Divergent Series and Analytical Continuation (LONG post)

[Caleb]

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

OOHH I see, your right that this is much more interesting for arbitrary functions! 

So... a simple condition on the fourier coefficents, implies a strong coherence between the two components? That IS insane! I really have to go back and read all the work you've done so far-- I've been busy so I'm far behind on all the progress you've made but soon I'll have time to read it.

Anyway, it looks like you've done great work so far I'm excited to read up on all you've done! This is very exciting stuff! Also, my recent post on micks function makes me super super curious about what a function like his looks like outside its natural boundary, it will be interseting to test it out in that case.