Explicit evaluations of a level 13 analogue of the Rogers–Ramanujan continued fraction

Journal of Number Theory 139 (2014) 91–111

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Journal of Number Theory www.elsevier.com/locate/jnt

Explicit evaluations of a level 13 analogue of the Rogers–Ramanujan continued fraction Shaun Cooper ∗ , Dongxi Ye Institute of Natural and Mathematical Sciences, Massey University-Albany, Private Bag 102904, North Shore Mail Centre, Auckland, New Zealand

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 16 August 2013 Received in revised form 16 December 2013 Accepted 16 December 2013 Available online 8 February 2014 Communicated by David Goss

The Rogers–Ramanujan continued fraction has a representation as an infinite product given by

MSC: primary 11J70 secondary 11B65, 11F11, 33E05

where |q| < 1 and ( pj ) is the Legendre symbol. In his letters to Hardy and in his notebooks, Ramanujan recorded some exact numerical values of the Rogers–Ramanujan continued fraction for specific values of q. In this work, we give explicit evaluations of the level 13 analogue defined by

Keywords: Kronecker limit formula P –Q modular equation Ramanujan–Weber class invariant Reciprocity formula Rogers–Ramanujan continued fraction

q 1/5

∞  

1 − qj

( j ) 5

j=1

q

∞  

1 − qj

(

j ) 13

.

j=1

© 2014 Elsevier Inc. All rights reserved.

* Corresponding author. E-mail addresses: [email protected] (S. Cooper), [email protected] (D. Ye). http://dx.doi.org/10.1016/j.jnt.2013.12.015 0022-314X/© 2014 Elsevier Inc. All rights reserved.

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

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1. Introduction Throughout this paper, it is assumed that Im(τ ) > 0 and q = e2πiτ . Let R(q) =

q 1/5 q

1+

(1.1) q

1+ 1+

2

q3 1 + ···

denote the Rogers–Ramanujan continued fraction. In 1913, Ramanujan asserted in his first letter to Hardy [11, p. 29] that 

√ √ 5+ 5 5+1 − , R e = 2 2  √ √  −π  5− 5 5−1 −R −e − = 2 2 

 −2π

√

and R(e−π n ) can be found exactly if n is a positive rational number. These results particularly impressed and intrigued Hardy who responded by writing to Ramanujan [11, p. 77]: “What I should like above all is a definite proof of some of your results concerning continued fractions of the type (1.1); and I am quite sure that the wisest thing you can do, in your own interests, is to let me have one as soon as possible.” A few months later Hardy reiterated the request for a proof [11, p. 87]: “If you will send me your proof written out carefully (so that it is easy to follow), I will (assuming that I agree with it—of which I have very little doubt) try to get it published for you in England. Write it in the form of a paper . . . giving a full proof of the principal and most remarkable theorem, viz., that the fraction can be expressed in √ finite terms when q = e−π n , where n is rational.” More than 25 years later Hardy recalled the profound impact that Ramanujan’s evaluations of R(q) had had on him [17, p. 9]: “(They) defeated me completely; I had never seen anything in the least like them before. A single look at them is enough to show that they could only be written down by a mathematician of the highest class. They must be true because, if they were not true, no one would have had the imagination to invent them.”

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

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Evaluations of R(q) and −R(−q) for other values of q have been studied by various mathematicians, including Ramanujan [25,27], Watson [30,31] Ramanathan [20–23], Berndt and Chan [6,7], and Berndt, Chan and Zhang [9]. A large number of evaluations of R(q) and −R(−q) have been summarized by Kang [19, pp. 64–67] and further evaluations have been given by Yi [35], Vasuki and Shivashankara [29] and Baruah and Saikia [2]. The Rogers–Ramanujan continued fraction has other beautiful properties: R(q) = q 1/5

∞  (1 − q 5j−4 )(1 − q 5j−1 ) , (1 − q 5j−3 )(1 − q 5j−2 ) j=1

(1.2)

E(q 1/5 ) 1 − 1 − R(q) = 1/5 R(q) q E(q 5 )

(1.3)

E 6 (q) 1 5 − 11 − R , (q) = R5 (q) qE 6 (q 5 )

(1.4)

and

where E(q) =

∞  

 1 − qj .

j=1

The identity (1.2) was proved by Rogers [28], and the other two identities (1.3) and (1.4) were recorded by Ramanujan [3, pp. 85 and 267] and proved by Watson [30]. In his second notebook [3,25, Chapter 20, Entry 8(i)], Ramanujan stated a striking analogue of the identities (1.2)–(1.4): If R(q) = q

∞   ( j ) 1 − q j 13 j=1

=q

∞  (1 − q 13j−12 )(1 − q 13j−10 )(1 − q 13j−9 )(1 − q 13j−4 )(1 − q 13j−3 )(1 − q 13j−1 ) (1 − q 13j−11 )(1 − q 13j−8 )(1 − q 13j−7 )(1 − q 13j−6 )(1 − q 13j−5 )(1 − q 13j−2 ) j=1

(1.5) then E 2 (q) 1 − 3 − R(q) = . R(q) qE 2 (q 13 )

(1.6)

Proofs of (1.6) have been given in [16, Theorem 5.1] and [18, Theorem 2.2]. From [12, Lemma 2.3] or [18, Theorem 1.1], we know that R(q) is invariant under the congruence subgroup Γ1 (13), where

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

94

 Γ1 (N ) =

a b c d



∈ SL2 (Z): c ≡ 0 (mod N ) and d ≡ 1 (mod N ) .

Since R(q) is invariant under Γ1 (13) and the infinite product formula (1.5) is an analogue of (1.2), we shall say that R(q) is a level 13 analogue of the Rogers–Ramanujan continued fraction. Motivated by the analogy between R(q) and R(q), this work is concerned with finding explicit evaluations of R(q). In the next section, √we deduce two reciprocity formulas for √ −π/ 13 R(q) and use them to give evaluations of R(e−2π/ 13 ) and R(−e ). In Sections 3–5,



we apply a variety of methods to evaluate R(e−2π n/13 ) and R(−e−π n/13 ) for various integers n that are summarized in the following table:

n

Method

1 2, 3, 7 5, 9, 13, 69, 129 15, 31, 55, 231, 255

Reciprocity formulas (Section 2) P –Q modular equations (Section 3) Ramanujan–Weber class invariant (Section 4) Kronecker’s limit formula (Section 5)

We shall also prove some general results. In Theorem 3.10 it is shown that for any n 0 < q < 1, if R(q) or R(−q) can be evaluated in terms of radicals, then so can R(±q 2 ) for any integer n. In Theorem 3.11 it is shown that if |q| < 1 and R(q) can be evaluated n in terms of √radicals, then so can R(q 3 ) for any integer n. In Theorem 6.2 it is shown that R(e−π n ) is a unit for any positive rational number n. Following (1.6), let F be defined by

F = F (q) =

∞ E 2 (q) 1  (1 − q j )2 1 − 3 − R(q) = = . R(q) qE 2 (q 13 ) q j=1 (1 − q 13j )2

(1.7)

R(q)(1 − 3R(q) − R(q)2 ) F (q) = . F (q)2 + 6F (q) + 13 (1 + R(q)2 )2

(1.8)

Let T be defined by

T = T (q) =

The function T (q) was studied in [14]. The function F (q) was also studied in [14] where it was denoted by 1/S(q). The values of T (q) often turn out to have simpler expressions in terms of radicals than the corresponding values of R(q) or F (q). Clearly, if T can be evaluated at a particular value of q, then F and R can be evaluated at the same value of q by solving appropriate quadratic equations and using numerical values to determine the root. We will use the phrase “explicit evaluation of R(q)” to mean that we have evaluated any of R(q), F (q) or T (q) in terms of radicals.

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95

2. Reciprocity formulas In his second notebook [3, p. 83], [25, Ch. 16, Entry 39], Ramanujan stated two reciprocity formulas for the Rogers–Ramanujan continued fraction: if α, β > 0 and αβ = 1, then √ √  √  −2πα   −2πβ  5+ 5 5+1 5+1 +R e +R e (2.1) = 2 2 2 and √

  5−1 − R −e−πα 2

 √

  5−1 − R −e−πβ 2

=

√ 5− 5 . 2

(2.2)

The formula (2.1) also appears in Ramanujan’s second letter to Hardy [11, p. 57], [26, p. xxviii]. In his lost notebook [1, pp. 91–92], [27, p. 364], Ramanujan recorded analogues of (2.1) and (2.2) for the fifth power of R(q): if α, β > 0 and αβ = 1/5, then  √

5 5  √     5+1 5+1 + R5 e−2πα + R5 e−2πβ 2 2 5 √ √ 5+1 =5 5 2

(2.3)

and  √

5 5  √  −πα   −πβ  5−1 5−1 5 5 − R −e − R −e 2 2 5 √ √ 5−1 . =5 5 2

(2.4)

The following result provides level 13 analogues of (2.1)–(2.4). Theorem 2.1. Let R(q) be defined by (1.5) and let α, β > 0 and αβ = 1/13. Then √

√  √     13 + 3 13 13 + 3 13 + 3 + R e−2πα + R e−2πβ = 2 2 2

(2.5)

and √

√  √  −πα   −πβ  13 − 3 13 13 − 3 13 − 3 − R −e − R −e . = 2 2 2

(2.6)

We call (2.5) and (2.6) reciprocity formulas for R(q). They can be used to eval

√ √ −2π/ 13n −π/ 13n −2π n/13 uate R(e ) and R(−e ) if we know the values of R(e ) and

R(−e−π n/13 ), respectively.

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96

Proof. The identity (2.5) was proved in [14]; we shall prove (2.6). From (1.6), we have √

13 − 3 − R(q) 2

 √ R(q)E 2 (q) 13 + 3 + R(q) = , 2 qE 2 (q 13 )

(2.7)

so for any t > 0, √ −

  13 + 3 < R −e−πt < 0. 2

(2.8)

Let a and b be positive real numbers with ab = π 2 . Then [3, p. 43, Entry 27(iv)]     e−a/24 a1/4 E −e−a = e−b/24 b1/4 E −e−b .

(2.9)

Successively taking (a, b) = (πα, 13πβ) and (a, b) = (13πα, πβ) in (2.9), we deduce that     e−πα/24 (πα)1/4 E −e−πα = e−13πβ/24 (13πβ)1/4 E −e−13πβ

(2.10)

    e−13πα/24 (13πα)1/4 E −e−13πα = e−πβ/24 (πβ)1/4 E −e−πβ ,

(2.11)

and

respectively. On dividing (2.10) by (2.11) and squaring, we get E 2 (−e−πα ) √ = 13(−e−πα )E 2 (−e−13πα )

√ 13(−e−πβ )E 2 (−e−13πβ ) . E 2 (−e−πβ )

(2.12)

Now let q take the values −e−πα and −e−πβ in (2.7) and multiply the two resulting identities. Making use of (2.12) we get (A + ρ)(B + ρ)(A + ρ)(B + ρ) = 13AB, where 

−πα

A = R −e





−πβ

B = R −e

,

 ,

ρ=

3+

√

13

2

,

ρ=

3−

√ 2

13

.

This can be rearranged to give 

√  √  (A + ρ)(B + ρ) − ρ 13 (A + ρ)(B + ρ) + ρ 13 = 0.

(2.13)

From (2.8) we deduce that √ 0 < (A + ρ)(B + ρ) < ρ2 < ρ 13, so the first factor in (2.13) cannot be zero. It follows that the second factor in (2.13) is zero and this gives (2.6). 2

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

The reciprocity formulas in Theorem 2.1 lead to evaluations of R(e−2π/ √ R(−e−π/ 13 ).

97 √

13

) and

Theorem 2.2. Let R(q) be defined by (1.5). Then √   −(3 + R e−2π/ 13 =

√

13) + 2

√ 26 + 6 13

(2.14)

and 

√  −π/ 13

R −e

=

−(3 −

√

13) − 2



√ 26 − 6 13

.

(2.15)

√ Proof. Setting α = β = 1/ 13 in (2.5) and (2.6), we get √

√  −2π/√13  2 13 + 3 13 13 + 3 +R e = 2 2

and √

√ √  2  13 − 3 13 13 − 3 − R −e−π/ 13 . = 2 2

Taking square roots and using numerical values of R(e−2π/ determine the signs, we deduce (2.14) and (2.15). 2

√

13

) and R(−e−π/

√ 13

) to

3. P –Q modular equations



In this section we will find explicit evaluations of R(e−2π n/13 ) and R(−e−π n/13 ) for n = 2, 3 and 7. We will need six lemmas of a type that are called P –Q modular equations. Before stating them, we point out that Lemmas 3.1–3.3 will involve the function q −n/2 E(q n )/E(q 13n ), whereas Lemmas 3.4–3.6 are about the squared function q −n E 2 (q n )/E 2 (q 13n ). In his second notebook [25, p. 327], Ramanujan recorded the quadratic transformation formula: Lemma 3.1. Let P and Q be defined by P = P(q) =

E(q) q 1/2 E(q 13 )

and

Q = Q(q) =

E(q 2 ) . qE(q 26 )

Then 13 = PQ + PQ



P3 Q3 + Q3 P3





 P Q −4 + . Q P

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98

Proof. See [4, p. 211, Entry 57].

2

The analogous cubic transformation formula was also known to Ramanujan [25, p. 322]: Lemma 3.2. Let P and Q be defined by P = P(q) =

E(q) q 1/2 E(q 13 )

E(q 3 )

Q = Q(q) =

and

q 3/2 E(q 39 )

.

Then 13 = PQ + PQ Proof. See [4, p. 237, Entry 72].



Q2 P2 + Q2 P2





Q P −3 + Q P

 − 3.

2

The corresponding transformation formula of degree 7 has been proved in [34]: Lemma 3.3. Let P and Q be defined by P = P(q) =

Let

E(q) 1/2 q E(q 13 )

and

Q = Q(q) =

E(q 7 ) . q 7/2 E(q 91 )

Then (PQ)3 +

  2    P3 P2 Q3 Q Q P − 7 − 196 + + − P3 Q3 P2 Q2 P Q     3 3   P P Q Q + − 273 + 3 − 7 P 3 Q + Q3 P − 21 3 P Q P Q   1 1 728 − 1183 + . − 56PQ − P 3 Q Q3 P PQ

 =

133 (PQ)3

P4 Q4 − 4 4 P Q





− 21

(3.1)

Now we are ready to give analogues of Lemmas 3.1–3.3 that involve the squared functions. Lemma 3.4. If P and Q are defined by P = P (q) =

E 2 (q) qE 2 (q 13 )

and

Q = Q(q) =

E 2 (q 2 ) q 2 E 2 (q 26 )

,

then P 3 + Q3 = P 2 Q2 + 4P Q(P + Q) + 13P Q.

(3.2)

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99

Proof. Square both sides of the identity in Lemma 3.1 and write P = P 2 and Q = Q2 . The result simplifies to  3  P − 4P 2 Q − 13P Q − P 2 Q2 − 4P Q2 + Q3   × P 3 − 4P 2 Q + 13P Q + P 2 Q2 − 4P Q2 + Q3 = 0. The q-expansion of the second factor is not zero. It follows that the first factor is identically zero and this completes the proof. 2 Lemma 3.5. If P and Q are defined by P = P (q) =

E 2 (q) qE 2 (q 13 )

and

Q = Q(q) =

E 2 (q 3 ) , q 3 E 2 (q 39 )

then 132 = PQ + PQ



Q2 P2 + Q2 P2





Q P − 15 + Q P





1 1 − 6(P + Q) − 78 + P Q

 − 33. (3.3)

Proof. Move the terms that involve odd powers of P and Q in Lemma 3.2 to one side, and the terms that involve even powers to the other side. Then square both sides and write P = P 2 and Q = Q2 . 2 Lemma 3.6. If P and Q are defined by P = P (q) =

E 2 (q) qE 2 (q 13 )

and

Q = Q(q) =

E 2 (q 7 ) , q 7 E 2 (q 91 )

then (P Q)3 +

  3   2    P P P Q P4 Q4 Q3 Q2 − 455 − 19 649 − 124 348 + + + + Q4 P4 Q3 P3 Q2 P2 Q P      3  Q2 Q3 Q2 P2 P2 P − 882 + + 2 − 11 466 + 3 − 14 406 2 3 Q P Q P Q P     3   3 Q Q Q P P P − 124 215 − 735 + − 187 278 + + Q P Q2 P2 Q3 P3     1 1 1 1 + − 5 198 102 + 3 2 − 2 599 051 P 3 Q Q3 P P 3 Q2 Q P       1 1 1 1 1 1 − 1 003 184 + − 738 192 + + − 3 229 590 P 2 Q Q2 P P3 Q3 P2 Q2

 =

136 (P Q)3

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

100

 − 616 798

1 1 + P Q



    − 336 P 3 + Q3 − 5936 P 2 + Q2 − 47 446(P + Q)

      − 14 P 3 Q2 + Q3 P 2 − 91 P 3 Q + Q3 P − 1470 P 2 Q + Q2 P − 210P 2 Q2 −

5 997 810 1 852 578 − 10 962P Q − 219 800. − P 2 Q2 PQ

(3.4)

Proof. Follow the same procedure as for the proof of Lemma 3.5.

2

The next three theorems provide explicit evaluations of R(q) at q = e−2π n/13 and

of the evaluations varies considerably. q = −e−π n/13 for n = 2, 3, 7. The complexity

For example, the value of T (q) at q = −e−π 7/13 turns out to be a rational number, whereas the evaluation at q = −e−π 2/13 is quite a complicated algebraic number. Observe that in Lemmas 3.4–3.6 we have P (q) = F (q) and Q(q) = F (q n ) where n = 2, 3 or 7, respectively, and F (q) is defined by (1.7). Theorem 3.7. Let t and u be defined by t=

√ √ 1/3 1/3 1 17 1 − 624 78 + 5239 624 78 − 5239 + 24 24 24

and u=

1 − 6t +

√ 1 − 12t − 16t2 2t

and let v1 , v2 and v3 be the algebraic numbers defined by     x3 − u2 + 4u x2 − 4u2 + 13u x + u3 = (x − v1 )(x − v2 )(x − v3 )

(3.5)

with v1 < 0 < v2 < v3 . Let F (q) be defined by (1.7). Then



    F −e−π 2/13 = v1 ≈ −4.97, F e−2π 2/13 = u ≈ 9.69,



    F e−π 2/13 = v2 ≈ 1.34 and F e−4π 2/13 = v3 ≈ 136.22.

Proof. Recall the transformation formula for the Dedekind eta-function, e.g., [3, p. 43, Entry 27(iii)]:     e−a/12 a1/4 E e−2a = e−b/12 b1/4 E e−2b , Let q = e−2π/ find that

√

26

a, b > 0, ab = π 2 .

(3.6)

. By the definitions of P and Q in Lemma 3.4 together with (3.6) we

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111 √   Q(q) = Q e−2π/ 26 =



E 2 (e−2π 2/13 ) √ √ e−2π/ 13 E 2 (e−2π 26 ) √

E 2 (e−26π/ 26 ) √ E 2 (e−2π/ 26 )

√ −2π/ 26

= 13e =

101

13

√ P (e−2π/ 26 )

=

13 . P (q)

On substituting P = 13/Q into the result of Lemma 3.4 and simplifying, we get Q6 − 52Q4 − 338Q3 − 676Q2 + 2197 = 0. On dividing by (Q2 + 6Q + 13)3 and rearranging, we deduce that  8

Q Q2 + 6Q + 13

3

 − 17

Q Q2 + 6Q + 13

2

 + 18

Q Q2 + 6Q + 13

 − 1 = 0.

Since Q(q) = P (q 2 ) = F (q 2 ) it follows from (1.8) that  3  2   8T q 2 − 17T q 2 + 18T q 2 − 1 = 0. On solving the cubic and selecting the appropriate root we conclude that

    T e−2π 2/13 = T q 2

=

√ √ 1/3 1/3 1 17 1 − 624 78 + 5239 624 78 − 5239 + = t. 24 24 24



The value of F (e−2π 2/13 ) follows from this on using (1.8). The other three evaluations follow from Lemma 3.4. We substitute Q = u in (3.2)



and solve for P to determine that F (−e−π 2/13 ) and F (e−π 2/13 ) are both roots of the cubic equation (3.5). Noting that (3.2) is symmetric in P and Q, then substituting

−4π 2/13 P = u and solving for Q, we deduce that F (e ) is also a root of the cubic equation (3.5). Numerical calculations may be used to match the three evaluations with the corresponding roots of the cubic equation. 2 Theorem 3.8. Let T (q) be defined by (1.8). Then 

−2π

T e

 3/13

√ 7 13 − 25 = 6

and

  1− T −e−π 3/13 =

√ 6

13

.

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102

√

Proof. Let P and Q be as for Lemma 3.5 and let q = e−2π/ 39 . By the same method of proof as for Theorem 3.7 we find that P = 13/Q, and hence from Lemma 3.5 we deduce that 

  Q4 − 13Q3 − 91Q2 − 169Q + 169 Q4 + 13Q3 + 65Q2 + 169Q + 169 = 0. (3.7)

A numerical calculation shows that Q = Q(e−2π/ Therefore

√

39

) is not a root of the second factor.

Q4 − 13Q3 − 91Q2 − 169Q + 169 = 0, or equivalently, 

Q 3 Q2 + 6Q + 13

2



Q + 25 Q2 + 6Q + 13

 − 1 = 0.

Since Q(q) = P (q 3 ) = F (q 3 ) it follows from (1.8) that  2   3T q 3 + 25T q 3 − 1 = 0.

The explicit evaluation of T (e−2π 3/13 ) can be calculated by solving this quadratic equation and choosing the solution that is positive.

−π 3/13 The evaluation of T (−e ) is similar. Let P and Q be as for Lemma 3.5 and √ −π/ 39 let q = −e . By (2.9) and Lemma 3.5 we √may deduce that P = 13/Q. Therefore, the identity (3.7) holds for the value q = −e−π/ 39 . A numerical calculation shows that √ Q = Q(−e−π/ 39 ) is not a root of the first factor of (3.7). It follows that Q4 + 13Q3 + 65Q2 + 169Q + 169 = 0, or equivalently 

Q 3 Q2 + 6Q + 13

2

 −

Q Q2 + 6Q + 13

 − 1 = 0.

Hence,  2   3T q 3 − T q 3 − 1 = 0 and the claimed evaluation follows on choosing the solution that is negative. 2 Theorem 3.9. Let T (q) be defined by (1.8). Then

√ √   17 1/3 1/3 2  2  − 4563 + 273 273 4563 − 273 273 T e−2π 7/13 = − 63 189 189

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103

and

  1 T −e−π 7/13 = − . 7

Proof. The proof is similar to the proof of Theorem 3.8 so we will be brief.√The functions P = P (q) and Q = Q(q) in Lemma 3.6 evaluated at the point q = e−2π/ 91 satisfy the relation P = 13/Q, and therefore 

 Q6 − 91Q5 − 702Q4 − 3887Q3 − 9126Q2 − 15 379Q + 2197  2   × Q4 + 39Q3 + 260Q2 + 507Q + 169 Q2 + 13Q + 13 = 0.

(3.8)

A numerical calculation shows that neither the second nor third factors are zero for this particular value of q, and hence Q6 − 91Q5 − 702Q4 − 3887Q3 − 9126Q2 − 15 379Q + 2197 = 0. This is equivalent to  567

Q 2 Q + 6Q + 13

3

 − 459

Q 2 Q + 6Q + 13

2

 + 109

Q 2 Q + 6Q + 13

 − 1 = 0.



The claimed evaluation of T (e−2π 7/13 ) follows on solving the cubic equation. √ Next, the function Q = Q(q) in Lemma 3.6 evaluated at q = −e−π/ 91 also satisfies the polynomial equation (3.8). A numerical calculation shows that neither the first nor second factors are zero for this particular value of q, and hence Q2 + 13Q + 13 = 0. It follows that

  T −e−π 7/13 =

Q 1 Q = =− . Q2 + 6Q + 13 (Q2 + 13Q + 13) − 7Q 7

2

The next two theorems provide some general results. Theorem 3.10 (Duplication and dimidiation). Suppose 0 < q < 1 and let n be an integer. n Suppose R(q) or R(−q) can be evaluated in terms of radicals. Then R(±q 2 ) can also be expressed in terms of radicals. Proof. Let P and Q be as for Lemma 3.4. We make two observations. First, since R(q) or R(−q) can be evaluated in terms of radicals, so can P (q) or P (−q), respectively, by (1.6). Then P (q 2 ) can be expressed in terms of radicals by putting P = P (q) in (3.2) and solving for Q. Second, if R(q) can be evaluated in terms of radicals, then so can P (q). Then, on setting Q = P (q) in (3.2) and solving for P we obtain values of P (q 1/2 ) and P (−q 1/2 ). The claimed results follow by repeated application of the two observations. 2

104

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

Theorem 3.11 (Triplication and trimidiation). Suppose 0 < |q| < 1 and let n be any n integer. Suppose R(q) can be evaluated in terms of radicals. Then R(q 3 ) can also be evaluated in terms of radicals. Proof. Let P and Q be defined in Lemma 3.5. The identity (3.3) is equivalent to the quartic equation in Q:     Q4 − 15P + 6P 2 + P 3 Q3 − 78P + 33P 2 + 6P 3 Q2   − 169P + 78P 2 + 15P 3 Q + P 4 = 0.

(3.9)

On substituting the value P = P (q) and solving for Q we can obtain the value of P (q 3 ). Similarly, on substituting the value Q = P (q) and solving for P , we can deduce the value of P (q 1/3 ). 2 4. Evaluation of R(q) using Ramanujan–Weber class invariants Let χ = χ(q) be defined by χ(q) =

∞  

 1 + q 2j−1 .

(4.1)

j=1

The Ramanujan–Weber class invariant Gn is defined for any positive rational number n by  Gn = 2−1/4 q −1/24 χ(q) q=exp(−π√n) .

(4.2)

Explicit values of Gn have been sought by many authors over the course of more than a century. See [5, Chapter 34], [8–10,24,36] for more √ details. √ In this section we will show that each of F (e−2π n ) and F (−e−π n ) can be deduced from the value of G169n /Gn by solving appropriate quadratic equations. We shall also give values of G169n /Gn for n = 5, 9, 13, 69 and 129. The main tool required to establish the connection with the Ramanujan–Weber class invariant is: Theorem 4.1. Let f1 , f2 and V be defined by   1 E 2 (q 2 ) f1 = f1 (q) = F q 2 = 2 2 26 , q E (q ) f2 = f2 (q) = −F (−q) = and

1 E 2 (−q) q E 2 (−q 13 )

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

V = V (q) =

105

1 χ(q 13 ) . q 1/2 χ(q)

Then 13V 13 f1 − = = f2 V − V f1 f2 V



1 V − V

3

  1 . +7 V − V

(4.3)

Proof. By routine manipulations of infinite products we find that χ(q) =

E 2 (q 2 ) E(q)E(q 4 )

and E(−q) =

E 3 (q 2 ) . E(q)E(q 4 )

Hence, V =

E(q)E(q 4 )E 2 (q 26 ) q 1/2 E 2 (q 2 )E(q 13 )E(q 52 )

f2 =

1 E 6 (q 2 )E 2 (q 13 )E 2 (q 52 ) . q E 2 (q)E 2 (q 4 )E 6 (q 26 )

1

and

It follows that 1 E 4 (q 2 )E(q 13 )E(q 52 ) f1 = 3/2 = f2 V. V E(q)E(q 4 )E 4 (q 26 ) q

(4.4)

Therefore, the first identity in (4.3) holds. To prove the second identity in (4.3), replace q with −q in Lemma 3.4 and note that f1 (q) = Q(q) and f2 (q) = −P (−q), to get f13 − f23 = f12 f22 + 4f1 f22 − 4f12 f2 − 13f1 f2 .

(4.5)

By (4.4) we have f1 = f2 V 2 . Using this in (4.5) we deduce that f23 V 6 − f23 = f24 V 4 + 4f23 V 2 − 4f23 V 4 − 13f22 V 2 . On dividing by f23 V 3 and rearranging we obtain the second identity in (4.3).

2

For future reference we note that 

√   1 χ(q 13 ) G169n = 1/2 = V e−π n . √ Gn χ(q) q=exp(−π n) q

(4.6) √

Hence, if √G169n /Gn is known for a particular value√of n, then F (e−2π n ) and −F (−e−π n ) can be determined by letting V = V (e−π n ) = G169n /Gn in (4.3) and solving the resulting quadratic equations for f1 or f2 , respectively.

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

106

The next five theorems provide values of G169n /Gn for n = 5, 9, 13, 69 and 129. Theorem 4.2. The following explicit evaluation holds: 

−π

V e

 5/13

G65 = = G5/13

√

5+1 2

1/2  √

13 + 3 2

1/2 .

Proof. From the transformation formula (3.6) and the definitions (4.1) and (4.2) it is straightforward to deduce the identity [5, p. 223], [24], Gn = G1/n .

(4.7)

In addition, from [10, Theorem 4.1], we have G65 = G13/5

√

5+1 2

1/2  √

13 + 3 2

1/2 .

(4.8)

Hence, by (4.6), (4.7) and (4.8) we have

  G65 G65 V e−π 5/13 = = = G5/13 G13/5

√

5+1 2

1/2  √

13 + 3 2

1/2 .

2

Theorem 4.3. The following explicit evaluation holds:

  G117 V e−π 9/13 = = G9/13

√ √ 1/3 

11 + 6 3 + 9 + 6 3



. √ √ 11 + 6 3 − 9 + 6 3

Proof. From [8, Theorem 1] and [9, p. 149], we have, for any positive integer n,

 1/6 G9n = Gn p + p2 − 1  



1/3  p2 − 2 + (p2 − 1)(p2 − 4) p2 − 4 + (p2 − 1)(p2 − 4) + × 2 2 (4.9) and

 1/6 Gn/9 = Gn p + p2 − 1  



1/3  p2 − 2 + (p2 − 1)(p2 − 4) p2 − 4 + (p2 − 1)(p2 − 4) − × , 2 2 (4.10) where

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

107

p = G4n + G−4 n . From [5, p. 190], we have √ G13 = so p =

√

13 + 3 2

1/4 ,

(4.11)

13 when n = 13. Substituting this into (4.9) and (4.10), we deduce that

G117

√ 1/6 √ = G13 13 + 12





√ 11 + 6 3 + 2



√ 1/3 9+6 3 2

(4.12)

and G13/9 = G13

√

13 +

√

12

1/6





√ 11 + 6 3 − 2



√ 1/3 9+6 3 , 2

(4.13)

respectively. Dividing (4.12) by (4.13) yields G117 = G13/9

√ √ 1/3 

11 + 6 3 + 9 + 6 3



. √ √ 11 + 6 3 − 9 + 6 3

Using (4.6) and (4.7), we get 

−π

V e

 9/13

G117 G117 = = = G9/13 G13/9

√ √ 1/3 

11 + 6 3 + 9 + 6 3



. √ √ 11 + 6 3 − 9 + 6 3

2

Theorem 4.4. The following explicit evaluation holds:   G169 V e−π = G1 √ 1/3   1 √ 13 + 3 13 ( 13 + 2) + = 3 2 √ √

 1/3    √ √ 1/3 11 + 13 11 + 13 . +3 3 −3 3 × + 2 2 Proof. Use (4.6) and the values of G1 and G169 given in [5, pp. 189, 195]. 2 Theorem 4.5. The following explicit evaluation holds:

  G897 V e−π 69/13 = G69/13     √ √  √ √  1 = 60 + 9 39 + 56 + 9 39 8 + 39 + 4 + 39 . 4

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

108

Proof. Use (4.6) and the value of G897 /G69/13 given in [36, p. 388].

2

Theorem 4.6. The following explicit evaluation holds:

  V e−π 129/13

G1677 G129/13     √ √  √ √  1 = 355 + 54 43 + 351 + 54 43 17 + 2 43 + 13 + 2 43 . 4

=

Proof. Use (4.6) and the value of G1677 /G129/13 given in [36, p. 390].

2

5. Kronecker’s limit formula In [18, Theorem 4.1], without giving explicit evaluations, Horie and Kanou used Kro

necker’s limit formula to deduce a formula for evaluating R(q) at q = −e−π n/13 for n = 7, 15, 31, 55, 231 and 255. In this section, using Horie and Kanou’s formula, we

complete the calculations to deduce the values of R(−e−π n/13 ) for n = 15, 31, 55, 231 and 255; the case n = 7 has already been studied in Section 3. √ Lemma 5.1. (See [18, Theorem 4.1].) Let K = Q( −13m) be an imaginary quadratic field with odd discriminant −13m and assume that each genus of K has only one class. For each character χ of genus, there is a unique decomposition −13m = d1 d2 , where d1 and d2 are fundamental discriminants and d1 < 0, d2 > 0. Let B be the ideal √ [1, (1 + −13m/13)/2] in K, which is non-principal but its square is contained in the principal ideal class. Let F (q) be as defined in (1.7). Then

√   F −e−π m/13 = − 13



4h(d1 )h(d2 )

(εd2 ) ωd1 h(−13m) ,

(5.1)

χ(B)=−1

where χ runs through all genus characters with χ(B) = −1, and h(d), εd and ωd denote the class number (in the narrow sense), the fundamental unit and the number of roots √ of unity of the quadratic field Q( d), respectively. The following results, which we state without proof, may be obtained by applying (5.1) with m = 15, 31, 55, 231 and 255 and making use of the data in [13, pp. 515–519] and [33]:

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

109

Theorem 5.2. The following evaluations hold: √ 2 √   √ 1+ 5 3 + 13 F −e = − 13 , 2 2 √ 3  √  −π 31/13  3 + 13 = − 13 , F −e 2 √ 5 

√ √   1+ 5 , F −e−π 55/13 = − 13(8 + 65) 2 √ 5/2 

√ √ 1/2    5 + 21 23 + 4 33 F −e−π 231/13 = − 13 2 √   √ 1/2 145 + 7 429 1/2  × 727 + 44 273 2 

−π

 15/13

and √ 3 √ 4  

√   3 + 13 1+ 5 F −e−π 255/13 = − 13 2 2 √ √ 2    15 + 221 9 + 85 . × 2 2 The definitions (1.7) and (1.8) can be used

with the results of Theorem 5.2 to determine the values of R(−e−π m/13 ) or T (−e−π m/13 ) for m = 15, 31, 55, 231 and 255. The value of T turns out to be particularly simple for m = 31:

  −1 . T −e−π 31/13 = 124

This was used in [14] to produce the formula  n  ∞ 9  161 1 −1 √ = A(n) n + , π 750 124 31 n=0 where the coefficients A(n) satisfy a certain 6-term recurrence relation.

Other values of R(q), such as the values of R(e−2π m/13 ) for m = 15, 31, 55, 231 and 255, can then be determined by using Theorem 3.10. 6. An algebraic property of R(q) √

In [9, Section 6] and [15, Section 2], R(e−π n ) was proved to be a unit for any rational number n. This algebraic property also holds for R(q). In order to state the main results of this section, recall that the normalized Eisenstein series of weights 4 and 6 are defined by

S. Cooper, D. Ye / Journal of Number Theory 139 (2014) 91–111

110

M (q) = 1 + 240

∞  j 3 qj 1 − qj j=1

and N (q) = 1 − 504

∞  j 5 qj , 1 − qj j=1

respectively. The modular j-function is defined by j(τ ) =

1728M 3 (q) , M 3 (q) − N 2 (q)

q = e2πiτ .

Lemma 6.1. (See [14, Corollary 3.5].) Let j(τ ) denote the modular j-function and let R = R(q). Then  13   j(τ )R R2 + 3R − 1 + 1 − R + 5R2 + R3 + R4 1 + 235R + 1207R2 3 + 955R3 + 3840R4 − 955R5 + 1207R6 − 235R7 + R8 = 0. Using Lemma 6.1, we deduce the following theorem. Theorem 6.2. Let R(q) be defined by (1.5). Then R(e−π rational number n.

√ n

) is a unit for any positive

Proof. It is known that j(τ ) is an algebraic integer if τ is in an imaginary quadratic √ field, e.g., [32, p. 423]. It follows that R(e−π n ) is a unit, for any positive rational value √ of n, by taking τ = −n/2 in Lemma 6.1. 2 References [1] G.E. Andrews, B.C. Berndt, Ramanujan’s Lost Notebook, Part I, Springer, New York, 2005. [2] N.D. Baruah, N. Saikia, Some new explicit values of Ramanujan’s continued fractions, Indian J. Math. 46 (2004) 197–222. [3] B.C. Berndt, Ramanujan’s Notebooks: Part III, Springer-Verlag, New York, 1991. [4] B.C. Berndt, Ramanujan’s Notebooks: Part IV, Springer-Verlag, New York, 1994. [5] B.C. Berndt, Ramanujan’s Notebooks: Part V, Springer-Verlag, New York, 1998. [6] B.C. Berndt, H.H. Chan, Ramanujan’s explicit values for the classical theta-function, Mathematika 42 (1995) 278–294. [7] B.C. Berndt, H.H. Chan, Some values for the Rogers–Ramanujan continued fraction, Canad. J. Math. 47 (1995) 897–914. [8] B.C. Berndt, H.H. Chan, L.-C. Zhang, Ramanujan’s class invariants and cubic continued fraction, Acta Arith. 73 (1995) 67–85. [9] B.C. Berndt, H.H. Chan, L.-C. Zhang, Explicit evaluations of the Rogers–Ramanujan continued fraction, J. Reine Angew. Math. 480 (1996) 141–159. [10] B.C. Berndt, H.H. Chan, L.-C. Zhang, Ramanujan’s class invariants, Kronecker’s limit formula, and modular equations, Trans. Amer. Math. Soc. 349 (1997) 2125–2173. [11] B.C. Berndt, R.A. Rankin, Ramanujan: Letters and Commentary, AMS, Providence, RI, 1995. [12] H.H. Chan, H. Hahn, R. Lewis, S.L. Tan, New Ramanujan–Kolberg type partition identities, Math. Res. Lett. 9 (2002) 801–811. [13] H. Cohen, A Course in Computational Algebraic Number Theory, Springer-Verlag, New York, 1996. [14] S. Cooper, D. Ye, The Rogers–Ramanujan continued fraction and its level 13 analogue, J. Approx. Theory (2014), http://dx.doi.org/10.1016/j.jat.2014.01.008. [15] W. Duke, Continued fractions and modular functions, Bull. Amer. Math. Soc. 42 (2005) 137–162. [16] R.J. Evans, Theta function identities, J. Math. Anal. Appl. 147 (1990) 97–121.

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[17] G.H. Hardy, Ramanujan. Twelve Lectures on Subjects Suggested by His Life and Work, Third edition, AMS, Chelsea, Providence, Rhode Island, 1999, The first edition was published by Cambridge University Press, 1940. [18] T. Horie, N. Kanou, Certain modular functions similar to the Dedekind eta function, Abh. Math. Semin. Univ. Hambg. 72 (2002) 89–117. [19] S.Y. Kang, Ramanujan’s formulas for the explicit evaluation of the Rogers–Ramanujan continued fraction and theta-functions, Acta Arith. 90 (1999) 49–68. [20] K.G. Ramanathan, On Ramanujan’s continued fraction, Acta Arith. 43 (1984) 209–226. [21] K.G. Ramanathan, On the Rogers–Ramanujan continued fraction, Proc. Indian Acad. Sci. Math. Sci. 93 (1984) 67–77. [22] K.G. Ramanathan, Ramanujan’s continued fraction, Indian J. Pure Appl. Math. 16 (1985) 695–724. [23] K.G. Ramanathan, Some applications of Kronecker’s limit formula, J. Indian Math. Soc. 52 (1987) 71–89. [24] S. Ramanujan, Modular equations and approximations to π, Quart. J. Math. (Oxford) 45 (1914) 350–372. [25] S. Ramanujan, Notebooks (2 volumes), Tata Inst. Fund. Res, Bombay, 1957. [26] S. Ramanujan, Collected Papers, Chelsea, New York, 1962. [27] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 1988. [28] L.J. Rogers, Second memoir on the expansion of certain infinite products, Proc. Lond. Math. Soc. 25 (1894) 318–343. [29] K.R. Vasuki, K. Shivashankara, Some new values for the Rogers–Ramanujan continued fraction, J. Indian Math. Soc. 70 (2003) 87–95. [30] G.N. Watson, Theorems stated by Ramanujan (VII): Theorems on continued fractions, J. Lond. Math. Soc. 4 (1929) 39–48. [31] G.N. Watson, Theorems stated by Ramanujan (IX): Two continued fractions, J. Lond. Math. Soc. 4 (1929) 231–237. [32] H. Weber, Lehrbuch der Algebra, III, Vieweg, Braunschweig, 1908, reprinted by Chelsea, New York, 1961. [33] E. Weisstein, Wolfram MathWorld, http://mathworld.wolfram.com/ClassNumber.html and http:// mathworld.wolfram.com/FundamentalUnit.html, accessed August 3, 2013. [34] D. Ye, Level 13 modular forms and their applications, Massey University, Project Report. [35] J. Yi, Evaluations of the Rogers–Ramanujan continued fraction R(q) by modular equations, Acta Arith. 97 (2001) 103–127. [36] L.-C. Zhang, Ramanujan’s class invariants, Kronecker’s limit formula and modular equations. III, Acta Arith. 82 (1997) 379–392.