Okay, I'll stop blowing up this thread after this last identification.
Let:
\[
L(z) : \mathbb{C}/ U \to \mathbb{C}\\
\]
Where \(U\) is the unit circle. Let's assume that:
\[
\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}} = 0\\
\]
For all \(R > 1\), and for all \(k > 0\). Then call the sum:
\[
F(z) = -\sum_{k\in \mathbb{Z}} z^{-k}\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}}\\
\]
This just reduces to:
\[
F(z) = -A - \sum_{k=1}^\infty \frac{1}{2\pi i}\int_{|\zeta| = R} L(\zeta)z^k \zeta^{k-1}\,d\zeta\\
\]
Which is just:
\[
F(z) = -A + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]
The conjecture by Caleb; is that:
\[
F(z) = L(z) + C\\
\]
Where the constant \(C\) is easily discoverable. Where it can be written more compactly as:
\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]
For all \(|z| < 1\). This actually creates a reflection formula between \(L(1/z)\) and \(L(z)\). Where we can simply evaluate this integral as:
\[
\frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\ = L(1/z) + A\\
\]
Where:
\[
A = \frac{1}{2\pi i} \int_{|z| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]
Because:
\[
\frac{L(\zeta)}{\zeta - \frac{1}{z}} = L(\zeta) \sum_{k=1}^\infty z^{1-k}\zeta^{-k}
\]
When \(|z\zeta| > 1\); which is always true if \(|\zeta| = R\) is large enough. And the only non-zero term, is when \(k = 1\).
I HAVE CONFIRMED THIS NUMERICALLY FOR ALL GENERALIZED LAMBERT FUNCTIONS!!!!!!!
I think this might be confirmation of a much much deeper identity; but we have to prove Caleb's conjecture first. Which is that:
\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta-\frac{1}{z}}\,d\zeta\\
\]
For all \(1 < \left|\frac{1}{z}\right| < R\), \(\frac{1}{R} < |z| < 1\).
This would intimately relate both functions \(L(z)\) and \(L(1/z)\) that I can't even believe is possible. This is like striking straight up gold!
BEAUTIFUL INTUITION, CALEB! This is a reflection formula for the books! And I think it'll go a long way in doing what you want to do! But do it without needing to fall back on Fourier Analysis/ Ramanujan shit!
Also, for you modular nerds! If \(L(0) = 0\), then we have shown that:
\[
L(1/z) = L(z) + A\\
\]
If we also have that \(A =0\)... then \(L(1/z) = L(z)\)! So if we assume that \(L(\infty) =0\) and \(L(0) = 0\); then these functions are carbon copies.
Which is the first step in identifying a modular relationship. The next would be a periodic condition... But I can imagine this as "analytically continuing modular functions"
YOU ARE ONTO SOMETHING HERE CALEB!!!!!!!!!!!!!!!!
Sincere regards, SUPER PUMPED, James
Let:
\[
L(z) : \mathbb{C}/ U \to \mathbb{C}\\
\]
Where \(U\) is the unit circle. Let's assume that:
\[
\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}} = 0\\
\]
For all \(R > 1\), and for all \(k > 0\). Then call the sum:
\[
F(z) = -\sum_{k\in \mathbb{Z}} z^{-k}\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}}\\
\]
This just reduces to:
\[
F(z) = -A - \sum_{k=1}^\infty \frac{1}{2\pi i}\int_{|\zeta| = R} L(\zeta)z^k \zeta^{k-1}\,d\zeta\\
\]
Which is just:
\[
F(z) = -A + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]
The conjecture by Caleb; is that:
\[
F(z) = L(z) + C\\
\]
Where the constant \(C\) is easily discoverable. Where it can be written more compactly as:
\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]
For all \(|z| < 1\). This actually creates a reflection formula between \(L(1/z)\) and \(L(z)\). Where we can simply evaluate this integral as:
\[
\frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\ = L(1/z) + A\\
\]
Where:
\[
A = \frac{1}{2\pi i} \int_{|z| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]
Because:
\[
\frac{L(\zeta)}{\zeta - \frac{1}{z}} = L(\zeta) \sum_{k=1}^\infty z^{1-k}\zeta^{-k}
\]
When \(|z\zeta| > 1\); which is always true if \(|\zeta| = R\) is large enough. And the only non-zero term, is when \(k = 1\).
I HAVE CONFIRMED THIS NUMERICALLY FOR ALL GENERALIZED LAMBERT FUNCTIONS!!!!!!!
I think this might be confirmation of a much much deeper identity; but we have to prove Caleb's conjecture first. Which is that:
\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta-\frac{1}{z}}\,d\zeta\\
\]
For all \(1 < \left|\frac{1}{z}\right| < R\), \(\frac{1}{R} < |z| < 1\).
This would intimately relate both functions \(L(z)\) and \(L(1/z)\) that I can't even believe is possible. This is like striking straight up gold!
BEAUTIFUL INTUITION, CALEB! This is a reflection formula for the books! And I think it'll go a long way in doing what you want to do! But do it without needing to fall back on Fourier Analysis/ Ramanujan shit!
Also, for you modular nerds! If \(L(0) = 0\), then we have shown that:
\[
L(1/z) = L(z) + A\\
\]
If we also have that \(A =0\)... then \(L(1/z) = L(z)\)! So if we assume that \(L(\infty) =0\) and \(L(0) = 0\); then these functions are carbon copies.
Which is the first step in identifying a modular relationship. The next would be a periodic condition... But I can imagine this as "analytically continuing modular functions"
YOU ARE ONTO SOMETHING HERE CALEB!!!!!!!!!!!!!!!!
Sincere regards, SUPER PUMPED, James
