Monotonicity rule for the quotient of two functions and its application

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1 Yang et al. Journal of Inequalities and Applications :106 DOI /s RESEARCH Open Access Monotonicity rule for the quotient of two functions and its application Zhen-Hang Yang1, Wei-Mao Qian, Yu-Ming Chu1* and Wen Zhang3 * Correspondence: chuyuming005@16.com Department of Mathematics, Huzhou University, Huzhou, , China Full list of author information is available at the end of the article 1 Abstract In the article, we provide a monotonicity rule for the function [Px + Ax]/[Px + Bx], where Px is a positive and decreasing function defined on R, R differentiable n n R > 0, and Ax = n=n0 an x and Bx = n=n0 bn x are two real power series converging on R, R such that the sequence {an /bn } n=n0 is increasing decreasing. As applications, we present new bounds with an0 /bn0 1 and bn > 0 for all n n0 π / 1 r sin t dt 0 < r < 1 of the second for the complete elliptic integral Er = 0 kind. MSC: 6A48; 33E05 Keywords: monotonicity rule; complete elliptic integral; absolute error; relative error 1 Introduction The most commonly used monotonicity rule in elementary calculus is that f is increasing decreasing on [a, b] if f : [a, b] R is continuous on [a, b] and has a positive negative derivative on a, b, and it can be proved easily by the Lagrange mean value theorem. The functions whose monotonicity we prove in this way are usually polynomials, rational functions, or other elementary functions. But we often find that the derivative of a quotient of two functions is quite messy and the process is tedious. Therefore, the improvements, generalizations and refinements of the method for proving monotonicity of quotients have attracted the attention of many researchers. In, Biernacki and Krzyż [ ] see also [ ], Lemma., [ ] found an important criterion for the monotonicity of the quotient of power series as follows. k k Theorem. [ ] Let At = k= ak t and Bt = k= bk t be two real power series converging on r, r r > with bk > for all k. If the non-constant sequence {ak /bk } is increasing decreasing for all k, then the function t At/Bt is strictly increasing decreasing on, r. In [ ], Cheeger et al. presented the monotonicity rule for the quotient of two functions. Theorem. [ ] If f and g are positive integrable functions on R such that f /g is de x x creasing, then the function x f t dt/ gt dt is decreasing. The Authors 017. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original authors and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Yang et al. Journal of Inequalities and Applications :106 Page of 13 Unaware of Theorem 1., Anderson et al. [5], Lemma. see also [6], Theorem 1.5 established l Hôpital s monotone rule that can be applied to a wide class of quotients of functions. Theorem 1.3 [5] Let < a < b <, f, g :[a, b] R be continuous on [a, b] and differentiable on a, b, and g x 0on a, b. If f x/g x is increasing decreasing on a, b, then so are the functions f x f a gx ga, f x f b gx gb. If f x/g x is strictly monotone, then the monotonicity in the conclusion is also strict. Pinelis [7] provided the following monotonicity theorem. Theorem 1.4 [7] Let f and g be differentiable and g nevervanishonanopeninterval a, b R. Then the following statements are true: 1 If gg >0on a, b, lim sup x a g xf x/gx / g x 0 and f /g is increasing on a, b, then f /g >0on a, b. If gg >0on a, b, lim inf x a g xf x/gx / g x 0 and f /g is decreasing on a, b, then f /g <0on a, b. 3 If gg <0on a, b, lim inf x b g xf x/gx / g x 0 and f /g is increasing on a, b, then f /g <0on a, b. 4 If gg <0on a, b, lim sup x b g xf x/gx / g x 0 and f /g is increasing on a, b, then f /g >0on a, b. Recently, Yang et al. [8], Theorem 1., established a more general monotonicity rule for the ratio of two power series. Theorem 1.5 [8] Let At= k=0 a kt k and Bt= k=0 b kt k be two real power series converging on r, r and b k >0for all k, and H A,B = A B/B A. Suppose that for certain m N, the non-constant sequence {a k /b k } is increasing decreasing for 0 k manddecreasing increasing for k m. Then the function A/B is strictly increasing decreasing on 0, r if and only if H A,B r 0.Moreover, if H A,B r < >0, then there exists t 0 0, r such that the function A/B is strictly increasing decreasing on 0, t 0 and strictly decreasing increasing on t 0, r. The foregoing monotonicity rules have been used very effectively in the study of special functions [9 3], differential geometry [4, 4], probability [5] and approximation theory [6]. The main purpose of the article is to present the monotonicity rule for the function [Px+ a n x n ]/[Px+ b n x n ] and to provide new bounds for the complete elliptic integral of the second kind. Some complicated computations are carried out using Mathematica computer algebra system. Monotonicity rule Theorem.1 Let Px be a positive differentiable and decreasing function defined on 0,r r >0,let Ax= a n x n and Bx= b n x n be two real power series converging on

3 Yang et al. Journal of Inequalities and Applications :106 Page 3 of 13 r, r. If a n0 /b n0 1,b n >0for all n n 0 and the non-constant sequence {a n /b n } is increasing decreasing, then the function x [Px+Ax]/[Px+Bx] is strictly increasing decreasing on 0,r. Proof Let x 0, r, and [ I 1 = npx xp x ] a n b n x n 1, I = na n x n 1 b n x n a n x n nb n x n 1..1 Then differentiating [Px+Ax]/[Px+Bx] gives [ [ ] Px+Ax Px+Bx Px+Bx = P x+ na n x n 1 Px+ b n x n Px+ a n x n P x+ nb n x n 1 ] = I 1 + I.. Note that I can be rewritten as I = = = = 1 ja j x j 1 b i x i a i x i jb j x j 1 j=n 0 i=n 0 i=n 0 j=n 0 aj jb i b j a i x i+j 1 b j=n j b i 0 i=n 0 j=n 0 i=n 0 ib j b i i=n 0 ai j=n 0 b i b j i j a j x i+j 1 b i b j ai a j x i+j 1..3 b i b j If a n0 /b n0 1,b n >0foralln n 0 and the non-constant sequence {a n /b n } is increasing decreasing, then we clearly see that a n b n.4 for all n n 0 and i=n 0 j=n 0 b i b j i j for all x 0, r. a j x i+j 1 > < 0.5 b i b j ai

4 Yang et al. Journal of Inequalities and Applications :106 Page 4 of 13 It followsfrom Px is a positive differentiable and decreasing function on 0, r that npx xp x>0.6 for all x 0, r. Therefore, [Px+Ax/Px+Bx] > < 0 for all x 0, r follows easily from.1-.6, and the proof of Theorem.1 is completed. 3 Bounds for the complete elliptic integral of the second kind For r 0, 1, Legendre s complete elliptic integral [7] of the second kind is given by Er= π/ 0 1 r sin t dt. It is well known that E0 + =π/, E1 =1,andEr is the particular case of the Gaussian hypergeometric function Fa, b; c; x= n=0 a n b n c n x n 1 < x <1, where a n = Ɣa + n/ɣaandɣx= 0 tx 1 e t dt x > 0 is the gamma function. Indeed, we have Er= π F 1, 1 ;1;r = π n=0 1 n 1 n r n. 3.1 Recently, the bounds for the complete elliptic integral Er of the second kind have been the subject of intensive research. In particular, many remarkable inequalities for Ercan be found in the literature [8 41]. Vuorinen [4] conjectured that the inequality 1+r 3/ /3 3. Er π holds for all r 0, 1, where, and in what follows, r =1 r 1/.Inequality3.wasproved by Barnard et al. in [43]. Very recently, the accurate bounds for Er in terms of the Stolarsky mean S p,q 1, r were given in [44, 45]: π S 11/4,7/4 1, r < Er< 11 7 S 11/4,7/4 1, r, S 5/, 1, r < Er< π S 5/, 1, r, 3.4 where S p,q a, b=[qa p b p /pa q b q ] 1/p q. In this section, we shall use Theorem.1 to present new bounds for the complete elliptic integral Er of the second kind. In order to prove our main result, we need three lemmas, which we present in this section.

5 Yang et al. Journal of Inequalities and Applications :106 Page 5 of 13 Lemma 3.1 see [46], Lemma 7 Let n N and m N {0} with n > m, a i 0 for all 0 i n, a n a m >0and P n t= m a i t i + i=0 n a i t i. i=m+1 Then there exists t 0 0, such that P n t 0 =0,P n t<0for t 0, t 0 and P n t>0for t t 0,. Lemma 3. see [47, 48] Thedoubleinequality 1 Ɣx + a < x + a 1 a Ɣx +1 < 1 x 1 a holds for all x >0and a 0, 1. Lemma 3.3 Let n N, p 1 n, p n, p 3 n and w n be defined by p 1 n=15n 4 +1,48n 3 +1,601n 7,53n +1,944, 3.5 p n=4,375n 6 +13,47n 5 + 1,696,697n 4 4,433,167n 3 1,571,83n + 4,66,360n 1,555,00, 3.6 p 3 n=65n 6 +,583n 5 304,413n 4 +34,181n 3 +1,906,38n,918,064n + 675,360, w n = 648np 1 n + p n + 5np 3n 3, 4 n n 3 n 5 4 n respectively. Then w n 0 for all n 3. Proof Let p 0 n, p 4 n, p 5 n, α n and β n be defined by p 0 n=65n 4,018n 3 +3,985n 4,03n +1,080, 3.9 p 4 n=875n 7 81,18n 6,341,894n 5 10,10,98n ,839,719n 3,615,904n +73,667,340n 71,971,00, 3.10 p 5 n=50n 6 1,834n ,75n ,000n 3 7,770,415n + 7,980,844n +98,140, 3.11 α n = p 4n 1, 3.1 nn +1 n 3 β n = 10n 1p 5n 3, 3.13 n 3n 5 4 n 3 respectively. Then from and elaborated computations we get w 3 = w 4 = w 5 = w 6 =0, 3.14

6 Yang et al. Journal of Inequalities and Applications :106 Page 6 of 13 w 7 =,848,355 > 0, w n+1 n +1p 1 n +1 n 7/4w n np 1 n = p n +1 n +1p 1 n +1 1 n 7/4p n n np 1 n 1 5p 3 n n 7/4p 3n n 3p 1 n +1 4 n n 5p 1 n [ n 5/p n +1 = n 7/4p ] n 1 n +1p 1 n +1 np 1 n n 3 [ 5n 9/4p3 n n 7/4p ] 3n 3 n 3p 1 n +1 n 5p 1 n 4 = p 0 n +1p 4 n 4nn +1p 1 np 1 n n 3 5n 1p 0 n +1p 5 n n 3n 5p 1 np 1 n n 3 p 0 n +1 = 4p 1 np 1 n +1 α n + β n, 3.16 α n = n 3n 5p 4n 1 n 3 β n 10nn 1n +1p 5 n 3 4, n α 7 = 7,313,056 β 7 7,313,875, n 3 αn+1 1 α n n 1 β n+1 β n = n 1n 3n 5/p 4n +1 nn +1n +n 9/4p 5 n +1 n 3n 5p 4n nn 1n +1p 5 n = n 3n 5n 3n 415n4 +1,98n 3 +6,797n +616n,380 n4n 9n 1n +1n +p 5 np 5 n +1 n ,750n 8 80,64n 7 54,353,513n ,7,431n 5 4,748,597,075n 4 + 1,568,863,989n 3 80,185,046,0n + 144,08,55,644n 85,5,499,160 = n 3n 5n 3n 415n4 +1,98n 3 +6,797n +616n,380 n4n 9n 1n +1n +p 5 np 5 n +1 [ 40,331,641,10 714,515,,844n 7 475,749,995,856n 7 n 3 15,56,681,341n 7 3 5,64,55,865n 7 4,63,535,771n ,63,449n ,64n ,750n 7 8] n 3 From Lemma 3.1 and 3.19 together with the facts that p 5 n>0andp 5 n +1>0for n 7, we clearly see that there exists n 0 > 7 such that the sequence {α n /β n } n=7 is increasing

7 Yang et al. Journal of Inequalities and Applications :106 Page 7 of 13 for 7 n n 0 and decreasing for n n 0, which implies that { α n α7 min, lim β n β 7 n α n β n It follows from Lemma 3. that }. 3.0 Ɣ 3 4 n 5 1/4 Ɣ 1 < 1 n = Ɣ 3 4 Ɣn n 3 Ɣ 1 Ɣn < Ɣ 3 13 n 4 4 1/4 Ɣ for all n 7. From 3.10, 3.11, 3.17, 3.18, 3.0and3.1weget α n β n α 7 β 7 > 1 3. for all n 7. Therefore, Lemma 3.3 followseasily from and3.together with the facts that p 1 n>0,p 0 n +1>0andp 1 n +1>0forn 7. Theorem 3.4 Thedoubleinequality 40π J r 51π 160 < Er< π 9 58 J r 3.3 holds for all r 0, 1, where J r = 51r +0r r +50r +0 r r r +5 Proof Let r 0, 1, x = r, Px, f 1 r, f randfrbedefinedby Px = 1,536 1,48x, 3.5 f 1 r=,59 r 1 r 3/ r 4 64r r 1/ 50 r 4 96r r 1/4 105r 4 +3,600r +,880, 3.6 f r=65r 4 384r +384, 3.7 Fr= 1 Jr 1 Er/π, 3.8 respectively. Then from 3.1and wehave f 1 r=,880+3,600r 105r 4 +, n n! rn 64 n= n=1 n=0 3 4 n r n n=1 1 n 1 n 1! rn n 1 n 1! rn n=0 1 n r n

8 Yang et al. Journal of Inequalities and Applications :106 Page 8 of n= 1 4 n n! rn 96 n=1 = 6,144 7,680r 105r 4 +, n= n= 1 4 n 1 n 1! rn +96 n= 7n 367n 70 1 n r n n 11n n r n = 6,144 7,680r + 11,48r 4 1, f r f 1 r =1,536r 1,48r =1,536r 1,48r 4 + n=0 n n 1 r n 7n 367n 70 1 n 3 r n n 11n n 3 r n, n n r n n 11n n 3 r n 7n 367n 70 1 n 3 r n +1,944 u n r n, 1 4 n r n n n r n where u n = 1 π Er 97n n 157n 367n 70 1 = n 1 n r n = n=1 n=1 16f r 1 π Er 1 5n 11n +70 n 3 1 n 1 1 n r n, =8 65r 4 384r n 1 1 n =1,536r 1,48r 4 + n=1 v n r n, r n 3, n 3

9 Yang et al. Journal of Inequalities and Applications :106 Page 9 of 13 where v n = 8p 0n 1 n 3 1 n 3.30 and p 0 nisdefinedby3.9. J r = 51r +0r r +50r +0 r + 515r r + 55r 46r r + r + 55r r + 55r 46r +5 = f 1r 16f r, 16f r f 1 r Px+ n= Fr= 16f r1 = u n+1x n Er Px+ π n= v n+1x n It follows from 3.9, 3.9, 3.30 and elaborated computations that u 3 =1, v u n+1 v n+1 15n 5 u n = v n 8n +1p 0 nn +1! w n, 3.33 where w n is defined by 3.8. It is not difficult to verify that p 0 n=65n 4,018n 3 +3,985n 4,03n +1,080> for all n 3. From Lemma 3.3,3.30, 3.33and3.34weknowthat v n > for all n 3, and the sequence {u n /v n } is decreasing. Equation 3.5impliesthat Px> for x 0, 1, and Px is decreasing on 0, 1. It follows from Theorem.1, 3.31 and3.3 together with the monotonicity of the sequence {u n /v n } and the function Px on 0, 1 that the function Fr isstrictlydecreasing on 0, 1 and lim r 1 Fr<Fr< lim Fr r 0 for all r 0, 1. Note that 3.4, 3.8and3.31leadtotheconclusionthat lim Fr= 1 J0+ r 1 1 E1 /π = 1 51/80 1 /π = 9π 80π, lim Fr= r 0 +

10 Yang et al. Journal of Inequalities and Applications :106 Page 10 of 13 Therefore, Theorem 3.4 followsfrom 3.8, 3.37and3.38. Remark 3.5 Let λ 1 r= π S 11/4,7/4 1, r, μ 1 r= 11 7 S 11/4,7/4 1, r, 3.39 λ r= 5 16 S 5/, 1, r, μ r= π S 5/, 1, r, 3.40 λr= 40π J r 9 51π 160, μr= π 58 J r, 3.41 where Jr isdefinedby3.4. Then simple computations lead to λ 1 1 = 7π = , μ = 11 = , λ 0 + = 5 16 =1.565, μ 1 = π = , λ 0 + = π = μ 0 + = , 3.44 λ 1 =1, μ 1 = 51π = From 3.3, 3.4, 3.3 and we clearly see that there exists small enough δ 0, 1 such that the lower bound given in 3.3forEr is better than the lower bound given in 3.3 forr δ,1 δ, the lower bound given in 3.3 forer is better than the lowerbound given in 3.4 for r 0,δ, the upper bound given in 3.3 for Er is better than the upper bound given in 3.3for r 0, δ, and the upper bound given in 3.3for Er is better than the upper bound given in 3.4forr δ,1 δ. Corollary 3.6 Let Jr be defined by 3.4. Then the double inequality π J r 51π < Er< π J r 3.46 holds for all r 0, 1. Proof Let Frbedefinedby3.8and Ar=J r Er π Then we clearly see that A 0 + = J 1 π E 0 + =0, A 1 = J 0 + π E 1 51π 160 =, π Ar= [1 π ][ ] [ Er 1 1 Jr 1 π Er = 1 ] [1 Fr π Er ] From 3.49 and the proof of Theorem 3.4 we know that Fr is strictly decreasing on 0, 1 and Ar is strictly increasing on 0, 1. Therefore, inequality 3.46 follows from 3.47 and3.48 together with the monotonicity of Ar on the interval0,1.

11 Yang et al. Journal of Inequalities and Applications :106 Page 11 of 13 Corollary 3.7 Let Jr be defined by 3.4. Then the double inequality J r < Er< π J r 3.50 holds for all r 0, 1. Proof Let Arbedefinedby3.47and Br= Jr 3.51 Er. π Then we clearly see that B 0 + = J1 π E0+ =1, B 1 = J0+ π E1 = 51π 160, Br= [J r π ] Er +1= Ar Er π Er π From 3.53 and the proof of Corollary 3.6 we know that both Ar andbr arestrictly increasing on 0, 1. Therefore, inequality 3.50 follows from 3.51 and3.5 together with the monotonicity of Brontheinterval0,1. Remark 3.8 FromCorollaries3.6and3.7we have Er π J r < 51π 160 1= , Er π Jr Er < 51π 160 1= for all r 0, 1, which implies that both the absolute and relative errors using πjr / to approximate Erarelessthan0.14%. 4 Conclusions In this paper, we find a monotonicity rule for the function [Px + a n x n ]/[Px + b n x n ]. As applications, we present new bounds for the complete elliptic integral Er = π/ 0 1 r sin tdt 0 < r < 1 of the second kind, and we show that our bounds are sharper than the previously known bounds for some r 0, 1. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, Huzhou University, Huzhou, , China. School of Distance Education, Huzhou Broadcast and TV University, Huzhou, , China. 3 Friedman Brain Institute, Icahn School of Medicine at Mount Sinai, New York, 1009, USA. Acknowledgements The research was supported by the Natural Science Foundation of China Grants Nos , , , , the Tianyuan Special Funds of the National Natural Science Foundation of China Grant No and the Natural Science Foundation of the Department of Education of Zhejiang Province Grant No. Y

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