Gaussian Triangle Picking

Author

Eric W. Weisstein
September 6, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/ComputationalGeometry/GaussianTrianglePicking.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/GaussianTrianglePicking.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Code

GaussianTriangle[n_]:=((2 (3^(1-(n/2))) Gamma[n] Hypergeometric2F1[(n/2),n,1+(n/2),-(1/3)])/(n ((Gamma[(n/2)])^2)))

Formula

expr=3Gamma[n]/Gamma[n/2]^2 Integrate[x^((n-2)/2)/(1+x)^n,{x,0,1/3},Assumptions->{Re[n]>0}]

Simplify[Table[expr,{n,7}]//FunctionExpand]

FullSimplify[expr,n∈Integers&&n>1]

Simplify[Table[((3 (I^(-n)) Beta[-(1/3),(n/2),1-n] Gamma[n])/((Gamma[(n/2)])^2)),{n,7}]//FunctionExpand]

expr2=3((Gamma[n])/((2^(n-1))((Gamma[n/2])^2) ))Integrate[((Sin[θ])^(n-1)),{θ,0,(π/3)},Assumptions->n>1]

FullSimplify[expr2,n∈Integers&&n>1]

Simplify[Table[expr2,{n,7}]//FunctionExpand]

Sum

Table[Sum[Binomial[2k-1,j]1/4^j(3/4)^(2k-1-j),{j,k,2k-1}],{k,5}]

3 (∑_(j=k)^(2 k-1))(1/(4^j))((((3/4)))^(-j+2 k-1)) (Binomial[2 k-1, j])

V5

FullSimplify[3Sum[Binomial[2k-1,j]1/4^j(3/4)^(2k-1-j),{j,k,2k-1}],k>0&&k∈Integers]

V7

Sum[Binomial[2k-1,j]1/4^j(3/4)^(2k-1-j),{j,k,2k-1}]

FullSimplify[%,k>0&&k∈Integers]

Table[(3^(-1+k)) (4^(1-2 k)) Binomial[-1+2 k,k] Hypergeometric2F1[1,1-k,1+k,-(1/3)],{k,5}]

Enumeration

FullSimplify[FunctionExpand[Table[GaussianTriangle[n],{n,2,8}]]]

FullSimplify[FunctionExpand[Table[GaussianTriangle[n],{n,2,8,2}]]]//InputForm

Through[{Numerator,Denominator}[%]]

FullSimplify[Pi/Sqrt[3](1-FunctionExpand[Table[GaussianTriangle[n],{n,3,10,2}]])]//InputForm

Through[{Numerator,Denominator}[%]]