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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.

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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
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See here. –  David Mitra Jan 28 '13 at 14:35
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marked as duplicate by Did, hardmath, P.., David Mitra, Nate Eldredge Jan 28 '13 at 15:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

up vote 5 down vote accepted

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.

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