Citation: Ling Zhu. New inequalities of Wilker’s type for circular functions[J]. AIMS Mathematics, 2020, 5(5): 4874-4888. doi: 10.3934/math.2020311
[1] | Ling Zhu . New inequalities of Wilker's type for hyperbolic functions. AIMS Mathematics, 2020, 5(1): 376-384. doi: 10.3934/math.2020025 |
[2] | Ling Zhu, Zhengjie Sun . Refinements of Huygens- and Wilker- type inequalities. AIMS Mathematics, 2020, 5(4): 2967-2978. doi: 10.3934/math.2020191 |
[3] | Dojin Kim, Patcharee Wongsason, Jongkyum Kwon . Type 2 degenerate modified poly-Bernoulli polynomials arising from the degenerate poly-exponential functions. AIMS Mathematics, 2022, 7(6): 9716-9730. doi: 10.3934/math.2022541 |
[4] | Xiaojun Hu, Qihan Wang, Boyong Long . Bohr-type inequalities for bounded analytic functions of Schwarz functions. AIMS Mathematics, 2021, 6(12): 13608-13621. doi: 10.3934/math.2021791 |
[5] | Muhammad Ghaffar Khan, Sheza.M. El-Deeb, Daniel Breaz, Wali Khan Mashwani, Bakhtiar Ahmad . Sufficiency criteria for a class of convex functions connected with tangent function. AIMS Mathematics, 2024, 9(7): 18608-18624. doi: 10.3934/math.2024906 |
[6] | Li Chen, Dmitry V. Dolgy, Taekyun Kim, Dae San Kim . Probabilistic type 2 Bernoulli and Euler polynomials. AIMS Mathematics, 2024, 9(6): 14312-14324. doi: 10.3934/math.2024696 |
[7] | Sarah Elahi, Muhammad Aslam Noor . Integral inequalities for hyperbolic type preinvex functions. AIMS Mathematics, 2021, 6(9): 10313-10326. doi: 10.3934/math.2021597 |
[8] | Wei-Mao Qian, Tie-Hong Zhao, Yu-Pei Lv . Refinements of bounds for the arithmetic mean by new Seiffert-like means. AIMS Mathematics, 2021, 6(8): 9036-9047. doi: 10.3934/math.2021524 |
[9] | Sabila Ali, Shahid Mubeen, Rana Safdar Ali, Gauhar Rahman, Ahmed Morsy, Kottakkaran Sooppy Nisar, Sunil Dutt Purohit, M. Zakarya . Dynamical significance of generalized fractional integral inequalities via convexity. AIMS Mathematics, 2021, 6(9): 9705-9730. doi: 10.3934/math.2021565 |
[10] | Muhammad Amer Latif . Fejér type inequalities for harmonically convex functions. AIMS Mathematics, 2022, 7(8): 15234-15257. doi: 10.3934/math.2022835 |
Inequalities are ubiquitous in all branches of pure and applied mathematics, while the trigonometric and hyperbolic functions inequalities are the indispensable parts in the whole theory of inequality, there are numerous researchers devoted to the study of mathematical inequalities. Recently, there are many novel inequalities in different areas have been discovered, for example, the Hermite-Hadamard type inequalities [1,2,3,4,5,6], bivariate means inequalities [7,8,9,10,11], gamma function inequalities [12], complete elliptic integrals inequalities [13,14,15], Bessel functions inequalities [16], Jensen type inequalities [17,18,19], Ostrowski type inequalities [20], Cauchy-Schwarz inequality [21], reverse Minkowski inequality [22], Petrović-type inequalities [23], Pólya-Szegö and Ćebyšev type inequalities [24], delay dynamic double integral inequalities [25], integral majorization type inequalities [26], generalized convex functions inequalities [27], generalized proportional fractional integral operators inequalities [28], generalized trigonometric and hyperbolic functions inequalities [29], exponentially convex inequalities [30] and so on.
In 1989, Montgomery et al. [31] proposed two open problems as follows:
(1) Does the inequality
(sinxx)2+tanxx>2 | (1.1) |
holds for all 0<x<π/2?
(2) Is there a largest constant c such that the inequality
(sinxx)2+tanxx>2+cx3tanx | (1.2) |
holds for all 0<x<π/2?
Wilker et al. [32] gave positive answers to the inequalities (1.1) and (1.2), and proved that the double inequality
16π4x3tanx<(sinxx)2+tanxx−2<845x3tanx | (1.3) |
holds for all x∈(0,π/2) with the best constants 16/π4 and 8/45.
For different proofs of inequalities (1.1) and (1.3), we recommend the literature [33,34] to the interested readers.
In [35], Zhu proved that the Wilker type inequality
(sinhxx)2+tanhxx−2>845x3tanhx | (1.4) |
for hyperbolic functions holds whenever x≠0, and the constant 8/45 in the right of inequality (1.4) can not be replaced by any larger number.
Sun and Zhu [36] provided a new version of the Wilker type inequality for hyperbolic functions as follows:
(xsinhx)2+xtanhx−2<245x3sinhx | (1.5) |
for all x≠0, and the constant 2/45 in the right of inequality (1.5) can not be replaced by any smaller number.
Very recently, Zhu [37] established the following novel results (Propositions 1.1–1.3) involving the Wilker type inequality for hyperbolic functions by use of the power series formulas and monotonicity criteria of the quotient of two power series [38].
Proposition 1.1. The inequality
(sinhxx)2+tanhxx−2>845x4(tanhxx)6/7 | (1.6) |
takes place for all x≠0 with the best possible constant 8/45 in the right of inequality (1.6).
Proposition 1.2. x0=1.54471⋯ is the unique point on the interval (0,∞) such that the function
G(x)=(xsinhx)2+xtanhx−2x3tanhx |
attain its maximum
θ0=G(x0)=maxx∈(0,∞)G(x)=0.050244⋯, |
and θ0 is the best possible constant such that the following inequality (1.7)
(xsinhx)2+xtanhx−2<θ0x3tanhx | (1.7) |
holds for all x∈(0,∞).
Proposition 1.3. 2/45 is the best possible constant such that the inequality (1.8)
(xsinhx)2+xtanhx−2<245x4(tanhxx)4/7 | (1.8) |
holds for all x≠0.
This main purpose of the article is to find new Wilker type inequalities for trigonometric functions. Our main results are the following Theorems 1.1–1.3.
Theorem 1.1. The inequality
(sinxx)2+tanxx−2>845x4(tanxx)6/7 | (1.9) |
is valid for all x∈(0,π/2) with the best possible constant 8/45 in the right of inequality (1.9).
Theorem 1.2. α=0 and β=2/45 are the best possible constants such that the two-sided inequality (1.10)
αx3tanx<(xsinx)2+xtanx−2<βx3tanx | (1.10) |
takes place for all x∈(0,π/2).
Theorem 1.3. Let 0<x<π/2. Then the double inequality
αx4(tanxx)4/7<(xsinx)2+xtanx−2<βx4(tanxx)4/7 | (1.11) |
holds with the best constants α=0 and β=2/45 in the left and right sides of inequality (1.11), respectively.
In order to prove our main results, we need four lemmas which we present in this section.
Lemma 2.1. Let |x|<π/2. Then we have the following power series formulas
tanx=∞∑n=122n(22n−1)(2n)!|B2n|x2n−1, | (2.1) |
tan2x=∞∑n=222n(22n−1)(2n−1)(n)|B2n|x2n−2, |
tan3x=∞∑n=322n(22n−1)(2n−1)(2n−2)2(2n)!|B2n|x2n−3 |
−∞∑n=222n(22n−1)(2n)!|B2n|x2n−1, |
tan4x=∞∑n=422n(22n−1)(2n−1)(2n−2)(2n−3)|B2n|6(2n)!x2n−4 |
−∞∑n=322n+2(22n−1)(2n−1)|B2n|3(2n)!x2n−2 |
and
sec2x tanx=∞∑n=222n(22n−1)(2n−1)(2n−2)2(2n)!|B2n|x2n−3, |
sec2xtan2x=∞∑n=222n(22n−1)(2n−1)(2n−2)(2n−3)6(2n)!|B2n|x2n−4 |
−∞∑n=122n(22n−1)(2n−1)3(2n)!|B2n|x2n−2, |
sec2xtan3x=∞∑n=222n(22n−1)(2n−1)(2n−2)(2n−3)(2n−4)|B2n|24(2n)!x2n−5 |
−∞∑n=222n+2(22n−1)(2n−1)(2n−2)12(2n)!|B2n|x2n−3, |
where Bn is the Bernoulli number.
Proof. The power series formula (2.1) can be found in the literature [39, 1.3.1.4(2)]. Next, we prove the remain power series formulas.
It follows from (2.1) that
tan2x=sec2x−1=(tanx)′−1 |
=∞∑n=222n(22n−1)(2n−1)(2n)!|B2n|x2n−2. | (2.2) |
Making use of (2.1) and (2.2) we get
tan3x=12[(tan2x)′−2tanx]=12(tan2x)′−tanx |
=∞∑n=222n(22n−1)(2n−1)(2n−2)2(2n)!|B2n|x2n−3 |
−∞∑n=122n(22n−1)(2n)!|B2n|x2n−1 |
=∞∑n=322n(22n−1)(2n−1)(2n−2)2(2n)!|B2n|x2n−3 |
−∞∑n=222n(22n−1)(2n)!|B2n|x2n−1 |
and
tan4x=13[(tan3x)′−3tan2x] |
=13(12(tan2x)′−tanx)′−tan2x |
=16(tan2x)′′−13(tanx)′−tan2x |
=∞∑n=222n(22n−1)(2n−1)(2n−2)(2n−3)6(2n)!|B2n|x2n−4 |
−∞∑n=1(22n−1)(2n−1)22n3(2n)!|B2n|x2n−2 |
−∞∑n=2(22n−1)(2n−1)22n(2n)!|B2n|x2n−2 |
=∞∑n=322n(22n−1)(2n−1)(2n−2)(2n−3)6(2n)!|B2n|x2n−4 |
−∞∑n=222n+2(22n−1)(2n−1)3(2n)!|B2n|x2n−2 |
=∞∑n=422n(22n−1)(2n−1)(2n−2)(2n−3)6(2n)!|B2n|x2n−4 |
−∞∑n=322n+2(22n−1)(2n−1)3(2n)!|B2n|x2n−2. |
Note that
sec2xtankx=1k+1ddxtank+1x,k≠−1. |
The desired power series formulas for the functions sec2xtanx, sec2xtan2x and sec2xtan3x can be derived easily.
Lemma 2.2. (See [40]) The power series formula
1sinx=1x+∞∑n=122n−2(2n)!|B2n|x2n−1 | (2.3) |
holds for all x∈(−π,0)∪(0,π), where Bn is the Bernoulli number.
Lemma 2.3. (See [41]) The double inequality
22n−1−122n+1−1(2n+2)(2n+1)π2<|B2n+2||B2n|<22n−122n+2−1(2n+2)(2n+1)π2. | (2.4) |
holds for n=1,2,3,⋯, where Bn is the Bernoulli number.
Lemma 2.4. (See [38]) Let an and bn (n=0,1,2,3,⋯) be real numbers, and the power series A(x)=∑∞n=0anxn and B(x)=∑∞n=0bnxn be convergent for |x|<R (R≤∞). If bn>0 for all n≥0 and the sequence {an/bn}∞n=0 is increasing (decreasing), then the function A(x)/B(x) is strictly increasing (decreasing) on (0,R).
Proof of Theorem 1.1. Let 0<|x|<π/2 and
F(x)=7ln[(sinxx)2+tanxx−2]−ln [(845)7x22(tanx)6]. |
Then elaborated computations lead to
F′(x)=cos2xx3(tanx)[(sinxx)2+tanxx−2]f(x), |
where
f(x)=12x3sec2x+x2sec2xtan3x+12x3tan2x−6xtan2x+14xtan2x |
−29xsec2xtan2x+45x2sec2xtanx−36tan3x−6xtan4x. |
By substituting the power series formulas in Lemma 2.1 into f(x), we obtain
f(x)=∞∑n=6lnx2n−1, |
where
ln=6⋅22n(22n−2−1)(2n−3)|B2n−2|(2n)!+22n(266n2−245n+164)(22n−1)|B2n|3(2n)! |
+(2n)22n(2n−145)(2n+1)(n+1)(22n+2−1)3(2n+2)!|B2n+2|. |
It follows from (2.4) that
322n|B2n|ln=18(22n−2−1)(2n−3)(2n)!|B2n−2||B2n| |
+(266n2−245n+164)(22n−1)(2n)! |
+2n(2n−145)(2n+1)(n+1)(22n+2−1)(2n+2)!|B2n+2||B2n| |
>18(22n−2−1)(2n−3)(2n)!π2(22n−1)(2n)(2n−1)(22n−2−1) |
+(266n2−245n+164)(22n−1)(2n)! |
+2n(2n−145)(2n+1)(n+1)(22n+2−1)(2n+2)! |
×(2n+2)(2n+1)(22n−1−1)π2(22n+1−1) |
=18(2n−3)(2n)!π2(22n−1)(2n)(2n−1)+(266n2−245n+164)(22n−1)(2n)! |
+2n(2n−145)(2n+1)(n+1)(22n+2−1)(2n)!(22n−1−1)π2(22n+1−1), |
which is equivalent to
3(2n)!22n|B2n|ln>18π2(2n−3)(22n−1)(2n)(2n−1)+(266n2−245n+164)(22n−1) |
+2n(2n−145)(2n+1)(n+1)(22n+2−1)(22n−1−1)π2(22n+1−1) |
=h(n)n(22n+1−1)(2n−1)π2, |
where
h(n)=[u1(n)22n−v1(n)]22n+w(n), |
u1(n)=32n6−2288n5+(1064π−2328)n4−(1512π2−572)n3 |
+(580+1146π2)n2−(328π2−36π4)n−54π4, |
v1(n)=72n6−5148n5+(1596π2−5238)n4−(2268π2−1287)n3 |
+(1305+1719π2)n2−(492π2−54π4)n−81π4, |
w(n)=16n6−1144n5+(532π2−1164)n4−(756π2−286)n3 |
+(573π2+290)n2+(18π4−164π2)n−27π4>0. |
It is easy to prove that 22n>v1(n)/u1(n) for all n≥6 by using mathematical induction. Therefore, we deduce that
h(n)>0⇒ln>0⇒f(x)>0⇒F′(x)>0, |
F(x) is strictly increasing on (0,π/2) and F(x)>F(0+)=0. Note that
limx→0+(sinxx)2+tanxx−2x4(tanxx)6/7=845. |
Which completes the proofs of Theorem 1.1.
Proof of Theorem 1.2. Let 0<x<π/2, and
A(x)=xsin2x+x2tanx−2sin2xtanxsin2x,B(x)=x3tan2x, |
G(x)=(xsinx)2+xtanx−2x3tanx. |
Then we clearly see that
G(x)=A(x)B(x). |
It follows from Lemma 2.1 and inequality (2.3) that
A(x)=xsin2x+x2tanx−2sin2xtanxsin2x |
=x−2tanx+x2sin2xtanx |
=x−2tanx+2x2(1sin2x) |
=x−2∞∑n=122n(22n−1)(2n)!|B2n|x2n−1 |
+2x2[12x+∞∑n=122n(22n−2)(2n)!|B2n|x2n−1] |
=2∞∑n=222n(22n−2)(2n)!|B2n|x2n+1−2∞∑n=322n(22n−1)(2n)!|B2n|x2n−1 |
=∞∑n=222n+1(22n−2)(2n)!|B2n|x2n+1−∞∑n=222n+3(22n+2−1)(2n+2)!|B2n+2|x2n+1 |
=∞∑n=2[22n+1(22n−2)(2n)!|B2n|−22n+3(22n+2−1)(2n+2)!|B2n+2|]x2n+1=:∞∑n=2anx2n+1 |
and
B(x)=∞∑n=222n(22n−1)(2n−1)(2n)!|B2n|x2n+1=:∞∑n=2bnx2n+1. |
Note that
anbn=22n+1(22n−2)(2n)!|B2n|−22n+3(22n+2−1)(2n+2)!|B2n+2|22n(22n−1)(2n−1)(2n)!|B2n| |
=2(22n−2)(22n−1)(2n−1)−8(22n+2−1)(22n−1)(2n−1)(2n+2)(2n+1)|B2n+2||B2n|. |
Next, we prove that the sequence {an/bn} is decreasing for n≥2. Indeed, from Lemma 2.3 we clearly see that
anbn>2(22n−2)(22n−1)(2n−1)−8(22n+2−1)(2n−1)(22n+2−1)π2 |
=2(22nπ2−22n+2−2π2+4)π2(2n−1)(22n−1) |
and
an+1bn+1<2(22n+2−2)(22n+2−1)(2n+1)−8(22n+4−1)(22n+1−1)(22n+2−1)(2n+1)(22n+3−1)π2 |
=4(22n+1−1)(22n+3π2−22n+5−π2+2)π2(22n+3−1)(22n+2−1)(2n+1). |
Therefore, the sequence {an/bn} is decreasing for n≥2 follows from the fact that
(22nπ2−22n+2−2π2+4)(22n+3−1)(22n+2−1)(2n+1) |
−2(2n−1)(22n−1)(22n+1−1)(22n+3π2−22n+5−π2+2) |
=[(64π2−256)22n−(48n+48π2n+128π2−376)]42n |
+22n(48n+6π2n+47π2−128)−(4π2−8)>0 |
for n≥2.
Therefore, it follows from Lemma 2.4 that the function G(x)=A(x)/B(x) is strictly increasing on (0,π/2), and Theorem 1.2 follows from the limit values
limx→0+G(x)=245,limx→(π2)−G(x)=0. |
Proof of Theorem 1.3. Let 0<x<π/2 and
H(x)=ln[(245)7x24tan4x]−7ln[(xsinx)2+xtanx−2]. |
Then elaborated computations lead to
H′(x)=p(x)x(tanx)2[(xsinx)2+xtanx−2], |
where
p(x)=11x2tan2x−8xtan3x+9xtanx+11x2−48tan2x+10x2sec2x |
+4x3sec2xtanx+8x3sin2x+28x3sin2x. |
It follows from Lemmas 2.1 and 2.2 that
p(x)=∞∑n=536⋅22n−2(22n−2−2)(2n−2)!|B2n−2|x2n |
+∞∑n=5(2n)22n(22n−1)(4n+15)(2n)!|B2n|x2n |
−∞∑n=522n+5(n+6)(2n+1)(22n+2−1)(2n+2)!|B2n+2|x2n=:∞∑n=5cnx2n, |
where
cn=36⋅22n−2(22n−2−2)(2n−2)!|B2n−2|+2n22n(22n−1)(4n+15)(2n)!|B2n| |
−22n+5(n+6)(2n+1)(22n+2−1)(2n+2)!|B2n+2|. |
From Lemma 2.3 we have
cn|B2n|=9⋅22n(22n−2−2)(2n−2)!|B2n−2||B2n|+(2n)22n(22n−1)(4n+15)(2n)! |
−22n+5(n+6)(2n+1)(22n+2−1)(2n+2)!|B2n+2||B2n| |
>9⋅22n(22n−2−2)(2n−2)!22n−122n−2−1π2(2n+2)(2n+1) |
+(2n)22n(22n−1)(4n+15)(2n)! |
−22n+5(n+6)(2n+1)(22n+2−1)(2n+2)!22n−122n+2−1(2n+2)(2n+1)π2 |
=9⋅22n(22n−2−2)(2n)!22n−122n−2−1π2(2n−1)(2n)(2n+2)(2n+1) |
+(2n)22n(22n−1)(4n+15)(2n)! |
−22n+5(n+6)(2n+1)(22n+2−1)(2n)!22n−122n+2−11π2, |
that is
(2n)!cn|B2n|>9⋅22n(22n−2−2)(22n−1)π2(2n−1)(2n)(22n−2−1)(2n+2)(2n+1) |
+(2n)22n(22n−1)(4n+15) |
−32⋅22n(n+6)(2n+1)(22n+2−1)(22n−1)(22n+2−1)π2 |
and
(2n)!cn22n(22n−1)|B2n|>9⋅(22n−2−2)π2(2n−1)(2n)(22n−2−1)(2n+2)(2n+1) |
+2n(4n+15)−32(n+6)(2n+1)π2 |
=q(n)π2(n+1)(2n+1)(22n−4), |
where
q(n)=u2(n)22n−v2(n), |
u2(n)=(16π2−128)n4−(1024n−84π2)n3 |
+(98π2+18π4−1696)n2−(9π4−30π2+992)n−192, |
v2(n)=(64π2−512)n4−(4096−336π2)n3 |
+(144π4+392π2−6784)n2−(3968−120π2+72π4)n−768. |
It is not difficulty to prove that q(n)>0 for all n≥5 by use of the mathematical induction. Therefore, H′(x)>0, F(x) is strictly increasing on (0,π/2) and H(x)>H(0+)=0. Note that
limx→0+(xsinx)2+xtanx−2x4(tanxx)4/7=245. |
Which completes the proofs of Theorem 1.3.
Remark 4.1. The first inequality of (1.3) and our inequality (1.9) are not comparable. Indeed, experiments and numerical simulation show that our inequality (1.9) is better than the first inequality of (1.3) for x∈(1.0828,π/2), but the first inequality of (1.3) is stronger than the inequality (1.9) for x∈(0,1.0828).
Remark 4.2. Let α=0. Then inequality (1.10) or (1.11) leads to
(xsinx)2+xtanx>2 |
for x∈(0,π/2), which also was proved by Zhu [42]. In addition, the second inequality of (1.11) is stronger than that of (1.10) due to the fact tanx>x for all x∈(0,π/2).
Remark 4.3. It is worth pointing out that it is very meaningful and important to improve the existing inequalities. We recommend the interested readers to read the literature [43,44,45] where one can found many new inequalities for the circular, hyperbolic, inverse circular, inverse hyperbolic and exponential functions as well as their related special functions.
We have established three sharp inequalities of Wilker type for trigonometric functions:
βx4(tanxx)6/7<(sinxx)2+tanxx−2,αx3tanx<(xsinx)2+xtanx−2<βx3tanx,αx4(tanxx)4/7<(xsinx)2+xtanx−2<βx4(tanxx)4/7, |
where 0<x<π/2, α=0, and β=8/45. The above three inequalities improve and develop the known famous results.
The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.
The research is supported by the Natural Science Foundation of China (Grant No. 61772025).
The authors declare that they have no competing interests.
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