I closed 29nov with: --------------- Thought for today (simplfied by Julian) sum(sech(((%pi * n)/(2 * sqrt(3)))),n, - inf,inf) = ((3 * 2^(1/6) * Gamma((1/3))^3)/(2 * 3^(1/4) * (sqrt(3) + 1)^(5/2) * sqrt(5 * sqrt(3) - 4 * sqrt(2) - 3) * (%pi)^2)) inf ==== \ %pi n
sech(---------) =
/ 2 sqrt(3) ==== n = - inf 1/6 3 1 3 2 gamma (-) 3 ------------------------------------------------------------ 1/4 5/2 2 2 3 (sqrt(3) + 1) sqrt(5 sqrt(3) - 4 sqrt(2) - 3) %pi ------------------- This was subseqently *radically* simplified by Corey's denester: Sum[Sech[(n*Pi)/(2*Sqrt[3])], {n, -Infinity, Infinity}] == (3^(3/4)*(1 + 2*Sqrt[2] + Sqrt[3])*Gamma[1/3]^3)/(4*2^(5/6)*Pi^2) which also completely poofed the 8th roots from my tabulated eta(e^-(2 pi rt 19), 1/3 8 1/8 (d148) ((3 sqrt(57) + 1) - -------------------) = 1/3 (3 sqrt(57) + 1) 1/3 3/8 4 (3 sqrt(57) + 1) sqrt(57) - 5 2 (--------------------- + ------------------- - 2) sqrt(57) - 5 1/3 (3 sqrt(57) + 1) ------------------------------------------------------ 6 (as with the 19th singular value) and eta(e^-(2 pi rt 11) 1 ------------------------------------------------------------------ = 1/3 1/3 1/3 1/3 1/8 (2 (21 sqrt(33) + 283) + 2 (283 - 21 sqrt(33)) + 8) 1/3 1/3 2/3 1/4 7/8 2 (21 sqrt(33) + 283) 2 (sqrt(33) - 1) 2 3 (--------------------------- - ---------------------- + 1) sqrt(33) - 1 1/3 (21 sqrt(33) + 283) -------------------------------------------------------------------- 9 (Gammastorms omitted for brevity.) This leads one to ask why we tabulate singular values instead of etas. I can get the former from the latter, but vice versa? 23Feb:Victor (Miller)> Hi Simon, I would bet that your values are related to this: http://www.mathstat.carleton.ca/~williams/papers/pdf/220.pdf (rwg)>YOW! So evaluating etas (and thetas and lambdas and ...) is reduced to evaluating L series! Presumably, a solved problem. I had no idea. YOWed too soon. This (truly impressive) paper solves only a fraction of the problem: It only does integers, not rationals, and only 0 or -1 mod 4. And it only produces magnitudes, not phases. I thought it might be possible to get the phase directly from the infinite product, but how the heck do you get j 1/2 (1/2 + j) Pi Sum[(-1) ArcCot[E ], {j, 0, Infinity}] == 1/4 1 4 2 - (Pi - ArcTan[---------------]) ? 8 3 (3 + Sqrt[2]) Eta triples permit extension to rationals, but is there a science to finding them? I just mine the powerseries for empirical relations. 3Mar: Integrating a (simpler but harder to justify) variation on the W&W Mittag-Leffler derivation, 2 r 2 y y -2 Pi Sum[(-1) r (ArcTan[----] + ArcTanh[----]) Csch[Pi r], Pi r Pi r 2 {r, 1, Infinity}] == Integrate[x Csc[x] Csch[x], {x, 0, y}] Unintegrated: inf ==== r 1 2 \ r (- 1) csch(%pi r) csc(x) csch(x) = -- - 4 %pi x > -------------------- 2 / 4 4 4 x ==== %pi r - x r = 1 which probably also comes out of convolving the separate csc and csch expansions. Expanding at x=0 gives the infinite set of identities r 3 (-1) Csch[Pi r] Pi {Sum[----------------, {r, 1, Infinity}] == -(---), 3 360 r r 7 (-1) Csch[Pi r] 13 Pi Sum[----------------, {r, 1, Infinity}] == -(------), 7 453600 r r 11 (-1) Csch[Pi r] 4009 Pi Sum[----------------, {r, 1, Infinity}] == -(-----------)} 11 13621608000 r ... These are less like Simon's results than I thought--he has r^3 instead of r^-3. I was hoping this constant would be in ISC: r 1 Pi Sum[(-1) r Csch[Pi r] Log[1 - --], {r, 2, Infinity}] == 4 r 3 1 4 t Csch[Pi] Integrate[-(-) - ------------- + t Csc[t] Csch[t], {t, 0, Pi}] t 4 3 t Pi (1 - ---) 4 Pi There's lot's more fun to be had here. --rwg Also, it should be possible to let sides->oo in that spherical triangle area formula and get a path integral for the area bounded by a piecewise smooth curve.