# Tetration Forum

Full Version: proof: Limit of self-super-roots is e^1/e. TPID 6
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http://math.eretrandre.org/tetrationforu...73#pid4073

First it is easy to see that for $1:

${^n b}=\exp_b^{\circ n}(1)\to a ($a$ is the lower fixed point of $b^x$)

Hence for $n_0 > 3$ we have for all $n\ge n_0$:
(*) ${^n b} < e < n$

We also know that for $b>\eta$, $\exp_b^{\circ n}\to\infty$ quite fast, particularly for each $b>\eta$ there is an $n_0$ such that for all $n\ge n_0$:
(**) ${^n b} > n$.

$\lim_{n\to\infty} b_n \neq \eta$ where ${^n b_n} = n,\quad b_n > 1$.

Then there must be a subsequence $b_{m},\quad m\in M\subseteq\mathbb{N}$ and $\epsilon>0$ such that this subsequence stays always more than $\eps$ apart from $\eta$:
$\left|b_{m}-\eta\right| \ge \epsilon$.
I.e. there is $B_1<\eta$ and $B_2>\eta$ such that
either $b_m \le B_1$ or $b_m \ge B_2$.

By (*) and (**) we have $n_0$ such that for all $m\ge n_0$:
${^m B_1} and ${^m B_2}>m$.
As ${^m x}$ is monotone increasing for $x>1$ we have also
${^m b_m} and ${^m b_m}>m$.

This particularly means ${^m b_m}\neq m$ and hence none of the $b_m$ can be the self superroot, in contradiction to our assumption.
Wow! Very nice! You make it seem so easy. I've been working on that one for while, ever since the xsrtx thread.
The same method of proof could possibly be used to easily prove that, possibly for all k>4, limit of self-hyper-k-root(x) as x -> infinity = $\eta_k$ (defined as the largest real x such that $x[k]\infty < \infty$, i.e. where the maximum of self-hyper-(k-1)-root function occurs; let's establish this notation); yeah I know, I only substituted the pentation-analogues into the proof and quickly checked.
(07/10/2010, 05:19 AM)Base-Acid Tetration Wrote: [ -> ]The same method of proof could possibly be used to easily prove that, possibly for all k>4

The thing is: to define the hyper k-self-root you need a hyper (k-1) operation defined on the reals.
And we still have several methods of doing this without equality proofs.