Is $\cot(\cot(\cot(\cdots\cot1)\cdots))$ always defined?

Asked Nov 05 '18 at 07:28

Active Nov 05 '18 at 08:08

Viewed 116 times

Consider sequence $a_1=1$, $a_{n+1}=\cot a_n$. Is $a_n$ always defined?

Numerical evaluation suggests this conjecture is true.

I have proved a weak version of this question: for a fixed $n$ and $a_1=x$, the measure of $x$ such that $a_n$ is not defined is 0.

Proof
There is a bijection between $\{x|a_n\text{ is not defined}\}$ and $\mathbb{N}^{n-1}$, the set is countable, hence the measure is 0.

Does anyone have some idea on the original question?

edited Nov 05 '18 at 08:08

asked Nov 05 '18 at 07:28

Kemono Chen

@Blue I think the brackets can be ignored? – Kemono Chen Nov 05 '18 at 07:40
1

What does it mean for a real number to be meaningful? – Carl Schildkraut Nov 05 '18 at 07:44
I mean the situation like $\cot0$ will never happen. – Kemono Chen Nov 05 '18 at 07:51
@KemonoChen I think this is called "defined" (?) – Tianlalu Nov 05 '18 at 08:00
@KemonoChen: In this context, I think it's better to use the parentheses. Otherwise, the expression looks like it might be an erroneously-formatted product of some kind. (That's actually what I thought it was, at first sight.) – Blue Nov 05 '18 at 08:02
@Tianlalu thanks for correcting word choosing. – Kemono Chen Nov 05 '18 at 08:09
2

I think it would be truly remarkable if $a_n$ just happens to hit $\pi\mathbb Z$. Therefore, I feel safe in asserting that this does not occur. But it also seems unlikely that there is a proof; unless perhaps there is some geometric interpretation for the sequence $a_n$. – GEdgar Nov 05 '18 at 12:30

0 Answers0