Lemma: If f(z) is analytic on a closed disk, then the average of f over the bounding circle equals its value at the center. [Proof: Expand f into a Taylor series, and make it into a Fourier series.] When r < 1, log(z) is analytic on and inside the circle, so <log(z)> = log(1) = 0. Since GM(f) = exp(<log(f)>), the answer is 1. When r > 1, we have a logarithmic branch point. Arbitrarily choose the branch cut so that the log is continuous for -pi < x < pi. We have log(1 + r exp(ix)) = log(r exp(ix)) + log(1 + exp(-ix)/r)) The lemma implies that the mean value of the last term is zero, so <log(1 + r exp(ix))> = log(r) + i <x> = log(r), since <x> = 0 by choice of branch cut. Then the geometric mean is r. -- Gene ________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, August 14, 2009 11:08:30 AM Subject: [math-fun] Geometric mean over a circle of complex numbers Puzzle: Find the geometric mean of the complex numbers lying on the circle {1 + r exp(ix) : -pi < x < pi}, as a function g(r) of r > 0. --Dan