# An explicit series representation for the analytic tetration with complex height

Tetration is the next hyperoperation after more familiar addition, multiplication and exponentiation. It can be seen as a repeated exponentiation, similar to how exponentiation can be seen as a repeated multiplication, and multiplication — as a repeated addition. Tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence $${^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$ so that $${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}},\tag2$$ where power towers are evaluated from top to bottom. To simplify matters, we restrict our attention to real bases in the interval $1<a<e^{1/e}$. Under this restriction, the sequence $\{^n a\}$ is strictly increasing and converges to a limit: $$^\infty a=\lim_{n\to\infty} {^n a} = \frac{W(-\ln a)}{-\ln a},\tag3$$ where $W(z)$ denotes the principal branch of the Lambert W-function. To simplify our following exposition it is convenient to introduce a new variable $q$ linked to $a$: $$q = \ln({^\infty a}) = {^\infty a} \cdot \ln a = -W(-\ln a).\tag4$$ The bounds on $a$ imply $0<q<1$. The value $q$ is closely related to the asymptotic growth of the tetration. In particular, it appears in remarkable limits: $$\lim_{n\to\infty} \, \frac{{^{n+k} a}-{^n a}}{{^{n-k} a}-{^n a}} = -q^k, \quad \lim_{n\to\infty} \, \frac{{^{n+k} a}-{^\infty a}}{{^n a}-{^\infty a}} = q^k .\tag5$$ Definitions of addition, multiplication and exponentiation for fractional, negative and complex arguments are well-established. A natural extension of the tetration to fractional and complex heights is a long-standing question. It appears to be much less straightforward problem, mainly because exponentiation, unlike addition and multiplication, lacks both commutativity and associativity, and so tetration does not obey simple laws similar to $a^m \cdot a^n = a^{m+n}$ or $(a^m)^n = a^{m\,n}$.

Despite these obstacles, several proposals aiming to extend the definition of the tetration to fractional and complex heights have been put forth. Their common goal is to define a function that agrees with the discrete tetration for all positive integer heights, and satisfies the functional equation $$^{z+1} a = a^{\left({^z a}\right)}\tag6$$ for all $z$ in its domain. These conditions alone are not sufficient to determine a unique function, so additional natural restrictions (such as continuity, smoothness or analyticity) have been proposed, aiming to narrow down the set of candidate functions to a single and, in some sense, "the most natural" extension. The Wikipedia article on tetration is very cautious and indicates that, apparently, so far there is no consensus about which extension is "the one". Some of the proposals seem to lead to the same function, although apparently no rigorous proof of their equivalence has been published yet. The Citizendum article is written in a much bolder style, and refers to tetration as a unique well-defined function. But, overall, the study of tetration seems to be a quite isolated area of mathematics, with few connections to other topics.

A few months ago I made some interesting conjectures related to tetration that I hoped I would prove soon. It appears that there are some connections between tetration and the theory of $q$-analogs. Unfortunately, I have not made much progress in proving them since then (although I have learnt a lot about $q$-series), so I finally decided to post them here and ask for your help. First, let's recall some definitions.

The $q$-Pochhammer symbol: $$(p;\,q)_\infty = \prod_{k=0}^\infty(1-p\,q^k)\tag7$$ $$(p;\,q)_z = \frac{(p;\,q)_\infty}{(p\,q^z;\,q)_\infty}, \quad z\in\mathbb C\tag8$$ It easily follows that $$(p;\,q)_n = \prod_{k=0}^{n-1}(1-p\,q^k), \quad n\in\mathbb N\tag9$$

The $q$-binomial coefficients (also known as the Gaussian binomial coefficients): $${r \brack s}_q = \frac{(q;\,q)_r}{(q;\,q)_s \, (q;\,q)_{r-s}}.\tag{10}$$

From now on we assume that $a$ and, consequently, $q$ are fixed. Let us define an analytic function $t(z)$ by the following series:

$$t(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\binom {n-k} 2} {z \brack n}_q {n \brack k}_q ({^k a}).\tag{11}$$

This formula can be seen as a combination of the direct and reverse $q$-binomial transforms. It can be proved by induction that for $n\in\mathbb N, \, t(n)={^n a}$, so $t(n)$ satisfies the functional equation for tetration $(6)$ at least for positive integer arguments. Indeed, partial sums of $(11)$ can be seen as Lagrange interpolating polynomials in powers of $q^z$ that exactly reproduce tetration values for progressively larger initial segments of $\mathbb N$. Numeric evidence suggests that adding new points to the interpolating polynomial of this form does not result in erratic oscillations of its arcs between existing points (i.e. Runge's phenomenon does not occur), but rather makes them converge to a monotone function satisfying the functional equation for the tetration $(6)$. So we have:

Conjecture 1. The series $t(z)$ converges at least in the half-plane $\Re(z)>-2$ to an analytic function that satisfies the functional equation for tetration $t(z+1)=a^{t(z)}$ for all arguments in its domain of analyticity.

Conjecture 2. The function $t(z)$ is periodic with a purely imaginary period $\tau = 2\pi n i/\ln q$, and admits an analytic continuation to the half-plane $\Re(z) \le -2$ by repeated backward application of the functional equation $(6)$. There it has a countable set of isolated singularities at $z = m + \tau\,n, \,m,n\in\mathbb Z, \, m \le -2$, and a countable set of branch cuts that can be made along horizontal rays at $z = x + \tau\,n, \, n\in\mathbb Z, \, x\in\mathbb R, \, x < -2$.

Conjecture 3. For real $x>-2$, the derivative $t'(x)$ is a completely monotone function.

Conjecture 4. There is only one analytic function that vanishes at $z=-1$ and satisfies both the functional equation for the tetration $(6)$, and the complete monotonicity condition from the Conjecture 3.

Related questions: [1][2][3].

• A numeric value of $^{1/2}\log(3)$ calculated with precision 300 decimal digits by application of Wynn's epsilon method to $(11)$: goo.gl/LVRkk7 — can be used for comparison with other approaches. – Vladimir Reshetnikov Jan 11 '17 at 0:07
• Fwiw: the value given in your comment agrees perfectly with that value which I get using the Schröder-mechanism – Gottfried Helms Jan 11 '17 at 8:28
• A couple of years ago I've found such a q-binomial-expression but for the "decremented exponentiation", meaning the tetration based on the iterates of $U(z)=t^z-1$ and $U^{°h+1}(z) = U^{°h}(U(z))$ (But I think my results can be translated somehow to the usual tetration with base $b=t^{1/t}$ , I've not looked at this a long time) In the surely very amateurish exercise in go.helms-net.de/math/tetdocs/… I've discussed my observations (plus the direct transformability into the Diagonalization/Schröder-mechanism(!)). I was relatively new to this then,... – Gottfried Helms Jan 11 '17 at 12:18
• A numeric value of ${^{1/2}\!\left(\sqrt2\right)}$: goo.gl/MLfEMC – Vladimir Reshetnikov Jan 11 '17 at 23:37
• @AlbertRenshaw Unlike with exponents, tetration with a fractional height does not correspond to a super-root. – Vladimir Reshetnikov Jun 7 '18 at 14:29

Not sure if this is an answer but a good suggestion of how to show (1) and (2) (and perhaps an actual solution), and a good start for (3). Consider the Newton series formula for tetration. This series converges for all $\Re(z) > -2$, and is very similar to your formula

$$^za = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \binom{z}{n}\binom{n}{k}(^k a)$$

(The op claims this expression diverges, an alternate expression is

$$\frac{1}{\Gamma(1-z)}\Big{(}\sum_{n=0}^\infty (^{n+1}a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty \big{(}\sum_{n=0}^\infty (^{n+1}a) \frac{(-x)^n}{n!}\big{)}x^{-z}\,dx\Big{)}$$

of which the rest of this discussion still applies.)

Now, since this solution is bounded in the half plane $\Re(z) > 0$ it does have a uniqueness criterion--quite a good one at that too. It is the only solution to the functional equation to do such. In fact, if a solution $F$ exists where $F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})$ for $\tau < \pi/2$ then $F$ is still the above solution.

What you have presented is a q-analog of the Newton series I just put above. Sadly I cannot find a rigorous paper detailing these types of series; I don't remember where I have seen them, but I have seen them. This suggests to me, that really you are just interpolating $(^na)$ in the same manner the Newton series does, except you are using q-analogs. It is necessary though that $q$ be fixed to the value you put (explanation below).

Consider the rather beautiful fact that your solution is bounded in the right half plane.

Observe if

$[z]_q = \frac{1-q^z}{1-q}$, $[z]_q$ is periodic in $z$.

$${z \brack n}_q = \frac{[z]_q!}{[n]_q![z-n]_q!} = \frac{[z]_q[z-1]_q...[z-n+1]_q}{[1]_q[2]_q...[n]_q}$$

And each of these functions have a period, namely $2\pi i/\log(q)$..

Therefore your final expansion has an imaginary period, it tends to $-W(\log(a))/\log(a)$ as $\Re(z) \to \infty$; thus, it is bounded in the right half plane.

From this it follows your function is the standard tetration function (If it converges).

So $t$ is bounded. The first obvious fact is $t(n) = (^n a)$. Next consider the wondrous power of a little known identity theorem for functions bounded in the right half plane. Namely, if $F(z)$ and $G(z)$ are bounded in the right half plane and $F\Big{|}_{\mathbb{N}} = G\Big{|}_{\mathbb{N}}$ then $F = G$. This follows because of Ramanujan's master theorem.

I'll explain the identity theorem for a second, and then explain how this applies to your case. If $F$ and $G$ are bounded, then surely

$$f(x) = \sum_{n=0}^\infty F(n+1)\frac{(-x)^n}{n!}$$ $$g(x) = \sum_{n=0}^\infty G(n+1)\frac{(-x)^n}{n!}$$

are both entire functions, and by Ramanujan both satisfy

$$\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(1-z)$$ $$\int_0^\infty g(x)x^{z-1}\,dx = \Gamma(z)G(1-z)$$

But $f = g$ and therefore $F = G$.

Look at $t(z+1)$ and $a^{t(z)}$, both of these functions are bounded and agree on the naturals ($t(n+1) = a^{t(n)}$), therefore they are equal everywhere.

Now as to the fact that this function is completely monotone: On the tetration forum Sheldon made a great point. I'm not sure of this fact, but I'm betting it's true. If $0 < \lambda < 1$ then $f(z) = \lambda^z$ for $\Re(z) > 0$ is the only completely monotone function on $\mathbb{R}^+$ such that $\lambda f(z) = f(z+1)$. Another function would have to be $f(z+\theta(z))$ for $\theta$ a 1-periodic function, but this will most assuredly destroy the completely monotone structure.

What's great about the standard solution (the one that is periodic, the one that is the Schroder iteration, the one that is bounded), is that it can be expanded in an exponential series for $\Re(z) > 0$. This is just Fourier series, essentially.

$$^za = F(z) = \sum_{k=0}^\infty a_k \lambda^{kz}$$

$$F'(x) \sim C\lambda^x$$

Since another solution is just $F(z+\theta(z))$ for $\theta$ periodic, we can most likely form a contradiction by using our knowledge of the uniqueness of the exponential function under these conditions.

All in all, I mostly just had to weigh in my two cents, but I think all the conjectures you put are true and that your solution is in fact the 'standard' solution, which is: the periodic one, the bounded one, the Schroder one, and hopefully, the completely monotone one. Hopefully the rough proof I gave of (1) and (2) is good enough for you.

• Hmm, it diverges does it? Anixx has a link in the comments to the very same expression mathoverflow.net/questions/20688/…. Are you using the standard Gaussian q-binomials, or the expression you wrote (which I'm not sure what to call.) If it's the former I'll rewrite everything I wrote to answer the question more clearly. if you use the Gaussian q-binomials you linked to (not the ones you wrote) then I am sure I can prove your function is tetration, and that it is the standard Schroder iteration. – user78249 Jan 11 '17 at 21:56
• @VladimirReshetnikov I edited my response to be clearer that your series is in fact tetration. – user78249 Jan 11 '17 at 22:17
• I realize that now, they just looked so different I was confused. Not very good at q analogs. Thanks for the clarification, though. Your series is definitely tetration, quite an odd formula for it I must say! – user78249 Jan 11 '17 at 22:24
• nice answer James. I think I proved the complete monotonicity of the Schroeder solution, so if you proved Vladimir's solution is the same solution, then we have complete monotonicity as well. This is because the S(z) Schroeder function has odd Taylor series coefficients negative and even Taylor series coefficients positive and the desired superfunction is $S(z) = \sum a_n (-1)^n \lambda^{nz}\;$ see my mathstack answer: math.stackexchange.com/questions/1987944/… – Sheldon Jan 12 '17 at 14:45
• Wow, so just because the Fourier coefficients oscillate it implies the function is completely monotone? That is really something! All we need to do is show that if the solution is completely monotone it is the Schroder solution and we have a really good uniqueness criterion talking only about the real positive line. Another great aspect is that your proof shows well behaved superfunctions about functions with multipliers $0 < \lambda < 1$ are completely monotone. This seems like it should imply that the derivatives of pentation, hexation, sexation... are completely monotone too. – user78249 Jan 12 '17 at 18:11

Here is a proof sketch that the only completely monotone tetration is in fact the standard tetration. I am not certain, as I'm a little fuzzy on Sheldon's details on how to show tetrations derivative is completely monotone. As Sheldon wrote in the comments to my previous answer, the standard tetration (the Schroder iteration) is completely monotone.

He shows it here https://math.stackexchange.com/questions/1987944/complete-monotonicity-of-a-sequence-related-to-tetration/1996129#1996129. I think using similar tricks we can show this function is unique under these conditions. By all means my proof is not complete, but I think it's on the right track.

First we start with the wondrous identity that

$$^z a = L + \sum_{n=1}^\infty a_n(-1)^n \lambda^{nz}$$

where $a_n(-1)^n < 0$ for all $n$.

Then if

$$F(z) = \frac{d}{dz}(^za)$$

it is completely monotone. We wish to show this is the sole function to satisfy this criterion.

This will be our starting point. The next fact to consider is that $\lambda^{x + \theta(x)}$ for $\theta$ 1-periodic is only completely monotone if $\theta$ is constant. I.e: a completely monotone function multiplied by a periodic function which isn't constant is not completely monotone.

Now assume we have another solution $A(x)$ such that $a^{A(x)} = A(x+1)$, $A(0) = 1$ and $A'(x)$ is completely monotone. It follows with very little work, I'll leave it to the reader, that $A(x) = (^{x+\theta(x)}a)$ for some 1-periodic function $\theta$. So that $A'(x) = (1+\theta'(x))F(x+\theta(x))$. We also know that $1 + \theta'(x) \neq 0$ as this would destroy complete monotonicity.

$A(x)$ can be written as

$$A(x) = L + \sum_{n=1}^\infty a_n (-1)^n\lambda^{nx}\phi(x)^{n}$$

where $\phi$ is one periodic. By just looking at this function it is rather clear its derivative can't be completely monotone. The $\phi$ is obviously going to kill the proper behaviour. To show this though, we'll need to use a good trick.

For the moment, let $\mu = -\lambda^x \phi(x)$, then

$$G(\mu) = L + \sum_{n=0}^\infty a_n \mu^n$$

This function is "fully monotonic," which is something that Sheldon showed in his post. The values $(-1)^na_n < 0$ (kind of like the opposite of completely monotone). It is in fact the inverse Schroder function of $l(\mu) = a^\mu$. Where $l(G(\lambda \mu)) = G(\mu)$.

Now, $-\mu = \lambda^x\phi(x)$ is not completely monotone (unless $\phi$ is constant), and therefore composing it with $G(\mu)$ breaks the monotonic structure. Therefore $\phi = C$ is constant. Therefore $A(x) = (^{x+C}a)$ but $A(0) = 1$, therefore $C = 0$ or is an imaginary period of $(^z a)$. Either way, $A(x) = (^xa)$.

I think I can say, QED.

• I need sometime to think about it -- but I upvoted you. – Sheldon Jan 12 '17 at 22:51
• Yeah, I wasn't certain, but I think this is at least a good layout on how to show it. I tried three different ways and they all seemed to break down, but this one seemed the most promising. – user78249 Jan 12 '17 at 22:59
• I like your usage of Ramanujan's master theorem, which provides a clear uniqueness condition. I'm not sure about your proof above. btw, I think I only proved that the inverse Schroder function is sort of monotonic at x=0, not necessarily at other values of x; with $a_n(-1)^n<0$. But the super-function is fully monotonic at the real axis. – Sheldon Jan 15 '17 at 4:57
• How do you prove the simplest case $\lambda^{z+\theta(z)}$ is only fully monotone if $\theta(z)$ is a constant? I know that the product of two monotone functions is monotone, but that doesn't tell you that the product of a monotone with a non-monotone is non-monotone. Do you have that as a known theorem? – Sheldon Jan 15 '17 at 5:03
• Never mind, I had a flaw in my reasoning, I can't even show that :/ I over extended my reach a bit. Essentially I was switching to $\lambda^{-z}\phi(-z)$ and showing this can't have all positive Taylor coefficients. But I think now that this might be possible :S – user78249 Jan 15 '17 at 17:47

I have written and submitted for publication The Existence and Uniqueness of the Taylor Series of Iterated Functions. It has cleared an initial peer review with three knowledgeable mathematicians and now is in further peer review for publication. The intent is to have a single authoritative work on extending tetration and all the higher hyperoperators to the complex numbers. The uniqueness and existence theorems in the paper guarantee a unique solution to the problems. So at best, other extensions of tetration must be consistent with the extension provided in the paper. Basing the mathematics on iterated functions provides the generality needed to extend all the higher hyperoperators to complex numbers, not just tetration.

• This is really cool. I always thought using a Taylor series construction would fall apart computationally, or even conceptually. This seems rather clear. I'm not sure how it applies to the question, but I guess you're talking about the Schroder iteration for $1 \le a \le e^{1/e}$ and its uniqueness through Taylor series? There are proposed solutions to hyper operators when $1 < a < e^{1/e}$. The standard Schroder iteration has a solution for all hyper-operators (minus a few qualms), and the boundedness condition I laid out above is its uniqueness. – user78249 Jan 14 '17 at 1:31
• This extension of tetration to the complex numbers is deeply consistent with the classification of complex fixed points. Both Schroeder's and Abel's functional equations can be derived from my paper, it not merely consistent with the classification of complex fixed points, it explains why it is the way that it is. – Daniel Geisler Jan 14 '17 at 4:04
• nice paper Daniel. Do you still maintain your tetration.org/Tetration/index.html website? And do you discuss Kneser's real valued solution for bases>exp(1/e) at all? To me, the most interesting problem is extending Kneser's solution to complex bases. Anyway, thanks for writing a peer reviewed paper. Perhaps someday I will publish something somewhere else other than Henryk Trapmann's Tetration website. – Sheldon Jan 15 '17 at 5:14
• My website is superseded by my paper. My research has focused on iterated functions since the early nineties. I don't both with working out techniques confined to extending tetration when extending iterated functions gives tetration and all the higher hyperoperators. Also my paper is completely self contained. I don't have good access to rare mathematical journals, so I avoid basing my work on techniques I have not mastered from the ground up. – Daniel Geisler Jan 17 '17 at 4:28
• Tetration.org is dead do u know when it will be back up? – frogeyedpeas Nov 2 '17 at 3:55