Dynamic Properties of Quantity Adjustment Process Under Demand Forecast Formed by Moving Average of Past Demands

Abstract

In this chapter, we shall analyze a family of multi-sector dynamic models of the quantity adjustment process in which firms determine their productions and raw material orders based on inventories and demand forecasts under constant prices and final demands. Special attention is paid to the role of demand forecast by a moving average of past demands. Section 4.1 describes the assumptions common to all the models and explains how and in what sequence firms determine productions and raw material orders. For the demand forecast method, we assume the simple moving average and the geometric moving average. Sections 4.2 and 4.3 investigate the dynamic properties of the process generated through the interactions between sectors under the above two demand forecast methods. A series of theorems will elucidate how the stability of the process depends on the input structure, the extent of averaging, and the buffer inventory coefficients. It will be shown that averaging of past demands in the demand forecast formation is essential for stability if the input matrix has negative or complex eigenvalues reflecting the interactions between sectors. Section 4.4 closely examines the mechanism of stabilization by averaging of past demands in forecast formation. Stability conditions corresponding to the forecast by the finite geometric moving average will indicate that stability depends on the allocation of weights to each past demand in addition to the number of past periods referred for averaging.

Appendix

1. Proof of Theorem  1 ²⁴

(i) Let ϕ ≠ 0 be an eigenvalue of Φ, and let ξ = [ξ _τ, ξ _τ−1, …, ξ ₁, η] be an associated eigenvector, where ξ _i and η are 1 × n vectors. Then, from ϕξ = ξΦ we obtain

$$ \xi =\left[{\phi}^{\tau}\eta, {\phi}^{\tau -1}\eta, \dots, \phi \eta, \eta \right],\quad \phi \eta =\eta Q\left(\phi \right)A, \vspace*{-8pt} $$

$$ Q\left(\phi \right)\equiv\left(\phi -1\right)\left({\phi}^{-1}{U}_1+\dots +{\phi}^{-\tau }{U}_{\tau}\right)B+I. $$

Therefore, ϕ is also an eigenvalue of Q(ϕ)A. Conversely, if ϕ is an eigenvalue of Q(ϕ)A and η is the corresponding eigenvector, then it is also an eigenvalue of Φ and the above ξ is the corresponding eigenvector. Consequently, ϕ ≠ 0 is an eigenvalue of Φ if and only if ϕ is an eigenvalue of Q(ϕ)A.

To prove the contraposition, assume ρ(Φ) ≥ 1 and ϕ is an eigenvalue of Φ such that ϕ ≥ 1. Then, from the above fact, it follows $1\le \rho \left(Q\left(\phi \right)A\right).\vspace*{-3pt}$ Let q _i be the i-th diagonal element of matrix Q(ϕ), then

$$\begin{array}{cc}\left|{\mathit{q}}_{\mathit{i}}\right|=& \left|\left(1-\frac{1}{{\varphi }^{{\tau }_{\mathit{i}}}}\right)\frac{{\mathit{b}}_{\mathit{i}}}{{\tau }_{\mathit{i}}}+1\right|\le \frac{2{\mathit{b}}_{\mathit{i}}}{{\tau }_{\mathit{i}}}+1\end{array}$$

(4.47)

Let C ⁺ denote a matrix obtained from a matrix C by replacing its elements with their absolute values. It is known that ρ(C) ≤ λ ^F(C ⁺) holds for any complex square matrix C.²⁵ Since Q(ϕ) is diagonal and A is nonnegative, (4.47) leads to

$$ {(QA)}^{+}\le {Q}^{+}A\le \left(2{T}^{-1}B+I\right)A. $$

Since ϕ is an eigenvalue of QA, these inequalities lead to

$$ 1\le \rho (QA)\le {\lambda}^{\mathrm{F}}\left({(QA)}^{+}\right)\le {\lambda}^{\mathrm{F}}\left({Q}^{+}A\right)\le {\lambda}^{\mathrm{F}}\left(\left(2{T}^{-1}B+I\right)A\right).\vspace*{-8pt} $$

(ii) To prove the contraposition, assume that ω λ ^F(A) is an eigenvalue of A satisfying (4.22) and ${\lambda}^{\ast }$  = λ ^F((2T ⁻¹ B + I)A) ≥ 1. Let ϕ ≠ 0, ψ = ω ⁻¹ ϕ, and $\tilde{A}={\omega}^{-1}A$; then since ${\omega}^{\tau_i}=-1$ for any i, ϕ is an eigenvalue of Φ if and only if ψ is an eigenvalue of $F\left(\psi \right)\tilde{A}$, where

$$ F\left(\psi \right)=\left[\begin{array}{ccc}1+{\psi}^{-{\tau}_1}& \cdots & 0\\ {}\vdots & \ddots & \vdots \\ {}0& \cdots & 1+{\psi}^{-{\tau}_n}\end{array}\right]{T}^{-1}B+I. $$

For any ψ > 0, matrix F(ψ)A is a nonnegative and identically cyclic with matrix A; hence ωλ ^F(F(ψ)A) is an eigenvalue of F(ψ)A, and hence λ ^F(F(ψ)A) is an eigenvalue of $F\left(\psi \right)\tilde{A}$. Let

$$ h\left(\psi \right)=\psi -{\lambda}^{\mathrm{F}}\left(F\left(\psi \right)A\right)\ \mathrm{for}\kern0.50em \psi >0, $$

then h(ψ) is an eigenvalue of $\psi I-F\left(\psi \right)\tilde{A}$; hence h(ψ) = 0 implies that ϕ = ωψ is an eigenvalue of Φ. It follows from F(λ $^{\ast}$) ≤ F(1) that ${\lambda}^{\mathrm{F}}\left(F\left({\lambda}^{\ast}\right)A\right)\le {\lambda}^{\mathrm{F}}\left(F(1)A\right)={\lambda}^{\ast };$hence

$h\left({1}\right)=1-{\lambda}^{\mathrm{F}}\left(F(1)A\right)=1-{\lambda}^{\ast}\le 0,\hspace*{2pt}$ $h\left({\lambda}^{\ast}\right)={\lambda}^{\mathrm{F}}\left(F(1)A\right)-{\lambda}^{\mathrm{F}}\left(F\left({\lambda}^{\ast}\right)A\right)\ge 0.\vspace*{8pt}$

Since polynomial h is continuous with respect to ψ, there exists an α such that

$$ 1\le \alpha \le {\lambda}^{\mathrm{F}}\left(\left(2{T}^{-1}B+I\right)A\right),\kern0.5em h\left(\alpha \right)=\alpha -{\lambda}^{\mathrm{F}}\left(F\left(\alpha \right)A\right)=0; \vspace*{-2pt} $$

hence αω is an eigenvalue of Φ. Therefore,

$$ \rho \left(\Phi \right)\ge \left|\alpha \omega \right|\ge 1. $$

2. Proof of Theorem 2

(i) The characteristic equation of Φ′ is

$$ F\left(\phi \right)=\left|{\phi}^{\tau +1}I-{\phi}^{\tau}\left(\frac{b}{\tau }+1\right)A+\frac{b}{\tau }A\right|=0. $$

Let Λ be the Jordan normal form of matrix A, then there exists a regular matrix X satisfying XAX ⁻¹ = Λ and

$$ {\displaystyle \begin{array}{c}F\left(\phi \right)=\left|X\left({\phi}^{\tau +1}I-{\phi}^{\tau}\left(\frac{b}{\tau }+1\right)A+\frac{b}{\tau }A\right){X}^{-1}\right|\\ {}=\prod \limits_{i=1}^{\nu }{\left({\phi}^{\tau +1}-{\phi}^{\tau}\left(\frac{b}{\tau}+1 \right){\lambda}_i+\frac{b}{\tau}{\lambda}_i\right)}^{\kappa_i},\end{array}} $$

where ν is the number of A’s different eigenvalues and κ _i is the multiplicity of an eigenvalue ${\lambda}_i={\mu}_i{\mathrm{e}}^{\mathrm{i}{\theta}_i}$ (thus ${\kappa}_1 + \cdots + {\kappa}_n=n$). Thus, ρ(Φ′) < 1 is equivalent to $\overline{\phi}\left({\mu}_i,{\theta}_i,\tau, b\right)<1$ for any i ∈ {1, …, ν}.

(ii) Let ξe^iω = ξ(cosω + i sin ω) (ξ ≥ 0) be a root of (4.24). Then, it satisfies

$$ f\left(\xi, \omega, \theta \right) \equiv {\xi}^{\tau +1}{\mathrm{e}}^{\mathrm{i}\left(\tau +1\right)\omega }-\left(\frac{b}{\tau}+1 \right)\mu {\mathrm{e}}^{\mathrm{i}\left(\tau \omega +\theta \right)}{\xi}^{\tau }+ \frac{b}{\tau}\mu {\mathrm{e}}^{\mathrm{i}\theta }=0. $$

Let ${\theta}^{\prime }=\theta +{2\pi}/{\tau },{\omega}^{\prime }=\omega +{2\pi}/{\tau },{\theta}^{{\prime\prime} }={2\pi}/{\tau }-\theta, {\omega}^{{\prime\prime} }= {2\pi}/{\tau }-\omega,$ then

$$ f\left(\xi, {\omega}^{\prime },{\theta}^{\prime}\right)={\mathrm{e}}^{2\mathrm{i}\frac{\pi }{\tau }}f\left(\xi, \omega, \theta \right)=0,\ f\left(\xi, {\omega}^{{\prime\prime} },{\theta}^{{\prime\prime}}\right)={\mathrm{e}}^{2\mathrm{i}\frac{\pi }{\tau }}f\left(\xi, -\omega, -\theta \right)=0. $$

The second equation is obtained by observing that f(ξ, −ω, −θ) is the conjugate of f(ξ, ω, θ). Therefore, when μe^iθ rotates by 2π/τ radians on the complex plane, each of the solutions of (4.24) also rotates by 2π/τ radians. Furthermore, if θ ₁ and θ ₂ are symmetric on the complex plane about the direction π/τ, two sets of the roots of (4.24) corresponding θ ₁ and θ ₂ are also symmetric about this direction.

(iii) This statement is obvious for μ = 0, so let us assume μ > 0. Let

$$ {{g}}_{+}\left(\xi; \tau, {b},\mu \right)\equiv -{{e}}^{\mathrm{-i}\frac{\pi}{\tau}}{f}\left(\xi, \frac{\pi}{\tau},\frac{\pi}{\tau}\right)={\xi}^{\tau +1}-{\xi}^{\tau}\left(\frac{{b}}{\tau}+1\right)\mu -\frac{{b}}{\tau}\mu,\vspace*{-4pt} $$

and ${\mathit{\varphi }}_{+}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)$be the unique positive$\mathit{\xi }$such that ${\mathit{g}}_{+}\left(\mathit{\xi };\mathit{\tau },\mathit{b},\mathit{\mu }\right)=0$.²⁶ By a theorem about the upper bound on roots of an algebraic equation²⁷, it follows $\overline{\phi}\left(\mu, \theta, \tau, b\right)\le {\phi}_{+}\left(\mu, \tau, b\right)$, where the equality sign holds if and only if θ = π/τ. Hence, $\overline{\phi}\left(\mu, {\pi}/{\tau },\tau, b\right)<1$ holds if and only if

$$ {g}_{+}\left(1;\tau, b,\mu \right)=\left(2b+\tau \right)\left(\overline{\mu}-\mu \right)>0, $$

where $\overline{\mu} \equiv {\tau }/{(2b+\tau)}$. Therefore, $\overline{\phi}\left(\overline{\mu},{\pi }/{\tau },\tau, b\right)=1$ and

$$ \overline{\phi}\left(\mu, \theta, \tau, b\right)\le \overline{\phi}\left(\mu, {\pi }/{\tau },\tau, b\right)<1\ \mathrm{for}\kern0.5em \mu <\overline{\mu}. $$

Let $\mathit{\alpha }\equiv -\left(\mathit{b}/\mathit{\tau }+1\right)\mathit{\mu },\beta \equiv \mathit{b\mu }/\mathit{\tau }$, then $1-\mathit{\alpha }>\mathit{\beta }$. According to Schur’s condition,²⁸ a necessary and sufficient condition for $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)<1$is that all the following inequalities hold:

$$ {C}_u\equiv \left|\begin{array}{cc}{I}_u+\alpha {\mathrm{e}}^{\mathrm{i}\theta }{J}_u& \beta {\mathrm{e}}^{\mathrm{i}\theta }{I}_u\\ {}\beta {\mathrm{e}}^{-\mathrm{i}\theta }{I}_u& {I}_u+\alpha {\mathrm{e}}^{-\mathrm{i}\theta }{J}_u^{\mathrm{T}}\end{array}\right|>0\ \left(u \in \{1,\dots, \tau\} \right),\vspace*{-3pt} $$

$$ \tilde{C}_{\tau +1}\equiv \left|\begin{array}{cc}{I}_{\tau +1}+\alpha {\mathrm{e}}^{\mathrm{i}\theta }{J}_{\tau +1}& \beta {\mathrm{e}}^{\mathrm{i}\theta }{I}_{\tau +1}+\alpha {\mathrm{e}}^{\mathrm{i}\theta }{K}_{\tau +1}\\ {}\beta {\mathrm{e}}^{-\mathrm{i}\theta }{I}_{\tau +1}+\alpha {\mathrm{e}}^{-\mathrm{i}\theta }{K}_{\tau +1}^{\mathrm{T}}& {I}_{\tau +1}+\alpha {\mathrm{e}}^{\mathrm{-i}\theta }{J}_{\tau +1}^{\mathrm{T}}\end{array}\right|>0,\vspace*{-3pt} $$

(4.48)

Where${\mathit{I}}_{\mathit{u}}$is the unit matrix of order$\mathit{u}$, ${\mathit{J}}_{\mathit{u}}\equiv \left[\begin{array}{ll}0& {\mathit{I}}_{\mathit{u}-1}\\ 0& 0_{\mathit{u}-1}\end{array}\right]$, ${\mathit{K}}_{\mathit{\tau }+1}\equiv \left[\begin{array}{ll}0& {\mathrm{O}}_{\mathit{\tau }}\\ 1& 0_{\mathit{\tau }}\end{array}\right]$($0_{\mathit{u}}$ is the zero row vector of order $\mathit{u}$, ${\mathrm{O}}_{\tau }$ is the zero matrix of order $\mathit{\tau }$), and $T$denotes transposition. Let ${\mathit{C}}_0=1$, $\mathit{\delta }\equiv 1+{\mathit{\alpha }}^2-{\mathit{\beta }}^2$. Then, since

$$ {{C}}_{{u}}=\left|\delta {{I}}_{{u}}+{{L}}_{{u}}-{\alpha}^2\left[\begin{array}{cc}{{\mathrm{O}}}_{{u}-1}& 0\\ {}{0}_{u-1}& 1\end{array}\right]\right|,\kern0.75em {{L}}_{{u}}\equiv \alpha {{e}}^{{i}\theta}{{J}}_{{u}}+\alpha {{e}}^{-{i}\theta}{{J}}_{{u}}^{{T}}, $$

${\mathit{C}}_{\mathit{u}}$satisfies the following recursion formula:

$$ {C}_u=\delta {C}_{u-1}-{\alpha}^2{C}_{u-2}. $$

(4.49)

Similarly, from

$${\tilde{C}}_{\mathit{\tau }+1}=\left|\mathit{\delta }{\mathit{I}}_{\mathit{\tau }+1}+{\mathit{L}}_{\mathit{\tau }+1}-{\mathit{\alpha }}^2\left[\begin{array}{lll}1& 0_{\mathit{\tau }-1}& 0\\ 0_{\mathit{\tau }-1}^{\mathrm{T}}& {\mathrm{O}}_{\mathit{\tau }-1}& 0_{\mathit{\tau }-1}^{\mathrm{T}}\\ 0& 0_{\mathit{\tau }-1}& 1\end{array}\right]-\mathit{\alpha \beta }\left({\mathit{K}}_{\mathit{\tau }+1}+{\mathit{K}}_{\mathit{\tau }+1}^{\mathrm{T}}\right)\right|,$$

we have

$$ \tilde{C}_{\tau +1}=\left(1-{\beta}^2\right){C}_{\tau }-{\alpha}^2\left(1+{\beta}^2\right){C}_{\tau -1}-{\alpha}^2{\beta}^2{D}_{\tau -1}+2\left(-\alpha \right)^{\tau+1}\beta \cos \tau \theta, \vspace*{-18pt} $$

$$ D_0 = 0, {D}_u\equiv \left|\begin{array}{cc}\delta {I}_{u-1}+{L}_{u-1}& 0\\ {}0& {\alpha}^2\end{array}\right|. $$

It can be shown that ${\mathit{D}}_{\mathit{u}}$ satisfies the same recursion formula as that of ${\mathit{C}}_{\mathit{u}}$. Let ${\mathit{C}}_{-1}=1,{\mathit{D}}_{-1}=-1$, and

$${\mathit{E}}_{\mathit{\tau }}\equiv \left(1 - {\mathit{\beta }}^{2}\right){\mathit{C}}_{\mathit{\tau }} - {\mathit{\alpha }}^{2}\left(1 + {\mathit{\beta }}^{2}\right){\mathit{C}}_{\mathit{\tau } - 1} - {\mathit{\alpha }}^{2}{\mathit{\beta }}^{2}{\mathit{D}}_{\mathit{\tau } - 1}, $$

then ${\tilde{C}}_{\mathit{\tau }+1}$can be written as

$$ \tilde{C}_{\tau +1}={E}_{\tau }+2\left(-\alpha \right)^{\tau + 1}\beta \cos \tau \theta, $$

(4.50)

where ${\mathit{E}}_{\mathit{u}}$satisfies the recursion formula:

$$ {E}_0=1-{\alpha}^2-{\beta}^2,\kern0.75em {E}_1={\left(1-\beta^2 \right)}^2-{\alpha}^2{\left(1+\beta^2 \right)},\kern0.75em {E}_u=\delta {E}_{u-1}-{\alpha}^2{E}_{u-2}. $$

Assume $\mu >\overline{\mu}$, then $-\alpha +\beta =\mu /\overline{\mu}>1$, hence

$$ {\delta}^2-4{\alpha}^2=\left({\left(1+\alpha \right)}^2-{\beta}^2\right)\left({\left(1-\alpha \right)}^2-{\beta}^2\right)<0. $$

Let ε = α ² + β ² − 1, $\eta = \sqrt{{4\alpha}^{2}-\delta^{2}}$. From (4.49) and (4.50), we have²⁹

$$\begin{array}{c}{\mathit{C}}_{\mathit{u}}=2\beta {\left(-\alpha \right)}^{\mathit{u}+1}{\eta }^{-1}cos\mathit{F}\left(\mu ,\mathit{u},\mathit{b}\right)\left(\mathit{u}\in \{1,\dots ,\tau \}\right),\\ {\tilde{\mathit{C}}}_{\tau +1}=2\beta {\left(-\alpha \right)}^{\tau +1}\left(cos\mathit{G}\left(\mu ,\tau ,\mathit{b}\right)+cos\mathit{\tau 𝜃}\right)\end{array}$$

where

$$ F\left(\mu, u,b\right)\equiv u\arctan\ \left(\eta /\delta \right)+\arctan\ \left(\varepsilon /\eta \right), \vspace*{-18pt} $$

$$ G\left(\mu, \tau, b\right)\equiv \tau \arctan\ \left({\eta}/{\delta}\right)+\arctan\ \left({\varepsilon}/{\eta}\right)+\pi /2=F\left(\mu, \tau, b\right)+\pi /2. \vspace*{-3pt} $$

and $\arctan$takes values in interval $\left[-\mathit{\pi }/2,\mathit{\pi }/2\right]$. Under the assumption$0\le \mathit{𝜃}\le \mathit{\pi }/\mathit{\tau }$ and $\overline{\mathit{\mu }}<\mathit{\mu }$, ${\mathit{C}}_{\mathit{\tau }}$ and ${\tilde{C}}_{\mathit{\tau }+1}$ can be simultaneously positive if and only if

$$ \begin{aligned} F\left(\mu, \tau, b\right) &=\tau\; \mathrm{arctan}\frac{\sqrt{\left(1{-}{\mu}^2\right)\left({\mu}^2{-}{\overline{\mu}}^2\right)}}{\mu^2+\overline{\mu}} \nonumber \\ &\quad +\arctan \frac{\mu^2\left(1+2{b}^2{\tau}^{-2}\overline{\mu}\right){-}\overline{\mu}}{\sqrt{\left(1{-}{\mu}^2\right)\left({\mu}^2{-}{\overline{\mu}}^2\right)}}{\,<\,}\frac{\pi }{2}{\,-\,}\tau \theta . \end{aligned} $$

(4.51)

Note that $\mathit{F}\left(\overline{\mathit{\mu }},\mathit{\tau },\mathit{b}\right)=-\mathit{\pi }/2$, $\mathit{F}\left(1,\mathit{\tau },\mathit{b}\right)=\mathit{\pi }/2$. Since

$$ -\pi{/}2\le {F}\left(\mu, 1,{b}\right)<{F}\left(\mu, 2,{b}\right)<\dots <{F}\left(\mu, \tau, {b}\right), $$

(4.51) guarantees ${\mathit{C}}_1>0$, ${\mathit{C}}_2>0$, …, ${\mathit{C}}_{\mathit{\tau }-1}>0.$ Thus, providing that $\mathit{\mu }>\overline{\mathit{\mu }}$ and$0\le \mathit{𝜃}<\mathit{\pi }/\mathit{\tau }$, (4.48) holds if and only if

$$ -\pi /2<F\left(\mu, \tau, b\right)<\pi /2-\tau \theta . $$

Let ${\mathit{\mu }}^{†}\equiv \sqrt{\frac{\mathit{\tau }+2}{\mathit{\tau }+2\mathit{b}}}$, then ${\mathit{\mu }}^{†2}-{\overline{\mathit{\mu }}}^2=\frac{2\left(1+\mathit{b}\overline{\mathit{\mu }}\right)}{2\mathit{b}+\mathit{\tau }}>0$. Thus, differentiating $\mathit{F}$with respect to $\mathit{\mu }$, we have

$$ {F}_{\mu}\left(\mu, \tau, b\right)=\frac{\tau \left({\mu}^{\dagger 2}-{\mu}^2\right)}{\mu \sqrt{\left(1-{\mu}^2\right)\left({\mu}^2-{\overline{\mu}}^2\right)}}>0 $$

for$\kern0.5em \overline{\mu}<\mu <{\mu}^{\dagger }$. Since $\mathit{F}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)$increase from $-\mathit{\pi }/2$to $\mathit{F}\left({\mathit{\mu }}^{†},\mathit{\tau },\mathit{b}\right)>\mathit{\pi }/2$as $\mathit{\mu }$ increases from $\overline{\mathit{\mu }}$to ${\mathit{\mu }}^{†}$, there exists a unique$\mathit{\mu }$such that

$$ \overline{\mu}<\mu <{\mu}^{\dagger },\kern0.5em F\left(\mu, \tau, b\right)=\pi /2-\tau \theta . $$

Let us denote this unique μ by μ ^∗(θ, τ, b), then (4.51) is equivalent to μ < μ ^∗(θ, τ, b) for $0\le \theta < {\pi }/{\tau }$. When $\theta = {\pi }/{\tau }$, (4.51) is never satisfied, and from the above-mentioned fact, (4.48) is equivalent to $\mu <\overline{\mu}$. However, by definition we have $F\left({\mu}^{\ast}\left({\pi}/{\tau },\tau, b\right),\tau, b\right)=-{\pi }/{2}$, hence, ${\mu}^{\ast}\left({\pi }/{\tau },\tau, b\right)=\overline{\mu}$, and consequently, (4.48) is equivalent to $\mu < \mu^{\ast}\left(\theta, \tau, b \right)$ for $0\le \theta \le {\pi }/{\tau }.$

(iv) For $0\le \theta <{\theta}^{\prime }< {\pi }/{\tau }$, it holds

$$ F\left({\mu}^{\ast}\left({\theta}^{\prime},\tau, b\right),\tau, b\right)={\pi }/{2}-\tau {\theta}^{\prime }< {\pi }/{2}-\tau \theta =F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau, b\right), $$

which implies μ ^∗(θ′, τ, b) < μ ^∗(θ, τ, b).

(v) The first part of this statement directly follows from statement (iii). As for the second part, μ ^∗(0, 1, b) and μ ^∗(0, 2, b) can be directly calculated from (4.48).

(vi) Let ${\mu}_1\equiv\sqrt{\frac{\tau \left(1+\tau \right)}{2{b}^2+\left(2b+\tau \right)\left(\tau -1\right)}}$ and ${\xi}_{1 }\equiv{\left(\frac{b\tau +b}{\tau +b}\right)}^{\frac{1}{\tau }}$, then we have

$$ {\displaystyle \begin{array}{l}F\left({\mu}_1,\tau, b\right)={\pi }/{2}- \mathrm{arctan} \sqrt{\xi^{2\tau}_{1}-1} + \tau\ \mathrm{arctan}\ \tau^{-1}\sqrt{\xi^{2\tau}_{1}-1}\\ {}\qquad \qquad \qquad \ge {\pi }/{2}\ge {\pi }/{2}-\tau\theta =F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau, b\right),\end{array}} $$

here we used the fact that x arctan $a/x \geq \text{arctan}\ a\ \text{holds \ for}\ \text{any} \ x \geq 1 \ \text{ and}\ a > 0$. Hence μ ₁(τ, b) ≥ μ ^∗(θ, τ, b), where the equality sign holds if and only if τ = 1, θ = 0 (note that $\mu_1 < \mu^{\dag}$). Treating τ as a continuous variable and differentiating F with respect to τ, for $\overline{\mu} < \mu < 1$ we obtain

$$ F_{\tau} (\mu, \tau, b) = \frac{-(1-\mu^{2})(1+b\overline{\mu})}{(b+\tau)\sqrt{(1-\mu^{2}){(\mu^{2}-\overline{\mu}^{2}})}}+ \mathrm{arctan} \frac{\sqrt{(1-\mu^{2})(\mu^{2}-\overline{\mu}^{2})}}{\overline{\mu}+ \mu^{2}}\vspace*{-12pt} $$

$$ < \sqrt{\frac{1-{\mu}^2}{\mu^2-{\overline{\mu}}^2}}\frac{\left(2{b}^2+\left(2b+\tau \right)\left(\tau -1\right)\right)\left({\mu}^2-{\mu}_1^2\right)\ }{\left(b+\tau \right)\left(2b+\tau \right)(\overline{\mu}+ \mu^{2})}. $$

Here we used the fact that $arctan\mathit{x}<x$ for x $>0$. Assume $0\le \mathit{𝜃}<\mathit{\pi }/\mathit{\tau }$, then ${\mathit{\mu }}_1>{\mathit{\mu }}^{{}_{\ast }}>\overline{\mathit{\mu }}$, hence ${\mathit{F}}_{\mathit{\tau }}<0$ at $\mathit{\mu }={\mathit{\mu }}^{{}_{\ast }}$. If ${\mathit{\mu }}^{{}_{\ast}} (\theta, \tau, b) < (\tau + 1)/(2b + \tau + 1)$, then ${\mathit{\mu }}^{{}_{\ast}} (\theta, \tau, b) < {\mathit{\mu }}^{{}_{\ast}} \left({\pi}/{(\tau+1)}, \tau + 1, b\right) \leq {\mathit{\mu }}^{{}_{\ast}} \left({\tau\theta}/{(\tau + 1)}, \tau + 1, b\right)$. Otherwise, since $\mu_1$ is increasing with respect to $\tau$, we have

$$ F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau +1,b\right)<F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau, b\right)=\frac{\pi }{2}-\tau \theta \vspace*{-12pt} $$

$$ =\frac{\pi }{2}-(\tau +1)\frac{\tau \theta}{\tau +1}=F\left({\mu}^{\ast}\left(\frac{\tau }{\tau +1}\theta, \tau +1,b\right),\tau +1,b\right). $$

Thus, $\mu^{\ast} (\theta, \tau, b)<\mu^{\ast}(\tau \theta/(\tau + 1), \tau + 1,b),$ and this holds also for $\theta = \pi/\tau$.

Next, let${\mathit{\mu }}_2\left(\mathit{\tau },\mathit{b}\right)\equiv \left(\mathit{\tau }+1\right)/\left(\mathit{\tau }+2\mathit{b}-1\right)$. Note that F(μ ₂(1, b), 1, b) = π/2 holds because ${\mathit{\mu }}^{{}_{\ast }}\left(0,1,\mathit{b}\right)=1/\mathit{b}={\mathit{\mu }}_2\left(1,\mathit{b}\right)$. Since

$$ \begin{aligned} F\left({\mu}_2,\tau, b\right) &=\tau \arctan \frac{2\sqrt{\left(b-1\right)\left(b+2 b\tau +{\tau}^2\right)}}{1+2 b\tau +{\tau}^2}\\ &\quad +\arctan \frac{b+2 b\tau +2{\tau}^2-b{\tau}^2}{2\tau\sqrt{\left(b-1\right)\left(b+2 b\tau +{\tau}^2\right)}}, \end{aligned} $$

$$ \begin{aligned} {F}_{\tau}\left({\mu}_2,\tau, b\right) &=-\sqrt{\frac{b-1}{b+2 b\tau +{\tau}^2}}\frac{2\left(\tau +2b-1+{\tau}^2+ b\tau \right)}{\left(\tau +2b-1\right)\left(\tau +1\right)}\\ & \quad +\arctan \frac{2\sqrt{\left(b-1\right)\left(b+2 b\tau +{\tau}^2\right)}}{1+2 b\tau +{\tau}^2}, \end{aligned} $$

$${\mathit{F}}_{\mathit{\tau b}}\left({\mathit{\mu }}_2,\mathit{\tau },\mathit{b}\right)=\sqrt{\frac{\mathit{b}-1}{\mathit{b}+2\mathit{b\tau }+{\mathit{\tau }}^2}}\frac{\left(\mathit{\tau }-1\right)\left(3{\mathit{\tau }}^2+2\mathit{\tau }+1\right)+2\left(\mathit{b}\left(3{\mathit{\tau }}^2+\mathit{\tau }-1\right)+1\right)}{\left(\mathit{b}+2\mathit{b\tau }+{\mathit{\tau }}^2\right){\left(\mathit{\tau }+2\mathit{b}-1\right)}^2}>0,$$

we have ${\mathit{F}}_{\mathit{\tau }}\left({\mathit{\mu }}_2\left(\mathit{\tau },\mathit{b}\right),\mathit{\tau },\mathit{b}\right)\ge {\mathit{F}}_{\mathit{\tau }}\left({\mathit{\mu }}_2\left(\mathit{\tau },2\right),\mathit{\tau },2\right)>0$ (the last inequality sign can be confirmed by some mechanical calculation). Therefore,

$$ \pi /2={F}\left({\mu}_2\left(1,{b}\right),1,{b}\right)<{F}\left({\mu}_2\left(\tau, {b}\right),\tau, {b}\right), $$

which implies $\mu_{2}(\tau, b) > \mu^{\ast}(0,\tau, b)$ for $\tau > 1$

(vii) Differentiating $\mathit{F}$with respect to $\mathit{b}$, we have

$$ {F}_b\left(\mu, \tau, b\right)=\sqrt{\frac{1-{\mu}^2}{\mu^2-{\overline{\mu}}^2}}\frac{\tau \left(1+b\overline{\mu}\right)}{b\left(b+\tau \right)}>0 $$

for $\overline{\mathit{\mu }}<\mathit{\mu }<1$. Hence, if $\mathit{b}<{\mathit{b}}^{\prime }$, then

$$ F\left({\mu}^{\ast}\left(\theta, \tau, {b}^{\prime}\right),\tau, {b}^{\prime}\right)=F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau, b\right)<F\left({\mu}^{\ast}\left(\theta, \tau, b\right),\tau, {b}^{\prime}\right), $$

which implies μ ^∗(θ, τ, b′) < μ ^∗(θ, τ, b).

3. Proof of Theorem 3

(i) By virtue of Theorem 4.2(ii), we can assume 0 ≤ θ ≤ π/τ without losing generality. Let ${\overline{\psi}}_m\left(\mu, \theta, \tau, b\right)$ denote, for $m> 0$, the dominant root of the equation:

$$ {\psi}^{\tau +1}-\left(\frac{b}{\tau}+1\right)m^{-1}\mu {\mathrm{e}}^{\mathrm{i}\theta} {\psi}^{\tau }+ \frac{b}{\tau}m^{-(\tau + 1)}{\mu{e}}^{\mathrm{i}\theta}=0.\vspace*{2pt}$$

(4.52)

Obviously, ψ is a solution of (4.52) if and only if mψ is a solution of (4.24), therefore, ${\overline{\psi}}_m\left(\mu, \theta, \tau, b\right)\lessgtr 1$ is equivalent to $\overline{\phi}\left(\mu, \theta, \tau, b\right)\lessgtr m$. Let

$$ B \equiv b+\tau, M\equiv {m}^{\tau },\alpha \equiv -\frac{B\mu}{\tau m},\beta \equiv \frac{b\mu}{\tau m M}, X \equiv \frac{b}{B}, {\xi}_0\equiv X^{\frac{1}{\tau }}, \vspace*{-9pt}$$

$$ Y \equiv X(1+\tau), \kern0.75em {\mathit{\mu }}^{{}_{\prime}}= \frac{\tau mM}{\mathit{BX}},\kern0.75em {\mu}_{-}\equiv \frac{\tau mM}{B\left(M-X\right)},\kern0.75em {\mu}_{+}\equiv \frac{\tau mM}{B\left(M+X\right)}, \vspace*{-9pt} $$

$$ \tilde{\mu}\equiv\frac{m M}{B}\sqrt{\frac{\tau \left(\tau +2\right)}{M^2-{X}^2}},\kern0.5em {\mu}_0\equiv \frac{m\left(1+\tau \right)}{B},\kern0.5em {\mu}_1\equiv\frac{m\tau M}{B\sqrt{M^2-{Y}^2}}. $$

Note that $X<1<Y=\left(\mathit{b}+\mathit{b\tau }\right)/\left(\mathit{b}+\mathit{\tau }\right)$. Define ${\mathit{E}}_{\mathit{\tau }}$ similarly to the case of $\mathit{m}=1$, then ${\overline{\mathit{\psi }}}_{\mathit{m}}<1$requires

$$0<1-{\mathit{\beta }}^2,0<{\mathit{E}}_{\mathit{\tau }}+2{\left(-a\right)}^{\tau +1}\mathit{\beta }cos\mathit{\tau 𝜃}\le {\mathit{E}}_{\mathit{\tau }}+2{(-a)}^{\tau +1}\mathrm{\beta .}$$

By induction, it can be shown that ${\mathit{E}}_{\mathit{\tau }}+2{(-a)}^{\tau +1}\beta$is a product of

$$ \left(1+\alpha +\beta \right)\left(1-\alpha \pm \beta \right)q_\tau{\left(\alpha, \beta \right)}^2, $$

where ${\mathit{q}}_{\tau }\left(\mathit{\alpha },\mathit{\beta }\right)$is a $\mathit{\tau }$-th order polynomial of $\mathit{\alpha },\mathit{\beta }$ and the double sign corresponds to odd $\mathit{\tau }$and even $\mathit{\tau }$ in this order. Since $1-\mathit{\alpha }\pm \mathit{\beta }>0$for $\mathrm{\beta }<1$, ${\overline{\mathit{\psi }}}_{\mathit{m}}<1$implies

$$ 1 - \beta =1 - \mu/{\mathit{\mu }}^{{}_{\prime}} > 0,\quad 1+\alpha +\beta =1 - \mu{/}{\mathit{\mu}}_{-} >0 $$

By the Gauss’ theorem about the bound on the solution of an algebraic equation³⁰, ${\overline{\mathit{\psi }}}_{\mathit{m}}<1$also requires $\mathit{\mu }<{\mathit{\mu }}_0$. Since ${\mathit{\mu }}_{-}-{\mathit{\mu }}_0=\mathit{m}\left(\mathit{M}-\mathit{Y}\right)/\left(\mathit{B}\left(\mathit{M}-\mathit{X}\right)\right)$, ${\mathit{\mu }}_0$ becomes more restrictive than $\mu_-$ when ${\mathit{\xi }}_0<{\mathit{\xi }}_1<\mathit{m}.$

Like the case of $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$, if $\mathit{\mu }<{\mathit{\mu }}_{+}$, then $1+\mathit{\alpha }-\mathit{\beta }>0$, which ensures ${\overline{\mathit{\psi }}}_{\mathit{m}}<1$. If ${\mathit{\mu }}_{+}<\mathit{\mu }<{\mathit{\mu }}_{-}$, then $1+\mathit{\alpha }-\mathit{\beta }<0$, and ${\overline{\mathit{\psi }}}_{\mathit{m}}<1$ holds if and only if $1+\mathit{\alpha }+\mathit{\beta }>0$and $-\mathit{\pi }/2<{\mathit{F}}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)<\mathit{\pi }/2-\mathit{\tau 𝜃}$(hence $\mathit{𝜃}<\mathit{\pi }/\mathit{\tau }$), where

$$ {F}^m\left(\mu, \tau, b\right)=\tau \arctan H/{G}_1+\arctan {G}_2/H,\kern0.5em {G}_1\equiv 1+{\alpha}^2-{\beta}^2>0, \vspace*{-8pt} $$

$$ {G}_2\equiv {\alpha}^2+{\beta}^2-1,H\equiv \sqrt{-\left({\left(1+\alpha \right)}^2-{\beta}^2\right)\left({\left(1-\alpha \right)}^2-{\beta}^2\right)}>0. $$

Note that $\mu_+ < {\mathit{\mu }}^{{}_{\prime}}$ and $F^m (\mu_+, \tau, b) = \pi/2$ for any $m >0$. Let ${\mathit{F}}_{\mathit{\mu }}^{\mathit{m}}$be the partial derivative of ${\mathit{F}}^{\mathit{m}}$with respect to $\mathit{\mu }$. If $\mathit{M}\le X$, then $1+\alpha +\beta >0$, $F^m ({\mathit{\mu}}^{{}_{\prime}}, \tau, b) \geq \pi/2$, and

$$ {F}_{\mu}^m\left(\mu, \tau, b\right)=\frac{m^2{\tau}\left(\tau +2\right){M}^2-{B}^2\left({M}^2-{X}^2\right){\mu}^2}{\mu \tau m^2 M^2H}>0. $$

If $\mathit{M}>\mathit{X}$, then $1+\mathit{\alpha }+\mathit{\beta }>0$is equivalent to $\mathit{\mu }<{\mathit{\mu }}_{-}$, and by definition, we have${\tilde{\mu }}^2-{\mathit{\mu }}_{+}^2=\frac{2\mathit{\tau }{m}^2{M}^2\left(\mathit{M}+\mathit{Y}\right)}{{\mathit{B}}^2{\left(\mathit{M}+\mathit{X}\right)}^2\left(\mathit{M}-\mathit{X}\right)}>0$. Accordingly,

$$ {F}_{\mu}^m\left(\mu, \tau, b\right)= {B}^2\left({M}^2-{X}^2\right)\left(\tilde{\mu}^2-{\mu}^2\right)/\left(\mu \tau m^2 M^2H\right)>0 $$

for ${\mathit{\mu }}_{+}<{\mathit{\mu }}^{\prime }<\tilde{\mu }$. Since ${\mathit{G}}_2\left({\mathit{\mu }}_{-}\right)=\frac{2\mathrm{MX}}{{(\mathit{M}-\mathit{X})}^2}>0,$ we have ${\mathit{F}}^{\mathit{m}}\left({\mathit{\mu }}_{ - },\mathit{\tau },\mathit{b}\right) = \mathit{\pi }/2.$ Therefore, there exists a unique $\mathit{\mu }$such that $\mu_+ \leq \mu \leq \mu^\prime \mbox{ for } M \leq X, {\mathit{\mu }}_{ + } \leq \mathit{\mu } \leq \min\left\{{\mathit{\mu }}_{ - },\tilde{\mu}\right\} \mbox{ for } X < M, \mbox{ and }$

$$ {\mathit{F}}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right) + \mathit{\tau \theta } = \mathit{\pi }/2.$$

Let us denote this unique $\mathit{\mu }$by${\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$, then ${\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{\pi }/\mathit{\tau },\mathit{\tau },\mathit{b}\right)={\mathit{\mu }}^{+}$; therefore, like the case of $\mathit{m}=1$, ${\overline{\mathit{\psi }}}_{\mathit{m}}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)<1$ is equivalent to $\mathit{\mu }<{\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right).$

Because of the continuity of ${\overline{\mathit{\psi }}}_{\mathit{m}}$with respect to its arguments, $\mu ={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$ is equivalent to $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)=\mathit{m}$. Assume $\mathit{\mu }<{\mathit{\mu }}^{\prime }<1$ and $\overline{\mathit{\varphi }}\left({\mathit{\mu }}^{\prime },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)=\mathit{m}$, then $\mathit{\mu }<{\mathit{\mu }}^{\prime }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$, which implies $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)<\mathit{m}=\overline{\mathit{\varphi }}\left({\mathit{\mu }}^{\prime },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$.³¹

(ii) Assume $0\le \theta <{\theta}^{\prime}\le {\pi }/{\tau }$ and $\overline{\phi}\left(\mu, {\theta}^{\prime},\tau, b\right)=m$. Then, since

$$ {F}^{m}\left(\mu, \tau, b\right)={\pi }/{2}-{\theta}^{\prime}\tau <{\pi }/{2}-\theta \tau, $$

we have $\overline{\phi}\left(\mu, \theta, \tau, b\right)<m=\overline{\phi}\left(\mu, {\theta}^{\prime},\tau, b\right).$

(iii) Assume m > ξ ₀ $(M >X)$ and $Z \equiv X\sqrt{2\tau+1}.$ Note that $X < Z < Y.$ Differentiating F ^m with respect to $\tau$, we have

$$ {F}_{\tau}^m\left(\mu, \tau, b\right)=\frac{K_1+B{K}_2\log m}{m^2 \tau^2 M^2 BH}+\arctan \frac{H}{G_1}, \vspace*{-6pt}$$

$$ {K}_1\equiv {\mu}^2\left(\left(1+b\right){B}^2{M}^2-{b}^2\left(1+\tau +B\right)\right)-\left(B+b+ b\tau \right)\tau {m}^2{M}^2, \vspace*{-6pt} $$

$$ {K}_2\equiv {B}^2\left({M}^2-{Z}^2\right){\mu}^2-{m}^2{\tau}^2{M}^2. $$

First, we consider the case $X<M<Y$. If $\mathit{M}\le \mathit{Z}$, then ${\mathit{K}}_2<0$. Assume $\mathit{Z}<\mathit{M}<\mathit{Y}$, then since ${\mathit{\mu }}_1^2-{\mathit{\mu }}_{-}^2=\frac{2{\tau }^2{\mathit{m}}^2{\mathit{M}}^2\left(\mathit{Y}-\mathit{M}\right)}{{\mathit{B}}^2{\left(\mathit{M}-\mathit{X}\right)}^2\left({\mathit{M}}^2-{\mathit{Z}}^2\right)}>0,$we have

$$ {K}_2={B}^2\left({M}^2-{Z}^2\right)\left({\mu}^2-{\mu}_1^2\right)<0 $$

for $\mathit{\mu }<{\mathit{\mu }}_{-}$. Therefore, in these cases, we have

$$ {F}_{\tau}^m\left(\mu, \tau, b\right)<\frac{K_{\mathrm{1}}}{m^2\tau^2 M^2 BH}+\frac{H\ }{G_1}=-\frac{L_1\left({\mu}^2-{\mu}_2^2\right)\left({\mu}^2-{\mu}_3^2\right)}{m^4\tau^4M^4G_1 H}, \vspace*{-9pt}$$

$$ {L}_1\equiv {B}\left(\left(\tau -1\right){B}{M}^2+{b}^2\left(1+\tau \right)\right)\left({M}^2-{X}^2\right)>0, \vspace*{-9pt} $$

$${\mathit{\mu }}_2\equiv \frac{\mathit{\tau m}M\sqrt{\mathit{\tau }+1}}{\sqrt{\left(\mathit{\tau }-1\right){\mathit{B}}^2{M}^2+{\mathit{b}}^2\left(\mathit{\tau }+1\right)}},{\mathit{\mu }}_3\equiv \frac{\mathit{mM}\sqrt{\tau (\mathit{b}+\mathit{B})}}{\mathit{B}\sqrt{{\mathit{M}}^2-{\mathit{X}}^2}}.$$

Because${\mathit{\mu }}_0^2-{\mathit{\mu }}_2^2=\frac{\left(\mathit{\tau }+1\right)m^2{B}^2\left({\mathit{Y}}^2-{\mathit{M}}^2\right)}{\left(\mathit{\tau }-1\right){\mathit{M}}^2{\mathit{B}}^2+{\mathit{b}}^2\left(\mathit{\tau }+1\right)},{\mathit{\mu }}_3^2-{\mathit{\mu }}_0^2=\frac{m^2\left(\left({\mathit{Y}}^2-{\mathit{M}}^2\right)+2{\mathit{M}}^2\mathit{\tau }\left(\mathit{b}-1\right)\right)}{{\mathit{B}}^2\left({\mathit{M}}^2-{\mathit{X}}^2\right)},$ and $\mu_0 - \mu_{-} = \frac{m(M-Y)}{B(M-X)}$, we obtain ${\mathit{\mu }}_2<min\{{\mu }_{-},{\mu }_3\}$, hence ${\mathit{F}}_{\mathit{\tau }}^{\mathit{m}}<0$for $\mathit{\mu }<{\mathit{\mu }}_2.$ On the other hand, since

$$ {F}^m\left({\mu}_2,\tau, b\right)=\frac{\pi }{2}+\tau \arctan \frac{\sqrt{Y^2-{M}^2}}{\tau M}\hbox{--} \arctan \frac{\sqrt{Y^2-{M}^2}}{M}, $$

we have ${\mathit{F}}^{\mathit{m}}\left({\mathit{\mu }}_2,\mathit{\tau },\mathit{b}\right)>\mathit{\pi }/2\ge \mathit{\pi }/2-\mathit{\tau 𝜃}={\mathit{F}}^{\mathit{m}}\left({\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right),\mathit{\tau },\mathit{b}\right)$, which implies ${\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)<{\mathit{\mu }}_2$. Therefore, ${\mathit{F}}_{\mathit{\tau }}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)<0$at $\mathit{\mu }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$.

Next, we turn to the case $Y<M$. In this case, it follows that ${\mathit{\mu }}_1\le {\mathit{\mu }}_{-}\le {\mathit{\mu }}_0\le {\mathit{\mu }}_2$.If $\mathit{\mu }<{\mathit{\mu }}_1$, then ${\mathit{K}}_2<0$ because $Z<\mathit{Y}\le \mathit{M}$, hence ${\mathit{F}}_{\mathit{\tau }}^m<0.$ If ${\mathit{\mu }}_1\le \mathit{\mu }<{\mathit{\mu }}_{-}$, then ${\mathit{K}}_2\ge 0$ and $0<log\mathit{m}\le \mathit{m}-1$because $1<Y\le M$. Thus,

$$ {F}_{\tau}^m<\frac{K_1+\left(m-1\right)B{K}_2}{m^2\tau^2M^2 BH}+\frac{H}{G_1}=\frac{L_2\left(\mu \right)}{m^2{\tau}^2{M}^2{G}_1H}, \vspace*{-9pt}$$

$$ {L}_2\left(\mu \right)\equiv -{L}_1\left({\mu}^2-{\mu}_2^2\right)\left({\mu}^2-{\mu}_3^2\right)+ \left(m-1\right)G_1{K}_2. $$

Since ${\mathit{\mu }}_3^2-{\mathit{\mu }}_{-}^2=\frac{2\mathit{b\tau }{\mathit{m}}^2{\mathit{M}}^2\left(\mathit{M}-1\right)}{{\mathit{B}}^2\left({\mathit{M}}^2-{\mathit{X}}^2\right){\left(\mathit{M}-\mathit{X}\right)}^2}$, we have ${{\mu }_1\le \mathit{\mu }}_{-}<min\{{\mu }_2,{\mu }_3\}$. Therefore, ${\mathit{L}}_2\left(\mathit{\mu }\right)$is increasing in${\mathit{\mu }}_1\le \mathit{\mu }<{\mathit{\mu }}_{-}$. Hence,

$$ {L}_2\left(\mu \right)\le {L}_2\left({\mu}_{-}\right)=-\frac{4b\tau m^2 M^3\left({m}^{\tau }- m\tau +\left(\tau -1\right)\right)\left(M-Y\right)}{B{\left(M-X\right)}^3}\le 0, $$

where the second inequality sign holds because

$$ {m}^{\tau }- m\tau +\left(\tau -1\right)={\left(m-1\right)}^2\sum_{j=1}^{\tau -1}j{m}^{\tau -1-j}>0 \vspace*{-8pt} $$

for $\mathit{m}>1,\mathit{\tau }\ge 2$. Consequently, ${\mathit{F}}_{\mathit{\tau }}^{\mathit{m}}<0$holds for any $\mathit{\mu }<{\mathit{\mu }}_{-}$. Hence, in this case also${\mathit{F}}_{\mathit{\tau }}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)<0$at $\mathit{\mu }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$.

In sum, $F_{\tau }^m\left(\mathit{\mu },\tau ,\mathit{b}\right)<0$ at $\mathit{\mu }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$for any$\mathit{m}>{\mathit{\xi }}_0$, and it also holds for any $\mu <\mu_2$ (when $X < M < Y$) and $\mu < \mu_{-}$ (when $Y \leq M$). Let $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)=\mathit{m}>{\mathit{\xi }}_0$. Then, since $\mathit{\mu }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\tau },\mathit{b}\right)$ and both $\mu_2$ and $\mu_{-}$ are increasing with respect to $\tau$ in their relevant ranges, we have

$$ {F}^m\left(\mu, \tau +1,b\right)<{F}^m\left(\mu, \tau, b\right)=\frac{\pi }{2}-\tau \theta =\frac{\pi }{2}-\left(\tau +1\right)\frac{\tau \theta}{\tau +1}, $$

which implies $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{\tau 𝜃}/\left(\mathit{\tau }+1\right),\mathit{\tau }+1,\mathit{b}\right)<\mathit{m}=\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right).$

(iv) Differentiating ${\mathit{F}}^{\mathit{m}}$with respect to $\mathit{b}$, we have

$$\mathit{F}_{\mathit{b}}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right) = \mathit{P}\left(\mathit{\mu }\right){/}(\mathit{bB}\tau m^2{\mathit{M}}^{2}\mathit{H}), \hspace*{0.5em}\mathit{P}\left(\mathit{\mu }\right)\equiv {\mathit{N}}_{1} + {\mathit{\mu }}^{2}{\mathit{N}}_{2}, \vspace*{-3pt}$$

${\mathit{N}}_1\equiv \tau m^2{\mathit{M}}^2\left(\mathit{B}+\mathit{b\tau }+\mathit{b}\right),{\mathit{N}}_2\equiv -\left(1+\mathit{b}\right){\mathit{B}}^2{\mathit{M}}^2+{\mathit{b}}^2\left(\mathit{B}+\mathit{\tau }+1\right)$.

Considering that the sign of ${\mathit{\mu }}_0-{\mathit{\mu }}_{-}$ is equal to that of $\mathit{M}-\mathit{Y}$, and

$$ P\left({\mu}_{+}\right)=2b {\tau} m^2 M^2\left(1+M\right)\left({M}+{Y}\right)/{\left(M+X\right)}^2>0, \vspace*{-10pt} $$

$$ P\left({\mu}_0\right)=-\left(1+\tau +B\right){m}^2\left({M}^2-{Y}^2\right),\vspace*{-10pt} $$

$$ P\left({\mu}_{-}\right)=2b\tau m^2 M^2\left(M-1\right)\left({M}-{Y}\right)/{\left(M-X\right)}^2, $$

${\mathit{F}}_{\mathit{b}}^{\mathit{m}}>0$ holds for ${\mathit{\mu }}_{+}<\mathit{\mu }<min\left\{{\mathit{\mu }}_0,{\mathit{\mu }}_{-}\right\}$. Let $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },{\mathit{b}}^{\prime }\right)=\mathit{m}$, then since $\mu = \mu_m^* (\theta, \tau, b^\prime)$ and both $\mu_2$ and $\mu_-$ are increasing with respect to $\tau$ in their relevant ranges, we have

$${\mathit{F}}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },\mathit{b}\right)< {\mathit{F}}^{\mathit{m}}\left(\mathit{\mu },\mathit{\tau },{\mathit{b}}^{{\prime}}\right) = \mathit{\pi }/2 - \mathit{\tau \theta }, $$

which implies $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },\mathit{b}\right)<\mathit{m}=\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\tau },{\mathit{b}}^{\prime }\right).$

4. Proof of Theorem 4

(i) To prove the contraposition of statement (i), let us assume ρ(Ψ) ≥ 1. Let ψ be an eigenvalue of Ψ such that |ψ| ≥ 1 and ξ = [ξ _I, ξ _II] be an associated eigenvector, where both ξ _I and ξ _II are 1 × n vectors. Then, from ψξ = ξ Ψ we obtain

$$ \xi =\left[{\xi}_{\mathrm{I}},{\xi}_{\mathrm{I}}\Gamma {\left(\psi I-\left(I-\Gamma \right)\right)}^{-1}\right],\kern0.5em {\psi \xi}_{\mathrm{I}}={\xi}_{\mathrm{I}}Q\left(\psi \right)A \vspace*{-12pt}$$

$$ Q\left(\psi \right) \equiv\left(I+\Gamma B\right)-\Gamma {\left(\psi I-\left(I-\Gamma \right)\right)}^{-1}\Gamma B. $$

Therefore, ψ is also an eigenvalue of matrix Q(ψ)A, hence 1 ≤ |ψ| ≤ ρ(Q(ψ)A). Let q _i be the i-th diagonal element of matrix Q(ψ), then

$$ \left|{q}_i\right|=\left|1+{\gamma}_i{b}_i-\frac{\gamma_i^2{b}_i}{\psi -\left(1-{\gamma}_i\right)}\right|=\left(1+{\gamma}_i{b}_i\right)\left|\frac{\psi -\frac{1-{\gamma}_i+{\gamma}_i{b}_i}{1+{\gamma}_i{b}_i}}{\psi -\left(1-{\gamma}_i\right)}\right|\le \frac{2{-}{\gamma}_i+2{\gamma}_i{b}_i}{2{-}{\gamma}_i}. $$

Since Q is diagonal and A is nonnegative, we have

$$ {(QA)}^{+}\le {Q}^{+}A\le \left(2\Gamma B{\left(2I-\Gamma \right)}^{-1}+I\right)A. $$

Because $\mathit{\psi }$ is an eigenvalue of $\mathit{QA}$, these inequalities imply

$$ {\displaystyle \begin{array}{c}1\le \rho (QA)\le {\lambda}^{\mathrm{F}}\left({(QA)}^{+}\right)\le {\lambda}^{\mathrm{F}}\left({Q}^{+}A\right)\\ {}\le {\lambda}^{\mathrm{F}}\left(\left(2\Gamma B{\left(2I-\Gamma \right)}^{-1}+I\right)A\right)\end{array}}\vspace*{-7pt} $$

(ii) To prove the contraposition of statement (ii), assume that –λ ^F(A) is an eigenvalue of A and $\lambda^* = \lambda^{\rm F} ((2\Gamma B(2I - \Gamma)^{-1} + I)A) \geq 1$. Let ψ ≠ 0, then ψ is an eigenvalue of Ψ if and only if

$$ \left|\psi I-Q\left(\psi \right)A\right|=0.\vspace*{-4pt} $$

For any ψ > 0, matrix Q(−ψ)A is nonnegative and identically cyclic with matrix A. Hence, −λ ^F(Q(−ψ)A) is an eigenvalue of Q(−ψ)A. Let

$$ h\left(\psi \right) \equiv \psi -{\lambda}^{\mathrm{F}}\left(Q\left(-\psi \right)A\right)\ $$

for ψ > 0, then h(ψ) is an eigenvalue of ψI + Q(−ψ)A =  − ((−ψ)I − Q(−ψ)A). Hence, h(ψ) = 0 implies that −ψ is an eigenvalue of Ψ. Since $Q(-\lambda^\ast) \leq Q(-1)$, we obtain $\lambda^{\mathrm{F}} (Q(-\lambda^\ast)A) \leq \lambda^{\mathrm{F}} (Q(-1)A) = \lambda^\ast$. Consequently,

$$ h(1)=1-{\lambda}^{\mathrm{F}}\left(Q\left(-1\right)A\right)=1-{\lambda}^{\ast}\le 0,\vspace*{-12pt} $$

$$ h\left({\lambda}^{\ast}\right)={\lambda}^{\ast }-{\lambda}^{\mathrm{F}}\left(Q\left(-{\lambda}^{\ast}\right)A\right)\ge 0. $$

Because of the continuity, there exists an α such that

$$ 1\le \alpha \le {\lambda}^{\ast },\kern0.5em h\left(\alpha \right)= \alpha -{\lambda}^{\mathrm{F}}\left(Q\left(-\alpha \right)A\right)=0. $$

Therefore, −α is an eigenvalue of Ψ, hence ρ(Ψ) ≥ 1.

5. Proof of Theorem 5

(i) Like the case of $\Phi^\prime$, the characteristic equation of $\Psi^\prime$ is

$$ \prod \limits_{i=1}^{\nu }{\left({\psi}^2I-\psi \left(\left(\gamma b+1\right){\lambda}_i+\left(1-\gamma \right)\right)+\left(1-\gamma +\gamma b\right){\lambda}_i\right)}^{\kappa_i}=0, \vspace*{-2pt} $$

where ν is the number of A’s different eigenvalues and κ _i is the multiplicity of an eigenvalue λ _i = ${\mu}_i{\mathrm{e}}^{\mathrm{i}{\theta}_i}$. Therefore, ρ $(\Psi^\prime) < 1$ is equivalent to $\overline{\psi}\left({\mu}_i,{\theta}_i,\gamma, b\right)<1$ for any i ∈ {1, …, ν}.

(ii) Let ξe^iω = ξ(cosω + i sin ω) (0 ≤ ξ, 0 ≤ ω < 2π) be a root of (4.33), then

$$ {\xi}^2{\mathrm{e}}^{2\mathrm{i}\omega }-\left(\left(\gamma b+1\right)\mu {\mathrm{e}}^{\mathrm{i}\theta }+\left(1-\gamma \right)\right)\xi {\mathrm{e}}^{\mathrm{i}\omega }+\left(1-\gamma +\gamma b\right)\mu {\mathrm{e}}^{\mathrm{i}\theta }=0. \vspace*{-2pt} $$

Let f(ξ, ω, θ) be the left-hand side of this equation; then f(ξ, −ω, −θ) = 0, because it is the conjugate of f(ξ, ω, θ). Therefore, $\overline{\psi}\left(\mu, \theta, \gamma, b\right)=\overline{\psi}\left(\mu, -\theta, \gamma, b\right)$.

(iii) From Schur’s condition, $\overline{\mathit{\psi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)<1$ holds if and only if

$$\begin{array}{cc}{\mathit{E}}_1& \equiv \left|\begin{array}{ll}1& \mathit{D}\mu {\mathrm{e}}^{\mathrm{i}\theta }\\ \mathit{D}\mu {\mathrm{e}}^{-\mathrm{i}\theta }& 1\end{array}\right|=1-D^2{\mu }^2>0,\\ {\mathit{E}}_2& \equiv \left|\begin{array}{llll}1& -\left(\mathit{B}\mu {\mathrm{e}}^{\mathrm{i}\theta }+\mathit{C}\right)& \mathit{D}\mu {\mathrm{e}}^{\mathrm{i}\theta }& 0\\ 0& 1& -\left(\mathit{B}\mu {\mathrm{e}}^{\mathrm{i}\theta }+\mathit{C}\right)& \mathit{D}\mu {\mathrm{e}}^{\mathrm{i}\theta }\\ \mathit{D}\mu {\mathrm{e}}^{-\mathrm{i}\theta }& -\left(\mathit{B}\mu {\mathrm{e}}^{-\mathrm{i}\theta }+\mathit{C}\right)& 1& 0\\ 0& \mathit{D}\mu {\mathrm{e}}^{-\mathrm{i}\theta }& -\left(\mathit{B}\mu {\mathrm{e}}^{-\mathrm{i}\theta }+\mathit{C}\right)& 1\end{array}\right|>0,\end{array}$$

(4.53)

where $\mathit{B}\equiv 1+\mathit{\gamma b},\mathit{C}\equiv 1-\mathit{\gamma },\mathit{D}\equiv 1-\mathit{\gamma }+\mathit{\gamma b}$. Note that

$$ B- CD=\gamma \left(B+C\right)>0,\kern0.75em BC-D=-{\gamma}^2b<0,\vspace*{-12pt} $$

$$ B-C=\gamma \left(b+1\right)>0,\kern1em B-D=1-C=\gamma, \kern0.75em 1-D=-\gamma \left(b-1\right)<0, $$

and E ₁ > 0 is equivalent to Dμ = (1 − γ + γb)μ < 1. Let

$$ {\mu}_1=\sqrt{\frac{C}{BD}},{\mu}_2=\frac{1+C}{B+D}=\frac{2-\gamma }{2-\gamma +2\gamma b},{\mu}_3=\frac{1}{D}=\frac{1}{1-\gamma +\gamma b}, $$

then μ ₁ < μ ₂ < μ ₃ because

$$ {\mu}_2^2-{\mu}_1^2=\frac{\gamma^3b\left(B+C\right)}{BD{\left(B+D\right)}^2}>0,{\mu}_3-{\mu}_2=\frac{\gamma \left(D+1\right)}{D\left(B+D\right)}>0. $$

By direct calculation, we have

$$ {E}_2=F\left(\mu \right)-\left(1-\cos \theta \right)G\left(\mu \right), \vspace*{-12pt} $$

$$ F\left(\mu \right)=\gamma D^2\left(1-\mu \right){\left({\mu}_{3}-\mu \right)}^2\left(1+C+\mu \left(B+D\right)\right)\ge 0, \vspace*{-12pt} $$

$$ G\left(\mu \right)=2\mu \left( BD{\mu}^2-C\right)\left(B- CD\right)=2\gamma \mu \left(B+C\right) BD\left({\mu}^2-{\mu}_1^2\right). $$

Let $\mathit{H}\left(\mathit{\mu }\right)=\mathit{F}\left(\mathit{\mu }\right)/\mathit{G}\left(\mathit{\mu }\right)$, then $\mathit{H}\left({\mathit{\mu }}_2\right)=2,\mathit{H}\left({\mathit{\mu }}_3\right)=0.$ Differentiating $\mathit{H}$ with respect to $\mathit{\mu }$, we have

$$ {H}_{\mu }=\frac{\left({\mu}_3^2-{\mu}^2\right)K\left(\mu \right)}{2{\mu}^2 B^2{\left({\mu}^2-{\mu}_1^2\right)}^2\left(B+C\right)}, \vspace*{-6pt} $$

$$ K\left(\mu \right)\equiv BD\left(B+D\right){\mu}^4-\left(3D- BC\right)\left(B+C\right){\mu}^2+C\left(1+C\right). $$

Since $\mathit{G}\left(\mathit{\mu }\right)$has the same sign as that of $\mathit{\mu }-{\mathit{\mu }}_1$, ${\mathit{E}}_2>0$ is equivalent to either $\mathit{\mu }\le {\mathit{\mu }}_1$ or ${\mathit{\mu }}_1<\mathit{\mu }$, $\mathit{H}\left(\mathit{\mu }\right)>1-cos\mathit{𝜃}.\mathit{K}\left(\mathit{\mu }\right)$is a quadratic expression of${\mathit{\mu }}^2$and

$$ K\left({\mu}_1\right)=-2b{\gamma}^2C\left(B+C\right)/(BD)<0,\kern0.75em K\left({\mu}_3\right)=-\gamma \left(b-1\right)P/{D}^3<0, $$

where $P\equiv \left(\mathit{B}+\mathit{C}\right)\left(\mathit{B}+2\mathit{D}+\mathit{CD}\right)$. Thus, $\mathit{H}\left(\mathit{\mu }\right)$is decreasing in ${\mathit{\mu }}_1<\mathit{\mu }<{\mathit{\mu }}_3$. Hence, there exists a unique$\mathit{\mu }$such that${\mathit{\mu }}_2\le \mathit{\mu }\le {\mathit{\mu }}_3$ and $\mathit{H}\left(\mathit{\mu }\right)=1-cos\mathit{𝜃}$. Let${\mathit{\mu }}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)$be this unique$\mathit{\mu }$, then (4.53) is equivalent to $\mathit{\mu }<{\mathit{\mu }}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)$.

(iv) From the above results, for 0 < θ < θ′ ≤ π we have

$$ H\left({\mu}^{\ast}\left(\theta, \gamma, b\right)\right)=1-\cos \theta <1-\cos {\theta}^{\prime}, $$

which implies ${\mathit{\mu }}^{{}_{\ast }}\left({\mathit{𝜃}}^{\prime },\mathit{\gamma },\mathit{b}\right)<{\mathit{\mu }}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right).$ The latter part of the statement directly follows from $\mathit{H}\left({\mathit{\mu }}_2\right)=2,\mathit{H}\left({\mathit{\mu }}_3\right)=0$.

(v) This statement directly follows from statement (iii).

(vi) Assume γ < γ′ < 1. Differentiating H with respect to γ, we have

$$ H_{\gamma}=\frac{\left(1-{\mu}^2\right)\left({\mu}_3^2-{\mu}^2\right)L\left(\mu \right)}{2\mu {\left(B+C\right)}^2 B^2\left({\mu}^2-{\mu}_1^2\right)},L\left(\mu \right)\equiv{P}_1-{\mu}^2{P}_2, \vspace*{-16pt}$$

$$ {P}_1=2\left(b-2\right)-\left(b-1\right)\gamma \left(4-\gamma \right), \vspace*{-16pt} $$

$$ {P}_2=2\left(b-2\right)+4\gamma {\left(b-1\right)}^2+{\gamma}^2\left(b-1\right)\left(1+2b\left(b-1\right)\right)>0, \vspace*{2pt} $$

Note that the sign of ${H}_{\gamma }$ is the same as that of L(μ) for μ ₂ < μ < μ ₃. From

$$ L\left({\mu}_3\right)<L\left(\mu \right)<L\left({\mu}_2\right)=-\frac{2\gamma b\left(B+C\right)\left({\left(\gamma -2\right)}^2+\gamma b\left(4-\gamma \right)\right)}{{\left(B+D\right)}^2}<0, \vspace*{2pt}$$

we have ${H}_{\gamma }<0$ for μ ₂ < μ < μ ₃. Assume $\mu = \mu^\ast (\theta, \gamma,b)$ and $0 \leq \theta < 2\pi$. If $\mu_3 (\gamma^\prime) =1/(1-\gamma^\prime+ \gamma^{\prime} b) < \mu$, then $\mu^\ast (\theta, \gamma^\prime, b) \leq \mu^\ast (0, \gamma^\prime, b) \leq \mu^\ast (\theta, \gamma, b)$. Otherwise, since

$$ 1-\cos \theta =H\left(\mu, \gamma, b\right)>H\left(\mu, \gamma^\prime, b\right), $$

we have μ ^∗(θ, γ, b) = μ > μ ^∗(θ, γ′, b), and this also holds for θ = 0.

(vii) Differentiating H with respect b, we have

$$ {H}_{b}=\frac{\gamma \left(1-{\mu}^2\right)\left({\mu}_3^2-{\mu}^2\right)M(\mu)}{2\mu {\left(B+C\right)}^2 B^2\left({\mu}^2-{\mu}_1^2\right)},M\left(\mu \right)\equiv{Q}_1-{\mu}^2{Q}_2,\vspace*{-8pt} $$

$$ Q_{1} \equiv \left(\gamma -1\right)\left(\gamma -2\right),\kern0.5em Q_{2} \equiv 2 + (1+4b)\gamma + (2b^2 + 2b -3)\gamma^2 + (b-1)^2\gamma^3. \vspace*{2pt} $$

Considering that

$$ M\left({\mu}_2\right)=-\frac{\gamma(2-\gamma)(B+C){\left({\left(\gamma -2\right)}^2+\gamma b\left(4-\gamma \right)\right)}^2}{(B+D)^2}<0, $$

we can apply the same argument as the case of (vi).

6. Proof of Theorem 6

(i) B, C, and D are the same as those in the proof of the previous theorem, and let $\overline{\phi}_{m}\left(\mu, \theta, \tau, b\right)$ be the dominant root of the equation:

$$ {\phi}^2-{m}^{-1}\left( B\mu {\mathrm{e}}^{\mathrm{i}\theta }+C\right)\phi +{m}^{-2} D\mu {\mathrm{e}}^{\mathrm{i}\theta }=0. $$

Then, it follows that $\overline{\psi}\left(\mu, \theta, \gamma, b\right)=m\overline{\phi}_m\left(\mu, \theta, \gamma, b\right).$ Let

$$ \kern0.5em {\psi}_0\equiv\sqrt{\frac{CD}{B}},\enspace {\mu}_0\equiv\frac{C}{B}=\frac{1-\gamma }{1+ b\gamma},\kern0.37em {\mu}_{\mathrm{I}}\equiv m \sqrt{\frac{C}{BD}}, \vspace*{-12pt}$$

$$ {\mu}_{\mathrm{I}\mathrm{I}}\equiv\frac{m\left(m+C\right)}{mB+D},\enspace {\mu}_{\mathrm{III}} \equiv \frac{m^2}{D},\enspace {\mu}_{\mathrm{IV}}\equiv\frac{m\left(m-C\right)}{mB-D} $$

As before, by Schur’s condition, ${\overline{\mathit{\varphi }}}_m\left(\mathit{\mu },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)<1$ holds if and only if

$$ \begin{aligned}&\mu <{\mu}_{\mathrm{III}},\kern0.5em {F}^m\left(\mu \right)-\left(1-\mathit{\cos}\theta \right){G}^m\left(\mu \right)>0,\\& {G}^m\left(\mu \right)\equiv 2\mu {m}^{-6}{B}^2D\left({m}^2-{\psi}_0^2\right)\left({\mu}^2-{\mu}_{\mathrm{I}}^2\right),\\&{F}^m\left(\mu \right)\equiv {m}^{-8}{D}^2\left({m}^2{B}^2-{D}^2\right){\left({\mu}_{\mathrm{I}\mathrm{I}\mathrm{I}}-\mu \right)}^2\left({\mu}_{\mathrm{I}\mathrm{I}}+\mu \right)\left({\mu}_{\mathrm{I}\mathrm{V}}-\mu \right)\end{aligned} $$

(4.54)

By the assumption $\mathit{\mu }\ge {\mathit{\mu }}_0$, it is sufficient to consider the case $\mathit{m}\ge \sqrt{{\mathit{\mu }}_0\mathit{D}}={\mathit{\psi }}_0$.Assume $\mathit{m}>{\mathit{\psi }}_0$ and let $\mathit{r}\left(\mathit{m}\right)\equiv B{\mathit{m}}^2-\mathit{CD},s\left(m\right)\equiv m^2{B}^2-D^2$, then from

$$ {\mu}_{\mathrm{I}}^2-{\mu}_0^2=\frac{Cr(m)}{BD}>0,\kern0.5em {\mu}_{\mathrm{I}\mathrm{I}}^2-{\mu}_{\mathrm{I}}^2=\frac{m^2{\gamma}^2 br(m)}{BD{\left( mB+D\right)}^2}>0,\vspace*{-6pt} $$

$$ {\mu}_{\mathrm{II}\mathrm{I}}-{\mu}_{\mathrm{II}}=\frac{m r(m)}{D\left( mB+D\right)}>0,\kern0.5em {\mu}_{\mathrm{IV}}-{\mu}_{\mathrm{II}}=\frac{2m^2{\gamma}^2b}{s(m)}, \vspace*{-6pt} $$

$$ {\mu}_{\mathrm{IV}}^2-{\mu}_{\mathrm{III}}^2=-\frac{m^2 r(m)q(m)}{D^2{\left( mB-D\right)}^2},\kern0.5em q(m)\equiv {m}^2B-2 mD+ \mathit{CD}, $$

we have ${\mathit{\mu }}_0<{\mathit{\mu }}_{\mathrm{I}}<{\mathit{\mu }}_{\mathrm{II}}<{\mathit{\mu }}_{\mathrm{III}}$. If $\mathit{\mu }\le {\mathit{\mu }}_{\mathrm{I}},\mathit{mB}>\mathit{D}$, then $\mathit{\mu }<{\mathit{\mu }}_{\mathrm{I}}<{\mathit{\mu }}_{\mathrm{II}}<{\mathit{\mu }}_{\mathrm{IV}}$. If $\mathit{\mu }\le {\mathit{\mu }}_{\mathrm{I}},\mathit{mB}<\mathit{D}$, then ${\mathit{\mu }}_{\mathrm{IV}}<0$. Hence, (4.54) holds when $\mathit{\mu }\le {\mathit{\mu }}_{\mathrm{I}}$.

Assume $\mathit{\mu }>{\mathit{\mu }}_{\mathrm{I}}$, then ${\mathit{G}}^{\mathit{m}}\left(\mathit{\mu }\right)>0.$ Since$\mathit{q}\left(0\right)>0$ and $\mathit{q}\left(\mathit{D}/\mathit{B}\right)=-{\mathit{\gamma }}^2\mathit{bD}/\mathit{B}<0$, equation$\mathit{q}\left(\mathit{m}\right)=0$has two positive roots. Let ${\mathit{\psi }}_{\mathrm{I}}$ be its larger root, then $\mathit{D}/\mathit{B}<1<{\mathit{\psi }}_{\mathrm{I}}$ because $\mathit{q}\left(1\right)=-\left(\mathit{b}-1\right){\mathit{\gamma }}^2<0$. Observe that $\mathit{q}\left({\mathit{\psi }}_0\right)=2\mathit{D}\left(\mathit{C}-{\mathit{\psi }}_0\right)<0$; ${\mathit{\mu }}_{\mathrm{III}}<\left|{\mathit{\mu }}_{\mathrm{IV}}\right|$ for ${\mathit{\psi }}_0<\mathit{m}<{\mathit{\psi }}_{\mathrm{I}}$; and ${\mathit{\mu }}_{\mathrm{II}}<{\mathit{\mu }}_{\mathrm{VI}}<{\mathit{\mu }}_{\mathrm{III}}$ for ${\mathit{\psi }}_{\mathrm{I}}<\mathit{m}.$ Let ${\mathit{H}}^{\mathit{m}}\left(\mathit{\mu }\right)={\mathit{F}}^{\mathit{m}}\left(\mathit{\mu }\right)/{\mathit{G}}^{\mathit{m}}\left(\mathit{\mu }\right),$ then

$$ {H}^m\left({\mu}_{\mathrm{II}}\right)=2,\kern0.75em {H}^m\left({\mu}_{\mathrm{II}\mathrm{I}}\right)={H}^m\left({\mu}_{\mathrm{VI}}\right)=0. $$

Differentiating ${\mathit{H}}^{\mathit{m}}$with respect to $\mathit{\mu }$, we have

$$ {H}_{\mu}^m=\frac{\left({\mu}_{\mathrm{I}\mathrm{II}}^2-{\mu}^2\right)K\left(\mu \right)}{2{m}^2{\mu}^2{B}^2{\left({\mu}^2-{\mu}_{\mathrm{I}}^2\right)}^2r(m)}, $$

$$ K\left(\mu \right)\equiv BD s(m) {\mu}^4-{m}^2\left(3D- BC\right)r(m){\mu}^2+{m}^4C\left({m}^2-{C}^2\right). $$

Let $\mathit{p}\left(\mathit{m}\right)\equiv {\mathit{m}}^2\mathit{B}+2\mathit{mD}+\mathit{CD}$. $\mathit{K}\left(\mathit{\mu }\right)$is a quadratic expression of${\mathit{\mu }}^2$, and

$$ K\left({\mu}_{\mathrm{I}}\right)=-2b{\gamma}^2{m}^4 Cr(m)/(BD),\kern1em {K}\left({\mu}_{\mathrm{I}\mathrm{II}}\right)={m}^4 r(m)p(m)q(m)/{D}^3, \vspace*{-12pt} $$

$$ {K}\left({\mu}_{\mathrm{IV}}\right)=-2b{\gamma}^2{m}^4 \left(m-C\right)r(m)q(m)/{\left( mB-D\right)}^3. $$

Hence, if ${\mathit{\psi }}_0<\mathit{m}<{\mathit{\psi }}_{\mathrm{I}},$ then ${\mathit{\mu }}_{\mathrm{I}}<\mathit{\mu }<{\mathit{\mu }}_{\mathrm{III}}<\left|{\mathit{\mu }}_{\mathrm{IV}}\right|$ and ${\mathit{K}}_{\mathit{m}}\left(\mathit{\mu }\right)<0$in ${\mathit{\mu }}_{\mathrm{I}}<\mathit{\mu }<{\mathit{\mu }}_{\mathrm{III}}$; if ${\mathit{\psi }}_{\mathrm{I}}<\mathit{m}$, then ${\mathit{\mu }}_{\mathrm{I}}<\mathit{\mu }<{\mathit{\mu }}_{\mathrm{IV}}<{\mathit{\mu }}_{\mathrm{III}}$and ${\mathit{K}}_{\mathit{m}}\left(\mathit{\mu }\right)<0$in ${\mathit{\mu }}_{\mathrm{I}}<\mathit{\mu }<{\mathit{\mu }}_{\mathrm{IV}}$. Thus, ${\mathit{H}}_{\mathit{\mu }}^{\mathit{m}}<0$ in ${\mathit{\mu }}_{\mathrm{I}}<\mathit{\mu }<min\left\{{\mathit{\mu }}_{\mathrm{III}},\left|{\mathit{\mu }}_{\mathrm{IV}}\right|\right\}$, so there exists a unique$\mathit{\mu }$such that

$$ {\mu}_{\mathrm{II}}\le \mu \le \min \left\{{\mu}_{\mathrm{II}\mathrm{I}},|{\mu}_{\mathrm{IV}}|\right\},{H}^m\left(\mu \right)=1-\cos \theta . $$

Let us denote this unique$\mathit{\mu }$by${\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)$, then, it satisfies

$$ {\mu}_m^{\ast}\left(0,\gamma, b\right)=\left\{\begin{array}{l}{\mu}_{\mathrm{I}\mathrm{I}\mathrm{I}}\kern1em {\psi}_0<m\le {\psi}_{\mathrm{I}},\\ {\mu}_{\mathrm{I}\mathrm{V}}\kern1em {\psi}_{\mathrm{I}}<m\end{array}\right.,\kern0.5em {\mu}_m^{\ast}\left(\pi, \gamma, b\right)={\mu}_{\mathrm{I}\mathrm{I}}, $$

and (4.54) is simplified into $\mathit{\mu }<{\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right).$

Assume ${\mu }_0<\mathit{\mu }<{\mathit{\mu }}^{\prime }<1$. Let $m=\overline{\mathit{\psi }}\left({\mathit{\mu }}^{\prime },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)$ then $\mathit{m}>{\mathit{\psi }}_0\left(\mathit{\gamma },\mathit{b}\right)$, hence ${\mathit{\mu }}^{\prime }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)$, which implies $\overline{\mathit{\psi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)<\mathit{m}=\overline{\mathit{\psi }}\left({\mathit{\mu }}^{\prime },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right).$ ³²

(ii) If $\mathit{m}={\mathit{\psi }}_0$, then ${\mathit{\mu }}_0={\mathit{\mu }}_{\mathrm{I}}={\mathit{\mu }}_{\mathrm{II}}={\mathit{\mu }}_{\mathrm{III}}$ and $\overline{\mathit{\psi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)<{\mathit{\psi }}_0$is equivalent to $\mathit{\mu }<{\mathit{\mu }}_0$, which implies that $\psi ={\mathit{\psi }}_0$holds irrespective of $\mathit{𝜃}$when $\mathit{\mu }={\mathit{\mu }}_0$.

Assume $\mu_0 < \mu, 0\leq \theta < \theta^\prime \leq \pi$, and $\psi\left(\mu, {\theta}^{\prime},\gamma, b\right)=m$.

Then, it follows that

$$ {H}^m\left(\mu, \gamma, b\right)={H}^{m}\left(\mu_m^\ast \left(\mathit{\theta^\prime },\mathit{\gamma },\mathit{b}\right), \gamma, b\right) = 1-\cos {\theta}^{\prime }>1-\cos \theta, $$

which implies $\overline{\phi}\left(\mu, \theta, \gamma, b\right)<m=\overline{\phi}\left(\mu, {\theta}^{\prime},\gamma, b\right).$

(iii) Differentiating ${\mathit{H}}^{\mathit{m}}$ with respect $\mathrm{\gamma }$, we have

$$ {H}_{\gamma}^m=-\frac{\left({\mu}_{\mathrm{I}\mathrm{II}}^2-{\mu}^2\right)L\left(\mu \right)}{2\mu {m}^2{B}^2{r}^2(m){\left({\mu}^2-{\mu}_{\mathrm{I}}^2\right)}^2},\kern0.75em L\left(\mu \right)\equiv {S}_1{\mu}^4+2{S}_2{m}^2{\mu}^2+{S}_3{m}^4,\vspace*{-6pt} $$

$$ {S}_1\equiv \left(b-1\right){B}^4{m}^4-2B{D}^2\left( bB-1\right){m}^2+{D}^3\left( bB-C\right), \vspace*{-12pt} $$

$$ {S}_2\equiv - B\Big(B\left(b-1\right)-\gamma b\Big)\ {m}^4+ BC\left(b-1\right)\left(3D- BC\right){m}^2-C{D}^2\left(b-C\right),\vspace*{-12pt} $$

$$ {S}_3\equiv \left( bC-B\right){m}^4-2C\left( bC-D\right){m}^2+{C}^4\left(b-1\right). $$

Note that $\mathit{L}\left(\mathit{\mu }\right)$is a quadratic expression of ${\mathit{\mu }}^2$. Since

$$ L\left({\mu}_{\mathrm{II}}\right)=-2{b}^2{\gamma}^2{m}^4{\left(1+m\right)}^2{r}^2(m)p(m)/{\left( mB+D\right)}^4,\vspace*{-10pt} $$

$$ \vspace*{-12pt}L\left({\mu}_{\mathrm{III}}\right)=\left(b-1\right){m}^4 {r}^2(m)p(m)q(m)/{D}^4, $$

$$ L\left({\mu}_{\mathrm{IV}}\right)=-2{b}^2{\gamma}^2{m}^4{\left(m-1\right)}^2{r}^2(m)q(m)/{\left( mB-D\right)}^4, $$

${\mathit{H}}_{\mathit{\gamma }}^{\mathit{m}}<0$ holds for ${\mathit{\mu }}_{\mathrm{II}}<\mathit{\mu }<min\left\{{\mathit{\mu }}_{\mathrm{III}},\left|{\mathit{\mu }}_{\mathrm{IV}}\right|\right\}$. Assume ${\mathit{\mu }}_0\left(\mathit{\gamma },\mathit{b}\right)\le \mathit{\mu },\mathit{\gamma }<{\mathit{\gamma }}^{\prime }\le 1$, and let $\overline{\mathit{\psi }}\left(\mathit{\mu },\mathit{𝜃},{\mathit{\gamma }}^{\prime },\mathit{b}\right)=\mathit{m}$, that is $\mathit{\mu }={\mathit{\mu }}_{\mathit{m}}^{{}_{\ast }}\left(\mathit{𝜃},{\mathit{\gamma }}^{\prime },\mathit{b}\right)$. Then, applying the same argument as in the proof of Theorem 5(vi), we have

$$ 1-\cos \theta ={H}^m\left(\mu, {\gamma}^{\prime },b\right)<{H}^m\left(\mu, \gamma, b\right), $$

which implies $\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},\mathit{\gamma },\mathit{b}\right)<\mathit{m}=\overline{\mathit{\varphi }}\left(\mathit{\mu },\mathit{𝜃},{\mathit{\gamma }}^{\prime },\mathit{b}\right).$

(iv) Differentiation H ^m with respect to b, we have

$$ {H}_b^{m}=\frac{-\gamma \left({\mu}_{\mathrm{I}\mathrm{II}}^2-{\mu}^2\right){M}_m\left(\mu \right)}{2{\mu}{m}^2{B}^2 r^2(m) {\left({\mu}^2-{\mu}_{\mathrm{I}}^2\right)}^2},\kern0.5em {M}_m\left(\mu \right) \equiv {T}_1{\mu}^4+{T}_2{m}^2{\mu}^2+{T}_3{m}^4, \vspace*{-6pt} $$

$$ {T}_1 \equiv {B}^4{m}^4+{BD}^2\left(\gamma B-2D\right){m}^2+{CD}^3\left(B+\gamma \right)\vspace*{-12pt} $$

$$ {T}_2 \equiv -B\left(2D+\gamma B\right){m}^4+2 BC\left(3D- BC\right){m}^2-{C}^2{D}^2\left(2+\gamma \right),\vspace*{-12pt} $$

$$ {T}_3 \equiv {Cm}^4-{C}^2\left(C+1\right){m}^2+{C}^4. $$

Note that $\mathit{M}\left(\mathit{\mu }\right)$is a quadratic expression of ${\mathit{\mu }}^2$. From

$$ M\left({\mu}_{\mathrm{II}}\right)=-2b{\gamma}^2{m}^4\left(1+m\right)\left(m+C\right)p(m){r}^2(m)/{\left( mB+D\right)}^4<0,\vspace*{-12pt} $$

$$ M\left({\mu}_{\mathrm{III}}\right)= {m}^4{r}^2(m)p(m)q(m)/{D}^4,\vspace*{-12pt} $$

$$ M\left({\mu}_{\mathrm{IV}}\right)=-2b{\gamma}^2{m}^4\left(m-1\right)\left(m-C\right){r}^2(m)q(m)/{\left( mB-C\right)}^4, $$

we have ${\mathit{H}}_b^{\mathit{m}}<0$for ${\mathit{\mu }}_{\mathrm{II}}\le \mathit{\mu }<min\left\{{\mathit{\mu }}_{\mathrm{III}},\left|{\mathit{\mu }}_{\mathrm{IV}}\right|\right\}$. Thus, we can repeat a similar argument to the case of (iii).

7. Proof of Theorem 7

(i) Let Ψ be the transitive matrix of FGMA and ϕ be one of its eigenvalues such that |ϕ| ≥ 1. Then, like the case of GMA, ϕ is also an eigenvalue of QA, where Q is an n × n diagonal matrix whose i-th diagonal element is

$$ {q}_i=\frac{1}{1+\dots +{r}_i^{\tau_i-1}}\frac{\phi -1}{\phi -{r}_i}\left(1-\frac{r^{\tau_i}}{\phi^{\tau_i}}\right){b}_i+1. $$

Let ψ = ϕ ⁻¹, then

$$ \frac{\phi -1}{\phi -{r}_i}\left(1-\frac{r^{\tau_i}}{\phi^{\tau_i}}\right)=\left(1-\psi \right)\left(1+ r\psi +\dots +{\left( r\psi \right)}^{\tau_i-1}\right). $$

Let us denote this polynomial of ψ by $\mathit{F}\left(\mathit{\psi }\right)$. Since $\mathit{F}\left(\mathit{\psi }\right)$is a complex integer expression, for any complex number $\mathit{\alpha }$and$\mathit{𝜖}>0$, there exists a complex number $\mathit{\beta }$such that $\left|\mathit{\beta }-\mathit{\alpha }\right|<\mathit{𝜖},\left|\mathit{F}\left(\mathit{\beta }\right)\right|>\left|\mathit{F}\left(\mathit{\alpha }\right)\right|.$ ³³ Therefore, under the constraint $\left|\mathit{\varphi }\right|\le 1$, $\mathit{F}\left(\mathit{\psi }\right)=\mathit{F}\left(1/\mathit{\varphi }\right)$takes the maximum value at a certain $\mathit{\varphi }$satisfying |$\mathit{\varphi }\mid =1$. In this case, observing

$$ \left|\frac{\phi -1}{\phi -{r}_i}\right|\le \frac{2}{1+{r}_i},\kern0.5em \left|1-\frac{r^{\tau_i}}{\phi^{\tau_i}}\right|\le 1+{r}_i^{\tau }, $$

we have $\mathit{F}\left(1/\mathit{\varphi }\right)\le 2\left(1+{\mathit{r}}_{\mathit{i}}^{\mathit{\tau }}\right)/\left(1+{\mathit{r}}_{\mathit{i}}\right)$. Consequently,

$$ \left|{{q}}_{\mathrm{i}}\right|\le \frac{2\left(1+{{r}}^{\tau_{{i}}}\right){{b}}_{{i}}}{\left(1+\dots +{{r}}_{{i}}^{\tau_{{i}}-1}\right)\left(1+{{r}}_{{i}}\right)}+1=\frac{2\left(1+{{r}}^{2}+\dots +{{r}}_{{i}}^{2\left({\tau}_{{i}}-1\right)}\right)}{{\left(1+\dots +{{r}}_{{i}}^{\tau_{{i}}-1}\right)}^2}{{b}}_{{i}}+1 $$

which implies $1\le \mathit{\rho }\left(\mathit{QA}\right)\le {\mathit{\lambda }}^{\mathrm{F}}\left({\mathit{Q}}^{+}\mathit{A}\right)\le {\mathit{\lambda }}^{\mathrm{F}}\left(\left(2\mathit{HB}+\mathit{I}\right)\mathit{A}\right).$

(ii) This can be shown in a similar manner to the proof of Theorem 4(ii).

8. Proof of Theorem 8

(i) By definition, it follows that

$$ \eta \left(\tau, {r}\right)-\eta \left(\tau +1,{r}\right)=\frac{2{{r}}^{\tau}}{\left(1+{r}\right)\left({\sum}_{\upsilon =0}^{\tau -1}{{r}}^{\upsilon}\right)\left({\sum}_{\upsilon =0}^{\tau}{{r}}^{\upsilon}\right)}>0 $$

(ii) Differentiating $\mathrm{\eta }\left(\mathit{\tau },\mathit{r}\right)$with respect to $\mathit{r}$, we have

$$ {\eta}_r\left(\tau, r\right)=\frac{2f\left(\tau, r\right)}{{\left(1+r\right)}^2{\left(1{-}r\right)}^2{\left({\sum}_{\upsilon =0}^{\tau -1}{r}^{\upsilon}\right)}^2},\kern0.5em f\left(\tau, r\right)\equiv {r}^{2\tau }{-}1+\tau {r}^{\tau -1}\!\left(\!1{-}{r}^2\right). $$

Since $\mathit{f}\left(1,\mathit{r}\right)=0$and

$$ {f}\left(\tau +1,{r}\right)-{f}\left(\tau, {r}\right)={{r}}^{\tau -1}\left({r}+1\right){\left({r}-1\right)}^3{\sum}_{\upsilon =1}^{\uptau}\upsilon {{r}}^{\tau -\upsilon}, $$

${\mathrm{\eta }}_{\mathit{r}}\left(\mathit{\tau },\mathit{r}\right)$ can be written as:

$$ {\eta}_r\left(\tau, r\right)=\frac{2\left(r-1\right){\sum}_{\upsilon =1}^{\tau -1}{r}^{\upsilon -1}{\sum}_{u=1}^{\upsilon}u {r}^{\upsilon -u}}{\left(1+r\right){\left({\sum}_{\upsilon =0}^{\tau -1}{r}^{\upsilon}\right)}^2}. $$

9. Proof of Theorem 9

(i) By Schur’s condition, the dominant root of a cubic equation with real coefficient ${\mathit{x}}^3+\mathit{\alpha }{\mathit{x}}^2+\mathit{\beta x}+\mathit{\gamma }=0$is less than unity if and only if

$$ \left|\gamma \right|<1,\kern0.5em {\left(1-{\gamma}^2\right)}^2>{\left|\beta +\alpha \gamma \right|}^2,\kern0.5em {\left|1+\beta \right|}^2>{\left|\alpha +\gamma \right|}^2. $$

Therefore, $\overline{\mathit{\varphi }}\left(2,\mathit{r},\mathit{\mu },0,\mathit{b}\right)<1$ holds if and only if

$$ \begin{aligned}\left| Rr\mu \right|<1, &\kern0.75em {\left(1-{R}^2{r}^2{\mu}^2\right)}^2>{\left(\left(1-r\right) R\mu +r{\mu}^2R\left(R+1\right)\right)}^2, \nonumber \\ & {\left(1+\left(1-r\right) R\mu \right)}^2>{\left(\left(r-1\right)R-1\right)}^2{\mu}^2, \end{aligned} $$

(4.55)

where $\mathit{R}\equiv b/(1+r)$. Let $\mathit{\beta }\left(\mathit{b}\right)\equiv \left(2\mathit{b}+1\right)/\left(\mathit{b}-1\right)$ (do not confuse this $\beta$ with the secondary coefficient of the above equation). Then, the pair of first two inequalities are equivalent to $\mathit{\mu }<{\mathit{\mu }}_1$ for $0<\mathit{r}<\mathit{\beta }$, and

$$ \mu <\nu \left(r,b\right)\equiv \left(1+r\right)\frac{1-r+\sqrt{5{r}^2-6r+1-4r\left(1+r\right)/b}}{2r\left(b\left(r-1\right)-\left(1+r\right)\right)} $$

for $\mathit{\beta }\le \mathit{r}$. The third inequality is equivalent to$\mathit{\mu }<{\mathit{\mu }}_2$ for $\left(\mathit{b}+1\right)/\left(\mathit{b}-1\right)<\mathit{r}$. Since ${\mathit{\mu }}_2<\mathit{\nu }$for $\mathit{\beta }<\mathit{r}$ and ${\mathit{\mu }}_1(\beta ,b)={\mathit{\mu }}_2(\beta ,b)=\mathrm{\nu }(\beta ,b)=3/\left(2\mathit{b}+1\right),$ these conditions are summarized into (4.44).

(ii) By replacing $\mathit{\mu }$ in (4.55) by $-\mathit{\mu }$, we have

$$ \left| Rr\mu \right|<1,\kern0.75em {\left(1-{R}^2{r}^2{\mu}^2\right)}^2>{\left(-\left(1-r\right) R\mu +r{\mu}^2R\left(R+1\right)\right)}^2, \vspace*{-14pt}$$

$$ {\left(1-\left(1-r\right) R\mu \right)}^2>{\left(\left(r-1\right)R-1\right)}^2{\mu}^2. $$

as the necessary and sufficient condition for $\overline{\mathit{\varphi }}\left(2,\mathit{r},\mathit{\mu },\mathit{𝜃},\mathit{b}\right)<1$. (4.45) can be derived from these condition in the similar way to (4.44).