# Solutions of $f(f(z)) = e^z$

It is my impression that if we find a function f(z) that satisfies

$$f(f(z)) = e^z$$

there is only one point z that satisfies the relation.

This dawned on me when I noticed that the pesky z that kept popping up in my attempts to look at the problem was the one my book proposed I start with, to wit: $z_o = 0.318 + 1.337i.$ So the joke was on me.

Now I would like to prove this. I would instinctively begin by assuming there was a $z \neq z_o$ and deriving a contradiction. Hopefully I will make some progress before an answer is posted but I am sure I will miss nuances. Maybe it's as simple as showing that $\log^nz$ has a fixed point, which I don't know to be true.

Thanks for any insights.

• 1. Must $f$ be holomorphic? (I say that because of the complex-analysis tag.) If yes, on an open set? On the whole of $\mathbf C$? 2. Can you put quantifiers in your equation? Must $f(f(z))=e^z$ be true for all $z$, or for some value(s) of $z$? 3. What do you mean by "start with" $z_0$? Start what? – jathd Jan 21 '13 at 15:40
• $z_0$ seems to be a fixed point of the exponential. – Hagen von Eitzen Jan 21 '13 at 15:42
• There has been quite a bit of discussion of this and similar questions on MathOverflow. For instance this question. – Old John Jan 21 '13 at 15:46
• Look for an article of Kneser; he developed a method for any real fractional self-composition which arrives at $\exp(z)$; in the tetration-forum (math.eretrandre.org/tetrationforum/index.php) you find even Pari/GP-code for an implementation based on Kneser's method (search for user "Sheldonison" and "Pari/GP") – Gottfried Helms Jan 21 '13 at 16:34
• See also mathoverflow.net/questions/4347/… and mathoverflow.net/questions/12081/… and mathoverflow.net/questions/45477/… and math.stackexchange.com/questions/1118/… and the links from those discussions. – Gerry Myerson Jan 21 '13 at 23:35

Daniel,

One solution would be the half iterate generated from real valued tetration, but one can also start with the $z_0\approx0.318+1.337i$ fixed point, and develop the half iterate directly from there. Of course, such a solution is not real valued at the real axis. Then, $z_0$ is defined such that $\exp(z_0)=z_0$. Then there is a Schroeder function $s(z)$, and its inverse, $s^{-1}(z)$. The Schroeder function for $\exp(z)$ has $\lambda=z_0\approx0.318+1.337i$; where $s(z_0)=0$, and s(z) is a taylor series developed in the neighborhood of $z_0$. And the inverse of the Schroder function, $s^{-1}(0)=z_0$.

$s(\exp(z))=\lambda s(z)\;\;\;\;\;\;\;\;\;\;\;s(z) = (z-z_0) + a_2(z-z_0)^2 + a_3(z-z_0)^3 + ...$

$s^{-1}(\lambda z)=\exp(s^{-1}(z))\;\;\;\;s^{-1}(z) = z_0 + z + b_2 z^2 + b_3 z^3 + ...$

Finally, the half iterate of f(z) that you might be looking for would be $h(z) = s^{-1}(s(z) \times \lambda^{0.5} )$ $h(h(z)) = \exp(z)$

The solution for h(z) isn't real valued at the real axis, and h(z) has a singularity at z=0. Nonetheless, the half iterate of say a number like 1/2 is defined. $h(0.5)\approx 0.99303919280011+0.1311428457124i$

A completely different solution involves real valued tetration, which actually involves both $z_0$ and $\overline{z_0}$. This is Kneser's solution, which involves both fixed points, and a Riemann mapping. I wrote a pari-gp program that purports to calculates Kneser's solution. You can download the pari-gp code from http://math.eretrandre.org/tetrationforum/showthread.php?tid=486 Call the abel function for the slog, or inverse of tetration, $\alpha(z)$, and the inverse abel function $\alpha^{-1}(z)$ is tetration itself.

Then $h(z)=\alpha^{-1}(\alpha(z)+0.5)$. With this solution, $h(0)$ is defined, and the half iterate of 0 is: $h(0)\approx 0.498563287941114434679619$

$h(h(z))=\exp(z)$

h(z)=
0.498563287941114434679619
+z^ 1*  0.876336132224813093948089
+z^ 2*  0.247552187310897996180728
+z^ 3*  0.0245718116969028132878499
+z^ 4* -0.000952136380204205567746147
+z^ 5*  0.000253339819008525443204593
+z^ 6*  0.0000709275516366955520540078
+z^ 7* -0.0000481808433402200348719782
+z^ 8*  0.00000263228465405932187872481
+z^ 9*  0.00000596598826774285620512586
+z^10* -0.00000130879479719985814669383
+z^11* -0.000000747165552015528631238906
+z^12*  0.000000268510892327234834856927
+z^13*  0.000000112440534247328702573054
+z^14* -0.0000000480789869461352605312404
+z^15* -0.0000000220118629742873551964336
+z^16*  0.00000000817933994010676141640719
+z^17*  0.00000000530688749879414641441033
+z^18* -0.00000000123819700193839015397174
+z^19* -0.00000000141844961463076076766377
+z^20*  1.05287927108075115225173 E-10
+z^21*  0.000000000389632939104117575233834
+z^22*  3.51707444355648805383811 E-11
+z^23* -1.04753098725701245543137 E-10
+z^24* -2.89321996209623932705115 E-11
+z^25*  2.62480364845324234254238 E-11
+z^26*  1.38848625050719277138625 E-11
+z^27* -5.58405052292307415587597 E-12
+z^28* -5.57754465436342011963328 E-12
+z^29*  6.94355445214550947800341 E-13
+z^30*  2.00345639417984951165738 E-12

• Thank you. The function $h(z)=s^{-1}(\sqrt{\lambda}~s(z))$ is very likely the one I intended to ask about, but the question was not as straightforward as I thought. – daniel Jan 22 '13 at 0:05