04/07/2009, 01:35 AM
(This post was last modified: 04/07/2009, 02:00 AM by nuninho1980.)
bo198214 Wrote: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.slog_b (x) = x <=> b ^^ x = x => b ^^^ oo = x
Of course there maybe always the dependency of the values from the chosen method of tetration.
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?
bo198214 Wrote: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