A GENERAL GENERALIZATION OF JORDAN’S INEQUALITY AND A REFINEMENT OF L. YANG’S INEQUALITY

FENG QI, DA-WEI NIU, AND JIAN CAO

Abstract. In this article, for t ≥ 2, a general generalization of Jordan’s inequality ∑_{k=1}^{n} μ_{k}(θ^{t} − x^{t})^{k} ≤ \frac{sin x}{x} − \frac{sin θ}{θ} ≤ ∑_{k=1}^{n} ω_{k}(θ^{t} − x^{t})^{k} for n ∈ ℕ and θ ∈ (0, π] is established, where the coefficients μ_{k} and ω_{k} defined by recursing formulas (11) and (12) are the best possible. As an application, L. Yang’s inequality is refined.

1. Introduction

The well known Jordan’s inequality (see [2, 6], [4, p. 143], [8, p. 269] and [11, p. 33]) states that

\frac{2}{π} ≤ \frac{sin x}{x} < 1  (1)

for 0 < |x| ≤ \frac{π}{2} . The equality in (1) is valid if and only if x = \frac{π}{2} .

Jordan’s inequality has important applications in analysis and other branches of mathematics. Therefore, many mathematicians have struggled to refine, generalize and apply it. For more detailed information, please refer to [8, pp. 274–275] and [1, 5, 6, 7, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 28, 31, 32, 33], especially [15], and the references therein.

In [1, 10, 14, 16, 17, 18, 19], among other things, Jordan’s inequality had been refined as

\frac{1}{π^{3}} x(π^{2} − 4x^{2}) ≤ sin x − \frac{2}{π} x ≤ \frac{π − 2}{π^{3}}x(π^{2} − 4x^{2}).  (2)

In [33], a stronger sharp double inequality for x ∈ (0, \frac{π}{2} ] was obtained:

\frac{12 − π^{2}}{16π^{5}} (π^{2} − 4x^{2})^{2} ≤ \frac{sin x}{x} − \frac{2}{π} − \frac{1}{π^{3}} (π^{2} − 4x^{2}) ≤ \frac{π − 3}{π^{5}}(π^{2} − 4x^{2})^{2}. (3)

Recently in [12], the following general refinement of Jordan’s inequality was showed:

\frac{2}{π} + ∑_{k=1}^{n} α_{k}(π^{2} − 4x^{2})^{k} ≤ \frac{sin x}{x} ≤ \frac{2}{π} + ∑_{k=1}^{n} β_{k}(π^{2} − 4x^{2})^{k},  (4)

where the constants

α_{k} = \frac{(−1)^{k}}{(4π)^{k}k!} ∑_{i=1}^{k+1}(\frac{2}{π} )^{i}c_{i−1}^{k} sin(\frac{k + i}{2}π)  (5)

2000 Mathematics Subject Classification. 26D05, 26D15.

Key words and phrases. Jordan’s inequality, L. Yang’s inequality, L’Hôspital’s rule, refinement, application.

This paper was typeset using AMS-L^{A}TEX.

and

β_{k} = \begin{cases}\frac{1 − \frac{2}{π} − ∑_{i=1}^{n−1} α_{i}π^{2i}}{π^{2n}} , & k = n \\ α_{k}, & 1 ≤ k < n \end{cases}  (6)

with

c_{i}^{k} = \begin{cases}(i + k − 1)c_{i−1}^{k−1} + c_{i}^{k−1}, & 0 < i ≤ k \\ 1, & i = 0 \end{cases}  (7)

in (4) are the best possible.

In [26], as a generalization of Jordan’s inequality (1), the following sharp inequality

\frac{1}{2τ^{2}} [(1 + λ)(\frac{sin θ}{θ} − cos θ) − θ sin θ](1 − \frac{x^{τ}}{θ^{τ}} )^{2}

≤ \frac{sin x}{x} − \frac{sin θ}{θ} − \frac{1}{λ}(\frac{sin θ}{θ} − cos θ)(1 − \frac{x^{λ}}{θ^{λ}} )

≤ [1 − \frac{sin θ}{θ} − \frac{1}{λ}(\frac{sin θ}{θ} − cos θ)](1 − \frac{x^{τ}}{θ^{τ}} )^{2} (8)

was obtained for 0 < x ≤ θ ∈ (0, \frac{π}{2} ], τ ≥ 2 and τ ≤ λ ≤ 2τ. The equalities in (8) holds if and only if x = θ. The coefficients of the term (1 − \frac{x^{τ}}{θ^{τ}} )^{2} are the best possible. If 1 ≤ τ ≤ \frac{5}{3} and either λ ≠ 0 or λ ≥ 2τ then inequality (8) is reversed. Specially, when θ = \frac{π}{2} , inequality (8) becomes

\frac{4λ + 4 − π^{2}}{4τ^{2}π^{2τ+1}} (π^{τ} − 2^{τ} x^{τ} )^{2} ≤ \frac{sin x}{x} − \frac{2}{π} − \frac{2}{λπ^{λ+1}} (π^{λ} − 2^{λ}x^{λ})

≤ \frac{λπ − 2λ − 2}{λπ^{2τ+1}}(π^{τ} − 2^{τ} x^{τ} )^{2} (9)

for 0 < x ≤ \frac{π}{2} , τ ≥ 2 and τ ≤ λ ≤ 2τ. If 1 ≤ τ ≤ \frac{5}{3} and either λ ≠ 0 or λ ≥ 2τ then inequality (9) is reversed. If taking (τ, λ) = (2, 2) in (9), then inequality (3) can be deduced.

For recent developments of the refinements, generalizations and applications of Jordan’s inequality, please refer to the expository and summary article [15].

The first aim of this paper is to generalize inequalities (4) and (8). One of the main results of this paper is the following Theorem 1.

Theorem 1. For 0 < x ≤ θ < π, n ∈ ℕ and t ≥ 2, inequality

∑_{k=1}^{n} μ_{k}(θ^{t} − x^{t})^{k} ≤ \frac{sin x}{x} − \frac{sin θ}{θ} ≤ ∑_{k=1}^{n} ω_{k}(θ^{t} − x^{t})^{k}  (10)

holds with the equalities if and only if x = θ, where the constants

μ_{k} = \frac{(−1)^{k}}{k!t^{k}} ∑_{i=1}^{k+1} a_{i−1}^{k}θ^{k−i−kt} sin(θ + \frac{k + i − 1}{2}π)  (11)

A GENERAL GENERALIZATION OF JORDAN’S INEQUALITY AND AN APPLICATION 3

and

ω_{k} = \begin{cases}\frac{1 − \frac{sin θ}{θ} − ∑_{i=1}^{n−1} μ_{i}θ^{ti}}{θ^{tn}} , & k = n \\ μ_{k}, & 1 ≤ k < n \end{cases}  (12)

with

a_{i}^{k} = \begin{cases}a_{i}^{k−1} + [i + (k − 1)(t − 1)]a_{i−1}^{k−1}, & 0 < i ≤ k \\ 1, & i = 0 \\ 0, & i > k \end{cases}  (13)

in (10) are the best possible.

Remark 1. Taking t = 2 in (10) yields inequality (4). Letting n = 2 in (10) leads to (8) for λ = τ = 2.

The second aim of this paper is to apply Theorem 1 to refine L. Yang’s inequality [27] as follows.

Theorem 2. Let 0 ≤ λ ≤ 1, 0 < x ≤ θ < π, t ≥ 2 and A_{i} > 0 with ∑_{i=1}^{n} A_{i} ≤ π for n ∈ ℕ. If m ∈ ℕ and n ≥ 2, then

L_{m}(n, λ) ≤ H(n, λ) ≤ R_{m}(n, λ),  (14)

where

L_{m}(n, λ) = \binom{n}{2}λ^{2}π^{2}[\frac{sin θ}{θ} + ∑_{k=1}^{m} 2^{−kt}μ_{k}(2^{t}θ^{t} − λ^{t}π^{t})^{k}]^{2} cos^{2}(\frac{λ}{2}π),  (15)

H(n, λ) = (n − 1) ∑_{k=1}^{n} cos^{2}(λA_{k}) − 2 cos(λπ) ∑_{1≤i<j≤n} cos(λA_{i}) cos(λA_{j}), (16)

R_{m}(n, λ) = \binom{n}{2}λ^{2}π^{2}[\frac{sin θ}{θ} + ∑_{k=1}^{m} 2^{−kt}ω_{k}(2^{t}θ^{t} − λ^{t}π^{t})^{k}]^{2} cos^{2}(\frac{λ}{2}π),  (17)

and μ_{k} and ω_{k} are defined by (11).

2. Lemmas

To prove our main results, the following lemmas are necessary.

Lemma 1. For x > 0, let u_{0}(x) = \frac{sin x}{x} and u_{k}(x) = \frac{u′_{k−1}(x)}{x^{r}} for k ∈ ℕ and r ≥ 1. Then

u_{k}(x) = ∑_{i=1}^{k+1} \frac{a_{i−1}^{k} sin(x + \frac{i+k−1}{2}π)}{x^{kr+i}},  (18)

where a_{i}^{k} is defined by (13).

Proof. It is apparent that u_{1}(x) = x^{−r}(\frac{sin x}{x})′ = x^{−1−r} cos x − x^{−2−r} sin x, which tells us that formula (18) is valid for k = 1.

Now assume formula (18) holds for some given k > 1. Direct computation by using (13) gives

u_{k+1} = ∑_{i=1}^{k+1} a_{i−1}^{k}[\frac{1}{x^{kr+i+r}} cos(x + \frac{k + i − 1}{2}π)

− \frac{1}{x^{kr+i+r+1}} sin(x + \frac{k + i − 1}{2}π)]

= \frac{a_{0}^{k}}{x^{kr+r+1}} cos(x + \frac{k}{2} π) − \frac{(kr + k + 1)a_{k}^{k}}{x^{kr+r+k+2}} sin(x + kπ)

− ∑_{i=0}^{k−1} \frac{a_{i}^{k}(kr + 1 + i) + a_{i+1}^{k}}{x^{kr+r+i+2}} sin(x + \frac{k + i}{2}π)

= \frac{a_{0}^{k+1}}{x^{kr+r+1}} sin(x + \frac{k + 1}{2}π) + \frac{a_{k+1}^{k+1}}{x^{kr+r+k+2}} sin[x + (k + 1)π]

+ ∑_{i=0}^{k−1} \frac{a_{i+1}^{k+1}}{x^{kr+r+i+2}} sin(x + \frac{k + i + 2}{2}π)

= ∑_{i=1}^{k+2} \frac{a_{i−1}^{k+1}}{x^{kr+i+r}} sin(x + \frac{k + i}{2}π).

By mathematical induction, Lemma 1 is proved. □

Lemma 2. For x > 0 and k ∈ ℕ, let v_{1}(x) = ∑_{i=1}^{k+1} a_{i−1}^{k}x^{k−i+1} sin(x + \frac{k+i−1}{2}π) and v_{j+1}(x) = \frac{1}{x}v′_{j}(x) for j ∈ ℕ. Then

v_{j}(x) = ∑_{i=0}^{k−j+1} b_{i}^{j}x^{k−i−j+1} sin(x + \frac{k + i + j − 1}{2}π)  (19)

is valid for j ∈ ℕ, where b_{i}^{1} = a_{i}^{k}, b_{0}^{j} = 1 and

b_{i}^{j} = b_{i}^{j−1} − (k − i − j + 3)b_{i−1}^{j−1},  0 < i ≤ k − j + 1, j > 1.  (20)

Proof. When j = 1, formula (19) is valid clearly.

By induction, suppose that formula (19) holds for some j > 1. Since k − j + 1 > k − (j + 1) + 1, it deduced from (20) that b_{k−j+1}^{j+1} = b_{k−j+1}^{j} − b_{k−j}^{j} = 0. Thus,

v_{j+1}(x) = \frac{1}{x}\{∑_{i=0}^{k−j} b_{i}^{j}[(k − i − j + 1)x^{k−i−j} sin(x + \frac{k + i + j − 1}{2}π)

+ x^{k−i−j+1} cos(x + \frac{k + i + j − 1}{2}π)] + b_{k−j+1}^{j} cos(x + kπ)\}

= b_{0}^{j}x^{k−j} sin(x + \frac{k + j}{2} π)

+ ∑_{i=0}^{k−j−1}[b_{i+1}^{j} − (k − i − j + 1)b_{i}^{j}]x^{k−i−j+1} sin(x + \frac{k + i + j + 1}{2}π)

= b_{0}^{j}x^{k−j} sin(x + \frac{k + j}{2} π) + ∑_{i=0}^{k−j−1} b_{i+1}^{j+1}x^{k−i−j+1} sin(x + \frac{k + i + j + 1}{2}π)

= ∑_{i=0}^{k−j} b_{i}^{j+1}x^{k−i−j} sin(x + \frac{k + i + j}{2} π).

By mathematical induction, formula (19) is proved. □

Lemma 3 ([3]). Let f and g be continuous on [a, b] and differentiable in (a, b) such that g′(x) ≠ 0 in (a, b). If \frac{f′(x)}{g′(x)} is increasing (or decreasing) in (a, b), then the

functions \frac{f(x)−f(b)}{g(x)−g(b)} and \frac{f(x)−f(a)}{g(x)−g(a)} are also increasing (or decreasing) in (a, b).

Lemma 4. Let 0 < x ≤ θ < π and t ≥ 2, then inequality

\frac{1}{t}(\frac{sin θ}{θ^{1+t}} − \frac{cos θ}{θ^{t}} )(θ^{t} − x^{t}) ≤ \frac{sin x}{x} − \frac{sin θ}{θ} ≤ (\frac{1}{θ^{t}} − \frac{sin θ}{θ^{1+t}} )(θ^{t} − x^{t})  (21)

holds with the equalities if and only if x = θ, where the constants

\frac{1}{t}(\frac{sin θ}{θ^{1+t}} − \frac{cos θ}{θ^{t}} ) and (\frac{1}{θ^{t}} − \frac{sin θ}{θ^{1+t}} )

are the best possible.

Proof. Let f(x) = \frac{sin x}{x} − \frac{sin θ}{θ}, g(x) = θ^{t} − x^{t}, f_{1}(x) = x cos x − sin x and g_{1}(x) = −tx^{1+t}. Then

\frac{f(x)}{g(x)} = \frac{f(x) − f(0)}{g(x) − g(0)},  \frac{f′(x)}{g′(x)} = \frac{f_{1}(x) − f_{1}(0)}{g_{1}(x) − g_{1}(0)},  \frac{f′_{1}(x)}{g′_{1}(x)} = \frac{sin x}{t(1 + t)x^{t}} .

Since \frac{sin x}{x^{t}} is decreasing in (0, π], then \frac{f′_{1}(x)}{g′_{1}(x)} is decreasing, and then, in virtue of

Lemma 3, the function \frac{f′(x)}{g′(x)} is decreasing, further \frac{f(x)}{g(x)} is decreasing in (0, π], thus,

\frac{1}{t}(\frac{sin θ}{θ^{1+t}} − \frac{cos θ}{θ^{t}} ) = lim_{x→θ^{−}} \frac{f(x)}{g(x)} ≤ \frac{f(x)}{g(x)} ≤ lim_{x→0^{+}} \frac{f(x)}{g(x)} = \frac{1}{θ^{t}} (1 − \frac{sin θ}{θ} )

and the two constants are the best possible. □

3. Proofs of theorems

Proof of Theorem 1. If n = 1, inequality (10) becomes (21) in Lemma 4. For n ≥ 2, let t = r + 1,

φ(x) = \frac{sin x}{x} − \frac{sin θ}{θ} − ∑_{k=1}^{n−1} μ_{k}(θ^{r+1} − x^{r+1})^{k}, ψ(x) = (θ^{r+1} − x^{r+1})^{n},

φ_{1}(x) = \frac{φ(x)}{x^{r}}, φ_{i+1}(x) = \frac{φ′_{i}(x)}{x^{r}}, ψ_{1}(x) = \frac{ψ′(x)}{x^{r}}, ψ_{i+1}(x) = \frac{ψ′_{i}(x)}{x^{r}},

where 2 ≤ i ≤ n. Then for 1 ≤ k ≤ n − 2,

φ_{k}(x) = u_{k}(x) − [−(r + 1)]^{k}k!μ_{k} − ∑_{i=1}^{n−k−1} \frac{(i + k)!}{i!}μ_{i+k}(θ^{1+r} − x^{1+r})^{i},

φ_{n−1}(x) = u_{n−1}(x) − (n − 1)![−(r + 1)]^{n−1}μ_{n−1} and φ_{n}(x) = u_{n}(x),

where u_{k}(x) for 1 ≤ k ≤ n is defined by (18).

In view of (18), it is deduced that [−(1+r)]^{k}k!μ_{k} = u_{k}(θ) for 1 ≤ k ≤ n−1, hence φ_{i}(θ) = 0 for 1 ≤ i ≤ n−1. A simple calculation gives ψ_{i}(x) = [−(1+r)]^{i} ∏_{ℓ=0}^{i−1}(n− ℓ)(θ^{r+1} − x^{r+1})^{n−i} for 1 ≤ i ≤ n, consequently ψ_{i}(θ) = 0 for 1 ≤ i ≤ n − 1. As a result, for 1 ≤ i ≤ n − 1,

\frac{φ(x)}{ψ(x)} = \frac{φ(x) − φ(θ)}{ψ(x) − ψ(θ)},  \frac{φ′(x)}{ψ′(x)} = \frac{φ_{1}(x) − φ_{1}(θ)}{ψ_{1}(x) − ψ_{1}(θ)},
\frac{φ′_{i}(x)}{ψ′_{i}(x)} = \frac{φ_{i+1}(x) − φ_{i+1}(θ)}{ψ_{i+1}(x) − ψ_{i+1}(θ)},  \frac{φ′_{n−1}(x)}{ψ′_{n−1}(x)} = \frac{φ_{n}(x)}{ψ_{n}(x)} = \frac{u_{n}(x)}{n![−(r + 1)]^{n}} .

Let h_{1}(x) = x^{nr+n+1} and h_{i+1}(x) = \frac{1}{x}h′_{i}(x) for 1 ≤ i ≤ n and n ∈ ℕ. Then it is easy to see that h_{i+1}(x) = ∏_{ℓ=1}^{i}(nr + n − 2ℓ + 3)x^{nr+n−2i+1} for 1 ≤ i ≤ n. Utilization of Lemma 1 and Lemma 2 leads to

\frac{φ′_{n−1}(x)}{ψ′_{n−1}(x)} = \frac{∑_{i=1}^{n+1} a_{i−1}^{n}x^{n−i+1} sin(x + \frac{n+i−1}{2}π)}{n![−(1 + r)]^{n}x^{rn+n+1}} = \frac{v_{1}(x)}{n![−(1 + r)]^{n}h_{1}(x)}

and, since v_{i}(0) = h_{i}(0) = 0 for 1 ≤ i ≤ n + 1,

\frac{v_{1}(x)}{h_{1}(x)} = \frac{v_{1}(x) − v_{1}(0)}{h_{1}(x) − h_{1}(0)}, \frac{v′_{j}(x)}{h′_{j}(x)} = \frac{v_{j+1}(x) − v_{j+1}(0)}{h_{j+1}(x) − h_{j+1}(0)},

\frac{v′_{n}(x)}{h′_{n}(x)} = \frac{v_{n+1}(x) − v_{n+1}(0)}{h_{n+1}(x) − h_{n+1}(0)} = \frac{(−1)^{n} sin x}{∏_{ℓ=1}^{i}(nr + n − 2ℓ + 3)x^{nr−n+1}}

for 1 ≤ j ≤ n − 1. Since \frac{sin x}{x} and x^{−n(r−1)} is decreasing on (0, π), then the function \frac{sin x}{x^{nr−n+1}} is decreasing and \frac{(−1)^{n}v′_{n}(x)}{h′_{n}(x)} is deceasing. Accordingly, from Lemma

3, it follows that the functions \frac{(−1)^{n}v′_{i}(x)}{h′_{i}(x)} and \frac{(−1)^{n}v′_{i−1}(x)}{h′_{i−1}(x)} for 2 ≤ i ≤ n are de-

creasing. Thus, the functions \frac{(−1)^{n}v′_{1}(x)}{h′_{1}(x)} and \frac{(−1)^{n}v_{1}(x)}{h_{1}(x)} are decreasing, and then

\frac{φ′_{n−1}(x)}{ψ′_{n−1}(x)} is decreasing in (0, π). Utilizing Lemma 3 again reveals that the functions

\frac{φ′_{j}(x)}{ψ′_{j}(x)} and \frac{φ′_{j−1}(x)}{ψ′_{j−1}(x)} for 2 ≤ j ≤ n − 1 are decreasing, which implies the deceas-

ingly monotonicity of \frac{φ(x)}{ψ(x)} in (0, π). By L’Hôspital’s rule, it is easy to deduce that

lim_{x→θ−} \frac{φ(x)}{ψ(x)} = lim_{x→θ−} \frac{φ′(x)}{ψ′(x)} = lim_{x→θ−} \frac{φ′_{i}(x)}{ψ′_{i}(x)} = \frac{u_{n}(θ)}{n![−(1+r)]^{n}} = μ_{n} for 1 ≤ i ≤ n−1

and lim_{x→0+} \frac{φ(x)}{ψ(x)} = ω_{n}, which implies μ_{n} ≤ \frac{φ(x)}{ψ(x)} ≤ ω_{n} and the constants μ_{k} and ω_{k} are the best possible.

By the mathematical induction, inequality (10) is proved. □

Proof of Theorem 2. It was proved in [29] and [30, (2.13)] that

sin^{2}(λπ) ≤ cos^{2}(λA_{i}) + cos^{2}(λA_{j}) − 2 cos(λA_{i}) cos(λA_{j}) cos(λπ)

≜ H_{ij} ≤ 4 sin^{2}(\frac{λ}{2}π). (22)

Summing up (22) for 1 ≤ i < j ≤ n yields

\binom{n}{2} sin^{2}(λπ) ≤ ∑_{1≤i<j≤n} H_{ij} = H(n, λ) ≤ 4\binom{n}{2} sin^{2}(\frac{λ}{2}π).  (23)

By virtue of inequality (10) in Theorem 1,

4 sin^{2}(\frac{λ}{2}π) ≤ λ^{2}π^{2}[\frac{sin θ}{θ} + ∑_{k=1}^{m} 2^{−kt}ω_{k}(2^{t}θ^{t} − λ^{t}π^{t})^{k}]^{2},  (24)

sin^{2}(λπ) = 4 cos^{2}(\frac{λ}{2}π) sin^{2}(\frac{λ}{2}π)

≥ λ^{2}π^{2}[\frac{sin θ}{θ} + ∑_{k=1}^{m} 2^{−kt}μ_{k}(2^{t}θ^{t} − λ^{t}π^{t})^{k}]^{2} cos^{2}(\frac{λ}{2}π).

(25)

Substituting (24) and (25) into (23) leads to (14). The proof of Theorem 2 is complete. □

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(F. Qi) Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: qifeng618@gmail.com, qifeng618@hotmail.com, qifeng618@msn.com, qifeng618@qq.com,
qifeng@hpu.edu.cn, fengqi618@member.ams.org
   URL: http://rgmia.vu.edu.au/qi.html

(D.-W. Niu) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: nnddww@163.com, nnddww@hotmail.com, nnddww@gmail.com, nnddww@tom.com

(J. Cao) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: 21caojian@163.com, goodfriendforeve@163.com