[math-fun] Dick Askey always chides me about how easy
these Gamma identities are: Gamma[-1/12] == (3^(3/8) ((2 + Sqrt[3])/Pi)^(1/4) Gamma[-1/4] Sqrt[-Gamma[-1/6]])/2^(2/3) I guess FullSimplify and I flunk his course. --rwg
This is a matter of taste, but I lean toward replacing (2+sqrt3)^(1/4) --> sqrt(1+sqrt3)/2^(1/4) since sqrt(2+sqrt3) = (1+sqrt3)/sqrt2. by analogy with 75^(1/4) --> sqrt(5) 3^(1/4). And the Gamma[-1/6] can be converted to something with Gamma[1/3], I think losing the sqrt in sqrt[-Gamma[-1/6]]. Does Askey have a different preferred basis for Gammas? Or a solution to HAKMEM #1? Rich ----- Quoting Bill Gosper <billgosper@gmail.com>:
these Gamma identities are: Gamma[-1/12] == (3^(3/8) ((2 + Sqrt[3])/Pi)^(1/4) Gamma[-1/4] Sqrt[-Gamma[-1/6]])/2^(2/3) I guess FullSimplify and I flunk his course. --rwg
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Bill Gosper -
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