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.

  • $\begingroup$ 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? $\endgroup$ – jathd Jan 21 '13 at 15:40
  • $\begingroup$ $z_0$ seems to be a fixed point of the exponential. $\endgroup$ – Hagen von Eitzen Jan 21 '13 at 15:42
  • 2
    $\begingroup$ There has been quite a bit of discussion of this and similar questions on MathOverflow. For instance this question. $\endgroup$ – Old John Jan 21 '13 at 15:46
  • 3
    $\begingroup$ 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") $\endgroup$ – Gottfried Helms Jan 21 '13 at 16:34
  • 2
    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jan 21 '13 at 23:35


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: $h(0)\approx 0.498563287941114434679619$


+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
  • $\begingroup$ 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. $\endgroup$ – daniel Jan 22 '13 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.