cp's OEIS Frontend

A275790 Triangle T(n, m) appearing in the expansion of the scaled phase space coordinate qhat of the plane pendulum in terms of the Jacobi nome q and sin(v) multiplying even powers of 2*cos(v), with v = u/((2/Pi)*K(k)).

%I A275790 #15 Dec 19 2017 02:15:31
%S A275790 1,8,1,-32,11,3,-736,-92,9,15,2816,-593,-249,-65,35,48976,6122,1581,
%T A275790 -970,-1295,315,-951424,61252,67791,46030,18515,-21735,3465,-1045952,
%U A275790 -130744,-92082,-30445,14455,53928,-25179,3003,26933248,1069361,-1666047,-634255,-1167740,-1258236,1562253,-471471,45045,634836808,79354601,24881793,17914550,30289840,12635028,-71064609,42480438,-9594585,765765
%N A275790 Triangle T(n, m) appearing in the expansion of the scaled phase space coordinate qhat of the plane pendulum in terms of the Jacobi nome q and sin(v) multiplying even powers of 2*cos(v), with v = u/((2/Pi)*K(k)).
%C A275790 The dimensionless scaled phase space coordinates of the plane pendulum are (qtilde(tau, k), ptilde(tau, k)) with tau = omega_0*t, omega^2 = g/L (L is the  length of the pendulum, g the acceleration), and the energy variable E = 2*k^2 = 2*sin^2(Theta_0/2), with the maximal deflection angle Theta_0 (from [0, Pi/2]). qtilde = Theta/(2*k)) with the deflection angle Theta. Similarly ptilde = (d(Theta)/d(tau))/(2*k).
%C A275790 The exact solution is qtilde(tau, k) = (1/k)*arcsin(k*sn(tau, k)) with Jacobi's elliptic sn function, and ptilde(tau,k) = cn(tau, k) with the elliptic cn function.
%C A275790 Here the expansion in new variables v and q is used where v = tau/((2/Pi)*K(k)) and q = exp(-Pi*K'(k)/K(k)) with the real and imaginary quarter periods K and K'. This leads to qhat(v, q) = qtilde(tau(v, q), k(q)) with tau(v, q) = theta_3^2(0, q)*v. (For theta_3^2(0, q) see A004018.) Because k is actually a function of k^2 one uses the q expansion of (k/4)^2 given in A005798.
%C A275790 Using the result for the sn expansion in q and v from A274662 one obtains qhat(v, q) = sin(v)*Sum_{n >= 0} q^n/L(n)*Sum_{m=0..n} T(n, m)*(2*cos(v))^(2*m) with L(n) = A025547(n+1) = lcm{1, 3, ..., (2*n+1)}.
%C A275790 This entry is inspired by a paper of Bradley Klee giving an approximation to the phase space solution of the plane pendulum (see A273506). Thanks for discussions via e-mail go to him.
%F A275790 T(n, m)*(2*cos(v))^(2*m)), n >= 0, m = 0, 1, ..., n, gives the contribution to q^n/L(n) (L(n) = A025547(n+1)) in the rescaled phase space coordinate qhat(v, q) expansion of the plane pendulum. See a comment above for details.
%e A275790 The triangle T(n, m) begins:
%e A275790 n\m    0    1    2    3     4   5 ...
%e A275790 0:     1
%e A275790 1:     8    1
%e A275790 2:   -32   11    3
%e A275790 3:  -736  -92    9   15
%e A275790 4:  2816 -593 -249  -65    35
%e A275790 5: 48976 6122 1581 -970 -1295 315
%e A275790 ...
%e A275790 row n=6: -951424 61252 67791 46030 18515 -21735 3465,
%e A275790 row n=7: -1045952 -130744 -92082 -30445 14455 53928 -25179 3003,
%e A275790 row n=8: 26933248 1069361 -1666047 -634255 -1167740 -1258236 1562253 -471471 45045,
%e A275790 row n=9: 634836808 79354601 24881793 17914550 30289840 12635028 -71064609 42480438 -9594585 765765.
%e A275790 ...
%e A275790 The corresponding L(n) = A025547(n+1) numbers are 1, 3, 15, 105, 315, 3465, 45045, 45045, 765765, 14549535,...
%e A275790 n=4: the contribution to qhat(v, q) of order q^4 is (q^4/315)*(2816 - 593*(2*cos(v))^2 - 249*(2*cos(v))^4 - 65*(2*cos(v))^6 + 35*(2*cos(v))^8).
%Y A275790 Cf. A004018, A005798, A025547, A274662.
%K A275790 sign,tabl
%O A275790 0,2
%A A275790 _Wolfdieter Lang_, Aug 09 2016