A Generalization on Product Degree Distance of Strong Product of Graphs

International J.Math. Combin. Vol.2(2018), 67-79

A Generalization on Product Degree Distance of Strong Product of Graphs K.Pattabiraman (Department of Mathematics, Annamalai University, Annamalainagar 608 002, India) E-mail: pramank@gmail.com

Abstract: In this paper, the exact formulae for the generalized product degree distance, reciprocal product degree distance and product degree distance of strong product of a connected graph and the complete multipartite graph with partite sets of sizes m0 , m1 , · · · , mr−1 are obtained.

Key Words: Reciprocal product degree distance, product degree distance, strong product.

AMS(2010): 05C12, 05C76

§1. Introduction All the graphs considered in this paper are simple and connected. For vertices u, v ∈ V (G), the distance between u and v in G, denoted by dG (u, v), is the length of a shortest (u, v)-path in G and let dG (v) be the degree of a vertex v ∈ V (G). The strong product of graphs G and H, denoted by G ⊠ H, is the graph with vertex set V (G) × V (H) = {(u, v) : u ∈ V (G), v ∈ V (H)} and (u, x)(v, y) is an edge whenever (i) u = v and xy ∈ E(H), or (ii) uv ∈ E(G) and x = y, or (iii) uv ∈ E(G) and xy ∈ E(H).

A topological index of a graph is a real number related to the graph; it does not depend on labeling or pictorial representation of a graph. In theoretical chemistry, molecular structure descriptors (also called topological indices) are used for modeling physicochemical, pharmacologic, toxicologic, biological and other properties of chemical compounds [12]. There exist several types of such indices, especially those based on vertex and edge distances. One of the most intensively studied topological indices is the Wiener index.

Let G be a connected graph. Then W iener index of G is defined as W (G) =

1 2

X

dG (u, v)

u, v ∈ V (G)

with the summation going over all pairs of distinct vertices of G. This definition can be further

1 Received

September 9, 2017, Accepted May 20, 2018.

68

K.Pattabiraman

generalized in the following way: Wλ (G) =

1 2

X

dλG (u, v),

u, v ∈ V (G)

where dλG (u, v) = (dG (u, v))λ and λ is a real number [13, 14]. If λ = −1, then W−1 (G) = H(G), where H(G) is Harary index of G. In the chemical literature also W 12 [29] as well as the general case Wλ were examined [10, 15].

Dobrynin and Kochetova [6] and Gutman [11] independently proposed a vertex-degreeweighted version of Wiener index called degree distance, which is defined for a connected graph G as 1 X DD(G) = (dG (u) + dG (v))dG (u, v), 2 u,v∈V (G)

where dG (u) is the degree of the vertex u in G. Similarly, the product degree distance or Gutman index of a connected graph G is defined as DD∗ (G) =

1 2

X

dG (u)dG (v)dG (u, v).

u,v∈V (G)

The additively weighted Harary index(HA ) or reciprocal degree distance(RDD) is defined in [3] as 1 X (dG (u) + dG (v)) HA (G) = RDD(G) = .

2 dG (u, v) u,v∈V (G)

Similarly, Su et al. [28] introduce the reciprocal product degree distance of graphs, which can be seen as a product-degree-weight version of Harary index RDD∗ (G) =

1 2

X

u,v∈V (G)

dG (u)dG (v) . dG (u, v)

In [16], Hamzeh et al. recently introduced generalized degree distance of graphs. Hua and Zhang [18] have obtained lower and upper bounds for the reciprocal degree distance of graph in terms of other graph invariants. Pattabiraman et al. [22, 23] have obtained the reciprocal degree distance of join, tensor product, strong product and wreath product of two connected graphs in terms of other graph invariants. The chemical applications and mathematical properties of the reciprocal degree distance are well studied in [3, 20, 27].

The generalized degree distance, denoted by Hλ (G), is defined as Hλ (G) =

1 2

X

(dG (u) + dG (v))dλG (u, v),

u,v∈V (G)

where λ is a real number. If λ = 1, then Hλ (G) = DD(G) and if λ = −1, then Hλ (G) =

A Generalization on Product Degree Distance of Strong Product of Graphs

69

RDD(G). Similarly, generalized product degree distance, denoted by Hλ∗ (G), is defined as Hλ∗ (G) =

1 2

X

dG (u)dG (v)dλG (u, v).

u,v∈V (G)

If λ = 1, then Hλ∗ (G) = DD∗ (G) and if λ = −1, then Hλ∗ (G) = RDD∗ (G). Therefore the study of the above topological indices are important and we try to obtain the results related to these indices. The generalized degree distance of unicyclic and bicyclic graphs are studied

by Hamzeh et al. [16, 17]. Also they are given the generalized degree distance of Cartesian product, join, symmetric difference, composition and disjunction of two graphs. The generalized degree distance and generalized product degree distance of some classes of graphs are obtained in [24, 25, 26]. In this paper, the exact formulae for the generalized product degree distance, reciprocal product degree distance and product degree distance of strong product G ⊠ Km0 , m1 , ··· , mr−1 , where Km0 , m1 , ··· , mr−1 is the complete multipartite graph with partite sets of sizes m0 , m1 , · · · , mr−1 are obtained.

The first Zagreb index is defined as M1 (G) =

X

dG (u)2

u∈V (G)

and the second Zagreb index is defined as M2 (G) =

X

dG (u)dG (v).

uv∈E(G)

In fact, one can rewrite the first Zagreb index as M1 (G) =

X

(dG (u) + dG (v)).

uv∈E(G)

The Zagreb indices were found to be successful in chemical and physico-chemical applications, especially in QSPR/QSAR studies, see [8, 9].

For S ⊆ V (G), hSi denotes the subgraph of G induced by S. For two subsets S, T ⊂ V (G), not necessarily disjoint, by dG (S, T ), we mean the sum of the distances in G from each vertex P of S to every vertex of T, that is, dG (S, T ) = dG (s, t). s ∈ S, t ∈ T

§2. Generalized Product Degree Distance of Strong Product of Graphs

In this section, we obtain the Generalized product degree distance of G ⊠ Km0 , m1 , ··· , mr−1 . Let G be a simple connected graph with V (G) = {v0 , v1 , · · · , vn−1 } and let Km0 , m1 , ··· , mr−1 , r ≥ 2, be the complete multiparite graph with partite sets V0 , V1 , · · · , Vr−1 and let |Vi | = mi , 0 ≤ i ≤ r − 1. In the graph G ⊠ Km0 , m1 , ··· , mr−1 , let Bij = vi × Vj , vi ∈ V (G) and 0 ≤ j ≤ r − 1.

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K.Pattabiraman

For our convenience, the vertex set of G ⊠ Km0 , m1 , ··· , mr−1 is written as

V (G) × V (Km0 , m1 , ..., mr−1 ) =

Let B = {Bij }i = 0,1,··· , n−1 . Let Xi = j = 0,1,··· , r−1

r−1 S

r−1 n−1 [

Bij .

i=0 j =0

Bij and Yj =

j =0

n−1 S

Bij ; we call Xi and Yj as layer

i=0

and column of G ⊠ Km0 , m1 , ..., mr−1 , respectively. If we denote V (Bij ) = {xi1 , xi2 , · · · , ximj } and V (Bkp ) = {xk1 , xk2 , · · · , xk mp }, then xiℓ and xkℓ , 1 ≤ ℓ ≤ j, are called the corresponding S vertices of Bij and Bkp . Further, if vi vk ∈ E(G), then the induced subgraph hBij Bkp i of G ⊠ Km0 , m1 , ··· , mr−1 is isomorphic to K|Vj ||Vp | or, mp independent edges joining the corresponding vertices of Bij and Bkj according as j 6= p or j = p, respectively.

The following remark is follows from the structure of the graph Km0 , m1 , ··· , mr−1 .

Remark 2.1 Let n0 and q be the number of vertices and edges of Km0 , m1 , ··· , mr−1 . Then the sums r−1 X

mj mp

= 2q,

j, p = 0 j 6= p r−1 X

m2j

= n20 − 2q,

j=0

r−1 X

j, p = 0 j 6= p

m2j mp = n0 q − 3t r−1 X

=

r−1 X j=0

mj m2p ,

j, p = 0 j 6= p

m3j

= n30 − 3n0 q + 3t

j=0

and

r−1 X

m4j = n40 − 4n20 q + 2q 2 + 4n0 t − 4τ, ′

where t and τ are the number of triangles and K4s in Km0 , m1 , ··· , mr−1 .

The proof of the following lemma follows easily from the properties and structure of G ⊠ Km0 , m1 , ··· , mr−1 .

Lemma 2.2 Let G be a connected graph and let Bij , Bkp ∈ B of the graph G′ = G ⊠ Km0 , m1 , ··· , mr−1 , where r ≥ 2. Then (i) If vi vk ∈ E(G) and xit ∈ Bij , xkℓ ∈ Bkj , then  1, if t = ℓ, dG′ (xit , xkℓ ) = 2, if t = 6 ℓ,

A Generalization on Product Degree Distance of Strong Product of Graphs

71

and if xit ∈ Bij , xkℓ ∈ Bkp , j 6= p, then dG′ (xit , xkℓ ) = 1.

(ii) If vi vk ∈ / E(G), then for any two vertices xit ∈ Bij , xkℓ ∈ Bkp , dG′ (xit , xkℓ ) = dG (vi , vk ). (iii) For any two distinct vertices in Bij , their distance is 2.

The proof of the following lemma follows easily from Lemma 2.2, which is used in the proof of the main theorems of this section.

Lemma 2.3 Let G be a connected graph and let Bij , Bkp ∈ B of the graph G′ = G ⊠ Km0 , m1 , ..., mr−1 , where r ≥ 2. (i) If vi vk ∈ E(G), then  m m , if j 6= p, j p dH (B , B ) = ′ ij kp G  mj (mj +1) , if j = p, 2

(ii) If vi vk ∈ / E(G), then

dH G′ (Bij , Bkp ) =

 

mj mp dG (vi ,vk ) , 2  mj , dG (vi ,vk )

if j 6= p, if j = p.

 m m , if j 6= p, j p (iii) dH (B , B ) = ′ ij ip G  mj (mj −1) , if j = p.

2

Lemma 2.4 Let G be a connected graph and let Bij in G′ = G ⊠ Km0 , m1 , ··· , mr−1 . Then the degree of a vertex (vi , uj ) ∈ Bij in G′ is dG′ ((vi , uj )) = dG (vi ) + (n0 − mj ) + dG (vi )(n0 − mj ), where n0 =

r−1 P

mj .

j=0

Now we obtain the generalized product degree distance of G ⊠ Km0 , m1 , ··· , mr−1 .

Theorem 2.5 Let G be a connected graph with n vertices and m edges. Then Hλ∗ (G ⊠ Km0 , m1 , ··· , mr−1 )

n = (4q 2 + n20 + 4n0 q)Hλ∗ (G) + 4q 2 Wλ (G) + (4q 2 + 2n0 q)Hλ (G) + (4q 2 − n0 q − 3t) 2 i M1 (G) h 2 + 4n0 q − 2q 2 + 4n0 t + 9t + 7n0 q − n0 − 3n20 − 2n30 + 8τ 2 h i +m 3n0 q + 2n0 t − 2q 2 − 3t − 4q + 4τ h i +2λ M1 (G)(2q 2 − 2n0 t − 6t − 2q − 4τ ) + m(2q 2 − 2n0 t − n0 q − 3t − 4τ ) i

+(2λ − 1)M2 (G) 2q 2 − 2n0 t − 3n30 + 10n0 q + n20 − 18t − 6q − n0 − 4τ .

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K.Pattabiraman

Proof Let G′ = G ⊠ Km0 , m1 , ..., mr−1 . Clearly, Hλ∗ (G′ )

1 2

=

1 2

=

+

X

dG′ (Bij )dG′ (Bkp )dλG′ (Bij , Bkp )

Bij , Bkp ∈ B n−1 r−1 X X

dG′ (Bij )dG′ (Bip )dλG′ (Bij , Bip )

i = 0 j, p = 0 j 6= p n−1 X

r−1 X

dG′ (Bij )dG′ (Bkj )dλG′ (Bij , Bkj )

i, k = 0 j = 0 i 6= k

+

n−1 X

r−1 X

dG′ (Bij )dG′ (Bkp )dλG′ (Bij , Bkp )

i, k = 0 j, p = 0 i 6= k j 6= p

+

n−1 r−1 XX

!

dG′ (Bij )dG′ (Bij )dλG′ (Bij , Bij )

i=0 j =0

.

(2.1)

We shall obtain the sums of (2.1) are separately.

First we calculate A1 =

n−1 P r−1 P

i = 0 j, p = 0 j 6= p

By Lemma 2.4, we have T1′

dG′ (Bij )dG′ (Bip )dλG′ (Bij , Bip ). For that first we find T1′ .

= dG′ (Bij )dG′ (Bip )

= dG (vi )(n0 − mj + 1) + (n0 − mj ) dG (vi )(n0 − mp + 1) + (n0 − mp )

= (n0 + 1)2 − (n0 + 1)mj − (n0 + 1)mp + mj mp d2G (vi )

+ 2n0 (n0 + 1) − (2n0 + 1)mj − (2n0 + 1)mp + 2mj mp dG (vi )

+ n20 − n0 mp − n0 mj + mj mp .

From Lemma 2.3, we have dλG′ (Bij , Bip ) = mj mp . Thus T1′ dλG′ (Bij , Bip ) = T1′ mj mp

= (n0 + 1)2 mj mp − (n0 + 1)m2j mp − (n0 + 1)mj m2p + m2j m2p d2G (vi )

+ 2n0 (n0 + 1)mj mp − (2n0 + 1)m2j mp − (2n0 + 1)mj m2p + 2m2j m2p dG (vi )

+ n20 mj mp − n0 m2j mp − n0 mj m2p + m2j m2p .

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73

By Remark 2.1, we have T1

r−1 X

=

T1′ dλG′ (Bij , Bip )

j, p = 0 j 6= p

=

2q 2 + 2qn0 + 2n0 t + 2q + 4τ + 6t d2G (vi )

+ 2qn0 + 4n0 t − 4q 2 + 6t + 8τ dG (vi )

+ 2n0 t + 2q 2 + 4τ .

From the definition of the first Zagreb index, we have A1

=

n−1 X

T1

i=0

=

Next we obtain A2 =

2q 2 + 2qn0 + 2n0 t + 2q + 4τ + 6t M1 (G)

+2m 2qn0 + 4n0 t − 4q 2 + 6t + 8τ

+n 2n0 t + 2q 2 + 4τ .

n−1 P

r−1 P

i, k = 0 j = 0 i 6= k

By Lemma 2.4, we have T2′

= = =

dG′ (Bij )dG′ (Bkj )dλG′ (Bij , Bkj ). For that first we find T2′ .

dG′ (Bij )dG′ (Bkj )

dG (vi )(n0 − mj + 1) + (n0 − mj ) dG (vk )(n0 − mj + 1) + (n0 − mj )

(n0 − mj + 1)2 dG (vi )dG (vk ) + (n0 − mj )(n0 − mj + 1)(dG (vi ) + dG (vk ))

+(n0 − mj )2 .

Thus A2

=

r−1 n−1 X X

T2′ dλG′ (Bij , Bkj )

j = 0 i, k = 0 i 6= k

=

r−1 X

j =0

n−1 X

i, k = 0 i 6= k vi vk ∈E(G)

T2′ dλG′ (Bij , Bkj ) +

r−1 X

j =0

n−1 X

i, k = 0 i 6= k vi vk ∈E(G) /

T2′ dλG′ (Bij , Bkj )

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K.Pattabiraman

By Lemma 2.3, we have A2

r−1 X

=

=

=

n−1 X

j=0

i, k = 0 i 6= k vi vk ∈E(G)

r−1 X

n−1 X

j=0

T2′

j=0

i, k = 0 i 6= k vi vk ∈E(G)

r−1 X

n−1 X

j=0

r−1

X T2′ 1 − 2λ + 2λ mj mj +

i, k = 0 i 6= k vi vk ∈E(G)

n−1 X

T2′ m2j dλG (vi , vk ),

i, k = 0 i 6= k vi vk ∈E(G) /

r−1

X 1 − 2λ + 2λ mj mj + m2j − m2j +

j =0

n−1 X

T2′ m2j dλG (vi , vk )

i, k = 0 i 6= k vi vk ∈E(G) /

r−1 n−1

X X T2′ m2j dλG (vi , vk ) T2′ (2λ − 1) m2j − mj + j = 0 i, k = 0 i 6= k

= S1 + S2 ,

(2.2)

where S1 and S2 are the sums of the terms of the above expression, in order.

Now we calculate S1 . For that first we find the following.

h (2λ − 1)T2′ m2j − mj = (2λ − 1) m4j − (2n0 + 3)m3j + (n20 + 4n0 + 3)m2j

−(n0 + 1)2 mj dG (vi )dG (vk )

+ m4j − (2n0 + 2)m3j + (n20 + 3n0 + 1)m2j − (n20 + n0 )mj (dG (vi ) + dG (vk ))

i + m4j − (2n0 + 1)m3j + (n20 + 2n0 )m2j − n2o mj .

By Remark 2.1, we have T2′′

r−1 X

=

(2λ − 1)T2′ m2j − mj

j =0

=

h

(2λ − 1) 2q 2 − 2n0 t − 4τ − 3n30 + 10n0 q − 18t + n20 − 6q − n0 dG (vi )dG (vk )

+ 2q 2 − 4τ − 2n0 t − 6t − 2q (dG (vi ) + dG (vk ))

i + 2q 2 − 4τ − 2n0 t − n0 q − 3t .

Hence S1

=

n−1 X

T2′′

i, k = 0 i 6= k vi vk ∈E(G)

=

h

(2λ − 1) 2q 2 − 2n0 t − 4τ − 3n30 + 10n0 q − 18t + n20 − 6q − n0 2M2 (G)

+ 2q 2 − 4τ − 2n0 t − 6t − 2q 2M1 (G)

i +2m 2q 2 − 4τ − 2n0 t − n0 q − 3t .

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75

Next we calculate S2 . For that we need the following.

T2′ m2j

m4j − (2n0 + 2)m3j + (n0 + 1)2 m2j dG (vi )dG (vk )

+ m4j − (2n0 + 1)m3j + (n20 + n0 )m2j (dG (vi ) + dG (vk ))

+ m4j − 2n0 m3j + n20 m2j .

=

By Remark 2.1, we have T2

=

r−1 X

T2′ m2j

j =0

=

2q 2 − 4τ − 2n0 t − 6t + 2n0 q − 2q + n20 dG (vi )dG (vk )

+ 2q 2 − 4τ − 2n0 t − 3t + n0 q (dG (vi ) + dG (vk ))

+ 2q 2 − 4τ − 2n0 t .

From the definitions of Hλ∗ ,Hλ and Wλ , we obtain S2

n−1 X

=

T2 dλG (vi , vk )

i, k = 0 i 6= k

=

2 2q 2 − 4τ − 2n0 t − 6t + 2n0 q − 2q + n20 Hλ∗ (G)

+2 2q 2 − 4τ − 2n0 t − 3t + n0 q Hλ (G)

+2 2q 2 − 4τ − 2n0 t Wλ (G).

Now we calculate A3 =

n−1 P

r−1 P

i, k = 0 j, p = 0 i 6= k j 6= p

pute T3′ . By Lemma 2.4, we have T3′

dG′ (Bij )dG′ (Bkp )dλG′ (Bij , Bkp ). For that first we com-

= dG′ (Bij )dG′ (Bkp )

= dG (vi )(n0 − mj + 1) + (n0 − mj ) dG (vk )(n0 − mp + 1) + (n0 − mp )

= dG (vi )dG (vk )(n0 − mj + 1)(n0 − mp + 1) + dG (vi )(n0 − mj + 1)(n0 − mp ) +dG (vk )(n0 − mp + 1)(n0 − mj ) + (n0 − mj )(n0 − mp ).

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K.Pattabiraman

Since the distance between Bij and Bkp is mj mp dλG (vi , vk ). Thus T3′ mj mp

=

dG (vi )dG (vk ) (n20 + 2n0 + 1)mj mp − (n0 + 1)m2j mp − (n0 + 1)mj m2p + m2j m2p

+dG (vi ) (n20 + n0 )mj mp − (n0 + 1)mj m2p − n0 m2j mp + m2j m2p

+dG (vk ) (n20 + n0 )mj mp − n0 mj m2p − (n0 + 1)m2j mp + m2j m2p

+ n20 mj mp − n0 mj m2p − n0 m2j mp + m2j m2p .

By Remark 2.1, we obtain T3

=

r−1 X

j, p = 0, j 6= p

T3′ mj mp = dG (vi )dG (vk ) 2n0 q + 2n0 t + 2q + 2q 2 + 6t + 4τ

+(dG (vi ) + dG (vk )) qn0 + 2n0 t + 3t + 2q 2 + 4τ

+ 2n0 t + 2q 2 + 4τ .

Hence A3

=

n−1 X

i, k = 0 i 6= k

T3 dλG (vi , vk ) = 2Hλ∗ (G) 2n0 q + 2n0 t + 2q + 2q 2 + 6t + 4τ

+2Hλ (G) qn0 + 2n0 t + 3t + 2q 2 + 4τ

+2Wλ (G) 2n0 t + 2q 2 + 4τ .

Finally, we obtain A4 =

n−1 P r−1 P

i=0 j =0

T4′ . By Lemma 2.4, we have T4′

dG′ (Bij )dG′ (Bij )dλG′ (Bij , Bij ). For that first we calculate

= dG′ (Bij )dG′ (Bij )

2 = dG (vi )(n0 − mj + 1) + (n0 − mj )

= d2G (vi )(n0 − mj + 1)2 + 2dG (vi )(n0 − mj )(n0 − mj + 1) + (n0 − mj )2 .

From Lemma 2.3, the distance between (Bij and (Bij is mj (mj − 1). Thus

T4′ mj (mj − 1) = d2G (vi ) m4j − (2n0 + 3)m3j + ((n0 + 1)2 + 2)m2j − (n0 + 1)2 mj

+2dG (vi ) m4j − (2n0 + 2)m3j + (n20 + 3n0 + 1)m2j − (n20 + n0 )mj

+ m4j − (2n0 + 1)m3j + (n20 + 2n0 )m2j − n20 mj .

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77

By Remark 2.1, we obtain T4

=

r−1 X

j =0

T4′ mj (mj − 1)

= d2G (vi ) 4n20 q − 2n30 − 3n20 − 2n0 t + 5n0 q − 9t − 6q − n0 − 4τ

+2dG (vi ) 2q 2 − 2n0 t − 2q − 6t − 4τ

+ 2q 2 − 2n0 t − n0 q − 3t − 4τ .

Hence A4

=

n−1 X

T4 dλG′ (Bij , Bij )

i=0

= M1 (G) 4n20 q − 2n30 − 3n20 − 2n0 t + 5n0 q − 9t − 6q − n0 − 4τ

+4m 2q 2 − 2n0 t − 2q − 6t − 4τ

+n 2q 2 − 2n0 t − n0 q − 3t − 4τ .

Adding A1 ,S1 ,S2 ,A3 and A4 we get the required result.

2

If we set λ = 1 in Theorem 2.5, we obtain the product degree distance of G⊠Km0 , m1 , ··· , mr−1 .

Theorem 2.6 Let G be a connected graph with n vertices and m edges. Then DD∗ (G ⊠ Km0 , m1 , ··· , mr−1 ) = (4q 2 + n20 + 4n0 q)DD∗ (G) + 4q 2 W (G) n +(4q 2 + 2n0 q)DD(G) + (4q 2 − n0 q − 3t) 2 i M1 (G) h 2 2 + 4n0 q + 6q − 4n0 t − 15t + 7n0 q − n0 − 3n20 − 2n30 − 8τ h2 i +m n0 q − 2n0 t + 2q 2 − 9t − 4q − 4τ h i +M2 (G) 2q 2 − 2n0 t − 3n30 + 10n0 q + n20 − 18t − 6q − n0 − 4τ

for r ≥ 2.

Setting λ = −1 in Theorem 2.5, we obtain the reciprocal product degree distance of G ⊠ Km0 , m1 , ··· , mr−1 .

Theorem 2.7 Let G be a connected graph with n vertices and m edges. Then RDD∗ (G ⊠ Km0 , m1 , ··· , mr−1 ) = (4q 2 + n20 + 4n0 q)RDD∗ (G) + 4q 2 H(G) n +(4q 2 + 2n0 q)RDD(G) + (4q 2 − n0 q − 3t) 2

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K.Pattabiraman

i M1 (G) h 2 4n0 q + 2n0 t + 3t + 7n0 q − n0 − 3n20 − 2n30 − 2q + 4τ 2 h 5n q i 9t 0 +m + n0 t − q 2 − − 4q + 2τ 2 2 i M2 (G) h 2 − 2q − 2n0 t − 3n30 + 10n0 q + n20 − 18t − 6q − n0 − 4τ 2 +

for r ≥ 2.

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