# What does recursive cosine sequence converge to?

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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It converges to the unique solution to $\cos x=x$. AFAIK, it's probably transcendental and not expressible in terms of $e, \pi$, etc. Also, this sounds like it'd be a duplicate for sure, but I can't find any. –  anon Aug 10 '11 at 20:44
The limit does not seem to have a closed form expression in terms of the standard functions, and even not so standard ones. –  André Nicolas Aug 10 '11 at 20:45
The sequence will converge to solution of $x=\cos(x)$ which can be found as point of intersection of two curves $y=x$ and $y=\cos(x)$. See Wolfram-Alpha for plot. –  Sasha Aug 10 '11 at 20:48
@Karel: you wrote it definitely does have limit by looking at the graph but I have no idea what you mean by that. If what you mean is that the equation x=cos(x) has one and only one solution on [0,1], that is true--but hardly the end of the story! See my comment on NKS's (accepted...) answer. –  Did Aug 10 '11 at 22:38
This is related. –  Guess who it is. Aug 11 '11 at 2:53

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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What negative solution ? –  Sasha Aug 10 '11 at 20:45
Oops. I had the picture wrong in my head. There is no negative solution. (For context, I fixed an error in the answer, as I had originally claimed there was also a negative solution). –  NKS Aug 10 '11 at 20:47
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 $x_0=\frac12$.) –  Did Aug 10 '11 at 22:34

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

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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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Always be careful with Wolfram Links. Dottie was a Professor of French, that is someone who teaches French, not a French Math Professor as the article would make you believe... –  Eric Naslund Aug 10 '11 at 21:08
@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) - –  mixedmath Aug 10 '11 at 21:13

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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