Download New sharp Cusa--Huygens type inequalities for trigonometric and

arXiv:1408.2243v1 [math.CA] 10 Aug 2014
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES FOR
TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS
Abstract. We prove that for p ∈ (0, 1], the double inequality
sin x
1
cos px + 1 − 3p12 <
< 3q12 cos qx + 1 − 3q12
3p2
x
√
holds for x ∈ (0, π/2) if and only if 0 < p ≤ p0 ≈ 0.77086 and 15/5 = p1 ≤
q ≤ 1. While its hyperbolic version holds for x > 0 if and only if 0 < p ≤ p1 =
√
15/5 and q ≥ 1. As applications, some more accurate estimates for certain
mathematical constants are derived, and some new and sharp inequalities for
Schwab-Borchardt mean and logarithmic means are established.
1. Introduction
The Cusa and Huygens (see, e.g., [1]) states that for x ∈ (0, π/2), the inequality
sin x
2 + cos x
(1.1)
<
x
3
holds true. Its version of hyperbolic functions refers to (see [2]) the inequality
2 + cosh x
sinh x
<
(1.2)
x
3
holds for x > 0, and it is know as hyperbolic Cusa–Huygens inequality (see [2]).
There are many improvements, refinements and generalizations of (1.1) and (1.2),
see [3], [4], [5], [6], [7], [8], [9]; [11], [12], [13], [14], [15], [16], [17], [18], [19], [20],
[21].
Now we focus on the bounds for (sin x) /x in terms of cos px, where x ∈ (0, π/2),
p ∈ (0, 1]. In 1945, Iyengar [22] (also see [23, subsection 3.4.6] ) proved that for
x ∈ (0, π/2),
sin x
(1.3)
cos px ≤
≤ cos qx
x
holds with the best possible constants
2
2
1
p = √ and q = arccos .
π
π
3
Moreover, the following chain of inequalities hold:
sin x
cos x
x
x
(1.4)
cos x ≤
≤ (cos x)1/3 ≤ cos √ ≤
≤ cos qx ≤ cos ≤ 1.
2
1 − x /3
x
2
3
Qi et al. [24] showed that
(1.5)
cos2
x
sin x
<
2
x
Date: April 10, 2014.
2010 Mathematics Subject Classification. Primary 26D05, 26D15; Secondary 33B10, 26E60 .
Key words and phrases. Trigonometric function, hyperbolic function, inequality, mean.
This paper is in final form and no version of it will be submitted for publication elsewhere.
1
2
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
holds for x ∈ (0, π/2). Klén et al. [25, Theorem 2.4]
out that the function
pointed
p
p
1/p
p 7→ (cos px)
is decreasing on (0, 1) and for x ∈ − 27/5, 27/5
sin x
x
2 + cos x
x
≤
≤ cos3 ≤
2
x
3
3
are valid. Subsequently, Yang [8] (also see [26]) gave a refinement of (1.6), which
states that for p, q ∈ (0, 1) the double inequality
sin x
1/p
1/q
(1.7)
(cos px)
<
< (cos qx)
x
holds for x ∈ (0, π/2) if and only if p ∈ [p∗0 , 1) and q ∈ (0, 1/3], where p∗0 ≈ 0.3473.
Moreover, the double inequality
x α
sin x x 3
<
cos
< cos
3
x
3
with the best exponents α = 2 (ln π − ln 2) / (ln 4 − ln 3) ≈ 3.1395 and 3. Also, he
pointed out that the value range of variable x such that (1.6) holds can be extended
to (0, π). Very recently, Yang [21] gave another improvement of (1.6), that is, for
x ∈ (0, π/2) the inequalities
2
sin x
2 + cos x
x
(1.8)
< 32 cos x2 + 31 < cos3 <
x
3
3
are true.
An important improvement for the inequality in (1.5) is due to Neuman [2]:
2/3
x
sin x
1 + cos x
(1.9)
cos4/3 =
<
, x ∈ 0, π2 .
2
2
x
(1.6)
cos2
Lv et al. [27] showed that for x ∈ (0, π/2) inequalities
x 4/3
sin x x θ
(1.10)
cos
<
< cos
2
x
2
hold, where θ = 2 (ln π − ln 2) / ln 2 = 1.3030... and 4/3 are the best possible con1/ 3p2
stants. By constructing a decreasing function p 7→ (cos px) ( ) (p ∈ (0, 1]), Yang
[9] showed that the double inequality
(1.11)
∗2
sin x
x
(cos p∗1 x)1/(3p1 ) <
< cos5/3 √
x
5
√
is true for x ∈ (0, π/2) with the best constants p∗1 ≈ 0.45346 and 1/ 5 ≈ 0.44721.
It follows that
√
∗2
x
(cos x)1/3 < cos1/2 36x < cos2/3 √x2 < cos √ < cos4/3 x2 < (cos p∗1 x)1/(3p1 )
3
2
sin x
5/3 √
2 √
3 x
x
x
x
< cos
< cos 6 < cos 3 < cos16/3 x4 < e−x /6 < 2+cos
(1.12)
3
5
x
are valid for x ∈ (0, π/2).
For the bounds for (sinh x) /x in terms of cosh px, it is known that the inequalities
sinh x
x
2 + cosh x
< cosh3 <
x
3
3
holds true for x > 0 (see [18]), which is exactly derived by the inequalities for means
L < A1/3 <
2G + A
3
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
3
(see e.g. [28], [29], [30]), where L, Ap , G and A stand for the logarithmic mean,
power mean of order p, geometric mean and arithmetic mean or positive numbers
a and b defined by
a−b
if a 6= b and L (a, a) = a,
ln a − ln b
p
1/p
√
a + bp
if p 6= 0 and A = A0 (a, b) = ab,
Ap (a, b) =
2
L ≡
Ap
L (a, b) =
≡
G = A0 and A = A1 , respectively. Zhu in [31] proved that for p > 1 or p ≤ 8/15,
and x ∈ (0, ∞), the inequality
q
sinh x
> p + (1 − p) cosh x
x
is true if and only if q ≥ 3 (1 − p). It follows by letting p = 1/2 and q = 3/2 that
sinh x
x
> cosh4/3
x
2
holds for x > 0 (also see [2, (2.8)]). Yang [19] showed that the inequality
sinh x
1/ 3p2
> (cosh px) ( )
x
√
holds for all x > 0 if and only if p ≥ 1/ 5 and its reverse holds if and only if
2
0 < p ≤ 1/3. And, the function p 7→ (cosh px)1/(3p ) is decreasing on (0, ∞).
The aim of this paper is to determine the best p such that the inequalities
(1.13)
sin x
x
sinh x
x
< (>) 3p12 cos px + 1 −
1
3p2 ,
< (>) 3p12 cosh px + 1 −
p ∈ (0, 1) , x ∈ (0, π/2) ,
1
3p2 ,
p, x ∈ (0, ∞)
hold true.
Our main results are contained in the following theorems.
Theorem 1. For p ∈ (0, 1] and x ∈ (0, π/2), the double inequality
1
3p2
cos px + 1 −
1
3p2
sin x
<
x
cos qx + 1 − 3q12
√
holds if and only if 0 < p ≤ p0 ≈ 0.77086 and 0.77460 ≈ 15/5 = p1 ≤ q ≤ 1,
where p0 is the unique root of the equation
2 1
(1.15)
Fp π2 − = − 3p12 cos pπ
=0
+
1
−
2
2
3p
π
on (0, 1). And, the bound for (sin x) /x given in (1.14) is increasing with respect to
parameters p or q.
(1.14)
<
1
3q2
Theorem 2. For p, x > 0, the double inequality
(1.16)
1
3p2
cosh px + 1 −
1
3p2
<
sinh x
<
x
1
3q2
cosh qx + 1 −
1
3q2
√
holds if and only if 0 < p ≤ p1 = 15/5 and q ≥ 1. And, the bound for (sinh x) /x
given in (1.16) is increasing with respect to parameters p or q.
4
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
Remark 1. The weighted basic inequality of two positive numbers of a and b tell
us that for α ∈ [0, 1], the inequality αa + (1 − α) b ≥ aα b1−α . It is reversed if and
only if α ≥ 1 or α ≤ 0 (see [32]). Hence, taking into account (1.11) and (1.14) we
see that
√
(i) if p ∈ [p∗1 , 1/ 3), where p∗1 ≈ 0.45346, then
sin x
1/ 3p2
> (cos px) ( ) > 3p12 cos px + 1 − 3p12 ;
x
√
(ii) if p ∈ (1/ 3, p0 ), where p0 ≈ 0.77086, then
sin x
1/ 3p2
> 3p12 cos px + 1 − 3p12 > (cos px) ( ) .
x
In the same way,√(1.13)√together with (1.16) leads us to
(iii) if p ∈ [1/ 5, 1/ 3), then
sinh x
1/ 3p2
> (cosh px) ( ) > 3p12 cosh px + 1 − 3p12 ;
x
√ √
(iv) if p ∈ (1/ 3, 15/5), then
sinh x
1/ 3p2
> 3p12 cosh px + 1 − 3p12 > (cosh px) ( ) .
x
p
p
√
√
√
Taking p = 3/4, 1/ 2, 2/3, 1/ 3 and q = 3/5, 2/3, 3/2, 1 in Theorem 1,
we have
Corollary 1. For x ∈ (0, π/2), the inequalities
(1.17) cos √x3
<
3
4
cos 2x
3 +
1
4
<
2
3
cos √x2 +
1
3
<
16
27
√
cos 3x
4 +
√
sin x
x
< 31 cos x + 23 .
11
27
<
+ 94 < cos2 √x6 < 49 cos 23x + 59
< 95 cos 15x
5
p
√
√
√
Putting p = 3/5, 3/4, 1/ 2, 2/3, 1/ 3 and q = 1, 2/ 3 in Theorem 2 we have
Corollary 2. For x > 0, the inequalities
(1.18)cosh √x3
<
3
4
cosh 2x
3 +
<
5
9
cosh
√
15x
5
1
4
<
2
3
cosh √x2 +
+
4
9
<
sinh x
<
x
1
3
1
3
<
16
27
cosh 3x
4 +
cosh x +
2
3
<
1
2
11
27
cosh2
x
√
3
+ 21 .
2. Proof of Theorem 1
In order to prove Theorem 1, we need some lemmas.
Lemma 1. For x ∈ (0, π/2), the function p 7→ Up (x) defined on [0, 1] by
Up (x) =
1
x2
1
cos px + 1 − 2 if p ∈ (0, 1] and U0 (x) = 1 −
2
3p
3p
6
is increasing.
Proof. Differentiation yields
∂Up
1
(2 − 2 cos px − px sin px)
=
∂p
3p3
px
px
px
12px sin 2
− cos
sin
> 0,
=
px
3
3p
2
2
2
which completes the proof.
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
5
Lemma 2. Let the function Fp be defined on (0, π/2) by
sin x
x
− ( 3p12 cos px + 1 − 3p12 ) if p ∈ (0, 1] and F0 (x) =
2
− 1 + x6 .
√
(i) If Fp (x) < 0 for all x ∈ (0, π/2), then p ∈ [p1 , 1], where p1 = 15/5 ≈
0.77460.
(ii) If Fp (x) > 0 for all x ∈ (0, π/2), then p ∈ [0, p0 ], where p0 ≈ 0.77086.
(2.1) Fp (x) =
sin x
x
Proof. (i) If Fp (x) < 0 for all x ∈ (0, π/2), then we have
sin x
1
1
−
cos
px
+
1
−
2
2
x
3p
3p
1
5p2 − 3 ≤ 0,
=−
(2.2)
lim
x→0
x4
360
√
which leads to p ∈ ( 15/5, 1].
(ii) If Fp (x) > 0 for all x ∈ (0, π/2), then we have
π−
2
pπ
1
1
Fp
= −
cos
+ 1 − 2 > 0.
2
π
3p2
2
3p
From Lemma 1 we see that the function p 7→ Fp (π/2− ) is decreasing on [0, 1], which
together with the facts
2
2
π−
2√
2
π−
1
= −
= − <0
F1/2
2 + > 0 and F1
2
π
3
3
2
π 3
gives that there is a unique number p0 ∈ (1/2, 1) such that Fp (π/2− ) > 0 for
p ∈ (0, p0 ) and Fp (π/2− ) < 0 for p ∈ (p0 , 1). Solving the equation Fp (π/2− ) = 0
for p by mathematical computer software we find that p0 ≈ 0.77086.
This completes the proof.
Lemma 3. Let c ∈ (0, 3/5] and let the sequence (an (c)) be defined by
an (c) = 3 − (2n + 1) cn−1 .
(2.3)
Then (i) an (c) ≥ 0 for n ∈ N; (ii) for n ≥ 3, we have
1<
an+1 (3/5)
11
an+1 (c)
≤
≤
.
an (c)
an (3/5)
5
Proof. (i) We first show that an (c) ≥ 0 for n ∈ N. A simple computation leads to
an+1 (c) − an (c)
=
≥
(2n + 1) cn−1 − (2n + 3) cn = cn−1 ((2n + 1) − (2n + 3) c)
3
4
cn−1 (2n + 1) − (2n + 3)
= cn−1 (n − 1) ≥ 0,
5
5
which implies that an+1 (c) ≥ an (c) ≥ a1 (c) = 0.
(ii) Since a1 (c) = 0, a2 (c) = 3 − 5c ≥ 0, an (c) > 0 for n ≥ 3, if we can show that
the function
3 − (2n + 3) cn
an+1 (c)
=
(c, n) 7→
an (c)
3 − (2n + 1) cn−1
is increasing in c on (0, 3/5] and decreasing in n ≥ 3, then we have
1=
an+1 (0)
an+1 (c)
an+1 (3/5)
a4 (3/5)
11
<
<
≤
=
,
an (0)
an (c)
an (3/5)
a3 (3/5)
5
6
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
which is the desired results. Now we prove that (c, n) 7→ an+1 (c) /an (c) is increasing in c on (0, 3/5] for n ≥ 3. Differentiation yields
′
an+1 (c)
= 4n2 + 8n + 3 cn −3n (2n + 3) c+ 6n2 − 3n − 3 := hn (c) ,
c2−n a2n (c)
an (c)
h′n (c) = −n (2n + 3) 3 − (2n + 1) cn−1 = −n (2n + 3) an (c) < 0,
where the last inequality holds due to an (c) > 0 for n ≥ 3. Therefore, we have
n
3
3
hn (c) ≥ hn (3/5) =
4n2 + 8n + 3 +
4n2 − 14n − 5 ,
5
5
which is clearly positive due to that h3 (3/5) = 876/125 > 0 and 4n2 − 14n − 5 =
n(4n − 14) − 5 ≥ 3 for n ≥ 4. This reveals that h′n (c) > 0, that is, (c, n) 7→
an+1 (c) /an (c) is increasing in c on (0, 3/5].
On the other hand, we have
an+1 (c) an+2 (c)
−
an (c)
an+1 (c)
4cn+1 + 6 (c − 1)2 n + 15c2 − 18c + 3
an (c) an+1 (c)
cn−1
2
4cn+1 + 6 (c − 1) × 3 + 15c2 − 18c + 3
an (c) an+1 (c)
cn−1
7
− c (1 − c) > 0,
4cn+1 + 33 11
an (c) an+1 (c)
= cn−1
≥
=
where the first inequality holds due to an (c) > 0 for n ≥ 3, while the last one
holds since c ∈ (0, 3/5]. This means that (c, n) 7→ an+1 (c) /an (c) is decreasing
with n ≥ 3.
Thus we complete the proof of this assertion.
Now we are in a position to prove Theorem 1.
Proof of Theorem 1. Expanding in power series gives
Fp (x)
=
=
=
=
sin x
x
∞
X
− ( 3p12 cos px + 1 −
x2n
(−1)
−
(2n + 1)!
n=0
∞
X
n
1
3p2 )
∞
2n
1
1 X
n (px)
+1− 2
(−1)
2
3p n=0
(2n)!
3p
!
3 − (2n + 1) p2n−2 2n
x
3 (2n + 1)!
n=2
∞
2
3 − 5p2 4 X
n an p
x +
(−1)
x2n ,
360
3
(2n
+
1)!
n=3
n
(−1)
where an (c) is defined by (2.3). Considering the function fp (x) = x−4 Fp (x), we
have
∞
an p 2
3 − 5p2 X
n
−4
(2.4)
fp (x) = x Fp (x) =
+
(−1)
x2n−4 ,
360
3
(2n
+
1)!
n=3
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
7
and differentiation yields
fp′
(x)
∞
X
=
(2n − 4) an p2 2n−5
x
(−1)
3 (2n + 1)!
n=3
:
=
∞
X
n
(−1)n un (x),
n=3
where
(2n − 4) an p2 2n−5
x
.
un (x) =
3 (2n + 1)!
Utilizing Lemma 3 we get that for p2 ∈ (0, 3/5] and n ≥ 3,
un+1 (x)
un (x)
=
(2n−2)an+1 (p2 ) 2n−3
x
3(2n+3)!
(2n−4)an (p2 ) 2n−5
3(2n+1)! x
(2n − 2) an+1 p2 2
1
x
=
(2n − 4) (2n + 3) (2n + 2) an (p2 )
11 π 2
11π 2
1
×1×
×
=
< 1,
(2 × 3 − 4) (2 × 3 + 3)
5
4
360
P∞
n
which implies that the power series n=3 (−1) un (x) is a Leibniz type alternating
one, and so fp′ (x) < 0 for p2 ∈ (0, 3/5].
√
(i) We first prove the second inequality in (1.14) holds, where p1 = 15/5 is the
best. As shown previously, we see that fp1 is decreasing on (0, π/2), and therefore,
<
fp1 (x) < fp1 0+ = lim
x→0+
1
3 − 5p21 = 0,
x−4 Fp (x) =
360
which together with (2.4)
√ yields Fp1 (x) < 0 for x ∈ (0, π/2).
Next we prove p1 = 15/5 is the best. If there is another p∗1 < p1 such that
the second inequality in (1.14) holds for x ∈ (0, π/2), then by Lemma
2 there must
√
∗
be p1 ∈ [p1 , 1], which yields a contradiction. Therefore, p1 = 15/5 can not be
replaced with other smaller ones.
(ii) Now we prove the first inequality in (1.14) holds with the best constant
p0 ≈ 0.77088. Since p20 ∈ (0, 3/5], fp0 is also decreasing on (0, π/2), and so
π −4
π−
π−
−4
fp0 (x) > fp0
x Fp0 (x) =
= lim
= 0,
Fp0
2
2
2
x→π/2−
where the last equality is true due to p0 is the unique root of the equation (1.15)
on (0, 1). It together with (2.4) gives Fp0 (x) > 0 for x ∈ (0, π/2).
Lastly, we show that p0 is the best. Assume that there is another p∗0 > p0 such
that Fp∗0 (x) > 0 for x ∈ (0, π/2). Then by Lemma 2 there must be p∗0 ∈ [0, p0 ],
which is clear a contradiction. Consequently, p0 can not be replaced by other larger
numbers.
Thus the proof is complete.
Remark 2. Application of the conclusion that fp′ (x) < 0 for x ∈ (0, π/2) if p2 ∈
(0, 3/5] gives fp (π/2− ) < fp (x) < fp (0+ ), that is,
π −4 π − 3 − 5p2
< x−4 Fp (x) < lim+ x−4 Fp (x) =
Fp
,
2
2
360
x→0
8
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
which can be changed into
sin x
< ( 3p12 cos px + 1 − 3p12 ) + c1 (p) x4 ,
x
−4
where c0 (p) = (π/2) Fp (π/2− ) and c1 (p) = 3 − 5p2 /360 are the best constants.
Then
√
(i) when p = p1 = 15/5, we have
sin x
< ( 3p12 cos p1 x + 1 − 3p12 ),
c0 (p1 ) x4 + ( 3p12 cos p1 x + 1 − 3p12 ) <
1
1
1
1
x
(2.5) ( 3p12 cos px + 1 −
1
3p2 ) + c0
(p) x4 <
−4
where c0 (p1 ) = (π/2− ) Fp1 (π/2− ) ≈ −7.2618 × 10−5 and c1 (p1 ) = 0 are the best
possible constants;
(ii) when p = p0 ≈ 0.77086, we get
< ( 3p12 cos p0 x + 1 − 3p12 ) + c1 (p0 ) x4 ,
0
0
where c0 (p0 ) = 0 and c1 (p0 ) = 3 − 5p20 /360 ≈ 8.0206 × 10−5 are the best constants.
( 3p12 cos p0 x + 1 −
0
1
)
3p20
<
sin x
x
3. Proof of Theorem 2
For proving Theorem 2, we first give the following lemmas.
Lemma 4. For x ∈ (0, ∞), the function p 7→ Vp (x) defined on [0, ∞) by
Vp (x) =
1
x2
1
cosh px + 1 − 2 if p 6= 0 and V0 (x) = 1 +
2
3p
3p
6
is increasing.
Proof. Differentiation yields
∂Vp
1
(px sinh px − 2 cosh px + 2)
=
∂p
3p3
px px sinh 2
2x
px
cosh
=
− px
> 0,
sinh
2
3p
2
2
2
which completes the proof.
Lemma 5. Let the function Gp be defined on (0, ∞) by
Gp (x) =
sinh x
x
− ( 3p12 cosh px + 1 −
1
3p2 )
if p 6= 0 and G0 (x) =
sinh x
x
−1−
x2
6 .
(i) If Gp (x) < 0 for all x ∈ (0, ∞), then p ≥ 1.
√
(ii) If Gp (x) > 0 for all x ∈ (0, π/2), then p ≤ p1 = 15/5 ≈ 0.77460.
Proof. In order to prove the desired results, we need the following two relations:
Gp (x)
1
(3.1)
5p2 − 3 ,
= −
lim
4
x→0
x
360

− 6p12 if p > 1,



Gp (x)
if p = 1,
− 61
=
(3.2)
lim
x→∞ epx

∞
if
0 < p < 1,


∞
if p = 0.
The first one follows by expanding in power series:
1
Gp (x) = −
5p2 − 3 x4 + o x6 .
360
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
9
To obtain the second one, it needs to note that
e−px Gp (x) = e(1−p)x 1−e2x
−2x
−
which gives (3.2).
(i) If Gp (x) < 0 for all x ∈ (0, ∞),
limx→∞ e−px Gp (x) ≤ 0. These together
(ii) If Gp (x) > 0 for all x ∈ (0, ∞),
limx→∞ e−px Gp (x) ≥ 0. These together
1
3p2
1 − e−2px − 1−
2
1
3p2
e−px ,
then we have limx→0 x−4 Gp (x) ≤ 0 and
with (3.1) and (3.2) give p ≥ 1.
then we have limx→0 x−4 Gp (x) ≥ 0 and
with (3.1) and (3.2) indicate p ≤ p1 .
We now can prove Theorem 2.
Proof of Theorem 2. The necessity follows by Lemma 5. To prove the sufficiency,
we expanding Gp (x) in power series to get
sinh x
− ( 3p12 cosh px + 1 − 3p12 )
x
!
∞
∞
X
x2n
1
1 X (px)2n
−
+1− 2
=
(2n + 1)!
3p2 n=0 (2n)!
3p
n=0
∞
∞
X 3 − (2n + 1) p2n−2
X an p 2
2n
=
x =
x2n .
3
(2n
+
1)!
3
(2n
+
1)!
n=2
n=2
√
2
It is derived
from Lemma 3 that an p ≥ 0 if 0 < p ≤ 15/5, and clearly,
an p2 < 0 if p ≥ 1.
Gp (x)
=
4. Applications
As simple applications of main results, we present some precise estimations for
certain special functions and constants in this section.
The sine integral is defined by
Z t
sin x
dx.
Si (t) =
x
0
Some estimates for sine integral can be seen [33], [34], [35], [26], [9]. Now we give
a new result.
√
Proposition 1. For t ∈ (0, π/2) and p ∈ (0, 15/5], we have
c0 (p) 5
c1 (p) 5
sin pt
sin pt
1
1
(4.1)
3p3 + 1 − 3p2 t + 5 t < Si (t) < 3p3 + 1 − 3p2 t + 5 t ,
where c0 (p) = (π/2)−4 Fp (π/2− ) and c1 (p) = 3 − 5p2 /360, here Fp (π/2− ) is
defined by (1.15). Particularly, putting p = 0+ , 2/3, we have
1
1 5
1 3 2π 3 − 48π + 96 5
t < Si (t) < t − t3 +
t +
t ,
18
15π5
18
600
9
2t 1
2 (16 − 5π) 5
2t 1
7 5
9
t < Si (t) < sin + t +
(4.3)
sin + t +
t ,
8
3
4
5π5
8
3
4
16200
and then,
1
1 3 1
1 3
1
2
π−
π + < Si π2 < π −
π +
π 5 ≈ 1.3714,
1.3705 ≈
5
360
5
2
144
19
200
√
√
1
1
9 3 1
7
9 3
1.3706 ≈
π+
+ < Si π2 < π +
π5 +
≈ 1.3711.
16
16
5
8
518 400
16
(4.2)
t−
10
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
Proof. Integrating each sides in (2.5) over [0, t] yields (4.1). Taking the limits of
the left and right hand sides in (4.1) as p → 0+ gives (4.2), and putting p = 2/3 in
(4.1) leads to (4.3). Substituting t = π/2 into (4.2) and (4.3) we get the last two
approximations of Si (π/2).
It is known that
Z
∞
0
′
1
π2
x
dx = ψ ′ ( 12 ) =
,
sinh x
2
4
where ψ is the tri-gamma function defined by
Z ∞
xe−tx
ψ ′ (t) =
dx.
1 − e−x
0
We define
Z
Sh (t) =
0
Then by (1.16) we have
t
x
dx.
sinh x
x
9
3
√
<
<
.
cosh x + 2
sinh x
5 cosh( 15x/5) + 4
Integrating over [0, t] and calculating lead to
Proposition 2. For t > 0, we have
√
√
√
√ t
√3+2 −
3 ln eet −
3
ln
2
−
3
+ 3+2
√
√
√
(4.4)
< Sh (t) < 2 15 arctan 53 e 15t/5 + 43 − 2 15 arctan 3.
In particular, we have
√
√
√ √
4.5621 ≈ 2 3 ln 2 + 3 < ψ ′ ( 21 ) < 2 15π − 4 15 arctan 3 ≈ 4.9845.
The Catalan constant [36]
G=
∞
X
n=0
(−1)n
2
(2n + 1)
= 0.9159655941772190...
is a famous mysterious constant appearing in many places in mathematics and
physics. Its integral representations contain the following [37]
Z π/4
Z
Z 1
1 π/2 x
π2
π
x2
arctan x
dx =
dx =
− ln 2 +
(4.5)
G=
dx.
x
2 0
sin x
16
4
sin2 x
0
0
We present an estimation for G below.
Proposition 3. We have
(4.6)
√
0.91586 ≈
√
15
4
ln
4 cos
4 cos
√
15π
15π
+3 sin 10
+5
√10
√
15π
15π
10 −3 sin
10 +5
<G<
√
15
5
ln
√ √ √ √√
√2−√2+3√15√√2+2+32 ≈ 0.91675,
11
11
2− 2−3 15
2+2+32
Proof. From the fourth and fifth inequalities in (1.17) we obtain that for x ∈
(0, π/2), the two-side inequality
1
5
9
cos
√
15x
5
+
4
9
<
x
<
sin x
1
16
27
cos 3x
4 +
11
27
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
holds true. Integrating both sides over [0, π/2] yields
Z π/2
Z π/2
Z π/2
x
dx
√
<
dx <
5
15x
sin x
0
0
0
+ 94
9 cos 5
Direct computations give
Z π/2
dx
5
9
0
Z
0
cos
π/2
16
27
√
15x
5
+
dx
cos 3x
4 +
4
9
11
27
=
=
√
√
16
27
11
dx
cos 3x
4 +
11
27
.
√
15π
15π
15 4 cos 10
+3 sin 10
+5
√
√
ln
≈ 1.8317,
15π
15π
4
cos
−3
sin
2
10
10 +5
√
√
√
2 15 11 2−√2+3√15 √2+2+32
ln √ √ √ √√
≈ 1.8335
11 2− 2−3 15
2+2+32
5
Utilizing the the second formula in (4.5) (4.4) follows.
We close this paper by giving some inequalities for bivariate means.
The Schwab-Borchardt mean of two numbers a ≥ 0 and b > 0, denoted by
SB (a, b), is defined as [38, Theorem 8.4], [39, 3, (2.3)]
 √2 2
b −a

 arccos(a/b) if a < b,
a
if a = b,
SB (a, b) =
√


a2 −b2
if
a > b.
arccosh(a/b)
The properties and certain inequalities involving Schwab-Borchardt mean can be
found in [40], [41]. We now establish a new inequality for this mean.
For a < b, letting x = arccos (a/b) in the fourth inequality of (1.17) and using
half-angle and triple-angle formulas for cosine function, and multiplying two sides
by b, we get
v
q
u
3
u
−3a/b
t 1 + 1+4(a/b)
2
11
16
+ 27
SB (a, b) ≥ 27 b
b
2
1/2
q
8
b1/4 + 11
2 (b − 2a)2 (a + b) + 2b3/2
= 27
27 b.
For a > b, letting x = arccosh (a/b) in the inequality connecting the fourth and sixth
members of (1.17) and using half-angle and triple-angle formulas for hyperbolic
cosine function, and multiplying two sides by b, we get the same inequality as
above.
Proposition 4. For a, b > 0, we have
(4.7)
SB (a, b) ≥
√
8 2
27
|b − 2a|
r
a+b
+ b3/2
2
!1/2
b1/4 +
11
27 b.
Remark 3. From the inequality (4.7), it is easy to get
√
11 + 8 2
SB (a, b) ≥
b ≈ 0.82643 × b
27
due to |b − 2a| ≥ 0. It seems to new and interesting.
For a, b > 0, with x = (1/2) ln (a/b), we have
L(a, b)
sinh x
=
,
x
G (a, b)
cosh px =
App (a, b)
(ap + bp ) /2
√ p = p
,
G (a, b)
ab
12
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
and by Theorem 2 we immediately get the following
Proposition 5. For a, b > 0 with a 6= b, the double inequality
√
√
5 √15/5 1− 15/5
G
A
9
15/5
(4.8)
+ 94 G < L < 13 A + 32 G
√
holds with the best constants p1 = 15/5 and 1. And, the function
(ln b − ln a)2
p 7→ 3p12 App G1−p + 1 − 3p12 G if p 6= 0 and G +
G if p = 0
24
is increasing on R.
Remark 4. Accordingly, Corollary 2 can be changed into the chain of inequalities
for means:
√
1/ 3
√
A1/√3 G1−1/
3
<
3 2/3 1/3
4 A2/3 G
<
16 3/4 1/4
27 A3/4 G
<
1
3A
√
1/ 2
√
+ 14 G < 32 A1/√2 G1−1/
+ 23 G <
2
+ 31 G
√
√
11
5 √15/5 1− 15/5
27 G < 9 A 15/5 G
√
√
1 2/√3 1−2/ 3
+ 21 G.
A
G
2 1/ 3
+
+ 94 G < L
1/(3q )
Remark 5. In [19], Yang obtained a sharp
lower bound Aq0 0 G1−1/(3q0 ) for
√
the logarithmic mean L, where q0 = 1/ 5, and pointed out that this one seems
to superior to most of known ones.√ Now, we derive a new sharp lower bound
5 p1 1−p1
+ 94 G for L, where p1 = 15/5. We claim that the latter is better than
9 Ap1 G
the former. In fact, we have
L > 59 App11 G1−p1 + 94 G > Aq1/(3q) G1−1/(3q)
√
if and only if q ≥ q0 = 1/ 5. In order for the second inequality in (4.9) to hold, it
suffices that for x > 0
1/ 3q2
D (x) = 3p12 cosh p1 x + 1 − 3p12 − (cosh qx) ( ) > 0
1
1
√
if and only if q ≥ q0 = 1/ 5.
The necessity can be obtained by limx→0 x−4 D (x) ≥ 0, which follows by expanding in power series
1 4 2
x p1 + 2q 2 − 1 + o x6 .
D (x) =
72
p
√
2
Then, q ≥ (1 − p1 ) /2 = 1/ 5.
1/ 3q2
Since q 7→ (cosh qx) ( ) is decreasing on (0, ∞) proved in [19, Lemma 2],
(4.9)
to prove D (x) ≥ 0 if q ≥ q0 , it suffices to show that D (x) ≥ 0 when q = q0 .
Differentiation yields
2
1
1
sinh p1 x −
cosh1/(3q0 )−1 q0 x sinh q0 x
D′ (x) =
3p1
3q0
2
sinh q0 x q0 sinh p1 x
=
− cosh1/(3q0 )−1 q0 x
3q0
p1 sinh q0 x
2
sinh q0 x
q0 sinh p1 x
=
×L
, cosh1/(3q0 )−1 q0 x × D1 (x) ,
3q0
p1 sinh q0 x
where
1
q0 sinh p1 x
−
− 1 ln cosh q0 x.
D1 (x) = ln
p1 sinh q0 x
3q02
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
13
Differentiating D1 (x) gives
D1′ (x) =
D2 (x)
,
6q0 sinh 2q0 x sinh p1 x
where
= 4(− sinh p1 x sinh2 q0 x − 3q02 cosh2 q0 x sinh p1 x
D2 (x)
+3q02 sinh p1 x sinh2 q0 x + 3p1 q0 cosh p1 x cosh q0 x sinh q0 x).
Utilizing ”product into sum” formulas and expanding in power series lead to
D2 (x) = −2 6q02 − 1 sinh p1 x + (3p1 q0 − 1) sinh (p1 x + 2q0 x)
+ (3p1 q0 + 1) sinh (2q0 x − p1 x)
∞
X
p2n−1 x2n−1
=
dn 1(2n−1)! ,
n=1
where
dn = (3p1 q0 − 1) 1 +
2q0
p1
2n−1
+ (3p1 q0 + 1)
2q0
p1
−1
2n−1
− 2 6q02 − 1
Now we show that dn ≥ 0 for n ≥ 1. A simple verification yields d1 = d2 = 0,
d3 = 64/45 > 0. Suppose that dn > 0 for n > 3, that is,
2n−1
2n−1
2q0
2q0
2
−1
.
> 2 6q0 − 1 − (3p1 q0 − 1) 1 +
(3p1 q0 + 1)
p1
p1
Then,
dn+1
=
>
(3p1 q0 − 1) 1 +
(3p1 q0 − 1) 1 +
×
2q0
−1
p1
2
2n+1
2n+1
0
+ (3p1 q0 + 1) 2q
−
1
− 2 6q02 − 1
p1
2n+1 2n−1 2q0
2q0
2
+ 2 6q0 − 1 − (3p1 q0 − 1) 1 + p1
p1
2q0
p1
− 2 6q02 − 1
!
2n−1
2q0
2
p1 (3p1 q0 − 1) 1 +
− 6q0 − 1 (p1 − q0 ) .
=
p1
√
Since (3p1 q0 − 1) = 3 3 − 5 /5 > 0, using binomial expansion we get
p21
2q0
− 6q02 − 1 (p1 − q0 )
dn+1 > p1 (3p1 q0 − 1) 1 + (2n − 1)
8q0
p1
= 2q0 (3p1 q0 − 1) (2n − 1) + q0 3p21 + 6q02 − 6p1 q0 − 1
8q0
p21
=
=
2q0 (3p1 q0 − 1) (2n − 1) − 2q0 (3p1 q0 − 1)
4q0 (3p1 q0 − 1) (n − 1) > 0,
where the third equality holds due to 3p21 + 6q02 = 3. By mathematical induction, we
have proven D2 (x) ≥ 0 for n ≥ 1. It follows that D1′ (x) > 0, which means that
D1 is increasing on (0, ∞), and then D1 (x) > limx→0+ D (x) = 0. This in turn
implies that D is increasing on (0, ∞), and therefore, D (x) ≥ D (0+ ) = 0, which
proves the sufficiency.
14
NEW SHARP CUSA–HUYGENS TYPE INEQUALITIES
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Power Supply Service Center, ZPEPC Electric Power Research Institute, Hangzhou,
Zhejiang, China, 310009
E-mail address: [email protected]