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

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?

• By what method do you want to prove this? Probabilistically your argument is true since $|sin(x)|$ is in the bounds $[0,1]$ and the values of your function are dense in the reals. – Brevan Ellefsen Feb 19 '17 at 18:09
• @BrevanEllefsen I just want to prove it mathematically. Do you know how to do that? – user19405892 Feb 19 '17 at 19:20
• you can actually turn my comment above into a decent proof if you imagine stretching the values around the unit circle. Imagine your function as a point rotating around the unit circle. You now just need to prove that all points will be hit (with some additional justifications on top of that). Is there any reason for the $\alpha$ you chose? – Brevan Ellefsen Feb 19 '17 at 19:57
• @user19405892: I know, I just find it easier to think as travelling the binary representation of $\beta$. Is there some particular reason for expecting arbitrarily long runs of zeroes or ones? Almost every number has such a property, of course, but why $\beta$ is one of such numbers? – Jack D'Aurizio Feb 25 '17 at 22:59
• @user19405892 JackD'Aurizio seems right; but perhaps you can just post the actual problem as a new, well specified question (without the character limits of comments, and in fact not buried under 20+ other comments)? – Anonymous Feb 28 '17 at 12:42

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.

• The periodic points of the map $x_{n+1} = 4x_n(1-x_n)$ over $[0,1]$ are dense in $[0,1]$ from here: di.univr.it/documenti/OccorrenzaIns/matdid/matdid044153.pdf. – user19405892 Feb 21 '17 at 2:06
• @user19405892 OK, so what does this tell us? – Brevan Ellefsen Feb 22 '17 at 19:19
• @user19405892 sorry, let me clarify - between abnry's post and your link, can we confirm the OP's conjecture, or do we need something tighter? Mappings and finite differences in general are not my forte. – Brevan Ellefsen Feb 22 '17 at 20:22
• @user19405892 The link shows that periodic points (the initial points $x_0$ that make $x_n$ periodic) are dense, not the points $x_n$'s. So, the question is still not confirmed. – Sungjin Kim Feb 22 '17 at 21:57
• Are you sure that $\sin(m \alpha) = T_m(\sin(\alpha))$? That would mean that $\sin(2x) = 2\sin^2(x)-1$, which isn't always true. – user19405892 Feb 27 '17 at 3:07

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.

• Nice observation. It looks like that you proved $\alpha/\pi$ is irrational. This can be proved via simpler way using algebraic integers. – Sungjin Kim Feb 27 '17 at 6:10
• @i707107 yes, I agree. Proving irrationality of $\frac{\alpha}{\pi}$ came as a side effect of studying the structure of the sequence. – rtybase Feb 27 '17 at 21:12

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 $a=\tan^{-1}(\frac{2}{3})$).

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.

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

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).

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 $\frac{\alpha}{\pi}$.