Python同情不能解决多项式函数

时间:2016-03-29 14:11:39

标签: python sympy polynomials

通过使用sympy求解多项式函数,我遇到了问题。以下示例显示了一个案例,该案例提供了我无法管理的错误消息。如果多项式变得更简单,则求解器正常工作。请复制并粘贴代码以检查系统上的错误。

 import sympy
 from sympy import I
 omega = sympy.symbols('omega')

 def function(omega):
    return - 0.34*omega**4      \
           + 7.44*omega**3      \
           + 4.51*I*omega**3    \
           + 87705.64*omega**2  \
           - 53.08*I*omega**2   \
           - 144140.03*omega    \
           - 22959.95*I*omega   \
           + 42357.18 + 50317.77*I
 sympy.solve(function(omega), omega)

你知道我怎么能取得成果吗? 谢谢你的帮助。

编辑:

这是错误消息:

TypeError                                 Traceback (most recent call last)
<ipython-input-7-512446a62fa9> in <module>()
      1 def function(omega):
      2     return - 0.34*omega**4                 + 7.44*omega**3                 + 4.51*I*omega**3               + 87705.64*omega**2             - 53.08*I*omega**2              - 144140.03*omega               - 22959.95*I*omega              + 42357.18 + 50317.77*I
----> 3 sympy.solve(function(omega), omega)

C:\Anaconda\lib\site-packages\sympy\solvers\solvers.pyc in solve(f, *symbols, **flags)
   1123     # restore floats
   1124     if floats and solution and flags.get('rational', None) is None:
-> 1125         solution = nfloat(solution, exponent=False)
   1126 
   1127     if check and solution:  # assumption checking

C:\Anaconda\lib\site-packages\sympy\core\function.pyc in nfloat(expr, n, exponent)
   2463             return type(expr)([(k, nfloat(v, n, exponent)) for k, v in
   2464                                list(expr.items())])
-> 2465         return type(expr)([nfloat(a, n, exponent) for a in expr])
   2466     rv = sympify(expr)
   2467 

C:\Anaconda\lib\site-packages\sympy\core\function.pyc in nfloat(expr, n, exponent)
   2497     return rv.xreplace(Transform(
   2498         lambda x: x.func(*nfloat(x.args, n, exponent)),
-> 2499         lambda x: isinstance(x, Function)))
   2500 
   2501 

C:\Anaconda\lib\site-packages\sympy\core\basic.pyc in xreplace(self, rule)
   1085 
   1086         """
-> 1087         value, _ = self._xreplace(rule)
   1088         return value
   1089 

C:\Anaconda\lib\site-packages\sympy\core\basic.pyc in _xreplace(self, rule)
   1093         """
   1094         if self in rule:
-> 1095             return rule[self], True
   1096         elif rule:
   1097             args = []

C:\Anaconda\lib\site-packages\sympy\core\rules.pyc in __getitem__(self, key)
     57     def __getitem__(self, key):
     58         if self._filter(key):
---> 59             return self._transform(key)
     60         else:
     61             raise KeyError(key)

C:\Anaconda\lib\site-packages\sympy\core\function.pyc in <lambda>(x)
   2496 
   2497     return rv.xreplace(Transform(
-> 2498         lambda x: x.func(*nfloat(x.args, n, exponent)),
   2499         lambda x: isinstance(x, Function)))
   2500 

C:\Anaconda\lib\site-packages\sympy\core\function.pyc in nfloat(expr, n, exponent)
   2463             return type(expr)([(k, nfloat(v, n, exponent)) for k, v in
   2464                                list(expr.items())])
-> 2465         return type(expr)([nfloat(a, n, exponent) for a in expr])
   2466     rv = sympify(expr)
   2467 

C:\Anaconda\lib\site-packages\sympy\core\function.pyc in nfloat(expr, n, exponent)
   2463             return type(expr)([(k, nfloat(v, n, exponent)) for k, v in
   2464                                list(expr.items())])
-> 2465         return type(expr)([nfloat(a, n, exponent) for a in expr])
   2466     rv = sympify(expr)
   2467 

TypeError: __new__() takes exactly 3 arguments (2 given)

2 个答案:

答案 0 :(得分:6)

正如我在评论中提到的,我不熟悉同情,但这里是如何使用任意精度mpmath模块找到等式的根。

为了避免Python浮点数的精度问题,通常以字符串形式将浮点数传递给mpmath,除非从整数构造它们很方便。我猜你的等式不是一个问题,因为你的系数精度相当低,但无论如何......

这里将您的等式直接翻译成mpmath语法:

from mpmath import mp

I = mp.mpc(0, 1)

def func(x):
    return (-mp.mpf('0.34') * x ** 4
        + mp.mpf('7.44') * x ** 3
        + mp.mpf('4.51') * I * x ** 3
        + mp.mpf('87705.64') * x ** 2
        - mp.mpf('53.08') * I * x ** 2
        - mp.mpf('144140.03') * x
        - mp.mpf('22959.95') * I * x
        + mp.mpf('42357.18') + mp.mpf('50317.77') * I)

mpf是任意精度的浮点构造函数,mpc是复数构造函数。有关如何调用这些构造函数的信息,请参阅mpmath文档。

但是,我们不需要像I这样混淆:我们可以直接将系数定义为复数。

from __future__ import print_function
from mpmath import mp

# set precision to ~30 decimal places
mp.dps = 30

def func(x):
    return (mp.mpf('-0.34') * x ** 4
        + mp.mpc('7.44', '4.51') * x ** 3
        + mp.mpc('87705.64', '-53.08') * x ** 2
        + mp.mpc('-144140.03', '-22959.95') * x
        + mp.mpc('42357.18', '50317.77'))

x = mp.findroot(func, 1)
print(x)
print('test:', func(x))

<强>输出

(1.35717989161653180236360985534 - 0.202974596285109153971324419197j)
test: (-3.2311742677852643549664402034e-26 + 6.4623485355705287099328804068e-27j)

但我们怎样才能找到其他根源呢?简单!

u 成为 f(x)的根。然后让 f(x)= g(x)(x - u) g(x)的任何根也是 f(x)的根。我们可以通过使用for循环方便地多次执行此操作,该循环将每个找到的根保存到列表中,然后从前一个函数构建一个新函数,将此新函数存储在另一个列表中。

在这个版本中,我使用&#34; muller&#34;解决者,在寻找复杂的根时推荐的,但它实际上给出了与使用默认&#34;割线&#34;相同的答案。解算器。

from __future__ import print_function
from mpmath import mp

# set precision to ~30 decimal places
mp.dps = 30

def func(x):
    return (mp.mpf('-0.34') * x ** 4
        + mp.mpc('7.44', '4.51') * x ** 3
        + mp.mpc('87705.64', '-53.08') * x ** 2
        + mp.mpc('-144140.03', '-22959.95') * x
        + mp.mpc('42357.18', '50317.77'))

x = mp.findroot(func, 1)
print(x)
print('test:', func(x))

funcs = [func]
roots = []

#Find all 4 roots
for i in range(4):
    x = mp.findroot(funcs[i], 1, solver="muller")
    print('%d: %s' % (i, x))
    print('test: %s\n%s\n' % (funcs[i](x), funcs[0](x)))
    roots.append(x)
    funcs.append(lambda x,i=i: funcs[i](x) / (x - roots[i]))    

<强>输出

(1.35717989161653180236360985534 - 0.202974596285109153971324419197j)
test: (-3.2311742677852643549664402034e-26 + 6.4623485355705287099328804068e-27j)
0: (1.35717989161653180236360985534 - 0.202974596285109153971324419197j)
test: (-3.2311742677852643549664402034e-26 + 6.4623485355705287099328804068e-27j)
(-3.2311742677852643549664402034e-26 + 6.4623485355705287099328804068e-27j)

1: (0.2859569222439674364374376897 + 0.465618100581394597267702975072j)
test: (2.70967991111831205485691272044e-27 - 4.34146435347156317045282996313e-27j)
(0.0 + 6.4623485355705287099328804068e-27j)

2: (-497.86487129703641182172086688 + 6.49836193448774263077718499855j)
test: (1.25428695883356196194609388771e-26 - 3.46609896266051486795576778151e-28j)
(3.11655159180984988723836070362e-21 - 1.65830325771275337225587644119e-22j)

3: (518.104087424352383171155113452 + 6.50370044356891310239702468087j)
test: (-4.82755073209873082528016484574e-30 + 7.38353321095804877623117526215e-31j)
(-1.31713649147437845007238988587e-21 + 1.68350641700147843422461467477e-22j)

lambda x,i=i: funcs[i](x) / (x - roots[i])

我们将i指定为默认关键字参数,以便使用i在定义函数时所具有的值。否则,调用函数时会使用{{1>}的当前值,这不是我们想要的。

这种用于查找多个根的技术可用于任意函数。但是,当我们想要求解多项式时,mpmath有一种更好的方法可以同时找到所有根:polyroots函数。我们只需要给它一个多项式系数的列表(或元组)。

i

<强>输出

from __future__ import print_function
from mpmath import mp

# set precision to ~30 decimal places
mp.dps = 30

coeff = (
    mp.mpf('-0.34'),
    mp.mpc('7.44', '4.51'),
    mp.mpc('87705.64', '-53.08'),
    mp.mpc('-144140.03', '-22959.95'),
    mp.mpc('42357.18', '50317.77'),
)

roots = mp.polyroots(coeff)
for i, x in enumerate(roots):
    print('%d: %s' % (i, x))
    print('test: %s\n' % mp.polyval(coeff, x))

如您所见,结果与前一技术获得的结果非常接近。使用0: (1.35717989161653180236360985534 - 0.202974596285109153971324419197j) test: (6.4623485355705287099328804068e-27 - 6.4623485355705287099328804068e-27j) 1: (0.2859569222439674364374376897 + 0.465618100581394597267702975072j) test: (0.0 + 0.0j) 2: (-497.86487129703641182172086688 + 6.49836193448774263077718499855j) test: (-2.27689218423463552142807161949e-21 + 7.09242751778865525915133624646e-23j) 3: (518.104087424352383171155113452 + 6.50370044356891310239702468087j) test: (-7.83663157514495734538720675411e-22 - 1.08373584941517766465574404422e-23j) 不仅更准确,它具有很大的优势,我们不需要为根指定起始近似值,mpmath足够聪明,可以为自己创建一个。

答案 1 :(得分:1)

感谢PM2Ring的帮助,我创建了一个完整的代码示例,它提取任意给定的四阶同情多项式的系数并解决它。

import sympy
from sympy import I
import mpmath as mp
import numpy as np

omega = sympy.symbols('omega')

# Define polynomial in sympy
def function(omega):
    return ( -  0.34                   *omega**4      \
             + (7.44 + 4.51*I)         *omega**3      \
             + (87705.64 - 53.08*I)    *omega**2  \
             - (144140.03 + 22959.95*I)*omega \
             + 42357.18 + 50317.77*I)

# Show defined polynomial
print''+ str(function(omega).expand()) +'\n'
result = sympy.Poly(function(omega).expand(), omega)

# Obtain coefficients of the polynomial
coeff = (
    mp.mpc(str(np.real(sympy.lambdify( (I),result.coeffs()[0])(1j))) , str(np.imag(sympy.lambdify( (I),result.coeffs()[0])(1j)))),
    mp.mpc(str(np.real(sympy.lambdify( (I),result.coeffs()[1])(1j))) , str(np.imag(sympy.lambdify( (I),result.coeffs()[1])(1j)))),
    mp.mpc(str(np.real(sympy.lambdify( (I),result.coeffs()[2])(1j))) , str(np.imag(sympy.lambdify( (I),result.coeffs()[2])(1j)))),
    mp.mpc(str(np.real(sympy.lambdify( (I),result.coeffs()[3])(1j))) , str(np.imag(sympy.lambdify( (I),result.coeffs()[3])(1j)))),
    mp.mpc(str(np.real(sympy.lambdify( (I),result.coeffs()[4])(1j))) , str(np.imag(sympy.lambdify( (I),result.coeffs()[4])(1j))) ),
  )

# Calculate roots of the polynomial
roots = mp.polyroots(coeff)
for no, frequency in enumerate(roots):
    print frequency

<强>输出

-0.34*omega**4 + 7.44*omega**3 + 4.51*I*omega**3 + 87705.64*omega**2 - 53.08*I*omega**2 - 144140.03*omega - 22959.95*I*omega + 42357.18 + 50317.77*I

(1.35717989161653 - 0.202974596285109j)
(0.285956922243967 + 0.465618100581395j)
(-497.864871297036 + 6.49836193448774j)
(518.104087424352 + 6.50370044356891j)