Hmm, I see that since gamma(x) gamma(1-x) = pi/sin(pi x) that means Don's suggestion of pi/gamma(1-x) satisfies pi/gamma(1-x) = sin(pi x) gamma(x) so we are just multiplying gamma by sin(pi x). It coincides with |gamma(x)| for x < 0 exactly where |sin(pi x)| = 1, i.e. at all the half-integers. (And it also coincides with gamma(x) for x > 0 at all the "even" half-integers, i.e. at 1/2, 5/2, 9/2, ....) And clearly since (d/dx)sin(pi x) = 0 at all these points, it follows that pi/gamma(1-x) (=sin(pi x) gamma(x)) is also *tangent* to |gamma(x)| at all these points of coincidence. sin(pi x)'s zeroes get rid of gamma(x)'s poles at negative integers, and sin(pi x)'s wiggles tame the ups and downs of gamma(x) for x < 0. Cool! ((( It's also amusing that when you multiply gamma(x) by sin(pi x) to tame the ups and downs of gamma(x) for negative x, you introduce massive wiggles with the same frequency for positive x. ))) ((( Also: It's also quite clear that the function I sought could not have been convex for *all* negative x. For, gamma on [-1,0] is not consistent with the convex "envelope" I asked about. This is because it's clear visually that the unique line tangent to the graph of gamma(x) at a point in [-1,0] and also at a point in [-3,-2] must lie *below* the graph of |gamma(x)| for x in [-2,-1]. It does appear clear that Don's suggestion of g(x) = pi/gamma(1-x) (=sin(pi x) gamma(x)) is indeed convex for x < -1. ))) —Dan

On Feb 3, 2017, at 11:42 AM, Dan Asimov <asimov@msri.org> wrote:

Yowza! Pi/gamma(1-x) is the one!

How did you arrive at that?

—Dan

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On Feb 3, 2017, at 9:59 AM, Don Reble <djr@nk.ca <mailto:djr@nk.ca>> wrote:

...

y = |gamma(x)| for negative x has one tangency in each interval (-n, -n+1) (n > 0) with the graph of some nice convex function y = g(x) ... Sort of an "envelope". If so, is there a unique "nicest" g that works here?

Is g = pi / gamma(1-x) nice enough?