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\lhead{\rmfamily \textsc{Alphonse P. Magnus.  Freud equations \& Gr\"unbaum \& al. problem  \today .} }
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\begin{center}
{\Large \textbf{
 Freud equations for Legendre polynomials on a circular arc
and solution of the Gr\"unbaum-Delsarte-Janssen-Vries problem.
}}


\bigskip

Alphonse P. Magnus
 \footnote{e-mail: \href{mailto:magnus@anma.ucl.ac.be}{\tt magnus@anma.ucl.ac.be}\\
       web: \href{http://www.math.ucl.ac.be/~magnus/}{\tt http://www.math.ucl.ac.be/\~{}magnus/}
       }

     Institut Math\'ematique,  Universit\'e Catholique de Louvain\\
      Chemin du Cyclotron 2\\
       B-1348 Louvain-la-Neuve\\
       Belgium\\

\end{center}
\bigskip
This pre-preprint ( \today \qquad ,  incomplete and unfinished), is
available as


 http://www.math.ucl.ac.be/\~{}magnus/freud/frgrde.pdf

\bigskip

\begin{flushright}
  \textsl{Although general methods led me to a complete solution, \\
      I soon saw that the result is obtained faster  \\ when the
      general procedure is left, and when one follows the path    \\
      suggested by the particular problem at hand.}   \\
                                    S.~Bernstein \qquad\qquad  \\
                            first lines of \cite{Bern}   \qquad\qquad
\end{flushright}

{\bf Abstract.} One establishes inequalities for the coefficients of orthogonal
polynomials
   $$  \Phi_n(z) = z^n +\xi_n z^{n-1} +\dotsb  + \Phi_n(0),
                                            \quad n=0,1,\dotsc  $$

 which are orthogonal with respect to a constant weight on the
 arc of the unit circle $S= \{e^{i\theta}, \alpha\pi<\theta
          < 2\pi-\alpha\pi\} $, with $0<\alpha<1$.
Recurrence relations (Freud equations), and differential relations
are used. Among other results, it is shown that $\Phi_n(0)>0, n=1,2,\dotsc$


\section{Introduction and statement of results.}

\subsection{Introduction}
The analysis of orthogonal polynomials on the unit circle has been limited
for a long time to measures supported on the whole circle (theories of
Szeg\H o, and, later on, of Rakhmanov).
Orthogonal polynomials on circular arcs were only known through special
cases (Geronimus, Akhiezer). They now enter a general theory as an
important subclass, as can be seen in
Khrushchev's paper \cite{Khrush}.

Actually, only a very special set of such orthogonal polynomials will be
studied here, namely the Legendre polynomials on an arc, i.e.,
$\Phi_0, \Phi_1,\dotsc$ are polynomials, with $\Phi_n$ of degree $n$,
and
$$
   \int_{\alpha\pi}^{2\pi-\alpha\pi} \Phi_n(e^{i\theta})
         \overline{\Phi_m(e^{i\theta})}\;d\theta =0
$$
when $n\neq m$, and where $\alpha$ is given ($0<\alpha<1$).

A property of these polynomials is needed in the solution of the
following problem:

\medskip

          ``3. The following Toeplitz matrix arises in  several  applications.
Define for $i\neq j$,  $A_{i,j}(\alpha) = \displaystyle{
  \frac{ \sin \pi\alpha (i-j) }{\pi(i-j)}  }$
and set $A_{i,i}=\alpha$.
Conjecture:  the matrix $M=(I-A)^{-1}$
has positive entries. A proof is known for $1/2\leqslant\alpha<1$.
Can one extend this to $0<\alpha<1$? Submitted by Alberto Gr\"unbaum, November 3, 1992.
(grunbaum@math.berkeley.edu)''  \cite{Grunb}.\\ \  \\
$I-A$ is the Gram matrix $[\langle z^i,z^j\rangle], i,j=0,1,\dotsc, N$
of the weight $w=1$ on the circular arc
$\alpha\pi <\theta < 2\pi-\alpha\pi$.
 For all the entries of all the
 $(I-A)^{-1}$ to be positive, it is necessary that all the
 coefficients $\Phi_n(0) >0$, $n=1,\dotsc,N$,
and the condition is known to be sufficient
\cite[p.~645]{DelGru}.

\medskip

In \cite{DelGru}, Delsarte \& {\it al}. study the robustness
of a signal recovery procedure amounting to find the polynomial
$p=p_0+\dotsb + p_N z^N$ minimizing the
integral of $|f(\theta)-p(e^{i\theta})|^2$ on the circular arc
shown above. This elementary least-squares problem involves the
Gram matrix $I-A$ of the problem above, and the stability of the
recovery procedure is related to the size of the smallest
eigenvalue of the matrix. The corresponding eigenvector is
shown to have elements of the same sign. The theory of this
eigenvalue-eigenvector pair sould be more complete if it could be
shown that $(I-A)^{-1}$ has only positive elements, for
any $N=1,2,\dotsc$, and any $\alpha\in(0,1)$. It is also
reported in \cite[p.~644]{DelGru} that Gr\"unbaum stated this
conjecture as early as 1981.

Now, the elements of $(I-A)^{-1}$ are positive combinations of
coefficients of the polynomials $\Phi_n$, and it is sufficient to show that
the sequence $\{\Phi_n(0)\}$ is positive (the opposite of $\Phi_n(0)$ is
the reflection coefficient $a(n+1,n+1)$ of \cite[p.~645]{DelGru}).

\smallskip

If $\alpha\geqslant 1/2$, all the zeros of $\Phi_n$ have negative real
part (Fej\'er), so $\Phi_n(0) = (-1)^n$ times the product of
all the zeros must be $>0$ (conjugate pairs have no influence on
the sign, and the number of real zeros is $n-$ an even number).

From continuity of the zeros with respect to $\alpha$, we are trying to
show that the \emph{\textbf{real zeros}} of $\Phi_n$ all remain
negative for all $0<\alpha<1$. Most zeros are close to the
support anyhow, and there are probably  only a small number of
 real zeros which are not close to $-1$.

\medskip

Here are some results containing the solution of the problem:

\subsection{{\bf Theorem.}\label{thmg}} {\sl The monic  polynomials

   $$  \Phi_n(z) = z^n +\xi_n z^{n-1} +\dotsb  + \Phi_n(0),
                                            \quad n=0,1,\dotsc  $$

 which are orthogonal with respect to a constant weight on the
 arc of the unit circle $S= \{e^{i\theta}, \alpha\pi<\theta
          < 2\pi-\alpha\pi\} $, with $0<\alpha<1$,
  have real coefficients  satisfying the
  following inequalities:
 \begin{enumerate}
   \item $0< \Phi_n(0) < \sigma$, \  $n=1,2,\dotsc$
   \item $ n\sigma^2 < \xi_n < (n-1)\sigma^2+\sigma$, \  $n=1,2,\dotsc$,
   \item $n\Phi_n(0) < (n+1)\Phi_{n+1}(0) \;,\qquad n=1,2,\dotsc  $,
   \item  $\Phi_n(0)$ is an increasing function
                  of $\alpha$, for any integer $n>0$.
 \end{enumerate}
where $\sigma=\sin(\pi\alpha/2)$.
 }

\subsection{{\bf Conjecture.}\label{conj}} {\sl Under the same
  conditions as above,}

$$ \Phi_n(0) < \Phi_{n+1}(0) \;,\qquad n=1,2,\dotsc  $$


\subsection{Method of proof of the theorem} \   \\

The proof mimics an algorithm of numerical calculation of
the sequence $\{\Phi_n(0)\}$ through a (non linear) recurrence
relation. It happens that a naive calculation based on an
approximate value of $\Phi_1(0)$ produces unsatisfactory values,
and that such numerical instabilities in recurrence
calculations can be fixed

Wimp

\begin{itemize}
 \item In section 2, a recurrence relation for the $\Phi_n(0)$'s
   (\emph{\textbf{Freud equations}}) will be produced,
 \item in section 3, the set of solutions of the latter recurrence
   relations will be shown to be a one-parameter set of
  sequences $\{ \boldsymbol{x} = \{x_1,x_2,\dotsc\}\;\}$, each
  solution $\boldsymbol{x}$ being completely determined by $x_1$.\\
  It will also be shown that there is at most one positive solution.
 \item In section 4, for each $N=1,2,\dotsc$, one will show
    how to construct the unique solution $\boldsymbol{x}^{(N)}$
   satisfying $0<x_n^{(N)} <\sigma$ for $n=1,2,\dotsc, N$ and
   $x_{N+1}^{(N)}=\sigma$.
 \item Finally, in section 5, we will see that, for each $n=1,2,\dotsc$,
   $x_n^{(N)}$ decreases when $N$ increases and reaches therefore
   a limit $x^*_n$ with which we build a nonnegative solution
   $\boldsymbol{x}^*$. This solution will finally be shown
   to be positive,
  ensuring the long sought existence of the positive solution!
\end{itemize}

\subsection{Known results}  \    \\

There are many results on asymptotic behaviour
\cite[etc.]{Golal1,Golakh,Golal2}

More subtle asymptotic estimates are also of interest in random
matrix theory \cite{AdvM,Wit}

\subsection{General identities of unit circle orthogonal polynomials}\  \\

Monic polynomials orthogonal on the unit circle with respect
to any valid measure $d\mu$:
$$
\Phi_n(z)=z^n +\xi_n z^{n-1}+\dotsb + \Phi_n(0)\;,
\langle \Phi_n,\Phi_m\rangle = \int_0^{2\pi} \Phi_n(z)
  \overline{\Phi_m(z)}\;d\mu(\theta) =0 \  \text{if}\;m\neq n,\   (z=
    e^{i\theta})
$$
satisfy quite a number of remarkable identities, most of them
stated by Szeg\H o in his book \cite[\S~11.3-11.4]{Sze}. The central one is that,
with
$$ \Phi_n^*(z) = \overline{\Phi_n(0)}\,z^n +\dotsb +
    \overline{\xi_n}\,z+1,
$$
 $\Phi_n^*/\|\Phi_n\|^2$ is the kernel polynomial with
respect to the origin:
\begin{equation} \label{kerneq}
 \dfrac{\Phi_n^*(z)}{\|\Phi_n\|^2} =  K_n(z;0)=
 \sum_{k=0}^n \dfrac{\overline{\Phi_k(0)}}{\|\Phi_k\|^2}\; \Phi_k(z)
\end{equation}
implying

\begin{equation} \label{norms}
   \|\Phi_{n+1}\|^2 = (1-|\Phi_{n+1}(0)|^2)\,\|\Phi_n\|^2
\end{equation}
\begin{equation} \label{recur3}
 \Phi_{n+1}(z) = z\Phi_n(z) +\Phi_{n+1}(0) \Phi^*_n(z)
\end{equation}
\begin{equation} \label{scal2}
  \langle \Phi_n, z^n\rangle = \|\Phi_n\|^2\;;
  \langle \Phi_n, z^{-1}\rangle = -\Phi_{n+1}(0)\,\|\Phi_n\|^2.
\end{equation}
For the last one:
$  \langle \Phi_n, z^{-1}\rangle =
  \langle z\Phi_n, 1\rangle =
  \Phi_{n+1}(0)  \langle \Phi^*_n, 1\rangle $, and
$\langle \Phi^*_n(z), P(z)\rangle =
   \|\Phi_n\|^2\langle K_n,P\rangle =
   \|\Phi_n\|^2\,\overline{P(0)}$
if $P$ is a polynomial of degree $\leqslant n$.

\begin{equation} \label{recur1}
 \Phi^*_{n+1}(z) =
 \dfrac{\|\Phi_{n+1}\|^2}{\|\Phi_n\|^2}   \Phi^*_n(z)
 +\overline{\Phi_{n+1}(0)}   \Phi_{n+1}(z)
\end{equation}
\begin{equation} \label{recur2}
 \Phi_{n+1}(z) =
 \dfrac{\|\Phi_{n+1}\|^2}{\|\Phi_n\|^2}   z\Phi_n(z)
 +\Phi_{n+1}(0)   \Phi^*_{n+1}(z)
\end{equation}

Identities for
the general kernel polynomial
$$
K_n(z;a)= \sum_{k=0}^n   \dfrac{\overline{\Phi_k(a)}\;\Phi_k(z) }
                                {\|\Phi_k\|^2}
$$
which is the only polynomial of degree $\leqslant n$ such that
$$
   \langle f,K_n \rangle = \int_{|z|=1}f(z)\overline{K_n(z;a)} \,d\mu(\theta)
       = \int_{|z|=1}f(z)\,K_n(a;z) \,d\mu(\theta) =
   f(a)
$$
for any $f$ of degree $\leqslant n$,
are best introduced through the determination of the polynomial
$F_n$ of degree $n$ of minimal norm with $F_n(a)=1$. As $F_n$ is
orthogonal to any polynomial $g$ of degree $\leqslant n$ vanishing
at $a$, it must be a scalar multiple of $K_n$, i.e.,
$$
    F_n(z;a) = \dfrac{K_n(z;a)}{K_n(a;a)}.
$$
Moreover, with $g(z)=(z-a)h(z)$,
$0=\langle g, F_n \rangle =
   \langle (z-a)h(z), F_n(z;a) \rangle =
   \langle zh(z), (1-\overline{a}\,z)F_n(z;a) \rangle,$
so that $(1-\overline{a}\,z)F_n(z;a)$ is orthogonal to $z,z^2,\dotsc, z^n$,
there is a constant $C$ such that
 $(1-\overline{a}\,z)F_n(z;a)-CK_n(0;z)$ is orthogonal to
  $1,z,\dotsc, z^n$, so is a constant multiple of $\Phi_{n+1}(z)$.
The final formula is

\begin{equation}\label{kernza}
 K_n(z;a) =\dfrac{  \overline{\Phi_n^*(a)}\, \Phi_n^*(z)
             -\overline{a} z\,  \overline{\Phi_n(a)}\, \Phi_n(z)  }
                 {(1-\overline{a}\,z)\|\Phi_n\|^2 }.
\end{equation}

N.B. The norm $\|F_n\| =1/\sqrt{K_n(a;a)} = \omega_n(\mu;a)$, the
famous Christoffel function \cite{NevChr}.

This latter piece of argument about $K_n(z;a)$ will not be needed in
the proof of the Theorem~\ref{thmg}, but 1) we will use similar
constructions, and 2) the formula may be useful in going further
with conjecture~\ref{conj}.



Finally, \eqref{recur3} yields expressions for the coefficients of
$z^{n-1}$ and $z$ in $\Phi_n(z)$:
\begin{equation} \label{xin}
 \xi_n = \xi_{n-1} +\Phi_n(0)\overline{\Phi_{n-1}(0)} =
 \Phi_1(0)
  +\Phi_2(0)\overline{\Phi_1(0)} +\dotsb
                   +\Phi_n(0)\overline{\Phi_{n-1}(0)}
\end{equation}
\begin{equation} \label{Phip}
                    \Phi'_n(0)=
  \Phi_{n-1}(0)+\Phi_n(0)\overline{\xi_{n-1}}
 =(1-|\Phi_n(0)|^2)  \Phi_{n-1}(0)+\Phi_n(0)\overline{\xi_n}
\end{equation}


\section{Recurrence relations (\emph{Freud equations}).}

\subsection{The Laguerre-Freud equations}
In looking for special non classical orthogonal polynomials related
to continued fractions satisfying differential equations, Laguerre
found some families of recurrence relations for the unknown coefficients.
Among the people who rediscovered some of these relations, G.~Freud
showed how to achieve progress in analysis by
 deriving from these relations a proof of inequalities and asymptotic properties, see
\cite{belron,Fr76,MagCam,NevChr} for more.

For orthogonal polynomials on the unit circle,
 the crux of the matter is that the weight function satisfies

\begin{equation} \label{PQ}
dw/d\theta   = R w,
\end{equation}
where $R$ is a rational function
of $z=\exp(i\theta)$, the same rational function $iP/Q$ on the whole
unit circle, up to a finite number of points \cite{AM}.  One shall
also need that $Qw=0$ at the endpoints of the support.

\subsection{The family of Legendre measures}  \   \\

Let us consider the measure $\displaystyle{ d\mu(\theta) =
   w(\theta) \dfrac{d\theta}{2\pi}  }$, with the following weight
  function:

\begin{equation} \label{weightAB} \begin{split}
  w(\theta) &= A \;,
       \qquad \alpha\pi < \theta < 2\pi-\alpha\pi, \\
            &= B \;,
       \qquad -\alpha\pi < \theta < \alpha\pi,
\end{split}\end{equation}
with $A$ and $B \geqslant 0$, $A+B>0$.

Our problem deals only with $B=0$, but we will need the full
family \eqref{weightAB} in a further discussion.

From symmetry with respect to the real axis, the polynomials
$\Phi_n$ have real coefficients.

Let $Q(z)=(z-e^{i\alpha\pi})(z-e^{-i\alpha\pi}) =z^2
                 -2\cos(\alpha\pi)z+1=2z(\cos\theta-\cos(\alpha\pi))$.

\subsection{The differential relation for the orthogonal polynomials}

We show that $Q\Phi'_n$ is a remarkably short linear combination
of some $\Phi$s and $\Phi^*$s \cite{AM}. To this end, we
 look at the integral of
$\displaystyle{ \dfrac{d}{dz}
   [ z^{-1} Q(z)f(z)\Phi_n(z^{-1}) ]  }$ on the two arcs
of \eqref{weightAB} for various polynomials $f$.
 Of course, the two integrals vanish, as $Q$
vanishes at the endpoints. So,

\begin{equation*}\begin{split}
0 &= A \int_{e^{i\alpha\pi}}^{e^{-i\alpha\pi}}
   d[z^{-1}  Q(z)f(z)\Phi_n(z^{-1}) ]
+B \int_{e^{-i\alpha\pi}}^{e^{i\alpha\pi}}
   d[ z^{-1} Q(z)f(z)\Phi_n(z^{-1}) ]  \\
 &= 2\pi i \int_0^{2\pi}
  z    \dfrac{d}{dz}
   [  z^{-1}Q(z)f(z)\Phi_n(z^{-1}) ] w(\theta)d\theta ,
\end{split}\end{equation*}
as $dz=de^{i\theta}=iz\,d\theta$.

The value is also
$$
\langle z(z^{-1}Qf)', \Phi_n\rangle
-\langle z^{-2}Qf, \Phi'_n\rangle =0.
$$
 The second scalar product is
also $\langle f, Q\Phi'_n\rangle$, as $z^{-2}Q(z)=Q(z^{-1})$, so
$$ \langle f, Q\Phi'_n \rangle =
  \langle z(z^{-1}Qf)', \Phi_n\rangle,$$
showing already that $Q\Phi'_n$ is a polynomial of degree $n+1$
which is orthogonal to $z,\dotsc, z^{n-2}$.    \\
By subtracting a suitable multiple of the kernel polynomial
$Q\Phi'_n -X_n  K_{n-1}$
is orthogonal to all the polynomials of degree $\leqslant n-2$,
 where
$X_n =\langle Q\Phi'_n,1\rangle  =
\langle z-z^{-1}, \Phi_n\rangle = \Phi_{n+1}(0)\|\Phi_n\|^2$.
\begin{equation} \label{diffrel0}
%Q\Phi'_n = (1-\Phi_n(0)^2)\Phi_{n+1}(0))\Phi^*_{n-1}
Q\Phi'_n = X_n \|\Phi_n\|^{-2}\Phi^*_{n-1}
   +n\Phi_{n+1} +Y_n\Phi_n+Z_n\Phi_{n-1},
\end{equation}
with the value of $X_n$ found above, even when $n=1$, as there is
no other orthogonality constraint.
The coefficient of $\Phi_{n+1}$ is obvious from the
leading coefficient of $Q\Phi'_n$. By looking at the coefficient
of $z^n$ in the expansion of $Q\Phi'_n$,  we get
$$ Y_n = (n-1)\xi_n -2n\cos(\alpha\pi)
 -n\xi_{n+1}
  =      -\xi_n -2n\cos(\alpha\pi)
         - n\Phi_{n+1}(0)\Phi_n(0)    .
$$
For $Z_n$,
\begin{equation*}\begin{split}
 Z_n\|\Phi_{n-1}\|^2 &=   \langle Q\Phi'_n, \Phi_{n-1}\rangle
    -X_n
   \langle K_{n-1} , \Phi_{n-1}\rangle \\
 &=
  \langle z(z^{-1}Q\Phi_{n-1})', \Phi_n\rangle
    -X_n  \Phi_{n-1}(0)  \\
  &=  \langle n z^n +\dotsb  -\Phi_{n-1}(0) z^{-1},\Phi_n  \rangle
     -X_n  \Phi_{n-1}(0)   \\
  &= n\|\Phi_n\|^2.
\end{split}\end{equation*}

$$
Q\Phi'_n = (1-\Phi_n(0)^2)\Phi_{n+1}(0)\Phi^*_{n-1} +n\Phi_{n+1}
  -[\xi_n+2n\cos(\alpha\pi)+n\Phi_n(0)\Phi_{n+1}(0)]\Phi_n
 +n(1-\Phi_n(0)^2)\Phi_{n-1}
$$
or also
\begin{equation} \label{diffrel}
Q\Phi'_n = (n+1)(1-\Phi_n(0)^2)\Phi_{n+1}(0)\Phi^*_{n-1}
  +[nz-\xi_n-2n\cos(\alpha\pi)]\Phi_n
 +n(1-\Phi_n(0)^2)\Phi_{n-1}
\end{equation}
which we evaluate at $z=0$:

\subsection{Recurrence relation for  $\Phi_n(0)$ } \   \\

\begin{equation} \label{recx1}
 (n+1) \Phi_{n+1}(0) - 2
\frac{ \xi_n
  +n \cos(\alpha\pi)  }{1-\Phi_n(0)^2}\, \Phi_n(0)
  +(n-1)  \Phi_{n-1}(0)=0,
\end{equation}
for $n=1,2,\dotsc$, and where
$ \xi_n =
 \Phi_1(0)
  +\Phi_1(0)\Phi_2(0) +\dotsb
                   +\Phi_{n-1}(0)\Phi_n(0)$.

Which is the recurrence relation determining $\Phi_{n+1}(0)$ from
$\Phi_1(0),\dotsc, \Phi_n(0)$, and which will be discussed in more detail
 in the next section.


\subsection{Differential equation for $\Phi_n$.}

Now, \eqref{diffrel} can be transformed into a differential system
for $\Phi_n$ and $\Phi^*_n$:
\begin{equation} \label{diffsys} \begin{split}
 zQ(z)\Phi'_n(z) &= [nQ(z)-(\xi_n+(n+1)\Phi_n(0)\Phi_{n+1}(0))z]\Phi_n(z)
               +[(n+1)\Phi_{n+1}(0)z-n\Phi_n(0)]\Phi_n^*(z)  \\
 Q(z)(\Phi^*_n)'(z) &= [n\Phi_n(0)z-(n+1)\Phi_{n+1}(0)]\Phi_n(z)
     +[\xi_n+(n+1)\Phi_n(0)\Phi_{n+1}(0)]\Phi^*_n(z)
\end{split}\end{equation}
Remark that, when $Q(z)=0$,
$$
\dfrac{\Phi_n(e^{\pm i\alpha\pi})}
      {\Phi^*_n(e^{\pm i\alpha\pi})} =
 \exp[\mp in\alpha\pi +2i\arg\Phi_n(e^{\pm i \alpha\pi})] =
 \dfrac{ (n+1)\Phi_{n+1}(0) -n\Phi_n(0)e^{\mp i\alpha\pi})  }
       { \xi_n +(n+1)\Phi_n(0)\Phi_{n+1}(0) },
$$
which makes sense if
$$ |\xi_n+(n+1)\Phi_n(0)\Phi_{n+1}(0)|=
       | (n+1)\Phi_{n+1}(0) -n\Phi_n(0)e^{\pm i\alpha\pi})  |,$$
another interesting indentity about the $\Phi_n(0)$'s.
By squaring\footnote{Squaring yields a proof by  induction: take the identity
at $n-1$ and add $2\{\xi_n +\Phi_n(0)[(n+1)\Phi_{n+1}(0)
                                     +(n-1)\Phi_{n-1}(0)]\}\Phi_n(0)
                 [(n+1)\Phi_{n+1}(0)-(n-1)\Phi_{n-1}(0)]$, so,
     \eqref{firsti} appears as a kinf of first integral of \eqref{recx1}.
   The form \eqref{firsti} appears essentially in Adler and van Moerbeke
   \cite{AdvM}, and in Forrester and Witte \cite{ForWit}.
}, one has
\begin{equation} \label{firsti}
 [\xi_n+(n+1)\Phi_n(0)\Phi_{n+1}(0)]^2 =
 (n+1)^2\Phi^2_{n+1}(0)
   -2n(n+1)\Phi_n(0)\Phi_{n+1}(0) \cos(\alpha\pi)
  +n^2\Phi^2_n(0).
\end{equation}
Also that, if one writes the system \eqref{diffsys} as
$\displaystyle{  \begin{bmatrix} zQ\Phi'_n \\  Q(\Phi^*)'_n
                 \end{bmatrix} =
               \begin{bmatrix} A  &  B \\  C & D
                 \end{bmatrix}
                 \begin{bmatrix} \Phi_n \\  \Phi^*_n
                 \end{bmatrix} }$, then
$AD-BC=n\xi_n Q$. useful in the construction of the scalar
differential equation for $\Phi_n$. Although this differential
equation will not be needed here, it would be a sin to neglect
to state it.    AM  Ismail

From \eqref{diffsys} a linear differential equation
 of second order for $\Phi_n$
follows
$$
zQ\Phi''_n +
\left[  zQ' -(n-1)Q
   -\dfrac{(n+1)\Phi_{n+1}(0)zQ}
          {(n+1)\Phi_{n+1}(0)z-n\Phi_n(0)} \right] \Phi'_n
$$
$$
+\left[
-nQ'+(n+1)\xi_{n+1}
 +[nQ-(\xi_n+(n+1)\Phi_n(0)\Phi_{n+1}(0))z]
    \dfrac{(n+1)\Phi_{n+1}(0)}
          {(n+1)\Phi_{n+1}(0)z-n\Phi_n(0)}  \right]
\Phi_n =0
$$

\subsection{Other weights:
semi classical orthogonal polynomials on the unit circle}

As already stated, similar relations hold whenever
$\displaystyle{\dfrac{dw/d\theta}{w}}$ is a rational function
of $z=\exp(i\theta)$, the same rational function $iP/Q$ on the whole
unit circle, up to a finite number of points. Then, $Q(z)\Phi'_n(z)$
is a remarkably short combination of some $\Phi$s and
$\Phi^*$s \cite{AM}. We just do as before, with
$$ 0= 2\pi i \int_0^{2\pi}
  z    \dfrac{d}{dz}
   [f(z)\overline{ Q(z) \Phi_n(z)}\,  w(\theta)]\,d\theta ,
$$
where $f$ is a polynomial,
$$
 \langle z(\overline{Q(z)}\; f)', \Phi_n \rangle
 -\langle z^{-1} f, Q\Phi'_n \rangle
  -\langle   f \overline{P(z)} , \Phi_n \rangle =0,
$$
using the pure imaginarity of $P(z)/Q(z)$ on the unit circle. We
see that $Q\Phi'_n$ is orthogonal to $z^{r-1},\dotsc, z^{n-2}$,
where $r$ is the maximum of the degrees of $P$ and $Q$ (so that
$\overline{P(z)}$ and
$\overline{Q(z)}$ are polynomials of degree $\leqslant r$ in
$z^{-1}$ on the unit circle).

As an exercise, consider the Gegenbauer case
$$ w(\theta) = A \text{\ or\ } B |\cos(\alpha\pi)-\cos\theta|^\beta$$
on the same arcs as in \eqref{weightAB}.Then,
$$
\dfrac{dw/d\theta}{w} =
\dfrac{-\beta \sin\theta}{\cos\theta-\cos(\alpha\pi)} =
\dfrac{i\beta (z^2-1)}{Q(z)},$$
with the same $Q$ as before. We still have \eqref{diffrel0},
but with the coefficient of $\Phi_{n-1}^*$ which is now
$$ -\|\Phi_{n-1}\|^{-2}\langle 1, Q\Phi'_n\rangle =
 -\|\Phi_{n-1}\|^{-2}\langle z(z\overline{Q})'-z\overline{P}, \Phi_n\rangle =
       (1+\beta) (1-\Phi_n(0)^2)\Phi_{n+1}(0)),
$$
 and with the same
$Y_n$ (i.e., the same \emph{formula}), and
$$
Z_n\|\Phi_{n-1}\|^2 = \langle Q\Phi'_n -X_n K_{n-1}, \Phi_{n-1}\rangle
$$
\begin{equation} \label{diffrelb}
Q\Phi'_n = (n+1+\beta)(1-\Phi_n(0)^2)\Phi_{n+1}(0)\Phi^*_{n-1}
  +[nz-\xi_n-2n\cos(\alpha\pi)]\Phi_n
 +(n+\beta)(1-\Phi_n(0)^2)\Phi_{n-1}
\end{equation}
which we evaluate at $z=0$:


\begin{equation} \label{recx1b}
 (n+1+\beta) \Phi_{n+1}(0) - 2
\frac{ \xi_n
  +n \cos(\alpha\pi)  }{1-\Phi_n(0)^2}\, \Phi_n(0)
  +(n-1+\beta)  \Phi_{n-1}(0)=0.
\end{equation}



\section{Properties of the solutions of the recurrence relations.}

\subsection{The set of solutions} \   \\

We now want to investigate all the solutions of the recurrence
relation
\begin{equation} \label{recx}
 (n+1) x_{n+1} -
2\frac{ \xi_n
  +n \cos(\alpha\pi) }{1-x_n^2}\, x_n
  +(n-1)  x_{n-1}=0,
\end{equation}
for $n=1,2,\dotsc$, where $\xi_n = x_1+x_1x_2+x_2x_3+\dotsb
   +x_{n-1}x_n$.

Each solution is a sequence $\{x_1,x_2,\dotsc\}$ completely determined
by the initial value $x_1$ (the value $x_0=1$ is common to all the
solutions considered here).

The particular solution we are interested in is determined by
$$ x_1 = \Phi_1(0) = -\dfrac{
   \displaystyle{\int_{\alpha\pi}^{(2-\alpha)\pi} e^{\pm i\theta}\;d\theta }}
 {\displaystyle{ \int_{\alpha\pi}^{(2-\alpha)\pi} d\theta }}   =
  \dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi}.
$$

But as \eqref{recx1} is valid for all the weights \eqref{weightAB}, we
find that $x_n$ is the related $\Phi_n(0)$, and that $x_1$ is the
ratio of moments
\begin{equation} \label{x1AB}
 x_1 =  -\dfrac{
   \displaystyle{A \int_{\alpha\pi}^{(2-\alpha)\pi} e^{\pm i\theta}\;d\theta
     +B \int_{-\alpha\pi}^{\alpha\pi} e^{\pm i\theta}\;d\theta }}
  {\displaystyle{ A\int_{\alpha\pi}^{(2-\alpha)\pi} d\theta
        +B \int_{-\alpha\pi}^{\alpha\pi} d\theta }}   =
  \dfrac{(A-B)\sin(\alpha\pi)}{A(1-\alpha)\pi+B\alpha\pi},
\end{equation}
relating $A/B$ to any $x_1$ (and even negative values of $A/B$ if
$x_1\notin [-\sin(\alpha\pi)/(\alpha\pi),
             \sin(\alpha\pi)/((1-\alpha)\pi)]$).

\subsection{Monotonicity with respect to $x_1$. {\bf Proposition.}\label{monot}}
{\sl While $x_1$, $x_2,\dotsc x_{n-1}$ are positive and less than 1,
and while $x_n$ is positive,
  $x_n$ is a continuous increasing function of $x_1$.}

Indeed, let us write the $i^{\text{th}}$ equation of \eqref{recx} as
$$ \dfrac{(i+1)x_{i+1}}{ix_i} =
 2\dfrac{x_1+x_1 x_2 +\dotsb +x_{i-1}x_i +i\cos(\alpha\pi)}{i(1-x_i^2)}
 -\dfrac{1}{
       \ \dfrac{ix_i}{(i-1)x_{i-1}}\   } ,
$$
for $i=1,2,\dotsc, n-1$.
As $x_1,\dotsc, x_n$ are positive, and $1-x_1^2,\dotsc, 1-x_{n-1}^2$
are positive too, the numerators $\xi_i+i\cos(\alpha\pi)$ are
positive too up to $i=n-1$. When $i=1$, we see that $x_2/x_1$, and
therefore $x_2$, is an increasing function of $x_1$.

If  $2x_2/x_1,\dotsc, ix_i/((i-1)x_{i-1})$ are continuous positive increasing
functions of $x_1$, then so is $x_{i+1}/x_i$, and therefore
$x_{i+1}$, as the two
terms of the right-hand side are increasing.  \qed

We look at the evolution of a solution with respect to $x_1\in (0,1)$.
We guess that if $x_1$ is too small, some $x_n$ will be negative, and
that if $x_1$ is too large,
 some $x_n$ will be larger than 1.

\subsection{Unicity of positive solution. \label{unicp} {\bf Proposition.}}
 {\sl The recurrence \eqref{recx} has at most one positive solution.}

Indeed, we consider four possibilities for $x_1$, according to the
ratio $A/B$ in \eqref{x1AB}:

\begin{enumerate}
  \item $x_1 =
      \displaystyle{\dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi} }$,
      corresponding to $B=0$. This is the solution we hope to
      show to be positive.
  \item
  $\displaystyle{ -\dfrac{\sin(\alpha\pi)}{\alpha\pi} <x_1 <
                    \dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi} }$,
   corresponding to $A>0$ and $B>0$. We then have a Szeg\H o
   weight, with $x_n \to 0$ and $\xi_n$ remaining bounded when
   $n\to \infty$. For $n$ large, and $p=0,1,2,\dotsc, P$ fixed, we have
  $$
   \dfrac{x_{n+p+1}}{x_{n+p}} \sim 2\cos(\alpha\pi)
      -\dfrac{1}{2\cos(\alpha\pi) -\dfrac{1}{\ddots -
               \dfrac{x_{n-1}}{x_n}  }  } =
     \dfrac{\sin((p+1)\alpha\pi+\rho_n)}
           {\sin(p\alpha\pi+\rho_n)},
  $$
   so that $x_{n+p} \sim C_n \sin(\rho_n +p\alpha\pi)$,
  $p=0,1,\dotsc, P-1$. We now choose $P$ so that $P\alpha$ is close
   to an even integer. The sines must change their signs, as
   the sum of these $P$ values is close to zero (actually, is
   $o(C_n)$).
  \item
  $\displaystyle{ x_1 =-\dfrac{\sin(\alpha\pi)}{\alpha\pi} }$,
   corresponds to $A=0$, and has of course no chance, as $x_1$
   is already negative! The asymptotic behaviour of $x_n$
   is known to be
  \item
  $\displaystyle{ x_1\notin \left[-\dfrac{\sin(\alpha\pi)}{\alpha\pi},
                    \dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi}\right] }$,
  corresponds to a non positive weight $A/B<0$, and we will either
   encounter a negative $x_n$, or $x_n>1$, but then $x_{n+1}<0$
  \footnote{
  If $x_{n-2}, x_{n-1}$, and $x_n$ are positive, with $x_{n-1}<1$,
    then $\xi_{n-1}+(n-1)\cos(\alpha\pi) > x_n/x_{n-1} -
     x_{n-1}x_n$, using \eqref{recx} with $n-1$. So,
     $\xi_n +n\cos(\alpha\pi) =
      \xi_{n-1} +(n-1)\cos(\alpha\pi) +x_{n-1}x_n +\cos(\alpha\pi)
      > x_n/x_{n-1}+\cos(\alpha\pi) >0$, and $x_{n+1} <0$}.
\end{enumerate}

That means that if we succeed in constructing a positive solution
of \eqref{recx}, this solution will have to be of the type 1)
above, and that will be the proof of positivity of the
sought solution.

One may also consider for each value of $x_1$ the smallest index
$\nu(x_1)$ where $x_\nu <0$. The propositions \ref{monot}
and \ref{unicp} above amount to stating that $\nu(x_1)$ is an
increasing function of $x_1\in
  \displaystyle{  \left(0,
                    \dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi}\right) }
$. The problem is to know if the limit of $\nu(x_1)$ will be
finite or infinite.


\section{Construction of a positive solution for $n=1,2,..., N+1$.  }

\subsection{Iteration of positive sequences} \  \\

 As it is so difficult to ``push'' a positive solution through an
 starting value $x_1$, we try to build a positive solution of
 \eqref{recx} through an iterative process keeping positive
 sequences. A good start is to write \eqref{recx} as

\begin{equation} \label{iter1}
  x_n = \sqrt{A_n^2(\boldsymbol{x})+1 }\,-
              A_n(\boldsymbol{x}) \,
      = \dfrac{1}{
        \sqrt{A_n^2(\boldsymbol{x})+1 }\,+
              A_n(\boldsymbol{x})    },\quad n=1,2,\dotsc
\end{equation}
where
\begin{equation*}
              A_n(\boldsymbol{x}) =
  \dfrac{x_1 +x_1x_2 +\dotsb + x_{n-1}x_n +n\cos(\alpha\pi)  }
        {(n-1)x_{n-1}+(n+1)x_{n+1} }.
\end{equation*}

Indeed, consider \eqref{recx} as an equation of degree two
for $x_n$
$$ x_n^2  +2A_n(\boldsymbol{x})\,x_n -1=0,$$
and take the unique positive root, which is \eqref{iter1}.

Therefore, the positive solution of \eqref{recx}, if it exists,
must satisfy \eqref{iter1}, and if we find a (positive, of course)
sequence satisfying \eqref{iter1}, we will have found the unique
positive solution of \eqref{recx}.

One may then consider to iterate \eqref{iter1}, hoping to see
it to converge towards the long sought positive solution.

Heavy numerical experiments (see \cite[\S~4.2]{Mag99}) suggest that
convergence indeed holds, but that no easy proof is at hand.
Moreover, some inequalities of Theorem~\ref{thmg} do not hold
for intermediate steps of application of \eqref{iter1}.

A modified iterative scheme will be much more satisfactory:


\subsection{An iteration of finite positive sequences.Proposition}
 \begin{itemize} \item
\textsl{
For any $\alpha\in (0,1)$, the function
 $\boldsymbol{F}^{(N,\varepsilon)}$ acting on a sequence
  $\boldsymbol{x} =\{x_n\}_1^\infty$ by
\begin{equation} \label{iter2}   \begin{split}
  F_n^{(N,\varepsilon)} &=
   \sqrt{[A_n^{(N,\varepsilon)}\boldsymbol{x})]^2+1 }\,-
              A_n^{(N,\varepsilon)}(\boldsymbol{x})\,
      = \dfrac{1}{
        \sqrt{[A_n^{(N,\varepsilon)}(\boldsymbol{x})]^2+1 }\,+
              A_n^{(N,\varepsilon)}(\boldsymbol{x})   },\quad n=1,2,
              \dotsc,N   \\
       &= \sigma,\qquad n=N+1,N+2,\dotsc
\end{split}\end{equation}
where $\sigma=\displaystyle{\sin\dfrac{\alpha\pi}{2}}$, and
\begin{equation}  \label{Anx}
              A_n^{(N,\varepsilon)}(\boldsymbol{x}) =
  \dfrac{N\sigma^2 +\varepsilon
   -x_n x_{n+1} - \dotsb - x_{N-1}x_N +n\cos(\alpha\pi)  }
        {(n-1)x_{n-1}+(n+1)x_{n+1} }, \quad n=1,2,\dotsc, N,
\end{equation}
transforms a positive sequence into a positive sequence; \\
if $\boldsymbol{x}\geqslant \boldsymbol{F}^{N,\varepsilon)}
(\boldsymbol{x})$
  (element-wise), then,
$\boldsymbol{F}^{(N,\varepsilon)}(\boldsymbol{x})    \geqslant
 \boldsymbol{F}^{(N,\varepsilon)}(
 \boldsymbol{F}^{(N,\varepsilon)}(\boldsymbol{x}))$
    when $\varepsilon\geqslant 0$.
} %end textsl
\item
\textsl{
Iterations of
$\boldsymbol{F}^{(N,\varepsilon)}$, starting with the constant
sequence $x_n=\sigma, n=1,2,\dotsc$,
  converge to a positive
fixed point $\boldsymbol{x}^{(N,\varepsilon)}$ of
$\boldsymbol{F}^{(N,\varepsilon)}$, i.e., a positive solution of
\begin{equation} \label{recxeps}
 (n+1) x_{n+1} -
2\frac{ N\sigma^2 +\varepsilon -x_n x_{n+1} -\dotsb - x_{N-1}x_N
  +n \cos(\alpha\pi) }{1-x_n^2}\, x_n
  +(n-1)  x_{n-1}=0,
\end{equation}
for $n=1,2,\dotsc,N$, and $x_n=\sigma$ for $n>N$.
} %end textsl
\item
\textsl{
 For any $\varepsilon\geqslant 0$, we now consider the function
  $$  f_N(\varepsilon)= N\sigma^2 +\varepsilon -x_1 -x_1x_2
    -\dotsb -x_{N-1}x_N $$
built with the sequence $\{x_1^{(N,\varepsilon)},\dotsc,
 x_N^{(N,\varepsilon)}\}$ found above.
The set of equations \eqref{recxeps} can also be written as
\begin{equation} \label{recxeps2}
 (n+1) x_{n+1} -
2\frac{ f_N(\varepsilon) +\xi_n
  +n \cos(\alpha\pi) }{1-x_n^2}\, x_n
  +(n-1)  x_{n-1}=0,
\end{equation}
Then, $f_N$ is an increasing function,
$f_N(0)=\sigma^2-\sigma <0$, $f_N(\varepsilon)\geqslant
  \sigma^2-\sigma+\varepsilon$, so that there is a unique positive
  zero  $\varepsilon_N$ of $f_N$, and the
  found positive solution $\boldsymbol{x}^{(N,\varepsilon_N)}$
  of \eqref{recxeps} is then the positive
  solution   $\boldsymbol{x}^{(N)}$
  of the equations \eqref{recx} for $n=1,2,\dotsc, N$, and
  $x_{N+1}=\sigma$.
}  % end textsl
\end{itemize}


Indeed, whenever $\boldsymbol{x}$ is a positive sequence,
each $A_n^{(N,\varepsilon)}(\boldsymbol{x})$ is a decreasing function
of the $x_i$'s, therefore,
 $F_n^{(N,\varepsilon)}(\boldsymbol{x})$ is an increasing function
of $\boldsymbol{x}$.

\smallskip

Next, the constant  positive sequence $x_n=\sigma$, $n=1,2,\dotsc$
satisfies
 $\boldsymbol{x}\geqslant \boldsymbol{F}^{(N,\varepsilon)}
(\boldsymbol{x})$, as
$   A_n^{(N,\varepsilon)}(\boldsymbol{x}) =   \displaystyle{
  \dfrac{n\sigma^2 +\varepsilon
    +n\cos(\alpha\pi)  }
        {2n\sigma  }, \geqslant \dfrac{\sigma^{-1}-\sigma}{2},
        n=1,2,\dotsc, N,   }$
from \eqref{Anx}, and $\cos(\alpha\pi)=1-2\sigma^2$.

Each $x_n$ will therefore decrease at each new iteration of
 $F_n^{(N,\varepsilon)}$, and will reach a nonnegative limit
called  $x_n^{(N,\varepsilon)}$, which satisfies \eqref{recxeps2}, as
stated above.
Remark that this
limit is not non only nonnegative, but actually positive:
if   $x_1^{(N,\varepsilon)}=0$, then
     $x_n^{(N,\varepsilon)}=0$ for all $n>0$;
if   $x_{n-1}^{(N,\varepsilon)}>0$, and
     $x_n^{(N,\varepsilon)}=0$, with $n>0$,then
     $x_{n+1}^{(N,\varepsilon)}<0$, and we could not have
$x_{N+1}=\sigma$.

We also have
     $x_n^{(N,\varepsilon)}<\sigma$ if $\varepsilon>0$.

Finally, we compare the values of some $x_n$ when the iterations
(\ref{iter2}-\ref{Anx}) are
performed with two different values of $\varepsilon$,
and find that $x_n$ is a decreasing function of $\varepsilon$,
whence the increasing character of the function $f_N$.

 \qed

Much more general
iterations with monotonocity properties are worked in Chapter~3
of Collatz' book \cite{Col}.

\section{Final limit process.}

\subsection{Proposition \label{finall}} \textsl{The sequence
$\boldsymbol{x}^{(N)}$ built above as the unique positive solution
of \eqref{recx} for $n=1,2,\dotsc,N$ with $x_{N+1}=\sigma$, decreases
when $N$ increases and converges to the unique positive solution
$\boldsymbol{x}$ of \eqref{recx}, whose existence had to be
established.}

Indeed, from $x_{N+1}^{(N)}=\sigma$, and
             $x_{N+1}^{(N+1)}<\sigma$,
       $x_1^{(N+1)}< x_1^{(N)}$ must follow, from
Proposition~\ref{monot}, and then
       $x_n^{(N+1)}< x_n^{(N)}$ for all $n\leqslant N+1$.

Moreover, $\boldsymbol{x}$ is actually positive, and not merely
nonnegative, as
$x_n <\sigma$ and $\varepsilon_N >0 \Rightarrow
0 > N\sigma^2 +\varepsilon_N -x_1^{(N)} -(N-1)\sigma^2$:
$x_1>\sigma^2$. And, as we saw above, we can not have $x_{n-1}>0,
x_n=0$, and $x_{n+1}\geqslant 0$.


This achieves the proof of (1-3) of Theorem~\ref{thmg}.

\medskip

\subsection{Numerical illustration and software}  \   \\

we choose $\alpha=1/4$, then
$\sigma=\sin(\alpha\pi/2)= 0.382683...$,


We iterate $F^{(5,0.01)}$, starting with the constant sequence
$x_n=\sigma:$

{\Small \begin{verbatim}
 it.   res.      x1       x2       x3       x4      x5        x6
  1   0.01306  0.38268  0.38268  0.38268  0.38268  0.38268  0.38268
  2   0.01053  0.37937  0.38102  0.38157  0.38185  0.38201  0.38268
  3   0.00960  0.37673  0.37939  0.38060  0.38118  0.38176  0.38268
  4   0.00804  0.37436  0.37803  0.37975  0.38076  0.38157  0.38268
  5   0.00679  0.37239  0.37686  0.37913  0.38041  0.38144  0.38268
  6   0.00542  0.37074  0.37594  0.37860  0.38017  0.38134  0.38268
  7   0.00445  0.36943  0.37517  0.37820  0.37996  0.38126  0.38268
  8   0.00352  0.36837  0.37457  0.37787  0.37980  0.38120  0.38268
  9   0.00285  0.36753  0.37408  0.37761  0.37968  0.38116  0.38268
 10   0.00226  0.36685  0.37370  0.37740  0.37958  0.38112  0.38268
\end{verbatim} }

where ``res'' is the norm of the residue at the particular iteration
step, i.e., the largest absolute value of the left-hand sides of
\eqref{recxeps}, $n=1,2,\dotsc, N$. This error norm decreases
rather slowly, and we proceed up to the reception of a reasonably
small value:

{\Small \begin{verbatim}
 it.   res.      x1       x2       x3       x4      x5        x6
 50   0.00000  0.36420  0.37218  0.37659  0.37918  0.38097  0.38268
\end{verbatim} }

one finds $f_5(0.01)=-0.18493$. We already knew that
$f_5(0)=\sigma^2-\sigma =-0.23623...$


We start the whole process again with various values of $\varepsilon$:


{\Small \begin{verbatim}
 eps.    f(eps)     x1       x2       x3       x4      x5        x6
  0    -0.23623  0.38268  0.38268  0.38268  0.38268  0.38268  0.38268
  0.01 -0.18493  0.36420  0.37218  0.37659  0.37918  0.38097  0.38268
  0.02 -0.13634  0.34700  0.36206  0.37061  0.37571  0.37927  0.38268
  0.03 -0.09021  0.33097  0.35231  0.36474  0.37228  0.37758  0.38268
  0.04 -0.04633  0.31600  0.34291  0.35898  0.36889  0.37591  0.38268
  0.05 -0.00450  0.30200  0.33384  0.35333  0.36552  0.37424  0.38268
  0.06  0.03544  0.28888  0.32509  0.34778  0.36220  0.37259  0.38268
\end{verbatim} }


we find  $\varepsilon_5=0.0511$, and perform the whole thing again
for several values of $N$:

{\Small \begin{verbatim}
  N    eps       x1       x2       x3       x4      x5        x6       x7       x8       x9      x10
  5  0.05110  0.30051  0.33286  0.35271  0.36516  0.37406  0.38268
  6  0.04124  0.30024  0.33242  0.35194  0.36370  0.37118  0.37682  0.38268
  7  0.03443  0.30015  0.33227  0.35167  0.36319  0.37019  0.37482  0.37853  0.38268
  8  0.02953  0.30012  0.33221  0.35157  0.36301  0.36984  0.37411  0.37707  0.37962  0.38268
  9  0.02585  0.30011  0.33219  0.35154  0.36295  0.36971  0.37385  0.37654  0.37852  0.38034  0.38268
 10  0.02299  0.30011  0.33219  0.35152  0.36292  0.36967  0.37376  0.37634  0.37810  0.37948  0.38084
\end{verbatim} }

And we see that we have indeed reconstructed
$x_1= \Phi_1(0)=\displaystyle{ \dfrac{\sin(\alpha\pi)}{(1-\alpha)\pi}
= 0.3001054...}$.

The gp-pari \cite{PARI} program used here can be found at

 \href{http://www.math.ucl.ac.be/~magnus/freud/grunbd.gp}
    {\tt http://www.math.ucl.ac.be/\~{}magnus/freud/grunbd.gp}.

A more experimental program, allowing $\beta\neq 0$ is at

 \href{http://www.math.ucl.ac.be/~magnus/freud/grunb2.gp}
    {\tt http://www.math.ucl.ac.be/\~{}magnus/freud/grunb2.gp}.

There is also a java program available at

 \href{http://www.math.ucl.ac.be/~magnus/freud/grunbd.htm}
    {\tt http://www.math.ucl.ac.be/\~{}magnus/freud/grunbd.htm}.


 The numerical efficiency of this demonstration
is close to zero! Should somebody really need a long subsequence
of the $\Phi_n(0)$'s,

\subsection{Proof of (4) of Theorem \ref{thmg}}
We show that, if $\boldsymbol{x}$ is a positive sequence bounded by
$\sigma$, and with $nx_n$ increasing with $n$, then the same holds
for
  $\boldsymbol{F}^{(N,\varepsilon)}(\boldsymbol{x})$. Indeed, by
\eqref{iter2},
$$ nF_n = \dfrac{1}{ \dfrac{A_n}{n} +
               \sqrt{ \left( \dfrac{A_n}{n}\right)^2 +
                      \dfrac{1}{n^2}  }  }$$
is increasing if $A_n/n$ is decreasing. Now, by \eqref{Anx},
$$ \dfrac{A_n}{n}=  \dfrac{ y_n +\cos(\alpha\pi)  }
        {(n-1)x_{n-1}+(n+1)x_{n+1} }, $$
$$\text{where\  } y_n =
                    \dfrac{N\sigma^2 +\varepsilon
   -x_n x_{n+1} - \dotsb - x_{N-1}x_N}{n} ,$$
has an increasing denominator, and a decreasing numerator. Indeed,
$$ y_{n+1}-y_n = \dfrac{(n+1)y_{n+1}-ny_n -y_{n+1} }{n} =
                 \dfrac{x_n x_{n+1} -y_{n+1} }{n} <0,$$
as $x_n<\sigma$ and $\varepsilon>0 \Rightarrow y_n>\sigma^2$ .\qed

\section{Differential equations with respect to $\alpha$.}

Let $\Phi_n$ and $\tilde{\Phi}_n$ be the monic orthogonal polynomials
of degree $n$ with respect to the measures $d\mu$ and $d\tilde{\mu}$.
As any polynomial of degree $n-1$, $\tilde{\Phi}_n-\Phi_n$ is represented
through the kernel polynomial $K_{n-1}$:
$$
%\tilde{\Phi}_n(z)-\Phi_n(z) = \int_\mathbb{T}
\tilde{\Phi}_n(z)-\Phi_n(z) = \int_{|t|=1}
(\tilde{\Phi}_n(t)-\Phi_n(t))K_{n-1}(z,t)\,d\mu.$$
We may suppress in the integral $\Phi_n$, which is orthogonal to
$K_{n-1}$; and replace $d\mu$ by $d\mu-d\tilde{\mu}$, as
$\tilde{\Phi}_n$ is orthogonal to $K_{n-1}$ with respect to
$d\tilde{\mu}$:
$$
\tilde{\Phi}_n(z)=\Phi_n(z) - \int_{|t|=1}
\tilde{\Phi}_n(t)K_{n-1}(z,t)\,(d\tilde{\mu}-d\mu).$$
sometimes called the Bernstein integral equation for
$\tilde{\Phi}_n$.  [...]

$$
\dfrac{\partial\Phi_n(z)}{\pi\partial\alpha}=
(A-B)[\Phi_n(e^{i\alpha\pi})K_{n-1}(z,e^{i\alpha\pi})
     +\Phi_n(e^{-i\alpha\pi})K_{n-1}(z,e^{-i\alpha\pi})]
$$
At $z=0$:
$$
\dfrac{d\Phi_n(0)}{\pi d\alpha}=
(A-B)[\Phi_n(e^{i\alpha\pi})K_{n-1}(0,e^{i\alpha\pi})
     +\Phi_n(e^{-i\alpha\pi})K_{n-1}(0,e^{-i\alpha\pi})]
$$
$$
=(A-B)\|\Phi_{n-1}\|^{-2}[\Phi_n(e^{i\alpha\pi})
     \overline{\Phi^*_{n-1}(e^{i\alpha\pi})}
     +\overline{\Phi_n(e^{-i\alpha\pi})}
     \Phi^*_{n-1}(e^{-i\alpha\pi})]
$$
relating $\Phi_n(0)$ to values at $e^{\pm i \alpha\pi}$, which may not
be easier. However,
$$
\dfrac{d\Phi_n(0)}{\pi d\alpha}= (A-B)
   \dfrac{ |\Phi_{n-1}(e^{i\alpha\pi})|^2}{\|\Phi_{n-1}\|^2}
  \left[
        \dfrac{\Phi_n(e^{i\alpha\pi})}{\Phi^*_{n-1}(e^{i\alpha\pi})}+
     \dfrac{\overline{\Phi_n(e^{i\alpha\pi})}}
           {\overline{\Phi^*_{n-1}(e^{i\alpha\pi})}}  \right],
$$
and we know that
$$
  \dfrac{\Phi_n(e^{i\alpha\pi})}{\Phi^*_{n-1}(e^{i\alpha\pi})}=
  e^{i\alpha\pi}
  \dfrac{\Phi_{n-1}(e^{i\alpha\pi})}{\Phi^*_{n-1}(e^{i\alpha\pi})}
  +\Phi_n(0)
$$
$$
=
 \dfrac{ n\Phi_n(0)e^{i\alpha\pi} -(n-1)\Phi_{n-1}(0)  }
       { \xi_n +(n-1)\Phi_{n-1}(0)\Phi_n(0) }+\Phi_n(0),
$$
and
$$
\dfrac{d\Phi_n(0)}{d\alpha}=\pi (A-B)
 [1-\Phi_n^2(0)]
   \dfrac{ |\Phi_{n-1}(e^{i\alpha\pi})|^2}{\|\Phi_{n-1}\|^2}
 \dfrac{ (n+1)\Phi_{n+1}(0) -(n-1)\Phi_{n-1}(0)  }
       { \xi_n +(n-1)\Phi_{n-1}(0)\Phi_n(0) }
$$
which achieves the proof of (4) of Theorem~\ref{thmg}. \qed

We certainly would like more explicit differential relations and
equations (Painlev\'e!) with respect to $\alpha$ here!

According to a formula in the proof of Prop. 5.3 of \cite{ForWit},

\begin{equation*} \begin{split}
\dfrac{d\Phi_n(0)}{d\alpha}&=-\pi
    \dfrac{ [\xi_n +n\cos(\alpha\pi)]\Phi_n(0)
            -(n-1)[1-\Phi_n^2(0)]\Phi_{n-1}(0)  }{\sin(\alpha\pi)} \\
  &= \pi \dfrac{ [\xi_n +n\cos(\alpha\pi)]\Phi_n(0)
            +(n+1)[1-\Phi_n^2(0)]\Phi_{n+1}(0)  }{\sin(\alpha\pi)}
\end{split}\end{equation*}

\section{Conclusion: new problems.}

We could establish the inequalities of Theorem~\ref{thmg} as
far as they are related to the unique positive solution of
the recurrence relations \eqref{recx1}. The method is to design
an iterative scheme converging towards this positive solution,
and to ensure that the required inequalities hold at each
intermediate step.

 Such a method may fail very easily: for
instance, the scheme \eqref{iter1} may have seemed very promising, but
produced sometimes unsatisfactory intermediate iterates.

Also, the conjecture~\ref{conj} cannot be proved by merely feeding
the iteration \eqref{iter2} with arbitrary increasing sequences: if
$\sigma$ is very small, we see that the sequence $\{A_n(\boldsymbol{x})\}_n$
is decreasing only if the sequence
$\displaystyle{  \dfrac{(n+1)x_{n+1}+(n-1)x_{n-1} }{n}  }$ is
increasing, which complels us to look for further inequalities. So,
something smarter is needed.

Final example of drawback of the method: if we want to
investigate the \emph{Gegenbauer} polynomials on a unit circle arc,
we only have to replace $(n+1)x_{n+1}+(n-1)x_{n-1}$ in the
denominator of \eqref{Anx} by
 $(n+\beta+1)x_{n+1}+(n+\beta-1)x_{n-1}$,
and the results of Section~1 are probably still true, at least if
$\beta\geqslant 0$. But we will now have to include the initial
condition $x_0=1$ explicitly, and have a lot of troubles with
the inequalities on the $x_n$'s. The conjecture \ref{conj} does not
hold for any $n$ and $\beta$ anyhow, as
$x_n \to \max(-1,\beta/(n+\beta))$ when $\alpha$ is small
(and $\xi_n \to n\beta/(n+\beta)$ if $\beta>-1/2$).

\section{Acknowledgements.}

Many thanks for ideas, references, and kind words to
M.~Adler,
F.A.~Gr\"unbaum,
J.~Meinguet,
P.~Nevai,
A.~Ronveaux,
W.~Van Assche,
P.~van Moerbeke,
N.S.~Witte


This paper presents research results of the Belgian Programme on
Interuniversity Attraction Poles, initiated by the Belgian State, Prime
Minister's Office for Science, Technology and Culture.  The scientific
responsibility rests with its author.



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\end{document}
