I know that $\lim_{n\to\infty}\sin(n)$ does not exist. But can $\sin(n)$ be arbitrary small? More formally: Let $\varepsilon>0$. Is there exist a positive integer $n$ such that $|\sin(n)|<\varepsilon$ ?
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You can prove that $\underset{n\to \infty}{\lim}\sin(n)$ is dense on it's value space meaning it will assume values arbitrarily close to any value that it takes. I think Jack has the details of what is required for that, I don't remember them right now. – mathreadler Aug 02 '17 at 07:33
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@TheSubstitute, he mentioned $n$ is an integer (Otherwise, trivial) – Covvar Aug 02 '17 at 07:37
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Would you guys please look up the meaning of "positive integer $n$"? It's written clearly enough in the question. So $n$ is neither continuous not passing through $0$. – Aug 02 '17 at 07:38
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Because the zeros of $\sin$ are the elements of $\pi\mathbb{Z}$ and $\sin$ is continuous, your question is equivalent to have for every $\epsilon>0$ positive integers $p$ and $q$ with $|q\pi-p|\leq \epsilon$. But this is possible by Dirichlet's approximation theorem.
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Thanks @sigmabe. Can you please elaborate your answer? Why it is equivalent? – boaz Aug 02 '17 at 07:50
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Weyl's equidistribution theorem implies that the fractional part of $M\pi$ is uniformly distributed in $(0,1)$.
Thus if $M\pi = n + f$ with $f \to 0$, then $f \gt \sin f= |\sin(M\pi -f)| = |\sin(n)|$
This shows that $|\sin(n)|$ can get arbitrarily close, infinitely often to any value in $(0, \sin 1)$.
Aryabhata
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