Modular Identities of New Ramanujan Continued ... - Semantic Scholar

International Mathematical Forum, Vol. 8, 2013, no. 14, 685 - 695 HIKARI Ltd, www.m-hikari.com

Modular Identities of New Ramanujan Continued Fraction and their Explicit Values B. N. Dharmendra*, M. R. Rajesh Kanna* and H. L. Parashivamurthy** *Post Graduate Department of Mathematics Maharani’s Science College for Women J. L. B. Road, Mysore-570 001, India [email protected], [email protected] **Department of Mathematics, BGS Institute of Technology, B.G. Nagar, Bellur, Nagamangala Thalque Mandya District-571448, India [email protected] c 2013 B. N. Dharmendra et al. This is an open access article distributed unCopyright  der the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper, we study new Ramanujan continued fraction and establish some new modular identities of their and some explicit values. Mathematics Subject Classification: Primary 33D10, 40A15, 11A55, 30B70 Keywords: Continued fraction, Theta functions 1. Introduction In Chapter 16 of his second notebook [2], Ramanujan develops the theory of theta-function and is defined by ∞  n(n+1) n(n−1) (1.1) a 2 b 2 , | ab |< 1, f (a, b) := n=−∞

= (−a; ab)∞ (−b; ab)∞ (ab; ab)∞ where (a; q)0 = 1 and (a; q)∞ = (1 − a)(1 − aq)(1 − aq 2 ) · · · .

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Following Ramanujan, we defined ∞ 

ϕ(q) := f (q, q) =

(1.2)

2

qn =

n=−∞ 3

ψ(q) := f (q, q ) =

(1.3)

∞ 

q

n(n+1) 2

n=0

(1.4)

2

f (−q) := f (−q, −q ) =

∞ 

(−q; −q)∞ , (q; −q)∞ (q 2 ; q 2 )∞ = , (q; q 2)∞

(−1)n q

n(3n−1) 2

= (q; q)∞

n=−∞

and χ(q) := (−q; q 2 )∞ .

(1.5)

Ramanujan recorded many q-continued fractions and some of their explicit values in his second notebook [9] and in his lost notebook [10]. The following beautiful continued fraction identity was recorded by Ramanujan in his second notebook and can be found in [1, p. 11, Entry 11]: (−a; q)∞ (b; q)∞ − (a; q)∞ (−b; q)∞ (−a; q)∞ (b; q)∞ + (a; q)∞ (−b; q)∞ (1.6) (a − bq)(aq − b) q(a − bq 2 )(aq 2 − b) a−b 1−q + 1 − q3 + 1 − q5 + ··· where either q, a, and b are complex numbers with mod q < 1, or q, a, and b are complex numbers with a = bq m for some integer m. Several elegant q-continued fractions can be expressed in terms of Ramanujan’s thetafunctions. The most famous of them is the celebrated Rogers-Ramanujan continued fraction R(q) is defined as (1.7)

R(q) :=

q 1/5 q 1/5 f (−q, −q 4 ) q q2 q3 = · · · , |q| < 1, f (−q 2 , −q 3 ) 1 + 1 + 1 + 1 +

On page 365 of his Lost Notebook [10], Ramanujan recorded five identities showing the relationships between R(q) and five continued fractions R(−q), R(q 2 ), R(q 3 ), R(q 4 ), and R(q 5 ). He also recorded these identities at the scattered places of his Notebooks [9]. L. J. Rogers [11] established the modular equations relating R(q) and R(q n ) for n=2,3,5, and 11. The last of these equations cannot be found in Ramanujan’s works. The Ramanujan’s cubic continued fraction G(q) is defined as (1.8)

G(q) :=

q 1/3 q + q2 q2 + q4 q 1/3 f (−q, −q 5 ) = · · · , |q| < 1, f (−q 3 , −q 3 ) 1 + 1 + 1 +

The continued fraction (1.8) was first introduced by Ramanujan in his second letter to G. H. Hardy [7]. He also recorded the continued fraction (1.8) on page 365 of his Lost Notebook [10] and claimed that there are many results for

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G(q) similar the results obtained for the famous Rogers-Ramanujan continued fraction (1.7). The Ramanjuan G¨ollnitz-Gordon continued fraction [7, p. 44], [5], [10] is defined as follows: q 1/2 f (−q 3 , −q 5 ) q 1/2 q2 q4 (1.9) L(q) := = · · · , |q| < 1, f (−q, −q 7 ) 1 + 1 + q3 + 1 + q5 +

Motivated by the above cited works on the continued fractions, changing q to q 2 in (1.7), we obtain

(1.10)

(−a; q 2 )∞ (b; q 2 )∞ − (a; q 2 )∞ (−b; q 2 )∞ (−a; q 2 )∞ (b; q 2 )∞ + (a; q 2 )∞ (−b; q 2 )∞ (a − bq 2 )(aq 2 − b) q 2 (a − bq 4 )(aq 4 − b) a−b = 1 − q2 + 1 − q6 + 1 − q 10 + ···

If a = q and b = 0 in the above equation (1.10), we get (1.11)

(−q; q 2 )∞ − (q; q 2)∞ q4 q8 q = (−q; q 2 )∞ + (q; q 2 )∞ 1 − q 2 + 1 − q 6 + 1 − q 10 + ···

In this paper, we study the Ramanujan’s continued fraction U(q) defined by (1.12)

U(q) =

q4 q8 q 1 − q 2 + 1 − q 6 + 1 − q 10 + ···

which is the right hand side of the equality (1.11). In section 3, we establish some modular relations connecting U(q) and U(q n ). In section 4, we find the some explicit evaluations of U(q). 2. Preliminary results In this section, we collect the necessary results required to prove our main results. Lemma 2.1. (2.1)

χ(q) = 21/6 {x(1 − x)q −1 }−1/24 ,

and (2.2)

χ(−q) = 21/6 (1 − x)1/12 (xq −1 )−1/24 .

For the proofs of (2.1) and (2.2), see [2, Entry 12(v),(vi), p.124].

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Lemma 2.2. If β is of degree 2 over α, then   √ (1 − 1 − α)(1 − β) = 2 β(1 − α). (2.3) For a proof of (2.3), see [1, Entry 17.3.1, p.385]. Lemma 2.3. If β has degree 3 over α, then (2.4)

(αβ)1/4 + {(1 − α)(1 − β)}1/4 = 1.

For a proof of (2.4), see [2, Entry 5(ii), p.230]. Lemma 2.4. If β has degree 4 over alpha, then   √ (2.5) (1 − 4 1 − α)(1 − 4 β) = 2 4 β(1 − α). For a proof of (2.5), see [1, Entry 17.3.2, p.385]. Lemma 2.5. If β has degree 5 over α, then (2.6)

(αβ)1/2 + {(1 − α)(1 − β)}1/2 + 2{16αβ(1 − α)(1 − β)}1/6 = 1.

For a proof of (2.6), see [2, Entry 13(i), p.280]. Lemma 2.6. If β has degree 7 over α, then (2.7)

(αβ)1/8 + {(1 − α)(1 − β)}1/8 = 1.

For a proof of (2.7), see [2, Entry 19(i), p.314]. Lemma 2.7. If β has degree 8 over α, then (2.8)

√ (1 − (1 − α)1/4 )(1 − β 1/4 ) = 2 2(β(1 − α))1/8 .

For a proof of (2.8), see [2]. Lemma 2.8. If β has degree 9 over α, then  1/8  1/8  1/8 √ β 1−β β(1 − β) (2.9) + − = m. α 1−α α(1 − α) (2.10)

 1/8  1/8  1/8 1−α α(1 − α) α 3 + − =√ . β 1−β β(1 − β) m

For the proofs of (2.9) and (2.10), see [2, Entry 3(x),(xi), p.352]. Lemma 2.9. If β has degree 11 over α, then (2.11)

(αβ)1/4 + {(1 − α)(1 − β)}1/4 + 2{16αβ(1 − α)(1 − β)}1/12 = 1.

For a proof of (2.11), see [2, Entry 7, p.363].

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3. Relation Between U(q) and U(q n ) Theorem 3.1. We have χ(−q) χ(q) U(q) := χ(−q) 1+ χ(q) 1−

(3.1)

Proof. From the equations (1.11) and (1.12), weget (−q, q 2 )∞ −1 (q, q 2 )∞ (3.2) U(q) = (−q, q 2 )∞ +1 (q, q 2 )∞ Employing the equation (1.5) in the above equation (3.2), we obtain χ(q) −1 χ(−q) (3.3) U(q) := χ(q) +1 χ(−q) the above equation can written as (3.1). Corollary 3.1. We have (3.4)

χ(−q) 1 − U(q) := χ(q) 1 + U(q)

Proof. Easily from the equations (3.1). Theorem 3.2. If x := U(q), and y := U(q 2 ), then 2    1 1 1 (3.5) y+ = y 2 + 2 + 6. x+ x y y Proof. From the equations (2.1) and (2.2), we get χ(−q) = (1 − x)1/8 , 0 < x < 1. (3.6) χ(q) The equation (2.3) can be written as √  2 1− 1−α √ (3.7) . β= 1+ 1−α Employing the equations (3.4) and (3.6) in the above equation (3.7), we obtain (y 4 x2 − y 3 − y − 2y 3 x2 + 6y 2x2 − 2yx2 + x2 − y 3 x4 − yx4 ) (3.8)

(y 4 + 2y 4x2 + y 4 x4 + 6y 2 + 1 − 16y 3x2 + 12y 2x2 − 16yx2 + 2x2 + 6y 2x4 + x4 ) = 0

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By examining the behavior of the above factors near q = 0, we can find a neighborhood about the origin, where the first factor is zero; whereas other factor are not zero in this neighborhood. By the Identity Theorem first factor vanishes identically. This completes the proof. Theorem 3.3. If x := U(q), and y := U(q 3 ), then         1 y2 1 x 1 2 (3.9) 3 x+ + + =3 y+ + x y+ 2 . x y2 x y xy Proof. The equation (2.4) can be written as (3.10)

αβ = [1 − {(1 − α)(1 − β)}1/4 ]4 .

Employing the equations (3.4) and (3.6) in the above equation (3.10), we obtain (3.11)

(−x3 y 4 + y 3 − 3y 2x − x + 3y 3x2 + 3yx2 − 3x3 y 2 + yx4 ) (−y 4 x + y − 3y 2x + 3y 3x2 + 3yx2 − 3x3 y 2 − x3 + y 3x4 ) = 0.

By examining the behavior of the above factors near q = 0, we can find a neighborhood about the origin, where the second factor is zero; whereas other factor are not zero in this neighborhood. By the Identity Theorem second factor vanishes identically. This completes the proof. Theorem 3.4. If x := U(q), and y := U(q 4 ), then         1 1 1 1 4 2 3 y + 3 +7 y+ x + 4 +4 x + 2 −2 x x y y     (3.12) 1 1 = y 4 + 4 + 28 y 2 + 2 + 70. y y Proof. The equation (2.5) can be written as √  4 1− 41−α √ β= (3.13) . 1+ 41−α Employing the equations (3.4) and (3.6) in the above equation (3.13), we obtain x4 − y + 28y 2x4 − 7y 3 + 14y 3x4 + 2yx4 − 28y 3x2 − 4yx2 + 70y 4x4 (3.14) − yx8 − y 7x8 + y 8 x4 − 4yx6 − 28y 3x6 − 7y 3x8 − 28y 5x2 + 14y 5x4 − 28y 5x6 − 7y 5x8 + 28y 6x4 − 4y 7x2 + 2y 7x4 − 4y 7x6 − 7y 5 − y 7 = 0. Theorem 3.5. If x := U(q), and y := U(q 5 ), then       1 1 1 2 3 2 + 5 xy + 2 − 2 x y + 3 2 10 x + x xy xy    2    2 (3.15) 3 y 1 y x x + 2 + = 10 y + + . +5 y y x y 3 x2

Modular identities of new Ramanujan continued fraction

691

Proof. The equation (2.6) can be written as (3.16)

6 16αβ(1 − α)(1 − β) = 2−1/6 1 − (αβ)1/2 − {(1 − α)(1 − β)}1/2 .

Employing the equations (3.4) and (3.6) in the above equation (3.16), we obtain (3.17) (−y + 5xy 4 − 5x2 y − 10x2 y 3 + 10x3 y 2 + 10x3 y 4 − 10x4 y 3 + x5 + y 6x − 5x4 y 5 + 5x5 y 2 − x6 y 5)(x + 5xy 2 − 10x2 y 3 + 10x3 y 2 + 10x3 y 4 − 5x4 y − 10x4 y 3 − y 5 − 5y 5 x2 + 5x5 y 4 + x5 y 6 − x6 y)(1 + 6x2 + 15x4 + 6y 2 + x12 y 12 + 200xy 3 + 276x2 y 2 + 1050x2 y 4 + 200x3 y − 280x3 y 3 + 1050x4 y 2 + 7905x4 y 4 + 40xy + 1560x2 y 6 + 13740x4 y 6 + 1560x6 y 2 + 13740x6y 4 + 28496x6 y 6 − 624xy 5 − 6960x5 y 3 − 17760x5y 5 − 6960x3 y 5 + 1050x2 y 8 − 6960x3 y 7 + 7905x4 y 8 − 17760x5 y 7 − 624x5 y − 624y 7x + 276y 10x2 − 280y 9x3 + 200x3 y 11 + 1050x4 y 10 − 6960x5 y 9 + 13740x6 y 8 + 15x4 y 12 − 624x5 y 11 + 1560x6 y 10 − 6960x7 y 3 − 17760x7y 5 − 17760x7y 7 − 6960x7 y 9 + 7905x8 y 4 + 13740x8 y 6 + 7905x8 y 8 + 20x6 y 12 − 624x7 y 11 + 1050x8 y 10 + 15x8 y 12 − 624x7 y + 200y 9x + 1050x8 y 2 + 6y 12 x2 − 280x9 y 3 − 6960x9 y 5 − 6960x9 y 7 − 280x9 y 9 + x12 + 1050x10 y 4 + 1560x10 y 6 + 1050x10 y 8 + 200x9 y 11 + 276x10 y 10 + 6x10 y 12 − 624x11 y 5 − 624x11 y 7 + 200x11 y 9 + 40x11 y 11 + 20x12 y 6 + 15x12 y 8 + 15y 4 + 6x12 y 10 + 200x9 y + 40y 11x + 276x10 y 2 + 200x11 y 3 + 15x12 y 4 + 40x11 y + 6x12 y 2 + 20x6 + 15x8 + 20y 6 + 15y 8 + 6y 10 + y 12 + 6x10 ) = 0. By examining the behavior of the above factors near q = 0, we can find a neighborhood about the origin, where the first factor is zero; whereas other factors are not zero in this neighborhood. By the Identity Theorem first factor vanishes identically. This completes the proof. Theorem 3.6. If x := U(q), and y := U(q 7 ), then (3.18)  3   2      x y3 y2 1 x x y x4 y 4 3 3 + +7 + + + 28 +7 + + 70 = x y + 3 3 y 4 x4 y 3 x3 y 2 x2 y x xy         1 1 1 1 1 + 49 xy + y+ x+ +7 x3 + 3 + y3 + 3 . xy x y y x Proof. The equation (2.7) can be written as (3.19)

8

αβ = 1 − {(1 − α)(1 − β)}1/8 .

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Employing the equations (3.4) and (3.6) in the above equation (3.19), we obtain x8 + y 8 + 28x2 y 6 + 70x4 y 4 + 28x6 y 2 − 7xy 3 − 7xy 5 + 7xy 7 (3.20)

− 7x3 y − 49x3 y 3 + 7x3 y 5 − 7x3 y 7 − 7x5 y + 7x5 y 3 − 49x5 y 5 − 7x5 y 7 + 7x7 y − 7x7 y 3 − 7x7 y 5 − x7 y 7 − xy = 0.

Theorem 3.7. If x := U(q), and y := U(q 8 ), then (3.21) (273y 11 + y 15 + y + 800y 8 + 504y 6 + 8y 2 + 273y 5 + 112y 4 + 8y 14 + 504y 10 + 112y 12 + 715y 7 + 715y 9 + 35y 3 + 35y 13)x16 + (5720y 7 + 8y + 5720y 9 + 64y 2 + 896y 12 + 4032y 6 + 2184y 11 + 4032y 10 + 896y 4 + 280y 3 + 6400y 8 + 8y 15 + 64y 14 + 280y 13 + 2184y 5)x14 + (9600y 8 + 12y + 12y 15 + 3276y 11 + 96y 2 + 96y 14 + 1344y 12 + 3276y 5 + 1344y 4 + 420y 3 + 420y 13 + 6048y 6 + 6048y 10 + 8580y 7 + 8580y 9)x12 − (5720y 9 + 6400y 8 + 280y 3 + 64y 2 + 8y 15 + 896y 12 + 2184y 5 + 4032y 6 + 4032y 10 + 8y + 896y 4 + 280y 13 + 64y 14 + 2184y 11 + 5720y 7)x10 − (21112y 6 + 10y + 10y 15 + 328y 2 + y 16 + 350y 3 + 1 + 2730y 11 + 33670y 8 + 4732y 12 + 7150y 9 + 21112y 10 + 7150y 7 + 4732y 4 + 350y 13 + 328y 14 + 2730y 5)x8 − (5720y 9 + 6400y 8 + 280y 3 + 896y 12 + 2184y 5 + 64y 2 + 4032y 6 + 8y 15 + 4032y 10 + 8y + 896y 4 + 280y 13 + 64y 14 + 2184y 11 + 5720y 7)x6 + (9600y 8 + 12y + 12y 15 + 3276y 11 + 96y 2 + 96y 14 + 1344y 12 + 3276y 5 + 1344y 4 + 420y 3 + 420y 13 + 6048y 6 + 6048y 10 + 8580y 7 + 8580y 9)x4 + (5720y 7 + 8y + 5720y 9 + 896y 12 + 4032y 6 + 2184y 11 + 4032y 10 + 896y 4 + 280y 3 + 6400y 8 + 8y 15 + 64y 14 + 64y 2 + 280y 13 + 2184y 5)x2 + 273y 11 + y 15 + y + 800y 8 + 504y 6 + 8y 2 + 273y 5 + 112y 4 + 8y 14 + 504y 10 + 112y 12 + 715y 7 + 715y 9 + 35y 3 + 35y 13 = 0. Proof. The equation (2.8) can be written as (3.22)

8

4096β(1 − α) = {(1 − α)1/4 }(1 − β 1/4 ) .

Employing the equations (3.4) and (3.6) in the above equation (3.22), we obtain(3.21).

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Theorem 3.8. If x := U(q), and y := U(q 9 ), then (3.23) x12 y 2 + (y − y 11 − 18y 7 + 9y 3 − 9y 9 + 6y 5 )x11 + (48y 8 + 1 − 3y 4 + 8y 10 + 12y 2)x10 + (33y 3 − 66y 5 − 114y 7 − 73y 9 − 9y 11 + 9y)x9 + (210y 6 − 3y 2 − 48y 4 + 48y 10 + 288y 8)x8 + (6y − 114y 9 − 66y 3 − 288y 7 − 312y 5 − 18y 11 )x7 + (504y 6 + 210y 4 + 210y 8)x6 + (6y 11 − 18y − 66y 9 − 114y 3 − 288y 5 − 312y 7)x5 + (48y 2 + 210y 6 − 48y 8 − 3y 10 + 288y 4)x4 + (9y 11 − 66y 7 − 114y 5 − 9y + 33y 9 − 73y 3)x3 + (48y 4 + 12y 10 + 8y 2 + y 12 − 3y 8)x2 + (6y 7 − 18y 5 − y − 9y 3 + 9y 9 + y 11 )x + y 10 = 0. Proof. The equations (2.9) and (2.10) can be written as

(3.24)

 1/4  1/4  1/8  1/8 1−α α 1−α α + − − β 1−β β 1−β  1/8  1/4  1/4  1/8 α α 1−α 1−α − − = 0. β 1−β β 1−β

Employing the equations (3.4) and (3.6) in the above equation (3.24), we obtain(3.23). Theorem 3.9. If x := U(q), and y := U(q 11 ), then (3.25) x12 y 11 + (33y 8 − 11y 6 − 33y 4 − 1 + 11y 10 − 11y 2)x11 + (−66y 3 + 11y 11 + 66y 9 − 11y + 66y 7)x10 + (−55y 6 − 396y 4 − 66y 2 + 231y 8 + 66y 10 )x9 + (−33y + 33y 11 + 396y 9 + 528y 7 − 198y 5 − 231y 3)x8 + (198y 8 − 66y 2 − 528y 4 − 396y 6)x7 + (11y + 55y 9 + 55y 3 + 396y 7 + 396y 5 + 11y 11)x6 + (−528y 8 − 396y 6 + 198y 4 − 66y 10)x5 + (528y 5 − 33y 11 − 198y 7 − 231y 9 + 33y + 396y 3)x4 + (−396y 8 − 55y 6 + 231y 4 + 66y 2 − 66y 10)x3 + (66y 3 − 66y 9 + 11y − 11y 11 + 66y 5)x2 + (−y 12 + 11y 2 − 33y 8 − 11y 6 + 33y 4 − 11y 10)x + y = 0. Proof. The equation (2.11) can be written as (3.26)



12 65536αβ(1 − α)(1 − β) = 1 − (αβ)1/4 − {(1 − α)(1 − β)}1/4 .

Employing the equations (3.4) and (3.6) in the above equation (3.26), we obtain(3.25).

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4. Explicit Evaluation of U(q) Theorem 4.1. We have

χ(−e−nπ ) χ(e−nπ ) U(e−nπ ) = χ(−e−nπ ) 1+ χ(e−nπ ) 1−

(4.1)

Proof. Put q := e−nπ in the equation (3.1), we obtain (4.1). In his first notebook, Ramanujan’s second notebook [9] recorded many elementary values of χ(q). In particularly, he recorded χ(e−nπ ) and χ(−e−nπ ) for n = 1, 2, and4. Noting from [3, Entry 2(v), (vi), (viii) and (ix), p.326], we have (4.2)

χ(e−π ) = 21/4 e−π/24

χ(−e−π ) = 21/8 e−π/24 .

and

Employing (4.2) in the equation (4.1), we get U(e−π ) =

(4.3) (4.4)

21/8 − 1 . 21/8 + 1

√ χ(e−2π ) = 21/16 ( 2 + 1)1/4 e−π/12

and χ(−e−2π ) = 23/8 e−π/12 .

Employing (4.4) in the equation (4.1), we obtain (4.5)

−2π

U(e

√ 1 − 25/16 ( 2 − 1)1/4 √ )= . 1 + 25/16 ( 2 − 1)1/4 References

[1] [2] [3] [4]

G. E. Andrews, B. C. Berndt, Ramanujan’s Lost Notebook, Part I. New York (2005). B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York (1991). B. C. Berndt, Ramanujan’s Notebooks, Part V, Springer-Verlag, New York (1998). H.H. Chan, S.-S. Haung, On the Ramanujan-G¨ollnitz-Gordon continued fraction. Ramanujan J. 1, 75-90 (1997). [5] H. G¨ ollnitz, Partition mit Differnzebendingungen, J. Reine Angew. Math., 25, 154-190 (1967). [6] B. Gordon, Some continued fractions of the Rogers-Ramanujan type, Duke Math. J., 32, 741-748 (1965). [7] G. H. Hardy, Ramanujan, Chelsea, New York (1978). [8] Nipen Saikia, Modular Idendities and Explicit Values of a New Continued Fraction of Ramanujan, Global Journal of Mathematical Sciences, Volume 4, Number 3, pp. 245248.India (2012). [9] S. Ramanujan, Notebooks (2 volumes). Bombay. Tata Institute of Fundamental Research (1957). [10] S. Ramanujan, The ‘lost’ notebook and other unpublished papers. New Delhi. Narosa (1988).

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[11] L. J. Rogers, On a type of modular relation. Proc. Lond. Math. Soc., 19, 387–397 (1921).

Received: January 2, 2013