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$x_1 = \sin x_0 > 0$

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

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.

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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.2012-04-23
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    I know that, but that does split the regions of values (poorly worded) (0,$\frac{\pi}{2}$) and $(\frac{\pi}{2}, \pi)$2012-04-23
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    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...)2012-04-23
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    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 $1/x_{n+1}^2-1/x_n^2\to1/3$.2012-04-23
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    @MaoYiyi , if you are confused, try to take $x_1$ as the start value.2012-04-23
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    [Found it.](http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n)2012-04-23
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    @RagibZaman your good, I tried checking before I posted. Now I feel bad. Thanks2012-04-23

1 Answers 1

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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 $x_n\to0$.2012-04-23
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    That $x_n$ goes to zero is trivial, and can be showed by a geometric argument (among others).2012-04-23
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    @nbubis how would you show the geometric argument?2012-04-25
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    Started to describe the image, when in turns out [wikipedia already had it](http://en.wikipedia.org/wiki/File:Sine_fixed_point.svg).2012-04-25
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    @MaoYiyi As an alternative, it is easy to show that $0, thus $\{x_n\}$ has a limit, which is the solution of $x=\sin x\,(0.2012-04-25
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    @ziyuang thanks. that really makes it easier.2012-04-27