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I've been experimenting with recursive sequences lately and I've come up with this problem:

Let $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$

Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$



I know of $3$ ways of examining whether a set is dense:

$i)$The definition, that is, whether its closure is the set on which it is dense, in our case this means if: $\bar L=[-1,1]$.

ii)$(\forall x \in [-1,1])(\forall \epsilon>0)(\exists b \in L):|x-b|<\epsilon$

$iii)$ $(\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow x$

So far I haven't been able to use these to answer the question. I tried plugging in different values of $a_0$ and see where that leads but I have not found any corresponding promising "pattern" for $a_n$. Any ideas on how to approach this?

MathematicianByMistake
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    Neat question. A small comment: you can make your sequence $a_{n} = \cos(a_{n-1})$, since $\cos(x)$ is a continuous function. – Alex Wertheim Mar 17 '16 at 17:13
  • @AlexWertheim Edited as you suggested! Thanks again! – MathematicianByMistake Mar 17 '16 at 17:31
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    The main problem is: $$a_0 \in \mathbb{R} \implies a_1 \in [-1,1] \implies a_2 \in [0.54,1].$$ Thus, this will never be dense in the whole interval. –  Mar 17 '16 at 19:56
  • @menag Thanks for your observation! So, should I look for a subset of $[-1,1]$ in which $L$ is dense? – MathematicianByMistake Mar 17 '16 at 20:15
  • @MathematicianByMistake if you look closely at my answer, you will have sufficient evidence to conduct a rigorous proof that L is not dense and can't be. – rtybase Mar 17 '16 at 20:17
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    In the meantime, you can play with https://www.wolframalpha.com/input/?i=f(n)%3Dcos(f(n-1)),+f(1)%3D5, with various values for $f(1)$ – rtybase Mar 17 '16 at 20:18

2 Answers2

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Ok, if you study recursive sequences, then you probably heard about "Lamere Ladder".

According to Lamere Ladder for $\cos(x)$ (which is also a contraction because of MVT), this function has a fixed (stationary) point. So, regardless of $a_0$, $a_1$ will end in between $[-1, 1]$ and from there on, the sequence $\{a_n\}$ will tend to the fixed point of $\cos(x)$. Which makes $L$ a converging sequence, so $L$ can't be dense, because the only point satisfying ii) (in your question) is its limit.

On another note, Kronecker's approximation theorem is quite an useful tool too. For example $\{n+ m \cdot 2 \cdot \pi \space | \space m,n \in \mathbb{Z} \}$ is dense on $\mathbb{R}$ and $cos(x)$ is a continuous function, making $\{cos(n)\}_{n \in \mathbb{Z}}$ dense on $[-1,1]$.

rtybase
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  • Anyway, could anyone explain why was this post downvoted? It's disappointing! – rtybase Mar 17 '16 at 18:54
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    Small comment: One neat way to add hyperlinks is this: "[some text](some link)". It looks better than adding the link. For instance, if you write "$[$Kronecker's approximation theorem$]$(http://mathworld.wolfram.com/KroneckersApproximationTheorem.html)" it will look like this: Kronecker's approximation theorem. (pd: I didn't downvote you). – Nate River Mar 17 '16 at 20:06
  • @NateRiver thank you for the advice, I fixed my answer! ;) – rtybase Mar 17 '16 at 20:14
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    Thanks for taking the time to answer! I am not formally-as part of a course-studying recursive functions, just checking some properties myself, so Lamere Ladder is not known to me. I am obviously aware of Banach's fixed point Theorem, but I have some difficulty understanding how it applies here. If you could answer in a more rigorous manner I would be obliged but in any case thank you for the usefull contribution! – MathematicianByMistake Mar 17 '16 at 20:19
  • Just read your comment-I guess I will try it myself after reading more about Lamere Ladder.. – MathematicianByMistake Mar 17 '16 at 20:20
  • @rtybase You are welcome. – Nate River Mar 17 '16 at 20:20
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    Banach's fixed point Theorem guarantees that sequence ${a_n}$ doesn't become periodic from some $n$ onwards. You will have to be careful around $x=\frac{\pi }{2}$ because $\left | \cos^{'}x \right |=\left | \sin x \right |=1$ in this point. – rtybase Mar 17 '16 at 20:29
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By continuity of cosine function, $a_n=\cos a_{n-1}$, which for any $a_0\in [-1,1]$ satisfies $a_n>0$ for all $n=1,2,...$, so can not intersect the open set $[-1,0)-\{a_0\}$.

Singh
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