Caleb! You are absolutely correct! I'm not trying to show my hand too much right now; but this is what's so confusing!
\[
p(z) = \sum_{k=0}^\infty p_k z^k\\
\]
And let's assume that:
\[
b_k = \sum_{ d \vert k} (-1)^{k/d} p_d \mu(k/d)
\]
Then \(p(z)\) is also written as:
\[
p(z) = \sum_{k=0}^\infty b_k \frac{z^k}{1+z^k}\\
\]
BUT! this tells us nothing about \(p(z)\) for \(|z| > 1\)... Just take \(p(z)\) to be a polynomial and you can identify this.
My theory is evolving constantly; but we have to make two additional claims to the above statement. We must ensure that \(p(z)\) only has poles at \(q^n = -1\); and we must use a generalized Lambert series...
If I write:
\[
p_k = \sum_{d \vert k} H(d, M(k/d)) P(k/d)\\
\]
Where:
\[
H(d,m) = \frac{(-1)^{m+1}}{(m-1)!} \prod_{i=0}^{m-2} (d+i)\\
\]
Then the expression \(p(z)\) is just:
\[
p(z) = \sum_{n=0}^\infty \frac{P(n)z^n}{(1+z^n)^{M(n)}}\\
\]
BUT NOW THE REFLECTION FORMULA HOLDS!
The trouble becomes; how do we find \(P(n)\) and \(M(n)\), as arithmetic functions, solely using the function \(p(z)\) and it's taylor coefficients \(p_k\)???? Ironically enough; this is solvable for polynomials... And the "generalized Lambert series" is just a polynomial. Which satisfies the reflection.
For example, take \(p(z) = z\). Then \(P(1) = 1\) and \(P(n) = 0\) everywhere else; and \(M(1) = 0\) and \(M(n)\) is arbitrary...
Voila! We've done our reflection formula.... \(M(n) : \mathbb{N} \to \mathbb{N}\) and \(P(n) : \mathbb{N} \to \mathbb{C}\); so there really is no problem. And the mechanics I use to write the reflection formula still follow; even if we choose the trivial \(M(n) = 0\)--but it can be a bit of a fucking headache to write everything perfectly....
I'm excited to release my program soon... I'm still developing some of the ideas; but finding a flawless reflection for arbitrary functions is appearing clearer and clearer.
EDIT:
I'm going to explain the polynomial interpolation; because this has to do with extending near the boundary. Additionally it is something you have mentioned.
Let's define:
\[
\begin{align}
P(1) &= 1\\
P(n) &= \epsilon\,\,\text{if}\,\, n \ge 0\,\,n\neq 1\\
M(1) &= 0\\
\end{align}
\]
Then the function:
\[
\sum_{n=0}^\infty \frac{P(n)z^n}{\left(1+z^n\right)^{M(n)}} = z + O(\epsilon)\\
\]
We can do this for all polynomials. And my number theoretic tricks will solve this generally...
The reflection formula becomes:
\[
L(z) = 1/\left(1/z + O(\epsilon)\right) = z + O(\epsilon)\\
\]
\[
p(z) = \sum_{k=0}^\infty p_k z^k\\
\]
And let's assume that:
\[
b_k = \sum_{ d \vert k} (-1)^{k/d} p_d \mu(k/d)
\]
Then \(p(z)\) is also written as:
\[
p(z) = \sum_{k=0}^\infty b_k \frac{z^k}{1+z^k}\\
\]
BUT! this tells us nothing about \(p(z)\) for \(|z| > 1\)... Just take \(p(z)\) to be a polynomial and you can identify this.
My theory is evolving constantly; but we have to make two additional claims to the above statement. We must ensure that \(p(z)\) only has poles at \(q^n = -1\); and we must use a generalized Lambert series...
If I write:
\[
p_k = \sum_{d \vert k} H(d, M(k/d)) P(k/d)\\
\]
Where:
\[
H(d,m) = \frac{(-1)^{m+1}}{(m-1)!} \prod_{i=0}^{m-2} (d+i)\\
\]
Then the expression \(p(z)\) is just:
\[
p(z) = \sum_{n=0}^\infty \frac{P(n)z^n}{(1+z^n)^{M(n)}}\\
\]
BUT NOW THE REFLECTION FORMULA HOLDS!
The trouble becomes; how do we find \(P(n)\) and \(M(n)\), as arithmetic functions, solely using the function \(p(z)\) and it's taylor coefficients \(p_k\)???? Ironically enough; this is solvable for polynomials... And the "generalized Lambert series" is just a polynomial. Which satisfies the reflection.
For example, take \(p(z) = z\). Then \(P(1) = 1\) and \(P(n) = 0\) everywhere else; and \(M(1) = 0\) and \(M(n)\) is arbitrary...
Voila! We've done our reflection formula.... \(M(n) : \mathbb{N} \to \mathbb{N}\) and \(P(n) : \mathbb{N} \to \mathbb{C}\); so there really is no problem. And the mechanics I use to write the reflection formula still follow; even if we choose the trivial \(M(n) = 0\)--but it can be a bit of a fucking headache to write everything perfectly....
I'm excited to release my program soon... I'm still developing some of the ideas; but finding a flawless reflection for arbitrary functions is appearing clearer and clearer.
EDIT:
I'm going to explain the polynomial interpolation; because this has to do with extending near the boundary. Additionally it is something you have mentioned.
Let's define:
\[
\begin{align}
P(1) &= 1\\
P(n) &= \epsilon\,\,\text{if}\,\, n \ge 0\,\,n\neq 1\\
M(1) &= 0\\
\end{align}
\]
Then the function:
\[
\sum_{n=0}^\infty \frac{P(n)z^n}{\left(1+z^n\right)^{M(n)}} = z + O(\epsilon)\\
\]
We can do this for all polynomials. And my number theoretic tricks will solve this generally...
The reflection formula becomes:
\[
L(z) = 1/\left(1/z + O(\epsilon)\right) = z + O(\epsilon)\\
\]
