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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    $\begingroup$ For the fast growing hierarchies, isn't $f_3(n)$ typically tetration hy definition? $\endgroup$
    – Mark S.
    Commented Mar 22, 2021 at 11:55
  • $\begingroup$ @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. $\endgroup$ Commented Mar 22, 2021 at 15:55
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    $\begingroup$ 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. $\endgroup$ Commented Mar 25, 2021 at 7:26
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    $\begingroup$ ... 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: math.stackexchange.com/questions/65876/thoughts-about-ffx-ex/… $\endgroup$ Commented Mar 25, 2021 at 7:29
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    $\begingroup$ 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) $\endgroup$ Commented Mar 27, 2021 at 23:39

2 Answers 2


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}


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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    $\begingroup$ It's interesting to see Faà Di Bruno on the scene. $\endgroup$ Commented Mar 27, 2021 at 3:15
  • $\begingroup$ 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? $\endgroup$ Commented Mar 28, 2023 at 6:01

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.

  • $\begingroup$ Can you provide published references for your tetration expressions? $\endgroup$
    – user63188
    Commented Mar 28, 2021 at 16:10
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    $\begingroup$ @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. $\endgroup$
    – Anixx
    Commented Mar 28, 2021 at 16:14
  • $\begingroup$ Thanks, very helpful. $\endgroup$
    – user63188
    Commented Mar 28, 2021 at 16:40

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