逆Broyden方法python实现

发篇博客纪念一下，折磨了我一个小时。。。

利用逆Broyden方法求方程组的数值解，迭代到某误差范围内结束，精确解x=（0.5,1）转秩

import numpy as np
import pprint

f_l = []
h_l = []
x_l = []
for i in range(12):
    if i == 0:
        x1 = 0.75
        x2 = 0.75
        # print(x1, x2)
        x_l.append([x1, x2])
        A = np.array([[8 * x1 + 2 * x2, 2 * x2 + 2 * x1 - 1], [4 * x1 + 3 * x2, 3 * x1 + 2 * x2]])  # F(X)的导数  A0
        H = np.linalg.inv(A)  # H0
        h_l.append(H)
    X = np.array([[x1], [x2]])  # Xk
    F = np.array([[x1 * x1 * 4 + x2 * x2 + 2 * x1 * x2 - x2 - 2], [2 * x1 * x1 + 3 * x1 * x2 + x2 * x2 - 3]])  # F_Xk
    f_l.append(F)
    if i >= 1:
        Y = f_l[-1] - f_l[-2]
        H = h_l[-1] + (
                (R - np.dot(h_l[-1], Y)) * np.dot(R.T, h_l[-1]) / (np.dot(np.dot(R.T, h_l[-1]), Y)))
        h_l.append(H)
    X_new = X - np.dot(H, F)  # x1
    R = X_new - X  # r0
    x1 = float(X_new[0])
    x2 = float(X_new[1])
    x_l.append([x1, x2])
    # print(x1, x2)
pprint.pprint(x_l)

for i in range(10):
    a = max(abs(x_l[i + 1][0] - x_l[i][0]), abs(x_l[i + 1][1] - x_l[i][1]))
    print(a, bool(a <= 0.5 / 100000), i + 1)