解渐近的有理不等式

Question

我不懂同情，但我想解决以下问题：

有多个不等式约束的形式

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，我可以让以下内容发挥作用：

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，g(x) = x/130.000。所以这是可行的。

现在，对于另一种情况，函数f( x ) = 10100.00 /x。

如果我把上面的食谱应用到上面，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

我只是用多项式搅乱了有理函数。我在寻找的是solve_poly_inequalities

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

顺便说一句:谁能告诉我(作为phython初学者)他们为什么对列表使用((...))，尽管手册将方括号[...]解释为列表语法？

无论如何，我终于解决了使用solveset命令解决不平等的实际问题，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，因此(1/x)在这个语法中是poly(1,x), poly(x)：

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

(-oo, 0)