# G Holms

Fractional iteration of the function f(x) = 1/(1+x) by

# Will Jagy

Below is an example by Will Jagy
"First, an example. Begin with
which has derivative 1 at $z=0$ but, along the positive real axis, is slightly less than $x$ when $x>0$.
We want to find a Fatou coordinate, which Milnor (page 107) denotes $\alpha$ that is infinite at $0$ and otherwise solves what is usually called the Abel functional equation,
There is only one holomorphic Fatou coordinate up to an additive constant. We take
To get fractional iterates $f_{s}(z)$ of $f(z)$, with real $0\leq s\leq 1$ we take
and finally
The desired semigroup homomorphism holds,
with $f_{0}(z)=z$ and $f_{1}(z)=f(z)$
Alright, the case of $\sin z$ emphasizing the positive real axis is not terribly different, as long as we restrict to the interval $0<x\leq {\frac {\pi }{2}}.$ For any such $x,$ define $x_{0}=x,\;x_{1}=\sin x,\;x_{2}=\sin \sin x,$ and in general $x_{n+1}=\sin x_{n}.$ This sequence approaches 0, and in fact does so for any $z$ in a certain open set around the interval $0<x\leq {\frac {\pi }{2}}$ that is called a petal.
Now, given a specific $x$ with $x_{1}=\sin x$ and $x_{n+1}=\sin x_{n}$ it is a result of Jean Ecalle at Orsay that we may take $\alpha (x)=\lim _{n\rightarrow \infty }\;\;\;{\frac {3}{x_{n}^{2}}}\;+\;{\frac {6\log x_{n}}{5}}\;+\;{\frac {79x_{n}^{2}}{1050}}\;+\;{\frac {29x_{n}^{4}}{2625}}\;-\;n.$
Note that $\alpha$ actually is defined on $0<x<\pi$ with $\alpha (\pi -x)=\alpha (x),$ but the symmetry also means that the inverse function returns to the interval $0<x\leq {\frac {\pi }{2}}.$
Before going on, the limit technique in the previous paragraph is given in pages 346-353 of *Iterative Functional Equations* by Marek Kuczma, Bogdan Choczewski, and Roman Ger. The solution is specifically Theorem 8.5.8 of subsection 8.5D, bottom of page 351 to top of page 353. Subsection 8.5A, pages 346-347, about Julia's equation, is part of the development.
As before, we define ( at least for $0<x\leq {\frac {\pi }{2}}$) the parametrized interpolating functions, $f_{s}(x)=\alpha ^{-1}\left(s+\alpha (x)\right)$
In particular $f_{1/2}(x)=\alpha ^{-1}\left({\frac {1}{2}}+\alpha (x)\right)$
I calculated all of this last night. First, by the kindness of Daniel Geisler, I have a pdf of the graph of this at:
Note that we use the evident symmetries $f_{1/2}(-x)=-f_{1/2}(x)$ and $f_{1/2}(\pi -x)=f_{1/2}(x)$
The result gives an interpolation of functions $f_{s}(x)$ ending at $f_{1}(x)=\sin x$ but beginning at the continuous periodic sawtooth function, $x$ for $-{\frac {\pi }{2}}\leq x\leq {\frac {\pi }{2}},$ then $\pi -x$ for ${\frac {\pi }{2}}\leq x\leq {\frac {3\pi }{2}},$ continue with period $2\pi .$ We do get $f_{s}(f_{t}(z))=f_{s+t}(z),$ plus the holomorphicity and symmetry of $\alpha$ show that $f_{s}(x)$ is analytic on the full open interval $0<x<\pi .$
** EDIT, TUTORIAL**: Given some $z$ in the complex plane in the interior of the equilateral triangle with vertices at $0,{\sqrt {3}}+i,{\sqrt {3}}-i,$ take $z_{0}=z,\;\;z_{1}=\sin z,\;z_{2}=\sin \sin z,$ in general $z_{n+1}=\sin z_{n}$ and $z_{n}=\sin ^{[n]}(z).$ It does not take long to show that $z_{n}$ stays within the triangle, and that $z_{n}\rightarrow 0$ as $n\rightarrow \infty .$
Second, say $\alpha (z)$ is a true Fatou coordinate on the triangle, $\alpha (\sin z)=\alpha (z)+1,$ although we do not know any specific value. Now, $\alpha (z_{1})-1=\alpha (\sin z_{0})-1=\alpha (z_{0})+1-1=\alpha (z_{0}).$ Also $\alpha (z_{2})-2=\alpha (\sin(z_{1}))-2=\alpha (z_{1})+1-2=\alpha (z_{1})-1=\alpha (z_{0}).$ Induction, given $\alpha (z_{n})-n=\alpha (z_{0}),$ we have $\alpha (z_{n+1})-(n+1)=\alpha (\sin z_{n})-n-1=\alpha (z_{n})+1-n-1=\alpha (z_{0}).$
So, given $z_{n}=\sin ^{[n]}(z),$ we have $\alpha (z_{n})-n=\alpha (z).$
Third , let $L(z)={\frac {3}{z^{2}}}+{\frac {6\log z}{5}}+{\frac {79z^{2}}{1050}}+{\frac {29z^{4}}{2625}}$. This is a sort of asymptotic expansion (at 0) for $\alpha (z),$ the error is $|L(z)-\alpha (z)|<c_{6}|z|^{6}.$ It is unlikely that putting more terms on $L(z)$ leads to a convergent series, even in the triangle.
Fourth, given some $z=z_{0}$ in the triangle. We know that $z_{n}\rightarrow 0$. So $|L(z_{n})-\alpha (z_{n})|<c_{6}|z_{n}|^{6}.$ Or $|(L(z_{n})-n)-(\alpha (z_{n})-n)|<c_{6}|z_{n}|^{6},$ finally $|(L(z_{n})-n)-\alpha (z)|<c_{6}|z_{n}|^{6}.$ Thus the limit being used is appropriate.
Fifth, there is a bootstrapping effect in use. We have no actual value for $\alpha (z),$ but we can write a formal power series for the solution of a Julia equation for $\lambda (z)=1/\alpha '(z),$ that is $\lambda (\sin z)=\cos z\;\lambda (z).$ The formal power series for $\lambda (z)$ begins (KCG Theorem 8.5.1) with $-z^{3}/6,$ the first term in the power series of $\sin z$ after the initial $z.$ We write several more terms, $\lambda (z)\asymp -{\frac {z^{3}}{6}}-{\frac {z^{5}}{30}}-{\frac {41z^{7}}{3780}}-{\frac {4z^{9}}{945}}\cdots .$ We find the formal reciprocal, ${\frac {1}{\lambda (z)}}=\alpha '(z)\asymp -{\frac {6}{z^{3}}}+{\frac {6}{5z}}+{\frac {79z}{525}}+{\frac {116z^{3}}{2625}}+{\frac {91543z^{5}}{6063750}}\cdots .$ Finally we integrate term by term, $\alpha (z)\asymp {\frac {3}{z^{2}}}+{\frac {6\log z}{5}}+{\frac {79z^{2}}{1050}}+{\frac {29z^{4}}{2625}}+{\frac {91543z^{6}}{36382500}}\cdots .$ and truncate where we like, $\alpha (z)={\frac {3}{z^{2}}}+{\frac {6\log z}{5}}+{\frac {79z^{2}}{1050}}+{\frac {29z^{4}}{2625}}+O(z^{6})$
Numerically, let me give some indication of what happens, in particular to emphasize

# References

* ↑ Fractional iteration of the function f(x) = 1/(1+x) by Gottfried Helms
* ↑ mathoverflow question (45608) : does-the-formal-power-series-solution-to-ffx-sin-x-converge
* ↑ Dynamics in one complex variable: introductory lectures John W. Milnor
* ↑ wikipedia: Abel equation