Timeline for Does this expression always vanish?
Current License: CC BY-SA 4.0
7 events
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| Sep 30, 2023 at 6:51 | comment | added | Markus Scheuer | @IlyaBogdanov: I see. Many thanks for your reply and thank you very much for this elegant answer. I have taken the liberty of writing a slightly more detailed version of your response to MSE. | |
| Sep 27, 2023 at 8:43 | comment | added | Ilya Bogdanov | @epi163sqrt When dealing with the $i$th summand, you apply all $\partial/\partial x_j$ with $j\neq i$ first, that’s easy. Then it remains to apply $\partial/\partial x_i$, and, perhaps, it would be more clear if I wrote the result using the logarithmic derivative, as $\displaystyle \left(\frac{n-1}{A_i}-2\sum_{j\neq i}\frac1{A_i-A_j}\right) A_i^{n-1} \prod_{j\neq i}\frac1{(A_i-A_j)^2}$. | |
| Sep 26, 2023 at 19:39 | comment | added | Markus Scheuer | @IlyaBogdanov: Very nice answer and instructive approach. (+1) May I ask how you arrived at the expression with factor $n-1$? Is there a known formula, or is the expression tool based? | |
| Sep 24, 2023 at 3:17 | comment | added | Silly Point | Thanks a lot for the answer. | |
| Sep 24, 2023 at 3:14 | vote | accept | Silly Point | ||
| Sep 24, 2023 at 3:13 | vote | accept | Silly Point | ||
| Sep 24, 2023 at 3:14 | |||||
| Sep 23, 2023 at 9:24 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |