Timeline for Determinant of the "quantum" version of the group $\mathbb{Z}_n$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2017 at 12:54 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Apr 21, 2017 at 3:28 | vote | accept | T. Amdeberhan | ||
| Apr 21, 2017 at 3:28 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Apr 18, 2017 at 5:10 | answer | added | Gjergji Zaimi | timeline score: 4 | |
| Apr 17, 2017 at 17:46 | answer | added | Fedor Petrov | timeline score: 2 | |
| Apr 17, 2017 at 17:06 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Apr 17, 2017 at 14:56 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 305 characters in body
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| Apr 17, 2017 at 13:58 | comment | added | T. Amdeberhan | That's nice. By the way, $[1]_q+\cdots+[n-1]_q=\frac{[n]_q-n}{q-1}$. | |
| Apr 17, 2017 at 6:55 | comment | added | მამუკა ჯიბლაძე | (the sign above is $(-1)^{\frac{n(n-1)}2}$) | |
| Apr 17, 2017 at 6:47 | comment | added | მამუკა ჯიბლაძე | ...which by the way I find more natural - why should $[0]_q=[1]_q$ hold?? | |
| Apr 17, 2017 at 6:46 | comment | added | მამუკა ჯიბლაძე | Oops that was with the $[0]_q=0$ convention | |
| Apr 17, 2017 at 6:27 | comment | added | მამუკა ჯიბლაძე | Without a proof it is $\pm[n]_q^{n-2}([1]_q+[2]_q+...+[n-1]_q)$ | |
| Apr 17, 2017 at 6:05 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |