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A younger student asked me:

What are accumulation points of the following set?

$$ \{x_n \in \mathbb{R}, n \in \mathbb{N} \ \ | \ x_n = n\sin(n) \}$$

I really can't answer this question, could anyone help me?

Ivan Di Liberti
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2 Answers2

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The irrationality measure $\mu(\pi)$ of $\pi$ is not known better than $2\le\mu(\pi)\le7.6063$, it seems. If we knew $\mu(\pi)>2$ then there are infinitely many fractions with $\left|\frac ab-\pi\right|<\frac1{b^\kappa}$ for some $\kappa>2$. Then $|\sin a\|<|a-b\pi|<\frac 1{b^{\kappa-1}}$ and so $|a\sin a|$ becomes arbitrarily small, making $0$ an accumulation point. Unfortunately, I can only confirm that $\left|\frac ab-\pi\right|<\frac c{b^2}$ for infinitely many fractions, which results in infinitely many $n$ with $|n\sin n|<c\pi$. So at least some accumulation point exists.

Martin Sleziak
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Hagen von Eitzen
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  • I am quite convinced that through Hurwitz theorem and the sum formulas for the sine function we can prove much more, i.e. $\bar{E}=\mathbb{R}$ depending on some parity property of the convergents of $\pi$. – Jack D'Aurizio Jun 02 '15 at 15:18
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Let $E=\{n\sin n:n\in\mathbb{N}\}$. For first, let we prove that a quite big neighbourhood of the origin belongs to $\bar{E}$ (the closure of $E$). By Hurwitz' theorem there are an infinite number of rational numbers $\frac{p_n}{q_n}$ such that: $$ \left|\pi-\frac{p_n}{q_n}\right|<\frac{1}{\sqrt{5} q_n^2}, $$ so $\left|\pi q_n-p_n\right|<\frac{1}{q_n\sqrt{5}}$ and: $$\begin{eqnarray*} p_n \sin p_n &=& p_n \left(\sin(\pi q_n)-\cos(\pi q_n)\cdot|\pi q_n-p_n|+O\left(\frac{1}{q_n^2}\right)\right)\\&=& p_n (-1)^{q_n}|\pi q_n-p_n|+O\left(\frac{1}{q_n}\right)\end{eqnarray*}$$ so there are an infinite number of elements of $E$ in the interval $I=\left(-\frac{\pi}{\sqrt{5}},\frac{\pi}{\sqrt{5}}\right)$.

Assuming that an infinite number of $q_n$s is odd (always true) and an infinite number of $q_n$s is even (it looks reasonable), by evaluating $n\sin n$ in $p_n+p_m$ we have that both $(E-E)$ and $(E+E)$ are dense in $E$.

By recalling the lemma for which an infinite set of points of an interval that is closed by difference is dense in the interval, it follows that $E$ is dense in $\mathbb{R}$.

Jack D'Aurizio
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  • "so a neighbourhood of the origin belongs to \overline{E}": how does that follow from the previous line, exactly? I only see that there is some accumulation point in $[-\pi/\sqrt{5}, \pi/\sqrt{5}]$. – D. Thomine Jun 02 '15 at 15:04
  • @D.Thomine: it wasn't exactly rigourous before. Now it is much better: we have that $\bar{E}=\mathbb{R}$ under the assumption that an infinite number of $q_n$s is even and an infinite number of $q_n$s is odd. – Jack D'Aurizio Jun 02 '15 at 15:16
  • First, all I see is that there is at least one accumulation point in $[-\pi/\sqrt{5}, \pi/\sqrt{5}]$. I don't see why there should be infinitely many. And I still don't follow your argument about $p_n+p_m$ or how any density ensues. – D. Thomine Jun 02 '15 at 16:12
  • @D.Thomine: for short, assuming that $q_n$ is even and $q_m$ is odd, $(p_n+p_m)\sin(p_n+p_m)\approx(p_n+p_m)(\sin p_n-\sin p_m)\approx\frac{p_m}{q_n}-\frac{p_n}{q_m}=\frac{p_m q_m-p_n q_n}{q_n q_m}.$ – Jack D'Aurizio Jun 02 '15 at 16:26
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    I still don't see why $E$ should be infinite, and even if it is, I still don't follow your arguement (I am not sure I would get the same equivalences, and even if I did, nothing tell me that the "new" accumulation points you would get would be different from $0$ or infinity. As it stands, your answer is not enough justified, and frankly, I think that your reasoning is false. – D. Thomine Jun 02 '15 at 19:20