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Remark on Gottfried's "problem with an infinite product" power tower variation
#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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Messages In This Thread
RE: Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 05/06/2014, 09:47 PM

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