Taylor series of cheta

[sheldonison]

I am interested in that how we could expand cheta function and its inverse to reals and complexes.
cheta(x) = e[x]e
It is important for me because of the rational and complex hyper-operators.
Please, help me.

There is a formal asymptotic series for the Abel \( \alpha(z) \) function solution for the parabolic case. Iterating \( f(z)=\exp(z)-1 \), is congruent through a simple linear transformation to iterating cheta \( f(y)=\eta^y \;\;\; \eta=\exp(\frac{1}{e}) \), by mapping \( z \mapsto \frac{y}{e}-1 \)



\(
\alpha(z) = \frac{-1}{2z} + \frac{\ln(z)}{3} - \frac{z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac{-71z^4}{435456} + ....
\alpha(\exp(z)-1) = \alpha(z)+1
\)

The formal series will work in either half plane, by changing the ln(z) to ln(-z). But it will not work in both at the same time. It helps to iterate (exp(z)-1) or ln(z+1) a few times to get closer to the fixed point of zero before evaulating the asymptotic series. See G Edgar's post in mathoverflow for some theoretical background on the parabolic case. http://mathoverflow.net/questions/45608/...x-converge

I don't really understand it. What is the connection between Abel and Cheta functions?

The cheta function was named by Jay Daniels in this post. It is the upper superfunction for base \( b=\eta=\exp(1/e) \). When Jay made his post, he did not know about the formal asymptotic series; but numerical results for his algorithm as well as an earlier algorithm of mine are identical to the formal asymptotic series. Jay's cheta function would be the inverse of the Abel function; with an additional arbitrary additive constant k. Then the superfunction base eta would be defined as follows using the inverse of the Abel function defined in my earlier post.
\( S_\eta(z) = \text{superfunction}_\eta(z) = e \cdot \left(\alpha^{-1}(z+k) + 1\right) \)
\( S_\eta(z+1) = \eta^{\left(S_\eta(z)\right)} \)

As far as the function e[x]e, where x is the operator sequence (1=addition, 2=multiplication, 3=exponentiation, 4=tetration), no analytic version of that is known since piecemeal definitions are almost never analytic. So all of this is probably not relevant to your post.

The only problem I understand is using tetration to get to the fractional exponentiation operators like \( f(z)=\exp^{(1/n)}(z)\;\;\; f^{\circ n}(z)=\exp(z)\;\;f(z)=\text{sexp}(\text{slog}(z)+1/n) \). This is directly analogous to \( f(z)=\exp(\log(z)+1/n)=z\cdot \exp(1/n) \). But this does not lead to an analytic e[x]e.