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A274659
Triangle entry T(n, m) gives the m-th contribution T(n, m)*sin((2*m+1)*v) to the coefficient of q^n in the Fourier expansion of Jacobi's elliptic sn(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.
4
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, -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, 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
OFFSET
0,8
COMMENTS
If one takes the row polynomials as R(n, x) = Sum_{m=0..n} T(n, m)*x^(2*m+1), n >= 0, Jacobi's elliptic sn(u|k) function in terms of the new variables v and q becomes sn(u|k) = Sum_{n>=0} R(n, x)*q^n, if one replaces in R(n, x) x^j by sin(j*v).
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.
A test for sn(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 0.8030018002 to be compared with sn(1|sqrt(1/2)) approximately 0.8030018249.
For the derivation of the Fourier series formula of sn 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.
For the cn expansion see A274661.
See also the W. Lang link, equations (34) and (35).
REFERENCES
J. V. Armitage and W. F. Eberlein, Elliptic Functions, London Mathematical Society, Student Texts 67, Cambridge University Press, 2006.
E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, fourth edition, reprinted, 1958, Cambridge at the University Press.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 375, 16.23.1.
FORMULA
T(n, m) = [x^(2*m+1)]Sum_{j=0..n} b(j, x)*a(n-j), with a(k) = A274621(k/2) if k is even and a(k) = 0 if k is odd, and b(j, x) = Sum_{r | 2*j+1} x^r = Sum_{k=1..A099774(j+1)} x^(A274658(j, k)), for j >= 0.
EXAMPLE
The triangle T(n, m) begins:
m 0 1 2 3 4 5 6 7 8 9 10 11
n\ 2m+1 1 3 5 7 9 11 13 15 17 19 21 23
0: 1
1: 1 1
2: -1 0 1
3: -1 -2 0 1
4: 2 1 -2 0 1
5: 2 3 0 -2 0 1
6: -4 -2 3 0 -2 0 1
7: -4 -5 1 3 0 -2 0 1
8: 7 3 -6 0 3 0 -2 0 1
9: 7 9 -2 -6 0 3 0 -2 0 1
10: -11 -5 11 1 -6 0 3 0 -2 0 1
11: -11 -15 3 11 0 -6 0 3 0 -2 0 1
...
T(4, 0) = 2 from the x^1 term in b(0, x)*a(4) + b(2, x)*a(2) + b(4, x)*a(0), that is x^1*3 + x^1*(-2) + x^1*1 = +2*x^1.
n=4: R(4, x) = 2*x^1 + 1*x^3 - 2*x^5 + 0*x^7 + 1*x^9, that is the sn(u|k) contribution of order q^4 in the new variables v and q is (2*sin(1*v) + 1*sin(3*v) - 2*sin(5*v) + 1*sin(9*v))*q^4.
KEYWORD
sign,tabl,easy
AUTHOR
Wolfdieter Lang, Jul 18 2016
STATUS
approved