Interesting! I encountered the LerchPhi function recently, too. Take the standard Gregory series for Pi/4 = 1 - 1/3 + 1/5 ... and introduce powers of Sinc into each term: Define f[k_, x_] := Sum[ Sinc[(2n-1)x]^k * (-1)^(n-1)/(2n-1), {n, 1, Infinity}] Then f[0, x] = Pi/4 for all x. MMA 7 expresses f[1, x], f[2, x] etc., in terms of the Lerch function. I can prove that f[1, x] = Sum[ Sinc[(2n-1)x] * (-1)^(n-1)/(2n-1) ] equals Pi/4 for x in [-Pi/2 , Pi/2]. This means that, for those x, we can multiply each term of the Gregory series by Sinc[(2n-1)x] without changing the sum. I conjecture that for k = 1, 2, 3, ..., f[k, x] equals Pi/4 for x in [-Pi/(2k) , Pi/(2k)]. (Was Lerch in the Addams family, or was it the Munsters?) Bob Baillie -------------------- rwg@sdf.lonestar.org wrote:
Mma 7.0 just startled me by turning the Fourier series for the line
Pi -- (4 Pi - 3 t + I Sqrt[3] t), 0 <= t <= 2 Pi, 3 into -I t 2 LerchPhi[E , 2, -] 3 (I t)/3 I t 1 L(t):= --------------------- + E LerchPhi[E , 2, -]. (2 I t)/3 3 E
But L(t + 2 Pi) = E^(2 Pi/3) L(t). I.e., translating by 2 Pi *rotates* by 120 degrees! Eh? Sure enough, plotting L(t), 0 < t < 6 Pi, draws a perfect equilateral triangle. There seems to be such a relation among n-1 Lerchs for each regular n-gon. Some simple consequence of n-secting the series? Psychoanalytic continuation. --rwg PS, Veit Elser's difference map algorithm, http://en.wikipedia.org/wiki/Difference_map_algorithm , has become only the second entity to solve the 82% Arnold Dozenegger disk packing puzzle completely unaided. (Not counting Emma Cohen, who got massive clues from Emma Cohen.) Also, it's clear that Rod Stephenson's clustering algorithm will do it, probably in ~1 hr --way longer than Veit's, who clearly has something dangerous. ------- Merriam-Webster's Unabridged: Main Entry: prince albert Usage: usually capitalized P&A Etymology: after Prince Albert Edward (later Edward VII king of England) [...] 2 : a man's house slipper with a low counter and goring on each side
ALGORISMIC MICROGLIAS
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