Monotonicity properties and inequalities related to generalized Grötzsch ring functions

Theorem 1

For r ∈ (0, 1), a ∈ (0, 12 ], b = 1 – a, and C = R(a)2 , define

where artanh denotes the inverse of the hyperbolic tangent function. Then the function F(r) is strictly increasing from (0, 1) onto (3(a2+b2)4,C). In particular, the double inequality

holds for A1=1,A2=3(a2+b2)4C,an=24n2−1, and r ∈ (0, 1).

Theorem 2

For B1=R(a)−ln⁡162,B2=B1ln⁡4,B3=3(1−2a)28,B4=eB1,B5=eB2,B6=B3B1,a∈(0,12), and r ∈ (0, 1), the following conclusions hold true:

1. The function

   G1(r)=μa(r)−μ(r)s2ln⁡(4/s) (1.10)

   is strictly increasing from (0, 1) onto (B₂, ∞). In particular, for r ∈ (0, 1),

   B2s2ln⁡4s<μa(r)−μ(r)<B1. (1.11)

2. The function

   G2(r)=B1−[μa(r)−μ(r)]1−(s2 artanh r)/r

   is strictly increasing from (0, 1) onto (B₃, B₁). In particular, for r ∈ (0, 1),

   B1s2 artanh rr<μa(r)−μ(r)<B1[1−B6(1−s2artanhrr)]. (1.12)

3. Let r0=s,rn=2rn−11+rn−1=φ2n(s) for n∈N,A(r)=s2 artanh rr,B(r)=s2ln⁡4s, and P(r)=∏n=0∞(1+rn)2−n. For a ∈ (0, 12 ] and r ∈ (0, 1), we have

   P(r)max{B4A(r),B5B(r)}≤exp⁡[μa(r)+ln⁡r]≤P(r)B41−B6[1−A(r)]. (1.13)

Theorem 3

For C₁ = R(a)−3ln⁡22ln⁡4 , the function

is strictly increasing from (0, 1) onto (C₁, ∞). In particular, for r ∈ (0, 1),

Lemma 1

([7, Theorem 1.25]). For –∞ < a < b < ∞, let g, h : [a, b] → ℝ be continuous on [a, b] and differentiable on (a, b) and let h′(x) ≠ 0 on (a, b). If g′(x)h′(x) is increasing (or decreasing respectively) on (a, b), then so are

Lemma 2

([2, Lemma 2]). Let r_n and s_n for n ∈ ℕ be real numbers and the power series

be convergent for |x| < 1. If s_n > 0 for n ∈ ℕ and if rnsn is strictly increasing (or decreasing respectively) for n ∈ ℕ, then the function R(x)S(x) is strictly increasing (or decreasing respectively) on (0, 1).

Lemma 3

([6, Lemmas 5.2 and 5.4] and [25, Theorem 2.2]). For r ∈ (0, 1) and b = 1 – a, the following conclusions hold true:

1. the function Ea−s2Kar2 is strictly increasing from (0, 1) onto (aπ2,sin⁡(aπ)2b) ;

2. the function Ka−Ealn⁡(1/s) is strictly decreasing from (0, 1) onto (sin(aπ), (1 – a)π);

3. the function π2/4−s2Ka2r2 is strictly increasing from (0, 1) onto ((a2+b2)π24,π24) ;

4. the function s^c𝓚_a is decreasing from (0, 1) onto (0, π2 ) if and only if c ≥ 2a(1 – a); the function s 𝓚_a(r) is decreasing for each a ∈ (0, 12 ];

5. the function E−1s2ln⁡(4/s) is strictly increasing from (0, 1) onto (π−22ln⁡4,12).

Lemma 4

([6, Theorem 5.5] and [24, Theorems 1 and 2]). Let R(a) be the Ramanujan constant defined in (1.7). Then

1. the function μ_a(r) + ln r is strictly decreasing from (0, 1) onto (0, R(a)2 );

2. the function μ_a(r) – μ(r) is strictly decreasing from (0, 1) onto (0,R(a)−ln⁡162) ;

3. the function μ_a(r) – artanh s is strictly decreasing from (0, 1) onto the interval (0,R(a)−3ln⁡22).

Lemma 5

For r ∈ (0, 1) and b = 1 – a, the following conclusions hold true:

1. the function

   I1(r)=ln⁡(1/s)(1+r2)(artanhr)/r−1

   is strictly increasing from (0, 1) onto (38,12) ;

2. the function

   I2(r)=Ka−Ea−(1−2a)(Ea−s2Ka)r2

   is strictly increasing from (0, 1) onto ((a2+b2)π2,∞) ;

3. the function

   I3(r)=π2/(4s2Ka2)−1ln⁡(1/s)

   is strictly increasing from (0, 1) onto (2(a² + b²), ∞).

Proof

Utilizing (1.1) and [29, 2.2.5] and using power series expansion lead to

and

Applying (2.5) yields

from which and (2.4), it follows that

where

Let c1(n)=a1(n)b1(n). Then

Hence, the inequality (2.6) implies that c₁(n) is strictly increasing in n. From Lemma 2, it follows that I₁(r) is strictly increasing in (0, 1).

By virtue of L’Hôpital’s rule and Lemma 5, we easily obtain the limits I1(0+)=38 and I1(1−)=12.

From (1.1) to (1.4), it is easy to verify that

and

for r ∈ (0, 1). Hence, after simplifying and utilizing (1.2), the function I₂(r) can be rewritten as

where J_n = [a² + b² + n](a)_n(1 – a)_n and b = 1 – a. A conclusion in [6, Lemma 7.1] states that the function (x)_n+1(1 – x)_n+1 for n ≥ 0 is positive, increasing on [0, 12 ], and decreasing on [ 12 , 1]. This implies that J_n > 0. Thus, the monotonicity of I₂(r) follows immediately.

The limits limx→0+I2(x)=(a2+b2)π2 and lim_x→1^– I₂(x) = ∞ are straightforward.

Let I3(r)=I4(r)I5(r), where I4(r)=π24s2Ka2−1 and I5(r)=ln⁡1s. By (1.3), it follows that lim_x→0⁺ I₄(x) = lim_x→0⁺ I₅(x) = 0.

From the formula (2.1) and elementary computation, it follows that

From Lemma 1, the fourth item in Lemma 3, and the second item in Lemma 5, the monotonicity of I₃(r) follows immediately.

By L’Hôpital’s rule and Lemmas 3 and 5, we arrive at

while lim_r→1^– I₃(r) = ∞ directly. The proof of Lemma 5 is complete.□

Lemma 6

For r ∈ (0, 1) and b = 1 – a, we have the following conclusions:

1. the function L1(r)=K−Kaln⁡(1/s) is strictly increasing from (0, 1) onto the interval (π(1−2a)24,1−sin⁡(aπ)) ;

2. the function L2(r)=K−KaKa−Ea is strictly increasing from (0, 1) onto the interval ((1−2a)24b,1sin⁡(aπ)−1) ; the function K−KaK−E is strictly increasing from (0, 1) onto ((1−2a)22,1−sin⁡(aπ)) ;

3. the function L3(r)=K−Ka(1+r2)(artanhr)/r−1 is strictly increasing from (0, 1) onto (3π(1−2a)232,1−sin⁡(aπ)2).

Proof

Let ℓ₁(r) = 𝓚 – 𝓚_a and ℓ₂(r) = ln 1s . It is obvious that L1(r)=ℓ1(r)ℓ2(r) and lim_r→0⁺ℓ₁(r) = lim_r→0⁺ ℓ₂(r) = 0.

It follows from (1.3) and (2.1) that

where

It is clear that lim_r→0⁺ ℓ₃(r) = lim_r→0⁺ ℓ₄(r) = 0. Applying (2.2) and (1.3) and differentiating give

where Tn=(12)n(12)n−4a(1−a)(a)n(1−a)n. From [6, Lemma 7.1], it is easy to see that T_n ≥ 0. Therefore, the monotonicity of L₁(r) follows from Lemma 2.

By L’Hôpital’s rule and Lemmas 1 and 3, we have

It is known [4, (1.6)] that F(a, b; a + b; x) satisfies the Ramanujan asymptotic relation

for a, b ∈ (0, ∞), where R(a, b) = –2γ – ψ(a) – ψ(b) and

where

Hence, the limit lim_r→1^– L₁(r) = 1 – sin(aπ) follows from (1.3), (2.7), and (2.8).

The function L₂(r) can be rewritten as

Hence, the monotonicity property of the function L₂(r) follows from the second item in Lemma 3 and the first item in 6. Furthermore, the limits

are easily obtained. Similarly, we can prove that the function K−KaK−E is strictly increasing from (0, 1) onto ((1−2a)22,1−sin⁡(aπ)).

The function L₃(r) can be rewritten as

From the first items in Lemmas 5 and 6, the monotonicity of L₃(r) follows immediately. Moreover, the limits

can be obtained from the first items in Lemmas 5 and 6. The proof of Lemma 6 is complete.□

Proof of Theorem 1

Let

Then F(r)=f1(r)f2(r) and, by Lemma 4, f₁(0⁺) = f₂(0⁺) = 0.

Differentiating and making use of (2.3) give

From Lemmas 1 and the first and third conclusions in Lemma 5, we see that the function F(r) is strictly increasing on (0, 1).

By L’Hôpital’s rule and the first and third items in Lemma 5, we obtain

Clearly, the limit F(1^–) = R(a)2 follows from the first item in Lemma 4.

Finally, by (2.5), the double inequality in (1.9) follows from the monotonicity property of F(r). The proof of Theorem 1 is complete.□

Corollary 1

For r ∈ (0, 1) and K ∈ (1, ∞), the inequality

holds true, where C=R(a)2 and an=24n2−1.

Proof

This follows from combining (1.8) with the double inequality (1.9).□

Remark 1

The upper and lower bounds in (1.9) are better than corresponding bounds in

obtained in [34, Theorem 2].

The inequality (3.1) gives an elementary and infinite series estimates for φ1/Ka (r) and, consequently, the bound of solutions to the Ramanujan generalized modular equations is refined.

Proof of Theorem 2

Write G₁(r) as

where

Let g₃(r) = μ_a(r) – μ(r) and g₄(r) = 𝓔 – 1. By (1.4) and the second item in Lemma 4, we obtain

Direct computation and utilization of (2.1) and (2.3) result in

Hence, from the fourth item in Lemma 3 and the second item in Lemma 6, it follows that the function g₁(r) is strictly increasing on (0, 1). Using L’Hôpital’s rule together with the fifth item in Lemma 3 and the second item in Lemma 6, the limits g1(0)=R(a)−ln⁡16π−2 and g₁(1^–) = ∞ follows readily.

By (3.2), the function G₁(r) is a product of two positive and strictly increasing functions, so the monotonicity of G₁(r) follows from the fifth item in Lemma 3. From the fifth item in Lemma 3 and the limit of g₁(r), we gain G1(0+)=R(a)−ln⁡162ln⁡4 and G₁(1^–) = ∞. Moreover, the double inequality (1.11) is obvious.

Let g₅(r) = B₁ – [μ_a(r) – μ(r)] and g6(r)=1−s2 artanh rr. Then G2(r)=g5(r)g6(r) and g₅(0) = g₆(0) = 0. By (2.3), simple computation leads to

Hence, by Lemma 1, the monotonicity of G₂(r) follows from the fourth item in Lemma 3 and the third item in Lemma 6.

Clearly, the limit G2(1−)=R(a)−ln⁡162 is valid. By L’Hôpital’s rule and the third item in Lemma 6, we readily obtain

By the monotonicity of G₂(r), the double inequality (1.12) follows immediately.

By the formula (1.11) in [28, Theorem 1], we have

Consequently, the third item in Theorem 2 follows from (1.11) and (1.12). The proof of Theorem 2 is complete.□

Corollary 2

For r ∈ (0, 1) and K ∈ (1, ∞), the inequality

holds true, where A(r)=s2 artanh rr and B(r)=s2ln⁡4s.

Proof

This follows from combining the double inequality (1.8), the equality (3.3), and the inequality (1.13).□

Remark 2

The lower bound in (1.11) is better than the corresponding bound in the equation (11) in [24, Theorem 1] which is referenced in item (2) of Lemma 4.

The upper and lower bounds in (1.12) are better than corresponding bounds in the equation (11) in [24, Theorem 1] which is referenced in item (2) of Lemma 4.

The inequality (3.4) provides an elementary and an infinite product estimates for φ1/Ka (r) and a new bound of solutions to the Ramanujan generalized modular equations is given.

Proof of Theorem 3

It is easy to see that the function H(r) can be written as

where G₁(r) is defined by (1.10) and

which can be equivalently written as the product of two functions

Denote

where h₂(r) = μ(r) – artanh s and h₃(r) = 𝓔 – 1. By the third item in Lemma 4 and (1.4), we obtain h₂(1^–) = h₃(1^–) = 0. Applying (2.1) and (2.3) and simply computing yield

Let

and s=1−r2. Using the substitution

Then u=1−t1+t. By Landen’s transformation formula

in [30] and (3.9), we have

By (3.10), the identity (3.8) is equivalent to

It is easy to show that the first factor in the right hand side of (3.11) is strictly increasing in t on (0, 1). Hence, by virtue of the third item in Lemma 3 and the relation between r and t, the function h₄(r) is strictly increasing on (0, 1).

It was given in [2, Theorem 15] that the function r→s(K−E)r2 is strictly decreasing from (0, 1) onto (0,π4). Therefore, by (3.7) and Lemma 1, the function h₁(r) is positive and strictly increasing.

From the fifth item in Lemma 3 and (3.6), we conclude that the function H₁(r) is strictly increasing on (0, 1). Hence, the monotonicity of H(r) follows from the first item in Theorem 2 and (3.5).

It is clear that the limits H₁(0⁺) = 14 and H₁(1^–) = ∞ follow from item (5) in Lemma 3, item (1) in Lemma 4, and item (1) in Theorem 2. Additionally we note that H(0⁺) = G₁(o⁺) + H₁(o⁺) = R(a)−ln⁡162ln⁡4 . The double inequality (1.14) follows immediately. The proof of Theorem 3 is complete.□

Remark 3

The lower bound in (1.14) is better than corresponding bounds in the equation (14) in [24, Theorem 2] which is presented in item (3) of Lemma 4.