(08/20/2010, 05:08 PM)sheldonison Wrote: When I was working with eta, I noticed the simmilarity between \( \text{sexpUpper}_\eta(z) \) and \( \text{sexpLower}_\eta(z) \). As I remember, in the complex plane, sexpLower(z) looked a lot like sexpUpper(z+i), especially as imag(i) increased. I'll try to dig up an old contour graph.... Anyway, there are two solutions at base eta, one, the regular superfunction developed from the fixed point of "e", that goes to infinity as real(z) increases. This is the sexpUpper(z) function which has no singularities, and real(z)>e for all z at the real axis. The sexpLower(z) function of eta has a singularity at sexp(z,z=-2), and approaches "e" as real(z) increases to infinity. What you are pointing out is that the sexpLower(z) is a conformal/Kneser map of the sexpUpper(z) which is pretty cool, since it can also be developed from the attracting fixed point at +real infinity=e. Neat!
The two functions \( \check{\eta}(z) \) (what you call "SexpUpper") and \( ^z \eta \) have similar asymptotic behavior, but along the real line, \( \check{\eta}(z) \) grows toward infinity, and is chaotically-behaved (shows the "fractal structure") in a half-strip containing that line. It's quite a weird function, and because of the tetrational growth I don't see how Carlson's theorem implies anything about it. How does this show the \( ^z \eta \) is a Kneser map of \( \check{\eta}(z) \)?
Quote:I wonder if Carlson's theorem can be used to say anything about bases >eta?
- Sheldon
Probably not. Those bases diverge, so the tetrational \( ^z b \) is not exponentially bounded in the right half-plane.