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Kouznetsov Wrote:
bo198214 Wrote:..
Kouznetsov Wrote:I doubt about "isolated". There should be a branchpoint.
So thats something left to find out
I work with slog. I have rotated the cutlines, and I found out the $2\pi \mathrm{i}$ periodicity and set of logarithmic singularities (with set of cutlines, of course) at the left hand side from the cutline
[attachment=423]
I do not copypast here desctiption from the previous figure; the only changes: I have rorated the cutline for 90 degrees, and I add levels $\Re(\sqrt{\exp}(z))=-3,-4,..-9$

But thats not the second fixed point of exp, is it?
Do you mean singularities at $z\approx -0.4+\pi \mathrm{i}+2\pi m), ~m\in \mathbb{N}$ ?
They are not fixed point of exp; $\exp(-0.4+\pi \mathrm{i}+2\pi m)\approx -0.7$ is real.
Why do you expect $\sqrt{\exp}$ to have many fixed points?

Another question: Could you deduce analytically the coefficients in the first terms of the asymptotic expanstion of slog in civinity of the fixed point?
Kouznetsov Wrote:Why do you expect $\sqrt{\exp}$ to have many fixed points?

As I already said, if $\exp$ has a fixed point then also $\exp^n$ has that fixed point for every integer $n$, at least on some branch. So I would expect that it has that fixed point also for every real $n$ on some branch.

Additionally for each singularity $s$ of $\text{slog}$ there should be singularities at $\log(s)+2\pi i k$ (on some branch), because:
$\text{slog}(s)=\text{slog}(\exp(\log(s)+2\pi i k))=\text{slog}(\log(s)+2\pi i k))+1$

A similar evidence was brought up by jdfox already
here
and by Andrew here

Those singularities are also singularities of $\sqrt{\exp}$ by
$\sqrt{\exp}(z)=\text{sexp}(0.5+\text{slog}(z))$
(as long as $\Re(\text{slog}(z))>-2$)

Quote:Another question: Could you deduce analytically the coefficients in the first terms of the asymptotic expanstion of slog in civinity of the fixed point?

Asymptotic from where?
bo198214 Wrote:As I already said, if $\exp$ has a fixed point then also $\exp^n$ has that fixed point for every integer $n$, at least on some branch. So I would expect that it has that fixed point also for every real $n$ on some branch.
Every fixed point of $\sqrt{f}$ is fixed point of $f$.
Some of fixed points of $f$ can be also fixed points of $\sqrt{f}$.

bo198214 Wrote:Additionally for each singularity $s$ of $\text{slog}$ there should be singularities at $\log(s)+2\pi i k$ (on some branch), because:
$\text{slog}(s)=\text{slog}(\exp(\log(s)+2\pi i k))=\text{slog}(\log(s)+2\pi i k))+1$
Yes, I see these singularities if I direct the cuts to the right hand side.
However, the higest fixed points of exp cannot be also those of $\sqrt{exp}$, at least, in the branch, where the function is $2\pi\mathrm{i}$ periodic: the fixed points of exp are not equidistant.

bo198214 Wrote:
Kouznetsov Wrote:Another question: Could you deduce analytically the coefficients in the first terms of the asymptotic expanstion of slog in civinity of the fixed point?
Asymptotic from where?
$
\mathrm{slog}(z)=\frac{1}{L}\left( \log(z-L) + r - (z-L)a+(z-L)^2 B +\mathcal{O}(z-L)^3 \right)
$

where $r\approx 1.07796 - 0.94654 \mathrm{i}$,
$a=0.5/(L-1)$ and so on.. $L$ is fixed point of log.
Kouznetsov Wrote:Every fixed point of $\sqrt{f}$ is fixed point of $f$.
Some of fixed points of $f$ can be also fixed points of $\sqrt{f}$.
So it remains still open whether the fixed point of $\log(z)+2\pi i$ is a fixed point of $\sqrt{\exp}$ on some branch.

Quote:$
\mathrm{slog}(z)=\frac{1}{L}\left( \log(z-L) + r - (z-L)a+(z-L)^2 B +\mathcal{O}(z-L)^3 \right)
$

where $r\approx 1.07796 - 0.94654 \mathrm{i}$,
$a=0.5/(L-1)$ and so on.. $L$ is fixed point of log.

That looks ingenious, Dmitrii! As for my first observation, this is an extension of Andrew's slog method - which does not work at fixed points - to fixed points. It seems that the coefficients can be computed directly without limits! This approach is also similar to the one of regular iteration where the Abel function is the logarithm at the fixed point plus some power series. As my time is currently really limited, I hope I can make a more concrete reply (or better a new thread as it is no more this topic) tomorrow.
bo198214 Wrote:I can make a more concrete reply
The "more concrete reply" (and the maple code that calculate the coefficients), is at
http://math.eretrandre.org/tetrationforu...hp?tid=215

Here I continue about branchpoints. Cutlines of the superlogarithm transfer to cuts of $\sqrt{\exp}$. Therefore, if one is interested to look behind the cutlines,
first, first look, that is happening with slog. I attach the contourplot of real and imaginary parts of slog in the left figure. It was already posted there, but now I post it with impeover desolution. At the right hand side, the similar figure with cut moved down.
[attachment=429]
Technically, for moving the cuts down, instead of slog, I plot the function $\mathrm{slog}_{\mathrm{u}}$ defined with $\mathrm{slog}_{\mathrm{u}}(z)=\mathrm{slog}(\log(z))+1$.
In this case, the cutlines remain straight, but the additional cutline along the negative part of the real axis appears.
Define the new operation cutter, which maked from a function $f$ function
$\mathrm{cutter}(f)(z)= f(log(z))+1$
In the right hand side of the picture above, cutter(slog) is plotted. Applying this cutter sequentially, one may cut the complex plane to the upper part and the lower part, which are holomorphically connected only at the right hand side of the real axis.

In the similar way, define the operation shroeder with equation $\mathrm{schroeder}(f)(z)= f(\exp(z))-1$. In the two pictures below I plot $\mathrm{shcroeder}(\mathrm{slog})$ and $\mathrm{shcroeder}^2(\mathrm{slog})=\mathrm{shcroeder}\Big(\mathrm{shcroeder}(\mathrm{slog})\Big)$ in the same notations.
[attachment=430]
The schroeder cuts the (simple) complex plane to the complicated (really "complex") labirinth with walls created by the cutlines. The cutlines (and singularities) of the sexp also contribute to the cutlines (and singulatities) of the generalized exponential.
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