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Remark on Gottfried's "problem with an infinite product" power tower variation - tommy1729 - 07/31/2013
Remark on Gottfried's "problem with an infinite product".
http://math.eretrandre.org/tetrationforum/showthread.php?tid=785
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A similar case occurs with power towers.
g(x) = x ^ f(x) ^ f^[2](x) ^ ...
An intresting case is ln(x) ^ ln(ln(x)) ^ ln^[3](x) ^ ... = A(x).
Let x >> 10.
Define f(x) = arcsinh(x/2)
( YES I USE THIS AGAIN , THE INVERSE OF 2SINH !!! )
Now there exists (small) constants C1,C2,C3,C4 such that
f(x) ^ f^[2](x) ^ f^[3](x) ^ ... ~ C1 (x + C2)^C3 + C4
Nice.
Return of the master
tommy1729
RE: Remark on Gottfried's "problem with an infinite product" power tower variation - tommy1729 - 05/02/2014
H(exp(x)) = x^(H(x)^a)
Such equations seem related.
Have they occured here before ?
regards
tommy1729
*edited*
RE: Remark on Gottfried's "problem with an infinite product" power tower variation - tommy1729 - 05/03/2014
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Conjecture B : This thread relates to another : http://math.eretrandre.org/tetrationforum/showthread.php?tid=825
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.
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regards
tommy1729
RE: Remark on Gottfried's "problem with an infinite product" power tower variation - tommy1729 - 05/05/2014
I referred to this thread in http://math.eretrandre.org/tetrationforum/showthread.php?tid=854&pid=6943#pid6943
For simplicity lets replace arc2sinh(x/2) with ln(x).
This is a bit informal handwaving, but for the time being it might be good enough.
Let f(x) = ln(x) ^ ln(ln(x)) ^ ln^[3](x) ^ ...
As often in math we get the weird result that something is true IFF something else is true.
Keeping this in mind :
This might be handwaving but lets go :
Let eps > 0
( the famous epsilon )
Assume for sufficiently large x :
x^eps < f(x) < x^A
for some real A > eps.
(the epsilon assumption)
This is more powerfull than you might expect ! :
We know ln(x^b(x)) / ln(x) = b(x)
Based on that :
f(x) = ln(x) ^ ln(ln(x)) ^ ln^[3](x) ^ ...
ln(f(x)) / ln(x)
= ln( ln(x) ^ ln(ln(x)) ^ ln^[3](x) ^ ... ) / ln(x)
= (ln(ln(x)) * ln(ln(x)) ^ ln^[3](x) ^ ...) / ln(x)
use f(x) = x^b(x)
= ln(ln(x)) * ln(x)^b(ln(x)) * (ln(x))^-1
=> the best constant fit for b(x) = b(ln(x)) = b => ( existance of b follows from the epsilon assumption )
=> b = 1 because ln(ln(x)) * ln(x) / ln(x) = ln(ln(x))
and ln(ln(x)) is smaller than a power of ln(x).
By a similar logic we can improve : f(x) = x^b(x) = x^b / ln(x)
This gives us :
= ln(ln(x)) * ln(x)^b(ln(x)) * (ln(x))^-1
= ln(ln(x)) * f(ln(x)) * (ln(x))^-1
= ln(ln(x)) * ln(x)^b / ln(ln(x)) * (ln(x))^-1
= 1.
= b.
QED.
Hence if x^eps < f(x) < x^A
then f(x) = x / (ln(x))^(1+o(1)).
WOW.
THIS IS THE PNT FOR POWER TOWERS !!
However like I said this is a bit handwaving.
Because ... " replace arc2sinh(x/2) with ln(x) " ?
Is x^eps < f(x) < x^A TRUE ???
Funny thing is also that currently the mellin transform is considered for tetration.
The same mellin transform used in the proof for PNT !
So maybe a mellin transform can be used for this power tower problem as well ???
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Remark :
Its not even clear f(x) is analytic ?
I considered Taylor series , four series and dirichlet.
And taking derivatives.
But if f(x) is not analytic ...
That might also be problematic for the mellin transform.
On the other hand f(x) is C^oo.
So I believe taking derivatives is justified.
This brings me to another idea :taking the logarithmic derivative of f(x).
That might help decide if the epsilon assumption is true !!
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regards
tommy1729
RE: Remark on Gottfried's "problem with an infinite product" power tower variation - tommy1729 - 05/06/2014
Lets dig a bit deeper.
H(exp(x)) = x^H(x)
I use =\(= notation here , =\)= compares two functions and concludes <,=,> for sufficiently large x.
What is special about this notation is that we do not switch LHS with RHS.
Say H is about x^2 :
=>
exp(2x) =\(= x^(x^2)
=>
exp(2x) =\)= exp(ln(x) x^2)
=>
exp(2x) < exp(ln(x) x^2)
Likewise say H is about x^0.5 :
=>
exp(0.5 x) =\(= x^(x^0.5)
=>
exp(0.5 x) =\)= exp(ln(x) x^0.5)
=>
exp(0.5 x) > exp(ln(x) x^0.5)
=> x^0.5 < H(x) < x^2
( It should be clear to you WHY =$= does not switch LHS with RHS now. )
It is easy to generalize this to :
x^a < H(x) < x^(1/a)
for all 0<a<1.
Hence H(x) = f(x) = x^(1+o(1)).
So far the semiformal part.
A quick estimate for f(x) gives me :
x/(ln(x)^ln(ln(x))^ln(ln(ln(x)))) < H(x) < x*(ln(x)^ln(ln(x))^ln(ln(ln(x))))
Notice I still do not know about using a derivative ...
IN FACT :
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H(exp(x)) = x^H(x)
suggests H(x) is not entire.
***
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Note : Power towers have the strange property of giving preference to positive nth derivatives.
This relates to a recent remark by sheldon about the first 14 derivatives of exp^[1/2] ...
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Thanks for your intrest
regards
tommy1729