tetration limit ??

[bo198214]

now the fixed point more smooth
1.6353244967 ^^^ oo ~= 3.0885549441 (200 dimensions of matrix)

b ^^^ oo = x => slog_b (y) = x, y=x, if b<=1.6353244967 and b=/=1.6353244968 then x<y and x>y.

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.

"tetration and slog" original by Andrew Robbins is as smoother as "new regular slog"

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