Let $\{a_{n}\}_{n=0}$ be defined recursively by $a_{0}=\pi/4$, and $a_{n+1}=\sin a_{n}$ for $n=0,1,\ldots$. How prove $\frac{1}{\sqrt{2006}}<a_{2006}<\frac{2}{\sqrt{2006}}$?
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What is the value of $a_0$? – Bumblebee Jul 23 '14 at 09:07
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This is non-trivial enough so that I will not vote to close. You still should list some of the available tools. What course does this exercise come from? Have you covered Taylor series of sine?... – Jyrki Lahtonen Jul 23 '14 at 09:23
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the value of $a_0=\frac{1}{4}\pi$ – piteer Jul 23 '14 at 09:25
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yes, covered Taylor series of sine – piteer Jul 23 '14 at 09:39
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For $x \in \left( 0, \sqrt{3} \right)$ we have $x^2 - \frac{x^4}{3} \le \sin^2 x \le x^2 - \frac{x^4}{3} + \frac{2}{45} x^6$, and what about this? – piteer Jul 23 '14 at 09:48
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But you could have searched the site also. See this related answer to get an idea of what this entails. That does use Taylor series (and may be only heuristic), but there are alternatives. Anyway, you need to estimate the difference between $x$ and $\sin x$, when $x$ is relatively small. See how that compares with similar difference for $f(x)=1/\sqrt x$ and $2/\sqrt x$. Or if that doesn't work, then you need to ask someone else for advice :-) – Jyrki Lahtonen Jul 23 '14 at 09:49
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Looks like that (or something similar) will be helpful here. Try it out! – Jyrki Lahtonen Jul 23 '14 at 09:54
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My suggestion is to check whether you can prove the inequalities $$a_n<\frac2{\sqrt n}$$ and $$\frac1{\sqrt n}<a_n$$ by induction using the inequality that you were given in the inductive step. Remember that sine is an increasing function in a relevant interval. As is square root. The inductive step may only work for a range of values of $n$, so I don't know what would be a good base case for the two inductions. – Jyrki Lahtonen Jul 23 '14 at 10:08
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ok will try to solve – piteer Jul 23 '14 at 11:22