06/27/2022, 11:57 PM

That's a very deep question, Catullus. I honestly don't know the answer to it.

But to point out, when you use \(1 \le \alpha \le \eta\), you have a very fine growth in the hyper-operations. In that, they tend to \(\alpha\) for \(\Re(z) > 0\). So doing this in this interval is about the closest I ever got. Where, as I've written before, I'll write again.

$$

\vartheta(w,u) = \sum_{n=0}^\infty\sum_{k=0}^\infty \alpha \uparrow^{n+2} (k+1)\frac{u^nw^k}{n!k!}\\

$$

If we differentiate repeatedly in \(w\), we get:

$$

\frac{d^j}{dw^j} \vartheta(w,u) \approx e^w\sum_{n=0}^\infty \omega_{n+2} \frac{u^n}{n!}\\

$$

Where \(\omega_{n+2} = \alpha \uparrow^{n+2} \infty\), which are the fixed points of \(\alpha \uparrow^{n+1} z\). And if we differentiate repeatedly in \(u\), we get:

$$

\frac{d^l}{du^l} \vartheta(w,u) \approx \alpha e^{w}e^{u}\\

$$

Now as asymptotic limits, both of these things are differintegrable. Whereby:

$$

\frac{d^{s}}{du^{s}} \frac{d^{z}}{dw^z}\Big{|}_{u=0,w=0} \vartheta(w,u) = \alpha \uparrow^{s+2} z+1\\

$$

This will satisfy the functional equation, but only if you can turn this heuristic into a proof. Which, I was unable to do. But I was only off by a few lemmas.

Not sure what else you can do here... Your question is very fucking hard, don't expect an honest and easy answer to it.

Regards, James

But to point out, when you use \(1 \le \alpha \le \eta\), you have a very fine growth in the hyper-operations. In that, they tend to \(\alpha\) for \(\Re(z) > 0\). So doing this in this interval is about the closest I ever got. Where, as I've written before, I'll write again.

$$

\vartheta(w,u) = \sum_{n=0}^\infty\sum_{k=0}^\infty \alpha \uparrow^{n+2} (k+1)\frac{u^nw^k}{n!k!}\\

$$

If we differentiate repeatedly in \(w\), we get:

$$

\frac{d^j}{dw^j} \vartheta(w,u) \approx e^w\sum_{n=0}^\infty \omega_{n+2} \frac{u^n}{n!}\\

$$

Where \(\omega_{n+2} = \alpha \uparrow^{n+2} \infty\), which are the fixed points of \(\alpha \uparrow^{n+1} z\). And if we differentiate repeatedly in \(u\), we get:

$$

\frac{d^l}{du^l} \vartheta(w,u) \approx \alpha e^{w}e^{u}\\

$$

Now as asymptotic limits, both of these things are differintegrable. Whereby:

$$

\frac{d^{s}}{du^{s}} \frac{d^{z}}{dw^z}\Big{|}_{u=0,w=0} \vartheta(w,u) = \alpha \uparrow^{s+2} z+1\\

$$

This will satisfy the functional equation, but only if you can turn this heuristic into a proof. Which, I was unable to do. But I was only off by a few lemmas.

Not sure what else you can do here... Your question is very fucking hard, don't expect an honest and easy answer to it.

Regards, James