Is There a Natural Way to Extend Repeated Exponentiation Beyond Integers?

Question

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.)

Answer 1

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.

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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.

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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.

Answer 2

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.

Answer 3

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.

Answer 4

If the base is  b = e^1/e,  which is  1.44466786 ..., there is one and only only one solution for its fractional exponential iterates.  That is because  b^x  has exactly one fixed point, at  x = e.  The technique is described in  The Fourth Operation  except it is applied to  e^x - 1  (where the fixed point is 0) in an attempt to find the iterates when  b = e.  The technique does provide a reasonable solution in that case but it is not unique.  However, to repeat, the technique gives a unique solution in the case  b = e^1/e.