Timeline for how to pass from algebraic power series to the analytic ones
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jul 18, 2019 at 10:14 | vote | accept | Dmitry Kerner | ||
Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
Apr 11, 2017 at 5:54 | history | edited | Todd Leason | CC BY-SA 3.0 |
Added answer to the second question.
|
Apr 4, 2017 at 9:03 | comment | added | Dmitry Kerner | Sorry for the confusion, $C^{\infty}(\Bbb{R}^1,0)$ is the ring of germs of infinitely differentiable functions (defined on some neighborhoods of $0\in \Bbb{R}^1$) | |
Apr 4, 2017 at 8:15 | comment | added | Todd Leason | So $C^\infty(\mathbb{R},0)$ is the (one-element) set of functions $\mathbb{R}\to \{0\}$ ? | |
Apr 4, 2017 at 8:06 | comment | added | Dmitry Kerner | Well, I meant a local ring with the non-trivial maximal ideal. (0 denotes the zero ideal) | |
Apr 4, 2017 at 7:58 | comment | added | Todd Leason | In my example $R$ is a field and hence local. BTW: What's the meaning of the $0$ in $C^\infty(\mathbb{R},0)$ ? | |
Apr 3, 2017 at 19:34 | comment | added | Dmitry Kerner | I meant mostly the local rings, and your example is not local, right? If the local ring has no flat functions, i.e. $\mathfrak{m}^\infty=\{0\}$, then the power series can be controlled by $\mathfrak{m}$-adic topology. But for the ring $C^\infty(\Bbb{R}^1,0)$ the situation seems to be tricky. | |
Apr 3, 2017 at 0:05 | history | answered | Todd Leason | CC BY-SA 3.0 |