Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!


Please don't use too much advanced techniques.

share|cite|improve this question

marked as duplicate by Brian M. Scott, Potato, rschwieb, Ittay Weiss, Stefan Hansen Jan 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.

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
up vote 7 down vote accepted

Baby Rudin has this as an example.

See Baby Rudin page 145

share|cite|improve this answer
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. – zrbecker 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 $\lim_{k\rightarrow \infty} [\cos (n! \pi x)]^{2k} = 0$.

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.