Assume real root in limit

[proy87]

Limit(((x - 6)**(Integer(1)/3) + 2)/(x + 2),x,-2,'-').doit() returns -oo*sign(2+2*(-1)^(1/3)). How to make it real-valued?

[gschintgen]

This is independent of limit.

You can use real_root:
https://docs.sympy.org/1.5.1/modules/functions/elementary.html#sympy.functions.elementary.miscellaneous.real_root

In [1]: limit((real_root(x - 6, 3) + 2)/(x + 2), x, -2, '-')
Out[1]: 1/12

[proy87]

@gschintgen , got some weird answer for right-hand limit: limit((real_root(x - 6, 3) + 2)/(x + 2), x, -2, '+') is 2*Subs(Derivative(sign(_x - 8), _x), _x, 0) + 1/12

[gschintgen]

Manually rewriting the root does the trick:

In [6]: l = Limit((-(6 - x)**(Integer(1)/3) + 2)/(x + 2),x,-2,'-'); l
Out[6]:
      ⎛    3 _______⎞
      ⎜2 - ╲╱ 6 - x ⎟
 lim  ⎜─────────────⎟
x─→-2⁻⎝    x + 2    ⎠

In [7]: _.doit()
Out[7]: 1/12

So the piecewise expression returned by real_root is not quite amenable to all operations, in particular derviatives and sign don't seem to mix well. For reference:

In [8]: real_root(x - 6, 3)
Out[8]:
⎧3 _________
⎪╲╱ │x - 6│ ⋅sign(x - 6)  for im(x) = 0
⎨
⎪       3 _______
⎩       ╲╱ x - 6            otherwise

I think the underlying issue is has to do with this: if x is real computing the diff is no problem:

In [16]: diff(sign(x-8))
Out[16]: 2⋅δ(x - 8)

In [17]: diff(sign(x-8)).subs(x, 0)
Out[17]: 0

But if I set x to be real before calling limit, this assumption is lost (or discarded on purpose?) when the Dummy in the diff.subs is created. And then we're left with the behavior for a general complex variable:

In [18]: diff(sign(y-8))
Out[18]:
d
──(sign(y - 8))
dy

In [19]: _.subs(y, 0)
Out[19]:
⎛d              ⎞│
⎜──(sign(y - 8))⎟│
⎝dy             ⎠│y=0

Since I'm not at all familiar with this part of the codebase that's as far as I'm going with this.

[gschintgen]

Also noteworthy:

In [26]: var('x', real=True)
Out[26]: x

In [27]: l = Limit((real_root(x - 6, 3) + 2)/(x + 2), x, -2, '+'); l
Out[27]:
      ⎛3 _________                ⎞
      ⎜╲╱ │x - 6│ ⋅sign(x - 6) + 2⎟
 lim  ⎜───────────────────────────⎟
x─→-2⁺⎝           x + 2           ⎠

A simple substitution of x = -2 would reveal that both abs and sign could be simplified even before the actual limit computation. But substitution is a dangerous idea in a limit computation, so I'm not sure if this will lead anywhere...

[proy87]

@gschintgen , thank you for the thorough answer.

[sachin-4099]

@gschintgen @jksuom Is there anything left to handle in this issue?

[gschintgen]

@gschintgen @jksuom Is there anything left to handle in this issue?

Yes, this:

In [1]: limit((real_root(x - 6, 3) + 2)/(x + 2), x, -2, '+')
Out[1]:
  ⎛d              ⎞│      1
2⋅⎜──(sign(x - 8))⎟│    + ──
  ⎝dx             ⎠│x=0   12

The first term shouldn't be there: it equals zero.
(Edit: same output if x is explicitly created with real=True.)