# Convergent subsequence of $\sin n$

It is well known (not to me -- ed.) that for every real number $\theta \in [0, 1]$ there exists a sequence $(k_i)$ such that $\lim\sin k_i = \theta,$ but there appear to be no explicit such (infinite) sequences, even for $\theta=0.$ Does anyone know of such?

• This is a nice question, really horribly phrased (which is why some have voted to close) -- I will try to rewrite. – Igor Rivin Aug 8 '13 at 21:53
• Presumably the $k_i$ are to be distinct positive integers. – Gerry Myerson Aug 8 '13 at 23:51
• Why is this not research level? Does any of the closers know the answer for any $\theta \neq 0?$ – Igor Rivin Aug 9 '13 at 0:14
• @IgorRivin: As you said, it is "horribly formulated". The OP asks for an explicit $(k_i)$ for each $\theta$ which is impossible: there are too many $\theta$'s and too few descriptions. For a given $\theta$ the question is about finding $k$ such that approximating $\arcsin(\theta)+2\pi k$ is close to an integer. An algorithm to find such a $k$ (for a good $\theta$) is obvious. – Mark Sapir Aug 9 '13 at 1:21
• I do not understand the question or what? Why $\sin \frac1n$ does not satisfy you? It seems it is infinite and tends to zero. Or do you mean $k_n$ to be integer? – Anixx Sep 4 '14 at 20:36

As to the convergenge to zero: note that the convergents of the continuous fraction for $\pi$ provide a rational approximation $|\pi - p_n/q_n| < 1/q_nq_{n+1}$ so that $\sin p_n\to 0$.

The sequence of numerators $0, 1, 3, 22, 333, 355,\dots$ is OEIS' A002485.

• (stil there could be subsequences with a simpler expression than the $p_n$'s, which have no known closed formula) – Pietro Majer Aug 8 '13 at 22:07

If you allow negative integers in your sequence, Pietro Majer's answer can be extended to the general case $\sin n \rightarrow \theta$. First, remark that, since $\sin$ is continuous it suffices to find a sequence of integer pairs $(u_n,v_n)$ such that $u_n+2\pi v_n \rightarrow \arcsin(\theta).$

Compute a continued fraction for $2\pi$ (instead of $\pi$ in the previous answer) with convergents $p_n/q_n$ and form the invertible matrix: $$M_n=\left(\begin{array}{cc} q_n & p_n \\ q_{n+1} & p_{n+1}\end{array}\right).$$

Let $V_n$ denotes the vector $(\lfloor q_n\arcsin(\theta)\rceil; \lfloor q_{n+1}\arcsin(\theta)\rceil).$ Consider the vector $W_n=M_n^{-1}V_n$, note that $W_n$ is an integer vector because $\det(M_n)=\pm 1$. If you rename the coordinates of $W_n$ by writing $W_n=(u_n;v_n)$ then $u_n/v_n \rightarrow \arcsin(\theta)$ and $\sin(u_n)\rightarrow \theta.$

Heuristically, you can probably expect the $u_n$ to contain an infinite subsequence of positive integers which would answer your initial question in the affirmative.

Example:

Take $\theta=1/10$ and $\alpha=\arcsin(\theta)\approx 0.1001674211615597963455231$. For $n=30$, we have: $$M_{30}=\left(\begin{array}{cc} 842468587426513207 & 5293386250278608690 \\ 68438367733593670 & 430010946591069243 \end{array} \right)$$ and $V_{30}=(6855294804380582;84387905812135908)$. We get $u_{30}=-1544426340857779936$ and $v_{30}=245803086388844114$.

And indeed: $$\sin(-1544426340857779936)\approx 0.099999999999999999558$$

Since $\pi$ is irrational, the sequence $n\bmod 2\pi$ is dense in $[0,2\pi]$. Now use the fact that $\sin$ is continuous.

• How does this give an explicit sequence? – Malik Younsi Aug 9 '13 at 11:56
• Let $\theta\in [0,2\pi]$. Pick a decreasing sequence of intervals $I_k\subset [0,2\pi]$ collapsing on $\theta$. For each $k$ find $n_k$ such that $n_k\bmod 2\pi \in I_k$. Then $\sin n_k\to \sin\theta$ as $k\to\infty$. – Liviu Nicolaescu Aug 9 '13 at 12:05