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slog(superfactorial(x)) = ?

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slog(superfactorial(x)) = ?
tommy1729 Offline
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#3
11/17/2012, 11:47 PM
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 ...

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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