(06/02/2011, 09:12 PM)tommy1729 Wrote: if i get it correctly :
the uppersuperfunction of eta^z is called cheta(z)
the lowersuperfunction of eta^z is called sexp(z)
you claim both cheta(z) and sexp(z) are not analytic at z = e.
No, not the superfunctions, of course is sexp analytic at e, its analytic on \( (-2,\infty) \) dont you remember?
We talk about the half iterate of \( \eta^x \), i.e. the function \( f \) such that
\( f(f(x))=\eta^x \). Though Sheldon's wording was indeed a bit misleading.
And even the explanation \( \eta^{0.5}(z) \) is not really clear, because \( \eta \) is not a function but a constant.
Better may be the expression \( \exp_\eta ^{0.5}(z) \) where \( \exp_\eta(z)=\eta^z \).
