Further observations on fractional calc solution to tetration

[tommy1729]

Hi, everyone. This is a continuation of my last thread http://math.eretrandre.org/tetrationforu...hp?tid=847

I don't have the time to explain too much now, but I realize a mistake I made and it causes for an inaccurate result. Take \( 0<\sigma <1 \)

\( \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \G(s) \frac{w^{-s}}{(^{-s} e)}\,ds = \vartheta(-w) \)

Quite remarkably:

\( \vartheta(-w) = \sum_{n=0}^\infty \frac{(-w)^n}{(^n e)n!} + k(w) \)

where I've recently calculated that:

\( k(w) = \lim_{n\to\infty} \frac{w^n}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i \infty} \G(s-n) \frac{w^{-s}}{(^{n-s} e)}\,ds \)

Where I have been unable to calculate if this converges to a limit. If it does, (and I think it does), we are good.

I'm not sure if I can show recursion with this new transform but I'll try my best to work on this.



We recall the important property, of recovering tetration: \( [\frac{d^z}{dw^z} \vartheta(w)]_{w=0} = \frac{1}{^z e} \), for \( \Re(z) > -1 \)

But every integral integrates a tetration component ... while we do not have that function yet ...

It feels a bit like when someone asks for a proof that a function is analytic , someone shouts : cauchy's theorem.

Which does nothing ... at least not by itself.

regards

tommy1729