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 Superlog with exact coefficients Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 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 « Next Oldest | Next Newest »

 Messages In This Thread Superlog with exact coefficients - by andydude - 06/11/2008, 05:46 AM RE: Superlog with exact coefficients - by andydude - 06/11/2008, 06:02 AM RE: Superlog with exact coefficients - by Gottfried - 06/11/2008, 07:18 AM RE: Superlog with exact coefficients - by Gottfried - 06/13/2008, 06:38 AM RE: Superlog with exact coefficients - by bo198214 - 06/20/2008, 01:26 PM RE: Superlog with exact coefficients - by Gottfried - 03/10/2009, 09:59 PM

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