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$$\lim_{n\to 0} n^{i} = \lim_{n\to 0} e^{i\log(n)} $$ I know that $0^{0}$ is generally undefined, but can equal one in the context of the empty set mapping to itself only one time. I realize that in terms of the equation above, the limit does not exist, but can $0^{i}$ be interpreted in a way to assign it a value? For the curious, I ran in to this when trying to calculate the imaginary-derivative of $\sin(x)$.

Martin Sleziak
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soultrane
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  • Well this might be related, but wolframalpha currently gives $\frac{d}{dx} 0^x= 0^x(-\infty)$ – Teoc Jan 12 '15 at 03:48
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    In the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $0^0=1$. Extending this to a complex arithmetic context is fraught with risks, as is the ambition to justify limits of this form generally by analogy to the value of a particular limit of this form. The derivative of the complex-valued sine function is everywhere well-defined. – hardmath Jan 12 '15 at 03:59
  • I was not taking the derivative of the complex sine function, but the imaginary-order derivative of sine: $$ \frac{d^{i}}{dx^{i}} \sin(x) $$ – soultrane Jan 12 '15 at 04:15
  • @Lubin that's pretty much the answer, why don't you post it as it? – hjhjhj57 Jan 12 '15 at 04:16
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    @hjhjhj57 the topic starter already indicated that the limit does not exist, so this comment does not add anything. He was asking what value can be assigned to it nevertheless. The mean value of the sequence is $0$ (see my ansswer). – Anixx Jan 12 '15 at 04:31
  • @soultrane i-th order derivative of sine is $i \sinh (\frac \pi 2-ix)$: http://tinyurl.com/kdmdkvf – Anixx Jan 12 '15 at 05:55
  • @Anixx I am familiar with the Fourier Inversion Theorem, but I do not see how what you linked to is equivalent to the i-th order derivative; can you explain? – soultrane Jan 12 '15 at 16:00
  • @soultrane the number after "^" determines the derivative order. Win $0$ it gives sin, with $1$ it gives cosine, with -1 it gives -cosine etc. But it is off-topic here, I can explain with more details as a separate answer to your question. – Anixx Jan 12 '15 at 16:04
  • I understand that, but I do not think you can necessarily generalize it to an imaginary power. I would assume that Wolfram has a formula it uses, and putting $i$ in there does not make anything inconsistent in Wolfram's formula, but that does not make it true. – soultrane Jan 12 '15 at 16:07
  • @soultrane there are multiple formulas, all give the same result, why you do not like it? I can explain with details. Wolfram specifics are not involved. – Anixx Jan 12 '15 at 16:08
  • Deriving the expression you're referencing involves integration by parts. I will make another post regarding the imaginary derivative, since that is really the main point of the original post I made. – soultrane Jan 12 '15 at 16:10
  • New post made: http://math.stackexchange.com/questions/1101432/imaginary-order-derivative – soultrane Jan 12 '15 at 16:38

4 Answers4

5

This was a comment, but @hjhjhj57 suggested that it might serve as an answer.

If you write the right-hand side of your equation as $\lim_{t\to−∞}e^{it}=\lim_{t\to−∞}(e^i)^t$, it’s completely clear that the limit doesn’t exist: you’re taking the number $e^i$, which is on the unit circle, and raising it to a large (but negative) power. You have a point that runs around the unit circle infinitely many times as $t\to−∞$, no limit at all.

Lubin
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    Hope this isn't too late of a response. I was looking at the problem again, and I am now wondering if it is relevant that $|{0^{i}}| = 1$, so the magnitude is convergent, but the phase is not. – soultrane Jan 15 '15 at 03:40
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    @soultrane, what you say is surely correct, and maybe we can say that it isolates somewhat the reason for nonconvergence. – Lubin Feb 14 '16 at 21:27
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It is possible to interpret such expressions in many ways that can make sense. The question is, what properties do we want such an interpretation to have?

$0^i = 0$ is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention $0^x = 0$. On the other hand, $0^{-1} = 0$ is clearly false (well, almost—see the discussion on goblin's answer), and $0^0=0$ is questionable, so this convention could be unwise when $x$ is not a positive real.

Digging deeper: One generally defines complex exponentiation as a multi-valued function: if $e^c = a$, then we can define $a^x = e^{cx}$. This is not unique, since it depends on the choice of $c$, but it's a good way to think about quantities like $i^i$ (this is sometimes claimed to be $e^{-\pi/2}$, but it can be interpreted as $e^{-\pi/2 + 2\pi n}$ for any $n\in\mathbb{Z}$)

This approach breaks down for $0^i$, because $0$ has no natural logarithm in the complex numbers. However, if we're comfortable calling $-\infty$ (or $-\infty + 2\pi i n$) a natural log of $0$, then we can say that $0^x = e^{-\infty \cdot x} = 0$ when $x$ has positive real part.

When $x$ has negative real part, this leads us to regard $0^x$ as a quantity with infinite magnitude and undefined argument. When $x$ is imaginary, the argument is still undefined, but the magnitude is multi-valued rather than infinite.

My conclusion is that we should avoid assigning meaning to $0^i$.

Writing $|0^i| = 1$ may be sensible, however, under some circumstances.

In a general setting, I would be comfortable saying that $|0^i| = e^{2\pi n}$, for any $n\in\mathbb{Z}$.

Andrew Dudzik
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    "follows the convention $0^x=0$" - where u got this "convention" from? "and $0^0=0$ is questionable" - it is not questionable, it it plainly wrong, it is usually define to be 1 or left undefined. – Anixx Jan 12 '15 at 07:28
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    @Anixx $0^0 = 1$ is also plainly wrong, but may be used in some circumstances (like power series evaluation) as a consequence of the formal identity $a^0 = 1$, which is valid for $a\neq 0$. By the same token, $0^x = 0$, which is valid for $x>0$, may be used in circumstances where it is regarded as a formal identity. – Andrew Dudzik Jan 12 '15 at 07:38
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    No, it is not wrong, it is widely accepted defginition. Prove it is wrong. – Anixx Jan 12 '15 at 07:48
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    @Anixx Where to begin? Most introductory calculus texts have a section on the problems with $0^0$, because it leads to some of the more common mistakes (from my experiences teaching calculus, anyway). For example, $\lim_{x\to 0^+} e^{-1/x} = 0$ and $\lim_{x\to 0^+} x = 0$, but $\lim_{x\to 0^+} (e^{-1/x})^x \neq 1$. – Andrew Dudzik Jan 12 '15 at 10:57
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    @Anixx Of course, $0^0 = 1$ is quite useful when doing "standard" algebra, e.g. power series, the binomial theorem, etc. But these things all have their limitations. In the comments to goblin's answer, there is even a link to a paper with $0^{-1} = 0$, which turns out to be a completely reasonable definition in some scenarios of interest. – Andrew Dudzik Jan 12 '15 at 11:05
  • For the case of $0^{0}$, couldn't you do something along the lines of $ \lim_{a \to 0} a^{0} = lim_{a \to 0} e^{0log(a)}$? I know strictly formally speaking there still may be an issue, but that quantity goes to $1$ since $log(a) \to -\infty$. – soultrane Jan 12 '15 at 14:42
  • Also, you say we should not assign meaning in your original comment, but then do so in the next few lines...The last line looks like you are saying the magnitude is equal to 1. – soultrane Jan 12 '15 at 14:44
  • @soultrane Depending on the choice of branch for $\ln$, the magnitude could be $1$, or $e^{2\pi}$, or $e^{-2\pi}$, etc. Speaking more precisely, the (multi-valued) function $z\mapsto |z^i|$ has a nice continuation at $0$, while the function $z\mapsto z^i$ does not. I do agree that $0^0=1$ makes sense under some circumstances, but it is possible to interpret $0^0$ as a limit so that it takes any value (including $\infty$). – Andrew Dudzik Jan 12 '15 at 19:59
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Wait.

The first reasonable way to understand if a good definition of $0^i$ can be given, is to define $$ 0^i:=\lim_{z\to0}z^i=\lim_{r\to0}(re^{i\theta})^i=e^{-\theta}\lim_{r\to0}r^i $$ But now $r^i=e^{i\log r}=\cos(\log r)+i\sin(\log r)$, and being $\log r\stackrel{r\to0}{\longrightarrow}-\infty$, we conclude the limit $\lim_{z\to0}z^i$ can't exist, thus $0^i$ can't be defined.

Joe
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We have:

$$\lim_{x\to0} e^{i \log x}=\lim_{x\to\infty} e^{-ix}=\lim_{x\to\infty} (-1)^{-x}=\lim_{x\to\infty} (-1)^{x}$$

The limit as you noticed, does not exist... But if you want to assign a value nevertheless... well, the mean value of $(-1)^n$ will be zero:

$$\lim _{x\to\infty} \frac 1{x}\int_0^x (-1)^x =0$$

Anixx
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  • Why the downvote? – Anixx Jan 12 '15 at 04:06
  • It's quite informal to use the infinity sign and the third equality appears to be wrong. I've cancelled it, since I may as well be wrong. If not, I'll downvote again. – hjhjhj57 Jan 12 '15 at 04:10
  • @hjhjhj57 what inequality? – Anixx Jan 12 '15 at 04:13
  • Oh, sorry! I meant equallity.. I've corrected it. What I mean is that neither $$\cos \lim_{x\to-\infty} x\quad\text{nor}\quad \sin \lim_{x\to-\infty}x$$ exist, but that doesn't mean they are equal to another limit that doesn't exist. – hjhjhj57 Jan 12 '15 at 04:13
  • @hjhjhj57 $e^{-i \pi}=-1$, so $e^{-i \infty}=e^{-i \pi \infty}=(-1)^{-\infty}$. Also, $(-1)^x=\cos (\pi x) + i \sin (\pi x)$ – Anixx Jan 12 '15 at 04:18
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    In that case $$ e^{-i\infty} = e^{-i\infty \frac{2\pi}{5}} = \left(\cos \left(-\frac{2\pi}{5} \right) + i\sin \left(-\frac{2\pi}{5} \right)\right)^\infty . $$ – hjhjhj57 Jan 12 '15 at 04:20
  • @hjhjhj57 yes, and what? – Anixx Jan 12 '15 at 04:21
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    By writing $e^{-i\infty}$ you assume that the object $e^{-i\infty}$ exists. Then you assume that De Moivre's formula can be used when $x=\infty$ twice. It's not a very clear answer. By rephrasing it as a limit then all of these objections goes away. – Winther Jan 12 '15 at 04:24
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    That expression isn't even defined, it's plain and simple abuse of notation. We all know the dangers of handling limits as ordinary numbers. – hjhjhj57 Jan 12 '15 at 04:26
  • @hjhjhj57 why u do not criticize Slade for the same? – Anixx Jan 12 '15 at 08:22
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    Because, even if your conclusion is the same (or similar), and you've edited your post: 1.- I still don't like the "$(-1)$" thing. 2.- He was very careful (or careful enough) when dealing with infinity, and the idea he's defending is $0^i=0$, which is nowhere mentioned in your answer. – hjhjhj57 Jan 12 '15 at 14:53
  • @hjhjhj57 I mention that the mean of this diverging series is $0$. – Anixx Jan 12 '15 at 15:00