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Expansion of e.g.f. tan(x)*sin(x)/2 (even powers only).
8

%I #39 Mar 01 2023 11:00:46

%S 0,1,2,31,692,25261,1351382,99680491,9695756072,1202439837721,

%T 185185594118762,34674437196568951,7757267081778543452,

%U 2043536254646561946181,626129820701814932734142,220771946624511552276841411,88759695789769644718332394832

%N Expansion of e.g.f. tan(x)*sin(x)/2 (even powers only).

%C From _Peter Bala_, Nov 10 2016: (Start)

%C This sequence gives the coefficients in an asymptotic expansion related to the constant Pi/8. Recall the Madhava-Gregory-Leibniz series Pi/4 = Sum_{k = 1..inf} (-1)^(k-1)/(2*k - 1). Borwein et al. gave an asymptotic expansion for the tails of this series: Pi/2 - 2*Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1) ~ 1/N - 1/N^3 + 5/N^5 - 61/N^7 + ..., where N is an integer divisible by 4 and the sequence of unsigned coefficients [1, 1, 5, 61,...] is the sequence of Euler numbers A000364.

%C Similarly, we have the series representation Pi/8 = Sum_{k = 1..inf} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)): using the approach of Borwein et al. we can show the associated asymptotic expansion for the tails of the series is Pi/4 - 2*Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/N^3 + 2/N^5 - 31/N^7 + 692/N^9 - ..., where N is divisible by 4 and where the sequence of unsigned coefficients [1, 2, 31, 692,...] forms the present sequence. A numerical example is given below. Cf. A278080 and A278195. (End)

%H Vincenzo Librandi, <a href="/A024235/b024235.txt">Table of n, a(n) for n = 0..50</a>

%H J. M. Borwein, P. B. Borwein, K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EulerPolynomial.html">Euler Polynomial</a>.

%F G.f.: 1/2*(G(0) - 1/(1+x)) where G(k) = 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/G(k+1) ); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Feb 09 2013

%F a(n) ~ (2*n)! * (2/Pi)^(2*n+1). - _Vaclav Kotesovec_, Jan 23 2015

%F From _Peter Bala_, Nov 10 2016: (Start)

%F a(n) = 1/2*(A000364(n) - (-1)^n).

%F a(n) = 1/8*(-4)^n*( -E(2*n,3/2) + 2*E(2*n,1/2) - E(2*n,-1/2) ), where E(n,x) is the Euler polynomial of order n.

%F G.f. 1/2!*sin^2(x)/cos(x) = x^2/2! + 2*x^4/4! + 31*x^6/6! + 692*x^8/8! + ....

%F O.g.f. for a signed version of the sequence: Sum_{n >= 0} ( 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n, k)/((1 - (2*k - 1)*x)*(1 - (2*k + 1)*x)*(1 - (2*k + 3)*x)) ) = 1 - 2*x^2 + 31*x^4 - 692*x^6 + .... (End)

%e tan(x)*sin(x)/2 = 1/2*x^2 + 1/12*x^4 + 31/720*x^6 + 173/10080*x^8 + ...

%e From _Peter Bala_, Nov 10 2016: (Start)

%e Asymptotic expansion at N = 100000.

%e The truncated series 2*Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) = 0.78539816339744(9)309615660(6)4581987(603) 104929(1657)84377... to 50 digits. The bracketed digits show where this decimal expansion differs from that of Pi/4. The numbers -1, 2, -31, 692 must be added to the bracketed numbers to give the correct decimal expansion to 50 digits: Pi/4 = 0.78539816339744(8)309615660(8)4581987(572)104929(2349)84377.... (End)

%p A000364 := proc(n)

%p abs(euler(2*n));

%p end proc:

%p seq(1/2*(A000364(n) - (-1)^n), n = 0..20); # _Peter Bala_, Nov 10 2016

%t With[{nn=30},Take[CoefficientList[Series[Tan[x]*Sin[x]/2,{x,0,nn}], x]Range[0,nn]!,{1,-1,2}]] (* _Harvey P. Dale_, Apr 27 2012 *)

%Y Cf. A009744, A000364, A004174, A019675, A278080, A278195.

%K nonn,easy

%O 0,3

%A _R. H. Hardin_

%E Extended and signs tested Mar 15 1997.

%E More terms from _Harvey P. Dale_, Apr 27 2012