Generalized discrete Lotka-Volterra equation, orthogonal polynomials and generalized epsilon algorithm

Abstract

In this paper, we propose a generalized discrete Lotka-Volterra equation and explore its connections with symmetric orthogonal polynomials, Hankel determinants and convergence acceleration algorithms. Firstly, we extend the fully discrete Lotka-Volterra equation to a generalized one with a sequence of given constants $\{u_{0}^{(n)}\}$ and derive its solution in terms of Hankel determinants. Then, it is shown that the discrete equation of motion is transformed into a discrete Riccati system for a discrete Stieltjes function, hence leading to a complete linearization. Besides, we obtain its Lax pair in terms of symmetric orthogonal polynomials by generalizing the Christoffel transformation for the symmetric orthogonal polynomials. Moreover, a generalization of the famous Wynn’s ε-algorithm is also derived via a Miura transformation to the generalized discrete Lotka-Volterra equation. Finally, the numerical effects of this generalized ε-algorithm are discussed by applying to some linearly, logarithmically convergent sequences and some divergent series.

Appendix A: Technical details for the main results

This appendix is devoted to providing some linear and bilinear relations which serve for the proofs of the theorems in Sections 2–4. In Appendix A.1, some notations and auxiliary determinants are introduced. Appendix A.2 contains some linear relations, while Appendix A.3 includes some bilinear relations. Based on these relations, one can prove the main results in Sections 2–4.

A.1. Notations

For integers k ≥ 1, j,r ≥− 1, s,n ≥ 0, and l = 0, 1, we first define two sets of sequences, $\{a_{s}^{(n)}\}$ and $\{d_{r}^{(n)}\}$ satisfying

$$a_{s}^{(n)}=c_{s}^{(n)}-\frac{h\alpha^{(n)}}{\beta^{(n)}}\sum\limits_{i=0}^{s-1}a_{i}^{(n)} c_{s-i}^{(n)},$$

and

$$d_r^{(n)}=c_r^{(n)}-\frac{h\alpha^{(n)}}{\beta^{(n)}}\sum\limits_{i=-1}^{r-1}d_i^{(n)} c_{r-i}^{(n)},$$

where α⁽ⁿ⁾ and β⁽ⁿ⁾ are defined by (3.12). Then, in order to express determinants in this appendix conveniently, we introduce the following vectors,

$$\begin{array}{@{}rcl@{}} &\boldsymbol{A}_{k,s}^{(n)}= \left(\begin{array}{c} a_{s}^{(n)} \\ a_{s+1}^{(n)} \\ {\vdots} \\ a_{s+k-1}^{(n)} \end{array}\right) , \quad \boldsymbol{B}_{k,j}^{(n+1)} = \left(\begin{array}{c} \frac{\beta^{(n)} c_{j}^{(n+1)}}{h}\\ \frac{\beta^{(n)} c_{j+1}^{(n+1)}}{h}+\alpha^{(n)} c_{1}^{(n)} c_{j}^{(n+1)}\\ \vdots\\ \frac{\beta^{(n)} c_{j+k-1}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{k-1}c_{i}^{(n)} c_{j+k-1-i}^{(n+1)} \end{array}\right),\ \ \end{array}$$

(A.1a)

$$\begin{array}{@{}rcl@{}} & \boldsymbol{C}_{k,j}^{(n)} = \left(\begin{array}{c} c_{j}^{(n)} \\ c_{j+1}^{(n)} \\ {\vdots} \\ c_{j+k-1}^{(n)} \end{array}\right) , \ \ \boldsymbol{D}_{k,j}^{(n)} = \left(\begin{array}{c} d_{j}^{(n)} \\ d_{j+1}^{(n)} \\ {\vdots} \\ d_{j+k-1}^{(n)} \end{array}\right), \ \ \boldsymbol{\xi}_{k,s}^{(n+1)} = \left(\begin{array}{c} {\sum}_{i=1}^{s}c_{i}^{(n)} c_{s-i}^{(n+1)}\\ {\sum}_{i=2}^{s+1}c_{i}^{(n)} c_{s+1-i}^{(n+1)}\\ \vdots\\ {\sum}_{i=k}^{s+k-1}c_{i}^{(n)} c_{s+k-1-i}^{(n+1)} \end{array}\right). \end{array}$$

(A.1b)

with the convention $\boldsymbol {\xi }_{k,0}^{(n+1)}=\boldsymbol {0}$.

Use these vectors, we define some auxiliary determinants as follows,

$$\begin{array}{@{}rcl@{}} F_{k,j,s}^{(n+1)}&=&|\boldsymbol{A}_{k,s}^{(n)},\boldsymbol{C}_{k,j}^{(n+1)},\ldots,\boldsymbol{C}_{k,j+k-3}^{(n+1)},\boldsymbol{C}_{k,j+k-2}^{(n+1)}|,\\ {\Gamma}_{k,j,r}^{(n+1)}&=&|\boldsymbol{D}_{k,r}^{(n)},\boldsymbol{C}_{k,j}^{(n+1)},\ldots,\boldsymbol{C}_{k,j+k-3}^{(n+1)},\boldsymbol{C}_{k,j+k-2}^{(n+1)}|,\\ E_{k,j,r}^{(n+1)}&=&|\boldsymbol{C}_{k,r}^{(n)},\boldsymbol{B}_{k,j}^{(n+1)},\ldots,\boldsymbol{B}_{k,j+k-3}^{(n+1)},\boldsymbol{B}_{k,j+k-2}^{(n+1)}|,\\ \tilde{E}_{k,s,r}^{(n+1)}&=&|\boldsymbol{B}_{k,r}^{(n+1)},\boldsymbol{B}_{k,s}^{(n+1)},\ldots,\boldsymbol{B}_{k,s+k-3}^{(n+1)},\boldsymbol{B}_{k,s+k-2}^{(n+1)}|,\\ K_{k,j,1}^{(n+1)}&=&\left| \begin{array}{cccccc} 1&-\alpha^{(n)} c_j^{(n+1)}&-\alpha^{(n)} c_{j+1}^{(n+1)}&\ldots&-\alpha^{(n)} c_{j+k-2}^{(n+1)}\\ \boldsymbol{C}_{k-1,1}^{(n)}&\boldsymbol{B}_{k-1,j+1}^{(n+1)}&\boldsymbol{B}_{k-1,j+2}^{(n+1)}&\ldots&\boldsymbol{B}_{k-1,j+k-1}^{(n+1)} \end{array}\right|,\\ G_{2k,l}^{(n+1)}(x)&=&\left| \begin{array}{cccccc} \boldsymbol{C}_{k,l}^{(n)}&\boldsymbol{B}_{k,l}^{(n+1)}&\ldots&\boldsymbol{B}_{k,l+k-2}^{(n+1)}&\boldsymbol{B}_{k,l+k-1}^{(n+1)}\\ 1&y_{l,2}&\ldots&y_{l,2k-2}&y_{l,2k} \end{array}\right|,\\ \tilde{G}_{2k,l}^{(n+1)}(x)&=&\left| \begin{array}{ccccc} \boldsymbol{B}_{k-1,l}^{(n+1)}&\boldsymbol{B}_{k-1,l+1}^{(n+1)}&\ldots&\boldsymbol{B}_{k-1,l+k-1}^{(n+1)}\\ y_{l,2}&y_{l,4}&\ldots&y_{l,2k} \end{array}\right|, \end{array}$$

with the conventions

$$\begin{array}{@{}rcl@{}} F_{0,j,s}^{(n+1)}&=&0,\quad F_{1,j,s}^{(n+1)}=a_{s}^{(n)},\quad {\Gamma}_{0,j,r}^{(n+1)}=0,\quad {\Gamma}_{1,j,r}^{(n+1)}=d_{r}^{(n)}, \end{array}$$

(A.2a)

$$\begin{array}{@{}rcl@{}} E_{1,j,r}^{(n+1)}&=&c_{r}^{(n)},\ \tilde{E}_{0,s,r}^{(n+1)}=1,\ \tilde{E}_{1,s,r}^{(n+1)}=\boldsymbol{B}_{1,r}^{(n+1)}, \end{array}$$

(A.2b)

$$\begin{array}{@{}rcl@{}} K_{1,j,1}^{(n+1)}&=&1,\ G_{0,l}^{(n+1)}(x)=1,\ \tilde{G}_{2,l}^{(n+1)}(x)=y_{l,2}, \end{array}$$

(A.2c)

where

$$y_{l,2k}=(x^2+\frac{1}{h})x^{2k-2}-\alpha^{(n)}\sum\limits_{i=1-l}^{k-1}c_{k-1-i}^{(n+1)}x^{2i}.$$

(A.3)

Using these notations, it is easy to see that the Hankel determinant $H_{k,j}^{(n)}$ can be expressed in terms of vector $\boldsymbol {C}_{k,j}^{(n)}$ as follows

$$H_{k,j}^{(n)}=|\boldsymbol{C}_{k,j}^{(n)},\boldsymbol{C}_{k,j+1}^{(n)},\ldots,\boldsymbol{C}_{k,j+k-1}^{(n)}|.$$

Now, let us present the relations among the vectors defined in (A.1). First, we rewrite (2.5) as follows

$$c_j^{(n)}=\frac{\beta^{(n)} c_{j-1}^{(n+1)}}{h}-\frac{c_{j-1}^{(n)}}{h}+\alpha^{(n)}\sum\limits_{i=1}^{j-1}c_i^nc_{j-1-i}^{(n+1)},\qquad j=1,2,3,\ldots.$$

(A.4)

Then, we have the following lemma.

Lemma A.1.1

Assume $\{c_{j}^{(n)}\}$ defined by the recursion relation (A.4) and (4.6), then

$$\begin{array}{@{}rcl@{}} {\boldsymbol{C}}_{k,0}^{(n)}&=&\frac{h{\boldsymbol{B}}_{k,-1}^{(n+1)}-\boldsymbol{C}_{k,-1}^{(n)}}{h\gamma^{(n)}}, \end{array}$$

(A.5a)

$$\begin{array}{@{}rcl@{}} {\boldsymbol{C}}_{k,j}^{(n)}&=&{\boldsymbol{B}}_{k,j-1}^{(n+1)}+\alpha^{(n)} {\boldsymbol{\xi}}_{k,j-1}^{(n+1)}-\frac{1}{h}{\boldsymbol{C}}_{k,j-1}^{(n)},\quad j=1,2,\ldots, \end{array}$$

(A.5b)

where γ⁽ⁿ⁾ is defined by

$$\gamma^{(n)}=1+\alpha^{(n)} c_{-1}^{(n+1)}.$$

Proof

The relation (A.5b) is obtained directly by applying (A.4) to $\boldsymbol {C}_{k,j}^{(n)}$. For (A.5a), noting the definition of recursion relation (4.6) for $c_{0}^{(n)}$, and using (A.4) to $\boldsymbol {C}_{k,0}^{(n)}$, we first have

$$\gamma^{(n)}\boldsymbol{C}_{k,0}^{(n)}=\alpha^{(n)} c_{-1}^{(n+1)}\boldsymbol{C}_{k,0}^{(n)}+\left(\begin{array}{c} \frac{c_{-1}^{(n+1)}}{h}\\ \frac{\beta^{(n)} c_{0}^{(n+1)}}{h}\\ \frac{\beta^{(n)} c_{1}^{(n+1)}}{h}+\alpha^{(n)} c_1^{(n)} c_0^{(n+1)}\\ \vdots\\ \frac{\beta^{(n)} c_{k-2}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{k-2}c_{i}^{(n)} c_{k-2-i}^{(n+1)} \end{array}\right)-\frac{1}{h}\boldsymbol{C}_{k,-1}^{(n)}.$$

Then, it is easy to find that the summation of the first two terms in the above right hand of equation identifies with $\boldsymbol {B}_{k,-1}^{(n+1)}$, by noting its definition in (A.1a). Thus, (A.5a) gets proved. 

A.2. Linear relations

In Lemma A.2.1, we list some important linear relations, which serve to deduce Corollary A.2.2. The proof of Lemma A.2.1 is implemented by performing some row and column transformations on determinants based on the determinant properties. Due to the lengthy details, we place its proof in Appendix A.4.

Lemma A.2.1

Assume $\{c_{j}^{(n)}\}$ satisfy the recursion relations (2.5) and (4.6), then for integers k ≥ 1, j,r ≥− 1, and l = 0, 1, the following formulae hold

$$\begin{array}{@{}rcl@{}} H_{k+1,-1}^{(n)}&=&\frac{1}{\gamma^{(n)}}E_{k+1,-1,-1}^{(n+1)}, \end{array}$$

(A.6a)

$$\begin{array}{@{}rcl@{}} H_{k,l}^{(n)}&=&E_{k,l,l}^{(n+1)}, \end{array}$$

(A.6b)

$$\begin{array}{@{}rcl@{}} H_{k,2}^{(n)}&=&K_{k+1,0,1}^{(n+1)}-\frac{1}{h}E_{k,2,1}^{(n+1)}, \end{array}$$

(A.6c)

$$\begin{array}{@{}rcl@{}} E_{k,j,-1}^{(n+1)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}{\Gamma}_{k,j,-1}^{(n+1)}, \end{array}$$

(A.6d)

$$\begin{array}{@{}rcl@{}} E_{k,j,0}^{(n+1)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}F_{k,j,0}^{(n+1)}, \end{array}$$

(A.6e)

$$\begin{array}{@{}rcl@{}} E_{k,j,0}^{(n+1)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}{\Gamma}_{k,j,0}^{(n+1)}+\frac{h\alpha^{(n)} c_{-1}^{(n)}}{{\beta^{(n)}}}E_{k,j,1}^{(n+1)}, \end{array}$$

(A.6f)

$$\begin{array}{@{}rcl@{}} E_{k,j,0}^{(n+1)}&=&\frac{h\tilde{E}_{k,j,-1}^{(n+1)}-E_{k,j,-1}^{(n+1)}}{h\gamma^{(n)}}, \end{array}$$

(A.6g)

$$\begin{array}{@{}rcl@{}} E_{k,j,1}^{(n+1)}&=&\frac{(\beta^{(n)})^k}{{h}^{k-1}}F_{k,j,1}^{(n+1)}, \end{array}$$

(A.6h)

$$\begin{array}{@{}rcl@{}} E_{k,j,1}^{(n+1)}&=&\tilde{E}_{k,j,0}^{(n+1)}-\frac{1}{h}E_{k,j,0}^{(n+1)}, \end{array}$$

(A.6i)

$$\begin{array}{@{}rcl@{}} \tilde{E}_{k,r+1,r}^{(n+1)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k}H_{k,r}^{(n+1)}, \end{array}$$

(A.6j)

$$\begin{array}{@{}rcl@{}} K_{k,j,1}^{(n+1)}&=&\frac{(\beta^{(n)})^k}{{h}^{k-1}}H_{k-1,j+1}^{(n+1)}-\frac{\alpha^{(n)}(\beta^{(n)})^{k-1}}{{h}^{k-2}}F_{k,j,0}^{(n+1)}, \end{array}$$

(A.6k)

$$\begin{array}{@{}rcl@{}} G_{2k,l}^{(n+1)}(x)&=&\frac{H_{k,l}^{(n)} P_{2k+l}^{(n)}(x)}{x^l}, \end{array}$$

(A.6l)

$$\begin{array}{@{}rcl@{}} \tilde{G}_{2k,l}^{(n+1)}(x)\\ \quad&=&\frac{(\beta^{(n)})^{k-1}H_{k-1,l}^{(n+1)}\Big((hx^2+1) P_{2k-2+l}^{(n+1)}(x)-hx\alpha^{(n)} Q_{2k-3+l}^{(n+1)}(x)\Big)}{h^kx^l}. \end{array}$$

(A.6m)

Using Lemma A.2.1, it is not hard to derive the following corollary, whose detail is omitted.

Corollary A.2.2

Under the recursion relations (2.5) and (4.6) satisfied by $\{c_{j}^{(n)}\}$, there hold the following important identities for k = 1, 2,…,

$$\begin{array}{@{}rcl@{}} H_{k+1,-1}^{(n)}&=&\frac{1}{\gamma^{(n)}}\Big(\frac{\beta^{(n)}}{h}\Big)^{k}{\Gamma}_{k+1,-1,-1}^{(n+1)}, \end{array}$$

(A.7a)

$$\begin{array}{@{}rcl@{}} H_{k,0}^{(n)}&=&\frac{(\beta^{(n)})^{k-1}(\beta^{(n)} H_{k,-1}^{(n+1)}-{\Gamma}_{k,0,-1}^{(n+1)})}{h^{k}\gamma^{(n)}}, \end{array}$$

(A.7b)

$$\begin{array}{@{}rcl@{}} H_{k,0}^{(n)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}F_{k,0,0}^{(n+1)}, \end{array}$$

(A.7c)

$$\begin{array}{@{}rcl@{}} H_{k,1}^{(n)}&=&\frac{(\beta^{(n)})^{k}}{h^{k-1}}{F}_{k,1,1}^{(n+1)}, \end{array}$$

(A.7d)

$$\begin{array}{@{}rcl@{}} H_{k,1}^{(n)}&=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k}H_{k,0}^{(n+1)}-\frac{(\beta^{(n)})^{k-1}}{h^{k}}F_{k,1,0}^{(n+1)}, \end{array}$$

(A.7e)

$$\begin{array}{@{}rcl@{}} H_{k,1}^{(n)}&=&\frac{(\beta^{(n)})^{k}(\beta^{(n)} H_{k,0}^{(n+1)}-{\Gamma}_{k,1,0}^{(n+1)})}{h^{k}(\beta^{(n)}+\alpha^{(n)} c_{-1}^{(n)})}, \end{array}$$

(A.7f)

$$\begin{array}{@{}rcl@{}} H_{k,2}^{(n)}&=&\frac{(\beta^{(n)})^{k+1}}{h^{k}}H_{k,1}^{(n+1)}-h\alpha^{(n)} H_{k+1,0}^{(n)}-\Big(\frac{\beta^{(n)}}{h}\Big)^{k}F_{k,2,1}^{(n+1)}. \end{array}$$

(A.7g)

Moreover, it is easy to check that relations (A.7e) and (A.7g) also hold for k = 0, by using the conventions (2.4) and (A.2a).

A.3. Bilinear relations

Now, we apply the Jacobi identity to produce some bilinear relations, which play crucial roles in the main results. For any determinant D, the Jacobi determinant identity [1] reads

$$D\cdot D\left[\begin{array}{cc}i_{1}&i_{2}\\j_{1}&j_{2} \end{array}\right]=D\left[\begin{array}{c}i_{1}\\j_{1} \end{array}\right]\cdot D\left[\begin{array}{c}i_{2}\\j_{2} \end{array}\right]-D\left[\begin{array}{c}i_{1}\\j_{2} \end{array}\right]\cdot D\left[\begin{array}{c}i_{2}\\j_{1} \end{array}\right] ,$$

(A.8)

where

$$D\left[\begin{array}{cccc}i_{1}&i_{2}&\cdots&i_{k}\\j_{1}&j_{2}&\cdots&j_{k} \end{array}\right] , \quad i_{1}<i_{2}<\cdots<i_{k} , \quad j_{1}<j_{2}<\cdots<j_{k} ,$$

denotes the determinant obtained from D by removing the rows at positions i₁,i₂,…,i_k, and the columns at positions j₁,j₂,…,j_k, in the respective matrix.

Lemma A.3.1

For k = 0, 1, 2,…, and j = 0, 1, the following bilinear relations hold,

$$\begin{array}{@{}rcl@{}} &&F_{k+2,0,0}^{(n+1)}H_{k,1}^{(n+1)}-F_{k+1,0,0}^{(n+1)}H_{k+1,1}^{(n+1)}+F_{k+1,1,1}^{(n+1)}H_{k+1,0}^{(n+1)}=0, \end{array}$$

(A.9a)

$$\begin{array}{@{}rcl@{}} &&F_{k+1,1,0}^{(n+1)}H_{k,0}^{(n+1)}-H_{k+1,0}^{(n+1)}F_{k,1,0}^{(n+1)}-F_{k+1,0,0}^{(n+1)}H_{k,1}^{(n+1)}=0, \end{array}$$

(A.9b)

$$\begin{array}{@{}rcl@{}} &&H_{k,1}^{(n+1)}F_{k,1,0}^{(n+1)}-F_{k,2,1}^{(n+1)}H_{k,0}^{(n+1)}-F_{k+1,0,0}^{(n+1)}H_{k-1,2}^{(n+1)}=0, \end{array}$$

(A.9c)

$$\begin{array}{@{}rcl@{}} &&F_{k+1,1,0}^{(n+1)}H_{k,1}^{(n+1)}-H_{k+1,0}^{(n+1)}F_{k,2,1}^{(n+1)}-F_{k+1,0,0}^{(n+1)}H_{k,2}^{(n+1)}=0, \end{array}$$

(A.9d)

$$\begin{array}{@{}rcl@{}} &&{\Gamma}_{k+2,-1,-1}^{(n+1)}H_{k,1}^{(n+1)}-H_{k+1,0}^{(n+1)}{\Gamma}_{k+1,0,-1}^{(n+1)}+{\Gamma}_{k+1,1,0}^{(n+1)}H_{k+1,-1}^{(n+1)}=0, \end{array}$$

(A.9e)

$$\begin{array}{@{}rcl@{}} &&{\Gamma}_{k+1,0,-1}^{(n+1)}H_{k,0}^{(n+1)}-H_{k+1,-1}^{(n+1)}{\Gamma}_{k,1,0}^{(n+1)}-{\Gamma}_{k+1,-1,-1}^{(n+1)}H_{k,1}^{(n+1)}=0, \end{array}$$

(A.9f)

$$\begin{array}{@{}rcl@{}} &&G_{2k+2,j}^{(n+1)}(x)\tilde{E}_{k,j+1j}^{(n+1)}-\tilde{G}_{2k+2,j}^{(n+1)}(x)E_{k+1,j,j}^{(n+1)}+G_{2k,j}^{(n+1)}(x)\tilde{E}_{k+1,j+1,j}^{(n+1)}=0. \end{array}$$

(A.9g)

Proof

For k = 0, all these equations are easily checked. For k ≥ 1, (A.9a) is obtained from the Jacobi identity (A.8) with $D=F_{k+2,0,0}^{(n+1)}$, and noticing

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}1&k+2\\1&k+2 \end{array}\right]=H_{k,1}^{(n+1)}, \quad D\left[\begin{array}{c}1\\1 \end{array}\right]=H_{k+1,1}^{(n+1)}, \\ && D\left[\begin{array}{c}k+2\\k+2 \end{array}\right]=F_{k+1,0,0}^{(n+1)}, \quad D\left[\begin{array}{c}1\\k+2 \end{array}\right]=F_{k+1,1,1}^{(n+1)}, \quad D\left[\begin{array}{c}k+2\\1 \end{array}\right]=H_{k+1,0}^{(n+1)} . \end{array}$$

(A.9b) follows from the Jacobi identity with

$$D=\left| \begin{array}{cccccc} a_{0}^{(n)}&c_{0}^{(n+1)}&\cdots&c_{k-1}^{(n+1)}&c_{k}^{(n+1)}\\ a_{1}^{(n)}&c_{1}^{(n+1)}&\cdots&c_{k}^{(n+1)}&c_{k+1}^{(n+1)}\\ \vdots&\vdots&{\ddots} &\vdots&\vdots\\ a_{k}^{(n)}&c_{k}^{(n+1)}&\cdots&c_{2k-1}^{(n+1)}&c_{2k}^{(n+1)}\\ 0&0&\cdots&0&1 \end{array}\right|=F_{k+1,0,0}^{(n+1)},$$

and

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}k+1&k+2\\1&2 \end{array}\right]=H_{k,1}^{(n+1)}, \quad D\left[\begin{array}{c}k+1\\1 \end{array}\right]=H_{k,0}^{(n+1)}, \\ && D\left[\begin{array}{c}k+2\\2 \end{array}\right]=F_{k+1,1,0}^{(n+1)}, \quad D\left[\begin{array}{c}k+1\\2 \end{array}\right]=F_{k,1,0}^{(n+1)}, \quad D\left[\begin{array}{c}k+2\\1 \end{array}\right]=H_{k+1,0}^{(n+1)} . \end{array}$$

(A.9c) is derived by employing the Jacobi identity with

$$D=\left| \begin{array}{ccccccc} 1&1&1&0&\cdots&0\\ 0&a_{0}^{(n)}&c_{0}^{(n+1)}&c_{1}^{(n+1)}&\cdots&c_{k-1}^{(n+1)}\\ 0&a_{1}^{(n)}&c_{1}^{(n+1)}&c_{2}^{(n+1)}&\cdots&c_{k}^{(n+1)}\\ \vdots&\vdots&\vdots&{\ddots} &\vdots\\ 0&a_{k}^{(n)}&c_{k}^{(n+1)}&c_{k+1}^{(n+1)}&\cdots&c_{2k-1}^{(n+1)} \end{array}\right|=F_{k+1,0,0}^{(n+1)},$$

and

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}2&k+2\\1&2 \end{array}\right]=H_{k-1,2}^{(n+1)}, \quad D\left[\begin{array}{c}k+2\\2 \end{array}\right]=H_{k,0}^{(n+1)}, \quad D\left[\begin{array}{c}2\\2 \end{array}\right]=H_{k,1}^{(n+1)}, \\ && D\left[\begin{array}{c}2\\1 \end{array}\right]=\left| \begin{array}{cccccc} 1&1&0&\cdots&0\\ a_{1}^{(n)}&c_{1}^{(n+1)}&c_{2}^{(n+1)}&\cdots&c_{k}^{(n+1)}\\ \vdots&\vdots&{\ddots} &\vdots\\ a_{k}^{(n)}&c_{k}^{(n+1)}&c_{k+1}^{(n+1)}&\cdots&c_{2k-1}^{(n+1)} \end{array}\right|=H_{k,1}^{(n+1)}-F_{k,2,1}^{(n+1)}, \\ && D\left[\begin{array}{c}k+2\\1 \end{array}\right]=\left| \begin{array}{cccccc} 1&1&0&\cdots&0\\ a_{0}^{(n)}&c_{0}^{(n+1)}&c_{1}^{(n+1)}&\cdots&c_{k-1}^{(n+1)}\\ \vdots&\vdots&{\ddots} &\vdots\\ a_{k-1}^{(n)}&c_{k-1}^{(n+1)}&c_{k}^{(n+1)}&\cdots&c_{2k-2}^{(n+1)} \end{array}\right|=H_{k,0}^{(n+1)}-F_{k,1,0}^{(n+1)}, \end{array}$$

where we subtract the first columns of the determinants from the second ones in the calculations of $D\left [\begin {array}{c}2\\1 \end {array}\right ]$ and $D\left [\begin {array}{c}k+2\\1 \end {array}\right ]$, and use the determinant property.

To prove (A.9d), we apply the Jacobi identity to

$$D=\left| \begin{array}{ccccccc} 0&0&0&\cdots&0&1\\ a_{0}^{(n)}&c_{0}^{(n+1)}&c_{1}^{(n+1)}&\cdots&c_{k-1}^{(n+1)}&c_{k}^{(n+1)}\\ a_{1}^{(n)}&c_{1}^{(n+1)}&c_{2}^{(n+1)}&\cdots&c_{k}^{(n+1)}&c_{k+1}^{(n+1)}\\ \vdots&\vdots&\vdots&{\ddots} &\vdots&\vdots\\ a_{k}^{(n)}&c_{k}^{(n+1)}&c_{k+1}^{(n+1)}&\cdots&c_{2k-1}^{(n+1)}&c_{2k}^{(n+1)} \end{array}\right|=(-1)^{k+3}F_{k+1,0,0}^{(n+1)},$$

with

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}1&2\\1&2 \end{array}\right]=H_{k,2}^{(n+1)}, \quad D\left[\begin{array}{c}1\\1 \end{array}\right]=H_{k+1,0}^{(n+1)}, \quad D\left[\begin{array}{c}2\\2 \end{array}\right]=(-1)^{k+2}F_{k,2,1}^{(n+1)},\\ && D\left[\begin{array}{c}1\\2 \end{array}\right]=F_{k+1,1,0}^{(n+1)}, \quad D\left[\begin{array}{c}2\\1 \end{array}\right]=(-1)^{k+2}H_{k,1}^{(n+1)} . \end{array}$$

(A.9e) is obtained from the Jacobi identity with $D={\Gamma }_{k+2,-1,-1}^{(n+1)}$, and

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}1&k+2\\1&2 \end{array}\right]=H_{k,1}^{(n+1)}, \quad D\left[\begin{array}{c}1\\1 \end{array}\right]=H_{k+1,0}^{(n+1)}, \\ && D\left[\begin{array}{c}k+2\\2 \end{array}\right]={\Gamma}_{k+1,0,-1}^{(n+1)}, \quad D\left[\begin{array}{c}1\\2 \end{array}\right]={\Gamma}_{k+1,1,0}^{(n+1)}, \quad D\left[\begin{array}{c}k+2\\1 \end{array}\right]=H_{k+1,-1}^{(n+1)} . \end{array}$$

(A.9f) is proved by applying the Jacobi identity to

$$D=\left| \begin{array}{ccccccc} 0&1&0&\cdots&0&0\\ d_{-1}^{(n)}&c_{-1}^{(n+1)}&c_{0}^{(n+1)}&\cdots&c_{k-2}^{(n+1)}&c_{k-1}^{(n+1)}\\ d_{0}^{(n)}&c_{0}^{(n+1)}&c_{1}^{(n+1)}&\cdots&c_{k-1}^{(n+1)}&c_{k}^{(n+1)}\\ \vdots&\vdots&\vdots&{\ddots} &\vdots&\vdots\\ d_{k-1}^{(n)}&c_{k-1}^{(n+1)}&c_{k}^{(n+1)}&\cdots&c_{2k-2}^{(n+1)}&c_{2k-1}^{(n+1)} \end{array}\right|=-{\Gamma}_{k+1,0,-1}^{(n+1)},$$

with

$$\begin{array}{@{}rcl@{}} && D\left[\begin{array}{cc}k+1&k+2\\1&k+2 \end{array}\right]=\tilde{E}_{k,j+1,j}^{(n+1)}, \quad D\left[\begin{array}{c}k+1\\1 \end{array}\right]=\tilde{G}_{2k+2,j}^{(n+1)}(x),\quad D\left[\begin{array}{c}k+2\\k+2 \end{array}\right]=E_{k+1,j,j}^{(n+1)}, \\ && D\left[\begin{array}{c}k+1\\k+2 \end{array}\right]=G_{2k,j}^{(n+1)}(x), \quad \quad D\left[\begin{array}{c}k+2\\1 \end{array}\right]=\tilde{E}_{k+1,j+1,j}^{(n+1)}. \end{array}$$

Corollary A.3.2

If we let

$$\tau_{2k}^{(n)}=H_{k,0}^{(n)},\qquad \tau_{2k+1}^{(n)}=H_{k,1}^{(n)}, \qquad k,n=0,1,\ldots,$$

with the elements of $H_{k,j}^{(n)}$ satisfying (2.5) as well as arbitrary constants $c_{0}^{(n)}\neq 0$, then $\tau _{k}^{(n)}$ satisfy the bilinear relations

$$\beta^{(n)}\tau_{k+2}^{(n+1)}\tau_{k+1}^{(n)}-\tau_{k+2}^{(n)}\tau_{k+1}^{(n+1)}-h \tau_{k+3}^{(n)}\tau_{k}^{(n+1)}=0,\qquad k,n=0,1,\ldots.$$

(A.10)

Proof

We prove (A.10) with respect to its odd and even parts, respectively.

For k = 2j + 1, j = 0, 1, 2,…, the relation (A.10) can be obtained from the identity (A.9a) and eliminating $F_{k+2,0,0}^{(n+1)},~F_{k+1,0,0}^{(n+1)}$ and $F_{k+1,1,1}^{(n+1)}$ via the relations (A.7c) and (A.7d) respectively.

For k = 2j, j = 0, 1, 2,…, the (A.10) immediately follows from the identities (A.9b) and eliminating $F_{k+1,0,0}^{(n+1)}$, and $F_{k,1,0}^{(n+1)},~F_{k+1,1,0}^{(n+1)}$ via the relations (A.7c) and (A.7e) respectively. 

Corollary A.3.3

For k = 0, 1, 2,…, the Hankel determinants $H_{k,j}^{(n)}$ with the elements restricted by (2.5) and (4.6) satisfy the following bilinear relations

$$\begin{array}{@{}rcl@{}} &&\mu^{(n)} H_{k+1,1}^{(n)} H_{k+1,-1}^{(n+1)}- \beta^{(n)} H_{k+1,0}^{(n+1)} H_{k+1,0}^{(n)}-H_{k+2,-1}^{(n)} H_{k,1}^{(n+1)}=0, \end{array}$$

(A.11a)

$$\begin{array}{@{}rcl@{}} &&\mu^{(n)} H_{k+1,-1}^{(n+1)} H_{k,1}^{(n)}- H_{k+1,-1}^{(n)} H_{k,1}^{(n+1)}-h H_{k+1,0}^{(n)} H_{k,0}^{(n+1)}=0, \end{array}$$

(A.11b)

$$\begin{array}{@{}rcl@{}} &&H_{k,2}^{(n)} H_{k,0}^{(n+1)}- \beta^{(n)} H_{k,1}^{(n+1)} H_{k,1}^{(n)}-H_{k+1,0}^{(n)}H_{k-1,2}^{(n+1)}\\ &&\quad+h\alpha^{(n)} H_{k,0}^{(n+1)} H_{k+1,0}^{(n)}=0, \end{array}$$

(A.11c)

$$\begin{array}{@{}rcl@{}} &&H_{k+1,0}^{(n+1)} H_{k,2}^{(n)}-H_{k+1,0}^{(n)} H_{k,2}^{(n+1)}- h H_{k+1,1}^{(n)} H_{k,1}^{(n+1)}\\ &&\quad+h\alpha^{(n)} H_{k+1,0}^{(n+1)} H_{k+1,0}^{(n)}=0, \end{array}$$

(A.11d)

where

$$\alpha^{(n)}=\frac{u_0^{(n)}}{c_0^{(n)}},\quad \beta^{(n)}=1+hu_0^{(n)} ,\quad \mu^{(n)}=\frac{\beta^{(n)}+\alpha^{(n)} c_{-1}^{(n)}}{1+\alpha^{(n)} c_{-1}^{(n+1)}}.$$

Proof

The proof can be achieved by employing the determinant relations given in Lemma A.3.1 and linear relations of determinants presented in Corollary A.2.2.

In particular, (A.11a) and (A.11b) are direct consequences of applying the relations (A.9e) and (A.9f) and replacing ${\Gamma }_{k+1,-1,-1}^{(n+1)},~{\Gamma }_{k,0,-1}^{(n+1)},~{\Gamma }_{k,1,0}^{(n+1)},$ in terms of $H_{k,j}^{(n)}$ via (A.7a), (A.7b) and (A.7f).

Similarly, (A.11c) and (A.11d) can be obtained by the bilinear relations (A.9c), (A.9d) and replacing $F_{k+1,0,0}^{(n+1)},~F_{k,1,0}^{(n+1)},~F_{k,2,1}^{(n+1)}$ in (A.9c) and (A.9d) via (A.7c), (A.7e) and (A.7g). 

A.4. Proof of Lemma A.2.1

For k = 1, all these relations (A.6a)–(A.6m) can be easily confirmed. We are going to prove that these relations also hold for k ≥ 2.

First of all, with the help of determinant properties, (A.6g) and (A.6i) can be easily checked by substituting (A.5a) and (A.5b) into the first column of $E_{k,j,0}^{(n+1)}$ and $E_{k,j,1}^{(n+1)}$, respectively, where the convention $\boldsymbol {\xi }_{k,0}^{(n+1)}=\boldsymbol {0}$ is used in the proof of (A.6i).

For the remaining relations, we will prove them by performing row or column transformation to a certain determinant and using the recursion relations (A.4) and (4.6) for $\{c_{j}^{(n)}\}$.

(1) Column transformation.

The proofs of (A.6a)–(A.6c) and (A.6l) are conducted by performing a series of column transformations on determinants.

We first consider (A.6b). To begin with, we deal with the last column of $H_{k,l}^{(n)}$ by substituting the relation (A.5b) into it, which yields

$$H_{k,l}^{(n)}=|\boldsymbol{C}_{k,l}^{(n)},\boldsymbol{C}_{k,l+1}^{(n)},\ldots,\boldsymbol{C}_{k,l+k-2}^{(n)},(\boldsymbol{B}_{k,l+k-2}^{(n+1)}+\alpha^{(n)} \boldsymbol{\xi}_{k,l+k-2}^{(n+1)}-\frac{1}{h}\boldsymbol{C}_{k,l+k-2}^{(n)})|.$$

Now, we perform some column transformations in order to eliminate the appeared vector $\alpha ^{(n)} \boldsymbol {\xi }_{k,l+k-2}^{(n+1)}$ and $\boldsymbol {C}_{k,l+k-2}^{(n)}/h$. Concretely, we add the (k − 1)-th column of $H_{k,l}^{(n)}$ multiplied by 1/h to the k-th column, then $\boldsymbol {C}_{k,l+k-2}^{(n)}/h$ is eliminated. Next, noticing l = 0, 1, adding the i-th column multiplied by $-\alpha ^{(n)} c_{k-1-i}^{(n+1)}$ to the k-th column for i = (2 − l), (3 − l),…, (k − 1), it is not hard to check that $\alpha ^{(n)} \boldsymbol {\xi }_{k,l+k-2}^{(n+1)}$ also disappears.

Then, we dispose the (k − 1)-th, (k − 2)-th, …, 2nd columns of $H_{k,l}^{(n)}$, successively. For each of these columns, after replacing the column vector $\boldsymbol {C}_{k,j}^{(n)}$ by (A.5b), we conduct similar column operations in order to eliminate the appeared vector $\alpha ^{(n)} \boldsymbol {\xi }_{k,j-1}^{(n+1)}$ and $\boldsymbol {C}_{k,j-1}^{(n)}/h$ in the corresponding column. Finally, it leads to

$$H_{k,l}^{(n)}=|\boldsymbol{C}_{k,l}^{(n)},\boldsymbol{B}_{k,l}^{(n+1)},\ldots,\boldsymbol{B}_{k,l+k-3}^{(n+1)},\boldsymbol{B}_{k,l+k-2}^{(n+1)}|,$$

which gives (A.6b).

Now, we turn to prove (A.6a). Based on the above calculations, we see that

$$H_{k+1,-1}^{(n)}=|\boldsymbol{C}_{k+1,-1}^{(n)},\boldsymbol{C}_{k+1,0}^{(n)},\boldsymbol{B}_{k+1,0}^{(n+1)},\ldots,\boldsymbol{B}_{k+1,k-3}^{(n+1)},\boldsymbol{B}_{k+1,k-2}^{(n+1)}|.$$

Then using (A.5a) and adding the first column multiplied by 1/(hγ⁽ⁿ⁾) to the second one, (A.6a) is immediately derived by eliminating $\boldsymbol {C}_{k+1,-1}^{(n)}/(h\gamma ^{(n)})$.

To prove (A.6c), we first rewrite the determinant as

$$H_{k,2}^{(n)}=\left| \begin{array}{cccccc} 1&0&0&\ldots&0\\ \boldsymbol{C}_{k,1}^{(n)}&\boldsymbol{C}_{k,2}^{(n+1)}&\boldsymbol{C}_{k,3}^{(n+1)}&\ldots&\boldsymbol{C}_{k,k+1}^{(n+1)} \end{array}\right|.$$

By following the similar column operations to the proof of (A.6b), we obtain

$$\begin{array}{@{}rcl@{}} H_{k,2}^{(n)}&=&\left| \begin{array}{cccccc} 1&\frac{1}{h}-\alpha^{(n)} c_0^{(n+1)}&-\alpha^{(n)} c_1^{(n+1)}&\ldots&-\alpha^{(n)} c_{k-1}^{(n+1)}\\ \boldsymbol{C}_{k,1}^{(n)}&\boldsymbol{B}_{k,1}^{(n+1)}&\boldsymbol{B}_{k,2}^{(n+1)}&\ldots&\boldsymbol{B}_{k,k}^{(n+1)} \end{array}\right|\\ &=&\left| \begin{array}{cccccc} 1&-\alpha^{(n)} c_0^{(n+1)}&-\alpha^{(n)} c_1^{(n+1)}&\ldots&-\alpha^{(n)} c_{k-1}^{(n+1)}\\ \boldsymbol{C}_{k,1}^{(n)}&\boldsymbol{B}_{k,1}^{(n+1)}&\boldsymbol{B}_{k,2}^{(n+1)}&\ldots&\boldsymbol{B}_{k,k}^{(n+1)} \end{array}\right|\\ &&+\left| \begin{array}{cccccc} 1&\frac{1}{h}&-\alpha^{(n)} c_1^{(n+1)}&\ldots&-\alpha^{(n)} c_{k-1}^{(n+1)}\\ \boldsymbol{C}_{k,1}^{(n)}&0&\boldsymbol{B}_{k,2}^{(n+1)}&\ldots&\boldsymbol{B}_{k,k}^{(n+1)} \end{array}\right|, \end{array}$$

which obviously leads to (A.6c).

Now, we proceed to prove (A.6l). Noticing first the determinant expression (3.1) of $P_{k}^{(n)}(x)$, it is easy to see that

$$\mathcal{H}_{2k,l}^{(n)}(x)=\frac{H_{k,l}^{(n)}P_{2k+l}^{(n)}(x)}{x^l},\qquad l=0,1.$$

Thus, to prove (A.6l), we only need to prove ${\mathscr{H}}_{2k,l}^{(n)}(x)=G_{2k,l}^{(n+1)}(x)$. This can be proved by performing some column transformations to ${\mathscr{H}}_{2k,l}^{(n)}(x)$ in a similar way to prove (A.6b). We omit the details.

(2) Row transformation.

The strategy to prove (A.6d)-(A.6f), (A.6h), (A.6j)-(A.6m) is based on a series of row transformations performed on determinants. The crucial observation is that, the column vectors with $\boldsymbol {B}_{k,j}^{(n+1)}$ can be converted to $\beta ^{(n)}\boldsymbol {C}_{k,j}^{(n+1)}/h$ after performing appropriate row transformations so that the summation term of $\boldsymbol {B}_{k,j}^{(n+1)}$ is eliminated.

Moreover, it should be noted that, since the determinants $\tilde {E}_{k,r+1,r}^{(n+1)},~E_{k,j,r}^{(n+1)},~K_{k,j,1}^{(n+1)}$, and $\tilde {G}_{2k,l}^{(n+1)}(x)$ have the majority columns in common in terms of $\boldsymbol {B}_{k,j}^{(n+1)}$, we will conduct the same row operations to transform them. In order to illustrate these consistent row transformations performed on determinants uniformly and clearly, we now take the following determinant as an example.

Noting that the types of the columns of the determinants in question mainly involve vectors $\boldsymbol {C}_{k,s}^{(n)},~\boldsymbol {B}_{k,j}^{(n+1)}$ with s = − 1, 0, 1 and j ≥− 1, we set

$$M_{k+2,j,-1}^{(n)}=|\boldsymbol{C}_{k+2,-1}^{(n)},\boldsymbol{C}_{k+2,0}^{(n)},\boldsymbol{B}_{k+2,j}^{(n+1)},\ldots,\boldsymbol{B}_{k+2,j+k-2}^{(n+1)},\boldsymbol{B}_{k+2,j+k-1}^{(n+1)}|,\qquad j=-1,0,1,\ldots.$$

The row transformations will be conducted on $M_{k+2,j,-1}^{(n)}$ from the second row until to the last row, successively. More precisely, for fixed p = 2, 3,…,k + 2, we add the i-th row multiplied by $-h\alpha ^{(n)} c_{p-i}^{(n)}/\beta ^{(n)}$ with i = 1, 2, 3,…,p − 1 to the p-th row of $M_{k+2,j,-1}^{(n)}$, which leads to

$$M_{k+2,j,-1}^{(n)}=\left| \begin{array}{cccccc} d_{-1}^{(n)}&a_0^{(n)}&\frac{\beta^{(n)} c_j^{(n+1)}}{h}&\cdots&\frac{\beta^{(n)} c_{j+k-1}^{(n+1)}}{h}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ d_{p-2}^{(n)}&a_{p-1}^{(n)}&\frac{\beta^{(n)} c_{j+p-1}^{(n+1)}}{h}&\cdots&\frac{\beta^{(n)} c_{j+p+k-2}^{(n+1)}}{h}\\ c_{p-1}^{(n)}&c_{p}^{(n)}&\frac{\beta^{(n)} c_{j+p}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{p}c_{i}^{(n)} c_{j+p-i}^{(n+1)}&\cdots&\frac{\beta^{(n)} c_{j+p+k-1}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{p}c_{i}^{(n)} c_{j+p+k-1-i}^{(n+1)}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ c_{k}^{(n)}&c_{k+1}^{(n)}&\frac{\beta^{(n)} c_{j+k+1}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{k+1}c_{i}^{(n)} c_{j+k+1-i}^{(n+1)}&\cdots&\frac{\beta^{(n)} c_{j+2k}^{(n+1)}}{h}+\alpha^{(n)}{\sum}_{i=1}^{k+1}c_{i}^{(n)} c_{j+2k-i}^{(n+1)} \end{array}\right|.$$

Obviously, for p = k + 2, we have

$$M_{k+2,j,-1}^{(n)}=\Big(\frac{\beta^{(n)}}{h}\Big)^{k}|\boldsymbol{D}_{k+2,-1}^{(n)},\boldsymbol{A}_{k+2,0}^{(n)},\boldsymbol{C}_{k+2,j}^{(n+1)},\ldots,\boldsymbol{C}_{k+2,j+k-1}^{(n+1)}|.$$

(A.5)

Then, (A.6d), (A.6e) and (A.6j) follow immediately since the involved columns of the determinants $E_{k,j,-1}^{(n+1)},~E_{k,j,0}^{(n+1)}$ and $\tilde {E}_{k,r+1,r}^{(n+1)}$, are only part of the determinant $M_{k+2,j,-1}^{(n)}$.

In order to prove the relations (A.6f), (A.6h) and (A.6k), we first conduct equivalent deformations on the determinants $E_{k,j,0}^{(n+1)}$, $E_{k,j,1}^{(n+1)}$ and $K_{k,0,1}^{(n+1)}$, before performing row transformations.

Note that

$$1-\frac{h u_0^{(n)}}{\beta^{(n)}}=\frac{1}{\beta^{(n)}},$$

and

$$\begin{array}{@{}rcl@{}} \boldsymbol{C}_{k,0}^{(n)}&=&(\boldsymbol{C}_{k,0}^{(n)}-\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}\boldsymbol{C}_{k,1}^{(n)})+\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}\boldsymbol{C}_{k,1}^{(n)},\\ \boldsymbol{C}_{k,1}^{(n)}&=&\beta^{(n)}(1-\frac{h u_0^{(n)}}{\beta^{(n)}})\boldsymbol{C}_{k,1}^{(n)}, \end{array}$$

from which, we immediately have

$$\begin{array}{@{}rcl@{}} E_{k,j,0}^{(n+1)}&=&|\Big(\boldsymbol{C}_{k,0}^{(n)}-\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}\boldsymbol{C}_{k,1}^{(n)}\Big),\boldsymbol{B}_{k,j}^{(n+1)},\ldots,\boldsymbol{B}_{k,j+k-3}^{(n+1)},\boldsymbol{B}_{k,j+k-2}^{(n+1)}|+\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}E_{k,j,1}^{(n+1)},\\ E_{k,j,1}^{(n+1)}&=&\beta^{(n)}|\Big(1-\frac{h u_0^{(n)}}{\beta^{(n)}}\Big)\boldsymbol{C}_{k,1}^{(n)},\boldsymbol{B}_{k,j}^{(n+1)},\ldots,\boldsymbol{B}_{k,j+k-3}^{(n+1)},\boldsymbol{B}_{k,j+k-2}^{(n+1)}|,\\ K_{k,j,1}^{(n+1)}&=&\beta^{(n)}\left| \begin{array}{cccccc} \frac{1}{\beta^{(n)}}&-\alpha^{(n)} c_j^{(n+1)}&-\alpha^{(n)} c_{j+1}^{(n+1)}&\ldots&-\alpha^{(n)} c_{j+k-2}^{(n+1)}\\ \Big(1-\frac{hu_0^{(n)}}{\beta^{(n)}}\Big)\boldsymbol{C}_{k-1,1}^{(n)}&\boldsymbol{B}_{k-1,j+1}^{(n+1)}&\boldsymbol{B}_{k-1,j+2}^{(n+1)}&\ldots&\boldsymbol{B}_{k-1,j+k-1}^{(n+1)} \end{array}\right|. \end{array}$$

Now, we operate the same row transformations as those on $M_{k+2,j,-1}^{(n)}$ upon the above determinants. For the first two, we have

$$\begin{array}{@{}rcl@{}} E_{k,j,0}^{(n+1)}&=&|\boldsymbol{D}_{k,0}^{(n)},\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k,j}^{(n+1)},\ldots\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k,j+k-2}^{(n+1)}|+\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}E_{k,j,1}^{(n+1)}\\ &=&\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}{\Gamma}_{k,j,0}^{(n+1)}+\frac{h\alpha^{(n)} d_{-1}^{(n)}}{\beta^{(n)}}E_{k,j,1}^{(n+1)},\\ E_{k,j,1}^{(n+1)}&=&\beta^{(n)}|\boldsymbol{A}_{k,1}^{(n)},\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k,j}^{(n+1)},\ldots,\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k,j+k-2}^{(n+1)}|=\frac{(\beta^{(n)})^{k}}{h^{k-1}}F_{k,j,1}^{(n+1)}, \end{array}$$

which are nothing but (A.6f) and (A.6h). While for the third one, it is not hard to obtain the following result by conducting the row transformations from the third row to the last one,

$$\begin{array}{@{}rcl@{}} K_{k,j,1}^{(n+1)}&=&\beta^{(n)}\left| \begin{array}{cccccc} \frac{1}{\beta^{(n)}}&-\alpha^{(n)} c_j^{(n+1)}&-\alpha^{(n)} c_{j+1}^{(n+1)}&\ldots&-\alpha^{(n)} c_{j+k-2}^{(n+1)}\\ \boldsymbol{A}_{k-1,1}^{(n+1)}&\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,j+1}^{(n+1)}&\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,j+2}^{(n+1)}&\ldots&\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,j+k-1}^{(n+1)} \end{array}\right|\\ &=&-\frac{\alpha^{(n)}(\beta^{(n)})^{k-1}}{h^{k-2}}\left| \begin{array}{cccccc} -\frac{1}{h\alpha^{(n)}}&c_j^{(n+1)}&c_{j+1}^{(n+1)}&\ldots&c_{j+k-2}^{(n+1)}\\ \boldsymbol{A}_{k-1,1}^{(n+1)}&\boldsymbol{C}_{k-1,j+1}^{(n+1)}&\boldsymbol{C}_{k-1,j+2}^{(n+1)}&\ldots&\boldsymbol{C}_{k-1,j+k-1}^{(n+1)} \end{array}\right|\\ &=&-\frac{\alpha^{(n)}(\beta^{(n)})^{k-1}}{h^{k-2}}\left| \begin{array}{cccccc} (A_{1,0}^{(n)}-\frac{\beta^{(n)}}{h\alpha^{(n)}})&c_j^{(n+1)}&c_{j+1}^{(n+1)}&\ldots&c_{j+k-2}^{(n+1)}\\ \boldsymbol{A}_{k-1,1}^{(n+1)}&\boldsymbol{C}_{k-1,j+1}^{(n+1)}&\boldsymbol{C}_{k-1,j+2}^{(n+1)}&\ldots&\boldsymbol{C}_{k-1,j+k-1}^{(n+1)} \end{array}\right|\\ &=&-\frac{\alpha^{(n)}(\beta^{(n)})^{k-1}}{h^{k-2}}\Big(F_{k,j,0}^{(n+1)}-\frac{\beta^{(n)}}{h\alpha^{(n)}}H_{k-1,j+1}^{(n+1)}\Big), \end{array}$$

which identifies with (A.6k) by observing the fact

$$-\frac{1}{h\alpha^{(n)}}=A_{1,0}^{(n)}-\frac{\beta^{(n)}}{h\alpha^{(n)}}.$$

Thus, (A.6f), (A.6h) and (A.6k) all get verified.

Now, we remain to prove (A.6m). For determinant $\tilde {G}_{2k,l}^{(n+1)}(x),~l=0,1$, from the second row to the last second row, after conducting several row transformations the same as those on $M_{k+2,j,-1}^{(n)}$, and noticing the expressions (A.3), (3.1) and (3.10b) of y_l,2k, $P_{k}^{(n)}$ and $Q_{k}^{(n)}$, respectively, we obtain that

$$\begin{array}{@{}rcl@{}} \tilde{G}_{2k,l}^{(n+1)}(x)&=&\left| \begin{array}{ccccc} \frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,l}^{(n+1)}&\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,l+1}^{(n+1)}&\ldots&\frac{\beta^{(n)}}{h}\boldsymbol{C}_{k-1,l+k-1}^{(n+1)}\\ y_{l,2}&y_{l,4}&\ldots&y_{l,2k} \end{array}\right|\\ &=&\frac{(\beta^{(n)})^{k-1}(hx^2+1)}{h^k}\left| \begin{array}{ccccc} \boldsymbol{C}_{k-1,l}^{(n+1)}&\boldsymbol{C}_{k-1,l+1}^{(n+1)}&\ldots&\boldsymbol{C}_{k-1,l+k-1}^{(n+1)}\\ 1&x^2&\ldots&x^{2k-2} \end{array}\right|\\ &&-\alpha^{(n)}\Big(\frac{\beta^{(n)}}{h}\Big)^{k-1}\left| \begin{array}{ccccc} \boldsymbol{C}_{k-1,l}^{(n+1)}&\boldsymbol{C}_{k-1,l+1}^{(n+1)}&\ldots&\boldsymbol{C}_{k-1,l+k-1}^{(n+1)}\\ {\sum}_{i=1-l}^0c_{-i}^{(n+1)}x^{2i}&{\sum}_{i=1-l}^1c_{1-i}^{(n+1)}x^{2i}&\ldots&{\sum}_{i=1-l}^{k-1}c_{k-1-i}^{(n+1)}x^{2i} \end{array}\right|\\ &=&\frac{(\beta^{(n)})^{k-1}(hx^2+1)H_{k-1,l}^{(n+1)} P_{2k-2+l}^{(n+1)}(x)}{h^kx^l}-\frac{\alpha^{(n)} (\beta^{(n)})^{k-1} H_{k-1,l}^{(n+1)}Q_{2k-3+l}^{(n+1)}(x)}{h^{k-1}x^{l-1}}, \end{array}$$

which is exactly (A.6m).

Therefore, we complete the proof of Lemma A.2.1.