I quite randomly stumbled upon the following phenomenon:
Let $ f:\mathbb R^+\to\mathbb R^+, x\mapsto x^{-2^{3^{-4^{\cdot^{\cdot^{\cdot}}}}}} $, then in the interval of $[1,20)$ the plot of $f$ looks like the following:
This looks surprisingly similar to the plot of $\frac{1}{x}$, plotted in red:
Why is that? And why is $f$ even well defined (i.e. why does the sequence converge)? It seems as if there exists some $\xi\in[1,2]$ for which $$\forall x\in\mathbb R^+:f(x)=x^{-\xi}\qquad\xi\simeq 1.2$$
$f$ is undefined for negative values and diverges at $0$: $$\lim\limits_{x\searrow0}f(x)\to\infty $$
Can anyone explain those properties / link to some proof? I'd be quite curious about the exact value of $\xi$ too.
Also, if you're interested or don't trust me, I've created the plots with this program.
So to clarify this a bit, the problem can be formulated the following way:
Show that $$\xi:=-\lim\limits_{n\to\infty}-2^{3^{-4^{\cdot^{\cdot^{\cdot^{\sigma(n)\cdot n}}}}}}\simeq1.1982330602188767$$ $$\sigma:\mathbb N\to\{-1,1\},\ n\mapsto\begin{cases}-1& n\mod2=0\\1&n\mod2=1\end{cases}$$
We can formulate this mathematically precisely:
Let $\sigma$ be as defined previously, then define $$e:\mathbb N^2\to\mathbb R,\ (m,n)\mapsto\begin{cases}\sigma(m)\cdot m^{e(m+1,n)}&m<n\\\sigma(m)\cdot m&m=n\end{cases}$$
then $\xi$ is defined as $$\xi:=\lim\limits_{n\to\infty}e(2,n)$$
For this we can proof that $\xi\in(-2,-1)$ by something like the following: $$\forall n\in\mathbb N:\ e(4,n)=-4^{r(n)}\quad r(n)>0\implies e(4,n)<-1$$ $$\implies e(3,n)=3^{e(4,n)}=\left|3^{e(4,n)}\right|<1$$ $$\implies e(2,n)=-2^{e(3,n)}\in(-2,-1)$$
However, this is not a full proof of convergence (but merely of limitedness). I feel like the proof of convergence should contain the value it converges to. But if the value of $\xi$ only appears by this construction, that's rather difficult.
It would also suffice to proof the monotony of $e(2,\cdot)$ - however, it clearly isn't monotonous.
Maybe, similarly to the proof I presented, the interval for $\xi$ can be minimized bit by bit, but I don't quite know how I'd proceed there.
The above proof can easily be generified to the following ($\forall n\in\mathbb N$): $$\forall m\in 2\mathbb N:\ e(m,n)<-1$$ $$\implies\forall o\in\mathbb N\setminus2\mathbb N:\ e(o,n)=|e(o,n)|<1$$ $$\implies\forall p\in2\mathbb N:\ e(p,n)\in(-m,-1)$$
However, a quick computer program indicates that $$\forall m\in2\mathbb N,\ n\gg m:\ e(m,n)\in(-2,-1)$$ I don't know if that's of any use though.
Let $t_n=e(2,n)$; i.e., $t_2=-2, t_3=-2^{3}, t_4=-2^{3^{-4}}, t_5=-2^{3^{-4^5}}, t_6=-2^{3^{-4^{5^{-6}}}}$, etc. Now, although the sequence $(t_n)_n$ is not monotone, it is composed of the following four monotone subsequences: $$\begin{align} &t_2<t_6<t_{10}<t_{14}<\cdots\tag{A}\\ &t_3<t_7<t_{11}<t_{15}<\cdots\tag{B}\\ &t_4>t_8>t_{12}>t_{16}>\cdots\tag{C}\\ &t_5>t_9>t_{13}>t_{17}>\cdots\tag{D} \end{align}$$
(A)-(D) are straightforward (but tedious) to prove by comparing the towers $t_n$ and $t_{n+4}$; similarly, it can be shown that for all $n\ge 6$, $$ \min(t_{n-4},t_{n-1})<t_n<\max(t_{n-4},t_{n-1}).\tag{E} $$ Since $-8\le t_n< -1$ the monotone convergence theorem gives that all four of the subsequences converge, and by (E) they all converge to the same limit. Computations (in Sage) show this to be $\xi=\lim\limits_{n\to\infty}t_n=-1.198233060218876592050411767\dots$
Here are some pictures:
Aside:
We can extend the OP's domain of definition of $e(m,n)$ to $\mathbb{Z}\times\mathbb{Z}$ and define $$\xi_m:=\lim\limits_{n\to\infty}e(m,n).$$ Then the doubly-infinite sequence $$(\ldots,\xi_{-2},\xi_{-1},\xi_{0},\xi_{1},\xi_{2},\ldots)$$ is well-defined and satisfies the recursion $$\xi_m=\sigma(m)\,m^{\xi_{m+1}}$$ where $\sigma(m) = -1 \text{ if $m$ is even, else $+1$ if $m$ is odd}.$
In particular, we find $$\begin{align} \vdots\\ \xi_{-p}&= p^{{(p-1)}^{ {(p-2)}^{.^{{.^{.}}^1}}} }, p\ge 1\\ \vdots\\ \xi_{-4}&=4^{3^{2^1}}\\ \xi_{-3}&=3^{2^1}\\ \xi_{-2}&=2^1\\ \xi_{-1}&=1\\ \xi_0&=0\\ \xi_1&=1\\ \xi_2&=-2^{3^{{-4}^{5^{.^{.^.}}}}}=−1.198233060218876592050411767\ldots\\ \xi_3&=\log_2(-\xi_2)=0.260908544287382650344471086\ldots\\ \xi_4&=\log_3(\xi_3)=-1.222984079054583379571125508\ldots\\ \xi_5&=\log_4(-\xi_4)=0.145202811409957592071338045\ldots\\ \xi_6&=\log_5(\xi_5)=-1.198942686476802521519469705\ldots\\ \xi_7&=\log_6(-\xi_6)=0.101263633269622829392559457\ldots\\ \xi_8&=\log_7(\xi_7)=-1.176841558483650376687654327\ldots\\ \vdots\\ \end{align}$$ and $$\begin{align}\lim\limits_{k\to\infty}\xi_{2k}&=-1\tag{F}\\ \lim\limits_{k\to\infty}\xi_{2k-1}&=0.\tag{G}\end{align} $$
Here are some pictures:
(The SageMath code that I used is available here.)