Back on Scott Aaronson's blog, I gave [an argument][1] that $e^z+z-1$ should have an analytic compositional square root. The important difference between this function and $e^z-1$ was that the fixed point at $0$ has derivative $>1$, not $=1$. This should warn us that arguments based on the growth rate near infinity are inadequate. (Or else it should point out that my argument was broken!) See comments below, my argument may have been broken. But, if so, I want to figure out why! UPDATE: OK, I'm looking for some empirical data myself now. Let $e(z)=e^z+z-1$. My argument claimed that there should be an analytic and invertible $u$ (near 0) such that $u(e(z)) = 2 u(z)$. If such a $u$ exists, then $u^{-1}(2^{1/2} u(z))$ should have the desired property. The nice thing about the equation $u(e(z)) = 2 u(z)$ is that it is linear in the coefficients of $u$. Here are the first 10 coefficients, computed with exact arithmetic. {1, -(1/4), 1/18, -(1/96), 17/10800, -(47/267840), 4069/354352320, -(24907/102863416320), 475411/2893033584000, -(108314387/ 1314080143488000)} And the numerical versions of the above {1., -0.25, 0.0555556, -0.0104167, 0.00157407, -0.000175478, 0.0000114829, -2.42137*10^-7, 1.6433*10^-7, -8.2426*10^-8} They seem to be converging rapidly. [1]: http://scottaaronson.com/blog/?p=263#comment-13954