[<< wikibooks] Fractals/Iterations in the complex plane/Fatou coordinate for f(z)=z+z^2)
= Will Jagy =
http://math.stackexchange.com/questions/208996/half-iterate-of-x2c?
"
Let

f
(
x
)
=

−
1
+

1
+
4
x

2

,

x
>
0

{\displaystyle f(x)={\frac {-1+{\sqrt {1+4x}}}{2}},\;\;x>0}

We use a technique of Ecalle to solve for the Fatou coordinate

α

{\displaystyle \alpha }
that solves

α
(
f
(
x
)
)
=
α
(
x
)
+
1

{\displaystyle \alpha (f(x))=\alpha (x)+1}

For any

x
>
0

{\displaystyle x>0}

let

x

0

=
x
,

x

1

=
f
(
x
)
,

x

2

=
f
(
f
(
x
)
)
,

x

n
+
1

=
f
(

x

n

)

{\displaystyle x_{0}=x,\;x_{1}=f(x),\;x_{2}=f(f(x)),\;x_{n+1}=f(x_{n})}

Then we get the exact

α
(
x
)
=

lim

n
→
∞

1

x

n

−
log
⁡

x

n

+

x

n

2

−

x

n

2

3

+

13

x

n

3

36

−

113

x

n

4

240

+

1187

x

n

5

1800

−

877

x

n

6

945

−
n

{\displaystyle \alpha (x)=\lim _{n\rightarrow \infty }{\frac {1}{x_{n}}}-\log x_{n}+{\frac {x_{n}}{2}}-{\frac {x_{n}^{2}}{3}}+{\frac {13x_{n}^{3}}{36}}-{\frac {113x_{n}^{4}}{240}}+{\frac {1187x_{n}^{5}}{1800}}-{\frac {877x_{n}^{6}}{945}}-n}

The point is that this expression converges far more rapidly than one would expect, and we may stop at a fairly small

n

{\displaystyle n}
.
It is fast enough that we may reasonably expect to solve numerically for

α

−
1

(
x
)

{\displaystyle \alpha ^{-1}(x)}
.
We have

f

−
1

(
x
)
=
x
+

x

2

{\displaystyle f^{-1}(x)=x+x^{2}}

Note

α
(
x
)
=
α
(

f

−
1

(
x
)
)
+
1

{\displaystyle \alpha (x)=\alpha (f^{-1}(x))+1}

α
(
x
)
−
1
=
α
(

f

−
1

(
x
)
)

{\displaystyle \alpha (x)-1=\alpha (f^{-1}(x))}

α

−
1

(

α
(
x
)
−
1

)

=

f

−
1

(
x
)

{\displaystyle \alpha ^{-1}\left(\alpha (x)-1\right)=f^{-1}(x)}

It follows that if we define

g
(
x
)
=

α

−
1

(

α
(
x
)
−

1
2

)

{\displaystyle g(x)=\alpha ^{-1}\left(\alpha (x)-{\frac {1}{2}}\right)}

we get the miraculous

g
(
g
(
x
)
)
=

α

−
1

(

α
(
x
)
−
1

)

=

f

−
1

(
x
)
=
x
+

x

2

{\displaystyle g(g(x))=\alpha ^{-1}\left(\alpha (x)-1\right)=f^{-1}(x)=x+x^{2}}

...
Note that

α

{\displaystyle \alpha }
is actually holomorphic in an open sector that does not include the origin, such as real part positive. That is the punchline here,

α

{\displaystyle \alpha }
cannot be extended around the origin as single-valued holomorphic.
So, since we are finding a power series around

0

{\displaystyle 0}
, not only are there a

1

/

z

{\displaystyle 1/z}
term, which would not be so bad, but there is also a

log
⁡
z

{\displaystyle \log z}
term.
So the

…
−
n

{\displaystyle \ldots -n}
I give a complete worked example at my question http://mathoverflow.net/questions/45608/formal-power-series-convergence
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.htmlThe Julia equation is Theorem 8.5.1 on page 346 of KCG.
It would be no problem to produce, say, 50 terms of

α
(
x
)

{\displaystyle \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

α
=

stuff

−
n

{\displaystyle \alpha ={\mbox{stuff}}-n}

when

f
≤
x

{\displaystyle f\leq x}

As I said in comment, the way to improve this is to take a few dozen terms in the expansion of

α
(
x
)

{\displaystyle \alpha (x)}
so as to get the desired decimal precision with a more reasonable number of evaluations of

f
(
x
)
.
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α
(
x
)

{\displaystyle f(x).