# Tetration Forum

Full Version: Kneser's Super Logarithm
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In order to understand it myself and as a service for the non-German community, in the following posts I will provide the construction of Kneser (for an analytic super logarithm) which he gives in his paper:

Hellmuth Kneser, Reelle analytische Lösungen der Gleichung und verwandter Funktionalgleichungen. J.Reine Angew. Math. 187, (1949), 56-67

I will proceed in several posts because my time is currently quite limited, and so I will give account of succeeding portions of his construction.
As mentioned in the title of his paper Kneser originally seeks for an analytic solution of the functional equation (This functional equation was considerably discussed at the conference of the German mathematician's association, October 1941 in Jena. So there the idea was born to construct an analytic solution.)

As is very well-known the construction of a half-iterate can be reduced to the solution of the Abel equation, which Kneser gives here in a bit more generality:
.

If one has such a solution then it is easy to construct the half iterate by:

because then

Closely related to the Abel equation is the Schroeder equation:

If one has a solution of the Schroeder equation, then is a solution of the Abel equation with , because:

And one can also derive the fractional iterates directly from the Schroeder function by:
.

The good thing is now that for holomorphic functions with fixed point and there is always the so called principal Schroeder function (holomorphic in a vicinity of ) which satisfies the Schroeder equation for and which satisfies .

There are two possibilities to construct this principal Schroeder function. First by development of the powerseries in and chosing the coefficients of such that they satisfy the Schroeder equation.
And second by the limit

As you can easily verify if is a solution of the Schroeder equation then is also a solution for any constant . They are called "regular" (by Szekeres) if is the principal Schroeder function.
Those regular solutions are characterized by that is holomorphic in .

As a first step Kneser computes this principal Schroeder function of at 's first fixed point in the upper half plane, the fixed point nearest to the real axis. is however not real on the real axis, so he determines some mapping properties of and later manipulates to become real and analytic on the real axis.
So up to now we have a Schroeder function in a vicinity of the fixed point . This function satisfies:

The next thing Kneser does is to analytically continue from this small vicinity to the whole upper halfplane without the points , that is:
.
He first verifies that for each (where the cut of the logarithm is the usual one, i.e. .)

And then he continues the function along the increasing sets , which contain all the points such that is contained in that initial vicinity of .

The inverse function (in a vicinity of ) satisifies:
and hence can be continued to the whole complex plane, i.e. is an entire function.

To see the properties of on he considers the following lines and areas. Each increasing index number indicates the application of exponentiation, for example , . The lines are without end points, the areas are without boundaries.

[attachment=420]

These areas are mapped by to (the letters indicate the source area):

[attachment=421]

Now is simply connected and does not contain 0 (only in the boundary). Hence it is possible to define a holomorphic logarithm on that domain, he defines
on . Which then satisfies
.

And this is the image under , considering that is biholomorphic:
[attachment=422]

Define and define , then we have and . Define . So one can see that the boundary of consists only of the cyan and violet arcs, hence the points of the real axis in are mapped to the boundary of . This property is used in the final step of Kneser's construction, which follows in my next post.
As the last step we use the Riemann mapping theorem, to map the area biholomorphically to the upper halfplane, which maps the boundary to the real axis. By some reason (to be explained) the corresponding function indeed satisfies the Abel equation.

But we do that in two steps first we map to the unit disk via and then we map to the upper (open) halfplane via .

The existence of is guarantied by the Riemann mapping theorem. The translation maps to without having a fixed point, hence the corresponding mapping in , maps the unit disk into the unit disk without a fixed point. It is well known that such a function needs to be linear fractional, i.e. of the form . Kneser shows that must be parabolic, that means it has only one fixed point at the boundary of .

To map the unit disk to the upper half plane we can again use a linear fractional transformation . There are enough parameters to chose it such that is mapped to infinity and such that .

Now we can define
which has the property
.
and is biholomorphic on the interior of .

The interior of is mapped to some area bordering on the real line. The boundary on the real axis is hence mapped to the real axis. More precisely the interval is mapped to the real axis. By the Schwarz reflection principle it can be continued to the complex conjugate of especially it is analytic on and can from there be continued to a vicinity of the whole real axis by and .
(06/02/2009, 10:04 AM)Kouznetsov Wrote: [ -> ]
(06/02/2009, 09:24 AM)bo198214 Wrote: [ -> ]
(06/02/2009, 03:10 AM)Kouznetsov Wrote: [ -> ]About Kneser's expansion: it would be good to check, that we evaluate the same function expanding it at fixed point L and doing that at L^*....
Yes, there are singularities at the real axis at . But thats only an intermediate step of Kneser's construction.
It is sufficient to show that the two expansions coincide along any segment of positive length. Then the two functions coincide in the whole range of the holomorphizm.

They dont coincide, because they are not real at the real axis.
Kneser's trick is to prepend a conformal mapping, that makes it real at the real axis.

Moderator's note: I moved the two previous posts from thread overview paper co-author inviation to here.
It occured to me that complex fourier analysis is at the heart of a lot of key topics for tetration. I finally have gotten to the point where I understand Kouznetsov's Cauchy integral, and implemented parts of it. I also think I might understand Kneser's approach, based on the Riemann mapping, starting with the entire analytic super-exponential, developed from the fixed point, c~=0.318+i*1.337. This is the inverse Abel function, where the Abel function is the slog developed from the fixed point c. , at least I think that is what is meant by . I tried really hard to understand Jay's post, and this post, but there are gaps.... I tried to fill in the gaps, but I have a feeling I'm going down a slightly different approach here.

Is the real valued sexp at the real axis connected to the Schroeder equation, chi, by a complex 1-cyclic function,

Where is the Riemann mapping of contour of the limit of iterating the natural logarithm, starting with the real interval, 0..1 (which corresponds to sexp(z) from z=0 to z=1). Then
subtracting c, which is the fixed point, ~= 0.318+i*1.337? Then the Riemann mapping unit circle is scaled by 2*pi and converted to a repeating unit length at the real axis. Is this equation correct, and is this Kneser's approach?

This equation is developed from the inverse sexp function, or the Abel function, . Assuming the equation is correct, here is my rough calculation of the resulting limit, whose contour is what I think is to be Riemann mapped, and then used to generate . For double precision numbers >0 and <1, the real part of the contour never exceeds 1.78, so most of the contour graph is only for values of z super-exponentially close to zero or one, but I thought I would include the entire contour for completeness. Also, I removed the scaling, and added the next iteration, which is developed from sexp(z), from z=1 to z=2, which shows that , and that is 1-cyclic. Contour values from this graph should have imaginary values of zero for . Also, the outermost for the contour is multi-valued, and as n increases, sometimes multiples of -2pi*i have to be added so that the contour equation converges. The other values correspond to other identical parts of the function, which repeats infinitely many times in the complex plane. Once the Riemann mapping is generated, and converted to a unit length at the real axis, to generate , I assume that the goes to a small constant (the center of the Riemann mapping unit circle), as the imaginary axis grows, and has a singularities at unit intervals at the real axis. Presumably, the singularity in the Riemann mapping is at one point on the edge of the unit circle, but not inside the unit circle.

The contour is valid before and after the Riemann mapping. But after the Riemann mapping, it would seem that is a 1-cyclic function, and it is the only analytic 1-cyclic function that converts from to the sexp_e such that will be a constant plus the sum of a complex fourier series, with complex coefficients, of . Because of the unique Riemann mapping, all Fourier terms of the form are zero. Because the Riemann mapping to drive these terms to zero is unique, there is only one sexp solution approaching the fixed point as i increases, with the complex fourier series decaying to zero as i increases.

Given that Kouznetsov's solution also assumes the limiting value of f as imaginary increases, it would seem that complex Fourier analysis would allow one to prove why the solution is unique, why Kouznetsov's solution is equal to Kneser's solution, and perhaps even why Kouznetsov's iterated Cauchy integral converges.
- Sheldon
What do you mean by ?
(01/25/2010, 06:35 AM)mike3 Wrote: [ -> ]What do you mean by ?
Typo, that should've been
f=fixed point of "e", 0.318131505+1.337235701i,

For clairty, I'm also changing this in the original reply I made.
- Sheldon
(01/25/2010, 07:42 AM)sheldonison Wrote: [ -> ]
(01/25/2010, 06:35 AM)mike3 Wrote: [ -> ]What do you mean by ?
Typo, that should've been
f=fixed point of "e", 0.318131505+1.337235701i,

For clairty, I'm also changing this in the original reply I made.
- Sheldon

Ah. So how would you use the given equation to calculate ?
(01/26/2010, 06:24 AM)mike3 Wrote: [ -> ]Ah. So how would you use the given equation to calculate ?
Update: fixed notation to use the Abel function, which is the inverse of the super-exponential developed from the fixed point "c".
Well, again, I was sort of trying to follow the graphs that Jay made, and Henryk's description of Kneser's approach, and it was not quite accessible to me. It appears they were iterating the logarithm on a unit length, [0..1], which is the first step in the limit towards the inverse super-exponential, or the Abel function, . The sequence of logarithms converges towards the fixed point, so the logical next step is the equation for , or the inverse super-exponential developed from the complex fixed point c. And the equation I got for is

Of course, I'm assuming that is the entire sexp_e developed from the fixed point, developed from the approximation for small (large negative) values of z, , which has complex values at the real axis. But I don't think that matches what Kneser is doing, but I think it must be closely related. Then I graphed the contour of , and I was wondering if that was indeed the contour that Kneser Reimann mapped, . But I couldn't tell from Jay's posts, or Henryk's post. It seemed more complicated than that. Anyway, If you already had the sexp_e to start with than

The 1-cyclic function is . Plugging into the equation for .

Of course, we don't know what sexp_e is, so we start with the contour generated over a unit length from [0..1]. A linear approximation for sexp_e is probably a reasonbly starting point, since it has a continuous first derivative. Then do a Riemann mapping of the contour of of the unit length, to a unit circle. After the Riemann mapping, unroll the the unit circle, divide by 2*pi, and we're back to a unit length. Only now is a 1-cyclic function that can be expressed as the unique sum of a complex fourier series exponentially decaying to a constant as i increases. I gather the computational difficulty comes in actually doing the Riemann mapping...

- Shel
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