Deriving tetration from selfroot?

[Ivars]

@Ivars -
The role of the selfroot is essential. Nevertheless, it is important as a solution of the y = b[s]y functional equation. But, unfortunately (so to say) it is not its unique "functional root". Another solution can be found by considering y = b[s](b[s]y), or [/b]s-log(y) = y = b[s]y.

E.g. for rank 3 (exponentiation), we should consider (supposing base b) an expression such: [/b]log x = x = b^x. I. e. also other intersection points between the exp and the log, which will give other "branches" of the functional roots, are relevant. And, the same should be valid for other hyperops ranks.

GFR

HI Gianfranco,

But if You only look at infinite tetration for time being. Do we need those intermediary heights that add solutions?

x= y^(1/y) does mean y=x^y but does NOT mean y=x^x^y. because it does not include all solutions.

However, in infinite tetration, h( x^(1/x) ) does mean x, and h(1/x^x) does mean 1/x. The intermediates disappear. Are You sure they reappear in branches of W, ln as solutions for infinite tetration?

Or am I wrong again

Ivars