Critical points for finite Fibonacci chains of point delta-interactions and orthogonal polynomials

By Laplace expansion of the determinant Q^τ_n along its last column, one has the recursion relation (12a)

$$\begin{eqnarray} &&Q_n^{\tau } \left(x\right)=2\left(1+x \tau ^{u_n}\right) Q_{n-1}^{\tau }\left(x\right) -Q_{n-2}^{\tau }\left(x\right), \end{eqnarray} \tag{ 12a }$$

$$\begin{eqnarray} &&Q_{-1}^{\tau } \left(x\right) =0,\quad \ Q_0^{\tau }\left(x\right)=1. \end{eqnarray} \tag{ 12b }$$

It is then easy to verify by explicit calculation of Q^τ₁(x) that the recursion relation (12a) together with definitions (12b) corresponds to the determinant of μT⁽ⁿ⁾. (12a) gives for the coefficient of xⁿ in Q^τ_n, 2ⁿ(u₁ + u₂ + ⋅⋅⋅ + u_n) = 2ⁿτ^{⌊(n + 1)/g⌋} where the last equality comes from (A.13). Also (12a) gives for the coefficient of x⁰ in Q^τ_n, n + 1.

The first explicit expressions (n = 1, 2, 3) are

$$\begin{eqnarray} &&Q_1^{\tau } \left(x\right)=2+2\tau x, \end{eqnarray} \tag{ 13a }$$

$$\begin{eqnarray} &&Q_2^{\tau } (x)=3+(4+4 \tau )x+4 \tau x^{2}, \end{eqnarray} \tag{ 13b }$$

$$\begin{eqnarray} &&Q_3^{\tau } (x)=4+(8+12 \tau ) x+(16 \tau +8 \tau ^2) x^2 +8\tau ^2 x^{3}. \end{eqnarray} \tag{ 13c }$$

The polynomials Q^τ_n(x) have been introduced and studied previously [8–10]. In the notations of [9], Q^τ_n(x) is denoted as S^(r)_n( − x) or S_n(2(1 + rx), 2(1 + x)) with r = τ. The viewpoint of polynomials in the two variables x and xτ is of particular interest when looking for a generating function of these polynomials [8–10]. In the present paper, we have tried to use notations similar to the general notation of [17]. The polynomial proportional to Q^τ_n(x), whose coefficient of the highest degree xⁿ is unity, i.e. the corresponding monic polynomials, will be denoted as $\hat{Q}_n^{\tau }\left(x\right)$. It is then easy to show that the monic polynomials are completely determined by

$$\begin{eqnarray} &&\hat{Q}_n^{\tau }(x)=(x +\tau ^{-u_n})\hat{Q}_{n-1}^{\tau }(x)-\lambda _n^{\tau }\hat{Q}_{n-2}^{\tau }(x), \end{eqnarray} \tag{ 14a }$$

$$\begin{eqnarray} &&\hat{Q}_{-1}^{\tau }\left(x\right)=0,\qquad \hat{Q}_0^{\tau }\left(x\right)=1, \end{eqnarray} \tag{ 14b }$$

$$\begin{eqnarray} \lambda _n^{\tau }=\left\lbrace \begin{array}{@{}ll@{}} 1 & {\rm if}\quad n=1, \\\ms \dsty\frac{\tau ^{-(u_{n}+u_{n-1})}}{4} & {\rm if} \quad n>1. \end{array}\right. \end{eqnarray} \tag{ 14c }$$

In fact, λ^τ₁ can be chosen arbitrarily without changing $\hat{Q}_n^{\tau }\left(x\right)$, but the choice λ₁ = 1 will be convenient for subsequent normalization of a measure. The sequence $\tau ^{-u_n}$, {n = 1, 2, ...} is a Fibonacci sequence ABAA... with $A=\frac{1}{\tau }$ and B = 1. The sequence λ^τ_n, n = 2, 3, ... (note that n starts at 2, not 1), is a Fibonacci sequence ABAA⋅⋅⋅ with $A=\frac{1}{4\tau }\frac{1}{4\tau }$ (the juxtaposition, not the product), and $B=\frac{1}{4\tau ^2}$ (see appendix A). It is a theorem known as the Favard theorem [17, 18] that if the coefficient $-\tau ^{-u_n}$ of $-\hat{Q}_{n-1}^{\tau }$ in (14a) is real and λ^τ_n > 0 for n ⩾ 1, the $\hat{Q}_n^{\tau }\left(x\right)$ form an orthogonal polynomial sequence (OPS) with respect to a unique, positive definite functional $\mathcal {L}$,

$$\begin{eqnarray} &&\mathcal {L}(1)=\lambda _1^{\tau }, \end{eqnarray} \tag{ 15 }$$

$$\begin{eqnarray} &&\mathcal {L}\big(\hat{Q}_n^{\tau }(x)\hat{Q}_m^{\tau }(x)\big)=K_n\delta _{mn},\qquad K_m\ne 0, m,n=0,1,2,\ldots , \end{eqnarray} \tag{ 16 }$$

with δ_mn the Kronecker symbol equal to 1 if m = n, 0 otherwise. The present $-\tau ^{-u_n}$ and λ^τ_n clearly satisfy the conditions required in the Favard theorem since τ > 0. As with every OPS, one can associate with the polynomials $\hat{Q}_n^{\tau }\left(x\right)$ other monic polynomials $\hat{R}_n^{\tau }\left(x\right)$, known as associated or numerator polynomials defined by

$$\begin{eqnarray} &&\hat{R}_n^{\tau }(x)=(x+\tau ^{-u_{n+1}})\hat{R}_{n-1}^{\tau }(x)-\lambda _{n+1}^{\tau }\hat{R}_{n-2}^{\tau }(x), \end{eqnarray} \tag{ 17a }$$

$$\begin{eqnarray} &&\hat{R}_{-1}^{\tau }(x)=0,\qquad \hat{R}_0^{\tau }(x)=1, \end{eqnarray} \tag{ 17b }$$

which also constitute an OPS, but relative to a different functional, $\mathcal {L}$'. It will be seen in section 5.1 that both Q^τ_n and R^τ_n are involved in the expression of the transmission coefficient at zero energy. Introducing the matrix [2–12]

$$\begin{equation} T_k^{\tau }=\left( \begin{array}{@{}cc@{}} 2(1+x \tau ^{u_k} ) & -1 \\ 1 & 0 \end{array} \right), \end{equation} \tag{ 18 }$$

it will be of interest to put (12a) and (12b) into the matrix form,

$$\begin{equation} \Phi _n^{\tau }\equiv T_n^{\tau }T_{n-1}^{\tau }\cdots T_1^{\tau }=\left( \begin{array}{@{}cc@{}} Q_n^{\tau }(x) & -R_{n-1}^{\tau }(x) \\\ms Q_{n-1}^{\tau }(x) & -R_{n-2}^{\tau }(x) \end{array} \right), \end{equation} \tag{ 19 }$$

where R^τ_n(x) is a polynomial proportional to $\hat{R}_n^{\tau } \left(x\right)$, defined by

$$\begin{eqnarray} &&R_n^{\tau } \left(x\right)=2\left(1+x \tau ^{u_{n+1}} \right) R_{n-1}^{\tau }\left(x\right) -R_{n-2}^{\tau }\left(x\right), \end{eqnarray} \tag{ 20a }$$

$$\begin{eqnarray} &&R_{-1}^{\tau } \left(x\right)=0,\qquad R_0^{\tau } \left(x\right)=1. \end{eqnarray} \tag{ 20b }$$

The determinant of Φ^τ_n is unity since it is the product of T^τ_k matrices whose determinant is unity. Different relations between the polynomials Q^τ_n and R^τ_n are given in [9] and [11]. Another one, which will be useful in section 5, is

$$\begin{equation} R_{F_{k}-3}^{\tau } \left(x\right)=Q_{F_{k}-3}^{\tau } \left(x\right) \; \mbox{if} \; k\ge 3. \end{equation} \tag{ 21 }$$

A proof of (21) relies on the fact that the Fibonacci words $W_{F_{k}-2}$ of F_k − 2 letters are palindromes. A proof of this palindromic property based on concatenations is given in [20]. Another proof, based on substitutions (1a) and (1b), is given in appendix B. Palindromes are words equal to the reversed words. For example, $W_{F_{3}-2}=\emptyset$, $W_{F_{4}-2}=A$, $W_{F_{5}-2}=ABA$, $W_{F_{6}-2}=ABAABA ,\ldots \,$. The palindromic property can equivalently be formulated algebraically:

$$\begin{equation} u_{F_{k}-1-j}=u_{j},\qquad j=1,\ldots , F_{k}-2. \end{equation} \tag{ 22 }$$

We now prove (21). Introducing the Pauli matrix

$$\begin{equation} \sigma _1=\left( \begin{array}{@{}cc@{}} 0 & 1 \\ 1 & 0 \end{array} \right) \end{equation} \tag{ 23 }$$

and considering (19), (21) is equivalent to Tr(σ₁Φ^τ_{Fk − 2}) = 0 with the symbol Tr for the trace. Due to the palindromic property,

$$\begin{equation} \Phi _{F_{k}-2}^{\tau }=T_1^\tau T_2^\tau \ldots T_j^\tau X T_j^\tau \ldots T_2^\tau T_1^\tau , \end{equation} \tag{ 24 }$$

where X is the identity matrix if F_k − 2 is even and T^τ_i otherwise. Since σ²₁ is the identity matrix,

$$\begin{equation} \mathrm{Tr}\left(\sigma _1\Phi _{F_{k}-2}^{\tau }\right)=\mathrm{Tr}\left(\sigma _1 T_1^\tau \sigma _1 \sigma _1 T_2^\tau \ldots T_j^\tau X T_j^\tau \ldots T_2^\tau T_1^\tau \right). \end{equation} \tag{ 25 }$$

Since σ₁T^τ_mσ₁ = (T^τ_m)⁻¹, by using the invariance of the trace by circular permutations, one obtains

$$\begin{equation} \mathrm{Tr}\left(\sigma _1\Phi _{F_{k}-2}^{\tau }\right)=\mathrm{Tr}\left(\sigma _1 T_2^\tau \ldots T_j^\tau X T_j^\tau \ldots T_2^\tau \right). \end{equation} \tag{ 26 }$$

The process can be iterated up to Tr(σ₁Φ^τ_{Fk − 2}) = Tr(σ₁T^τ_jXT^τ_j). If X is the identity matrix, this reduces to Tr(σ₁Φ^τ_{Fk − 2}) = Tr(σ₁) = 0. If X = T^τ_i, iteration of the process gives Tr(σ₁Φ^τ_{Fk − 2}) = Tr(σ₁T^τ_i) = 0. Thus, (21) is finally proved.

For τ = 1, (19) reduces to

$$\begin{equation} \left( \begin{array}{@{}cc@{}} 2+2 x & -1 \\ 1 & 0 \end{array} \right)^n=\left( \begin{array}{@{}cc@{}} U_n(1+x) & -U_{n-1}(1+x) \\ U_{n-1}(1+x) & -U_{n-2}(1+x) \end{array} \right), \end{equation} \tag{ 27 }$$

with U_n the Chebyshev polynomial of second kind defined by U_n(cos θ) = sin [(n + 1)θ]/sin θ.

We also give defining relations for orthonormal (with respect to the functional $\mathcal {L}$) polynomials q^τ_n(x) proportional to $\hat{Q}_n^{\tau }\left(x\right)$, and orthonormal (with respect to the functional $\mathcal {L}$') polynomials r^τ_n(x) proportional to $\hat{R}_n^{\tau }\left(x\right)$ [17],

$$\begin{equation} q_{n}^{\tau }\left(x\right)= \left(\lambda _1^{\tau }\lambda _2^{\tau }\cdots \lambda _{n+1}^{\tau }\right)^{-1/2}\hat{Q}_{n}^{\tau }(x), \end{equation} \tag{ 28 }$$

$$\begin{equation} r_{n}^{\tau }\left(x\right)= \left(\lambda _2^{\tau }\lambda _3^{\tau }\cdots \lambda _{n+2}^{\tau }\right){}^{-1/2}\hat{R}_{n}^{\tau }(x). \end{equation} \tag{ 29 }$$

The orthonormal polynomials q^τ_n(x) will be used later (60) for numerical investigation of the measure of orthogonality. As a result, these orthonormal polynomials are completely determined by

$$\begin{eqnarray} &&q_n^{\tau }\left(x\right)=\big(\lambda _{n+1}^{\tau }\big)^{-1/2}\big[\big(x+\tau ^{-u_{n}}\big)q_{n-1}^{\tau }(x)-\big(\lambda _n^{\tau }\big) ^{1/2}q_{n-2}^{\tau }(x) \big], \end{eqnarray} \tag{ 30a }$$

$$\begin{eqnarray} &&q_{-1}^{\tau }\left(x\right)=0,\qquad q_0^{\tau }\left(x\right)=\frac{1}{\sqrt{\lambda _1}}, \end{eqnarray} \tag{ 30b }$$

$$\begin{eqnarray} &&r_n^{\tau }\left(x\right)=\left(\lambda _{n+2}^{\tau }\right)^{-1/2}\big[(x+\tau ^{-u_{n+1}})r_{n-1}^{\tau }(x)-\big(\lambda _{n+1}^{\tau }\big) ^{1/2}r_{n-2}^{\tau }(x) \big], \end{eqnarray} \tag{ 31a }$$

$$\begin{eqnarray} &&r_{-1}^{\tau }\left(x\right)=0, \qquad r_0^{\tau }\left(x\right)=\frac{1}{\sqrt{\lambda _2}}. \end{eqnarray} \tag{ 31b }$$

Let us define the function t by

$$\begin{equation} t\left(k\right)=\left\lfloor \frac{k+2}{g}\right\rfloor +\left\lfloor \frac{k+1}{g}\right\rfloor -1. \end{equation} \tag{ 32 }$$

The Christoffel Darboux identities [17, 18] are

$$\begin{eqnarray} \fl \sum _{k=0}^n \hat{Q}_k^{\tau }\left(x\right)\hat{Q}_k^{\tau }\left(y\right) 4^{k} \tau ^{t\left(k\right)}&=&4^{n} \tau ^{t\left(n\right)}\frac{\hat{Q}_{n+1}^{\tau }\left(x\right)\hat{Q}_n^{\tau }\left(y\right)-\hat{Q}_n^{\tau }\left(x\right)\hat{Q}_{n+1}^{\tau }\left(y\right)}{x-y}, \end{eqnarray} \tag{ 33a }$$

$$\begin{eqnarray} \fl \sum _{k=0}^n q_k^{\tau }\left(x\right)q_k^{\tau }\left(y\right)&=&\left(\lambda _{n+2}^{\tau }\right)^{1/2} \frac{q_{n+1}^{\tau }\left(x\right)q_n^{\tau }\left(y\right)-q_n^{\tau }\left(x\right)q_{n+1}^{\tau }\left(y\right)}{x-y}, \end{eqnarray} \tag{ 33b }$$

$$\begin{eqnarray} \fl \sum _{k=0}^n \hat{R}_k^{\tau }\left(x\right)\hat{R}_k^{\tau }\left(y\right) 4^{k+1} \tau ^{t\left(k+1\right)}&=&4^{n+1} \tau ^{t\left(n+1\right)}\frac{\hat{R}_{n+1}^{\tau }\left(x\right)\hat{R}_n^{\tau }\left(y\right)-\hat{R}_n^{\tau }\left(x\right)\hat{R}_{n+1}^{\tau }\left(y\right)}{x-y}, \end{eqnarray} \tag{ 33c }$$

$$\begin{eqnarray} \fl \sum _{k=0}^n r_k^{\tau }\left(x\right)r_k^{\tau }\left(y\right)&=&\left(\lambda _{n+3}^{\tau }\right){}^{1/2} \frac{r_{n+1}^{\tau }\left(x\right)r_n^{\tau }\left(y\right)-r_n^{\tau }\left(x\right)r_{n+1}^{\tau }\left(y\right)}{x-y}. \end{eqnarray} \tag{ 33d }$$

Equations (33a) and (33c) were obtained using (A.18).

Let us consider the continued fractions $C_n=\frac{A_n}{B_n}$ defined by $C_1= b_0+\frac{a_1}{b_1}, C_2=b_0+\frac{a_1}{b_1+\frac{a_2}{b_2}}, C_3=b_0+\frac{a_1}{b_1+\frac{a_2}{b_2+\frac{a_3}{b_3}}},\ldots$ and A_n = b_nA_{n − 1} + a_nA_{n − 2}, A₋₁ = 1, B_n = b_nB_{n − 1} + a_nB_{n − 2}, B₋₁ = 0 with b_i ≠ 0 for i ⩾ 1. Then with b₀ = 0, a₁ = λ₁, and a_{n + 1} = −λ_{n + 1}, b_n = x − c_n for n > 1, we have q^τ_n(x) = B_n, r^τ_n(x) = (λ₁^τ)⁻¹A_{n + 1} (hence the other name numerator polynomial for the associated polynomial). For more details and other properties shared by all orthonormal polynomials, see [17–19].

The so-called Fibonacci operator [2–12] is a discrete one-dimensional Schrödinger-type operator, which occurs in the context of tight binding Hamiltonians. Let |n〉 be an orthonormal basis in some Hilbert space and let H be the Hamiltonian defined by H|n〉 = |n + 1〉 + |n − 1〉 + αu_n|n〉 with u_n defined by (A.7) and α > 0. Then the Schrödinger equation H|ψ〉 = z|ψ〉 with |ψ〉 = Σ_na_n|n〉 gives for the sequence $a_n\in \ell ^2\left(\mathbb {Z}\right)$, a_{n + 1} + a_{n − 1} + αu_na_n = za_n. By comparison with (10) it is clear that −μT^{(n − 1)} has the same matrix elements as H − z with z = 2(1 + x) and α = −2x(τ − 1). Therefore, some results obtained for the Fibonacci operator [2–12, 16] can be translated in the present context. In particular, defining Ξ^τ_n ≡ Φ_Fn^τ and $s_n^{\tau }\equiv \frac{1}{2}\mathrm{Tr}\left(\Xi _n^{\tau }\right)$, one has Ξ^τ_{n + 1} = Ξ_{n − 1}^τΞ^τ_n, s^τ_{n + 1} = 2s_n^τs^τ_{n − 1} − s_{n − 2}^τ, (s^τ_{n + 1})² + (s_n^τ)² + (s^τ_{n − 1})² − 2s_{n + 1}^τs^τ_ns_{n − 1}^τ − 1 = x²( − 1 + τ)².

The zeros of q^τ_n and those of r^τ_n are simple. Throughout this paper, the n zeros x^τ_{n, i}, i = 1, ...n, of q^τ_n will be noted in increasing order:

$$\begin{equation} x_{n,1}^{\tau }<x_{n,2}^{\tau }<\cdots <x_{n,n}^{\tau }. \end{equation} \tag{ 34 }$$

Then we have the interlacing properties [17] x^τ_{n + 1, j} < x_{n, j}^τ < x^τ_{n + 1, j + 1}. The same inequality holds for the zeros y^τ_{n, i} of r^τ_n [17]. Moreover [17], x^τ_{n + 1, j} < y_{n, j}^τ < x^τ_{n + 1, j + 1}.

Subtract Q^τ_{n − 1}(x) from both members of (12a) and sum over n. One obtains

$$\begin{equation} Q_k^{\tau } \left(x\right)=1+Q_{k-1}^{\tau } \left(x\right)+2x\sum _{j=1}^{k} \tau ^{u_{j}} Q_{j-1}^{\tau } \left(x\right). \end{equation} \tag{ 35 }$$

With k = 1, (35) shows that the coefficients of every power of x are positive for the polynomials Q^τ₁(x) since Q^τ₀(x) = 1 (recall that τ > 0). Then, by recursion on k, (35) shows that the coefficients of every power of x are positive for Q^τ_k(x) whatever the value of k. Recalling that the zeros are real, the zeros of a polynomial with positive coefficients cannot be positive. Since Q^τ_k(0) ≠ 0, the zeros are negative.

It is known [15] that μ in (5) should be negative in order that the Hamiltonian supports bound states. What has been done in this section is the proof using only the definition of the polynomials.

The zeros x^τ_{n, i} of q^τ_n are the eigenvalues of a tridiagonal symmetric real matrix M^τ_n of order n whose non-zero elements are [17] $\left(M_n^{\tau }\right)_{j, j}=-\tau ^{-u_{j}},\left(M_n^{\tau }\right)_{j, j+1}=\left(M_n^{\tau }\right)_{j+1, j}=\sqrt{\lambda _{j+1}^{\tau }}$. Indeed, the Laplace expansion along the last column of $\det \left(x\mathbb {I}-M_n^{\tau }\right)$ corresponds to (14a) and (14b). This gives another proof that the zeros are real. For example,

$$\begin{equation} M_4^{\tau }=\left( \begin{array}{@{}cccc@{}} -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} & 0 & 0 \\\ms \dsty\frac{1}{2 \sqrt{\tau }} & -1 & \dsty\frac{1}{2 \sqrt{\tau }} & 0 \\\ms 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \tau } \\\ms 0 & 0 & \dsty\frac{1}{2 \tau } & -\dsty\frac{1}{\tau } \end{array} \right). \end{equation} \tag{ 36 }$$

The zeros y^τ_{n, i} of r^τ_n are the eigenvalues of a tridiagonal symmetric real matrix N^τ_n of order n whose non-zero elements are

$$\begin{equation*} \left(N_n^{\tau }\right)_{j, j}=-\tau ^{-u_{j+1}}, \left(N_n^{\tau }\right)_{j, j+1}=\left(N_n^{\tau }\right)_{j+1, j}=\sqrt{\lambda _{j+2}^{\tau }}. \end{equation*}$$

The important property in the title of this section can be derived along the framework of the Hellmann–Feynman theorem. Let the ith zero x^τ_{n, i} be considered as a function of τ. Let w be a normalized real column eigenvector belonging to the zero eigenvalue of the real symmetric matrix μT⁽ⁿ⁾ (10), and take the total derivative of w^tμT⁽ⁿ⁾w = 0 with respect to τ (w^t is the transpose of w). The sum $\left(\frac{{\rm d}}{{\rm d}\tau }w^{t}\right)\mu T^{\left( n \right)}w+w^{t}\mu T^{\left( n \right)}\left(\frac{{\rm d}}{{\rm d}\tau }w\right)$ yields zero because T⁽ⁿ⁾w is the null vector (another argument is that w^tw = 1 and w is an eigenvector), and it remains

$$\begin{equation} w^{t}\left(\frac{\partial }{\partial x}\mu T^{\left( n \right)}\right)\frac{{\rm d}x}{{\rm d}\tau }w+w^{t}\left(\frac{\partial }{\partial \tau }\mu T^{\left( n \right)}\right)w=0. \end{equation} \tag{ 37 }$$

Now $\frac{\partial }{\partial x}\mu T^{\left( n \right)}$ is a diagonal matrix with diagonal elements equal to 2 or 2τ and therefore $w^{t}\left(\frac{\partial }{\partial x}\mu T^{\left( n \right)}\right)w>0$. Similarly, $\frac{\partial }{\partial \tau }\mu T^{\left( n \right)}$ is a diagonal matrix with diagonal elements equal to 0 or 2x^τ_{n, i} and therefore $w^{t}\left(\frac{\partial }{\partial \tau }\mu T^{\left( n \right)}\right)w<0$. One therefore deduces from (37) that

$$\begin{equation} \frac{{\rm d}}{{\rm d} \tau }x_{n,i}^\tau >0. \end{equation} \tag{ 38 }$$

Two different zeros x^τ_{n, i}, x^τ_{n, j} (i ≠ j) cannot cross since all the zeros are simple.

For the sum Σ^τ_n := ∑_{k = 1}ⁿx^τ_{n, k} of the zeros, i.e. minus the coefficient of x^{n − 1} in $\hat{Q}_n^{\tau }(x)$, one obtains either directly from (14a), or by noting that this sum is the trace of M^τ_n,

$$\begin{equation} \Sigma _n^{\tau }=-\sum _{k=1}^n \tau ^{-u_k}=-\left\lfloor \frac{n+1}{g}\right\rfloor \tau ^{-1}-\left\lfloor \frac{n+1}{g^2}\right\rfloor . \end{equation} \tag{ 39 }$$

The evaluation of the last sum is obtained using (A.13) and (A.14). For other results concerning sums of powers of roots, see [21].

The product Π^τ_n of the zeros can be obtained from the expression of the coefficients of highest power xⁿ and constant term x⁰ indicated above (13a). Therefore,

$$\begin{equation} \Pi _n^\tau =(-1)^n\frac{n+1}{2^n\tau ^{\left\lfloor \frac{n+1}{g}\right\rfloor }}. \end{equation} \tag{ 40 }$$

Note that (40) is also the determinant of M^τ_n.

The sum of the absolute values of matrix elements belonging to the same line of the tridiagonal symmetric matrix M^τ_n can only be one of the following: $\big\lbrace 1+\frac{1}{2\sqrt{\tau }},1+\frac{1}{\sqrt{\tau }},\frac{1}{\tau }+\frac{1}{\sqrt{\tau }},\frac{2+\sqrt{\tau }}{2 \tau },\frac{3+\sqrt{\tau }}{2 \tau }\big\rbrace$. Then by a general theorem [22, 23], the spectral radius of M^τ_n is less than or equal to the maximum of these numbers, and therefore

$$\begin{equation} \left\lbrace \begin{array}{@{}ll@{}} -\dsty\frac{3}{2 \tau }-\dsty\frac{1}{2\sqrt{\tau }}\le x_{n,1}^\tau & {\rm if} \quad \tau \le 1, \\\ms -1-\dsty\frac{1}{\sqrt{\tau }}\le x_{n,1}^\tau & {\rm if} \quad \tau \ge 1.\end{array}\right. \end{equation} \tag{ 41 }$$

Figure 1 shows the zeros as a function of τ for n = 100. The dashed line curve at the bottom represents the lower bound (41). There is only two different diagonal matrix elements for M^τ_n, namely −1/τ and −1. The maximum radius of the Geršgorin disk centered at −1 is $1/\sqrt{\tau }$, and the maximum radius of the Geršgorin disk centered at −1/τ is $(1+\sqrt{\tau })/(2 \tau )$ if τ ⩽ 1 and $1/\sqrt{\tau }$ if τ ⩾ 1. Therefore, for τ ⩽ 1 the union of all Geršgorin disks is a disconnected set if and only if $-\frac{1}{\tau } + \frac{1+\sqrt{\tau }}{2 \tau }<-1-\frac{1}{\sqrt{\tau }}$, i.e. if $\tau <\frac{1}{8} (13-3 \sqrt{17})\simeq 0.078\,8354$. Also, for τ ⩾ 1 the union of all Geršgorin disks is a disconnected set if and only if $-1 +( 1/\sqrt{\tau }) < -(1/\tau )-1/\sqrt{\tau }$, i.e. if $\tau >3+2 \sqrt{2}\simeq 5.828\,43$. It remains to count the number of centers −1 and the number of centers −1/τ to find the number of zeros of Q^τ_n in each component of the disconnected set. Since the diagonal elements of M^τ_n form a Fibonacci word W_n( − 1/τ, −1), (A.13) and (A.14) give the following.

• Case $\tau <\frac{1}{8} (13-3 \sqrt{17})$: in the interval $\big[{-}\frac{1}{\tau }- \frac{1}{2}\big(\frac{1}{\tau }+ \frac{1}{\sqrt{\tau }}\big),-\frac{1}{\tau }+ \frac{1}{2}\big(\frac{1}{\tau }+ \frac{1}{\sqrt{\tau }}\big)\big]$, there are $\big\lfloor \frac{n+1}{g}\big\rfloor$ zeros. In the non-overlapping interval $\big[-1- \frac{1}{\sqrt{\tau }},-1+ \frac{1}{\sqrt{\tau }}\big]$, there are $\big\lfloor \frac{n+1}{g^2}\big\rfloor$ zeros.
• Case $\tau >3+2 \sqrt{2}$: in the interval $\big[{-}\frac{1}{\tau }-\frac{1}{\sqrt{\tau }},-\frac{1}{\tau }+ \frac{1}{\sqrt{\tau }}\big]$, there are $\big\lfloor \frac{n+1}{g}\big\rfloor$ zeros. In the non-overlapping interval $\big[{-}1- \frac{1}{\sqrt{\tau }},-1+ \frac{1}{\sqrt{\tau }}\big]$, there are $\big\lfloor \frac{n+1}{g^2}\big\rfloor$ zeros.

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The case τ = 1 corresponds to an equidistant linear chain. The zeros at τ = 1 are exactly known

$$\begin{equation} x_{n,k}^1=-1+\cos \left(\frac{(n+1-k)\pi }{n+1}\right), \end{equation} \tag{ 42 }$$

with k = 1, ..., n. They are the zeros of the Chebyshev polynomials U_n(1 + x). We shall see in section 4.6.2 that the series of x^τ_{n, i} in integer powers of τ − 1 can be expressed in a compact form and calculated to any finite order k. The first-order perturbation correction is first calculated in the traditional way in section 4.6.1.

4.6.1. First-order correction around τ = 1

For a finite periodic chain, i.e. for τ = 1, the critical points (zeros of Q¹_n or equivalently eigenvalues of M¹_n) are known analytically (42), and the same holds for the eigenvectors of M¹_n. Indeed, let e_{n, k}, k = 1, ..., n, denote the normalized column eigenvector corresponding to the eigenvalue x¹_{n, k} .They are orthogonal as eigenvectors of a real symmetric matrix. It can be verified that e_{n, k} has, within arbitrary global phase factors, the components e^j_{n, k}, j = 1, ..., n:

$$\begin{equation*} e_{n,k}^j=\sqrt{\frac{2}{n+1}}(-1)^{k+1}\sin \left\lbrace \left(\frac{kj}{n+1}\right)\pi \right\rbrace , \end{equation*}$$

$$\begin{equation*} \sum _{k=1}^n e_{n,k}^je_{n,k}^i=\delta _{ji}. \end{equation*}$$

To first order in , one has M^{1 +} _n = M_n¹ + C_n with C_n a tridiagonal real symmetric matrix whose main diagonal elements form a Fibonacci word W_n(, 0), and whose elements of the two other non-zero diagonals form a Fibonacci sequence with $A=-\frac{\epsilon }{4}-\frac{\epsilon }{4}$ (juxtaposition) and $B=-\frac{\epsilon }{2}$ (see appendix A). For example,

$$\begin{equation*} C_4=\left( \begin{array}{@{}cccc@{}} \epsilon & -\dsty\frac{\epsilon }{4} & 0 & 0 \\\ms -\dsty\frac{\epsilon }{4} & 0 & -\dsty\frac{\epsilon }{4} & 0 \\\ms 0 & -\dsty\frac{\epsilon }{4} & \epsilon & -\dsty\frac{\epsilon }{2} \\\ms 0 & 0 & -\dsty\frac{\epsilon }{2} & \epsilon \end{array} \right). \end{equation*}$$

The first-order perturbation theory yields to first order in

$$\begin{equation*} x_{n,i}^{\tau }=x_{n,i}^1+\sum _{k=1}^n \sum _{j=1}^n e_{n,k}^i\left(C_n\right){}_{kj}e_{n,j}^{i}. \end{equation*}$$

The tridiagonal symmetric structure of C_n allows us to break these sums into two parts: one concerning the main diagonal and the other concerning the two other diagonals. In order to minimize the number of summation terms for the main diagonal, we first evaluate it for the case where all diagonal elements are , which gives a contribution of , and then subtract the contribution of only the zero diagonal elements which are replaced by . The largest integer k such that ⌊kg²⌋ ⩽ n is $k= \big\lfloor \frac{n+1}{g^2}\big\rfloor$ (see appendix A). The first part is then

$$\begin{equation*} \epsilon \bigg\lbrace 1-\sum _{k=1}^{ \lfloor \frac{n+1}{g^2}\rfloor } \big(e_{n,\lfloor {kg}^2\rfloor }^i\big)^2\bigg\rbrace . \end{equation*}$$

The same procedure can be applied to the two other diagonals by first considering all elements equal to $-\frac{\epsilon }{4}$ and then correcting for the contribution of the elements equal to $-\frac{\epsilon }{2}$. Since the largest integer k such that ⌊⌊kg²⌋g⌋ ⩽ n is $k=\big\lfloor \frac{\lfloor \frac{n+1}{g}\rfloor +1}{g^{2}}\big\rfloor$ (see appendix A), one obtains

$$\begin{equation*} 2\bigg\lbrace \bigg(-\frac{\epsilon }{4}\sum _{k=1}^{n-1} e_{n,k}^ie_{n,k+1}^i\big)-\frac{\epsilon }{4}\sum _{k=1}^{\lfloor \frac{\lfloor \frac{n}{g}\rfloor +1}{g^2}\rfloor } e_{n,\lfloor \lfloor {kg}^2\rfloor g\rfloor }^ie_{n,\lfloor \lfloor {kg}^2\rfloor g\rfloor +1}^i\bigg\rbrace . \end{equation*}$$

Using the trigonometrical relation,

$$\begin{equation*} \fl \left.-\frac{2}{n+1}\sum _{k=1}^{n-1} \sin \left\lbrace \left(\frac{(k+1)j}{n+1}\right)\pi \right\rbrace \sin \left\lbrace \left(\frac{kj}{n+1}\right)\pi \right\rbrace =\cos \left\lbrace \pi \left(1-\frac{j}{n+1}\right)\right\rbrace \right), \end{equation*}$$

and putting all results together, one finally obtains to first order in

$$\begin{equation} x_{n,i}^{\tau }=-1+\cos \left(\frac{j\pi }{n+1}\right) +c_{1,n,i}\left(\tau -1\right), \end{equation} \tag{ 43 }$$

$$\begin{eqnarray} &&\fl c_{1,n,i}= \left\lbrace \begin{array}{@{}c@{}} 1-\left[\dsty\frac{2}{n+1}\dsty\sum _{k=1}^{ \lfloor \frac{n+1}{g^2}\rfloor } \left(\sin \left(\dsty\frac{\lfloor {kg}^2\rfloor i}{n+1}\pi \right)\right)^2\right]-\dsty\frac{1}{2}\cos \left[ \left(1-\dsty\frac{i}{n+1}\right)\pi \right] \\\ms +\dsty\frac{1}{n+1}\dsty\sum _{k=1}^{\lfloor \frac{\lfloor \frac{n}{g}\rfloor +1}{g^2}\rfloor } \sin \left(\dsty\frac{\lfloor \lfloor {kg}^2\rfloor g\rfloor i}{n+1}\pi \right)\sin \left(\dsty\frac{(\lfloor \lfloor {kg}^2\rfloor g\rfloor +1)i}{n+1}\pi \right)\end{array}\right\rbrace . \end{eqnarray} \tag{ 44 }$$

The presence of the floor function in (44) is at the origin of the onset of fractal aspect as soon as τ departs from unity.

4.6.2. Higher order corrections to the critical point around τ = 1

Let us first give a relation that allows us to calculate the derivative of the zeros x^τ_{n, i} with respect to τ using only the knowledge of the zeros. Taking the total derivative with respect to τ of Q^τ_n(x_{n, i}^τ) = 0, one obtains

$$\begin{equation} \frac{{\rm d}}{{\rm d} \tau }x_{n,i}^\tau =-\frac{\frac{\partial Q_n^\tau }{\partial \tau }}{\frac{\partial Q_n^\tau }{\partial x}}\equiv f\left(\tau ,x_{n,i}^\tau \right), \end{equation} \tag{ 45 }$$

where $\frac{\partial Q_n^\tau }{\partial \tau }$ and $\frac{\partial Q_n^\tau }{\partial x}$ are evaluated at x = x^τ_{n, i}. This is a first-order differential equation satisfied by x^τ_{n, i}. Since the value of x = x¹_{n, i} is known, (45) is also a sufficient relation that also allows us to determine x^τ_{n, i}. Now, the first-order coefficient c_{1, n, i} (44) can also be computed as

$$\begin{equation} c_{1,n,i}=f\left(1,x_{n,i}^1\right). \end{equation} \tag{ 46 }$$

More generally, the coefficients c_{k, n, i} of the power series

$$\begin{equation} x_{n,i}^\tau =\sum _{k=0}^\infty c_{k,n,i}\left(\tau -1\right)^k \end{equation} \tag{ 47 }$$

can be expressed as

$$\begin{equation} c_{k,n,i}=\frac{{\rm d}^{k-1} f\left(\tau ,x_{n,i}^\tau \right)}{k! {\rm d} \tau ^{k-1}}, \end{equation} \tag{ 48 }$$

where the right member is to be evaluated at τ = 1. It is stressed that in (48), the derivatives are the total derivatives with respect to τ. For example,

$$\begin{equation} \frac{{\rm d} f\left(\tau ,x\right)}{{\rm d} \tau }=\frac{\partial f\left(\tau ,x\right)}{\partial x}\frac{{\rm d} x}{{\rm d} \tau }+\frac{\partial f\left(\tau ,x\right)}{\partial \tau }=\frac{\partial f\left(\tau ,x\right)}{\partial x} f\left(\tau ,x\right)+\frac{\partial f\left(\tau ,x\right)}{\partial \tau }. \end{equation} \tag{ 49 }$$

Equation (48) provides a direct algorithm for the computation of the coefficients of the Rayleigh–Schrödinger perturbation series (47). Figure 2 shows the power series (48) truncated with a maximum value of k equal to 5 for the case n = 10.

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There are only two values for the non-zero nondiagonal matrix elements of M⁽ⁿ⁾, namely 1/(2τ) and $1/(2\sqrt{\tau })$. For τ near zero, it is therefore natural to consider as a zero-order starting point the matrix M⁽ⁿ⁾₀ obtained from the matrix M⁽ⁿ⁾ by replacing the nondiagonal matrix elements $1/(2\sqrt{\tau })$ by zero. In the limit n → ∞, the resulting matrix M⁽ⁿ⁾₀ is block diagonal, with block submatrices of order 1 or 2. The submatrices of order 1 have for matrix elements either −1/τ or −1, and the block submatrices of order 2 are all equal to

$$\begin{equation} \left( \begin{array}{@{}cc@{}} -1 /\tau & 1/(2\tau ) \\ 1/(2\tau ) & -1/\tau \end{array} \right), \end{equation} \tag{ 50 }$$

whose eigenvalues are −3/(2τ) and −1/(2τ). If n is such as the block submatrix of order 2 is truncated, it becomes a one-order matrix of matrix element equal to −1/τ. To summarize, this zero-order approximation yields only four zeros, namely

$$\begin{equation} x_1,x_2,x_3,x_4=\frac{-3}{2\tau },\frac{-1}{\tau },\frac{-1}{2\tau },-1. \end{equation} \tag{ 51 }$$

Since this zero-order approximation M⁽ⁿ⁾₀ leads to only four eigenvalues, it remains to give the degrees of degeneracy y_i of each eigenvalue x_i. The number of −1 diagonal elements of M⁽ⁿ⁾₀ is the same as M⁽ⁿ⁾, and therefore $y_4=\big\lfloor \frac{n+1}{g^2}\big\rfloor$ (A.14). The number of elements of the upper diagonal of Mⁿ is n − 1 and using (A.22), $y_{1}=y_{3}=\big\lfloor \frac{\lfloor \frac{n+1}{g}\rfloor }{g^2}\big\rfloor$. The difference between n and the previous degrees gives the degeneracy degree of the remaining value −1/τ: $y_{2}=n-\big\lfloor \frac{n+1}{g^2}\big\rfloor -2\big\lfloor \frac{\lfloor \frac{n+1}{g}\rfloor }{g^2}\big\rfloor$. Figure 3 shows the four zero-order approximation zeros (51) together with the numerical results for x^τ_{n, i} for n = 100. It is seen that x₁, x₂ and x₃ are good zero-order approximations, but x₄ = −1 is a rather poor zero-order approximation. The reason is that for an eigenvalue close to zero, the relative error introduced by a perturbation is of course greater than for eigenvalues with large moduli. The number of exact zeros near each zero-order approximations is equal to the zero-order degree of degeneracy y_i.

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It is of interest to note that the limit τ → 0 maps to another chain model, namely one with uniform spacings, smaller n, but non-uniform couplings.

From the defining relations (14a) and (14b), the limit of the monic polynomial $\hat{Q}_n^{\tau }\left(x\right)$ when τ → ∞ can be determined:

$$\begin{equation} \lim _{\tau \rightarrow \infty } \hat{Q}_n^{\tau }(x)=x^{\lfloor \frac{n+1}{g}\rfloor }(1+x)^{\lfloor \frac{n+1}{g^2}\rfloor } \end{equation} \tag{ 52 }$$

and therefore, among the n zeros, $\big\lfloor \frac{n+1}{g}\big\rfloor$ tend to 0 and $\big\lfloor \frac{n+1}{g^2}\big\rfloor$ tend to −1. Using the same line of arguments as in section 4.7, it is natural for the case τ → ∞, to consider as a zero-order starting point the matrix M⁽ⁿ⁾_∞ obtained from the matrix M⁽ⁿ⁾ by replacing the nondiagonal matrix elements 1/(2τ) by zero. In the limit n → ∞, the resulting matrix M⁽ⁿ⁾_∞ is block diagonal, with two submatrices, one of order 3,

$$\begin{equation} b_3=\left( \begin{array}{@{}ccc@{}} -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} & 0 \\\ms \dsty\frac{1}{2 \sqrt{\tau }} & -1 & \dsty\frac{1}{2 \sqrt{\tau }} \\\ms 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -\dsty\frac{1}{\tau } \end{array} \right), \end{equation} \tag{ 53 }$$

whose eigenvalues are $\big\lbrace {-}\frac{1}{\tau },\frac{-\sqrt{\tau ^2+1}-\tau -1}{2 \tau },\frac{\sqrt{\tau ^2+1}-\tau -1}{2 \tau }\big\rbrace$, and one of order 5

$$\begin{equation} b_5=\left( \begin{array}{@{}ccccc@{}} -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} & 0 & 0 & 0 \\\ms \dsty\frac{1}{2 \sqrt{\tau }} & -1 & \dsty\frac{1}{2 \sqrt{\tau }} & 0 & 0 \\\ms 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} & 0 \\\ms 0 & 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -1 & \dsty\frac{1}{2 \sqrt{\tau }} \\\ms 0 & 0 & 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -\dsty\frac{1}{\tau } \end{array} \right), \end{equation} \tag{ 54 }$$

whose eigenvalues are $\big\lbrace \frac{-\sqrt{\tau ^2+\tau +1}-\tau -1}{2 \tau },\frac{-\sqrt{\tau ^2-\tau +1}-\tau -1}{2 \tau },-\frac{1}{\tau },\frac{\sqrt{\tau ^2-\tau +1}-\tau -1}{2 \tau },\frac{\sqrt{\tau ^2+\tau +1}-\tau -1}{2 \tau }\big\rbrace$. Thus, in the case where the last block diagonal submatrix is not truncated, there are seven eigenvalues, X_i < X_{i + 1}:

$$\begin{eqnarray} &&\fl X_{1\ldots 7} = \frac{-\sqrt{\tau ^2+\tau +1}-\tau -1}{2 \tau }, \frac{-\sqrt{\tau ^2+1}-\tau -1}{2 \tau },\frac{-\sqrt{\tau ^2-\tau +1}-\tau -1}{2 \tau }, \nonumber \\ &&-\frac{1}{\tau }, \frac{\sqrt{\tau ^2-\tau +1}-\tau -1}{2 \tau }, \frac{\sqrt{\tau ^2+1}-\tau -1}{2 \tau },\frac{\sqrt{\tau ^2+\tau +1}-\tau -1}{2 \tau }. \end{eqnarray} \tag{ 55 }$$

The asymptotic behaviors of the X_i are −1 − 3/(4τ), −1 − 1/(2τ), −1 − 1/(4τ), −1/τ, −3/(4τ), −1/(2τ), −1/(4τ). We then have to consider successively all the different cases where n is such that the last block diagonal matrix is truncated. If the last diagonal submatrix (53) is truncated, it becomes

$$\begin{equation} b_2=\left( \begin{array}{@{}cc@{}} -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} \\\ms \dsty\frac{1}{2 \sqrt{\tau }} & -1 \end{array} \right), \end{equation} \tag{ 56 }$$

whose eigenvalues are X₃ and X₅. A further truncation leads to the one-order matrix of eigenvalue X₄.

If the last diagonal submatrix (54) is truncated, it becomes

$$\begin{equation} \left( \begin{array}{@{}cccc@{}} -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} & 0 & 0 \\\ms \dsty\frac{1}{2 \sqrt{\tau }} & -1 & \frac{1}{2 \sqrt{\tau }} & 0 \\\ms 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -\dsty\frac{1}{\tau } & \dsty\frac{1}{2 \sqrt{\tau }} \\\ms 0 & 0 & \dsty\frac{1}{2 \sqrt{\tau }} & -1 \end{array} \right), \end{equation} \tag{ 57 }$$

whose eigenvalues are

$$\begin{eqnarray} &&\fl X_i, i=8,\ldots ,11\nonumber\\ &&\hphantom{\fl X_i, i}=\frac{-\sqrt{2} \sqrt{2 \tau ^2+\sqrt{5} \tau -\tau +2}-2 \tau -2}{4 \tau },\qquad \frac{-\sqrt{2} \sqrt{2 \tau ^2-\sqrt{5} \tau -\tau +2}-2 \tau -2}{4 \tau }, \nonumber \\ &&\hphantom{\fl X_i, i=}\ \frac{\sqrt{2} \sqrt{2 \tau ^2-\sqrt{5} \tau -\tau +2}-2 \tau -2}{4 \tau },\qquad \frac{\sqrt{2} \sqrt{2 \tau ^2+\sqrt{5} \tau -\tau +2}-2 \tau -2}{4 \tau }. \nonumber\\ \end{eqnarray} \tag{ 58 }$$

A further truncation leads to (53) whose eigenvalues are X₂, X₄ and X₆. A further truncation leads to (56) whose eigenvalues are X₃ and X₅.

Let us now consider the degeneracy degree Y_i of each eigenvalues X_i. We consider here only the case where the last block matrix is not truncated. Then the upper diagonal elements (M^τ_n)_{j, j + 1} = 2^{s_{j + 1}/2}, j = 1, ...n − 1, are the Fibonacci word $W_{n-1}\big(\frac{1}{2 \sqrt{\tau }}\frac{1}{2\sqrt{\tau }},\frac{1}{2\tau }\big)$ (see appendix A). The positions of 1/(2τ) in the upper diagonal are ⌊⌊kg²⌋g⌋ with integer k (see appendix A). Thus, n can be expressed as n = ⌊⌊mg²⌋g⌋ with integer m. In that case, m can be expressed as $m=\big\lfloor \frac{\lfloor \frac{n+1}{g}\rfloor +1}{g^2}\big\rfloor$ (see appendix A). The number of 1/(2τ) in the upper diagonal element of M^τ_n is m − 1, and therefore the total number of blocks matrices is m. As a result, the degeneracy degree Y₄ of −1/τ which is an eigenvalue common to both block matrices (53) and (54) is m. Consider the letter $\frac{1}{2 \sqrt{\tau }}\frac{1}{2\sqrt{\tau }}$. Its number in the upper diagonal is equal to the half of n − 1 − (m − 1) where n − 1 is the number of elements in the upper diagonal and m − 1 is the number of 1/(2τ) in that diagonal. The total number of letters in $W_{n-1}\big(\frac{1}{2 \sqrt{\tau }}\frac{1}{2\sqrt{\tau }},\frac{1}{2\tau }\big)$ therefore is equal to $\frac{n-m}{2}+m-1$. Using (A.23), one obtains the number n₅ of block matrices of order 5, $n_5=\big\lfloor \frac{\lfloor \frac{n+m}{2g} \rfloor }{g^2}\big\rfloor =\big\lfloor \frac{\lfloor \frac{\lfloor \lfloor m g^2\rfloor g\rfloor +m}{2g} \rfloor }{g^2}\big\rfloor$. Let us show that $\frac{\lfloor \lfloor m g^2\rfloor g\rfloor +m}{2} =\lfloor m g^2\rfloor$. Using g = 1 + (1/g), this is equivalent to showing that $\lfloor m g^{2}\rfloor =m+\big\lfloor \frac{\lfloor m g^2\rfloor }{g}\big\rfloor$. Using (A.3), one has $\lfloor m g^{2}\rfloor =\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g}\big\rfloor +\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g^2}\big\rfloor$. Therefore, the relation to be proved is equivalent to $m=\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g}\big\rfloor -\big\lfloor \frac{\lfloor m g^2\rfloor }{g}\big\rfloor +\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g^2}\big\rfloor =u_{\lfloor m g^2\rfloor }+\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g^2}\big\rfloor$. Since $u_{\lfloor m g^2\rfloor }=0$, we have to prove that $m=\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g^2}\big\rfloor$. This is true since $\big\lfloor \frac{\lfloor m g^2\rfloor +1}{g^2}\big\rfloor =\big\lfloor m+\frac{1-\lbrace m g^2\rbrace }{g^2}\big\rfloor =m$. Therefore, we have obtained $n_{5}=\big\lfloor \frac{\lfloor \frac{\lfloor m g^2\rfloor }{g}\rfloor }{g^2}\big\rfloor \stackrel{(A.2)}{=}\big\lfloor \frac{\lfloor \frac{\lfloor \lfloor m g^2\rfloor g\rfloor }{g^2}\rfloor }{g^2}\big\rfloor =\big\lfloor \frac{\lfloor \frac{n}{g^2}\rfloor }{g^2}\big\rfloor$. Since the number of block matrices (53) is m − n₅, the difference between the total number of blocks matrices and the number of block matrices (54), we finally obtain for the degeneracy degrees Y_i

$$\begin{eqnarray} &&Y_{1}=Y_{3}=Y_{5}=Y_{7}=\left\lfloor \frac{\big\lfloor \frac{n}{g^2}\big\rfloor }{g^2}\right\rfloor , \end{eqnarray} \tag{ 59a }$$

$$\begin{eqnarray} &&Y_{4}=\left\lfloor \frac{\big\lfloor \frac{n+1}{g}\big\rfloor +1}{g^2}\right\rfloor , \end{eqnarray} \tag{ 59b }$$

$$\begin{eqnarray} &&Y_{2}=Y_{6}=\left\lfloor \frac{\big\lfloor \frac{n+1}{g}\big\rfloor +1}{g^2}\right\rfloor -\left\lfloor \frac{\big\lfloor \frac{n}{g^2}\big\rfloor }{g^2}\right\rfloor . \end{eqnarray} \tag{ 59c }$$

The cases where the last block diagonal submatrix is truncated can then easily be deduced by inspection, and we do not detail all the results here.

Since ⌊⌊24g²⌋g⌋ = 100, the last block diagonal submatrix is not truncated for n = 100 and figure 4 shows the numerical results for x^τ_{n, i} together with the seven zero-order approximations X_i. It is seen again that these zero-order approximations are not sufficient for those tending to zero (X₄, X₅, X₆, X₇) but are much better for those tending to −1 (X₁, X₂, X₃).

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Let us define the two functions f₀ and f₁ by $f_{0}(k)=\big\lfloor \frac{k+1}{g^2}\big\rfloor , f_{1}(k)=\big\lfloor \frac{k+1}{g}\big\rfloor$. Then it has been seen that a gap in the distributions of zeros occurs at position f₀(n). In view of the structure of the matrices μT^{(n − 1)} (8), of the construction of Fibonacci words by substitution rules (1a), (1b) and of the asymptotics results obtained previously, one expects that the zeros below and above this first gap each will present a gap at about the same position. This determines a hierarchy on the gaps. For example, if hierarchy number 1 is attributed to the gap at position f₀(n), then the gaps with hierarchy number 2 have positions f²₀(n) and f₀(n) + f₀f₁(n). The gaps with hierarchy number 3 have positions f³₀(n), f₀²(n) + f₀f₁f₀(n), f₀(n) + f²₀f₁(n), f₀(n) + f₀f₁(n) + f₀f²₁(n) and so on. Hierarchy number i gaps thus can be labeled by a number in the binary system with i − 1 digits and with the order of the digits reversed. This number then reflects the structure of the terms of higher degree in the functions f₀ and f₁. (Note that each term always begins with f₀, so that this first f₀ can be omitted in the labeling process.) For example, the above gaps with hierarchy number 3 can be labeled successively as 00, 10, 01 and 11, and the gaps with hierarchy number 4 can be labeled successively as 000, 100, 010, 110, 001, 101, 011 and 111.

For n = 500, the hierarchy numbers and positions (thereafter called hierarchical positions) of the gaps computed according to the previous rules are {1, {191}}{2, {73, 309}}{3, {28, 118, 236, 382}}{4, {11, 45, 90, 146, 208, 264, 337, 427}}. Figure 5 shows the numerically computed zeros for n = 500 in the cases τ = 1/2 and τ = 2, together with vertical lines at the previous hierarchical positions.

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The Gauss quadrature formula gives [17] $\mathcal {L}(p(x))=\sum _{k=1}^n A_{n,k}^{\tau }p\left(x_{n,k}^{\tau }\right)$ with p(x) a polynomial of degree at most 2n − 1, and the weigths A^τ_nk positive [17],

$$\begin{equation} A_{n,k}^{\tau }=-\frac{\lambda _1^{\tau }\lambda _2^{\tau }\cdots \lambda _{n+1}^{\tau }}{\hat{Q}_{n+1}^{\tau }\big(x_{n,k}^{\tau }\big)\hat{Q}_{n}^{\tau \prime } \big(x_{n,k}^{\tau }\big)}=\frac{1}{\sum _{j=0}^n \big[q_j^{\tau }\big(x_{n,k}^{\tau }\big)\big]^2}. \end{equation} \tag{ 60 }$$

The prime in $\hat{Q}_{n}^{\tau \prime }$ means the derivative with respect to x of $\hat{Q}_{n}^{\tau }(x)$. The positive weights satisfy [17] ∑ⁿ_{k = 1}A_{n, k}^τ = 1. Let ψ^τ_n a bounded, right continuous, non-decreasing step function defined by [17]

$$\begin{equation} \psi _n^{\tau }(x)=\left\lbrace \begin{array}{@{}ll@{}} 0&{\rm if} \quad x<x_{n,1}^{\tau }, \\\ms A_{n,1}^{\tau }+A_{n,2}^{\tau }+\cdots + A_{n,p}^{\tau }& {\rm if} \quad x_{n,p}^{\tau }\le x<x_{n,p+1}^{\tau }, \\\ms 1& {\rm if} \quad x\ge x_{n,n}^{\tau }.\end{array} \right. \end{equation} \tag{ 61 }$$

Then [17] there is a subsequence of the sequence ψ^τ_n that converges to a distribution ψ^τ for which

$$\begin{equation} \int _{-\infty }^{\infty } x^{k}\, {\rm d}\psi ^{\tau }(x)=\mathcal {L}(x^{k}) \end{equation} \tag{ 62 }$$

for k = 0, 1, 2, ... . One says [17] that ψ^τ is a representative of $\mathcal {L}$.

We shall now verify explicitly that for τ = 1, the sequence ψ¹_n converges and that dψ¹ is indeed the measure of orthogonality for Chebyshev polynomials U_j. For the case τ = 1, $q_j^1(x)=U_j(1+x)=\frac{\sin [(j+1)\arccos (1+x)]}{\sin [\arccos (1+x)]}$ and (42) gives the zeros. Explicit computation from the last equality in (60) gives $A_{n,k}^1=\frac{2}{n+1}\big(\sin \big[\frac{(n+1-k)\pi }{n+1}\big]\big)^2=\frac{2}{n+1}\big[1-\big(1+x_{n,k}^{1}\big)^2\big]$. It is clear that the maximum jump of the step function ψ¹_n(x) is less than or equal to $\frac{2}{n+1}$ and thus decreases to zero as n tends to infinity. The orthonormalization relation for Chebyshev polynomials reads $\frac{2}{\pi }\int _{-2}^0U_i(1+x)U_j(1+x)\sqrt{1-(1+x)^2}\,{\rm d}x=\delta _{ij}$. One therefore has to verify that in the limit n → ∞, A¹_{n, k} is equal to $\frac{2}{\pi }\sqrt{1-\big(1+x_{n,k}^1\big){}^2}\big(x_{n,k+1}^1-x_{n,k}^1\big)$. Indeed, $x_{n,k+1}^1-x_{n,k}^1=2\sin \big[\frac{\pi }{2+2 n}\big]\sin \big[\frac{\pi -2 k \pi +2 n \pi }{2+2 n}\big]$ $\simeq \frac{\pi }{n+1}\sin \big[\frac{(n+1-k)\pi }{n+1}\big] =\frac{\pi }{n+1}\sqrt{1-\big(1+x_{n,k}^{1}\big){}^2}$ as n → ∞. To summarize, it has been verified explicitly that dψ¹_n(x) converges pointwise to $\frac{2}{\pi }\sqrt{1-(1+x)^2}\,{\rm d}x$ when n → ∞. Thus, the measure of orthogonality for τ = 1 has been deduced from the only knowledge of the zeros (42). The function whose derivative is $\sqrt{1-(1+x)^2}$ and which takes the values 0 for x = −2 and 1 for x = 0 is

$$\begin{equation} \frac{1}{2}+\frac{(1+x) \sqrt{1-(1+x)^2}+\arcsin (1+x)}{\pi }. \end{equation} \tag{ 63 }$$

It is therefore natural to also investigate numerically the sequence ψ^τ_n for τ ≠ 1. Results are reported in figures 6(a) and (c). The fact that results for n = 100 (thin curve) appear as refinement of the results for n = 20 (thick gray curve) indicates that ψ^τ_n should indeed converge in some sense when n → ∞. However, in contrast with the case τ = 1, some discontinuities of the functions ψ^τ_n do not decrease as n increases, and one has evidence that ψ^τ_n converges toward a so-called devil staircase. The same numerical calculations made for τ = 1 give for n = 100 a step curve which almost coincides with (63) at the scale of the graph (see figure 6(b)). Figure 6 also indicates that the region where the ψ^τ_n are constant is clearly related to the gaps between the zeros, as should be the case from the defining relation (61). To conclude, the measure of orthogonality dψ^τ should involve, for τ ≠ 1, Dirac distributions whose supports are presently unknown functions of τ, and should exhibit self-similar structures.

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It will now be shown explicitly that the transmission coefficient throughout a Fibonacci chain in the limit of zero-energy incident plane wave is generically zero, except for a critical Fibonacci chain, i.e. a Fibonacci chain with the critical point.

It is well known [24] that the transmission coefficient t(k) for an incident plane wave with wave number k has the expression

$$\begin{equation} t(k)=\frac{4 k^2}{[k^{2}(S_{n}(k))_{12}-(S_{n}(k))_{21}]^{2}+k^{2}[(S_{n}(k))_{11}+(S_{n}(k))_{22}]^2}, \end{equation} \tag{ 64 }$$

with the second-order matrix S_n defined for a Fibonacci chain with n centers by

$$\begin{eqnarray} &&S_{n}^{\tau }\left(k\right)=M\left(0,k\right)\mathcal {S}_{n-1}^{\tau }\left(k\right), \end{eqnarray} \tag{ 65 }$$

$$\begin{eqnarray} &&\mathcal {S}_{n-1}^{\tau }\left(k\right)=M\left(d_{n},k\right)M\left(d_{n-1},k\right)\ldots M\left(d_{2},k\right), \end{eqnarray} \tag{ 66 }$$

$$\begin{eqnarray} M(d,k)=\left( \begin{array}{@{}cc@{}} \cos (k d)+2m \mu \dsty\frac{\sin (k d)}{k} & \dsty\frac{\sin (k d)}{k} \\\ms -k \sin (k d)+2m \mu \cos (k d) & \cos (k d) \end{array} \right), \end{eqnarray} \tag{ 67 }$$

with d_i the intercenter distance (6). Now at the limit of zero energy,

$$\begin{equation} M\left(d,0\right)=\left( \begin{array}{@{}cc@{}} 2 d m \mu +1 & d \\ 2 m \mu & 1 \end{array} \right). \end{equation} \tag{ 68 }$$

From (64), it is clear that lim_{k → 0}t(k) = 0 for (S_n(0)₂₁) ≠ 0. From the relations

$$\begin{eqnarray} &&S_{n}^{\tau }(0)=M(0,0)M(d_{n},0)M(d_{n-1},0)\mathcal {S}_{n-3}^{\tau }(0), \end{eqnarray} \tag{ 69 }$$

$$\begin{eqnarray} &&S_{n-1}^{\tau }(0)=M(0,0)M(d_{n-1},0)\mathcal {S}_{n-3}^{\tau }(0), \end{eqnarray} \tag{ 70 }$$

$$\begin{eqnarray} &&S_{n-2}^{\tau }(0)=M(0,0)\mathcal {S}_{n-3}^{\tau }(0), \end{eqnarray} \tag{ 71 }$$

one obtains

$$\begin{eqnarray} &&\fl S_{n}^{\tau }\left(0\right)-2\left(1+x \tau ^{u_{n-1}} \right)S_{n-1}^{\tau }\left(0\right)+S_{n-2}^{\tau }\left(0\right) \nonumber \\ \fl\qquad= M(d,0)\left\lbrace M(d_{n},0)M(d_{n-1},0)-2(1+x \tau ^{u_{n-1}} )M(d_{n-1},0)+\left( \begin{array}{@{}cc@{}} 1 & 0 \\ 0 & 1 \end{array} \right)\right\rbrace \mathcal {S}_{n-3}^{\tau }(0). \nonumber\\ \end{eqnarray} \tag{ 72 }$$

The left member has the same form as the one of recursion relations (12a). It can be verified by explicit computation that the whole factor in front of $\mathcal {S}_{n-3}\left(0\right)$ in the right member of (72), i.e. M(d, 0) times the second-order matrices in the curly braces, is a matrix whose second line is 0, 0. Therefore, both (S_n(0)₂₁) and (S_n(0)₂₂) satisfy the recursion relation (12a). By explicit computation of S₁(0) and S₂(0) and comparison with (12b) and (13a), one deduces that

$$\begin{equation} \left( S_{n}^{\tau }\left(0\right)_{21}\right)=2 m \mu Q_{n-1}^{\tau } \left(x\right), \end{equation} \tag{ 73 }$$

and therefore (S^τ_n(0)₂₁) = 0 at critical Fibonacci chains. To summarize, the zero-energy transmission coefficient for a Fibonacci chain is zero except the cases of critical Fibonacci chains where it is

$$\begin{equation} t_{n}^{\tau }(0)=\left(\frac{2}{\mathrm{Tr}\big(S_{n}^{\tau }(0)\big)}\right)^{2}. \end{equation} \tag{ 74 }$$

Let us further determine the matrix S^τ_n(0). From (12a) and (20a), one deduces that Q^τ_n − R_{n − 1}^τ satisfies the recursion relation (12a). By explicit calculation of S^τ₁(0) and S^τ₂(0) and from (12b) and (20b), one obtains (S^τ_n(0)₂₂) = Q^τ_{n − 1} − R_{n − 2}^τ. Now it is clear from (65), (67) and (66) that

$$\begin{equation} S_{n}^{\tau }(0)=M(0,0)M(d_{n},0)(M(0,0))^{-1}S_{n-1}^{\tau }(0). \end{equation} \tag{ 75 }$$

Since

$$\begin{equation} M\left(0,0\right)M\left(d_{n},0\right)\left(M\left(0,0\right)\right)^{-1}= \left( \begin{array}{@{}cc@{}} 1 & d_n \\ 2m\mu & 1+2m\mu d_n \end{array} \right), \end{equation} \tag{ 76 }$$

from (75), (76) and (73), one obtains

$$\begin{equation} \left(S_{n}^{\tau }\left(0\right)\right)_{21}=2 m \mu Q_{n-1}^{\tau } \left(x\right)=2m\mu \left(S_{n}^{\tau }\left(0\right)\right)_{21}+\left(1+2m\mu d_{n}\right)2m\mu Q_{n-2}^{\tau } \end{equation} \tag{ 77 }$$

and, using (12a), (S^τ_n(0))₁₁ = Q^τ_{n − 1} − Q_{n − 2}^τ. Finally, since $\det \left(S_n^{\tau }\left(0\right)\right)=1$,

$$\begin{equation} \fl S_n^{\tau }\left(0\right)= \left( \begin{array}{@{}cc@{}} Q_{n-1}^\tau (x)- Q_{n-2}^\tau (x)& \dsty\frac{\big(Q_{n-1}^\tau (x)- Q_{n-2}^\tau (x)\big)\big(Q_{n-1}^\tau (x)- R_{n-2}^\tau (x)\big)-1}{2m \mu Q_{n-1}^\tau (x)} \\\ms 2m\mu Q_{n-1}^\tau (x) & Q_{n-1}^\tau (x)- R_{n-2}^\tau (x) \end{array} \right), \end{equation} \tag{ 78 }$$

with x defined by (9). The matrix element S^τ_n(0)₁₂ can also be computed recursively with (75):

$$\begin{equation} S_{n}^{\tau }\left(0\right)_{12}=\sum _{j=1}^{n-1} f \tau ^{j}\left(Q_{j-1}^{\tau }\left(x\right)-R_{j-2}^{\tau }\left(x\right)\right). \end{equation} \tag{ 79 }$$

Equation (79) proves that Q^τ_{n − 1}(x) divides (Q^τ_{n − 1}(x) − Q_{n − 2}^τ(x))(Q^τ_{n − 1}(x) − R_{n − 2}^τ(x)) − 1. The transmission coefficient for the critical point i, i.e. for x equal to the zero x^τ_{n − 1, i} (34) of Q^τ_{n − 1} will be denoted as t_{n, i}(0). Then (74) can be expressed as

$$\begin{equation} t_{n,i}^{\tau }(0)=\left(\frac{2}{Q_{n-2}^{\tau }\big(x_{n-1,i}^{\tau }\big)+R_{n-2}^{\tau }\big(x_{n-1,i}^{\tau }\big)}\right)^{2}. \end{equation} \tag{ 80 }$$

This proves that |Q^τ_{n − 2}(x_{n − 1, i}^τ) + R^τ_{n − 2}(x_{n − 1, i}^τ)| ⩾ 2.

The critical transmission coefficients (80) are not equal to zero but not necessarily equal to unity. For example, for n = 3,

$$\begin{eqnarray} &&t_{3,1}^{\tau }(0)=\frac{4}{\big(\tau +\sqrt{(\tau -1) \tau +1}+\frac{1}{\tau +\sqrt{(\tau -1) \tau +1}-1}-1\big)^2}, \end{eqnarray} \tag{ 81a }$$

$$\begin{eqnarray} &&t_{3,2}^{\tau }(0)=\frac{4}{\big(\tau -\sqrt{(\tau -1) \tau +1}+\frac{1}{\tau -\sqrt{(\tau -1) \tau +1}-1}-1\big)^2}, \end{eqnarray} \tag{ 81b }$$

and these transmission coefficients are equal to unity only for τ = 1. For the special cases where n is a Fibonacci number minus 1, and therefore the n − 1 intervals display palindromic property, it is now shown that all the critical transmission coefficients are equal to unity. (For spectral properties of infinite Fibonacci chains with palindromic properties, see [25].)

First note that $Q_{F_{k}-3}^{\tau }$ evaluated at the zeros of $Q_{F_{k}-2}^{\tau }$ is equal to ±1. This property is obtained by taking the determinant of $\Phi _{F_{k}-2}^{\tau }$ (19) at the zeros $x_{F_{k}-2,i}^\tau$. Indeed this determinant is unity and equal to $\big(Q_{F_{k}-3}^{\tau }\big(x_{F_{k}-2,i}^\tau \big)\big)^2$ by (19) and (21). Then, from (21) and (80), one obtains

$$\begin{equation} t_{F_{k}-1,i}^{\tau }\left(0\right)=1 \end{equation} \tag{ 82 }$$

for all the F_k − 2 values of i.