04/25/2010, 09:11 AM
(04/25/2010, 08:22 AM)bo198214 Wrote: Another one:
\( f(x)=x^2+(1+\sqrt{5})x+1 \)
\( F(x)=e^{2^x} - \frac{1+\sqrt{5}}{2} \)
\( f \) has two fixed points with the derivations:
\( f'(\frac{1-\sqrt{5}}{2}) = 2 \) and
\( f'(\frac{-1-\sqrt{5}}{2}) = 0 \).
The above superfunction \( F \) is the regular iteration at the upper fixed point \( \frac{1-\sqrt{5}}{2} \)
Henryk, it seems to me that such a case can be obtained from the example 5 of the Table 1 of our article
D.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12 (English version), p.8-14 (Russian version;
http://www.ils.uec.ac.jp/~dima/PAPERS/2009superfae.pdf
http://www.ils.uec.ac.jp/~dima/PAPERS/2010superfar.pdf
with transformation at the bottom of that table at
\( P(z)=z - \frac{1+\sqrt{5}}{2} \)
I suspect, with that short table we have covered the most of simple elementary superfunctions...

