# Limit of $\sin (a^{n}\theta\pi)$ as $n \to \infty$ where $a$ is an integer greater than $2$

In Hardy's Pure Mathematics, Hardy discusses the limit $$\lim_{n\to\infty}\sin (2^{n}\theta\pi)$$ and says that if this limit exists it must be zero and then $\theta$ must be a rational number whose denominator is a power of $2$. Then he asks the reader to consider the limit $$\lim_{n\to\infty}\sin (a^{n}\theta\pi)$$ where $a$ is an integer greater than $2$.

It is also mentioned that when $a > 2$ then the limit can be non-zero and as an example Hardy states that if $a = 9$ and $\theta = 1/2$ then the above limit is $1$. In this case I reasoned (based on Hardy's technique for the case $a = 2$) that $\lim_{n\to\infty}\sin((9^{n}\pi)/2)$ will be $1$ only when we can write $$\frac{9^{n}\pi}{2} = 2b_{n}\pi + \frac{\pi}{2} + c_{n}$$ for all sufficiently large values of $n$, where $\{b_{n}\}$ is a sequence which takes only integer values and $\{c_{n}\}$ is a sequence which tends to zero as $n \to \infty$. Thus we get

\displaystyle \begin{aligned}9^{n}\pi &= 4b_{n}\pi + \pi + 2c_{n}\\ \Rightarrow (9^{n} - 1)\pi &= 4b_{n}\pi + 2c_{n}\end{aligned}

Since $9^{n} - 1$ is divisible by $9 - 1 = 8$ and hence also by $4$, it follows that we can take $b_{n} = (9^{n} - 1)/4$ and $c_{n} = 0$. The same logic could be applied when $a$ is any integer of the form $a = 4m + 1$ and $\theta = 1/2$.

But, and this is my question, how does one handle the case for general integer $a > 2$ and any real $\theta$ rational or irrational?

EDIT: I should also provide more details so that readers get the full context. For the specific case when $a = 2$ Hardy reasons that if the limit exists and, say is equal to $l$, then we must have $|l| \leq 1$ so that we have a constant $\alpha = \sin^{-1} l$ lying in interval $[-\pi/2, \pi/2]$. Since the solution of $\sin x = \sin \alpha$ is given by $x = m\pi + (-1)^{m}\alpha$ for all integers $m$, we can see that the existence of limit $l$ entails that $$2^{n}\theta\pi = b_{n}\pi + (-1)^{b_{n}}\alpha + c_{n}$$ where $b_{n}$ takes integer values and $c_{n} \to 0$ as $n\to\infty$. We can now see that $$2^{n}\theta = b_{n} + (-1)^{b_{n}}\beta + d_{n}$$ where $\beta = \alpha/\pi$ so that $\beta \in [-1/2, 1/2]$ and $d_{n} = c_{n}/\pi \to 0$ as $n \to\infty$. Hardy somehow assumes that $b_{n}$ will always be even and then proceeds as follows:

$\displaystyle 2^{n}\theta = b_{n} + \beta + d_{n}$ so that multiplication by $2$ gives

$\displaystyle 2^{n + 1}\theta = 2b_{n} + 2\beta + 2d_{n}$ and also we have

$\displaystyle 2^{n + 1}\theta = b_{n + 1} + \beta + d_{n + 1}$ and therefore we get

$\displaystyle (b_{n + 1} - 2b_{n}) - \beta + (d_{n + 1} - 2d_{n}) = 0$

After this Hardy uses simple arguments to show that $\beta = 0$ and $d_{n}$ should be identically zero from a certain value of $n = n_{0}$ so that $2^{n_{0}}\theta = b_{n_{0}}$ and thus $\theta$ is rational with denominator a power of $2$.

In the above I don't understand why he assumes $b_{n}$ as even. Also I think the argument can be continued without assuming parity of $b_{n}$ but I am not sure. I wonder how this can be carried forward for higher values of $a$.

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Nice question! However, my intuition says it is really difficult and is related to the irrationality measure of $\pi$. I'd love to be totally wrong, though. – Ian Mateus Jul 20 '13 at 14:04
Hello Ian, I suppose that probably the case when $\theta$ is irrational might be very difficult so I am first trying to figure how if we can handle the rational $\theta$ and also start with small values of $a$ for example $a = 3$. But I have not been able to solve even the case for $a = 3$. – Paramanand Singh Jul 20 '13 at 14:24
The only idea I had was computing $a^n\theta\pi\pmod{2\pi}.$ This line of reasoning seems to work with the $a=2$ case ($\equiv 0$), but I don't know if it works in general. – Ian Mateus Jul 20 '13 at 14:58
I didn't understood your initial implicit question and thought that this reference completely answers it. So I move my answer to comments. See lemma 2.15 in this article – Norbert Jul 20 '13 at 15:23
To Norbert: My question is find the limit of $\sin(a^{n}\theta\pi)$ whenever it exists and also to find values of $\theta$ for which it exists. The lemma in your referred article handles the case when this limit is zero. By the way thanks for that excellent article on non-differentiable functions. I need to solve the case when limit is non-zero. – Paramanand Singh Jul 20 '13 at 15:34

Here is a very partial answer, but at least it contains some ideas

If $\theta=m/a^p$ for integers $m,p$ then $\sin a^n\theta\pi=\sin a^{n-p}m\pi=0$ as soon as $n\geq p$.

The article cited by Norbert shows that this is the only case where the limit is zero.

Note that $\sin a^n\pi/2=\sin (a+4k)^n\pi/2$ for any integer $k$. In particular, $\sin 9^n\pi/2=\sin\pi/2=1$ for any $n$.

The same idea occurs with other denominators. For example $$\sin 7^n\pi/3=\sin(6+1)^n\pi/3=\sin\pi/3=\frac{\sqrt3}2\ \ \mbox{ for all } n.$$

More generally, if we consider $\sin a^n\,m\pi/q$, we can write $a=2kq+r$ with $k$ integer and $0\leq r<q$, and in that case $$\sin \frac{a^n\,m\pi}q=\sin \frac{r^n\,\pi}q.$$ This last expression will usually oscillate when $r\ne0,1$.

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Thanks Martin, this seems to be a good approach towards solution. At least it handles some rational values of $\theta$ and some corresponding values of $a$. My best guess is if there is a limit then $\theta$ must be rational. But a strict proof of this does not seem easy. Considering this was a problem set in elementary book written in 1908, I think Hardy used to give research problems in his book. – Paramanand Singh Jul 20 '13 at 16:37