11
$\begingroup$

I'm trying to find this limit:

$$\lim_{n \to \infty} \underbrace{\sin \sin \ldots \sin }_{\text{$n$ times}}x$$

Thank you

$\endgroup$
  • 2
    $\begingroup$ Please use LaTeX in the future, just as Jonas did for you on this question. $\endgroup$ – Raphael Aug 20 '11 at 16:18
  • 23
    $\begingroup$ "Forgive me, father, for I have sinned. Many many times..." $\endgroup$ – Mark Aug 20 '11 at 16:18
  • 10
    $\begingroup$ @user14829 $\Large \textrm{Thi}\int_{s}^{i} \LaTeX.$ $\endgroup$ – Mateen Ulhaq Aug 20 '11 at 18:04
  • 7
    $\begingroup$ $\top \mathbb{N} \chi $ $\endgroup$ – Jozef Aug 20 '11 at 20:26
  • 3
    $\begingroup$ This question is not a duplicate. In the other question, $n$ and $x$ are the same. $\endgroup$ – Qiaochu Yuan Aug 23 '11 at 1:19
36
$\begingroup$

First, you have to prove there is a limit. the first application of $\sin x$ will get us into $[-1,1]$ If $x \in [0,1]$ we have $0 \le \sin x \lt x$, so the sequence (after the first term, maybe) is monotonically decreasing and bounded below by $0$. Then, if there is a limit, you must have $\sin x=x$. Where is that true? The case below zero is for you.

$\endgroup$
7
$\begingroup$

Here is a different approach that doesn't lead as directly to a rigorous proof as Ross Millikan's, but is more concrete and shows the rate of convergence. Let the $n$th member of the sequence be given by the function $x(n)$. Using the Taylor series of the sine function, and approximating the discrete function $x$ by a continuous one, we have $dx/dn\approx -(1/6)x^3$. Separation of variables and integration gives $x\approx \pm\sqrt{3/n}$ for large $n$.

$\endgroup$

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.