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 tetration limit ?? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/01/2009, 05:49 PM maybe this is the dumbest tetration question ever , but here goes. i was thinking about a " tetration limit ". as you all know , im particular intrested in half iterates of exponential functions. we all know lim n -> oo ( 1 + f(n) ) ^ n = e if f(n) = 1/n now this is the typical exponential limit. but what is the typical half iterate exponential limit ? lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ??? where F stands for the half iterate exponential of base ( 1 + f(n) ) and f(n) and Q are the unknown function and unknown value. regards tommy1729 nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009 04/01/2009, 08:15 PM (This post was last modified: 04/04/2009, 02:25 PM by nuninho1980.) Hi! (e^(1/e))^^oo = e 1.6353 "pentate" oo ~= 3.0886 (30 dimensions of matrix) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/02/2009, 02:50 PM (This post was last modified: 04/02/2009, 02:50 PM by bo198214.) tommy1729 Wrote:lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ??? where F stands for the half iterate exponential of base ( 1 + f(n) ) and what is "; n"? Power, iteration, or, or, or? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/02/2009, 09:24 PM bo198214 Wrote:tommy1729 Wrote:lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ??? where F stands for the half iterate exponential of base ( 1 + f(n) ) and what is "; n"? Power, iteration, or, or, or? n is an integer F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n F has two arguments seperated by " ; " and has the form F[ base ; z ] and is the half iterate of base ^ z. clear ? regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/02/2009, 09:56 PM tommy1729 Wrote:n is an integer F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n F has two arguments seperated by " ; " and has the form F[ base ; z ] and is the half iterate of base ^ z. Perhaps then you should start with the simpler case of the double iterate. And look what a suitable function f you would find that: $\lim_{n\to\infty} (1+f(n))^{(1+f(n))^{n}} = Q$ I dont see what useful function f that could be. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/02/2009, 09:58 PM nuninho1980 Wrote:Hi! (e^(1/e))^^oo = e 1.6353 "pentate" oo ~= 3.0886 (30 terms) Hello! How do you tetrate 1.6353^^1.6353? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/02/2009, 10:39 PM bo198214 Wrote:tommy1729 Wrote:n is an integer F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n F has two arguments seperated by " ; " and has the form F[ base ; z ] and is the half iterate of base ^ z. Perhaps then you should start with the simpler case of the double iterate. And look what a suitable function f you would find that: $\lim_{n\to\infty} (1+f(n))^{(1+f(n))^{n}} = Q$ I dont see what useful function f that could be. first of all , i have to comment that i find you are changing subject a little bit. what you call a simpler case might be a harder case or an unrelated case. no offense. ( i think your limit like question is to " sensitive ". i wont explain that ) but i will try to give it a go anyways ... (1 + f(n)) ^ (1 + f(n)) ^ n = Q i try to use the identity lim (1 + 1 / g(n)) ^ g(n) = e ( for many g(n) ) so : 1 + 1/ ((1 + f(n)) ^n) = 1 + f(n) so f(n) = (1 + f(n)) ^ - n which leads to f(z) = ( 1 + f(z) ) ^ - z and Q = e ... maybe ... however note that ( 1 + 1/n + 1/n^3 ) ^ n = e too , so some flexibility can be added to the equation for f(z) ... at first sight id estimate f(z) around a / log(z) + b/ log(z)^c for some reals a, b and c. another method might be taking the log of both sides getting an expression for log(Q) and then using l'hospital. maybe that is more succesfull. maybe both give a working result but different !?! this might be bo's objection somewhat hidden. but i dont think such an issue occurs in my OP. ( informally : in general limits with double exponential speed tends to converge to a finite numbers less then limits of slower functions ) regards tommy1729 nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009 04/03/2009, 12:53 AM (This post was last modified: 04/03/2009, 12:54 AM by nuninho1980.) bo198214 Wrote:Hello! How do you tetrate 1.6353^^1.6353? my pc evaluated it, running "maple 12" w/ code v1.1 about "tetration and slog". x^^y = z if the "z" isn't to close the point fixed (may to infinity) then you reduce any case decimal "x" and/or "y" but it's bit difficult to close the point fixed (3.08~3.10). the tetration is slower than slog. but I will try more smooth for calcuate slog. my cpu c2d e6600 is slow. but the future the gpu (CUDA ) will help to the cpu for calculate very fast. my gpu geforce 9800gt maybe will support! if you don't know CUDA then you see http://www.nvidia.com/object/cuda_what_is.html. note: sorry for bad english. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/03/2009, 12:49 PM nuninho1980 Wrote:bo198214 Wrote:Hello! How do you tetrate 1.6353^^1.6353? my pc evaluated it, running "maple 12" w/ code v1.1 about "tetration and slog". Which code did you use? Quote: note: sorry for bad english. No problem, as long as I understand you nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009 04/03/2009, 05:54 PM (This post was last modified: 04/03/2009, 05:55 PM by nuninho1980.) bo198214 Wrote:Which code did you use? "tetration and slog" original by Andrew Robbins is as smoother as "new regular slog". « Next Oldest | Next Newest »

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