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I have the following function f(x) = $\lim_{m \to \infty} \lim_{n \to \infty} (\cos(\pi m!x))^{2n}$

I have to prove that f(x) exists and calculate its value. I was wondering if using $e^{i\pi x}$ to direct calculate it was enough or I need the definition with $\epsilon$.

After that, I need, as well, to find {x$\in\mathbb{R}$| f is continuous at x}.

I decided to consider the série $e^u = \sum_{k=0}^{n}\frac{u^k}{k!}$. Here, u = $i\pi x m!$ I choose r > 0 and x$\in$[-r,r], so we can $i\pi x m! \leq i \pi r m!$ and using the Abel's theorem (don't know how it's called in english) then the serie is normally convergent.

Thus, f is continuous when x $\in \mathbb{R}$

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  • $\begingroup$ I am not sure why you need to go to $e^{i\pi x}$. Just consider rational and irrational points separately. Once you have the limit, continuity should be more clear. $\endgroup$
    – Dunham
    Nov 7, 2020 at 16:49
  • $\begingroup$ consider $m=2$ fixed. $0 \leq \cos(2\pi x)^2 \leq 1$. Only the points where the value is 1 will converge to 1 as $n\rightarrow \infty$. At all other points, it converges to 0. For all values of $m$, there is a similar form. You then need to look at convergence of these functions at rational and irrational points. $\endgroup$
    – Dunham
    Nov 7, 2020 at 19:50
  • $\begingroup$ @Dunham Re($(e^{i x \pi})^{m!}$) = cos($\pi x m!$). We know that $-1 \leq e^{I x \pi} \leq 1$ so f(x) = 1. Am I wrong? $\endgroup$
    – Moh18
    Nov 7, 2020 at 19:52
  • $\begingroup$ $e^{i \pi x}$ is not real valued, and $f(x)$ is not identically 1. $\endgroup$
    – Dunham
    Nov 7, 2020 at 19:52
  • $\begingroup$ I don't think that expanding the exponential is a good idea. $\endgroup$
    – user65203
    Nov 9, 2020 at 20:27

1 Answer 1

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Define $f_m(x) = \lim_{n\to \infty} (\cos(\pi m! x))^{2n}$.

If $x$ is rational, then $x=p/q$ for some integers $p,q$. If $2q$ divides $m!$, then $(\cos(\pi m! x))^2 = 1$, which will be the case for $m$ sufficiently large. Hence $f_m(x)=1$ eventually for all $m$ after some point.

If $x$ is irrational, then $0<(\cos(\pi m! x))^2<1$, so $f_m(x)=0$ for all $m$.

Therefore $f(x) = 0$ for irrational $x$ and $f(x)=1$ for rational $x$. This function is not continuous anywhere.

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  • $\begingroup$ Thanks! It helps verify what I have done. Actually, when I understood that $\cos(m!\pi x)$ was like $\cos(kx)$ here $k = \pi m!$. It's very easy to do after that thinking. $\endgroup$
    – Moh18
    Nov 9, 2020 at 20:59

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