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I have a sequence defined as follow:

$a_0 = 1, a_n=\cos\left(a_{n-1}\right)$.

I want to count $\lim_{n\rightarrow\infty} a_n$ - it definitely does have limit by looking at the graph, the first few numbers of the limit are 0.7390851, but I have no idea, if that number is related to some other real number ($\pi$ or something like that).

The sequence is from this site, but they don't provide actual result for their own sequence.

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    [This](http://math.stackexchange.com/questions/46934) is related.2011-08-11

4 Answers 4

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This is a standard trick worth knowing.

Supposing the limit does exist, call it $x$. If $x$ is the limit of the sequence, it has the property that $x = \mbox{cos}(x)$. From a graph, we can see that there is exactly one solution. Lastly, Wolfram Alpha tells us that $x = \mbox{cos}(x)$ has the solution $x = 0.739085$ as you said.

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    *Supposing the limit does exist*: good idea. Soooo... one should read a proof that the limit does exist--as opposed to, say, a cycle of length 2. (For a specific example of the behaviour one does not want to meet, one can consider the sequence $x_{n+1}=(x_n-1)^2$ with $x_0=\frac12$.)2011-08-10
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This happens to be a relatively well-known number. It is called the Dottie Number, named after the not-at-all famous Professor of French, Dottie. I should also point out that the number is transcendental (also pointed out in the link).

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    @Eric - I didn't know that! I knew it was the Dottie number, but I'm not sure why. Can you think of some place where one might come across the Dottie number? Did Gardner write on it? Ian Stewart? (I don't think I would have come across it in a very technical setting, although maybe I came across it from the Fixed Point Theorem and an imaginative professor) -2011-08-10
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Here is an elementary proof of convergence of the sequence:

Notice that $0 \leq a_n \leq 1$ for all $n$.

Consider the function $f(x) = x - \cos \cos x$.

This is increasing in $[0,1]$.

Now since $f(0) \lt 0$ and $f(1) \gt 0$, $f(x) = 0$ has a unique root (say $D$) in $(0,1)$, which is also the root of $x = \cos x$.

Now if $a_n \lt D$, then $a_n - a_{n+2} = f(a_n) \lt 0$

if $a_n \gt D$, then $a_n - a_{n+2} = f(a_n) \gt 0$

We also have that $g(x) = \cos x - D$ is decreasing in $[0,1]$ and thus if $a_n \lt D$ then $a_{n+1} \gt D$ and if $a_n \gt D$, then $a_{n+1} \lt D$.

Since $a_0 = 1 \gt D$

The sub-sequence $a_0, a_2, a_4, \dots$ is monotonically decreasing and bounded below and hence is convergent (to $D$).

Similarly, the sub-sequence $a_1, a_3, a_5, \dots$ is monotonically increasing and bounded above, and is convergent (to $D$).

Thus $\lim a_n = D$

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The limit is the (unique) solution of the equation $\cos x = x$. This can't be expressed in any simpler way.