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slog(superfactorial(x)) = ? - Printable Version

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slog(superfactorial(x)) = ? - tommy1729 - 11/14/2012

Define the superfactorial as superfactorial(x) = super!(x) and
super!(x+1) = factorial(super!(x)) = super!(x)!

Where the factorial is computed from the gamma function and super!(-oo) = 2 and super!(0) = 3.

Now I wonder how slog(super!(x)) looks like for large real x ?
We know from basechange and the fact that gamma grows faster than exp that slog(super!(x)) must be strictly increasing for x > y for some real y.

I wonder how fast this is.
Could it be O(x/ln(x)) ? or O(ln(x) sqrt(x)) ?
I have no theoretical reasons to assume anything apart from lim sup slog(super!(x)) < 2x.

I could make up some arguments for this or that based upon asymptotics of gamma , but formal analysis instead of dubious arguments seems not easy.

There are superfunctions known of the factorial but we only need integer iterations for the general behaviour ... on the other hand maybe the other values are needed in some proof ...

I wonder what you guys think.

I was thinking about a plot , but im not sure how we do this since superfactorial grows faster than tetration !!

Maybe this requires some sort of ' simplify ' ( as math apps and books call it ) *command* towards slog(super!(x)) for which i have no idea how to do it.

It looks a bit like basechange , but to me it looks harder.

I think this is intresting because it might learn us alot about iteration in general and slog in particular.



RE: slog(superfactorial(x)) = ? - sheldonison - 11/15/2012

(11/14/2012, 08:30 PM)tommy1729 Wrote: Define the superfactorial as superfactorial(x) = super!(x) and
super!(x+1) = factorial(super!(x)) = super!(x)!
Hey Tommy,
Interesting question, I'll post more later. The fixed point=2, super! has a well defined schroder and inverse schroder function. I think it comes down to how big is and it seems like its not going to be that much bigger than . So should converge to a constant for integer values of n. I would expect a constant plus a 1-cyclic function real values of n.
updateThe , for large enough x, the log(x) multiplier term becomes arbitrarily accurate. .

The important thing is that slog((x+1)log(x)), if x is big enough, approaches arbitrarily close slog(x), so slog(super!(n))-n approaches a constant as n increases.
- Sheldon

RE: slog(superfactorial(x)) = ? - tommy1729 - 11/17/2012

Lim x-> +oo slog(super!(x)) < x(1+eps).

The reason is gamma(x) grows slower than exp(sexp(slog(x)+eps)).

Thus Lim x-> +oo [slog(super!(x))-x]/x = 0

However statements as slog(super!(x)) = x + O(ln(x)) are still mysterious.

I guess slog(super!(x)) = x + O(sqrt(slog(x log(x)))).

Although my guess looks strong , sheldon's statement is even stronger : slog(super!(x)) = x + O(1).

In other words , Im not convinced yet ...



RE: slog(superfactorial(x)) = ? - tommy1729 - 06/02/2014

Sheldon is right.

This follows from the " growth rate " insights recently discovered (posts around may 2014).

I agreed with him for much longer but forgot to reply that.

So for completeness and avoiding confusion , my reply now.