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slog(superfactorial(x)) = ?
#1
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.

regards

tommy1729
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Messages In This Thread
slog(superfactorial(x)) = ? - by tommy1729 - 11/14/2012, 08:30 PM
RE: slog(superfactorial(x)) = ? - by sheldonison - 11/15/2012, 06:34 PM
RE: slog(superfactorial(x)) = ? - by tommy1729 - 11/17/2012, 11:47 PM
RE: slog(superfactorial(x)) = ? - by tommy1729 - 06/02/2014, 11:29 PM

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