I gave [...] Amazingly worse: n ==== \ pi j pi j pi j
cot(---- + f) cot(---- + g) cot(---- + h) = / n n n ==== j = 1 2 sin(h - g) cot(f n) sin(h n - g n) n (--------------------------------------- + ...)+ <hairy mess> sin(g - f) sin(h - f) sin(g n) sin(h n)
from a harrowing derivation via 16-page intermediate results, including a quadrivariate generating fcn to circumvent a *nasty* cubic. Here's the stringout if you want to test or simplify it: 'SUM(COT(%PI*J/N+F)*COT(%PI*J/N+G)*COT(%PI*J/N+H),J,1,N) = N^2*(SIN(H-G)*COT(F*N)*SIN(... I found a neat way to drastically simplify it: Lim(n->oo, Sum(cotcotcot)/n) = Integral(cotcotcot,0,1) is finite (f#g#h#f), so the coeff of n^2 is zero! When the smoke clears, sum(cot(%pi*j/n+f)*cot(%pi*j/n+g)*cot(%pi*j/n+h),j,1,n) = -n*((sin(2*g-2*f)*cot(h*n)+sin(2*f-2*h)*cot(g*n)+sin(2*h-2*g)*cot(f*n)) /(2*sin(g-f)*sin(f-h)*sin(h-g))+cot(h*n)+cot(g*n)+cot(f*n))); n ==== \ %pi j %pi j %pi j
cot(----- + f) cot(----- + g) cot(----- + h) = / n n n ==== j = 1 - n ((sin(2 g - 2 f) cot(h n) + sin(2 f - 2 h) cot(g n) + sin(2 h - 2 g) cot(f n))/(2 sin(g - f) sin(f - h) sin(h - g)) + cot(h n) + cot(g n) + cot(f n)).
Now what *is* that integral? This limit doesn't exist! Empirically, it's 0. Anyway, the sum answer is verifiably equal to the mess. --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com