Does anyone have taylor series approximations for tetration and slog base e^(1/e)?

[sheldonison]

I hate to be a bit of a dunce but:
\( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z-2e)^n \) is the correct formula for the first series correct? I only ask because this is the code I'm using (and I've also tried\( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z+2e)^n \)) and neither seem to converge anywhere?

The only convergence I do get, is \( \text{cheta}(0) = 2e \) when I remove the plus/minus 2e.

Hey James,

Hopefully, this helps. The correct formula is \( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z^n) \)

Cheta(z) is the upper entire superfunction of eta, which grows superexponentially as z increases. Unfortunately, how to define cheta(0) is a somewhat arbitrary, since cheta(z) is always bigger than e, as z goes to minus infinity. Jay suggested defining cheta(0)=2e. Then \( \text{cheta}(1)=\eta^{2e}=e^2 \), and cheta(2)=e^e, which seems like a reasonable choice for how to define cheta(0). Centered around 0, the series I posted will generate these values. More recently, Henryk and Dimitrii have written a paper where the upper \( \text{SuperFunction}_\eta(0)=3 \). Perhaps there will eventually be a reason to pick a definitive value, but that hasn't happened yet.

There is also a lower superfunction for base eta, that I usually refer to as \( \text{sexp}_\eta(z) \), since \( \text{sexp}_\eta(0)=1 \), \( \text{sexp}_\eta(-1)=0 \), and there is a singularity at \( \text{sexp}_\eta(-2) \). \( \text{sexp}_\eta(z) \) does not grow superexponentially, but converges towards e as z increases. I recently posted the Taylor series for that function, centered at z=0, here:
http://math.eretrandre.org/tetrationforu...17#pid5817
- Sheldon