Let $x_{0}=0$. Define $x_{n+1}=cos (x_{n})$ for $n \geq 0$. then prove that sequence $x_{n}$ is convergent and $x_{2n}< \lim_{n\to\infty} x_{n} < x_{2n+1}$ for every $n \in \mathbb{N}$. Sequence is 0,1,cos(1), cos(cos(1)) then i am not getting the how to show that $x_{2n}< \lim_{n\to\infty} x_{n} < x_{2n+1}$. Please help.Thanks for precise time.
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First of all have to prove convergence... – neelkanth Jun 06 '15 at 18:28
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See also this question. – dafinguzman Sep 30 '15 at 13:25
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Define $f : [0, 1] \to [0, 1]$ by $f(x) = \cos x$. Prove that $f$ is a contraction, hence the sequence $x_n$ is convergent by the contraction mapping theorem.
For the second part it suffices to show that
$$x_0 < x_2 < \ldots < x_{2n} < \ldots < x_{2n+1} < \ldots < x_3 < x_1.$$
$f$ is decreasing, so $f^2 = f \circ f$ is increasing. Therefore from $x_0 < x_2$ follows $f^2(x_0) < f^2(x_2)$ i.e. $x_2 < x_4$, then $x_4 < x_6$, and so on. It also follows that $f(x_2) < f(x_0)$ i.e. $x_3 < x_1$, so $x_5 < x_3$ and so further. By the same argument, $x_0 < x_1$ so $x_2 < x_3$ and generally $x_{2n} < x_{2n+1}$. That completes the proof.
Adayah
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In contraction map theorem we have k such that k is non negative and k<1 – neelkanth Jun 06 '15 at 18:37
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1On $[0, 1]$ the constant can be made smaller. There is $K < 1$ s.t. for every $x, y \in [0, 1]$ we have $|\cos x - \cos y| \leqslant K \cdot |x-y|$. – Adayah Jun 06 '15 at 18:46
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First you show the cosine function maps $[0,1]\,$ to $\,[0,1]$ and is decreasing on this interval. Then observe $x_{2n}=\cos(\cos x_{2n-2})$ and similarly $x_{2n+1}=\cos(\cos x_{2n-1})$.
Initially $x_0=0<x_1=1$. A simple induction shows $x_{2n}<x_{2n+1}$ (note $\cos\circ\cos$ is the composition of two decreasing functions, hence is increasing on $[0,1]$).
Moreover, the subsequence of even rank terms is increasing while the subsequence of odd rank terms is decreasing.
Indeed, $x_0=0<x_2=\cos(\cos1)$ and the same easy induction shows $x_{2n}<x_{2n+2}$ for all $n$.
For the odd rank terms, we first note $0<\cos 1$ hence $\cos 0=x_1 > \cos(\cos1) = x_3$. Same induction as above.
Finally, we prove $x_{2n+1}-x_{2n}<k(x_{2n-1}-x_{2n-2})$ for some $0<k<1$. This results from the Mean Value theorem, which guarantees there exists $\theta\in [x_{2n-1},x_{2n-2}]\subset [0,1]$ such that $$(x_{2n-1}-x_{2n-2})\cdot \sin(\cos\theta)\cdot\sin\theta$$ Observe that $0<\sin(\cos\theta)\cdot\sin\theta<\sin(cos1)\sin1=k<1$ so that $$0<x_{2n+1}-x_{2n}<k(x_{2n-1}-x_{2n-2})$$ This proves $\lim\limits_{n\to\infty}(x_{2n+1}-x_{2n})= 0$
Conclusion: The subsequences of even and odd terms are adjacent sequences, hence they converge to the same limit, which is between them.
Bernard
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