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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jul 24, 2012 at 18:15 vote accept Malik Younsi
Jul 24, 2012 at 18:15 history bounty ended Malik Younsi
Jul 24, 2012 at 15:17 answer added Malik Younsi timeline score: 0
Jul 23, 2012 at 16:53 answer added GH from MO timeline score: 6
Jul 22, 2012 at 19:46 answer added Gerald Edgar timeline score: 4
Jul 22, 2012 at 17:24 answer added juan timeline score: 16
Jul 22, 2012 at 13:24 history bounty started Malik Younsi
Jul 22, 2012 at 13:24 history edited Malik Younsi CC BY-SA 3.0
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Jul 20, 2012 at 15:49 comment added Malik Younsi For information, $K(k)=EllipticK(k)$ in Maple.
Jul 20, 2012 at 15:47 comment added Malik Younsi Yes, according to reference.wolfram.com/mathematica/ref/EllipticK.html. Weird...
Jul 20, 2012 at 15:38 comment added GH from MO I clarified: K(k)=EllipticK[k^2] in Mathematica.
Jul 20, 2012 at 15:20 comment added GH from MO Are you sure you typed in the function all right? In Mathematica I get a very different plot. For example, at k=4/5 it tells me K(3/5)=1.94957, K(4/5)=2.25721, so f(4/5)=0.8*2.25721*Sinh[(pi/2)*1.94957/2.25721]=3.27375, while your graph suggests a value below 3.
Jul 20, 2012 at 13:44 comment added Malik Younsi @Suvrit : Good idea, I'll check this out.. Thank you!
Jul 20, 2012 at 13:35 history edited Malik Younsi CC BY-SA 3.0
added 236 characters in body
Jul 19, 2012 at 19:50 comment added Suvrit I have a feeling that looking at Laplace transforms and Bernstein's theorem on completely monotonic functions along with some inverse Laplace transforms might help...but translating this feeling into a proof might be non-trivial...
Jul 19, 2012 at 17:13 history asked Malik Younsi CC BY-SA 3.0