If

$$p(x,y) = x^N + a_{N-1}(y)x^{N-1} + \ldots + a_0(y), \quad x,y \in \mathbf{C}$$

is a monic polynomial in $x$, and the coefficients $a_j$ are analytic functions of $y$, then the roots of $p$ have expansions in Puiseux series (in powers of $y^{1/m}$ for some $m$) which are convergent for $y$ sufficiently close to 0.

Is this true in an asymptotic sense when the $a_j$ are only assumed to be smooth? i.e. Do the roots have asymptotic expansions which are formal Puiseux series (not necessarily convergent)?

For $A$ to have an asymptotic expansion $A \sim B_1 + B_2 + \ldots$ with respect to some grading $\mathcal{O}(n)$ means that $B_n \in \mathcal{O}(n)$, and for each $N$, $A - \sum_{n=1}^N B_n \in O(N+1)$. Here $\mathcal{O}(n)$ means $\mathcal{O}(|y|^{n/m})$ in the usual big-O notation, as $y\to 0$.

formalPuiseux series expansion of $x$. – Torsten Ekedahl Aug 13 '10 at 8:29computeda Puiseux series, but I suppose the basic algorithm is just to backsolve for the coefficients. If the algorithm works, I guess it has to work asymptotically. I'm a little surprised at this outcome as, combined with the Malgrange preparation theorem, this should imply that the germ of zeros ofanysmooth function has such an expansion. – Mike Hall Aug 13 '10 at 16:36