Tutte Polynomial of Pseudofractal Scale-Free Web

Abstract

The Tutte polynomial of a graph is a 2-variable polynomial which is quite important in both Combinatorics and Statistical physics. It contains various numerical invariants and polynomial invariants, such as the number of spanning trees, the number of spanning forests, the number of acyclic orientations, the reliability polynomial, chromatic polynomial and flow polynomial. In this paper, we study and obtain a recursive formula for the Tutte polynomial of pseudofractal scale-free web (PSFW), and thus logarithmic complexity algorithm to calculate the Tutte polynomial of the PSFW is obtained, although it is NP-hard for general graph. By solving the recurrence relations derived from the Tutte polynomial, the rigorous solution for the number of spanning trees of the PSFW is obtained. Therefore, an alternative approach to determine explicitly the number of spanning trees of the PSFW is given. Furthermore, we analyze the all-terminal reliability of the PSFW and compare the results with those of the Sierpinski gasket which has the same number of nodes and edges as the PSFW. In contrast with the well-known conclusion that inhomogeneous networks (e.g., scale-free networks) are more robust than homogeneous networks (i.e., networks in which each node has approximately the same number of links) with respect to random deletion of nodes, the Sierpinski gasket (which is a homogeneous network), as our results show, is more robust than the PSFW (which is an inhomogeneous network) with respect to random edge failures.

Appendices

Appendix 1: A Proof of Theorem 1

In order to derive recursive formula of $T_{n}(x,y)$, we must analyze the relation of spanning subgraphs between $G(n+1)$ and its three subgraphs $\Gamma _i(i=1,2,3)$ which are a copy of $G(n)$ respectively.

For any $H\in D_{n+1}$, it is a spanning subgraphs of $G(n+1)$. Let

$$\begin{aligned} E_i=E(H)\cap E(\Gamma _i),V_i=V(H)\cap V(\Gamma _i),i=1,2,3. \end{aligned}$$

(30)

We find that $H_i=(V_i,E_i)$ is a spanning subgraphs of $\Gamma _i$ which is a copy of $G(n)$, thus $H_i \in D_n$, for $i=1,2,3$.

On the contrary, given three spanning subgraphs $H_1$, $H_2$ and $H_3$ of $\Gamma _1, \Gamma _2$ and $\Gamma _3$. Let

$$\begin{aligned} E=E(H_1)\cup E(H_2)\cup E(H_3),V=V(H_1)\cup V(H_2)\cup V(H_3). \end{aligned}$$

(31)

We find that $H=(V, E)$ is a spanning subgraphs of $G(n+1)$ and there exists a bijection between spanning subgraphs of $G(n+1)$ and spanning subgraphs of $\Gamma _1, \Gamma _2, \Gamma _3$ inside $G(n+1)$. Therefore

$$\begin{aligned} T_{n+1}(x,y)=\sum _{H\in D_{n+1}}\varPhi _{n+1}(H)&= \sum _{H_i\in D_{n},i=1,2,3 }\varPhi _{n+1}(H), \end{aligned}$$

(32)

where the relation between $H=(V,E)$ and $H_i=(V_i,E_i)$ ($i=1,2,3$) are shown in Eqs. (30) and (31). We denote this relation always holds when they appear in the following paragraphs.

We first derive the relations between $\varPhi _{n+1}(H)$ and $\varPhi _{n}(H_i)$ ($i=1,2,3$), and then the recursive formula of $T_{n}(x,y)$.

1.1 Relations Between $\varPhi _{n+1}(H)$ and $\varPhi _{n}(H_i)$

Now, we first derive recursive formula of $r(G(n))$, and the relations between $r(H)$ and $r(H_i)$, $n(H)$ and $n(H_i)$. Then we can obtain relations between $\varPhi _{n+1}(H)$ and $\varPhi _{n}(H_i)$.

Because the PSFW is a connected graph, therefore $k(G(n))=1$, for any $n\ge 0$. Note that $G(n+1)$ is composed of three copies of $G(n)$ denoted as $\Gamma _1$, $\Gamma _2$, $\Gamma _3$, we have

$$\begin{aligned} |V(G(n+1))|=|V(\Gamma _1)|+|V(\Gamma _2)|+|V(\Gamma _3)|-3=3|V(G(n))|-3. \end{aligned}$$

Thus

$$\begin{aligned} r(G(n+1))&= |V(G(n+1))|-k(G(n+1)) \nonumber \\&= (3|V(G(n))|-3) -1 \nonumber \\&= 3(|V(G(n))|-1)-1 \nonumber \\&= 3 r(G(n))-1. \end{aligned}$$

(33)

According to the second construction algorithm of the PSFW, we have

$$\begin{aligned} |V(H)|&= |V(H_1)|+|V(H_2)|+|V(H_3)|-3, \end{aligned}$$

(34)

$$\begin{aligned} |E(H)|&= |E(H_1)|+|E(H_2)|+|E(H_3)|. \end{aligned}$$

(35)

As for the relation between $r(H)$ and $r(H_i)$, $n(H)$ and $n(H_i)$, they can be divided into two cases. Thus relations between summation term of $T_{n+1}(x,y)$ and that of $T_n(x,y)$ can also be divided into two cases.

Case 1 The hub nodes of $G({n+1})$ belong to the same connected component in the spanning subgraph $H_i$ of $G_i$, for any $i=1, 2, 3$ (i.e., hub nodes $A,C$ of $G({n+1})$ belong to the same connected component in $H_1$, hub nodes $B, C$ of $G({n+1})$ belong to the same connected component in $H_2$, hub nodes $A, B$ of $G({n+1})$ belong to the same connected component in $H_3$).

According to the second construction algorithm of the PSFW, we have

$$\begin{aligned} k(H) = k(H_1) +k(H_2) + k(H_3) -2. \end{aligned}$$

(36)

Substituting $|V(H)|$, $|k(H)|$ from Eqs. (34) and (36) respectively in Eq. (1),

$$\begin{aligned} r(H)&= |V(H)|-k(H) \nonumber \\&= |V(H_1)|-k(H_1)+|V(H_2)|-k(H_2)+|V(H_3)|-k(H_3)-1 \nonumber \\&= r(H_1) + r(H_2) + r(H_3) -1. \end{aligned}$$

(37)

Substituting $|E(H)|$, $|r(H)|$ from Eqs. (35) and (37) respectively in Eq. (2),

$$\begin{aligned} n(H)=|E(H)|-r(H)=n(H_1) + n(H_2) + n(H_3) + 1. \end{aligned}$$

Let Eq. (33) minus Eq. (37), we obtain

$$\begin{aligned} r(G(n+1))-r(H) = \sum _{i=1}^3 (r(G(n))-r(H_i)). \end{aligned}$$

Hence

$$\begin{aligned} \varPhi _{n+1}(H)&= (x-1)^{r(G(n+1))-r(H)}(y-1)^{n(H)} \nonumber \\&= (y-1) \prod _{i=1}^3(x-1)^{r(G(n))-r(H_i)}(y-1)^{n(H_i)}\nonumber \\&= (y-1) \prod _{i=1}^3\varPhi _{n}(H_i). \end{aligned}$$

(38)

Case 2 The conditions for case $1$ are not satisfied (i.e., for certain $i=1, 2, 3$, the hub nodes of $G({n+1})$ do not belong to the same connected component in the spanning subgraph $H_i$), we have

$$\begin{aligned} k(H) = k(H_1) +k(H_2) + k(H_3) -3. \end{aligned}$$

(39)

Substituting $|V(H)|$, $|k(H)|$ from Eqs. (34) and (39) respectively in Eq. (1), we have

$$\begin{aligned} r(H)&= |V(H)|-k(H) \nonumber \\&= |V(H_1)|-k(H_1)+|V(H_2)|-k(H_2)+|V(H_3)|-k(H_3) \nonumber \\&= r(H_1) + r(H_2) + r(H_3). \end{aligned}$$

(40)

Thus

$$\begin{aligned} n(H)=|E(H)|-r(H)= n(H_1) + n(H_2) + n(H_3), \end{aligned}$$

and

$$\begin{aligned} r(G(n+1))-r(H) = \sum _{i=1}^3 (r(G(n))-r(H_i))-1. \end{aligned}$$

Hence

$$\begin{aligned} \varPhi _{n+1}(H)&= (x-1)^{r(G(n+1))-r(H)}(y-1)^{n(H)}\nonumber \\&= \frac{1}{(x-1)} \prod _{i=1}^3 (x-1)^{r(G(n))-r(H_i)}(y-1)^{n(H_i)}\nonumber \\&= \frac{1}{(x-1)} \prod _{i=1}^3 \varPhi _{n}(H_i). \end{aligned}$$

(41)

1.2 Recursive Formulas of $T_{1, n}(x,y)$ , $T_{2, n}(x,y)$ and $T_{3, n}(x,y)$

Here, we proof $T_{1,n}(x,y)$, $T_{2,n}(x,y)$,$T_{3,n}(x,y)$ satisfy the following recurrence relations:

with initial conditions

$$\begin{aligned} T_{1,0}(x,y)=y+2, \qquad T_{2,0}(x,y)=x-1, \qquad T_{3,0}(x,y)=(x-1)^2, \end{aligned}$$

where $T_{1,n}$, $T_{2,n}$, $T_{3,n}$ are the abbreviations of $T_{1,n}(x,y)$, $T_{2,n}(x,y)$, $T_{3,n}(x,y)$ respectively.

The initial conditions are easy to be verified according to the definition of $T_{1,n}(x,y)$, $T_{2,n}(x,y)$ and $T_{3,n}(x,y)$. As for the recurrence relations, the strategy of proof is to study all the possible structure of spanning subgraphs $H$ of $G(n+1)$ and analyze which kind of contribution they give to $T_{1,n+1}(x,y)$, $T_{2,n+1}(x,y)$ and $T_{3,n+1}(x,y)$.

As defined in Sect. 3, the spanning subgraphs of $G(n)$ has five different structures (i.e., $D_{1,n}, D_{2,n}^A, D_{2,n}^B, D_{2,n}^C$, $D_{3,n}$), thus $H_i$ also has five different structures, for any $i=1, 2, 3$, and $H$ has $5^3$ kinds of structure.

In order to depict the relations between the structure of $H_i$ and $H$ clearly, we introduce five corresponding notations for the five different structures of $H_i$ which are shown in Fig. 5. Each structure is denoted by a triangle whose three vertices denote the three hub nodes of $G(n)$. The hub nodes which are in the same connected component are connected by a solid line, and the hub nodes which are not in the same connected component connected by a dotted line.

Fig. 5

The notation of $D_{1,n}, D_{2,n}^A, D_{2,n}^B, D_{2,n}^C$ and $D_{3,n}$. The hub nodes of them are denoted by the three vertices of the triangle. The hub nodes which are in the same connected component are connected by a solid line, and the hub nodes which are not in the same connected component connected by a dotted line

Now, we will study all the $5^3$ kinds of structures of $H$, and analyze which kind of contributions they give to $T_{1,n+1}(x,y)$, $T_{2,n+1}(x,y)$ and $T_{3,n+1}(x,y)$.

First, we analyze the possible structures of $H$ which give contribution to $T_{1,n+1}(x,y)$. According to the definition of $T_{1,n+1}(x,y)$, the condition $H\in D_{1,n+1}$ holds. We find it has ten kinds of possible structures. The first three of them are shown in Fig. 6, the rest seven kinds of structures are shown in Fig. 7. For the first structure (left one of Fig. 6), $H_i\in D_{1,n}$ for any $i=1,2,3$, the hub nodes $A, B$ of $G_{n+1}$ are connected by a solid line in $H_3$, the hub nodes $A, C$ of $G_{n+1}$ are connected by a solid line in $H_1$, the hub nodes $B, C$ of $G_{n+1}$ are connected by a solid line in $H_2$. Thus the conditions for the case $1$ of Appendix 1 are satisfied. According to Eq. (38), we have

$$\begin{aligned}&\sum _{H_i\in D_{1,n},i=1,2,3 }\varPhi _{n+1}(H) \nonumber \\&\qquad \quad = \sum _{H_i\in D_{1,n},i=1,2,3 }\left[ (y-1) \prod _{i=1}^3\varPhi _{n}(H_i)\right] \nonumber \\&\qquad \quad = (y-1) \prod _{i=1}^3\sum _{H_i\in D_{1,n}}\left[ (y-1) \prod _{i=1}^3\varPhi _{n}(H_i)\right] \nonumber \\&\qquad \quad = (y-1)T_{1,n}^3. \end{aligned}$$

(45)

Thus, it contributes to $T_{1,n+1}(x,y)$ by a term $(y-1)T_{1,n}^3$.

For the second structure (center one of Fig. 6), $H_1\in D_{1,n}$, $H_2\in D_{1,n}$ and $H_3\in D_{2,n}^C$, the hub nodes $A, B$ of $G_{n+1}$ are connected by a solid line in $H_3$, the hub nodes $A, C$ of $G_{n+1}$ are connected by a solid line in $H_1$, the hub nodes $B, C$ of $G_{n+1}$ are connected by a solid line in $H_2$. Thus the conditions for the case $1$ of Appendix 1 are also satisfied. According to Eq. (38), we have

$$\begin{aligned}&\sum _{H_1\in D_{1,n}, H_2\in D_{1,n}, H_3\in D_{2,n}^C }\varPhi _{n+1}(H) \nonumber \\&\qquad \qquad \quad = (y-1) \prod _{i=1}^2\sum _{H_i\in D_{1,n}}\varPhi _{n}(H_i) \cdot \sum _{H_3\in D_{2,n}^C}\varPhi _{n}(H_3) \nonumber \\&\qquad \qquad \quad = (y-1)T_{1,n}^2T_{2,n}. \end{aligned}$$

(46)

Noticing the possible rotations, we find it has $3$ equivalent structures. Thus this kind of spanning subgraphs contributes to $T_{1,n+1}(x,y)$ by a term $3(y-1)T_{1,n}^2T_{2,n}$. Computing the contributions of all possible structures and adding them together, we obtain Eq. (42).

Fig. 6

Three kinds of structures which satisfy $H\in D_{1,n+1}$. Each one of them corresponds to a kind of spanning subgraph $H$ of $G(n+1)$. The hub nodes $A, B, C$ of $G(n+1)$ are connected by a path of solid line which shows they are in the same connected component

Fig. 7

Seven kinds of structures which satisfy $H\in D_{1,n+1}$. The hub nodes $A, B, C$ are connected by a path of solid line which shows the are in the same connected component

Next, we analyze the possible structures of $H$ which give contribution to $T_{2,n+1}(x,y)$. By symmetry, we only study $T_{2,n+1}^C(x,y)$. Then the condition $H\in D_{2,n+1}^C$ holds. We find it has $6$ possible structures which are shown in Fig. 8. For the first structure, $H_3\in D_{1,n}$, $H_1\in D_{2,n}^B$ and $H_2\in D_{2,n}^A$. Therefore the hub nodes $A, C$ of $G_{n+1}$ are connected by a dotted line in $H_1$ which shows that they are not in the same connected component. Thus the conditions for case $2$ are satisfied. According to Eq. (41), it contributes to $T_{2,n+1}^C(x,y)$ by a term $\frac{4}{x-1}T_{1,n}T_{2,n}^2$ (we have consider $4$ equivalent structures: $H_1\in D_{2,n}^C$ (or $D_{2,n}^B$), $H_2\in D_{2,n}^A$ (or $D_{2,n}^C$) ).

Fig. 8

All possible structures which satisfy $H \in D_{2,n+1}^C$. Each one of them corresponds to a kind of spanning subgraph $H$ of $G(n+1)$. The hub nodes $A, B$ are connected by solid line, but $C$ are not connected with $A, B$ by a path of solid line which shows $C$ are not in the same connected component with $A, B$

Computing the contributions of all the $6$ possible structures and adding them together, we obtain Eq. (43).

Finally, we analyze the possible structures of $H$ which give contribution to $T_{3,n+1}(x,y)$. Then the condition $H\in D_{3,n+1}$ holds. We find it has $4$ possible structures which is shown in Fig. 9. For the first structure, $H_1\in D_{2,n}^B$ (or $D_{2,n}^C)$, $H_2\in D_{2,n}^A$ (or $D_{2,n}^C$), $H_3\in D_{2,n}^A$ (or $D_{2,n}^B)$. It contributes to $T_{3,n+1}(x,y)$ by a term $\frac{8}{x-1}T_{2,n}^3$ according to Eq. (41). Computing the contributions of all the $4$ possible structures and adding them together, we obtain Eq. (44).

Fig. 9

All possible structures which satisfy $H\in D_{3,n+1}$. The hub nodes $A, B, C$ are not connected by a path of solid line which shows $A, B, C$ belong to three different connected components

Appendix 2: Proof of Lemma 1

In this appendix, we prove the following result by mathematical induction.

For any $n \ge 0$, $T_{2,n}(x,y),T_{3,n}(x,y)$ can be factored as

$$\begin{aligned} \left\{ \begin{array}{l} T_{2,n}(x,y)= (x-1)P_n(x,y), \\ T_{3,n}(x,y)= (x-1)^2Q_n(x,y), \end{array} \right. \end{aligned}$$

(47)

where $P_n(x,y)$, $Q_n(x,y)$ are polynomials of $x$ and $y$.

Initial step: for $n=0$, $T_{2,0}(x,y)=x-1, T_{3,0}(x,y)=(x-1)^2$, let $P_0(x,y)=Q_0(x,y)=1$, we know Eq. (47) holds.

Inductive step: assuming Eq. (47) holds for certain $n \ge 0$, we prove Eq. (47) holds for $n+1$ as follows.

For convince, we abbreviate $P_{n}(x,y)$, $Q_{n}(x,y)$ as $P_{n}$ and $Q_{n}$ respectively. Substituting $T_{2,n}(x,y),T_{3,n}(x,y)$ from Eq. (47) in Eqs. (43) and (44), we have

$$\begin{aligned} T_{2,n+1}&= \frac{1}{x-1}(4T_{1,n}T_{2,n}^2+4T_{1,n}T_{2,n}T_{3,n}+4T_{2,n}^3 \\&\quad +\,4T_{2,n}^2T_{3,n}+T_{1,n}T_{3,n}^2+T_{2,n}T_{3,n}^2)\\&= \frac{1}{x-1}(4T_{1,n}((x-1)P_n)^2+4T_{1,n}((x-1)P_n)((x-1)^2Q_n)\\&\quad +\,4((x-1)P_n)^3+4((x-1)P_n)^2((x-1)^2Q_n)\\&\quad +\,T_{1,n}((x-1)^2Q_n)^2+((x-1)P_n)((x-1)^2Q_n)^2) \\&= (x-1)[4T_{1,n}P_{n}^2+4(x-1)T_{1,n}P_{n}Q_{n}+4(x-1)P_{n}^3 \\&\quad +\,4(x-1)^2P_{n}^2Q_{n}+(x-1)^2T_{1,n}Q_{n}^2+(x-1)^3P_{n}Q_{n}^2], \end{aligned}$$

and

$$\begin{aligned} T_{3,n+1}&= \frac{1}{x-1}\left( 8T_{2,n}^3+ 12T_{2,n}^2T_{3,n}+6T_{2,n}T_{3,n}^2+T_{3,n}^3\right) \\&= \frac{1}{x-1}( 8((x-1)P_n)^3+ 12((x-1)P_n)^2(x-1)^2Q_n\\&\quad +\,6(x-1)P_n((x-1)^2Q_n)^2+((x-1)^2Q_n)^3)\\&= (x-1)^2(8P_{n}^3+ 12(x-1)P_{n}^2Q_{n}+6(x-1)^2P_{n}Q_{n}^2+(x-1)^3Q_{n}^3). \end{aligned}$$

Let

$$\begin{aligned} P_{n+1}&= 4T_{1,n}P_{n}^2+4(x-1)T_{1,n}P_{n}Q_{n}+4(x-1)P_{n}^3 \nonumber \\&\quad +\,4(x-1)^2P_{n}^2Q_{n}+(x-1)^2T_{1,n}Q_{n}^2+(x-1)^3P_{n}Q_{n}^2,\end{aligned}$$

(48)

$$\begin{aligned} Q_{n+1}&= 8P_{n}^3+ 12(x-1)P_{n}^2Q_{n}+6(x-1)^2P_{n}Q_{n}^2+(x-1)^3Q_{n}^3. \end{aligned}$$

(49)

We find

$$\begin{aligned} \left\{ \begin{array}{l} T_{2,n+1}(x,y)= (x-1)P_{n+1}(x,y)\\ T_{3,n+1}(x,y)= (x-1)^2Q_{n+1}(x,y) \end{array} \right. . \end{aligned}$$

(50)

Thus Eq. (47) holds for $n+1$ which end the proof.

Appendix 3: Comparison of $R_{s}(n)$ and $R(n)$

For the Sierpinski gasket $SG(n)$ ($n\ge 0$), Alfredo Donno [20] found that the all-terminal reliability polynomial is given by

$$\begin{aligned} R_{s}(n) =p^{V_{n}-1}(1-p)^{E_{n}-V_{n}+1}T_{1,n}\left( 1,\frac{1}{1-p}\right) , \end{aligned}$$

(51)

and $T_{1,n}\left( 1,\frac{1}{1-p}\right)$ satisfy the following recurrence relation

$$\begin{aligned} \left\{ \begin{aligned} \textstyle T_{1,n+1}\left( 1,\frac{1}{1-p}\right)&\textstyle =\frac{p}{1-p}T_{1,n}^3 +6T_{1,n}^2N_{n} \\ \textstyle N_{n+1}\left( 1,\frac{1}{1-p}\right)&\textstyle = \frac{p}{1-p}T_{1,n}^2N_{n}+T_{1,n}^2M_{n}+7T_{1,n}N_{n}^2\\ \textstyle M_{n+1}\left( 1,\frac{1}{1-p}\right)&\textstyle =\frac{3p}{1-p}T_{1,n}N_{n}^2+ 12T_{1,n}N_{n}M_{n}+14N_{n}^3 \end{aligned} \right. , \end{aligned}$$

(52)

with initial conditions

$$\begin{aligned} T_{1,0}\left( 1,\frac{1}{1-p}\right) =\frac{3-2p}{1-p}, \qquad N_0\left( 1,\frac{1}{1-p}\right) = M_0\left( 1,\frac{1}{1-p}\right) =1. \end{aligned}$$

Let

$$\begin{aligned} B_s(n)= p^{(V_{n}-2)}(1-p)^{(E_{n}-V_{n}+2)}N_{n},\end{aligned}$$

(53)

$$\begin{aligned} C_s(n)= p^{(V_{n}-3)}(1-p)^{(E_{n}-V_{n}+3)}M_{n}. \end{aligned}$$

(54)

We obtain the following recurrence relation from Eqs. (51), (52), (53) and (54).

with initial conditions

$$\begin{aligned} R_{s}(0) =p^2(3-2p), \qquad B_{s}(0) = p(1-p)^2, \qquad C_{s}(0) = (1-p)^3. \end{aligned}$$

Now, we compare all-terminal reliability between the Sierpinski gasket and the PSFW while $p\in (0,1)$.

For $n=0$, the Sierpinski gasket and the PSFW have the same topology (i.e., a triangle). Thus

$$\begin{aligned} R_{s}(0) =R(0), \qquad B_{s}(0) =B(0). \end{aligned}$$

According to Eqs. (25), (55), (26) and (56), we have

$$\begin{aligned} R_{s}(1)=R(1), B_{s}(1)> B(1) \end{aligned}$$

But for $n \ge 2$, we have

$$\begin{aligned} \left\{ \begin{array}{l} R_{s}(n)-R(n)>0 \\ B_{s}(n)-B(n)>0 \end{array} \right. , \end{aligned}$$

(57)

which is proved by mathematical induction as follows.

Initial step: for $n=2$,

$$\begin{aligned} R_{s}(2)-R(2)&= R_{s}^3(1)-R^3(1)+6R_{s}^2(1)B_s(1)-6R^2(1)B(1) \\&= 6R^2(1)[B_s(1)-B(1)] \\&> 0, \end{aligned}$$

and

$$\begin{aligned} B_{s}(2)-B(2)&= R_{s}^2(1)B_s(1)+R_{s}^2(1)C_s(1)+7R_{s}(1)B_s^2(1)-4R(1)B^2(1)\\&\ge R_{s}^2(1)B_s(1)+R_{s}^2(1)C_s(1)+3R_{s}(1)B_s^2(1)\\&> 0. \end{aligned}$$

Thus Eq. (57) holds for $n=2$.

Inductive step: assuming that Eq. (57) holds for certain $n \ge 2$, we prove Eq. (57) holds for $n+1$.

Let Eq. (55) minus Eq. (25) and Eq. (56) minus Eq. (26), we have

$$\begin{aligned} R_{s}(n+1)-R(n+1)&= R_{s}^3(n)-R^3(n)+6R_{s}^2(n)B_s(n)-6R^2(n)B(n)\\&> 6R_{s}^2(n)B_s(n)-6R^2(n)B(n) \\&> 0, \end{aligned}$$

and

$$\begin{aligned} B_s(n+1)-B(n+1)&= R_{s}^2(n)B_s(n)+R_{s}^2(n)C_s(n)+7R_{s}(n)B_s^2(n)-4R(n)B^2(n)\\&> R_{s}^2(n)B_s(n)+R_{s}^2(n)C_s(n)+3R_{s}(n)B_s^2(n) \\&> 0. \end{aligned}$$

Thus Eq. (57) holds for $n+1$ which end the proof.