entire function close to sexp ??
#3
I got a function that is very close to tetration:

Let us take the function:

\( \vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^ne)} \)

Then \( \vartheta \) is exponential order zero.

If \( \int_0^\infty |\vartheta(-w)|w^{\sigma-1}\,dw<\infty \) for \( 0 < \sigma < 1 \)

Then
\( \int_0^\infty \vartheta(-w)w^{s-1} \,dw = \frac{\Gamma(s)}{(^{-s} e)} \)

Which IS TETRATION and satisfies the recursion. However, I haven't been able to prove absolute convergence of that integral.

BUT! We can take:


\( \phi(s) = \frac{1}{\Gamma(s)}\int_0^\infty e^{-\lambda w}\vartheta(-w)w^{s-1}\,dw \)

for \( 0<\lambda<\epsilon \) and \( \phi(s) \approx \frac{1}{(^{-s}e)} \) for small \( \epsilon \) hopefully not too small to blow up.

Furthermore this can be made even better for \( \forall s \in \mathbb{C} \)

\( \frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw) \)

These functions are so close to tetration it makes me want to punch a hole in the wall that I can' prove absolute convergence. -_- lol. It's Real to real and interpolates tetration too. The limit as \( \lambda \to 0 \) is tetration in both cases if it converges. but both functions are not the same \( \psi \neq \phi \).
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Messages In This Thread
entire function close to sexp ?? - by tommy1729 - 04/26/2014, 12:23 PM
RE: entire function close to sexp ?? - by JmsNxn - 04/29/2014, 05:47 PM
RE: entire function close to sexp ?? - by JmsNxn - 04/29/2014, 11:21 PM
RE: entire function close to sexp ?? - by JmsNxn - 04/30/2014, 03:49 PM

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