Timeline for How to find the limit of this recursion
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2023 at 12:25 | comment | added | fedja | @IosifPinelis And inequalities for polynomials proved by Mathematica? :lol:? I posted the formal derivation. Sorry for the delay: had some other fish to fry yesterday. | |
| Jan 4, 2023 at 12:23 | answer | added | fedja | timeline score: 1 | |
| Jan 4, 2023 at 4:57 | review | Close votes | |||
| Jan 22, 2023 at 3:08 | |||||
| Jan 4, 2023 at 3:11 | comment | added | Iosif Pinelis | @fedja : A formal, complete proof certainly seems within reach now (reducing to two certain inequalities for elementary functions of one variable, which in turn are reducible to inequalities for polynomials in one variable). However, my proof is going to be long and messy, and at this point I am unsure if it is worth the effort. | |
| Jan 3, 2023 at 21:19 | comment | added | Christophe Leuridan | It is still not clear. Is $n$ a parameter or the index of the sequence. First, you say that it is a parameter, and you define a sequence $(a_k)$ by recursion. The recursion formula involves $n$ as a parameter. Then you ask for the behavior of $a_n$ as $n$ goes to infinity. Should we read $a_k$ as $k$ goes to infinity? Is there a typo somewhere? Do you have a sequence $(a_k)$ depending on a parameter $n$? Be clear with the notations, the assumptions and the question! | |
| Jan 3, 2023 at 21:01 | comment | added | Iosif Pinelis | @fedja : So, it's a spelling problem :-) I have this asymptotics only heuristically at this point. Do you have a formal proof of this? | |
| Jan 3, 2023 at 20:13 | comment | added | fedja | @IosifPinelis It is fairly clear that "limit" is the OP's way to spell "asymptotics", so, if I don't mistake, the answer is $\asymp n\log(2e-1)$... | |
| Jan 3, 2023 at 18:58 | comment | added | Iosif Pinelis | If $n\ge1$, then $a_k\ge a_{k-1}+1$ for all $k\ge1$ and hence $a_n\ge n$, so that the limit of $a_n$ as $n\rightarrow\infty$ is $\infty$. | |
| Jan 3, 2023 at 18:54 | comment | added | user497270 | Sorry, $a_0=0$ , the parameter $n$ goes to infinity, essentially I'm taking the limit of the series $b_n$ where $b_n$ is defined as $a_n$ (the $n$th element in the defined recursion) with the parameter $n$. | |
| Jan 3, 2023 at 18:29 | comment | added | Christophe Leuridan | The question is not clear. What goes to infinity? $k$ or the parameter $n$? | |
| Jan 3, 2023 at 17:55 | comment | added | fedja | What is $a_0$ (or, rather, does it depend on $n$?). | |
| S Jan 3, 2023 at 17:38 | review | First questions | |||
| Jan 4, 2023 at 11:27 | |||||
| S Jan 3, 2023 at 17:38 | history | asked | user497270 | CC BY-SA 4.0 |