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 Kneser's Super Logarithm bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/22/2008, 06:11 PM (This post was last modified: 11/23/2008, 09:55 AM by bo198214.) So up to now we have a Schroeder function $\chi$ in a vicinity $H_0$ of the fixed point $c$. This function satisfies: $\chi(\log(z))=\chi(z)/c$ $\chi'( c)=1$ The next thing Kneser does is to analytically continue $\chi$ from this small vicinity $H_0$ to the whole upper halfplane without the points $\exp^n(0)$, that is: $H=\{z\in\mathbb{C}: \Im(z)\ge 0, z\neq \exp^n(0), n= 0,1,\dots\}$. He first verifies that $\lim_{n\to\infty} \log^n(z) = c$ for each $z\in H$ (where the cut of the logarithm is the usual one, i.e. $-\pi<\Im(\log(z))\le \pi$.) And then he continues the function along the increasing sets $H_n$, which contain all the points $z$ such that $\log^n(z)$ is contained in that initial vicinity $H_0$ of $c$. The inverse function (in a vicinity of $c$) $\chi^{-1}$ satisifies: $\chi^{-1}(cz)=\exp(\chi^{-1}(z))$ and hence can be continued to the whole complex plane, i.e. is an entire function. To see the properties of $\chi$ on $H$ he considers the following lines and areas. Each increasing index number indicates the application of exponentiation, for example $A_1=\exp(A_0)$, $H_{-1}=\log(H_0)$. The lines are without end points, the areas are without boundaries.     These areas are mapped by $\chi$ to (the letters indicate the source area):     Now $\chi(H_{-1})\cup \chi(H_0)\cup \chi(H_1)$ is simply connected and does not contain 0 (only in the boundary). Hence it is possible to define a holomorphic logarithm $\underline{\log}$ on that domain, he defines $\psi(z)=\underline{\log}(\chi(z))$ on $H_{-1}\cup H_0\cup H_1$. Which then satisfies $\psi(e^z)=\psi(z)+\underline{\log}( c)=\psi(z)+c$. And this is the image under $\psi$, considering that $\underline{\log}$ is biholomorphic:     Define $L_0=\psi(H_0)$ and define $L_n=L_0+nc$, then we have $L_{-1}=\psi(H_{-1})$ and $L_1=\psi(H_1)$. Define $L=\bigcup_{n=-\infty}^\infty L_n$. So one can see that the boundary of $L$ consists only of the cyan and violet arcs, hence the points of the real axis in $H$ are mapped to the boundary of $L$. This property is used in the final step of Kneser's construction, which follows in my next post. « 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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