Dmitrii Kouznetsov's Tetration Extension

[Kouznetsov]

Dear Bo. I am glad to see your interest. I cut your post for pieces and try to answer one by one.

2. There are two regular superexponentials at base \( b \) such that \( 1<b<\exp(1/{\rm e}) \).

What do you mean by regular? I would deprecate the term regular superexponential except in the sense of "the superexponential retrieved by regular iteration": \( \text{sexp}_b(x)=\exp_b^{\circ x}(1) \).
Otherwise we impose major confusion on this forum.
I think your use of "regular" can be replaced by "analytic", or something similar?

Yes. Sorry about slang. Regular means that \( F(z) \) has no singularities at \( \Re(z)>-2 \).

I have plotted the only one, \( F \) such that \( F(0)=1 \).
At \( b=\sqrt{2} \), for example,
\( \lim_{x \rightarrow \infty} F_{\sqrt{2}}(x+{\rm i}y)=2 \);
\( \lim_{x \rightarrow -\infty} F_{\sqrt{2}}(x+{\rm i}y)=4 \).

This contradicts \( F(z^\ast)=F(z)^\ast \)?

No. \( F_{\sqrt{2}}(x+{\rm i}y) \) approaches the limiting values at any real \( y \). (in spite the cutlines!)

How does it come anyway that you now can compute tetration also for bases \( <e^{1/e} \)?

Yes, I can. You can too. I tried to post the code as "source" together with the picture, but it was not accepted...
I shall post it in different way.

There is another one, \( G \) that grows up along the real axis faster than any exponential and aproaches its limiting values in the opposite direction.

But if we use base \( \sqrt{2} \) it can not grow to infinity, it has to be limited by \( 2 \) on the real axis. Or what exactly do you mean?

Yes, it is limited, you can see, it approaches 2 at the picture. But if we withdraw the equation \( F(0)=1 \) and replace it with requirement \( F(z) = 2 + \mathcal O (\exp(Qz)) \) at \( \Re(z)\rightarrow -\infty \), then the "second" tetration can be plotted, let us call it \( G \); and it should be so beautiful, as tetration \( F \) for \( b=2 \) and \( b=\rm e \).
It is that I expect, I did not yet plot it.

I am writing source for its evaluation.

Did you read my previous post? I asked you to send me some code of your computations.

Oops... I see, I used to read it "by diagonal". I should stop to write new codes and share that I already have. I know that usually the "self-made" codes dislike to run at another computer, so, let us do it with few steps, reproducing the figures one by one. Let us begin with very simple code. Please, reproduce first Figure 1 from the source posted at http://en.citizendium.org/wiki/Image:Exa...nLog01.png
please, tell me how does it run at your computer and send me (or post) the resulting picture. It is supposed to be <b>identical</b> with the eps file I got at my computer. I begin with so simple picture because its source is in "one piece" and does not require any input files.

3. Then we have covered the ranges \( 1<b<\exp(1/\rm e) \) and \( b>\exp(1/\rm e) \); and I think about cases \( b=\exp(1/\rm e}) \) and \( b<1 \).

Yes, we would have covered them in one way. However if you read through this forum we have established several other (at least 3) ways to compute analytic tetration. And we even dont know yet which are equal or which are equal to yours.

Yes, but they seem to be singular.. What example would you suggest to begin with?
"Equal to mine" is tetration that has no singularities at the right hand side of the complex halfplane. I did not see any map of real and imaginary parts, nor those of modulus and phase. Tell me if I am wrong.

I suggest that you use the same idea: first, find the asymptotics and periodicity (if any); then recover the analytic function with these properties. Could you calculate some pictures (similar to those I have posted) for these cases?

As I said: for the case of tetration by regular iteration, which is applicable in the range \( e^{-e}<b<e^{1/e} \), I can (and did) compute pictures.

Ah! very good! Will you plot the real and imaginary parts in the complex plane? Then any difference (even exponentially-small) leads to singularities which are easy to see.

Apropos pictures: I think you have to explain something more about your second posted picture.

Yes. But let us begin with Fig.1. I expect, is trivial and you can easy reproduce it. Please, confirm, that you can do it namely with my code; if we need any adjustment, it is important to reveal this with simplest possible case. While I prepare the description of Figure 2 and post its source.