# Some Refinements of Inequalities for Circular Functions

Authors: Zhengjie Sun; Ling Zhu
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We give new lower bound and upper bound for Papenfuss-Bach inequality and improve Ruehr-Shafer inequality by providing a new lower bound.


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## Body

## 1. Introduction

Papenfuss [1] proposes an open problem described as follows.Theorem 1.1.
Let0≤x<π/2. Then
(1.1)xsec2x-tanx≤8π2x3(π2-4x2)2.Bach [2] prove Theorem 1.1 and obtain a further result.Theorem 1.2.
Let0≤x<π/2. Then
(1.2)xsec2x-tanx≤2π23x3(π2-4x2)2.Ge gives a lower bound of the above inequality in [3] as follows.Theorem 1.3.
Let0<x<π/2. Then
(1.3)64x3(π2-4x2)2<xsec2x-tanx<2π43x3(π2-4x2)2.
Furthermore, 64 and 2π4/3 are the best constants in (1.3).Besides, Bach [2] obtain the improvement of the Papenfuss-Bach inequality as another one.Theorem 1.4.
Let0≤x<π/2. Then
(1.4)xsec2x-tanx≤2π2(tanx-x)π2-4x2.In this note, we firstly obtain better bounds for Papenfuss-Bach inequality as in the following statement.Theorem 1.5.
Let0<x<π/2. Then
(1.5)(2π4/3)x3+((8π4/15)-16π2/3)x5(π2-4x2)2<xsec2x-tanx<(2π4/3)x3+((256/π2)⋅(513/511)-(8π2/3))x5(π2-4x2)2.And then we give the refinement of the Ruehr-Shafer inequality as follows.Theorem 1.6.
Let0<x<π/2. Then
(1.6)1018π2511(tanx-x)π2-4x2<xsec2x-tanx<2π2(tanx-x)π2-4x2.Remark 1.7.
Since(256/π2)·(513/511)-(8π2/3)≈-0.2792<0, we know that the upper bound in the inequality (1.5) is better than the one in (1.3). At the same time, we find that the lower estimate in (1.5) is larger than the one in (1.3) on the interval (0,1.169880805) meanwhile the lower estimate in (1.3) is larger than the one in (1.5) on (1.169880805,π/2).


## 2. Lemmas

Lemma 2.1 (see [4–7]).
LetB2n be the even-indexed Bernoulli numbers. Then
(2.1)2(2n)!(2π)2n11-2-2n<|B2n|<2(2n)!(2π)2n11-21-2n,n=1,2,3,….Lemma 2.2 (see [7–9]).
Let|x|<π/2.  Then
(2.2)tanx=∑n=1∞22n(22n-1)(2n)!(-1)n-1B2nx2n-1=∑n=1∞22n(22n-1)(2n)!|B2n|x2n-1,tanx-x=∑n=2∞22n(22n-1)(2n)!|B2n|x2n-1.Lemma 2.3.
LetF(x)=(π2-4x2)(xsec2x-tanx) and  |x|<π/2. Then
(2.3)F(x)=2π23x3+∑n=2∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1.Proof.
By using Lemma2.2, we have
(2.4)sec2x=(tanx)′=∑n=1∞22n(22n-1)(2n)!(2n-1)|B2n|x2n-2,|x|<π2,xsec2x-tanx=∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1,|x|<π2.
We calculate
(2.5)(π2-4x2)(xsec2x-tanx)=π2∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1=2π23x3+π2∑n=3∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1=2π23x3+π2∑n=2∞22n+2(22n+2-1)(2n+2)!(2n)|B2n+2|x2n+1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1=2π23x3+∑n=2∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1.
The proof of Lemma 2.3 is completed.Lemma 2.4.
Let|x|<π/2. Then
(2.6)1π2-4x2=1π2∑n=0∞(2π)2nx2n.


## 3. The Proof of Theorem1.5

The proof of Theorem1.5 is completed when proving the truth of the following double inequality: (3.1)(2π4/3)x3+(8π4/15-16π2/3)x5π2-4x2<(π2-4x2)(xsec2x-tanx)<(2π4/3)x3+((256/π2)⋅(513/511)-(8π2/3))x5π2-4x2.We firstly process the left-hand side of the above inequality. We compute that(3.2)(π2-4x2)(xsec2x-tanx)-(2π4/3)x3+(8π4/15-16π2/3)x5π2-4x2=2π23x3+∑n=2∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1-[2π23x3+2π23∑n=2∞22n-2π2n-2x2n+1+(8π215-163)x5+(8π215-163)∑n=3∞22n-4π2n-4x2n+1]=∑n=3∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1-[2π23∑n=3∞22n-2π2n-2x2n+1+(8π215-163)∑n=3∞22n-4π2n-4x2n+1]=a3x7+a4x9+∑n=5∞anx2n+1,
where
(3.3)an=2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|-2π2322n-2π2n-2-(8π215-163)22n-4π2n-4
for n≥3 and n∈N+.We calculatea3≈0.0100>0 and a4≈0.0049>0. For n≥5, by using Lemma 2.1, we can estimate an as follows:(3.4)an>2π222n+2(22n+2-1)(2n+2)!n2(2n+2)!(2π)2n+2(1-2-2n-2)-422n(22n-1)(2n)!(2n-2)2(2n)!(2π)2n(1-21-2n)-2π2322n-2π2n-2-(8π215-163)22n-4π2n-4=4n⋅22n+2π2n-(16n-16)22nπ2n⋅22n-122n-2-2π2322n-2π2n-2-(8π215-163)22n-4π2n-4=22n-4π2n-4[4n⋅26π4-24(16n-16)π4(1+122n-2)-2π2322π2-8π215+163]=22n-4π2n-4[256π4-256(n-1)π4(22n-2)-8π215+83].Since, forn=5,6,…, we have
(3.5)256π4-256(n-1)π4(22n-2)-8π215+83≥256π4-256(5-1)π4(210-2)-8π215+83≈0.0207>0,
so an>0 for n≥5 and n∈N+. Combining with the results of a3 and a4, we finish the proof of the left-hand side of inequality (3.1).Now, Let’s discuss the right-hand side of inequality (3.1) by taking the same method. We compute that (3.6)(π2-4x2)(xsec2x-tanx)-(2π4/3)x3+((256/π2)⋅(513/511)-8π2/3)x5π2-4x2=2π23x3+∑n=2∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1-[2π23x3+2π23∑n=2∞22n-2π2n-2x2n+1+(256π4⋅513511-83)x5+(256π4⋅513511-83)∑n=3∞22n-4π2n-4x2n+1]=b2x5+∑n=3∞[2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|]x2n+1-[2π23∑n=3∞22n-2π2n-2x2n+1+(256π4⋅513511-83)∑n=3∞22n-4π2n-4x2n+1]=b2x5+b3x7+∑n=4∞bnx2n+1,
where
(3.7)b2=(8π215-163)-(256π4⋅513511-83)≈-0.0413<0,b3=[2π228(28-1)38!⋅30-426(26-1)46!⋅42-2π2324π4-(256π4⋅513511-83)22π2]≈-0.0068<0,
and, for n≥4(3.8)bn=2π222n+2(22n+2-1)(2n+2)!n|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|-2π2322n-2π2n-2-(256π4⋅513511-83)22n-4π2n-4.
We can reckon that, for n=4,5,…(3.9)bn<2π222n+2(22n+2-1)(2n+2)!n2(2n+2)!(2π)2n+2(1-21-2n-2)-422n(22n-1)(2n)!(2n-2)2(2n)!(2π)2n(1-2-2n)-2π2322n-2π2n-2-(256π4⋅513511-83)22n-4π2n-4=4π2n⋅22n+2(22n+2-1)π2n(22n+2-2)-(16n-16)22nπ2n-2π2322n-2π2n-2-(256π4⋅513511-83)22n-4π2n-4=22n-4π2n-4(256nπ4(1+122n+2-2)-24(16n-16)π4-2π2322π2-256π4⋅513511+83)=22n-4π2n-4(256nπ4122n+2-2+256π4-83-256π4⋅513511+83)=22n-4π2n-4(256nπ4122n+2-2-256π4⋅2511).As we can see,(3.10)256nπ4122n+2-2≤256⋅4π4122⋅4+2-2=256π42511
holds for n=4,5,…, so we conclude that bn<0 for n≥4 and n∈N+. Observing the value of b2 and b3, we complete the proof of the right-hand side of inequality (3.1).


## 4. The Proof of Theorem1.6

Now we prove the left-hand side of the inequality in Theorem1.6, which is equivalent to (4.1)(π2-4x2)(xsec2x-tanx)-1018π2511(tanx-x)>0.
By using the power series expansions of (xsec2x-tanx) and (tanx-x), we can rewrite the above inequality as follows: (4.2)π2∑n=2∞22n(22n-1)(2n)!(2n-2-1018511)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1>0.Now we simplify the left expression and prove that it is positive:(4.3)π2∑n=2∞22n(22n-1)(2n)!(2n-2-1018511)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1=π2∑n=1∞22n+2(22n+2-1)(2n+2)!(2n-1018511)|B2n+2|x2n+1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1=4π23⋅511x3+c2x5+c3x7+c4x9+c5x11+∑n=6∞cnx2n+1,
where (4.4)cn=π222n+2(22n+2-1)(2n+2)!(2n-1018511)|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|
for n=2,3,…. Particularly, c2≈-0.02447,c3≈0.00142, and c4≈0.00151,c5≈6.74143×10-4. We also use Lemma 2.1 in order to give the lower bound of cn for n=6,7,…: (4.5)cn=π222n+2(22n+2-1)(2n+2)!(2n-1018511)|B2n+2|-422n(22n-1)(2n)!(2n-2)|B2n|>π222n+2(22n+2-1)(2n+2)!(2n-1018511)2(2n+2)!(2π)2n+2(1-2-2n-2)-422n(22n-1)(2n)!(2n-2)2(2n)!(2π)2n(1-21-2n)=2⋅22n+2(2n-1018/511)π2n-(16n-16)22n(22n-1)π2n(22n-2)=22n+3π2n[2n-1018511-(2n-2)(1+122n-2)]=22n+3π2n(4511-2n-222n-2)>22n+3π2n(4511-2⋅6-222⋅6-2)>0.We denote that(4.6)G(x)≡4π23⋅511x3+c2x5+c3x7+c4x9+c5x11=x3(4π23⋅511+c2x2+c3x4+c4x6+c5x8)≡x3H(x),
where (4.7)H(x)=4π23⋅511+c2x2+c3x4+c4x6+c5x8.Lett=x2, and (4.8)P(t)≡4π23⋅511+c2t+c3t2+c4t3+c5t4.
We compute that (4.9)P′(t)=c2+2c3t+3c4t2+4c5t3,0<t<π24,P′′(t)=2+c3+6c4t+12c5t2>0,0<t<π24.
So P′(t) is increasing on (0,π2/4). Since P′(0)<0 and P′(π2/4)>0, we have that P(t) is decreasing firstly and then increasing. Let t0 be only one point of minimum of the function P(t). Then t0≈1.5262621 and P(t0)≈7.33921×10-4>0; this implies that P(t)>0, so H(x)>0 and G(x)>0 for 0<x<π/2.Combining withcn>0 for n≥6, we have proved Theorem 1.6.


## 5. Open Problem

In the last section we pose a problem as follows: Let0<x<π/2. Then(5.1)(2π4/3)x3+(8π4/15-16π2/3)x5(π2-4x2)2<xsec2x-tanx<(2π4/3)x3+(256/π2-8π2/3)x5(π2-4x2)2
hold, where (8π4/15-16π2/3) and (256/π2-8π2/3) are the best constants in (5.1).



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*Source: 869261-2011-11-24.xml*