A019675 Decimal expansion of Pi/8.

3, 9, 2, 6, 9, 9, 0, 8, 1, 6, 9, 8, 7, 2, 4, 1, 5, 4, 8, 0, 7, 8, 3, 0, 4, 2, 2, 9, 0, 9, 9, 3, 7, 8, 6, 0, 5, 2, 4, 6, 4, 6, 1, 7, 4, 9, 2, 1, 8, 8, 8, 2, 2, 7, 6, 2, 1, 8, 6, 8, 0, 7, 4, 0, 3, 8, 4, 7, 7, 0, 5, 0, 7, 8, 5, 7, 7, 6, 1, 2, 4, 8, 2, 8, 5, 0, 4, 3, 5, 3, 1, 6, 7, 7, 6, 4, 6, 3, 3

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• cons

Comments

Equals Integral_{x>=0} sin(4*x)/(4*x) dx. - Jean-François Alcover, Feb 28 2013

Consider 4 circles inscribed in a square. Inscribe a square in each circle. And finally, inscribe 4 circles inside each four small squares. Totally we get 16 small circles. Pi/8 is the ratio of the area of the 16 small circles to the area of initial square. See the link. - Kirill Ustyantsev, Apr 30 2020

Examples

			Pi/8 = 0.392699081698724154807830422909937860524646174921888227621868... - _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009
		

References

• Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 492.

Links

• Ivan Panchenko, Table of n, a(n) for n = 0..1000
• Kirill Ustyantsev, Geometric sense of Pi/8.
• Index entries for transcendental numbers.

Crossrefs

Cf. A195909, A195913, A195697. - Mohammad K. Azarian, Oct 11 2011

Cf. A000796, A003881, A019670, A024235, A244978, A334422 (Pi/128).

Cf. A001539, A061550.

Programs

Magma

pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^100*(pi)/8))); // Vincenzo Librandi, Oct 07 2015

Mathematica

RealDigits[N[Pi/8,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

PARI

default(realprecision, 1002);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+2)))), "3..-2"))  \\ Gheorghe Coserea, Oct 06 2015

Formula

From Peter Bala, Nov 15 2016: (Start)

Pi/8 = Sum_{k >= 1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)).

More generally, for n >= 0 we have 1/(2*n)! * Pi/4 = Sum_{k >= 1} (-1)^(k+n-1) * 1/Product_{j = -n..n} (2*k + 2*j - 1): when n = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.

For N divisible by 4, we have the asymptotic expansion Pi/8 - Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/2*(1/N^3 - 2/N^5 + 31/N^7 - 692/N^9 + ...), where the sequence of unsigned coefficients [1, 2, 31, 692, ...] equals A024235. (End)

Equals Integral_{x = 0..1} x*sqrt(1 - x^4) dx. - Peter Bala, Oct 27 2019

Equals Integral_{x = 0..oo} sin(x)^6/x^4 dx = Sum_{n >= 1} sin(n)^6/n^4, by the Abel-Plana formula. - Peter Bala, Nov 04 2019

From Amiram Eldar, Jul 12 2020: (Start)

Equals arctan(sqrt(2) - 1).

Equals Sum_{k>=0} (-1)^k/(4*k+2).

Equals Sum_{k>=0} 1/((4*k+1)*(4*k+3)) = Sum_{k>=0} 1/A001539(k).

Equals Integral_{x=0..oo} dx/(x^2 + 16).

Equals Integral_{x=0..oo} dx/(x^4 + 4) = Integral_{x=0..oo} x/(x^4 + 4) dx.

Equals Integral_{x=0..oo} x/(x^4 + 1)^2 dx = Integral_{x=0..1} x/(x^4 + 1) dx.

Equals Integral_{x=0..1} x * arcsin(x) dx. (End)

From Kritsada Moomuang, Jun 18 2025: (Start)

Equals Integral_{x=0..oo} (x*log(x + 1))/((x^2 + 1)^2) dx.

Equals Integral_{x=0..oo} (x^3 - 3*x + 3*arctan(x))/(3*x^5) dx. (End)