OFFSET
0,1
COMMENTS
The constant represents the area of a circular segment bounded by an arc of Pi/2 radians (the right angle) of a unit circle and by a chord of the length of sqrt(2). Four such segments result when a square with the side length of sqrt(2) is circumscribed by a unit circle. The area of each segment is:
A = (R^2 / 2) * (theta - sin(theta))
A = (1^2 / 2) * (Pi/2 - sin(Pi/2))
A = (1 / 2) * (Pi/2 - 1)
A = (Pi - 2) / 4 = 0.28539816...
where Pi = 3.14159265... (A000796) is the area bounded by the unit circle, and 2 is the area of the inscribed square.
Apart from the first digit this is the same as Pi/4 = 0.78539816... (A003881), the area of a circular sector bounded by the arc of Pi/2 = 1.57079632... (A019669) radians of the unit circle and by two radii of unit length, and 1/2 = 0.5 (A020761) is one-quarter of the area of the inscribed square.
The constant is close to 2/7 = 0.28571428... (2 * A020806) and Pi/11 = 0.28559933... (A019678). The equation (x - 2)/4 = x/11 has a solution x = 22/7 = 3.14285714... (A068028), which is an approximation of Pi.
The best rational approximation of the constant using small positive integers (less than 1000) is 129/452 = 0.28539823..., the next best approximation is 4771/16717 = 0.28539809...
The reciprocal of the constant is:
1/A = 4 / (Pi - 2) = 3.50387678... (A309091).
The sagitta (height) of the circular segment is:
h = R * (1 - cos(theta/2))
h = 1 * (1 - cos(Pi/4))
h = 1 - sqrt(2) / 2
h = 1 - 1 / sqrt(2) = 0.29289321... (A268682).
FORMULA
Equals Integral_{x=-sqrt(2)/2..sqrt(2)/2} Integral_{y=sqrt(2)/2..sqrt(1-x^2)} dy dx.
Equals Sum_{k>=1} (-1)^(k + 1)/(4*k^2 - 1). - Amiram Eldar, Jun 08 2021
Continued fraction: 1/(3 + 3/(4 + 15/(4 + 35/(4 + ... + (4*n^2 - 1)/(4 + ...). - Peter Bala, Feb 22 2024
EXAMPLE
0.2853981633974483...
MATHEMATICA
RealDigits[Pi/4 - 1/2, 10, 100][[1]] (* Amiram Eldar, Jun 08 2021 *)
PROG
(PARI) (Pi - 2) / 4
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Michal Paulovic, Apr 01 2021
STATUS
approved