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Possible Duplicate:
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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marked as duplicate by Brian M. Scott, Potato, rschwieb, Ittay Weiss, Stefan Hansen Jan 11 '13 at 20:02

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  • $\begingroup$ is that $\pi$, or really $\Pi$? $\endgroup$ – Maesumi Jan 11 '13 at 19:10
  • $\begingroup$ just Pi = 3.1415... $\endgroup$ – Steve Jan 11 '13 at 19:12
  • $\begingroup$ +1 Nice question. Also, the absolute value isn't really necessary, since $\cos$ is an even function. $\endgroup$ – Calvin Lin Jan 11 '13 at 19:14
  • $\begingroup$ I know that I’ve answered this or a very similar question here before, but it may take a while to find it. $\endgroup$ – Brian M. Scott Jan 11 '13 at 19:16
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Baby Rudin has this as an example.

See Baby Rudin page 145

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  • $\begingroup$ I don't understand still, why the lim lim is equal 0 for irrational x $\endgroup$ – Steve Jan 11 '13 at 19:19
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    $\begingroup$ because $cos(m!\pi x) < 1$ when $x$ is irrational, so if you raise it to large powers, it goes to 0. $\endgroup$ – zrbecker Jan 11 '13 at 19:21
  • $\begingroup$ Ok, now understand! $\endgroup$ – 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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  • $\begingroup$ So then x = p/q. When N > 2q, $2\mid n!\pi x$. Could You advice more about the second part? $\endgroup$ – Steve Jan 11 '13 at 19:17

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