We give an elementary proof of an interesting combinatorial identity which is of particular interest in graph theory and its applications

Dorlas, T. C.

Rebenko, Alexei L.

Savoie, Baptiste

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arXiv:1902.03470v2  [math.CO]  5 Apr 2019A combinatorial identityTony Dorlas∗, Alexei Rebenko†, Baptiste Savoie∗,‡AbstractWe give an elementary proof of an interesting combinatorial identity which is of partic-ular interest in graph theory and its applications.The aim of this note is to give a new, elementary proof of the combinatorial identity (1) inTheorem 1. Done by induction, the proof we give is simple in that it only requires the useof the binomial formula along with the derivation operator. This identity finds an interest ingraph theory (for enumeration of forests) and its applications. For instance, we came across(1) when evaluating, within the framework of rigorous statistical mechanics, contributionsfrom forest graphs to a cluster expansion for classical gas correlation functions in [1]. Thecombinatorial identity (1) provides a means to directly derive a closed-form expression forthe number of distinct forests on a collection of sets of vertices, see formula (2) in Theorem2. We point out that the proof of formula (2) is self-contained and does not rely on graphtheory related techniques as, e.g., in [2, Chap. 5].1 The combinatorial identity.Theorem 1. Given m, p ∈ N with 1 ≤ p ≤ m, and a collection (xi)mi=1 of (complex) numbersxi ∈ C, the following identity holds∑{I1,...,Ip}∈Πp({1,...,m})p∏j=1∑i∈Ijxi|Ij |−1=(m− 1p− 1)( m∑i=1xi)m−p, (1)where the sum on the left-hand side is over the set Πp({1, . . . ,m}) of all partitions (Ij)pj=1of {1, . . . ,m} into p non-empty subsets.The proof of Theorem 1 is postponed to Sec. 3.2 An application to enumeration of forests.A forest is defined on a collection (Vi)mi=1 of sets of vertices Vi as a graph on V = V1∪· · ·∪Vmwith connected components which are trees, such that there are no lines between vertices ofany individual Vj , and moreover, if each set Vj is reduced to a single point, then the graphreduces to a single (connected) tree. Further, if in the reduced tree, Vj (j ≥ 2) is connectedto Vi in the path from Vj to V1, then all lines between Vi and Vj eminate from a singlevertex in Vi (a root). Deriving a closed-form expression for the number of (distinct) forestsas described above becomes straightforward by using Theorem 1∗Dublin Institute for Advanced Studies, School of Theoretical Physics, Dublin, Ireland†Institute of Mathematics, Ukrainian National Academy of Sciences, Kyiv, Ukraine‡Corresponding author - e-mail: baptiste.savoie@gmail.com1Theorem 2. Given m ∈ N, the number Nm of distinct forests on a collection of sets ofvertices (Vi)mi=1 is given byNm = |V1|(m∑i=1|Vi|)m−2 m∏i=2(2|Vi| − 1). (2)Proof. For m = 2 we can choose the root in |V1| ways and connect it to k = 1, . . . , |V2| pointsin V2. This results in |V1|(2|V2|−1) different forests (in fact they are trees). Suppose now thatthe statement is true for m. Adding an additional set of vertices Vm+1, it can be connectedto p = 1, . . . ,m other sets Vj in the reduced tree on V1, . . . , Vm. Omitting the connections toVm+1 we obtain p separate forests given by subsets I1, . . . , Ip of {1, . . . ,m}. If I1 contains 1,then this yields a factor|V1| ∏i∈I1\{1}(2|Vi| − 1)(∑i∈I1|Vi|)|I1|−2(∑i∈I1|Vi|)(2|Vm+1| − 1),the additional factor(∑i∈I1|Vi|)being due to the choice of root for the connection to Vm+1.Similarly, the other branches yield factors|Vm+1|(∑i∈Ir|Vi|)|Ir |−1 ∏i∈Ir(2|Vi| − 1), r = 2, . . . ,m.The resulting expression for Nm+1 is|V1|m∑p=1|Vm+1|p−1∑{I1,...,Ip}∈Πp({1,...,m})p∏r=1(∑i∈Ir|Vi|)|Ir |−1 ∏i∈Iri6=1(2|Vi| − 1) .Identity (1) yieldsNm+1 = |V1| m∑p=1(m− 1p− 1)|Vm+1|p−1(m∑i=1|Vi|)m−pm+1∏i=2(2|Vi| − 1)= |V1|(m+1∑i=1|Vi|)m−1 m+1∏i=2(2|Vi| − 1).This concludes the proof of Theorem 2. 3 Proof of Theorem 1.The proof is done by induction on m and p. Note that, for m = p both sides are equal to 1and for p = 1 both sides are equal to (∑mi=1 xi)m−1. Now assume that identity (1) holds truefor a given m ≥ 1 and all p ≤ m. Let Lm and Rm denote the left-hand side and right-handside of (1) respectively. We may assume that 1 ∈ I1 and expand the factor(∑i∈I1xi)|I1|−1by the binomial formula in powers of x1(∑i∈I1xi)|I1|−1=|I1|−1∑n=0(|I1| − 1n)xn1 ∑i∈I1\{1}xi|I1|−1−n.2Inserting this into Lm+1 and denoting I˜1 := I1 \ {1}, we haveLm+1 =m+1−p∑n=0xn1∑{I1,...,Ip}∈Πp({1,...,m+1})1∈I1, |I1|≥n+1(|I˜1|n)∑i∈I˜1xi|I˜1|−np∏j=2∑i∈Ijxi|Ij |−1.If n = 0, we separate out I˜1 = ∅, for which {I2, . . . , Ip} ∈ Πp−1({2, . . . ,m + 1}). If I˜1 6= ∅then {I˜1, I2, . . . , Ip} ∈ Πp({2, . . . ,m + 1}). Conversely, given a partition {I˜1, I˜2, . . . , I˜p} ∈Πp({2, . . . ,m+ 1}) we obtain a unique partition of {1, . . . ,m+ 1} by adding 1 to any of thesets I˜l with l ∈ {1, . . . , p}. We can therefore writeLm+1 =∑{I2,...,Ip}∈Πp−1({2,...,m+1})p∏j=2∑i∈Ijxi|Ij |−1+m+1−p∑n=0xn1p∑l=1∑{I˜1,...,I˜p}∈Πp({2,...,m+1})|I˜l|≥n(|I˜l|n)∑i∈I˜lxi|I˜l|−np∏j=1j 6=l∑i∈I˜jxi|I˜j |−1.Note that in the second term on the right-hand side, |I˜l| 6= 0 since {I˜1, . . . , I˜p} ∈ Πp({2, . . . ,m+1}). Expanding the quantity Rm+1 (the right-hand side of (1) but with m + 1) in powersof x1 and then replacing {2, . . . ,m+ 1} by {1, . . . ,m}, it follows that it suffices to prove theequivalent identities∑{I1,...,Ip−1}∈Πp−1({1,...,m})p−1∏j=1∑i∈Ijxi|Ij |−1+p∑l=1∑{I˜1,...,I˜p}∈Πp({1,...,m})∑i∈I˜lxi|I˜l|p∏j=1j 6=l∑i∈I˜jxi|I˜j |−1=(mp− 1)( m∑i=1xi)m+1−p(3)for n = 0, andp∑l=1∑{I˜1,...,I˜p}∈Πp({1,...,m})|I˜l|≥n(|I˜l|n)∑i∈I˜lxi|I˜l|−np∏j=1j 6=l∑i∈I˜jxi|I˜j |−1=(mp− 1)(m+ 1− pn)( m∑i=1xi)m+1−p−n(4)for all 1 ≤ n ≤ m+1− p. We start with the case of n = 0. By the induction hypothesis, thefirst term on the left-hand side of (3) is equal to(m− 1p− 2)( m∑i=1xi)m+1−p.The second term on the left-hand side of (3) can be rewritten as∑{I˜1,...,I˜p}∈Πp({1,...,m})p∏j=1∑i∈I˜jxi|I˜j |−1p∑l=1∑i∈I˜lxi = (m− 1p− 1)( m∑i=1xi)m+1−p,3again by the induction hypothesis. Adding the two contributions for n = 0 yields the right-hand side of (3). For the case n = 1, the left-hand side of (4) can simply be rewrittenas∑{I˜1,...,I˜p}∈Πp({1,...,m})(p∑l=1|I˜l|)p∏j=1∑i∈I˜jxi|I˜j |−1= m(m− 1p− 1)( m∑i=1xi)m−p=(mp− 1)(m+ 1− p)(m∑i=1xi)m−p.Next, we prove the case where 2 ≤ n ≤ m + 1 − p. The key idea is to apply the derivationoperator∑mk=1∂n−1∂xn−1kto the left-hand side of (1). This givesm∑k=1∂n−1∂xn−1k∑{I1,...,Ip}∈Πp({1,...,m})p∏j=1∑i∈Ijxi|Ij |−1=p∑j′=1∑{I1,...,Ip}∈Πp({1,...,m})p∏j=1j 6=j′∑i∈Ijxi|Ij |−1 ∑k∈Ij′∂n−1∂xn−1k∑i∈Ij′xi|Ij′ |−1. (5)The j′ term on the right-hand side of (5) is equal to zero unless |Ij′ | ≥ n. In that case,∑k∈Ij′∂n−1∂xn−1k∑i∈Ij′xi|Ij′ |−1= |Ij′ |n−1∏r=1(|Ij′ | − r)∑i∈Ij′xi|Ij′ |−n,independently of k ∈ Ij′ . Hence, the left-hand side of (5) can be rewritten asp∑j′=1∑{I1,...,Ip}∈Πp({1,...,m})|Ij′ |≥nn−1∏r=0(|Ij′ | − r)∑i∈Ij′xi|Ij′ |−np∏j=1j 6=j′∑i∈Ijxi|Ij |−1,which is nothing but n! times the left-hand side of (4). On the other hand, applying thederivation operator∑mk=1∂n−1∂xn−1kto the right-hand side of (1) givesm(m− 1p− 1) n−2∏r=0(m− p− r)(m∑i=1xi)m+1−p−n= n!(mp− 1)(m+ 1− pn)( m∑i=1xi)m+1−p−n,which is just n! times the right-hand side of (4). This concludes the proof of (4), and hencealso the proof of the theorem. Acknowledgment.The authors thank Adrien Kassel for having brought to their attention reference [2]. The sec-ond author gratefully acknowledges the financial support of the Ukrainian Scientific Project”III-12-16 Research of models of mathematical physics describing deterministic and stochasticprocesses in complex systems of natural science”.4References[1] T.C. Dorlas, A.L. Rebenko, B. Savoie, Correlation of clusters: Partially truncated correlationfunctions and their decay, e-print arXiv:1811.12342[2] R.P. Stanley, Enumerative Combinatorics, Volume 2, Cambridge Studies in Advanced Mathe-matics, Series 62, Cambridge University Press, Cambridge, 19995

A combinatorial identity

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A combinatorial identity

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