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Remark on Gottfried's "problem with an infinite product" power tower variation
#1
Remark on Gottfried's "problem with an infinite product".


http://math.eretrandre.org/tetrationforu...hp?tid=785

***

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
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#2
H(exp(x)) = x^(H(x)^a)

Such equations seem related.

Have they occured here before ?


regards

tommy1729

*edited*
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#3
------------------------------------------------------------------------------
Conjecture B : This thread relates to another : http://math.eretrandre.org/tetrationforu...hp?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.
------------------------------------------------------------------------------

regards

tommy1729
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#4
I referred to this thread in http://math.eretrandre.org/tetrationforu...43#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 ???


------------------------------------------------------------------------------

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 !!

------------------------------------------------------------------------------

regards

tommy1729
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#5
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 :

***

H(exp(x)) = x^H(x)
suggests H(x) is not entire.

***

------
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] ...
------

Thanks for your intrest

regards

tommy1729
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