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# Prototype of N-R for a system of two non-linear equations
#evaluating  functions of two variables
# f(x,y)=1.6 * x ** 2 + 3.6 * x * y - 7.8 * x - 2.6 * y + 5.2
# g(x,y)=0.9 * y ** 2 + 3.1 * x **2 - 6.2 * x + 6.2 * y

# x = 0.5
# y =0.4

from math import *

eq1 = raw_input('Enter the equation 1: ')
eq2 = raw_input('Enter the equation 2: ')
x0 = float(input('Enter x: '))
y0 = float(input('Enter y: '))

def f(x,y):
    return eval(eq1)

def g(x,y):
    return eval(eq2)

Ea_X = 1
x = x0
y = y0

for n in range(1, 8):

    a = (f(x + 1e-06, y) - f(x,y)) / 1e-06   #in this one start the trouble
    b = (f(x, y + 1e-06) - f(x,y)) / 1e-06
    c = 0 - f(x,y)
    d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
    eE = (g(x, y + 1e-06) - g(x,y)) / 1e-06
    f = 0 - g(x,y)


    print "f(x, y)= ", eq1
    print "g(x, y)= ", eq2
    print """x   y """
    print x, y
    print """a   b   c   d   e   f """
    print a, b, c, d, e, f

    print """
    a * x + b * y = c
    d * x + e * y = f
    """

    print a," * x  +  ",b," * y  =  ",c
    print d," * x  +  ",eE," * y  =  ",f

    _Sy = (c - a * f / d) / (b - a * eE / d)
    _Sx = (f / d) - (eE / d) * _Sy

    Ea_X = (_Sx ** 2 + _Sy ** 2)**0.5


    x = x + _Sx
    y = y + _Sy

    print "Sx = ", _Sx
    print "Sy = ", _Sy

    print "x = ", x
    print "y = ", y

    print "|X_1 - X_0| = ", Ea_X

我一直在为两个非线性方程测试 Newton-Rapson 方法，原型代码有效，但后来我想让它更有用，因为原型是关于 2 个方程的输入和第一次猜测，它最好实现一个 for 循环，而不是启动像 6 或 10 这样的过程来解决我正在使用的众多方程中的一个

# Prototype of N-R for a system of two non-linear equations
# f(x,y)=1.6 * x ** 2 + 3.6 * x * y - 7.8 * x - 2.6 * y + 5.2
# g(x,y)=0.9 * y ** 2 + 3.1 * x **2 - 6.2 * x + 6.2 * y

# x = 0.5
# y =0.4


# evaluating  functions of two variables

from math import *


eq1 = raw_input('Enter the equation 1: ')
eq2 = raw_input('Enter the equation 2: ')
x0 = float(input('Enter x: '))
y0 = float(input('Enter y: '))

def f(x,y):
    return eval(eq1)

def g(x,y):
    return eval(eq2)

Ea_X = 1
x = x0
y = y0

a = (f(x + 1e-06, y) - f(x,y)) / 1e-06
b = (f(x, y + 1e-06) - f(x,y)) / 1e-06
c = 0 - f(x,y)
d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
eE = (g(x, y + 1e-06) - g(x,y)) / 1e-06
f = 0 - g(x,y)


print "f(x, y)= ", eq1
print "g(x, y)= ", eq2
print """x   y """
print x, y
print """a   b   c   d   e   f """
print a, b, c, d, e, f

print """
a * x + b * y = c
d * x + e * y = f
"""

print a," * x  +  ",b," * y  =  ",c
print d," * x  +  ",eE," * y  =  ",f

_Sy = (c - a * f / d) / (b - a * eE / d)
_Sx = (f / d) - (eE / d) * _Sy

Ea_X = (_Sx ** 2 + _Sy ** 2)**0.5


x = x + _Sx
y = y + _Sy

print "Sx = ", _Sx
print "Sy = ", _Sy

print "x = ", x
print "y = ", y

print "|X_1 - X_0| = ", Ea_X
4

2 回答 2

3

这是问题所在：

f = 0 - g(x,y)

您正在f从一个函数重新绑定到一个float.

于 2013-01-27T20:34:27.330 回答
3

在行

f = 0 - g(x,y)

您为名称分配一个数字f。由于函数和其他变量在 Python 中共享一个命名空间（函数只是一个可调用对象，绑定到任何变量），这使得未来的迭代失败。为您在上述行中分配的值选择另一个名称。

于 2013-01-27T20:35:24.353 回答