|
%I #96 Nov 20 2023 08:18:12
%S 4,4,4,2,8,8,2,9,3,8,1,5,8,3,6,6,2,4,7,0,1,5,8,8,0,9,9,0,0,6,0,6,9,3,
%T 6,9,8,6,1,4,6,2,1,6,8,9,3,7,5,6,9,0,2,2,3,0,8,5,3,9,5,6,0,6,9,5,6,4,
%U 3,4,7,9,3,0,9,9,4,7,3,9,1,0,5,7,5,3,2,6,9,3,4,7,6,4,7,6,5,2,3
%N Decimal expansion of Pi * sqrt(2).
%C Hypotenuse of the right triangle with legs Pi and Pi. - _Zak Seidov_, May 04 2005
%C Circumference of the circumcircle of the unit square. - _Jonathan Sondow_, Nov 23 2017
%C Half-perimeter of the closed curve with implicit Cartesian equation x^2 + y^2 = abs(x) + abs(y). - _Stefano Spezia_, Oct 20 2020
%H Harry J. Smith, <a href="/A063448/b063448.txt">Table of n, a(n) for n = 1..20000</a>
%H Peter Bala, <a href="/A063448/a063448.txt">Series for sqrt(2)*Pi</a>
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/GaussMultiplicationFormula.html">Gauss Multiplication Formula</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Equals Gamma(1/4)*Gamma(3/4). - _Jean-François Alcover_, Nov 24 2014
%F From _Amiram Eldar_, Aug 06 2020: (Start)
%F Equals Integral_{x=0..oo} log(1 + 1/x^4) dx.
%F Equals Integral_{x=0..oo} log(1 + 2/x^2) dx.
%F Equals Integral_{x=-oo..oo} exp(x/4)/(exp(x) + 1) dx.
%F Equals Integral_{x=0..2*Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..2*Pi} 1/(sin(x)^2 + 1) dx. (End)
%F From _Andrea Pinos_, Jul 03 2023: (Start)
%F Equals (Product_{k=1..4} Gamma(k/8)*Gamma(1 - k/8))^(1/4).
%F General result: 2*Pi/(4*y)^(1/(2*y)) = (Product_{k=1..y} Gamma(k/(2*y))*Gamma(1 - k/(2*y)) )^(1/y). (End)
%F From _Peter Bala_, Oct 22 2023: (Start)
%F sqrt(2)*Pi = 4 + 8*Sum_{n >= 0} (-1)^n/(16*n^2 + 32*n + 15). See A141759.
%F In the following the Eisenstein summation convention is assumed: that is,
%F Sum_{n = -oo..oo} means Limit_{N -> oo} Sum_{n = -N..N}:
%F sqrt(2)*Pi = 4*Sum_{n = -oo..oo} (-1)^n/(4*n + 1).
%F More generally, it appears that for k >= 0, k not of the form 4*m + 1,
%F sqrt(2)*Pi = -sign(cos(Pi*(k - 3)/4)) * 4*(2^floor(k/2))*k! * Sum_{n = -oo..oo} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 2*k + 1)) (verified up to k = 50).
%F sqrt(2)*Pi = (2^4)*Sum_{n >= 0} (-1)^n * (2*n + 1)/((4*n + 1)*(4*n + 3)) = 512/105 - (2^6)*4!*Sum_{n >= 0} (-1)^n * (2*n + 3)/((4*n + 1)*(4*n + 3)*...*(4*n + 11)).
%F sqrt(2)*Pi = 4 + (2^3)*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*(4*n + 5)) = 1408/315 - (2^5)*5!*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*...*(4*n + 13)).
%F sqrt(2)*Pi = 16/3 - (2^4)*3!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*(4*n + 5)*(4*n + 7)) = 14848/3465 + (2^6)*7!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 15)). (End)
%F From _Peter Bala_, Nov 19 2023: (Start)
%F sqrt(2)*Pi = 512*Sum_{k >= 1} (-1)^(k+1) * k^2/((16*k^2 - 1)*(16*k^2 - 9)).
%F This is the case n = 1 of the more general result: for n >= 1,
%F sqrt(2)*Pi = (-1)^(n+1) * 2^(n+7) * (2*n)!/(2*n - 1) * Sum_{k >= 1} (-1)^(k+1) * k^2/( Product_{i = 0..n} (16*k^2 - (2*i+1)^2) ). Cf. A334422. (End)
%e 4.4428829381583662470158809900606936986146216893756902230853...
%t RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* _Vladimir Joseph Stephan Orlovsky_, Mar 21 2011*)
%o (PARI) \p 400; Pi * sqrt(2)
%o (PARI) default(realprecision, 20080); x=Pi*sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b063448.txt", n, " ", d)) \\ _Harry J. Smith_, Aug 21 2009
%o (Python) # Use some guard digits when computing.
%o # BBP formula (1/8) P(1, 64, 12, (32, 0, 8, 0, 8, 0, -4, 0, -1, 0, -1, 0))
%o from decimal import Decimal as dec, getcontext
%o def BBPpisqrt2(n: int) -> dec:
%o getcontext().prec = n
%o s = dec(0); f = dec(1); g = dec(64)
%o for k in range(int(n * 0.5536546824812272) + 1):
%o twk = dec(12 * k)
%o s += f * ( dec(32) / (twk + 1) + dec(8) / (twk + 3)
%o + dec(8) / (twk + 5) - dec(4) / (twk + 7)
%o - dec(1) / (twk + 9) - dec(1) / (twk + 11))
%o f /= g
%o return s / dec(8)
%o print(BBPpisqrt2(200)) # _Peter Luschny_, Nov 03 2023
%Y Cf. A063447 (continued fraction), A093954, A153799, A193887, A244976, A247719.
%K cons,nonn
%O 1,1
%A _Jason Earls_, Jul 24 2001
%E Edited by _N. J. A. Sloane_, May 05 2007
%E Corrected by Neven Juric, Apr 24 2008
|
