2

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 $0&lt\theta&lt1$. Since $|\phi(n)| &lt 1$ 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 &lt \theta &lt 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.

Comic Book Guy
  • 5,748
  • 1
    (i) There's already a term for "oscillate finitely": periodic; (ii) it's clear $\phi(n)$ has period at most twice the denominator for $\theta$ rational, simply because $\sin$ and $\cos$ have period $2\pi$, so a large section of your question seems superfluous and distracting to me; (iii) "Since $|\phi(n)|<1$ either it oscillates finitely or tends to a limit." What makes you think that? – anon Apr 22 '12 at 01:50
  • @anon since we assumed $\theta$ is a constant between 0 and 1 hence $|cos\theta n \pi| = |\phi(n)| < 1$. Also purpose of the part 1 proof attempt was just to get feedback on the irrational case, would u recommend splitting this question into 2 questions ? – Comic Book Guy Apr 22 '12 at 02:28
  • I have made the edits as suggested. – Comic Book Guy Apr 22 '12 at 03:01
  • It's fine as one question. The issue is, you think that $ |\phi(n)|<1$ implies $\phi$ either has a limit or finite period. This is not the case. In fact, even restricting our attention to prime $n$, the range of $\phi$ is dense in $[-1,1]$. (Its being dense in the interval demolishes the idea that $\phi$ could have a finite range or a limit.) – anon Apr 22 '12 at 03:11
  • 3
    $\cos n\pi\theta$ certainly is periodic if $\theta$ is an even integer. The sequence $1,1,1,\dots$ is periodic with period 1. – Gerry Myerson Apr 22 '12 at 03:53
  • @anon, sorry buddy i just saw your comment i ended up deciding to break it up so my thread did n't suffer from tldr. – Comic Book Guy Apr 22 '12 at 05:04
  • @GerryMyerson you may be right, but on that point a very reputable author stated to prove that $\cos n\pi\theta$ oscillates finitely unless $\theta$ is an even integer, which was improved in terminology based on Anon's comment above, so i am unsure where the fault lies, as in with their statement or your definition of periodic. Please do n't take this as an insult or take offence as my intentions is not to do that. – Comic Book Guy Apr 22 '12 at 05:10
  • Unless the "reputable author" gave a definition for the phrase "oscillate finitely," we can't tell whether she meant to include or exclude the constant case, but it's for certain that constant functions qualify as periodic. Anyway, no offense taken. – Gerry Myerson Apr 22 '12 at 05:43
  • @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. – Comic Book Guy Apr 22 '12 at 07:37

1 Answers1

2

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?

Gerry Myerson
  • 179,216
  • Firstly thank you very much for your answer. Yes G.H.Hardy does indeed define the term here it goes. "If $\phi(n)$ oscillates as n tends to $\inf$, then $\phi(n)$ be said to oscillate finitely or infinitely according as it is or is not possible to assign, a number $K$ such that all the values of $\phi(n)$ are numerically less than K, i.e. $|\phi(n)| < K$ for all values of n. – Comic Book Guy Apr 22 '12 at 09:02
  • Also buddy I am not convinced with the statement that $\cos n \pi \theta$ is not periodic for irrational $\theta$ values. Could you provide a proof of this or point me to a source for it i went through the post anon referred to in his comment but i do not think it was about the same question it was talking about range of such a function being densely filled with irrationals. – Comic Book Guy Apr 22 '12 at 09:02
  • 2
    The original version of your question asked to show $\cos n\pi\theta$ oscillates finitely unless $\theta$ is an even integer. Now that statement is true, and easy: it's clearly bounded in absolute value, and it does oscillate (that is, it's not constant) unless $\theta$ is an even integer. Periodicity is a red herring; if I understand what GHH wrote, it has nothing to do with oscillation. And what is this "buddy" business? – Gerry Myerson Apr 22 '12 at 09:39
  • 1
    I guess i was trying to be nice, obviously i am hogging your time primarily for my educational benefit.Thanks a lot for your help. – Comic Book Guy Apr 22 '12 at 10:10
  • 3
    You're very welcome. You're not taking any more of my time than I'm willing to give, so don't worry about that. And it seems I read something into "buddy" that wasn't actually there. It seemed patronizing, but I accept that that was not the intention. – Gerry Myerson Apr 22 '12 at 10:49
  • 2
    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. – Robert Israel Jun 24 '12 at 18:56