True or false: For every $N\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $\sin{(k^n)}<0$ for $n=1,2,3,...,N$.

Question

I made up this question. I suspect the statement is true. I have considered complex numbers ( $\sin{(k^n)}=\text{Im}(e^{k^n i})$ ), proof by contradiction, and induction, to no avail.

As an example, for $N\le10$ we have $k=539$.

EDIT

As @durianice commented, $\sin{(539^7)}>0$. (I was using desmos, which is not precise enough to calculate the sine of huge numbers.)

$\sin(539^7)>0$. You need to use numerical methods that allow arbitrary precision. — Paprika7191, Apr 23 '23 at 11:35
@bluebril No. Let $m=2103495018943925748$. Then $2m\pi <539^7<(2m+1)\pi$. Again, use a numerical library that allows arbitrary precision (such as sage or mpmath). To double-check, see wolframalpha. — Paprika7191, Apr 23 '23 at 11:44
For $N\le 17$, the smallest solution is $k=32928$. For $N=18$, we have $k>10^5$. — Paprika7191, Apr 23 '23 at 12:10
Also, don't actually calculate the value of $\sin(k^n)$. Note that $\sin(k^n)<0$ if and only if $\lfloor \frac{k^n}{\pi} \rfloor$ is not divisible by $2$. — Paprika7191, Apr 23 '23 at 12:23
It seems likely to be true just by chance... if $k^n$ is effectively random modulo $2\pi$, then you should expect to find such a $k$ after checking something like $2^N$ arbitrary integers, for any fixed $N$. — mjqxxxx, Apr 27 '23 at 22:23

score 4 · Accepted Answer · answered Apr 27 '23 at 23:10

It is true. In fact, the natural density of such $k$ is $2^{-N}$. There's probably a nicer way to show this, but everything we need is given here.

A natural number $k$ satisfies $\sin(k^n)<0$ if and only if $\operatorname{frac}(\frac{k^n}{2\pi})>\frac{1}{2}$. We show that the sequence of vectors $\{(\frac{k^1}{2\pi}, \dots, \frac{k^N}{2\pi})\}_{k\in \mathbb{N}}$ is uniformly distributed mod 1.

By the multidimensional version of Weyl's Criterion it suffices to show that for all $\ell\in\mathbb{Z}^N$ the sequence $p(k) := \{\sum_{n=1}^N \ell_n \frac{k^n}{2\pi}\}_{k\in \mathbb{N}}$ is uniformly distributed mod 1. Since any integer multiple of $\frac{1}{2\pi}$ is irrational, $p(k)$ is a polynomial in $k$ with an irrational coefficient, and therefore by result of Weyl (Thm 7.2 in the link) $p(k)$ is uniformly distributed mod 1.

True or false: For every $N\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $\sin{(k^n)}<0$ for $n=1,2,3,...,N$.

1 Answers1