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Consider the polynomial $K(z,u)= u^e- z (p_0+p_1 u+...+p_k u^k)$, where: $k,e$ are nonnegative integers and the coefficients $p_i$ are nonnegative reals such that $0< \sum_i p_i \leq 1$ with $p_0>0$. We say $u_s(z)$ is a small root of $K(z,u)$ if $K(z,u_s(z))=0$ identically, and $|u_s(z)|$ is bounded near $z=0$ (in other words, the Puiseux expansion at $z=0$ of $u_s(z)$ has no negative exponent).

Question: under what conditions (on $k,e,p_i,...$) every small root $u_s(z)$ of $K$ can be analytically continued along the interval $[0,1)$ and $\lim_{z\rightarrow 1^-} u_s(z)$ exists finite ($\lim$ taken along $[0,1)$)?

Context. The polynomial $K(z,u)$ arises in connection with a counting problem of random walks. Cf. directed lattice paths in Flajolet's Analytic Combinatorics; the "small root" terminology comes from there.

An elementary proof / reference would be particularly appreciated.

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  • $\begingroup$ Your conditions do not define the function $u_s(z)$ on $(0,1)$ unambiguously. First of all, the multi-valued function $u(z)$ has no poles at all for finite $z$, so your condition that $|u(z)|$ is bounded is redundant. Second, the branch with the property $u(0)=0$ has a ramification point at $z=0$ when $e>1$. So you have to specify whether your question is about SOME branch on $(0,1)$ or about ALL branches. $\endgroup$ Commented Sep 14 at 12:22
  • $\begingroup$ @AlexandreEremenko. I am not an expert in complex analysis, but let me try to clarify with an example. Consider $K(z,u)=u-z(1/2+(1/2)u^2)$. This has two branches: $u_1(z)=2z^{-1}-(1/2)z-(1/8)z^3-(1/16)z^5+...$ and $u_2(z)=(1/2)z+(1/8)z^3+(1/16)z^5+...$. Out of these, only $u_2(z)$ is a small root, because $|u_1(z)|$ is not bounded around $z=0$. In this case, I am only interested in what happens to $u_2(z)$ as $z$ approaches 1 from the left (so not in $u_1(z)$). I edited my question to make this clearer, hopefully. $\endgroup$ Commented Sep 16 at 7:26
  • $\begingroup$ NB: I am identifying Puiseux series with functions on a suitable subset of the complex plane. $\endgroup$ Commented Sep 16 at 7:27
  • $\begingroup$ I was talking of the case $e>k$. In this case, no branch becomes infinite, and your boundedness condition is not sufficient to distinguish branches. In your example, only one branch is finite at 0, but in general, this is not so. $\endgroup$ Commented Sep 16 at 13:23

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