05/26/2022, 10:49 PM
Oh okay, Yes, I understand your concern now.
I actually had a rough outline of how it could work. This is a little tricky but I'll try my best.
Consider
\[
\alpha \uparrow^s z+1\\
\]
Assume it satisfies Ramanujan's bounds, so the "gamma interpolation" works. Now, here is the real kicker. We also need:
\[
\alpha \uparrow^{s-1} \left( \alpha \uparrow^s z\right)\\
\]
To be a holomorphic function for \(\Re(s) > A\) for some \(A\) and satisfy Ramanujan's bounds. Where they are defined similarly. IF (and that's a big IF) you can show this, then we're done.
This is because:
\[
\left(\alpha \uparrow^s z+1\right) - \left(\alpha \uparrow^{s-1} \left( \alpha \uparrow^s z\right)\right)\Big{|}_{s \in \mathbb{N}} = 0\\
\]
So Ramanujan pretty much takes care of everything, if you can show it's appropriately bounded. It's still a big if though, but I made a good amount of head way.
I actually had a rough outline of how it could work. This is a little tricky but I'll try my best.
Consider
\[
\alpha \uparrow^s z+1\\
\]
Assume it satisfies Ramanujan's bounds, so the "gamma interpolation" works. Now, here is the real kicker. We also need:
\[
\alpha \uparrow^{s-1} \left( \alpha \uparrow^s z\right)\\
\]
To be a holomorphic function for \(\Re(s) > A\) for some \(A\) and satisfy Ramanujan's bounds. Where they are defined similarly. IF (and that's a big IF) you can show this, then we're done.
This is because:
\[
\left(\alpha \uparrow^s z+1\right) - \left(\alpha \uparrow^{s-1} \left( \alpha \uparrow^s z\right)\right)\Big{|}_{s \in \mathbb{N}} = 0\\
\]
So Ramanujan pretty much takes care of everything, if you can show it's appropriately bounded. It's still a big if though, but I made a good amount of head way.