Sharp Cusa and Becker-Stark inequalities

Abstract

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

$${\left(\frac{2+cosx}{3}\right)}^{\theta }<\frac{sinx}{x}<{\left(\frac{2+cosx}{3}\right)}^{\vartheta }$$

and

$${\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}^{\alpha }<\frac{tanx}{x}<{\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}^{\beta }$$

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

Mathematics Subject Classification (2000): 26D05.

1. Introduction

For 0 < × < π/ 2, it is known in the literature that

$$\frac{sinx}{x}<\frac{2+cosx}{3}.$$

(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 [2–5]. 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,

$${\left(\frac{2+cosx}{3}\right)}^{\theta }<\frac{sinx}{x}<{\left(\frac{2+cosx}{3}\right)}^{\vartheta }$$

(2)

with the best possible constants

$$\theta =\frac{ln\left(\pi ∕2\right)}{ln\left(3∕2\right)}=1.11373998\dots \mathsf{\text{and}}\vartheta =1.$$

Becker and Stark [6] obtained the inequalities

$$\frac{8}{{\pi }^2-4x^2}<\frac{tanx}{x}<\frac{{\pi }^2}{{\pi }^2-4x^2}\left(0<x<\frac{\pi }{2}\right).$$

(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,

$${\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}^{\alpha }<\frac{tanx}{x}<{\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}^{\beta }$$

(4)

with the best possible constants

$$\alpha =\frac{{\pi }^2}{12}=0.822467033\dots \mathsf{\text{and}}\beta =1.$$

Remark 1. There is no strict comparison between the two lower bounds$\frac{8}{{\pi }^2-4x^2}$ and

${\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}^{{\pi }^2∕12}$in (3) and (4).

The following lemma is needed in our present investigation.

Lemma 1 ([9–11]). Let - ∞ < a < b < ∞, and f, g : [a, b] → ℝ 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

$$\left[f\left(x\right)-f\left(a\right)\right]∕\left[g\left(x\right)-g\left(a\right)\right]\mathsf{\text{and}}\left[f\left(x\right)-f\left(b\right)\right]∕\left[g\left(x\right)-g\left(b\right)\right].$$

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

$$\begin{array}{c}F\left(x\right)=\frac{\mathsf{\text{ln}}\left(\frac{sinx}{x}\right)}{\mathsf{\text{ln}}\left(\frac{2+cosx}{3}\right)},0<x<\frac{\pi }{2},\\ F\left(0\right)=1\mathsf{\text{and}}\mathsf{\text{F}}\left(\frac{\pi }{2}\right)=\frac{\mathsf{\text{ln(}}\pi \mathsf{\text{/2)}}}{\mathsf{\text{ln(3/2)}}}.\end{array}$$

For 0 < x < π/2, let

$$F_1\left(x\right)=ln\left(\frac{sinx}{x}\right)\mathsf{\text{and}}{F}_2\left(x\right)=ln\left(\frac{2+cosx}{3}\right).$$

Then,

$$\frac{{F^{\prime }}_1\left(x\right)}{{F^{\prime }}_2\left(x\right)}=\frac{-2xcosx-x{cos}^{2}x+2sinx+sinxcosx}{x{sin}^{2}x}=\frac{F_3\left(x\right)}{F_4\left(x\right)},$$

where

$$F_3\left(x\right)=-2xcosx-x{cos}^{2}x+2sinx+sinxcosx\mathsf{\text{and}}{F}_4\left(x\right)=x{sin}^{2}x.$$

Differentiating with respect to x yields

$$\frac{{F^{\prime }}_3\left(x\right)}{{F^{\prime }}_4\left(x\right)}=\frac{2x+2xcosx-sinx}{sinx+2xcosx}\triangleq F_5\left(x\right).$$

Elementary calculations reveal that

$$F_5^{\prime }\left(x\right)=\frac{2F_6\left(x\right)}{2xsin\left(2x\right)+4x^2{cos}^{2}x+{sin}^{2}x},$$

where

$$F_6\left(x\right)=sin\left(2x\right)+\left(2x^2+1\right)sinx-2x-xcosx.$$

By using the power series expansions of sine and cosine functions, we find that

$$F_6\left(x\right)=x^3-\frac{1}{10}{x}^5-\frac{19}{2520}{x}^7+2\sum _{n=4}^{\infty }{(-1)}^n{u}_n\left(x\right),$$

where

$$u_n\left(x\right)=\frac{4^n-4n^2-3n}{\left(2n+1\right)!}{x}^{2n+1}.$$

Elementary calculations reveal that, for 0 < × < π/ 2 and n ≥ 4,

$$\begin{array}{cc}\hfill \frac{u_{n+1}\left(x\right)}{u_n\left(x\right)}& =\frac{x^2}{2}\frac{2^{2n+2}-4n^2-11n-7}{\left(n+1\right)\left(2n+3\right)\left(4^n-4n^2-3n\right)}\hfill \\ <\frac{1}{2}{\left(\frac{\pi }{2}\right)}^2\frac{2^{2n+2}-4n^2-11n-7}{\left(n+1\right)\left(2n+3\right)\left(4^n-4n^2-3n\right)}\hfill \\ =\frac{{\pi }^2}{8\left(n+1\right)}\frac{4^{n+1}-4n^2-11n-7}{\left(2n+3\right)\left(4^n-4n^2-3n\right)}\hfill \\ <\frac{{\pi }^2}{8\left(n+1\right)}<1.\hfill \end{array}$$

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,

$$F_6\left(x\right)=x^3-\frac{1}{10}{x}^5-\frac{19}{2520}{x}^7>0\left(0<x<\frac{\pi }{2}\right),$$

and therefore, the functions F₅(x) and $\frac{{F^{\prime }}_3\left(x\right)}{{F^{\prime }}_4\left(x\right)}$are both strictly increasing on (0, π/2).

By Lemma 1, the function

$$\frac{{F^{\prime }}_1\left(x\right)}{{F^{\prime }}_2\left(x\right)}=\frac{F_3\left(x\right)}{F_4\left(x\right)}=\frac{F_3\left(x\right)-F_3\left(0\right)}{F_4\left(x\right)-F_4\left(0\right)}$$

is strictly increasing on (0, π/2). By Lemma 1, the function

$$F\left(x\right)=\frac{F_1\left(x\right)}{F_2\left(x\right)}=\frac{F_1\left(x\right)-F_1\left(0\right)}{F_2\left(x\right)-F\left(0\right)}$$

is strictly increasing on (0, π/2), and we have

$$1=F\left(0\right)<F\left(x\right)=\frac{ln\left(\frac{sinx}{x}\right)}{ln\left(\frac{2+cosx}{3}\right)}<F\left(\frac{\pi }{2}\right)=\frac{ln\left(\pi ∕2\right)}{ln\left(3∕2\right)}\forall x\in \left(0,\frac{\pi }{2}\right).$$

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

Proof of Theorem 2. Consider the function f(x) defined by

$$\begin{array}{ll}\hfill f\left(x\right)& =\frac{ln\left(\frac{tanx}{x}\right)}{ln\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)},0<x<\frac{\pi }{2},\\ \hfill f\left(0\right)& =\frac{{\pi }^2}{12}\mathsf{\text{and}}f\left(\frac{\pi }{2}\right)=1.\end{array}$$

For 0 < x < π/2, let

$$f_1\left(x\right)=ln\left(\frac{tanx}{x}\right)\mathsf{\text{and}}{f}_2\left(x\right)=ln\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right).$$

Then,

$$\frac{{f^{\prime }}_1\left(x\right)}{{f^{\prime }}_2\left(x\right)}=\frac{\left({\pi }^2-4x^2\right)\left(2x-sin\left(2x\right)\right)}{8x^{2}sin\left(2x\right)}\triangleq g\left(x\right).$$

Elementary calculations reveal that

$$4x^3{sin}^2\left(2x\right)g^{\prime }\left(x\right)=-\left({\pi }^2+4x^2\right)xsin\left(2x\right)-2\left({\pi }^2-4x^2\right)x^{2}cos\left(2x\right)+{\pi }^2{sin}^2\left(2x\right)\triangleq h\left(x\right).$$

Motivated by the investigations in [12], we are in a position to prove h(x) > 0 for x ∈ (0, π/2).Let

$$H\left(x\right)=\left\{\begin{array}{cc}\lambda ,\hfill & x=0,\hfill \\ \frac{h\left(x\right)}{x^6{\left(\frac{\pi }{2}-x\right)}^2}\hfill & 0<x<\frac{\pi }{2},\hfill \\ \mu ,\hfill & x=\frac{\pi }{2},\hfill \end{array}\right.$$

Where λ and μ are constants determined with limits:

$$\begin{array}{cc}\hfill \lambda & =\underset{x\to 0^{+}}{lim}\frac{h\left(x\right)}{x^6{\left(\frac{\pi }{2}-x\right)}^2}=\frac{224{\pi }^2-1920}{45{\pi }^2}=0.654740609...,\hfill \\ \hfill \mu & =\underset{t\to {\left(\pi ∕2\right)}^{-}}{lim}\frac{h\left(x\right)}{x^6{\left(\frac{\pi }{2}-x\right)}^2}=\frac{128}{{\pi }^4}=1.31404572....\hfill \end{array}$$

Using Maple, we determine Taylor approximation for the function H(x) by the polynomial of the first order:

$$P_1\left(x\right)=\frac{32\left(7{\pi }^2-60\right)}{45{\pi }^2}+\frac{128\left(7{\pi }^2-60\right)}{45{\pi }^3}x,$$

which has a bound of absolute error

$${\epsilon }_1=\frac{-1920-1920{\pi }^2+224{\pi }^4}{15{\pi }^4}=0.650176097...$$

for values x ∈ [0,π/2]. It is true that

$$H\left(x\right)-\left(P_1\left(x\right)-{\mathit{ℰ}}_1\right)\ge 0,P_1\left(x\right)-{\mathit{ℰ}}_1=\frac{64\left(60{\pi }^2+90-7{\pi }^4\right)}{45{\pi }^4}+\frac{128\left(7{\pi }^2-60\right)}{45{\pi }^3}x>0$$

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 $\frac{{f^{\prime }}_1\left(x\right)}{{f^{\prime }}_2\left(x\right)}$is strictly increasing on. (0, π/ 2).By Lemma 1, the function

$$f\left(x\right)=\frac{f_1\left(x\right)}{f_2\left(x\right)}$$

is strictly increasing on (0, π/ 2), and we have

$$\frac{{\pi }^2}{12}=f\left(0\right)<f\left(x\right)=\frac{ln\left(\frac{tanx}{x}\right)}{ln\left(\frac{{\pi }^2}{{\pi }^2-4x^2}\right)}<f\left(\frac{\pi }{2}\right)=1.$$

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