Timeline for Low-level proof of identity related to Weierstrass P-function
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7 at 14:47 | vote | accept | Kevin Buzzard | ||
| Apr 18 at 19:57 | answer | added | user516477 | timeline score: 6 | |
| Apr 17 at 14:21 | comment | added | François Brunault | We have the identity $(\partial X/\partial u)/(2Y+X) = 1/u$, which expresses the fact that the invariant differential is $du/u$ (since $u=e^{2\pi iz}$). It seems easy to prove. | |
| Apr 17 at 13:37 | comment | added | Kevin Buzzard | @i9Fn thanks -- fixed | |
| Apr 17 at 13:37 | history | edited | Kevin Buzzard | CC BY-SA 4.0 |
Fix formula for Y as per comment
|
| Apr 15 at 13:39 | comment | added | Peter Taylor | It might be worth rewriting the double sums to instead be of the form $\sum_{d \ge 1} f(d) \frac{q^d}{1-q^d}$. | |
| Apr 15 at 9:19 | comment | added | i9Fn | A minor typo, it should be $Y=\frac{u^{\color{red}{2}}}{(1-u)^3}+\ldots$ | |
| Apr 15 at 8:38 | comment | added | Chris Wuthrich | I would even be interested in any argument avoiding the complex parametrisation, but high-brow algebraic geometry would be ok. Though Kevin might not want that. | |
| Apr 15 at 7:57 | comment | added | Kevin Buzzard | I've removed the word "combinatorial" from the question because it has connotations which I didn't mean to imply. | |
| Apr 15 at 7:56 | history | edited | Kevin Buzzard | CC BY-SA 4.0 |
remove the word "combinatorial" from the question as per comments
|
| Apr 15 at 7:51 | comment | added | Kevin Buzzard | Sorry for the ambiguity -- I just mean "proof which involves goofing around with power series and no complex analysis" | |
| Apr 14 at 3:08 | comment | added | Ira Gessel | Some people use the word "combinatorial" for anything to do with (formal) power series. Other people call such things "analytic." | |
| Apr 14 at 2:59 | comment | added | Timothy Chow | Combinatorialists sometimes reserve the term "purely combinatorial" for situations involving only positive integers (because then you can hope that the integers count some kind of combinatorial object). Is there a formulation of the identity that doesn't involve minus signs? Alternatively, what about a proof that doesn't use any analysis, and involves only manipulation of formal power series? Combinatorialists might not call such an argument "combinatorial" but maybe you'd be satisfied with it. | |
| Apr 14 at 2:33 | comment | added | Steven Stadnicki | By 'combinatorial' are you thinking expanding out $X$ and $Y$ into formal power series in $u$ and $q$ and finding a way of equating terms? Do you have a combinatorial interpretation for either of them? (Very much looking forward to seeing if anyone has an answer to your question, incidentally; it's an excellent one!) | |
| Apr 14 at 0:33 | history | asked | Kevin Buzzard | CC BY-SA 4.0 |