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arXiv:1408.2243v1 [math.CA] 10 Aug 2014 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES FOR TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS Abstract. We prove that for p ∈ (0, 1], the double inequality sin x 1 cos px + 1 − 3p12 < < 3q12 cos qx + 1 − 3q12 3p2 x √ holds for x ∈ (0, π/2) if and only if 0 < p ≤ p0 ≈ 0.77086 and 15/5 = p1 ≤ q ≤ 1. While its hyperbolic version holds for x > 0 if and only if 0 < p ≤ p1 = √ 15/5 and q ≥ 1. As applications, some more accurate estimates for certain mathematical constants are derived, and some new and sharp inequalities for Schwab-Borchardt mean and logarithmic means are established. 1. Introduction The Cusa and Huygens (see, e.g., [1]) states that for x ∈ (0, π/2), the inequality sin x 2 + cos x (1.1) < x 3 holds true. Its version of hyperbolic functions refers to (see [2]) the inequality 2 + cosh x sinh x < (1.2) x 3 holds for x > 0, and it is know as hyperbolic Cusa–Huygens inequality (see [2]). There are many improvements, refinements and generalizations of (1.1) and (1.2), see [3], [4], [5], [6], [7], [8], [9]; [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21]. Now we focus on the bounds for (sin x) /x in terms of cos px, where x ∈ (0, π/2), p ∈ (0, 1]. In 1945, Iyengar [22] (also see [23, subsection 3.4.6] ) proved that for x ∈ (0, π/2), sin x (1.3) cos px ≤ ≤ cos qx x holds with the best possible constants 2 2 1 p = √ and q = arccos . π π 3 Moreover, the following chain of inequalities hold: sin x cos x x x (1.4) cos x ≤ ≤ (cos x)1/3 ≤ cos √ ≤ ≤ cos qx ≤ cos ≤ 1. 2 1 − x /3 x 2 3 Qi et al. [24] showed that (1.5) cos2 x sin x < 2 x Date: April 10, 2014. 2010 Mathematics Subject Classification. Primary 26D05, 26D15; Secondary 33B10, 26E60 . Key words and phrases. Trigonometric function, hyperbolic function, inequality, mean. This paper is in final form and no version of it will be submitted for publication elsewhere. 1 2 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES holds for x ∈ (0, π/2). Klén et al. [25, Theorem 2.4] out that the function pointed p p 1/p p 7→ (cos px) is decreasing on (0, 1) and for x ∈ − 27/5, 27/5 sin x x 2 + cos x x ≤ ≤ cos3 ≤ 2 x 3 3 are valid. Subsequently, Yang [8] (also see [26]) gave a refinement of (1.6), which states that for p, q ∈ (0, 1) the double inequality sin x 1/p 1/q (1.7) (cos px) < < (cos qx) x holds for x ∈ (0, π/2) if and only if p ∈ [p∗0 , 1) and q ∈ (0, 1/3], where p∗0 ≈ 0.3473. Moreover, the double inequality x α sin x x 3 < cos < cos 3 x 3 with the best exponents α = 2 (ln π − ln 2) / (ln 4 − ln 3) ≈ 3.1395 and 3. Also, he pointed out that the value range of variable x such that (1.6) holds can be extended to (0, π). Very recently, Yang [21] gave another improvement of (1.6), that is, for x ∈ (0, π/2) the inequalities 2 sin x 2 + cos x x (1.8) < 32 cos x2 + 31 < cos3 < x 3 3 are true. An important improvement for the inequality in (1.5) is due to Neuman [2]: 2/3 x sin x 1 + cos x (1.9) cos4/3 = < , x ∈ 0, π2 . 2 2 x (1.6) cos2 Lv et al. [27] showed that for x ∈ (0, π/2) inequalities x 4/3 sin x x θ (1.10) cos < < cos 2 x 2 hold, where θ = 2 (ln π − ln 2) / ln 2 = 1.3030... and 4/3 are the best possible con1/ 3p2 stants. By constructing a decreasing function p 7→ (cos px) ( ) (p ∈ (0, 1]), Yang [9] showed that the double inequality (1.11) ∗2 sin x x (cos p∗1 x)1/(3p1 ) < < cos5/3 √ x 5 √ is true for x ∈ (0, π/2) with the best constants p∗1 ≈ 0.45346 and 1/ 5 ≈ 0.44721. It follows that √ ∗2 x (cos x)1/3 < cos1/2 36x < cos2/3 √x2 < cos √ < cos4/3 x2 < (cos p∗1 x)1/(3p1 ) 3 2 sin x 5/3 √ 2 √ 3 x x x x < cos < cos 6 < cos 3 < cos16/3 x4 < e−x /6 < 2+cos (1.12) 3 5 x are valid for x ∈ (0, π/2). For the bounds for (sinh x) /x in terms of cosh px, it is known that the inequalities sinh x x 2 + cosh x < cosh3 < x 3 3 holds true for x > 0 (see [18]), which is exactly derived by the inequalities for means L < A1/3 < 2G + A 3 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES 3 (see e.g. [28], [29], [30]), where L, Ap , G and A stand for the logarithmic mean, power mean of order p, geometric mean and arithmetic mean or positive numbers a and b defined by a−b if a 6= b and L (a, a) = a, ln a − ln b p 1/p √ a + bp if p 6= 0 and A = A0 (a, b) = ab, Ap (a, b) = 2 L ≡ Ap L (a, b) = ≡ G = A0 and A = A1 , respectively. Zhu in [31] proved that for p > 1 or p ≤ 8/15, and x ∈ (0, ∞), the inequality q sinh x > p + (1 − p) cosh x x is true if and only if q ≥ 3 (1 − p). It follows by letting p = 1/2 and q = 3/2 that sinh x x > cosh4/3 x 2 holds for x > 0 (also see [2, (2.8)]). Yang [19] showed that the inequality sinh x 1/ 3p2 > (cosh px) ( ) x √ holds for all x > 0 if and only if p ≥ 1/ 5 and its reverse holds if and only if 2 0 < p ≤ 1/3. And, the function p 7→ (cosh px)1/(3p ) is decreasing on (0, ∞). The aim of this paper is to determine the best p such that the inequalities (1.13) sin x x sinh x x < (>) 3p12 cos px + 1 − 1 3p2 , < (>) 3p12 cosh px + 1 − p ∈ (0, 1) , x ∈ (0, π/2) , 1 3p2 , p, x ∈ (0, ∞) hold true. Our main results are contained in the following theorems. Theorem 1. For p ∈ (0, 1] and x ∈ (0, π/2), the double inequality 1 3p2 cos px + 1 − 1 3p2 sin x < x cos qx + 1 − 3q12 √ holds if and only if 0 < p ≤ p0 ≈ 0.77086 and 0.77460 ≈ 15/5 = p1 ≤ q ≤ 1, where p0 is the unique root of the equation 2 1 (1.15) Fp π2 − = − 3p12 cos pπ =0 + 1 − 2 2 3p π on (0, 1). And, the bound for (sin x) /x given in (1.14) is increasing with respect to parameters p or q. (1.14) < 1 3q2 Theorem 2. For p, x > 0, the double inequality (1.16) 1 3p2 cosh px + 1 − 1 3p2 < sinh x < x 1 3q2 cosh qx + 1 − 1 3q2 √ holds if and only if 0 < p ≤ p1 = 15/5 and q ≥ 1. And, the bound for (sinh x) /x given in (1.16) is increasing with respect to parameters p or q. 4 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES Remark 1. The weighted basic inequality of two positive numbers of a and b tell us that for α ∈ [0, 1], the inequality αa + (1 − α) b ≥ aα b1−α . It is reversed if and only if α ≥ 1 or α ≤ 0 (see [32]). Hence, taking into account (1.11) and (1.14) we see that √ (i) if p ∈ [p∗1 , 1/ 3), where p∗1 ≈ 0.45346, then sin x 1/ 3p2 > (cos px) ( ) > 3p12 cos px + 1 − 3p12 ; x √ (ii) if p ∈ (1/ 3, p0 ), where p0 ≈ 0.77086, then sin x 1/ 3p2 > 3p12 cos px + 1 − 3p12 > (cos px) ( ) . x In the same way,√(1.13)√together with (1.16) leads us to (iii) if p ∈ [1/ 5, 1/ 3), then sinh x 1/ 3p2 > (cosh px) ( ) > 3p12 cosh px + 1 − 3p12 ; x √ √ (iv) if p ∈ (1/ 3, 15/5), then sinh x 1/ 3p2 > 3p12 cosh px + 1 − 3p12 > (cosh px) ( ) . x p p √ √ √ Taking p = 3/4, 1/ 2, 2/3, 1/ 3 and q = 3/5, 2/3, 3/2, 1 in Theorem 1, we have Corollary 1. For x ∈ (0, π/2), the inequalities (1.17) cos √x3 < 3 4 cos 2x 3 + 1 4 < 2 3 cos √x2 + 1 3 < 16 27 √ cos 3x 4 + √ sin x x < 31 cos x + 23 . 11 27 < + 94 < cos2 √x6 < 49 cos 23x + 59 < 95 cos 15x 5 p √ √ √ Putting p = 3/5, 3/4, 1/ 2, 2/3, 1/ 3 and q = 1, 2/ 3 in Theorem 2 we have Corollary 2. For x > 0, the inequalities (1.18)cosh √x3 < 3 4 cosh 2x 3 + < 5 9 cosh √ 15x 5 1 4 < 2 3 cosh √x2 + + 4 9 < sinh x < x 1 3 1 3 < 16 27 cosh 3x 4 + cosh x + 2 3 < 1 2 11 27 cosh2 x √ 3 + 21 . 2. Proof of Theorem 1 In order to prove Theorem 1, we need some lemmas. Lemma 1. For x ∈ (0, π/2), the function p 7→ Up (x) defined on [0, 1] by Up (x) = 1 x2 1 cos px + 1 − 2 if p ∈ (0, 1] and U0 (x) = 1 − 2 3p 3p 6 is increasing. Proof. Differentiation yields ∂Up 1 (2 − 2 cos px − px sin px) = ∂p 3p3 px px px 12px sin 2 − cos sin > 0, = px 3 3p 2 2 2 which completes the proof. NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES 5 Lemma 2. Let the function Fp be defined on (0, π/2) by sin x x − ( 3p12 cos px + 1 − 3p12 ) if p ∈ (0, 1] and F0 (x) = 2 − 1 + x6 . √ (i) If Fp (x) < 0 for all x ∈ (0, π/2), then p ∈ [p1 , 1], where p1 = 15/5 ≈ 0.77460. (ii) If Fp (x) > 0 for all x ∈ (0, π/2), then p ∈ [0, p0 ], where p0 ≈ 0.77086. (2.1) Fp (x) = sin x x Proof. (i) If Fp (x) < 0 for all x ∈ (0, π/2), then we have sin x 1 1 − cos px + 1 − 2 2 x 3p 3p 1 5p2 − 3 ≤ 0, =− (2.2) lim x→0 x4 360 √ which leads to p ∈ ( 15/5, 1]. (ii) If Fp (x) > 0 for all x ∈ (0, π/2), then we have π− 2 pπ 1 1 Fp = − cos + 1 − 2 > 0. 2 π 3p2 2 3p From Lemma 1 we see that the function p 7→ Fp (π/2− ) is decreasing on [0, 1], which together with the facts 2 2 π− 2√ 2 π− 1 = − = − <0 F1/2 2 + > 0 and F1 2 π 3 3 2 π 3 gives that there is a unique number p0 ∈ (1/2, 1) such that Fp (π/2− ) > 0 for p ∈ (0, p0 ) and Fp (π/2− ) < 0 for p ∈ (p0 , 1). Solving the equation Fp (π/2− ) = 0 for p by mathematical computer software we find that p0 ≈ 0.77086. This completes the proof. Lemma 3. Let c ∈ (0, 3/5] and let the sequence (an (c)) be defined by an (c) = 3 − (2n + 1) cn−1 . (2.3) Then (i) an (c) ≥ 0 for n ∈ N; (ii) for n ≥ 3, we have 1< an+1 (3/5) 11 an+1 (c) ≤ ≤ . an (c) an (3/5) 5 Proof. (i) We first show that an (c) ≥ 0 for n ∈ N. A simple computation leads to an+1 (c) − an (c) = ≥ (2n + 1) cn−1 − (2n + 3) cn = cn−1 ((2n + 1) − (2n + 3) c) 3 4 cn−1 (2n + 1) − (2n + 3) = cn−1 (n − 1) ≥ 0, 5 5 which implies that an+1 (c) ≥ an (c) ≥ a1 (c) = 0. (ii) Since a1 (c) = 0, a2 (c) = 3 − 5c ≥ 0, an (c) > 0 for n ≥ 3, if we can show that the function 3 − (2n + 3) cn an+1 (c) = (c, n) 7→ an (c) 3 − (2n + 1) cn−1 is increasing in c on (0, 3/5] and decreasing in n ≥ 3, then we have 1= an+1 (0) an+1 (c) an+1 (3/5) a4 (3/5) 11 < < ≤ = , an (0) an (c) an (3/5) a3 (3/5) 5 6 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES which is the desired results. Now we prove that (c, n) 7→ an+1 (c) /an (c) is increasing in c on (0, 3/5] for n ≥ 3. Differentiation yields ′ an+1 (c) = 4n2 + 8n + 3 cn −3n (2n + 3) c+ 6n2 − 3n − 3 := hn (c) , c2−n a2n (c) an (c) h′n (c) = −n (2n + 3) 3 − (2n + 1) cn−1 = −n (2n + 3) an (c) < 0, where the last inequality holds due to an (c) > 0 for n ≥ 3. Therefore, we have n 3 3 hn (c) ≥ hn (3/5) = 4n2 + 8n + 3 + 4n2 − 14n − 5 , 5 5 which is clearly positive due to that h3 (3/5) = 876/125 > 0 and 4n2 − 14n − 5 = n(4n − 14) − 5 ≥ 3 for n ≥ 4. This reveals that h′n (c) > 0, that is, (c, n) 7→ an+1 (c) /an (c) is increasing in c on (0, 3/5]. On the other hand, we have an+1 (c) an+2 (c) − an (c) an+1 (c) 4cn+1 + 6 (c − 1)2 n + 15c2 − 18c + 3 an (c) an+1 (c) cn−1 2 4cn+1 + 6 (c − 1) × 3 + 15c2 − 18c + 3 an (c) an+1 (c) cn−1 7 − c (1 − c) > 0, 4cn+1 + 33 11 an (c) an+1 (c) = cn−1 ≥ = where the first inequality holds due to an (c) > 0 for n ≥ 3, while the last one holds since c ∈ (0, 3/5]. This means that (c, n) 7→ an+1 (c) /an (c) is decreasing with n ≥ 3. Thus we complete the proof of this assertion. Now we are in a position to prove Theorem 1. Proof of Theorem 1. Expanding in power series gives Fp (x) = = = = sin x x ∞ X − ( 3p12 cos px + 1 − x2n (−1) − (2n + 1)! n=0 ∞ X n 1 3p2 ) ∞ 2n 1 1 X n (px) +1− 2 (−1) 2 3p n=0 (2n)! 3p ! 3 − (2n + 1) p2n−2 2n x 3 (2n + 1)! n=2 ∞ 2 3 − 5p2 4 X n an p x + (−1) x2n , 360 3 (2n + 1)! n=3 n (−1) where an (c) is defined by (2.3). Considering the function fp (x) = x−4 Fp (x), we have ∞ an p 2 3 − 5p2 X n −4 (2.4) fp (x) = x Fp (x) = + (−1) x2n−4 , 360 3 (2n + 1)! n=3 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES 7 and differentiation yields fp′ (x) ∞ X = (2n − 4) an p2 2n−5 x (−1) 3 (2n + 1)! n=3 : = ∞ X n (−1)n un (x), n=3 where (2n − 4) an p2 2n−5 x . un (x) = 3 (2n + 1)! Utilizing Lemma 3 we get that for p2 ∈ (0, 3/5] and n ≥ 3, un+1 (x) un (x) = (2n−2)an+1 (p2 ) 2n−3 x 3(2n+3)! (2n−4)an (p2 ) 2n−5 3(2n+1)! x (2n − 2) an+1 p2 2 1 x = (2n − 4) (2n + 3) (2n + 2) an (p2 ) 11 π 2 11π 2 1 ×1× × = < 1, (2 × 3 − 4) (2 × 3 + 3) 5 4 360 P∞ n which implies that the power series n=3 (−1) un (x) is a Leibniz type alternating one, and so fp′ (x) < 0 for p2 ∈ (0, 3/5]. √ (i) We first prove the second inequality in (1.14) holds, where p1 = 15/5 is the best. As shown previously, we see that fp1 is decreasing on (0, π/2), and therefore, < fp1 (x) < fp1 0+ = lim x→0+ 1 3 − 5p21 = 0, x−4 Fp (x) = 360 which together with (2.4) √ yields Fp1 (x) < 0 for x ∈ (0, π/2). Next we prove p1 = 15/5 is the best. If there is another p∗1 < p1 such that the second inequality in (1.14) holds for x ∈ (0, π/2), then by Lemma 2 there must √ ∗ be p1 ∈ [p1 , 1], which yields a contradiction. Therefore, p1 = 15/5 can not be replaced with other smaller ones. (ii) Now we prove the first inequality in (1.14) holds with the best constant p0 ≈ 0.77088. Since p20 ∈ (0, 3/5], fp0 is also decreasing on (0, π/2), and so π −4 π− π− −4 fp0 (x) > fp0 x Fp0 (x) = = lim = 0, Fp0 2 2 2 x→π/2− where the last equality is true due to p0 is the unique root of the equation (1.15) on (0, 1). It together with (2.4) gives Fp0 (x) > 0 for x ∈ (0, π/2). Lastly, we show that p0 is the best. Assume that there is another p∗0 > p0 such that Fp∗0 (x) > 0 for x ∈ (0, π/2). Then by Lemma 2 there must be p∗0 ∈ [0, p0 ], which is clear a contradiction. Consequently, p0 can not be replaced by other larger numbers. Thus the proof is complete. Remark 2. Application of the conclusion that fp′ (x) < 0 for x ∈ (0, π/2) if p2 ∈ (0, 3/5] gives fp (π/2− ) < fp (x) < fp (0+ ), that is, π −4 π − 3 − 5p2 < x−4 Fp (x) < lim+ x−4 Fp (x) = Fp , 2 2 360 x→0 8 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES which can be changed into sin x < ( 3p12 cos px + 1 − 3p12 ) + c1 (p) x4 , x −4 where c0 (p) = (π/2) Fp (π/2− ) and c1 (p) = 3 − 5p2 /360 are the best constants. Then √ (i) when p = p1 = 15/5, we have sin x < ( 3p12 cos p1 x + 1 − 3p12 ), c0 (p1 ) x4 + ( 3p12 cos p1 x + 1 − 3p12 ) < 1 1 1 1 x (2.5) ( 3p12 cos px + 1 − 1 3p2 ) + c0 (p) x4 < −4 where c0 (p1 ) = (π/2− ) Fp1 (π/2− ) ≈ −7.2618 × 10−5 and c1 (p1 ) = 0 are the best possible constants; (ii) when p = p0 ≈ 0.77086, we get < ( 3p12 cos p0 x + 1 − 3p12 ) + c1 (p0 ) x4 , 0 0 where c0 (p0 ) = 0 and c1 (p0 ) = 3 − 5p20 /360 ≈ 8.0206 × 10−5 are the best constants. ( 3p12 cos p0 x + 1 − 0 1 ) 3p20 < sin x x 3. Proof of Theorem 2 For proving Theorem 2, we first give the following lemmas. Lemma 4. For x ∈ (0, ∞), the function p 7→ Vp (x) defined on [0, ∞) by Vp (x) = 1 x2 1 cosh px + 1 − 2 if p 6= 0 and V0 (x) = 1 + 2 3p 3p 6 is increasing. Proof. Differentiation yields ∂Vp 1 (px sinh px − 2 cosh px + 2) = ∂p 3p3 px px sinh 2 2x px cosh = − px > 0, sinh 2 3p 2 2 2 which completes the proof. Lemma 5. Let the function Gp be defined on (0, ∞) by Gp (x) = sinh x x − ( 3p12 cosh px + 1 − 1 3p2 ) if p 6= 0 and G0 (x) = sinh x x −1− x2 6 . (i) If Gp (x) < 0 for all x ∈ (0, ∞), then p ≥ 1. √ (ii) If Gp (x) > 0 for all x ∈ (0, π/2), then p ≤ p1 = 15/5 ≈ 0.77460. Proof. In order to prove the desired results, we need the following two relations: Gp (x) 1 (3.1) 5p2 − 3 , = − lim 4 x→0 x 360 − 6p12 if p > 1, Gp (x) if p = 1, − 61 = (3.2) lim x→∞ epx ∞ if 0 < p < 1, ∞ if p = 0. The first one follows by expanding in power series: 1 Gp (x) = − 5p2 − 3 x4 + o x6 . 360 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES 9 To obtain the second one, it needs to note that e−px Gp (x) = e(1−p)x 1−e2x −2x − which gives (3.2). (i) If Gp (x) < 0 for all x ∈ (0, ∞), limx→∞ e−px Gp (x) ≤ 0. These together (ii) If Gp (x) > 0 for all x ∈ (0, ∞), limx→∞ e−px Gp (x) ≥ 0. These together 1 3p2 1 − e−2px − 1− 2 1 3p2 e−px , then we have limx→0 x−4 Gp (x) ≤ 0 and with (3.1) and (3.2) give p ≥ 1. then we have limx→0 x−4 Gp (x) ≥ 0 and with (3.1) and (3.2) indicate p ≤ p1 . We now can prove Theorem 2. Proof of Theorem 2. The necessity follows by Lemma 5. To prove the sufficiency, we expanding Gp (x) in power series to get sinh x − ( 3p12 cosh px + 1 − 3p12 ) x ! ∞ ∞ X x2n 1 1 X (px)2n − +1− 2 = (2n + 1)! 3p2 n=0 (2n)! 3p n=0 ∞ ∞ X 3 − (2n + 1) p2n−2 X an p 2 2n = x = x2n . 3 (2n + 1)! 3 (2n + 1)! n=2 n=2 √ 2 It is derived from Lemma 3 that an p ≥ 0 if 0 < p ≤ 15/5, and clearly, an p2 < 0 if p ≥ 1. Gp (x) = 4. Applications As simple applications of main results, we present some precise estimations for certain special functions and constants in this section. The sine integral is defined by Z t sin x dx. Si (t) = x 0 Some estimates for sine integral can be seen [33], [34], [35], [26], [9]. Now we give a new result. √ Proposition 1. For t ∈ (0, π/2) and p ∈ (0, 15/5], we have c0 (p) 5 c1 (p) 5 sin pt sin pt 1 1 (4.1) 3p3 + 1 − 3p2 t + 5 t < Si (t) < 3p3 + 1 − 3p2 t + 5 t , where c0 (p) = (π/2)−4 Fp (π/2− ) and c1 (p) = 3 − 5p2 /360, here Fp (π/2− ) is defined by (1.15). Particularly, putting p = 0+ , 2/3, we have 1 1 5 1 3 2π 3 − 48π + 96 5 t < Si (t) < t − t3 + t + t , 18 15π5 18 600 9 2t 1 2 (16 − 5π) 5 2t 1 7 5 9 t < Si (t) < sin + t + (4.3) sin + t + t , 8 3 4 5π5 8 3 4 16200 and then, 1 1 3 1 1 3 1 2 π− π + < Si π2 < π − π + π 5 ≈ 1.3714, 1.3705 ≈ 5 360 5 2 144 19 200 √ √ 1 1 9 3 1 7 9 3 1.3706 ≈ π+ + < Si π2 < π + π5 + ≈ 1.3711. 16 16 5 8 518 400 16 (4.2) t− 10 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES Proof. Integrating each sides in (2.5) over [0, t] yields (4.1). Taking the limits of the left and right hand sides in (4.1) as p → 0+ gives (4.2), and putting p = 2/3 in (4.1) leads to (4.3). Substituting t = π/2 into (4.2) and (4.3) we get the last two approximations of Si (π/2). It is known that Z ∞ 0 ′ 1 π2 x dx = ψ ′ ( 12 ) = , sinh x 2 4 where ψ is the tri-gamma function defined by Z ∞ xe−tx ψ ′ (t) = dx. 1 − e−x 0 We define Z Sh (t) = 0 Then by (1.16) we have t x dx. sinh x x 9 3 √ < < . cosh x + 2 sinh x 5 cosh( 15x/5) + 4 Integrating over [0, t] and calculating lead to Proposition 2. For t > 0, we have √ √ √ √ t √3+2 − 3 ln eet − 3 ln 2 − 3 + 3+2 √ √ √ (4.4) < Sh (t) < 2 15 arctan 53 e 15t/5 + 43 − 2 15 arctan 3. In particular, we have √ √ √ √ 4.5621 ≈ 2 3 ln 2 + 3 < ψ ′ ( 21 ) < 2 15π − 4 15 arctan 3 ≈ 4.9845. The Catalan constant [36] G= ∞ X n=0 (−1)n 2 (2n + 1) = 0.9159655941772190... is a famous mysterious constant appearing in many places in mathematics and physics. Its integral representations contain the following [37] Z π/4 Z Z 1 1 π/2 x π2 π x2 arctan x dx = dx = − ln 2 + (4.5) G= dx. x 2 0 sin x 16 4 sin2 x 0 0 We present an estimation for G below. Proposition 3. We have (4.6) √ 0.91586 ≈ √ 15 4 ln 4 cos 4 cos √ 15π 15π +3 sin 10 +5 √10 √ 15π 15π 10 −3 sin 10 +5 <G< √ 15 5 ln √ √ √ √√ √2−√2+3√15√√2+2+32 ≈ 0.91675, 11 11 2− 2−3 15 2+2+32 Proof. From the fourth and fifth inequalities in (1.17) we obtain that for x ∈ (0, π/2), the two-side inequality 1 5 9 cos √ 15x 5 + 4 9 < x < sin x 1 16 27 cos 3x 4 + 11 27 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES holds true. Integrating both sides over [0, π/2] yields Z π/2 Z π/2 Z π/2 x dx √ < dx < 5 15x sin x 0 0 0 + 94 9 cos 5 Direct computations give Z π/2 dx 5 9 0 Z 0 cos π/2 16 27 √ 15x 5 + dx cos 3x 4 + 4 9 11 27 = = √ √ 16 27 11 dx cos 3x 4 + 11 27 . √ 15π 15π 15 4 cos 10 +3 sin 10 +5 √ √ ln ≈ 1.8317, 15π 15π 4 cos −3 sin 2 10 10 +5 √ √ √ 2 15 11 2−√2+3√15 √2+2+32 ln √ √ √ √√ ≈ 1.8335 11 2− 2−3 15 2+2+32 5 Utilizing the the second formula in (4.5) (4.4) follows. We close this paper by giving some inequalities for bivariate means. The Schwab-Borchardt mean of two numbers a ≥ 0 and b > 0, denoted by SB (a, b), is defined as [38, Theorem 8.4], [39, 3, (2.3)] √2 2 b −a arccos(a/b) if a < b, a if a = b, SB (a, b) = √ a2 −b2 if a > b. arccosh(a/b) The properties and certain inequalities involving Schwab-Borchardt mean can be found in [40], [41]. We now establish a new inequality for this mean. For a < b, letting x = arccos (a/b) in the fourth inequality of (1.17) and using half-angle and triple-angle formulas for cosine function, and multiplying two sides by b, we get v q u 3 u −3a/b t 1 + 1+4(a/b) 2 11 16 + 27 SB (a, b) ≥ 27 b b 2 1/2 q 8 b1/4 + 11 2 (b − 2a)2 (a + b) + 2b3/2 = 27 27 b. For a > b, letting x = arccosh (a/b) in the inequality connecting the fourth and sixth members of (1.17) and using half-angle and triple-angle formulas for hyperbolic cosine function, and multiplying two sides by b, we get the same inequality as above. Proposition 4. For a, b > 0, we have (4.7) SB (a, b) ≥ √ 8 2 27 |b − 2a| r a+b + b3/2 2 !1/2 b1/4 + 11 27 b. Remark 3. From the inequality (4.7), it is easy to get √ 11 + 8 2 SB (a, b) ≥ b ≈ 0.82643 × b 27 due to |b − 2a| ≥ 0. It seems to new and interesting. For a, b > 0, with x = (1/2) ln (a/b), we have L(a, b) sinh x = , x G (a, b) cosh px = App (a, b) (ap + bp ) /2 √ p = p , G (a, b) ab 12 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES and by Theorem 2 we immediately get the following Proposition 5. For a, b > 0 with a 6= b, the double inequality √ √ 5 √15/5 1− 15/5 G A 9 15/5 (4.8) + 94 G < L < 13 A + 32 G √ holds with the best constants p1 = 15/5 and 1. And, the function (ln b − ln a)2 p 7→ 3p12 App G1−p + 1 − 3p12 G if p 6= 0 and G + G if p = 0 24 is increasing on R. Remark 4. Accordingly, Corollary 2 can be changed into the chain of inequalities for means: √ 1/ 3 √ A1/√3 G1−1/ 3 < 3 2/3 1/3 4 A2/3 G < 16 3/4 1/4 27 A3/4 G < 1 3A √ 1/ 2 √ + 14 G < 32 A1/√2 G1−1/ + 23 G < 2 + 31 G √ √ 11 5 √15/5 1− 15/5 27 G < 9 A 15/5 G √ √ 1 2/√3 1−2/ 3 + 21 G. A G 2 1/ 3 + + 94 G < L 1/(3q ) Remark 5. In [19], Yang obtained a sharp lower bound Aq0 0 G1−1/(3q0 ) for √ the logarithmic mean L, where q0 = 1/ 5, and pointed out that this one seems to superior to most of known ones.√ Now, we derive a new sharp lower bound 5 p1 1−p1 + 94 G for L, where p1 = 15/5. We claim that the latter is better than 9 Ap1 G the former. In fact, we have L > 59 App11 G1−p1 + 94 G > Aq1/(3q) G1−1/(3q) √ if and only if q ≥ q0 = 1/ 5. In order for the second inequality in (4.9) to hold, it suffices that for x > 0 1/ 3q2 D (x) = 3p12 cosh p1 x + 1 − 3p12 − (cosh qx) ( ) > 0 1 1 √ if and only if q ≥ q0 = 1/ 5. The necessity can be obtained by limx→0 x−4 D (x) ≥ 0, which follows by expanding in power series 1 4 2 x p1 + 2q 2 − 1 + o x6 . D (x) = 72 p √ 2 Then, q ≥ (1 − p1 ) /2 = 1/ 5. 1/ 3q2 Since q 7→ (cosh qx) ( ) is decreasing on (0, ∞) proved in [19, Lemma 2], (4.9) to prove D (x) ≥ 0 if q ≥ q0 , it suffices to show that D (x) ≥ 0 when q = q0 . Differentiation yields 2 1 1 sinh p1 x − cosh1/(3q0 )−1 q0 x sinh q0 x D′ (x) = 3p1 3q0 2 sinh q0 x q0 sinh p1 x = − cosh1/(3q0 )−1 q0 x 3q0 p1 sinh q0 x 2 sinh q0 x q0 sinh p1 x = ×L , cosh1/(3q0 )−1 q0 x × D1 (x) , 3q0 p1 sinh q0 x where 1 q0 sinh p1 x − − 1 ln cosh q0 x. D1 (x) = ln p1 sinh q0 x 3q02 NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES 13 Differentiating D1 (x) gives D1′ (x) = D2 (x) , 6q0 sinh 2q0 x sinh p1 x where = 4(− sinh p1 x sinh2 q0 x − 3q02 cosh2 q0 x sinh p1 x D2 (x) +3q02 sinh p1 x sinh2 q0 x + 3p1 q0 cosh p1 x cosh q0 x sinh q0 x). Utilizing ”product into sum” formulas and expanding in power series lead to D2 (x) = −2 6q02 − 1 sinh p1 x + (3p1 q0 − 1) sinh (p1 x + 2q0 x) + (3p1 q0 + 1) sinh (2q0 x − p1 x) ∞ X p2n−1 x2n−1 = dn 1(2n−1)! , n=1 where dn = (3p1 q0 − 1) 1 + 2q0 p1 2n−1 + (3p1 q0 + 1) 2q0 p1 −1 2n−1 − 2 6q02 − 1 Now we show that dn ≥ 0 for n ≥ 1. A simple verification yields d1 = d2 = 0, d3 = 64/45 > 0. Suppose that dn > 0 for n > 3, that is, 2n−1 2n−1 2q0 2q0 2 −1 . > 2 6q0 − 1 − (3p1 q0 − 1) 1 + (3p1 q0 + 1) p1 p1 Then, dn+1 = > (3p1 q0 − 1) 1 + (3p1 q0 − 1) 1 + × 2q0 −1 p1 2 2n+1 2n+1 0 + (3p1 q0 + 1) 2q − 1 − 2 6q02 − 1 p1 2n+1 2n−1 2q0 2q0 2 + 2 6q0 − 1 − (3p1 q0 − 1) 1 + p1 p1 2q0 p1 − 2 6q02 − 1 ! 2n−1 2q0 2 p1 (3p1 q0 − 1) 1 + − 6q0 − 1 (p1 − q0 ) . = p1 √ Since (3p1 q0 − 1) = 3 3 − 5 /5 > 0, using binomial expansion we get p21 2q0 − 6q02 − 1 (p1 − q0 ) dn+1 > p1 (3p1 q0 − 1) 1 + (2n − 1) 8q0 p1 = 2q0 (3p1 q0 − 1) (2n − 1) + q0 3p21 + 6q02 − 6p1 q0 − 1 8q0 p21 = = 2q0 (3p1 q0 − 1) (2n − 1) − 2q0 (3p1 q0 − 1) 4q0 (3p1 q0 − 1) (n − 1) > 0, where the third equality holds due to 3p21 + 6q02 = 3. By mathematical induction, we have proven D2 (x) ≥ 0 for n ≥ 1. It follows that D1′ (x) > 0, which means that D1 is increasing on (0, ∞), and then D1 (x) > limx→0+ D (x) = 0. 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