# Limit points of (signless) Laplacian spectral radii of linear trees

Francesco Belardo  Department of Mathematics and Applications, University of
Naples Federico II, Italy<EMAIL_ADDRESS>,  Elismar R. Oliveira  Instituto
de Matemática e Estatística, UFRGS, Porto Alegre, Brazil
<EMAIL_ADDRESS>and  Vilmar Trevisan  Instituto de Matemática e
Estatística, UFRGS, Porto Alegre, Brazil<EMAIL_ADDRESS>
###### Abstract.

We study limit points of the spectral radii of Laplacian matrices of graphs.
We adapted the method used by J. B. Shearer in 1989, devised to prove the
density of adjacency limit points of caterpillars, to Laplacian limit points.
We show that this fails, in the sense that there is an interval for which the
method produces no limit points. Then we generalize the method to Laplacian
limit points of linear trees and prove that it generates a larger set of limit
points. The results of this manuscript may provide important tools for proving
the density of Laplacian limit points in $[4.38+,\infty)$.

###### Key words and phrases:

spectral radius; tree; limit points;

###### 1991 Mathematics Subject Classification:

05C50, 05C05, 15A18

##  1\. Introduction

The 1972 seminal paper of A. J. Hoffman [4] introduced the concept of limit
points of eigenvalues of graphs. Let $\mathcal{A}$ be the set of all symmetric
matrices of all orders, every entry of which is a non-negative integer and
$R=\\{\rho:\rho=\rho(A)\mbox{ for some }A\in\mathcal{A}\\}$ where $\rho(A)$ is
the largest eigenvalue of $A$. He asked which real number can be in $R$ and
showed that it is sufficient to consider matrices of $\mathcal{A}$ having only
entries in $\\{0,1\\}$ and 0 diagonal, e.g. adjacency matrices of graphs.
Additionally, he determined all limit points of $R\leq\sqrt{2+\sqrt{5}}$. More
precisely, let $\tau=\frac{1+\sqrt{5}}{2}$ (the golden mean). For
$n=1,2,\ldots$, let $\overline{\beta}_{n}$ be the positive root of

$Q_{n}(x)=x^{n+1}-\left(1+x+x^{2}+\cdots+x^{n-1}\right).$

Let $\bar{\alpha}_{n}=\bar{\beta}_{n}^{1/2}+\bar{\beta}_{n}^{-1/2}\cdot$ Then

$2=\bar{\alpha}_{1}<\bar{\alpha}_{2}<\cdots$

are all the limit points of $R$ smaller than
${\displaystyle\lim_{n\to\infty}\bar{\alpha}_{n}=\tau^{1/2}+\tau^{-1/2}=\sqrt{2+\sqrt{5}}(=2.05+)}$.

In 1989, a remarkable result due to J. B. Shearer [13] extended the work of
Hoffman. He showed that every real number larger than $\sqrt{2+\sqrt{5}}$ is a
limit point of $R$.

In his original paper, Hoffman asks about limit points of other eigenvalues of
graphs, and this originated several interesting results on this topic
culminating with the fact that every real number is a limit point of some
eigenvalue of a graph (see Zhang & Chen [16]). We remark that Estes [2] proved
in 1992 that every totally real algebraic integer – which are roots of monic
integral polynomials having only real roots – is an eigenvalue of a graph.
Since eigenvalues are totally real algebraic integers (roots of the
characteristic polynomial), it follows that this set is dense in the real
line. What is perhaps surprising, is that only eigenvalues of trees are needed
to be considered for these results. Salez [12], extending Estes’ result,
showed that any totally real algebraic integer is an eigenvalue of a tree,
while both Hoffman and Shearer found sequences of trees whose spectral radii
were limit points. It seems remarkable that the set composed only by the
largest eigenvalue of the adjacency matrix of trees is dense in the interval
$[\sqrt{2+\sqrt{5}},\infty)$.

It is worth noticing that in 2003, Kirkland [8] showed that any nonnegative
real number is a limit point for the algebraic connectivity (the second
smallest Laplacian eigenvalue). The same author [9] considered limit points
for the positive eigenvalues of the normalized Laplacian matrix of a graph.
Specifically, he shows that the set of limit points for the $j$-th smallest
such eigenvalues is equal to $[0,\,1]$, while the set of limit points for the
$j$-th largest such eigenvalues is equal to $[1,\,2]$. Wang et al. [14] proved
that any nonnegative real number is a limit point for some eigenvalue of the
signless Laplacian matrix of graphs.

In this paper, we are interested in the problem originated by Hoffman’s
question, which deals only with limit points of the spectral radius of graphs.
More specifically, we want to study the Laplacian version of Hoffman and
Shearer’s results, that is, what real numbers are limit points of the spectral
radii of Laplacian matrices of graphs. The converse of this problem may also
be viewed as _which sequences of graphs have Laplacian spectral radius with
limit points_ (we refer to the next section for the precise definitions).

In order to explain our results, we recall the work of Guo [3]. Let

$\omega=\frac{1}{3}(\sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}+1),$

$\leavevmode\nobreak\ \beta_{0}=1$ and $\beta_{n},\;n\geqslant 1$ be the
largest positive root of

$P_{n}(x)=x^{n+1}-\left(1+x+\cdots+x^{n-1}\right)(\sqrt{x}+1)^{2}.$

Let $\alpha_{n}=2+\beta_{n}^{\frac{1}{2}}+\beta_{n}^{-\frac{1}{2}}$. Then

$4=\alpha_{0}<\alpha_{1}<\alpha_{2}<\cdots$

are all the limit points of Laplacian spectral radii of graphs smaller than
$\lim_{n\rightarrow\infty}\alpha_{n}=2+\omega+\omega^{-1}$ (= $4.38+)$.

By analogy to the adjacency case, it is natural to ask whether any real number
$\mu\geq 2+w+w^{-1}=4.38+$ is the limit point of the Laplacian spectral radii
of graphs. We refer to Figure 1 for an illustration of the current state of
knowledge.

Figure 1. Current status of (Laplacian) limit points

Our contribution in this paper is the development of some analytical tools
allowing one to study the density of Laplacian spectral radius in
$[4.38+,\infty)$. We first adapt Shearer’s method to the Laplacian case,
verifying that it is not sufficient to prove density. In spite of the fact
that the method produces sequence of caterpillars having a limit point, we
show that this limit point is not always the _desired_ number. As a
consequence, we find a whole interval where the method produces no Laplacian
limit points. We then extend Shearer’s method to the class of linear trees,
e.g. we define sequences of linear trees whose Laplacian spectral radius has
limit points. This provides a generalization of Shearer’s process, since
caterpillars is a subclass of linear trees.

It is tempting to conjecture that any $\mu\geq 4.38+$ is a limit point of
Laplacian spectral radii of graphs, however, it seems that Laplacian spectral
radii are a bit slick! Our generalization improves Shearer’s process in the
sense that we are able to find a larger set of limit points. We do not know
whether sequences of linear trees suffice to prove the density conjecture (if
true).

The rest of the paper is organized as follows. In the next section, we present
the necessary notation, definitions and the main tools used in the paper. In
particular, we explain Shearer’s procedure using our tools. In Section 3, we
apply Shearer’s approach to Laplacian case, verifying that it fails. In
Section 4 we study the convergence of the Laplacian spectral radius of
sequences of linear trees. In Section 5 we use these results to generalize
Shearer’s work: we obtain a method that produces a sequence of linear trees
for a given number $\mu$ in a way that the Laplacian spectral radius has a
limit point $\leq\mu$. We call this _generalized Shearer’s sequence_ and show
that the process provide a larger set of limit points. In Section 6 we study
the Laplacian limit points given by the method. We present some numerical
evidence of the density of these limit points. We also prove a density result,
showing that a truncated sequence of linear trees have algebraic limit points
that is dense in a subset of the limit points. In Section 7, we apply some
analytical tools to obtain approximations, providing quicker ways to obtain
sequence of linear trees having a specified limit point. Finally, in Section
8, we suggest a few open problems and make final considerations.

######  Remark 1.1.

The results of this paper are stated for the Laplacian matrix, however, as all
the graphs involved are trees, they naturally extend to signless Laplacian
matrix, since the spectrum of these two matrices are equal for bipartite
graphs.

##  2\. Notation and preliminaries

Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. If
$|V|=n$, then its adjacency matrix $A(G)=[a_{ij}]$ is the $n\times n$ matrix
of zeros and ones such that $a_{ij}=1$ if and only if $v_{i}$ is adjacent to
$v_{j}$ (that is, there is an edge between $v_{i}$ and $v_{j}$). A value
$\lambda$ is an eigenvalue if ${\rm det}(\lambda I_{n}-A)=0$, and since $A$ is
real symmetric its eigenvalues are real. In this setting, we denote by
$\rho_{A}(G)$ the largest eigenvalue of $A(G)$, which is called the spectral
radius of $G$. One can also consider the Laplacian matrix $L(G)$ of a graph,
which is given by $L(G):=D(G)-A(G)$, where $D(G)=[d_{ij}]$ is the $n\times n$
diagonal matrix with $d_{ii}={\rm degree}(v_{i})$. A value $\mu$ is an
eigenvalue if ${\rm det}(\mu I_{n}-L)=0$, and since $L$ is positive semi-
definite, its eigenvalues are non-negative. We denote by $\rho_{L}(G)$ the
largest eigenvalue of $L(G)$ which is called the Laplacian spectral radius of
$G$. One can also consider the signless Laplacian associated to a graph, which
is denoted $Q(G):=D(G)+A(G)$.

Generalizing the concept of limit point of graphs, we say that a real number
$\gamma$ is an $M$-limit point of the $M$-spectral radius of graphs if there
exists a sequence of graphs $\\{G_{k}\;|\;k\in\mathbb{N}\\}$ such that

$\lim_{k\to\infty}\rho_{M}(G_{k})=\gamma,$

where $\rho_{M}(G_{i})\neq\rho_{M}(G_{j}),\;i\neq j$ and $M$ is a class of
matrices associated with a graph such as adjacency, Laplacian, signless
Laplacian, etc. (See Wang & Brunetti [15]).

In this note we will extend Shearer’s ideas for the adjacency matrix to the
Laplacian version. The background of this process can be recovered by the use
of modern methods such as the Jacobs-Trevisan diagonalization algorithm [5],
and properties of some recurrence equations from [10]. Those are our main
tools and we briefly explain here, so that the note is self contained.

Given a symmetric matrix $A$ whose underlying graph is a tree, the following
algorithm of Figure 2 outputs a diagonal matrix that is congruent to $A+xI$
and hence, the sign of its diagonal values determine the number of eigenvalues
that greater than/equal to/larger than $x$. More precisely, we state the
following result for future reference.

######  Lemma 2.1.

([5]) Let $A$ be a symmetric matrix whose underlying graph is a tree and let
$D$ be the matrix output by $\rm{Diagonalize}(A,-x)$. The number of
eigenvalues of $A$ smaller than/equal to/larger than $x$ is, respectively, the
number of negative/zero/positive diagonal values of $D$

Input: matrix $A=(a_{ij})$ and underlying tree $T$ whose vertices
$v_{1},\ldots,v_{n}$
---
are ordered bottom-up and real $x$
Output: diagonal matrix $D=\mbox{diag}(d_{1},\ldots,d_{n})$ congruent to
$A+xI$
Algorithm $\mbox{Diagonalize}(A,x)$
|  initialize $d_{i}\longleftarrow a_{ii}+x$, for all $i$
|  for  $k=1$ to $n$
|  |  if $v_{k}$ is a leaf then continue
|  |  else if $d_{c}\neq 0$ for all children $c$ of $v_{k}$ then
|  |  |  $d_{k}\longleftarrow d_{k}-\sum\frac{(a_{ck})^{2}}{d_{c}}$, summing over all children of $v_{k}$
|  |  else
|  |  |  select one child $v_{j}$ of $v_{k}$ for which $d_{j}=0$
|  |  |  $d_{k}\longleftarrow-\frac{(a_{jk})^{2}}{2}$
|  |  |  $d_{j}\longleftarrow 2$
|  |  |  if $v_{k}$ has a parent $v_{\ell}$, delete the edge $\\{v_{k},v_{\ell}\\}$.
|  end loop
Figure 2.  Diagonalizing $A$ for a symmetric matrix $A$ with an underlying
tree.

Applying the tree algorithm on a path of a tree, produces certain numerical
rational sequences. They all may be seen in a unified elementary form given by

(1) $x_{j+1}=\varphi(x_{j}),j\geq 1$

where $\varphi(t)=\alpha+\frac{\gamma}{t}$, for $t\neq 0$,
$\alpha,\gamma\in\mathbb{R}$ are fixed numbers ($\gamma\neq 0$) and $x_{1}$ is
a given initial condition. The explicit solution is a function $j\to f(j)$
such $x_{j}=f(j)$ for $j\geq 1$ and has been studied in [10].

###  2.1. Revisiting Shearer’s work for adjacency limit points

In [13] Shearer has proved the density of the limit points for adjacency
matrices in the interval $[\sqrt{2+\sqrt{5}},\infty)$. The proof is based on
an explicit construction of a sequence of graphs. For any given
$\lambda\geq\sqrt{2+\sqrt{5}}$, Shearer constructs a sequence of caterpillars
whose spectral radii converge to $\lambda$. We now explain, in our framework,
why this method works.

Recall that a _caterpillar_ is a tree in which the removal of all its leaves
transforms it into a path. An arbitrary caterpillar with $k$ back nodes
$v_{1},\ldots,v_{k}$, where $v_{i}$ has $r_{i}$ leaves may be represented by

$G_{k}:=[r_{1},r_{2},...,r_{k}],\;k\geq 2.$

We refer to Figure 3 for an illustration.

Figure 3. Representation of $G_{k}$

We notice that, if we replace each $r_{i}$ by a star tree with $r_{i}$ rays,
we have an alternative representation for $G$. For instance

$[3,1,0,2,2]=[[1,1,1],[1],[0],[1,1],[1,1]]$

are different representations of the same tree.

For each $\lambda\in[\sqrt{2+\sqrt{5}},\infty)$ Shearer finds a caterpillar
$G_{k}(\lambda)=[r_{1},r_{2},...,r_{k}],\;k\geq 2$, in such a way that the
number of leaves in each vertex satisfies a recurrence related to the number
$\frac{-\lambda\pm\sqrt{\lambda^{2}-4}}{2}$.

Consider a given number $\lambda\in[\sqrt{2+\sqrt{5}},\infty)$ and an
arbitrary caterpillar $G_{k}(\lambda):=[r_{1},r_{2},...,r_{k}],\;k\geq 2.$ We
apply the algorithm ${\rm Diagonalize}(A(G_{k}),-\lambda)$, the adjacency
matrix of $G_{k}$. Each leaf is initialized with $-\lambda<0$. In the vertex
$v_{1}$ we obtain the value

$R_{1}:=-\lambda-\frac{r_{1}}{-\lambda}=-\lambda+\frac{r_{1}}{\lambda},$

and in the next vertex $v_{2}$ we obtain the value

$R_{2}:=-\lambda-\frac{r_{2}}{-\lambda}-\frac{1}{R_{1}}=-\lambda-\frac{1}{R_{1}}+\frac{r_{2}}{\lambda},$

and so on.

We observe that the sequence of equations may be seen as the recurrence
relation

(2) $\left\\{\begin{array}[]{ll}x_{j+1}=\varphi(x_{j})\\\
x_{1}=-\lambda,\end{array}\right.$

where $\varphi(t)=-\lambda-\frac{1}{t},\;t\neq 0$, disturbed by a drift factor
$\delta_{j}:=\frac{r_{j}}{\lambda}\geq 0$.

From [10] we know that the iterates $x_{j}$ obeying Equation (2) are
asymptotic to $\theta=\frac{-\lambda-\sqrt{\lambda^{2}-4}}{2}$, which is a
fixed point $\varphi(t)=t$, satisfies $t^{2}+\lambda t+1=0$. It is shown that
$\theta:=\frac{-\lambda-\sqrt{\lambda^{2}-4}}{2}$ is an attracting point and
$\theta^{\prime}:=\frac{-\lambda+\sqrt{\lambda^{2}-4}}{2}=\theta^{-1}>\theta$
is a repelling point, with respect to the iteration process (2).

From Lemma 2.1, it follows that $\rho_{A}(G_{k})<\lambda$ if and only if
$R_{j}<0$ for all $j$. As the attracting interval for $\theta$ is
$(-\infty,\theta^{\prime}]$ we need to require that $R_{j}\leq\theta^{\prime}$
otherwise $R_{j^{\prime}}>0$ for some $j^{\prime}>j$ contradicting
$\rho_{A}(G_{k})<\lambda$.

This is the basis for the Shearer’s construction (see [13] equations (1)-(4)),
that is, the sequential choice of $r_{1},r_{2},...,r_{k}$:
If $-\lambda+\frac{r_{1}}{\lambda}\leq\theta^{\prime}$ then $R_{1}<0$, thus we
choose

$r_{1}:=\max_{r\geq
0}\\{-\lambda+\frac{r}{\lambda}\leq\theta^{\prime}\\}\Leftrightarrow
r_{1}:=\lfloor\lambda(\theta^{\prime}-(-\lambda))\rfloor.$

If $\varphi(R_{1})+\frac{r_{2}}{\lambda}\leq\theta^{\prime},$ then $R_{2}<0$,
thus we choose

$r_{2}:=\max_{r\geq
0}\\{\varphi(R_{1})+\frac{r_{2}}{\lambda}\leq\theta^{\prime}\\}\Leftrightarrow
r_{2}:=\lfloor\lambda(\theta^{\prime}-\varphi(R_{1}))\rfloor,$

and so on.

######  Definition 2.2.

Given $\lambda\geq\sqrt{2+\sqrt{5}}$ we define the adjacency Sharer’s sequence
as the sequence of caterpillars
$G_{k}(\lambda):=[r_{1},r_{2},...,r_{k}],\;k\geq 2$ where

$\left\\{\begin{array}[]{ll}r_{1}:=\lfloor\lambda(\theta^{\prime}-(-\lambda))\rfloor;\\\
R_{1}=-\lambda+\frac{r_{1}}{\lambda};\\\
r_{j}:=\lfloor\lambda(\theta^{\prime}-\varphi(R_{j-1}))\rfloor,2\leq j\leq
k;\\\ R_{j}=\varphi(R_{j-1})+\frac{r_{j}}{\lambda},2\leq j\leq
k.\end{array}\right.$

The adjacency Sharer’s sequence $G_{k},\;k\geq 2$ is such that
$\rho_{A}(G_{k})<\rho_{A}(G_{k+1})<\lambda$. Indeed, the part
$\rho_{A}(G_{k})<\lambda$ is a trivial consequence of the construction because
$R_{1},...,R_{k}<\theta^{\prime}<0$ then the spectral radius is necessarily
smaller than $\lambda$ (see Theorem 3, from [5]). The part
$\rho_{A}(G_{k})<\rho_{A}(G_{k+1})$ is a consequence of the interlacing
property because $G_{k}$ is always a subgraph of $G_{k+1}$ since the choice of
$r_{k}:=\lfloor\lambda(\theta^{\prime}-\varphi(R_{k-1}))\rfloor$ is the same
for both graphs.

In particular there exists

$\lim_{k\to\infty}\rho_{A}(G_{k})=\gamma(\lambda)\leq\lambda.$

Shearer [13] showed that for every real number $\lambda\geq\sqrt{2+\sqrt{5}}$,
the limit point $\gamma(\lambda)=\lambda$, that is
$\displaystyle\lim_{k\to\infty}\rho_{A}(G_{k})=\lambda$.

In this note we will adapt Shearer’s construction to the Laplacian matrix. For
this, we set next some notation that is going to be used throughout the paper.

###  2.2. Notation for Laplacian Matrices

From Guo’s result [3], the problem for limit points of spectral radii of
Laplacian matrices that remains to study is the interval $\mu\geq 4.38+$. More
precisely, the question is whether any real number $\mu\in[4.38+,\infty)$ is a
limit point of Laplacian spectral radius of graphs.

In the next section, we extend Shearer’s method of caterpillar construction
for Laplacian matrices. We will see the necessity to consider other trees.
Here, we set the notation for this larger set of trees.

We recall that a _linear tree_ is a tree obtained by attaching a starlike tree
(that has at most one vertex of degree larger than 2) at each vertex of a
path. We denote a linear tree by $G=[T_{1},\ldots,T_{k}]$, for $k\geq 1$,
where all high degree vertices (HDVs) are located in one single path
$P=\\{v_{1}\to v_{2}\to\cdots\to v_{k}\\}$, called the main path. At each
vertex $v_{j}$ (we call then back nodes), we attach a number of paths
$P_{q_{1}^{j}},\ldots,P_{q_{r_{j}}^{j}}$ forming a starlike tree, which is
denoted by $T_{j}=[q_{1}^{j},...,q_{r_{j}}^{j}],q_{i}^{j}\leq q_{i+1}^{j}$,
for $1\leq j\leq k$ (see Figure 4 for an illustration).

The class of linear trees was introduced by Johnson, Li & Walker in [6] and
further studied by Johnson & Saiago (see [7] Chapter 10). Some special cases
of linear trees are caterpillars, where all the HDVs have a number of pendant
leafs $P_{1}$, and the open quipus, where all the HDVs have degree at most
three.

Figure 4. Linear tree $G_{k}=[[1,2,2,3],[1,1,2],\ldots,[1,1],[1,1,2]]$

We notice that the representation $T_{j}=[0]$ means that the graph $T_{j}$ is
formed just by the vertex $v_{j}$, or that $T_{j}$ is empty.

We denote by $\ell(G_{k})=k$, the _length_ of $G=[T_{1},\ldots,T_{k}]$, as the
number of back nodes in the main path, by $\omega(T_{j})=r_{j}$, the _width_
of $T_{j}=[q_{1}^{j},\ldots,q_{r_{j}}^{j}]$, as the number of paths it is
composed, and by $h(T_{j})=q_{r_{j}}^{j}$, the height of
$T_{j}=[q_{1}^{j},...,q_{r_{j}}^{j}]$, as the maximum length of paths in it.

The representation $G_{k}=[T_{1},...,T_{k}]$ is almost never unique unless
$T_{1}=T_{k}=[0]$ and $k$ is fixed, because we may choose another main path by
picking paths from $T_{1}$ and/or $T_{k}$. For instance
$[[2,2,2],[1,2],[1,1]]$ and $[[0],[0],[2,2],[1,2],[1,1]]$ are representations
of the same linear tree. Although, in any new representation the length could
increase at most by $h(T_{1})+h(T_{k})$.

We consider applying ${\rm Diagonalize}(L,-\mu)$, for $\mu\geq 4.38+$ and $L$
the Laplacian matrix of a linear tree with $k$ back nodes
$G_{k}=[T_{1},T_{2},\ldots,T_{k}]$.

When executing ${\rm Diagonalize}(L,-\mu)$, we first apply in
$T_{i},\leavevmode\nobreak\ i=1,\ldots,k$, towards the vertex $v_{k}$, which
is the root. We remark that $T_{i}$ may be empty and in this case there is
nothing to be done. If $T_{i}$ is a non-empty starlike, for each path of
$T_{i}$, we have that leaves receive the value $b_{1}=1-\mu$, while the
remaining values satisfy $b_{j}=2-\mu-\frac{1}{b_{j-1}}$, for $j>1$. We refer
to Figure 5 for an illustration.

Figure 5. Linear tree with the values obtained by ${\rm Diagonalize}(L,-\mu)$
depicted in each vertex.

It will be convenient to denote, for future reference, the function
$\psi:\mathbb{R}\rightarrow\mathbb{R}$ by

(3) $\psi(t):=2-\mu-\frac{1}{t}.$

With this notation, the values in each path of $T_{i}$ satisfy

(4) $\left\\{\begin{array}[]{ll}b_{1}=1-\mu\\\
b_{j+1}=\psi(b_{j}),\leavevmode\nobreak\ \leavevmode\nobreak\
j>1.\end{array}\right.$

From [10] we know that the iterates $b_{j}$ obeying Equation (4) have the
following properties. Let $\theta$ and $\theta^{\prime}$ be fixed points of
$\psi(t)=t$. They are given by

(5) $\theta:=\frac{-(\mu-2)-\sqrt{(\mu-2)^{2}-4}}{2}\mbox{ and
}\theta^{\prime}:=\frac{-(\mu-2)+\sqrt{(\mu-2)^{2}-4}}{2}=\theta^{-1}>\theta.$

They are attracting and repelling points, respectively, meaning that for
$b_{1}\leq\theta^{\prime}$, we have $b_{j}$ converging to $\theta$. Since
$\mu>4$, we have that $b_{1}<\theta^{\prime}<0$ and, in particular, that all
$b_{j}<0$. We process each path of $T_{i}$ until the vertices adjacent to
$v_{i}$. We notice that to process the vertex $v_{i}$ we need the value
resultant from processing $T_{i}$. We denote by $\omega(T_{i})=r_{i}$ the
number of paths of $T_{i}$ and, for $m=1,\ldots,r_{i}$, by $b_{i,m}$ the last
value output by the algorithm at the $m$-th path of $T_{i}$. The net result of
the starlike $T_{i}$ is given by

(6)
$\delta(T_{i},\mu):=\left\\{\begin{array}[]{ll}\sum_{j=1}^{\ell_{i}}\left(1-\frac{1}{b_{i,j}}\right),\mbox{
for non-empty }T_{i}\\\ 0\mbox{ if }T_{i}\mbox{ is empty. }\\\
\end{array}\right.$

Continuing applying the algorithm, we go from $v_{1}$ towards $v_{k}$, the
root of $G_{k}$. We denote $S_{i}$, for $i\in\\{1,\ldots,k\\}$, the value
output by ${\rm Diagonalize}(L,-\mu)$. We have
$S_{1}=1-\mu+\delta(T_{1},\mu),\leavevmode\nobreak\
S_{2}=2-\mu+\delta(T_{2},\mu)-\frac{1}{S_{1}},\leavevmode\nobreak\
S_{3}=2-\mu+\delta(T_{2},\mu)-\frac{1}{S_{2}}$ and so on. For $S_{k}$ the
expression is $S_{k}:=1-\mu-\frac{1}{S_{k-1}}+\delta(T_{k},\mu)$ (see Figure 5
for an illustration). Using the function $\psi$, we notice that $S_{i}$ obeys
the following recurrence.

(7) $\left\\{\begin{array}[]{ll}S_{1}=1-\mu+\delta(T_{1},\mu),\\\
S_{i}=\psi(S_{i-1})+\delta(T_{2},\mu),\;2\leq i\leq k-1,\\\
S_{k}=-1+\psi(S_{k-1})+\delta(T_{k},\mu).\end{array}\right.$

We are interested in determining whether $\mu$ is smaller than, equal to, or
larger than $\rho_{L}(G_{k})$, the Laplacian spectral radius of $G_{k}$. Since
the values of all $b_{j}$’s are negative, by Lemma 2.1, only the signs of the
values $S_{j}$ suffice to determine this. More precisely, consider the
function

$\Pi:=(G_{k},\mu)\to\left(S_{1},S_{2},\ldots,S_{k}\right),$

and let
$\mbox{sign}(\Pi)(G_{k},\mu)=\left(\mbox{sign}(S_{1}),\mbox{sign}(S_{2}),\ldots,\mbox{sign}(S_{k})\right)$.
From Lemma 2.1 we can state the following result.

######  Lemma 2.3.

With the established notation, we have

  * $(a)$

$\rho_{L}(G_{k})<\mu\Leftrightarrow\mbox{sign}(\Pi)(G_{k},\mu)=(-,-,\ldots,-,-)$;

  * $(b)$

$\rho_{L}(G_{k})=\mu\Leftrightarrow\mbox{sign}(\Pi)(G_{k},\mu)=(-,-,\ldots,-,0)$;

  * $(c)$

$\rho_{L}(G_{k})>\mu\Leftrightarrow\mbox{sign}(\Pi)(G_{k},\mu)$ has a + entry.

##  3\. Applying Shearer’s approach for Laplacian limit points

In the previous section, we set a framework to apply ${\rm
Diagonalize}(L(G_{k}),-\mu)$, where $G_{k}$ is a linear tree and $\mu$ is any
real number larger than 4. We are going to use the notation established there,
but we observe a subtle difference. Here, from a given number $\mu\geq 4.38+$,
we construct a prescribed caterpillar
$G_{k}(\mu):=[r_{1},r_{2},...,r_{k}],\;k\geq 2$ (see Figure 3) by adapting
Shearer’s method to Laplacian matrices. Additionally, we will study the
convergence of $\rho_{L}(G_{k}(\mu))$.

Let us then fix a real number $\mu\in[4.28+,\infty)$. Shearer’s approach needs
to determine numbers $r_{1},r_{2},...,r_{k},$ such that
$G_{k}(\mu)=[r_{1},r_{2},...,r_{k}]$ has $\rho_{L}(G_{k})<\mu$. In the
language of Section 2.2, the linear trees $T_{i}$ are simply stars, whose
paths have length one. As $T_{i}=[1,1,\ldots,1]$, we have
$\delta(T_{i},\mu):=\sum_{j=1}^{r_{i}}\left(1-\frac{1}{b_{1}}\right)=\sum_{j=1}^{r_{i}}\left(1-\frac{1}{1-\mu}\right)=r_{i}\frac{\mu}{\mu-1}=\delta_{i}.$
We recall that, from Lemma 2.3, $\rho_{L}(G_{k})<\mu$ if and only if $S_{j}<0$
for all $j<k$. As the attracting interval for $\theta$, given by Equation (5),
is $(-\infty,\theta^{\prime}]$ we need to require that
$S_{j}\leq\theta^{\prime}$ otherwise $S_{j^{\prime}}>0$ for some
$j^{\prime}>j$ contradicting $\rho_{L}(G_{k})<\mu$.

Following the reasoning of Shearer’s construction, we determine the sequential
choice of $r_{1},r_{2},\ldots,r_{k}$ as follows.
Since $S_{1}<0\text{ if }1-\mu+r_{1}\frac{\mu}{\mu-1}\leq\theta^{\prime}$, we
choose

$r_{1}:=\max_{r\geq
0}\left\\{1-\mu+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}\Leftrightarrow
r_{1}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-(1-\mu))\right\rfloor.$

We point out that here we deal with real inequalities, such as $S_{1}<0$, or
equivalently $1-\mu+r_{1}\frac{\mu}{\mu-1}\leq\theta^{\prime}$. Of course,
this inequality is true for $r_{1}=0$. Moreover, if we increase $r$ by one,
then the left hand side of the inequality increases by $\frac{\mu}{\mu-1}>0$.
Thus, there will be a maximum number $r$ satisfying the inequality. If we
assume that $r$ could be a real number, then the equality happens for the real
number $\tilde{r}:=\frac{\mu-1}{\mu}(\theta^{\prime}-(1-\mu))$. Clearly, the
largest positive integer satisfying this property is
$r_{1}:=\left\lfloor\tilde{r}\right\rfloor.$
And this is the choice for $r_{1}$. As $S_{2}<0\text{ if
}\psi(S_{1})+r_{2}\frac{\mu}{\mu-1}\leq\theta^{\prime}$ we choose

$r_{2}:=\max_{r\geq
0}\left\\{\psi(S_{1})+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}\Leftrightarrow
r_{2}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-\psi(S_{1}))\right\rfloor,$

and so on. We notice that $r_{k}$ has a distinct expression. Since
$S_{k}<0\text{ if
}-1+\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}\leq\theta^{\prime}$, we choose

$r_{k}:=\max_{r\geq
0}\left\\{-1+\psi(S_{k-1})+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}\Leftrightarrow
r_{k}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}+1-\psi(S_{k-1}))\right\rfloor.$

######  Definition 3.1.

Given $\mu\geq 4.38+$, let
$\theta^{\prime}:=\frac{-(\mu-2)+\sqrt{(\mu-2)^{2}-4}}{2}$. For
$\psi(t)=2-\mu-\frac{1}{t}$, we define the classic Laplacian Sharer’s sequence
as the sequence of caterpillars $G_{k}(\mu):=[r_{1},r_{2},...,r_{k}],\;k\geq
2$ where

(8)
$\left\\{\begin{array}[]{ll}r_{1}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-(1-\mu))\right\rfloor.\\\
S_{1}=1-\mu+r_{1}\frac{\mu}{\mu-1}\\\
r_{j}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-\psi(S_{j-1}))\right\rfloor\\\
S_{j}=\psi(S_{j-1})+r_{j}\frac{\mu}{\mu-1}\\\
r_{k}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}+1-\psi(S_{k-1}))\right\rfloor\\\
S_{k}=-1+\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}.\end{array}\right.$

######  Example 3.2.

Consider applying Definition 3.1 for $\mu=5.4$. We have
$\theta:=\frac{-(\mu-2)-\sqrt{(\mu-2)^{2}-4}}{2}\approx-3.074772708$ and
$\theta^{\prime}:=\frac{-(\mu-2)+\sqrt{(\mu-2)^{2}-4}}{2}=\theta^{-1}\approx-0.32522729>\theta$.
We obtain by computation the first 11 terms, as follows.

$r_{1}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-(1-\mu))\right\rfloor$,
gives $r_{1}=3$
$S_{1}=1-\mu+r_{1}\frac{\mu}{\mu-1}$, gives $S_{1}\approx-0.718181818$
$r_{2}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-\psi(S_{1}))\right\rfloor$
, gives $r_{2}=1$
$r_{j}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}-\psi(S_{j-1}))\right\rfloor$
, gives $r_{j}=1$ for $j=2,\ldots,10$, while
$r_{11}:=\left\lfloor\frac{\mu-1}{\mu}(\theta^{\prime}+1-\psi(S_{10}))\right\rfloor$,
where $S_{10}\approx-1.503801894,$ giving $r_{11}=2$.

Hence for $k=11$ our $G_{k}(5.4)$ is given by

$G_{k}:=[[1,1,1],[1],[1],[1],[1],[1],[1],[1],[1],[1],[1,1]].$

Before we proceed with the analysis of this process, it is of great importance
to understand how it evolves, in particular we want to obtain a relation
between the sequence $S_{j}$ for
$G_{\mu}(\mu):=[r_{1},r_{2},...,r_{k-1},r_{k}]$ comparatively with the one for
$G_{k+1}(\mu)$. Let us denote it by $\tilde{S}_{j}$ and
$G_{k+1}(\mu):=[\tilde{r}_{1},\tilde{r}_{2},...,\tilde{r}_{k-1},\tilde{r}_{k},\tilde{r}_{k+1}]$.

We claim that, if $r_{k}\geq 1$ then

$\tilde{r}_{k}=r_{k}-1.$

To prove this, we first notice that, by construction $r_{j}=\tilde{r}_{j}$ and
$S_{j}=\tilde{S}_{j}$ for $1\leq j\leq k-1$. We also recall that

(9) $r_{k}:=\max_{r\geq
0}\left\\{-1+\psi(S_{k-1})+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}.$

We notice that this is equivalent to

(10)
$\theta^{\prime}-\frac{\mu}{\mu-1}<-1+\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}\leq\theta^{\prime}.$

Equation (10) is equivalent to

$-1+\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}\leq\theta^{\prime}\text{ and
}-1+\psi(S_{k-1})+(r_{k}+1)\frac{\mu}{\mu-1}>\theta^{\prime},$

which are equivalent to

$\psi(S_{k-1})+(r_{k}-1)\frac{\mu}{\mu-1}+\frac{1}{\mu-1}\leq\theta^{\prime}\text{
and }\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}+\frac{1}{\mu-1}>\theta^{\prime}.$

Rewriting these, we have

$\psi(S_{k-1})+(r_{k}-1)\frac{\mu}{\mu-1}\leq\theta^{\prime}-\frac{1}{\mu-1},\text{
and }\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}>\theta^{\prime}-\frac{1}{\mu-1}.$

We claim that these inequalities imply

(11) $\psi(S_{k-1})+(r_{k}-1)\frac{\mu}{\mu-1}\leq\theta^{\prime},\text{ and
}\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}>\theta^{\prime}.$

Indeed, as $\theta-\frac{1}{\mu-1}\leq\theta^{\prime}$, the first inequality
follows. Now if $\psi(S_{k-1})+r_{k}\frac{\mu}{\mu-1}\leq\theta^{\prime}$, the
first inequality implies that $\frac{\mu}{\mu-1}\leq 0$, which is a
contradiction. It follows that, by comparing Equation (11) with Equation (10)
we can write

(12) $r_{k}-1=\max_{r\geq
0}\left\\{\psi(S_{k-1})+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}.$

Now we notice that, because $\psi(\tilde{S}_{k-1})=\psi(S_{k-1})$, the
expression for $\tilde{r}_{k}$ is

(13) $\tilde{r}_{k}:=\max_{r\geq
0}\left\\{\psi(S_{k-1})+r\frac{\mu}{\mu-1}\leq\theta^{\prime}\right\\}=r_{k}-1,$

in view of Equation (12).

Analogously to the adjacency case, the procedure given in Definition 3.1,
always works. In the sense that it produces a sequence of caterpillars

$G_{k}(\mu):=[r_{1},r_{2},...,r_{k}],\;k\geq 2$

such that the spectral radius $\rho_{L}(G_{k}(\mu)):=\rho_{L}(G_{k})$ has a
limit point smaller than $\mu$. More precisely, the sequence $\rho_{L}(G_{k})$
is increasing and bounded by $\mu$. Indeed, the part $\rho_{L}(G_{k})<\mu$ is
a trivial consequence of the construction because
$S_{1},\ldots,S_{k}<\theta^{\prime}<0$ then the spectral radius for the
Laplacian matrix of $G_{k}$ is necessarily smaller than $\mu$ by Lemma (2.3).
The part $\rho_{L}(G_{k})<\rho_{L}(G_{k+1})$ is not quiet obvious because the
choice of $r_{k}$ for $G_{k}$ is possibly different from the one for
$G_{k+1}$, unless $r_{k}=0$. In that case $G_{k}$ is necessarily a subgraph of
$G_{k+1}$ and $\rho_{L}(G_{k})<\rho_{L}(G_{k+1})$. Thus, we assume that
$r_{k}\geq 1$. In order to verify the inequality
$\rho_{L}(G_{k})<\rho_{L}(G_{k+1})$, we recall that, from Equation (13), we
have $\tilde{r}_{k}=r_{k}-1$ meaning that $G_{k}$ is necessarily a subgraph of
$G_{k+1}$ and then $\rho_{L}(G_{k})<\rho_{L}(G_{k+1})$ again. In particular
there exists
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\gamma(\mu)\leq\mu$. We
summarize the findings of this section, stating the following result.

######  Theorem 3.3.

Let $\mu\in[4.38+,\infty)$. The sequence $G_{k}(\mu)$ of caterpillars given by
Equation (8) produces a limit point of $L$-spectral radii: there exists

$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k}(\mu))=\gamma(\mu)\leq\mu.$

The question is: what are the limit points produced by this process?

######  Definition 3.4.

We denote by $\Omega_{1}\subseteq[4.38+,\infty)$ the set of all Laplacian
limit points produced by Shearer’s method

$\Omega_{1}:=\\{\gamma\;|\;\lim_{k\to\infty}\rho_{L}(G_{k}(\mu))=\gamma\\}.$

Similar to the adjacency case studied by Shearer [13], where he shows that
this method produces an $A$-limit point for any real number in
$[\sqrt{2+\sqrt{5}},\infty)$, it is natural to conjecture that
$\Omega_{1}=[4.38+,\infty)$.

Unlike the adjacency case, however, we show in the sequel that there are
points where $\gamma(\mu)<\mu$. In fact, we will provide a whole interval $I$,
for which the sequence of caterpillars $G_{k}(\mu)$, given by Definition 3.1
is the same for every $\mu\in I$. This means $G_{k}(\mu)$ has Laplacian
spectral radius converging to the same value $\gamma<\mu$.

This only shows that the classic Shearer’s caterpillar construction for the
Laplacian version is not powerful enough to prove the density of Laplacian
spectral radii in $[4.38+,\infty)$. In subsequent sections of this note, we
extend this method and show how this extension obtains more accumulation
points.

###  3.1. A nasty interval

In this section we study the spectral radius of the caterpillar
$G_{k},\leavevmode\nobreak\ \leavevmode\nobreak\ k\geq 3$ given by

$G_{k}:=[[1,1,1],[1],[1],\ldots,[1],[1,1]].$

We notice this was obtained in Example 5.3 as the classic Laplacian Shearer’s
construction with $\mu=5.4$ and $k=11$. We first consider applying
Diagonalize$(G_{k},-\mu)$, for $\mu>4.38+$, using the notation established in
Section 2.2, disregarding the fact that this sequence is generated by some
actual $\mu$.

Since the starlike trees $T_{i}$ are actually stars whose number of rays are
$r_{1}=3,r_{j}=1,\leavevmode\nobreak\ 2\leq j\leq k-1$, and $r_{k}=2$, the
application of Equation (7) gives
$S_{1}=1-\mu+3\frac{\mu}{\mu-1},\leavevmode\nobreak\ \leavevmode\nobreak\
S_{j}=2-\mu-\frac{1}{S_{j-1}}+\frac{\mu}{\mu-1},\leavevmode\nobreak\ 2\leq
j\leq k-1$ and $S_{k}=1-\mu-\frac{1}{S_{k-1}}+2\frac{\mu}{\mu-1}$. For
convenience we let $\tilde{\psi}(t)=(2-\mu+\frac{\mu}{\mu-1})-\frac{1}{t}$,
and rewrite as the following recurrence.

(14) $\left\\{\begin{array}[]{ll}S_{1}=1-\mu+3\frac{\mu}{\mu-1},\\\
S_{j}=\tilde{\psi}(S_{j-1}),\leavevmode\nobreak\ 2\leq j\leq k-1,\\\
S_{k}=\tilde{\psi}(S_{k-1})+\frac{1}{\mu-1}\end{array}\right.$

The fixed points $\psi(t)=t$ are $\theta$ and $\theta^{\prime}$ given by
Equation (5), while the fixed points of $\tilde{\psi}(t)=t$ are given by

(15)
$\left\\{\begin{array}[]{ll}\sigma=\frac{(2-\mu+\frac{\mu}{\mu-1})-\sqrt{(2-\mu+\frac{\mu}{\mu-1})^{2}-4}}{2}\\\
\sigma^{\prime}=\frac{(2-\mu+\frac{\mu}{\mu-1})+\sqrt{(2-\mu+\frac{\mu}{\mu-1})^{2}-4}}{2}=\sigma^{-1}.\end{array}\right.$

######  Lemma 3.5.

Let $4<\mu<6$. Let $\mu_{*}$ and $\mu^{*}$ be, respectively, the solutions of
$S_{1}=\sigma^{\prime}$ and $\theta^{-1}-\frac{\mu}{\mu-1}=\sigma$.

  * $(a)$

$\mu_{*}=\frac{5+\sqrt{33}}{2}=5.3722+$ is the largest root of the polynomial
$x^{2}-5x-2$;

  * $(b)$

$\mu^{*}=5.4207+$ is the largest root of
$2\,{x}^{4}-14\,{x}^{3}+19\,{x}^{2}-10\,x-1.$

###### Proof.

$(a)$ $\leavevmode\nobreak\ S_{1}=\sigma^{\prime}$ is equivalent to

$1-\mu+3\frac{\mu}{\mu-1}=\frac{(2-\mu+\frac{\mu}{\mu-1})+\sqrt{(2-\mu+\frac{\mu}{\mu-1})^{2}-4}}{2},$

which can be transformed into the polynomial $\mu^{2}-5\mu-2$. In the range
$4<\mu<6$, $\mu_{*}$ is the only root of the polynomial.
$(b)$ Now, for $\theta^{-1}-\frac{\mu}{\mu-1}=\sigma$ we can perform the same
computations transforming

$\frac{(2-\mu)+\sqrt{(2-\mu)^{2}-4}}{2}-\frac{\mu}{\mu-1}=\frac{(2-\mu+\frac{\mu}{\mu-1})-\sqrt{(2-\mu+\frac{\mu}{\mu-1})^{2}-4}}{2}$

into a polynomial equation as prescribed. Performing that we obtain that
$\mu^{*}$ is the largest root of
$2\,{\mu}^{4}-14\,{\mu}^{3}+19\,{\mu}^{2}-10\,\mu-1.$

∎

######  Lemma 3.6.

Let $G_{k}$ be the caterpillar $G_{k}:=[[1,1,1],[1],[1],\ldots,[1],[1,1]]$ and
$\mu_{k}=\rho_{L}(G_{k})$ be the spectral radius of $G_{k}$. Then

$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu_{*}.$

###### Proof.

We consider the auxiliary graphs
$H_{k}(\mu):=[[1,1,1],[1],\ldots,[1],[1]],\;k\geq 3$. By applying ${\rm
Diagonalize}(H_{k},-\mu_{k})$ with $\mu_{k}=\rho_{L}(G_{k})$, we obtain the
sequence of values $W_{1},\ldots,W_{k-1},W_{k}$,

$W_{1}=1-\mu_{k}+3\frac{\mu_{k}}{\mu_{k}-1}<0$
$W_{j+1}:=\tilde{\psi}(W_{j}),\;1\leq j,$

where
$\tilde{\psi}(t):=(2-\mu_{k}+\frac{\mu_{k}}{\mu_{k}-1})-\frac{1}{t},\;t\neq
0$. From [10] we know that a closed formula $W_{j}$ is as follows.

$W_{j}:=\sigma_{k}+\frac{\sigma_{k}^{-1}-\sigma_{k}}{\beta_{k}\sigma_{k}^{2j-2}+1},\;1\leq
j,$

where $\beta_{k}:=\frac{\sigma_{k}^{-1}-\sigma_{k}}{W_{1}-\sigma_{k}}-1$ and
$\sigma_{k}=\frac{(2-\mu_{k}+\frac{\mu_{k}}{\mu_{k}-1})-\sqrt{(2-\mu_{k}+\frac{\mu_{k}}{\mu_{k}-1})^{2}-4}}{2}$.
Since $H_{k}$ and $G_{k}$ only differ in the last term, applying ${\rm
Diagonalize}(G_{k},-\mu_{k})$, we obtain a sequence of values
$S_{1}=W_{1},\ldots,S_{k-1}=W_{k-1}$ and $S_{k}=W_{k}+\frac{1}{\mu_{k}-1}=0$
because $\mu_{k}:=\rho_{L}(G_{k})$. An easy computation transforming this
equation into a polynomial with respect to $\sigma_{k}$ shows that taking
limit when $k\to\infty$ in $W_{k}+\frac{1}{\mu_{k}-1}=0$, we obtain that the
limit point $\mu:=\lim_{k\to\infty}\rho_{L}(G_{k})$ satisfies
$(\sigma+\frac{1}{\mu-1})\cdot(S_{1}-\sigma^{-1})=0$.

Indeed,

$S_{k}=W_{k}+\frac{1}{\mu_{k}-1}=0\Leftrightarrow W_{k}+\frac{1}{\mu_{k}-1}=0$
$\sigma_{k}+\frac{\sigma_{k}^{-1}-\sigma_{k}}{\beta_{k}\sigma_{k}^{2j-2}+1}+\frac{1}{\mu_{k}-1}=0$
$\sigma_{k}+\frac{\sigma_{k}^{-1}-\sigma_{k}}{\left(\frac{\sigma_{k}^{-1}-\sigma_{k}}{W_{1}-\sigma_{k}}-1\right)\sigma_{k}^{2j-2}+1}+\frac{1}{\mu_{k}-1}=0.$

Now we just transform this into a polynomial and use that
$\sigma_{k}^{2j-2}\to\infty$. The remaining equation is
$(\sigma_{k}+\frac{1}{\mu-1})\cdot(W_{1}-\sigma_{k}^{-1})=0$. As $W_{1}=S_{1}$
and since $\sigma_{k}+\frac{1}{\mu-1}$ has no solution for $\mu>4$, we obtain
the equation $S_{1}=\sigma_{k}^{-1}$, which, by Lemma 3.5 is $\mu_{*}$. ∎

######  Lemma 3.7.

Let $\mu_{*}=5.3722+$ and $\mu^{*}=5.4207+$ be as in Lemma 3.5 and let
$I=[\mu_{*},\mu^{*}]$. Then for all $\mu\in I$, the classic Laplacian
Shearer’s sequence given by Definition 3.1 is
$G_{k}(\mu):=[[1,1,1],[1],[1],\ldots,[1],[1,1]].$

###### Proof.

We consider two facts, the first one is

$\theta^{-1}-\frac{\mu}{\mu-1}<1-\mu+3\frac{\mu}{\mu-1}<\theta^{-1}$

for $5.09807+<\mu<6.05505+$ (see Figure 6), meaning that in this interval
every $G_{k}$ begins with $[[1,1,1],?,?,\ldots]$ because
$I\subset[5.09807+,6.05505+]$.

Figure 6. Inequality
$\theta^{-1}-\frac{\mu}{\mu-1}<1-\mu+3\frac{\mu}{\mu-1}<\theta^{-1}.$

By applying ${\rm Diagonalize}(G_{k},-\mu)$ with $\mu\in I$, we obtain a
sequence of values $S_{1},...,S_{k-1},S_{k}$ where
$S_{1}=1-\mu+3\frac{\mu}{\mu-1}\in(\theta^{-1}-\frac{\mu}{\mu-1},\theta^{-1}]$,
that is $r_{1}=3$.

The second fact is the inequality

$\theta^{-1}-\frac{\mu}{\mu-1}<\sigma<1-\mu+3\frac{\mu}{\mu-1}<\sigma^{-1}<\theta^{-1},$

where $\sigma$ and $\sigma^{-1}$ are defined in Equation (15). This is easily
verified from the graphics given in Figure 7, since those are functions of
$\mu$.

Figure 7. Inequality
$\theta^{-1}-\frac{\mu}{\mu-1}<\sigma<1-\mu+3\frac{\mu}{\mu-1}<\sigma^{-1}<\theta^{-1}$.

We know that $S_{2}=2-\mu-\frac{1}{S_{1}}+r_{2}\frac{\mu}{\mu-1}$ and we claim
that $r_{2}=1$ because
$2-\mu-\frac{1}{S_{1}}+\frac{\mu}{\mu-1}=\tilde{\psi}(S_{1})\in(\sigma,\sigma^{-1}]\subset(\theta^{-1}-\frac{\mu}{\mu-1},\theta^{-1}]$
meaning that $r_{2}=1$. In each step, $S_{j+1}=\tilde{\psi}(S_{j})$ obeys the
recurrence generated by
$\tilde{\psi}(t):=(2-\mu+\frac{\mu}{\mu-1})-\frac{1}{t},\;t\neq 0$. From [10]
we know that, since $\Delta=(2-\mu+\frac{\mu}{\mu-1})^{2}-4>0$, we have

$S_{j}:=\sigma+\frac{\sigma^{-1}-\sigma}{\beta\sigma^{2j-2}+1},\;1\leq j\leq
k-1,$

where $\beta:=\frac{\sigma^{-1}-\sigma}{S_{1}-\sigma}-1$ and
$\sigma=\frac{(2-\mu+\frac{\mu}{\mu-1})-\sqrt{(2-\mu+\frac{\mu}{\mu-1})^{2}-4}}{2}$.
We notice that
$\psi(S_{j})\in(\sigma,\sigma^{-1}]\subset(\theta^{-1}-\frac{\mu}{\mu-1},\theta^{-1}]$,
meaning that $r_{j}=1$ for $j\leq k-1$. We could extrapolate for
$\psi(S_{k-1})\in(\sigma,\sigma^{-1}]\subset(\theta^{-1}-\frac{\mu}{\mu-1},\theta^{-1}]$
obtaining $r_{k}=1+1$.
Thus $G_{k}(\mu):=[[1,1,1],[1],...,[1],[1,1]],\;k\geq 3$, as claimed.

∎

######  Remark 3.8.

We are not claiming that the points $\mu\in I$ are not Laplacian limit points,
only that they can not be achieved by using caterpillar graphs according to
the Shearer’s process. This motivates us to consider a more general class of
trees in the next section.

##  4\. Sequences of linear trees

Inspired by the Shearer’s procedure from [13], we will consider increasing
sequences of Laplacian spectral radii of graphs
$\\{G_{k}\;|\;k\in\mathbb{N}\\}$, where $G_{k}$ are linear trees, as
introduced in Section 2.2.

We need some technical results. The first one is:

######  Lemma 4.1.

[1, 17] Let $G=(V,E)$ be a graph on $n$ vertices and $m$ edges. Let
$V=\\{v_{1},v_{2},\ldots,v_{n}\\}$ and let $d(v_{i})$ be the degree of the
vertex $v_{i}$. Then

$\rho_{L}(G)\leq\max\left\\{d(v_{i})+d(v_{j})\leavevmode\nobreak\
|\leavevmode\nobreak\ v_{i},v_{j}\in E\right\\}.$

In particular, we can say that $\rho_{L}(G)\leq 2\Delta(G)$, where $\Delta(G)$
is the largest degree of $G$.

The second one is a bound on the spectral radii of
$\\{G_{k}\;|\;k\in\mathbb{N}\\}$. We recall that $\omega(T_{j})$ is the number
of paths that compose the starlike $T_{j}$.

######  Lemma 4.2.

Consider a sequence of linear trees $\\{G_{k}\;|\;k\in\mathbb{N}\\}$. Then,

$\exists M>0,\;\rho_{L}(G_{k})\leq M,\forall
k\in\mathbb{N}\Leftrightarrow\max_{k}\max_{1\leq j\leq
k}\omega(T_{j})<+\infty.$

###### Proof.

Assume that $\exists M>0,\;\rho_{L}(G_{k})\leq M,\forall k\in\mathbb{N}$, then
necessarily ${\displaystyle\max_{k}\max_{1\leq j\leq k}\omega(T_{j})<+\infty}$
otherwise if there exists a sequence of $T_{j}$ with $\omega(T_{j})\to\infty$
then, since $\rho_{L}(G_{k})\geq 1+\omega(T_{j})$, would be a contradiction.

Reciprocally, let ${\displaystyle r_{0}:=\max_{k}\max_{1\leq j\leq
k}\omega(T_{j})<+\infty}$ and ${\displaystyle h_{0}:=\max_{1\leq j\leq
k}h(T_{j})}$ then for any $k$, $G_{k}$ is a subgraph of
$F_{k}=[[h_{0},\ldots,h_{0}],\ldots,[h_{0},\ldots,h_{0}]],$ where
$w([h_{0},\ldots,h_{0}])=r_{0}$ and $\ell(F_{k})=k$. Thus,
$\rho_{L}(G_{k})\leq\rho_{L}(F_{k})$, for any $k$.

A well known upper bound for the $\rho_{L}(F_{k})$ is $\rho_{L}(F_{k})\leq
2\Delta(F_{k})$ (see Lemma 4.1), where $\Delta(F_{k})$ is the largest degree
of $F_{k}$. As the $\Delta(F_{k})\leq r_{0}+2$, we have
$\rho_{L}(G_{k})\leq\rho_{L}(F_{k})\leq 2(r_{0}+2)$ and the result follows.

∎

######  Definition 4.3.

We denote by $\mathcal{L}_{\infty}$ the set of all sequences of starlike trees
with bounded width, that is,

$\mathcal{L}_{\infty}:=\left\\{(T_{1},T_{2},...)\;|\;T_{j}^{\prime}s\text{ are
starlike trees and }\max_{k\geq 1}\max_{1\leq j\leq
k}\omega(T_{j})<+\infty\right\\}.$

######  Definition 4.4.

For each $\mathbb{T}=(T_{1},T_{2},...)\in\mathcal{L}_{\infty}$ and
$C=(C_{2},C_{3},...)\in\mathcal{L}_{\infty}$ we can define a sequence of
linear trees

$\left\\{\begin{array}[]{ll}G_{1}(\mathbb{T}):=[T_{1}]\\\
G_{k}(\mathbb{T}):=[T_{1},T_{2},...,T_{k-1},C_{k}],\;k\geq
2,\end{array}\right.$

called (Laplacian) _generalized Shearer sequence_ associated to $\mathbb{T}$
and $C$ if $\rho_{L}(G_{k}(\mathbb{T}))<\rho_{L}(G_{k+1}(\mathbb{T}))$ for all
$k\geq 1$.

We actually should use the notation $G_{j}(\mathbb{T},C)$, but we will omit
the dependence in $C$ when it is clear. A sufficient condition to have
$\rho_{L}(G_{k}(\mathbb{T}))<\rho_{L}(G_{k+1}(\mathbb{T}))$ is
$G_{k}(\mathbb{T})$ to be a subgraph of $G_{k+1}(\mathbb{T})$ for all $k\geq
1$, called subgraph condition.

######  Remark 4.5.

The canonical case we may deal is,
$\mathbb{T}=(T_{1},T_{2},...)\in\mathcal{L}_{\infty}$ and
$C=(T_{2},T_{3},...)$ a shift of $\mathbb{T}$ (thus
$C\in\mathcal{L}_{\infty}$), then $G_{1}=[T_{1}]$, $G_{2}=[T_{1},T_{2}]$ and
$G_{3}=[T_{1},T_{2},T_{3}]$, and so on. This sequence naturally satisfies the
subgraph condition so it is a generalized Shearer sequence.

######  Example 4.6.

Let us consider some cases for generalized Shearer sequences:

  * a)

Consider $\mathbb{T}=([1,1,1],[1],[1],[1],[1],...)\in\mathcal{L}_{\infty}$ and
$C=([1,1],[1,1],[1,1],[1,1],...)\in\mathcal{L}_{\infty}$, then
$G_{1}=[[1,1,1]]$, $G_{2}=[[1,1,1],[1,1]]$, $G_{3}=[[1,1,1],[1],[1,1]]$,
$G_{4}=[[1,1,1],[1],[1],[1,1]]$, and so on, is obviously a generalized Shearer
sequence coincident with the one in Section 3.1.

  * b)

Consider $\mathbb{T}=([1,1],[1],[0],[1],[0],...)\in\mathcal{L}_{\infty}$ and
$C=([0],[0],[0],[0],...)\in\mathcal{L}_{\infty}$ then $G_{1}=[[1,1]]$,
$G_{2}=[[1,1],[0]]$, $G_{3}=[[1,1],[1],[0]]$, $G_{4}=[[1,1],[1],[0],[0]]$,
$G_{5}=[1,1],[1],[0],[1],[0]$, and so on, is a generalized Shearer sequence.

  * c)

Consider $\mathbb{T}=([1,1],[1,1],[1],[1],[1],...)\in\mathcal{L}_{\infty}$ and
$C=([1,1,1,1],[0],[0],[0],...)\in\mathcal{L}_{\infty}$ then $G_{1}=[[1,1]]$,
$G_{2}=[[1,1],[1,1,1,1]]$ and $G_{3}=[[1,1],[1,1],[0]]$. Note that $G_{2}$ is
not a subgraph of $G_{3}$, moreover,

$\rho_{L}(G_{2})=6.141336+>5.261802+=\rho_{L}(G_{3})$

thus it is not a generalized Shearer sequence.

  * d)

Consider $\mathbb{T}=([1],[1,1],[1,1,1],[1,1,1,1],...)$ and
$C=([1],[1],[1],[1],...)\in\mathcal{L}_{\infty}$ then $G_{1}=[[1]]$,
$G_{2}=[[1],[1]]$ and $G_{3}=[[1],[1,1],[1]]$,
$G_{4}=[[1],[1,1],[1,1,1],[1]]$, and so on. This sequence satisfies the
subgraph condition however it is not a generalized Shearer sequence because
$\omega(T_{j})=j-1\to\infty$, that is, $C\in\mathcal{L}_{\infty}$ but
$\mathbb{T}\not\in\mathcal{L}_{\infty}$.

By construction, each generalized Shearer sequence satisfies the property that
$\rho_{L}(G_{k})\leq\rho_{L}(G_{k+1})$ and by Lemma 4.2 we also know that
$k\to\rho_{L}(G_{k})$ is bounded. Thus, there exists
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k}):=\gamma(\mathbb{T})$.

######  Definition 4.7.

We denote by $\mathcal{S}$ the set of all limit points of (Laplacian)
generalized Shearer sequences, that is,

$\mathcal{S}:=\\{\gamma\;|\;\gamma=\gamma(\mathbb{T})\text{ for some
}\mathbb{T},C\in\mathcal{L}_{\infty}\\}.$

##  5\. Applications: Generalized random Shearer process

As the set $\Omega_{1}\subseteq\mathcal{S}\subseteq[4.38+,\infty)$ (see
Section 3), it is natural to believe that $\mathcal{S}$ is a larger set of
Laplacian limit points. Indeed, we will show this. Neverthless, it will be
useful to have a structured way to construct linear trees as generalized
Shearer sequences, for a given parameter $\mu$. The main goal of this section
is provide one, which is an adaptation of the classical process used by
Shearer to produce caterpillars. We denominate it as Generalized random
Shearer process.

Fixed $\mu>4.38+$, we will construct a generalized Shearer sequence in the
following way. Let $\mathbb{T}=(T_{1},T_{2},\ldots)$ and
$C=(C_{2},C_{3},\ldots)$ be arbitrary sequences of starlike trees. From that
we define linear trees

$G_{k}:=[T_{1},T_{2},,T_{k-1},C_{k}],\;k\geq 2,$

with $G_{1}:=[T_{1}]$. We would like to have $\rho_{L}(G_{k})<\mu$ and $G_{k}$
a subgraph of $G_{k+1}$.

To achieve this property, we use ${\rm Diagonalize}(G_{k},-\mu)$ obtaining
$\Pi(G_{k},\mu)=(S_{1},\ldots,S_{k})$ (see Section 2.2). We recall that
$\psi(t)=2-\mu-\frac{1}{t},\;t\neq 0$,
$\theta:=\frac{-(\mu-2)-\sqrt{(\mu-2)^{2}-4}}{2}$ is an attracting point and
$\theta^{\prime}:=\frac{-(\mu-2)+\sqrt{(\mu-2)^{2}-4}}{2}=\theta^{-1}>\theta$
is a repelling one.

Thus, in order to obtain $S_{j}<0$ for all $j$ we successively choose the
following starlike trees:

$\left\\{\begin{array}[]{ll}T_{1}\text{ such that
}1-\mu+\delta(T_{1},\mu)<\theta^{-1};\\\ T_{j}\text{ such that
}\psi(S_{j-1})+\delta(T_{j},\mu)<\theta^{-1},\;2\leq j\leq k;\\\ C_{k}\text{
such that }C_{k}:=T_{k}.\end{array}\right.$

Since $\psi(S_{k-1})+\delta(T_{k},\mu)<\theta^{-1}$, $\theta^{-1}<0<1$ and
$C_{k}=T_{k}$ we obtain

$\psi(S_{k-1})+\delta(T_{k},\mu)<1$ $-1+\psi(S_{k-1})+\delta(C_{k},\mu)<0$
$S_{k}<0.$

Thus $\rho_{L}(G_{k})<\mu$ because $\mbox{sign}(\Pi)(G_{k},\mu)=(-,\ldots,-)$.

This procedure is random since we can choose different trees in each step and
it will always work because $1-\mu<\theta^{-1}$ so, in the worst case, we can
choose $T_{1}=[0]$ and the same is true for the next ones since the previous
$S_{j-1}<\theta^{-1}$ implies that $\psi(S_{j-1})<\theta^{-1}$, by
construction.

We notice that necessarily $\mathbb{T},C\in\mathcal{L}_{\infty}$ because for
each choice of $T_{j}$ or $C_{k}$ we must have subgraphs of $G_{k}$ which has
spectral radius smaller than $\mu$, a fixed number, this means that the width
$\omega(T_{j})$ or $\omega(C_{k})$ can not exceed $\mu-4$ (recall that
$\rho_{L}(G_{k})\geq\max\\{\omega(T_{j})+3,\omega(C_{k})+2\\})$.

From the way we built the $G_{k}$’s, we always end up with a generalized
Shearer sequence since $C$ is a shift of $\mathbb{T}$. We then have
$\rho_{L}(G_{k})<\rho_{L}(G_{k+1})$ and, in particular, there exists

$\lim_{k\to\infty}\rho_{L}(G_{k})=\gamma(\mu)\leq\mu.$

######  Definition 5.1.

We denote by $\Omega\subseteq[4.38+,\infty)$ the set of all Laplacian limit
points produced by the Generalized random Shearer’s method

$\Omega:=\\{\gamma\;|\;\lim_{k\to\infty}\rho_{L}(G_{k}(\mu))=\gamma\\}.$

######  Remark 5.2.

We notice that, if in addition to the above requirements, we restrict our
choices to $\max\\{h(T_{j}),h(C_{k})\\}=1$ and we maximize the value of each
$S_{j}<\theta^{-1}$ we recover the caterpillars from the classic Shearer’s
paper. Thus $\Omega_{1}\subset\Omega$.

######  Example 5.3.

Consider $\mu=5.4$ and a random Shearer process running up computationally
producing a generalized Shearer sequence

$\mathbb{T}:=([1],[1,2],[1],[0],[0],[1],[1],[0],[1],[2],[2],[2],[1],[1],[2],[2],[2],[1],[2],[1],[2],$
$[1],[2],[2],[1],[2],[1],[2],[2],[2],[2],[2],[2],[2],[1],[1],[2],[1],[2],[2],[1],[2],[2],[2],[1],$
$[2],[2],[2],[2],[2],[1],[2],[2],[2],[2],[2],[2],[1],[2],[2],[1],[1],[2],[2],[2],[2],[2],[2],[2],$
$[1],[2],[2],[1],[2],[1],[1],[2],[2],[2],[1],[1],[1],[1],[1],[1],[2],[2],[1],[1],[2],[2],[1],[2],$
$[2],[1],[2],[1],[2],[2],[2],[0]^{\infty}).$

By a direct computation we obtain
$G_{1}:=[[1]]$ and $\rho_{L}(G_{1})=2$;
$G_{2}:=[[1],[1,2]]$ and $\rho_{L}(G_{2})=4.302775+$;
$G_{3}:=[[1],[1,2],[1]]$ and $\rho_{L}(G_{3})=5.236067+$;
$G_{4}:=[[1],[1,2],[1],[0]]$ and $\rho_{L}(G_{4})=5.3817984+$;
$G_{5}:=[[1],[1,2],[1],[0],[0]]$ and $\rho_{L}(G_{4})=5.397488988+$
and so on. For instance $\rho_{L}(G_{100})=5.39999504+$, proving that by
continuing the process we obtain some Laplacian limit point in the interval
$(\mu_{*},\mu^{*})$. This proves that the generalized random Shearer process
generates Laplacian limit points which are not achieved by classical approach
using caterpillars, that is, $\Omega_{1}\subsetneq\Omega$.

It is worth mentioning that, for the same $\mu=5.4$, we could have made
different choices in each step, obtaining a very different sequence. We will
compute the first terms using the maximum criteria (as the classical Shearer’s
method) and leave to the reader to check that these are optimal choices once
we assume $h(T_{j})\leq 2$. This, however, result that is worse in terms of
approximation to $\mu=5.4$:
Take $T_{1}:=[2,2,2]$ because
$S_{1}=1-\mu-\delta([2,2,2],\mu)=-0.8843+<-0.3252+=\theta^{-1}$;
Take $T_{2}:=[0]$ because
$S_{2}=\psi(S_{1})-\delta([0],\mu)=-1.1994+<-0.3252+=\theta^{-1}$;
Take $T_{3}:=[2]$ because
$S_{3}=\psi(S_{2})-\delta([2],\mu)=-1.2511+<-0.3252+=\theta^{-1}$;
Take $T_{4}:=[2]$ because
$S_{4}=\psi(S_{3})-\delta([2],\mu)=-1.2855+<-0.3252+=\theta^{-1}$ and so on.
Thus we have the sequence $\mathbb{T}:=([2,2,2],[0],[2],[2],[2]^{\infty}).$ By
a direct computation we obtain
$G_{1}:=[[2,2,2]]$ and $\rho_{L}(G_{1})=4.4142+$
$G_{2}:=[[2,2,2],[0]]$ and $\rho_{L}(G_{2})=5.2360+$
$G_{3}:=[[2,2,2],[0],[2]]$ and $\rho_{L}(G_{3})=5.3105+$
$G_{4}:=[[2,2,2],[0],[2],[2]]$ and $\rho_{L}(G_{4})=5.3325+$
$G_{5}:=[[2,2,2],[0],[2],[2],[2]]$ and $\rho_{L}(G_{4})=5.3423+$.
Finally, notice that $\rho_{L}(G_{100})=5.3554+<\mu_{*}$ indicating that the
limit does not reach the interval $(\mu_{*},\mu^{*})$ as the previous one!

In some sense, a generalized Shearer sequence $G_{k}(\mathbb{T},C)$ is always
the realization of a generalized random Shearer process, for a given $\mu$.

######  Lemma 5.4.

Consider the generalized Shearer sequence $G_{k}(\mathbb{T},C)$

$\left\\{\begin{array}[]{ll}G_{1}(\mathbb{T}):=[T_{1}]\\\
G_{k}(\mathbb{T}):=[T_{1},T_{2},\ldots,T_{k-1},C_{k}],\;k\geq
2,\end{array}\right.$

where $\mathbb{T}=(T_{1},T_{2},\ldots)\in\mathcal{L}_{\infty}$ and
$C=(C_{2},C_{3},\ldots)\in\mathcal{L}_{\infty}$. Let
$\mu:=\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})$ and
$\theta:=\frac{-(\mu-2)-\sqrt{(\mu-2)^{2}-4}}{2}$, then

$\left\\{\begin{array}[]{ll}S_{1}=1-\mu+\delta(T_{1},\mu)<\theta^{-1};\\\
S_{j}=\psi(S_{j-1})+\delta(T_{j},\mu)<\theta^{-1},\;2\leq j\leq
k,\end{array}\right.$

where $\Pi(G_{k},\mu)=(S_{1},\ldots,S_{k})$ is obtained from ${\rm
Diagonalize}(G_{k},-\mu)$. In particular, if $C$ is a shift of $\mathbb{T}$
then we have a generalized random Shearer process.

###### Proof.

We know that $\rho_{L}(G_{k})$ is strictly increasing, thus
$\rho_{L}(G_{k})<\mu$. Applying ${\rm Diagonalize}(G_{k},-\mu)$ we should
obtain only negative values $S_{j}$. Let us suppose, by contradiction, that
$S_{j_{0}}>\theta^{-1}$, for some $j_{0}\geq 1$, that is,
$S_{j_{0}}\in(\theta^{-1},0)$, then

$S_{j_{0}+1}=\psi(S_{j_{0}})+\delta(T_{j_{0}+1},\mu)\geq\psi(S_{j_{0}})>S_{j_{0}}.$

If $S_{j_{0}+1}>0$ we have a contradiction, otherwise
$S_{j_{0}+1}\in(\theta^{-1},0)$ and

$S_{j_{0}+2}=\psi(S_{j_{0}+1})+\delta(T_{j_{0}+2},\mu)\geq\psi^{2}(S_{j_{0}}).$

In this way, after $m$ steps not reaching a contradiction we obtain

$S_{j_{0}+m}\geq\psi^{m}(S_{j_{0}}).$

As $\psi$ is uniformly expanding in the interval $(\theta^{-1},0)$ (actually
$\psi^{\prime}(t)>\psi^{\prime}(\theta^{-1})=\frac{1}{\left(\theta^{-1}\right)^{2}}=\theta^{2}>1$)
we must have $S_{j_{0}+m}>0$ for a large enough iterate $m$ (see Figure 8), a
contradiction with $S_{j}<0,j\geq 1$.

Figure 8. Representation of $\psi$. In this case $S_{j_{0}+m}>0$ for $m=1$.

∎

##  6\. Density of Laplacian limit points

The main goal of this section is to investigate, for a generalized Shearer
sequence $G_{k}(\mathbb{T},C)$

$\left\\{\begin{array}[]{ll}G_{1}(\mathbb{T}):=[T_{1}]\\\
G_{k}(\mathbb{T}):=[T_{1},T_{2},...,T_{k-1},C_{k}],\;k\geq
2,\end{array}\right.$

where $\mathbb{T}=(T_{1},T_{2},...)\in\mathcal{L}_{\infty}$ and
$C=(C_{2},C_{3},...)\in\mathcal{L}_{\infty}$, the Laplacian limit points
$\mu:=\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})$, which we denoted by
$\mathcal{S}$ (see Definition 4.7).

To justify our choices, we start by pointing out that for a starlike
$T_{j}\in\mathbb{T}$, its position, that is dependent on the index $j$,
affects $\rho_{L}(G_{k})$ in different ways. In fact, if $j=1$ then ${\rm
deg}(v_{1})=1+\omega(T_{j})$ and $\rho_{L}(G_{k})\geq 2+\omega(T_{j})$.
However, if $j>1$ then ${\rm deg}(v_{j})=2+\omega(T_{j})$ and
$\rho_{L}(G_{k})\geq 3+\omega(T_{j})$. Thus, in order to stay within an
interval $[m,m+1]$ we must control $\omega(T_{j})$ across the entire sequence.
For instance for $\mathbb{T}=([1,1],T_{2},...)\in\mathcal{L}_{\infty}$ we can
have limit points in $[4,5]$ but we need to require $\omega(T_{j})\leq 1$ for
$j\geq 2$, otherwise all the limit points will be in $[5,\infty)$. In other
words every $G_{k}$ must be a quipu!

Another important remark is that if
$\mathbb{T}=(T_{1},T_{2},...)\in\mathcal{L}_{\infty}$ is constant or pre-
periodic, we can always compute explicitly the limit point, often being an
algebraic number. Recall that we already show that for
$\mathbb{T}=([1,1,1],[1]^{\infty},\ldots)\in\mathcal{L}_{\infty}$ and
$C=([1,1],[1,1]^{\infty},\ldots)\in\mathcal{L}_{\infty}$ we obtain

$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu_{*}$

where $\mu_{*}=\frac{5+\sqrt{33}}{2}(=5.3722+)$ is the larger root of the
polynomial $x^{2}-5x-2$.

Other remarkable case is the sequence
$\mathbb{T}=([0]^{\infty},\ldots)\in\mathcal{L}_{\infty}$ and

$C=([1,1],[1,2],[1,3],\ldots)\in\mathcal{L}_{\infty}.$

A close inspection shows that
$G_{1}:=[[0]]$;
$G_{2}:=[[0],[1,1]]=[1,1,1]$;
$G_{3}:=[[0],[0],[1,2]]=[1,2,2]$
and so on. Thus $G_{k}:=[[0]^{k-1},[1,k-1]]=[1,k-1,k-1]$. Obviously $G_{k}$
satisfy the subgraph condition constituting a generalized Shearer sequence. By
[10], Theorem 4.2, we know that

$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=2+\epsilon=4.382975+$

where $\epsilon$ is the real root of $x^{3}-4x-4$. It is worth mentioning that
the convergence is quite fast, for instance
$\rho_{L}([[0],[0],[0],[0],[0],[0],[0],[0],[1,8]])=4.3829331+$ is already
$10^{-4}$ close to the actual value!

######  Example 6.1.

Motivated by the above considerations we may take a closer look at the
following family of generalized Shearer sequences:

(16)
$\mathcal{F}_{1}:=\\{G_{k}(\mathbb{T},C)\;|\;T_{1}=[1,1,1],\;T_{j}\in\\{[0],[1],[1,1]\\},j\geq
2,\;\text{ and }C\text{ is a shift of }\mathbb{T}\\}$

As we observed before $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})>5$ for
any $G_{k}\in\mathcal{F}_{1}$, thus the question is the distribution of these
limit points in the interval $[5,5+m],\;m\geq 1$.

To give some idea on that distribution we run a numerical experiment sampling
3000 sequences for $G_{k}\in\mathcal{F}_{1}$ and plot, in the vertical
coordinate the value $\rho_{L}(G_{100})$ in Figure 9.

Figure 9. Limit points distribution (in the left) and a zoom of it (in the
right).

Figure 9 suggests an abundance of limit points, but we need a rigorous study
to quantify it. In the experiment $\\{[0],[1],[1,1]\\}$ were sorted with equal
probability, which may affect the variability (see the zoom on the right of
Figure 9). The smallest gap is $1.4\times 10^{-8}$ while the largest is
$0.01+$. We limit in 3000 the number of samples, because larger simulations
are computationally prohibitive and pointless to establish density results.

In order to compare different limit points, we start with some general
properties regarding the perturbation of a generalized Shearer sequence. In
the next result we will use a truncation process, where a sequence
$\mathbb{T}=(T_{1},T_{2},T_{3},\ldots)$ is replaced by
$\tilde{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{k_{0}},[0]^{\infty})\in\mathcal{L}_{\infty}$
always having a smaller limit point because, after $k_{0}$ there is no drift
in the recurrence relation used to compute ${\rm
Diagonalize}(\hat{G}_{k},-\mu_{0})$ for
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu_{0}$. Then,
$S_{k_{0}}<\theta^{-1}$ means that
$S_{k_{0}+i}=\psi(S_{k_{0}})<\psi(\theta^{-1})=\theta^{-1}$, for all $i\geq
1$. Perhaps more interesting is the fact that this truncation process provides
a set of algebraic limit points that is dense in a subset of $\mathcal{S}$
(see Definition 5.1).

######  Theorem 6.2.

Let $G_{k}(\mathbb{T},C)$ be a generalized Shearer sequence where
$C=(T_{2},T_{3},\ldots)$ is a shift of $\mathbb{T}=(T_{1},T_{2},T_{3},\ldots)$
and $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu_{0}\in\mathcal{S}$.
Consider $\varepsilon>0$ and $k_{\varepsilon}\geq 1$ such that
$\rho_{L}(G_{k_{\epsilon}})>\mu_{0}-\varepsilon$.

  1. $(a)$

If
$\hat{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{j_{0}-1},\hat{T}_{j_{0}},T_{j_{0}+1},\ldots)$,
defines the sequence $\hat{G}_{k}:=G_{k}(\hat{\mathbb{T}})$ and
$\delta(\hat{T}_{j_{0}},\mu_{0})>\delta(T_{j_{0}},\mu_{0})$ then
$\displaystyle\lim_{k\to\infty}\rho_{L}(\hat{G}_{k})\geq\mu_{0}$;

  2. $(b)$

If
$\tilde{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{k_{\varepsilon}},[0]^{\infty})$,
defines the sequence $\tilde{G}_{k}:=G_{k}(\tilde{\mathbb{T}})$ and

$\displaystyle\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})=\mu_{1}$

then $\mu_{1}\in(\mu_{0}-\varepsilon,\mu_{0}]$;

  3. $(c)$

The set $\mathcal{A}$ of Laplacian limit points of sequences defined by

$\tilde{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{k_{0}},[0]^{\infty})\in\mathcal{L}_{\infty}$

is algebraic;

  4. $(d)$

The set $\mathcal{A}$ is dense in the set $\mathcal{B}$ of Laplacian limit
points of sequences defined by
$\mathbb{T}=(T_{1},T_{2},\ldots)\in\mathcal{L}_{\infty}$ and
$C=(T_{2},T_{3},\ldots)\in\mathcal{L}_{\infty}$.

###### Proof.

(a) Since $\hat{G}_{k}=G_{k}$ for $k<j_{0}$, we may assume $k>j_{0}$. This
time we will apply ${\rm Diagonalize}(G_{k},-\mu)$ and ${\rm
Diagonalize}(\hat{G}_{k},\mu)$ with $\mu:=\rho_{L}(G_{k})$, take $v_{j_{0}}$
as root, $f_{G_{k}}(v_{j_{0}})=0$ the final value in $G_{k}$, and
$f_{\hat{G}_{k}}(v_{j_{0}})$ the final value in $\hat{G}_{k}$. By Lemma 2.1,
we know that $\rho_{L}(\hat{G}_{k})>\rho_{L}(G_{k})$ if
$f_{\hat{G}_{k}}(v_{j_{0}})>0$.

By direct computation, we obtain

$0=f_{G_{k}}(v_{j_{0}})=2-\mu-\frac{1}{f_{G_{k}}(v_{j_{0}-1})}-\frac{1}{f_{G_{k}}(v_{j_{0}+1})}+\delta(T_{j_{0}},\mu),$

or

$2-\mu-\frac{1}{f_{G_{k}}(v_{j_{0}-1})}-\frac{1}{f_{G_{k}}(v_{j_{0}+1})}=-\delta(T_{j_{0}},\mu).$

On the other hand, using the above equation we get

$f_{\hat{G}_{k}}(v_{j_{0}})=2-\mu-\frac{1}{f_{G_{k}}(v_{j_{0}-1})}-\frac{1}{f_{G_{k}}(v_{j_{0}+1})}+\delta(\hat{T}_{j_{0}},\mu)=\delta(\hat{T}_{j_{0}},\mu)-\delta(T_{j_{0}},\mu)>0,$

by hypothesis.

(b) The easy part is to prove that $\mu_{1}>\mu_{0}-\varepsilon$. We just need
to observe that, by hypothesis,
$\rho_{L}(G_{k_{\varepsilon}})>\mu_{0}-\varepsilon$ and for
$k>k_{\varepsilon}$ we have
$\tilde{G}_{k}=[T_{1},T_{2},T_{3},\ldots,T_{k_{\varepsilon}},[0]^{k-k_{\varepsilon}}]$,
meaning that, $G_{k_{\varepsilon}}$ is a subgraph of $\tilde{G}_{k}$. Thus,
$\displaystyle\mu_{1}=\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})\geq\rho_{L}(G_{k_{\varepsilon}})>\mu_{0}-\varepsilon$.

In order to see that $\mu_{1}\leq\mu_{0}$ we need a more sophisticated
argument. Again we apply ${\rm Diagonalize}(G_{k},-\mu_{0})$ and ${\rm
Diagonalize}(\hat{G}_{k},-\mu_{0})$ taking $v_{k}$ as root in both graphs. Let
$S_{1},\ldots,S_{k-1},S_{k}$ the output of ${\rm Diagonalize}(G_{k},-\mu_{0})$
at the vertices $v_{1},\ldots,v_{k-1},v_{k}$ and
$\tilde{S}_{1},\ldots,\tilde{S}_{k-1},\tilde{S}_{k}$ the output of ${\rm
Diagonalize}(\tilde{G}_{k},-\mu_{0})$ at the vertices
$v_{1},\ldots,v_{k-1},v_{k}$.

From Lemma 5.4 we know that if
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu_{0}$ and
$\theta_{0}:=\frac{-(\mu_{0}-2)-\sqrt{(\mu_{0}-2)^{2}-4}}{2}$, then

$\left\\{\begin{array}[]{ll}S_{1}=1-\mu_{0}+\delta(T_{1},\mu_{0})<\theta^{-1}_{0};\\\
\psi(S_{j-1})+\delta(T_{j},\mu_{0})<\theta^{-1}_{0},\;2\leq j\leq
k.\end{array}\right.$

Assuming $k>k_{\varepsilon}$ (if $k\leq k_{\varepsilon}$ then
$\rho_{L}(\tilde{G}_{k})=\rho_{L}(G_{k}))$, we have
$\tilde{S}_{j}=S_{j}<\theta^{-1}_{0}<0,\;1\leq j\leq k_{\varepsilon}$. For the
next index, we obtain

$S_{k_{\varepsilon}+1}=\psi(S_{k_{\varepsilon}})+\delta(T_{k_{\varepsilon}+1},\mu_{0})<\theta^{-1}_{0}$

and

$\tilde{S}_{k_{\varepsilon}+1}=\psi(S_{k_{\varepsilon}})+\delta([0],\mu_{0})=\psi(S_{k_{\varepsilon}})<\theta^{-1}_{0}.$

Proceeding in this way we obtain

$\tilde{S}_{k_{\varepsilon}+m}=\psi(S_{k_{\varepsilon}+m-1})+\delta([0],\mu_{0})=\psi^{m-1}(S_{k_{\varepsilon}})<\theta^{-1}_{0}.$

In particular, as $\psi(S_{k-1})<\theta^{-1}_{0}$ implies that
$\tilde{S}_{k}=-1+\psi(S_{k-1})<-1+\theta^{-1}_{0}<0$. Thus
$\rho_{L}(\tilde{G}_{k})<\mu_{0}$ for $k\geq k_{\varepsilon}$ and
$\displaystyle\mu_{1}=\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})\leq\mu_{0}$.

(c) Take $\mu\in\mathcal{A}$, then there exists
$\mu_{k}:=\rho_{L}(\tilde{G}_{k})$, defined by
$\tilde{\mathbb{T}}=(T_{1},T_{2},T_{3},...,T_{k_{0}},[0]^{\infty})\in\mathcal{L}_{\infty}$
such that $\displaystyle\mu=\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})$.

Consider $S_{1},\ldots,S_{k-1},S_{k}$ the output of ${\rm
Diagonalize}(\tilde{G}_{k},-\mu_{k})$ at the vertices
$v_{1},\ldots,v_{k-1},v_{k}$. We know that
$S_{1},\ldots,S_{k_{0}-1},S_{k_{0}}$ are negative rational functions with
respect to the variable $\mu_{k}$. Now,
$S_{k_{0}+1}=\psi(S_{k_{0}})+\delta([0],\mu_{k})=\psi(S_{k_{0}})$,
$S_{k_{0}+2}=\psi(S_{k_{0}+1})+\delta([0],\mu_{k})=\psi(S_{k_{0}+1})$ until we
reach $S_{k}=-1+\psi(S_{k-1})+\delta([0],\mu_{k})=-1+\psi(S_{k-1})=0,$ by
construction, or equivalently $\mu_{k}$ must solve the equation

$\psi(S_{k-1})=1.$

Defining $a_{1}=S_{k_{0}}$, $a_{2}=S_{k_{0}+1}=\psi(a_{1})$, …,
$a_{k-k_{0}}=S_{k_{0}+(k-k_{0})}=\psi(S_{k-1})$, where
$\psi(t)=2-\mu_{k}-\frac{1}{t},t\neq 0$. From [10] we know that
$\Delta=(2-\mu_{k})^{2}-4>0$ thus

$a_{j}:=\theta_{k}+\frac{\theta_{k}^{-1}-\theta_{k}}{\beta_{k}\theta_{k}^{2(j-1)}+1},\;1\leq
j,$

where $\beta_{k}:=\frac{\theta_{k}^{-1}-\theta_{k}}{a_{1}-\theta_{k}}-1$ and
$\theta_{k}=\frac{(2-\mu_{k})-\sqrt{(2-\mu_{k})^{2}-4}}{2}$.

Now we are able to rewrite the equation $\psi(S_{k-1})=1$ as $a_{k-k_{0}}=1$
or

$\theta_{k}+\frac{\theta_{k}^{-1}-\theta_{k}}{\beta_{k}\theta_{k}^{2(k-k_{0}-1)}+1}=1$
$\theta_{k}^{-1}-\theta_{k}=(1-\theta_{k})\left(\beta_{k}\theta_{k}^{2(k-k_{0}-1)}+1\right)$
$\theta_{k}^{-1}-\theta_{k}=(1-\theta_{k})\left(\left(\frac{\theta_{k}^{-1}-\theta_{k}}{S_{k_{0}}-\theta_{k}}-1\right)\theta_{k}^{2(k-k_{0}-1)}+1\right)$
$(\theta_{k}^{-1}-\theta_{k})(S_{k_{0}}-\theta_{k})=(1-\theta_{k})\left(\left(\theta_{k}^{-1}-S_{k_{0}}\right)\theta_{k}^{2(k-k_{0}-1)}+S_{k_{0}}-\theta_{k}\right)$
$\frac{(\theta_{k}^{-1}-\theta_{k})(S_{k_{0}}-\theta_{k})}{\theta_{k}^{2(k-k_{0}-1)}}=(1-\theta_{k})\left(\left(\theta_{k}^{-1}-S_{k_{0}}\right)+\frac{S_{k_{0}}-\theta_{k}}{\theta_{k}^{2(k-k_{0}-1)}}\right).$

Notice that $(\theta_{k}^{-1}-\theta_{k})=\sqrt{(2-\mu_{k})^{2}-4}\not\to 0$
when $k\to\infty$. Thus, or $S_{k_{0}}-\theta_{k}\to 0$ when $k\to\infty$ or
taking the limit in the above equation we obtain $\theta_{k}^{-1}-S_{k_{0}}\to
0$ when $k\to\infty$. Thus
$\displaystyle\mu=\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})$ must solve one of
the equations

$\theta^{-1}=S_{k_{0}}\text{ or }\theta=S_{k_{0}}.$

All in all, as $S_{k_{0}}=\frac{P(\mu)}{Q(\mu)}$ we conclude that $\mu$ must
solve

$\frac{(2-\mu)\pm\sqrt{(2-\mu)^{2}-4}}{2}=\frac{P(\mu)}{Q(\mu)}$
$\pm\sqrt{(2-\mu)^{2}-4}=\frac{2P(\mu)}{Q(\mu)}-(2-\mu)$
$(2-\mu)^{2}-4=\left(\frac{2P(\mu)}{Q(\mu)}-(2-\mu)\right)^{2}$
$(2-\mu)^{2}-4=4\left(\frac{P(\mu)}{Q(\mu)}\right)^{2}-4(2-\mu)\left(\frac{P(\mu)}{Q(\mu)}\right)+(2-\mu)^{2}$
$-4=4\left(\frac{P(\mu)}{Q(\mu)}\right)^{2}-4(2-\mu)\left(\frac{P(\mu)}{Q(\mu)}\right)$
$\left(\frac{P(\mu)}{Q(\mu)}\right)^{2}-(2-\mu)\left(\frac{P(\mu)}{Q(\mu)}\right)+1=0$
$P^{2}(\mu)-(2-\mu)\left(P(\mu)Q(\mu)\right)+Q^{2}(\mu)=0.$

After a simplification we conclude that $\mu$ is a real root of the polynomial
at the numerator, meaning that $\mu$ is an algebraic number.

(d) Consider $\mu_{0}\in\mathcal{B}$ and $\varepsilon>0$ arbitrarily small. By
(b) we can find $k_{\varepsilon}$ such that
$\tilde{\mathbb{T}}_{\varepsilon}=(T_{1},T_{2},T_{3},...,T_{k_{\varepsilon}},[0]^{\infty})$
defines a sequence $\tilde{G}_{k}:=G_{k}(\tilde{\mathbb{T}_{\varepsilon}})$
such that
$\displaystyle\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})=\mu_{1}\in(\mu_{0}-\varepsilon,\mu_{0}]$.
As $\mu_{1}\in\mathcal{A}$, this proves the density. ∎

The next general result is on the set of possible limit points _dominated_ by
$\mu$.

######  Definition 6.3.

Let $\mu\geq 4.38+$ be a fixed real number. A generalized Shearer sequence
$G_{k}(\mathbb{T},C)$ is dominated by $\mu$ if
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})\leq\mu$. We denote by
$\Gamma(\mu)$ the set of all generalized Shearer sequences
$G_{k}(\mathbb{T},C)$ which are dominated by $\mu$.

We notice that if $\Gamma(\mu)$ is known, this means we completely understand
the approximation of $\mu$ by Laplacian limit points originated by linear
trees. In other words, if

$\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})<\mu$

then either $\mu$ is not a Laplacian limit point or the approximation must be
taken through a wider class such as general trees or general graphs. On the
other hand if

$\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$

then $\mu$ is certainly a Laplacian limit point, but we do not know whether
$\mu\in\mathcal{S}$ that is, whether the supremum is attained by a sequence of
linear trees.

######  Lemma 6.4.

Let $\mu\geq 4.38+$ be a fixed real number and let $G_{k}(\mathbb{T},C)$ be a
generalized Shearer sequence where $C=(T_{2},T_{3},\ldots)$ is a shift of
$\mathbb{T}=(T_{1},T_{2},T_{3},\ldots)$.

  1. $(a)$

If $G_{k}(\mathbb{T},C)\in\Gamma(\mu)$ then

$\omega(T_{j})\leq\delta(T_{j},\mu)<\mu,$

for all $j\geq 1$. The same is true for $C$, that is,
$\omega(C_{k})\leq\delta(C_{k},\mu)<\mu$;

  2. $(b)$

Consider $G_{k}(\mathbb{T},C)\in\Gamma(\mu)$ and a new sequence

$\tilde{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{j_{0}-1},\tilde{T}_{j_{0}},T_{j_{0}+1},\ldots)$

defining $\tilde{G}_{k}:=G_{k}(\tilde{\mathbb{T}})$. If
$\delta(\tilde{T}_{j_{0}},\mu)<\delta(T_{j_{0}},\mu)$ then
$\tilde{G}_{k}\in\Gamma(\mu)$;

  3. $(c)$

Consider $G_{k}(\mathbb{T},C)\in\Gamma(\mu)$ and starlike trees
$\hat{T}_{j_{0}},\hat{T}_{j_{0}+1}$ such that
$\delta(\hat{T}_{j_{0}},\mu)\leq\delta(T_{j_{0}},\mu)$ and
$\displaystyle\theta^{-1}-\psi(\hat{S}_{j_{0}})>\delta(\hat{T}_{j_{0}+1},\mu)\geq\delta(T_{j_{0}+1},\mu).$
There exist some $m\in\mathbb{N}$ and starlike trees
$\hat{T}_{j_{0}+2},\ldots,\hat{T}_{j_{0}+m}$ such that, defining

$\hat{\mathbb{T}}=(T_{1},T_{2},T_{3},\ldots,T_{j_{0}-1},\hat{T}_{j_{0}},\hat{T}_{j_{0}+1},\ldots,\hat{T}_{j_{0}+m},T_{j_{0}+m+1},\ldots),$

we obtain a generalized Shearer sequence
$\hat{G}_{k}:=G_{k}(\hat{\mathbb{T}},C)$ with $\hat{G}_{k}\in\Gamma(\mu)$.

###### Proof.

(a) Recall that $\Pi(G_{k},\mu)=\left(S_{1},S_{2},\ldots,S_{k}\right)$ where

$\left\\{\begin{array}[]{ll}S_{1}=1-\mu+\delta(T_{1},\mu),\\\
S_{j}=\psi(S_{j-1})+\delta(T_{j},\mu),\;2\leq j\leq k-1,\\\
S_{k}=-1+\psi(S_{k-1})+\delta(C_{k},\mu).\end{array}\right.$

As $G_{k}(\mathbb{T},C)\in\Gamma(\mu)$, we have from Lemma 5.4 that

$1-\mu+\delta(T_{1},\mu)<\theta^{-1},$

but

$\theta^{-1}-(1-\mu)=\frac{2-\mu+\sqrt{(2-\mu)^{2}-4}}{2}-(1-\mu)=$
$=\frac{2-\mu-2(1-\mu)+\sqrt{(2-\mu)^{2}-4}}{2}=\frac{\mu+\sqrt{(2-\mu)^{2}-4}}{2}$

thus

$\omega(T_{1})\leq\delta(T_{1},\mu)<\frac{\mu+\sqrt{(2-\mu)^{2}-4}}{2}.$

By construction, $S_{j}>S_{1}>1-\mu$ for all $j$ and
$S_{j}<\theta^{-1}\Rightarrow\psi(S_{j-1})<\theta^{-1}$ thus

$1-\mu+\delta(T_{j},\mu)<S_{j}=\psi(S_{j-1})+\delta(T_{j},\mu)<\theta^{-1}$
$\omega(T_{j})\leq\delta(T_{j},\mu)<\theta^{-1}-(1-\mu)=\frac{\mu+\sqrt{(2-\mu)^{2}-4}}{2}.$

Notice that

$\frac{\mu+\sqrt{(2-\mu)^{2}-4}}{2}=\frac{\mu}{2}\left(1+\sqrt{\left(\frac{2}{\mu}-1\right)^{2}-\left(\frac{2}{\mu}\right)^{2}}\right)=$
$=\frac{\mu}{2}\left(1+\sqrt{\left(\frac{2}{\mu}-1+\frac{2}{\mu}\right)\left(\frac{2}{\mu}-1-\frac{2}{\mu}\right)}\right)=\frac{\mu}{2}\left(1+\sqrt{\left(1-\frac{4}{\mu}\right)}\right)\leq\frac{\mu}{2}2=\mu,$

because $\frac{4}{\mu}<1$.

Finally, $S_{k}=-1+\psi(S_{k-1})+\delta(C_{k},\mu)<0$ and
$1-\mu<S_{k-1}<\theta^{-1}\Rightarrow 1-\mu<\psi(S_{k-1})<\theta^{-1}$ thus
$-\mu<-1+\psi(S_{k-1})<-1+\theta^{-1}$. So,

$-1+\psi(S_{k-1})+\delta(C_{k},\mu)<0$
$\delta(C_{k},\mu)<-(-1+\psi(S_{k-1}))<\mu$
$\omega(C_{k})\leq\delta(C_{k},\mu)<-(-1+\psi(S_{k-1}))<\mu.$

(b) Consider $\Pi(G_{k},\mu)=\left(S_{1},S_{2},\ldots,S_{k}\right)$ where (by
Lemma 5.4)

$\left\\{\begin{array}[]{ll}S_{1}=1-\mu+\delta(T_{1},\mu)<\theta^{-1},\\\
S_{j}=\psi(S_{j-1})+\delta(T_{j},\mu)<\theta^{-1},\;2\leq j\leq k-1,\\\
S_{k}=-1+\psi(S_{k-1})+\delta(C_{k},\mu)<0.\end{array}\right.$

Now we compute
$\Pi:=(\tilde{G}_{k},\mu)\mapsto\left(\tilde{S}_{1},\tilde{S}_{2},\ldots,\tilde{S}_{k}\right)$.
We know that $\tilde{S}_{j}=S_{j}<\theta^{-1}<0$ for $1\leq j_{0}-1$ and

$\tilde{S}_{j_{0}}=\psi(S_{j_{0}-1})+\delta(\tilde{T}_{j_{0}},\mu)-\delta(T_{j_{0}},\mu)+\delta(T_{j_{0}},\mu)=$
$=S_{j_{0}}+\delta(\tilde{T}_{j_{0}},\mu)-\delta(T_{j_{0}},\mu)<S_{j_{0}}+0<\theta^{-1}<0.$

Since $\tilde{S}_{j_{0}}<S_{j_{0}}$,
$\tilde{T}_{j_{0}+m}=T_{j_{0}+m},\leavevmode\nobreak\ \forall m\geq 1$ and
because $\psi$ is order preserving, we have
$\tilde{S}_{j_{0}+m}<S_{j_{0}+m}<\theta^{-1}<0$,

$\tilde{S}_{j_{0}+m}<S_{j_{0}+m}<\theta^{-1}$
$\psi(\tilde{S}_{j_{0}+m})<\psi(S_{j_{0}+m})<\psi(\theta^{-1})$
$\psi(\tilde{S}_{j_{0}+m})+0<\psi(S_{j_{0}+m})+0<\theta^{-1}$
$\tilde{S}_{j_{0}+(m+1)}<S_{j_{0}+(m+1)}<\theta^{-1}.$

In particular $\tilde{S}_{j_{0}+1}<S_{j_{0}+1}<\theta^{-1}<0$. This concludes
the proof because
$\mbox{sign}(\Pi):=(\tilde{G}_{k},\mu)=\left(-,-,\ldots,-\right)$ thus
$\rho_{L}(\tilde{G}_{k})<\mu$ and
$\displaystyle\lim_{k\to\infty}\rho_{L}(\tilde{G}_{k})\leq\mu$ or
$\tilde{G}_{k}\in\Gamma(\mu)$.

(c) We have $\Pi(G_{k},\mu)=\left(S_{1},S_{2},\ldots,S_{k}\right)$ where

$\left\\{\begin{array}[]{ll}S_{1}=1-\mu+\delta(T_{1},\mu)<\theta^{-1},\\\
S_{j}=\psi(S_{j-1})+\delta(T_{j},\mu)<\theta^{-1},\;2\leq j\leq k-1,\\\
S_{k}=-1+\psi(S_{k-1})+\delta(C_{k},\mu)<0.\end{array}\right.$

At the same time we have
$\Pi(\hat{G}_{k},\mu)=\left(\hat{S}_{1},\hat{S}_{2},\ldots,\hat{S}_{k}\right)$.
For $1\leq j_{0}-1$ we have $\hat{S}_{j}=S_{j}<\theta^{-1}$ and for
$\hat{S}_{j_{0}}$ we have

$S_{j_{0}}=\psi(S_{j_{0}-1})+\delta(T_{j_{0}},\mu)<\theta^{-1}$

and by hypothesis

(17)
$\delta(\hat{T}_{j_{0}},\mu)<\delta(T_{j_{0}},\mu)<\theta^{-1}-\psi(S_{j_{0}-1})$

On the other hand

$\hat{S}_{j_{0}}=\psi(S_{j_{0}-1})+\delta(\hat{T}_{j_{0}},\mu)<S_{j_{0}}$
$\psi(\hat{S}_{j_{0}})<\psi(S_{j_{0}})$
$\theta^{-1}-\psi(\hat{S}_{j_{0}})>\theta^{-1}-\psi(S_{j_{0}}).$

Using this inequality in Equation (17) we obtain

$\delta(\hat{T}_{j_{0}},\mu)<\delta(T_{j_{0}},\mu)<\theta^{-1}-\psi(S_{j_{0}-1})<\theta^{-1}-\psi(\hat{S}_{j_{0}})$
$\hat{S}_{j_{0}}=\psi(\hat{S}_{j_{0}})+\delta(\hat{T}_{j_{0}},\mu)<\theta^{-1}.$

The next is
$\hat{S}_{j_{0}+1}=\psi(\hat{S}_{j_{0}})+\delta(\hat{T}_{j_{0}+1},\mu)$. In
order to have $\hat{S}_{j_{0}+1}<\theta^{-1}$ we need
$\delta(\hat{T}_{j_{0}+1},\mu)<\theta^{-1}-\psi(\hat{S}_{j_{0}})$.

Notice that

$\theta^{-1}-\psi(\hat{S}_{j_{0}})>\theta^{-1}-\psi(S_{j_{0}})>\delta(T_{j_{0}+1},\mu).$

It is always possible
$\theta^{-1}-\psi(\hat{S}_{j_{0}})>\delta(\hat{T}_{j_{0}+1},\mu)$ (or
equivalently $\hat{S}_{j_{0}+1}<\theta^{-1}$) provided that

$\theta^{-1}-\psi(\hat{S}_{j_{0}})>\delta(\hat{T}_{j_{0}+1},\mu)\geq\delta(T_{j_{0}+1},\mu),$

in the worst case one may take $\hat{T}_{j_{0}+1}=T_{j_{0}+1}$.

Assuming our hypothesis, we know that $\hat{S}_{j_{0}+1}<\theta^{-1}$ we may
take, in the worst case, $\hat{T}_{j_{0}+2}=[0]$, $\hat{T}_{j_{0}+3}=[0]$, …,
$\hat{T}_{j_{0}+m+1}=[0]$. This means that
$\hat{S}_{j_{0}+2}=\psi(\hat{S}_{j_{0}+1})<\theta^{-1}$,
$\hat{S}_{j_{0}+3}=\psi^{2}(\hat{S}_{j_{0}+1})<\theta^{-1}$, …,
$\hat{S}_{j_{0}+m+1}=\psi^{m}(\hat{S}_{j_{0}+1})<\theta^{-1}$. As $m$
increases
$\hat{S}_{j_{0}+m+1}=\psi^{m}(\hat{S}_{j_{0}+1})\to\theta<\theta^{-1}$. Thus,
unless the original sequence is formed only by $[0]$’s after $T_{j_{0}+2}$
will exist such a $m$ in the way that
$\hat{S}_{j_{0}+m+1}<S_{j_{0}+m+1}<\theta^{-1}$ for all the next indices.
Finally we conclude that
$\mbox{sign}(\Pi)(\hat{G}_{k},\mu)=\left(-,-,\ldots,-\right)$ for all $k$ thus
$\hat{G}_{k}\in\Gamma(\mu)$.
∎

######  Remark 6.5.

In item (c) of Lemma 6.4 we can take $\hat{T}_{j_{0}}=T_{j_{0}}$ and the proof
would still be valid. At the same time, taking
$\delta(\hat{T}_{j_{0}},\mu)<\delta(T_{j_{0}},\mu),$ we will be able to choose
$\hat{T}_{j_{0}+1}$ in a wider range. Moreover, one can take
$\hat{T}_{j_{0}+1}$ in such a way that we maximize the quantity
$\delta(\hat{T}_{j_{0}+1},\mu)$ in the range

$[\delta(T_{j_{0}+1},\mu),\;\theta^{-1}-\psi(\hat{S}_{j_{0}})).$

Finally, instead of taking the easier choice $\hat{T}_{j_{0}+2}=[0]$,
$\hat{T}_{j_{0}+3}=[0]$, …, $\hat{T}_{j_{0}+m+1}=[0]$, one can simply take

$\delta(\hat{T}_{j_{0}+m+1},\mu)<\delta(T_{j_{0}+m+1},\mu),\;m\geq 1,$

with $m$ such that $\hat{S}_{j_{0}+m+1}<S_{j_{0}+m+1}$ may be reached much
later.

##  7\. Variational technique

The goal of this section is to investigate, for a generalized Shearer sequence
$G_{k}(\mathbb{T},C)$

$\left\\{\begin{array}[]{ll}G_{1}(\mathbb{T}):=[T_{1}]\\\
G_{k}(\mathbb{T}):=[T_{1},T_{2},...,T_{k-1},C_{k}],\;k\geq
2,\end{array}\right.$

where $\mathbb{T}=(T_{1},T_{2},...)\in\mathcal{L}_{\infty}$ and
$C=(C_{2},C_{3},...)\in\mathcal{L}_{\infty}$, when the Laplacian limit points
$\mu_{0}:=\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})$ is close to a given
number $\mu\geq 4.38+$.

In order to do that, we assume $G_{k}\in\Gamma(\mu)$ and consider a variation
$\mu_{\varepsilon}:=\mu-\varepsilon$. We ask whether
$\mu_{0}>\mu_{\varepsilon}$. We already know that $\mu_{0}>\mu_{\varepsilon}$
if, and only if, there exists some $k\in\mathbb{N}$ such that the vector

$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})$

has some positive entry, where
$\Pi(G_{k},\mu_{\varepsilon})=\left(S_{1}(\varepsilon),S_{2}(\varepsilon),\ldots,S_{k}(\varepsilon)\right)$
is given by

$\left\\{\begin{array}[]{ll}S_{1}(\varepsilon)=1-\mu_{\varepsilon}+\delta(T_{1},\mu_{\varepsilon}),\\\
S_{j}(\varepsilon)=\psi_{\varepsilon}(S_{j-1}(\varepsilon))+\delta(T_{j},\mu_{\varepsilon}),\;2\leq
j\leq k-1,\\\
S_{k}(\varepsilon)=-1+\psi_{\varepsilon}(S_{k-1}(\varepsilon))+\delta(C_{k},\mu_{\varepsilon}).\end{array}\right.$

We need to take a closer look at some terms at the above formula:

  * •

The function $\psi_{\varepsilon}$ is given by
$\psi_{\varepsilon}(t)=2-\mu_{\varepsilon}-\frac{1}{t}=2-\mu-\frac{1}{t}+\varepsilon=\psi(t)+\varepsilon$.
Of course, $\psi_{0}(t)=\psi(t)$.

  * •

The processing of each path in $T_{i}$ satisfies the formula

(18) $\left\\{\begin{array}[]{ll}b_{1}(\varepsilon)=1-\mu_{\varepsilon}\\\
b_{j+1}(\varepsilon)=\psi_{\varepsilon}(b_{j}(\varepsilon)),\leavevmode\nobreak\
\leavevmode\nobreak\ j>1.\end{array}\right.$

By a reasoning analogous to the case $\varepsilon=0$, one can prove that for a
small $\varepsilon$, the iterates $b_{j}(\varepsilon)$ obeying Equation (18)
has the following properties. Let $\theta_{\varepsilon}$ and
$\theta^{\prime}_{\varepsilon}$ be fixed points of $\psi_{\varepsilon}(t)=t$.
They are given by

(19)
$\theta_{\varepsilon}:=\frac{-(\mu_{\varepsilon}-2)-\sqrt{(\mu_{\varepsilon}-2)^{2}-4}}{2}\mbox{
and
}\theta^{\prime}_{\varepsilon}:=\theta_{\varepsilon}^{-1}>\theta_{\varepsilon}.$

They are attracting and repelling points, respectively, meaning that for
$b_{1}(\varepsilon)\leq\theta^{\prime}_{\varepsilon}$, we have
$b_{j}(\varepsilon)$ converging to $\theta_{\varepsilon}$. Since $\mu>4.38+$
we have $\mu_{\varepsilon}>4$, so that

$1-\mu_{\varepsilon}<b_{1}(\varepsilon)\leq
b_{j}(\varepsilon)<\theta_{\varepsilon}<0$

and, in particular, that all $b_{j}(\varepsilon)<0$.

  * •

The function $\varepsilon\to b_{j}(\varepsilon)$ is differentiable and
$\frac{d\,b_{j}(\varepsilon)}{d\,\varepsilon}\geq 1$. In particular,
$\varepsilon\to b_{j}(\varepsilon)$ is strictly increasing. Define
$b^{\prime}_{j}(\varepsilon):=\frac{d\,b_{j}(\varepsilon)}{d\,\varepsilon}$.
Then, the sequence $(b^{\prime}_{j}(\varepsilon))$ satisfy the recurrence
relation

(20) $\left\\{\begin{array}[]{ll}b^{\prime}_{1}(\varepsilon)=1\\\
b^{\prime}_{j}(\varepsilon)=1+\frac{1}{\left(b_{j-1}(\varepsilon)\right)^{2}}b^{\prime}_{j-1}(\varepsilon),\;j\geq
2.\end{array}\right.$

Indeed,
$b^{\prime}_{1}(\varepsilon)=\frac{d\,(1-\mu_{\varepsilon})}{d\,\varepsilon}=1>0$.
For $j>1$ we have
$b^{\prime}_{j}(\varepsilon)=\frac{d\,\psi_{\varepsilon}(b_{j-1}(\varepsilon))}{d\,\varepsilon}=1+\frac{1}{b_{j-1}(\varepsilon)^{2}}b^{\prime}_{j-1}(\varepsilon)$.
By induction we get $b^{\prime}_{j}(\varepsilon)\geq 1$, for $j\geq 1$.
Additionally, we can also set a upper bound to $b^{\prime}_{j}(\varepsilon)$:

$b_{j}(\varepsilon)<\theta_{\varepsilon}<0\Rightarrow
0<\frac{1}{-b_{j}(\varepsilon)}<\frac{1}{-\theta_{\varepsilon}}\Rightarrow\frac{1}{(b_{j}(\varepsilon))^{2}}<\theta_{\varepsilon}^{-2}.$

Thus

$b^{\prime}_{j}(\varepsilon)=1+\frac{1}{\left(b_{j-1}(\varepsilon)\right)^{2}}b^{\prime}_{j-1}(\varepsilon)\leq
1+\theta_{\varepsilon}^{-2}b^{\prime}_{j-1}(\varepsilon),$

which means that $b^{\prime}_{j}(\varepsilon)\leq
1+\theta_{\varepsilon}^{-2}+\ldots+\theta_{\varepsilon}^{-2j}\leq\frac{1}{1-\theta_{\varepsilon}^{-2}}<\infty$.

  * •

The function $\varepsilon\to b_{j}(\varepsilon)$ is twice differentiable and
$\frac{d^{2}\,b_{j}(\varepsilon)}{d\,\varepsilon^{2}}\geq 0$. In particular,
if $j>1$ then $\varepsilon\to b_{j}(\varepsilon)$ is strictly concave, because
$\frac{d\,b_{j}(\varepsilon)}{d\,\varepsilon}$ is not constant. Define
$b^{\prime\prime}_{j}(\varepsilon):=\frac{d^{2}\,b_{j}(\varepsilon)}{d\,\varepsilon^{2}}$.
Then, the sequence $(b^{\prime\prime}_{j}(\varepsilon))$ satisfies the
recurrence relation

(21) $\left\\{\begin{array}[]{ll}b^{\prime\prime}_{1}(\varepsilon)=0\\\
b^{\prime\prime}_{j}(\varepsilon)=-\frac{2}{b_{j-1}(\varepsilon)^{3}}\left(b^{\prime}_{j}(\varepsilon)\right)^{2}+\frac{1}{b_{j-1}(\varepsilon)^{2}}\;b^{\prime\prime}_{j-1}(\varepsilon),\;j\geq
2.\end{array}\right.$

Indeed, we already computed $b^{\prime}_{1}(\varepsilon)=1$ thus
$b^{\prime\prime}_{1}(\varepsilon)=\frac{d\,}{d\,\varepsilon}(1)=0$.
For $j>1$ we have

$b^{\prime\prime}_{j}(\varepsilon)=\frac{d\,}{d\,\varepsilon}\left(1+\frac{1}{b_{j-1}(\varepsilon)^{2}}b^{\prime}_{j-1}(\varepsilon)\right)=$
$=-\frac{2}{b_{j-1}(\varepsilon)^{3}}\left(b^{\prime}_{j}(\varepsilon)\right)^{2}+\frac{1}{b_{j-1}(\varepsilon)^{2}}\;b^{\prime\prime}_{j-1}(\varepsilon).$

By induction we get $b^{\prime\prime}_{j}(\varepsilon)\geq 0$, for $j\geq 1$.

  * •

Let $T_{j}$ be one starlike tree attached to $G_{k}$. Recall that
$\omega(T_{j})$ denote the number of paths of
$T_{j}=[q_{1}^{j},\ldots,q_{\omega(T_{j})}]$ and, for
$i=1,\ldots,\omega(T_{j})$, by $b_{q_{i}^{j}}$ the last value output by the
algorithm ${\rm Diagonalize}(G_{k},-\mu_{\varepsilon})$ at the $i$-th path of
$T_{j}$. We denote the net result of the starlike $T_{j}$ by

(22)
$\delta(T_{j},\mu_{\varepsilon}):=\left\\{\begin{array}[]{ll}\sum_{i=1}^{\omega(T_{j})}\left(1-\frac{1}{b_{q_{i}^{j}}(\varepsilon)}\right),\mbox{
for non-empty }T_{j}\\\ 0\mbox{ if }T_{j}\mbox{ is empty. }\end{array}\right.$

From the previous items we can define the differential drift,
$\delta^{\prime}(T_{j},\mu_{\varepsilon}):=\frac{d\,}{d\,\varepsilon}\delta(T_{j},\mu_{\varepsilon})$
which satisfies the formula for $T_{j}\neq[0]$

(23)
$\delta^{\prime}(T_{j},\mu_{\varepsilon})=\sum_{i=1}^{\omega(T_{j})}\frac{1}{\left(b_{q_{i}^{j}}(\varepsilon)\right)^{2}}b^{\prime}_{q_{i}^{j}}(\varepsilon)\geq
0.$

We can also find an upper bound for
$\delta^{\prime}(T_{j},\mu_{\varepsilon})$. Recall that
$\frac{1}{(b_{j}(\varepsilon))^{2}}<\theta_{\varepsilon}^{-2}$ and
$b^{\prime}_{j}(\varepsilon)<\frac{1}{1-\theta_{\varepsilon}^{-2}}$, thus

$\delta^{\prime}(T_{j},\mu_{\varepsilon})=\sum_{i=1}^{\omega(T_{j})}\frac{1}{\left(b_{q_{i}^{j}}(\varepsilon)\right)^{2}}b^{\prime}_{q_{i}^{j}}(\varepsilon)\leq\sum_{i=1}^{\omega(T_{j})}\theta_{\varepsilon}^{-2}\frac{1}{1-\theta_{\varepsilon}^{-2}}=\frac{\omega(T_{j})}{\theta_{\varepsilon}^{2}-1}<\infty$

because $\omega(T_{j})<\infty$ (see Lemma 4.2).

######  Theorem 7.1.

Let $G_{k}:=G_{k}(\mathbb{T},C)$ be a generalized Shearer sequence such that
$(G_{k})\in\Gamma(\mu)$. Then
$\mu:=\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})$ if and only if the
sequence $(\varepsilon_{j})$ converges to zero, where $\varepsilon_{j}$ is the
smallest positive solution of $S_{j}(\varepsilon)=0$, for
$\Pi(G_{k},\mu_{\varepsilon})=\left(S_{1}(\varepsilon),S_{2}(\varepsilon),\ldots,S_{k}(\varepsilon)\right)$.

###### Proof.

First, we will investigate the sign of
$S_{1}(\varepsilon)=1-\mu_{\varepsilon}+\delta(T_{1},\mu_{\varepsilon})$. In
this case,

$\mbox{sign}(S_{1}(\varepsilon))=+\Leftrightarrow
1-\mu_{\varepsilon}+\delta(T_{1},\mu_{\varepsilon})>0\Leftrightarrow\varepsilon>\mu-1-\delta(T_{1},\mu_{\varepsilon}).$

To study the above inequality we define the function
$g_{1}:[0,\infty)\to\mathbb{R}$ given by

$g_{1}(\varepsilon)=S_{1}(\varepsilon))=1-\mu_{\varepsilon}+\delta(T_{1},\mu_{\varepsilon}).$

It is easy to see that $g_{1}(0)=1-\mu+\delta(T_{1},\mu)<\theta^{-1}<0$
$(G_{k})\in\Gamma(\mu)$. Also, $g_{1}(\varepsilon)$ is differentiable in some
neighborhood of $\varepsilon=0$. Indeed, for
$T_{1}=[q_{1}^{1},\ldots,q_{\omega(T_{1})}^{1}]$ we get

$\frac{d\,g_{1}(\varepsilon)}{d\,\varepsilon}=1+\delta^{\prime}(T_{1},\mu_{\varepsilon})\geq
1.$

Since $\frac{d\,g_{1}(\varepsilon)}{d\,\varepsilon}\geq 1$ and $g_{1}(0)<0$ we
obtain an unique root $\varepsilon_{1}$ such that $g_{1}(\varepsilon)<0$ for
$\varepsilon<\varepsilon_{1}$ and $g_{1}(\varepsilon)>0$ for
$\varepsilon>\varepsilon_{1}$. In this way, we conclude that
$\mbox{sign}(S_{1}(\varepsilon))=+$ for $\varepsilon>\varepsilon_{1}$, in
other words,
$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})=\left(+,?,\ldots,?\right)$, meaning
that, $\rho_{L}(G_{k})>\mu_{\varepsilon}$.

In order to get a better approximation of $\mu$ we need to look for the
possibility of having $\mbox{sign}(S_{2}(\varepsilon))=+$ for
$\varepsilon<\varepsilon_{1}$. Analogously to the previous case, we define

$g_{2}(\varepsilon)=S_{2}(\varepsilon)=\psi_{\varepsilon}(S_{1}(\varepsilon))+\delta(T_{2},\mu_{\varepsilon}),\;0\geq\varepsilon<\varepsilon_{1}.$

It is easy to see that $g_{2}(0)=S_{2}(0))=S_{2}<\theta^{-1}<0$ because
$(G_{k})\in\Gamma(\mu)$. For $T_{2}=[q_{1}^{2},\ldots,q_{\omega(T_{2})}^{2}]$
we get

$\frac{d\,g_{2}(\varepsilon)}{d\,\varepsilon}=1+\frac{1}{\left(g_{1}(\varepsilon)\right)^{2}}\frac{d\,g_{1}(\varepsilon)}{d\,\varepsilon}+\delta^{\prime}(T_{2},\mu_{\varepsilon})\geq
1.$

Again, as $\frac{d\,g_{2}(\varepsilon)}{d\,\varepsilon}\geq 1$ and
$g_{2}(0)<0$ we obtain a unique root $\varepsilon_{2}$ such that
$g_{2}(\varepsilon)<0$ for $\varepsilon<\varepsilon_{2}$ and
$g_{2}(\varepsilon)>0$ for $\varepsilon>\varepsilon_{2}$.
We claim that $\varepsilon_{2}<\varepsilon_{1}$. To see that, we analyze

$\lim_{\varepsilon\to\varepsilon_{1}^{-}}g_{2}(\varepsilon)=\lim_{\varepsilon\to\varepsilon_{1}^{-}}2-\mu-\frac{1}{g_{1}(\varepsilon)}+\varepsilon+\delta(T_{2},\mu_{\varepsilon})=+\infty$

because $g_{1}(\varepsilon)<0$ for $\varepsilon<\varepsilon_{1}$ and
$\displaystyle\lim_{\varepsilon\to\varepsilon_{1}^{-}}g_{1}(\varepsilon)=0$.
Thus, $g_{2}(\varepsilon)>0$ in some left neighborhood of $\varepsilon_{1}$,
that is, the root $\varepsilon_{2}<\varepsilon_{1}$.
In this way, we conclude that $\mbox{sign}(S_{2}(\varepsilon))=+$ for
$\varepsilon_{2}<\varepsilon<\varepsilon_{1}$, in other words,
$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})=\left(-,+,?,\ldots,?\right)$,
meaning that, $\rho_{L}(G_{k})>\mu_{\varepsilon}$.

If we continue this procedure for $2\leq j\leq k-1$ we obtain a sequence
$0<\varepsilon_{j}<\ldots\varepsilon_{2}<\varepsilon_{1}$ such that, for
$\varepsilon_{j}<\varepsilon<\varepsilon_{j-1}$ we have
$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})=\left(-,\ldots,-,+,?,\ldots,?\right)$
(obviously, by construction,
$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})=\left(-,\ldots,-,-,?,\ldots,?\right)$
for $\varepsilon<\varepsilon_{j}$), meaning that,
$\rho_{L}(G_{k})>\mu_{\varepsilon}$.

The last step is to look for the possibility of having
$\mbox{sign}(S_{k}(\varepsilon))=+$ for $\varepsilon<\varepsilon_{k-1}$.
Analogously to the previous case, we define

$g_{k}(\varepsilon)=S_{k}(\varepsilon)=-1+\psi_{\varepsilon}(S_{k-1}(\varepsilon))+\delta(C_{k},\mu_{\varepsilon}),\;0\geq\varepsilon<\varepsilon_{k-1}.$

It is easy to see that $g_{k}(0)=S_{k}(0))=S_{k}<0$ because
$(G_{k})\in\Gamma(\mu)$. For $C_{k}=[p_{1}^{k},\ldots,p_{\omega(C_{k})}^{k}]$
we get

$\frac{d\,g_{k}(\varepsilon)}{d\,\varepsilon}=1+\frac{1}{\left(g_{k-1}(\varepsilon)\right)^{2}}\frac{d\,g_{k-1}(\varepsilon)}{d\,\varepsilon}+\delta^{\prime}(C_{k},\mu_{\varepsilon})\geq
1.$

Again, as $\frac{d\,g_{k}(\varepsilon)}{d\,\varepsilon}\geq 1$ and
$g_{k}(0)<0$ we obtain a unique root $\varepsilon_{k}$ such that
$g_{k}(\varepsilon)<0$ for $\varepsilon<\varepsilon_{k}$ and
$g_{k}(\varepsilon)>0$ for $\varepsilon>\varepsilon_{k}$.
We claim that $\varepsilon_{k}<\varepsilon_{k-1}$. To see that, we analyze

$\lim_{\varepsilon\to\varepsilon_{k-1}^{-}}g_{k}(\varepsilon)=\lim_{\varepsilon\to\varepsilon_{k-1}^{-}}2-\mu-\frac{1}{g_{k-1}(\varepsilon)}+\varepsilon+\delta(C_{k},\mu_{\varepsilon})=+\infty$

because $g_{k-1}(\varepsilon)<0$ for $\varepsilon<\varepsilon_{k-1}$ and
$\displaystyle\lim_{\varepsilon\to\varepsilon_{k-1}^{-}}g_{k-1}(\varepsilon)=0$.
Thus, $g_{k}(\varepsilon)>0$ in some left neighborhood of $\varepsilon_{1}$,
that is, the root $\varepsilon_{k}<\varepsilon_{k-1}$.
In this way, we conclude that $\mbox{sign}(S_{k}(\varepsilon))=+$ for
$\varepsilon_{k}<\varepsilon<\varepsilon_{k-1}$, in other words,
$\mbox{sign}(\Pi)(G_{k},\mu_{\varepsilon})=\left(-,\ldots,-,+\right)$, meaning
that, $\rho_{L}(G_{k})>\mu_{\varepsilon}$.

The conclusion is that, $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$
if, and only if, the sequence $(\varepsilon_{j})$ converges to zero.
Otherwise, if $\varepsilon_{j}\geq\varepsilon_{\infty}>0$ we get
$\rho_{L}(G_{k})<\mu-\varepsilon_{\infty}$ thus
$\mu_{0}=\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})<\mu$. ∎

######  Remark 7.2.

Notice that $\varepsilon_{k}$ is the solution of

$S_{k}(\varepsilon_{k})=-1+\psi_{\varepsilon}(S_{k-1}(\varepsilon_{k}))+\delta(C_{k},\mu_{\varepsilon_{k}})=0$

meaning that

$-\varepsilon_{k}:=-1+\psi(S_{k-1}(\varepsilon_{k}))+\delta(C_{k},\mu_{\varepsilon_{k}})$

If $\varepsilon_{j}\to 0$ then
$\mu_{\varepsilon_{k}}=\mu-\varepsilon_{k}\to\mu$ and
$S_{k}=-1+\psi(S_{k-1})+\delta(C_{k},\mu)=[-1+\psi(S_{k-1})+\delta(C_{k},\mu)-(-1+\psi(S_{k-1}(\varepsilon_{k}))+\delta(C_{k},\mu_{\varepsilon_{k}}))]-\varepsilon_{k}\to
0$ .

For practical purposes, it is feasible, from a theoretical point of view, but
computationally very hard, to obtain the numbers $(\varepsilon_{j})$, because
each function $g_{j}(\varepsilon)$ is obtained recursively from the previous
one. We notice, however, that we can compute another sequence $(\alpha_{j})$
such that $\varepsilon_{j}<\alpha_{j}$. Thus, if $(\alpha_{j})$ converges to
zero then $(\varepsilon_{j})$ converges to zero, providing a sufficient
condition to ensure that $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$.
The numbers $(\alpha_{j})$ will be the root of the linear approximation
$h_{j}(\varepsilon):=g_{j}(0)+\varepsilon\frac{d\,g_{j}}{d\,\varepsilon}(0)$
of $g_{j}(\varepsilon)$ in $\varepsilon=0$, see Figure 10.

Figure 10. Representation of $\alpha_{j}$.

######  Corollary 7.3.

Under the hypothesis of Theorem 7.1, consider the sequence
$(\varepsilon_{j})$. Let $(\alpha_{j})$ be the sequence of roots of the
functions

$h_{j}(\varepsilon):=g_{j}(0)+\varepsilon\frac{d\,g_{j}}{d\,\varepsilon}(0).$

Then, $\varepsilon_{j}<\alpha_{j}$ and, if $(\alpha_{j})$ converges to zero,
then $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$. Moreover,
$\displaystyle\alpha_{1}:=\frac{-S_{1}}{1+\delta^{\prime}(T_{1},\mu_{\varepsilon})(0)}$
and, for $2\leq j\leq k$,

(24)
$\alpha_{j}=\frac{-S_{j}}{\beta_{j}-\frac{1}{S_{j-1}}\frac{1}{\alpha_{j-1}}},$

where $\beta_{j}:=1+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0)$, and for
$2\leq j\leq k-1$, $\beta_{k}:=1+\delta^{\prime}(C_{k},\mu_{\varepsilon})(0)$.

###### Proof.

By definition, $\alpha_{j}$ is the solution of $h_{j}(\alpha_{j})=0$, which is
given by

$\alpha_{j}=\frac{-g_{j}(0)}{\frac{d\,g_{j}}{d\,\varepsilon}(0)}.$

We recall that $g_{j}(0)=S_{j}$,

$\frac{d\,b_{j}}{d\,\varepsilon}(0)=1+\frac{1}{b_{j-1}^{2}}\frac{d\,b_{j-1}}{d\,\varepsilon}(0)$

and

$\frac{d\,g_{j}}{d\,\varepsilon}(\varepsilon)=\left\\{\begin{array}[]{ll}1+\delta^{\prime}(T_{1},\mu_{\varepsilon})(0),&j=1,\;T_{1}=[q_{1}^{1},\ldots,q_{\omega(T_{1})}^{1}]\\\
1+\frac{1}{\left(g_{j-1}(\varepsilon)\right)^{2}}\frac{d\,g_{j-1}(\varepsilon)}{d\,\varepsilon}+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0),&2\leq
j\leq k-1,\;T_{j}=[q_{1}^{j},\ldots,q_{\omega(T_{j})}^{j}]\\\
1+\frac{1}{\left(g_{k-1}(\varepsilon)\right)^{2}}\frac{d\,g_{k-1}(\varepsilon)}{d\,\varepsilon}+\delta^{\prime}(C_{k},\mu_{\varepsilon})(0),&j=k,\;C_{k}=[p_{1}^{k},\ldots,p_{\omega(C_{k})}^{k}]\end{array}\right.$

so

(25)
$\frac{d\,g_{j}}{d\,\varepsilon}(0)=1+\frac{1}{\left(S_{j-1}\right)^{2}}\frac{d\,g_{j-1}}{d\,\varepsilon}(0)+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0).$

Substituting the above formulas, for $2\leq j\leq k-1$, we obtain

$\alpha_{j}=\frac{-S_{j}}{1+\frac{1}{\left(S_{j-1}\right)^{2}}\frac{d\,g_{j-1}}{d\,\varepsilon}(0)+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0)}.$

Notice that
$\displaystyle\alpha_{j-1}=\frac{-g_{j-1}(0)}{\frac{d\,g_{j-1}}{d\,\varepsilon}(0)}=\frac{-S_{j-1}}{\frac{d\,g_{j-1}}{d\,\varepsilon}(0)}$
so

(26)
$\alpha_{j}=\frac{-S_{j}}{\beta_{j}-\frac{1}{S_{j-1}}\frac{1}{\alpha_{j-1}}},$

where $\beta_{j}:=1+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0)$. The above
formula provides a way to obtain $\alpha_{j}$ from $\alpha_{j-1}$ without
compositions, only the computation of $S_{j}$, which is easy to perform, and
the knowledge of the derivatives $b^{\prime}_{j}(0)$, which is a very standard
recursion problem (see Equation (20), for $\varepsilon=0$), independent from
this particular construction.
In order to conclude our proof we just need to show that
$\varepsilon_{j}<\alpha_{j}$. We claim that it is true if $g_{j}(\varepsilon)$
is a concave function ($\frac{d^{2}\,}{d\,\varepsilon^{2}}(g_{j})\geq 0$).
Indeed, $h_{j}(0)=g_{j}(0)$ and for $\varepsilon>0$ we have

$\frac{d\,}{d\,\varepsilon}(g_{j}(\varepsilon)-h_{j}(\varepsilon))=\frac{d\,g_{j}}{d\,\varepsilon}-\frac{d\,g_{j}}{d\,\varepsilon}(0)\geq
0.$

Thus $g_{j}(\alpha_{j})-h_{j}(\alpha_{j})=g_{j}(\alpha_{j})\geq
g_{j}(0)-h_{j}(0)=0$, so that $g_{j}(\alpha_{j})\geq 0$. Since $g_{j}$ is
increasing we get $\varepsilon_{j}<\alpha_{j}$.
In this way, we only need to show that
$\frac{d^{2}\,}{d\,\varepsilon^{2}}(g_{j})\geq 0$:

$\frac{d^{2}\,g_{j}(\varepsilon)}{d\,\varepsilon^{2}}=\frac{d\,}{d\,\varepsilon}\left(1+\frac{1}{\left(g_{j-1}(\varepsilon)\right)^{2}}\frac{d\,g_{j-1}(\varepsilon)}{d\,\varepsilon}+\sum_{i=1}^{\omega(T_{j})}\frac{1}{\left(b_{q_{i}^{j}}(\varepsilon)\right)^{2}}\frac{d\,b_{q_{i}^{j}}(\varepsilon)}{d\,\varepsilon}\right)=$
$=\frac{-2}{\left(g_{j-1}(\varepsilon)\right)^{3}}\left(\frac{d\,g_{j-1}(\varepsilon)}{d\,\varepsilon}\right)^{2}+\frac{1}{\left(g_{j-1}(\varepsilon)\right)^{2}}\frac{d^{2}\,g_{j-1}(\varepsilon)}{d^{2}\,\varepsilon}+$
$\sum_{i=1}^{\omega(T_{j})}\left(\frac{-2}{\left(b_{q_{i}^{j}}(\varepsilon)\right)^{3}}\left(\frac{d\,b_{q_{i}^{j}}(\varepsilon)}{d\,\varepsilon}\right)^{2}+\frac{1}{\left(b_{q_{i}^{j}}(\varepsilon)\right)^{2}}\frac{d^{2}\,b_{q_{i}^{j}}(\varepsilon)}{d\,\varepsilon^{2}}\right)\geq
0,$

because $b_{j}<0$ for all $j$, $\frac{d^{2}\,}{d\,\varepsilon^{2}}(g_{1})=0$
and we already proved that $b^{\prime\prime}_{j}(\varepsilon)\geq 0$. ∎

######  Example 7.4.

For $\mu=\frac{5+\sqrt{33}}{2}$ and the generalized Shearer sequence
$\mathbb{T}=([1,1,1],[1],\ldots)$ and $C=([1,1],\ldots)$ we know that
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$. We can check, from the
formula (24) that
$\alpha_{1}=0.5930703308$
$\alpha_{10}=0.0003726377$
$\ldots$
$\alpha_{100}=1.33485599\times 10^{-33}$
$\alpha_{150}=5.84453701\times 10^{-50}$
$\alpha_{190}=4.78412668\times 10^{-63}$. Thus, $|\rho_{L}(G_{190})-\mu|\leq
10^{-63}$.
On the other hand, using for $\mu=5.4$ the same sequence (we can do that since
$G_{k}\in\Gamma(5.4)$) we get $\alpha_{1}=0.62182468$,
$\alpha_{10}=0.670219903$, $\alpha_{50}=0.807268557543450193961813121639$,
$\alpha_{100}=0.807268557543452357547180905158$,
$\alpha_{150}=0.80726855754345235754718090515772416691821457170280270632003$
and so on, as expected, since
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\frac{5+\sqrt{33}}{2}=5.37+<5.4$.
Finally, we consider a sequence selected from a genetic algorithm to fit to
$\mu=5.4$:

$\mathbb{T}=([0],[1,1],[1],[7],[5],[6],[7],[7],[2],[3],[5],[6],[2],[5],[4],[4],[6],$
$[6],[6],[0],[6],[1,1],[0],[6],[0],[3],[4],[4],[7],[1,1],...)$

and $C$ a shift of $\mathbb{T}$. In this case $G_{k}\in\Gamma(5.4)$ and
$\alpha_{29}=0.0001005914$, suggesting that $\rho_{L}(G_{29})$is at least
$10^{-4}$ close to $\mu=5.4$. Actually, a direct computation shows that
$\rho_{L}(G_{29})=5.399999999963451$, which is $3.65\times 10^{-11}$ close to
$\mu=5.4$. That is natural since the convergence of $\alpha_{j}\to 0$ is only
a sufficient condition and $\varepsilon_{j}$ could be much smaller than
$\alpha_{j}$.

An alternative way to prove that $\alpha_{j}\to 0$ and so
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$, without a direct
computation, is the following:

######  Theorem 7.5.

If

(27)
$\sum_{m=1}^{j}\frac{1+\delta^{\prime}(T_{m},\mu_{\varepsilon})(0)}{\left(\prod_{n=m}^{j-1}S_{n}\right)^{2}}$

is unbounded, then $\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$.

###### Proof.

Let $\alpha_{j}$ be the sequence given by Corollary 7.3. By definition,
$\alpha_{j}$ is given by

$\alpha_{j}=\frac{-S_{j}}{\frac{d\,g_{j}}{d\,\varepsilon}(0)}.$

From Equation (25) we get

$\frac{d\,g_{j}}{d\,\varepsilon}(0)=1+\frac{1}{\left(S_{j-1}\right)^{2}}\frac{d\,g_{j-1}}{d\,\varepsilon}(0)+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0),$

and
$\frac{d\,g_{1}}{d\,\varepsilon}(0)=1+\delta^{\prime}(T_{1},\mu_{\varepsilon})(0)$.
Let us introduce the following auxiliary sequences

  * •

$A_{j}:=1+\delta^{\prime}(T_{j},\mu_{\varepsilon})(0)$;

  * •

$B_{j}:=\frac{1}{\left(S_{j}\right)^{2}}$;

  * •

$X_{j}:=\frac{d\,g_{j}}{d\,\varepsilon}(0)$.

Then, we obtain the standard difference equation problem

(28) $\left\\{\begin{array}[]{ll}X_{j}=A_{j}+B_{j-1}X_{j-1}\\\
X_{1}=A_{1}.\end{array}\right.$

This equation has an explicit solution:

$X_{2}=A_{2}+B_{1}X_{1}=A_{2}+B_{1}A_{1},$
$X_{3}=A_{3}+B_{2}(A_{2}+B_{1}A_{1})=A_{3}+A_{2}B_{2}+A_{1}B_{1}B_{2},$
$X_{4}=A_{4}+A_{3}B_{3}+A_{2}B_{2}B_{3}+A_{1}B_{1}B_{2}B_{3},$

and so on, obtaining

$\frac{d\,g_{j}}{d\,\varepsilon}(0)=X_{j}=\sum_{m=1}^{j}A_{m}\prod_{n=m}^{j-1}B_{n}=\sum_{m=1}^{j}\left(1+\delta^{\prime}(T_{m},\mu_{\varepsilon})(0)\right)\prod_{n=m}^{j-1}\frac{1}{\left(S_{n}\right)^{2}}.$

Notice that $S_{1}<S_{j}<\theta^{-1}$ so $-S_{1}>-S_{j}>-\theta^{-1}$. Thus,
if $\frac{d\,g_{j}}{d\,\varepsilon}(0)$ is unbounded when $j\to\infty$, then
$\alpha_{j}\to 0$. By Corollary 7.3 we obtain
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$. ∎

######  Remark 7.6.

The formula

$\frac{d\,g_{j}}{d\,\varepsilon}(0)=\sum_{m=1}^{j}\frac{1+\delta^{\prime}(T_{m},\mu_{\varepsilon})(0)}{\left(\prod_{n=m}^{j-1}S_{n}\right)^{2}},$

has an analogous for adjacency limit points, and the fact that it is unbounded
proves Shearer’s result in [13]. Although the approach is a little different,
the basis for Shearer’s argument is to show that the product
$\prod_{n=m}^{j-1}\left(S_{n}\right)$ has absolute values smaller than 1. This
is somehow hidden in the proof of the main theorem (actually he proves that
$R_{j-1}R_{j}<1,R_{j-2}R_{j+1}<1,...$ thus $R_{j-2}R_{j-1}R_{j}R_{j+1}<1$,
etc, where the $R_{j}^{\prime}s$ are the values obtained in the adjacency
case). In our Laplacian case, if this was true, we would obtain
$\frac{d\,g_{j}}{d\,\varepsilon}(0)\to\infty$ and
$\displaystyle\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$ according to Theorem 7.5.
Since it is not true that Equation (27) is unbounded for any $\mu$, this
explains why the classical Laplacian Shearer fails to prove convergence in
general.

##  8\. Final considerations and future work

In this section we discuss possible consequences of the previous results and
additional investigations one could carry on in the future, towards the proof
of the conjecture that all the points in the interval $[4.38+,\infty)$ are
Laplacian limit points.

We mention here an ongoing research work [11], where we study limit points for
the spectral radius of $A_{\alpha}$-matrix of a graph $G$, where
$A_{\alpha}(G):=\alpha D(G)+(1-\alpha)A(G)$, for $0\leq\alpha\leq 1$. It is
easy to see that $A_{0}(G)=A(G)$, the adjacency matrix of $G$ and
$A_{1}(G)=D(G)$ the degree matrix of $G$. Also,
$A_{1/2}(G)=1/2(D(G)+A(G))=1/2Q(G)$ where $Q$ is the signless Laplacian matrix
of $G$. Our main result is that for any $\alpha\in[0,1/2)$ there exists a
positive number $\tau_{2}(\alpha)>2$ such that any value
$\lambda>\tau_{2}(\alpha)$ is an $A_{\alpha}$-limit point. Notice that this
generalization of the Shearer’s work for the adjacency matrix seems to stretch
full power (which is $\alpha<\frac{1}{2}$ ). Beyond that, the technique used
does not work. As for $\alpha=\frac{1}{2}$ we have the signless Laplacian
matrix, these two manuscripts show how difficult it is to study Laplacian
limit points .

Question 1

Can one use Theorem 7.1, or Corollary 7.3, or Theorem 7.5 to prove that

$\displaystyle\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})=\mu$

at least for some class of points $\mu\geq 4.38$?
The answer of this question is definitive for approximation by linear trees
(which include caterpillars), using increasing sequences of spectral radii.

Question 2

Are there are any points where
$\displaystyle\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})<\mu?$
If yes, then $\mu$ is not a Laplacian limit point or at least it can not be
limit of increasing sequences of Laplacian spectral radii of linear trees,
because $\Gamma(\mu)$ contains all linear trees with spectral radius smaller
than $\mu$.

Question 3

In Question 2, can one find (from the previous lemmas and theorems) a maximum,
where the supremum is attained, despite the fact that
$\displaystyle\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})<\mu$
?
If yes, this would be a special number in some way!

Question 4

Is there some particular subinterval of $[4.38+,\infty)$ regarding properties
of the product $S_{j}S_{j+1}<1$, provided that $S_{j}>-1$ and $S_{j+1}<-1$?
If the answer is yes, then one may be able to use Theorem 7.5 to prove that
$\displaystyle\sup_{G_{k}\in\Gamma(\mu)}\lim_{k\to\infty}\rho_{L}(G_{k})=\mu.$

Question 5

A natural future problem would be to investigate the application of Theorem
7.5. Can one prove that $X_{k}>m_{k}\to\infty$?
In order to do that, one needs to control weather $S_{j}S_{j+1}<1$, at least
for $j\geq k$ because (for $k\geq 2$)

$X_{4}=A_{4}+A_{3}B_{3}+A_{2}B_{2}B_{3}+A_{1}B_{1}B_{2}B_{3},$

we start with $A_{2}B_{2}B_{3}$ where
$A_{2}=1+\delta^{\prime}(T_{2},\mu_{\varepsilon})(0)>1$ and
$B_{2}B_{3}=\frac{1}{S_{2}^{2}S_{3}^{2}}>1$, provided that $S_{2}S_{3}<1$.
This does not depend on $S_{1}$, same as for the next additives, except for
$A_{1}B_{1}B_{2}B_{3}$. Recall that the goal is to show that $X_{k}\to\infty$,
in this case one may have $X_{4}>3$ if $B_{3}>1$ because $A_{4}>1$,
$A_{2}B_{2}B_{3}>1$, but we do not have control of $A_{1}B_{1}B_{2}B_{3}$
since $S_{1}$ could be negative and as large as $1-\mu$. In the worst case
$X_{4}>2$.

Question 6

Despite the success of the Shearer’s method for adjacency limit points, one
may ask if part of our difficulties when we translate to the Laplacian limit
points, comes from the choice of an iterative process that builds sequences of
increasing spectral radii, as we did. One possible line of research is to
consider Lapalacian limit points arbitrarily constructed, but still using
treatable trees like caterpillars, linear trees and others. If one succeeds,
then one may expect a proof of the conjecture or to conclude that one should
look at a more general class of trees or general graphs.

## Acknowledgments

This work is partially supported by MATH-AMSUD under project GSA, brazilian
team financed by CAPES under project 88881.694479/2022-01. It was executed
while F. Belardo was visiting UFRGS, under the financial of CAPES - PRINT,
project 88887.716953/2022-00. The research of Francesco Belardo and Vilmar
Trevisan was supported by a grant from the group GNSAGA of INdAM (Italy).
Elismar Oliveira and Vilmar Trevisan acknowledge the support of CNPq grants
408180/2023-4 and 310827/2020-5.

## References

  * [1] Anderson, William N.; Morley, T. D. Eigenvalues of the laplacian of a graph.  Linear and Multilinear Algebra 1985-oct vol. 18 iss. 2 18 (oct 1985).
  * [2] Estes, D. R. Eigenvalues of symmetric integer matrices.  J. Number Theory 42, 3 (1992), 292–296.
  * [3] Guo, J.-M. On limit points of Laplacian spectral radii of graphs.  Linear Algebra Appl. 429, 7 (2008), 1705–1718.
  * [4] Hoffman, A. J. On limit points of spectral radii of non-negative symmetric integral matrices.  In Graph theory and applications (Proc. Conf., Western Michigan Univ., Kalamazoo, Mich., 1972; dedicated to the memory of J. W. T. Youngs) (1972), Lecture Notes in Math., Vol. 303, Springer, Berlin, pp. 165–172.
  * [5] Jacobs, D. P., and Trevisan, V. Locating the eigenvalues of trees.  Linear Algebra Appl. 434, 1 (2011), 81–88.
  * [6] Johnson, C. R., Li, A. A., and Walker, A. J. Ordered multiplicity lists for eigenvalues of symmetric matrices whose graph is a linear tree.  Discrete Math. 333 (2014), 39–55.
  * [7] Johnson, C. R., and Saiago, C. M. Eigenvalues, multiplicities and graphs, vol. 211 of Cambridge Tracts in Mathematics.  Cambridge University Press, Cambridge, 2018.
  * [8] Kirkland, S. A note on limit points for algebraic connectivity.  Linear Algebra Appl. 373 (2003), 5–11.  Combinatorial Matrix Theory Conference (Pohang, 2002).
  * [9] Kirkland, S. Limit points for normalized Laplacian eigenvalues.  Electron. J. Linear Algebra 15 (2006), 337–344.
  * [10] Oliveira, E. R., and Trevisan, V. Applications of rational difference equations to spectral graph theory.  J. Difference Equ. Appl. 27, 7 (2021), 1024–1051.
  * [11] Oliveira, E. R., and Trevisan, V. Limit points of ${A}_{\alpha}$-matrices of graphs.  arXiv preprint arXiv:2404.10953 (2024).
  * [12] Salez, J. Every totally real algebraic integer is a tree eigenvalue.  J. Combin. Theory Ser. B 111 (2015), 249–256.
  * [13] Shearer, J. B. On the distribution of the maximum eigenvalue of graphs.  Linear Algebra Appl. 114/115 (1989), 17–20.
  * [14] Wang, J., Huang, Q., An, X., and Belardo, F. Some results on the signless Laplacians of graphs.  Appl. Math. Lett. 23, 9 (2010), 1045–1049.
  * [15] Wang, J., Wang, J., and Brunetti, M. The hoffman program of graphs: old and new.  arXiv preprint arXiv:2012.13079 (2020).
  * [16] Zhang, F., and Chen, Z. Limit points of eigenvalues of (di)graphs.  Czechoslovak Math. J. 56(131), 3 (2006), 895–902.
  * [17] Zhang, X.-D. The laplacian eigenvalues of graphs: a survey.  arXiv preprint arXiv:1111.2897 (2011).