Maximum Modulus of Independence Roots of Graphs and Trees
aa r X i v : . [ m a t h . C O ] D ec Maximum Modulus of Independence Roots ofGraphs and Trees
Jason I. Brown ∗ and Ben CameronDepartment of Mathematics and StatisticsDalhousie University, Halifax, NS B3H 3J5, Canada Abstract
The independence polynomial of a graph is the generating polynomialfor the number of independent sets of each size and its roots are called in-dependence roots . We bound the maximum modulus, maxmod ( n ) , of anindependence root over all graphs on n vertices and the maximum modulus,maxmod T ( n ) , of an independence root over all trees on n vertices in termsof n . In particular, we show thatlog ( maxmod ( n )) n = + o ( ) and log ( maxmod T ( n )) n = + o ( ) . The independence number of a graph G , denoted α ( G ) , is the maximum size ofan independent of G . The independence polynomial of G , denoted i ( G , x ) , is thegenerating polynomial for the number of independence sets of each size: i ( G , x ) = α ( G ) ∑ k = i k x k , ∗ Corresponding author. [email protected] i k denotes the number of independent sets of size k in G . (When dealingwith the independence polynomials of multiple graphs, we will distinguish thecoefficients with a superscript to avoid confusion, so that i Gk is the number ofindependent sets of size k in G .) The roots of i ( G , x ) are called the independenceroots of G .The independence polynomial was first introduced by Gutman and Harary in1983 [15] and has been a fascinating object of study ever since (see Levit andMandrescu’s survey [18]). One topic that has generated much research is on theindependence roots [1, 2, 4, 5, 7, 9, 10, 19, 23].The roots of other graph polynomials have also been of interest and the natureand location in C of these roots can vary considerably depending on the polyno-mial (see [20]). Determining bounds on the moduli of these roots is an importantquestion. In 1992, the first author and Colbourn [3] conjectured that the rootsof reliability polynomials lie in the unit disk. The Brown-Colbourn conjecturestood for 12 years until it was shown to be false (although just barely) in [25]. Itwas later shown that if G is a connected graph on n vertices and q is a reliabilityroot, then | q | ≤ n −
1, yet the largest known reliability root has modulus approx-imately 1 . | z | < ( + √ ) + √ [11]. In contrast, the collection of all roots of independence polyno-mials [5], domination polynomials [8], and chromatic polynomials [28] are eachdense in C .Although these polynomials have roots with arbitrarily large moduli, an in-teresting question to ask is: for fixed n , how large can the modulus of a root ofone of these polynomials be for a graph in n vertices? Sokal [27] showed thatall simple graphs on n vertices have their chromatic roots contained in the disk | z | ≤ . ( n − ) , so that the maximum moduli of chromatic roots grows atmost linearly in n . The growth rate of domination roots is unknown. There hasbeen work done on bounding the independence roots; for example, it was shownin [7] that for fixed α , the largest modulus of an independence root of a graphwith independence number α on n vertices is (cid:0) n α − (cid:1) α − + O ( n α − ) . Althoughthis bound is tight, the O ( n α − ) term hides enough information to make it unclearif the maximum moduli of independence roots is a polynomial in n or exponen-tial in n . In this paper, we consider the problem of fixing n as the number ofvertices and determining the maximum modulus of an independence root over all2raphs on n vertices. We will show that the growth rate is indeed exponential. Tothat end, let maxmod ( n ) denote the maximum modulus of an independence rootover all graphs on n vertices and maxmod T ( n ) denote the maximum modulus ofan independence root over all trees on n vertices. We show that, in contrast toSokal’s linear bound for chromatic roots, maxmod ( n ) and maxmod T ( n ) are bothexponential in n : in Section 2, we prove that3 n − r + ≤ maxmod ( n ) ≤ n + n − , where 1 ≤ r ≤
3, while in Section 3, we prove that2 n − ≤ maxmod T ( n ) ≤ n − + n − n is odd and 2 n − ≤ maxmod T ( n ) ≤ n − + n n is even.We shall need some notation. The number of maximum independent sets in G is denoted by ξ ( G ) . The number of maximal independent sets in G is denoted µ ( G ) . Note that ξ ( G ) = i G α ( G ) , the leading coefficient of the independence poly-nomial of G . For S ⊆ V ( G ) , let G − S be the graph obtained from G by deleting allvertices of S as well as their incident edges. If S = { v } , we will use the shorthand, G − v to denote G − { v } . To bound the roots of independence polynomials, we will make extensive useof the classical Enestr¨om-Kakeya Theorem which uses the ratios of consecutivecoefficients of a given polynomial to describe an annulus in C that contains all itsroots. Theorem 2.1 (Enestr¨om-Kakeya [12, 17])
If f ( x ) = a + a x + · · · + a n x n haspositive real coefficients, then all complex roots of f lie in the annulus r ≤ | z | ≤ Rwherer = min (cid:26) a i a i + : 0 ≤ i ≤ n − (cid:27) and R = max (cid:26) a i a i + : 0 ≤ i ≤ n − (cid:27) .
3e will also need to make use of two basic results on computing the indepen-dence polynomial.
Proposition 2.2 ([15])
If G and H are graphs and v ∈ V ( G ) , then:i) i ( G , x ) = i ( G − v , x ) + x · i ( G − N [ v ] , x ) .ii) i ( G ∪ H , x ) = i ( G , x ) i ( H , x ) . Note that from Proposition 2.2, ξ ( G ∪ H ) = ξ ( G ) · ξ ( H ) . Our proofs are in-ductive and often require upper bounds ξ ( G ) for all graphs on n vertices, a col-lection of which can be found in [16]. Theorem 2.3 ([22])
If G is a graph of order n ≥ , then ξ ( G ) ≤ µ ( G ) ≤ g ( n ) = n if n ≡ · n − if n ≡ · n − if n ≡ . Note that an easy corollary of this is that for a graph on n vertices, ξ ( G ) ≤ µ ( G ) ≤ n , since ξ ( K ) = µ ( K ) = ≤ , 3 n ≥ · n − , and 3 n ≥ · n − for all n ≥ Proposition 2.4
For all n ≥ , maxmod ( n ) ≥ n − if n ≡ n − if n ≡ n − if n ≡ . Proof
The proof is in three cases depending on n mod 3. Each relies on in-dependence polynomials of the graphs G , G , and G , respectively, in Figure 1where G is obtained by joining a central vertex to all but one vertex in each of k copies of K , G is obtained by joining one vertex in K to the central vertex in G , and G is obtained by joining one vertex in another copy of K to the centralvertex in G . Note that the orders of G , G , and G are congruent to 0, 1, and 2,4 k y y . . . (a) G y k y y . . . (b) G y k y y . . . (c) G Figure 1: Graphs with independence roots of large moduli.respectively, mod 3. We then use the Intermediate Value Theorem (IVT) to findthat each has a real root of large modulus.From Proposition 2.2, it easily follows that i ( G , x ) = ( + x ) k ( + x ) + x ( + x ) k + i ( G , x ) = ( + x ) k + x ( + x ) k i ( G , x ) = ( + x ) k ( + x ) + x ( + x ) k + .
5t is now straightforward to determine thatsign (cid:18) lim x →− ∞ i ( G , x ) (cid:19) = ( − ) k sign (cid:18) lim x →− ∞ i ( G , x ) (cid:19) = ( − ) k + sign (cid:18) lim x →− ∞ i ( G , x ) (cid:19) = ( − ) k + . We now prove the lower bounds for maxmod ( n ) by exhibiting, in each oneof the cases, a graph with a real independence root with modulus larger than thebound. Case 0: n ≡ n =
3, then we can use the quadratic formula to find that P has a realindependence root with modulus approximately 2 . >
1. So we may assumethat n ≥ k ≥ G . For all k ≥
1, we have that i ( G , − k ) = (cid:16) − k + (cid:17) k (cid:16) − · k + (cid:17) − k (cid:16) − k (cid:17) k = ( − ) k h ( − k ) (cid:16) ( k + − ) k − ( k + − ) k (cid:17) − k ( k − ) k i which has the same sign as ( − ) k + since 0 > ( k + − ) k − ( k + − ) k . Thus, i ( G , x ) alternates sign on ( − ∞ , − k ] and by the IVT and since k = n − , i ( G , x ) has a root in the interval ( − ∞ , − n − ) . Case 1: n ≡ n =
1, then K is the only graph to consider and the result clearly holds.So we may assume that n ≥ k ≥ G . Since i ( G , − k ) = ( − ) k (( k + − ) k − ( k + − ) k ) , it follows that i ( G , − k ) has thesame sign as ( − ) k . Thus i ( G , x ) alternates sign on ( − ∞ , − k ] and by IVT itmust have a root in the interval ( − ∞ , − n − ) . Case 2: n ≡ n =
2, then the graph K has and − | − | = = . If n =
5, then P has a real independence root of modulus approximately5 . n ≥ k ≥ G . We now have, i ( G , − k ) = ( − ) k h ( − · k ) ( k + − ) k − k ( − k ) ( k − ) k i = ( − ) k h ( − · k + · k )( k + − ) k − ( − k ) ( k + − ) k i = ( − ) k h ( − · k + k + k + )( k + − ) k − ( − k ) ( k + − ) k i = ( − ) k h ( − k ) (cid:16) ( k + − ) k − ( k + − ) k (cid:17) +( k + − · k )( k + − ) k i which has sign ( − ) k since ( k + − ) k − ( k + − ) k > ( − k ) >
0, and ( k + − · k )( k + − ) k > i ( G , x ) must have a root in the interval ( − ∞ , − n − ) .This completes the proof.Therefore, maxmod ( n ) is at least exponential in n . We require the next twolemmas to put an upper bound on maxmod ( n ) . Lemma 2.5
For all graphs G with at least one edge, there exists a non-isolatedvertex v such that α ( G ) = α ( G − v ) ≥ α ( G − N [ v ]) + . Proof
Let G be a graph with at least one edge. It is clear that for any vertex v of G , α ( G ) ≥ α ( G − N [ v ]) +
1, since any maximum independent set in G − N [ v ] will still be independent in G with the addition of v . Suppose that for all vertices v ∈ V ( G ) , that α ( G ) > α ( G − v ) . Then every vertex belongs to every maximumindependent set. However, G has at least one edge, so the vertices incident withthis edge cannot belong to the same independent set, which contradicts both ofthese vertices being in every maximum independent set. Therefore, there existssome v ∈ V ( G ) incident with some edge such that α ( G ) = α ( G − v ) ≥ α ( G − N [ v ]) + . emma 2.6 If G is a graph on n vertices such that ξ ( G ) = , then i α ( G ) − ≤ n + n − . Proof
Let G be a graph on n vertices such that ξ ( G ) =
1. Every independent setof size α ( G ) − α ( G ) . Therefore, i α ( G ) − ≤ µ ( G ) − + α ( G ) ≤ n + n − µ ( G ) to account for the one maximum independent set) by the note followingTheorem 2.3. Theorem 2.7
For all n ≥ , maxmod ( n ) ≤ n + n − . Proof
We actually prove the stronger result that for a graph on n vertices, theratios of coefficients, given by R in the statement of Theorem 2.1 (the Enestr¨om-Kakeya Theorem), of its independence polynomial are bounded above by 3 n + n −
1. It then follows directly from the Enestr¨om-Kakeya Theorem that the rootsare bounded by this value. We proceed by induction on n .The results hold for graphs on n ≤ ≤ k < n , and let G be a graphon n vertices. If G has no edges, then we are done, since G has only − G has at least one edge. Let v be a nonisolated vertex in G such that α ( G ) = α ( G − v ) ≥ α ( G − N [ v ]) + v exists by Lemma 2.5. Now, by Proposition 2.2, i ( G , x ) = i ( G − v , x ) + x · i ( G − N [ v ] , x )= α ( G − v ) ∑ k = i G − vk x k + x α ( G − N [ v ]) ∑ k = i G − N [ v ] k x k = + α ( G − v ) ∑ k = i G − vk x k + α ( G − N [ v ])+ ∑ k = i G − N [ v ] k − x k . (1)We now have two cases. Case 1: α ( G ) = α ( G − v ) = α ( G − N [ v ]) + i ( G , x ) = + α ( G − v ) ∑ k = (cid:16) i G − vk + i G − N [ v ] k − (cid:17) x k . This gives the following ratios between coefficients,1 n and i G − vk + i G − N [ v ] k − i G − vk + + i G − N [ v ] k for k = , , . . . , α ( G − N [ v ]) . For all n ≥ n < n + n −
1, and by the inductive hypothesis, i G − vk + i G − N [ v ] k − i G − vk + + i G − N [ v ] k < (cid:16) n − + n − (cid:17) i G − vk + + (cid:16) n −| N [ v ] | + n − − | N [ v ] | (cid:17) i G − N [ v ] k i G − vk + + i G − N [ v ] k ≤ (cid:16) n − + n − (cid:17) (cid:16) i G − vk + + i G − N [ v ] k (cid:17) i G − vk + + i G − N [ v ] k = n − + n − < n + n − . Case 2: α ( G ) = α ( G − v ) > α ( G − N [ v ]) + i ( G , x ) = + α ( G − N [ v ])+ ∑ k = (cid:16) i G − vk + i G − N [ v ] k − (cid:17) x k + α ( G − v ) ∑ α ( G − N [ v ])+ i G − vk x k . This gives four different forms for i Gk i Gk + . The first two, namely n and i G − vk + i G − N [ v ] k − i G − vk + + i G − N [ v ] k ,are less than or equal to 3 n + n − k = , , . . . , α ( G − N [ v ]) by the sameargument as Case 1. This leaves, i G − v α ( G − N [ v ])+ + i G − N [ v ] α ( G − N [ v ]) i G − v α ( G − N [ v ])+ , and i G − vk i G − vk + for k ≥ α ( G − N [ v ]) +
29y the inductive hypothesis, i G − vk i G − vk + ≤ n − + n − < n + n −
1, so we are left onlywith i G − v α ( G − N [ v ])+ + i G − N [ v ] α ( G − N [ v ]) i G − v α ( G − N [ v ])+ .In this case, we first show that | N [ v ] | ≥
3. As v is not isolated, | N [ v ] | ≥ | N [ v ] | =
2, then v is a leaf, and since v was chosen such that α ( G ) = α ( G − v ) ≥ α ( G − N [ v ]) + v is not in every maximum independent set in G . Butevery maximum independent set in G must contain either v or its neighbour, so α ( G − v ) = α ( G − N [ v ]) + | N [ v ] | ≥
3. We also note that i G − v α ( G − N [ v ])+ + i G − N [ v ] α ( G − N [ v ]) i G − v α ( G − N [ v ])+ = i G − v α ( G − N [ v ])+ i G − v α ( G − N [ v ])+ + ξ ( G − N [ v ]) i G − v α ( G − N [ v ])+ . There are three subcases to consider.
Case 2a: α ( G − N [ v ]) + < α ( G − v ) = α ( G ) .If α ( G − N [ v ]) + < α ( G − v ) , then G − v has an independent set of size α ( G − N [ v ]) +
3. Therefore, i G − v α ( G − N [ v ])+ ≥ α ( G − N [ v ]) + ≥
3, since any inde-pendent set of size k , contains at least (cid:0) kk − (cid:1) = k independent sets of size k − i G − v α ( G − N [ v ])+ i G − v α ( G − N [ v ])+ + ξ ( G − N [ v ]) i G − v α ( G − N [ v ])+ ≤ n − + n − + ξ ( G − N [ v ]) ≤ n − + n − + n −| N [ v ] |− ≤ n − + n − + n − = n (cid:18) − + (cid:19) + n − ≤ n + n − . Case 2b: α ( G − N [ v ]) + = α ( G − v ) = α ( G ) and | N [ v ] | ≥ G − v α ( G − N [ v ])+ i G − v α ( G − N [ v ])+ + ξ ( G − N [ v ]) i G − v α ( G − N [ v ])+ ≤ n − + n − + ξ ( G − N [ v ]) ≤ n − + n − + n −| N [ v ] | ≤ n − + n − + n − = n (cid:16) − + − (cid:17) + n − < n + n − . Case 2c: α ( G − N [ v ]) + = α ( G − v ) and | N [ v ] | = i G − v α ( G − N [ v ])+ .First, if i G − v α ( G − N [ v ])+ ≥
2, then by the inductive hypothesis and the note followingTheorem 2.3, i G − v α ( G − N [ v ])+ i G − v α ( G − N [ v ])+ + ξ ( G − N [ v ]) i G − v α ( G − N [ v ])+ ≤ n − + n − + ξ ( G − N [ v ]) ≤ n − + n − + n − = n (cid:18) − + (cid:19) + n − ≤ n + n − . Note if some maximum independent set in G contained v , then this set with v removed would be an independent set of size α ( G ) − = α ( G − N [ v ]) + G − N [ v ] , which is a contradiction. Therefore, the maximum independent sets in G and G − v are exactly the same sets and, in particular, ξ ( G ) = ξ ( G − v ) . Now,if 1 = i G − v α ( G − N [ v ])+ = ξ ( G − v ) = ξ ( G ) , then Lemma 2.6 applied to G gives a bound on i G α ( G ) − in the last line of thefollowing, 11 G − v α ( G − N [ v ])+ + i G − N [ v ] α ( G − N [ v ]) i G − v α ( G − N [ v ])+ = i G α ( G ) − i G α ( G ) = i G α ( G ) − ≤ n + n − . Now, if z is an independence root of G , then, by the Enestr¨om-Kakeya Theo-rem, | z | ≤ n + n − Corollary 2.8 log ( maxmod ( n )) n = + o ( ) . n Now that we have determined bounds on maxmod ( n ) , a natural extension of thisis to determine the largest modulus an independence root can obtain among allgraphs of order n in a specific family of graphs. In particular, the bound weobtained for maxmod ( n ) seems to be much too large when we restrict our attentionto trees. In this section, we consider maxmod T ( n ) , the maximum modulus of anindependence root over all trees on n vertices.Let T k be the tree obtained by gluing k copies of P together at a leaf (see Fig-ure 2). This tree is known [26] to have the largest number of maximal independentsets among trees on 2 k + T ( n ) . Theorem 3.1 ([29])
If G is a tree of order n ≥ , then ξ ( G ) ≤ t ′ ( n ) = n − if n is odd2 n − + n is even . k x k y x y x z ... Figure 2: The tree T k on 2 k + [ − k − k , − k ) .We need to extend our notation to maxmod F ( n ) , the maximum modulus of anindependence root of a forest of order n ; clearly maxmod T ( n ) ≤ maxmod F ( n ) .The following lemma will be required in proving an upper bound on the ratio ofconsecutive coefficients of the independence polynomials of forests. Lemma 3.2
If F is a forest on n vertices, n ≥ , and v ∈ V ( F ) , then ξ ( F ) ξ ( F − v ) ≤ n − + n is odd2 n − + n is even . Proof
Let F = H ∪ H ∪ · · · ∪ H k , where k ≥
1, and each H i is a connectedcomponent of F . Suppose, without loss of generality, that v ∈ H k . Then F − v = H ∪ H ∪ · · · H k − ∪ F ′ , where F ′ is the forest obtained from deleting v from H k (note that if v was anisolated vertex in F , then F ′ may have no vertices and ξ ( H k ) = ( F ) ξ ( F − v ) = ξ ( H ) · ξ ( H ) · · · ξ ( H k ) ξ ( H ) · ξ ( H ) · · · ξ ( H k − ) · ξ ( F ′ )= ξ ( H k ) ξ ( F ′ ) ≤ ξ ( H k ) ≤ max { t ′ ( i ) : 1 ≤ i ≤ n } (from Theorem 3.1) = max n n − ( i + ) + i = , , . . . , n o if n is oddmax n n − i + i = , , . . . , n o if n is even = n − + n is odd2 n − + n is even . Theorem 3.3
For n ≥ , maxmod F ( n ) ≤ n − + n − if n is odd2 n − + n if n is even . Proof
As in the proof of Theorem 2.7, we actually prove a stronger result,bounding the ratios of consecutive coefficients. The Enestr¨om-Kakeya Theoremthen applies to obtain the bound the roots. We proceed by induction on n .For n = , , , ≤ k ≤ n − F be a forest on n vertices.Note that if F = K n , then the largest ratio of consecutive coefficients of i ( F , x ) caneasily be verified to be n which is less than the result in either case, so suppose F has at least one edge and therefore at least one leaf. Let v be a leaf of F andlet u be adjacent to v . Note that α ( F − { u , v } ) ≤ α ( F − v ) ≤ α ( F − { u , v } ) + α ( F − v ) = α ( F − { u , v } ) as our argumentswill hold (and be even shorter) when α ( F − v ) = α ( F − { u , v } ) +
1. To simplify14otation, let α = α ( F − v ) = α ( F − { u , v } ) . By Proposition 2.2, we have i ( F , x ) = i ( F − v , x ) + x · i ( F − { u , v } , x )= α ∑ k = i F − vk x k + x α ∑ k = i F −{ u , v } k x k = + α ∑ k = (cid:16) i F − vk + i F −{ u , v } k − (cid:17) x k + i F −{ u , v } α x α + . (2)We need to show that i Fk i Fk + is bounded above by the desired value and from (2),we see that i Fk i Fk + can take on the following forms,1 n , i F − vk + i F −{ u , v } k − i F − vk + + i F −{ u , v } k for k = , , . . . α ( F − { u , v } ) −
1, and i F − v α + i F −{ u , v } α − i F −{ u , v } α . The first ratio, n , clearly satisfies the desired bound regardless of the parity of n . We now only need to verify the remaining two forms of i Fk i Fk + . We will do this intwo cases depending on the parity of n . Case 1: n is odd.We apply the inductive hypothesis to get, i F − vk + i F −{ u , v } k − i F − vk + + i F −{ u , v } k ≤ (cid:16) n − + n − (cid:17) i F − vk + + (cid:16) n − + n − (cid:17) i F −{ u , v } k i F − vk + + i F −{ u , v } k ≤ (cid:16) n − + n − (cid:17) (cid:16) i F − vk + + i F −{ u , v } k (cid:17) i F − vk + + i F −{ u , v } k = n − + n − < n − + n − . i F − v α + i F −{ u , v } α − i F −{ u , v } α = i F − v α i F −{ u , v } α + i F −{ u , v } α − i F −{ u , v } α ≤ ξ ( F − v ) ξ ( F − { u , v } ) + n − + n − (by the inductive hypothesis) ≤ n − + + n − + n − (by the Lemma 3.2) = n − + n − . Therefore, the result holds when n is odd by the Enestr¨om-Kakeya Theorem. Case 2:
Suppose that n is even.Then we apply the inductive hypothesis to get, i F − vk + i F −{ u , v } k − i F − vk + + i F −{ u , v } k ≤ (cid:16) n − + n − (cid:17) i F − vk + + (cid:16) n − + n − (cid:17) i F −{ u , v } k i F − vk + + i F −{ u , v } k ≤ (cid:16) n − + n − (cid:17) (cid:16) i F − vk + + i F −{ u , v } k (cid:17) i F − vk + + i F −{ u , v } k = n − + n − < n − + n . For the last ratio, we have, i F − v α + i F −{ u , v } α − i F −{ u , v } α = i F − v α i F −{ u , v } α + i F −{ u , v } α − i F −{ u , v } α ≤ ξ ( F − v ) ξ ( F − { u , v } ) + n − + n − (by the inductive hypothesis) ≤ n − + + n − + n − (by the Lemma 3.2) = n − + n . Therefore,the result holds when n is even by the Enestr¨om-Kakeya Theorem.16 orollary 3.4 For n ≥ , maxmod T ( n ) ≤ n − + n − if n is odd2 n − + n if n is even . We remark that, at least in terms of the bounds on the ratio of consecutivecoefficients, this is best possible as there are forests that achieve these bounds.Let n be odd, and consider the graph T n − as previously defined and pictured inFigure 2. The independence polynomial of this tree has 2 n − + n − as its last ratioof consecutive coefficients. If n is even then look at the forest T n − ∪ K , whoseindependence polynomial has 2 n − + n as its last ratio of consecutive coefficients.We have shown that the bounds on the ratio of consecutive coefficients aretight, but are these bounds tight on the roots? It is not always the case that theupper bound on the moduli of the roots of a polynomial is tight, even for treesand forests. Take for example, the tree K , which has 30 as an upper bound onthe roots from Enestr¨om-Kakeya but its actual root of largest modulus is approx-imately 2 . k copies of K , . This forest will have the same root of maximum modu-lus but the bound on the root from Enestr¨om-Kakeya is 30 k , which is unbounded.Fortunately, it turns out that the bound we found in Theorem 3.3 is asymptoticallytight when n is odd.For the case where n is even in the next proof we require the definition of thetree T ′ k as shown in Figure 3. Let T ′ k be the graph obtained by adding two leaves toeach vertex in K and then gluing a leaf of the resulting graph to the central vertexof T k . Proposition 3.5
For all n ≥ , maxmod T ( n ) ≥ n − if n is odd2 n − if n is even . dcbay k x k y x y x z ... Figure 3: The tree T ′ k on 2 ( k + ) vertices that has an independence root in [ − k + − k − , − k ) . Proof
The proof is similar to the proof of Proposition 2.4, finding trees that havereal independence roots of large modulus.
Case 1: n is odd.If n =
1, then the result clearly holds so we may assume n ≥
1. Let n = k + k ≥ k = n − , and set T = T k as in Figure 2. A simple calculationvia Proposition 2.2 shows that i ( T , x ) = ( + x ) k + x ( + x ) k . We will use theIntermediate Value Theorem to show that i ( T , x ) has a real root to the left of − k .Now, i ( T , − k ) = ( − k + ) k − k ( − k ) k = ( − ) k (cid:16) ( k + − ) k − ( k + − ) k (cid:17) , so i ( T , − k ) has the same sign as ( − ) k . On the other hand, i ( T , x ) has sign ( − ) k + as x tends to ∞ We now compute the limit as x tends to − ∞ . Thus, i ( T k , x ) alternates sign on ( − ∞ , − k ] , so by IVT it must have a real root in the interval ( − ∞ , − k ) . We remark that from Theorem 3.3, that i ( T , x ) actually has a real rootin the interval [ − k − k , − k ) . Case 2: n is even.For n = n ≥
6, we will show that T ′ k , the graph inFigure 3, has a real root to the left of − k . Let n = ( k + ) for k ≥
0. If k =
0, then i ( T ′ k , x ) = ( + x ) ( + x + x ) which has roots − − + √
3, and − − √
3, with − − √ − − = − ) . If k =
1, then i ( T ′ k , x ) = x + x + x + x + x +
1, which has its largest root at approximately − . − − = −
4. Since the result holds for k = ,
1, we maynow assume that k ≥ i ( T ′ k , x ) = ( + x ) k ( + x + x + x ) + x ( + x ) k ( + x + x + x ) . Let g ( x ) = + x + x + x and h ( x ) = + x + x + x . We can easily verifythat g ( x ) < x ≤ − h ( x ) < x ≤ −
3. Moreover, h ( x ) = g ( x ) − x ( x + ) . We consider the function f ( x ) = ( − x − ) k ( + x + x + x ) + x ( − x − ) k ( + x + x + x ) , so that i ( T ′ k , x ) = ( − ) k f ( x ) . Now, f ( − k ) = ( k + − ) k g ( − k ) − k ( k − ) k h ( − k )= ( k + − ) k g ( − k ) − k ( k − ) k ( g ( − k ) + k ( − k ) )= g ( − k )(( k + − ) k − ( k + − ) k ) − k ( k − ) k + and since g ( − k ) and − k ( k − ) k + are both negative for k ≥
2, it follows that f ( − k ) <
0. Therefore, i ( T ′ k , x ) has sign ( − ) k ( − ) = ( − ) k + . On the otherhand, i ( T ′ k , x ) has sign ( − ) k + = ( − ) k as x tends to ∞ . Thus, by the IVT, i ( T ′ k , x ) has a real root to the left of − k . From, Theorem 3.3 and the Enestr¨om-KakeyaTheorem, i ( T , x ) has no roots in ( − ∞ , − k + / − k − ) , so i ( T ′ k , x ) has a root inthe interval [ − k + / − k − , − k ) .Tables 1 and 2 show values of maxmod T ( n ) for small values of n in compari-son to our bounds.Although the bounds on maxmod T ( n ) are not as tight for even n as for odd n ,Corollary 3.4 and Proposition 3.5 give the following corollary for all n . Corollary 3.6 log ( maxmod T ( n )) n = + o ( ) . n − maxmod T ( n ) n − + n − . . . . . . . . T ( n ) toour bounds for odd n n n − maxmod T ( n ) n − + n .
25 0 . . . . . . . . . T ( n ) toour bounds for even n We were able to prove upper and lower bounds on maxmod ( n ) and maxmod T ( n ) for all n but questions remain about tightening our bounds for all graphs and aboutthe growth rate of the moduli of independence roots for other families of graphs.One highly structured and highly interesting family of graphs is the family ofwell-covered graphs [13, 14, 24], that is, graphs with all maximal independentsets of the same size. For each well-covered graph with independence number α , it is known that all of its independence roots lie in the disk | z | ≤ α and thereare well-covered graphs with independence roots arbitrarily close to the boundary[4]. This difference between the independence roots of graphs and well-coveredgraphs begs the question of what happens for well-covered trees? Finbow et al.[14] showed that every well-covered tree is obtained by attaching a leaf to everyvertex of another tree. This construction of attaching a leaf to every vertex in agraph G is know as the graph star operation, the resulting graph denoted G ∗ . Levitand Mandrescu [19] proved a formula for i ( G ∗ , x ) in terms of i ( G , x ) for all graphs G . Using Maple and nauty [21], we were able exploit this formula to verify thatall well-covered trees on n ≤
40 vertices have their independence roots containedin the unit disk!This makes it extremely tempting to conjecture that the independence rootsof all well-covered trees are contained in the unit disk. However, the relationshipbetween the independence roots of a tree and the independence roots of its well-covered extension are bound by the properties of M ¨obius transformations (thisrelationship was used by the authors in [1, 2]). Drawing on the theory of these20ransfomrations, it can be shown that any tree T with independence roots to theright of the line Re ( z ) = , will yield a well-covered tree T ∗ with independenceroots outside of the unit disk. It was shown in [1], that there are trees with indepen-dence roots arbitrarily far in the right half of C , therefore, there are well-coveredtrees with independence roots outside of the unit disk. The tantalizing questionremains: Question 4.1
What is the maximum modulus of an independence root of a well-covered tree on n vertices?
Our bounds on log ( maxmod ( n )) and log ( maxmod T ( n )) are very good asymp-totically and a fairly good estimate for all n . Nevertheless, from computations withMaple and nauty [21], we have the following conjectures. Conjecture 4.2
If G is a graph on n vertices, then for n ≥ , maxmod ( n ) ≤ · n − + n if n ≡ n − + n − if n ≡ · n − + n + if n ≡ Conjecture 4.3
The graphs G , G , and G are the only graphs to achieve maxmod ( n ) . Conjecture 4.4
If T is a tree on n vertices with n ≥ even , then, maxmod T ( n ) ≤ n − + n + Conjecture 4.5
The trees T n − and T ′ n − (see Figures 2 and 3) are the only treesto achieve maxmod T ( n ) for n odd and even respectively. Acknowledgements
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