Monotonicity criterion for the quotient of power series with applications

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Monotonicity criterion for the quotient of power series with applications Zhen-Hang Yang a , Yu-Ming Chu a,∗ , Miao-Kun Wang b a b

School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China Department of Mathematics, Huzhou University, Huzhou 313000, China

a r t i c l e

i n f o

Article history: Received 4 January 2015 Available online xxxx Submitted by J. Xiao Keywords: Quotient of power series Piecewise monotonicity Hypergeometric function Landen inequality

a b s t r a c t In this paper, we present the necessary and sufficient condition for the monotonicity of the quotient of power series. As applications, some gaps and misquotations in certain published articles are pointed out and corrected, and some known results involving the Landen inequalities for zero-balanced hypergeometric functions are improved. © 2015 Elsevier Inc. All rights reserved.

1. Introduction In 1955, Biernacki and J. Krzyż [12] (see also [16, Lemma 2.1], [15]) found an important criterion for the monotonicity of the quotient of power series as follows. ∞ ∞ Theorem 1.1. (See [12].) Let A(t) = k=0 ak tk and B(t) = k=0 bk tk be two real power series converging on (−r, r) (r > 0) with bk > 0 for all k. If the non-constant sequence {ak /bk } is increasing (resp. decreasing) for all k, then the function t → A(t)/B(t) is strictly increasing (resp. decreasing) on (0, r). Theorem 1.1 has been widely used to find analytic inequalities [2,5,7–9,22,24,27,28]. The polynomial version of Theorem 1.1 can be found in the literature [15]. n n Theorem 1.2. (See [15, Theorem 4.4].) Let An (t) = k=0 ak tk and Bn (t) = k=0 bk tk be two real polynomials with bk > 0 for all k. If the sequence {ak /bk } is increasing (decreasing), then so is the function t → An (t)/Bn (t) for all t > 0. More results involving the quotient of power series can be found in [6] and the references therein. * Corresponding author. Fax: +86 572 2321163. E-mail addresses: [email protected] (Z.-H. Yang), [email protected] (Y.-M. Chu), [email protected] (M.-K. Wang). http://dx.doi.org/10.1016/j.jmaa.2015.03.043 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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Next, we deal with the case of the sequence {ak /bk } being piecewise monotone, that is, for certain m ∈ N, {ak /bk } is increasing (resp. decreasing) for 0 ≤ k ≤ m and decreasing (resp. increasing) for k ≥ m. In this case, how to determine the monotonicity of the function t → A(t)/B(t)? In 2007, Belzunce, Ortega and Ruiz [11] offered a criterion when A(t) and B(t) have the radius of convergence R = ∞, but without giving the details of the proof. Proposition 1.3. (See [11, Lemma 6.4].) Suppose that {ak /bk , k = 0, 1, . . .} is a real non-constant sequence with the property that ak bk

 is increasing in k, is decreasing in k,

for k ≤ k0 , for k ≥ k0

∞ for some positive integer k0 , also assume that the power series defined by A(t) = k=0 ak tk and B(t) = ∞ k k=0 bk t , t ≥ 0, converge absolutely for all t ≥ 0. Then there exists a real number t0 > 0 such that A(t) B(t)

 is strictly increasing in t, is strictly decreasing in t,

for for

t < t0 , t ≥ t0 .

In [10], Proposition 1.3 was rewritten by Baricz, Vesti and Vuorinen in different form as follows.  Proposition 1.4. (See [10, Theorem 4.6].) Suppose that the power series f (x) = n≥0 an xn and g(x) =  n n≥0 bn x have the radius of convergence r > 0, and bn > 0 for all n ∈ {0, 1, 2, . . .}. If the sequence {an /bn } satisfies a0 /b0 ≤ a1 /b1 ≤ · · · ≤ an0 /bn0 and an0 /bn0 ≥ an0 +1 /bn0 +1 ≥ · · · ≥ an /bn ≥ · · · for some n0 ≥ 1, then there exists x0 ∈ (0, r) such that the function x → f (x)/g(x) is increasing on (0, x0 ) and decreasing on (x0 , r). Proposition 1.3 was also rewritten by Simić and Vuorinen [19] in another different form.  Proposition 1.5. (See [19, Lemma 1.1(3)].) Suppose that the power series f (x) = n≥0 an xn and g(x) =  n n≥0 bn x have the radius of convergence r > 0 and bn > 0 for all n ∈ {0, 1, 2, . . .}. If the sequence {an /bn } is monotone increasing (resp. decreasing) for 0 < n ≤ n0 and monotone decreasing (resp. increasing) for n > n0 , then there exists x0 ∈ (0, r) such that f (x)/g(x) is increasing (resp. decreasing) on (0, x0 ) and decreasing (resp. increasing) on (x0 , r). Recently, Proposition 1.5 has been cited and used in the literature [26, Lemma 2.1], [25, Lemma 1], [17, Lemma 6] and [20, Lemma 2.2]. Since without giving the details in the proof of Proposition 1.3, its correctness is questionable. Moreover, Propositions 1.4 and 1.5 may be not valid even if Proposition 1.3 is true, since Proposition 1.3 only contains the case of the radius of convergence r = ∞, but both Proposition 1.4 and Proposition 1.5 also contain the case of the radius of convergence r ∈ (0, ∞). In fact, we easily find the following counter-examples to refute Propositions 1.4 and 1.5. Example 1.6. Let f (x) = 3 +

∞  22k − 2 k=1

(2k)!

|B2k |xk

and g(x) = 1 +

∞  22k |B2k |xk , (2k)!

k=1

where Bk is the Bernoulli number. Then we clearly see that f (x) and g(x) have the radius of convergence R = π 2 . Making use of the well-known formulas [13, pp. 227–229]

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∞

1  22k − 2 1 = + |B2k |x2k−1 , sin x x (2k)!

|x| < π,

k=1 ∞

cot x =

1  22k − |B2k |x2k−1 , x (2k)!

|x| < π,

k=1

we know that f (x)/g(x) converges to √

x √ +2 f (x) sin x √ √ , = g(x) 2 − x cot x

0 < x < π2 .

Clearly, bk > 0 for all k, a1 1 a0 =3> = , b0 2 b1 that is, ak /bk is decreasing for k = 0, 1, and 22k − 2 ak = = 1 − 21−2k bk 22k is increasing for k ≥ 1. If Propositions 1.4 and 1.5 hold, then there exists x0 ∈ (0, π 2 ) such that f (x)/g(x) is decreasing on (0, x0 ) and increasing on (x0 , π 2 ). However, simple computations lead to 

f (x) g(x)

 =−

x sin

√ √ √ √ x + 2(1 + cos x)( x − sin x) √ √ √ √ <0 2 x(2 sin x − x cos x)2

(1.1)

for x ∈ (0, π 2 ), where the inequality holds due to x − sin x > 0 for x > 0 and sin x − x cos x > 0 for x ∈ (0, π). Remark 1.7. If we replace f (x) with −f (x) in Example 1.6, then Propositions 1.4 and 1.5 lead to the conclusion that there exists x0 ∈ (0, π 2 ) such that −f (x)/g(x) is increasing on (0, x0 ) and decreasing on (x0 , π 2 ). But (1.1) implies that −f (x)/g(x) is strictly increasing on (0, π 2 ). Inspired by the above discussions, in this paper, we will establish a criterion for the monotonicity of quotient of two power series on the finite interval of convergence so as to modify the wrong conclusion in Propositions 1.4 and 1.5, and give a strict proof of Proposition 1.3. The paper is organized as follows. In Section 2, we state and prove a necessary and sufficient condition for the monotonicity of quotient of two power series on (0, r), where r ∈ (0, ∞) is their radius of convergence. As a corollary, a strict proof of Proposition 1.3 is given. The monotonicity of quotient of two polynomials are also investigated. In Section 3, we apply our main result (Theorem 2.1) to correct and improve lemmas and main theorems concerning Landen inequalities for zero-balanced hypergeometric functions in [19]. 2. Main results In order to prove our main results, we introduce an auxiliary function at first. Let −∞ ≤ a < b ≤ ∞, f and g be differentiable on (a, b), and g  = 0 on (a, b). Then the function Hf,g is defined by Hf,g ≡

f g − f. g

(2.1)

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The auxiliary function Hf,g has the following properties [23, Property 1]: (i) If g = 0 on (a, b), then   f g = 2 Hf,g , g g

(2.2)

and therefore, sgn

  f = sgng  sgnHf,g . g

(2.3)

(ii) If f and g are twice differentiable on (a, b), then  Hf,g

 =

f g

 g.

(2.4)

Next we state and prove the correct assertion corresponding to Propositions 1.4 and 1.5. ∞ ∞ Theorem 2.1. Let A(t) = k=0 ak tk and B(t) = k=0 bk tk be two real power series converging on (−r, r) and bk > 0 for all k. Suppose that for certain m ∈ N, the non-constant sequence {ak /bk } is increasing (resp. decreasing) for 0 ≤ k ≤ m and decreasing (resp. increasing) for k ≥ m. Then the function A/B is strictly increasing (resp. decreasing) on (0, r) if and only if HA,B (r− ) ≥ (resp. ≤)0. Moreover, if HA,B (r− ) < (>)0, then there exists t0 ∈ (0, r) such that the function A/B is strictly increasing (resp. decreasing) on (0, t0 ) and strictly decreasing (resp. increasing) on (t0 , r). Proof. Since A/B = −(−A)/B, it is sufficient to prove the desired result in the case of the sequence {ak /bk } is increasing for 0 ≤ k ≤ m and decreasing for k ≥ m. From (2.2) and (2.4) we have 

A B



 HA,B

B HA,B , B2    A = B. B =

The necessity in the first statement follows easily from HA,B = (A/B) B 2 /B  ≥ 0. Therefore, in order to complete the proof we only need to prove the following fact: Let the non-constant sequence {ak /bk } is increasing for 0 ≤ k ≤ m and decreasing for k ≥ m, then A/B is strictly increasing on (0, r) if HA,B (r− ) ≥ 0, and there exists t0 ∈ (0, r) such that the function A/B is strictly increasing on (0, t0 ) and strictly decreasing on (t0 , r) if HA,B (r− ) < 0. Next, we prove the fact by mathematical induction for m ∈ N. (1) When m = 1, that is, {ak /bk } is increasing for k = 0, 1 and decreasing for k ≥ 1. It follows from Theorem 1.1 and the sequence {ak /bk } is decreasing for k ≥ 1 that the function A /B  is strictly decreasing,  that is, (A /B  ) ≤ 0. Therefore, HA,B = (A /B  ) B ≤ 0 for t ∈ (0, r) and HA,B is strictly decreasing on (0, r). We divide the proof into two cases. Case 1.1 HA,B (r− ) ≥ 0. Then the monotonicity of HA,B leads to the conclusion that HA,B (t) > HA,B (r− ) ≥ 0. Therefore, (A/B) = B −2 B  HA,B > 0. Case 1.2 HA,B (r− ) < 0. We first assert that HA,B (0+ ) > 0. Indeed, the non-constant sequence {ak /bk } is increasing for k = 0, 1 and decreasing for k ≥ 1 implies a1 /b1 > a0 /b0 . Otherwise, the non-constant

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sequence {ak /bk } is decreasing for all k ≥ 0. On the other hand, simple computation leads to HA,B (0+ ) =

A (0+ ) a1 B(0+ ) − A(0+ ) = b0 − a0 = b0  + B (0 ) b1



a0 a1 − b1 b0

 .

It follows from HA,B (r− ) < 0 and HA,B (0+ ) > 0 together with the monotonicity of HA,B we clearly see that there exists t0 ∈ (0, r) such that HA,B (t) > 0 for t ∈ (0, t0 ) and HA,B (t) < 0 for t ∈ (t0 , r). Therefore, the desired statement follows easily from the identity (A/B) = B −2 B  HA,B . (2) Suppose that the required assertion is true for m = n. We will prove it is also true for m = n + 1. Differentiation gives ∞ ∞ ∞ ∗ k kak tk−1 (k + 1)ak+1 tk ak t A (t) k=1 k=0 = ∞ = ∞ ≡ k=0 ∞ ∗ k. k−1 k B  (t) kb t (k + 1)b t k k+1 k=1 k=0 k=0 bk t

(2.5)

Clearly, the sequence {a∗k /b∗k } is increasing for 0 ≤ k ≤ n and decreasing for k ≥ n. By the inductive assumption, the function A /B  is strictly increasing on (0, r) if HA ,B  (r− ) ≥ 0, and there exists t0 ∈ (0, r) such that the function A /B  is strictly increasing on (0, t0 ) and strictly decreasing on (t0 , r) if HA ,B  (r− ) < 0. We divide the proof into three cases. Case 2.1 HA ,B  (r− ) ≥ 0. Then A /B  is strictly increasing on (0, r), this leads to the conclusion that = (A /B  ) B ≥ 0, HA,B (t) > HA,B (0+ ) ≥ 0 and (A/B) = B −2 B  HA,B > 0 for t ∈ (0, r).

 HA,B

Case 2.2 HA,B (r− ) ≥ 0, HA ,B  (r− ) < 0. Then there exists t0 ∈ (0, r) such that A /B  is strictly  increasing on (0, t0 ) and strictly decreasing on (t0 , r). It follows from HA,B = (A /B  ) B we know that   HA,B is strictly increasing on (0, t0 ) and strictly decreasing on (t0 , r). Therefore, (A/B) = B −2 B  HA,B > 0 follows from HA,B (r− ) ≥ 0 and HA,B (0+ ) ≥ 0 together with the piecewise monotonicity of HA,B . Case 2.3 HA,B (r− ) < 0. Then it follows from Case 2.1 that HA ,B  (r− ) < 0. It follows from Case 2.2 that there exists t0 ∈ (0, r) such that HA,B is strictly increasing on (0, t0 ) and strictly decreasing on (t0 , r). From HA,B (r− ) < 0 and HA,B (0+ ) ≥ 0 together with the piecewise monotonicity of HA,B we clearly see that there exists t0 ∈ (t0 , r) ⊂ (0, r) such that HA,B (t) > 0 for t ∈ (0, t0 ) and HA,B (t) < 0 for t ∈ (t0 , r). Therefore, the desired assertion follows easily from the identity (A/B) = B −2 B  HA,B . 2 Remark 2.2. Let us return to Example 1.6, it has shown that {ak /bk } is decreasing for k = 0, 1 and increasing for k ≥ 1. It is easy to check that √

x √ ( sin √ + 2) √ √ f  (x) x x √ √  (2 − x cot x) − √ −2 g(x) − f (x) = Hf,g (x) =  g (x) (2 − x cot x) sin x √ √ √ √ √ x(1 − cos x) + 2 x sin x − 2 sin2 x √ √ √ √ =− (1 + cos x), sin x( x − cos x sin x) √ √ √ √ √ x(1 − cos x) + 2 x sin x − 2 sin2 x 2− √ √ √ √ Hf,g (π ) = − lim2 (1 + cos x) = 0. x→π sin x( x − cos x sin x)   By our Theorem 2.1 we know f /g is strictly decreasing on 0, π 2 .

Letting r = ∞ in Theorem 2.1, we obtain the following corollary, which gives a strict proof of Proposition 1.3. ∞ ∞ Corollary 2.3. Let A(t) = k=0 ak tk and B(t) = k=0 bk tk be two real power series converging on R with bk > 0 for all k. If for certain m ∈ N, the non-constant sequence {ak /bk } is increasing (decreasing) for 0 ≤ k ≤ m and decreasing (increasing) for k ≥ m, then there exists t0 ∈ (0, ∞) such that the function A/B is strictly increasing (decreasing) on (0, t0 ) and strictly decreasing (increasing) on (t0 , ∞).

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Proof. Likewise, we only prove the desired result in the case of the sequence {ak /bk } is increasing for 0 ≤ k ≤ m and decreasing for k ≥ m. By Theorem 2.1, it suffices to check that HA,B (∞) < 0. We first prove that ∞ ak t k an lim k=0 = lim . ∞ k t→∞ n→∞ bn b t k=0 k

(2.6)

It follows from the decreasing property of the non-constant sequence {ak /bk } for k ≥ m that there exists λ (λ is a finite number or −∞) such that limn→∞ (an /bn ) = λ. We divide two cases to prove (2.6). Case 1 λ is a finite number. Let n ak t−k a0 tn + a1 tn−1 + · · · + an−1 t + an Cn (t) = k=0 = , n −k b0 tn + b1 tn−1 + · · · + bn−1 t + bn k=0 bk t  A(1/t) , t = 0, C (t) = B(1/t) λ, t = 0. Then we clearly see that the function sequence {Cn (t)} converges uniformly to C(t) on [a, b] for any −∞ < a < b < +∞. Therefore, (2.6) follows from ∞ ak t k lim k=0 = lim lim Cn (t) = lim lim Cn (t). ∞ k t→∞ n→∞ t→0 t→0 n→∞ k=0 bk t Case 2 λ = −∞. Then from Case 1 we know that ∞ k bn k=0 bk t lim ∞ = lim . k t→∞ n→∞ an a t k k=0 Next, we prove that HA,B (∞) < 0. To this end, we write B  (t) HA,B (t) as B  (t) HA,B (t) = A (t) B (t) − A (t) B  (t) =

∞ 

kak tk−1

k=0

∞ 

bk tk −

k=0

∞ 

ak t k

k=0

∞ 

kbk tk−1 =

∞  ∞ 

(i − j) ai bj ti+j−1 .

(2.7)

i=0 j=0

k=0

The symmetry of i and j in (2.7) leads to 

B (t) HA,B (t) =

∞  ∞ 

(j − i) aj bi tj+i−1 .

(2.8)

j=0 i=0

It follows from (2.7) and (2.8) that 2B  (t) HA,B (t) =

∞  ∞ 

(i − j) ai bj ti+j−1 +

i=0 j=0

=

∞  ∞ 

(j − i) aj bi tj+i−1

j=0 i=0

 bi bj (i − j)

i=0 j=0

=

∞  ∞ 

m−1  m−1  i=0 j=0

aj ai − bi bj

ci,j ti+j−1 + 2

 ti+j−1 =

m−1 ∞   i=0 j=m

∞  ∞ 

ci,j ti+j−1

i=0 j=0

ci,j ti+j−1 +

∞  ∞  i=m j=m

ci,j ti+j−1 ,

(2.9)

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where  ci,j = bi bj (i − j)

aj ai − bi bj



≥0 ≤0

if i, j ≤ m, if i, j ≥ m.

It follows from the decreasing property of the sequence {ak /bk } for k ≥ m that aj /bj ≤ am /bm for j ≥ m and 2

m−1 ∞  

ci,j t

i+j−1

=2

i=0 j=m

m−1 ∞  

 bi bj (j − i)

i=0 j=m

≤2

m−1 ∞  

 bi bj (j − i)

i=0 j=m

ai aj − bj bi



am ai − bm bi

ti+j−1  ti+j−1 = 2

m−1 ∞  

bi bj (j − i) λi ti+j−1 ,

(2.10)

i=0 j=m

where λi = am /bm − ai /bi ≥ 0 for all i = 0, 1, 2, . . . . Eq. (2.9) and inequality (2.10) lead to 2B  (t) HA,B (t) ≤

m−1  m−1 

ci,j ti+j−1 + 2

m−1 ∞  

i=0 j=0

= tm

bi bj (j − i) λi ti+j−1 +

i=0 j=m

m−1  m−1  i=0 j=0

ci,j ti+j−1

i=m j=m

∞ 

ci,j ti+j−m−1 + tm

∞  ∞ 

pj (t) tj−m

j=m

and ⎛ 2HA,B (t) < tm ⎝

1 B  (t)

m−1  m−1  i=0 j=0

⎞ ∞  pj (t) j−m ⎠ t ci,j ti+j−m−1 + ,  (t) B j=m

(2.11)

where pj (t) =

m−1 

2bi bj (j − i) λi ti−1 +

i=0

∞ 

ci,j ti−1 .

(2.12)

i=m

Note that m−1 m−1 1   ci,j ti+j−m−1 = 0. t→∞ B  (t) i=0 j=0

lim

(2.13)

From (2.11) and (2.13) we know that it suffices to prove that lim

pj (t) ≤0 B  (t)

(2.14)

lim

pm (t) < 0. B  (t)

(2.15)

t→∞

for j > m and

t→∞

Indeed, if j ≥ m, then (2.6) and (2.12) lead to

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 i−1 c t i,j i=m B  (t) m−1 ∞ i−1 i−1 i=0 2bi bj (j − i) λi t i=m ci,j t = lim + lim   t→∞ t→∞ B (t) B (t)  ∞ ∞ m−1 m i−m−1 i−1 cm,j t +t i=m+1 ci,j t i=m ci,j t  = lim = lim ∞ k t→∞ t→∞ B  (t) k=0 (k + 1) bk+1 t ∞ ∞  m ∞  k k tm cm+k+1,j tk t k=0 cm+k+1,j t k=m (k + 1) bk+1 t   = lim ∞ k=0 = lim × ∞ ∞ k k k t→∞ t→∞ k=0 (k + 1) bk+1 t k=m (k + 1) bk+1 t k=0 (k + 1) bk+1 t ∞    ∞ k k k=0 cm+k+1,j t k=m (k + 1) bk+1 t  = lim ∞ × ∞ k k t→∞ k=0 (m + k + 1) bm+k+1 t k=0 (k + 1) bk+1 t

 m−1 ∞ cm+k+1,j tk (k + 1) bk+1 tk k=0 k=0 × lim 1 − ∞ = lim ∞ k k t→∞ t→∞ k=0 (m + k + 1) bm+k+1 t k=0 (k + 1) bk+1 t   aj am+k+1 b b (m + k + 1 − j) − m+k+1 j bm+k+1 bj cm+k+1,j = lim = lim k→∞ (m + k + 1) bm+k+1 k→∞ (m + k + 1) bm+k+1   ak aj . (2.16) = bj lim − k→∞ bk bj

pj (t) = lim lim t→∞ B  (t) t→∞



m−1 i=0

2bi bj (j − i) λi ti−1 + B  (t)

∞

From the decreasing property of the non-constant sequence {ak /bk } for j ≥ m we clearly see that lim

aj ak − ≤0 bk bj

(2.17)

am ak − < 0, bk bm

(2.18)

k→∞

for j ≥ m. In particular, we have lim

k→∞

otherwise, limk→∞ ak /bk − am /bm = 0 leads to the conclusion that the non-constant sequence {ak /bk } is increasing for k ≤ m and am /bm = am+1 /bm+1 = · · · = an /bn = · · · = 0, which means that {ak /bk } is a increasing sequence. Therefore, (2.14) and (2.15) follows from (2.16)–(2.18). 2 Remark 2.4. Although [19, Lemma 1.1(3)] was wrongly cited and used in [17,20,25,26], but the validity of the given results did not be influenced, because the authors in fact used our Corollary 2.3. Making use of the similar method in proving Theorem 2.1 and Corollary 2.3, we easily get the criterion for the monotonicity of quotient of two polynomials on (0, r) (r > 0). n n k k Theorem 2.5. Let An (t) = k=0 ak t and Bn (t) = k=0 bk t be two real polynomials defined on (0, r) (r > 0) with bk > 0 for all k. Suppose that for certain m ∈ N with m < n, the non-constant sequence {ak /bk } is increasing (resp. decreasing) for 0 ≤ k ≤ m and decreasing (resp. increasing) for m ≤ k ≤ n. Then the function An /Bn is strictly increasing (resp. decreasing) on (0, r) if and only if HAn ,Bn (r− ) ≥ (≤)0, and there exists t0 ∈ (0, r) such that the function A/B is strictly increasing (resp. decreasing) on (0, t0 ) and strictly decreasing (resp. increasing) on (t0 , r) if HAn ,Bn (r− ) < (>)0. n n Corollary 2.6. Let An (t) = k=0 ak tk and Bn (t) = k=0 bk tk be two real polynomials with bk > 0 for all k. If for certain m ∈ N with m < n, the non-constant sequence {ak /bk } is increasing (resp. decreasing) for

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0 ≤ k ≤ m and decreasing (resp. increasing) for m ≤ k ≤ n, then there is a unique t0 ∈ (0, ∞) such that the function An /Bn is strictly increasing (resp. decreasing) on (0, t0 ) and strictly decreasing (resp. increasing) on (t0 , ∞). 3. Applications In this section, we introduce some necessary knowledge for Gaussian hypergeometric function and Landen inequalities in Subsection 3.1. In Subsection 3.2, we refine Lemmas 3.1 and 3.2 in [19] by use of Theorem 2.1. In Subsection 3.3, we improve the main results involving the Landen inequalities for zero-balanced hypergeometric function given by Simić and Vuorinen in [19]. 3.1. Gaussian hypergeometric function and Landen inequalities For real numbers a, b, and c with c = 0, −1, −2, . . . , the Gaussian hypergeometric function is defined by F (a, b; c; x) =2 F1 (a, b; c; x) =

∞  (a)n (b)n xn (c)n n! n=0

for x ∈ (−1, 1), where (a)n = a(a + 1) · · · (a + n − 1), for n = 1, 2, . . . , and (a)0 = 1 for a = 0. It is well-known that the complete elliptic integral of the first kind K(x) can be expressed as π K(x) = F 2



1 1 , ; 1; x2 2 2



π/2 = (1 − x2 sin2 t)−1/2 dt,

x ∈ (0, 1).

0

Generally, F (a, b; c; x) is called zero-balanced if c = a + b. In the zero-balanced case, there is a logarithmic singularity at x = 1 and Gauss proved the asymptotic formula (see [14]) F (a, b; a + b; x) ∼ −

1 ln(1 − x) as B(a, b)

x → 1,

(3.1)

where B(z, w) =

Γ(z)Γ(w) , Γ(z + w)

(z) > 0,

(w) > 0

is the classical beta function. Ramanujan found a much sharper asymptotic formula (see [14]) B(a, b)F (a, b; a + b; x) + ln(1 − x) = R(a, b) + O((1 − x) ln(1 − x)) as

x → 1,

(3.2)

where  R(a, b) = −ψ(a) − ψ(b) − 2γ,

R

1 1 , 2 2



ψ(z) = Γ (z)/Γ(z), (z) > 0 and γ is the Euler–Mascheroni constant.

= ln 16,

(3.3)

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∗ ∗ Fig. 1. The regions D00 , D00 , D11 , D10 , D10 .

For the complete elliptic integral of the first kind K(x), the Landen identities proved in 1771 (see [1], [21, p. 507]) are interesting and important, which state that K

 √  2 r = (1 + r)K(r), 1+r

 K

1−r 1+r

 =

1+r  K (r), 2

(3.4)

√ where K (r) ≡ K( 1 − r2 ), r ∈ (0, 1). Naturally, a similar problem was proposed by Anderson et al. in [3, p. 79], that is: Problem 3.1. Find an analog of Landen’s transformation formulas in (3.4) for F (a, b; a + b; x). In particular, if k(r) = F (a, b; a + b; r2 ) and a, b ∈ (0, 1), is it true that k

 √  2 r ≤ Ck(r) 1+r

(3.5)

for some constant C and all r ∈ (0, 1)? ∗ ≡ {(a, b)|a, b > 0, a + b ≤ 1} (the regions Qiu and Vuorinen [18, Theorem 1.2] proved that for (a, b) ∈ D00 ∗ D00 , D11 , D10 , D10 , see Fig. 1), the Landen inequality

∗ D00 ,

k

 √  2 r ≤ (1 + r)k(r) 1+r

(3.6)

∗ holds for all r ∈ (0, 1). Thus we obtained the best constant C = 2 in Problem 3.1 for (a, b) ∈ D00 . ∗ Recently, Simić and Vuorinen [19, Corollary 2.3, Lemma 3.2] have extended the region D00 to D00 ≡ {(a, b)|a, b > 0, ab ≤ 1/4}. Moreover, they presented a reverse inequality of (3.6) for (a, b) ∈ D11 ≡ {(a, b)|a, b > 0, 1/a + 1/b ≤ 4}.

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∗ Fig. 2. The regions E00 , E11 , E10 , E10 .

In [18, Theorem 1.2 (1)], Qiu and Vuorinen tried to proved the function s(r) ≡ (1 +

√

 √ 4 r √ a, b; a + b; (1 + r)2

 r)F (a, b; a + b; r) − F

(3.7)

∗ is increasing in r ∈ (0, 1) for (a, b) ∈ D00 . Unfortunately, there is a gap in their proof. Simić and Vuorinen [19, Lemma 3.2] gave a correct proof, and pointed out that the function r → s(r) is decreasing on (0, 1) for (a, b) ∈ D11 . As mentioned in Section 1, Simić and Vuorinen [19] incorrectly cited Proposition 1.3 as a lemma, and used it to prove Theorem 2.1 in [19]. In Theorem 2.1 [19], Simić and Vuorinen claimed that neither of (3.6) nor its reverse inequality holds in the remaining region D10 ≡ {(a, b)|a, b > 0, 1/4 < ab < (a + b)/4} for each r ∈ (0, 1). Indeed, by our Theorem 2.1 in Section 2, we find that inequality (3.6) also holds in ∗ D10 ≡ {(a, b)|a, b > 0, 1/4 < ab < (a + b)/4} ∩ {(a, b)|a, b > 0, R(a, b) ≥ ln 16} ⊂ D10 . ∗ ∗ , D00 , D11 , D10 , D10 again, and introduce some new regions in For convenience, we list the regions D00 2 {(a, b) ∈ R |a > 0, b > 0}: (see Figs. 1, 2 and 3)

D00 = {(a, b)|a, b > 0, ab ≤ 1/4} ,

∗ D10

∗ D00 = {(a, b)|a, b > 0, a + b ≤ 1},

D11 = {(a, b)|a, b > 0, 1/a + 1/b ≤ 4} ,   a+b , D10 = (a, b)|a, b > 0, 1/4 < ab < 4   a+b ∩ {(a, b)|a, b > 0, R(a, b) ≥ ln 16} , = (a, b)|a, b > 0, 1/4 < ab < 4

E00 = {(a, b)|a, b > 0, a + b ≤ 5/4 − ab} , E11 = {(a, b)|a, b > 0, a + b ≤ 8ab − 1} ,   a+b+1 , E10 = (a, b)|a, b > 0, 5/4 − (a + b) < ab < 8

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Fig. 3. The regions Ω1 , Ω2 .

∗ E10 =

 (a, b)|a, b > 0, 5/4 − (a + b) < ab <

∗ ∗ ] ∩ [D00 ∪ D10 ], Ω1 = [E00 ∪ E10

1 4

 ,

Ω2 = E11 ∩ D11 .

Clearly, D11 ∪ D00 ∪ D10 = {(a, b) ∈ R2 |a > 0, b > 0}, E11 ∪ E00 ∪ E10 = {(a, b) ∈ R2 |a > 0, b > 0}, ∗ ∗ ∗ D00 ⊂ D00 , D10 ⊂ D10 , E10 ⊂ E10 , and Ω1 = D00 , Ω2 = D11 . To end this subsection, we shall denote the following four functions, which are frequently used in Subsections 3.2 and 3.3. Denoted by

F (x) = F (a, b; a + b; x) =

∞ 

Fˆn xn ,

G(x) = F (a, b; a + b + 1; x) =

n=0

 F0 (x) = F

∞ 

ˆ n xn , G

n=0

  ∞ 1 1 , ; 1; x = Fˆ0n xn , 2 2 n=0

 G0 (x) = F

  ∞ 1 1 ˆ 0n xn , , ; 2; x = G 2 2 n=0

where a, b > 0 and (a, b) = (1/2, 1/2). Simple computations yield (a)n+1 (b)n+1 /(a + b)n+1 (a)n (b)n /(a + b)n Fˆn+1 Fˆn / −1= / −1 [(1/2)n+1 ]2 /(1)n+1 [(1/2)n ]2 /(1)n Fˆ0n+1 Fˆ0n =

4 (2n +

1)2 (a

+ b + n)

Tn ,

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ˆn ˆ n+1 G (a)n+1 (b)n+1 /(a + b + 1)n+1 (a)n (b)n /(a + b + 1)n G / = / −1 ˆ 0n+1 G ˆ 0n [(1/2)n+1 ]2 /(2)n+1 [(1/2)n ]2 /(2)n G =

4 T, (2n + 1)2 (a + b + n + 1) n

where  a+b 1 n + ab − ≡ nC1 (a, b) + C2 (a, b), 4 4     a+b 1 5  Tn = ab + a + b − n + 2ab − − ≡ nC3 (a, b) + C4 (a, b). 4 4 4 

Tn =

ab −

(3.8) (3.9)

3.2. Improvements of Lemma 3.1 and 3.2 in [19] The basic results in [19] were deduced from the following lemma. Lemma 3.1. (See [19, Lemma 3.1].) (1) The function r → F (r)/F0 (r) is strictly decreasing in r ∈ (0, 1) on D00 and strictly increasing on D11 . (2) The function r →  G(r)/G0 (r) is strictly decreasing on E00 and strictly increasing on E11 . Remark 3.2. In Lemma 3.1, the monotonicity of F/F0 on (0, 1) in the remaining region D10 was not considered. Evidently, if (a, b) ∈ D10 , then the sequence {Fˆn /Fˆ0n } decreases and then increases, and by our Theorem 2.1, the monotonicity of the function F/F0 relies on the sign of HF,F0 (1− ). Therefore, there may be a region in D10 such that F/F0 is strictly decreasing on (0, 1). In fact, a complete assertion is included in the following proposition. Proposition 3.3. Let a, b > 0 and R(a, b) be defined by (3.3). Then the following statements are true: (i) If (a, b) ∈ D11 , that is, a + b ≤ 4ab, then the function r → F (r)/F0 (r) is strictly increasing on (0, 1). (ii) If (a, b) ∈ D00 ∪ (D10 ∩ {R(a, b) ≥ ln 16}), that is, R(a, b) ≥ ln 16, then the function r → F (r)/F0 (r) is strictly decreasing on (0, 1). (iii) If (a, b) ∈ D10 ∩ {R(a, b) < ln 16}, that is, 1 ≤ 4ab ≤ a + b and R(a, b) < ln 16, then there exists r0 ∈ (0, 1) such that F/F0 is strictly decreasing for r ∈ (0, r0 ) and strictly increasing for r ∈ (r0 , 1). Remark 3.4. Similar, the second assertion in Lemma 3.1 can be improved since the case of (a, b) ∈ E10 ˆ n /G ˆ 0n } decreases and then was not considered in [19, Theorem 2.1]. If (a, b) ∈ E10 , then the sequence {G increases, by our Theorem 2.1 we know that the function G/G0 is decreasing on (0, 1) if HG,G0 (1− ) ≤ 0. As a matter of fact, a supplement to the second assertion in Lemma 3.1 can be stated as follows. Proposition 3.5. Let a, b > 0. Then the following statements are true: (i) If (a, b) ∈ E11 , that is, a +b ≤ 8ab −1, then the function r → G(r)/G0 (r) is strictly increasing on (0, 1). (ii) If (a, b) ∈ E00 ∪ (E10 ∩ {4ab ≤ 1}), that is, ab ≤ 1/4, then the function r → G(r)/G0 (r) is strictly decreasing on (0, 1). (iii) If (a, b) ∈ E10 ∩ {4ab > 1}, that is, a + b ≥ 8ab − 1 > 1, then there exists r0 ∈ (0, 1) such that G/G0 is strictly decreasing on (0, r0 ) and strictly increasing on (r0 , 1).

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14

To prove Theorems 2.4 and 2.5 in [19], Simić and Vuorinen presented Lemma 3.2, which states that Lemma 3.6. (See [19, Lemma 3.2].) The function r → s(r) defined by (3.7) is strictly increasing in r ∈ (0, 1) ∗ and strictly decreasing on D11 . on D00 Remark 3.7. From the proof of Lemma 3.2 in [19], we find that the increasing (decreasing) property of s(r) on (0, 1) depends on the decreasing (increasing) property of the functions G/G0 and F/F0 . Consequently, ∗ by our Propositions 3.3 and 3.5, after replacing D00 and D11 by ∗ ∗ Ω1 = [E00 ∪ E10 ] ∩ [D00 ∪ D10 ] = D00

and Ω2 = E11 ∩ D11 = D11 , respectively, Lemma 3.2 in [19] still holds. Thus Simić and Vuorinen’s Lemma 3.2 can be improved as follows. Proposition 3.8. Let a, b > 0. Then the function s(r) defined by (3.7) is strictly increasing in r ∈ (0, 1) for (a, b) ∈ D00 and strictly decreasing for (a, b) ∈ D11 . As mentioned previously, to prove our propositions, it is crucial to determine the signs of HF,F0 (1− ) and HG,G0 (1− ). For this, let us recall the following formulas: F (k) (a, b; c; x) =

(a)k (b)k F (a + k, b + k; c + k; x), (c)k

(1 − x)F  (a, b; a + b; x) = F (a, b; c; 1) =

ab F (a, b; a + b + 1; x), a+b

Γ(c)Γ(c − a − b) , Γ(c − a)Γ(c − b)

(c) > (a + b).

(3.10) (3.11) (3.12)

We also need the following Lemma 3.9. Lemma 3.9. Let Hf,g be defined by (2.1). Then one has lim HF,F0 (x) =

1 [ln 16 − R(a, b)], B(a, b)

(3.13)

lim HG,G0 (x) =

Γ(a + b + 1) (4ab − 1). Γ(a + 1)Γ(b + 1)

(3.14)

x→1−

x→1−

Proof. First of all, by the formula (3.12) we get G(1− ) = F (a, b; a + b + 1; 1) = −



G0 (1 ) = F (i) It follows from (3.11) and (3.2) that

Γ(a + b + 1)Γ(1) , Γ(a + 1)Γ(b + 1)

 Γ(2)Γ(1) 4 1 1 , ; 2; 1 = 2 = . 2 2 Γ (3/2) π

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lim

x→1−

15

F  (x) F  (x) (1 − x)F  (x) F  (x) 4ab G(x) = lim = lim F0 (x) x→1− F0 (x) (1 − x)F0 (x) x→1− F0 (x) a + b G0 (x) = lim− x→1

=π

4ab Γ(a + b + 1)Γ(1)/[Γ(a + 1)Γ(b + 1)] a+b Γ(2)Γ(1)/Γ(3/2)2

Γ(a + b) , Γ(a)Γ(b)

lim HF,F0 (x) = lim

x→1−

x→1−

F  (x) F0 (x) − F (x) F0 (x)

= lim π x→1−

1 Γ(a + b) [R(1/2, 1/2) − ln(1 − x)] Γ(a)Γ(b) B(1/2, 1/2)

1 [R(a, b) − ln(1 − x)] + O((1 − x) ln(1 − x)) B(a, b) 1 = [ln 16 − R(a, b)], B(a, b) −

which implies (3.13). (ii) From (3.10) and (3.1) we have ab F (a + 1, b + 1; a + b + 2; x) G (x) a + b+1 = lim− lim−  1 x→1 G0 (x) x→1 F (3/2, 3/2; 3; x) 8   1 1 ln + O(1) 8ab B(a + 1, b + 1) 1−x   = lim 1 1 x→1− a + b + 1 ln + O(1) B(3/2, 3/2) 1−x =

B(3/2, 3/2) Γ(a + b + 1) 8ab =π , a + b + 1 B(a + 1, b + 1) Γ(a)Γ(b)

lim− HG,G0 (x) = lim−

x→1

x→1

=π = which implies (3.14).

G (x) G0 (x) − G(x) G0 (x)

Γ(a + b + 1) 4 Γ(a + b + 1) − Γ(a)Γ(b) π Γ(a + 1)Γ(b + 1)

Γ(a + b + 1) (4ab − 1), Γ(a + 1)Γ(b + 1)

2

Proof of Proposition 3.3. Since Lemma 3.1 has shown that the function r → F (r)/F0 (r) is strictly increasing in r ∈ (0, 1) for (a, b) ∈ D11 and strictly decreasing for (a, b) ∈ D00 , we only discuss the monotonicity of the function r → F (r)/F0 (r) on (0, 1) for (a, b) ∈ D10 . Evidently, for fixed (a, b) ∈ D10 , from (3.8) we clearly see that there exists n0 ∈ N such that Tn = (ab − 1/4)n + ab − (a + b)/4 ≤ 0 for 0 ≤ n ≤ n0 and Tn ≥ 0 for n ≥ n0 , which implies that the sequence {Fˆn /Fˆ0n } is decreasing for 0 ≤ n ≤ n0 and increasing for n ≥ n0 . Therefore, Proposition 3.3 directly follows from Theorem 2.1 and (3.13). 2 Proof of Proposition 3.5. In the cases of (a, b) ∈ E11 and (a, b) ∈ E00 , we clearly see that the sequence ˆ n /G ˆ 0n } is increasing and decreasing, respectively, and so is the function G/G0 by Theorem 1.1. {G

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If (a, b) ∈ E10 , that is, C3 (a, b) = ab + a + b − 5/4 ≥ 0 and C4 (a, b) = 2ab − (a + b)/4 − 1/4 ≤ 0, then there exists n0 ∈ N such that Tn = nC3 (a, b) + C4 (a, b) ≤ 0 for 0 ≤ n ≤ n0 and Tn ≥ 0 for n ≥ n0 . This ˆ n /G ˆ 0n } is decreasing for 0 ≤ n ≤ n0 and increasing for n ≥ n0 . indicates that the sequence {G Therefore, by Theorem 2.1 and (3.14), Proposition 3.5 follows. 2 3.3. Improvements of Theorems 2.1, 2.2 and 2.4 Since Corollary 2.3 in [19] can be deduced by their Theorem 2.2, while Theorem 2.5 in [19] is a consequence of Theorems 2.2 and 2.4 in [19], hence we only need to improve Theorems 2.1, 2.2 and 2.4. Employing our propositions, Theorem 2.1 in [19] can be corrected and improved as follows. Theorem 3.10. Let a, b > 0, then the following statements are true: ∗ , that is R(a, b) ≥ ln 16, then the inequality (i) If (a, b) ∈ D00 ∪ D10

F

 a, b; a + b;

4r (1 + r)2

 < (1 + r)F (a, b; a + b; r2 )

(3.15)

holds for each r ∈ (0, 1). (ii) The inequality (3.15) is reversed if (a, b) ∈ D11 . ∗ (iii) In the remain region, that is (a, b) ∈ D10 \ D10 , neither (3.15) nor its reversed inequality holds for all r ∈ (0, 1). Remark 3.11. Theorem 3.10 shows that the inequality (3.15) also holds on the region {1 ≤ 4ab ≤ a + b, R(a, b) ≥ ln 16}, which improve Simić and Vuorinen’s Theorem 2.1. Up to now, for the zero-balanced Gaussian hypergeometric functions F (a, b; a + b; x) (a, b > 0), we present the maximal regions in ab-plane where the Landen identities for the complete elliptic integral of the first kind turn to corresponding inequalities for each x ∈ (0, 1). Theorem 2.2 in [19] can be improved as follows. Theorem 3.12. Let a, b > 0. Then we have ∗ , that is, R(a, b) ≥ ln 16, then the double inequality (i) If (a, b) ∈ D00 ∪ D10

1≤

B(a, b) (1 + r)F (a, b; a + b; r2 ) ≤ F (a, b; a + b; 4r/(1 + r)2 ) π

(3.16)

holds for r ∈ (0, 1), where 1 and B(a, b)/π are the best constants and equalities hold if and only if a = b = 1/2. (ii) If (a, b) ∈ D11 , that is, a + b ≤ 4ab, then the double inequality (3.16) is reversed. ∗ (iii) If (a, b) ∈ D10 \ D10 , that is, 1 ≤ 4ab ≤ a + b and R(a, b) < ln 16, then it holds that  c0 min



B(a, b) ,1 π

<

1 (1 + r)F (a, b; a + b; r2 ) < max F (a, b; a + b; 4r/(1 + r)2 ) c0



 π ,1 , B(a, b)

where c0 = F (r0 )/F0 (r0 ), r0 is the unique solution of the equation (F (r)/F0 (r)) = 0 on (0, 1).

(3.17)

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17

Proof. The double inequality (3.16) and its reverse are the direct consequences of parts (i) and (ii) of Proposition 3.3. Now we use part (iii) to show the inequality (3.17). In the case of 1 < 4ab < a +b and R(a, b) < ln 16, we clearly see that there exists r0 ∈ (0, 1) such that F/F0 is strictly decreasing in r ∈ (0, r0 ) and strictly increasing in r ∈ (r0 , 1). Therefore, c0 =

F (r) F (r0 ) F (r) < < lim = 1 for r ∈ (0, r0 ), F0 (r0 ) F0 (r) r→0+ F0 (r)

F (r) π F (r) F (r0 ) < < lim = − F0 (r0 ) F0 (r) r→1 F0 (r) B(a, b)

for r ∈ (r0 , 1),

that is, F (r) < max c0 ≤ F0 (r)





π ,1 B(a, b)

for r ∈ (0, 1).

Let r ∈ (0, 1), x(r) = r2 and y(r) = 4r/(1 + r)2 . Then  max F (x(r)) F0 (y(r)) (1 + r)F (x(r))  < = < π F (y(r)) F0 (x(r)) F (y(r)) max B(a,b) ,1 c0



π B(a,b) , 1

c0

follows from the Landen identity, which implies (3.17). 2 Remark 3.13. In [19, Theorem 2.2] it only considered the regions D1 = {ab ≤ 1/4} and D2 = {a + b ≤ 4ab}, but in our Theorem 3.12 all the first quadrant in ab-plane is considered. By using our Proposition 3.8, Theorem 2.4 in [19] becomes more perfect. Theorem 3.14. Let a, b > 0, R(a, b) be defined by (3.3), and (a, b) ∈ D00 , that is ab ≤ 1/4. Then one has 0 ≤ (1 +

√

 √ R(a, b) − ln 16 4 r √ 2 ≤ . a, b; a + b; B(a, b) (1 + r)

 r)F (a, b; a + b; r) − F

(3.18)

The above inequalities are reversed if (a, b) ∈ D11 (that is, a, b > 0, a + b ≤ 4ab). Remark 3.15. In [19, Theorem 2.4], it was proved that inequality (3.18) holds for all a, b > 0 with a + b ≤ 1. But our Theorem 3.14 indicates that inequality (3.18) also holds on the region {1 < a + b} ∩ {ab ≤ 1/4}. Remark 3.16. In the same way, using our Theorem 2.1, Theorem 1 in [9] and Theorem 1.4 in [4] also can be improved. Acknowledgments The research was supported by the National Natural Science Foundation of China under Grants 61374086 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

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