TITLE: How to calculate $f(x)$ in $f(f(x)) = e^x$?
QUESTION [11 upvotes]: 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?

REPLY [15 votes]: Look at this answer:
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.
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