As a note on a similar technique you are applying Mike but trying to keep the vibe much more fractional calculus'y (since it's what I am familiar with) We will try the following function:
\( 0 < \lambda \) We want \( \lambda \) fairly small
\( \beta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)} \)
We know that \( |\beta(w)| < C e^{\kappa |w|} \) for \( \kappa > 0 \) because \( \frac{1}{(^n e)} < C_\kappa \kappa^n \)
So that \( \Re(z) > 0 \).
\( F(-z) = \frac{1}{\Gamma(z)}\int_0^\infty e^{-\lambda x}\beta(-x)x^{z-1} \,dx \)
This function should be smaller then tetration at natural values.
\( F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)} \)
We would get the entire expression for \( F(z) \) by Lemma 3 of my paper:
\( F(z) = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty F(n)\frac{(-1)^n}{n!(n-z)} + \int_1^\infty e^{-\lambda x} \beta(-x)x^{z-1}\,dx) \)
Now F(z) will be susceptible to alot of the techniques I have in my belt involving fractional calculus. This Idea just popped into my head but I'm thinking working with a function like this will pull down the imaginary behaviour and pull down the real behaviour.
We also note that
\( e^{F(z) }\approx F(z+1) \). Which again will be more obvious if you look at the paper, but it basically follows because:
\( F(n) \approx (^n e) \)
\( 0 < \lambda \) We want \( \lambda \) fairly small
\( \beta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)} \)
We know that \( |\beta(w)| < C e^{\kappa |w|} \) for \( \kappa > 0 \) because \( \frac{1}{(^n e)} < C_\kappa \kappa^n \)
So that \( \Re(z) > 0 \).
\( F(-z) = \frac{1}{\Gamma(z)}\int_0^\infty e^{-\lambda x}\beta(-x)x^{z-1} \,dx \)
This function should be smaller then tetration at natural values.
\( F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)} \)
We would get the entire expression for \( F(z) \) by Lemma 3 of my paper:
\( F(z) = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty F(n)\frac{(-1)^n}{n!(n-z)} + \int_1^\infty e^{-\lambda x} \beta(-x)x^{z-1}\,dx) \)
Now F(z) will be susceptible to alot of the techniques I have in my belt involving fractional calculus. This Idea just popped into my head but I'm thinking working with a function like this will pull down the imaginary behaviour and pull down the real behaviour.
We also note that
\( e^{F(z) }\approx F(z+1) \). Which again will be more obvious if you look at the paper, but it basically follows because:
\( F(n) \approx (^n e) \)
