# Analytic continuation of $f(x)=\sqrt{\frac{1-x}{1-x^p}}$ over the p-adics

Consider the power series $$f(x)=\sqrt{\frac{1-x}{1-x^p}}$$ over the algebraic closure of $$\mathbb{Q}_p$$, defined by $$f(0)=1$$. What can be said about an analytic continuation "in the form of Mittag-Leffler"? By that I mean that $$f(x)$$ can be expressed as $$f(x)=g(x)+\sum_{n=0}^{\infty} \frac{a_n(x)}{h(x)^n}$$, where $$g(x)$$ converges in a disk greater than the unit ball, $$h(x)$$ is a polynomial of norm $$\le 1$$, and the $$a_n(x)$$ are polynomials satisfying $$\text{ord}_p(a_n)\ge rn-m$$ for some $$r>0$$ and a constant $$m \in \mathbb{R}$$. Here the order of a polynomial is the minimum order of its coefficients.

I am particuarly interested in the values $$f(\lambda)$$, for Teichmuller representatives $$\lambda$$, which satisfy $$\lambda^q=\lambda$$ for some $$q=p^r$$. Thus I would like $$h(x)$$ not to have a zero at the Teichmuller representatives. Apparently the values at the Teichmuller representatives are related to classical results on Gauss Sums but I couldn't find anything in the literature.

• By "analytic continuation", you mean that the power series converges also outside the unit ball, or analytic in the sense of Krasner? – EFinat-S Feb 13 at 15:09
• @EFinat-S, I have clarified what I mean by analytic continuation, I think it corresponds to the latter. – Martin Ortiz Feb 13 at 16:11
• Ok, that makes an important difference. I'll delete my answer. – EFinat-S Feb 13 at 16:16
• Sorry for another question: is the condition $\operatorname{ord}_p(a_n) \ge r n$ a condition on some coefficient of $a_n(x)$, all coefficients of $a_n(x)$, some value of $a_n(x)$, or all coefficients of $a_n(x)$, or is it something else? – LSpice Feb 13 at 18:54
• I don't know if this will help you, but the book "Analytic elements in $p$-adic analysis" by Alain Escassut has a chapter called "Composition of anaytic elements". Perhaps this is useful since you are composing an analytic function with a rational function. – EFinat-S Feb 13 at 23:38