Timeline for Puiseux series for roots of polynomials with smooth coefficients
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2010 at 2:07 | vote | accept | Mike Hall | ||
| Aug 14, 2010 at 18:41 | vote | accept | Mike Hall | ||
| Aug 15, 2010 at 2:07 | |||||
| Aug 14, 2010 at 14:42 | answer | added | quim | timeline score: 3 | |
| Aug 13, 2010 at 16:36 | comment | added | Mike Hall | @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. | |
| Aug 13, 2010 at 8:29 | comment | added | Torsten Ekedahl | 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 $x$. | |
| Aug 13, 2010 at 6:05 | history | asked | Mike Hall | CC BY-SA 2.5 |