Open Access

Sharp Cusa and Becker-Stark inequalities

Journal of Inequalities and Applications20112011:136

DOI: 10.1186/1029-242X-2011-136

Received: 8 June 2011

Accepted: 7 December 2011

Published: 7 December 2011

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
(2)
Department of Mathematics, the University of Hong Kong

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Copyright

© Chen and Cheung; licensee Springer. 2011

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