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$x_1 = \sin x_0 > 0$

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


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

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

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marked as duplicate by Martin Sleziak, mau, naslundx, Najib Idrissi, Mark Bennet Apr 24 '14 at 9:28

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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 '12 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 '12 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...) –  Gottfried Helms Apr 23 '12 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 '12 at 6:49
@MaoYiyi , if you are confused, try to take $x_1$ as the start value. –  ziyuang Apr 23 '12 at 7:37

1 Answer 1

up vote 7 down vote accepted

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 $n\to\infty$

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Which assumes (as I mentioned in a comment) that $x_n\to0$. –  Did Apr 23 '12 at 9:31
That $x_n$ goes to zero is trivial, and can be showed by a geometric argument (among others). –  nbubis Apr 23 '12 at 10:27
@nbubis how would you show the geometric argument? –  yiyi Apr 25 '12 at 5:47
Started to describe the image, when in turns out wikipedia already had it. –  nbubis Apr 25 '12 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 $x=\sin x\,(0<x<1)$. –  ziyuang Apr 25 '12 at 9:10

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