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Nov 28, 2012 at 23:06 comment added Daniel Thank you for the compliments. I think there is some inconsistency in the question by Leitz, but I simply chose to start the indices at 1 in my answer and I tried to be consistent with it. Probably 0 would have been a better choice as that is the choice in the referenced paper.
Nov 28, 2012 at 20:16 comment added The Bridge @Daniel Marahrens and Leitz: There is a little typo I think, $k_1$ should start at 0 and not at 1, but this comes from a typo in the question itself, where the definition of $\bar{s}_k$ should include $s_0$. Best regards
Nov 27, 2012 at 21:20 comment added Leitz Thank you for the answer, I just recognized it is also possible by using integrals.
Nov 27, 2012 at 21:15 vote accept Leitz
Nov 27, 2012 at 20:48 comment added The Bridge @Daniel Marahrens : Very nice proof. Best regards
Nov 27, 2012 at 20:24 history answered Daniel CC BY-SA 3.0