Divergent Series and Analytical Continuation (LONG post)

[JmsNxn]

This is mainly aimed at Caleb because it might be new to him , but ofcourse to all people


Although it is not a natural boundary at all I can not help think about this 

pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)...

where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there.

But pxp(z) is actually just exp(z).

This is a product expansion for exp(z) valid for |z| < 1.

I will put the pxp paper by Gottfried in an attachement.

Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation :

exp(- x) = 1/exp(x)

hence

pxp(- z) = 1/pxp(z).

 
Then again it does not map the inside to the outside of the boundary and vice versa.

Just a comment ...

And a reminder that we have not even considered products yet.

I haven't posted much on products yet; but I have considered them in my work upto now. Essentially; from my perspective; this becomes a problem of where there are FINITE singularities along \(|z| = 1\). Where the finite number happens to be zero.

A large part of what I am trying to do is to extend products with natural boundaries to a larger domain by considering a reflection formula that is universal.

So when you write:

\[
\text{pxp}(-z) = \prod_{n=0}^\infty \left(1+ a_n z^n\right)^{-1}\\
\]

I want to use the residues (which are zero in this case) at the poles, to construct the Taylor series or the Laurent Series; or both of them as one. And in order to do that, I want to write them as a Lambert series first. Another point to observe is that if:

\[
\frac{(-1)^k}{k!} = \sum_{d | k} b_d\\
\]

Where then:

\[
b_n = \sum_{k|n} \mu(k/n) \frac{(-1)^k}{k!}\\
\]

Then:

\[
\exp(-z) = \sum_{n=0}^\infty b_n \frac{z^n}{1-z^n}\\
\]

Things get more complicated though; once we allow for second order poles; rather than just first order poles. This switch becomes more intricate. Not to mention, when we allow arbitrary order poles on the boundary.

Now, we can still express it as a Lambert series; but it hides the actual complexity a good amount. My goal so far; has been to relate the sequence \(b_n\), To the values at \(q^n = -1\):

\[
L(z) = \sum_{j=1}^m \frac{r_j}{(z-q)^j} + O(1)\\
\]

This has been unbelievably effective! But the key is to include more general Lambert series; so that we can talk about arbitrary order poles.

For example; let's take Euler's series:

\[
E(z) = \sum_{n=0}^\infty \#n z^n = \prod_{n=1}^\infty \frac{1}{(1-z^n)}\\
\]

with the partition function \(\# n\). It's been long known that the residues; and singular part of the poles at \(z^n = 1\) create an exact equality to \(\# n\)--a la circle method. I'm basically trying to reprove this same result; but in turn, by trying to reprove it, I am constructing \(E(1/z)\).

Ultimately the goal is: If you give me the singular part of \(L(z)\) at each \(q^n = -1\), I can construct a generalized Lambert series, and vice versa. The amount of number theory that I'm using is minimal for the moment; but I am seeing huge applications to Caleb's original idea of "analytically continuing" a modular function.

I'm still a little spooked by this. And I'm not trying to overstep the deepness of this. But this is looking really really fucking cool!   There is definitely something very deep going on behind the scenes!

I'm sticking for the moment to the case where the singularities happen at \(q^n = -1\) (Rather than \(q^n = 1\) because it makes the "gateway" at \(z= 1\)); but this can be altered to \(q^n = e^{i\theta}\) with little change to the fundamentals. I do not think this will generalize to when the singularities aren't measure zero on the boundary. I think the singularities need to be measure zero... Additionally, I'm only working on rationally dense sequences \(q^n = e^{i\theta}\)...

I'm still wishy washy though. I'm at page 15 of a rough draft; and I'm only starting to get to the deep stuff!