Evaluate this function
$$ f(x) = \lim_{n \to \infty} \left(\tfrac12 + \tfrac12\cos(\pi \, n! \, x)\right)^n $$
for every $x \in \mathbb{R}$.
It's not (present day) homework. I'm too old for that.
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${\bf Theorem:}\text{ For any $a \in [0,1]$ there exists $x$ such that $f(x) = a$.}$
Proof: As other users have shown, if $x \in \mathbb{Q}$, then $f(x) = 1$. Find $x$ such that the limit is $0$ is pretty straightforward (I expect this is the case for almost all $x$ but I'm not sure how that could be proven). Now suppose $0< a < 1$. Let $(x_n) \in \mathbb{Z}$ be a sequence such that $\lim\limits_{n\rightarrow\infty}\frac{x_n}{\sqrt{n}} = \frac{\sqrt{2}}{\pi}\sqrt{\log\frac{1}{\sqrt{a}}}$. Then, we claim $$ f\left(2\sum_{k=0}^\infty \frac{x_k}{k!}\right) = a $$ Observe that by the half angle formula, $$ f(x) = \lim_{n\rightarrow\infty} \left(\cos(n!\pi x /2)\right)^{2n} $$ We also note that as $t$ approaches $0$, we have asymptotically $\cos(n\pi + t)^2 \sim (1 - \frac12t^2)^2$. Now: $$ f\left(2\sum_{k=0}^\infty \frac{x_k}{k!}\right) = \lim_{n\rightarrow\infty} \cos\left(n! \pi \sum_{k=0}^\infty \frac{x_k}{k!}\right)^{2n} = \lim_{n\rightarrow\infty} \cos\left(\pi \sum_{k=n+1}^\infty \frac{n! x_k}{k!}\right)^{2n} $$ since the first $n$ terms of the series are integer multiples of $\pi$. Then, we have: $$ f\left(2\sum_{k=0}^\infty \frac{x_k}{k!}\right) =\lim_{n\rightarrow\infty} \cos\left(\pi \sum_{k=n+1}^\infty \frac{n! x_k}{k!}\right)^{2n} = \lim_{n\rightarrow\infty} \left(1 - \left(\pi \sum_{k=n+1}^\infty \frac{n! x_k}{k!}\right)^2\right)^{2n} = \lim_{n\rightarrow\infty} \left(1 - \frac{1}{2}\left(\frac{\pi x_{n+1}}{n+1}\right)^2\right)^{2n} = \lim_{n\rightarrow\infty} \left(1 - \frac12\left(\frac{\pi \sqrt{n+1} \frac{\sqrt{2}}{\pi}\sqrt{\log\frac{1}{\sqrt{a}}}}{n+1}\right)^2\right)^{2n} = \lim_{n\rightarrow\infty} \left(1 - \frac{\log{\frac1{\sqrt{a}}}}{n+1}\right)^{2n} = \left(\lim_{n\rightarrow\infty} \left(1 + \frac{\log{\sqrt{a}}}{n+1}\right)^{n}\right)^2 = \left(\exp(\log\sqrt{a})\right)^2 = a $$ (We can ignore all other terms of the series because they go to $0$ significantly faster than the first.)
This shows that $x = 2\sum_{k=0}^\infty \frac{x_k}{k!}$ indeed satisfies $f(x) =a$.
A mildly interesting fact is that since this proof only depends on the asymptotic behavior of the sequence $x_n$, the set of points $x$ such that $f(x) = a$ will actually be dense in $\mathbb{R}$.
Dark Malthorp
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Why is there a sequence $(x_n)_n$ in $\mathbb{Z}$ such that $x_n/\sqrt{n}\to \sqrt{2}/\pi(\log(a)^{-1/2})^{1/2}$? – Vinícius Novelli Sep 27 '18 at 23:13
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Just take $x_n $ to be the floor of $\sqrt{2n}\frac{1}\pi (\log{a^{-1/2}})^{1/2}$ – Dark Malthorp Sep 27 '18 at 23:28
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$f(x) = \lim_{n \to \infty} \left(\tfrac12 + \tfrac12\cos(\pi \, n! \, x)\right)^n $
If $x$ is rational, $n!x$ will be an even integer for large enough $n$ so $\cos(\pi \, n! \, x) =1$ and the result is $1$.
If $x$ is not rational, it depends on how close $n!x$ is to an even integer. If it is often not close, then the limit for those such $n$ is zero.
My guess is that the limit does not exist for irrational $x$, but I don't know how to prove it.
It may be related to work on equidistributed sequences.
marty cohen
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if $x$ is irrational, is $n! , x$ *ever* equal to an integer? is $\pi n! , x$ ever equal to an integer multiple of $\pi$? – robert bristow-johnson Sep 27 '18 at 18:19
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If it gets close enough to an integer often enough, then the limit won't converge. – Dark Malthorp Sep 27 '18 at 18:25
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and remember, as $n$ gets sufficiently large, there will always be a leftover factor of $2$ in $n!$ that is not canceled by anything in a rational $x$. – robert bristow-johnson Sep 27 '18 at 18:26
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@DarkMalthorp, i might downvote your comment if i could. is the content of $( \cdot )^n$ ever bigger in magnitude than 1? – robert bristow-johnson Sep 27 '18 at 18:29
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No, obviously. But it can get arbitrarily close to $1$, so it is plausible that $\limsup (\cdot)^n = 1$ even if $x$ is irrational. – Dark Malthorp Sep 27 '18 at 18:32
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1It reminds me a lot of https://math.stackexchange.com/questions/8690/is-there-a-limit-of-cos-n – Dark Malthorp Sep 27 '18 at 18:43
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it does look related, but that question doesn't raise the $\cos(n!)$ to the $n$th power. – robert bristow-johnson Sep 27 '18 at 18:46
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1Well, using the ideas mentioned there by David E Spayer in his answer, you can show that if $x = e/2$ the limit is 1. – Dark Malthorp Sep 27 '18 at 19:00
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Correction: $x = 2e$, I got my half- and double- angle formulas mixed up. – Dark Malthorp Sep 27 '18 at 19:15
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