## Does the exponential function have a square root?

(asked by Nathaniel Hellerstein on the Q&A board at JMM)

Is there a "half-exponential" function h(x) such that h(h(x))=ex? Is it unique? Is it analytic?

Related question: Is there an invertible smooth function E such that E(x+1)=eE(x)? Is it unique? If so, then we can take h(x)=E(E-1(x)+1/2).

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A related question: mathoverflow.net/questions/4347/… – Gil Kalai Jan 17 2010 at 9:38
See the tag fractional-iteration for other related questions. – Daniel Geisler Nov 6 2010 at 6:52
Having read the answers $1-7$, for me the most interesting problem here is the one posed by Gil Kalai in MO question 4347: Find a function definable in finite terms that for $x\to\infty$ satisfies $f(f(x)\sim e^x$ in some sense. – Christian Blatter Nov 7 2010 at 19:10

Yes, there exists real-analytic injective $E$.

Its inverse $E^{-1}$ is constructed in the paper of Kneser [1].

Uniqueness: $E^{-1}$ is the only holomorphic and injective Abel function (up to an additive constant) on the sickel $G$ given in the following way: Let $z_0$ and its conjugate $z_0^\ast$ be the both fixpoints of $e^z$ closest to the real axis, let $\ell$ be the straight line connecting $z_0$ and $z_0^\ast$ and let $\ell_1$ be the image of $\ell$ under $e^z$. Now let $G$ be the region bounded by $\ell$ and $\ell_1$ inclusive and $z_0$, $z_0^\ast$ exclusive. See also [2].

For the general theory of Abel functions between two complex fixed points (which is a relatively new development in holomorphic dynamics which is unaware of its application to the exponential) [3].

[1] Kneser, H. (1949). Reelle analytische Lösungen der Gleichung $\phi(\phi(x))=e^x$ und verwandter Funktionalgleichungen. J. Reine Angew. Math., 187, 56–67.

[2] Trappmann, H., & Kouznetsov, D. (2010). Uniqueness of holomorphic Abel functions at a complex fixed point pair. Aequationes Math.

[3] Shishikura, M. (2000). Bifurcation of parabolic fixed points. In Lei, Tan, The Mandelbrot set, theme and variations. Cambridge: Cambridge University Press. Lond. Math. Soc. Lect. Note Ser. 274, 325-363.

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To avoid repetition see What’s a natural candidate for an analytic function that interpolates the tower function? for my general explanation of tetration.

A distinguishing feature of the exponential function is that it has an infinite number of complex fixed points, for example $0.318132 + 1.33724 i$. The lack of real fixed points results in $f: \mathbb{R} \rightarrow \mathbb{C}$, where $f(f(x))=e^x$.

• An Introduction to Chaotic Dynamical Systems; Devaney
• Complex Dynamics; Carleson, Gamelin
• Iterative Functional Equations; Kuczma

In fact, a picture of the exponential map is on the cover of Devaney's book. Both dynamics and functional equations address fractional iteration. Elements of dynamics are covered in functional equations and the reverse is also true, at least when discussing fractional iteration.

One way to validate or invalidate a proposed solution is to see if it is consistent or inconsistent with the theorems of complex dynamics. If a solution doesn't treat a fixed point as a fixed point, then it is defective. If the solution is inconsistent with the linearization theorem in the neighborhood of a fixed point, then the solution is defective. If the solution is inconsistent with the classification of fixed points and violates topological conjugacy the solution is defective.

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If you take exponential with a base $a < e^{1/e}$ finding the iterative square root of $f(x)=a^x$ is not so difficult.

You can use one of the following formulas:

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m \binom mk (-1)^{m-k}\exp_a^{[k]}(x)$$

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

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

$$f^{[1/2]}(x)=\lim_{n\to\infty} \log_a^{[n]}\left(\left(1-\left(\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\right)^{1/2}\right)\frac{W(-\ln a)}{-\ln a}+\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\exp_a^{[n]}(x)\right)$$

Here is a graph of an iterative square root of a function $f(x)=(\sqrt{2})^x$:

Regarding second part of the question, such specially constructed function is called superfunction or flow. For base $a \le e^{1/e}$ superfunction can be easily constructed from the above formulas by substituting variable instead of 1/2. Of course the superfunction is not unique but there are certain more "natural" solutions.

The question becomes more difficult when speaking about the base $a>e^{1/e}$. But in this case the superfunction can also be constructed, see this article.

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For many references see http://reglos.de/lars/ffx.html.

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Great collection on the topic of iterative functional equations! – ex falso quodlibet Jan 19 2010 at 2:58

Re: second question.

Yes, there are many of them on a half-line. Just take any smooth increasing function on $[0,1]$ with the property that $E(1)$ and $E'1(1)$ have desired values and extend it by the rule above. Since $E(x) > x$ the function will be increasing, thus invertible.

No, there are none on the whole line. For any number $x$, the sequence $\mathop{\text{ln}} \mathop{\text{ln}} \dots \mathop{\text{ln}} x$ cannot be continued indefinitely; at some point you encounter negative numbers.

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There is no entire (holomorphic everywhere) function $f(z)$ with $f(f(z)) = e^z$. To see this, one can use Picard's little theorem, and do case analysis. For example, $f(z)$ cannot take all values, for then so does $f(f(z))$, while $e^z$ omits zero. The case that $f(z)$ has an omitted value can also be excluded, this is not difficult.

EDIT: Perhaps I should be more explicit. If $f(z)$ omits a value, that has to be zero, for $e^z$ takes all other values. Thus there is an entire function $h(z)$ such that $f(z) = e^{h(z)}$ (the complex plane is simply connected). Now

$e^{h(e^{h(z)})} = e^z$

and so $h(e^{h(z)}) = z + 2{\pi}ik$ for some fixed integer $k$. Since the right hand side takes all values, so does the left hand side. So $h(z)$ takes the two values $0$ and $2{\pi}i$, say $h(a) = 0$ and $h(b) = 2{\pi}i$. Now

$a + 2{\pi}ik = h(e^{h(a)}) = h(e^0) = h(e^{2{\pi}i}) = h(e^{h(b)}) = b + 2{\pi}ik$

and so $a = b$. Contradiction!

Alternatively, use the theorem of Polya that if $f(z)$ and $g(z)$ are entire functions, then $f(g(z))$ is of infinite order unless (i) $f(z)$ is of finite order and $g(z)$ is a polynomial, or (ii) $f(z)$ has order zero and $g(z)$ is of finite order. The exponential function has order $1$.

EDIT: Here $f(f(z)) = e^z$ yields the absurd conclusion that $e^z$ is a polynomial in case (i). While in case (ii) we find that $f(z)$ is an entire function of order zero with $f(f(z)) = e^z$. But an entire function of order zero that is not a polynomial takes every value infinitely often (by the Hadamard factorization theorem), so we are led to the absurd conclusion that $e^z$ takes the value zero infinitely often.

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If f(f(x))=exp(x) then f(x)=log(f(exp(x)), so it is not reasonable to take f any more analytic than log. – Ron Maimon Jul 31 2011 at 0:47

For any nonpositive $x$, define the sequence $(x_0, x_1, ...)$ by $x_0 = x$, $x_{i+1} = e^{x_i}$. If we let $y$ be another nonpositive real number, and define $(y_0, ...)$ similarly, then if we define $h$ on the union of these two sequences by $h(x_i) = y_i$, $h(y_i) = x_{i+1}$, we get $h(h(x_i)) = x_{i+1}$, $h(h(y_i)) = y_{i+1}$. So, to define $h$ everywhere, we just pair off the nonpositive reals any old way and define $h$ as above for each pair (every number can be written as an $x_i$ for a unique $i \ge 0, x \le 0$, so this defines $h$ on the entire real line). Thus, there are actually uncountably many square roots of the exponential function, almost all of which are horribly discontinuous.

Edit: in fact, any such $h$ must be of this form - if $h(x_0) = y_i$, then we must have $h(y_i) = x_1, h(x_1) = y_{i+1}, ..., h(y_{i+j}) = x_{j+1}, h(x_j) = y_{i+j}$, if $y = x$ then we have a problem when we plug in $j = i$, so we must have $y \ne x$, and by a similar argument if $h(y_0) = z_k$ for some $k$, then $x_1 = h(y_i) = z_{i+k}$ so $x = z, i+k = 1$, so either $h(x_0) = y_0, h(y_0) = x_1$ or vice-versa.

To find such a pairing producing a continuous $h$, you can take any homeomorphism $f : (-\infty, -1] \rightarrow (-1, 0]$, such as $f(x) = e^{x+1}-1$, and have your pairs be of the form $(x, f(x))$.

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