Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2011, Article ID 869261, 9 pages
doi:10.1155/2011/869261
Research Article
Some Refinements of Inequalities for
Circular Functions
Zhengjie Sun and Ling Zhu
Department of Mathematics, Zhejiang Gongshang University, Zhejiang, Hangzhou 310018, China
Correspondence should be addressed to Ling Zhu, zhuling0571@163.com
Received 22 August 2011; Accepted 4 October 2011
Academic Editor: Morteza Rafei
Copyright q 2011 Z. Sun and L. Zhu. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We give new lower bound and upper bound for Papenfuss-Bach inequality and improve RuehrShafer inequality by providing a new lower bound.
1. Introduction
Papenfuss 1 proposes an open problem described as follows.
Theorem 1.1. Let 0 ≤ x < π/2. Then
x sec2 x − tan x ≤
8π 2 x3
π 2 − 4x2 2
.
1.1
Bach 2 prove Theorem 1.1 and obtain a further result.
Theorem 1.2. Let 0 ≤ x < π/2. Then
x sec2 x − tan x ≤
x3
2π 2
.
3 π 2 − 4x2 2
Ge gives a lower bound of the above inequality in 3 as follows.
1.2
2
Journal of Applied Mathematics
Theorem 1.3. Let 0 < x < π/2. Then
64x3
π 2 − 4x2 2
< x sec2 x − tan x <
x3
2π 4
.
3 π 2 − 4x2 2
1.3
Furthermore, 64 and 2π 4 /3 are the best constants in 1.3.
Besides, Bach 2 obtain the improvement of the Papenfuss-Bach inequality as another
one.
Theorem 1.4. Let 0 ≤ x < π/2. Then
x sec2 x − tan x ≤
2π 2 tan x − x
.
π 2 − 4x2
1.4
In this note, we firstly obtain better bounds for Papenfuss-Bach inequality as in the
following statement.
Theorem 1.5. Let 0 < x < π/2. Then
2π 4 /3 x3 8π 4 /15 − 16π 2 /3 x5
2
π 2 − 4x2 < x sec2 x − tan x <
2π 4 /3 x3 256/π 2 · 513/511 − 8π 2 /3 x5
1.5
π 2 − 4x2 2
.
And then we give the refinement of the Ruehr-Shafer inequality as follows.
Theorem 1.6. Let 0 < x < π/2. Then
1018π 2 tan x − x
2π 2 tan x − x
< x sec2 x − tan x <
.
2
2
511
π − 4x
π 2 − 4x2
1.6
Remark 1.7. Since 256/π 2 ·513/511−8π 2 /3 ≈ −0.2792 < 0, we know that the upper bound
in the inequality 1.5 is better than the one in 1.3. At the same time, we find that the lower
estimate in 1.5 is larger than the one in 1.3 on the interval 0, 1.169880805 meanwhile the
lower estimate in 1.3 is larger than the one in 1.5 on 1.169880805, π/2.
2. Lemmas
Lemma 2.1 see 4–7. Let B2n be the even-indexed Bernoulli numbers. Then
22n!
2π
2n
1
1
22n!
< |B2n | <
,
−2n
2n
1−2
2π 1 − 21−2n
n 1, 2, 3, . . . .
2.1
Journal of Applied Mathematics
3
Lemma 2.2 see 7–9. Let |x| < π/2. Then
∞ 22n 22n − 1
∞ 22n 22n − 1
n−1
2n−1
tan x −1 B2n x
|B2n |x2n−1 ,
2n!
2n!
n1
n1
∞ 22n 22n − 1
tan x − x |B2n |x2n−1 .
2n!
n2
2.2
Lemma 2.3. Let Fx π 2 − 4x2 x sec2 x − tan x and |x| < π/2. Then
∞
22n2 22n2 − 1
22n 22n − 1
2π 2 3 2
2π
x Fx n|B2n2 | − 4
2n − 2|B2n | x2n1 .
3
2!
2n
2n!
n2
2.3
Proof. By using Lemma 2.2, we have
∞ 22n 22n − 1
π
sec x tan x 2n − 1|B2n |x2n−2 , |x| < ,
2
2n!
n1
∞ 22n 22n − 1
π
xsec2 x − tan x 2n − 2|B2n |x2n−1 , |x| < .
2
2n!
n2
2
2.4
We calculate
π 2 − 4x2 xsec2 x − tan x
∞ 22n 22n − 1
∞ 22n 22n − 1
2n−1
−4
π
2n − 2|B2n |x
2n − 2|B2n |x2n1
2n!
2n!
n2
n2
∞ 22n 22n − 1
∞ 22n 22n − 1
2π 2 3
2
2n−1
x π
−4
2n − 2|B2n |x
2n − 2|B2n |x2n1
3
2n!
2n!
n3
n2
2n2
2n2
∞ 22n 22n − 1
∞
2
2
−1
2
2π 3
2
2n1
x π
−4
2n|B2n2 |x
2n − 2|B2n |x2n1
3
2!
2n
2n!
n2
n2
∞
22n2 22n2 − 1
22n 22n − 1
2π 2 3 2
2π
x n|B2n2 | − 4
2n − 2|B2n | x2n1 .
3
2!
2n
2n!
n2
2.5
2
The proof of Lemma 2.3 is completed.
Lemma 2.4. Let |x| < π/2. Then
∞ 2n
1
1 2
x2n .
2
2
2
π − 4x
π n0 π
2.6
4
Journal of Applied Mathematics
3. The Proof of Theorem 1.5
The proof of Theorem 1.5 is completed when proving the truth of the following double
inequality:
2π 4 /3 x3 8π 4 /15 − 16π 2 /3 x5
π 2 − 4x2
< π 2 − 4x2 x sec2 x − tan x
3.1
2π 4 /3 x3 256/π 2 · 513/511 − 8π 2 /3 x5
.
<
π 2 − 4x2
We firstly process the left-hand side of the above inequality. We compute that
2π 4 /3x3 8π 4 /15 − 16π 2 /3x5
π 2 − 4x2 x sec2 x − tan x −
π 2 − 4x2
∞
22n2 22n2 − 1
22n 22n − 1
2π 2 3 2
2π
x n|B2n2 | − 4
2n − 2|B2n | x2n1
3
2n 2!
2n!
n2
∞ 2n−2
∞ 2n−4
2
2
2π 2 3 2π 2 8π 2 16
8π 2 16 2n1
5
2n1
x −
x −
−
x
x
3
3 n2 π 2n−2
15
3
15
3 n3 π 2n−4
∞
22n2 22n2 − 1
22n 22n − 1
2
2π
n|B2n2 | − 4
2n − 2|B2n | x2n1
2!
2n
2n!
n3
∞ 2n−2
∞ 2n−4
2
2
2π 2 8π 2 16 2n1
2n1
−
−
x
x
3 n3 π 2n−2
15
3 n3 π 2n−4
a3 x7 a4 x9 ∞
an x2n1 ,
n5
3.2
where
22n2 22n2 − 1
22n 22n − 1
8π 2 16 22n−4
2π 2 22n−2
an 2π
−
−
n|B2n2 |−4
2n−2|B2n |−
3 π 2n−2
15
3 π 2n−4
2n 2!
2n!
3.3
2
for n ≥ 3 and n ∈ N .
Journal of Applied Mathematics
5
We calculate a3 ≈ 0.0100 > 0 and a4 ≈ 0.0049 > 0. For n ≥ 5, by using Lemma 2.1, we
can estimate an as follows:
22n2 22n2 − 1
22n 2!
n
an > 2π
2n2
2n 2!
2π
1 − 2−2n−2 22n 22n − 1
22n!
8π 2 16 22n−4
2π 2 22n−2
−4
−
−
2n − 2
−
3 π 2n−2
15
3 π 2n−4
2n!
2π2n 1 − 21−2n
4n · 22n2 16n − 1622n 22n − 1 2π 2 22n−2
8π 2 16 22n−4
−
−
−
· 2n
−
3 π 2n−2
15
3 π 2n−4
π 2n
π 2n
2 −2
2π 2 22 8π 2 16
22n−4 4n · 26 24 16n − 16
1
−
2n−4
−
−
1 2n
3 π2
15
3
2 −2
π
π4
π4
22n−4 256 256n − 1 8π 2 8
.
− 4 2n
−
2n−4
15
3
π
π4
π 2 − 2
2
3.4
Since, for n 5, 6, . . ., we have
256 256n − 1 8π 2 8 256
8π 2 8
2565 − 1
≥
≈ 0.0207 > 0,
−
−
−
−
15
3
15
3
π4
π 4 22n − 2
π4
π 4 210 − 2
3.5
so an > 0 for n ≥ 5 and n ∈ N . Combining with the results of a3 and a4 , we finish the proof
of the left-hand side of inequality 3.1.
Now, Let’s discuss the right-hand side of inequality 3.1 by taking the same method.
We compute that
2π 4 /3x3 256/π 2 · 513/511 − 8π 2 /3x5
π 2 − 4x2 x sec2 x − tan x −
π 2 − 4x2
∞
22n2 22n2 − 1
22n 22n − 1
2π 2 3 2
2π
x n|B2n2 | − 4
2n − 2|B2n | x2n1
3
2n 2!
2n!
n2
∞ 2n−4
∞ 2n−2
2
2
2π 2 3 2π 2 256 513 8 5
256 513 8 2n1
2n1
x −
−
−
x
·
·
x
x 3
3 n2 π 2n−2
π 4 511 3
π 4 511 3 n3 π 2n−4
∞
22n 22n − 1
22n2 22n2 − 1
5
2
2π
n|B2n2 | − 4
b2 x 2n − 2|B2n | x2n1
2!
2n
2n!
n3
∞ 2n−4
∞ 2n−2
2
2
2π 2 256 513 8 2n1
2n1
−
−
x
·
x
3 n3 π 2n−2
π 4 511 3 n3 π 2n−4
b2 x5 b3 x7 ∞
n4
bn x2n1 ,
3.6
6
Journal of Applied Mathematics
where
256 513 8
−
−
≈ −0.0413 < 0,
b2 ·
π 4 511 3
3.7
2
8 8
6 6
2 4
2
2
−
1
3
−
1
4
2
2
2π
2
8
256
513
2
b3 2π 2
≈ −0.0068 < 0,
−4
−
−
−
·
8! · 30
6! · 42
3 π4
π 4 511 3 π 2
8π 2 16
−
15
3
and, for n ≥ 4
22n2 22n2 − 1
22n 22n − 1
bn 2π
n|B2n2 | − 4
2n − 2|B2n |
2n 2!
2n!
2π 2 22n−2
256 513 8 22n−4
−
−
−
·
.
3 π 2n−2
π 4 511 3 π 2n−4
2
3.8
We can reckon that, for n 4, 5, . . .
22n 22n − 1
22n2 22n2 − 1
22n 2!
22n!
n
bn < 2π
2n − 2
−4
2n2
2n
1−2n−2
2n 2!
2n!
1−2
2π
2π 1 − 2−2n 2
2π 2 22n−2
256 513 8 22n−4
−
−
·
3 π 2n−2
π 4 511 3 π 2n−4
4π 2 n · 22n2 22n2 − 1
256 513 8 22n−4
16n − 1622n 2π 2 22n−2
−
−
−
−
·
3 π 2n−2
π 2n 22n2 − 2
π 2n
π 4 511 3 π 2n−4
24 16n − 16 2π 2 22 256 513 8
1
22n−4 256n
−
1 2n2
2n−4
−
− 4 ·
3 π2
511 3
2
−2
π
π4
π4
π
−
22n−4
π 2n−4
256 8 256 513 8
1
256n
4 − − 4 ·
2n2
4
3 π
511 3
−2 π
π 2
22n−4 256n
256 2
1
2n−4
.
−
·
π
π 4 22n2 − 2 π 4 511
3.9
As we can see,
256n
1
1
256 · 4
256 2
≤
4
2n2
4
4
2·42
−2
π 2
π 2
−2
π 511
3.10
holds for n 4, 5, . . ., so we conclude that bn < 0 for n ≥ 4 and n ∈ N . Observing the value
of b2 and b3 , we complete the proof of the right-hand side of inequality 3.1.
Journal of Applied Mathematics
7
4. The Proof of Theorem 1.6
Now we prove the left-hand side of the inequality in Theorem 1.6, which is equivalent to
π 2 − 4x2
1018π 2
xsec2 x − tan x −
tan x − x > 0.
511
4.1
By using the power series expansions of x sec2 x − tan x and tan x − x, we can rewrite the
above inequality as follows:
∞ 22n 22n − 1
∞ 22n 22n − 1 1018
2n−1
π
−4
2n − 2 −
2n − 2|B2n |x2n1 > 0.
|B2n |x
511
2n!
2n!
n2
n2
2
4.2
Now we simplify the left expression and prove that it is positive:
π2
∞ 22n 22n − 1
∞ 22n 22n − 1 1018
2n − 2 −
2n − 2|B2n |x2n1
|B2n |x2n−1 − 4
511
2n!
2n!
n2
n2
∞ 22n 22n − 1
∞ 22n2 22n2 − 1
1018
2
2n1
π
−4
2n −
2n − 2|B2n |x2n1
|B2n2 |x
511
2n 2!
2n!
n2
n1
∞
4π 2 3
x c2 x5 c3 x7 c4 x9 c5 x11 cn x2n1 ,
3 · 511
n6
4.3
where
22n2 22n2 − 1
22n 22n − 1
1018
cn π
2n −
2n − 2|B2n |
|B2n2 | − 4
511
2n 2!
2n!
2
4.4
for n 2, 3, . . .. Particularly, c2 ≈ −0.02447, c3 ≈ 0.00142, and c4 ≈ 0.00151, c5 ≈ 6.74143 × 10−4 .
We also use Lemma 2.1 in order to give the lower bound of cn for n 6, 7, . . .:
22n2 22n2 − 1
cn π
2n −
2n 2!
22n2 22n2 − 1
2
>π
2n −
2n 2!
2
22n 22n − 1
1018
2n − 2|B2n |
|B2n2 | − 4
511
2n!
22n 2!
1018
2n2
511 2π
1 − 2−2n−2 8
Journal of Applied Mathematics
22n 22n − 1
22n!
2n − 2
2n 2n!
2π 1 − 21−2n
2 · 22n2 2n − 1018/511 16n − 1622n 22n − 1
−
π 2n
π 2n 22n − 2
22n3
1
1018
2n 2n −
− 2n − 2 1 2n
511
π
2 −2
4
4
2n − 2
2·6−2
22n3
22n3
− 2n
− 2·6
2n
> 2n
> 0.
511 2 − 2
511 2 − 2
π
π
−4
4.5
We denote that
4π 2 3
x c2 x5 c3 x7 c4 x9 c5 x11
3 · 511
4π 2
3
2
4
6
8
c2 x c3 x c4 x c5 x
x
3 · 511
Gx ≡
4.6
≡ x3 Hx,
where
Hx 4π 2
c2 x 2 c3 x 4 c4 x 6 c5 x 8 .
3 · 511
4.7
4π 2
c 2 t c 3 t2 c 4 t3 c 5 t4 .
3 · 511
4.8
Let t x2 , and
P t ≡
We compute that
P t c2 2c3 t 3c4 t2 4c5 t3 ,
P t 2 c3 6c4 t 12c5 t > 0,
2
0<t<
π2
,
4
π2
.
0<t<
4
4.9
So P t is increasing on 0, π 2 /4. Since P 0 < 0 and P π 2 /4 > 0, we have that P t is
decreasing firstly and then increasing. Let t0 be only one point of minimum of the function
P t. Then t0 ≈ 1.5262621 and P t0 ≈ 7.33921×10−4 > 0; this implies that P t > 0, so Hx > 0
and Gx > 0 for 0 < x < π/2.
Combining with cn > 0 for n ≥ 6, we have proved Theorem 1.6.
Journal of Applied Mathematics
9
5. Open Problem
In the last section we pose a problem as follows: Let 0 < x < π/2. Then
2π 4 /3 x3 8π 4 /15 − 16π 2 /3 x5
π 2 − 4x2 2
< x sec2 x − tan x
2π 4 /3 x3 256/π 2 − 8π 2 /3 x5
<
π 2 − 4x2 5.1
2
hold, where 8π 4 /15 − 16π 2 /3 and 256/π 2 − 8π 2 /3 are the best constants in 5.1.
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4 W. Scharlau and H. Opolka, From Fermat to Minkowski, Undergraduate Texts in Mathematics, Springer,
New York, NY, USA, 1985.
5 M. Abramowitz and I. A. Stegun, Eds., Handbook of Mathematical Functions with Formulas, Graphs and
Mathematical Tables, Dover, New York, NY, USA, 1965.
6 C. D’Aniello, “On some inequalities for the Bernoulli numbers,” Rendiconti del Circolo Matematico di
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