Orthogonal sequences constructed from quasi-orthogonal ultraspherical polynomials

TL;DR

This paper explores the construction of orthogonal sequences from interlacing zeros of ultraspherical polynomials, providing an algorithm and analyzing the zeros' behavior for different parameter ranges.

Contribution

It introduces a novel algorithm for generating orthogonal sequences from ultraspherical polynomials' interlacing zeros, addressing a question posed by Mourad Ismail.

Findings

Zeros of constructed polynomials differ significantly for certain parameter ranges.

The shape of zero curves varies with the parameter λ, showing distinct behaviors.

The method extends the understanding of orthogonal sequences related to ultraspherical polynomials.

Abstract

Let $\displaystyle \{x_{k,n-1}\} _{k=1}^{n-1}$ and $\displaystyle \{x_{k,n}\} _{k=1}^{n},$ $n \in \mathbb{N}$, be two sets of real, distinct points satisfying the interlacing property $ x_{i,n}<x_{i,n-1}< x_{i+1,n}, \, \, \, i = 1,2,\dots,n-1.$ Wendroff proved that if $p_{n-1}(x) = \displaystyle \prod _{k=1}^{n-1} (x-x_{k,n-1})$ and $p_n(x) = \displaystyle \prod _{k=1}^n (x-x_{k,n})$, then $p_{n-1}$ and $p_n$ can be embedded in a non-unique monic orthogonal sequence $\{p_{n}\} _{n=0}^\infty. $ We investigate a question raised by Mourad Ismail at OPSFA 2015 as to the nature and properties of orthogonal sequences generated by applying Wendroff's Theorem to the interlacing zeros of $C_{n-1}^{\lambda}(x)$ and $ (x^2-1) C_{n-2}^{\lambda}(x)$, where $\{C_{k}^{\lambda}(x)\} _{k=0}^\infty$ is a sequence of monic ultraspherical polynomials and $-3/2 < \lambda < -1/2,$ $\lambda \neq -1.$ We construct an algorithm for generating infinite monic orthogonal sequences $\{D_{k}^{\lambda}(x)\} _{k=0}^\infty$ from the two polynomials $D_n^{\lambda} (x): = (x^2-1) C_{n-2}^{\lambda} (x)$ and $D_{n-1}^{\lambda} (x): = C_{n-1}^{\lambda} (x)$, which is applicable for each pair of fixed parameters $n,\lambda$ in the ranges $n \in \mathbb{N}, n \geq 5$ and $\lambda > -3/2$, $\lambda \neq -1,0, (2k-1)/2, k=0,1,\ldots$. We plot and compare the zeros of $D_m^{\lambda} (x)$ and $C_m^{\lambda} (x)$ for several choices of $m \in \mathbb{N}$ and a range of values of the parameters $\lambda$ and $n$. For $-3/2 < \lambda < -1,$ the curves that the zeros of $D_m^{\lambda} (x)$ and $C_m^{\lambda} (x)$ approach are substantially different for large values of $m.$ When $-1 < \lambda < -1/2,$ the two curves have a similar shape while the curves are almost identical for $\lambda >-1/2.$

Figures

Equations

$$C_n^{\lambda }(x)=x{C}_{n-1}^{\lambda }(x)-b_n^{\lambda }{C}_{n-2}^{\lambda }(x),\lambda \ne 0,-1,\dots ;n=1,2,\dots ,$$

$$C_{-1}^{\lambda }(x)\equiv 0,C_0^{\lambda }(x)=1,b_n^{\lambda }=\frac{(n-1)(n-2+2\lambda )}{4(n-2+\lambda )(n-1+\lambda )},\lambda \ne 0,-1,\dots ;n=1,2,\dots$$

$$-1<x_{1,n}<x_{1,n-1}<\cdots <x_{n-1,n}<x_{n-1,n-1}<x_{n,n}<1.$$

$$x_{1,n-1}<x_{1,n}<-1<x_{2,n}<x_{2,n-1}<\cdots <x_{n-2,n-1}<x_{n-1,n}<1<x_{n,n}<x_{n-1,n-1}.$$

$$p_{n+k}(x)=(x-a_k)p_{n+k-1}(x)-b_k{p}_{n+k-2}(x),a_k\in \mathbb{R},b_k>0,k=1,2,\dots .$$

$$D_{n-1}^{\lambda }(x):=C_{n-1}^{\lambda }(x),D_n^{\lambda }(x):=(x^2-1)C_{n-2}^{\lambda }(x).$$

$$C_m^{\lambda }(x)=x^m+\sum _{j=1}^m{\alpha }_{j,m}{x}^{m-j}=\prod _{j=1}^m(x-x_{j,m}),c_m=\sum _{j=1}^m{x}_{j,m}.$$

$$D_m^{\lambda }(x)=x^m+\sum _{j=1}^m{\beta }_{j,m}{x}^{m-j}=\prod _{j=1}^m(x-y_{j,m}),d_m=\sum _{j=1}^m{y}_{j,m}.$$

$$y_{1,n}=x_{1,n-2},y_{2,n}=-1,y_{3,n}=x_{2,n-2},\dots ,y_{n-1,n}=1,y_{n,n}=x_{n-2,n-2}$$

$$y_{k,n-1}=x_{k,n-1},k=1,2,..\dots n-1.$$

$$a_n:=x_{n-2,n-2}+ϵ$$

$$D_n^{\lambda }(x)=(x^2-1)C_{n-2}^{\lambda }(x),D_{n-1}^{\lambda }(x)=C_{n-1}^{\lambda }(x),$$

$$D_{n-j}^{\lambda }(x)=-\frac{1}{{\ell }_{n-j+2}}\left[D_{n-j+2}^{\lambda }(x)-x{D}_{n-j+1}^{\lambda }(x)\right],j=2,3,\dots ,n$$

$$D_{n+j}^{\lambda }(x)=x{D}_{n+j-1}^{\lambda }(x)-{\ell }_{n+j}{D}_{n+j-2}^{\lambda }(x),j=1,2,\dots$$

$${\ell }_{n-j}={\beta }_{2,n-j-1}-{\beta }_{2,n-j}>0,j=0,1,\dots ,n-2,$$

$${\ell }_{n+j}\in \left(0,\frac{a_n{D}_{n+j-1}^{\lambda }(a_n)}{D_{n+j-2}^{\lambda }(a_n)}\right),j=1,2,\dots ,$$

$$D_n^{\lambda }(x)-x{D}_{n-1}^{\lambda }(x)=-{\ell }_n{D}_{n-2}^{\lambda }(x)$$

$$D_{n+j}(x)=x{D}_{n+j-1}^{\lambda }(x)-{\ell }_{n+j}{D}_{n+j-2}(x),j=1,2,\dots ,$$

$$D_{n+1}^{\lambda }(y_{n,n})=-{\ell }_{n+1}{D}_{n-1}^{\lambda }(y_{n,n})<0.$$

$$D_{n+1}^{\lambda }(a_n)=a_n{D}_n^{\lambda }(a_n)-{\ell }_{n+1}{D}_{n-1}^{\lambda }(a_n)>a_n{D}_n^{\lambda }(a_n)-\frac{a_n{D}_n^{\lambda }(a_n)}{D_{n-1}^{\lambda }(a_n)}{D}_{n-1}^{\lambda }(a_n)=0.$$

$$a=A_1(\lambda ):={\left(\frac{2}{2\lambda +3}\right)}^{1/2}$$

$$a=A_2(\lambda ):=\frac{4(2+\lambda )}{3(3+2\lambda )}.$$

$$D_0^{\lambda }(x)=1,$$

$$D_1^{\lambda }(x)=x,$$

$$D_2^{\lambda }(x)=x^2-\frac{2{\lambda }^2+7\lambda +9}{2(2{\lambda }^3+7{\lambda }^2+9\lambda +6)},$$

$$D_3^{\lambda }(x)=x^3-\frac{3(2\lambda +5)}{2(2{\lambda }^2+7\lambda +9)}x,$$

$$D_4^{\lambda }(x)=x^4-\frac{3}{\lambda +3}{x}^2+\frac{3}{4({\lambda }^2+5\lambda +6)},$$

$$D_5^{\lambda }(x)=x^5-\frac{(2\lambda +7)}{2\lambda +4}{x}^3+\frac{3}{2\lambda +4}x,$$

$$D_6^{\lambda }(x)=x^6$$

$$+\frac{-26624{\lambda }^8-315136{\lambda }^7-1452096{\lambda }^6-3030464{\lambda }^5-1350544{\lambda }^4+6634848{\lambda }^3+14325052{\lambda }^2+11993936\lambda +3814971)}{18(\lambda +2)(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}{x}^4$$

$$-\frac{(-8192{\lambda }^7-55552{\lambda }^6-93408{\lambda }^5+238480{\lambda }^4+1249616{\lambda }^3+2167224{\lambda }^2+1773274\lambda +577325)}{6(\lambda +2)(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}{x}^2$$

$$+\frac{4(2\lambda +1{)}^2(80{\lambda }^3+426{\lambda }^2+753\lambda +442)}{3(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)},$$

$$D_7^{\lambda }(x)=x^7$$

$$+\frac{(-10240{\lambda }^8-121088{\lambda }^7-561408{\lambda }^6-1198624{\lambda }^5-656672{\lambda }^4+2255736{\lambda }^3+5154212{\lambda }^2+4387984\lambda +1408809)}{6(\lambda +2)(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}{x}^5$$

$$+\frac{4096{\lambda }^8+78848{\lambda }^7+458688{\lambda }^6+1074016{\lambda }^5+295376{\lambda }^4-3734688{\lambda }^3-8130836{\lambda }^2-7213054\lambda -2452023}{18(\lambda +2)(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}{x}^3$$

$$-\frac{2(512{\lambda }^6+3328{\lambda }^5+7048{\lambda }^4+828{\lambda }^3-19042{\lambda }^2-28747\lambda -13742)}{3(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}x,$$

$$D_8^{\lambda }(x)=x^8$$

$$+\frac{-34816{\lambda }^8-411392{\lambda }^7-1916352{\lambda }^6-4161280{\lambda }^5-2589488{\lambda }^4+6899568{\lambda }^3+16600220{\lambda }^2+14333968\lambda +4637883}{18(\lambda +2)(2\lambda +3{)}^2(512{\lambda }^5+2944{\lambda }^4+5208{\lambda }^3+4{\lambda }^2-8638\lambda -6429)}{x}^6$$

$$\displaystyle+\Big{[}\frac{253952\lambda^{10}+4967424\lambda^{9}+36636672\lambda^{8}+134987136\lambda^{7}+240904128\lambda^{6}+28003872\lambda^{5}-813980016\lambda^{4}}{162(\lambda+2)(2\lambda+3)^{4}(512\lambda^{5}+2944\lambda^{4}+5208\lambda^{3}+4\lambda^{2}-8638\lambda-6429)}$$

$$\displaystyle-\frac{1823644104\lambda^{3}+1962244068\lambda^{2}+1102018370\lambda+259653399}{162(\lambda+2)(2\lambda+3)^{4}(512\lambda^{5}+2944\lambda^{4}+5208\lambda^{3}+4\lambda^{2}-8638\lambda-6429)}\Big{]}x^{4}$$

Full text

MSC: primary 33C50; secondary 42C05.

Keywords: Ultraspherical polynomials, Wendroff’s Theorem, interlacing of zeros, quasi-orthogonal polynomials.

1 Introduction

The monic ultraspherical polynomial $C_{n}^{\lambda}(x)$ is defined by the three term recurrence relation [8, eqn.(8.18)]

[TABLE]

where

[TABLE]

For each $\lambda>-\frac{1}{2},$ the sequence $\{C_{n}^{\lambda}(x)\}_{n=0}^{\infty}$ is orthogonal on $(-1,1)$ with respect to the weight function $(1-x^{2})^{\lambda-\frac{1}{2}}$ and for each $n\in\mathbb{N},$ $n\geq 1,$ the zeros of $C_{n}^{\lambda}(x)$ are real, simple, symmetric, lie in $(-1,1)$ and the zeros of $C_{n-1}^{\lambda}(x)$ interlace with the zeros of $C_{n}^{\lambda}(x),$ $n\geq 2,$   ( see [9, Theorem 3.3.2]) namely,

[TABLE]

where $\displaystyle\{x_{i,n}\}_{i=1}^{n}$ are the zeros of $C_{n}^{\lambda}(x)$ in increasing order.

As $\lambda$ decreases below $-1/2,$ two (symmetric) zeros of $C_{n}^{\lambda}(x)$ leave the interval $(-1,1)$ through the endpoints $-1$ and $1$ ( see [5, p. 296] ) and remain real with absolute value $>1$ for each $n\in\mathbb{N},$ $n\geq 3,$ and $-\frac{3}{2}<\lambda<-\frac{1}{2},\lambda\neq-1.$ For $-\frac{3}{2}<\lambda<-\frac{1}{2},\lambda\neq-1,$ the sequence $\{C_{n}^{\lambda}(x)\}_{n=0}^{\infty}$ is quasi-orthogonal of order $2$ with respect to the weight function $(1-x^{2})^{\lambda+\frac{1}{2}}$ , (see [2, Theorem 6] and [3, p.144]) and, for any $n\in\mathbb{N},n\geq 4,$ ( see [6, Theorem 3.1]),

[TABLE]

It follows from $(4)$ that the zeros of $C_{n-1}^{\lambda}(x)$ and $C_{n}^{\lambda}(x)$ are not interlacing for any $n\in\mathbb{N},n\geq 4$ and $-\frac{3}{2}<\lambda<-\frac{1}{2},\lambda\neq-1,$ but we see from $(3)$ and $(4)$ that the zeros of $C_{n}^{\lambda}(x)$ interlace with the zeros of $(x^{2}-1)C_{n-1}^{\lambda}(x)$ for each $n\in\mathbb{N},n\geq 4$ and each $\lambda$ with $\lambda>-\frac{3}{2},\lambda\neq-1.$

In 1961, Wendroff [10, p. 554] proved that, for any fixed positive integer $n,$ $n\geq 2,$ if $\displaystyle\{x_{i}\}_{i=1}^{n}$ and $\displaystyle\{y_{i}\}_{i=1}^{n-1}$ are two sets of real, distinct points satisfying the interlacing property $x_{1}<y_{1}<x_{2}<y_{2}<\dots<x_{n-1}<y_{n-1}<x_{n},$ there exist infinitely many sequences $\{p_{k}(x)\}_{k=0}^{\infty}$ of monic orthogonal polynomials with $p_{n}(x)=\displaystyle\prod_{k=1}^{n}(x-x_{k})$ and $p_{n-1}(x)=\displaystyle\prod_{k=1}^{n-1}(x-y_{k}).$ His proof is constructive and for a given, fixed $n\in\mathbb{N},$ $n\geq 2,$ each polynomial $p_{k}(x)$ of degree $k\leq n-2$ is uniquely determined by $p_{n}(x)$ and $p_{n-1}(x).$ In contrast, the monic polynomials of degree $n+1,n+2,\dots$ in any orthogonal sequence that includes $p_{n}(x)$ and $p_{n-1}(x)$ are only constrained by the requirement that any infinite sequence of (monic) orthogonal polynomials satisfies a three term recurrence relation of the form

[TABLE]

Since there are infinitely many choices of the coefficients $a_{k}$ and $b_{k}$ with $a_{k}\in\mathbb{R}$ and $b_{k}>0$ for $k=1,2,\dots,$ there are infinitely many distinct monic orthogonal sequences $\{p_{k}(x)\}_{k=0}^{\infty}$ that include $p_{n}(x)$ and $p_{n-1}(x).$

Here, we fix $n\in\mathbb{N},n\geq 5$, fix $\lambda,$ $\lambda>-3/2$, $\lambda\neq-1,0,(2k-1)/2,k=0,1,\ldots$, and define

[TABLE]

We investigate the properties of the zeros of polynomials in monic orthogonal sequences $\{D_{m}^{\lambda}\}_{m=0}^{\infty}$ generated by the Wendroff process. It is important to emphasize the dependence on the “starting value” of $n\in\mathbb{N}$ when generating each monic orthogonal sequence $\{D_{m}^{\lambda}\}_{m=0}^{\infty}$ that includes $D_{n-1}^{\lambda}(x)$ and $D_{n}^{\lambda}(x).$ If $n\in\mathbb{N},$ $n\geq 5$ is large, the number of polynomials (namely, $n-2$) that are uniquely determined in every monic orthogonal sequence that includes $D_{n-1}^{\lambda}(x)$ and $D_{n}^{\lambda}(x)$ is correspondingly large whereas, for example, if $n=5$ we have two degrees of freedom when generating each of the monic polynomials of degree $>5$ and exactly $4$ of the polynomials of lower degree are completely determined. The restriction $n\geq 5$ arises from the fact that when $-\frac{3}{2}<\lambda<-1,$ the quadratic ultraspherical polynomial $C_{2}^{\lambda}(x)$ has two pure imaginary zeros (see [3, p. 144]). If we restrict $\lambda$ to the range $\lambda>-1,$ the results proved here apply for $n\geq 3.$

2 Notation

For each $m\in\mathbb{N},$ denote

[TABLE]

[TABLE]

The zeros $\{x_{j,m}\}_{j=1}^{m}$ of $C_{m}^{\lambda}(x)$ are distinct, real and symmetric with respect to the origin for $m\geq 3$ so that $c_{m}=0$ for all $m\in\mathbb{N},m\geq 3$ while the zeros $\{y_{j,m}\}_{j=1}^{m}$ of $D_{m}^{\lambda}(x)$ are distinct, real and symmetric when $m=n$ or $m=n-1$ so that $d_{n}=d_{n-1}=0.$ Note that

[TABLE]

and

[TABLE]

3 Orthogonal sequences generated by $C_{n-1}^{\lambda}(x)$ and $(x^{2}-1)C_{n-2}^{\lambda}(x),\,n\geq 5$

Our main result is the following theorem.

Theorem 3.1

Let $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ be the sequence of monic ultraspherical polynomials defined by $(1)$ and $(2).$ Fix $n\in{\mathbb{N}},$ $n\geq 5,$ fix $\lambda\in(-\frac{3}{2},-\frac{1}{2}),$ $\lambda\neq-1,$ and suppose that $\epsilon>0$ is abitrary.

Define

[TABLE]

where $x_{n-2,n-2}>1$ is the largest zero of $C_{n-2}^{\lambda}(x).$

Let the sequence of monic polynomials $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ be defined by:

[TABLE]

where

[TABLE]

and $\beta_{2,m}$ is the coefficient of $x^{m-2}$ in $D_{m}^{\lambda}(x)$, see (8). Then the sequence $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ is symmetric and orthogonal with respect to a positive measure supported on the interval $(-a_{n},a_{n})$.

Proof of Theorem 3.1

Fix $n\in\mathbb{N},n\geq 5.$ The monic polynomial $D_{n-2}^{\lambda}(x)$ is uniquely determined by

[TABLE]

where $D_{n}^{\lambda}(x)$ and $D_{n-1}^{\lambda}(x)$ are defined by $(12)$ and the coefficient $\ell_{n}$ is chosen so that $D_{n-2}^{\lambda}(x)$ is monic. The positivity of $\ell_{n}$ follows from the interlacing property of the zeros of $D_{n}^{\lambda}(x)$ and $D_{n-1}^{\lambda}(x).$ In the same way, for each $j=3,4,\ldots,n$, the polynomial $D_{n-j}^{\lambda}(x)$ is constructed from the polynomials $D_{n-j+1}^{\lambda}(x)$ and $D_{n-j+2}^{\lambda}(x.)$ The process is repeated until we obtain $D_{0}^{\lambda}(x)=1.$ The polynomials $D_{n+j}^{\lambda}(x),$ $j=1,2,\dots$ are constructed recursively using the three-term recurrence relation

[TABLE]

choosing positive coefficients $\ell_{n+j}$ for $j=1,2,\dots.$ This ensures (Favard’s Theorem) that the infinite sequence $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ is orthogonal with respect to a positive measure. Wendroff mentions in his proof [10, p. 554] that the coefficients $\ell_{n+j}$ can be chosen in such a way that all zeros of $D_{n+j}^{\lambda}(x)$ lie in the interval $(-a_{n},a_{n})$ for each $j\geq 1$ but does not indicate how to choose the coefficients to achieve this outcome. Here, we prove that the choice of $\ell_{n+1}$ given in $(16)$ ensures that all the zeros of $D_{n+1}^{\lambda}(x)$ lie in the interval $(-a_{n},a_{n}).$ Applying the same argument iteratively, it can be shown that the choice of $\ell_{n+j}$ given in $(16)$ ensures that the zeros of $D_{n+j}^{\lambda}(x)$ lie in the interval $(-a_{n},a_{n})$ for each $j\in\mathbb{N}.$

Suppose $\ell_{n+1}\in\left(0,\frac{a_{n}\,D_{n}^{\lambda}(a_{n})}{D_{n-1}^{\lambda}(a_{n})}\right).$ We show that the zeros of $D_{n+1}^{\lambda}$ lie in the interval $(-a_{n},a_{n})$. From $(4)$ with $n$ replaced by $n-1,$ it follows immediately that the zeros of $D_{n}^{\lambda}(x)=(x^{2}-1)C_{n-2}^{\lambda}(x)$ and $D_{n-1}^{\lambda}(x)=C_{n-1}^{\lambda}(x)$ lie in the open interval $(-a_{n},a_{n}).$ In addition, $D_{n}^{\lambda}(x)$ and $D_{n-1}^{\lambda}(x)$ are monic polynomials with no zeros greater than $x_{n-2,n-2}$ so $D_{n}^{\lambda}(a_{n})>0$ and $D_{n-1}^{\lambda}(a_{n})>0.$ Since $a_{n}>0,$ it follows that $\frac{a_{n}\,D_{n}^{\lambda}(a_{n})}{D_{n-1}^{\lambda}(a_{n})}>0$. By construction, the zeros of $D_{n+1}^{\lambda}$ and $D_{n}^{\lambda}$ interlace and the zeros of $D_{n}^{\lambda}$ lie in the interval $(-a_{n},a_{n})$ so it follows that $y_{1,n+1}<y_{1,n}<y_{2,n+1}<y_{2,n}<\cdots<y_{n,n+1}<y_{n,n}<y_{n+1,n+1}$ and $-a_{n}<y_{1,n}<y_{n,n}<a_{n}.$ We need to show that the largest zero $y_{n+1,n+1}$ of $D_{n+1}^{\lambda}$ satisfies $y_{n+1,n+1}<a_{n}$.

From $(14)$ with $j=1,$ we have $D_{n+1}^{\lambda}(x)=xD_{n}^{\lambda}(x)-\ell_{n+1}D_{n-1}^{\lambda}(x)$ so that

[TABLE]

On the other hand, since we are assuming that $\ell_{n+1}\in\left(0,\frac{a_{n}\,D_{n}^{\lambda}(a_{n})}{D_{n-1}^{\lambda}(a_{n})}\right),$ we have

[TABLE]

Therefore, $D_{n+1}^{\lambda}(y_{n,n})<0$ and $D_{n+1}^{\lambda}(a_{n})>0$ so that $y_{n+1,n+1}<a_{n}$ as required.

Remark 3.1. The sequences of polynomials $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ defined in Theorem 3.1 with $D_{n-1}^{\lambda}(x)=C_{n-1}^{\lambda}(x)$ and $D_{n}^{\lambda}(x)=(x^{2}-1)C_{n-2}^{\lambda}(x)$ are (by construction) orthogonal for each $n\geq 5$ and $\lambda>-\frac{1}{2}$ satisfying $\lambda\neq 0,\lambda\neq\frac{2k-1}{2},$ $k=1,2\ldots$. It is therefore natural to compare orthogonal sequences $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ generated by the Wendroff process with the sequences $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ of ultraspherical polynomials orthogonal on $(-1,1)$. In this case, we can choose $a_{n}=1$ so the interval $(-1,1)$ contains the zeros of all the polynomials in the sequence $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ as well as the sequence $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$.

Remark 3.2. For $-3/2<\lambda<-1/2,$ $\lambda\neq-1,$ and $n\geq 5$ fixed, the largest (real) zero $x_{n-2,n-2}$ of $C_{n-2}^{\lambda}(x)$ is bounded above by $\left(\frac{n-3}{2\lambda+n-2}\right)^{1/2},$ see [7, (4)]. An alternative upper bound for $x_{n-2,n-2}$ is given by

$1-\frac{2\lambda+1}{(n-2)(n+2\lambda-2)},$ see [7, (15)]. These bounds give estimates for the interval of orthogonality $(-a_{n},a_{n})$ of the sequences $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$, where $D_{n-1}^{\lambda}(x)=C_{n-1}^{\lambda}(x)$ and $D_{n}^{\lambda}(x)=(x^{2}-1)C_{n-2}^{\lambda}(x)$.

Remark 3.3. For $-3/2<\lambda<-1/2,$ $\lambda\neq-1,$ and $n\geq 5$ fixed, we can choose $\epsilon>0$ in Theorem 3.1 in such a way that the interval $(-a_{n},a_{n})$ contains all the zeros of the polynomials $\{C_{m}^{\lambda}(x)\}_{m=3}^{\infty}$. To this end, we use the estimates $x_{m,m}<\left(\frac{m-1}{2\lambda+m}\right)^{1/2}$ and $x_{m,m}<1-\frac{2\lambda+1}{m(m+2\lambda)}$, $m\geq 3$, see [7], where $x_{m,m}$ is the largest zero of $C_{m}^{\lambda}$. Because $\max\{\left(\frac{m-1}{2\lambda+m}\right)^{1/2}:m\geq 3,\lambda>-3/2\}=\left(\frac{2}{2\lambda+3}\right)^{1/2}$ and $\max\{1-\frac{2\lambda+1}{m(m+2\lambda)}:m\geq 3,\lambda>-3/2\}=\frac{4(2+\lambda)}{3(3+2\lambda)}$, we may choose $a_{n}:=a$ independent of $n\in\mathbb{N}$, namely

[TABLE]

or

[TABLE]

To choose the sharper of the two bounds $A_{1}(\lambda)$ and $A_{2}(\lambda)$, we use the following comparisons: $A_{1}(-5/4)=A_{2}(-5/4)$, $A_{1}(\lambda)<A_{2}(\lambda)$ if $\lambda\in(-3/2,-5/4)$ and $A_{1}(\lambda)>A_{2}(\lambda)$ if $\lambda\in(-5/4,-1/2)$. Note that $A_{1}(-1/2)=A_{2}(-1/2)$ and $\lim\limits_{\lambda\to(-3/2)+}A_{k}(\lambda)=+\infty$ for $k=1,2$.

Remark 3.4. When developing an algorithm for generating orthogonal sequences, we can choose $\ell_{n+j}=\frac{aD_{n+j-1}^{\lambda}(a)}{\sigma D_{n+j-2}^{\lambda}(a)}$, $j=1,2,\ldots$, where $\sigma>1$. Using this expression for $\ell_{n+j}$ and putting $x=a$ into (14), we obtain $\ell_{n+j+1}=\frac{(\sigma-1)}{\sigma^{2}}a^{2}$ for all $j=1,2,\ldots$. The advantage of choosing $\ell_{n+j}=\frac{aD_{n+j-1}^{\lambda}(a)}{\sigma D_{n+j-2}^{\lambda}(a)}$, $j=1,2,\ldots$, with $\sigma>1$, is that all coefficients $\ell_{n+j+1},$ $j=1,2,\ldots$ are equal, namely, $\ell_{n+j+1}=\frac{(\sigma-1)}{\sigma^{2}}a^{2}$ for $j=1,2,\ldots$. This results in a significant reduction in the computational complexity of the algorithm.

4 Algorithm for construction of $\{D_{m}^{\lambda}(x)\}_{m=0}^{n+k},$ $n$ fixed, $n\geq 5$, $k\in\mathbb{N}$

Using Theorem 3.1 and Remarks 1,3,4, we present an algorithm for construction of the first $n+k+1$ terms of orthogonal sequences $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$.

Choose integers $n\geq 5$ and $k\geq 1$. 2. 2.

Choose $\lambda\in(-\frac{3}{2},+\infty)$ with $\lambda\neq-1,0,(2k-1)/2$, $k=0,1,2,\ldots.$ 3. 3.

If $-3/2<\lambda<-5/4$, define $a=A_{1}(\lambda)=\left(\frac{2}{2\lambda+3}\right)^{1/2}$. If $-5/4\leq\lambda<-1/2$, define $a=A_{2}(\lambda)=\frac{4(2+\lambda)}{3(3+2\lambda)}$. If $\lambda>-\frac{1}{2}$, define $a=1$. 4. 4.

Choose $\sigma>1$. 5. 5.

Let $D_{n}^{\lambda}(x)=(x^{2}-1)C_{n-2}^{\lambda}(x),\;\;\;D_{n-1}^{\lambda}(x)=C_{n-1}^{\lambda}(x)$. 6. 6.

For $j=0,1$, let $\beta_{2,n-j}$ be the coefficient of $x^{n-j-2}$ in $D_{n-j}^{\lambda}(x)$. 7. 7.

For $j=2,3,\ldots,n$,

let $\ell_{n-j+2}=\beta_{2,n-j+1}-\beta_{2,n-j+2}$;

let $D_{n-j}^{\lambda}(x)=-\frac{1}{\ell_{n-j+2}}\left[D_{n-j+2}^{\lambda}(x)-xD_{n-j+1}^{\lambda}(x)\right]$;

let $\beta_{2,n-j}$ be the coefficient of $x^{n-j-2}$ in $D_{n-j}^{\lambda}(x)$. 8. 8.

Let $\ell_{n+1}=\frac{a\,D_{n}^{\lambda}(a)}{\sigma D_{n-1}^{\lambda}(a)}$ and $D_{n+1}^{\lambda}(x)=xD_{n}^{\lambda}(x)-\ell_{n+1}D_{n-1}^{\lambda}(x)$. 9. 9.

For $j=2,\ldots,k$, let $D_{n+j}^{\lambda}(x)=xD_{n+j-1}^{\lambda}(x)-\frac{(\sigma-1)}{\sigma^{2}}a^{2}D_{n+j-2}^{\lambda}(x)$.

The above algorithm generates the first $n+k+1$ terms of a sequence of symmetric polynomials $\{D_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ orthogonal with respect to some positive measure supported on the interval $(-a,a)$, which contains all the zeros of the symmetric polynomials $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$. If $\lambda\in(-3/2,-1/2)$ and $\lambda\neq-1$, the sequence $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ is quasi-orthogonal of order $2$ on $(-1,1)$ with respect to the weight function $(1-x^{2})^{\lambda+1/2}$; if $\lambda\in(-1/2,+\infty)$ and $\lambda\neq 0$, $\lambda\neq\frac{2k-1}{2}$ for $k=1,2,\ldots$, the sequence $\{C_{m}^{\lambda}(x)\}_{m=0}^{\infty}$ is orthogonal with respect to the weight function $(1-x^{2})^{\lambda-1/2}$ on the interval $(-1,1)$.

Example 4.1. In this example we present the first 11 terms of the sequence $\{D_{m}^{\lambda}\}_{m=0}^{\infty}$ using our algorithm with $n=5$, $k=5$, $\sigma=2$, and $a=\frac{4(2+\lambda)}{3(3+2\lambda)}$, where $\lambda\in(-\frac{3}{2},+\infty)$, $\lambda\neq{-1,0}$ and $\lambda\neq(2k-1)/2$, $k=0,1,2,\ldots$:

[TABLE]

5 The zeros of $D_{m}^{\lambda}(x)$ and $C_{m}^{\lambda}(x)$

In this section, we plot and compare the zeros of $D_{m}^{\lambda}(x)$ constructed using the algorithm in Section 4 with the zeros of $C_{m}^{\lambda}(x)$, where $m=3,4,\ldots,n+k$ and $n\geq 5$, $k\geq 1$ are fixed integers.

Example 5.1. Let $n=5$ and $\sigma=2.$ Choose $k=5$, $\lambda=-5/4$ and $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=2.$ The polynomials $D_{m}^{\lambda},0\leq m\leq 10,$ are listed below with the approximate values of their zeros in curly brackets $\{\}$ :

[TABLE]

Note that the zeros of $D_{5}^{-5/4}(x)$ are $-\sqrt{2};-1,0,1,\sqrt{2}.$ The largest and smallest zeros of $D_{10}^{-5/4}(x)$ are close to the limits $-2$ and $2.$

In Figures 1 through 4, the $y$-coordinates of the plotted points are the zeros of $D_{m}^{-5/4}(x)$ (diamond, brown) and $C_{m}^{-5/4}(x)$ (round, blue) for $m=3,4,5,10$. The figures suggest that the greatest difference between the zeros of $D_{m}^{-5/4}(x)$ and $C_{m}^{-5/4}(x)$ are at the extreme zeros.

Example 5.1 provides numerical confirmation that the relative ordering of the zeros of $D_{n+1}^{-5/4},D_{n}^{-5/4},$ and $D_{n-1}^{-5/4}$, is consistent with [1, Theorem 4]. Replacing $n$ by $n-1$ and putting $b_{n}=0$ in (7) and (8) in [1, Theorem 4], the negative zeros of $D_{m}^{-5/4},$ $m\in\{n-1,n,n+1\},$ should satisfy

[TABLE]

while the positive zeros of $D_{m}^{-5/4},$ $m\in\{n-1,n,n+1\},$ should satisfy

[TABLE]

From Example 5.1, we see that the zeros of $D_{4}^{-5/4}$, $D_{5}^{-5/4}$, and $D_{6}^{-5/4}$ satisfy

[TABLE]

and

[TABLE]

as expected.

In the examples and figures that follow, we plot the zeros of $D_{m}^{\lambda}(x)$ and $C_{m}^{\lambda}(x)$ for selected values of $n$ (the “starting value” ), $m$ and $\lambda.$

Example 5.2. Let $n=5$, $k=5$, and $\sigma=2$, as in the previous example. In Figures 5 and 6, the $y$-coordinates of the plotted points are the zeros of $D_{10}^{\lambda}$ (diamond, brown) and $C_{10}^{\lambda}$ (round, blue) respectively for $\lambda=-3/4$, $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=\frac{10}{9}$ and $\lambda=-1/4$, $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=\frac{14}{15}$.

Example 5.3. Let $n=10$, $k=58$, $\sigma=2.$ Choose $\lambda=-5/4$ and $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=2$. In Figures 7 through 14 the $y$-coordinates of the plotted points are the zeros of $D_{m}^{\lambda}$ (diamond, brown) and $C_{m}^{\lambda}$ (round, blue) for selected integer values of $m$ between $3$ and $68$. The figures suggest that, as $m$ increases, the curves that fit the zeros of $D_{m}^{-5/4}$ and $C_{m}^{-5/4}$ are significantly different.

Example 5.4. Let $n=10$, $k=58$, $\sigma=2.$ Choose $\lambda=-3/4$ and $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=\frac{10}{9}$. In Figures 15 through 20 the $y$-coordinates of the plotted points are the zeros of $D_{m}^{\lambda}$ and $C_{m}^{\lambda}$ for selected integer values of $m$ between $8$ and $68$.

Example 5.5. Let $n=10$, $k=58$, $\sigma=2.$ Choose $\lambda=-1/4$ and $a=\frac{4(2+\lambda)}{3(3+2\lambda)}=\frac{14}{15}$. In Figures 21 through 26 the $y$-coordinates of the plotted points are the zeros of $D_{m}^{\lambda}$ and $C_{m}^{\lambda}$ for several integer values of $m$ between $8$ and $68$.

Remark 5.2. As $m$ increases, the zeros of $D_{m}^{\lambda}(x)$ and $C_{m}^{\lambda}(x)$ appear to be asymptotically equal for $\lambda=-1/4$. This is not unexpected since $\lambda=-1/4$ lies in the orthogonal range $\lambda>-1/2$ for ultraspherical polynomials. Note that the interval of orthogonality is $(-a,a)$, where $a<1.$

Example 5.6. Let $n=5$, $k=18$, $\sigma=2.$ In Figures 27 through 34, the $y$-coordinates of the plotted points are the zeros of $D_{m}^{\lambda}$ and $C_{m}^{\lambda}$ for a selection of values of $\lambda\in(-\frac{3}{2},+\infty)$ with $\lambda\neq-1,0,(2k-1)/2$, $k=0,1,2,\ldots$ where $m=23$ is fixed. We choose $a=\frac{4(2+\lambda)}{3(3+2\lambda)}$ if $\lambda<-1/2$ and $a=1$ if $\lambda>-1/2$; the zeros of $D_{m}^{\lambda}$ and $C_{m}^{\lambda}$ are contained in $(-a,a)$.

For $-3/2<\lambda<-1,$ the curves to which the zeros of $D_{m}^{\lambda}(x)$ and $C_{m}^{\lambda}(x)$ can be fitted are substantially different for some values of $m.$ As $\lambda$ approaches $-1/2$ from the left, the two curves are very similar, and, as $\lambda>$ increases further, the curves are almost identical.

6 Acknowledgements

Kathy Driver would like to express her thanks to the Mathematics Department, University of Colorado, Colorado Springs, for its hospitality during her visit in Spring 2018, during which the work on this paper began.

7 Bibliography