%%
%%  This is AmSLaTeX    m3xxx00.tex
%%
%% Here is a character listing to check to be sure that no
%% unwanted translations took place within the bowels of the net.
%% Upper case letters: ABCDEFGHIJKLMNOPQRSTUVWXYZ
%% Lower case letters: abcdefghijklmnopqrstuvwxyz
%% round parentheses, square brackets, curly braces: ()  []  {}
%% Exclaim, at, sharp, dollar, percent: ! @ # $ %
%% Caret, ampersand, star, underscore, hyphen: ^ & * _ -
%% vertical bar, backslash, tilde, backprime, plus: | \ ~ ` +
%% plus, equal, prime, quote, colon: + = ' " :
%% less than, greater than, slash, question, comma: < > / ? ,
%% period, semicolon: . ;
%%
%%
 \documentclass[11pt,reqno,twoside,final]{amsart}

%page size
\hsize=7.7677truein\vsize=10.5truein

% LaTeX's page layout parameters   Pedro Quaresma de Almeida
%                                   1995

\textwidth=6.4in
\textheight=9.00in   %9.5in
\oddsidemargin=.2in
\evensidemargin=.2in
\headheight=0.5in
\topmargin=-.05in
\headsep=0.25in
\voffset=-0.5in

\input epsf.tex    \epsfverbosetrue


\makeatletter
%\def\ddots{\mathinner{\mkern1mu\raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern2mu
%    \raise4\p@\hbox{.}\mkern2mu\raise\p@\hbox{.}\mkern1mu}}
\def\secddots{\mathinner{\mkern1mu\raise\p@\vbox{\kern7\p@\hbox{.}}\mkern2mu
    \raise4\p@\hbox{.}\mkern2mu\raise7\p@\hbox{.}\mkern1mu}}
\makeatother

\def\pain{\boldsymbol{\mathfrak{P}}}

\usepackage{amssymb}
\usepackage{mathrsfs}
\usepackage{array}
\usepackage{cite}
%\usepackage{epic}
%\usepackage{eepic}
\usepackage{picins}
\usepackage{fancyhdr}
\usepackage{pslatex}

\def\mylmark{\ mylmark initial}
\markboth{}{} \markright{}

\pagestyle{fancy}
\lhead{\rmfamily \textsf{MAPA3xxxA\ 2000-2001-2002\ }-- Complex rational approx. \ \textbf{\thesection}\ --\ \mylmark\  --} \chead{} \rhead{\rmfamily\thepage}
\lfoot{}\cfoot{}\rfoot{}


\usepackage{pst-plot}

% tables des matieres avec ... : extrait de amsbook.cls avec dotfill au
%   de hfil
\makeatletter
\def\@tocline#1#2#3#4#5#6#7{\relax
  \ifnum #1>\c@tocdepth % then omit
  \else
    \par \addpenalty\@secpenalty\addvspace{#2}%
    \begingroup \hyphenpenalty\@M
    \@ifempty{#4}{%
      \@tempdima\csname r@tocindent\number#1\endcsname\relax
    }{%
      \@tempdima#4\relax
    }%
    \parindent\z@ \leftskip#3\relax \advance\leftskip\@tempdima\relax
    \rightskip\@pnumwidth plus1em \parfillskip-\@pnumwidth
    #5\leavevmode\hskip-\@tempdima #6\relax
    \ \ \dotfill\hbox to\@pnumwidth{\@tocpagenum{#7}}\par
    \nobreak
    \endgroup
  \fi}

% police de chapitre
\def\@makechapterhead#1{\global\topskip 7.5pc\relax
  \begingroup
  \fontsize{\@xivpt}{18}\bfseries\centering
    \ifnum\c@secnumdepth>\m@ne
      \leavevmode \hskip-\leftskip
      \rlap{\vbox to\z@{\vss
%         \centerline{\normalsize\bfseries
          \centerline{\Large    \bfseries
                        \@xp{\chaptername}\enspace\thechapter}
          \vskip 3pc}}\hskip\leftskip\fi
     #1\par \endgroup
  \skip@34\p@ \advance\skip@-\normalbaselineskip
  \vskip\skip@ }
\newcommand{\@chapapp}{\chaptername}

\def\section{\@startsection{section}{1}%
  \z@{.7\linespacing\@plus\linespacing}{.5\linespacing}%
% {\normalfont\bfseries\centering}}
  {\Large\normalfont\bfseries\centering}}


\makeatother


% comp.text.tex #141312 (21 + 1180 more)                      (1)+-(1)+-(1)
% From: J Scott Berg <jsberg@indiana.edu>                        |    \-(1)--(1)
% [1] Re: Cauchy principal value integral                        \-[1]
% Date: Fri May 22 19:36:45 EDT 1998

% There seems to be some feeling that you shouldn't use the integral
% sign with the bar through it.  If you want to anyhow, here it is:

\newcommand\PVint{\mathop{\setbox0\hbox{$\displaystyle\intop$}%
    \hskip0.2\wd0%
    \vcenter{\hrule width0.6\wd0height0.5pt depth0.5pt}%
    \hskip-0.8\wd0%
    }\mskip-\thinmuskip\intop\nolimits}

%                                -Scott Berg
% --
% J. Scott Berg                                 Phone: (812) 855-5465
% Indiana University Cyclotron Facility         FAX:   (812) 855-6645
% 2401 Milo B. Sampson Lane                     email: jsberg@indiana.edu
% Bloomington, IN  47408-1398
% End of article 141312 (of 141732) -- what next? [npq]

\renewcommand{\boldsymbol}{\pmb}
\renewcommand{\refname}{ }

%Comment these:
%\usepackage[dvips,colorlinks=true]{hyperref}
\usepackage{color}
\definecolor{dred}{cmyk}{0.0,0.5,0,0.5}\definecolor{dgree}{cmyk}{0.25,0,0.25,0.5}
\makeatletter\def\@linkcolor{dred}\def\@citecolor{dgree}\def\@pagecolor{dgree}\def\@urlcolor{dred}
 \makeatother
% or these:
\newcommand{\href}[2]{{#2}}
%  Destroy toc and aux files!

\begin{document}

\thispagestyle{empty}

\begin{center} \textbf{\Large
  MAPA 3071 Special topics in approximation theory.}

\smallskip

 \textbf{2000-2002: Asymptotic estimates in complex rational approximation.}

\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
        (0)(10)473157 ,
   \href{mailto:magnus@anma.ucl.ac.be}
        {\tt magnus@anma.ucl.ac.be} ,
  \href{http://www.math.ucl.ac.be/\~magnus/ }
       {\tt http://www.math.ucl.ac.be/\~{}magnus/ }
%Paul Van Dooren, EULER, a119,  478040 , vandooren@anma.ucl.ac.be
\end{center}
\bigskip
$1^{\text{st}}$ lect.: Friday 13 Oct. 2000, 14h30, room b101.

 Friday 14 Oct. 2001, 14h30, room b101.

%pp.~1-23: October 15, 1999. \\
This version: \today \qquad  (incomplete and unfinished)

\bigskip

Abstract: several ways to build rational approximations to various
 species of analytic functions are examined. Special emphasis is put
 on strong asymptotic estimates of the form
  $ f(z) -R_n(z) \sim \sigma(z) \rho^n(z)$, when such estimates are
available.

\bigskip


\tableofcontents

\bigskip



\section{Complex approximation theory and potential theory.}
\renewcommand{\mylmark}{Complex \& potential.}
\pagestyle{fancy}

\subsection{Taylor expansions} \   \\

The Taylor series expansion of a function with finite convergence
domain, say
$$
  \frac{1}{1-z/R} = \sum_0^\infty \frac{z^k}{R^k}\;, \qquad
  \text{or} \qquad \sqrt{1-z/R} =
   \sum_0^\infty \left( -\,\frac{1}{2}\right)\, \frac{1}{2}
      \dotsb \left(k-\frac{3}{2}\right) \frac{z^k}{k! R^k}
$$
shows ``typically'' almost circular level lines of equal
approximation:
\psset{unit=1.0cm}\begin{pspicture}(0,1)(2.2,2.2)
 \pscircle*(1,1){0.1}
 \pscircle(1,1){0.5}
 \pscircle(1,1){0.73}
 \pscircle(1,1){0.9}
 \pscircle[linewidth=0.05](1,1){1}
\end{pspicture}
, explained by a convenient representation of the error
\begin{equation} \label{errtayl}
  f(z) - \sum_0^n  c_k z^k = K_n(z) (z/R)^n,
\end{equation}
where $K_n$ is ``typically'' slowly variable in $n$. What is meant
by ``typical'' must be estimated on particular classes of functions.
The only general truth here is that $K_n$ is bounded by a slowly
variable\footnote{slowly variable $=$ less than exponentially
variable.} function of $n$ when $|z|<R$:
$$
  K_n(z) = \left(\frac{R}{z}\right)^n
    \frac{1}{2\pi i} \int_{|t|=r} f(t) \frac{z^{n+1}}{t^n(t-z)}\;dt,
$$
with $|z|<r<R$, and $r$ arbitrarily close to $R$. For some functions
$f$, an infinite subset $\{K_{n_i}\}_i$ may be much smaller than
expected, that's why some special classes of functions $f$ will be
described (sometimes a \emph{single} function\dots) when accurate
asymptotic estimates of \eqref{errtayl} will be needed.
For the examples above,
$$
\frac{1}{1-z/R} -\sum_0^n (z/R)^k = \frac{ (z/R)^{n+1}}{1-z/R}\;,
\sqrt{1-z/R}-\sum_0^n c_k (z/R)^k = \frac{ d_{n+1}^{(n)} (z/R)^{n+1}
       +\dotsb + d_{2n}^{(n)} (z/R)^{2n} }
       {\sqrt{1-z/R}\;+\;\sum_0^n c_k (z/R)^k},
$$
with suitable $d_{n+1}^{(n)}$, etc. A special asymptotic analysis
is needed to show that $d_{n+1}^{(n)} =2c_{n+1} \sim
 -1/\sqrt{\pi n^3}$.

The circle $|z|=R$ is the boundary of the convergence region. It is often
conveniently approached by the zeros of the truncated expansion, here
with $n=10$:

\psset{unit=0.9cm}\begin{pspicture}(-1.50,-2.6)(4.2,4.2)
 \pscircle(0,0){1.0}
 \psdots[dotstyle=*,dotsize=4pt](0.42,0.91)(0.42,-0.91)(0.84,0.54)%
       (0.84,-0.54)(-0.96,0.28)(-0.96,-0.28)(-0.66,0.76)(-0.66,-0.76)%
       (-0.14,0.99)(-0.14,-0.99)
 \psdots[dotstyle=*,dotsize=4pt](-1.1,-2.1)
 \uput[0](-0.8,-2.1){: zeros of approximation}
 \psdots[dotstyle=square*,linecolor=red,dotsize=4pt](0,0)(-1.1,-2.4)
 \uput[0](-0.8,-2.4){: zeros of error}
\end{pspicture}
\hspace*{5cm}
\psset{unit=0.9cm}\begin{pspicture}(-2.00,-2.6)(4.2,4.2)
 \pscircle(0,0){1.0}
 \psdots[dotstyle=*,dotsize=4pt](-1.77,0.0)(1.09,0.0)(0.50,1.57)%
       (0.50,-1.57)(1.23,0.89)(1.23,-0.89)(-1.43,1.03)(-1.43,-1.03)%
       (-0.54,1.63)(-0.54,-1.63)
 \psdots[dotstyle=square*,linecolor=red,dotsize=4pt](0,0)(-1.89,0)%
     (0.54,1.68)(0.54,-1.68)(1.33,0.93)(1.33,-0.93)(-1.53,1.1)(-1.53,-1.1)%
     (-0.58,1.75)(-0.58,-1.75)
 \psdots[dotstyle=*,dotsize=4pt](-1.1,-2.1)
 \uput[0](-0.8,-2.1){: zeros of approximation}
 \psdots[dotstyle=square*,linecolor=red,dotsize=4pt](-1.1,-2.4)
 \uput[0](-0.8,-2.4){: zeros of error}
\end{pspicture}

For $f(z)=1/(1-z/R)$, the truncated Taylor expansion is
$p_n(z)=(1-(z/R)^{n+1})/(1-z/R)$, therefore vanishes at the $n$ nontrivial
 $(n+1)^{\text{th}}$ roots of unity. For $f(z)=\sqrt{1-z/R}$, the
picture is not so simple, but the zeros of $p_n$ appear to have the circle as
limit set as well. It is not essential to take zeros, similar results
hold for $z: p_n(z)=\xi$, for any $\xi$: as a ``typical'' function $f$
is expected to be almost univalent, or not very many-valent, in the
convergence disk, most of the roots of $p_n(z)=\xi$ will be outside the
disk interior.

\bigskip

Darboux

\subsection{Interpolatory (Jacobi) expansions} \   \\

\begin{flushright} \dots, tout entier \`a une id\'ee \\
                        qui lui \'etait venue sur les \textit{potentiels}. \  \\
                                  \   \\
                        Alphonse Allais (from \textsl{Madrigal manqu\'e}\ \ \ %
\end{flushright}

\subsection{General polynomial interpolation} \   \\

\subsection{Pad\'e approximation} \   \\

Pad\'e approximation is closely linked to continued fractions, also
to orthogonal polynomials. The latter aspect appears especially with
expansion about $\infty$.

Just as the simplest explicit arithmetic continued fractions are
the periodic ones giving square roots, we have square roots of
polynomials:

$$
 \sqrt{z^2-1} = z-\dfrac{1}{2z} -\dfrac{1}{8z^3} -\dotsb
  = z-\displaystyle{   \frac{1/2}
             {\displaystyle{z-   \frac{1/4}
                 {\displaystyle{z-   \frac{1/4}{z-\dotsb} } } } } }
$$
and the $[n+1/n]$ Pad\'e approximant is the ratio of
\emph{\textbf{Chebyshev}} polynomials
$$\displaystyle{  \frac{T_{n+1}(z)}{U_n(z)}
 = \sqrt{z^2-1} \frac{ (z+\sqrt{z^2-1})^{n+1}
                      +(z-\sqrt{z^2-1})^{n+1}    }
                     { (z+\sqrt{z^2-1})^{n+1}
                      -(z-\sqrt{z^2-1})^{n+1}    }
}$$

Square roots of polynomials of degree $>2$ will usually not yield
periodic continued fractions, but a kind of quasi-periodicity which
will be described further on. It will also be seen that the arrangement
$f=\displaystyle{ \frac{X-\sqrt{Y}}{Z} }$ is almost pure periodic
\cite{Mag4}. So,

\begin{multline*}
\dfrac{z^2-\alpha z -\beta\sqrt{1-\alpha^2}-
     \sqrt{(z^2-1)(z-\alpha-\beta)(z-\alpha+\beta)} }{z-\alpha}=
 \frac{\gamma}{z} +\alpha\frac{\gamma -(1-\beta^2)/2}{z^2} +\dotsb
 \\
   = \displaystyle{
            \frac{\gamma}
           {\displaystyle{z-\alpha\beta(\beta-\sqrt{1-\alpha^2})/\gamma-
          \frac{\gamma'} {z- \dotsb}  } }\qquad%
        }
\end{multline*}

with $\gamma=(1+\beta^2)/2-\beta\sqrt{1-\alpha^2}$, and
$\gamma'= ...$
turns into an (apparent) mess.
Special periodic cases are

\begin{itemize}

  \item Two-periodic $\alpha=0$:
    $$ \frac{ z^2-\beta -\sqrt{(z^2-1)(z^2-\beta^2)}}{z} =
    \displaystyle{
            \frac{(\beta-1)^2/2 }
           {\displaystyle{z-
          \frac{(\beta+1)^2/4} {z-
                  {\displaystyle{ \frac{(\beta-1)^2/4}
                       {z-\dotsb}} }   } }
                  }}
 $$
Approximants $q_n/p_n=[n-1/n]$:
 $$  p_{n+2} = (z^2- (1+\beta^2)/2) p_n - \left(\frac{1-\beta^2}{4}
                      \right)^2 p_{n-2}.$$
  $$
    \frac{q_{2n}}{p_{2n}} = \frac{
           z {\displaystyle{\frac{(\beta-1)^2}{2}   }}
        \;   {\displaystyle{\frac{
                                 \rho^{2n}-\rho^{-2n} }
                          {\sqrt{(z^2-1)(z^2-\beta^2)}}  }}
       \;    {\displaystyle{\left(\frac{1-\beta^2}{4}\right)^n   }}
                                }{
             (A\rho^{2n}+B\rho^{-2n})
       \;    {\displaystyle{\left(\frac{1-\beta^2}{4}\right)^n   }}
                                  } \;,
 $$
  $$
    \frac{q_{2n+1}}{p_{2n+1}} = ((\beta-1)^2/2)\frac{
             (C\rho^{2n+1}+D\rho^{-2n-1})
       \;    {\displaystyle{\left(\frac{1-\beta^2}{4}\right)^n   }}
                                     }{
            z(\rho^{2n}+\rho^{-2n} )
       \;    {\displaystyle{\left(\frac{1-\beta^2}{4}\right)^n   }}
                                  } \;
 $$
 with $\rho=\displaystyle{\sqrt{\frac{z^2-(1+\beta^2)/2+
                                 \sqrt{(z^2-1)(z^2-\beta^2)}}{(1-\beta^2)/2}
                         }}$, and
     $A,B=\displaystyle{ \frac{1}{2}\pm \frac{z^2-\beta}
                                 {2\sqrt{(z^2-1)(z^1-\beta^2)}} }$,
   $C,D=\displaystyle{\rho^{\mp 1} \left(\frac{1}{2}\pm \frac{z^2+\beta}
                         {2\sqrt{(z^2-1)(z^2-\beta^2)}}\right) }$.

The approximants converge (check that the limit is $f$) outside
the singular locus $S=\{ z: |\rho(z)|=1\}$. All the zeros and
poles but one are on, or tend when $n\to\infty$ towards, $S$.
There is a supplementary zero at 0 for the even approximants, a
supplementary pole at 0 for the odd approximants.

 The numerator and the denominator have the forms
 $C_n\rho^n+D_n\rho^{-n}$, and $A_n\rho^n+B_n\rho^{-n}$, with
 $A_n$, etc., $D_n$ two-periodic in $n$.

  \item Period three: $\alpha=(1-\beta^2)/2$.
\begin{multline*}
\dfrac{z^2-\dfrac{1-\beta^2}{2}z +\beta\dfrac{1-2\beta-\beta^2}{2} -
  \sqrt{\displaystyle{(z^2-1)\left(z-\frac{1-2\beta-\beta^2}{2}\right)
              \left(z-\frac{1+2\beta-\beta^2}{2}\right)}}}
  {z-\dfrac{1-2\beta-\beta^2}{2}}=   \\
 \displaystyle{
     \frac{(1-\beta)(1+\beta)^2/2}
          {\displaystyle{z+\beta(1+\beta)/2 -
                 \frac{(1-\beta)^2/4}
                      {\displaystyle{z+\beta(1+\beta)/2 -
                            \frac{(1-\beta)(1+\beta)^2/4}
                                 {\displaystyle{z-\beta -
                                       \frac{(1-\beta)(1+\beta)^2/4}
                                            {z+\beta(1+\beta)/2 -\dotsb}
                                 }}
                       }}
             }}
     }
\end{multline*}
Approximants $[n-1/n] = q_n/p_n :$
$$ p_{n+3} = \left( z^3 +\beta^2 z^2
                +\frac{\beta^4 -2\beta^2-3}{4}\, z -\beta^2\right) p_n
         -\left( \frac{ (1-\beta^2)^2 }{8} \right)^2 p_{n-3}.$$
$$ \frac{q_{3n}}{p_{3n}} = \frac{(1-\beta)(1+\beta)^2}{2}
    \frac{  \displaystyle{\left(z-\frac{\beta+1}{2}\right)
                      \frac{\rho^{3n}-\rho^{-3n}}{2\sqrt{Y}}
                      \left( \frac{(\beta^2-1)^2}{8} \right)^n
         }}
         {  \displaystyle{
                  (A \rho^{3n} +B\rho^{-3n} )
                      \left( \frac{(\beta^2-1)^2}{8} \right)^n
         }}
$$
$$ \frac{q_{3n+1}}{p_{3n+1}} = \frac{(1-\beta)(1+\beta)^2}{2}
    \frac{  (C \rho^{3n+1} +D \rho^{-3n-1})
                      \left( \dfrac{(\beta^2-1)^2}{8} \right)^n
         }
         {  (A' \rho^{3n+1} +B' \rho^{-3n-1})
                      \left( \dfrac{(\beta^2-1)^2}{8} \right)^n
         }
$$
$$ \frac{q_{3n+2}}{p_{3n+2}} = \frac{(1-\beta)(1+\beta)^2}{2}
    \frac{  (C' \rho^{3n+2} +D' \rho^{-3n-2})
                      \left( \dfrac{(\beta^2-1)^2}{8} \right)^n
         }
         {  \displaystyle{\left(z+\frac{\beta^2+2\beta-1}{2}\right)
                      \frac{\rho^{3n+3}-\rho^{-3n-3}}{2\sqrt{Y}}
                      \left( \frac{(\beta^2-1)^2}{8} \right)^n
         }}
$$
with
  $Y=\displaystyle{(z^2-1)\left(z-\frac{1-2\beta-\beta^2}{2}\right)
              \left(z-\frac{1+2\beta-\beta^2}{2}\right)}  $,

  $\rho=\displaystyle{\sqrt[3]{\frac{
          z^3+\beta^2 z^2 -
             {\displaystyle{\frac{3+2\beta^2-\beta^4}{4}\, z}}-\beta^2
  + \left(z+\dfrac{1+\beta^2}{2}\right)  \sqrt{Y} }
                 {(1-\beta^2)^2/4}
                         }}$,\\
$ A,B= \displaystyle{ \frac{1}{2} \pm \frac{2z^2 +(\beta^2-1)z
            +\beta(1-2\beta-\beta^2)}{4\sqrt{Y}} }$,
$C,D= \rho^{\mp 1} \displaystyle{ \left(\frac{1}{2} \pm \frac{ 2z^2
            +(\beta-1)^2 z -2\beta}{4\sqrt{Y}}\right) }$,
$A'\rho = C'\rho^2, B'\rho^{-1}=D'\rho^{-2} =
       \displaystyle{ \frac{z+\beta(1+\beta)/2}{2}
             \pm \frac{4z^3+2(2\beta^2+\beta-1)z^2 +
         (\beta^4+\beta^3-3\beta^2-\beta-2)z
            -(2\beta^3+6\beta^2+2\beta-2)}{8\sqrt{Y}} }$.

 \item General quadratic case \cite{Mag4}

  We consider $f(z):=f_0(z) = \displaystyle{
                                 \frac{\beta_0(z)}
                                      {\alpha_0(z) -
                                        \displaystyle{
                                           \frac{\beta_1(z)}
                                                {\alpha_1(z) - \dotsb}
                                         }}
                           = \frac{X_0(z)-\sqrt{Y(z)}}{Z_0(z)}
                                             }$,
  where $X_0,Y,$ and $Z_0$ are given polynomials, and where the continued
fraction arrangement (Jacobi, or Stieltjes, or $C-$fraction, etc.)
keeps $\alpha_n$ and $\beta_n$ to be polynomials of small degrees.

Then,
              $f_n(z):= \displaystyle{
                                 \frac{\beta_n(z)}
                                      {\alpha_n(z) -
                                        \displaystyle{
                                           \frac{\beta_{n+1}(z)}
                                                {\alpha_{n+1}(z) - \dotsb}
                                         }}
                           = \frac{X_n(z)-\sqrt{Y(z)}}{Z_n(z)}
                                             }$, or
\begin{equation} \label{fn}
 f_n = \frac{\beta_n}{\alpha_n-f_{n+1}},
\end{equation}
with polynomial $X_n$ and $Z_n$ of ultimate degrees related to the
degree of $Y$. For Jacobi continued fractions (Pad\'e at $\infty$),
the degrees are $g+1$ and $g$ if the degree of $Y$ is $2g+2$.

\begin{equation} \label{ZZ}
X_n^2-Y=\beta_n Z_n Z_{n+1}, \qquad X_{n+1}= -X_n-\alpha_n Z_{n+1}.
\end{equation}
Also, the $X_n$'s and $Z_n$'s have a quasi-periodic behaviour with
respect to $n$.

For the two and three periodic examples above,
$$ X_{2n}(z)= z-\beta,
   X_{2n+1}(z)= z+\beta;\qquad Z_n(z)=z; $$
$$ X_{3n}(z)= X_{3n+2}(z)=z^2 -\frac{1-\beta^2}{2}z +\beta
       \frac{1-2\beta-\beta^2}{2},
   X_{3n+1}(z)= z^2 -\frac{1-\beta^2}{2}z -\beta, $$
 $$ Z_{3n}(z)= z-\frac{1-2\beta-\beta^2}{2},
 Z_{3n+1}(z)= Z_{3n+2}(z)=z-\frac{1+\beta}{2}.
$$

The recurrence $p_{n+1} = \alpha_n p_n -\beta_n p_{n-1}$ of $p_n$ and
$q_n$ of the approximants $q_n/p_n =1/0 , \beta_0/\alpha_0,$ \\
$\beta_0\alpha_1/(\alpha_0\alpha_1-\beta_1),\dotsc$, has also the solution
$ f_0 f_1 \dotsc f_n$ (from \eqref{fn},
  $\alpha_n f_0\dotsc f_n - f_0 \dotsc f_n f_{n+1} = \beta_n
       f_0\dotsc f_{n-1}$), and we have
\begin{equation} \label{prfn}
 f_0 \dotsc f_n = q_n -f p_n.
\end{equation}

Considering the two possible choices of the square root of $Y$ in $f$
and the $f_n$'s,
\begin{equation} \label{pq}
  p_n= - \frac{f_{0,\text{conj.}}\dotsb f_{n,\text{conj.}} -
               f_0\dotsb f_n  }{f_{\text{conj.}}-f},\qquad%
  q_n=  \frac{f_{1,\text{conj.}}\dotsb f_{n,\text{conj.}} -
               f_1\dotsb f_n  }{1/f_{\text{conj.}}-1/f}.
\end{equation}

\begin{equation*}\begin{split}
(f_0 \dotsc f_n)^2 &=\frac{ (X_0-\sqrt{Y})^2 \dotsb  (X_n-\sqrt{Y})^2\; Z_{n+1}}
   {Z_0  \underbrace{Z_0Z_1}_{(X_0^2-Y)/\beta_0}\; \dotsb \;%
        \underbrace{Z_nZ_{n+1}}_{(X_n^2-Y)/\beta_n}
   }  \\
 &= \beta_0 \dotsc \beta_n \; \frac{Z_{n+1}}{Z_0}\;%
   \frac{ X_0-\sqrt{Y} }{ X_0+\sqrt{Y} }\,%
     \dotsb\,\frac{ X_n-\sqrt{Y} }{ X_n+\sqrt{Y} }
\end{split}\end{equation*}
\emph{\textbf{Definition.}} \textsl{The average complex rate of growth
 $=$ exponential of the complex Green function is}
$$
\rho = e^{\mathcal{G}} =  \lim_{n\to\infty}\left(
   \frac{ X_0+\sqrt{Y} }{ X_0-\sqrt{Y} }\,%
     \dotsb\,\frac{ X_n+\sqrt{Y} }{ X_n-\sqrt{Y}
     }\right)^{1/(2n)}\;,
$$
with the square root of $Y$ such that $|\rho|\geqslant 1$ ($=$ the
real Green function $G:= \text{Re}\, \mathcal{G} \geqslant 0$)
everywhere.
\end{itemize}

\subsection{Strong asymptotics of Pad\'e terms for functions with
  branch points}  \    \\

Two branch points. Szeg\H o: denominator $p_n$ is orthogonal to $w$,
if $f(z)= \displaystyle{ \int_a^b \frac{w(t)}{z-t}\,dt }$.
Then, $p_n(z) \sim \alpha(z) \rho^n(z)$,


\subsection{General rational interpolation. Condenser capacity} \   \\



\section{Asymptotic features of rational interpolation.}
\renewcommand{\mylmark}{Asymptotics.}
\pagestyle{fancy}

\subsection{According to Gon\v car-Stahl (a sloppy rendering)}  \   \\

Interpolation to $\displaystyle{ f_n(z)= \int_{C_f} \dfrac{
  \varphi_0(t)\varphi^n(t) }{z-t}\;dt }$ at $z_0,\dotsc, z_{m+n}$
by $p_m/q_n$ yields
$$
f_n(z) -\dfrac{p_m(z)}{q_n(z)} = \dfrac{ \prod_0^{m+n} (z-z_j)}{q_n^2(z)}
  \int_{C_f}
        \dfrac{q_n^2(t)}{ \prod_0^{m+n} (t-z_j)} \; %
    \dfrac{\varphi_0(t)\varphi^n(t) }{z-t}\;dt       ,
$$
where $q_n$ is (formally) orthogonal with respect to
$\displaystyle{ \dfrac{\varphi_0(t)\varphi^n(t) }
        { \prod_0^{m+n} (t-z_j)}  }$ on $C_f$.

Well, we expect that most of the poles of $q_n$ will tend to a set
of arcs $S$, with a limit distribution $\mu_p$, that $C_f$ may be
modified within the closure of the domain where $\varphi_0$ and
$\varphi$ are analytic, so that $S\subseteq C_f$. On the support of $\mu_p$, $q_n$ is almost
a Szeg\H o orthogonal polynomial! which means that
$\pm\displaystyle{ q_n(t) \dfrac{\varphi^{n/2}(t)}
        {\sqrt{ \prod_0^{m+n} (t-z_j)}}  }$ has  slowly varying
phase and absolute value there.

Sloppy asymptotic explanation with (complex) potentials
$\displaystyle{ \mathcal{V}_p(z) := \int_{\text{supp\ } \mu_p}
         \log(z-t)\;d\mu_p(t)  }$  and     \\
$\displaystyle{ \mathcal{V}_i(z) := \int_{\text{supp\ } \mu_i}
         \log(z-t)\;d\mu_i(t)  }$  (interpolation points), so that
$q_n(z)\sim \exp(n\mathcal{V}_p(z))$ when $z\notin $ supp$(\mu_p)$,
$\prod_0^{m+n} (z-z_j) \sim \exp(2n\mathcal{V}_i(z))$, and
$ q_n(z) \approx \exp(n\mathcal{V}_{p,+}(z))
                +\exp(n\mathcal{V}_{p,-}(z))   $ on supp$(\mu_p)$.
Then,
\begin{equation*}\begin{split}
   q_n(t) \dfrac{\varphi^{n/2}(t)}
        {\sqrt{ \prod_0^{m+n} (t-z_j)}}  &\approx
[ \exp(n\mathcal{V}_{p,+}(t))
                +\exp(n\mathcal{V}_{p,-}(t)) ] \; %
  \exp( n\log\varphi(t)/2 -n\mathcal{V}_i(t)) \\    &\approx
  \exp\left( n\left[\dfrac{\log\varphi(t)}{2}
   -\mathcal{V}_i(t) +\dfrac{\mathcal{V}_{p,+}(t)+\mathcal{V}_{p,-}(t)}
                          {2} \right]\right)\;\; \cos n\left(
 \dfrac{\mathcal{V}_{p,+}(t)-\mathcal{V}_{p,-}(t)}{i} \right)
\end{split}\end{equation*}
on $S$, or:
\begin{equation} \label{VCf}
  \log\varphi(t)/2 -\mathcal{V}_i(t)+[\mathcal{V}_{p,+}(t)
            +\mathcal{V}_{p,-}(t)]/2 = \text{constant},
\end{equation}
       the same real constant on all the arcs of $S$, has a real part smaller
     than this constant on $C_f\setminus S$.

For derivatives:


\begin{equation} \label{pvCf}
   (\log\varphi(z))'/2 -\mathcal{V}'_i(z)+
    \PVint_S \frac{d\mu_p(t)}{z-t} = 0 \text{\ on\ } z\in S.
\end{equation}


\medskip

Remark that the (complex conjugate of) the derivative
$  (\log\varphi(z))'/2 -\mathcal{V}'_i(z)+ \mathcal{V}'_p(z) $
on the two sides of $S$ gives the gradient of the real
potential
Re$\,[\log\varphi(z)/2 -\mathcal{V}_i(z)+ \mathcal{V}_p(z)] $,
and has \emph{\textbf{opposite}} values $\pm \pi i \overline{\mu'(z)}$
on the two sides of $S$, from \eqref{pvCf} and the
\emph{\textbf{Sokhotskyi-Plemelj}} formulas (see \S~\ref{severalarcs})
 for $\mathcal{V}'_p$:
\emph{\textbf{symmetry property}} \cite[etc.]{Apt2001,GR,Sta,St}.

\subsection{According to Nuttall (in construction)}  \   \\


\subsection{The Riemann-Hilbert way (not even started)}  \   \\


\subsection{\label{singlearc}Conditions on a single arc} \  \\

Let the function (often associated to a distribution of poles)
$\displaystyle{  \mathcal{V}'_p(z) = \int_\alpha^\beta \frac{d\mu_p(t)}{z-t} }$.
Suppose that we know that
\begin{equation} \label{pvab}
   = \PVint_\alpha^\beta \frac{d\mu_p(t)}{z-t} = g(z),
\end{equation}
with $g$ analytic in some domain (the arc $[\alpha, \beta]$ is not yet
known). The trick is to multiply $\mathcal{V}'_p$
by a function $[(z-\alpha)(z-\beta)]^{\gamma/2}$ taking \emph{opposite}
values on the two sides of $[\alpha,\beta]$. We consider only
$\gamma=1$ and $\gamma=-1$. Also,
              $[(z-\alpha)(z-\beta)]^{\gamma/2}$
is defined to be continuous outside the arc, and behaves like
$z^\gamma$ for large $z$. As
 $\mathcal{V}'_p(z) [(z-\alpha)(z-\beta)]^{\gamma/2} -\delta_{\gamma,1} $
has a Laurent expansion with only negative powers at $\infty$,
$$
 \mathcal{V}'_p [(z-\alpha)(z-\beta)]^{\gamma/2} -\delta_{\gamma,1} =
 \frac{1}{2\pi i} \oint
 \frac{
 \mathcal{V}'_p(t) [(t-\alpha)(t-\beta)]^{\gamma/2} -\delta_{\gamma,1}}
   {z-t}\;dt
$$
on a big counterclockwise contour having the arc $[\alpha,\beta]$ inside
and $z$ outside. Making the contour shrink to a neighbourhood of the
arc $[\alpha,\beta]$, we get
$\displaystyle{
 \frac{1}{2\pi i} \int_\alpha^\beta
 \frac{  \Delta\{
 \mathcal{V}'_p(t) [(t-\alpha)(t-\beta)]^{\gamma/2} -\delta_{\gamma,1}\}}
   {z-t}\;dt,
}$
where $\Delta\{F\}$ means the difference $F_--F_+$ between the limit
values of $F$ on the ``lower'' side of the arc (from which the arc is seen
at left), and the ``upper'' side. The difference is here
$2 g(t) [(t-\alpha)(t-\beta)]^{\gamma/2}_-$, whence quite explicit
solutions
\begin{equation} \label{pvsol}
 \mathcal{V}'_p [(z-\alpha)(z-\beta)]^{\gamma/2} -\delta_{\gamma,1} =
 \frac{1}{\pi i} \int_\alpha^\beta
\frac{g(t)[(t-\alpha)(t-\beta)]^{\gamma/2}_-}{z-t}\;dt,\qquad \gamma=\pm 1.
\end{equation}
It may help to realize that the phase of
$\displaystyle{
 \frac{  \beta-\alpha}
          {[(t-\alpha)(t-\beta)]^{1/2}_- }
}$  is exactly the one of $+i$ on the rectilinear segment
$[\alpha,\beta]$.\\
Some questions: the $-1$ in the left-hand side of \eqref{pvsol} when
$\gamma=1$ is needed from $\mathcal{V}'_p(z) = 1/z +O(1/z^2)$ for
large $z$. But the two sides of \eqref{pvsol} when $\gamma=-1$
should be $\sim 1/z^2$ for large $z$, everything works only if
\begin{equation} \label{pvcond}
  \int_\alpha^\beta
 \frac{  g(t)\,dt    }
          {[(t-\alpha)(t-\beta)]^{1/2}_- } \; =0,\  \
  \int_\alpha^\beta
 \frac{t g(t)\,dt    }
          {[(t-\alpha)(t-\beta)]^{1/2}_- } \; =\pi i.
\end{equation}
The two forms of \eqref{pvsol} then agree, either with $\gamma=-1$, or
$\gamma=1$.
It will also be useful to check that, as \eqref{pvab} is a plain
integral when $z=\alpha$ and $z=\beta$, one has
$\mathcal{V}'_p(\alpha)=g(\alpha)$, and
$\mathcal{V}'_p(\beta)=g(\beta)$.

\subsubsection{A little bit of Chebyshev polynomials calculus.}

N.B. Ullman

Let us consider the Chebyshev polynomials expansion of a generic function
$F$ on $[\alpha,\beta]$:
$$
 F(t) = \frac{c_0}{2} +\sum_1^\infty c_n
             T_n\left( \frac{2t-\alpha-\beta}{\beta-\alpha}\right).
$$
Then, we have the integral
$$
  \int_\alpha^\beta
 \frac{  F(t)\,dt    }
          {[(t-\alpha)(t-\beta)]^{1/2}_- } \; =\pi i \frac{c_0}{2}.
$$
Therefore, from \eqref{pvsol} with $\gamma=-1$,
$\mathcal{V}'_p [(z-\alpha)(z-\beta)]^{-1/2}$ is the constant
term of the Chebyshev expansion of $g(t)/(z-t)$.

Let $g_0/2 +\sum_1^\infty g_n T_n$ be the expansion of $g$.
Remark that \eqref{pvcond} becomes
\begin{equation} \label{pvcond2}
  g_0=0,\qquad g_1 = \frac{4}{\beta-\alpha}.
\end{equation}

We need the expansion of $1/(z-t) = X_0/2 +\sum_1^\infty X_n T_n$, which we
multiply by $\displaystyle{ \frac{2(z-t)}{\beta-\alpha} =
  \frac{2z-\alpha-\beta}{\beta-\alpha} -
  \frac{2t-\alpha-\beta}{\beta-\alpha}  }$:
$$
\frac{2}{\beta-\alpha} = \frac{X_0}{2} \left(
  \frac{2z-\alpha-\beta}{\beta-\alpha} - T_1((2t\alpha-\beta)/(\beta-\alpha))\right)
  +\sum_{n=1}^\infty X_n \left(
  \frac{2z-\alpha-\beta}{\beta-\alpha}  T_n -\frac{ T_{n-1}+T_{n+1}}{2})\right)
$$
whence the recurrence
$\displaystyle{
X_{n+1}
-2  \frac{2z-\alpha-\beta}{\beta-\alpha}  X_n +X_{n-1} =0 }$ for
$n=1,2,\dotsc$ solved by $X_n = X_0 \rho^n$, where $\rho$ is a
root of
\begin{equation} \label{rhoz}
    \frac{ \rho + \rho^{-1} }{2} =
    \frac{2z-\alpha-\beta}{\beta-\alpha}  \;,
\end{equation}
normally with $|\rho|<1$, but this will have to be discussed later.
The value of $X_0$ comes from $n=0$:
$4/(\beta-\alpha)= X_0((\rho+\rho^{-1})/2)-X_1 =
 X_0(\rho^{_1}-\rho)/2$, so
$$
 X_0 = \frac{8}{(\beta-\alpha)(\rho^{-1}-\rho)}.
$$
Remark that $[(z-\alpha)(z-\beta)]^{1/2} = (\beta-\alpha)^2 (1-\rho^2)^2/
 (16\rho^2)$, so that
\begin{equation} \label{VpT}
   \mathcal{V}'_p(z) = \sum_{n=1}^\infty g_n \rho^n.
\end{equation}
The two determinations of $\mathcal{V}'_p$ on the two sides of the
cut $[\alpha,\beta]$ are obtained with the \emph{two} roots
$\rho$ and $1/\rho$ of \eqref{rhoz}. One checks that the arithmetic mean
is indeed
$$
 (\mathcal{V}'_{p,+} (z) +    \mathcal{V}'_{p,-} (z) )/2 =
  \sum_1^\infty g_n (\rho^n +\rho^{-n})/2
= \sum_1^\infty g_n T_n  = g(z).
$$
As for the discontinuity along the cut,
\begin{equation} \label{mupT}
\pm \pi i \mu'_p(z) =
 \mathcal{V}'_{p,-} (z) +    \mathcal{V}'_{p,+} (z) =
  \sum_1^\infty g_n (\rho^{-n} -\rho^n) =
  \frac{4}{\beta-\alpha}[(z-\alpha)(z-\beta)]^{1/2}
   \sum_1^\infty g_n U_{n-1}(z),
\end{equation}
it appears as a kind of harmonic conjugate to $g$.


\subsection{\label{severalarcs}Conditions on several  arcs} \  \\

Let \eqref{pvCf} be
\begin{equation} \label{pvCf2}
    \PVint_S \frac{d\mu_p(t)}{z-t} = g(z)\;,\ \ z\in S,
\end{equation}
with
 $g(z)= \mathcal{V}'_i(z)-  (\log\varphi(z))'/2 $,
whereas $\displaystyle{\mathcal{V}'_p(z)=
    \PVint_S \frac{d\mu_p(t)}{z-t} \pm \pi i \mu'_p(z) }$
(\emph{\textbf{Sokhotskyi-Plemelj}} formulas \cite{Hen}, chap.~14, etc.).

If the (unknown) endpoints are $\alpha_1,\dotsc, \alpha_{2m}$,
and $P(z)=(z-\alpha_1)\dotsb (z-\alpha_{2m})$,

$\displaystyle{\dfrac{\mathcal{V}'_p(z)}{\sqrt{P(z)}} =
 \pm \dfrac{g(z)}{\sqrt{P(z)}}+ }$ the same
function on the two sides of $S$, so,

$$
\dfrac{\mathcal{V}'_p(z)}{\sqrt{P(z)}} = \dfrac{1}{\pi i} \int_S
  \dfrac{g(t)}{\sqrt{P(t)}}\;\dfrac{dt}{z-t}
$$
(Sokhotskyi-Plemelj again).

As $\mathcal{V}'_p(z) = z^{-1}+\dotsb$ for large $z$,
$m+1$ equations follow
$$
 \int_S t^k\; \dfrac{g(t)}{\sqrt{P(t)}}\;dt =0, \ \ \ k=0,\dotsc, m-1;\ \ %
 \int_S t^m\; \dfrac{g(t)}{\sqrt{P(t)}}\;dt =\pi i,
$$
and the $m-1$ further equalities, from \eqref{VCf}
$$
  \log\varphi(\alpha_{2k})/2 -\mathcal{V}_i(\alpha_{2k})
  \mathcal{V}_p(\alpha_{2k} =
  \log\varphi(\alpha_{2k+1})/2 -\mathcal{V}_i(\alpha_{2k+1})
  \mathcal{V}_p(\alpha_{2k+1} \ \ k=1,\dotsc,m-1.
$$

\subsubsection{Differential equation for $\mathcal{V}'_p$ \cite{GR} .}

If there is a polynomial $Q$ such that $Qg'/g$ is a polynomial,
$$
Q(z)\sqrt{P(z)}\;\mathcal{V}'_p(z)= \text{\ polynomial\ }
+\dfrac{1}{\pi i} \int_S
  Q(t)g(t)\sqrt{P(t)}\;\dfrac{dt}{z-t}
$$
we derive in $z$ and perform an integration by parts
$$
[Q(z)\sqrt{P(z)}\;\mathcal{V}'_p(z)]'= \text{\ another\ polynomial\ }
+\left(\dfrac{P'(z)Q(z)}{2} +\dfrac{P(z)Q(z)g'(z)}{g(z)} +P(z)Q'(z)\right)
\dfrac{1}{\pi i} \int_S
  \dfrac{g(t)}{\sqrt{P(t)}}\;\dfrac{dt}{z-t},
$$
or $\displaystyle{ Q\sqrt{P} \left(\dfrac{\mathcal{V}_p'}{g}\right)' =
  \dfrac{\text{polynomial}}{g}  }$.

\medskip

\subsubsection{An example from \cite[p.~404]{DKM} (J.~Meinguet).}

We have $g(x)=-2\delta x^3$ on $[-1,1]$ with $\delta>0$ and we suspect $S$ to
be $[-1,-\alpha]\bigcup [-\beta,\beta]\bigcup [\alpha,1]$.

The formula for $\mathcal{V}'_p$ is surprisingly elementary:
$$
\mathcal{V}'_p(z) = -2\delta z^3 +\delta (2z^2+\alpha^2+\beta^2-1)
  \sqrt{ \dfrac{(z^2-\alpha^2)(z^2-\beta^2)}{z^2-1} }
$$
we have indeed $(\mathcal{V}'_{p,+}(x)+
                 \mathcal{V}'_{p,-}(x))/2 = -2\delta x^3$ on $S=$ the
set of points where the square root is pure imaginary. Let us look
at the argument of this square root just above and just below the
real axis:

\begin{center}
\psset{unit=1cm}\begin{pspicture}(-5,-1)(5,1.5)
 \uput[90](-4,0){$-$}
 \uput[270](-4,0){$-$}
 \uput[90](-2.5,0){$-i$}
 \uput[270](-2.5,0){$+i$}
 \uput[90](-1.5,0){$-$}
 \uput[270](-1.5,0){$-$}
 \uput[90](0,0){$+i$}
 \uput[270](0,0){$-i$}
 \uput[90](1.5,0){$+$}
 \uput[270](1.5,0){$+$}
 \uput[90](2.5,0){$-i$}
 \uput[270](2.5,0){$+i$}
 \uput[90](4,0){$+$}
 \uput[270](4,0){$+$}
 \psline{-}(-5,0)(5,0)
 \psline[linewidth=0.09]{-}(-3,0)(-2,0)
 \psline[linewidth=0.09]{-}(-1,0)(1,0)
 \psline[linewidth=0.09]{-}(2,0)(3,0)
\psline[linestyle=dashed, dash=2mm 2mm](-3,0)(-3,0.7) \uput[90](-3,0.7){$-1$}
\psline[linestyle=dashed, dash=2mm 2mm](-2,0)(-2,0.7) \uput[90](-2,0.7){$-\alpha$}
\psline[linestyle=dashed, dash=2mm 2mm](-1,0)(-1,0.7) \uput[90](-1,0.7){$-\beta$}
\psline[linestyle=dashed, dash=2mm 2mm](1,0)(1,0.7) \uput[90](1,0.7){$\beta$}
\psline[linestyle=dashed, dash=2mm 2mm](2,0)(2,0.7) \uput[90](2,0.7){$\alpha$}
\psline[linestyle=dashed, dash=2mm 2mm](3,0)(3,0.7) \uput[90](3,0.7){$1$}
\end{pspicture}
\end{center}

 The coefficient of $z^{-1}$ in
the expansion of $\mathcal{V}'_p(z)$ about $\infty$ is 1 if
$4=\delta[ 3-2(\alpha^2+\beta^2)-(\alpha^2-\beta^2)^2
  -(\alpha^2+\beta^2-1)^2]$. Finally, Re$\;\mathcal{V}_p(x) +\delta x^4/2$ must
be the same constant on the intervals of $S$, which holds if
$\displaystyle{  \int_\beta^\alpha
 (2t^2+\alpha^2+\beta^2-1)
  \sqrt{ \dfrac{(t^2-\alpha^2)(t^2-\beta^2)}{t^2-1} } \;dt =0 }$.
This latter equation gives, say, $\alpha$ as a function of $\beta$,
and the former equation gives the corresponding $\delta$:

\begin{center}\begin{tabular}{|l|c|c|}   \hline
      &       &               \\
 $\beta^2$  & $\alpha^2$  &   $\delta$ \\
      &       &             \\
\hline      &       &             \\
0  &  0.5402057601  &    3.319654336   \\
0.0001  &  0.5398986321  &  3.318351926 \\
0.01  &  0.5221272482  &  3.237132862 \\
0.02  &  0.5073410955  &  3.172076626 \\
0.03  &  0.4936208059  &  3.115410709 \\
0.04  &  0.4805403767  &  3.064945359 \\
0.05  &  0.4679087488  &  3.019526908 \\
0.06  &  0.4556185329  &  2.978410767 \\
0.07  &  0.4436016321  &  2.941064425 \\
0.08  &  0.4318114417  &  2.907083036 \\
0.09  &  0.4202143377  &  2.876146324 \\
0.10 &  0.4087851073  &  2.847993877 \\
0.11  &  0.3975042842  &  2.822409762 \\
0.12  &  0.3863564943  &  2.799212342 \\
0.13  &  0.3753293774  &  2.778247201 \\
0.14  &  0.3644128565  &  2.759382044 \\
0.15  &  0.3535986256  &  2.742502906 \\
0.16  &  0.3428797816  &  2.727511262 \\
0.17  &  0.3322505543  &  2.714321783 \\
0.18  &  0.3217061032  &  2.702860566 \\
0.19  &  0.3112423637  &  2.693063714 \\
0.20 &  0.3008559284  &  2.684876183 \\
0.21  &  0.2905439552  &  2.678250852 \\
0.22  &  0.2803040951  &  2.673147746 \\
0.23  &  0.2701344357  &  2.669533405 \\
0.24  &  0.2600334576  &  2.667380353 \\
0.25  &  0.2500000000  &  2.666666667 \\
      &       &            \\
\hline
\end{tabular} \end{center}

\noindent Inequalities: \\
We want $\mu'(x)\geqslant 0$ on the whole of $S$, OK if
$3\alpha^2+\beta^2-1\geqslant 0$ and
$\alpha^2+3\beta^2-1\leqslant 0$. The real part of $\mathcal{V}_p(x)+
\delta x^4/2$ must be smaller on $(\beta,\alpha)$ than its constant
value on $S$, OK, as
$$ \mathcal{V}_p(x)+ \delta x^4/2 -
   \mathcal{V}_p(\beta)- \delta \beta^4/2 =
\delta  \int_\beta^x
 (2t^2+\alpha^2+\beta^2-1)
  \sqrt{ \dfrac{(t^2-\alpha^2)(t^2-\beta^2)}{t^2-1} } \;dt
$$
is the integral of a function which is negative on a first part
of $(\beta, \alpha)$, positive on another part, but the integral
remains negative, as it must vanish when $x=\alpha$ (the square root
is positive in the whole interval $(\beta,\alpha)$).

The set $S$ is reduced to two intervals when $\delta>3.319\dotsc$,
and to a single interval when $\delta<8/3$. It is conjectured in
\cite{DKM} that $S$ has at most 3 parts for any $g(x)=-\delta x^{2q-1}$.

\section{The exponential function $e^{Az}$.}
\renewcommand{\mylmark}{Exponential function.}
\pagestyle{fancy}

\subsection{Taylor} \  \\

\subsection{Pad\'e} \  \\

for $e^z$,
\begin{equation} \label{padexp1}
   [m/n] = \frac{ \displaystyle{
                       1 +\frac{m}{m+n}\, \frac{z}{1!}
                +\frac{m(m-1)}{(m+n)(m+n-1)}\, \frac{z^2}{2!}
        +\dotsb  +\frac{m(m-1)\dotsb 2.1}
                             {(m+n)(m+n-1)\dotsb (n+1)}\, \frac{z^m}{m!}
                               }}
                { \displaystyle{
                       1 -\frac{n}{m+n}\, \frac{z}{1!}
                +\frac{n(n-1)}{(m+n)(m+n-1)}\, \frac{z^2}{2!}
        -\dotsb  +(-1)^n \frac{n(n-1)\dotsb 2.1}
                             {(m+n)(m+n-1)\dotsb (m+1)}\, \frac{z^n}{n!}
                               }}
\end{equation}
\begin{equation}\begin{split}    \label{padeint}
 e^z\;\text{den}. - \text{num.} &=
  \frac{(-1)^n}{(m+1)\dotsb (m+n)} \sum_{k=m+n+1}^\infty
     \frac{(k-m-1)\dotsb (k-m-n)}{k!} z^k  \\
  &=
  \frac{(-1)^n}{(m+n)!}  \left[
   \int_{-\infty}^z -\int_{-\infty}^0
  = \int_0^z e^t (z-t)^m t^n\,dt    \right]
\end{split}\end{equation}
\cite{Per}


Exponential behaviour of numerator and denominator has been much
worked, especially the distribution of zeros and poles.
Saff \& Varga remark \cite[II]{SVe} that, when $m\sim n$, these distributions
had already been examined by Olver\cite{Olv54}  in a study of Bessel functions.

Integrals of the form \eqref{padeint} behave for large $n$ as value at
saddlepoint. With $m\sim n$, saddlepoint is a root of
$$
  1 +\frac{n}{t-z}+\frac{n}{t} =0, \  \text{whence\  }
    t=\frac{z}{2}  -n +\sqrt{ \frac{z^2}{4}+n^2 },
$$
with some choice for the square root (see later), and
$ (z-t)t = \displaystyle{ 2n^2 \left(
    \sqrt{1+\frac{z^2}{4n^2}} \;-1 \right)  }$.
\medskip

Denominator behaves like $n^{\text{th}}$ power of

\begin{equation} \label{paddenexp1}
  \frac{1+\sqrt{ \displaystyle{1+\frac{z^2}{4n^2}} }  }{2}
 \;\exp\left( 1-\frac{z}{2n}-\sqrt{ 1+\frac{z^2}{4n^2}}\right),
\end{equation}

where the value of the square root is 1 at $z=0$. When $z$ increases,
the value of \eqref{paddenexp1} becomes very small and must be replaced
by the same formula with the other choice of the square root as soon as
the new formula has an absolute value which is larger than the former one.
This happens near $z/n=1.3255$.
In the $z/n-$plane:
\psset{unit=1cm}
\begin{pspicture}(0,2.0)(4,4.2)
  \psline{->}(0.0,2)(4,2)
  \psline{->}(2.0,0.0)(2,4.2)
 \pscurve[linewidth=0.05,linecolor=red](2.0,4.0)(3.3,2.0)(2.0,0.0)
 \uput[0](2.0,4.0){$2i$}  \uput[0](2.00,0.0){$-2i$}
 \uput[45](3.3,2.0){$1.325$}
\end{pspicture}
\ \  the formula \eqref{paddenexp1} holds with a continuous square root
outside

\vspace*{2cm}

the shown arc, which is the locus where the two formulas have
the same absolute value, also the limit of the \emph{\textbf{poles}}
of the approximant. The equation of the arc is $|w(z)|=1$, where
$$
w(z) = \sqrt{
              \frac{\sqrt{ 1+\frac{z^2}{4n^2}}-1  }
                   {\sqrt{ 1+\frac{z^2}{4n^2}}+1  }   }
      \;\exp\left( \sqrt{ 1+\frac{z^2}{4n^2}}\right)  =
       \frac{  (z/2n) \exp\left(\sqrt{1+z^2/(4n^2)}\right) }
            {1+\sqrt{ 1+\frac{z^2}{4n^2}}  }
$$

This $w$ looks like the $\rho$ of before, but is not a plain
quadratic algebraic function.

Remark the square root behaves like $-z/(2n)$ for large $z$ in
\eqref{paddenexp1}.

Numerator:
$\displaystyle{  \frac{1+\sqrt{ \displaystyle{1+\frac{z^2}{4n^2}} }  }{2}
 \;\exp\left( 1+\frac{z}{2n}-\sqrt{ 1+\frac{z^2}{4n^2}}\right) }$,
\ \ \ \  \psset{unit=1cm}
\begin{pspicture}(0,2.0)(4,4.2)
  \psline{->}(0.0,2)(4,2)
  \psline{->}(2.0,0.0)(2,4.2)
 \pscurve[linewidth=0.05,linecolor=green](2.0,4.0)(0.7,2.0)(2.0,0.0)
 \uput[0](2.0,4.0){$2i$}  \uput[0](2.00,0.0){$-2i$}
 \uput[135](0.5,2.0){$-1.325$}
\end{pspicture}    Here the square

\vspace*{2cm}

\noindent root behaves like $z/(2n)$ for large $z$.  The formula
holds with a continuous square root outside another arc ( the limit set
of the \emph{\textbf{zeros}} of the approximant) which is another
part of the locus $|w(z)|=1$.

And the remainder (or residual) $e^z$ den. $-$ num. behaves
 like the $n^{\text{th}}$ power of  \\
$\displaystyle{  \frac{1-\sqrt{ \displaystyle{1+\frac{z^2}{4n^2}} }  }{2}
 \;\exp\left( 1+\frac{z}{2n}-\sqrt{ 1+\frac{z^2}{4n^2}}\right) }$
to the right of the latter arc, together with the part of the imaginary
axis outside $[-2ni,2ni]$ (which is also a part of the locus
$|w(z)|=1$)
\ \ \ \ \ \  \psset{unit=1cm}
\begin{pspicture}(0,2.5)(4,5.2)
  \psline{->}(0.0,2.5)(4,2.5)
  \psline{->}(2.0,2.0)(2,3.2)
 \pscurve[linewidth=0.05,linecolor=green](2.0,4.5)(0.7,2.5)(2.0,0.5)
 \psline[linewidth=0.05,linecolor=green](2.0,4.5)(2.0,5.1)
 \psline[linewidth=0.05,linecolor=green](2.0,0.5)(2.0,0.1)
 \uput[0](2.0,4.5){$2i$}  \uput[0](2.00,0.5){$-2i$}
 \uput[135](0.6,2.5){$-1.325$}
\end{pspicture}

\vspace*{2.5cm}

Finally, for the error $e^z-$ approximant, the powers are

\begin{center}
\psset{unit=1cm}
\begin{pspicture}(0,2.5)(4,5.2)
  \psline{->}(1.5,2.5)(2.5,2.5)
  \psline{->}(2.0,2.0)(2,3.2)
 \pscurve[linewidth=0.05,linecolor=green](2.0,4.5)(0.7,2.5)(2.0,0.5)
 \pscurve[linewidth=0.05,linecolor=red](2.0,4.5)(3.3,2.5)(2.0,0.5)
 \psline[linewidth=0.05](2.0,4.5)(2.0,5.1)
 \psline[linewidth=0.05](2.0,0.5)(2.0,0.1)
 \uput[0](2.0,4.5){$2i$}  \uput[0](2.00,0.5){$-2i$}
 \uput[135](0.6,2.5){$-1.325$}       \uput[45](3.3,2.5){$1.325$}
  \uput[0](3.5,4){$e^{z/n}$}
  \uput[0](1.2,1.6){$-w^2e^{z/n}$}
  \uput[0](-6,2){$ \displaystyle{  \dfrac{1+\sqrt{1+z^2/(4n^2)}}
                                         {1-\sqrt{1+z^2/(4n^2)}}\;%
                           e^{z/n-2\sqrt{1+z^2/(4n^2)}}  }$}
\end{pspicture}   \label{padefig}
\end{center}

\vspace*{2.5cm}

\medskip

The example just seen shows the interest of a technique much used by
J.~Nuttall, for instance in  \cite[\S~3.2]{NuttJAT},
consisting in building asymptotic
approximations $\chi_1$, $\chi_2$, and $\chi_3$
to the remainder, denominator, and numerator\footnote{In
  \cite[\S~3.2]{NuttJAT}, $\chi_1$, $\chi_2$, and $\chi_3$ are
called $R$, $\chi_1$, and $\chi_2$.}.
 These $\chi$ functions,
which are here these $n^{\text{th}}$ powers of exponentials and
algebraic functions, are expected to be piecewise analytic,



\subsection{Pad\'e approximation to $\exp(nB_1 z+n B_2 z^2)$ and a bit
of Painlev\'e  \label{Padexp2} functions}  \   \\


Laguerre wrote quite a number of papers dealing with continued
fraction expansion of solutions of linear differential equations
of first order with polynomial coefficients. His theory culminates in
\cite{Lag2}, but he considered separately in \cite{Lag1} exponentials
of polynomials $\exp(F(z))$. Laguerre starts immediately with the
Pad\'e property (here, about 0):
$$ e^{F(z)}-\frac{q_n(z)}{p_n(z)} = O(z^{m+n+1})
  =\epsilon_n z^{m+n+1} +\eta_n z^{m+n+2}+\dotsb
,$$
if the degrees of $q_n$ and $p_n$ are $m$ and $n$ ($m-n$ a small fixed
integer). Divide by $e^F$ and derivate:
$$
  -\dfrac{d}{dz}\left( \dfrac{q_n}{p_n e^F}\right) =
  e^{-F(0)} \left[
  (m+n+1)\epsilon_n  z^{m+n} +(m+n+2)(\eta_n-\epsilon_n F'(0))
   z^{m+n+1}+\dotsb \right]
$$
The left-hand side must be a polynomial divided by $p_n^2 e^F$. The
left-hand side numerator is $-q_n p_n(q_n'/q_n - p'_n/p_n -F') =
q_n p_n F' -p_n q_n' +p_n' q_n$ of degree $m+n+g$ if the degree
of $F$ is $g+1$ (yes, $g$ will be the genus of something\dots) And has
the right-hand side has order $m+n$ at 0, we have
\begin{equation}  \label{Thetan}
  -\dfrac{d}{dz}\left( \dfrac{q_n(z)}{p_n(z) e^F(z)}\right) =
 \frac{ z^{m+n} \epsilon_n \Theta_n(z) }{p_n^2(z) e^{F(z)} }
\end{equation}
with $\Theta_n$ of degree $g$. Remark that $\Theta_n(0)=m+n+1$ if
we choose $p_n(0)=1$. As the right-hand side is the
derivative of a meromorphic function, it must have vanishing
residues: $\displaystyle{ \frac{m+n}{z} +\frac{\Theta_n'(z)}{\Theta_n(z)}
-\frac{p_n''(z)}{p_n'(z)} -F'(z) =0}$ at the zeros of $p_n$, whence
\begin{equation}  \label{Thetan2}
z\Theta_n(z) p_n''(z) +[zF'(z)\Theta_n(z)-z\Theta_n'(z)-(m+n)
   \Theta_n(z)]p_n'(z) +M_n(z) p_n(z)=0,
\end{equation}
where $M_n$ is a new polynomial of degree $2g$.

We need to know a little more on $\Theta_n$ and $M_n$ for large $n$, in
order to extract the asymptotic behaviour of $p_n$.

Any other solution $y$ of \eqref{Thetan2} satisfies
$\displaystyle{ \left( \frac{m+n}{z} +\frac{\Theta_n'(z)}{\Theta_n(z)}
 -F'(z)\right)\left( \frac{p_n'(z)}{p_n(z} - \frac{y'}{y}\right) =
                     \frac{p_n''(z)}{p_n(z} - \frac{y''}{y}  }$, or
\\
$\displaystyle{ p'_n y-p_n y' = \text{const.\ } z^{m+n} e^{-F} \Theta_n }$,
whence $\displaystyle{ \left(\frac{y}{p_n}\right)' = \text{const.\ }
 \frac{ z^{m+n} \Theta_n}{e^F \; p_n^2} }$, showing that
$$
   A p_n(z) + B q_n(z) e^{-F(z)}
$$
is the general solution of \eqref{Thetan2}.

Such differential equations with apparent singular points (as all
solutions are entire functions) have been much worked by W.~Hahn
\cite{ha49,ha76,ha1,ha2}. The coefficients must be Painlev\'e-like
functions (see \cite{MagP}, containing references to R.~Fuchs,
D. \& G. Chudnovsky, etc.).

We use $q_n p_{n-1} - p_n q_{n-1} = \epsilon_{n-1} z^{m+n-1}$ to
write \eqref{Thetan} as
$$
  -\dfrac{d}{dz}\left( \dfrac{q_n(z)}{p_n(z) e^{F(z)}}\right) =
 \frac{ z \epsilon_n
       [q_n p_{n-1} - p_n q_{n-1}]
 \Theta_n(z) }{\epsilon_{n-1} p_n^2(z) e^{F(z)} },
$$
or
$$
q_n [p_n' +F' p_n] -q_n' p_n = z  \frac{\epsilon_n}{\epsilon_{n-1}}
       [q_n p_{n-1} - p_n q_{n-1}]  \Theta_n,
$$

$$
q_n \left[p_n' - z  \frac{\epsilon_n}{\epsilon_{n-1}} \Theta_n p_{n-1}
    \right] =
p_n \left[q_n' -F'q_n
               - z  \frac{\epsilon_n}{\epsilon_{n-1}} \Theta_n q_{n-1}
    \right]
$$
must be a polynomial of the form $p_n q_n \Omega_n$, with $\Omega_n$
of degree $g$. Whence the \emph{\textbf{differential system}} of
first order
$$
  p_n' = \Omega_n p_n +z \frac{\epsilon_n}{\epsilon_{n-1}}  \Theta_n  p_{n-1}\  ,
  q_n' = (F'+\Omega_n) q_n +z \frac{\epsilon_n}{\epsilon_{n-1}}  \Theta_n  q_{n-1}.
$$
Conditions at $n=0$: $p_0=1$, $p_{-1}=0$, $q_0=$ Taylor-Maclaurin
expansion of $e^F$ truncated to degree $m-n$.
 $q_n p_{n-1} - p_n q_{n-1} = \epsilon_{n-1} z^{m+n-1} \Rightarrow$
$q_{-1}= -\epsilon_{-1} z^{m-n-1}$. Also, $\Omega_0=0$.

The same equations for $p_{n-1}$ and $q_{n-1}$ involve $p$ and $q_{n-2}$,
which we eliminate through the three-terms recurrence relation
$p_n(z) = (1-\alpha_n z) -\dfrac{\epsilon_{n-1}}{\epsilon_{n-2}}
  z^2 p_{n-2}(z)$, and we arrive at
\begin{equation} \label{YA}
 \boldsymbol{Y_n}' = \boldsymbol{A}_n\boldsymbol{Y}_n\  ,\  %
\boldsymbol{Y}_n = \begin{bmatrix} p_n  &  q_n e^{-F} \\
                                   p_{n-1}  &  q_{n-1} e^{-F}
                   \end{bmatrix} \  ,  \   %
\boldsymbol{A}_n = \begin{bmatrix} \Omega_n  &  z \dfrac{\epsilon_n}
                     {\epsilon_{n-1}} \Theta_n \\
            -z^{-1}\Theta_{n-1}    & \Omega_{n-1}+(z^{-1}-\alpha_n)
                  \Theta_{n-1}
                   \end{bmatrix} .
\end{equation}
Remark that $\boldsymbol{Y}_0$ being upper triangular, so is
$\boldsymbol{A}_0 = (\boldsymbol{Y}'_0)^{-1}
                     \boldsymbol{Y}_0 \Rightarrow \Theta_{-1}=0$.

From trace $\boldsymbol{A}_n = (\log\det\boldsymbol{Y})'$,
\begin{equation} \label{Omegann1}
  \Omega_n(z)+\Omega_{n-1}(z) = - \frac{ (1-\alpha_n z)\Theta_{n-1}(z)
       +zF'(z) -m-n+1}{z}.
\end{equation}
When we reconstruct the scalar second order differential
equation for $p_n$ using now

$\displaystyle{
\boldsymbol{A}_n = \begin{bmatrix} \Omega_n  &  z \dfrac{\epsilon_n}
                     {\epsilon_{n-1}} \Theta_n \\
            -z^{-1}\Theta_{n-1}    & z^{-1}(m+n-1)-F'-\Omega_n
                   \end{bmatrix}
            }$,
we recover the two first terms of \eqref{Thetan2}, and conclude that
$M_n = z(\Omega_n \Theta'_n - \Omega'_n \Theta_n) +(m+n)\Omega_n
  \Theta_n -z\Theta_n[\Omega_n^2 +F'\Omega_n -(\epsilon_n/\epsilon_{n-1})
   \Theta_n\Theta_{n-1}]$.
This complicated formula makes not even clear if $M_n$ has indeed
degree $2g$ (the degree seems to be $3g+1$). We still need more
relations by looking at the first order differential equation
for $p_{n+1}(z)= (1-\alpha_{n+1} z)p_n(z)-(\epsilon_n/ \epsilon_{n-1})
z^2 p_{n-1}(z)$ to find
$$
 (1-\alpha_{n+1} z)(\Omega_n-\Omega_{n+1}) +(\epsilon_n/\epsilon_{n-1})
  z\Theta_{n-1} -\alpha_{n+1} = z (\epsilon_{n+1}/\epsilon_n)
  \Theta_{n+1},
$$
or, at $n-1$,
\begin{equation} \label{Omegann12}
 (1-\alpha_n z)(\Omega_n-\Omega_{n-1}) =
          \dfrac{\epsilon_{n-1}}{\epsilon_{n-2}}
  z\Theta_{n-2} -\alpha_n - \dfrac{\epsilon_n}{\epsilon_{n-1}}
  z\Theta_n,
\end{equation}
leading to, by multiplying \eqref{Omegann1} by \eqref{Omegann12},
and by summation in $n$:
\begin{equation} \label{Thetann1}
\Omega_n^2 -\frac{\epsilon_n}{\epsilon_{n-1}} \Theta_n\Theta_{n-1}
  +\Omega_n\frac{ zF' -m-n}{z} =
 \frac{ \displaystyle{ \sum_0^{n-1} \Theta_k \; -mn}}{z^2}
  +\dfrac{nF'}{z}\;.
\end{equation}

\smallskip

It is then possible to advance the recurrence for the $\Theta$'s and
the $\Omega$'s: if $\Theta_{n-1}$, $\Omega_{n-1}$, and
$\epsilon_{n-1}/\epsilon_{n-2}$ are known, we find
\begin{equation} \label{alphaomega}
 \alpha_n = \frac{ 2\Omega_{n-1}(0) +\Theta'_n(0)+F'(0)}{m+n}
\end{equation}
from the difference of \eqref{Omegann1} and \eqref{Omegann12}
at $z=0$; then \eqref{Omegann1} yields $\Omega_n$, and \eqref{Omegann12}
gives $\Theta_n$ and $\epsilon_n/\epsilon_{n-1}$, knowing that
$\Theta_n(0)=m+n+1$.

With $F(z)=Az$, one finds the explicit solution $\Theta_n=m+n+1$,
$\Omega_n= -\dfrac{nA}{m+n}$, $\alpha_n = \dfrac{(m-n)A}{(m+n)(m+n-2)}$,
($\alpha_1=\dfrac{A}{m-n+2}$),
$\epsilon_n/\epsilon_{n-1} = -\dfrac{mnA^2}{(m+n-1)(m+n)^2(m+n+1)}$.

See later for a much more difficult example of degree 2 for $F$.

\medskip

The recurrence equations \eqref{Omegann1} and \eqref{Thetann1} have
a structure somewhat similar to the equations \eqref{ZZ}! This is an
important remark already made by Gammel and Nuttall
in \cite{GaN} also about the Laguerre theory.

We are very close to \eqref{ZZ} by taking $X_n = \Omega_n +F'/2
  -(m+n)/(2z)$ and $ \Theta_n = Z_{n+1}$. But what is $Y$?

\medskip

Ah, Painlev\'e. Let us introduce a parameter $t$ in $F$. Then, the
coefficients of $p_n$, $q_n$, etc. will be functions of $t$. The
Painlev\'e-like equations are differential equations in $t$ for the
coefficients of the polynomials $\Theta_n$ and $\Omega_n$, for each
$n$.

Let $\boldsymbol{H}_n := \dot{\boldsymbol{Y}}_n \boldsymbol{Y}_n^{-1}$.
From \eqref{YA},
$$
\boldsymbol{H}_n = \frac{-1}{\epsilon_{n-1}z^{m+n-1}} \begin{bmatrix}
 \dot{p}_n q_{n-1} -\dot{q}_n p_{n-1}    +\dot{F} q_n p_{n-1} &
 -\dot{p}_n q_{n} +\dot{q}_n p_{n}    -\dot{F} q_n p_{n} \\
 \dot{p}_{n-1} q_{n-1} -\dot{q}_{n-1} p_{n-1} +\dot{F} q_{n-1} p_{n-1} &
 -\dot{p}_{n-1} q_{n} +\dot{q}_{n-1} p_{n} -\dot{F} q_{n-1} p_{n}
 \end{bmatrix}
$$
is a matrix of rational functions in $z$. Remark that
trace~$\boldsymbol{H}_n =
\dfrac{\dot{\epsilon}_{n-1}}{\epsilon_{n-1}} -\dot{F}$.
 Is there such a big pole
at the origin? No: from $q_n(z)e^{-F(z)} -p_n(z) = -\epsilon_n
 z^{m+n+1} +\dotsb$,
\begin{equation*}\begin{split}
 \dot{p}_{n-r} q_{n+s-1} -\dot{q}_{n-r} p_{n+s-1} +\dot{F} q_{n-r} p_{n+s-1}
  &= e^F\left\{
 \dot{p}_{n-r} e^{-F}q_{n+s-1} - \frac{\partial}{\partial t}\left[
 e^{-F}{q}_{n-r} \right] p_{n+s-1} \right\}  \\
  &= e^F\left\{ p_{n+s-1} \frac{\partial}{\partial t}\left[
     p_{n-r}-e^{-F} q_{n-r} \right] +\dot{p}_{n-r}
      (e^{-F}q_{n+s-1}-p_{n+s-1}) \right\}
\end{split}\end{equation*}
showing that the orders in $\boldsymbol{H}_n$ are
${1\ 2}\choose{0\ 0}$ ($r,s=0,1$), so, polynomials of degrees not higher
than $g+1$, much less if only coefficients of low degree in $F$ actually
depend on $t$: if the degree of $\partial F/\partial t$ is $\delta$,
then the degrees of the elements
                        of $\boldsymbol{H}_n$ are
${\delta\ \  , \delta+1}\choose{\delta-1,\delta}$.
 More precisely, the coefficients of lowest order in
$\boldsymbol{H}_n$ are
$\displaystyle{  \begin{bmatrix}
    \dot{p}_n'(0) z   &  \dfrac{\dot{\epsilon}_n}{\epsilon_{n-1}} z^2 \\
    -\dfrac{\dot{\epsilon}_{n-1}}{\epsilon_{n-1}}  &
     \dfrac{\dot{\epsilon}_{n-1}}{\epsilon_{n-1}}   \end{bmatrix}
     }$.

When $\delta=1$ (and when $F(0)$ does not depend on $t$), we have
$$ \boldsymbol{H}_n=
    \begin{bmatrix}
    \dot{p}_n'(0) z   &  \dfrac{\dot{\epsilon}_n}{\epsilon_{n-1}} z^2 \\
    -\dfrac{\dot{\epsilon}_{n-1}}{\epsilon_{n-1}}  &
     \dfrac{\dot{\epsilon}_{n-1}}{\epsilon_{n-1}} -\dot{p}_n'(0)z-\dot{F}
     \end{bmatrix}  .
$$

And the differential equation in $t$ for $\boldsymbol{A}_n$ comes
from
$$
\frac{\partial}{\partial z} \frac{\partial \boldsymbol{Y}_n}{\partial t}=
\frac{\partial \boldsymbol{H}_n\boldsymbol{Y}_n}{\partial z}=
 \frac{\partial \boldsymbol{H}_n}{\partial z} \boldsymbol{Y}_n
 + \boldsymbol{H}_n \boldsymbol{A}_n \boldsymbol{Y}_n
$$
$$  =
\frac{\partial}{\partial t} \frac{\partial \boldsymbol{Y}_n}{\partial z}=
\frac{\partial \boldsymbol{A}_n\boldsymbol{Y}_n}{\partial t}=
 \frac{\partial \boldsymbol{A}_n}{\partial t} \boldsymbol{Y}_n
 + \boldsymbol{A}_n \boldsymbol{H}_n \boldsymbol{Y}_n
$$
whence
\begin{equation} \label{AHHA}
 \frac{\partial \boldsymbol{A}_n}{\partial t} =
 \frac{\partial \boldsymbol{H}_n}{\partial z}
 + \boldsymbol{H}_n \boldsymbol{A}_n
 - \boldsymbol{A}_n \boldsymbol{H}_n
\end{equation}

\medskip

We apply to the Pad\'e approximations $[n-1/n$ to
$\exp(\alpha x +\beta x^2)$. It is not bad to look at low degrees,
with the \texttt{GP-PARI} program \cite{PARI}:

\begin{verbatim}
{
/* padee21.gp:  launch gp and type   \r padee21

Pade [n-1 / n]  de exp(ax+b x2)

*/
default(seriesprecision,10);

Yn=matrix(2,2); An=matrix(2,2) ; Hn=matrix(2,2) ;

F=a*x+b*x^2;  Fp=deriv(F,x);        eF=exp(F);

/* n=0:  */   thn1=0+0*x;  omn1=0+0*x ;   epn1=0;    thn2=thn1;
      ep=1;  pn=1; pn1=0 ;  qn=0 ; qn1=0;
 for(n=1,3,
  print(" n=",n);
  m=n-1;
  aln=(2*polcoeff(omn1,0)+polcoeff(thn1,1)+polcoeff(Fp,0))/(m+n);
  print("    alphan=",aln);
  omn=-omn1-(1-aln*x)*thn1/x-Fp+(m+n-1)/x;
  thn=((1-aln*x)*(omn1-omn)+epn1*x*thn2-aln)/x;
  epn=polcoeff(thn,0)/(m+n+1);
  pnp=(1-aln*x)*pn-epn*x^2*pn1;
  qnp=(1-aln*x)*qn-epn*x^2*qn1;
  pn1=pn; pn=pnp; qn1=qn; qn=qnp;
  if(n==1,qn=1);
  print(" qn= ",qn);
  print(" pn= ",pn);
  thn=thn/epn;
  ep=ep*epn;
  print("   epsilon n=",ep);
  print(" Theta n =",thn);
  print(" Omega n =",omn);
    Yn[1,1]=pn;Yn[1,2]=qn/eF;
    Yn[2,1]=pn1;Yn[2,2]=qn1/eF;
    An=deriv(Yn,x)*Yn^(-1);
    Hn=deriv(Yn,a)*Yn^(-1);
   print("check: ",deriv(An,a)-deriv(Hn,x)-Hn*An+An*Hn);
  omn1=omn; epn1=epn; thn2=thn1;thn1=thn;
    );
}

Script V1.1 session started Wed Oct 11 16:07:49 2000

C:\calc\pari>pari217

                    GP/PARI CALCULATOR Version 2.0.17 (beta)
                  Windows NT ix86 (ix86 kernel) 32-bit version
                (readline disabled, extended help not available)

                           Copyright (C) 1989-1999 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

? \r padee2
 n=1
    alphan=a
 qn= 1
 pn= -a*x + 1
   epsilon n=-1/2*a^2 + b
 Theta n =-4*a*b/(-a^2 + 2*b)*x + 2
 Omega n =-2*b*x - a
check: [O(x^9), O(x^9); 0, 0]
 n=2
    alphan=(a^3 - 6*a*b)/(-3*a^2 + 6*b)
 qn= ((-a^3 + 6*a*b)/(-3*a^2 + 6*b))*x + 1
\end{verbatim}

etc. In a more readable writing:
$$
 \dfrac{q_1}{p_1}= \dfrac{1}{1-\alpha z}\  ,  \; \alpha_1 =\alpha\;,\;%
  \epsilon_1= \dfrac{2\beta-\alpha^2}{2},
$$
$$ \Theta_1=2+\dfrac{4\alpha\beta}{\alpha^2-2\beta}\,z\  \quad , \quad  %
   \Omega_1 = -F' = -\alpha -2\beta z;$$
$$
 \dfrac{q_2}{p_2}= \dfrac{
    1+\alpha\dfrac{\alpha^2-6\beta}{3(\alpha^2-2\beta)}\,z }
                         {
    1-\dfrac{2\alpha^3}{3(\alpha^2-2\beta)}\,z
     +\dfrac{\alpha^4+12\beta^2}{6(\alpha^2-2\beta)}\,z^2 }\ ,\ %
$$
$$
\alpha_2 = -\alpha\dfrac{\alpha^2-6\beta}{3(\alpha^2-2\beta)}\ ,\ %
\epsilon_2 = \dfrac{
  \alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3}
         {72(\alpha^2-2\beta)}\ ,
$$
$$ \Theta_2 = 4+ 8\alpha\beta
     \dfrac{(\alpha^2 -6\beta)( \alpha^4 +12\beta^2)}
{(\alpha^2-2\beta) (\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3)}
z\;,\  \Omega_2 = -\dfrac{2\alpha^3}{3(\alpha^2-2\beta)}
  -\dfrac{4\beta\alpha^2}{3}\;
  \dfrac{\alpha^2-6\beta}{(\alpha^2-2\beta)^2}\;z;
$$
$$\hspace*{-32pt}
 \dfrac{q_3}{p_3}= \dfrac{ 1+   \dfrac{  \dfrac{2\alpha^3}{5}
        (\alpha^4 -12\beta\alpha^2 +60\beta^2) z +\dfrac{1}{20}
  (\alpha^8 -16\beta\alpha^6 +120 \beta^2\alpha^4 +720\beta^4) z^2
                                      }
               {\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3}
                         }
                         { 1+ \dfrac{ \dfrac{-3\alpha}{5}
      (\alpha^6 -2\beta\alpha^4 +20\beta^2\alpha^2+120\beta^3)z
 +\dfrac{3}{20}(\alpha^8+40\beta^2\alpha^4 -240\beta^4) z^2
 -\dfrac{\alpha}{60}(\alpha^8+72\beta^2\alpha^4 -2160\beta^4)z^3
                         }
               {\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3}
                         }
\  ,
$$
$$
\alpha_3= -\dfrac{\alpha}{15}\, \dfrac{\alpha^8-24\beta\alpha^6
           +144\beta^2\alpha^4+2160\beta^4}{(\alpha^2-2\beta)
                 (\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3)}
$$
$$
 \epsilon_3=-\dfrac{\alpha^{12} -12\beta\alpha^{10}
 +180\beta^2\alpha^8 -480\beta^3 \alpha^6 -3600\beta^4\alpha^4
    -43200\beta^5\alpha^2 +43200\beta^6}
        {7200 (\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3)}
$$
$$\hspace*{-32pt}
\Theta_3 = 6+
\dfrac{12\alpha\beta(\alpha^8-16\beta\alpha^6+120\beta^2\alpha^4+720\beta^4)
       (\alpha^8+72\beta^2\alpha^4-2160\beta^4)}
       { (\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3)
       (\alpha^{12} -12\beta\alpha^{10}
 +180\beta^2\alpha^8 -480\beta^3 \alpha^6 -3600\beta^4\alpha^4
    -43200\beta^5\alpha^2 +43200\beta^6)} z,
$$
$$
\Omega_3 = -\dfrac{3\alpha}{5}\, %
    \dfrac{\alpha^6-2\beta\alpha^4+20\beta^2\alpha^2 +120\beta^3}
    {\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3}
    -\dfrac{6\beta}{5}\, \dfrac{ (\alpha^4+12\beta^2)
    (\alpha^8-16\beta\alpha^6+120\beta^2\alpha^4+720\beta^4)}{
    (\alpha^6-6\beta\alpha^4+36\beta^2\alpha^2+72\beta^3)^2} z.
$$

\medskip

 Hmm\ldots these functions contain more and more complicated
terms. For each $n$, there are basically four non obvious terms
to consider: $\epsilon_n$, $\Omega_n(0)$, and the coefficients of $z$
of $\Theta_n$ and $\Omega_n$, i.e., $\Theta'_n$ and $\Omega'_n$.
It happens that, if one looks for differential equations (in $t=\alpha$)
for these four terms through \eqref{AHHA}, what comes out is a real
mess (I tried).

\medskip

One finds more tractable expressions by considering the even
function $\exp(\widetilde{F}(Z))$, with
$\widetilde{F}(Z)= F(Z^2) = \alpha Z^2+\beta Z^4$. Nothing is lost,
as the $[2n-1/2n]$ Pad\'e approximants to $\exp(F(Z^2))$ is related to
the $[n-1/n]$ Pad\'e approximant to $\exp(F)$:
with $z=Z^2$,
$$ e^{F(z)}-\dfrac{q_n(z)}{p_n(z)} = \epsilon_n z^{2n} +\dotsb
  \Rightarrow
 e^{\widetilde{F}(Z)}-\dfrac{\tilde{q}_{2n}(z)}{\tilde{p}_{2n}(z)} =
  \tilde{\epsilon}_{2n} Z^{4n} +\dotsb ,
$$
so, $p_n(z)=\tilde{p}_{2n}(Z) =  \tilde{p}_{2n}(\sqrt{z}) $,
    $q_n(z)=\tilde{q}_{2n}(Z) =  \tilde{q}_{2n}(\sqrt{z}) $,
 $\tilde{\epsilon}_{2n} = \epsilon_n$. Remark that $\tilde{q}_{2n}$ is
an even polynomial of true degree $2n-2$.

The Laguerre theory above (p.~\pageref{Padexp2}) builds the $\Theta$'s
and the $\Omega$'s for this $\exp(\alpha Z^2+\beta Z^4)$ function, which
one writes $\widetilde{\Theta}_N$ and $\widetilde{\Omega}_N$ (and which
may be interesting on their own right):

\smallskip

\begin{tabular}{|c|c|c|c|c|}   \hline
      &       &        &     &       \\
 N & $\dfrac{\tilde{q}_N(Z)}{\tilde{p}_N(Z)}$   &  $\widetilde{\Theta}_N(Z)$
    &  $\widetilde{\Omega}_N(Z)$  &  $\tilde{\epsilon}_N$  \\
      &       &        &     &       \\
\hline      &       &        &     &       \\
    0   &  $\dfrac{0}{1}$  &    0    &    0   &  1   \\
      &       &        &     &       \\
\hline      &       &        &     &       \\
   1   &  $\dfrac{1}{1}$  & $2\left(1+2\dfrac{\beta}{\alpha}Z^2\right)$
    & $-2\alpha Z -4\beta Z^3$  &  $\alpha$  \\
      &       &        &     &       \\
\hline      &       &        &     &       \\
 2 &  $\dfrac{1}{1-\alpha Z^2}$ &
  $4\left( 1+\dfrac{2\alpha\beta}{\alpha^2-2\beta} Z^2\right)$ &
  $ -\dfrac{4\beta}{\alpha}Z$   &  $-\dfrac{\alpha^2-2\beta}{2}$ \\
      &       &        &     &       \\
\hline      &       &        &     &       \\
 3  &  $\dfrac{1+\displaystyle{\dfrac{\alpha^2-2\beta}{2\alpha}Z^2}}
                    {1-\displaystyle{\dfrac{\alpha^2+2\beta}{2\alpha}Z^2}}
  $  &
 $6\left( 1+\dfrac{2\beta(\alpha^4-4\beta^2)}{\alpha(\alpha^4+12\beta^2)}
          Z^2\right)$  &
 $-\dfrac{2(\alpha^4+4\beta^2)}{\alpha(\alpha^2-2\beta)}Z -4\beta Z^3$  &
  $-\dfrac{\alpha^4+12\beta^2}{12\alpha}$  \\
      &       &        &     &       \\
\hline
\end{tabular}


\smallskip
\medskip

Now, $\widetilde{\Theta}_N$ is an even polynomial of degree 2, whose
constant coefficient is $2N$; $\widetilde{\Omega}_N$ is an odd
polynomial of degree 3, whose coefficient of $Z^3$ is given by a
simple rule, easily deduced from \eqref{Omegann1} and
$\widetilde{\Omega}_0=0,
 \widetilde{\Omega}_1=-\widetilde{F}' = -2\alpha Z-4\beta Z^3$:
this coefficient of $Z^3$ is $2\beta((-1)^N-1)$.

So, we now have three unknown terms for each $N$: $\epsilon_N$, and,
say $\xi_N$ and $\chi_N$ in
$\widetilde{\Theta}_N(Z)= 2N(1+\xi_N Z^2)$ and
$\widetilde{\Omega}_N(Z)= 2\beta((-1)^N-1)Z^3 -\chi_N Z$.

We know apply the differential relations \eqref{AHHA} with $t=\alpha$,
remarking first that the matrix $\widetilde{\boldsymbol{H}}_N$ is
made of even polynomials, therefore of degrees
${2  , 2}\choose{0,2}$, (as $\delta=$ degree of $\partial\widetilde{F}/
 \partial\alpha = 2$), and orders
${2  , 2}\choose{0,0}$, (as seen before): we have monomials, excepting
the second diagonal element,
$$
 \widetilde{\boldsymbol{H}}_N(Z) =
  \begin{bmatrix}  \zeta_N Z^2  &  \dfrac{\dot{\tilde{\epsilon}}_N}
                                   {\tilde{\epsilon}_{N-1}} Z^2 \\

 -\dfrac{\dot{\tilde{\epsilon}}_{N-1}}{\tilde{\epsilon}_{N-1}} &
 \dfrac{\dot{\tilde{\epsilon}}_{N-1}}{\tilde{\epsilon}_{N-1}}
 -\dot{\widetilde{F}}(Z) -\zeta_N Z^2  \end{bmatrix},
$$
as we don't yet know what the first diagonal could be, but we know the
trace.

\eqref{AHHA} yields
$$
\dfrac{\partial \widetilde{\Omega}_N}{\partial\alpha} =
 2  \zeta_N Z
+ Z \dfrac{\tilde{\epsilon}_N \dot{\tilde{\epsilon}}_{N-1}}
                          {\tilde{\epsilon}^2_{N-1}} \widetilde{\Theta}_N
- Z \dfrac{ \dot{\tilde{\epsilon}}_N}{\tilde{\epsilon}_{N-1}}
                        \widetilde{\Theta}_{N-1},
$$
for the two diagonal elements of $\widetilde{\boldsymbol{A}}_N$, leading to
the coefficient of $Z^2$ of $\widetilde{\Theta}_N$
$$
2N\xi_N =
\widetilde{\Theta}_{N,2} = 4\beta \dfrac{\dot{\tilde{\epsilon}}_N}
                 {\tilde{\epsilon}_n}\ ,  \qquad%
\dot{\chi}_N = 1-(-1)^N
-2N \dfrac{\tilde{\epsilon}_N \dot{\tilde{\epsilon}}_{N-1}}
                          {\tilde{\epsilon}^2_{N-1}} \widetilde{\Theta}_N
+2(N-1)\dfrac{ \dot{\tilde{\epsilon}}_N}{\tilde{\epsilon}_{N-1}}
,
$$
knowing that the coefficient of $Z^3$ in $\widetilde{\Omega}_N$
does not depend on $\alpha$, and
using $\widetilde{\Theta}_N(Z)=2N(1+\xi_N Z^2)$;
the off-diagonal yield, after cleaning a little bit,

\begin{multline*}
\dot{\widetilde{\Theta}}'_N =
    \dfrac{\dot{\tilde{\epsilon}}_N}{\tilde{\epsilon_N}}
       [2N-\widetilde{\Theta}_N -Z(\widetilde{F}'+2\widetilde{\Omega}_N)]
        +(1+2\zeta_N)Z^2\widetilde{\Theta}_N
       \;,\ \\
\dot{\widetilde{\Theta}}'_{N-1} =
    \dfrac{\dot{\tilde{\epsilon}}_{N-1}}{\tilde{\epsilon}_{N-1}}
       [\widetilde{\Theta}_N-(2N-2) +Z(\widetilde{F}'+2\widetilde{\Omega}_N)]
        -(1+2\zeta_N)Z^2\widetilde{\Theta}_{N-1}
       \;,\ %
\end{multline*}
whence $\zeta_N = ((-1)^N-1)/2$, as the coefficient of $Z^3$ of
$\widetilde{F}'+2\widetilde{\Omega}_N$ is $4(-1)^N\beta$, and as
the coefficient of $Z^2$ of $\widetilde{\Theta}_N$ is
       $ 4\beta \dot{\tilde{\epsilon}}_N /\tilde{\epsilon}_n$,
as seen above;
\begin{multline*}
2N\dot{\xi}_N =
  \dfrac{\dot{\tilde{\epsilon}}_N}{\tilde{\epsilon}_N}
     [-2N\xi_N -2\alpha +2\chi_N ] +2N(-1)^N\;,\;\\
2(N-1)\dot{\xi}_{N-1} =
 \dfrac{\dot{\tilde{\epsilon}}_{N-1}}{\tilde{\epsilon}_{N-1}}
    [2(N-1)\xi_{N-1} +2\alpha -2\chi_N] -2(N-1)(-1)^N\;,
\end{multline*}
amounting to the differential system
\begin{equation} \label{pain4}
  4\beta \dfrac{\partial}{\partial\alpha}
 \begin{bmatrix} \tilde{\epsilon}_N/\tilde{\epsilon}_{N-1} \\
            \chi_N \\   \xi_N \\  \xi_{N-1}
 \end{bmatrix} =
 \begin{bmatrix}
    (\tilde{\epsilon}_N/\tilde{\epsilon}_{N-1}) (2N\xi_N -2(N-1)\xi_{N-1}) \\
   4\beta [1-(-1)^N] - 4N(N-1)
   (\tilde{\epsilon}_N/\tilde{\epsilon}_{N-1}) (\xi_N - \xi_{N-1} ) \\
  \xi_N [-2N\xi_N -2\alpha +2\chi_N] +4\beta (-1)^N \\
  \xi_{N-1} [ 2(N-1)\xi_{N-1} +2\alpha-2\chi_N ] +4\beta (-1)^{N-1}
 \end{bmatrix}
\end{equation}
a system of 4 equations, but the coefficients of $Z^4$ and  $Z^2$
of \eqref{Thetann1} yield 2 first integrals
$$
(-1)^N \beta \chi_N + N(N-1)
 \dfrac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}}
  \xi_N \xi_{N-1} = \alpha\beta [(-1)^N-1],
$$
$$
 \chi_N^2 -4N(N-1)
 \dfrac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}}
  (\xi_N+ \xi_{N-1}) -2 \alpha \chi_N = 2\beta[(2N-1)(-1)^N +1].
$$
There must remain a system of two first order differential equations,
or scalar second order differential equations.

Well, let us try with $\xi_N$ first:

from the third row of \eqref{pain4},
$\chi_N = \alpha+N\xi_N+2\beta [\dot{\xi}_N -(-1)^N]/\xi_N$;
use the first integrals: \\
$\displaystyle{N(N-1)\dfrac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}}
  \xi_{N-1} = -\dfrac{ \alpha\beta}{\xi_N}
-(-1)^N\beta\dfrac{N\xi_N+2\beta [\dot{\xi}_N -(-1)^N]/\xi_N
                  }{\xi_N}     }$
$\displaystyle{ = -\frac{\alpha\beta}{\xi_N} -N\beta(-1)^N
  -2\beta^2\frac{ (-1)^N\dot{\xi}_N -1}{\xi_N^2}  }$,
$$
\frac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}} =
\frac{N}{4(N-1)}\xi_N
  +\frac{\beta}{N-1} \frac{\dot{\xi}_N}{\xi_N}
  -\beta\frac{(2N-1)(-1)^N+1}{2N(N-1)\xi_N}
  +\frac{\beta^2}{N(N-1)} \frac{1}{\xi_N} \left(\frac{\dot{\xi}_N}
                                            {\xi_N}\right)^2
  -\frac{1}{4N(N-1)\xi_N}\left( \frac{2\beta}{\xi_N}-\alpha\right)^2
$$
gives $\tilde{\epsilon}_N/\tilde{\epsilon}_{N-1}$
 as a function of $\xi_N$ and its
first derivative $\dot{\xi}_N = \partial\xi_N/\partial\alpha$.
We enter all that in the first equation of \eqref{pain4}, and get a
second order differential equation for $\xi_N$
\begin{equation} \label{painxi}
\frac{\ddot{\xi}_N}{\xi_N} -\frac{3}{2} \left( \frac{\dot{\xi}_N}
      {\xi_N}\right)^2 +\frac{1+(-1)^N}{4\beta}
  +\frac{\alpha^2}{8\beta^2}
  +\frac{3}{2\xi_N^2} -\frac{\alpha}{\beta\xi_N}
 -\frac{N^2\xi_N^2}{8\beta^2} =0
\end{equation}
which relates $\xi_N$ to a solution of the Painlev\'e-IV equation
by
\begin{equation} \label{painxid}
  \xi_N = \frac{ \sqrt{-4\beta} }{\pain_{IV,N}\left(\dfrac{\alpha}
                     {\sqrt{-4\beta}}\right) },
\end{equation}
$\pain_{IV,N}(x)$ being a solution of the Painlev\'e IV equation
\begin{equation} \label{painxit}
 p'' = \frac{p'{}^2}{2p} +p\frac{(3p+2x)(p+2x)}{2}
  -[1+(-1)^N]p -\frac{2N^2}{p}
\end{equation}
of parameters $[1+(-1)^N]/2$ and $-2N^2$ (see  \cite[chap.~3]{Iwasa} for the
Painlev\'e-IV equation, also Ince \cite[\S~14.4]{Ince} of course).

Remark that, as $\xi_N$ is rational in $\beta$, \eqref{painxid} shows
that $\pain_{IV,N}$ is an \emph{odd} function. From the instances of
$\Theta_n$ and $\widetilde{\Theta}_N$ seen above, we have some
samples

\smallskip

\begin{tabular}{|c|c|c|c|c|}
\hline  &   &   &   &   \\
     $N$&  1&  2&  3&  4 \\
        &   &   &   &   \\
\hline  &   &   &   &   \\
  $\pain_{IV,N}(x)$ & $-2x$ &  $-\dfrac{2x^2+1}{x}$   &
    $-\dfrac{2x(4x^4+3)}{4x^4-1}$  &
  $-\dfrac{(2x^2+1)(8x^6+12x^4+18x^2-9)}{x(2x^2+3)(4x^4+3)}$ \\
        &   &   &   &   \\
\hline\end{tabular}

\smallskip

For the other functions of interest,
$$ \chi_N = \sqrt{-4\beta}\left[ x+\frac{N}{\pain_{IV,N}}
   +\dfrac{\pain_{IV,N}'}{2\pain_{IV,N}}
   +\dfrac{(-1)^N}{2\pain_{IV,N}}\right], $$
$$ N(N-1)\dfrac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}} \xi_{N-1}=
 \beta\left[ -x\pain_{IV,N} -N(-1)^N -\dfrac{(-1)^N\pain_{IV,N}'}{2}
  +\dfrac{\pain_{IV,N}^2}{2}\right],
$$
$$ 8N(N-1)\dfrac{\tilde{\epsilon}_N}{\tilde{\epsilon}_{N-1}}=
   \sqrt{-4\beta}\left[ \dfrac{2N^2}{\pain_{IV,N}}
          +\dfrac{2N\pain_{IV,N}'}{\pain_{IV,N}}
          +[ (2N-1)(-1)^N+1]\pain_{IV,N}
   +\dfrac{\pain_{IV,N}'{}^2}{2\pain_{IV,N}}
  -\dfrac{\pain_{IV,N}}{2} (\pain_{IV,N}+2x)^2 \right]
$$

but what may be the use of all this?

To be continued\dots

\subsection{Rational interpolation}\  \\

 \subsubsection{Equidistant points.}  \cite{iser1}

As far as we only need $e^{Az}$ at $z=z_0, z_0+h,\dotsc, z_0+(m+n)h$,

\begin{equation*} \begin{split}
e^{Az} &= (\boldsymbol{I}+\boldsymbol{\Delta})^{(z-z_0)/h} e^{Az_0} \\
 &= \sum_{k=0}^{m+n}  { (z-z_0)/h  \choose  k} \boldsymbol{\Delta}^k
     e^{Az_0} \\
 &= \sum_{k=0}^{m+n} \left( \frac{ e^{Ah}-1}{h}\right)^k
       \frac{1}{k!} (z-z_0) (z-z_0-h) \dotsb (z-z_0-(k-1)h),
\end{split}\end{equation*}
which we multiply by the denominator $\displaystyle{ Q(z) =
\sum_{j=0}^n q_j (z-z_0)\dotsb (z-z_0-(j-1)h) }$, using
\begin{multline*}
 (z-z_0) (z-z_0-h) \dotsb (z-z_0-(j-1)h) e^{Az} = \\
   e^{A(z_0+jh)}\sum_{k=0}^{m+n} \left(\frac{e^{Ah}-1}{h}\right)^{k-j} \frac{1}{(k-j)!}
   (z-z_0)(z-z_0-h) \dotsb (z-z_0-(k-1)h),
\end{multline*}

\begin{equation*}
  Q(z) e^{Az} =
   e^{Az_0}\sum_{k=0}^{m+n} \left(\frac{e^{Ah}-1}{h}\right)^{k} \frac{C(k)}{k!}
   (z-z_0)(z-z_0-h) \dotsb (z-z_0-(k-1)h),
\end{equation*}

where $\displaystyle{ C(k)=\sum_{j=0}^n q_j e^{Ajh}
      \left(\frac{e^{Ah}-1}{h}\right)^{-j} \frac{1}{(k-j)!} }$ is a polynomial of
degree $n$ in $k$, which must vanish at $k=m+1, m+2,\dotsc, m+n$,

\begin{equation} \label{Pqrat} \begin{split}
 P(z) &= e^{Az_0} \sum_{k=0}^m \left(\frac{e^{Ah}-1}{h}\right)^k {m \choose k}
          (m+n-k)! (z-z_0)(z-z_0-h)\dotsb (z-z_0-(k-1)h), \\
 Q(z) &=  \sum_{k=0}^n \left(\frac{e^{-Ah}-1}{h}\right)^k {n \choose k}
          (m+n-k)! (z-z_0)(z-z_0-h)\dotsb (z-z_0-(k-1)h), \\
       &  \text{and,\ formally:}  \\
     Q(z)e^{Az}&-P(z) =   \\
 \hspace*{-33pt}
 e^{Az_0} m! (-1)^n&\sum_{k=m+n+1}^\infty
 \left(\frac{e^{Ah}-1}{h}\right)^k
      \frac{(k-m-1)(k-m-2)\dotsb (k-m-n)  }{k!}
         (z-z_0)(z-z_0-h)\dotsb (z-z_0-(k-1)h),
\end{split}\end{equation}

We look at the performance of some examples of the region of good
approximation in the complex plane, coloured in light gray:

\epsfxsize=477bp \epsfbox[0  200 477 350]{expval.ps}

colouring is made with respect of the average of
$\log| \exp(A z)-P(z)/Q(z)|$ in the square $[-2,2]\times [-2,2]$.
The degrees of $P$ and $Q$ are here 5 and 4. When $A$ is small, the
region is an oval around the locus of the interpolation points
(here, the interval $[-1,1]$ shown by a thin horizontal black
line).

\smallskip

The interpolation points should appear as bright white dots, but they
are hardly visible in somewhat big pixels, if colouring  is made according
to an arbitrary point of the pixel. This chosen point happens to be an
actual interpolation point only for the endpoints, whence the rightmost
interpolation point looking like a beacon in a dark environment.

The graph of the error function in the interpolation interval also
looks like its envelope:

 exp(    5(0.250z)),    -1     1

 2.6e-010    -2.3e-009

\hspace*{1cm}
\psset{xunit=0.5cm,yunit=3cm}\begin{pspicture}(-10,-2.6)(11,0.3)
\psline{->}(-10,0)(11,0)
\psline{->}(0,-2.6)(0,0.5)
\uput[90](-10,0){$-1$}      \uput[90](10,0){$1$}
\uput[60](11,0){$\zeta$}
\psline(-0.2,0.25)(0.2,0.25)   \uput[0](0.2,0.25){$2.5\ 10^{-10}$}
\psline(-0.2,-2.50)(0.2,-2.5)   \uput[0](0.2,-2.5){$-2.5\ 10^{-9}$}
\psline%
(-10,0)(-9.80,-0.195)(-9.60,-0.301)(-9.40,-0.343)(-9.20,-0.340)(-9.00,-0.308)(-8.80,-0.259)(-8.60,-0.202)(-8.40,-0.144)%
(-8.20,-0.0906)(-8.00,-0.0430)(-7.80,-0.00379)(-7.60,0.0262)(-7.40,0.0473)(-7.20,0.0599)(-7.00,0.0651)(-6.80,0.0642)(-6.60,0.0585)%
(-6.40,0.0494)(-6.20,0.0381)(-6.00,0.0258)(-5.80,0.0136)(-5.60,0.00232)(-5.40,-0.00748)(-5.20,-0.0153)(-5.00,-0.0209)%
(-4.80,-0.0243)(-4.60,-0.0254)(-4.40,-0.0245)(-4.20,-0.0219)(-4.00,-0.0179)(-3.80,-0.0129)(-3.60,-0.00746)(-3.40,-0.00183)(-3.20,0.00352)%
(-3.00,0.00827)(-2.80,0.0121)(-2.60,0.0148)(-2.40,0.0163)(-2.20,0.0165)(-2.00,0.0155)(-1.80,0.0133)(-1.60,0.0102)(-1.40,0.00633)%
(-1.20,0.00199)(-1.00,-0.00249)(-0.800,-0.00683)(-0.600,-0.0107)(-0.400,-0.0138)(-0.200,-0.0159)(0,-0.0169)%
(0.200,-0.0166)%
(0.400,-0.0149)(0.600,-0.0120)(0.800,-0.00802)(1.00,-0.00305)(1.20,0.00253)(1.40,0.00838)(1.60,0.0140)(1.80,0.0191)(2.00,0.0232)%
(2.20,0.0257)(2.40,0.0264)(2.60,0.0250)(2.80,0.0212)(3.00,0.0150)(3.20,0.00669)(3.40,-0.00362)(3.60,-0.0153)(3.80,-0.0278)%
(4.00,-0.0400)(4.20,-0.0509)(4.40,-0.0593)(4.60,-0.0640)(4.80,-0.0637)(5.00,-0.0571)(5.20,-0.0434)(5.40,-0.0220)(5.60,0.00715)%
(5.80,0.0436)(6.00,0.0860)(6.20,0.131)(6.40,0.178)(6.60,0.219)(6.80,0.250)(7.00,0.264)(7.20,0.253)(7.40,0.208)(7.60,0.120)%
(7.80,-0.0181)%
(8.00,-0.213)(8.20,-0.467)(8.40,-0.778)(8.60,-1.13)(8.80,-1.51)(9.00,-1.86)(9.20,-2.14)(9.40,-2.25)(9.60,-2.05)%
(9.80,-1.38)(10.0,0)
\end{pspicture}

\smallskip

This suggests that we shall have valuable asymptotic estimates containing
something like a nonvanishing power $\rho^n$, even up to the interpolation
locus.

Somewhat similarly
poles may even enter the locus of interpolation points. Here,
the poles of $P/Q$ (degrees $20/19$) interpolating $\exp(20az)$,
with $a=1$ and $a=2$:

\epsfxsize=477bp \epsfbox[0  600 477 815]{expra4.ps}

So, the locus of poles of $P/Q$ of degrees $m\sim n$ and $n$ approximating
$\exp(naz)$ enters the locus $[-1,1]$ of interpolation points when
$a$ becomes larger than a number slightly smaller than 2. Such
features will be explained.

\textbf{Integral form. }\label{jacobiint}

%For orthogonal polynomials, see Chen \& Ismail \cite{che}, Gawronski
%\& Shawyer \cite{Gaw}, also Perron

The sums and series \eqref{Pqrat} are special hypergeometric expansions
formally related to Jacobi polynomials with large symbols, say
$P_n^{(n\alpha+\beta, n\gamma+\delta)}(x)$. Detailed asymptotics of these
polynomials are achieved in \cite{che} and \cite{Gaw}. These authors used either
generating functions or integral forms. We adapt the integral form here:

the polynomials of \eqref{Pqrat} have the same form
\begin{equation}  \label{somjacobi}
   S =  \sum_{k=0}^M X^k {M\choose k} (P-k)! Y(Y-1)\dotsb (Y-k+1),
\end{equation}
where $P$ is an integer larger than $M$( in \eqref{Pqrat}, $X=e^{Ah}-1$ or
$e^{-Ah}-1$, $M=m$ or $n$, $P=m+n$, $Y=(z-z_0)/h )$).

As a hypergeometric function, $S=$ const. ${}_2F_1(-M, -Y; -P; X)$, which
is the Jacobi polynomial \\ $P_M^{(-P-1,-Y-M+P)}(1+2X)$ with parameters
$-P-1$ and $-Y-M+P$ which are never both positive, so that we cannot
just quote \cite{che,Gaw}, but follow their methods.

We try to write $(P-k)! Y(Y-1)\dotsb (Y-k+1)$ as an integral involving
a $k^{\text{th}}$ power, in order to achieve a closed form of
\eqref{somjacobi} through the binomial theorem.

The special Beta integral $\displaystyle{ \int_C u^{P-k} (1-u)^{k-Y-1}du}$,
where $C$ is an arc starting from the origin, turning around $u=1$, and
returning to the origin   \qquad %
\psset{unit=1cm}
\begin{pspicture}(0,0.4)(2,1)
 \psline[linewidth=0.05,linecolor=red](0.0,0.5)(0.5,0.40)   \psline[linewidth=0.05,linecolor=red](0.5,0.40)(1.5,0.40)
 \psline[linewidth=0.05,linecolor=red](0.0,0.5)(0.5,0.60)   \psline[linewidth=0.05,linecolor=red](0.5,0.60)(1.5,0.60)
 \psarc[linewidth=0.05,linecolor=red](1.75,0.5){0.25}{-160}{160}
%\psarc[linewidth=0.05,linecolor=red]{*-*}(2.15,2.5){2}{50}{310}
  \psline(0,0.5)(1.75,0.5)
  \psline{->}(0.6,0.3)(1,0.3)
  \psline{<-}(0.6,0.7)(1,0.7)
 \uput[0](2.0,0.5){$1$}  \uput[180](0,0.5){$0$}
\end{pspicture}
\qquad is what we need:

\medskip
 $\displaystyle{ \int_C u^{P-k} (1-u)^{k-Y-1}du = \dotsb =
 \frac{(-1)^{P-k}(P-k)!}{(Y-k)(Y-k-1)\dotsb (Y-P+1)}\int_C  (1-u)^{P-Y-1}du }$, or

 $\displaystyle{  (P-k)! Y(Y-1)\dotsb (Y-k+1) = (-1)^{P-k}
                         Y(Y-1)\dotsb (Y-P+1)
 \frac{ \int_C u^{P-k} (1-u)^{k-Y-1}du }
      {\int_C  (1-u)^{P-Y-1}du }  }$, then \eqref{somjacobi} becomes

$$
  S = (-1)^P
                         Y(Y-1)\dotsb (Y-P+1)
 \frac{ \displaystyle{\int_C u^{P-M} (1-u)^{-Y-1} [u(1+X)-X]^M\,du }}
      {\displaystyle{\int_C  (1-u)^{P-Y-1}du }} .
$$
(Check: if $X=-1$, $S=(P-M)! (Y-P)(Y-P+1)\dotsb (Y-P+M-1)$).

The formula is not intended to be of any use if $Y=P-1, P-2,\dotsc$ where
it yields $0/0$. The integral of the denominator
is $(1-e^{-2\pi i Y})/(P-Y)$ if the phase of $1-u$ is 0 in the first part
of the integral. Then, another form of the formula is
$$
  S = (-1)^{P-1} \frac{ \Gamma(Y+1)\Gamma(P-Y+1) e^{\pi i Y} }
                       {2\pi i}
 \int_C u^{P-M} (1-u)^{-Y-1} [u(1+X)-X]^M\,du
$$

{\bf Rough asymptotics.}
 Let $E:= \exp(Ah)$ and
 $\zeta:= \displaystyle{\frac{z-z_0 -(m+n)h/2}{nh} =
                        \frac{Y  -(m+n)/2}{n}
   }$. Then, with $m\sim n$,
we intend to follow things at constant $E$ and $\zeta$, i.e., a fixed
exponential and $z$ expanding linearly with $n$, or $A$ increasing linearly
with $n$, and $2n$ interpolating points filling a fixed segment
$[z_0, z_0+2nh]$. Remark that this segment of interpolation points
is $-1\leqslant \zeta\leqslant 1$.

Remark also that $e^{Az} = e^{Az_0} E^Y$, so
$$
 e^{Az}Q(z)-P(z) =
   (-1)^{m+n-1} e^{Az_0}\frac{ \Gamma(Y+1)\Gamma(m+n-Y+1) e^{\pi i (m+n)} }
                       {2\pi i}
$$
$$
 \int_C \{ u^{m} E^Y (1-u)^{-Y-1} [uE^{-1}-E^{-1}+1]^n
          -u^{n} (1-u)^{-Y-1} [uE-E+1]^m \}
 \,du
$$

...

Saddlepoint \& Stirling:

$$
S \approx e^{-P} P^P \left\{
  \frac{1}{4}  (1+\zeta)^{1+\zeta}   (1-\zeta)^{1-\zeta}
  u (1-u)^{-\zeta-1} [u(1+X)-X]  \right\}^n,
$$
where
$$
  \frac{1}{u}
  -\frac{\zeta+1}{u-1}
  +\frac{1}{u-\displaystyle{ \frac{X}{1+X} } }  =0,
$$
or
$$
  (\zeta-1)u^2 +\left(2-\frac{\zeta X}{1+X}\right)u -\frac{X}{1+X} =0
$$
so,
\begin{equation} \label{saddu}
  u= \frac{ 1+X -\zeta X/2 \pm
            \sqrt{ 1+X+\zeta^2 X^2/4 }  }
          {(1-\zeta)(1+X)}.
\end{equation}
One of these saddlepoints, or both, enter an asymptotic expression
of the denominator and the numerator. It is easy to get lost in
intractable, although elementary, formulas.

Strangely enough,
a cruder method ends up with almost readable expressions:

At least when each term has the same phase, the sum is roughly
given by the term reached when the preceding ratio has unit value:

\begin{equation} \label{kappa}
  \frac{X  \kappa (\zeta+\kappa)}{ 1-\kappa^2} =1,
\end{equation}
where $\kappa:= 1-k/n$. Remark that the ratio $\to 0$ when $\kappa\to 0$,
and $\to\infty$ when $\kappa\to 1$.
The roots are
$$
  \kappa= \frac{  -\zeta X/2 \pm
            \sqrt{ 1+X+\zeta^2 X^2/4 }  }
          {1+X},
$$
clearly related to the saddlepoints of \eqref{saddu} by
$$
  \kappa = (1-\zeta)u -1.
$$

The dominant term (there will sometimes be two dominant terms%, see
%also in \S~\ref{jacobiint}
), is, still roughly,
$$
  \frac{ X^{n-n\kappa}  n!  (n+n\kappa)!
             \Gamma(n\zeta+n) }
       {  (n-n\kappa)! (n\kappa)! \Gamma(n\zeta+n\kappa) },
$$
where $\kappa$ is one of the roots of \eqref{kappa}. Tedious
application of Stirling's formula yields, keeping only the
wildest factors,
\begin{align}
  & e^{-2n} n^{2n} \left[
     \frac{X^{1-\kappa} (1+\kappa)^{1+\kappa}
           (1+\zeta)^{1+\zeta}  }
         { (1-\kappa)^{1-\kappa} \kappa^\kappa
               (\zeta+\kappa)^{\zeta+\kappa} } \right]^n  \notag\\
 =&
   e^{-2n} n^{2n} \left[  \underbrace{ \left(
       \frac{1-\kappa^2}{X\kappa(\zeta+\kappa)}
         \right)^\kappa              }_1\;
     \frac{X (1+\kappa)
           (1+\zeta)^{1+\zeta}  }
         { (1-\kappa)
               (\zeta+\kappa)^\zeta } \right]^n  \notag\\
 =&
   e^{-2n} n^{2n} \left[
     \frac{X (1+\kappa)
           (1+\zeta)^{1+\zeta}  }
         { (1-\kappa)
               (\zeta+\kappa)^\zeta } \right]^n \notag \\
 =&
   e^{-2n} n^{2n} \left[
       \frac{(1+(1+X)\kappa)^{1+\zeta} (1+\kappa)^{1-\zeta} }
            {\kappa}  \right]^n, \label{ratio}
\end{align}
using $\displaystyle{  \zeta =
   \frac{ \kappa^{-1} -(1+X)\kappa }{X} }$, from
\eqref{kappa}.


Remark that, if $\kappa$ is one of the roots of \eqref{kappa},
for the numerator ($X=E-1$), the
other root is $-1/(E\kappa)$, and that the values of $\kappa$ for
the denominator expansion are the inverses of the ones for the
numerator:
$$ E^{-1}\kappa_{\text{den}}^2 +(E^{-1}-1) \zeta \kappa_{\text{den}} -1=0
 \Rightarrow
 E +(E-1) \zeta \kappa_{\text{den}} -\kappa_{\text{den}}^2 =0.
$$
 Finally, the last expansion of \eqref{Pqrat} yields the
same equation \eqref{kappa} as for the numerator (i.e., with
$X=E-1$), and we have the following possible
exponential behaviours:

\smallskip

\begin{tabular}{|c|c|c|c|}   \hline
      &       &        &          \\
 num. & $\displaystyle{ \frac{P(z)}{e^{-2n} n^{2n}  } }$ &
  $\displaystyle{ \left[  \frac{(1+E\kappa)^{1+\zeta} (1+\kappa)^{1-\zeta} }
                               {\kappa}  \right]^n }$&
  $\displaystyle{ \left[-E^\zeta\frac{(\kappa-1)^{1+\zeta} (E\kappa-1)^{1-\zeta} }
                               {\kappa}  \right]^n }$    \\
      &       &        &          \\
 \hline     &       &        &          \\
 den. & $\displaystyle{ \frac{Q(z)}{e^{-2n} n^{2n}  } }$ &
  $\displaystyle{ \left[ E^{-1-\zeta} \frac{(1+E\kappa)^{1+\zeta} (1+\kappa)^{1-\zeta} }
                               {\kappa}  \right]^n }$&
  $\displaystyle{ \left[-\frac{(\kappa-1)^{1+\zeta} (E\kappa-1)^{1-\zeta} }
                              {E \kappa}  \right]^n }$    \\
      &       &        &          \\
 \hline     &       &        &          \\
 ratio& $\displaystyle{ \frac{P(z)}{Q(z)} }$ &
  $\displaystyle{ \left[ E^{1+\zeta}   \right]^n }$&
  $\displaystyle{ E^{n(\zeta+1)}
  \left[-E^\zeta \frac{(\kappa-1)^{1+\zeta} (E\kappa-1)^{1-\zeta} }
        { (1+\kappa)^{1-\zeta} (1+E\kappa)^{1+\zeta} }    \right]^{\pm n} }$    \\
      &       &        &          \\
 \hline     &       &        &          \\
      & $\displaystyle{ \frac{Q(z)e^{Az}-P(z)}{e^{-2n} n^{2n}  } }$ &
  $\displaystyle{ \left[  \frac{(1+E\kappa)^{1+\zeta} (1+\kappa)^{1-\zeta} }
                               {\kappa}  \right]^n }$&
  $\displaystyle{ \left[-E^\zeta\frac{(\kappa-1)^{1+\zeta} (E\kappa-1)^{1-\zeta} }
                               {\kappa}  \right]^n }$    \\
      &       &        &          \\
 \hline     &       &        &          \\
 err.      & $\displaystyle{ e^{Az}-\frac{P(z)}{Q(z)}  }$ &
  $\displaystyle{ \left[ E^{1+\zeta}   \right]^n }$&
  $\displaystyle{ E^{n(\zeta+1)}
  \left[-E^\zeta \frac{(\kappa-1)^{1+\zeta} (E\kappa-1)^{1-\zeta} }
        { (1+\kappa)^{1-\zeta} (1+E\kappa)^{1+\zeta} }    \right]^{\pm n} }$    \\
      &       &        &          \\
  \hline
\end{tabular}


\smallskip

Of course, the integral formula above gives more weight to these
asymptotic constructions,
% (see later, \S~\ref{jacobiint})
with $\kappa = (1-\zeta)u-1$.

\smallskip

\smallskip

We expect the numerator, denominator, ratio $=$ the approximation,
and the error to behave as one of the given $n^{\text{th}}$ powers, or
a combination of both. At the present level of formal manipulation,
we don't yet know which power(s) to use.

 At least for large $z$, we
know that we should have $(E^{\pm 1}-1)^n n! (z/h)^n \approx
(E^{\pm 1}-1)^n e^{-n} n^{2n} \zeta^n$ for the numerator and the
denominator. Indeed, with the root $\kappa = (E^{-1}-1)\zeta+\dotsb$
one finds, with the second power, $\displaystyle {
\left[\frac{P(z)}{(e^{-2n} n^{2n}} \right]^{1/n} \approx
 -E\kappa \left( \frac{\kappa-1}{\kappa-1/E}\right)^\zeta \approx
(E-1)\zeta
  \left(1+\frac{1}{\zeta}\right)^\zeta \approx e(E-1)\zeta  }$.
The first power behaves for large $z$ as $[e(1-E)\zeta E^{1+\zeta}]^n$,
such a behaviour is impossible with a polynomial.
  Similarly, one
has for the denominator, with the first power,
 $\displaystyle  {
\left[\frac{Q(z)}{(e^{-2n} n^{2n}} \right]^{1/n} \approx
 e(E^{-1}-1)\zeta  }$.

The region where the asymptotic behaviours of $P$ and $Q$ are such that
$(P/Q)^{1/n} \approx E^{1+\zeta}$ is simply the region of good
approximation! Indeed, $E^{n(1+\zeta)} = e^{Ahn(1+\zeta)} =e^{A(z-z_0)}$.

When $E$ is close to 1, we almost have the Pad\'e situation of the
figure of p.~\pageref{padefig}

\medskip

Various interesting situations occur, the wildest situation
being $E=-1$: we then interpolate merely the sequence $1, -1, 1,
-1,\dotsc$ at $z_0, z_1,\dotsc, z_{m+n}$ by $q/p$ of degrees
$m$ and $n$, without any reference to an exponential function anymore!

Moreover, the solution of the Cauchy problem (in the sense of
\cite{Me}) is then immediate: $q+p$ must vanish at $z_1, z_3,\dotsc$,
and $q-p$ vanishes at $z_0, z_2,\dotsc$:
$$
\frac{q(z)}{p(z)} = \frac
 {c (z-z_1)(z-z_3)\dotsb   +c' (z-z_0)(z-z_2)\dotsb  }
 {c (z-z_1)(z-z_3)\dotsb   -c' (z-z_0)(z-z_2)\dotsb  }  ,
$$
where one of the two numbers $c$ or $c'$ may very well vanish if it is
the only way to achieve degrees $\leqslant m$ and $n$!

\medskip

The integral of the numerator is $\displaystyle{ \int_C F(u)
  e^{n\Phi(u)}\,du }$ estimated, through deformation of the contour,
by steepest descent contributions of the neighbourhoods of saddle
points (there will be at most two of them) $u_s$ such that
$\Phi'(u_s)=0$:

$\displaystyle{\Phi(u)=\Phi(u_s) + \frac{(u-u_s)^2}{2}\Phi''(u_s)
 +\dotsb, \qquad      \int_C F(u)
  e^{n\Phi(u)}\,du \sim F(u_s) e^{n\Phi(u_s)}
    \sqrt{\frac{2\pi}{\Phi''(u_s)}}    }$.

Here, when $m\sim n$,

$F(u)= e^{Az_0}(1-u)^{-1} [u(1+E)-E]^{m-n}$,
 $\Phi(u)=\log\{u(1-u)^{-\zeta-1}[u(1+E)-E]\}$ (numerator $P$),

$F(u)= u^{m-n}(1-u)^{-1}$,
 $\Phi(u)=\log\{u(1-u)^{-\zeta-1}[u(1+E^{-1})-E^{-1}]\}$ (denominator $Q$).


\subsubsection{Check with \S~\ref{singlearc}.\label{check1}}


\  \\

From \eqref{ratio},
$$
\mathcal{V}_p(z) = \lim_{n\to\infty} \frac{\log Q(z)}{n} =
  (1+\zeta)\log(1+E^{-1}\kappa)+(1-\zeta)\log(1+\kappa)-\log(\kappa),
$$
where $\zeta$ is basically our $z$ (managed so that the interpolation
points are in $[-1,1]$). Then, the derivative in $\zeta$ simplifies into
$$
\frac{d\mathcal{V}_p(z)}{d\zeta} = \log \frac{1+E^{-1}\kappa}{1+\kappa},
$$
where $\kappa$ is related to $\zeta$ through
$\displaystyle{  \zeta = \frac{ \kappa^{-1} -E^{-1}\kappa }{E^{-1}-1} }$,
from \eqref{kappa}. This matches \eqref{rhoz} provided
$$
\rho= iE^{-1/2}\kappa, \alpha = -\beta = \frac{2i}{E^{1/2}-E^{-1/2}},
$$
and
$$
\frac{d\mathcal{V}_p(z)}{d\zeta} =
 \log \frac{1-iE^{-1/2}\rho}{1-iE^{1/2}\rho}
= -\sum_{n=1}^\infty \frac{i^n (E^{-n/2}-E^{n/2})}{n} \rho^n.
$$
Remark that $g_1 = -i(E^{-1/2}-E^{1/2}) = 4/(\beta-\alpha)$ as it
should.

\section{Rational interpolation to $\exp(nB_1 z +nB_2 z^2)$. }\  \\

This very interesting rational interpolation appears in
special nonlinear Schr\"odinger problems (   \cite{KM} and remarks
by J.~Nuttall)  . The Pad\'e approximation (\S~\ref{Padexp2})
already told that there may be several arcs in the discussion.

\subsection{The single arc case} \  \\

Let the interpolation points be equidistant on $[I_1, I_2]$. Then,
\begin{equation} \label{gKM}
 g(z) = \int_{I_1}^{I_2} \frac{(I_2-I_1)^{-1}\,dt}{z-t} -\frac{B_1}{2}
     -B_2 z =
  \frac{\log  {\displaystyle{\frac{z-I_1}{z-I_2}}} }{I_2-I_1}
    -\frac{B_1}{2} -B_2\left(  \frac{\beta-\alpha}{2}
      \frac{2z-\alpha-\beta}{\beta-\alpha}
     +\frac{\alpha+\beta}{2} \right)
\end{equation}
The logarithms have the expansions
$$
  \log (z-I_k) = \log\frac{\alpha-\beta}{4\rho_k}
       -2 \sum_1^\infty \frac{\rho_k^n}{n} T_n,
$$
where $\rho_k$ is now a root of
\begin{equation} \label{rhok}
    \frac{ \rho_k + \rho_k^{-1} }{2} =
    \frac{2I_k-\alpha-\beta}{\beta-\alpha}  \;, k=1,2,
\end{equation}
where $|\rho_k|<1$ should be the orthodox choice, but which will not
be kept in the final formula. Precisely, the closed form is now
\begin{equation} \label{Vp2}
\mathcal{V}'_p(z) = \sum_1^\infty g_n \rho^n =
  \frac{2}{I_2-I_1} \log\frac{1-\rho_1\rho}{1-\rho_2\rho}
     -B_2 \frac{\beta-\alpha}{2}\rho,
\end{equation}
with the conditions \eqref{pvcond2} on $g_0$ and $g_1$
\begin{equation} \label{B1cond}
\frac{\log(\rho_2/\rho_1)}{I_2-I_1} -\frac{B_1}{2}
      -B_2 \frac{\alpha+\beta}{2} =0,
\end{equation}
\begin{equation} \label{B2cond}
 g_1 = 2 \frac{\rho_2-\rho_1}{I_2-I_1} -B_2 \frac{\beta-\alpha}{2} =
 \frac{4}{\beta-\alpha},
\end{equation}

If $\rho_1$ and $\rho_2$ are known, $\alpha$ and $\beta$ are got by
\eqref{rhok}:
\begin{equation} \label{abr1r2}
 \alpha = \frac{ \rho_1(1+\rho_2)^2 I_1 -\rho_2 (1+\rho_1)^2 I_2}
               { (\rho_2-\rho_1)(\rho_1\rho_2-1)}\;,
 \beta = \frac{ \rho_1(1-\rho_2)^2 I_1 -\rho_2 (1-\rho_1)^2 I_2}
               { (\rho_2-\rho_1)(\rho_1\rho_2-1)}\;.
\end{equation}
(As for $\rho_1$ and $\rho_2$, they are simply found to be, if
$B_2=0$, $\rho_1= i\exp(-B_1(I_2-I_1)/4)$ and
         $\rho_2= i\exp(B_1(I_2-I_1)/4)$. Remark that $\rho_1\rho_2 =-1$:
no chance to have the comfortable $|\rho_k|<1$\dots)

Also,
$$
\frac{\beta-\alpha}{2} = \frac{ 2\rho_1\rho_2(I_2-I_1)}{
         (\rho_2-\rho_1)(\rho_1\rho_2-1)}\,,
\frac{\alpha+\beta}{2} = \frac{I_1+I_2}{2} -\frac{
         (\rho_1+\rho_2)(\rho_1\rho_2+1)(I_2-I_1)}{
        2(\rho_2-\rho_1)(\rho_1\rho_2-1)}\,.
$$
and \eqref{B1cond} and \eqref{B2cond} become
$$
\frac{\log(\rho_2/\rho_1)}{I_2-I_1} -\frac{B_1}{2}
  -B_2\frac{I_1+I_2}{2} +\frac{B_2}{2}
    \frac{(\rho_2+\rho_1)(\rho_1\rho_2+1)}
         {(\rho_2-\rho_1)(\rho_1\rho_2-1)} (I_2-I_1)=0,
B_2(I_2-I_1)^2 = \frac{(\rho_2-\rho_1)^2 (\rho_1^2\rho_2^2-1) }
                      { 2\rho_1^2\rho_2^2},
$$
or this form emphasizing $\rho_1\rho_2$ and $\rho_2/\rho_1$:
\begin{equation} \label{B2rr}
 2B_2 (I_2-I_1)^2 = \left( \frac{\rho_2}{\rho_1} -2
                          +\frac{\rho_1}{\rho_2}   \right)
                    \left( \rho_1\rho_2-\frac{1}{\rho_1\rho_2}\right),
\end{equation}
\begin{equation} \label{logrr}
\log\left( \frac{\rho_2}{\rho_1}\right) +\frac{1}{4}
                    \left( \frac{\rho_2}{\rho_1} -
                           \frac{\rho_1}{\rho_2}   \right)
           \left( \rho_1\rho_2+2+\frac{1}{\rho_1\rho_2}\right)=
           \frac{B_1}{2}(I_2-I_1)
  +B_2\frac{I_2^2-I_1^2}{2}.
\end{equation}

For a given $B_2$ and various ratios $\rho_2/\rho_1$, we find
valid values for $B_1$, etc. For instance, $I_1=-Ai$, $I_2=Ai$,
$B_2$ negative imaginary
and $\rho_2/\rho_1$ negative real, which is of interest in \cite{KM}:

{\Small

\begin{verbatim}

Script V1.1 session started Mon Jan 17 14:32:31 2000

C:\calc\pari>gp
                   GP/PARI CALCULATOR Version 2.0.12 (alpha)
                           Copyright (C) 1989-1998 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

? \r expr1r2
\end{verbatim}

$A=1,  B_1=\pi-2ix, x= 0.5,  B_2=-2iAt$,

\begin{tabular}{cccccc}
$At$  &  $-\rho_2/\rho_1$   & $Q$ &  $\rho_1$ & $\rho_2$  & $\alpha=\overline{\beta}$   \\
$0.10$ & $ 2.758$ & $0.02455$ & $0.6002 + 0.04717i$ & $ -1.655 - 0.1301i$ & $ 0.03677 + 0.8865i$ \\
$0.10$ & $ 0.439$ & $3.971  $ & $0.1284 + 1.502i$ & $ -0.05646 - 0.6606i$ & $ -4.554 + 10.81i$ \\
$0.20$ & $ 2.888$ & $0.09570$ & $0.5812 + 0.09100i$ & $ -1.679 - 0.2629i$ & $ 0.07604 + 0.8847i$ \\
$0.20$ & $ 0.429$ & $3.883  $ & $0.2604 + 1.504i$ & $ -0.1117 - 0.6454i$ & $ -2.307 + 5.374i$ \\
$0.30$ & $ 3.138$ & $0.2038 $ & $0.5499 + 0.1274i$ & $ -1.725 - 0.3998i$ & $ 0.1197 + 0.8788i$ \\
$0.30$ & $ 0.411$ & $3.737  $ & $0.3997 + 1.507i$ & $ -0.1643 - 0.6197i$ & $ -1.574 + 3.545i$ \\
$0.40$ & $ 3.553$ & $0.3275 $ & $0.5082 + 0.1518i$ & $ -1.806 - 0.5394i$ & $ 0.1674 + 0.8641i$ \\
$0.40$ & $ 0.384$ & $3.533  $ & $0.5507 + 1.514i$ & $ -0.2119 - 0.5830i$ & $ -1.221 + 2.622i$ \\
$0.50$ & $ 4.176$ & $0.4363 $ & $0.4618 + 0.1616i$ & $ -1.928 - 0.6749i$ & $ 0.2147 + 0.8365i$ \\
$0.50$ & $ 0.351$ & $3.277  $ & $0.7171 + 1.527i$ & $ -0.2517 - 0.5363i$ & $ -1.023 + 2.064i$ \\
$0.60$ & $ 5.006$ & $0.5082 $ & $0.4175 + 0.1593i$ & $ -2.090 - 0.7976i$ & $ 0.2544 + 0.7974i$ \\
$0.60$ & $ 0.311$ & $2.988  $ & $0.9009 + 1.548i$ & $ -0.2805 - 0.4824i$ & $ -0.9027 + 1.692i$ \\
$0.70$ & $ 6.007$ & $0.5430 $ & $0.3792 + 0.1503i$ & $ -2.278 - 0.9031i$ & $ 0.2832 + 0.7524i$ \\
$0.70$ & $ 0.270$ & $2.692  $ & $1.099 + 1.578i$ & $ -0.2972 - 0.4266i$ & $ -0.8243 + 1.431i$ \\
$0.80$ & $ 7.139$ & $0.5518 $ & $0.3474 + 0.1390i$ & $ -2.480 - 0.9924i$ & $ 0.3017 + 0.7071i$ \\
$0.80$ & $ 0.232$ & $2.414  $ & $1.307 + 1.612i$ & $ -0.3033 - 0.3742i$ & $ -0.7690 + 1.241i$ \\
$0.90$ & $ 8.372$ & $0.5452 $ & $0.3211 + 0.1275i$ & $ -2.689 - 1.068i$ & $ 0.3124 + 0.6643i$ \\
$0.90$ & $ 0.198$ & $2.166  $ & $1.517 + 1.650i$ & $ -0.3019 - 0.3282i$ & $ -0.7263 + 1.099i$ \\
$1.0$ & $  9.692$ & $0.5303 $ & $0.2991 + 0.1169i$ & $ -2.899 - 1.133i$ & $ 0.3178 + 0.6252i$ \\
$1.0$ & $  0.171$ & $1.953  $ & $1.727 + 1.687i$ & $ -0.2962 - 0.2893i$ & $ -0.6908 + 0.9883i$ \\
$1.1$ & $  11.08$ & $0.5113 $ & $0.2804 + 0.1073i$ & $ -3.109 - 1.190i$ & $ 0.3195 + 0.5899i$ \\
$1.1$ & $  0.149$ & $1.770  $ & $1.933 + 1.723i$ & $ -0.2882 - 0.2568i$ & $ -0.6599 + 0.9000i$ \\
$1.2$ & $  12.55$ & $0.4906 $ & $0.2643 + 0.09885i$ & $ -3.318 - 1.240i$ & $ 0.3187 + 0.5582i$ \\
$1.2$ & $  0.130$ & $1.613  $ & $2.136 + 1.757i$ & $ -0.2791 - 0.2296i$ & $ -0.6323 + 0.8278i$
\end{tabular}

\begin{verbatim}

? quit
Good bye!

C:\calc\pari>exit

Script completed Mon Jan 17 14:34:12 2000

\end{verbatim}

}   %end Small

We integrate \eqref{Vp2} along the lines suggested by the
exercises of section \ref{check1}, p.~\pageref{check1}:
$$
  \mathcal{V}_p(z) = \frac{2}{I_2-I_1} [ (z-I_1)\log(1-\rho_1\rho)
                                        -(z-I_2)\log(1-\rho_2\rho)
  +X(\rho) ],
$$
which yields indeed, using from \eqref{rhoz} and \eqref{rhok}
$\displaystyle{ z-I_k = \frac{\beta-\alpha}{4} \left(1-\frac{\rho_k}{\rho}\right)
                    \left(\rho-\frac{1}{\rho_k}\right)  }$,
$$
\frac{d\mathcal{V}_p(z)}{dz} =
  \frac{2}{I_2-I_1}\left\{ \log\frac{1-\rho_1\rho}{1-\rho_2\rho}
+\left[  \frac{z-I_1}{\rho-\rho_1^{-1} } -  \frac{z-I_2}{\rho-\rho_1^{-2} }
    +\frac{dX(\rho)}{d\rho}\right] \frac{d\rho}{dz} \right\}
$$
$$
=
  \frac{2}{I_2-I_1}\left\{ \log\frac{1-\rho_1\rho}{1-\rho_2\rho}
+\left[  \frac{\beta-\alpha}{4} \frac{\rho_2-\rho_1}{\rho}
    +\frac{dX(\rho)}{d\rho}\right] \frac{4}{(\beta-\alpha)(1-\rho^{-2})} \right\}
$$
One must have $\displaystyle{  \frac{dX}{d\rho} =
 - \frac{\beta-\alpha}{4} \frac{\rho_2-\rho_1}{\rho}
 -B_2 \frac{(\beta-\alpha)^2}{16} (I_2-I_1) \left( \rho-\frac{1}{\rho}\right) }$,
finally:
$$
  \frac{dX}{d\rho} =
  - \frac{I_2-I_1}{2}  \frac{\rho_1\rho_2+1}
                                  {\rho_1\rho_2-1}  \rho +\frac{1}{\rho},
$$
\begin{equation}     \label{V2}
  \mathcal{V}_p(z) = \frac{2}{I_2-I_1} [ (z-I_1)\log(1-\rho_1\rho)
                                        -(z-I_2)\log(1-\rho_2\rho) ]
  -  \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}  \frac{\rho^2}{2}
  -\log\rho.
\end{equation}

The two determinations of $\mathcal{V}_p$ on the two sides of the
cut are found with the two roots $\rho$ and $1/\rho$ of
\eqref{rhoz}. In particular, the \emph{arithmetic mean} of the two
values of the derivative
must give  \eqref{pvab} again, with $g$ given by  \eqref{gKM}.
Indeed, one finds
$$
    \frac{1}{I_2-I_1} \{ \log[(1-\rho_1\rho)(1-\rho_1/\rho)]
                            -\log[(1-\rho_2\rho)(1-\rho_2/\rho)] \}
  -  B_2 \frac{\beta-\alpha}{2}  \frac{\rho+\rho^{-1}}{2},
$$
which is
$$
   \frac{1}{I_2-I_1} \left[ \log\frac{z-I_1}{z-I_2}
                           +\log\frac{\rho_1}{\rho_2} \right]
   -B_2 z +B_2 \frac{\alpha+\beta}{2},
$$
  \

The \emph{difference} of the two determinations of
$\mathcal{V}_p'$ must be $\pm 2\pi i \mu'$:
\begin{equation} \label{mup}
    \pm 2\pi i \mu'(z) =
    \frac{2}{I_2-I_1} \left[ \log\frac{1-\rho_1\rho}{1-\rho_1/\rho}
                      -\log\frac{1-\rho_2\rho}{1-\rho_2/\rho}  \right]
  -  B_2 \frac{\beta-\alpha}{2}  (\rho-\rho^{-1}),
\end{equation}
(Nuttall's $\Delta\Psi_2$)

and the cut itself is the locus $\{ z: \mu'(z)\,dz $ real $\}$,
which is integrated as $\{z: \mathcal{V}_{p,+}(z)-
                             \mathcal{V}_{p,-}(z) $ pure imaginary $\}$,
\begin{equation} \label{intmup}
 \frac{2}{I_2-I_1} \left[ (z-I_1)\log\frac{1-\rho_1\rho}{1-\rho_1/\rho}
                     -(z-I_2)\log\frac{1-\rho_2\rho}{1-\rho_2/\rho} \right]
 -\frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}  \frac{\rho^2-\rho^{-2}}{2}
  -2\log\rho  \ \text{pure imaginary}.
\end{equation}

 Writing \eqref{intmup} as a function of $\rho$ (using \eqref{rhoz} and
\eqref{rhok}), we have
$$
 F(\rho) = \frac{2}{(\rho_2-\rho_1)(1-1/(\rho_1\rho_2))}\left[
     (\rho-\rho_1)\left( 1-\frac{1}{\rho\rho_1} \right) L_1
    -(\rho-\rho_2)\left( 1-\frac{1}{\rho\rho_2} \right) L_2 \right]
 -\frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}  \frac{\rho^2-\rho^{-2}}{2}
,$$
with $\displaystyle{ L_1 = \log\frac{1-\rho_1\rho}{\rho-\rho_1}\;,
                     L_2 = \log\frac{1-\rho_2\rho}{\rho-\rho_2}\;, }$
and where, for given $B_1, B_2, I_1, I_2$, one must determine $\rho_1$
and $\rho_2$ from \eqref{B2rr} and \eqref{logrr}.

\bigskip

\subsection{First caustic}  \  \\

 The present setting of the limit set of poles as a single arc joining
$z=\alpha$ to $z=\beta$ (or $\rho=-1$ to $\rho=1$) holds as long as
$\mu_p'(z)dz$ remains positive on the cut.  A critical situation
occurs when $\mu_p'$ happens to vanish right on the cut, i.e., if
$dF/dz$ vanishes at a point where the real part of $F$ vanishes too.

\epsfxsize=477bp \epsfbox[50   50 527 470]{m3001.ps}

The locus of $(x,At)$ with $B_1=\pi-2ix$, $B_2=-2iAt$, where this
happens is called the (first) caustic in \cite{KM}. We then have
$\rho_1= R^{-1/2} e^{i\theta}$,
$\rho_2= -R^{1/2} e^{i\theta}$, with real $R$ and $\theta$.
For a trial value of $At$, we look for $R$ and $\theta$ such that
$(R+1/R)/2 = 2At/\sin 2\theta\;-1$ (from \eqref{B2rr}) and
$2x=\log R +(1/R-R)\sin^2\theta$ (from \eqref{logrr}). Knowing
$\rho_1$ and $\rho_2$, one looks for the zero of the analytic
function $dF/dz$, or $dF/d\rho$. This yields the equation $\mu'_p=0$
in \eqref{mup} as
\begin{equation} \label{mup1}
 L_1-L_2 =
     \left( 1+\frac{1}{\rho_1\rho_2} \right) (\rho_2-\rho_1)
    \frac{ \rho-\rho^{-1} }{2}.
\end{equation}
One then manages to have the real part of $F=0$ as well.

Some values:


\begin{tabular}{ccccccc}
$x$&$At$&$\sin\theta$&$\rho_2$  &  $R=-\rho_2/\rho_1$   &  $\rho$  & $F$ \\
 0.001 &   0.500973 &  0.6990 &   -0.749  -0.733\ $i$ &      1.098 &   -1.27846  -0.24352\ $i$ & 6.282\ $i$\\
 0.010 &   0.509711 &  0.6781 &   -0.854  -0.788\ $i$ &      1.350 &   -1.52506  -0.51989\ $i$ & 6.261\ $i$ \\
 0.050 &   0.548383 &  0.6288 &   -1.095  -0.885\ $i$ &      1.982 &   -1.84975  -0.94099\ $i$ & 6.123\ $i$\\
 0.100 &   0.596697 &  0.5838 &   -1.324  -0.952\ $i$ &      2.660 &   -2.06555  -1.23283\ $i$ & 5.922\ $i$ \\
 0.250 &   0.744301 &  0.4821 &   -1.928  -1.061\ $i$ &      4.842 &   -2.48476  -1.76294\ $i$ & 5.301\ $i$\\
 0.500 &   1.009193 &  0.3636 &   -2.919  -1.139\ $i$ &      9.817 &   -2.98988  -2.26747\ $i$ & 4.371\ $i$\\
 0.750 &   1.312145 &  0.2794 &   -4.017  -1.169\ $i$ &     17.504 &   -3.41534  -2.56407\ $i$ & 3.587\ $i$\\
 1.000 &   1.672677 &  0.2167 &   -5.311  -1.179\ $i$ &     29.600 &   -3.78112  -2.72755\ $i$ & 2.923\ $i$\\
 1.250 &   2.117750 &  0.1686 &   -6.897  -1.180\ $i$ &     48.954 &   -4.06939  -2.79596\ $i$ & 2.359\ $i$\\
 1.500 &   2.682341 &  0.1314 &   -8.888  -1.178\ $i$ &     80.381 &   -4.27061  -2.81082\ $i$ & 1.884\ $i$\\
 1.750 &   3.408138 &  0.1024 &  -11.423  -1.176\ $i$ &    131.867 &   -4.39913  -2.80582\ $i$ & 1.493\ $i$\\
 2.000 &   4.344519 &  0.0798 &  -14.669  -1.174\ $i$ &    216.553 &   -4.47798  -2.79719\ $i$ & 1.175\ $i$\\
 2.500 &   7.107960 &  0.0484 &  -24.179  -1.172\ $i$ &    585.997 &   -4.55461  -2.78484\ $i$ & 0.721\ $i$\\
 3.000 &  11.684073 &  0.0294 &  -39.858  -1.171\ $i$ &   1590.003 &   -4.58262  -2.77939\ $i$ & 0.439\ $i$
\end{tabular}

We see that $\theta\to \pi/4$ when $x\to 0$, and that $\theta\to 0$ when $x\to\infty$, but
many features are still unexplained\dots

Here is a tentative explanation of the behaviour for large $x$: as it
seems that $|\rho_1| << |\rho| << |\rho_2|$, the logarithms
are approximated by
$L_1 \approx \rho_1 (\rho^{-1}-\rho) -\log\rho$,
$L_2 \approx (\rho-\rho^{-1})/\rho_2 +\log\rho$, the equation
\eqref{mup1} becomes
$\displaystyle{   \frac{\log\rho}{\rho-\rho^{-1}} \approx  i\xi }$,
with $\xi = \theta\sqrt{R}\,/2$. Also,
$F/\theta \approx (\rho-\rho^{-1}) [i (\rho+\rho^{-1})/2 -\xi^{-1}-2\xi]$
must be pure imaginary, making a second equation for $\xi$ and $\rho$,
whence fixed solutions. And
$$
 \frac{At}{e^x} \approx \frac{ \theta R/4}{ \sqrt{R}\; \exp(-\theta^2 R/2)} =
            \frac{\xi}{2} \exp( 2\xi^2).
$$

\begin{verbatim}

Script V1.1 session started Thu Feb 17 11:24:23 2000

C:\calc\pari>gp

                   GP/PARI CALCULATOR Version 2.0.12 (alpha)
                           Copyright (C) 1989-1998 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

? \r expcaus2
...
    rho=-4.59885439353246016460 - 2.77599828040642492631*I

 F/theta= -1.14496588753536343500 E-40 + 14.9829543045360004158*I
? xi
  0.585318492448534646977
? (xi/2)*exp(2*xi*xi)
  0.580682668039111487078
? quit
Good bye!

C:\calc\pari>exit

Script completed Thu Feb 17 11:29:16 2000

\end{verbatim}

\bigskip

\emph{\textbf{Numerator and interpolation.}}

Remind that  $\mathcal{V}_p(z)$ is the limit when $n\to\infty$ of
$n^{-1}\log Q_n(z) = n^{-1} \sum \log(z-$ poles$)$. It must
behave like $\log(z) +O(1/z)$ for large $z$, compatible with
\eqref{V2} if one adds a constant:
$$
  \mathcal{V}_p(z) = \frac{2}{I_2-I_1} [ (z-I_1)\log(1-\rho_1\rho)
                                        -(z-I_2)\log(1-\rho_2\rho) ]
  - \frac{\rho^2}{2}\, \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}
  -\log\rho +C_p,
$$
which behaves for large $z$ as
$\displaystyle{  \log z  -log\frac{\beta-\alpha}{4}
    +\frac{(\beta-\alpha)(\rho_2-\rho_1)}{2(I_2-I_1)} +C_p }$,
from $\rho \sim (\beta-\alpha)/(4z)$. Therefore,
$$
  C_p =   log\frac{\beta-\alpha}{4}
    -\frac{(\beta-\alpha)(\rho_2-\rho_1)}{2(I_2-I_1)} .
$$

The numerator of the interpolant to $\exp(n(B_1z+B_2z^2))$
is the denominator of the interpolant to $\exp(n(-B_1z-B_2z^2))$,
so that the calculations made before apply with $(B_1,B_2)
\longrightarrow (-B_1,-B_2)$. The equations \eqref{B1cond} and
\eqref{B2cond} are now satisfied by
$(\rho_1,\rho_2)\longrightarrow (1/\rho_1,1/\rho_2)$. And the
values for $\alpha$ and $\beta$ are the same as before. Let
$\mathcal{V}_{\text{num}}(z)$ be the (presumed to exist) limit when
$n\to\infty$ of $n^{-1}\log P_n(z)$, where $P_n$ is the numerator.
We expect a formula similar to \eqref{V2}, but with another
constant:
$$
  \mathcal{V}_{\text{num}}(z) =
   \frac{2}{I_2-I_1} [ (z-I_1)\log(1-\rho/\rho_1)
                                        -(z-I_2)\log(1-\rho/\rho_2) ]
  + \frac{\rho^2}{2} \,\frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}
  -\log\rho +C_{\text{num}}.
$$
The remaining constant $C_{\text{num}}$ is determined by
$  \mathcal{V}_{\text{num}}(z) -\mathcal{V}_p(z)=  B_1z+B_2 z^2$ in a
neighbourhood of the set of the interpolation points. Everything
works if one determination, say with $\rho$, is used for
$\mathcal{V}_p$, while the determination with $1/\rho$ is used
for  $  \mathcal{V}_{\text{num}}$:
$$
  \mathcal{V}_{\text{num}}(z) -\mathcal{V}_p(z)=
   \frac{2}{I_2-I_1} \left[ (z-I_1)\log
                      \frac{1-1/(\rho\rho_1)}{1-\rho\rho_1}
            -(z-I_2)\log\frac{1-1/(\rho/\rho_2)}{1-\rho\rho_2} \right]
 + \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}  \frac{\rho^{-2}+\rho^2}{2}
  +2\log\rho +C_{\text{num}}-C_p,
$$
$$
=   \frac{2}{I_2-I_1} [ (z-I_2)\log(-\rho_2) -(z-I_1)\log(-\rho_1) ]
 + \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}
   \left[2\left(\frac{2z-\alpha-\beta}{\beta-\alpha}\right)^2-1\right]
   +C_{\text{num}}-C_p,
$$
$$
 = [B_1+B_2(\alpha+\beta)]z
  +2 \frac{ I_1\log(-\rho_1)-I_2\log(-\rho_2)}{I_2-I_1}
 +\frac{B_2}{4} (2z-\alpha-\beta)^2
 - \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}
   +C_{\text{num}}-C_p,
$$
whence
\begin{equation} \label{Cnump}
   C_{\text{num}}-C_p=
  2 \frac{ I_2\log(-\rho_2)-I_1\log(-\rho_1)}{I_2-I_1}
 -\frac{B_2}{4} (\alpha+\beta)^2
 + \frac{\rho_1\rho_2+1}{\rho_1\rho_2-1}.
\end{equation}




\section{Best rational approximation to $e^{-(An+B)z}$ on a real interval} \   \\


\subsection{}
Best rational approximation to $\exp(-z)$ on a given real interval,
say $[0,c]$ has a strict equioscillating error function, as seen
here with $e^z-p_n(z)/q_n(z)$ on $[0,1]$, for $n=1,2$:

\psset{unit=15cm}
\begin{pspicture}(0,-0.1)(1,0.1)
\psline{->}(0,0)(1.1,0) \uput[90](0.7,0.079){$0.00158$}
\psline{->}(0,-0.079)(0,0.10) % \uput[0](0,3.5){$y$}
\psplot[linewidth=0.003,plotstyle=curve]{0.0}{1.0}{x dup neg %
    2.718281828 exch exp exch  %   now we have exp(-x),  x
 dup   %
  -0.63217546610245 mul  1.57418925342864   add   exch %
     1.57170561775033   add   div sub 50.0  mul}
\psline[linestyle=dashed, dash=3mm 3mm](0,0.079)(1,0.079)
\psline[linestyle=dashed, dash=3mm 3mm](0,-0.079)(1,-0.079)
\psline[linestyle=dashed, dash=3mm 3mm](1,-0.079)(1,0.079)
\end{pspicture}

\psset{unit=15cm}
\begin{pspicture}(0,-0.1)(1,0.1)
\psline{->}(0,0)(1.1,0) \uput[90](0.7,0.082){$1.64\  10^{-6}$}
\psline{->}(0,-0.082)(0,0.09)  \uput[0](0,3.5){$y$}
\psplot[linewidth=0.003,plotstyle=curve,plotpoints=20]{0.0}{1.0}{x dup neg %
    2.718281828459 exch exp exch  %   now we have exp(-x),  x
 dup dup  %
   0.62186183854470  mul -4.34141740064031 add mul  9.42324963553348 %
  add exch    dup %
   5.08091346772400  add mul 9.42326514088167  add div sub  50000.0 mul}
\psline[linestyle=dashed, dash=3mm 3mm](0,0.082)(1,0.082)
\psline[linestyle=dashed, dash=3mm 3mm](0,-0.082)(1,-0.082)
\psline[linestyle=dashed, dash=3mm 3mm](1,-0.082)(1,0.082)
\end{pspicture}

\bigskip

\parpic{
\psset{unit=0.5cm}
\begin{pspicture}(-6,-6)(0,6)
\psline{->}(-6,0)(0.8,0) \uput[90](-5,0){$-5$}
\psline{->}(0,-6)(0,6)  \uput[0](0,5){$5$}
\psdots[dotsize=0.25](-1.57,0)  \uput[0](-1.57,0){$1$}
\psdots[dotsize=0.25](-2.54,1.72)  \uput[0](-2.54,1.72){$2$}
\psdots[dotsize=0.25](-2.54,-1.72)  \uput[0](-2.54,-1.72){$2$}
\psdots[dotsize=0.25](-4.17,0)  \uput[0](-4.17,0){$3$}
\psdots[dotsize=0.25](-3.24,3.52)  \uput[0](-3.24,3.52){$3$}
\psdots[dotsize=0.25](-3.24,-3.52)  \uput[0](-3.24,-3.52){$3$}
\psdots[dotsize=0.25](-3.74,5.32)  \uput[0](-3.74,5.32){$4$}
\psdots[dotsize=0.25](-3.74,-5.32)  \uput[0](-3.74,-5.32){$4$}
\psdots[dotsize=0.25](-5.34,1.72)  \uput[0](-5.34,1.72){$4$}
\psdots[dotsize=0.25](-5.34,-1.72)  \uput[0](-5.34,-1.72){$4$}
\pscurve[linestyle=dashed, dash=3mm 3mm](-3.74,-5.32)(-5.34,-1.72)(-5.34,1.72)(-3.74,5.32)
\psdots[dotsize=0.25](-6.81,0)  \uput[0](-6.81,0){$5$}
\psdots[dotsize=0.25](-6.22,3.48)  \uput[0](-6.22,3.48){$5$}
\psdots[dotsize=0.25](-6.22,-3.48)  \uput[0](-6.22,-3.48){$5$}
\psdots[dotsize=0.25](-4.16,-7.14)  \uput[0](-4.16,-7.14){$5$}
\psdots[dotsize=0.25](-4.16,7.14)  \uput[0](-4.16,7.14){$5$}
\pscurve[linestyle=dashed, dash=3mm 3mm](-4.16,-7.14)(-6.22,-3.48)(-6.81,0)(-6.22,3.48)(-4.16,7.14)
\end{pspicture}
}
For varying degrees, we have a now familiar scaling effect
best seen through the poles:

Sets of poles expand and tend to follow the Pad\'e poles;
errors decrease factorially fast with $n$( here, the error
is about $\displaystyle{ \dfrac{e^{-1/2} n! (n-1)! }{4^{2n+1}
(2n+1)! (2n-1)!}}$. For an
accurate asymptotic picture, see Braess' proof of Meinardus'
conjecture \cite{Braess}).

\picskip{1}

\   \\

\vspace*{3cm}

\parpic{
\psset{unit=3cm}
\begin{pspicture}(-1.5,-1.7)(0,1.7)
\psline{->}(-2,0)(0.1,0) \uput[90](-1,0){$-1$}
\psline{->}(0,-1.4)(0,1.8)  \uput[0](0,1){$1$}
\psdots[dotsize=0.05](-1.57,0)  \uput[135](-1.57,0){$1$}
\psdots[dotsize=0.05](-1.08,0.85)  \uput[0](-1.08,0.85){$2$}
\psdots[dotsize=0.05](-1.08,-0.85)  \uput[0](-1.08,-0.85){$2$}
\psdots[dotsize=0.05](-1.14,0)  \uput[135](-1.14,0){$3$}
\psdots[dotsize=0.05](-0.805,-1.146)  \uput[0](-0.805,-1.146){$3$}
\psdots[dotsize=0.05](-0.805,1.146)  \uput[0](-0.805,1.146){$3$}
\psdots[dotsize=0.05](-0.63,1.30)  \uput[0](-0.63,1.30){$4$}
\psdots[dotsize=0.05](-0.63,-1.30)  \uput[0](-0.63,-1.30){$4$}
\psdots[dotsize=0.05](-1.04,0.42)  \uput[0](-1.04,0.42){$4$}
\psdots[dotsize=0.05](-1.04,-0.42)  \uput[0](-1.04,-0.42){$4$}
\psdots[dotsize=0.05](-1.05,0)  \uput[135](-1.05,0){$5$}
\psdots[dotsize=0.05](-0.93,-0.68)  \uput[0](-0.93,-0.68){$5$}
\psdots[dotsize=0.05](-0.93,0.68)  \uput[0](-0.93,0.68){$5$}
\psdots[dotsize=0.05](-0.506,-1.40)  \uput[0](-0.506,-1.40){$5$}
\psdots[dotsize=0.05](-0.506,1.40)  \uput[0](-0.506,1.40){$5$}
\end{pspicture}
}
We find a stable picture if we look at the poles of the
best approximants of degree $n$ to $\exp(-nz)$. Moreover,
the norms $E_n$ of the errors tend to decrease in an exponential
way with $n$:

\begin{tabular}{cccccc}
$n=1$ & 2 & 3 & 4 & 5 & 6  \\
1.58 E-3 &
3.197 E-5 &
5.921 E-7 &
1.068 E-8 &
1.91 E-10 &
3.383 E-12
\end{tabular}

The ratio of two successive errors seems to tend towards a
limit of about $1/60$. The exact value, as it will be shown
later (in \eqref{rho}, p.~\pageref{rho}), is
$\rho= 1/57.0699681\cdots $. Could we have $E_n \sim C \rho^n$, and what is the value
of $C$? I can't wait: here are the products $E_n\rho^{-n}$:

\begin{tabular}{cccccc}
$n=1$ & 2 & 3 & 4 & 5 & 6  \\
0.090 &
0.104 &
0.110 &
0.113 &
0.116 &
0.117
\end{tabular}

Hmmm, what could it be?   The numbers follow the approximate formula
$0.125 -0.05/(n+1/2)$. The limit $0.125$ is reasonably close to
an estimate which will be given in \S~\ref{strongas}.

Ah, an obscure insight (hindsight?) coming from long and
painful experiments with the $'1/9'$ problem \cite{Mag1,Mag2} tells me to
try $\exp(-(n+1/2)z)$ instead of $\exp(-nz)$, and to multiply
the errors by $\rho^{-n+1/2}$:

\begin{tabular}{ccccccc}
$n=0$ & 1 & 2 & 3 & 4 & 5 & 6  \\
 0.197 &
4.161 E-3 &
7.610 E-5 &
1.356 E-6 &
2.406 E-8 &
4.244 E-10 &
7.470 E-12 \\
1.488 &
1.794 &
1.872 &
1.904 &
1.928 &
1.941 &
1.950
\end{tabular}

\textbf{Aha!} Now, the limit seems to be 2.
This phenomenon will also be explained in \S~\ref{strongas}.


\picskip{1}

\  \\

Ah, the numerical experiments with best real rational approximation
on a bounded interval have been performed with the truly remarkable
MATLAB CF program of Trefethen  \cite{TrM}. Here is how it is
adapted to the present problem of approximation to $\exp(-(An+B)z)$
on $[0,c]$. Of course, one must enter $c$, $n$, $A$, and $B$, also
nfft $=$ preferably a power of 2, say, 512, or 1024,  and $K=$ order
of an auxiliary Hankel matrix, say, between 20 and 30.

{\small  \begin{verbatim}
%cfneuf3.m :
%
%  Quasi best rational approximation of degree n to  exp(- nf z)
%
%        on 0 <=  z  <=  c.
%
% enter:  c   n   nfft (a power of 2, say 1024)  K ( order of Hankel
%             matrix in CF construction, say 20 or more)
%
% Output:  plot of error function,    restart with higher nfft and/or K
%               if not nice;
%          two estimates of error norm. They should of course be
%              very close.
%          zeros &  poles.
%
%  cfneuf.m  adaptation de
% cftref.m
%  Approximation Theory V
% RCF -- REAL RATIONAL CF APPROXIMATION ON THE UNIT INTERVAL
%
% Loyd N. Trefethen, Dept. of Math., M.I.T., March 1986
% Reference: L.N.T. and M.H. Gutknecht,
%            SIAM J. Numer. Anal. 20 (1983), pp. 420-436
%   L.N.~Trefethen,  MATLAB programs for CF approximation,  pp.599-602
%     {\it in}  {\sl Approximation Theory V\ },
%     ( C.K.Chui, L.L.Schumaker, J.D.Ward, eds.),
%      Academic Press, Orlando 1986.
%
%
%     Fx(x) - function to be approximated by R(x)=P(x)/Q(x)
%      m,n  - degree of P,Q
%     nfft  - number of points in FFT (power of 2)
%        K  - degree at which Chebyshev series is truncated
%   F,P,Q,R - functions evaluated on FFT mesh (Chebyshev points)
%    Pc,Qc  - Chebyshev coefficients of P and Q
%
%  If Fx is even, take (m,n)=( odd, even).
%  If Fx is odd , take (m,n)=(even,even ).
%
  diary cfneuf3.txt
%
%  CONTROL PARAMETERS
   format long;
   c=input('c in interval 0..c ? ');
     m=input('degree n? ') ; n=m; np=n+1;
    aa=input(' A  in exp(-(An+B)z) ? ');bb=input(' B  in exp(-(An+B)z) ? ');
       nf=aa*n+bb;
     nfft=input('nfft? ');nfft2=nfft/2;
     K=input('K? '); dim=K+n-m;
%
%  CHEBYSHEV COEFFICIENTS OF Fx
     z = exp(2*pi*sqrt(-1)*(0:nfft-1)/nfft);
     x = real(z); F = exp(nf*c*(x-1)/2); Fc = real(fft(F))/nfft2;
%    disp(Fc(dim:dim+2));
%
%  SVD OF HANKEL MATRIX H
     H = toeplitz(Fc(1+rem((dim:-1:1)+nfft+m-n,nfft)));
        H = triu(H); H=H(:,(dim:-1:1));
     [u,s,v] = svd(H);
     s = s(np,np); u = u((dim:-1:1),np)';v = v(:,np)';
%
%  DENOMINATOR POLYNOMIAL Q
     zr = roots(v); qout = []; for i=1:dim-1;
         if abs(zr(i))>1 qout = [qout, zr(i)];end; end;
     qc = real(poly(qout)); qc = qc/qc(np); q = polyval(qc,z);
     Q = q.*conj(q); Qc = real(fft(Q))/nfft2;
     Qc(1)= Qc(1)/2; Q=Q/Qc(1); Qc = Qc(1:np)/Qc(1);
%
%  NUMERATOR POLYNOMIAL P
     b = fft([u zeros(1,nfft-dim)])./fft([v zeros(1,nfft-dim)]);
     Rt = F-real(s*z.^K.*b); Rtc = real(fft(Rt))/nfft2;
     gam = real(fft((1)./Q))/nfft2; gam = toeplitz(gam(1:2*m+1));
     if m==0 Pc = 2*Rtc(1)/gam;
       else Pc = 2*[Rtc(m+1:-1:2) Rtc(1:m+1)]/gam; end;
     Pc = Pc(m+1:2*m+1); Pc(1) = Pc(1)/2;
     P = real(polyval(Pc(m+1:-1:1),z)); R = P./Q;
%
% RESULTS
%    plot(x,F-R,'-',x,[s;0;-s]*ones(1,nfft),':'); pause;
%   s, err = norm(F-R,'inf'), Pc, Qc
       err = norm(F-R,'inf');
    serr=[s,err],
  disp(' zeros');
  qth=zeros(1,2*n+1);qth(1:n)=Pc(np:-1:2);
                  qth(n+2:2*n+1)=Pc(2:np);
                  qth(n+1)=2*Pc(1);
  polth=roots(qth);
  polx=[];
  ki=0;for kz=1:2*n;
        polxx=( polth(kz)+1/polth(kz) )/2;
         if kz==1, polx=[polx,polxx]; ki=1;kip=0;end;
         if kz>1 , kip=1;
             for kz2=1:ki;  if abs(polxx-polx(kz2))<0.0000001 , kip=0;end;
             end;
         end;
         if kip==1 , ki=ki+1; polx=[polx,polxx] ; end;
       end;
  polz= c*(1-polx)/2;
%  real(polz'), imag(polz')
  Pcc=real(poly(polz));
  polz'
  disp(' poles');
  qth=zeros(1,2*n+1);qth(1:n)=Qc(np:-1:2);
                  qth(n+2:2*n+1)=Qc(2:np);
                  qth(n+1)=2;
  polth=roots(qth);
  polx=zeros(1,n+1);
  ki=0;for kz=1:2*n;if abs(polth(kz))>1 , ki=ki+1;
                               polx(ki)=( polth(kz)+1/polth(kz) )/2;
                      end;
       end;
  polx=polx(1:ki);
  polz= c*(1-polx)/2;
%  real(polz'), imag(polz')
  polz'
  Qcc=real(poly(polz));
  Pcc=Pcc*R(1)*Qcc(np)/Pcc(np), Qcc
    tx=0:(c/100):c;er2=polyval(Pcc,tx)./polyval(Qcc,tx)-exp(-nf*tx);
    plot(tx,er2,'-',tx,[s;0;-s]*ones(1,101),':');  pause;

\end{verbatim}    }  %end small

%\vspace*{4cm}

\subsection{Root asymptotics}  \   \\


We expect the poles to tend to be ultimately distributed on a fixed arc $F$
with a limit distribution $d\mu_p$, and the interpolation points on
$E=[0,c]$ with a limit distribution $d\mu_i$, so that the complex
potential
\begin{equation}  \label{defV}
  \mathcal{V}(z) := \int_F \log\frac{1}{z-t}\;d\mu_p(t)
                -\int_E \log\frac{1}{z-t}\;d\mu_i(t)
\end{equation}
satisfies
\begin{equation}  \label{VE}
  V:={\text Re}\,\mathcal{V}= \text{\ a constant}\;= \rho \text{\ on}\; E,
\end{equation}
\begin{equation}  \label{VF}
  V(z) +\frac{ A\text{Re}\;z}{2} =
                           \text{\ a constant}\;= \sigma \text{\ on}\; F,
\end{equation}

\begin{equation}  \label{VnF}
  V(z) +\frac{A \text{Re}\;z}{2} \text{\  has equal normal derivatives
    on the two sides of\ }F,
\end{equation}

\begin{equation}  \label{unitc}
  \int_E d\mu_i(t) = \int_F d\mu_p(t) =1
\end{equation}
 (charges on $E$ and $F$), equivalent to $\mathcal{V}$ bounded
at $\infty$, actually, $\mathcal{V}'(z) \sim$ constant $z^{-2}$ for
large $z$, and $\displaystyle{ \int_C \frac{\partial
V(t)}{\partial n}\,|dt| = -2\pi }$ on any contour containing $F$
but not $E$, or also, that the imaginary part of
$\mathcal{V}$ increases by $\pi$ on $[0,c]$.

Conditions \eqref{VF} and \eqref{VnF} amount to the realization that
$V+A\text{Re}\,z/2$ has opposite \emph{gradients} along the normal
on the two sides of $F$:

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(0,0)(7,3)
\pscurve[linewidth=0.05](2,0)(4,2)(6,3)
\psline{->}(4.1,1.8)(4.9,1.0)\psline{->}(3.8,2.1)(3,2.9)
\uput[180](1,0.2){$F$}
\end{pspicture}
}
As the derivative of an analytic function has real and imaginary parts building the gradient
of its real part (Cauchy-Riemann: grad Re$\,F = \overline{F'}$), it follows that
$\mathcal{V}'+A/2$ takes opposite values on the two sides of $F$.

\smallskip

Now, limit values of such functions are given by
\emph{\textbf{Sokhotskyi-Plemelj}} formulas \cite{Hen}, chap.~14, etc.
\begin{equation}  \label{SP}
  \mathcal{V}'(z) = -\int_{E\bigcup F} \frac{d\mu(t)}{z-t}
   = -\PVint_{E\bigcup F} \frac{d\mu(t)}{z-t}
     \pm \pi i \mu'(z)
\end{equation}
when $z$ tends to a point of $E$ or $F$, and where $\displaystyle{\PVint}$ is the
Cauchy principal value. We therefore have
\begin{equation}  \label{VnF2}
  \PVint_{E\bigcup F} \frac{d\mu(t)}{z-t}
 = \PVint_F \frac{d\mu_p(t)}{z-t}
   -\int_E \frac{d\mu_i(t)}{z-t} = \frac{A}{2}\,,\  z\in F,
\end{equation}
which is an integral equation for the distribution $\mu_p$, to be
considered with \eqref{VE} as another equation for $\mu_p$ and
$\mu_i$\dots

Now, there are various ways to go further, and to conclude with
more or less neat expressions. There may be wrong turns, which may
however yield a useful piece of information.

We get rid of the condition \eqref{VE} by using complex Green functions
\footnote{used by Gonchar in several works\dots}
of $E$: first, let
\begin{equation}  \label{varphi1}
  \varphi(z) := \frac{2z}{c} -1
    +\sqrt{\left(\frac{2z}{c} -1\right)^2-1}
\end{equation}
with the square root such that $|\varphi(z)|>1$ for $z\notin E$:
$\varphi$ maps $\mathbb{C}\setminus E$ on the exterior of the unit
disk, with $\varphi(\infty)=\infty$.
Remark that $\displaystyle{ \varphi(z) +\frac{1}{\varphi(z)} =
        \frac{4z}{c} -2  }$.

We now build $\varphi(z,t)$,
with $\varphi(t,t)=\infty$:
\begin{equation} \label{varphi2}
\varphi(z,t) = \frac{\varphi(z) \overline{\varphi(t)}-1}
                    {\varphi(z)-\varphi(t)}, \qquad t\notin E,
\end{equation}
and reconsider a formula for $\mathcal{V}$:
\begin{equation}  \label{defV2}
  \mathcal{V}(z) := \int_F \log\varphi(z,t)\;d\mu(t),
\end{equation}
which automatically satisfies \eqref{VE}, with $\rho=0$, as
Re$\,\log\varphi(z,t) = \log|\varphi(z,t)| =0$ when $z\in E$.

As $\displaystyle{ \frac{d}{dz}\log\varphi(z,t) =
  \frac{\varphi'(z)}{\varphi(z)-1/\overline{\varphi(t)} }
 -\frac{\varphi'(z)}{\varphi(z)-\varphi(t) }  }$,
\begin{equation}  \label{defVn2}
  \frac{d\mathcal{V}(z)}{d\varphi(z)} =
   \int_F \frac{d\mu(t)}{\varphi(z)-1/\overline{\varphi(t)}}
  - \int_F \frac{d\mu(t)}{\varphi(z)-\varphi(t)}
\end{equation}
corresponding to charges and their images spread on $\varphi(F)$
and $1/\overline{\varphi(F)}$ in the $\varphi-$plane.

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(0,0)(7,5)
 \pscircle[linestyle=dashed, dash=3mm 3mm](4.5,2.5){2}
\pscurve[linewidth=0.05](1,1)(0,2.5)(1,4)
\pscurve[linewidth=0.05](3,1.5)(3.75,2.5)(3,3.5)
\pscurve(0.5,2.3)(1,0.7)(-0.3,2.5)(1,4.3)(0.5,2.7)
\pscurve(3.2,2.7)(2.7,3.9)(4,2.5)(2.7,1.1)(3.2,2.3)
\psline(0.5,2.3)(3.2,2.3)  \psline(0.5,2.7)(3.2,2.7)
%\psline{->}(6.0,1.7)(6.8,0.9)\psline{->}(5.8,2.1)(5,2.9)
\psline{->}(1.2,2.1)(2.1,2.1)    \psline{->}(2.1,2.9)(1.2,2.9)
\psline{->}(0.2,2.8)(0.2,2.4)
\uput[180](0.2,2.2){$\varphi(F)$}
\uput[0](2.7,2.2){$1/\varphi(F)$}
\end{pspicture}
}
??
As an argument of validity of the form \eqref{defV2}, let us show how
to recover, at least partially,  the derivative of \eqref{defV2}:

$\displaystyle{\mathcal{V}'(z) =
-\int_{E\bigcup F} \frac{d\mu(t)}{z-t} }$ is analytic in a neighbourhood
of $\infty$, and can be written for large $z$ as
$\mathcal{V}'(z) = \alpha z^{-1} +\beta z^{-2} +\dotsb$ (actually,
$\alpha=0$), with the contour integrals
$\alpha = (2\pi i)^{-1}\displaystyle{ \int_C \mathcal{V}'(t)\,dt }$,
$\beta = (2\pi i)^{-1}\displaystyle{ \int_C t\mathcal{V}'(t)\,dt }$,\dots
so,
$\displaystyle{\mathcal{V}'(z) = \frac{1}{2\pi i}
\int_C \frac{\mathcal{V}'(t)}{z-t}\;dt }$
on a large contour $C$ containing the singular loci $E$ and $F$, for
$z$ outside $C$.  We may as well consider the Laurent series in
powers of $\varphi(z)$

$$
\mathcal{V}'(z) = \mathcal{V}'\left(c\frac{\varphi+1/\varphi+2}{4}\right)=
 \frac{1}{2\pi i}
\int_D \frac{\mathcal{V}'((c/4)(u+1/u+2)}{\varphi-u}\;du \;,
$$
and we make the contour $D$ shrink about the singular loci $\varphi(F)$ and
$1/\varphi(F)$

The contributions about $\varphi(F)$ and $1/\varphi(F)$ sum as
$$
\mathcal{V}'((c/4)(\varphi+1/\varphi+2)) = \frac{1}{2\pi i}
\int_{\varphi(F)} \mathcal{V}' \left[ \frac{1}{\varphi-u}
 -\frac{1/u^2}{\varphi-1/u}\right]du
$$

It seems simpler to return now to the original variable $z$, but we
learned at least that $\mathcal{V}'$ has the same singularities than
$\varphi'$ at $0$ and $c$, i.e.,
$$ \mathcal{V}'(z) \sim \ \text{constant}\; z^{-1/2}, z\to 0\;;\qquad %
   \mathcal{V}'(z) \sim \ \text{constant}\; (z-c)^{-1/2}, z\to c. $$

We turn to a classical way to deal with
the Sokhotskyi-Plemelj formulas \eqref{SP}-\eqref{VnF2} in the
$z-$plane, by considering $\sqrt{z(z-c)(z-a)(z-b)}\; \mathcal{V}'(z)$
which is meromorphic outside $F$, even holomorphic, as the product
remains bounded near 0 and $c$. The product has therefore a Laurent
expansion about $\infty$, say
$$ \sqrt{z(z-c)(z-a)(z-b)}\; \mathcal{V}'(z) = \delta_0 +\delta_1
     z^{-1} +\dotsb $$
with $\displaystyle{ \delta_k = \dfrac{1}{2\pi i}\int_C
   t^{k-1}\sqrt{t(t-c)(t-a)(t-b)}\; \mathcal{V}'(t)\;dt }$, on
a contour $C$ around $F$, or
$$ \sqrt{z(z-c)(z-a)(z-b)}\; \mathcal{V}'(z) = \delta_0 +
                                \dfrac{1}{2\pi i}\int_C
  \dfrac{\sqrt{t(t-c)(t-a)(t-b)}}{z-t}\; \mathcal{V}'(t)\;dt
$$
which, considering that $\mathcal{V}'(t)= -A/2 \pm \pi i \mu'_p(t)$
on the two sides of $F$, where the square root takes opposite values,
yields
$$ \sqrt{z(z-c)(z-a)(z-b)}\; \mathcal{V}'(z) = \delta_0 +
                                \dfrac{A}{2\pi i}\int_a^b
  \dfrac{\sqrt{t(t-c)(t-a)(t-b)}}{z-t}\;dt.
$$
Now, everything is known up to the three numbers $a$, $b$, and
$\delta_0$! The clumsy move leaving the $\delta_0$ term comes from
$\mathcal{V}'(z)= O(z^{-2})$ at $\infty$, so that the product of
$\mathcal{V'}$ and the big square root has an unknown nonzero
limit there. Multiplying by
$ \sqrt{z^{-1}(z-c)(z-a)(z-b)}$ instead of
$ \sqrt{z(z-c)(z-a)(z-b)}$ removes this problem, but introduces an
unwanted residue at the origin! After several trials, I came on the--
 Bingo!
\begin{equation} \label{Vpellip1}
 \sqrt{\dfrac{z(z-c)}{(z-a)(z-b)}}\; \mathcal{V}'(z) =
                                \dfrac{A}{2\pi i}\int_a^b
  \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\; \dfrac{dt}{z-t},
\end{equation}
where one not only got rid of unwanted constants, but, as
$\mathcal{V}'(z)$ is only $O(z^{-2})$ at $\infty$, leaves
\begin{equation} \label{Vpellip2}
 \int_a^b  \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\; dt =0
\end{equation}
as a bonus!! \eqref{Vpellip2} gives one equation for $a$ and $b$,
knowing $c$ (and another equation will be worked later on, from
the unit charge condition \eqref{unitc}).
 For instance, when $c=0$, we have indeed a vanishing
integral of an odd function if $a=-b$, but, as we know (or suspect)
that $a$ and $b$ are complex conjugates, we see that $a$ and $b$ must
be opposite pure imaginary numbers, as they are indeed in the Pad\'e
case. To work \eqref{Vpellip2} a bit further, we see that it is a
complete elliptic integral of the third  kind (complete because one
integrates on a arc joining two branchpoints; of the third kind
because the incomplete integral behaves like a logarithm somewhere
[near $\infty$, the square root is $1 +(a+b-c)/(2t) +\dotsb$]).

A convenient transformation sending the four branchpoints $0$, $c$,
$a$, and $b$ on and from a symmetric set is
$$
  t = \frac{\alpha+iv}{1+i\gamma v}.
$$
So, $v=i\alpha$ is mapped on $t=0$, one must have, for $v=-i\alpha$,
$\displaystyle{ \frac{2\alpha}{1+\gamma \alpha}=c }$, and
$\displaystyle{ \frac{\alpha\pm i\beta}{1\pm i\gamma \beta} = a,b }$. As neither
$a$ nor $b$ is known, we may as well take $\alpha$ and $\beta$, keeping in
mind that $\displaystyle{ \gamma = \frac{2}{c} -\frac{1}{\alpha} }$
(for given $\alpha$, $a$ and $b$ are on a circle of diametral points
$\alpha$ and $1/\gamma$).

\begin{center}
\psset{unit=1cm}
\begin{pspicture}(16,5)
% \psline(0.0,2.5)(4.8,2.5)    \psline(2.5,0.1)(2.5,4.8)
 \psarc[linewidth=0.1,linecolor=red]{*-*}(2.15,2.5){2}{50}{310}
 \pscircle[linestyle=dashed, dash=3mm 3mm](2.15,2.5){2}
  \uput[40](3.35,4.0){$b=v(\beta)$}
  \uput[310](3.35,1.0){$a=v(-\beta)$}
  %\psdots[dotsize=0.15,dotstyle=o](12,4.5)
 \psline[linewidth=0.1,linecolor=red]{*-*}(2.65,2.5)(5.6,2.5)
   \uput[90](2.65,2.5){$0=v(i\alpha)$}       \uput[90](5.6,2.5){$c=v(-i\alpha)$}
   \uput[0](0.15,2.5){$\alpha$}  \uput[325](4.15,2.5){$1/\gamma$}
% \psline[linewidth=0.1,linecolor=green]{-*}(7.65,2.5)(10.6,2.5)
% \psline[linewidth=0.1,linecolor=green]{*-*}(11.5,2.5)(12.5,2.5)
% \psline[linewidth=0.1,linecolor=green]{*-}(13.4,2.5)(15.5,2.5)
% \uput[90](10.6,2.5){$-1/k$} \uput[90](11.5,2.5){$-1$}
% \uput[90](12.5,2.5){$1$} \uput[90](13.4,2.5){$1/k$}
 \psline[linewidth=0.1,linecolor=green]{-*}(11,4.5)(11,3)
 \psline[linewidth=0.1,linecolor=green]{-*}(11,0.5)(11,2)
 \psline[linewidth=0.1,linecolor=green]{*-*}(9,2.5)(13,2.5)
 \uput[0](11,3){$-i\alpha$} \uput[0](11,2){$i\alpha$}
 \uput[90](9,2.5){$-\beta$} \uput[90](13,2.5){$\beta$}
\end{pspicture}


 The $z-$ plane and the $v-$plane.
\end{center}

Now, \eqref{Vpellip2} becomes
$\displaystyle{ \int_{-\beta}^{\beta} \dfrac{1}{(1+i\gamma v)^2}
  \sqrt{\dfrac{\alpha^2+v^2}{\beta^2-v^2}}\; dv =0}$, or, with $v=\beta\cos\theta$,
$$
   \int_0^\pi \dfrac{ \sqrt{1+(\beta^2/\alpha^2)\cos^2\theta}}
                    {(1+i\gamma \beta\cos\theta)^2}\;d\theta=
   \int_0^\pi  \dfrac{1-\gamma^2 \beta^2 \cos^2\theta}
                     {(1+\gamma^2 \beta^2 \cos^2\theta)^2}
    \sqrt{1+(\beta^2/\alpha^2)\cos^2\theta} =0.
$$

The best way to study this integral is to look how it may
be computed efficiently. Consider first the Fourier expansion
of $(1+i\gamma\beta\cos\theta)^{-2}$:
$$
\dfrac{1}{(1+i\gamma\beta\cos\theta)^2} = \left[
\dfrac{2}{(\xi-\xi^{-1})i\gamma\beta}\left(\dfrac{\xi e^{-i\theta}}
  {1-\xi e^{-i\theta}} +\dfrac{1}{1-\xi e^{i\theta}}\right)\right]^2
= \dfrac{-4}{(\xi-\xi^{-1})^2\gamma^2\beta^2}
  \sum_{p=-\infty}^\infty \left( |p|+\dfrac{1+\xi^2}{1-\xi^2}\right)
    \xi^{|p|} e^{ip\theta},
$$
\hspace*{-15pt}where $\xi$ is the root of
$\displaystyle{ \xi^2 +\dfrac{2}{i\gamma\beta}
\xi+1=0 }$, with $|\xi|<1$. The integral is therefore
$\displaystyle{ \dfrac{1}{1+\gamma^2\beta^2} \sum_{n=-\infty}^\infty
 \left(2|n|+\dfrac{1+\xi^2}{1-\xi^2}\right)\xi^{2|n|}\tau_n }$,
where $\tau_n =\displaystyle{ \int_0^\pi e^{2in\theta}
  \sqrt{1+(\beta/\alpha)^2\cos^2\theta}\;d\theta }$, computed from
the recurrence relation
%\begin{multline*}
$$0=\int_0^\pi d\left[e^{2in\theta} \left(1+\dfrac{\beta^2}{\alpha^2}
 \cos^2\theta\right)^{3/2}\right]  \Rightarrow %\\
(2n+3)\tau_{n+1}+4n(1+2\alpha^2/\beta^2)\tau_n +(2n-3)\tau_{n-1}=0.
$$%\end{multline*}
The recurrence must be performed backwards (Miller's algorithm), as
$\tau_n$ tends to zero exponentially fast when $n\to\infty$, see
Gautschi \cite{Gau67} for a survey of these matters. The $\tau_n$'s
are particular Legendre functions, by the way. And their generating
function $F(Z)=\sum_0^\infty \tau_n Z^n$ satisfies
$$
\left(Z+Z^{-1}+2+4\dfrac{\alpha^2}{\beta^2}\right) \dfrac{dF(Z)}{dZ}
+\dfrac{Z^{-2}-1}{2} F(Z) =
                           \dfrac{\tau_0 +3\tau_1 Z}{2Z^2}
$$
i.e. $\displaystyle{  F(Z)=
\left(Z+Z^{-1}+2+4\dfrac{\alpha^2}{\beta^2}\right)^{1/2}
 \int_0^Z
                           \dfrac{\tau_0 +3\tau_1 Y}{2Y^2}
       (Y+Y^{-1}+2+4\alpha^2/\beta^2)^{-3/2} \,dY  }$,
a way to see how complete elliptic integrals of the thitd kind
turn into incomplete elliptic of first and second kind. Indeed, the
integral is $G(\xi^2)=\displaystyle{ 4\xi^2 F'(\xi^2) +\dfrac{1+\xi^2}{1-\xi^2}
[2F(\xi^2)-\tau_0] }$, and it satisfies
$$ \dfrac{d}{dZ} \left[ \dfrac{1-Z}{1+Z} (Z+Z^{-1} +2+4\alpha^2/\beta^2)
  G(Z)\right] +\dfrac{Z^{-2}-1}{2}\; \dfrac{1-Z}{1+Z}  G(Z)=
\dfrac{(1-Z)^2}{2Z(1+Z)^2} [6\tau_1 -(Z+4+Z^{-1})\tau_0], $$
\begin{multline*}
(Z+Z^{-1}+2+4\alpha^2/\beta^2)^{1/2}\dfrac{1-Z}{1+Z}G(Z)=
  \int_0^Z (t+t^{-1}+2+4\alpha^2/\beta^2)^{-1/2}\dfrac{(1-t)^2}
   {2t(1+t)^2} [6\tau_1-(t+4+t^{-1})\tau_0]\,dt \\
 = -i\int_{+i\infty}^{(2i)^{-1}\log Z}
            \dfrac{\tan^2\theta}{\sqrt{\cos^2\theta+\alpha^2/\beta^2}}
      [3\tau_1-(1+2\cos^2\theta)\tau_0]\,d\theta \ \ \ \ \ \ (t=e^{2i\theta})
\end{multline*}


Some points of the locus:

\parpic{
\begin{tabular}{|c|c|c|c|}   \hline
      &       &        &            \\
 $a/c=\overline{b}/c$ & $\alpha/c$  &  $\beta/c$ & $\gamma c$ \\
      &       &        &            \\
\hline      &       &             &       \\
$ 0.023671 - 0.0280i$&$ -0.0390$&$  0.0809$&$  27.614   $\\
$ 0.061784 - 0.0757i$&$ -0.1159$&$  0.2206$&$  10.628   $\\
$ 0.091190 - 0.1152i$&$ -0.1910$&$  0.3385$&$   7.2343  $\\
$ 0.114713 - 0.1488i$&$ -0.2647$&$  0.4412$&$   5.7773  $\\
$ 0.134055 - 0.1781i$&$ -0.3371$&$  0.5327$&$   4.9659  $\\
$ 0.150306 - 0.2040i$&$ -0.4084$&$  0.6156$&$   4.4482  $\\
$ 0.204616 - 0.3022i$&$ -0.7512$&$  0.9493$&$   3.3310  $\\
$ 0.241306 - 0.3830i$&$ -1.1399$&$  1.2531$&$   2.8772  $\\
$ 0.358549 - 0.8401i$&$ -6.3630$&$  3.7085$&$ 2.1571    $\\
$ 0.406333 - 1.3084i$&$ -20.529$&$  7.8098$&$ 2.0487    $\\
$ 0.425044 - 1.6475i$&$ -39.115$&$  11.848$&$ 2.0255    $\\
$ 0.439587 - 2.0532i$&$ -73.476$&$  17.878$&$ 2.0136    $\\
$ 0.445607 - 2.2839i$&$ -100.04$&$  21.890$&$ 2.0099    $\\
      &       &        &            \\
\hline
\end{tabular}
}
\psset{unit=1cm}
\begin{pspicture}(0,-4)(5,0)
 \psline{->}(0,0)(4,0)  \psline{->}(0,-4)(0,0.5)
 \uput[90](1,0){$0.5$}     \uput[90](2,0){$1$}
 \psline[linestyle=dashed, dash=3mm 3mm](1,0)(1,-4)
 \pscurve[linewidth=0.07](0,0)(0.18,-0.23)(0.30,-0.41)(0.81,-2.61)(0.88,-4.1)
  \uput[180](0.8,-2.5){$c=1$}
 \psline[linestyle=dashed, dash=3mm 3mm](2,0)(2,-4)
 \pscurve[linewidth=0.07](0,0)(0.36,-0.46)(0.60,-0.82)(1.43,-3.36)
  \uput[0](1.5,-3.5){$c=2$}
 \psline[linewidth=0.1,linecolor=red](0,0)(4,-4.65)
  \uput[0](3,-3.4){$c=\infty$}
\end{pspicture}

\vspace*{1cm}

For each $c$, the locus of $a$ is a curve with vertical asymptote
of abscissa $c/2$, and of tangent at the origin matching the $c=\infty$
locus, given by arg$\;a = -0.860274\dotsc$ (see \cite[end of \S~3.2]{Mag3}).

\medskip
\picskip{2}
Constants already encountered appear as similar integrals, as
$$
 \gamma_0 = \lim_{z\to\infty} z^2 \mathcal{V}'(z)
      = \dfrac{A}{2\pi i}\int_a^b t \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\; dt
$$
using $\displaystyle{ \dfrac{z^2}{z-t} = z+t +O(z^{-1} }$ in
\eqref{Vpellip1}. More integrals appear in a sequence of
transformations needed in order to get a convenient incomplete
elliptic integral form for $\mathcal{V}'$. We first multiply
\eqref{Vpellip1} by $z(z-c)(z-a)(z-b)$:
$\sqrt{z^3(z-c)^3(z-a)(z-b)}\; \mathcal{V}'(z) = $
$$
\gamma_0 z^2 +[\gamma_1-(a+b+c)\gamma_0]z
           +\gamma_2-(a+b+c)\gamma_1 +(ab+ac+bc)\gamma_0 +
                                \dfrac{A}{2\pi i}\int_a^b
  \sqrt{t^3(t-c)^3(t-a)(t-b)}\; \dfrac{dt}{z-t},
$$

using $\displaystyle{ \dfrac{z(z-c)(z-a)(z-b)}{z-t} =
   (z-c)(z-a)(z-b)+tz^2 +(t-c-a-b)tz +t^3-(a+b+c)t^2
  +(ab+ac+bc)t + }$
        $ \displaystyle{\dfrac{t(t-c)(t-a)(t-b)}{z-t} }$, and
where
\begin{equation} \label{gammak}
 \gamma_k
      = \dfrac{A}{2\pi i}\int_a^b t^{k+1} \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\; dt
\end{equation}
Now, we take the derivative in $z$, and perform an integration by
parts in the integral:
$$
\sqrt{P(z)}\;\mathcal{V}''(z) + \dfrac{P'(z)}{2\sqrt{P(z)}}
 \mathcal{V} = 2\gamma_0 z +\gamma_1-(a+b+c)\gamma_0
-\dfrac{A}{2\pi i}\int_a^b \dfrac{P'(t)}{2\sqrt{P(t)}}\;\dfrac{dt}{z-t},
$$
where $P(z)=z^3(z-c)^3(z-a)(z-b)$. Now, $\displaystyle{
  \dfrac{P'(t)}{2\sqrt{P(t)}} }$ is
$\displaystyle{ \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)} } }$ (well, who's
there!) times $[4t^3 -(7a+7b+5c)t^2/2 +(3ab+2ac+2bc)t-3abc/2]/(z-t)=$
$-4z^2 +((7a+7b+5c)/2-4t)z -4t^2 +(7a+7b+5c)t/2 -3ab-2ac-2bc
+[4z^3 -(7a+7b+5c)z^2/2 +(3ab+2ac+2bc)z-3abc/2]/(z-t)$, so that
the integral is $-4\gamma_0 z -4\gamma_1 +(7a+7b+5c)\gamma_0/2 +$
what turns out to be just
$\displaystyle{  \dfrac{P'(z)}{2\sqrt{P(z)}} \mathcal{V}'(z)}$, so,

$$  \sqrt{P(z)}\; \mathcal{V}''(z) = -2\gamma_0 z -\gamma'_1,$$
where $\gamma'_1=3\gamma_1 -(5a+5b+3c)\gamma_0/2$.

\smallskip

As remarked by J.~Nuttall (22 Oct. 1999), we could have adapted
 Gonchar and Rakhmanov \cite{GR}, and use the form
$$
   \mathcal{V}''(z) = \frac{\text{constant} + \text{constant}\;z}{\sqrt{z^3 (z-c)^3 (z-a)(z-b)}}
$$
but without knowledge of the two constants\dots

Check: from \eqref{Vpellip1}, $\displaystyle{ \left(1+\dfrac{a+b-c}{2z}
   +\dotsb\right) \mathcal{V}'(z) = \dfrac{\gamma_0}{z^2}+
                                          {\gamma_1}{z^3}+  \dotsb,
   \mathcal{V}'(z) = \dfrac{\gamma_0}{z^2}+
                  {2\gamma_1+(c-a-b)\gamma_0}{2z^3}+  \dotsb,   }$ \\
$\displaystyle{
\sqrt{P(z)}\mathcal{V}''(z)= \left( z^4 -\dfrac{a+b+3c}{2}z^3+\dotsb
            \right)\left(\dfrac{-2\gamma_0}{z^3} -3
            \dfrac{2\gamma_1+(c-a-b)\gamma_0}{2z^4} +\dotsb\right) =
-2\gamma_0 z -\gamma'_1 }$.

\medskip

\begin{equation} \label{Vvp} \begin{split}
     \mathcal{V}'(z) &= -
 \int_\infty^z \frac{(\gamma'_1 + 2\gamma_0 t)dt}{\sqrt{t^3 (t-c)^3 (t-a)(t-b)}}\;,  \\
    \mathcal{V}(z) &= \text{constant}\;-
 \int_\infty^z \frac{(z-t)(\gamma'_1 + 2\gamma_0 t)dt}{\sqrt{t^3 (t-c)^3 (t-a)(t-b)}}
 \end{split}\end{equation}
 are elliptic integrals of first and second kind (there is no more logarithmic behaviour).

 The standard forms of the elliptic integrals of $1^{\text{st}}$
 and $2^{\text{nd}}$ kinds are \cite{MilneT}

 $$
 F(x,k^2) = \int_0^{\arcsin x}
     (1-k^2 \sin^2\theta)^{-1/2}\,d\theta
     = \int_0^x [(1-u^2)(1-k^2u^2)]^{-1/2}\,du,
 $$
  $$
 E(x,k^2) = \int_0^{\arcsin x}
     (1-k^2\sin^2\theta)^{1/2}\,d\theta
     = \int_0^x (1-u^2)^{-1/2}(1-k^2u^2)^{1/2}\,du.
 $$

 Elementary change of variable will not easily lead to these
 forms, but what is closest to our needs appears to be
 \cite[17.4.51]{MilneT}
 \begin{equation} \label{FE} \begin{split}
    F(y,k^2) &= (\alpha^2+\beta^2)^{1/2}\int_0^x
       \frac{dv}{[(v^2+\alpha^2)(\beta^2-v^2)]^{1/2}}\;, \\
    E(y,k^2) &= \alpha^2(\alpha^2+\beta^2)^{1/2}\int_0^x
       \frac{1}{v^2+\alpha^2}\,
       \frac{dv}{[(v^2+\alpha^2)(\beta^2-v^2)]^{1/2}}\;,
 \end{split}\end{equation}
where  $k^2=\beta^2/(\alpha^2+\beta^2)$,
$\displaystyle{y^2= \frac{x^2(\alpha^2+\beta^2)}{\beta^2(\alpha^2+x^2)}
}$. One may check that $dF/dy$ and $dE/dy$ are
what they should be (using
$\displaystyle{  1-y^2 =\frac{\alpha^2(\beta^2-x^2)}{\beta^2(\alpha^2+x^2)}}$and
$\displaystyle{  1-k^2y^2 =\frac{\alpha^2} {\alpha^2+x^2}}$ ).

Also, for complete elliptic integrals,
\begin{equation} \label{FE3} \begin{split}
0=&\int_{-\beta}^{\beta} d[v(\alpha^2+v^2)^{p+1/2}(\beta^2-v^2)^{1/2}] \Rightarrow \\
&\int_{-\beta}^{\beta} (\alpha^2+v^2)^p
    \dfrac{-(2p+3)(v^2+\alpha^2)^2 +2(p+1)(\beta^2+2\alpha^2)(v^2+\alpha^2)
         -(2p+1)\alpha^2(\alpha^2+\beta^2)}
         {\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}}dv =0.
\end{split}\end{equation}
In particular, with $p=-1$, we find two equivalent formulas for the
complete elliptic integral of second kind:
$$  \mathsf{E}=
    \alpha^2(\alpha^2+\beta^2)^{1/2}\int_0^{\beta}
       \frac{1}{v^2+\alpha^2}\,
       \frac{dv}{[(v^2+\alpha^2)(\beta^2-v^2)]^{1/2}}\; =
    (\alpha^2+\beta^2)^{-1/2}\int_0^{\beta}
      \left[ \frac{v^2+\alpha^2}{\beta^2-v^2}\right]^{1/2}\,dv.
$$

We need integrals
\begin{equation}\label{ellR}
 \int_\infty^z R(t)\frac{dt}{\sqrt{t (t-c) (t-a)(t-b)}}\;,
\end{equation}
where $R$ is a rational function with at most simple poles at 0
and/or $c$. One finds
$$
 \sqrt{(1+\gamma^2 \beta^2)\frac{1+\gamma \alpha}{1-\gamma \alpha}}
 \int_{i/\gamma}^{v(z)} R\left( \frac{\alpha+iv}{1+i\gamma v}\right)
 \frac{dv}{\sqrt{ (\alpha^2+v^2)(\beta^2-v^2)}},
$$
where $\displaystyle{ x=v(z)= i\frac{\alpha-z}{1-\gamma z}}$.

So, the constant and the simple fractions $1/t$ and $1/(t-c)$ lead
to the (indefinite) integrals
$$ R(t)=1:
\sqrt{\frac{(1+\gamma^2\beta^2)(1+\gamma \alpha)}{(1-\gamma \alpha)(\alpha^2+\beta^2)}}
 F(y,k^2),
$$
$$ R(t)=\frac{1}{t}:
\sqrt{\frac{(1+\gamma^2\beta^2)(1+\gamma \alpha)}{(1-\gamma \alpha)(\alpha^2+\beta^2)}}
\left\{ \gamma F(x,k^2)
        +i\frac{1-\alpha\gamma}{\sqrt{\alpha^2+\beta^2}}\sqrt{\frac{\beta^2-x^2}{x^2+\alpha^2}}
      +(1-\alpha\gamma) E(y,k^2)/\alpha \right\},
$$
$$ R(t)=\frac{1}{t-c}:
\sqrt{\frac{(1+\gamma^2\beta^2)(1+\gamma \alpha)^3}{(1-\gamma \alpha)^3(\alpha^2+\beta^2)}}
\left\{ \gamma F(y,k^2)
        +i\frac{1+\alpha\gamma}{\sqrt{\alpha^2+\beta^2}}\sqrt{\frac{\beta^2-x^2}{x^2+\alpha^2}}
      -(1+\alpha\gamma) E(y,k^2)/\alpha \right\},
$$
For $\mathcal{V}'(z), R(t)=-\displaystyle{
   \frac{\gamma'_1+2\gamma_0 t}{t(t-c)} =
   \frac{\gamma'_1/c}{t} - \frac{\gamma''_1/c}{t-c} }$,

\begin{equation} \label{Vp} \begin{split}
\mathcal{V}'(z) &= \text{constant}\ +
\sqrt{\frac{(1+\gamma^2\beta^2)(1+\gamma \alpha)}{(1-\gamma \alpha)^3(\alpha^2+\beta^2)}}
\\
 & \left\{
  \dfrac{\gamma}{c} \left[
     \gamma'_1 (1-\gamma \alpha) -\gamma''_1 (1+\gamma \alpha)\right]
                             F(y,k^2)  \right. \\
  &+\frac{i}{c\sqrt{\alpha^2+\beta^2}}  \left[
     \gamma'_1 (1-\gamma \alpha)^2 -\gamma''_1 (1+\gamma \alpha)^2\right]
          \sqrt{\frac{\beta^2-x^2}{x^2+\alpha^2}}    \\
 & + \left. \left[
  \gamma'_1 (1-\gamma \alpha)^2 +\gamma''_1 (1+\gamma \alpha)^2\right]
               \dfrac{E(y,k^2)}{\alpha c}  \right\}.
\end{split}\end{equation}

Remark that everything but the constant vanishes in \eqref{Vp} at
$x=\pm \beta$, i.e., at $z=a$ and $z=b$, so that this constant is
$\mathcal{V}'(a)=  \mathcal{V}'(b)= -A/2$.

The function $\mathcal{V}'$ must be single-valued in
$\mathbb{C}\setminus\{E\bigcup F\}$, i.e., have vanishing periods about
the sets $E$ and $F$.
% As the periods of the functions $F$ and $E$ are
%$\{2\mathsf{K}, 2i\mathsf{K}'\}$ and
%$\{2\mathsf{E}, 2i(\mathsf{K}'-\mathsf{E}')\}$ \cite[\S~31, formula (4)]{Akh},
\begin{equation} \label{KE} %\begin{split}
  \gamma \left[
     -\gamma'_1 (1-\gamma \alpha) -\gamma''_1 (1+\gamma \alpha)\right]
             \mathsf{K}
  +  \left[
  \gamma'_1(1-\gamma \alpha)^2 +\gamma''_1 (1+\gamma \alpha)^2\right]
             \mathsf{E}/\alpha = 0,
\end{equation}

What a mess! Wait! The integrals $\gamma_0$ and $\gamma_1$ entering
$\gamma'_1$ and $\gamma''_1$ in \eqref{KE} are complete elliptic of
first and second kind too:  \eqref{gammak} and \eqref{Vpellip2}
 mean that $\gamma_{-1}=0$, which will be useful in the calculation
of the subsequent $\gamma_k$'s, from the recurrence relation
\begin{equation} \label{recgamm}     \hspace*{-20pt}
(k+2)\gamma_k -[k(a+b+c)+3(a+b)/2+c/2]\gamma_{k-1}
  +[k(ab+ac+bc)+ab]\gamma_{k-2}  -(k-1/2)abc\;\gamma_{k-3}=0,
\end{equation}
found by
$$
0= \int_a^b d[t^{k-1/2}(t-c)^{3/2}(t-a)^{1/2}(t-b)^{1/2} \Rightarrow
\qquad\qquad%
$$
$$   \hspace*{-25pt}
 \int_a^b t^{k-2}
     \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\;
 [(k-1/2)(t-a)(t-b)(t-c) +3t(t-a)(t-b)/2   +t(t-c)(t-b)/2
                               +t(t-c)(t-a)/2 ]dt =0.
$$
When $k$ is negative enough, $\gamma_k$ in \eqref{gammak} is
immediately an elliptic integral of second kind at most:
$$
 \gamma_k
      = \dfrac{A}{2\pi i}\int_a^b t^{k+1} \sqrt{\dfrac{t(t-c)}{(t-a)(t-b)}}\; dt
=    \dfrac{A}{2\pi } \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}
                           {\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{(\alpha+iv)^{k+1}}{(1+i\gamma v)^{k+3}}
     \sqrt{\dfrac{v^2+\alpha^2}{\beta^2-v^2}}\; dv
$$
So,
\begin{equation*}\begin{split}
\gamma_{-4}&=
   \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{1+i\gamma v}{(\alpha+iv)^3}
     \sqrt{\dfrac{v^2+\alpha^2}{\beta^2-v^2}}\; dv     \\
&=   \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{(1+i\gamma v)(\alpha-iv)^3}
     {(v^2+\alpha^2)^2\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}}\; dv     \\
&=   \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{-\gamma v^4 +3\alpha(\alpha\gamma-1)v^2+\alpha^3}
     {(v^2+\alpha^2)^2\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}}\; dv
           \qquad\qquad\text{(keep only even terms)}         \\
&=   \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{[(\alpha^2-3\beta^2)\gamma-4\alpha](v^2+\alpha^2)+\alpha[\alpha\gamma(7\beta^2-\alpha^2)+7\alpha^2-\beta^2]}
     {3(\alpha^2+\beta^2)(v^2+\alpha^2)\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}}\; dv    \\
 &   \qquad\qquad\qquad\qquad\text{(subtract a suitable multiple of \eqref{FE3}
         with $p=-2$)}         \\
&=   \dfrac{A}{2\pi }
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
     \dfrac{[(\alpha^2-3\beta^2)\gamma-4\alpha] \mathsf{K} +
       [\gamma(7\beta^2-\alpha^2)+7\alpha-\beta^2/\alpha] \mathsf{E}}{3(\alpha^2+\beta^2)^{3/2}}
\end{split}\end{equation*}
$$
\gamma_{-3}
=    \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
    \int_{-\beta}^{\beta}
    \dfrac{(\alpha-iv)^2}{\alpha^2+v^2}
     \dfrac{dv}{\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}}
$$
$$
=
  \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
  \dfrac{4\mathsf{E}-2\mathsf{K}}{\sqrt{\alpha^2+\beta^2}}
$$


The relation \eqref{recgamm} allows to find the next $\gamma_k$'s\dots
excepting $\gamma_{-2}$, which is precisely the first elliptic integral
of third kind. However, knowing that $\gamma_{-1}=0$, \eqref{recgamm}
at $k=-1$ actually gives
$$
\gamma_{-2}=  c\dfrac{-2(a+b)\gamma_{-3} +3ab\gamma_{-4} }
                    {a+b-c},
$$
amounting to
$$
\gamma_{-2} =
  \dfrac{A}{2\pi}
   \sqrt{\dfrac{(1+\gamma^2\beta^2)(\alpha\gamma-1)}{\alpha\gamma+1}}
  \dfrac{2\alpha\mathsf{K}-2(\alpha+1/\gamma)\mathsf{E}}{\sqrt{\alpha^2+\beta^2}}.
$$
For $\gamma_0$, use \eqref{recgamm} at $k=0$:
$\gamma_0 = -ab(c\gamma_{-3} +2\gamma_{-2})/4$,
$$
\gamma_0 =
  \dfrac{A}{2\pi}
   \sqrt{\dfrac{(\alpha\gamma-1)(\alpha^2+\beta^2)}
               {(\alpha\gamma+1)^3  (1+\gamma^2\beta^2)  }}
  \left[\alpha^2\gamma (\mathsf{K}-\mathsf{E})-\dfrac{\mathsf{E}}{\gamma}
                              \right]
$$
and finally, $\gamma_1= \displaystyle{ \dfrac{ (5(a+b)+3c)\gamma_0
                +abc \gamma_{-2} }{6}}$,
 $\gamma'_1= 3\gamma_1-\displaystyle{ \dfrac{ (5(a+b)+3c)\gamma_0
                 }{2}} = abc\gamma_{-2}/2$,
                   $\gamma''_1=\gamma'_1+2c\gamma_0= -abc(\gamma_{-2}+c\gamma_{-3})/2$:
\begin{equation*}\begin{split}
\gamma'_1 &=
  \dfrac{A}{2\pi}
   \sqrt{\dfrac{(\alpha\gamma-1)(\alpha^2+\beta^2)}
               {(\alpha\gamma+1)^3  (1+\gamma^2\beta^2)  }}
  \;2\alpha^2\left[\mathsf{K}-\mathsf{E}-\dfrac{\mathsf{E}}{\alpha\gamma}
                              \right], \\
\gamma''_1 &=
  \dfrac{A}{2\pi}
   \sqrt{\dfrac{(\alpha\gamma-1)(\alpha^2+\beta^2)}
               {(\alpha\gamma+1)^3  (1+\gamma^2\beta^2)  }}
  \;\; \dfrac{1-\alpha\gamma}{1+\alpha\gamma}   \;2\alpha^2
  \left[\mathsf{K}-\mathsf{E} +\dfrac{\mathsf{E}}{\alpha\gamma}
                       \right],
\end{split}\end{equation*}
and \eqref{Vp} becomes
$$
\mathcal{V}'(z)= -\dfrac{A}{2} +\dfrac{A}{\pi}\left\{
    \mathsf{K}E(y,k^2)-
    \mathsf{E}F(y,k^2)
  -\dfrac{i\alpha^2}{c\sqrt{\alpha^2+\beta^2}}\left[
      \dfrac{2\alpha\gamma}{1+\alpha\gamma}\mathsf{K}
     +\dfrac{2}{\alpha\gamma(1-\alpha\gamma)}\mathsf{E}  \right]
      \sqrt{\dfrac{\beta^2-x^2}{x^2+\alpha^2}}  \right\}
$$
which confirms \eqref{KE}.

For $\mathcal{V}$, use \eqref{Vvp} with the discussion following
\eqref{ellR}: $\displaystyle{R(t)= \dfrac{(t-z)(\gamma'_1+2\gamma_0t)}
  {t(t-c)} = 2\gamma_0 +\dfrac{z\gamma'_1/c}{t}
                       -\dfrac{(z-c)\gamma''_1/c}{t-c}  }$,
\begin{equation} \label{V} \begin{split}
\mathcal{V}(z) &= \text{constant}\ +
\sqrt{\frac{(1+\gamma^2\beta^2)(1+\gamma \alpha)}{(1-\gamma \alpha)^3(\alpha^2+\beta^2)}}
\\
 & \left\{
   2\gamma_0 (1-\gamma \alpha)F(y,k^2)
  +\dfrac{\gamma}{c} \left[
    z \gamma'_1 (1-\gamma \alpha) -(z-c)\gamma''_1 (1+\gamma \alpha)\right]
                             F(y,k^2)  \right. \\
  &+\frac{i}{c\sqrt{\alpha^2+\beta^2}}  \left[
     z\gamma'_1 (1-\gamma \alpha)^2 -(z-c)\gamma''_1 (1+\gamma \alpha)^2\right]
          \sqrt{\frac{\beta^2-x^2}{x^2+\alpha^2}}    \\
 & + \left. \left[
  z\gamma'_1 (1-\gamma \alpha)^2 +(z-c)\gamma''_1 (1+\gamma \alpha)^2\right]
               \dfrac{E(y,k^2)}{\alpha c}  \right\}.
\end{split}\end{equation}

Now we find, at last!, a second equation for $a$ and $b$, from
$\mathcal{V}(b)-\mathcal{V}(a)= i\pi$: at $z=a$ and $z=b$,
$x=\pm \beta$, $F=\pm \mathsf{K}$,
            $E=\pm \mathsf{E}$,
\begin{equation*}\begin{split}
\mathcal{V}(a) \  \text{and\ } \mathcal{V}(b) &= \text{const.\ }
\pm \dfrac{iA}{2\pi(1-\alpha^2\gamma^2)}2\alpha^2\left\{ \gamma
  \left[ \mathsf{K}-\mathsf{E}- \dfrac{\mathsf{E}}{\alpha^2\gamma^2}\right]
   (1-\gamma \alpha)\mathsf{K}  \right. \\
  +z(1-\gamma \alpha)&2\alpha^2
  \left[ \mathsf{K}-\mathsf{E} -\dfrac{\mathsf{E}}{\alpha\gamma}\right]
   \dfrac{\gamma}{c}
  \left[ \mathsf{K}-\mathsf{E} +\dfrac{\mathsf{E}}{\alpha\gamma}\right]
     \\
  -(z-c)(1+\gamma \alpha)& \left. 2\alpha^2  \dfrac{1-\alpha\gamma}{1+\alpha\gamma}
  \left[ \mathsf{K}-\mathsf{E} +\dfrac{\mathsf{E}}{\alpha\gamma}\right]
   \dfrac{\gamma}{c}
  \left[ \mathsf{K}-\mathsf{E} -\dfrac{\mathsf{E}}{\alpha\gamma}\right]
 \right\}
\end{split}\end{equation*}
$$
= \text{const.} \pm \dfrac{iAc}{2\pi} \left\{
    \left[ \alpha\gamma(\mathsf{K}-\mathsf{E}) -\dfrac{\mathsf{E}}{\alpha\gamma}
          \right] \mathsf{K}
   -   \left[ \alpha\gamma(\mathsf{K}-\mathsf{E})^2
    -\dfrac{\mathsf{E}^2}{\alpha\gamma}    \right] \right\},
$$
so that, $\mathcal{V}(b)-\mathcal{V}(a) =i\pi \Rightarrow$
\begin{equation} \label{eqV2}
  Ac = \dfrac{\pi^2}{ \left( \dfrac{1}{\alpha\gamma}-\alpha\gamma  \right)
  \mathsf{E}(\mathsf{K}-\mathsf{E})   }
\end{equation}
which, together with  \eqref{Vpellip2}, gives $\alpha$, $\gamma$,
$\mathsf{K}$, etc., from $Ac$ (no wonder that everything depends
essentially on the product $Ac$: remember that we approximate
$\exp(-nAz)$ on $0\leqslant z\leqslant c$, equivalent up to a scaling
to the approximation of $\exp(-nAct)$ on $[0,1]$.

Check: when $c\to\infty$, $\alpha\gamma = -1+2\alpha/c \to -1$,
$1/(\alpha\gamma)-\alpha\gamma \sim -4\alpha/c$, and we should check that
$-4\alpha \mathsf{E}(\mathsf{K}-\mathsf{E}) \to \pi^2$. Yes:
$\displaystyle{ |a|^2 = ab = \dfrac{\alpha^2+\beta^2}{1+\gamma^2 \beta^2}
\to \alpha^2}$, and we know that $\mathsf{E}\to \mathsf{K}/2$ when
$c\to\infty$, so that the limit of \eqref{eqV2} is $|a|\mathsf{K}^2 =
\pi^2$, confirmed by $|\xi_1|\mathsf{K}= \pi$ in
\cite[\S~2]{Mag2} (where $\xi_1 = \sqrt{-a}$), and
by $|a|=\pi/\omega$ of \cite[\S~3.4, eq.~(34)]{Mag3}.

For $\mathcal{V}(0)$ and $\mathcal{V}(c)$,
it is better to show $\mathcal{V}$ to be bounded at 0 and $c$ by adapting
\eqref{FE3} to incomplete elliptic integrals:
$$
    E(y,k^2) = \dfrac{x}{\sqrt{\alpha^2+\beta^2}}
               \sqrt{\dfrac{\beta^2-x^2}{\alpha^2+x^2}} +
                     \dfrac{1}{\sqrt{\alpha^2+\beta^2}}
         \int_0^x\sqrt{\dfrac{\alpha^2+v^2}{\beta^2-v^2}}\;dv,
$$
and the two last lines of \eqref{V} become
$$  \hspace*{-25pt}
 \dfrac{i(1-\gamma \alpha)}{c\alpha (1+i\gamma x)}
 [\gamma'_1 (1-\gamma \alpha) +\gamma''_1 (1+\gamma \alpha)]
 \dfrac{\sqrt{(\alpha^2+x^2)(\beta^2-x^2)}   }{\sqrt{\alpha^2+\beta^2}}
+\dfrac{
  z\gamma'_1 (1-\gamma \alpha)^2 +(z-c)\gamma''_1 (1+\gamma \alpha)^2}
 {\alpha c\sqrt{\alpha^2+\beta^2} }
         \int_0^x\sqrt{\dfrac{\alpha^2+v^2}{\beta^2-v^2}}\;dv,
$$
using $\displaystyle{z= \dfrac{\alpha+ix}{1+i\gamma x}}$ and
 $\displaystyle{z-c= z-\dfrac{2\alpha}{1+\gamma \alpha}=
  \dfrac{(\gamma \alpha-1)(\alpha-ix)}{(1+\gamma \alpha)(1+i\gamma x)}}$

$\mathcal{V}(c)-\mathcal{V}(0)= \pi i$: at $z=0$ and $z=c$,
$x=i(\alpha-z)/(1-\gamma z) = \pm i\alpha$, where
$$
F(y,k^2)= \sqrt{\alpha^2+\beta^2} \int_0^{\pm i\alpha}
 \dfrac{dv}{\sqrt{(v^2+\alpha^2)(\beta^2-v^2)}} =
                           \pm i\sqrt{\alpha^2+\beta^2} \int_0^{\alpha}
 \dfrac{dw}{\sqrt{(\alpha^2-w^2)(\beta^2+w^2)}} =   \pm i \mathsf{K}', $$
as $\alpha$ and $\beta$ have been interchanged in $\mathsf{K}$.

Also, $\displaystyle{ \dfrac{1}{\sqrt{\alpha^2+\beta^2}}
  \int_0^{\pm i \alpha} \sqrt{\dfrac{\alpha^2+v^2}{\beta^2-v^2}}\,dv =   }$
$  \pm i\displaystyle{ \dfrac{1}{\sqrt{\alpha^2+\beta^2}}
  \int_0^{\alpha} \dfrac{\alpha^2+\beta^2 -(\beta^2+w^2)}{\sqrt{(\alpha^2-w^2)(\beta^2+w^2)}}\,dw =   }$
$\pm i(\mathsf{K}'-\mathsf{E}')$.

What a mess again. Well, the important thing now is the rate of
decrease of the error with $n$, which is
$\rho = \exp\{-2[ $ Re $(\mathcal{V}(z) +Az/2)$ on $F  -$ Re $\mathcal{V}(z)$
on $E ]\}$. From $\mathcal{V}'(a)+A/2=0$, one has
$ \mathcal{V}'(z)+A/2 =    $ \\
$\displaystyle{  -\int_a^z
  \dfrac{\gamma'_1 +2\gamma_0 t = \gamma''_1 +2\gamma_0(t-c) }
        {\sqrt{t^3(t-c)^3 (t-a)(t-b)}}\,dt  }$,
$\displaystyle{ \mathcal{V}(z)+Az/2 = \text{const.} -\int_a^z   (z-t)
  \dfrac{\gamma''_1 +2\gamma_0(t-c) }
        {\sqrt{t^3(t-c)^3 (t-a)(t-b)}}\,dt  }$,
and what is needed is the complete elliptic integral
$$
\log\rho = 2\,\text{Re}\; [\mathcal{V}(0) -\mathcal{V}(a)-Aa/2] =
 -2\;\text{Re}\; \int_0^a
  \dfrac{ \gamma''_1 +2\gamma_0(t-c) }
        {\sqrt{t(t-c)^3 (t-a)(t-b)}}\,dt  ,
$$
which turns as
\begin{equation} \label{rho}
 \log\dfrac{1}{\rho} = \pi \dfrac{ \alpha\gamma (\mathsf{K}-\mathsf{E})
                                           (\mathsf{K}'-\mathsf{E}')
                                    -\mathsf{E}\mathsf{E}' }
              {(\alpha\gamma-1)\mathsf{E}  (\mathsf{K}-\mathsf{E})  }
\end{equation}

The Legendre relation
$\mathsf{E}\mathsf{E}'  -(\mathsf{K}-\mathsf{E})
            (\mathsf{K}'-\mathsf{E}') =\pi/2$ may be useful here, but
does not give a much nicer formula.

Check when $c\to\infty$: $\alpha\gamma \to -1$ and $
                                    \mathsf{K}-2\mathsf{E} \to 0$,
and we recover the $\rho=\exp(-\pi\mathsf{K}'/\mathsf{K})$
of \cite{Mag1,Mag2,Mag3}!

In the limit case $c\to 0$, $\mathcal{V} \sim \mathcal{V}_{p,\text{Pad\'e}}
-\mathcal{V}_{[0,c]}$, where
$\mathcal{V}_{p,\text{Pad\'e}}$ is the potential of the poles of
the Pad\'e approximant, and where
$\mathcal{V}_{[0,c]}$ is the equilibrium potential of $[0,c]$:
$$
\mathcal{V}(z) \sim -\log\left( 1+\sqrt{1+\dfrac{A^2 z^2}{4}} \right)
      +\sqrt{1+\dfrac{A^2 z^2}{4}}  -\dfrac{Az}{2}
  +\log\left[ \dfrac{2z}{c}-1 +\sqrt{
       \left( \dfrac{2z}{c}-1 \right)^2 -1 } \right]
$$
where the square roots behave like $Az/2$ and $2z/c$ for large $z$.
Then,
$$
\rho \sim \exp(2\;\text{Re}\;( \mathcal{V}(0) -\mathcal{V}(a)-Aa/2))
\sim \exp\left(2-2\log 2 -2\log\dfrac{8}{Ac} \right)
 = \left(\dfrac{eAc}{16}\right)^2.
$$


Some instances:

\medskip

{\small

\hspace*{-22pt}\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}   \hline
      &       &        &    & & & & & &        \\
 $Ac$  & $a/c=\overline{b}/c$ & $\alpha/c$  &  $\beta/c$ & $\gamma c$ &
 $\mathsf{K}$  &  $\mathsf{E}$ & $\mathsf{K}'$ & $\mathsf{E}'$ &
  $1/\rho$  \\
      &       &        &    & & &  & & &        \\
\hline      &       &             &  & & &  & & &     \\
$\infty$ & 0 & $ 0$ & $0$ & $\infty$ &
$2.32105$ & $1.16052$ & $1.64669$ & $1.50011$ &  $9.2890255$  \\
      &       &        &    & & &  & & &        \\
      &       &        &    &  & $=2\mathsf{E}$  &  & & &  $=1/'1/9'$      \\
\hline      &       &             &  & & &  & & &     \\
10 & $0.11626-0.15112i$ & $-0.27010$ & $0.44836$ & $5.70233$ &
$2.12833$ & $1.22121$ & $1.69467$ & $1.46033$ &  $10.55420$  \\
      &       &        &    & & &  & & &        \\
\hline      &       &             &  & & &  & & &     \\
5 & $0.21759-0.32920i$ & $-0.86874$ & $1.04722$ & $3.15109$ &
$1.94139$ & $1.30256$ & $1.78286$ &  $1.39577$ &
$12.43300$ \\
      &       &        &    & & &  & & &        \\
\hline      &       &             &  & & &  & & &     \\
2 & $0.37281-0.94489i$ & $-8.58507$ & $4.47929$ & $2.11648$ &
$1.66674$ & $1.48306$ & $2.22657$ & $1.18782$ & $23.22870$ \\
      &       &        &    & & &  & & &        \\
\hline      &       &             &  & & &  & & &     \\
1 & $0.43707-1.96963i$ & $-65.1743$ & $16.5290$ & $2.01534$ &
$1.59537$ & $1.54679$ & $2.81732$ & $1.07039$ & $57.069968$ \\
      &       &        &    & & &  & & &        \\
\hline      &       &             &  & & &  & & &     \\
$0.5$ & $0.46869-3.98448i$ & $-514.5877$ & $64.57005$ & $2.00194$ &
$1.57694$ & $1.56469$ & $3.47937$ & $1.02313$ & $177.934379$ \\
      &       &        &    & & &  & & &    \\  \hline
\end{tabular}  }

The modulus of the elliptic integrals is $k=\beta/\sqrt{\alpha^2
+\beta^2}$. The limit value when $Ac\to\infty$ is $0.90890856$.
Why don't we have a ``decent'' limit for $k$ when $Ac\to\infty$,
such as $k\to 1$? Well, if we extrapolate the locus of $a$, $b$, etc.
``above'' $Ac=\infty$, we simply get\dots negative values of $Ac$
which  repeat already known results for positive values, up to simple
reflections: if $c$ unchanged and $A\leftrightarrow -A$, approximating
$\exp(-nAz) \leftrightarrow$ approximating $\exp(-nAc)$ times
$\exp(nAz)$ with $z\leftrightarrow c-z$, so that $a,b \leftrightarrow
c-a,c-b$,etc. (yes, but what about $k$?).

\subsection{Strong asymptotics  \label{strongas}} \    \\

Consider rational approximants to functions $f^ng$, and suppose that
the Hermite-Walsh error formula can already be written as
$$
f^n(z)g(z) -\dfrac{p_n(z)}{q_n(z)} \sim e^{\mathcal{W}_n(z)}
\dfrac{1}{2\pi i} \int_C f^n(t)g(t) e^{-\mathcal{W}_n(t)}
  \dfrac{dt}{z-t},
$$
where $\mathcal{W}_n$ is a ``smoothed'' approximation of the
discrete potential created by the poles and the interpolation
points. The function $\exp(\mathcal{W}_n)$ (corresponding to
Nuttall's $\chi_1$ and/or $\chi_2$ \cite{cafe,N1105,NuttJAT}) has
branch points, even if $f$ and $g$ are entire. What is this function?
The influence of $f$ is overwhelming in the determination of the
branchpoints and other main features when $n$ is large. So, we solve
first with $f$, and find the active part $F\subset C$ and the main
behaviour $(\exp(\mathcal{W}_n))^{1/n} \to \exp(2\mathcal{V})$ (root
asymptotics, also called zero order asymptotics by Nuttall).

Aptekarev \cite{Apt2001} established in some cases a more accurate
picture $\mathcal{W}_n = 2n\mathcal{V} + \widetilde{\mathcal{V}} +o(1)$
(strong asymptotics, also called first order asymptotics by Nuttall).
I give here a probably very sloppy account of Aptekarev's
wonderful results (to be available soon):


\parpic{  \psset{unit=1cm}  \begin{pspicture}(1,3)  \end{pspicture} }
\textsl{Also sprache Aptekarev}:
$\widetilde{\mathcal{V}} $ is (multivalued) analytic outside
$E\bigcup F$,
with a period $2\pi i$ about $F$, and $-2\pi i$ about $E$,
corresponding to a positive unit charge on $F$, and a negative unit
charge on $E$, with
$\widetilde{\mathcal{V}}_+   +  \widetilde{\mathcal{V}}_-$
constant on $E$,
 $\widetilde{\mathcal{V}}(z)_+ +\widetilde{\mathcal{V}}(z)_- +2\log g(z)=$
  another constant on
$F$,
and finally $\widetilde{\mathcal{V}}(z) = $ const. $+o(1)$
when $z\to\infty$ (if $E$ and $F$ are bounded).

 Moreover, the error norm is
$ E_n \sim 2 \rho^n \tilde{\rho}$, where $2\log\tilde{\rho} =
\text{Re}\,\{
 (\widetilde{\mathcal{V}}_+(z)+\widetilde{\mathcal{V}}_-(z))_E
  -[\widetilde{\mathcal{V}}_+(z)+\widetilde{\mathcal{V}}_-(z) +2\log g(z)]_F \}$.

\picskip{1}  \     \\

\bigskip

This means also that $\widetilde{\mathcal{V}}'$ is analytic outside $E$ and
$F$, taking opposite values on the two sides of $E$, and with
 $\widetilde{\mathcal{V}}'+g'(z)/g(z)$ taking opposite values on the two sides
of $F$.

\medskip

\textbf{\emph{Important special case:}} if $g=\sqrt{f}$, the
conditions on $\widetilde{\mathcal{V}}$ are exactly the conditions
 \eqref{VE}-\eqref{unitc} which we already saw for $\mathcal{V}$
itself! So, $\widetilde{\mathcal{V}} = \mathcal{V}$, and
$\tilde{\rho} = \sqrt{\rho}$ in this case.

\medskip

Remark: the real part of $\widetilde{\mathcal{V}}+\log g$ need not, and will
normally not be a constant on $F$. However, the cut on which the
boundary conditions for $\widetilde{\mathcal{V}}$ are set may be
modified (keeping the endpoints as the endpoints of $F$), and one may
dream to find the locus $\widetilde{F}$ where $\widetilde{\mathcal{V}}+\log g$
has a constant real part.
 The use and even the existence of $\widetilde{F}$
seem questionable (Aptekarev). It may be wiser and more useful to look
for a locus $F_n$ where the whole complex potential
 $\mathcal{V}_n = 2n\mathcal{V} + \widetilde{\mathcal{V}} +n\log f +\log g$ has
a constant real part, as this locus may be a fair approximation to the
set of poles for a given value of $n$ (Nuttall).


\bigskip

\textbf{\emph{Application to best approximation to}}
$\exp(-(nA+B)z)$ on $[0,c]$:

$ E_n \sim 2 \rho^n \rho_B$, where $2\log\rho_B =
\text{Re}\,\{
 (\mathcal{V}_{B,+}+(z)+\mathcal{V}_{B,-}(z))_E
  -[\mathcal{V}_{B,+}(z)+\mathcal{V}_{B,-}(z) +2Bz]_F \}$,
  $\mathcal{V}'_B=\widetilde{\mathcal{V}}'$ being analytic
    outside $E\bigcup F$, taking opposite values on the two sides
of $E=[0,c]$, $\mathcal{V}'_B(z)+B$ taking opposite values on the
two sides of $F$, or any arc of endpoints $a$ and $b$,
and
corresponding to a positive unit charge on $F$, and a negative unit
charge on $E$, and finally $\mathcal{V}'_B(z) =$ const. $z^{-2}+\dotsb$
when $z\to\infty$.


The problem is solved by $\mathcal{V}_{A/2} = \mathcal{V}$ if
$B=A/2$.

And if $B=0$? Then, $\mathcal{V}'_0$ is the simple algebraic function
 $\displaystyle{\mathcal{V}'_0(z) = \dfrac{ \text{constant} }
                                     {\sqrt{z(z-c)(z-a)(z-b)}} }$
associated to
the potential of a plain (and plane)
condenser $(E,\widetilde{F})$, although we do not need to know what
$\widetilde{F}$ is. The capacity is $2\mathsf{K}/(\pi\mathsf{K}')$, and

$\displaystyle{ \rho_0 = \exp\left(-\dfrac{\pi}{2}\,
    \dfrac{\mathsf{K}'}{\mathsf{K}}\right) }$.

And for any $B$,
\begin{equation} \label{anyB}
  \mathcal{V}_B = \dfrac{2B}{A} \mathcal{V} +
  \left(1-\dfrac{2B}{A}\right)\mathcal{V}_0
\end{equation}
does the trick,
see Meinguet \cite{Me2} for such relations. So,
$$
\rho_B = \rho^{B/A} \rho_0^{(1-2B/A)}\;.
$$
and we just have to get $\rho_0 = \exp(-1/C)$, where $C$ is the
plain condenser capacity of $(E,\widetilde{F})$.

\medskip

Now, we look at some error norms $E_n=\displaystyle{\| e^{-nz}
  -\dfrac{p_n(z)}{q_n(z)} \|_\infty }$ on $[0,c]$, and
the products $\rho^{-n} E_n/2$ which should tend towards
$\rho_0$:


{\Small

\hspace*{-42pt}\begin{tabular}{|c|cc|cc|cc|cc|cc|}   \hline
  &    &       &        &    & & & & & &        \\
$n$ & $c=$  & $0.5$ & $c=$  &  $1$ & $c=$ &
 $2$  &  $c=$ & $5$ & $c=$ &  $\infty$  \\
  &    &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$$ & $E_n$ & $\rho^{-n}E_n/2$ & $E_n$ & $\rho^{-n}E_n/2$ & $E_n$ &
$\rho^{-n}E_n/2$ & $E_n$ & $\rho^{-n}E_n/2$ & $E_n$ &  $\rho^{-n}E_n/2$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$1$ & $2.5352E-4$ & $0.02255$ & $1.5802E-3$ & $0.04509$ & $7.7144E-3$ &
$0.08960$ & $3.13E-2$ & $0.1946$ & $6.6831E-2$ &  $0.3104$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$2$ & $1.6454E-6$ & $0.02605$ & $3.1969E-5$ & $0.05206$ & $3.8218E-4$ &
$0.1031$ & $2.7691E-3$ & $0.2140$ & $7.3587E-3$ &  $0.3175$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$3$ & $9.7750E-9$ & $0.02753$ & $5.9205E-7$ & $0.05502$ & $1.7358E-5$ &
$0.1088$ & $2.301E-4$ & $0.2211$ & $7.9938E-4$ &  $0.3204$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$4$ & $5.658E-11$ & $0.02836$ & $1.0684E-8$ & $0.05667$ & $7.6871E-7$ &
$0.1119$ & $1.8815E-5$ & $0.2248$ & $8.6522E-5$ &  $0.3221$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$5$ & $3.240E-13$ & $0.02888$ & $1.907E-10$ & $0.05771$ & $3.3678E-8$ &
$0.1139$ & $1.5284E-6$ & $0.2270$ & $9.3457E-6$ &  $0.3232$  \\
 &     &       &        &    & & &  & & &        \\
\hline &     &       &             &  & & &  & & &     \\
$\lim$ & $$ & $0.03123$ & $$ & $0.06241$ & $$ &
$0.1227$ & $$ & $0.236$ & $$ &  $0.328$  \\
&      &       &        &    & & &  & & &    \\
\hline &     &       &             &  & & &  & & &     \\
$e^{-\frac{\pi\mathsf{K}'}{2\mathsf{K}}}$ & $$ & $0.03126$ & $$ & $0.06240$ & $$ &
$0.1226$ & $$ & $0.2362$ & $$ &  $0.328$  \\
&      &       &        &    & & &  & & &    \\
 \hline
\end{tabular}  }

\smallskip

The last rows are: the limit when $n\to\infty$ estimated through
a simple step of Thiele interpolatory continued fraction, i.e.,
$\lambda$ from $\lambda+\mu/(n+\nu)$ interpolation three
values, $=$ first nontrivial step of $\rho-$algorithm \cite{Bre}; \\
and the  formula
$\exp(-\pi\mathsf{K}'/(2\mathsf{K}))$.

%%\parpic{
%\psset{unit=2cm}
%\begin{pspicture}(-1,-2)(7,2)
% \psline{->}(-1,0)(1.25,0)  \psline{->}(0,-1.5)(0,2)
% \psline[linewidth=0.05](0.0,0.0)(1.0,0)
%%\psdots[dotsize=0.07,dotstyle=o](12,4.5)
%\psdots(0.437,1.967)(0.437,-1.967)(-0.932,0)(-0.296,1.416)(-0.296,-1.416)%
%(-0.679,0.912)(-0.679,-0.912)(-0.872,0.45)(-0.872,-0.45)
%\pscurve[linecolor=red](0.437,1.967)(-0.296,1.416)(-0.679,0.912)(-0.872,0.45)%
%(-0.932,0)(-0.872,-0.45)(-0.679,-0.912)(-0.296,-1.416)(0.437,-1.967)
%\uput[0](0.437,1.967){$b$}       \uput[0](0.437,-1.967){$a$}
%\uput[90](-0.93,0){$\alpha'$}
%\uput[0](1.8,0){\Huge $\leftrightarrow$}
% \psline{->}(2.5,0)(5,0)  \psline{->}(3,-1.5)(3,2)
% \psline[linewidth=0.05,linecolor=red](3.5,2)(3.5,0.6)
% \psline[linewidth=0.05,linecolor=red](3.5,-1.6)(3.5,-0.6)
% \psline[linewidth=0.07](3.75,0)(4.6,0)
%\uput[0](5.8,0){\Huge $\leftrightarrow$}
% \pscircle[linewidth=0.05](7,0){1}
% \pscircle[linewidth=0.05](7,0){0.6}
%\end{pspicture} %}
%A quick and dirty estimation is found by replacing $F$ by a circular
%arc (see the poles of $p_7/q_7$, together with the known endpoints
%$a$ and $b$, for $c=1$ on the figure),
% and considering a small value of $c$, so that $[0,c]$ is ``seen''
%from $F$ almost as a point charge at the origin, and a first
%approximation of the potential function is the Green function of
%$F$ with a singularity at the origin, i.e., for a circular arc
%of endpoints $a$ and $b$, and of midpoint $\alpha'$
%\cite{Hille}\footnote{Also, from Nehari \cite[Exercise 4 of V.7]{Neh},
%$\displaystyle{ G(z;w) = \log\left| \dfrac{ \overline{X} -i
%  \sqrt{\dfrac{\alpha'-b}{\alpha'-a}\, \dfrac{z-a}{z-b} }  }
% {X+i \sqrt{\dfrac{\alpha'-b}{\alpha'-a}\, \dfrac{z-a}{z-b} }  } \right| }$,
%and we take $X$ so that the denominator vanishes at $z=w$.}
%%$$ \widetilde{\mathcal{V}}(z) \approx
%$$ \qquad -G(z;0) \approx
%-\log \left| \dfrac{(z-\alpha')|a| -\alpha'\sqrt{(z-a)(z-b)} }{z}
% \right|,
%$$
%where the square root is continuous outside $F$, such that the
%numerator does not vanish at $z=0$.
%
%For $V_0$, replace the denominator $z$ by
%the exponential of the Green function of $[0,c]$:
%$$ V_0(z) \approx
%-\log \left| \dfrac{(z-\alpha')|a| -\alpha'\sqrt{(z-a)(z-b)} }
%{z-c/2 +\sqrt{z(z-c)} }\right|,
%$$
%and $\rho_0 \approx \exp(\widetilde{V}_E -\widetilde{V}_F)$:
%$$ \rho_0 \approx \dfrac{c}{4|a|}\, \dfrac{|\alpha'-a|}
%       {|\alpha'-c/2+\sqrt{\alpha'(\alpha'-c)}|}
%   \approx \dfrac{c}{4|a|}\, \dfrac{|\alpha'-a|}
%       {2|\alpha'|}
%$$
%when $c\to 0$. (Check: for a full large circle, $a=b=R$ and $\alpha'=-R$,
%whence $\rho_0 \approx c/(4R)$ which is OK: logarithmic capacity/$R$).
%
%For the Pad\'e locus of poles, we have $a\approx -2i$ and
%$\alpha'\approx -1.325$, should give $\rho_0 \approx c/8$.
%
%Hmm, numerical tests with $A=1$, $c=1$ and $c=0.5$ suggest about
%$c/16$, two times smaller! Well, I do it again: for small $c$,
%$\rho_0 \sim c/4$ times the logarithmic capacity of $1/F$, which is
%an arc of endpoints $\pm i/2$, hence with a log. capacity a bit larger
%than $1/4$, so $\rho_0$ a bit larger than $c/16$: promising!
%
%To do: better estimation of condenser capacity.
%
%
%\picskip{1} \   \\
%
%%\vspace*{4cm}

\section{Best rational approximation to other exponential functions} \   \\

Not even started!   (things from \cite{Mag1})

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

\begin{thebibliography}{99}

\bibitem{Akh} N.I.~Akhiezer, {\sl Elements of the Theory
  of Elliptic Functions}, $2^{\rm nd}$~edition, ``Nauka'', Moscow,
  1970 =    Translations Math. Monographs {\bf 79}, AMS,
  Providence, 1990.

\bibitem{And} V.~Andrievskii, On the approximation of a continuum
   by lemniscates, \textsl{J.\ Approx.\ Theory} \textbf{105} (2000)
    292-304.

\bibitem{Apt} A.I.~Aptekarev, Asymptotic properties of polynomials
  orthogonal on a system of contours, and periodic motions of Toda
  lattices, \textsl{Mat. Sb.} \textbf{125} (\textbf{167}) (1984)
  231-258 (Russian) = \textsl{Math. USSR Sb.} \textbf{53} (1986)
  233-260.

\bibitem{Apt2001} A.I.~Aptekarev, new results, 2001.

 \bibitem{AptB}
   A.I.~Aptekarev, A.I., A.~Branquinho, F.~Marcellan,
   Toda-type differential equations for the recurrence coefficients of
   orthogonal polynomials and Freud transformation,
 \textsl{J. Comput. Appl. Math.} \textbf{78} (1997), No.1, 139-160.

\bibitem{bak} G.A.~Baker, Jr., {\sl Essentials of Pad\'e Approximants},
  Ac. Press, new York, 1975.


\bibitem{PARI} C.~Batut, K.~Belabas, D.~Bernardi, H.~Cohen,
  M.~Olivier, {\sl The PARI-GP calculator}, guides and software at
  \href{ftp://megrez.math.u-bordeaux.fr/pub/pari/}
  {\texttt{ftp://megrez.math.u-bordeaux.fr/pub/pari/}}  \ ,\  %
  \href{http://www.parigp-home.de/}
  {\texttt{http://www.parigp-home.de/}}

\bibitem{Beck}
   B. Beckermann, On a conjecture of E.A. Rakhmanov,
  \textsl{Constr. Approx.} \textbf{16} (2000) 427-448.


\bibitem{Braess}
  D.~Braess,
  On the conjecture of Meinardus on rational approximation of $e\sp x$, II.
   \textsl{J. Approximation Theory} \textbf{40} (1984) 375-379.

\bibitem{Bre} C.~Brezinski,

\bibitem{BSV} M.G.~de Bruin, E.B.~Saff, R.S.~Varga, On the zeros
  of generalized Bessel polynomials. I. \textsl{Proc. Koninklijke
  Nederlandse Akademie van Wetenschappen Amsterdam} \textbf{A 84}(1),
  (1981)      ; II, \emph{ibid.} \textbf{A 84}(1), (1981)

\bibitem{CRV} A.J.~Carpenter, A.~Ruttan, R.S.~Varga, Extended numerical
    computations on the "1/9" conjecture in rational approximation theory,
    pp.~383-411 {\it in} \textsl{Rational Approximation and Interpolation,
     Proceedings, Tampa, Florida, 1983} (P.R.~Graves-Morris, E.B.~Saff
    and R.S.~Varga, editors), \textsl{Springer Lecture Notes Math.}
    \textbf{1105}, Springer, Berlin, 1984.

\bibitem {che} Li-C.~Chen, M.E.H.~Ismail. On asymptotics of Jacobi
      polynomials. {\sl SIAM J.\ Math.\ Anal.\ } {\bf 22\/} (1991)
       1442-1449.

\bibitem{Cour} R.~Courant, \textsl{Dirichlet's Principle},
        Interscience 1950.

\bibitem{Deift} P.~Deift, \textsl{Orthogonal Polynomials and
   Random Matrices: A Riemann-Hilbert Approach}, Courant Inst.
   Math. Sci. \& Amer. Math . Soc. 1998.

\bibitem{DKM} P.~Deift, T.~Kriecherbauer, K.T.-R.~McLaughlin,
   New results on the equilibrium measure for logarithmic
   potentials in the presence of an external field,
   \textsl{J. Approx. Theory} \textbf{95} (1998) 388-475.

\bibitem{ESV} A.~Edrei, E.B.~Saff, R.S.~Varga, \textsl{Zeros of
   Sections of Power Series}, Springer Lecture Notes Math.
  \textbf{1002}, Springer, Berlin, 1983.

\bibitem{Thiele} R.W. Floyd: Algorithm 18. Rational interpolation by continued
  fraction, \textsl{Comm. ACM}, Sept. 1960 =
  \href{http://www.netlib.org/tomspdf/18.pdf}
       {\texttt{http://www.netlib.org/tomspdf/18.pdf}}

\bibitem{Gaier}
               D.~Gaier, {\sl Vorlesungen \"uber Approximation im
    Komplexen\/}, Birkh\"auser, 1980 = {\sl Lectures on Complex
    Approximation\/}, {\it ibid.}, 1987.

\bibitem{GaN} J.L.~Gammel, J.~Nuttall, Note on generalized Jacobi
     polynomials, {\it in\/} ``The Riemann Problem, Complete
     Integrability and Arithmetic Applications'' (D.~Chudnovsky and
     G.~Chudnovski, Eds.), pp.258-270, Springer-Verlag (Lecture
     Notes Math. {\bf 925\/}), Berlin, 1982.

\bibitem{Ga} T.~Ganelius, Degree of rational approximation, pp.~9-78
    {\it in\/} T.~Ganelius {\it et al.\/} {\sl Lectures on
    Approximation and Value Distribution, S\'em.\ Math.\
    Sup.\ - S\'em.\ Sci.\ OTAN\/}, Universit\'e de Montr\'eal,
   Qu\'ebec, 1982.

\bibitem{Gau67} W.~Gautschi,
 Computational aspects of three-term recurrence relations, \textsl{SIAM
       Rev.} \textbf{9} (1967) 24-82.

\bibitem{Gaw} W.~Gawronski, B.~Shawyer,
    Strong asymptotics and the limit distribution
    of the zeros of Jacobi polynomials $P_n^{(an+\alpha, bn+\beta)}$,
     pp.~379-404 {\it in}
       \textsl{Progress in Approximation Theory}
       (P.~Nevai \& A.~Pinkus, editors), Academic Press, 1991.

\bibitem{GV} J.S. Geronimo, W. Van Assche, Orthogonal polynomials on
   several intervals via a polynomial mapping, \textsl{Trans. Amer.
   Math. Soc.} \textbf{308} (1988), 559-581.

\bibitem{GR} A.A.~Gonchar, E.A.~Rakhmanov, Equilibrium distribution and
     the degree of rational approximation of analytic functions,
    {\sl Mat.\ Sb.\ } {\bf 134\/} (176) (1987) 306-352 =
    {\sl Math.\ USSR Sbornik\/} {\bf 62\/} (1989) 305-348.

\bibitem{ha49} W.~Hahn,
   Ueber lineare Differentialgleichungen, deren Loesungen einer
   Rekursionsformel genuegen.
   \textsl{Math. Nachr.} \textbf{4}, 1-11 (1951);
   Ueber lineare Differentialgleichungen, deren Loesungen einer
   Rekursionsformel genuegen. II.
   \textsl{Math. Nachr.} \textbf{7}, 85-104 (1952).


\bibitem{ha76} W.~Hahn,
   Kennzeichnung der linearen Differentialgleichungen, die durch
   Orthogonalpolynome befriedigt werden.
    Bericht No.54. Graz: Mathematisch-statistische Sektion im
   Forschungszentrum Graz. 16 S. (1976).

\bibitem{ha1} W.~Hahn, On differential equations for orthogonal
  polynomials, \textsl{Funk.\ Ekv.} \textbf{21} (1978), 1-9.

\bibitem{ha2} W.~Hahn, \"Uber Orthogonalpolynome, die linearen
  Functionalgleichungen gen\"ugen, pp.~16-35 \emph{in}
  \textsl{Polyn\^omes Orthogonaux et Applications, Proceedings,
  Bar-le-Duc 1984},
(C.BREZINSKI , A.RONVEAUX , A.DRAUX , A.P.MAGNUS , P.MARONI editors),
\textsl{Lecture Notes in Mathematics} {\bf 1171\/}, Springer, Berlin 1985.



 \bibitem{Ha} G.H.~Halphen, {\sl Trait\'e des fonctions elliptiques
 et de leurs applications. Premi\`ere partie. Th\'eorie des fonctions
 elliptiques et de leurs d\'eveloppements en s\'eries.}
 Gauthier-Villars, Paris, 1886. \\
 Also available at
 \href{http://gallica.bnf.fr/}{\tt http://gallica.bnf.fr}\ \ \ ,\ \ \ %
 \href{http://moa.cit.cornell.edu/}{\tt http://moa.cit.cornell.edu/}

\bibitem{Hen} P.~Henrici, \textsl{Applied and Computational Complex
  Analysis}, vol.~\textbf{3}, Wiley, 1986.

\bibitem{Hille} Hille

\bibitem{Hu} Chenglie Hu, Algorithm 785: a software package for
 computing Schwarz-Christoffel conformal transformation for
  doubly connected polygonal regions, \textsl{ACM Trans. Math.
  Soft.} \textbf{24} (1998) 317-333; FORTRAN 77 program
  available in
  \href{http://www.netlib.org/toms/785}
  {\texttt{http://www.netlib.org/toms/785}}

\bibitem{Ince}   Ince

 \bibitem{iser1} A.~Iserles,  Rational interpolation to exp(-x) with application to certain stiff
     systems,  \textsl{SIAM J. Numer. Anal.} \textbf{18} (1981), 1-12.

%\bibitem{iser2} A.~Iserles, M.J.D.~Powell,  On the A-acceptability of rational approximations
% that interpolate the  exponential function,
%  \textsl{IMA J. Numer. Anal.} \textbf{1} (1981), 241-251.

\bibitem{Iwasa} K.~Iwasaki, H.~Kimura, S.~Shimomura, M.~Yoshida,
 \textsl{From Gauss to Painlev\'e. A Modern Theory of Special
  Functions}, Vieweg, Braunschweig, 1991.

\bibitem{JE} Jahnke, Emde, L\"osch, \textsl{Tables of Higher Functions-Tafeln
  h\"oherer Funktionen}, $6^{\text{th}}$~ed., Teubner, 1960.

\bibitem{KMcM} S.~Kamvissis, K.T.-R.~McLaughlin, P.D.~Miller,
   \textsl{Semiclassical soliton ensembles for the focusing
   nonlinear Schr\"odinger equation}, preprint

\bibitem{KR}
     Kuijlaars,A.B.J., Rakhmanov, E.A.,
   Zero distributions for discrete orthogonal polynomials,
  \textsl{J. Comput. Appl. Math.} \textbf{99}, No.2, 255-274 (1998);
   corrigendum ibid. \textbf{104}, No.2, 213 (1999).

\bibitem{Lag1}  E. Laguerre, Sur la r\'eduction en fraction continue
   de $e^{F(x)}$, $F(x)$ d\'esignant un polynome entier,
  \textsl{Journal de math\'ematiques pures et appliqu\'ees},
 \textbf{VI}, 1880 =
      pp. 325-335 {\it in\/}
      {\sl Oeuvres\/}, Vol.I, Chelsea, New-York 1972.


\bibitem{Lag2}  E. Laguerre, Sur la r\'eduction en fractions continues d'une
      fraction qui satisfait \`a une \'equation diff\'erentielle lin\'eaire
      du premier ordre dont les coefficients sont rationnels,
      {\sl J. Math. Pures Appl. (4)\/} {\bf 1\/} (1885), 135-165 =
      pp. 685-711 {\it in\/}
      {\sl Oeuvres\/}, Vol.II, Chelsea, New-York 1972.

\bibitem{Li} Ninghua Li, \textsl{Strong Asymptotics of Pad\'e
  Polynomials}, Ph.D. Univ. Western Ontario, London, Ontario, 1991.

\bibitem{Lopez} G. L\'opez Lagomasino, Survey of multipoint Pad\'e
  approximation to Markov type meromorphic functions and asymptotic
  properties of the orthogonal polynomials generated by them,
  pp.~309-316 {\it in} \textsl{Polyn\^omes orthogonaux et applications,
  Proceedings, Bar-le-Duc 1984} (C.~Brezinski {\it et al.}, editors),
  \textsl{Springer Lecture Notes Math.} \textbf{1171}, 1985.

\bibitem{Lub} D.S.~Lubinsky, Asymptotics of orthogonal polynomials:
some old, some new, some identities, \textsl{Acta Applicand\ae
Mathematic\ae} \textbf{61} (2000) 207-256.

\bibitem{Mag1}  A.P.Magnus, On the use of Carath\'eodory-Fej\'er method for
investigating '1/9' and similar constants, pp. 105-132 {\it in\/} A.Cuyt, Editor:
{\sl Nonlinear Numerical Methods and Rational Approximation\/}, D.Reidel,
Dordrecht 1988.

\bibitem{Mag2} A.P.Magnus, Asymptotics  and super asymptotics of
     best rational approximation error norms
     for the exponential function (the '1/9' problem)
     by the Carath\'eodory-Fej\'er's method,
     pp.~173-185 {\it in\/}
     A.\ Cuyt, editor:
     {\sl Nonlinear Methods and Rational Approximation~II\/},
     Kluwer, Dordrecht, 1994.


\bibitem{MagP} A.P.Magnus,
           Painlev\'e-type differential equations for the recurrence
           coefficients of semi-classical orthogonal polynomials.
           {\sl J.\ Comp.\ Appl.\ Math.} {\bf 57\/} (1995) 215-237.

\bibitem{Mag3}
  A.P. Magnus, J. Meinguet,
The elliptic functions and integrals of the `1/9' problem,
 \textsl{Numerical Algorithms}, \textbf{24} (2000) 117-139.
See in \href{http://www.math.ucl.ac.be/~magnus}
    {\texttt{http://www.math.ucl.ac.be/\~{}magnus}}

\bibitem{Mag4}
  A.P.~Magnus,
  Pad\'e approximation to functions with branch points.
 Lecture notes 1998-1999,
 \href{http://www.math.ucl.ac.be/~magnus/num3/m3xxx98.ps}
 {\texttt{http://www.math.ucl.ac.be/\~{}magnus/num3/m3xxx98.ps}}

\bibitem{cafe} A.P.Magnus, J.~Nuttall, On the constructive rational
   approximation of certain entire functions, preliminary notes in
 \href{http://www.crosswinds.net/~jnuttall/approx.html}
 {\texttt{http://www.crosswinds.net/\~{}jnuttall/approx.html}}=
 \href{http://www.math.ucl.ac.be/~magnus/cafe.pdf}
 {\texttt{http://www.math.ucl.ac.be/\~{}magnus/cafe.pdf}}

\bibitem{McL99}
K. McLaughlin, On the semi-classical limit of the focusing nonlinear
 Schroedinger equation, {\it in} \textsl{Random Matrix, Statistical
 Mechanics, and Integrable Systems Workshop}, MSRI, February 22-26,
 1999. Slides and video at
\href{http://msri.org/publications/ln/msri/1999/random/mclaughlin/1/title.html}
{\texttt{http://msri.org/publications/ln/msri/1999/random/mclaughlin/1/title.html}}

 \bibitem{Me} J.~Meinguet, On the solubility of the Cauchy problem,
   pp.~137-163
\emph{in} \textsl{Approximation Theory, Proceedings of a Symposium held at
Lancaster, July 1969} (A. Talbot, editor), Academic Press, London, 1970.

\bibitem{Me2} J.~Meinguet, An electrostatic approach to the determination
  of extremal measures,  \textsl{Mathematical Physics, Analysis and
  Geometry} \textbf{3} (2000) 323-337.

 \bibitem{KM} P.D.~Miller, S.~Kamvissis, On the semiclassical limit of
 the focusing nonlinear Schr\"odinger equation, \textsl{Physics
 Letters A} \textbf{247} (1998) 75-86.

\bibitem{MilneT} L.M.~Milne-Thomson, Jacobian elliptic functions
and theta functions; Elliptic integrals, chapters 16 \& 17 {\it
in} \textsl{Handbook of Mathematical Functions} (M.~Abramowitz \&
I.A.~Stegun, editors), National Bureau of Standards, 1964, =
Dover, 1965, etc.

\bibitem{Mus} Mushkelichvili

\bibitem{Neh} Z.~Nehari, \textsl{Conformal Mapping}, 1952, =Dover 1975.

\bibitem{NuttS} J.~Nuttall, S.R.~Singh, Orthogonal polynomials and
  Pad\'e approximants associated to a sustem of arcs,
   \textsl{J. Approx. Theory}
     \textbf{21} (1977), 1-42.

\bibitem{NuttSV} J.~Nuttall, The convergence of Pad\'e approximants
   to functions with branchpoints,
   pp.~101-109 {\it in} E.B.~Saff and R.S.~Varga, editors,
  \textsl{Pad\'e and Rational Approximation, Theory and Applications},
  Academic Press, New York, 1977.

\bibitem {Nuttmin} J.~Nuttall, Sets of minimum capacity, Pad\'e approximants
     and the bubble problem, pp. 185--201 \textit{in} C.~Bardos,
    D.~Bessis, eds: \textsl{Bifurcation Phenomena in Mathematical Physics
    and Related Topics, Proceedings NATO ASI Carg\`ese, 1979}, NATO
    ASI Ser.~C \textbf{54}, D.~Reidel, Dordrecht, 1980.

\bibitem{N1105} J.~Nuttall, Location of poles of Pad\'e approximants
    to entire functions,
    pp.~354-363 {\it in} \textsl{Rational Approximation and Interpolation,
     Proceedings, Tampa, Florida, 1983} (P.R.~Graves-Morris, E.B.~Saff
    and R.S.~Varga, editors), \textsl{Springer Lecture Notes Math.}
    \textbf{1105}, Springer, Berlin, 1984.

\bibitem{NuttJAT} J.~Nuttall, Asymptotics of diagonal
  Hermite-Pad\'e polynomials, \textsl{J. Approx. Theory}
  \textbf{42} (1984) 299-386.

\bibitem{NuttCA} J.~Nuttall, Asymptotics of generalized Jacobi
   polynomials, \textsl{Constr. Approx.} \textbf{2} (1986) 59-77.

\bibitem{Olv54} F.W.J.~Olver, The asymptotic expansion of Bessel
     functions of large order, \textsl{Phil. Trans. Roy. Soc.
      London Ser.A} \textbf{247} (1954), 328-368.

\bibitem{Olver} F.W.J.~Olver, \textsl{Asymptotics and Special Functions},
    Academic Press, 1974.

\bibitem{Per} O.~Perron, \textsl{Die Lehre von den Kettenbr\"uchen},
    $2^{\text{nd}}$ edition, Teubner 1929 = Chelsea.

\bibitem{SVe} E.B.~Saff, R.S.~Varga, On the zeros and poles of Pad\'e
   aproximants to $e^z$, \textsl{Numer. Math.} \textbf{25} (1975) 1-14;
   II:
   pp.~195-213 {\it in} E.B.~Saff and R.S.~Varga, editors,
  \textsl{Pad\'e and Rational Approximation, Theory and Applications},
  Academic Press, New York, 1977;
  III:
              \textsl{Numer. Math.} \textbf{30} (1978) 241-266.

\bibitem{SaTo} E.B.~Saff, V.~Totik, \textsl{Logarithmic Potentials
   with External Fields}, Springer, Berlin, 1997.

\bibitem{Sha1} D. Browne Shaffer, Distortion theorems for lemniscates
   and level loci of Green's functions, \textsl{J. d'Analyse Math.}
   \textbf{17} (1966) 59-70.

\bibitem {sta1} H.~Stahl, Three different approaches to a proof of
    convergence for Pad\'e approximants, pp. 79-124 \textit{in}
    J.~Gilewicz, M.~Pindor, W.~Siemaszko, eds: \textsl{Rational
    Approximation and its Applications in Mathematics and Physics,
    Proceedings, \L ancut 1985, Lect.\ Notes Math.} \textbf{1237},
    Springer, Berlin, 1987.

\bibitem{Sta1} H.~Stahl, On the convergence of generalized Pad\'e
  approximants, {\sl Constr. Approx.} {\bf 5} (1989) 221-240.

\bibitem{St2} H.~Stahl, Uniform rational approximation of $|x|$,
      p.\ 110-130 {\it in\/} {\sl Methods of Approximation Theory in
     Complex Analysis and Mathematical Physics, Euler Institute, 1991.\/}
     (A.A.\ GONCHAR and E.B.\ SAFF, editors), ``Nauka'', Moscow, 1992
     and Springer ({\sl Lecture Notes Math.\ } {\bf 1550\/}), Berlin,
     1993.

\bibitem{St3} H.~Stahl, Best uniform rational approximation of $x^\alpha$
     on $[0,1]$, {\sl Bull.\ AMS\/}, {\bf 28\/} (1993) 116-122.

\bibitem{Sta} H.~Stahl, Convergence of rational interpolants,
  {\sl  Bull. Belg. Math. Soc. - Simon Stevin Suppl.}, 11-32 (1996).

\bibitem{St}
  H.~Stahl, The convergence of Pad\'e approximants to functions with
  branch points, \textsl{J. Approx. Th.} \textbf{91} (1997) 139-204.

\bibitem{Sze} G.~Szeg\H o, \textsl{Orthogonal Polynomials}, $3^{\text{rd}}$
 ed., Amer.\ Math.\ Soc.\ , Colloquium Publications, vol.~23,
   Providence, 1967.

\bibitem{Takas} K.~Takasaki, Painlev\'e equations,
 \href{http://www.math.h.kyoto-u.ac.jp/~takasaki/soliton-lab/chron/painleve.html}
 {\texttt{http://www.math.h.kyoto-u.ac.jp/\~{}takasaki/soliton-lab/chron/painleve.html}}

\bibitem{TG} L.N.~Trefethen, M.~Gutknecht, The Carath\'eodory-Fej\'er
    method for real rational approximation, {\sl SIAM J.\ Numer.\
    Anal.\ } {\bf 20\/} (1983) 420-436.

\bibitem{TrM}  L.N.~Trefethen,  MATLAB programs for CF approximation,
      pp.599-602
      {\it in}  {\sl Approximation Theory V\ },
      ( C.K.Chui, L.L.Schumaker, J.D.Ward, eds.),
       Academic Press, Orlando 1986.

\bibitem{U} J.~Ullman,

\bibitem{VA} W. Van Assche, \textsl{Asymptotics for Orthogonal
           Polynomials}, Lect. Notes Math. \textbf{1265}, Springer, 1987.

\bibitem{Va} R.S.~Varga, {\sl Scientific Computation on Mathematical
        Problems and Conjectures,\/} CBMS-NSF Reg.\ Conf.\ Series in
        Appl.\ Math.\ {\bf 60\/}, SIAM, Philadelphia, 1990.

\bibitem{Vas} A. Almendral V\'asquez, The spectrum of periodic
   complex Jacobi matrix revisited,
  \textsl{J.\ Approx.\ Theory} \textbf{105} (2000) 344-351.

\bibitem{Wal2} J.L.~Walsh, \textsl{The Location of Critical Points of
  Analytic and Harmonic Functions},
             Amer.\ Math.\ Soc., New York, 1950.

\bibitem {wal0} J.L.~Walsh, Pad\'e approximants as limits of rational
   functions of best approximation, {\sl J.\ Math.\ Mech.\/} {\bf 13}
   (1964) 305-312.

\bibitem {wal} J.L.~Walsh, {\sl Interpolation and Approximation by
    Rational Functions in the Complex Domain\/}, $4^{\textrm{th}}$
    edition, Amer.\ Math.\ Soc., Providence, 1965.


\bibitem {wid1} H.~Widom, Extremal polynomials associated with a
   system of curves in the complex plane, {\sl Adv.\ Math.} {\bf 3}
   (1969) 127--232.


\end{thebibliography}


\end{document}
