.cos(x))

import numpy as np
np.array(list(Newton(f,dfdx, 1,10e-4)))

产生以下输出：

array([[0.00000000e+00, 4.74646213e-03],
   [1.00000000e+00, 1.78222779e-08]])

最后：

import numpy as np
import matplotlib.pyplot as plt

data = np.array(list(Newton(f,dfdx, 1, 10e-14)))
plt.plot(data[:,0], data[:,1])
plt.yscale('log')
plt.show()

产生图形：

Answer 1

您的Newton函数不应同时屈服并返回
使用更慢的收敛函数来测试结果

这就是我要做的：

import math
import sys
import numpy as np
import matplotlib.pyplot as plt


def newton(f, dfdx, x, eps):
    f_value = f(x)
    iteration_counter = 0
    while abs(f_value) > eps and iteration_counter < 100:
        try:
            x = x - float(f_value)/dfdx(x)
            yield iteration_counter, x, abs(f(x))
        except ZeroDivisionError:
            print ("Error! - derivative zero for x = ", x)
            sys.exit(1)     # Abort with error
        f_value = f(x)
        iteration_counter += 1

def f(x):
    return x ** 2 - 1.34

def dfdx(x):
    return 2 * x

data = np.array(list(newton(f, dfdx, 10, 10e-14)))

# plt.plot(data[:, 0], data[:, 1]) # x-axis: iteration, y-axis: x values
plt.plot(data[:, 0], data[:, 2]) # x-axis: iteration, y-axis: f(x) values
plt.yscale('log')
plt.show()

Python中的图形图

1 个答案: