A well-known limit with sine, but with $P(n)$ instead of $n$

Asked Jul 12 '19 at 21:03

Active Jul 13 '19 at 00:08

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If $\lim_{n\to\infty}\sin(nx\pi)$ exists for some $x\in\mathbb{R}$, then $x$ has to be integer (this is easy).

Le $P$ be a nonconstant polynomial with integer coefficients.

Assume $\lim_{n\to\infty}\sin(P(n)x\pi)$ exists for some $x\in\mathbb{R}$.

Then $x$ doesn't have to be integer (counterexample: $P(t)=t(t+1),\ x=\frac{1}{2}$, then $xP(n)\in\mathbb{Z}$ for all $n\in\mathbb{Z}$).

What about $x$ being rational?

Is this thesis true?

This is a very natural question, I'm sure somebody thought about this before, but I couldn't find this result anywhere.

asked Jul 12 '19 at 21:03

larry01

I would suggest defining $L_P={x\in\mathbb{R}: \lim_{n\to\infty}\sin(P(n)x\pi)\text{ exists}}$. Then for $P(t)=t$, $L_P=\mathbb{Z}$ while for $P(t)=t(t+1)$, $L_P={1\over2}\mathbb{Z}$. It sounds like you're asking if $L_P$ is always a subset of the rationals, $\mathbb{Q}$. – Barry Cipra Jul 12 '19 at 22:24
Yes, exactly, this is the problem. My question contains also some background (why to use polynomials there). – larry01 Jul 12 '19 at 23:01
@larry01 Note also that $\mathbb Q \subseteq \cup_P L_P$. – eyeballfrog Jul 13 '19 at 00:13
If $x$ is not rational the polynomial $Q=xP$ has (at least one) irrational coefficients, so $Q(n)$ is uniformly distributed modulo $1$ by Van Der Corput's theorem and that is obviously enough to show that $\sin{\pi Q(n)}$ cannot converge – Conrad Jul 13 '19 at 00:21
Note this is closely related to an about $3$ hours later question by the OP at A problem with the existence of limits of $\sin(P(n)\pi)$ and $\cos(P(n)\pi)$. – John Omielan Jul 13 '19 at 02:28

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