标签：方程解 return 迭代 Aitken theta x0 x1 def 法求

埃特肯加速法求方程解Aitken!

import numpy as np
import matplotlib.pyplot as plt

def f(x):
    return x**3 - x**2 - 1

构造迭代函数

\[\begin{aligned} x^{3} - x^{2} - 1 &= 0 \\ x = \sqrt{x^{3} - 1} \\ \end{aligned} \]

令\(\varphi(x) = \sqrt{x^{3} - 1}\)为迭代公式

def phi(x):
    return (x**3 - 1)**(1/2)

x0 = 1.5 # 初始值
while True:
    x1 = phi(x0)
    if abs(x1 - x0) < 1e-12:
        break
    x2 = phi(x1)
    theta = (x2 - x1)/(x1 - x0)
    x0 = (phi(x0) - theta * x0)/(1 - theta) 
    print(x0)

1.465365431478883
1.4655712202898679
1.4655712318767682

选取其它迭代函数

\[\begin{aligned} x^{3} - x^{2} - 1 &= 0 \\ x = {(x^{2} + 1)} ^ {\frac{1}{3}} \\ \end{aligned} \]

令\(\psi(x) = {(x^{2} + 1)} ^ {\frac{1}{3}}\)为迭代公式

# define other iteration methods
def psi(x):
    return (x**2 + 1)**(1/3)

x0 = 1.5 # 初始值
while True:
    x1 = psi(x0)
    if abs(x1 - x0) < 1e-12:
        break
    x2 = psi(x1)
    theta = (x2 - x1)/(x1 - x0)
    x0 = (psi(x0) - theta * x0)/(1 - theta) 
    print(x0)

1.4655584829667796
1.4655712318748688
1.4655712318767682

我们可以看到，\(\varphi(x)\)和 \(\psi(x)\) 的选取对迭代次数并没有明显的改善，个人认为应该是初值选取的太好了，导致迭代次数很少。下面我们来看一下不同初值的迭代次数。

选取初值

重新构造埃特肯 aitken函数

# aitken method
def aitken(x0, func = lambda x0 : x0 , eps = 1e-4 ,isprint=True):
    x0 = 1.5 # 初始值
    while True:
        x1 = func(x0)
        if abs(x1 - x0) < eps:
            break
        x2 = func(x1)
        theta = (x2 - x1)/(x1 - x0)
        x0 = (func(x0) - theta * x0)/(1 - theta) 
        print(x0)

aitken(1.5, phi, 1e-12)

1.465365431478883
1.4655712202898679
1.4655712318767682

观察不同初值的迭代次数

aitken(1, psi, 1e-12)

1.4655584829667796
1.4655712318748688
1.4655712318767682

aitken(1, phi, 1e-12)

1.465365431478883
1.4655712202898679
1.4655712318767682

更多迭代函数（free time~~~）

\(x^{3} - x^{2} - 1 = 0\)经过简单的变换可以得到下面的迭代函数

\[\begin{aligned} x &= x^{2} - \frac{1}{x} \\ x &= \frac{1}{(x - 1)^{\frac{1}{2}}} \\ x &= 1 + \frac{1}{x^{2}} \\ \end{aligned} \]

def f1(x):
    return x**2 - 1/x
def f2(x):
    return 1/((x-1)**0.5)
def f3(x):
    return 1 + 1/x**2

def f1(x):
    return x**2 - 1/x
def f2(x):
    return 1/((x-1)**0.5)
def f3(x):
    return 1 + 1/x**2

aitken(1, f1, 1e-12)

1.466725043782837
1.4655725197594736
1.465571231878372
1.465571231876768

aitken(1, f2, 1e-12)

1.4673422863283745
1.4655760852065471
1.4655712319132883
1.465571231876768

aitken(1, f3, 1e-12)

1.465858585858586
1.4655712527301703
1.4655712318767682

总结

虽然这只是数值分析的一道题但总体来说，觉得这门课如果设置成一门实践课可能也有好处，不过现在这样自己探索真的好有意思，也让我对这门课有了更深的理解。

标签：方程解,return,迭代,Aitken,theta,x0,x1,def,法求
From： https://www.cnblogs.com/Linkdom/p/16729110.html

[数值分析]埃特肯加速法求方程解Aitken!