Timeline for Сlosed formula for $(g\partial)^n$
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2019 at 20:00 | comment | added | Tom Copeland | Related history mathoverflow.net/questions/287742/… | |
| Aug 20, 2019 at 17:34 | history | edited | Tom Copeland |
Added relevant tags
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| Aug 1, 2019 at 20:59 | answer | added | Tom Copeland | timeline score: 4 | |
| Aug 1, 2019 at 12:13 | comment | added | Abdelmalek Abdesselam | I did not know about the work of Scherk in the reference given by Dan. It is certainly relevant. I looked at it and it does not have explicit drawings of trees. I also looked at Cayley's paper and he does not mention Scherk's thesis which definitely preceded him in this investigation. | |
| Aug 1, 2019 at 11:54 | comment | added | Wakabaloola | @AbdelmalekAbdesselam many thanks for adding further insight | |
| Aug 1, 2019 at 11:47 | comment | added | Abdelmalek Abdesselam | The natural objects for indexing the terms of the expansion are trees. This goes back to Cayley who called expressions like $g(z)\partial_z$ "operandators" because they are at the same time operators and operands. See mathoverflow.net/questions/168888/… | |
| Jul 31, 2019 at 14:48 | answer | added | Max Alekseyev | timeline score: 11 | |
| Jul 31, 2019 at 14:30 | comment | added | Wakabaloola | @MaxAlekseyev A recurrence formula for $C_{n,p}(m_1,\dots)$? Perhaps you could write it as an answer, it might accelerate progress towards a more explicit answer. | |
| Jul 31, 2019 at 14:19 | comment | added | Max Alekseyev | @Wakabaloola: There is no much to say, besides a recurrence formula for the coefficients. | |
| Jul 31, 2019 at 14:18 | comment | added | Wakabaloola | @Andrew thanks you for the comment, I'm not quite sure what you have in mind, could you elaborate? | |
| Jul 31, 2019 at 14:17 | comment | added | Wakabaloola | @MaxAlekseyev thank you for your comment, perhaps you can be a bit more explicit, maybe write-up an answer? | |
| Jul 31, 2019 at 14:15 | history | edited | Wakabaloola | CC BY-SA 4.0 |
fixed typos and simplified discussion
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| Jul 31, 2019 at 14:14 | comment | added | Max Alekseyev | The unknown coefficients are essentially given by oeis.org/A124796 | |
| Jul 31, 2019 at 11:55 | comment | added | Wakabaloola | @DanFox Thanks for the useful reference. It seems this is a difficult problem for general $g(z)$. And Scherk back in 1823 apparently stated after thinking about precisely the above question: `... the process has become so un-tractable that we could not succeed with an investigation of all the numerical coefficients, based on the sole discovery of some individual terms.' | |
| Jul 31, 2019 at 11:33 | history | edited | Wakabaloola | CC BY-SA 4.0 |
added more flesh
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| Jul 31, 2019 at 11:27 | comment | added | Dan Fox | The paper arxiv.org/abs/1010.0354 of Blasiak and Flajolet treats this sort of question in a more general setting. In particular, see the appendix. | |
| Jul 31, 2019 at 10:55 | history | edited | Wakabaloola | CC BY-SA 4.0 |
elaboration
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| Jul 31, 2019 at 10:54 | comment | added | Andrew | If to put all $g^k$ to one, unsigned Stirling numbers of the first kind appear oeis.org/A132393 May be it's connected to incomplete Bell polynomials, after making change of variables $u=\int g(z)$. | |
| Jul 31, 2019 at 10:24 | comment | added | Wakabaloola | many thanks for the comment. it's not clear to me how Faa di Bruno's formula might be implemented here, but maybe you see a way? | |
| Jul 31, 2019 at 10:19 | comment | added | Dirk | Maybe this is helpful en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula?wprov=sfla1 | |
| Jul 31, 2019 at 9:46 | history | edited | user64494 | CC BY-SA 4.0 |
Typos in the title.
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| Jul 31, 2019 at 8:53 | history | asked | Wakabaloola | CC BY-SA 4.0 |