Bill Gosper <rwmgosper@yahoo.com> wrote: rwg>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. The key identity is just prod(a-b*%e^(2*%i*%pi*k/n),k,1,n) = a^n-b^n; n 2 %i %pi k /===\ ---------- | | n n n | | (a - b %e ) = a - b , | | k = 1 from which follows prod(%e^(4*%i*%p i*k/n)*r-%e^(2*%i*%pi*k/n)*q+p,k,1,n) = -(sqrt(q^2-4*p*r)+q)^n/2^n-(q -sqrt(q^2-4*p*r))^n/2^n+r^n+p^n; n 4 %i %pi k 2 %i %pi k /===\ ---------- ---------- | | n n | | (%e r - %e q + p) = | | k = 1 2 n 2 n (sqrt(q - 4 p r) + q) (q - sqrt(q - 4 p r)) n n - ----------------------- - ----------------------- + r + p . n n 2 2 Identities so constructed can then be extended via my nonlocal derangement formulae, which blend each term with all the rest, along with an additional, essentially arbitrary sequence, which, if also trigonometric, invites reduction to closed form via the regular polygon formulae. The devil is in the details. Using just c_k := cos(a + 2 pi k/n) for the arbitrary sequence, the derangement of the easy result sum(sec(2*%pi*k/n+a),k,1,n) = n*sin(%pi*n/2)^2*cot((%pi/2-a)*n/2)*cs c((a*n+%pi*n/2)/2)^2/2; n ==== \ 2 %pi k > sec(------- + a) = / n ==== k = 1 %pi %pi n (--- - a) n a n + ----- 2 %pi n 2 2 2 n sin (-----) cot(-----------) csc (-----------) 2 2 2 ------------------------------------------------ 2 required deriving the identity prod(cos(4*%pi*j/n+2*a)/2-b*cos(2*%pi*j/n+a)+3/2,j,1,n) = cos(2*a *n)/2^(2*n-1)-chebyshev_t[n]((sqrt(2-b)-sqrt(b+2))/2) *chebyshev_t[n]((sqrt(b+2)+sqrt(2- b))/2)*cos(a*n)/2^(2*n-3)+ (chebyshev_t[n](sqrt(4-b^2)+1)+chebyshev_t[n](1-sqrt(4-b^2))+ 1)/2^(2*n-1) n 4 %pi j /===\ cos(------- + 2 a) | | n 2 %pi j 3 | | (------------------ - b cos(------- + a) + -) = | | 2 n 2 j = 1 cos(2 a n) sqrt(2 - b) - sqrt(b + 2) ---------- - T (-------------------------) 2 n - 1 n 2 2 sqrt(b + 2) + sqrt(2 - b) 2 n - 3 T (-------------------------) cos(a n)/2 n 2 2 2 T (sqrt(4 - b ) + 1) + T (1 - sqrt(4 - b )) + 1 n n + -----------------------------------------------, 2 n - 1 2 which, at least in my state of dotage, was *ridiculously* hard. (Three Macsyma notebooks, thousands of steps, mostly into intractable algebra, spread over several weeks.) Yet it is clear that this result could be generalized to arbitrary constants replacing, 1/2, 3/2, 2a, and even to more addends with 6 pi j and 8 pi j ... . Much less clear is how, while avoiding tsunamis of algebraic uglitude on the rhs. Anyway, applying this to the sec sum gives, for n > 1, the somewhat homely sum(sin(2*%pi*k/n+a)*(4* chebyshev_t[n](-(sqrt(cos(2*%pi*k/n+a)+2)-sqrt(2-cos(2*%pi*k/n+a)))/2) *chebyshev_t[n]((sqrt(cos(2*%pi*k/n+a)+2) +sqrt(2-cos(2*%pi*k/n+a)))/2)*cos(a*n) -chebyshev_t[n](sqrt(4-cos(2*% pi*k/n+a)^2)+1) -chebyshev_t[n](1-sqrt(4-cos(2*%pi*k/n+a)^2))),k,1,n) = n^2*sin(%pi*n/2)^2*sin(2*a*n)*cot((a-%pi/2)*n/2)*csc((a+%pi/2)*n/2)^2/2 n ==== \ 2 %pi k

sin(------- + a) (4 cos(a n) / n ==== k = 1 2 %pi k 2 %pi k sqrt(cos(------- + a) + 2) - sqrt(2 - cos(------- + a)) n n T (- -------------------------------------------------------) n 2 2 %pi k 2 %pi k sqrt(cos(------- + a) + 2) + sqrt(2 - cos(------- + a)) n n T (-------------------------------------------------------) n 2 2 2 %pi k - T (sqrt(4 - cos (------- + a)) + 1) n n 2 2 %pi k - T (1 - sqrt(4 - cos (------- + a)))) n n %pi %pi (a - ---) n (a + ---) n 2 2 %pi n 2 2 2 n sin (-----) sin(2 a n) cot(-----------) csc (-----------) 2 2 2 = ------------------------------------------------------------. 2

So we get sums/prods over periods of some algebraic functions of trigs as well as rational. --rwg (Reminder: T_n(x) = cos(n acos x) = cosh(n acosh x) = 2 x T_n-1 - T_n-2, T_0(x)=1, T_1(x)=x. T_n(T_m(x)) = T_nm(x), so, e.g. T_(n/2)(x) := T_n(sqrt((x+1)/2)) = sqrt((T_n(x)+1)/2) .) For general n, inf i %pi (n - i) n + i i ==== 2 cos(-----------) binomial(-----, i) x \ 2 2 T = n > ----------------------------------------- n / n + i ==== i = 0 (Technically, not hypergeometric, but the sum of two.) To test that the deranged series is actually nontrivial, n=5, a = pi/6 gives 1653.8323176939d0 + 2934.61886896907d0 - 3275.80964540526d0 4447 - 225.891541257714d0 - ---- = - 25 4 The irrational summands persist with the removal of the sin(a + 2 pi k/n) factor. Too bad it's so ugly. From an even uglier, more general result it might be possible to specialize something nicer. --rwg --------------------------------- You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost.