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Sharp Cusa and Becker-Stark inequalities

Journal of Inequalities and Applications20112011:136

https://doi.org/10.1186/1029-242X-2011-136

  • Received: 8 June 2011
  • Accepted: 7 December 2011
  • Published:

Abstract

We determine the best possible constants θ,ϑ,α and β such that the inequalities

2 + cos x 3 θ < sin x x < 2 + cos x 3 ϑ

and

π 2 π 2 - 4 x 2 α < tan x x < π 2 π 2 - 4 x 2 β

are valid for 0 < × < π/ 2. Our results sharpen inequalities presented by Cusa, Becker and Stark.

Mathematics Subject Classification (2000): 26D05.

Keywords

  • Inequalities
  • trigonometric functions

1. Introduction

For 0 < × < π/ 2, it is known in the literature that
sin x x < 2 + cos x 3 .
(1)

Inequality (1) was first mentioned by the German philosopher and theologian Nicolaus de Cusa (1401-1464), by a geometrical method. A rigorous proof of inequality (1) was given by Huygens [1], who used (1) to estimate the number π. The inequality is now known as Cusa's inequality [25]. Further interesting historical facts about the inequality (1) can be found in [2].

It is the first aim of present paper to establish sharp Cusa's inequality.

Theorem 1. For 0 < × < π/ 2,
2 + cos x 3 θ < sin x x < 2 + cos x 3 ϑ
(2)
with the best possible constants
θ = ln ( π 2 ) ln ( 3 2 ) = 1 . 11373998 and ϑ = 1 .
Becker and Stark [6] obtained the inequalities
8 π 2 - 4 x 2 < tan x x < π 2 π 2 - 4 x 2 0 < x < π 2 .
(3)

The constant 8 and π2 are the best possible.

Zhu and Hua [7] established a general refinement of the Becker-Stark inequalities by using the power series expansion of the tangent function via Bernoulli numbers and the property of a function involving Riemann's zeta one. Zhu [8] extended the tangent function to Bessel functions.

It is the second aim of present paper to establish sharp Becker-Stark inequality.

Theorem 2. For 0 < × < π/ 2,
π 2 π 2 - 4 x 2 α < tan x x < π 2 π 2 - 4 x 2 β
(4)
with the best possible constants
α = π 2 12 = 0 . 822467033 and β = 1 .

Remark 1. There is no strict comparison between the two lower bounds 8 π 2 - 4 x 2 and

π 2 π 2 - 4 x 2 π 2 12 in (3) and (4).

The following lemma is needed in our present investigation.

Lemma 1 ([911]). Let -< a < b < ∞, and f, g : [a, b] → be continuous on [a, b] and differentiable in (a, b). Suppose g' ≠ 0 on (a; b). If f'(x)/g' (x) is increasing (decreasing) on (a, b), then so are
[ f ( x ) - f ( a ) ] [ g ( x ) - g ( a ) ] and [ f ( x ) - f ( b ) ] [ g ( x ) - g ( b ) ] .

If f'(x) = g'(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

2. Proofs of Theorems 1 and 2

Proof of Theorem [1]. Consider the function f(x) defined by
F ( x ) = ln sin x x ln 2 + cos x 3 , 0 < x < π 2 , F ( 0 ) = 1 and F π 2 = ln( π /2) ln(3/2) .
For 0 < x < π/2, let
F 1 ( x ) = ln sin x x and F 2 ( x ) = ln 2 + cos x 3 .
Then,
F 1 ( x ) F 2 ( x ) = - 2 x cos x - x cos 2 x + 2 sin x + sin x cos x x sin 2 x = F 3 ( x ) F 4 ( x ) ,
where
F 3 ( x ) = - 2 x cos x - x cos 2 x + 2 sin x + sin x cos x and F 4 ( x ) = x sin 2 x .
Differentiating with respect to x yields
F 3 ( x ) F 4 ( x ) = 2 x + 2 x cos x - sin x sin x + 2 x cos x F 5 ( x ) .
Elementary calculations reveal that
F 5 ( x ) = 2 F 6 ( x ) 2 x sin ( 2 x ) + 4 x 2 cos 2 x + sin 2 x ,
where
F 6 ( x ) = sin ( 2 x ) + ( 2 x 2 + 1 ) sin x - 2 x - x cos x .
By using the power series expansions of sine and cosine functions, we find that
F 6 ( x ) = x 3 - 1 10 x 5 - 19 2520 x 7 + 2 n = 4 ( - 1 ) n u n ( x ) ,
where
u n ( x ) = 4 n - 4 n 2 - 3 n ( 2 n + 1 ) ! x 2 n + 1 .
Elementary calculations reveal that, for 0 < × < π/ 2 and n ≥ 4,
u n + 1 ( x ) u n ( x ) = x 2 2 2 2 n + 2 - 4 n 2 - 11 n - 7 ( n + 1 ) ( 2 n + 3 ) ( 4 n - 4 n 2 - 3 n ) < 1 2 π 2 2 2 2 n + 2 - 4 n 2 - 11 n - 7 ( n + 1 ) ( 2 n + 3 ) ( 4 n - 4 n 2 - 3 n ) = π 2 8 ( n + 1 ) 4 n + 1 - 4 n 2 - 11 n - 7 ( 2 n + 3 ) ( 4 n - 4 n 2 - 3 n ) < π 2 8 ( n + 1 ) < 1 .
Hence, for fixed x (0, π/ 2), the sequence n u n (x) is strictly decreasing with regard to n ≥ 4. Hence, for 0 < × < π/2,
F 6 ( x ) = x 3 - 1 10 x 5 - 19 2520 x 7 > 0 0 < x < π 2 ,

and therefore, the functions F5(x) and F 3 ( x ) F 4 ( x ) are both strictly increasing on (0, π/2).

By Lemma 1, the function
F 1 ( x ) F 2 ( x ) = F 3 ( x ) F 4 ( x ) = F 3 ( x ) - F 3 ( 0 ) F 4 ( x ) - F 4 ( 0 )
is strictly increasing on (0, π/2). By Lemma 1, the function
F ( x ) = F 1 ( x ) F 2 ( x ) = F 1 ( x ) - F 1 ( 0 ) F 2 ( x ) - F ( 0 )
is strictly increasing on (0, π/2), and we have
1 = F ( 0 ) < F ( x ) = ln sin x x ln 2 + cos x 3 < F π 2 = ln ( π 2 ) ln ( 3 2 ) x 0 , π 2 .

By rearranging terms in the last expression, Theorem 1 follows.

Proof of Theorem 2. Consider the function f(x) defined by
f ( x ) = ln tan x x ln π 2 π 2 - 4 x 2 , 0 < x < π 2 , f ( 0 ) = π 2 12 and f π 2 = 1 .
For 0 < x < π/2, let
f 1 ( x ) = ln tan x x and f 2 ( x ) = ln π 2 π 2 - 4 x 2 .
Then,
f 1 ( x ) f 2 ( x ) = ( π 2 - 4 x 2 ) ( 2 x - sin ( 2 x ) ) 8 x 2 sin ( 2 x ) g ( x ) .
Elementary calculations reveal that
4 x 3 sin 2 ( 2 x ) g ( x ) = - ( π 2 + 4 x 2 ) x sin ( 2 x ) - 2 ( π 2 - 4 x 2 ) x 2 cos ( 2 x ) + π 2 sin 2 ( 2 x ) h ( x ) .
Motivated by the investigations in [12], we are in a position to prove h(x) > 0 for x (0, π/2).Let
H ( x ) = λ , x = 0 , h ( x ) x 6 π 2 - x 2 0 < x < π 2 , μ , x = π 2 ,
Where λ and μ are constants determined with limits:
λ = lim x 0 + h ( x ) x 6 ( π 2 - x ) 2 = 224 π 2 - 1920 45 π 2 = 0 . 654740609 . . . , μ = lim t ( π 2 ) - h ( x ) x 6 ( π 2 - x ) 2 = 128 π 4 = 1 . 31404572 . . . .
Using Maple, we determine Taylor approximation for the function H(x) by the polynomial of the first order:
P 1 ( x ) = 32 ( 7 π 2 - 60 ) 45 π 2 + 128 ( 7 π 2 - 60 ) 45 π 3 x ,
which has a bound of absolute error
ε 1 = - 1920 - 1920 π 2 + 224 π 4 15 π 4 = 0 . 650176097 . . .
for values x [0,π/2]. It is true that
H ( x ) - ( P 1 ( x ) - 1 ) 0 , P 1 ( x ) - 1 = 64 ( 60 π 2 + 90 - 7 π 4 ) 45 π 4 + 128 ( 7 π 2 - 60 ) 45 π 3 x > 0
for x [0, π/ 2]. Hence, for x [0, π/ 2], it is true that H (x) > 0 and, therefore, h (x) > 0 and g'(x) > 0 for x [0, π/ 2]. Therefore, the function f 1 ( x ) f 2 ( x ) is strictly increasing on. (0, π/ 2).By Lemma 1, the function
f ( x ) = f 1 ( x ) f 2 ( x )
is strictly increasing on (0, π/ 2), and we have
π 2 12 = f ( 0 ) < f ( x ) = ln tan x x ln π 2 π 2 - 4 x 2 < f π 2 = 1 .

By rearranging terms in the last expression, Theorem 2 follows.

Declarations

Acknowledgements

Research is supported in part by the Research Grants Council of the Hong Kong SAR, Project No. HKU7016/07P.

Authors’ Affiliations

(1)
School of Mathematics and Informatics, Henan Polytechnic, University, Jiaozuo City, Henan Province, 454003, People's Republic of China
(2)
Department of Mathematics, the University of Hong Kong, Pokfulam Road, Hong Kong, China

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Copyright

© Chen and Cheung; 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.

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