# Extremal Trees for the Exponential of Forgotten Topological Index

Authors: Akbar Jahanbani; Murat Cancan; Ruhollah Motamedi
Journal: Journal of Mathematics
 (2022)

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## Abstract

LetF be the forgotten topological index of a graph G. The exponential of the forgotten topological index is defined as eFG=∑x,y∈Stx,yGex2+y2, where tx,yG is the number of edges joining vertices of degree x and y. Let Tn be the set of trees with n vertices; then, in this paper, we will show that the path Pn has the minimum value for eF over Tn.


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## Body

## 1. Introduction

In this paper, letV=VG and EG be the vertex set and edge set, respectively. Let dv=dGv be the degree of a vertex v in graph G. A vertex of degree one is a pendant vertex or a leaf. A branching vertex v of a tree T is a vertex of degree dv≥3.LetP=w0w1,…,wq−1wq be the path graph with length of rP=q; if dTw0≥3,dTw1=…=dTwq−1=2 and dTwq=1, then we called P is a pendant path.Recently, topological indices have been considered by many researchers due to their many applications in various sciences. The forgotten topological index is defined in [1] as follows:(1)FG=∑uv∈EGdu2+dv2.For applications of the forgotten topological index, see [2–4].Before starting a new definition, we consider the setS=x,y∈N×N:1≤x≤y≤n−1, and let tx,yG be the number of edges joining vertices of degree x and y in a graph G. Therefore, the new definition will be as follows:(2)F=FG=∑x,y∈Stx,yGx2+y2.The exponential of the forgotten topological indexF, denoted by eF, is defined as(3)eF=eFG=∑x,y∈Stx,yGex2+y2.Recently, the exponential topological indices have attracted the attention of many researchers. In [5], the exponential Randić index is characterized. In [6], the authors have characterized the exponential atom bond connectivity and the exponential augmented Zagreb index. In [7], the problem maximal value of trees for the exponential second Zagreb index is solved. Then, in this paper, we solve the problem with finding the minimal value of eF among trees.


## 2. Trees with Minimum Exponential of the Forgotten Topological Index

In this section, we will show that the pathPn has the minimal value of the exponential forgotten topological index among all trees.Lemma 1.
LetT and T1 be the trees in Figure 1 and A be a subtree of T. If s≥3, then eFT>eFT1.Figure 1
The treesT and T1.Proof.
By settingk=dTu, hence, we can write(4)eFT−eFT1=es2+k2+s−1e1+s2−e4+k2+s−2e8+e5=es2+k2−e4+k2+s−2e1+s2−e8+e1+s2−e5≥e9+k2−e4+k2+e10−e8+e10−e5>e9+k2−e4+k2>0.Lemma 2.
LetT be a tree with minimum value of eF in Tn and let u be a pendant vertex T, v∈T. If uv∈ET, then dTv=2.Proof.
SupposedTv=g and P be the largest path of T and contains v. Let t be an end vertex of P and o a vertex in P, where ot∈ET; hence, by applying Lemma 1, we have dTo=2.
We continue the proof with the following two cases.Case 1.
IfdTv=g≥4.
Assuming thatT1 be the tree in Figure 2 and Av=∑i=1g−1eg2+yi2, where y1,…,yg−1 are the degrees of the adjacent vertices to v different from u. Hence, we can write(5)eFT−eFT1=Av+e1+g2+e5−∑i=1g−1eg−12+yi2−e8−e5=Av−∑i=1g−1eg−12+yi2+e1+g2−e8≥Av−∑i=1g−1eg−12+yi2+e17−e8>Av−∑i=1g−1eg−12+yi2>0.
This contradicts the minimality ofT.Figure 2
The treesT and T1.Case 2.
IfdTv=3.
Supposeh,e∈VT and hv,ev∈ET, where h,e≠u, dTh=a, and dTe=b. By applying Lemma 1, we get a≥2 and b≥2. Let T2 be the tree described in Figure 3. Therefore, we can write(6)eFT−eFT2=e9+a2+e9+b2+e10−ea2+b2−2e8=e9+a2+e9+b2−ea2+b2+e10−2e8.
Here, we show(7)fa,b=e9+a2+e9+b2+e10>ea2+b2+2e8.
Since(8)fa,2=e9+a2+e13+e10>ea2+4+2e8and(9)fa,3=e9+a2+e18+e10>ea2+9+2e8,the above inequality holds for a≥2. Therefore, for a,b≥2, we have fa,b>0. Hence, we get eFT>eFT2. That is a contradiction; hence, we get dTv=2.
Letv be a branching vertex of degree y of a tree T; hence, T can be viewed as the coalescence of y subtrees of T at the vertex v. We call T1,…,Ty are the y branches of T at v (see Figure 4).Figure 3
The treesT and T2.Figure 4
Branches of the treeT at v.Definition 1.
(see [5]). A branching vertex x of tree T is an outer branching vertex of T if all branches of T at x except for possibly one are paths.Lemma 3 (see [5])..
A treeT∈Tn has no outer branching vertex if and only if T≅Pn.Lemma 4.
LetT,T1∈Tn be the trees in Figure 5 and R be a subtree of T, z≥3 and x=dTu≥2. Then, eFT>eFT1.Figure 5
The treesT and T1.Proof.
By direct calculation, it is not difficult to see3≤z≤8. Here, we let z≥9; therefore, we have(10)eFT−eFT1=ez+12+4+z−1ez+12+4−ez2+4+ez+12+x2−ez2+x2+e5−2e8>ez+12+4+e5−2e8≥e104+e5−2e8>0.Lemma 5.
LetT,T1∈Tn be the trees in Figure 6 and t≥3. Then, eFT>eFT1, where sPu=−2+∑i=1tsPi.Figure 6
The treesT and T1.Proof.
We describe the graph in Figure7; let Pt be the path of length sPt=−2t+2+∑i=1tsPi. It is not difficult to see eFT=eFT3. Then, by repeated of Lemma 4, we get eFT>eFT1.Figure 7
The treeT3.Corollary 1.
LetT∈Tn be a tree with a unique outer branching vertex and every pendant path has length at least 2. Then, eFT>eFPn.Proof.
IfT has a unique outer branching vertex, hence, T has the form trees in Figure 6. Let M be the tree in Figure 8. If s=2, then eFT=eFM. If s≥3, then by using Lemma 5, we have eFT>eFM. Hence, we can write(11)eFT−eFPn=3e5+3e13+n−7e8−2e5−n−3e8=e5+3e13−4e8>0.Figure 8
The treeM.Lemma 6.
LetT,T1∈Tn be the trees in Figure 9, such that 1≤dTv≤3; then, eFT>eFT1.Figure 9
The treesT and T1.Proof.
By settingx=dTv, we can write(12)eFT−eFT1=2e5+2e13+e9+x2−e5−3e8−e4+x2=e5+2e13−3e8+e9+x2−e4+x2>e5+2e13−3e8>0.Lemma 7.
LetT∈Tn be the tree in Figure 10, s≥1 and z≥0; then, T is not minimal in Tn.Figure 10
The treeT.Proof.
Setx=dTu≥2, and let T1 be tree in Figure 11. Hence, we have eFT>eFT1 if the following conditions are hold:(1)
s≥1,z≥2.(2)
s≥4,z≥0.It is not difficult to see that our result holds fors+z≤11. Therefore, we let z+s≥12. Then,(13)eFT−eFT1=2e5+2e13−5e8+s−1e9+s+z+12−e9+s+z2+ze4+s+z+12−e4+s+z2+ex2+s+z+12−ex2+s+z2+e9+s+z+12>2e5+2e13−5e8>0.To continue the proof, we must consider the following conditions:(3)
s=1 and z=0.(4)
s=1 and z=1.(5)
s=2 and z=0.(6)
s=3 and z=1.(7)
s=2 and z=1.(8)
s=3 and z=0.
Note that, in (3), (4), and (5),, we have2≤dTw≤3; therefore, by Lemma 6, we can obtain trees with the minimum value of eF.
Here, if (6) holds, then we consider graphT2 in Figure 12. Hence, we can write(14)eFT−eFT2=e5−2e8+3e34−e25+e29−e20+ex2+25−ex2+16+e20>e5−2e8+3e34−e25+e29−e20+e20>e5−2e8+e20>0.
If (7) holds, then we consider graphB in Figure 13. So, we have(15)eFT−eFB=6e5+4e13+2e25+e20+ex2+16−3e5−2e13−8e8−ex2+9=ex2+16−ex2+9+3e5+2e13+2e25+e20−8e8>e20−8e8>0.
Finally, if (8) holds, then we consider graphC in Figure 14. Hence, we can write(16)eFT−eFC=7e5+6e13+3e25+ex2+16−3e5−2e13−11e8−ex2+9=ex2+16−ex2+9+4e5+4e13+3e25−11e8>4e5+4e13+3e25−11e8>0.
Therefore,T is not minimal in Tn.Figure 11
The treeT1.Figure 12
The treesT and T2.Figure 13
The treesT and B.Figure 14
The treesT and C.Theorem 1.
LetT∈Tn and T≇Pn; then, T is not minimal for eF.Proof.
By using Lemma3, we know that T has an outer branching vertex u. Using Lemmas 2, 5, and 6, we let all pendant paths of T have length at least 2 and T has the form in Figure 15, such that dTv≥4, and otherwise, T is not minimal. If u is the unique outer branching vertex of T, then the result obtained by Corollary 1. Otherwise, among all outer branching vertices of T, choose u as the farthest from u. From Lemma 5, we let T is the form in Figure 16, such that dTv1≥4. Note that u1 is the farthest outer branching vertex from u; it is clear if Ti is not a path; then, wi is an outer branching vertex of T, and by Lemma 5, we let Ti have the form in Figure 17. Therefore, we get eFT=eFE, where E is described in Figure 18 and s+z=q+1. The result follows from Lemma 7.Figure 15
The treeT.Figure 16
The treeT.Figure 17
The treeTi.Figure 18
The treeT and E.



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