# Problem 43 **/***

Evaluate \displaystyle \lim_{m \to \infty} \left(\lim_{n \to \infty} \cos^{2 n} (m! \pi x)\right).

[originally posted by Shamika on TSR]

will post a solution when I’ve posted 44-49 and then we’ll be up to date.

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Note that -1 \le \cos x \le 1.

If \cos (m! \pi x) = \pm 1, then \cos^{2 n} (m! \pi x) = 1 \to 1 as n \to \infty. Note that this is the case iff m! x is an integer. (since \cos (m! \pi x) = \pm 1 iff m! \pi x is a multiple of \pi)

Otherwise, if \cos (m! \pi x) = c \ne \pm 1, then \cos^{2 n} (m! \pi x) = c^{2 n} \to 0 as n \to \infty since |c| < 1.

So:

\displaystyle \lim_{n \to \infty} \cos^{2 n} (m! \pi x) = \begin{cases}1 & m!x \in \mathbb Z \\ 0 & m!x \not\in \mathbb Z\end{cases}

Note that if x is rational, then x = p/q for integers p,q with q non-zero. Then q! x is clearly an integer and m! x is an integer for m \ge q since for m \ge q, m! is a multiple of q!. Similarly, if m! x is an integer for some m, then x = \dfrac n {m!} for some integer n, rational. So x is rational iff m! x is an integer for sufficiently large m.

That is x is rational iff for all sufficiently large m, \displaystyle \lim_{n \to \infty} \cos^{2 n} (m! \pi x) = 1. Otherwise, \displaystyle \lim_{n \to \infty} \cos^{2 n} (m! \pi x) = 0 for all m. So:

\displaystyle \lim_{m \to \infty} \left(\lim_{n \to \infty} \cos^{2 n} (m! \pi x)\right) = \begin{cases}1 & x \in \mathbb Q \\ 0 & x \not \in \mathbb Q\end{cases}