MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi 10.1155/2018/4178629 4178629 Research Article Refined Estimates and Generalizations of Inequalities Related to the Arctangent Function and Shafer’s Inequality http://orcid.org/0000-0002-4963-4149 Malešević Branko 1 Rašajski Marija 1 Lutovac Tatjana 1 López-Ochoa Luis M. University of Belgrade School of Electrical Engineering Department of Applied Mathematics Serbia bg.ac.rs 2018 10102018 2018 10 11 2017 11 09 2018 10102018 2018 Copyright © 2018 Branko Malešević et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We give some sharper refinements and generalizations of inequalities related to Shafer’s inequality for the arctangent function, stated in Theorems  1, 2,  and  4 in Mortici and Srivastava, 2014, by C. Mortici and H.M. Srivastava.

Ministarstvo Prosvete, Nauke i Tehnološkog Razvoja ON 174032 III 44006 ON 174033 TR 32023
1. Introduction

Inverse trigonometric functions play an important role and have many applications in engineering . In particular, the arctangent function and various related inequalities have been studied and effectively applied to problems in fundamental sciences and many areas of engineering, such as electronics, mechanics, and aeronautics [3, 5, 6]; see also .

Various approximations of the arctangent function can be found in [46, 821]; see also [22, 23]. One of the inequalities that attracted attention of many authors is Shafer’s inequality :(1)3x1+21+x2<arctanx,which holds for x>0; see also .

Recently, in , Mortici and Srivastava proved the following results, cited here as Statements 1, 2, and 3, related to the above inequality. These results are the starting point of our research.

Statement 1 (Theorem 1, [<xref ref-type="bibr" rid="B1">8</xref>]).

For every x>0, the following two-sided inequality holds:(2)3x1+21+x2+ax<arctanx<3x1+21+x2+bx,where a(x)=1/180x5-13/1512x7 and b(x)=1/180x5.

Statement 2 (Theorem 2, [<xref ref-type="bibr" rid="B1">8</xref>]).

For every x>0, it is asserted that(3)3x+cx1+21+x2<arctanx<3x+dx1+21+x2,where c(x)=1/60x5-17/840x7 and d(x)=1/60x5.

Statement 3 (Theorem 4, [<xref ref-type="bibr" rid="B1">8</xref>]).

For every x>0, it is asserted that(4)-112x3<arctanx-2x1+1+x2<-112x3+340x5.

The main results of this paper are refined estimates and generalizations of the inequalities given in Statements 1, 2, and 3. Although inequalities (2), (3), and (4) hold for x>0, considering them in a neighborhood of zero is of primary importance, as noted in .

2. Main Results

First, let us recall some well-known power series expansions that will be used in our proofs.

For x1,(5)arctanx=m=0-1mAmx2m+1,where(6)Am=12m+1.

For x1,(7)1+x2=1+m=0-1mKmx2m+2,where(8)Km=2m!m!m+1!22m+1.

The following power series expansion holds:(9)3x1+21+x2=m=0Bmx2m+1, where |x|3/2, with B(0)=1 and B(1)=-1/3, and for m2,(10)Bm=-1m4m-13m1-8i=2m2i-2!i-1!i!22i-134i.Power series coefficients are calculated by applying Cauchy’s product to the power series expansions arising from the following transformation of the corresponding function:(11)3x1+21+x2=-x1-21+x21+4/3x2.

It is easy to prove that sequence BmmN0 for m1 satisfies the recurrence equation:(12)Bm+1+43Bm=-1m2m-1!22m-1m-1!m+1!=2-1m2m-1!!2m+2!!.

2.1. Refinements of the Inequalities in Statement <xref ref-type="statement" rid="state1">1</xref>

Before we proceed to Theorem 9, which represents an improvement and generalization of Statement 1, we need the following lemmas.

Lemma 4.

Let β(0)=1, β(1)=-1/3, and β(m)=3-1m/22m+1k=02k+2m-1!/k+m-1!k+m+1!3/16k, for m2. The sequence β(m)mN0 for m1 satisfies the recurrence relation (12).

Proof.

In the proof of this lemma we use the Wilf- Zeilberger method . (The same approach we used in .)

The assertion is obviously true for m=1.

Let m2 and (13)gk,m=2k+2m-1!k+m-1!k+m+1!andϕm=3-1m22m+1.Then we have (14)βm=ϕmk=0gk,m316k.Further we have(15)βm+1+43βm=k=0ϕm+1gk,m+1+43ϕmgk,m316k=-1m22m-1·k=02k+2m+132k+2m-1!8k+8m+16k+m-1!k+m+1!316k.Consider now the sequence S(m)mN,m2, where(16)Sm=k=0m-1Fm,kandFm,k=2m+2k+132m+2k-1!8m+8k+16m+k-1!m+k+1!316k.Consider the function (an algorithm for determining function G(m,k) for a given function F(m,k) is described in . Note that the pair of discrete functions   Fm,k,Gm,k is the so-called Wilf- Zeilberger pair)(17)Gm,k=-8k-8m-162k+2m-1!8k+8m+16k+m-1!k+m+1!316k,where mN0 and kN0. It is not hard to verify that functions F(m,k) and G(m,k) satisfy the following relation:(18)Fm,k=Gm,k+1-Gm,k.If we sum both sides of (18) over all kN0, we get the following relation: (19)Sm=-Gm,0. Finally, as (20)Gm,0=-2m-1!m-1!m+1! we have(21)Sm=2m-1!m-1!m+1!.Therefore from (15) and (21) we conclude that (22)βm+1+43βm=-1m22m-1·Sm=-1m22m-1·2m-1!m-1!m+1!.

Corollary 5.

Given that the sequences B(m)mN0 and β(m)mN0 satisfy the same recurrence relation and as they agree for m=0 and m=1, we conclude that(23)Bm=βm,for  mN0.

Let us introduce the notation:(24)Cm=Am-Bm, where C(0)=C(1)=0, and for m2 the following holds:(25)Cm=12m+1-4m-13m1-8i=2m2i-2!i-1!i!22i-134i.

Thus, we have the power series expansion:(26)fx=arctanx-3x2+1+x2=m=0-1mCmx2m+1,for mN0 and |x|3/2.

Let us introduce the notation:(27)β+m=322m+1k=02k+2m-1!k+m-1!k+m+1!22k34k,mN0β1m=k=0+n=0k-1m+1/2+nn=0k-1m+2+n34k,β1km=n=0k-1m+1/2+nn=0k-1m+2+n,mN0.

Lemma 6.

For mN0 the following holds:(28)β+m=322m-1!!2m+2!!β1m.

Proof.

(29) 3 2 2 m + 1 2 m + 2 k - 1 ! m + k - 1 ! m + k + 1 ! 2 2 k = 3 2 1 2 m + 2 ! ! 2 m + 2 k - 1 ! ! m + k + 1 ! m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! 2 m + 1 2 m + 3 2 m + 2 k - 1 m + 1 ! m + 2 m + 3 m + k + 1 m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! n = 0 k - 1 m + 1 / 2 + n n = 0 k - 1 m + 2 + n .

Lemma 7.

For mN0 the following holds:(30)8m+22m+13=k=0m+1/2m+2k34k<β1m<k=034k=4.

Proof.

The statement immediately follows from the inequalities:(31)m+1/2m+2k<β1km<1.

Lemma 8.

For mN0 the following holds:(32)β1m+1>β1mm+2m+1/21-321m+2.

Proof.

(33) β 1 k m + 1 = β 1 k m m + 2 m + 1 / 2 + k 1 / 2 + m m + 2 + k = β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 + k > β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 .

Theorem 9.

For the real analytic function(34)fx=arctanx-3x2+1+x2the following inequalities hold for kN and x(0,3/2]:(35)m=02k+1-1mCmx2m+1<fx<m=02k-1mCmx2m+1.where C(0)=C(1)=0, and for m2 the following holds: (36)Cm=12m+1-4m-13m1-8i=2m2i-2!i-1!i!22i-134i.

Proof.

We will prove that the sequence {C(m)}mN0 is positive and monotonically decreasing and tends to zero as m tends to infinity. We will use Lemmas 7 and 8. (37)Cm=12m+1-322m-1!!2m+2!!β1m>12m+1-4322m-1!!2m+2!!=12m+11-62m+1!!2m+2!!.It is easy to verify that 6(2m+1)!!/(2m+2)!!<1 for m11; therefore C(m)>0. Let us note that 0<C(m)<1/2m+1, so we can conclude that limm+C(m)=0.

Let us now prove that {C(m)}mN0 is a monotonically decreasing sequence. (38)Cm+1-Cm=-22m+12m+3-322m+1!!2m+4!!β1m+1+322m-1!!2m+2!!β1m=-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4β1m+1<-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4·m+2m+1/21-321m+2β1m=-22m+12m+3+β1m942m-1!!m+22m+2!!<-22m+12m+3+4942m-1!!m+22m+2!!=-22m+12m+31-92m+3!!2m+4!!.It is easy to prove that C(m+1)-C(m)<0 for m8, i.e., the sequence is monotonically decreasing. Since CmmN0 is positive for m2 and monotonically decreasing (for m8) and tends to zero, the same holds true for the sequence C(m)x2m+1mN0 for a fixed x(0,3/2] (noting that it is decreasing for m3 ), so we can apply Leibniz’s theorem for alternating series , thus proving the claim of Theorem 9: (39)m=02k+1-1mCmx2m+1<fx<m=02k-1mCmx2m+1,kN.

Examples. For k=1 and x(0,3/2] we get Statement 1.

For k=2 and x(0,3/2],(40)x5180-13x71512+53x95184-3791x11342144<arctanx-3x1+21+x2<x5180-13x71512+53x95184.For k=3 and x(0,3/2],(41)x5180-13x71512+53x95184-3791x11342144+55801x134852224-130591x1511197440<arctanx-3x1+21+x2<x5180-13x71512+53x95184-3791x11342144+55801x134852224,etc.

2.2. Refinements of the Inequalities in Statement <xref ref-type="statement" rid="state2">2</xref>

We propose the following improvement and generalization of Statement 2.

Theorem 10.

For every x(0,1] and kN, it is asserted that(42)3x+m=22k+1-1mEmx2m+11+21+x2<arctanx<3x+m=22k-1mEmx2m+11+21+x2,where (43)Em=32m+1-i=0m-12m-2i-2!22m-2i-22i+1m-i-1!m-i!.

Examples. For x(0,1] and k=1 we get inequality (3) from Statement 2.

For x(0,1] and k2 inequality (42) refines inequality (3) from Statement 2 and we have the following new results:

Taking k=2 in (42) gives (44)3x+1/60x5-17/840x7+139/6720x9-8947/443520x111+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x91+21+x2.

Taking k=3 in (42) gives (45)3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x13-851677/46126080x151+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x131+21+x2,

etc.

Proof of Theorem <xref ref-type="statement" rid="thm2">10</xref>.

Based on Cauchy’s product of power series (7) and (5), the real analytical function,(46)fx=1+21+x2·arctanx-3x,for x(0,1] has the following power series:(47)fx=m=2-1mEmx2m+1,where(48)Em=32m+1-i=0m-12m-2i-2!2i+1m-i-1!m-i!22m-2i-2.

We aim to show that sequence E(m)mN,m2 decreases monotonically and that limm+E(m)=0. It is easy to verify that sequence E(m)mN,m2 satisfies the following recurrence relation:(49)-2mEm+2m+3Em+1=12m+1-m+12m!m+1!24m.Consider the sequence e(m)mN,m2 where (50)em=gm·Sm and (51)Sm=j=1m-12j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2 and (52)gm=m!m-1!22m+12m+1!.

It is easy to verify that sequence e(m)mN,m2 satisfies the recurrence relation (49). Given that sequences E(m)mN,m2 and e(m)mN,m2 agree for m=2 and m=3, we conclude that(53)Em=em,formN,m2.

We prove that sequence e(m)mN,m2 is a monotonically decreasing sequence and limm+e(m)=0.

By the principle of mathematical induction, it follows that (54)2j+1!<2j!!2j+1 is true for all jN. Therefore S(m)>0 for m2, i.e.,(55)em>0,form2.To prove that emmN,m2 is a monotonically decreasing sequence, let us use the following notation: (56)Sm=j=1m-1hj where (57)hj=2j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2.

Consider the following equivalences for m2:(58)em+1em<1gm+1Sm+1gmSm<1gm+1gm·Sm+hmSm<12m2m+3·Sm+hmSm<12mhm<3Sm.

Consider the last inequality. It is easy to verify that it is true for m=2. Observing that(59)3Sm+1=3Sm+hm=3Sm+3hmand using the induction hypothesis 3Sm>2mhm for some positive integer m2, we conclude that (60)3Sm+1>2m+3hm>2m+1hm. Therefore, by the principle of mathematical induction, the inequality(61)2mhm<3Sm is true for m2, i.e.,(62)em+1em<1,form2.

Let us further consider the positive addend of S(m), i.e.,(63)S+m=j=1m-12j+2!2j2j+122j+32j+1j+1!22j2=j=1m-12j!42j!!2.By the principle of mathematical induction, it follows that(64)S+m=m2m!22m!!2-14.

Finally, given that for m2(65)0<emgm·S+m=12m+1 we have limm+e(m)=0.

Finally, based on (53) we conclude that EmmN,m2 is a positive monotonically decreasing sequence and that it tends to zero. The same holds true for the sequence E(m)x2m+1mN,m2 for a fixed x(0,1] so we can apply Leibniz’s theorem for alternating series , thus proving the claim of Theorem 10.

2.3. Refinements of the Inequalities in Statement <xref ref-type="statement" rid="state3">3</xref>

We propose the following improvement and generalization of Statement 3.

Theorem 11.

For every x(0,1] and kN, it is asserted that(66)m=12k-1-1mCmx2m+1<arctanx-2x1+1+x2<m=12k-1mCmx2m+1,where (67)Cm=12m+1-2m-1!!m+1!2m.

Examples. For x(0,1] and k=1 we get inequality (4) from Statement 3.

For x(0,1] and k2 inequality (66) from Theorem 11 refines inequality (4) from Statement 3 and we have the following new results:

Taking k=2 in (66) gives (68)-112x3+340x5-29448x7<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9.

Taking k=3 in (66) gives (69)-112x3+340x5-29448x7+651152x9-2815632x11<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9-2815632x11+59513312x13,

etc.

Proof of Theorem <xref ref-type="statement" rid="thm3">11</xref>.

For x(0,1] the following power series expansion holds: (70)arctanx-2x1+1+x2=arctanx+21-x2+1x=m=1-1mCmx2m+1 where(71)Cm=12m+1-2m-1!!m+1!2m.

We prove that the sequence C(m)mN is positive and monotonically decreasing and tends to zero as m tends to infinity.

It is easy to verify that (2m+2)!!>2(2m+1)!! is true for mN. Thus, the following equivalences hold true for every mN: (72)Cm>012m+1>2m-1!!m+1!2m2m+2!!>22m+1!!,and we conclude that C(m)>0 for every mN.

Let us now prove that C(m)mN is a monotonically decreasing sequence. We have (73)Cm-Cm+1>022m+12m+3-32m-1!!2m+2!!m+2>02m+4!!>32m+3!!.

As it is easy to show (by the principle of mathematical induction) that the last inequality holds true for mN, we may conclude that C(m)mN is a monotonically decreasing sequence.

Finally, as 0<C(m)<1/2m+1, we conclude that limm+C(m)=0.

Since CmmN is a positive monotonically decreasing sequence, and it tends to zero, the same holds true for the sequence C(m)x2m+1mN for a fixed x(0,1]. So we can apply Leibniz’s theorem for alternating series  and thus prove the claim of Theorem 11.

3. Conclusion

In Theorems 9, 10, and 11 of this paper we proved some new inequalities related to Shafer’s inequality for the arctangent function. These inequalities represent sharpening and generalization of the inequalities given in  (Theorems1,2,and4).

Let us mention that it is possible to prove inequality (35), for any fixed kN and x(0,3/2], by substituting x=tant for t(0,π/3] using the algorithms and methods (see also [29, 30]) developed in [31, 32]. Also, inequalities (42) and (66) for any fixed kN and x0,1 can be proved by substituting x=tant for t0,π/4 using the algorithms and methods (see also [29, 30]) developed in [31, 32].

Conflicts of Interest

The authors would like to state that they do not have any conflicts of interest in the subject of this research.

Authors’ Contributions

All the authors participated in every phase of the research conducted for this paper.

Acknowledgments

Research of the first, second, and third author was supported in part by the Serbian Ministry of Education, Science and Technological Development, under Projects ON 174032, III 44006, ON 174033, and TR 32023, respectively.

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