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A recurrent sequence $a_{n+1}=\arcsin (a_n)$, $a_1=a$ $(a\in [-1,1])$ is given. If, for example, $a=1$, then $a_3$ does not exist. But does the sequence exist and converge if $a$ is a small enough positive number?

My Attempt (copied from answer)

Given sequence is bounded because range of function $y=\arcsin(x)$ is $[-\pi/2,\pi/2]$. This sequence is strictly increasing when $a_1\gt0$ and strictly decreasing when $a_1\lt0$ (for example, because for any $t$ from $(0,\pi/2)$, $\sin(t)$ is less than $t$ and so if $a_1\gt0$, $a_{n+1}\gt \sin(a_{n+1})=a_n$ for any $n$). For $a_1=0$ this sequence is constant. So the sequence is monotonic. So this sequence, being bounded and monotonic, converges for any $a_1$.

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  • $\begingroup$ Next time you ask a question be sure to show how you attempted to solve the problem. $\endgroup$ – Arbuja Dec 11 '15 at 23:35
  • $\begingroup$ @Arbuja: good advice. However, the OP did proceed to give an answer, which does show where they were having difficulty. $\endgroup$ – robjohn Dec 12 '15 at 2:25
  • $\begingroup$ The OP has written an answer. Although this context might better be added to the question, perhaps we could reopen the question. $\endgroup$ – robjohn Dec 12 '15 at 3:38
  • $\begingroup$ I had edited the question. $\endgroup$ – Boris I. Model Dec 12 '15 at 4:48
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    $\begingroup$ @BorisModel : You might consider also adding the work you did in one of the (now deleted) answer to the question. $\endgroup$ – user99914 Dec 12 '15 at 11:35
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Assuming $a_1\gt0$, $$ \begin{align} a_{n+1} &=\arcsin(a_n)\\ &=\int_0^{a_n}\frac{\mathrm{d}t}{\sqrt{1-t^2}}\\ &\ge\int_0^{a_n}\left(1+\frac{t^2}2\right)\mathrm{d}t\\ &=a_n+\frac{a_n^3}6\\ &=a_n\left(1+\frac{a_n^2}6\right) \end{align} $$ Thus, $$ \begin{align} a_n &\ge a_1\prod_{k=1}^{n-1}\left(1+\frac{a_k^2}6\right)\\ &\ge a_1\left(1+\frac{a_1^2}6\right)^{n-1}\\ \end{align} $$ and therefore, for some $n$ no greater than $$ \frac{\log\left(\frac1{a_1}\right)}{\log\left(1+\frac{a_1^2}6\right)}+2 $$ we have $a_n\gt1$.

Thus, for any $a_1\gt0$, there is an $n$ so that $a_n\gt1$. Since $\sin(-x)=-\sin(x)$, for any $a_1\lt0$, there is an $n$ so that $a_n\lt-1$.

Thus, only for $a_1=0$ does the sequence continue without reaching a value outside of $[-1,1]$.

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  • $\begingroup$ Thank you very very much. Your proof is just beautiful. Finely I undeleted my wrong proof because you needed it. I think at the very end of your proof there is some insignificant slip of hand. Probably you ment "for some n greater than ..." instead of "for some n no greater than ...". $\endgroup$ – Boris I. Model Dec 12 '15 at 12:42
  • $\begingroup$ If $n=\frac{\log\left(\frac1{a_1}\right)}{\log\left(1+\frac{a_1^2}6\right)}+1$, then $a_1\left(1+\frac{a_1^2}6\right)^{n-1}=1$, so for some $n$ no greater than $\frac{\log\left(\frac1{a_1}\right)}{\log\left(1+\frac{a_1^2}6\right)}+1$, $a_n$ will be greater than $1$, and $a_{n+1}$ won't exist. $\endgroup$ – robjohn Dec 12 '15 at 17:18
  • $\begingroup$ It's correct if ...+2 instead of ...+1 in above expression because expression with ...+1 can be not integer. You had already corrected slightly your proof. But actually these details do not matter. $\endgroup$ – Boris I. Model Dec 12 '15 at 18:10
  • $\begingroup$ Once more your proof is just beautiful. $\endgroup$ – Boris I. Model Dec 12 '15 at 18:17
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$\{a\}_{i=1}^\infty$ is converging for only $a_1=0$. If $a_1=0$, $a_1$ is a fixed point of $\arcsin$, so we are done.

If $a_1>0$, then the sequence is increasing as $\arcsin(x)\geq x$ for positive $x$. $\arcsin(x)>x+\frac{x^3}{6}$ (Maclaurin Series), as $a_n$ is increasing, we have

$$a_{1+\left\lceil\frac{6}{a_1^3}\right\rceil}>a_1+\left\lceil\frac{6}{a_1^3}\right\rceil\frac{a_1^3}{6}>1$$

If $a_1<0$, then the sequence is decreasing as $\arcsin(x)\leq x$ for negative $x$. $\arcsin(x)<x+\frac{x^3}{6}$ (Maclaurin Series), as $a_n$ is decreasing, we have

$$a_{1+\left\lceil\frac{6}{|a_1|^3}\right\rceil}>a_1+\left\lceil\frac{6}{|a_1|^3}\right\rceil\frac{a_1^3}{6}>-1$$

So $a_n$ is eventually not a real number, hence the sequence is not converging in this case.

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  • $\begingroup$ I have a question: and how about the Imaginary component of this function? $\endgroup$ – Guilherme Thompson Dec 11 '15 at 22:45
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    $\begingroup$ Fixed error in proof. I'm assuming that $\{a\}^n_{i=1}$ is a real-valued sequence. $\endgroup$ – Element118 Dec 11 '15 at 22:57
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The sequence is convergent only for $a_1=0$ because $\arcsin(x)$ for $x>0$ is convex and $\arcsin(x)>x$, and for $x<0$ concave and $\arcsin(x)<x$.

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  • $\begingroup$ What is correct in this is answer is that the given sequence is converging when a(1)=0. But in any case thank you very much for your attention and effort. $\endgroup$ – Boris I. Model Dec 11 '15 at 22:35
  • $\begingroup$ @Element118 edited $\endgroup$ – Kamil Jarosz Dec 11 '15 at 22:47
  • $\begingroup$ @BorisModel: both answers are correct. The sequence is not convergent. In fact there are only finitely many terms that are defined unless $a_1=0$. Your answer is the only one that is not correct. $\endgroup$ – robjohn Dec 12 '15 at 2:22
  • $\begingroup$ Maybe. But still it's to be proved that for some n a(n)>1 if a(1) is small enough. $\endgroup$ – Boris I. Model Dec 12 '15 at 3:28
  • $\begingroup$ @BorisModel: If the question were not closed, perhaps someone could expand on that. Perhaps some context might help to get the question reopened. $\endgroup$ – robjohn Dec 12 '15 at 3:36

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