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 Unique Holomorphic Super Logarithm bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/18/2008, 06:18 PM When I continued to find a suitable domain of definition for the slog such that the it is unique there by the universal uniqueness criterion II, I followed a line of thoughts I will describe later and came up with the following condition: Proposition. There is at most one holomorphic super logarithm $\text{slog}$ that has a convergence radius of at least $|L|$ when developed at 0 and that maps an open set $G^o$ containing $G$ - which is defined below - (or $\log(G)$) biholomorphically to some set $T$, that contains for each real value $y$ a horizontal line of length $>1$ with imaginary part $y$. Here $L$ is the first fixed point of $\exp$ and $G$ can be roughly seen on the following picture:     The idea behind is the following. If we look at the straight line between $L$ and $L^\ast$ which can be given by $\el(t)=\Re(L)+i\Im(L)t$ for $-1, then the area $G$ bounded by $\el$ and $e^{\el}$ can be considered as an initial area from which you can derive for example the values on $e^G$ or on $\log(G)$ by $\text{slog}(\exp(z))=\text{slog}(z)+1$, or $\text{slog}(\log(z))=\text{slog}(z)-1$. You can see very well on the picture that $e^{\el}$ lies on the circle with radius $|L|$ (red dashed line). This can be easily derived: By $e^L=L$ we know that $e^{\Re(L)}=|e^L|=|L|$ and hence $e^{\el(t)}=e^{\Re(L)+i\Im(L)t}=|L|e^{i\Im(L)t}$ which is an arc with radius |L| around 0. To be more precise we define exactly what we mean: Let $G$ be the set enclosed by $\el(t)$ and $e^{\el(t)}$ for $-1, $\el$ included but $e^\el$ excluded. This set is not open and hence not a domain, but if we move $\el$ slightly to the left we get a domain $G^o$ containing $G$. We call a function $\text{sexp}$ defined on the domain $D$ a super exponential iff it satisfies $\text{sexp}(0)=1$ and $\text{sexp}(z+1)=\exp(\text{sexp}(z))$ for all $z$ such that $z, z+1\in D$. We call a function $\text{slog}$ defined on the domain $H$ a super logarithm iff $\text{sexp}(\text{slog}(z))=z$ for each $z\in H$ (but not necessarily $\text{slog}(\text{sexp}(z))$ for each $z\in\text{slog}(H)$). With those specifications we can come to the proof. Proof. Assume there are two holomorphic super logarithms $f^{-1}: G_1^o\leftrightarrow T_1$ and $g^{-1}: G_2^o\leftrightarrow T_2$ . Then both are defined on the domain $G^o=G_1^o\cap G_2^o$ and map it bihomorphically, say $f^{-1}:G^o\leftrightarrow T_1'$ and $g^{-1}:G^o\leftrightarrow T_2'$. $\delta:=g^{-1}\circ f$ is holomorphic on the domain $T_1'$. By the condition on $T_1$ and by $\delta(z+1)=\delta(z)+1$ it can be continued to an entire function. The same is true for $\delta_2:=f^{-1}\circ g$, which is the inverse of $\delta$ and hence must $\delta(z)=z$ as it was shown in the proof in universal uniqueness criterion II.$\boxdot$ « Next Oldest | Next Newest »

 Messages In This Thread Unique Holomorphic Super Logarithm - by bo198214 - 11/18/2008, 06:18 PM RE: Unique Holomorphic Super Logarithm - by Kouznetsov - 11/19/2008, 03:24 AM RE: Unique Holomorphic Super Logarithm - by bo198214 - 11/19/2008, 01:13 PM RE: Unique Holomorphic Super Logarithm - by Kouznetsov - 11/24/2008, 06:23 AM

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