Home Ramanujan’s function k(τ)=r(τ)r2(2τ) and its modularity
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Ramanujan’s function k(τ)=r(τ)r2(2τ) and its modularity

  • Yoonjin Lee and Yoon Kyung Park EMAIL logo
Published/Copyright: December 31, 2020

Abstract

We study the modularity of Ramanujan’s function k ( τ ) = r ( τ ) r 2 ( 2 τ ) , where r ( τ ) is the Rogers-Ramanujan continued fraction. We first find the modular equation of k ( τ ) of “an” level, and we obtain some symmetry relations and some congruence relations which are satisfied by the modular equations; these relations are quite useful for reduction of the computation cost for finding the modular equations. We also show that for some τ in an imaginary quadratic field, the value k ( τ ) generates the ray class field over an imaginary quadratic field modulo 10; this is because the function k is a generator of the field of the modular function on Γ 1 ( 10 ) . Furthermore, we suggest a rather optimal way of evaluating the singular values of k ( τ ) using the modular equations in the following two ways: one is that if j ( τ ) is the elliptic modular function, then one can explicitly evaluate the value k ( τ ) , and the other is that once the value k ( τ ) is given, we can obtain the value k ( r τ ) for any positive rational number r immediately.

MSC 2010: 11F03; 11R37; 11R04; 14H55

1 Introduction

Let Γ be a congruence subgroup of S L 2 ( Z ) , be the complex upper half plane, and { } . We consider the compact Riemann surface Γ \ and the field K ( Γ ) of meromorphic functions which are invariant under Γ . If the congruence subgroup Γ has genus zero, there is a function f ( τ ) which satisfies K ( Γ ) = ( f ( τ ) ) . For a positive integer N, we call the function j ( N ) ( τ ) a Hauptmodul of level N if it is a generator for K ( Γ 0 ( N ) ) which has a simple pole at the cusp at infinity. Clearly, it is unique up to a constant. Note that the genus of Γ 0 ( N ) is zero only for N = 1 , 2 , , 10 , 12 , 13 , 16 , 18 , and 25, and Table 1 describes Hauptmoduln j ( N ) ( τ ) in terms of η -quotients for the levels N 2 [1,2], where η ( τ ) = q 1 / 24 n = 1 ( 1 q n ) with q = e 2 π i τ .

Table 1

Hauptmoduln of levels 2 , , 10 , 12 , 13 , 16 , 18 , and 25

N 2 3 4 5 6 7 8
j ( N ) ( τ ) η ( τ ) 24 η ( 2 τ ) 24 η ( τ ) 12 η ( 3 τ ) 12 η ( τ ) 8 η ( 4 τ ) 8 η ( τ ) 6 η ( 5 τ ) 6 η ( 2 τ ) 3 η ( 3 τ ) 9 η ( τ ) 3 η ( 6 τ ) 9 η ( τ ) 4 η ( 7 τ ) 4 η ( τ ) 4 η ( 4 τ ) 2 η ( 2 τ ) 2 η ( 8 τ ) 4
N 9 10 12 13 16 18 25
j ( N ) ( τ ) η ( τ ) 3 η ( 9 τ ) 3 η ( 2 τ ) η ( 5 τ ) 5 η ( τ ) η ( 10 τ ) 5 η ( 4 τ ) 4 η ( 6 τ ) 2 η ( 2 τ ) 2 η ( 12 τ ) 4 η ( τ ) 2 η ( 13 τ ) 2 η ( τ ) 2 η ( 8 τ ) η ( 2 τ ) η ( 16 τ ) 2 η ( 6 τ ) η ( 9 τ ) 3 η ( 3 τ ) η ( 18 τ ) 3 η ( τ ) η ( 25 τ )

We consider the following equation:

(1.1) 1 X N X N = j ( N ) ( τ ) + c N

with a solution ( X N , c N ) , where X N is the modular function and c N is a constant. Then the solutions ( r 5 ( τ ) , 11 ) , ( r 13 ( τ ) , 3 ) and ( r ( 5 τ ) , 1 ) are found for N = 5 , 13 , and 25, respectively. In detail,

(1.2) 1 r 5 ( τ ) r 5 ( τ ) = j ( 5 ) ( τ ) + 11 ,

(1.3) 1 r 13 ( τ ) r 13 ( τ ) = j ( 13 ) ( τ ) + 3 ,

and

(1.4) 1 r ( 5 τ ) r ( 5 τ ) = j ( 25 ) ( τ ) + 1 ,

where r ( τ ) is the Rogers-Ramanujan continued fraction

r ( τ ) = q 1 / 5 1 + q 1 + q 2 1 + q 3 1 + = q 1 5 n = 1 ( 1 q n ) n 5 ,

and r 13 ( τ ) is the level 13 analogue of the Rogers-Ramanujan continued fraction

r 13 ( τ ) = q n = 1 ( 1 q n ) n 13

with the Legendre symbol n p . Identities (1.2) and (1.4) were stated by Ramanujan [3, pp. 85 and 267] and proved by Watson [4], and Identity (1.3) was proved by Cooper and Ye [5] and Lee and Park [6].

On the other hand, Cooper [7, Theorem 3.5] showed that ( X 10 , c 10 ) = ( k ( τ ) , 0 ) is a solution to the following equation:

(1.5) 1 X 10 X 10 = j ( 10 ) ( τ ) + c 10

by using the following identity:

(1.6) k ( τ ) = r ( τ ) r 2 ( 2 τ ) .

In fact, Ramanujan used the function k ( τ ) as a parameter for expressing r 5 ( τ ) and r 5 ( 2 τ ) in [8, p. 326]:

if r 5 ( τ ) = k 1 k 1 + k 2 , then r 5 ( 2 τ ) = k 2 1 + k 1 k .

Therefore, it is clear that there is a significant meaning of the function k ( τ ) , but there has been no investigation yet for k ( τ ) as a modular function. It is thus certainly worthy of studying the modularity of the function k ( τ ) . This is one of the motivations for this paper: we study the modularity of Ramanujan’s function k ( τ ) = r ( τ ) r 2 ( 2 τ ) .

The field of modular functions on Γ ( 5 ) and Γ 0 ( 13 ) is generated by r ( τ ) and 1 / r 13 ( τ ) r 13 ( τ ) , respectively. Koo and Shin found all the generators of Γ 1 ( N ) whose genus is zero [9, p. 161], and they chose 1 / k ( τ ) as a generator of Γ 1 ( 10 ) . Therefore, the modular function k ( τ ) is a generator of the function field K ( Γ 1 ( 10 ) ) . Furthermore, using (1.5), we see that 1 / k ( τ ) k ( τ ) is a generator of K ( Γ 0 ( 10 ) ) . One can get the relation between k ( τ ) and k ( n τ ) for any positive integer n using the previous facts as above. For τ = τ + 1 / 2 , 2 τ + 1 / 2 , 3 τ , and 5 τ , the relations of this type were found by Xia and Yao [10]. These relations are called modular equations which satisfy Kronecker’s congruence.

In this paper, we investigate the modularity of Ramanujan’s function k ( τ ) = r ( τ ) r 2 ( 2 τ ) . We first find the modular equation of k ( τ ) of “any” level, and we obtain some symmetry relations and some congruence relations which are satisfied by the modular equations; these relations are quite useful for reduction of the computation cost for finding the modular equations (Theorem 1.1). We also show that for some τ in an imaginary quadratic field, the value k ( τ ) generates the ray class field over an imaginary quadratic field modulo 10; this is because the function k is a generator of the field of the modular function on Γ 1 ( 10 ) (Theorem 1.2). Furthermore, we suggest a rather optimal way of evaluating the singular values of k ( τ ) using the modular equations in the following two ways: one is that if j ( τ ) is the elliptic modular function, then one can explicitly evaluate the value k ( τ ) (Theorem 1.3), and the other is that once the value k ( τ ) is given, we can obtain the value k ( r τ ) for any positive rational number r immediately (Theorem 1.4). For any congruence subgroup Γ such that Γ 1 ( 10 ) Γ SL 2 ( Z ) , any function f K ( Γ ) can be written as the rational function of k ( τ ) since k ( τ ) is the generator of K ( Γ 1 ( 10 ) ) . For example, combining the first formula in [11, Corollary 3.40] with (1.5), we get the same formula in Theorem 1.3. In Theorem 1.3, we focus on finding the explicit relation of k ( τ ) with the modular j-function directly. If one is interested in expressing j ( d τ ) in terms of k ( τ ) for d = 1 , 2 , 5 , 10 , then one can use the result in [11, Theorems 10.8 and 10.13] in order to write η ( d τ ) and Q ( q d ) in terms of k ( τ ) since j ( τ ) = Q 3 ( q ) / η 24 ( τ ) , where Q ( q ) = 1 / 240 + m = 1 m 3 q m / ( 1 q m ) .

Now we state four theorems, which are our main results in detail:

Theorem 1.1

Let k ( τ ) be a solution to (1.1) for N = 10 and c 10 = 0.

  1. One can explicitly obtain the modular equation of k ( τ ) of level n for any positive integer n.

  2. For every positive integer n with ( n , 10 ) = 1 , the modular equation F n ( X , Y ) of k ( τ ) has the following symmetry:

    F n ( X , Y ) = F n ( Y , X ) i f n ± 1 ( mod 10 ) , F n ( X , Y ) = Y d F n ( Y 1 , X ) i f n ± 3 ( mod 10 ) ,

    where d = n p | n ( 1 + p 1 ) .

  3. For any odd prime p 5 , the modular equation F p ( X , Y ) of k ( τ ) of level p is congruent to

( X p Y ) ( X Y p ) mod p Z [ X , Y ] i f p ± 1 ( mod 10 ) , ( X p Y ) ( X Y p + 1 ) mod p Z [ X , Y ] i f p ± 3 ( mod 10 ) .

Using the modularity, one can construct some ray class fields over an imaginary quadratic field. Theorem 1.2 shows that k ( τ ) generates the ray class field modulo 10 over an imaginary quadratic field.

Theorem 1.2

Let K be an imaginary quadratic field with discriminant d K and τ K be a root of the primitive equation a x 2 + b x + c = 0 such that b 2 4 a c = d K and ( a , 10 ) = 1 , where a , b , c Z . Then K ( k ( τ ) ) is the ray class field modulo 10 over K.

It is mentioned that the value k ( τ ) is a unit for an imaginary quadratic quantity τ [9, Theorem 6.7]. We prove that this value k ( τ ) can be written in the following two ways. Theorem 1.3 shows that we can evaluate the value k ( τ ) by using its relation with j ( τ ) , and the other way is given in Theorem 1.4.

Theorem 1.3

Let k be the modular function k ( τ ) . Then j ( τ ) = G 1 ( 1 / k k ) 3 / G 2 ( 1 / k k ) , where

G 1 ( X ) = X 6 + 236 X 5 + 1,440 X 4 + 1,920 X 3 + 3,840 X 2 + 256 X + 256 , G 2 ( X ) = X 2 ( X + 1 ) 5 ( X 4 ) 10

and j ( τ ) is a generator of K ( S L 2 ( Z ) ) with q-expansion q 1 + 744 + 196,884 q + O ( q 2 ) .

Theorem 1.4

When k ( τ ) is expressed in terms of radicals, we can express k ( r τ ) in terms of radicals for any positive rational number r.

We present some examples (Examples 4.4 and 4.5) for Theorems 1.3 and 1.4, respectively, in Section 4.

This paper is organized as follows. We briefly mention some definitions and lemmas which are used to prove our theorems in Section 2. In Section 3, we focus on the modular equations of k ( τ ) . We use the fact that k ( τ ) generates the field of modular functions on Γ 1 ( 10 ) , and we find the modular equations which satisfy Kronecker’s congruence (Theorem 1.1). It is possible to reduce the computation cost for finding modular equations by using useful properties for the coefficients C i , j ’s of modular equations given in Theorem 1.1 (2) and Proposition 3.5. Modular equations of levels 2, 3, 4, 5, 7, 8, 11, and 19 are found, and they are presented in Appendix A. The proofs of Theorems 1.2–1.4 are given in Section 4. First of all, the value k ( τ ) generates the ray class field modulo 10 over an imaginary quadratic field (Theorem 1.2). Some relation between k ( τ ) and j ( τ ) is obtained (Theorem 1.3), and it is used to get the value k ( τ ) in terms of radicals (Theorem 1.4). Using the modular equations, we find an explicit method for evaluating the value k ( τ ) in terms of radicals. Some calculations in this paper are performed by the MAPLE program.

2 Preliminaries

Ramanujan’s function k ( τ ) can be written as an infinite product

k ( τ ) = q n = 1 ( 1 q 10 n 9 ) ( 1 q 10 n 8 ) ( 1 q 10 n 2 ) ( 1 q 10 n 1 ) ( 1 q 10 n 7 ) ( 1 q 10 n 6 ) ( 1 q 10 n 4 ) ( 1 q 10 n 3 )

by (1.6) and

r ( τ ) = q 1 / 5 n = 1 ( 1 q 5 n 4 ) ( 1 q 5 n 1 ) ( 1 q 5 n 3 ) ( 1 q 5 n 2 ) .

Hence, in [9, Table 2, p. 161] the generator of K ( Γ 1 ( 10 ) ) is G 10 ( τ ) = 1 / k ( τ ) , and we get the following:

Proposition 2.1

The field K ( Γ 1 ( 10 ) ) is generated by k ( τ ) , that is,

K ( Γ 1 ( 10 ) ) = ( k ( τ ) ) .

We assume that two modular functions f 1 ( τ ) and f 2 ( τ ) satisfy the relation F ( f 1 ( τ ) , f 2 ( τ ) ) = 0 , where F ( X , Y ) is a two-variable polynomial with complex coefficients. The following lemma which is proved by Ishida and Ishii [12] provides some information on the coefficients of F ( X , Y ) .

Lemma 2.2

For any congruence subgroup Γ , let f 1 ( τ ) and f 2 ( τ ) be nonconstants such that ( f 1 ( τ ) , f 2 ( τ ) ) = K ( Γ ) with the total degree D k of poles of f k ( τ ) for k = 1 , 2 . Let

F ( X , Y ) = 0 i D 2 0 j D 1 C i , j X i Y j [ X , Y ]

be such that F ( f 1 ( τ ) , f 2 ( τ ) ) = 0 . Let S Γ be a set of all the inequivalent cusps of Γ , and for k = 1 , 2 ,

S k , 0 = { s S Γ : f k ( τ ) h a s z e r o s a t s } ,

S k , = { s S Γ : f k ( τ ) h a s p o l e s a t s } ,

a = s S 1 , S 2 , 0 o r d s f 1 ( τ ) a n d b = s S 1 , 0 S 2 , 0 o r d s f 1 ( τ ) .

We assume that a (resp. b) is zero if S 1 , S 2 , 0 (resp. S 1 , 0 S 2 , 0 ) is empty. Then we obtain the following assertions:

  1. C D 2 , a 0 . If further S 1 , S 2 , S 2 , 0 , then C D 2 , j = 0 for any j a .

  2. C 0 , b 0 . If further S 1 , 0 S 2 , S 2 , 0 , then C 0 , j = 0 for any j b .

If we interchange the roles of f 1 and f 2 , then we obtain further properties similar to (1) and (2).

3 Modular equations: Proof of Theorem 1.1

Let Γ = Γ 1 ( N ) Γ 0 ( m N ) for positive integers N and m. We denote the subgroup Δ by

Δ { ± ( 1 + N k ) ¯ ( Z / m N Z ) × : k = 0 , , m 1 } .

We understand that ± 1 / 0 = . Then we use the following lemmas to obtain useful properties.

Lemma 3.1

Let a , c , a , c Z be such that ( a , c ) = ( a , c ) = 1 . Then with the notation Δ as above, a/c and a / c are equivalent under Γ 1 ( N ) Γ 0 ( m N ) if and only if there exist s ¯ Δ ( Z / m N Z ) × and n Z such that ( a , c ) ( s ¯ 1 a + n c , s ¯ c ) ( mod m N ) .

Proof

See [13, Lemma 1].□

Let K ( Γ ) be the subfield of K ( Γ ) which consists of all modular functions whose Fourier coefficients are in .

Lemma 3.2

Let n be a positive integer. Then we have

( k ( τ ) , k ( n τ ) ) = K ( Γ 1 ( 10 ) Γ 0 ( 10 n ) ) .

Proof

For a convenience, let Γ Γ 1 ( 10 ) Γ 0 ( 10 n ) . Since ( k ( τ ) ) = K ( Γ 1 ( 10 ) ) , we know that for any α GL 2 + ( ) , k ( α τ ) = k ( τ ) if and only if α × Γ 1 ( 10 ) . Let β = n 0 0 1 . Note that

Γ 1 ( 10 ) β 1 Γ 1 ( 10 ) β = Γ 1 ( 10 ) Γ 0 ( 10 n ) = Γ .

Hence, we get k ( τ ) , k ( n τ ) K ( Γ ) . Now we show that ( k ( τ ) , k ( n τ ) ) contains K ( Γ ) . We choose M i Γ 1 ( 10 ) satisfying

(3.1) Γ 1 ( 10 ) = i Γ M i ,

which is a disjoint union.

Let f ( τ ) be k ( n τ ) = ( k β ) ( τ ) . For distinct indices i and j, we assume that f M i = f M j . Then k β M i = k β M j ; so, k β M i M j 1 β 1 = k . This means that β M i M j 1 β 1 × Γ 1 ( 10 ) and M i M j 1 β 1 Γ 1 ( 10 ) β . Since M i M j 1 Γ 1 ( 10 ) , M i M j 1 Γ 1 ( 10 ) β 1 Γ 1 ( 10 ) β = Γ , which is a contradiction to (3.1). Therefore, all functions f M i are distinct with distinct indices and ( k ( τ ) , k ( n τ ) ) = K ( Γ 1 ( 10 ) Γ 0 ( 10 n ) ) .□

The following lemma tells us the behavior of k ( τ ) .

Lemma 3.3

Let a , c , a , c Z . Then the functions k ( τ ) and k ( n τ ) have the following properties:

  1. k ( τ ) has a pole at a / c { } with ( a , c ) = 1 if and only if ( a , c ) = 1 , a ± 3 ( mod 10 ) , and c 0 ( mod 10 ) .

  2. k ( n τ ) has a pole at a / c { } if and only if there exist a , c Z such that a / c = n a / c , ( a , c ) = 1 , a ± 3 ( mod 10 ) , and c 0 ( mod 10 ) .

  3. k ( τ ) has a zero at a / c { } with ( a , c ) = 1 if and only if ( a , c ) = 1 , a ± 1 ( mod 10 ) , and c 0 ( mod 10 ) .

  4. k ( n τ ) has a zero at a / c { } if and only if there exist a , c Z such that a / c = n a / c , ( a , c ) = 1 , a ± 1 ( mod 10 ) , and c 0 ( mod 10 ) .

Proof

First we prove that k ( τ ) has a simple zero at and a simple pole at 3/10. From Table 1, j ( 10 ) ( τ ) has a pole at x if and only if k ( τ ) has a pole or zero at x. Since j ( 10 ) ( τ ) has a pole at , if ord x k ( τ ) is nonzero, then x is equivalent to under Γ 0 ( 10 ) . Among the elements of S Γ 1 ( 10 ) , and 3/10 are equivalent under Γ 0 ( 10 ) by Lemma 3.1. It is easy to see that ord k ( τ ) = 1 , thus we know that ord 3 / 10 k ( τ ) 0 . Since x Γ 1 ( 10 ) \ ord x k ( τ ) = 0 , ord 3 / 10 k ( τ ) = 1 .

By Lemma 3.1, k ( τ ) has a simple pole at a/c if and only if ( a , c ) ± ( 3 , 0 ) ( mod 10 ) . Similarly, k ( τ ) has a simple zero at a/c if and only if ( a , c ) ± ( 1 , 0 ) ( mod 10 ) . Hence, we proved (1) and (3).

We easily obtain (2) and (4) by (1) and (3).□

Hereafter, we fix the sets S Γ 0 ( 10 ) and S Γ 1 ( 10 ) of inequivalent cusps of Γ 0 ( 10 ) and Γ 1 ( 10 ) , respectively, as follows:

S Γ 0 ( 10 ) , 0 , 1 2 , 1 5

and

S Γ 1 ( 10 ) , 0 , 1 2 , 1 3 , 1 5 , 1 6 , 3 5 , 3 10 .

Before proving Theorem 1.1 we check the action of Γ 0 ( 10 ) on the function k ( τ ) in the following lemma.

Lemma 3.4

Let γ Γ 0 ( 10 ) Γ 1 ( 10 ) . Then k γ ( τ ) = 1 / k ( τ ) .

Proof

Note that K ( γ 1 Γ 1 ( 10 ) γ ) = K ( Γ 1 ( 10 ) ) . Since

( ( k γ ) ( τ ) ) = K ( γ 1 Γ 1 ( 10 ) γ ) = K ( Γ 1 ( 10 ) ) = ( k ( τ ) ) ,

there are four constants a , b , c , and d satisfying that

( k γ ) ( τ ) = a k ( τ ) + b c k ( τ ) + d and a b c d 2 = ± I .

By Lemma 3.3 (1), k ( τ ) has a pole at 3/10 and the q-expansion of ( k γ ) ( τ ) = q 1 + . This means that we may assume that a = 0 , c = 1 , and b 0 . Then

0 b 1 d 2 = b b d d d 2 + b = ± 1 0 0 1 ,

so we have d = 0 and b = ± 1 . Hence, k γ = ± 1 / k . Suppose that k γ = 1 / k . Using γ Γ 0 ( 10 ) and (1.5),

j ( 10 ) ( τ ) = ( j ( 10 ) γ ) ( τ ) = 1 k k γ ( τ ) = k ( τ ) 1 k ( τ ) = j ( 10 ) ( τ ) ;

this is a contradiction to the fact that j ( 10 ) ( τ ) is a generator of the field of modular functions on Γ 0 ( 10 ) . Therefore, ( k γ ) ( τ ) = 1 / k ( τ ) .□

Until the proof of Theorem 1.1 ends, assume that n is a positive integer relatively prime to 10. We take σ a SL 2 ( Z ) such that σ a a 1 0 0 a ( mod 10 ) for any integer a relatively prime to 10. Then by [14, Proposition 3.36] we have a disjoint union such as

Γ 1 ( 10 ) 1 0 0 n Γ 1 ( 10 ) = 0 < a | n 0 b < n / a ( a , b , n / a ) = 1 Γ 1 ( 10 ) σ a a b 0 n / a .

Let d = n p | n ( 1 + p 1 ) for prime p dividing n. Then [ Γ 1 ( 10 ) 1 0 0 n Γ 1 ( 10 ) : Γ 1 ( 10 ) ] = d . It is clear that σ a depends on a ( mod 10 ) , and so we may take σ a as follows:

σ ± 1 = ± 1 0 0 1 , σ ± 3 = ± 3 10 10 33 .

Then k σ ± 1 = k and k σ ± 3 = 1 / k by Lemma 3.4.

We denote α a , b σ a a b 0 n / a for integers a and b such that a is a positive divisor of n, 0 b < n / a , and ( a , b , n / a ) = 1 . Consider the polynomial Ψ n ( X , τ ) having an indeterminate X

Ψ n ( X , k ( τ ) ) 0 < a | n 0 b < n / a ( a , b , n / a ) = 1 ( X ( k α a , b ) ( τ ) ) .

We just write Ψ n ( X , k ( τ ) ) instead of Ψ n ( X , τ ) because Ψ n ( X , τ ) ( k ( τ ) ) [ X ] from the fact that all the coefficients of Φ n ( X , τ ) are elementary symmetric functions of k α a , b (i.e., they are invariant under the action of Γ 1 ( 10 ) ).

Let S m , 0 be the set of cusps which k ( m τ ) has a zero and S m , is the set of cusps which k ( m τ ) . We denote by r n the nonnegative integer

r n s S 1 , S n , 0 ord s k ( τ ) ,

where S m , 0 is the set of cusps which k ( m τ ) has a zero and S m , is the set of cusps which k ( m τ ) . Here we consider k ( τ ) as a function on K ( Γ 1 ( 10 ) Γ 0 ( 10 n ) ) and r n = 0 if the set S n , 0 S 1 , is empty. Then define the polynomial

F n ( X , k ( τ ) ) = k ( τ ) r n Ψ n ( X , k ( τ ) ) ,

that is, F n ( X , Y ) = Y r n Ψ n ( X , Y ) .

Now we are ready to prove our first theorem.

Proof of Theorem 1.1

Let F n ( X , Y ) be the polynomial defined as before. Then it is easy to see that F n ( X , Y ) Z [ X , Y ] , deg X F n ( X , Y ) = n p | n (1 + p 1 ) , and F n ( X , Y ) is irreducible as a polynomial in X (resp. Y) over ( Y ) (resp. ( X ) ); this is obtained in a similar way to that of [12, Theorem 10], so we omit the proof.

  1. First, we show that the modular equations of k ( τ ) exist for all levels. Let ( f 1 ( τ ) , f 2 ( τ ) ) be the field of modular functions on a certain congruence subgroup Γ for nonconstants f 1 ( τ ) and f 2 ( τ ) . The degree [ ( f 1 ( τ ) , f 2 ( τ ) ) : ( f 1 ( τ ) ) ] (resp. [ ( f 1 ( τ ) , f 2 ( τ ) ) : ( f 2 ( τ ) ) ] ) of field of extension is the total degree d 1 (resp. d 2 ) of poles f 1 ( τ ) (resp. f 2 ( τ ) ) in Riemann surface Γ \ .

    Assume that f 1 ( τ ) = k ( τ ) and f 2 ( τ ) = k ( n τ ) for any positive integer n. By Lemma 3.2, ( f 1 ( τ ) , f 2 ( τ ) ) = K ( Γ 1 ( 10 ) Γ 0 ( 10 n ) ) . From Lemma 2.2, one can have a polynomial F ( X , Y ) such that F ( k ( τ ) , k ( n τ ) ) = 0 , deg X F ( X , Y ) is the total degree of poles k ( n τ ) , and deg Y F ( X , Y ) is the total degree of poles of k ( τ ) . So we obtain the modular equation F ( X , Y ) of k ( τ ) of level n.

  2. Assume that n ± 1 ( mod 10 ) . As before, F n ( X , k ( τ ) ) is an irreducible polynomial in X over ( k ( τ ) ) with F n ( k ( τ / n ) , k ( τ ) ) = 0 . Using Ψ n ( k ( n τ ) , k ( τ ) ) = 0 , it is easy to see that k ( τ / n ) is a root of F n ( k ( τ ) , X ) Z [ X , k ( τ ) ] . Hence, there is a polynomial G ( X , k ( τ ) ) Z [ X , k ( τ ) ] such that F n ( k ( τ ) , X ) = G ( X , k ( τ ) ) F n ( X , k ( τ ) ) . By interchanging the places of X and k ( τ ) , we have that F n ( X , k ( τ ) ) = G ( k ( τ ) , X ) F n ( k ( τ ) , X ) and F n ( k ( τ ) , X ) = G ( X , k ( τ ) ) G ( k ( τ ) , X ) F n ( k ( τ ) , X ) . So G ( X , k ( τ ) ) should be ± 1 . Suppose that G G ( X , k ( τ ) ) = 1 . Then F n ( X , k ( τ ) ) + F n ( k ( τ ) , X ) = 0 . When X = k ( τ ) , F n ( k ( τ ) , k ( τ ) ) = 0 and k ( τ ) is a root of F n ( X , k ( τ ) ) . Since X k ( τ ) divides F n ( X , k ( τ ) ) , F n ( X , k ( τ ) ) is not irreducible by considering d > 1 ; this is a contradiction to the irreducibility of F n ( X , Y ) . Hence, we get

    F n ( X , Y ) = F n ( Y , X ) .

    Now assume that n ± 3 ( mod 10 ) . We already know that Ψ n ( 1 / k ( n τ ) , k ( τ ) ) = 0 and Ψ n ( 1 / k ( τ ) , k ( τ / n ) ) = 0 . Then k ( τ ) d F n ( 1 / k ( τ ) , X ) Z [ X , k ( τ ) ] also has a root k ( τ / n ) . Hence, k ( τ ) d F n ( 1 / k ( τ ) , X ) is written as follows:

    k ( τ ) d F n 1 k ( τ ) , X = G ( X , k ( τ ) ) F n ( X , k ( τ ) )

    for some G ( X , k ( τ ) ) Z [ X , k ( τ ) ] by using the fact that F n ( X , k ( τ ) ) is an irreducible polynomial with root k ( τ / n ) .

    Then

    deg X F n ( X , Y ) + deg X G ( X , Y ) = deg X Y d F n 1 Y , X = deg Y F n ( X , Y )

    and

    deg Y F n ( X , Y ) + deg Y G ( X , Y ) = deg Y Y d F n 1 Y , X = deg X F n ( X , Y ) ,

    where the second identity is obtained because Y d F n ( 1 / Y , X ) is written as

    1 C d n , r ( ( 1 ) d n C d n , r X r Y d d n + C 0 , s X s Y d + ( 1 ) r C r , d 1 X d 1 Y d r + ( 1 ) s C s , 0 Y d s + ( lower degree terms ) ) ,

    where d m is the total degrees of pole of k ( m τ ) , r r n ,

    r s S n , S 1 , 0 ord s k ( n τ ) , s s S 1 , 0 S n , 0 ord s k ( τ ) , and s s S 1 , 0 S n , 0 ord s k ( n τ ) .

    Hence, deg X G ( X , Y ) + deg Y G ( X , Y ) = 0 and G G ( X , Y ) is constant. Moreover, we obtain that d = deg X F n ( X , Y ) = deg Y F n ( X , Y ) . By using that F n ( X , Y ) is a primitive polynomial, G = ± 1 . Since F n ( 1 / Y , X ) = G Y d F n ( X , Y ) ,

    F n ( X , Y ) = G ( X ) d F n Y , 1 X = G ( X ) d ( Y d ( X ) r + ) = G ( 1 ) d r X d r Y d + .

    Noting that Ψ n ( X , k ( τ ) ) can be written as the following product:

    0 < a | n a ± 1 ( mod 10 ) 0 b < n / a ( a , b , n / a ) = 1 ( X ζ n a b q a 2 / n + ) 0 < a | n a ± 3 ( mod 10 ) 0 b < n / a ( a , b , n / a ) = 1 ( X + ζ n a b q a 2 / n + ) ,

    we can get that the coefficient of X d r Y d in F n ( X , Y ) is given by

    (3.2) 0 < a | n a ± 3 ( mod 10 ) 0 b < n / a ( a , b , n / a ) = 1 ζ n a b .

    Hence, (3.2) should be ( 1 ) d r G . Denote by Π Π the double product

    0 < a | n a ± 3 ( mod 10 ) 0 b < n / a ( a , b , n / a ) = 1 .

    By [15, Lemma 6.7], in (3.2) ζ n a b = 1 . Since ( 1 ) = ( 1 ) r , we have G = ( 1 ) d . Hence,

    F n ( X , Y ) = Y d F n 1 Y , X

    because n is odd and d is even.

  3. Now we focus on the congruence properties which the modular equation of k ( τ ) satisfies. Let p be an odd prime not 5. We denote x y ( mod α ) for x , y R if x y α R .

Write k ( τ ) = n = 1 c m a m with integers c m . Since

( k α 1 , b ) ( τ ) = k σ 1 1 b 0 p ( τ ) = k τ + b p = m 1 c m ζ p b m q m p m 1 c m q m p = ( k α 1 , 0 ) ( τ ) ( mod 1 ζ p ) ,

b = 0 p 1 ( X ( k α 1 , b ) ( τ ) ) ( X ( k α 1 , 0 ) ( τ ) ) p ( mod 1 ζ p ) ( X p ( k α 1 , 0 ) ( τ ) p ) ( mod 1 ζ p ) ( X p k ( τ ) ) ( mod 1 ζ p ) ;

in last congruence we use that

( k α 1 , 0 ) ( τ ) p = k τ p p = m 1 c m q m p p m 1 c m p q m m 1 c m q m ( mod 1 ζ p ) .

On the other hand,

r = s S 1 , S n , 0 ord s k ( τ ) = 0 if p ± 1 ( mod 10 ) , p if p ± 3 ( mod 10 ) .

Hence,

F p ( X , k ( τ ) ) = k ( τ ) r a = 1 , p 0 b p / a ( X ( k α a , b ) ( τ ) ) k ( τ ) r ( X p k ( τ ) ) ( X ( k α p , 0 ) ( τ ) ) k ( τ ) r ( X p k ( τ ) ) ( X ( k σ p ) ( p τ ) ) ( X p k ( τ ) ) ( X k ( p τ ) ) if p ± 1 ( mod 10 ) , k ( τ ) p ( X p k ( τ ) ) X + 1 k ( p τ ) if p ± 3 ( mod 10 ) ( X p k ( τ ) ) ( X k ( τ ) p ) if p ± 1 ( mod 10 ) , ( X p k ( τ ) ) ( k ( τ ) p X + 1 ) if p ± 3 ( mod 10 ) ,

where a b means that a b ( mod 1 ζ p ) , because k ( p τ ) = c m q p m c m p q p m c m q m p = k ( τ ) p ( mod p ) . We have that 1 ζ p divides F p ( X , Y ) H p ( X , Y ) in Z [ ζ p ] and F p ( X , Y ) Z [ X , Y ] , where

H p ( X , Y ) = ( X p Y ) ( X Y p ) if p ± 1 ( mod 10 ) , ( X p Y ) ( X Y p + 1 ) if p ± 3 ( mod 10 ) .

Our result follows since p is the smallest integer which is divisible by 1 ζ p in Z [ ζ p ] .□

We study the property of F p ( X , Y ) of odd prime level p 5 in Proposition 3.5. It is helpful for finding the modular equations in practice; it reduces the computation cost.

Proposition 3.5

Let p be an odd prime with p 5 and F p ( X , Y ) = 0 i , j p + 1 C i , j X i Y j be a modular equation of level p. Then F p ( X , Y ) satisfies the following:

  1. If p ± 1 ( mod 10 ) , then

    1. C p + 1 , 0 0 and C 0 , p + 1 0 ,

    2. C p + 1 , j = C j , p + 1 = 0 for j = 1 , 2 , , p + 1 ,

    3. C 0 , j = C j , 0 = 0 for j = 0 , 1 , , p .

  2. If p ± 3 ( mod 10 ) , then

    1. C p + 1 , p 0 , C 0 , 1 0 , C 1 , p + 1 0 and C p , 0 0 ,

    2. C 0 , j = C j , p + 1 = 0 for j = 0 , 2 , , p + 1 ,

    3. C p + 1 , j = C j , 0 = 0 for j = 0 , 1 , , p 1 , p + 1 .

Proof

Let Γ = Γ 1 ( 10 ) Γ 0 ( 10 p ) . Since all statements below also hold for the case p = 3 , we deal with the only case of the prime p 7 .

Assume that the cusp a/c of Γ with ord x k ( τ ) or o r d x k ( p τ ) is nonzero. Then c is a multiple of 10. Let f 1 ( τ ) = k ( τ ) and f 2 ( τ ) = k ( p τ ) . Define the sets S j , and S j , 0 for j = 1 , 2 as Lemma 2.2. Then we know that

S 1 , = 3 10 , 3 10 p , S 1 , 0 = 1 10 , 1 10 p

and

S 2 , = 3 10 , 3 10 p if p ± 1 ( mod 10 ) , 1 10 , 3 10 p if p ± 3 ( mod 10 ) , S 2 , 0 = 1 10 , 1 10 p if p ± 1 ( mod 10 ) , 3 10 , 1 10 p if p ± 3 ( mod 10 ) .

Moreover,

ord 3 / 10 p f 1 ( τ ) = 1 , ord 3 / 10 f 1 ( τ ) = p , o r d 1 / 10 p f 1 ( τ ) = 1 , o r d 1 / 10 f 1 ( τ ) = p .

We calculate o r d x f 2 ( τ ) ( x = 1 / 10 , 3 / 10 , 1 / 10 p , 3 / 10 p ) as follows:

  1. ord 1 / 10 p f 2 ( τ ) = ord k ( p τ ) = p .

  2. ord 3 / 10 p f 2 ( τ ) = ord q k p 0 0 1 3 b 10 p c ( τ ) = ord q k 3 b p 10 c ( p τ ) = ord q 1 k ( p τ ) = p ,

    where b and c are integers such that 3 c 10 b p = 1 .

  3. ord 1 / 10 f 2 ( τ ) = 1 if p ± 1 ( mod 10 ) , 1 if p ± 3 ( mod 10 ) ,

    because the width of 1/10 is p and

    f 2 1 0 10 1 ( τ ) = k p 0 0 1 1 0 10 1 ( τ ) = k p 0 10 1 ( τ ) = k p t 10 y 1 t 0 p ( τ ) = k 1 t 0 p ( τ ) = ζ p t q 1 / p + , if p ± 1 ( mod 10 ) , 1 / k 1 t 0 p ( τ ) = ζ p t q 1 / p + if p ± 3 ( mod 10 ) ,

    where t , y Z such that 10 t + y p = 1 and 0 t p 1 .

  4. ord 3 / 10 f 2 ( τ ) = 1 if p ± 1 ( mod 10 ) , 1 if p ± 3 ( mod 10 ) ,

because the width of 3/10 is p and

f 2 3 2 10 7 ( τ ) = k p 0 0 1 3 2 10 7 ( τ ) = k 3 p 2 p 10 7 ( τ ) = k 3 p 2 3 t 10 y 1 t 0 p ( τ ) = 1 / k 1 t 0 p ( τ ) = ζ p t q 1 / p + if p ± 1 ( mod 10 ) , k 1 t 0 p ( τ ) = ζ p t q 1 / p + , if p ± 3 ( mod 10 ) ,

where t , y Z such that 10 t + y p = 7 and 0 t p 1 .

Hence, the total degree of poles of f i ( τ ) is p + 1 for i = 1 , 2 , and the modular equation F p ( X , Y ) of k ( τ ) of level p is written as:

F p ( X , Y ) = 0 i , j p + 1 C i , j X i Y j .

Now we have the information on S j , 0 and S j , for j = 1 , 2 as follows:

  1. S 1 , S 1 , 0 = S 2 , S 2 , 0 .

  2. S 1 , S 2 , 0 = ϕ if p ± 1 ( mod 10 ) , 3 10 if p ± 3 ( mod 10 ) .

  3. S 1 , 0 S 2 , 0 = 1 10 , 1 10 p if p ± 1 ( mod 10 ) , 1 10 p if p ± 3 ( mod 10 ) .

Thus, we have

C p + 1 , 0 0 , C 0 , p + 1 0 ,

C p + 1 , 1 = C p + 1 , 2 = = C p + 1 , p + 1 = 0 ,

and

C 0 , 0 = C 0 , 1 = = C 0 , p = 0

if p ± 1 ( mod 10 ) . Moreover,

C p + 1 , p 0 , C 0 , 1 0 ,

C p + 1 , 0 = = C p + 1 , p 1 = C p + 1 , p + 1 = 0 ,

and

C 0 , 0 = C 0 , 2 = = C 0 , p + 1 = 0

if p ± 3 ( mod 10 ) .

By letting f 1 ( τ ) = k ( p τ ) and f 2 ( τ ) = k ( τ ) and repeating this similar calculation for S j , and S j , 0 , the result follows.□

We find the modular equations of levels 2 and 5 explicitly in Proposition 3.6 using the method presented in Theorem 1.1 (1).

Proposition 3.6

Let U k ( τ ) , V k ( 2 τ ) , and W k ( 5 τ ) . Then

  1. (modular equation of level 2)

    U 2 ( 1 2 U U 2 ) V + V 2 = 0 .

  2. (modular equation of level 5)

U 5 ( 1 5 U + 15 U 3 2 U 5 ) W ( 2 5 U 15 U 2 10 U 3 + 5 U 4 + U 5 ) W 2 + ( 1 5 U 10 U 2 + 15 U 3 5 U 4 2 U 5 ) W 3 + ( 2 15 U 2 + 5 U 4 + U 5 ) W 4 W 5 .

Proof

  1. We consider Γ 1 ( 10 ) Γ 0 ( 20 ) . If a / c satisfies a ± 3 ( mod 10 ) and c 0 ( mod 20 ) , then a/c is equivalent to either 3/10 or 3/20 under Γ 1 ( 10 ) Γ 0 ( 20 ) . Similarly, if a / c satisfies a ± 1 ( mod 10 ) and c 0 ( mod 20 ) , then a/c is equivalent to either 1/10 or 1/20 under Γ 1 ( 10 ) Γ 0 ( 20 ) . Moreover, the widths of 1 / 20 , 3 / 20 , 1 / 10 , and 3/10 are all 1. Hence, we have

    ord 1 / 10 k ( τ ) = ord 1 / 20 k ( τ ) = 1 , ord 3 / 10 k ( τ ) = 1 , ord 3 / 20 k ( τ ) = 1

    and

    ord 1 / 20 k ( 2 τ ) = 2 , ord 3 / 20 k ( 2 τ ) = 2 .

    By Lemma 2.2, the modular equation F 2 ( X , Y ) is given by

    F 2 ( X , Y ) = 0 i , j 2 C i , j X i Y j .

    Since there is no point x such that o r d x k ( τ ) < 0 and o r d x k ( 2 τ ) > 0 , by Lemma 2.2 (1), C 2 , 0 is nonzero, so we may assume that C 2 , 0 is 1. By substituting X k ( τ ) and Y k ( 2 τ ) as q-expansions, we get that

    C 0 , 1 = 1 , C 1 , 1 = 2 , C 2 , 1 = 1 , C 0 , 2 = 1 .

  2. Under the group Γ 1 ( 1 ) Γ 0 ( 50 ) , we may consider the points

x = 1 10 , 3 10 , 7 10 , 9 10 , 1 20 , 3 20 , 7 20 , 9 20 , 1 50 , 3 50

in such that o r d x k ( τ ) is nonzero. Note that all widths of these points are 1. Hence,

ord 3 / 10 k ( τ ) = ord 7 / 10 k ( τ ) = ord 3 / 20 k ( τ ) = ord 7 / 20 k ( τ ) = ord 3 / 50 k ( τ ) = 1 ,

ord 1 / 10 k ( τ ) = ord 9 / 10 k ( τ ) = ord 1 / 20 k ( τ ) = ord 9 / 20 k ( τ ) = ord 1 / 50 k ( τ ) = 1 ,

and

ord 3 / 50 k ( 5 τ ) = 5 , ord 1 / 50 k ( 5 τ ) = 5 .

Thus, we get the modular equation F 5 ( X , Y ) = 0 i , j 5 C i , j X i Y j and C 5 , 0 is nonzero because there is no point x such that ord x k ( τ ) < 0 and ord x k ( 5 τ ) > 0 by Lemma 2.2 (1). When we put C 5 , 0 = 1 and substitute X k ( τ ) and k ( 5 τ ) in F 5 ( X , Y ) , we conclude that

F 5 ( X , Y ) = X 5 ( 1 5 X + 15 X 3 2 X 5 ) Y ( 2 5 X 15 X 2 10 X 3 + 5 X 4 + X 5 ) Y 2 + ( 1 5 X 10 X 2 + 15 X 3 5 X 4 2 X 5 ) Y 3 + ( 2 15 X 2 + 5 X 4 + X 5 ) Y 4 Y 5 .

4 Class fields and evaluations: Proofs of Theorems 1.2, 1.3, and 1.4

For an imaginary quadratic field K with discriminant d K and positive integer N, let K ( N ) be the ray class field modulo N over K. Let τ K be a root of a primitive equation a x 2 + b x + c = 0 satisfying b 2 4 a c = d K , where a , b , and c are integers. In this section, we show that k ( τ ) generates K ( 10 ) over K, and we work on evaluation of k ( τ ) .

Lemma 4.1

Let K be an imaginary quadratic field with discriminant d K and τ K be a root of a primitive equation a x 2 + b x + c = 0 such that b 2 4 a c = d K and a , b , c Z . Let Γ be a congruence subgroup such that Γ ( N ) Γ Γ 1 ( N ) . Suppose that ( N , a ) = 1 . Then the field generated over K by all the values h ( τ ) , where h K ( Γ ) is defined and finite at τ , is the ray class field modulo N over K.

Proof

See [16, Corollary 5.2].□

Proof of Theorem 1.2

By Proposition 1.1, we have K ( Γ 1 ( 10 ) ) = ( k ( τ ) ) . Hence, k ( τ ) satisfies the condition of Lemma 4.1 when N = 10 , and so we can conclude that for an imaginary quadratic field K with discriminant d K , the field K ( k ( τ ) ) is the ray class field modulo 10 over K when τ K satisfies a τ 2 + b τ + c = 0 , b 2 4 a c = d K , and ( a , 10 ) = 1 .□

Now we evaluate k ( τ ) for an imaginary quadratic quantity τ by proving Theorems 1.3 and 1.4.

Proof of Theorem 1.3

Let j ( 5 ) and j ( 10 ) be the Hauptmoduln which are given in Table 1. We claim that the following two identities are true:

(4.1) j ( 5 ) ( τ ) = j ( 10 ) ( τ ) 3 7 j ( 10 ) ( τ ) 2 + 8 j ( 10 ) ( τ ) + 16 j ( 10 ) ( τ ) 2

and

(4.2) j ( τ ) = ( j ( 5 ) ( τ ) 2 + 250 j ( 5 ) ( τ ) + 3,125 ) 3 j ( 5 ) ( τ ) 5 .

First we show (4.1). Since [ Γ 0 ( 5 ) : Γ 0 ( 10 ) ] = [ Γ 0 ( 1 ) : Γ 0 ( 10 ) ] / [ Γ 0 ( 1 ) : Γ 0 ( 5 ) ] = 18 / 6 = 3 , a Hauptmodul j ( 5 ) ( τ ) of level 5 is written as the fraction P 1 ( j 10 ( τ ) ) / P 2 ( j ( 10 ) ( τ ) ) , where P 1 ( X ) and P 2 ( X ) are polynomials and max { deg X P 1 ( X ) , deg X P 2 ( X ) } = 3 . Note that both j ( 10 ) ( τ ) and j ( 5 ) ( τ ) have a simple pole at . Hence deg X P 1 ( X ) > deg X P 2 ( X ) and deg X P 1 ( X ) = 3 . This implies that there is a polynomial A ( X , Y ) = P 2 ( X ) Y P 1 ( X ) such that A ( j ( 10 ) ( τ ) , j ( 5 ) ( τ ) ) = 0 , P 2 ( X ) = a 2 X 2 + a 1 X + a 0 , and P 1 ( X ) = a 3 X 3 + a 2 X 2 + a 1 X + a 0 .

Since

j ( 10 ) ( τ ) = q 1 + 1 + q + 2 q 2 + 2 q 3 2 q 4 q 5 4 q 7 2 q 8 + O ( q 9 )

and

j ( 5 ) ( τ ) = q 1 6 + 9 q + 10 q 2 30 q 3 + 6 q 4 25 q 5 + 96 q 6 + 60 q 7 250 q 8 + O ( q 9 ) ,

by substituting the q-expansions of X j ( 5 ) ( τ ) and Y j ( 10 ) ( τ ) to A ( X , Y ) , we get the relation between j ( 10 ) ( τ ) and j ( 5 ) ( τ ) .

Now we show (4.2). Note that [ ( j ( 5 ) ( τ ) ) : ( j ( τ ) ) ] = [ K ( Γ 0 ( 1 ) ) : K ( Γ 0 ( 5 ) ) ] = 6 ; so, we can take the polynomials of degree less than or equal to 6 similar to P 1 ( X ) and P 2 ( X ) in the case of (4.1). Since

j ( τ ) = q 1 + 744 + 196,884 q + 21,493,760 q 2 + 864,299,970 q 3 + 20,245,856,256 q 4 + 333,202,640,600 q 5 + O ( q 6 ) ,

(4.2) follows.

Combining (1.5), (4.1), and (4.2), we obtain our assertion.□

Remark 4.2

Identity (4.1) can be obtained by using the third formula in [11, Theorem 10.5]

η ( 5 τ ) 6 η ( τ ) 6 = k 1 + k k 2 1 k 2 1 4 k k 2 2

and (1.5) because j ( 5 ) ( τ ) = η ( τ ) 6 / η ( 5 τ ) 6 . The function field K ( Γ 0 ( 10 ) ) is generated by j ( 10 ) ( τ ) and K ( Γ 0 ( 5 ) ) K ( Γ 0 ( 10 ) ) . Thus, j ( 5 ) ( 2 τ ) is contained in K ( Γ 0 ( 10 ) ) , and so it is also written in terms of j ( 10 ) ( τ ) :

j ( 5 ) ( 2 τ ) = ( j ( 10 ) ( τ ) + 1 ) 2 j ( 10 ) ( τ ) 4 j ( 10 ) ( τ ) .

Furthermore, j ( 5 ) ( 2 τ ) belongs to the bigger field K ( Γ 1 ( 10 ) ) which is generated by k = k ( τ ) , and thus it can be written as the rational function in k as follows (this appears in [11, Theorem 10.5]):

j ( 5 ) ( 2 τ ) = 1 + k k 2 k 2 1 4 k k 2 1 k 2 .

Remark 4.3

[11, Theorem 5.26] contains (4.2). It also contains the relation between j ( 5 τ ) and j ( 5 ) ( τ ) :

j ( 5 τ ) = ( 5 + 10 j ( 5 ) ( τ ) + j ( 5 ) ( τ ) 2 ) 3 j ( 5 ) ( τ )

because j ( 5 τ ) is a modular function on Γ 0 ( 5 ) .

Example 4.4

By Theorem 1.3, we can compute k 1 + 3 2 and k ( i ) .

  1. k 1 + 3 2 = 1 3 α 0 ( 5 α 0 2 + 1,320 + 590 5 + 25 5 α 0 + 59 α 0 10,491,900 + 455,760 α 0 + 203,820 5 α 0 + 8,850 5 α 0 2 + 4,692,120 5 + 19,815 α 0 2 ) , where α 0 2,728 5 + 6,100 3 by using j 1 + 3 2 = 0 , j ( 5 ) 1 + 3 2 = 125 50 5 , and j ( 10 ) 1 + 3 2 = 10 α 0 3 2,640 + 1,180 5 3 α 0 50 5 3 118 3 .

  2. k ( i ) = 67 30 5 5 β 0 + 17,990 + 8,045 5 + 670 β 0 + 300 5 β 0 ,

    where β 0 360 + 161 5 by using j ( i ) = 1728 , j ( 5 ) ( i ) = 250 + 125 5 , and j ( 10 ) ( i ) = 134 + 60 5 + 10 360 + 161 5 .

  3. k 1 + i 2 = 1 2 ( 11 + 5 5 250 110 5 )

by using j 1 + i 2 = 1728 , j ( 5 ) 1 + i 2 = 250 125 5 , and j ( 10 ) 1 + i 2 = 11 5 5 .

Proof of Theorem 1.4

We choose positive integers a and b such that r = a / b and ( a , b ) = 1 . When p is a prime factor of r, let τ 0 = p τ (resp. τ / p ) if p | a (resp. p | b ). By finding the modular equation F p ( X , Y ) in Appendix A, let the polynomial P ( T ) F p ( k ( τ ) , T ) (resp. F p ( T , k ( τ ) ) ) if p | a (resp. p | b ). Then the solutions t 1 , , t m of the equation P ( T ) = 0 which can be written in terms of radicals are candidates for the value k ( τ 0 ) . Among them, we choose t j to have the smallest absolute value

t j q 0 n = 1 M ( 1 q 0 10 n 9 ) ( 1 q 0 10 n 8 ) ( 1 q 0 10 n 2 ) ( 1 q 0 10 n 1 ) ( 1 q 0 10 n 7 ) ( 1 q 0 10 n 6 ) ( 1 q 0 10 n 4 ) ( 1 q 0 10 n 3 )

for q 0 e 2 π i τ 0 ; then t k is the value k ( τ 0 ) . One can repeat this procedure until getting k ( r τ ) .□

Example 4.5

By Theorem 1.4, we have the value

k ( 3 ) = 1 2 α α 2 2 1 4 α 2 α 2 + 4 α 3 + α 4 2 ,

where α k 1 + 3 2 is the value found in Example 4.4

α = 1 3 α 0 ( 5 α 0 2 1,320 590 5 25 5 α 0 59 α 0 + 10,491,900 + 455,760 α 0 + 203,820 5 α 0 + 8,850 5 α 0 2 + 4,692,120 5 + 19,815 α 0 2 ) ,

and

α 0 2,728 5 + 6,100 3 .

Acknowledgments

The authors would like to thank the referees for their valuable comments which were helpful for improving the clarity of our paper. The authors also would like to thank Nayoung Han for computing a modular equation of levels 4, 8, 11, and 19. Yoonjin Lee was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (Grant No. 2019R1A6A1A11051177) and also by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (NRF-2017R1A2B2004574). Yoon Kyung Park was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2017R1D1A1B03029519).

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Received: 2020-04-07
Revised: 2020-10-21
Accepted: 2020-10-26
Published Online: 2020-12-31

© 2020 Yoonjin Lee and Yoon Kyung Park, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  56. Asymptotic relations for the products of elements of some positive sequences
  57. Jordan {g,h}-derivations on triangular algebras
  58. A systolic inequality with remainder in the real projective plane
  59. A new characterization of L2(p2)
  60. Nonlinear boundary value problems for mixed-type fractional equations and Ulam-Hyers stability
  61. Asymptotic normality and mean consistency of LS estimators in the errors-in-variables model with dependent errors
  62. Some non-commuting solutions of the Yang-Baxter-like matrix equation
  63. General (p,q)-mixed projection bodies
  64. An extension of the method of brackets. Part 2
  65. A new approach in the context of ordered incomplete partial b-metric spaces
  66. Sharper existence and uniqueness results for solutions to fourth-order boundary value problems and elastic beam analysis
  67. Remark on subgroup intersection graph of finite abelian groups
  68. Detectable sensation of a stochastic smoking model
  69. Almost Kenmotsu 3-h-manifolds with transversely Killing-type Ricci operators
  70. Some inequalities for star duality of the radial Blaschke-Minkowski homomorphisms
  71. Results on nonlocal stochastic integro-differential equations driven by a fractional Brownian motion
  72. On surrounding quasi-contractions on non-triangular metric spaces
  73. SEMT valuation and strength of subdivided star of K 1,4
  74. Weak solutions and optimal controls of stochastic fractional reaction-diffusion systems
  75. Gradient estimates for a weighted nonlinear parabolic equation and applications
  76. On the equivalence of three-dimensional differential systems
  77. Free nonunitary Rota-Baxter family algebras and typed leaf-spaced decorated planar rooted forests
  78. The prime and maximal spectra and the reticulation of residuated lattices with applications to De Morgan residuated lattices
  79. Explicit determinantal formula for a class of banded matrices
  80. Dynamics of a diffusive delayed competition and cooperation system
  81. Error term of the mean value theorem for binary Egyptian fractions
  82. The integral part of a nonlinear form with a square, a cube and a biquadrate
  83. Meromorphic solutions of certain nonlinear difference equations
  84. Characterizations for the potential operators on Carleson curves in local generalized Morrey spaces
  85. Some integral curves with a new frame
  86. Meromorphic exact solutions of the (2 + 1)-dimensional generalized Calogero-Bogoyavlenskii-Schiff equation
  87. Towards a homological generalization of the direct summand theorem
  88. A standard form in (some) free fields: How to construct minimal linear representations
  89. On the determination of the number of positive and negative polynomial zeros and their isolation
  90. Perturbation of the one-dimensional time-independent Schrödinger equation with a rectangular potential barrier
  91. Simply connected topological spaces of weighted composition operators
  92. Generalized derivatives and optimization problems for n-dimensional fuzzy-number-valued functions
  93. A study of uniformities on the space of uniformly continuous mappings
  94. The strong nil-cleanness of semigroup rings
  95. On an equivalence between regular ordered Γ-semigroups and regular ordered semigroups
  96. Evolution of the first eigenvalue of the Laplace operator and the p-Laplace operator under a forced mean curvature flow
  97. Noetherian properties in composite generalized power series rings
  98. Inequalities for the generalized trigonometric and hyperbolic functions
  99. Blow-up analyses in nonlocal reaction diffusion equations with time-dependent coefficients under Neumann boundary conditions
  100. A new characterization of a proper type B semigroup
  101. Constructions of pseudorandom binary lattices using cyclotomic classes in finite fields
  102. Estimates of entropy numbers in probabilistic setting
  103. Ramsey numbers of partial order graphs (comparability graphs) and implications in ring theory
  104. S-shaped connected component of positive solutions for second-order discrete Neumann boundary value problems
  105. The logarithmic mean of two convex functionals
  106. A modified Tikhonov regularization method based on Hermite expansion for solving the Cauchy problem of the Laplace equation
  107. Approximation properties of tensor norms and operator ideals for Banach spaces
  108. A multi-power and multi-splitting inner-outer iteration for PageRank computation
  109. The edge-regular complete maps
  110. Ramanujan’s function k(τ)=r(τ)r2(2τ) and its modularity
  111. Finite groups with some weakly pronormal subgroups
  112. A new refinement of Jensen’s inequality with applications in information theory
  113. Skew-symmetric and essentially unitary operators via Berezin symbols
  114. The limit Riemann solutions to nonisentropic Chaplygin Euler equations
  115. On singularities of real algebraic sets and applications to kinematics
  116. Results on analytic functions defined by Laplace-Stieltjes transforms with perfect ϕ-type
  117. New (p, q)-estimates for different types of integral inequalities via (α, m)-convex mappings
  118. Boundary value problems of Hilfer-type fractional integro-differential equations and inclusions with nonlocal integro-multipoint boundary conditions
  119. Boundary layer analysis for a 2-D Keller-Segel model
  120. On some extensions of Gauss’ work and applications
  121. A study on strongly convex hyper S-subposets in hyper S-posets
  122. On the Gevrey ultradifferentiability of weak solutions of an abstract evolution equation with a scalar type spectral operator on the real axis
  123. Special Issue on Graph Theory (GWGT 2019), Part II
  124. On applications of bipartite graph associated with algebraic structures
  125. Further new results on strong resolving partitions for graphs
  126. The second out-neighborhood for local tournaments
  127. On the N-spectrum of oriented graphs
  128. The H-force sets of the graphs satisfying the condition of Ore’s theorem
  129. Bipartite graphs with close domination and k-domination numbers
  130. On the sandpile model of modified wheels II
  131. Connected even factors in k-tree
  132. On triangular matroids induced by n3-configurations
  133. The domination number of round digraphs
  134. Special Issue on Variational/Hemivariational Inequalities
  135. A new blow-up criterion for the Nabc family of Camassa-Holm type equation with both dissipation and dispersion
  136. On the finite approximate controllability for Hilfer fractional evolution systems with nonlocal conditions
  137. On the well-posedness of differential quasi-variational-hemivariational inequalities
  138. An efficient approach for the numerical solution of fifth-order KdV equations
  139. Generalized fractional integral inequalities of Hermite-Hadamard-type for a convex function
  140. Karush-Kuhn-Tucker optimality conditions for a class of robust optimization problems with an interval-valued objective function
  141. An equivalent quasinorm for the Lipschitz space of noncommutative martingales
  142. Optimal control of a viscous generalized θ-type dispersive equation with weak dissipation
  143. Special Issue on Problems, Methods and Applications of Nonlinear analysis
  144. Generalized Picone inequalities and their applications to (p,q)-Laplace equations
  145. Positive solutions for parametric (p(z),q(z))-equations
  146. Revisiting the sub- and super-solution method for the classical radial solutions of the mean curvature equation
  147. (p,Q) systems with critical singular exponential nonlinearities in the Heisenberg group
  148. Quasilinear Dirichlet problems with competing operators and convection
  149. Hyers-Ulam-Rassias stability of (m, n)-Jordan derivations
  150. Special Issue on Evolution Equations, Theory and Applications
  151. Instantaneous blow-up of solutions to the Cauchy problem for the fractional Khokhlov-Zabolotskaya equation
  152. Three classes of decomposable distributions
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