Timeline for What is the "best" good kernel?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2023 at 20:31 | comment | added | Iosif Pinelis | Your post is clearer now. Yet, as Alexei said, you can just take any subsequence $(k_{n_j})$ of $(k_n)$ and thus make the convergence however fast. | |
| Apr 16, 2023 at 20:24 | comment | added | Dr. Pi | Thank you Iosif Pinelis and Aleksei Kulikov for the questions! I added some detail and an example to explain a little bit better what I meant. | |
| Apr 16, 2023 at 20:23 | history | edited | Dr. Pi | CC BY-SA 4.0 |
added example and more detail to question
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| Apr 16, 2023 at 19:25 | comment | added | Aleksei Kulikov | The problem is, if you take a subsequence $k_{n_i}$ then it will also be a good kernel, and this way you can accelerate the convergence however you want by a diagonal process. So it is not clear what exactly are you asking for? | |
| Apr 16, 2023 at 19:25 | comment | added | Iosif Pinelis | What do you mean here? Would every function whatsoever of $n$ and $\epsilon$ be considered super fast by you? | |
| Apr 16, 2023 at 19:23 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting
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| S Apr 16, 2023 at 19:21 | history | suggested | CommunityBot | CC BY-SA 4.0 |
proper spacing separating dx
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| Apr 16, 2023 at 19:13 | comment | added | Dr. Pi | As a function of $n$ and $\epsilon$. | |
| Apr 16, 2023 at 19:13 | review | Suggested edits | |||
| S Apr 16, 2023 at 19:21 | |||||
| Apr 16, 2023 at 18:40 | comment | added | Iosif Pinelis | What do you mean by "super fast"? | |
| Apr 16, 2023 at 18:34 | history | asked | Dr. Pi | CC BY-SA 4.0 |