i wanted to add that z = 0 or z = 1 are probably the only solutions such that
z^^z = 1
that means apart from those 2 branches there are no other branches for f(1) unless ALL branches intersect at one of those 2 branches of f(1).
the taylor series at that point , if existing , is thus either 1 + az + bz^2 + ... or (z-1) ( a' + b' z + c' z^2 + ...)
i find that fascinating.
also intresting is the question what is z^^z a superfunction of ?
you could define it as (f(z) + 1)^^(f(z) + 1).
and one could wonder about its fixed points.
furthermore , is z^^z actually garantueed to converge for all complex z ??
and of course it might depend alot on what kind of tetration we work with ...
tommy1729
z^^z = 1
that means apart from those 2 branches there are no other branches for f(1) unless ALL branches intersect at one of those 2 branches of f(1).
the taylor series at that point , if existing , is thus either 1 + az + bz^2 + ... or (z-1) ( a' + b' z + c' z^2 + ...)
i find that fascinating.
also intresting is the question what is z^^z a superfunction of ?
you could define it as (f(z) + 1)^^(f(z) + 1).
and one could wonder about its fixed points.
furthermore , is z^^z actually garantueed to converge for all complex z ??
and of course it might depend alot on what kind of tetration we work with ...
tommy1729