CC BY
65B. A. Bhayo, J. Sandor
ON CARLSON’S AND SHAFER’S INEQUALITIES
Abstract. In this paper the authors refine the Carlson’s inequalities for inverse cosine function, and the Shafer’s inequalities for inverse tangent function.
Key words: Carlson’s inequality, Shafer’s inequality, inverse trigonometric functions.
2010 Mathematical Subject Classification: 26D05, 26D07,
26D99.
§ 1. Introduction
During the past fifteen years, numerous authors have studied various inequalities for trigonometric functions [1-12]. Thus, some classical and also more recent inequalities, such as inequalities of Jordan, Cusa-Huygens, Shafer-Fink, and Wilker have been refined and generalized. One of the key methods in these studies has been so called monotone l’Hospital Rule from [1] and an extensive survey of the related literature is given in
[13]. This Rule is formulated as Lemma 1 and it will be also applied here. Motivated by these studies, in this paper we make a contribution to the topic by sharpening Carlson’s and Shafer’s inequalities, and our inequalities refine the existing results in literature.
We start our discussion with the following well-known inequalities,
cos(t)i/3 < sinW < co^a+l, (1)
t 3
for 0 < \t\ < n/2. The first inequality is due to by D. D. Adamovic and D. S. Mitrinovic [14, p. 238], while the second inequality was obtained by N. Cusa and C. Huygens [15]. These inequalities can be written as
3 sin(t) sin(t)
2 + cos(t) < ^ < cos(t)1/3
© Bhayo B. A., Sandor J., 2014
For the further studies and refinements of inequalities in (1), e.g., see [2, 5-8,13,16] and the references therein. For the easy references we recall the following inequalities
/1 \ 1 sin(t) (t\
"H^) <T< “Ha) ’ (2)
the first inequality holds for t E (0, n/2) [6], while the second one is valid for t E (—\/27/5, 27/5), and was proved by Klen et al. [2]. The first
inequality in (2) refines the following one
2sin(t) n
t < ----------—, 0 < x < —,
1 + cos(t) 2
which was constructed in [17] by using Chebyshev’s integral inequality.
Oppenheim’s problem [14, 18, 19] states to determine the largest a2
and the least a3 as a function of a1 > 0, such that the following inequalities
a2 sin(x) a3 sin(x) ^
< X < ~ z r (3)
1 + a1 cos(x) 1 + a1 cos(x)
hold for all x E (0, n/2). A partial solution of this problem was given by Oppenheim and Carver [19], they showed that (3) holds for all a1 E E (0,1/2) and x E (0, n/2) when a2 = 1 + a1 and a3 = n/2. In 2007, Zhu [20, Theorem 7] solved Oppenheim’s problem completely by proving that the inequalities in (3) hold if a1, a2 and a3 are as follows:
1) if a1 E (0,1/2), then a2 = 1 + a1 and a3 = n/2,
2) if a1 E (1/2, n/2 — 1), then a2 = 4a1(1 — a2) and a3 = n/2,
3) if a1 E (n/2 — 1, 2/n), then a2 = 4a1 (1 — af) and a3 = 1 + a1,
4) if a1 > 2/n, then a2 = n/2 and a3 = 1 + a1,
where a2 and a3 are the best possible constants in (1) and (4), while a3 is the best possible constant in (2) and (3). Thereafter, Carver’s solution was extended to the Bessel functions for the further results by Baricz
[21, 22]. On the basis of computer experiments we came up that the following lower and upper bounds for x,
are the best possible bounds, and can be obtained from case (4) and (3), respectively.
Recently, Qi et al. [23] have given a new proof of Oppenheim’s problem, and deduced the following inequalities,
This implies that the second inequality of (4) is better than the corresponding inequality of (5).
Our first main result, which refines the inequalities in (4), reads as follows.
Theorem 1. For x E (0,n/2), we have
with the best possible constants a = 3 and 3 = (8^ — 2)/n ~ 2.96465.
By using Mathematica Software® [24], one can see that Theorem 1 refines the inequalities in (4) as follows:
Zi(x) < Ca(x), for x E (0,1.28966),
(n/2)sin(x) n sin(x)
< x <
(4)
1 + (2/n) cos(x) 2 + (n — 2) cos(x)
(n/2)sin(x) < < (n + 2) sin(x)
(5)
for x E (0,n/2). It is obvious that
((n — 2) — 4)(1 — cos(x)) < 0,
which is equivalent to
2 + (n — 2) cos(x) n + 2 cos(x)
n sin(x) < (n + 2) sin(x)
Ca < x < Cp,
(6)
where
8sin(x/2) — sin(x) 8sin(x/2) — sin(x)
Zu(x) < Cp(x), for x E (0, 0.980316),
where Zi and Zu denote the lower and upper bound of (4), respectively. It is worth to mention that the first inequality in Theorem 1 was discovered heuristically by Huygens [25], here we have given a proof.
In 1970, Carlson [26] established the following inequalities,
6(1 — x)1/2 . , 41/3(1 — x)1/2 , ,
< arccos(x) < — ----------—j6—, (7)
2\[2 + (1 + x)1/2 (1 + x)1/6
0 < x < 1. These inequalities are known as Carlson’s inequalities in literature. Thereafter, several authors studied these inequalities, and gave some generalization and partial refinement, e.g., see [27-30]. It is interesting to observe that the Adamovic-Mitrinovic and Cusa-Huygens inequality
(1) implies the second and the first inequality of (7), respectively, with the transformation x = arccos(t), 0 < t < n/4.
For 0 < x < 1, Guo and Qi [28, 29] gave the following inequalities,
n (1 — x)1/2 (1/2 + V2)(1 — x)1/2
“7-------< arccos(x) < _-----—----------, (8)
2(1 + x)1/6 2\[2 + (1 + x)1/2 ’
41/n (1 — x)1/2 n(1 — x)1/2
------ry.—' ' < arccos(x) < —-------------y-,—, ,.
(1 + x)(4-n)/(2n) 2(1 + x)(4-n)/(2n)
(9)
They concluded that these inequalities don’t refine (7) in the whole interval (0, 1) of x.
Chen et al. [27] established the lower bound for arccos(x) as follows,
n (1 — x)(n+2)/n2
— -r?—on / 2 < arccos(x), 0 < x < 1. (10)
2(1 + x)(n-2)/n2 W V 7
The inequality (10) refines the first inequality of (7) for x E (0, 0.345693). In [30], Zhu proved that for p > 1 and x E (0,1)
2 • 31/p^T—x . . 2nV1—x
— < arccos(x) <
((2V2)p + (V1 + x)p)1 ((2V2)P + (np — 2p)^1 + x)p)p ’
(11)
inequalities reverse for p E [0, 4/5].
We give the following theorem, which refines Carlson’s inequality, see Figure 1.
Theorem 2. For x E (0,1)
1 (V2 - < arccos(x) < (J^6^)2/3. (12)
We see that Theorem 2 refines the inequalities in (11) by using the Mathematica Software® [24].
In 1967, Shafer [31] proposed the following elementary inequality
3x
1 2^1 = < arctan(x), x > 0. (13)
This inequality was proved by Grinstein, Marsh and Konhauser by different ways in [32].
In 2009, Qi et al. [33] refined the inequality (13) as follows,
(1 + a)x . . (n/2)x . ,
< arctan(x) < ------------------------------------ , x > 0, —1 < a < 1/2, (14)
a + y/1 + x2 4 + y/1 + x2
4a(1 + a2 )x max{n/2,1 + a}x
._______ < arctan(x) < ------------/ —, x > 0, 1/2 < a < 2/n.
a + V1 + x2 a W1 + x2
Recently, Alirezaei [34] has sharpened Shafer’s inequality (13) by giving the following bounds for arctan(x),
< arctan(x) < (15)
4/n2 + ^/(1 — 4/n2)2 + 4x2/n2
x
<
1 — 6/n2 + \J(6/n2)2 + 4x2/n2 ’
for x E R. Graphically, it is shown that the maximum relative errors of the obtained bounds are approximately smaller than 0.27% and 0.23% for the lower and upper bound, respectively.
Our next result refines the bounds given in (15), which is illustrated in Figure 2.
Theorem 3. For x E (0,1), we have
22/3x
V1 + x2 (1 + 1/V1 + x2) 2
§ 2. Preliminaries
For easy reference, we recall the the following Monotone l’Hopital rule due to Anderson et al. [1, Theorem 2], which has been extremely used in literature.
Lemma 1. For —^ < a < b < x>, let f,g : [a, b] ^ R be continuous on
[a, b], and be differentiable on (a, b). Let g (x) = 0 on (a, b). If f (x)/g (x)
is increasing (decreasing) on (a, b), then so are
f (x) — f (a) f (x) — f (b)
g(x) — g(a) an g(x) — g(b)'
Iff (x) /g (x) is strictly monotone, then the monotonicity in the conclusion is also strict.
Lemma 2. The function
f (x) = 4x sin(x) + (4 — x2) cos(x) — x2
is strictly decreasing from (0, n/2) onto (a, 4), a = n(8 — n)/4 ~ 3.81578. In particular,
n(8 — n)/4 + x2 — (4 — x2) cos(x) sin(x) 4 + x2 — (4 — x2) cos(x)
4x2
x
4x2
for x E (0, n/2).
Proof. By differentiating and using the indentities sin(x) = 2sin(x/2) x x cos(x/2) and 1 — cos(x) = 2sin(x/2)2 we get
f'(x) = x(2cos(x) + x sin(x) — 2) =
= 2sin(x/2)(xcos(x/2) — 2sin(x/2)) < 0.
Hence f is strictly decreasing in x E (0, n/2), and the limiting values can be obtained easily. □
Lemma 3. The following function
sin(x) — x cos(x)
f (x) =
2 sin(x/2) — x cos(x/2)
is strictly decreasing from (0, n/2) onto (b, 4), b = 2^/2/(4 — n) & 3.81578. In particular,
2V2 (2 • (x) (x
^2 sin ^x j — x cos ^x^ < sin(x) — x cos(x) < <4 (2sin (2)—x cos (x)),
4 — n V \2J V2,
(x\ (x'
2J — x cos ^2
for x E (0, n/2).
Proof. We get
x sin(x) x sin(x/2)(sin(x) — x cos(x))
f (x) =
2sin(x/2) — x cos(x/2) 2(2sin(x/2) — x cos(x/2))2
x sin(x/2)(x(2 + cos(x)) — 3sin(x))
4 — (4x sin(x) + (4 — x2) cos(x) — x2) ’
which is negative by the second inequality of (1) and Lemma 2. This implies that f is strictly decreasing in x E (0, n/2), and by applying l’Hopital rule we get the limiting values. □
Lemma 4. The following function
8 sin(x/2) — sin(x)
g(x) =
x
is strictly decreasing from (0, n/2) onto (3, 3), 3 = (8\/2 —2)/n & 2.96465. Also, the function
f (z) = 8sin(z)
6z + sin(2z)
is strictly decreasing from (0,n/4) onto (1,j),j = 8v/2/(2 + 3n) & & 0.99028.
Proof. We get
4cos(x/2) — cos(x) 8sin(x/2) — sin(x)
g'(x) =
sin(x) — x cos(x) — 4(2 sin(x/2) — x cos(x/2))
x2
which is negative by Lemma 3. Thus, g is strictly decreasing in x E E (0, n/2) , and the limiting values follow from the l’Hopital rule.
Next, let f = fi(z)/f2(z), z E (0, n/4), where fi(z) = 8sin(z) and f2(z) = 6z + sin(2z). We get
f1(z) = 4cos(z) = f ( )
f2(z) 1 + cos(z)2 m
One has,
>30- — < *
Clearly, f1 (0) = f2 (0) = 0, hence by Lemma 1 f is strictly decreasing, and we get
lim f (z) = 8^/(2 + 3n) & 0.99028 < f (z) < lim f (z) = 1, this implies the proof. □
§ 3. Proof of Theorems
Proof of Theorem 1. The proof follows easily from Lemma 4. □
Corollary. For x E (0,n/2), we have
8 sin(x/2) — sin(x) 8 sin(x/2) — 3 sin(x)
3 < Y ,
where 3 and y are as in Lemma 4.
Proof. For x E (0, n/2), let f (x) = sin(x/2)/sin(x). One has
f (x) = > 0.
sin(x)2
Hence, f is strictly increasing, and
1 = hm f (x) <f (x) < hm f (x) = T.
2 x^0 x^n/2 y2
We observe that
sin(x/2) < 1 2 — 8^/2 + 3n
sin(x) V2 16 — 2V2 — 3^’
which is equivalent to
(16 — 2.2 — 3.2n) sin(x/2) + (2 — 8.2 + 3n) sin(x)
24.2 — 6
This is equivalent to the desired inequality. □
> 0.
Proof of Theorem 2. Let x = cos(2t) for 0 < t < n/4. Then arccos(x)/2 = t, and clearly 0 < x < 1. From (2) and (6) we have
8 sin(t/2) — sin(t) 22/3 sin(t)
--------------- <t<TTT-------------, (17)
3 (1 + cos(t))2/3
for t E (0, n/2). Replacing cos(t), sin(t) and t by \/(1 + x)/2, ^/(1 — x)/2 and arccos(x)/2, respectively, in (17), we get
8((1 — \J(1 + x)/2)/2)1/2 — a/(1 — x)/2 < arccos(x) <
3 < 2 <
22/V (1 — x)/2
<
(1 + ^(1 + x)/2)2/3'
After simplification we get the desired inequality. □
Proof of Theorem 3. Next, let x = tan(t), t E (0,n/2) and x E (0, 1). Then t = arctan(x), and by using the identity 1 + tan(t)2 = sec(t)2 we get
x d • (t) (VlTx2 — 1 ^
sin(t) = . = m, and sin - = ----. = n.
w WTx2 \2J \ 2VTTx )
.1 + x2 ’ \2J y 2.1 +x
We get the desired inequalities if we replace, t, sin(t), sin(t/2) by arctan(x), m, n, respectively, in (17). □
For the comparison of the bounds of arccos(x) and arctan(x) given in
(7) and (12) with the corresponding bounds appear in Theorem 2 and 3, we use the the graphical method, see Figure 1 and 2.
Figure 1: We denote the left-hand sides of (7) and (12) by Ciow and Niow, respectively, while the right-hand sides by Cup and Nup, respectively. It is clear that (12) refines the Carlson’s inequality (7)
0.0014
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
: 1 1 1 1 : 1 1 1 ^ 1 ^ 1
: 1 1 : Blow ~Alowj f1 1 1- 1
- 1 4 J-
: r , 1 , , , r , , , 1
0.00012
0.00010
0.00008
0.00006
0.00004
0.00002
0.0 0.2 0.4 0.6 0.8 1.0
upj
Figure 2: We denote the lower and upper bound of (16) by Biow and B respectively, while the corresponding bounds of (15) are denoted by Aiow and Aup. The differences Biow — Aiow, Aup — Bup are positive, this implies that the inequalities in (16) are better than the corresponding inequalities of (15)
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The work is received on May 7, 2014.
University of Jyvaaskylaa,
Department of Mathematical Information Technology,
40014 Jyvaskyla, Finland.
E-mail: [email protected]
Babes-Bolyai University, Department of Mathematics,
Str. Kogalniceanu nr. 1, 400084 Cluj-Napoca, Romania.
E-mail: [email protected]
