#  $a$-potent Schwarz matrices and Bessel-like Jacobi polynomials

Alexander Dyachenko  Keldysh Institute of Applied Mathematics, Russian Academy
of Sciences, 125047, Moscow, Russia<EMAIL_ADDRESS>,  Carlos M. da
Fonseca  Kuwait College of Science and Technology, Doha District, Safat 13133,
Kuwait  Chair of Computational Mathematics, University of Deusto, 48007
Bilbao, Spain<EMAIL_ADDRESS>and  Mikhail Tyaglov  Department of
Mathematics and Computer Sciences, St.Petersburg State University,
St.Petersburg, 199178, Russia<EMAIL_ADDRESS>
###### Abstract.

We consider the problem of the reconstruction of a Schwarz matrix from exactly
one given eigenvalue. This inverse eigenvalue problem leads to the Jacobi
orthogonal polynomials $\\{P_{k}^{(-n,n)}\\}_{k=0}^{n-1}$ that can be treated
as a discrete finite analogue of Bessel polynomials.

###### Key words and phrases:

Inverse eigenvalue problem, Schwarz matrices, unipotent matrices, orthogonal
polynomials

###### 1991 Mathematics Subject Classification:

15B05, 33C45, 15A18, 15A29, 15A15

##  1\. Introduction

The matrix inverse eigenvalue problem consists in reconstructing a specific
structured matrix from a prescribed spectral data. This topic is of great
interest in different branches of mathematics, continually providing new and
surprising results (see for example [10, 22, 33, 34] and the references
there).

From the celebrated work [36] by Hubert S. Wall, it follows that there exists
a _unique_ matrix

(1.1) $J_{n}=\begin{bmatrix}-b_{0}&1&&&\\\ -b_{1}&0&1&&\\\
&-b_{2}&\ddots&\ddots&\\\ &&\ddots&\ddots&1\\\ &&&-b_{n-1}&0\\\ \end{bmatrix}$

with $b_{k}>0$, for $k=0,1,\ldots,n-1$, whose characteristic polynomial is
$(x+1)^{n}$ (see also [32, 35, 37] for more details). Wall’s argumentation is
based on the stability of the characteristic polynomial, i.e. on that all its
zeros have negative real parts — which is of course true for $(x+1)^{n}$.

By analogy with nilpotent matrices, one introduces unipotent matrices: an
$n\times n$ matrix is called unipotent if its characteristic polynomial is
$(x-1)^{n}$, or, equivalently, if all its eigenvalues are equal to $1$.
Various disciplines deal with unipotent matrices, cf. e.g. [6, 21, 28, 38]. We
consider an extension of this definition, namely matrices with only a single
eigenvalue labelled by $a$. In brief, we are looking for an “$a$-potent”
matrix of the form (1.1) where $a\in\mathbb{C}\setminus\\{0\\}$.

Matrices of the form (1.1) are called Schwarz matrices. Such matrices are very
important in automatic control theory and signal processing (see, for example,
[8]), and several general inverse problems have been previously considered in
[3, 7, 11, 13, 16]. So, here we consider the inverse eigenvalue problem for a
Schwarz matrix whose characteristic polynomial is

(1.2) $p(z)=(z-a)^{n}=\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}a^{n-i}z^{i},$

where $a$ is a given nonzero complex number. That is, we want to construct a
Schwarz matrix (1.1) with only one eigenvalue $a$.

Since $J_{k}$ is the leading principal $k\times k$ submatrix of $J_{n}$, with
$k\in\\{1,\ldots,n\\}$, the polynomials

$P_{k}(z)=\det(zI_{k}-J_{k})\,,\quad\mbox{for}\;k=1,\ldots,n\,,$

where $I_{k}$ is the identity matrix of order $k$ and $P_{n}(z)=p(z)$, satisfy
the following three-term recurrence relation:

$\displaystyle P_{0}(z)$ $\displaystyle=$ $\displaystyle 1\,,$ (1.3)
$\displaystyle P_{1}(z)$ $\displaystyle=$ $\displaystyle z+b_{0}\,,$
$\displaystyle P_{k+1}(z)$ $\displaystyle=$ $\displaystyle
zP_{k}(z)+b_{k}P_{k-1}(z)\,,\quad\mbox{for}\;k=1,\ldots,n-1\,.$

By the Favard theorem, see111In [27], this extended version is attributed to
Chihara, although in essence it already existed in [17]. [9, pp. 21–22], the
sequence $\\{P_{k}(z)\\}_{k=0}^{n-1}$ is orthogonal w.r.t. a linear
functional. Observe that, given arbitrary $n>1$ and
$a\in\mathbb{C}\setminus\\{0\\}$, this functional is not positive and cannot
be written as an integral w.r.t. a measure on the real line (otherwise $J_{n}$
would have $n$ distinct real eigenvalues). Accordingly, for $n>1$ the numbers
$b_{1},\dots,b_{n-1}$ in (1.3) cannot be chosen negative while $b_{0}$ remains
real. As is seen from Section 3.2, the support of the orthogonality functional
is natural to take consisting of a unique point $a$.

We find the aforementioned matrix $J_{n}$, see the values of $b_{0}$ and
$b_{1},\dots,b_{n-1}$ in, resp., (2.9) and (2.23). The sequence
$\\{P_{k}(z)\\}_{k=0}^{n-1}$ satisfying (1.3) reduces to a finite chunk of the
Jacobi polynomials with non-standard parameters. The explicit expression of
the polynomial $P_{k}(z)$ for $a=1$ is given in (3.3), while the relation of
the polynomials with distinct $a$ is in (3.2).

It turns out that three-term recurrence relations for the Bessel polynomials
[24] have a form similar to (1.3), so Bessel polynomials are also
characteristic polynomials of Schwarz matrices, and for $a=-\dfrac{1}{n}$ the
polynomials $P_{k}(z)$ approximate Bessel polynomials uniformly on the compact
sets of the complex plane as $n\to+\infty$. Moreover, the support of the
functional of orthogonality for Bessel polynomials on the real line may also
be treated as consisting of one point of essential singularity. Thus, we can
treat polynomials $\\{P_{k}(z)\\}_{k=0}^{n-1}$ as a discrete finite analogue
of Bessel polynomials.

The paper is organised as follows. Section 2 is devoted to finding the entries
of the matrix $J_{n}$. In Sections 3.1 and 3.2, we investigate properties of
the orthogonal polynomials $P_{k}(z)$ and find their functional of
orthogonality on the real line. In Section 3.3, we compare our polynomials
with Bessel polynomials and provide some calculations showing that the zeroes
of $P_{k}(z)$ approach the zeroes of Bessel polynomials for $a=-\dfrac{1}{n}$
as $n\to+\infty$. In Section 3.4, we calculate zeroes of $P_{k}(z)$ for
varying $a$ and $k$ to estimate numerically their asymptotic behaviour as $n$
is large.

##  2\. Explicit formula for Schwarz matrix

To find the entries of the matrix $J_{n}$ or, similarly, the coefficients of
the three-term recurrence relations (1.3), we consider the rational function

(2.1) $R(z)=b_{0}((zI_{n}-J_{n})^{-1}e_{1},e_{1})=\dfrac{q(z)}{p(z)},$

where $e_{1}$ is the first coordinate vector. Here the polynomial $q(z)$,
multiplied by $b_{0}$, is the characteristic polynomial of the principal
submatrix of $J_{n}$ (1.1) constructed from $J_{n}$ by deleting its first row
and the first column:

$q(z)=b_{0}\det\begin{bmatrix}z&-1&0&&\\\ b_{2}&z&-1&&\\\
&b_{3}&\ddots&\ddots&\\\ &&\ddots&\ddots&-1\\\ &&&b_{n-1}&z\\\ \end{bmatrix}.$

Observe that the polynomial $p(z)$, the characteristic polynomial of the
matrix $J_{n}$, can be represented as follows

(2.2) $p(z)=Q(z)+q(z)\,,$

where

$Q(z)=\det\begin{bmatrix}z&-1&0&&\\\ b_{1}&z&-1&&\\\ &b_{2}&\ddots&\ddots&\\\
&&\ddots&\ddots&-1\\\ &&&b_{n-1}&z\\\ \end{bmatrix}\,.$

It is easy to see that $q(z)$ is an even polynomial whenever $n$ is odd, and
is an odd polynomial otherwise, while the polynomial $Q(z)$ is odd if $n$ is
odd, and is even if $n$ is even. Therefore, by (2.2), the polynomial $q(z)$ is
the even (odd) part of the polynomial $p(z)$ whenever $n$ is odd (even). This
means that $q(z)$ can be obtained from $p(z)$ in the following way

$q(z)=\dfrac{p(z)-(-1)^{n}p(-z)}{2}\,,$

so that

(2.3) $\displaystyle q(z)$ $\displaystyle=$
$\displaystyle-\binom{n}{1}az^{n-1}-\binom{n}{3}a^{3}z^{n-3}-\binom{n}{5}a^{5}z^{n-5}-\cdots$
$\displaystyle=$
$\displaystyle-\sum_{j=0}^{\left\lfloor\tfrac{n-1}{2}\right\rfloor}\binom{n}{2j+1}a^{2j+1}z^{n-2j-1}.$

For the calculation of the entries of the matrix $J_{n}$ we first study the
rational function $R(z)$ defined in (2.1).

######  Theorem 2.1.

The rational function $R(z)=\dfrac{q(z)}{p(z)}$ has the form

$R(z)=\dfrac{\alpha_{1}}{z-a}+\dfrac{\alpha_{1}}{(z-a)^{2}}+\cdots+\dfrac{\alpha_{n}}{(z-a)^{n}}=\dfrac{s_{0}}{z}+\dfrac{s_{1}}{z^{2}}+\dfrac{s_{2}}{z^{3}}+\cdots,$

where

(2.4)
$\alpha_{k}=-2^{k-1}a^{k}\binom{n}{k}\,,\quad\mbox{for}\;k=1,\ldots,n\,,$

and

(2.5)
$s_{m}=\sum\limits_{k=1}^{n}a^{m-k+1}\binom{m}{k-1}\alpha_{k}=-a^{m+1}\sum\limits_{k=1}^{n}2^{k-1}\binom{m}{k-1}\binom{n}{k}.$

###### Proof.

Expanding the polynomial $q(z)$ into the Taylor series at the point $a$, we
obtain

$R(z)=\dfrac{q(a)}{(z-a)^{n}}+\dfrac{q^{\prime}(a)}{1!(z-a)^{n-1}}+\dfrac{q^{\prime\prime}(a)}{2!(z-a)^{n-2}}+\cdots+\dfrac{q^{(n-1)}(a)}{(n-1)!(z-a)}\,,$

so

$\alpha_{k}=\dfrac{q^{(n-k)}(a)}{(n-k)!}\,\,,\quad\mbox{for}\;k=1,\ldots,n\,.$

From (2.3), we have

$\dfrac{q^{(n-k)}(z)}{(n-k)!}=-\sum_{j=1}^{\left\lfloor\tfrac{k-1}{2}\right\rfloor}\binom{n}{2j+1}\binom{n-2j-1}{n-k}a^{2j+1}z^{k-2j-1}\,,\quad\mbox{for}\;k=1,\ldots,n,$

which implies

(2.6)
$\alpha_{k}=\dfrac{q^{(n-k)}(a)}{(n-k)!}=-a^{k}\binom{n}{k}\sum_{j=0}^{\left\lfloor\tfrac{k-1}{2}\right\rfloor}\binom{k}{2j+1}\,,\quad\mbox{for}\;k=1,\ldots,n.$

Now substituting

$2^{k-1}=\sum_{j=0}^{\left\lfloor\tfrac{k-1}{2}\right\rfloor}\binom{k}{2j+1}$

into (2.6), we get (2.4).

To establish (2.5), we recall that, for any nonnegative integer number $\ell$,

$\displaystyle\dfrac{1}{(z-a)^{\ell}}=\sum_{m=0}^{\infty}\binom{m+\ell-1}{\ell-1}\dfrac{a^{m}}{z^{m+\ell}}\,,$

so

$\displaystyle R(z)$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\dfrac{\alpha_{k}}{(z-a)^{k}}=\sum_{m=0}^{\infty}\sum_{k=1}^{n}\dfrac{\binom{m+k-1}{k-1}a^{m}\alpha_{k}}{z^{m+k}}$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\dfrac{1}{z^{m+1}}\sum_{k=1}^{n}a^{m-k+1}\binom{m}{k-1}\alpha_{k}\,.$

This formula together with (2.4) gives us (2.5). ∎

######  Remark 2.2.

Note that the numbers $s_{m}$ satisfy the following recurrence relation

$(1+m)a^{2}s_{m}+2nas_{m+1}-(3+m)s_{m+2}=0\,.$

A careful consideration shows that the moments $s_{m}$ are polynomials in $n$,
and can be represented as

$s_{m}(n)=-\dfrac{a^{m+1}n}{(m+1)!}\,p_{m}(n)\,,\quad\mbox{for}\;m=0,1,2,\ldots,$

where $p_{m}(x)$ are the polynomials satisfying the following three-term
recurrence relations:

$p_{m+1}(x)=xp_{m}(x)+m(m+1)p_{m-1}(x)\,,\quad\mbox{for}\;m=1,2,\ldots$

with the initial conditions $p_{0}(x)=1$, $p_{1}(x)=x$. These recurrence
relations generate the following continued fraction (see, e.g., [26, p.231]):

$\Phi(x)=x\int\limits_{0}^{\infty}e^{-xt}\tanh(t)\,dt=\cfrac{1}{x+\cfrac{1\cdot
2}{x+\cfrac{2\cdot 3}{x+\cdots}}}$

convergent for $x>0$. In fact, the integral itself converges in the right
half-plane, so it generates a measure of orthogonality for the polynomials
$p_{m}(x)$ with the support on the imaginary axis.

It is not hard to identify this system of orthogonal polynomials: on comparing
the recurrence coefficients one observes that they are the Meixner–Pollaczek
polynomials $P_{m}^{(1)}\big{(}\>\cdot\>;\frac{\pi}{2}\big{)}$ up to rotation
and renormalisation. Namely,

$p_{m}(n)=-\frac{(m+1)!}{a^{m+1}n}s_{m}(n)=(m+1)!\>{}_{2}F_{1}\left(\begin{matrix}-m,1-n\\\
2\end{matrix}\,\Big{|}\,2\right)=\frac{m!}{i^{m}}P_{m}^{(1)}\left(in;\frac{\pi}{2}\right),$

so the corresponding orthogonality measure is $\left|\Gamma(1-x)\right|^{2}$
with $x\in i\mathbb{R}$, see222Note a typo in formula (3.22) on that page: the
third parameter of ${}_{2}F_{1}$ is actually $2\lambda$, not $2$. [9, p. 180].

The entries $b_{m}$ of the matrix $J_{n}$ defined in (1.1) can be found by the
formula (see, e.g., [20, 35] and references therein)

(2.7)
$b_{m}=-\dfrac{D_{m-1}(R)D_{m+1}(R)}{D_{m}^{2}(R)}\,,\quad\mbox{for}\;m=1,2,\ldots,n-1\,,$

where $D_{0}(R)=1$, and $D_{m}(R)$, for $m=1,\ldots,n$, are the Hankel
determinants defined as follows

(2.8) $D_{m}(R)=\begin{vmatrix}s_{0}&s_{1}&s_{2}&\cdots&s_{m-1}\\\
s_{1}&s_{2}&s_{3}&\cdots&s_{m}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
s_{m-1}&s_{m}&s_{m+1}&\cdots&s_{2m-2}\end{vmatrix}\,,\quad\mbox{for}\;m=1,2,\dots,n.$

Recall that

$D_{m}(R)=0\,,\qquad\mbox{for $m=n+1,n+2,\ldots,$}$

since the rational function $R(z)$ has exactly $n$ poles, counting
multiplicities (see, e.g., [15, Ch. XV, §10, Theorem 8] or, alternatively,
[20, Theorem 1.2]).

Moreover,

(2.9) $b_{0}=-an,$

since $-b_{0}$ equals the trace of $J_{n}$.

To find the exact formula for the minors $D_{m}(R)$ we need the following
general result that can be obtained using elementary determinant properties.

######  Lemma 2.3.

The minors $D_{m}(R)$ can be represented in the following form:

(2.10) $D_{m}(R)=\begin{vmatrix}\alpha_{1}&\alpha_{2}&\dots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\dots&\alpha_{m+1}\\\ \vdots&\vdots&\ddots&\vdots\\\
\alpha_{m}&\alpha_{m+1}&\dots&\alpha_{2m-1}\end{vmatrix}\,,$

for $m=1,2,\dots,n$, where we set $\alpha_{j}=0$, for $j>n$.

###### Proof.

Indeed, the $k$th column $\mathbf{C}_{k}$ of the minor $D_{m}(R)$ has the form

(2.11) $\mathbf{C}_{k}=\begin{bmatrix}s_{k-1}\\\ \vdots\\\
s_{k+m-2}\end{bmatrix}\,,\quad\mbox{for}\;k=1,\dots,m\,.$

If we change columns as follows

(2.12)
$\widetilde{\mathbf{C}}_{k}=\sum_{j=1}^{k}(-a)^{k-j}\binom{k-1}{j-1}\mathbf{C}_{j}\,,$

for $k=1,\dots,m$, where $\widetilde{\mathbf{C}}_{k}$ is the $k$th new column,
then the minor $D_{m}(R)$ does not change its value. From (2.11)–(2.12), we
obtain that the $\ell$-entry $(\widetilde{\mathbf{C}}_{k})_{\ell}$ of the new
$k$th column has the form

$\displaystyle(\widetilde{\mathbf{C}}_{k})_{\ell}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{k}(-a)^{k-j}\binom{k-1}{j-1}s_{j+\ell-2}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{k}(-a)^{k-j}\binom{k-1}{j-1}\sum\limits_{i=1}^{n}a^{j+\ell-1-i}\binom{j+\ell-2}{i-1}\alpha_{i}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{n}a^{k+\ell-1-i}\alpha_{i}\sum_{j=0}^{k-1}(-1)^{k-1-j}\binom{k-1}{j}\binom{j+\ell-1}{i-1}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}a^{k+\ell-1-i}\binom{\ell-1}{i-k}\alpha_{i}=\sum_{i=1}^{\ell}a^{\ell-i}\binom{\ell-1}{i-1}\alpha_{k+i-1}\,,$

where $k,\ell=1,\dots,m$ and $\alpha_{j}=0$, for $j>n$. Here we used the
identity

$\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}\binom{k+\ell}{i}=\binom{\ell}{i-m}$

which is true for any nonnegative integer numbers $m,\ell,i$ (cf., e.g., [31,
Section 4.2.5, no. 55] or [18, Chapter 5, Table 169]).

Thus, we get the following formula for the minors $D_{m}(R)$:

(2.13) $D_{m}(R)=\begin{vmatrix}\alpha_{1}&\alpha_{2}&\cdots&\alpha_{m}\\\
a\alpha_{1}+\alpha_{2}&a\alpha_{2}+\alpha_{3}&\cdots&a\alpha_{m}+\alpha_{m+1}\\\
\vdots&\vdots&\ddots&\vdots\\\
\displaystyle\sum_{r=1}^{m}a^{m-r}\binom{m}{r}\alpha_{r}&\displaystyle\sum_{r=1}^{m}a^{m-r}\binom{m}{r}\alpha_{r+1}&\cdots&\displaystyle\sum_{r=1}^{m}a^{m-r}\binom{m}{r}\alpha_{r+m-1}\end{vmatrix}\,,$

for $m=1,2,\dots,n$.

Let us consider now the $\ell$th row of the matrix (2.13). It has the form

(2.14)
$\mathbf{R}_{\ell}=\begin{bmatrix}v_{\ell,1},v_{\ell,2},\ldots,v_{\ell,m}\end{bmatrix}\,\,,\quad\mbox{for}\;\ell=1,\dots,m\,,$

where

(2.15) $v_{\ell
k}=\sum_{i=1}^{l}a^{\ell-i}\binom{\ell-1}{i-1}\alpha_{i+k-1}\,,\quad\mbox{for}\;\ell,k=1,2,\ldots,m\,.$

Now we change the rows as follows

(2.16)
$\widetilde{\mathbf{R}}_{\ell}=\sum_{j=1}^{\ell}(-a)^{\ell-j}\binom{\ell-1}{j-1}\mathbf{R}_{j}\,,\quad\mbox{for}\;\ell=1,\dots,m\,,$

where $\widetilde{\mathbf{R}}_{\ell}$ is the $\ell$th new row. This operation
does not change the value of the minor $D_{m}(R)$. From (2.14)–(2.16), we
conclude that the $k$th entry $(\widetilde{\mathbf{R}}_{\ell})_{k}$ of the new
$\ell$th row has the form

(2.17) $\displaystyle(\widetilde{\mathbf{R}}_{\ell})_{k}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{\ell}(-a)^{\ell-j}\binom{\ell-1}{j-1}v_{j,k}$
$\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{\ell}\sum\limits_{i=1}^{j}a^{\ell-i}(-1)^{\ell-j}\binom{\ell-1}{j-1}\binom{j-1}{i-1}\alpha_{i+k-1}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{\ell}a^{\ell-i}\alpha_{i+k-1}\sum\limits_{j=i}^{\ell}(-1)^{\ell-j}\binom{\ell-1}{j-1}\binom{j-1}{i-1}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{\ell}a^{\ell-i}\alpha_{i+k-1}\delta_{\ell
i}=\alpha_{k+\ell-1},$

for $k,\ell=1,\dots,m$. Now (2.17) implies (2.10). ∎

Now we are in a position to calculate the values of the Hankel minors
$D_{m}(R)$.

######  Theorem 2.4.

The minors $D_{m}(R)$ have the form

(2.18)
$D_{m}(R)=(-1)^{\tfrac{m(m+1)}{2}}a^{m^{2}}2^{m(m-1)}\binom{n}{m}\prod_{k=0}^{m-1}\dfrac{\binom{n+k}{n-k}}{\binom{m+k}{m-k}}\,,$

for $m=1,2,\dots,n$.

###### Proof.

From (2.4) and (2.10) we have

(2.19) $\displaystyle D_{m}(R)$ $\displaystyle=$
$\displaystyle\begin{vmatrix}-a\binom{n}{1}&-2a^{2}\binom{n}{2}&\dots&-2^{m-1}a^{m}\binom{n}{m}\\\
-2a^{2}\binom{n}{2}&-2^{2}a^{3}\binom{n}{3}&\dots&-2^{m}a^{m+1}\binom{n}{m+1}\\\
\vdots&\vdots&\ddots&\vdots\\\
-2^{m-1}a^{m}\binom{n}{m}&-2^{m}a^{m+1}\binom{n}{m+1}&\dots&-2^{2m-2}a^{2m-1}\binom{n}{2m-1}\end{vmatrix}$
$\displaystyle=$
$\displaystyle(-1)^{m}2^{\tfrac{m(m-1)}{2}}a^{\tfrac{m(m+1)}{2}}\begin{vmatrix}\binom{n}{1}&2a\binom{n}{2}&\dots&2^{m-1}a^{m-1}\binom{n}{m}\\\
\binom{n}{2}&2a\binom{n}{3}&\dots&2^{m-1}a^{m-1}\binom{n}{m+1}\\\
\vdots&\vdots&\ddots&\vdots\\\
\binom{n}{m}&2a\binom{n}{m+1}&\dots&2^{m-1}a^{m-1}\binom{n}{2m-1}\end{vmatrix}$
$\displaystyle=$
$\displaystyle(-1)^{m}2^{\tfrac{m(m-1)}{2}}a^{\tfrac{m(m+1)}{2}}(2a)^{\tfrac{m(m-1)}{2}}\begin{vmatrix}\binom{n}{1}&\binom{n}{2}&\dots&\binom{n}{m}\\\
\binom{n}{2}&\binom{n}{3}&\dots&\binom{n}{m+1}\\\
\vdots&\vdots&\ddots&\vdots\\\
\binom{n}{m}&\binom{n}{m+1}&\dots&\binom{n}{2m-1}\end{vmatrix}$
$\displaystyle=$ $\displaystyle(-1)^{m}2^{m(m-1)}a^{m^{2}}T_{m}\,,$

where

(2.20) $\displaystyle T_{m}$ $\displaystyle=$
$\displaystyle\begin{vmatrix}\binom{n}{1}&\binom{n}{2}&\dots&\binom{n}{m}\\\
\binom{n}{2}&\binom{n}{3}&\dots&\binom{n}{m+1}\\\
\vdots&\vdots&\ddots&\vdots\\\
\binom{n}{m}&\binom{n}{m+1}&\dots&\binom{n}{2m-1}\end{vmatrix}$
$\displaystyle=$
$\displaystyle\begin{vmatrix}\tfrac{n}{1!}&\tfrac{n(n-1)}{2!}&\dots&\tfrac{n(n-1)\cdots(n-m+1)}{m!}\\\
\tfrac{n(n-1)}{2!}&\tfrac{n(n-1)(n-2)}{3!}&\dots&\tfrac{n(n-1)\cdots(n-m)}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{n(n-1)\cdots(n-m+1)}{m!}&\tfrac{n(n-1)\cdots(n-m)}{(m+1)!}&\dots&\tfrac{n(n-1)\cdots(n-2m+2)}{(2m-1)!}\end{vmatrix}$

for $m=1,2,3,\dots,n$. Taking into account that the $j$th, $(j+1)$th,…, $m$th
rows have the common factor $n(n-1)\cdots(n-j+1)$, for $j=1,\ldots,m$, we get

$\displaystyle T_{m}$ $\displaystyle=$ $\displaystyle
n^{m}(n-1)^{m-1}\cdots(n-m)^{2}(n-m+1)\begin{vmatrix}\tfrac{1}{1!}&\tfrac{n-1}{2!}&\dots&\tfrac{(n-1)\cdots(n-m+1)}{m!}\\\
\tfrac{1}{2!}&\tfrac{n-2}{3!}&\dots&\tfrac{(n-2)\cdots(n-m)}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{n-m}{(m+1)!}&\dots&\tfrac{(n-m)\cdots(n-2m+2)}{(2m-1)!}\end{vmatrix}$
$\displaystyle=$
$\displaystyle\prod_{k=1}^{m}k!\binom{n}{k}\,\begin{vmatrix}\tfrac{1}{1!}&\tfrac{n-1}{2!}&\dots&\tfrac{(n-1)\cdots(n-m+1)}{m!}\\\
\tfrac{1}{2!}&\tfrac{n-2}{3!}&\dots&\tfrac{(n-2)\cdots(n-m)}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{n-m}{(m+1)!}&\dots&\tfrac{(n-m)\cdots(n-2m+2)}{(2m-1)!}\end{vmatrix}$

Now we add the $(m-1)$th column to the $m$th column, then the $(m-2)$th column
to the $(m-1)$th column, etc. The identity

(2.21)
$\begin{array}[]{l}\dfrac{(n-i)(n-i-1)\cdots(n-i-j+3)}{(i+j-2)!}+\dfrac{(n-i)(n-i-1)\cdots(n-i-j+2)}{(i+j-1)!}=\\\
\\\ =(n+1)\dfrac{(n-i)(n-i-1)\cdots(n-i-j+3)}{(i+j-1)!}\end{array}$

is used repeatedly for each element of the $j$th column, for
$j=m,m-1,\ldots,2$. Thus, we conclude that the determinant does not change its
value, while its $2$nd, $3$rd, …, $m$th columns have the common factor $n+1$.
Therefore, we have

$T_{m}=(n+1)^{m-1}\prod_{k=1}^{m}k!\binom{n}{k}\,\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\dots&\tfrac{(n-1)\cdots(n-m+2)}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\dots&\tfrac{(n-2)\cdots(n-m+1)}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\dots&\tfrac{(n-m)\cdots(n-2m+3)}{(2m-1)!}\end{vmatrix}$

Now we use the similar procedure for the $3$rd, $4$th,$\ldots$, $(m+1)$th
columns. Namely, we add the $(m-1)$th column to the $m$th, then the $(m-2)$th
column to the $(m-1)$th, etc., and use the identity

$\begin{array}[]{l}\dfrac{(n-i)(n-i-1)\cdots(n-i-j+4)}{(i+j-2)!}+\dfrac{(n-i)(n-i-1)\cdots(n-i-j+3)}{(i+j-1)!}=\\\
\\\ =(n+2)\dfrac{(n-i)(n-i-1)\cdots(n-i-j+4)}{(i+j-1)!}\end{array}$

repeatedly for each element of the $j$th column, for $j=3,\ldots,m$. Again we
conclude that the determinant does not change its value, while its $3$rd,
$4$th,$\ldots$, $m$th columns have the common factor $n+2$. So we have

$T_{m}\displaystyle=(n+1)^{m-1}(n+2)^{m-2}\prod_{k=1}^{m}k!\binom{n}{k}\,\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\cdots&\tfrac{(n-1)\cdots(n-m+3)}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\cdots&\tfrac{(n-2)\cdots(n-m+2)}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\dots&\tfrac{(n-m)\cdots(n-2m+4)}{(2m-1)!}\end{vmatrix}$

Consequently, from a straightforward inductive argument, we get

$\displaystyle T_{m}$ $\displaystyle=$
$\displaystyle\prod_{\ell=1}^{m-1}(n+\ell)^{m-\ell}\prod_{k=1}^{m}k!\binom{n}{k}\,\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\cdots&\tfrac{1}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\cdots&\tfrac{1}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\cdots&\tfrac{1}{(2m-1)!}\end{vmatrix}$
$\displaystyle=$
$\displaystyle\prod_{k=1}^{m}k!\binom{n}{k}\prod_{j=1}^{m-1}j!\binom{n+j}{j}\,\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\cdots&\tfrac{1}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\cdots&\tfrac{1}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\cdots&\tfrac{1}{(2m-1)!}\end{vmatrix}$
$\displaystyle=$ $\displaystyle
m!\binom{n}{m}\prod_{k=1}^{m-1}(2k)!\binom{n+k}{n-k}\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\cdots&\tfrac{1}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\cdots&\tfrac{1}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\cdots&\tfrac{1}{(2m-1)!}\end{vmatrix}\,.$

Note that the determinant $T_{m}$ equals $(-1)^{\tfrac{m(m-1)}{2}}$, if $n=m$,
as it follows from (2.20). Therefore,

(2.22) $\begin{vmatrix}\tfrac{1}{1!}&\tfrac{1}{2!}&\cdots&\tfrac{1}{m!}\\\
\tfrac{1}{2!}&\tfrac{1}{3!}&\cdots&\tfrac{1}{(m+1)!}\\\
\vdots&\vdots&\ddots&\vdots\\\
\tfrac{1}{m!}&\tfrac{1}{(m+1)!}&\cdots&\tfrac{1}{(2m-1)!}\end{vmatrix}=\displaystyle\dfrac{(-1)^{\tfrac{m(m-1)}{2}}}{m!\displaystyle\prod_{k=1}^{m-1}(2k)!\binom{m+k}{m-k}},$

and, consequently,

$\displaystyle\displaystyle T_{m}$ $\displaystyle=$
$\displaystyle(-1)^{\tfrac{m(m-1)}{2}}\dfrac{m!\displaystyle\binom{n}{m}\prod_{k=1}^{m-1}(2k)!\binom{n+k}{n-k}}{m!\displaystyle\prod_{k=1}^{m-1}(2k)!\binom{m+k}{m-k}}$
$\displaystyle=$
$\displaystyle(-1)^{\tfrac{m(m-1)}{2}}\binom{n}{m}\prod_{k=1}^{m-1}\dfrac{\binom{n+k}{n-k}}{\binom{m+k}{m-k}}\,.$

Substituting this expression for $T_{m}$ into (2.19), we obtain (2.18), as
required. ∎

######  Remark 2.5.

The determinant (2.22) is connected to the Padé approximations to the
exponential, and its value can be found, e.g., in [2, Section 1.2, p.9].

######  Remark 2.6.

On summing up the matrix transformations behind Theorem 2.4, one additionally
arrives at the following factorisation of the matrix from (2.20):

$\left[\binom{n}{i+j-1}\right]_{i,j=1}^{m}=\operatorname{diag}\left[(n-j)_{j+1}\right]_{j=0}^{m-1}\cdot\left[\frac{1}{(i+j-1)!}\right]_{i,j=1}^{m}\cdot\operatorname{diag}\left[(n)_{j}\right]_{j=0}^{m-1}\cdot\left[(-1)^{i+j}\binom{j}{i}\right]_{i,j=0}^{m-1}\,,$

where

$\left[(-1)^{i+j}\binom{j}{i}\right]_{i,j=0}^{m-1}=\left(\left[\binom{j}{i}\right]_{i,j=0}^{m-1}\right)^{-1}$

is the inverse of the upper-triangular Pascal matrix. This inverse of the
Pascal matrix is a product of $m-1$ bidiagonal terms stemming from column
operations and application of the Pascal identity starting (2.21). The Pascal
matrix itself also has an analogous bidiagonal factorisation (but the terms
are unsigned). The remaining term, whose determinant appears on the left-hand
side of (2.22), can be expressed from the matrix
$\left[\binom{m}{i+j-1}\right]_{i,j=1}^{m}$ by the reverse transformations
with $n$ set to be $m$, that is

$\left[\frac{1}{(i+j-1)!}\right]_{i,j=1}^{m}\\!\\!\\!\\!\\!\\!=\operatorname{diag}\left[\frac{1}{(m-j)_{j+1}}\right]_{j=0}^{m-1}\cdot\left[\binom{m}{i+j-1}\right]_{i,j=1}^{m}\\!\\!\\!\\!\\!\\!\cdot\operatorname{diag}\left[\frac{1}{(m)_{j}}\right]_{j=0}^{m-1}\cdot\left[\binom{j}{i}\right]_{i,j=0}^{m-1}\\!\\!\\!.$

At the same time, the Hankel matrix
$\left[\binom{m}{i+j-1}\right]_{i,j=1}^{m}$ is built from the coefficients of
the polynomial $(x+1)^{m}$, and hence it may be further factorised into
$m\times m$ factors

$\left[\binom{m}{i+j-1}\right]_{i,j=1}^{m}=\begin{bmatrix}0&\ldots&0&0&1\\\\[-3.0pt]
\vdots&\iddots&0&1&0\\\\[-3.0pt] \vdots&\iddots&1&0&0\\\\[-3.0pt]
0&\iddots&\iddots&\iddots&\vdots\\\ 1&0&0&\ldots&0\\\
\end{bmatrix}\cdot\begin{bmatrix}1&0&0&\ldots&0\\\\[-3.0pt]
1&1&0&\ddots&\vdots\\\\[-3.0pt] 0&1&1&\ddots&0\\\\[-3.0pt]
\vdots&\ddots&\ddots&\ddots&0\\\ 0&\ldots&0&1&1\\\
\end{bmatrix}^{m}\\!\\!\\!.$

Now (2.7) and (2.18) give us the exact formula for the entries of the matrix
(1.1) whose characteristic polynomial is (1.2).

######  Corollary 2.7.

The coefficients $b_{m}$ of the relations (1.3) have the form

(2.23)
$b_{m}=a^{2}\,\dfrac{n^{2}-m^{2}}{4m^{2}-1}\,,\quad\mbox{for}\;m=1,2,\ldots,n-1.$

###### Proof.

Let us consider first the relation

(2.24) $\displaystyle\dfrac{D_{m+1}(R)}{D_{m}(R)}$ $\displaystyle=$
$\displaystyle\dfrac{(-1)^{\tfrac{(m+1)(m+2)}{2}}2^{(m+1)m}a^{(m+1)^{2}}\binom{n}{m+1}\displaystyle\prod_{k=1}^{m}\dfrac{\binom{n+k}{n-k}}{\binom{m+1+k}{m+1-k}}}{(-1)^{\tfrac{m(m+1)}{2}}2^{m(m-1)}a^{m^{2}}\binom{n}{m}\displaystyle\prod_{k=1}^{m-1}\dfrac{\binom{n+k}{n-k}}{\binom{m+k}{m-k}}}$
$\displaystyle=$
$\displaystyle\displaystyle(-1)^{m+1}4^{m}a^{2m+1}\dfrac{n-m}{m+1}\dfrac{\binom{n+m}{n-m}}{2m+1}\prod_{k=1}^{m-1}\dfrac{\binom{m+k}{m-k}}{\binom{m+1+k}{m+1-k}}$
$\displaystyle=$
$\displaystyle(-1)^{m+1}4^{m}a^{2m+1}\dfrac{n-m}{m+1}\dfrac{\binom{n+m}{n-m}}{2m+1}\prod_{k=1}^{m-1}\dfrac{m+1-k}{m+1+k}$
$\displaystyle=$
$\displaystyle(-1)^{m+1}4^{m}a^{2m+1}\dfrac{n-m}{(2m+1)!}\displaystyle\binom{n+m}{n-m}(m!)^{2}.$

Now, from (2.7) and (2.24), we have

$\displaystyle b_{m}$ $\displaystyle=$
$\displaystyle-\dfrac{D_{m+1}(R)/D_{m}(R)}{D_{m}(R)/D_{m-1}(R)}$
$\displaystyle=$
$\displaystyle-\dfrac{(-1)^{m+1}4^{m}a^{2m+1}\dfrac{n-m}{(2m+1)!}\displaystyle\binom{n+m}{n-m}(m!)^{2}}{(-1)^{m}4^{m-1}a^{2m-1}\dfrac{n-m+1}{(2m-1)!}\displaystyle\binom{n+m-1}{n-m+1}((m-1)!)^{2}}$
$\displaystyle=$ $\displaystyle
4m^{2}a^{2}\,\dfrac{n-m}{2m(2m+1)}\dfrac{(n+m)!}{(n-m)!(2m)!}\dfrac{(n-m)!(2m-2)!}{(n+m-1)!}$
$\displaystyle=$ $\displaystyle
a^{2}\,\dfrac{n-m}{(2m-1)(2m+1)}\dfrac{(n+m)!}{(n+m-1)!}=a^{2}\,\dfrac{n^{2}-m^{2}}{4m^{2}-1}\,.$

∎

As a conclusion, up to similarity, the only Schwarz matrix $J_{n}$ whose
characteristic polynomial is $(z-a)^{n}$ is

(2.25) $J_{n}=\begin{bmatrix}an&1&&&\\\ -a^{2}\dfrac{n^{2}-1}{3}&0&1&&\\\
&-a^{2}\dfrac{n^{2}-4}{15}&\ddots&\ddots&\\\ &&\ddots&\ddots&1\\\
&&&-a^{2}\dfrac{1}{2n-3}&0\\\ \end{bmatrix}.$

Clearly, the spectrum of the matrix

$\tilde{J}_{n}=a\begin{bmatrix}n&\dfrac{n-1}{1}&&&\\\
-\,\dfrac{n+1}{3}&0&\dfrac{n-2}{3}&&\\\ &-\,\dfrac{n+2}{5}&\ddots&\ddots&\\\
&&\ddots&\ddots&\dfrac{1}{2n-3}\\\ &&&-1&0\\\ \end{bmatrix}$

is reduced to $a$ as well.

##  3\. Orthogonal polynomials

In this section, we consider the sequence $\\{P_{k}(z)\\}_{k=0}^{n-1}$ of the
characteristic polynomials of the leading principal minors of the matrix
(2.25) satisfying the three-term recurrence relation (1.1)–(1.3):

$P_{0}(z)=1,\quad P_{1}(z)=z-an,$ (3.1)
$P_{k+1}(z)=zP_{k}(z)+\dfrac{a^{2}(n^{2}-k^{2})}{(2k-1)(2k+1)}P_{k-1}(z)\,,\quad\mbox{for}\;k=1,\ldots,n-1.$

We give their explicit representations and study their even and odd parts.
Then we show that $\\{P_{k}(z)\\}_{k=0}^{n-1}$ are orthogonal w.r.t. a linear
functional supported at the only point $z=a$. We also find the inverse Fourier
transform of the corresponding distributional weight and depict the zeroes of
the polynomials $P_{k}(z)$ for various $a$ and relations between $k$ and $n$
as $n\to+\infty$.

###  3.1. Basic properties

For the current subsection we assume $a=1$ keeping in mind that the general
case $a\neq 0$ may be obtained by

(3.2) $a^{-k}P_{k}(az)=P_{k}(z)\big{|}_{a=1},$

which immediately follows from (3.1). Observe that the monic Jacobi
polynomials

$\widehat{P}_{k}^{(-n,n)}(z)=\frac{2^{k}}{\binom{2k}{k}}P_{k}^{(-n,n)}(z)\qquad\left(=\frac{(-2)^{k}}{\binom{2k}{k}}P_{k}^{(n,-n)}(-z)\right)$

satisfy for $k\geq 1$ the three-term recurrence relation [9, p. 153, eq.
(2.29)]

$\widehat{P}_{k+1}^{(-n,n)}(z)=z\widehat{P}_{k}^{(-n,n)}(z)-\frac{4k(k-n)(k+n)k}{(2k-1)(2k+1)(2k)^{2}}\widehat{P}_{k-1}^{(-n,n)}(z),$

which on reduction of similar terms turns into (3.1) with $a=1$. Similarly,
$\widehat{P}_{0}^{(-n,n)}(z)\equiv 1$ and $\widehat{P}_{1}^{(-n,n)}(z)=z-n$.
Therefore, our $P_{k}(z)$ are exactly $\widehat{P}_{k}^{(-n,n)}(z)$, namely,
for all $k=0,\dots,n,$

(3.3)
$P_{k}(z)=\widehat{P}_{k}^{(-n,n)}(z)=(-2)^{k}\frac{(n+1)_{k}}{(k+1)_{k}}\,{}_{2}F_{1}\left(\genfrac{.}{|}{0.0pt}{0}{-k,\
k+1}{n+1}\dfrac{1+z}{2}\right).$

Also,

$P_{k}(z)\big{|}_{a=1}=\dfrac{(-1)^{k}}{(k+1)_{k}}\left(\dfrac{1-z}{1+z}\right)^{n}\dfrac{d^{k}}{dz^{k}}\left[(1-z)^{n-k}(1+z)^{n+k}\right],$
$(1-z^{2})P_{k}^{\prime\prime}(z)+2(n-z)P_{k}^{\prime}(z)+k(k+1)P_{k}(z)=0.$

Although the three-term recurrence degenerates to a two-term recurrence for
$k=n$, the sequence $\\{P_{k}\\}_{k=0}^{n}$ may be further extended to an
infinite sequence. Properties of infinite sequences of the monic polynomials
$\\{\widehat{P}_{k}^{(-n,\beta)}\\}_{k=0}^{\infty}$ with
$\beta\notin\\{-1,-2,\dots\\}$ were studied in [1], where orthogonality was
introduced via a bilinear form.

Now, on introducing two auxiliary families of polynomials via

(3.4) $f_{0}(z)=1\,,\quad f_{1}(z)=z\,,\quad\text{and}\quad
f_{k+1}(z)=zf_{k}(z)+\dfrac{n^{2}-k^{2}}{4k^{2}-1}f_{k-1}(z)$

and

(3.5) $g_{0}(z)=0\,,\quad g_{1}(z)=1\,,\quad\text{and}\quad
g_{k+1}(z)=zg_{k}(z)+\dfrac{n^{2}-k^{2}}{4k^{2}-1}g_{k-1}(z)$

with $k=1,\dots,n-1$, one immediately obtains that the decomposition

$P_{k}(z)=f_{k}(z)-ng_{k}(z)$

is valid for all $k$. The formula (3.4) determines monic orthogonal
polynomials supported on the imaginary axis, and (3.5) gives exactly the
corresponding monic numerator polynomials. In particular, for each $k$ the
zeroes of $f_{k}$ and $g_{k}$ are all simple and pure imaginary, they also
interlace on the imaginary axis, see [9, p. 86]. Therefore, the Hermite-
Biehler theorem implies that all zeroes of the polynomial
$P_{k}(z)=f_{k}(z)-ng_{k}(z)$ have positive real parts, [25] (see also [20]
and references there). That is to say, all zeroes of the polynomial $P_{k}$
for arbitrary $a\neq 0$ lie in the half-plane
$\\{z\in\mathbb{C}\,:\,\operatorname{Re}\left(z/a\right)>0\\}$, which is
clearly seen on the figures in Sections 3.3 and 3.4.

In fact, the polynomials $f_{k}$ and $g_{k}$ may also be expressed via
classical families of orthogonal polynomials. The three-term relations (3.4)
and (3.5) are known to determine monic versions of the so-called associated
Jacobi polynomials $\widehat{P}_{k}^{(\alpha,\beta)}(z;c)$, namely,

$f_{k}(z)=\widehat{P}_{k}^{(-n,-n)}(z;n)\quad\text{and}\quad
g_{k}(z)=\widehat{P}_{k-1}^{(-n,-n)}(z;n+1).$

The explicit representation of the associated Jacobi polynomials is due to
Wimp [23, p. 182], however our specific case allows to write these polynomials
using the simple expressions

$f_{k}(z)=\frac{(-1)^{k}P_{k}(-z)+P_{k}(z)}{2}\quad\text{and}\quad
g_{k}(z)=\frac{(-1)^{k}P_{k}(-z)-P_{k}(z)}{2n}.$

Note that the sequence $\\{f_{k}\\}_{k=0}^{n-1}$ consists of the so-called
(monic) exceptional Jacobi polynomials [23, p. 97], although the parameter
$\alpha=-n$ is non-standard for them. Observe also that $f_{k}$ and $-ng_{k}$
turn into $Q$ and $q$ from (2.2) on letting $k=n$.

###  3.2. Orthogonality

Now, let us introduce the following linear functional
$\mathcal{L}_{n}^{a}(f(x))$ acting on $(n-1)$th differentiable functions $f$
and depending on the positive integer number $n$ and the (complex) nonzero
number $a$:

$\mathcal{L}_{n}^{a}(f(x))=\sum_{k=1}^{n}\dfrac{f^{(k-1)}(a)\alpha_{k}}{(k-1)!}=\sum_{k=1}^{n}\dfrac{f^{(k-1)}(a)}{(k-1)!}\dfrac{q^{(n-k)}(a)}{(n-k)!}=\dfrac{1}{(n-1)!}\dfrac{d^{n-1}[f(x)q(x)]}{dx^{n-1}}\Big{|}_{x=a}\,.$

where the numbers $\alpha_{k}$ are defined in (2.4).

Alternatively, assuming that $f$ is analytic near the point $a$ this
functional may be expressed in a more usual form — as an integral with a
continuous (complex-valued) Jacobi weight $(z-a)^{-n}(z+a)^{n}$ over the
circle $|z-a|=2|a|$, which is typical for the Jacobi polynomials with such
parameters, cf. [17, p. 222]. The weight is analytic in
$\mathbb{C}\setminus\\{0\\}$, so the circle may be replaced with any Jordan
curve enclosing the point $z=a$. Due to Cauchy’s integral formula, this
integral will essentially express the same functional $\mathcal{L}_{n}^{a}$,
namely:

$\begin{split}\mathcal{L}_{n}^{a}(f)&=\dfrac{1}{(n-1)!}\dfrac{d^{n-1}[f(z)q(z)]}{dz^{n-1}}\Big{|}_{z=a}=\frac{1}{2\pi
i}\oint\limits_{|z-a|=2|a|}\frac{f(z)q(z)}{(z-a)^{n}}dz\\\ &=\frac{1}{2\pi
i}\oint\limits_{|z-a|=2|a|}f(z)\cdot\frac{1}{2}\left(\left(\frac{z-a}{z-a}\right)^{n}-\left(\frac{z+a}{z-a}\right)^{n}\right)dz\\\
&=\frac{i}{4\pi}\oint\limits_{|z-a|=2|a|}f(z)\left(\frac{z+a}{z-a}\right)^{n}dz.\end{split}$

With this functional we can represent the rational function $R(z)$ defined in
(2.1) as follows

(3.6)
$R(z)=\mathcal{L}_{n}^{a}\left(\dfrac{1}{z-x}\right)=\sum_{k=1}^{n}\dfrac{\alpha_{k}}{(z-a)^{k}}\,.$

Moreover, from (3.2) we have

$\mathcal{L}_{n}^{a}(x^{m})=s_{m}\,,\quad\mbox{for}\;m=0,1,\ldots,$

where $s_{m}$ are the moments defined in (2.5). In fact,

$\displaystyle\mathcal{L}_{n}^{a}(x^{m})$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\dfrac{(x^{m})^{(k-1)}\Big{|}_{x=a}}{(k-1)!}\,\alpha_{k}=a^{m}\alpha_{1}+\sum_{k=2}^{n}\dfrac{m(m-1)\cdots(m-k+2)a^{m-k+1}}{(k-1)!}\,\alpha_{k}$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\binom{m}{k-1}a^{m-k+1}\alpha_{k}=s_{m}\,.$

We now prove that the polynomials $P_{k}(z)$, for $k=0,1,\ldots,n-1$, defined
in (3.1) are orthogonal w.r.t. the functional (3.2), that is,

$\mathcal{L}_{n}^{a}(P_{m}(x)P_{k}(x))=C_{m}\delta_{km}\ \
\quad\text{for}\quad m,k=0,1,\ldots,n-1\,,$

where $\delta_{km}$ is the Kronecker symbol. To this end, we find the three-
term recurrence relations for monic polynomials $Q_{m}(z)$ orthogonal w.r.t.
the functional $\mathcal{L}_{n}^{a}(f)$ and compare the obtained relations
with (3.1). Firstly, let us note that all the minors $D_{m}(R)$,
$m=1,\ldots,n$, defined in (2.8) are nonzero, i.e.,

$D_{m}(R)\neq 0\,,\quad\mbox{for}\;m=1,\ldots,n,$

by (2.18), so such a sequence exists (see, e.g., [9, Theorem 3.1]). These
orthogonal polynomials have the form [23, p. 21]

(3.7)
$Q_{m}(z)=\dfrac{1}{D_{m}(R)}\begin{vmatrix}s_{0}&s_{1}&s_{2}&\dots&s_{m}\\\
s_{1}&s_{2}&s_{3}&\dots&s_{m+1}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
s_{m-1}&s_{m}&s_{m+1}&\dots&s_{2m-1}\\\ 1&z&z^{2}&\dots&z^{m}\\\
\end{vmatrix}\,,$

for $m=1,\ldots,n$. The roots of the last polynomial $Q_{n}(z)$ (partially)
define the functional $\mathcal{L}_{n}^{a}(f)$. If $\mathcal{L}_{n}^{a}(f)$ is
a positive or negative functional, $Q_{n}(z)$ would have exactly $n$ real
distinct roots that are the growth point of the measure generated by
$\mathcal{L}_{n}^{a}(f)$, see, e.g., [23, Section 2.2]. But the formula (2.18)
shows that $\mathcal{L}_{n}^{a}(f)$ is neither a positive nor a negative
functional, since the minors $D_{m}(R)$ are not all positive nor of
alternating signs $(\mbox{sgn}\,D_{m}(R)\neq(-1)^{m}$, for $m=1,\ldots,n$).

The polynomials $Q_{m}(z)$ satisfy a three-term recurrence relation

$Q_{m+1}(z)=(z-c_{m+1})Q_{m}(z)-d_{m+1}Q_{m-1}(z).$

It is well known, e.g., [9, 20, 23, 37], that

(3.8) $d_{m+1}=\dfrac{D_{m-1}(R)D_{m+1}(R)}{D^{2}_{m}(R)}=-b_{m},$

so the polynomials $Q_{m}(z)$ satisfy the recurrence relation

(3.9)
$Q_{m+1}(z)=(z-c_{m+1})Q_{m}(z)+b_{m}Q_{m-1}(z)\,,\quad\mbox{for}\;m=0,1,\cdots,n-1,$

where $Q_{-1}(z)=0$ and $Q_{0}(z)=1$.

Furthermore, if

$Q_{m}(z)=z^{m}+\displaystyle\sum_{i=1}^{m}A_{im}z^{m-i}\,,$

then (cf., e.g., [9])

(3.10) $c_{1}=A_{11}\,,\quad
c_{m+1}=A_{1m}-A_{1,m+1}\,,\quad\mbox{for}\;m=1,\ldots,n-1\,,$

and, from (3.7), we have

(3.11)
$A_{1,m}=-\dfrac{D^{\prime}_{m}(R)}{D_{m}(R)}\,,\quad\mbox{for}\;m=1,\ldots,n.$

where

$D^{\prime}_{m}(R)=\begin{vmatrix}s_{0}&s_{1}&s_{2}&\cdots&s_{m-2}&s_{m}\\\
s_{1}&s_{2}&s_{3}&\cdots&s_{m-1}&s_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
s_{m-1}&s_{m}&s_{m+1}&\cdots&s_{2m-3}&s_{2m-1}\\\
\end{vmatrix}\,,\quad\mbox{for}\;m=1,\ldots,n\,.$

To find the formula for the minors $D^{\prime}_{m}(R)$, we need the following
lemma which can be proved in the same way as Lemma 2.3.

######  Lemma 3.1.

The minors $D^{\prime}_{m}(R)$ can be represented in the following form:

$\displaystyle D^{\prime}_{m}(R)$ $\displaystyle=$
$\displaystyle\begin{vmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\cdots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\alpha_{4}&\cdots&\alpha_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\alpha_{m-1}&\alpha_{m}&\alpha_{m+1}&\cdots&\alpha_{2m-2}\\\
\alpha_{m+1}&\alpha_{m+2}&\alpha_{m+3}&\cdots&\alpha_{2m}\end{vmatrix}$
$\displaystyle\;+\,am\begin{vmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\cdots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\alpha_{4}&\cdots&\alpha_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\alpha_{m-1}&\alpha_{m}&\alpha_{m+1}&\cdots&\alpha_{2m-2}\\\
\alpha_{m}&\alpha_{m+1}&\alpha_{m+2}&\cdots&\alpha_{2m-1}\end{vmatrix},$

for $m=1,\dots,n$, and setting $\alpha_{j}=0$, for $j>n$.

Furthermore, by the same technique that we used in proof of Theorem 2.4, we
may conclude the next theorem (here, we omit the proof to avoid long and
unnecessary repetitions).

######  Theorem 3.2.

The following identity holds for $m=1,\dots,n$,

(3.13) $\begin{vmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\cdots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\alpha_{4}&\cdots&\alpha_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\alpha_{m-1}&\alpha_{m}&\alpha_{m+1}&\dots&\alpha_{2m-2}\\\
\alpha_{m+1}&\alpha_{m+2}&\alpha_{m+3}&\dots&\alpha_{2m}\end{vmatrix}=a(n-m)\begin{vmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\cdots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\alpha_{4}&\cdots&\alpha_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\alpha_{m-1}&\alpha_{m}&\alpha_{m+1}&\cdots&\alpha_{2m-2}\\\
\alpha_{m}&\alpha_{m+1}&\alpha_{m+2}&\cdots&\alpha_{2m-1}\end{vmatrix},$

where we set $\alpha_{j}=0$, for $j>n$, and for $m=1$ we mean that the left
determinant equals $\alpha_{2}$.

So, from (3.1) and (3.13) we have

$D^{\prime}_{m}(R)=an\begin{vmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\dots&\alpha_{m}\\\
\alpha_{2}&\alpha_{3}&\alpha_{4}&\dots&\alpha_{m+1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\alpha_{m-1}&\alpha_{m}&\alpha_{m+1}&\dots&\alpha_{2m-2}\\\
\alpha_{m}&\alpha_{m+1}&\alpha_{m+2}&\dots&\alpha_{2m-1}\end{vmatrix},$

i.e.,

$D^{\prime}_{m}(R)=anD_{m}(R)\,.$

Now, (2.10) and (3.11) imply

$A_{1,m}=-an\,,\quad\mbox{for}\;m=1,2,\ldots,n\,.$

Hence

(3.14) $c_{1}=an\quad\mbox{and}\quad
c_{m}=0\,,\quad\mbox{for}\;m=2,\ldots,n\,.$

by (3.10).

Thus, from (3.8)–(3.9) and (3.14) we have

$Q_{m+1}(z)=zQ_{m}(z)+\dfrac{a^{2}(n^{2}-m^{2})}{(2m-1)(2m+1)}Q_{m-1}(z)\,,\quad\mbox{with}\;m=1,\ldots,n\,,$

with initial conditions $Q_{0}(z)=1$, and $Q_{1}(z)=z-an$. Therefore, the
monic polynomials $P_{k}(z)$ coincide with the monic polynomials $Q_{m}(z)$
which are orthogonal with respect to the functional
$\mathcal{L}_{n}^{a}(f(x))$ defined in (3.2) and

$\displaystyle C_{m}$ $\displaystyle=$
$\displaystyle\mathcal{L}_{a}^{n}(P^{2}_{m}(x))=b_{0}\prod_{k=1}^{m}(-b_{k})=(-1)^{m+1}a^{2m+1}n\prod_{k=1}^{m}\dfrac{(n^{2}-k^{2})}{4k^{2}-1}$
$\displaystyle=$
$\displaystyle(-1)^{m+1}a^{2m+1}\binom{n+m}{2m+1}\,\dfrac{(2m)!!}{(2m-1)!!}\,,$

or, equivalently,

$C_{m}=(-1)^{m+1}a^{2m+1}\sqrt{\pi}\binom{n+m}{2m+1}\,\dfrac{\Gamma(m+1)}{\Gamma\left(m+\tfrac{1}{2}\right)},$

for $m=0,1,\ldots,n-1$.

For example, setting $m=n-1$, we have the following interesting fact:

$\displaystyle C_{n-1}$ $\displaystyle=$
$\displaystyle\mathcal{L}_{a}^{n}(P^{2}_{n-1}(x))(-1)^{n}a^{2n-1}\sqrt{\pi}\,\dfrac{\Gamma(n)}{\Gamma\left(n-\tfrac{1}{2}\right)}\sim(-1)^{n}a^{2n-1}\sqrt{\pi
n}\quad\text{as}\quad n\to+\infty.$

Having the moments $s_{m}$ defined in (2.5) which are related to the
orthogonal polynomials $\\{P_{m}(z)\\}_{m=0}^{n-1}$, one can introduce the
distributional weight function $w_{a}$ characterised by those moments (see
[29]) such that

(3.15) $\langle
w_{a},f\rangle=\sum\limits_{m=0}^{\infty}\dfrac{s_{m}f^{(m)}(0)}{m!}$

for infinitely differentiable function $f$ with a compact support. Therefore,
the distributional weight function $\omega_{a}$ of this functional has the
form [29]

(3.16)
$\omega_{a}(x)=\sum\limits_{m=0}^{\infty}\dfrac{(-1)^{m}s_{m}\delta^{(m)}(x)}{m!}\,,$

where $\delta(x)=\delta^{(0)}(x)$ is the Dirac delta-function, and
$\delta^{(m)}(x)$, for $m=1,2,\ldots$, are its derivatives. From (2.5) it
easily follows that the moments $s_{m}$ satisfy the inequality

$|s_{m}|\leqslant C\,|a|^{m}\,m!\,,$

for certain constants $C=C(a,n)>0$, and the distributional weight $w_{a}(x)$
has the inverse Fourier transform [29]

$F^{-1}w_{a}(t)=\dfrac{1}{2\pi}\int\limits_{-\infty}^{+\infty}e^{-ixt}\omega_{a}(x)dx=\dfrac{1}{2\pi}\sum\limits_{m=0}^{+\infty}\dfrac{s_{m}(-it)^{m}}{m!}\,.$

The sum of this series can be found explicitly.

######  Theorem 3.3.

The inverse Fourier transform of the distributional weight (3.16)
corresponding to the functional $w_{a}$ has the form

$F^{-1}w_{a}(t)=\dfrac{-ae^{-iat}}{2\pi}\,L^{(1)}_{n-1}(2iat),\quad
t\in\mathbb{C}\,,$

where $L^{(\alpha)}_{n}(x)$ denotes the (generalised) Laguerre polynomial.

###### Proof.

From (2.5) one has

$\displaystyle\displaystyle F^{-1}w_{a}(t)$ $\displaystyle=$
$\displaystyle\dfrac{-a}{2\pi}\sum\limits_{m=0}^{\infty}\sum\limits_{k=1}^{n}2^{k-1}a^{m}\binom{m}{k-1}\binom{n}{k}\dfrac{(-it)^{m}}{m!}$
$\displaystyle=$
$\displaystyle\dfrac{-a}{2\pi}\sum\limits_{k=1}^{n}\dfrac{2^{k-1}}{(k-1)!}\binom{n}{k}\sum\limits_{m=0}^{\infty}\dfrac{(-ait)^{m}}{(m-k+1)!}$
$\displaystyle=$
$\displaystyle\dfrac{-ae^{-ait}}{2\pi}\sum\limits_{k=0}^{n-1}\dfrac{1}{k!}\binom{n}{k+1}(-2ait)^{k}=\dfrac{-ae^{-iat}}{2\pi}\,L^{(1)}_{n-1}(2iat)\,,$

where the series converges for $t\in\mathbb{C}$. ∎

In (3.15) we defined the functional $w_{a}$ for infinitely differentiable
functions with compact support. In fact it will be extended to a larger class
of functions. Indeed, let us find the functional $w_{a}$ by applying the
Fourier transform to $F^{-1}w_{a}(t)$.

$\begin{split}F\left(F^{-1}w_{a}(t)\right)&=-\dfrac{a}{2\pi}F\left(e^{-iat}\,L^{(1)}_{n-1}(2iat)\right)\\\\[3.0pt]
&=\sum\limits_{k=1}^{n}\dfrac{-2^{k-1}a^{k}\binom{n}{k}}{(k-1)!}\,\dfrac{(-i)^{k-1}F(e^{-iat}\,t^{k-1})}{2\pi}=\sum\limits_{k=1}^{n}\dfrac{\alpha_{k}\delta^{(k-1)}(x-a)}{(k-1)!}\,,\end{split}$

where $\alpha_{k}$ are defined in (2.4). Thus, we can continuously extend the
functional $w_{a}$ to the set of all $(n-1)$-times differentiable functions.
From (3.2) it follows that the continuously extended functional $w_{a}$
coincides with $\mathcal{L}_{a}^{n}$.

Note that since the matrix $J_{n}$ defined in (1.1) is tridiagonal, it is non-
derogatory. Thus, the normalized (by the first unit entry) eigenvector
corresponding to the unique eigenvalue of $J_{n}$ is
$\mathbf{u}_{0}:=\mathbf{u}(a)$ where
$\mathbf{u}(z)=(P_{0}(z),P_{1}(z),\ldots,P_{n-2}(z),P_{n-1}(z))^{T}$. The
corresponding generalised eigenvectors are
$\mathbf{u}_{k}:=\dfrac{d^{k}\mathbf{u}(z)}{dz^{k}}\Bigg{|}_{z=a}$,
$k=1,\ldots,n-1$.

###  3.3. Relation to Bessel polynomials

The Bessel polynomials $\\{B_{n}(x)\\}_{n=0}^{\infty}$ is a sequence of
polynomials satisfying the differential equation

$x^{2}\dfrac{d^{2}y}{dx^{2}}+(2x+2)\dfrac{dy}{dx}=n(n+1)y.$

They occur naturally in the theory of traveling spherical waves [24]. These
(monic) polynomials satisfy the following three-term recurrence relations

(3.17) $\begin{array}[]{l}B_{0}(z)=1,\quad B_{1}(z)=z+1,\\\ \\\
B_{k+1}(z)=zB_{k}(z)+\dfrac{1}{4k^{2}-1}B_{k-1}(z)\,,\quad\mbox{for}\;k=1,2,\ldots\end{array}$

The Bessel polynomials were introduced by H.L. Krall and O. Frink in [24].
They satisfy complex orthogonality conditions

$\int\limits_{C}B_{n}(z)B_{m}(z)e^{-2/z}dz=0,\qquad\quad m\neq n,$

where $C$ is a closed path in the complex plane enclosing the origin. Their
orthogonality on the real axis was a subject of discussion for many years
until this problem was resolved in a few different ways, see [12, 14, 29]. We
note that on the real line, the support of the functional of orthogonality for
the Bessel polynomials consists of a unique point $z=0$ which is an essential
singularity. Moreover, from the recurrence relations (3.17), it follows that
every monic Bessel polynomial is the characteristic polynomial of a Schwarz
matrix defined by these relations and, therefore, is stable. Since the
polynomials $\\{P_{k}\\}_{k=0}^{n-1}$ satisfying (3.1) and orthogonal w.r.t.
the functional $\mathcal{L}_{a}^{n}$ defined in (3.2) are characteristic
polynomials of Schwarz matrices stable for $a<0$, and the support of the
functional $\mathcal{L}_{a}^{n}$ consists of one point, we may say that
$\\{P_{k}\\}_{k=0}^{n-1}$ can be treated as a finite discrete analogue of
Bessel polynomials. Additionally, if one sets $a=-\dfrac{1}{n}$, then the
three-term recurrence relations (3.1) for the polynomials $P_{k}(z)$ take the
form

$\begin{array}[]{l}P_{0}(z)=1,\quad P_{1}(z)=z+1,\\\\[5.0pt]
P_{k+1}(z)=zP_{k}(z)+\dfrac{P_{k-1}(z)}{4k^{2}-1}-\dfrac{1}{n^{2}}\,\dfrac{k^{2}}{4k^{2}-1}P_{k-1}(z)\,,\quad\mbox{for}\;k=1,\ldots,n-1,\end{array}$

from which it follows that as $n\to+\infty$, these recurrence relations are
close to (3.17). Thus, the polynomials $P_{k}(z)$ approach $B_{k}(z)$
coefficient-wise, and, therefore, on the compact subsets of the complex plane.
So for $a=-\dfrac{1}{n}$, sufficiently large $n$, and relatively small $k$
(comparing with $n$) one can consider the polynomials $P_{k}(z)$ as an
approximation of the monic Bessel polynomials $B_{k}(z)$. Thus, we can expect
that for a fixed $k$ and large $n$ the zeroes of $P_{k}(z)$ will approximate
zeroes of $B_{k}(z)$, and as $k\to+\infty$ and $n\to+\infty$ with
$\dfrac{k}{n}\to 0$ the zeroes of $P_{k}(z)$ will approach the same curve as
the zeroes of $B_{k}(z)$ for $k\to+\infty$, see [4, 5] and references there.
Figures 3.2 and 3.2 show the zeroes of the polynomials $B_{50}(z)$ and
$B_{75}(z)$, respectively, and the zeroes of $P_{50}(z)$ and $P_{75}(z)$
approaching them as $n\to+\infty$.

Figure 3.1. Zeroes of $B_{50}$ and $P_{50}$ for $n=50,\dots,150$

Figure 3.2. Zeroes of $B_{75}$ and $P_{75}$ for $n=75,\dots,175$

Figures 3.4 and 3.4 represent the zeroes of the polynomials $P_{k}(z)$ and
$B_{k}(z)$ for $n=k^{2}$ and $k=5,\ldots,24$. There is a visibly good
approximation of $B_{k}(z)$ by $P_{k}(z)$ in such a case.

Figure 3.3. Zeroes of $P_{k}$ and $B_{k}$ for $n=k^{2}$ and $k=5,\ldots,19$

Figure 3.4. Zeroes of $P_{k}$ and $B_{k}$ for $n=k^{2}$ and $k=20,\ldots,24$

Note also that if $a=-\dfrac{1}{n}$, then as $n\to+\infty$ the limits of the
moments $s_{m}(n)$, $m=0,1,2,\ldots$, are close to the moments of the Bessel
polynomials (but not the same). At the same time, the inverse Fourier
transform of the distributional weight function corresponding to the Bessel
polynomials has the form

$F^{-1}w(t)=-\dfrac{1}{2\pi}\sum\limits_{k=0}^{+\infty}\dfrac{2^{k+1}(it)^{k}}{k!(k+1)!}=-\dfrac{I_{1}(\sqrt{8it})}{\pi\sqrt{8it}},$

where $I_{1}(x)$ is the modified Bessel function [29].

###  3.4. Zeroes of polynomials $P_{k}(z)$ for fixed $a$

Apart of approximation of Bessel polynomials, it is interesting to study how
the zeroes the polynomials $P_{k}$ behave for a fixed
$a\in\mathbb{C}\setminus\\{0\\}$. As it was already observed in Section 3.1,
the results on the polynomials $P_{k}$ for a certain fixed
$a\in\mathbb{C}\setminus\\{0\\}$ immediately extend to other values of $a$ due
to

$a^{-k}P_{k}(az)=P_{k}(z)\big{|}_{a=1}\quad\text{or,
equivalently,}\quad(-a)^{-k}P_{k}(-az)=P_{k}(z)\big{|}_{a=-1}.$

Accordingly, varying values of $a$ only scales and/or rotates the zeroes of
these polynomials in $\mathbb{C}$, see Figure 3.6. We shall concentrate on the
case $a=-1$ which is related to the previous section.

Figure 3.5. Zeroes of $P_{k}$ for $a=e^{i\varphi}$ with $n=k+1=71$, and
$\varphi=-\pi,\dots,-\frac{\pi}{2}$

Figure 3.6. Zeroes of $P_{k}$ for $a=-1$, $k=17,\ldots,50$, and $n=50$

Figures 3.8 and 3.8 represent the zeroes of the polynomials $P_{n/10}(z)$ and
$P_{n-1}(z)$ for $a=-1$. According to calculations, the zeroes tend to a
certain curve in the left half-plane that depends on the limiting ratio
$\dfrac{k}{n}$ as $k\to\infty$, $n>k$.

Figure 3.7. Zeroes of $P_{k}$ for $a=-1$, $n=10k$, and $k=10,\ldots,60$

Figure 3.8. Zeroes of $P_{k}$ for $a=-1$, $n=k+1$, and $k=50,\ldots,150$

If one changes linear dependence between $n$ and $k$ to higher powers, the
zeroes of $P_{k}$ move to infinity as $k\to+\infty$, see Figures 3.10 and
3.10.

Figure 3.9. Zeroes of $P_{k}$ for $a=-1$, $n=k^{2}$, and $k=9,\ldots,144$

Figure 3.10. Zeroes of $P_{k}$ for $a=-1$, $n=\lfloor k^{5/4}\rfloor$, and
$k=9,\ldots,144$

This phenomenon may be explained as follows. As $\dfrac{k}{n}\to 0$ we can
rescale the polynomials $P_{k}(z)$ to have the zeroes bounded. The proper
rescaling may be achieved by letting $a=-\dfrac{1}{n}$ and study the
asymptotic behaviour of the zeroes of $P_{k}\left(\dfrac{z}{k}\right)$, which
is equivalent to letting $a=-\dfrac{k}{n}$ and studying the zeroes of
$P_{k}(z)$, see Figures 3.12 and 3.12. From the discussion in Section 3.3 it
follows that the zeroes of $P_{k}\left(\dfrac{z}{k}\right)$ for
$a=-\dfrac{1}{n}$ should approximate those of $B_{k}\left(\dfrac{z}{k}\right)$
as $k\to\infty$, $n\to\infty$, $\dfrac{k}{n}\to 0$. At the same time, the
latter zeroes are known to remain bounded [19, 30], see also [4, 5].

Figure 3.11. Zeroes of $P_{k}$ for $a=-k/n$, $n=k^{2}$, and $k=9,\ldots,144$

Figure 3.12. Zeroes of $P_{k}$ for $a=-k/n$, $n=\lfloor k^{5/4}\rfloor$, and
$k=9,\ldots,144$

## Acknowledgement

The results of Theorems 2.1 and 2.4, Corollary 2.7 and Sections 3.2 and 3.3
(except Theorem 3.2) were obtained with the support of the Russian Science
Foundation grant 19-71-30002. The work on Remarks 2.2 and 2.6, Section 3.1 and
all numerical calculations were supported by the state assignment,
registration number 122041100132-9.

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