(11/06/2009, 04:15 AM)mike3 Wrote: I've also been toying with this, too. It appears, however, that it continues to real values, not complex values, for \( b > e^{1/e} \).
that would be awesome. And this would mean that there is no singularity at \( b=e^{1/e} \)?
Quote:For \( z = 1.5 = \frac{3}{2} \), we can use this get \( ^{\frac{1}{2}} \left(\frac{3}{2}\right) \approx 1.28087727794 \), which is real, not complex.
with "use this" you mean the powerseries development you just derived at \( z_0 \)? But how do you know that it converges and that there is no branchpoint at \( z=e^{1/e} \)?
Quote:I'm not sure of a formal proof of the "continuability", though one approach may be to try and differentiate the regular iteration formula, then prove that the limit of the derivative as \( b \rightarrow e^{1/e} \) converges -- in order for it to switch to non-real complex values as \( b = e^{1/e} \) is passed, that point would have to be some sort of singularity, like a branch point, and so the function would not be differentiable there, and if it is, then that is not the case.
*nods* but at least it is already known that the regular iteration \( f(z)=\exp_{e^{1/e}}^{\circ t}(z) \) is not analytic at \( z=e \). However it is currently not clear to me what this states about the regularity of \( f(z)=\exp_z^{\circ t}(1) \) at \( z=e^{1/e} \).
Quote:I'll see if maybe I can get some graphs on the complex plane but calculating the regular iteration is a bear as it requires lots of numerical precision, at least for the limit formula. Maybe that series formula would be better?
Ya I will try it with the series formula (or perhaps a mixture with limit formulas).
Actually it seems that no-one posted pictures of tetra-powers yet!!!
(yes Bat I mean tetra-powers.)
So it will be time that we have some pictures at least, as the theoretic consideration seems utmost complicated to me.
