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Research article

On simultaneous characterizations of partner-ruled surfaces in Minkowski 3-space

  • Received: 16 May 2023 Revised: 21 June 2023 Accepted: 29 June 2023 Published: 13 July 2023
  • MSC : 53A04, 53A05

  • In this study, the partner-ruled surfaces in Minkowski 3-space, which are defined according to the Frenet vectors of non-null space curves, are introduced with extra conditions that guarantee the existence of definite surface normals. First, the requirements of each pair of partner-ruled surfaces to be simultaneously developable and minimal (or maximal for spacelike surfaces) are investigated. The surfaces also simultaneously characterize the asymptotic, geodesic and curvature lines of the parameter curves of these surfaces. Finally, the study provides examples of timelike and spacelike partner-ruled surfaces and includes their graphs.

    Citation: Yanlin Li, Kemal Eren, Soley Ersoy. On simultaneous characterizations of partner-ruled surfaces in Minkowski 3-space[J]. AIMS Mathematics, 2023, 8(9): 22256-22273. doi: 10.3934/math.20231135

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  • In this study, the partner-ruled surfaces in Minkowski 3-space, which are defined according to the Frenet vectors of non-null space curves, are introduced with extra conditions that guarantee the existence of definite surface normals. First, the requirements of each pair of partner-ruled surfaces to be simultaneously developable and minimal (or maximal for spacelike surfaces) are investigated. The surfaces also simultaneously characterize the asymptotic, geodesic and curvature lines of the parameter curves of these surfaces. Finally, the study provides examples of timelike and spacelike partner-ruled surfaces and includes their graphs.



    In 1998, Kahlig and Matkowski [1] proved in particular that every homogeneous bivariable mean M in (0,) can be represented in the form

    M(x,y)=A(x,y)fM,A(xyx+y),

    where A is the arithmetic mean and fM,A: (1,1)(0,2) is a unique single variable function (with the graph laying in a set of a butterfly shape), called an A-index of M.

    In this paper we consider Seiffert function f:(0,1)R which fulfils the following condition

    t1+tf(t)t1t.

    According to the results of Witkowski [2] we introduce the mean Mf of the form

    Mf(x,y)={|xy|2f(|xy|x+y)xy,xx=y. (1.1)

    In this paper a mean Mf:R2+R is the function that is symmetric, positively homogeneous and internal in sense [2]. Basic result of Witkowski is correspondence between a mean Mf and Seiffert function f of Mf is given by the following formula

    f(t)=tMf(1t,1+t),  (1.2)

    where

    t=|xy|x+y. (1.3)

    Therefore, f and Mf form a one-to-one correspondence via (1.1) and (1.2). For this reason, in the following we can rewrite f=:fM.

    Throughout this article, we say xy, that is, t(0,1). For convenience, we note that M1<M2 means M1(x,y)<M2(x,y) holds for two means M1 and M2 with xy. Then there is a fact that the inequality fM1(t)>fM2(t) holds if and only if M1<M2. That is to say,

    1fM1<1fM2M1<M2. (1.4)

    The above relationship (1.4) inspires us to ask a question: Can we transform the means inequality problem into the reciprocal inequality problem of the corresponding Seiffert functions? Witkowski [2] answers this question from the perspective of one-to-one correspondence. We find that these two kinds of inequalities are equivalent in similar linear inequalities. We describe this result in Lemma 2.1 as a support of this paper.

    As we know, the study of inequalities for mean values has always been a hot topic in the field of inequalities. For example, two common means can be used to define some new means. The recent success in this respect can be seen in references [3,4,5,6,7,8]. In [2], Witkowski introduced the following two new means, one called sine mean

    Msin(x,y)={|xy|2sin(|xy|x+y)xyxx=y, (1.5)

    and the other called hyperbolic tangent mean

    Mtanh(x,y)={|xy|2tanh(|xy|x+y)xyxx=y. (1.6)

    Recently, Nowicka and Witkowski [9] determined various optimal bounds for the Msin(x,y) and Mtanh(x,y) by the arithmetic mean A(x,y)=(x+y)/2 and centroidal mean

    Ce(x,y)=23x2+xy+y2x+y

    as follows:

    Proposition 1.1. The double inequality

    (1α)A+αCe<Msin<(1β)A+βCe

    holds if and only if α 1/2 and β(3/sin1)30.5652.

    Proposition 1.2. The double inequality

    (1α)A+αCe<Mtanh<(1β)A+βCe

    holds if and only if α (3/tanh1)30.9391 and β1.

    Proposition 1.3. The double inequality

    (1α)C1e+αA1<M1sin<(1β)C1e+βA1

    holds if and only if α 4sin130.3659 and β1/2.

    Proposition 1.4. The double inequality

    (1α)C1e+αA1<M1tanh<(1β)C1e+βA1

    holds if and only if α 0 and β4tanh130.0464.

    Proposition 1.5. The double inequality

    (1α)A2+αC2e<M2sin<(1β)A2+βC2e

    holds if and only if α 1/2 and β(9cot21)/70.5301.

    Proposition 1.6. The double inequality

    (1α)A2+αC2e<M2tanh<(1β)A2+βC2e

    holds if and only if α (9(coth211))/70.9309 and β1.

    Proposition 1.7. The double inequality

    (1α)C2e+αA2<M2sin<(1β)C2e+βA2

    holds if and only if α(16sin219)/70.3327 and β1/2.

    Proposition 1.8. The double inequality

    (1α)C2e+αA2<M2tanh<(1β)C2e+βA2

    holds if and only if α0 and β(16tanh219)/70.0401.

    In essence, the above results are how the two new means Msin and Mtanh are expressed linearly, harmoniously, squarely, and harmoniously in square by the two classical means Ce(x,y) and A(x,y). In this paper, we study the following two-sided inequalities in exponential form for nonzero number pR

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep, (1.7)
    (1λp)Ap+λpCep<Mptanh<(1μp)Ap+μpCep (1.8)

    in order to reach a broader conclusion including all the above properties. The main conclusions of this paper are as follows:

    Theorem 1.1. Let x,y>0, xy, p0 and

    p=3cos2+sin2+13sin2cos234.588.

    Then the following are considered.

    (i) If pp, the double inequality

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep (1.9)

    holds if and only if αp3p(1sinp1)/[(sinp1)(4p3p)] and βp1/2.

    (ii) If 0<p12/5, the double inequality

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep (1.10)

    holds if and only if αp1/2 and βp3p(1sinp1)/[(sinp1)(4p3p)].

    (iii) If p<0, the double inequality

    (1βp)Ap+βpCep<Mpsin<(1αp)Ap+αpCep (1.11)

    holds if and only if αp1/2 and βp3p(1sinp1)/[(sinp1)(4p3p)].

    Theorem 1.2. Let x,y>0, xy, p0 and

    p=16cosh23cosh4+4sinh2+3cosh412sinh2+153.4776.

    Then the following are considered:

    (i) If p>0, the double inequality

    (1λp)Ap+λpCep<Mptanh<(1μp)Ap+μpCep (1.12)

    holds if and only if λp((coth1)p1)/((4/3)p1) and μp1.

    (ii) If pp<0,

    (1μp)Ap+μpλpCep<Mptanh<(1λp)Ap+λpCep (1.13)

    holds if and only if λp((coth1)p1)/((4/3)p1) and μp1.

    We first introduce a theoretical support of this paper.

    Lemma 2.1. ([10]) Let K(x,y),R(x,y), and N(x,y) be three means with two positive distinct parameters x and y; fK(t), fR(t), and fN(t) be the corresponding Seiffert functions of the former, ϑ1,ϑ2,θ1,θ2,pR, and p0. Then

    ϑ1Kp(x,y)+ϑ2Np(x,y)Rp(x,y)θ1Kp(x,y)+θ2Np(x,y) (2.1)
    ϑ1fpK(t)+ϑ2fpN(t)1fpR(t)θ1fpK(t)+θ2fpN(t). (2.2)

    It must be mentioned that the key steps to prove the above results are following:

    Mf(u,v)=Mf(λ2xx+y,λ2yx+y)=λMf(2xx+y,2yx+y)=λMf(1t,1+t)=λtfM(t), (2.3)

    where

    {u=λ2xx+yv=λ2yx+y, 0<x<y, λ>0.

    and 0<t<1,

    t=yxx+y.

    In order to prove the main conclusions, we shall introduce some very suitable methods which are called the monotone form of L'Hospital's rule (see Lemma 2.2) and the criterion for the monotonicity of the quotient of power series (see Lemma 2.3).

    Lemma 2.2. ([11,12]) For <a<b<, let f,g:[a,b]R be continuous functions that are differentiable on (a,b), with f(a)=g(a)=0 or f(b)=g(b)=0. Assume that g(t)0 for each x in (a,b). If f/g is increasing (decreasing) on (a,b), then so is f/g.

    Lemma 2.3. ([13]) Let an and bn (n=0,1,2,) be real numbers, and let the power series A(x)=n=0anxn and B(x)=n=0bnxn be convergent for |x|<R (R+). If bn>0 for n=0,1,2,, and if εn=an/bn is strictly increasing (or decreasing) for n=0,1,2,, then the function A(x)/B(x) is strictly increasing (or decreasing) on (0,R) (R+).

    Lemma 2.4. ([14,15]) Let B2n be the even-indexed Bernoulli numbers. Then we have the following power series expansions

    cotx=1xn=122n(2n)!|B2n|x2n1, 0<|x|<π, (2.4)
    1sin2x=csc2x=(cotx)=1x2+n=122n(2n1)(2n)!|B2n|x2n2, 0<|x|<π.  (2.5)

    Lemma 2.5. ([16,17,18,19,20]) Let B2n the even-indexed Bernoulli numbers, n=1,2,. Then

    22n1122n+11(2n+2)(2n+1)π2<|B2n+2||B2n|<22n122n+21(2n+2)(2n+1)π2.

    Lemma 2.6. Let l1(t) be defined by

    l1(t)=s1(t)r1(t),

    where

    s1(t)=6t2+2t412sin2t2t3costsint+6tcostsintsin2t,r1(t)=8t2sin2t+2t4sin2t6t22t46sin2t+12tcostsintsin2t.

    Then the double inequality

    125<l1(t)<p=3cos2+sin2+13sin2cos234.588 (2.6)

    holds for all t(0,1), where the constants 12/5 and (3cos2+sin2+1)/(3sin2cos23)4.588 are the best possible in (2.6).

    Proof. Since

    1l1(t)=r1(t)s1(t),

    and

    r1(t)=8t2sin2t+2t4sin2t6t22t46sin2t+12tcostsintsin2t=8t22t41sin2t6t21sin2t+2t4+12tcostsint6=8t22t4[1t2+n=122n(2n1)(2n)!|B2n|t2n2]6t2[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+2t4+12t[1tn=122n(2n)!|B2n|t2n1]6=23t4n=3[22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|]t2n=:n=2ant2n,

    where

    a2=23,an=[22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|], n=3,4,,
    s1(t)=6t2+2t412sin2t2t3costsint+6tcostsintsin2t=6t21sin2t+2t41sin2t+6tcostsint2t3costsint12=6t2[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+2t4[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+6t[1tn=122n(2n)!|B2n|t2n1]2t3[1tn=122n(2n)!|B2n|t2n1]12=n=21222n(n1)(2n)!|B2n|t2n+n=14n22n(2n)!|B2n|t2n+2=n=21222n(n1)(2n)!|B2n|t2n+n=2(n1)22n(2n2)!|B2n2|t2n=n=2[1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|]t2n=85t4+n=3[1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|]t2n=:n=2bnt2n,

    where

    b2=85,bn=1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|>0, n=3,4,.

    Setting

    qn=anbn, n=2,3,,

    we have

    q2=512=0.41667,qn=22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|, n=3,4,.

    Here we prove that the sequence {qn}n2 decreases monotonously. Obviously, q2>0>q3. We shall prove that for n3,

    qn>qn+122n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|>22n+1(2n1)(2n)!|B2n|+622n+2(2n+3)(2n+2)!|B2n+2|1222n+2n(2n+2)!|B2n+2|+n22n+2(2n)!|B2n|22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|<22n+1(2n1)(2n)!|B2n|+622n+2(2n+3)(2n+2)!|B2n+2|1222n+2n(2n+2)!|B2n+2|+n22n+2(2n)!|B2n|,

    that is,

    2(2n)!(2n2)!|B2n2||B2n|+24(4n3)(2n2)!(2n+2)!|B2n+2||B2n2||B2n||B2n|>24(4n1)((2n)!)2+864(2n)!(2n+2)!|B2n+2||B2n|. (2.7)

    By Lemma 2.5 we have

    2(2n)!(2n2)!|B2n2||B2n|+24(4n3)(2n2)!(2n+2)!|B2n+2||B2n2||B2n||B2n|>2(2n)!(2n2)!22n122n21π2(2n)(2n1)+24(4n3)(2n2)!(2n+2)!22n1122n+11(2n+2)(2n+1)π222n122n21π2(2n)(2n1)=2π2(2n)!(2n)!22n122n21+24(4n3)(2n)!(2n)!22n1122n+1122n122n21,

    and

    24(4n1)(2n)!2+864(2n)!(2n+2)!|B2n+2||B2n|<24(4n1)(2n)!2+864(2n)!(2n+2)!22n122n+21(2n+2)(2n+1)π2=24(4n1)(2n)!2+864(2n)!(2n)!22n122n+211π2.

    So we can complete the prove (2.7) when proving

    2π2(2n)!(2n)!22n122n21+24(4n3)(2n)!(2n)!22n1122n+1122n122n21>24(4n1)((2n)!)2+864(2n)!(2n)!22n122n+211π2

    or

    2π2(22n1)22n21+24(4n3)22n1122n+1122n122n21>24(4n1)+22n122n+21864π2.

    In fact,

    2π2(22n1)22n21+24(4n3)22n1122n+1122n122n21[24(4n1)+22n122n+21864π2]=:8H(n)π2(22n+21)(22n4)(22n+11),

    where

    H(n)=826n(π+3)(π3)(π2+3)+224n(72π2n+60π27π4+594)22n(36π2n+123π27π4+1404)+(24π2π4+432)>0

    for all n3.

    So the sequence {qn}n2 decreases monotonously. By Lemma 2.3 we obtain that r1(t)/s1(t) is decreasing on (0,1), which means that the function l1(t) is increasing on (0,1). In view of

    lim

    the proof of this lemma is complete.

    Lemma 2.7. Let l_{2}(t) be defined by

    \begin{equation*} l_{2}\left( t\right) = 2\cdot \frac{ 3\cosh 4t-12t^{2}\cosh 2t-4t^{4}\cosh 2t+2 t^{3}\sinh 2t-6t\sinh 2t-3}{ t^{2}\cosh 4t-3\cosh 4t+24t\sinh 2t-25t^{2}- 8t^{4}+3} = :2\frac{ B(t)}{A(t)}, \text{ }0 < t < \infty , \end{equation*}

    where

    \begin{eqnarray*} A(t) & = & t^{2}\cosh 4t-3\cosh 4t+24t\sinh 2t-25t^{2}- 8t^{4}+3 , \\ B(t) & = &3\cosh 4t-12t^{2}\cosh 2t-4t^{4}\cosh 2t+2 t^{3}\sinh 2t-6t\sinh 2t-3. \end{eqnarray*}

    Then l_{2}(t) is strictly decreasing on \left(0, \infty \right) .

    Proof. Let's take the power series expansions

    \begin{equation*} \sinh kt = \sum\limits_{n = 0}^{\infty }\frac{k^{2n+1}}{\left( 2n+1\right) !}t^{2n+1}, \text{ }\cosh kt = \sum\limits_{n = 0}^{\infty }\frac{k^{2n}}{\left( 2n\right) !}t^{2n} \end{equation*}

    into A(t) and B(t) , and get

    \begin{equation*} A(t) = \sum\limits_{n = 2}^{\infty }c_{n}t^{2n+2}, \text{ }B(t) = \sum\limits_{n = 2}^{\infty }d_{n}t^{2n+2}, \end{equation*}

    where

    \begin{eqnarray*} c_{2} & = &0, \\ c_{n} & = &\left[ \frac{ 2\left( 3n+2n^{2}-23\right) 2^{2n}+48\left( 2n+2\right) }{\left( 2n+2\right) !}\right] 2^{2n}, \text{ } n = 3, 4, \ldots , \\ d_{n} & = &\left[ \frac{48\cdot 2^{2n} -8\left( n+1\right) \left( 5n-n^{2}+2n^{3}+6\right) }{\left( 2n+2\right) !}\right] 2^{2n}, \text{ } n = 2, 3, \ldots , \end{eqnarray*}

    Setting

    \begin{equation*} k_{n} = \frac{c_{n}}{d_{n}} = \frac{ 48\left( n+1\right) + 2^{2n}\left( 3n+2n^{2}-23\right) }{ 4\left( 6\cdot 2^{2n}-11n-4n^{2}-n^{3}-2n^{4}-6\right) }, \text{ }n = 2, 3, \ldots , \end{equation*}

    Here we prove that the sequence \{k_{n}\}_{n\geq 2} decreases monotonously. Obviously, k_{2} = 0 < k_{3} . For n\geq 3,

    \begin{eqnarray*} k_{n} & < &k_{n+1} \\ &\Longleftrightarrow & \\ &&\frac{ 48\left( n+1\right) + 2^{2n}\left( 3n+2n^{2}-23\right) }{ 4\left( 6\cdot 2^{2n}-11n-4n^{2}-n^{3}-2n^{4}-6\right) } \\ & < &\frac{ 48\left( n+2\right) + 2^{2n+2}\left( 3\left( n+1\right) +2\left( n+1\right) ^{2}-23\right) }{ 4\left( 6\cdot 2^{2n+2}-11\left( n+1\right) -4\left( n+1\right) ^{2}-\left( n+1\right) ^{3}-2\left( n+1\right) ^{4}-6\right) } \\ &\Longleftrightarrow & \\ &&\frac{ 48\left( n+1\right) + 2^{2n}\left( 3n+2n^{2}-23\right) }{ 6\cdot 2^{2n}-11n-4n^{2}-n^{3}-2n^{4}-6} \\ & < & \frac{48n+96+ 2^{2n+2}\left( 7n+2n^{2}-18\right) }{ 6\cdot 2^{2n+2}-30n-19n^{2}-9n^{3}-2n^{4}-24} \end{eqnarray*}

    follows from \Delta (n) > 0 for all n\geq 2 , where

    \begin{eqnarray*} \Delta (n) & = &\left( 48n+96+ 2^{2n+2}\left( 7n+2n^{2}-18\right) \right) \left( 6\cdot 2^{2n}-11n-4n^{2}-n^{3}-2n^{4}-6\right) \\ &&-\left( 48\left( n+1\right) + 2^{2n}\left( 3n+2n^{2}-23\right) \right) \left( 6\cdot 2^{2n+2}-30n-19n^{2}-9n^{3}-2n^{4}-24\right) \\ & = & 24\cdot 2^{4n}\left( 4n+5\right) -2^{2n}\left( 858n+367n^{2}+218n^{3}-103n^{4}+40n^{5}+12n^{6}+696\right) \\ &&+1248n+1440n^{2}+1056n^{3}+288 n^{4}+576 \\ & = :&2^{2n}\left[ j(n)2^{2n}- i(n)\right] +w(n) \end{eqnarray*}

    with

    \begin{eqnarray*} j(n) & = & 24\left( 4n+5\right) , \\ i(n) & = &858n+367n^{2}+218n^{3}-103n^{4}+40n^{5}+12n^{6}+696, \\ w(n) & = &1248n+1440n^{2}+1056n^{3}+288 n^{4}+576 > 0. \end{eqnarray*}

    We have that \Delta (2) = 5376 > 0 and shall prove that

    \begin{eqnarray} j(n)2^{2n}- i(n) & > &0\Longleftrightarrow \\ 2^{2n} & > &\frac{ i(n)}{j(n)} \end{eqnarray} (2.8)

    holds for all n\geq 3 . Now we use mathematical induction to prove (2.8) . When n = 3 , the left-hand side and right-hand side of (2.8) are 2^{6} = 64 and i(3)/j(3) = 941/17\thickapprox 55.\, 353 , which implies (2.8) holds for n = 3 . Assuming that (2.8) holds for n = m , that is,

    \begin{equation} 2^{2m} > \frac{i(m)}{j(m)}. \end{equation} (2.9)

    Next, we prove that (2.8) is valid for n = m+1 . By (2.9) we have

    \begin{equation*} 2^{2\left( m+1\right) } = 4\cdot 2^{2m} > 4\frac{i(m)}{j(m)}, \end{equation*}

    in order to complete the proof of (2.8) it suffices to show that

    \begin{equation*} 4\frac{i(m)}{j(m)} > \frac{i(m+1)}{j(m+1)}\Longleftrightarrow 4i(m)j(m+1)-i(m+1)j(m) > 0. \end{equation*}

    In fact,

    \begin{eqnarray*} &&4i(m)j(m+1)-i(m+1)j(m) \\ & = & 17\, 280m^{7}+90\, 720m^{6}-60\, 000m^{5}-97\, 176m^{4}+ 1169\, 232m^{3}+2266\, 104m^{2} \\ &&+3581\, 136m+2154\, 816 \\ & = &146\, 337\, 408+234\, 401\, 616\left( m-3\right) +189\, 746\, 328 \left( m-3\right) ^{2}+92\, 580\, 720\left( m-3\right) ^{3} \\ &&+27\, 579\, 624\left( m-3\right) ^{4}+ 4838\, 880\left( m-3\right) ^{5}+453\, 600\left( m-3\right) ^{6}+17\, 280 \left( m-3\right) ^{7} \\ & > &0 \end{eqnarray*}

    for m\geq 3 due to the coefficients of the power square of \left(m-1\right) are positive.

    By Lemma 2.3 we get that A(t)/B(t) is strictly increasing on \left(0, \infty \right). So the function l_{2}(x) is strictly decreasing on \left(0, \infty \right) .

    The proof of Lemma 2.7 is complete.

    Via (1.3) and (1.2) we can obtain

    \begin{eqnarray*} f_{\mathbf{A}}(t) & = &t, \\ f_{\mathbf{C}_{e}}(t) & = &\frac{3t}{3+t^{2}}, \\ f_{\mathbf{M}_{\sin }}(t) & = &\sin t, \\ f_{\mathbf{M}_{\tanh }}(t) & = &\tanh t. \end{eqnarray*}

    Then by Lemma 2.1 and \left(2.3\right) we have

    \begin{eqnarray*} \alpha _{p} & < &\frac{\mathbf{M}_{\sin }^{p}-\mathbf{A}^{p}}{\mathbf{Ce}^{p}- \mathbf{A}^{p}} < \beta _{p}\Longleftrightarrow \alpha _{p} < \frac{\left( \frac{ 1}{\sin t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}}{\left( \frac{3+t^{2}}{ 3t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}} < \beta _{p}, \\ \lambda _{p} & < &\frac{\mathbf{M}_{\tanh }^{p}-\mathbf{A}^{p}}{\mathbf{Ce} ^{p}-\mathbf{A}^{p}} < \mu _{p}\Longleftrightarrow \lambda _{p} < \frac{\left( \frac{1}{\tanh t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}}{\left( \frac{ 3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}} < \mu _{p}. \end{eqnarray*}

    So we turn to the proof of the following two theorems.

    Theorem 3.1. Let t\in (0, 1) and

    \begin{equation*} p^{\clubsuit } = \frac{3\cos 2+\sin 2+1}{3\sin 2-\cos 2-3}\thickapprox 4.\, 588. \end{equation*}

    Then,

    (i) if p\geq p^{\clubsuit } , the double inequality

    \begin{equation} \alpha _{p} < \frac{\left( \frac{1}{\sin t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}}{\left( \frac{3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}} < \beta _{p} \end{equation} (3.1)

    holds if and only if \alpha _{p}\leq 3^{p}\left(1-\sin ^{p}1\right) /\left[\left(\sin ^{p}1\right) \left(4^{p}-3^{p}\right) \right] and \beta _{p}\geq 1/2 ;

    (ii) if 0\neq p\leq 12/5 = 2.\, 4 and p\neq 0 the double inequality

    \begin{equation} \beta _{p} < \frac{\left( \frac{1}{\sin t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}}{\left( \frac{3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}} < \alpha _{p} \end{equation} (3.2)

    holds if and only if \alpha _{p}\leq 1/2 and \beta \geq 3^{p}\left(1-\sin ^{p}1\right) /\left[\left(\sin ^{p}1\right) \left(4^{p}-3^{p}\right) \right].

    Theorem 3.2. Let t\in (0, 1) and

    \begin{equation*} p^{\ast } = -\frac{16\cosh 2-3\cosh 4+4\sinh 2+3}{\cosh 4-12\sinh 2+15} \thickapprox - 3.\, 477\, 6. \end{equation*}

    If 0\neq p\geq - 3.\, 477\, 6 , the double inequality

    \begin{equation} \lambda _{p} < \frac{\left( \frac{1}{\tanh t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}}{\left( \frac{3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t} \right) ^{p}} < \mu _{p} \end{equation} (3.3)

    holds if and only if \lambda _{p}\leq \left(\left(\coth 1\right) ^{p}-1\right) /\left(\left(4/3\right) ^{p}-1\right) and \mu _{p}\geq 1 .

    Let

    \begin{eqnarray*} F(t) & = &\frac{\left( \frac{1}{\sin t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}}{\left( \frac{3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}} = \frac{\left( \frac{t}{\sin t}\right) ^{p}-1}{\left( \frac{3+t^{2}}{3}\right) ^{p}-1} \\ & = :&\frac{f(t)}{g(t)} = \frac{f(t)-f(0^{+})}{g(t)-g(0^{+})}, \end{eqnarray*}

    where

    \begin{eqnarray*} f(t) & = &\left( \frac{t}{\sin t}\right) ^{p}-1, \\ g(t) & = &\left( \frac{3+t^{2}}{3}\right) ^{p}-1. \end{eqnarray*}

    Then

    \begin{eqnarray*} f^{\prime }(t) & = & \frac{p}{\sin ^{2}t}\left( \sin t-t\cos t\right) \left( \frac{t}{\sin t}\right) ^{p-1}, \\ g^{\prime }(t) & = & \frac{2}{3}\left( \frac{1}{3}\right) ^{p-1}pt\left( t^{2}+3\right) ^{p-1}, \end{eqnarray*}
    \begin{equation*} \frac{f^{\prime }(t)}{g^{\prime }(t)} = \frac{3^{p}}{2}\frac{1}{ t\sin ^{2}t}\left( \sin t-t\cos t\right) \left( \frac{t}{\left( t^{2}+3\right) \sin t}\right) ^{p-1} , \end{equation*}

    and

    \begin{eqnarray*} \left( \frac{f^{\prime }(t)}{g^{\prime }(t)}\right) ^{\prime } & = &\frac{1}{4} \left( \frac{3t}{\left( \sin t\right) \left( t^{2}+3\right) }\right) ^{p} \frac{r_{1}(t)}{t^{3}\sin ^{2}t}\left[ \frac{s_{1}(t)}{r_{1}(t)}-p\right] \\ & = :&\frac{1}{4}\left( \frac{3t}{\left( \sin t\right) \left( t^{2}+3\right) } \right) ^{p}\frac{r_{1}(t)}{t^{3}\sin ^{2}t}\left[ l_{1}(t)-p\right] , \end{eqnarray*}

    where the three functions s_{1}(t) , r_{1}(t) , and l_{1}(t) are shown in Lemma 2.6 .

    By Lemma 2.6 we can obtain the following results:

    (a) When p\geq \max_{t\in (0, 1)}l_{1}(t) = :p^{\clubsuit } = \left(3\cos 2+\sin 2+1\right) /\left(3\sin 2-\cos 2-3\right) \thickapprox 4.\, 588 ,

    \begin{equation*} \left( \frac{f^{\prime }(t)}{g^{\prime }(t)}\right) ^{\prime }\leq 0\Longrightarrow \frac{f^{\prime }(t)}{g^{\prime }(t)}~\text{ is decreasing on }~(0, 1)\text{, } \end{equation*}

    this leads to F(t) = f(t)/g(t) is decreasing on (0, 1) by Lemma 2.1 . In view of

    \begin{equation} F(0^{+}) = \frac{1}{2}, \text{ }F(1^{-}) = \frac{3^{p}\left( 1-\sin ^{p}1\right) }{\left( \sin ^{p}1\right) \left( 4^{p}-3^{p}\right) }, \end{equation} (3.4)

    we have that \left(3.1\right) holds.

    (b) When 0\neq p\leq 12/5 = \min_{t\in (0, 1)}l_{1}(t),

    \begin{equation*} \left( \frac{f^{\prime }(t)}{g^{\prime }(t)}\right) ^{\prime }\geq 0\Longrightarrow \frac{f^{\prime }(t)}{g^{\prime }(t)}\text{ is increasing on }(0, 1)\text{, } \end{equation*}

    this leads to F(t) = f(t)/g(t) is increasing on (0, 1) by Lemma 2.2 . In view of \left(3.4\right) we have that \left(3.2\right) holds.

    The proof of Theorem 3.1 is complete.

    Let

    \begin{eqnarray*} G(t) & = &\frac{\left( \frac{1}{\tanh t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}}{\left( \frac{3+t^{2}}{3t}\right) ^{p}-\left( \frac{1}{t}\right) ^{p}} = \frac{\left( \frac{t}{\tanh t}\right) ^{p}-1}{\left( \frac{3+t^{2}}{3} \right) ^{p}-1} \\ & = :&\frac{u(t)}{v(t)} = \frac{u(t)-u(0^{+})}{v(t)-v(0^{+})}. \end{eqnarray*}

    Then

    \begin{eqnarray*} u^{\prime }(t) & = & \frac{p}{\tanh ^{2}t}\left( \frac{t}{\tanh t} \right) ^{p-1}\left( t\tanh ^{2}t+\tanh t-t\right) , \\ v^{\prime }(t) & = & \frac{2}{3}pt\left( \frac{t^{2}+3}{3}\right) ^{p-1}, \end{eqnarray*}
    \begin{equation*} \frac{u^{\prime }(t)}{v^{\prime }(t)} = \frac{3}{2}\frac{t\tanh ^{2}t+\tanh t-t }{t\tanh ^{2}t}\left[ \frac{3t}{\left( t^{2}+3\right) \tanh t}\right] ^{p-1} , \end{equation*}

    and

    \begin{eqnarray*} \left( \frac{u^{\prime }(t)}{v^{\prime }(t)}\right) ^{\prime } & = &-\frac{1}{ 16}\left[ \frac{3t\cosh t}{\left( 3+t^{2}\right) \sinh t}\right] ^{p}\frac{ A(t)}{t^{3}\cosh ^{2}t\sinh ^{2}t}\left[ p+\frac{2B(t)}{A(t)}\right] \\ & = :&-\frac{1}{16}\left[ \frac{3t\cosh t}{\left( 3+t^{2}\right) \sinh t} \right] ^{p}\frac{A(t)}{t^{3}\cosh ^{2}t\sinh ^{2}t}\left[ p+l_{2}\left( t\right) \right] , \end{eqnarray*}

    where the three functions A(t) , B(t) , and l_{2}(t) are shown in Lemma 2.7 . By Lemma 2.7 we see that l_{2}(x) is strictly decreasing on \left(0, 1\right) . Since

    \begin{eqnarray*} \lim\limits_{t\rightarrow 0^{+}}l_{2}(t) & = &\infty , \\ \lim\limits_{t\rightarrow 1^{-}}l_{2}(t) & = &\frac{16\cosh 2-3\cosh 4+4\sinh 2+3}{ \cosh 4-12\sinh 2+15} = :p_{\#}\thickapprox 3.\, 477\, 6, \end{eqnarray*}

    we obtain the following result:

    When p\geq \max_{t\in \left(0, 1\right) }\left\{ -l_{2}(t)\right\} = -p_{\#} = :p^{\ast }\thickapprox -3.\, 477\, 6,

    \begin{equation*} \left( \frac{u^{\prime }(t)}{v^{\prime }(t)}\right) ^{\prime }\leq 0\Longrightarrow \frac{u^{\prime }(t)}{v^{\prime }(t)}~\text{ is decreasing on }~(0, 1)\text{, } \end{equation*}

    this leads to G(t) = u(t)/v(t) is decreasing on (0, 1) by Lemma 2.2 . Since

    \begin{equation*} G(0^{+}) = 1, G(1^{-}) = \frac{\left( \frac{\cosh 1}{\sinh 1}\right) ^{p}-1}{ \left( \frac{4}{3}\right) ^{p}-1}, \end{equation*}

    we have

    \begin{equation*} G(1^{-}) < G(t) < G(0^{+}), \end{equation*}

    which completes the proof of Theorem 3.2.

    Remark 4.1. Letting p = 1, -1, 2, -2 in Theorems 1.1 and 1.2 respectively, one can obtain Propositions 1.1–1.8.

    From Theorems 1.1 and 1.2, we can also get the following important conclusions:

    Corollary 4.1. Let x, y > 0 , x\neq y , and

    \begin{eqnarray*} p^{\clubsuit } & = &\frac{3\cos 2+\sin 2+1}{3\sin 2-\cos 2-3}\thickapprox 4.\, 588, \\ \alpha & = &\frac{3^{p^{\clubsuit }}\left( 1-\sin ^{p^{\clubsuit }}1\right) }{ \left( \sin ^{p^{\clubsuit }}1\right) \left( 4^{p^{\clubsuit }}-3^{p^{\clubsuit }}\right) }\thickapprox 0.440\, 25, \\ \beta & = &\frac{1}{2}. \end{eqnarray*}

    Then the double inequality

    \begin{equation} (1-\alpha )\mathbf{A}^{p^{\clubsuit }}+\alpha \mathbf{Ce}^{p^{\clubsuit }} < \mathbf{M}_{\sin }^{p^{\clubsuit }} < (1-\beta )\mathbf{A}^{p^{\clubsuit }}+\beta \mathbf{Ce}^{p^{\clubsuit }} \end{equation} (4.1)

    holds, where the constants \alpha and \beta are the best possible in (4.1).

    Corollary 4.2. Let x, y > 0 , x\neq y , and

    \begin{eqnarray*} \theta & = &\frac{1}{2}, \\ \vartheta & = &\frac{3^{12/5}\left( 1-\sin ^{12/5}1\right) }{\left( \sin ^{12/5}1\right) \left( 4^{12/5}-3^{12/5}\right) }\thickapprox 0.516\, 03. \end{eqnarray*}

    Then the double inequality

    \begin{equation} (1-\theta )\mathbf{A}^{12/5}+\theta \mathbf{Ce}^{12/5} < \mathbf{M}_{\sin }^{12/5} < (1-\vartheta )\mathbf{A}^{12/5}+\vartheta \mathbf{Ce}^{12/5} \end{equation} (4.2)

    holds, where the constants \theta and \vartheta are the best possible in (4.2).

    Corollary 4.3. Let x, y > 0 , x\neq y , and

    \begin{eqnarray*} p^{\ast } & = &-\frac{16\cosh 2-3\cosh 4+4\sinh 2+3}{\cosh 4-12\sinh 2+15} \thickapprox - 3.\, 477\, 6, \\ \lambda & = &\frac{\left( \coth 1\right) ^{p^{\ast }}-1}{\left( 4/3\right) ^{p^{\ast }}-1}\thickapprox 0.968\, 13, \\ \mu & = &1. \end{eqnarray*}

    Then the double inequality

    \begin{equation} (1-\mu)\mathbf{A}^{p^{\ast }}+\mu\mathbf{Ce}^{p^{\ast }} < \mathbf{M}_{\tanh }^{p^{\ast }} < (1-\lambda)\mathbf{A}^{p^{\ast }}+\lambda\mathbf{Ce}^{p^{\ast }} \end{equation} (4.3)

    holds, where the constants \lambda and \mu are the best possible in (4.3).

    In this paper, we have studied exponential type inequalities for \mathbf{M} _{\sin } and \mathbf{M}_{\tanh } in term of \mathbf{A} and \mathbf{Ce} for nonzero number p\in \mathbb{R} :

    \begin{eqnarray*} (1-\alpha _{p})\mathbf{A}^{p}+\alpha _{p}\mathbf{Ce}^{p} & < &\mathbf{M}_{\sin }^{p} < (1-\beta _{p})\mathbf{A}^{p}+\beta _{p}\mathbf{Ce}^{p}, \\ (1-\lambda _{p})\mathbf{A}^{p}+\lambda _{p}\mathbf{Ce}^{p} & < &\mathbf{M} _{\tanh }^{p} < (1-\mu _{p})\mathbf{A}^{p}+\mu _{p}\mathbf{Ce}^{p}, \end{eqnarray*}

    obtained a lot of interesting conclusions which include the ones of the previous similar literature. In fact, we can consider similar inequalities for dual means of the two means \mathbf{M}_{\sin } and \mathbf{M}_{\tanh } , and we can replace \mathbf{A} and \mathbf{Ce} by other famous means. Therefore, the content of this research is very extensive.

    The authors are grateful to editor and anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

    The first author was supported by the National Natural Science Foundation of China (no. 61772025). The second author was supported in part by the Serbian Ministry of Education, Science and Technological Development, under projects ON 174032 and III 44006.

    The authors declare that they have no conflict of interest.



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