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 Question about tetration limit mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/13/2011, 03:17 AM Hi. Consider the limit of tetration $^z b$ as $b \rightarrow \eta$ from the right, along the real axis (Here, we use the Kneser/etc. tetrational). The tetrational looks to converge to the parabolic attracting regular iteration at base $\eta$. This suggests that the resulting fractional iterates of $\exp_b(x)$ converge to the attracting parabolic regular iterates of base $\eta$ when $x < e$. But what about $x > e$? Does it converge to the iterates obtained from the "cheta" function $\check{\eta}(x)$? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 07/13/2011, 12:08 PM (This post was last modified: 07/13/2011, 12:11 PM by tommy1729.) if eta^eta^...x >> e then no. this resembles the base change idea , if two bases are close it works better unless one base is smaller than eta and the other is bigger. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 07/13/2011, 12:24 PM (This post was last modified: 07/13/2011, 01:15 PM by sheldonison.) (07/13/2011, 03:17 AM)mike3 Wrote: Hi. Consider the limit of tetration $^z b$ as $b \rightarrow \eta$ from the right, along the real axis (Here, we use the Kneser/etc. tetrational). The tetrational looks to converge to the parabolic attracting regular iteration at base $\eta$. This suggests that the resulting fractional iterates of $\exp_b(x)$ converge to the attracting parabolic regular iterates of base $\eta$ when $x < e$. But what about $x > e$? Does it converge to the iterates obtained from the "cheta" function $\check{\eta}(x)$? I think if you recenter the tetrations for different bases>eta, all at the same point, and if the value chosen to recenter at, f(x), is large enough, then the answer may be that it does converge to $\check{\eta}(x)$. For example the upper superfunction at base eta, $\check{\eta}(x)$, can be centered so that $\check{\eta}(0)=2e$. If you take the limit for f(x)=sexp(x) for bases>eta, but approaching eta, all equivalently recentered so that f(0)=2e, then it seems that the sequence of functions would converge arbitrarily closely, so that in the limit, f(x) converges to the $\check{\eta}(x)$ function. I think this would hold in the complex plane as well. The key is that the real part of the pseudo period for sexp for bases approaching but greater than eta gets arbitrarily large. Recentering at 2e, convergence might hold for a little less than half a psuedo period on the left, which is approximately where the inflection point is for sexp(x). This is mostly speculation, btw. - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 07/13/2011, 12:51 PM and i agree with sheldon. « Next Oldest | Next Newest »

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