I wonder whether anyone ever tried to introduce an extension of real numbers by adding an element $\nu$ which would signify the behavior of the function $(-1)^x$ as $x$ goes to infinity?

In other words, in some generalization of limit, $$\operatorname{gen}\lim_{x\to\infty} (-1)^x =\nu$$

This value $\nu$ should characterize the particular way in which $(-1)^x$ behaves at infinity, so that if a function oscillates at infinity in a different way, for instance, with different frequency, it has a different "limit" but it still may be expressable in terms of $\nu$, if we can do some operations on the function to make it behave like $(-1)^x$ at infinity. For instance, $\operatorname{gen} \lim_{x\to\infty} \cos^2 (\pi x/2)=(\nu+\frac1\nu)/2+1/2$. I think, due to Fourier series we can represent the behavior of any periodic function at infinity in terms of $\nu$, though the series could turn infinite.

Introduction of such element would also allow to represent the series like $\sum_{k=0}^\infty (-1)^k$ more precisely than just giving the Cesaro-regularized value.

The functions like $\cos (1/x)$ can be made continous at zero...

Particularly we can define

$$\operatorname{gen}\lim_{x\to\infty} \cos ax = \frac{\nu^{a/\pi}+\nu^{-a/\pi}}2$$


$$\operatorname{gen}\lim_{x\to\infty} \sin ax = i\frac{\nu^{-a/\pi}-\nu^{a/\pi}}2$$

This way sine and cosine would obey the Pythagorean trigonometric identity even at infinity, despite the Cesaro-regularized (mean) value of the both is zero.

Moreover, we can represent some non-periodic functions this way as well, for instance we would formally have

$$\operatorname{gen}\lim_{x\to\infty} \exp x = \nu^{-i/\pi}$$

We also would be able to define Dirac Delta function as an elementary function: $$\delta (z)=\frac{\nu ^{z/\pi }-\nu ^{-z/\pi }}{2 \pi i z}$$

That said, I wonder whether anyone came up with an idea of oscillating unity, and what would be the algebraic properties of such "number"?

  • $\begingroup$ What does $(-1)^x$ mean? $\endgroup$ Dec 22, 2019 at 14:25
  • $\begingroup$ @Steven Landsburg : Since he has not specified whether the domain is $\mathbb{Z}$ or $\mathbb{R}$, I suspect it is either two of: $(-1)^n = \pm 1$ depending on the parity of $n$, for $n \in \mathbb{Z}$, otherwise it is $(-1)^x = e^{(2k + 1) i \pi x}$ for $x \in \mathbb{R}$, and some set $k \in \mathbb{Z}$ (presumably $k = 0$ but ideally would be a further piece of information to be included.). $\endgroup$ Dec 23, 2019 at 3:17
  • 1
    $\begingroup$ @The_Sympathizer : It somehow failed to occur to me that the domain might be anything other than ${\mathbb R}$. Mea culpa. $\endgroup$ Dec 23, 2019 at 3:42

3 Answers 3


Do you know the Ultrafunction theory by Benci et al. https://arxiv.org/pdf/1405.4152.pdf

If I remember well from the seminars, once you define the $\Lambda$ limit, the sequence $(-1)^n$ has a limit, and you have limits also for $(-1)^x$.

The interesting thing about the theory of ultrafunctions is that in some sense permits us to do analysis and solve some functional analytic problems that were intractable without it.

If you look at section 3.2 they define the $\delta$ ultrafunctions, which are ultrafunctions that play the exact role of the $\delta$ distributions.


There are, actually, a number of ways one can talk of the limiting behavior of a function for which the limit in the "classical" sense you are referencing does not exist. But pretty much all of them do not make the generalized "limit" a number: instead, the limit must be a type of object that is has more "degrees of freedom" or is more "expressive" than a number, because there are a vast (infinite, uncountably infinite) number of possible limiting behaviors once you get beyond just the usual concept of limit and perhaps also its simple extension to unbounded growth by adding the numbers $\pm \infty$ to the number system.

For example, if you are treating $x \mapsto (-1)^x$ as a function with domain $\mathbb{Z}$, so perhaps better we'd write $n$, one would have to note that its limiting behavior - to oscillate between two values, i.e. -1 and +1 - is far from the only possible such. We could, say, easily create an oscillation between three values: take

$$f: \mathbb{Z} \rightarrow \mathbb{Z},\ f(n) := n\ \mathrm{mod}\ 3$$

which cycles through 0, 1, and 2. A natural place where these appear non-trivially as a "limiting behavior" is in dynamical systems: consider the sequence given by

$$f: \mathbb{N} \rightarrow \mathbb{R}, f(n) := \left[x \mapsto 3.839x(1 - x)\right]^n \left(\frac{1}{2}\right)$$

where the exponent iterates the anonymous function taking $x$ to $3.839x(1 - x)$ - this is a special case of the famous logistic map with parameter $r = 3.839$. The (nontrivial, with changing values throughout) limit behavior is a cycle between three real numbers that are approximately 0.149888, 0.489172, and 0.959299. Note that these numbers aren't even equally spaced - so would it make sense to categorize this as the same as $n \mapsto (-1)^n$? This is likely why that no single "number" exists in the sense you're thinking of to accomplish this: there's just far too many possibilities for the limiting behaviors.

So what do you do? Well, one way to do it is to introduce the idea of a limit point, and you can talk of a limiting set: in particular, you could say

$$\mbox{gen lim}_{x \rightarrow \infty} f(x)$$

is the set $S_f$ of real numbers (or perhaps also extended real numbers) such that for each $L \in S_f$, there is a subsequence $x_0, x_1, x_2, \cdots$ of inputs such that

$$\lim_{n \rightarrow \infty} x_n = \infty$$


$$\lim_{n \rightarrow \infty} f(x_n) = L$$

. You can see easily then that

$$\mbox{gen lim}_{n \rightarrow \infty} (-1)^n = \left\{ -1, +1 \right\}$$

as you'd expect. The relevant sequence of inputs $n$ for the limit point $-1$ are the odd values, and for $+1$, are the even values.

Likewise, if you take (see comment below your question post)

$$f: \mathbb{R} \rightarrow \mathbb{C}, f(x) := (-1)^x$$

with a prior choice of branch, then you can see the generalized limit set is the unit circle, in the same way: the relevant sequence is found by taking any $x \in \mathbb{R}$ and incrementing by 2s. Which, of course, is "intuitively" what you'd think - it circles around and around on just that.

And if we take

$$f: \mathbb{R} \rightarrow \mathbb{C}, f(x) := (-2)^x$$

instead, then we see the generalized limit set is empty, i.e. no limit exists, or else we go to the Riemann sphere and it contains the single element complex infinity, as this function spirals out to larger and larger complex values (I believe it describes a loxodrome [line of constant compass bearing] on the sphere, as understood through the usual stereographic mapping). In particular, if the generalized limit set contains a single element, that single element will be the usual limit, so the idea of this as being worthy of being called a generalization is sensible.

  • $\begingroup$ No, I am treating $(-1)^x$ as function of reals. By definition, $\operatorname{gen}\lim_{x\to\infty} f(x)=\nu$ if and only if $\lim_{x\to\infty} (f(x)-(-1)^x)=0$ $\endgroup$
    – Anixx
    Dec 23, 2019 at 9:23
  • $\begingroup$ @Anixx : I covered that case further down, too. $\endgroup$ Dec 23, 2019 at 9:24
  • $\begingroup$ I am not considering the generalized limit as a set of the values of the function at infinity, but rather the characteristic of the behavior of the function. In other words, the functions which take values from $[-1,1]$ at infinity may have different generalized limits depending on the frequency, the phase and the shape on their graphic. $\endgroup$
    – Anixx
    Dec 23, 2019 at 9:34
  • $\begingroup$ For instance, since $\cos ax=\frac{(-1)^{a/\pi}+(-1)^{-a/\pi}}2$ it follows that $\operatorname{gen}\lim_{x\to\infty} \cos ax = \frac{\nu^{a/\pi}+\nu^{-a/\pi}}2$ while the generalized limit of sine would be different. Only functions that approach each other at infinity have the same generalized limit. $\endgroup$
    – Anixx
    Dec 23, 2019 at 9:45
  • 2
    $\begingroup$ Sounds to me like you are looking for the Stone-Cech compactification of the real line or something very similar. $\endgroup$ Jun 1, 2020 at 17:20

There is nothing stopping you from constructing a theory of numbers and analysis rigorously by which the generalized limit of the function $exp(πix)$ exists. In fact, for all I know, maybe this has already been considered. The question is: how is this useful? Are there any new insights you can gain by doing this? Are you solving any previously unsolved problems with this theory? Are there any situations that would incentivize me applying this theory instead of simply using classical analysis? If the answer to these questions is never "Yes," then maybe that would explain why you have not heard of anyone trying to do this. It certainly is an interesting concept for the sake of exploration itself, but unless there is anything else to gain from such a theory in applications, you will not see anything like this becoming the next big thing, definitely nothing akin to the invention of complex numbers or quaternions. But I applaud the fact that you at least arrived at this exploration on your own. Hey, maybe you're actually onto something ;)

  • 3
    $\begingroup$ Sorry to ask, bot how does this answer the actual question? I am a bit confused. $\endgroup$ Nov 22, 2020 at 8:57
  • $\begingroup$ @AndrásBátkai Looking back, this is a bit of a non-answer, but I also think it has to be that way, since the question really has no answer. I do not believe there has been a serious consideration of the concept being proposed in the mathematical community, which I think is implied in my comment, and that gets as close to an answer as it gets. But I also think it is equally important to remind the asker and make them think, why would that be the case? $\endgroup$
    – Angel
    Oct 19, 2021 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.