01/13/2016, 10:55 PM
(This post was last modified: 01/13/2016, 10:55 PM by sheldonison.)
(01/13/2016, 09:21 PM)andydude Wrote:\( \alpha^{-1}(z)=S^{-1}(L^z) \)(01/13/2016, 05:36 PM)sheldonison Wrote: Lets say we have the Schroeder function, and its inverse \( S(z)\;\;S^{-1}(z) \) which have corresponding Abel and super functions, \( \alpha(z)=\log_L(S(z))\;\;\alpha^{-1}(z)=S^{-1}(L_0+L^z)\;\; \) Notice that for b=e, \( \exp(z)\; L_0=L\;\; \) but this is not the case for other bases.
So
\( \alpha^{-1}(z)=S^{-1}(L_0+L^z) \)
\( S(\alpha^{-1}(z))=L_0+L^z \)
\( S(z)=L_0+L^{\alpha(z)} \)
\( S(z)=L_0+S(z) \)
\( S(z) - S(z) = L_0 = 0 \)
which means \( L_0 = L = 0 \)?
I don't understand.
That was a typo; working from the top of my head. Anyway, the inverse of the Schroeder function of L^z is the superfunction. For base-e, it will be around the fixed point, so \( S^{-1}(0)=L \). Anyway, the formal Schroeder equation is what you use, at the fixed point of the exponential for base=b.
- Sheldon

