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Dec 11, 2014 at 7:18 comment added larry I can't possibly express my gratitude for your help. Thx u!
Dec 11, 2014 at 7:00 vote accept larry
Dec 11, 2014 at 6:34 vote accept larry
Dec 11, 2014 at 6:34
Dec 10, 2014 at 17:44 comment added Pietro Majer If we interpret $K^{1/2}$ as (any) square root of $K$ the same computation holds true, therefore $e^{-F}$ is a solution too.
Dec 10, 2014 at 16:38 history answered Bazin CC BY-SA 3.0