On Friday 15 February 2008, Dan Asimov wrote:

As happens so often, I blew it again. Sorry.

Here is the corrected version: the coefficient of the cosine term is 4 (and not 2):

. . . . . . . . . . ... spoiler space ... . . . . . . . . . . On a moderate amount of numerical evidence, I believe I know the answer to your original and corrected puzzles, and to a common generalization of both: if n=2k+1 and f(k) = product {1..k} of (a-2b.cos(2j.pi/n)) then f(0)=1, f(1) = a+b, f(k+2) = a.f(k+1)-b^2.f(k) from which we derive that if u,v = a/2 +- sqrt(a^2/4-b^2) and A,B = (1 +- sqrt[(a+2b)/(a-2b)]) / 2 then f(k) = Au^k + Bv^k. Can we prove this? Thinking aloud ... Clearly multiplying a,b by C multiplies f(k) by C^k, both for the given product formula and for my alleged general expression, so if we're trying to prove the latter we can scale a,b however we like. Let's take a=2, so f(k) = 2^k product {1..k} of (1-b.cos(2j.pi/n)) u,v = 1 +- sqrt(1-b^2) A,B = 1/2 (1 +- sqrt[(1+b)/(1-b)]) = 1/2(1-b) (1 +- sqrt(1-b^2) - b) and our alleged value for f(k) is g(k) = [(u-b)u^k + (v-b)v^k] / 2(1-b). Let's throw in a factor of 2(1-b); we want to prove 2^(k+1) product {0..k} of (1-b.cos(2j.pi/n)) = u^(k+1) + v^(k+1) - b(u^k+v^k). Now it's apparent that expanding out the powers on the RHS will make the square roots go away, so we get a polynomial in b on both sides. We know the roots of the LHS polynomial; the degrees are equal; so let's substitute a root of the LHS into the RHS and see what happens. So, if b = sec t where t = j/n.2pi then 1-b^2 = 1-sec^2 t = - tan^2 t so u,v = 1 +- i tan t, which we might prefer to write as sec t exp(+-it). So (u-b)u^k = sec^(k+1) t (exp(it)-1) exp(ikt) (v-b)v^k = sec^(k+1) t (exp(-it)-1) exp(-ikt) and their sum is sec^(k+1) t [exp(i(k+1)t) + exp(-i(k+1)t) - exp(ikt) - exp(-ikt)] = -2 sec^(k+1) t (cos (k+1)t - cos kt) = 4 sec^(k+1) t sin (2k+1)t/2 sin t/2 and now the first sin() factor is zero. I can't be bothered to check the constant coefficient, but obviously it'll come out right; then we're done. (It would probably have been cleaner to take b=1/2 instead of a=2, in hindsight.) There's probably a much more elegant way to see this. -- g