Some New Explicit Values of Quotients of Ramanujan’s Theta Functions and Continued Fractions

Abstract

We evaluate some new explicit values of quotients of Ramanujan’s theta functions and use them to find explicit values of Ramanujan’s continued fractions.

1. Introduction

Ramanujan’s general theta function 𝑓(𝑎,𝑏) is defined by 𝑓(𝑎,𝑏)∶=^∞_𝑘=−∞𝑎^{𝑘(𝑘+1)/2}𝑏^{𝑘(𝑘−1)/2},|𝑎𝑏|<1. (1) Two important special cases of 𝑓(𝑎,𝑏) are the theta functions 𝜙(𝑞) and 𝜓(𝑞) [1, page 36, Entry 22] defined by, for |𝑞|<1, 𝜙𝑞∶=𝑓𝑞,𝑞=^∞_𝑛=−∞𝑞^𝑛2=−𝑞;−𝑞_∞𝑞;−𝑞_∞,𝜓𝑞∶=𝑓𝑞,𝑞³=^∞_𝑛=0𝑞^{𝑛(𝑛+1)/2}=𝑞²;𝑞²_∞𝑞;𝑞²_∞, (2) where (𝑎;𝑞)_∞=∏^∞_𝑛=0(1−𝑎𝑞^𝑛).

In his notebooks [2], Ramanujan recorded many explicit values of theta functions 𝜙(𝑞) and 𝜓(𝑞). All these values were proved by Berndt [3, page 325] and Berndt and Chan [4]. Yi [5] introduced the parameter ℎ_𝑘,𝑛 for positive real numbers 𝑘 and 𝑛 defined by ℎ_𝑘,𝑛=𝜙𝑞𝑘^1/4𝜙𝑞^𝑘;𝑞=𝑒^{−𝜋𝑛/𝑘} (3) and used the particular case ℎ_𝑛,𝑛 to find explicit values 𝜙(𝑞). Baruah and Saikia [6] defined the parameter 𝑔_𝑘,𝑛 for positive real numbers 𝑘 and 𝑛 as 𝑔_𝑘,𝑛=𝜓−𝑞𝑘^1/4𝑞^(𝑘−1)/8𝜓−𝑞^𝑘;𝑞=𝑒^{−𝜋𝑛/𝑘} (4) and used the particular case 𝑔_𝑛,𝑛 to find explicit values 𝜓(𝑞). Saikia [7] also established some explicit values 𝜓(𝑞).

In this paper, we consider the particular cases ℎ_3,𝑛 and 𝑔_3,𝑛 of the parameters ℎ_𝑘,𝑛 and 𝑔_𝑘,𝑛, respectively. By using theta function identities, we find some new explicit values of the parameters ℎ_3,𝑛 and 𝑔_3,𝑛. Particularly, we evaluate ℎ_3,𝑛 and 𝑔_3,𝑛 for 𝑛= 3/2, 2/3, 6, 1/6, 5/2, 2/5, 10, and 1/10. Previously, Yi [5] evaluated ℎ_3,𝑛 for 𝑛= 1, 3, 1/3, 9, 1/9, 5, 1/5, 25, and 1/25. Saikia [8] evaluated ℎ_3,𝑛 for 𝑛= 2, 1/2, 4, 1/4, 7, 1/7, 49, and 1/49. Baruah and Saikia [6] evaluated 𝑔_3,𝑛 for 𝑛= 1, 3, 1/3, 9, 1/9, 5, 1/5, 25, 1/25, 7, 1/7, 13, 1/13, 49, and 1/49. As an application to our new values, we evaluate some old and new explicit values of Ramanujan’s cubic continued fraction 𝐺(𝑞) and a continued fraction of order twelve 𝐻(𝑞) which are, respectively, defined by, for |𝑞|<1, 𝐺𝑞𝑞∶=^1/31₊𝑞+𝑞²1₊𝑞²+𝑞⁴1₊𝑞³+𝑞⁶1_+⋯,𝐻𝑞∶=𝑞𝑓−𝑞,−𝑞¹¹𝑓−𝑞⁵,−𝑞⁷=𝑞(1−𝑞)1−𝑞³₊𝑞³(1−𝑞²)(1−𝑞⁴)(1−𝑞³)(1+𝑞⁶)₊𝑞³(1−𝑞⁸)(1−𝑞¹⁰)(1−𝑞³)(1+𝑞¹²)_+⋯. (5) The continued fraction 𝐺(𝑞) was recorded by Ramanujan on page 366 of lost notebook [9]. We refer to [10–14] for explicit evaluations of 𝐺(𝑞). The continued fraction 𝐻(𝑞) was introduced by Naika et al. [15]. We refer to [8, 15] for explicit evaluations of 𝐻(𝑞).

The presentation of the paper is as follows. In Section 2 we record some preliminary results for ready references in this paper. Section 3 is devoted to explicit evaluations of the parameters ℎ_3,𝑛 and 𝑔_3,𝑛. In Section 4 we use new explicit values of ℎ_3,𝑛 and 𝑔_3,𝑛 to evaluate some explicit values of the continued fractions 𝐺(𝑞) and 𝐻(𝑞).

We end this introduction by noting the following remarks regarding 𝑔_3,𝑛 and ℎ_3,𝑛 from [6, page 1764, Theorem 4.1] and [5, page 385, Remark 2.3], respectively.

Remark 1. The parameter 𝑔_3,𝑛 has positive real value with 𝑔_3,1=1 and 𝑔_3,𝑛 increases as 𝑛 increases.

Remark 2. The parameter ℎ_3,𝑛 has positive real value with ℎ_3,1=1 and the values of ℎ_3,𝑛 decrease as 𝑛 increases.

2. Preliminary Results

Lemma 3 (see [16, Theorem 3.2]). If 𝑃∶=√𝑞𝜓(−𝑞)𝜓(−𝑞⁹)/𝜓²(−𝑞³) and 𝑄∶=𝑞𝜓(−𝑞²)𝜓(−𝑞¹⁸)/𝜓²(−𝑞⁶), then 𝑃²𝑄²+𝑄²𝑃²+𝑃²𝑄²+𝑄²𝑃²²=13𝑃𝑄−1𝑃𝑄3𝑃𝑄−+𝑃𝑄3𝑃𝑄+3𝑄𝑃+𝑃³𝑄³+𝑄³𝑃³+14. (6)

Lemma 4 (see [16, Theorem 3.1]). If 𝐿∶=𝜙(𝑞)𝜙(𝑞⁹)/𝜙²(𝑞) and 𝑀∶=√𝑞𝜓(−𝑞)𝜓(−𝑞⁹)/𝜓²(−𝑞³), then 𝐿²=𝑀²+11−3𝑀². (7)

Lemma 5 (see [16, Theorem 3.2]). If 𝑃∶=𝜓(−𝑞³)𝜓(−𝑞⁵)/𝑞𝜓(−𝑞)𝜓(−𝑞¹⁵) and 𝑄∶=𝜓(−𝑞⁶)𝜓(−𝑞¹⁰)/𝑞²𝜓(−𝑞²)𝜓(−𝑞³⁰), then 𝑃²𝑄²+𝑄²𝑃²+𝑃𝑄+𝑄𝑃−1𝑃𝑄++𝑃𝑄1𝑃𝑄−𝑃𝑄𝑃³𝑄³+𝑄³𝑃³+𝑃𝑄+𝑄𝑃=0. (8)

Lemma 6 (see [16, Theorem 3.9]). If 𝐿∶=𝜙(𝑞³)𝜙(𝑞⁵)/𝜙(𝑞)𝜙(𝑞¹⁵) and 𝑀∶=𝜓(−𝑞³)𝜓(−𝑞⁵)/𝑞𝜓(−𝑞)𝜓(−𝑞¹⁵), then 𝑀=1+𝐿.1−𝐿 (9)

Lemma 7 (see [5, page 385, Theorem 2.2]). For all positive real numbers 𝑘 and 𝑛, one has ℎ_𝑘,1/𝑛=1ℎ_𝑘,𝑛. (10)

Lemma 8 (see [6, page 1764, Theorem 4.1]). For all positive real numbers 𝑘 and 𝑛, one has 𝑔_𝑘,1/𝑛=1𝑔_𝑘,𝑛. (11)

Lemma 9 (see [5, page 393, Theorem 4.9(i)]). For any positive real number 𝑛, one has √3ℎ_3,𝑛ℎ_3,9𝑛+1ℎ_3,𝑛ℎ_9,𝑛=ℎ_3,9𝑛ℎ_3,𝑛²+3. (12)

Lemma 10 (see [5, page 394, Theorem 4.12(i)]). For any positive real number 𝑛, one has 3ℎ_3,𝑛ℎ_3,25𝑛²+1ℎ_3,𝑛ℎ_3,25𝑛²=ℎ_3,25𝑛ℎ_3,𝑛³−ℎ_3,𝑛ℎ_3,25𝑛³ℎ+5^3,25𝑛ℎ_3,𝑛²+ℎ_3,𝑛ℎ_3,25𝑛²ℎ+5^3,25𝑛ℎ_3,𝑛−ℎ_3,𝑛ℎ_3,25𝑛. (13)

Lemma 11 (see [6, page 1769, Theorem 5.1(i) and (iii)]). For any positive real number 𝑛, one has(i)(1+√3𝑔_3,𝑛𝑔_3,9𝑛)³=1+3𝑔⁴_3,9𝑛, (ii)3{(𝑔_3,𝑛𝑔_3,25𝑛)²+1/(𝑔_3,𝑛𝑔_3,25𝑛)²} + 5{(𝑔_3,25𝑛/𝑔_3,𝑛)²+(𝑔_3,𝑛/𝑔_3,25𝑛)²} = {(𝑔_3,25𝑛/𝑔_3,𝑛)³−(𝑔_3,𝑛/𝑔_3,25𝑛)³} + 5{(𝑔_3,25𝑛/𝑔_3,𝑛)−(𝑔_3,𝑛/𝑔_3,25𝑛)}.

3. Explicit Evaluations of the Parameters 𝑔_𝑘,𝑛 and ℎ_𝑘,𝑛

In this section, we prove some general theorems for the explicit evaluations of the parameters 𝑔_𝑘,𝑛 and ℎ_𝑘,𝑛 and evaluate some new explicit values therefrom.

Theorem 12. One has 𝑔_3,𝑛𝑔_3,36𝑛𝑔_3,4𝑛𝑔_3,9𝑛²+𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,𝑛𝑔_3,36𝑛²+𝑔_3,𝑛𝑔_3,36𝑛𝑔_3,4𝑛𝑔_3,9𝑛²+𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,𝑛𝑔_3,36𝑛²²=3𝑔−14_3,𝑛𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,36𝑛−𝑔_3,9𝑛𝑔_3,36𝑛𝑔_3,𝑛𝑔_3,4𝑛×3𝑔_3,𝑛𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,36𝑛−𝑔_3,9𝑛𝑔_3,36𝑛𝑔_3,𝑛𝑔_3,4𝑛𝑔+3^3,𝑛𝑔_3,36𝑛𝑔_3,4𝑛𝑔_3,9𝑛+3𝑔_3,𝑛𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,36𝑛−𝑔_3,9𝑛𝑔_3,36𝑛𝑔_3,𝑛𝑔_3,4𝑛×3𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,𝑛𝑔_3,36𝑛+𝑔_3,𝑛𝑔_3,36𝑛𝑔_3,4𝑛𝑔_3,9𝑛³+𝑔_3,4𝑛𝑔_3,9𝑛𝑔_3,𝑛𝑔_3,36𝑛³. (14)

Proof. We set 𝑞∶=𝑒^{−𝜋𝑛/3} in Lemma 3 and use the definition of 𝑔_3,𝑛.

Corollary 13. One has 𝑔_3,3/2=32+2√2−6+6√2√2,𝑔_3,6=61+√232+2√2−6+6√2,𝑔_3,2/3=√232+2√2−6+6√2,𝑔_3,6=32+2√2−6+6√261+√2. (15)

Proof. Setting 𝑛=1/6 in Theorem 12 and simplifying using Lemma 8, we obtain 3𝑔_3,3/2𝑔_3,6²−𝑔_3,3/2𝑔_3,6²×3𝑔_3,3/2𝑔_3,6²−𝑔_3,3/2𝑔_3,6²+8+8=0. (16) Solving (16) for 𝑔_3,3/2𝑔_3,6 and noting the fact in Remark 1, we find that 𝑔_3,3/2𝑔_3,6=31+√2. (17) Again, setting 𝑛=1/6 in Lemma 11(i) and simplifying using Lemma 8, we deduce that 1+√3𝑔_3,3/2𝑔_3,6³=1+3𝑔⁴_3,3/2. (18) Eliminating 𝑔_3,6 from (18) using (17) and simplifying, we obtain 𝑔⁴_3,3/2+31+√2−1+√2^3/2𝑔²_3,3/2+3+3√2=0. (19) Solving (19) and noting the fact in Remark 1, we obtain 𝑔_3,3/2=32+2√2−6+6√2√2. (20) Employing (20) in (17) and simplifying, we obtain 𝑔_3,6=61+√232+2√2−6+6√2. (21) Now the values of 𝑔_3,2/3 and 𝑔_3,1/6 follow from the values of 𝑔_3,3/2 and 𝑔_3,6, respectively, and Lemma 8.

Theorem 14. One has ℎ_3,𝑛ℎ_3,9𝑛²𝑔1−3^3,𝑛𝑔_3,9𝑛²−𝑔_3,𝑛𝑔_3,9𝑛²−1=0. (22)

Proof. We set 𝑞∶=𝑒^{−𝜋𝑛/3} in Lemma 4 and use the definitions of ℎ_3,𝑛 and 𝑔_3,𝑛.

Corollary 15. One has ℎ_3,3/2=2^−5/83√2−4^3/46+6√2+√6^1/2,ℎ_3,6=2^1/83^1/23√2+4^1/46+6√2+√6^−1/2,ℎ_3,2/3=2^5/83√2−4^−3/46+6√2+√6^−1/2,ℎ_3,1/6=2^−1/83^−1/23√2+4^−1/46+6√2+√6^1/2. (23)

Proof. Setting 𝑛=1/6 in Theorem 14 and simplifying using Lemma 7, we obtain 1ℎ_3,6ℎ_3,3/2²31−𝑔_3,6𝑔_3,3/2²−1𝑔_3,6𝑔_3,3/2²−1=0. (24) Employing (17) in (24), solving the resulting equation, and noting the fact in Remark 2, we deduce that ℎ_3,3/2ℎ_3,6=2^1/434+3√2. (25) Again setting 𝑛=1/6 in Lemma 9 and simplifying using Lemma 7, we obtain √3ℎ_3,3/2ℎ_3,6+ℎ_3,6ℎ_3,3/2=ℎ_3,3/2ℎ_3,6²+3. (26) Employing (25) in (26), solving the resulting equation, and noting the fact in Remark 2, we deduce that ℎ_3,3/2ℎ_3,6=6+6√2+√6√34+3√2. (27) Multiplying (25) and (27), we evaluate the value of ℎ_3,3/2, and dividing (25) by (27) and simplifying, we evaluate the value of ℎ_3,6. Now the values of ℎ_3,2/3 and ℎ_3,1/6 easily follow from the values of ℎ_3,3/2 and ℎ_3,6, respectively, and Lemma 7.

Theorem 16. One has 𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,𝑛𝑔_3,100𝑛²+𝑔_3,𝑛𝑔_3,100𝑛𝑔_3,4𝑛𝑔_3,25𝑛²+𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,𝑛𝑔_3,100𝑛+𝑔_3,𝑛𝑔_3,100𝑛𝑔_3,4𝑛𝑔_3,25𝑛−𝑔_3,25𝑛𝑔_3,100𝑛𝑔_3,𝑛𝑔_3,4𝑛−𝑔_3,𝑛𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,100𝑛+𝑔_3,25𝑛𝑔_3,100𝑛𝑔_3,𝑛𝑔_3,4𝑛^1/2−𝑔_3,𝑛𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,100𝑛^1/2×𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,𝑛𝑔_3,100𝑛^3/2+𝑔_3,𝑛𝑔_3,100𝑛𝑔_3,4𝑛𝑔_3,25𝑛^3/2+𝑔_3,4𝑛𝑔_3,25𝑛𝑔_3,𝑛𝑔_3,100𝑛^1/2+𝑔_3,𝑛𝑔_3,100𝑛𝑔_3,4𝑛𝑔_3,25𝑛^1/2=0. (28)

Proof. We set 𝑞∶=𝑒^{−𝜋𝑛/3} in Lemma 5 and use the definition of 𝑔_3,𝑛.

Corollary 17. One has 𝑔_3,5/2=1+√3+2+√3^1/2𝑎+√5𝑏^1/4𝑐^−1/4,𝑔_3,10=1+√3+2+√3^1/2𝑎+√5𝑏^−1/4𝑐^1/4,𝑔_3,2/5=1+√3+2+√3^−1/2𝑎+√5𝑏^−1/4𝑐^1/4,𝑔_3,1/10=1+√3+2+√3^−1/2𝑎+√5𝑏^1/4𝑐^−1/4, (29) where 𝑎=63+35√3+302+√3+206+3√√3,𝑏=2574+14853+13302+√3+7706+3√√3,𝑐=25+153+142+√3+76+3√3. (30)

Proof. Setting 𝑛=1/10 in Theorem 16 and simplifying using Lemma 8, we obtain 𝑔_3,5/2𝑔_3,10²+1𝑔_3,5/2𝑔_3,10²−4𝑔_3,5/2𝑔_3,10−1𝑔_3,5/2𝑔_3,10−4=0. (31) Solving (31) and noting the fact in Remark 2, we find that 𝑔_3,5/2𝑔_3,10=1+√3+2+√3. (32) Again, setting 𝑛=1/10 in Lemma 11 (ii) and simplifying using Lemma 8, we obtain 3𝑔_3,5/2𝑔_3,10²+𝑔_3,5/2𝑔_3,10+5𝑔_3,5/2𝑔_3,10²+1𝑔_3,5/2𝑔_3,10²−𝑔_3,5/2𝑔_3,10³−1𝑔_3,5/2𝑔_3,10³−5𝑔_3,5/2𝑔_3,10−1𝑔_3,5/2𝑔_3,10=0. (33) Employing (32) in (33), solving the resulting equation, and noting the fact in Remark 1, we obtain 𝑔_3,10𝑔_3,5/2=𝑎+√5𝑏𝑐^1/2, (34) where 𝑎=63+35√3+302+√3+206+3√3, 𝑏=2574+1485√3+13302+√3+7706+3√3, and 𝑐=25+15√3+142+√3+76+3√3.
Combining (32) and (34), we calculate the values of 𝑔_3,5/2 and 𝑔_3,10. Then, the values of 𝑔_3,2/5 and 𝑔_3,1/10 follow from the values of 𝑔_3,5/2 and 𝑔_3,10, respectively, and Lemma 8.

Theorem 18. One has 𝑔_3,25𝑛𝑔_3,𝑛ℎ1−^3,25𝑛ℎ_3,𝑛ℎ−1−^3,25𝑛ℎ_3,𝑛=0. (35)

Proof. We set 𝑞∶=𝑒^{−𝜋𝑛/3} in Lemma 6 and use the definitions of ℎ_3,𝑛 and 𝑔_3,𝑛.

Corollary 19. One has ℎ_3,5/2=√3+2+√3^1/22+√3+2+√3^−1/2×𝑓−4√5𝑏^1/4𝑑^−1/4,ℎ_3,10=√3+2+√3^1/22+√3+2+√3^−1/2×𝑓−4√5𝑏^−1/4𝑑^1/4,ℎ_3,2/5=√3+2+√3^−1/22+√3+2+√3^1/2×𝑓−4√5𝑏^−1/4𝑑^1/4,ℎ_3,1/10=√3+2+√3^−1/22+√3+2+√3^1/2×𝑓−4√5𝑏^1/4𝑑^−1/4, (36) where 𝑏 is given in Corollary 17, 𝑓=1143+655√3+5822+√3+3466+3√3, and 𝑑=1117+645√3 + 5782+√3+3346+3√3.

Proof. Setting 𝑛=1/10 in Theorem 18 and simplifying using Lemma 7, we get 𝑔_3,5/2𝑔_3,101−ℎ_3,5/2ℎ_3,10−1−ℎ_3,5/2ℎ_3,10=0. (37) Employing (32) and solving the resulting equation, we obtain ℎ_3,5/2ℎ_3,10=√3+2+√32+√3+2+√3. (38) Again setting 𝑛=1/10 in Lemma 10 and simplifying using Lemma 7, we find that 3ℎ_3,5/2ℎ_3,10²+ℎ_3,10ℎ_3,5/2²−ℎ_3,5/2ℎ_3,10³−1ℎ_3,5/2ℎ_3,10³−5ℎ_3,5/2ℎ_3,10²+1ℎ_3,5/2ℎ_3,10²−5ℎ_3,5/2ℎ_3,10−1ℎ_3,5/2ℎ_3,10=0. (39) Employing (38) in (39), solving the resulting equation, and noting the fact in Remark 2, we obtain ℎ_3,5/2ℎ_3,10=𝑓−4√5𝑏𝑑, (40) where 𝑏 is given in Corollary 17, 𝑓=1143+655√3+5822+√3+3466+3√3, and 𝑑=1117+645√3 + 5782+√3 + 3346+3√3.
Combining (38) and (40), we evaluate the values of ℎ_3,5/2 and ℎ_3,10. Then the values of ℎ_3,2/5 and ℎ_3,1/10 follow from ℎ_3,5/2 and ℎ_3,10, respectively, and Lemma 7.

4. Explicit Evaluations of Continued Fractions

This section is devoted to finding some explicit values of the continued fractions 𝐺(𝑞) and 𝐻(𝑞) by using new values of 𝑔_3,𝑛 and ℎ_3,𝑛 evaluated in Section 3.

Lemma 20 (see [6, page 1788, Theorem 9.1(i)]). One has 𝐺³𝑒^{−𝜋𝑛/3}=−11+3𝑔⁴_3,𝑛. (41)

Theorem 21. One has 𝐺³−𝑒^√−𝜋/2=2936+42√2−38+36√2,𝐺³−𝑒^√−𝜋2=32+2√2−6+6√2²62+2√26+6√2−58+40√2,𝐺³−𝑒^{√−𝜋2/3}=32+2√2−6+6√2²62+2√26+6√2−6+4√2,𝐺³−𝑒^{√−𝜋/32}=23+2√236+24√2−10+8√2,𝐺³−𝑒^−𝜋5/6=−𝑐𝑐+3𝑎+√5𝑏1+√3+2+√3²,𝐺³−𝑒^−𝜋10/3=−𝑎+√5𝑏𝑎+√5𝑏+3𝑐1+√3+2+√3²,𝐺³−𝑒^−𝜋2/15=−𝑎+√5𝑏1+√3+2+√3²𝑎+√5𝑏1+√3+2+√3²,𝐺+3𝑐₃−𝑒^{√−𝜋/30}=−𝑐1+√3+2+√3²𝑐1+√3+2+√3²+3𝑎+√,5𝑏 (42) where 𝑎, 𝑏, and 𝑐 are given in Corollary 17.

Proof. We set 𝑛= 3/2, 6, 2/3, 1/6, 5/2, 10, 2/5, and 1/10 in Lemma 20 and use the corresponding values of 𝑔_3,𝑛 from Corollaries 13 and 17 to complete the proof.

The explicit values of 𝐺³(−𝑒^{√−𝜋/32}), 𝐺³(−𝑒^−𝜋5/6), 𝐺³(−𝑒^−𝜋10/3), 𝐺³(−𝑒^−𝜋2/15), and 𝐺³(−𝑒^{√−𝜋/30}) are new.

Lemma 22 (see [8, page 144, Theorem 5.1]). One has 𝐻𝑒^{−𝜋𝑛/3}=3^1/4ℎ_3,𝑛−13^1/4ℎ_3,𝑛.+1 (43)

By setting 𝑛= 3/2, 6, 2/3, 1/6, 5/2, 10, 2/5, and 1/10 in Lemma 22 and employing the corresponding values of ℎ_3,𝑛 from Corollaries 15 and 19, we calculate the following new explicit values of the continued fraction 𝐻(𝑞).

Theorem 23. One has 𝐻𝑒^√−𝜋/2=−2+2^3/83^1/43√2−4^3/41+6√2+√62+2^3/83^1/43√2−4^3/41+6√2+√6,𝐻𝑒^√−𝜋2=2^1/83^3/44+3√2^1/4−1+6√2+√62^1/83^3/44+3√2^1/4+1+6√2+√6,𝐻𝑒^{√−𝜋2/3}=2^5/83^1/4−3√2−4^3/41+6√2+√62^5/83^1/4+3√2−4^3/41+6√2+√6,𝐻𝑒^{√−𝜋/18}=2^7/81+6√2+√6−212+9√2^1/42^7/81+6√2+√6+212+9√2^1/4,𝐻𝑒^−𝜋5/6=3^1/43+2+√3^1/2𝑓−4√5𝑏^1/4−2+√3+2+√3^1/2𝑑^1/4×3^1/43+2+√3^1/2𝑓−4√5𝑏^1/4+2+√3+2+√3^1/2𝑑^1/4−1,𝐻𝑒^−𝜋10/3=3^1/43+2+√3^1/2𝑑^1/4−2+√3+2+√3^1/2𝑓−4√5𝑏^1/4×3^1/43+2+√3^1/2𝑑^1/4+2+√3+2+√3^1/2𝑓−4√5𝑏^1/4−1,𝐻𝑒^−𝜋2/15=3^1/42+√3+2+√3^1/2𝑑^1/4−3+2+√3^1/2𝑓−4√5𝑏^1/4×3^1/42+√3+2+√3^1/2𝑑^1/4+3+2+√3^1/2𝑓−4√5𝑏^1/4−1,𝐻𝑒^{√−𝜋/30}=3^1/42+√3+2+√3^1/2𝑓−4√5𝑏^1/4−3+2+√3^1/2𝑑^1/4×3^1/42+√3+2+√3^1/2𝑓−4√5𝑏^1/4+3+2+√3^1/2𝑑^1/4−1, (44) where 𝑏 is given in Corollary 17 and 𝑑 and 𝑓 are given in Corollary 19.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author is thankful to the University Grants Commission, New Delhi, India, for partially supporting the research work under Grant F. no. 41-1394/2012(SR).