Question

我试图用wxMaxima和sympy解决以下无限积分：

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。我不懂为什么。我是Python的重要用户，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。

我做错了还是Maxiam更好？（我也在Mathematica中解决了同样问题）

我在wxMaxima中得到了以下答案：

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

给出了这个答案：

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

它看起来很难看，但忘掉了。这里的观点是，同情不能解决这个问题。试着用这段代码解决同样问题：

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

给出此错误消息：

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

当你略微改写你的等式时，你会得到一个解决方案：

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()

因此，我使用expr**0.5而不是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的结果相同很难分辨，因为sympy为您提供了两个部分的解决方案，这取决于sqrt中的argumnt是大于还是小于0.您可以尝试使用实际界限并检查你是否得到了与Maxima解决方案相同的结果，并且结果的第二部分会给你。

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本：

 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在有理数上的表现要好于浮点数，尤其是，当这些数字是幂时。

这是因为“好的”封闭式解决方案通常只存在于确切的权力。考虑一下这个
之间的区别
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和这个

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

功率几乎为0.5，但答案需要使用特殊功能来表示。 SymPy可能没有注意到0.5应该是1/2（并且浮点数的不精确性对此没有帮助）。

话虽如此，我认为这是一个SymPy错误，特别是因为它可以计算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 ⎠。

Sympy失败了，wxMaxima没有

2 个答案: