Michael Reid wrote: <<

Let N = 2K+1 for some positive integer K. Evaluate explicitly

f(N) := Product (5 - 4*cos(2*j*pi/N))

where j runs through 1,2,...,K, and prove your answer.

This is one of those regular polygon sums I was blathering about. It's pretty clear that any product or sum over a period of a rational function of trigs comes out in closed form. You can interconvert such products and sums via trig, calculus, and generating function hacks. (This would make a nice computer algebra project.) A handy generalization of your product is n /===\ | | 2 %pi j 2 %pi j | | (x + b cos(------- + g) + a cos(------- + f)) = | | n n j = 1 2 2 n/2 2 (2 a b cos(g - f) + b + a ) x (chebyshev_t (--------------------------------) n 2 2 sqrt(2 a b cos(g - f) + b + a ) b cos(g) + a cos(f) n - chebyshev_t (- --------------------------------))/2 . n 2 2 sqrt(2 a b cos(g - f) + b + a ) Log-differentiating dt the special case n 1 2 ------------------------------ = -------------------------------- n 1 n /===\ 2 (cos(n acos(-)) - cos(f n)) t | | 2 %pi j t | | (1 - cos(------- + f) t) | | n j = 1 and using the standard Chebyshev series with x=1/t, n ==== \ 1

---------------------- = n / 2 %pi j ==== sec(------- + f) x - 1 j = 1 n

inf j n j n + j j ==== j 2 cos(%pi (- - -)) binomial(-----, j) x \ 2 2 2 (( > -------------------------------------------) / n + j ==== j = 1 inf j n j n + j j ==== 2 cos(%pi (- - -)) binomial(-----, j) x \ 2 2 2 /(n > ----------------------------------------- - cos(f n) / n + j ==== j = 1 %pi n + cos(-----)) - 1), 2 which is the generating function of the identities for sum(sec^k(...)). Expanding thru x^3 gives n %pi n ==== n sin(-----) \ 2 %pi j 2

sec(------- + f) = ---------------------, / n %pi n ==== cos(f n) - cos(-----) j = 1 2

n 2 %pi n ==== n (1 - cos(-----) cos(f n)) \ 2 f n + 2 %pi j 2

sec (-------------) = ----------------------------, / n %pi n 2 ==== (cos(-----) - cos(f n)) j = 1 2

and n ==== \ 3 2 %pi j %pi n

sec (------- + f) = n sin(-----) / n 2 ==== j = 1

2 %pi n 2 2 2 2 (((n + 2) cos(f n) - 2 cos(-----)) - n ((n + 8) cos (f n) - 8)) 2 %pi n 3 /(8 (cos(f n) - cos(-----)) ). 2 The phase coincident case of that messy cotcotcot sum is just n ==== \ 3 %pi j 2 2

cot (----- + f) = n cot(f n) (n csc (f n) - 1). / n ==== j = 1

Note the half as long period. --rwg --------------------------------- Never miss a thing. Make Yahoo your homepage.