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What is the domain of $(-1)^x$ as a real function?

We have $(-1)^{1/3}=-1$ and $(-1)^{1/2}$ undefined.

I'm confuse.

Mary
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    This is another way of asking the same question: "For what $x$ does the expression $(-1)^x$ make sense?" – Arthur Oct 10 '15 at 22:00
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    @Arthur that's not at all the same. – rewritten Oct 10 '15 at 22:10
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    How is $(-1)^x$ defined? – Cameron Buie Oct 10 '15 at 22:23
  • @CameronBuie I think that's the main issue. If it's a definition that uses polynomials, of course it makes sense only on rationals, if on the other way it's an actual function (with analytical continuation etc) then its value at $\frac{1}{2n+1}$ is not $-1$ but the principal root $e^\frac{i\pi}{n}$. – rewritten Oct 10 '15 at 22:27

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It's strange as a precalculus question, anyway, $(-1)^x$ is a function defined in the whole complex plane, but the only points on the real axis where it takes real values are the integers, where it takes value $1$ or $-1$ depending on parity.

To calculate it, start from $-1=e^{(2n+1)i\pi}$, so $$ (-1)^x = e^{(2n+1)i\pi x} = \cos((2n+1)\pi x)+i \sin((2n+1)\pi x) $$

For it to be real, you need $$ \sin((2n+1)\pi x) = 0 $$ so $$ (2n+1)x \in \mathbb Z $$ for some $n\in\mathbb Z$, that is, $x$ can be represented as a fraction with odd denominator.

Simpler solution.

After a long discussion in the comments, it seems clear that the OP needs an elementary solution that does not involve complex numbers. Exactly as @Arthur says in his comment on the question.

So let's suppose that we "define" $(-1)^{\frac n m}$ ($m$ and $n$ coprime), as the only real number (if any) $r$ that satisfy $$ r^m = (-1)^n. $$ There is no way to define anything similar on irrational numbers anyway.

In this case the function is defined whenever $m$ is odd. If $m$ is even, then $n$ must be odd and the LHS of the equation ($r^m$) is positive where the RHS is negative.

So this function is defined on "some" dense subset of rationals (as said, those whose reduction has odd denominator), and it takes values $-1$ and $1$ in two co-dense subsets of its domain.

rewritten
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  • We have $(-1)^{1/3}=-1$ therefore $1/3 \in Dom (-1)^x$ – Mary Oct 10 '15 at 22:14
  • Do we just completely leave out all the fractions that also give real asnwers? (Because there is an infinite amount of fractions that give a real answer and an infinite amount that do not?) – user265554 Oct 10 '15 at 22:14
  • Okay, if you are willing to define $(-1)^{1/3}=-1$, what about $(-1)^{1/5}$? – dbanet Oct 10 '15 at 22:21
  • The problem here is that $(-1)^x$ is multivalued if $x$ is a rational number with an odd denominator. For example, $(-1)^{1/3}$ has three possible values which are $\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})$, $\cos(\pi)+i\sin(\pi)$, and $\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$. If we are talking about real valued functions and there is a real number present we define the value to be it. So as a real valued function $(-1)^{1/3}=\cos(\pi)+i\sin(\pi)=1$. As such, I don't know that using euler's formula is the best approach here. – Eric Oct 10 '15 at 22:22
  • This is incorrect. To see why, observe that we also have $e^{3\pi i}=-1,$ so that $x=\frac13$ is part of the domain of definition. – Cameron Buie Oct 10 '15 at 22:26
  • @Eric $x^3=-1$ has got three solutions, one of which is real yes, but $(-1)^{1/3}$ can and does often mean different things, depending on what is considered more useful, the intentions, preferences of the author and the overall context of usage. I personally do not prefer the multivaluedness viewpoint. I either do not prefer the real viewpoint, but that's what is taught in high school. Go figure what do they want from the OP. – dbanet Oct 10 '15 at 22:28
  • Also, I think you meant $\Bbb Z,$ and not $\Bbb N.$ – Cameron Buie Oct 10 '15 at 22:34
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    @CameronBuie that's true, I'm changing that, thanks. On the rest, I still think that there is no definition of $(-1)^x$ as a function of $x$ that gives meaningful values outside of the complex one. Using tricks like "a real solution of the polynomials" or extracting roots is not fair in proofs, as overly demostrated by $(-1)^2 = (1)^2 \nRightarrow -1 = 1$. – rewritten Oct 10 '15 at 22:38
  • @CameronBuie about $e^{3πi}=−1$, this proves that $3\in \Dom(f)$, nothing to say about $\frac 1 3$ – rewritten Oct 10 '15 at 22:40
  • I think it should be pretty obvious that the question is asked in the low-level calculus context where complex numbers are explicitly avoided. Then the expression $(-1)^{1/n}$ is the real root if $n$ is odd and undefined if $n$ is even. The intended answer is a proper subset of the rational numbers. – James Staff Oct 10 '15 at 22:47
  • It's true that $(-1)^2=1^2\nRightarrow-1=1,$ because $x\mapsto x^2$ is not a one-to-one function on the reals. However, for example $x\mapsto x^3$ is a one-to-one function on the reals, so there is no particular problem in general with extracting appropriate roots. Also, since $e^{3\pi i}=-1,$ then reasoning in exactly the same way as in your answer, we find that $(-1)^x=\cos(3\pi x)+i\sin(3\pi x),$ whence $x=\frac13$ is a solution. – Cameron Buie Oct 10 '15 at 22:48
  • With equal validity one could say that $(-1)^x = e^{-i\pi x}$, so the assertion that it's $e^{i\pi x}$ should have some qualification. ${}\qquad{}$ – Michael Hardy Oct 10 '15 at 23:12
  • @MichaelHardy why so? I am missing something. – rewritten Oct 10 '15 at 23:14
  • I have altered my downvote to an upvote, in light of your edit. Still, I recommend either fixing the portion using Euler's formula, or removing it altogether. As it stands, it suggests that different approaches would yield different domains, which is not the case (and could further confuse the OP)! – Cameron Buie Oct 10 '15 at 23:20
  • This is relevant: http://math.stackexchange.com/questions/1332412/is-the-set-of-all-rational-numbers-with-odd-denominators-a-subring-of-bbb-q?rq=1 and so is this: http://math.stackexchange.com/questions/106465/ring-of-rational-numbers-with-odd-denominator – James Staff Oct 10 '15 at 23:43
  • @rewritten : Because it's only one possibility; the other is equally valid. – Michael Hardy Oct 11 '15 at 00:20
  • What you are missing is that the function $f(z)=e^z$ is periodic, with fundamental period $2\pi i.$ So, $-1=e^z$ if and only if $z=(2n+1)\pi i$ for some $n\in\Bbb Z.$ From here, we can derive the same answer as was found in your edit, using similar reasoning as in you original answer. – Cameron Buie Oct 11 '15 at 16:06
  • :) of course, I didn't think about that. I'll edit the first part too, thank you very much for shading light onto this. – rewritten Oct 11 '15 at 16:12
  • @CameronBuie after editing, it's absolutely clear that I had entirely missed the point, the first answer was plainly wrong, not just confusing (and confused). – rewritten Oct 11 '15 at 16:17
  • You are very welcome. Note that you need $n\in\Bbb Z,$ rather than $n\in\Bbb N.$ Once you've made that adjustment, it's all fixed! – Cameron Buie Oct 12 '15 at 00:14
  • Actually, I think $n\in\Bbb N$ suffices, but it still helps to point out that it can be any $n\in\Bbb Z.$ – Cameron Buie Oct 12 '15 at 00:22
  • in fact it suffices, because $(2n+1)x∈ℤ \iff (2(-n-1)+1)x∈ℤ$ . – rewritten Oct 12 '15 at 06:57