On the largest real root of independence polynomials of graphs, an ordering on graphs, and starlike trees
Combinatorics
2013-03-14 v1
Abstract
Let G be a simple graph of order n. An independent set in a graph is a set of pairwise non-adjacent vertices. The independence polynomial of G is the polynomial I(G,x)=∑k=0ns(G,k)xk, where s(G,k) is the number of independent sets of G of size k and s(G,0)=1. Clearly all real roots of I(G,x) are negative. Let ξ(G) be the largest real root of I(G,x). Let H be a simple graph. By G⪰H we mean that I(H,x)≥I(G,x) for every x in the interval [ξ(G),0]. We note that G⪰H implies that ξ(G)≥ξ(H). Also we let G≻H if and only if G⪰H and I(G,x)=I(H,x). We prove that for every tree T of order n, Sn⪰T⪰Pn, where Sn and Pn are the star and the path of order n, respectively. By T=T(n1,…,nk) we mean a tree T which has a vertex v of degree k such that T∖v=Pn1−1+⋯+Pnk−1, that is T∖v is the disjoint union of the paths Pn1−1,…,Pnk−1. Let X=(x1,…,xk) and Y=(y1,…,yk), where x1≥⋯≥xk and y1≥⋯≥yk are real. By X≻Y, we mean x1=y1,…,xt−1=yt−1 and xt>yt for some t∈{1,…,k}. We let X≻dY, if X=Y and for every j, 1≤j≤k, ∑i=1jxi≥∑i=1jyi. Among all trees with fixed number of vertices, we show that if (m1,…,mk)≻d(n1,…,nk), then T(n1,…,nk)≻T(m1,…,mk). We conjecture that T(n1,…,nk)≻T(m1,…,mk) if and only if (m1,…,mk)≻(n1,…,nk), where ∑i=1kni=∑i=1kmi.
Cite
@article{arxiv.1303.3222,
title = {On the largest real root of independence polynomials of graphs, an ordering on graphs, and starlike trees},
author = {Mohammad Reza Oboudi},
journal= {arXiv preprint arXiv:1303.3222},
year = {2013}
}
Comments
16 pages,5 figures