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volume 7, issue 2, article 46, 2006.

Received 12 July, 2005;

accepted 21 January, 2006.

Communicated by:A.G. Babenko

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Journal of Inequalities in Pure and Applied Mathematics

A NOTE ON A PAPER OF H. ALZER AND S. KOUMANDOS

KUNYANG WANG

School of Mathematical Science Beijing Normal University EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 208-05

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006

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Abstract

In the paper “ Sharp inequalities for trigonometric sums in two variables,” (Illi- nois Journal of Mathematics, Vol. 48, No.3, (2004), 887–907) Alzer and Kouman- dos investigated some special trigonometric sums. One of them is the sum

An(x, y) :=

n

X

k=1

cos(k−12)xsin(k−12)y

k−12 .

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:

An(x, y) =

n

X

k=1

coskxsinky

k , An(x, y) =

n

X

k=1

cos k− 12

xsin k−12 y

k− 12 ,

Bn(x, y) =

n

X

k=1

sinkxsinky

k .

Their results can be restated as follows:

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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(A) |An(x, y)|<sup{An(x, y) : x, y ∈[0, π], n∈N}=Rπ 0

sint t dt;

(B) min{Bn(x, y) : x, y ∈[0, π], n∈N}=−18, Bn(x, y) = −1

8 ⇐⇒n= 2and(x, y) = 5π

6 ,π 6

or(x, y) = π

6,5π 6

; (C) −23(√

2−1)≤An(x, y)≤2, An(x, y) = −2

3(√

2−1)⇐⇒n= 2,(x, y) = 3π

4 ,π 4

, An(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 functionf onD:= [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)

max(An) = An

0, π n+ 1

=

Z 2(n+1)π

0

2 cos(n+ 1)tsinnt sint dt, min(An) = −max(An) =An

π, π− π n+ 1

, max(An)< lim

n→∞max(An) = Z π

0

sint t dt.

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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(B0) Forn≥2

min(Bn) = Bn

(2n+ 1)π n(n+ 1) , π

n(n+ 1)

= Z πn

π n+1

sin(n+ 1)tsinnt

sint dt <min(Bn+1),

n→∞lim min(Bn) = 0.

(C0) For alln

max(An) = An 0,π

n

= Z 2nπ

0

sin 2nt sint dt

>max(An+1)→ Z π

0

sint

t dt, (n → ∞), and forn≥2

min(An) = An

2n, π 2n

= Z πn

π 2n

sin 2nt

2 sint dt <min(An+1)→0 (n→ ∞).

In particular,min(A2) = 23(1−√ 2).

The results (A), (B), (C) are easy consequences of (A0), (B0) and (C0) respec- tively.

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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Proof of (A0). We have An(x, y) =

n

X

k=1

sink(x+y)−sink(x−y) 2k

=

n

X

k=1

1 2k

Z k(x+y)

k(x−y)

cost dt

=

n

X

k=1

Z x+y2

x−y 2

cos 2kt dt

= Z x+y2

x−y 2

n

X

k=1

2 cos 2ktsint 2 sint

= Z x+y2

x−y 2

sin(2n+ 1)t−sint 2 sint dt=

Z x+y2

x−y 2

cos(n+ 1)tsinnt sint dt.

Then we get

max(An) =An

0, π n+ 1

=

Z 2(n+1)π

−π 2(n+1)

cos(n+ 1)tsinnt sint dt

=

Z 2(n+1)π

0

2 cos(n+ 1)tsinnt

sint dt

<

Z 2(n+1)π

0

2 cos(n+ 1)tsin(n+ 1)t

t dt=

Z π

0

sint t dt;

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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J. Ineq. Pure and Appl. Math. 7(2) Art. 46, 2006

http://jipam.vu.edu.au n→∞lim max(An) =

Z π

0

sint t dt;

min(An) = min{An(π−x, π−y) : (x, y)∈D}

=−max(An) = An

π, π− π n+ 1

.

Proof of (B0) and (C0). We have Bn(x, y) =

n

X

k=1

cosk|x−y| −cosk(x+y) 2k

=

n

X

k=1

Z x+y2

|x−y|

2

sin 2kt dt

= Z x+y2

|x−y|

2

cost−cos(2n+ 1)t 2 sint dt

= Z x+y2

|x−y|

2

sin(n+ 1)tsinnt sint dt, An(x, y) = 1

2

n

X

k=1

Z x+y

x−y

cos

k− 1 2

t dt=

Z x+y2

x−y 2

sin 2nt 2 sint dt.

Then we get (B0) and (C0).

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A Note on a Paper of H. Alzer and S. Koumandos

Kunyang Wang

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References

[1] H. ALZER AND S. KOUMANDOS, Sharp inequalities for trigonometric sums in two variables, Illinois J. Math., 48(3) (2004), 887–907.

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