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 Kneser's Super Logarithm bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 01/26/2010, 11:19 PM (This post was last modified: 01/26/2010, 11:20 PM by bo198214.) I can not yet follow you: (01/23/2010, 01:01 PM)sheldonison Wrote: $\chi(z)=\lim_{n\to\infty}( \exp^{\circ n}(f+f^{(z-n)})$, at least I think that is what is meant by $\chi(z)$. I think the formula is corrupt, the correct formula should be: $\chi(z)=\lim_{n\to\infty} \frac{\log^{\circ n}(z) -c}{c^{-n}}$. edit: oh now I see: you call the fixed point $f$! Please avoid, $f$ is reserved for functions. Kneser calls it $c$, in this thread I keep with his convention. edit: well now I also see that you mean the inverted Schröder function. As we can derive from the previous formula we get: $\chi^{-1}(z)=\lim_{n\to\infty}\exp^{\circ n}(z c^{-n} +c )$ No, this still is not your formula, I guess you mean the inverse of the Abel function $\psi(z)=\log_c(\chi(z))$ this would have the formula: $\psi^{-1}(z)=\lim_{n\to\infty}\exp^{\circ n}(c^{z-n} +c )$ which is finally your formula Quote:Is the real valued sexp at the real axis connected to the Schroeder equation, chi, by a complex 1-cyclic function, $\theta(z)-z$ $\text{sexp}_e(z)=\chi(\theta(z))$ The Abel function $\psi$ maps $H_0$ to some region $X_0$. But our final Abel function $\Psi$ shall map $H_0$ into the upper halfplane, correspondingly also $H_1=\exp(H_0)$ to $X_0+1$ and $H_{-1}=\log(H_0)$ to $X_0-1$. Thatswhy we consider the Riemann mapping $\rho$ that biholomorphically maps the union of the by integer translated $X_0$ to the upper halfplane. (though is this marginally different from Kneser's original approach I think it is better accessible.) Then $\Psi(z)=\rho(\psi(z))$ is the wanted Abel function (slog), and we obtain sexp as: $\operatorname{sexp}(z)=\psi^{-1}(\rho^{-1}(z))$ Quote:Where $\theta(z)$ is the Riemann mapping of contour of the limit of iterating the natural logarithm, starting with the real interval, 0..1 [color=#0000CD](which corresponds to sexp(z) from z=0 to z=1). Not sure what contour exactly you mean, but I think we are close already. Your theta should be the Riemann mapping that maps the upper halfplane to the union of the by integer translated images $\psi(H_0)$. Please continue from here with unified denotation. « Next Oldest | Next Newest »

 Messages In This Thread Kneser's Super Logarithm - by bo198214 - 11/19/2008, 02:20 PM RE: Kneser's Super Logarithm - by bo198214 - 11/19/2008, 03:25 PM RE: Kneser's Super Logarithm - by sheldonison - 01/23/2010, 01:01 PM RE: Kneser's Super Logarithm - by mike3 - 01/25/2010, 06:35 AM RE: Kneser's Super Logarithm - by sheldonison - 01/25/2010, 07:42 AM RE: Kneser's Super Logarithm - by mike3 - 01/26/2010, 06:24 AM RE: Kneser's Super Logarithm - by sheldonison - 01/26/2010, 01:22 PM RE: Kneser's Super Logarithm - by mike3 - 01/27/2010, 06:28 PM RE: Kneser's Super Logarithm - by sheldonison - 01/27/2010, 08:30 PM RE: Kneser's Super Logarithm - by mike3 - 01/28/2010, 08:52 PM RE: Kneser's Super Logarithm - by sheldonison - 01/28/2010, 10:08 PM RE: Kneser's Super Logarithm - by mike3 - 01/29/2010, 06:43 AM RE: Kneser's Super Logarithm - by bo198214 - 01/26/2010, 11:19 PM RE: Kneser's Super Logarithm - by sheldonison - 01/27/2010, 07:51 PM RE: Kneser's Super Logarithm - by bo198214 - 11/22/2008, 06:11 PM RE: Kneser's Super Logarithm - by bo198214 - 11/23/2008, 01:00 PM

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