Refinements of Jordan's inequality

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Abstract

A method of sharpening Jordan's inequality proposed by Li-Li would be improved. Increasing lower bounds and decreasing upper bounds for strengthened Jordan's inequality can be constructed and the errors of lower-upper bounds for strengthened Jordan's inequality can be estimated.

(2010) Mathematics Subject Classification: 26D05; 26D15.

Keywords

Jordan's inequality lower bound upper bound 

1 Introduction

The well-known Jordan's inequality [1, p. 33] reads that
2 π sin x x < 1 , 0 < x π 2 , Open image in new window
(1.1)
with equality holds if and only if x = π/ 2. This inequality plays an important role in many areas of pure and applied mathematics. Jordan's inequality (1.1) has been refined, generalized, and applied by many mathematicians. It was first extended to the following:
sin x x 2 π + 1 12 π ( π 2 - 4 x 2 ) , 0 < x π 2 , Open image in new window
(1.2)
and then, it was further refined to inequality (1.3),
sin x x 2 π + 1 π 3 ( π 2 - 4 x 2 ) , 0 < x π 2 . Open image in new window
(1.3)

For detailed information, please refer to the expository and survey articles [2] and related references therein.

In [3, Theorem 2.1] or [2, (2.26)], a new method of sharpening Jordan's inequality was established by Li-Li which shows that one can obtain strengthened Jordan's inequalities from old ones. This result may be stated as follows.

Theorem 1.1 (Li-Li) Let g: [0, π/2] → [0, 1] be a continuous function. If
sin x x g ( x ) , 0 < x π 2 , Open image in new window
(1.4)
then the double inequality
2 π - h π 2 + h ( x ) sin x x 1 + h ( x ) , 0 < x π 2 , Open image in new window
(1.5)
holds with equality if and only if x = π/ 2, where
h ( x ) = - 0 x 1 u 2 0 u v 2 g ( v ) d v d u , 0 x π 2 . Open image in new window
(1.6)
However, the new ones are not always stronger than the old ones. For example, by taking g(x) as the function in the right-hand side of (1.3), it was obtained that
sin x x 2 π + 1 10 π ( π 2 - 4 x 2 ) + 1 80 π 3 ( π 2 - 4 x 2 ) 2 , 0 < x π 2 . Open image in new window
(1.7)

The right-hand side inequality in (1.3) is slightly stronger than the right-hand side inequality in (1.7) at x = ( 5 ( π 2 - 8 ) 4 ) 1 2 Open image in new window but less than it at x = 0. These two lower bounds for sin x/x can not be included each other. In fact, this obstacle can be improved and we can obtain better results (see (2.9)). Moreover, we can estimate the errors of lower bounds and upper bounds for strengthened Jordan's inequality (see (2.12)).

Let g: [0, π/2] → [0, 1] be a continuous function. Define the following functions, h g (x), L g (x) and U g (x), on the interval [0, π/2] by
h g ( x ) : = - 0 x 1 u 2 0 u v 2 g ( v ) d v d u , Open image in new window
(1.8)
L g ( x ) : = 2 π - h g π 2 + h g ( x ) , U g ( x ) : = 1 + h g ( x ) . Open image in new window
(1.9)
For a positive integer m, we define
L g m ( x ) : = L L g m - 1 ( x ) , L g 1 ( x ) : = L g ( x ) , L g 0 ( x ) : = g ( x ) . Open image in new window
(1.10)

Corollary 2.2 leads us to know that the lower bounds L g m ( x ) Open image in new window and upper bounds U L g m ( x ) Open image in new window for sin x/x are increasing and decreasing with respect to positive integers m, respectively, and we obtain the double inequality (2.9). Corollary 2.4, equality (2.12), leads us to know that the errors of lower-upper bounds for strengthened Jordan's inequality can be estimated and we give two examples such as (2.14) and (2.16) to estimate errors.

2 Lower-upper bounds for Jordan's inequality

With the help of Theorem 1.1, we find the important relations among lower-upper bounds for strengthened Jordan's inequality.

Theorem 2.1 Let g1, g2: [0, π/2] → [0, 1] be continuous functions. If g1(x) ≤ g2(x), 0 ≤ x ≤ π/2, then
L g 1 ( x ) L g 2 ( x ) , 0 x π 2 , Open image in new window
(2.1)
U g 1 ( x ) U g 2 ( x ) , 0 x π 2 , Open image in new window
(2.2)
and
sup 0 x π 2 ( L g 2 ( x ) - L g 1 ( x ) ) = L g 2 ( 0 ) - L g 1 ( 0 ) = 0 π 2 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u , Open image in new window
(2.3)
sup 0 x π 2 ( U g 1 ( x ) - U g 2 ( x ) ) = U g 1 π 2 - U g 2 π 2 = 0 π 2 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u . Open image in new window
(2.4)
Proof. Let 0 ≤ xπ/2. It follows from (1.8) and (1.9) that we have
L g 2 ( x ) - L g 1 ( x ) = - h g 2 π 2 + h g 1 π 2 + h g 2 ( x ) - h g 1 ( x ) = 0 π 2 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u - 0 x 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u = x π 2 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u 0 , Open image in new window
(2.5)
and (2.1) is obtained. The last term of (2.5) leads us to know that L g 2 ( x ) - L g 1 ( x ) Open image in new window is decreasing to 0 as x approaches to π/ 2 and its supremum occurs at x = 0 so that we obtain (2.3).
U g 1 ( x ) - U g 2 ( x ) = h g 1 ( x ) - h g 2 ( x ) = 0 x 1 u 2 0 u v 2 ( g 2 ( v ) - g 1 ( v ) ) d v d u 0 . Open image in new window
(2.6)

(2.6) is nonnegative and increasing so that we obtain (2.2) and (2.4) and we complete the proof.

If we set g2(x) = sin x/x in Theorem 2.1, then we would obtain important results in the following corollary.

Corollary 2.2 Let g: [0, π/2] → [0, 1] be a continuous function. if sin x/xg(x) and L g (x) ≥ g(x), 0 ≤ xπ/2, then for all positive integers m,
L g m - 1 ( x ) L g m ( x ) , 0 x π 2 , Open image in new window
(2.7)
and
U L g m - 1 ( x ) U L g m ( x ) , 0 x π 2 . Open image in new window
(2.8)
Moreover,
L g m ( x ) sin x x U L g m - 1 ( x ) , 0 < x π 2 . Open image in new window
(2.9)

Proof. It follows from Theorem 2.1, if replacing g1(x) and g2(x) by g(x) and L g (x), respectively, then we have L g ( x ) L g 2 ( x ) Open image in new window. Repeating (2.1) in this inequality, we get that L g m ( x ) Open image in new window is increasing with respect to positive integers m. A similar way, repeating (2.2), we have that U L g m ( x ) Open image in new window is decreasing with respect to positive integers m. With the help of Theorem 1.1, we have L g (x) ≤ sin x/xU g (x). Replacing g(x) by L g (x), we have L g 2 ( x ) sin x x U L g ( x ) Open image in new window. Repeating this method, we obtain (2.9) and we complete our proof.

The following lemma will be used in the next corollary.

Lemma 2.3 Let g(x) = sin x/x, 0 < xπ/2 and g(0) = 1. Then
L g ( x ) = sin x x , 0 < x π 2 , Open image in new window
(2.10)
U g ( x ) = sin x x , 0 < x π 2 , Open image in new window
(2.11)

L g (0) = 1 and U g (0) = 1.

Proof. For 0 < xπ/2. It follows from elementary calculus and Fubini's Theorem that we have
0 x 1 u 2 0 u v 2 sin v v d v d u = 0 x v sin v v x 1 u 2 d u d v = 0 x v sin v 1 v - 1 x d v = 1 - sin x x , Open image in new window
and
L g ( x ) = 2 π + 0 π 2 1 u 2 0 u v 2 sin v v d v d u - 0 x 1 u 2 0 u v 2 sin v v d v d u = 2 π + 1 - 2 π sin π 2 - 1 - sin x x = sin x x , Open image in new window
also we have
U g ( x ) = 1 - 0 x 1 u 2 0 u v 2 sin v v d v d u = sin x x . Open image in new window

By approximation, we set L g (0) = 1 and U g (0) = 1 and we get our desired results.

The errors of lower-upper bounds for strengthened Jordan's inequality can be estimated in the following corollary.

Corollary 2.4 Let g: [0, π/2] → [0, 1] be a continuous function. If
sin x x g ( x ) , 0 < x π 2 , Open image in new window
then
sup 0 x π 2 sin x x - L g ( x ) = 1 - 2 π - 0 π 2 1 u 2 0 u v 2 g ( v ) d v d u = sup 0 x π 2 U g ( x ) - sin x x . Open image in new window
(2.12)
Proof. It follows from Lemma 2.3 and equalities (2.3) and (2.4), if replacing g1(x) and g2(x) by g(x) and sin x/x, respectively, then we have
sup 0 x π 2 sin x x - L g ( x ) = 1 - L g ( 0 ) = 1 - 2 π - h g π 2 + h g ( 0 ) = 1 - 2 π - 0 π 2 1 u 2 0 u v 2 g ( v ) d v d u , Open image in new window
sup 0 x π 2 U g ( x ) - sin x x = U g π 2 - 2 π = 1 + h g π 2 - 2 π = 1 - 0 π 2 1 u 2 0 u v 2 g ( v ) d v d u - 2 π . Open image in new window
We give some examples of (2.9) and (2.12) here. Let g(x) = π/2. It follows from (1.9) that L g ( x ) = 2 π + π 12 - 1 3 π x 2 . L g ( x ) - g ( x ) = π 12 - 1 3 π x 2 Open image in new window is decreasing on [0, π/2] and inf0≤x≤π/2L g (x)-g(x) = L g (π/2) - g(π/2) = 0, so that L g (x) ≥ g(x). It follows from (1.1), L g (x) ≥ g(x) and choosing m = 2 in (2.9) that we have L g 2 ( x ) sin x x U L g ( x ) Open image in new window,
2 π + π 12 + 7 π 3 2 , 880 - 1 3 π + π 72 x 2 + 1 60 π x 4 sin x x 1 - 1 3 π + π 72 x 2 + 1 60 π x 4 . Open image in new window
(2.13)
Moreover, by (2.12), we have
sup 0 x π 2 sin x x - L g 2 ( x ) = sup 0 x π 2 U L g ( x ) - sin x x = 1 - 2 π - π 12 - 7 π 3 2 , 880 0 . 0262 . Open image in new window
(2.14)
The lower bound of (2.13) is stronger than the lower bound of (1.2) and equal at x = π/2. But the lower bounds of (2.13) and (1.3) cannot be included each other. In this case, when x ∈ [0, β], the left-hand side inequality in (2.13) is stronger than the right-hand side inequality in (1.3) but less than it as x is close to π/2, where
β = 5 π 4 + 120 π 2 - 1 , 440 - 2 ( π 8 + 120 π 6 + 2 , 160 π 4 - 86 , 400 π 2 + 518 , 400 ) 1 2 12 π 2 1 2 1 . 2 . Open image in new window
For L g 3 ( x ) sin x x U L g 2 ( x ) Open image in new window, we have
2 π + π 12 + 7 π 3 2 , 880 + 31 π 5 483 , 840 - 1 3 π + π 72 + 7 π 3 17 , 280 x 2 + 1 60 π + π 1 , 440 x 4 - 1 2 , 520 π x 6 sin x x 1 - 1 3 π + π 72 + 7 π 3 17 , 280 x 2 + 1 60 π + π 1 , 440 x 4 - 1 2 , 520 π x 6 , Open image in new window
(2.15)
and
sup 0 x π 2 sin x x - L g 3 ( x ) = sup 0 x π 2 U L g 2 ( x ) - sin x x = 1 - 2 π - π 12 - 7 π 3 2 , 880 - 31 π 5 483 , 480 0 . 0066 Open image in new window
(2.16)

It is clear that the error in (2.16) is much smaller than that in (2.14), and it seems to us that L g m ( x ) Open image in new window and U L g m ( x ) Open image in new window are convergent to sin x/x uniformly. However, we have not proved our conjecture yet.

Notes

Acknowledgements

The author expresses his sincere thanks to the referees for careful reading of the article and several helpful suggestions.

References

  1. 1.
    Mitrinović DS, Vasić PM: Analytic Inequalities. Springer, New York; 1970.CrossRefGoogle Scholar
  2. 2.
    Qi F, Niu D-W, Guo B-N: Refinements, generalizations, and applications of Jordan's inequality and related problems. J Inequal Appl 2009., 52: 2009(Article ID 271923)Google Scholar
  3. 3.
    Li J-L, Li Y-L: On the strengthened Jordan's inequality. J Inequal Appl 2007., 8: 2007(Article ID 74328)Google Scholar

Copyright information

© Kuo; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Center of General EducationJen-Teh Junior College of Medicine, Nursing and ManagementMiaoli CountyRepublic of China

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