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It appears that there is a new extension related to complex heights.

Andrew Robbins
andydude Wrote:See Analytic solution of F(z+1)=exp(F(z)) in complex z-plane for more information.

About the uniqueness: It is well known that if we have a solution $\alpha$ of the Abel equation $\alpha(z+1)=f(\alpha(z))$ then for any 1-periodic function $\phi$ also $\beta(z)=\alpha(z+\phi(z))$ is a solution to the Abel equation. (Because $\beta(z+1)=\alpha(z+1+\phi(z+1))=\alpha(z+\phi(z)+1)=f(\alpha(z+\phi(z))=f( \beta( z))$).

So let $F$ be one solution of
(*) $F(z+1)=\exp(F(z))$ with
(**) $\lim_{y\to\infty} F(x+iy) = L$ and $\lim_{y\to -\infty} F(x+iy)=L^\ast$
then $G(z)=G(z+\sin(2\pi z))$ is another solution of (*). Let us now consider (**). We know that $\sin(z)=-i\frac{e^{iz}-e^{-iz}}{2}$ and $\sin(x+iy)=i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2}$

$G(x+iy)=F\left(x+iy+i\frac{-e^{2\pi ix}e^{-2\pi y}+e^{-2\pi ix}e^{2\pi y}}{2}\right)$. As $e^{-2\pi y}\to 0$
at least for x=0 also
$\lim_{y\to\infty} G(iy)=\lim_{y\to\infty} F(i(y+e^{2\pi y}))=L$

 fixed some negligences. [/edit]
Hello, bo198214; you have cathced the important point! I see, in my paper, I have to write

Theorem 0. There exist function $F(z)$, analytic in the whole complex $z$ plane except $z<-2$, satisfying
(100) $~~~~ \exp(F(z))=F(z+1)$,
(101) $~~~~ F(0)=1$
and
(102) $~~~~ F(z)=L + {\mathcal O}\Big( \exp( L z ) \Big)~$ at any fixed $\Re(z)$ and $\Im(z) \rightarrow +\infty$

Theorem 1. There exist only one such function.

I hope, the reviewer catches this point, and I already have the corresponding correction above.
By the way, your deduction gives the hint, how to prove the Theorem 1. (However, we have to scale the argument of sin function.) How about the collaboration?
Kouznetsov Wrote:By the way, your deduction gives the hint, how to prove the Theorem 1.

Perhaps. I am not convinced yet that it is true, but if it was, this would be great. We even know vice versa if we have two solutions $F$ and $G$ of the Abel equation then $G^{-1}(F(x+1))-(x+1)=G^{-1}(\exp(F(x)))-(x+1)=G^{-1}(\exp(G(G^{-1}(F(x)))))-(x+1)=G^{-1}(F(x))+1-(x+1)=G^{-1}(F(x))-x$, meaning that $\phi=G^{-1}\circ F-\text{id}$ is a 1-periodic function, so we know already that each other solution $G$ of the Abel equation must be of the form $F(z+\phi(z))$ for some 1-periodic $\phi$. To prove theorem 1 everything depends on the behaviour of those 1-periodic functions for $\Im(z)\to\infty$. For uniqueness roughly the real value must be go to infinity for the imaginary argument going to infinity.

Quote:(However, we have to scale the argument of sin function.)

Corrected in the original post.

That would be great.

Quote:(P.S. Does the number of replies I should answer grow as the Ackermann function of time, or just exponentially?)

I would guess its rather logarithmically (much questions at the start but then slowly ebbing away)! At least here is another question: Can you please compute values on the real axis for bases $b for example for $b=\sqrt{2}$? I would like to compare your solution with the regular tetration developed at the lower real fixed point (which would be 2 in the case of $b=\sqrt{2}$) of $b^x$.

@Andrew
Can you post a comparison graph with your slog/sexp? The values are in Dmitrii's paper. How does the periodicity of your slog (or was it sexp?) at the imaginary axis compare with Dmitrii's limit $\lim_{y\to\infty} F(x+iy)=L$ (where $L$ is a fixed point of $\exp$)?
bo198214 Wrote:
andydude Wrote:See Analytic solution of F(z+1)=exp(F(z)) in complex z-plane for more information.
About the uniqueness: It is well known that if we have a solution $\alpha$ of the Abel equation $\alpha(z+1)=f(\alpha(z))$ then for any 1-periodic function $\phi$ also $\beta(z)=\alpha(z+\phi(z))$ is a solution to the Abel equation. (Because $\beta(z+1)=\alpha(z+1+\phi(z+1))=\alpha(z+\phi(z)+1)=f(\alpha(z+\phi(z))=f( \beta( z))$).

So let $F$ be one solution of
(*) $F(z+1)=\exp(F(z))$ with
(**) $\lim_{y\to\infty} F(x+iy) = L$ and $\lim_{y\to -\infty} F(x+iy)=L^\ast$
then $G(z)=G(z+\sin(z))$ is another solution of (*). Let us now consider (**). We know that $\sin(z)=-i\frac{e^{iz}-e^{-iz}}{2}$ and $\sin(x+iy)=i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2}$

$G(x+iy)=F\left(x+iy+i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2}\right)$. As $e^{-y}\to 0$
at least for x=0 also
$\lim_{y\to\infty} G(iy)=\lim_{y\to\infty} F(i(y+e^y))=L$

bo198214 Wrote:
Kouznetsov Wrote:By the way, your deduction gives the hint, how to prove the Theorem 1.

Perhaps. I am not convinced yet that it is true, but if it was, this would be great. We even know vice versa if we have two solutions $F$ and $G$ of the Abel equation then $G^{-1}(F(x+1))-(x+1)=G^{-1}(\exp(F(x)))-(x+1)=G^{-1}(\exp(G(G^{-1}(F(x)))-(x+1)=G^{-1}(F(x))+1-(x+1)=G^{-1}(F(x))-x$, meaning that $\phi=G^{-1}\circ F-\text{id}$ is a 1-periodic function, so we know already that each other solution $G$ of the Abel equation must be of the form $F(z+\phi(z))$ for some 1-periodic $\phi$. To prove theorem 1 everything depends on the behaviour of those 1-periodic functions for $\Im(z)\to\infty$. For uniqueness roughly the real value must be go to infinity for the imaginary argument going to infinity.

Quote:(However, we have to scale the argument of sin function.)

Yes, my negligence. The $z$ in $\sin(z)$ has to be replaced by $\frac{z}{2\pi}$ in the previous post.

That would be great.

Can you please compute values on the real axis for bases $b for example for $b=\sqrt{2}$? I would like to compare your solution with the regular tetration developed at the lower real fixed point (which would be 2 in the case of $b=\sqrt{2}$) of $b^x$.

Bo, I got your message about base b=e^(1/e) and b=sqrt(2). In these cases, the real part of quasiperiod is zero, and I cannot run my algorithm as is. I need to adopt it. It will take time. I do not think that b=sqrt(2) is of specific interest (just integer L(b)=4); we need to consider the general case.
You may advance faster than I do. You may begin with the plot of the asymptotic period T(b) and analysis of its limiting behavior in vicinity of b=1 and b=e^(1/e). Please, provide the good approximation for (at least) the leading terms.

P.S. you may also correct misprints in your post:
invert the scaling factor for the argument of sin, and
delete the expression with unmatched parenthesis.
Kouznetsov Wrote:Bo, I got your message about base b=e^(1/e) and b=sqrt(2). In these cases, the real part of quasiperiod is zero, and I cannot run my algorithm as is. I need to adopt it. It will take time. I do not think that b=sqrt(2) is of specific interest (just integer L(b)=4); we need to consider the general case.
$b=\sqrt{2}$ is indeed only of interest because it has a simple (integer) fixed point 2. So that is our standard reference (on this forum) base to compare two different methods of computing a tetration. As for example real regular iteration/tetration is no more possible for $b>e^{1/e}$ because there is no real fixed point.

Quote:You may advance faster than I do. You may begin with the plot of the asymptotic period T(b) and analysis of its limiting behavior in vicinity of b=1 and b=e^(1/e). Please, provide the good approximation for (at least) the leading terms.
yeah, I am also not that richly blessed with time. I will see, what I can do.

Quote:P.S. you may also correct misprints in your post:
invert the scaling factor for the argument of sin, and
delete the expression with unmatched parenthesis.
blush (again!).
Bo, when you make the first step, please, plot the asymptotics in the complex plane, and we compare our results. It will be the second step.

bo198214 Wrote:$b=\sqrt{2}$ is indeed only of interest because it has a simple (integer) fixed point 2. So that is our standard reference (on this forum) base to compare two different methods of computing a tetration.
At 1<b<e^(1/e), there are two real fixed points; each of them should correspond to the analytic tetration. Now I try to plot them both; then hope to provide the algorithm for the precize evaluation. Then I shall run it at b=sqrt(2).
In such a way, my theorem is wrong at b < e^(1/e); and, perhaps, at b=e^(1/e); so, it should be reformulated: the equirement b > e^(1/e) should be included into the conditions of the Theorem.
bo198214 Wrote:As for example real regular iteration/tetration is no more possible for $b>e^{1/e}$ because there is no real fixed point.
I am not sure if I understand you well. At b=2 and b=e, tetration F(z) looks pretty regular (except $z\le-2$), and it is real at z>-2. Complex fixed points are easy to work with.
Kouznetsov Wrote:
bo198214 Wrote:As for example real regular iteration/tetration is no more possible for $b>e^{1/e}$ because there is no real fixed point.
I am not sure if I understand you well. At b=2 and b=e, tetration F(z) looks pretty regular (except $z\le-2$), and it is real at z>-2. Complex fixed points are easy to work with.

Oh I use "regular" in the sense of "regular iteration" this is a well studied (mostly by Szekeres and Ecalle) way to compute arbitrary real or complex iterates of a function at a fixed point. There is only one solution for the iterates such that the fixed point still remains analytic or at least asymptotically analytic, this is called regular iteration. You will find the iterational formulas as well the formulas for the coefficients of the powerseries of regular iteration throughout the forum (keywords: hyperbolic and parabolic iteration).

For tetration we have $b[4]t=\exp_b^{\circ t}(1)$. So if we have a fixed point of $\exp_b$ then we can just consider $\exp_b^{\circ t}$ to be the regular iteration, which gives us the regular tetration. In almost all cases the regular iteration at different fixed points give different solutions.

As I now see those regular tetration (at the lower real fixed point, which is btw the only attracting fixed point of $b^x$) is cyclic along the imaginary axis: For regular iteration we have the iterational formula:

$\exp_b^{\circ t}(1)=\lim_{n\to\infty} \log_b^{\circ n}(a(1-c^t) + c^t \exp_b^{\circ n}(1)),\quad c=\exp_b'(a)=\ln(a)$ where $a$ is the fixed point.

We see that the regular tetration $b[4](it)=\exp_b^{\circ it}(1)$ is periodic with $2\pi/\ln(\ln(a))$ so it can not have a limit for $t->\infty$.
Dmitrii, I am currently programming your tetration extension.
I just want to mention some misprints, that shouldnt go into your published paper.
In formulas (3.2), (3.3), (3.4), (3.6), (4.4) you always omit the minus sign in front of the 1 below the log. Only in the computation formula 4.2 the minus sign is at the right place.

There may be also some simplifications (avoidance of doublification with $L$ and $\overline{L}$) if you would put $F(\overline{z})=\overline{F(z)}$ into your assumption, which is quite reasonable and which you are also using in (4.7).

In formula (3.4),(4.4) either the $A$ in ${\mathcal K}_A$ should be omitted, or it should be appended whenever you use ${\mathcal K}$
bo198214 Wrote:Dmitrii, I am currently programming your tetration extension.
I just want to mention some misprints, that shouldnt go into your published paper.
In formulas (3.2), (3.3), (3.4), (3.6), (4.4) you always omit the minus sign in front of the 1 below the log. Only in the computation formula 4.2 the minus sign is at the right place.
Do you mean paper at http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp.pdf ?
bo198214 Wrote:There may be also some simplifications (avoidance of doublification with $L$ and $\overline{L}$) if you would put $F(\overline{z})=\overline{F(z)}$ into your assumption, which is quite reasonable and which you are also using in (4.7).
Yes, I assume $F(z^*)=F(z)^*$, but it follows from the analyticity and the assumption that $F(z)$ is real at z>-2.
bo198214 Wrote:In formula (3.4),(4.4) either the $A$ in ${\mathcal K}_A$ should be omitted, or it should be appended whenever you use ${\mathcal K}$