Timeline for Wanted: Positivity certificate for the AM-GM inequality in low dimension
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2019 at 12:16 | comment | added | Michael Rozenberg | Yes, of course! | |
| Oct 3, 2019 at 11:29 | comment | added | vidyarthi | Shouldnt it be $\sum_{cyc}a^2(b+c+d+e)=\frac1{6}\sum_{sym}a^2b$ | |
| Oct 3, 2019 at 11:05 | comment | added | Michael Rozenberg | For example $\sum\limits_{sym}a^2b=\frac{1}{6}\sum\limits_{cyc}a^2(b+c+d+e)$ because in the expression $\sum\limits_{cyc}a^2(b+c+d+e)$ we have $20$ terms and $\frac{5!}{20}=6.$ This thinking is very useful for the work with symmetric polynomials! | |
| Oct 3, 2019 at 7:56 | comment | added | vidyarthi | sorry, I didnt get you. You mean, you apply the same faormula again and again, is it? | |
| Oct 3, 2019 at 3:09 | comment | added | Michael Rozenberg | @vidyarthi I do it every time again. | |
| Oct 2, 2019 at 21:33 | comment | added | vidyarthi | thanks! So that seems something like the relation between cyclic and symmetric groups. By the way, could you refer a proper repository of such formulae and factorizations, that is a work dealing with symmetric and cyclic inequalities | |
| Oct 2, 2019 at 21:00 | comment | added | Michael Rozenberg | @vidyarthi I used $\sum\limits_{sym}a^5=\frac{5!}{5}\sum\limits_{cyc}a^5=24\sum\limits_{cyc}a^5.$ | |
| Oct 2, 2019 at 20:20 | comment | added | vidyarthi | could you add some more explanations, like how the constant $\frac1{48}$ came in the first step? | |
| Jan 30, 2019 at 8:21 | history | answered | Michael Rozenberg | CC BY-SA 4.0 |