How do I prove that iteration $x_{n+1}=\cos x_n$ converges for any $x_0\in \Bbb R$ ?
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1use the fixed point theorem of Banach – Dr. Sonnhard Graubner Dec 25 '16 at 11:08
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There is a lot of duplicates, about 5 or so. – mvw Dec 25 '16 at 11:21
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3Come on folks, down votes without giving a comment what is wrong on christmas? – mvw Dec 25 '16 at 11:22
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Yep, here is another duplicate http://math.stackexchange.com/questions/1701935/let-a-n-cosa-n-1-l-a-1-a-2-a-n-is-there-an-a-0-such-that-l/ – rtybase Dec 25 '16 at 12:28
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Hint
WLOG, we can suppose $0\leq x_0\leq \pi$. then $\forall n\geq 2 \;\; |x_n|\leq 1$ and $\forall c\in[-1,1]\;| \sin(c)|\leq \sin(1)<1$. thus, by MVT
$$|x_{n+1}-x_n|<\sin(1)|x_n-x_{n-1}|.$$
From here, you prove that $(x_n)$ is Cauchy and converges to the fixed point.
hamam_Abdallah
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$x_0\in\Bbb R$ gives $x_1\in[-1,1]$ and $x_2\in[\cos 1,1]$, which gives a little, but insignificantly so, more information than what you wrote. – Lutz Lehmann Dec 25 '16 at 12:34