11/19/2008, 03:24 AM
Your figure looks similar to Fig.3 by Kneser.
\( H=\mathrm{sexp}(D) \) ?
\( D=\mathrm{slog}(H) \) ?
bo198214 Wrote:We call a function \( \text{slog} \) defined on the domain \( H \) a super logarithm iff \( \text{sexp}(\text{slog}(z))=z \) for each \( z\in H \) (but not necessarily \( \text{slog}(\text{sexp}(z)) \) for each \( z\in\text{slog}(H) \)).Why "non necessarily"?
\( H=\mathrm{sexp}(D) \) ?
\( D=\mathrm{slog}(H) \) ?

