bo198214 Wrote:Argh, sorry, I meant \( \sqrt{2} \). \( t\mapsto \exp_{\sqrt{2}}^{\circ t}(1.0) \) maps \( (-2,\infty) \) to \( (-\infty,2) \), so the slog is defined on \( (-\infty,2) \)Quote:Also, there's a question about whether you consider the slog function to only apply to the inverse of the function of iterated exponentials/logarithms from 1. For b=2, for example, the domain of slog is negative infinity to 2.Dont understand this, \( t\mapsto \exp_2^{\circ t}(1.0) \) maps \( (-2,\infty) \) to \( (-\infty,\infty) \), so the slog is defined on \( (-\infty,\infty) \)?
~ Jay Daniel Fox

