Timeline for Constructing a vector consisting of nonnegative entries
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2021 at 16:16 | vote | accept | vidyarthi | ||
| Feb 9, 2021 at 16:17 | |||||
| Feb 10, 2020 at 21:25 | vote | accept | vidyarthi | ||
| Feb 9, 2021 at 16:16 | |||||
| Feb 10, 2020 at 21:17 | answer | added | Max Alekseyev | timeline score: 2 | |
| Feb 10, 2020 at 17:52 | history | edited | vidyarthi | CC BY-SA 4.0 |
added 100 characters in body
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| Feb 10, 2020 at 17:02 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 11 characters in body
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| S Feb 10, 2020 at 1:08 | history | suggested | RobPratt | CC BY-SA 4.0 |
changed curly braces to parentheses
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| Feb 9, 2020 at 22:16 | comment | added | vidyarthi | @MaxAlekseyev yes, suppose I specify any $n$ and $m$ | |
| Feb 9, 2020 at 22:08 | comment | added | Max Alekseyev | What is given? Both $n$ and $m$? | |
| Feb 9, 2020 at 21:08 | review | Suggested edits | |||
| S Feb 10, 2020 at 1:08 | |||||
| Feb 9, 2020 at 20:59 | answer | added | RobPratt | timeline score: 0 | |
| Feb 9, 2020 at 19:57 | comment | added | vidyarthi | @RobPratt well observed, edited again now. Actually the condition is derived from the condition of being able to obtain a symmetric anti-circulant matrix by shifting the vector $v$. | |
| Feb 9, 2020 at 19:55 | history | edited | vidyarthi | CC BY-SA 4.0 |
added 11 characters in body
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| Feb 9, 2020 at 18:39 | comment | added | RobPratt | Your second example fails for $j=6$ because $5\not\equiv 2+6-1 \pmod{5}$. | |
| Feb 9, 2020 at 0:39 | comment | added | vidyarthi | @RobPratt sorry, again edited hope its correct now. | |
| Feb 9, 2020 at 0:38 | history | edited | vidyarthi | CC BY-SA 4.0 |
edited body
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| Feb 8, 2020 at 23:43 | comment | added | RobPratt | For the first example, $n=5$, and $j=2$ yields $a_{n-j+1}=a_{5-2+1}=a_4=5\not= 3$. | |
| Feb 8, 2020 at 22:40 | comment | added | vidyarthi | @RobPratt no, we do have $4\equiv 3+2-1 \pmod 5$. In fact you misread $a_{n-j+1}$ for $n$ | |
| Feb 8, 2020 at 22:37 | history | edited | LSpice | CC BY-SA 4.0 |
TeX fix
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| Feb 8, 2020 at 22:37 | comment | added | RobPratt | Still not right. For $j=2$, $4\not\equiv 5+2-1 \pmod 5$. | |
| Feb 8, 2020 at 22:35 | history | edited | LSpice | CC BY-SA 4.0 |
TeX fix
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| Feb 8, 2020 at 22:29 | comment | added | vidyarthi | @RobPratt thanks! edited. see now. | |
| S Feb 8, 2020 at 22:28 | history | edited | vidyarthi | CC BY-SA 4.0 |
Corrected spelling in title
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| S Feb 8, 2020 at 22:28 | history | suggested | RobPratt | CC BY-SA 4.0 |
Corrected spelling in title
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| Feb 8, 2020 at 22:24 | comment | added | RobPratt | Your two examples are nonnegative, have $a_1=1$, and have distinct nonzero entries, but the mod condition fails for both. | |
| Feb 8, 2020 at 22:17 | review | Suggested edits | |||
| S Feb 8, 2020 at 22:28 | |||||
| Feb 8, 2020 at 21:58 | history | asked | vidyarthi | CC BY-SA 4.0 |