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

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 $y\to 0$.

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How would you expand the two roots of $x^2=exp(-1/y^2)$? – Piero D'Ancona Aug 13 '10 at 7:10
I would have thought that you get an asymptotic expansion by replacing the $a_i$ by their asymptotic expansions, i.e., their Taylor series (which are formal power series), and then do the usual formal Puiseux series expansion of $x$. – Torsten Ekedahl Aug 13 '10 at 8:29
@Piero I believe the expansions would simply be 0 for both roots. @Torsten Ah, I guess I've never actually computed a 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 of any smooth function has such an expansion. – Mike Hall Aug 13 '10 at 16:36
up vote 3 down vote accepted

The constructive proof of the Newton-Puiseux theorem works formally, and a posteriori one can show that if the original coefficients are convergent, then the Puiseux series are convergent (see the details in Casas-Alvero's book, "Singularities of Plane Curves"). So Torsten Ekedahl's comment is right, of course. However, I wonder how useful this would be when the power series don't converge to the original smooth functions.

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That was exactly my point. You can not hope to represent the solutions by a Puiseux series, even in the simplest examples. And you can use $C^\infty$ functions to produce much weirder examples (e.g., infinite number of oscillations near many points) – Piero D'Ancona Aug 14 '10 at 17:10
The Taylor series gives the asymptotic expansion of a smooth function in terms of powers of $y$ and as such is uniquely determined and does give a lot of information about the behaviour of the function around the point. Of course the flat functions (those which are $O(y^n)$ for all $n$) get ignored so that $\exp(-1/y^2) \sim 0$ and then also so is any solution of $x^2=\exp(-1/y^2)$. Whether or not an asymptotic expansion of this type is useful or not will of course depend on your needs but the OP seems to have found it so as the question is about asymptotic expansions. – Torsten Ekedahl Aug 14 '10 at 17:44
@Torsten: sure. I am questioning that the Puiseux expansion may be given any reasonable meaning. I'm thinking of examples like Glaeser's non negative, smooth function, flat at 0, whose square root fails to be $C^2$. In that case what is the meaning of the expansion? which would be a 'Taylor' expansion of any order of a non smooth function? I'm just perplexed and slightly curious about the usefulness of Mike's efforts. – Piero D'Ancona Aug 14 '10 at 19:18
@Piero: Interesting, I didn't know that you had that kind of complications. I agree that Puiseux expansions as asymptic expansions for functions on the real line look a little fishy. For one thing it seems safer to have them defined only for positive reals as otherwise $y^{m/n}$ may not even be well defined. – Torsten Ekedahl Aug 15 '10 at 22:12
I think the expansions certainly tell you something meaningful about where the zeros are, namely that they are within $\mathcal{O}(y^\infty)$ of several curves with the given Puiseux expansions. It's true that the $\mathcal{O}(y^\infty)$ error means you have essentially no control over derivatives. Still, I think this gives a nice picture of what the zero set looks like near a point. – Mike Hall Aug 16 '10 at 17:51

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