Timeline for How to work with this power series? [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 13, 2016 at 5:58 | vote | accept | Radmir Sultamuratov | ||
| Oct 12, 2016 at 22:00 | history | closed |
Franz Lemmermeyer Michael Albanese Alex Degtyarev Stefan Kohl♦ Myshkin |
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| Oct 12, 2016 at 21:10 | comment | added | darij grinberg | See web.mit.edu/~darij/www/algebra/va3.pdf (Sections 1-3) for some fundamentals of these "doubly-infinite" formal series. (This is all folklore material, but few people have bothered writing it down. Note that my writeup is complicated by the extra lengths I went to make things work natively in positive characteristic.) | |
| Oct 12, 2016 at 19:47 | comment | added | vltava | This Laurent series DOES converge---on the unit circle (the limit is the Dirac delta function, with singularity at the point "1"). Convergence is, of course, not in the classical but in the distributional sense. The details of this were worked out by Gottfried Köthe in 1952---for a reference, see my answer below. | |
| Oct 12, 2016 at 17:35 | review | Close votes | |||
| Oct 12, 2016 at 22:00 | |||||
| Oct 12, 2016 at 17:04 | comment | added | Gerald Edgar | This is known as a "Laurent series". The region of the complex plane where it converges is, in general, an annulus centered at $0$. Of course the annulus may be (as in your example) empty. | |
| Oct 12, 2016 at 16:40 | answer | added | vltava | timeline score: 2 | |
| Oct 12, 2016 at 16:22 | answer | added | R.P. | timeline score: 4 | |
| Oct 12, 2016 at 16:15 | answer | added | Fedor Petrov | timeline score: 7 | |
| Oct 12, 2016 at 16:13 | comment | added | Radmir Sultamuratov | thank you, but i know it, my question is:Is there ANY WAY to work with such generating functions? | |
| Oct 12, 2016 at 16:12 | comment | added | Alan | The radius of convergence is not the same for the two series. For the series $1/(1-z)$ the radius of convergence is $|z|<1$ and for $1/(1-1/z))$ is $|z|>1$. | |
| Oct 12, 2016 at 16:08 | history | asked | Radmir Sultamuratov | CC BY-SA 3.0 |