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 Infinite Pentation (and x-srt-x) andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/12/2009, 08:19 AM nuninho1980 Wrote:yes. but I don't get value "x" of a^^x if x<1 using your code "tetration and slog". Don't be sad, that's to be expected. My code only applies when (a>1), which would require that (x>1) for this function. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/12/2009, 08:58 AM So this is my attempt at a proof, and there are a bunch of inequalities, which always confuse me , so let me know if there are any mistakes. Let $\eta = e^{1/e}$ as usual. I will use k (instead of n) to avoid confusion with $\eta$. Lemma 1 (Knoebel). $(a < b)$ iff $({}^{k}a < {}^{k}b)$ for all $a,b,k > 1$. Lemma 2. $\eta < {}^{x}\eta < e$ for all real $x > 1$. Proof. The base-$\eta$ tetrational function is continuous and monotonic? Lemma 3. $b^x > x + 1$ for all positive real $x \ge 5$ and real $b > \eta$. Proof. The function $(x+1)^{1/x} < \eta$ for all $x \ge 5$, thus $\eta^x > x+1$. If $b > \eta > 1$, then $b^x > \eta^x$, so $b^x > x+1$. Lemma 4 (lower bound). For all integer $k \ge 3$, $\eta < a_k = \text{srt}_k(k)$. Proof. Since $k \ge 3$, then obviously $k > e$. Together with lemma (2), this implies that ${}^{k}\eta < e < k$. Substituting $k = {}^{k}(a_k)$ (hypothesis), this can be written as ${}^{k}\eta < {}^{k}(a_k)$ which implies $\eta < a_k$ by lemma (1). Lemma 5 (decreasing). For all integer $k \ge 5$ and $a_k = \text{srt}_k(k)$, $a_{k} > a_{k+1}$. Proof. We have ${}^{k}\left(a_{k}\right) = k$ for all integer $k$ by definition. It follows that $ \begin{tabular}{rl} a_{k+1}^{\left({}^{k}\left(a_{k+1}\right)\right)} & = {}^{k+1}\left(a_{k+1}\right) \\ 1 + {}^{k}\left(a_{k+1}\right) & < {}^{k+1}\left(a_{k+1}\right) \\ 1 + {}^{k}\left(a_{k+1}\right) & < k + 1 \\ {}^{k}\left(a_{k+1}\right) & < k \end{tabular}$ by lemma (3). Thus ${}^{k}\left(a_{k+1}\right) < {}^{k}\left(a_{k}\right)$ which implies $a_{k+1} < a_k$ by lemma (1). Theorem. $\lim_{x\to\infty} \text{srt}_x(x) = \eta$ Proof. $\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ by lemma (4) and lemma (5). ... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to $\eta$. Is this right? Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/13/2009, 05:27 AM So I tried using regular and natural tetration to fill out the rest of the graph. I almost ignored using bases where $1 < a < \eta$ because these bases would have 2 fixed points: (1) the upper fixed point would overlap the "blue" region which I already calculated with interpolation (but a good area for comparison), and (2) the lower fixed point would overlap the "green" region which I already calculated with natural iteration (also a good area for comparison). The only part I used regular tetration for was the "red" region, which corresponds to bases where $e^{-e} < a < 1$. Using pure regular iteration from the primary fixed point of $a^z$ the base-a tetrational is real-valued only for integers, and complex-valued everywhere else. So I fudged a little bit to make the graph, I solve the equation $\text{Re}({}^{x}a) = x$ and these points are what you see in the "red" region of the graph. I'm pretty sure these are wrong, and that the graph would have to be complex here (if tetrationals for bases less than 1 form complex outputs for real inputs), so it's more of a heuristic than a continuation of the blue curve.     Andrew Robbins Attached Files   x-srt-x-more.pdf (Size: 10.29 KB / Downloads: 326) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/13/2009, 05:01 PM (This post was last modified: 04/13/2009, 05:01 PM by bo198214.) andydude Wrote:$\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ by lemma (4) and lemma (5). ... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to $\eta$. Is this right? I dont think so. When the sequence is decreasing and bounded it has a limit. But this limit could be probably bigger than $\eta$. Or did I overlook something? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/03/2009, 10:42 PM (04/13/2009, 05:01 PM)bo198214 Wrote: andydude Wrote:$\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ by lemma (4) and lemma (5). ... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to $\eta$. Is this right? I dont think so. When the sequence is decreasing and bounded it has a limit. But this limit could be probably bigger than $\eta$. Or did I overlook something? $\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?! or did i miss something ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 10:51 PM (05/03/2009, 10:42 PM)tommy1729 Wrote: $\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?! or did i miss something ? yes, take a similar example: $-1 < \frac{1}{k+1} < \frac{1}{k}$ by your argumentation $\lim_{k\to\infty} \frac{1}{k} = -1$. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/03/2009, 10:57 PM (05/03/2009, 10:51 PM)bo198214 Wrote: (05/03/2009, 10:42 PM)tommy1729 Wrote: $\eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k)$ for all $k \ge 5$ so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?! or did i miss something ? yes, take a similar example: $-1 < \frac{1}{k+1} < \frac{1}{k}$ by your argumentation $\lim_{k\to\infty} \frac{1}{k} = -1$. thats not the same , the limit value cannot possibly be smaller than eta , so eta is the ultimate boundary. and because any infinite tetration of real base (eta + q) = oo our resulting limit must be between eta < " limit "< eta + q taking lim q = 0+ => " limit " = eta. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 11:20 PM (05/03/2009, 10:57 PM)tommy1729 Wrote: thats not the same , the limit value cannot possibly be smaller than eta ,Of course not the limit in the above case is 0. It is *greater* than -1. Thats what I already said in my original objection: the limit can possibly be *greater* than $\eta$. Quote:and because any infinite tetration of real base (eta + q) = oo proof? BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 05/04/2009, 08:20 PM (05/03/2009, 10:57 PM)tommy1729 Wrote: and because any infinite tetration of real base (eta + q) = oo I thought this was about infinite pentation, not infinite tetration. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/04/2009, 08:57 PM (05/04/2009, 08:20 PM)BenStandeven Wrote: (05/03/2009, 10:57 PM)tommy1729 Wrote: and because any infinite tetration of real base (eta + q) = oo I thought this was about infinite pentation, not infinite tetration. Well depends how you interpret it. Surely the title "infinite pentation" means $\lim_{n\to\infty} b [5] n$ which is the same as infinitely repeated tetration, which tommy refers to. « Next Oldest | Next Newest »

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