Refinements of Certain Hyperbolic Inequalities Via the Padé

REFINEMENTS OF CERTAIN HYPERBOLIC INEQUALITIES VIA THE PADE´ APPROXIMATION METHOD

GABRIEL BERCU AND SHANHE WU

Abstract. The aim of this paper is to deal with the reﬁnements of certain inequalities for hyperbolic functions using Pad´eapproximation method. We provide a useful way of improving the inequalities for trigonometric functions and hyperbolic functions.

1. Introduction The famous Wilker inequality for trigonometric functions asserts that sin x2 tan x π (1.1) + > 2, 0 < x < , x x 2 while the Huygens trigonometric inequality states sin x tan x π (1.2) 2 + > 3, 0 < x < . x x 2 The following double inequality √ sin x 2 + cos x π (1.3) 3 cos x < < , 0 < x < x 3 2 has also attracted the attention of many authors in the recent past. The left-hand side of (1.3) was discovered by I. Lazarovi´c(see, [5, p. 238]), while the right- hand side of (1.3) is the trigonometric Cusa-Huygens inequality (see [10]). For more connections between these inequalities see [8]. The hyperbolic counterparts of the Wilker and Huygens trigonometric inequalities have been introduced by L. Zhu (see [21],[22]), E. Neumann and J. S´andor (see [8]) as follows: sinh x2 tanh x (1.4) + > 2, x 6= 0 x x and respectively sinh x tanh x (1.5) 2 + > 3, x 6= 0. x x

The hyperbolic Cusa-Huygens inequality (see [8]) and the hyperbolic Lazarevi´cinequality (see [6]) state that √ sinh x cosh x + 2 (1.6) 3 cosh x < < , x 6= 0 x 3 These inequalities were proved using the variation of some functions and their derivatives. Recently, some of the above inequalities have been improved using the Taylor’s expansions for hyperbolic functions (see [7]) as follows sinh x tanh x 3 3 (1.7) 2 + > 3 + x4 − x6, x > 0 x x 20 56

1991 Mathematics Subject Classification. 41A21; 26D05; 26D15; 33B10; 11J10. Key words and phrases. Padéapproximation; Taylor expansion; continued fraction; hyperbolic inequalities; refinements. 1 2 GABRIEL BERCU AND SHANHE WU

2 x 2 sin x tan x sin x tan 2 (1.8) + > + x > 2 x x x 2 and its hyperbolic counterpart (see [9]) x 2 sinh x tanh 2 (1.9) + x > 2. x 2

The following inequality (see [11], [12]) sin x π2 − x2 (1.10) ≥ , for all x ∈ x π2 + x2 R is known as Redheffer’s inequality. Using mathematical induction and infinite product representation of the hyperbolic functions sinh x and cosh x, C. P. Chen, J. W. Zhao and F. Qi (see [2]) found the hyperbolic analogue of inequality (1.10), by showing that sinh x π2 + x2 (1.11) ≤ , for all |x| < π. x π2 − x2 They also proved that π2 + 4x2 π (1.12) cosh x ≤ , for all |x| ≤ . π2 − 4x2 2 Recently, J. Sándor and B.A.Bhayo (see [13]) improved the inequality (1.11), by showing that sinh x 12 + x2 (1.13) ≤ , for all |x| < π. x 12 − x2 The relevant papers on the topic are also [1], [3] - [5], [10], [20] and [14] - [19]. The aim of this paper is to prove and refine the aforesaid hyperbolic inequalities using Padé approximation method. It is known that a Padéapproximation is the best approximation of a function by a rational function of given order. The rational approximation is particularly good for series with alter- nating terms and poor polynomial convergence. Optimal rational polynomials are frequently used in computer calculations, because they provide a good compromise in accuracy, size and speed. The Padéapproximation [L/M] corresponds to the Taylor series. When it exists, the [L/M] ∞ P j Padéapproximation to any power series A (x) = ajx is unique. If A (x) is a transcendental j=0 1 function, then the terms are given by the Taylor series about x , a = A(n) (x ). 0 n n! 0 L p0 + p1x + ... + pLx The coefficients are found by setting A (x) = M . These give the set of 1 + q1x + ... + qM x equations  p0 = a0   p1 = a0q1 + a1   p2 = a0q2 + a1q1 + a2  . .   pL = a0qL + ... + aL−1q1 + aL   0 = aL−M+1qM + ... + aLq1 + aL+1  0 = aLqM + ... + aL+M−1q1 + aL+M . x2 x4 x6 x8 For example, let us consider the Taylor series for cosh : cosh x = 1 + 2 + 24 + 720 + 40320 + 10 x2 x4 x6 x8 O (x ) and its associate polynomial: 1 + 2 + 24 + 720 + 40320 . REFINEMENTS OF CERTAIN HYPERBOLIC INEQUALITIES VIA THE PADE´ APPROXIMATION 3

The Padéapproximation 2 3 4 5 6 p0 + p1x + p2x + p3x + p4x + p5x + p6x cosh[6/4](x) = 2 3 4 1 + q1x + q2x + q3x + q4x satisfies x2 x4 x6 x8 1 + + + + 1 + q x + q x2 + q x3 + q x4 = p +p +p x2+p x3+p x4+p x5+p x6. 2 24 720 40320 1 2 3 4 0 1 2 3 4 5 6 We find 1 1 6 101 17 p = p = p = q = q = 0, q = − , q = , p = 1, p = , p = , p = . 1 3 5 1 3 2 26 4 1456 0 2 13 4 4368 6 131040 Therefore we obtain 1 + 6 x2 + 101 x4 + 17 x6 cosh (x) = 13 4368 131040 [6/4] 1 1 − x2 + 1 x4 26 1456 131040 + 60480x2 + 3030x4 + 17x6 = . 131040 − 5040x2 + 90x4 Similar calculations lead us to the following results x3 + 15x 7x3 + 60x tanh (x) = and sinh (x) = . [3/2] 6x2 + 15 [3/2] −3x2 + 60 2. Some Lemmas In order to obtain our main results, we first prove several lemmas.

Lemma 2.1. For every x ∈ R, one has 17x6 + 3030x4 + 60480x2 + 131040 (2.1) cosh x > . 90x4 − 5040x2 + 131040 Proof. First we remark that the denominator 90x4 − 5040x2 + 131040 is positive for all x ∈ R. We consider the function f : (0, ∞) → R, f (x) = (90x4 − 5040x2 + 131040) cosh x − 17x6 − 3030x4 − 60480x2 − 131040. Elementary calculations reveal that f 0 (x) = 360x3 − 10080x cosh x+90x4 − 5040x2 + 131040 sinh x−102x5−12120x3−120960x, f (2) (x) = 90x4 − 3960x2 + 120960 cosh x+720x3 − 20160x sinh x−510x4−36360x2−120960, f (3) (x) = 1080x3 − 28080x cosh x + 90x4 − 1800x2 + 100800 sinh x − 2040x3 − 72720x, f (4) (x) = 90x4 + 1440x2 + 72720 cosh x + 1440x3 − 31680x sinh x − 6120x2 − 72720, f (5) (x) = 1800x3 − 28800x cosh x + 90x4 + 5760x2 + 41040 sinh x − 12240x, f (6) (x) = 90x4 + 11160x2 + 12240 cosh x + 2160x3 − 17280x sinh x − 12240, f (7) (x) = 2520x3 + 5040x cosh x + 90x4 + 17640x2 − 5040 sinh x, f (8) (x) = 90x4 + 25200 cosh x + 2880x3 + 40320x sinh x. We see that f (8) (x) > 0 for all x > 0. Then f (7) is strictly increasing on (0, ∞). As f (7) (0) = 0, we get f (7) > 0 on (0, ∞). Continuing the algorithm, ﬁnally we obtain f (x) > 0 for all x ∈ (0, ∞). Due to the form of function f, it follows that the inequality f (x) > 0 holds also for x < 0. The proof is completed. Lemma 2.2. For every x 6= 0, one has 10x2 + 105 tanh x x2 + 15 (2.2) < < . x4 + 45x2 + 105 x 6x2 + 15 4 GABRIEL BERCU AND SHANHE WU

Proof. We introduce the function g : (0, ∞) → R, g (x) = (6x2 + 15) sinh x − (x3 + 15x) cosh x. Its derivative is g0 (x) = x[−x2 − 3 sinh x + 3x cosh x]. Then we consider the function r : (0, ∞) → R, r (x) = (−x2 − 3) sinh x + 3x cosh x and its derivative r0 (x) = x (sinh x − x cosh x) . The function p (x) = sinh x−x cosh x has the derivative p0 (x) = −x sinh x, therefore p0 (x) < 0 for all x ∈ (0, ∞). Then p is strictly decreasing on (0, ∞). As p (0) = 0, it follows p < 0 on (0, ∞), hence r0 (x) < 0 for every x ∈ (0, ∞). Due to similar arguments, ﬁnally it results that g (x) < 0 for all x ∈ (0, ∞). For proving the ﬁrst part of Lemma 2.2, we consider the function s : (0, ∞) → R, s (x) = (x4 + 45x2 + 105) sinh x − (10x3 + 105x) cosh x and its derivative s0 (x) = x[−6x2 − 15 sinh x + x3 + 15x cosh x] = −xg (x). Since g (x) < 0 for all x ∈ (0, ∞), it follows that s0 > 0 on (0, ∞), therefore s is strictly increasing on (0, ∞). As s (0) = 0, we get s (x) > 0 for all x ∈ (0, ∞). We remark that if the inequality (2.2) is true for x > 0, then it holds clearly also for x < 0. This completes the proof. 10x3 + 105x Remark 2.1. The idea to compare the function tanh x with the function is x4 + 45x2 + 105 given by the continued fraction representation of the hyperbolic tangent function: x tanh x = . x2 1 + x2 3 + x2 5 + 7 + ... We also have Lemma 2.3. a) For every x 6= 0, the inequality sinh x 170x8 + 32085x6 + 922950x4 + 7660800x2 + 13759200 (2.3) > x 90x8 − 990x6 − 86310x4 + 5367600x2 + 13759200 holds. √ b) For every |x| < 20, one has sinh x 7x2 + 60 (2.4) < . x −3x2 + 60 Proof. a) Multiplying the inequalities of positive functions (2.1) and (2.2), we obtain the desired lower rational bound for the hyperbolic√ sine function. 2 3 b) We consider the function t : 0, 20 → R, t (x) = (−3x + 60) sinh x − 7x − 60x. Its derivatives are t0 (x) = −6x sinh x + −3x2 + 60 cosh x − 21x2 − 60,

t(2) (x) = 54 − 3x2 sinh x − 12x cosh x − 42x, t(3) (x) = −18x sinh x + 42 − 3x2 cosh x − 42, t(4) (x) = 24 − 3x2 sinh x − 24x cosh x, t(5) (x) = −30x sinh x − 3x2 cosh x. REFINEMENTS OF CERTAIN HYPERBOLIC INEQUALITIES VIA THE PADE´ APPROXIMATION 5

We notice that t(5) < 0 on (0, ∞). Then the function t(4) is strictly decreasing for all x ∈ (0, ∞). As t(4) (0) = 0, we have t(4) < 0 on (0, ∞). Using the same arguments, ﬁnally we conclude that t (x) < 0 for all x ∈ (0, ∞). √ √ Since the inequality (2.4) is true for x ∈ 0, 20, then it holds also for x ∈ − 20, 0. The proof is completed.

3. Main results In this section we will formulate and prove the rational refinements of the aforesaid hyperbolic inequalities. Firstly we will refine the Mortici’s improved version of hyperbolic Huygens inequality as follows: Theorem 3.1. For every x 6= 0, the rational inequality 170x12 + 40185x10 + 2384400x8 + 52078950x6 + 477711675x4+ sinh x tanh x 1774143000x2 + 2167074000 2 + > x x 45x12 + 1530x10 − 60705x8 + 689850x6 + 123119325x4+ 591381000x2 + 722358000 3 3 > 3 + x4 − x6 20 56 holds. Proof. We remark that if the inequality is true for x > 0, then it holds clearly also for x < 0, so it is sufficient to show only for x > 0. The first inequality is an easy consequence of Lemma 2.2 and respectively Lemma 2.3. The second inequality has the equivalent form 135 1053 194967 52222365 63118965 x8 x10 + x8 − x6 + 46097x4 + x2 + > 0. 56 14 56 8 4 The polynomial function from the left - hand side has no real non - zero roots, hence the last inequality holds true for every x 6= 0. Using Padéapproximation method, we also improve the hyperbolic version of Wilker inequality as follows: Theorem 3.2. For every x 6= 0, one has sinh x2 tanh x A (x) + > > 2, x x B (x) where A (x) = 28900x20 + 12290400x18 + 1836253725x16 + 123257800125x14 + +4281358717125x12 + 82293438024000x10 + 892105608933000x8 + +5736412528560000x6 + 22757784325920000x4 + +48057033024000000x2 + 39756272774400000 and B (x) = 8100x20 + 186300x18 − 21724200x16 + 463344300x14 + +48937656600x12 − 865986754500x10 − 16558594573500x8 + +24028516512000000x2 + 19878136387200000. Proof. We need to prove only for x > 0. 6 GABRIEL BERCU AND SHANHE WU

The inequalities from Lemma 2.2 and respectively Lemma 2.3 lead us to the following rational inequality sinh x2 tanh x (170x8 + 32085x6 + 922950x4 + 7660800x2 + 13759200)2 10x2 + 105 + > + . x x (90x8 − 990x6 − 86310x4 + 5367600x2 + 13759200)2 x4 + 45x2 + 105 A (x) The right - hand side of the above inequality takes the equivalent form . B (x) A (x) On the other hand, the inequality > 2 becomes B (x) 12700x20 + 11917800x18 + 1879702125x16 + 122331111525x14 + +4183483403925x12 + 84025411533000x10 + 925222798080000x8 + +3262053150720000x6 + 3533890913280000x4 > 0, which is obviously true for all x 6= 0. Since the hyperbolic version of Lazarevi´c’sinequality (1.6) can be re - written as sinh x2 tanh x > 1, x 6= 0, x x we will improve this result as follows: Theorem 3.3. For every x 6= 0, one has sinh x2 tanh x 10x6 + 120x4 + 360x2 + 105x4 + 1260x2 + 3780 > > 1. x x 36x4 + 1620x2 + 3780 Proof. We remark that if the inequality is true for x > 0, then it holds also for x < 0, so it is enough to prove only for x > 0. Arising from Taylor’s expansion for the hyperbolic sine, the sinh x x2 following estimate > 1 + holds for x > 0. x 6 Using also the estimate for hyperbolic tangent function obtained in Lemma 2.2, we have sinh x2 tanh x · > E (x) , x x where x2 2 10x2 + 105 E (x) = 1 + · 6 x4 + 45x2 + 105 10x6 + 225x4 + 1620x2 + 3780 = . 36x4 + 1620x2 + 3780 The inequality E (x) > 1 has the equivalent form 10x6 + 225x4 > 0, which is obviously true for all x 6= 0. We also will sharp the hyperbolic Cusa - Huygens inequality as follows: Theorem 3.4. For every x 6= 0, one has sinh x x2 + 15 cosh x + 2 < cosh x < . x 6x2 + 15 3 Proof. Let us consider x > 0. From inequality (2.2) we have tanh x x2 + 15 < x 6x2 + 15 or equivalent sinh x x2 + 15 < cosh x. x 6x2 + 15 REFINEMENTS OF CERTAIN HYPERBOLIC INEQUALITIES VIA THE PADE´ APPROXIMATION 7

We transform the inequality x2 + 15 cosh x + 2 cosh x < 6x2 + 15 3 into the inequality 10 − x2 cosh x − 4x2 − 10 < 0 for all x > 0. We introduce the function 2 2 h : (0, ∞) → R, h (x) = 10 − x cosh x − 4x − 10. Its derivatives are h0 (x) = −2x cosh x + 10 − x2 sinh x − 8x, h(2) (x) = 8 − x2 cosh x − 4x sinh x − 8, h(3) (x) = 4 − x2 sinh x − 6x cosh x, h(4) (x) = −2 − x2 cosh x − 8x sinh x. We have h(4) < 0 on (0, ∞), hence h(3) is strictly decreasing on (0, ∞). As h(3) (0) = 0, it follows that h(3) < 0 on (0, ∞). Using the same arguments, finally we conclude that h (x) < 0 on (0, ∞). Since the inequalities from Theorem 3.4 are true for x > 0, then they hold clearly also for x < 0. In order to refine the inequality (1.9), we will prove the following Theorem 3.5. For all x 6= 0, one has  x 2 sinh x tanh x10 + 366x8 + 37920x6 + 828960x4 + 7257600x2 + 33868800 + 2 > > 2. x  x  6x8 + 2160x6 + 214560x4 + 3628800x2 + 16934400 2 Proof. It is sufficient only to prove for x > 0. From Taylor’s expansion for the hyperbolic sine and from Lemma 2.2, we have  x 2  x2 2 sinh x tanh x2 + 6 10 · + 105 + 2 > +  2  x  x  6 x4 x2  + 45 · + 105 2 2 2 x10 + 366x8 + 37920x6 + 828960x4 + 7257600x2 + 33868800 = . 6x8 + 2160x6 + 214560x4 + 3628800x2 + 16934400 The last rational expression is greater than 2 since the inequality x10 + 354x8 + 33600x6 + 4 399840x > 0 is obviously true for all x 6= 0. Finally, we also will improve the Redheffer - type inequalities (1.11), (1.12) and (1.13) as follows: Theorem 3.6. a) For all |x| < π, one has sinh x 60 + 7x2 12 + x2 < < . x 60 − 3x2 12 − x2 π b) For all |x| < , one has 2 10 + 4x2 π2 + 4x2 cosh x < < . 10 − x2 π2 − 4x2 8 GABRIEL BERCU AND SHANHE WU

Proof. a) First inequality is a consequence of Lemma 2.2. We remark that the denominator 60 − 3x2 is positive for all |x| < π. It is easy to see that the difference 12 + x2 7x2 + 60 C (x) = − 12 − x2 60 − 3x2 4x4 has the equivalent positive form . (12 − x2) (60 − 3x2) b) In the demonstration of Theorem 3.5, we showed that 2 2 10 − x cosh x − 4x − 10 < 0 for all x ∈ R. 10 + 4x2 π √ Therefore we get cosh x < for all |x| < < 10. 10 − x2 2 10 + 4x2 π2 + 4x2 The inequality < can be rewritten as the following true inequality 10 − x2 π2 − 4x2 π 0 < 12x4 + 80 − 5π2 x2 for all |x| < . 2 4. Final Remarks Let us emphasize that the Padéapproximation method was here applied for proving the refinements of some remarkable hyperbolic inequalities. We are convinced that Padéapproxi- mation method is suitable to establish many other similar inequalities.

Acknowledgements The present investigation was supported by the Natural Science Foundation of Fujian Province of China (No.2016J01023). Competing interests The authors declare that they have no competing interests.

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“Dun˘areade Jos” University of Galat¸i, Department of Mathematics and Computer Sciences, 111 Domneasc˘aStreet, Galat¸i, 800201, ROMANIA E-mail address: [email protected]

Longyan University, Department of Mathematics, Longyan, Fujian, 364012, P. R. China E-mail address: [email protected]