4

$x_1 = \sin x_0 > 0$

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

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.

Martin Sleziak
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yiyi
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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 05:54
  • 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 06:17
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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...) – Gottfried Helms Apr 23 '12 at 06:30
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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 $1/x_{n+1}^2-1/x_n^2\to1/3$. – Did Apr 23 '12 at 06:49
  • @MaoYiyi , if you are confused, try to take $x_1$ as the start value. – Ziyuan Apr 23 '12 at 07:37
  • Found it. – Ragib Zaman Apr 23 '12 at 10:51
  • @RagibZaman your good, I tried checking before I posted. Now I feel bad. Thanks – yiyi Apr 23 '12 at 23:28

1 Answers1

8

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 $n\to\infty$

Community
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Ziyuan
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