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\begin{document}
\hfill {\small June 1997}
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\begin{center}
{\bf Asymptotic Distance Between Zeros of Orthogonal Polynomials
Satisfying Second-Order Differential Equations}\\
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I.V.Krasovsky\\
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Max-Planck-Institut f\"ur Physik Komplexer Systeme\\
Bayreuter Str. 40, Haus 16, D-01187, Dresden, Germany\\
(present address)\\
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B.I.Verkin Institute for Low Temperature Physics and Engineering\\
47 Lenina Ave., Kharkov 310164, Ukraine.\\
E-mail: ivk@igorvk.kharkov.ua
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\noindent
{\bf Abstract.} There is a simple way to find the distance between zeros of
orthogonal polynomials (in the limit as the degree of a polynomial
becomes large) if the latter satisfy second-order differential
equations. The proposed approach is based on application of a polynomial
ansatz to the differential equation. As an example,
the classical orthogonal polynomials are considered.
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As is well known (see, e.g., [\ref{Szego},\ref{BE},\ref{NSU}]),
some orthogonal
polynomials (including the classical ones: Jacobi, Laguerre, and
Hermite) satisfy differential equations of the following form:
\begin{equation}
a(x)y^{''}_n+b(x)y^{'}_n+c(x,n)y_n=0.
\end{equation}
Here $y_n(x)$ is a polynomial of exactly degree $n$, functions $a(x)$,
$b(x)$, $c(x,n)$, where only the last one depends on $n$, are real and
continuous.
Although for the known to author systems of orthogonal polynomials these
functions are themselves polynomials in $x$ and ($c(x,n)$) in $n$
(see [\ref{Geron_in_Szego},\ref{Bochner},\ref{Brenke}] for related questions),
this fact is not important for what follows. We shall only assume that there
is such a constant $\gamma>0$ that
\[
\lim_{n\to\infty}\frac{c(x,n)}{n^{2\gamma}}=c_\infty(x),
\]
where $c_\infty(x)$ is continuous.

Now fix the system of polynomials $\{y_n(x)\}_{n=0}^\infty$ satisfying
equations (1) for $n=0,1,\dots$ We further assume that these polynomials
are orthogonal with a continuous weight function on a set
$x\in\cal A$.
Let $x_1^{(n)}<x_2^{(n)}<\cdots<x_n^{(n)}$ be zeros of $y_n(x)$. (Recall
that zeros of orthogonal polynomials are real and simple.) Since the
weight function of the polynomials is continuous, it follows from the general
principles [\ref{Geron_in_Szego}] that the set of zeros $x_k^{(n)}$,
$k=1,\dots,n$
becomes everywhere dense in $\cal A$ as $n\to\infty$.

We shall be interested in the quantities $x^{(n)}_i-x^{(n)}_k$ when the value
of $n$ becomes large.

Consider a small
$\varepsilon$-neighbourhood of a point $x$. If $a(x)c_\infty(x)>0$ then,
starting with a sufficiently large number $n$, the characteristic values
of the equation (1) (considered as an equation with constant coefficients),
$\omega_\pm=\{-b(x)\pm\sqrt{b(x)^2-4a(x)c(x,n)}\}/(2a(x))$, have an
imaginary part growing as $n^{\gamma}$. Hence, it is easy to conclude
that the asymptotic distance between consecutive zeros is (henceforth I omit
the upper index $n$ of zeros)\footnote{
Note that it follows from (\ref{de}) that
${\cal B}=\{x|\quad a(x)c_\infty(x)>0\}$
{\it coincides} with the orthogonality set $\cal A$.

Note also that (\ref{de}) should look obvious to the reader who is familiar
with the quasiclassical approximation in the theory of Schr\"odinger
equation [\ref{LL3}].}
\begin{equation}
\eqalign{
x_{k+1}-x_k\sim\gamma_k/n^{\gamma},\\
\gamma_k=\pi\sqrt{\frac{a(x)}{c_\infty(x)}}+o(1).}
\label{de}
\end{equation}

We see from here that in the limit as $n\to\infty$ (i.e. in the main order
in $n$), the consecutive zeros
are separated equidistantly, that is $x_i-x_k\sim(i-k)\gamma_k/n^{\gamma}$.
This result is a consequence of the fact that we replaced the coefficients
of the equation (1) by constants in the $\varepsilon$-neighbourhood of the
point $x$.

In the next order in $n$, it is reasonable to suppose a linear growth
of the value $x_{i+1}-x_i$ with $i$, that is
\begin{equation}
x_i-x_k=(i-k)\left(\frac{\gamma_k}{n^{\gamma}}+
\frac{(i-k-1)}{2}\frac{\delta_k}{n^{\delta}}+
o(n^{-\delta})\right),\qquad\delta>\gamma.
\label{xixk}
\end{equation}

The {\it purpose} of this paper is to present a way to find $\delta_k$,
$\gamma_k$ (in the limit as $n\to\infty$),
and $\delta$ from the differential equation.

Up to a constant factor, $y_n(x)$ can be written in the
form
\begin{equation}
y_n(x)=\prod_{k=1}^n (x-x_k).\label{pa}
\end{equation}
Substituting this expression in (1)
and dividing the equation by $y_n(x)$, we get
\begin{equation}
a(x)\sum_{k=1}^n\sum_{i=1\atop i\neq k}^n\frac1{(x-x_k)(x-x_i)}+
b(x)\sum_{k=1}^n\frac1{x-x_k}+c(x,n)=0.\label{4}
\end{equation}
Obviously, the singularities of the function at the left-hand side must vanish.
Equating the residues in the simple poles $x_k$, $k=1,\dots,n$, to zero,
we obtain the following system:
\begin{equation}
2a(x_k)\sum_{i=1\atop i\neq k}^n\frac1{x_k-x_i}+b(x_k)=0,\qquad
k=1,\dots,n.\label{d0}
\end{equation}
A procedure similar to the deriving of (\ref{4}) and
(\ref{d0}) from (1) by application
of the polynomial ansatz (\ref{pa}) is used
in the context of exactly and quasi-exactly solvable
spectral problems for one-dimensional Shr\"odinger operators
(see [\ref{Shifman},\ref{WZ}]).

Let us now differentiate (1) with respect to $x$ and apply the
same polynomial-ansatz technique to {\it the resulting equation}.
Then instead of (\ref{d0}) we get
\begin{eqnarray}
\eqalign{
3a(x_k)\left\{\sum_{i=1\atop i\neq k}^n\frac1{x_k-x_i}
\sum_{j=1\atop j\neq k}^n\frac1{x_k-x_j}-
\sum_{i=1\atop i\neq k}^n\frac1{(x_k-x_i)^2}\right\}+\\
2(a'(x_k)+b(x_k))\sum_{i=1\atop i\neq k}^n\frac1{x_k-x_i}+
c(x_k,n)+b'(x_k)=0,\qquad k=1,\dots,n.}\label{d1}
\end{eqnarray}
Now divide (\ref{d1}) by $n^{2\gamma}$ and take the limit as $n\to\infty$ on
condition that the point $x=\lim_{n\to\infty}x_{k(n)}$, $x\in\cal A$,
is fixed. Then,
using the fact that, as follows from (\ref{d0}),
\begin{equation}
\lim_{n\to\infty}\frac1{n^\varepsilon}\sum_{i=1\atop i\neq k}^n
\frac1{x_k-x_i}=0\label{inf}
\end{equation}
for any $\varepsilon>0$, we obtain from (\ref{d1})
\begin{equation}
\lim_{n\to\infty}\frac1{n^{2\gamma}}\sum_{i=1\atop i\neq k}^n
\frac1{(x_k-x_i)^2}=\frac{c_\infty(x)}{3a(x)}.\label{n2}
\end{equation}
On the other hand, applying (\ref{xixk}), we can write
($\gamma(x)=\lim_{n\to\infty}\gamma_{k(n)}$)
\begin{equation}
\lim_{n\to\infty}\frac1{n^{2\gamma}}\sum_{i=1\atop i\neq k}^n
\frac1{(x_k-x_i)^2}=
\frac1{\gamma(x)^2}\sum_{i=1\atop i\neq k}^\infty\frac1{(k-i)^2}=
2\frac1{\gamma(x)^2}\sum_{i=1}^\infty\frac1{i^2}=
\frac{\pi^2}3\frac1{\gamma(x)^2},\label{9}
\end{equation}
where the identity $\sum_{i=1}^\infty i^{-2}=\pi^2/6$ has been used.

So putting (\ref{n2}) and (\ref{9}) together, we have
\begin{equation}
\gamma(x)=\pi\sqrt{\frac{a(x)}{c_\infty(x)}}\label{gamma}
\end{equation}
on the set ${\cal A}=\{x|\quad a(x)c_\infty(x)>0\}$, the result which
coincides with (\ref{de}) as $n\to\infty$.

In order to find $\delta$ and $\delta_k$ in (\ref{xixk}), let us
consider {\it the second derivative} of (1). Using the same technique as above,
we obtain a set of equations analogous to (\ref{d0}) and
(\ref{d1}). From this set of equations,
using the formulas (\ref{d0}), (\ref{inf}), and (\ref{n2}) we can derive
\begin{equation}
\lim_{n\to\infty}\frac1{n^{2\gamma}}\sum_{i=1\atop i\neq k}^n
\frac1{(x_k-x_i)^3}=\frac{a'(x)c_\infty(x)}{4a(x)^2}.\label{n3}
\end{equation}

On the other hand, using (\ref{xixk}), we obtain the following expansion in
powers of $n$:
\begin{equation}
\frac1{n^{2\gamma}(x_k-x_i)^3}=-\frac{n^\gamma}{(i-k)^3\gamma_k^3}+
\frac{3\delta_k(i-k-1)n^{2\gamma-\delta}}{2\gamma_k^4(i-k)^3}+
o(n^{2\gamma-\delta}).\label{13}
\end{equation}
After the summation over $i\neq k$, the first term at the right hand side
yields zero and hence we see that in order to obtain in the limit as
$n\to\infty$ a finite and (for $a'(x)\neq 0$) nonzero sum,
there should be $\delta=2\gamma$. (If $a'(x)=0$ then, provided
(\ref{xixk}) holds, $\delta>2\gamma$.
This case needs some further investigation, which is not presented here.)
Then performing the summation in (\ref{13})
and comparing the result with (\ref{n3}), we get
\begin{equation}
\delta(x)=\frac{\pi^2 a'(x)}{2c_\infty(x)},\label{delta}
\end{equation}
where $\delta(x)=\lim_{n\to\infty}\delta_{k(n)}$.





Consider some particular cases. If $a(x)$ and $b(x)$ are polynomials of
no greater than the second and first degree, respectively, and $c(n)$
does not depend on $x$, then (1) is an equation of the hypergeometric
type. By differentiating it $n$ times, it is easy to
prove [\ref{NSU}] that the necessary and sufficient
condition for such an equation to have a polynomial solution of degree
$n$ is that $c(n)=-nb'(x)-n(n-1)a''(x)/2$. There is only one such
polynomial solution (up to a constant factor).

As is well known, the classical orthogonal polynomials satisfy the
hypergeometric-type equations (1). Namely:

1) $a(x)=1-x^2$, $b(x)=-(\alpha+\beta+2)x+\beta-\alpha$,
$c(n)=n(n+\alpha+\beta+1)$. Then $y_n(x)$ is the Jacobi polynomial.
(For certain particular values of the parameters $\alpha$ and $\beta$,
$y_n(x)$ is called the ultraspherical (Gegenbauer), Legendre, and
Chebyshev polynomial).
The system $\{y_n(x)\}_{n=0}^\infty$
is  orthogonal on the interval $(-1,1)$ with
the weight function $w(x)=(1-x)^\alpha(1+x)^\beta$ for $\alpha>-1$,
$\beta>-1$. Equations (\ref{gamma}) and (\ref{delta}) yield
expressions for the asymptotic distance between consecutive zeros
and its linear increase in the main order in $n$:
\begin{equation}
\gamma(x)/n=\frac{\pi\sqrt{1-x^2}}{n},\qquad
\delta(x)/n^2=-\frac{x\pi^2}{n^2},\qquad x\in(-1,1).
\end{equation}

2) $a(x)=x$, $b(x)=-x+\alpha+1$, $c(n)=n$. In this case $y_n(x)$ is the
Laguerre polynomial.
The system $\{y_n(x)\}_{n=0}^\infty$ is orthogonal on the interval
$(0,\infty)$ with the weight function $w(x)=x^\alpha e^{-x}$ for $\alpha>-1$.
In this case we have:
\begin{equation}
\gamma(x)/\sqrt{n}=\pi\sqrt{\frac{x}{n}},\qquad
\delta(x)/n=\frac{\pi^2}{2n},\qquad x>0.
\end{equation}

3) $a(x)=1$, $b(x)=-2x$, $c(n)=2n$. $y_n(x)$ is the
Hermite polynomial.
In this case the orthogonality interval is
$(-\infty,\infty)$, and the weight function is $w(x)=e^{-x^2}$.
The asymptotic distance between consecutive zeros,
according to (\ref{gamma}), is constant
$\gamma(x)/\sqrt{n}=\pi/\sqrt{2n}$. In this case if (\ref{xixk}) holds then
$\delta>2\gamma$,
and to understand the asymptotic behaviour of the distance between zeros,
we would need a more careful analysis.


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\end{document}

