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If $x\in \Bbb R$ ; then is the sequence $\{a_n\}$ where $a_1=x$; $a_{n+1}=\cos (a_n)$ convergent?

Obviously $|a_n|\le1$ and hence $(a_n)$ is bounded.

Also $f(x)=\cos x$ is decreasing for $x>0$.

But here $x\in \Bbb R$ .How to proceed here?Please help.

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    Fixed point theorems? Google? – 2017-02-27
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    See http://mathworld.wolfram.com/DottieNumber.html – 2017-02-27
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    Hint : Use the Banach-fixpoint-theorem – 2017-02-27
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    This has been asked before, probably à propos the similar recursion based on the sine since I remember having explained the differences between these two situations... – 2017-02-27

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Note that since $|a_n| \le 1$ we have $a_n \in [0,1]$ since $\cos$ is positive on $[-1,1]$.

Since $\cos$ is decreasing on $[0,1]$ we see that $a_n \ge \cos 1 >0$ for all $n \ge 2$.

In particular, $a_n \in [\cos 1, 1]$ for $n \ge 2$ and we have $|\cos'x| \le \sin 1 < 1$ for $x \in [\cos 1, 1]$ and so $\cos$ is a contraction map on $[\cos 1, 1]$. Hence $a_n$ converges to the unique fixed point.

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    I don't get one thing.Take $x=(\pi+\frac{\pi}{3})$;then $\cos x=\frac{-1}{2}$;Why do you say that $a_n\in [0,1]$?? – 2017-02-28
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    I meant for $n \ge 2$. As you noted, $|a_n | \le 1$ for $n \ge 2$ (the first value is arbitrary), but there after $|a_n| \le1$. If $|x| \le 1$, then $\cos x \ge \cos 1$Also, i – 2017-02-28
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The sequence $\{a_n\} $ is an infinite subset of $[-1,1] $. Therefore it has a convergent subsequence since $[-1,1]$ is compact. Now check that "is the subsequences converge to a unique point or not"!