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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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+tanxx−2<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+tanhxx−2>845x3tanhx, x≠0, | (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+xtanhx−2<245x3sinhx, x≠0, | (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 x≠0. Then
(sinhxx)2+tanhxx−2>αx4(tanhxx)6/7 | (1.6) |
holds if and only if α≤8/45.
Theorem 2. The function
G(x)=(xsinhx)2+xtanhx−2x3tanhx |
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 x≠0,
(xsinhx)2+xtanhx−2<θx3tanhx | (1.7) |
holds if and only if θ≥θ0.
Theorem 3. Let x≠0. Then
(xsinhx)2+xtanhx−2<βx4(tanhxx)4/7 | (1.8) |
holds if and only if β≥2/45.
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 g′≠0 on (a,b). Then the function Hf,g is defined by
Hf,g:=f′g′g−f. |
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 m∈N, the non-constant sequence {ak/bk} is increasing (resp. decreasing) for 0≤k≤m and decreasing (resp. increasing) for k≥m. 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+tanhxx−2>845x4(tanhxx)6/7 |
⇔[(sinhxx)2+tanhxx−2]7>[845x4(tanhxx)6/7]7= 2097152373 669453125x22tanh6x, |
we can let
F(x)=7ln[(sinhxx)2+tanhxx−2]−ln[2097152373 669453125x22tanh6x], x>0. |
Then
F′(x)= h(x)8x3(tanhx)(cosh3x)[(sinhxx)2+tanhxx−2] , |
where
h(x)=18sinh3x−18sinh5x+36sinhx+88x2sinh3x+56xcoshx−63xcosh3x+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)=18∞∑n=0(3x)2n+1(2n+1)!−18∞∑n=0(5x)2n+1(2n+1)!+36∞∑n=0x2n+1(2n+1)!+88x2∞∑n=0(3x)2n+1(2n+1)!+56x∞∑n=0x2n(2n)!−63x∞∑n=0(3x)2n(2n)!+7x∞∑n=0(5x)2n(2n)!+96x3∞∑n=0x2n(2n)!+96x2∞∑n=0x2n+1(2n+1)! |
=18∞∑n=532n+1(2n+1)!x2n+1−18∞∑n=552n+1(2n+1)!x2n+1+36∞∑n=51(2n+1)!x2n+1+88∞∑n=432n+1(2n+1)!x2n+3+56∞∑n=51(2n)!x2n+1−63∞∑n=532n(2n)!x2n+1+7∞∑n=552n(2n)!x2n+1+96∞∑n=41(2n)!x2n+3+96∞∑n=41(2n+1)!x2n+3 |
=∞∑n=5 18(32n+1−52n+1+2)+(2n+1)(56−63⋅32n+7⋅52n)(2n+1)!x2n+1+∞∑n=4 24(11⋅32n+4)+96(2n+1)(2n+1)!x2n+3=∞∑n=5 18(32n+1−52n+1+2)+(2n+1)(56−63⋅32n+7⋅52n)(2n+1)!x2n+1+∞∑n=5 24(11⋅32n−2+4)+96(2n−1)(2n−1)!x2n+1=∞∑n=5l(n)3(2n+1)!x2n+1, |
where
l(n)= 3(14n−83)52n+(352n2−202n−27)32n+(2304n3+1152n2+336n+276). |
Since l(5)=77856768>0 and for n≥6,
14n−83≥14⋅6−83= 1>0,352n2−202n−27≥352⋅62−202⋅6−27= 11433>0,2304n3+1152n2+336n+276>0, |
we have that l(n)>0 for n≥5. 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
limx→0+(sinhxx)2+tanhxx−2x4(tanhxx)6/7=845. |
Then the proof of Theorem 1 is complete.
Let
G(x)=(xsinhx)2+xtanhx−2x3tanhx, 0<x<∞. |
Rewrite G(x) as
G(x)= coshxx3sinhx(x2sinh2x+xcoshxsinhx−2)= (x2+xcoshxsinhx−2sinh2x)(coshx)x3sinh3x=14(xsinh3x−2cosh3x+2coshx+xsinhx+4x2coshx) 14x3(sinh3x−3sinhx)= xsinh3x−2cosh3x+2coshx+xsinhx+4x2coshxx3(sinh3x−3sinhx):=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)!−2∞∑n=032nx2n(2n)!+2∞∑n=0x2n(2n)!+∞∑n=0x2n+2(2n+1)!+4∞∑n=0x2n+2(2n)!=∞∑n=232n+1x2n+2(2n+1)!−2∞∑n=332nx2n(2n)!+2∞∑n=3x2n(2n)!+∞∑n=2x2n+2(2n+1)!+4∞∑n=2x2n+2(2n)!=∞∑n=232n+1+1+4(2n+1)(2n+1)!x2n+2−∞∑n=32(32n−1)(2n)!x2n=∞∑n=232n+1+1+4(2n+1)(2n+1)!x2n+2−∞∑n=22(32n+2−1)(2n+2)!x2n+2=∞∑n=2(2n+2)(32n+1+1+4(2n+1))−2(32n+2−1)(2n+2)!x2n+2=∞∑n=2 2[(3n−6)32n+8n2+13n+6](2n+2)!x2n+2:=∞∑n=2anx2n+2, |
and
g(x)=x3(∞∑n=0(3x)2n+1(2n+1)!−3∞∑n=0x2n+1(2n+1)!)=∞∑n=0(32n+1−3)x2n+4(2n+1)!=∞∑n=1(32n+1−3)x2n+4(2n+1)!=∞∑n=2(32n−1−3)(2n−1)!x2n+2:=∞∑n=2bnx2n+1. |
Let
cn=anbn= 2[(3n−6)32n+8n2+13n+6](2n+2)!(32n−1−3)(2n−1)!= 2[(3n−6)32n+8n2+13n+6](2n+2)(2n+1)2n(32n−1−3). |
We can calculate that
c2=245<c3=235>c4=2064095, |
and show that {cn}n≥4 is decreasing:
cn≥cn+1⇔CD:= 2((3n−6)32n+8n2+13n+6)(2n+2)(2n+1)2n(32n−1−3)≥ 2((3n−3)32n+2+8(n+1)2+13(n+1)+6)(2n+4)(2n+3)(2n+2)(32n+1−3):=EF. |
In fact,
CF−DE= 23p(n) , |
where
p(n)=(108n2−189n−324)34n+(128n4+1104n3+952n2+1026n+648)32n−(144n3+612n2+837n+324). |
Since for n≥4, we have 108n2−189n−324≥648>0. By mathematical induction it is easy to prove that
32n>144n3+612n2+837n+324128n4+1104n3+952n2+1026n+648 |
holds for n≥4. So p(n)>0 for n≥4. This leads to {cn}n≥4 is decreasing. Therefore
c2<c3>c4>c5>⋯. |
We compute to get
Hf,g(∞)=limx→∞(f′g′g−f)= −∞, |
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(15cosh2x−6cosh4x+cosh6x−10) , |
where
r(x)=−10x−15sinh2x+12sinh4x−3sinh6x+8x3cosh2x+4x3cosh4x−12x2sinh2x+6x2sinh4x+15xcosh2x−6xcosh4x+xcosh6x−12x3. |
We find out that
G′(1.54471)=1. 1086×10−7, G′(1.54472)=−2. 2292×10−9. |
So x0=1.54471… and
G(x0)=maxx∈(0,∞)G(x)=0.050244…. |
Considering the reasons
limx→0+G(x)= 245=0.044444…, limx→∞G(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+xtanhx−2], |
where 0<x<∞. Then
H′(x)= q(x)8x( coshxsinh3x) [(xsinhx)2+xtanhx−2] , |
where
q(x)=15x+96sinh2x−48sinh4x+56x3cosh2x+84x2sinh2x−32xcosh2x+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+96∞∑n=022n+1x2n+1(2n+1)!−48∞∑n=042n+1x2n+1(2n+1)!+56x3∞∑n=022nx2n(2n)!+84x2∞∑n=022n+1x2n+1(2n+1)!−32x∞∑n=022nx2n(2n)!+17x∞∑n=042nx2n(2n)!+88x3=96∞∑n=522n+1x2n+1(2n+1)!−48∞∑n=542n+1x2n+1(2n+1)!−32∞∑n=522nx2n+1(2n)!+17∞∑n=542nx2n+1(2n)!+56∞∑n=422nx2n+3(2n)!+84∞∑n=422n+1x2n+3(2n+1)!+56∞∑n=422nx2n+3(2n)!+84∞∑n=422n+1x2n+3(2n+1)! |
=∞∑n=5(34n−175)42n−(64n−160)22n(2n+1)!x2n+1+∞∑n=4112(n+2)22n(2n+1)!x2n+3=∞∑n=5(34n−175)42n−(64n−160)22n(2n+1)!x2n+1+∞∑n=528(n+1)22n(2n−1)!x2n+1=∞∑n=5[(34n−175)42n−(64n−160)22n(2n+1)!+28(n+1)22n(2n−1)!]x2n+1=∞∑n=5(34n−175)42n−(64n−160)22n+28(2n+1)(2n)(n+1)22n(2n+1)!x2n+1=∞∑n=5(34n−175)42n+8(14n3+21n2−n+20)22n(2n+1)!x2n+1=∞∑n=5dn(2n+1)!x2n+1, |
where
dn=(34n−175)42n+814n3+21n2−n+20, n≥5. |
We can find d5= 13516800>0 and dn>0 holds for n≥6 due to
34n−175≥34×6−175= 29>0,14n3+21n2−n+20≥14⋅63+21⋅62−6+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
limx→0+(xsinhx)2+xtanhx−2x4(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.
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[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. |
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