Prove that $\sin(2^n \alpha)$ can be arbitrarily small

Question

Prove that $\sin(2^n \alpha)$ can be arbitrarily small where $\alpha = \tan^{-1}\left(\frac23\right)$ and $n$ is a positive integer.

I was wondering how I should go about solving this. Can we prove this by contradiction?

Answer 1

This isn't really a complete answer, but it is of interest since it converts the problem into a well known one. First, convert problem to $\cos (2^n \alpha)$ being made arbitrarily close to 1.

The Chebychev polynomials $T_m$ have the property that $\cos (m \alpha) = T_m(\cos \alpha)$. With $\alpha = \arctan(5/12)$ we have $\cos(\alpha) = 5/13$.

So we are interested in $$a_n = T_{2^n}(5/12).$$

Now we have this wonderful identity for Chebychev polynomials that $T_m(T_n(x)) = T_{mn}(x)$. Using this, we obtain:

$$a_{n+1}= T_2(T_{2^n}(5/13)) = T_2(a_n).$$

Hence, this turns into studying the orbit of the iterated map $a_{n+1} = T_2(a_n)$ with $a_0 = 5/13$. In particular, $T_2(x) = 2x^2 -1$.

Take $y_n = -(x_n+1)/2$ so $x_n=-2y_n+1$ and $$f(y)=(2y+1)^2-1=4y^2-4y=4y(1-y)$$ where $y_n=f(y_n)$ and $y_0=-9/13$.

This is the famous logistic map with parameter 4. Thus, $x_n$ being close to 1 (remember we switch is equivalent to $\cos$) $y_n$ being close to $1$. I'm pretty sure that for almost all initial points the orbits under the logistic map are dense. However, how this is proved may as just as well use facts about the original question.

Answer 2

This is not a complete answer.

If we note $a_n=\sin{(2^n\alpha)}$ then $$a_n=\sin{(2^n\alpha)}=2\sin{(2^{n-1}\alpha)}\cos{(2^{n-1}\alpha)}=2a_{n-1}\cos{(2^{n-1}\alpha)}$$ Initial value is irrational: $$a_0=\sin{\alpha}=\frac{1}{\sqrt{1+\frac{9}{4}}}=\frac{2}{\sqrt{13}}$$ however: $$|a_1|=|\sin{2\alpha}|=2\frac{2}{\sqrt{13}}\frac{3}{\sqrt{13}}=\frac{12}{13}$$ $$|a_2|=|\sin{4\alpha}|=2\frac{12}{13}\frac{\sqrt{13^2-12^2}}{13}=\frac{120}{169}$$ $$|a_3|=|\sin{8\alpha}|=2\frac{120}{169}\frac{\sqrt{169^2-120^2}}{169}=\frac{28560}{28561}$$ $$|a_4|=|\sin{16\alpha}|=2\frac{28560}{28561}\frac{\sqrt{28561^2-28560^2}}{28561}=\frac{13651680}{815730721}$$ all the subsequent values are rational, because $$|a_{n+1}|=2|a_n|\left|\sqrt{1-a_n^2}\right|=2|a_n|\sqrt{(1-|a_n|)(1+|a_n|)}=\\ 2|a_n|\sqrt{\left(1-2|a_{n-1}||\cos{(2^{n-1}\alpha)}|\right)\left(1+2|a_{n-1}||\cos{(2^{n-1}\alpha)}|\right)}=\\ 2|a_n|\sqrt{\left(a_{n-1}-\cos{(2^{n-1}\alpha)}\right)^2\left(a_{n-1}+\cos{(2^{n-1}\alpha)}\right)^2}=\\ 2|a_n|\left|a_{n-1}-\cos{(2^{n-1}\alpha)}\right|\left|a_{n-1}+\cos{(2^{n-1}\alpha)}\right|=\\ 2|a_n|\left|a_{n-1}^2-\cos^2{(2^{n-1}\alpha)}\right|=2|a_n|\left|2a_{n-1}^2-1\right|$$ and by induction $a_n \in \mathbb{Q}, \forall n>0$. It's impossible not to mention the striking appearance of Pythagorean Triples due to this fact, e.g. for $|a_2|$ - $(5, 12, 13)$ and for $|a_3|$ - $(119, 120, 169)$: $$|a_n|=\frac{2mn}{m^2+n^2}$$ then $$|a_{n+1}|=2|a_n|\left|\sqrt{1-a_n^2}\right|=2\frac{2mn}{(m^2+n^2)^2}|m^2-n^2|=...$$ where by noting $m_1=2mn$, $n_1=|m^2-n^2|$ we have $m_1^2+n_1^2=(m^2+n^2)^2$ $$...=\frac{2m_1n_1}{m_1^2+n_1^2}$$ These fractions don't seem to be reducible and repeating (again, by induction and using $\gcd(m,n)=1$, $m$-even, $n$-odd), for example $$|a_{n+2}|=\frac{2m_2n_2}{m_2^2+n_2^2}=\frac{4m_1n_1|m_1^2-n_1^2|}{(m_1^2+n_1^2)^2}=\frac{8mn|m^2-n^2||(2mn)^2-(m^2-n^2)^2|}{(m^2+n^2)^4}$$ thus, there is no periodicity.

I should also mention this result from the wikipedia article about logistic maps, saying that

$$a_n^2=\sin^2{(2^n\pi \theta)}$$ for rational $\theta$, after a finite number of iterations $a_{n}^2$ maps into a periodic sequence.

by contradiction, $\theta=\frac{\alpha}{\pi}$ is not rational.

This may not answer the original question, but, at least, reveals some structure.

Answer 3

This is not an answer - just a brief argument of why obtaining the proof, or the proof of the converse, appears extremely difficult and likely to elude casual attempts.

Note that $|\sin(2^n a)|$ can be made arbitrarily small if and only if $2^n a$ can be made arbitrarily close to an integer multiple of $\pi$; i.e. if and only if the fractional part of $\frac{2^n a}{\pi}$ can be made arbitrarily close to $0$ or to $1$; i.e. if and only if the binary representation of $\frac{a}{\pi}$ contains arbitrarily long sequences of $0$s, or arbitrarily long sequences of $1$s.

It is currently unknown if the binary representation of $\pi$ or $\pi^{-1}$ contains arbitrarily long sequences of $0$s or of $1$s (however, note that this is a weaker property than normality in base $2$, implied by, but not equivalent to it). Thus proving or disproving that the binary representation of $\frac{a}{\pi}$ contains arbitrarily long sequences of $0$s or of $1$s appears beyond the reach of current mathematics, unless one can prove that $a$ has some “special property” that “cancels out the $\pi-$ness” from $\frac{a}{\pi}$ (which seems unlikely for $a=\tan^{-1}(\frac{2}{3})$).

Answer 4

The above claim means:

$\hspace{1cm}$ For each $\, \delta>0\, $ exists $\, n\in\mathbb{N}\, $ and $\, m\in\mathbb{N}_0\, $ such that $\enspace\displaystyle |m-2^n \frac{\arctan\frac{2}{3}}{\pi} |<\delta$ .

The problem now is to find out if for $\frac{\arctan\frac{2}{3}}{\pi}=:0.a_1 a_2 a_3 ...$ with all $a_j\in\{0;1\}$

the length of all binary squences $a_{n+k+1}...a_{n+k+l}=0$ are limited or if there can be

always found a longer sequence than any previous one (with which we can proof the claim).

There is no regularity in the sequence for $\frac{\arctan\frac{2}{3}}{\pi}$ and therefore it seems to be that

the claim cannot be proofed (or disproofed).

Answer 5

Thoughts on using $\tan$ formulas... not an answer, yet.

Note that if $\sin(\theta)$ is very close to zero, $\tan(\theta)$ is also very close to zero and and $|\tan(\theta)| \approx |\sin(\theta)|$ to increasing accuracy closer to zero; but in particular $|\tan(\theta)| > |\sin(\theta)|$, so if we can show that $\tan(2^n\alpha)$ gets arbitrarily close to zero, that guarantees the same for $\sin(2^n\alpha)$.

So we can use the double angle formula for $\tan$ :

$$\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} $$

to determine $\tan(2^n\alpha)$ for successive values of $n$, without converting to either $\sin$ values or angles.

Converting this into recurrence formulas for a rational number with integer numerator $N(n)$ and denominator $D(n)$, we have:

$N(0)=2, D(0)=3$
$N(n+1)=2N(n)\cdot D(n)$
$D(n+1)=D(n)^2-N(n)^2$

Unfortunately these get too large too quickly to be of particular value.

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I'm still thinking about how to carry this forward.....

Answer 6

This is not an answer, but it greatly simplifies the problem.

$sin(2^n\alpha)$ can get arbitrarily small if and only if $2^n\alpha$ can get arbitrarily close to $0$ or $\pi$ (mod $\pi$).

It can be proven that $2^n\alpha \equiv \pi \cdot (0.b_nb_{n+1}...)_{2} $ (mod $\pi$) , where $0.b_0b_1 ...$ is the binary representation of the fractional part of $\frac{\alpha}{\pi}$. The proof is done by induction:

$(0.b_0b_1 ...)_2$ = {$\frac{\alpha}{\pi}$} $\iff \alpha \equiv \pi \cdot (0.b_0b_1 ...)_2$ (mod $\pi$) as the starting case.

$2^n\alpha \equiv \pi \cdot (0.b_nb_{n+1}...)_{2} $ (mod $\pi$) $\iff (0.b_nb_{n+1} ...)_2$ = {$\frac{2^n\alpha}{\pi}$} $\iff (0.b_{n+1}b_{n+2} ...)_2$

= {$2${$\frac{2^n\alpha}{\pi}$}} = {$\frac{2^{n+1}\alpha}{\pi}$} $\iff 2^{n+1}\alpha \equiv \pi \cdot (0.b_{n+1}b_{n+2} ...)_2$ (mod $\pi$) as the induction.

So the conjecture is true if and only if you can find arbitrarily long sequences of $1$'s or $0$'s in the binary representation of $\frac{\alpha}{\pi}$.