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While reading other topics, e,g, Is $n \sin n$ dense on the real line? or Is $\{ \sin n^m \mid n \in \mathbb{N} \}$ dense in $[-1,1]$ for every natural number $m$?, the following problem appeared in my head:

  • is $\{\sin(x^n)|n\in\mathbb{N}\}$ dense in $[-1,1]$ for all $x>1$?

or a weaker problem:

  • if $x>1$, then $\lim_{n\to\infty} \sin(x^n)$ does not exist?

I proved the second one for $x=2$ and $x=3$ (with use of sine/cosine multiple angle formulas) and have some thoughts for $x\in\mathbb{N}$, but I have completely no idea how to deal e.g. with $x=e$.

Khashayar Baghizadeh
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tong_nor
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    The first problem depends on the distribution of the digits of $\pi$ in base $x$ and it is essentially intractable with the current technology (Van Der Corput's trick and Weyl's inequality). The weaker problem is related to the finiteness of the irrationality measure of $\pi$. – Jack D'Aurizio Apr 05 '18 at 23:53
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    I can share that I wondered if BBP-type formulas can assess something on the distribution of the binary substrings of $\pi$ (hence tackle the density problem of $\sin(2^n)$ or $\sin(16^n)$), but the arithmetics does not seem easy to handle. – Jack D'Aurizio Apr 05 '18 at 23:59

1 Answers1

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The equality $x^n=e^{n\ln x}$ (for $x>1$) and Zeldich’ attraction theorem applied to a function $f(t)=e^t$, any number $0\le x_0\le 2\pi$, and any open neighborhood $A$ of the set $2\pi\Bbb Z+x_0$ imply that a set $X$ of $x>1$ for which the first problem has a negative answer, is meager. On the other hand, Proposition from the same answer implies that $|X|=2^{\aleph_0}$.

Alex Ravsky
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