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A003881
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Decimal expansion of Pi/4.
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91
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7, 8, 5, 3, 9, 8, 1, 6, 3, 3, 9, 7, 4, 4, 8, 3, 0, 9, 6, 1, 5, 6, 6, 0, 8, 4, 5, 8, 1, 9, 8, 7, 5, 7, 2, 1, 0, 4, 9, 2, 9, 2, 3, 4, 9, 8, 4, 3, 7, 7, 6, 4, 5, 5, 2, 4, 3, 7, 3, 6, 1, 4, 8, 0, 7, 6, 9, 5, 4, 1, 0, 1, 5, 7, 1, 5, 5, 2, 2, 4, 9, 6, 5, 7, 0, 0, 8, 7, 0, 6, 3, 3, 5, 5, 2, 9, 2, 6, 6, 9, 9, 5, 5, 3, 7
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OFFSET
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0,1
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COMMENTS
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Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023
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REFERENCES
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Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
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LINKS
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FORMULA
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Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
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EXAMPLE
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0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - Peter Bala, Nov 15 2016
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MAPLE
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evalf(Pi/4) ;
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MATHEMATICA
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(* PROGRAM STARTS *)
(* Define the nested radicals a_k by recurrence *)
a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
(* Example of Pi/4 approximation at K = 100 *)
Print["The actual value of Pi/4 is"]
N[Pi/4, 40]
Print["At K = 100 the approximated value of Pi/4 is"]
K := 100; (* the truncating integer *)
N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
(* Error terms for Pi/4 approximations *)
Print["Error terms for Pi/4"]
k := 1; (* initial value of the index k *)
K := 10; (* initial value of the truncating integer K *)
sqn := {}; (* initiate the sequence *)
AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
While[K <= 30,
AppendTo[sqn, {K,
N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
N}]; K++]
Print[MatrixForm[sqn]]
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PROG
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(Haskell) -- see link: Literate Programs
import Data.Char (digitToInt)
a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
machin = 4 * arccot 5 unity - arccot 239 unity
unity = 10 ^ (len + 10)
arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
arccot' x unity summa xpow n sign
| term == 0 = summa
| otherwise = arccot'
x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
where term = xpow `div` n
(SageMath) # Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
def FastLeibniz(n):
b = 2^(2*n-1); c = b; s = 0
for k in range(n-1, -1, -1):
t = 2*k+1
s = s + c/t if is_even(k) else s - c/t
b *= (t*(k+1))/(2*(n-k)*(n+k))
c += b
return s/c
A003881 = RealField(3333)(FastLeibniz(1330))
(Magma) R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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