Search: a244644 -id:a244644
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A093954
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Decimal expansion of Pi/(2*sqrt(2)).
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+10
32
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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, 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, 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
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OFFSET
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1,5
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COMMENTS
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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
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
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
The area of a circle circumscribing a unit-area regular octagon. - Amiram Eldar, Nov 05 2020
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REFERENCES
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J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), p. 149.
L. B. W. Jolley, Summation of Series, Dover (1961), eq 76 page 16.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 53.
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LINKS
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Eric Weisstein's World of Mathematics, Bifoliate.
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FORMULA
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Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
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)
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). See Wells.
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)
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.
c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
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.
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.
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)
Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - Peter Bala, Nov 01 2019
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
Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
Equals Integral_{x=0..oo} dx/(x^2 + 2).
Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - Bernard Schott, May 19 2022
Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - Peter Bala, Jul 22 2022
Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - Terry D. Grant, Mar 17 2023
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
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EXAMPLE
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1.11072073453959156175397...
Asymptotic expansion at n = 5000.
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)
Case m = 1, n = 1:
Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
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.
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)
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MAPLE
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simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) ); # Peter Bala, Mar 09 2015
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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STATUS
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approved
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A014549
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Decimal expansion of 1 / M(1,sqrt(2)) (Gauss's constant).
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+10
17
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8, 3, 4, 6, 2, 6, 8, 4, 1, 6, 7, 4, 0, 7, 3, 1, 8, 6, 2, 8, 1, 4, 2, 9, 7, 3, 2, 7, 9, 9, 0, 4, 6, 8, 0, 8, 9, 9, 3, 9, 9, 3, 0, 1, 3, 4, 9, 0, 3, 4, 7, 0, 0, 2, 4, 4, 9, 8, 2, 7, 3, 7, 0, 1, 0, 3, 6, 8, 1, 9, 9, 2, 7, 0, 9, 5, 2, 6, 4, 1, 1, 8, 6, 9, 6, 9, 1, 1, 6, 0, 3, 5, 1, 2, 7, 5, 3, 2, 4, 1, 2, 9, 0, 6, 7, 8, 5
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OFFSET
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0,1
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COMMENTS
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On May 30, 1799, Gauss discovered that this number is also equal to (2/Pi)*Integral_{t=0..1} 1/sqrt(1-t^4).
M(a,b) is the limit of the arithmetic-geometric mean iteration applied repeatedly starting with a and b: a_0 = a, b_0 = b, a_{n+1} = (a_n + b_n)/2, b_{n+1} = sqrt(a_n*b_n).
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REFERENCES
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J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, page 5.
J. R. Goldman, The Queen of Mathematics, 1998, p. 92.
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LINKS
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FORMULA
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Equals (lim_{k->oo} p(k))/(1+i) and (lim_{k->oo} q(k))/(1+i), where i is the imaginary unit, p(0) = 1, q(0) = i, p(k+1) = 2*p(k)*q(k)/(p(k)+q(k)) and q(k+1) = sqrt(p(k)*q(k)) for k >= 0. - A.H.M. Smeets, Jul 26 2018
Equals the infinite quotient product (3/4)*(6/5)*(7/8)*(10/9)*(11/12)*(14/13)*(15/16)*... . - James Maclachlan, Jul 28 2019
Equals (9/15)*hypergeom([1/2, 3/4], [9/4], 1). - Peter Bala, Mar 03 2022
Equals theta4(exp(-Pi))^2.
Equals Sum_{k>=0} (-1)^k * binomial(2*k,k)^2/16^k. - Amiram Eldar, Jul 04 2023
Equals 2*Gamma(5/4)/(sqrt(Pi)*Gamma(3/4)).
Equals hypergeom([1/4, -2/4], [1], 1). (End)
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EXAMPLE
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0.8346268416740731862814297327990468...
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MAPLE
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MATHEMATICA
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RealDigits[Gamma[1/4]^2/(2*Pi^(3/2)*Sqrt[2]), 10, 105][[1]] (* or: *)
First[RealDigits[N[EllipticTheta[4, Exp[-Pi]]^2, 90]]] (* Stefano Spezia, Sep 29 2022 *)
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PROG
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(PARI) default(realprecision, 20080); x=10*agm(1, sqrt(2))^-1; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b014549.txt", n, " ", d)); \\ Harry J. Smith, Apr 20 2009
(Python)
from mpmath import mp, agm, sqrt
mp.dps=105
print([int(z) for z in list(str(1/agm(sqrt(2)))[2:-1])]) # Indranil Ghosh, Jul 11 2017
(Magma) SetDefaultRealField(RealField(100)); R:= RealField(); Sqrt(Pi(R)/2)/Gamma(3/4)^2; // G. C. Greubel, Aug 17 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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A188615
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Decimal expansion of Brocard angle of side-silver right triangle.
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+10
9
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3, 3, 9, 8, 3, 6, 9, 0, 9, 4, 5, 4, 1, 2, 1, 9, 3, 7, 0, 9, 6, 3, 9, 2, 5, 1, 3, 3, 9, 1, 7, 6, 4, 0, 6, 6, 3, 8, 8, 2, 4, 4, 6, 9, 0, 3, 3, 2, 4, 5, 8, 0, 7, 1, 4, 3, 1, 9, 2, 3, 9, 6, 2, 4, 8, 9, 9, 1, 5, 8, 8, 8, 6, 6, 4, 8, 4, 8, 4, 1, 1, 4, 6, 0, 7, 6, 5, 7, 9, 2, 5, 0, 0, 1, 9, 7, 6, 1, 2, 8, 5, 2, 1, 2, 9, 7, 6, 3, 8, 0, 7, 4, 0, 2, 2, 9, 4, 4, 7, 4, 1, 5, 2, 3, 9, 3, 5, 7, 5, 6
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OFFSET
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0,1
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COMMENTS
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The Brocard angle is invariant of the size of the side-silver right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(silver ratio)=(1+sqrt(2)). This is the unique right triangle matching the continued fraction [2,2,2,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there are exactly 2 removable subtriangles at each stage. (This is analogous to the removal of 2 squares at each stage of the partitioning of the silver rectangle as a nest of squares.)
Archimedes's-like scheme: set p(0) = 1/(2*sqrt(2)), q(0) = 1/3; 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
This angle is also the half-angle at the summit of the Kelvin wake pattern traced by a boat. - Robert FERREOL, Sep 27 2019
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LINKS
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FORMULA
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(Brocard angle) = arccot((a^2+b^2+c^2)/(4*area(ABC))) = arccot(sqrt(8)).
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EXAMPLE
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Brocard angle: 0.3398369094541219370963925133917640663882 approx.
Brocard angle: 19.471220634490691369245999 degrees, approx.
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MATHEMATICA
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r=1+2^(1/2);
b=1; a=r*b; c=(a^2+b^2)^(1/2);
area=(1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);
brocard=ArcCot[(a^2+b^2+c^2)/(4area)];
N[brocard, 130]
RealDigits[N[brocard, 130]][[1]]
N[180 brocard/Pi, 130] (* degrees *)
RealDigits[ArcCos[Sqrt[8/9]], 10, 50][[1]] (* G. C. Greubel, Nov 18 2017 *)
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CROSSREFS
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approved
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A244645
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Decimal expansion of the sum of the reciprocals of the octagonal numbers (A000567).
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+10
8
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1, 2, 7, 7, 4, 0, 9, 0, 5, 7, 5, 5, 9, 6, 3, 6, 7, 3, 1, 1, 9, 4, 9, 5, 3, 4, 9, 2, 1, 0, 2, 4, 3, 3, 2, 1, 1, 5, 5, 6, 6, 3, 4, 4, 8, 0, 3, 9, 0, 2, 4, 7, 2, 3, 2, 6, 9, 3, 4, 9, 1, 9, 8, 4, 0, 7, 5, 1, 5, 1, 5, 1, 5, 1, 9, 5, 5, 4, 5, 1, 9, 6, 0, 7, 6, 2, 4, 3, 0, 6, 3, 1, 6, 3, 3, 1, 4, 1, 0, 8, 8, 0, 5, 0, 3
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OFFSET
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1,2
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FORMULA
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Equals Sum_{n>=1} 1/(3*n^2 - 2*n).
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EXAMPLE
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1.2774090575596367311949534921024332115566344803902472326934919840751515151955452...
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MATHEMATICA
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RealDigits[ Sum[1/(3n^2 - 2n), {n, 1 , Infinity}], 10, 111][[1]]
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PROG
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(PARI) sumpos(n=1, 1/(3*n^2 - 2*n)) \\ Michel Marcus, Sep 12 2016
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CROSSREFS
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approved
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A244646
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Decimal expansion of the sum of the reciprocals of the 9-gonal (or enneagonal or nonagonal) numbers (A001106).
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+10
7
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1, 2, 4, 3, 3, 2, 0, 9, 2, 6, 1, 5, 3, 7, 1, 2, 9, 8, 9, 2, 0, 6, 6, 0, 7, 7, 3, 9, 6, 3, 1, 0, 1, 4, 2, 8, 2, 1, 3, 5, 8, 4, 4, 1, 0, 1, 0, 3, 0, 0, 9, 9, 6, 2, 4, 4, 1, 5, 2, 8, 1, 7, 5, 2, 5, 3, 8, 6, 6, 0, 7, 4, 3, 8, 4, 4, 0, 8, 5, 1, 9, 7, 8, 6, 9, 0, 0, 1, 3, 2, 3, 2, 5, 8, 8, 3, 2, 8, 6, 0, 0, 7, 3, 6, 8
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OFFSET
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1,2
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LINKS
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FORMULA
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Equals Sum_{n>=1} 2/(7n^2 - 5n).
Equals (2*log(14) + 4*(cos(Pi/7)*log(cos(3*Pi/14)) + log(sin(Pi/7))*sin(Pi/14) - log(cos(Pi/14)) * sin(3*Pi/14)) + Pi*tan(3*Pi/14))/5. - Vaclav Kotesovec, Jul 04 2014
Equals 14/25 - (2/5)*(gamma + psi(-5/7)), where gamma is Euler's constant (A001620) and psi(x) is the digamma function (Agarwal, 2021), psi(-5/7) = psi(2/7)+7/5 = -2.285517..., see A354628. - Amiram Eldar, Nov 12 2021
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EXAMPLE
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1.2433209261537129892066077396310142821358441010300996244152817525...
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MATHEMATICA
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RealDigits[ Sum[2/(7n^2 - 5n), {n, 1 , Infinity}], 10, 111][[1]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A244649
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Decimal expansion of the sum of the reciprocals of the Dodecagonal numbers (A051624).
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+10
7
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1, 1, 7, 7, 9, 5, 6, 0, 5, 7, 9, 2, 2, 6, 6, 3, 8, 5, 8, 7, 3, 5, 1, 7, 3, 9, 6, 8, 0, 9, 1, 8, 8, 7, 4, 1, 8, 4, 4, 5, 8, 5, 7, 2, 3, 4, 5, 6, 6, 6, 7, 9, 8, 0, 2, 8, 4, 2, 5, 2, 2, 8, 5, 7, 3, 2, 6, 6, 8, 9, 2, 5, 6, 8, 2, 8, 4, 8, 8, 7, 4, 5, 4, 0, 2, 4, 0, 7, 6, 9, 0, 2, 5, 6, 9, 5, 5, 9, 0, 3, 2, 2, 4, 4, 4
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OFFSET
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1,3
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COMMENTS
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In the Downey et al. link this is the instance k = 5 of the formula given there for S_{2*k+2}. A simpler formula is given in the Koecher reference as (5/4)*v_5(1) on p. 192. See the Kotesovec formula given below.
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REFERENCES
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Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, pp. 189 - 193.
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LINKS
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FORMULA
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Equals Sum_{n>=1} 1/(5n^2 - 4n).
Equals Pi/8*sqrt(1+2/sqrt(5)) + (5*log(5) + sqrt(5)*log((3+sqrt(5))/2))/16. - Vaclav Kotesovec, Jul 04 2014
This is the value given in the Koecher reference (see a comment above), and rewritten with the golden section phi = (1 + sqrt(5))/2 this becomes
((5/2)*log(5) + (2*phi - 1)*(log(phi) + (Pi/5)*sqrt(3 + 4*phi)))/8. - Wolfdieter Lang, Nov 09 2017
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EXAMPLE
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1.1779560579226638587351739680918874184458572345666798028425228573...
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MATHEMATICA
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RealDigits[ Sum[1/(5n^2 - 4n), {n, 1 , Infinity}], 10, 111][[1]]
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CROSSREFS
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KEYWORD
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STATUS
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approved
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A244648
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Decimal expansion of the sum of the reciprocals of the hendecagonal numbers (A051682).
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+10
5
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1, 1, 9, 5, 4, 3, 4, 1, 1, 6, 5, 2, 9, 6, 2, 7, 9, 7, 4, 3, 5, 2, 4, 9, 9, 2, 3, 4, 6, 9, 8, 4, 9, 9, 3, 5, 4, 8, 8, 4, 6, 8, 2, 6, 2, 7, 0, 8, 4, 6, 5, 8, 0, 6, 2, 3, 8, 6, 0, 2, 1, 6, 0, 3, 0, 1, 7, 3, 5, 8, 4, 7, 3, 3, 7, 0, 3, 1, 7, 6, 0, 1, 4, 6, 4, 4, 8, 4, 1, 7, 5, 4, 8, 5, 5, 1, 1, 2, 3, 1, 8, 5, 5, 4, 7
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OFFSET
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1,3
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LINKS
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FORMULA
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Sum_{n=1..infinity} 2/(9n^2 - 7n).
Equals (5*log(3) + Pi*cot(2*Pi/9) - 4*cos(2*Pi/9)*log(cos(Pi/18)) + 4*cos(Pi/9)*log(sin(2*Pi/9)) - 4*log(sin(Pi/9))*sin(Pi/18))/7. - Vaclav Kotesovec, Jul 04 2014
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EXAMPLE
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1.195434116529627974352499234698499354884682627084658062386021603017...
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MATHEMATICA
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RealDigits[ Sum[2/(9n^2 - 7n), {n, 1 , Infinity}], 10, 111][[1]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A129200
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Decimal expansion of arcsinh(1/4).
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+10
4
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2, 4, 7, 4, 6, 6, 4, 6, 1, 5, 4, 7, 2, 6, 3, 4, 5, 2, 9, 4, 4, 7, 8, 1, 5, 4, 9, 7, 8, 8, 3, 5, 9, 2, 8, 9, 2, 5, 3, 7, 6, 6, 9, 0, 3, 0, 9, 8, 5, 6, 7, 6, 9, 6, 4, 6, 9, 1, 1, 7, 3, 5, 7, 9, 4, 4, 3, 6, 5, 1, 7, 9, 4, 4, 3, 6, 6, 6, 3, 6, 4, 9, 7, 4, 7, 5, 4, 8, 8, 3, 3, 2, 9, 3, 9, 8, 5, 9, 6
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OFFSET
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0,1
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COMMENTS
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Archimedes's-like scheme: set p(0) = 1/sqrt(17), q(0) = 1/4; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (arithmetic mean of reciprocals, 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
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LINKS
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FORMULA
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EXAMPLE
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.24746646154726345294478154978835928925376690309856769646911...
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MATHEMATICA
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PROG
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(Magma) SetDefaultRealField(RealField(100)); Argsinh(1/4); // G. C. Greubel, Nov 11 2018
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KEYWORD
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AUTHOR
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STATUS
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approved
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A195621
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Decimal expansion of arccsc(4).
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+10
4
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2, 5, 2, 6, 8, 0, 2, 5, 5, 1, 4, 2, 0, 7, 8, 6, 5, 3, 4, 8, 5, 6, 5, 7, 4, 3, 6, 9, 9, 3, 7, 1, 0, 9, 7, 2, 2, 5, 2, 1, 9, 3, 7, 3, 3, 0, 9, 6, 8, 3, 8, 1, 9, 3, 6, 3, 3, 9, 2, 3, 7, 7, 8, 7, 4, 0, 5, 7, 5, 0, 6, 0, 4, 8, 1, 0, 2, 1, 2, 2, 2, 4, 1, 1, 7, 4, 8, 7, 4, 2, 2, 2, 8, 0, 1, 4, 6, 0, 1, 6
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OFFSET
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0,1
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COMMENTS
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Archimedes's-like scheme: set p(0) = 1/sqrt(15), q(0) = 1/4; 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
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LINKS
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FORMULA
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Equals arccos(sqrt(15)/4) = arctan(1/sqrt(15)). - Amiram Eldar, Jul 11 2023
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EXAMPLE
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arccsc(4) = arcsin(1/4) = 0.25268025514207865...
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MATHEMATICA
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PROG
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(Magma) SetDefaultRealField(RealField(100)); Arcsin(1/4); // G. C. Greubel, Nov 11 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A129187
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Decimal expansion of arcsinh(1/3).
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+10
3
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3, 2, 7, 4, 5, 0, 1, 5, 0, 2, 3, 7, 2, 5, 8, 4, 4, 3, 3, 2, 2, 5, 3, 5, 2, 5, 9, 9, 8, 8, 2, 5, 8, 1, 2, 7, 7, 0, 0, 5, 2, 4, 5, 2, 8, 9, 9, 0, 7, 6, 7, 4, 5, 1, 2, 7, 5, 6, 2, 9, 5, 1, 5, 4, 2, 7, 1, 7, 6, 5, 6, 2, 9, 4, 9, 3, 2, 7, 2, 1, 4, 1, 1, 9, 8, 2, 4, 7, 7, 3, 0, 6, 3, 2, 3, 1, 9, 5, 5
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OFFSET
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0,1
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COMMENTS
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Archimedes's-like scheme: set p(0) = 1/sqrt(10), q(0) = 1/3; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (arithmetic mean of reciprocals, 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
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LINKS
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FORMULA
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EXAMPLE
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0.32745015023725844332253525998825812770052452899076745127562...
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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