Journal of Inequalities in Pure and
Applied Mathematics
A NOTE ON A PAPER OF H. ALZER AND S. KOUMANDOS
KUNYANG WANG
volume 7, issue 2, article 46,
2006.
School of Mathematical Science
Beijing Normal University
Received 12 July, 2005;
accepted 21 January, 2006.
EMail: wangky@bnu.edu.cn
Communicated by: A.G. Babenko
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
208-05
Quit
Abstract
In the paper “ Sharp inequalities for trigonometric sums in two variables,” (Illinois Journal of Mathematics, Vol. 48, No.3, (2004), 887–907) Alzer and Koumandos investigated some special trigonometric sums. One of them is the sum
A∗n (x, y) :=
n
X
cos(k − 1 )x sin(k − 1 )y
2
k=1
k − 12
2
.
In the present note we show that the results of [1] can be easily obtained by a
very simple elementary argument. And the results we obtained are more exact.
2000 Mathematics Subject Classification: 26D05.
Key words: Inequalities, Trigonometric sums.
Supported by NSF of China, Grant # 10471010.
In a recent long paper [1], Alzer and Koumandos investigated the trigonometric
sums:
n
n
X
X
cos k − 21 x sin k − 12 y
cos kx sin ky
∗
An (x, y) =
, An (x, y) =
,
1
k
k
−
2
k=1
k=1
Bn (x, y) =
n
X
k=1
sin kx sin ky
.
k
A Note on a Paper of H. Alzer
and S. Koumandos
Kunyang Wang
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
Their results can be restated as follows:
http://jipam.vu.edu.au
(A) |An (x, y)| < sup{An (x, y) : x, y ∈ [0, π], n ∈ N} =
Rπ
0
sin t
t
dt;
(B) min{Bn (x, y) : x, y ∈ [0, π], n ∈ N} = − 18 ,
1
5π π
π 5π
Bn (x, y) = − ⇐⇒ n = 2 and (x, y) =
,
or (x, y) =
,
;
8
6 6
6 6
√
(C) − 23 ( 2 − 1) ≤ A∗n (x, y) ≤ 2,
2 √
3π π
∗
,
,
An (x, y) = − ( 2 − 1) ⇐⇒ n = 2, (x, y) =
3
4 4
A∗n (x, y) = 2 ⇐⇒ n = 1, (x, y) = (0, π).
The purpose of the present note is to give more exact results by very much
simpler proof.
For a continuous function f on D := [0, π] × [0, π] we define
min(f ) = min{f (x, y) : (x, y) ∈ D},
max(f ) = max{f (x, y) : (x, y) ∈ D}.
Our results are
(A0 )
π
0,
n+1
Z
π
2(n+1)
2 cos(n + 1)t sin nt
max(An ) = An
=
dt,
sin t
0
π
min(An ) = − max(An ) = An π, π −
,
n+1
Z π
sin t
dt.
max(An ) < lim max(An ) =
n→∞
t
0
A Note on a Paper of H. Alzer
and S. Koumandos
Kunyang Wang
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 3 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
http://jipam.vu.edu.au
(B0 ) For n ≥ 2
min(Bn ) = Bn
π
n
Z
=
π
n+1
(2n + 1)π
π
,
n(n + 1) n(n + 1)
sin(n + 1)t sin nt
dt < min(Bn+1 ),
sin t
lim min(Bn ) = 0.
n→∞
A Note on a Paper of H. Alzer
and S. Koumandos
0
(C ) For all n
max(A∗n )
Kunyang Wang
π
=
0,
n
Z π
2n sin 2nt
dt
=
sin t
0
A∗n
> max(A∗n+1 ) →
Title Page
Contents
Z
0
π
sin t
dt, (n → ∞),
t
and for n ≥ 2
min(A∗n ) = A∗n
JJ
J
II
I
Go Back
3π π
,
2n 2n
Z
π
n
=
In particular, min(A∗2 ) = 32 (1 −
π
2n
√
sin 2nt
dt < min(A∗n+1 ) → 0 (n → ∞).
2 sin t
2).
The results (A), (B), (C) are easy consequences of (A0 ), (B0 ) and (C0 ) respectively.
Close
Quit
Page 4 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
http://jipam.vu.edu.au
Proof of (A0 ). We have
n
X
sin k(x + y) − sin k(x − y)
An (x, y) =
2k
k=1
Z k(x+y)
n
X
1
=
cos t dt
2k k(x−y)
k=1
n Z x+y
X
2
cos 2kt dt
=
x−y
2
k=1
Z
x+y
2
n
X
2 cos 2kt sin t
=
x−y
2
Z
A Note on a Paper of H. Alzer
and S. Koumandos
k=1
x+y
2
sin(2n + 1)t − sin t
dt =
2 sin t
=
x−y
2
Kunyang Wang
2 sin t
Z
x+y
2
x−y
2
cos(n + 1)t sin nt
dt.
sin t
Then we get
max(An ) = An
π
0,
n+1
Z
π
2(n+1)
Z
−π
2(n+1)
π
2(n+1)
=
=
0
Z
<
0
π
2(n+1)
cos(n + 1)t sin nt
dt
sin t
2 cos(n + 1)t sin nt
dt
sin t
Z π
2 cos(n + 1)t sin(n + 1)t
sin t
dt =
dt;
t
t
0
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 5 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
http://jipam.vu.edu.au
π
Z
sin t
dt;
t
lim max(An ) =
n→∞
0
min(An ) = min{An (π − x, π − y) : (x, y) ∈ D}
π
= − max(An ) = An π, π −
.
n+1
A Note on a Paper of H. Alzer
and S. Koumandos
Proof of (B0 ) and (C0 ). We have
Bn (x, y) =
=
n
X
cos k|x − y| − cos k(x + y)
k=1
n Z
X
Z
x+y
2
Z
|x−y|
2
x+y
2
=
=
n
1X
A∗n (x, y) =
2 k=1
Then we get (B0 ) and (C0 ).
Z
|x−y|
2
x+y
x−y
2k
Title Page
x+y
2
|x−y|
2
k=1
Kunyang Wang
sin 2kt dt
Contents
JJ
J
cos t − cos(2n + 1)t
dt
2 sin t
Go Back
sin(n + 1)t sin nt
dt,
sin t
1
cos k −
2
Z
x+y
2
t dt =
x−y
2
II
I
Close
Quit
sin 2nt
dt.
2 sin t
Page 6 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
http://jipam.vu.edu.au
References
[1] H. ALZER AND S. KOUMANDOS, Sharp inequalities for trigonometric
sums in two variables, Illinois J. Math., 48(3) (2004), 887–907.
A Note on a Paper of H. Alzer
and S. Koumandos
Kunyang Wang
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 7
J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006
http://jipam.vu.edu.au