jaydfox Wrote:Following up on the \( \Im({}^{i}{e}) = \mathrm{slog}_e'(0) \) conjecture, my initial estimate puts \( \Im({}^{i}{e}) \) closer to 0.9163, give or take. It would have been very cool if it had been true.andydude Wrote:Super-logarithm DerivativeIn computing the solutions (for base e) to moderately large systems with your proposed slog, I think the first derivative is closer to 0.9159460564995..., which puts Catalan's constant out of the running (or puts your slog out of the running?).
From my approximations, \( \text{slog}_e'(0) \approx 0.916 \), and \( {}^{i}{e} \approx 0.786 + i 0.916 \).
My first conjecture is that \( \text{Im}({}^{i}{e}) = \text{slog}_e'(0) \), and my second conjecture is that \( \text{slog}_e'(0) = 0.915965594177\cdots \) otherwise known as Catalan's constant.
PS: Which looks better, \( \text{Im}(z) \) or \( \Im(z) \)?
~ Jay Daniel Fox

