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+--- Thread: about power towers and base change (/showthread.php?tid=825)
about power towers and base change - tommy1729 - 10/23/2013
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.
RE: about power towers and base change - tommy1729 - 05/02/2014
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
RE: about power towers and base change - sheldonison - 05/03/2014
(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
RE: about power towers and base change - tommy1729 - 05/03/2014
(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
RE: about power towers and base change - sheldonison - 05/03/2014
(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
RE: about power towers and base change - tommy1729 - 05/03/2014
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/tetrationforum/showthread.php?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
RE: about power towers and base change - sheldonison - 05/04/2014
(05/03/2014, 08:24 PM)tommy1729 Wrote: ....Im fascinated by this f(x).you're welcome, and you are correct.
My x_n is now given by f^[-1]( exp^[n] f(x) ).
SO x_n does grow superexponential !
Thank you Sheldon !
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 :This is the equation for f
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) ??
\( f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n \)
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
RE: about power towers and base change - tommy1729 - 05/04/2014
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