06/13/2008, 06:38 AM
andydude Wrote:This means that \( \beta(x) = \text{slog}_{(b^{1/b})}(x) \) which relates back to the super-logarithm as follows:
\( \text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left(
(x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)}
- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots
\right) \)
andydude Wrote:Also, I wonder if this is the same as regular slog (rslog)?
Hmm, I give up here - I couldn't reproduce these coefficients yet; I had to determine the coefficients of \( ln(\frac{\sigma(x)}{\sigma(1)}) \) ... I'll try another time. So, if your coefficients give the correct powerseries (and values): congratulation! (and I put it aside for a later reconsideration)
Gottfried
Gottfried Helms, Kassel

