OFFSET
0,1
COMMENTS
Considering the isosceles right triangle x>0, y>0, x+y=1; there exists a unique least [electrostatic] capacity point (0.301..., 0.301...).
It can be observed that this constant is close to but different from the analog for the center of the incircle, which is the inradius 1-1/sqrt(2) = 0.29289...
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..10000
Steven R. Finch, Least Capacity Point of Triangles, arXiv:1407.4105 [math.HO] 2014, p. 6.
FORMULA
2*sqrt(Pi)*re(F(arcsin(sqrt(1 + sqrt(3))), 1/2))/gamma(1/4)^2, where F is the elliptic integral of the first kind.
EXAMPLE
0.30112161084132208155382545585018896837090055390464695504633924...
MATHEMATICA
t0 = 2*Sqrt[Pi]*Re[EllipticF[ArcSin[Sqrt[1 + Sqrt[3]]], 1/2]]/Gamma[1/4]^2;
RealDigits[t0, 10, 100][[1]]
(* second program: *)
digits = 100;
s[z_] := WeierstrassSigma[z, {Gamma[1/4]^8/(256 Pi^2), 0}];
h[x_] := Abs[(s[(2+I) - (1+I)x] s[1-(1-I)x] s[2I*x + (1-I)] s[(2+2I)x] s[1 + (1-I)x] s[(1+I)x-I] s[2x+I-1])/(s[1+I] s[1-(1+I)x] s[(2-I) - (1-I)x] s[2I*x] s[2x] s[(1-I)x + I] s[1+(1+I)x] s[(1+I)(2x-1)])]
t0 = x /. FindMinimum[h[x], {x, 1/3}, WorkingPrecision -> 3 digits][[2]];
RealDigits[t0, 10, digits][[1]]
RealDigits[2 Sqrt[Pi] EllipticF[ArcTan[Sqrt[1 + Sqrt[3]]], 1/2]/Gamma[1/4]^2, 10, 100][[1]] (* Jan Mangaldan, Jan 04 2017 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Jean-François Alcover, Jul 23 2016
STATUS
approved