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A085565
Decimal expansion of lemniscate constant A.
23
1, 3, 1, 1, 0, 2, 8, 7, 7, 7, 1, 4, 6, 0, 5, 9, 9, 0, 5, 2, 3, 2, 4, 1, 9, 7, 9, 4, 9, 4, 5, 5, 5, 9, 7, 0, 6, 8, 4, 1, 3, 7, 7, 4, 7, 5, 7, 1, 5, 8, 1, 1, 5, 8, 1, 4, 0, 8, 4, 1, 0, 8, 5, 1, 9, 0, 0, 3, 9, 5, 2, 9, 3, 5, 3, 5, 2, 0, 7, 1, 2, 5, 1, 1, 5, 1, 4, 7, 7, 6, 6, 4, 8, 0, 7, 1, 4, 5, 4
OFFSET
1,2
COMMENTS
This number is transcendental by a result of Schneider on elliptic integrals. - Benoit Cloitre, Jan 08 2006
The two lemniscate constants are A = Integral_{x = 0..1} 1/sqrt(1 - x^4) dx and B = Integral_{x = 0..1} x^2/sqrt(1 - x^4) dx. See A076390. - Peter Bala, Oct 25 2019
Also the ratio of generating curve length to diameter of a "Mylar balloon" (see Paulsen). - Jeremy Tan, May 05 2021
REFERENCES
B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
Th. Schneider, Transzendenzuntersuchungen periodischer Funktionen (1934).
Th. Schneider, Arithmetische Untersuchungen elliptischer Integrale (1937).
LINKS
John Maxwell Campbell, WZ proofs for lemniscate-like constant evaluations, Integers 21 (2021), Article A107, 15.
S. Khrushchev, Orthogonal polynomials and continued fractions from Euler’s point of view, Encyclopedia of Mathematics and its Applications 122.
W. H. Paulsen, What Is the Shape of a Mylar Balloon?, Amer. Math. Monthly 101 (10), (Dec. 1994), pp. 953-958.
J. Todd, The lemniscate constants, Comm. ACM, 18 (1975), 14-19; 18 (1975), 462.
J. Todd, The lemniscate constants, in Pi: A Source Book, pp. 412-417.
Eric Weisstein's World of Mathematics, Lemniscate Constant
Wikipedia, Mylar balloon
FORMULA
Equals (1/4)*(2*Pi)^(-1/2)*GAMMA(1/4)^2.
Equals Integral_{x>=1}dx/sqrt(4x^3-4x). - Benoit Cloitre, Jan 08 2006
Equals Product_(k>=0, [(4k+3)(4k+4)] / [(4k+5)(4k+2)] ) (Gauss). - Ralf Stephan, Mar 04 2008 [corrected by Vaclav Kotesovec, May 01 2020]
Equals Pi/sqrt(8)/agm(1,sqrt(1/2)).
Equals Pi/sqrt(8)*hypergeom([1/2,1/2],[1],1/2).
Product_{m>=1} ((2*m)/(2*m+1))^(-1)^m. - Jean-François Alcover, Sep 02 2014, after Steven Finch
From Peter Bala, Mar 09 2015: (Start)
Equals Integral_{x = 0..1} 1/sqrt(1 - x^4) dx.
Continued fraction representations: 2/(1 + 1*3/(2 + 5*7/(2 + 9*11/(2 + ... )))) due to Euler - see Khrushchev, p. 179.
Also equals 1 + 1/(2 + 2*3/(2 + 4*5/(2 + 6*7/(2 + ... )))). (End)
From Peter Bala, Oct 25 2019: (Start)
Equals 1 + 1/5 + (1*3)/(5*9) + (1*3*5)/(5*9*13) + ... = hypergeom([1/2,1],[5/4],1/2) by Gauss's second summation theorem.
Equivalently, define a sequence of rational numbers r(n) recursively by r(n) = (2*n - 3)/(4*n - 3)*r(n-1) with r(1) = 1. Then the constant equals Sum_{n >= 1} r(n) = 1 + 1/5 + 1/15 + 1/39 + 7/663 + 1/221 + 11/5525 + 11/12325 + 1/2465 + .... The partial sum of the series to 100 terms gives 32 correct decimal digits for the constant.
Equals (1*3)/(1*5) + (1*3*5)/(1*5*9) + (1*3*5*7)/(1*5*9*13) + ... = (3/5) * hypergeom([5/2,1],[9/4],1/2). (End)
Equals (3/2)*A225119. - Peter Bala, Oct 27 2019
Equals Integral_{x=0..Pi/2} 1/sqrt(1 + cos(x)^2) dx = Integral_{x=0..Pi/2} 1/sqrt(1 + sin(x)^2) dx. - Amiram Eldar, Aug 09 2020
From Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 4*k + 1 for k >= 0.
For example, taking k = 0 and k = 1 yields
A = 2/(1 + (1*3)/(2 + (5*7)/(2 + (9*11)/(2 + (13*15)/(2 + ... + (4*n + 1)*(4*n + 3)/(2 + ... )))))) and
A = (1/4)*(5 + (1*3)/(10 + (5*7)/(10 + (9*11)/(10 + (13*15)/(10 + ... + (4*n + 1)*(4*n + 3)/(10 + ... )))))). (End)
EXAMPLE
1.3110287771460599052324197949455597068413774757158115814084108519...
MATHEMATICA
RealDigits[ Gamma[1/4]^2/(4*Sqrt[2*Pi]), 10, 99][[1]]
(* or *)
RealDigits[ EllipticK[-1], 10, 99][[1]] (* Jean-François Alcover, Mar 07 2013, updated Jul 30 2016 *)
PROG
(PARI) gamma(1/4)^2/4/sqrt(2*Pi)
(PARI) K(x)=Pi/2/agm(1, sqrt(1-x))
K(-1) \\ Charles R Greathouse IV, Aug 02 2018
(Magma) C<i> := ComplexField(); [Gamma(1/4)^2/(4*Sqrt(2*Pi(C)))]; // G. C. Greubel, Nov 05 2017
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
N. J. A. Sloane, Jul 06 2003
STATUS
approved