Wilker and Huygens type inequalities for mixed trigonometric-hyperbolic functions
Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo, Vinay M. Raut
aa r X i v : . [ m a t h . G M ] S e p Wilker and Huygens type inequalities for mixedtrigonometric-hyperbolic functions
Yogesh J. Bagul , Ramkrishna M. Dhaigude , Barkat A. Bhayo ∗ , Vinay M. Raut Department of Mathematics,K. K. M. College, Manwath,Dist: Parbhani(M. S.) - 431505, India Department of Mathematics,Government Vidarbha Institute of Scienceand Humanities, Amravati(M. S.)-444604, India ∗ Department of Mathematics, Sukkur IBA University,Sindh, Pakistan Department of Mathematics,Shri. Shivaji Science College, Amravati(M. S.)-444603, India
Abstract
Motivated by the work of Sándor [19], in this paper we establish a new Wilker type andHuygens type inequalities involving the trigonometric and hyperbolic functions. Moreover,in terms of hyperbolic functions, the upper and lower bounds of sin( x ) /x and tan( x ) /x are given. Mathematics Subject Classification (2010).
Keywords.
Wilker-type inequality, Huygens-type inequality, trigonometric-hyperbolicfunctions.
1. Introduction
In the last two decades, the refinements of the inequalities involving trigonometricand hyperbolic functions such as Wilker type inequalities and Huygens type inequalitieshave been studied by several authors e.g., see [4–7, 10, 17, 19, 22–24] and the referencestherein. Motivated by the work of Sándor [19], and above studies, in this paper wemake a contribution to the subject by establishing a new Wilker type and Huygens typeinequalities involving the trigonometric and hyperbolic functions. In all cases, we give theupper and lower bounds of sin( x ) /x and tan( x ) /x in terms of elementary functions.For 0 < x < π/
2, Wilker [20] and Huygens [9] proposed the following inequalities (cid:18) sin xx (cid:19) + tan xx > , (1.1)2 sin xx + tan xx > , (1.2)respectively. In literature, inequality (1.1) and (1.2) are known as Wilker’s and Huygens’inequalities, respectively.In [24], Zhu proved the the hyperbolic version of (1.1) as follows, (cid:18) sinh xx (cid:19) + tanh xx > , x > , (1.3) ∗ Corresponding Author.Email addresses: [email protected], [email protected], [email protected],[email protected]
Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo
Similarly, Neuman and Sándor [16] established2 sinh xx + tanh xx > x > , (1.4)as a hyperbolic version of (1.2).Recently, Wu and Srivastava [21] proved that for 0 < x < π/ (cid:18) x sin x (cid:19) + x tan x > , (1.5)and the following hyperbolic version of (1.5) (cid:18) x sinh x (cid:19) + x tanh x > x > , (1.6)was proved by Wu and Debnath in [22].For 0 < x < π/
2, the following relations were established by Neuman and Sándor2 sin xx + tan xx > x sin x + x tan x > , (1.7)2 sinh xx + tanh xx > x sinh x + x tanh x > , (1.8)in [16] and [17], respectively. For 0 < x < π/
2, one can write thatsin xx + 2 tan xx > xx + tan xx , (1.9)because cos x < < x < π/ (cid:18) sinh xx (cid:19) + sin xx > , (1.10) (cid:18) x sin x (cid:19) + x sinh x > , (1.11)2 sinh xx + sin xx > , (1.12)and 2 x sin x + x sinh x > . (1.13)In [19], the inequalities (1.10), (1.11) are called as trigonometric-hyperbolic Wilker typeinequalities and the inequalities (1.12), (1.13) are called as trigonometric-hyperbolic Huy-gens type inequalities. The main purpose of this paper is to obtain some trigonometric-hyperbolic Wilker type and Huygens type inequalities.The paper is organized as follows. In this section, we give an introduction and highlightthe concerned previous inequalities together with the statements of main results in theform of theorems. Section 2 consists of preliminary ancestor and lemmas, which will beused in the proving procedures sequel. Section 3 closes the paper by proving the mainresults from Section 1.Theorems 1.1-1.2 and Theorems 1.3-1.4 are dealing with Wilker-type inequalities andHuygens-type inequalities, respectively. ilker and Huygens type inequalities for mixed trigonometric-hyperbolic functions Theorem 1.1.
For < x < π/ it is asserted that − x < x sin x + (cid:18) tanh xx (cid:19) < < x < (cid:18) sin xx (cid:19) + x tanh x . (1.14) Theorem 1.2.
The inequalities sinh xx + (cid:18) x tan x (cid:19) < − x < < x
45 tan x < (cid:18) sinh xx (cid:19) + x tan x (1.15) hold for all x ∈ (0 , π/ . Theorem 1.3.
Let < x < π/ . Then the following inequalities hold: xx + x tan x < − x < < x < x sinh x + tan xx (1.16) and sinh xx + 2 x tan x < − x < < x < x sinh x + 2 tan xx . (1.17) Theorem 1.4. If < x < π/ then the following inequalities hold: − x < xx + x tanh x < < x sin x + tanh xx < x , (1.18) and − x < x sin x + 2 tanh xx < < sin xx + 2 x tanh x < x . (1.19)
2. Preliminaries and Lemmas
In this section, we give a few series expansion formulas and lemmas, which will be usedin the proofs of our main result.For the following series expansions, we refer to [8, 1.411].cos x = ∞ X k =0 ( − k (2 k )! x k , cosh x = ∞ X k =0 x k (2 k )! , (2.1)sin xx = ∞ X k =0 ( − k (2 k + 1)! x k , sinh xx = ∞ X k =0 x k (2 k + 1)! , x = 0 , (2.2) x sin x = 1 + ∞ X k =1 k − − k )! | B k | x k , | x | < π, (2.3) x sinh x = 1 − ∞ X k =1 k − − k )! B k x k , | x | < π, (2.4) x cot x = 1 − ∞ X k =1 k | B k | (2 k )! x k , | x | < π, (2.5) x coth x = 1 + ∞ X k =1 k B k (2 k )! x k , | x | < π, (2.6) Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo tan x = ∞ X k =1 k (2 k − k )! | B k | x k − , | x | < π/ x + x x
15 + 17 x
315 + 62 x · · · (2.7)tanh x = ∞ X k =1 k (2 k − k )! B k x k − , | x | < π/ x − x x − x
315 + 62 x − · · · (2.8)Differentiating (2.8) we getsech x = 1 − tanh x = ∞ X k =1 k (2 k − k − k )! B k x k − , | x | < π/ . Hence (cid:18) tanh xx (cid:19) = − ∞ X k =2 k (2 k − k − k )! B k x k − , | x | < π/ . (2.9)The following lemmas are also necessary. Lemma 2.1. [10] For < x < π/ , we have − x < sin xx < x sinh x . Lemma 2.2. [4] For every x = 0 , we have x x + 15 < x tanh x . Lemma 2.3.
For < x < π/ , it is true that x tanh x < tan xx . Proof.
The famous Adamović-Mitrinović inequality [13]cos x < (cid:18) sin xx (cid:19) (0 < x < π/ (cid:18) x sin x (cid:19) < tan xx (0 < x < π/ . Combining this with the inequality [2, Proposition 1.2] x tanh x < (cid:18) x sin x (cid:19) (0 < x < π/ , we get the required inequality. (cid:3) Lemma 2.4. [1, p. 805] k )!(2 π ) k < | B k | < k )!(2 π ) k k − k − − , ( k = 1 , , , · · · ) where B , B , B , · · · are Bernoulli numbers. Lemma 2.5.
The inequality | B k | > k − (2 k + 1) = 2 − k (2 k + 1) holds for all integers k ≥ . ilker and Huygens type inequalities for mixed trigonometric-hyperbolic functions Proof.
It is evident that (2 k + 1)! > π k ( k = 2 , , · · · )i.e., 1 π k > k + 1)! . This yields | B k | > (2 k )!2 k − π k > (2 k )!2 k − (2 k + 1)! = 2 − k (2 k + 1)due to Lemma 2.4. (cid:3) Lemma 2.6.
The inequality | B k | > k (2 k + 1) = 2 − k (2 k + 1) holds for all integers k ≥ . Proof.
Since, 2(2 k + 1)! > π k ( k = 1 , , , · · · ) , we get the desired result by applying the same argument as in the proof of Lemma 2.5. (cid:3) Lemma 2.7.
The inequality | B k +2 || B k | > ( k + 1)(2 k + 1)(2 k − − k − (2 k +2 − is true for all integers k ≥ . Proof.
From Lemma 2.4, we can write | B k +2 || B k | > k + 2)!(2 π ) k +2 (2 π ) k (2 k − − k )!2 k − = ( k + 1)(2 k + 1)(2 k − − k π > ( k + 1)(2 k + 1)(2 k − − k − (2 k +2 − , as 2 k π < k − (2 k +2 −
1) i.e., 2 π + 1 < k +2 for all integers k ≥ . (cid:3)
3. Proofs and corollaries
Proof of Theorem 1.1.
For the leftmost double inequality of (1.14), we add (2.3) and(2.9) to get x sin x + (cid:18) tanh xx (cid:19) = 1 + ∞ X k =1 k − − k )! | B k | x k − ∞ X k =2 k (2 k − k − k )! B k x k − , | x | < π/
2= 2 + ∞ X k =1 " k − − k )! | B k | − k +4 (2 k + 3)(2 k +4 − k + 4)! B k +4 x k = 2 + ∞ X k =1 a k x k := A ( x ) ! , | x | < π/ a k = k − − k )! | B k | − k +4 (2 k +3)(2 k +4 − k +4)! B k +4 ( k = 1 , , , · · · ) . Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo
Clearly a k > k = 2 , , , · · · . For k = 1 , , , , · · · we claim that a k < | B k || B k +4 | < k +1 (2 k +4 − k + 1)( k + 2)(2 k + 1) 1(2 k − −
1) = L ( k ) . From Lemma 2.4, we write | B k | < k )!2 k π k k − (2 k − −
1) and 1 | B k +4 | < k +4 π k +4 k + 4)! . Therefore | B k || B k +4 | < k + 1)( k + 2)(2 k + 1) 1((2 k − − π k +1 (2 k + 3) = M ( k ) . Since π < (2 k + 3)(2 k +4 −
1) ( k = 1 , , , · · · ) implies M ( k ) < L ( k ) , so a k < k = 1 , , , · · · and we conclude that A ( x ) is an alternating convergent series whose firstterm is negative. The required double inequality follows by the truncation of A ( x ) . For the rightmost double inequality of (1.14), we have by Lemmas 2.1 and 2.2 that (cid:18) sin xx (cid:19) + x tanh x > − x ! + x x + 15 ! (0 < x < π/ x − x x + 45= 2 + x x + 3 x + 15 ! > x x + 35 x + 15 ! = 2 + x > . This completes the proof. (cid:3)
Remark 3.1. If n is any odd positive integer and 0 < x < π/ n X k =1 a k x k < x sin x + (cid:18) tanh xx (cid:19) < n +1 X k =1 a k x k where a k is as defined in the proof of Theorem 1.1. In particular, we get2 − x < x sin x + (cid:18) tanh xx (cid:19) < − x x . Proof of Theorem 1.2. As x tan x < < x < π/ , we can writesinh xx + (cid:18) x tan x (cid:19) < sinh xx + x tan x (0 < x < π/ . Utilizing series expansions from (2.2) and (2.5),sinh xx + (cid:18) x tan x (cid:19) < ∞ X k =0 x k (2 k + 1)! + 1 − ∞ X k =1 k | B k | (2 k )! x k = 2 + ∞ X k =1 k )! (cid:20) k + 1) − k | B k | (cid:21) x k = 2 + ∞ X k =1 b k x k := B ( x ) ! (0 < x < π/ , ilker and Huygens type inequalities for mixed trigonometric-hyperbolic functions b k = k )! h k +1) − k | B k | i < k ≥ xx + (cid:18) x tan x (cid:19) < b x (0 < x < π/ b = − / x sinh x − x tan x + x > x > < x < π/ . Let f ( x ) = tan x sinh x − x tan x + x (0 < x < π/ . It can be written as f ( x ) = x − x tan x + tan x x − tan x . Using known series expansions of cosh x (see (2.1)) and tan x (see (2.7)), we write f ( x ) = x − x tan x + ∞ X k =2 k − (2 k )! x k tan x = x − x (cid:18) x + 13 x + 215 x + 17315 x + 622835 x + · · · (cid:19) + 13 x tan x + 245 x tan x + 1315 x tan x + 214175 x tan x + · · · = x + (cid:18) − x − x − x − x − x − · · · (cid:19) + (cid:18) x + 245 x + 1315 x + 214175 x + · · · (cid:19) × (cid:18) x + 13 x + 215 x + 17315 x + 622835 x + · · · (cid:19) = (cid:18) − x − x − x − x − · · · (cid:19) + (cid:18) x + 745 x + 59945 x + 249299225 x + · · · (cid:19) = 145 x + 8945 x + 4614175 x + · · · > x > . The proof is completed. (cid:3)
Remark 3.2.
The rightmost inequality of (1.15) can be put as (cid:18) sinh xx (cid:19) + x tan x − x ! > < x < π/ . Corollary 3.3.
For < x < π/ , we have the following chains of inequalities: (cid:18) x sin x (cid:19) + tanh xx > (cid:18) sinh xx (cid:19) + tanh xx > (cid:18) sinh xx (cid:19) + x tan x > x
45 tan x > and (cid:18) x sinh x (cid:19) + tan xx > (cid:18) sin xx (cid:19) + tan xx > (cid:18) sin xx (cid:19) + x tanh x > . (3.2) Proof.
The chain of inequalities in (3.1) immediately follows by applying Theorem 1.2with Lemmas 2.1 and 2.3. Similarly (3.2) follows due to Theorem 1.1 and Lemmas 2.1and 2.3. (cid:3)
Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo
Again the following corollary is an easy consequence of Theorems 1.1 and 1.2.
Corollary 3.4.
For < x < π/ , we have following chain of inequalities: (cid:18) x sin x (cid:19) + x tan x > (cid:18) sinh xx (cid:19) + x tan x > x
45 tan x > and (cid:18) x sinh x (cid:19) + x tanh x > (cid:18) sin xx (cid:19) + x tanh x > . (3.4)As the applications of Theorems 1.1 and 1.2, we have in fact obtained/refined theinequalities (1.1)-(1.6) in Corollaries 3.3 and 3.4. Proof of Theorem 1.3.
We first prove the inequalities in (1.16). For the leftmostinequality, we have by (2.2) and (2.5) that2 sinh xx + x tan x = ∞ X k =0 k + 1)! x k + 1 − ∞ X k =1 k | B k | (2 k )! x k = 3 + ∞ X k =1 k + 1)! x k − ∞ X k =1 k | B k | (2 k )! x k = 3 + ∞ X k =2 k )! (cid:18) k + 1 − k − | B k | (cid:19) x k < − x < x sinh x + tan xx = 2 − ∞ X k =1 (2 k − − k )! B k x k + ∞ X k =1 k (2 k − k )! | B k | x k − = 2 − ∞ X k =1 (2 k − − k )! B k x k + ∞ X k =0 k +2 (2 k +2 − k + 2)! | B k +2 | x k = 3 + ∞ X k =2 (2 k )! " k − (2 k +2 − k + 1)(2 k + 1) | B k +2 | − ( − k +1 (2 k − − | B k | x k = 3 + ∞ X k =2 c k x k := C ( x ) ! , where c k = (2 k )! h k − (2 k +2 − k +1)(2 k +1) | B k +2 | − ( − k +1 (2 k − − | B k | i ( k ≥ . Now c k > k = 2 , , , · · · and c k = 2 (2 k )! " k − (2 k +2 − k + 1)(2 k + 1) | B k +2 | − (2 k − − | B k | > k = 3 , , , · · · by Lemma 2.7. Thus all the terms of C ( x ) are positive and hence by truncating C ( x ) weget 2 x sinh x + tan xx > c x > c = 31180 . The inequalities (1.17) can be proved easily by applying similar arguments. (cid:3) ilker and Huygens type inequalities for mixed trigonometric-hyperbolic functions Proof of Theorem 1.4.
We give proof of the inequalities in (1.18) only. The inequalitiesin (1.19) will follow similarly. From (2.2) and (2.6) we have2 sin xx + x tanh x = 2 ∞ X k =0 ( − k (2 k + 1)! x k + 1 + ∞ X k =1 k B k (2 k )! x k = 3 + ∞ X k =1 − k (2 k + 1)! x k + ∞ X k =1 k | B k | (2 k )! x k = 3 + ∞ X k =1 k )! " ( − k k + 1 − ( − k k − | B k | x k = 3 + ∞ X k =2 − k (2 k )! (cid:20) k + 1 − k − | B k | (cid:21) x k = 3 + ∞ X k =2 d k x k := D ( x ) ! , where d k = − k (2 k )! h k +1 − k − | B k | i ( k ≥ . By Lemma 2.5, the series D ( x ) is analternating convergent series and its first term is negative. Consequently3 + d x < xx + x tanh x < d = − . This gives leftmost double inequality of (1.18).For the rightmost double inequality of (1.18), we make use of (2.3) and (2.8) to obtain2 x sin x + tanh xx = 2 + ∞ X k =1 (2 k − − k )! | B k | x k + ∞ X k =1 k (2 k − k )! B k x k − = 2 + ∞ X k =1 (2 k − − k )! | B k | x k + ∞ X k =1 k +2 (2 k +2 − k + 2)! B k +2 x k = 3 + ∞ X k =2 (2 k )! " (2 k − − | B k | + ( − k k − (2 k +2 − k + 1)(2 k + 1) | B k +2 | x k = 3 + ∞ X k =2 e k x k = E ( x ) ! , where e k = (2 k )! h (2 k − − | B k | + ( − k k − (2 k +2 − k +1)(2 k +1) | B k +2 | i ( k ≥ . Clearly e k > k = 2 , , , · · · and e k < k = 3 , , , · · · by Lemma 2.7. Thus E ( x ) is an alternatingconvergent series whose first term is positive, i.e. x . The desired inequality follows bythe truncation of E ( x ) . This ends the proof. (cid:3)
Remark 3.5. If n is even positive integer and 0 < x < π/ n X k =2 d k x k < xx + x tanh x < n +1 X k =2 d k x k , and 3 + n +1 X k =2 e k x k < x sin x + tanh xx < n X k =2 e k x k , where d k and e k are as defined in the proof of Theorem 1.4.Before concluding this section, we must emphasize that the Huygens-type inequalities inTheorems 1.3 and 1.4 give sharp bounds to the much-discussed functions in the literature0 Yogesh J. Bagul, Ramkrishna M. Dhaigude, Barkat A. Bhayo such as sin( x ) /x, x/ tan( x ) etc. For example, the inequalitiessin xx < (cid:18) − x tanh x (cid:19) (0 < x < π/ , (3.5)and sinh xx < (cid:18) − x tan x (cid:19) (0 < x < π/
2) (3.6)are very sharp and interesting for further studies.
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