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I used

FunctionDomain[1/(x^3 + 1), x, Complexes]

and got

1 + x^3 != 0

Is this a bug in FunctionDomain (V.10)?

Can I receive $x \neq -1$, $x \neq \dfrac{1-\sqrt{3}I}{2}$ and $x \neq \dfrac{1+\sqrt{3}I}{2}$? With Maple, I tried enter image description here

If I used

FunctionDomain[1/(x^3 + 1), x]

I got true answer

x < -1 || x > -1

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I don't think your Maple comparison is quite fair. You've explicitly asked Maple to solve an inequality, rather than asked for the domain of a function. That doesn't account for more complicated examples, as I've placed in my answer. –  Mark McClure Jul 16 at 16:52

1 Answer 1

up vote 12 down vote accepted

This is not a bug; it is an implicit representation of the domain. Note that condition $1-x^3 \neq 0$ cannot be simplified further over the complexes or, at least, Reduce doesn't simplify it further:

Reduce[1 - x^3 != 0, x, Reals] 
(* Out: x < 1 || x > 1 *)

Reduce[1 - x^3 != 0, x, Complexes] 
(* Out: -1 + x^3 != 0 *)

A similar thing can happen over the reals:

FunctionDomain[1/(2 + Sin[x] + Cos[Pi*x]), x]
(* Out: Cos[Pi*x] + Sin[x] != -2 *)

The expression $\cos(\pi x) + \sin(x)$ is never equal to $-2$, by the way, but Reduce can't prove that and I don't think it's reasonable to expect it to do so. Is this a bug? Probably not; the answer is correct.

Here's another example:

FunctionDomain[1/Sqrt[Sin[1/x] - x], x]
(* Out: x != 0 && x - Sin[1/x] < 0 *)

How might we improve this result? The graph of $x-\sin(1/x)$ crosses the $x$-axis infinitely many times near $x=0$ and at values that are not expressible in closed form.

Plot[Sin[1/x] - x, {x, 0, 1/2},
  PlotPoints -> 200, MaxRecursion -> 12]

enter image description here

Considering these types of examples, I think that implicit representations are quite reasonable - even preferable. In simple cases, it's quite easy to extract out Unequal expressions, solve the corresponding equations, and come to the conclusion that you did with Maple.

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Mclure First, thanks to your answer. I think, we want to find the domain of the given function, we must solve the inequality x^3+1<>0. –  minthao_2011 Jul 17 at 2:01

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