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Unique Holomorphic Super Logarithm - Printable Version

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Unique Holomorphic Super Logarithm - bo198214 - 11/18/2008

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 that has a convergence radius of at least when developed at 0 and that maps an open set containing - which is defined below - (or ) biholomorphically to some set , that contains for each real value a horizontal line of length with imaginary part .

Here is the first fixed point of and can be roughly seen on the following picture:

The idea behind is the following. If we look at the straight line between and which can be given by for , then the area bounded by and can be considered as an initial area from which you can derive for example the values on or on by , or .

You can see very well on the picture that lies on the circle with radius (red dashed line). This can be easily derived:
By we know that and hence
which is an arc with radius |L| around 0.

To be more precise we define exactly what we mean:
Let be the set enclosed by and for , included but excluded. This set is not open and hence not a domain, but if we move slightly to the left we get a domain containing .

We call a function defined on the domain a super exponential iff it satisfies and for all such that .

We call a function defined on the domain a super logarithm iff for each (but not necessarily for each ).

With those specifications we can come to the proof.
Proof. Assume there are two holomorphic super logarithms and .
Then both are defined on the domain and map it bihomorphically, say and . is holomorphic on the domain . By the condition on and by it can be continued to an entire function. The same is true for , which is the inverse of and hence must as it was shown in the proof in universal uniqueness criterion II.

RE: Unique Holomorphic Super Logarithm - Kouznetsov - 11/19/2008

Your figure looks similar to Fig.3 by Kneser.
bo198214 Wrote:We call a function defined on the domain a super logarithm iff for each (but not necessarily for each ).
Why "non necessarily"?

RE: Unique Holomorphic Super Logarithm - bo198214 - 11/19/2008

Kouznetsov Wrote:Your figure looks similar to Fig.3 by Kneser.

Ya, now that I go again through his paper do find out what he really does, I see that his picture is indeed the upper half of mine. I will start a thread about the Kneser construction, step by step how I understand it. I think there maybe broad interest in it, especially because not everyone interested in his paper is able to read German.

bo198214 Wrote:We call a function defined on the domain a super logarithm iff for each (but not necessarily for each ).
Why "non necessarily"?

Well if you define on then I guess the image of is whole . While has a singularity at .

I mean you do it the same for the logarithm. You chose a branch for the logarithm and then for all complex . But of course . A branching similar to the logarithm at 0 takes place at and for the .

RE: Unique Holomorphic Super Logarithm - Kouznetsov - 11/24/2008

Henryk, I am still with your deduction. I hope to confirm it soon. While, it seems to me that you make the slog more basic than sexp, and I begin to like it. If we calculate few tens of derivatives of sexp at minus unity, invert the series and calculate some tens of terms of the Tailor expansion for slog around zero, it will be precize and fast representation. If close to the branch point, we can use the asymptotic expansion, which is even faster. We may get the same 14 significant digits with only a hundred of flops. (Now it takes 10^4 flops per value). Then, we can evaluate tetration as inverse of slog, and it will be also fast and precize representation. We get two orders of magnitude in the speed of evaluation of sexp and slog. With these representations, we can evaluate the holomorphic pentation in the similar maner, define sslog and so on. Do you plan to create the pentationforum?

P.S. Henryk, I hope to have the simplification of your proof. It leads to the linear integral equation of Fredholm of 2d kind with the smooth kernel; smooth Fredholm-2 always has one solution, and the only one.
I need to verify it with the numerical implementation; then I post it together with figures. The resulting slog can be much faster than sexp I have for now; it can be so fast as other special functions already implemented in the conventional software (C++, Fortran, Mathematica, etc.).