Sympy 失败，wxMaxima 没有

Question

我正在尝试用 wxMaxima 和 sympy 解决以下不定积分：

integrate(r^2*sqrt(R^2-r^2),r)

在千里马我确实得到了答案，但没有同情。我不理解为什么。我是 Python 的重度用户，并且喜欢在 Python 中进行符号数学，但由于 sympy 没有解决这个问题，我仍然坚持使用 Maxima。

是我做错了什么还是Maxiam更好？ （我也在 Mathematica 中解决了同样的问题）

我在 wxMaxima 中得到了以下答案：

f:r^2*sqrt(R^2-r^2);
g:integrate(f,r);

给出这个答案：

g:(R^4*asin(r/abs(R)))/8-(r*(R^2-r^2)^(3/2))/4+(r*R^2*sqrt(R^2-r^2))/8  

它看起来很丑，但忘了这一点。这里的重点是 sympy 无法解决这个积分。尝试用这段代码解决同样的问题：

import sympy as sy
import math
R,r = sy.symbols('R r')
g = sy.integrate(r**2*(R**2-r**2)**0.5,r)
print g

给出以下错误消息：

Traceback (most recent call last):
  File "E:\UD\Software\BendStiffener\calculate-moment\moment-calculation-equations\sympy-test.py", line 4, in <module>
    g = sy.integrate(r**2*(R**2-r**2)**0.5,r)
  File "C:\Python27\lib\site-packages\sympy\utilities\decorator.py", line 35, in threaded_func
    return func(expr, *args, **kwargs)
  File "C:\Python27\lib\site-packages\sympy\integrals\integrals.py", line 1232, in integrate
    risch=risch, manual=manual)
  File "C:\Python27\lib\site-packages\sympy\integrals\integrals.py", line 487, in doit
    conds=conds)
  File "C:\Python27\lib\site-packages\sympy\integrals\integrals.py", line 876, in _eval_integral
    h = meijerint_indefinite(g, x)
  File "C:\Python27\lib\site-packages\sympy\integrals\meijerint.py", line 1596, in meijerint_indefinite
    res = _meijerint_indefinite_1(f.subs(x, x + a), x)
  File "C:\Python27\lib\site-packages\sympy\integrals\meijerint.py", line 1646, in _meijerint_indefinite_1
    r = hyperexpand(r.subs(t, a*x**b))
  File "C:\Python27\lib\site-packages\sympy\simplify\hyperexpand.py", line 2482, in hyperexpand
    return f.replace(hyper, do_replace).replace(meijerg, do_meijer)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 1351, in replace
    rv = bottom_up(self, rec_replace, atoms=True)
  File "C:\Python27\lib\site-packages\sympy\simplify\simplify.py", line 4082, in bottom_up
    rv = F(rv)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 1336, in rec_replace
    new = _value(expr, result)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 1280, in <lambda>
    _value = lambda expr, result: value(*expr.args)
  File "C:\Python27\lib\site-packages\sympy\simplify\hyperexpand.py", line 2479, in do_meijer
    allow_hyper, rewrite=rewrite)
  File "C:\Python27\lib\site-packages\sympy\simplify\hyperexpand.py", line 2375, in _meijergexpand
    t, 1/z0)
  File "C:\Python27\lib\site-packages\sympy\simplify\hyperexpand.py", line 2335, in do_slater
    resid = residue(integrand, s, b_ + r)
  File "C:\Python27\lib\site-packages\sympy\series\residues.py", line 57, in residue
    s = expr.series(x, n=0)
  File "C:\Python27\lib\site-packages\sympy\core\expr.py", line 2435, in series
    rv = self.subs(x, xpos).series(xpos, x0, n, dir, logx=logx)
  File "C:\Python27\lib\site-packages\sympy\core\expr.py", line 2442, in series
    s1 = self._eval_nseries(x, n=n, logx=logx)
  File "C:\Python27\lib\site-packages\sympy\core\mul.py", line 1446, in _eval_nseries
    terms = [t.nseries(x, n=n, logx=logx) for t in self.args]
  File "C:\Python27\lib\site-packages\sympy\core\expr.py", line 2639, in nseries
    return self._eval_nseries(x, n=n, logx=logx)
  File "C:\Python27\lib\site-packages\sympy\functions\special\gamma_functions.py", line 168, in _eval_nseries
    return super(gamma, self)._eval_nseries(x, n, logx)
  File "C:\Python27\lib\site-packages\sympy\core\function.py", line 598, in _eval_nseries
    term = e.subs(x, S.Zero)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 892, in subs
    rv = rv._subs(old, new, **kwargs)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 1006, in _subs
    rv = fallback(self, old, new)
  File "C:\Python27\lib\site-packages\sympy\core\basic.py", line 983, in fallback
    rv = self.func(*args)
  File "C:\Python27\lib\site-packages\sympy\core\function.py", line 382, in __new__
    return result.evalf(mlib.libmpf.prec_to_dps(pr))
  File "C:\Python27\lib\site-packages\sympy\core\evalf.py", line 1317, in evalf
    result = evalf(self, prec + 4, options)
  File "C:\Python27\lib\site-packages\sympy\core\evalf.py", line 1217, in evalf
    re, im = x._eval_evalf(prec).as_real_imag()
  File "C:\Python27\lib\site-packages\sympy\core\function.py", line 486, in _eval_evalf
    v = func(*args)
  File "C:\Python27\lib\site-packages\sympy\mpmath\ctx_mp_python.py", line 993, in f
    return ctx.make_mpf(mpf_f(x._mpf_, prec, rounding))
  File "C:\Python27\lib\site-packages\sympy\mpmath\libmp\gammazeta.py", line 1947, in mpf_gamma
    raise ValueError("gamma function pole")
ValueError: gamma function pole

Answer 1

稍稍改写方程即可得到解：

import sympy as sy
import math
R, r = sy.symbols('R r')

g = sy.integrate(r**2 * sy.sqrt((R**2 - r**2)), r)

print g.simplify()

所以我没有使用expr**0.5，而是使用sy.sqrt(expr)。这给了

Piecewise((I*R*(-R**3*acosh(r/R) - R**2*r*sqrt((-R**2 + r**2)/R**2) + 2*r**3*sqrt((-R**2 + r**2)/R**2))/8, Abs(r**2/R**2) > 1), (R*(R**3*asin(r/R) - R**2*r*sqrt(1 - r**2/R**2) + 2*r**3*sqrt(1 - r**2/R**2))/8, True))

这是否与 Maxima 的结果相同很难说，因为 sympy 分两部分为您提供解决方案，这取决于 sqrt 中的参数是大于还是小于 0。您可以尝试使用实际边界并检查您是否得到与 Maxima 解决方案相同的结果以及结果 sympy 的第二部分给您的结果。

您可以像这样访问解决方案的第二部分：

g.args[1][0]

给你：

 R**4*asin(r/R)/8 - R**3*r/(8*sqrt(1 - r**2/R**2)) + 3*R*r**3/(8*sqrt(1 - r**2/R**2)) - r**5/(4*R*sqrt(1 - r**2/R**2))

您也可以通过以下方式获得简化版：

 g.args[1][0].simplify()

给你：

R*(R**3*asin(r/R) - R**2*r*sqrt(1 - r**2/R**2) + 2*r**3*sqrt(1 - r**2/R**2))/8

现在看起来与 Maxima 获得的结果非常相似。

Answer 2

一般来说，SymPy 处理有理数比处理浮点数表现更好，尤其是当这些数字是幂时。

这是因为“好的”封闭形式解决方案通常只存在于精确的幂。例如考虑一下这之间的区别

In [39]: integrate(r**2*sqrt(R**2-r**2), r)
Out[39]:
⎧     4      ⎛r⎞
⎪  ⅈ⋅R ⋅acosh⎜─⎟            3                      3                  5             │ 2│
⎪            ⎝R⎠         ⅈ⋅R ⋅r             3⋅ⅈ⋅R⋅r                ⅈ⋅r              │r │
⎪- ───────────── + ───────────────── - ───────────────── + ───────────────────  for │──│ > 1
⎪        8                 _________           _________             _________      │ 2│
⎪                         ╱       2           ╱       2             ╱       2       │R │
⎪                        ╱       r           ╱       r             ╱       r
⎪                  8⋅   ╱   -1 + ──    8⋅   ╱   -1 + ──    4⋅R⋅   ╱   -1 + ──
⎪                      ╱          2        ╱          2          ╱          2
⎪                    ╲╱          R       ╲╱          R         ╲╱          R
⎨
⎪     4     ⎛r⎞
⎪    R ⋅asin⎜─⎟          3                     3                 5
⎪           ⎝R⎠         R ⋅r              3⋅R⋅r                 r
⎪    ────────── - ──────────────── + ──────────────── - ──────────────────       otherwise
⎪        8                ________           ________             ________
⎪                        ╱      2           ╱      2             ╱      2
⎪                       ╱      r           ╱      r             ╱      r
⎪                 8⋅   ╱   1 - ──    8⋅   ╱   1 - ──    4⋅R⋅   ╱   1 - ──
⎪                     ╱         2        ╱         2          ╱         2
⎩                   ╲╱         R       ╲╱         R         ╲╱         R

还有这个

In [40]: integrate(r**2*(R**2-r**2)**0.5001, r)
Out[40]:
                                  ⎛             │  2  2⋅ⅈ⋅π⎞
                   1.0002  3  ┌─  ⎜-0.5001, 3/2 │ r ⋅ℯ     ⎟
0.333333333333333⋅R      ⋅r ⋅ ├─  ⎜             │ ─────────⎟
                             2╵ 1 ⎜    5/2      │      2   ⎟
                                  ⎝             │     R    ⎠ 

幂几乎是0.5，但是答案需要用一个特殊的函数来表示。 SymPy 可能没有注意到 0.5 应该是 1/2（浮点数的不精确性对此无济于事）。

话虽如此，我认为这是一个 SymPy 错误，尤其是因为它可以计算 integrate(r**2*(R**2-r**2)**0.5001, r)。