A new generalized and sharp version of Jordan’s inequality and its applications to the improvement of the Yang Le inequality, II

Applied Mathematics Letters 20 (2007) 532–538 www.elsevier.com/locate/aml

A new generalized and sharp version of Jordan’s inequality and its applications to the improvement of the Yang Le inequality, II Shanhe Wu a , Lokenath Debnath b,∗ a Department of Mathematics, Longyan College, Longyan Fujian 364012, People’s Republic of China b Department of Mathematics, University of Texas-Pan American, Edinburg, TX 78539, USA

Received 15 May 2006; accepted 20 May 2006

Abstract This paper deals with a new generalized and sharp version of Jordan’s inequality which is a unified generalization of some known results. An application of our results to the improvement of the Yang Le inequality is provided in this paper. c 2006 Elsevier Ltd. All rights reserved.  Keywords: Jordan’s inequality; Yang Le inequality; Sharpness; Generalization; Best coefficient

1. Introduction The Jordan’s inequality (see Mitrinovi´c and Vasi´c [1]) 2 sin x ≤ < 1, π x

0
π 2

(1)

has important applications in many areas of pure and applied mathematics. This simple inequality has motivated a large number of research papers concerning its new proofs, various generalizations, sharpness and applications (e.g., ¨ see Zhu [2,3], Ozban [4], Debnath and Zhao [5], Yuefeng [6], Wu [7], Mercer et al. [8]). In a recent paper Wu and Debnath [9] established the following sharp lower and upper bounds for the function ( sinx x ) as an improvement of Jordan’s inequality max {N1 (θ, x), N2 (θ, x)} ≤ where N1 (θ, x) =

1 sin θ + θ 2



sin x ≤ min {M1 (θ, x), M2 (θ, x)} , 0 < x ≤ θ ≤ π, x

sin θ − cos θ θ

    x2 sin θ  x 2 3 2 cos θ 1− 2 + + − 1− , θ 2 3 3 θ θ

∗ Corresponding author. Tel.: +1 956 381 3459; fax: +1 956 384 5091.

E-mail addresses: [email protected] (S. Wu), [email protected] (L. Debnath). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter  doi:10.1016/j.aml.2006.05.022

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S. Wu, L. Debnath / Applied Mathematics Letters 20 (2007) 532–538

533

   2  sin θ x2 3 1 x2 sin θ 1− 2 , − cos θ 1− 2 − θ sin θ + cos θ − θ 8 3 θ θ θ       2 x 2 sin θ 1 sin θ x sin θ 3 1 1− , M1 (θ, x) = + − cos θ 1− 2 − θ sin θ + cos θ − θ 2 θ 2 3 θ θ θ   2    x2 1 sin θ sin θ sin θ 3 2 cos θ x2 + − cos θ 1− 2 + + − M2 (θ, x) = 1− 2 . θ 2 θ θ 2 3 3 θ θ sin θ 1 N2 (θ, x) = + θ 2



Motivated by the above inequality, we consider further improvement of Jordan’s inequality. In this paper, we give a new generalized and sharp version of Jordan’s inequality by introducing exponential parameters, our result is a further generalization of inequality (2) as well as a unified generalization of several results of previous papers [2–5]. Finally, we provide an application of our results to the improvement of the Yang Le inequality. 2. Generalization and sharpness of Jordan’s inequality Theorem 1. Let 0 < x ≤ θ, θ ∈ (0, π2 ], and let τ ≥ 2, τ ≤ λ ≤ 2τ . Then we have the inequality         1 sin θ xλ xτ 2 1 sin θ sin θ 1 − λ + 2 (1 + λ) 1− τ + − cos θ − cos θ − θ sin θ θ λ θ θ 2τ θ θ         λ sin θ 1 sin θ x 1 sin θ xτ 2 sin x sin θ 1− λ + 1− 1− τ ≤ + − cos θ − − cos θ ≤ , (3) x θ λ θ θ θ λ θ θ     where the equality holds if and only if x = θ . Furthermore, 2τ1 2 (1 + λ) sinθ θ − cos θ − θ sin θ and 1 − sinθ θ −   1 sin θ λ θ − cos θ are the best coefficients in (3). Further, the inequality (3) is reversed for 1 ≤ τ ≤ 5/3, λ ≤ τ and λ = 0 (or λ ≥ 2τ ). In order to prove the Theorem 1, we will use the following lemma (see Anderson et al. [10] and [11]). Lemma 1. Let f , g : [a, b] → R be two continuous functions which are differentiable on (a, b). Further, let g = 0 on (a, b). If ( f /g ) is increasing (or decreasing) on (a, b), then the functions f (x) − f (b) g (x) − g (b)

and

f (x) − f (a) g (x) − g (a)

are also increasing (or decreasing) on (a, b). Proof of Theorem 1. When x = θ , (3) becomes an identity. We suppose 0 < x < θ ≤ π2 below.      λ τ 2 Let f 1 (x) = sinx x − sinθ θ − λ1 sinθ θ − cos θ 1 − θx λ , f 2 (x) = 1 − θx τ , f 3 (x) = x −τ cos x −   x −τ −1 sin x + θ1λ sinθ θ − cos θ x λ−τ , f 4 (x) = 2τ θ −2τ (x τ − θ τ ), f5 (x) = (τ + 1) sin x − (τ + 1)x cos x −

2 −2τ 2τ +1 , f (x) = (τ − 1) sin x − x cos x, f (x) = 2τ 2 (2τ + 1)θ −2τ x 2τ −1 , and let x 2 sin x, f 6 (x) 7 8  = 2τ θ  x λ−τ sin θ 2τ −λ x λ−2τ . Then, we have F(x) = 2τ 2 − cos θ θ θ    sin x sin θ 1 sin θ xλ − − − cos θ 1 − λ f 1 (x) − f 1 (θ ) x θ λ θ f 1 (x) θ = = ,   xτ 2 f 2 (x) f 2 (x) − f 2 (θ ) 1 − θτ   x −τ cos x − x −τ −1 sin x + θ1λ sinθ θ − cos θ x λ−τ f 1 (x) f 3 (x) − f 3 (θ ) = = , −2τ τ τ f 2 (x) 2τ θ f 4 (x) − f 4 (θ ) (x − θ )   f 3 (x) λ − τ sin θ (τ + 1) sin x − (τ + 1)x cos x − x 2 sin x − cos θ θ 2τ −λ x λ−2τ + = f 4 (x) θ 2τ 2 θ −2τ x 2τ +1 2τ 2 f 5 (x) − f 5 (0) = + F(x), f 6 (x) − f 6 (0)

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f 5 (x) (τ − 1) sin x − x cos x f 7 (x) − f 7 (0) = 2 , = −2τ 2τ −1 f 6 (x) f 8 (x) − f 8 (0) 2τ (2τ + 1)θ x f 7 (x) (τ − 2) cos x + x sin x = 2 = H (x). f 8 (x) 2τ (2τ + 1)(2τ − 1)θ −2τ x 2τ −2 Differentiating H (x) and F(x) with respect to x gives (5 − 3τ )x sin x − 2(τ − 1)(τ − 2) cos x + x 2 cos x , 2τ 2 (2τ + 1)(2τ − 1)θ −2τ x 2τ −1   (λ − τ )(λ − 2τ ) sin θ − cos θ θ 2τ −λ x λ−2τ −1 . F (x) = 2τ 2 θ

H (x) =

Case (I). When τ ≥ 2 and τ ≤ λ ≤ 2τ . By the well-known inequality (see Mitrinovic and Vasic [1]):  π sin x − cos x > 0, x ∈ 0, , x 2 we find   −3(τ − 2)x sin x − 2(τ − 1)(τ − 2) cos x − sinx x − cos x x 2 < 0, H (x) = 2τ 2 (2τ + 1)(2τ − 1)θ −2τ x 2τ −1 and (λ − τ )(λ − 2τ ) F (x) = 2τ 2



 sin θ − cos θ θ 2τ −λ x λ−2τ −1 ≤ 0 θ

for 0 < x < θ ≤

π . 2

These show that H (x) and F(x) are decreasing on (0, θ ). Further, applying Lemma 1, we can deduce in order from the above relation chains that In view of the fact that f 1 (x) = f (x) = f 2 (x)

sin x x

f 7 (x) f 5 (x) f 3 (x) f 1 (x) f 1 (x) , , , , f 8 (x) f 6 (x) f 4 (x) f 2 (x) f 2 (x)

−

sin θ θ

−

1 λ





sin θ θ

1−

− cos θ  xτ 2

are all decreasing on (0, θ ).

 1−

xλ θλ



θτ

is decreasing on (0, θ ), this result together with   1 sin θ sin θ lim f (x) = 1 − − − cos θ , θ λ θ x→0+

lim f (x) =

x→θ −

1 2τ 2

    sin θ − cos θ − θ sin θ , (1 + λ) θ

gives the following inequality       1 sin θ 1 sin θ sin θ − cos θ − θ sin θ < f (x) < 1 − − − cos θ + λ) (1 θ θ λ θ 2τ 2     This yields inequality (3) and shows that 2τ1 2 (1 + λ) sinθ θ − cos θ − θ sin θ and 1 − are the best coefficients in (3). Case (II). When 1 ≤ τ ≤ 5/3, λ ≤ τ and λ = 0 (or λ ≥ 2τ ), we have H (x) =

3( 53 − τ )x sin x + 2( 53 − τ )(τ − 1) cos x + 23 (τ − 1) cos x + x 2 cos x 2τ 2 (2τ + 1)(2τ − 1)θ −2τ x 2τ −1

and (λ − τ )(λ − 2τ ) F (x) = 2τ 2



 sin θ − cos θ θ 2τ −λ x λ−2τ −1 ≥ 0 θ

for x ∈ (0, θ ). sin θ θ

−

1 λ



sin θ θ

− cos θ



> 0,

for 0 < x < θ ≤

π , 2

this means that H (x) and F(x) are increasing on (0, θ ). Making use of Lemma 1, we deduce in order from the above relation chains that

f 7 (x) f 5 (x) f 3 (x) f 1 (x) f 1 (x) , , , , f 8 (x) f 6 (x) f 4 (x) f 2 (x) f 2 (x)

are all increasing on (0, θ ). Consequently, we conclude that

S. Wu, L. Debnath / Applied Mathematics Letters 20 (2007) 532–538

limx→0+ 1−

f 1 (x) f 2 (x)

<

f 1 (x) f 2 (x)

1 sin θ − θ λ



< limx→θ − sin θ − cos θ θ

535

f 1 (x) f 2 (x)



for 0 < x < θ . Consequently,     f 1 (x) 1 sin θ < < 2 (1 + λ) − cos θ − θ sin θ , f 2 (x) 2τ θ

which leads to the reversed inequality of (3). This completes the proof of Theorem 1.



We give here some direct consequences from Theorem 1. Putting τ = λ and τ = 1 in Theorem 1 respectively, we obtain Corollary 1. If 0 < x ≤ θ, θ ∈ (0, π2 ] and λ ≥ 2. Then max {N1 (λ, θ, x), N2 (λ, θ, x)} ≤ where

sin x ≤ min{M1 (λ, θ, x), M2 (λ, θ, x)}, x

(4)

      x 2 xλ 1 sin θ sin θ 1− 1− λ + 1− − − cos θ , θ θ λ θ θ       2  1 sin θ xλ xλ sin θ 1 sin θ N2 (λ, θ, x) = + − cos θ 1 − λ + 2 (1 + λ) − cos θ − θ sin θ 1 − λ , θ λ θ θ 2λ θ θ        x 2 1 sin θ xλ sin θ 1 sin θ 1− λ + + − cos θ − cos θ − θ sin θ 1 − M1 (λ, θ, x) = , (1 + λ) θ λ θ θ 2 θ θ        2 sin θ 1 sin θ xλ 1 sin θ xλ sin θ M2 (λ, θ, x) = + − cos θ 1− λ + 1− − − cos θ 1− λ . θ λ θ θ θ λ θ θ 1 sin θ + N1 (λ, θ, x) = θ λ



sin θ − cos θ θ

In particular, taking λ = 2 in Corollary 1 gives inequality (2). Putting θ = π2 in Theorem 1, we get the following generalized and sharp version of Jordan’s inequality Corollary 2. If 0 < x ≤

π 2,τ

≥ 2 and τ ≤ λ ≤ 2τ . Then

2 2 4λ + 4 − π 2  τ 2 + π − 2τ x τ (π λ − 2λ x λ ) + λ+1 2 2τ +1 π λπ 4τ π 2 sin x λπ − 2λ − 2  τ 2 2 π − 2τ x τ . ≤ (π λ − 2λ x λ ) + ≤ + x π λπ λ+1 λπ 2τ +1 Inequality (5) is reversed for 1 ≤ τ ≤ 5/3, λ ≤ τ and λ = 0 (or λ ≥ 2τ ).

(5)

Specially, putting (τ, λ) = (2, 2) and (τ, λ) = (1, 2) in Corollary 2 respectively yields the main results of previous ¨ papers due to Zhu [2,3] and Ozban, [4], that is Corollary 3. If 0 < x ≤

π 2,

then

1 2 1 12 − π 2 2 sin x π −3 2 2 + 3 (π 2 − 4x 2) + ≤ + 3 (π 2 − 4x 2) + (π − 4x 2)2 ≤ (π − 4x 2)2 , π π 16π 5 x π π π5 π 2 sin x π 2 1 2 1 2 4(π − 3)  12 − π 2  x− x− + 3 (π 2 − 4x 2) + ≤ + 3 (π 2 − 4x 2 ) + ≤ . 3 3 π 2 x π 2 π π π π

(6) (7)

3. Application to the improvement of the Yang Le inequality It is well-known that the Yang Le inequality plays an important role in the theory of distribution of values of functions (see Yang [12]). This inequality is stated below: If A1 > 0, A2 > 0, A1 + A2 ≤ π and 0 ≤ μ ≤ 1. Then cos2 μA1 + cos2 μA2 − 2 cos μπ cos μA1 cos μA2 ≥ sin2 μπ.

(8)

Considerable attention has been given to the Yang Le inequality by many authors, and it has been generalized and extended in different directions (see [2–5]). We give here a new improvement of the Yang Le inequality.

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S. Wu, L. Debnath / Applied Mathematics Letters 20 (2007) 532–538

Theorem 2. Let Ai ≥ 0 (i = 1, 2, . . . , n, n ≥ 2), with max{K 1 (λ, θ ), K 2 (λ, θ )} ≤ (n − 1)

n

n i=1

Ai ≤ θ, θ ∈ [0, π], and let λ ≥ 2. Then

cos2 Ak − 2 cos θ

cos Ai cos A j

1≤i< j ≤n

k=1

≤ min{Q 1 (λ, θ ), Q 2 (λ, θ )}, where K 1 (λ, K 2 (λ, Q 1 (λ, Q 2 (λ,

   θλ n θ) = λ+1− λ 2 π    θλ n θ) = λ+1− λ 2 π    θλ n θ) = λ+1− λ 2 π    θλ n θ) = λ+1− λ 2 π

(9)

2   λπ − 2λ − 2 θ 2 2θ θ + , 1− cos 2 π λπ 2 2 2  2θ 4λ + 4 − π 2 θλ θ + , 1− λ cos 8λ π λπ 2 2   4λ + 4λ2 − λπ 2 θ 2 2θ + , 1− 8 π λπ 2  2 2θ θλ λπ − 2λ − 2 1− λ + . 2 π λπ

We first state the following lemma (see Wu and Debnath [9]): Lemma 2. Let A ≥ 0, B ≥ 0 and let A + B ≤ θ , θ ∈ [0, π]. Then sin2 θ ≤ cos2 A + cos2 B − 2 cos θ cos A cos B ≤ 4 sin2

θ . 2

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Proof of Theorem 2. When θ = 0, (9) is an identity. We suppose 0 < θ ≤ π below. Let Hi j = cos2 Ai + cos2 A j − 2 cos θ cos Ai cos A j . It follows from Lemma 2 that sin2 θ ≤ Hi j ≤ 4 sin2

θ 2

(1 ≤ i < j ≤ n).

Taking the sum of all inequalities in (11), we obtain



θ sin2 θ ≤ Hi j ≤ 4 sin2 . 2 1≤i< j ≤n 1≤i< j ≤n 1≤i< j ≤n A simple calculation of (12) leads to the following inequality:     n



θ θ θ n n cos2 Ak − 2 cos θ cos Ai cos A j ≤ 4 4 sin2 . sin2 cos2 ≤ (n − 1) 2 2 2 2 2 k=1 1≤i< j ≤n

(11)

(12)

(13)

On the other hand, applying Corollary 1 together with 0 < θ2 ≤ π2 , gives           sin θ2 π θ π θ π θ π θ max N1 λ, , , N2 λ, , ≤ θ ≤ min M1 λ, , , M2 λ, , . 2 2 2 2 2 2 2 2 2 This yields the following result 0 < max {N1 (λ, θ ), N2 (λ, θ )} ≤ sin where θ N1 (λ, θ ) = λπ



θ ≤ min {M1 (λ, θ ), M2 (λ, θ )} , 2

   λπ − 2λ − 2 θλ θ 2 , λ+1− λ + 1− π 2 π

(14)

S. Wu, L. Debnath / Applied Mathematics Letters 20 (2007) 532–538

θ N2 (λ, θ ) = λπ M1 (λ, θ ) =

θ λπ

M2 (λ, θ ) =

θ λπ



537

 λ 2

 θλ 4λ + 4 − π 2 θ λ+1− λ + , 1− λ π 8λ π     θλ 4λ + 4λ2 − λπ 2 θ 2 , λ+1− λ + 1− π 8 π  2   λπ − 2λ − 2 θλ θλ . λ+1− λ + 1− λ π 2 π

Combining inequalities (13) and (14) yields inequality (9). This completes the proof of Theorem 2.  Note that when ni=1 Ai ≤ π and 0 ≤ μ ≤ 1, it implies ni=1 μAi ≤ μπ and μπ ∈ [0, π]. Now, using the Theorem 2 and substituting Ai → μAi (i = 1, 2, . . . , n) and θ → μπ in inequality (9) yields the following: Corollary 4. Let Ai > 0 (i = 1, 2, . . . , n, n ≥ 2), with ni=1 Ai ≤ π, and let 0 ≤ μ ≤ 1, λ ≥ 2. Then max {K 1 (λ, μ), K 2 (λ, μ)} ≤ (n − 1)

n



cos2 μAk − 2 cos μπ

cos μAi cos μA j

1≤i< j ≤n

k=1

≤ min {Q 1 (λ, μ), Q 2 (λ, μ)} , where

(15)

     μπ 2 λπ − 2λ − 2 n λ 2 2μ K 1 (λ, μ) = cos , λ+1−μ + (1 − μ) 2 2 λ 2 2      μπ 4λ + 4 − π 2  n λ λ 2 2μ cos K 2 (λ, μ) = , 1−μ λ+1−μ + 2 8λ λ 2    2  4λ + 4λ2 − λπ 2 n λ 2 2μ Q 1 (λ, μ) = , λ+1−μ + (1 − μ) 2 8 λ      2 2μ 2 λπ − 2λ − 2  n λ + 1 − μλ + Q 2 (λ, μ) = . 1 − μλ 2 2 λ

In particular, when λ = 2 in Corollary 4, we obtain Corollary 5. Let Ai > 0 (i = 1, 2, . . . , n, n ≥ 2), with max {K 1 (μ), K 2 (μ)} ≤ (n − 1)

n

n i=1

Ai ≤ π, and let 0 ≤ μ ≤ 1. Then

cos2 μAk − 2 cos μπ

k=1



≤ min {Q 1 (μ), Q 2 (μ)} , where

cos μAi cos μA j

1≤i< j ≤n

(16)

  2  μπ  n , 3 − μ2 + (π − 3) (1 − μ)2 μ2 cos2 2 2   2 2  μπ  12 − π 2  n 1 − μ2 , μ2 cos2 K 2 (μ) = 3 − μ2 + 2 16 2   2 12 − π 2 n Q 1 (μ) = 3 − μ2 + (1 − μ)2 μ2 , 2 4    2 2 n 2 2 Q 2 (μ) = μ2 . 3 − μ + (π − 3) 1 − μ 2

K 1 (μ) =

Inequality (16) is the main result of our recent paper [9], which is also a unified generalization of the earlier results of many authors (see [2–5,13].

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S. Wu, L. Debnath / Applied Mathematics Letters 20 (2007) 532–538

Acknowledgment The research was supported by the Natural Science Foundation of Fujian Province Education Department of China under grant No. JA05324. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

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