# A New Part-Metric-Related Inequality Chain and an Application

Authors: Xiaofan Yang; Fangkuan Sun; Yuan Yan Tang
Journal: Discrete Dynamics in Nature and Society
 (2008)

---

## Abstract

Part-metric-related (PMR) inequality chains are elegant and are useful in the study of difference equations. In this paper, we establish a new PMR inequality chain, which is then applied to show the global asymptotic stability of a class of rational difference equations.


---

## Body

## 1. Introduction

A  part-metric related (PMR) inequality
chain is a chain of inequalities of the form(1.1)
min
1
≤
i
≤
n
{
a
i
,
1
a
i
}
≤
f
(
a
1
,
…
,
a
n
)
≤
max
⁡
1
≤
i
≤
n
{
a
i
,
1
a
i
}
,
which
is closely related to the well-known part metric [1] and has important
applications in the study of difference equations [2–13]. Below are three
previously known PMR inequality chains:
(1.2)
min
⁡
1
≤
i
≤
4
{
a
i
,
1
a
i
}
≤
a
1
+
a
2
+
a
3
a
4
a
1
a
2
+
a
3
+
a
4
≤
max
⁡
1
≤
i
≤
4
{
a
i
,
1
a
i
}

(
s
e
e
⁢
[
5
]
)
,
(1.3)
min
⁡
1
≤
i
≤
k
{
a
i
,
1
a
i
}
≤
a
1
+
⋯
+
a
k
−
2
+
a
k
−
1
a
k
a
1
a
2
+
a
3
+
⋯
+
a
k
≤
max
⁡
1
≤
i
≤
k
{
a
i
,
1
a
i
}

(
s
e
e
⁢
[
11
]
)
,
(1.4)
min
⁡
1
≤
i
≤
5
{
a
i
,
1
a
i
}
≤
(
1
+
w
)
a
1
a
2
a
3
+
a
4
+
a
5
a
1
a
2
+
a
1
a
3
+
a
2
a
3
+
w
a
4
a
5
≤
max
⁡
1
≤
i
≤
5
{
a
i
,
1
a
i
}
,

1
≤
w
≤
2

(
s
e
e
⁢
[
13
]
)
.
In this article, we establish the following PMR inequality chain:
(1.5)
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
≤
h
w
(
a
1
,
…
,
a
2
p
−
1
)
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
,
where h
w
will be defined in the next section, p
−
2
≤
w
≤
p
−
1. When p
=
3, chain (1.5) reduces to chain (1.4). On this basis, we prove that the
difference equation
(1.6)
x
n
=
h
w
(
x
n
−
2
p
+
1
,
…
,
x
n
−
1
)
,

n
=
1
,
2
,
…
,
with positive initial conditions admits
a globally asymptotically stable equilibrium c
=
1.


## 2. Main Results

This section establishes the main results of
this paper. For a functionf
(
a
1
,
…
,
a
2
p
−
1
),
let
(2.1)
f
(
a
1
,
…
,
a
2
p
−
1
)
|
i
1
~
i
r
=
f
(
a
1
,
…
,
a
2
p
−
1
)
|
a
i
j
=
m
,
1
≤
j
≤
r
.Lemma 2.1.
Leta
1
,
…
,
a
n
,
b
1
,
…
,
b
n
>
0. Then min
1
≤
i
≤
n
{
a
i
/
b
i
}
≤
(
a
1
+
⋯
+
a
n
)
/
(
b
1
+
⋯
+
b
n
)
≤
max
1
≤
i
≤
n
{
a
i
/
b
i
}.
One equality in the chain holds if and only if a
1
/
b
1
=
⋯
=
a
n
/
b
n.Forp
≥
3 and w
>
0, define a function h
w
:
(
ℜ
+
)
2
p
−
1
→
ℜ
+ as follows:
(2.2)
h
w
(
a
1
,
…
,
a
2
p
−
1
)
=
(
1
+
w
)
∏
i
=
1
p
a
i
+
∏
i
=
p
+
1
2
p
−
1
a
i
×
∑
i
=
p
+
1
2
p
−
1
(
1
/
a
i
)
∏
i
=
1
p
a
i
×
∑
i
=
1
p
(
1
/
a
i
)
+
w
∏
i
=
p
+
1
2
p
−
1
a
i
.
Below are two examples of this function:
(2.3)
h
w
(
a
1
,
…
,
a
5
)
=
(
1
+
w
)
a
1
a
2
a
3
+
a
4
+
a
5
a
1
a
2
+
a
1
a
3
+
a
2
a
3
+
w
a
4
a
5
,
h
w
(
a
1
,
…
,
a
7
)
=
(
1
+
w
)
a
1
a
2
a
3
a
4
+
a
5
a
6
+
a
5
a
7
+
a
6
a
7
a
1
a
2
a
3
+
a
1
a
2
a
4
+
a
1
a
3
a
4
+
a
2
a
3
a
4
+
w
a
5
a
6
a
7
.For brevity, leth
w
=
h
w
(
a
1
,
…
,
a
2
p
−
1
).
Note that, for each a
r, h
w
is linear fractional in a
r. As a consequence, h
w is monotone in a
r. Through simple
calculations, we get the following two lemmas.Lemma 2.2.
Letp
≥
3, a
1
,
…
,
a
2
p
−
1
>
0, m
=
min
1
≤
i
≤
2
p
−
1
{
a
i
}, 1
≤
r
≤
p.
(1)
Ifh
p
−
2 is
increasing in ar, then h
p
−
2
≤
(
p
−
1
)
/
∑
i
=
1
,
i
≠
r
p
(
1
/
a
i
).
The equality holds if and only if  h
p
−
2 is constant
in a
r.
(2)
Ifh
p
−
2 is
strictly decreasing in a
r,
then h
p
−
2
≤
h
p
−
2
|
a
r
=
m.
The equality holds if and only if a
r
=
m.Lemma 2.3.
Letp
≥
3, a
1
,
…
,
a
2
p
−
1
>
0, m
=
min
1
≤
i
≤
2
p
−
1
{
a
i
}, p
+
1
≤
r
≤
2
p
−
1.
(1)
Ifh
p
−
2 is
increasing in a
r, then h
p
−
2
≤
∑
i
=
p
+
1
,
i
≠
r
2
p
−
1
(
1
/
a
i
)
/
(
p
−
2
).
The equality holds if and only if h
p
−
2 is constant
in a
r.
(2)
Ifh
p
−
2 is
strictly decreasing in a
r,
then h
p
−
2
≤
h
p
−
2
|
a
r
=
m.
The equality holds if and only if a
r
=
m.Theorem 2.4.
Letp
≥
3, a
1
,
…
,
a
2
p
−
1
>
0.
Then min
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
≤
h
p
−
2
≤
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}.
One of the two equalities holds if and only if a
1
=
⋯
=
a
2
p
−
1
=
1.Proof.
Letm
=
min
1
≤
i
≤
2
p
−
1
{
a
i
}, M
=
max
1
≤
i
≤
2
p
−
1
{
a
i
}.
We prove onlyh
p
−
2
≤
max
{
M
,
1
/
m
} because min
{
M
,
1
/
m
}
≤
h
p
−
2 can be proved similarly. We proceed by
distinguishing two possible cases.
Case 1.
There is a permutationi
1
,
…
,
i
2
p
−
1 of 1
,
2
,
…
,
2
p
−
1 such that for each 1
≤
k
≤
2
p
−
1, either a
i
k
=
m or h
p
−
2
|
i
1
~
i
k
−
1 is strictly decreasing in a
i
k.
Then(2.4)
h
p
−
2
≤
h
p
−
2
|
i
1
≤
⋯
≤
h
p
−
2
|
i
1
~
i
2
p
−
1
=
1
2
(
m
+
1
m
)
≤
max
⁡
{
m
,
1
m
}
≤
max
⁡
{
M
,
1
m
}
.
Case 2.
There is a partial permutationi
1
,
…
,
i
r of 1
,
2
,
…
,
2
p
−
1 (1
≤
r
≤
2
p
−
2) such that (a) for each 1
≤
k
≤
r,
either a
i
k
=
m or h
p
−
2
|
i
1
~
i
k
−
1 is strictly decreasing in a
i
k,
and (b) for each t
∈
{
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}, a
i
t
≠
m and h
p
−
2
|
i
1
~
i
r is increasing in a
t.
Then
(2.5)
m
<
M
,


h
p
−
2
≤
h
p
−
2
|
i
1
≤
h
p
−
2
|
i
1
~
i
2
≤
⋯
≤
h
p
−
2
|
i
1
~
i
r
.
Sincer
≤
2
p
−
2,
there is t
∈
{
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}.
If t
∈
{
1
,
…
,
p
}
−
{
i
1
,
…
,
i
r
},
it  follows from (2.5) and Lemma 2.2 that
(2.6)
h
p
−
2
≤
h
p
−
2
|
i
1
~
i
r
≤
(
p
−
1
)
∑
i
=
1
,
i
≠
t
p
(
1
/
a
i
)
|
i
1
~
i
r
≤
max
⁡
1
≤
i
≤
p
,
i
≠
t
{
a
i
}
|
i
1
~
i
r
≤
M
≤
max
⁡
{
M
,
1
m
}
.
Whereas if t
∈
{
p
+
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
},
it follows from (2.5) and Lemma 2.3 that
(2.7)
h
p
−
2
≤
h
p
−
2
|
i
1
~
i
r
≤
∑
i
=
p
+
1
,
i
≠
t
2
p
−
1
(
1
/
a
i
)
(
p
−
2
)
|
i
1
~
i
r
≤
max
⁡
p
+
1
≤
i
≤
2
p
−
1
,
i
≠
t
{
1
a
i
}
|
i
1
~
i
r
≤
1
m
≤
max
⁡
{
M
,
1
m
}
.
Hence, h
p
−
2
≤
max
{
M
,
1
/
m
} is proven.
Second, we prove thata
1
=
⋯
=
a
2
p
−
1
=
1 if h
p
−
2
=
max
{
M
,
1
/
m
}.
The claim of “a
1
=
⋯
=
a
2
p
−
1
=
1 if h
p
−
2
=
min
{
M
,
1
/
m
}” can be treated similarly. To this end, we need
to prove the following.Claim 1.
Ifh
p
−
2
=
max
{
M
,
1
/
m
},
then there is a permutation i
1
,
…
,
i
2
p
−
1 of 1
,
…
,
2
p
−
1 such that for each 1
≤
k
≤
2
p
−
1, either a
i
k
=
m or h
p
−
2
|
i
1
~
i
k
−
1 is strictly decreasing in a
i
k.Proof of Claim1.
On the
contrary, assume that
Claim1 is not true. Then there is a partial permutation i
1
,
…
,
i
r of 1
,
2
,
…
,
2
p
−
1 (1
≤
r
≤
2
p
−
2) such that (a) for each 1
≤
k
≤
r,
either a
i
k
=
m or h
p
−
2
|
i
1
~
i
k
−
1 is strictly decreasing in a
i
k,
and (b) for each t
∈
{
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}, a
i
t
≠
m and h
p
−
2
|
i
1
~
i
r is increasing in a
t.
One of the following two cases must occur.
Case 1.
There ist
∈
{
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
} such that h
p
−
2
|
i
1
~
i
r is strictly increasing in a
t. If t
∈
{
1
,
…
,
p
}
−
{
i
1
,
…
,
i
r
},
it follows by (2.5), (2.6), and Lemma 2.2 that
(2.8)
h
p
−
2
≤
h
p
−
2
|
i
1
~
i
r
<
(
p
−
1
)
∑
i
=
1
,
i
≠
t
p
(
1
/
a
i
)
|
i
1
~
i
r
≤
max
⁡
1
≤
i
≤
p
,
i
≠
t
{
a
i
}
|
i
1
~
i
r
≤
max
⁡
{
M
,
1
m
}
.
A contradiction occurs. Whereas if t
∈
{
p
+
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
},
it follows by (2.5), (2.7), and Lemma 2.3 that
(2.9)
h
p
−
2
≤
h
p
−
2
|
i
1
~
i
r
<
∑
i
=
p
+
1
,
i
≠
t
2
p
−
1
(
1
/
a
i
)
(
p
−
2
)
|
i
1
~
i
r
≤
max
⁡
p
+
1
≤
i
≤
2
p
−
1
,
i
≠
t
{
1
a
i
}
|
i
1
~
i
r
≤
max
⁡
{
M
,
1
m
}
.
Again a contradiction occurs.
Case 2.
For eacht
∈
{
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}, h
p
−
2
|
i
1
~
i
r is constant in a
t.
First, let us show that{
1
,
…
,
p
}
⊆
{
i
1
,
…
,
i
r
}.
Otherwise, there is t
∈
{
1
,
…
,
p
}
−
{
i
1
,
…
,
i
r
}.
By Lemma 2.2, we have
(2.10)
h
p
−
2
|
i
1
~
i
r
=
(
p
−
1
)
∑
i
=
1
,
i
≠
t
p
(
1
/
a
i
)
|
i
1
~
i
r
.If there is s
∈
{
1
,
…
,
p
}
−
{
i
1
,
…
,
i
r
,
t
},
it follows from (2.10) that h
p
−
2
|
i
1
~
i
r is strictly increasing in a
s,
a contradiction occurs. So, {
1
,
…
,
p
}
−
{
i
1
,
…
,
i
r
}
=
{
t
} and thus
(2.11)
max
{
M
,
1
m
}
=
h
p
−
2
≤
h
p
−
2
|
i
1
~
i
r
=
h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
|
a
i
1
=
⋯
=
a
i
r
=
m
=
m
<
M
,
from which a contradiction follows.
So, {
1
,
…
,
p
}
⊆
{
i
1
,
…
,
i
r
}.According to the previous argument, there ist
∈
{
p
+
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}.
By Lemma 2.3, we get
(2.12)
h
p
−
2
|
i
1
~
i
r
=
∑
i
=
p
+
1
,
i
≠
t
2
p
−
1
(
1
/
a
i
)
(
p
−
2
)
|
i
1
~
i
r
.
If there is s
∈
{
p
+
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
,
t
},
it follows from (2.12) that h
p
−
2
|
i
1
~
i
r is strictly decreasing in a
s,
a contradiction. So, {
p
+
1
,
…
,
2
p
−
1
}
−
{
i
1
,
…
,
i
r
}
=
{
t
} and thus
(2.13)
a
1
=
⋯
=
a
t
−
1
=
a
t
+
1
=
⋯
=
a
2
p
−
1
=
m
.
By (2.13) and (2.2), we get
(2.14)
h
p
−
2
=
h
p
−
2
|
i
1
~
i
r
=
(
p
−
1
)
m
3
+
m
+
(
p
−
2
)
a
t
p
m
2
+
(
p
−
2
)
m
a
t
.
Since h
p
−
2
|
i
1
~
i
r is constant in a
t,
and (
d/
d
a
t
)
h
p
−
2
|
i
1
~
i
r
=
(
(
p
−
1
)
(
p
−
2
)
m
2
(
1
−
m
2
)
)
/
[
p
m
2
+
(
p
−
2
)
m
a
t
]
2,
we derive m
=
1. From (2.12) and (2.13),
we get h
p
−
2
|
i
1
~
i
r
=
1
/
m.
Since h
p
−
2
=
max
{
M
,
1
/
m
},
all equalities in chains (2.5) and (2.7) hold. These plus m
=
1 yield h
p
−
2
|
i
1
~
i
r
=
1
/
m
=
1
≥
M,
from which we derive M
=
m
=
1. So, a
t
=
1
=
m.
This is a contradiction. Claim 1 is proved.By Claim1 and h
p
−
2
=
max
{
M
,
1
/
m
},
all equalities in (2.4) must hold. This plus Lemma 2.2 yields a
1
=
⋯
=
a
2
p
−
1
=
m and h
p
−
2
(
m
,
…
,
m
)
=
(
m
+
1
/
m
)
/
2
=
m.
This implies a
1
=
⋯
=
a
2
p
−
1
=
1.Theorem 2.5.
Letp
≥
3, a
1
,
…
,
a
2
p
−
1
>
0.
Then, min
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
≤
h
p
−
1
≤
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}.
One of the two equalities holds if and only if a
1
=
⋯
=
a
p
=
1
/
a
p
+
1
=
⋯
=
1
/
a
2
p
−
1.Proof.
By Lemma2.1 and (2.2), we get
(2.15)
h
p
−
1
≤
max
⁡
{
a
1
,
…
,
a
p
,
1
a
p
+
1
,
…
,
1
a
2
p
−
1
}
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
,
h
p
−
1
≥
min
⁡
{
a
1
,
…
,
a
p
,
1
a
p
+
1
,
…
,
1
a
2
p
−
1
}
≥
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.
The second claim follows immediately
from Lemma 2.1.
We are ready to present the main result of this paper.Theorem 2.6.
Letp
≥
3, p
−
2
≤
w
≤
p
−
1, a
1
,
…
,
a
2
p
−
1
>
0.
Let
(2.16)
a
k
=
h
w
(
a
k
−
2
p
+
1
,
…
,
a
k
−
1
)
,

k
=
2
p
,
2
p
+
1
,
….
Then  (2.16)
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
≤
a
k
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
},  (2.16)
k
=
2
p
,
2
p
+
1
,
….
If (2.16)
k
≥
2
p
+
1, then one of the two equalities holds if and only if (2.16)
a
1
=
⋯
=
a
2
p
−
1
=
1.Proof 4
Regard (2.16)h
w
as a linear fractional function in w,
which is monotone in w. By Theorems
and , we obtain
(2.16)
a
2
p
≥
min
⁡
{
h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
,
h
p
−
1
(
a
1
,
…
,
a
2
p
−
1
)
}
≥
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
,
a
2
p
≤
max
⁡
{
h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
,
h
p
−
1
(
a
1
,
…
,
a
2
p
−
1
)
}
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
,
a
2
p
+
1
≥
min
⁡
{
h
p
−
2
(
a
2
,
…
,
a
2
p
)
,
h
p
−
1
(
a
2
,
…
,
a
2
p
)
}
≥
min
⁡
2
≤
i
≤
2
p
{
a
i
,
1
a
i
}
≥
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
,
a
2
p
+
1
≤
max
⁡
{
h
p
−
2
(
a
2
,
…
,
a
2
p
)
,
h
p
−
1
(
a
2
,
…
,
a
2
p
)
}
≤
max
⁡
2
≤
i
≤
2
p
{
a
i
,
1
a
i
}
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.
Working inductively, we conclude that
for k
=
2
p, 2
p
+
1
,
…,
(2.18)
a
k
≥
min
⁡
{
h
p
−
2
(
a
k
−
2
p
+
1
,
…
,
a
k
−
1
)
,
h
p
−
1
(
a
k
−
2
p
+
1
,
…
,
a
k
−
1
)
}
≥
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
},
(2.18)
a
k
≤
max
⁡
{
h
p
−
2
(
a
k
−
2
p
+
1
,
…
,
a
k
−
1
)
,
h
p
−
1
(
a
k
−
2
p
+
1
,
…
,
a
k
−
1
)
}
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.Claim 2.
Ifa
2
p
+
1
=
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
,
then a
1
=
⋯
=
a
2
p
−
1
=
1.Proof of Claim2.
By (2.19),
we get
(2.20)
a
2
p
+
1
=
max
⁡
{
h
p
−
2
(
a
2
,
…
,
a
2
p
)
,
h
p
−
1
(
a
2
,
…
,
a
2
p
)
}
=
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.
Here, we encounter two possible cases.
Case 1.
a
2
p
+
1
=
h
p
−
2
(
a
2
,
…
,
a
2
p
)
=
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}.
By Theorem 2.4, we get a
2
=
⋯
=
a
2
p
=
1 and,
hence, a
2
p
+
1
=
1.
Then 1
=
a
2
p
+
1
=
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
=
max
{
a
1
,
1
/
a
1
},
implying a
1
=
1.
Case 2.
a
2
p
+
1
=
h
p
−
1
(
a
2
,
…
,
a
2
p
)
=
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}.
By Theorem 2.5, we get
(2.21)
a
2
=
⋯
=
a
p
+
1
=
1
a
p
+
2
=
⋯
=
1
a
2
p
,
and consequently,
(2.22)
a
2
p
+
1
=
h
p
−
1
(
a
2
,
…
,
a
2
p
)
=
a
2
.
Then,
(2.23)
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
=
a
2
p
+
1
=
1
a
2
p
≤
1
min
⁡
{
h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
,
h
p
−
1
(
a
1
,
…
,
a
2
p
−
1
)
}
≤
max
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.
Hence, all equalities in this chain
hold. In particular, we have
(2.24)
min
⁡
{
h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
,
h
p
−
1
(
a
1
,
…
,
a
2
p
−
1
)
}
=
min
⁡
1
≤
i
≤
2
p
−
1
{
a
i
,
1
a
i
}
.
If h
p
−
2
(
a
1
,
…
,
a
2
p
−
1
)
=
min
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
},
it follows from Theorem 2.4 that a
1
=
⋯
=
a
2
p
−
1
=
1.
Now, assume that h
p
−
1
(
a
1
,
…
,
a
2
p
−
1
)
=
min
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}.
By Theorem 2.5, we get
(2.25)
a
1
=
⋯
=
a
p
=
1
a
p
+
1
=
⋯
=
1
a
2
p
−
1
.
Equations (2.21) and (2.25) imply that a
1
=
⋯
=
a
2
p
−
1
=
1.
Claim 2 is proven.By Claim2 and working inductively, we get that if  a
k
=
max
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}
for somek
≥
2
p
+
1, then a
1
=
⋯
=
a
2
p
−
1
=
1.Similarly, we can show thata
1
=
⋯
=
a
2
p
−
1
=
1 if  a
k
=
min
1
≤
i
≤
2
p
−
1
{
a
i
,
1
/
a
i
}  holds for some k
≥
2
p
+
1.As an application of Theorem2.6, we
have the following theorem.Theorem 2.7.
Letp
≥
3, p
−
2
≤
w
≤
p
−
1. The difference equation
(2.26)
x
n
=
h
w
(
x
n
−
2
p
+
1
,
…
,
x
n
−
1
)
,

n
=
1
,
2
,
...
,
with positive initial conditions
admits the globally asymptotically stable equilibrium c = 1.The proof of this theorem is similar to those in [11, 13], and hence is
omitted.



---

*Source: 193872-2008-02-25.xml*