Timeline for Asymptotics of a recurrence relation
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2015 at 10:19 | comment | added | r9m | @BrendanMcKay could you explain how we get to the quasi-periodic component, it'd be nice if you elaborate over an answer! Thanks!! :) | |
| Jul 31, 2015 at 13:11 | comment | added | Brendan McKay | The solution of $\sum t^{2^i} 2^i=n$ is $t=2^{-R(n)/n}$, where $R(n)$ is a quasi-periodic function that oscillates around $1/\ln(2)^2$ without converging. If I didn't get it wrong. There will be a quasi-periodic component to the asymptotics. | |
| Jul 31, 2015 at 12:39 | comment | added | r9m | Indeed I tried Hayman's method but couldn't have an exact asymptotics for $a_n$, (+1) Thanks! :) | |
| Jul 31, 2015 at 12:39 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
added 56 characters in body
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| Jul 31, 2015 at 12:25 | history | answered | Ofir Gorodetsky | CC BY-SA 3.0 |