使用sympy的求解器对具有复系数的多项式（第4度）的误差

  • Z时代
  • 2024-01-10
  • 分类：问答

尝试用sympy求解四次多项式方程，我遇到了一些困难。我的代码和公式我试图解决：使用sympy的求解器对具有复系数的多项式（第4度）的误差

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)

我真的不能做任何事情出来。我不太确定是什么导致了这一点，我“测试”这个求解器以获得更紧凑的多项式，并且运行良好。

回答：

看起来您可以使用solve(expr2, k, rational=False)解决该问题。

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