EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE

OF THE ROGERS-RAMANUJAN CONTINUED FRACTION

SHAUN COOPER AND DONGXI YE

Abstract. The Rogers-Ramanujan continued fraction has a represen-tation as an infinite product given by

q1/5∞∏j=1

(1− qj

)( j5 )

where |q| < 1 and(

jp

)is the Legendre symbol. In his letters to Hardy

and in his notebooks, Ramanujan recorded some exact numerical valuesof the Rogers-Ramanujan continued fraction for specific values of q. Inthis work, we give explicit evaluations of the level 13 analogue definedby

q

∞∏j=1

(1− qj

)( j13 )

.

1. Introduction

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

(1.1) R(q) =q1/5

1 +q

1 +q2

1 +q3

1 + · · ·denote the Rogers-Ramanujan continued fraction. In 1913, Ramanujan as-serted in his first letter to Hardy [11, p. 29] that

R(e−2π

)=

√5 +√

5

2−√

5 + 1

2,

−R(−e−π

)=

√5−√

5

2−√

5− 1

2

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 bywriting to Ramanujan [11, p. 77]:

2010 Mathematics Subject Classification. Primary—11J70; Secondary—11B65, 11F11,33E05. Date: December 19, 2013. Status: Journal of Number Theory, to appear.

1

2 SHAUN COOPER AND DONGXI YE

“What I should like above all is a definite proof of some of your resultsconcerning continued fractions of the type (1.1); and I am quite sure thatthe wisest thing you can do, in your own interests, is to let me have one assoon 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 easyto follow), I will (assuming that I agree with it—of which I have very littledoubt) try to get it published for you in England. Write it in the form of apaper · · · 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 Ra-manujan’s evaluations of R(q) had had on him [17, p. 9]:

“(They) defeated me completely; I had never seen anything in the leastlike them before. A single look at them is enough to show that they couldonly be written down by a mathematician of the highest class. They must betrue because, if they were not true, no one would have had the imaginationto invent them.”

Evaluations of R(q) and −R(−q) for other values of q have been studiedby various mathematicians, including Ramanujan [25, 27], Watson [30, 31]Ramanathan [20, 21, 22, 23], Berndt and Chan [6, 7], and Berndt, Chanand Zhang [9]. A large number of evaluations of R(q) and −R(−q) havebeen summarized by Kang [19, pp. 64–67] and further evaluations have beengiven by Yi [35], Vasuki and Shivashankara [29] and Baruah and Saikia [2].

The Rogers-Ramanujan continued fraction has other beautiful properties:

(1.2) R(q) = q1/5∞∏j=1

(1− q5j−4)(1− q5j−1)(1− q5j−3)(1− q5j−2)

,

1

R(q)− 1−R(q) =

E(q1/5

)q1/5E (q5)

(1.3)

and

1

R5(q)− 11−R5(q) =

E6 (q)

qE6 (q5),(1.4)

where

E(q) =

∞∏j=1

(1− qj).

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 provedby Watson [30].

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 3

In his second notebook [3, 25, Chapter 20, Entry 8(i)], Ramanujan stateda striking analogue of the identities (1.2)–(1.4): If

R(q) = q

∞∏j=1

(1− qj)(j13 )

(1.5)

= q

∞∏j=1

(1− q13j−12)(1− q13j−10)(1− q13j−9)(1− q13j−4)(1− q13j−3)(1− q13j−1)

(1− q13j−11)(1− q13j−8)(1− q13j−7)(1− q13j−6)(1− q13j−5)(1− q13j−2)

then

(1.6)1

R(q)− 3−R(q) =

E2(q)

qE2(q13).

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 invariantunder the congruence subgroup Γ1(13), where

Γ1(N) =

{(a bc 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 theRogers-Ramanujan continued fraction.

Motivated by the analogy between R(q) and R(q), this work is con-cerned with finding explicit evaluations of R(q). In the next section, wededuce two reciprocity formulas for R(q) and use them to give evaluations

of R(e−2π/

√13)

and R(−e−π/

√13)

. 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 Reciprocity formulas (Section 2)

2, 3, 7 P–Q modular equations (Section 3)

5, 9, 13, 69, 129 Ramanujan-Weber class invariant (Section 4)

15, 31, 55, 231, 255 Kronecker’s limit formula (Section 5)

4 SHAUN COOPER AND DONGXI YE

We shall also prove some general results. In Theorem 3.10 it is shown thatfor any 0 < q < 1, if R(q) or R(−q) can be evaluated in terms of radicals,then so can R(±q2n) for any integer n. In Theorem 3.11 it is shown thatif |q| < 1 and R(q) can be evaluated in terms of radicals, then so can R(q3

n)

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

(1.7) F = F (q) =1

R(q)− 3−R(q) =

E2(q)

qE2(q13)=

1

q

∞∏j=1

(1− qj)2

(1− q13j)2.

Let T be defined by

(1.8) T = T (q) =F (q)

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

(1 +R(q)2)2.

The function T (q) was studied in [14]. The function F (q) was also studiedin [14] where it was denoted by 1/S(q).

The values of T (q) often turn out to have simpler expressions in termsof radicals than the corresponding values of R(q) or F (q). Clearly, if T canbe evaluated at a particular value of q, then F and R can be evaluated atthe same value of q by solving appropriate quadratic equations and usingnumerical values to determine the root. We will use the phrase “explicitevaluation of R(q)” to mean that we have evaluated any of R(q), F (q) orT (q) in terms of radicals.

2. Reciprocity formulas

In his second notebook [3, p. 83], [25, Ch. 16, Entry 39], Ramanujanstated two reciprocity formulas for the Rogers-Ramanujan continued frac-tion: if α, β > 0 and αβ = 1, then{√

5 + 1

2+R

(e−2πα

)}{√5 + 1

2+R

(e−2πβ

)}=

5 +√

5

2(2.1)

and {√5− 1

2−R

(−e−πα

)}{√5− 1

2−R

(−e−πβ

)}=

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 + 1

2

)5

+R5(e−2πα

)}{(√5 + 1

2

)5

+R5(e−2πβ

)}(2.3)

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 5

= 5√

5

(√5 + 1

2

)5

and {(√5− 1

2

)5

−R5(−e−πα

)}{(√5− 1

2

)5

−R5(−e−πβ

)}(2.4)

= 5√

5

(√5− 1

2

)5

.

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

2+R

(e−2πα

)}{√13 + 3

2+R

(e−2πβ

)}=

13 + 3√

13

2(2.5)

and {√13− 3

2−R

(−e−πα

)}{√13− 3

2−R

(−e−πβ

)}=

13− 3√

13

2.(2.6)

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

to evaluate R(e−2π/

√13n)

and R(−e−π/

√13n)

if we know the values of

R(e−2π√n/13

)and R

(−e−π

√n/13

), respectively.

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

(2.7)

{√13− 3

2−R(q)

}{√13 + 3

2+R(q)

}=R(q)E2(q)

qE2(q13),

so for any t > 0,

(2.8) −√

13 + 3

2< R

(−e−πt

)< 0.

Let a and b be positive real numbers with ab = π2. Then [3, p. 43, En-try 27(iv)]

(2.9) e−a/24a1/4E(−e−a

)= e−b/24b1/4E

(−e−b

).

Successively taking (a, b) = (πα, 13πβ) and (a, b) = (13πα, πβ) in (2.9), wededuce that

e−πα/24(πα)1/4E(−e−πα

)= e−13πβ/24(13πβ)1/4E

(−e−13πβ

)(2.10)

6 SHAUN COOPER AND DONGXI YE

and

e−13πα/24(13πα)1/4E(−e−13πα

)= e−πβ/24(πβ)1/4E

(−e−πβ

),(2.11)

respectively. On dividing (2.10) by (2.11) and squaring, we get

(2.12)E2 (−e−πα)√

13 (−e−πα)E2 (−e−13πα)=

√13(−e−πβ

)E2(−e−13πβ

)E2 (−e−πβ)

.

Now let q take the values −e−πα and −e−πβ in (2.7) and multiply the tworesulting identities. Making use of (2.12) we get

(A+ ρ)(B + ρ)(A+ ρ )(B + ρ ) = 13AB,

where

A = R(−e−πα

), B = R

(−e−πβ

), ρ =

3 +√

13

2, ρ =

3−√

13

2.

This can be rearranged to give

(2.13)(

(A+ ρ) (B + ρ)− ρ√

13)(

(A+ ρ ) (B + ρ ) + ρ√

13)

= 0.

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 factorin (2.13) is zero and this gives (2.6). �

The reciprocity formulas in Theorem 2.1 lead to evaluations ofR(e−2π/

√13)

and R(−e−π/

√13)

.

Theorem 2.2. Let R(q) be defined by (1.5). Then

R(e−2π/

√13)

=−(3 +

√13) +

√26 + 6

√13

2(2.14)

and

R(−e−π/

√13)

=−(3−

√13)−

√26− 6

√13

2.(2.15)

Proof. Setting α = β = 1/√

13 in (2.5) and (2.6), we get{√13 + 3

2+R

(e−2π/

√13)}2

=13 + 3

√13

2

and {√13− 3

2−R

(−e−π/

√13)}2

=13− 3

√13

2.

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 7

Taking square roots and using numerical values of R(e−2π/

√13)

and

R(−e−π/

√13)

to determine the signs, we deduce (2.14) and (2.15). �

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 outthat Lemmas 3.1–3.3 will involve the function q−n/2E(qn)/E(q13n), whereasLemmas 3.4–3.6 are about the squared function q−nE2(qn)/E2(q13n).

In his second notebook [25, p. 327], Ramanujan recorded the quadratictransformation formula:

Lemma 3.1. Let P and Q be defined by

P = P(q) =E(q)

q1/2E(q13)and Q = Q(q) =

E(q2)

qE(q26).

Then

PQ+13

PQ=

(P3

Q3+Q3

P3

)− 4

(PQ

+QP

).

Proof. See [4, p. 211, Entry 57]. �

The analogous cubic transformation formula was also known to Ramanu-jan [25, p. 322]:

Lemma 3.2. Let P and Q be defined by

P = P(q) =E(q)

q1/2E(q13)and Q = Q(q) =

E(q3)

q3/2E(q39).

Then

PQ+13

PQ=

(P2

Q2+Q2

P2

)− 3

(PQ

+QP

)− 3.

Proof. See [4, p. 237, Entry 72]. �

The corresponding transformation formula of degree 7 has been provedin [34]:

Lemma 3.3. Let P and Q be defined by

Let P = P(q) =E(q)

q1/2E(q13)and Q = Q(q) =

E(q7)

q7/2E(q91).

Then

(PQ)3 +133

(PQ)3

(3.1)

8 SHAUN COOPER AND DONGXI YE

=

(Q4

P4− P

4

Q4

)− 21

(Q3

P3+P3

Q3

)− 7

(Q2

P2− P

2

Q2

)− 196

(QP

+PQ

)− 7

(P3Q+Q3P

)− 21

(Q3

P+P3

Q

)− 273

(QP3

+PQ3

)− 1183

(1

P3Q+

1

Q3P

)− 56PQ− 728

PQ.

Now we are ready to give analogues of Lemmas 3.1–3.3 that involve thesquared functions.

Lemma 3.4.If P and Q are defined by

P = P (q) =E2(q)

qE2(q13)and Q = Q(q) =

E2(q2)

q2E2(q26),

then

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

Proof. Square both sides of the identity in Lemma 3.1 and write P = P2

and Q = Q2. The result simplifies to

(P 3 − 4P 2Q− 13PQ− P 2Q2 − 4PQ2 +Q3)

× (P 3 − 4P 2Q+ 13PQ+ P 2Q2 − 4PQ2 +Q3) = 0.

The q-expansion of the second factor is not zero. It follows that the firstfactor is identically zero and this completes the proof. �

Lemma 3.5.If P and Q are defined by

P = P (q) =E2(q)

qE2(q13)and Q = Q(q) =

E2(q3)

q3E2(q39),

then

PQ+132

PQ(3.3)

=

(P 2

Q2+Q2

P 2

)− 15

(P

Q+Q

P

)− 6 (P +Q)− 78

(1

P+

1

Q

)− 33.

Proof. Move the terms that involve odd powers of P and Q in Lemma 3.2to one side, and the terms that involve even powers to the other side. Thensquare both sides and write P = P2 and Q = Q2. �

Lemma 3.6.If P and Q are defined by

P = P (q) =E2(q)

qE2(q13)and Q = Q(q) =

E2(q7)

q7E2(q91),

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 9

then

(PQ)3 +136

(PQ)3

(3.4)

=

(P 4

Q4+Q4

P 4

)− 455

(P 3

Q3+Q3

P 3

)− 19649

(P 2

Q2+Q2

P 2

)− 124348

(P

Q+Q

P

)− 882

(P 3

Q2+Q3

P 2

)− 11466

(P 2

Q3+Q2

P 3

)− 14406

(P 2

Q+Q2

P

)− 735

(P 3

Q+Q3

P

)− 187278

(P

Q2+

Q

P 2

)− 124215

(P

Q3+

Q

P 3

)− 2599051

(1

P 3Q+

1

Q3P

)− 5198102

(1

P 3Q2+

1

Q3P 2

)− 3229590

(1

P 2Q+

1

Q2P

)− 738192

(1

P 3+

1

Q3

)− 1003184

(1

P 2+

1

Q2

)− 616798

(1

P+

1

Q

)− 336

(P 3 +Q3

)− 5936

(P 2 +Q2

)− 47446 (P +Q)

− 14(P 3Q2 +Q3P 2

)− 91

(P 3Q+Q3P

)− 1470

(P 2Q+Q2P

)− 210P 2Q2

− 5997810

P 2Q2− 1852578

PQ− 10962PQ− 219800.

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

The next three theorems provide explicit evaluations of R(q) at q =

e−2π√n/13 and q = −e−π

√n/13 for n = 2, 3, 7. The complexity of the evalu-

ations varies considerably. 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 (qn)

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 =17

24− 1

24

(624√

78 + 5239)1/3

+1

24

(624√

78− 5239)1/3

and

u =1− 6t+

√1− 12t− 16t2

2tand let v1, v2 and v3 be the algebraic numbers defined by

(3.5) x3 − (u2 + 4u)x2 − (4u2 + 13u)x+ u3 = (x− v1)(x− v2)(x− v3)with v1 < 0 < v2 < v3. Let F (q) be defined by (1.7). Then

F(e−2π√

2/13)

= u ≈ 9.69, F(− e−π

√2/13

)= v1 ≈ −4.97,

10 SHAUN COOPER AND DONGXI YE

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)]:

(3.6) e−a/12a1/4E(e−2a

)= e−b/12b1/4E

(e−2b

), a, b > 0, ab = π2.

Let q = e−2π/√26. By the definitions of P and Q in Lemma 3.4 together

with (3.6) we find that

Q(q) = Q(e−2π/

√26)

=E2(e−2π√

2/13)

e−2π/√13E2

(e−2π

√26)

= 13e−2π/√26 E

2(e−26π/

√26)

E2(e−2π/

√26)

=13

P (e−2π/√26)

=13

P (q).

On substituting P = 13/Q into the result of Lemma 3.4 and simplifying, weget

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 (q2) = F (q2) it follows from (1.8) that

8T (q2)3 − 17T (q2)2 + 18T (q2)− 1 = 0.

On solving the cubic and selecting the appropriate root we conclude that

T(e−2π√

2/13)

= T(q2)

=17

24− 1

24

(624√

78 + 5239)1/3

+1

24

(624√

78− 5239)1/3

= t.

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 symmetricin P and Q, then substituting P = u and solving for Q, we deduce that

F(e−4π√

2/13)

is also a root of the cubic equation (3.5). Numerical calcu-lations may be used to match the three evaluations with the correspondingroots of the cubic equation. �

Theorem 3.8. Let T (q) be defined by (1.8). Then

T(e−2π√

3/13)

=7√

13− 25

6

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 11

and

T(− e−π

√3/13

)=

1−√

13

6.

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 fromLemma 3.5 we deduce that(3.7)(Q4 − 13Q3 − 91Q2 − 169Q+ 169

) (Q4 + 13Q3 + 65Q2 + 169Q+ 169

)= 0.

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

√39)

is not a root of thesecond factor. Therefore

Q4 − 13Q3 − 91Q2 − 169Q+ 169 = 0,

or equivalently,

3

(Q

Q2 + 6Q+ 13

)2

+ 25

(Q

Q2 + 6Q+ 13

)− 1 = 0.

Since Q(q) = P (q3) = F (q3) it follows from (1.8) that

3T (q3)2 + 25T (q3)− 1 = 0.

The explicit evaluation of T(e−2π√

3/13)

can be calculated by solving thisquadratic equation and choosing the solution that is positive.

The evaluation of T(− e−π

√3/13

)is similar. Let P and Q be as for

Lemma 3.5 and let q = −e−π/√39. By (2.9) and Lemma 3.5 we may de-

duce 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)

isnot a root of the first factor of (3.7). It follows that

Q4 + 13Q3 + 65Q2 + 169Q+ 169 = 0,

or equivalently

3

(Q

Q2 + 6Q+ 13

)2

−(

Q

Q2 + 6Q+ 13

)− 1 = 0.

Hence,

3T (q3)2 − T (q3)− 1 = 0

and the claimed evaluation follows on choosing the solution that is negative.�

Theorem 3.9. Let T (q) be defined by (1.8). Then

T(e−2π√

7/13)

=17

63− 2

189

(4563 + 273

√273)1/3− 2

189

(4563− 273

√273)1/3

and

T(− e−π

√7/13

)= −1

7.

12 SHAUN COOPER AND DONGXI YE

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 − 15379Q+ 2197

)(3.8)

×(Q4 + 39Q3 + 260Q2 + 507Q+ 169

)2 (Q2 + 13Q+ 13

)= 0.

A numerical calculation shows that neither the second nor third factors arezero for this particular value of q, and hence

Q6 − 91Q5 − 702Q4 − 3887Q3 − 9126Q2 − 15379Q+ 2197 = 0.

This is equivalent to

567

(Q

Q2 + 6Q+ 13

)3

− 459

(Q

Q2 + 6Q+ 13

)2

+ 109

(Q

Q2 + 6Q+ 13

)− 1 = 0.

The claimed evaluation of T(e−2π√

7/13)

follows on solving the cubic equa-tion.

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 showsthat neither the first nor second factors are zero for this particular valueof q, and hence Q2 + 13Q+ 13 = 0. It follows that

T(− e−π

√7/13

)=

Q

Q2 + 6Q+ 13=

Q

(Q2 + 13Q+ 13)− 7Q= −1

7.

�

The next two theorems provide some general results.

Theorem 3.10 (Duplication and dimidiation). Suppose 0 < q < 1 and let nbe an integer. Suppose R(q) or R(−q) can be evaluated in terms of radicals.Then R(±q2n) 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) orP (−q), respectively, by (1.6). Then P (q2) can be expressed in terms ofradicals 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 (q1/2) and P (−q1/2).The claimed results follow by repeated application of the two observations.

�

Theorem 3.11 (Triplication and trimidiation). Suppose 0 < |q| < 1 and letn be any integer. Suppose R(q) can be evaluated in terms of radicals. ThenR(q3

n) can also be evaluated in terms of radicals.

Proof. Let P andQ be defined in Lemma 3.5. The identity (3.3) is equivalentto the quartic equation in Q:

Q4 −(15P + 6P 2 + P 3

)Q3 −

(78P + 33P 2 + 6P 3

)Q2(3.9)

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 13

−(169P + 78P 2 + 15P 3

)Q+ P 4 = 0.

On substituting the value P = P (q) and solving for Q we can obtain thevalue of P (q3). Similarly, on substituting the value Q = P (q) and solving

for P , we can deduce the value of P (q1/3). �

4. Evaluation of R(q) using Ramanujan-Weber class invariants

Let χ = χ(q) be defined by

(4.1) χ(q) =∞∏j=1

(1 + q2j−1).

The Ramanujan-Weber class invariant Gn is defined for any positive rationalnumber n by

(4.2) Gn = 2−1/4[q−1/24χ(q)

]q=exp(−π

√n).

Explicit values of Gn have been sought by many authors over the course ofmore than a century. See [5, Chapter 34], [8, 9, 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 quadraticequations. We shall also give values of G169n/Gn for n = 5, 9, 13, 69and 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

f1 = f1(q) = F (q2) =1

q2E2(q2)

E2(q26),

f2 = f2(q) = −F (−q) =1

q

E2(−q)E2(−q13)

and

V = V (q) =1

q1/2χ(q13)

χ(q).

Then

(4.3)f1V− 13V

f1= f2V −

13

f2V=

(V − 1

V

)3

+ 7

(V − 1

V

).

Proof. By routine manipulations of infinite products we find that

χ(q) =E2(q2)

E(q)E(q4)and E(−q) =

E3(q2)

E(q)E(q4).

Hence,

V =1

q1/2E(q)E(q4)E2(q26)

E2(q2)E(q13)E(q52)

14 SHAUN COOPER AND DONGXI YE

and

f2 =1

q

E6(q2)E2(q13)E2(q52)

E2(q)E2(q4)E6(q26).

It follows that

(4.4)f1V

=1

q3/2E4(q2)E(q13)E(q52)

E(q)E(q4)E4(q26)= f2V.

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

(4.5) f31 − f32 = f21 f22 + 4f1f

22 − 4f21 f2 − 13f1f2.

By (4.4) we have f1 = f2V2. Using this in (4.5) we deduce that

f32V6 − f32 = f42V

4 + 4f32V2 − 4f32V

4 − 13f22V2.

On dividing by f32V3 and rearranging we obtain the second identity in (4.3).

�

For future reference we note that

(4.6)G169n

Gn=

[1

q1/2χ(q13)

χ(q)

]q=exp(−π

√n)

= V(e−π√n).

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.

The next five theorems provide values of G169n/Gn for n = 5, 9, 13, 69and 129.

Theorem 4.2. The following explicit evaluation holds:

V(e−π√

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

V(e−π√

5/13)

=G65

G5/13=

G65

G13/5=

(√5 + 1

2

)1/2(√13 + 3

2

)1/2

.

�

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 15

Theorem 4.3. The following explicit evaluation holds:

V(e−π√

9/13)

=G117

G9/13=

(√11 + 6

√3 +

√9 + 6

√3√

11 + 6√

3−√

9 + 6√

3

)1/3

.

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

G9n = Gn

(p+

√p2 − 1

)1/6(4.9)

×

√p2 − 2 +

√(p2 − 1)(p2 − 4)

2+

√p2 − 4 +

√(p2 − 1)(p2 − 4)

2

1/3

and

Gn/9 = Gn

(p+

√p2 − 1

)1/6(4.10)

×

√p2 − 2 +

√(p2 − 1)(p2 − 4)

2−

√p2 − 4 +

√(p2 − 1)(p2 − 4)

2

1/3

,

where

p = G4n +G−4n .

From [5, p. 190], we have

(4.11) G13 =

(√13 + 3

2

)1/4

,

so p =√

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

G117 = G13

(√13 +

√12)1/6√11 + 6

√3

2+

√9 + 6

√3

2

1/3

(4.12)

and

G13/9 = G13

(√13 +

√12)1/6√11 + 6

√3

2−

√9 + 6

√3

2

1/3

,(4.13)

16 SHAUN COOPER AND DONGXI YE

respectively. Dividing (4.12) by (4.13) yields

G117

G13/9=

(√11 + 6

√3 +

√9 + 6

√3√

11 + 6√

3−√

9 + 6√

3

)1/3

.

Using (4.6) and (4.7), we get

V(e−π√

9/13)

=G117

G9/13=

G117

G13/9=

(√11 + 6

√3 +

√9 + 6

√3√

11 + 6√

3−√

9 + 6√

3

)1/3

.

�

Theorem 4.4. The following explicit evaluation holds:

V(e−π

)=G169

G1

=1

3

(√13 + 2)

+

(13 + 3

√13

2

)1/3

×

(11 +√

13

2+ 3√

3

)1/3

+

(11 +

√13

2− 3√

3

)1/3 .

Proof. Use (4.6) and the values of G1 and G169 given in [5, pp. 189, 195]. �

Theorem 4.5. The following explicit evaluation holds:

V(e−π√

69/13)

=G897

G69/13

=1

4

(√60 + 9

√39 +

√56 + 9

√39

)(√8 +√

39 +

√4 +√

39

).

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

Theorem 4.6. The following explicit evaluation holds:

V(e−π√

129/13)

=G1677

G129/13

=1

4

(√355 + 54

√43 +

√351 + 54

√43

)(√17 + 2

√43 +

√13 + 2

√43

).

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

5. Kronecker’s limit formula

In [18, Theorem 4.1], without giving explicit evaluations, Horie and Kanouused Kronecker’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.

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 17

Lemma 5.1. [18, Theorem 4.1] Let K = Q(√−13m

)be an imaginary

quadratic field with odd discriminant −13m and assume that each genus ofK has only one class. For each character χ of genus, there is a uniquedecomposition

−13m = d1d2,

where d1 and d2 are fundamental discriminants and d1 < 0, d2 > 0. LetB 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

(5.1) F(−e−π

√m/13

)= −√

13∏

χ(B)=−1

(εd2)4h(d1)h(d2)ωd1

h(−13m) ,

where χ runs through all genus characters with χ(B) = −1, and h(d), εdand 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 byapplying (5.1) with m = 15, 31, 55, 231 and 255 and making use of the datain [13, pp. 515–519] and [33]:

Theorem 5.2. The following evaluations hold:

F(−e−π

√15/13

)= −√

13

(3 +√

13

2

)(1 +√

5

2

)2

,

F(−e−π

√31/13

)= −√

13

(3 +√

13

2

)3

,

F(−e−π

√55/13

)= −√

13(

8 +√

65)(1 +

√5

2

)5

,

F(−e−π

√231/13

)= −√

13

(5 +√

21

2

)5/2 (23 + 4

√33)1/2

×(

727 + 44√

273)1/2(145 + 7

√429

2

)1/2

and

F(−e−π

√255/13

)= −√

13

(1 +√

5

2

)4(3 +√

13

2

)3

×

(9 +√

85

2

)2(15 +

√221

2

).

18 SHAUN COOPER AND DONGXI YE

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 simplefor m = 31:

T(−e−π

√31/13

)=−1

124.

This was used in [14] to produce the formula

1

π=

9√31

∞∑n=0

A(n)

(n+

161

750

)(−1

124

)n,

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

M(q) = 1 + 240

∞∑j=1

j3qj

1− qjand N(q) = 1− 504

∞∑j=1

j5qj

1− qj,

respectively. The modular j-function is defined by

j(τ) =1728M3(q)

M3(q)−N2(q), q = e2πiτ .

Lemma 6.1. [14, Corollary 3.5] Let j(τ) denote the modular j-function andlet R = R(q). Then

j(τ)R(R2 + 3R− 1

)13+(1−R+ 5R2 +R3 +R4

) (1 + 235R+ 1207R2

+ 955R3 + 3840R4 − 955R5 + 1207R6 − 235R7 +R8)3

= 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√n)

is a unit for

any positive rational number n.

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. �

EXPLICIT EVALUATIONS OF A LEVEL 13 ANALOGUE 19

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Institute of Natural and Mathematical Sciences, Massey University-Albany,Private Bag 102904, North Shore Mail Centre, Auckland, New Zealand, E-mail: s.cooper@massey.ac.nz, lawrencefrommath@gmail.com