Pictures of some generalized analytical continuations

[JmsNxn]

that last picture seems to have an analytic opening ? 

truncated result or optical illusion ?

regards

tommy1729

That opening is a bit of an optical illusion. What's happening is that at \(z=1\) this function is essentially infinitely differentiable. So it is \(\mathcal{C}^\infty\) here. This causes a kind of regularity near here. It is still absolutely not holomorphic here, or near here. But the singularities are measure zero on the unit disk; so because it is smooth at this point; it looks like things are "cancelling out" and the poles are almost negligible.

It's definitely not a truncation error because I took 50 digit precision, and 100 terms to the sum:

\[
L(z) = \sum_{n=0}^{100} \frac{z^n}{(1+z^n)^n} \frac{1}{2^n}\\
\]

So any truncation error would be absolutely invisible to the naked eye. My hypothesis is the fact that this function is smooth at \(1\). There's probably an asymptotic series here; which I'm sure you could figure out; doesn't look too hard to manipulate out.

Also additionally, the values \(|q-1| < \delta\) is small and \(q^n = -1\) implies that \(n\) is very very large (how well we are rationally approximating \(\pi\)); which then implies that it is a miniscule term that is added to the value. I think all functions:

\[
L(z) = \sum_{n=0}^\infty P(n) \frac{z^n}{(1+z^n)^{M(n)}}\\
\]

Should have a similar result near \(1\)--a kind of "gateway" near \(z= 1\). Provided that:

\[
\limsup_{n\to\infty} |P(n)|^{1/n} \le 1\\
\]

But the "gateway" can look worse or better depending on how fast \(M(n)\) and \(P(n)\) grows.

I'm not certain this is the reasoning; but it's my best guess at the moment

So, to be to the point. There are still a bunch of poles near this gateway; they are just \(\frac{1}{2^N}* \frac{1}{z}\) where \(N\) is so ridiculously large, that the naked eye can't see the pole...