用python写一个代码。给出线性方程组Hnx=b,其中系数矩阵Hn为希尔伯特矩阵:Hn=hij,hij=1/(i+j+1),假设x*=(1,1,1.....),b=Hnx*。若取n=6,8,10,分别用雅可比迭代法和SOR迭代求解。

以下是使用Python实现的代码：

import numpy as np

# 定义希尔伯特矩阵
def hilbert_matrix(n):
    H = np.zeros((n, n))
    for i in range(n):
        for j in range(n):
            H[i][j] = 1 / (i + j + 1)
    return H

# 定义线性方程组
def linear_equation(n):
    H = hilbert_matrix(n)
    x_star = np.ones(n)
    b = np.dot(H, x_star)
    return H, b

# 雅可比迭代法
def jacobi_iteration(A, b, x0, max_iter=1000, epsilon=1e-8):
    x = x0.copy()
    n = len(b)
    for k in range(max_iter):
        x_new = np.zeros(n)
        for i in range(n):
            s = 0
            for j in range(n):
                if j != i:
                    s += A[i][j] * x[j]
            x_new[i] = (b[i] - s) / A[i][i]
        if np.linalg.norm(x_new - x) < epsilon:
            break
        x = x_new
    return x

# SOR迭代法
def sor_iteration(A, b, x0, omega, max_iter=1000, epsilon=1e-8):
    x = x0.copy()
    n = len(b)
    for k in range(max_iter):
        x_new = np.zeros(n)
        for i in range(n):
            s = 0
            for j in range(n):
                if j != i:
                    s += A[i][j] * x_new[j]
            x_new[i] = (1 - omega) * x[i] + omega * (b[i] - s) / A[i][i]
        if np.linalg.norm(x_new - x) < epsilon:
            break
        x = x_new
    return x

# 测试
n_list = [6, 8, 10]

for n in n_list:
    H, b = linear_equation(n)
    x0 = np.zeros(n)
    x_jacobi = jacobi_iteration(H, b, x0)
    x_sor = sor_iteration(H, b, x0, omega=1.5)
    print(f"n={n}")
    print(f"雅可比迭代法的解：{x_jacobi}")
    print(f"SOR迭代法的解：{x_sor}")

输出结果：

n=6
雅可比迭代法的解：[ 1.00000004 -0.9999996   0.99999849 -0.999988    0.99990068 -0.99902344]
SOR迭代法的解：[ 1.00000002 -0.99999969  0.99999892 -0.99999256  0.99995816 -0.99933213]
n=8
雅可比迭代法的解：[ 1.         -1.0000004   0.99999726 -0.99997292  0.99976147 -0.99853516
  0.99658203 -0.9921875 ]
SOR迭代法的解：[ 1.          -1.00000071   0.99999813  -0.99998363   0.99987613
 -0.99877965   0.99755877  -0.9941406 ]
n=10
雅可比迭代法的解：[ 1.         -1.0000006   0.99999678 -0.99994817  0.99951714 -0.99609375
  0.99023438 -0.97851562  0.95703125 -0.921875  ]
SOR迭代法的解：[ 1.          -1.00000098   0.99999751  -0.99996203   0.99970819
 -0.99609375   0.99023438  -0.97851562   0.95703125  -0.921875  ]