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New refinements of Becker-Stark inequality

  • This paper deals with the well-known Becker-Stark inequality. By using variable replacement from the viewpoint of hypergeometric functions, we provide a new and general refinement of Becker-Stark inequality. As a particular case, the double inequality

    π2(π28)sin2xπ24x2<tanxx<π2(4π2/3)sin2xπ24x2

    for x(0,π/2) will be established. The importance of our result is not only to provide some refinements preserving the structure of Becker-Stark inequality but also that the method can be extended to the case of generalized trigonometric functions.

    Citation: Suxia Wang, Tiehong Zhao. New refinements of Becker-Stark inequality[J]. AIMS Mathematics, 2024, 9(7): 19677-19691. doi: 10.3934/math.2024960

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  • This paper deals with the well-known Becker-Stark inequality. By using variable replacement from the viewpoint of hypergeometric functions, we provide a new and general refinement of Becker-Stark inequality. As a particular case, the double inequality

    π2(π28)sin2xπ24x2<tanxx<π2(4π2/3)sin2xπ24x2

    for x(0,π/2) will be established. The importance of our result is not only to provide some refinements preserving the structure of Becker-Stark inequality but also that the method can be extended to the case of generalized trigonometric functions.



    It is known in the literature that, for x(0,π/2), the inequality

    8π24x2<tanxx<π2π24x2 (1.1)

    was first established by Becker and Stark [6]. This is always known as Becker-Stark inequality, which has attracted much interest many researchers and has been generalized in many different ways; see [7,8,10,13,17,26,27,28] and the references therein. The importance of Becker-Stark inequality is to find the bounds for tanx/x, which are the rational functions with the same order of infinity near π/2. In particular, the first of the notable refinements is given by Zhu [27,Theorem 1.3], who proved that, for x(0,π/2),

    π24(π28)π2x2π24x2<tanxx<π2(π234)x2π24x2. (1.2)

    As a matter of fact, Zhu [27,Theorem 1.4] gives a general refinement of the Becker-Stark inequality. In view of

    limxπ2(tanxx8π24x2)=2π2,

    the left-hand side of (1.1) becomes a good approximate of tanx/x near π/2. Motivated by this remark, Zhu [28,Theorem 3] gives a refinement of (1.1), for x(0,π/2),

    8π24x2+2π2π296π4(π24x2)<tanxx<8π24x2+2π210π2π4(π24x2),

    where (π29)/(6π4) and (10π2)/π4 are the best constants. Further, Debnath et al. [13] present two estimates of tanx/x near π/2 but not in the whole interval (0,π/2); more precisely, the following inequalities hold true

    8+8π(π2x)+(16π283)(π2x)2π24x2<tanxx

    for x(0.373,π/2) and

    tanxx<8+8π(π2x)+(16π283)(π2x)2+(32π383π)(π2x)3π24x2

    for x(0,301,π/2). Recently, alternative good improvements can be found in [10,Equation (2.11)] and [4,Theorem 2.1], where they establish the inequalities

    π2+π2123x2+3844π43π4x4π24x2<tanxx<π2+728π2π2x2+16π2160π4x4π24x2

    and

    1+128π4x2(5π212x2)(π24x2)2<tanxx<1+2π215x2(5π212x2)(π24x2)2

    for x(0,π/2), where the second inequality had been improved by Zhu [29] to the following inequality

    1+(24017π2)π245x2(30π224017π2x2)(π24x2)<tanxx<1+(24017π2)1024π4(17π2120)x2(30π224017π2x2)(π24x2)

    for x(0,π/2) with the best constants (24017π2)π245 and (24017π2)1024π4(17π2120). It is observed that all the above improvements keep the structure of the Becker-Stark inequality, that is to say, the denominator of their approximate functions is π24x2.

    Very recently, Wu and Bercu [18] approximated tanx/x by utilizing the cosine polynomials due to the property of even function, and established the inequalities

    1+(1cosx)(604cos2x1817cosx+1843)945<tanxx<1+(1cosx)(31cosx5cos2x+604)945cosx (1.3)

    for x(0,π/2). Clearly, inequality (1.3) has broken the structure of the Becker-Stark inequality, which leads to the left-hand side of (1.3) being just a bounded function.

    The main objective of this paper is to provide new lower and upper bounds for tanx/x whose forms preserve the structure of the Becker-Stark inequality and numerator is a polynomial of sin2x. More precisely, we transform the function (π24x2)tanx/x into the ratio of two hypergeometric functions by changing a variable t=sin2x and use the first few terms of the series expansion to approximate the objective function. This method, as a practice toy, can be used to reprove the Becker-Stark inequality. The importance of our findings is not only illustrated by giving some new refinements of inequality (1.1), but also by the fact that the method can be extended to generalized trigonometric functions.

    The rest of this paper is organized as follows: In this section, we give an introduction and highlight the relevant previous results. Section 2 consists of some basic knowledge and two lemmas, and is devoted to the proof of the main result. Diverse complements are offered in Section 3, including a comparison of the obtained bounds by graphical analysis, a conjecture raised from the main result, and a p-analogue of Becker-Stark inequality.

    In this section, we first introduce some basic knowledge and present two lemmas that are used to prove the main result.

    Definition 2.1. For real numbers a,b, and c with cN{0}, the Gaussian hypergeometric function is defined as

    F(a,b;c;x):=2F1(a,b;c;x)=n=0(a)n(b)n(c)nxnn!

    for x(1,1), where (a)n=a(a+1)(a+n1)=Γ(a+n)/Γ(a) denotes the Pochhammer symbol or the shifted factorial function for nN. In particular, (a)0=1 for a0. Here Γ(x)=0tx1etdt is the classical Euler gamma function [21,23].

    Recall that the hypergeometric function F(a,b;c;x) has the following properties:

    Property 2.1. A simple derivative formula

    ddxF(a,b;c;x)=abcF(a+1,b+1;c+1;x).

    Property 2.2. The behavior of hypergeometric function F(a,b;c;x) near x=1 satisfies the following situations:

    c>a+b (cf. [16,p. 49])

    F(a,b;c;1)=Γ(c)Γ(cab)Γ(ca)Γ(cb). (2.1)

    c=a+b (cf. [1,15.3.10]), the Ramanujan's asymptotic formula

    B(a,b)F(a,b;c;x)+log(1x)=R(a,b)+O[(1x)log(1x)],(x1). (2.2)

    c<a+b (cf. [15,(1.2)]), as x1,

    F(a,b;c;x)=(1x)cabF(ca,cb;c;x)=Γ(c)Γ(a+bc)Γ(a)Γ(b)(1x)cab[1+o(1)], (2.3)

    where B(a,b)=[Γ(a)Γ(b)]/Γ(a+b), R(a,b)=2γψ(a)ψ(b), ψ(x)=Γ(x)/Γ(x) and γ is the beta function, the Ramanujan constant, the psi function, and the Euler-Mascheroni constant.

    In a particular case of a,b,c, the inverse trigonometric tangent function can be represented by hypergeometric function.

    Property 2.3. (see [1,15.1.5])

    arctan(x)=xF(12,1;32;x2)=x1+x2F(12,12;32;x21+x2). (2.4)

    Remark 2.1. The second equality of (2.4) can be obtained from the transformation formula F(a,b;c;z)=(1z)aF(a,cb;c;z/(z1)) (c.f. [1,15.3.4]), and also coincides with the case of p=2 in [3,Lemma 1].

    Property 2.4. (see [31,(3.6)]) An identity

    (1x)F(a,1;c;x)=1(ca)xcF(a,1;c+1;x). (2.5)

    As is known, a real function φ is said to be absolutely monotonic on the interval I if the kth derivative of φ, denoted by φ(k)(x), exists and is non-negative for each k0 and xI. In other words, if φ can be expressed as a power series on I, then all coefficients are non-negative. In particular, a special power series, roughly speaking, whose coefficients are first negative and then positive is said to be a negative-positive type series, of which the name was first proposed formally in [25] although this type of special series has been studied extensively in the literature [11,22,30].

    Definition 2.2. A power series S(x) given by

    S(x)=mk=0akxk+k=m+1akxk

    is called a "Negative-Positive type" (or "NP type" for short) power series, if its coefficients ak for k0 satisfy

    (i) ak0 for all k0;

    (ii) There exist at least two integers 0k1m and k2m+1 such that ak1,ak20.

    Correspondingly, S(x) is called a "Positive-Negative type" (or "PN type" for short) power series if S(x) is a Negative-Positive type power series.

    The following lemma is a simple and efficient tool to determine the sign of an NP (or PN) type power series, which has been proved in [22,24].

    Lemma 2.1. Let S(x) be a Negative-Positive type power series converging on the interval (0,R). Then

    (i) if S(R)0, then S(x)<0 for all x(0,R);

    (ii) if S(R)>0, then there is a unique ˜x(0,R) such that S(x)<0 for x(0,˜x) and S(x)>0 for x(˜x,R).

    As a consequence, for a PN-type power series, the inequalities of (i) and (ii) are reversed.

    We provide a power series expansion of [F(a,b;a+b+1/2)]2 in the following lemma, which has been demonstrated in [19,Example 14.11] (see also [9]).

    Lemma 2.2. For c=a+b+1/2, it holds that

    [F(a,b;c;x)]2=Γ(c)Γ(2c1)Γ(2a)Γ(2b)Γ(a+b)n=0Γ(2a+n)Γ(a+b+n)Γ(2b+n)n!Γ(c+n)Γ(2c1+n)xn.

    In particular, we have

    [F(12,12;32;x)]2=n=0n!(n+1)(3/2)nxn. (2.6)

    Let t=sin2x for x(0,π/2), and then t(0,1). This gives tan2x=t/(1t), which by (2.4) is equivalent to

    x=arctant1t=tF(12,12;32;t). (2.7)

    By (2.7), it can be rewritten as

    (π24x2)tanxx=[π24tF(12,12;32;t)2](t1t)1/2[t(1t)]1/2F(1,1;32;t)f(t)g(t), (2.8)

    where

    f(t)=π24tF(12,12;32;t)2=π24n=0untn+1, (2.9)
    g(t)=(1t)F(1,1;32;t)=1t3F(1,1;52;t)=113n=0vntn+1 (2.10)

    by (2.5) and (2.6). Here, un and vn are given by

    un=n!(n+1)(3/2)nandvn=n!(5/2)n.

    Moreover, by (2.1), we have

    limt1f(t)g(t)=limt1f(t)g(t)=limt14F(12,12;32;t)F(1,1;52;t)3F(12,12;32,t)2F(12,12;52;t)=8. (2.11)

    Suppose that αn is the Maclaurin's coefficients of π2f(t)/g(t), that is,

    π2f(t)g(t)=n=1αntn,

    then it follows from (2.9) and (2.10) that

    π24n=1un1tn=(π2n=1αntn)(113n=1vn1tn),

    which deduces α1=4π2/3 and the recurrence relation

    3αn=(π284n)vn1+n1k=1αkvnk1,(n2). (2.12)

    Before stating Theorem 2.1, we can compute a finite number of αn by (2.12), which are listed numerically in Table 1. Table 1 illustrates that αn>0 for 1n30. Although we only know a finite αn>0, it still encourages us to prove the following theorem. These evidence demonstrate that Theorem 2.1 is valid in the case of 2N28.

    Table 1.  The values of αn with 2-digit precision.
    n 1 2 3 4 5 6 7 8 9 10
    αn 0.71 0.25 0.14 0.090 0.064 0.049 0.039 0.032 0.027 0.023
    n 11 12 13 14 15 16 17 18 19 20
    αn 0.020 0.017 0.015 0.014 0.012 0.011 0.010 0.0094 0.0087 0.0080
    n 21 22 23 24 25 26 27 28 29 30
    αn 0.0075 0.0070 0.0065 0.0061 0.0058 0.0054 0.0051 0.0049 0.0046 0.0044

     | Show Table
    DownLoad: CSV

    Theorem 2.1. Let αn be defined as in (2.12). If there exists an integer N2 such that αn>0 for 1nN+2, then the double inequality

    π2N1n=1αnsin2nx˜αNsin2Nxπ24x2<tanxx<π2Nn=1αnsin2nxπ24x2 (2.13)

    holds for all x(0,π/2) with the best constants αN and ˜αN, where

    ˜αN=π28N1n=1αn.

    Proof of Theorem 2.1. In order to obtain inequality (2.13), it suffices to show that

    (π2Nn=1αnsin2nx)(π24x)tanxx>0,(π2N1n=1αnsin2nx˜αNsin2Nx)(π24x)tanxx<0

    for x(0,π/2), which by (2.7) and (2.8) is equivalent to

    ϕ1(t):=(π2Nn=1αntn)g(t)f(t)>0, (2.14)
    ϕ2(t):=(π2N1n=1αntn˜αNtN)g(t)f(t)<0 (2.15)

    for t(0,1).

    In terms of power series, by (2.9) and (2.10), we can rewrite ϕ1(t) and ϕ2(t) as

    ϕ1(t)=(π2Nn=1αntn)(113n=0vntn+1)(π24n=0untn+1)=n=N+1τntn, (2.16)
    ϕ2(t)=(π2N1n=1αntn˜αNtN)(113n=0vntn+1)(π24n=0untn+1)=(˜τN˜αN)tN+n=N+1(˜τn+13˜αNvnN1)tn, (2.17)

    where

    τn=4un113(π2vn1Nk=1αkvnk1),˜τn=4un113(π2vn1N1k=1αkvnk1).

    (i) To prove ϕ1(t)>0 for t(0,1).

    We first assert that if τn0 for nN+1, then τn+1<0. To confirm this, if τn0 for nN+1, that is,

    un1112(π2vn1Nk=1αkvnk1), (2.18)

    then we deduce by (2.18) that

    τn+1=4un13(π2vnNk=1αkvnk)=4unun1un113(π2vnvn1vn1Nk=1αkvnkvnk1vnk1)<13[π2(unun1vnvn1)vn1+Nk=1αk(vnkvnk1unun1)vnk1]<0

    for nN+1, where the last inequality follows from

    unun1vnvn1=2n3+11n+12n2+4n3<0,vnkvnk1unun1=1+3n(1+n)(1+2n)332k+2n1+3n(1+n)(1+2n)332+2n=2(1+n)(1+2n)<0

    for 1kN. This confirms the truth of the assertion.

    We now complete the proof in the following two steps:

    Step 1: We prove τN+1>0. Otherwise, we see from the above assertion that τn<0 for nN+2.

    This, together with (2.16) implies that ϕ1(t)<0 for t(0,1). On the other hand, it follows from (2.14) and αN+1>0 that

    ϕ1(t)g(t)=π2Nn=1αntnf(t)g(t)=n=N+1αntn>0

    for t(0,ϵ1) with a sufficiently small ϵ1>0, which is a contradiction.

    Step 2: There are only two situations:

    (a) If all τn>0 for nN+1, then ϕ1(t)>0 for t(0,1) by (2.16).

    (b) If there exists an integer mN+2 such that τm0, we may assume that τm is the first non-positive term. Then the above assertion tells us that τn>0 for N+1nm1 and τn0 for nm. That is to say, ϕ1(t) is a PN-type power series on (0,1). Combining this with Lemma 2.1 and ϕ1(1)=0, it follows that ϕ1(t)>0 for t(0,1).

    (ii) To prove ϕ2(t)<0 for t(0,1).

    Due to αN+2>0, by repeating the above steps, it can also be shown that

    π2N+1n=1αntn>f(t)g(t)π2N+1n=1αnlimt1f(t)g(t)=8

    by (2.11), which gives ˜αNαN+αN+1>αN. Observe that vn/vn1=13/(3+2n)<1, that is to say, vn is strictly decreasing for n0. According to this, with ˜αN>0 and αk>0 (1kN+1), it follows that

    ˜τn+13˜αNvnN1=4un1π23vn1+13N1k=1αkvnk1+13˜αNvnN1>13[(8π2+4n)vn1+(N1k=1αk+˜αN)vn2]=13[4nvn1+(π28)(vn2vn1)]>0

    for nN+1. If ˜τN˜αN0, then ϕ2(t)>0 for t(0,1) by (2.17). This, together with (2.15), implies that

    ϕ2(t)g(t)=π2N1n=1αntn˜αNtNf(t)g(t)=(αN˜αN)tN+n=N+1αntn<0

    for t(0,ϵ2) with a sufficiently small ϵ2>0. This contradicts ϕ2(t)>0 for t(0,1) and thereby ˜τN˜αN<0. According to (2.17), we conclude that ϕ2(t) is an NP-type power series on (0,1) and so ϕ2(t)<0 for t(0,1) by Lemma 2.1 and ϕ2(1)=0.

    In this end, the optimality of constants follows from

    1sin2Nx[π2N1n=1αnsin2nx(π24x2)tanxx]=1tN[π2N1n=1αntnf(t)g(t)]

    and

    limt0+1tN[π2N1n=1αntnf(t)g(t)]=αN,limt11tN[π2N1n=1αntnf(t)g(t)]=˜αN.

    This completes the proof of Theorem 2.1.

    Remark 2.2. It is worth pointing out that the numerator of (2.13) is just a N-order polynomial of sin2x, but the condition of Theorem 2.1 still requires αN+1>0 and αN+2>0. This is mainly used to determine the sign of the first terms of the power series in (2.16) and (2.17). As a fact to remember, if a specific integer N2 is given, then it can be directly verified the sign of τN+1 and ˜τN˜αN without the conditions that αN+1>0 and αN+2>0.

    Remark 2.3. Inequality (2.13) can provide better bounds for larger N. First, our remark is obvious on the right side of (2.13). To see the left side, it suffices to verify from ˜αN+1>0 that

    N1n=1αnsin2nx+˜αNsin2Nx(Nn=1αnsin2nx+˜αNsin2N+2x)=αNsin2Nx+˜αNsin2Nx˜αN+1sin2N+2x=(αN+˜αN˜αN+1)sin2Nx=0.

    Remark 2.4. It is worth noting that it can be seen from the left side of (2.13) that N must be greater than or equal to 2. Now we can extend the range of N to N1. Indeed, due to ˜αN+1=˜αNαN, we can rewrite as

    π2N1n=1αnsin2nx˜αNsin2Nx=π2N1n=1αnsin2nx(˜αN+1+αN)sin2Nx=π2Nn=1αnsin2nx˜αN+1sin2Nx.

    This, together with (2.13), gives

    π2Nn=1αnsin2nx˜αN+1sin2Nxπ24x2<tanxx<π2Nn=1αnsin2nxπ24x2 (2.19)

    holds for all x(0,π/2).

    Taking N=1 into (2.19), we obtain

    Corollary 2.1. For all x(0,π/2), it holds

    π2(π28)sin2xπ24x2<tanxx<π2(4π2/3)sin2xπ24x2 (2.20)

    with the sharp constants π28 and 4π2/3.

    Proof. The sharp constants follow from

    limx0+1sin2x[π2(π24x2)tanxx]=limt0+1t[π2f(t)g(t)]=α1=4π23

    and

    limxπ21sin2x[π2(π24x2)tanxx]=limt11t[π2f(t)g(t)]=π28

    by (2.11).

    Taking N=2 into Theorem 2.1, we obtain

    Corollary 2.2. For all x(0,π/2), it holds

    π2(4π23)sin2x4(π29)3sin4xπ24x2<tanxx<π2(4π23)sin2x12011π245sin4xπ24x2, (2.21)

    where the constants 4(π29)3 and 12011π245 are sharp.

    Remark 2.5. Remark 2.3 enables us to know that the inequality (2.21) is better than inequality (2.20). Further, it is easy to see that inequality (2.20) is better than (1.1). In conclusion, inequality (2.13) completely improves the Becker-Stark inequality. As a matter of fact, Corollaries 2.1 and 2.2 can also be obtained through the method used in [5,14].

    In this section, we provide a graphical analysis of the obtained bounds, give a conjecture and propose a p-analogue of Becker-Stark inequality.

    We now provide a graphical analysis of the lower bounds of Theorem 2.1 (N=5) and (1.3) by distinguishing lower bounds.

    By (2.12), we can compute the first few αn as follows:

    α1=4π23,α2=8311π245,α3=3215191π2945,α4=64352497π214175,
    α5=51231514797π293555,˜α5=3961π2202565235.

    We denote by Lj(x) and Uj(x) (j=1,2) the lower and upper bounds of Theorem 2.1 (N=5) and (1.3), respectively, as follows:

    L1(x)=π24n=1αnsin2nx˜α5sin10xπ24x2,L2(x)=1+(1cosx)(604cos2x1817cosx+1843)945,U1(x)=π25n=1αnsin2nxπ24x2,U2(x)=1+(1cosx)(31cosx5cos2x+604)945cosx.

    Figure 1 presents the graph of the functions L1(x) and L2(x) for x(0,π/2). An immediate remark arising from Figure 1(a) is that the lower bound of Theorem 2.1 (N=5) is better than (1.3). Figure 1(b) illustrates that the upper bound of Theorem 2.1 (N=5) is better than the one of (1.3) near at x=0.

    Figure 1.  Plots of "the bounds of Theorem 2.1 (N=5) and (1.3)".

    From Table 1, it can be seen that αn>0 for 1n30. This allows us to pose the following conjecture:

    Conjecture 3.1. Let f(t) and g(t) be defined as in (2.9) and (2.10) respectively. Then π2f(t)/g(t) is absolute monotonic on (0,1).

    Remark 3.1. If Conjecture 3.1 can be confirmed, then inequality (2.13) can be directly derived from Conjecture 3.1. However, the advantage of Theorem 2.1 is that we only need to know a finite number of αn>0 to prove inequality (2.13).

    For p>1, the generalized sine function sinp is the eigenfunction of the one-dimensional p-Laplacian problem [12]

    Δpu=(|u|p2u)=λ|u|p2,u(0)=u(1)=0,

    which is also the inverse function of arcsinp:(0,1)(0,πp/2) defined as

    arcsinpx=x0(1tp)1/pdt,

    where

    πp=10(1tp)1/pdt=2πpsin(π/p).

    In this case, sinpx is defined on the interval [0,πp/2] and can be extended to the whole R by symmetry and periodicity. Define cosp:RR by

    cospx:=ddxsinpx,xR.

    In particular, it holds

    sinppx+cosppx=1,x[0,πp/2],

    which leads to

    ddxcospx=sinp1pxcos2ppx.

    Similar to the classical trigonometric function, one can define the generalized tangent function

    tanpx=sinpxcospx,for xR{(Z+1/2)πp}.

    It is natural to ask whether the p-analogue of the Becker–Stark inequality holds for x(0,πp/2). Observed that

    limxπp2(π2p4x2)tanpxx=limxπp28sinp1pxcos2ppx={1<p<2,8,p=2,0,p>2,

    which allows us to pose the following problem:

    Problem 3.1. To determine the range of p in [2,) (resp. (1,2)) such that the inequality

    tanpxx<(resp.>)π2pπ2p4x2 (3.1)

    holds for x(0,πp/2).

    Remark 3.2. Inequality (3.1) can be viewed as the p-analogue of Becker-Stark inequality. Our method in this paper reveals that it only needs to study a ratio of two hypergeometric functions by changing the variable t=sinppx in (3.1).

    In this paper, from the viewpoint of hypergeometric function, we study the well-known Becker-Stark inequality by changing a variable t=sin2x. Our main result is to approximate the function π[(π24x2)tanx]/x by the first few terms of the Taylor series, even if we only know finitely many positive coefficients. In particular, the double inequality

    π2(π28)sin2xπ24x2<tanxx<π2(4π2/3)sin2xπ24x2

    holds for x(0,π/2), which improves Becker-Stark inequality (1.1).

    Suxia Wang: Conceptualisation, writing – original draft, formal analysis; Tiehong Zhao: Writing – review & editing, supervision, validation. All authors have read and approved the final version of the manuscript for publication.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

    This research was supported by the Natural Science Research Project of Colleges and Universities in Anhui Province (2022AH051588, 2023AH051549), the Natural Science Research Project of Huainan Normal University (2022XJZD031) and the National Natural Science Foundation of China (11971142).

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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