Gonzalo Muga, Andreas Ruschhaupt, Adolfo del Campo. where T(q, q') depends on ... 1 v. ) T( 0-(-; * + v. 0) T( o -i. XC 211 ôq2 q (1, (1 21 9q” q (1, (1 ×sgn ... =#1;± (vo-voto-0 go and satisfies the condition dT(q, ôT(q, q' (4.4). (q, q ...
... q-deformation of U(sl(2)). After this we describe the general case U, (g). Example 2.1 Let k = C and q e C", not a root of unity. Let U,(sl(2)) be the non-commutative - H + H. algebra generated by 1 ... =# - 1 (the notation is meant to ...
... Q, = "Tl A # (a)da 0 For Q = const , the integrals in the numerators are readily calculated: l in knl } Q?n (x) dx =# [1 – cos knl +# (1 – ch K.)] - For the computation of the integrals in the denominator, we may again use Table l. In ...
... ), n + m . Ot, " ; q + q =#1 r . |=0 " (" "), , , ;", a +4 j – 1 % 3 = r-i-1 - -(?) , .m. 1 -: XDa'(-1)^2(3) (q ;4), | r+ w | '" * j=0 J J (by (Andrews, 1976, p. 35, eq. (3.3.3))). Hence Proposition 3.4 follows by induction on r. D ...
... q"(1- 2bx°), [Q =# (1 - bx*] o This is the pressure treated in the previous section; see (4.1). Problem II: (5.2) q = q x", [Q = + x'] . sk In (5.1) and (5.2), the constant q is the amplitude of the pressure. We define a new variable y ...
... q| < 1, then # k2 (-aid").(a d'). (i) % (q) TE f(q, q) = 1 + 2 Q 2: 2 2 2 3. k=l (q; q"). (-q"; q"). OC) (g";a") (ii) ... =# n=l n=l 1 + q 2n l 2) 5 CO n' (1 - "in a ") (dia”). Go",a which is a famous identity of Euler. Substituting (22.2) ...
... q), 201(x, y – 6 yla, Ya") (10.89) 10.3.3 Three formulas of Andrews We continue ... 1(a; 6, 6', yla, q', q') (a; q)<(6 + x; q)<(6' + y, q)< (Y; q).2c (x;q) oc ... =#XX. (a + m + n, q).2e (1; q), ( m=0 n=0 _(a; q)< \\ \\ \\ q"q"q"(y – a q) ...
... 1-simulate it Theorem 1. For k > 1, T € {o, di, de, f}, for any A, the following containments hold: 1. – C =#1 ... [q] denote the equivalence class of q e Q with respect to ~. The quotient of A with respect to ~ is the automaton A/~ = (X ...
Kevin W. J. Kadell. and [1] # w(t1,...,tn) abcn (a,b, kit 1,...,tn) (6.14) = (a" - 1)[1].”(t1,...,tn) ben(a,b, kiti,...,tn) + q " [1] : w(t1,...,tn) 1bcn (a, b, kiti,...,tn). Proof. Observe that (615) (#) (, ) = (1 ...
... 1 kak :5C, nEN. k=1 Thus the desired result follows from Theorem 3.3.1 (ii). Another direct consequence of Theorem 3.3.1 is the construction of some special function's suggesting that Qp is a large subspace of Dp. Example 3.3.1. Let p E n( ...