On the relevance of the property $\exp^{\circ a}(\exp^{\circ b}(z))=\exp^{\circ a+b}(z)$ for the *fractional* iteration ("tetration")

Question

In the concept of fractional iteration of the exponential function ("tetration") the property of $$\exp^{\circ a}(\exp^{\circ b}(z))=\exp^{\circ b}(\exp^{\circ a}(z))=\exp^{\circ a+b}(z) \tag 1$$ has been/is a very basic, even a foundational one, for my understanding.

In some recent consideration of n-periodic points it occured to me, that this cannot hold (independently of the style of interpolation, be it "regular"/"Schroeder-" or "Kneser-" or any other method), where for example with the 3-periodic points $p_1$, $p_2=\exp(p_1)$, $p_3=\exp(p_2)$, $p_1=\exp(p_3)=\exp^{\circ 3}(p_1)$ we have with some point $p_{1.1} = \exp^{\circ 0.1}(p_1)$ the following

$$ p_{1.1}=\exp^{\circ 0.1}(\exp^{\circ 3}(p_1))=\exp^{\circ 0.1}(p_1) \tag {2.1} $$ $$ \text{ but } $$ $$ \exp^{\circ 3}(\exp^{\circ 0.1}(p_1)) \ne p_{1.1} \tag {2.2} $$

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Example data: $$ \small { \begin{array} {rll} p_1&=0.90866853431523997218 + 0.67624988547121701164*I \\ p_2&=1.9350078633658531684 + 1.5528005817432416377*I \\ p_3&=0.12459758600926988160 + 6.9229773584312693320*I \\ p_4 &= \exp^{\circ 3}(p_1) = p_1 \end{array} }$$ _{Remark 1: the value of $p_1$ can be made arbitrarily precise by initializing $\small{p_1=1+I}$ and then iterating $\small{p_1=\ln(\ln(\ln(p_1)+2\pi I))}$ to sufficient convergence}
$$ \small{ \begin{array} {rll} p_{1.1} &= \text{rtet}(p_1,0.1) &= 0.99523184831984789219 + 0.70987078655389452780*I \\ p_{4.1 \, a}&=\exp^{\circ 3}(\text{rtet}(p_1,0.1) ) & = 0.048217006014677061506 + 0.22062947724802093650*I \\ p_{4.1\,b}&=\text{rtet}(\exp^{\circ 3}(p_1),0.1)&= 0.99523184831984789219 + 0.70987078655389452780*I \\ p_{4.1\,a} & \neq p_{4.1\,b} \qquad \qquad\text{(!)} \end{array} } $$ _{Remark 2: here, rtet(z,height) is an implementation of fractional iteration, which is numerically approximate to the Kneser-solution as given by the implementation of S. Levenstein in the tetrationforum but works simply by diagonalization of a finite truncated Carlemanmatrix of size $\small {16 \times 16}$}

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That the two iterations in different orders cannot be equal can easily be seen by the argument,

• that the infinity of 3-periodic points (as well as in general n-periodic points) is countable,
• a continuous curve connecting $p_1 \to p_2 \to p_3 \to p_1 \to ...$ if it were itself periodic would represent uncountably many 3-periodic points,

but which is a contradiction.

Another, perhaps better known, argument comes from the allegory of a "hair" coined by R. L. Devaney in his study of periodic points in the exponential function. It says (paraphrased here)

• that in an epsilon neighbourhood of a periodic point (say $p_1$) the iteration of any point (except of $p_1$ itself) diverges, even chaotically, towards infinity.

Of course this implies as well that there cannot be a trajectory of fractional iterates, whose partial curves $l_{12}=[p_1,p_2]$, $l_{23}=[p_2,p_3]$ and $l_{31}=[p_3,p_1]$ have the 3-periodicity $\exp^{\circ 3}(l_{12}) \ne l_{12}$. (Actually that iterations blow the partial curves out quickly and few such iterations chaotize completely any graphical plot).

This inequality in (2.1) breaks the -in my view- foundational equality (1) such that I now question at all the meaningfulness of the fractional iteration in such cases.

• Do I possibly misrepresent/misinterpret the foundational role of (1)?

• Has tetration been developed so far with well knowing and possibly answering the problem in (2.2)?

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A discussion starting at some initial irritation because of existence of 6-periodic points, the chaotizing of the interpolated trajectories, towards more precise graphical display and towards finding the thoughts presented here can be checked at tetrationforum

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A picture illustrates the partial trajectories and their non-periodicity. The coordinates of the periodic points can be approximated to arbitrary precision by simple fixpoint-iteration: P1=log(log(log(P1)+2*Pi*I)) until satisfactory precision (I use, by default, 200 internal decimal digits with Pari/GP), or Newton-iteration starting at the given coordinates.

A short article discussing the initial observation of existence of periodic points is here This does not arrive at the discussion of the problem of non-periodicity of the fractional iterated trajectories along the n-periodic points.

Answer 1

I believe I can answer this, and it's a little tricky. Obviously, I don't think Gottfried is declaring that Kneser's solution to tetration is incorrect; and I don't believe one can show this from what Gottfried has stated. The trick; Gottfried has ignored a very important discussion of domains.

To elaborate I'll start with the simple case of $b = \sqrt{2}$ and the function $\phi(\xi) = b^{\xi}$. And then I'll extrapolate this to the case of Kneser's tetration when $b=e$.

To begin, let's focus on the fixed point $2$ in which $\phi(2) = 2$ and $0 < \phi'(2) = log(2) < 1$. Then there is an immediate basin of attraction $\mathcal{A}_0$ about $2$, which is connected, in which,

$\lim_{n\to\infty} \phi^{\circ n}(\xi) = 2$

If we take a neighborhood about $2$; we have a Schroder function which is injective on the neighborhood, in which,

$ \Psi( \phi(\xi)) = \log(2) \Psi(\xi)$

And we can construct a fractional iteration,

$\phi^{\circ z}(\xi) = \Psi^{-1} (\log(2)^z \Psi(\xi))$

This fractional iteration can be extended to at least a function,

$\phi^{\circ z}(\xi) : \mathbb{C}_{\Re(z) > 0} \times \mathcal{A}_0 \to \mathcal{A}_0$

Which satisfies the group law Gottfried mentioned. Now, particular to this case, is that $\mathcal{A}_0$ has no periodic points within it. This is by first principle, it's the immediate basin. So trying to do what Gottfried suggests, is in fact impossible.

So, what does this tell us about $e$? It's actually pretty obvious,

$ exp^{\circ z}(\xi)$ cannot be holomorphic in $\xi$ about any periodic point. This will actually arise as a type of branching error. To justify this, we'll stick to Kneser's Tetration, and Kneser's superlogarithm.

So the function $\text{tet}_K(z) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}$ and satisfies,

$ \text{tet}_K(0) = 1$

$\text{tet}_K(z+1) = e^{\text{tet}_K(z)}$

And the function,

$\text{slog}(1) = 0$

$\text{slog}(e^z) = \text{slog}(z) +1$

Kneser's tetration is bijective from $(-2,\infty) \to \mathbb{R}$, so let's choose that super-logarithm. Then in a neighborhood of the real line, it's holomorphic. Call a neighborhood $\mathcal{N} = \{z \in \mathbb{C}\,|\, |z -1| < \delta\}$ for an appropriately small $\delta>0$.

Then, by the functional equation $\text{slog}$ exhibits, this can be extended to the domain,

$\mathcal{S} = \bigcup_{n=0}^\infty \exp^{\circ n}(\mathcal{N})$

By properties of the exponential, it's well known (see Milnor's Dynamics in One complex variable, for instance) that,

$\overline{\mathcal{S}} = \mathbb{C}$

But, the only points not in $\mathcal{S}$, yes you guessed it, are the points of normality, which are all the periodic points. So Gottfried's suggested function,

$\exp^{\circ z}(\xi) = \text{tet}_K(z+\text{slog}(\xi))$

IS not holomorphic at these points of periodicity; at least, no one's proved that. In fact, you can prove that they can't be; which, we might as well do for completeness.

Assume that,

$\text{slog}(p_1) = A$

When then,

$\text{slog}(p_1) = A+3$

too...

This can't work. All in all, Tetration is safe, and Kneser is still correct. We just have to pay very close attention when we talk about iterating. Very close.

Hope this answers the question, Gottfried.

EDIT:

I thought I'd add that to iterate about periodic points, we don't just take the regular iteration. If we go back to $\phi(\xi) = \sqrt{2}^\xi$ and choose a sequence of periodic points $p_1,p_2,p_3$, then we certainly can iterate about these points. But, inherently it'll be a different iteration.

To clarify, the function $\phi^{\circ 3}(\xi)$ has a fixed point at each point $p_1,p_2,p_3$. And we can construct an iteration $\mu_j(z,\xi)$ about each of these points $p_j$. However, none of these commute $\mu_j(z,\mu_i(z',\xi)) \neq \mu_i(z',\mu_j(z,\xi))$. These functions are different iterations about a different fixed point.

In such a sense, we don't get the nice clean identity we'd want. It follows that,

$(\phi^{\circ 3}(\xi))^{\circ z/3} \neq \phi^{\circ z}(\xi)$

And these certainly aren't reconcilable as a single function about all three fixed points. I think the best way to explain this, is that there exists ANOTHER $\phi^{\circ z}(\xi)$ about the fixed point $4$; and in absolutely no way are the two iterations (about $2$ and about $4$) the same thing, or commute, or satisfy a group law. Nicely enough, these functions are holomorphic in $\xi$ on different domains. Similarly to each $\mu_j$, they are defined for different domains in $\xi$. And furthermore, there will be a jump discontinuity if we try to move $\xi$ to being about a different fixed point because we will cross the Julia set somewhere here; where no nice iteration exists (except for the iteration about $4$).

It's one of these moments that we should take a step back and really remember what $\phi^{\circ z}$ means. It's about a fixed point, and it's specific to a fixed point. And different fixed points give different answers. And these iterations don't like other fixed points, or other periodic points; they'll blow up either at the other fixed point (like with Kneser), or they'll blow up before they even get close (we hit the Julia set in our path towards the other fixed point).

What you are doing is similar to taking three fixed points $p_1,p_2,p_3$ of $\exp^{\circ 3}$ and expecting the iteration,

$\rho_j(z,\xi)$

which satisfies $\rho(m,\xi) = \exp^{\circ 3m}(\xi)$ and $\rho_j(z,p_j) = p_j$ to be equal. Or even worse, that $\rho_j(z/3,\xi)$ will be Kneser's iteration...spoiler alert, it won't be.