andydude Wrote:This means that \( \beta(x) = \text{slog}_{(b^{1/b})}(x) \) which relates back to the super-logarithm as follows:
\( \text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left(
(x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)}
- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots
\right) \)
Hmm, at least it looks somehow similar to the "regular"-formula.
If I replace a by u, and introduce t for exp(u) so that \( \text{slog}_{(e^{a(e^{-a})})}(x) = \text{slog}_{(e^{u/t})}(x) \)
and \( (x(e^{-a})-1) = \frac{x}{t}-1 = x' \)
and factorize the denominators differently, for instance
\( \hspace{24}
(a-1)^3(a+1)(1+a+a^2)
= (a-1)(a-1)(a-1)*(a+1)*(1+a+a^2)
= (a-1)(a^2-1)(a^3-1)
\)
to get a more familiar looking formula for me, this is then, using "rsdxplog" as rslog for the dxp-function:
\( \text{slog}_{e^{u/t}}(x) = \text{rsdxplog}_t(x') = C + \frac{1}{\ln(u)}\left(
(x' - \frac{ux'^2}{4(u-1)}
\hspace{12}
+ \frac{u^2(1+5u)x'^3}{36(u-1)(u^2-1)}
\hspace{12}
- \frac{u^4(2+u+3u^2)x'^4}{32(u-1)(u^2-1)(u^3-1)}
\hspace{12}
+ \cdots
\right) \)
where especially the denominators-products are the same as in my Ut-formulae, and also the numerators look very familiar. I'll see, whether we have the same coefficients later today.
I had the rsdxplog as logarithm of the Schroeder-function, assuming x' as h'th (continuous) iteration of \( \text{dxp}_t^{{^o}h}(1) \)
then
\(
\\ \\[12pt]
\hspace{24}
x' = \text{dxp}_t^{{^o}h}(1) = \sigma_t^{-1}(u^h \sigma_t(1))
\\ \\[12pt]
\hspace{24}
\sigma_t(x') = u^h \sigma_t(1)
\\ \\[12pt]
\hspace{24}
\frac {\sigma_t(x')}{\sigma_t(1)} = u^h
\\ \\[12pt]
\hspace{24}
\text{rsdxplog}(x')= h = \log_u(\sigma_t(x')) - \log_u(\sigma_t(1))
\\ \\[12pt]
\hspace{24}
\text{rsdxplog}(x')= h = C + \log_u(\sigma_t(x')) =C + \frac1{ln(u)}\log(\sigma_t(x'))
\)
where I got the coefficients of the sigma-function by the eigenmatrices of Ut - and the structure of these coefficients look very similar to yours above
Gottfried
Gottfried Helms, Kassel