# How to calculate $f(x)$ in $f(f(x)) = e^x$?

How would I calculate the power series of $f(x)$ if $f(f(x)) = e^x$? Is there a faster-converging method than power series for fractional iteration/functional square roots?

• Are you looking for an analytic function? Such a $f$ does not exist in the analytic category. Aug 22, 2011 at 16:03
• I wasn't aware that $f$ wasn't analytic, but I'm looking for whatever I can graph.
– jnm2
Aug 22, 2011 at 16:05
• Aside: $f$ cannot be a function composed of basic arithmetic operations, exponentials, and logs: Scott Aaronson has a proof here (see Comment #52), that for any such function, $f(f(x))$ is either subexponential or superexponential. See also Mathoverflow where it is said that "there are analytic solutions in a neighborhood of the real line, but they are known not to be entire". Aug 22, 2011 at 16:06
• Also an answer at the above Mathoverflow question links to these discussions which are about precisely your question. Aug 22, 2011 at 16:07
• @Henri: An explanation of why no such analytic function can exist would probably be a good answer, even though it does not completely answer the question. Aug 22, 2011 at 17:27

https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/44727#44727

In short, the analytic solution is

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

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

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

Insert here $g(x)=a^x$ The same way you can find not only square iterative root but iterative root of any order. Unfortunately this does not converge for $g(x)=a^x$ where $a > e^{1/e}$.

Here is a graphic for iterative root of $g(x)=(\sqrt{2})^x$ The question becomes more difficult when speaking about the base $a>e^{1/e}$. But in this case the solution can also be constructed, see this article.

• How can you compute a solution for $a=\sqrt2$ while you wrote " (...) does not converge where $a \le e^{1/e}$ (...)" ? And how can we approximate a solution when in the third formula high iterates of $g(x)$ are required and the base is greater than $e^{1/e}$ ? Dec 25, 2014 at 12:37
• @Gottfried Helms typo. It does not converge when $a > e^{1/e}$, of course. Dec 25, 2014 at 13:24
• I see. <sigh> (big relief!) Thank you for clarification. P.s. Did you know that I made a comparision between the computation which you called the Newton-series (when I did it I didn't know that my computations were equivalent to that) and an approach which involves diagonalization? In that discussion the diagonalization-approach was characterized by the property that it converges to the same solution, but much faster (with less terms). If you're interested: go.helms-net.de/math/tetdocs/index.htm , see the third entry "comparision of two methods" Dec 25, 2014 at 15:02

Here's the proof of a theorem due to Thron (1956), extracted from a article of Laurent Bonavero (available at his webpage).

Theorem. There is no entire function $f$ (that is $f:\mathbb C \to \mathbb C$ holomorphic) such that $\exp = f \circ f$.

Proof. If such a function $f$ exists, then $f(\mathbb C)= \mathbb C^*$. Indeed, $f(\mathbb C) \supset \exp(\mathbb C) = \mathbb C^*$, but $0$ can't be in the image of $f$: if $f(x)=0$, then as $x \neq 0$, there exists $y$ such that $x=f(y)$ so that $\exp(y)=0$, absurd.

Therefore $f$ can be lifted by the exponential, $f=\exp \circ g \,$ for $g$ entire. So $\exp = \exp(g \circ f)$, and there must exist a constant $C$ such that $g \circ f(z)=z+C$ for all $z\in \mathbb C$. It follows that $f$ is injective, so $\exp$ should be injective too, which is absurd!

• Seems this post does not answer the question. The questioner did not look for an entire function and even for funtions on the complex plane at all. Dec 20, 2012 at 14:40
• @Anixx I agree. Nov 1, 2021 at 13:52

There is a lot of material about this question here and in mathoverflow. There is also a "Tetration forum", where someone has implemented a version of tetration due to Hellmuth Kneser, see some forum entries there: http://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=8" also in citizendium there is an extensive article of Dmitri Kousznetzov who claims he has a usable interpretation (and implementation) see http://en.citizendium.org/wiki/Tetration