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 $a_1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $[-1,1]$.

  • $\begingroup$ 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

$\{a\}_{i=1}^\infty$ 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


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


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

  • $\begingroup$ I have a question: and how about the Imaginary component of this function? $\endgroup$ – Guilherme Thompson Dec 11 '15 at 22:45
  • 1
    $\begingroup$ 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

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 $\arcsin(x)<x$.

  • $\begingroup$ 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

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.