(06/16/2010, 11:48 AM)bo198214 Wrote: Hm, I think that the intuitive method converges for bases \( b>1 \) does not necessarily mean that it is holomorphic in b (in the trumpet area). Why cant there be an invisible boundary at which it is merely continuous and not holomorphic? I really dont know, but do you have some more evidence that the intuitive iteration is holomorphic in b?
Hmm. I dug up this:
http://www.math.wustl.edu/~sk/limits.pdf
It says that if a function converges pointwise in a region, the convergent need only be holomorphic on a dense, open subset. The region without its boundary and with a slit in it that cuts it into two pieces would still be a dense, open subset, dense since even the points in the slit have points in the set "near" them (namely, "right next to" them, i.e. in any neighborhood no matter how small), and open because it's the union of two open sets (those on each side of the slit). So it could conceivably be holomorphic on all but a slit. But there's no proof, or proof of the opposite...
Another theorem is if it converges uniformly on compact subsets of the region (e.g. consider closed disks \( D \subset \Omega \), where \( \Omega \) is the region), the convergent is holomorphic on the whole region, also called "compact convergence". This is the manner in which a Taylor series converges. Compact convergence is stronger than just convergence, and more difficult to prove. (Is the AM even proven convergent?) I suppose one would need an closed-form-for-the-terms formula for the AM convergents, no (i.e. for the inverses of each partial matrix), to analyze this problem and determine the character of the convergence (and also, perhaps, the precise region, too...)?
Failing that, could examining the derivatives on the part of the STB where it converges be useful? If the convergent is not holomorphic on the STB, then wouldn't these derivatives explode, go nuts, etc.?
