Timeline for Evaluate $\sum_{\sigma} (2\pi i)^{-n}\oint \frac{f_{\sigma(1)}(u)\dots f_{\sigma_n(1)}(u)}{(u_2 - u_1)\dots (u_n - u_{n-1})}du_1\dots du_n$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2017 at 8:51 | comment | added | Fedor Petrov | I do not understand what the identity is already for $n=2$. If the integrand is $(f_1(u_1)f_2(u_2)-f_1(u_2)f_2(u_1))/(u_2-u_1)$, the identity does not hold: for pairs $(f_1,f_2)=(f,1)$ and $(f_1,f_2)=(1,f)$ the values of the integral have opposite sign | |
| Oct 27, 2017 at 19:53 | comment | added | Fedor Petrov | There should be $\sigma$ in the denominator too, else we get something wrong for $n=2$ (the difference $u_2-u_1$ does not cancel) | |
| Oct 27, 2017 at 15:47 | comment | added | john mangual | Hope I wrote it correctly now. Usually they just wrote $\Sigma_{cyc}$ as you might see in Olympiad. These integral identities were convenient as they evaluated $n$-point correlations. | |
| Oct 27, 2017 at 15:34 | comment | added | Fedor Petrov | I would add that $S^{cyc}$ means the cyclic subgroup of $S_n$ generated by a long cycle. | |
| Oct 27, 2017 at 14:34 | comment | added | Jules Lamers | Just a comment: I encountered this too, see Lemma 7 of arxiv.org/abs/1510.00342. Though not lengthy, the proof I give there is still by induction. | |
| Oct 27, 2017 at 14:10 | history | edited | john mangual | CC BY-SA 3.0 |
hopefully integral is more suggestive
|
| Oct 27, 2017 at 14:03 | history | asked | john mangual | CC BY-SA 3.0 |