On the largest real root of independence polynomials of graphs, an ordering on graphs, and starlike trees

preprint2013arXivOpen access

Abstract

Let GG be a simple graph of order nn. An independent set in a graph is a set of pairwise non-adjacent vertices. The independence polynomial of GG is the polynomial I(G,x)=∑k=0ns(G,k)xkI(G,x)=\sum_{k=0}^{n} s(G,k) x^{k}, where s(G,k)s(G,k) is the number of independent sets of GG of size kk and s(G,0)=1s(G,0)=1. Clearly all real roots of I(G,x)I(G,x) are negative. Let ξ(G)ξ(G) be the largest real root of I(G,x)I(G,x). Let HH be a simple graph. By G⪰HG \succeq H we mean that I(H,x)≥I(G,x)I(H,x)\geq I(G,x) for every xx in the interval [ξ(G),0][ξ(G),0]. We note that G⪰HG\succeq H implies that ξ(G)≥ξ(H)ξ(G)\geq ξ(H). Also we let G≻HG\succ H if and only if G⪰HG\succeq H and I(G,x)≠I(H,x)I(G,x)\neq I(H,x). We prove that for every tree TT of order nn, Sn⪰T⪰PnS_n\succeq T\succeq P_n, where SnS_n and PnP_n are the star and the path of order n, respectively. By T=T(n1,…,nk)T=T(n_1,\ldots,n_k) we mean a tree TT which has a vertex vv of degree kk such that T∖v=Pn1−1+⋯+Pnk−1T\setminus v=P_{n_1-1}+\cdots+P_{n_k-1}, that is T∖vT\setminus v is the disjoint union of the paths Pn1−1,…,Pnk−1P_{n_1-1},\ldots,P_{n_k-1}. Let X=(x1,…,xk)X=(x_1,\ldots,x_k) and Y=(y1,…,yk)Y=(y_1,\ldots,y_k), where x1≥⋯≥xkx_1\geq \cdots\geq x_k and y1≥⋯≥yky_1\geq\cdots\geq y_k are real. By X≻YX\succ Y, we mean x1=y1,…,xt−1=yt−1x_1=y_1,\ldots,x_{t-1}=y_{t-1} and xt>ytx_t>y_t for some t∈{1,…,k}t\in\{1,\ldots,k\}. We let X≻dYX\succ_{d}Y, if X≠YX\neq Y and for every jj, 1≤j≤k1\leq j\leq k, ∑i=1jxi≥∑i=1jyi\sum_{i=1}^{j}x_i\geq \sum_{i=1}^{j}y_i. Among all trees with fixed number of vertices, we show that if (m1,…,mk)≻d(n1,…,nk)(m_1,\ldots,m_k)\succ_{d}(n_1,\ldots,n_k), then T(n1,…,nk)≻T(m1,…,mk)T(n_1,\ldots,n_k)\succ T(m_1,\ldots,m_k). We conjecture that T(n1,…,nk)≻T(m1,…,mk)T(n_1,\ldots,n_k)\succ T(m_1,\ldots,m_k) if and only if (m1,…,mk)≻(n1,…,nk)(m_1,\ldots,m_k)\succ (n_1,\ldots,n_k), where ∑i=1kni=∑i=1kmi\sum_{i=1}^kn_i=\sum_{i=1}^km_i.

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