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+--- Thread: On extension to "other" iteration roots (/showthread.php?tid=1351)
RE: On extension to "other" iteration roots - JmsNxn - 08/15/2022
Bo, again. I think you are misinterpreting me.
Let's say that \(z_0 = \inf_{z \in \mathbb{C}} \{|\Im(z_0)|\,|\, b^{z_0} = z_0\}\). Then within the shell thron region for \(b > e^{-e}\) the infimum is \(z_0\) and it is real valued. When you let \(b < e^{-e}\), the value \(z_0\) is split into fixed point pairs \(z_0^{\pm}\) where \(|\Im(z_0)|> 0\).
This is what I believed Leo was referring to, when I made my typo; that we no longer can take the P-iteration--or the Schroder iteration with \(\cos(\pi x)\) (as you did), because we have to map between two fixed points; as opposed to the single real valued fixed point when \(z_0 \in \mathbb{R}^+\).
I mean, I'm not sure what you mean by that question. Primary, was as I defined it in the post; which is the infimum I just drew out. I guess I thought that was apparent. I really am just saying, that it's going to be a bit more involved outside of the Shell thron region, and I agree with everything you've said and within the thread. I apologize if I've said something off.
Regards.
RE: On extension to "other" iteration roots - bo198214 - 08/15/2022
(08/15/2022, 04:46 AM)JmsNxn Wrote: Let's say that \(z_0 = \inf_{z \in \mathbb{C}} \{|\Im(z_0)|\,|\, b^{z_0} = z_0\}\). Then within the shell thron region for \(b > e^{-e}\) the infimum is \(z_0\) and it is real valued. When you let \(b < e^{-e}\), the value \(z_0\) is split into fixed point pairs \(z_0^{\pm}\) where \(|\Im(z_0)|> 0\).
We really have a problem here. I thought from the picture I gave it was clear that there is no split at eta minor - it is a totally continuous behaviour regarding the fixed points. The \(\inf \{|\Im(z_0)|\,|\, b^{z_0} = z_0, z_0 \in \mathbb{C}\}\) is 0 because of the fixed point on the real line (in and out of the STR). I really have no clue what split you are referring to or what essentially changes in the fixed points when passing eta minor. Inside and outside we have one real fixed point and many conjugated fixed point pairs.
RE: On extension to "other" iteration roots - JmsNxn - 08/17/2022
(08/15/2022, 07:47 AM)bo198214 Wrote:(08/15/2022, 04:46 AM)JmsNxn Wrote: Let's say that \(z_0 = \inf_{z \in \mathbb{C}} \{|\Im(z_0)|\,|\, b^{z_0} = z_0\}\). Then within the shell thron region for \(b > e^{-e}\) the infimum is \(z_0\) and it is real valued. When you let \(b < e^{-e}\), the value \(z_0\) is split into fixed point pairs \(z_0^{\pm}\) where \(|\Im(z_0)|> 0\).
We really have a problem here. I thought from the picture I gave it was clear that there is no split at eta minor - it is a totally continuous behaviour regarding the fixed points. The \(\inf \{|\Im(z_0)|\,|\, b^{z_0} = z_0, z_0 \in \mathbb{C}\}\) is 0 because of the fixed point on the real line (in and out of the STR). I really have no clue what split you are referring to or what essentially changes in the fixed points when passing eta minor. Inside and outside we have one real fixed point and many conjugated fixed point pairs
I realized that that infimum appears meaningless.
There's a split in the sense that \(b^z\) no longer has an attracting fixed point on the real line; there is not a discontinuity in the Fixed points. The fixed points follow a continuous curve; I know this. There's no splitting of fixed points. I said they are not "primary"; and once you have \(b < \eta^-\), we no longer have an attracting fixed point on the real line. Okay, so to maintain real valued; we need our "primary" fixed point to be the up/down fixedpoints in the complex plane. And in doing such, we won't have the orbit about \(0\) in our image; so it won't have \(0,1,b,...\) in its image.
What I meant by "split" I meant: not a split in the continuity of the fixed points; there's a split in our choice of my words to describe what primary is. By which we don't have \(0,1,b\) in the orbit, but it is still a real valued super function.
I think I just misspoke one too many times and I keep digging a hole here. I'm happy to take the L, but when you switch from \(b > \eta^-\) to \(b < \eta^-\) there is a "split" which I used in my own internal language. I apologize bo, if I can appear thick.
RE: On extension to "other" iteration roots - Leo.W - 08/17/2022
Quote:Actually your expertise and technical arsenal in creating superfunctions is a bit wasted on me.
I am still puzzled how a superfunction can not come out of a iteration (semi)group.
In fact all superfunctions can come out of an iteration group, but the price is that it almost always is multivalued, we mostly speak and talk about a superfunction and automatically take it as singlevalued and avoid talking about branch cuts, but the natural way is to consider the superfunction as a multivalued function, or almost equivalently a riemann surface.
I don't think it's expertise or something or even "wasted", it's very nice to have a simplication as you did.
I pointed out that your \(tet_{e^{-2e}}\) is also generalized by P, only to remind you that it's different from a tetration. It's because
if we have a superfunction \(F(z+1)=f(F(z))\), then F(-z), is superfunction of the inverse of f, that is, \(F(-(z+1))=f^{-1}(F(-z))\), so this time it's not tetration, since it oscillate and eventually converges at 1 limit at negative infinity.
Quote:The name of the method is in your hand, you even called it re-construction method once.Sure. I just personally don't have a brain to memorize all different names. I think it's pretty good to call it P method.
I would say to choose a name wisely in the beginning and don't say things like "call it what you want".
We have enough confusion about naming, so that would really improve things to start with good names in the first place.
Quote:James: that the nearest fixed points are the conjugate pairs (which are still there for b=η−+δ, but aren't the "primary fixed point/points"). But once you go outside of the shell thron region, you get (when forcing real valued solutions) two complex conjugate fixed points.I think it's my fault I didnt clarify the "2 limits" or "2 fixed points".
Bo: We really have a problem here. I thought from the picture I gave it was clear that there is no split at eta minor - it is a totally continuous behaviour regarding the fixed points. The inf{|ℑ(z0)||bz0=z0,z0∈C} is 0 because of the fixed point on the real line (in and out of the STR). I really have no clue what split you are referring to or what essentially changes in the fixed points when passing eta minor. Inside and outside we have one real fixed point and many conjugated fixed point pairs.
The 2 fixed points refers to the 3 real fixed points of \(f(z)=b^{b^z}\), and excluded the one real fixed point of \(f^{1/2}(z)=b^z\). It's exactly the bifurcation in main branch of chaos theory.
When b fits \(0<b<e^{-e}=\eta_-\), the function \(f(z)\) will always have 3 real fixed points, one of them are the real fixed point of \(b^z\). Because the excluded one fixed point is repelling, the tetration of b (to integer heights) won't converge and will oscillate between the other 2 values, or have 2 limits at infinity.
For example, when \(b=0.06\), we have 3 fixed points of \(b^{b^z}\), approx are 0.216898, 0.36158, 0.54323
where 0.36158 is also the fixed point of \(b^z\), this time the tetration to integer heights will oscillate between 0.216898 and 0.54323 like the pic shown below.
By James, it's correct about the splitting choices, when you take the split 2 fixed points and merge their superfunction it's liable to have a real tetration. It's just too hard to compute and make it converge. I'm trying to do this. It makes me feel that we should take a different contour integral as double-dagger method or others do, we should take a broader contour in real-axis-direction.
RE: On extension to "other" iteration roots - JmsNxn - 08/17/2022
THANK YOU, LEO! I KNEW I WASN'T TOTALLY WRONG.
You can make a super function at \(\eta^-\), through the bifurcation between both the upper and lower fixed points. It won't be tetrational, which is the final point you pull out. It won't be real valued and equal \(1\). It'll equal \(0\) somewhere in the complex plane when it's real valued. But it won't be "tetrational" as Bo, and Kouznetsov, described; when we consider it on the real line.
RE: On extension to "other" iteration roots - bo198214 - 08/17/2022
But, Leo.W, I really have to say, you set me on a totally wrong track here, with statements like
(08/12/2022, 05:21 PM)Leo.W Wrote: Well bo, if you take abelian property, it'll fail then
Leo.W Wrote: Albeit these superfunctions would oscillate around the fixed point as a limit at infty, thus uninvertible, and thus would not grant you for \(f^s\circ f^t=f^{s+t}\), they're contradicts.
Leo.W Wrote: And meanwhile the superfunction only guarantees \(F(z+1)=T(F(z))\) for some T, not \(F(z+t)=T^t(F(z))\) for all real or even complex t. So these examples indeed are superfunctions but wont always allow you to have \(f^s\circ f^t=f^{s+t}\)Well, in the end it lead me to find a way to handle the multi-valued case, so one could say you did something good
I interpret your post as to find a superfunction that is real-to-real and also preserve the property \(f^s\circ f^t=f^{s+t}\), and in that sense it's impossible though,
RE: On extension to "other" iteration roots - Leo.W - 08/17/2022
(08/17/2022, 07:46 AM)bo198214 Wrote: Well, in the end it lead me to find a way to handle the multi-valued case, so one could say you did something good
hey bo, it's my own fault to have said it's impossible to do so and led you to a wrong track, but we in everyday talk and in forum it's default to say something taking singlevalued-ness for granted, I'd refer to multivalued-ness when I have to just on my own
In fact, we talked a lot way before in my previous post thread
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1318
in which I defined ways to talk about multivalued iterations and compositions and so on, not very rigorously.
But in modern days, we still cannot compute a multivalued function easily, so we'd better looking and researching in the world of singlevalued-ness and mentioning multivalued-ness when possible or pertinent.
Also, it is for sure impossible to do so, to harvest such a superfunction with both abelian-ity and singlevalued-ness
RE: On extension to "other" iteration roots - bo198214 - 08/17/2022
(08/17/2022, 08:08 AM)Leo.W Wrote: but we in everyday talk and in forum it's default to say something taking singlevalued-ness for granted,
On the other hand, in the complex plane, it is rather the default to consider multivaluedness. E.g. log or sqrt, when you continue from one point around the singularity back to the same point, you arrive at a different value and that's also the secret with the iteration group, it is not so much having multiple separate values, they just come naturally into existence by analytic continuation (and that's why the graph is one line not many separate lines).
Apropos Abelian property: just want to remind you to be cautious with the term, because \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) does not automatically imply \(f^{s}\circ f^{t} = f^{s+t}\) if the latter is what you actually mean.
I say that because I just encountered that case with the real valued Fibonacci extension
\begin{align}
\phi'_t &:=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}\\
f^{t}(z) &:= \frac{\phi'_t + \phi'_{t-1}z}{\phi'_{t+1} + \phi'_t z}\\
\end{align}
There we have \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) but we don't have \(f^{s}\circ f^{t} = f^{s+t}\) for most s,t and hence it is not a continous iteration group.
And that would actually be your counterexample to have a real single valued function family, with the Abelian property at a fixed point with negative multiplier
RE: On extension to "other" iteration roots - tommy1729 - 08/17/2022
(08/17/2022, 10:46 AM)bo198214 Wrote:(08/17/2022, 08:08 AM)Leo.W Wrote: but we in everyday talk and in forum it's default to say something taking singlevalued-ness for granted,
On the other hand, in the complex plane, it is rather the default to consider multivaluedness. E.g. log or sqrt, when you continue from one point around the singularity back to the same point, you arrive at a different value and that's also the secret with the iteration group, it is not so much having multiple separate values, they just come naturally into existence by analytic continuation (and that's why the graph is one line not many separate lines).
Apropos Abelian property: just want to remind you to be cautious with the term, because \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) does not automatically imply \(f^{s}\circ f^{t} = f^{s+t}\) if the latter is what you actually mean.
I say that because I just encountered that case with the real valued Fibonacci extension
\begin{align}
\phi'_t &:=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}\\
f^{t}(z) &:= \frac{\phi'_t + \phi'_{t-1}z}{\phi'_{t+1} + \phi'_t z}\\
\end{align}
There we have \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) but we don't have \(f^{s}\circ f^{t} = f^{s+t}\) for most s,t and hence it is not a continous iteration group.
And that would actually be your counterexample to have a real single valued function family, with the Abelian property at a fixed point with negative multiplier
I call the continuous iteration group valid within a jordan curve on the complex plane and being analytic there a " flow " and the related ideas " flow theory ".
which bring me to the natural question
if we have a continuous iteration group for real s and t within real interval [a,b] , meaning s + t is within [a,b] and s and t being positive, does that mean that we have a " flow " near the interval [a,b].
In other words does the property hold for complex s and t near [a,b] and is it analytic there ?
and this is the same as the semi-group homom by simply adding the z in f^[s] ( f^[s](z) ).
If the answer is yes , this implies 2sinh method is analytic.
regards
tommy1729
RE: On extension to "other" iteration roots - Leo.W - 08/18/2022
(08/17/2022, 10:46 AM)bo198214 Wrote:Not to disappoint ya, bo, but you'd missed one thing is that, you proved the \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) identity taken its singlevaluedness, this will be the same as:(08/17/2022, 08:08 AM)Leo.W Wrote: but we in everyday talk and in forum it's default to say something taking singlevalued-ness for granted,
On the other hand, in the complex plane, it is rather the default to consider multivaluedness. E.g. log or sqrt, when you continue from one point around the singularity back to the same point, you arrive at a different value and that's also the secret with the iteration group, it is not so much having multiple separate values, they just come naturally into existence by analytic continuation (and that's why the graph is one line not many separate lines).
Apropos Abelian property: just want to remind you to be cautious with the term, because \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) does not automatically imply \(f^{s}\circ f^{t} = f^{s+t}\) if the latter is what you actually mean.
I say that because I just encountered that case with the real valued Fibonacci extension
\begin{align}
\phi'_t &:=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}\\
f^{t}(z) &:= \frac{\phi'_t + \phi'_{t-1}z}{\phi'_{t+1} + \phi'_t z}\\
\end{align}
There we have \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) but we don't have \(f^{s}\circ f^{t} = f^{s+t}\) for most s,t and hence it is not a continous iteration group.
And that would actually be your counterexample to have a real single valued function family, with the Abelian property at a fixed point with negative multiplier
Q: Find an abelian realvalued (singlevalued) superfunction for \(f(z)=z+1\)
A: It's okay you write \(f^t(z)=z+g(t)\) where g is any function with \(g(n)=n\) for integers because \(g(s)+g(t)=g(t)+g(s)\)
Thus there lies such a superfunction \(F(z+1)=f(F(z))=F(z)+g(t)\), \(F(z)=z+G(t)\) is a superfunction where \(G(t)=F(0)+\sum_{n=0}^{t-1}{g(n)}\), choose a real-valued arbitrary function g(n) with \(g(n)=n\) for integers n, will always promise you such a superfunction and abelian-ity.
But the core is, Abelian-ity for iterations must refers to \(f^{s+t}=f^s\circ f^t=f^t\circ f^s\), otherwise we're just doing almost the same thing as finding "All matrices A that commutes with B", it not suffices to call itself an iteration, just an Abelian group of functions that commutes with each other.
If you look into the multivaluedness of your superfunction, then your superfunction can preserve an Abelian-ity by some branch cuts.
Assume your superfunction \(F\) has a multivalued inverse \(H\), then since \(H(f)=H+1, F(z+1)=f(F)\), by conjugacy \(H(F(z)+1)=f(z)\), thus a naturally extended iterations \(H(F(z)+t)=f^t(z)\) and satisfy Abelianity. All equalities here represents "There lies such a branch which fits this equality".
Another thing, your extended fibonacci is used for so long, it's derived from a complex decomposition by Re and Im:
Assume \(f(z+1)=L(f(z),f(z-1),f(z-2),\cdots,f(z-n))\) has a complexvalued solution f where L is a linear function, then
\(\Re(f(z+1))=\Re(L(f(z),f(z-1),f(z-2),\cdots,f(z-n)))=L(\Re(f(z)),\Re(f(z-1)),\cdots,\Re(f(z-n)))\)
also true for replacing \(\Re\) with \(\Im\), So \(\Re(f(z))\) and \(\Im(f(z))\) are also solutions, your superfunction can be derived from \(\Re\) case.
It may be possible for you to get a real singlevalued/multivalued superfunction, try on this:
(since the hidden eigen equation is \(f(z+1)=f(z)+f(z-1)\) or fibonacci, it must follows any \(c_tf(z+t)\) satisfies the equation. We even can write \(G(z)=\int_{a}^{b}{f(z+t)a(t)\mathrm{d}t}\) or \(H(z)=\sum_{n=a}^{b-1}(c_nf(z+n))\) as extended solution, where a(t) arbitrary function, a,b arbitrary constant. It's almost the same as writing a series, or almost equivalently a thetamapping. But intrinsically it's only) assuming \(G(z)=a_1(t)r_1^z+a_2(t)r_2^z\). where \(r_1,r_2\) solves \(r^2-r-1=0\)