Research article

New inequalities of Wilker's type for hyperbolic functions

  • Received: 24 October 2019 Accepted: 19 November 2019 Published: 22 November 2019
  • MSC : 26D05, 26D15

  • Using power series expansions of functions and the monotonicity criterion for the quotient of power series, we establish some new Wilker-type inequalities for hyperbolic functions.

    Citation: Ling Zhu. New inequalities of Wilker's type for hyperbolic functions[J]. AIMS Mathematics, 2020, 5(1): 376-384. doi: 10.3934/math.2020025

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  • Using power series expansions of functions and the monotonicity criterion for the quotient of power series, we establish some new Wilker-type inequalities for hyperbolic functions.


    Wilker [1] proposed two open problems as the following statements:

    (a) If 0<x<π/2, then

    (sinxx)2+tanxx>2. (1.1)

    (b) There exists a largest constant c such that

    (sinxx)2+tanxx>2+cx3tanx (1.2)

    for 0<x<π/2.

    Sumner et al. [2] affirmed the truth of two problems above and obtained a further results as follows

    16π4x3tanx<(sinxx)2+tanxx2<845x3tanx, 0<x<π2, (1.3)

    where 16/π4 and 8/45 are the best constants in (1.3).

    Some refreshing proofs of inequalities (1.1) and (1.3) can be found in Pinelis [3]. In 2007, the author of this paper [4] established a new Wilker-type inequality involving hyperbolic functions as follows:

    (sinhxx)2+tanhxx2>845x3tanhx, x0,  (1.4)

    where 8/45 can not be replaced by any larger number.

    In 2011, Sun and Zhu [5] showed another new Wilker-type inequality involving hyperbolic functions and obtained the following result.

    (xsinhx)2+xtanhx2<245x3sinhx, x0, (1.5)

    where 2/45 can not be replaced by any smaller number.

    The purpose of this paper is to give three new inequalities of Wilker-type for hyperbolic functions.

    Theorem 1. Let x0. Then

    (sinhxx)2+tanhxx2>αx4(tanhxx)6/7 (1.6)

    holds if and only if  α8/45.

    Theorem 2. The function

    G(x)=(xsinhx)2+xtanhx2x3tanhx

    has only one maximum point x0=1.54471 on (0,), so the function G(x) has the maximum value

    G(x0)=maxx(0,)G(x)=0.050244=θ0.

    Specifically, for x0,

    (xsinhx)2+xtanhx2<θx3tanhx (1.7)

    holds if and only if θθ0.

    Theorem 3. Let x0. Then

    (xsinhx)2+xtanhx2<βx4(tanhxx)4/7 (1.8)

    holds if and only if  β2/45.

    In order to prove Theorem 2, we need the following lemma. We introduce a useful auxiliary function Hf,g. For a<b, let f and g be differentiable on (a,b) and g0 on (a,b). Then the function Hf,g is defined by

    Hf,g:=fggf.

    The function Hf,g has some well properties and play an important role in the proof of a monotonicity criterion for the quotient of power series.

    Lemma 1. ([6]) Let f(t)=k=0aktk and g(t)=k=0bktk be two real power series converging on (r,r) and bk>0 for all k. Suppose that for certain mN, the non-constant sequence {ak/bk} is increasing (resp. decreasing) for 0km and decreasing (resp. increasing) for km. Then the function f/g is strictly increasing (resp. decreasing) on (0,r) if and only if Hf,g(r)0 (resp. 0). Moreover, if Hf,g(r)<0 (resp. >0), then there exists t0(0,r) such that the function f/g is strictly increasing (resp. decreasing) on (0,t0) and strictly decreasing (resp. increasing) on (t0,r).

    Since

    (sinhxx)2+tanhxx2>845x4(tanhxx)6/7
    [(sinhxx)2+tanhxx2]7>[845x4(tanhxx)6/7]7= 2097152373 669453125x22tanh6x,

    we can let

    F(x)=7ln[(sinhxx)2+tanhxx2]ln[2097152373 669453125x22tanh6x], x>0.

    Then

    F(x)= h(x)8x3(tanhx)(cosh3x)[(sinhxx)2+tanhxx2] ,

    where

    h(x)=18sinh3x18sinh5x+36sinhx+88x2sinh3x+56xcoshx63xcosh3x+7xcosh5x+96x3coshx+96x2sinhx .

    By substituting the power series expansions of all hyperbolic functions involved in the above formula into h(x), we obtain that

    h(x)=18n=0(3x)2n+1(2n+1)!18n=0(5x)2n+1(2n+1)!+36n=0x2n+1(2n+1)!+88x2n=0(3x)2n+1(2n+1)!+56xn=0x2n(2n)!63xn=0(3x)2n(2n)!+7xn=0(5x)2n(2n)!+96x3n=0x2n(2n)!+96x2n=0x2n+1(2n+1)!
    =18n=532n+1(2n+1)!x2n+118n=552n+1(2n+1)!x2n+1+36n=51(2n+1)!x2n+1+88n=432n+1(2n+1)!x2n+3+56n=51(2n)!x2n+163n=532n(2n)!x2n+1+7n=552n(2n)!x2n+1+96n=41(2n)!x2n+3+96n=41(2n+1)!x2n+3
    =n=5 18(32n+152n+1+2)+(2n+1)(566332n+752n)(2n+1)!x2n+1+n=4 24(1132n+4)+96(2n+1)(2n+1)!x2n+3=n=5 18(32n+152n+1+2)+(2n+1)(566332n+752n)(2n+1)!x2n+1+n=5 24(1132n2+4)+96(2n1)(2n1)!x2n+1=n=5l(n)3(2n+1)!x2n+1,

    where

    l(n)= 3(14n83)52n+(352n2202n27)32n+(2304n3+1152n2+336n+276).

    Since l(5)=77856768>0 and for n6,

    14n8314683= 1>0,352n2202n2735262202627= 11433>0,2304n3+1152n2+336n+276>0,

    we have that l(n)>0 for n5. So h(x)>0 and F(x)>0 for x>0. Then the function F(x) is strictly increasing on (0,). Therefore, F(x)>F(0+)=0. At the same time, we find that

    limx0+(sinhxx)2+tanhxx2x4(tanhxx)6/7=845.

    Then the proof of Theorem 1 is complete.

    Let

    G(x)=(xsinhx)2+xtanhx2x3tanhx, 0<x<.

    Rewrite G(x) as

    G(x)= coshxx3sinhx(x2sinh2x+xcoshxsinhx2)= (x2+xcoshxsinhx2sinh2x)(coshx)x3sinh3x=14(xsinh3x2cosh3x+2coshx+xsinhx+4x2coshx) 14x3(sinh3x3sinhx)= xsinh3x2cosh3x+2coshx+xsinhx+4x2coshxx3(sinh3x3sinhx):=f(x)g(x).

    By substituting the power series expansions of the hyperbolic functions involved in the above formula into f(x) and g(x) we have

    f(x)=n=032n+1x2n+2(2n+1)!2n=032nx2n(2n)!+2n=0x2n(2n)!+n=0x2n+2(2n+1)!+4n=0x2n+2(2n)!=n=232n+1x2n+2(2n+1)!2n=332nx2n(2n)!+2n=3x2n(2n)!+n=2x2n+2(2n+1)!+4n=2x2n+2(2n)!=n=232n+1+1+4(2n+1)(2n+1)!x2n+2n=32(32n1)(2n)!x2n=n=232n+1+1+4(2n+1)(2n+1)!x2n+2n=22(32n+21)(2n+2)!x2n+2=n=2(2n+2)(32n+1+1+4(2n+1))2(32n+21)(2n+2)!x2n+2=n=2 2[(3n6)32n+8n2+13n+6](2n+2)!x2n+2:=n=2anx2n+2,

    and

    g(x)=x3(n=0(3x)2n+1(2n+1)!3n=0x2n+1(2n+1)!)=n=0(32n+13)x2n+4(2n+1)!=n=1(32n+13)x2n+4(2n+1)!=n=2(32n13)(2n1)!x2n+2:=n=2bnx2n+1.

    Let

    cn=anbn= 2[(3n6)32n+8n2+13n+6](2n+2)!(32n13)(2n1)!= 2[(3n6)32n+8n2+13n+6](2n+2)(2n+1)2n(32n13).

    We can calculate that

    c2=245<c3=235>c4=2064095,

    and show that {cn}n4 is decreasing:

    cncn+1CD:= 2((3n6)32n+8n2+13n+6)(2n+2)(2n+1)2n(32n13) 2((3n3)32n+2+8(n+1)2+13(n+1)+6)(2n+4)(2n+3)(2n+2)(32n+13):=EF.

    In fact,

    CFDE= 23p(n) ,

    where

    p(n)=(108n2189n324)34n+(128n4+1104n3+952n2+1026n+648)32n(144n3+612n2+837n+324).

    Since for n4, we have 108n2189n324648>0. By mathematical induction it is easy to prove that

    32n>144n3+612n2+837n+324128n4+1104n3+952n2+1026n+648

    holds for n4. So p(n)>0 for n4. This leads to {cn}n4 is decreasing. Therefore

    c2<c3>c4>c5>.

    We compute to get

    Hf,g()=limx(fggf)= ,

    By Lemma 1 we obtain that there exists x0(0,) such that the function G(x)=f/g is strictly increasing on (0,x0) and strictly decreasing on (x0,). That is, x0 is the only maximum point of the function G(x) on (0,). Let us determine the maximum point x0 and the maximum value G(x0). We compute

    G(x)=2r(x)x4(15cosh2x6cosh4x+cosh6x10) ,

    where

    r(x)=10x15sinh2x+12sinh4x3sinh6x+8x3cosh2x+4x3cosh4x12x2sinh2x+6x2sinh4x+15xcosh2x6xcosh4x+xcosh6x12x3.

    We find out that

    G(1.54471)=1. 1086×107, G(1.54472)=2. 2292×109.

    So x0=1.54471 and

    G(x0)=maxx(0,)G(x)=0.050244.

    Considering the reasons

    limx0+G(x)= 245=0.044444, limxG(x)= 0,

    we have

    minx(0,)G(x)=0.

    The proof of Theorem 2 is completed.

    Let

    H(x)=ln[128373 669453125x24cosh4xsinh4x]7ln[(xsinhx)2+xtanhx2],

    where 0<x<. Then

    H(x)= q(x)8x( coshxsinh3x) [(xsinhx)2+xtanhx2] ,

    where

    q(x)=15x+96sinh2x48sinh4x+56x3cosh2x+84x2sinh2x32xcosh2x+17xcosh4x+88x3.

    By substituting the power series expansions of the hyperbolic functions involved in the above formula into q(x), we have

    q(x)=15x+96n=022n+1x2n+1(2n+1)!48n=042n+1x2n+1(2n+1)!+56x3n=022nx2n(2n)!+84x2n=022n+1x2n+1(2n+1)!32xn=022nx2n(2n)!+17xn=042nx2n(2n)!+88x3=96n=522n+1x2n+1(2n+1)!48n=542n+1x2n+1(2n+1)!32n=522nx2n+1(2n)!+17n=542nx2n+1(2n)!+56n=422nx2n+3(2n)!+84n=422n+1x2n+3(2n+1)!+56n=422nx2n+3(2n)!+84n=422n+1x2n+3(2n+1)!
    =n=5(34n175)42n(64n160)22n(2n+1)!x2n+1+n=4112(n+2)22n(2n+1)!x2n+3=n=5(34n175)42n(64n160)22n(2n+1)!x2n+1+n=528(n+1)22n(2n1)!x2n+1=n=5[(34n175)42n(64n160)22n(2n+1)!+28(n+1)22n(2n1)!]x2n+1=n=5(34n175)42n(64n160)22n+28(2n+1)(2n)(n+1)22n(2n+1)!x2n+1=n=5(34n175)42n+8(14n3+21n2n+20)22n(2n+1)!x2n+1=n=5dn(2n+1)!x2n+1,

    where

    dn=(34n175)42n+814n3+21n2n+20, n5.

    We can find d5= 13516800>0 and dn>0 holds for n6 due to

    34n17534×6175= 29>0,14n3+21n2n+201463+21626+20= 3794>0.

    So H(x)> 0. Then H(x) is increasing on (0,). Therefore, H(x)>H(0+)=0. At the same time, we find that

    limx0+(xsinhx)2+xtanhx2x4(tanhxx)4/7=245.

    Then the proof of Theorem 3 is complete.

    This paper is supported by the Natural Science Foundation of China grants No.61772025.

    The author declares no conflict of interest in this paper.



    [1] J. B. Wilker, Problem E 3306, Amer. Math. Monthly, 96 (1989), 55.
    [2] J. S. Sumner, A. A. Jagers, M. Vowe, et al. Inequalities involving trigonometric functions, Amer. Math. Monthly, 98 (1991), 264-267.
    [3] I. Pinelis, L'Hospital rules for monotonicity and the Wilker-Anglesio inequality, Amer. Math. Monthly, 111 (2004), 905-909.
    [4] L. Zhu, On Wilker-type inequalities, Math. Inequal. Appl., 10 (2007), 727-731.
    [5] Z. J. Sun and L. Zhu, On New Wilker-type inequalities, ISRN Mathematical Analysis, 2011 (2011), 1-7.
    [6] Zh. H. Yang, Y. M. Chu and M. K. Wang, Monotonicity criterion for the quotient of power series with applications, J. Math. Anal. Appl., 428 (2015), 587-604.
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