A275336 Decimal expansion of the location of the least capacity point of a unit isosceles right triangle.

3, 0, 1, 1, 2, 1, 6, 1, 0, 8, 4, 1, 3, 2, 2, 0, 8, 1, 5, 5, 3, 8, 2, 5, 4, 5, 5, 8, 5, 0, 1, 8, 8, 9, 6, 8, 3, 7, 0, 9, 0, 0, 5, 5, 3, 9, 0, 4, 6, 4, 6, 9, 5, 5, 0, 4, 6, 3, 3, 9, 2, 4, 1, 1, 4, 9, 4, 8, 1, 5, 1, 5, 5, 9, 2, 5, 6, 2, 0, 2, 3, 2, 9, 0, 8, 6, 8, 6, 9, 2, 3, 4, 3, 3, 6, 6, 8, 3, 6, 3

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

Sequence in context: A334739 A131802 A069023 * A373949 A091614 A350829

Adjacent sequences: A275333 A275334 A275335 * A275337 A275338 A275339

Keyword

nonn,cons

Author

Jean-François Alcover, Jul 23 2016

Status

approved