paradox, accurate taylor series half iterate of eta not analytic at e

[sheldonison]

Since we're dealing with half-iterates of \( \text{sexp}_\eta(z) \), does this tell us anything about pentation? If \( \text{pent}_\eta(z) \) is pentation base \( \eta \),

if \( \text{sexp}_\eta^{\frac{1}{2}}(z) = \text{pent}_\eta(\text{pent}_\eta^{-1}(z) + \frac{1}{2}) \) holds up,
I would assume since \( \text{sexp}_\eta^{\frac{1}{2}}(z) \) isn't analytic about e, this says \( \text{pent}_\eta(z) \) and \( \text{pent}_\eta^{-1}(z) \) aren't analytic about e also right?
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This thread was all very beautiful. Thank you for posting it.

you're welcome, but fyi, the wording may have been misleading, in that the half iterate I was generating was \( \exp_\eta^{[0.5]}(z) \), where \( \eta \)=1/e. This is the half iterate of an exponential function, not the half iterate of a superexponential function.

But one way to calculate the half iterate is this equation, which works for other bases than \( \eta \) as well. \( \exp_\eta^{[0.5]}(z)=\text{sexp}^_\eta(\text{sexp}^{-1}_\eta(z)+0.5) \). Also, your equation for the half iterate of sexp(z) in terms of pent(z) is correct. But notice, that we're only saying that the half iterate of eta^z is not analytic at a single point, when z=e. At other points, it is analytic, and at z=e, all of its derivatives are apparently defined, though with a zero radius of convergence.

As you may have noticed from other posts, the two superfunctions of \( \eta^z \) are a bit of an oddball, and this is just one more example. As a consequence, I don't think anyone has ever thought about or calculated pentation for base eta. Also the lower sexp(z) for eta never grows super-exponentially. The upper superfunction for eta^z does grow superexponentially, but never gets smaller than e. So that might take away some of the appeal for doing pentation base eta.
- Sheldon