Infinite tetration and superroot of infinitesimal
#97
I agree that it has a real limit in the Eta-Beta range [Eta = e^(1/e), Beta = e^(-e) !!]. Nevertheless, we must admit that there is also another real upper branch. Think of the famous example of b = sqrt(2). We have two heights h(sqrt(2)) = {2, 4}.
In fact, by applying b = h^(1/h), we have: 2^(1/2) = 4^(1/4) = sqrt(2).

The lower branch is reachable starting from points (-1,0) and (0,1) by applying b#(x+1) = b^(b#x). The upper branch is unreachablke (big mystery), but still it satisfies the same conditions. Always in the case of b = sqrt(1/2), if the upper branch is 4, we have that:
if: sqrt(2) # x = 4,
then: sqrt(2) ^ (sqrt(2) # x) = sqrt(2) ^ 4 = 4 , for any "x" !!
I mean that the upper branch defines (according to my point of view) a second constant real numerical value in y = b^x, in the 1-to-Eta domain of b.

I might be wrong. But, I still believe it. We always have the same problems with two-valued "functions", one of which, in this particular case, should have a "constant" value. I think we already discussed about the example y = x^2. A lot of orthodox mathematicians say that it is not inversible (!?!), and that sqrt(4) = 2, like all the pocket calculators indicate. Neverthaless, we know that sqrt(4) = {-2, +2}.

This time, I go and drink a "koke", or a "cola", as you say in Germany.

GFR
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