Does $\lim_{n\to\infty}\underset{n}{\underbrace{\cos(\cos(...\cos x))}}$ exist?

Question

Possible Duplicate: Explaining $\cos^\infty$

Does the following limit exist?

$$\lim_{n\to\infty}\underset{n}{\underbrace{\cos(\cos(...\cos x))}}$$

If yes, find the limit.

If no, please explain why the limit doesn't exist.

I think, the limit exists.

So, I tried to use Squeeze theorem but didn't work.

You should guess what the limit should be, then prove that it actually is the limit. — Tunococ, Jan 28 '13 at 14:32
yes, I also try to use a sequence instead the limit. $a_1=\cos x$ and $a_{n+1}=\cos(a_n)$ , but my friend said that was wrong — cwk709394, Jan 28 '13 at 14:35

Trismegistos · Accepted Answer · 2013-01-28T15:16:53.090

6

You may treat is as dynamical system with state transition function $f(x) = cos(x)$. After first two iterations $f^{n > 2}(x)$ will lay in interval $I = [cos(1), 1]$. Line $g(x) = x$ will intercept $cos(x)$ in interval $I$ exactly once so $cos(x)$ has unique fixed point in $I$. Because of unique fixed point and because $|f'(x)| < 1$ sequence $f^n(x)$ will converge to this fixed point.

edited Jan 28 '13 at 15:16

answered Jan 28 '13 at 14:57

Trismegistos

2,410

Consider adding this answer to the earlier duplicate, linked above. Rather than calling it a "dynamical system", though, it would be a little more self- contained to call it a fixed-point iteration. – hardmath Jan 28 '13 at 15:11
1

I have copied this answer to duplicate post. – Trismegistos Jan 28 '13 at 15:18

Does $\lim_{n\to\infty}\underset{n}{\underbrace{\cos(\cos(...\cos x))}}$ exist?

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