Timeline for A truncated divisor function sum
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2020 at 12:53 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added 4 characters in body; edited tags
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| Dec 26, 2020 at 12:37 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag (the question was bumped anyway)
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| Dec 26, 2020 at 7:04 | history | edited | kodlu | CC BY-SA 4.0 |
deleted 125 characters in body
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| Dec 26, 2020 at 6:38 | history | edited | kodlu | CC BY-SA 4.0 |
added 127 characters in body
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| Apr 12, 2012 at 4:07 | vote | accept | kodlu | ||
| Apr 8, 2012 at 9:14 | answer | added | Dr. Pi | timeline score: 1 | |
| Apr 6, 2012 at 5:22 | answer | added | Dimitris Koukoulopoulos | timeline score: 6 | |
| Apr 3, 2012 at 23:56 | comment | added | kodlu | Of course I meant $A(x)$ is the same value as the first sum, i.e., is asymptotically $\sum_{n\leq x} d(n)$, for $x$ large enough. | |
| Apr 3, 2012 at 23:47 | comment | added | kodlu | @Woett: Thanks, you're right. So $A(x)$ is the same value as the second sum for $x$ large enough. @GH: Are you able to comment some more on the case $f(x)$ a power of $\log x$? | |
| Apr 3, 2012 at 17:49 | comment | added | Woett | We definitely have $d(n) = o(n^{\epsilon})$. So $min[d(n), f(x)] = d(n)$ if $f(x) \ge cx^{\alpha}$. Am I missing something? | |
| Apr 3, 2012 at 7:00 | comment | added | kodlu | @GH: I am sorry, I meant to put in min, so Large values of d(n) get truncated. | |
| Apr 3, 2012 at 7:00 | history | edited | kodlu | CC BY-SA 3.0 |
Corrected by replacing Max with Min
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| Apr 3, 2012 at 6:52 | comment | added | GH from MO | The difference between the two sums comes from small values of $d(n)$, namely from those less than $f(x)$. Your candidates $f(x)$ make the second sum very simple, namely $xf(x)$. Things become interesting for much smaller $f(x)$ such as a power $\log x$. | |
| Apr 3, 2012 at 6:15 | history | asked | kodlu | CC BY-SA 3.0 |