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\lhead{\rmfamily Complex rational approx. \ \textbf{\thesection}\ --\ \mylmark\  --} \chead{} \rhead{\rmfamily\thepage}
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% 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

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%                                -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
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\begin{document}

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\begin{center}  Seminar series on \\
\textbf{\Large
  Rational  approximations and systems theory.}    \\
    February-March 2002

\smallskip

 \textbf{\large Asymptotic convergence rates of rational interpolation
             to exponential functions.}

\end{center}

\medskip
\noindent The present file is
 \texttt{\textbf{http://www.math.ucl.ac.be/\~{}magnus/num3/rsummary.ps}}
  and \texttt{\textbf{pdf}}
For more,
see \href{http://www.math.ucl.ac.be/~magnus/num3/m3xxx00.pdf}
 {\texttt{\textbf{http://www.math.ucl.ac.be/\~{}magnus/num3/m3xxx00.pdf}}}
and
 \href{http://www.math.ucl.ac.be/~magnus/num3/m3xxx00.pdf}
 {\texttt{\textbf{ps}}}   and references therein.

\bigskip
\begin{center}
  Alphonse Magnus,    \\
  Institut de  Math\'ematique Pure et Appliqu\'ee,  \\
 Universit\'e Catholique de Louvain, \\
 Chemin du Cyclotron,2, \\
  B-1348 Louvain-la-Neuve \\
      (Belgium)\\% Research address for author one
   \href{mailto: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

Various instances of behaviour of error of rational approximations
in the complex plane.

\tableofcontents

\bigskip



\section{Taylor, polynomial, and rational interpolation.}
\renewcommand{\mylmark}{Taylor \& polynomial \& rational.}
\pagestyle{fancy}

\subsection{Taylor expansions} \   \\

The Taylor series expansion of a function with finite convergence
domain
shows ``typically'' almost circular level lines of equal
approximation:
\parpic{
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 \pscircle(1,1.5){0.9}
 \pscircle[linewidth=0.05](1,1.5){1}
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}
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/Z)^n, \qquad |z|<|Z|=R,
\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 $|z|<|Z|=R$:
$$
  K_n(z) = \left(\frac{Z}{z}\right)^n
    \frac{1}{2\pi i} \int_{|t|=r} f(t) \frac{z^{n+1}}{t^n(t-z)}\;dt,
$$
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.

\medskip

So, we stretch integration contours as far as possible, so to have
smallest possible integrands, in order to catch an idea of the integral.
This is a way to cope with the additive logic of integration, where the
same result may be achieved with many terms of different phases. But if there
are not many terms, each of them has to cooperate in the same
direction, more or less.

\bigskip

Darboux: $K_n(z)$ depends on the behaviour of $f_+(t)-f_-(t)$ in
neighbourhoods of the singular points
\parpic{
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 of $f$ on the convergence circle.
The main part of the error integral is
$$\dfrac{z^{n+1}}{2\pi i}\int_0^\epsilon \dfrac{f_+(Z+u)-f_-(Z+u)}
                               {(Z+u)^n (Z+u-z)}\,du $$
$$\sim\dfrac{z^{n+1}}{2\pi i Z^n (Z-z)}
   \int_0^\epsilon [f_+(Z+u)-f_-(Z+u)] \exp(-nu/Z)\,du $$
\hspace*{3cm}(Laplace transform)

\hspace*{3cm}For instance, if $f(t)\sim A(t-Z)^\alpha$ near
$Z$,
$$ f(z)-\sum_0^n c_k z^k \sim \dfrac{ z^{n+1} A (1-e^{2\pi i \alpha})
   Z^{\alpha+1} \Gamma(\alpha+1) }{2\pi i Z^n (Z-z) n^{\alpha+1} }$$
(Watson's lemma for Laplace transforms).

So, refined asymptotics will have to deal with differences
$f_+(t)-f_-(t)$ on the the two sides of cuts.

For the \emph{\textbf{exponential}} function, there is no
finite convergence radius! Stretching the contour as far as possible
will put us in orbit! Of course,  the size of $e^t$ will now have to
be considered too.
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                                                 (0,-0.8)(2,0)
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}
The error integral of $z^{n+1} e^t t^{-n}(t-z)^{-1}
dt$ is best analysed on a contour of radius $|t|=n$, whence the main
behaviour $z^{n+1} (e/n)^n$ (when $|z|$ is much
smaller than $n$). Remark the famous 100\% relative error curve
$|e^z|=|z|^n(e/n)^n$ where zeros of the approximants can be found
(Szeg\H o).

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


We consider repeated (confluent) interpolation at a finite number
of points $z_1,\dotsc, z_k$, amounting in a modified polynomial
series (Jacobi expansion)
$$ \sum_{m=0}^\infty d_m(z) [(z-z_1)\dotsc (z-z_k)]^m, $$
where $d_m$ is a polynomial of degree $<k$.
\parpic{
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What we have is similar to a power expansion, with powers of
$(z-z_1)\dotsc (z-z_k)$. And the relevant contour integral
is examined after stretching the contour so to have
$|(t-z_1)\dotsc (t-z_k)|$ as large as possible on it:
\ \ \ \ $f(z)- \text{\ interpolant\ of\ degree\ } n =  $  \\
$\hspace*{-12pt}\displaystyle{\dfrac{[(z-z_1)\dotsc (z-z_k)]^{n'} \rho_n(z)}{2\pi i}
 \int_C
  \dfrac{(t-z)^{-1}f(t)\,dt}{[(t-z_1)\dotsc (t-z_k)]^{n'} \rho_n(t)} }$  \\
where $n'=\lfloor n/k\rfloor$, and where $\rho_n$ is the product
of less than $k$ factors $z-z_j$. The main $n^{\text{th}}$ power
involved in the error at $z$ is
$\displaystyle{\left( \dfrac{\Phi(z)}{\Phi(Z)}\right)^n}$,
with $\Phi(z)=[(t-z_1)\dotsc (t-z_k)]^{1/k}$, and where
$|\Phi(Z)|=\min |\Phi(t)|$ on $t\in C$. We also see that
$|\Phi(t)|=|\Phi(Z)|$ is the largest \emph{\textbf{lemniscate}}
${\red L_2}$ within the contour $C$. \\
Let ${\blue L_1}$ be the lemniscate $\{u: |\Phi(u)|= \text{\ constant\ }
= |\Phi(z)|\}$ containing $z$. The error level lines are
the lemniscates $|\Phi(z)|=$ constant.

\subsection{General rational interpolation with given poles.
 Condenser capacity} \   \\

\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}

At the end of this line of thought, one puts $m+1$ interpolation
points on a locus $E$, and $n$ poles on a locus $F$.

The error at $z$ is
\begin{equation} \label{errat}
 f(z)-\dfrac{p(z)}{q(z)} =\dfrac{ \prod(z-z_j) }{2\pi i q(z)}
  \int_C
     \dfrac{q(t)f(t)\;dt}{(t-z) \prod(t-z_j) }
\end{equation}
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\noindent
involving mainly $\displaystyle{ \left(\dfrac{\Phi(z)}{\Phi(t^*)}\right)^n}$,
with $\Phi(z) = \exp \mathcal{V}(z) =
             \exp[\mathcal{V}_p(z) - \mathcal{V}_i(z) ]$, where
$\mathcal{V}_p$ and
$\mathcal{V}_i$ are the (complex) \emph{\textbf{potentials}} of the
distributions of the poles and interpolation points:
$\displaystyle{ \mathcal{V}_p(z) := \int_C \log\dfrac{1}{z-t}\,d\mu_p(t)}$,
$\displaystyle{ \mathcal{V}_i(z) := \int_I \log\dfrac{1}{z-t}\,d\mu_i(t)}$.
We now have flexibility enough \cite{wal} to make the set of interpolation
points ${\blue L_i}$ and the set of poles ${\red L_p}$ sets where
Re $\mathcal{V} = \log|\Phi|$ take two constant values (Dirichlet
boundary-value problem for Re $\mathcal{V}$). Then,
$$\dfrac{|\Phi(z)|}{|\Phi(Z)|} =
  \exp[\text{Re}\;(\mathcal{V}(z)- (\text{Re}\mathcal{V}\text{\ on \ }
    L_i)]= \exp\left(\dfrac{-1}{\text{cap\ }(L,L_i)}\right),$$
where cap$\;(L,L_i)$ is the \emph{\textbf{capacity}} of
the \emph{\textbf{condenser}} $(L,L_i)$.

\subsection{The problem of rational interpolation at $m+n+1$ points,
  orthogonal polynomials} \   \\

The rational interpolation setting above, with given poles and
interpolation points, is a well-conditioned problem, somewhat like the
boundary value problems with Dirichlet data on a part of the boundary,
and Neumann data on the complementary part.

The seemingly inocuous alternative interpolation problem, to construct
a numerator of degree $\leqslant m$, and a denominator of degree
$\leqslant n$, such that the ratio interpolates at $m+n+1$ points,
is an ill-conditioned problem (when $m$ and $n$ are large), somewhat
like an elliptic boundary value problem with Dirichlet
\emph{\textbf{and}} Neumann data on the same part of the boundary.

\smallskip

Then, why venture such dangerous hasards? Because we don't have to
look for the boundary of a region of analyticity, we expect that the
interpolant will find it through the locus of its poles! So, full
interpolation is a tool of discovery. This was expecially clear
when \emph{\textbf{Pad\'e}} approximants (the full confluent case)
were heavily used some decades ago.

\smallskip

We look for $p_m$ and $q_n$ such that
$\displaystyle{ \dfrac{p_m(z_i)}{q_n(z_i)} =f(z_i) }$ for
$i=0,\dotsc, m+n$. OK, suppose the denominator $q_n$ known (that's the
hard part), then $p_m$ interpolates $q_nf$. But a polynomial interpolant
at $m+n+1$ points will normally have a degree as high as $m+n$! That the degree
of $p_m$ is actually $\leqslant m$ represents the set of conditions
for the denominator $q_n$. For analytic functions, the interpolant
is
$$ p_m(z)=\dfrac{1}{2\pi i}
  \int_{C_f} q_n(t) \sum_{j=0}^{m+n} \dfrac{L_j(z)}{t-z_j} f(t)\;dt =
   \dfrac{1}{2\pi i}\int_{C_f} q_n(t) \left[\dfrac{1}{t-z}-
        \dfrac{ \prod_0^{m+n} (z-z_j)}
              {(t-z)  \prod_0^{m+n} (t-z_j)} \right] f(t)\;dt,
$$
where $C_f$ is a valid contour with the $z_j$'s as interior points.
So, $p_m$ is normally $O(z^{m+n})$ for large $z$, it only be
$O(z^m)$ as it should if
$\displaystyle{  \int_{C_f}
        \dfrac{ q_n(t)f(t)}
              {(t-z)  \prod_0^{m+n} (t-z_j)} \;dt }$
is $O(z^{-n-1})$, so, as $(t-z)^{-1} = -z^{-1} -tz^{-2} -\dotsb
  +t^nz^{-n}(z-t)^{-1}$, if
\begin{equation} \label{orthogf}
   \int_{C_f} q_n(t) t^j w_n(t)\;dt =0,\qquad j=0,\dotsc, n-1,
\end{equation}
where $\displaystyle{ w_n(t)= \dfrac{f(t)}{\prod_0^{m+n} (t-z_j) }  }$.

What we have are conditions of \emph{\textbf{orthogonality}}, although
the complex facors in \eqref{orthogf} represent by no means ``nice''
orthogonality.

\section{Known rational interpolations to the exponential function.}
\renewcommand{\mylmark}{Exponential function.}
\pagestyle{fancy}

\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}


Exponential behaviour of numerator and denominator has been much
worked, especially the distribution of zeros and poles.
Saff \& Varga remark that, when $m\sim n$, these distributions
had already been examined by Olver 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:
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  \psline{->}(0.0,2)(4,2)
  \psline{->}(2.0,0.0)(2,4.2)
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 \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
$$
w(z) = \sqrt{
              \frac{\sqrt{ 1+\frac{z^2}{4n^2}}-1  }
                   {\sqrt{ 1+\frac{z^2}{4n^2}}+1  }   }
      \;\exp\left( \sqrt{ 1+\frac{z^2}{4n^2}}\right)  =
       \frac{  (z/2n) \exp\left(\sqrt{1+z^2/(4n^2)}\right) }
            {1+\sqrt{ 1+\frac{z^2}{4n^2}}  }
$$

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}


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



\subsection{Rational interpolation at equidistant points (Iserles)}
 \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.


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.


{\bf Rough asymptotics.}
 Let
 $\zeta:= \displaystyle{\frac{z-z_0 -(m+n)h/2}{nh}   }$.
  Then, with $m\sim n$,
we intend to follow things at constant $Ah$ 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 $-1\leqslant \zeta\leqslant 1$.

\smallskip

One finds that the numerator, denominator, and the error behave
like the $n^{\text{th}}$ powers of
\begin{equation}  \label{ratio} \begin{split}
 \text{num.}\qquad e^{-Az_0}P(z)  &\qquad  \exp \left[ \zeta \log \left( e^{Ah/2}
     \dfrac{ \gamma\zeta-\sqrt{\sigma^2\zeta^2+1} }{\zeta-1} \right)
      +\dfrac{Ah}{2}+\log( \gamma+\sqrt{\sigma^2\zeta^2+1}) \right] \\
 \text{den.} \qquad Q(z)  &\qquad  \exp \left[ \zeta \log \left( e^{-Ah/2}
     \dfrac{ \gamma\zeta-\sqrt{\sigma^2\zeta^2+1} }{\zeta-1} \right)
    -\dfrac{Ah}{2}  +\log( \gamma+\sqrt{\sigma^2\zeta^2+1}) \right] \\
 \text{err.}\  e^{-Az_0}\left(e^{Az}-\dfrac{P(z)}{Q(z)}\right) & \qquad
      \exp \left[ \zeta \log \left( e^{Ah}
     \dfrac{ \gamma\zeta-\sqrt{\sigma^2\zeta^2+1} }
           { \gamma\zeta+\sqrt{\sigma^2\zeta^2+1} }   \right)
    +Ah  +\log\left( \dfrac{\gamma+\sqrt{\sigma^2\zeta^2+1} }
                           {\gamma-\sqrt{\sigma^2\zeta^2+1} } \right)
     \right]  \\
   &  \qquad\qquad\qquad \text{in the convergence region}
\end{split}\end{equation}
where $\gamma=\cosh Ah/2$, $\sigma=\sinh Ah/2$, and where
determinations of the logarithms and of the square root have to be
chosen appropriately.

 At least for large $z$, we
know that $P$ and $Q$ must behave like $z^n$, possible only if
different choices of the square root are taken in the first two rows
of \eqref{ratio}, with $\sqrt{\sigma^2\zeta^2+1}\sim \sigma\zeta$
for $P$, $-\sigma\zeta$ for $Q$.

The region where the same determination of the square root holds in
the asymptotic behaviours of $P$ and $Q$
 is simply the region of good
approximation! Indeed, one finds for the ration the $n^{\text{th}}$
power of $\exp(Ah(\zeta+1))$, which makes $\exp(A(z-z_0))$.

When $Ah$ is small, we almost have the Pad\'e situation of the
figure of p.~\pageref{padefig}   (take $\zeta\to \infty$ and
 $\sigma\zeta \sim z/(2n)$).

\medskip

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

Moreover, the solution of the Cauchy problem (i.e., find $p$ and $q$
such that $q(z)f(z)-p(z)=0$ at $z_0,\dotsc, z_{m+n}$) is then immediate:
 $q+p$ must vanish at $z_1, z_3,\dotsc$,
and $q-p$ vanishes at $z_0, z_2,\dotsc$:
$$
\frac{p(z)}{q(z)} = \frac
 {c (z-z_1)(z-z_3)\dotsb   -c' (z-z_0)(z-z_2)\dotsb  }
 {c (z-z_1)(z-z_3)\dotsb   +c' (z-z_0)(z-z_2)\dotsb  }  ,
$$
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 $n$!



\section{Asymptotic features of rational interpolation.}
%\qquad \qquad {\blue \#\noire}  }
\renewcommand{\mylmark}{Asymptotics.}
\pagestyle{fancy}

\subsection{According to Gon\v car-Rahmanov-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
$\displaystyle{ w_n(t):=\dfrac{\varphi_0(t)\varphi^n(t) }
        { \prod_0^{m+n} (t-z_j)}  }$ on $C_f$, as in \eqref{orthogf}.

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 $S\subseteq C_f$.

\smallskip

{\blue\noire} \textsl{
On the support of $\mu_p$, $q_n$ is almost
a Szeg\H o orthogonal polynomial! which means that
$\pm\displaystyle{ q_n(t) \sqrt{w_n(t)} }$ has  slowly varying
phase and absolute value there.  }

%\smallskip

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(0,0.5)(3.2,3.5)
 \pscurve[linewidth=0.025](0.5,1)(1,1.5)(1.5,2)(2,2.3)(2.5,2.5)(2.9,2.7)
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 \pscircle[linestyle=dashed, dash=2mm 2mm](2.5,2.5){0.7}
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 \psline{-*}(0.5,1)(0.9,1)
 \psline{-*}(1,1.5)(1.1,2.3)
%\psline{-*}(1.5,2)(0.8,2.1)
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 \psline{-*}(2.5,2.5)(2.8,2.2)
\uput[180](2.9,2.7){$S$}
\uput[0](2,1.9){$\rho_n\sqrt{w_n}$}
\end{pspicture}
}
Indeed, a monic polynomial $\rho_n$ orthogonal in Szeg\H o's sense with
respect to $|w_n|$ on $C_f$ minimizes its (square) norm
$\displaystyle{ \int_{C_f}  |\rho_n(t)|^2
  |w_n(t)| |dt| }$. The absolute value $|\rho_n \sqrt{w_n}|$ will remain
almost constant along $S$, but the argument of
     $\rho_n \sqrt{w_n}$ will normally have fast variation.
So, $|\rho_n^2 w_n|$ is normally not the same thing as
$\rho_n^2 w_n$.

\parpic{
\psset{unit=1.0cm}\begin{pspicture}(0,0.5)(3.2,3.5)
 \pscurve[linewidth=0.025](0.5,1)(1,1.5)(1.5,2)(2,2.3)(2.5,2.5)
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 \psline{-*}(1,1.5)(1.8,1.5)
 \psline{-*}(1.5,2)(1.6,2)
 \psline{-*}(2,2.3)(1.3,2.3)
 \psline{-*}(2.5,2.5)(2.2,2.5)
\uput[315](2.2,2.5){$S$}
\uput[0](1.8,1.5){$\rho_n\sqrt{w_n}$}
\end{pspicture}
}
 \textbf{But} if there is a particular $C_f$ where
$\rho_n \sqrt{w_n}$ happens to be --almost-- \textbf{real}, then
$q_n$ may be expected to be close to $\rho_n$.

Sloppy asymptotic explanation with (complex) potentials
$\displaystyle{ \mathcal{V}_p(z) := \int_{\text{supp\ } \mu_p}
         \log\dfrac{1}{z-t}\;d\mu_p(t)  }$  and     \\
$\displaystyle{ \mathcal{V}_i(z) := \int_{\text{supp\ } \mu_i}
         \log\dfrac{1}{z-t}\;d\mu_i(t)  }$  (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
 for $\mathcal{V}'_p$:
\emph{\textbf{symmetry property}} \cite[,etc.]{Apt2001,GR,Sta}.


\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 $z$.


The solution is
\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

$\mathcal{V}'_p [(z-\alpha)(z-\beta)]^{-1/2}$ 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 $1/(z-t) = X_0/2 +\sum_1^\infty X_n T_n$:
$X_n = X_0 \rho^n$, 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 $n=0$:
$$
 X_0 = \frac{8}{(\beta-\alpha)(\rho^{-1}-\rho)}.
$$
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 $g$.




\subsubsection{Check with \eqref{ratio}.\label{check1}}


\  \\

From \eqref{ratio},
\begin{equation*}\begin{split}
-\mathcal{V}_p(z) = \lim_{n\to\infty} \frac{\log Q(z)}{n} &=
  \zeta \log \left( e^{-Ah/2}
     \dfrac{ \gamma\zeta-\sqrt{\sigma^2\zeta^2+1} }{\zeta-1} \right)
    -\dfrac{Ah}{2}  +\log( \gamma+\sqrt{\sigma^2\zeta^2+1})  \\
   &= \zeta \log \dfrac{1-i e^{-Ah/2} \rho }
                       {1-i e^{Ah/2} \rho }
    -\dfrac{Ah}{2}  +\log( \gamma+ (\rho+\rho^{-1})/(2i)\;)
\end{split}\end{equation*}
where $\zeta$ is basically our $z$ (managed so that the interpolation
points are in $[-1,1]$), and where $\rho+\rho^{-1}=2i\sigma\zeta$.
 Then, the derivative in $\zeta$ simplifies into
$$
\frac{d\mathcal{V}_p(z)}{d\zeta} =
         -\log \dfrac{1-i e^{-Ah/2} \rho }
                       {1-i e^{Ah/2} \rho }
$$
This matches \eqref{rhoz} provided
$
\alpha = -\beta = i/\sigma = 2i/[\exp(Ah/2) -\exp(-Ah/2)]
$
so that
$\displaystyle{
\frac{d\mathcal{V}_p(z)}{d\zeta} =
 \log \frac{1-ie^{-Ah/2}\rho}{1-ie^{Ah/2}\rho}     }$ \\
$\displaystyle{
= -\sum_{n=1}^\infty \frac{i^n (e^{-nAh/2}-e^{nAh/2})}{n} \rho^n.
}$
Remark that $g_1 = -i(e^{-Ah/2}-e^{Ah/2}) = 2i\sigma=4/(\beta-\alpha)$ as it
should.

\subsection{Strong asymptotics (Aptekarev)} \   \\

Aptekarev \cite{Apt2001} has the following accurate
asymptotic formulas for formal orthogonal polynomials:

Let the set of complex functions $w_n(z):= \tilde{w}_n(z) \exp(-2nQ(z))$,
with $\tilde{w}_n \to \tilde{w}_\infty$ when $n\to\infty$ be such that
the boundary value problem for complex potentials
  $$ Q(t) +[\mathcal{V}_{p,+}(t)
            +\mathcal{V}_{p,-}(t)]/2 =  \Gamma,
             \qquad t\in S $$
where $\Gamma$ is a constant,
has a solution with $S=$ a single analytic arc of endpoints $\alpha$
and $\beta$. Then the monic formal orthogonal polynomials related to
$w_n$ on $S$ satisfy
$$ q_n(z) \sim C_n \Phi^n(z) \psi(z) ,\qquad z\notin S,$$
when $n\to\infty$,
where $\Phi(z)=\exp(-\mathcal{V}_p(z)+\Gamma)$,
$\psi$ is analytic nonvanishing outside $S$ such that
$$ \psi_+(t)\psi_-(t) \tilde{w}_\infty(t) = \dfrac{i}{\sqrt{(t-\alpha)
    (t-\beta)}_+ },\qquad t\in S,$$
and $C_n=[e^{n\Gamma} \psi(\infty)]^{-1}$.
This is exactly the extension of the famous Szeg\H o's theory.
Also,    \\
$\displaystyle{
 q_n(z) \sim C_n [\Phi^n_+(z) \psi_+(z)
                     +\Phi^n_-(z) \psi_-(z)]   }$ on $S$.
For the functions of the second kind
$\displaystyle{  R_n(z):= \dfrac{1}{2\pi i}\int_S \dfrac{q_n(t) w_n(t)}
       {t-z}\;dt }$,
$$ R_n(z) \sim \dfrac{iC_n}{\Phi^n(z) \psi(z) \sqrt{(z-\alpha)(z-\beta)}}
$$
for $z\notin S$, and up to the interior of $S$.

There are also accurate uniform estimates on the whole arc $S$ if
the product $\mu'_p(t) \sqrt{(t-\alpha)(t-\beta)}$ is known to
be regular.

\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,McL99} and remarks
by J.~Nuttall in \cite{cafe})  .

\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}



We integrate \eqref{Vp2} along the lines suggested by the
exercises of section \ref{check1}, p.~\pageref{check1}:
\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 \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 $\rho_2$.

\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
$2x=\log R +(1/R-R)\sin^2\theta$ ). 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\ $i$ \\
 0.100 &   0.596697 &  0.5838 &   -1.324  -0.952\ $i$ &      2.660 &   -2.06555  -1.23283\ $i$ & 5.922\ $i$ \\
 0.500 &   1.009193 &  0.3636 &   -2.919  -1.139\ $i$ &      9.817 &   -2.98988  -2.26747\ $i$ & 4.371\ $i$\\
 1.000 &   1.672677 &  0.2167 &   -5.311  -1.179\ $i$ &     29.600 &   -3.78112  -2.72755\ $i$ & 2.923\ $i$\\
 2.000 &   4.344519 &  0.0798 &  -14.669  -1.174\ $i$ &    216.553 &   -4.47798  -2.79719\ $i$ & 1.175\ $i$\\
 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

C:\calc\pari>exit        Script completed Thu Feb 17 11:29:16 2000

\end{verbatim}

\bigskip



\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 $n=1,2$:

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    2.718281828459 exch exp exch  %   now we have exp(-x),  x
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\bigskip

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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).

\picskip{1}

\   \\

\vspace*{2.7cm}

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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  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}

\  \\


\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) = \mathcal{V}_p(z) -   \mathcal{V}_i(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 \eqref{VCf} with $\varphi(z)=\exp(-Az)$:
\begin{equation}  \label{VF}
 ( \mathcal{V}_+(z) +
  \mathcal{V}_-(z))/2 +\frac{ Az}{2} =
                       \text{\ a real constant}\;= \sigma \text{\ on}\; F,
\end{equation}
and equioscillation on $E$:
\begin{equation}  \label{VE}
  (\mathcal{V}_+ +
  \mathcal{V}_-)/2= \text{\ another real constant}\;= \rho \text{\ on}\; E,
\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 $[0,c]$.

From \eqref{pvCf},
$\mathcal{V}'+A/2$ takes opposite values on the two sides of $F$,
and also
\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 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 $c$.
 Best combination is
\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}


complete elliptic integrals of first and second kind:
$\displaystyle{  \mathsf{K}=
    (\alpha^2+\beta^2)^{1/2}\int_0^{\beta}
       \frac{dv}{[(v^2+\alpha^2)(\beta^2-v^2)]^{1/2}} }$, \\
$\displaystyle{  \mathsf{E}=
    \alpha^2(\alpha^2+\beta^2)^{1/2}\int_0^{\beta}
       \frac{1}{v^2+\alpha^2}\,
       \frac{dv}{[(v^2+\alpha^2)(\beta^2-v^2)]^{1/2}}\; =
    (\alpha^2+\beta^2)^{-1/2}\int_0^{\beta}
      \left[ \frac{v^2+\alpha^2}{\beta^2-v^2}\right]^{1/2}\,dv.
}$


\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
 $|a|=\pi/\omega$ of \cite[\S~3.4, eq.~(34)]{Mag3}.

Now, \eqref{Vpellip2} becomes
$\displaystyle{ \int_{-\beta}^{\beta} \dfrac{1}{(1+i\gamma v)^2}
  \sqrt{\dfrac{\alpha^2+v^2}{\beta^2-v^2}}\; dv =0}$.


Some points of the locus:

\begin{minipage}[t]{90mm}   %deux colonnes
\  \par

   \smallskip


\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}
\end{minipage}
\begin{minipage}[t]{70mm}   %deux colonnes
\  \par

   \bigskip

\psset{unit=1cm}
\begin{pspicture}(0,-3)(5,0)
 \psline{->}(0,0)(4,0)  \psline{->}(0,-5)(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*{2cm}

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}).
\end{minipage}

Finally, the error decreases like $\rho^n$, with
\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}


\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}) 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


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

\begin{thebibliography}{99}


\bibitem{Apt2001} A.I.~Aptekarev,
Sharp constants for rational approximation of analytic
functions (in Russian),
\textsl{Mathematical Sbornik}, Vol \textbf{193}(1), 2002,  pp. 3-72,\\
the english translation will soon be published in
the \textsl{Russ. Acad. Sci. Sb. Math.}


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

\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{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/}
{\texttt{http://msri.org/publications/ln/msri/1999/random/mclaughlin/1/}}

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



\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.

\end{thebibliography}


\end{document}
