目录
- 一元函数到多元函数的牛顿迭代法
- python代码实现过程
一元函数到多元函数的牛顿迭代法
一元函数的牛顿迭代公式:
多元函数的牛顿迭代公式:
其中Hession矩阵为:
; python代码实现过程
- 计算梯度函数
使用了sympy库的diff函数计算导数,再代入具体数值进行计算,返回X处的梯度grad
def get_grad(f, X):
f1 = diff(f, x1)
f2 = diff(f, x2)
grad = np.array([[f1.subs([(x1, X[0]), (x2, X[1])])],
[f2.subs([(x1, X[0]), (x2, X[1])])]])
return grad
- 计算Hession矩阵函数
同样使用diff函数计算二次偏导,组成Hession矩阵(里面的括号、中括号要盯对清楚)
def get_hess(f, X):
f1 = diff(f, x1)
f2 = diff(f, x2)
f11 = diff(f,x1,2)
f22 = diff(f,x2,2)
f12 = diff(f1,x2)
f21 = diff(f2,x1)
hess = np.array([[f11.subs([(x1,X[0]), (x2,X[1])]),
f12.subs([(x1,X[0]), (x2,X[1])])],
[f21.subs([(x1,X[0]), (x2,X[1])]),
f22.subs([(x1,X[0]), (x2,X[1])])]])
hess = np.array(hess, dtype = 'float')
return hess
- 牛顿迭代过程
迭代公式里使用的是Hession矩阵的逆,所以还要求个逆
def newton_iter(X0, err):
count = 0
X1 = np.array([[0],[0]])
while True:
X2 = X0 - X1
if sqrt(X2[0]**2 + X2[1]**2) err:
break
else:
hess = get_hess(f, X0)
hess_inv = np.linalg.inv(hess)
grad = get_grad(f, X0)
X1 = X0
X0 = X1 - np.dot(hess_inv, grad)
count += 1
print('第',count,'次迭代:','x1=',X0[0],'x2=',X0[1])
print('迭代次数为:',count)
print('迭代结果为',X0)
全部代码如下
请使用1.4版本的sympy
from sympy import *
import numpy as np
x1,x2 = symbols('x1, x2')
f = 60 - 10*x1 - 4*x2 + x1**2 + x2**2 - x1*x2
def get_grad(f, X):
f1 = diff(f, x1)
f2 = diff(f, x2)
grad = np.array([[f1.subs([(x1, X[0]), (x2, X[1])])],
[f2.subs([(x1, X[0]), (x2, X[1])])]])
return grad
def get_hess(f, X):
f1 = diff(f, x1)
f2 = diff(f, x2)
f11 = diff(f,x1,2)
f22 = diff(f,x2,2)
f12 = diff(f1,x2)
f21 = diff(f2,x1)
hess = np.array([[f11.subs([(x1,X[0]), (x2,X[1])]),
f12.subs([(x1,X[0]), (x2,X[1])])],
[f21.subs([(x1,X[0]), (x2,X[1])]),
f22.subs([(x1,X[0]), (x2,X[1])])]])
hess = np.array(hess, dtype = 'float')
return hess
def newton_iter(X0, err):
count = 0
X1 = np.array([[0],[0]])
while True:
X2 = X0 - X1
if sqrt(X2[0]**2 + X2[1]**2) err:
break
else:
hess = get_hess(f, X0)
hess_inv = np.linalg.inv(hess)
grad = get_grad(f, X0)
X1 = X0
X0 = X1 - np.dot(hess_inv, grad)
count += 1
print('第',count,'次迭代:','x1=',X0[0],'x2=',X0[1])
print('迭代次数为:',count)
print('迭代结果为',X0)
X0 = np.array([[1],[1]])
err = 1e-10
newton_iter(X0, err)
Original: https://blog.csdn.net/qq_45726331/article/details/115804812
Author: 空空7
Title: 【python】牛顿迭代法求解多元函数的最小值–以二元函数为例
原创文章受到原创版权保护。转载请注明出处:https://www.johngo689.com/694508/
转载文章受原作者版权保护。转载请注明原作者出处!