06/04/2011, 09:08 AM
(06/04/2011, 08:22 AM)bo198214 Wrote:(06/03/2011, 10:57 PM)mike3 Wrote:(06/02/2011, 02:04 PM)bo198214 Wrote: The kneser tetration \( b\mapsto b[4]p \) is real on the real axis \( b>\eta \), which implies that \( \overline{b} [4] p = \overline{b[4]p} \) (conjugation).
So approaching from above or below is just conjugate to each other.
So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at \( b<\eta \).
_Or_ show that it is not conjugate-symmetric there.
Maybe I was not insistent enough on that:
If we have a function \( f \) that is real-analytic on any interval (a,b) and you continue it through any path \( \gamma(t) \) in the upper halfplane.
Then for the conintuation in the lower halfplane \( \overline{\gamma} \) we have:
\( f(\overline{\gamma(t)}) = \overline{f(\gamma(t))} \),
i.e. in simple words: f is conjugate-symmetric *everywhere*
Correct. So showing that it is not conjugate-symmetric at all would seem to work, no? Thus if it behaves like the regular iteration at fixed point 2 in the upper halfplane, and like that at fixed point 4 in the lower halfplane, then it would seem it would not be conjugate-symmetric (note the difference in behaviors implied: the function would be bounded in the upper-right quadrant, while not so in the lower-right), thus not real-valued for real heights greater than -2. However, I wonder if the deviation from real-valuedness may be relatively small, which is why the pure regular "seems to work so well" for \( 1 < b < \eta \).

