Talk:Reinhardt domain: Difference between revisions
Content deleted Content added
m Maintain {{WPBS}} and vital articles: 1 WikiProject template. Create {{WPBS}}. Tag: |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
{{WikiProject banner shell| |
|||
{{WikiProject Mathematics}} |
|||
}} |
|||
{{merged-from|Logarithmically convex set|2017-07-03}} |
{{merged-from|Logarithmically convex set|2017-07-03}} |
||
{{merged-to|Several complex variables|2020-11-20}} |
|||
== Example in 1 dimensional case == |
== Example in 1 dimensional case == |
||
It seems to me that the following sentence is not exact: |
It seems to me that the following sentence is not exact: |
||
Latest revision as of 18:02, 8 February 2024
 | This redirect does not require a rating on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | |||||||
| ||||||||
 | The contents of the Logarithmically convex set page were merged into Reinhardt domain on 2017-07-03. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see its talk page. |
 | The contents of the Reinhardt domain page were merged into Several complex variables on 2020-11-20. For the contribution history and old versions of the merged article please see its history. |
Example in 1 dimensional case
[edit]It seems to me that the following sentence is not exact: "Note that in one complex variable, a logarithmically convex Reinhardt domain is simply a disc." I think it should be anulus instead of disc. But if we add the condition of containing the origin then it has to be a disc. What do you think? That is actually the first time I study Reinhardt domain, so I don't want to do any change myself for the moment. Bongilles (talk) 22:15, 30 January 2013 (UTC)
- I basically thought the same thing. Based on the definition presented here, the Reinhardt domain appears to be up to n open annuluses containing the zn. This could include a disc if one of the zn is the origin. It might be useful to mention annuluses (annuli?) in the article, if this interpretaion is correct (I've first encountered Reinhardt domains just now by reading this article). It may be that to be "logarithmically convex" a 1d Reinhardt domain must be a disk, which would make the article correct (I don't know what logarithmically convex means).--Wikimedes (talk) 19:56, 8 August 2014 (UTC)
- I see the same issue, but I also don't know this subject well. Looking at the referenced EoM article, it looks like what's missing the concept of a "complete" Reinhardt domain. Also, Reinhardt domains are "domains of holomorphy" for functions, which seems to include Laurent series. Taylor series (I guess) have complete Reinhardt domains as their domain of convergence, and the statement in question should read (I think) "...a logarithmically convex, complete Reinhardt domain is simply a disc." Baccala@freesoft.org (talk) 18:30, 2 February 2020 (UTC)