inbound2948046473811376725

290 13 Hadamard Products for Two q-Series θ2m+1,1 = (−1)mq−m2−mθ1,1, (13.4.7) θ2m,2 = (−1)mq−m2+1θ4 + (−1)mm(m + 1)q−m2 θ4, (13.4.8) θ2m+1,2 = 2(−1)m+1(m + 1)q−m2−mθ1,1, (13.4.9) θ2m,3 = 3(−1)m−1(m + 1)q−m2+1θ4 + (−1)m−1m(m + 1)(m + 2)q−m2 θ4, (13.4.10) θ2m+1,3 = (−1)mq1−m−m2 θ1,1 + 3(−1)m(m + 1)(m + 2)q−m2−mθ1,1, (13.4.11) θ2m+1,4 = 4(−1)m+1(m + 2)q1−m2−mθ1,1 (13.4.12) + 4(−1)m+1(m + 1)(m + 2)(m + 3)q−m2−mθ1,1, where the prime indicates differentiation with respect to q. Each of these is proved in the same manner. We illustrate two proofs; the remainder are similar. We will see that four of the sums that arise are equal to 0. In each case, we can demonstrate this by taking the terms with negative index and replacing n by −n − 1. First, ∞ (−1)nn(n − 1)qn2+n+2mn θ2m+1,2 = n=−∞ = (−1)mq−m2−m ∞ (−1)n(n − m)(n − m − 1)qn2+n n=−∞ = (−1)mq−m2−m ∞ (−1)n n2 + n − (2m + 2)n + m(m + 1) qn2+n n=−∞ = 0 + 2(−1)m+1(m + 1)q−m2−mθ1,1 + 0, and second, ∞ (−1)nn(n − 1)(n − 2)qn2+2mn θ2m,3 = n=−∞ = (−1)mq−m2 ∞ (−1)n(n − m)(n − m − 1)(n − m − 2)qn2 n=−∞ = (−1)mq−m2 ∞ (−1)n n3 − 3(m + 1)n2 n=−∞ +(3m2 + 6m + 2)n − m(m + 1)(m + 2) qn2 = (−1)mq−m2 (0 − 3(m + 1)qθ4 + 0 − m(m + 1)(m + 2)θ4) . Comparable formulas can be found for all θm,k, and elegant formulas for the coefficients can be produced using the methods of [44, Section 2].

13.5 A Formal Power Series 291 Theorem 13.4.1. Both qm2 θ2m,k and qm2+mθ2m+1,k are analytic functions of q inside |q| < 1. Proof. We have qm2 θ2m,k = ∞ (−1)nn(n − 1) · · · (n − k + 1)q(n+m)2 n=−∞ ∞ (−1)n(n − m)(n − m − 1) · · · (n − m − k + 1)qn2 = (−1)m n=−∞ and qm2+mθ2m+1,k = ∞ (−1)nn(n − 1) · · · (n − k + 1)q(n+m)2+(n+m) n=−∞ ∞ (−1)n(n − m)(n − m − 1) · · · (n − m − k + 1)qn2+n, = (−1)m n=−∞ from which the desired conclusions immediately follow. Corollary 13.4.1. qm(m+1)/2θm+1,k is analytic in q for |q| < 1. Proof. We have qµ(2µ+1)θ2µ+1,k = qµ2 qµ2+µθ2µ+1,k and qµ(2µ−1)θ2µ,k = qµ2−µ qµ2 θ2µ,k . So our assertion follows from Theorem 13.4.1. 13.5 A Formal Power Series Theorem 13.5.1. Let yi = yi(q), i = 1, 2, 3, . . ., satisfy, for each N ≥ 0, N (−1)N−j q(N−j)(N−j+1)/2 j=0 (q; q)N−j ⎫ (−1)k1+···+kj y1k1 · · · yjkj θN −j+1,k1+···+kj ⎬ = 0, (13.5.1) k1! · · · kj! ⎭ k1+2k2+···+jkj =j where ki ≥ 0, 1 ≤ i ≤ j. Then each yi is a uniquely defined function of q analytic inside |q| < 1, and the following identity holds as a formal power series identity in z: ∞ (−1)n(1 − zy1 − z2y2 − z3y3 − · · · )n(zqn+1; q)∞qn2+n = 0. (13.5.2) n=−∞

292 13 Hadamard Products for Two q-Series Proof. We begin with N = 1 in (13.5.1) which asserts that 0 = −qθ2,0 − y1θ1,1. 1−q Hence, by (13.4.4), y1 = θ4 = 1 (13.5.3) (1 − q)(q2; q2)∞3 (1 − q)ψ2(q) , which we observe in passing is identical with the y1 appearing in (13.1.2). For N = 2, in (13.5.1), we see that 0 = q3 + (1 q q) y1θ2,1 + y12θ1,2 − y2θ1,1, (q; q)2 θ3,0 − 2 and by (13.5.1), (13.4.9), and (13.4.6), y2 = y1 y1θ1,2 + qθ2,1 θ1,1 2 (1 − q) = y1 − θ4θ1,1 + θ4 q) = 0, θ1,1 (1 − q)θ1,1 (1 − which coincides with (13.1.3). For N = 3, we find that 1 − q6θ4,0 − q3y1θ3,1 − qy12θ2,2 − y13θ1,3 y3 = θ1,1 (q; q)3 (q; q)2 2(1 − q) 6 and, after simplification, with the use of (13.4.4), (13.4.7), (13.4.8), and (13.4.11), q + q3 q 1 θ4 − 1 θ1,1 y3 = (q; q)3ψ2(q) + (1 − q)3ψ6(q) 2 θ4 6 θ1,1 ∞ (2n − 1)q2n−1 = q + q3 − n=1 1 − q2n−1 , (13.5.4) (q; q)3ψ2(q) − q)3ψ6(q) (1 where the final step requires logarithmic differentiation of the product rep- resentations of θ4 and θ1,1. We therefore have a result coinciding with the formula for y3 in (13.1.4). Putting N = 4 in (13.5.1), we find that 1 q10θ5,0 + q6y1θ4,1 + q3y12θ3,2 y4 = (q; q)4 (q; q)3 2(q; q)2 θ1,1 + qy13θ2,3 + qy3θ2,1 + y14θ1,4 + y1y3θ1,2 . 6(1 − q) 1−q 24

13.5 A Formal Power Series 293 Using (13.4.5), (13.4.6), (13.4.7), (13.4.9), (13.4.10), and (13.4.12), we may simplify this latter expression to one involving θ4, θ4, θ1,1, and θ1,1. Thus, after simplification and the use of (13.5.3) and (13.5.4), we deduce that y4 = 2(q + q3)y12 − y1y3 + y14 qθ4 − q θ1,1 = y1y3, (1 − q2)(1 − q3) θ4 3 θ1,1 which is in agreement with (13.1.5). For larger N , we see that yN always appears uniquely in (13.5.1). Indeed, the only term containing yN arises from k1 = · · · = kN−1 = 0 and kN = 1, j = N . This term is therefore −yN θ1,1. Consequently, for N > 1, we see that 1 N (−1)N−j q(N−j)(N−j+1)/2 yN = θ1,1 j=0 (q; q)N−j × (−1)ky1k1 . . . yjkj θN−j+1,k , k1!k2! · · · kj! k1+2k2+···+jkj =j k:=k1+k2+···+kj , ki≥0 excluding k1=···=kN−1=0,kN =1 and proceeding by mathematical induction on N with the use of Corollary 13.4.1, we see that each yN is analytic in q inside the unit circle. Now we turn to (13.5.2). Clearly, the left-hand side of (13.5.2) defines a formal power series in z and q. While it first appears that (zqn+1; q)∞qn2+n might contribute a negative power of q, we observe that by Euler’s series [21, p. 19, equation (2.2.6)], (zqn+1; q)∞ = ∞ (−1)mqm(m+1)/2+mnzm . m=0 (q; q)m Consequently, the exponent on q is n2 + n + m + 1 + mn = m + n + 1 + n + 1 ≥ |n| > 0. 2 2 22 Now (13.5.2) is equivalent to 0 = [z0] 1 dN ∞ N! dzN (−1)n(1 − zy1 − z2y2 − · · · )n(zqn+1; q)∞qn2+n, n=−∞ for every N ≥ 0, where

294 13 Hadamard Products for Two q-Series ∞ [z0] amzm := a0. m=0 To find the formal N th derivative, we need several facts. First [z0] dH (z qn+1 ; q)∞ = ∞ (−1)ii(i − 1) · · · (i − H + 1)zi−H qi(i+1)/2+in dzH (q; q)i [z0] i=H (−1)H H !qH(H+1)/2+Hn =. (q; q)H Next, by the Faa` di Bruno formula, [z0] dM (1 − y1z − y2z2 − · · · )n dzM = M !n(n − 1) · · · (n − k + 1)(−1)ky1k1 y2k2 · · · yMkM . k1!k2! · · · kM ! k1+2k2+···+M kM =M k:=k1+···+kM ,ki≥0 (For an excellent historical exposition of Faa` di Bruno’s formula, see W. John- son’s article [168].) Also, by Leibniz’s rule, dN N N f (N−j)(z)g(j)(z). dzN f (z)g(z) = j=0 j Therefore, [z0] 1 dN ∞ N! dzN (−1)n(1 − zy1 − z2y2 − · · · )n(zqn+1; q)∞qn2+n n=−∞ 1 ∞N N = (−1)nqn2+n N! j n=−∞ j=0 × j!n(n − 1) · · · (n − k + 1)(−1)ky1k1 . . . yjkj k1!k2! · · · kj! k1+2k2+···+jkj =j k:=k1+···+kj , ki≥0 × (−1)N−j (N − j)! q(N−j)(N−j+1)/2+(N−j)n (q; q)N−j N (−1)N−j q(N−j)(N−j+1)/2 (−1)ky1k1 . . . yjkj θN−j+1,k = (q; q)N−j k1!k2! · · · kj! j=0 k1+2k2+···+jkj =j k:=k1+···+kj , ki≥0 = 0, and (13.5.2) is established.

13.6 The Zeros of K∞(zx) 295 13.6 The Zeros of K∞(zx) We know from Section 13.3 that K∞(z) has real, simple, negative zeros z1, z2, z3, . . . with −q1−2n > zn > −q−1−2n, (13.6.1) provided that 0 < q < 1/4. We shall sharpen this inequality. Theorem 13.6.1. For 0 < q < 1/4, −q1−2n > zn > −q−2n. (13.6.2) Proof. We know that (−1)iK∞ − q−2i−1 > 0 from the proof of Theorem 13.2.1. We also need to show that (−1)iK∞ − q−2i > 0. We require Jacobi’s triple product identity from Lemma 1.2.2 of Chapter 1, namely, ∞ (13.6.3) qn2 zn = q2; q2 ∞ −zq; q2 ∞ −z−1q; q2 ∞ . n=−∞ Now, ∞ qn2+nzn K∞(z) = (q; q)n n=0 = 1∞ qn2+nzn qn+1; q ∞ (q; q)∞ n=−∞ 1 ∞ qn2+nzn ∞ (−1)mqm(m+1)/2+mn = (q; q)m (q; q)∞ n=−∞ m=0 1 ∞ qm(2m+1)−m2−mz−m ∞ = (q; q)2m q(n+m)2+n+mzn+m (q; q)∞ m=0 n=−∞ ∞ qm(2m−1)−m2 z−m ∞ − (q; q)2m−1 q(n+m)2 zn+m m=1 n=−∞ = (−q; q)∞(−zq2; q2)∞(−z−1; q2)∞ ∞ qm2 z−m m=0 (q; q)2m − (−q; q)∞(−zq; q2)∞(−z−1q; q2)∞ ∞ qm2−mz−m , (13.6.4) m=1 (q; q)2m−1 by (13.6.3). Hence, for any positive integer i, (−1)iK∞ − q−2i = −(−1)i(−q; q)∞ q1−2i; q2 ∞ q1+2i; q2 ∞ (−1)mqm2−m+2im (q; q)2m−1 ∞ m=1 = q−i2 (q; q2)∞ q2i q2+4i . (13.6.5) 1 − q − (1 − q)(1 − q2)(1 − q3) + · · ·

296 13 Hadamard Products for Two q-Series Now observing that i ≥ 1 and 0 < q < 1/4, we see that the infinite series above is an alternating series and consequently has the lower bound q2i 1 − (1 − q2+2i q3) > 0. 1−q q2)(1 − Hence, for any positive integer i, (−1)iK∞ − q−2i > 0. (13.6.6) We already know that zn lies in the open interval −q−1−2n, −q1−2n . Furthermore, we have just established that K∞(−q1−2n) and K∞(−q−2n) are of opposite signs. Hence, invoking Theorem 13.2.1, we find that K∞(z) must have a zero in −q−2n, −q1−2n , and zn is the only candidate for this role. Thus, Theorem 13.6.1 is proved. Next, we note that the zeros of each Kn(z) are algebraic functions of the coefficients of Kn(z), which are in turn polynomials in q. Thus, in turn, the zeros of Kn(z) are analytic functions of q for 0 < q < 1/4 (the domain specified in Theorem 13.2.1). This is an immediate corollary of the implicit function theorem [283], whose hypotheses are fulfilled here. Finally, for n ≥ N the zeros of Kn(z), say ζn,N , form a decreasing sequence in n (by the interlacing theorem [124, p. 28]). Consequently, ζ∞,N = zN is analytic in q. So, by (13.3.7) and Theorem 13.6.1, 1 > ωn(q) > q. Therefore, we conclude that for some sequence {an,j}, j ≥ 0, ∞ zn = −q−2n an,j qj . j=0 Hence, the equality K∞(zn) = 0 implies that ∞ qh2+h−2nh(−1)h ∞ h 0= an,j qj h=0 (q; q)h j=0 (−1)nq−n2+n ∞ ∞ h (q; q)∞ = (qh+1; q)∞q(h−n)2 (−q)h−n an,j qj h=−∞ j=0 (−1)nq−n2+n ∞ n an,j qj ∞ ∞ h = j=0 (−1)h(qh+n+1; q)∞qh2+h an,j qj . (q; q)∞ h=−∞ j=0 Therefore, ∞ ∞ h (13.6.7) 0 = (−1)h(qh+n+1; q)∞qh2+h an,j qj , h=−∞ j=0

13.8 A New Polynomial Sequence 297 and the an,j are uniquely determined from (13.6.7) in the same way that the yi were determined in Theorem 13.5.1; however, this time we know in ∞ advance that the series n=0 an,j qj converges in 0 < q < 1/4. The change from Theorem 13.5.1 is that now (13.6.7) is valid as an analytic assertion for 0 < q < 1/4 as well as a formal power series identity in q; recall that the derivation of (13.6.7) guarantees that there are no negative powers of q. The reduction of (13.5.2) to a formal power series in q by the replacement of z by qn means that we must have the formal series identity ∞∞ (13.6.8) an,j qj = 1 − yiqni. j=0 i=1 Substituting (13.6.8) and (13.3.7) back into (13.3.6), we conclude the proof of our primary theorem. Entry 13.6.1 (p. 57). Identity (13.1.1) holds for all complex a and real q with 0 < q < 1/4. 13.7 Small Zeros of K∞(z) The primary consequences of our work are the remarkable formulas (13.1.2)– (13.1.5), which provide the series expansions for the zeros zn of K∞(z). For example, z1 = −q−2(1 − q + q2 − 2q3 + 4q4 − · · · ) z2 = −q−4(1 − q2 + q3 − 2q4 + 4q5 − 7q6 + 11q7 − 18q8 + 33q9 − · · · ) z3 = −q−6(1 − q3 + q4 − 2q5 + 4q6 − 7q7 + 11q8 − 17q9 + 27q10 − 43q11 + 68q12 − 112q13 + 196q14 − · · · ) z4 = −q−8(1 − q4 + q5 − 2q6 + 4q7 − 7q8 + 11q9 − 17q10 + 27q11 − 42q12 + 62q13 − 91q14 + 138q15 − 213q16 + 334q17 − 549q18 + 957q19 − · · · ) z5 = −q−10(1 − q5 + q6 − 2q7 + 4q8 + 7q9 + 11q10 − 17q11 + 27q12 − 42q13 + 62q14 − 90q15 + 132q16 − 192q17 + 275q18 − 398q19 + 591q20 − 900q21 + 1417q22 − 2327q23 + 3971q24 − · · · ). 13.8 A New Polynomial Sequence In order to prove (13.1.7), we must study a new sequence of polynomials