Timeline for An asymptotic formula for a sum involving powers of floor functions
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| S Aug 16, 2019 at 7:45 | history | suggested | CommunityBot | CC BY-SA 4.0 |
sizes of delimiters
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| Aug 16, 2019 at 4:08 | review | Suggested edits | |||
| S Aug 16, 2019 at 7:45 | |||||
| Aug 15, 2019 at 19:44 | vote | accept | Inequalityforall | ||
| Aug 15, 2019 at 19:32 | answer | added | Harry Richman | timeline score: 5 | |
| Aug 15, 2019 at 19:08 | comment | added | Gerhard Paseman | It seems your question was about independence. Is your question really about the value of c(theta)? Gerhard "Perhaps I Do Not Understand" Paseman, 2019.08.15. | |
| Aug 15, 2019 at 19:06 | comment | added | Inequalityforall | @GerhardPaseman Sorry if I am misunderstanding, but does this approach lead to an expression for $c(\theta)?$ Or why the leading term must be of this form? | |
| Aug 15, 2019 at 19:00 | comment | added | Gerhard Paseman | To me, it answers the question of why it is true that the O(1) does not depend on theta, because the "shape" of the sum (and thus a bound on the error) does not depend on theta. You can get a better error that depends on theta, but it is unclear whether this is worth the effort. Gerhard "And The Equation Needs Fixing" Paseman, 2019.08.15. | |
| Aug 15, 2019 at 18:59 | comment | added | Inequalityforall | So this suggests the answer should involve generalized harmonic numbers in some way. How is not entirely clear to me. | |
| Aug 15, 2019 at 18:51 | comment | added | Inequalityforall | I can easily see that this sum looks as follows: for $n$ less than or equal to $x$ but strictly larger than $x/2,$ the summands are $1.$ For $x/3< n \leq x/2$ they are $2^{-\theta}$ and so on. But I am having trouble using this to get anywhere | |
| Aug 15, 2019 at 18:48 | comment | added | Inequalityforall | @HarryRichman It is from the book ”A course in analytic number theory” by Marius Overholt. | |
| Aug 15, 2019 at 18:47 | comment | added | Inequalityforall | @GerhardPaseman It is obvious to me that most of the summands are between $0$ and $1$. Does this observation make the answer straightforward? | |
| Aug 15, 2019 at 18:40 | comment | added | Harry Richman | Out of curiosity, where have you seen this claim? | |
| Aug 15, 2019 at 18:34 | comment | added | Gerhard Paseman | Even without theta, most of the summands are 0 or 1. With theta positive, the rest are between 0 and 1. If you are having trouble seeing this part, you should ask this on math.stackexchange. Gerhard "The Equations Will Cost Extra" Paseman, 2019.08.15. | |
| Aug 15, 2019 at 18:25 | review | First posts | |||
| Aug 15, 2019 at 18:31 | |||||
| Aug 15, 2019 at 18:25 | history | asked | Inequalityforall | CC BY-SA 4.0 |