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about power towers and base change
#1
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.
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#2
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
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#3
(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.



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 , so we calculate . 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,
- Sheldon
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#4
(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.



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 . 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,
- 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
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#5
(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 , and to generate .

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
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#6
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
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#7
(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


For numerical results for , I iterate using Newton's approximations and it works pretty well.

for n=2, you get a sense of how f(x) grows:

for n=3

for n=4,

.... for n=5, there are three nested logarithms ....
- Sheldon
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#8
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
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