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\lhead{\rmfamily \textsf{MAPA3xxxA\ 2002-03\ }-- Semi-classical orth. pol. on unit circle. \ \textbf{\thesection}\ --\ \mylmark\  --} \chead{} \rhead{\rmfamily\thepage}
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\begin{document}

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\begin{center} \textbf{\Large
  MAPA 3072A Special topics in approximation theory.}

\smallskip

 \textbf{1999-2000, 2002-2003, 2012-2013:
  Semi-classical orthogonal polynomials on the unit circle.}

\bigskip
  Alphonse Magnus,    \\
  Institut de  Math\'ematique Pure et Appliqu\'ee,  \\
 Universit\'e Catholique de Louvain, \\
 Chemin du Cyclotron,2, \\
  B-1348 Louvain-la-Neuve \\
      (Belgium)\\% Research address for author one
       \href{mailto:alphonse.magnus@uclouvain.be}
        {\framebox{\tt alphonse.magnus@uclouvain.be}} ,
  \href{http://perso.uclouvain.be/alphonse.magnus}
       {\framebox{\tt http://perso.uclouvain.be/alphonse.magnus }}
%Paul Van Dooren, EULER, a119,  478040 , vandooren@anma.ucl.ac.be
\end{center}
\bigskip

%pp.~1-23: October 15, 1999. \\
%pp.~24-28: November 17, 1999. \\
\textbf{{\red New} in 2002-2003:} essentially the linear differential
equation, more and better ordered experimental evidence about the
generalized Jacobi polynomials, and various corrections. Thanks for
remarks and kind words: L.~Golinskii, M.~Ismail, F.~Marcell\'an,
N.~Witte. Earlier thanks to P.~Nevai for sending me a copy of
the Badkov paper \cite{Badk} in a former century.

Er, most of these ``new'' items are still in construction,
but there are new facts about the Gr\"unbaum-Delsarte-Janssen-Vries
problem in \S~\ref{grunl}.

\textbf{{\red New, new}} in 2012-2013: final asymptotic of unit circle gen. Jacobi
\begin{equation*} 
   w(\theta) =
    \begin{cases}
        r_1 e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_1<\theta<\theta_2\\
        r_2e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_2-2\pi <\theta <\theta_1]
    \end{cases}
\end{equation*}
where $r_1$ and $r_2$ are positive,  see \S~\ref{finalgjac}, p.~\pageref{finalgjac}.
 
This version: 
\today 
%May 1, 2003 
\qquad  (incomplete and unfinished)

\medskip

\begin{flushright}
\textit{ Rien n'est plus agr\'eable que de savoir
quelqu'un aux prises avec des difficult\'es insurmontables, surtout si l'on
y voit du travail g\^ach\'e, de l'argent perdu et des crises de
 nerfs\footnote{There is nothing like knowing somebody having to
   struggle with a hopeless task, especially if it represents
   bad work, loss of money, and nervous tantrums.}.}\\
     \hspace*{72pt} J.~Giono\\
\end{flushright}

Abstract: Semi-classical orthogonal polynomials on the unit circle
are examined. Special care is given to generalized Jacobi
polynomials.

%\medskip
%\enlargethispage{33pt}

\tableofcontents

\bigskip



\section{Complex orthogonal polynomials.}
\renewcommand{\mylmark}{Complex orthogonal polynomials.}
\pagestyle{fancy}

\subsection{General scalar product.}

For any scalar product (positive definite sesquilinear\footnote{i.e., $(f,g)$ involves
the \emph{\textbf{complex conjugate}} of $g$.}
 hermitian symmetric form) defined at least on
the polynomials of degree $\leqslant N \leqslant\infty$, there is exactly one
sequence $\{\Phi_0, \Phi_1,\dotsc, \Phi_N\}$ of \emph{monic} polynomials,
degree $\Phi_n=n, n=0,1,\dotsc, n$, such that $(\Phi_n, \Phi_m)=0, 0\leqslant m\neq n
\leqslant N$.

The $n$ coefficients of $\Phi_n$ are determined by $(\Phi_n,p)=0$ for any $p$ of
degree $<n$ :
$\Phi_n(z) = z^n +\displaystyle{ \sum_0^{n-1} \Phi_{n,z^k} z^k }$,
$$
\boldsymbol{G}_{n-1}
   \begin{bmatrix} \Phi_n(0) \\ \Phi'_n(0) \\ \vdots \\  \Phi_{n,z^{n-1}} \end{bmatrix}
  =
  \begin{bmatrix}
    (z^0, z^0) & (z^1, z^0) & \cdots & (z^{n-1},z^0) \\
    (z^0, z^1) & (z^1, z^1) & \cdots & (z^{n-1},z^1) \\
      \vdots   &  \vdots    &        &  \vdots     \\
    (z^0, z^{n-1}) & (z^1, z^{n-1}) & \cdots & (z^{n-1},z^{n-1})
  \end{bmatrix}
   \begin{bmatrix} \Phi_n(0) \\ \Phi'_n(0) \\ \vdots \\  \Phi_{n,z^{n-1}} \end{bmatrix}
=-
  \begin{bmatrix}
   (z^n, z^0) \\ (z^n,z^1) \\  \vdots  \\  (z^n,z^{n-1})
  \end{bmatrix} \;.
$$
$\boldsymbol{G}_{n-1}$ is the transposed of the \emph{Gram matrix}
of $\{ z^0, z^1,\dotsc, z^{n-1}\}$. The matrix is hermitian and
positive definite (cf. \framebox{\cite{Dav}}).

$  \Phi_0=1\; \qquad
    \Phi_1(z) = z- \displaystyle{\frac{ (z^1, z^0)}{(z^0, z^0)}  }$.

Also,
$$
\boldsymbol{G}_n \begin{bmatrix} \Phi_n(0) \\
                                 \Phi'_n(0) \\
                                 \vdots \\
                                \Phi_{n,z^{n-1}} \\
                                    1  \end{bmatrix} =
                 \begin{bmatrix} 0 \\
                                 0 \\
                                 \vdots \\
                                  0 \\
                                 (\Phi_n,z^n)=\|\Phi_n\|^2  \end{bmatrix}
$$
i.e., the last column of $\boldsymbol{G}_n^{-1}$ is made of the
coefficients of $\Phi_n/\|\Phi_n\|^2$.

$\Phi_n$ is the only monic $n^{\text{th}}$ degree polynomial of minimal norm:
any $n^{\text{th}}$ degree monic polynomial is $\Phi_n +p$ with degree $p < n$, and
$$
\| \Phi_n +p\|^2 = (\Phi_n+p, \Phi_n+p) = \|\Phi_n\|^2 +
2\;\underbrace{\text{Re}\;(\Phi_n,p)}_{0} +\|p\|^2.
$$

One has
$$
  \|\Phi_n\|^2  = \frac{ \det \boldsymbol{G}_n}
                       { \det \boldsymbol{G}_{n-1}}  \;.
$$

\medskip

\subsection{Playing with the idea.}

\subsubsection{} The simplest case occurs when $\boldsymbol{G}_n$
happens to be a \emph{\textbf{diagonal matrix}}, such as for the
simplest $L^2$ scalar products on a circle or on a disk:
$$
(f,g)=\int_{|z=R\exp(i\theta)|=R} f(z)\overline{g(z)}\dfrac{d\theta}{2\pi R},\  %
(f,g)=\int_{|z|<R} f(z)\overline{g(z)}\;\dfrac{dx\,dy}{\pi R^2}.
$$
This also happens with \emph{\textbf{Sobolev}} scalar products
$\displaystyle{ \int  f(z)\overline{g(z)}
                     + f'(z)\overline{g'(z)}   }$

\subsubsection{} The next simplest situation seems to be
\emph{\textbf{binomial}} polynomials: when do we have
$\Phi_m(z)=z^m-\alpha_m$, $m=1,2,\dotsc$?
Then one must have $(\Phi_m, z^k)=0$, i.e.,
$(z^m,z^k)= \alpha_m (z^0,z^k)$ for $k=0,1,\dotsc, m-1\geqslant 0$.
 In particular, for
$k=0$: $(z^m,z^0)=\alpha_n (z^0,z^0)$ for all $m>0$. Whence
\begin{equation} \label{mombin}
 (z^m,z^0)=\alpha_m(z^0,z^0),
 (z^0,z^m)=\overline{\alpha_m}(z^0,z^0),\   %
 (z^m,z^k)=\alpha_m\overline{\alpha_k}(z^0,z^0),\   %
  m,k=1,2,\dotsc, m\neq k.
\end{equation}
Oh, I see, this is just a familiar perturbation of rank one to
a diagonal matrix, easily explainable by the addition of a
masspoint to an elementary $L^2$ or Sobolev measure, and
we invert $\boldsymbol{G}_n = \boldsymbol{D}+\boldsymbol{u}
\boldsymbol{v}^T$ by Sherman-Morrison formula
\framebox{\cite[p.~3]{GolVL}}
 $\displaystyle{\boldsymbol{G}_n^{-1} = \boldsymbol{D}^{-1}
      -\dfrac{  \boldsymbol{D}^{-1}
                \boldsymbol{u}
                \boldsymbol{v}^T
                \boldsymbol{D}^{-1}      }
      {1+\boldsymbol{v}^T  \boldsymbol{D}^{-1} \boldsymbol{u}  }}$
and\dots hey! this is not sparse!   Indeed, \eqref{mombin} is
\emph{\textbf{not}} a rank one perturbation to a diagonal matrix
$\boldsymbol{D}$! From the nondiagonal elements of
$\boldsymbol{G}_n$,
$\alpha_m\overline{\alpha_k}(z^0,z^0) = u_m v_k$, (with $\alpha_0=1$),so,
$u_m = \gamma \alpha_m, v_m =\gamma^{-1} (z^0,z^0)\overline{\alpha_m}$.
But then the first diagonal element should be
$(z^0,z^0)=d_0 +u_0 v_0 =   d_0 + (z^0,z^0)$, or $d_0=0$, aaargh.  \\
Well, there is nothing wrong with $d_0=0$, such as in
$$
 (f,g)= \int_0^{2\pi} f'(e^{i\theta})\, \overline{g'(e^{i\theta})}\,
    \dfrac{d\theta}{2\pi} +f(re^{ia})\overline{g(re^{ia})},
$$
so that \framebox{\eqref{mombin}} holds with
$\alpha_m = r^m e^{ima}$, $m=0,1,\dotsc$

Remark that, if $r>1$, all the zeros but one of $\Phi_m$ are outside
the convex hull of the support of the measures involved in the
scalar product, a known feature of Sobolev orthogonal polynomials.
Actually, I do not know if these zeros are useful in any respect.

\smallskip

$z^m-\alpha_m z^{m'}$

\subsubsection{} Here is an example where the
$(z^\ell, z^k)$ satisfy special  relations which make  orthogonal
polynomials immediately appear:    \label{dioph}

let $\alpha$ be an \emph{irrational} positive number, and the
scalar product be $\displaystyle{ (f,g) = \frac{4}{\pi^2}
\sum_{\substack{m=-\infty
         \\m\;\text{odd}}}^\infty  \frac{1}{m^2} f(e^{im\alpha\pi})
                                \overline{g(e^{im\alpha\pi})}  }$,
so that
$(z^\ell, z^k)
=\displaystyle{  \frac{8}{\pi^2} \sum_{\substack{m=1
  \\m\;\text{odd}}}^\infty  \frac{1}{m^2} \cos[ (k-\ell)m\alpha\pi]
  = F((k-\ell)\alpha)  }$,
where $F$ is the even periodic function of period $2$, with
value $F(u)= 1-2|u|$ on $[-1, 1]$.
Remark that $F(u+k) = (-1)^k F(u)$ if $k\in\mathbb{Z}$.

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(2.2,2.2)
\psline[linewidth=0.04](0.25,1)(0.5,0)(1,2)(1.5,0)
 \pscircle*(1.2,1.2){0.1}
% \pscircle[linewidth=0.05](1,1.5){1}
\end{pspicture}
}
Now, let $m'/n'$ be the best rational approximant to $\alpha$ in the sense that
$1\leqslant n'\leqslant n$ and $|n'\alpha-m'|< $ any other $p\alpha-q$
with $1\leqslant p\leqslant n$; $m''/n''$ the second best rational
approximant to $\alpha$. Then, $\Phi_n(z) = z^n+$ a combination of
$z^{n-n'}$ and $z^{n-n''}$. Indeed, $(\Phi_n, z^{n-k})= F(k\alpha)+$
the same combination of $F((k-n')\alpha)$ and $F((k-n'')\alpha)$.

more on ``second best approx.''

About this second best rational approximation
\framebox{\cite{Smith}}
which most of us would ignore completely, if it were not
available in B.~Casselman's home page
\href{http://www.math.ubc.ca/people/faculty/cass/smith/smith.html}
{\framebox{\texttt{http://www.math.ubc.ca/people/faculty/cass/smith/smith.html}}}

$F((k-n')\alpha) = (-1)^{m'} F(k\alpha -(n'\alpha-m'))$,
$F((k-n'')\alpha) = (-1)^{m''} F(k\alpha -(n''\alpha-m''))$.

As long as $k\alpha$, $k\alpha-(n'\alpha-m')$, and
                      $k\alpha-(n''\alpha-m'')$ have the \emph{same}
integer (floor) part:
$$
  \frac{F(k\alpha)-F(k\alpha-(n'\alpha-m')) }{n'\alpha-m'}
 -\frac{F(k\alpha)-F(k\alpha-(n''\alpha-m'')) }{n''\alpha-m''}  =0,
$$
which is interpreted as $(\Phi_n, z^{n-k})=0$, where
$$
\Phi_n(z) = \frac{
   \displaystyle{ \frac{z^n-z^{n-n'}}{(-1)^{m'}(n'\alpha-m')}  }
  - \displaystyle{ \frac{z^n-z^{n-n''}}{(-1)^{m''}(n''\alpha-m'')}  }
                 }{
   \displaystyle{ \frac{(-1)^{m'}}{n'\alpha-m'}  }
  - \displaystyle{ \frac{(-1)^{m''}}{n''\alpha-m''}  }
                  }
$$


Check with the program used in  p.~\pageref{prpari}:

\begin{verbatim}
{
 Foucoeff(k)=

   -(-1)^(floor(k*sqrt(2)))*( 2*frac(k*sqrt(2))-1 )
}
...
Script V1.1 session started Mon Sep 20 17:54:35 1999

C:\calc\pari>gp
                   GP/PARI CALCULATOR Version 2.0.12 (alpha)
                 ix86 running emx (ix86 kernel) 32-bit version
                (readline enabled, extended help not available)

                           Copyright (C) 1989-1998 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

   realprecision = 28 significant digits
   seriesprecision = 16 significant terms
   format = g0.28

parisize = 4000000, primelimit = 500000

? \r reflec
\end{verbatim}

\noindent  \hspace*{3cm} $\Phi_n(z)$ \hspace*{6cm} $\|\Phi_n\|^2$

\begin{verbatim}
 1                                           1.0000000000
 z    + 0.1715728752                         0.9705627484
 z^2  + 0.2928932188*x   + 0.7071067811      0.4852813742
 z^3  + 0.5857864376*z   - 0.4142135623      0.4020202535
 z^4  + 0.5857864376*z^2 - 0.4142135623*z    0.4020202535
 z^5  + 0.2928932188*z^3 + 0.7071067811      0.2010101267
 z^6  + 0.2928932188*z^4 + 0.7071067811*z    0.2010101267
 z^7  + 0.5857864376*z^2 - 0.4142136237      0.1665222413
...
 z^12 + 0.2928932188*z^7 + 0.7071067811      0.0832611206
...
 z^17 + 0.5857864376*z^5 - 0.4142135623      0.0689757708
? quit
Good bye!

C:\calc\pari>exit

Script completed Mon Sep 20 17:54:57 1999
\end{verbatim}

\subsection{Cholesky factors}  \  \\

Considering all the monic orthogonal polynomials
$\Phi_k(z) = z^k + \xi_{k-1,k}z^{k-1} +\dotsb + \xi_{0,k}$,
$k=0,\dotsc, n$:

$$
\boldsymbol{G}_n
   \begin{bmatrix}
     \xi_{0,0}=1&  \xi_{0,1} & \cdots & \xi_{0,n} \\
                &  \xi_{1,1}=1 & \cdots &  \xi_{1,n} \\
                &            & \ddots &  \vdots   \\
                &            &        &  \xi_{n,n}=1 \end{bmatrix}
   =
   \begin{bmatrix}
   (\Phi_0, z^0)=\|\Phi_0\|^2 &              &            &            \\
   (\Phi_0,z^1)  &  (\Phi_1,z^1)=\|\Phi_1\|^2 &           &            \\
    \vdots       &  \vdots      &  \ddots    &            \\
   (\Phi_0,z^n)  &  (\Phi_1,z^n) & \cdots    &  (\Phi_n,z^n)=\|\Phi_n\|^2
  \end{bmatrix}
$$
i.e., the \emph{Cholesky} factorization of $\boldsymbol{G}_n$ is
$\boldsymbol{L}\, \overline{\boldsymbol{L}}^T$, with
$\boldsymbol{L}_{k,\ell} = (\varphi_\ell,z^k)$, where
$$
   \varphi_\ell = \frac{\Phi_\ell}{\|\Phi_\ell\|}
$$
is the $\ell^{\text{th}}$ degree \emph{orthonormal} polynomial.

Check (Riesz-Fisher- generalized Parseval):
$$
 (z^\ell,z^k) = \sum_{m=0}^{\min(k,\ell)} (z^\ell,\varphi_m)
                                          (\varphi_m,z^k).
$$

And, as
$\overline{\boldsymbol{L}}^T_{k,\ell}= (z^\ell,\varphi_k)$ is the coefficient of $z^\ell$
 in the
$\varphi-$basis: $[z^0,\dotsc, z^n] =[\varphi_0,\dotsc, \varphi_n]
\overline{\boldsymbol{L}}^T $,
we also confirm that $(\overline{\boldsymbol{L}}^{-1})^T$ contains the coefficients
of the $\varphi$'s, and that
  $\xi_{k,l} =(\overline{\boldsymbol{L}}^{-1})^T_{k,\ell}/\|\Phi_\ell\|$:

$$
\boldsymbol{G}_n =
\boldsymbol{L}\, \overline{\boldsymbol{L}}^T =
   \begin{bmatrix}
   (\varphi_0, z^0)  &              &            &            \\
   (\varphi_0,z^1)  &  (\varphi_1,z^1) &           &            \\
    \vdots       &  \vdots      &  \ddots    &            \\
   (\varphi_0,z^n)  &  (\varphi_1,z^n) & \cdots    &  (\varphi_n,z^n)
  \end{bmatrix}
   \begin{bmatrix}
   (z^0,\varphi_0)  & (z^1,\varphi_0) & \cdots     & (z^n,\varphi_0) \\
                    & (z^1,\varphi_1) & \cdots     & (z^n,\varphi_1) \\
                    &                 & \ddots     &    \vdots \\
                    &                 &            & (z^n,\varphi_n)
  \end{bmatrix}
    \;,$$

 $$
(\boldsymbol{G}_n)^{-1} =
( \overline{\boldsymbol{L}}^{-1})^T\, \boldsymbol{L}^{-1} =
   \begin{bmatrix}
     \xi_{0,0}&  \xi_{0,1} & \cdots & \xi_{0,n} \\
                &  \xi_{1,1} & \cdots &  \xi_{1,n} \\
                &            & \ddots &  \vdots   \\
                &            &        &  \xi_{n,n} \end{bmatrix}
   \begin{bmatrix} \|\Phi_0\|^{-2} &     &    &   \\
                   &  \|\Phi_1\|^{-2} &  &  \\  &  &  \ddots &   \\
                & & &    \|\Phi_n\|^{-2} \end{bmatrix}
 \begin{bmatrix}    \overline{\xi_{0,0}}&            &        &           \\
   \overline{\xi_{0,1}}& \overline{\xi_{1,1}}&  &        \\
        \vdots           &   \vdots   & \ddots   &      \\
\overline{\xi_{0,n}}& \overline{\xi_{1,n}}& \cdots & \overline{\xi_{n,n}}
  \end{bmatrix}  .
$$
In particular, the \emph{last} column of
$(\boldsymbol{G}_n)^{-1}$ is
\begin{equation} \label{lastco}
 \frac{1}{\|\Phi_n\|^2} \begin{bmatrix}
     \xi_{0,n}&  \xi_{1,n} & \cdots & \xi_{n,n} \end{bmatrix}^T .
\end{equation}

\medskip

\noindent\textbf{Example: Jacobi $(\alpha,0)$ and Gegenbauer
 polynomials on the unit circle.}
There are not so many explicit scalar products allowing a complete
description of the Cholesky factors. Most known cases are related to
combinatorial identities (Knuth? Wilf?). Here is such a case (Delsarte
\& Genin?):

The scalar product is $\displaystyle{ (f,g) = \frac{1}{2\pi}
 \int_{\mathbb{T}} f(e^{i\theta}) \overline{ g(e^{i\theta})}
  |\sin \theta/2|^{2\alpha}\,d\theta }$ on the unit circle
 $\mathbb{T}$, $\alpha> -1/2$. The elements of $\boldsymbol{G}_n$ are
$c_{k-\ell}$, with $c_k = \displaystyle{ (-1)^k
                \frac{ \Gamma(\alpha+1)\Gamma(\alpha+1/2)}
          {\sqrt{\pi} \Gamma(\alpha+k+1)\Gamma(\alpha-k+1)}
           =   \frac{ (-1)^k  \Gamma(2\alpha+1)}
          {4^\alpha  \Gamma(\alpha+k+1)\Gamma(\alpha-k+1)}
           }$
(see later on, \S~\ref{sectToep} for unit circle orthogonality)

One finds $\|\Phi_n\|^2 =
                          \displaystyle{
                \frac{ n! \Gamma(2\alpha+n+1)}
          {4^\alpha[ \Gamma(\alpha+n+1)]^2}    }$,

          $\boldsymbol{L}_{k,\ell} = \|\Phi_\ell\| (-1)^{k-\ell}
          \displaystyle{  {k  \choose \ell}
                \frac{  \Gamma(\alpha+1) \Gamma(\alpha+\ell+1)}
          { \Gamma(\alpha-k+\ell+1) \Gamma(\alpha+k+1)}    }$,
     $\boldsymbol{L}^{-1}_{k,\ell} = \|\Phi_k\|^{-1}
              \displaystyle{  {k  \choose \ell}
                \frac{  \Gamma(\alpha+k-\ell) \Gamma(\alpha+\ell+1)}
          { \Gamma(\alpha+k+1) \Gamma(\alpha)}    }$,

 \begin{equation} \label{jacob0}
  \Phi_n(z) = \sum_{\ell=0}^n {n \choose \ell}
    \frac{\Gamma(\alpha+n-\ell)\Gamma(\alpha+\ell+1)}
         {\Gamma(\alpha+n+1)\Gamma(\alpha)}\,z^\ell.
 \end{equation}

 Gegenbauer: sieved polynomials $\dotsc, \Phi_n(z^2), z\Phi_n(z^2),\dotsc $

\medskip

{\Small \noindent Indeed, first, the product of the two latter
matrices is, for $k\geqslant \ell$, \\
$ \displaystyle{  \sum_{m=\ell}^k \|\Phi_m\| (-1)^{k-m}
    {k\choose m} \frac{ \Gamma(\alpha+1) \Gamma(\alpha+m+1)}
                  { \Gamma(\alpha-k+m+1) \Gamma(\alpha+k+1)}
                                  \|\Phi_m\|^{-1}
    {m\choose \ell} \frac{ \Gamma(\alpha+m-\ell) \Gamma(\alpha+\ell+1)}
                  { \Gamma(\alpha+m+1) \Gamma(\alpha)}
              }$ \\
$ \displaystyle{\ \ = \alpha \frac{\Gamma(\alpha+\ell+1)}
                                  {\Gamma(\alpha+k+1)}
  \sum_{m=\ell}^k  (-1)^{k-m}
    {k\choose m}  {m \choose \ell}
    \frac{ \Gamma(\alpha+m-\ell)}{ \Gamma(\alpha-k+m+1)}
              }$ \\
$ \displaystyle{\ \ = \alpha \frac{\Gamma(\alpha+\ell+1)}
                                  {\Gamma(\alpha+k+1)}   {k\choose\ell}
  \sum_{m=\ell}^k  (-1)^{k-m}
    {k-\ell\choose m-\ell}
    \frac{ \Gamma(\alpha+m-\ell)}{ \Gamma(\alpha-k+m+1)}
              }$ \\
$ \displaystyle{\ \ = \alpha \frac{\Gamma(\alpha+\ell+1)}
                                  {\Gamma(\alpha+k+1)}   {k\choose\ell}
  \left[\Delta^{k-\ell}
        (\alpha+m-\ell-1)(\alpha+m-\ell-2)\dotsb (\alpha-k+m+1)
  \right]_{m=\ell}
              }$ \\
$\ \ = \delta_{k,\ell}$, as
$    (\alpha+m-\ell-1)(\alpha+m-\ell-2)\dotsb (\alpha-k+m+1) $ is
a polynomial of degree $k-\ell-1$ if $k>\ell$.

\medskip

Next, the $(k,\ell)^{\text{th}}$ element of $\boldsymbol{L}
\overline{\boldsymbol{L}}^T$ is \\ $ \displaystyle{
\sum_{m=0}^{\min(k,\ell)} \|\Phi_m\| (-1)^{k-m}
    {k\choose m} \frac{ \Gamma(\alpha+1) \Gamma(\alpha+m+1)}
                  { \Gamma(\alpha-k+m+1) \Gamma(\alpha+k+1)}
                                  \|\Phi_m\|  (-1)^{\ell-m}
    {\ell\choose m} \frac{ \Gamma(\alpha+1) \Gamma(\alpha+m+1)}
                  { \Gamma(\alpha-\ell+m+1) \Gamma(\alpha+\ell+1)}
              }$ \\
$ \displaystyle{\ \ = \frac{ (-1)^{k-\ell} \Gamma(2\alpha+1)}
    {4^\alpha\Gamma(\alpha+k-\ell+1)\Gamma(\alpha-k+\ell+1)}
  \frac{1}{(\alpha+k)\dotsb (\alpha+1)
           (\alpha+\ell)\dotsb (\alpha+\ell-k+1) }  }$ \\  $\displaystyle{\  \  \  %
  \sum_{m=0}^k
    \frac{(2\alpha+m)\dotsb (2\alpha+1)}{m!}
   k\dotsb (k-m+1) \ell \dotsb (\ell-m+1)
    \alpha\dotsb (\alpha-k+m+1)
    (\alpha+k-\ell)\dotsb (\alpha-\ell+m+1)
              }$, if $k\leqslant \ell$. \\
The sum must be  \label{gegen}
  $(\alpha+k)\dotsb (\alpha+1)
           (\alpha+\ell)\dotsb (\alpha+\ell-k+1)$. It is indeed a
polynomial $\alpha^{2k}+\dotsb$ of degree $2k$ in $\alpha$. The
polynomial vanishes at $\alpha=-1,\dotsc, -k$, as it contains  then
the $(-2\alpha-1)^{\text{th}}$ difference of $(k-m-\alpha-1)\dotsc (k-m+1)
(\ell-m-\alpha-1)\dotsc (\ell-m+1)$ which has degree $-2\alpha-2$ in $m$, etc.
} %end small

\bigskip

It is current practice to accept that the scalar product may fail to be
positive definite on polynomials of exact degree $N$, i.e., that
some nonzero polynomials of degree $N$, but not $N-1$, may have zero
norm: $\|\Phi_{N-1}\|>0$, $\|\Phi_N\|=0$, as $\Phi_N$ achieves least
possible norm. If this happens, $\Phi_N$, which is still orthogonal
to any polynomial of degree $<N$, is the only monic polynomial
of degree $N$ with zero norm: any other monic polynomial of degree
$N$ is $\Phi_N +\Psi_{N-1}$, and $\|\Phi_N +\Psi_{N-1}\|^2 =
\|\Phi_N\|^2 + 2 \text{Re}\;(\Phi_N, \Psi_{N-1}) +\|\Psi_{N-1}\|^2 =
\|\Psi_{N-1}\|^2 >0$. See for instance \cite{Zhed2}.

What is stated above is still true, the positive semidefinite
$\boldsymbol{G}_N$ still has a Cholesky factorization.

\medskip

\subsection{Decreasing sequences}  In the semidefinite case just seen
the sequence of norms $\{\|\Phi_n\|\}$ stops at $\|\Phi_N\|=0$.
Can we measure such a situation by producing a positive decreasing
sequence up to the index $N-1$? The sequence of norms $\{\|\Phi_n\|\}$
is generally \emph{not} decreasing (it is in the unit circle case\dots),
but an interesting decreasing sequence is certainly the sequence of
smallest eigenvalues of $\boldsymbol{G}_n$, from the minimizing
property:
$$
  \lambda_{\min}(\boldsymbol{G}_n) = \min_{\|\boldsymbol{\eta}\|=1}
    \overline{\boldsymbol{\eta}}^T \boldsymbol{G}_n
              \boldsymbol{\eta},
$$
where $\|\boldsymbol{\eta}\|$ is the Euclidian norm on $\mathbb{C}^n$.

\smallskip

Any item constructed as a minimum on larger and larger subspaces will
do, such as norms of \emph{kernel polynomials}:
$$
   \min \|p\|, \  \text{on}\; p\in \mathscr{P}_n, \ \text{with}\  %
    p(z_0)=1.
$$
Solution \cite{Dav}:
$$
  \frac{K_n(z,z_0)}{K_n(z_0,z_0)} = \frac
  {\displaystyle{ \sum_{k=0}^n \overline{\varphi_k(z_0)}\, \varphi_k(z) }}
  {\displaystyle{ \sum_{k=0}^n |\varphi_k(z_0)|^2  }} \;,
$$
and the norm is
$\displaystyle{ \frac{1}{\sqrt{\displaystyle{\sum_{k=0}^n
                        |\varphi_k(z_0)|^2 }}}   }$, ($n<N$).

\medskip

A very interesting feature of the coefficients of the kernel
polynomial $K_n(z) := K_n(z,0)$ is that they build the \emph{first}
column of $\boldsymbol{G}_n^{-1}$:
\begin{equation} \label{kern0}
  \boldsymbol{G}_n
 \begin{bmatrix} K_n(0) \\  K'_n(0) \\  \vdots \\  K_{n,n} \end{bmatrix}
=
 \begin{bmatrix} (K_n,z^0) \\  (K_n,z) \\  \vdots \\  (Kn,z^n) \end{bmatrix}
=
 \begin{bmatrix} 1 \\  0 \\  \vdots \\  0 \end{bmatrix}
\end{equation}
from the reproducing property $(K_n,f)=\overline{f(0)}$ for all $f$ of
degree $\leqslant n$.

\medskip

A hint of Szeg\H o's theory: the sequence $\{K_n(0)\}$ is of course
increasing. Should it be bounded, the \emph{functions} $K_n$ make a
Cauchy sequence: if $m\leqslant n$,
 $\|K_n-K_m\|^2 = \|K_n\|^2 -2\;\text{Re}\;(K_n,K_m) +\|K_m\|^2 =
 K_n(0)-K_m(0)$. What can $\displaystyle{ \lim_{n\to\infty} K_n }$ be?

\subsection{Most general scalar product} on polynomials is

$$
(\sum a_k z^k, \sum b_\ell z^\ell) =
 \begin{bmatrix} \overline{b_0} & \overline{b_1} & \cdots & \overline{b_L} \end{bmatrix}
  \boldsymbol{G}^T_{(K,L)}
  \begin{bmatrix} a_0 \\ a_1 \\ \vdots \\ a_K \end{bmatrix}
$$
with any positive definite Hermitian tableau $\boldsymbol{G}$, whether it can be
interpreted as a functional integral or not.

If one does not go further than the degree $N$, one may always write
$(f,g)$ through functional values:
$(f,g) = \displaystyle{ \sum_{k,\ell=0}^N h_{k,\ell} f(a_\ell)
  \overline{g(a_k)}  }$, where $h_{k,\ell} =
    (\ell_\ell, \ell_k) $ (Lagrange interpolation).

A kind of ``true support'' could be deduced as the set of $z_0$ such that the sequence
$\{K_n(z_0, z_0)\}$ is bounded or mildly increasing, for what the meaning of this may be.

\subsection{$\displaystyle{\int_S f(t)\overline{g(t)}\,d\mu(t) }$
   scalar product.} \  \\

Here, $d\mu$ is a positive measure supported on a closed complex set
$S$ containing at least $N$ points.

The two arguments scalar product form is now a simple linear
form with the single argument $f\overline{g}$, written (Maroni, Marcellan):
\begin{equation} \label{formu}
   (f,g) = u(f\overline{g}).
\end{equation}

\noindent\textbf{Theorem. (Fej\'er) } \cite{Dav} \textsl{All the zeros of $\Phi_n$ are in
the convex closure of the support $S$}.

Indeed,
let $z_k$ be a zero of $\Phi_n$, and $\Psi(z) = \displaystyle{
  \frac{\Phi_n(z)}{z-z_k} }$. As $\Psi$ has degree $n-1$,
$$
0=\int_S \Phi_n(t)\overline{\Psi(t)}\,d\mu(t) =
  \int_S (t-z_k)|\Psi(t)|^2\,d\mu(t) \Rightarrow
 z_k= \frac{ \int_S t|\Psi(t)|^2\,d\mu(t) }
          { \int_S |\Psi(t)|^2\,d\mu(t) }
$$
is therefore in the convex closure of $S$, as center of mass of the
non-negative distribution $|\Psi|^2\,d\mu$ on $S$. \qed

See other proof in \cite{AM}

A very interesting example of orthogonal polynomials on the unit circle are
the \emph{\textbf{Rogers-Szeg\H o}} polynomials \cite{LubSaff}
\label{RSzeg}
$$
  \Phi_n(z) := \sum_{j=0}^n \frac{(1-q^n)(1-q^{n-1})\dotsb (1-q^{n-j+1}) }
                              {(1-q)(1-q^2)\dotsb (1-q^j) } \,z^j.
$$
Zeros have strange properties if $|q|=1$ and the argument of $q$ is a
\emph{rational} multiple of $\pi$ \cite{LubSaff}\dots

\section{$\displaystyle{\int_\mathbb{T} f(t)\overline{g(t)}\,
   \frac{d\mu(t)}{2\pi} }$ scalar product on the unit circle $\mathbb{T}$.}
\renewcommand{\mylmark}{Unit circle orthogonal polynomials.}
\pagestyle{fancy}

\subsection{Toeplitz matrix and Fourier series} \   \label{sectToep}\\

Now, the Gram matrix is the \emph{\textbf{Toeplitz matrix}}
$$
  \boldsymbol{G}_n =
\begin{bmatrix}
  c_0 & c_{-1}  &  \cdots & c_{-n} \\
  c_1 &  c_0    &  \cdots & c_{-n+1} \\
  \vdots & \vdots &     &  \vdots \\
  c_n & c_{n-1} & \cdots & c_0
\end{bmatrix}\;,
$$
with
\begin{equation} \label{ck}
 c_k = \frac{1}{2\pi}
                        \int_\mathbb{T} t^{-k}\,d\mu(t)   ,
\end{equation}
$k\in\mathbb{Z}$ (or $k=-N, -N+1,\dotsb, N-1, N$).

We shall only consider measures $d\mu$ without singularly continuous part, then
$d\mu(t) = w(\theta)d\theta $, where $w$ is $2\pi-$ periodic positive integrable, together
with a countable set of positive Dirac masses. Then, $c_k$ is the \emph{Fourier} coefficient
of $w$:
$$
   w(\theta) = \sum_{-\infty}^\infty c_k e^{ik\theta} .
$$

Simplest Fourier series is the square wave
%\setlength{\unitlength}{\textwidth}
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 \put(2.5,0.0){\vector(0,1){1.02}} \put(2.5,0.625){1}
 \put(3.0,-0.3){$\pi/2$}   \put(4.4,-0.3){$3\pi/2$}
 \put(1.6,-0.3){$-\pi/2$}   \put(-0.2,-0.3){$-3\pi/2$}
 \thicklines
\put(0,0.5){\line(1,0){0.3}}
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\put(4.7,0.0){\line(0,1){0.5}}
\put(4.7,0.5){\line(1,0){0.3}}
\end{picture}

\medskip    \medskip

The Fourier coefficients are $c_0= 1/2$, $c_k= (-1)^{(k-1)/2}/(\pi k)$
for odd $k$, $c_k=0$ for even $k\neq 0$.

Another well known expansion is $w(\theta) =|\theta|$ on
$[-\pi,\pi]$: \setlength{\unitlength}{1cm}
\begin{picture}(4,1.8)
 \put(0.0,0.0){\vector(1,0){4.025}} \put(4.20,0.165){$\theta$}
 \put(2.0,0.0){\vector(0,1){1.52}}%\put(2.5,0.625){1}
 \put(0.4,-0.3){-$\pi$}   \put(3.4,-0.3){$\pi$}
 \thicklines
\put(0,1.0){\line(1,1){0.5}}
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\put(2.0,0.0){\line(1,1){1.5}}
\put(3.5,1.5){\line(1,-1){0.5}}
\end{picture}
with $c_0=\pi/2$,

\smallskip \noindent
 $c_k= -2/(\pi k^2)$ for odd $k$, $c_k=0$ for
even $k\neq 0$. Will these simple Fourier series yield easy
orthogonal polynomials?

\subsection{Recurrence relation} \  \\

An Hermitian Toeplitz matrix turns into its conjugate when elements
are moved left-to-right and top-to-bottom

$ \overline{\boldsymbol{G}_n} = \boldsymbol{SG}_n \boldsymbol{S}\ \ ,
\boldsymbol{S} = \begin{bmatrix}
             &           &       &      1  \\
             &           &  1    &         \\
             & \secddots &       &         \\
       1     &           &       &      \end{bmatrix}  $.

So,
$ \boldsymbol{G}^{-1}_n =
\boldsymbol{S}\;\overline{\boldsymbol{G}^{-1}_n}\, \boldsymbol{S}$.

Remember from \eqref{kern0} that the first column is made of the
coefficients of $K_n$. It is therefore the same than the
conjugate upside-down \emph{last} column:
$\displaystyle{
  \begin{bmatrix} K_n(0) \\ K'_n(0) \\ \vdots  \\  K_{n,n}\end{bmatrix}=
 \frac{1}{\|\Phi_n\|^2}
  \begin{bmatrix} 1 \\ \overline{\xi_{n-1,n}} \\ \vdots  \\
    \overline{\xi_{0,n}} = \overline{\Phi_n(0)} \end{bmatrix} }$
(from \eqref{lastco}), i.e.,
\begin{equation} \label{phistar}
  K_n(z) := \sum_{k=0}^n \frac{ \overline{\Phi_k(0)}\; \Phi_k(z) }
                              {\|\Phi_k\|^2}  =
 \frac{ z^n \overline{ \Phi_n(1/\overline{z})} }{\|\Phi_n\|^2} :=
 \frac{  \Phi_n^*(z) }{\|\Phi_n\|^2}.
\end{equation}

\medskip

As $\Phi_n$ leads to the knowledge of $K_n =$ constant $\Phi^*_n$, we
use the obvious recurrence for the $K$'s:
$$
  \frac{\Phi^*_n}{\|\Phi_n\|^2} -
  \frac{\Phi^*_{n-1}}{\|\Phi_{n-1}\|^2} =
  \frac{\overline{\Phi_n(0)}\, \Phi_n}{\|\Phi_n\|^2},
$$
which is the  ${}^*$ operator applied to
\begin{equation} \label{phinm1}
  \Phi_n(z) - z \frac{\|\Phi_n\|^2}{\|\Phi_{n-1}\|^2}\, \Phi_{n-1}(z) =
  \Phi_n(0) \Phi_n^*(z),
\end{equation}
and combine the two latter equations:


\begin{equation} \label{recur}
\Phi_n(z) = z \Phi_{n-1}(z) + \Phi_n(0) \Phi^*_{n-1}(z),
\end{equation}
together with an interesting identity
\begin{equation}   \label{rnormphi}
  \frac{\|\Phi_n\|^2}{\|\Phi_{n-1}\|^2} = 1-|\Phi_n(0)|^2.
\end{equation}

Of course, $\Phi_n(0)$ is not yet known when we try to deduce
$\Phi_n$ from $\Phi_{n-1}$!

In matrix-vector form, \eqref{recur} tells that
$\displaystyle{
 \begin{bmatrix} \Phi_n(0) \\  \Phi'_n(0) \\    \vdots  \\  1 \end{bmatrix}
  =
 \begin{bmatrix} 0  \\ \Phi_{n-1}(0) \\     \vdots  \\  1 \end{bmatrix}
 + \Phi_n(0)
 \begin{bmatrix} 1  \\ \overline{\xi_{n-2,n-1}} \\     \vdots  \\ 0\end{bmatrix}
 }$,
which is not surprising, as a left-multiplication by $\boldsymbol{G}_n$
shows
$\displaystyle{
 \begin{bmatrix} 0 \\ 0 \\    \vdots  \\ \|\Phi_n\|^2  \end{bmatrix}
  =
 \begin{bmatrix} \eta_{n-1} \\ 0 \\     \vdots  \\ \|\Phi_{n-1}\|^2 \end{bmatrix}
 + \Phi_n(0)
 \begin{bmatrix} \|\Phi_{n-1}\|^2  \\ 0 \\ \vdots  \\ \overline{\eta_{n-1}}\end{bmatrix}
 }$,
where $\eta_{n-1}$ can be computed from $\Phi_{n-1}$:    \label{etan}

$\displaystyle{
  \eta_{n-1} = c_{-1} \Phi_{n-1}(0) + c_{-2} \xi_{1,n-1} +\dotsb +
      c_{-n+1} \xi_{n-2,n-1} +c_{-n} = (z\Phi_{n-1}(z), z^0)  }$.



Let us look at more rows of the matrix-vector relation:

$$
  \hspace*{-25pt}\begin{bmatrix}
   c_{-2} & c_{-3} &  \cdots & c_{-n-2} & c_{-n-3}  \\
   c_{-1} & c_{-2} &  \cdots & c_{-n-1} & c_{-n-2}  \\
   c_{0} & c_{-1} &  \cdots & c_{-n} & c_{-n-1}  \\
  \vdots & \vdots &         & \vdots & \vdots   \\
   c_{n} & c_{n-1} &  \cdots & c_{0} & c_{-1}  \\
   c_{n+1} & c_{n} &  \cdots & c_{1} & c_{0}  \\
   c_{n+2} & c_{n+1} &  \cdots & c_{2} & c_{1}  \\
  \end{bmatrix}
  \begin{bmatrix}
    \xi_{0,n+1} &  0  & \overline{\xi_{n,n}} \\
    \xi_{1,n+1} &  \xi_{0,n}  & \overline{\xi_{n-1,n}} \\
  \vdots & \vdots &    \vdots    \\
  \vdots & \vdots &    \vdots    \\
    \xi_{n,n+1} &  \xi_{n-1,n}  & \overline{\xi_{0,n}} \\
    \xi_{n+1,n+1} &  \xi_{n,n}  &  0 \\
  \end{bmatrix}
=
  \begin{bmatrix}
    u(z^2\Phi_{n+1}(z)) & u(z^3\Phi_n(z)) & \overline{u(z^{-n-2}\Phi_n(z))} \\
    u(z\Phi_{n+1}(z)) & u(z^2\Phi_n(z)) & \overline{u(z^{-n-1}\Phi_n(z))} \\
      0         & u(z\Phi_n(z)) & \|\Phi_n\|^2 \\
      0         &    0          &   0          \\
  \vdots & \vdots &    \vdots    \\
  \|\Phi_{n+1}\|^2 & \|\Phi_n\|^2 & \overline{u(z\Phi_n(z))} \\
  u(z^{-n-2}\Phi_{n+1}(z)) & u(z^{-n-1}\Phi_n(z)) & \overline{u(z^2\Phi_n(z))}
  \end{bmatrix}
$$
and, as $1^{\text{st}}$ column $= 2^{\text{nd}}$ column
 $+ \Phi_{n+1}(0)$ times $3^{\text{rd}}$ column,  \\
$u(z^2\Phi_n(z)) = u(z\Phi_{n+1}(z)) -\Phi_{n+1}(0)
                 \overline{u(z^{-n-1}\Phi_n(z))}$ \\
$ u(z^{-n-2}\Phi_{n+1}(z)) = u(z^{-n-1}\Phi_n(z)) +\Phi_{n+1}(0)
[ \overline{u(z\Phi_{n+1}(z))} -\overline{\Phi_{n+1}(0)}
                 u(z^{-n-1}\Phi_n(z))]$, \\
finally:
\begin{align}     \label{uuu}
  u(z\Phi_n(z))=\eta_n  &= -\Phi_{n+1}(0) \|\Phi_n\|^2, \\
 u(z^{-n-1}\Phi_n(z)) &= -(\Phi_0(0)\overline{\Phi_1(0)} +
                           \Phi_1(0)\overline{\Phi_2(0)} +  \dotsb +
                      \Phi_n(0)\overline{\Phi_{n+1}(0)} )\|\Phi_n\|^2,
 % \\  u(z^2\Phi_n(z)) &=
\end{align}

\medskip

Other forms of \eqref{recur}:

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

Also,
\begin{equation} \label{recur3}
  \begin{bmatrix} \Phi_{n+r}(z)  \\  \Phi^*_{n+r}(z) \end{bmatrix}
 =
  \begin{bmatrix}
            z U_{r-1}(z;n) &   V_{r-1}(z;n)    \\
              z V^*_{r-1}(z;n) & U^*_{r-1}(z;n)
             \end{bmatrix}
  \begin{bmatrix} \Phi_n(z)  \\  \Phi^*_n(z) \end{bmatrix}
\end{equation}
where $U_{r-1}$ and $V_{r-1}$ are polynomials of degrees
$\leqslant r-1$:

\begin{center} \begin{tabular}{c|c|c}
           $r$ &      $U_r(z;n)$  &  $V_r(z;n)$         \\
\hline     $0$ &       $1$        &    $x_{n+1}$         \\
     $1$ &     $z+x_{n+2}\overline{x_{n+1}}$    &  $x_{n+1}z+x_{n+2}$   \\
     $2$ &  $z^2+(x_{n+2}\overline{x_{n+1}}+x_{n+3}\overline{x_{n+2}})z
                +x_{n+3}\overline{x_{n+1}}$
          &
         $x_{n+1}z^2 +
                 (x_{n+3}\overline{x_{n+2}}x_{n+1}+x_{n+2})z
           +x_{n+3}$
\end{tabular}  \end{center}

where $x_n=\Phi_n(0)$, etc.

One has $U_r(z;n)=zU_{r-1}(z;n)+x_{n+r-1}V^*_{r-1}(z;n),
         V_r(z;n)=zV_{r-1}(z;n)+x_{n+r-1}U^*_{r-1}(z;n)$.


\medskip


 The Cholesky factorization
becomes
\begin{equation} \label{cholt}
 \boldsymbol{G}_n = \boldsymbol{L} \overline{\boldsymbol{L}}^T
  \  \Rightarrow \
 \boldsymbol{G}_n^{-1} =
  (\boldsymbol{S} \overline{\boldsymbol{L}}^T \boldsymbol{S})^{-1}
  (\boldsymbol{SLS})^{-1},
\end{equation}
which is the Cholesky factorization of $\boldsymbol{G}_n^{-1}$!

\bigskip

\subsection{Multiplication operator: Toeplitz matrices in
 Grenander and Szeg\H o's sense} \  \\



A deep remark (Bultheel, Marcellan): the operator $\mathscr{M}$
of multiplication by
$z$ is represented in any polynomial basis by a Hessenberg matrix
$\displaystyle{
  \begin{bmatrix} x & x & x& x& x& x& x& x& \cdots \\
                  x & x & x& x& x& x& x& x& \cdots \\
                    & x & x&x & x& x& x& x& \cdots \\
                    &   & x&x &x & x& x& x& \cdots \\
                    &   &  &x &x &x & x& x& \cdots  \\
                    &   &  &  &\ddots  &\ddots  & \ddots &\ddots &\ddots  \end{bmatrix}  }$.
On the real interval, the operator is \emph{symmetric}, hence the
matrix representation is tridiagonal $\Rightarrow$ the three terms
recurrence relation; on the circle, the operator is \emph{unitary}.

On curves $|P(z)|=$ constant (\emph{lemniscates}), or Re$\,P(z)=$
constant, one has to consider the polynomial $P(\mathscr{M})$
(Vigil, Marcellan).

\medskip

Let $\{\varphi_0, \varphi_1,\dotsc\}$ be a complete orthonormal
sequence in a Hilbert function space $X$, then

\begin{equation} \label{GSToep}
 \mathscr{M}(f) = \begin{bmatrix}
    \langle \varphi_0 f,\varphi_0\rangle  &
    \langle \varphi_1 f,\varphi_0\rangle  &
    \langle \varphi_2 f,\varphi_0\rangle  &
    \dots \\
    \langle \varphi_0 f,\varphi_1\rangle  &
    \langle \varphi_1 f,\varphi_1\rangle  &
    \langle \varphi_2 f,\varphi_1\rangle  &
    \dots \\
    \dots & \dots & \dots & \dots  \end{bmatrix}
\end{equation}

represents the operator of multiplication by $f$ if an element $F$
of $X$ is reprensented by its sequence of coefficients
$\{c_k(F)\}$ in the expansion $F=\sum_0^\infty c_K(F)\varphi_k$.
Indeed\cite{GrS,MagT},
\begin{equation*}
\mathscr{f} \begin{bmatrix} c_0(F) \\  c_1(F) \\ \vdots \end{bmatrix} =
     \begin{bmatrix} c_0(fF) \\  c_1(fF) \\ \vdots \end{bmatrix} .
\end{equation*}

One also have $\mathscr{M}(f)
               \mathscr{M}(g)=
               \mathscr{M}(fg)$.

\medskip

The $\mathscr{M}$ of above is simply $\mathscr{M}(z)$. On the circle,
we have $\displaystyle{ \mathscr{M}_{m,n} =
 \int_\mathbb{T} z\varphi_m(z)\,\overline{\varphi_n(z)}\;d\mu }$, so
that $\mathscr{M}(z^{-1})$ is the transpose of the complex cojugate
of $\mathscr{M}$, and is also $\mathscr{M}^{-1}$: $\mathscr{M}$ is
a unitary matrix.

The $n^{\text{th}}$ column (starting at 0) of $\mathscr{M}$ is
made with the coefficients of $z\varphi_n(z)$. From the
recurrence relation \eqref{recur},
$$ z\varphi_n(z) =\dfrac{z\Phi_n(z)}{\|\Phi_n\|} =
   =\dfrac{\Phi_{n+1}(z)-\Phi_{n+1}(0)\Phi_n^*(z)}{\|\Phi_n\|}$$
$$= \dfrac{\|\Phi_{n+1}\|}{\|\Phi_n\|}\varphi_{n+1}(z)
  -\Phi_{n+1}(0)\|\Phi_n\|\sum_{k=0}^n \overline{\varphi_k(0)}
     \varphi_k(z) $$
from \eqref{phistar}. Whence
$$
\mathscr{M}_{n+1,n}=
 \dfrac{\|\Phi_{n+1}\|}{\|\Phi_n\|}\;,
\mathscr{M}_{k,n}=
 -\dfrac{\|\Phi_{n+1}\| \Phi_{n+1}(0) \overline{\Phi_k(0)} }
 {\|\Phi_k\|}\;, k\leqslant n.
$$
found in B.Simon's preprint \cite{Simon}, with reference to
Nevai \cite{} for some identities involving the $\Phi_n^*$s.


\subsection{First and last columns of
       \label{firstlast} $\boldsymbol{G}_n^{-1}$} \  \\

\subsubsection{First column}
We already saw that the
first column of $\boldsymbol{G}_n^{-1}$ is made of
the coefficients of the kernel polynomial $K_n$, which is here
$\Phi_n^*/\|\Phi_n\|^2$, so is

$$ [ \overline{\xi_{n,n}} ,
     \overline{\xi_{n-1,n}} , \dotsc,
     \overline{\xi_{1,n}} ,
     \overline{\xi_{0,n}} ]^T/\|\Phi_n\|^2,
$$

\subsubsection{Second column} Let us try $z\Phi_{n-1}^*(z)$, i.e.,
$$
  \hspace*{-5pt}\begin{bmatrix}
   c_{0} & c_{-1} &  \cdots & c_{-n}   \\
   c_{1} & c_{0} &  \cdots & c_{-n+1}   \\
  \vdots & \vdots &         & \vdots    \\
   c_{n} & c_{n-1} &  \cdots & c_{0}   \\
  \end{bmatrix}
  \begin{bmatrix}
      0   \\
     \overline{\xi_{n-1,n-1}} \\
     \overline{\xi_{n-2,n-1}} \\
      \vdots     \\
    \overline{\xi_{1,n}} \\
    \overline{\xi_{0,n}}
  \end{bmatrix}.
$$
Thanks to the Toeplitz structure, the $n$ last elements of the product are
$\|\Phi_{n-1}\|^2, 0,\dotsc, 0$, and the first element of the product
is  $u(z\Phi^*_{n-1}(z)) = \overline{u(z^{-n}\Phi_{n-1}(z))}$, known from
\eqref{uuu}. So, second column is
$$ \{
 [ 0, \overline{\xi_{n-1,n-1}} ,
     \overline{\xi_{n-2,n-1}} , \dotsc,
     \overline{\xi_{1,n-1}} ,
     \overline{\xi_{0,n-1}} ]^T -
                           \overline{u(z^{-n}\Phi_{n-1}(z))}
 [ \overline{\xi_{n,n}} ,
     \overline{\xi_{n-1,n}} , \dotsc,
     \overline{\xi_{1,n}} ,
     \overline{\xi_{0,n}} ]^T/\|\Phi_n\|^2\}/\|\Phi_{n-1}\|^2.
$$

\subsubsection{Last column}    Known to be
$$ [\xi_{0,n} , \xi_{1,n}, \dotsc, \xi{n,n}]/\|\Phi_n\|^2.
$$

\subsubsection{First before last column}
$$ \{
 [ \xi_{0,n-1} ,
     \xi_{1,n-1} , \dotsc,
     \xi_{n-1,n-1} ,   0  ]^T -
                           u(z^{-n}\Phi_{n-1}(z))
 [ \xi_{0,n} ,
     \xi_{1,n} , \dotsc,
     \xi_{n-1,n} ,
     \xi_{n,n} ]^T/\|\Phi_n\|^2\}/\|\Phi_{n-1}\|^2.
$$

\subsection{Behaviour of reflection coefficients}  \  \\

With the program PARI
\cite{PARI},      \label{prpari}
\begin{verbatim}

{
\\  reflec.gp    :  launch gp and make  \r reflec
\\
\\  Reflection coefficients given Fourier coefficients
\\
N=22;
}
\\ vector of Fourier coeff.  c_k = c[k+N+1]
{
 Foucoeff(k)=

   if(k==0,  1.0/2 ,
             (I^k-(-I)^k)/(2*Pi*I*k)
     )
}
{
c=vector(2*N+1,k,Foucoeff(k-N-1));
Phin=x^0;normPhin2=c[N+1];
for(n=1,N-1,
    print(n-1," ",polcoeff(Phin,0),"   ",normPhin2);
    scalPhizm1 = sum(k=1,n, c[N+1-k]*polcoeff(Phin,k-1) );
  \\print(scalPhizm1);
    Phin = x*Phin -(scalPhizm1/normPhin2)* polrecip(conj(Phin));
    normPhin2  -=  scalPhizm1*conj(scalPhizm1)/normPhin2;
   )
}


Script V1.1 session started Wed Aug 18 11:19:38 1999

C:\calc\pari>gp
                   GP/PARI CALCULATOR Version 2.0.12 (alpha)
                 ix86 running emx (ix86 kernel) 32-bit version
                (readline enabled, extended help not available)

                           Copyright (C) 1989-1998 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.

   realprecision = 28 significant digits
   seriesprecision = 16 significant terms
   format = g0.28

parisize = 4000000, primelimit = 500000
? \r reflec
\end{verbatim}

One finds more and more complicated expressions,
          $\Phi_1(0)= \displaystyle{- \frac{2}{\pi}}$,
          $\Phi_2(0)= \displaystyle{ \frac{4}{\pi^2-4}}$,
          $\Phi_3(0)= \displaystyle{-\frac{2}{3\pi}\,\frac{16-\pi^2}{\pi^2-8}}$, etc.
 which we can perhaps explore further through numerical values:

\noindent $n$ \hspace*{3cm} $\Phi_n(0)$ \hspace*{6cm} $\|\Phi_n\|^2$

\begin{verbatim}
 0  1                                 0.5000000000000000000000000000
 1 -0.6366197723675813430755350535    0.2973576327153244571122410735
 2  0.6814769321178829752922525447    0.1592615339410585123538737276
 3 -0.6958211852068310273669299707    0.08215233544212813372178866790
 4  0.7010703573176610826474364438    0.04177449166168262201071826460
 5 -0.7033737451875978766777148650    0.02110720417736141209933540938
 6  0.7045635263840126108977529709    0.01062938196019166045331220303
 7 -0.7052581523381021267708256617    0.005342443643331533145252544298
 8  0.7057004468702704665764798408    0.002681836612283204103389461465
 9 -0.7060001718947297527963793848    0.001345112047680199190623848503
10  0.7062129569336616331861078971    0.0006742552493577655387959406246
...
19 -0.7068612070337355176642206532    0.000001331509045263803822704034814
20  0.7068852157013753769732944692    0.0000006661716735277939346122326538
? quit
Good bye!

C:\calc\pari>exit

Script completed Wed Aug 18 11:20:08 1999
\end{verbatim}

And the $|\theta|$ example:
\begin{verbatim}
...
{
 Foucoeff(k)=

   if(k==0,  Pi/2 ,
             ((-1)^k-(1)^k)/(Pi*k^2)
     )
}
Script V1.1 session started Mon Sep 20 17:03:52 1999

C:\calc\pari>gp
? \r reflec
\end{verbatim}

\noindent $n$ \hspace*{3cm} $\Phi_n(0)$ \hspace*{6cm} $\|\Phi_n\|^2$

\begin{verbatim}
 0 1                                1.570796326794896619231321691
 1 0.4052847345693510857755178528   1.312784051329300705755945270
 2 0.1965382464879410164306311434   1.262074771136928879634464159
 3 0.1551854679951381617834429861   1.231680818259009766792617169
 4 0.1098125062480936823071229883   1.216828242200391418572744386
 5 0.09591351266518851754113126190  1.205634150142790986549470868
 6 0.07623828544046405724091782636  1.198626671505913288215311961
 7 0.06937466183732452782488437927  1.192857868675259251133628348
 8 0.05839893203204895727424656476  1.188789704136797789141563113
 9 0.05432890225589902022644379786  1.185280837233723055616638076
10 0.04732939132232474162137501120  1.182625723668014639762979918
...
19 0.02605076565085498126742871016  1.170344822783165775227484266
20 0.02430644869083064612989477131  1.169653379026563365817484145
? quit
Good bye!

C:\calc\pari>exit

Script completed Mon Sep 20 17:04:26 1999
\end{verbatim}

The description and understanding of the behaviour of these
coefficients $\Phi_n(0)$ is the \emph{main subject} of the present
lecture. The following completely explicit cases are known:

\begin{itemize}
  \item Special relations, as in p.~\pageref{dioph}, also
     in \cite{Zhed,Zhed2},
  \item Rogers-Szeg\H o polynomials, p.~\pageref{RSzeg},
  \item Bernstein-Szeg\H o, \S~\ref{BerSze}, p.~\pageref{BerSze},
  \item Bernstein-Szeg\H o pol. on a circular arc \cite{Golakh},
  \item Gegenbauer, p.~\pageref{gegen}, and more generally:
  \item Jacobi, \S~\ref{jacob},
  \item and others from known interval cases, from \S~\ref{cirint},
    as in \cite{Zhed}.
\end{itemize}

\subsection{Circle versus interval \label{cirint}} \  \\

Among other feats, Szeg\H o achieved the description of the connection
between orthogonal polynomials on $[-1,1]$ and orthogonal polynomials
on the unit circle \cite[\S~11.5]{Sze}. The connection is usually interpreted as a reduction
of interval polynomials to the ``more basic'' circle polynomials, but
wait.

Ah, let $P_n$ be the $n^{\text{th}}$ degree monic polynomial orthogonal
with respect to a measure $d\nu$ on $[-1,1]$, i.e., a (real) scalar
product $\langle f,g\rangle = \displaystyle{\int_{-1}^1 f(x)g(x)\,d\nu(x)
}$. We express orthogonality to lower degree polynomials through
\emph{Chebyshev} polynomials of \emph{first kind}: let
$$
  P_n(x) = \frac{d_0}{2} +d_1 T_1(x) + \dotsb + d_n T_n(x),
$$
which must be $\langle , \rangle-$orthogonal to $T_0$,\dots,
$T_{n-1}$:
$$
 \begin{bmatrix} 0 \\  \vdots \\  0  \\  \langle P_n, T_n\rangle \end{bmatrix}
 =
 \begin{bmatrix} \frac{1}{2}\langle T_0, T_0\rangle & \cdots  &
                            \langle T_n, T_0\rangle \\
                 \vdots  &          &  \vdots  \\
                 \frac{1}{2}\langle T_0, T_{n-1}\rangle & \cdots  &
                            \langle T_n, T_{n-1}\rangle \\
                 \frac{1}{2}\langle T_0, T_n\rangle & \cdots  &
                            \langle T_n, T_n\rangle
   \end{bmatrix}
  \begin{bmatrix}  d_0 \\  \vdots  \\  d_{n-1} \\  d_n \end{bmatrix}.
$$
As $\langle T_\ell, T_k\rangle =
    \langle T_\ell T_k,1\rangle$, and
$\displaystyle{ T_\ell T_k = \frac{T_{\ell+k} +T_{\ell-k}}{2} }$, let
$c_k = \langle T_k,1\rangle$. Remark that $c_{-k} = c_k$, also that
        $c_k =  \displaystyle{\int_{-1}^1 T_k(x)  \,d\nu(x)
             =  \int_\pi^0 \cos k\theta  \,d\nu(\cos\theta)
             = \frac{1}{2} \int_\pi^{-\pi} e^{-ik\theta}  \,d\nu(\cos\theta)
             = \frac{1}{2\pi} \int_{-\pi}^\pi e^{-ik\theta}  \,d\mu(e^{i\theta})
         }$,
with \\ $d\mu(e^{i\theta}) = -\pi d\nu(\cos\theta)$.
 Then, the equations above may be
rearranged as
$$
 \begin{bmatrix} 2\langle P_n, T_n\rangle \\
 0 \\  \vdots \\  0  \\  \vdots \\  0 \\ 2\langle P_n, T_n\rangle
  \end{bmatrix}
 =
 \begin{bmatrix}
  c_0 & \cdots & c_{1-n} &  c_{-n}   & c_{-n-1} &\cdots  &  c_{-2n}  \\
  c_1 & \cdots & c_{2-n} &  c_{1-n}   & c_{-n} &\cdots  &  c_{-2n+1}  \\
\vdots&        &\vdots   & \vdots &  \vdots    &       &  \vdots    \\
  c_n & \cdots & c_1 &  c_0   & c_{-1} &\cdots  &  c_{-n}  \\
\vdots&        &\vdots   & \vdots &  \vdots    &       &  \vdots    \\
  c_{2n-1} & \cdots & c_n &  c_{n-1}   & c_{n-2} &\cdots  &  c_{-1}  \\
  c_{2n} & \cdots & c_{n+1} &  c_n   & c_{n-1} &\cdots  &  c_0
   \end{bmatrix}
  \begin{bmatrix}  d_n \\  \vdots  \\  d_1  \\  d_0  \\  d_1  \\
        \vdots  \\  d_n \end{bmatrix},
$$
which is exactly a Toeplitz system! The solution is $\langle P_n, T_n\rangle$
times the sum of the first and the last columns of
$\boldsymbol{G}_{2n}^{-1}$, so, $2z^n P_n(x) =$
$$
d_n z^{2n} +\dotsb + d_1 z^{n+1} +d_0 z^n +d_1 z^{n-1} + \dotsb d_n =
 2\langle P_n, T_n\rangle \left(K_{2n}(z) +
                  \frac{\Phi_{2n}(z)}{\|\Phi_{2n}\|^2} \right)
=
 2\frac{\langle P_n, T_n\rangle}{\|\Phi_{2n}\|^2}
      (\Phi_{2n}(z) + \Phi_{2n}^*(z)),
$$
with $d_n=2^{1-n} \Rightarrow \|\Phi_{2n}\|^2 = 2^n \langle P_n, T_n
  \rangle (1+\Phi_{2n}(0))$. (N.B., $\Phi_{2n}(0)$ is real).

\medskip

\emph{\textbf{Conversely}}, can we recover the $\Phi$'s from the
$P$'s??  We need the \emph{difference} of the first and last columns
of $\boldsymbol{G}_{2n}^{-1}$\dots Let $Q_{n-1}$ be the monic orthogonal
polynomial of degree $n-1$ with respect to some scalar product
$\{\ ,\ \}$ on $[-1,1]$. We use the basis of Chebyshev polynomials
of \emph{second} kind:  $\displaystyle{ Q_{n-1} = \sum_{k=0}^{n-1}
   e_k U_k }$,
$$
 \begin{bmatrix} 0 \\  \vdots \\  0  \\  \{ Q_{n-1}, U_{n-1}\} \end{bmatrix}
 =
 \begin{bmatrix} \{ U_0, U_0\} & \cdots  &
                            \{ U_{n-1}, U_0\}  \\
                 \vdots  &          &  \vdots  \\
                 \{ U_0, U_{n-2}\rangle & \cdots  &
                            \{ U_{n-1}, U_{n-2}\} \\
                 \{ U_0, U_{n-1}\} & \cdots  &
                            \{ U_{n-1}, U_{n-1}\}
   \end{bmatrix}
  \begin{bmatrix}  e_0 \\  \vdots  \\  e_{n-2} \\  e_{n-1} \end{bmatrix}.
$$
Now, $U_\ell(\cos\theta) U_k(\cos\theta) = \displaystyle{
  \frac{\sin(\ell+1)\theta\; \sin(k+1)\theta}{\sin^2\theta} =
  \frac{\cos(\ell-k)\theta -\cos(k+\ell+2)\theta}{\sin^2\theta} =
  \frac{T_{\ell-k}(x) -T_{k+\ell+2}(x)}{1-x^2} }$,
and we recover the $c_k$'s if $\{f(x),1\} = \langle (1-x^2)f(x),1\rangle$,
or
$\displaystyle{   \{f,g\} = \int_{-1}^1 f(x)g(x)\,(1-x^2)\,d\nu(x)  }$.
The convenient rearrangement of the equations is
$$
 \begin{bmatrix} 2\{ Q_{n-1}, U_{n-1}\} \\
 0 \\  \vdots \\  0  \\  0  \\  0  \\  \vdots \\   0 \\ -2\{ Q_{n-1}, U_{n-1}\}
  \end{bmatrix}
 =
 \begin{bmatrix}
  c_0 & \cdots & c_{1-n} &  c_{-n}   & c_{-n-1} &\cdots  &  c_{-2n}  \\
  c_1 & \cdots & c_{2-n} &  c_{1-n}   & c_{-n} &\cdots  &  c_{-2n+1}  \\
\vdots&        &\vdots   & \vdots &  \vdots    &       &  \vdots    \\
  c_{n-1} & \cdots & c_0 &  c_{-1}   & c_{-2} &\cdots  &  c_{-n-1}  \\
  c_n & \cdots & c_1 &  c_0   & c_{-1} &\cdots  &  c_{-n}  \\
  c_{n+1} & \cdots & c_2 &  c_1   & c_0 &\cdots  &  c_{-n+1}  \\
\vdots&        &\vdots   & \vdots &  \vdots    &       &  \vdots    \\
  c_{2n-1} & \cdots & c_n &  c_{n-1}   & c_{n-2} &\cdots  &  c_{-1}  \\
  c_{2n} & \cdots & c_{n+1} &  c_n   & c_{n-1} &\cdots  &  c_0
   \end{bmatrix}
  \begin{bmatrix}  e_{n-1} \\ \vdots \\ \vdots  \\  e_0  \\  0  \\  -e_0  \\
        \vdots \\  \vdots \\  -e_{n-1} \end{bmatrix},
$$
so,  $-z^n (z-z^{-1}) Q_{n-1}(x) =$
$$
e_{n-1} +\dotsb + e_0 z^{n-1} +0 z^n -e_0 z^{n+1} - \dotsb -e_{n-1}z^{2n} =
 2\{ Q_{n-1}, U_{n-1}\} \left(K_{2n}(z) -
                  \frac{\Phi_{2n}(z)}{\|\Phi_{2n}\|^2} \right)
$$
$$
=
 2\frac{\{ Q_{n-1}, U_{n-1}\} }{\|\Phi_{2n}\|^2}
      (-\Phi_{2n}(z) + \Phi_{2n}^*(z)),
$$
with $e_{n-1}=2^{1-n} \Rightarrow \|\Phi_{2n}\|^2 = 2^n \{ Q_{n-1},
 U_{n-1}  \} (1-\Phi_{2n}(0))$.

\medskip

Finally,
$$
 \Phi_{2n}(z) = \frac{\|\Phi_{2n}\|^2}{2\langle P_n,T_n\rangle }
   z^n P_n(x)
              +\frac{\|\Phi_{2n}\|^2}{4\{Q_{n-1},U_{n-1}\}  }
   z^n (z-z^{-1}) Q_{n-1}(x),
$$
$$
 \Phi^*_{2n}(z) = \frac{\|\Phi_{2n}\|^2}{2\langle P_n,T_n\rangle }
   z^n P_n(x)
              -\frac{\|\Phi_{2n}\|^2}{4\{Q_{n-1},U_{n-1}\}  }
   z^n (z-z^{-1}) Q_{n-1}(x),
$$
$$
\Phi_{2n}\  \text{monic\  } \Rightarrow
\|\Phi_{2n}\|^2 = \frac{2^{n+1} }{
   \displaystyle{  \frac{1}{\langle P_n,T_n\rangle }
                  +\frac{1}{\{Q_{n-1},U_{n-1}\}    }    }    },
$$
$$
\Phi_{2n}(0) =
  \frac{\{Q_{n-1},U_{n-1}\} -\langle P_n,T_n\rangle }
       {\{Q_{n-1},U_{n-1}\} +\langle P_n,T_n\rangle },
$$
$$
\|\Phi_{2n-1}\|^2\; \|\Phi_{2n}\|^2\, =
   2^{n-1}     \{Q_{n-1},U_{n-1}\} \langle P_n,T_n\rangle\,,
$$
$$
\Phi_{2n-1}(z) =  \frac{\|\Phi_{2n-1}\|^2}{\|\Phi_{2n}\|^2}\,
  \frac{\Phi_{2n}(z)-\Phi_{2n}(0)\Phi^*_{2n}(z)}{z}
=   2^{n-1}
 z^{n-1} \left[ P_n(x) + \displaystyle{\frac{z-z^{-1}}{2}}
              Q_{n-1}(x)\right].
$$

See also \cite{Zhed}

\subsection{Jacobi polynomials on the unit circle\label{jacob}} \  \\

Here, $d\mu(e^{i\theta}) = \displaystyle{ \left(\cos\frac{\theta}{2}
                   \right)^{2\beta}
            \left|\sin\frac{\theta}{2}
                   \right|^{2\alpha} \,d\theta }$\  \   %
\ \  ($-\pi <\theta <\pi$, $\alpha$ and $\beta >-1/2$).

$ d\nu(\cos\theta) = (-1/\pi) d\mu(e^{i\theta}) $, so
$ d\nu(x) =  (1/\pi)
 \displaystyle{ \left(\frac{1+x}{2}
                   \right)^{\beta}
            \left(\frac{1-x}{2}
                   \right)^{\alpha} (1-x^2)^{-1/2}\,dx }$,

$P_n = P_{n,\text{monic}}^{\alpha-1/2, \beta-1/2} =
 \displaystyle{x^n +\frac{n(\alpha-\beta)}{2n+\alpha+\beta-1}x^{n-1}
  +\dotsb  },
 Q_{n-1} = P_{n-1,\text{monic}}^{\alpha+1/2, \beta+1/2}
= \displaystyle{x^{n-1} +\frac{(n-1)(\alpha-\beta)}{2n+\alpha+\beta-1}
       x^{n-2} +\dotsb  }$,

$\langle P_n,T_n\rangle = 2^{n-1}  \langle P_n,P_n\rangle  =
\frac{ 2^{3n+\alpha+\beta-1} \Gamma(n+\alpha+1/2)
                  \Gamma(n+\beta+1/2)\Gamma(n+\alpha+\beta) n!}
    {\Gamma(2n+\alpha+\beta)   \Gamma(2n+\alpha+\beta+1)}  $, \\
$\{ Q_{n-1},U_{n-1}\} = 2^{n-1}  \{ Q_{n-1},Q_{n-1}\}  =
\frac{ 2^{3n+\alpha+\beta-1} \Gamma(n+\alpha+1/2)
                  \Gamma(n+\beta+1/2)\Gamma(n+\alpha+\beta+1) (n-1)!}
    {\Gamma(2n+\alpha+\beta)   \Gamma(2n+\alpha+\beta+1)}  $,

$$
 \Phi_{2n}(0) = \frac{ \alpha+\beta }{2n+\alpha+\beta },
$$
$$
\hspace*{-22pt}\Phi_{2n-1}(z) = 2^{n-1}z^{n-1} \left[
    \left(\frac{z+z^{-1}}{2}\right)^n
      +\frac{n(\alpha-\beta)}{2n+\alpha+\beta-1}
       \left(\frac{z+z^{-1}}{2}\right)^{n-1}
  + \frac{z-z^{-1}}{2}   \left(\frac{z+z^{-1}}{2}\right)^{n-1}
  + \frac{z-z^{-1}}{2}
      \frac{(n-1)(\alpha-\beta)}{2n+\alpha+\beta-1}
    \left(\frac{z+z^{-1}}{2}\right)^{n-2}
 +\dotsb\right]
$$
$$
  = z^{2n-1} +\frac{(2n-1)(\alpha-\beta)}{2n+\alpha+\beta-1} z^{2n-2}
    +\dotsb + \frac{\alpha-\beta}{2n+\alpha+\beta-1}
$$
so,

\begin{equation}  \label{recjacobi}
 \Phi_n(0) = \frac{\alpha +(-1)^n \beta}{n+\alpha+\beta}
\end{equation}
\cite{Badk}

\subsection{The Szeg\H o-Geronimus theory} \  \\

Szeg\H o \cite{Sze} investigated the cases where the increasing
sequence $\{K_n(0)\}$ remains bounded,
the sequence of functions
$$
 K_n(z;0) = \sum_{k=0}^n \overline{\varphi_k(0)} \varphi_k(z)
       = \sum_{k=0}^n \overline{\Phi_k(0)} \Phi_k(z) /\|\Phi_k\|^2
$$
is a Cauchy sequence. What about the limit?

\subsubsection{The Bernstein-Szeg\H o polynomials \label{BerSze}}

If $p(z) = p_0 z^{-d} + \dotsb + p_{2d} z^d$ is a Laurent polynomial,
with real values on the unit circle, $p_k = \overline{p_{2d-k}}$:
$p_{2d-k} = (2\pi)^{-1}\int_{\mathbb{T}} p(e^{i\theta}) e^{i(-d+k)\theta}
d\theta$ is the complex conjugate of
$p_k = (2\pi)^{-1}\int_{\mathbb{T}} p(e^{i\theta}) e^{i(d-k)\theta}
d\theta$. The $2d$ zeros of $p$ come in inverse pairs: if $p(\zeta)=0$,
$p(1/\overline{\zeta}) = \sum_k p_k (1/\overline{\zeta})^{-d+k}
= \overline{ \sum_k p_{2d-k} \zeta^{d-k} }  =0$.

So, $p(z) = q(z) r(z)$ (Fej\'er), where $q(z)= q_0 +q_1 z^{-1}+\dotsb +q_d z^{-d}$
has its zeros in the unit disk  (in the open unit disk if
$p$ does not vanish on the unit circle), and $r(z) = r_0 +\dotsb +r_d z^d$ has
its zeros outside the unit disk.

When $w(\theta) = 1/p(e^{i\theta})$, where $p$ is a positive Laurent
polynomial of degree $d$,
$$\Phi_n(z) = \frac{1}{q_0} q(z) z^n$$
from $n=d$ onwards.\\
Indeed, $(\Phi_n(z), z^k) =  (1/2\pi i  q_0) \displaystyle{ \int_{\mathbb{T}} q(z)
z^{n-k} \frac{1}{q(z) r(z) }\, \frac{dz}{z}   =0}$ for
$k=0,\dotsc, n-1$, as there is no residue. When $k=n$, we find the residue
$\|\Phi_n\|^2 = 1/(q_0 r_0)$.

And, when $n\geqslant d$, $K_n$ is the polynomial
$$
  K_n(z) = \frac{\Phi_n^*(z)}{\|\Phi_n\|^2} =
  q_0 r(z).
$$

\subsubsection{The Szeg\H o theory} \  \\

A necessary condition to get a generalization of the factorization
of $w$ is $\log w \in L^1$, so that $\log w$ has a Fourier series,
and we take
$$
\log w = \underbrace{\dotsb + \lambda_{-2} z^{-2}
           +\lambda_{-1}z^{-1} +\lambda_0}_{\log q(z)}
    +   \underbrace{ \lambda_{0}
           +\lambda_{1}z +\lambda_2 z^2 +\dotsb}_{\log r(z)}\ ,
$$
and we expect $\Phi_n^*(z) \to r(z)/exp(\lambda_0)$ when $n\to\infty$, for
$z\in \mathbb{D}$. For the coefficients:
$$
  \Phi_n^*(z) = 1 +\overline{\xi_{n-1,n}}\; z
                  + \overline{\xi_{n-2,n}}\; z^2 +\dotsb
  \to_{n\to\infty}
      \exp( \lambda_1 z + \lambda_2 z^2 +\dotsb ) .
$$

Proofs and extensions: \cite[etc.]{Sze,Fre}.

\subsubsection{Some theorems by Geronimus} cf. \cite{Gol2}\  \\

\textsl{The following statements are equivalent}
 \begin{itemize}
   \item $\displaystyle{ \sum_{n=0}^\infty |\Phi_n(0)|^2 <\infty }$,
   \item $\displaystyle{ \sum_{n=0}^\infty |\varphi_n(z)|^2 <\infty }$
        for at least one point $z\in\mathbb{D}$,
   \item there is a subsequence $n_k$ such that $\displaystyle{
        \varphi^*_{n_k}(z) }$ converges
         at least for one point $z\in\mathbb{D}$,
   \item ${\displaystyle \lim_{n\to\infty} \varphi^*_n(z) = S(z) }$
        uniformly inside $\mathbb{D}$.
 \end{itemize}

\subsection{Formal unit circle orthogonal polynomials: non hermitian
  Toeplitz matrices} \  \\

\section{Semi-classical orthogonal polynomials on unit circle.}
\renewcommand{\mylmark}{Semi-classical orth. pol.}
\pagestyle{fancy}

\subsection{Definition, forms, and differential equation for weight
function}  \  \\

Cf. \cite{AM} and references therein

We can not guess easily the behaviour, especially the asymptotic
behaviour of the reflection coefficients $-\Phi_n(0)$ from the
entries $c_k$ with the recurrence relation \eqref{recur}: this relation
is a universal relation for all unit circle orthogonal polynomials.
We only need $\Phi_n(0)= \xi_{0,n}$, but \eqref{recur} compels us to
get all the two-dimensional tableau of the $\xi_{k,n}$'s.

It is very useful to select classes of orthogonal polynomials whose
coefficients $\Phi_n(0)$ satisfy a kind of one-dimensional recurrence
relation. This will happen with the class defined hereafter, where such
recurrence relations will be associated to \emph{differential relations}.

\smallskip

Speaking of differential relations, the famous Sonine-Hahn
characterization of \emph{classical} orthogonal polynomials as
having orthogonal derivatives falls short on the unit circle, as it
only works for $z^n$ \cite{AM,MarMar}.

\medskip

In the language of forms defined on Laurent polynomials
\cite{AM}, one defines the product of a function
and a form $fu$ as the form such that $\forall p, (fu)(p) = u(fp)$; and
the derivative $Du$ of $u$ as the form such that
$\forall p, (Du)(p) = -iu(t p'(t))$. This latter strange definition
corresponds actually to a simple relation for the \emph{weight
functions}: let $d\mu(t=e^{i\theta}) = w(\theta)d\theta$, where $w$
has at most a finite number of Dirac masses, and let $P$ be a
Laurent polynomial vanishing at these masses, then
if $\displaystyle{  u(p) = \frac{1}{2\pi} \int_0^{2\pi} p(e^{i\theta})
  d\mu(e^{i\theta}) }$, \\
$\displaystyle{
\forall p:
 (Pu)(p) =
 \frac{1}{2\pi } \int_0^{2\pi} P(e^{i\theta})  p(e^{i\theta})
 w(\theta)\,d\theta\;; }$
 $$
  (D(Pu))(p) = -i(Pu)(tp'(t)) =
 \frac{1}{2\pi i} \int_0^{2\pi} P(e^{i\theta}) e^{i\theta}
 \frac{d p(e^{i\theta})}{d e^{i\theta}}
  w(\theta)\,d\theta =
 \frac{1}{2\pi } \int_0^{2\pi}
  p(e^{i\theta})
 \frac{dP(e^{i\theta})w(\theta)}{d\theta}\,d\theta,
 $$
i.e., $D(Pu)$ has the integral representation involving $d[P(e^{i\theta})
w(\theta)]/d\theta$. Remark that, in order to get rid of annoying
boundary terms
in the integration by parts, $P$ must vanish at all Dirac points and
other discontinuities of $w$.

\textbf{Definition. A semi-classical form} on the circle satisfies

\begin{equation} \label{defsemi}
D( A(z)u) = B(z) u,
\end{equation}
with Laurent polynomials $A$ and $B$,
which means, if $u$ has an integral representation involving
$w(\theta)d\theta$, that
$$
   \frac{d[A(e^{i\theta})w(\theta)]}{d\theta} = B(e^{i\theta})w(\theta)\;,
$$
or
$$
\frac{dw(\theta)/d\theta}{w(\theta)} = \frac{B(e^{i\theta})
                -dA(e^{i\theta})
         /d\theta}{A(e^{i\theta})}  =
               \frac{B(e^{i\theta}) -i e^{i\theta}\,dA(e^{i\theta})
         /de^{i\theta}}{A(e^{i\theta})}\;
$$
with $A(e^{i\theta})=0$ at the singular points of $w$.


Jacobi:
$$
  u(p) = \frac{1}{2\pi} \int_{-\pi}^\pi p(e^{i\theta})
        \left( \cos \frac{\theta}{2} \right)^{2\beta}
        \left| \sin \frac{\theta}{2} \right|^{2\alpha}  \;d\theta\;,
$$
$$
  \frac{dw(\theta)/w(\theta)}{w(\theta)} = \alpha\cot\frac{\theta}{2}
                                -\beta\tan\frac{\theta}{2}
  = i\alpha \frac{z+2+z^{-1}}{z-z^{-1}}
    +i\beta \frac{z-2+z^{-1}}{z-z^{-1}}       \,, \  (z=e^{i\theta})\,,
$$
so, $A(z) = z-z^{-1}$,
 $B(z) = A(z)(dw(\theta)/d\theta)/w(\theta) +izdA(z)/dz =
  i(\alpha+\beta+1)z +2i(\alpha-\beta)
 +i(\alpha+\beta-1)z^{-1}$, or, if we prefer polynomials without
negative powers,
$$
 A(z)=z^2-1; B(z) = i\{ (\alpha+\beta+2)z^2 +2(\alpha-\beta)z
   +\alpha+\beta \}
$$

\smallskip
\noindent\textbf{Exercise.} Show that the square wave exemple
is semi-classical, but that $w(\theta) = |\theta|$ is not semi-classical.

\subsection{Recurrence relation for the moments} \   \\

From \eqref{ck} (p.~\pageref{ck}), the moment $c_k$ is just
$u(t^{-k})$ when $u$ has the integral representation
$u(p)=(2\pi)^{-1}\int_{\mathbb{T}} p(e^{i\theta})d\mu$. So,
we define $c_k=u(t^{-k})$, $k\in\mathbb{Z}$ for any form
defined on Laurent polynomials.

Let us apply \eqref{defsemi} to momomials $t^{-k}$, $k\in\mathbb{Z}$:

$$ iu(A(t)kt^{-k}) = u(B(t)t^{-k}),\qquad k\in\mathbb{Z}.  $$

If $\displaystyle{A(t)=\sum_{p=0}^d a_p t^{p-p_0}  \  \text{and\  }
                  B(t)=\sum_{p=0}^d b_p t^{p-p_0}  }$, then

\begin{equation} \label{recsemi}
  \sum_{p=0}^d  (ika_p-b_p) c_{k-p+p_0}=0, \qquad k\in\mathbb{Z} .
\end{equation}

A linear recurrence relation  of the form  \eqref{recsemi} is
another way to recognize a semi-classical functional.

Without more information than \eqref{recsemi}, we may find differential
equations for the generating functions
$G_{\pm(z)} = \displaystyle{ \sum_{m=0}^\infty c_{\pm m} z^m}$,
and contour integral representations for $c_k$

to be continued

\subsection{Differential relations for the orthogonal polynomials and
 recurrence relations for the reflection coefficients} \   \\

Let $A$ and $B$ be (plain, i.e., without negative powers) polynomials
of degree $\leqslant d$. Then,
the product $A(z)d\Phi_n(z)/dz$ is a remarkably short combination of the
$\Phi$ and the $\Phi^*$s.

The matrix $\boldsymbol{G}_{n+d-1}$ times the vector
$[A(0)\Phi_n'(0),A(0)\Phi_n''(0)+A'(0)\Phi_n'(0),\dotsc,
(n-1)A_d\xi_{n-1,n}+nA_{d-1}, nA_d]^T$ of the coefficients of $A\Phi_n'$
is the vector of
$$
(A(z)\Phi_n'(z), z^k) =
 u(A(z)\Phi_n'(z)z^{-k}) =
 u(zA(z)[\Phi_n(z)z^{-k-1}]')
- u(zA(z)\Phi_n(z)[z^{-k-1}]')
$$
$$
=
i(D( Au))(\Phi_n(z)z^{-k-1})
+ (k+1)u(\Phi_n(z)A(z)z^{-k-1})
=
iu(B(z)\Phi_n(z) z^{-k-1})+(k+1)u(A(z)\Phi_n(z)z^{-k-1})
$$
$$
= u\left(
\Phi_n(z)\frac{iB(z) +(k+1)A(z)}{z^{k+1}}\right)
$$
for $k=0,\dotsc, n+d-1$. The latter form vanishes as soon as
$(iB(z)+(k+1)A(z)) z^{-k-1} $ contains only nonpositive powers,
downto $z^{-n+1}$, which makes $k=d-1,\dotsc, n-2$. Therefore,
$A\Phi_n'$ is a combination of $\Phi_{n+d-1},\dotsc, \Phi_{n-1}$, and
$\Phi^*_{n+d-1}, z\Phi^*_{n+d-2},\dotsc, z^{d-2}\Phi^*_{n+1}$.

A more detailed look:


$$
\boldsymbol{G}_{n+d-1}
  \begin{bmatrix} A(0)\Phi'_n(0) \\  A(0)\Phi''_n(0)+A'(0)\Phi'_n(0) \\
         \vdots  \\
       (n-1)A_d\xi_{n-1,n}+nA_{d-1}  \\ nA_d  \end{bmatrix} =
 \begin{bmatrix}
  \vdots  \\  u((iB_d +(d-1)A_d)z \Phi_n(z)) \\  0  \\ \vdots \\ 0 \\
  u( (iB(0)+nA(0))\Phi_n(z)/z^n) \\  \vdots  \end{bmatrix}
 \begin{matrix}      k=d-2   \\  \\   \\    \\   k=n-1  \\  \end{matrix}
$$

\begin{equation} \label{diffrel} \begin{split}
A(z)\Phi'_n(z) =& \dotsb \Phi^*_{n+d-1}(z) + \dotsb +
   (iB_d+(d-1)A_d)\eta_n z^{d-2}\frac{\Phi^*_{n+1}(z)}{\|\Phi_{n+1}\|^2}
        \\
 &+ (iB(0)+nA(0)) \|\Phi_n\|^2 \frac{\Phi_{n-1}(z)}{\|\Phi_{n-1}\|^2}
 + \dotsb + \dotsb \Phi_{n+d-1}(z),
\end{split}\end{equation}
where we already encountered $\eta_n = -\Phi_{n+1}(0)\|\Phi_n\|^2$
p.~\pageref{etan}

\medskip

\noindent\textbf{Manifesto.}

\textsl{ As all the polynomials $\Phi_n$ depend, from the recurrence relations
 \eqref{recur}, on the reflection coefficients $-\Phi_m(0)$, the semi-classical
 identities will yield equations for these reflection coefficients.}

\smallskip

How $\Phi_n$ depends on the $\Phi_m(0)$:
with $x_n:=\Phi_n(0)$,

$$    \hspace*{-15pt}
\Phi_n(z) = z^n +\xi_{n-1,n}z^{n-1} +\dotsb + \Phi'_n(0)z+\Phi_n(0) \  \text{with\  }
 \xi_{n-1,n} = x_1\overline{x_0}+ x_2\overline{x_1} +
   \dotsb +x_n\overline{x_{n-1}}\;,
$$
$$
 \Phi'_n(0) = \Phi_{n-1}(0) +\Phi_n(0)\overline{\xi_{n-2,n-1}}.
$$

\medskip

Simplest case: Jacobi $(\alpha,0)$, $d=1$, $A(z)=z-1$,
$B(z)= i(\alpha+1)z +i\alpha$, $iB(0)+nA(0) = -(n+\alpha)$,
one finds, from \eqref{jacob0}
$$
   (z-1)\Phi'_n(z) = n\Phi_n(z) -\frac{n(n+2\alpha)}{n+\alpha}
     \Phi_{n-1}(z).
$$

remember \eqref{rnormphi}: $\|\Phi_n\|^2/\|\Phi_{n-1}\|^2 =
 1-|\Phi_n(0)|^2$.

\subsection{More Jacobi polynomials on the unit circle}   \  \\


Let us see how the features of the polynomials $\Phi_n$, in particular the reflection
coefficients $-\Phi_n(0)$, can be recovered (and, why not, \emph{discovered}) from the
semi-classical identities:

$d=2$,
we know $A(z)=z^2-1$, $B(z)= i(\alpha+\beta+2)z^2 +2i(\alpha-\beta)z +i(\alpha+\beta)$.

$(z^2-1)\Phi'_n(z) = nz^{n+1} +(n-1)\xi_{n-1,n}z^n
           +[(n-2)\xi_{n-2,n}-n]z^{n-1} +\dotsb =  $\\
$\ \  X_n \Phi^*_{n+1}(z) + Y_n \Phi_{n+1}(z) +Z_n \Phi_n(z) +W_n \Phi_{n-1}(z),
 $

 We know that $X_n=-(\alpha+\beta+1)\Phi_{n+1}(0)\|\Phi_n\|^2
          /\|\Phi_{n+1}\|^2 = \displaystyle{
       -(\alpha+\beta+1)\frac{\Phi_{n+1}(0)}{1-|\Phi_{n+1}(0)|^2}   }$,\\
 $W_n = -(n+\alpha+\beta)\|\Phi_n\|^2/\|\Phi_{n-1}\|^2 =
         -(n+\alpha+\beta)(1-|\Phi_n(0)|^2) $.

Comparing the terms in
\begin{itemize}
   \item $z^{n+1}$: $X_n\overline{\Phi_{n+1}(0)}+Y_n=n$,
   \item $z^n$:
       $X_n\overline{\Phi'_{n+1}(0)}+Y_n\xi_{n,n+1}+Z_n=(n-1)\xi_{n-1,n}$,
  \item $z^{n-1}$:
       $X_n\overline{\xi_{2,n+1}}+Y_n\xi_{n-1,n+1}+Z_n
              \xi_{n-1,n}  =(n-2)\xi_{n-2,n}-n$.
\end{itemize}

The two first equations merely yield expressions for $Y_n$ and
 $Z_n$, and the third equation is actually an equation for the
 $\Phi_m(0)$\dots after all the $\xi$'s have been expanded!

We avoid to have to bother with the $\xi_{2,n}$'s by

$Z_n \|\Phi_n\|^2 +W_n u(z^{-n}\Phi_{n-1}(z)) =
        (iB(0)+(n+1)A(0)) u(z^{-n-1}\Phi_n(z))
       + (iB'(0)+nA'(0)) \|\Phi_n\|^2$

 \medskip

$x_n:= \Phi_n(0)$:
$$
 (n+\alpha+\beta+1)x_{n+1} =
 (n+\alpha+\beta-1)x_{n-1}
  +\frac{ 2x_n\,\text{Im}\,(\overline{x_0} x_1
                           +\overline{x_1} x_2  +\dotsb
                           +\overline{x_{n-1}} x_n )
        }{1-|x_n|^2}
$$

if $x_1$ is real, so are the next $x_n$'s, and one recovers
\eqref{recjacobi}

what if $x_1$ is \emph{not} real?

%\subsection{Summary of the method} \  \\

\subsection{The second order differential equation} \   \\

The most attractive feature for most people is obviously the
linear second order differential equation

\begin{equation} \label{eqdiff2}
 zA(z)A^*(z)\Theta_n(z) \Phi''_n(z)= R_n(z)\Phi'_n(z)
  +S_n(z)\Phi_n(z)
\end{equation}
with $\Theta_n$ of degree $\leqslant d$, $R_n$ and $S_n$

Indeed, we put \eqref{recur3} in \eqref{diffrel}, to have
$A\Phi'_n=$ polynomials times $\Phi_n$ and $\Phi^*_n$, and also
a constant times $\Phi_{n-1}$, which is eliminated from the
second equation of \eqref{recur2}, the net result is
\begin{equation} \label{eqdiff21}
  zA(z)\Phi'_n(z)= \mathfrak{A}_n(z) \Phi_n(z) +
                   \mathfrak{B}_n(z) \Phi^*_n(z)
\end{equation}
with $\mathfrak{A}_n$ and $\mathfrak{B}_n$ of degree $\leqslant d$.
Considering the contribution of $\Phi_{n-1}$ in
\eqref{diffrel}, one has
$\mathfrak{A}_n(0)=iB(0)+nA(0)$ and
$\mathfrak{B}_n(0)=-\Phi_n(0)[iB(0)+nA(0)]$.

With
\begin{equation} \label{eqdiff22}
  zA^*(z)(\Phi^*_n(z))'= -\mathfrak{B}^*_n(z) \Phi_n(z) +
                (nA^*(z)- \mathfrak{A}^*_n(z)) \Phi^*_n(z),
\end{equation}
using\footnote{If $\Phi(z)= az^n+ b z^{n-1} +\dotsb +uz+v$,
 $\Phi^*(z)= \overline{v}z^n
            +\overline{u}z^{n-1} +\dotsb +
             \overline{b}z  +\overline{a}$,
 $(\Phi^*)'(z)= n\overline{v}z^{n-1}
            +(n-1)\overline{u}z^{n-2} +\dotsb +
             \overline{b}$,
 $(z\Phi')^*(z)= \overline{u}z^{n-1}  +\dotsb
            +(n-1)\overline{b}z  +  n \overline{a}$.}$(z\Phi_n')^*(z)=
n\Phi^*_n(z)-z(\Phi^*)'(z)$, we have a linear differential
system of the first order for $[\Phi_n,\Phi^*_n]$. And we
eliminate $\Phi_n^*$ between \eqref{eqdiff21}-\eqref{eqdiff22}

\subsection{Linear differential equation for Jacobi polynomials on
the unit circle: a bigger flop} \  \\

Of course we want to see the
                   differential equation for Jacobi polynomials on
the unit circle, to compare with the differential equation for
plain Jacobi polynomials. What a thrill.

...

to be completed

...

So, a unit circle Jacobi polynomial is a more composite object than
a plain Jacobi polynomial. Thats's why I do not believe that
reduction to unit circle is always the simplest thing to do.

\section{Generalized Jacobi polynomials on the unit circle, with
    two singular points.}
\renewcommand{\mylmark}{Gen. Jacobi.}
\pagestyle{fancy}


\subsection{The recurrence relation for the $\Phi_n(0)$'s} \   \\

We must have $\displaystyle{ \frac{dw/d\theta}{w} =
   \frac{a z^2 +bz+c}{(z-e^{i\theta_1})(z-e^{i\theta_2}) }  }$, real
on the unit circle. As denominator $/(z \exp(i(\theta_1+\theta_2)/2))$
is real, $(az+b+cz^{-1})/\exp(i(\theta_1+\theta_2)/2)$ must remain
real for $z$ on the unit circle
$\Rightarrow |a|=|c|, ac\exp(-i(\theta_1+\theta_2)) >0$ and
$            b\exp(-i(\theta_1+\theta_2)/2)$ real.

We solve now for $w$:
$$
\frac{dw/dz}{w} = \frac{dw/d\theta}{izw} =
  \frac{az+b +cz^{-1}}{i(z-e^{i\theta_1})(z-e^{i\theta_2})}
$$
$$
 = \frac{2\alpha}{z-e^{i\theta_1}}
 + \frac{2\beta}{z-e^{i\theta_2}}
 + \frac{\tilde{\gamma}}{z},
$$
with the residues
$$
 2\alpha = \frac{ae^{i\theta_1} +b +ce^{-i\theta_1} }
               {i(e^{i\theta_1}-e^{i\theta_2})}
        =  \frac{ae^{i\theta_1} +b +ce^{-i\theta_1} }
               {2e^{i(\theta_1+\theta_2)/2}\sin((\theta_2-\theta_1)/2)},
 2\beta =  \frac{ae^{i\theta_2} +b +ce^{-i\theta_2} }
               {2e^{i(\theta_1+\theta_2)/2}\sin((\theta_1-\theta_2)/2)},
 \tilde{\gamma} = \frac{c}{i e^{i(\theta_1+\theta_2)}},
$$

so, $w = $ constant $ z^{\tilde{\gamma}} (z-e^{i\theta_1})^{2\alpha}
                               (z-e^{i\theta_2})^{2\beta}
 =$ constant $\exp(i(\tilde{\gamma}+\alpha+\beta)\theta)
    |\sin((\theta-\theta_1)/2)|^{2\alpha}
    |\sin((\theta-\theta_2)/2)|^{2\beta} $,
with two constants on the two arcs of endpoints $\exp(i\theta_1)$ and
                                                $\exp(i\theta_2)$.

$\alpha$ and $\beta$ are real,
$\tilde{\gamma} +\alpha+\beta = \displaystyle{  \frac{a}{2i}
    +\frac{c}{2i e^{i(\theta_1+\theta_2)}}  = i\gamma} $ is a pure
imaginary number. So,

$w=$ constants  $\exp(-\gamma\theta)
    |\sin((\theta-\theta_1)/2)|^{2\alpha}
    |\sin((\theta-\theta_2)/2)|^{2\beta} $,

$d=2$, $A(z) = (z-e^{i\theta})(z-e^{i\theta_2})
           = z^2 -(e^{i\theta_1}+e^{i\theta_2})z +e^{i(\theta_1+\theta_2)},
$

$B(z) = izdA/dz +A [(dw/d\theta)/w]
       = izdA/dz +A iz[(dw/dz)/w]
      = iz(2z-e^{i\theta_1} -e^{i\theta_2})
  +i[2\alpha z(z-e^{i\theta_2})
    +2\beta z(z-e^{i\theta_1})
     (i\gamma-\alpha-\beta) (z-e^{i\theta_1})(z-e^{i\theta_2})]$,

$B(z)= i(\alpha+\beta+2+i\gamma)z^2
 +[i(\alpha-\beta)(e^{i\theta_1} -e^{i\theta_2})
   +(\gamma -i)(e^{i\theta_1} +e^{i\theta_2}) ]z
  -(i(\alpha+\beta)+\gamma)e^{i(\theta_1+\theta_2)} $.

The building of the equations proceeds as before,
$$
 A(z)\Phi'_n(z) = X_n\Phi_{n+1}^*(z) +Y_n\Phi_{n+1}(z)+Z_n\Phi_n(z)
  +W_n\Phi_{n-1}(z)
 = X'_n x_{n+1}\Phi_n^*(z) +n\Phi_{n+1}(z)+Z_n\Phi_n(z)
  +W_n\Phi_{n-1}(z),
$$
$X'_n x_{n+1}= X_n (1-|x_{n+1}|^2) = iB_2+A_2 = -(\alpha+\beta+1
     +i\gamma)$,

$W_n = (iB(0)+nA(0))(1-|x_n|^2) =
   (n+\alpha+\beta-i\gamma)e^{i(\theta_1+\theta_2)}(1-|x_n|^2)$,

and we look at the coefficients of $A\Phi'_n$:

$z^n$: $(n-1)\xi_{n-1,n} -n(e^{i\theta_1}+e^{i\theta_2}) =
 X'_n x_{n+1} \overline{x_n} +n \xi_{n,n+1} +Z_n$,

$z^0$: $A(0)\Phi'_n(0) = X'_n x_{n+1} +nx_{n+1} +Z_n x_n +W_n
  x_{n-1}$,

elimination of $Z_n$:

$A(0)[ x_{n-1}+x_n \overline{\xi_{n-2,n-1}} ] =
 (n+X'_n) x_{n+1}
      +  (n-1)\xi_{n-1,n}x_n -n(e^{i\theta_1}+e^{i\theta_2})x_n
 -X'_n x_n x_{n+1} \overline{x_n} -n x_n \xi_{n,n+1} +W_n x_{n-1}$,

which is a linear equation for $x_{n+1}$ if we know all the $x_m$
with $m\leqslant n$,

$(n+X'_n)(1-|x_n|^2)x_{n+1} =
[A(0)\overline{\xi_{n-1,n}} +\xi_{n-1,n} +n(e^{i\theta_1}
     +e^{i\theta_2})]x_n +A(0)(1-|x_n|^2)x_{n-1} -W_n x_{n-1}$,



\begin{equation}  \label{jacobg2}
\hspace*{-15pt} (n+1+\alpha+\beta+i\gamma) x_{n+1} =
\frac{ e^{i(\theta_1+\theta_2)} \overline{\xi_{n-1,n}} +\xi_{n-1,n}
  +n(e^{i\theta_1} +e^{i\theta_2})  }{1-|x_n|^2}\, x_n
  -(n-1+\alpha+\beta-i\gamma) e^{i(\theta_1+\theta_2)} x_{n-1},
\end{equation}
with
$ \xi_{n-1,n} = x_1\overline{x_0}+ x_2\overline{x_1} +
   \dotsb +x_n\overline{x_{n-1}}$.



\subsection{Numerical experiments with the asymptotics of
 \eqref{jacobg2}}   \   \\

\subsubsection{}.  We now proceed to extract the last asymptotic
secrets of the $\Phi_n(0)=x_n$'s of \eqref{jacobg2}.

One obviously expects $\Phi_n(0)$ to behave like a combination of
$e^{in\theta_1}$ and  $e^{in\theta_2}$ with slowly varying
coefficients. What works is

 

$$
  \Phi_n(0) \sim A_1 n^{\kappa_1} e^{in\theta_1}
            + A_2 n^{\kappa_2} e^{in\theta_2}
$$
as guessed from \cite[etc.]{Badk,GolJac,Ver}, also \cite{Maggjac}. The powers of $n$
are deduced from
  \eqref{jacobg2} looking as a linear recurrence relation when $|x_n|^2$ is neglected,
and where $\xi_{n-1,n}$ is replaced by its limit, say $\xi$. Then each
particular solution $A_j n^{\kappa_j} e^{in\theta_j}$ satisfies
$\displaystyle \dfrac{x_{n+1}}{x_n} \sim (1+\kappa_j /n)e^{i\theta_j}$, so
 \eqref{jacobg2} becomes
$ [1+( 1+\alpha+\beta+i\gamma)/n] (1+\kappa_j/n) e^{i\theta_j}\sim 
[e^{i(\theta_1+\theta_2)} \overline{\xi} +\xi]/n
  +e^{i\theta_1} +e^{i\theta_2})  
  -[1+(-1+\alpha+\beta-i\gamma)/n] e^{i(\theta_1+\theta_2)} (1-\kappa_j/n) e^{-i\theta_j}$,
match the $n^{-1}$ contributions: 
$ ( 1+\alpha+\beta+i\gamma+\kappa_j) e^{i\theta_j} = 
e^{i(\theta_1+\theta_2)} \overline{\xi} +\xi
  +  (1-\alpha-\beta+i\gamma+\kappa_j) e^{-i\theta_j}$, or

$\kappa_1$ and $\kappa_2= -1-i\gamma \pm i\delta$, with
$\displaystyle i\delta= -(\alpha+\beta)\dfrac{e^{i\theta_1}+e^{i\theta_2}} {e^{i\theta_1}-e^{i\theta_2}} 
 + \dfrac{  \overline{\xi} } {e^{-i\theta_2}-e^{-i\theta_1}}
 +\dfrac{  \xi} {e^{i\theta_1}-e^{i\theta_2}}$. 
 



What can $A_1$ and $A_2$ be? We perform numerical explorations.
First with known cases: when $\theta_2-\theta_1=\pi$, and when
$w(\theta)/[|\sin(\theta-\theta_1)/2|^{2\alpha}
|\sin(\theta-\theta_2)/2|^{2\beta}]$ is the same constant on the whole
circle, one must have $A_1=\alpha$, $A_2=\beta$, and $\delta=0$.

When $\gamma=0$, and where is no jump,
more experiments lead to

\begin{equation} \label{a1a2}
A_1 = \alpha \exp(i\beta(\pi-\theta_2+\theta_1)),
A_2 = \beta \exp(-i\alpha(\pi-\theta_2+\theta_1)),
\end{equation}

where $\pi -\theta_2+\theta_1$ must be understood modulo
$2\pi$, with a value between $-\pi$ and $\pi$.



%\newpage

\subsection{Final (so far) experiments\label{finalgjac}}
\renewcommand{\mylmark}{Final?}
\pagestyle{fancy}




\subsubsection{Discontinuous case.}


Now,
 $w(\theta)/[|\sin(\theta-\theta_1)/2|^{2\alpha}
|\sin(\theta-\theta_2)/2|^{2\beta}]=$ discontinuous, i.e.:

\begin{equation} \label{wdisc}
   w(\theta) =
    \begin{cases}
        r_1 e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_1<\theta<\theta_2\\
        r_2e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_2-2\pi <\theta <\theta_1]
    \end{cases}
\end{equation}
where $r_1$ and $r_2$ are positive, possibly different.  Of course, it's
the ratio $r_2/r_1$ which is important.

The multiplicative jump is $\dfrac{r_1}{r_2}$ at $\theta_1$, and 
$\displaystyle \dfrac{r_2 e^{-\gamma(\theta_2-2\pi)}  }{r_1 e^{-\gamma\theta_2}} =
e^{2\pi\gamma}\dfrac{r_2}{r_1}$ at $\theta_2$. 

We proceed with examining the output of \eqref{jacobg2}, but where
some tricks must be explained: \eqref{jacobg2} needs a complex
number $x_1=\Phi_1(0)$ which is not so easy to get, as it is the
ratio $-c_{-1}/c_0$ of the two Fourier coefficients
$-\int_0^{2\pi} w(\theta)\exp(i\theta)\,d\theta /
  \int_0^{2\pi} w(\theta)\,d\theta $.

However, from the already encountered differential equation for
   $w = $ constants $ z^{\tilde{\gamma}} (z-e^{i\theta_1})^{2\alpha}
                               (z-e^{i\theta_2})^{2\beta}$,
$$
\frac{dw/dz}{w} = \frac{\sum_{-\infty}^\infty k\,c_k z^{k-1} }
                       {\sum_{-\infty}^\infty \,c_k z^{k} }
  = \frac{\tilde{\gamma}}{z}
   +\frac{2\alpha}{z-e^{i\theta_1} }
   +\frac{2\beta}{z-e^{i\theta_2} },
$$
(almost) everywhere on $(0,2\pi)$,
with $\tilde{\gamma} =i\gamma-\alpha-\beta$:
$\displaystyle
(\sum_{-\infty}^\infty k\,c_k z^{k} )(z^2-(e^{i\theta_1}+e^{i\theta_2})z
+e^{i(\theta_1+\theta_2)}) =\\
(\sum_{-\infty}^\infty \,c_k z^{k} )[(i\gamma+\alpha+\beta)z^2+
[(\beta-\alpha)
  (e^{i\theta_2}-e^{i\theta_1})-i\gamma(e^{i\theta_2}+e^{i\theta_1})]z+(i\gamma -\alpha-\beta)e^{i(\theta_1+\theta_2)}]
$
whence the \emph{\textbf{linear recurrence relation}} for the $c_k$'s
(see also \eqref{recsemi}, p.~\pageref{recsemi})
$$
(k-1-i\gamma-\alpha-\beta)c_{k-1}
-[k(e^{i\theta_1}+e^{i\theta_2})+
        (\beta-\alpha)(e^{i\theta_2}-e^{i\theta_1})-i\gamma(e^{i\theta_2}+e^{i\theta_1})]c_k
 +(k+1+\alpha+\beta-i\gamma)  e^{i(\theta_1+\theta_2)} c_{k+1} =0.
$$
At $k=0$, knowing that $c_1=\overline{c_{-1}}$:
$ (\alpha+\beta+1+i\gamma)x_1
       -(\beta-\alpha)(e^{i\theta_2}-e^{i\theta_1})+i\gamma(e^{i\theta_2}+e^{i\theta_1})
 -(\alpha+\beta+1-i\gamma)  e^{i(\theta_1+\theta_2)} \overline{x_1} =0$,
or $\displaystyle (\alpha+\beta+1+i\gamma)\dfrac{x_1}{e^{i\theta_2}-e^{i\theta_1}} 
+\alpha-\beta +\gamma\cot ((\theta_2-\theta_1)/2) 
-(\alpha+\beta+1-i\gamma)\dfrac{\overline{x_1}}{e^{-i\theta_1}-e^{-i\theta_2}}$
amounting to
$$
 x_1 = (e^{i\theta_2}-e^{i\theta_1}) \left[ -\dfrac{ \alpha-\beta +\gamma\cot ((\theta_2-\theta_1)/2) }
         {2(\alpha+\beta+1+i\gamma)} +(\gamma +(\alpha+\beta+1)i )t \right],\ \ \ \  t\in{\mathbb R} 
$$
so that we only have to experiment with various (real) values of $t$, which
correspond to various real ratios $r_2/r_1$, which we recover in
the Christoffel function-aided weight function reconstruction.

Check that $|x_1|<1$ when $t=0$: $|x_1|=\displaystyle \dfrac{ |(\alpha-\beta)\sin((\theta_2-\theta_1)/2)
+\gamma\cos((\theta_2-\theta_1)/2) | }{|\alpha+\beta+1+i\gamma|}$,
OK if $|\alpha-\beta| < |\alpha+\beta+1|$.


Simple check with $\alpha=\beta=\gamma=0$: we immediately have
$x_1=\Phi_1(0)=-c_{-1}/c_0 = i(e^{i\theta_2}-e^{i\theta_1})(r_1-r_2)/
[r_1(\theta_2-\theta_1)+r_2(2\pi-\theta_2+\theta_1)]$.
When $\gamma\neq 0$, $\displaystyle x_1= \dfrac
{ (i-\gamma)^{-1} [r_1 (e^{(i-\gamma)\theta_2} -e^{(i-\gamma)\theta_1} )
     +r_2(e^{(i-\gamma)\theta_1} - e^{(i-\gamma)(\theta_2-2\pi)}  )] }
{ \gamma^{-1} [ r_1( e^{-\gamma\theta_2} -e^{-\gamma\theta_1} )
   +r_2 (e^{-\gamma\theta_1} -e^{-\gamma(\theta_2-2\pi)} )] }
= e^{i(\theta_1+\theta_2)/2} \dfrac{\gamma}{i-\gamma} \left[ \cos((\theta_2-\theta_1)/2)
   +i\sin((\theta_2-\theta_1)/2)   \underbrace{\dfrac{
     r_1( e^{-\gamma\theta_2} +e^{-\gamma\theta_1} )
   -r_2 (e^{-\gamma\theta_1} +e^{-\gamma(\theta_2-2\pi)}  )} 
   {  r_1( e^{-\gamma\theta_2} -e^{-\gamma\theta_1} )
   +r_2 (e^{-\gamma\theta_1} -e^{-\gamma(\theta_2-2\pi)})  }  }_{\displaystyle -2(\gamma^{-1}+\gamma)t}\right] 
$

Check with $r_1/r_2=\exp(2\pi\rho)$ with $\rho=0.25$:
\begin{verbatim}

reflecj2 unit circle gen Jacobi  Sat 9 Mar 2013 19:26

 alpha=0 beta=0 , gamma=0.50000; theta1,2/pi=-0.35000 0.25000
 Phi1(0)= -0.063355 - 0.27785*I
A1=-0.0089576 - 0.24984*I A2=-0.0089729 - 0.24984*I 
 xi=-0.045993 - 0.29027*I i delta= - 0.000022164*I
r1=0.79468 r2=0.16520 rho=0.25000 
A1/(alpha-rho i)=0.99936 - 0.035830*I A2/(beta+(rho-gamma) i)=0.99936 - 0.035892*I
 abs: 1.0000 1.0000
 








2.4313888 |"'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''|
          |:"                                                            |
          |: "                                                           |
          |:  "_                                                         |
          |:    _                                                        |
          :      x                                                       |
          :       x                                                      |
          :        "_                                                    |
          :          x                                                   |
          :           "_                                                 |
          "             x_                             "_                "
          |               x_                           : "_              :
          |                 x_                         :   "_            :
          |                   x_                       :     "x_        :|
          |                     "x_                    :        x_      :|
          |                        "x_                :           "x__  :|
          |                           "x__            :               "x_|
          |                               "xx_        :                  |
          |                                   ""xx___ :                  |
          |                                          ""                  |
          |                                                              |
        0 ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
          -5.497787                                              0.7853982
(19:26) gp > t
%7 = -0.060843
(19:27) gp > quit
Goodbye!
\end{verbatim}

For general $\alpha$ and $\beta$,
we could  use that, as $c_0$ and $c_{-1}$ are linear in $r_1$ and
$r_2$, $\displaystyle{ x_1 = \frac{ X +Y (r_2/r_1) }{1+Z (r_2/r_1)} }$,
so that we could extract $X, Y$, and $Z$ from three different trials,
without having to compute $X, Y$, and $Z$, which are complicated
integrals, and arrive at a full algorithm for computing long sequences
of reflection coefficients for \eqref{wdisc}, given $\alpha$, $\beta$,
$\theta_1$, $\theta_2$, $r_1$, and $r_2$, but we just finish now
asymptotic experiments.

In the recurrence \eqref{jacobg2}, 
 we know that $x_n\to 0$, and even that
 $ \overline{\xi_{n-1,n}}   \to_{n\to\infty}  \lambda_1$,
from Szeg\H o-Geronimus theory, where
$$
\log w = \underbrace{\dotsb + \lambda_{-2} z^{-2}
           +\lambda_{-1}z^{-1} +\lambda_0}_{\log q(z)}
    +   \underbrace{ \lambda_{0}
           +\lambda_{1}z +\lambda_2 z^2 +\dotsb}_{\log r(z)}\ ,
$$
We  recover $A_1$ and $A_2$ from
numerical runs and guess:

$$
  A_1= (\alpha-\rho_1 i) e^{i\beta(\pi-\theta_2+\theta_1)+i\psi_1}\;,
 \qquad A_2= (\beta-\rho_2 i) e^{-i\alpha(\pi-\theta_2+\theta_1)+i\psi_2}\;,
$$
with $\displaystyle{ \rho_1 = \frac{\log(r_1/r_2)}{2\pi} }$ and $\rho_2=\gamma-\rho_1$,
the jumps of $(2\pi)^{-1}\log w$ at $\theta_1$ and $\theta_2$, as seen above.

 The absolute
values of $A_1$ and $A_2$ are probably correct, as they agree with what
is needed in the Hartwig-Fisher formula of next section; the phases
$\psi_1$ and $\psi_2$ are still unknown, they
will be examined later on.


\  

\  

\subsubsection{Relation with Fisher-Hartwig determinants} \   \\

We may \emph{\textbf{conjecture}} that the main behaviour of
$\Phi_n(0)$ for a weight which is smooth except at $\theta_1,\dotsc,
\theta_p$ where it behaves like $r_{k,1} |\theta-\theta_k|^{2\alpha_k}$
for $\theta\to\theta_k, \theta<\theta_k$, and like
                                $r_{k,2} |\theta-\theta_k|^{2\alpha_k}$
for $\theta\to\theta_k, \theta>\theta_k$, will be
$$
  x_n=\Phi_n(0) = \sum_{k=1}^p \frac{A_k n^{i\delta_k}}{n} e^{in\theta_k}
  +o(1/n),
$$
with $|A_k|^2 = \alpha_k^2 + \rho_k^2$, where
$\rho_k = \displaystyle{ \frac{\log(r_{k,1}/r_{k,2})}{2\pi} }$.
From \eqref{rnormphi},
$$
  \|\Phi_n\|^2  = \prod_{m=1}^n (1-|\Phi_m(0)|^2) =
  \text{const.} \left( 1 +\frac{ \sum_{k=1}^p |A_k|^2}{n} +\dotsb \right),
$$
as the only non-oscillating terms in the expansion of
$|\Phi_m(0)|^2 =\Phi_m(0)\overline{\Phi_m(0)} =
  \sum_k\sum_j A_k \overline{A_j} m^{i(\delta_k-\delta_j)}
\exp(im(\theta_k-\theta_j))/m^2$ are the $|A_k|^2/m^2$ terms.
The constant is actually very well known from the Szeg\H o
theory, as it must be $\exp(2\lambda_0)$.

Finally, the product of these square norms yields the determinant
of the Gram matrix (here, a Toeplitz matrix)
$\displaystyle 
   \det \boldsymbol{G}_n = \prod_{m=1}^n \|\Phi_m\|^2
 = \text{const.}\; n^{\sum_k |A_k|^2} \exp(2n\lambda_0),
$
(with another constant). This formula is exactly the Hartwig-Fisher
asymptotic formula  for such Toeplitz determinants, see \cite{Bot} for
details and full history!

\vspace*{-9pt}

\subsubsection{Last last calculations, Spring 2012 and March 2013.}  \   \\

We look at the influence of the jump $\rho=\rho_1$ in
$ A_1= (\alpha-\rho_1 i) e^{i\beta(\pi-\theta_2+\theta_1)+i\psi_1}\;,\\
 \qquad A_2= (\beta-\rho_2 i) e^{-i\alpha(\pi-\theta_2+\theta_1)+i\psi_2}$,
when $\alpha=\beta=\gamma=0$ and $\theta_1, \theta_2 =\mp\pi/2$.

\vspace*{-1pt}

\parpic{
\begin{tabular}{c|c|c|c|c}
\hline     $\rho$ &      $\psi_1$  &  $d\psi_1/d\rho$ & $\rho\dfrac{ d^2\psi_1}{d\rho^2}$  &
   $\begin{matrix}  2 \arg\Gamma(1+\rho i) \\  -\rho\log 4\end{matrix}$      \\
\hline     -0.10347  &  0.26208  & &  & 0.26201\\
          0    &  0  &    & & 0\\
      0.10347  & - 0.26208  & -2.533 &  & -0.26201\\
     0.23504   & - 0.58723  &  &  & -0.58705   \\
   0.55998     & - 1.3012   &  &   &  -1.30086 \\
0.68822    &- 1.5363  &   &  & 1.53590\\
0.98394  &- 1.9705  &  &  &  -1.97015\\
1.8739 &- 2.5110&   &  & -2.51096\\
1.9103 &- 2.5133&  &  &  -2.51325\\
1.9576 &- 2.5142&  &  & -2.51425\\
2.0253 &- 2.5119& &  &  -2.51191\\
4.4543 &- 0.23977&  &  &  -0.24197\\
4.8284 & 0.39333&  &  &  0.39062\\
5.3055 & 1.2836&  &  &  1.28040\\
5.9850 & 2.6974& &  & 2.69334\\
7.0797 &  5.2921&  &  &  5.28652\\
8.1195 & 8.071&  &  & 8.06407\\
8.9465 & 10.4736&  &   & 10.4649\\
9.9983 & 13.7476 & &  & 13.7374\\ 
11.009 & 17.1021 & 3.3190 & 2.006 & 17.0908 \\
11.999  & 20.5684 & 3.5013 & 2.021 & 20.5555\\
13.015 & 24.2974 & 3.6703 & 1.954&  24.2810 \\
14.005 & 28.0801   & 3.8209 & 2.043 & 28.0638\\
15.000 & 32.0260  &  3.9657 & & 32.0068
\end{tabular}  
}
Various values
of $t$, so of $x_1$ are entered in the recurrence \eqref{jacobg2}, $A_1$ and $A_2$
are extracted from $x_n$ with large $n$, and $\rho$ from weight reconstruction.
One finds $\psi_2=-\psi_1$, and we look at $\psi_1$ as a continuous function
of $\rho$, which may ask for adding integer multiples of $2\pi$. \\
The second order divided difference on almost uniformly spaced sets of three values of
$\rho$ yield a good estimate of (half the) second derivative at the middle point.
The formula $d^2\psi_1/d\rho^2 \sim  2/\rho$ seems to hold.\\
This means a main behaviour $2\rho\log\rho$ for the
function itself. The simplest special function with this behaviour is the Gamma function.
More precisely, an exploration of the pages of Abramowitz \& Stegun's masterpiece, where
even the pages with tables are wonderfully inspiring, shows that the imaginary part
of $\log\Gamma(1+i\rho)$ is the needed odd continuous function of $\rho$. From Stirling
formula, Abr 6.1.44, the behaviour for large positive $\rho$ is
$\rho\log\rho-\rho +\pi/4+o(1)$.So the first and second derivatives are close to $\log\rho$
and $1/\rho$. \\
The derivative at $\rho=0$ of two times the argument of the Gamma function of
$1+i\rho$ is $-1.15443=-2$ times the Euler constant. The derivative of our function
there is estimated to be $-2.533$. The difference is $-1.379$, close enough to
$-\log 4$. GOT IT! (Spring 2012). Music, at last!

For general $\theta_1$ and $\theta_2$, we find readily the term 
$-2\rho\log\sin((\theta_2-\theta_1)/2)$ to be present n $\psi_1$.

\    \\

\picskip{1}

\ \hspace*{-4pt}\vspace*{3pt}\includegraphics[angle=0, scale=0.75]{abrgam.ps}

We now must consider the influence of $\alpha$, $\beta$, and $\gamma$.


\subsubsection{The final conjecture.} \textsl{Consider the special weight on the unit circle} 

\begin{equation*} 
   w(\theta) =
    \begin{cases}
        r_1 e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_1<\theta<\theta_2\\
        r_2e^{-\gamma\theta}|\sin(\theta-\theta_1)/2|^{2\alpha}
   |\sin(\theta-\theta_2)/2|^{2\beta}\;,\qquad & \theta_2-2\pi <\theta <\theta_1
    \end{cases}
\end{equation*}
\textsl{where $\theta_1 < \theta_2 < \theta_1+2\pi$, $r_1$ and $r_2$ are positive,
$\alpha$, $\beta$, and $\gamma$ real, $\alpha$ and $\beta >-1/2$. Then} 
$$
  \Phi_n(0) = A_1 n^{\kappa_1} e^{in\theta_1}
            + A_2 n^{\kappa_2} e^{in\theta_2}+o(n^{-1}),
$$
\textsl{with $\kappa_1$ and $\kappa_2= -1-i\gamma \pm i\delta$, where }
$\displaystyle i\delta= -(\alpha+\beta)\dfrac{e^{i\theta_1}+e^{i\theta_2}} {e^{i\theta_1}-e^{i\theta_2}} 
 + \dfrac{  \overline{\xi} } {e^{-i\theta_2}-e^{-i\theta_1}}
 +\dfrac{  \xi} {e^{i\theta_1}-e^{i\theta_2}}$, $\xi$
\textsl{being the coefficient of $e^{-i\theta}$ in the Fourier expansion}
 $\log w(\theta) = \dotsb + \xi e^{-i\theta} +\lambda_0
    +   \overline{\xi}e^{i\theta}  +\dotsb$,
\textsl{$\xi$ is also the limit when $n\to\infty$ of $\xi_n$ in 
$\Phi_n(z)=z^n+\xi_n z^{n-1}+\dotsb$.
And the big conjecture is 
}  % end sl



\   \\
$A_1= (\alpha-\rho_1\; i)\exp[ i[ 2\arg\Gamma(1+\alpha+\rho_1\; i)+2(\gamma-\rho_1)\log(2\sin((\theta_2-\theta_1)/2))+\beta(\pi-\theta_2+\theta_1)\  ]],\\
A_2= (\beta-\rho_2\; i)\exp[ i[ 2\arg\Gamma(1+\beta+\rho_2\; i)+2(\gamma-\rho_2)\log(2\sin((\theta_2-\theta_1)/2))-\alpha(\pi-\theta_2+\theta_1)\  ]]
$,\\
\textsl{where  $\displaystyle{ \rho_1 = \frac{\log(r_1/r_2)}{2\pi} }$ and $\rho_2=\gamma-\rho_1$,
the jumps of $(2\pi)^{-1}\log w$ at $\theta_1$ and $\theta_2$.
} 
 

\subsubsection{The GP-PARI program.}

Given $\alpha, \beta, \gamma, \theta_1$, and $\theta_2$, the program 
computes $x_1=\Phi_1(0)=\\
\displaystyle (e^{i\theta_2}-e^{i\theta_1}) \left[ -\dfrac{ \alpha-\beta +\gamma\cot ((\theta_2-\theta_1)/2) }
         {2(\alpha+\beta+1+i\gamma)} +(\gamma +(\alpha+\beta+1)i )t \right],\ \ \ \  t\in{\mathbb R} 
$, with various values of $t$ which will allow experiments with various values of $\rho$.
Then, several thousands of the next $x_n$s are computed with
\eqref{jacobg2}. At each power of two, $A_1$ and $A_2$ are estimated from
$x_n\sim A_1 n^{\kappa_1} e^{in\theta_1} +A_2 n^{\kappa_2} e^{in\theta_2} $ and
$x_{n+1}\sim A_1 (n+1)^{\kappa_1} e^{i(n+1)\theta_1} +A_2 (n+1)^{\kappa_2} e^{i(n+1)\theta_2}
\Rightarrow n^{-\kappa_2} e^{i\theta_2} x_n - (n+1)^{-\kappa_2} x_{n+1} \sim
[n^{\kappa_1-\kappa_2} e^{i\theta_2} - (n+1)^{\kappa_1-\kappa_2}e^{i\theta_1}]e^{in\theta_1}A_1 $,etc.
A further simple acceleration is performed with $A_1$ replaced by $2A_1$ minus the former
$A_1$ from the step $n/2$. Several thousands of $\Phi_n(0)$s allow an accurate reconstruction
of the weight function through Christoffel function (P.Nevai, the big 1986 paper, p.~26)
$\displaystyle w(\theta)=  \lim_{n\to\infty} n \omega_n(e^{i\theta}) = \dfrac{n}{\sum_0^{n-1} |\varphi_k(e^{i\theta})|^2  }$, from which $\rho=\rho_1= (2\pi)^{-1}\log(r_1/r_2)$ is found, and
the formulas for $A_1$ and $A_2$ are checked.

{\Small

\begin{verbatim}

{
/*
  reflecj2.gp    :  launch gp and make  \r reflecj2

  Reflection coefficients and reconstruction of gen. Jacobi
  weight on unit circle with 2 singular points exp(i theta1)
     and exp(i theta2).

  w(theta) = A or B   exp(-gamma theta) (sin(theta-theta1)/2)^(2alpha)
                                        (sin(theta-theta2)/2)^(2 beta)


*/
sextern=extern("date +%a%e%b20%y__%H-%M");print("reflecj2 unit circle gen Jacobi ",sextern);
\\default(realprecision,55);
default(format,"g4.5");
N=2^15+1; 
Nr=2^13+1;  \\ only for reconstruction
tp1=-0.15; tp2=0.275;  \\  theta1/pi and theta2/pi    albe=alpha+beta
al=1.6;be=0.9;albe=al+be;ga=2;   \\  alpha, beta, gamma
et1=exp(I*Pi*tp1);et2=exp(I*Pi*tp2);print(" alpha=",al," beta=",be," , gamma=",ga,"; theta1,2/pi=",tp1," ",tp2);
st1t2=sin(Pi*(tp2-tp1)/2);L12=2*log(2*st1t2);ct1t2=cos(Pi*(tp2-tp1)/2);

\\ P.Nevai's algorithm of weight reconstruction through Christoffel function
}
{
 Chris(z)=

       locp=1;locps=1;locsom=1;
       for(n=1,Nr,locp=z*locp+ref[n+1]*locps;
                 locps=(1-(abs(ref[n+1]))^2)*locps+conj(ref[n+1])*locp;
                 locsom=locsom+(abs(locp))^2/norm2[n+1]
          );
   1/locsom
}

{
ref=vector(Nr+1,k,0);
\\cas alpha=beta=0
\\rr=exp(2*Pi*0.25);t=-( (rr-1)*exp(-ga*Pi*tp1)+(rr-exp(2*ga*Pi))*exp(-ga*Pi*tp2)  )/(2*(ga+1/ga)*(  (1-rr)*exp(-ga*Pi*tp1)+(rr-exp(2*ga*Pi))*exp(-ga*Pi*tp2) )  );  
t=0.1;ref[2]=(et2-et1)*( (be-al-ga*ct1t2/st1t2)/(2*(albe+1+ga*I)) +(ga+(albe+1)*I)*t );
\\ vector of reflection coeff.
ref[1]=1;xi=ref[2];
print1(" Phi1(0)= ",xi);default(format,"g4.30");print(" , t=",t);default(format,"g4.5");
stop=0;
p2=1;
refn=ref[1]; refn1=ref[2];refn1o=refn1;cr1o=0;cr2o=0;
for(n=1,N-1,      
 if( abs(refn1)>1,stop=1;return);
 refn2=(  refn1*(et1*et2*conj(xi)+xi+n*(et1+et2))/(1-(abs(refn1))^2)
             -(n-1+albe-I*ga)*et1*et2*refn
          )/(n+1+albe+I*ga);
 \\   xn = A1 n^(-1+i delta) e^(i n theta1) +A2 n^(-1-i delta) e^(i n theta2)
 xi=xi+conj(refn1)*refn2;la1=conj(xi);id1=la1/( 1/et2-1/et1 )+conj(la1)/( et1-et2 )
    -albe*( et1+et2 )/( et1-et2 );
 if(n+2<= Nr+1,  ref[n+2]=refn2);
 \\ asymptotic behaviour  cr1 et1^n/n^(1+i gamma -i delta)  + cr2 et2^n /n^(1+i gamma +i delta)
 if(n==p2,dn=n;dn1=n+1;
 cr1=(dn^(1+ga*I+id1)*refn1*et2-dn1^(1+ga*I+id1)*refn2)/(et1^n*(dn^(2*id1)*et2-dn1^(2*id1)*et1));
 cr2=(dn^(1+ga*I-id1)*refn1*et1-dn1^(1+ga*I-id1)*refn2)/(et2^n*(dn^(-2*id1)*et1-dn1^(-2*id1)*et2));
     cr12=2*cr1-cr1o; cr22=2*cr2-cr2o;errx=refn1-cr12*et1^n/(dn^(1+ga*I-id1))-cr22*et2^n/(dn^(1+ga*I+id1));
     print(n," A1=",cr12," A2=",cr22," relerr=10^",log(abs(errx/refn1))/2.3026  );
      p2=p2*2;cr1o=cr1;cr2o=cr2;
     ); \\ end if  p2
 refn=refn1; refn1=refn2
   );    \\  end for n
if(stop,return);
cr1=cr12;cr2=cr22;
\\  norms
norm2=vector(Nr+1,k,0);norm2[1]=1;
for(n=2,Nr+1,  norm2[n]= (1 -(abs(ref[n]))^2)*norm2[n-1] );
if(stop,return);
\\  B/A, al, be:
\\    dw/d theta =w[  alpha cot((theta-theta1)/2)+beta cot((theta-theta2)/2) -gamma  ]
  th1=Pi*(tp1+tp2)/2;w1=Nr*Chris(exp(I*th1));
  tans=tan( Pi*(tp2-tp1)/4 ); cots=1/tans;
 dw1=Nr*( Chris(exp(I*(th1+0.0001)))-Chris(exp(I*(th1-0.0001))))/(0.0002*w1)+ga;
al1=(dw1*tans+albe)/2;be1=(-dw1*tans+albe)/2;print1(" check alpha, beta= ",al1," ",be1);
th2=Pi*(-2+tp1+tp2)/2;w2=Nr*Chris(exp(I*th2));
dw2=Nr*( Chris(exp(I*(th2+0.0001)))-Chris(exp(I*(th2-0.0001))))/(0.0002*w2) +ga;
al2=(-dw2*cots+albe)/2;be2=(dw2*cots+albe)/2;print(" ;  ",al2," ",be2);
\\al1=(albe-(abs(cr2)^2-abs(cr1)^2)/albe)/2;be1=(albe+(abs(cr2)^2-abs(cr1)^2)/albe)/2;
print(" xi=",xi," i delta=",id1);
r1=w1/( abs(sin((th1-Pi*tp1)/2))^(2*al) *
        abs(sin((th1-Pi*tp2)/2))^(2*be) *
        exp(-ga*th1)   );
r2=w2/( abs(sin((th2-Pi*tp1)/2))^(2*al) *
        abs(sin((th2-Pi*tp2)/2))^(2*be) *
        exp(-ga*th2) );
rho1=log(r1/r2)/(2*Pi);rho2=ga-rho1;errho=r1/r2;
Astart=cr1/(al-rho1*I);Bstart=cr2/(be-rho2*I) ;
print("r1=",r1," r2=",r2, " rho=",rho1 , " A1/(alpha-rho i)=",Astart, " A2/(beta-(gamma-rho) i)=",Bstart," abs: ",abs(Astart)," ",abs(Bstart) );
psi1=-rho1*L12+2* arg(gamma(1+al+rho1*I)) ;psi11=be*Pi*(1-tp2+tp1)+psi1+ga*L12;
psi11=psi11-2*Pi*round(0.5*psi11/Pi);
psi2=-rho2*L12+2* arg(gamma(1+be+rho2*I)) ;psi12= -al*Pi*(1-tp2+tp1)+psi2+ga*L12;
psi12=psi12-2*Pi*round(0.5*psi12/Pi);
print(" logarithmes: ",log(Astart)," ",log(Bstart)," ",psi11," ",psi12    ); 
plot(th=Pi*tp2-2*Pi,Pi*tp2,Nr*Chris(exp(I*th)));

}

          GP/PARI CALCULATOR Version 2.2.8 (development CHANGES-1.887)
                i686 running cygwin (ix86 kernel) 32-bit version
               compiled: Jan 13 2004, gcc-3.3.1 (cygming special)
                (readline v4.3 enabled, extended help available)

                       Copyright (C) 2003 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and 
comes WITHOUT ANY WARRANTY WHATSOEVER.

Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.

   realprecision = 28 significant digits
   seriesprecision = 16 significant terms
   format = g0.28

parisize = 4000000, primelimit = 500000

reflecj2 unit circle gen Jacobi Tue 19 Mar 2013 15:15
 alpha=1.6000 beta=0.90000 , gamma=2; theta1,2/pi=-0.15000 0.27500
 Phi1(0)= -0.63105 - 0.31312*I , t=0.100000000000000000000000000000

    1 A1=-5.6514 + 2.8024*I A2=-2.7015  - 5.5964*I  relerr=10^4.6045 E-28
    2 A1=1.0605  + 1.4044*I A2= 0.24594 + 0.78057*I relerr=10^-1.3150
    4 A1=-1.6032 - 3.7685*I A2= 0.72380 - 1.1202*I  relerr=10^0.20871
    8 A1=2.7833  - 2.5577*I A2=-1.4613  + 0.20793*I relerr=10^-0.32309
   16 A1=2.7643  - 1.2968*I A2=-0.58543 - 0.23763*I relerr=10^-0.50384
   32 A1=2.0948  - 2.0138*I A2=-0.68737 - 0.78442*I relerr=10^-0.93469
   64 A1=1.5325  - 2.5385*I A2=-0.42672 - 0.90166*I relerr=10^-0.56313
  128 A1=1.0397  - 2.8077*I A2=-0.31776 - 0.98557*I relerr=10^-0.56055
  256 A1=0.78972 - 2.8517*I A2=-0.22935 - 0.98711*I relerr=10^-0.64465
  512 A1=0.66794 - 2.8593*I A2=-0.18537 - 0.99123*I relerr=10^-0.80112
 1024 A1=0.61129 - 2.8605*I A2=-0.16559 - 0.99087*I relerr=10^-1.5224
 2048 A1=0.58541 - 2.8612*I A2=-0.15641 - 0.99045*I relerr=10^-1.7537
 4096 A1=0.57276 - 2.8622*I A2=-0.15201 - 0.99067*I relerr=10^-2.0813
 8192 A1=0.56663 - 2.8629*I A2=-0.14989 - 0.99080*I relerr=10^-2.3460
16384 A1=0.56362 - 2.8633*I A2=-0.14885 - 0.99090*I relerr=10^-2.0060
32768 A1=0.56213 - 2.8635*I A2=-0.14833 - 0.99096*I relerr=10^-2.4234
 check alpha, beta= 1.6000 0.89998 ;  1.6000 0.90002
 xi=0.56702 - 1.9303*I i delta=4.1709 E-29 - 2.8812*I
r1=2934.6 r2=0.00064375 rho=2.4402 A1/(alpha-rho i)=0.92629 - 0.37698*I A2/(beta-(gamma-rho) i)=-0.56760 - 0.82341*I abs: 1.0001 1.0001
 logarithmes: 0.000059741 - 0.38651*I 0.000087415 - 2.1743*I -0.38733 -2.1739

7.4349977 |'''''''''''''''''''''''''''''''''''''''''''''''''''''''"x'''''|
          |                                                        :     |
          |                                                      "  :    |
          |                                                      :  "    |
          |                                                     :   :    |
          |                                                     :    :   |
          |                                                     x    :   |
          |                                                     :    x   |
          |                                                     :    :   |
          |                                                    :     :   |
          |                                                    :      :  |
          |                                                    _      :  |
          |                                                    :      x  |
          |                                                    :      :  |
          |                                                   :        : |
          |                                                   :        : |
          |                                                   :        _ |
          |                                                   "        : |
          |                                                   :         :|
          |      _x"""""xx__                                 :          :|
          |   _x"           "xx__                            _          x|
        0 __x",,,,,,,,,,,,,,,,,,,""xxxxx_____________________,,,,,,,,,,,,_
          -5.419247                                               0.863938
(15:16) gp > quit
Goodbye!

\end{verbatim} 


} % end Small

\renewcommand{\mylmark}{Grunbaum problem.}
\pagestyle{fancy}
\hypertarget{grunt}{ }
\section{The Gr\"unbaum-Delsarte-Janssen-Vries problem.  \label{grunl}} \  \\

\subsection{The problem\   }\   \\.

         \begin{minipage}[t]{5cm}
         \  \par

          \vspace*{-5mm}
\psset{unit=1cm}
\begin{pspicture}(5,5)
 \psline(0.0,2.5)(4.8,2.5)    \psline(2.5,0.1)(2.5,4.8)
 \psarc[linewidth=0.1,linecolor=red]{*-*}(2.5,2.5){2}{50}{310}
 \pscircle[linestyle=dashed, dash=3mm 3mm](2.5,2.5){2}
  \uput[40](3.7,4.0){$e^{i\pi\alpha}$}
  \uput[310](3.7,1.0){$e^{-i\pi\alpha}$}
  %\psdots[dotsize=0.15,dotstyle=o](12,2.5)
\end{pspicture}

         \end{minipage}
         \begin{minipage}[t]{10.7cm}
         \  \par

          \vspace*{-5mm}
          ``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 $\boldsymbol{G}_N$ of the weight $w=1$ on the circular arc
shown left. For all the entries of all the
 $\boldsymbol{G}_N^{-1}$ to be positive, it is necessary that all the
reflection coefficients $\Phi_n(0) >0$, $n=1,\dotsc,N$,
and the condition is sufficient:
from \eqref{recur}, all the $\Phi_n$'s have positive coefficients,
so does $\boldsymbol{L}^{-1}$, and $\boldsymbol{G}_N^{-1}$, from
\eqref{cholt} \cite[p.645]{DelGru}.
              \end{minipage}

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 $\boldsymbol{G}_N$, 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 $\boldsymbol{G}_N^{-1}$ 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.

\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

In order to remove the insufferable tension, here is some numerical
evidence, where the relevant $\Phi_n(0)$'s are computed either
with the all-purpose algorithm  of p.~\pageref{prpari}, or from
the formula \eqref{jacobg2} which will be further worked in
\eqref{eqgrun}:

\noindent
\ $\alpha \qquad  \Phi_1(0)\qquad   \Phi_2(0)\qquad   \Phi_3(0) \qquad   \Phi_4(0)
 \qquad    \Phi_5(0)\qquad \  \Phi_6(0)\qquad  \Phi_7(0)\qquad  \  \Phi_8(0)
 \qquad  \Phi_9(0)\qquad   \Phi_{10}(0)\qquad   \Phi_{11}(0)  $

\begin{verbatim}
0.1 0.109292 0.117289 0.124056 0.129708 0.134380 0.138214 0.141343 0.143889 0.145959 0.147640 0.149008
0.2 0.233872 0.258015 0.274428 0.285306 0.292452 0.297160 0.300299 0.302429 0.303905 0.304953 0.305715
0.3 0.367883 0.406603 0.427202 0.438032 0.443863 0.447146 0.449098 0.450327 0.451143 0.451711 0.452123
0.4 0.504551 0.550672 0.569686 0.577835 0.581659 0.583657 0.584812 0.585537 0.586023 0.586365 0.586616
0.5 0.636619 0.681477 0.695821 0.701070 0.703373 0.704563 0.705258 0.705700 0.706000 0.706213 0.706369
0.6 0.756826 0.793314 0.802464 0.805497 0.806813 0.807503 0.807911 0.808173 0.808352 0.808479 0.808573
0.7 0.858393 0.882688 0.887579 0.889138 0.889826 0.890192 0.890410 0.890551 0.890647 0.890716 0.890766
0.8 0.935489 0.947570 0.949601 0.950254 0.950547 0.950704 0.950798 0.950859 0.950901 0.950930 0.950952
0.9 0.983631 0.986853 0.987333 0.987491 0.987563 0.987601 0.987625 0.987639 0.987650 0.987657 0.987662

? quit
Good bye!

C:\calc\pari>exit    Script completed Wed Oct 13 14:09:47 1999
\end{verbatim}

\medskip

All  the $\Phi_n(0)$'s  seem indeed to be positive. Moreover,
it is known that $\Phi_n(0) \to \sin(\pi\alpha/2)$ when $n\to \infty$
\cite[etc.]{Golal1,Golakh,Golal2}. N.B.: the capacity of the support
is $\cos(\pi\alpha/2)$ \cite{ThDe}.  A possible strategy, although not
extremely elegant, is to look for a more accurate asymptotic behaviour,
so to prove the positivity for $n>$ some finite $n_0$, and to show
positivity of the finite number of remaining $\Phi_n(0)$, $n\leqslant
n_0$\dots But perhaps we may learn something from a (nonlinear) recurrence
relation between the $x_n=\Phi_n(0)$:

\subsubsection{A recurrence relation for the $\Phi_n(0)$'s}.

\eqref{jacobg2} becomes
\begin{equation}  \label{eqgrun}
 (n+1) x_{n+1} =
\frac{ 2(x_1x_0 +\dotsb + x_n x_{n-1}) +2n\cos\pi\alpha }
   {1-x_n^2}\, x_n
  -(n-1)  x_{n-1},
\end{equation}
with $x_0=1$ and $x_1= \displaystyle{ \frac{\sin\pi\alpha}
                                           {(1-\alpha)\pi} }$.

          Question: are all the  $x_n$'s positive??

The most elegant proof should establish that the sequence of the
$\Phi_n(0)$'s is \emph{\textbf{increasing}}, as suggested by the
numerical tests, but how to achieve that?

OK, it will be achieved in section \ref{grfinal}, p.~\pageref{grfinal}!

 \medskip

\subsubsection{Trying to solve \eqref{eqgrun}}.

From numerical runs of \eqref{eqgrun}, the following empiric asymptotic
formula:
$$
  x_n = \sin(\pi\alpha/2) - \dfrac{\cos^2(\pi\alpha/2)}
                                  {8n^2 \sin(\pi\alpha/2)}
      +O(n^{-4}).
%   \dfrac{ (\sigma^{-1}-\sigma)(13\sigma^{-2}-9) }{128n^4}
$$

quite in agreement with formula (56) of Golinskii, Nevai, and
Van Assche \cite{Golal1} stating that  \\
$\displaystyle{ \Phi_n(0) =
   \sin(\pi\alpha/2)  + (-1)^n \dfrac{\delta \cos^2(\pi\alpha/2)}{2n}
  }$\\
$\displaystyle{ \hspace*{55pt}
   - \dfrac{\cos^2(\pi\alpha/2)}
                                  {8n^2 \sin(\pi\alpha/2)}
   \left[ 1+\delta^2 \sin^2(\pi\alpha/2)-4\gamma^2
          +2(-1)^n \delta\sin(\pi\alpha/2)(1+\delta+2\gamma
         -\sin\pi\alpha/2)\right]
      +O(n^{-3})  }$ for the weight function
$(\cos\pi\alpha-\cos\theta)^\gamma |\cos\theta/2|^\delta \sin\theta/2$
on $(\pi\alpha,2\pi-\pi\alpha)$. The formula is based on values of
Jacobi polynomials, as the problem is reduced to the
weight $(\cos\pi\alpha-x)^\gamma (1+x)^{(\delta-1)/2}$ on the real
interval $(-1,\cos\pi\alpha)$. Unfortunately, the same technique
would lead here to $(1-x^2)^{-1/2}$, which is not related to
known real line orthogonal polynomials. Actually, we are struggling here like
mad to make these polynomials known!!!


 \medskip

What can be shown from \eqref{eqgrun} are -probably useless-
expansions in powers of  $\alpha$:

$$
 x_n = \alpha +n\alpha^2 -n^2(\pi^2-6)\alpha^3/6
  -[2(\pi^2-9)n^3+\pi^2 n]\alpha^4/18 +\dotsb
$$


 \medskip

\subsubsection{First exploration of the solutions.}

Making some numerical runs of \eqref{eqgrun} with a definite precision
 leads to troubles in the long run:

{\Small \begin{verbatim}
ubasic
 20   ' grunbaum
 25   print=print+"grunbaum.l"
 30   point 2:print point
 40   A=0.25
201   X0=1:X1=sin(A*#pi)/((1-A)*#pi):C=cos(A*#pi):N=100
202   ' recurrence xi
205   Xi=X1:for M=1 to N-1:Xp1=(2*(Xi+M*C)*X1/(1-X1^2)-(M-1)*X0)/(M+1)
207   Xi=Xi+X1*Xp1
208   print M;" ";X1
210   X0=X1:X1=Xp1:next M
                                          POINT 20:N=1000
Words for fractionals 2          Words for fractionals 20
 (Decimals for display 9)        (Decimals for display 96)

   1   0.300105438    0.30010543871903535651839973033558019422152471665174020852381249041048624357301544833269425925
   2   0.332187784    0.33218778472931665288245731180806017658179317346382917680179154875483312187176135233108180410
   3   0.351521621    0.35152162294422913023077917777756810454770623402302924991862850016080478323045805931585155286
   4   0.362911325    0.36291132841133703559148099728672089327661037086490658695601024932418242378075626620349744161
   5   0.369642298    0.36964230414617100082455879714538699957111424116368282234713049524225202661075914125700868793
...
  10   0.379705895    0.37970610732170667627690919215696544125235169168336493512833090267521617687995515802880861153
...
  15   0.38140294     0.38141260538836773527973794641115525299704926868489141173147263080761372750575709626124897350
...
  20   0.381507291    0.38197709531954745708872462188088469085131034342993947565188373696463195928218581338364145094
...
  25   0.359328655    0.38223363113152422566539446313925294718174035952089916592661769607185779367707722953576139206
  26   0.33380412     0.38226783587997069754501344133199300414889805882502962439425175103144000554173338854464982793
  27   0.283680342    0.38229827224178648248897756342088539583120704553237426022769614013079833313901849890267256446
  28   0.19583214     0.38232547558253590090583512789915126966979241634998984803814280158620168118659115728618094801
  29   0.06894989     0.38234988919997318707066772121869458362935550034600147036957873706274887837963850442405177831
  30  -0.069915877    0.38237188274696440864757885812660868563700744456975329172516834746939873310358047138879475398
  31  -0.178450092    0.38239176647877182260828221401760111929952098525858090585509367995015773326020324459415969305
  32  -0.232336628    0.38240980237615231100928484254239109991190175909042196733162348821472604125307250818526323131
  33  -0.228388816    0.38242621290583003217892565150355728581237614963997901075587048950143154352696536133052332830
  34  -0.169224556    0.38244118797653500191397562375656644882021538492111033982941207639018837735166389285083144606
  35  -0.06452273     0.38245489050425285349841140056469392251245493330925299372873853561638663630740083712202534368
  36   0.056045639    0.38246746089634594236881186435008946389041960246681232513269030025603005075975327201287546768
  37   0.150789831    0.38247902068856021809542526624567517164608107160067233419768268463102476757695614041602471949
  38   0.192559004    0.38248967551332896042000271238809590502281674099493684721299554818511437818292493373748237581
  39   0.174702805    0.38249951753651672968156468259575246726644840148792253730571079200644531605083678549433684531
  40   0.102928098    0.38250862746884118758802777537257439941615157615474177408716915704407258161586912556317338089
  41  -0.00150102     0.38251707623486851027186672908137167366417593635412494409608333257617189561148172785924684577
  42  -0.100396835    0.38252492636470828130111112092949665986452357740717420125401438705769567876753379880205431437
  43  -0.158318881    0.38253223315990433815802515153331897925942421237549662074542351498305157692084071705241339428
  44  -0.159476973    0.38253904567449028763179388703498779736063673103023585505165204569309473098715206680150057840
  45  -0.105601602    0.38254540754399199970670582104074210039074420829458282464158064325046969281025221079348111888
...
 270                  0.382744697429545214205569710252972145011105886234432141000123131448570516044686415992395419733862
 271                  0.382825171593994643573964872452942989072548889230770044858518762152261603290217814319805929378455
 272                  0.383005115178603908297729659252774132437806148545073750205937561125762025283647943345680819959853
 273                  0.383407693123411762781081308864926981882854078652404908016236701849905142868079449206644266862794
 274                  0.384309260277383199740966886773247087076501732863160406974422029456305768323279768754385346890067
 275                  0.386332645419367229398555894946266358286055717424823017373820224935700131437280929943159426794277
 276                  0.390895472236394426226318869995802256570317686718730469630040862009891598218750444520423067016592
 277                  0.401296382804890155001167568833596412927958758090652961405462166619546285423623051527368852850507
 278                  0.42559870839830301078566465334668442878147630940739219473815370962571400040489793264993458763346
 279                  0.485817896654772081180182592943949119791027973866130902939426856629992033664413377027008310128195
 280                  0.659591632013078076495279692160117728029109732045336735815917620092431931955707678680374202692895
 281                  1.505981932218524693248679285080550185985168274952391079436473794522035120244896473701415524363117
 282                 -2.684730905253404076857008359913051690030802859469963700444726916544630820508025116920734662853239
 283                 -0.769027600578075813998542996552764162552927499420269710407386875385048969210487027191600119423235
 284                 -0.520646107978621653729441890047411402229182714785934046677082857067925770404980285328513467963788
 285                 -0.446940562352144823593956486331798507581505504129963913856443519445741931859240726200496361835672
 286                 -0.429940300494259475782118920216492570518288717983891969828601753099268185005737512251683829560596
 287                 -0.450631234197750035739348291117722745560011099125023026520076929306020186970129527878417319998356
 288                 -0.532189975665453428670239748584646710962486153427092388646093042830706109578642228439649984829031
...
\end{verbatim} }

That happens because the general solution of \eqref{eqgrun}, keeping only $x_0=1$, is
related to the piecewise constant weight function $w=A$ on
$\pi\alpha <\theta < \pi(2-\alpha)$, and $w=B$ on $-\pi\alpha <\theta
<\pi\alpha$.
The overwhelming majority of solutions have $A$ and $B\neq 0$, are
related to a Szeg\H o weight if $A$ and $B>0$,
 and have therefore $x_n\to 0$ when
$n\to\infty$. Taking $x_1$ just a trifle above or below the
ideal value $\sin(\pi\alpha)/((1-\alpha)\pi)$ will therefore end up
with unsatisfactory values of $x_n$ for large $n$. More precisely,
$x_1$ is the $\Phi_1(0)$ related to the $(A,B)$ weight function:
$$ x_1 = -\dfrac{\mu_1}{\mu_0} =
  -\dfrac{ \displaystyle{A\int_{\text{left}}\cos\theta\,d\theta
                        +B\int_{\text{right}}\cos\theta\,d\theta  }}
         { \displaystyle{A\int_{\text{left}}\,d\theta
                        +B\int_{\text{right}}\,d\theta  }}
 =\dfrac{(1-B/A)\sin\pi\alpha}
        {\pi[1-\alpha+(B/A)\alpha]}. $$
Values of $x_1$ smaller than $\sin\pi\alpha/((1-\alpha)\pi)$
correspond indeed to a Szeg\H o weight (as long as
$x_1> -\sin\pi\alpha/(\pi\alpha)$); larger values of $x_1$ does
not even correspond to positive weights and will have some
$x_n\notin (-1,1)$!

These numerical difficulties show that a direct study of the
recurrence relation \eqref{eqgrun}, trying to get an induction
of the form ``if some inequality is valid for $x_n$, it is also
valid for $x_{n+1}$'', will lead nowhere.

\medskip


{\Small \begin{verbatim}

Script V1.1 session started Tue Mar 18 14:45:45 2003

type grunbg.gp

{
/*
  grunbg.gp    :  launch gp and make  \r grunbg

  Reflection coefficients for Grunbaum- Delsarte et al.  problem

 by modified iteration,   January 2003.

  weight= (cos pi a - cos theta)^(beta)  on pi a < theta < 2pi -pi a
*/
default(format,"g2.7");

for(bigN=50,50,print("N= ",bigN);
for(ia=1,1,a=0.25*ia;
\\  a=2*asin(a)/Pi;
print1("alpha=",a," ");    sa=sin(Pi*a/2);
print1(" sigma=sin pi a/2= ",sa);
ca=cos(a*Pi);
print(" cos pi a= ",ca);
\\ vector of reflection coeff.
v=vector(bigN+2,k,0);
print("\\psline(0,-5)(0,0)(10,0)(0,0)(0,5)(10,5)(10,-5)(0,-5)");
v[1]=1;
for(iba=1,25,
   ba=10*(iba-20);
   if(iba<=20,ba=0.25*(iba-10));
   if(iba<=10,ba=0.01*(iba-5));
v[2]=(1-ba)*sin(a*Pi)/(Pi*(1-a+ba*a)); \\print(v[2]);
print("\\psline(0,",5*v[2],")(-0.25,",5*v[2],")");
print("\\uput[180](-0.25,",5*v[2],"){$",ba,"$}");
print("\\psline%");
som=0; kpr=1;
for(n=1,bigN,
   som=som+v[n+1]*v[n];
  v[n+2]=(  2*(som+n*ca)*v[n+1]/(1-v[n+1]^2) -(n-1)*v[n] )/(n+1);
  if(kpr==1,print1("(",0.2*n,",",5*v[n+1],")"));
  if(divrem(n,5)[2]==0,print("%"));
  if(abs(v[n+1])>1.2,kpr=0);
   );
)))
}

C:\calc\pari>gp-sta
                  GP/PARI CALCULATOR Version 2.1.3 (released)
                      Copyright (C) 2000 The PARI Group PARI/GP
 \r grunbg
N= 50
alpha=0.250000003  sigma=sin pi a/2= 0.382683438 cos pi a= 0.707106791

-0.200000005 0.385849855
-0.100000002 0.341499297
 0           0.300105443
 0.100000001 0.261382160
 0.200000002 0.225079082
\end{verbatim}    }

 \psset{unit=1cm}
 \begin{pspicture}(-4,-5)(10,5)

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%
%
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%
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\psline(0,1.520601)(-0.25,1.520601)
\uput[180](-0.25,1.520601){$-0.010000000$}
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\uput[180](-0.25,1.500527){$ 0$}
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\psline(0,0.643083092)(-0.25,0.643083092)
\uput[180](-0.25,0.643083092){$0.500000007$}
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\psline(0,0.300105443)(-0.25,0.300105443)
\uput[180](-0.25,0.300105443){$0.750000011$}
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\psline(0,0)(-0.25,0)
\uput[180](-0.25,0){$1.000000$}
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(5.200000,0)(5.400000,0)(5.600000,0)(5.800000,0)(6.000000,0)%
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\psline(0,-0.264798924)(-0.25,-0.264798924)
\uput[180](-0.25,-0.264798924){$1.250000$}
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\psline(0,-0.500175746)(-0.25,-0.500175746)
\uput[180](-0.25,-0.500175746){$1.500000$}
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\psline(0,-0.710776060)(-0.25,-0.710776060)
\uput[180](-0.25,-0.710776060){$1.750000$}
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\psline(0,-0.900316342)(-0.25,-0.900316342)
\uput[180](-0.25,-0.900316342){$2.000000$}
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\uput[180](-0.25,-1.071805){$2.250000$}
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\uput[180](-0.25,-1.227704){$2.500000$}
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\uput[180](-0.25,-3.116479){$10$}
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\uput[180](-0.25,-3.718697){$20$}
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\uput[180](-0.25,-3.955935){$30$}
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\psline(0,-4.082829)(-0.25,-4.082829)
\uput[180](-0.25,-4.082829){$40$}
\psline%
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\end{pspicture}

{\Small \begin{verbatim}
? quit
Good bye!

C:\calc\pari>exit

Script completed Tue Mar 18 14:54:01 2003
\end{verbatim}}

\medskip

 Check that $x_n \to \sigma\neq 0$ in \eqref{eqgrun} yields
  $\sigma=\pm \sin(\pi \alpha/2)$. There is
 also an alternating solution $(-1)^n x_n \to \sigma$ which yields
  $\sigma=\pm \cos(\pi\alpha/2)$,
 corresponding to the arc of circle joining $e^{i\pi\alpha}$ to $e^{-i\pi\alpha}$ and
 containing $+1$.  These two exceptional solutions correspond to $B=0$
and $A=0$.
So, we are looking for the only solution of \eqref{eqgrun}
satisfying $x_n\to \sin \pi\alpha/2$ when $n\to\infty$, and we are
wondering if all the $x_n$'s are positive.



Of course, each $x_n$ is a rational function of $x_1$:
$$ x_2=\dfrac{x_1(\cos\pi\alpha+x_1)}{1-x_1^2}\;,
   x_3=\dfrac{x_1(4\cos^2\pi\alpha-1+6\cos\pi\alpha x_1
       +(4-\cos^2\pi\alpha)x_1^2)}{3(1-(2+\cos^2\pi\alpha)x_1^2
       -2\cos\pi\alpha x_1^3) }\;\dotsc
$$

Here is an information on the variation of each $x_n$ with respect
to $x_1$, while $x_n$ remains positive:

We  look at the influence of $x_1$ on $x_n$,
i.e., at $\partial x_n/\partial x_1$, which we write
$\dot{x}_n$.

\subsubsection{Influence of $x_1$ on $x_n$\label{influ1n}}

\noindent\textbf{Proposition.}  \textsl{If $x_1, x_2,\dotsc, x_n
 \in (0,1)$, then
$\dot{x}_n >0$.}

Derivating \eqref{eqgrun} for $i=1,2,\dotsc, n-1$:
$$ (i+1)\dot{x}_{i+1} = \dfrac{2x_i}{1-x_i^2}\left[\dot{x}_1
 +\sum_{j=1}^{i-1}(x_j\dot{x}_{j+1}+x_{j+1}\dot{x}_j)\right]
+2\dfrac{\text{num.}}{(1-x_i^2)^2}(1+x_i^2)\dot{x}_i
 -(i-1)\dot{x}_{i-1}, $$
where ``num.'' is the numerator in the right-hand side of
\eqref{eqgrun}. We now use precisely this equation \eqref{eqgrun}
to replace ``2num.''$/(1-x_i^2)$ by $((i+1)x_{i+1}+(i-1)x_{i-1})/x_i$:

$$ (i+1)\dot{x}_{i+1} = \dfrac{2x_i}{1-x_i^2}\left[\dot{x}_1
 +\sum_{j=1}^{i-1}(x_j\dot{x}_{j+1}+x_{j+1}\dot{x}_j)\right]
+\dfrac{(i+1)x_{i+1}+(i-1)x_{i-1}}{x_i(1-x_i^2)}(1+x_i^2)\dot{x}_i
 -(i-1)\dot{x}_{i-1}, $$

All the $\dot{x}$'s are positive: if true up to $i$,
$$ (i+1)\dot{x}_{i+1} > \dfrac{(i+1)x_{i+1}+(i-1)x_{i-1}}{x_i}
   \dot{x}_i-(i-1)\dot{x}_{i-1},$$

$$ (i+1)x_{i+1}\left[ \dfrac{\dot{x}_{i+1}}{x_{i+1}} -
                      \dfrac{\dot{x}_{i}}{x_{i}} \right] >
   (i-1)x_{i-1}\left[ \dfrac{\dot{x}_{i}}{x_{i}} -
                      \dfrac{\dot{x}_{i-1}}{x_{i-1}} \right] .
$$
whence indeed $\dot{x}_{i+1}>0$ too.   \qed

Incidentally, we also have $\dot{x}_i/x_i$ increasing with $i$,
as also
$   i(i-1)[ x_{i-1}\dot{x}_{i} - x_i \dot{x}_{i-1}]$.


Remark that, if $x_1$ is very small,
$$
   x_n \sim   \dfrac{\sin n\pi\alpha}{n\,\sin\pi\alpha}\,x_1,
$$
seeing that \eqref{eqgrun} is
$(n+1)x_{n+1}\sim 2nx_n\cos\pi\alpha -(n-1)x_{n-1}$ at first order.
So, the first $x_n$'s are positive as long as $n\alpha <1$ if $x_1$
is a small positive number. And when $x_1$ increases, the length of
the initial sequence of positive $x_n$'s must increase too, and we hope
to have the whole infinite sequence at $x_1=\sin\alpha\pi/((1-\alpha)\pi)$.
If this is true, what happens when $x_1$ still increases a little bit,
may we ask? According to the Proposition above, all the $\dot{x}_n$,s
are positive, so that all the $x_n$'s will increase by an
infinitesimal amount and still keep a valid value $<1$. What about the
unicity of the positive solution? Answer: for any small but nonzero
increase of $x_1$, there is an $n$ large enough where $x_n$
will increase so much as to be $>1$. A strong hint is the fast
exponential increase of $\dot{x}_n$ with $n$:


{\Small \begin{verbatim}

ubasic

 20 ' grunbaum
 25 print=print+"grunbaum.l"
 30 point 21:word 74
 40 A=0.25
201 X0=1:X1=sin(A*#pi)/((1-A)*#pi):Dx0=0:Dx1=1:C=cos(A*#pi):N=100
202 ' recurrence xi  and dxi/x1
205 Xi=X1:Dxi=1:for M=1 to N-1:Xp1=(2*(Xi+M*C)*X1/(1-X1^2)-(M-1)*X0)/(M+1)
206 Dxp1=((2*DXi*X1+((M+1)*Xp1+(M-1)*X0)*(1+X1^2)*Dx1/X1)/(1-X1^2)-(M-1)*Dx0)/(M+1)
207 Xi=Xi+X1*Xp1:Dxi=Dxi+Dx1*Xp1+X1*Dxp1
208 print M;" ";X1;" ";Dx1
210 X0=X1:X1=Xp1:Dx0=Dx1:Dx1=Dxp1:next M
\end{verbatim}

\begin{tabular}{c|rrrrrrrrrr}
$n$         &    1   &    2   &   3    &   4    &    5    &    6    &    7    &     8   &     9    &     10    \\
$x_n$       & 0.3001 & 0.3322 & 0.3515 & 0.3629 &  0.3696 &  0.3737 &  0.3762 &  0.3779 &   0.3789 &   0.3797  \\
$\dot{x}_n$ & 1.0000 & 1.6558 & 2.9312 & 5.5059 & 10.7885 & 21.7452 & 44.6731 & 93.0072 & 195.5287 & 414.1064
\end{tabular}  }

The lesson here is that the above Proposition is useful only
with respect to a finite sequence.

\medskip

More on the nonpositivity of the Szeg\H o's solutions:

When $A$ and $B>0$, we know that $x_n\to 0$ and $\xi_n$ has a bounded
limit, hence \eqref{eqgrun} ``looks like''
\\ $(n+1)x_{n+1} = 2n\cos(\pi\alpha)x_n-(n-1)x_{n-1}$, and we expect
a $\sin(n\pi\alpha)$ behaviour as seen before. What is true is that
in $\displaystyle{  \begin{bmatrix} x_n \\ x_{n+1} \end{bmatrix} =
     A_n
                  \begin{bmatrix} x_{n-1} \\ x_n \end{bmatrix} }$,
the matrix $A_n$ is close to
 $\displaystyle{  A_\infty:=
 \begin{bmatrix} 0 & 1 \\ -1 & 2\cos\pi\alpha \end{bmatrix} }$
for large $n$, so that, for a given $p$,
 $\displaystyle{  \begin{bmatrix} x_{n+p} \\ x_{n+p+1} \end{bmatrix} =
     A_{n+p}A_{n+p-1}\dotsb A_{n+1}
                   \begin{bmatrix} x_n \\ x_{n+1} \end{bmatrix} }$,
with
  $  A_{n+p}A_{n+p-1}\dotsb A_{n+1}   $ close to
\\ $\displaystyle{  A_\infty^p=  \dfrac{1}{\sin\pi\alpha}
 \begin{bmatrix} -\sin(p-1)\pi\alpha  & \sin p\pi\alpha \\
  -\sin p\pi\alpha & \sin(p+1)\pi\alpha \end{bmatrix} }$.
So, $x_{n+p}$ is close to $C_n \sin(p\pi\alpha+\zeta_n)$,

\medskip

We shall need a finite, non infinitesimal, version of the
Proposition:

\subsubsection{Difference of two solutions\label{influ2n}}

\noindent\textbf{Proposition.}  \textsl{If $x'_1, x'_2,\dotsc$ and
                                           $x''_1, x''_2,\dotsc$ are
two solutions of \eqref{eqgrun}, with $0<x'_i<1$ and $0<x''_i<1$
for $i=1,2,\dotsc,n$, then $x'_i=x''_i$ or all the $x''_i-x'_i$ have
the same sign for $i=1,2,\dotsc,n$. }

Indeed, let us write \eqref{eqgrun} as
\begin{equation} \label{recCi}
  (i+1)x_{i+1}= C_i x_i -(i-1)x_{i-1}, i=1,2,\dotsc
\end{equation}
where $C_i= \displaystyle{ 2\dfrac{x_1+x_2x_1+\dotsb +x_i x_{i-1}
  +i\cos\pi\alpha}{1-x_i^2} }$.

Suppose $x''_1>x'_1$, and that $x''_j>x'_j$ for $j=1,2,\dotsc,i$.

\begin{equation*}\begin{split}
 (i+1)[x''_{i+1}-x'_{i+1}] &=
    C''_i[x''_i-x'_i] +[C''_i-C'_i]x'_i -(i-1)[x''_{i-1}-x'_{i-1}] \\
  &> C''_i[x''_i-x'_i]  -(i-1)[x''_{i-1}-x'_{i-1}] \  \   \text{as \ }
                    C''_i>C'_i \\
  &> \displaystyle{\dfrac{(i+1)x''_{i+1}+(i-1)x'_{i-1}}{x''_i} }
  [x''_i-x'_i]
    -(i-1)[x''_{i-1}-x'_{i-1}] \  \   \text{from \ } \eqref{recCi}
\end{split}\end{equation*}
and $x''_{i+1}>x'_{i+1}$ follows. \qed


\smallskip

As we still do not know if the $\Phi_n(0)$'s are all positive, we
shall try to build a positive solution of \eqref{eqgrun}.

\begin{flushright}
  \textsc{Chico}: Go to the house next door. \\
  \textsc{Groucho}: That's great. Suppose there isn't any house\\
         next door. \\
  \textsc{Chico}: Well, then of course we gotta build one!\\
    from \textsl{Animal Crackers}\footnote{considered by
   J.~Adamson, in \textsl{Groucho, Harpo, Chico and sometimes Zeppo},
   Simon \& Schuster 1973 = Pocket Book 1976, as an overestimated
   sample of the Marx Brothers humor. Perhaps a better instance
   is the much repeated scene where a handshake produces a flood
   of knives falling from Harpo sleeves\dots when one should have
   expected mere spoons.}.
\end{flushright}


\subsubsection{It seems that a smart idea}
 is to look at \eqref{eqgrun} as a
relation between  positive sequences.
We use \eqref{eqgrun} to extract $x_n$ through the
only positive root of

\begin{equation} \label{griter}
\dfrac{x_n^{-1}-x_n}{2} =
\dfrac{ x_1x_0 +\dotsb + x_n x_{n-1} +n\cos\pi\alpha }
       {(n+1)x_{n+1}+(n-1)  x_{n-1}},
  n=1,2,\dotsc   \quad x_0=1,
\end{equation}
recomputing new estimates of $x_1, x_2,\dots, x_N$ from old
ones in the right-hand side of \eqref{griter}, with $x_0=1$
and $x_{N+1}=$ the limit $\sin(\pi\alpha/2)$. The first estimate
of the sequence $\{x_n\}$ may simply be $x_n=\sin(\pi\alpha/2)$
for all $n>0$. Remark that, as a bonus, $x_n<1$ if all the $x_i$'s
on the right-hand side are positive.

It works!! Here is a test with $\alpha=0.3$, applying first
\eqref{eqgrun} directly with $x_1=(\sin \alpha\pi)/((1-\alpha)\pi)=
0.367883098...$, and using several iterations of \eqref{griter}
with $x_n=\sin(\alpha\pi/2)=0.453990607...$, $n=1,2,\dotsc$
 as starting sequence:


{\Small \begin{verbatim}

{
/*
  reflecjg.gp    :  launch gp and make  \r reflecjg

  Reflection coefficients for Delsarte et al. and Grunbaum problem

*/
default(realprecision,50);
flone=1.0;pr=precision(flone);ep=10^(-pr);
default(format,"g1.6");
N=20;

for(ia=3,3,alpha=0.1*ia;print1(alpha," ");ca=cos(alpha*Pi);
\\ vector of reflection coeff.
ref=vector(N+1,k,0);
ref[1]=1;ref[2]=sin(alpha*Pi)/(Pi*(1-alpha));
xi=ref[2];
 for(n=1,N-1,
  print1(ref[n+1]," ");
  ref[n+2]=(  ref[n+1]*(2*xi+2*n*ca)/(1-(ref[n+1])^2)
             -(n-1)*ref[n]     )/(n+1);
  xi=xi+ref[n+1]*ref[n+2]
    ) ;
print(" ");

   sa=sin(Pi*alpha/2);    print(" sin pi alpha/2= ",sa);
  \\  par iteration de suites
   newref=vector(N+1,k,0);
   for(n=2,N,ref[n]=sa);ref[1]=1;
    for(ns=1,99,   newref[1]=1; residm=0; xi=0;  print(ns," ");
     for(n=2,N-1,
            print1(ref[n]," ");
             xi=xi+ref[n-1]*ref[n];
      resid=
      ref[n+1]-(  ref[n]*(2*xi+2*(n-1)*ca)/(1-(ref[n])^2)
                        -(n-2)*ref[n-1]     )/n  ;
         residm=max(residm,abs(resid));
         newref[n]=
         (xi+(n-1)*ca) / (  n*ref[n+1]+(n-2)*ref[n-1] )  ;
         newref[n]= 1/( sqrt(  1+newref[n]^2 )+newref[n]  );
        ) ;  print(" ",residm);
       for(n=2,N-1,ref[n]=newref[n]);
       );
 );
}

Script V1.1 session started Thu Dec 19 11:38:53 2002

C:\calc\pari>gp-2-1
                  GP/PARI CALCULATOR Version 2.1.0 (released)
             ix86 running Windows 3.2 (ix86 kernel) 32-bit version
              (readline v4.0 enabled, extended help not available)

                       Copyright (C) 2000 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and
comes WITHOUT ANY WARRANTY WHATSOEVER.

? \r reflecjc
alpha= 0.3000000  sin pi alpha/2= 0.453990607
\end{verbatim}

\noindent
\ iter. $\hspace*{23pt}  \Phi_1(0)\hspace*{23pt}   \Phi_2(0)\hspace*{23pt}   \Phi_3(0) \hspace*{23pt}   \Phi_4(0)
 \hspace*{23pt}    \Phi_5(0)\hspace*{23pt} \  \Phi_6(0)\hspace*{23pt}  \Phi_7(0)\hspace*{23pt}  \  \Phi_8(0)
 \hspace*{23pt}  \Phi_9(0)  $


\begin{verbatim}
   0.367883098 0.406603001 0.427202438 0.438032120 0.443863125 0.447146049 0.449098425 0.450327477 0.451143664
\end{verbatim}

\noindent
\ iter. $\hspace*{33pt}  x_1\hspace*{33pt}   x_2\hspace*{33pt}   x_3 \hspace*{33pt}   x_4
 \hspace*{33pt}    x_5\hspace*{33pt} \  x_6\hspace*{33pt}  x_7\hspace*{33pt}  \  x_8
 \hspace*{33pt}  x_9  $

\begin{verbatim}
1  0.453990607 0.453990607 0.453990607 0.453990607 0.453990607 0.453990607 0.453990607 0.453990607 0.453990607
2  0.374625426 0.411055959 0.424578085 0.431625072 0.435948566 0.438871514 0.440979521 0.442571680 0.443816684
3  0.368965781 0.404493014 0.423836837 0.433240534 0.438634789 0.442072607 0.444425731 0.446121465 0.447391852
4  0.366110087 0.404880268 0.424682308 0.435405794 0.441214990 0.444727398 0.447021691 0.448606494 0.449748319
5  0.367214055 0.405488905 0.426180133 0.436916193 0.442878899 0.446324018 0.448484612 0.449923774 0.450926738
...
10 0.367870659 0.406611437 0.427205887 0.438050902 0.443884096 0.447176753 0.449133606 0.450367787 0.451186558
...
20 0.367883024 0.406603045 0.427202390 0.438032144 0.443863094 0.447146049 0.449098403 0.450327458 0.451143670
...
30 0.367883097 0.406603002 0.427202438 0.438032121 0.443863126 0.447146052 0.449098433 0.450327496 0.451143714
...
50 0.367883098 0.406603001 0.427202438 0.438032121 0.443863127 0.447146052 0.449098433 0.450327497 0.451143714

? quit   Good bye!   C:\calc\pari>exit    Script completed Thu Dec 19 11:39:56 2002

\end{verbatim}        }

Hmm, we should now show that \eqref{griter} leads to a
\emph{\textbf{contraction}} of positive sequences, so to allow a
proof of a unique fixed point. We can see \eqref{griter} as
$$
  \boldsymbol{x} = \boldsymbol{F}(\boldsymbol{x}),
$$
acting on positive sequences $\boldsymbol{x}=\{x_1,x_2\dotsc\}$,
with $F_n(\boldsymbol{x})=\sqrt{
      A_n(\boldsymbol{x})^2+1 }-
      A_n(\boldsymbol{x}) $, and where
$A_n(\boldsymbol{x}) $ is the right-hand side of
\eqref{griter}.

We estimate the contraction in the $\|\ \|_\infty$ norm by looking
at the sum of all the $|\partial F_n/\partial x_i|$. For $i=1,2,\dotsc,
n-2$ and $i=n$, one finds
$\displaystyle{  \dfrac{\partial A_n}{\partial x_i} =
   \dfrac{x_{i-1}+x_{i+1}}{(n+1)x_{n+1}+(n-1)x_{n-1}} }$. For
$i=n\pm 1$, one must subtract
$\displaystyle{
 (n\pm 1)  \dfrac{ \text{num.}}{[(n+1)x_{n+1}+(n-1)x_{n-1}]^2 } }$,
where ``num'' is the numerator of the right-hand side of \eqref{griter}.
The sum of absolute values is bounded by
$$
  \dfrac{  1+2x_1+2x_2+\dotsb +2x_{n-2} +x_{n-1}+x_n
         +n(F_n^{-1}-F_n) }
          {(n+1)x_{n+1}+(n-1)x_{n-1}},
$$
having introduced $F_n^{-1}-F_n$ from the left-hand side of
\eqref{griter}. For the derivatives of $F_n$, we multiply by
$dF_n/dA_n = -F_n/\sqrt{A_n^2+1} = -2F_n/(F_n^{-1}+F_n)$. The sum of
the
 $|\partial F_n/\partial x_i|$ is therefore bounded by
$$
 2F_n \dfrac{  1+2x_1+2x_2+\dotsb +2x_{n-2} +x_{n-1}+x_n
         +n(F_n^{-1}-F_n) }
          {(F_n^{-1}+F_n)[(n+1)x_{n+1}+(n-1)x_{n-1}]}.
$$
For large $n$, when most of the $x_i$'s and $F_n$ itself are close
to the limit $\sigma$, we find asymptotically
$\displaystyle{ 2\sigma \dfrac{ 2n\sigma +n(\sigma^{-1}-\sigma)}
                    {(\sigma^{-1}+\sigma)2n\sigma}  }$
which is\dots 1!

\smallskip

If the positive sequence $\boldsymbol{x}=\{x_n\}$ has a limit
$\sigma$ with $\sigma^2+\cos\pi\alpha >0$,  then
$\displaystyle{ A_n(\boldsymbol{x}) \to
   \dfrac{\sigma^2+\cos\pi\alpha}{2\sigma} }$, and
$$
F_n(\boldsymbol{x}) \to f(\sigma):=
   \sqrt{\left(
     \dfrac{\sigma}{2}+\dfrac{\cos\pi\alpha}{2\sigma}
       \right)^2+1 } -
     \dfrac{\sigma}{2}-\dfrac{\cos\pi\alpha}{2\sigma}
$$
which has $\sin(\pi\alpha/2)$ as attractive fixed point, as
$f'(\sigma)=\displaystyle{
  \dfrac{3\cos\pi\alpha-1}{3-\cos\pi\alpha}}$ at
the fixed point $\sigma=\sin(\pi\alpha/2)$.

\psset{unit=1cm}
\begin{pspicture}(-0.1,-0.3)(15,4)
\psline{->}(1,0)(4.5,0) \uput[90](4.5,0){$\sigma$}
\psline{->}(1,0)(1,3.5)  \uput[180](1,3){$1$}
  \psline[linestyle=dashed,dash=3mm 3mm](1,3)(4,3)(4,0)
    \uput[90](2.5,3.25){$\cos\pi\alpha<0$}
\psplot[linewidth=0.075,plotstyle=curve]{2}{4}{6 x div }
  \psline[linestyle=dashed,dash=3mm 3mm](2,3)(2,0)
%\psplot[linewidth=0.075,plotstyle=curve]{2}{4}{x 4 div dup 0.398 mul 0.961 add exch dup mul 1 add sqrt div 1 sub 50 mul}
\psline(1,0)(4,3)  \uput[270](2,0){$\sqrt{-\cos\pi\alpha}$}
\psdots[dotsize=0.25](3,2)
%
\psline{->}(6,0)(9.5,0) \uput[90](9.5,0){$\sigma$}
\psline{->}(6,0)(6,3.5)  \uput[180](6,3){$1$}
\psplot[linewidth=0.075,plotstyle=curve]{6}{7}{x 7 sub dup mul neg 1 add 2.4 mul}
\psplot[linewidth=0.075,plotstyle=curve]{7}{9}{1 x 7 sub dup mul 1 add div 2.4 mul}
    \uput[90](7.5,3.25){$0<\cos\pi\alpha<1/3$}
  \psline[linestyle=dashed,dash=3mm 3mm](6,3)(9,3)(9,0)
\psline(6,0)(9,3)  \uput[270](7,0){$\sqrt{\cos\pi\alpha}$}
\psdots[dotsize=0.25](7.665,1.665)
  \psline[linestyle=dashed,dash=3mm 3mm](7,2.4)(7,0)
%
\psline{->}(11,0)(14.5,0) \uput[90](14.5,0){$\sigma$}
\psline{->}(11,0)(11,3.5)  \uput[180](11,3){$1$}
    \uput[90](12.5,3.25){$1/3<\cos\pi\alpha<1$}
\psplot[linewidth=0.075,plotstyle=curve]{11}{14}{x 13 sub dup mul neg 4 add 0.4 mul}
  \psline[linestyle=dashed,dash=3mm 3mm](11,3)(14,3)(14,0)
  \psline[linestyle=dashed,dash=3mm 3mm](13,1.6)(13,0)
\psline(11,0)(14,3)  \uput[270](13,0){$\sqrt{\cos\pi\alpha}$}
\psdots[dotsize=0.25](12.5,1.5)
\end{pspicture}


\smallskip

However, contraction cannot be established on the set of all the
positive sequences. A convenient subset must be found. Let us
try the already suspected set of the \emph{\textbf{increasing}}
positive sequences. Does   $\boldsymbol{F}$ transforms an increasing
sequence into an increasing sequence? Alas no, some iterated sequences
may show decreasing episodes:

\hspace*{-15pt}{\Small
\begin{verbatim}

alpha= 0.4000  sin pi alpha/2= 0.587785261    cos pi alpha= 0.309016998

     x1               x23         x24        x25          x26        x27         x28         x29          x30

0.587785261 ...  0.587785261 0.587785261 0.587785261 0.587785261 0.587785261 0.587785261 0.587785261 0.587785261
0.494895897 ...  0.583209880 0.583399460 0.583573954 0.583735095 0.583884358 0.584023011 0.584152146 0.584272710
0.501288967 ...  0.589828341 0.589838555 0.589844230 0.589846167 0.589845023 0.589841333 0.589835538 0.589828002
0.504031732 ...  0.588181205 0.588167246 0.588152427 0.588137067 0.588121407 0.588105632 0.588089882 0.588074262
0.504781340 ...  0.587432119 0.587439806 0.587446203 0.587451582 0.587456152 0.587460079 0.587463494 0.587466499
...
0.504551159 ...  0.587521237 0.587542850 0.587561911 0.587578807 0.587593854 0.587607313 0.587619400 0.587630295

\end{verbatim}     }

\subsubsection{Anticlimax}.  Moreover, some intermediate $x_i$'s
are larger that the limit $\sigma=\sin\pi\alpha/2$, whereas all the final
values appear to be $<\sigma$. Finally, asymptotic behaviour of
intermediate $x_n$'s is not as expected: \\
with $x_n=\sigma$, $n=1,2,\dotsc$ as first iterate, the next one is
$F_n(x)$ such that
$\displaystyle{  \dfrac{F_n(x)^{-1}-F_n(x)}{2} =    }$ \\
$\displaystyle{
 \dfrac{ \xi_n +n\cos\pi\alpha = x_1x_0+\dotsb + x_nx_{n-1} +n\cos\pi\alpha}
     {(n+1)x_{n+1}+(n-1)x_{n-1}} =
  \dfrac{ \sigma+(n-1)\sigma^2+n\cos\pi\alpha}{2n\sigma}
  \sim \dfrac{\sigma^{-1}-\sigma}{2} +\dfrac{1-\sigma}{2n} }$,
whence $F_n(x)= \sigma+O(n^{-1})$, whereas
$O(n^{-2}$ is expected.

Obviously, we have a problem with $\xi_n=x_1x_0+\dotsb +x_nx_{n-1}$
which is numerically found to be $n\sigma^2+O(n^{-1})$, a feature
which is impossible to catch.

\smallskip

\subsubsection{More smart ideas.}

\emph{\textbf{Aha!}} In order to be sure of the sought feature for $\xi_n$,
I intend to compute $x_1$, $x_2,\dotsc, x_N$ with boundary values
$x_0=1$ and $x_{N+1}=\sigma$, and where $\xi_n$ is approximated by
$N\sigma^2 - x_{n+1}x_n -x_{n+2}x_{n+1} -\dotsb - x_Nx_{N-1}$, which
has the expected right behaviour.

So, the iteration is $\boldsymbol{x}=
                      \boldsymbol{F}(
                      \boldsymbol{x})$, with
\begin{equation}   \label{iter20}
  \dfrac{F_n(\boldsymbol{x})^{-1} -
         F_n(\boldsymbol{x}) }{2}  =
 A_n(\boldsymbol({x})  =
\dfrac{\xi_n+n\cos\pi\alpha =
  N\sigma^2 - x_{n+1}x_n -x_{n+2}x_{n+1} -\dotsb - x_Nx_{N-1}
   +n\cos\pi\alpha}{(n+1)x_{n+1}+(n-1)x_{n-1}}
\end{equation}
for $n=1,2,\dotsc, N$, with $x_0=1$, $x_{N+1}=\sigma$.

Let us try a run, I can't wait:

{\Small  \begin{verbatim}

{
/*
  grunb2.gp    :  launch gp and make  \r grunb2

  Reflection coefficients for Grunbaum- Delsarte et al.  problem

 by modified iteration,   January 2003.

*/
default(realprecision,75);
flone=1.0;pr=precision(flone);ep=10^(-pr);
default(format,"g1.7");
N=50;
a=0.2;
print1(a," ");    sa=sin(Pi*a/2);
print(" sigma=sin pi a/2= ",sa);
ca=cos(a*Pi);
print(" cos pi a= ",ca);
\\ vector of reflection coeff.
ref=vector(N+2,k,0);
newref=vector(N+2,k,0);
   print(" iteration");
   for(n=2,N+2,ref[n]=sa);ref[1]=1;
    for(ns=1,5,   newref[1]=1; xi=N*sa^2;  print(ns," ");
          for(ni=1,N,
            n=N+1-ni;
    An= (xi+n*ca) / (  (n+1)* ref[n+2]+(n-1)*ref[n] )  ;
    newref[n+1]= 1/( sqrt(  1+An^2 )+An  );
             xi=xi-ref[n+1]*ref[n];
            ) ;
   print(" final xi=", xi);
   resid=0;  for(n=2,N+1, print1(ref[n]," ");
               resid=max(resid,abs(newref[n]-ref[n]));ref[n]=newref[n]);
              print(" ",resid);
       );
}

Script V1.1 session started Mon Jan 13 09:55:38 2003

? \r grunb2
0.200000 sigma=sin pi a/2= 0.309016998     cos pi a= 0.809017006

iteration x1       x2          x3         x4           x5          x6
1 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998
2 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998
3 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998
...
 final xi=-0.213525497

Script completed Mon Jan 13 10:01:06 2003

\end{verbatim} }

This is ridiculous! Indeed, $x_n=\sigma$ for all $n\geqslant 1$ is a
solution of \eqref{iter20}. The mistake was to input the wrong
value $\xi_N =N\sigma^2$ and, even as the error is small when $N$
is large, it induced a big error on $\xi_0=0$.

I now try $\xi_N=N\sigma^2+\epsilon$, with a small value of
$\epsilon$:

\begin{equation}   \label{iter21}
  \dfrac{F_n(\boldsymbol{x})^{-1} -
         F_n(\boldsymbol{x}) }{2}  =
 A_n(\boldsymbol{x})  =
\dfrac{\xi_n+n\cos\pi\alpha =
  N\sigma^2+\epsilon  - x_{n+1}x_n -x_{n+2}x_{n+1} -\dotsb - x_Nx_{N-1}
   +n\cos\pi\alpha}{(n+1)x_{n+1}+(n-1)x_{n-1}}
\end{equation}
for $n=1,2,\dotsc, N$, with $x_0=1$, and $x_{N+1}$ given in $(0,\sigma]$.


{\Small  \begin{verbatim}
   ...
   epsN=0.01;
   for(n=2,N+2,ref[n]=sa);ref[1]=1;
    for(ns=1,99,   newref[1]=1; xi=N*sa^2+epsN;  print(ns," ");
   ...
Script V1.1 session started Mon Jan 13 10:15:37 2003

C:\calc\pari>gp-2-1
? \r grunb2
0.20000  sigma=sin pi a/2= 0.309016998     cos pi a= 0.809017006
iteration  x1       x2          x3         x4           x5          x6          x7         x8
 1 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998
 2 0.306219520 0.307612449 0.308079335 0.308313263 0.308453776 0.308547516 0.308614505 0.308664765
 3 0.303157786 0.305671680 0.306848166 0.307423936 0.307763642 0.307987009 0.308144695 0.308261757
 4 0.299463531 0.303418696 0.305337763 0.306365638 0.306962124 0.307349253 0.307619605 0.307818446
 5 0.295361338 0.300807905 0.303625347 0.305154810 0.306064180 0.306647176 0.307050272 0.307344243
10 0.270967997 0.284787140 0.292876381 0.297689581 0.300651827 0.302561102 0.303851791 0.304765798
20 0.225981032 0.252575546 0.270358158 0.281962161 0.289514290 0.294493259 0.297852035 0.300183570
50 0.179686381 0.215487507 0.242549751 0.261914544 0.275338719 0.284519575 0.290798615 0.295135010

 final xi=0.174535940
...
? quit      Good bye!    C:\calc\pari>exit
Script completed Mon Jan 13 10:25:40 2003

\end{verbatim}  }

Now, we get a slightly better estimate, and we try other values of
$\epsilon$, trying to have a vanishing $\xi_0$. As the iteration
has become so slow, only the $50^{\text{th}}$ and the
                             $75^{\text{th}}$ are shown:

{\Small  \begin{verbatim}

Script V1.1 session started Mon Jan 13 10:51:51 2003

epsN  it.      xi0          x1         x2          x3           x4          x5          x6         x7          x8
0.001 50 -0.163917929  0.291013681 0.297533706 0.301534030 0.304008217 0.305561057 0.306554982 0.307206219 0.307644146
      75 -0.161816339  0.290153919 0.296967949 0.301165472 0.303769074 0.305405851 0.306453912 0.307140014 0.307600427

0.002 50 -0.117394143  0.274474296 0.286600667 0.294248505 0.299064334 0.302122190 0.304094074 0.305392139 0.306267487
      75 -0.113484834  0.272896542 0.285524758 0.293531093 0.298591648 0.301812283 0.303890901 0.305258463 0.306178964

0.003 50 -0.073636237  0.259241672 0.276185619 0.287156206 0.294186150 0.298701449 0.301634917 0.303575083 0.304887172
      75 -0.068164090  0.257063455 0.274649366 0.286108540 0.293485440 0.298237418 0.301328662 0.303372693 0.304752768

0.004 50 -0.032368428  0.245179644 0.266258421 0.280252791 0.289374295 0.295299816 0.299178134 0.301755372 0.303503350
      75 -0.025538800  0.242498971 0.264306568 0.278892477 0.288451021 0.294682317 0.298767874 0.301483053 0.303321994

0.005 50  0.006649610  0.232169733 0.256790922 0.273533841 0.284629255 0.291918212 0.296724326 0.299933321 0.302116170
      75  0.014663500  0.229068764 0.254463778 0.271877573 0.283488829 0.291147982 0.296209200 0.299589886 0.301886801

0.006 50  0.043628206  0.220108412 0.247756840 0.266994894 0.279951380 0.288557493 0.294274072 0.298109238 0.300725777
      75  0.052679462  0.216656357 0.245090801 0.265058525 0.278599174 0.287635359 0.293653287 0.297693533 0.300447346

...

Script completed Mon Jan 13 11:00:21 2003
\end{verbatim}  }

Still further detail, keeping now
 $100^{\text{th}}$ and the
                             $200^{\text{th}}$ iterations:

{\Small  \begin{verbatim}

Script V1.1 session started Mon Jan 13 12:05:34 2003


  epsN  it.       res.          xi0          x1         x2          x3           x4          x5          x6         x7          x8
0.0044 100 0.00003326434 -0.008321663 0.236659067 0.260060575 0.275886322 0.286337328 0.293185851 0.297691239 0.300691562 0.302725829
       200 0.00000000397 -0.008217685 0.236617948 0.260030122 0.275864899 0.286322753 0.293176147 0.297684867 0.300687413 0.302723143

0.0045 100 0.00003413821 -0.004276811 0.235306088 0.259069462 0.275180283 0.285838125 0.292830474 0.297434110 0.300501426 0.302581778
       200 0.00000000408 -0.004171089 0.235264338 0.259038445 0.275158416 0.285823227 0.292820544 0.297427584 0.300497175 0.302579024

0.0046 100 0.00003501818 -0.000255118 0.233963922 0.258083236 0.274476246 0.285339652 0.292475309 0.297177003 0.300311254 0.302437682
       200 0.00000000420 -0.000147672 0.233921551 0.258051661 0.274453938 0.285324430 0.292465153 0.297170323 0.300306901 0.302434861

0.0047 100 0.00003590429  0.003743652 0.232632453 0.257101867 0.273774206 0.284841907 0.292120358 0.296919919 0.300121047 0.302293540
       200 0.00000000431  0.003852803 0.232589470 0.257069738 0.273751457 0.284826361 0.292109975 0.296913085 0.300116592 0.302290652

0.0048 100 0.00003679661  0.007719737 0.231311564 0.256125325 0.273074157 0.284344892 0.291765620 0.296662858 0.299930806 0.302149354
       200 0.00000000442  0.007830572 0.231267979 0.256092646 0.273050970 0.284329022 0.291755010 0.296655870 0.299926247 0.302146398

0.0049 100 0.00003769518  0.011673364 0.230001143 0.255153580 0.272376095 0.283848606 0.291411097 0.296405822 0.299740529 0.302005123
       200 0.00000000454  0.011785865 0.229956965 0.255120356 0.272352470 0.283832413 0.291400260 0.296398679 0.299735867 0.302002099

0.0050 100 0.00003860006  0.015604763 0.228701077 0.254186601 0.271680013 0.283353050 0.291056790 0.296148811 0.299550218 0.301860848
       200 0.00000000466  0.015718911 0.228656316 0.254152836 0.271655952 0.283336534 0.291045725 0.296141513 0.299545452 0.301857755

? quit  Good bye! C:\calc\pari>exit   Script completed Mon Jan 13 13:54:13 2003

\end{verbatim}  }

where ``res.'' is the residue norm
$\max_n |(x_n^{-1}-x_n)/2-A_n(\boldsymbol{x})|$. It seems indeed that we get
the correct sequence when $\xi_0=0$, if one reminds that one should
find


{\Small  \begin{verbatim}

      x1         x2          x3          x4          x5          x6          x7          x8          x9         x10
0.233872324 0.258015427 0.274428042 0.285306081 0.292452074 0.297160853 0.300299896 0.302429553 0.303905700 0.304953308
\end{verbatim}  }

\medskip

The iteration based on \eqref{iter21} is numerically rather
poor, but has very interesting monotony properties:

\subsubsection{\textbf{Lemma.}} For any $\alpha\in (0,1)$, the map
 $\boldsymbol{F}$ acting on $\mathbb{R}^N$ through \eqref{iter21}
 has the properties:

 \begin{enumerate}
   \item For any $\epsilon \geqslant 0$, a positive %increasing
   sequence bounded by $\sigma=\sin \pi\alpha/2$ is transformed
    by $\boldsymbol{F}$ in
   a sequence with the same properties.
   \item If $\boldsymbol{x}$ is a positive   %increasing
    sequence
     bounded by $\sigma$, and if $\boldsymbol{F}(\boldsymbol{x})
            \leqslant \boldsymbol{x}$, then
   $\boldsymbol{F}(\boldsymbol{F}(\boldsymbol{x})) \leqslant
      \boldsymbol{F}(\boldsymbol{x})$, where
      $\boldsymbol{x} \leqslant \boldsymbol{y}$ means
    $x_n\leqslant y_n$, $n=1,2,\dotsc, N$.
  \item $\xi_0$ is an increasing function of $\epsilon$ when one
    starts \eqref{iter21} with the constant sequence
    $x_n=\sigma$, $n=1,\dotsc, N$.
  \item $\xi_0 = \sigma^2-\sigma <0$ if $\epsilon=0$ and
    $x_{N+1}=\sigma$; $\xi_0\geqslant \epsilon+\sigma^2-\sigma$.
   There is therefore exactly one $\epsilon>0$ such that $\xi_0=0$
   if $x_{N+1}=\sigma$.
 \end{enumerate}

Indeed,

\begin{enumerate}
  \item If all the $x_n$'s are positive and
$\leqslant \sigma=\sin\pi\alpha/2$,
the numerator of $A_n$ in \eqref{iter21} is $\geqslant
 n\sigma^2+n\cos\pi\alpha = n\cos^2(\pi\alpha/2)$, so that
 $A_n\geqslant \cos^2(\pi\alpha/2)/(2\sin\pi\alpha/2 =
  (\sigma^{-1}-\sigma)/2$, and $0<F_n(\boldsymbol{x})\leqslant
 \sigma$.
% \\
%  If all the $x_i$'s are positive, the numerator of $A_n$
%  decreases when $n$ increases, and the denominator increases
%  with $n$ if $x_{i+1}\geqslant x_i$ for all $i\geqslant 1$, so
% $A_n$ decreases with $n$, and $F_n =\sqrt{A_n^2+1}-A_n$ increases with $n$.
  \item We show $\boldsymbol{y}\leqslant \boldsymbol{x} \Rightarrow
     \boldsymbol{F}(\boldsymbol{y})\leqslant
        \boldsymbol{F}(\boldsymbol{x})$ on positive sequences
    such that $A_n$ is positive. Indeed: $A_n(\boldsymbol{y})
   \geqslant A_n(\boldsymbol{x})$, as $A_n$ is a decreasing
   function of its $x_i$'s arguments.
  \item At each iteration, $x'_n\geqslant x''_n$ if $x'_n$ and
   $x''_n$ correspond to $\epsilon'$ and $\epsilon''$, with
                         $\epsilon' \leqslant \epsilon''$.
  This is true for the first step; if true at some iteration,
   $A'_n\leqslant A''_n$, and $x'_n\geqslant x''_n$ again at the
   next iteration. Finally, $\xi'_0 = N\sigma^2+\epsilon'
   -x'_1 -x'_1x'_2 -\dotsb -x'_{N-1}x'_N \leqslant \xi''_0$.
\end{enumerate}

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(2.2,2.2)
\psline{->}(0,0.5)(2,0.5) \uput[90](2,0.5){$\epsilon_N$}
\psline{->}(0,0)(0,2)  \uput[0](0,2){$\xi_0$}
  \psline[linestyle=dashed,dash=3mm 3mm](0,1)(1,2)
\pscurve[linewidth=0.04](0,0)(0.2,0.5)(0.5,1.2)(1,1.7)
% \pscircle*(1.2,1.2){0.1}
% \pscircle[linewidth=0.05](1,1.5){1}
\uput[0](0,0){$\sigma^2-\sigma$}
\uput[90](1,1.9){$N\sigma^2+\epsilon$}
\end{pspicture}
}
That there is always an $\epsilon_N>0$ ensuring $\xi_0=0$ follows
from 1) that $\xi_0= N\sigma^2-x_1 -x_1x_2 -\dotsb -x_{N-1}x_N =
  \sigma^2-\sigma <0$ at $\epsilon=0$ (as all the $x_i$'s $=\sigma$);
2) when $\epsilon$ is large, the $x_n$'s are very small, so that
$\xi_0$ is close to $N\sigma^2+\epsilon$.


See here a computer run with $N=50$ and $\sigma=0.309...$, so that
$\xi_0$ is expected near $4.774+\epsilon$:

{\Small  \begin{verbatim}

Script V1.1 session started Wed Jan 15 13:51:32 2003

C:\calc\pari>gp-2-1
? \r grunb2
0.200000002 0.200000002  sigma=sin pi a/2= 0.309016998   cos pi a= 0.809017006

it.    res.     xi0        x1             x2          x3             x4             x5
 epsN= 1
100 43632.74 4.311780 1.020088 E-07 5.177575 E-07 1.995604 E-06 6.411419 E-06 1.801285 E-05
200 611.2635 4.316579 8.342916 E-08 4.275806 E-07 1.663710 E-06 5.395415 E-06 1.530021 E-05
300 7.097863 4.316634 8.323656 E-08 4.266411 E-07 1.660235 E-06 5.384725 E-06 1.527153 E-05
400 0.082361 4.316635 8.323433 E-08 4.266302 E-07 1.660194 E-06 5.384602 E-06 1.527120 E-05
500 0.000955 4.316635 8.323430 E-08 4.266301 E-07 1.660194 E-06 5.384600 E-06 1.527119 E-05

 epsN= 2
400 3010.520 6.101422 5.711563 E-11 3.946940 E-10 2.012176 E-09 8.383047 E-09 3.010333 E-08
500 75.78560 6.101422 5.711510 E-11 3.946905 E-10 2.012159 E-09 8.382978 E-09 3.010309 E-08

 epsN= 3
400 1665142. 7.455424 2.227993 E-13 1.841310 E-12 1.106376 E-11 5.374799 E-11 2.232201 E-10
500 44603.46 7.455424 2.227948 E-13 1.841274 E-12 1.106355 E-11 5.374702 E-11 2.232162 E-10

? quit    Good bye!     C:\calc\pari>exit     Script completed Wed Jan 15 13:58:43 2003
\end{verbatim}  }

The amazing feature is the small size of the first $x_n$'s
when $\epsilon$ is large.

\medskip

We are far enough to get the

\subsubsection{\textbf{Theorem}\label{grfinal}} {\sl For any
 $\alpha\in(0,1)$, the positive
solution -known to be unique- of \eqref{eqgrun} with the sole
boundary condition $x_0=1$ is the limit (in the $\ell_\infty$ norm)
when $N\to\infty$
of the unique positive solution of the system of $N$ equations from \eqref{griter}
with $n=1,2,\dotsc, N$, where $x_{N+1}, x_{N+2},\dotsc$ are replaced
by $\sigma=\sin(\pi\alpha/2)$.\\
One also has  $\Phi_n(0) < \sigma=\sin(\pi\alpha/2),\  n=1,2,\dotsc$ \\
  $\Phi_1(0)+\Phi_2(0)\Phi_1(0)+\dotsb +\Phi_n(0)\Phi_{n-1}(0)
    > n\,\sigma^2,\ \ \ n=1,2,\dotsc$
 }

Of course, all this means that the sequence $\{\Phi_n(0)\}$ is
positive% and increasing!!

I would be more at ease with a lower bound of $x_n$, to be sure
that we do not fall towards the black hole which is the trivial
solution $x_n=0, n=1,2,\dotsc$ of \eqref{eqgrun}!
The example above shows that the $x_n$'s may become very small.

No problem if $\xi_0=0$: then,
$ x_1+x_1x_2+x_2x_3+\dotsb + x_{n-1}x_n >n\sigma^2$ and all the
$x_n$'s $<\sigma \Rightarrow x_1 > \sigma^2$.

Now, the part ``limit when $N\to\infty$'':

\noindent\textbf{The step $N\to N+1$.}

To show: $x_i^{(N+1)} < x_i^{(N)}, i=1,2,\dotsc, N+1$.

$\{x_1^{(N)},\dotsc, x_N^{(N)}\}$ is the  unique
positive solution of the $N$ first equations of
\eqref{eqgrun}, when $x_0=1$ and $x_{N+1}=\sigma$ are given.

Let $\{x_i^{(N+1)}\}$, $i=1,\dotsc, N+1$, be as above with
$x_{N+2}=\sigma$. We know that $x_{N+1}^{(N+1)} <\sigma$, as all the
other $x_i^{(N+1)}$ for $i=1,\dotsc, N$.

A first point is that $x_i^{(N+1)}$ for $i=1,2,\dotsc,N$ is another
positive solution of the $N$ first equations of \eqref{eqgrun}, with
the boundary conditions $x_0^{(N+1)}=1$ and
                        $x_{N+1}^{(N+1)}<\sigma$.

Then, $x_i^{(N+1)} < x_i^{(N)}$ for $i=1,2,\dotsc, N+1$ from
Proposition~\ref{influ2n}, p.~\pageref{influ2n}!

\bigskip

\subsubsection{\textbf{Extended conjecture.}}

Another generalized Jacobi weight giving rise to real $\Phi_n(0)$'s
is\\ $\displaystyle{2^{2\beta}\left|\sin\dfrac{\theta-\theta_1}{2}\right|^{2\beta}
                  \left|\sin\dfrac{\theta+\theta_1}{2}\right|^{2\beta}=
(\cos\theta_1 -\cos\theta)^{2\beta} }$ on the same arc
$\theta_1 <\theta <2\pi-\theta_1$ as above.

Then we apply \eqref{jacobg2} with $\alpha=\beta$ (the $\alpha$ of
\eqref{jacobg2}, of course), and $\theta_1=\pi\alpha $:

\begin{equation}  \label{eqgrum2}
\hspace*{-1pt} (n+1+2\beta) x_{n+1} =
\frac{ 2\xi_{n-1,n}
  +2n\cos\pi\alpha  }{1-x_n^2}\, x_n
  -(n-1+2\beta)  x_{n-1},
\end{equation}
with
$ \xi_{n-1,n} = x_1 x_0+ x_2 x_1 +
   \dotsb +x_n x_{n-1}$.

Everything should follow as previously, at least when $\beta\geqslant 0$.
If $-1/2<\beta<0$, $x_1$ may be negative. Of course: when $\alpha\to 0$,
$x_n=\Phi_n(0)\to \displaystyle{\dfrac{2\beta}{n+2\beta+1}}$,
from the Jacobi case \eqref{jacob0}. Check that it is a solution of
 \eqref{eqgrum2}, and that
$\displaystyle{ \xi_n\to    \dfrac{2n}{n+2\beta} }$.

The algorithm based on \eqref{iter21} still works, but monotonicity
is lost if $\beta<0$:


{\Small  \begin{verbatim}

Script V1.1 session started Wed Jan 15 19:26:42 2003

C:\calc\pari>gp-2-1
? \r grunb2
 beta= -0.4000
a=0.200  sigma=sin pi a/2= 0.309016998    cos pi a= 0.809017006

  epsN   it.   res.         xi0           x1          x2          x3          x4          x5          x6          x7          x8
-0.7300   1 1.870135   -0.943525519  0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998 0.309016998
          2 0.7042453   0.338622633 -0.672895643 0.310526529 0.309838715 0.309581500 0.309446924 0.309364162 0.309308119 0.309267656
          3 0.3774387  -0.049577267 -0.383801812 0.239853530 0.310968644 0.310318867 0.309987500 0.309787483 0.309654043 0.309558889
          4 0.1884398   0.111600434 -0.506704566 0.259264601 0.295726171 0.311221992 0.310629146 0.310277695 0.310046423 0.309883345
          5 0.0789971   0.050028948 -0.463622974 0.237411815 0.300642846 0.307529406 0.311362442 0.310825839 0.310477387 0.310234251
          6 0.0719641   0.088081151 -0.493017577 0.244316002 0.293604070 0.309517456 0.310682941 0.311423509 0.310939671 0.310605480
          7 0.0340438   0.068994451 -0.481645602 0.236455395 0.296619959 0.307394269 0.311786747 0.311564930 0.311426647 0.310991532
          8 0.0248932   0.081868112 -0.492970739 0.240113465 0.293811476 0.309078412 0.311378238 0.312301968 0.311760261 0.311387545
          9 0.0254564   0.073029980 -0.489236169 0.237428235 0.296032722 0.308307282 0.312454837 0.312461032 0.312298826 0.311728218
         10 0.0120816   0.077756786 -0.494995718 0.240174885 0.295322860 0.309772801 0.312470898 0.313205727 0.312608697 0.312138140
        ...
        250 2.157 E-11  0.015164409 -0.513154068 0.305572573 0.341790084 0.339335893 0.332603304 0.326763176 0.322394200 0.319244409
        500 1.025 E-21  0.015164409 -0.513154068 0.305572573 0.341790084 0.339335893 0.332603304 0.326763176 0.322394200 0.319244409

-0.7305 250 2.236 E-11  0.006866933 -0.516129642 0.315181023 0.347903715 0.343117240 0.335033848 0.328400153 0.323546940 0.320089279
        500 1.027 E-21  0.006866933 -0.516129642 0.315181023 0.347903715 0.343117240 0.335033848 0.328400153 0.323546940 0.320089279
-0.7310 250 2.306 E-11 -0.001575852 -0.519134554 0.325188111 0.354143600 0.346933581 0.337471851 0.330036888 0.324697764 0.320932273
        500 1.025 E-21 -0.001575852 -0.519134554 0.325188111 0.354143600 0.346933581 0.337471851 0.330036888 0.324697764 0.320932273

? quit  Good bye! C:\calc\pari>exit  Script completed Wed Jan 15 19:32:49 2003
\end{verbatim}  }

The true values are

{\Small  \begin{verbatim}
       x1         x2          x3          x4          x5          x6          x7          x8          x9          x10
-0.518575525 0.323302616 0.352977636 0.346223740 0.337019515 0.329733610 0.324484652 0.320776200 0.318145256 0.316253055
\end{verbatim}  }

\section{Generalized Jacobi polynomials on the unit circle, with
    $>2$ singular points.} \  \\

\cite{Vanl}



\bigskip


\section{References.}
\renewcommand{\mylmark}{References.}
\pagestyle{fancy}

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\end{thebibliography}

\end{document}
