# How to prove that $\lim(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k}))=\chi_\mathbb{Q}$ [duplicate]

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How is this called? Rationals and irrationals

Please help me prove, that $$\underset{n\rightarrow\infty}{\lim}\left(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k})\right)=\begin{cases} 1 & \iff x\in\mathbb{Q}\\ 0 & \iff x\notin\mathbb{Q} \end{cases}$$

Seems very complicated, but it's on calc I. I've tried use series expansions of cos, but it don't lead to answer. Thanks in advance!

Edit

Please don't use too much advanced techniques.

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is that $\pi$, or really $\Pi$? –  Maesumi Jan 11 '13 at 19:10
just Pi = 3.1415... –  Steve Jan 11 '13 at 19:12
+1 Nice question. Also, the absolute value isn't really necessary, since $\cos$ is an even function. –  Calvin Lin Jan 11 '13 at 19:14
I know that I’ve answered this or a very similar question here before, but it may take a while to find it. –  Brian M. Scott Jan 11 '13 at 19:16
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## marked as duplicate by Brian M. Scott, Potato, rschwieb, Ittay Weiss, Stefan HansenJan 11 '13 at 20:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

Baby Rudin has this as an example.

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I don't understand still, why the lim lim is equal 0 for irrational x –  Steve Jan 11 '13 at 19:19
because $cos(m!\pi x) < 1$ when $x$ is irrational, so if you raise it to large powers, it goes to 0. –  Danikar Jan 11 '13 at 19:21
Ok, now understand! –  Steve Jan 11 '13 at 19:22
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Hint: Show that if $x \in \mathbb{Q}$, then there exists some $N$ such that for $n > N$, $n! \pi x$ is an integer multiple of $2\pi$. Conclude that it tends to 1.

Show that if $x \not \in \mathbb{Q}$, then $\lim_{k\rightarrow \infty} [\cos (n! \pi x)]^{2k} = 0$.

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So then x = p/q. When N > 2q, $2\mid n!\pi x$. Could You advice more about the second part? –  Steve Jan 11 '13 at 19:17
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