Research Article On Jordan Type Inequalities for Hyperbolic Functions

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1 Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 010, Article ID 6548, 14 pages doi: /010/6548 Research Article On Jordan Type Inequalities for Hyperbolic Functions R. Klén, M. Visuri, and M. Vuorinen Department of Mathematics, University of Turku, 0014 Turku, Finland Correspondence should be addressed to R. Klén, Received 8 January 010; Accepted 9 April 010 Academic Editor: Andrea Laforgia Copyright q 010 R. Klén et al. 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. This paper deals with some inequalities for trigonometric and hyperbolic functions such as the Jordan inequality and its generalizations. In particular, lower and upper bounds for functions such as sin / and / sinh are proved. 1. Introduction During the past several years there has been a great deal of interest in trigonometric inequalities 1 7. The classical Jordan inequality 8, page 1 π sin, 0 <<π 1.1 has been in the focus of these studies and many refinements have been proved for it by Wu and Srivastava 9, 10, Zhang et al. 11, J.-L. Li and Y.-L. Li 5, 1,WuandDebnath1 15, Özban 16,Qietal.17,Zhu18 9,Sándor 0, 1,BariczandWu,, Neuman and Sándor 4, Agarwal et al. 5,Niuetal.6, Pan and Zhu 7, and Qi and Guo 8. For a long list of recent papers on this topic see 7 and for an etensive survey see 17. The proofs are based on familiar methods of calculus. In particular, a method based on a l Hospital type criterion for monotonicity of the quotient of two functions from Anderson et al. 9 is a key tool in these studies. Some other applications of this criterion are reviewed in 40, 41. Pinelis has found several applications of this criterion in 4 and in several other papers. The inequality 1 cos sin cos, 1. where π/,π/ is well-known and it was studied recently by Baricz in 4, page 111.

2 Journal of Inequalities and Applications The second inequality of 1. is given in 8, page 54,.9. for 0 π.for a refinement of the first inequality in 1. see Remark 1.1 and of the second inequality see Theorem.4. This paper is motivated by these studies and it is based on the Master Thesis of Visuri 44. Some of our main results are the following theorems. Theorem 1.1. For 0,π/ sinh < sin < sinh. 1. Theorem 1.. For 0, 1 ( ) 1 1/ ( ) 1 1/4 < cosh sinh <. 1.4 cosh We will consider quotients sin / and /sinh at origin as limiting values lim 0 sin / 1 and lim 0 / sinh 1. Remark Let g 1 1 cos Then g 1 g > 0on0,π/ because sin, g 6 1 cos. 1.5 d ( g1 g ) cos > d 6 In 45, 7 it is proved that for 0,π/ sin g Hence 1 cos / sin /6 is a better lower bound for sin / than 1. for 0,π/. Observe that cos cos / 1 cos, 1.8 which holds true as equality if and only if cos/ ± 1/4. In conclusion, 1.8 holds for all π/, π/. Together with 1. we now have cos 1 cos > cos, 1.9

3 Journal of Inequalities and Applications and by 1.8. Jordan s Inequality cos < sin < cos In this section we will find upper and lower bounds for sin / by using hyperbolic trigonometric functions. Theorem.1. For 0,π/ 1 cosh < sin < sinh..1 Proof. The lower bound of sin / holds true if the function f sin cosh is positive on 0,π/. Since f cos sinh,. we have f > 0for 0,π/ and f is increasing on 0,π/. Therefore f cos cosh sin sinh 1 >f 0 0,. and the function f is increasing on 0,π/. Nowf >f0 0for 0,π/. The upper bound of sin / holds true if the function g sin sinh is positive on 0,π/. Let us denote h tan tanh. Since cos <1 < cosh for 0,π/ we have h cosh cos >0andh >h0 0for 0,π/. Now g cos cosh h,.4 which is positive on 0,π/, because cos cosh >0andh > 0for 0,π/. Therefore g 1 cos cosh >g0 0, g cos sinh sin cosh >g for 0,π/. Nowg >g0 0for 0,π/. Proof of Theorem 1.1. The upper bound of sin / is clear by Theorem.1. The lower bound of sin / holds true if the function f sin sinh is positive on 0,π/. Let us assume 0,π/. Since sin > /6 6 /6 we have f > 6 sinh /6. We will show that g 6 sinh 6.6 is positive which implies the assertion.

4 4 Journal of Inequalities and Applications Now g > 0 is equivalent to sinh > Since 1 sinh > 1 /6itissufficient to show that 1 /6 > 6/ 6, which is equivalent to ( ) > 0..8 Let us denote h Now h 4 and therefore h / 0and h > min{h0,hπ/} > 0. Therefore inequality.8 holds for 0,π/ and the assertion follows. We net show that for 0, 1 the upper and lower bounds of 1. are better than the upper and lower bounds in Theorem.1. Theorem.. i For 1, 1 cos sinh..9 ii For π/,π/ 1 1 cos cos cosh..10 iii For π/,π/ 1 1 sin 1 cos 1 cos..11 Proof. i The claim holds true if the function f sinh sinh cos is nonnegative on 0, 1. By a simple computation we obtain f cosh sin sinh. Inequality f 0 is equivalent to sin tanh. By the series epansions of sin and tanh we obtain sin tanh n,n 1 mod n,n 1 mod 1 n 1/ n 1 n1( n1 1 ) B n1 n n 1! c n n,.1

5 Journal of Inequalities and Applications 5 where B j is the jth Bernoulli number. By the properties of the Bernoullin numbers c 1 1/6, c 1/8, coefficients c n,forn 1, form an alternating sequence, c n n 0asn and c m1 > c m for m 1. Therefore by Leibniz Criterion sin tanh ( 4 ).1 4 and sin tanh for all 0, 1. Nowf is a conve function on 0, 1 and f is nondecreasing on 0, 1 with f 0 0. Therefore f is nondecreasing and f f0 0. ii The claim holds true if the function g cosh 1 cos is nonnegative on 0,π/. By the series epansion of cos we have cos 1 / 0 and therefore by the series epansion of cosh ( ) g 1 1 cos cos 1 cos.14 cos 1 0, and the assertion follows. iii Clearly we have ( )( ) 1 cos 1 sin sin cos,.15 which implies the first inequality of the claim. The second inequality is trivial since cos 0, 1. Theorem.. Let 0,π/. Then i the function ft cos t t.16 is increasing on 1,, ii the function gt sin t t.17 is decreasing on 1,, iii the functions ft cosh t /t and gt sinh t /t are decreasing on 0,.

6 6 Journal of Inequalities and Applications Proof. i Let us consider instead of f the function ( ) f 1 y log cos y.18 y for y 0,. Notethatft epf 1 /t and therefore the claim is equivalent to the function f 1 y being decreasing on 0,. We have f ( ) ( ) 1 y log cos y y tan y, y.19 and f 1 y 0 is equivalent to f y log cos y y tan y 0. Since f y y/cos y 0we have f y f 0 0. Therefore ft is increasing on 1,. ii We will consider instead of g the function ( ) g 1 y log sin y.0 y for y 0,.Notethatgt epg 1 /t and therefore the claim is equivalent to the function g 1 y being increasing on 0,. We have ( ) g ( ) y 1 y log cos y, y.1 tan y and g 1 y 0 is equivalent to g y y/ tan y log cos y 0. Since g y 1/ cos y y/ sin y/ sin y 0 we have g y f 0 1. Therefore g 1 y 0 and the assertion follows. iii We will show that h 1 y /y log cosh y is increasing on 0,. Nowh 1y y tanh y log cosh y/y, d ( y tanh y log cosh y ) dy y cosh y > 0,. and y tanh y log cosh y 0. Therefore the function h 1 y is increasing on 0, and ft is decreasing on 0,. We will show that h y /y log sinh y is increasing on 0,. Nowh y y coth y log sinh y/y, d ( y coth y log sinh y ) y dy sinh y < 0,. and coth y log sinh y/y lim y coth y log sinh y/y 0. Therefore the function h y is increasing on 0, and gt is decreasing on 0,. We net will improve the upper bound of 1..

7 Journal of Inequalities and Applications 7 Theorem.4. For 7/5, 7/5 cos sin cos ( ) cos..4 Proof. The first inequality of.4 follows from 1.. By the series epansions of sin and cos sin 1 6 ( ) ( ) cos,.5 18 where the second inequality is equivalent to / and the second inequality of.4 follows. By the identity cos cos cos /4 the upper bound of.4 is equivalent to 0 8 cos 9 cos/. By the series epansion of cos ( ) 8 cos 9 cos 1 n n 9 n n! n,.6 n and by the Leibniz Criterion the assertion follows.. Hyperbolic Jordan s Inequality In this section we will find upper and lower bounds for the functions /sinh and cosh. Theorem.1. For π/,π/ 1 6 sin 1 π..1 Proof. We obtain from the series epansion of sin sin 1 6,. which proves the lower bound. By using the identity 1 cos sin / the chain of inequalities 1. gives sin 1 sin /. and the assertion follows from inequality sin / /π.

8 8 Journal of Inequalities and Applications Remark.. J.-L. Li and Y.-L. Li have proved 1, 4.9 that sin ( ) <p 1 π < 1, 0 <<π,.4 π where p 1/ 1 /π 4 < 1. This result improves Theorem.1. Lemma.. For 0, 1 i sinh < /5, ii cosh <15 /9, iii 1/ cosh <1 /. Proof. i For 0, 1 we have 1 > 0 which is equivalent to 1 < 1 1 / By Theorems.1,.1,and.5 sinh sin < 1 / ii Since n! > 6 n for n we have cosh 18 n n! n 18 n! n ( ) n! n ( ) n n.7 > 0. iii By the series epansion of cosh we have ( ) ( )( ) cosh >

9 Journal of Inequalities and Applications 9 Proof of Theorem 1.. The lower bound of / sinh follows from Lemma. and Theorem.1 since ( ) ( ) 1 cosh < 1 < sinh The upper bound of / sinh holds true if the function g sinh 4 4 cosh is positive on 0, 1. By the series epansion it is clear that By Lemma. and.10 sinh > ( ) 4 ( ) g > ( ) > 0, and the assertion follows. Theorem.4. For 0,π/4 cosh < cos cos sin..1 Proof. The upper bound of cosh holds true if the function f cos cosh cos sin is positive on 0,π/4. Since f 4sin sinh > 0,.1 we have f sin sinh cos cosh >f Therefore f >f0 0 and the assertion follows. Theorem.5. For 0,π/4 1 cos / < cosh < 1 cos..15 Proof. The upper bound of cosh holds true if the function f 1 cos cosh is positive on 0,π/4. Since f sin sinh >0 the function f cosh sin cos sinh is increasing. Therefore f >f 0 0andf >f0 0.

10 10 Journal of Inequalities and Applications The lower bound of cosh holds true if the function g cos cosh 1 is positive on 0,π/4. By the series epansions we have ( ) ( g > 1 1 ) 1 ( ) By a straightforward computation we see that the polynomial h is strictly decreasing on 0,π/4. Therefore h >hπ/4 16 π π4 π6 π > > 0, and the assertion follows. Remark.6. Baricz and Wu have shown in, page that the right hand side of Theorem.1 is true for 0,π and the right hand side of Theorem.5 is true for 0,π/. Their proof is based on the infinite product representations. Note that for 0,π/4 1 cos cos. cos sin.18 Hence, the upper bound in Theorem.5 is better that in Theorem Trigonometric Inequalities Theorem 4.1. For 0, 1 the following inequalities hold i / arcsin sin /, ii /arcsinh sinh /, iii / arctan tan /, iv /arctanh tanh /. Proof. i By setting sin t the assertion is equivalent to sinc t sinc sin t, 4.1 which is true because sinc t sin t/t is decreasing on 0,π/ and sin t t.

11 Journal of Inequalities and Applications 11 ii By the series epansions of sinh and arcsinh we have by Leibniz Criterion ( )( sinh arcsinh 6 6 ) ( ) , and since > 77 on 0, 1 the assertion follows. iii By the series epansions of tan and arctan we have by Leibniz Criterion ( tan arctan ( ), 945 )( ) and since > 4on0, 1 the assertion follows. iv By the series epansions of tanh and arctanh we have by Leibniz Criterion ( )( tanh arctanh 6 ( 4 ), 45 ) and since 4 > 1on0, 1 the assertion follows. Remark 4.. Similar inequalities to Theorem 4.1 have been considered by Neuman in 46, page 4-5. Theorem 4.. Let k 0, 1. Then i for 0,π sin sink, 4.5 k ii for >0 sinh sinhk, 4.6 k iii for 0, 1 tanh tanhk. 4.7 k

12 1 Journal of Inequalities and Applications Proof. i The claim follows from the fact that sinc is decreasing on 0,π. ii The claim is equivalent to saying that the function f sinh / is increasing for >0. Since f cosh / sinh / 0andf 0 is equivalent to tanh the assertion follows. iii The claim is equivalent to tanhk k tanh 0. By the series epansion of tanh we have 4 n1( 4 n1 1 ) B n1 n1 ( ) tanhk k tanh k k n 1, 4.8 n! n1 where B j is the jth Bernoulli number B 0 1, B 1 1/, B 1/6,... The assertion follows from the Leibniz Criterion, if k k ( k k 5) 5 > for all 0, 1. Since 4.9 is equivalent to < 5 1 k, 4.10 the assertion follows from the assumptions k 0, 1 and 0, 1. References 1 Á. Baricz, Redheffer type inequality for Bessel functions, Journal of Inequalities in Pure and Applied Mathematics, vol. 8, no. 1, article 11, pp. 1 6, 007. Á. Baricz, Jordan-type inequalities for generalized Bessel functions, Journal of Inequalities in Pure and Applied Mathematics, vol. 9, no., article 9, pp. 1 6, 008. L. Debnath and C.-J. Zhao, New strengthened Jordan s inequality and its applications, Applied Mathematics Letters, vol. 16, no. 4, pp , W. D. Jiang and H. Yun, Sharpening of Jordan s inequality and its applications, Journal of Inequalities in Pure and Applied Mathematics, vol. 7, no., article 10, pp. 1 4, J.-L. Li, An identity related to Jordan s inequality, International Journal of Mathematics and Mathematical Sciences, vol. 006, Article ID 7678, 6 pages, S.-H. Wu, H. M. Srivastava, and L. Debnath, Some refined families of Jordan-type inequalities and their applications, Integral Transforms and Special Functions, vol. 19, no. -4, pp , L. Zhu and J. Sun, Si new Redheffer-type inequalities for circular and hyperbolic functions, Computers & Mathematics with Applications, vol. 56, no., pp. 5 59, D. S. Mitrinović, Analytic Inequalities, vol. 16 of Die Grundlehren der mathematischen Wissenschaften, Springer, New York, NY, USA, 1970, in cooperation with P. M. Vasić. 9 S.-H. Wu and H. M. Srivastava, A weighted and eponential generalization of Wilker s inequality and its applications, Integral Transforms and Special Functions, vol. 18, no. 7-8, pp , S.-H. Wu and H. M. Srivastava, A further refinement of a Jordan type inequality and its application, Applied Mathematics and Computation, vol. 197, no., pp , X. Zhang, G. Wang, and Y. Chu, Etensions and sharpenings of Jordan s and Kober s inequalities, Journal of Inequalities in Pure and Applied Mathematics, vol. 7, no., article 6, pp. 1, J.-L. Li and Y.-L. Li, On the strengthened Jordan s inequality, Journal of Inequalities and Applications, vol. 007, Article ID 748, 8 pages, 007.

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