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Does decimal expansion of $\pi$ contain blocks of zeroes of any integer length? I.e. $0$, $00$, $000$, $\ldots$

I discovered this question, when trying to prove $$\limsup_{n \to \infty} |\sin(n)| = 1.$$ Or is there any idea how to find subsequence $k_n$ such that $\sin(k_n) \rightarrow 1$?

Martin Sleziak
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joseph
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  • I think (correct me if I'm wrong) that would imply that $\pi$ is normal, but that has not been proven. – Eemil Wallin Mar 31 '16 at 12:47
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    That is, simply, not known – user247327 Mar 31 '16 at 12:50
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    @EemilWallin Why would that imply that $\pi$ was normal? The number $0.1010010001000001000001\dots$ is certainly not normal. – David C. Ullrich Mar 31 '16 at 12:55
  • Given any finite string of numbers, it is most likely to be somewhere in the expansion of $\pi$ (the probability of it happening, given that the expansion of $\pi$ is psuedo-random, is $1$), but we do not know whether it's always the case. – Arthur Mar 31 '16 at 12:59
  • I don't get how knowing anything about the decimal expansion of $\pi$ could help to prove $\limsup_n |\sin n| = 1$. – Crostul Mar 31 '16 at 13:10
  • If $\pi$ is multiplied by power of $10$, the block of zeroes can be shiftet right behind the decimal point. Then can be found $n$ and $k$ (power of 10) such that $|n-k\frac{\pi}{2}|$ is small. – joseph Mar 31 '16 at 13:15
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    @joseph Consider the approximations of $\pi /2$ given by the continued fraction of $\pi /2$. – Crostul Mar 31 '16 at 13:19
  • @Crostul Thanks I'll try.. – joseph Mar 31 '16 at 13:22
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    In general, truncated continued fractions give you (always) better approximations than truncated decimal expansion. You don't even need to compute them, since you know that they are good enough. See here for details. – Crostul Mar 31 '16 at 13:26
  • Thank you, that solved my problem:) – joseph Mar 31 '16 at 13:31
  • Some related posts: http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1, http://math.stackexchange.com/questions/63526/showing-sup-sin-n-mid-n-in-mathbb-n-1 and http://math.stackexchange.com/questions/438636/proof-of-limsup-sin-nx-1-n-rightarrow-infty-forall-x-in-mathbbr (You can find also other questions linked there.) – Martin Sleziak Mar 31 '16 at 15:41

1 Answers1

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See this article, a special case is that there is two increasing sequences of odd positive integers $(p_n),(q_n)$ such that $$ \left|\frac{\pi}{2} - \frac{p_n}{q_n} \right| \leq \frac{1}{q_n^2},\quad n>1.$$

Note that $\sin |x| = |\sin x |$ for $x\in [0,\pi]$, then $$\sin \left|\frac{q_n\pi }{2} - p_n\right|= \left|\cos p_n\right| <\frac{1}{q_n} \to 0.$$ therefore $|\sin p_n|\to 1$.

Tulip
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