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This question arose when I read a (very introductory) googology book. Tetration is essentially repeated exponentiation (right-associative), just like how multiplication is repeated addition, defined for integers $a>0,n\geq0$: $$a\uparrow\uparrow n=\begin{cases}1, n=0\\ a^{a\uparrow\uparrow(n-1)}, n>0\end{cases}$$

My question is whether there are nice extensions of this operation to the reals, like how the gamma function extends the factorial. Personally I don't expect it to be extended nicely to the complex numbers (given how $i^i$ is not even nicely defined), but maybe for reals we could use monotonicity (though very loose) to bound it somehow. I have seen an answer here but it seems to be a uniqueness theorem and an algorithm, and I wonder if there are more "explicit" constructions (using elementary functions, integrals, elliptic functions, etc). I may have overlooked something though.

I also considered a weaker version: can we find a simpler function that mimics the growth rate of tetration (and is also defined on the reals)? This should probably be an easier task, so in exchange I would like to see more "elementary" constructions. (By growth rate, I think big-O is too strict and maybe FGH is better, but I'm open to more reasonable options.)

At last, if we don't currently have an answer as satisfactory for tetrations as for factorial, what's the difficulty here? Is it because the recursive relation is more complicated? Or do we need stronger arithmetic to even prove that tetration is a function? (I don't know much about the latter though, and only listed it here as a possibility)

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    For the fast growing hierarchies, isn't $f_3(n)$ typically tetration hy definition? – Mark S. Mar 22 '21 at 11:55
  • @MarkS. Yes, but I thought $f_3$ is also defined by recursion, thus an integer function and not a real extension. I have edited my question to be clearer. – Tesla Daybreak Mar 22 '21 at 15:55
  • Perhaps https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-compositional-square-root is a good source for discussion of basic difficulties with the fractional iteration of the $\exp()$. – Gottfried Helms Mar 24 '21 at 14:19
  • @GottfriedHelms Yes, I see how that would be helpful. It confirms by suspicion that nice complex extensions are hard to have. But we can largely get around the question by only looking at the reals. – Tesla Daybreak Mar 24 '21 at 17:28
  • ... yes, which is already a conceptional decision. I've once done a discussion of the two interpolations of the fibonacchi-numbers: the real-only one and the complex one; and while the real-only one was preferred by ,say, wikipedia, for my taste the complex one is a more "natural" one. The same then with tetration: the Kneser- (parallel with the "polynomial" (as I called it)) solution as real-to-real might widely be preferred, but the more "natural" one seems to me the complex solution along the Schroeder-/Koenigs mechanism. – Gottfried Helms Mar 24 '21 at 18:58
  • ... for a quick reference one might look at my small essay on comparision of 5 interpolation methods http://go.helms-net.de/math/tetdocs/ComparisionOfInterpolations.pdf to see what I mean. The discussion of the Fibonacchi-interpolation occured accidentally, when I simply wanted to do an exercise with "an easy case" of function. See http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm . I think it is not trivial to declare the real-to-real or the complex one as "better" general solution! – Gottfried Helms Mar 24 '21 at 19:02
  • @GottfriedHelms After going through your links, I kind of see why the complex extension would have its appeal; being able to see the spirals is a great pro. However I'm still bothered by some unexpected behaviors near the real axis--guess that's personal preference. In any case, thank you very much for the illuminating stuff. I'll be looking for more "growth-rate" approximations, since "nice" exact extensions seem already hard enough. – Tesla Daybreak Mar 25 '21 at 05:01
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    Wish you a similar exciting and intellectual adventurous journey like I had with this all! I'm a bit tired after this 12-year quest and likely shall not step in deeper again. But perhaps there is some more of the older material helpful for improving intuition ... One which might still be interesting is a vizualisation of smoothness of the real-to-real case, where one can insert an initial guess for the fractional iteration interpolation to see, how the resulting curve would be smooth or edgy. This is so far only in a Excel (2000) file which I could make available if it is of interest. – Gottfried Helms Mar 25 '21 at 07:26
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    ... ah, here is a link where I've explained and showed what I mean with that vizualisation, but only a screenshot of the Excel-file: https://math.stackexchange.com/questions/65876/thoughts-about-ffx-ex/66042#66042 – Gottfried Helms Mar 25 '21 at 07:29
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    Did you think already of using a base $a$ from the interval $(1,\exp(\exp(-1)))$? This makes things much easier, because such bases have a real fixpoint and one can (via Schroeder-mechanism) construct a powerseries which converts $a\uparrow \uparrow h$ into $ q \log(t)^h$ (where for $t$ one uses the attracting fixpoint, and $q$ is some normed value), and then the fractional iterate can be computed by the series-reverse. (I don't want to make more text here in the comment, maybe this is trivial knowledge for you) – Gottfried Helms Mar 27 '21 at 23:39
  • The technique I use is completely general for tetration regardless of the base being non-zero. More importantly is the mathematics that presents itself. For example is the combinatorics of iterated functions. –  Mar 28 '21 at 00:12
  • @Daniel - upps, sorry, I meant to ask the OP ... – Gottfried Helms Mar 28 '21 at 00:34
  • @GottfriedHelms No I have not known this technique. I assume it is similar to the second answer here, which is fun but I have not seen before (given how much of a freshman I am). It's good to see something more explicit, but if the base cannot even be 2 then I guess that's not the best thing I'm looking for... – Tesla Daybreak Mar 28 '21 at 19:04
  • But still, already quite illuminating. – Tesla Daybreak Mar 28 '21 at 19:04

2 Answers2

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Extensions of tetration

  1. Counter-intuitively it appears that extending tetration to the complex numbers is easier than extending tetration to the real numbers. Likely because the complex numbers have more degrees of freedom.
  2. Special functions for extending tetration - Faà Di Bruno's formula and the Bell polynomials are invaluable in extending tetration. They provide exact solutions to tetration of whole numbers. Most extensions of tetration to the real and complex numbers are only approximations.
  3. As a member of the Ackermann function, tetration denotes a unique growth rate.
  4. The difficulty with finding special functions for constructing tetration lies with the powers being so successful in constructing functions. But the umbral calculus emphasizes other fundamental processes besides the powers, like composition.

Consider iterated functions as iterated composition.

Faà Di Bruno's formula is $$D^mf(g(z)) = \sum_{\pi(m)} \frac{m!}{k_1! \cdots k_m!} (D^kf)(g(z)) \left(\frac{Dg(z)}{1!}\right)^{k_1} \cdots \left(\frac{D^ng(z)}{m!}\right)^{k_m}$$

where $\pi(m)$ denotes a partition of $m$, usually denoted by $1^{k_1}2^{k_2}\cdots n^{k_m}$, with $k_1+2k_2+ \cdots mk_m=k$; where $k_i$ is the number of parts of size $i$. The partition function $p(m)$ is a decategorized version of $\pi(m)$, the function $\pi(m)$ enumerates the integer partitions of $m$, while $p(m)$ is the cardinality of the enumeration of $\pi(m)$.

Mathematica code for Faà Di Bruno's formula $$\small D[f[g[z]], \{z, m\}] = \ \underset{K[1]=0}{\overset{n}{\sum }}f^{(K[1])}[g[z]]\ \text{BellY}\left[m,K[1],\text{Table}\left[g^{(K[2])}[z],\{K[2],1,-K[1]+m+1\}\right]\right]$$


$$D^m \ a\uparrow \uparrow z = D^m \ a^{(a\uparrow \uparrow {(z-1)})} = \sum_{\pi(m)} \frac{m!}{k_1! \cdots k_m!} (D^k \ {a^z})(a^{(z-1)}) \left(\frac{D \ ^{z-1}a}{1!}\right)^{k_1} \cdots \left(\frac{D^n \ ^{z-1}a}{m!}\right)^{k_m}$$

$$a\uparrow \uparrow b=\sum_{m=1}^{\infty} \frac{1}{m!} \ D^m \ (a\uparrow \uparrow b) \ (b-a\uparrow \uparrow \infty)^m$$

Fixed Points and Infinite Solutions

When the Lyapunov multipliers are a root of unity, a special solution is called for. Since there are infinite roots of unity, there are infinite solutions.

$f(0)=f_0=0$ - Fixed Point

$q \in \mathbb{Q}$, $n \in \mathbb{N^+}$, $\lambda \in \mathbb{S1}$

$\lambda = e^{2 q \pi i}$ - root of unity (symmetry)


$f'(0)=1$ - Abel's Functional Equation produces series of polynomials

$$f^t(z)=z+\frac{1}{2}t f_2 (z-f_0)^2 +\frac{1}{12}(3(t^2-t)f_2^2+2tf_3)(z-f_0)^3+\ldots$$

$f'(0)\ne \lambda^n$ - Schröder's Functional Equation produces series of exponentials

\begin{eqnarray} f^t(z)=f_0 &+& \lambda ^t (z-f_0)+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) f_2}{2 (-1+\lambda )} (z-f_0)^2 \\ & + & \frac{1}{6} \left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right) f_2^2}{(-1+\lambda )^2 (1+\lambda )}+\frac{\lambda ^{-1+t} \left(-1+\lambda ^{2 t}\right) f_3}{-1+\lambda ^2}\right) (z-f_0)^3+\ldots \end{eqnarray}

$f'(0)= \lambda^n$ - Neutral Fixed Points


Neutral Fixed Points in Tetration

$e^{e^{2 \pi i x-{e^{2 \pi i x}}}}$

Neutral Fixed Points in Tetration


Tetration Mandelbrot Fractal

Tetration Mandelbrot Fractal


\begin{eqnarray} {}^t a = a_o & + & \lambda ^t\left(1-a_o\right)+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \text{Log}\left(a_o\right){}^2}{2 (-1+\lambda )}\left(1-a_o\right){}^2 \\ & + &\frac{1}{6}\text{ }\left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right)\text{ }\text{Log}\left(a_o\right){}^4}{(-1+\lambda )^2 (1+\lambda )}\\ +\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \left(1+\lambda ^t\right)\text{ }\text{Log}\left(a_o\right){}^3}{(-1+\lambda ) (1+\lambda )}\right)\left(1-a_o\right){}^3+\ldots \end{eqnarray}


Convergence

The convergence of each type of fixed point needs to be considered in order to prove tetration as a whole always converges.

Convergence can be proven by functional equation.

Points described by the Schröder's Functional Equation are dense in the complex plane. It's fractal spirals are exponential spirals, excluding negative powers like $z^{-1}$. But repeated exponentiation is convergent.

On the other hand are the roots of unity which manifest as the symmetry of the exponential function. For each root there are an infinite number of "larger" associated numbers from the Schröder's Functional Equation that are convergent.

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    It's interesting to see Faà Di Bruno on the scene. – Tesla Daybreak Mar 27 '21 at 03:15
  • Can you please clarify the iterating function you used for the mandelbrot set? For the "original" mandelbrot set, I expect that this is f(z) = z^2 + c. What iterating function did you use to draw the image you attached? – Vivian River Mar 28 '23 at 06:01
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What makes it difficult? The growth rate. The Taylor series at this growth rates usually do not converge.

If you relax the condition so to find a solution for $f(x+1)=a^{f(x)}$ such that $$a \le e^{1/e} $$ then there are multiple expressions for tetration:

$$f(x)=\sum_{m=0}^{\infty} \binom xm \sum_{k=0}^m \binom mk (-1)^{m-k}\exp_a^{[k]}(1)$$

$$f(x)=\lim_{n\to\infty}\binom xn\sum_{k=0}^n\frac{x-n}{x-k}\binom nk(-1)^{n-k}\exp_a^{[k]}(1)$$

$$f(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{2n} \frac{(-1)^k \exp_a^{[k]}(1)}{(x-k)k!(2n-k)!}}{\sum_{k=0}^{2n} \frac{(-1)^k }{(x-k) k!(2n-k)!}}$$

$$f(x)=\lim_{n\to\infty} \log_a^{[n]}\left(\left(1-\left(\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\right)^x\right)\frac{W(-\ln a)}{-\ln a}+\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\exp_a^{[n]}(1)\right)$$

Always here the number in square brackets designates n-th iteration and $W(x)$ is the Lambert's function.

There is also an expression for inverse function:

$$ f^{[-1]}(x)=\lim_{n\to\infty} \frac{\ln \left(\frac{\frac{W(-\ln a )}{\ln a}+\exp_a^{[n]}(x)}{\frac{W(-\ln a)}{\ln a}+\exp_a^{[n]}(1)}\right)}{\ln \ln \left(\frac{W(-\ln a)}{-\ln a}\right)}$$

As to for the case $a > e^{1/e}$, I know only the iterative solution with Cauchy integrals by Dmitrii Kouznetsov and Henryk Trappmann.

Anixx
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  • Can you provide published references for your tetration expressions? –  Mar 28 '21 at 16:10
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    @DanielGeisler the first one is Newton series expansion. Second one is Lagrange. Third... Is something like that as well. The fourth and fifth ones I have seen on Tetration Forum (tetration.org), you can ask there. Since you asked for general solution for any base, the only one is that I linked. – Anixx Mar 28 '21 at 16:14
  • Thanks, very helpful. –  Mar 28 '21 at 16:40