I am interested in solving the following functional equation:


Here $n\geq 2$ is a fixed integer and $F(z,w)$ is a power series in two variables with complex coefficients, which you may assume that converges in a neighborhood of the origin in $\mathbb C^2$.

My thoughts:

Defining $G(j)=F(z,w^j)$ for $j\geq 1$ and using the functional equation we obtain


Dividing both sides by $\prod_{k=0}^\infty a_{2^kj}=c_j$ and defining $H(j)=G(j)/c_j$ we get


which implies

$$\begin{align*} F(z,w)=&\,G(1)=c_1H(1)\\ =&\,c_1\sum_{r=0}^\infty\bigl[H(2^r)-H(2^{r+1})\bigr]+c_1\lim_{r\to\infty}H(2^r)\\ =&\,\sum_{r=0}^\infty b_{2^r}\frac{c_1}{c_{2^r}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ %=&\,\sum_{r=0}^\infty \frac{z-zw^{2^r}}{1-w^{2^r}-zw^{2^r}}\,\prod_{k=0}^{r-1}\frac{zw^{2^kn}}{1-w^{2^k}-zw^{2^k}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ \end{align*}$$

Since $F$ is holomorphic, then for $|w|<1$ we have $\lim\limits_{r\to\infty}G(2^r)=F(z,0)=z$ by the functional equation. Defining


we arrive to the following "explicit" formula, provided that the sequence $\boldsymbol{(d_r)_{r\geq1}}$ converges for $\boldsymbol{|z|,|w|}$ small:

$$F(z,w)=\sum_{r=0}^\infty b_{2^r}d_r+z\lim_{r\to\infty}d_r\,.$$

I personally believe that the sequence $(d_r)_{r\geq1}$ indeed converges. Ironically, I put a great effort into the partial solution above, but I am not feeling like solving the system of infinite linear equations satisfied by the coefficients of $F$. Who knows? some seasoned complex analyst can help me and obtain an explicit formula for both the series and the infinite product above.


In this answer, we make some observations that OP might find useful in his search for an explicit solution.

  1. Let us rewrite the double power series $$F(z,w)~=~\sum_{j,k=0}^{\infty}a_{jk}z^jw^k ~=~\sum_{k=0}^{\infty}F_k(z)w^k \tag{A}$$ in terms of coefficient functions $F_k(z)$. OP's functional equation leads to a recursion relation for the coefficient functions $$ F_k(z) ~=~(1+z) F_{k-1}(z) - z F_{\frac{k-n}{2}}(z) + z \delta_k^0 - z \delta_k^1, \qquad k~\in~\mathbb{N}_0.\tag{B} $$ [It is implicitly understood in eq. (B) that $F_{\ell}(z)\equiv 0$ if $\ell\notin\mathbb{N}_0$.] Hence the first few coefficient functions read $$ F_0(z)~=~z, \qquad F_1(z)~=~z^2, \qquad F_2(z)~=~z^3 + (1-\delta_n^2)z^2,\qquad \ldots.\tag{C} $$ It is not hard to see that the $k$'th coefficient function $F_k(z)$ is a $k\!+\!1$ order polynomial of the form $$F_k(z)~\stackrel{(B)}{=}~\sum_{j=1}^{k+1}a_{jk}z^j~\stackrel{(B)}{=}~~z^{k+1}+ \text{lower-order terms}.\tag{D}$$

  2. It becomes clear that there exists a unique solution to OP's functional equation.

  3. It follows that the power series $F(z,w)$ is absolutely convergent for at least $$|z|, |w|~<~\frac{1}{2},\tag{E}$$ thereby agreeing with OP's comment about the existence of an absolutely convergent neighborhood around the origin $(z,w)=(0,0)$. To prove this convergence claim, define recursively non-negative majorant coefficient functions $$ G_0(z)~:=~|z|, \qquad G_1(z)~:=~|z|^2,$$ $$ G_k(z)~:=~(1+|z|) G_{k-1}(z) + |z| G_{\frac{k-n}{2}}(z) , \qquad k~\in~\mathbb{N}\backslash\{0,1\}. \tag{F} $$ and note that by induction $$ |z|~<~\frac{1}{2}\qquad \stackrel{(B)+(F)}{\Rightarrow} \qquad \sum_{j=1}^{k+1} |a_{jk}||z|^j~\leq~|G_k(z)|~<~2^{k-1}. \tag{G}$$


We begin with defining a new function $$ f(z,q,r) = z \sum_{k\ge 0} (zq)^k a_k(q,r)$$ by power series where we define $$ a_0(q,r) = 1, a_{k+1}(q,r) = (a_k(q,r) - q^kra_k(q^2,r^2))/(1-q). $$ It is not obvious from the recursive definition that if $n\ge0$ then $a_k(q,q^n)$ is a polynomial in $q$ with positive integer coefficients. The connection with the original function is $$ F(z,w) = f(z,w,w^{n-1})$$ while the new function satisfies $$ (1-q-zq)f(z,q,r) + zqrf(z,q^2,r^2) = z(1-q). $$ As special cases of $F$ we have $F(0,w) = 0$ and also if $n=1$ or $w=0$ then $F(z,w) = z$ because if $k>0$ then $a_k(w,1) = 0$ or $(zw)^k = 0$ respectively.

Another special case is if $n=2$ and $w=1$, then $F(z,1) = z A(z)$ where $A(z)$ is the generating function of sequence A008934. If we look at the power series coefficients for $n\ge1$ with $w=1$ they form the table in sequence A093729.


Let us write ${\Bbb D}(0,r)=\{z\in {\Bbb C}: |z|<r\}$ for the open disk of radius $r>0$ centered at the origin in the complex plan. I claim that the functional equation defines a meromorphic function in the domain $(z,w)\in {\Bbb C} \times {\Bbb D}(0,1)$.

It is slightly easier to work with a different function than $F$. Note that since $F(0,w)\equiv 0$ if $F$ is analytic in a neighborhood of the origin it has the form $F(z,w)=z H(z,w)$ again with $H$ analytic and verifying $$ H(z,w) = \frac{1-w}{1-w(1+z)} - \frac{z w^n}{1-w(1+z)} H(z,w^2).$$

To find solutions of the latter consider for $0<\delta<1$, $0<R$ the polydisk $\Omega=\Omega_{R,\delta}= {\Bbb D}(0,R) \times {\Bbb D}(0,\delta) \subset {\Bbb C}^2$ and the space $$ A=A_{R,\delta} = C^\omega( \Omega_{R,\delta}) \cap C(\overline{\Omega}_{R,\delta}) $$ of analytic functions on $\Omega$ having a continuous continuation to the boundary. It is a Banach space under the sup norm $\|H\| = \sup_{(z,w)\in \Omega} |H(z,w)|$.

We will look at the map: $$ \Phi(H) (z,w) = \frac{1-w}{1-w(1+z)} - \frac{z w^n}{1-w(1+z)} H(z,w^2), \ \ \ H \in A$$ When $1-\delta(1+R)>0$ then $\Phi$ maps $A$ into $A$ since $(z,w)\in \Omega$ implies $(z,w^2)\in \Omega$ (and the RHS is analytic). If in addition the contraction condition $$ \eta=\frac{R \delta^n}{1 - \delta(1+R)} <1 $$ holds, then $$ \|\Phi (H_1) - \Phi(H_2) \| \leq \eta \|H_1 - H_2\|, \ \ \ H_1,H_2\in A.$$ This shows that $\Phi$ is a Lipschitz contraction, whence by the fixed point theorem has a unique fixed point, again denoted by $H$. As mentioned above $F(z,w)=zH(z,w)$ then solves the posed problem on the domain $\Omega$. Note that given any $0<|z|=R<+\infty$ you may find $\delta=\delta_R>0$ which satifies the contraction condition. Furthermore, when $|w|<1$ then iterating the map $\Phi$, $p$ times we get meromorphic factors times $H(z,w^{2^p})$ and for $p$ large enough we have $|w|^{2^p}< \delta_R$ so that $H(z,w^{2^p})$ is analytic. In other words, $H(z,w)$ is meromorphic (e.g. by viewing it for fixed $w$ as a function of $w$, or the other way around). Singularities appear precisely when $1-w^{2^k}(1+z)=0$, $k\geq 0$. We conclude that the functional equation defines a function which is analytic on the domain: $$ \{ (z,w)\in {\Bbb C}^2: |w|<1 \ \ \mbox{and} \ \ \forall k\geq 0 : w^{2^k}(1+z)\neq 1\}$$

The above map $\Phi$ also gives good power series developments in $\Omega$ (assuming the contraction condition). However, when $|w|=1$, the iteration fails, and I imagine that this constitutes a natural boundary for a meromorphic extension of $H^*$ (not sure about this, and anyway wouldn't know how to prove this off hand). The reason is that if $w=\exp(2\pi i u)$ with $u$ irrational the map $w\mapsto w^2$ becomes a rather erratic (ergodic) map of the circle, which probably induces a rather bad behaviour of any possible extension. This also suggests that there are no explicit solution to the problem, other than developping the infinite product/sum obtained by iteration.


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