This question has been in my mind since high school.

We can get multiplication of natural numbers by repeated addition; equivalently, if we define $f$ recursively by $f(1)=m$ and $f(n+1)=f(n)+m$, then $f(n) = m \times n$. Likewise, we get exponentiation by repeated multiplication. If $g(1)=m$ and $g(n+1)=mg(n)$, then $g(n) = m^n$. In my high school mind it was natural to imagine a new function defined by repeated exponentiation: $h(1)=m$ and $h(n+1)=m^{h(n)}$.

These definitions only make sense for $n$ a natural number, but of course there are standard very mathematically satisfying ways to define multiplication and exponentiation by any real number. My question is this:

Can the function $h$ defined above also be extended in a natural way to $\mathbb{R}^{>0}$?

The question is in the spirit of seeking an extension of $f(n)=n!$ to $\mathbb{R}$ and arriving at $\Gamma(x)$.

Let me focus the question, and attempt to make precise what I mean by "in a natural way." Take $h(1)=2$ and $h(n+1)=2^{h(n)}$. $h$ is now defined on $\mathbb{N}$, and $h(2)=4$, $h(3)=16$, $h(4)=2^{16}=65,536$ etc. Is it possible to extend the domain of definition of $h$ to all positive reals in such a way that

a) The functional equation $h(x+1)=2^{h(x)}$ continues to be satisfied for all $x$ in the domain.

b) $h$ is $C^\infty$. (Analytic would be even better but this seems maybe too much to hope for?)

c) All $h$'s derivatives are monotone.

These requirements are my attempt to codify what would count as "natural." I am open to suggestions about what would be a better list of requirements.

If such a function exists, I would like to know how to construct it; if it doesn't, I would like to know why (i.e. outline of proof), and if relaxing some of the requirements (e.g. just the first derivative monotone) would make it possible.

(If the function exists, I am also interested in the questions, "is it unique?" "Could we add some natural requirements to make it unique?" But my main query is about existence.)

  • $\begingroup$ LambertW through tetration: math.stackexchange.com/questions/531276/… $\endgroup$ Oct 21, 2013 at 15:23
  • 6
    $\begingroup$ @H.R.: It is not a mistake not to capitalize every word of a title. It is a matter of taste and style, and I ask that you stop making those style-only edits. If you have actual things to contribute to the post by editing it, that's great. If you only want to enforce your style choices over the OP's style choices, this is not okay. Not to mention that bumping questions from over four years ago just for capitalization of the title is a plain waste of front-page space. And the same can be said about adding boldface to Zev's answer below. Completely useless and out of line. $\endgroup$
    – Asaf Karagila
    Dec 3, 2015 at 18:48
  • $\begingroup$ @AsafKaragila: OK :) I just edited two posts and it got the whole space of front page? :) About the Zev's answer there was a typo that I fixed ("it" changed to "is") beside bolding. :) $\endgroup$ Dec 3, 2015 at 19:47
  • 6
    $\begingroup$ @H.R. Front page space aside (that's only significant when multiple posts are edited), please do not capitalise all words in a title when the title is already capitalised correctly as a sentence. If a title is all lower case, you have more freedom of choice. I'd still much prefer to capitalise only proper nouns and the start of sentences, but when it's all wrong to begin with, the editor may give their own preferences more weight. The same principle applies to adding emphasis by bolding or italicising. If the post is a mess, you can choose, otherwise, respect the author's choice. $\endgroup$ Dec 3, 2015 at 20:41
  • 1
    $\begingroup$ You can take a look at mathoverflow.net/q/259278/9550 $\endgroup$ Jan 11, 2017 at 0:22

3 Answers 3


This is not an answer to your question but a long comment on its motivation. Multiplication is at least two conceptually distinct things, only one of which can reasonably be described as repeated addition:

  • The natural map $\mathbb{Z} \times A \to A$ given by $(n, a) \mapsto na$ where $A$ is an abelian group; this really is repeated addition, and is in particular bilinear.
  • The composition $\text{End}(A) \times \text{End}(A) \to \text{End}(A)$ of endomorphisms of an abelian group.

What's confusing is that these two definitions agree in familiar cases. If $A = \mathbb{Z}$ (the abelian group), repeated addition gives a natural map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. On the other hand, $\text{End}(\mathbb{Z}) \cong \mathbb{Z}$ (the ring), and composition of endomorphisms gives a natural map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. These happen to be the same map, but this is an illusion caused by the fact that we're looking at such a fundamental abelian group $\mathbb{Z}$.

Similarly, if $A = \mathbb{R}$ (the abelian group), repeated addition gives a natural map $\mathbb{Z} \times \mathbb{R} \to \mathbb{R}$. On the other hand, the reasonable endomorphisms of $\mathbb{R}$ form a ring isomorphic to $\mathbb{R}$ (the ring), giving a natural map $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, and the restriction to $\mathbb{Z}$ in the first factor of the second map gives the first.

But when we think of multiplication of real numbers, the first picture is misleading: in what sense is $\pi \times \pi = \pi^2$ repeated addition? Simple: it isn't. It's better conceptualized as composition of scalings of the real line (the endomorphism definition).

The endomorphism definition generalizes immediately to multiplication of complex numbers and matrix multiplication, a context where "repeated addition" doesn't even begin to capture what multiplication is all about. This is probably a big reason why people have a difficult time with complex numbers: nobody's explained to them that they're just composing rotations and scalings of the plane.

Exponentiation is at least three conceptually distinct things, only one of which can reasonably be described as repeated multiplication:

  • The natural map $\mathbb{Z} \times G \to G$ given by $(n, g) \mapsto g^n$ where $G$ is a group; this really is repeated multiplication. Note that for fixed $n$ we don't get a homomorphism in general if $G$ is non-abelian, but for fixed $g$ we get a homomorphism $\mathbb{Z} \to G$.
  • The natural map $B \to B$ given by $x \mapsto e^x = \exp(x) = \sum \frac{x^k}{k!}$ where $B$ is a topological ring and the series converges (which for example is always true in a Banach algebra). If it exists for all $x \in B$, this map is a homomorphism from the additive group of $B$ to the multiplicative group of $B$; moreover, the homomorphism $t \mapsto e^{tx}$ from $\mathbb{R}$ to $B^{\times}$ is (in nice cases) uniquely determined by the fact that its derivative at $t = 0$ (in nice cases where this exists) is $x$. In other words, this is a very, very natural map.
  • Any map that extends or is analogous to one or both of the above two maps.

The reason maps in the third category exist is because of the nice homomorphism properties that anything behaving like an exponential ought to satisfy, which we often want to imitate in other settings (e.g. the exponential map in Riemannian geometry). Thus, for example, we have an exponential $(a, x) \mapsto a^x = e^{x \log a}$ where $a$ is a positive real and $x$ is an element of a topological ring, generalizing the second definition, such that

  • $a^{x+y} = a^x a^y$ (when $x, y$ commute) and
  • $(ab)^x = a^x b^x$

and if $x$ is chosen to be a scalar multiple of the identity, we get back a special case of the first map for $G = (\mathbb{R}_{>0}, \times)$.

But I still think this is misleading. One sign is that exponentials with arbitrary bases behave quite badly once $a$ is allowed to be anything other than a positive real. The first time I tried to graph the equation

$$y = (-10)^x$$

on my calculator impressed this point on me very strongly. (Try it and see what happens.) Of course this is due to the fact that logarithms aren't well-defined in general, which, while interesting, only further emphasises the point that instead of allowing both arbitrary bases and exponents we should stick to repeated multiplication, $e^x$, and logarithms, which are the real stars of the show.

So insofar as multiplication and exponentiation are repeated addition and multiplication, at least this is sensible because addition and multiplication are associative.

Exponentiation is not associative, and it shouldn't be, because in many more general cases its two inputs are different types of things.

Therefore, there's no reason to expect repeated exponentiation to have any reasonable properties along the lines of the natural and useful homomorphism properties of multiplication and exponentiation, and as far as I know, it doesn't.

  • 3
    $\begingroup$ I was going to write this up as a blog post at some point, but it might as well be here instead. See also math.stackexchange.com/questions/35598/… and math.stackexchange.com/questions/17711/… for my previous answers discussing this theme. $\endgroup$ Aug 10, 2011 at 15:39
  • 3
    $\begingroup$ @Ben: your last comment describes my attitude almost exactly. There is very little room for them not to coincide given that they are both bilinear and act as the identity when fed $1 \in \mathbb{Z} \subseteq \text{End}(A)$. The fact that they do is fine for dealing with the integers but very misleading once you get to the complex numbers, as I mentioned above. $\endgroup$ Aug 10, 2011 at 19:27
  • 3
    $\begingroup$ @Ben: as for your third comment, I don't know. The factorial is deceptively easy to define but makes its appearance in mathematics in several fundamental, nontrivial ways (Taylor series, the sizes of the symmetric groups, various probability distributions). The same is probably not true of an arbitrary "natural" function on $\mathbb{N}$. $\endgroup$ Aug 10, 2011 at 19:28
  • 1
    $\begingroup$ Now to decategorify: viewing ring as a category with one object, its two operations (multiplication by integers = repeated addition, and ring multiplication) coincide and are commutative on the characteristic subring because muliplication of integers is commutative, and because of Eckmann-Hilton. The two kinds of multiplication are unrelated, except when restricted to characteristic subring, where they must coincide by Eckmann-Hilton. What do you think? $\endgroup$
    – ziggurism
    Oct 31, 2017 at 15:02
  • 1
    $\begingroup$ On second thought, my proposal as stated doesn't make much sense. A ring, viewed as a category with one object, doesn't have a terminal object, and the inclusion of its characteristic subring is not due to any generating object. $\endgroup$
    – ziggurism
    Oct 31, 2017 at 18:09

What you're after is called tetration (the example you computed is given here), and it has an active community of people who are interested in it (though my sense is that it is not quite in the mainstream of mathematics research at the moment, for whatever reason). The Wikipedia page indicates that the problem of extending tetration to arbitrary real powers in a sufficiently regular/smooth way is still not satisfactorily solved, so I'm afraid I don't know the answer to your question about the existence of such a function $h(x)$.

Tetration is further generalized by Knuth's up-arrow notation, and then generalized even more by Conway's chained arrow notation.

  • 6
    $\begingroup$ Why the downvote? $\endgroup$ Aug 10, 2011 at 14:52
  • 4
    $\begingroup$ I don't understand the downvote. It seems a reasonable and helpful answer. Supplying the proper term (here tetration) can give OP the entry needed to find information-I have been on the receiving end and appreciate it greatly. $\endgroup$ Aug 10, 2011 at 15:31
  • $\begingroup$ I also don't understand the downvote. I find the answer extremely helpful. I didn't mark the question answered yet only because I hope to get more perspectives. $\endgroup$ Aug 10, 2011 at 17:22
  • $\begingroup$ Perhaps Qiaochu's answer offers some illumination about why it's not more mainstream... $\endgroup$ Aug 10, 2011 at 21:25

A function is analytic at $x$ for some open interval containing $x$, it can be expressed as a power series centered at $x$.

I once thought of a way to extend repeated exponentation for real numbers to any real base greater than 1. It's called tetration. $x ↑↑ 0$ is defined to be 1. For any integer $y$, $x ↑↑ (y + 1)$ is defined to be $x^{x ↑↑ y}$. We see that when ever $x$ > 1 and $x$ < $e^{\frac{1}{e}}$, the sequence $x ↑↑ 0$, $x ↑↑ 1$, $x ↑↑ 2$ ... approaches a fixed number and as it gets closer to that number, it approaches it very nearly exponentially. Now tetration to the base $x$ can uniquely be extended to all real numbers > -2 in such a way $\forall t > -2,x ↑↑ (t + 1) = x^{x ↑↑ t}$ and it gets closer and closer to exponentially approaching a fixed number. Using analytic continuation, we can also define for any $x \geq e^{\frac{1}{e}}$ and $y \geq -2$ what $x ↑↑ y$ is.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .