# Limit of ##i^\frac{1}{n}## as ##n \to \infty##

• A
• mathman
In summary, the conversation discusses the existence of a limit for ##i^\frac{1}{n}## and the importance of specifying a branch in order to have a well-defined function. It is noted that if the principal branch is chosen, the limit is 1, but if other branches are chosen, the limit may not exist or may have different values. The use of default assumptions in communication is also mentioned.

#### mathman

TL;DR Summary
Does a limit exist? I believe no!
##i^\frac{1}{n}## has n roots. If one is not careful, the limit as ##n \to \infty## is 1. Simple proof: ##i=e^\frac{\pi i}{2}## or ##i^\frac{1}{n}=e^\frac{\pi i}{2n} \to e^0=1##.

This does not take into account the n roots, since ##i=e^{(\pi i)(2k+\frac{1}{2})}##.. Here ##\frac{k}{n} ## can have any value with ##1\le k \le n## for the n roots. Therefore given any ##0\lt x \le 1##, it is easy to construct a sequence of ##\frac{k}{n}\to x## so any point on the unit circle will be the limit of a sub-sequence.

Am I missing something?

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Suppose that you specify the principle root? Does it have a limit then?

mathman said:
Summary: Does a limit exist? I believe no!

so any point on the unit circle will be the limit of a sub-sequence.
which means that any point on the circle is a cluster point (not a limit).

I suppose the distinction here is in defining $$f_n: \mathbb{C} \setminus\{0\} \to \mathbb{C} : z \mapsto \exp\left(\frac{\log z + 2ik\pi}{n}\right)$$ with $k$ fixed at the outset on the one hand, and defining $$f_n: \mathbb{C} \setminus\{0\}\to \mathbb{C} : z \mapsto \exp\left(\frac{\log z + 2ik_n\pi}{n}\right)$$ on the other. In the first, $k$ is fixed and necessarily $k/n \to 0$ as $n \to \infty$ and $f_n(z) \to 1$; in the second $k_n$ can be chosen such that $k_n/n \to x$ for any $x \in [0,1]$ and then $f_n(z) \to e^{2i\pi x}$. Alternatively, $k_n$ could be chosen such that $k_n/n$ does not converge to a limit.

The question is, therefore, which of these approaches do you understand $\lim_{n \to \infty} i^{1/n}$ to mean? I would take it to mean the result of using the principal branch ($k = 0$) throughout unless the author specified a different $(k_n)$ and a justification for doing so.

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pasmith said:
I suppose the distinction here is in defining $$f_n: \mathbb{C} \setminus\{0\} \to \mathbb{C} : z \mapsto \exp\left(\frac{\log z + 2ik\pi}{n}\right)$$ with $k$ fixed at the outset on the one hand, and defining $$f_n: \mathbb{C} \setminus\{0\}\to \mathbb{C} : z \mapsto \exp\left(\frac{\log z + 2ik_n\pi}{n}\right)$$ on the other. In the first, $k$ is fixed and necessarily $k/n \to 0$ as $n \to \infty$ and $f_n(z) \to 1$; in the second $k_n$ can be chosen such that $k_n/n \to x$ for any $x \in [0,1]$ and then $f_n(z) \to e^{2i\pi x}$. Alternatively, $k_n$ could be chosen such that $k_n/n$ does not converge to a limit.

The question is, therefore, which of these approaches do you understand $\lim_{n \to \infty} i^{1/n}$ to mean? I would take it to mean the result of using the principal branch ($k = 0$) throughout unless the author specified a different $(k_n)$ and a justification for doing so.
Taking the principal branch is your assumption, but it is NOT part of the original question.

It's kind of part of the original question. Limits normally apply to functions. If you don't pick a branch, you don't even have a function, so what the heck does the limit even mean? And if you're going to pick a branch, you normally take the principal one by default.

The trick is just taking advantage of a small amount of ambiguity to pretend someone made a mistake when the real issue is you failed to communicate properly.

It is not a small amount! I can set up a sequence of (k,n) such as (1,1),(1,2),(2,2),(1,3),(2,3),(3,3)... representing that part of the exponent of ##i^\frac{1}{n}##. There is nothing except arbitrary choice that k is fixed for all n.

I mean, let's try something else.

##\lim_{n\to \infty} \sqrt{1}##

Does it exist?

You can define the question so that there is no limit or so that there is a limit. You made up the question, so you can decide. I don't think that your decision is very interesting. On the other hand, if you specify the principle branch, you can even talk about derivatives and entire functions.

mathman said:
Taking the principal branch is your assumption, but it is NOT part of the original question.

I also now realize that I had assumed, entirely without justification, that in taking the limit I should do so with respect to a topology on $\mathbb{C}$ equivalent to that induced by the metric $|z - w|$. Was I correct, or did you have smoething different in mind? In fact, if I use the indiscrete topology instead then for any sequence of branches we have $\lim_{n \to \infty} i^{1/n} = 1726$.

Default assumptions are a thing in communication. If you don't spell something out explicitly, people are going to make assumptions about what you would have said, and the default asumption is the default because in most circumstances it is the simplest, most obvious, or most convenient, or poeple have previously been told to use it by other authority. If there's a reason in the specific circumstances that the default assumption does not work or is less convenient, then it is for you to call attention to it and specify your preferred alternative.

Office_Shredder
Office_Shredder said:
I mean, let's try something else.

##\lim_{n\to \infty} \sqrt{1}##

Does it exist?
Your expression is a constnt ##\sqrt{1}##. Where is n?

pasmith said:
I also now realize that I had assumed, entirely without justification, that in taking the limit I should do so with respect to a topology on $\mathbb{C}$ equivalent to that induced by the metric $|z - w|$. Was I correct, or did you have smoething different in mind? In fact, if I use the indiscrete topology instead then for any sequence of branches we have $\lim_{n \to \infty} i^{1/n} = 1726$.

Default assumptions are a thing in communication. If you don't spell something out explicitly, people are going to make assumptions about what you would have said, and the default asumption is the default because in most circumstances it is the simplest, most obvious, or most convenient, or poeple have previously been told to use it by other authority. If there's a reason in the specific circumstances that the default assumption does not work or is less convenient, then it is for you to call attention to it and specify your preferred alternative.
What topology? The limit is for a discrete sequence. Default asssumption is nice, but here there is no obvious one. Simpler example ##1^\frac{1}{2n}##. as ##n\to \infty##, assuming real. Here there are two limits 1 and -1.

mathman said:
Your expression is a constnt ##\sqrt{1}##. Where is n?
Are you sure it's a constant?

Office_Shredder said:
Are you sure it's a constant?
It has two values. WHERE IS n?

mathman said:
It has two values. WHERE IS n?
Nowhere. But for every choice of n, I can pick which of the two values it picks, right? Therefore the limit doesn't exist.

Sounds dumb? It is!