F(x)=[（x2）-（5.1/4*π**21）*（x1）**2+（5/π）*（x1）-6]**2+10*(1-1/8*π）*cos(x1）+10用梯度下降法求最小值

以下是使用梯度下降法求解F(x)最小值的Python代码：

import numpy as np

# 定义函数F(x)
def F(x):
    x1, x2 = x
    return (x2 - (5.1 / (4 * np.pi ** 2)) * x1 ** 2 + (5 / np.pi) * x1 - 6) ** 2 + 10 * (1 - 1 / (8 * np.pi)) * np.cos(x1) + 10

# 定义函数F(x)的梯度
def grad_F(x):
    x1, x2 = x
    grad_x1 = 2 * (x2 - (5.1 / (4 * np.pi ** 2)) * x1 ** 2 + (5 / np.pi) * x1 - 6) * ((5.1 / (2 * np.pi ** 2)) * x1 + (5 / np.pi) * np.sin(x1) - 10 / (8 * np.pi) * np.sin(x1))
    grad_x2 = 2 * (x2 - (5.1 / (4 * np.pi ** 2)) * x1 ** 2 + (5 / np.pi) * x1 - 6)
    return np.array([grad_x1, grad_x2])

# 定义梯度下降函数
def gradient_descent(F, grad_F, init_x, lr=0.01, max_iter=10000, tol=1e-6):
    x = init_x
    for i in range(max_iter):
        grad = grad_F(x)
        if np.linalg.norm(grad) < tol:
            break
        x -= lr * grad
    return x, F(x)

# 设置初始点并调用梯度下降函数求解最小值
init_x = np.array([0, 0])
min_x, min_val = gradient_descent(F, grad_F, init_x)

# 输出最小值及对应的x值
print("最小值为：", min_val)
print("对应的x值为：", min_x)