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 proof: Limit of self-super-roots is e^1/e. TPID 6 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/07/2009, 08:00 AM (This post was last modified: 10/07/2009, 08:01 AM by bo198214.) In reply to 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 $n\ge n_0$: (**) ${^n b} > n$. Now we lead proof by contradiction, suppose that $\lim_{n\to\infty} b_n \neq \eta$ 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. « Next Oldest | Next Newest »

 Messages In This Thread proof: Limit of self-super-roots is e^1/e. TPID 6 - by bo198214 - 10/07/2009, 08:00 AM RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - by andydude - 10/07/2009, 09:52 PM RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - by Base-Acid Tetration - 07/10/2010, 05:19 AM RE: proof: Limit of self-super-roots is e^1/e. TPID 6 - by bo198214 - 07/10/2010, 09:13 AM

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