Need help solving Recursive series defined by $x_1 = \sin x_0$ and $x_{n+1} = \sin x_n$ [duplicate]

$x_1 = \sin x_0 > 0$

$x_{n+1} = \sin x_n$

Prove

$\lim_{x \to \infty }$ $\sqrt{\frac{n}{3}}$ $x_n = 1$

having problem of trying to figure out what value for the $x_0$ starts at.

• It doesn't sound like it's supposed to matter what $x_0$ starts at, so long as $\sin x_0$ is positive.
– anon
Apr 23, 2012 at 5:54
• I know that, but that does split the regions of values (poorly worded) (0,$\frac{\pi}{2}$) and $(\frac{\pi}{2}, \pi)$
– yiyi
Apr 23, 2012 at 6:17
• in the limit-expression, is it really meant that x approaches infinity or isn't that n approaches infinity? Also, I vaguely remember we have a question(and answer) with the infinitely iterated sin-function and the square-root of 3 already, (but don't have the reference at hand...) Apr 23, 2012 at 6:30
• Like @anon said: initial values in $(0,\frac{\pi}2)$ or in $(\frac{\pi}2,\pi)$ make no difference. // About the asymptotics: the simplest route might be to show that $x_n\to0$ and that $1/x_{n+1}^2-1/x_n^2\to1/3$.
– Did
Apr 23, 2012 at 6:49
• @MaoYiyi , if you are confused, try to take $x_1$ as the start value. Apr 23, 2012 at 7:37

Use Stolz theorem:

$$nx_n^2=\frac{n}{\frac{1}{x_n^2}}\to\frac{n+1-n}{\frac{1}{x_{n+1}^2}-\frac{1}{x_{n}^2}}=\frac{x_{n+1}^2x_{n}^2}{x_{n}^2-x_{n+1}^2}=\frac{x_n^2\sin x_n^2}{x_n^2-\sin^2x_n}=\frac{\sin^2x_n}{1-\frac{\sin^2x_n}{x_n^2}}$$

By $$\sin x\sim x,\displaystyle\frac{\sin x}{x}\sim 1-\frac{x^2}{3!}$$ and $$x_n\to 0$$, you can obtain $$nx_n^2\to3$$

I assume $$x\to\infty$$ is a typo, which should be $$n\to\infty$$

• Which assumes (as I mentioned in a comment) that $x_n\to0$.
– Did
Apr 23, 2012 at 9:31
• That $x_n$ goes to zero is trivial, and can be showed by a geometric argument (among others). Apr 23, 2012 at 10:27
• @nbubis how would you show the geometric argument?
– yiyi
Apr 25, 2012 at 5:47
• Started to describe the image, when in turns out wikipedia already had it. Apr 25, 2012 at 6:52
• @MaoYiyi As an alternative, it is easy to show that $0<x_{n+1}<x_n<x_1\le 1$, thus $\{x_n\}$ has a limit, which is the solution of $x=\sin x\,(0<x<1)$. Apr 25, 2012 at 9:10