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It's clear that the set $\{\sin k \mid k \in \mathbb{N}\}$ is dense in $[0, 1)$. What I'm having trouble determining is how the sequence $\{(\sin k)^k \mid k \in \mathbb{N}\}$ behaves for large $k$.

My main question is: does there exist $N \in \mathbb{N}$ and $\epsilon \in \mathbb{R}, 0 \leq \epsilon < 1$ such that for all $k \geq N, k \in \mathbb{N}$, $(\sin k)^k \leq \epsilon$? Or equivalently, does there exist $N \in \mathbb{N}$ such that $\sup(\{(\sin k)^k \mid k \in \mathbb{N}, k \geq N\}) \neq 1$?

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Sully Chen
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  • Hint: use power series of $sin(k)^k$ – Ahmed Masud Apr 24 '18 at 18:32
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    @AhmedMasud How? It's very non-obvious to me, because that seems more relevant for $(\sin x)^k$ for a fixed argument. –  Apr 24 '18 at 18:33
  • This is no answer as I have no proof, but off-hand it does seem that there are arbitrarily large values of $k$ s.t. $(\sin k)^k$ is at least $e^{-2}$. I mean, there should be at least one $k \in [m,2m]$, where $m$ is an arbitrarily large integer, s.t. $k-2\pi\lfloor \frac{k}{2\pi} \rfloor$ lands in the interval $[\frac{\pi}{2} -\frac{1}{m}, \frac{\pi}{2} + \frac{1}{m}]$, which would imply $\sin k \geq 1 - \frac{1}{2k}$... – Mike Apr 24 '18 at 18:58
  • ...excuse me, I should have written $k - 2\pi\lfloor \frac{k}{2\pi} \rfloor$ lands in the interval $[\frac{\pi}{2} - \frac{2 \pi}{2m}, \frac{\pi}{2} + \frac{2\pi}{2m}]$ or $[\frac{3\pi}{2} - \frac{2 \pi}{2m}, \frac{3\pi}{2} + \frac{2\pi}{2m}]$which would give $|\sin k|$ $\ge 1-\frac{2\pi}{k}$ which would give something close to $|\sin k|^k \geq e^{-\pi}$ for at least some $k$ arbitrarily large. – Mike Apr 24 '18 at 23:42
  • It can be shown that $$\limsup_{n\to \infty} |\sin^n n| = 1$$ so the answer to your question is no. For a proof, see my answer here, https://math.stackexchange.com/questions/2684662/compute-limsup-sinn-n/2684948#2684948 – pisco Aug 04 '18 at 09:34

1 Answers1

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So I think this is actually an open question, because it's so closely related to the irrationality measure of $\pi$. We are essentially asking whether $k$ can be really close to a number of the form $n\pi + \pi/2$, because that's a necessary condition to have $\sin k \approx 1$. In fact, we need to have $\sin k \approx 1$ to an error that is controlled even after raising to power $k$. This is pretty delicate.

We can write down rational approximations of $\pi/2$. In particular, suppose we can find a sequence of rational numbers such that

$$\left|\frac{\pi}{2} - \frac{p_j}{q_j}\right| \le \frac{1}{q_j^{2 + \mu}}.$$ for some $\mu > 0$ (equivalently, that the irrationality measure of $\pi$ is strictly greater than $2$).

Without loss of generality, $q_j$ is odd. In that case, we have

$$\left|p_j - q_j \frac{\pi}{2}\right| \le \frac{1}{q_j^{1 + \mu}} \implies \left|p_j - (2n_j + 1)\frac{\pi}{2}\right| \le \frac{1}{q_j^{1 + \mu}}$$

for some integer $n_j$. Now $p_j$ is our candidate index; we have

$$1 - |\sin p_j|^{p_j} \le 1 - \left(1 - \frac 1 {q_j^{1 + \mu}}\right)^{p_j} \approx \frac{p_j}{q_j^{1 + \mu}} \approx \frac{1}{q_j^{\mu}} \to 0$$

as $j \to \infty$. In this case, the supremum of $\{(\sin p_j)^{p_j}\}$ is equal to $1$.

On the other hand, suppose that there were an $\epsilon > 0$ such that for all sufficiently large $p, q$ we had

$$\left|\frac{\pi}{2} - \frac p q\right| \ge \epsilon \frac{1}{q^2}.$$

Then we can play the same game, finding that

$$1 - |\sin p|^p$$

is bounded below, and so the supremum is less than $1$.

  • As a further comment on the last part, it basically boils down to whether you can improve the constant in Borel's theorem for the special case of $\pi$. –  Apr 24 '18 at 19:38
  • Ah I see. The only part I'm a little fuzzy on is how we can show that $1 - |\sin p_j|^{p_j} \leq 1 - \left ( 1 - \frac{1}{{q_j}^{1 + \mu}} \right ) ^ {p_j}$ is true. That implies that $\left ( 1 - \frac{1}{{q_j}^{1 + \mu}} \right ) \leq |\sin p_j|$, which makes sense, but I'm not sure how to verify that's true. – Sully Chen Apr 24 '18 at 23:57
  • Also, just to make sure I understand this correctly, this problem is the exact same rephrased as $(\cos k)^k$, in which case the inequality would be $\left |p_j - 2\pi q_j \right | \leq \frac{1}{{q_j}^{1 + \mu}}$ right? – Sully Chen Apr 25 '18 at 00:00
  • The inequality comes from knowing the difference between $p_j$ and the translate of $\pi/2$; since it's smaller than $1/q_j^{1 + \mu}$ and the derivative of sine is bounded by $1$, the difference in outputs is controlled by $1 / q_j^{1 + \mu}$. –  Apr 25 '18 at 00:01
  • Yes, for $\cos$ you'd look at approximations of $\pi$ instead. –  Apr 25 '18 at 00:02
  • I see. Thanks, phenomenal answer. – Sully Chen Apr 25 '18 at 00:05
  • Thank you! :) ${}$ –  Apr 25 '18 at 00:07
  • This is a fascinating subject that I'd like to write a paper on for one of my classes (I'm an undergraduate math major). The result you've shown is very useful to me, so I was wondering if I should cite you, and if so, how would you like to be cited? – Sully Chen Apr 25 '18 at 09:05
  • I'm glad that this was helpful; you can see information on citing posts on meta. –  Apr 25 '18 at 18:52
  • If you have the time, could you perhaps show the work on the second case where $\mu = 0$? I did the derivation but I didn’t find that it imposes a strong enough condition to show that the supremum is less than one; instead I found it shows the supremum is at least greater than a lower bound, meaning it could still be 1. – Sully Chen Apr 28 '18 at 04:43
  • @SullyChen Unfortunately, $\mu = 0$ is sort of indeterminate, and it really depends on the precise degree to which you can improve the constant in Borel's theorem. If you have some sort of quantitative decay, you would get that the supremum is $1$, and if you don't then you don't improve the result. Something like logarithmic decay would be enough (or any other decay to zero!), but not detectable through something as blunt as irrationality measure. –  Apr 28 '18 at 04:44
  • that makes more sense thanks! – Sully Chen Apr 28 '18 at 05:21
  • I think this is not an open problem. It can be shown that $$\limsup_{n\to \infty} |\sin^n n| = 1$$, see my answer here https://math.stackexchange.com/questions/2684662 – pisco Aug 04 '18 at 09:38
  • @pisco I don't think that resolves the open-ness of the originally stated problem, that is, the end behaviour of the sequence given. What you have shown is that there exists a subsequence with limit 1, so it has limit superior 1. That does not resolve the density of the sequence on the interval [-1,1] – Jack Tiger Lam Jun 13 '19 at 14:43