tetration limit ??

[nuninho1980]

Oh, you mean we have an upper fixed point of the tetrational for \( b\le 1.6353244967 \) and the fixed point can then be computed by \( \operatorname{slog}_b(x)=x \) or \( {^x b} = x \). Ya, interesting. I dont know whether we even have a thread on the forum that dealt with the topic of the fixed point of tetrationals.
Of course there maybe always the dependency of the values from the chosen method of tetration.

slog_b (x) = x <=> b ^^ x = x => b ^^^ oo = x
yeah!

to remember:
one - 1^oo = 1
Euler - (e^(1/e)) ^^ oo = e

now new fixed point
(1.63532...) ^^^ oo ~= (3.08855...) what is this new result? it's a Super-Euler?

So how big is the difference between both methods, with respect to the computed fixed point?

to remember - you follow "regular slog" - http://en.wikipedia.org/wiki/Talk:Tetrat...on_methods lol