(07/09/2010, 05:32 AM)bo198214 Wrote:(07/03/2010, 09:42 AM)mike3 Wrote: What do you think of this function? Especially the similarity between the shape of its graph and that of the selfroot. One could almost imagine a continuous spectrum of similar functions in between them -- "fractional-rank hyperoperations", anyone?
That indeed looks interesting (Andrew posted this self tetra root already somewhere on the forum and I proved somewhere that the limit is indeed e^(1/e)).
However I would expect to have a functional equation for the tetra self root.
As well as to know how you derive tetration from the tetra self root.
Yes, a functional equation would be good, but I'm not sure of one. As for deriving tetration, you can't really, except that evaluating this at some fractional number will give you the value of a specific base \( b \in [1, K) \) (where \( K = 1.6353... \)) -- i.e. the value of the function, tetrated to a certain tower, i.e. the point of evaluation, and the value of such tetration, also the point of evaluation (this means the point of evaluation is a fixed point of the tetrational of the base given by the result of evaluation). For bases \( b > L \), where \( L \) is the limit of the function at \( +\infty \) (which you are saying is \( e^{1/e} \) -- if that's so then it must converge to it at a hideously slow rate: do you know what the big-\( O \) of \( |\mathrm{selftetroot}(n) - e^{1/e}| \) is as \( n \rightarrow \infty \)?), this procedure yields fractional tetration values at two points.
If the fractional tetration values this yields agree with regular iteration for \( b \in (1, e^{1/e}] \), that may suggest that regular iteration does not have a natural boundary and can be analytically continued to the wider complex plane. If not, then the regular may still have such a boundary, and I'd be curious what this "superior-to-regular-iteration" solution for \( b \in (1, e^{1/e}] \) is that it is "pointing at" with the values for the fractional towers that are fixed points.
