Divergent Series and Analytical Continuation (LONG post)

[JmsNxn]

I apoligise for the late response, I should have mentioned this earlier- but I belive we have only stubled upon a trivial property  . In particular, its not that \( L(z) \) and \( L(1/z) \) are related in some interesting way-- they are actually just the same evaulation. In paritcular, notice that
\[ f(x) = \sum \frac{x^n}{1+x^n} a_n \]
Using the geometric series formula for \( |x|<1 \) gives
\[ f(x) = \sum_n a_n  \sum_k x^{n(k+1)}\]
Now take a look at 
\[f(1/x) = \sum \frac{(\frac{1}{x})^n}{1+(\frac{1}{x})^n} a_n = \sum a_n \frac{1}{x^n +1} \]
But now, for \(|x|<1\), we have that 
\[\sum a_n \frac{1}{x^n +1} = \sum_n a_n \sum_{k=0}^\infty (x)^{nk}\]
But now the only missing term is \( k=0\), so 
\[f(1/x) - f(x) = \sum_n a_n \sum_{k=0}^0 = \sum_n a_n \]
So, unforunately I don't think anything interseting is actually going on here, at least for this relationship.

I appreciate your candor, Caleb! This is definitely a trivial Fourier type identity. The point I am trying to make is this follows without this expansion! And that is definitely non-trivial!!!!

So, this result follows pretty trivially if we follow the coefficients on a generalized Lambert function. The real trouble would be to generalize this to arbitrary functions:

\[
\begin{align}
L(z) &: \mathbb{C} / U \to \mathbb{C}\\
L(z) &= L(0) + \frac{1}{2\pi i} \int_{|\zeta| =R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\end{align}
\]

Where we are given, if:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d \zeta}{\zeta^{k-1}} = 0\\
\]

For \( k \ge 1\); then:

\[
L(1/z) = L(z) - L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

This is HIGHLY non-trivial!

Yes for the Lambert function, this reduces to some fourier rearrangements and is trivial. But When we write this as arbitrary functions, this is absolutely not trivial!

This is a fucking crazy result, Caleb!

It is a generalized function result; for functions holomorphic on \(\mathbb{D}\) and on \(\mathbb{C}/\overline{\mathbb{D}}\). If they satisfy the above integral condition, they must satisfy this reflection formula.

For things like Lambert's functions, it is very obvious; just by rearranging the series. It is not obvious for general functions. We have gold here, bro

For example; if we have \(L(z) : \mathbb{D} \to \mathbb{C}\) with a natural boundary on \(|z| =1 \). There exists a function \(G(z) : \mathbb{C}/\mathbb{D} \to \mathbb{C}\) such that:

\[
G(z) = L(1/z) + C\\
\]

For a constant \(C\). This constant can be chosen uniquely such that:

\[
\int_{|\zeta| = R} \frac{G(\zeta)}{\zeta^{k+1}}\,d\zeta = 0\\
\]

For all \(k \ge 1\). And this gives our analytic continuation!!!!!!!!

Usually, finding \(C\) will be very hard; and in most cases won't exist. But if it does exist; we have continued a function \(L:\mathbb{D} \to\mathbb{D}\) to \(L:\mathbb{C}/U \to \mathbb{C}\). This is super fucking coool111