Some Refinements of Inequalities for Circular Functions

Abstract

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

1. Introduction

Papenfuss [1] proposes an open problem described as follows.

Theorem 1.1. Let 0≤𝑥<𝜋/2. Then 𝑥sec²𝑥−tan𝑥≤8𝜋²𝑥³𝜋²−4𝑥²². (1.1)

Bach [2] prove Theorem 1.1 and obtain a further result.

Theorem 1.2. Let 0≤𝑥<𝜋/2. Then 𝑥sec²𝑥−tan𝑥≤2𝜋²3𝑥³𝜋²−4𝑥²². (1.2)

Ge gives a lower bound of the above inequality in [3] as follows.

Theorem 1.3. Let 0<𝑥<𝜋/2. Then 64𝑥³𝜋²−4𝑥²²<𝑥sec²𝑥−tan𝑥<2𝜋⁴3𝑥³𝜋²−4𝑥²². (1.3) 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. Let 0≤𝑥<𝜋/2. Then 𝑥sec²𝑥−tan𝑥≤2𝜋²(tan𝑥−𝑥)𝜋²−4𝑥². (1.4)

In this note, we firstly obtain better bounds for Papenfuss-Bach inequality as in the following statement.

Theorem 1.5. Let 0<𝑥<𝜋/2. Then 2𝜋⁴/3𝑥³+8𝜋⁴/15−16𝜋²/3𝑥⁵𝜋²−4𝑥²²<𝑥sec²𝑥−tan𝑥<2𝜋⁴/3𝑥³+256/𝜋²⋅(513/511)−8𝜋²/3𝑥⁵𝜋²−4𝑥²². (1.5)

And then we give the refinement of the Ruehr-Shafer inequality as follows.

Theorem 1.6. Let 0<𝑥<𝜋/2. Then 1018𝜋²()511tan𝑥−𝑥𝜋²−4𝑥²<𝑥sec²𝑥−tan𝑥<2𝜋²(tan𝑥−𝑥)𝜋²−4𝑥². (1.6)

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]). Let 𝐵_2𝑛 be the even-indexed Bernoulli numbers. Then 2(2𝑛)!(2𝜋)^2𝑛11−2^−2𝑛<𝐵_2𝑛<2(2𝑛)!(2𝜋)^2𝑛11−2^1−2𝑛,𝑛=1,2,3,…. (2.1)

Lemma 2.2 (see [7–9]). Let |𝑥|<𝜋/2. Then tan𝑥=^∞_𝑛=12^2𝑛2^2𝑛−1()!()2𝑛−1^𝑛−1𝐵_2𝑛𝑥^2𝑛−1=^∞_𝑛=12^2𝑛2^2𝑛−1()!𝐵2𝑛^2𝑛𝑥^2𝑛−1,tan𝑥−𝑥=^∞_𝑛=22^2𝑛2^2𝑛−1()!𝐵2𝑛^2𝑛𝑥^2𝑛−1. (2.2)

Lemma 2.3. Let 𝐹(𝑥)=(𝜋²−4𝑥²)(𝑥sec²𝑥−tan𝑥) and |𝑥|<𝜋/2. Then 𝐹(𝑥)=2𝜋²3𝑥³+^∞_𝑛=22𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1. (2.3)

Proof. By using Lemma 2.2, we have sec²𝑥=(tan𝑥)^′=^∞_𝑛=12^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−1)_2𝑛𝑥^2𝑛−2,|𝑥|<𝜋2,𝑥sec²𝑥−tan𝑥=^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛−1,|𝑥|<𝜋2. (2.4) We calculate 𝜋²−4𝑥²𝑥sec²𝑥−tan𝑥=𝜋^2∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛−1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1=2𝜋²3𝑥³+𝜋^2∞_𝑛=32^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛−1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1=2𝜋²3𝑥³+𝜋^2∞_𝑛=22^2𝑛+22^2𝑛+2−1()!(𝐵2𝑛+22𝑛)_2𝑛+2𝑥^2𝑛+1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1=2𝜋²3𝑥³+^∞_𝑛=22𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1. (2.5) The proof of Lemma 2.3 is completed.

Lemma 2.4. Let |𝑥|<𝜋/2. Then 1𝜋²−4𝑥²=1𝜋^2∞_𝑛=02𝜋^2𝑛𝑥^2𝑛. (2.6)

3. The Proof of Theorem 1.5

The proof of Theorem 1.5 is completed when proving the truth of the following double inequality: 2𝜋⁴/3𝑥³+8𝜋⁴/15−16𝜋²/3𝑥⁵𝜋²−4𝑥²<𝜋²−4𝑥²𝑥sec²<𝑥−tan𝑥2𝜋⁴/3𝑥³+256/𝜋²⋅(513/511)−8𝜋²/3𝑥⁵𝜋²−4𝑥². (3.1)

We firstly process the left-hand side of the above inequality. We compute that 𝜋²−4𝑥²𝑥sec²𝑥−tan𝑥−2𝜋⁴/3𝑥³+8𝜋⁴/15−16𝜋²/3𝑥⁵𝜋²−4𝑥²=2𝜋²3𝑥³+^∞_𝑛=22𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1−2𝜋²3𝑥³+2𝜋²3^∞_𝑛=22^2𝑛−2𝜋^2𝑛−2𝑥^2𝑛+1+8𝜋²−15163𝑥⁵+8𝜋²−15163^∞_𝑛=32^2𝑛−4𝜋^2𝑛−4𝑥^2𝑛+1=^∞_𝑛=32𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1−2𝜋²3^∞_𝑛=32^2𝑛−2𝜋^2𝑛−2𝑥^2𝑛+1+8𝜋²−15163^∞_𝑛=32^2𝑛−4𝜋^2𝑛−4𝑥^2𝑛+1=𝑎₃𝑥⁷+𝑎₄𝑥⁹+^∞_𝑛=5𝑎_𝑛𝑥^2𝑛+1, (3.2) where 𝑎_𝑛=2𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛−2𝜋²32^2𝑛−2𝜋^2𝑛−2−8𝜋²−151632^2𝑛−4𝜋^2𝑛−4 (3.3) for 𝑛≥3 and 𝑛∈𝑁⁺.

We calculate 𝑎₃≈0.0100>0 and 𝑎₄≈0.0049>0. For 𝑛≥5, by using Lemma 2.1, we can estimate 𝑎_𝑛 as follows:𝑎_𝑛>2𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛2()!2𝑛+22𝑛+2(2𝜋)^2𝑛+21−2^−2𝑛−22−4^2𝑛2^2𝑛−1()!()2()!2𝑛2𝑛−22𝑛(2𝜋)^2𝑛1−2^1−2𝑛−2𝜋²32^2𝑛−2𝜋^2𝑛−2−8𝜋²−151632^2𝑛−4𝜋^2𝑛−4=4𝑛⋅2^2𝑛+2𝜋^2𝑛−(16𝑛−16)2^2𝑛𝜋^2𝑛⋅2^2𝑛−12^2𝑛−−22𝜋²32^2𝑛−2𝜋^2𝑛−2−8𝜋²−151632^2𝑛−4𝜋^2𝑛−4=2^2𝑛−4𝜋^2𝑛−44𝑛⋅2⁶𝜋⁴−2⁴(16𝑛−16)𝜋⁴11+2^2𝑛−−22𝜋²32²𝜋²−8𝜋²+15163=2^2𝑛−4𝜋^2𝑛−4256𝜋⁴−256(𝑛−1)𝜋⁴2^2𝑛−−28𝜋²+8153. (3.4)

Since, for 𝑛=5,6,…, we have 256𝜋⁴−256(𝑛−1)𝜋⁴2^2𝑛−−28𝜋²+8153≥256𝜋⁴−256(5−1)𝜋⁴2¹⁰−−28𝜋²+8153≈0.0207>0, (3.5) so 𝑎_𝑛>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 𝜋²−4𝑥²𝑥sec²𝑥−tan𝑥−2𝜋⁴/3𝑥³+256/𝜋²⋅(513/511)−8𝜋²/3𝑥⁵𝜋²−4𝑥²=2𝜋²3𝑥³+^∞_𝑛=22𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1−2𝜋²3𝑥³+2𝜋²3^∞_𝑛=22^2𝑛−2𝜋^2𝑛−2𝑥^2𝑛+1+256𝜋⁴⋅513−85113𝑥⁵+256𝜋⁴⋅513−85113^∞_𝑛=32^2𝑛−4𝜋^2𝑛−4𝑥^2𝑛+1=𝑏₂𝑥⁵+^∞_𝑛=32𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1−2𝜋²3^∞_𝑛=32^2𝑛−2𝜋^2𝑛−2𝑥^2𝑛+1+256𝜋⁴⋅513−85113^∞_𝑛=32^2𝑛−4𝜋^2𝑛−4𝑥^2𝑛+1=𝑏₂𝑥⁵+𝑏₃𝑥⁷+^∞_𝑛=4𝑏_𝑛𝑥^2𝑛+1, (3.6) where 𝑏₂=8𝜋²−15163−256𝜋⁴⋅513−85113𝑏≈−0.0413<0,₃=2𝜋²2⁸2⁸−1328!⋅30−4⁶2⁶−14−6!⋅422𝜋²32⁴𝜋⁴−256𝜋⁴⋅513−851132²𝜋²≈−0.0068<0, (3.7) and, for 𝑛≥4𝑏_𝑛=2𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛𝐵2𝑛+2^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛−2𝜋²32^2𝑛−2𝜋^2𝑛−2−256𝜋⁴⋅513−851132^2𝑛−4𝜋^2𝑛−4. (3.8) We can reckon that, for 𝑛=4,5,…𝑏_𝑛<2𝜋²2^2𝑛+22^2𝑛+2−1()!𝑛2()!2𝑛+22𝑛+2(2𝜋)^2𝑛+21−2^{1−2𝑛−2}2−4^2𝑛2^2𝑛−1()!()2()!2𝑛2𝑛−22𝑛(2𝜋)^2𝑛1−2^−2𝑛−2𝜋²32^2𝑛−2𝜋^2𝑛−2−256𝜋⁴⋅513−851132^2𝑛−4𝜋^2𝑛−4=4𝜋²𝑛⋅2^2𝑛+22^2𝑛+2−1𝜋^2𝑛2^2𝑛+2−()2−216𝑛−16^2𝑛𝜋^2𝑛−2𝜋²32^2𝑛−2𝜋^2𝑛−2−256𝜋⁴⋅513−851132^2𝑛−4𝜋^2𝑛−4=2^2𝑛−4𝜋^2𝑛−4256𝑛𝜋⁴11+2^2𝑛+2−2−2⁴(16𝑛−16)𝜋⁴−2𝜋²32²𝜋²−256𝜋⁴⋅513+85113=2^2𝑛−4𝜋^2𝑛−4256𝑛𝜋⁴12^2𝑛+2+−2256𝜋⁴−83−256𝜋⁴⋅513+85113=2^2𝑛−4𝜋^2𝑛−4256𝑛𝜋⁴12^2𝑛+2−−2256𝜋⁴⋅2.511 (3.9)

As we can see, 256𝑛𝜋⁴12^2𝑛+2≤−2256⋅4𝜋⁴12^2⋅4+2=−2256𝜋⁴2511 (3.10) holds for 𝑛=4,5,…, so we conclude that 𝑏_𝑛<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 Theorem 1.6

Now we prove the left-hand side of the inequality in Theorem 1.6, which is equivalent to 𝜋²−4𝑥²𝑥sec²𝑥−tan𝑥−1018𝜋²()511tan𝑥−𝑥>0. (4.1) By using the power series expansions of (𝑥sec²𝑥−tan𝑥) and (tan𝑥−𝑥), we can rewrite the above inequality as follows: 𝜋^2∞_𝑛=22^2𝑛2^2𝑛−1(2𝑛)!2𝑛−2−1018𝐵511^2𝑛𝑥^2𝑛−1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1>0. (4.2)

Now we simplify the left expression and prove that it is positive: 𝜋^2∞_𝑛=22^2𝑛2^2𝑛−1(2𝑛)!2𝑛−2−1018𝐵511^2𝑛𝑥^2𝑛−1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1=𝜋^2∞_𝑛=12^2𝑛+22^2𝑛+2−1(2𝑛+2)!2𝑛−1018𝐵511^2𝑛+2𝑥^2𝑛+1−4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1=4𝜋²𝑥3⋅511³+𝑐₂𝑥⁵+𝑐₃𝑥⁷+𝑐₄𝑥⁹+𝑐₅𝑥¹¹+^∞_𝑛=6𝑐_𝑛𝑥^2𝑛+1, (4.3) where 𝑐_𝑛=𝜋²2^2𝑛+22^2𝑛+2−1(2𝑛+2)!2𝑛−1018𝐵511^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛 (4.4) 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 𝑐_𝑛 for 𝑛=6,7,…: 𝑐_𝑛=𝜋²2^2𝑛+22^2𝑛+2−1(2𝑛+2)!2𝑛−1018𝐵511^2𝑛+22−4^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛>𝜋²2^2𝑛+22^2𝑛+2−1(2𝑛+2)!2𝑛−10182()!5112𝑛+2(2𝜋)^2𝑛+21−2^−2𝑛−22−4^2𝑛2^2𝑛−1()!()2()!2𝑛2𝑛−22𝑛(2𝜋)^2𝑛1−2^1−2𝑛=2⋅2^2𝑛+2(2𝑛−1018/511)𝜋^2𝑛−(16𝑛−16)2^2𝑛2^2𝑛−1𝜋^2𝑛2^2𝑛=2−2_2𝑛+3𝜋^2𝑛2𝑛−1018−()15112𝑛−21+2^2𝑛=2−2_2𝑛+3𝜋^2𝑛4−5112𝑛−22^2𝑛>2−2𝜋^2𝑛+32𝑛4−5112⋅6−22^2⋅6−2>0. (4.5)

We denote that 𝐺(𝑥)≡4𝜋²𝑥3⋅511³+𝑐₂𝑥⁵+𝑐₃𝑥⁷+𝑐₄𝑥⁹+𝑐₅𝑥¹¹=𝑥³4𝜋²3⋅511+𝑐₂𝑥²+𝑐₃𝑥⁴+𝑐₄𝑥⁶+𝑐₅𝑥⁸≡𝑥³𝐻(𝑥), (4.6) where 𝐻(𝑥)=4𝜋²3⋅511+𝑐₂𝑥²+𝑐₃𝑥⁴+𝑐₄𝑥⁶+𝑐₅𝑥⁸. (4.7)

Let 𝑡=𝑥², and 𝑃(𝑡)≡4𝜋²3⋅511+𝑐₂𝑡+𝑐₃𝑡²+𝑐₄𝑡³+𝑐₅𝑡⁴. (4.8) We compute that 𝑃^′(𝑡)=𝑐₂+2𝑐₃𝑡+3𝑐₄𝑡²+4𝑐₅𝑡³𝜋,0<𝑡<²4,𝑃^′′(𝑡)=2+𝑐₃+6𝑐₄𝑡+12𝑐₅𝑡²𝜋>0,0<𝑡<²4. (4.9) 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 with 𝑐_𝑛>0 for 𝑛≥6, we have proved Theorem 1.6.

5. Open Problem

In the last section we pose a problem as follows: Let 0<𝑥<𝜋/2. Then2𝜋⁴/3𝑥³+8𝜋⁴/15−16𝜋²/3𝑥⁵𝜋²−4𝑥²²<𝑥sec²<𝑥−tan𝑥2𝜋⁴/3𝑥³+256/𝜋²−8𝜋²/3𝑥⁵𝜋²−4𝑥²² (5.1) hold, where (8𝜋⁴/15−16𝜋²/3) and (256/𝜋²−8𝜋²/3) are the best constants in (5.1).