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Let $x \in \mathbb{R} \setminus \mathbb{Q}$. What is the value of $$\lim_{m \to \infty} \lim_{n \to \infty} \left[ \cos(n!\pi x) \right]^{2m}, \qquad (m,n \in \mathbb{N})$$


The answer given is $0$. I don't understand why it is so.

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  • $\begingroup$ Related post: Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals. Found using Approach0. Other questions linked there might be of interest, too. $\endgroup$ Dec 26 '16 at 14:23
  • $\begingroup$ @koolman, something is not right about your question: are you sure that it's $\lim_{m \to \infty} \lim_{n \to \infty}$ and not $\lim_{n \to \infty} \lim_{m \to \infty}$? Equivalently, are you sure that it's $[ \cos(n!\pi x) ]^{2m}$ and not $[ \cos(m!\pi x) ]^{2n}$? I bet you've inadvertently interchanged $m$ and $n$, because your question is not quite right. $\endgroup$
    – Alex M.
    Dec 26 '16 at 16:46
  • $\begingroup$ @AlexM. This is the original question imgur.com/a/3XjRM $\endgroup$
    – Koolman
    Dec 26 '16 at 16:53
  • $\begingroup$ @AlexM. And how does that matter $\endgroup$
    – Koolman
    Dec 26 '16 at 16:53
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[This answer has been rewritten thanks to the helpful criticism in Alex's comment.]

The statement is wrong. One has to take into consideration first the existence of the limit $$ \lim_{n\to\infty}\cos(n!\pi x)^{2m}, $$ which is a highly non-trivial problem.

When $x=\dfrac{1}{\pi}$, $ \cos(n!\pi x)=\cos(n!). $ But it seems that the existence of limit (which could be $1$) $$ \lim_{n\to\infty}\cos(n!)\qquad\text{(which could be $1$)}, $$ is an open problem (edited due to Fimpellizieri's comment) thanks to the answer of this question: Is there a limit of cos (n!)?.

[Added: Interestingly, there is related question in MO: On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational.]


If you were to ask the double limit related to the Dirichlet function on the other hand: $$ \lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}. $$ you could read the accepted answer to this question:

Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals.

Note carefully (again, thanks to Alex's comment) that the order of taking the double limit is different from the one in your question: $$ \lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}. $$

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  • $\begingroup$ How can we say this ?? $\endgroup$
    – Koolman
    Dec 25 '16 at 16:43
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    $\begingroup$ No , I don't know that $\endgroup$
    – Koolman
    Dec 25 '16 at 16:50
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    $\begingroup$ ... Then you might either ask a follow-up question in a new post or look it up in your calculus textbook. $\endgroup$
    – user9464
    Dec 25 '16 at 16:52
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    $\begingroup$ @AlexM.: I understand that you could have given your own answer without writing any comment to my previously stupid wrong hint. However you provided me a hint to learn something I didn't fully understand since I naively thought OP is asking about the Dirichlet function. Thank you very much and I apologize for my unreasonable complaint. $\endgroup$
    – user9464
    Dec 26 '16 at 0:53
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    $\begingroup$ The linked answer does not state that $\cos(n!)\to1$, but rather that there is an (unlikely) scenario, which we do not yet know if it is or not true, in which that limit holds. $\endgroup$ Dec 26 '16 at 0:59
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Let $e=\sum_{j=0}^{\infty}j!^{-1},$ which is irrational, because $n!e\not \in \mathbb Z$ for any $n\in \mathbb N.$

For $n\in \mathbb N$ we have $$n!e=A_n+\sum_{j=n+1}^{\infty}n!/j!=A_n+B_n,$$ where $A_n\in \mathbb N$ and $B_n\in (0,1)$.By induction on $n,$ if $n$ is even then $A_n$ is odd, and if $n$ is odd then $A_n$ is even.

So for infinitely many $n$ we have $\cos (\pi n!e)=\cos \pi B_n$ and for infinitely many $n$ we have $\cos (\pi n!e)=-\cos \pi B_n.$

Therefore, since $\lim_{n\to \infty }B_n=0,$ the sequence $(\cos n!\pi e)_{n\in \mathbb N}$ has a $\lim \sup$ of $+1$ and a $\lim \inf$ of $-1.$ Therefore $\lim_{n\to \infty}\cos(n!\pi e)$ does not exist.

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  • $\begingroup$ @koolman, if I had been you, then this is the answer that I would have accepted. $\endgroup$
    – Alex M.
    Dec 26 '16 at 16:39

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