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%I #143 Dec 17 2023 03:11:53
%S 1,1,1,0,7,2,0,7,3,4,5,3,9,5,9,1,5,6,1,7,5,3,9,7,0,2,4,7,5,1,5,1,7,3,
%T 4,2,4,6,5,3,6,5,5,4,2,2,3,4,3,9,2,2,5,5,5,7,7,1,3,4,8,9,0,1,7,3,9,1,
%U 0,8,6,9,8,2,7,4,8,6,8,4,7,7,6,4,3,8,3,1,7,3,3,6,9,1,1,9,1,3,0,9,3,4
%N Decimal expansion of Pi/(2*sqrt(2)).
%C The value is the length Pi*sqrt(2)/4 of the diagonal in the square with side length Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) = A003881. The area of the circumcircle of this square is Pi*(Pi*sqrt(2)/8)^2 = Pi^3/32 = A153071. - _Eric Desbiaux_, Jan 18 2009
%C This is the value of the Dirichlet L-function of modulus m=8 at argument s=1 for the non-principal character (1,0,1,0,-1,0,-1,0). See arXiv:1008.2547. - _R. J. Mathar_, Mar 22 2011
%C Archimedes's-like scheme: set p(0) = sqrt(2), q(0) = 1; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - _A.H.M. Smeets_, Jul 12 2018
%C The area of a circle circumscribing a unit-area regular octagon. - _Amiram Eldar_, Nov 05 2020
%D J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
%D George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), p. 149.
%D L. B. W. Jolley, Summation of Series, Dover (1961), eq 76 page 16.
%H Harry J. Smith, <a href="/A093954/b093954.txt">Table of n, a(n) for n = 1..20000</a>
%H J. M. Borwein, P. B. Borwein, and K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.
%H R. J. Mathar, <a href="http://arxiv.org/abs/1008.2547">Table of Dirichlet L-series and prime zeta modulo functions for small moduli</a>, arXiv:1008.2547 [math.NT], 2010-2015, table 7 and section 2.2, value of L(m=8,r=4,s=1).
%H Michael Penn, <a href="https://www.youtube.com/watch?v=qoQ2pQi7mvM">Newton's sum</a> (2023), YouTube video.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Bifoliate.html">Bifoliate</a>.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Equals 1/A112628.
%F Equals Integral_{x=0..oo} 1/(x^4+1) dx. - _Jean-François Alcover_, Apr 29 2013
%F From _Peter Bala_, Feb 05 2015: (Start)
%F Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
%F The integer sequences A(n) := 2^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k ) both satisfy the second order recurrence equation u(n) = (12*n^2 + 1)*u(n-1) - 4*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/(2*sqrt(2)) = 1 + 1/(12 - 4*3^3/(49 - 4*2*5^3/(109 - 4*3*7^3/(193 - ... - 4*(n - 1)*(2*n - 1)^3/((12*n^2 + 1) - ... ))))). Cf. A002388 and A019670. (End)
%F From _Peter Bala_, Mar 03 2015: (Start)
%F Pi/(2*sqrt(2)) = Sum_{k >= 0} (-1)^floor(k/2)/(2*k + 1) = limit (n -> infinity) Sum_{k = -n .. n - 1} (-1)^k/(4*k + 1).
%F We conjecture the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + ..., where n is a multiple of 4 and the sequence of unsigned coefficients [1, 3, 57, 2763, ...] is A000281. An example with n = 5000 is given below. (End)
%F From _Peter Bala_, Sep 21 2016: (Start)
%F c = 2 * Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)) = A181048 + A181049. The asymptotic expansion conjectured above follows from the asymptotic expansions given in A181048 and A181049.
%F c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
%F From _Peter Bala_, Nov 24 2016: (Start)
%F Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 2^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 2*m*j). Cf. A003881.
%F In the particular case m = 1 the result has the equivalent form: for n a nonnegative integer, Pi/(2*sqrt(2)) = 2^n*(2*n)!*Sum_{k >= 0} (-1)^(n+k)*(8*k + 4)* 1/Product_{j = -n..n+1} (4*k + 2*j + 1). The case m = 1, n = 1 is considered in the Example section below.
%F Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 4^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 4*m*j). (End)
%F Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - _Peter Bala_, Nov 01 2019
%F Equals Sum_{k>=1} A188510(k)/k = Sum_{k>=1} Kronecker(-8,k)/k = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - 1/15 + ... - _Jianing Song_, Nov 16 2019
%F From _Amiram Eldar_, Jul 16 2020: (Start)
%F Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
%F Equals Integral_{x=0..oo} dx/(x^2 + 2).
%F Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
%F Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - _Bernard Schott_, May 19 2022
%F Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - _Peter Bala_, Jul 22 2022
%F Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - _Terry D. Grant_, Mar 17 2023
%F Equals 1/Product_{p prime} (1 - Kronecker(-8,p)/p), where Kronecker(-8,p) = 0 if p = 2, 1 if p == 1 or 3 (mod 8) or -1 if p == 5 or 7 (mod 8). - _Amiram Eldar_, Dec 17 2023
%e 1.11072073453959156175397...
%e From _Peter Bala_, Mar 03 2015: (Start)
%e Asymptotic expansion at n = 5000.
%e The truncated series Sum_{k = 0..5000 - 1} (-1)^floor(k/2)/(2*k + 1) = 1.110(6)207345(42)591561(18)3970(5238)1.... The bracketed digits show where this decimal expansion differs from that of Pi/(2*sqrt(2)). The numbers 1, -3, 57, -2763 must be added to the bracketed numbers to give the correct decimal expansion to 30 digits: Pi/(2*sqrt(2)) = 1.110(7)207345(39)591561(75)3970 (2475)1.... (End)
%e From _Peter Bala_, Nov 24 2016: (Start)
%e Case m = 1, n = 1:
%e Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
%e We appear to have the following asymptotic expansion for the tails of this series: for N divisible by 4, Sum_{k >= N/2} (-1)^floor(k/2)/((2*k - 1)*(2*k + 1)*(2*k + 3)) ~ 1/N^3 - 14/N^5 + 691/N^7 - 62684/N^9 - ..., where the coefficient sequence [1, 0, -14, 0, 691, 0, -62684, ...] appears to come from the e.g.f. (1/2!)*cosh(x)/cosh(2*x)*sinh(x)^2 = x^2/2! - 14*x^4/4! + 691*x^6/6! - 62684*x^8/8! + .... Cf. A019670.
%e For example, take N = 10^5. The truncated series Sum_{k = 0..N/2 -1} (-1)^(1+floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.27768018363489(8)89043849(11)61878(80026)6163(351171)58.... The bracketed digits show where this decimal expansion differs from that of (1/4)*Pi/(2*sqrt(2)). The numbers -1, 14, -691, 62684 must be added to the bracketed numbers to give the correct decimal expansion: (1/4)*Pi/(2*sqrt(2)) = 0.27768018363489(7) 89043849(25)61878(79335)6163(413855)58... (End)
%p simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) ); # _Peter Bala_, Mar 09 2015
%t RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* _Michael De Vlieger_, Sep 23 2016 and slightly modified by _Robert G. Wilson v_, Jul 23 2018 *)
%o (PARI) default(realprecision, 20080); x=Pi*sqrt(2)/4; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093954.txt", n, " ", d)); \\ _Harry J. Smith_, Jun 17 2009
%Y Cf. A161684 (continued fraction).
%Y Cf. A002388, A019670.
%Y Cf. A000281, A063448, A247719, A193887, A244976, A181048, A181049.
%Y Cf. A003881, A251809, A188510.
%Y Cf. A352324, A248897, A019669.
%K nonn,cons,easy
%O 1,5
%A _Eric W. Weisstein_, Apr 19 2004
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