Timeline for Asymptotic analysis of $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2021 at 9:43 | comment | added | user164469 | Also 10 upvotes, congratulations ! | |
| Feb 16, 2021 at 17:40 | answer | added | juan | timeline score: 1 | |
| Feb 16, 2021 at 4:38 | history | edited | River Li | CC BY-SA 4.0 |
add what is done in [1]
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| Feb 16, 2021 at 3:03 | answer | added | River Li | timeline score: 2 | |
| Feb 15, 2021 at 23:50 | comment | added | River Li | @fedja :) I mean in MSE and AoPS, I saw many question of recurrence relations. | |
| Feb 15, 2021 at 23:48 | comment | added | fedja | @RiverLi Really? How about $x_{n+1}=1-x_n$? :lol: | |
| Feb 15, 2021 at 23:46 | comment | added | River Li | @fedja Thanks. It is the first time I see an recurrence relation has no common asymptotics for odd and even indices. | |
| Feb 15, 2021 at 23:41 | vote | accept | River Li | ||
| Feb 15, 2021 at 22:29 | answer | added | Iosif Pinelis | timeline score: 8 | |
| Feb 15, 2021 at 17:43 | history | edited | gmvh |
Added top-level tag
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| Feb 15, 2021 at 17:03 | comment | added | fedja | @Marcel Be careful: all that shows is that IF (and that is a big and false if) the asymptotics of the given type exists, then it must be with $5/8$. However, the OP is completely right that there is no common asymptotics for odd and even indices. Moreover, the two different coefficients he gets depend on $x_1$, so, unlike it is with the first two terms, they change if you start with some other number. | |
| Feb 15, 2021 at 16:21 | comment | added | Marcel | Substitute $x_n=n+1/2+a/n$ into the equation. Expand for large $n$. Leading order is $(2a-5/4)/n=0$, so indeed $a=5/8$. | |
| Feb 15, 2021 at 15:56 | history | asked | River Li | CC BY-SA 4.0 |