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Dec 14, 2017 at 16:35 vote accept Nick Dong
Dec 14, 2017 at 16:04 comment added Nick Dong $\langle k^n \rangle = \sum_a F(a)\langle k^n|a \rangle=\sum_a F(a) \sum_k k^n g(k|a) =\sum_a F(a) \sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}\lambda^m =\sum_a F(a)\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}[T(a+\langle a\rangle)]^m =\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}T^m\sum_a F(a)(a+\langle a\rangle)^m =\sum_m\genfrac{\{}{\}}{0pt}{}{n}{m}T^m\kappa_m$
Dec 14, 2017 at 15:51 history edited Carlo Beenakker CC BY-SA 3.0
added 151 characters in body
Dec 14, 2017 at 11:21 history answered Carlo Beenakker CC BY-SA 3.0