6
$\begingroup$

I'm trying to show that the series: $\sin(m) + \sin(\sin(m)) + \sin(\sin(\sin(m)) + \cdots$ converges for all real numbers $m$. To be specific, the series is defined as follows:

$\sum_1^\infty{a_k}$ where $a_{k+1}=\sin(a_k)$ for $k=1, 2, \ldots$ and $a_1=\sin(m)$.

I've concluded that the terms in the series tend to $0$ as $k\rightarrow\infty$. I also tried applying the ratio test:

$\dfrac{|a_k+1|}{|a_k|}=\dfrac{|\sin(\sin\cdots(\sin(m))|}{|\sin(\sin(\cdots(m)|} =\dfrac{|\sin(a_k)|}{|a_k|}\rightarrow1$ as $k\rightarrow\infty$, but since the limit is $1$ the test is inconclusive. I don't know an awful lot about series, but I do know about the the root test, the integral test and about doing comparisons. I can't see how I can apply any of those methods here.

I would be really grateful for any help with this problem!

$\endgroup$
  • $\begingroup$ I would not expect this series to converge. After getting below, say, $1/10$ quickly, the individual terms approach zero really, really slowly. I used the sequence of terms for finding fractional iterates of $\sin x;$ that's why i have an opinion... $\endgroup$ – Will Jagy Sep 23 '14 at 19:58
  • $\begingroup$ So did you use $\sin x < x$ and then iterate, to show the sequence tends to $0$? How did you know the ratio tended to $1$? This is just for my own curiosity $\endgroup$ – snulty Sep 23 '14 at 20:03
  • 5
    $\begingroup$ (This has been asked before on the site.) Except if $m$ is such that some $a_k$ is zero (and then the series converges), one can show that $a_k\sim\alpha\sqrt{3/k}$ when $k\to\infty$ where $\alpha=\pm1$ depends on $a_1$, hence the series diverges. $\endgroup$ – Did Sep 23 '14 at 20:14
  • $\begingroup$ @snulty No, I didn't really make a formal proof of the fact that the terms tend to zero, but I convinced myself drawing the unit circle. No matter what value m is, $\sin{m}$ will always be a number between -1 and 1. Let's say that the value is positive. Inserting this value (which is now an angle measured in radians) into $sin$ will give you something even smaller - I concluded this by drawing a unit circle and exploring the angles... I know that the ratio tends to 1 because of the $\frac{\sin{x}}{x}\rightarrow1$ when $x\rightarrow0$ and $a_k\rightarrow0$ when $k\rightarrow\infty$. $\endgroup$ – Jarvi79 Sep 23 '14 at 20:17
  • $\begingroup$ @Jarvi79 oh so it's like the sequential version of a limit, that's cool. $\endgroup$ – snulty Sep 23 '14 at 20:20
7
$\begingroup$

Heuristics: Say $a_n$ is near zero but not zero, then $\sin(x)=x-\frac16x^3+o(x^3)$ when $x\to0$ hence $$a_{n+1}=a_n-\frac16a_n^3+o(a_n^3).$$ If $a_n\sim c/n^b$, this imposes that $c^2=3$ and $b=\frac12$, hence the idea that $a_n\sim\alpha\sqrt{\alpha/n}$ with $\alpha=\pm1$ depending on $a_1$.

Full proof: Change variables and consider $$b_n=\frac1{a_n^2},$$ thus, one knows that $b_n\to+\infty$, and the same expansion of sine yields $$b_{n+1}^{-1}=\sin(b_n^{-1/2})^2=\left(b_n^{-1/2}-\frac16b_n^{-3/2}+o(b_n^{-3/2})\right)^2=b_n^{-1}-\frac13b_n^{-2}+o(b_n^{-2}),$$ which implies $$b_{n+1}=b_n+\frac13+o(1),$$ hence we are done.

$\endgroup$
  • $\begingroup$ Thanks for this answer. I'm gonna have to work through that thouroughly if I'm going to have a chance to understand it, I'm afraid. How about my idea about using Leibniz convergence test? $\endgroup$ – Jarvi79 Sep 23 '14 at 20:52
  • $\begingroup$ Explain it in detail and we will see if it works... $\endgroup$ – Did Sep 23 '14 at 20:53
  • $\begingroup$ Well, I need to show that the terms alternate in sign. This is easy to see since sin(-v) = -sin v so for example $-\sin(-\sin(m))=sin(sin(m))$. This way you can see that the signs alternate. I also need to show that $a_k\rightarrow0$ as $k\rightarrow\infty$. This is more difficult. I tried to convince myself using a geometric approach where I drew the unit circle and explored the angles. Do you have a better idea? I also need to show that $|a_{k+1}|<|a_k|$. Maybe this follows from $|\sin{x}|<|x|$? My reasoning is that that implies that $|\sin{a_k}|<|a_k|$ which shows that $|a_{k+1}<|a_k|$ $\endgroup$ – Jarvi79 Sep 23 '14 at 21:09
  • $\begingroup$ @Did: Nice! I wonder if you can get an asymptotic expansion for $a_n$. $\endgroup$ – Orest Bucicovschi Sep 23 '14 at 21:34
  • $\begingroup$ "Do you have a better idea?" To show convergence to zero? Well... this is kind of... like... in my answer, no? $\endgroup$ – Did Sep 23 '14 at 22:16
1
$\begingroup$

We will prove that $$\lim n \cdot a_n^2 \to 3$$ where $a_{n+1} = \sin a_n$ and $a_0 = a \ne k \pi$.

It is easy to see that $a_n \ne 0$ for all $n$. ( the only root of $\sin x$ in the interval $\sin (\mathbb{R}) = [-1,1]$ is $x=0$) Moreover, since $|\sin x| \le |x|$ for all $x \ne 0$ we conclude that $(a_n)_{n \ge 1}$ is strictly decreasing and $>0$ if $a_1 \in (0,1]$ and strictly increasing and $<0$ if $a_1 \in [-1,0)$. Therefore the sequence $(a_n)$ is convergent. Let $l$ be the limit. Since $x\mapsto \sin x$ is a continuous function from $a_{n+1} = \sin a_n$ we conclude $\sin l = l$ and therefore $l = 0$.

We will show that $n\cdot a_n^2 \to 3$, or, equivalently $$\lim \frac{1/a_n^2}{n} =1/3$$ We'll prove a stronger(use Stolz–Cesàro) $$\lim_{n\to \infty} (1/a^2_{n+1} - 1/a_n^2) = 1/3$$ Since $a_n\to 0$ we'll prove instead $$\lim_{x->0} (1/\sin^2 x- 1/x^2) = 1/3$$

Indeed we have the expansion $$ 1/\sin ^2 x =1/x^2 + 1/3 + x^2/15 + O(x^4)$$

and the statement is proved.

It is apparent that the speed of the convergence of a sequence of iterates $f^n(a)$ to the (locally) unique fixed point $0$ of $f(x)$ has to do with the expansion of $f(x)$ at $0$. Write: $$f(x) =x( 1 + c_s x^s + c_{s'}x^{s'}+ \ldots )$$ Then we have $$ \lim_{x\to 0} \frac{1}{f(x)^s} - \frac{1}{x^s} = -s \cdot c_s$$ Conclusion: If $0$ is an isolated fixed point and $a_n \ne 0$ is a sequence of iterates converging to $0$ then $$n\cdot a_n^s\to - s\cdot c_s$$

Observation: If the expansion at $0$ is more general $c_0 x + \ldots $ then for $|c_0|<1$ any sequence of iterates has convergence at least exponential while for $|c_0|>1$ it seems that $0$ will not be an attractive fixed point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.