# Will Jagy

http://math.stackexchange.com/questions/208996/half-iterate-of-x2c?
" This may be helpful.
Let
We use a technique of Ecalle to solve for the Fatou coordinate $\alpha$ that solves
For any
let
Then we get the exact
The point is that this expression converges far more rapidly than one would expect, and we may stop at a fairly small $n$.
It is fast enough that we may reasonably expect to solve numerically for $\alpha ^{-1}(x)$.
We have $f^{-1}(x)=x+x^{2}$
Note
It follows that if we define
we get the miraculous
Note that $\alpha$ is actually holomorphic in an open sector that does not include the origin, such as real part positive. That is the punchline here, $\alpha$ cannot be extended around the origin as single-valued holomorphic. So, since we are finding a power series around $0$, not only are there a $1/z$ term, which would not be so bad, but there is also a $\log z$ term. So the $\ldots -n$ business is crucial.
I give a complete worked example at my question http://mathoverflow.net/questions/45608/formal-power-series-convergence as my answer http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765
The Ecalle technique is described in English in a book, see
* [K_C_G PDF] http://zakuski.utsa.edu/~jagy/K_C_G_book_excerpts.pdf
* [BAKER] http://zakuski.utsa.edu/~jagy/other.html
The Julia equation is Theorem 8.5.1 on page 346 of KCG.
It would be no problem to produce, say, 50 terms of $\alpha (x)$ with some other computer algebra system that allows longer power series and enough programming that the finding of the correct coefficients, which i did one at a time, can be automated.
No matter what, you always get the
when
As I said in comment, the way to improve this is to take a few dozen terms in the expansion of $\alpha (x)$ so as to get the desired decimal precision with a more reasonable number of evaluations of $f(x).<math>SohereisabriefversionoftheGP-PARIsessionthatproduced<math>\alpha (x)$:

# Jonathan Lubin

From http://math.stackexchange.com/questions/911818/how-to-obtain-fx-if-it-is-known-that-ffx-x2x?
"Here’s a technique for finding the first few terms of a formal power series representing the fractional iterate of a given function like
$f(x)=x+x^{2}$.
I repeat that this is a formal solution to the problem, and leaves unaddressed all considerations of convergence of the series answer.
I’m going to find the first six terms of
$f^{\circ 1/2}(x)$
the “half-th” iterate of $f$, out to the $x^{5}$-term.
Let’s write down the iterates of $f$, starting with the zero-th.
${\begin{aligned}f^{\circ 0}(x)&=x\\f^{\circ 1}=f&=x&+x^{2}\\f^{\circ 2}&=x&+2x^{2}&+2x^{3}&+x^{4}\\f^{\circ 3}&\equiv x&+3x^{3}&+6x^{3}&+9x^{4}&+10x^{5}&+8x^{6}\\f^{\circ 4}&\equiv x&+4x^{2}&+12x^{3}&+30x^{4}&+64x^{5}&+118x^{6}\\f^{\circ 5}&\equiv x&+5x^{2}&+20x^{3}&+70x^{4}&+220x^{5}&+630x^{6}\\f^{\circ 6}&\equiv x&+6x^{2}&+30x^{3}&+135x^{4}&+560x^{5}&+2170x^{6}\\f^{\circ 7}&\equiv x&+7x^{2}&+42x^{3}&+231x^{4}&+1190x^{5}&+5810x^{6}\,,\end{aligned}}$
where the congruences are modulo all terms of degree $7$ and more.
Now look at the coefficients
* of the $x$-term: always $1$.
* Of the $x^{2}$-term? In $f^{\circ n}$, it’s $C_{2}(n)=n$.
* The coefficient of $x^{3}$ in $f^{\circ n}$ is $C_{3}(n)=n(n-1)=n^{2}-n$, as one can see by inspection.
Now, a moment’s thought (well, maybe several moments’) tells you that $C_{j}(n)$, the coefficient of $x^{j}$ in $f^{\circ n}$, is a polynomial in $n$ of degree $j-1$.
And a familiar technique of finite differences shows you that
${\begin{aligned}C_{4}(n)&={\frac {2n^{3}-5n^{2}+3n}{2}}\\C_{5}(n)&={\frac {3n^{4}-13n^{3}+18n^{2}-8n}{3}}\,,\end{aligned}}$
I won’t go into the details of that technique. The upshot is that, modulo terms of degree $6$ and higher, you have
$f^{\circ n}(x)\equiv x+nx^{2}+(n^{2}-n)x^{3}+{\frac {1}{2}}(2n^{3}-5n^{2}+3n)x^{4}+{\frac {1}{3}}(3n^{4}-13n^{3}+18n^{2}-8n)x^{5}$.
Now, you just plug in $n={\frac {1}{2}}$ in this formula to get your desired series.
And I’ll leave it to you to go one degree higher, using the iterates I’ve given you."