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函数求最值上课学生代码.py

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+# -*- coding: utf-8 -*-
+"""
+Created on Tue Jun  8 17:19:05 2021
+
+@author: hzh
+"""
+# 例2：求函数y=x2-10x-30  的最小值
+# 绘图
+import numpy as np
+import matplotlib.pyplot as plt
+
+
+def f(x):
+	return x ** 2 - 10 * x - 30
+
+
+x = np.arange(-10, 10, 0.1)
+plt.plot(x, f(x))
+plt.grid()
+plt.show()
+
+# 梯度下降法求函数最值
+np.random.seed(3)
+x = np.random.randn(1);  # 10代表数据的维度
+# print(x)
+learning_rate = 0.1
+err = 0.000001;
+max_iters = 10000
+
+
+def f(x):
+	return x ** 2 - 10 * x - 30
+
+
+def df(x):
+	return 2 * x - 10
+
+
+for i in range(max_iters):
+	print("第 %d 次迭代：x=%.8f y=%.8f" % (i, x, f(x)))
+	if abs(df(x)) < err:
+		break
+	x = x - df(x) * learning_rate  # xk+1=xk- η* f’(xk)  (迭代公式)
+
+# 练习题1：求函数y=x2-8x-25的最小值
+# 练习题2：
+# 已知函数y=x+sqrt(2*R*x-x**2), R为大于0的常数，x的取值范围在[R,2*R]之间。
+# 当R=10时，求该函数在x取值区间内函数的最大值。
+# 绘图
+
+# 梯度下降法求函数最大值
+
+
+# 给定一个二元函数f(x,y)= - exp(x-y)*(x**2-2*y**2)，
+# 用梯度下降法求其在x<0,y<0范围上的极小值
+
+x = -1;
+y = -1
+learning_rate = 0.1
+err = 0.000001;
+max_iters = 10000
+
+
+def f(x, y):
+	return -np.exp(x - y) * (x ** 2 - 2 * y ** 2)
+
+
+def dx(x, y):
+	return -(np.exp(x - y) * (2 * x) + (x ** 2 - 2 * y ** 2) * np.exp(x - y))
+
+
+def dy(x, y):
+	return -(np.exp(x - y) * (-4 * y) + (x ** 2 - 2 * y ** 2) * np.exp(x - y) * (-1))
+
+
+for t in range(max_iters):
+	if t % 100 == 0:
+		print("Iter %d, x=%.8f,y=%.8f,z=%.8f,dx=%.8f,dy=%.8f" % (t, x, y, f(x, y), dx(x, y), dy(x, y)))
+	if abs(dx(x, y)) < err and abs(dy(x, y)) < err:
+		print("Iter %d, x=%.8f,y=%.8f,z=%.8f,dx=%.8f,dy=%.8f" % (t, x, y, f(x, y), dx(x, y), dy(x, y)))
+		break
+	x = x - learning_rate * dx(x, y);
+	y = y - learning_rate * dy(x, y)  # 迭代公式
+
+'''#练习题：
+二元函数f(x,y)=x3-y3+3x2+3y2-9x求最值
+三元函数f(x,y,z)=x2+2y2+3z2+2x+4y-6z求最值
+'''