Skip to main content

Advertisement

You are viewing the new article page. Let us know what you think. Return to old version

Research | Open | Published:

Sharp lower bounds involving circuit layout system

Abstract

The circuit layout system CLS { Γ n ( A ) , Γ N ( A ) , δ } E in a Euclidean space is defined. By means of algebraic,analytic, geometric and inequality theories, we obtain several sharp lowerbounds involving the circuit layout system.

MSC: 51K05, 26D15.

1 Introduction

We first introduce a circuit layout problem as follows. Let Γ be a rectangular(or polygon) courtyard (or street). Five light poles (or street lamp), with a fixedminimal distance apart from each other, are proposed to be erected on the boundaryof Γ, and straight underground pipes are planned to connect these poles (seeFigure 1). Assuming that the major cost of theconstruction project is the price of the pipes, it is then important to find out theminimal total lengths required for the project, its purpose is toestimate the installation costs.

Figure 1
figure1

The graph of CLS { Γ 4 ( A ) , Γ 5 ( A ) , 2 } R 2 .

We can easily illustrate this problem by means of Figure 1in a later Example 4.3 in which the corners of the courtyard Γ areindicated by the points A 1 , A 2 , A 3 and A 4 , while the light poles are indicated by A 1 , A 2 , A 3 , A 4 and A 5 , respectively. The light poles are kept apart fromeach other for clear reasons so that we may assume the distances

A 2 A 1 , A 3 A 2 , A 4 A 3 , A 5 A 4 , A 5 A 1 δ>0.

We need to find among all possible locations of A 1 ,, A 5 such that the total length

A 2 A 1 + A 3 A 2 + A 4 A 3 + A 5 A 4 + A 5 A 1

is the minimal one.

The above problem can easily be generalized. To this end, we need to recall somebasic concepts as follows.

Let be a Euclideanspace, and α,βE. The inner product of α andβ is denoted by α,β and the norm of α is denoted byα. The dimension dimE of satisfiesdimEn if and only if there exist n linearlyindependent vectors ε 1 , ε 2 ,, ε n in (see [1]).

Let B, C be points in , the closed, openand closed-open segments joining them will respectively be denoted by

[BC],(BC),[BC)and(BC]

and defined as usual by

{ χ B , C ( t ) t [ 0 , 1 ] } , { χ B , C ( t ) t ( 0 , 1 ) } , { χ B , C ( t ) | t [ 0 , 1 ) } and { χ B , C ( t ) t ( 0 , 1 ] } ,

where

χ B , C (t):=(1t)B+tC.

Let dimE2, A=( A 1 , A 2 ,, A n ) E n , where

A i A i + 1 ,i=1,2,,n,n3,

be a sequence of points in and

A i = A j ij(modn),i,j=0,±1,±2,.

We call the set

Γ n (A)= i = 1 n [ A i A i + 1 )

an n-polygon, or a polygon if no confusion is caused. The angleaat A i and the angle A are defined as

A i :=( A i A i 1 , A i + 1 A i ),i=1,2,,nandA:= min 1 i n { A i }.

In case each A i is the same, we say that our polygon is equiangular.We will also denote the total length (or perimeter) of an n-gon by

| Γ n (A)|= i = 1 n a i = i = 1 n A i + 1 A i ,

where, and in the future,

a i := A i + 1 A i ,i=1,2,,n.

Now we give the definition of the circuit layout system in a Euclidean space asfollows.

Definition 1.1 Let Γ n (A) and Γ N ( A ), where Nn3, be two polygons in with the dimensiondimE2. We say that the set

CLS { Γ n ( A ) , Γ N ( A ) , δ } E := { Γ n ( A ) , Γ N ( A ) , δ }

is a circuit layout system (or CLS for short) if the following conditions aresatisfied:

(H1.1) A i (0,π), i=1,2,,n.

(H1.2) A j Γ n (A) for j{1,2,,N} and A 1 [ A 1 A 2 ).

(H1.3) If A j , A j + 1 [ A i A i + 1 ), then A j + 1 ( A j A i + 1 ) for i=1,2,,n and j=1,2,,N.

(H1.4) If A j [ A i A i + 1 ) and A k [ A i + 1 A i + 2 ) for j,k{1,2,,N} and i{1,2,,n}, then j<k.

(H1.5) For any i{1,2,,n}, there exists j{1,2,,N} such that A j [ A i A i + 1 ).

(H1.6) For any j{1,2,,N}, there is δ>0 such that

A j + 1 A j δ.

In this paper, we are concerned with the sharp lower bound (see [25]) of | Γ N ( A )|, its purpose is to estimate the installation costs ofthe circuit layout problem. In other words, we will mainly be concerned with thefollowing problem.

Problem 1.1 (Circuit layout problem)

Let CLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS. How can we determine the lower bound of| Γ N ( A )| by means of n, N, δand Γ n (A)?

In this paper, by means of algebraic, analytic, geometric and inequality theories,several sharp lower bounds of | Γ N ( A )| in Problem 1.1 are obtained. As applications of ourresults, in Section 4, we calculate that inf{| Γ N ( A )|} for the special circuit layout systemCLS { Γ n ( A ) , Γ N ( A ) , δ } R 2 by means of three effective examples.

2 Preliminaries

We provide in this section some basic terminologies and results which are necessaryfor the investigation of Problem 1.1.

We first recall the concept of parallel vectors for later use. Two vectors xand y in are said to be inthe same (opposite) direction if (i) x=0 or y=0, or (ii) x0 and y0 and x is a positive (respectively negative)constant multiple of y. Two vectors x and y in the same(opposite) direction are indicated by xy (respectively xy) (see [1]).

Next, we set that

x= ( x 1 , x 2 , , x n ) T , x p = ( | x 1 | p + | x 2 | p + + | x n | p ) 1 / p ,

where p(1,).

In order to study Problem 1.1, we need six lemmas as follows.

Lemma 2.1 (Minkowski’s inequality [6])

Ifx,y R n andp(1,), then

x + y p x p + y p .
(1)

Furthermore, the equality holds if and only ifxy.

According to Lemma 2.1 and the algebraic theory, we easily get the followinglemma.

Lemma 2.2 (Minkowski-type inequality [6])

LetA R n × n . If A T =A, A0andf(x)= x T Ax, then for anyx,y R n , we have

f ( x + y ) f ( x ) + f ( y ) .
(2)

Furthermore, ifA>0, then the equality in (2) holds if andonly ifxy.

Lemma 2.3 Let the function φ:[0,c](0,) be defined by

φ(u)= ( c u ) 2 + y 2 2 ( c u ) y cos θ +u,

wherec,u,y[0,), θ(0,π). Then the function φ is nondecreasing. If in addition, y>0, then φ is increasing.

Proof Note that

φ u = ( c u ) cos θ y ( c u ) 2 + y 2 2 ( c u ) y cos θ + 1 = ( c u ) 2 + y 2 2 ( c u ) y cos θ [ ( c u ) cos θ y ] ( c u ) 2 + y 2 2 ( c u ) y cos θ ( c u ) 2 + y 2 2 ( c u ) y cos θ | ( c u ) cos θ y | ( c u ) 2 + y 2 2 ( c u ) y cos θ = [ ( c u ) 2 + y 2 2 ( c u ) y cos θ ] 1 / 2 sin 2 θ y 2 ( c u ) 2 + y 2 2 ( c u ) y cos θ + | ( c u ) cos θ y | 0 .

Thus, φ:[0,c](0,) is nondecreasing. Furthermore, ify>0, then φ:[0,c](0,) is a strictly increasing function. This ends theproof. □

Lemma 2.4 LetB,CE. IfBCandD[BC], then

CB=CD+DB.
(3)

The result of Lemma 2.4 is well known.

By our assumptions (H1.2)-(H1.5), we may easily get the following result.

Lemma 2.5 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS. Then, for anyi{1,2,,n}, there existσ(i){1,2,,N}andτ(i){0,1,2,,Nn}such that

A σ ( i ) + k [ A i A i + 1 ),k=0,1,2,,τ(i).

Furthermore,

i = 1 n τ(i)=Nn.
(4)

Lemma 2.6 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS. If the infimum of| Γ N ( A )|can be attained, then for any

i{1,2,,n},k { 1 , 2 , , τ ( i ) } ,τ(i)1,

we have

A σ ( i ) + k A σ ( i ) + k 1 =δ,
(5)

whereσ(i)andτ(i)are defined in Lemma  2.5.

Proof Suppose to the contrary that there exist i{1,2,,n} and k{1,2,,τ(i)} such that

A σ ( i ) + k A σ ( i ) + k 1 >δ.
(6)

We construct a new CLS { Γ n ( A ) , Γ N ( A ) , δ } E as follows: If

jσ(i)+k,k=0,1,,τ(i)1,

then A j = A j . If there exists k{0,1,,τ(i)1} such that j=σ(i)+k, then

A j [ A i A σ ( i ) + τ ( i ) )

and

A σ ( i ) + 1 A σ ( i ) = A σ ( i ) + 2 A σ ( i ) + 1 == A σ ( i ) + τ ( i ) A σ ( i ) + τ ( i ) 1 =δ.
(7)

Now fix A σ ( i ) + τ ( i ) [ A i A i + 1 ). Denote

( c i , u i , y i ):= ( A σ ( i ) + τ ( i ) A i , A σ ( i ) + τ ( i ) A σ ( i ) , A σ ( i 1 ) + τ ( i 1 ) A i ) .
(8)

Without loss of generality, we can assume that y i >0, i=1,2,,n. By condition (H1.3) and Lemma 2.4, we obtainthat

A σ ( i ) A i = c i u i , c i u i = A σ ( i ) + τ ( i ) A σ ( i ) = k = 1 τ ( i ) A σ ( i ) + k A σ ( i ) + k 1 .
(9)

Since

αβ= α 2 + β 2 2 α β cos ( α , β ) ,α,βE,

we see that

| Γ N ( A ) | = j = 1 n ( A σ ( j ) A σ ( j 1 ) + τ ( j 1 ) + A σ ( j ) + τ ( j ) A σ ( j ) ) = j = 1 n [ ( A σ ( j ) A i ) ( A σ ( j 1 ) + τ ( j 1 ) A j ) + A σ ( j ) + τ ( j ) A σ ( j ) ] = j = 1 n [ ( c j u j ) 2 + y j 2 2 ( c j u j ) y j cos A j + u j ] = j = 1 n φ j ( u j ) = 1 j n , j i φ j ( u j ) + φ i ( u i ) ,
(10)

where

φ j ( u j )= ( c j u j ) 2 + y j 2 2 ( c j u j ) y j cos A j + u j ,j=1,2,,n.

By condition (H1.6), (6) and (9), we see that

u i = k = 1 τ ( i ) A σ ( i ) + k A σ ( i ) + k 1 >τ(i)δ.
(11)

According to Lemma 2.3, the function φ i :[0, c i ](0,) is increasing. Thus, by (10) and (11), we have

| Γ N ( A ) |= 1 j n , j i φ j ( u j )+ φ i ( u i )> 1 j n , j i φ j ( u j )+ φ i ( τ ( i ) δ ) =| Γ N ( A ) |.

This is contrary to the minimality of | Γ N ( A )|. The proof is completed. □

3 Study of Problem 1.1

3.1 The case where n is an odd number

We first study the case of Problem 1.1 where n is an odd number. In thissituation we have the following result.

Theorem 3.1 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS and n is an odd number. Then we have the following inequality:

| Γ N ( A ) || Γ n (A)|sin A 2 + ( 1 sin A 2 ) (Nn)δ.
(12)

Proof We construct another CLS { Γ n ( A ) , Γ N ( A ) , δ } E such that (7) holds for anyi{1,2,,n}. Set

A σ ( j 1 ) + τ ( j 1 ) A j = y j , A σ ( j ) A j = x j .

By equality (7) and Lemma 2.4, we see that

y j + 1 = A j + 1 A σ ( j ) + τ ( j ) = ( A j + 1 A j ) ( A σ ( j ) + τ ( j ) A j ) = A j + 1 A j A σ ( j ) + τ ( j ) A j = a j x j τ ( j ) δ ,

therefore,

x j + y j + 1 = a j τ(j)δ,j{1,2,,n}.
(13)

Since

i = 1 n x i + 1 = i = 1 n x i , i = 1 n y i + 1 = i = 1 n y i ,

by (13) and (14) we obtain that

j = 1 n x j + j = 1 n y j = j = 1 n a j δ j = 1 n τ(j)=| Γ n (A)|δ(Nn).
(14)

In view of Lemma 2.6 and equality (10), we see that

| Γ N ( A ) | | Γ N ( A ) | = j = 1 n ( A σ ( j ) A σ ( j 1 ) + τ ( j 1 ) + A σ ( j ) + τ ( j ) A σ ( j ) ) = j = 1 n [ ( A σ ( j ) A i ) ( A σ ( j 1 ) + τ ( j 1 ) A j ) + A σ ( j ) + τ ( j ) A σ ( j ) ] = j = 1 n [ x j 2 + y j 2 2 x j y j cos A j + τ ( j ) δ ] j = 1 n [ x j 2 + y j 2 2 x j y j cos A + τ ( j ) δ ] = j = 1 n x j 2 + y j 2 2 x j y j cos A + j = 1 n τ ( j ) δ = j = 1 n x j 2 + y j 2 2 x j y j cos A + δ ( N n ) = j = 1 n f ( x j , y j ) T + δ ( N n ) ,

i.e.,

| Γ N ( A ) | j = 1 n f ( x j , y j ) T +δ(Nn),
(15)

where

f ( x j , y j ) T := x j 2 + y j 2 2 x j y j cosA=( x j , y j ) [ 1 cos A cos A 1 ] ( x j y j ) .

Note that condition (H1.1) implies 0<A<π, thus, where the matrix is positive definite.According to inequality (15), Lemma 2.2 and equality (14), we obtain that

| Γ N ( A ) | j = 1 n f ( x j , y j ) T + δ ( N n ) = 1 2 [ j = 1 n f ( x j , y j ) T + j = 1 n f ( x j , y j ) T ] + δ ( N n ) = 1 2 [ j = 1 n f ( x j , y j ) T + j = 1 n f ( y j , x j ) T ] + δ ( N n ) 1 2 f { j = 1 n [ ( x j , y j ) T + ( y j , x j ) T ] } + δ ( N n ) = 1 2 f { [ j = 1 n ( x j + y j , y j + x j ) T ] } + δ ( N n ) = 1 2 f { [ j = 1 n ( x j + y j ) , j = 1 n ( y j + x j ) ] T } + δ ( N n ) = 1 2 f [ ( j = 1 n x j + j = 1 n y j ) ( 1 , 1 ) T ] + δ ( N n ) = 1 2 ( j = 1 n x j + j = 1 n y j ) f ( 1 , 1 ) T + δ ( N n ) = 1 2 [ | Γ n ( A ) | δ ( N n ) ] 2 2 cos A + δ ( N n ) = [ | Γ n ( A ) | δ ( N n ) ] sin A 2 + δ ( N n ) = | Γ n ( A ) | sin A 2 + ( 1 sin A 2 ) ( N n ) δ .

This means that inequality (12) holds.

In addition, from the above analysis we may easily see that the equality in (12)holds if

(H3.1) Γ n (A) is an equiangular n-gon;

(H3.2) for any i{1,2,,n}, τ(i)1, equality (5) holds;

(H3.3) equality (4) holds; and

(H3.4) there exist x,y [ 0 , ) n such that

{ ( x j , y j ) T ( x 1 , y 1 ) T ( y j , x j ) T , j = 1 , 2 , , n , x j + y j + 1 = a j τ ( j ) δ , j = 1 , 2 , , n , x j 2 + y j 2 2 x j y j cos A δ , j = 1 , 2 , , n ,

where

( x i , y i ) T = ( x j , y j ) T ij(modn),i,j=0,±1,±2,.

The proof is completed. □

3.2 The case where n is an even number

Now, we consider the case of Problem 1.1 where n is an even number.

Theorem 3.2 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS, n is even, and let

i = 1 n ( 1 ) j + 1 a j 0.

Then we have the following two assertions:

  1. (I)

    If

    δ(Nn)> j = 1 n ( 1 ) j + 1 a j ,

then we have

| Γ N ( A ) | { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 min 2 { { ω } , 1 { ω } } } 1 / 2 + δ ( N n ) ,
(16)

where

ω= j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ ,{ω}=ω[ω][0,1),

and[ω]is the Gaussian function.

  1. (II)

    If

    δ(Nn) j = 1 n ( 1 ) j + 1 a j ,

then we have

| Γ N ( A ) | { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + cos 2 A 2 [ j = 1 n ( 1 ) j + 1 a j δ ( N n ) ] 2 } 1 / 2 + δ ( N n ) .
(17)

Proof First we consider the case where

δ(Nn)> j = 1 n ( 1 ) j + 1 a j .

We show that for any i{1,2,,n}, equality (7) holds by constructing a newCLS { Γ n ( A ) , Γ N ( A ) , δ } E . Since n is an even number, by the proofof Theorem 3.1 and (13), we see that

j = 1 n / 2 ( x 2 j 1 + y 2 j )= j = 1 n / 2 a 2 j 1 δ i = 1 n / 2 τ(2j1)
(18)

and

j = 1 n / 2 ( y 2 j 1 + x 2 j ) = ( y 1 + x 2 ) + ( y 3 + x 4 ) + + ( y n 1 + x n ) = ( x 2 + y 3 ) + ( x 4 + y 5 ) + + ( x n 2 + y n 1 ) + ( x n + y 1 ) = j = 1 n / 2 ( x 2 j + y 2 j + 1 ) = j = 1 n / 2 a 2 j δ i = 1 n / 2 τ ( 2 j ) .
(19)

Inequality (15) is still valid where

f ( x j , y j ) T := x j 2 + y j 2 2 x j y j cosA=( x j , y j ) [ 1 cos A cos A 1 ] ( x j y j )

is a positive definite quadratic function. By means of inequality (15),Lemma 2.2, (18), (19) and (4), we then obtain that

| Γ N ( A ) | | Γ N ( A ) | j = 1 n f ( x j , y j ) T + δ ( N n ) = j = 1 n / 2 f ( x 2 j 1 , y 2 j 1 ) T + j = 1 n / 2 f ( x 2 j , y 2 j ) T + δ ( N n ) = j = 1 n / 2 f ( x 2 j 1 , y 2 j 1 ) T + j = 1 n / 2 f ( y 2 j , x 2 j ) T + δ ( N n ) f [ j = 1 n / 2 ( x 2 j 1 , y 2 j 1 ) T + j = 1 n / 2 ( y 2 j , x 2 j ) T ] + δ ( N n ) = f [ j = 1 n / 2 ( x 2 j 1 + y 2 j ) , j = 1 n / 2 ( y 2 j 1 + x 2 j ) ] T + δ ( N n ) = f [ j = 1 n / 2 a 2 j 1 δ i = 1 n / 2 τ ( 2 j 1 ) , j = 1 n / 2 a 2 j δ i = 1 n / 2 τ ( 2 j ) ] T + δ ( N n ) = f [ P n δ τ , Q n δ ( N n τ ) ] T + δ ( N n ) ,

i.e.,

| Γ N ( A ) | f [ P n δ τ , Q n δ ( N n τ ) ] T +δ(Nn),
(20)

where

P n := j = 1 n / 2 a 2 j 1 (0,), Q n := j = 1 n / 2 a 2 j (0,), P n Q n

and

τ:= i = 1 n / 2 τ(2j1){0,1,2,,Nn}.

Note that

f [ P n δ τ , Q n δ ( N n τ ) ] T = ( P n δ τ ) 2 + [ Q n δ ( N n τ ) ] 2 2 ( P n δ τ ) [ Q n δ ( N n τ ) ] cos A = [ P n + Q n δ ( N n ) ] 2 2 ( 1 + cos A ) ( P n δ τ ) [ Q n δ ( N n τ ) ] = [ P n + Q n δ ( N n ) ] 2 4 cos 2 A 2 ( P n δ τ ) [ Q n δ ( N n τ ) ] = sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 [ j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ τ ] 2 = sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 ( { ω } + [ ω ] τ ) 2 ,

i.e.,

f [ P n δ τ , Q n δ ( N n τ ) ] T = sin 2 A 2 [ | Γ n | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 ( { ω } + [ ω ] τ ) 2 ,
(21)

where

ω= j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ .
(22)

Set

τ 0 = { [ ω ] if  0 { ω } 1 2 , [ ω ] + 1 if  1 2 < { ω } < 1 .
(23)

Since

δ(Nn)> j = 1 n ( 1 ) j + 1 a j ,τ{0,1,,Nn},

we see that

j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ (Nn){ω}+[ω]τ{ω}+[ω].
(24)

Thus, if

0{ω} 1 2 ,

then

f [ P n δ τ , Q n δ ( N n τ ) ] T sin 2 A 2 [ | Γ n | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 { ω } 2 ,
(25)

where equality holds if and only if τ=[ω]= τ 0 , and if

1 2 <{ω}<1,

then

f [ P n δ τ , Q n δ ( N n τ ) ] T sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 ( 1 { ω } ) 2 .
(26)

The equalities in (25) and (26) hold if and only if τ=[ω]+1= τ 0 . From (20), (21), (25) and (26), we see that

| Γ N ( A ) | f [ P n δ τ , Q n δ ( N n τ ) ] T + δ ( N n ) { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 min 2 { { ω } , 1 { ω } } } 1 / 2 + δ ( N n ) .

Thus, inequality (16) is proved.

Second, we consider the case

δ(Nn) j = 1 n ( 1 ) j + 1 a j .

Since

δ(Nn) j = 1 n ( 1 ) j + 1 a j ,τ{0,1,,Nn},

we see that

0 j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ (Nn){ω}+[ω]τ{ω}+[ω]
(27)

and

( { ω } + [ ω ] τ ) 2 [ j = 1 n ( 1 ) j + 1 a j + δ ( N n ) 2 δ ( N n ) ] 2 = 1 4 δ 2 [ j = 1 n ( 1 ) j + 1 a j δ ( N n ) ] 2 ,
(28)

where equality holds in (28) if and only if τ=Nn. By (20), (21) and (28), we have

| Γ N ( A ) | f [ P n δ τ , Q n δ ( N n τ ) ] T + δ ( N n ) { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + cos 2 A 2 [ j = 1 n ( 1 ) j + 1 a j δ ( N n ) ] 2 } 1 / 2 + δ ( N n ) .

Thus, inequality (16) is proved.

Finally, the conditions for the equality in (16) to hold are as follows:

(H3.5) The n-gon Γ n (A) is an equiangular n-gon.

(H3.6) Equality (5) holds for any i{1,2,,n}, τ(i)1.

(H3.7) Equality (4) holds.

(H3.8) i = 1 n / 2 τ(2j1)= τ 0 {0,1,2,,Nn}, where τ 0 is defined by (23).

(H3.9) There exist x,y [ 0 , ) n such that

{ ( x 2 j 1 , y 2 j 1 ) T ( x 1 , y 1 ) T ( y 2 j , x 2 j ) T , j = 1 , 2 , , n / 2 , x j + y j + 1 = a j τ ( j ) δ , j = 1 , 2 , , n , x j 2 + y j 2 2 x j y j cos A δ , j = 1 , 2 , , n .

While the conditions for the equality in (17) to hold are as follows:

(H3.10) The conditions (H3.5)-(H3.7) and (H3.9) hold.

(H3.11) i = 1 n / 2 τ(2j1)=Nn.

Here,

( x i , y i ) T = ( x j , y j ) T ij(modn),i,j=0,±1,±2,.

This completes the proof of this theorem. □

3.3 The case Γ n (A)is an equiangular n-gon

Equiangular polygon is a special kind of polygons. Regular polygon in R 2 is an equiangular polygon. If is a Euclideanspace with dimE2, then there is an equiangular 4-gon Γ 4 (A) in . Indeed, in , there exist atleast two linearly independent vectors α,β. Then, by the Gram-Schmidt orthogonalizationprocess, we may obtain two orthogonal unit vectors i,jE from α, β. If we set

A 1 =a ( i cos π 3 + j sin π 3 ) , A 2 =b ( i cos π 3 + j sin π 3 ) , A 3 =bi, A 4 =ai,

where a,b(0,), then the 4-gon Γ 4 (A) is an equiangular polygon in with

A 1 = A 2 = A 3 = A 4 = π 3 .

Similarly, if dimE3, then the 8-gon

Γ 8 (A):= Γ 8 (0,i,i+j,j,j+k,i+j+k,i+k,k)

is equiangular with

A 1 = A 2 == A 8 = π 2 ,

where i, j, k are three mutually orthogonal unit vectors in .

We now turn to the calculation of inf{| Γ N ( A )|}.

Theorem 3.3 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS with n odd, and let Γ n (A)be an equiangular n-gon. Suppose that there existτ(i){0,1,,Nn}for eachi=1,2,,n, such that:

(H3.12) i = 1 n τ(i)=Nn.

(H3.13) 1 2 j = 1 n [ ( 1 ) j + 1 ( a j τ(j)δ)] 1 2 δcsc A 2 .

(H3.14) ( 1 ) k { j = 1 k 1 [ ( 1 ) j + 1 ( a j τ(j)δ)] 1 2 j = 1 n [ ( 1 ) j + 1 ( a j τ(j)δ)]} 1 2 δcsc A 2 , wherek=2,3,,n.

Then

inf { | Γ N ( A ) | } =| Γ n (A)|sin A 2 + ( 1 sin A 2 ) (Nn)δ.
(29)

Proof By the assumptions in Theorem 3.3, conditions (H3.1) and(H3.3) hold. Since dimE2, there exists Γ N ( A )E such that condition (H3.2) holds. Thus, we justneed to consider condition (H3.4).

From

( x j , y j ) T ( x 1 , y 1 ) T ( y j , x j ) T ,j=1,2,,n,

we have

x j = y j ,j=1,2,,n.
(30)

By

x j + y j + 1 = a j τ(j)δ,j=1,2,,n,

and (30), we get

y j + y j + 1 = a j τ(j)δ,j=1,2,,n.
(31)

We can rewrite (31) as

( 1 ) j + 1 y j + 1 ( 1 ) j y j = ( 1 ) j + 1 ( a j τ ( j ) δ ) ,j=1,2,,n.
(32)

By (30), (32), y n + 1 = y 1 and n is odd number, we obtain that

j = 1 k [ ( 1 ) j + 1 y j + 1 ( 1 ) j y j ] = j = 1 k [ ( 1 ) j + 1 ( a j τ ( j ) δ ) ] , ( 1 ) k + 1 y k + 1 + y 1 = j = 1 k [ ( 1 ) j + 1 ( a j τ ( j ) δ ) ] , k = 1 , 2 , , n , x 1 = y 1 = 1 2 j = 1 n [ ( 1 ) j + 1 ( a j τ ( j ) δ ) ] ,
(33)
x k = y k = ( 1 ) k { j = 1 k 1 [ ( 1 ) j + 1 ( a j τ ( j ) δ ) ] 1 2 j = 1 n [ ( 1 ) j + 1 ( a j τ ( j ) δ ) ] } ,
(34)

where k=2,3,,n. By (30), x,y [ 0 , ) n and

x j 2 + y j 2 2 x j y j cos A δ,j=1,2,,n,

we have

x j = y j 1 2 δcsc A 2 ,j=1,2,,n.
(35)

According to the assumption in Theorem 3.3, and (33)-(35), condition (H3.4)holds. Consequently, by Theorem 3.1, Theorem 3.3 holds.

This completes the proof of Theorem 3.3. □

Theorem 3.4 LetCLS { Γ n ( A ) , Γ N ( A ) , δ } E be a CLS. Assume that Γ n (A)is an equiangular n-gon where n is an even number, and

j = 1 n ( 1 ) j + 1 a j 0.

Suppose that there existτ(i){0,1,,Nn}for eachi=1,2,,nand free variable y n (0,)such that:

(H3.15) Condition (H3.12) holds.

(H3.16) The following inequalities hold:

y j λ 2 ( 1 ) j + 1 + 1 λ ( 1 ) j + 1 cos A δ,j=1,2,,n,

where

( y 1 y 2 y 3 y n 2 y n 1 ) = [ λ 1 0 0 0 0 0 λ ( 1 ) j + 1 1 0 0 0 0 0 λ 1 1 0 0 0 0 0 λ ] 1 ( a 1 τ ( 1 ) δ a j τ ( j ) δ a n 2 τ ( j 2 ) δ a n 1 τ ( n 1 ) δ y n ) ,
(36)
λ= j = 1 n / 2 [ a 2 j 1 τ ( 2 j 1 ) δ ] j = 1 n / 2 [ a 2 j τ ( 2 j ) δ ] .
(37)

(H3.17)

j = 1 n / 2 τ(2j1)= { [ ω ] , δ ( N n ) > j = 1 n ( 1 ) j + 1 a j , 0 { ω } 1 2 , [ ω ] + 1 , δ ( N n ) > j = 1 n ( 1 ) j + 1 a j , 1 2 < { ω } < 1 , N n , δ ( N n ) j = 1 n ( 1 ) j + 1 a j ,

where ω is defined by (22).

Then we have the following two assertions:

  1. (I)

    If

    δ(Nn)> j = 1 n ( 1 ) j + 1 a j ,

then

inf { | Γ N ( A ) | } = { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + 4 δ 2 cos 2 A 2 min 2 { { ω } , 1 { ω } } } 1 / 2 + δ ( N n ) .
(38)
  1. (II)

    If

    δ(Nn) j = 1 n ( 1 ) j + 1 a j ,

then

inf { | Γ N ( A ) | } = { sin 2 A 2 [ | Γ n ( A ) | δ ( N n ) ] 2 + cos 2 A 2 [ j = 1 n ( 1 ) j + 1 a j δ ( N n ) ] 2 } 1 / 2 + δ ( N n ) .
(39)

Proof We first look for the conditions for equality in (16)-(17) tohold. The conditions are either (H3.5)-(H3.9) or (H3.10)-(H3.11). By theassumptions in Theorem 3.4 and dimE2, conditions (H3.5)-(H3.8) and (H3.11) hold. If(H3.5)-(H3.9) hold, then (H3.10) hold. Therefore we just need to show that(H3.9) holds.

Form (18)-(19) and

( x 2 j 1 , y 2 j 1 ) T ( x 1 , y 1 ) T ( y 2 j , x 2 j ) T ,j=1,2,, n 2 ,

we see that

x 2 j 1 y 2 j 1 = y 2 j x 2 j =λ,j=1,2,, n 2 ,
(40)

where

λ= j = 1 n / 2 ( x 2 j 1 + y 2 j ) j = 1 n / 2 ( y 2 j 1 + x 2 j ) = j = 1 n / 2 [ a 2 j 1 τ ( 2 j 1 ) δ ] j = 1 n / 2 [ a 2 j τ ( 2 j ) δ ] .
(41)

Consequently,

x j = λ ( 1 ) j + 1 y j ,j=1,2,,n.
(42)

By

x j + y j + 1 = a j τ(j)δ,j=1,2,,n,

and (42), we have

λ ( 1 ) j + 1 y j + y j + 1 = a j τ(j)δ,j=1,2,,n1
(43)

and

λ 1 y n + y 1 = a n τ(n)δ.
(44)

Equalities (40)-(43) imply that if j=n, then (43) holds, i.e., (44) holds. From(43) we have get (36), where y n are free variables. By (42),x,y [ 0 , ) n ,

x j 2 + y j 2 2 x j y j cos A δ,j=1,2,,n,

we have

y j λ 2 ( 1 ) j + 1 + 1 λ ( 1 ) j + 1 cos A δ,j=1,2,,n.
(45)

This means that condition (H3.9) can be deduced from conditions (H3.15)-(H3.17).Thus Theorem 3.4 holds by applying Theorem 3.2.

This completes the proof of Theorem 3.4. □

4 Three effective examples

For a general CLS { Γ n ( A ) , Γ N ( A ) , δ } E , the equalities in (12), (16) and (17) may not hold,this is most probably because conditions (H3.1)-(H3.4) or conditions (H3.5)-(H3.11)cannot be met at the same time. We will discuss Problem 1.1 of a special CLS in R 2 .

Example 4.1 Consider the CLS { Δ A B C , Δ A B C , δ } R 2 , where

A [BC), B [CA), C [AB),0<δ<min{a,b,c},

and a, b, c is the length of the sides of the triangleΔABC. We will calculate that inf{|Δ A B C |}.

By Theorem 3.1 we have

|Δ A B C ||ΔABC|sin min { A , B , C } 2 ,
(46)

where A, B, C are three inner angles of the triangleΔABC. According to conditions (H3.1)-(H3.4), the equalityin (46) holds if and only if ΔABC is a normal triangle, and A , B , C are the midpoints of line segment[BC], [CA], [AB], respectively. Consequently, the equality ininequality (12) does not hold in general.

It is well known that ΔABC is an acute triangle if and only if

[ A A ] [BC], [ B B ] [CA], [ C C ] [AB]

implies that |Δ A B C | takes the minimum. By this we see that

inf { | Δ A B C | } =acosA+bcosB+ccosC.
(47)

If π/2A<π and δ=0, then

B = C =A, [ A A ] [BC]

is necessary and sufficient for |Δ A B C | to take the minimum. By this we see that

inf { | Δ A B C | } =2 | A A | =2bsinC=2csinB.
(48)

Example 4.2 Consider the CLS { Γ 3 ( A ) , Γ 4 ( A ) , 1 } R 2 (see Figure 2), where

A 1 A 2 =6, A 2 A 3 =8, A 3 A 2 =10

and

A 1 [ A 1 A 2 ), A 2 [ A 2 A 3 ), A 3 , A 4 [ A 3 A 1 ).

We will calculate that inf{| Γ 4 ( A )|}.

Figure 2
figure2

The graph of CLS { Γ 3 ( A ) , Γ 4 ( A ) , 1 } R 2 .

Note that

cos A 1 = 3 5 ,cos A 2 =0,cos A 3 = 4 5 .

Without loss of generality, we may assume that A 4 A 3 =1. By Lemma 2.6, we have

| Γ 4 ( A ) | ψ ( x , y , z ) = x 2 + ( 6 z ) 2 + ( 8 x ) 2 + y 2 8 5 ( 8 x ) y + ( 9 y ) 2 + z 2 6 5 ( 9 y ) z + 1 ,

where (x,y,z)[0,6]×[0,9]×[0,6], and

{ x 2 + ( 6 z ) 2 1 , ( 8 x ) 2 + y 2 8 5 ( 8 x ) y 1 , ( 9 y ) 2 + z 2 6 5 ( 9 y ) z 1 .
(49)

By the help of Mathematica software, we obtain that

inf { | Γ 4 ( A ) | } = ψ ( 0.621684 , 5.90012 , 5.1705 ) = 10.600001492661695 .

We note that (49) and

ψ ( x , y , z ) x = ψ ( x , y , z ) y = ψ ( x , y , z ) z =0

hold if

(x,y,z)=(0.621684,5.90012,5.1705).

Example 4.3 Consider the CLS { Γ 4 ( A ) , Γ 5 ( A ) , 2 } R 2 (see Figure 1), where Γ 4 (A) is a rectangle, and

A 2 A 1 = A 4 A 3 =6, A 3 A 2 = A 1 A 4 =5

and

A 1 [ A 1 A 2 ), A 2 [ A 2 A 3 ), A 3 , A 4 [ A 3 A 4 ), A 5 [ A 4 A 1 ).

We will calculate that inf{| Γ 5 ( A )|}.

We may assume that A 4 A 3 =2. By Lemma 2.6, we have

| Γ 5 ( A ) | φ ( x , y , z , w ) = x 2 + ( 5 w ) 2 + ( 6 x ) 2 + y 2 + ( 5 y ) 2 + z 2 + ( 4 z ) 2 + w 2 + 2 ,

where (x,y,z,w)[0,6]×[0,5]×[0,4]×[0,5], and

{ x 2 + ( 5 w ) 2 2 , ( 6 x ) 2 + y 2 2 , ( 5 y ) 2 + z 2 2 , ( 4 z ) 2 + w 2 2 .
(50)

By the help of Mathematica software, we obtain that

inf { φ ( x , y , z , w ) } = φ ( 2.75296 , 3.24704 , 1.75296 , 2.24704 ) = 16.1421356237309 = 10 2 + 2 .
(51)

On the other hand, by Theorem 3.2, we have

| Γ 5 ( A ) | | Γ n ( A ) | sin A 2 + ( 1 sin A 2 ) ( N n ) δ = 10 2 + 2 .

It should be noted that the apparent error is caused by the computer. Therefore

inf { | Γ 5 ( A ) | } =10 2 +2.
(52)

We note that (51) holds if

(x,y,z,w)=(2.75296,3.24704,1.75296,2.24704).

We can also give another intuitive proof of equation (52) as follows.

By Theorem 3.2, the equality in (51) holds if conditions (H3.5)-(H3.11) hold. Infact, there exist

(x,y,z,w)[0,6]×[0,5]×[0,4]×[0,5],

such that (50) holds. From (H3.9) we get

x 5 w = z 5 y = 6 x y = 4 z w = x + z + ( 6 x ) + ( 4 z ) ( 5 w ) + ( 5 y ) + y + w =1,

i.e.,

{ x = 5 w , y = 1 + w , z = 4 w .
(53)

Combined with (50), (53), we get

2 w4,
(54)

where w is a free variable. This means that (51) holds if and only if(53)-(54) hold. This proves equation (52).

In addition, we can also prove (52) by Theorem 3.4.

Example 4.3 is a geometry problem. However, we can see this example as a circuitlayout problem of a family. In addition, this example also means that the equalitiesin (12), (16) and (17) can hold.

Remark 4.1 Since a Euclidean space is an abstract space, we will find theapplications of CLS in theoretical fields such as statistics (see [1, 7]), matrix theory (see [8]), geometry (see [6, 912]) and space science (see [1, 9]), etc.

Remark 4.2 A large number of theories of algebra, analysis, geometry, computer(see [9, 1214]) with inequality are used in this paper, which can be found in the latestliterature [1, 616].

Endnote

The angle between two nonzero vectors B and C is defined to be(B,C)=:arccos(B,C/BC)[0,π].

References

  1. 1.

    Wen JJ, Han TY, Cheng SS: Inequalities involving Dresher variance mean. J. Inequal. Appl. 2013., 2013: Article ID 366.http://www.journalofinequalitiesandapplications.com/content/2013/1/366

  2. 2.

    Matejc̆ka L: Sharp bounds for the weighted geometric mean of the first Seiffert andlogarithmic means in terms of weighted generalized Heronian mean. Abstr. Appl. Anal. 2013., 2013: Article ID 721539

  3. 3.

    Gao HY, Guo JL, Yu WG: Sharp bounds for power mean in terms of generalized Heronian mean. Abstr. Appl. Anal. 2011., 2011: Article ID 679201

  4. 4.

    Chu YM, Long BY: Bounds of the Neuman-Sándor mean using power and identric means. Abstr. Appl. Anal. 2013., 2013: Article ID 832591

  5. 5.

    Chu YM, Hou SW: Sharp bounds for Seiffert mean in terms of contraharmonic mean. Abstr. Appl. Anal. 2012., 2012: Article ID 425175

  6. 6.

    Gardner RJ: The Brunn-Minkowski inequality. Bull. Am. Math. Soc. 2002, 39: 355–405. 10.1090/S0273-0979-02-00941-2

  7. 7.

    Wen JJ, Zhang ZH: Jensen type inequalities involving homogeneous polynomials. J. Inequal. Appl. 2010., 2010: Article ID 850215 10.1155/2010/850215

  8. 8.

    Wen JJ, Wang WL: Chebyshev type inequalities involving permanents and their applications. Linear Algebra Appl. 2007, 422(1):295–303. 10.1016/j.laa.2006.10.014

  9. 9.

    Gao CB, Wen JJ: Theory of surround system and associated inequalities. Comput. Math. Appl. 2012, 63: 1621–1640. 10.1016/j.camwa.2012.03.037

  10. 10.

    Gao CB, Wen JJ: Several identities and inequalities involving Jordan closed curves. Appl. Math. E-Notes 2008, 8: 148–158.

  11. 11.

    Wen JJ, Ke R, Lu T: A class of geometric inequalities involving k -Brocard distance. Chin. Q. J. Math. 2006, 21(2):210–219.

  12. 12.

    Wen JJ, Wang WL: The inequalities involving generalized interpolation polynomial. Comput. Math. Appl. 2008, 56(4):1045–1058. 10.1016/j.camwa.2008.01.032

  13. 13.

    Wen JJ, Han TY, Gao CB: Convergence tests on constant Dirichlet series. Comput. Math. Appl. 2011, 62(9):3472–3489. 10.1016/j.camwa.2011.08.064

  14. 14.

    Wen JJ, Cheng SS: Closed balls for interpolating quasi-polynomials. Comput. Appl. Math. 2011, 30(3):545–570.

  15. 15.

    Pečarić JE, Wen JJ, Wang WL, Tao L: A generalization of Maclaurin’s inequalities and its applications. Math. Inequal. Appl. 2005, 8(4):583–598.

  16. 16.

    Wen JJ, Wang WL: The optimization for the inequalities of power means. J. Inequal. Appl. 2006., 2006: Article ID 46782 10.1155/JIA/2006/46782

Download references

Acknowledgements

This work was supported in part by the Natural Science Foundation of China (No.61309015) and in part by the Foundation of Scientific Research Project of FujianProvince Education Department of China (No. JK2012049). The authors are deeplyindebted to Professor Sui Sun Cheng, Tsing Hua University, Taiwan, for manyuseful comments and keen observations which led to the present improved versionof the paper as it stands.

Author information

Correspondence to Shanhe Wu.

Additional information

Competing interests

The authors declare that they have no conflicts of interest to this work.

Authors’ contributions

All authors contributed equally and significantly in this paper. All authors read andapproved the final manuscript.

Authors’ original submitted files for images

Authors’ original file for figure 1

Authors’ original file for figure 2

Rights and permissions

Reprints and Permissions

About this article

Keywords

  • circuit layout system
  • Euclidean space
  • Minkowski’s inequality