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I know that $\{\sin n\}_{n=1}^\infty$ is dense in $[-1, 1]$. That is, $$ \forall x\in[-1, 1] \quad \forall\varepsilon>0 \quad \exists m\in\mathbb{N}: \quad |\sin m-x|<\varepsilon. $$

Here $\mathbb{N}=\{1,2,3,\ldots\}$.

I have a hypothesis.

Hypothesis $$ \forall n\in\mathbb{N} \quad \forall x_k\in[-1, 1] \quad (k=\overline{1,n}) \quad \forall\varepsilon>0 \quad \exists\alpha\in\mathbb{R} \quad \exists\beta\in\mathbb{R}: \quad |\sin(\alpha k+\beta)-x_k|<\varepsilon \quad \forall k=\overline{1,n}. $$

And as I see, $\alpha$ may be 'very large' in absolute value.

Is the hypothesis true or false? I cannot prove it using density of $\sin n$ in $[-1, 1]$. Could you help me please?

peter218
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2 Answers2

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Here is a proof sketch. $\sin$ is continuous taking inputs from the compact set $[0,2\pi]$, so it is uniformly continuous. Thus, if $\theta_k=\arcsin x_k$, then there exists a neighborhood $(\theta_k-\delta,\theta_k+\delta)\ni y$ (indep. of $k$) for which $\lvert\sin y-x_k\rvert<\varepsilon$. Thus, the objective is to find an arithmetic sequence $a_1,\dots,a_n$ where $a_k\in2\pi\mathbb{N}+(\theta_k-\delta,\theta_k+\delta)=S_k$. Put $a_1=\theta_1$. Letting $d_k=\theta_{k+1}-\theta_k$ for $k=1$ up to $n-1$, we want integers $m_k$ for those same $k$s such that there exists real $d$ with $\lvert d-(d_k+2\pi m_k)\rvert<\frac{\delta}{k}$ (to make sure that adding $d$ each time stays in the sets $S_k$). This is possible because of $2\pi$'s irrationality. Thus the desired arithmetic sequence exists.

zjs
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I think what you are really asking is this: Given $$g:\mathbb T^2\to \mathbb T^n\\ (\alpha,\beta)\mapsto (\alpha+\beta, 2\alpha+\beta,\ldots, n\alpha+\beta),$$ is the range of $g$ dense in $\mathbb T^n$? (Here $\mathbb T$ is the additive group of $\mathbb R$ modulo $2\pi$.) It seems the answer is trivially yes if $n=1$ or $n=2$, and (on dimension counting grounds) obviously no if $n>2$. Let $\gamma=(\gamma_1,\ldots,\gamma_n)\in\mathbb T^n\setminus g(\mathbb T^2)$ and take $x_k=\sin\gamma_k$ to construct a counterexample to the question as stated.

kimchi lover
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