Timeline for Convergence rate of a sequence
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2022 at 15:32 | vote | accept | Jean Legall | ||
| Oct 13, 2022 at 14:32 | comment | added | Saúl RM | I see. In the end I didn't need $\eta<\frac{1}{12}$, it can be improved to $\frac{1}{8}$ with one more line of argument. I wrote my answer supposing that $x_k>0$ for all $k$, although that is not obvious to me | |
| Oct 13, 2022 at 14:22 | answer | added | Saúl RM | timeline score: 1 | |
| Oct 13, 2022 at 4:03 | comment | added | Jean Legall | For $\eta=1/8$, I run numerical experiments and it seems that the whole sequence $x_k$ is positive and decreasing. Furthermore, as $k\to\infty$, I found that $kx_k^2\to 2$. It would still be very helpful if you can write your proof for $\eta<1/12$. @SaúlRM | |
| Oct 13, 2022 at 3:52 | comment | added | Jean Legall | Thanks for your reply. I can prove $x_{k+1}/x_k\in(0,1)$ for sufficiently large $k$, which means that eventually $x_k$ will not change sign and $|x_k|$ will be decreasing (it is possible to be negative). If we want $x_k=\Theta(1/\sqrt{k})$, then we need to show $x_{k+1}/x_k\to 1$. However, I can only show $x_{k+1}/x_k$ converges to either $1$ or $1/2$, using the result in mathoverflow.net/questions/432262/convergence-of-a-sequence. I don't know how to exclude the possibility that $x_{k+1}/x_k\to 1/2$, and it that is true, then the convergence rate of $x_k$ would be linear convergence. | |
| Oct 12, 2022 at 23:13 | comment | added | Saúl RM | In fact I think the delicate part is proving that $x_k>0$ for all $k$ and that $\frac{x_{k+1}}{x_k}\to 0$, which seem like they should be obvious at first glance. The condition $\eta<\frac{1}{8}$ is to ensure that the sequence $x_k$ doesn't become negative? | |
| Oct 12, 2022 at 22:57 | comment | added | Saúl RM | Are you interested in the case where $\eta$ is close to $\frac{1}{8}$? I'm pretty sure I have a proof but it works better if $\eta<\frac{1}{12}$, if $\eta$ is close to $\frac{1}{8}$ it seems like a more complicated argument would be needed | |
| Oct 12, 2022 at 2:38 | history | edited | Jean Legall | CC BY-SA 4.0 |
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| Oct 12, 2022 at 0:01 | history | asked | Jean Legall | CC BY-SA 4.0 |