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ISRN Computational Mathematics
VolumeΒ 2012Β (2012), Article IDΒ 169050, 14 pages
http://dx.doi.org/10.5402/2012/169050
Research Article

A Parameter for Ramanujan's Function Ο‡(q): Its Explicit Values and Applications

Department of Mathematics, Rajiv Gandhi University, Rono Hills, Doimukh 791112, India

Received 3 May 2012; Accepted 28 June 2012

Academic Editors: L.Β Hajdu, L. S.Β Heath, and H. J.Β Ruskin

Copyright Β© 2012 Nipen Saikia. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We define a new parameter πΌπ‘˜,𝑛 involving quotient of Ramanujan's function πœ’(π‘ž) for positive real numbers π‘˜ and 𝑛 and study its several properties. We prove some general theorems for the explicit evaluations of the parameter πΌπ‘˜,𝑛 and find many explicit values. Some values of πΌπ‘˜,𝑛 are then used to find some new and known values of Ramanujan's class invariant 𝐺𝑛.

1. Introduction

In Chapter 16 of his second notebook [1], Ramanujan develops the theory of theta-function. Ramanujan's general theta-function is defined by 𝑓(π‘Ž,𝑏)=βˆžξ“π‘›=βˆ’βˆžπ‘Žπ‘›(𝑛+1)/2𝑏𝑛(π‘›βˆ’1)/2,||||π‘Žπ‘<1.(1) After Ramanujan, for |π‘ž|<1, we define 𝑓(βˆ’π‘ž)∢=π‘“βˆ’π‘ž,βˆ’π‘ž2ξ€Έ=βˆžξ“π‘›=βˆ’βˆž(βˆ’1)π‘›π‘žπ‘›(3π‘›βˆ’1)/2=(π‘ž;π‘ž)∞,(2) where (π‘Ž;π‘ž)∞∏∢=βˆžπ‘›=0(1βˆ’π‘Žπ‘žπ‘›). If π‘ž=𝑒2πœ‹π‘–π‘§ with Im(𝑧)>0, then 𝑓(βˆ’π‘ž)=π‘žβˆ’1/24πœ‚(𝑧), where πœ‚(𝑧) denotes the classical Dedekind eta function.

Ramanujan's function πœ’(π‘ž) is defined by πœ’(π‘ž)∢=𝑓(π‘ž)π‘“ξ€·βˆ’π‘ž2ξ€Έ=ξ€·βˆ’π‘ž;π‘ž2ξ€Έβˆž.(3) The function πœ’(π‘ž) is intimately connected to Ramanujan's class invariants 𝐺𝑛 and 𝑔𝑛, which are defined by 𝐺𝑛=2βˆ’1/4π‘žβˆ’1/24πœ’(π‘ž),𝑔𝑛=2βˆ’1/4π‘žβˆ’1/24πœ’(βˆ’π‘ž),(4) where π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘› and 𝑛 is a positive rational number. Since from [2, page 124, Entry 12(v) & (vi)] πœ’(π‘ž)=21/6𝛼(1βˆ’π›Ό)π‘žξ‚Όβˆ’1/24,πœ’(βˆ’π‘ž)=21/6(1βˆ’π›Ό)1/12ξ‚΅π›Όπ‘žξ‚Άβˆ’1/24,(5) it follows from (4) that 𝐺𝑛={4𝛼(1βˆ’π›Ό)}βˆ’1/24,𝑔𝑛=2βˆ’1/12(1βˆ’π›Ό)1/12π›Όβˆ’1/24.(6) Also, if 𝛽 has degree π‘Ÿ over 𝛼, then πΊπ‘Ÿ2𝑛={4𝛽(1βˆ’π›½)}βˆ’1/24,π‘”π‘Ÿ2𝑛=2βˆ’1/12(1βˆ’π›½)1/12π›½βˆ’1/24.(7) In his notebooks [1] and paper [3], Ramanujan recorded a total of 116 class invariants or monic polynomials satisfied by them. The table at the end of Weber's book [4, page 721–726] also contains the values of 107 class invariants. Weber primarily was motivated to calculate class invariants so that he could construct Hilbert class fields. On the other hand Ramanujan calculated class invariants to approximate πœ‹ and probably for finding explicit values of Rogers-Ramanujan continued fractions, theta-functions, and so on. An account of Ramanujan's class invariants and applications can be found in Berndt's book [5]. For further references, see [6–12].

Ramanujan and Weber independently and many others in the literature calculated class invariants 𝐺𝑛 for odd values of 𝑛 and 𝑔𝑛 for even values of 𝑛. For the first time, Yi [13] calculated some values of 𝑔𝑛 for odd values of 𝑛 by finding explicit values of the parameter π‘Ÿπ‘˜,𝑛 (see [13, page 11, (2.1.1)] or [14, page 4, (1.11)]) defined by π‘Ÿπ‘˜,π‘›βˆΆ=𝑓(βˆ’π‘ž)π‘˜1/4π‘ž(π‘˜βˆ’1)/24π‘“ξ€·βˆ’π‘žπ‘˜ξ€Έ,π‘ž=π‘’βˆšβˆ’2πœ‹π‘›/π‘˜.(8) In particular, she established the result [13, page 18, Theorem 2.2.3] 𝑔𝑛=π‘Ÿ2,𝑛/2.(9) However, the values of 𝐺𝑛 for even values of 𝑛 have not been calculated. The main objective of this paper is to evaluate some new values of 𝐺𝑛 for even values of 𝑛. We also prove some known values of 𝐺𝑛. For evaluation of class invariant 𝐺𝑛 in this paper, we introduce the parameter πΌπ‘˜,𝑛, which is defined as πΌπ‘˜,π‘›βˆΆ=πœ’(π‘ž)π‘ž(βˆ’π‘˜+1)/24πœ’ξ€·π‘žπ‘˜ξ€Έ,π‘ž=π‘’βˆšβˆ’πœ‹π‘›/π‘˜,(10) where π‘˜ and 𝑛 are positive real numbers.

In Section 3, we study some properties of πΌπ‘˜,𝑛 and also establish its relations with Ramanujan's class invariant 𝐺𝑛. In Section 4, by employing Ramanujan's modular equations, we present some general theorems for the explicit evaluations of πΌπ‘˜,𝑛 and find several explicit values of πΌπ‘˜,𝑛. In Section 5, we establish some general theorems connecting the parameter πΌπ‘˜,𝑛 and the class invariant 𝐺𝑛. We also evaluate some explicit values of the product πΊπ‘›π‘˜πΊπ‘›/π‘˜ by employing some values of πΌπ‘˜,𝑛 evaluated in Section 4. Finally, in Section 6, we calculate new and known values of class invariant 𝐺𝑛 by combining the explicit values of πΌπ‘˜,𝑛 and the product πΊπ‘›π‘˜πΊπ‘›/π‘˜ evaluated in Sections 4 and 5, respectively. Section 2 is devoted to record some preliminary results.

Since Ramanujan's modular equations are key in our evaluations of πΌπ‘˜,𝑛 and 𝐺𝑛, we complete this introduction by defining Ramanujan's modular equation from Berndt's book [2]. The complete elliptic integral of the first kind 𝐾(π‘˜) is defined by ξ€œπΎ(π‘˜)∢=0πœ‹/2π‘‘πœ™βˆš1βˆ’π‘˜2sin2πœ™=πœ‹2βˆžξ“π‘›=0(1/2)2𝑛(𝑛!)2π‘˜2𝑛=πœ‹22𝐹1ξ‚€12,12;1;π‘˜2,(11) where 0<π‘˜<1, 2𝐹1 denotes the ordinary or Gaussian hypergeometric function, and (π‘Ž)𝑛=π‘Ž(π‘Ž+1)(π‘Ž+2)β‹―(π‘Ž+π‘›βˆ’1).(12) The number π‘˜ is called the modulus of 𝐾, and π‘˜ξ…žβˆšβˆΆ=1βˆ’π‘˜2 is called the complementary modulus. Let 𝐾,πΎξ…ž, 𝐿, and πΏξ…ž denote the complete elliptic integrals of the first kind associated with the moduli π‘˜, π‘˜ξ…ž, 𝑙, and π‘™ξ…ž, respectively. Suppose that the equality 𝑛𝐾′𝐾=𝐿′𝐿(13) holds for some positive integer 𝑛. Then, a modular equation of degree 𝑛 is a relation between the moduli π‘˜ and 𝑙, which is implied by (13).

If we set ξ‚΅πΎπ‘ž=expβˆ’πœ‹ξ…žπΎξ‚Ά,π‘žξ…žξ‚΅πΏ=expβˆ’πœ‹ξ…žπΏξ‚Ά,(14) we see that (13) is equivalent to the relation π‘žπ‘›=π‘žξ…ž. Thus, a modular equation can be viewed as an identity involving theta-functions at the arguments π‘ž and π‘žπ‘›. Ramanujan recorded his modular equations in terms of 𝛼 and 𝛽, where 𝛼=π‘˜2 and 𝛽=𝑙2. We say that 𝛽 has degree 𝑛 over 𝛼. The multiplier π‘š connecting 𝛼 and 𝛽 is defined by πΎπ‘š=𝐿.(15) Ramanujan also established many β€œmixed” modular equations in which four distinct moduli appear, which we define from Berndt's book [2, page 325].

Let 𝐾, πΎξ…ž, 𝐿1, πΏξ…ž1, 𝐿2, πΏξ…ž2, 𝐿3, and πΏξ…ž3 denote complete elliptic integrals of the first kind corresponding, in pairs, to the moduli βˆšπ›Ό, βˆšπ›½, βˆšπ›Ύ, and βˆšπ›Ώ and their complementary moduli, respectively. Let 𝑛1, 𝑛2, and 𝑛3 be positive integers such that 𝑛3=𝑛1𝑛2. Suppose that the equalities 𝑛1𝐾′𝐾=πΏξ…ž1𝐿1,𝑛2𝐾′𝐾=πΏξ…ž2𝐿2,𝑛3𝐾′𝐾=πΏξ…ž3𝐿3(16) hold. Then, a β€œmixed” modular equation is a relation between the moduli βˆšπ›Ό, βˆšπ›½, βˆšπ›Ύ, and βˆšπ›Ώ that is induced by (16). In such an instance, we say that 𝛽, 𝛾, and 𝛿 are of degrees 𝑛1, 𝑛2, and 𝑛3, respectively, over 𝛼 or 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 𝑛1, 𝑛2, and 𝑛3, respectively. Denoting π‘§π‘Ÿ=πœ™2(π‘žπ‘Ÿ), where ξ‚΅βˆ’π‘ž=expπœ‹πΎξ…žπΎξ‚Ά||π‘ž||,πœ™(π‘ž)=𝑓(π‘ž,π‘ž),<1(17) the multipliers π‘š and π‘šξ…ž associated with 𝛼, 𝛽, and 𝛾, 𝛿, respectively, are defined by π‘š=𝑧1/𝑧𝑛1 and π‘šβ€²=𝑧𝑛2/𝑧𝑛3.

2. Preliminary Results

Lemma 1 (see [2, page 43, Entry 27(v)]). If 𝛼 and 𝛽 are such that the modulus of each exponential argument is less than 1 and 𝛼𝛽=πœ‹2, then 𝑒𝛼/24πœ’(π‘’βˆ’π›Ό)=𝑒𝛽/24πœ’ξ€·π‘’βˆ’π›½ξ€Έ.(18)

Lemma 2 (see [15, page 241, Lemma 2.3]). Let π‘‹βˆΆ=π‘ž1/12(πœ’(π‘ž)/πœ’(π‘ž3)) and π‘ŒβˆΆ=π‘ž1/6(πœ’(π‘ž2)/πœ’(π‘ž6)); then ξ‚€π‘‹π‘Œξ‚12+ξ‚€π‘Œπ‘‹ξ‚12+ξ‚Έξ‚€π‘‹π‘Œξ‚6+ξ‚€π‘Œπ‘‹ξ‚6ξ‚ΉΓ—ξ€Ί(π‘‹π‘Œ)10+(π‘‹π‘Œ)βˆ’10ξ€½+16(π‘‹π‘Œ)6+(π‘‹π‘Œ)βˆ’6ξ€Ύξ€½+71(π‘‹π‘Œ)2+(π‘‹π‘Œ)βˆ’2ξ€Ύξ€»=(π‘‹π‘Œ)12+(π‘‹π‘Œ)βˆ’12ξ€½+25(π‘‹π‘Œ)8+(π‘‹π‘Œ)βˆ’8ξ€Ύξ€½+200(π‘‹π‘Œ)4+(π‘‹π‘Œ)βˆ’4ξ€Ύ+550.(19)

Lemma 3 (see [15, page 241, Lemma 2.8]). Let π‘‹βˆΆ=π‘ž1/6(πœ’(π‘ž)/πœ’(π‘ž5)) and π‘ŒβˆΆ=π‘ž1/3(πœ’(π‘ž2)/πœ’(π‘ž10)); then ξ‚Έξ‚€π‘‹π‘Œξ‚3+ξ‚€π‘Œπ‘‹ξ‚3ξ‚Ήξ€Ί(π‘‹π‘Œ)5+(π‘‹π‘Œ)βˆ’5ξ€·+8(π‘‹π‘Œ)3+(π‘‹π‘Œ)βˆ’3ξ€Έξ€·+19π‘‹π‘Œ+(π‘‹π‘Œ)βˆ’1+ξ‚€π‘‹ξ€Έξ€»π‘Œξ‚6+ξ‚€π‘Œπ‘‹ξ‚6=ξ€·(π‘‹π‘Œ)6+(π‘‹π‘Œ)βˆ’6ξ€Έ(+13ξ€·ξ€·π‘‹π‘Œ)4+(π‘‹π‘Œ)βˆ’4ξ€Έξ€Έ+52ξ€·ξ€·(π‘‹π‘Œ)2+(π‘‹π‘Œ)βˆ’2ξ€Έξ€Έ+82.(20)

Lemma 4 (see [15, page 252, Lemma 2.13]). Let π‘‹βˆΆ=π‘ž1/4(πœ’(π‘ž)/πœ’(π‘ž7)) and π‘ŒβˆΆ=π‘ž1/2(πœ’(π‘ž2)/πœ’(π‘ž14)); then ξ‚€π‘‹π‘Œξ‚12+ξ‚€π‘Œπ‘‹ξ‚12𝑋+16π‘Œξ‚10+ξ‚€π‘Œπ‘‹ξ‚10ξ‚Ήξ€Ί(π‘‹π‘Œ)2+(π‘‹π‘Œ)βˆ’2𝑋+8π‘Œξ‚8+ξ‚€π‘Œπ‘‹ξ‚8ξ‚Ήξ€Ί7ξ€½(π‘‹π‘Œ)4+(π‘‹π‘Œ)βˆ’4ξ€Ύξ€»+ξ‚Έξ‚€π‘‹βˆ’19π‘Œξ‚6+ξ‚€π‘Œπ‘‹ξ‚6ξ‚ΉΓ—ξ€Ί(π‘‹π‘Œ)10+(π‘‹π‘Œ)βˆ’10ξ€½+32(π‘‹π‘Œ)6+(𝑋Y)βˆ’6ξ€Ύξ€½(βˆ’81π‘‹π‘Œ)2+(π‘‹π‘Œ)βˆ’2+ξ‚Έξ‚€π‘‹ξ€Ύξ€»π‘Œξ‚4+ξ‚€π‘Œπ‘‹ξ‚4ξ‚Ήξ€Ίξ€½16(π‘‹π‘Œ)8+(π‘‹π‘Œ)βˆ’8ξ€Ύξ€½βˆ’288(π‘‹π‘Œ)4+(π‘‹π‘Œ)βˆ’4ξ€Ύξ€»+𝑋+352π‘Œξ‚2+ξ‚€π‘Œπ‘‹ξ‚2ξ‚Ήξ€Ίξ€½296(π‘‹π‘Œ)2+(π‘‹π‘Œ)βˆ’2ξ€Ύξ€½βˆ’256(π‘‹π‘Œ)6+(π‘‹π‘Œ)βˆ’6ξ€Ύξ€½βˆ’8(π‘‹π‘Œ)10+(π‘‹π‘Œ)βˆ’10=ξ€½ξ€Ύξ€»+1746(π‘‹π‘Œ)12+(π‘‹π‘Œ)βˆ’12ξ€Ύ+145ξ€Ίξ€½(π‘‹π‘Œ)8+(π‘‹π‘Œ)βˆ’8ξ€Ύξ€»+496ξ€Ίξ€½(π‘‹π‘Œ)4+(π‘‹π‘Œ)βˆ’4.ξ€Ύξ€»(21)

Lemma 5 (see [2, page 231, Entry 5(xii)]). Let 𝑃={16𝛼𝛽(1βˆ’π›Ό)(1βˆ’π›½)}1/8 and 𝑄=(𝛽(1βˆ’π›½)/𝛼(1βˆ’π›Ό))1/4; then 𝑄+π‘„βˆ’1√+22ξ€·π‘ƒβˆ’π‘ƒβˆ’1ξ€Έ=0,(22) where 𝛽 has degree 3 over 𝛼.

Lemma 6 (see [2, page 282, Entry 13(xiv)]). Let 𝑃={16𝛼𝛽(1βˆ’π›Ό)(1βˆ’π›½)}1/12 and 𝑄=(𝛽(1βˆ’π›½)/𝛼(1βˆ’π›Ό))1/8; then 𝑄+π‘„βˆ’1ξ€·+2π‘ƒβˆ’π‘ƒβˆ’1ξ€Έ=0,(23) where 𝛽 has degree 5 over 𝛼.

Lemma 7 (see [2, page 315, Entry 13(xiv)]). Let 𝑃={16𝛼𝛽(1βˆ’π›Ό)(1βˆ’π›½)}1/8 and 𝑄=(𝛽(1βˆ’π›½)/𝛼(1βˆ’π›Ό))1/6; then 𝑄+π‘„βˆ’1√+7=22𝑃+π‘ƒβˆ’1ξ€Έ=0,(24) where 𝛽 has degree 7 over 𝛼.

Lemma 8 (see [5, page 378, Entry 41]). Let 𝑃=21/6{𝛼𝛽(1βˆ’π›Ό)(1βˆ’π›½)}1/24 and 𝑄=(𝛽(1βˆ’π›½)/𝛼(1βˆ’π›Ό))1/24; then 𝑄7+π‘„βˆ’7𝑄+135+π‘„βˆ’5𝑄+523+π‘„βˆ’3ξ€Έξ€·+78𝑄+π‘„βˆ’1ξ€Έξ€·π‘ƒβˆ’8βˆ’6+𝑃6ξ€Έ=0,(25) where 𝛽 has degree 13 over 𝛼.

For Lemmas 9 to 15, we set π‘ƒβˆΆ=(256𝛼𝛽𝛾𝛿(1βˆ’π›Ό)(1βˆ’π›½)(1βˆ’π›Ύ)(1βˆ’π›Ώ))1/48,ξ‚΅π‘„βˆΆ=𝛼𝛿(1βˆ’π›Ό)(1βˆ’π›Ώ)𝛽𝛾(1βˆ’π›½)(1βˆ’π›Ύ)1/48,ξ‚΅π‘…βˆΆ=𝛾𝛿(1βˆ’π›Ύ)(1βˆ’π›Ώ)𝛼𝛽(1βˆ’π›Ό)(1βˆ’π›½)1/48,ξ‚΅π‘‡βˆΆ=𝛽𝛿(1βˆ’π›½)(1βˆ’π›Ώ)𝛼𝛾(1βˆ’π›Ό)(1βˆ’π›Ύ)1/48.(26)

Lemma 9 (see [5, page 381, Entry 50]). If 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 5, 7, and 35, respectively, then 𝑅4+π‘…βˆ’4βˆ’ξ€·π‘„6+π‘„βˆ’6𝑄+54+π‘„βˆ’4ξ€Έξ€·π‘„βˆ’102+π‘„βˆ’2ξ€Έ+15=0.(27)

Lemma 10 (see [5, page 381, Entry 51]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 13, 3, and 39, respectively, then 𝑄4+π‘„βˆ’4ξ€·π‘„βˆ’32+π‘„βˆ’2ξ€Έβˆ’ξ€·π‘‡2+π‘‡βˆ’2ξ€Έ+3=0.(28)

Lemma 11 (see [5, page 381, Entry 52]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 13, 5, and 65, respectively, then 𝑄6+π‘„βˆ’6ξ€·βˆ’5𝑄+π‘„βˆ’1ξ€Έ2𝑇+π‘‡βˆ’1ξ€Έ2βˆ’ξ€·π‘‡4+π‘‡βˆ’4ξ€Έ=0.(29)

Lemma 12 (see [16, page 277, Lemma 3.1]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 3, 7, and 21, respectively, then 𝑅2+π‘…βˆ’2βˆ’π‘„4βˆ’π‘„βˆ’4+3=0.(30)

Lemma 13 (see [15, page 243, Theorem 2.5]). If 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 2, 3, and 6, respectively, then 𝑅4+π‘…βˆ’4+𝑃2βˆ’2π‘ƒβˆ’2=0.(31)

Lemma 14 (see [15, page 248, Theorem 2.10]). If 𝛼, 𝛽, 𝛾 and 𝛿 have degrees 1, 2, 5, and 10, respectively, then 𝑅6+π‘…βˆ’6𝑅+52+π‘…βˆ’2ξ€Έ=4π‘ƒβˆ’4βˆ’π‘ƒ4.(32)

Lemma 15 (see [15, page 252, Theorem 2.12]). If 𝛼, 𝛽, 𝛾 and 𝛿 have degrees 1, 2, 7, and 14, respectively, then 2√2𝑇3+π‘‡βˆ’3ξ€Έ=4π‘ƒβˆ’1βˆ’2𝑃+4𝑃3βˆ’π‘ƒ5.(33)

3. Properties of πΌπ‘˜,𝑛

In this section, we study some properties of πΌπ‘˜,𝑛. We also establish a relation connecting πΌπ‘˜,𝑛 and Ramanujan's class invariants 𝐺𝑛.

Theorem 16. For all positive real numbers π‘˜ and 𝑛, one has (i)πΌπ‘˜,1=1,(ii)πΌπ‘˜,1𝑛=1πΌπ‘˜,𝑛,(iii)πΌπ‘˜,𝑛=𝐼𝑛,π‘˜.(34)

Proof . Using the definition of πΌπ‘˜,𝑛 and Lemma 1, we easily arrive at (i). Replacing 𝑛 by 1/𝑛 in πΌπ‘˜,𝑛 and using Lemma 1, we find that πΌπ‘˜,π‘›πΌπ‘˜,1/𝑛=1, which completes the proof of (ii). To prove (iii), we use Lemma 1 in the definition of πΌπ‘˜,𝑛 to arrive at (πΌπ‘˜,𝑛/𝐼𝑛,π‘˜)=1.

Remark 17. By using the definitions of πœ’(π‘ž) and πΌπ‘˜,𝑛, it can be seen that πΌπ‘˜,𝑛 has positive real value less than 1 and that the values of πΌπ‘˜,𝑛 decrease as 𝑛 increases when π‘˜>1. Thus, by Theorem 16(i), πΌπ‘˜,𝑛<1 for all 𝑛>1 if π‘˜>1.

Theorem 18. For all positive real numbers k, m, and n, one has πΌπ‘˜,𝑛/π‘š=πΌπ‘šπ‘˜,π‘›πΌβˆ’1π‘›π‘˜,π‘š.(35)

Proof. Using the definition of πΌπ‘˜,𝑛, we obtain πΌπ‘šπ‘˜,π‘›πΌπ‘›π‘˜,π‘š=πœ’ξ‚€π‘’βˆšβˆ’πœ‹π‘›/π‘šπ‘˜ξ‚π‘’βˆšπœ‹(βˆšπ‘š/π‘›π‘˜βˆ’π‘›/π‘šπ‘˜)/24πœ’ξ‚€π‘’βˆšβˆ’πœ‹π‘š/π‘›π‘˜ξ‚.(36) Using Lemma 1 in the denominator of the right-hand side of (36) and simplifying, we complete the proof.

Corollary 19. For all positive real numbers k and n, one has πΌπ‘˜2,𝑛=πΌπ‘›π‘˜,π‘›πΌπ‘˜,𝑛/π‘˜.(37)

Proof. Setting π‘˜=𝑛 in Theorem 18 and simplifying using Theorem 16(ii), we obtain πΌπ‘˜2,π‘š=πΌπ‘šπ‘˜,π‘˜πΌπ‘˜,π‘š/π‘˜.(38) Replacing π‘š by 𝑛, we complete the proof.

Theorem 20. Let k, a, b, c, and d be positive real numbers such that ab=cd. Then πΌπ‘Ž,π‘πΌπ‘˜π‘,π‘˜π‘‘=πΌπ‘˜π‘Ž,π‘˜π‘πΌπ‘,𝑑.(39)

Proof. From the definition of πΌπ‘˜,𝑛, we deduce that, for positive real numbers π‘˜,π‘Ž,𝑏,𝑐, and 𝑑, πΌπ‘˜π‘Ž,π‘˜π‘πΌβˆ’1π‘Ž,𝑏=πœ’ξ‚€π‘’βˆšβˆ’πœ‹π‘Žπ‘ξ‚π‘’βˆšπœ‹(π‘˜βˆšπ‘Žπ‘βˆ’π‘Žπ‘)/24πœ’ξ‚€π‘’βˆšβˆ’π‘˜πœ‹π‘Žπ‘ξ‚πΌπ‘˜π‘,π‘˜π‘‘πΌβˆ’1𝑐,𝑑=πœ’ξ‚€π‘’βˆšβˆ’πœ‹π‘π‘‘ξ‚π‘’βˆšπœ‹(π‘˜βˆšπ‘π‘‘βˆ’π‘π‘‘)/24πœ’ξ‚€π‘’βˆšβˆ’π‘˜πœ‹π‘π‘‘ξ‚.(40) Now the result follows readily from (40), and the hypothesis that π‘Žπ‘=𝑐𝑑.

Corollary 21. For any positive real numbers 𝑛 and 𝑝, we have 𝐼𝑛𝑝,𝑛𝑝=𝐼𝑛𝑝2,𝑛𝐼𝑝,𝑝.(41)

Proof. The result follows immediately from Theorem 20 with π‘Ž=𝑝2,𝑏=1,𝑐=𝑑=𝑝, and π‘˜=𝑛.

Now, we give some relations connecting the parameter πΌπ‘˜,𝑛 and Ramanujan's class invariants 𝐺𝑛.

Theorem 22. Let π‘˜ and 𝑛 be any positive real numbers. Then (i)πΌπ‘˜,𝑛=𝐺𝑛/π‘˜πΊβˆ’1π‘›π‘˜,(ii)𝐺1/𝑛=𝐺𝑛.(42)

Proof. Proof of (i) follows easily from the definitions of πΌπ‘˜,𝑛 and 𝐺𝑛 from (10) and (4), respectively. To prove (ii), we set π‘˜=1 in part (i) and use Theorem 16(i) and (iii).

4. General Theorems and Explicit Evaluations of πΌπ‘˜,𝑛

In this section, we prove some general theorems for the explicit evaluations of πΌπ‘˜,𝑛 and find its explicit values.

Theorem 23. One has 𝐼3,𝑛𝐼3,4𝑛12+𝐼3,𝑛𝐼3,4π‘›ξ‚Άβˆ’12+𝐼3,𝑛𝐼3,4𝑛6+𝐼3,𝑛𝐼3,4π‘›ξ‚Άβˆ’6×𝐼3,𝑛𝐼3,4𝑛10+𝐼3,𝑛𝐼3,4π‘›ξ€Έβˆ’10𝐼+163,𝑛𝐼3,4𝑛6+(𝐼3,𝑛𝐼3,4𝑛)βˆ’6𝐼+713,𝑛𝐼3,4𝑛2+𝐼3,𝑛𝐼3,4π‘›ξ€Έβˆ’2=𝐼3,𝑛𝐼3,4𝑛12+𝐼3,𝑛𝐼3,4π‘›ξ€Έβˆ’12𝐼+253,𝑛𝐼3,4𝑛8+𝐼3,n𝐼3,4π‘›ξ€Έβˆ’8𝐼+2003,𝑛𝐼3,4𝑛4+𝐼3,𝑛𝐼3,4π‘›ξ€Έβˆ’4+550.(43)

Proof. The proof follows easily from the definition of πΌπ‘˜,𝑛 and Lemma 2.

Corollary 24. One has (i)𝐼3,2=ξ‚΅βˆšβˆ’44+273βˆ’3√458βˆ’2643ξ‚Ά1/12,(ii)𝐼3,4=βŽ›βŽœβŽœβŽœβŽœβŽξ‚΅ξ”βˆšβˆ’2+36βˆ’βˆšβˆ’6+36ξ‚Ά2⎞⎟⎟⎟⎟⎠1/2,(iii)𝐼3,1/2=ξ‚΅βˆšβˆ’44+273+3√458βˆ’2643ξ‚Ά1/12,(iv)𝐼3,1/4=βŽ›βŽœβŽœβŽœβŽœβŽξ‚΅ξ”βˆšβˆ’6+36+βˆšβˆ’2+36ξ‚Ά2⎞⎟⎟⎟⎟⎠1/2.(44)

Proof. Setting 𝑛=1/2 in Theorem 23 and using Theorem 16(ii), we obtain 𝐼243,2+πΌβˆ’243,2𝐼+176123,2+πΌβˆ’123,2ξ‚βˆ’1002=0.(45) Equivalently, 𝐡2+176π΅βˆ’1004=0,(46) where 𝐡=𝐼123,2+πΌβˆ’123,2.(47) Solving (46) and using the fact in Remark 17, we obtain √𝐡=545βˆ’88.(48) Employing (48) in (47), solving the resulting equation for 𝐼3,2, and noting that 𝐼3,2<1, we arrive at 𝐼3,2=ξ‚΅βˆšβˆ’44+273βˆ’3√458βˆ’2643ξ‚Ά1/12.(49) This completes the proof of (i).
Again setting 𝑛=1 in Theorem 23 and using Theorem 16(i), we obtain 𝐼63,4+πΌβˆ’63,4I103,4+πΌβˆ’103,4𝐼+1663,4+πΌβˆ’63,4𝐼+7123,4+πΌβˆ’23,4𝐼=2583,4+πΌβˆ’83,4𝐼+20043,4+πΌβˆ’43,4+550.(50) Equivalently, 𝐷2ξ€Έ+12𝐷4+4𝐷2ξ€Έβˆ’50=0,(51) where 𝐷=𝐼23,4+πΌβˆ’23,4.(52) Since the first factor of (51) is nonzero, solving the second factor, we deduce that ξ‚€βˆšπ·=βˆ’2+361/2.(53) Employing (53) in (52), solving the resulting equation, and using the fact that 𝐼3,4<1, we obtain 𝐼3,4=βŽ›βŽœβŽœβŽœβŽœβŽξ‚΅ξ”βˆšβˆ’2+36βˆ’βˆšβˆ’6+36ξ‚Ά2⎞⎟⎟⎟⎟⎠1/2.(54) This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 25. One has 𝐼5,𝑛𝐼5,4𝑛6+𝐼5,𝑛𝐼5,4π‘›ξ‚Άβˆ’6+𝐼5,𝑛𝐼5,4𝑛3+𝐼5,𝑛𝐼5,4π‘›ξ‚Άβˆ’3×𝐼5,𝑛𝐼5,4𝑛5+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’5𝐼+85,𝑛𝐼5,4𝑛3+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’3𝐼+19ξ€·ξ€·5,𝑛𝐼5,4𝑛+𝐼5,𝑛𝐼5,4𝑛=𝐼5,𝑛𝐼5,4𝑛6+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’6𝐼+135,𝑛𝐼5,4𝑛4+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’4𝐼+525,𝑛𝐼5,4𝑛2+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’2+82.(55)

Proof. The proof follows from Lemma 3 and the definition of πΌπ‘˜,𝑛.

Corollary 26. One has (i)𝐼5,2=ξ‚΅βˆšβˆ’14+510βˆ’βˆš445βˆ’140ξ‚Ά101/6,(ii)𝐼5,4=ξƒ©ξ‚€βˆš11+551/4βˆ’ξ‚™ξ”βˆ’4+√11+55ξƒͺ2,(iii)𝐼5,1/2=ξ‚΅βˆšβˆ’14+510+√445βˆ’140ξ‚Ά101/6,(iv)𝐼5,1/4=ξƒ©ξ‚€βˆš11+551/4+ξ‚™ξ”βˆ’4+√11+55ξƒͺ2.(56)

Proof. Setting 𝑛=1/2 in Theorem 25 and using Theorem 16(ii), we obtain 𝐢2+56πΆβˆ’216=0,(57) where 𝐢=𝐼65,2+πΌβˆ’65,2.(58) Solving (57) and noting the fact in Remark 17, we obtain √𝐢=βˆ’28+1010.(59) Employing (59) in (58), solving the resulting equation, and noting that 𝐼5,2<1, we obtain 𝐼5,2=ξ‚΅βˆšβˆ’14+510βˆ’βˆš445βˆ’140ξ‚Ά101/6.(60) This completes the proof of (i).
Again, setting 𝑛=1 in Theorem 25 and using Theorem 16(i), we obtain 𝐡8βˆ’22𝐡4βˆ’4=0,(61) where 𝐡=𝐼5,4+πΌβˆ’14,5.(62) Solving (61), we obtain ξ‚€βˆšπ΅=11+551/4.(63) Using (63) in (62), solving the resulting equation, and noting that 𝐼5,4<1, we arrive at 𝐼5,4=ξƒ©ξ‚€βˆš11+551/4βˆ’ξ‚™ξ”βˆ’4+√11+55ξƒͺ2.(64) This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 27. One has 𝐼7,𝑛𝐼4,7𝑛12+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’12𝐼+167,𝑛𝐼4,7𝑛10+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’10𝐼7,𝑛𝐼7,4𝑛2+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’2𝐼+87,𝑛𝐼4,7𝑛8+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’8×7𝐼7,𝑛𝐼7,4𝑛4+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’4+ξƒ¬ξ‚΅πΌβˆ’197,𝑛𝐼4,7𝑛6+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’6×𝐼7,𝑛𝐼7,4𝑛10+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’10𝐼+327,𝑛𝐼7,4𝑛6+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’6ξ‚‡ξ‚†ξ€·πΌβˆ’817,𝑛𝐼7,4𝑛2+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’2+𝐼7,𝑛𝐼4,7𝑛4+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’4×𝐼167,𝑛𝐼7,4𝑛8+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’8ξ‚‡ξ‚†ξ€·πΌβˆ’2887,𝑛𝐼7,4𝑛4+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’4+𝐼+3527,𝑛𝐼4,7𝑛2+𝐼7,𝑛𝐼4,7π‘›ξ‚Άβˆ’2×𝐼2967,𝑛𝐼7,4𝑛2+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’2ξ‚‡ξ‚†ξ€·πΌβˆ’2567,𝑛𝐼7,4𝑛6+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’6ξ‚‡ξ‚†ξ€·πΌβˆ’87,𝑛𝐼7,4𝑛10+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’10=𝐼+17467,𝑛𝐼7,4𝑛12+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’12𝐼+1457,𝑛𝐼7,4𝑛8+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’8𝐼+4967,𝑛𝐼7,4𝑛4+𝐼7,𝑛𝐼7,4π‘›ξ€Έβˆ’4.(65)

Corollary 28. One has (i)𝐼7,2=2βˆ’1/4Γ—βŽ›βŽœβŽœβŽβˆšβˆ’8βˆ’52+√163+1162βˆ’ξƒŽξ‚΅βˆšβˆ’4+βˆ’8βˆ’52+√163+1162ξ‚Ά2⎞⎟⎟⎠1/4,(ii)𝐼7,4=ξ‚™3ξ”βˆšβˆ’6+52βˆ’βˆšβˆ’58+452√2,(iii)𝐼7,1/2=2βˆ’1/4,Γ—βŽ›βŽœβŽœβŽβˆšβˆ’8βˆ’52+√163+1162+ξƒŽξ‚΅βˆšβˆ’4+βˆ’8βˆ’52+√163+1162ξ‚Ά2⎞⎟⎟⎠1/4(iv)𝐼7,1/4=ξ‚™3ξ”βˆšβˆ’6+52+βˆšβˆ’58+452√2.(66)

Proof. Setting 𝑛=1/2 and simplifying using Theorem 16(ii), we obtain 𝐼247,2+πΌβˆ’247,2𝐼+32207,2+πΌβˆ’207,2ξ‚ξ‚€πΌβˆ’40167,2+πΌβˆ’167,2ξ‚ξ‚€πΌβˆ’96127,2+πΌβˆ’127,2ξ‚ξ‚€πΌβˆ’19287,2+πΌβˆ’87,2𝐼+6447,2+πΌβˆ’47,2+462=0.(67) Equivalently, 𝐴2π΄βˆ’4ξ€Έξ€·4+32𝐴3βˆ’42𝐴2ξ€Έβˆ’128π΄βˆ’191=0,(68) where 𝐴=𝐼47,2+πΌβˆ’47,2.(69) By using the fact in Remark 17, it is seen that the first factor of (68) is nonzero, and so from the second factor, we deduce that √𝐴=βˆ’8βˆ’52+√163+1162.(70) Combining (69) and (70) and noting that 𝐼7,2<1, we obtain 𝐼7,2=2βˆ’1/4βŽ›βŽœβŽœβŽβˆšβˆ’8βˆ’52+√163+1162βˆ’ξƒŽξ‚΅βˆšβˆ’4+βˆ’8βˆ’52+√163+1162ξ‚Ά2⎞⎟⎟⎠1/4.(71) This completes the proof of (i).

To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐸2𝐸2πΈβˆ’4ξ€Έξ€·4+108𝐸2ξ€Έβˆ’1134=0,(72)

where 𝐸=𝐼27,4+πΌβˆ’27,4.(73)

Using the fact in Remark 17 it is seen that the first two factors of (72) are nonzero, and so solving the third factor, we obtain 𝐸=3βˆšβˆ’6+52.(74)

Combining (73) and (74) and noting that 𝐼7,4<1, we deduce that 𝐼7,4=ξ‚™3ξ”βˆšβˆ’6+52βˆ’βˆšβˆ’58+452√2.(75) So the proof of (ii) is complete.

Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 29. One has 𝐼7,𝑛𝐼7,25𝑛2+𝐼7,𝑛𝐼7,25π‘›ξ€Έβˆ’2βˆ’ξƒ―ξ‚΅πΌ7,25𝑛𝐼7,𝑛3+𝐼7,25𝑛𝐼7,π‘›ξ‚Άβˆ’3𝐼+57,25𝑛𝐼7,𝑛2+𝐼7,25𝑛𝐼7,π‘›ξ‚Άβˆ’2ξƒ°πΌβˆ’10ξ‚»ξ‚΅7,25𝑛𝐼7,𝑛+𝐼7,25𝑛𝐼7,𝑛+15=0.(76)

Proof. Using (5) in Lemma 9, we find that ξ„Άξ„΅ξ„΅βŽ·π‘„=ξ€·π‘žπ‘žπœ’5ξ€Έπœ’ξ€·π‘ž7ξ€Έξ€·π‘žπœ’(π‘ž)πœ’35ξ€Έξ„Άξ„΅ξ„΅βŽ·,𝑅=π‘ž3/2ξ€·π‘žπœ’(π‘ž)πœ’5ξ€Έπœ’ξ€·π‘ž7ξ€Έπœ’ξ€·π‘ž35ξ€Έ.(77) Setting π‘ž=π‘’βˆšβˆ’πœ‹π‘›/7 and using the definition of πΌπ‘˜,𝑛 in (77), we get 𝐼𝑄=7,25𝑛𝐼7,𝑛1/2𝐼,𝑅=7,𝑛𝐼7,25𝑛1/2.(78) Employing (78) in (27), we complete the proof.

Corollary 30. One has (i)𝐼7,5=ξ”βˆšβ„Žβˆ’βˆ’36+β„Ž2√6,(ii)𝐼7,1/5=ξ”βˆšβ„Ž+βˆ’36+β„Ž2√6,(iii)𝐼7,25=ξ‚€βˆšπ‘‘βˆ’βˆ’144+𝑑2,12(iv)𝐼7,1/25=ξ‚€βˆšπ‘‘+βˆ’144+𝑑2,12(79) where βˆšβ„Ž=5+(62βˆ’6105)1/3√+(62+6105)1/3 and 𝑑=12+22/3√(135βˆ’1521)1/3+22/3√(135+1521)1/3.

Proof. Setting 𝑛=1/5 in Theorem 29 and simplifying using Theorem 16(ii), we obtain 𝐼67,5+πΌβˆ’67,5ξ‚€πΌβˆ’547,5+πΌβˆ’47,5𝐼+1027,5+πΌβˆ’27,5ξ‚βˆ’17=0.(80) Equivalently, 𝐻3βˆ’5𝐻2+7π»βˆ’7=0,(81) where 𝐻=𝐼27,5+πΌβˆ’27,5.(82) Solving (81) and noting the fact in Remark 17, we obtain ξ‚΅ξ‚€βˆšπ»=5+62βˆ’61051/3+ξ‚€βˆš62+61051/3ξ‚Ά3(83) Combining (82) and (83) and noting that 𝐼7,5<1, we deduce that 𝐼7,5=ξ”βˆšβ„Žβˆ’βˆ’36+β„Ž2√6,(84) where ξ‚€βˆšβ„Ž=5+62βˆ’61051/3+ξ‚€βˆš62+61051/3.(85) This completes the proof of (i).
Again setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at π‘ˆ3βˆ’6π‘ˆ2+7π‘ˆβˆ’3=0,(86) where π‘ˆ=𝐼7,25+πΌβˆ’17,25.(87) Solving (86) and using Remark 17, we get ξ‚΅ξ‚€ξ‚€βˆšπ‘ˆ=6+159βˆ’ξ‚ξ‚21/21/3+ξ‚€ξ‚€βˆš159+21/21/3ξ‚Ά3.(88) Combining (87) and (88) and noting that 𝐼7,25<1, we obtain 𝐼7,25=ξ‚€βˆšπ‘‘βˆ’βˆ’144+𝑑2,12(89) where 𝑑=12+22/3ξ‚€βˆš135βˆ’15211/3+22/3ξ‚€βˆš135+15211/3.(90) This completes the proof of (ii). Now (ii) and (iv) easily follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 31. One has 𝐼13,𝑛𝐼13,9𝑛2+𝐼13,𝑛𝐼13,9π‘›ξ‚Άβˆ’2ξƒ―ξ‚΅πΌβˆ’313,𝑛𝐼13,9𝑛+𝐼13,𝑛𝐼13,9π‘›ξ‚Άβˆ’1ξƒ°βˆ’ξ‚†ξ€·πΌ13,𝑛𝐼13,9𝑛+𝐼13,𝑛𝐼13,9π‘›ξ€Έβˆ’1+3=0.(91)

Proof. Proceeding as in the proof of Theorem 29, using (5) in Lemma 10, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/13, and using the definition of πΌπ‘˜,𝑛, we arrive at 𝐼𝑄=13,9𝑛𝐼13,𝑛1/2𝐼,𝑇=13,𝑛𝐼13,9𝑛1/2.(92) Employing (92) in (28), we complete the proof.

Corollary 32. One has (i)𝐼13,3=ξ‚΅βˆš1+13βˆ’βˆšβˆ’2+2ξ‚Ά134,(ii)𝐼13,9=ξ‚΅βˆš2+3βˆ’βˆš3+43ξ‚Ά2,(iii)𝐼13,1/3=ξ‚΅βˆš1+13+βˆšβˆ’2+2ξ‚Ά134,(iv)𝐼13,1/9=ξ‚΅βˆš2+3+√3+43ξ‚Ά2.(93)

Proof. Setting 𝑛=1/3 in Theorem 31 and simplifying using Theorem 16(ii), we obtain 𝑉2βˆ’3π‘‰βˆ’1=0,(94) where 𝑉=𝐼213,3+πΌβˆ’213,3.(95) Solving (94) and using Remark 17, we get βˆšπ‘‰=3+132.(96) Combining (95) and (96) and noting that 𝐼13,3<1, we obtain 𝐼13,3=ξ‚΅βˆš1+13βˆ’βˆšβˆ’2+2ξ‚Ά134.(97) So we complete the proof of (i).
Again setting 𝑛=1 and using Theorem 16(i), we obtain 𝐽2βˆ’4𝐽+1=0,(98) where 𝐽=𝐼13,9+πΌβˆ’113,9.(99) Solving (98) and using Remark 17, we get √𝐽=2+3.(100) Combing (99) and (100) and noting that 𝐼13,9<1, we deduce that 𝐼13,9=ξ‚΅βˆš2+3βˆ’βˆš3+43ξ‚Ά2.(101) So the proofs of (ii) is complete. Now the proof of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 33. One has 𝐼13,25𝑛𝐼13,𝑛3𝐼13,25𝑛𝐼13,π‘›ξ‚Άβˆ’3ξƒ―ξ‚΅πΌβˆ’513,25𝑛𝐼13,𝑛+𝐼13,25𝑛𝐼13,π‘›ξ‚Άβˆ’1×𝐼+213,25𝑛𝐼13,𝑛+𝐼13,25𝑛𝐼13,π‘›ξ€Έβˆ’1ξ‚‡βˆ’ξ‚†ξ€·πΌ+213,25𝑛𝐼13,𝑛2+𝐼13,25𝑛𝐼13,π‘›ξ€Έβˆ’2=0.(102)

Proof. Using (5) in Lemma 11, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/13, and using the definition of πΌπ‘˜,𝑛, we arrive at 𝐼𝑄=13,25𝑛𝐼13,𝑛1/2𝐼,𝑇=13,𝑛𝐼13,25𝑛1/2.(103) Employing (103) in (29), we complete the proof.

Corollary 34. One has (i)𝐼13,5=ξƒ©ξ‚™βˆš3+65βˆ’βˆš58+6ξƒͺ652,(ii)𝐼13,25=ξ‚€βˆšπ‘βˆ’βˆ’36+𝑐26,(iii)𝐼13,1/5=ξƒ©ξ‚™βˆš3+65+√58+6ξƒͺ652,(iv)𝐼13,1/25=ξ‚€βˆšπ‘+βˆ’36+𝑐26,(104) where βˆšπ‘=6+(1080βˆ’1539)1/3√+(1080+1539)1/3.

Proof. Setting 𝑛=1/5 and simplifying using Theorem 16(ii), we arrive at 𝐿3βˆ’23πΏβˆ’42=0,(105) where 𝐿=𝐼213,5+πΌβˆ’213,5.(106) Solving (105) and using the fact in Remark 17, we obtain √𝐿=3+652.(107) Employing (107) in (106), solving the resulting equation, and noting that 𝐼13,5<1, we obtain 𝐼13,5=ξƒ©ξ‚™βˆš3+65βˆ’βˆš58+6ξƒͺ652.(108) This completes the proof of (i).
To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐴3βˆ’6𝐴2βˆ’23π΄βˆ’18=0,(109) where 𝐴=𝐼13,25+πΌβˆ’113,25.(110) Solving (109) and using the fact in Remark 17, we obtain ξ‚΅ξ‚€βˆšπ΄=6+1080βˆ’15391/3+ξ‚€βˆš1080+15391/3ξ‚Ά3.(111) Employing (111) and (110), solving the resulting equation, and noting that 𝐼13,25<1, we obtain 𝐼13,25=ξ‚€βˆšπ‘βˆ’βˆ’36+𝑐26,(112) where βˆšπ‘=6+(1080βˆ’1539)1/3√+(1080+1539)1/3.
This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 35. One has 𝐼7,𝑛𝐼7,9𝑛+𝐼7,𝑛𝐼7,9π‘›ξ€Έβˆ’1βˆ’ξƒ―ξ‚΅πΌ7,9𝑛𝐼7,𝑛2+𝐼7,9𝑛𝐼7,π‘›ξ‚Άβˆ’2ξƒ°+3=0.(113)

Proof. Using (5) in Lemma 12, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/7, and using the definition of πΌπ‘˜,𝑛, we arrive at 𝐼𝑄=7,9𝑛𝐼7,𝑛1/2𝐼,𝑅=7,𝑛𝐼7,9𝑛1/2.(114) Employing (114) in (30), we complete the proof.

Corollary 36. One has (i)𝐼7,3=ξƒ©βˆš5βˆ’212ξƒͺ1/4,(ii)𝐼7,9=ξ‚΅βˆš1+21βˆ’βˆš6+2ξ‚Ά214,(iii)𝐼7,1/3=ξƒ©βˆš5+212ξƒͺ1/4,(iv)𝐼7,1/9=ξ‚΅βˆš1+21+√6+2ξ‚Ά214.(115)

Proof. Setting 𝑛=1/3 and simplifying using Theorem 16(ii), we arrive at 𝐼47,3+πΌβˆ’47,3βˆ’5=0.(116) Solving (116) and noting the fact in Remark 17, we obtain 𝐼7,3=ξƒ©βˆš5βˆ’212ξƒͺ1/4.(117) This completes the proof of (i).
To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐷2βˆ’π·βˆ’5=0,(118) where 𝐷=𝐼7,9+πΌβˆ’17,9.(119) Solving (118) and using the fact in Remark 17, we obtain ξƒ©βˆšπ·=1+212ξƒͺ.(120) Employing (120) in (119), solving the resulting equation, and noting that 𝐼7,9<1, we deduce that 𝐼7,9=ξ‚΅βˆš1+21βˆ’βˆš6+2ξ‚Ά214.(121) This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

5. General Theorems and Explicit Evaluations of πΊπ‘›π‘˜πΊπ‘›/π‘˜

In this section we evaluate some explicit values of the product πΊπ‘›π‘˜πΊπ‘›/π‘˜ by establishing some general theorems and employing the values of πΌπ‘˜,𝑛 obtained in Section 4. We recall from Theorem 22(ii) that 𝐺1/𝑛=𝐺𝑛 for ready references in this section.

Theorem 37. One has 𝐼(i)3,𝑛𝐼3,4𝑛2+𝐼3,𝑛𝐼3,4π‘›ξ€Έβˆ’2+𝐺𝑛/3𝐺4𝑛/3𝐺9𝑛/3𝐺36𝑛/3ξ€Έβˆ’1ξ€·πΊβˆ’2𝑛/3𝐺4𝑛/3𝐺9𝑛/3𝐺36𝑛/3ξ€Έ=0,(ii)𝐼63,𝑛+πΌβˆ’63,π‘›βˆš+22𝐺𝑛/3𝐺3π‘›ξ€Έβˆ’3βˆ’ξ€·πΊπ‘›/3𝐺3𝑛3=0.(122)

Proof. To prove (i), using (5) in Lemma 13, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/3, and employing the definitions of πΌπ‘˜,𝑛 and 𝐺𝑛, we obtain ξ„Άξ„΅ξ„΅βŽ·π‘…=π‘ž1/4ξ€·π‘žπœ’(π‘ž)πœ’2ξ€Έπœ’ξ€·π‘ž3ξ€Έπœ’ξ€·π‘ž6ξ€Έ=𝐼3,𝑛𝐼3,4𝑛1/2,𝐺𝑃=𝑛/3𝐺4𝑛/3𝐺9𝑛/3𝐺36𝑛/3ξ€Έβˆ’1/2.(123) Employing (123) in (31), we complete the proof. (ii) follows similarly from Lemma 5 and the definition of πΌπ‘˜,𝑛 and 𝐺𝑛 with π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/3.

Corollary 38. One has (i)𝐺6𝐺3/2=ξƒŽβˆš1+32,(ii)𝐺12𝐺4/3=2βˆ’13/6ξ‚΅ξ”βˆš6+36+βˆšβˆ’2+36ξ‚Ά,(iii)𝐺39𝐺13/3=21/6ξ‚€βˆš3+131/3.(124)

Proof. Setting 𝑛=1/2 in Theorem 37(i) and simplifying using Theorem 16(ii) and the result 𝐺1/𝑛=𝐺𝑛, we obtain 2𝐺6𝐺3/2ξ€Έ2βˆ’ξ€·πΊ6𝐺3/2ξ€Έβˆ’2βˆ’2=0.(125) Solving (125) and noting that 𝐺6𝐺3/2>1, we complete the proof of (i).
To prove (ii), setting 𝑛=1 in Theorem 37(i); using Theorem 16(i), and noting that 𝐺1/𝑛=𝐺𝑛, we obtain 𝐼23,4+πΌβˆ’23,4+𝐺23𝐺4/3𝐺12ξ€Έβˆ’1ξ€·πΊβˆ’223𝐺4/3𝐺12ξ€Έ=0.(126) Employing (53) in (126), solving the resulting equation, and noting that 𝐺23𝐺4/3𝐺12>1, we obtain 𝐺23𝐺4/3𝐺12=ξ‚΅ξ”βˆš6+36+βˆšβˆ’2+36ξ‚Ά4.(127) Using the value 𝐺3=21/12 from [5, p. 189] in (127), we complete the proof of (ii).
To prove (iii), setting 𝑛=13 in Theorem 37(ii), we obtain 𝐼63,13+πΌβˆ’63,13√+22𝐺3/13𝐺39ξ€Έβˆ’2βˆ’ξ€·πΊ3/13𝐺39ξ€Έ2=0.(128) Cubing (96) and then employing in (128) and solving the resulting equation, we complete the proof.

Theorem 39. One has 𝐼(i)5,𝑛𝐼5,4𝑛3+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’3𝐼+55,𝑛𝐼5,4𝑛+𝐼5,𝑛𝐼5,4π‘›ξ€Έβˆ’1𝐺=4𝑛/5𝐺4𝑛/5𝐺25𝑛/5𝐺100𝑛/5ξ€Έ2βˆ’ξ€·πΊπ‘›/5𝐺4𝑛/5𝐺25𝑛/5𝐺100𝑛/5ξ€Έβˆ’2,(ii)𝐼35,𝑛+πΌβˆ’35,𝑛𝐺+2𝑛/5𝐺5π‘›ξ€Έβˆ’2βˆ’ξ€·πΊπ‘›/5𝐺5𝑛2=0.(129)

Proof. Using (5) in Lemma 14, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/5, and employing the definitions of πΌπ‘˜,𝑛 and 𝐺𝑛, we obtain ξ„Άξ„΅ξ„΅βŽ·π‘…=π‘ž1/2ξ€·π‘žπœ’(π‘ž)πœ’2ξ€Έπœ’ξ€·π‘ž5ξ€Έπœ’ξ€·π‘ž10ξ€Έ=𝐼5,𝑛𝐼5,4𝑛1/2,𝐺𝑃=𝑛/5𝐺4𝑛/5𝐺25𝑛/5𝐺100𝑛/5ξ€Έβˆ’1/2.(130) Employing (130) in (32), we complete the proof of (i). Similarly, (ii) follows from Lemma 6 and the definition of πΌπ‘˜,𝑛 and 𝐺𝑛 with π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/5.

Corollary 40. One has (i)𝐺10𝐺5/2=ξƒ©βˆš3+102ξƒͺ1/4,(ii)𝐺20𝐺5/4=12ξ‚™ξ‚€βˆšβˆ’311+551/4+ξ‚€βˆš11+553/4+𝔐,(131) where 𝔐 denotes 16+√11+55(βˆ’3+√11+55)2.

Proof. Setting 𝑛=1/2 in Theorem 39(i) and simplifying using Theorem 16(ii) and the result 𝐺1/𝑛=𝐺𝑛, we obtain 4𝐺10𝐺5/2ξ€Έ4βˆ’ξ€·πΊ10𝐺5/2ξ€Έβˆ’4βˆ’12=0.(132) Solving (132) and noting that 𝐺10𝐺5/2>1, we complete the proof of (i).
For proof of (ii), setting 𝑛=1 in Theorem 39(ii) and simplifying using Theorem 18(i) and the result 𝐺1/𝑛=𝐺𝑛, we obtain 𝐼35,4+πΌβˆ’35,4𝐺+24/5𝐺20ξ€Έβˆ’2βˆ’ξ€·πΊ4/5𝐺20ξ€Έ2=0.(133) Employing the value of 𝐼5,4+πΌβˆ’15,4 from (63) in (133), solving the resulting equation, and noting that 𝐺4/5𝐺20>1, we get 𝐺4/5𝐺20=12ξ‚™ξ‚€βˆšβˆ’311+551/4+ξ‚€βˆš11+553/4+𝔐.(134) So the proof is complete.

Theorem 41. One has 2√2𝐼2,𝑛𝐼2,49𝑛3/2+𝐼2,𝑛𝐼2,49π‘›ξ€Έβˆ’3/2𝐺=4𝑛/2𝐺4𝑛/2𝐺47𝑛/2𝐺196𝑛/2ξ€Έ1/2ξ€·πΊβˆ’2𝑛/2𝐺4𝑛/2𝐺47𝑛/2𝐺196𝑛/2ξ€Έβˆ’1/2𝐺+4𝑛/2𝐺4𝑛/2𝐺47𝑛/2𝐺196𝑛/2ξ€Έβˆ’3/2βˆ’ξ€·πΊπ‘›/2𝐺4𝑛/2𝐺47𝑛/2𝐺196𝑛/2ξ€Έβˆ’5/2.(135)

Proof. Using (5) in Lemma 15, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/2, and employing the definitions of πΌπ‘˜,𝑛 and 𝐺𝑛, we obtain, ξ„Άξ„΅ξ„΅βŽ·π‘…=π‘ž1/3ξ€·π‘žπœ’(π‘ž)πœ’7ξ€Έπœ’ξ€·π‘ž2ξ€Έπœ’ξ€·π‘ž14ξ€Έ=𝐼2,𝑛𝐼2,49𝑛1/2,𝐺𝑃=𝑛/2𝐺4𝑛/2𝐺47𝑛/2𝐺196𝑛/2ξ€Έβˆ’1/2.(136) Employing (136) in (33), we complete the proof.

Corollary 42. One has 𝐺14𝐺7/2=2+√4+82ξ‚Ά4.(137)

Proof. Setting 𝑛=1/7 and simplifying using Theorem 16(ii) and the result 𝐺1/𝑛=𝐺𝑛, we get 4𝐺14𝐺7/2ξ€Έ6βˆšβˆ’42𝐺14𝐺7/2ξ€Έ5ξ€·πΊβˆ’214𝐺7/2ξ€Έ4𝐺+414𝐺7/2ξ€Έ2βˆ’1=0.(138) Solving (138) and noting that 𝐺14𝐺7/2>1, we complete the proof.

Theorem 43. One has 𝐼47,𝑛+πΌβˆ’47,π‘›βˆš+7=22𝐺7/𝑛𝐺7π‘›ξ€Έβˆ’3+𝐺7/𝑛𝐺7𝑛3.(139)

Proof. Using (5) in Lemma 7, setting π‘žβˆΆ=π‘’βˆšβˆ’πœ‹π‘›/7, and employing definitions of πΌπ‘˜,𝑛 and 𝐺𝑛, we arrive at 𝐺𝑃=𝑛/7𝐺7π‘›ξ€Έβˆ’3,ξƒ©π‘žπ‘„=1/4πœ’(π‘ž)πœ’ξ€·π‘ž7ξ€Έξƒͺ4=𝐼47,𝑛.(140) Using (140) in (24), we complete the proof.

Corollary 44. One has (i)𝐺7/4𝐺28=ξ‚€βˆš2+22,(ii)𝐺7/3𝐺21=2βˆ’1/6ξ‚€βˆš3+71/3,(iii)𝐺7/9𝐺63=21/6ξ‚€βˆš5+211/3.(141)

Proof. Setting 𝑛=4 in Theorem 43, we get 𝐼47,4+πΌβˆ’47,4√+7=22𝐺7/4𝐺28ξ€Έ3+𝐺7/4𝐺28ξ€Έβˆ’3.(142) Squaring (74) and simplifying, we obtain 𝐼47,4+πΌβˆ’47,4√=βˆ’56+452.(143) Employing (143) in (142), solving the resulting equation, and noting that 𝐺7/4𝐺28>1, we complete the proof of (i).
To prove (ii), we set 𝑛=3 in Theorem 43, and employing (116), we obatin 𝐺21𝐺7/3ξ€Έ3+𝐺21𝐺7/3ξ€Έβˆ’3βˆšβˆ’32=0.(144) Solving (144) and noting that 𝐺21𝐺7/3>1, we complete the proof.
To prove (iii), setting 𝑛=9 in Theorem 43, we get 𝐼47,9+πΌβˆ’47,9ξ‚βˆš+7=22𝐺7/9𝐺63ξ€Έ3+𝐺7/9𝐺63ξ€Έβˆ’3.(145) Squaring (120) twice and simplifying, we obtain