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%I A274661 #27 Aug 13 2023 17:32:49
%S A274661 1,-1,1,-1,0,1,1,-2,0,1,2,-1,-2,0,1,-2,3,0,-2,0,1,-4,2,3,0,-2,0,1,4,
%T A274661 -5,-1,3,0,-2,0,1,7,-3,-6,0,3,0,-2,0,1,-7,9,2,-6,0,3,0,-2,0,1,-11,5,
%U A274661 11,-1,-6,0,3,0,-2,0,1,11,-15,-3,11,0,-6,0,3,0,-2,0,1,17,-9,-17,2,11,0,-6,0,3,0,-2,0,1,-17,23,6,-18,-1,11,0,-6,0,3,0,-2,0,1
%N A274661 Triangle read by rows: T(n, m) gives the m-th contribution T(n, m)*cos((2*m+1)*v) to the coefficient of q^n in the Fourier expansion of Jacobi's elliptic cn(u|k) function when expressed in the variables v = u/(2*K(k)/Pi) and q, the Jacobi nome, written as series in (k/4)^2. K is the real quarter period of elliptic functions.
%C A274661 If one takes the row polynomials as P(n, x) = Sum_{m=0..n} T(n, m)*x^m, n >= 0, Jacobi's elliptic function cn(u|k) in terms of the new variables v and q becomes cn(u|k) = Sum_{n>=0} P(n, x)*q^n, if in P(n, x) one replaces x^j by cos((2*j+1)*v).
%C A274661 v=v(u,k^2) and q=q(k^2) are computed with the help of A038534/A056982 for (2/Pi)*K and A002103 for q expanded in powers of (k/4)^2.
%C A274661 A test for cn(u|k) with u = 1, k = sqrt(1/2), that is v approximately 0.8472130848 and q approximately 0.04321389673, with rows n=0..10 (q powers not exceeding 10) gives 0.5959766014 to be compared with cn(1|sqrt(1/2)) approximately 0.5959765676.
%C A274661 For the derivation of the Fourier series formula of cn given in Abramowitz-Stegun (but there the notation sn(u|m=k^2) is used for sn(u|k)) see, e.g., Whittaker and Watson, p. 511 or Armitage and Eberlein, Exercises on p. 55.
%C A274661 For sn see A274659 (differently signed triangle).
%C A274661 The sum of entries in row n is P(n, 1) = A000007(n): 1, repeat 0. Proof: due to the g.f. identity (from the convolution)
%C A274661 Sum_{n >= 0} x^n/(1 + x^(2*n+1)) = (Sum_{n >= 0} x^(n*(n+1)))^2.
%C A274661 This is proved by bisecting the g.f. on the l.h.s. which generates c(n, 1) = (-1)^n*Sum_{2*r+1 | 2*n+1} (-1)^n. The part with n = 2*k+1 vanishes due to r_2(4*k+1)/4 = 0, where r_2(n) is the number of solutions of n as a sum of two squares. See the Grosswald reference. The part with n = 2*k becomes Sum_{k >= 0} x^(2*k) r_2(4*k+1)/4 which is the r.h.s. See A008441, the Broadhurst Oct 20 2002 comment.
%C A274661 For another version of this expansion of cn see A275791.
%C A274661 See also the W. Lang link, eqs. (43) and (44). - _Wolfdieter Lang_, Aug 26 2016
%D A274661 J. V. Armitage and W. F. Eberlein, Elliptic Functions, London Mathematical Society, Student Texts 67, Cambridge University Press, 2006.
%D A274661 E. Grosswald, Representations of Integers as Sums of Squares. Springer-Verlag, NY, 1985, p. 15, Theorem 3.
%D A274661 E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, fourth edition, reprinted, 1958, Cambridge at the University Press.
%H A274661 M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 375, 16.23.2.
%H A274661 Wolfdieter Lang, <a href="/A273506/a273506_5.pdf">Expansions for phase space coordinates for the plane pendulum</a>
%F A274661 T(n, m) = [x^(2*m+1)]Sum_{j=0..n} c(j, x)*a(n-j), with a(k) = A274621(k/2) if k is even and a(k) = 0 if k is odd, and c(j, x) = (-1)^j*Sum_{2*r+1 | 2*j+1} (-1)^r*x^(2*r+1) = Sum_{k=1..A099774(j+1)} sign(A274660(j, k))*x^(abs(A274660(j, k))), for j >= 0.
%e A274661 The triangle T(n, m) begins:
%e A274661 m 0 1 2 3 4 5 6 7 8 9 10 11
%e A274661 n\ 2m+1 1 3 5 7 9 11 13 15 17 19 21 23
%e A274661 0: 1
%e A274661 1: -1 1
%e A274661 2: -1 0 1
%e A274661 3: 1 -2 0 1
%e A274661 4: 2 -1 -2 0 1
%e A274661 5: -2 3 0 -2 0 1
%e A274661 6: -4 2 3 0 -2 0 1
%e A274661 7: 4 -5 -1 3 0 -2 0 1
%e A274661 8: 7 -3 -6 0 3 0 -2 0 1
%e A274661 9: -7 9 2 -6 0 3 0 -2 0 1
%e A274661 10: -11 5 11 -1 -6 0 3 0 -2 0 1
%e A274661 11: 11 -15 -3 11 0 -6 0 3 0 -2 0 1
%e A274661 ...
%e A274661 n = 4: c(0, x)*a(4) + c(2, x)*a(2) + c(4, x)*a(0) = (+x^1)*3 + (+x^1 + x^5)*(-2) + (+x^1 - x^3 + x^9)*1 = +2*x^1 - x^3 - 2*x^5 + 0*x^7 + x^9. Hence row n=4 is 2, -1, -2, 0, 1.
%e A274661 From A274660, row n = 4: c(4, x) = +x^1 - x^3 +x^9.
%e A274661 n = 4: P(4, x) = 2 - 1*x^1 - 2*x^2 + 1*x^4, that is the contribution of order q^4 to cn in the new variables is (2*cos(v) - 1*cos(3*v) - 2*cos(5*v) + 1*cos(9*v))*q^4.
%Y A274661 Cf. A000007, A099774, A274621, A274659, A274660, A275791.
%K A274661 sign,easy,tabl
%O A274661 0,8
%A A274661 _Wolfdieter Lang_, Jul 27 2016