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 about power towers and base change tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/23/2013, 12:26 PM I was wondering about the following x^x = e^e^e x^x^x = e^e^e^e ... x^^n = e^^(n-1) if we solve for x we get (i have little time for research at the moment , but i will give a brute estimate , which might even be quite exact. ) e^e $$x take elog on both sides : e$$ elog(x) e^e^e $$x^x take elog on both sides : e^e$$ elog(x) * x take elog again : e $$elog(x) + elog(elog(x)) keep increasing tower hight and taking elogs : e^e$$ elog(x^x) + elog(elog(x^x)) = e^e $$elog(x) * x + elog(elog(x) * x) = e^e$$ elog(x) * x + elog(elog(x)) + elog(x) = e^e^e $$elog(x^x) * x^x + elog(elog(x^x) * x^x) = e^e^e$$ elog(x) * x * x^x + elog(elog(x) * x * x^x) = e^e^e $$elog(x) * x * x^x + elog(elog(x)) + elog(x) + elog(x) * x = e^e^e$$ elog(x)[1 + x + x^(x+1)] + elog(elog(x)) now replace '' with '=' and solve for real x > e : => q estimated around q = x = 6,6568558380496 in the limit e^^(n+1) grows slower than 6,6568558380496^^n. I have confidence in my digits BECAUSE it is known that the convergeance is superexponential. However we can continue 6,6568... ^^(n+1) = x_2 ^^n etc and we end up with a sequence : x_0 = e , x_1 = x = 6.65... , x_2 , x_3 , ... How fast does this x_n grow ?? it seems x_n is the superfunction of y^^(n-1) = x^^n where n goes to oo but is well approximated by small n FOR A SINGLE STEP AT LEAST. the behaviour of x_n though troubles me. this is clearly base change. But what is the behaviour of the super of y^y^y^y = x^x^x^x^x ? or even y^y^y^y = x^x^x^x^x ? is there a known series expansion for y^^(n-1) = x^^n ?? numeric overflow seems like an often encountered problem. it is easy to show x_n < C x_(n-1)^x_(n-1). But does x_n then grow faster / slower or equal to exponential ?? I know Lambert W and the Taylor series for a power tower. Maybe a certain limit with those would help. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/02/2014, 10:42 PM This is still unresolved. Despite my increasing understanding of sexp and slog, I do not know how to work with this. Im confident others here could solve the problem. regards tommy1729 sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 05/03/2014, 03:39 AM (This post was last modified: 05/03/2014, 11:16 PM by sheldonison.) (05/02/2014, 10:42 PM)tommy1729 Wrote: This is still unresolved. Despite my increasing understanding of sexp and slog, I do not know how to work with this. Im confident others here could solve the problem. regards tommy1729 Traveling -- little time right now. This is problem is closely related to my mathstack thread; ln-n-of-tetrationx-n-is-nowhere-analytic. My relevant function is the following equation. $f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n$ One of Tommy's original question was "is there a known series expansion for y^^(n-1) = x^^n ??" For the limiting answer, as n goes to infinity, given base base x, we use my function above. We desire $f(y)=e^{f(x)}$, so we calculate $y=f^{-1}(e^{f(x)})$. Of course, there is the small issue that f(x) is conjectured coo and nowhere analytic .... I plan to eventually post a proof. Using the above equation, if b~=7.28550781987618684208203148323, from this mathstack thread then f(b)=e, and as n goes to infinity, $b\uparrow\uparrow n = e \uparrow \uparrow (n+1)$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/03/2014, 12:22 PM (05/03/2014, 03:39 AM)sheldonison Wrote: (05/02/2014, 10:42 PM)tommy1729 Wrote: This is still unresolved. Despite my increasing understanding of sexp and slog, I do not know how to work with this. Im confident others here could solve the problem. regards tommy1729 Traveling -- little time right now. This is problem is closely related to my mathstack thread; ln-n-of-tetrationx-n-is-nowhere-analytic. My relevant function is the following equation. $f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n$ One of Tommy's original question was "is there a known series expansion for y^^(n-1) = x^^n ??" For the limiting answer, as n goes to infinity, given base base x, we use my function above. We desire f(y)=f(x)+1, so we calculate $y=f^{-1}(f(x)+1)$. Of course, there is the small issue that f(x) is conjectured coo and nowhere analytic .... I plan to eventually post a proof. Using the above equation, if b~=7.28550781987618684208203148323, from this mathstack thread then f(b)=f(e)+1, and as n goes to infinity, $b\uparrow\uparrow n = e \uparrow \uparrow (n+1)$ - Sheldon I wonder where you are travelling too. I have seen your mathstack thread. Its intresting and reminds me of base change. In fact isnt your function f(x) just the base change ? Could it be you made a typo ? Your value of 7.28 does not agree with my 6.65. I used paper and you probably a computer. But still I wonder, are we talking about the same things and if so , why is there such a difference ? Thanks for your intrest. I dont have much time either. regards tommy1729 sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 05/03/2014, 01:14 PM (This post was last modified: 05/03/2014, 01:18 PM by sheldonison.) (05/03/2014, 12:22 PM)tommy1729 Wrote: In fact isnt your function f(x) just the base change ? Could it be you made a typo ? Your value of 7.28 does not agree with my 6.65. I used paper and you probably a computer. But still I wonder, are we talking about the same things and if so , why is there such a difference ? Thanks for your intrest. I dont have much time either. regards tommy1729 The classic use of base change was to start with $\text{cheta}(z)=\exp_\eta^{oz}$, and to generate $\exp^{oz} = \lim_{n\to\infty}\log^{on}(\exp_\eta^{o(z+n)})$. The f(z) function isn't a superexponential, though it is related to the base change. It is just a function that tells you how fast any given base for tetration would grow; you can plug any pair of bases z=b1,b2 into f(b1), f(b2). And then if f(b2)=exp(f(b1)), you know b2^^n=b1^^(n+1). I did have a typo, in my previous post today, now fixed, but the value, ~=7.2855 is/was correct. -Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/03/2014, 08:24 PM Well according to me the value y was between 6.65 and e^2. This is still true. Yet I assumed it would be closer to 6.65 due to fast convergeance. But I guess the " fast " part comes later after more iterations. I assume sheldon used a computer to do the iterations. I could not improve the value 6.65 by hand. I havent given it much attention due to lack of time, but if anyone knows how to get 7.28 by hand that would be appreciated. Im fascinated by how close y is to e^2. I assume this is not coincidence. Im fascinated by this f(x). My x_n is now given by f^[-1]( exp^[n] f(x) ). SO x_n does grow superexponential ! Thank you Sheldon ! 2 remaining questions pop up immediately in my head : 1) how fast does f(x) grow ? Can we express f(x) in other functions ? 2) is there a limit form for f^[-1](x) ?? Question 2 reminds me of the idea that finding a limit form of a function that is not given by a sum and that is not analytic is somehow " hard " usually. Im not quite sure if and why that is true. With some luck this was already given in the many threads about base change. I did not find the time to investigate this yet. Thanks for the help !! And good luck on MSE. One more thing, ------------------------------------------------------------------------------ Conjecture B : This thread relates to another : http://math.eretrandre.org/tetrationforu...hp?tid=799 More specifically my g(x) and/or H(x) relates to sheldon's f(x) and/or sheldon's f^[-1](x). Its BTW no coincidence that I gave attention to them again at the same day. In other words I expected this connection for a long time. ------------------------------------------------------------------------------ regards tommy1729 sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 05/04/2014, 04:29 AM (This post was last modified: 05/04/2014, 04:37 AM by sheldonison.) (05/03/2014, 08:24 PM)tommy1729 Wrote: ....Im fascinated by this f(x). My x_n is now given by f^[-1]( exp^[n] f(x) ). SO x_n does grow superexponential ! Thank you Sheldon !you're welcome, and you are correct. The approximate value for b^^n = e^^(n+2) as n to infinity, is b~= 302263.280461758 Quote:2 remaining questions pop up immediately in my head : 1) how fast does f(x) grow ? Can we express f(x) in other functions ? 2) is there a limit form for f^[-1](x) ?? This is the equation for f $f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n$ For numerical results for $f^{-1}(x)$, I iterate using Newton's approximations and it works pretty well. for n=2, you get a sense of how f(x) grows: $f_2(x) = \ln^{o2}(x\uparrow \uparrow 2) = \ln(x) + \ln(\ln(x))$ for n=3 $f_3(x) = \ln^{o3}(x\uparrow \uparrow 3) = \ln(\exp(f_2(x))+\ln(\ln(x))) = f_2(x) + \ln(1+\frac{\ln(\ln(x))}{\exp(f_2(x))})$ for n=4, $f_4(x) = \ln^{o4}(x\uparrow \uparrow 4) = f_3(x) + \ln(1 + \frac{1}{\exp(f_3(x))}\ln(1+\frac{\ln(\ln(x))}{\exp^{o2}(f_3(x))} ))$ .... for n=5, there are three nested logarithms .... - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/04/2014, 08:30 AM I was about to post that f^[-1](x) for x > e is Always smaller than x^x. This is easy to show. Also f^[-1] = (approximate) = (ln(x)+ln^[2](x) + ln(1+1/x))^[-1]. But this now seems trivial after sheldon's post. (ln(x)+ln^[2](x) + ln(1+1/x))^[-1] < exp(x) and that can be improved. The approximate value for x_n is thus e^^(n) < x_n < e^^(n+1) regards tommy1729 « Next Oldest | Next Newest »

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