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I was working on an analysis problem when this question arose in one my proofs. I think it may be either $-1$ or $1$, but it seems like there can only be an arbitrary way to assign this.

So is there an agreed upon method for determining $(-1)^x$ for irrational values of $x$?

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    $\begingroup$ might be trickier than you think - what happens at $x = \frac{1}{2}$? $\endgroup$ – MikeO Mar 12 '14 at 13:53
  • $\begingroup$ not to mention for $1.5,-1.5$ or other weird fractions... +1 $\endgroup$ – imranfat Mar 12 '14 at 13:55
  • $\begingroup$ This isn't an answer, it's simply reinforcing what the OP already knows, that their problem is not a simple one. $\endgroup$ – Dan Rust Mar 12 '14 at 14:17
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    $\begingroup$ @DanielRust: The OP stated that the result can only be $1$ or $-1$... That's a very big oversight IMO. $\endgroup$ – Najib Idrissi Mar 12 '14 at 14:26
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    $\begingroup$ How is this a duplicate?! The linked post discusses nothing about $a^x$ where $a$ is negative... $\endgroup$ – 6005 Mar 12 '14 at 15:13
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Euler's Formula ($e^{ix}=\cos(x)+i\sin(x)$) is pretty useful here. Since $\cos((2n+1)\pi)=-1$ and $\sin((2n+1)\pi)=0$ for integer $n$, $$\cos((2n+1)\pi)+i\sin((2n+1)\pi)=e^{i(2n+1)\pi}=-1$$

$$ (-1)^z = (e^{i(2n+1)\pi})^z = \cos((2n+1)\pi z)+i\sin((2n+1)\pi z) \tag{$n\in\mathbb{Z}$} $$ (This equation can also be viewed as a Generalization of De Moivre's formula)

Since $n$ goes over all integers, $(-1)^z$ has infinite values if the exponent $z$ is irrational. If $z$ is rational, its values repeat after a certain value of $n$.

Take $(-1)^\pi$ as an example: $$ \begin{align} (-1)^\pi &= (e^{i(2n+1)\pi})^\pi\\ &=e^{i(2n+1)\pi^2} \\ &=\cos((2n+1)\pi^2)+i\sin((2n+1)\pi^2) \tag{$n\in\mathbb{Z}$} \end{align} $$

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    $\begingroup$ Except $-1$ is also equal to $e^{3i\pi}$, so the result isn't well defined. $\endgroup$ – Najib Idrissi Mar 12 '14 at 14:14
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    $\begingroup$ @nik This is the principal value though. $\endgroup$ – user85798 Mar 12 '14 at 15:06
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    $\begingroup$ @OJB: Principal values apply to the complex logarithm. It is no proper to call values computed by plugging in a principal value of the logarithm into some formula a principal value as well. $(-1)^\pi$ simply has no meaningful value. $\endgroup$ – Marc van Leeuwen Mar 12 '14 at 15:34
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    $\begingroup$ @MarcvanLeeuwen en.wikipedia.org/wiki/Principal_value#Exponential_function Is wikipedia wrong? $\endgroup$ – user85798 Mar 12 '14 at 17:32
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    $\begingroup$ @Vibhav: You're just making things up here. $\endgroup$ – TonyK Mar 15 '14 at 8:39
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The general idea is that $a^b$ is to be defined as $\exp(b\log a)$, where that’s the natural logarithm in the parentheses. The problem is that the log has no unambiguous definition for numbers that are not positive real; generally there’s an ambiguity of multiples of $2\pi i$. Since “the” log of $-1$ is $(2k+1)\pi i$ for all integers $k$, you get infinitely many values for $(-1)^x$ when $x$ is not a rational number.

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As others have pointed out, using $a^x=\exp(\ln(a)x)$ to define exponentials with irrational exponents, the expression $(-1)^x$ would designate $\exp(m\pi\mathbf ix)$ for some odd integer $m$ depending on how the natural logarithm of $-1$ is chosen, which can be done in infinitely many ways. What I want to add to other answers is that, while some value for $m$ can be chosen, the fact that this ambiguity exists removes all utility that using the expression $(-1)^x$ might be imagined to have.

The situation is not at all like that of for instance the complex logarithm or arctangent functions, which functions cannot be continuously defined everywhere in the complex plane except at their singularities, which forces making branch cuts and choosing a value somewhere in the complement of the cuts to determine the branch. In those cases there is an inevitable problem in passing from a locally well defined function a globally defined one, the different branches share all the essential properties required (being local inverses of the exponential respectively tangent function, having a specific derivative) and can be transformed into one another by the process of gradually moving branch cuts. For $(-1)^x$ none of this applies. Each individual function $x\mapsto \exp(m\pi\mathbf ix)$ is defined without problem on the whole complex plane, and they have quite different properties; if one did not know that they were obtained as candidates for representing $x\mapsto(-1)^x$, it would be hard to say what property they have in common that sets them apart from other (exponential) functions. So if one wants to make one specific choice for $m$, say $m=1$, then by all means use $\exp(\pi\mathbf ix)$ and write it like that; however nothing is gained (except a great potential for confusion) by calling this $(-1)^x$. And by the above discussion, the term Principal Value is misplaced.

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  • $\begingroup$ I have removed the part of my answer mentioning principal values. Thanks for pointing that out. $\endgroup$ – Vibhav Pant Mar 18 '14 at 10:55
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You could think in this: $$(-1)^x=e^{i\pi x}$$

After a brief commentary actually one really have is: $$-1=e^{i(\pi +2k\pi)}, k \in \mathbb{Z}$$ And then: $$(-1)^x=e^{i(\pi +2k\pi)x}, k \in \mathbb{Z} $$ And one would have more than one value for $(-1)^x$

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  • $\begingroup$ That is on the right track, but I think you need $e^{(2k+1)i\pi x}$, where $k$ can be any integer. For $e^{3i\pi}=-1$ also. $\endgroup$ – MJD Mar 12 '14 at 14:03
  • $\begingroup$ yes, in that I was thinking $\endgroup$ – rlartiga Mar 12 '14 at 14:05
  • $\begingroup$ but that could make more than one definition for the function $\endgroup$ – rlartiga Mar 12 '14 at 14:06
  • $\begingroup$ Treated as a problem in complex numbers it will have more than one solution. Treated as a problem in real numbers it has no solutions if x is irrational. This follows because pi(1 + 2k)x must = n.pi for a real solution, i.e. (1 + 2k)x = n so that x = n/(1+ 2k) and so x has to be rational for a real solution. $\endgroup$ – Tom Collinge Mar 13 '14 at 11:38
  • $\begingroup$ In fact, as a problem in complex numbers it will have an infinite number of solututions. This follows from the formula above, and that for a finite number of solutions the values of pi(1 + 2k)x have to repeat so that pi(1 + 2k)x = pi(1 + 2m)x + 2n.pi for some k, m, and n. This requires that x = n/(k-m) which cannot be if x is irrational. $\endgroup$ – Tom Collinge Mar 13 '14 at 11:54
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The usual agreed-apon method, is to imagine that the imaginary part of the exponent is a line wrapped around the circle, the circumference is $2\pi$. The agreed on position is some point either in the range $-\pi\lt x \le\pi$ or $0 \le x \lt 2\pi$.

So if one has a number like $i$, then its complex logrithm is $\pi/2$. If we mean to raise it to some power, like $\pi^{17}$, then one calculates first $(\pi^{17}.\pi/2) \mod 2\pi$, that is, the remainder when an integer multiple of $2\pi$ is subtracted from the calculated value. And this remainder gives the power of the number raised to that power.

The solution for equations like $x^7=1$ is effected by $e^{2\pi ni}x^7=1$ for $n=0$ to $6$, and using the same model as above. When $n$ is not rational, this can go on forever, progressively splattering the unit circle with answers.

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  • $\begingroup$ ...In other words (and more accurately), "when $n$ is not rational", $x^n$ is not defined. $\endgroup$ – Did Mar 15 '14 at 9:58

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