This paper studies the problem of estimating the spectral radiusof trees with the given number of vertices and maximum degree.We obtain the new upper bounds on the spectral radius of thetrees, and the results are the best upper bounds expressed by thenumber of vertices and maximum degree, at present.Let T = (V , E) be a tree on n vertices with maximum degree �,where 3 � � � n − 2. Denote by ρ(T ) the spectral radius of T .We prove that
(1) if n � 2�, then ρ(T ) �√
n−1+√
(n−2�)2+2n−32 , and equality
holds if and only if T is an almost completely full-degree treeof 3 levels;
(2) if 2� < n � �2 + 1, then ρ(T ) � √2� − 1, and equality holds
if and only if T is a completely full-degree tree of 3 levels;(3) if n > �2 + 1, then ρ(T ) < 2
Let T = (V , E) be a tree on n vertices with maximum degree �. Its adjacency matrix is definedto be the n × n matrix A(T ) = (aij), where aij = 1 if vi is adjacent to v j ; and aij = 0, otherwise. The
✩ This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural ScienceFoundation of Fujian Province (Z0511028).
2528 H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
characteristic polynomial of T is det(xI − A(T )) and is denoted by Φ(T , x). Since A(T ) is a symmetricmatrix, each of its eigenvalues is real. We assume, without loss of generality, that they are orderedin nonincreasing order, i.e., ρ1(T ) � ρ2(T ) � · · · � ρn(T ). We call them the eigenvalues of T . In par-ticular, the largest eigenvalue ρ1(T ) is called the spectral radius of T , denoted by ρ(T ). Since T is aconnected graph, A(T ) is irreducible, the spectral radius is simple and there is a unique positive uniteigenvector by the Perron–Frobenius Theorem, e.g. [1]. We shall refer to such an eigenvector as thePerron vector of T .
Let dv be the degree of v (v ∈ V ), � = max{dv : v ∈ V }. Let NT (v) denote the set of vertices adja-cent to v in T . A pendant vertex of T is a vertex of degree 1. Denote by ρ(A) the largest eigenvalueof the matrix A. The distance dist(v, u) from a vertex v to a vertex u is the length of the shortestpath joining v and u. We recall that the eccentricity eu of a vertex u is the largest distance from u toany other vertex of the graph. We recall some results on the upper bounds of the spectral radius oftrees.
In [2] Stevanovic proves that
ρ(T ) < 2√
� − 1 (1.1)
where T is a tree with maximum degree �.In [3] Oscar Rojo gives an improved upper bound on the spectral radius of a tree. Let T be a
tree with maximum degree �, u be a vertex of T such that du = �. Let k = eu + 1, where eu is theeccentricity of u. For j = 1,2, . . . ,k − 1, let δ j = max{dv : dist(v, u) = j}. Then
ρ(T ) < max{
max2� j�k−2
{√δ j − 1 + √δ j−1 − 1 },√δ1 − 1 + √
�}. (1.2)
In [4] Oscar Rojo gives another upper bound on the spectral radius of a tree. Let T be a tree withmaximum degree � and such that there exist two adjacent vertices u and v with du = dv = �. LetT ′ be the forest obtained from T by deleting the edge uv . Thus T ′ is the union of two disjoint treesTu = (V u, Eu) and T v = (V v , E v). Let ku = eu + 1, kv = ev + 1, where eu and ev are the eccentricitiesof u and v with respect to the trees Tu and T v respectively. Now, we define k = max{ku,kv }, γ j(u) =max{dx: x ∈ V u, dist(x, u) = j}, 1 � j � k − 2, γ j(v) = max{dy: y ∈ V v , dist(y, v) = j}, 1 � j � k − 2,and γ j = max{γ j(u), γ j(v)}, 1 � j � k − 2, where γ j(u) = 0 for j > ku − 1 or γ j(v) = 0 for j > kv − 1.Then
ρ(T ) < max{
max2� j�k−2
{√γ j − 1 + √γ j−1 − 1 },√γ1 − 1 + √
� − 1}. (1.3)
This paper studies the problem of estimating the spectral radius of trees with the given number ofvertices and maximum degree. When � = n − 1 or � = 2, the tree T is a star or a path, respectively,and it is easy to obtain the spectral radius of the trees. Thus in this paper we only study for 3 �� � n − 2. We obtain the new upper bounds on the spectral radius of trees, as the following twotheorems. The results are the best upper bounds expressed by the number of vertices and maximumdegree, at present.
Theorem 3.1. Let T = (V , E) be a tree on n vertices with maximum degree �, where 3 � � � n − 2. Denoteby ρ(T ) the spectral radius of T . We prove that
(1) if n � 2�, then ρ(T ) �√
n−1+√
(n−2�)2+2n−32 , and equality holds if and only if T is an almost completely
full-degree tree of 3 levels;(2) if 2� < n � �2 + 1, then ρ(T ) �
√2� − 1, and equality holds if and only if T is a completely full-degree
tree of 3 levels;(3) if n > �2 + 1, then ρ(T ) < 2
√� − 1 cos π
2k+1 , where k = �log�−1((�−2)(n−1)
�+ 1)� + 1.
Theorem 3.2. Let T = (V , E) be a tree on n vertices with maximum degree �, where � = 3 and n � � + 2.Let t be the cardinality of the vertices of degree 3. We prove that
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2529
Fig. 1. Completely full-degree tree.
Fig. 2. Almost completely full-degree tree.
(1) if t = 1, then ρ(T ) <√
4.5;(2) if t = 2, then ρ(T ) <
√5;
(3) if t = 3, then ρ(T ) < 4√
33 ;
(4) if t = 4, then ρ(T ) <√
2 + √13;
(5) if t � 5, then ρ(T ) < 2√
2 cos π2k+1 , where k = �log2
n+23 � + 1.
2. Some basic concepts and lemmas
First, we introduce some needed definitions and lemmas for getting the main results of this article.
Definition 2.1. Let T be a rooted tree on n vertices with maximum degree �, r be the root node of T .If u is a vertex of T and dist(r, u) = k − 1 (1 � k), then we call u is in the kth level, note that the rootvertex is in the first level. If v is a vertex of T and dist(r, v) = maxu∈V (T ) dist(r, u), then we call v isin the last level.
Definition 2.2. Let T be a rooted tree on n vertices with maximum degree �. If in the tree T , thedegrees of all the nodes in each level except for the last level are �. Then we call T is a completelyfull-degree tree.
Definition 2.3. Let T be a rooted tree on n vertices with maximum degree �. If T is a completelyfull-degree tree; or if in the tree T , the number of nodes in the last level has not reached the maximalvalue but only lacks some right nodes (nodes on the right-hand side), and the number of nodes ineach level except for the last level has reached the maximal value. Then we call T is an almostcompletely full-degree tree.
For example, Fig. 1 shows a completely full-degree tree with level 3 and degree 4, and Fig. 2 showsan almost completely full-degree tree with level 4 and degree 3.
Lemma 2.1. Let T = (V , E) be a tree on n vertices with maximum degree �, and let T ∗ be an almost completelyfull-degree tree on n vertices with maximum degree � corresponding to T . Then ρ(T ) � ρ(T ∗), and equalityholds if and only if T = T ∗ .
Proof. The proof is similar to Theorem 2.1 in [5]. �Lemma 2.2. Let T ∗ = (V ∗, E∗) be an almost completely full-degree tree on n vertices with maximum degree �.Let T ∗∗ be a tree defined as follows:
2530 H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
(1) if T ∗ is a completely full-degree tree, then T ∗ = T ∗∗;(2) if T ∗ is not completely full-degree tree, then T ∗∗ is the completely full-degree tree obtained from T ∗ by
supplying all the lacking vertices. Then ρ(T ∗) � ρ(T ∗∗).
Proof. The proof is trivial. �Lemma 2.3. (See [6].) Let T = (V , E) be a completely full-degree tree on n vertices with maximum degree �,and whose number of levels is k. Let dk− j+1 be the degree of vertices in the level j (1 � j � k), and ρ(T ) bethe spectral radius of T . If Rk is a symmetric tridiagonal matrix of order k × k, where
Rk =
⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
0√
d2 − 1 0 · · · · · · 0√
d2 − 1 0√
d3 − 1. . .
...
0√
d3 − 1. . .
. . .. . .
......
. . .. . .
. . .√
dk−1 − 1 0...
. . .√
dk−1 − 1 0√
dk0 · · · · · · 0
√dk 0
⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
k×k
,
then ρ(Rk) = ρ(T ).
Lemma 2.4. (See [7].) If b > 0, then the spectral radius of the m × m symmetric tridiagonal matrix
Dm(b) =
⎡⎢⎢⎢⎢⎢⎢⎢⎣
0 bb 0 b
b. . .
. . .. . . bb 0 b
b b
⎤⎥⎥⎥⎥⎥⎥⎥⎦
m×m
is ρ(Dm(b)) = 2b cos π2m+1 .
Lemma 2.5. (See [1].) Let A and B be both the nonnegative irreducible real symmetric matrices, and A � B,A �= B. Then ρ(A) < ρ(B).
Lemma 2.6. (See [8].) Let T = (V , E) be a tree, u ∈ V , du = 1, and v /∈ V . Denote by ρ(T ) the spectral radiusof T . If T1 = T + uv, then ρ(T ) < ρ(T1).
Lemma 2.7. (See [9].) Let T = (V , E) be a tree, u ∈ V , v ∈ V , du � 3 and dv � 3. Let w0 w1 w2 · · · wk+1 bethe only path from u to v, where w0 = u, wk+1 = v,k � 1, and dwi = 2 (1 � i � k). If T1 = T − wi−1 wi −wi wi+1 + wi−1 wi+1 (1 � i � k), then ρ(T ) � ρ(T1).
Lemma 2.8. (See [10,11].) Suppose that u is a vertex of a tree T . For positive integers k and l, let Tk,l be the treeobtained from T by adding two new paths Pk and Pl of lengths k and l at u. If k � l � 1, then ρ(Tk+1,l−1) <
ρ(Tk,l).
Lemma 2.9. (See [10,12].) Suppose that u and v are two adjacent vertices of the tree T , d(u) > 1,d(v) > 1.For nonnegative integers k and l, let T 1
k,l be the tree obtained from T by adding two new paths Pk and Pl of
lengths k and l at u and v, respectively. If k � l � 1, then ρ(T 1k+1,l−1) < ρ(T 1
k,l).
Lemma 2.10. (See [9,13].) Suppose that T ∗ is a tree and b is a pendant vertex of T ∗ . Let a be the vertex whichis adjacent to b, that is, NT ∗ (b) = a. Let Φ(T ∗, x), Φ(T ∗ − b, x), and Φ(T ∗ − b − a, x) be the characteristicpolynomials of T ∗ , T ∗ − b, and T ∗ − b − a, respectively, then Φ(T ∗, x) = xΦ(T ∗ − b, x) − Φ(T ∗ − b − a, x).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2531
Fig. 3. A tree T1.
Fig. 4. The star K1,s .
3. Main results and proofs
For the tree neither a path nor a star, we obtain a new upper bound on the spectral radius of thetree. As the following theorem.
Theorem 3.1. Let T = (V , E) be a tree on n vertices with maximum degree �, where 3 � � � n − 2. Denoteby ρ(T ) the spectral radius of T . We prove that
(1) if n � 2�, then ρ(T ) �√
n−1+√
(n−2�)2+2n−32 , and equality holds if and only if T is an almost completely
full-degree tree of 3 levels;(2) if 2� < n � �2 + 1, then ρ(T ) �
√2� − 1, and equality holds if and only if T is a completely full-degree
tree of 3 levels;(3) if n > �2 + 1, then ρ(T ) < 2
√� − 1 cos π
2k+1 , where k = �log�−1((�−2)(n−1)
�+ 1)� + 1.
Proof. (1) If � + 1 < n � 2�. Then by Lemma 2.1, we have ρ(T ) � ρ(T1), where T1 is an almostcompletely full-degree tree on n vertices with maximum degree � (Fig. 3). Let r be the root node,then dr = �. Let b1,b2, . . . ,bd be the pendent vertices of T1, and bi /∈ NT1 (r) (1 � i � d). Then d +� + 1 = n.
Denote by Φ(K1,s, x) the characteristic polynomial of the star K1,s (Fig. 4). By Lemma 2.10, wehave
Let Φ(T1, x) = 0. Then xd+�−3[x4 − (d + �)x2 + d(� − 1)] = 0. It is easy to prove thatρ(T ) > 1, so ρ(T ) is the largest root of the equation x4 − (d + �)x2 + d(� − 1) = 0. Thus,
x2 = (d+�)+√
(d+�)2−4d(�−1)2 .
Since d = n − 1 − �, we have x2 = n−1+√
(n−2�)2+2n−32 . So ρ(T1) =
√n−1+
√(n−2�)2+2n−3
2 .
Therefore, ρ(T ) �√
n−1+√
(n−2�)2+2n−32 , and equality holds if and only if T is an almost completely
full-degree tree of 3 levels.(2) If 2� < n � �2 + 1. Then by Lemma 2.1, we have ρ(T ) � ρ(T2), where T2 is an almost com-
pletely full-degree tree on n vertices with maximum degree �. Let T3 be a completely full-degreetree corresponding to T2 with maximum degree � and such that the number of levels in T3 is 3(may be T2 = T3). Then by Lemma 2.1 and Lemma 2.2, we have ρ(T ) � ρ(T2)� ρ(T3).
By Lemma 2.3, we have ρ(R3) = ρ(T3), where
R3 =⎡⎣ 0
√� − 1 0√
� − 1 0√
�
0√
� 0
⎤⎦ .
Then ρ(R3) = √2� − 1, that is, ρ(T3) = √
2� − 1. Thus, ρ(T ) �√
2� − 1.Therefore, ρ(T ) �
√2� − 1, and equality holds if and only if T is a completely full-degree tree
of 3 levels.(3) If n > �2 + 1. Then by Lemma 2.1, we have ρ(T ) � ρ(T4), where T4 is an almost completely
full-degree tree on n vertices with maximum degree �, and the number of levels in T4 is k. Let T5be a completely full-degree tree corresponding to T4 with maximum degree � and the level of T5 isalso k (may be T4 = T5). By Lemma 2.2, we have ρ(T4) � ρ(T5).
+ 1)� + 1. �Corollary 3.1. Let T = (V , E) be a tree on n vertices with maximum degree �, where � � 3. If n � � + 1,then ρ(T ) < 2
√� − 1 cos π
2k+1 , where k = �log�−1((�−2)(n−1)
�+ 1)� + 1.
Proof. The proof is similar to (3) in Theorem 3.1. �In the following, we give the upper bounds on the spectral radius of trees when � = 3. Since it
is easy to get the upper bounds on the spectral radius of trees when � = 3 and n is small, we onlystudy the upper bounds on the spectral radius of trees for � = 3 and n is large.
Theorem 3.2. Let T = (V , E) be a tree on n vertices with maximum degree �, where � = 3 and n � � + 2.Let t be the cardinality of the vertices of degree 3. Following results are established.
(1) if t = 1, then ρ(T ) <√
4.5;(2) if t = 2, then ρ(T ) <
√5;
(3) if t = 3, then ρ(T ) < 4√
33 ;
(4) if t = 4, then ρ(T ) <√
2 + √13;
(5) if t � 5, then ρ(T ) < 2√
2 cos π2k+1 , where k = �log2
n+23 � + 1.
Proof. Let T 3n be the set of the trees on n vertices with maximum degree 3. By Corollary 3.1, we
obtain that for ∀T ∈ T 3n , ρ(T ) < 2
√2 cos π
2k+1 , where k = �log2n+2
3 �+ 1. Obviously, Theorem 3.2 holdswhen ρ(T ) � 2. In the following, we assume that ρ(T ) > 2.
(1) If t = 1, then by Lemma 2.6, Lemma 2.7 and Lemma 2.8, for ∀T ∈ T 3n , there is a nature number
M1 such that ρ(T ) � ρ(Tm) for m � M1 and Tm is a tree as shown in Fig. 5. Denote ρ(Tm) by ρm .It is easy to prove that {ρm} (m � M1) is strictly monotone increasing and has a upper bound. Solimm→+∞ ρm exists. Let limm→+∞ ρm = α1, then ρm < α1.
Let x be the Perron vector of Tm , and xvi be a component of x corresponds to the vertex vi(1 � i � n). Then A(Tm)x = ρ(Tm)x. Therefore, we have ρmxvi = ∑
v ∈N (v ) xv j .
j Tm i
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2535
Fig. 6. A tree for t = 2.
It is easy to prove that xai = xbi = xci (i = 1,2, . . . ,m). Let xi = xai (i = 1,2, . . . ,m), xm+1 = xr . Thenρmxk = xk+1 + xk−1, k = 2,3, . . . ,m.
Let x0 = 0, then we have ρmxk = xk+1 + xk−1, k = 1,2, . . . ,m. Its characteristic equation is λ2 −ρmλ + 1 = 0. Then the characteristic roots are
λ1(m) = ρm +√
ρ2m − 4
2, λ2(m) = ρm −
√ρ2
m − 4
2.
So the general solution is
xk = C1λk1(m) + C2λ
k2(m) (k = 1,2, . . . ,m + 1),
where C1, C2 are constants, λ1(m)λ2(m) = 1, and λ1(m) > λ2(m).Since ρmxm+1 = 3xm , we have ρm = 3xm
xm+1. Then
α1 = limm→+∞
3xm
xm+1= 3 lim
m→+∞C1λ
m1 (m) + C2λ
m2 (m)
C1λm+11 (m) + C2λ
m+12 (m)
= 3 limm→+∞λ2(m).
For limm→+∞ λ2(m) = α1−√
α21−4
2 , we have α1 = 3α1−
√α2
1−4
2 . Then α1 = √4.5. So for ∀T ∈ T 3
n , ρ(T ) �ρ(Tm) <
√4.5.
However, for t = 1 and n � � + 2, we have k � 3. Then 2√
2 cos π2k+1 � 2
√2 cos π
7 ≈ 2.5483. Obvi-
ously, 2√
2 cos π2k+1 >
√4.5.
Therefore, if t = 1, then ρ(T ) <√
4.5.(2) If t = 2, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀T ∈ T 3
n , there is anature number M2 such that ρ(T ) � ρ(Tm) for m � M2 and Tm is a tree as shown in Fig. 6. Denoteρ(Tm) by ρm . It is easy to prove that {ρm} (m � M2) is strictly monotone increasing and has a upperbound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α2, then ρm < α2.
Similar to (1), we have ρmxvi = ∑v j∈NTm (vi)
xv j . It is easy to prove that xai = xbi = xci = xdi
(i = 1,2, . . . ,m), and xs = xt . Let xi = xai (i = 1,2, . . . ,m), xm+1 = xs . Then ρmxk = xk+1 + xk−1,k = 2,3, . . . ,m.
Let x0 = 0, then we have ρmxk = xk+1 + xk−1, k = 1,2, . . . ,m. Its characteristic equation is λ2 −ρmλ + 1 = 0. Then the characteristic roots are
λ1(m) = ρm +√
ρ2m − 4
2, λ2(m) = ρm −
√ρ2
m − 4
2.
So the general solution is
xk = C ′1λ
k1(m) + C ′
2λk2(m) (k = 1,2, . . . ,m + 1),
where C ′1, C ′
2 are constants, λ1(m)λ2(m) = 1, and λ1(m) > λ2(m).
2536 H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 7. A tree for t = 3.
Since ρmxm+1 = 2xm + xm+1, we have ρm = 2xmxm+1
+ 1. Then
α2 = limm→+∞
2xm
xm+1+ 1 = 2 lim
m→+∞C ′
1λm1 (m) + C ′
2λm2 (m)
C ′1λ
m+11 (m) + C ′
2λm+12 (m)
+ 1 = 2 limm→+∞λ2(m) + 1.
For limm→+∞ λ2(m) = α2−√
α22−4
2 , we have α2 = 2α2−
√α2
2−4
2 + 1. Then α2 = √5. So for ∀T ∈ T 3
n ,
ρ(T ) � ρ(Tm) <√
5.However, for t = 2 and n � � + 2, we have k � 3. Then 2
√2 cos π
2k+1 � 2√
2 cos π7 ≈ 2.5483. Obvi-
ously, 2√
2 cos π2k+1 >
√5.
Therefore, if t = 2, then ρ(T ) <√
5.(3) If t = 3, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀T ∈ T 3
n , there is anature number M3 such that ρ(T ) � ρ(Tm) for m � M3 and Tm is a tree as shown in Fig. 7. Denoteρ(Tm) by ρm . It is easy to prove that {ρm} (m � M3) is strictly monotone increasing and has a upperbound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α3, then ρm < α3.
Similar to (1), we have ρmxvi = ∑v j∈NTm (vi)
xv j . It is easy to prove that xai = xbi = xci = xdi
(i = 1,2, . . . ,m), and xs = xt . Let xi = xai (i = 1,2, . . . ,m), xm+1 = xs . Then ρmxk = xk+1 + xk−1,k = 2,3, . . . ,m.
Let x0 = 0, then we have ρmxk = xk+1 + xk−1, k = 1,2, . . . ,m. Its characteristic equation is λ2 −ρmλ + 1 = 0. Then the characteristic roots are
λ1(m) = ρm +√
ρ2m − 4
2, λ2(m) = ρm −
√ρ2
m − 4
2.
So the general solution is
xk = C3λk1(m) + C4λ
k2(m) (k = 1,2, . . . ,m + 1),
where C3, C4 are constants, λ1(m)λ2(m) = 1, and λ1(m) > λ2(m).Let yi = xei (i = 1,2, . . . ,m), ym+1 = xr , y0 = 0. Then ρm yk = yk+1 + yk−1, k = 1,2, . . . ,m. So the
general solution is
yk = C ′3λ
k1(m) + C ′
4λk2(m) (k = 1,2, . . . ,m + 1),
where C ′3, C ′
4 are constants; λ1(m), λ2(m) are the same as the above.For ρmxm+1 = 2xm + ym+1 and ρm ym+1 = 2xm+1 + ym , we have⎧⎪⎪⎨
⎪⎪⎩ρm = 2xm
xm+1+ ym+1
xm+1,
ρm = 2xm+1
ym+1+ ym
ym+1.
It has been proved that limm→+∞ ρm exists, and limm→+∞ xmxm+1
= limm→+∞ ymym+1
= limm→+∞ λ2(m)=α3−
√α2
3−4
2 exists. So limm→+∞ ym+1x exists. Let limm→+∞ ym+1
x = β1, then
m+1 m+1
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2537
Fig. 8. Four 3-degree vertices.
Fig. 9. A tree for t = 4.
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
α3 = 2α3 −
√α2
3 − 4
2+ β1,
α3 = 2
β1+
α3 −√
α23 − 4
2.
So α3 = 4√
33 . Then for ∀T ∈ T 3
n , ρ(T ) � ρ(Tm) < 4√
33 .
However, for t = 3 and n � � + 2, we have k � 3. Then 2√
2 cos π2k+1 � 2
√2 cos π
7 ≈ 2.5483. Obvi-
ously, 2√
2 cos π2k+1 > 4
√3
3 .
Therefore, if t = 3, then ρ(T ) < 4√
33 .
(4) If t = 4, then according to the connected relationships of the four 3-degree vertices, we discussthe following two cases.
Case 1. The connected relationships of the four 3-degree vertices as shown in Fig. 8.By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀T ∈ T 3
n , there is a nature number M4such that ρ(T ) � ρ(Tm) for m � M4 and Tm is a tree as shown in Fig. 9. Denote ρ(Tm) by ρm .It is easy to prove that {ρm} (m � M4) is strictly monotone increasing and has a upper bound. Solimm→+∞ ρm exists. Let limm→+∞ ρm = α4, then ρm < α4.
Similar to (1), we have ρmxvi = ∑v j∈NTm (vi)
xv j . It is easy to prove that xai = xbi = xci = xdi =xei = x fi (i = 1,2, . . . ,m), xs = xt = xr . Let xi = xai (i = 1,2, . . . ,m), xm+1 = xs . Then ρmxk = xk+1 +xk−1, k = 2,3, . . . ,m.
Let x0 = 0, then we have ρmxk = xk+1 + xk−1, k = 1,2, . . . ,m. Then
xk = C5λk1(m) + C6λ
k2(m) (k = 1,2, . . . ,m + 1),
where C5, C6 are constants; λ1(m), λ2(m) are the same as the above.
2538 H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 10. Four 3-degree vertices.
Fig. 11. A tree for t = 4.
For ρmxm+1 = 2xm + xw and ρmxw = 3xm+1, then
⎧⎪⎪⎨⎪⎪⎩
ρm = 2xm
xm+1+ xw
xm+1,
ρm = 3xm+1
xw.
It has been proved that limm→+∞ ρm exists, and limm→+∞ xmxm+1
= limm→+∞ ymym+1
= limm→+∞ λ2(m)=α4−
√α2
4−4
2 exists. So limm→+∞ xwxm+1
exists. Let limm→+∞ xwxm+1
= β2, then
⎧⎪⎪⎪⎨⎪⎪⎪⎩
α4 = 2α4 −
√α2
4 − 4
2+ β2,
α4 = 3
β2.
So α4 =√
2 + √13. Then for ∀T ∈ T 3
n , ρ(T ) <√
2 + √13.
Case 2. The connected relationships of the four 3-degree vertices as shown in Fig. 10.By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀T ∈ T 3
n , there is a nature num-ber M5 such that ρ(T ) � ρ(Tm) for m � M5 and Tm is a tree as shown in Fig. 11. Denote ρ(Tm)
by ρm . It is easy to prove that {ρm} (m � M5) is strictly monotone increasing and has a upper bound.So limm→+∞ ρm exists. Let limm→+∞ ρm = α5, then ρm < α5.
Similar to (1), we have ρmxvi = ∑v j∈NTm (vi)
xv j . It is easy to prove that xai = xbi = xci = xdi ,xei = x fi (i = 1,2, . . . ,m), xs = xr , xu = xv . Let xi = xai (i = 1,2, . . . ,m), xm+1 = xr . Then ρmxk =xk+1 + xk−1, k = 2,3, . . . ,m.
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2539
Let x0 = 0, then we have ρmxk = xk+1 + xk−1, k = 1,2, . . . ,m. Then
xk = C ′5λ
k1(m) + C ′
6λk2(m) (k = 1,2, . . . ,m + 1),
where C ′5, C ′
6 are constants; λ1(m), λ2(m) are the same as the above.Let yi = xei (i = 1,2, . . . ,m), ym+1 = xu , y0 = 0. Then ρm yk = yk+1 + yk−1, k = 1,2, . . . ,m. So the
general solution is
yk = C7λk1(m) + C8λ
k2(m) (k = 1,2, . . . ,m + 1),
where C7, C8 are constants; λ1(m), λ2(m) are the same as the above.For ρmxm+1 = 2xm + ym+1 and ρm ym+1 = ym+1 + xm+1 + ym , then⎧⎪⎪⎨
⎪⎪⎩ρm = 2xm
xm+1+ ym+1
xm+1,
ρm = xm+1
ym+1+ ym
ym+1+ 1.
It has been proved that limm→+∞ ρm exists, and limm→+∞ xmxm+1
= limm→+∞ ymym+1
= limm→+∞ λ2(m) =α5−
√α2
5−4
2 exists. So limm→+∞ ym+1xm+1
exists. Let limm→+∞ ym+1xm+1
= β3, then
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
α5 = 2α5 −
√α2
5 − 4
2+ β3,
α5 = 1
β3+
α5 −√
α25 − 4
2+ 1.
So α5 ≈ 2.3569. Then for ∀T ∈ T 3n , ρ(T ) < 2.357.
For√
2 + √13 > 2.357, by the two cases, we know if t = 4, then for ∀T ∈ T 3
n , ρ(T ) <√
2 + √13.
However, for t = 4 and n � � + 2, we have k � 3. Then 2√
2 cos π2k+1 � 2
√2 cos π
7 ≈ 2.5483. Obvi-
ously, 2√
2 cos π2k+1 >
√2 + √
13.
Therefore, if t = 4, then ρ(T ) <√
2 + √13.
(5) If t � 5, then by Corollary 3.1, we have ρ(T ) < 2√
2 cos π2k+1 , where k = �log2
n+23 � + 1. �
It is easy to prove that, when � = 3, the upper bounds in Theorem 3.2 are better than the upperbounds in Theorem 3.1, respectively.
4. The comparison of the results
In the following, we compare our upper bounds on the spectral radius of trees in Theorem 3.1with the previous results (1.1), (1.2), and (1.3).
4.1. Comparing with (1.1)
We know that (1.1) gives a result ρ(T ) < 2√
� − 1.
(1) When � + 1 < n � 2�. In Theorem 3.1, we get ρ(T ) �√
n−1+√
(n−2�)2+2n−32 .
Since � � 3, we have 9�2 − 28� + 13 = 9(� − 149 )2 − 79
9 > 0. Then we get 36�2 − 112� +52 + 12n > 0. Thus, 60�2 − 112� + 52 − 12�n + 12n > 0. So we can get
√(n − 2�)2 + 2n − 3 <
8(� − 1) − (n − 1). Therefore,√
n−1+√
(n−2�)2+2n−32 < 2
√� − 1. Thus, our result is better than (1.1).
(2) When 2� < n ��2 + 1. In Theorem 3.1, we get ρ(T ) �√
2� − 1.It is easy to prove that
√2� − 1 < 2
√� − 1 for �� 3. Thus, our result is better than (1.1).
2540 H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 12. A tree with 7 vertices.
Fig. 13. A tree with 6 vertices.
(3) When n > �2 + 1. In Theorem 3.1, we get ρ(T ) < 2√
� − 1 cos π2k+1 (k � 4).
Because 2√
� − 1 cos π2k+1 < 2
√� − 1, our result is better than (1.1).
Therefore, our upper bounds in Theorem 3.1 are better than (1.1).
4.2. Comparing with (1.2)
We know that (1.2) gives a result
ρ(T ) < max{
max2� j�k−2
{√δ j − 1 + √δ j−1 − 1 },√δ1 − 1 + √
�}.
(1) When � + 1 < n � 2�. Let T1 be an almost completely full-degree tree as shown in Fig. 12,
then we can get our upper bound is ρ(T1) =√
7−1+√
(7−8)2+14−32 =
√3 + √
3.
However, the upper bound in (1.2) is 2 + √2.
Because 2 + √2 >
√3 + √
3, our result is better than (1.2).(2) When 2� < n � �2 + 1. Let T be a tree with maximum degree �, and let u, v be two adjacent
vertices of T such that du = dv = �. Then the upper bound in (1.2) is√
� + √� − 1. Obviously,√
� + √� − 1 > 2
√� − 1 >
√2� − 1. Thus, our result is better than (1.2).
(3) When n > �2 + 1. Let T be a tree with maximum degree �, and let u, v be two adjacentvertices of T such that du = dv = �. Then the upper bound in (1.2) is
√� + √
� − 1. Obviously,√� + √
� − 1 > 2√
� − 1 > 2√
� − 1 cos π2k+1 . Thus, our result is better than (1.2).
Therefore, our upper bounds in Theorem 3.1 are better than (1.2) in some cases.
4.3. Comparing with (1.3)
We know that (1.3) gives a result
ρ(T ) < max{
max2� j�k−2
{√γ j − 1 + √γ j−1 − 1 },√γ1 − 1 + √
� − 1}.
(1) When � + 1 < n � 2�. Let T2 be an almost completely full-degree tree as shown in Fig. 13,
then we can get our upper bound is ρ(T2) =√
6−1+√12−3
2 = 2.
However, the upper bound in (1.3) is 2√
2.Because 2
√2 > 2, our result is better than (1.3).
(2) When 2� < n � �2 + 1. Let T be a tree with maximum degree �, and let u, v, w be thevertices of T such that du = dv = dw = � and uv ∈ E(T ), uw ∈ E(T ). Then the upper bound in (1.3)is 2
√� − 1. Obviously, 2
√� − 1 >
√2� − 1. Thus, our result is better than (1.3).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541 2541
(3) When n > �2 + 1. Let T be a tree with maximum degree �, and let u, v, w be the vertices ofT such that du = dv = dw = � and uv ∈ E(T ), uw ∈ E(T ). Then the upper bound in (1.3) is 2
√� − 1.
Obviously, 2√
� − 1 > 2√
� − 1 cos π2k+1 . Thus, our result is better than (1.3).
Therefore, our upper bounds in Theorem 3.1 are better than (1.3) in some cases.In conclusion, comparing our results with the previous results, we prove that our upper bounds
are better than (1.1). For (1.2) and (1.3), in order to get the upper bound on the spectral radius of atree T , we require to know the degrees of all the vertices of T . But in Theorem 3.1, we only require toknow the number of vertices and maximum degree, and we prove that our upper bounds are betterthan (1.2) and (1.3) in some cases.
Acknowledgements
The authors would like to thank the anonymous referees for valuable suggestions and corrections,which have improved the original manuscript.
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