Sharp inequalities for hyperbolic functions and circular functions

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R E S E A R C H

Open Access

Sharp inequalities for hyperbolic functions

and circular functions

Ling Zhu

1*

*_{Correspondence:}

[email protected] 1_{Department of Mathematics,}

Zhejiang Gongshang University, Hangzhou, China

Abstract

In this paper, we obtain some new sharp bounds for the exponential functions whose powers involve hyperbolic functions and circular functions, respectively.

MSC: Primary 26D05; 26D15; secondary 33B10

Keywords: Inequalities; Hyperbolic functions; Circular functions

1 Introduction

In [1], Stolarsky obtained the bounds for the exponential function whose power involves a hyperbolic function.

Theorem 1 Let x> 0.Then

1 <excothx–1<coshx. (1.1) On the other hand, Pittenger [2] and Stolarsky [3] got the lower bound for the function excothx–1_{as follows.}

cosh2x

3 3

2

<excothx–1. (1.2)

In fact, Zhu [4] and Kouba [5] showed a new sharp lower bound for the functionexcothx–1 as follows.

2(coshx)65 + 1

3

5 6

<excothx–1. (1.3)

Zhu [6] proved

cosh2x

3 3

2

<

2(coshx)65 _{+ 1}

3

5 6

, x> 0 (1.4)

to illustrate that the inequality (1.3) is stronger than (1.2).

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It should be pointed out that the paper of Yang et al. [7] has made a great deal of im-provement on inequality (1.2). The subject of the present paper is to further study the inequality (1.3), and to obtain the following results.

Theorem 4 Let p= 0,p1= (ln(3/2))/(ln(e/2)) = 1.3214 . . . ,and x∈(0, +∞).Then we have (i) whenp∈[2, +∞),the double inequality

ep(xcothx–1)<2cosh p_x_{+ 1}

3 < 2 3

e 2

p

ep(xcothx–1) (1.5)

holds,where the constants1and(2/3)(e/2)p_{are the best possible}_; (ii) whenp∈(–∞, 6/5],we have

(a) ifp∈(0, 6/5],the double inequality(1.5)reverses,that is,

2 3

e 2

p

ep(xcothx–1)<2cosh p_x_{+ 1}

3 <e

p(xcothx–1) _(1.6)

holds,where the constants(2/3)(e/2)p_and₁_{are the best possible}_; (b) ifp∈(–∞, 0),the left-hand side of the double inequality(1.5)holds too; (iii) whenp∈(6/5, 2),we have

(c) ifp∈[p1, 2),the left-hand side of inequality(1.5)holds; (d) ifp∈(6/5,p1),the left-hand side of inequality(1.6)holds.

As straightforward consequences of Theorem4, Theorem5which is due to Kouba [5] may be derived immediately.

Theorem 5 Let p= 0,and p1= (ln(3/2))/(ln(e/2)) = 1.3214 . . . .Then (1) the inequality

2coshpx+ 1 3 <e

p(xcothx–1) _(1.7)

holds for allx∈(0, +∞)if and only ifp∈(0, 6/5];

(2) the inequality(1.7)reverses for allx∈(0, +∞)if and only ifp∈(–∞, 0)∪[p1, +∞).

The analog of Theorem4for the circular functions is the following result.

Theorem 6 Let p= 0,p2=ln3 = 1.0986 . . . ,and x∈(0,π/2).Then we have (i) whenp∈[6/5, +∞),the double inequality

ep(xcotx–1)<2cos p_x_{+ 1}

3 <

ep 3

ep(xcotx–1) (1.8)

holds,where the constants1andep_/3_{are the best possible}_; (ii) whenp∈(–∞, 1],we have

(a) ifp∈(0, 1],the double inequality(1.8)reverses,that is,

ep 3

3 <e

p(xcotx–1) _(1.9)

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(b) ifp∈(–∞, 0),the double inequality(1.8)holds too; (iii) whenp∈(1, 6/5),we have

(c) ifp∈(p2, 6/5),the right-hand side of inequality(1.8)holds; (d) ifp∈(1,p2],the right-hand side of inequality(1.9)holds.

The following result which is due to Yang et al. [7] is a straightforward consequence of Theorem6.

Theorem 7 Let p= 0,and p2=ln3 = 1.0986 . . . .Then (A) the inequality

3 (1.10)

holds for allx∈(0,π/2)if and only ifp∈(–∞, 0)∪[6/5, +∞);

(B) the inequality(1.10)reverses for allx∈(0,π/2)if and only ifp∈(0,p2].

2 Lemmas

Lemma 1([8]) For–∞<a<b<∞,let f,g: [a,b]→Rbe continuous functions that are diﬀerentiable on(a,b),with f(a) =g(a) = 0or f(b) =g(b) = 0.Assume that g(x)= 0for each x in(a,b).If f/gis increasing(decreasing)on(a,b),then so is f/g.

Lemma 2([9]) Let A(x) =∞_n₌₀anxn_{and B(x) =}∞

n=0bnxnbe two real power series con-verging on(0,r) (r> 0)with bn> 0for all n.If the sequence{an/bn}is increasing(decreasing) for all n,then the function A(x)/B(x)is also increasing(decreasing)on(0,r).

Lemma 3 Let l(x)be deﬁned by

l(x) =ln

sinh2x–2x 4(xcoshx–sinhx)

ln coshx . (2.1)

Then l(x)is strictly increasing from(0, +∞)onto(1/5, 1).

Proof Let

l(x) =ln

ln coshx =: f(x) g(x)=

f(x) –f(0+₎ g(x) –g(0+₎,

where

f(x) =ln sinh2x– 2x

4(xcoshx–sinhx), g(x) =ln coshx. Then

f(x) = 2cosh2x– 2

sinh2x– 2x +

xsinhx

sinhx–xcoshx =sinhx

4sinhx

sinh2x– 2x+

x

sinhx–xcoshx

,

g(x) = sinhx

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We compute

f(x) g(x) =

sinhx(_sinh4sinh₂_x_–2x_x+_sinh_x_–x_x_cosh_x)

sinhx

coshx

=

x

sinhx–xcoshx+ 4

sinhx –2x+sinh2x

coshx

= (coshx)–4sinh

2_x_–_x_sinh_2x_{+ 2x}2_{+ 4x}_cosh_x_sinh_x (–2x+sinh2x)(–sinhx+xcoshx)

= (2x

2_cosh_x_{+ 2x}_cosh2_x_sinh_x_{– 4}_cosh_x_sinh2_x) –(2x2_cosh_x_{– 2x}_cosh2_x_sinh_x_{– 2x}_sinh_x_{+ 2}_cosh_x_sinh2_x) = (coshx–cosh3x+

1

2xsinhx+ 1

2xsinh3x+ 2x 2_cosh_x) –(1₂cosh3x–1₂coshx–5₂xsinhx–1₂xsinh3x+ 2x2_cosh_x) = 2coshx– 2cosh3x+xsinhx+xsinh3x+ 4x

2_cosh_x –cosh3x+coshx+ 5xsinhx+xsinh3x– 4x2_cosh_x :=A(x)

B(x),

where

A(x) = 2coshx– 2cosh3x+xsinhx+xsinh3x+ 4x2coshx, B(x) = –cosh3x+coshx+ 5xsinhx+xsinh3x– 4x2coshx.

Then

A(x) = +∞

n=2

6(n– 2)32n_{+ 16n}2_{+ 26n}_{+ 12} (2n+ 2)! x

2n+2_=: +∞

n=2

anx2n+2,

B(x) = +∞

n=2

3(2n– 1)32n_{– 16n}2_{– 14n}_{+ 3} (2n+ 2)! x

2n+2_=: +∞

n=2

bnx2n+2,

where

an=6(n– 2)3

2n_{+ 16n}2_{+ 26n}_{+ 12}

(2n+ 2)! ,

bn=3(2n– 1)3

2n_{– 16n}2_{– 14n}_{+ 3}

(2n+ 2)! .

We can obtain

an bn= 2

3(n– 2)32n_{+ 8n}2_{+ 13n}_{+ 6}

3(2n– 1)32n_{– 16n}2_{– 14n}_{+ 3}=: 2sn, n≥2.

Now we will prove that{sn}n≥2is strictly increasing, which means

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where

h(n) =: 81·32n–256n3+ 224n2+ 112n+ 162> 0

forn≥2 due to

h(n+ 1) – 9h(n) = 2048n3+ 1024n2– 320n+ 704

= 20,544 + 28,352(n– 2) + 13,312(n– 2)2+ 2048(n– 2)3> 0

forn≥2 andh(2) = 3231 > 0. This leads tosn<sn+1forn≥2. So{an/bn}n≥2is strictly increasing. By Lemma 2, we know that A(x)/B(x) =f(x)/g(x) is strictly increasing on (0, +∞). Thenl(x) =f(x)/g(x) is strictly increasing on (0, +∞) by Lemma1.

Since

lim

x→0+

ln₄₍_xsinh_cosh2_xx_––2_sinhx_x₎ ln coshx =

1

5, x→lim+∞

ln₄₍_xsinh_cosh2_xx_––2_sinhx _x₎ ln coshx = 1,

this completes the proof of Lemma3.

Lemma 4 Let x> 0,B2nbe the even-indexed Bernoulli numbers(see[10]).Then the follow-ing power series expansions:

tanx=

∞

n=1 22n_{– 1}

(2n)! 2 2n_|_B

2n|x2n–1, (2.2)

sec2x=

∞

n=1

22n₍₂2n_{– 1)(2n}_{– 1)} (2n)! |B2n|x

2n–2_, _(2.3)

tanxsec2x=

∞

n=2

22n₍₂2n_{– 1)(2n}_{– 1)(n}_{– 1)} (2n)! |B2n|x

2n–3 _(2.4)

hold for all x∈(–π/2,π/2).

Proof The power series expansion (2.2) can be found in [11, equations 1.3.1.4(3)]. By (2.2) we have

(secx)2= (tanx)=

∞

n=1

22n₍₂2n_{– 1)(2n}_{– 1)} (2n)! |B2n|x

2n–2_, _|_x_|_<π 2,

and

tanxsec2x=1 2

sec2x

=1 2

∞

n=2

22n₍₂2n_{– 1)(2n}_{– 1)(2n}_{– 2)} (2n)! |B2n|x

2n–3

=

∞

n=2

22n₍₂2n_{– 1)(2n}_{– 1)(n}_{– 1)} (2n)! |B2n|x

2n–3_, _|_x_|_<π

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Lemma 5([12,13]) Let B2nbe the even-indexed Bernoulli numbers.Then

(2n+ 2)(2n+ 1)(22n–1– 1) π2₍₂2n+1_{– 1)} <

|B2n+2|

|B2n|

<(2n+ 2)(2n+ 1)(2 2n_{– 1)}

π2₍₂2n+2_{– 1)} (2.5)

holds for n= 1, 2, . . . .

Lemma 6 Let z(x)be deﬁned by

z(x) =ln

2x–sin2x 4(sinx–xcosx)

ln cosx . (2.6)

Then z(x)is strictly decreasing from(0,π/2)onto(0, 1/5).

Proof Let

z(x) = ln

ln cosx = p(x) q(x)=

p(x) –p(0+₎ q(x) –q(0+₎,

where

p(x) =ln 2x–sin2x

4(sinx–xcosx), q(x) =ln cosx. Since

p(x) = (sinx)

4sinx 2x–sin2x–

x

sinx–xcosx , q

_{(x) = –}sinx

cosx, we have

p(x) q(x) =

xcosx

sinx–xcosx–

2sinxcosx x–cosxsinx

= x

2_cos_x₊_x_cos2_x_sin_x_{– 2}_cos_x_sin2_x –x2_cos_x₊_x_cos2_x_sin_x₊_x_sin_x_–_cos_x_sin2_x = (x

2_cos_x₊_x_cos2_x_sin_x_{– 2}_cos_x_sin2_x)/_cos3_x (–x2_cos_x₊_x_cos2_x_sin_x₊_x_sin_x_–_cos_x_sin2_x)/_cos3_x

= x

2_sec2_x₊_x_tan_x_{– 2}_tan2_x –x2_sec2_x₊_x_tan_x₊_x_tan_x_sec2_x_–_tan2_x =: C(x)

D(x).

From Lemma4we obtain

C(x) =x2sec2x+xtanx– 2tan2x =

∞

n=3

(2n)(22n– 1)|B2n| (2n)! –

8(22n+2– 1)(2n+ 1)|B2n+2| (2n+ 2)! 2

2n_x2n

=:

∞

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D(x) = –x2sec2x+xtanx+xtanxsec2x–tan2x =

∞

n=3

2(2n+ 1)(22n+2_{– 1)}_|_B 2n+2| (2n+ 2)! –

(22n_{– 1)}_|_B 2n|

(2n)! (n– 1)2 2n+1_x2n

=:

∞

n=3 dnx2n.

We consider the monotonicity ofC(x)/D(x), and compute that

cn dn=

1 2(n– 1)

(n+ 1)(2n)(22n_{– 1)}_|_B

2n|– 4(22n+2– 1)|B2n+2| (22n+2_{– 1)}_|_B

2n+2|– (n+ 1)(22n– 1)|B2n| .

Then forn≥3

cn dn>

cn+1 dn+1 ⇐⇒

L M=:

(n+ 1)(2n)(22n_{– 1)}_|_B

2n|– 4(22n+2– 1)|B2n+2| (n– 1)[(22n+2_{– 1)}|_B

2n+2|– (n+ 1)(22n– 1)|B2n|]

>(n+ 2)(2n+ 2)(2

2n+2_{– 1)}_|_B

2n+2|– 4(22n+4– 1)|B2n+4| n[(22n+4_{– 1)}|_B

2n+4|– (n+ 2)(22n+2– 1)|B2n+2|]

:=X Y.

Since

LY–MX 2|B2n+2|2

= –(n+ 2)n2– 2n– 122n+2– 12

–22n+2– 122n– 1(n+ 2)(n+ 1) |B2n|

|B2n+2|

+ (n+ 1)n2– 2n+ 222n+4– 122n– 1 |B2n|

|B2n+2|

|B2n+4|

|B2n+2|

– 222n+4– 122n+2– 1|B2n+4|

|B2n+2| ,

by Lemma5we have

LY–MX 2|B2n+2|2

> –(n+ 2)n2– 2n– 122n+2– 12

–(2

2n+2_{– 1)(2}2n_{– 1)(n}_{+ 2)(n}_{+ 1)π}2₍₂2n+1_{– 1)}

(2n+ 2)(2n+ 1)(22n–1_{– 1)}

+ (n+ 1)n2– 2n+ 222n+4– 122n– 1

× π2(22n+2– 1)

(2n+ 2)(2n+ 1)(22n_{– 1)}

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– 222n+4– 122n+2– 1(2n+ 4)(2n+ 3)(2 2n+2_{– 1)}

π2₍₂2n+4_{– 1)}

= (2

2n+2_{– 1)(n}_{+ 2)}

π2₍₂2n+3_{– 1)(2}2n_{– 2)(2n}_{+ 1)}r(n),

where

r(n) =u1(n)22n–v1(n)

24n+u2(n)22n–v2(n)

with

u1(n) =

64π2– 512n2–1024 – 64π2n+ 224π2– 16π4– 384,

v1(n) = 12π2n3+

146π2– 1216n2–2432 – 140π2n

+568π2– 26π4– 912,

u2(n) = 24π2n3–

400 – 38π2n2–800 – 26π2n

+ 247π2– 11π4– 300,

v2(n) =

4π2– 32n2+64 – 4π2n–14π2–π4– 24.

Then we havecn/dn>cn+1/dn+1forn≥3 when proving

u1(n)22n–v1(n) > 0 ⇐⇒ 22n> v1(n) u1(n)

, (2.7)

u2(n)22n–v2(n) > 0 ⇐⇒ 22n> v2(n) u2(n)

. (2.8)

Now we use mathematical induction to prove (2.7). Whenn= 3, (2.7) clearly holds. Assuming that (2.7) holds forn=m, that is,

22m> v1(m) u1(m)

. (2.9)

Next, we prove that (2.7) is valid forn=m+ 1. By (2.9) we have

22(m+1)= 4·22m> 4v1(m) u1(m) ,

in order to complete the proof of (2.7) it suﬃces to show that

4v1(m) u1(m)

> v1(m+ 1) u1(m+ 1) ⇐⇒

4v1(m)u1(m+ 1) –v1(m+ 1)u1(m) > 0.

In fact,

4v1(m)u1(m+ 1) –v1(m+ 1)u1(m) =k(m),

where

k(m) =13,452,288π4– 184,343,040π2– 230,976π6+ 1248π8+ 728,082,432

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+3,983,424π4– 59,525,376π2– 16,608π6+ 225,067,008(m– 3)2

+738,048π4– 10,274,304π2– 576π6+ 33,619,968(m– 3)3

+68,736π4– 801,792π2+ 1,867,776(m– 3)4

+2304π4– 18,432π2(m– 3)5

> 0

form≥3.

Similarly, we can prove (2.8). By (2.7) and (2.8) we ﬁnd that{cn/dn}n≥3is a monotonic decreasing sequence. Then we arrive at the conclusion thatp(x)/q(x) =C(x)/D(x) is de-creasing on (0,π/2) by Lemma2. By Lemma1we see thatz(x) is decreasing on (0,π/2).

Since

z0+=1 5, z

π 2

– = 0,

this completes the proof of Lemma6.

3 The proofs of main results 3.1 The proof of Theorem4

Proof Let

G(x) =2cosh p_x_{+ 1}

3ep(xcothx–1), x> 0.

Then

G(+∞) = ⎧ ⎨ ⎩ 2 3(

e 2)

p_, _p_{> 0,}

+∞, p< 0,

and

G(x) = p 3

Q(x)

ep(xcothx–1), (3.1)

where

Q(x) = 2(xcoshx–sinhx)

sinh2x cosh

p–1_x_–coshxsinhx–x

sinh2x = 2(xcoshx–sinhx)

sinh2x

coshp–1x– coshxsinhx–x 2(xcoshx–sinhx) = 2(xcoshx–sinhx)(ln coshx)

sinh2x

coshp–1x–₂₍cosh_x_coshxsinh_x_–_sinhx–x_x₎

ln(coshp–1x) –ln₂₍cosh_x_coshxsinh_x_–_sinhx–x_x₎

×

p– 1 –ln

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=: 2(xcoshx–sinhx)(ln coshx)

sinh2x

coshp–1x–₂₍cosh_x_coshxsinh_x_–_sinhx–x_x₎

ln(coshp–1x) –ln₂₍cosh_x_coshxsinh_x_–_sinhx–x_x₎

×p– 1 –l(x) (3.2)

with

l(x) =ln

ln coshx .

We consider the following three cases. Case1:p≥2.

From Lemma3, we getmaxx∈(0,+∞)l(x) = 1. Sop– 1 –l(x) > 0. This leads toQ(x) > 0 by (3.2), andG(x) > 0 by (3.1). Then

G0+<G(x) <G(+∞),

this is the double inequality (1.5). Case2:p≤6/5.

From Lemma3, we getminx∈(0,+∞)l(x) = 1/5. Sop– 1 –l(x) < 0. This leads toQ(x) < 0. Subcase2.1: 0 <p≤6/5. In this case,G(x) < 0 by (3.1). Then

G(+∞) <G(x) <G0+,

this is the double inequality (1.6).

Subcase2.2:p< 0. We haveG(x) > 0 by (3.1). In view ofG(+∞) = +∞, the left-hand side of inequality (1.5) holds too.

Case3: 6/5 <p< 2.

Letr(x) :=l(x) + 1 –p. Then

r0+=l0++ 1 –p=6 5–p< 0,

r(+∞) =l(+∞) + 1 –p= 2 –p> 0,

and there is the unique pointξ ∈(0, +∞) such that r(x) < 0 holds for all x∈(0,ξ) and r(x) > 0 holds for all forx∈(ξ, +∞). That is,p– 1 –l(x) > 0 holds for allx∈(0,ξ) and p– 1 –l(x) > 0 holds for allx∈(ξ, +∞). By (3.2) and (3.1), we haveG(x) > 0 for allx∈(0,ξ) andG(x) < 0 holds for allx∈(ξ, +∞). Then

min(G0+,G(+∞) <G(x) <G(ξ).

Subcase3.1:p1= (ln(3/2))/(ln(e/2)) <p< 2. In this case, 1 < (2/3)(e/2)p, that is,G(0+) < G(+∞) holds, somin(G(0+_),_G(+_∞_{)) =}_G(0+_{). This leads to the left-hand side of inequality} (1.5).

Subcase3.2: 6/5 <p<p1= (ln(3/2))/(ln(e/2)). In this case, 1 > (2/3)(e/2)p, that is,G(0+) > G(+∞) holds, somin(G(0+_),_G(+_∞_{)) =}_G(+_∞_{). This leads to the left-hand side of} inequal-ity (1.6).

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3.2 The proof of Theorem6

Proof Let

F(x) =2cos p_x_{+ 1}

3ep(xcotx–1), 0 <x< π

2.

Then

F

π 2

– =e

p

3

and

F(x) =p 3

2(sinx–xcosx)[–ln(cosx)] (sin2_x)_exp_(p(x_cot_x_{– 1))}

cosp–1_x_– x–cosxsinx 2(sinx–xcosx) (p– 1)ln cosx–ln₂₍(x_sin–cos_x_–x_xsin_cosx_x)₎

×

p– 1 –ln

ln cosx

=p 3

2(sinx–xcosx)[–ln(cosx)] (sin2x)exp(p(xcotx– 1))

cosp–1x– x–cosxsinx 2(sinx–xcosx) (p– 1)ln cosx–ln₂₍(x_sin–cos_x_–x_xsin_cosx_x)₎

×p– 1 –z(x), (3.3)

where

z(x) =ln

ln cosx .

We consider the following three cases. Case1:p≥6/5.

From Lemma6, we getmaxx∈(0,+∞)z(x) = 1/5. Sop– 1 –z(x) > 0 holds. This leads to F(x) < 0 by (3.3). Then

F0+<F(x) <F

π 2

– ,

which is the double inequality (1.8). Case2:p≤1.

From Lemma6, we getminx∈(0,+∞)z(x) = 0. Sop– 1 –z(x) < 0 holds. Subcase2.1: 0 <p≤1. In this case,F(x) < 0 by (3.3). Then

F

π 2

–

<F(x) <F0+,

which is the double inequality (1.9).

Subcase2.2:p< 0. We haveF(x) > 0 by (3.3), the double inequality (1.8) holds too. Case3: 1 <p< 6/5.

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andq(x) < 0 holds for all forx∈(η,π/2). That is,p– 1 –z(x) < 0 holds for allx∈(0,η) and p– 1 –z(x) > 0 holds for allx∈(η,π/2). By (3.3), we haveF(x) < 0 for allx∈(0,η) and F(x) > 0 for allx∈(η,π/2). Then

F(η) <F(x) <maxF0+,F(π/2)–.

Subcase3.1:p2=ln3 = 1.0986 <p< 6/5. In this case, 1 <ep/3, that is,F(0+) <F((π/2)–), somax(F(0+_),_F((π/2)–_{)) =}_F((π/2)–_{). This leads to the right-hand side of inequality (1.8).} Subcase3.2: 1 <p≤p2=ln3 = 1.0986. In this case, 1≥ep/3, that is,F(0+)≥F((π/2)–), somax(F(0+),F((π/2)–)) =F(0+). This leads to the right-hand side of inequality (1.9).

The proof of Theorem6is complete.

Acknowledgements

The author is thankful to reviewers for their careful corrections to and valuable comments on the original version of this paper.

Funding

The author was supported by the National Natural Science Foundation of China (no. 61772025).

Competing interests

The author declares that he has no competing interests.

Authors’ contributions

The author provided the questions and gave the proof for all results. He read and approved this manuscript.

Publisher’s Note

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Received: 6 March 2019 Accepted: 7 August 2019 References

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