Timeline for Interesting question about the Thomson problem for arbitrary number of electrons
Current License: CC BY-SA 4.0
14 events
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| Jun 14, 2023 at 17:25 | comment | added | Iosif Pinelis | @Rodrigo : $u''(0)$ is a random variable, whose values depend on the values of Rademacher random signs $z_0,\dots,z_{n-1}$. If the average $Eu''(0)$ of the values of $u''(0)$ (over all possible sequences of the signs) is $<0$, then there exist some values of the signs $z_0,\dots,z_{n-1}$ for which the value of the random variable $u''(0)$ is $<0$ -- because the average of nonnegative numbers is nonnegative, whereas our average $Eu''(0)$ is $<0$. | |
| Jun 14, 2023 at 16:33 | comment | added | Rodrigo | @IosifPinelis Sorry to disturb you again. I have one final doubt: How does $s_n > 0$ imply that $u''(0) < 0?$ I understand it implies $Eu''(0) < 0$ but I don't know how to reach $u''(0) < 0$. I am not really good at probability, perhaps I am missing something really obvious. | |
| Jun 14, 2023 at 15:44 | vote | accept | Rodrigo | ||
| Jun 14, 2023 at 15:44 | comment | added | Rodrigo | @IosifPinelis Got it... Thank you so much for your help. Really clarifying. | |
| Jun 14, 2023 at 15:42 | comment | added | Iosif Pinelis | @Rodrigo : I think you actually understood everything in the answer that I actually said. :-) The condition $n\ge12$ is needed and used only for the last inequality in that multiline display -- and I did not say that other inequalities are true only for $n\ge12$. | |
| Jun 14, 2023 at 15:34 | comment | added | Rodrigo | From the last inequality, I believe it follows that $$ s_n \geqslant a_{\star} \sum_{1 \leqslant k \leqslant k_n} (n-k) + a_{min} \sum_{1 \leqslant k \leqslant n-1} (n-k),$$ since $a_{min}$ is negative. The part I don't understand is how is this only valid for $n \geq 12$. | |
| Jun 14, 2023 at 15:33 | comment | added | Rodrigo | @IosifPinelis Thanks. I understood that part of the inequality originally. The one I didn't understand is how you get that $$ s_n \geqslant a_\star \sum_{1 \leqslant k \leqslant k_n} (n-k) + a_{min} \sum_{1 \leqslant k \leqslant n-1} (n-k), $$ for every $n \geqslant 12.$ I did the following: $$ s_n = \sum_{1 \leqslant k \leqslant k_n} (n-k)a_k + \sum_{k_n < k \leqslant n-1} (n-k)a_k.$$ From here, one gets $$ s_n \geqslant a_{\star} \sum_{1 \leqslant k \leqslant k_n} (n-k) + a_{min} \sum_{k_n < k \leqslant n-1} (n-k) $$. I will keep writing in a comment below (out of characters). | |
| Jun 14, 2023 at 15:25 | comment | added | Fawen90 | @IosifPinelis Thanks Iosif for the clarification. I get it | |
| Jun 14, 2023 at 15:24 | comment | added | Iosif Pinelis | @Rodrigo : The two sums in your comments are just the sums of arithmetic progressions. I have added details about that. | |
| Jun 14, 2023 at 15:22 | comment | added | Iosif Pinelis | @Fawen90 : I did not make or use such a claim, that the parametrization covers all possible points of the sphere (and it does not -- we are going along the meridian from each of the $n$ points on the equator towards one of the poles). Please look at the question, highlighted in the OP. | |
| Jun 14, 2023 at 15:15 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 317 characters in body
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| Jun 14, 2023 at 11:08 | comment | added | Rodrigo | Hello Iosif. Thanks for your answer to my problem. I have to say I understood mostly everything, except a few "details" ( important details, I would say). Basically, I guess I understand everything until the last inequality, i.e., until $$ s_n \geq a_{\star} \sum_{1 \leq k \leq n/8} (n-k) + a_{min} \sum_{1 \leqslant k \leqslant n-1} (n-k) > 0. $$ I really can't understand where this inequality comes from. Can you be a little bit more detailed in this part? Also, how does this only work for $n \geqslant 12?$ Thanks for your help. | |
| Jun 14, 2023 at 7:26 | comment | added | Fawen90 | Dear Iosif, I've a question about your solution. Could you justify that your parametrisation $x_1(t),\ldots, x_n(t)$ go through all possible points of $\mathbb S^2$ as $t$ varies in $[-\min_k|a_k|,\min_k|a_k|]$? | |
| Jun 14, 2023 at 5:08 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |