在具有复系数（4 度）的多项式上使用 sympy 的求解器时出错

Question

尝试用 sympy 求解 4 次多项式方程时，我遇到了一些困难。我的代码和我试图求解的方程式：

import sympy as sym
from sympy import I
sym.init_printing()

k = sym.Symbol('k')
t, sigma ,k0, L , V = sym.symbols('t, sigma, k0, L,V')

x4 = ( -t**2 + 2*I * t / sigma**2 + 1/sigma**4)
x3 = ( -2*I * t * k0 / sigma**2 - 2*k0 / sigma**4)
x2 = ( L**2 + k0 **2 / sigma **4 + t**2 * V - 2 * I * t * V / sigma**2 -V/sigma**4)
x1 = (2*I * V * k0 / sigma**2 + 2*k0 * V / sigma **4)
x0 = (2*I*k0*t*V / sigma**2 - k0 **2 *V / sigma**4)

expr = x4 * k**4 + x3 * k**3 + x2 * k**2 + x1 * k + x0
expr2 = expr.subs({k0 :2 , sigma : .2 , L : 1, V:1})

sym.solvers.solve(expr2,k)

输出：

Traceback (most recent call last):

  File "<ipython-input-4-e1ce7d8c9531>", line 1, in <module>
    sols = sym.solvers.solve(expr2,k)

  File "/usr/local/lib/python2.7/dist-packages/sympy/solvers    /solvers.py", line 1125, in solve
    solution = nfloat(solution, exponent=False)

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/function.py",     line 2465, in nfloat
    return type(expr)([nfloat(a, n, exponent) for a in expr])

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/function.py",     line 2499, in nfloat
    lambda x: isinstance(x, Function)))

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/basic.py",     line 1087, in xreplace
    value, _ = self._xreplace(rule)

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/basic.py",     line 1095, in _xreplace
    return rule[self], True

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/rules.py",     line 59, in __getitem__
    return self._transform(key)

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/function.py",     line 2498, in <lambda>
    lambda x: x.func(*nfloat(x.args, n, exponent)),

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/function.py",     line 2465, in nfloat
    return type(expr)([nfloat(a, n, exponent) for a in expr])

  File "/usr/local/lib/python2.7/dist-packages/sympy/core/function.py",     line 2465, in nfloat
    return type(expr)([nfloat(a, n, exponent) for a in expr])

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

而且我真的无法从中得到任何东西。我不太确定是什么原因造成的，我 "tested" 这个解算器用于更紧凑的多项式并且它运行良好。

Answer 1

看来您可以使用 solve(expr2, k, rational=False) 解决此问题。

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