Divergent Series and Analytical Continuation (LONG post)

[Caleb]

From \(t =0\) to \(t = 2 \pi\)... Then \(J(e^{it})\) graphs as:



This looks just like a very simple modular kind of function. We only need a single frequency to get it.
I suspect this is because it's only simple poles on the wall. If we have arbitrary order poles; we will get more complication.

Please remember the entire point is that:

\[
f(0,e^{it}) - f(J(e^{it})x,e^{it}) = \sum_{k=1}^{\infty} F_{-k}^{t}x^k\\
\]

Recalling that:

\[
F_{-k}^t = \frac{1}{2\pi i} \int_{|z| =2} f(z,e^{it})z^{k-1}\,dz\\
\]

To remind everyone:

\[
f(x,e^{it}) = \sum_{n=0}^\infty \frac{x^n}{1+e^{itn}x^n} \frac{1}{2^n}\\\
\]

I haven't read through all that you've written yet (that will take some time-- I'd like to think about all your points in detail). However, I can perhaps provide some reading if you aren't already aware of these things.

First, on your point that it looks modular, you can check out: the wiki page on lambert series. Part of my motivation for studying the series here is that they have really nice arithmetic properties. 

Second, to your point about it looking like some fourier series decomposition and the connection between \( \frac{f^{(-k)}}{(-k)!} \) and \(b_{-k}\), I have this MO post, which discusses the relationship between \( \sum a_n x^n\) and \( \sum a_{-n} x^{-n} \). Basically, I expect that these two series are always equal, up to the residues of \( a_n \) when we view \( a_n \) as a complex-analytic function defined on the plane. Since \( a_n = f^{(n)}{n!} \), then we should expect that \( \sum \frac{f^{(n)}}{n!} x^n  = -\sum \frac{f^{(-n)}}{(-n)!} x^{-n} \)

Also, a bit of a tangent, but its probably quite intersting to study the series
\[\sum \frac{x^p}{1-x^p}\]
where \(p\) is prime. If you view this as an arthemtic function, it becomes \( \sum \omega(n) x^n \) where \( \omega(n) \) counts the number of primes that divide a given n. Thus, if one found a different way to continue the series (besides just evaulting the series directly) it would provide a way to basically count the number of prime factors without actually computing the primes. The series also has the really nice property that all of its poles are 'unique' in some way, in the sense that the poles of different terms \( \frac{1}{1-x^p} \) never intersect, since to intersect they would have to share a prime factor (like, \( \frac{1}{1-x^2} \) and \( \frac{1}{1-x^4} \) have some of their poles intersect together).