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 '$$' with '=' and solve for real x > e :
=> q estimated around q = x = 6,6568558380496
in the limit
e^^(n+1) grows slower than 6,6568558380496^^n.
I have confidence in my digits BECAUSE it is known that the convergeance is superexponential.
However we can continue
6,6568... ^^(n+1) = x_2 ^^n
etc
and we end up with a sequence : x_0 = e , x_1 = x = 6.65... , x_2 , x_3 , ...
How fast does this x_n grow ??
it seems x_n is the superfunction of y^^(n-1) = x^^n where n goes to oo but is well approximated by small n FOR A SINGLE STEP AT LEAST.
the behaviour of x_n though troubles me.
this is clearly base change.
But what is the behaviour of the super of y^y^y^y = x^x^x^x^x ?
or even y^y^y^y = x^x^x^x^x ?
is there a known series expansion for y^^(n-1) = x^^n ??
numeric overflow seems like an often encountered problem.
it is easy to show x_n < C x_(n-1)^x_(n-1).
But does x_n then grow faster / slower or equal to exponential ??
I know Lambert W and the Taylor series for a power tower.
Maybe a certain limit with those would help.
x^x = e^e^e
x^x^x = e^e^e^e
...
x^^n = e^^(n-1)
if we solve for x we get
(i have little time for research at the moment , but i will give a brute
estimate , which might even be quite exact. )
e^e $$ 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.