Newton_iterative

0.背景

惊讶的发现，在过去的一年里，我的python编程技巧几乎没有进步。为了提高编程能力和拥抱AI时代，有必要加强python在真正的数值分析方面的能力，因此开始学习鸣凤老师的课程

第一节课主要讲解了牛顿迭代法求根的python实现，我们这里把实现的代码记录下来，方便后续的复习(本人记忆太差了)。

• 函数在整个定义域内最好是二阶可导的
• 起始点对求根计算影响重大，可以增加一些别的判断手段进行试错

1. 代码实现

python_advance_0001.py
#!/usr/bin/env python
from sys import argv, exit
from sympy import *


def function():
    """Equation of expression"""
    x = symbols('x')
    return x*exp(-x)*sin(x) + 2*cos(x) - 0.25


def diff_function():
    """The first derivative"""
    return function().diff()


def Newton_iterative(x0, eps, maxiter):
    """Newton_iterative"""
    x = symnols("x")
    fh = function()
    dfh = diff_function()
    x_n = x0

    # output header
    print("%5s %25s %25s" % ("Iter", "Approximate_Solution", "Error_Rate"))
    k = 0
    errval = 0
    for k in range(maxiter):
        x_b = x_n
        fx = fh.evalf(subs = {x : x_b})
        dfx = dfh.evalf(subs = {x : x_b})
        x_n = x_b - fx/dfx

        # 这里注意，求根嘛 看当前y值与X轴的差异
        errval = abs(fh.evalf(subs = {x: x_n}))
        print("%5d %25.15f %25.15f" % (k+1, x_n, errval))

        if errval <= eps:
            break
    
    if k+1 <= maxiter-1:
        print(f"Under the accuracy:{eps}, the approximate solution of the equation"+ 
            f" is {x_n} and the error rate is {errval}.")
    else:
        print("The maximum number of iterations is exceeded, convergence may not occur!")

    return None


if __name__ == "__main__":
    if len(argv) != 4:
        exit("Usage: python python_advance_0001.py [initial_x] [accuracy_requirement] [iter_count]")

    Newton_iterative(argv[1], argv[2], argv[3])