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

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

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,

Added a summary of what iterations about periodic points look like.
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Richard Diagram
Richard Diagram
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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 $\exp$ 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.

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 $\exp$ 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.

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Richard Diagram
Richard Diagram
  • 388
  • 1
  • 9

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.