OFFSET
1,2
COMMENTS
With offset 2, decimal expansion of 5*Pi. - Omar E. Pol, Oct 03 2013
Decimal expansion of the number of radians in a quadrant. - John W. Nicholson, Oct 07 2013
Not the same as A085679. First differing term occurs at 10^-49, as list -49, or 51st counting term (a(-49)= 5 and A085679(-49) = 4). - John W. Nicholson, Oct 07 2013
5*Pi is also the surface area of a sphere whose diameter equals the square root of 5. More generally x*Pi is also the surface area of a sphere whose diameter equals the square root of x. - Omar E. Pol, Dec 22 2013
Pi/2 is also the radius of a sphere whose surface area equals the volume of the circumscribed cube. - Omar E. Pol, Dec 27 2013
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..20000
David H. Bailey and Richard E. Crandall, On the Random Character of Fundamental Constant Expansions, Experimental Mathematics, Vol. 10 (2001), Issue 2, p. 185 (preprint draft).
Richard J. Mathar, Chebyshev approximation of x^m (-log x)^l in the interval 0 <= x <= 1, arXiv:2408.15212 [math.CA], 2024. See p. 2.
Michael Penn, A nice sum from the Harvard MIT math trust, YouTube video, 2022.
L. D. Servi, Nested Square Roots of 2, The American Mathematical Monthly 110:4 (Apr. 2003), pp. 326-330.
Johan Wästlund, An Elementary Proof of the Wallis Product Formula for pi, The American Mathematical Monthly 114:10 (Dec. 2007), pp. 914-917.
Eric W. Weisstein and Jonathan Sondow, Wallis Formula, MathWorld.
Wikipedia, Viète's formula
FORMULA
Pi/2 = log(i)/i, where i = sqrt(-1). - Eric Desbiaux, Jun 27 2009
Pi/2 = Product_{n>=1} (n/(n+1))^((-1)^n) = 2 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 * 10/9 * ... (Wallis formula). - William Keith and Alonso del Arte, Jun 24 2012
Equals Sum_{k>1} 2^k/binomial(2*k,k). - Bruno Berselli, Sep 11 2015
The previous result is the particular case n = 1 of the more general identity: Pi/2 = 4^(n-1) * n!/(2*n)! * Sum_{k >= 2} 2^(k+1)*(k + n - 1)!*(k + 2*n - 2)!/(2*k + 2*n - 2)! valid for n = 0,1,2,... . - Peter Bala, Oct 26 2016
Pi/2 = Product_{n>=1} (4*n^2)/(4*n^2-1). - Fred Daniel Kline, Oct 29 2016
Pi/2 = lim_{n->oo} F(2^(n+3))/2, with one half of the area of a regular 2^(n+3)-gon, for n >= 0, inscribed in the unit circle, written as iterated square roots of 2 as F(2^(n+3))/2 = 2^n*sqrt(2 + sq2(n)), with sq2(n) = sqrt(2 + sq2(n-1)), n >= 1, with input sq2(0) = 0 (2 appears n times in sq2(n)). Viète's infinite product formula works with the partial product F(2^(n+2))/2 = Product_{j=1..n} (2/sq2(j)), n >= 1, which corresponds to the above given formula. - Wolfdieter Lang, Jul 06 2018
Pi/2 = Integral_{x = 0..oo} sin(x)^2/x^2 dx = 1/2 + Sum_{n >= 1} sin(n)^2/n^2, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
From Amiram Eldar, Aug 15 2020: (Start)
Equals Sum_{k>=0} k!/(2*k + 1)!!.
Equals Sum_{k>=0} (-1)^k/(k + 1/2).
Equals Integral_{x=0..oo} 1/(x^2 + 1) dx.
Equals Integral_{x=0..oo} sin(x)/x dx.
Equals Integral_{x=0..oo} exp(x/2)/(exp(x) + 1) dx.
Equals Product_{p prime > 2} p/(p + (-1)^((p-1)/2)). (End)
Pi/2 = Integral_{x = 0..oo} 1/(1 - x^2 + x^4) dx = (1 + 2/3 + 1/5) - (1/7 + 2/9 + 1/11) + (1/13 + 2/15 + 1/17) - .... - Peter Bala, Jul 22 2022
Equals arcsin(9/10) + sqrt(19)*Sum_{k >= 1} A106854(k-1)/(k*10^k) (see Bailey and Crandall, 2001). - Paolo Xausa, Jul 15 2024
Equals 2F1(1/2,1/2 ; 3/2; 1). - R. J. Mathar, Aug 20 2024
EXAMPLE
Pi/2 = 1.570796326794896619231321691639751442098584699...
5*Pi = 15.70796326794896619231321691639751442098584699...
MAPLE
Digits:=100: evalf(Pi/2); # Wesley Ivan Hurt, Oct 26 2016
MATHEMATICA
RealDigits[N[Pi/2, 200]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
PROG
(PARI) default(realprecision, 20080); x=Pi/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b019669.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
STATUS
approved