FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK. INTRODUCTION On pages and 7 in his Lost Notebook [3], Ramanujan recorded four identities involving the rank generating function. Of course, Ramanujan would not have used this terminology, because the rank of a partition was not defined until 944 by F. J. Dyson []. He defined the rank of a partition to be the largest part minus the number of parts. For example, the rank of the partition 4 + is 4. Let N(m, n) denote the number of partitions of the positive integer n with rank m. Dyson showed that the generating function for N(m, n) is given by m N(m, n)q n z m q n (zq) n (q/z) n : G(z, q), q <. (.) Ramanujan s four identities involve special cases of G(z, q). Here, we use the standard notation (a) n : (a; q) n : ( a)( aq) ( aq n ), n, (a) 0 : (a; q) 0 :. In the sequel, we also use the notation (a, a,..., a m ; q) n : (a ; q) n (a ; q) n (a m ; q) n, n 0, (a) : (a; q) : lim n (a; q) n, (a, a,..., a m ; q) : (a ; q) (a ; q) (a m ; q), [a; q] n : (a; q) n (q/a; q) n, n 0, a 0, [a, a,..., a m ; q] n : [a ; q] n [a ; q] n [a m ; q] n, n 0, [a] : [a; q] : lim n [a; q] n, [a, a,..., a m ; q] : [a ; q] [a ; q] [a m ; q]. Throughout this paper, q <. To state the aforementioned four identities of Ramanujan, we need to define Ramanujan s theta function ψ(q), ψ(q) : q n(n+)/ ( q; q) (q; q) (q ; q ) (q; q ), (.) The third author was partially supported by the Singapore Ministry of Education Academic Research Fund, Tier, project number MOE04-T--05, ARC40/4.
GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK by the Jacobi triple product identity (given in its general form in (3.) below) and Euler s theorem. Appearing in each of the four identities are instances of q n f a (q) : ( + aq + q )( + aq + q 4 ) ( + aq n + q n ), (.3) where a is any real number. Observe that ( ) a ± a 4 f a (q) G, q, where G(z, q) is defined in (.). A focus in this paper is the special case when a, for which we can write Furthermore, define f (q) φ(q) : q n (e 3πi/4 q) n (e 5πi/4 q) n G(e 3πi/4, q). (.4) q n ( q ; q ) n q n (iq) n (q/i) n G(i, q), (.5) which is featured in Ramanujan s fourth identity. We are now ready to state the four identities of Ramanujan, which were first proved in a wonderful paper by H. Yesilyurt [6]. Entry.. [3, p. ] Suppose that a and b are real numbers such that a + b 4. Recall that f a (q) is defined by (.3). Then b a + 4 f a ( q) + b + a + f a ( q) b 4 f b(q) (q4 ; q 4 ) ( q; q ) n bq n + q n. (.6) + (a b )q 4n + q8n Entry.. [3, p. ] Let a and b be real numbers with a + ab + b 3. Then, with f a (q) defined by (.3), (a + )f a (q) + (b + )f b (q) (a + b )f a+b (q) 3 (q3 ; q 3 ). (.7) (q; q) + ab(a + b)q 3n + q6n n Entry.3. [3, p. 7] Let f a (q) and ψ(q) be defined by (.3) and (.), respectively. Then + 3 f ( q) + 3 + 3 q n f ( q) 6 ( + 3q + q ) ( + 3q n + q n ) + 3 ψ( q) (q4 ; q 4 ) (q 6 ; q 6 ) +. (.8) 3q n + qn n
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 3 Entry.4. [3, p. 7] Let φ(q) be defined by (.5) and ψ(q) be defined by (.). Then ( + eπi/4 ) φ(iq) + ( + e πi/4 ) φ( iq) f (q) + ψ( q)( q ; q 4 ) n +. (.9) q n + qn Yesilyurt s proofs [6] of Entries..4 depend upon the following famous lemma of A. O. L. Atkin and H. P. F. Swinnerton-Dyer. Lemma.5. Let q, q <, be fixed. Suppose that ϑ(z) is an analytic function of z, except for possibly a finite number of poles, in every annulus 0 < z z z. If ϑ(zq) Az k ϑ(z) for some integer k (positive, negative, or 0) and some constant A, then either ϑ(z) has k more poles than zeros in the region q < z, or ϑ(z) vanishes identically. Since it is very unlikely that Ramanujan would have given proofs of Entries..4 using complex analysis, in particular, using Lemma.5, the primary purpose of this paper is to give completely different proofs using q-series, perhaps more in line with what Ramanujan might have devised. However, although our proofs of Entries..3 are simple, our proof of Entry.4 is much more difficult. Our proof of Entry.4 relies on the following -dissections for two special cases of the rank generating function G(z, q), when z i and when z is a primitive eighth root of unity. These two -dissections of the rank, with their immediate consequences, comprise a second major focus of this paper. Theorem.6. The -dissection of the rank function G(i, q) is given by q n ( ) n q 4n +8n [q4 ; q 6 ] (q 6 ; q 6 ) (.0) (iq) n (q/i) n (q 6 ; q 6 ) + q 6n+ [ q, q, q 6 ; q 6 ] n ( ) n q 4n +4n+5 + + q [q4 ; q 6 ] (q 6 ; q 6 ). (q 6 ; q 6 ) + q 6n+6 [ q, q 6, q 6 ; q 6 ] n Theorem.7. Let a be a primitive eighth root of unity. Then q n a /a ( ) n q 4n +8n + (a + /a )[q4 ; q 6 ] (q 6 ; q 6 ) (aq) n (q/a) n (q 6 ; q 6 ) q 6n+ [q, q, q 6 ; q 6 ] n + q [q4 ; q 6 ] (q 6 ; q 6 ) [q, q 6, q 6 ; q 6 ] + a + /a (q 6 ; q 6 ) n The proofs of Theorems.6 and.7 will be given in Sections 4 6.. PROOFS OF ENTRIES..3 ( ) n q 4n +4n+5 q 6n+6. (.) Our starting point is a corollary of Lemma.3. from [3, p. 9]. (It is to be assumed in the sequel that parameters, such as z and ζ below, are chosen so that all relevant expressions are well defined.)
4 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK Theorem.. For any complex numbers z, ζ, Proof. Define R(z, ζ, q) : ζ S(z, ζ, q) : ζ 3 n ζ (ζ, q/ζ ; q) (ζ, q/ζ; q) ( ) n ζ 3n q n(3n+)/ zζq n + ζ n n ( ) n q n(3n+)/ zq n ( ) n ζ 3n q n(3n+)/ zq n /ζ z(ζ, q/ζ, ζ, q/ζ, q, q; q) (z/ζ, qζ/z, z, q/z, zζ, q/(zζ); q). (.) n ζ (ζ, q/ζ ; q) (ζ, q/ζ; q) Then from [3, p. 9, Lemma.3.], S(z, ζ, q) ( ) n ζ 3n q 3n(n+)/ zζq n + n n ( ) n q 3n(n+)/ zq n. (ζ, q/ζ, ζ, q/ζ, q, q; q) (z/ζ, qζ/z, z, q/z, zζ, q/(zζ); q). ( ) n ζ 3n q 3n(n+)/ zq n /ζ Hence, to conclude the proof of Theorem., we are required to show that S(z, ζ, q) R(z, ζ, q). z Using the pentagonal number theorem in the second equality below, we find that ( S(z, ζ, q) ζ 3 ( ) n ζ 3n q n(3n+)/ zζq n zζ ) zζ ( zζqn ) n ( ( ) n ζ 3n q n(3n+)/ ζ + zq n /ζ z ζ ( zζ )) z qn n ζ(ζ, q/ζ ; q) ( ) n q n(3n+)/ (ζ, q/ζ; q) zq n n ( z R(z, ζ, q) ζ ζ ( ) n ζ 3n q n(3n+)/ z n ) + ( ) n ζ 3n q n(3n+)/ (ζ, q/ζ ; q) (q; q) (ζ, q/ζ; q) n ( z ) z ( zqn ) z R(z, ζ, q), because the expression inside the large parentheses equals 0, as it is a formulation of the quintuple product identity [3, p., eq. (8..8)], [8, p. 8]. In particular, if we take the
formulation from [8, p. 8] FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 5 n q 3n +n (z 3n q 3n z 3n q 3n+ ) (q ; q ) (qz; q ) (q/z; q ) (z ; q 4 ) (q 4 /z ; q 4 ), (.) replace q by q, and then set z ζ q, we find that the sum of the expressions within large parentheses on the far right side above equals 0. We frequently use the observation that if a t /t, then q n t ( ) n q n(3n+)/, (.3) (tq) n (q/t) n (q) tq n by [, p. 63, eq. (..3)]. n Proof of Entry.. First we note that we may parameterize the circle a + b 4 by and with z e iθ, we find that Hence, b cos θ, a sin θ, b z z and a i(z z ). (.4) a b z 4 z 4. (.5) Let us now set ζ i ( ) in Theorem.. Thus, by (.3), (.4), and (.5), ( ) n i 3n q n(3n+)/ ( ) n i 3n q n(3n+)/ + i ziq n + ziq n n n i(, q; q) (q; q) (i, iq; q) ( z) f b(q) z(i, iq,, q, q, q; q) ( iz, qi/z, z, q/z, iz, iq/z; q) z( i)( q; q) (q 4 ; q 4 ) ( bq n + q n ) ( + z )( z)( q; q ) ( + (a b )q 4n + q 8n ). (.6) Multiply both sides of (.6) by n ( + z )( z) z( i)( q; q). Upon doing so, we then see that the right-hand side of (.6) becomes the right-hand side of (.6). The third expression on the left-hand side of (.6) then becomes ( + z )( z)( i)(, q; q) (q; q) z( i)( q; q) (i, iq; q) ( z) f b(q) b f b(q).
6 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK Therefore, we will complete the proof if we can show that ( ( + z )( z) ( ) n i 3n q n(3n+)/ z( i)( q; q) ziq n n ) ( ) n i 3n q n(3n+)/ +i + ziq n n b a + 4 b a + 4 + b + a + 4 f a ( q) + b + a + f a ( q) 4 ( + iz) ( q; q) n ( iz) ( q; q) n ( ) n ( q) n(3n+)/ + iz( q) n ( ) n ( q) n(3n+)/ iz( q) n, (.7) where we have twice used (.3). Proving (.7) is equivalent to proving that ( ) n i 3n q n(3n+)/ ( ) n i 3n q n(3n+)/ + i ziq n + ziq n i n n ( ) n ( q) n(3n+)/ + iz( q) n n n ( ) n ( q) n(3n+)/ iz( q) n. (.8) Combining sums on each side of (.8), we see that our task has been reduced to proving that ( ) n q n(3n+)/ ( i 3n ( + ziq n ) + i 3n ( ziq n ) ) + z q n n n and this follows immediately because and ( ) n q n(3n+)/ + z q n ( i( ) n(3n+)/ ( iz( q) n ) ( ) n(3n+)/ ( + iz( q) n ) ), i 3n + i 3n ( ) n(3n+)/ (i ) i +3n i 3n ( ) n(3n+)/+n ( + i). The last two assertions are most easily proved by noting that each expression is periodic with period 4, and that the assertions hold for n 0,,, 3. Proof of Entry.. First we note that we may parameterize the ellipse a + ab + b 3 by a cos ( θ + 3 π), b cos θ. So with z e iθ, we find that where ω e πi/3. Hence, b z + z and a zω + (zω), a + b zω (zω ) (.9)
and FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 7 ab(a + b) z 3 z 3. Therefore, we now set ζ ω in (.). Thus, the resulting right-hand side, by (.9), equals z( ω)( ω )(q 3 ; q 3 ) z(q; q) 3(q 3 ; q 3 ) ( z 3 )(z 3 q 3, z 3 q 3 ; q 3 ) ( z 3 ) (q; q) n ( + ab(a + b)q3n + q 6n ). (.0) We now observe that the latter quotient on the right-hand side of (.0) is the same as the right-hand side of (.7). We are thus led to multiply the left-hand side of (.) with ζ ω by ( z 3 ) z(q; q) to deduce, with the help of three applications of (.3), that ( ( z 3 ) ω ( ) n q n(3n+)/ ( ) n q n(3n+)/ + ω z(q; q) zωq n zω q n n ω( ω ) ( ω) n n ) ( ) n q n(3n+)/ zq n ( z3 )ω z( ωz) f a(q) + ( z3 )ω z( ω z) f a+b(q) ( z3 )ω( ω ) z( ω)( z) f b(q) (a + )f a (q) (a + b )f a+b (q) + (b + )f b (q), which is the left-hand side of (.7). This completes the proof. Proof of Entry.3. Let a and b 3 in Entry. to deduce that + 3 f ( q) + 3 + 3 3 f ( q) 4 4 f 3 (q) (q4 ; q 4 ) 3q n + q n. (.) ( q; q ) + q 4n + q8n n Now multiply both sides of (.) by / 3 to arrive at 3 + 3 6 f ( q) + + 3 f ( q) f 3 (q) 3 (q 4 ; q 4 ) ( q; q ) Examining (.8) and (.), we see that we are required to show that ψ( q) (q6 ; q 6 ) ( q; q ) n n 3q n + q n. (.) + q 4n + q8n ( 3q n + q n )( + 3q n + q n ) + q 4n + q 8n. (.3)
8 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK To that end, (q 6 ; q 6 ) ( q; q ) (q6 ; q 6 ) ( q; q ) (q6 ; q 6 ) ( q; q ) ( 3q n + q n )( + 3q n + q n ) + q 4n + q 8n n q n + q 4n + q 4n + q 8n n n ( q n ) ( + q n + q 4n )( q n ) (q6 ; q 6 ) ( q; q ) (q ; q ) (q 6 ; q 6 ) (q ; q ) ( q; q ) ψ( q), by (.). Thus, we have shown (.3), and so the proof of Entry.3 is finished. 3. PROOF OF ENTRY.4; PART We show in this section that Entry.4 follows from the two -dissections for two special cases of the rank generating function G(z, q) given in Theorems.6 and.7. Proof of Entry.4. We need knowledge of theta functions. After Ramanujan, set f(a, b) : a n(n+)/ b n(n )/ ( a; ab) ( b; ab) (ab; ab), ab <, (3.) n where the latter equality is the Jacobi triple product identity [7, p. 35, Entry 9]. To simplify the product on the right side of (.9), use (.) and (3.) to deduce that ψ( q)( q ; q 4 ) n + q n + q n (q 4 ; q 8 ) (qe πi/4, qe πi/4, q; q) (q ; q 4 ) (q 4 ; q 8 ) f( e πi/4, qe πi/4 ). (3.) (q ; q 4 ) ( e πi/4 ) We need a special case of an identity of Ramanujan for theta functions [7, p. 48, Entry 3]. To that end, if U n : a n(n+)/ b n(n )/ and V n : a n(n )/ b n(n+)/ for each integer n, then n ( ) Un+r f(u, V ) U r f. (3.3) r0 U r, V n r U r We apply (3.3) with a U e πi/4, b V qe πi/4, and n 4 to the theta function in the last equality of (3.). Thus, f( e πi/4, qe πi/4 ) f( q 6, q 0 ) e πi/4 f( q 0, q 6 ) + iqf( q 4, q ) + e πi/4 q 3 f( q 8, q ) ( e πi/4 )f( q 6, q 0 ) + (i e πi/4 )qf( q, q 4 ).
Hence, f( e πi/4, qe πi/4 ) ( e πi/4 ) FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 9 f( q 6, q 0 ) ( + )f( q 4, q ) (q 6 ; q 6 ) (q 0 ; q 6 ) (q 6 ; q 6 ) ( + )q(q ; q 6 ) (q 4 ; q 6 ) (q 6 ; q 6 ) [q 6 ; q 6 ] (q 6 ; q 6 ) ( + )q[q ; q 6 ] (q 6 ; q 6 ), (3.4) where we made two applications of (3.). Hence, inserting (3.4) into (3.), we deduce that ψ( q)( q ; q 4 ) (q 4 ; q 8 ) (q ; q 4 ) [q 4 ; q 6 ] (q 6 ; q 6 ) [q, q, q 6 ; q 6 ] + q n n + q n ([q 6 ; q 6 ] (q 6 ; q 6 ) ( + ) )q[q ; q 6 ] (q 6 ; q 6 ) ( + ) q [q4 ; q 6 ] (q 6 ; q 6 ). [q, q 6, q 6 ; q 6 ] Therefore, identity (.9) is equivalent to f (q) ( + ) ( φ(iq) + φ( iq)) [q 4 ; q 6 ] (q 6 ; q 6 ) [q, q, q 6 ; q 6 ] + ( i φ(iq) i φ( iq) ) ( + + ) q [q4 ; q 6 ] (q 6 ; q 6 ). (3.5) [q, q 6, q 6 ; q 6 ] We now apply Theorem.6 twice, with q replaced by iq and iq, obtaining, by (.5), φ(iq) and φ( iq), respectively. We next apply Theorem.7 with a e 3πi/4 thereby obtaining, by (.4), f (q). If we substitute these three representations into (3.5), we see indeed that (3.5) is valid. It therefore remains to prove Theorems.6 and.7, which we do in the following sections. 4. PROOF OF ENTRY.4; PART, IDENTITIES FOR THETA FUNCTIONS AND LAMBERT SERIES We offer the following lemmas that are needed to simplify the dissections in the proofs of Theorem.6 and.7. Lemma 4.. We have (q6 ; q 6 ) ( ) [ q 6 ; q 6 ] (q) (q ; q ) + q[ q ; q 6 ]. (4.) Proof. By (.) and (3.3) with a q, b q 3, and n, (q) (q ; q ) (q6 ; q 6 ) (q ; q ) (q ; q ) (q; q ) (q ; q ) n q n(n+) ( [ q 6 ; q 6 ] + q[ q ; q 6 ] ),
0 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK and so the proof is complete. Next, we state Halphen s identity [4, p. 87] in the form discovered and presented in [5], H(a, b, c, q) : [ab, bc, ca] (q) [a, b, c, abc] + F (a, q) + F (b, q) + F (c, q) F (abc, q), (4.) where F (x) : F (x, q) : k0 xq k xq q k /x, q <. (4.3) k q k /x Below are three identities that are consequences of Halphen s identity [0, Cor. 4.4]. Corollary 4.. We have k H(a, b, c, q ) H(a, b, d, q ) H(c, /d, abd, q ), (4.4) H(a, a, q/a, q ) + H(b, b, q/b, q ) H(a, q/a, b, q ), (4.5) H(a, a, q/a, q ) H(b, b, q/b, q ) H(a, q/a, b/q, q ). (4.6) Setting r 0 and s 3 in [9, Theorem.] and then replacing q by q 6, we derive the generalized Lambert series identity (q 6 ; q 6 ) [b, c, d; q 6 ] [c/b, d/b; q 6 ] + [b/c, d/c; q 6 ] + [b/d, c/d; q 6 ] n n n ( ) n q 4n(n+) bq 6n ( ) n q 4n(n+) cq 6n ( ) n q 4n(n+) dq 6n ( b ) n cd ( c ) n bd ( ) d n. bc Multiplying both sides by [c/b, d/b; q 6 ] and rearranging, we deduce the following lemma. Lemma 4.3. We have n Let S(z, ζ) : ( ) n q 4n(n+) bq 6n ( ) b n c [d/b; q 6 ] cd b [d/c; q 6 ] (q6 ; q 6 ) [c/b, d/b; q 6 ] [b, c, d; q 6 ] n ( ) n q 4n(n+) ζ n + d b n [c/b; q 6 ] [c/d; q 6 ] ( ) n q 4n(n+) cq 6n n zq 6n and S (ζ) : ( c bd ( ) n q 4n(n+) dq 6n n ) n ( ) d n. (4.7) bc ( ) n q 4n(n+) ζ n q 6n, (4.8)
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS where the prime in the second sum denotes the omission of the term n 0. Then by substituting (b, c, d) (q 8, q, q ) and (q 8, q, q 6 ) in Lemma 4.3, we find that, respectively, q 6 S(q 8, q ) q 4 S(q, q 4 ) (q6 ; q 6 ) [ q 6 ; q 6 ] [ q, q 8 ; q 6 ] q S(q 8, q 0 ) q S(q, q ) q 4 (q6 ; q 6 ) [ q ; q 6 ] [ q 6, q 8 ; q 6 ] [q4 ; q 6 ] [ q ; q 6 ] [q4 ; q 6 ] [ q 6 ; q 6 ] S( q, q 6 ), n ( ) n q 4n 8n + q 6n 6 (4.9) q 4 (q6 ; q 6 ) [ q ; q 6 ] [ q 6, q 8 ; q 6 ] + q 6 [q4 ; q 6 ] [ q ; q 6 ] S( q 0, q 6 ), (4.0) where we replaced n by n + in the previous line. Similarly, substituting (b, c, d) (, q 4, q 8 ), ( q 8, q, q ), (, q, q 6 ), and ( q 4, q 8, q 6 ) in Lemma 4.3, we obtain, respectively, S(, q ) q S( q 4, q 0 ) (q6 ; q 6 ) [ q, q 4, ; q 6 ] [, q, q 4 ; q 6 ] q 3 [q4 ; q 6 ] [ q ; q 6 ] S(q 8, q 3 ) (q6 ; q 6 ) [ q, q 4, ; q 6 ] [, q, q 4 ; q 6 ] + [q4 ; q 6 ] [ q ; q 6 ] S(q, q 6 ), (4.) q 6 S( q 8, q ) q 4 S( q, q 4 ) (q6 ; q 6 ) [q 4, q 6 ; q 6 ] [q, q 8, q ; q 6 ] [q4 ; q 6 ] [ q 6 ; q 6 ] S(q, q 6 ), (4.) S(, q 8 ) q 8 S( q, q 8 ) (q6 ; q 6 ) [q 4, q 6 ; q 6 ] [, q 4, q 6 ; q 6 ] q 6 [q4 ; q 6 ] [ q 6 ; q 6 ] S(q 6, ), S( q 4, q 6 ) q 6 S( q 8, q 6 ) (q6 ; q 6 ) [ q, q 4 ; q 6 ] [ q 4, q 6, q 8 ; q 6 ] q [q4 ; q 6 ] [ q ; q 6 ] S(q 6, ). Lemma 4.4. We have n ( ) n q 4n(n+) q 6n ( ) n c [d; q6 ] cd [d/c; q 6 ] n ( ) n q 4n(n+) cq 6n ( c ( cq6n cq + q6n+6 /c 6n q 6n+6 /c dq6n dq + q6n+6 /d 6n q 6n+6 /d ( ) d n, + d [c; q6 ] [c/d; q 6 ] n ( ) n q 4n(n+) dq 6n where the prime on the sum on the left-hand side denotes the omission of the term n 0. c d ) n ) (4.3) (4.4)
GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK Proof. Multiply both sides of (4.7) by (b ), differentiate with respect to b, and let b. We find that the only terms that remain are those in Lemma 4.4. Lemma 4.5. Recall that S(z, ζ) and S (ζ) are defined in (4.8). Then + S ( q ) S ( q 4 ) q S(q 4, q 0 ) + q 4 S(q 4, q ) (4.5) [q4 ; q 6 ] S( q, q 6 ) + q 6 [q4 ; q 6 ] S( q 0, q 6 ) [q 4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ). [ q ; q 6 ] [ q 6 ; q 6 ] [ q, q, q 6, q 6 ; q 6 ] Proof. Substituting c q 4 and d q 8 in Lemma 4.4, we find that S ( q ) q S(q 4, q 0 ) q q 3 [q4 ; q 6 ] [ q ; q 6 ] + ( q6n+4 q ( q6n+4 q6n+ q6n+8 q6n + + 6n+4 q6n+ + q6n+8 n ( ) n q 4n +56n + q 6n+8 q + q + q 6n q6n+ q6n+ q6n+4 + + 6n+4 6n+ 6n+ ) + q 6n+4 + [q4 ; q 6 ] [ q ; q 6 ] S( q, q 6 ). (4.6) Similarly, substituting c q 4 and d q 0 in Lemma 4.4 gives S ( q 4 ) q 4 S(q 4, q ) q ( q6n+4 q6n+ q6n+0 q6n+6 + + 6n+4 q6n+ + q6n+0 + q 6n+6 q 6 [q4 ; q 6 ] S( q 0, q 6 ). (4.7) [ q 6 ; q 6 ] Taking the difference between (4.6) and (4.7), we obtain S ( q ) S ( q 4 ) q S(q 4, q 0 ) + q 4 S(q 4, q ) ( ) q 6n+ q6n+6 q6n+0 q6n+4 + + + q6n+ + q6n+6 + q6n+0 + q 6n+4 + [q4 ; q 6 ] S( q, q 6 ) + q 6 [q4 ; q 6 ] S( q 0, q 6 ). (4.8) [ q ; q 6 ] [ q 6 ; q 6 ] Note that by Halphen s identity (4.) with q replaced by q 6 and (a, b, c) ( q, q, q 6 ), + ( ) q 6n+ q6n+6 q6n+0 q6n+4 + + q6n+ + q6n+6 + q6n+0 + q 6n+4 [q 4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ). [ q, q, q 6, q 6 ; q 6 ] ) )
Therefore, (4.8) is equivalent to FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 3 + S ( q ) S ( q 4 ) q S(q 4, q 0 ) + q 4 S(q 4, q ) [q4 ; q 6 ] S( q, q 6 ) + q 6 [q4 ; q 6 ] S( q 0, q 6 ) [q 4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ), [ q ; q 6 ] [ q 6 ; q 6 ] [ q, q, q 6, q 6 ; q 6 ] which is identical with (4.5). 5. PROOF OF ENTRY.4; PART 3, PROOF OF THEOREM.6 Proof of Theorem.6. Invoking the partial fraction identity (.3) with t i, we find that the left side of (.0) is q n i ( ) n q n(3n+)/ (iq) n (q/i) n (q) iq n n i ( ) n q n(3n+)/ ( + iq n ) (q) + q n n (q) n n ( ) n q n(3n+)/ [( + q n ) i( q n )] + q n. (5.) Replacing n by n and multiplying both the numerator and denominator by q n, we see that ( ) n q n(3n+)/ ( ) n q n(3n+3)/. + q n + q n Therefore, the identity (5.) simplifies to q n (iq) n (q/i) n (q) n n n ( ) n q n(3n+)/ + q n. (5.) We focus our attention on the series on the right side above. By subdividing the index of summation into residue classes modulo 4 and using the definitions (4.8), we find that ( ) n q n(3n+)/ q 4n +n + q n + q q 4n +4n+ 8n + q 8n+ + n + + n n n q 4n +6n+7 + q 8n+4 n q 4n +n ( q 8n ) q 6n n q 4n +38n+5 + q 8n+6 n q 4n +6n+7 ( q 8n+4 ) q 6n+8 q 4n +4n+ ( q 8n+ ) q 6n+4 n q 4n +38n+5 ( q 8n+6 ) q 6n+ + S ( q ) S ( q 4 ) q S(q 4, q 0 ) + q 4 S(q 4, q )
4 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK + q 7 S(q 8, q ) q S(q 8, q 0 ) q 5 S(q, q 4 ) + q S(q, q ). (5.3) Therefore, by (5.), (4.), (5.3), (4.9), (4.0), and (4.5), { } q n ( q 6, q 0, q 6 ; q 6 ) + q ( q, q 4, q 6 ; q 6 ) (iq) n (q/i) n (q ; q ) (q ; q ) { [q4 ; q 6 ] S( q, q 6 ) + q 6 [q4 ; q 6 ] S( q 0, q 6 ) [ q ; q 6 ] [ q 6 ; q 6 ] [q4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) [ q, q, q 6, q 6 ; q 6 ] + q (q6 ; q 6 ) [ q 6 ; q 6 ] [ q, q 8 ; q 6 ] q 3 (q6 ; q 6 ) [ q ; q 6 ] [ q 6, q 8 ; q 6 ] q 5 [q4 ; q 6 ] [ q ; q 6 ] S( q 0, q 6 ) q [q4 ; q 6 ] S( q, q 6 ) [ q 6 ; q 6 ] }. (5.4) We now prove two identities that we use in our collection of the even powers from (5.4). First, we prove the identity [ q 6 ; q 6 ] q [ q ; q 6 ] [q, q 4, q 4, q 6, q 8 ; q 6 ]. (5.5) Replacing q by q and noting that [ q, q 3 ; q 8 ] [q ; q 8 ] /[q, q 3 ; q 8 ], we see that (5.5) is equivalent to [q 3 ; q 8 ] + q[q; q 8 ] [q, q, q, q 4 ; q 8 ] [q, q 3 ; q 8 ]. (5.6) Dividing (5.6) throughout by [q, q, q 4 ; q 8 ] and rearranging, we see that (5.5) is in turn equivalent to [q 3, q 3, q ; q 8 ] [q, q, q ; q 8 ] q [q, q, q 5 ; q 8 ], [q, q, q, q 4 ; q 8 ] [q, q, q, q 3 ; q 8 ] [q 4 ; q 8 ] [q, /q, q 3, q 4 ; q 8 ] and this follows from (4.4) with q replaced by q 4 and (a, b, c, d) (q, q, q, q). Second, we establish the identity [q 8, q 8 ; q 6 ] q [ q, q, q, q 4, q 6 ; q 6 ] [q, q, q 4, q 4, q 4, q 6, q 6, q 8 ; q 6 ]. (5.7) Replacing q by q, we find that (5.7) is equivalent to [q 4, q 4 ; q 8 ] + q[q, q, q, q, q 3 ; q 8 ] [ q, q, q, q, q, q 3, q 3, q 4 ; q 8 ], which in turn is equivalent to [q 4, q 4 ; q 8 ] + q [q, q, q ; q 8 ] [q, q, q, q, q, q 4 ; q 8 ]. (5.8) [q 3 ; q 8 ] [q, q, q 3, q 3 ; q 8 ] We thus want to prove (5.8). We apply (4.6) with q replaced by q 4 and with (a, b) (q, q ) to obtain the identity. [q, q 4, q 4 ; q 8 ] (q 8 ; q 8 ) [q, q, q 3, q 3 ; q 8 ] [q4, q 4, q 4 ; q 8 ] (q 8 ; q 8 ) [q, q, q, q ; q 8 ] [q, q, q 4 ; q 8 ] (q 8 ; q 8 ) [q, q 3, q, q ; q 8 ] q [q, q4 ; q 8 ] (q 8 ; q 8 ) [q, q, q 3 ; q 8 ]. (5.9)
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 5 Multiplying both sides by [q ; q 8 ] 4 and then dividing both sides by [q 4 ; q 8 ] (q 8 ; q 8 ), we see that (5.9) is (5.8) upon rearrangement. Invoking (5.5) in the second equality below and (5.7) in the last equality below, we find that the even powers from (5.4) are equal to [ q 6 ; q 6 ] (q 6 ; q 6 ) (q ; q ) [q4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) [ q, q, q 6, q 6 ; q 6 ] { q (q6 ; q 6 ) [ q 6 ; q 6 ] [ q, q 8 ; q 6 ] { [q4 ; q 6 ] S( q, q 6 ) + q 6 [q4 ; q 6 ] S( q 0, q 6 ) [ q ; q 6 ] [ q 6 ; q 6 ] } + q [ q ; q 6 ] (q 6 ; q 6 ) (q ; q ) q [q4 ; q 6 ] [ q 6 ; q 6 ] S( q, q 6 ) q 3 (q6 ; q 6 ) [ q ; q 6 ] [ q 6, q 8 ; q 6 ] q 5 [q4 ; q 6 ] [ q ; q 6 ] S( q 0, q 6 ) (q 6 ; q 6 ) S( q, q 6 ) ( ) [ q 6 ; q 6 ] (q ; q ) (q 4 ; q 4 q [ q ; q 6 ] ) [q4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) 3 (q ; q ) [ q, q, q 6 ; q 6 ] + q (q 6 ; q 6 ) 3 ( [ q 6 ; q 6 ] (q ; q ) [ q 6, q 8 ; q 6 q [ q ; q 6 ] ] (q 6 ; q 6 ) S( q, q 6 )[q, q 4, q 4, q 6, q 8 ; q 6 ] (q ; q ) (q 4 ; q 4 ) [q4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) 3 + q (q 6 ; q 6 ) 3 [q, q 4, q 4, q 6, q 8 ; q 6 ] (q ; q ) [ q, q, q 6 ; q 6 ] (q ; q ) [ q 6, q 8 ; q 6 ] (q 6 ; q 6 ) S( q, q 6 )[q, q 4, q 4, q 6, q 8 ; q 6 ] (q ; q ) (q 4 ; q 4 ) [q 4 ; q 6 ] (q 6 ; q 6 ) 3 ( ) [q 8, q 8 ; q 6 ] (q ; q ) [ q, q, q 6 ; q 6 q [ q, q, q, q 4, q 6 ; q 6 ] ] S( q, q 6 ) [q4 ; q 6 ] (q 6 ; q 6 ). (5.0) (q 6 ; q 6 ) [ q, q, q 6 ; q 6 ] We now collect the odd powers of q from (5.4). In the analysis below, we need to use the identity [q, q 4, q 6, q 6, q 6 ; q 6 ] [q 8, q 8 ; q 6 ] [q, q, q 4, q 4, q 4, q 6, q 6, q 8 ; q 6 ], (5.) which we now prove. Replacing q by q and rearranging, we see that (5.) is equivalent to ) } [q 4 ; q 8 ] + [q ; q 8 ] 5 [q 4 ; q 8 ] [q, q 3 ; q 8 ] [q, q, q 3 ; q 8 ] [q; q 8 ]. (5.)
6 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK Multiplying both sides of (5.) by [q 4 ; q 8 ] (q 8 ; q 8 ) and dividing both sides by [q ; q 8 ] 4, we obtain [q 4 ; q 8 ] 3 (q 8 ; q 8 ) [q ; q 8 ] 4 + [q, q 4, q 4 ; q 8 ] (q 8 ; q 8 ) [q, q 3 ; q 8 ] [q3, q 4 ; q 8 ] (q 8 ; q 8 ) [q, q, q ; q 8 ]. (5.3) If in (4.5), we replace q by q 4 and set (a, b) (q, q), we arrive at (5.3). Hence, (5.) has been established. In the first equality below, we employ (5.5) twice, and in the penultimate equality below, we utilize (5.). Thus, collecting the odd powers of q from (5.4), we find that q 5 [q 4 ; q 6 ] (q 6 ; q 6 ) S( q 0, q 6 )([ q 6 ; q 6 ] (q ; q ) [ q, q 6 ; q 6 q [ q ; q 6 ] ] ) (q 6 ; q 6 ) 3 + q ([ q 6 ; q 6 ] (q ; q ) [ q, q 8 ; q 6 q [ q ; q 6 ] ] ) [q 4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) 3 q (q ; q ) [ q, q 6, q 6 ; q 6 ] q 5 S( q 0, q 6 ) (q 6 ; q 6 ) (q 6 ; q 6 ) 3 + q [q, q 4, q 4, q 6, q 8 ; q 6 [q 4, q 8, q 8 ; q 6 ] (q 6 ; q 6 ) 3 ] (q ; q ) [ q, q 8 ; q 6 q ] (q ; q ) [ q, q 6, q 6 ; q 6 ] q 5 S( q 0, q 6 ) (q 6 ; q 6 ) [q 4 ; q 6 ] (q 6 ; q 6 ) 3 + q ([q, q 4, q 6, q 6, q 6 ; q 6 ] (q ; q ) [ q, q 6, q 6 ; q 6 [q 8, q 8 ; q 6 ] ) ] q 5 (q 6 ; q 6 ) S( q 0, q 6 ) [q 4 ; q 6 ] (q 6 ; q 6 ) 3 + q [q, q, q 4, q 4, q 4, q 6, q 6, q 8 ; q 6 ] (q ; q ) [ q, q 6, q 6 ; q 6 ] q 5 (q 6 ; q 6 ) S( q 0, q 6 ) + q [q4 ; q 6 ] (q 6 ; q 6 ) [ q, q 6, q 6 ; q 6 ] q 5 (q 6 ; q 6 ) S( q 6, ) + q [q4 ; q 6 ] (q 6 ; q 6 ) [ q, q 6, q 6 ; q 6 ]. (5.4) We now return to (5.4). On the right side of (5.4), we substitute the expressions that we found for the even powers in (5.0) and the representation for the odd powers from (5.4). We immediately obtain the proposed identity (.0), thus completing the proof of Theorem.6.
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 7 6. PROOF OF ENTRY.4; PART 4, PROOF OF THEOREM.7 Proof of Theorem.7. Let a be a primitive eighth root of unity. Then by the partial fraction identity (.3), q n a (aq) n (q/a) n (q) a (q) (q) n n n ( ) n q n(3n+)/ aq n (6.) ( ) n q n(3n+)/ ( + aq n + a q n + a 3 q 3n ) + q 4n ( ) n q n(3n+)/ [( a 4 q 3n ) a( a q 3n ) + aq n ( a q n ) a q n ( q n )] + q 4n. Replacing n by n and multiplying both the numerator and denominator of the summands by q 4n, we see that n Therefore, we deduce that and n n n n ( ) n q n(3n+)/ q kn + q 4n n ( ) n q n(3n+)/ ( a 4 q 3n ) + q 4n ( ) n q n(3n+)/ q (3 k)n + q 4n. n ( ) n q n(3n+)/ ( a q 3n ) + q 4n ( a ) ( ) n q n(3n+)/ q n ( a q n ) + q 4n ( a ) ( ) n q n(3n+)/ n n + q 4n, (6.) ( ) n q n(3n+)/, (6.3) + q 4n ( ) n q n(3n+3)/, (6.4) + q 4n ( ) n q n(3n+)/ q n ( q n ) + q 4n 0. (6.5) Thus, noting that a 3 /a and putting (6.) (6.5) in (6.), we find that q n (aq) n (q/a) n a /a (q) n ( ) n q n(3n+)/ + q 4n + a + /a (q) n ( ) n q n(3n+3)/ + q 4n. (6.6) We now proceed as we did in the proof of Theorem.6. We first work out the dissections of the relevant Lambert series. For each of the two Lambert series below, we divide the index of summation n into residue classes modulo 4 and express each sum in terms of S(z, ζ), defined in (4.8). By applying (4.) and (4.) in the first case and (4.3) and (4.4) in the
8 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK second, we find that ( ) n q n(3n+)/ n S(, q ) q S( q 4, q 0 ) + q 4n + q 7 S( q 8, q ) q 5 S( q, q 4 ) (q6 ; q 6 ) [ q, q 4 ; q 6 ] [, q, q 4 ; q 6 ] + [q4 ; q 6 ] [ q ; q 6 ] S(q, q 6 ) and n ( ) n q (3n +3n)/ + q (q6 ; q 6 ) [q 4, q 6 ; q 6 ] [q, q 4, q 8 ; q 6 ] q [q4 ; q 6 ] [ q 6 ; q 6 ] S(q, q 6 ) (6.7) + q 4n S(, q 8 ) q 3 S( q 4, q 6 ) + q 9 S( q 8, q 6 ) q 8 S( q, q 8 ) (q6 ; q 6 ) [q 4, q 6 ; q 6 ] [, q 4, q 6 ; q 6 ] q 6 [q4 ; q 6 ] [ q 6 ; q 6 ] S(q 6, ) q 3 (q6 ; q 6 ) [ q, q 4 ; q 6 ] [ q 4, q 6, q 8 ; q 6 ] + q 5 [q4 ; q 6 ] [ q ; q 6 ] S(q 6, ). (6.8) To study (6.7), we first verify two theta function identities, both used in the second equality of (6.3) below. The first equality that we shall employ is [ q 8 ; q 6 ] q [ ; q 6 ] [q, q, q 4, q 6, q 6 ; q 6 ] (q 8 ; q 6 ), (6.9) which, after multiplying both sides by (q 6 ; q 6 ) and utilizing the Jacobi triple product identity (3.), is equivalent to the elementary identity q 8n q q 8n(n+) q (n) q (n+) ( ) n q n. n n The second equality that we shall use is given by n n [, q 6, q 6 ; q 6 ] [ q, q, q 8 ; q 6 ] [q, q, q 4, q 4 ; q 6 ] (q 8 ; q 6 ), (6.0) which we now prove. Multiplying both sides of (6.0) by [q 4 ; q 6 ] (q 6 ; q 6 ) and then dividing both sides by [, q, q, q 4, q 8 ; q 6 ], we see that (6.0) is equivalent to [q 4, q 6, q 6 ; q 6 ] (q 6 ; q 6 ) [q, q, q 4, q 8 ; q 6 ] [ q, q, q 4 ; q 6 ] (q 6 ; q 6 ) [, q, q, q 4 ; q 6 ] [q4, q 4, q 4 ; q 6 ] (q 8 ; q 6 ) (q 6 ; q 6 ) [, q 4, q 8 ; q 6 ] Note that [q4, q 4, q 8 ; q 6 ] (q 6 ; q 6 ) [, q 4, q 4, q 8 ; q 6 ]. (6.) [q 4 ; q 6 ] (q 8 ; q 6 ) (q8 ; q 6 ) (q 8 ; q 6 ) (q8 ; q 6 ), (6.) ( q 4 ; q 4 ) ( q 4 ; q 8 ) ( q 8 ; q 8 ) [ q 4 ; q 6 ] which we use on the right side of (6.). Then (6.) follows from (4.4) upon replacing q by q 8 and setting (a, b, c, d) (q, q, q 4, ).
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 9 We now focus on (6.7). By invoking (4.) and using (5.5) in the second equality below and, as we mentioned above, (6.9) and (6.0) as well, we find that (q) ( ) n q n(3n+)/ + q 4n n [q4 ; q 6 ] (q 6 ; q 6 ) (q ; q ) [ q, q 6 ; q 6 ] S(q, q 6 )([ q 6 ; q 6 ] q [ q ; q 6 ] ) (q6 ; q 6 ) 3 [ q, q 4, q 6 ; q 6 ] ([ q 8 ; q 6 ] (q ; q ) [, q, q 4, q 8 ; q 6 q [ ; q 6 ] ) ] (q 6 ; q 6 ) 3 + q [q 4 ; q 6 ] ([, q 6, q 6 ; q 6 ] (q ; q ) [, q, q 4, q 8 ; q 6 [ q, q, q 8 ; q 6 ] ) ] [q4 ; q 6 ] (q 6 ; q 6 ) (q ; q ) [ q, q 6 ; q 6 ] S(q, q 6 )[q, q 4, q 4, q 6, q 8 ; q 6 ] (q6 ; q 6 ) 3 [ q, q 4, q 6 ; q 6 ] [q, q, q 4, q 6, q 6 ; q 6 ] (q ; q ) [, q, q 4, q 8 ; q 6 (q 8 ; q 6 ) ] (q 6 ; q 6 ) 3 + q [q 4 ; q 6 ] [q, q, q 4, q 4 ; q 6 ] (q ; q ) [, q, q 4, q 8 ; q 6 (q 8 ; q 6 ) ] S(q, q 6 ) (q6 ; q 6 ) [q 4 ; q 6 ] + q (q6 ; q 6 ) [q 4 ; q 6 ]. (6.3) (q 6 ; q 6 ) [q, q, q 6 ; q 6 ] [q, q 6, q 6 ; q 6 ] To study (6.8), we again need to first establish a certain theta function identity, namely, [ q 6, q 6, q 8 ; q 6 ] q 4 [, q, q ; q 6 ] [q 4, q 4, q 6, q 6 ; q 6 ] (q 8 ; q 6 ). (6.4) Multiplying both sides of (6.4) by [q 4 ; q 6 ] (q 6 ; q 6 ) and then dividing both sides by [, q 4, q 6, q 6, q 8 ; q 6 ], we see that (6.4) is equivalent to [q 4, q 6, q 6 ; q 6 ] (q 6 ; q 6 ) [, q 6, q 6, q 4 ; q 6 ] [ q, q, q 4 ; q 6 ] (q 6 ; q 6 ) [q 6, q 6, q 4, q 8 ; q 6 ] [q4, q 4, q 8 ; q 6 ] (q 6 ; q 6 ) [, q 4, q 4, q 8 ; q 6 ], (6.5) where we applied (6.) and the elementary identity [ q 4 ; q 6 ] q 4 [ q 4 ; q 6 ]. In (4.4), replace q by q 8 and set (a, b, c, d) (q 6, q 6,, q 4 ). Then (6.5) follows immediately. We now give our attention to (6.8). As before, we utilize (4.), and then applying (5.5), (6.4), and (6.9) in the second equality below, we find that (q) ( ) n q n(3n+3)/ + q 4n n q 5 [q 4 ; q 6 ] (q 6 ; q 6 ) (q ; q ) [ q, q 6 ; q 6 ] S(q 6, )([ q 6 ; q 6 ] q [ q ; q 6 ] )
0 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK + (q 6 ; q 6 ) 3 [q 4 ; q 6 ] (q ; q ) [, q 4, q 6, q 8 ; q 6 ] ([ q 6, q 6, q 8 ; q 6 ] q 4 [, q, q ; q 6 ] ) + q (q6 ; q 6 ) 3 [ q, q 4, q 6 ; q 6 ] (q ; q ) [, q 4, q 6, q 8 ; q 6 ] ([ q 8 ; q 6 ] q [ ; q 6 ] ) q 5 [q 4 ; q 6 ] (q 6 ; q 6 ) (q ; q ) [ q, q 6 ; q 6 ] S(q 6, )[q, q 4, q 4, q 6, q 8 ; q 6 ] + (q 6 ; q 6 ) 3 [q 4 ; q 6 ] (q ; q ) [, q 4, q 6, q 8 ; q 6 ] [q 4, q 4, q 6, q 6 ; q 6 ] (q 8 ; q 6 ) + q (q6 ; q 6 ) [q 4 ; q 6 ] [q, q 6, q 6 ; q 6 ] q 5 (q 6 ; q 6 ) S(q 6, ) + (q6 ; q 6 ) [q 4 ; q 6 ] [q, q, q 6 ; q 6 ] + q (q6 ; q 6 ) [q 4 ; q 6 ] [q, q 6, q 6 ; q 6 ]. (6.6) We can now complete the proof of Theorem.7 by substituting (6.3) and (6.6) into (6.6). 7. SOME CONSEQUENCES OF THEOREMS.6 AND.7 In this final section, we give some immediate consequences of Theorems.6 and.7 related to ranks and to mock theta functions. First, let N(k, t, n) denote the number of partitions of n with rank congruent to k modulo t. Denote R b,c (t, l, d) [N(b, t, ln + d) N(c, t, ln + d)]q n. Then the following results follow from Theorems.6 and.7. Corollary 7.. We have i.e., for n 0, R 0, (4, 4, 0) R 0,4 (8, 4, 0), (7.) R 0, (4, 4, ) R 0,4 (8, 4, ), (7.) R 0, (4, 4, ) R 0,4 (8, 4, ) + R,3 (8, 4, ), (7.3) R 0, (4, 4, 3) R 0,4 (8, 4, 3) R,3 (8, 4, 3), (7.4) N(0, 4, n) N(, 4, n) ( ) n [N(0, 8, n) N(4, 8, n)], (7.5) N(0, 4, n + ) N(, 4, n + ) ( ) n [N(0, 8, n + ) + N(, 8, n + ) N(3, 8, n + ) N(4, 8, n + )]. (7.6) Proof. On page 7 of [], F. G. Garvan explains how one can obtain results on rank differences from the dissections of the rank generating functions. We follow his argument here. First, note that q n (iq) n (q/i) n k0 3 N(k, 4, n)i k q n (N(0, 4, n) N(, 4, n))q n,
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS since N(, 4, n) N(3, 4, n) [, eqs. (.09), (.0)] and i 3 i. By extracting the terms with even powers of q on both sides of (.0), we deduce that (N(0, 4, n) N(, 4, n))q n (q 6 ; q 6 ) n ( ) n q 4n +8n + q 6n+ [q4 ; q 6 ] (q 6 ; q 6 ) [ q, q, q 6 ; q 6 ]. (7.7) As above, let a denote a primitive eighth root of unity. Using the relations a a 6, N(, 8, n) N(6, 8, n), and a 3 + a 5 (a + a 7 ), we see that q n (aq) n (q/a) n k0 7 N(k, 8, n)a k q n { [N(0, 8, n) N(4, 8, n)] + (a + a 7 )[N(, 8, n) N(3, 8, n)] } q n. Since and a + a 7 are linearly independent over the set of integers, extracting the terms with even powers of q on both sides of (.), we obtain the two identities (N(0, 8, n) N(4, 8, n))q n (q 6 ; q 6 ) n (N(, 8, n) N(3, 8, n))q n (q 6 ; q 6 ) n ( ) n q 4n +8n q 6n+ [q4 ; q 6 ] (q 6 ; q 6 ) [q, q, q 6 ; q 6 ], (7.8) ( ) n q 4n +8n q 6n+ + [q4 ; q 6 ] (q 6 ; q 6 ) [q, q, q 6 ; q 6 ]. (7.9) We see that the right side of (7.7) is exactly (7.8) but with q replaced by q. This implies (7.), (7.), and (7.5). Equations (7.3), (7.4), and (7.6) are proved similarly by selecting the terms with odd powers of q on both sides of (.0) and (.). Remark 7.. N. Santa-Gadea and R. Lewis proved many results on ranks and cranks modulo 4 and 8. See, for example, [6, 7, 8, 9, 0,,, 4, 5]. Equation (4.5) of Santa- Gadea s Thesis [4] gives the generating function of the relation N(0, 4, n) N(, 4, n) N(0, 8, n) N(, 8, n) + N(4, 8, n). (7.0) Through (7.0), the relations given in equations (4.) (4.4) of [4] are immediately seen to be equivalent to (7.5) and (7.6). The relations (4.) (4.4) in [4] were originally conjectured by Lewis in two papers [6], [8]. They were proved again in a later paper by Santa-Gadea and Lewis [].
GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK Each of Ramanujan s mock theta functions satisfies a transformation formula involving the rank function G(z, q), defined in (.). For example, the famous mock theta conjectures, first proved by D. Hickerson [5, eqs. (0.9), (0.0)], are given by q n ( q) n q G(q, q 0 ) + (q5 ; q 5 ) (q 5 ; q 0 ), [q; q 5 ] q n +n ( q) n q q( q 4 ) G(q4, q 0 ) + (q5 ; q 5 ) (q 5 ; q 0 ) [q ; q 5 ]. By summing the odd and even indices n s separately, we can write down the -dissection of the rank function modulo 4 directly as q n q n q 4n q 4(n +n) + q. (iq) n (q/i) n ( q ; q ) n ( q ; q ) n ( q ; q ) n+ By applying (.3) and the identity [, eq. (7.0)] z ( ) n q 3n(n+)/ + G(z, q), (q) zq n z n in Theorem.6, we can then derive analogous identities. Corollary 7.3. We have q n ( q; q) n + q G( q, q8 ) [q, q ; q 8 ] (q 8 ; q 8 ), [ q, q, q 3 ; q 8 ] q (n +n) ( q; q) n+ q q( + q 3 ) G( q3, q 8 ) + [q, q ; q 8 ] (q 8 ; q 8 ). [ q, q 3, q 3 ; q 8 ] This is not the first time these two functions have been studied. For example, in [, eqs. (.4), (.5)]), the first author gave Hecke-type series representations for the two functions q n ( q; q) n q (n +n) ( q; q) n+ (q ; q ) (q ; q ) q 4n +n ( q 6n+3 ) q 4n +3n ( q n+ ) REFERENCES n ( ) j q j, j n n ( ) j q j. j n [] G. E. Andrews, q-orthogonal polynomials, Rogers-Ramanujan identities, and mock theta functions, Tr. Mat. Inst. Steklova 76 (0), 7 38. [] G. E. Andrews and B. C. Berndt, Ramanujan s Lost Notebook, Part I, Springer, New York, 005. [3] G. E. Andrews and B. C. Berndt, Ramanujan s Lost Notebook, Part III, Springer, New York, 0. [4] G. E. Andrews and R. Lewis, The ranks and cranks of partitions moduli, 3, and 4, J. Number Theory 85 (000), no., 74 84. [5] G. E. Andrews, R. Lewis, and Z. G. Liu, An identity relating a theta function to a sum of Lambert series, Bull. London Math. Soc. 33 (00), 5 3.
FOUR IDENTITIES FOR THIRD ORDER MOCK THETA FUNCTIONS 3 [6] A. O. L. Atkin and H. P. F. Swinnerton Dyer, Some properties of partitions, Proc. London Math. Soc. (3) 4 (954), 84 06. [7] B. C. Berndt, Ramanujan s Notebooks, Part III, Springer Verlag, New York, 99. [8] B. C. Berndt, Number Theory in the Spirit of Ramanujan, American Mathematical Society, Providence, RI, 006. [9] S. H. Chan, Generalized Lambert series identities, Proc. London Math. Soc. 9 (005), no. 3, 598 6. [0] S. H. Chan, Congruences for Ramanujan s φ function, Acta Arith. 53 (0), no., 6 89. [] F. J. Dyson, Some guesses in the theory of partitions, Eureka, Cambridge 8 (944), 0 5. [] F. G. Garvan, New combinatorial interpretations of Ramanujan s partition congruences mod 5, 7 and, Trans. Amer. Math. Soc. 305 (988), 47 77. [3] F. G. Garvan, The crank of partitions mod 8, 9 and 0, Trans. Amer. Math. Soc. 3 (990), no., 79 94. [4] G. H. Halphen, Traité des fonctions elliptiques et de leurs applications, Vol., Gauthier-Villars, Paris, 888. [5] D. Hickerson, A proof of the mock theta conjectures, Invent. Math. 94 (998), no. 3, 66 677. [6] R. Lewis, On the rank and the crank modulo 4, Proc. Amer. Math. Soc. (99), no. 4, 95 933. [7] R. Lewis, On the ranks of partitions modulo 9, Bull. London Math. Soc. 3 (99), no. 5, 47 4. [8] R. Lewis, On some relations between the rank and the crank, J. Combin. Thy., Ser. A 59 (99), no., 04 0. [9] R. Lewis, Relations between the rank and the crank modulo 9, J. London Math. Soc. () 45 (99), no., 3. [0] R. Lewis, The ranks of partitions modulo, 5-th British Combinatorial Conference (Stirling, 995), Discrete Math. 67/68 (997), 445 449. [] R. Lewis, The generating functions of the rank and crank modulo 8, Ramanujan J. 8 (009), no., 46. [] R. Lewis and N. Santa-Gadea, On the rank and the crank modulo 4 and 8, Trans. Amer. Math. Soc. 34 (994), no., 449 465. [3] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 988. [4] N. Santa-Gadea, On the rank and the crank moduli 8, 9, and, Ph.D. Thesis, The Pennsylvania State University, 990, 70 pp. [5] N. Santa-Gadea, On some relations for the rank moduli 9 and, J. Number Theory 40 (99), no., 30 45. [6] H. Yesilyurt, Four identities related to third order mock theta functions in Ramanujan s lost notebook, Adv. Math. 90 (005), 78 99.
4 GEORGE E. ANDREWS, BRUCE C. BERNDT, SONG HENG CHAN, SUN KIM, AND AMITA MALIK DEPARTMENT OF MATHEMATICS, PENNSYLVANIA STATE UNIVERSITY, UNIVERSITY PARK, PA 680, USA E-mail address: andrews@math.psu.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS, 409 WEST GREEN STREET, URBANA, IL 680, USA. E-mail address: berndt@illinois.edu DIVISION OF MATHEMATICAL SCIENCES, SCHOOL OF PHYSICAL AND MATHEMATICAL SCIENCES, NANYANG TECHNOLOGICAL UNIVERSITY, NANYANG LINK, SINGAPORE 63737, REPUBLIC OF SIN- GAPORE E-mail address: ChanSH@ntu.edu.sg DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS, 409 WEST GREEN STREET, URBANA, IL 680, USA E-mail address: sunkim@illinois.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS, 409 WEST GREEN STREET, URBANA, IL 680, USA E-mail address: amalik0@illinois.edu