Timeline for Who needs a symmetric upper asymptotic density on the integers?
Current License: CC BY-SA 3.0
15 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Jun 16, 2015 at 12:55 | answer | added | Salvo Tringali | timeline score: 1 | |
Jun 9, 2015 at 21:55 | vote | accept | Salvo Tringali | ||
Jun 9, 2015 at 21:38 | answer | added | Joe Silverman | timeline score: 3 | |
Jun 9, 2015 at 19:07 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
edited title
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Jun 9, 2015 at 18:59 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Changed the title; new update
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Jun 9, 2015 at 15:52 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added 79 characters in body
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Jun 9, 2015 at 15:42 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Update
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May 30, 2015 at 17:35 | comment | added | Anthony Quas | From my part of the world, where sequences come from invariant measures on $\{0,1\}^{\mathbb Z}$, the two quantities agree almost surely. | |
May 30, 2015 at 13:37 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added 193 characters in body
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May 30, 2015 at 13:29 | comment | added | Salvo Tringali | @Wojowu: I've hopefully answered your question in the OP. | |
May 30, 2015 at 13:26 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added 609 characters in body
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May 30, 2015 at 12:35 | comment | added | Wojowu | I have never seen upper asymptotic density (not the symmetric one) being used for $\Bbb Z$, because it completely ignores half of the set. May I ask where have you seen it being used? | |
May 30, 2015 at 10:25 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Fixed a typo
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May 30, 2015 at 10:16 | history | asked | Salvo Tringali | CC BY-SA 3.0 |