Timeline for Conjectured Somos-like closed form of recurrences with polynomial coefficients
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 12 at 14:06 | history | bounty ended | joro | ||
| S Aug 12 at 14:06 | history | notice removed | joro | ||
| Aug 6 at 14:51 | answer | added | joro | timeline score: 0 | |
| S Aug 5 at 13:35 | history | bounty started | joro | ||
| S Aug 5 at 13:35 | history | notice added | joro | Authoritative reference needed | |
| Aug 5 at 13:34 | history | edited | joro | CC BY-SA 4.0 |
Minor edit
|
| Aug 4 at 17:27 | comment | added | Max Alekseyev | @joro: I'm still unsure about the conjectures, but the search for recurrences in the rational function form can be streamlined - see my answer for detail. | |
| Aug 4 at 8:54 | history | edited | joro | CC BY-SA 4.0 |
addressed comments
|
| Aug 4 at 5:40 | history | edited | joro | CC BY-SA 4.0 |
Addressed the comments and the deleted answer
|
| Aug 3 at 17:33 | answer | added | Max Alekseyev | timeline score: 3 | |
| Aug 3 at 13:49 | comment | added | joro |
@MaxAlekseyev Thanks, I see your idea. I rediscovered something similar with resultants. Does it work if there is n^2 in the recurrence?
|
|
| Aug 3 at 10:30 | comment | added | Max Alekseyev |
@joro: Not quite that. See this sample Sage code - just replace in the result each variable f[i] with $f(n+i)$.
|
|
| Aug 3 at 6:44 | comment | added | joro | @MaxAlekseyev By "solve" I mean I can't find algebraic dependency using your original method. | |
| Aug 3 at 5:47 | comment | added | joro | @MaxAlekseyev I can't solve $f(n)=f(n-1)(n+2)+f(n-2)(3 n + 5)$ with your method. Maybe you need to add variables $y_n$,$y_{n+1}$ and take the resultant of $f(n)-y_n,f(n+1)-y_{n+1}$, but this doesn't seem linear, not sure. | |
| Aug 3 at 5:08 | comment | added | Max Alekseyev | What recurrence you cannot process with resultant? | |
| Aug 3 at 4:51 | comment | added | joro | @MaxAlekseyev The paper discusses non-linearity and based on experiments it suggests to try dependency on more f(n-i). I can't reproduce your resultant result, are you sure it works? | |
| Aug 2 at 23:08 | comment | added | Max Alekseyev | Not sure about the conjecture, but using algebraic dependency in the paper is an overkill. It is enough to take two polynomials representing the recurrence for two consecutive terms and compute their resultant with respect to $n$, which will give a polynomial ("algebraic dependency") not depending on $n$. It also shows that there is no option for failure in finding an algebraic dependency, however, I'm not about the "luck" of linearity for the highest term in general. | |
| Aug 2 at 14:10 | comment | added | Gerald Edgar | Remark. Recurrence $b_{n}' = n^2b_{n-1}' - \omega^2 (n-1)^2 b_{n-2}'$ appears here mathoverflow.net/a/475581/454 Maybe this method would help with it when $\omega$ is an integer.. | |
| Aug 2 at 12:32 | history | asked | joro | CC BY-SA 4.0 |