Seems about "twice as hard". (1/3)! == (2*2^(47/72)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1/2^(3/16) + 1/(2^(7/16)*Sqrt[1 + Sqrt[3]]) + Sqrt[1 + Sqrt[3]]/(3 - 4*Sqrt[2] + 5*Sqrt[3])^(1/8)))^(2/3))/(3*3^(1/4)) Again just the terms through first order: 2 (1 + ----------------) Pi (16 Pi)/Sqrt[3] 47/72 E 2/3 2 2 (----------------------------------------------------------------) 1 1 Sqrt[1 + Sqrt[3]] ----- + ----------------------- + ------------------------------ 3/16 7/16 1/8 1 2 2 Sqrt[1 + Sqrt[3]] (3 - 4 Sqrt[2] + 5 Sqrt[3]) {(-)!, ------------------------------------------------------------------------------} 3 1/4 3 3 N[%, 51] {0.892979511569249211218564313658225881376229792652434, 0.892979511569249211218564313658225881376229792652429} --rwg
For many of us, idiot GMail badly linewrapped the OutputForm display, but it's actually not hard to repair. Convert the display to a fixed pitch font (if it isn't already). Then simply go to the beginning of each line which shouldn't be left-adjusted, and rub out the preceding linebreak. Do the obvious ones first. --rwg On Wed, Dec 28, 2011 at 1:46 AM, Bill Gosper <billgosper@gmail.com> wrote:
Seems about "twice as hard". (1/3)! == (2*2^(47/72)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1/2^(3/16) + 1/(2^(7/16)*Sqrt[1 + Sqrt[3]]) + Sqrt[1 + Sqrt[3]]/(3 - 4*Sqrt[2] + 5*Sqrt[3])^(1/8)))^(2/3))/(3*3^(1/4))
Again just the terms through first order: 2 (1 + ----------------) Pi (16 Pi)/Sqrt[3] 47/72 E 2/3 2 2 (----------------------------------------------------------------) 1 1 Sqrt[1 + Sqrt[3]] ----- + ----------------------- + ------------------------------ 3/16 7/16 1/8 1 2 2 Sqrt[1 + Sqrt[3]] (3 - 4 Sqrt[2] + 5 Sqrt[3]) {(-)!, ------------------------------------------------------------------------------} 3 1/4 3 3
N[%, 51]
{0.892979511569249211218564313658225881376229792652434, 0.892979511569249211218564313658225881376229792652429} --rwg
That peculiar trinomial factors into binomials: Gamma[1/3] == (2*2^(7/9)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1 + 1/(2^(1/4)*Sqrt[1 + Sqrt[3]]) + (2^(7/16)*((-1 + Sqrt[2])/(-Sqrt[2] + Sqrt[3]))^(1/4))/(-1 + Sqrt[3])^(1/8)))^ (2/3))/3^(1/4) -((16 Pi)/Sqrt[3]) 7/9 Pi EllipticTheta[3, 0, E ] 2/3 2 2 (----------------------------------------------------------) 7/16 -1 + Sqrt[2] 1/4 2 (------------------) 1 -Sqrt[2] + Sqrt[3] 1 + ---------------------- + ----------------------------- 1/4 1/8 1 2 Sqrt[1 + Sqrt[3]] (-1 + Sqrt[3]) Gamma[-] == ---------------------------------------------------------------------- 3 1/4 3 Macsyma: GAMMA(1/3) = 4^(8/9)*%PI^(2/3)*THETA[3](0,%E^-(16*%PI/SQRT(3)))^(2/3)/(3^(1/4)*(2^(7/16)*(SQRT(2)-1)^(1/4)/((SQRT(3)-1)^(1/8)*(SQRT(3)-SQRT(2))^(1/4))+1/(2^(1/4)*SQRT(SQRT(3)+1))+1)^(2/3)) --rwg On Wed, Dec 28, 2011 at 1:46 AM, Bill Gosper <billgosper@gmail.com> wrote:
Seems about "twice as hard". (1/3)! == (2*2^(47/72)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1/2^(3/16) + 1/(2^(7/16)*Sqrt[1 + Sqrt[3]]) + Sqrt[1 + Sqrt[3]]/(3 - 4*Sqrt[2] + 5*Sqrt[3])^(1/8)))^(2/3))/(3*3^(1/4))
Again just the terms through first order: 2 (1 + ----------------) Pi (16 Pi)/Sqrt[3] 47/72 E 2/3 2 2 (----------------------------------------------------------------) 1 1 Sqrt[1 + Sqrt[3]] ----- + ----------------------- + ------------------------------ 3/16 7/16 1/8 1 2 2 Sqrt[1 + Sqrt[3]] (3 - 4 Sqrt[2] + 5 Sqrt[3]) {(-)!, ------------------------------------------------------------------------------} 3 1/4 3 3
N[%, 51]
{0.892979511569249211218564313658225881376229792652434, 0.892979511569249211218564313658225881376229792652429} --rwg
participants (1)
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Bill Gosper