# 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

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## marked as duplicate by Brian M. Scott, Potato, rschwieb, Ittay Weiss, Stefan HansenJan 11 '13 at 20:02

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

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

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