I can not yet follow you:
\( \chi(z)=\lim_{n\to\infty} \frac{\log^{\circ n}(z) -c}{c^{-n}} \).
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
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:
\( \operatorname{sexp}(z)=\psi^{-1}(\rho^{-1}(z)) \)
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.
(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.
