Timeline for Puiseux series for roots of polynomials with smooth coefficients
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Aug 16, 2010 at 20:17 | comment | added | Mike Hall | Also, @quim, thanks! That's a very nice reference. | |
| Aug 16, 2010 at 17:53 | comment | added | Mike Hall | @Torsten, when I said "powers of $y^{1/m}$" I was being lazy, as really it means that you get a root for each branch of the $m$th root function. Somewhat relatedly, it's true that I want to know about real roots, but in fact, I'm not interested in exact roots so much as roots modulo $\mathcal{O}(y^\infty)$, in which case we can ignore the error terms, and the zero is real modulo $\mathcal{O}(y^\infty)$ iff all coefficients in the Puiseux expansion are real. I guess it could be a sticky matter in general. | |
| Aug 16, 2010 at 17:51 | comment | added | Mike Hall | 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. | |
| Aug 15, 2010 at 22:12 | comment | added | Torsten Ekedahl | @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. | |
| Aug 15, 2010 at 2:07 | vote | accept | Mike Hall | ||
| Aug 14, 2010 at 19:18 | comment | added | Piero D'Ancona | @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. | |
| Aug 14, 2010 at 18:41 | vote | accept | Mike Hall | ||
| Aug 15, 2010 at 2:07 | |||||
| Aug 14, 2010 at 17:44 | comment | added | Torsten Ekedahl | 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. | |
| Aug 14, 2010 at 17:10 | comment | added | Piero D'Ancona | 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) | |
| Aug 14, 2010 at 14:42 | history | answered | quim | CC BY-SA 2.5 |