Complexity of Join and Corona graphs and Chebyshev polynomials

ABSTRACT

Boesh and Prodinger have shown how to use properties of Chebyshev polynomials to compute formulas for the number of spanning trees of some special graphs. In this paper, we extend this idea for some operations on graphs such as, Join and Corona of two graphs $G_1$ and $G_2$, if $G_1$ and $G_2$ are one of the following graphs: (i) Path graph $P_n$, (ii) Cycle graph $C_n$, (iii) Complete graph $K_n$, (iv) Complete bipartite $K_{m,n}$, (v) Hypercube graph $Q_n$, (vi) Fan graph $F_n$ and (vi) Wheel graph $W_n$.

KEYWORDS:

• Number of spanning trees
• Chebyshev polynomials
• Join
• Corona

Mathematics subject classifications:

• 05c05
• 05c50
• 05c99

1. Introduction

The study of the number of spanning trees in a graph has a long history and has been very active because computing this number is important: (1) in analysing energy of masers in investigating the possible particle transitions; (2) in estimating the reliability of a network; (3) in designing electrical circuits; (4) in enumerating certain chemical isomers; (5) in counting the number of Eulerian circuits in a graph [1–10].

The number of spanning trees $\tau \left(G\right)$ of a finite undirected graph $G$ is equal to the total number of distinct spanning subgraphs of $G$ that are trees. This quantity is also known as the complexity of $G$. There exist various methods of finding this number. Let us recall the famous Matrix tree theorem of Kichhoff [11]: The complexity $\tau \left(G\right)$ of a graph $G$ is equal any cofactor of Laplace matrix $L\left(G\right)=D\left(G\right)-A\left(G\right)$, where $D\left(G\right)=diag\left(d\right(u),u\in V)$ is the diagonal matrix of vertex degrees of $G$ and $A\left(G\right)$ is the adjacency matrix of $G$.

Another way to compute the number of spanning trees a graph, is to recursively degenerate the graph by deleting and contracting edges. In this way, the number of spanning trees of an unknown graph $G$ is reduced to a sum of the number of spanning trees in small graphs which is well known. Let $e$be an edge with endpoints $u$ and $v$in the graph $G$, the contraction $G\cdot e$ of $e$ from $G$ is the graph obtained by removing $e$ and identifying $u$ and $v$. The deletion $G-e$ of $e$ from $G$ is the graph obtained by removing $e$. The formula for counting spanning trees in a graph $G$ is $\tau \left(G\right)=\tau (G\cdot e)+\tau (G-e)$ (see [12,13]).

Another way to compute the number of spanning trees in a graph is using linear algebra. Let $G(V,E)$ be a graph with vertex set is $V=\{v_1,v_2,\dots ,v_n\}$ with non-increasing degree sequence $d_1\ge d_2\ge \dots \ge d_n$, where $d_i$ is the degree of vertex $v_i$ for $i=1,2,\dots ,n.$ The eigenvalues of $L\left(G\right)$ are called the Laplacian eigenvalues (Spectrum) and denoted by ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_n=0.$ It is well known that ${\lambda }_1=n.$

The following formula in terms of Laplacian eigenvalues of $G$ is well known by Kelmans and Chelnoknov [14] as: (1) $\tau \left(G\right)=\frac{1}{n}\prod _{i=1}^{n-1}{\lambda }_i.$(1) By $S\left(G\right)$, we denote the Laplacian spectrum of $G$. For a graph $G$ of order $n$, we write $S\left(G\right)=({\lambda }_1,\dots ,{\lambda }_n)$ with the understanding that ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_n.$

2. Some properties of Chebyshev polynomial

In this section, we introduce some relations concerning Chebyshev polynomials of the first and second kind which we use in our computations [15].

The Chebyshev polynomials of the first kind are defined by: (2) $T_n\left(x\right)=\cos (n\arccos x).$(2)

It is easily verified that (3) $T_{n+1}\left(x\right)=2x{T}_n\left(x\right)-T_{n-1}\left(x\right);T_0\left(x\right)=1,T_1\left(x\right)=x.$(3)

Furthermore by using standard methods for solving the recursion (3), one obtains the explicit formula: (4) $T_n\left(x\right)=\frac{1}{2}\left[{\left(x+\sqrt{x^2-1}\right)}^n+{\left(x-\sqrt{x^2-1}\right)}^n\right].$(4)

The Chebyshev polynomials of the second kind are defined by: (5) $U_n\left(x\right)=\frac{1}{n+1}\frac{\text{d}}{\text{d}x}{T}_{n+1}\left(x\right)=\frac{\sin \left(\right(n+1)\arccos x)}{\sin (\arccos x)}.$(5)

It is easily verified that (6) $\begin{array}{rl}{U}_{n+1}\left(x\right)& =2x{U}_n\left(x\right)-U_{n-1}\left(x\right);U_0\left(x\right)\\ & =1,U_1\left(x\right)=2x.\end{array}$(6)

Furthermore by using standard methods for solving the recursion (6), one obtains the explicit formula: (7) $\begin{array}{rl}{U}_n\left(x\right)& =\frac{1}{2\sqrt{x^2-1}}\left[{\left(x+\sqrt{x^2-1}\right)}^{n+1}\right.\\ & \left.-{\left(x-\sqrt{x^2-1}\right)}^{n+1}\right],n\ge 1.\end{array}$(7)

We have, $U_{n-1}\left(\cos \frac{i\pi }{n}\right)=0,i=1,2,\dots ,n-1.$ Hence (8) $U_{n-1}\left(x\right)=2^{n-1}\prod _{i=1}^{n-1}\left(x-\cos \frac{i\pi }{n}\right).$(8)

One further notes that: (9) $\begin{array}{rl}{U}_{n-1}(-x)& =(-1{)}^{n-1}{U}_{n-1}\left(x\right)\\ & =2^{n-1}\prod _{i=1}^{n-1}\left(x+\cos \frac{i\pi }{n}\right).\end{array}$(9)

These two results yield another formula for $U_n\left(x\right)$, (10) $U_{n-1}^2\left(x\right)=4^{n-1}\prod _{j=1}^{n-1}\left(x^2-{\cos }^2\frac{j\pi }{n}\right).$(10)

Simple manipulation of the above formula yields the following, which also will be extremely useful to us latter: (11) $U_{n-1}^2\left(\sqrt{\frac{x+2}{4}}\right)=\prod _{j=1}^{n-1}\left(x-2\cos \frac{2j\pi }{n}\right).$(11)

One further notes that: (12) $\prod _{i=1}^{n-1}\left(2-2\cos \frac{i\pi }{n}\right)=n,n\ge 2,$(12) (13) $\prod _{i=1}^{n-1}\left(2-2\cos \frac{2i\pi }{n}\right)=n^2,n\ge 2.$(13) Polynomials $T_n\left(x\right)$ and $U_{n-1}\left(x\right)$ are related by the following identity (14) $U_{n-1}^2\left(x\right)=\frac{1}{2(x^2-1)}\left[T_n\left(2x^2-1\right)-1\right].$(14) Fibonacci numbers, ${\text{F}}_n$.are defined by recursive relation (15) ${\text{F}}_n={\text{F}}_{n-1}+{\text{F}}_{n-2},n\ge 2.$(15) With initial conditions ${\text{F}}_0=0$ and ${\text{F}}_1={\text{F}}_2=1$ [16].

An interesting fact follows by comparing (7) with the well known closed form formula for the Fibonacci numbers ${\text{F}}_n$. (16) ${\text{F}}_n=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right],$(16) namely, (17) $\begin{array}{rl}{U}_{n-1}\left(\frac{3}{2}\right)& ={\text{F}}_{2n}\\ & =\frac{1}{\sqrt{5}}\left[{\left(\frac{3+\sqrt{5}}{2}\right)}^n-{\left(\frac{3-\sqrt{5}}{2}\right)}^n\right].\end{array}$(17)

The Lucas numbers, ${\text{L}}_n$ are defined by recursive relation (18) ${\text{L}}_n={\text{L}}_{n-1}+{\text{L}}_{n-2},n\ge 2,$(18) with initial conditions ${\text{L}}_0=2,{\text{L}}_1=1$ and ${\text{L}}_2=3$ [17].

Also we can prove that: (19) $2T_n\left(\frac{3}{2}\right)={\text{L}}_{2n}={\left(\frac{3+\sqrt{5}}{2}\right)}^n+{\left(\frac{3-\sqrt{5}}{2}\right)}^n.$(19)

3. Complexity of the join of two graphs

The join of two graphs $G_1$ and $G_2$ is the graph $G_1\vee G_2=\left(\overline{{\overline{G}}_1+\overline{G_2}}\right)$. This is the union of two graphs $G_1$ and $G_2$, with every vertex in $G_1$ connected to every vertex in $G_2$. We note that $G_1\vee G_2$ is always connected [18].

Theorem 3.1:

Let $G_1$ and $G_2$ be graphs on disjoint sets of $n_1$ and $n_2$ vertices respectively. If $S\left(G_1\right)=({\lambda }_1,\dots ,{\lambda }_{n_1})$ and $S\left(G_2\right)=({\sigma }_1,\dots ,{\sigma }_{n_2})$ are the eigenvalues of $L\left(G_1\right)$ and $L\left(G_2\right)$ arranged in non-increasing order, then the eigenvalues of $L(G_1\vee G_2)$ are $n_1+n_2;{\lambda }_1+n_2,\dots ,{\lambda }_{n_1-1}+n_2;{\sigma }_1+n_1,\dots ,{\sigma }_{n_2-1}+n_1$ and $0$.

Proof

The adjacency matrix of $G_1\vee G_2$ can be defined as: $A(G_1\vee G_2)=\left(\begin{array}{cc}A\left(G_1\right)& J_{n_1{n}_2}\\ J_{n_2{n}_1}& A\left(G_2\right)\end{array}\right)$

Let $1_n$ be all one column vector of length $n$, the degree sequence of $G_1\vee G_2$, $\begin{array}{rl}d(G_1\vee G_2)& =A(G_1\vee G_2)1_{|V(G_1\vee G_2)|}\\ & =\left(\begin{array}{cc}A\left(G_1\right)& J_{n_1{n}_2}\\ J_{n_2{n}_1}& A\left(G_2\right)\end{array}\right)\left(\begin{array}{c}1\\ ⋮\\ ⋮\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}{d}_1\left(G_1\right)+n_2\\ ⋮\\ d_{n_1}\left(G_1\right)+n_2\\ d_1\left(G_2\right)+n_1\\ ⋮\\ d_{n_2}\left(G_2\right)+n_1\end{array}\right)\end{array}$

Thus the degree matrix of $G_1\vee G_2$ can be defined as $D(G_1\vee G_2)=\left(\begin{array}{cc}D\left(G_1\right)+n_2{I}_{n{}_1}& 0\\ 0& D\left(G_2\right)+n_1{I}_{n{}_2}\end{array}\right)$ and hence the Laplacian matrix of $G_1\vee G_2$ is $L(G_1\vee G_2)=\left(\begin{array}{cc}L\left(G_1\right)+n_2{I}_{n{}_1}& -J_{n_1{n}_2}\\ -J_{n_2{n}_1}& L\left(G_2\right)+n_1{I}_{n{}_2}\end{array}\right)$.

From this we may readily deduce the Laplacian spectrum of $G_1\vee G_2$ by exhibiting a complete set of eigenvectors. If $w$ is an eigenvector of $L\left(G_1\right)$ corresponding to ${\lambda }_i$, $1\le i\le n_1-1$, then the vector $x_k=w_k$, $1\le k\le n_1$ and $x_k=0$ otherwise, can be seen to be an eigenvector of $L(G_1\vee G_2)$ with eigenvalue ${\lambda }_i+n_2$.

Similarly, we obtain the eigenvalue ${\sigma }_j+{n_1}_{}$ for $1\le j\le n_2-1$. The all ones vector gives as usual, the eigenvalue $0$, and the eigenvector whose value is $-n_1$ on the vertices of $G_1$ and $n_2$ on the vertices of $G_2$ gives the eigenvalue $n_1+n_2$.

Lemma 3.1:

Let $L\left(F_n\right)$ be the Laplacian matrix of the fan graph $F_n=K_1\vee P_n,n\ge 2$. Then its eigenvalues are $0,n+1,3-2\cos \frac{i\pi }{n},(1\le i\le n-1).$

Proof

The Laplacian matrix of the fan graph $F_n$ can be defined as: $\begin{array}{rl}& L\left(F_n\right)\\ & =\left(\begin{array}{ccccccc}n-\lambda & -1& \cdots & \cdots & \cdots & \cdots & -1\\ -1& 2-\lambda & -1& 0& \cdots & \cdots & 0\\ ⋮& -1& 3-\lambda & \ddots & \ddots & \ddots & ⋮\\ ⋮& 0& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & \ddots & \ddots & 0\\ ⋮& ⋮& \ddots & \ddots & \ddots & 3-\lambda & -1\\ -1& 0& \cdots & \cdots & 0& -1& 2-\lambda \end{array}\right),\end{array}$

Straightforward induction using properties of determinants, we have $\begin{array}{rl}& det\left(L\left(F_n\right)\right)=\frac{\lambda +n\lambda -{\lambda }^2}{\lambda -1}det\\ & \left(\begin{array}{cccccc}2-\lambda & -1& 0& \cdots & \cdots & 0\\ -1& 3-\lambda & \ddots & \ddots & \ddots & ⋮\\ 0& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & 0\\ ⋮& \ddots & \ddots & \ddots & 3-\lambda & -1\\ 0& \cdots & \cdots & 0& -1& 2-\lambda \end{array}\right).\end{array}$

Applying Lemma 3.1 by the author in [19], we get: $\begin{array}{rl}det\left(L\left(F_n\right)\right)& =\frac{\lambda +n\lambda -{\lambda }^2}{\lambda -1}(2-\lambda -1)U_{n-1}\\ & \left(\frac{1+2-\lambda }{2}\right).\end{array}$

Using identity (8), we have: $det\left(L\right(F_n\left)\right)=\lambda (\lambda -n-1)\prod _{k=1}^{n-1}\left(3-\lambda -\cos \frac{k\pi }{n}\right),$

Therefore, $det\left(L\right(F_n\left)\right)=0$ implies, $\lambda =0,\lambda =n+1,\lambda =3-2\cos \frac{i\pi }{n},1\le i\le n-1.$

Lemma 3.2

Let $L\left(w_n\right)$ be the Laplacian matrix of the wheel graph $W_n=K_1\vee C_n,n\ge 2$. Then its eigenvalues are $0,n-1,3-2\cos \frac{2j\pi }{n},(1\le j\le n-1).$

Proof

The Laplacian matrix of the wheel graph $W_n$ can be defined as: $\begin{array}{rl}& L\left(W_n\right)\\ & =\left(\begin{array}{ccccccc}n-\lambda & -1& \cdots & \cdots & \cdots & \cdots & -1\\ -1& 3-\lambda & -1& 0& \cdots & 0& -1\\ ⋮& -1& 3-\lambda & \ddots & \ddots & \ddots & 0\\ ⋮& 0& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & \ddots & \ddots & 0\\ ⋮& 0& \ddots & \ddots & \ddots & 3-\lambda & -1\\ -1& -1& 0& \cdots & 0& -1& 3-\lambda \end{array}\right).\end{array}$

Straightforward induction using properties of determinants, we get $\begin{array}{rl}& det\left(L\left(W_n\right)\right)=\frac{\lambda +n\lambda -{\lambda }^2}{\lambda -1}det\\ & \left(\begin{array}{cccccc}3-\lambda & -1& 0& \cdots & 0& -1\\ -1& 3-\lambda & \ddots & \ddots & \ddots & 0\\ 0& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & 0\\ 0& \ddots & \ddots & \ddots & 3-\lambda & -1\\ -1& 0& \cdots & 0& -1& 3-\lambda \end{array}\right).\end{array}$

Applying Lemma 3.2 by the author in [20] together with Equation (14), we get: $\begin{array}{rl}det\left(L\right(W_n\left)\right)& =\frac{2(\lambda +n\lambda -{\lambda }^2)}{\lambda -1}\left[T_n\left(\frac{3-\lambda }{2}\right)-1\right]\\ & =\frac{\lambda +n\lambda -{\lambda }^2}{\lambda -1}(1-\lambda )U_{n-1}^2\left(\sqrt{\frac{5-\lambda }{4}}\right)\\ & =\lambda (\lambda -n-1)\prod _{k=1}^{n-1}\left(3-\lambda -2\cos \frac{2k\pi }{n}\right).\end{array}$

Thus, $det\left(L\right(W_n\left)\right)=0,$ implies, $\lambda =0,\lambda =n+1,\lambda =3-2\cos \frac{2j\pi }{n},(1\le j\le n-1).$

Now we can introduce the following theorem:

Theorem 3.2

Let $G_1$ and $G_2$ be graphs on disjoint sets of $n_1$ and $n_2$ vertices respectively. If $S\left(G_1\right)=({\lambda }_1,\dots ,{\lambda }_{n_1})$ and $S\left(G_2\right)=({\sigma }_1,\dots ,{\sigma }_{n_2})$ are the eigenvalues of $L\left(G_1\right)$ and $L\left(G_2\right)$, then $\tau (G_1\vee G_2)=\prod _{i=1}^{n_1-1}({\lambda }_i+n_2)\prod _{j=1}^{n_2-1}({\sigma }_j+n_1).$

Theorem 3.5

For $m,n,r$ and $s\ge 1$, we have

1. $\begin{array}{rl}& \tau (P_m\vee P_n)\\ & =\frac{1}{2^{m+n}\sqrt{({(n+2)}^2-4)({(m+2)}^2-4)}}\\ & \times \left[{\left(n+2+\sqrt{{(n+2)}^2-4}\right)}^m\right.\\ & \left.-{\left(n+2-\sqrt{{(n+2)}^2-4}\right)}^m\right]\times \\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n\right].\end{array}$

2. $\begin{array}{rl}& \tau (P_m\vee C_n)\\ & =\frac{n}{2^{m+n}\times m\sqrt{{(n+2)}^2-4}}\\ & \times \left[{\left(n+2+\sqrt{{(n+2)}^2-4}\right)}^m\right.\\ & \left.-{\left(n+2-\sqrt{{(n+2)}^2-4}\right)}^m\right]\times \\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

3. $\begin{array}{rl}& \tau (C_m\vee C_n)=\frac{1}{2^{m+n}\times mn}\\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n-2^{n+1}\right]\\ & \times \left[{\left(n+2+\sqrt{{(n+2)}^2-4}\right)}^m\right.\\ & \left.+{\left(n+2-\sqrt{{(n+2)}^2-4}\right)}^m-2^{m+1}\right].\end{array}$

4. $\begin{array}{rl}& \tau (K_m\vee P_n)=\frac{{(m+n)}^{m-1}}{2^n\sqrt{{(m+2)}^2-4}}\\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n\right].\end{array}$

5. $\begin{array}{rl}& \tau (K_m\vee C_n)=\frac{{(m+n)}^{m-1}}{m\times 2^n}\\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

6. $\begin{array}{rl}& \tau (K_{r,s}\vee P_n)=\frac{(r+s+m){(r+m)}^{s-1}{(s+m)}^{r-1}}{2^m\sqrt{{(r+s+2)}^2-4}}\\ & \times \left[{\left(r+s+2+\sqrt{{(r+s+2)}^2-4}\right)}^m\right.\\ & \left.-{\left(r+s+2-\sqrt{{(r+s+2)}^2-4}\right)}^m\right].\end{array}$

7. $\begin{array}{rl}& \tau (K_{r,s}\vee C_m)=\frac{(r+s+m){(r+m)}^{s-1}{(s+m)}^{r-1}}{2^m\times (r+s)}\\ & \times \left[{\left(r+s+2+\sqrt{{(r+s+2)}^2-4}\right)}^m\right.\\ & \left.+{\left(r+s+2-\sqrt{{(r+s+2)}^2-4}\right)}^m-2^{m+1}\right].\end{array}$

8. $\begin{array}{rl}\tau (K_{r,s}\vee K_m)& =(r+m{)}^{s-1}(s+m{)}^{r-1}\\ & (r+s+m{)}^m.\end{array}$

9. $\begin{array}{rl}\tau (K_{m,n}\vee K_{r,s})& =(m+r+s{)}^{n-1}(n+r+s{)}^{m-1}\\ & (r+m+n{)}^{s-1}(s+m+n{)}^{r-1}\\ & (m+n+r+s{)}^2.\end{array}$

10. $\begin{array}{rl}& \tau \left(\overline{K_m}\vee P_n\right)=\frac{n^{m-1}}{2^n\sqrt{{(m+2)}^2-4}}\\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n\right].\end{array}$

11. $\begin{array}{rl}& \tau \left(\overline{K_m}\vee C_n\right)=\frac{n^{m-1}}{2^n\times m}\\ & \times \left[{\left(m+2+\sqrt{{(m+2)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+2-\sqrt{{(m+2)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

Proof

1. The eigenvalues of $L\left(P_m\right)$ and $L\left(P_n\right)$ are $2-2\cos \frac{i\pi }{m}(0\le i\le m-1)$ and $2-2\cos \frac{j\pi }{n}(0\le j\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (P_m\vee P_n)=\frac{1}{m+n}(m+n)\\ & \prod _{i=1}^{m-1}\left(n+2-2\cos \frac{i\pi }{m}\right)\prod _{j=1}^{n-1}\left(m+2-2\cos \frac{j\pi }{n}\right)\\ & =U_{m-1}\left(\frac{n+2}{2}\right)U_{n-1}\left(\frac{m+2}{2}\right).\end{array}$

The explicit formula follows from Equation (7).

1. The eigenvalues of $L\left(P_m\right)$ and $L\left(C_n\right)$ are $2-2\cos \frac{i\pi }{m}(0\le i\le m-1)$ and $2-2\cos \frac{2j\pi }{n}(0\le j\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}\tau (P_m\vee C_n)& =\frac{1}{m+n}(m+n)\\ & \prod _{i=1}^{m-1}\left(n+2-2\cos \frac{i\pi }{m}\right)\\ & \times \prod _{j=1}^{n-1}\left(m+2-2\cos \frac{2j\pi }{n}\right)\\ & =U_{m-1}\left(\frac{n+2}{2}\right)U_{n-1}^2\left(\sqrt{\frac{m+4}{4}}\right)\\ & =\frac{2}{m}{U}_{m-1}\left(\frac{n+2}{2}\right)\left[T_n\left(\frac{m+2}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equations (4) and (7).

1. The eigenvalues of $L\left(C_m\right)$ and $L\left(C_n\right)$ are $2-2\cos \frac{2i\pi }{m}(0\le i\le m-1)$ and $2-2\cos \frac{2j\pi }{n}(0\le j\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (C_m\vee C_n)=\frac{1}{m+n}(m+n)\\ & \prod _{i=1}^{m-1}\left(n+2-2\cos \frac{2i\pi }{m}\right)\\ & \times \prod _{j=1}^{n-1}\left(m+2-2\cos \frac{2j\pi }{n}\right)\\ & =U_{m-1}^2\left(\sqrt{\frac{n+4}{4}}\right)U_{n-1}^2\left(\sqrt{\frac{m+4}{4}}\right)\\ & =\frac{4}{mn}\left[T_m\left(\frac{n+2}{2}\right)-1\right]\left[T_n\left(\frac{m+2}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equation (4).

1. The eigenvalues of $L\left(K_m\right)$ and $L\left(P_n\right)$ are $0^1,m^{m-1}$ and $2-2\cos \frac{i\pi }{n}(0\le i\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_m\vee P_n)=(m+n{)}^{m-1}\\ & \prod _{i=1}^{n-1}\left(m+2-2\cos \frac{i\pi }{n}\right)\\ & =(m+n{)}^{m-1}{U}_{n-1}\left(\frac{m+2}{2}\right).\end{array}$

The explicit formula follows from Equation (7).

1. The eigenvalues of $L\left(K_m\right)$ and $L\left(C_n\right)$ are $0^1,m^{m-1}$ and $2-2\cos \frac{2i\pi }{n}(0\le i\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_m\vee C_n)=(m+n{)}^{m-1}\\ & \prod _{i=1}^{n-1}\left(m+2-2\cos \frac{2i\pi }{n}\right)\\ & =(m+n{)}^{m-1}{U}_{n-1}^2\left(\sqrt{\frac{m+4}{4}}\right)\\ & =\frac{2{(m+n)}^{m-1}}{m}\left[T_n\left(\frac{m+2}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equation (4).

1. The eigenvalues of $L\left(K_{r,s}\right)$ and $L\left(P_m\right)$ are $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $2-2\cos \frac{i\pi }{m}(0\le i\le m-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_{r,s}\vee P_m)=(r+m{)}^{s-1}(s+m{)}^{r-1}\\ & (r+s+m)\prod _{i=1}^{m-1}\left(r+s+2-2\cos \frac{i\pi }{m}\right)\\ & =(r+m{)}^{s-1}(s+m{)}^{r-1}(r+s+m)\\ & U_{m-1}\left(\frac{r+s+2}{2}\right).\end{array}$

The explicit formula follows from Equation (7).

1. The eigenvalues of $L\left(K_{r,s}\right)$ and $L\left(C_m\right)$ are $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $2-2\cos \frac{2i\pi }{m}(0\le i\le m-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_{r,s}\vee C_m)=(r+m{)}^{s-1}(s+m{)}^{r-1}(r+s+m)\\ & \prod _{i=1}^{m-1}\left(r+s+2-2\cos \frac{2i\pi }{m}\right)\\ & =(r+m{)}^{s-1}(s+m{)}^{r-1}(r+s+m)\\ & U_{m-1}^2\left(\sqrt{\frac{r+s+4}{4}}\right)\\ & =\frac{2{(r+m)}^{s-1}{(s+m)}^{r-1}(r+s+m)}{r+s}\\ & \left[T_m\left(\frac{r+s+2}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equation (4).

1. The eigenvalues of $L\left(K_{r,s}\right)$ and $L\left(K_m\right)$ are $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $0^1,m^{m-1}$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_{r,s}\vee K_m)=(r+m{)}^{s-1}(s+m{)}^{r-1}(r+s+m)\\ & (r+s+m{)}^{m-1}\\ & =(r+m{)}^{s-1}(s+m{)}^{r-1}(r+s+m{)}^m.\end{array}$

2. The eigenvalues of $L\left(K_{r,s}\right)$ and $L\left(K_m\right)$ are $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $0^1,m^{n-1},n^{m-1},(m+n{)}^1$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (K_{m,n}\vee K_{r,s})=(m+r+s{)}^{n-1}\\ & (n+r+s{)}^{m-1}(r+m+n{)}^{s-1}(s+m+n{)}^{r-1}\\ & (m+n+r+s{)}^2.\end{array}$

3. The eigenvalues of $L\left(\overline{K_m}\right)$ and $L\left(P_n\right)$ are $0^1,0^{m-1}$ and $2-2\cos \frac{i\pi }{n}(0\le i\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau \left(\overline{K_m}\vee P_n\right)=\frac{1}{m+n}(m+n)n^{m-1}\\ & \prod _{i=1}^{n-1}\left(m+2-2\cos \frac{i\pi }{n}\right)\\ & =n^{m-1}{U}_{n-1}\left(\frac{m+2}{2}\right).\end{array}$

The explicit formula follows from Equation (7).

1. The eigenvalues of $L\left(\overline{K_m}\right)$ and $L\left(C_n\right)$ are $0^1,0^{m-1}$ and $2-2\cos \frac{2i\pi }{n}(0\le i\le n-1)$, respectively. Then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau \left(\overline{K_m}\vee C_n\right)=\frac{1}{m+n}(m+n)n^{m-1}\\ & \prod _{i=1}^{n-1}\left(m+2-2\cos \frac{2i\pi }{n}\right)\\ & =n^{m-1}{U}_{n-1}^2\left(\sqrt{\frac{m+4}{4}}\right)=n^{m-1}\\ & \times \frac{2}{m}\left[T_n\left(\frac{m+2}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equation (4).

Theorem 3.4

For $m,n,r$ and $s\ge 1$, we have

1. $\tau (P_m\vee Q_n)=U_{m-1}(2^{n-1}+1)\prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.$

2. $\begin{array}{rl}& \tau (C_m\vee Q_n)=\frac{1}{2^{n-1}}\left[T_m(2^{n-1}+1)-1\right]\\ & \prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

3. $\tau (K_m\vee Q_n)=(2^n+m{)}^{m-1}\prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.$

4. $\begin{array}{rl}& \tau (K_{r,s}\vee Q_n)=(2^n+r{)}^{s-1}(2^n+s{)}^{r-1}\\ & (2^n+r+s)\prod _{i=1}^n{(r+s+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

5. $\begin{array}{rl}& \tau (Q_m\vee Q_n)=\prod _{i=1}^m{(2^n+2i)}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & \prod _{j=1}^n{(2^m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

Proof

1. The eigenvalues of $L\left(P_m\right)$ and $L\left(Q_n\right)$ are $2-2\cos \frac{j\pi }{m},(0\le j\le m-1)$ and $(2i{)}^{\left(\begin{array}{c}n\\ i\end{array}\right)},(0\le i\le n)$, respectively, then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (P_m\vee Q_n)=\prod _{j=1}^{m-1}\left(2^n+2-2\cos \frac{j\pi }{m}\right)\\ & \prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}=U_{m-1}(2^{n-1}+1)\\ & \prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

The explicit formula follows from Equation (7).

1. The eigenvalues of $L\left(C_m\right)$ and $L\left(Q_n\right)$ are $2-2\cos \frac{2j\pi }{m},(0\le j\le m-1)$ and $(2i{)}^{\left(\begin{array}{c}n\\ i\end{array}\right)},(0\le i\le n)$, respectively, then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (C_m\vee Q_n)=\prod _{j=1}^{m-1}\left(2^n+2-2\cos \frac{2j\pi }{m}\right)\\ & \times \prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}=U_{m-1}^2\left(\sqrt{\frac{2^n+4}{4}}\right)\\ & \times \prod _{j=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}=\frac{1}{2^{n-1}}\left[T_m\right(2^{n-1}+1)\\ & -1]\prod _{i=1}^n{(m+2i)}^{\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

The explicit formula follows from Equation (4).

The proofs of (iii–v) follows similarly, using the facts, the eigenvalues of $L\left(K_m\right)$ and $L\left(K_{r,s}\right)$ are $0^1,m^{m-1}$; and $0^1,r^{s-1},s^{r-1},(r+s{)}^1$, respectively.

Theorem 3.5

For $m,r,s\ge 1$ and $n,k\ge 2$, we have

1. $\begin{array}{rl}& \tau (P_m\vee F_n)\\ & =\frac{m+n+1}{2^{m+n}\sqrt{({(n+3)}^2-4)({(m+3)}^2-4)}}\\ & \times \left[{\left(n+3+\sqrt{{(n+3)}^2-4}\right)}^m\right.\\ & \left.-{\left(n+3-\sqrt{{(n+3)}^2-4}\right)}^m\right]\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n\right].\end{array}$

2. $\begin{array}{rl}& \tau \left(\right(C_m\vee F_n)\\ & =\frac{m+n+1}{2^{m+n}(n+1)\sqrt{{(m+3)}^2-4}}\\ & \times \left[{\left(n+3+\sqrt{{(n+3)}^2-4}\right)}^m\right.\\ & \left.+{\left(n+3-\sqrt{{(n+3)}^2-4}\right)}^m\right]-2^{m+1}]\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n\right].\end{array}$

3. $\begin{array}{rl}& \tau (K_m\vee F_n)\\ & =\frac{{(m+n+1)}^m}{2^n\sqrt{{(m+3)}^2-4}}\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.-{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n\right].\end{array}$

4. $\begin{array}{rl}& \tau (K_{r,s}\vee F_n)\\ & =\frac{{(r+n+1)}^{s-1}{(s+n+1)}^{r-1}{(r+s+n+1)}^2}{2^n\sqrt{{(r+s+3)}^2-4}}\\ & \times \left[{\left(r+s+3+\sqrt{{(r+s+3)}^2-4}\right)}^n\right.\\ & \left.-{\left(r+s+3-\sqrt{{(r+s+3)}^2-4}\right)}^n\right].\end{array}$

5. $\begin{array}{rl}& \tau (Q_m\vee F_n)\\ & =\frac{(2^m+n+1)\prod _{i=1}^m{(n+2i+1)}^{\left(\genfrac{}{}{0ex}{}{m}{i}\right)}}{2^n\sqrt{{(2^{m}s+3)}^2-4}}\\ & \times \left[{\left(2^m+3+\sqrt{{(2^m+3)}^2-4}\right)}^n\right.\\ & \left.-{\left(2^m+3-\sqrt{{(2^m+3)}^2-4}\right)}^n\right].\end{array}$

6. $\begin{array}{rl}& \tau (F_k\vee F_n)\\ & =\frac{{(k+n+2)}^2}{2^{k+n}\sqrt{({(n+4)}^2-4)({(k+4)}^2-4)}}\\ & \times \left[{\left(n+4+\sqrt{{(n+4)}^2-4}\right)}^k\right.\\ & \left.-{\left(n+4-\sqrt{{(n+4)}^2-4}\right)}^k\right]\\ & \times \left[{\left(k+4+\sqrt{{(k+4)}^2-4}\right)}^n\right.\\ & \left.-{\left(k+4-\sqrt{{(k+4)}^2-4}\right)}^n\right].\end{array}$

Proof

1. The eigenvalues of $L\left(P_m\right)$ and $L\left(F_n\right)$ are $2-2\cos \frac{i\pi }{m},(0\le i\le m-1)$ and $0,n+1,3-2\cos \frac{j\pi }{n},(1\le j\le n-1)$, then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (P_m\vee F_n)=(m+n+1)\prod _{i=1}^{m-1}\left(n+3-2\cos \frac{i\pi }{m}\right)\\ & \prod _{j=1}^{n-1}\left(m+3-2\cos \frac{i\pi }{n}\right)\\ & =(m+n+1)U_{m-1}\left(\frac{n+3}{2}\right)U_{n-1}\left(\frac{m+3}{2}\right).\end{array}$

The proofs of (ii–vi) follows similarly, using the facts, the eigenvalues of $L\left(C_n\right),L\left(K_m\right),L\left(K_{r,s}\right)$ and $L\left(Q_k\right)$ are $2-2\cos \frac{2i\pi }{n},(0\le i\le n-1)$; $0^1,m^{m-1}$; $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $(2i{)}^{\left(\begin{array}{c}k\\ i\end{array}\right)},(0\le i\le k)$, respectively, as follows:

1. $\begin{array}{rl}& \tau (C_m\vee F_n)=(m+n+1)\\ & \times \prod _{i=1}^{m-1}\left(n+1+2-2\cos \frac{2i\pi }{m}\right)\\ & \times \prod _{j=1}^{n-1}\left(m+3-2\cos \frac{i\pi }{n}\right)\\ & =(m+n+1)U_{m-1}^2\left(\sqrt{\frac{n+5}{4}}\right)U_{n-1}\left(\frac{m+3}{2}\right)\\ & =\frac{2(m+n+1)}{n+1}\left[T_m\left(\frac{n+3}{2}\right)-1\right]\\ & U_{n-1}\left(\frac{m+3}{2}\right).\end{array}$

2. $\begin{array}{rl}& \tau (K_m\vee F_n)=(m+n+1{)}^m\\ & \times \prod _{j=1}^{n-1}\left(m+3-2\cos \frac{j\pi }{n}\right)=(m+n+1{)}^m\\ & U_{n-1}\left(\frac{m+3}{2}\right).\end{array}$

3. $\begin{array}{rl}& \tau (K_{r,s}\vee F_n)=(r+n+1{)}^{s-1}(s+n+1{)}^{r-1}\\ & (r+s+n+1{)}^2\prod _{j=1}^{n-1}\left(r+s+3-2\cos \frac{i\pi }{n}\right)\\ & =(r+n+1{)}^{s-1}(s+n+1{)}^{r-1}(r+s+n+1{)}^2\\ & U_{n-1}\left(\frac{r+s+3}{2}\right).\end{array}$

4. $\begin{array}{rl}& \tau (Q_m\vee F_n)=\prod _{i=1}^m{(n+1+2i)}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & \times (2^m+n+1)\prod _{j=1}^{n-1}\left(2^m+3-2\cos \frac{j\pi }{n}\right)\\ & =\prod _{i=1}^m{(n+1+2i)}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\times (2^m+n+1)\\ & U_{n-1}\left(\frac{2^m+3}{2}\right).\end{array}$

5. $\begin{array}{rl}& \tau (F_k\vee F_n)=(k+1+n+1{)}^2\\ & \times \prod _{i=1}^{k-1}\left(n+4-2\cos \frac{i\pi }{k}\right)\\ & \times \prod _{j=1}^{n-1}\left(n+4-2\cos \frac{j\pi }{n}\right)\\ & =(k+n+2{)}^2\times U_{k-1}\left(\frac{n+4}{2}\right)\\ & \times U_{n-1}\left(\frac{k+4}{2}\right).\end{array}$

The explicit formula follows from Equations (4) and (7).

Theorem 3.6

For $m,r,s\ge 1$ and $n,k\ge 2$, we have:

1. $\begin{array}{rl}& \tau \left(P_m\vee W_n\right)=\frac{m+n+1}{2^{m+n}(m+1)\sqrt{{(n+3)}^2-4}}\\ & \times \left[{\left(n+3+\sqrt{{(n+3)}^2-4}\right)}^m\right.\\ & \left.-{\left(n+3-\sqrt{{(n+3)}^2-4}\right)}^m\right]\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

2. $\begin{array}{rl}& \tau \left(C_m\vee W_n\right)=\frac{m+n+1}{2^{m+n}\times (m+1)(n+1)}\\ & \times \left[{\left(n+3+\sqrt{{(n+3)}^2-4}\right)}^m\right.\\ & \left.+{\left(n+3-\sqrt{{(n+3)}^2-4}\right)}^m-2^{m+1}\right]\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

3. $\begin{array}{rl}& \tau \left(K_m\vee W_n\right)=\frac{{(m+n+1)}^m}{2^n\times (m+1)}\\ & \times \left[{\left(m+3+\sqrt{{(m+3)}^2-4}\right)}^n\right.\\ & \left.+{\left(m+3-\sqrt{{(m+3)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

4. $\begin{array}{rl}& \tau \left(K_{r,s}\vee W_n\right)\\ & =\frac{{(r+n+1)}^{s-1}{(s+n+1)}^{r-1}{(r+s+n+1)}^2}{2^n\times (r+s+1)}\\ & \times \left[{\left(r+s+3+\sqrt{{(r+s+3)}^2-4}\right)}^n\right.\\ & \left.+{\left(r+s+3-\sqrt{{(r+s+3)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

5. $\begin{array}{rl}& \tau \left(Q_m\vee W_n\right)=\frac{(2^m+n+1)\prod _{i=1}^m{(n+2i+1)}^{\left(\genfrac{}{}{0ex}{}{m}{i}\right)}}{2^n\times (2^m+1)}\\ & \times \left[{\left(2^m+1+\sqrt{{(2^m+1)}^2-4}\right)}^n\right.\\ & \left.+{\left(2^m+1-\sqrt{{(2^m+1)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

6. $\begin{array}{rl}& \tau \left(W_k\vee W_n\right)=\frac{1}{2^{k+n}\times (k+2)(n+2)}\\ & \times \left[{\left(n+4+\sqrt{{(n+4)}^2-4}\right)}^k\right.\\ & \left.+{\left(n+4-\sqrt{{(n+4)}^2-4}\right)}^k-2^{k+1}\right]\\ & \times \left[{\left(k+4+\sqrt{{(k+4)}^2-4}\right)}^n\right.\\ & \left.+{\left(k+4-\sqrt{{(k+4)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

7. $\begin{array}{rl}& \tau \left(F_k\vee W_n\right)=\frac{1}{2^n\times (k+2)\sqrt{{(n+4)}^2-4}}\\ & \times \left[{\left(n+4+\sqrt{{(n+4)}^2-4}\right)}^k\right.\\ & \left.-{\left(n+4-\sqrt{{(n+4)}^2-4}\right)}^k\right]\\ & \times \left[{\left(k+4+\sqrt{{(k+4)}^2-4}\right)}^n\right.\\ & \left.+{\left(k+4-\sqrt{{(k+4)}^2-4}\right)}^n-2^{n+1}\right].\end{array}$

Proof

1. The eigenvalues of $L\left(P_m\right)$ and $L\left(W_n\right)$ are $2-2\cos \frac{i\pi }{m},(0\le i\le m-1)$ and $0,n+1,3-2\cos \frac{2j\pi }{n},(1\le j\le n-1)$, then applying Equation (1) together with Theorem 3.2, we get $\begin{array}{rl}& \tau (P_m\vee W_n)=(m+n+1)\\ & \prod _{i=1}^{m-1}\left(n+1+2-2\cos \frac{i\pi }{m}\right)\\ & \prod _{j=1}^{n-1}\left(m+3-2\cos \frac{2j\pi }{n}\right)\\ & =(m+n+1)U_{m-1}\left(\frac{n+3}{2}\right)U_{n-1}^2\left(\sqrt{\frac{m+5}{4}}\right)\\ & =\frac{2(m+n+1)}{m+1}{U}_{m-1}\left(\frac{n+3}{2}\right)\\ & \left[T_n\left(\frac{m+3}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equations (4) and (7).

The proofs of (ii–vii) follows similarly, using the facts, the eigenvalues of $L\left(C_n\right),L\left(K_m\right),L\left(K_{r,s}\right)$ and $L\left(Q_k\right)$ are $2-2\cos \frac{2i\pi }{n},(0\le i\le n-1)$; $0^1,m^{m-1}$; $0^1,r^{s-1},s^{r-1},(r+s{)}^1$ and $(2i{)}^{\left(\begin{array}{c}k\\ i\end{array}\right)},(0\le i\le k)$, respectively, as follows:

1. $\begin{array}{rl}& \tau (C_m\vee W_n)=(m+n+1)\\ & \prod _{i=1}^{m-1}\left(n+3-2\cos \frac{2i\pi }{m}\right)\\ & \prod _{j=1}^{n-1}\left(m+3-2\cos \frac{2j\pi }{n}\right)\\ & =(m+n+1)U_{m-1}^2\left(\sqrt{\frac{n+5}{4}}\right)U_{n-1}^2\left(\sqrt{\frac{m+5}{4}}\right)\\ & =\frac{4(m+n+1)}{(m+1)(n+1)}\left[T_m\left(\frac{n+3}{2}\right)-1\right]\\ & \left[T_n\left(\frac{m+3}{2}\right)-1\right].\end{array}$

2. $\begin{array}{rl}& \tau (K_m\vee W_n)=(m+n+1{)}^{m-1}(m+n+1)\\ & \prod _{i=1}^{n-1}\left(m+3-2\cos \frac{2i\pi }{n}\right)\\ & =(m+n+1{)}^m{U}_{n-1}^2\left(\sqrt{\frac{m+5}{4}}\right)\\ & =\frac{4{(m+n+1)}^m}{(m+1)}\left[T_n\left(\frac{m+3}{2}\right)-1\right].\end{array}$

3. $\begin{array}{rl}& \tau (K_{r,s}\vee W_n)=(n+r+1{)}^{s-1}(n+s+1{)}^{r-1}\\ & (r+s+n+1{)}^2\prod _{i=1}^{n-1}\left(r+s+3-2\cos \frac{2i\pi }{n}\right)\\ & =(n+r+1{)}^{s-1}(n+s+1{)}^{r-1}(r+s+n+1{)}^2\\ & U_{n-1}^2\left(\sqrt{\frac{r+s+5}{4}}\right)\\ & =\frac{2{(n+r+1)}^{s-1}{(n+s+1)}^{r-1}{(r+s+n+1)}^2}{(r+s+1)}\\ & \left[T_n\left(\frac{r+s+3}{2}\right)-1\right].\end{array}$

4. $\begin{array}{rl}& \tau (Q_m\vee W_n)=\prod _{i=1}^m{(n+2i+1)}^{\left(\genfrac{}{}{0ex}{}{m}{i}\right)}\\ & \times (2^m+n+1)\\ & \prod _{j=1}^{n-1}\left(2^m+3-2\cos \frac{2j\pi }{n}\right)\\ & =\prod _{i=1}^m{(n+2i+1)}^{\left(\genfrac{}{}{0ex}{}{m}{i}\right)}\times (2^m+n+1)\\ & U_{n-1}^2\left(\sqrt{\frac{2^m+5}{4}}\right)\\ & =\frac{2}{2^m+1}\prod _{i=1}^m{(n+2i+1)}^{\left(\genfrac{}{}{0ex}{}{m}{i}\right)}\times (2^m+n+1)\\ & \left[T_n\left(\frac{2^m+3}{2}\right)-1\right].\end{array}$

5. $\begin{array}{rl}& \tau (W_m\vee W_n)=\prod _{i=1}^{m-1}\left(n+4-2\cos \frac{2i\pi }{m}\right)\\ & \prod _{j=1}^{n-1}\left(m+4-2\cos \frac{2j\pi }{n}\right)\\ & =U_{m-1}^2\left(\sqrt{\frac{n+6}{4}}\right)U_{n-1}^2\left(\sqrt{\frac{m+6}{4}}\right)\\ & =\frac{4}{(m+2)(n+2)}\left[T_m\left(\frac{n+4}{2}\right)-1\right]\\ & \left[T_n\left(\frac{m+4}{2}\right)-1\right].\end{array}$

6. $\begin{array}{rl}& \tau (F_k\vee W_n)=\prod _{i=1}^{m-1}\left(n+4-2\cos \frac{i\pi }{m}\right)\\ & \prod _{j=1}^{n-1}\left(m+4-2\cos \frac{2j\pi }{n}\right)\\ & =U_{m-1}\left(\frac{n+4}{2}\right)U_{n-1}^2\left(\sqrt{\frac{m+6}{4}}\right)\\ & \frac{2}{(m+2)}{U}_{m-1}\left(\frac{n+4}{2}\right)\\ & \left[T_n\left(\frac{m+4}{2}\right)-1\right].\end{array}$

The explicit formula follows from Equations(4) and (7).

4. Complexity of corona of two graphs

The corona $G_1\odot G_2$ of two graphs $G_1$ and $G_2$ is the graph obtained by taking one copy of $G_1$ (which has $n_1$ vertices) and $n_1$ copies of $G_2$ and then joining the i-th vertex of $G_1$ to every vertex in the copy of $G_2$ [21]. It is easy to see that $G_1\odot G_2$ is not isomorphic $G_2\odot G_1$.

Theorem 4.1

[22] Let $G_1$ and $G_2$ be graphs of order $n_1$ and $n_2$ respectively. If Let $G_1$ and $G_2$ be graphs of order $n_1$ and $n_2$ respectively. If $S\left(G_1\right)=({\lambda }_1,\dots ,{\lambda }_{n_{{}_1}})$ and $S\left(G_2\right)=({\sigma }_1,\dots ,{\sigma }_{n_{{}_2}})$ are the eigenvalues of $L\left(G_1\right)$ and $L\left(G_2\right)$ arranged in non-increasing order, then the eigenvalues of $L(G_1\odot G_2)$ is given by the eigenvalues${\sigma }_j+1$ with multiplicity $n_1$ for $j=2,\dots ,n_2$ and $\frac{({\lambda }_i+n_2+1)\pm \sqrt{{({\lambda }_i+n_2+1)}^2-4{\lambda }_i}}{2}$ for $i=1,2,\dots ,n_1.$

Now we can introduce the following theorem:

Theorem 4.2

Let $G_1$ and $G_2$ be graphs of order $n_1$ and $n_2$ respectively. If $S\left(G_1\right)=({\lambda }_1,\dots ,{\lambda }_{n_{{}_1}})$ and $S\left(G_2\right)=({\sigma }_1,\dots ,{\sigma }_{n_{{}_2}})$ are the eigenvalues of $L\left(G_1\right)$ and $L\left(G_2\right)$, then $\tau (G_1\odot G_2)=\tau \left(G_1\right)\prod _{j=1}^{n_2-1}{({\sigma }_j+1)}^{n_1}.$

Theorem 4.3

For $m,n,r\text{and}s\ge 1$, we have

1. $\begin{array}{rl}& \tau (P_m\odot P_n)\\ & ={\left[\frac{1}{2^n\sqrt{5}}\left[{\left(3+\sqrt{5}\right)}^n-{\left(3-\sqrt{5}\right)}^n\right]\right]}^m.\end{array}$

2. $\begin{array}{rl}& \tau (P_m\odot C_n)\\ & =\frac{1}{2^{mn}}{\left[{\left(3+\sqrt{5}\right)}^n+{\left(3-\sqrt{5}\right)}^n-2^{n+1}\right]}^m.\end{array}$

3. $\begin{array}{rl}& \tau (C_m\odot P_n)\\ & =m{\left[\frac{1}{2^n\sqrt{5}}\left[{\left(3+\sqrt{5}\right)}^n-{\left(3-\sqrt{5}\right)}^n\right]\right]}^m.\end{array}$

4. $\begin{array}{rl}& \tau (C_m\odot C_n)\\ & =\frac{m}{2^{mn}}{\left[{\left(3+\sqrt{5}\right)}^n+{\left(3-\sqrt{5}\right)}^n-2^{n+1}\right]}^m.\end{array}$

5. $\begin{array}{rl}& \tau (K_m\odot P_n)\\ & =m^{m-2}{\left[\frac{1}{2^n\sqrt{5}}\left[{\left(3+\sqrt{5}\right)}^n-{\left(3-\sqrt{5}\right)}^n\right]\right]}^m.\end{array}$

6. $\tau (P_m\odot K_n)=(n+1{)}^{m(n-1)}.$

7. $\begin{array}{rl}& \tau (K_m\odot C_n)\\ & =\frac{m^{m-2}}{2^{mn}}{\left[{\left(3+\sqrt{5}\right)}^n+{\left(3-\sqrt{5}\right)}^n-2^{n+1}\right]}^m.\end{array}$

8. $\tau (C_m\odot K_n)=m(n+1{)}^{m(n-1)}.$

9. $\begin{array}{rl}& \tau (K_{r,s}\odot P_m)\\ & =r^{s-1}{s}^{r-1}\left[\frac{1}{2^m\sqrt{5}}\left[{(3+\sqrt{5})}^n\right.\right.\\ & {\left.\left.-{(3-\sqrt{5})}^n\right]\right]}^{r+s}.\end{array}$

10. $\begin{array}{rl}& \tau (P_m\odot K_{r,s})\\ & =\left[\right(r+s+1)(r+1{)}^{s-1}(s+1{)}^{r-1}{]}^m.\end{array}$

11. $\begin{array}{rl}& \tau (K_{r,s}\odot C_m)\\ & =r^{s-1}{s}^{r-1}\left[\frac{1}{2^m}\left[{(3+\sqrt{5})}^m\right.\right.\\ & {\left.\left.{(3-\sqrt{5})}^m-2^{m+1}\right]\right]}^{r+s}.\end{array}$

12. $\begin{array}{rl}& \tau (C_m\odot K_{r,s})\\ & =m\left[\right(r+s+1)(r+1{)}^{s-1}(s+1{)}^{r-1}{]}^m.\end{array}$

13. $\begin{array}{rl}& \tau (K_{r,s}\odot K_m)\\ & =r^{s-1}{s}^{r-1}(m+1{)}^{(m-1)(r+s)}.\end{array}$

14. $\begin{array}{rl}& \tau (K_m\odot K_{r,s})\\ & =m^{m-2}\left[\right(r+s+1)(r+1{)}^{s-1}(s+1{)}^{r-1}{]}^m.\end{array}$

15. $\begin{array}{rl}& \tau (K_{m,n}\odot K_{r,s})\\ & =m^{n-1}{n}^{m-1}[(r+1{)}^{s-1}\\ & (s+1{)}^{r-1}(r+s+1){]}^{m+n}.\end{array}$

Proof

Applying Theorem 4.2, we get

1. $\begin{array}{rl}& \tau (P_m\odot P_n)=\prod _{i=1}^{n-1}{\left(3-2\cos \frac{i\pi }{n}\right)}^m\\ & ={\left[U_{n-1}\left(\frac{3}{2}\right)\right]}^m=\tau (P_m\odot P_n)={\left[U_{n-1}\left(\frac{3}{2}\right)\right]}^m\\ & ={\text{F}}_{2n}^m.\end{array}$

2. $\begin{array}{rl}& \tau (P_m\odot C_n)=\prod _{i=1}^{n-1}{\left(3-2\cos \frac{2i\pi }{n}\right)}^m\\ & ={\left[U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^m={\left(2\left[T_n\left(\frac{3}{2}\right)-1\right]\right)}^m\\ & =({\text{L}}_{2n}-2{)}^m.\end{array}$

3. $\begin{array}{rl}& \tau (C_m\odot P_n)=m\prod _{i=1}^{n-1}{\left(3-2\cos \frac{i\pi }{n}\right)}^m\\ & =m{\left[U_{n-1}\left(\frac{3}{2}\right)\right]}^m=m{\text{F}}_{2n}^m.\end{array}$

4. $\begin{array}{rl}& \tau (C_m\odot C_n)=m\prod _{i=1}^{n-1}{\left(3-2\cos \frac{2i\pi }{n}\right)}^m\\ & =m{\left[U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^m=m{\left(2\left[T_n\left(\frac{3}{2}\right)-1\right]\right)}^m\\ & =m(L_{2n}-2{)}^m.\end{array}$

5. $\begin{array}{rl}& \tau (K_m\odot P_n)=m^{m-2}\prod _{i=1}^{n-1}{\left(3-2\cos \frac{i\pi }{n}\right)}^m\\ & =m^{m-2}{\left[U_{n-1}\left(\frac{3}{2}\right)\right]}^m=m^{m-2}{\text{F}}_{2n}^m.\end{array}$

6. $\begin{array}{rl}& \tau (P_m\odot K_n)\\ & =(n+1{)}^{m(n-1)}.\end{array}$

7. $\begin{array}{rl}& \tau (K_m\odot C_n)=m^{m-2}\prod _{i=1}^{n-1}{\left(3-2\cos \frac{2i\pi }{n}\right)}^m\\ & =m^{m-2}{\left[U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^m\\ & =m^{m-2}{\left(2\left[T_n\left(\frac{3}{2}\right)-1\right]\right)}^m=m^{m-2}(L_{2n}-2{)}^m.\end{array}$

8. $\begin{array}{r}\tau (C_m\odot K_n)=m(n+1{)}^{m(n-1)}.\end{array}$

9. $\begin{array}{rl}& \tau (K_{r,s}\odot P_m)=r^{s-1}{s}^{r-1}\prod _{i=1}^{m-1}{\left(3-2\cos \frac{i\pi }{m}\right)}^{r+s}\\ & =r^{s-1}{s}^{r-1}{\left[U_{m-1}\left(\frac{3}{2}\right)\right]}^{r+s}=r^{s-1}{s}^{r-1}{\text{F}}_{2m}^{r+s}.\end{array}$

10. $\begin{array}{rl}& \tau (P_m\odot K_{r,s})\\ & =\left[\right(r+s+1)(r+1{)}^{s-1}(s+1{)}^{r-1}{]}^m.\end{array}$

11. $\begin{array}{rl}& \tau (K_{r,s}\odot C_m)=r^{s-1}{s}^{r-1}\prod _{i=1}^{m-1}{\left(3-2\cos \frac{2i\pi }{m}\right)}^{r+s}\\ & =r^{s-1}{s}^{r-1}{\left[U_{m-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^{r+s}\\ & =r^{s-1}{s}^{r-1}{\left(2\left[T_m\left(\frac{3}{2}\right)-1\right]\right)}^{r+s}\\ & =r^{s-1}{s}^{r-1}(L_{2m}-2{)}^{r+s}.\end{array}$

The explicit formula follows from Equations (17) and (19).

Theorem 4.4

For $m,n,r\text{and}s\ge 1$, we have

1. $\begin{array}{rl}& \tau (Q_m\odot P_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & {\left[\frac{1}{2^n\sqrt{5}}\left[{(3+\sqrt{5})}^n-{(3-\sqrt{5})}^n\right]\right]}^m.\end{array}$

2. $\begin{array}{rl}& \tau (Q_m\odot C_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & {\left[\frac{1}{2^n}\left[{(3+\sqrt{5})}^n+{(3-\sqrt{5})}^n-2^{n+1}\right]\right]}^m.\end{array}$

3. $\begin{array}{r}\tau (P_m\odot Q_n)=\prod _{i=1}^n{(2i+1)}^{m\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

4. $\begin{array}{r}\tau (C_m\odot Q_n)=m\prod _{i=1}^n{(2i+1)}^{m\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

5. $\begin{array}{rl}& \tau (Q_m\odot Q_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}n\\ i\end{array}\right)}\prod _{j=1}^n{(2j+1)}^{m\left(\begin{array}{c}n\\ j\end{array}\right)}.\end{array}$

6. $\begin{array}{r}\tau (K_m\odot Q_n)=m^{m-2}\prod _{i=1}^n{(2i+1)}^{m\left(\begin{array}{c}n\\ i\end{array}\right)}.\end{array}$

7. $\begin{array}{r}\tau (Q_m\odot K_n)=2^{2^m-m-1}(n+1{)}^{m(n-1)}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}.\end{array}$

8. $\begin{array}{r}\tau (K_{r,s}\odot Q_m)=r^{s-1}{s}^{r-1}\prod _{i=1}^m{(2i+1)}^{(r+s)\left(\begin{array}{c}m\\ i\end{array}\right)}.\end{array}$

9. $\begin{array}{rl}& \tau (Q_m\odot K_{r,s})=2^{2^m-m-1}(r+1{)}^{m(s-1)}\\ & (s+1{)}^{m(r-1)}(r+s+1{)}^m\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}.\end{array}$

Proof

Applying Theorem 4.2, we get:

1. $\begin{array}{rl}\tau (Q_m\odot P_n)& =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\prod _{j=1}^{n-1}\\ & \times {\left(3-2\cos \frac{j\pi }{n}\right)}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}[{\left[U_{n-1}\left(\frac{3}{2}\right)\right]}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}{\text{F}}_{2n}^m.\end{array}$

2. $\begin{array}{rl}\tau (Q_m\odot C_n)& =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\prod _{j=1}^{n-1}\\ & \times {\left(3-2\cos \frac{2j\pi }{n}\right)}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}{\left[U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & \times {\left(2\left[T_n\left(\frac{3}{2}\right)-1\right]\right)}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}({\text{L}}_{2n}-2{)}^m.\end{array}$

The explicit formula follows from Equations (17) and (19).

Theorem 4.5

For $m,r\text{,}s\ge 1$ and $k,n\ge 2$, we have

1. $\tau (P_m\odot F_n)={\left[\frac{n+2}{2\sqrt{3}}\left[{\left(2+\sqrt{3}\right)}^n-{(2-\sqrt{3})}^n\right]\right]}^m.$

2. $\tau (F_n\odot P_m)=\frac{1}{2^n}\left[{(3+\sqrt{5})}^n+{(3-\sqrt{5})}^n-\right.\left.2^{n+1}\phantom{{(3+\sqrt{5})}^n}\right]{\left[\frac{1}{2^m}\left[{(3+\sqrt{5})}^m+{(3-\sqrt{5})}^m-2^{m+1}\right]\right]}^n.$

3. $\tau (C_m\odot F_n)=m\left[\frac{n+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^n-\right.\right.{\left.\left.{(2-\sqrt{3})}^n\right]\right]}^m.$

4. $\tau (F_n\odot C_m)=\frac{1}{2^n\sqrt{5}}\left[{(3+\sqrt{5})}^n-{(3-\sqrt{5})}^n\right]{\left[\frac{1}{2^m}\left[{(3+\sqrt{5})}^m+{(3-\sqrt{5})}^m-2^{m+1}\right]\right]}^n.$

5. $\tau (K_m\odot F_n)=m^{m-2}\left[\frac{n+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^n-\right.\right.{\left.\left.{(2-\sqrt{3})}^n\right]\right]}^m.$

6. $\tau (F_n\odot K_m)=\frac{{(m+1)}^{n(m-1)}}{2^n\sqrt{5}}\left[{(3+\sqrt{5})}^n-{(3-\sqrt{5})}^n\right].$

7. $\tau (K_{r,s}\odot F_n)=r^{s-1}{s}^{r-1}{\left[\frac{n+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^n-{(2-\sqrt{3})}^n\right]\right]}^{r+s}.$

8. $\tau (F_n\odot K_{r,s})=\frac{{(r+1)}^{(s-1)}{(s+1)}^{(r-1)}{(r+s+1)}^n}{2\sqrt{5}}\left[{(3+\sqrt{5})}^n-{(3-\sqrt{5})}^n\right].$

9. $\tau (Q_m\odot F_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}{\left[\frac{n+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^n-{(2-\sqrt{3})}^n)\right]\right]}^m.$

10. $\tau (F_n\odot Q_m)=\prod _{i=1}^m{(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}\times \frac{1}{2^n\sqrt{5}}\left[{(3+\sqrt{5})}^n-{(3-\sqrt{5})}^n\right].$

11. $\tau (F_k\odot F_n)=\frac{1}{2^k\sqrt{5}}\left[{(3+\sqrt{5})}^k-{(3-\sqrt{5})}^k\right]\times {\left[\frac{n+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^n-{(2-\sqrt{3})}^n\right]\right]}^k.$

Proof

Applying Theorem 4.2, we get:

1. $\tau (P_m\odot F_n)=\prod _{i=1}^{n-1}{\left(4-2\cos \frac{i\pi }{n}\right)}^m=(n+2{)}^m\left[U_{n-1}\right(2){]}^m.$

2. $\tau (F_n\odot P_m)=\prod _{i=1}^{n-1}\left(3-2\cos \frac{i\pi }{n}\right)\prod _{j=1}^{m-1}{\left(3-2\cos \frac{j\pi }{m}\right)}^n=U_{n-1}\left(\frac{3}{2}\right){\left[U_{m-1}\left(\frac{3}{2}\right)\right]}^n={\text{F}}_{2n}{\text{F}}_{2m}^n.$

3. $\tau (C_m\odot F_n)=m(n+2{)}^m\prod _{i=1}^{n-1}{\left(4-2\cos \frac{i\pi }{n}\right)}^m=m(n+2{)}^m\left[U_{n-1}\right(2){]}^m.$

4. $\begin{array}{rl}\tau (F_n\odot C_m)& =\prod _{i=1}^{n-1}\left(3-2\cos \frac{i\pi }{n}\right)\prod _{i=1}^{m-1}\\ & \times {\left(3-2\cos \frac{2j\pi }{m}\right)}^n\\ & =U_{n-1}\left(\frac{3}{2}\right){\left[U_{m-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^n\\ & =U_{n-1}\left(\frac{3}{2}\right){\left[2\left(T_n\left(\frac{m+3}{2}\right)-1\right)!\right]}^n\\ & =F_{2n}({\text{L}}_{2m}-2{)}^n.\end{array}$

5. $\tau (K_m\odot F_n)=m^{m-2}(n+2{)}^m\prod _{i=1}^{n-1}{\left(4-2\cos \frac{i\pi }{n}\right)}^m=m^{m-2}(n+2{)}^m\left[U_{n-1}\right(2){]}^m.$

6. $\tau (F_n\odot K_m)=(m+1{)}^{n(m-1)}\prod _{i=1}^{n-1}\left(3-2\cos \frac{i\pi }{n}\right)=(m+1{)}^{n(m-1)}{U}_{n-1}\left(\frac{3}{2}\right)=(m+1{)}^{n(m-1)}{\text{F}}_{2n}$

7. $\tau (K_{r,s}\odot F_n)=r^{s-1}{s}^{r-1}(n+2{)}^{r+s}\prod _{i=1}^{n-1}{\left(4-2\cos \frac{i\pi }{n}\right)}^{r+s}=r^{s-1}{s}^{r-1}(n+2{)}^{r+s}[U_{n-1}\left(2\right){]}^{r+s}.$

8. $\begin{array}{rl}\tau (F_n\odot K_{r,s})& =(r+1{)}^{(s-1)n}(s+1{)}^{(r-1)n}\\ & \times (r+s+1{)}^n\prod _{i=1}^{n-1}\left(3-2\cos \frac{i\pi }{n}\right)\\ & =(r+1{)}^{(s-1)n}(s+1{)}^{(r-1)n}\\ & \times (r+s+1{)}^n{U}_{n-1}\left(\frac{3}{2}\right)\\ & =(r+1{)}^{(s-1)n}(s+1{)}^{(r-1)n}\\ & \times (r+s+1{)}^n{\text{F}}_{2n}.\end{array}$

9. $\tau (Q_m\odot F_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}\times (n+2{)}^m\prod _{j=1}^{n-1}\left(3-2\cos \frac{j\pi }{n}\right)=2^{2^m-m-1}(n+2{)}^m\left[U_{n-1}\right(2){]}^m\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}.$

10. $\tau (F_n\odot Q_m)=\prod _{i=1}^{n-1}\left(3-2\cos \frac{i\pi }{n}\right)\prod _{j=1}^m{(2j+1)}^{n\left(\begin{array}{c}m\\ j\end{array}\right)}=U_{n-1}\left(\frac{3}{2}\right)\prod _{i=1}^m{(2j+1)}^{n\left(\begin{array}{c}m\\ j\end{array}\right)}={\text{F}}_{2n}\prod _{i=1}^m{(2j+1)}^{n\left(\begin{array}{c}m\\ j\end{array}\right)}.$

11. $\begin{array}{rl}\tau (F_k\odot F_n)& =\prod _{i=1}^{k-1}\left(3-2\cos \frac{i\pi }{k}\right)\times (n+2{)}^k\\ & \times \prod _{j=1}^{n-1}{\left(4-2\cos \frac{j\pi }{n}\right)}^k\\ & =(n+2{)}^k{U}_{k-1}\left(\frac{3}{2}\right)\left[U_{n-1}\right(2){]}^k\\ & ={\text{F}}_{2k}(n+2{)}^k\left[U_{n-1}\right(2){]}^k.\end{array}$

Theorem 4.6

For $m,r\text{,}s\ge 1$ and $n,p,q\ge 2$, we have

1. $\tau (P_m\odot W_n)={\left[\frac{n+2}{2}\left[{(2+\sqrt{3})}^n\text{+ (}2-\sqrt{3}{)}^n-2\right]\right]}^m.$

2. $\tau (W_n\odot P_m)=\frac{1}{2^n}\left[{(3+\sqrt{5})}^n\text{+ (}3-\sqrt{5}{)}^n-2^{n+1}\right]{\left[\frac{1}{2^m\sqrt{5}}\left[{(3+\sqrt{5})}^m-(3-\sqrt{5}{)}^m\right]\right]}^n$

3. $\tau (C_m\odot W_n)=m{\left[\frac{n+2}{2}\left[{(2+\sqrt{3})}^n\text{+ (}2-\sqrt{3}{)}^n-2\right]\right]}^m.$

4. $\tau (W_n\odot C_m)=\frac{1}{2^n}\left[{(3+\sqrt{5})}^n\text{+ (}3-\sqrt{5}{)}^n-2^{n+1}\right]{\left[\frac{1}{2^m}\left[{(3+\sqrt{5})}^m\text{+ (}3-\sqrt{5}{)}^m-2^{m+1}\right]\right]}^n.$

5. $\tau (K_m\odot W_n)=m^{m-2}{\left[\frac{n+2}{2}\left[{(2+\sqrt{3})}^n\text{+ (}2-\sqrt{3}{)}^n-2\right]\right]}^m.$

6. $\tau (W_n\odot K_m)=\frac{{(n+1)}^{m(n-1)}}{2^n}\left[{(3+\sqrt{5})}^n\text{+ (}3-\sqrt{5}{)}^n-2^{n+1}\right].$

7. $\tau (K_{r,s}\odot W_n)=r^{s-1}{s}^{r-1}{\left[\frac{n+2}{2}\left[{(2+\sqrt{3})}^n\text{+ (}2-\sqrt{3}{)}^n-2\right]\right]}^{r+s}.$

8. $\tau (W_n\odot K_{r,s})=\frac{{(r+1)}^{n(s-1)}{(s+1)}^{n(r-1)}{(r+s+1)}^n}{2^n}\left[{(3+\sqrt{5})}^n\text{+ (}3-\sqrt{5}{)}^n-2^{n+1}\right].$

9. $\tau (Q_m\odot W_n)=2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}{\left[\frac{n+2}{2}\left[{(2+\sqrt{3})}^n\text{+ (}2-\sqrt{3}{)}^n-2\right]\right]}^m.$

10. $\tau (W_n\odot Q_m)=\prod _{i=1}^m{(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}\times \frac{1}{2^n}\left[{(3+\sqrt{5})}^n\text{+ (}3-\sqrt{5}{)}^n-2^{n+1}\right].$

11. $\tau (W_p\odot W_q)=\frac{1}{2^p}\left[{(3+\sqrt{5})}^p\text{+ (}3-\sqrt{5}{)}^p-2^{p+1}\right]\times {\left[\frac{q+2}{2}\left[{(2+\sqrt{3})}^q\text{+ (}2-\sqrt{3}{)}^q-2\right]\right]}^p.$

12. $\tau (F_p\odot W_q)=\frac{1}{2^n\sqrt{5}}\left[{(3+\sqrt{5})}^p-(3-\sqrt{5}{)}^p\right]{\left[\frac{q+2}{2}\left[{(2+\sqrt{3})}^q\text{+ (}2-\sqrt{3}{)}^q-2\right]\right]}^p.$

13. $\tau (W_p\odot F_q)=\frac{1}{2^p}\left[{(3+\sqrt{5})}^p\text{+ (}3-\sqrt{5}{)}^p-2^{p+1}\right]{\left[\frac{q+2}{2\sqrt{3}}\left[{(2+\sqrt{3})}^q-(2-\sqrt{3}{)}^q\right]\right]}^p.$

Proof

Applying Theorem 4.2, we get:

1. $\tau (P_m\odot W_n)=(n+2{)}^m\prod _{i=1}^{n-1}{\left(4-2\cos \frac{2i\pi }{n}\right)}^m=(n+2{)}^m{\left[U_{n-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^m=(n+2{)}^m[T_n\left(2\right)-1{]}^m.$

2. $\begin{array}{rl}\tau (W_n\odot P_m)& =\prod _{i=1}^{n-1}\left(3-2\cos \frac{2i\pi }{n}\right)\prod _{j=1}^{m-1}\\ & \times {\left(3-2\cos \frac{j\pi }{m}\right)}^n\\ & =U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right){\left[U_{m-1}\left(\frac{3}{2}\right)\right]}^n\\ & =2\left[T_n\left(\frac{3}{2}\right)-1\right]{\left[U_{m-1}\left(\frac{3}{2}\right)\right]}^n\\ & =(L_{2n}-2)F_{2m}^n.\end{array}$

3. $\tau (C_m\odot W_n)=m(n+2{)}^m\prod _{i=1}^{m-1}{\left(4-2\cos \frac{2i\pi }{n}\right)}^m=m(n+2{)}^m{\left[U_{n-1}^2\left(\frac{3}{2}\right)\right]}^m=m(n+2{)}^m\left[T_n\right(2)-1{]}^m.$

4. $\begin{array}{rl}\tau (W_n\odot C_m)& =\prod _{i=1}^{n-1}\left(3-2\cos \frac{2i\pi }{n}\right)\prod _{j=1}^{m-1}\\ & \times {\left(3-2\cos \frac{2j\pi }{m}\right)}^n\\ & =U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right){\left[U_{m-1}^2\left(\sqrt{\frac{5}{4}}\right)\right]}^n\\ & =2\left[T_n\left(\frac{3}{2}\right)-1\right]\\ & \times {\left(2\left[T_m\left(\frac{3}{2}\right)-1\right]\right)}^n\\ & =(L_{2n}-2)(L_{2m}-2{)}^n.\end{array}$

5. $\begin{array}{rl}\tau (K_m\odot W_n)& =m^{m-2}(n+2{)}^m\prod _{i=1}^{n-1}\\ & \times {\left(4-2\cos \frac{2i\pi }{n}\right)}^m\\ & =m^{m-2}(n+2{)}^m{\left[U_{n-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^m\\ & =m^{m-2}(n+2{)}^m\left[T_n\right(2)-1{]}^m.\end{array}$

6. $\begin{array}{rl}\tau (W_n\odot K_m)& =(m+1{)}^{n(m-1)}\prod _{i=1}^{n-1}\left(3-2\cos \frac{2i\pi }{n}\right)\\ & =(m+1{)}^{n(m-1)}{U}_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\\ & =2(m+1{)}^{n(m-1)}\left[T_n\left(\frac{3}{2}\right)-1\right]\\ & =(m+1{)}^{n(m-1)}(L_{2n}-2).\end{array}$

7. $\begin{array}{rl}\tau (K_{r,s}\odot W_n)& =r^{s-1}{s}^{r-1}(n+2{)}^{r+s}\prod _{i=1}^{n-1}\\ & \times {\left(4-2\cos \frac{2i\pi }{n}\right)}^{r+s}\\ & =r^{s-1}{s}^{r-1}(n+2{)}^{r+s}\\ & \times {\left[U_{n-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^{r+s}\\ & =r^{s-1}{s}^{r-1}(n+2{)}^{r+s}\left[T_n\right(2)-1{]}^{r+s}.\end{array}$

8. $\begin{array}{rl}\tau (W_n\odot K_{r,s})& =(r+1{)}^{n(s-1)}(s+1{)}^{n(r-1)}\\ & \times (r+s+1{)}^n\prod _{i=1}^{n-1}\\ & \times \left(3-2\cos \frac{2i\pi }{n}\right)\\ & =(r+1{)}^{n(s-1)}(s+1{)}^{n(r-1)}\\ & \times (r+s+1{)}^n{U}_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\\ & =2(r+1{)}^{n(s-1)}(s+1{)}^{n(r-1)}\\ & \times (r+s+1{)}^n\left[T_n\left(\frac{3}{2}\right)-1\right]\\ & =(r+1{)}^{n(s-1)}(s+1{)}^{n(r-1)}\\ & \times (r+s+1{)}^n(L_{2n}-2).\end{array}$

9. $\begin{array}{rl}\tau (Q_m\odot W_n)& =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}(n+2{)}^m\prod _{j=1}^{n-1}\\ & \times {\left(4-2\cos \frac{2j\pi }{n}\right)}^m=2^{2^m-m-1}\\ & \times \prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}(n+2{)}^m\\ & \times {\left[U_{n-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^m\\ & =2^{2^m-m-1}\prod _{i=1}^m{i}^{\left(\begin{array}{c}m\\ i\end{array}\right)}(n+2{)}^m\\ & \times \left[T_n\right(2)-1{]}^m.\end{array}$

10. $\begin{array}{rl}\tau (W_n\odot Q_m)& =\prod _{j=1}^{n-1}\left(3-2\cos \frac{2j\pi }{n}\right)\prod _{j=1}^m\\ & \times {(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}=U_{n-1}^2\left(\sqrt{\frac{5}{4}}\right)\\ & \times \prod _{i=1}^m{(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & =2\left[T_n\left(\frac{3}{2}\right)-1\right]\prod _{i=1}^m{(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}\\ & =(L_{2n}-2)\prod _{i=1}^m{(2i+1)}^{n\left(\begin{array}{c}m\\ i\end{array}\right)}.\end{array}$

11. $\begin{array}{rl}\tau (W_p\odot W_q)& =\prod _{i=1}^{p-1}\left(3-2\cos \frac{2i\pi }{p}\right)(q+2{)}^p\prod _{j=1}^{q-1}\\ & \times {\left(4-2\cos \frac{2j\pi }{q}\right)}^p=U_{p-1}^2\\ & \times \left(\sqrt{\frac{5}{4}}\right)(q+2{)}^p{\left[U_{q-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^p\\ & =2(q+2{)}^p\left[T_p\left(\frac{3}{2}\right)-1\right]\left[T_q\right(2)-1{]}^p\\ & =(L_{2p}-2)(q+2{)}^p\left[T_q\right(2)-1{]}^p.\end{array}$

12. $\begin{array}{rl}\tau (F_p\odot W_q)& =\prod _{i=1}^{p-1}\left(3-2\cos \frac{i\pi }{p}\right)(q+2{)}^p\prod _{j=1}^{q-1}\\ & \times {\left(4-2\cos \frac{2j\pi }{q}\right)}^p\\ & =U_{p-1}\left(\frac{3}{2}\right)(q+2{)}^p{\left[U_{q-1}^2\left(\sqrt{\frac{3}{2}}\right)\right]}^p\\ & =(q+2{)}^p{U}_{p-1}\left(\frac{3}{2}\right)\left[T_q\right(2)-1{]}^p\\ & =(q+2{)}^p{\text{F}}_{2p}\left[T_q\right(2)-1{]}^p.\end{array}$

13. $\begin{array}{rl}\tau (W_p\odot F_q)& =\prod _{i=1}^{p-1}\left(3-2\cos \frac{2i\pi }{p}\right)(q+2{)}^p\prod _{j=1}^{q-1}\\ & \times {\left(4-2\cos \frac{j\pi }{q}\right)}^p\\ & =(q+2{)}^p{U}_{p-1}^2\left(\sqrt{\frac{5}{4}}\right)\left[U_{q-1}\right(2){]}^p\\ & =2(q+2{)}^p\left[T_p\left(\frac{3}{2}\right)-1\right]\left[U_{q-1}\right(2){]}^p\\ & =(q+2{)}^p(L_{2p}-2)\left[U_{q-1}\right(2){]}^p.\end{array}$

The explicit formula follows from Equations (4), (7), (17) and (19).

5. Conclusions

In this work, we investigate the spectrum of the Laplacian matrix of some operations on graphs such as, Join and Corona product. Using spectral graph theory, we proved some interesting trigonometric identities, these identities are of special interest to engineers, physicists and mathematicians. We can also develop, using this technique another set of identities.

Acknowledgments

The author would like to record their indebtedness and thankfulness to the reviewers for their valuable and fruitful comments as well as for their powerful reading and suggestions.

Disclosure statement

No potential conflict of interest was reported by the author.

ORCID

S. N. Daoud http://orcid.org/0000-0003-3809-2521