12/21/2009, 08:57 PM
(12/20/2009, 06:31 AM)Ansus Wrote: It is possible to use this formula and substitute sexp as inverse function operator of slog.
\( \operatorname{slog}_z C=-\int \left( \frac{1}{z (\ln z)^2}\sum_{q=0}^{\operatorname{slog}_z C -1}\frac{1}{D_q \operatorname{sexp}_z(q-1)}\right) dz \)
Then it will be an iterating formula for slog.
But this approach is complicated.
Hmm. I noticed this when preparing for the Mueller sum:
\(
\sum_{q=0}^{\mathrm{slog}(x)-1} \frac{1}{\mathrm{tet}'(q-1)} = \sum_{q=-1}^{\mathrm{slog}(x)-2} \frac{1}{\mathrm{tet}'(q)} = \frac{1}{\mathrm{tet}'(-1)} + \frac{1}{\mathrm{tet}'(0)} + \left(\sum_{q=1}^{\mathrm{slog}(x)} \frac{1}{\mathrm{tet}'(q)}\right) - \frac{1}{\mathrm{tet}'(\mathrm{slog}(x))} - \frac{1}{\mathrm{tet}'(\mathrm{slog}(x)-1)}.
\)
But the last term has a singularity at 0! (note that \( \mathrm{slog}(0) - 1 = -2 \)) Does this help "complicate" things?
