If $\theta$ is linearly indep. to $\pi,$ then is $\liminf_{n\to\infty}f(n)\left\vert\cos(n\theta)\right\vert=+\infty$ where $f(n)=n^k?$ or $f(n)=2^n?$ If, over $\ \mathbb{Q},\ \theta\ $ is linearly independent to $\ \pi,\ $ then what is  $\ \displaystyle\liminf_{n\to\infty} \left( f(n) \left\vert \cos(n\theta) \right\vert \right)\ $ where:

*

*$f(n) = n^k,\ $ for each $k\geq 1?$


*$f(n) = 2^n?$
Maybe these questions have been asked before, but I couldn't find them...
The motivation behind the question is that maybe the answer would give some sort of measure of how quickly the smallest values of the sequence $\vert\cos(n\theta)\vert$ closely approaches $0$.
 A: What you really seem to want to understand is the growth of
$$a_n = \text{min}_{1 \le k \le n} |\cos k \theta|$$
so let's just talk about that directly. It will be more convenient to write $\theta = \frac{\pi \alpha}{2}$; then the condition we want for this sequence to not be periodic is that $\alpha$ is irrational.
The behavior of $a_n$ is controlled by the behavior of rational approximations to $\alpha$ in the following sense. For $\cos \frac{\pi \alpha n}{2}$ to be close to $0$, $\frac{\pi \alpha n}{2}$ needs to be close to $\frac{\pi}{2} + \pi k$ for some integer $k$, and so
$$\alpha \approx \frac{2k+1}{n}$$
needs to be close to a rational number with odd numerator (this awkward constraint would not be necessary if we were considering $\sin$ instead of $\cos$) and denominator $n$. So for $a_n$ to be close to $0$, $\alpha$ needs to be close to a rational number with odd numerator and denominator $\le n$. Let $\frac{p_n}{q_n}$ be the closest rational approximation to $\alpha$ among all rational numbers with odd numerator and denominator $\le n$. Then
$$\begin{align*} a_n &= \left| \cos \frac{\pi \alpha q_n}{2} \right| \\
 &= \left| \cos \frac{\pi}{2} \left( q_n - \alpha - p_n \right) + \frac{\pi}{2} p_n \right| \\
 &\sim \frac{\pi}{2} \left| q_n \alpha - p_n \right| \end{align*}$$
where we use $\cos \left( \frac{\pi}{2} + x \right) = \sin x$ and then the asymptotic $\sin x \sim x$ for $x$ small. We have $\sin x \le x$ for $x \ge 0$ so the $\sim$ above is also a $\le$.
So, what can we say about the sequence of closest rational approximations to an irrational number? The basic fact here is Dirichlet's approximation theorem, that among the rational numbers with denominator $\le n$ a rational approximation $\frac{p}{q}$ exists satisfying $|q \alpha - p| < \frac{1}{n}$; the additional constraint that $p$ is odd weakens this bound but probably by a factor of $2$ or so. Without the oddness constraint this would immediately imply $a_n \le \frac{\pi}{2n}$ and with it we should still get $a_n = O \left( \frac{1}{n} \right)$ but, again, with the constant worse by a factor of $2$ or so. (Edit: Actually the pigeonhole argument that proves Dirichlet doesn't obviously generalize to give odd numerators so I wonder if the bound could actually become substantially worse.)
Almost all irrational numbers $\alpha$ have the property that the growth rate of $a_n$ is $\Theta \left( \frac{1}{n} \right)$; this is equivalent to saying that almost all irrational numbers have irrationality measure $2$, although for some reason this is not stated in the Mathworld article. In other words, with probability $1$ it is not possible to rationally approximate an irrational number substantially more accurately than what Dirichlet's theorem tells you is possible. However, one can cook up specific irrational numbers which can be approximatedly more accurately, e.g. Liouville numbers, and correspondingly for those numbers $a_n$ will decay more quickly. We can get $a_n = \Theta \left( \frac{1}{n^k} \right)$ for any $k \ge 1$ from irrational numbers with irrationality measure $k + 1$, and $a_n$ decaying more quickly than $\frac{1}{n^k}$ for any $k$ from irrational numbers with irrationality measure $\infty$; I believe it's possible for $a_n$ to decay arbitrarily quickly using a generalization of the Liouville number construction but I haven't thought about the details.
