inbound2948046473811376725

194 7 Asymptotic Formulas for Continued Fractions Now, 1 √ 1 −1 + √ + 4a − √ 2ax +··· −1 + 1 + 4ae−x = 1 2 2 1 + 4a = ac0 − √ ax 4a + ·· · 1+ and ec1(ae−x)x = 1 + c1(ae−x)x + · · · = 1 + c1(a)x + O(x2), as x → 0. Using the two expansions above in (7.4.7) and displaying only the terms up to the first power of x, which are needed to obtain the next approximation, we set c0(1 + c1x + · · · ) + c0(1 + 2c1x + · · · ) ac0 − √ ax 4a + · ·· − 1 = 0. (7.4.8) 1+ If we equate constant coefficients in (7.4.8), we arrive at c0 + ac20 − 1 = 0, which again yields (7.4.4). If we equate coefficients of x in (7.4.8), we find that c1 + 2ac0c1 − √a = 0. 1 + 4a Solving for c1 and employing (7.4.4), we conclude that c1 = a , (7.4.9) 4a 1 + which is in agreement with what Ramanujan claims in (7.4.1). For the third approximation, set r(a, x) = c0(a)ec1(a)x+c2(a)x2 and use this approximation in (7.4.6). We repeat the procedure detailed above to calculate c2(a). In fact, at this point, we turn to Maple to effect the calcula- tions. After several iterations of (7.4.6), we deduce the asymptotic expansion √ r(a, x) = −1 + 1 + 4a 2a × exp ax − a(1 − a)x2 + a(1 − a)(1 − 14a)x3 + 4a 2(1 + 4a)5/2 6(1 + 4a)4 1 − a(1 − a)(1 − 66a + 378a2 − 20a3)x4 24(1 + 4a)11/2 a(1 − a)(1 − 230a + 4860a2 − 17000a3 + 1984a4)x5 + 120(1 + 4a)7 − a(1 − a)(1 − 726a + 40530a2 − 455740a3 + 1155960a4 − 211776a5)x6 720(1 + 4a)17/2 + a(1 − a)976a6x6 + O(x7) . (7.4.10) 720(1 + 4a)17/2

7.4 An Asymptotic Formula for R(a, q) 195 This establishes (7.4.1), gives further evidence that the coefficient of xn, n ≥ 2, has a(1 − a) as a factor, and indicates that finding a general formula for the coefficient of xn is a daunting task. We now prove the claims about the factors a and 1−a. The assertion about a is trivial to prove. Inducting on n, suppose that cj(0) = 0, 1 ≤ j ≤ n − 1. Then from (7.4.6), exp cn(0)xn + O(xn+1) = 1. It follows that cn(0) = 0. The assertion about the factor 1 − a is deeper, but it follows from Theo- rem 7.3.1. In fact, our proof in the previous section gives a slightly stronger result, which we now state for only the product representation for the Rogers– Ramanujan continued fraction. For every positive number N > 0, as x → 0+, R(1, e−x) = (e−x; e−5x)∞(e−4x; e−5x)∞ = √ 1 x + O(xN ) . (e−2x; e−5x)∞(e−3x; e−5x)∞ 5 − 1 exp 5 2 (7.4.11) Comparing (7.4.11) with (7.4.1), we conclude that cn(1) = 0 for every n ≥ 2, since N > 0 can be made arbitrarily large. This completes the proof of Entry 7.4.1. The ideas used to prove Entry 7.4.1 can be applied to the generalized cubic continued fraction C(a, e−x) := 1 ae−x + a2e−2x ae−2x + a2e−4x ae−3x + a2e−6x ···, 1 1 + 1 1 + + + (7.4.12) where a is any complex number and x > 0. The continued fraction (7.4.12) generalizes Ramanujan’s cubic continued fraction [112] C (q ) := 1 q + q2 q2 + q4 q3 + q6 = (q; q6)∞ (q5; q6 )∞ , 1 1 1 1 (q3; q6)2∞ + + + + ··· where |q| < 1. Theorem 7.4.1. As x → 0+, C(a, e−x) ax − a(1 − a)x2 + a(1 − a)(1 − 12a − 4a2)x3 − ··· . = 1 ×exp + 2a 2(1 + 2a)3 6(1 + 2a)5 1 a+1 Moreover, each term of the asymptotic expansion beginning with the second has a factor of a(1 − a). See the paper [84] by Berndt and Yee for more details.

8 Ramanujan’s Continued Fraction for (q2; q3)∞/(q; q3)∞ 8.1 Introduction In Chapter 6, we proved some general theorems on continued fractions from the lost notebook that yielded several beautiful examples as special cases, in particular, the Rogers–Ramanujan continued fraction, the Ramanujan– Go¨llnitz–Gordon continued fraction, and Ramanujan’s cubic continued frac- tion. In Chapter 7, we considered asymptotic formulas for continued fractions, but one of the examples on which we focused in that chapter does not fall under the purview of the general theorems in Chapter 6. Our goal in this chapter is to prove two remarkable theorems for this continued fraction (q2; q3)∞ = 1 − q − q3 − q5 − ···, |q| < 1. (8.1.1) (q; q3)∞ 1 1+q 1 + q2 1 + q3 The continued fraction (8.1.1) is due to Ramanujan and is found in his second notebook [227, p. 290]. Of the many q-continued fractions found by Ramanu- jan, (8.1.1) is, by far, the most difficult to prove. Up until recently, the only known proof was found by Andrews, Berndt, L. Jacobsen, and R.L. Lamphere [39], [63, p. 46, Entry 19] in 1992, which uses a deep theorem of Andrews [17]. However, a considerably shorter and more natural proof was recently given by Andrews, Berndt, J. Sohn, A.J. Yee, and A. Zaharescu [40]. On page 45 in his lost notebook, Ramanujan claims, in an unorthodox fashion, that a certain q-continued fraction possesses three limit points. More precisely, he asserts that as n tends to ∞ in the three residue classes modulo 3, the nth partial quotients tend, respectively, to three distinct limits, which he explicitly gives. In fact, Ramanujan claims that a more general continued fraction has three distinct limits under the broader concept of “general con- vergence,” which was not defined in the literature until about 70 years later. If ω = e2πi/3, then, except for the simplification of notation, Ramanujan [228, p. 45] claimed that for |q| < 1,

198 8 Ramanujan’s Continued Fraction for (q2; q3)∞/(q; q3)∞ lim 11 1 1 n→∞ 1 − 1 + q − 1 + q2 − · · · − 1 + qn + a = −ω2 Ω − ωn+1 · (q2; q3)∞ , (8.1.2) Ω − ωn−1 (q; q3)∞ where Ω := 1 − aω2 (ω2q; q)∞ . 1 − aω (ωq; q)∞ (8.1.3) After (8.1.2), Ramanujan appended the note, “Numerators and Denominators can be equated separately.” Of course, because of the appearance of the limiting variable n on the right side of (8.1.2), Ramanujan’s claim is meaningless as it stands. But after a few minutes of reflection, we readily conclude that Ramanujan was affirming that there are three distinct limits depending on the congruence class modulo 3 in which n → ∞. In the note after (8.1.3), Ramanujan evidently asserted that the limits can be obtained by determining separately the limits of both the partial numerators and denominators. Ramanujan’s claim is very interesting for several reasons. First, if a = 0, the left side of (8.1.2) is a continued fraction (in the normal sense) that diverges. We prove that the three partial quotients tend to the required limits if n is restricted to any one of the three residue classes modulo 3. This is in contrast to the classical result from the general theory of continued fractions, which asserts that if all the elements of a divergent continued fraction are positive, then the even and odd approximants approach distinct limits [182, pp. 96–97]. Second, if a = 0, we prove that the continued fraction in (8.1.2) converges “generally” in the sense that when n is confined to any one of the three residue classes modulo 3, the limit of the left side indeed exists and is equal to that claimed on the right side of (8.1.2) in each of the three cases. The concept of general convergence is due to L. Jacobsen [167] in 1986. See also her book with H. Waadeland [182, pp. 41–44]. For some results of Ramanujan of a different kind on general convergence, see Chapter 5. Thus, we have one further example of Ramanujan’s having discovered a fundamental concept long ahead of his time, before anyone else ever thought of it. Third, note that the continued fraction (8.1.1) can be written in the equiv- alent form (q2; q3)∞ = 1 − 1 1 − 1 − 1 − ···. (8.1.4) (q; q3)∞ 1 q−1 + q−2 + 1 q−3 + 1 Thus, when a = 0, the continued fraction on the left side of (8.1.2) is the same as the continued fraction of (8.1.4), but with q replaced by 1/q. Observe that, remarkably, (q2; q3)∞/(q; q3)∞ also appears in the three limits on the right side of (8.1.2). In this sense, Ramanujan’s result (8.1.2) is analogous to his theorem on the divergence of the Rogers–Ramanujan continued fraction found

8.2 A Proof of Ramanujan’s Formula (8.1.2) 199 on pages 374 and 382 in his third notebook [227], which was first proved by Andrews, Berndt, Jacobsen, and Lamphere [39], [63, p. 30, Entry 11]. In the latter result, Ramanujan explicitly determines the limits of the even and odd indexed approximants of the divergent Rogers–Ramanujan continued fraction for |q| > 1 and shows that these limits can be expressed in terms of the Rogers–Ramanujan continued fraction itself, but at different arguments. Thus, our first important goal in this chapter is to give a proof of (8.1.2), which we think is one of the most fascinating results in Ramanujan’s lost notebook. Our proof is taken from the paper of Andrews, Berndt, Sohn, Yee, and Zaharescu [41], in which the authors also establish general theorems pro- viding classes of continued fractions with three distinct limit points. However, Ramanujan’s result (8.1.2) is deeper and does not come under the umbrella of the general theorems of [41]. At the top of page 45 in his lost notebook [228], Ramanujan states separately the special case of (8.1.2) when a = ω. This can be proved in a more elementary fashion, and we do so in the section following our proof of (8.1.2). The second major purpose of this chapter is to prove another asymptotic formula for (8.1.1), which has a flavor different from that proved in Chapter 7 and which is also found on page 45 of the lost notebook. In fact, the con- tinued fraction examined by Ramanujan is slightly more general than (8.1.1). Although both (8.1.1) and its generalization do not converge for q > 1, Ra- manujan claims that his asymptotic formula is valid as q → 1 from both direc- tions. However, the continued fraction satisfies a simple difference equation, which is given by Ramanujan immediately preceding the asymptotic formula. Thus, Ramanujan’s asymptotic formula should be more properly interpreted as an asymptotic formula for solutions of this difference equation, which does not have a unique solution. Therefore, a sequence of arbitrary constants arises in Ramanujan’s asymptotic formula. If q > 1, as discussed above, the contin- ued fraction in (8.1.1) has three limit points, and so it would not be possible in any way to prescribe values to these arbitrary constants. 8.2 A Proof of Ramanujan’s Formula (8.1.2) We first introduce needed notation. Define P0(a) = 0, P1(a) = 1, Q0(a) = 1, Q1(a) = 1, (8.2.1) and for N ≥ 2, set PN (a) = 1 − 1 − 1 − ··· − 1 . (8.2.2) QN (a) 1 1+q 1 + q2 1 + qN−1 +a From the general theory of continued fractions [182, p. 9, equation (1.2.9)], for N ≥ 2, the partial numerators PN (0) and QN (0) satisfy the recurrence relations

200 8 Ramanujan’s Continued Fraction for (q2; q3)∞/(q; q3)∞ PN (0) = (1 + qN−1)PN−1(0) − PN−2(0), (8.2.3) QN (0) = (1 + qN−1)QN−1(0) − QN−2(0), where P0(0), P1(0), Q0(0), and Q1(0) are defined by (8.2.1). To prove (8.1.2), our first task will be to derive explicit formulas for PN (0) and QN (0). To do so, we need to recall the definition of the Gaussian polyno- mials and two versions of the q-binomial theorem [21, pp. 35–36]. Lemma 8.2.1. If [ n ] denotes the Gaussian polynomial defined by m n n ⎧ (q; q)n , if 0 ≤ m ≤ n, m m ⎨ otherwise, := := (q; q)m(q; q)n−m ⎩0, q then N N (−1)j zj qj(j−1)/2, (8.2.4) j (8.2.5) (z; q)N = j=0 1∞ N +j−1 zj. = j (z; q)N j=0 Lemma 8.2.2. Let N − 1 = 3v + , where = 0, ±1. Then (−1)vPN (0) = ∞ (−1)( −n−r)/3qn(n+1)/2+r(r−1)/2 n,r=0 n+r≡ (mod 3) × n+1 n+ 1 (N − 1 − (n + r)) . (8.2.6) r 3 n q3 Proof. Recall from (8.2.3) and (8.2.1) that PN (0) satisfies the recurrence relation PN (0) = (1 + qN−1)PN−1(0) − PN−2(0), N ≥ 2, (8.2.7) and the initial conditions P0(0) = 0 and P1(0) = 1. Define ∞ F (t) := PN (0)tN . N =1 Multiplying the recurrence relation (8.2.7) by tN and summing over N ≥ 2, we obtain F (t) − t = tF (t) + tF (tq) − t2F (t). So, tt F (t) = 1 − t + t2 + 1 − t + t2 F (tq).