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I am trying to show that $\cos n\pi\theta$ is periodic unless $\theta$ is an even integer.

I wish to provide a proof based on an example of the $\sin n\pi\theta$ case of the first result, I would appreciate if you could let me know if it is correct or what improvements could be made, I am somewhat doubt full about the last few steps of the irrational case.

If $\theta$ is rational then let $\theta = \frac {p} {q}$ and let $n = aq + b$ then $\phi(n) = \cos n \frac{p}{q} \pi = \cos(ap\pi + \frac{bp}{q} \pi) $ Since $a\in Z+$ so $\cos(ap\pi + \frac{bp}{q} \pi) = (-1)^{ap}\cos(\frac{bp}{q} \pi)$

Now if p is even as n increases from $0$ to $q-1$ then $\phi(n)$ takes the values $1, \cos\frac{p}{q} \pi, \cos\frac{2p}{q} \pi, \dots, \cos\frac{(q-1)p}{q} \pi$

These values repeat as n goes from $q$ to $2q -1$ hence $\phi(n)$ is cyclic.

Assuming if p is odd as n increases and is odd or even we get $np$ as odd or even determined by n and we observe cyclic values.

Let $\theta$ be irrational and without loss of generality be a constant in 0<\theta<1. Since |\phi(n)| < 1 either it periodically or tends to a limit.

If $\cos n \theta \pi \rightarrow l$ so $\cos (n+1) \theta \pi -\cos n\theta \pi = 2 \sin((n+\frac{1}{2})\theta\pi)\sin\frac{\theta\pi}{2} \rightarrow 0$ Hence $\sin((n+\frac{1}{2})\theta\pi) \rightarrow 0$ then $(n+\frac{1}{2})\theta = k_{n} + \frac{1}{2} + \epsilon_{n}$ where $k_{n} \in Z$ and $\epsilon_{n}\rightarrow 0$. Hence $\theta = k_{n} - k_{n-1} + \epsilon_{n} - \epsilon_{n-1} =l_{n}+\eta_{n}$ where $l_{n} \in Z$ and $\eta_{n}\rightarrow 0$. This is impossible since $\theta$ is a constant and lies between 0 < \theta < 1 . So we have reached a contradiction.

If you could shed some light on the arguments provided in the proof from the entry of $k_{n}$ forwards that would be much appreciated.

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    @GerryMyerson Glad to hear that buddy, Actually the question is from G.H.Hardy's textbook, would you be kind enough to shed some light on this proof. I was hoping that someone would provide a yes or no answer with some reason if the above proof establishes the result or not, but i am some what surprised that no one has as i thought this was trivial. I guess in the essence what i was trying to show is that there can be no limit for $\cos n \pi \theta$ unless $\theta$ is an even integer.2012-04-22

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I think anon has already pointed out the flaw, but as OP seems to disagree, here goes:

The proof can't be right, because what you are trying to prove is nonsense. $\cos n\pi\theta$ is periodic if and only if $\theta$ is rational. It is possible for a sequence to be bounded, not periodic, and not converge to a limit, and that's what happens to these sequences when $\theta$ is irrational.

As an example of a sequence that is bounded by 1, not periodic, and not convergent (though not arising from $\cos n\pi\theta$, consider the sequence $.1,.2,.1,.1,.2,.1,.1,.1,.2,.1,.1,.1,.1,.2,\dots$

Maybe the difficulty is in that term, "oscillate finitely." Is that the phrase G H Hardy uses? If so, does he define it somewhere?

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    If $\cos(n \pi \theta)$ is periodic, say with period $p$, that implies $\cos(p \pi \theta) = \cos(0) = 1$, so $p \pi \theta = 2 n \pi$ for some integer $n$, and then $\theta = 2 n/p$ is rational.2012-06-24