Newton法解非线性方程组

#P223 例7
import numpy as np

def f1(x1,x2):
    return (4*x1**2+3*x2**2-1)
def f2(x1,x2):
    return (x1**3-8*x2**3-1)

def fd11(x1):
    return 8*x1
def fd12(x2):
    return 6*x2

def fd21(x1):
    return 3*x1**2
def fd22(x2):
    return -24*x2**2

x1=0.25
x2=-0.5
for i in range(1,10):
    a=([fd11(x1),fd12(x2)],[fd21(x1),fd22(x2)])
    b=([-f1(x1,x2)],[-f2(x1,x2)])
    ans=np.linalg.solve(a,b)
    #print(ans)
    x1=round(x1+ans[0][0],8)
    x2=round(x2+ans[1][0],8)
    print(x1,x2)

习题八 18

import numpy as np

def f1(x1,x2):
    return (4*x1+0.1*np.e**x1-x2)
def f2(x1,x2):
    return (1/8*x1**2-x1+4*x2-4)

def fd11(x1):
    return 4+0.1*np.e**x1
def fd12(x2):
    return -1

def fd21(x1):
    return 1/4*x1-1
def fd22(x2):
    return 4

x1=0
x2=1
for i in range(1,10):
    a=([fd11(x1),fd12(x2)],[fd21(x1),fd22(x2)])
    b=([-f1(x1,x2)],[-f2(x1,x2)])
    #print(a,b)
    ans=np.linalg.solve(a,b)
    #print(ans)
    x1=round(x1+ans[0][0],6)
    x2=round(x2+ans[1][0],6)
    print(x1,x2)

19.(1)

import numpy as np

def f1(x1,x2):
    return (x1**2-x1+x2**2)
def f2(x1,x2):
    return (x1**2-x2**2-x2)

def fd11(x1):
    return 2*x1-1
def fd12(x2):
    return 2*x2

def fd21(x1):
    return 2*x1
def fd22(x2):
    return -2*x2-1

x1=0.8
x2=0.4
for i in range(1,10):
    a=([fd11(x1),fd12(x2)],[fd21(x1),fd22(x2)])
    b=([-f1(x1,x2)],[-f2(x1,x2)])
    #print(a,b)
    ans=np.linalg.solve(a,b)
    #print(ans)
    x1=round(x1+ans[0][0],6)
    x2=round(x2+ans[1][0],6)
    print(x1,x2)