I'd like to gather information and references on the following functional equation for power series $$f(f(x))=x+f(x)^2,$$$$f(x)=\sum_{k=1}^\infty c_k x^k$$

(so $c_0=0$ is imposed).

First things that can be established quickly:

- it has a unique solution in $\mathbb{R}[[x]]$, as the coefficients are recursively determined;
- its formal inverse is $f^{-1}(x)=-f(-x)$ , as both solve uniquely the same functional equation;
- since the equation may be rewritten $f(x)=f^{-1}(x)+x^2$, it also follows that $f(x)+f(-x)=x^2$, the even part of $f$ is just $x^2/2$, and $c_2$ is the only non-zero coefficient of even degree;
- from the recursive formula for the coefficients, they appear to be integer multiples of negative powers of $2$ (see below the recursive formmula).
**Rmk.**It seems (but I did not try to prove it) that $2^{k-1}c_k$ is an integer for all $k$, and that $(-1)^k c_{2k-1} > 0$ for all $k\geq 2$.

Question :how to see in a quick way that this series has a positive radius of convergence, and possibly to compute or to evaluate it?

[updated]A more reasonable question, after the numeric results and various comments, seems to be, rather: how to prove that this seriesdoes notconverge.

Note that the radius of convergence has to be finite, otherwise $f$ would be an automorphism of $\mathbb{C}$. Yes, of course I did evaluate the first coefficients and put them in OEIS, getting the sequence of numerators A107700; unfortunately, it has no further information.

**Motivation.** I want to understand a simple discrete dynamical system on $\mathbb{R}^2$, namely the diffeomorphism $\phi: (x,y)\mapsto (y, x+y^2)$, which has a unique fixed point at the origin. It is not hard to show that the stable manifold and the unstable manifold of $\phi$ are
$$W^s(\phi)=\mathrm{graph}\big( g_{|(-\infty,\ 0]}\big)$$
$$W^u(\phi)=\mathrm{graph}\big( g_{|[0, \ +\infty)}\big)$$

for a certain continuous, strictly increasing function $g:\mathbb{R}\to\mathbb{R}$, that solves the above functional equation. Therefore, knowing that the power series solution has a positive radius of convergence immediately implies that it coincides locally with $g$ (indeed, if $f$ converges we have $f(x)=x+x^2/2+o(x^2)$ at $x=0$ so its graph on $x\le0$ is included in $W^s$, and its graph on $x\ge0$ is included in $W^u$: therefore the whole graph of $f$ would be included in the graph of $g$,implying that $f$ coincides locally with $g$). If this is the case, $g$ is then analytic everywhere, for suitable iterates of $\phi$ give analytic diffeomorphism of any large portion of the graph of $g$ with a small portion close to the origin.

One may also argue the other way, showing directly that $g$ is analytic, which would imply the convergence of $f$. Although it seems feasible, the latter argument would look a bit indirect way, and in that case I'd like to make sure there is no easy direct way of working on the coefficients (of course, it may happen that $g$ is not analytic and $f$ is not convergent).

**Details**: equating the coefficients in both sided of the equation for $f$ one has, for the 2-Jet
$$c_1^2x+(c_1c_2+c_2c_1^2)x^2 =x + c_1^2x^2,$$
whence $c_1=1$ and $c_2=\frac 1 2;$ and for $n>2$
$$2c_n=\sum_{1\le j\le n-1}c_jc_{n-j}\,-\sum_{1 < r < n \atop \|k\|_1=n}c_rc_{k_1}\dots c_{k_r}.$$

**More details**: since it may be of interest, let me add the argument to see $W^s(\phi)$ and
$W^u(\phi)$ as graphs.

Since $\phi$ is conjugate to $\phi^{-1}=J\phi J $ by the linear involution $J:(x,y)\mapsto (-y,-x)$, we have $W^u(\phi):=W^s(\phi^{-1})=J\ W^s(\phi)$, and it suffices to study $\Gamma:=W^s(\phi)$. For any $(a,b)\in\mathbb{R}^2$ we have $\phi^n(a,b)=(x_n,x_{n+1})$, with $x_0=a$, $x_1=b$, and $x_{n+1}=x_n^2+x_{n-1}$ for all $n\in\mathbb{N}$. From this it is easy to see that $x_{2n}$ and $x_{2n+1}$ are both increasing; moreover, $x_{2n}$ is bounded above iff $x_{2n+1}$ is bounded above, iff $x_{2n}$ converges, iff $x_n\to 0$, iff $x_n\le 0 $ for all $n\in\mathbb{N}$.

As a consequence $(a,b)\in \Gamma$ iff $\phi^n(a,b)\in Q:=(-\infty,0]\times(-\infty,0]$, whence $ \Gamma=\cap_{ n\in\mathbb{N}} \phi^{-n}(Q)$. The latter is a nested intersection of connected unbounded closed sets containing the origin, therefore such is $\Gamma$ too.

In particular, for any $a\leq 0$ there exists at least a $b\leq 0$ such that $(a,b)\in \Gamma$: to prove that $b$ is unique, that is, that $\Gamma$ is a graph over $(\infty,0]$, the argument is as follows. Consider the function $V:\Gamma\times\Gamma\to\mathbb{R}$ such that $V(p,p'):=(a-a')(b-b')$ for all $p:=(a,b)$ and $p':=(a',b')$ in $\Gamma$.

Showing that $\Gamma$ is the graph of a strictly increasing function is equivalent to show that $V(p,p')>0$ for all pair of distinct points $p\neq p'$ in $\Gamma$.

By direct computation we have $V\big(\phi(p),\phi(p')\big)\leq V(p,p')$ and $\big(\phi(p)-\phi(p')\big)^2\geq \|p-p'\|^2+2V(p,p')(b+b')$. Now, if a pair $(p,p')\in\Gamma\times\Gamma$ has $V(p,p')\le0$, then also by induction $V\big(\phi^n(p),\phi^n(p')\big)\leq 0$ and $\big(\phi^n(p)-\phi^n(p')\big)^2\geq \|p-p'\|^2$ for all $n$, so $p=p'$ since both $\phi^n(p)$ and $\phi^n(p')$ tend to $0$. This proves that $\Gamma$ is a graph of a strictly increasing function $g:\mathbb{R}\to\mathbb{R}$: since it is connected, $g$ is also continuous. Of course the fact that $\Gamma$ is $\phi$-invariant implies that $g$ solves the functional equation.

Enumerative Combinatorics, vol. 1, second ed. $\endgroup$ – Richard Stanley Sep 25 '13 at 1:189more comments