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So, let $q$ be a generic complex number with $|q| = 1$, i.e. $q^n \ne 1$ for any $n \in \mathbb{N}$.

Then, does it hold that \begin{align} \lim_{n \to \infty} M_n |q^{M_n} -1| \to 0? \end{align} for any sequences of integer, $\{ M_n \}$ where $q^{M_n}$ converging to $1$?

If it isn't well-defined statement, are thery any similar well-defined or opposite statement(like, counterexamples) alternatives to this? Also, could you recommend the common way to approach such problems?

ChoMedit
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  • I don't think that converges for anything other than $q=1$. – march Mar 22 '22 at 16:19
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    $\liminf$ might converge though. – Testcase Mar 22 '22 at 16:55
  • Thanks for comments. I've edited the question. It's a little complicated to write down it rigourosuly, and now I found what I really want to ask. I think $\liminf$ is good, but It seems rough to cover every case of sequence. – ChoMedit Mar 23 '22 at 00:00
  • Isn't this similar to the closest the function $\lceil an \rceil - a n$ gets to $0$ for a transcendental real $a$ and an $n \in {1,\ldots, N}$? [function on $N$]. If it is $\theta(1/N)$ or larger, then I think the answer to your question is a no. – Mike Mar 23 '22 at 00:02
  • @Mike Hello, Mike. Thanks for comment. What does $\theta$ means? – ChoMedit Mar 23 '22 at 00:13
  • Write $q=e^{2\pi i a}$ where $a \in [0,1)$. Then for $\epsilon$ small enough, $|q^M-1| \in O(\epsilon)$ iff $\lceil an \rceil - an$ is $O(\epsilon)$. – Mike Mar 23 '22 at 00:16
  • Ah, I see you use theta notation. – ChoMedit Mar 23 '22 at 00:17
  • $\theta$ is asymptotic notation. For example, $2x=\theta(x)$. You probably will need to google this and become fluent in this to be able to work on this problem! – Mike Mar 23 '22 at 00:17
  • My apologies, I should have written this instead: "Write $q=e^{2\pi i a}$ where $a \in [0,1)$. Then for $\epsilon>0$ small enough, $|q^M-1| $ is $O(\epsilon)$ iff either $\lceil aM \rceil - aM$ or $\lfloor aM\rfloor -aM$ is also $O(\epsilon)$." – Mike Mar 23 '22 at 00:23

1 Answers1

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This topic is called Diophantine approximation.

The answer to your problem is: no, a counterexample is easily found.

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Firstly, for $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, the sequence $z_n=e^{2\pi i n\alpha}$ is dense on the unit circle. Every point on the circle is a limit point of the sequence. You can refer to this link.

Therefore, for any choice of $r>0$ and $\epsilon>0$, you can find arbitrarily large positive integers $M$ such that $|e^{2\pi i M\alpha}-e^{2\pi ir}|<\epsilon$.

$\\$

Now, let $r_n=\dfrac{1}{n}$. See, for fixed $n\in\mathbb{Z}^+$, the distance $|1-e^{2\pi ir_n}|$ is fixed.

$\\$

So choose $M_n$ such that

$|e^{2\pi i M_n\alpha}-e^{2\pi ir_n}|<\dfrac{1}{2}|1-e^{2\pi ir_n}|$

but $M_n|1-e^{2\pi ir_n}|$ is really large, e.g. greater than $n$.

These inequalities, by the way, ensure that

$|1-e^{2\pi i M_n\alpha}|$

$\leq|1-e^{2\pi ir_n}|+|e^{2\pi ir_n}-e^{2\pi i M_n\alpha}|$

$<(1+\frac{1}{2})|1-e^{2\pi ir_n}|\to0$,

but also,

$M_n|1-e^{2\pi i M_n\alpha}|$

$\geq M_n(|1-e^{2\pi ir_n}|-|e^{2\pi ir_n}-e^{2\pi i M_n\alpha}|)$

$>M_n(1-\frac{1}{2})|1-e^{2\pi ir_n}|$

$>\frac{n}{2}\to+\infty$.

Chris Sanders
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