Question

我很新见，但想解决以下问题：

我有多种形式的不等式约束

A < f(x) < B
C < g(x) < D
...

其中A，B，C，D只是数字。 f和g是有理函数。

例如，我可以使用以下方法：

solve_rational_inequalities([[
   ((Poly(x-10000.00), Poly(1, x)), '>'), 
   ((Poly(x-100000.00), Poly(1, x)), '<'), 
   ((Poly((x/130000.00)-0.00), Poly(1, x)), '>'), 
   ((Poly((x/130000.00)-0.19), Poly(1, x)), '<')]])
Interval.open(10000, 24700)

这里，A = 10.000，B = 100.000，C = 0，D = 0.19，f（x）= x并且g（x）= x / 130.000。所以这很有效。

现在，对于另一种情况，我有函数f（x）= 10100.00 / x。

如果我只是应用上面的配方，我会得到：

solve_rational_inequalities([[
    ((Poly((10100.00/x)-0.00), Poly(1, x)), '>'), 
    ((Poly((10100.00/x)-0.19), Poly(1, x)), '<')]])                     
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/oliverdemetz/Library/Python/2.7/lib/python/site-packages/sympy/solvers/inequalities.py", line 162, in solve_rational_inequalities
    numer_intervals = solve_poly_inequality(numer*denom, rel)
  File "/Users/oliverdemetz/Library/Python/2.7/lib/python/site-packages/sympy/solvers/inequalities.py", line 55, in solve_poly_inequality
    reals, intervals = poly.real_roots(multiple=False), []
  File "/Users/oliverdemetz/Library/Python/2.7/lib/python/site-packages/sympy/polys/polytools.py", line 3454, in real_roots
    reals = sympy.polys.rootoftools.CRootOf.real_roots(f, radicals=radicals)
  File "/Users/oliverdemetz/Library/Python/2.7/lib/python/site-packages/sympy/polys/rootoftools.py", line 196, in real_roots
    return cls._get_roots("_real_roots", poly, radicals)
  File "/Users/oliverdemetz/Library/Python/2.7/lib/python/site-packages/sympy/polys/rootoftools.py", line 565, in _get_roots
    raise PolynomialError("only univariate polynomials are allowed")
sympy.polys.polyerrors.PolynomialError: only univariate polynomials are allowed

Answer 1

我只是用多项式弄乱了有理函数。我在寻找的是solve_poly_inequalities：

solve_poly_inequalities(((
    (Poly((10100.00/x)-0.00),'>'),
    (Poly((10100.00/x)-0.19),'<')
)))

BTW：有人可以告诉我（作为初学者）为什么他们使用((...))作为列表，尽管手册将方括号[...]解释为列表语法???

无论如何，我终于使用solveset命令更轻松地解决了我的不平等问题的实际问题，请参阅this other SO post

Answer 2

在solve_rational_inequality()语法中第一个多项式（poly）是分子，第二个多项式是分母，所以（1 / x）在此语法中为poly(1,x), poly(x)：

solve_rational_inequalities([[((poly(-10100.00,x), poly(x)), '>'),((poly(-0.19*x+10100.00), poly(x)), '<')]])

(-oo, 0)

解决同情中的理性不平等

2 个答案: