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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
% End of article 141312 (of 141732) -- what next? [npq]


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\begin{center}
\textbf{\Large
  `1/9', summary \& afterthoughts.}    \\
    Nov.  2005

\smallskip

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

\end{center}

\medskip

\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@inma.ucl.ac.be}
        {\tt magnus@inma.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

\def\mylmark{\ contents}
\markboth{}{} \markright{}

\tableofcontents

\bigskip

\def\mylmark{\ measures \& potentials}
\markboth{}{} \markright{}

\section{Measures and potentials.}

\subsection{Algebra of rational interpolation, orthogonality} \  \\

Rational interpolant $p/q$ of degrees $m/n$ of $f$ at $m+n+1$ points
$z_0,\dotsc, z_{m+n} \in E$: if $f$ analytic in a domain
containing $E$,
\begin{equation}    \label{hermw}
 q(z)f(z)-p(z) = \dfrac{1}{2\pi i} \int_{C_n}
  \dfrac{(z-z_0)\dotsb (z-z_{m+n})}
  {(t-z_0)\dotsb (t-z_{m+n})}
  \dfrac{q(t)f(t)}{t-z}   \,dt \qquad\text{\ (Hermite, Walsh)},
\end{equation}
where $C_n$ is a contour containing $z_0,\dotsc,z_{m+n}$, and $z$ in its
interior. This leaves a numerator
\begin{equation*}
 p(z) = \dfrac{1}{2\pi i} \int_{C_n}
  \dfrac{(t-z_0)\dotsb (t-z_{m+n})-(z-z_0)\dotsb (z-z_{m+n})}{t-z}
 \dfrac{f(t)}{(t-z_0)\dotsb (t-z_{m+n})}
  q(t) \,dt
\end{equation*}
which is of degree $m+n$ unless $q$ is `orthogonal' with respect
to the `weight' \\  $w_n(t):= f(t)/[(t-z_0)\dotsb (t-z_{m+n})]$:
$\displaystyle{ \int_{C_n} q(t) t^k w_n(t)\,dt=0}$, for $k=0,\dotsb, n-1$.
Indeed, the big polynomial in $z$ and $t$ above contains terms
$t^a z^b$, with $a+b\leqslant m+n$. If only the $t^a$ with $a\geqslant n$
have to be considered, then only $z^b$ with $b\leqslant m$ are left.

Using orthogonality, \eqref{hermw} is left unchanged when one
susbtracts from $(t-z)^{-1}$ its interpolant at the zeros of
$q$, so that $\displaystyle{\dfrac{q(t)}{(t-z)q(z)}}$ is left,
and
\begin{equation}    \label{hermwo}
 f(z)-\dfrac{p(z)}{q(z)} = \dfrac{1}{2\pi i} \int_{C_n}
  \dfrac{(z-z_0)\dotsb (z-z_{m+n})q^2(t)}
  {(t-z_0)\dotsb (t-z_{m+n})q^2(z)}
  \dfrac{f(t)}{t-z}   \,dt.
\end{equation}

Remark also that the `scalar' product of two functions $u$ and $v$ is

\begin{equation}  \label{scalp}
  \langle u,v\rangle = \dfrac{1}{2\pi i} \int_{C_n}
  \dfrac{u(t)v(t)}
  {(t-z_0)\dotsb (t-z_{m+n})} f(t)\,dt
\end{equation}
is the \emph{\textbf{divided difference}} of $uvf$ at
$z_0,\dotsc, z_{m+n}$.


\subsection{Distributions of interpolation points, poles, and their
potentials} \   \\

Let $\mu_{n,i}$ and $\mu_{n,p}$ the distributions of interpolations points
on $E$, and poles on $C_n$, with unit total weight, i.e., such that
\begin{equation*}
  \int_E F(t)\,d\mu_{n,i}(t) = \dfrac{1}{m+n+1}\sum_{k=0}^{m+n}
                     F(z_k),\qquad
  \int_{C_n} F(t)\,d\mu_{n,p}(t) = \dfrac{1}{n}\sum_{k=1}^{n}
                     F(p_k).
\end{equation*}
These distributions can be seen as staircase functions, but they
will receive smoother approximations.

(Complex) logarithmic potentials: $\mathcal{V}(z)=\int \log(z-t)\,d\mu(t)$.
Then,
\begin{equation}    \label{hermwv}
 f(z)-\dfrac{p(z)}{q(z)} = \dfrac{1}{2\pi i}
    \exp((m+n+1)\mathcal{V}_{n,i}(z)-
           2n  \mathcal{V}_{n,p}(z))
     \int_{C_n}
       \exp(2n  \mathcal{V}_{n,p}(t))
          -(m+n+1)\mathcal{V}_{n,i}(t))
      \dfrac{f(t)}{t-z}   \,dt.
\end{equation}

When $m\sim n$, everything depends on
$\mathcal{V}_n(z):= \mathcal{V}_{n,i}(z)-\mathcal{V}_{n,p}(z)$.

Let $c_n$ be the largest absolute value of
 $f(t)\exp(-2n\mathcal{V}_n(t))$ on $C_n$,
then the error bound of \eqref{hermwv} is dominated by
\begin{equation} \label{errbound}
\left\{
 \begin{aligned}
c_n \exp(2n\,\text{Re}\mathcal{V}_n(z)) &\ \text{if $z$ is inside $C_n$},\\
\max\left[c_n \exp(2n\,\text{Re}\mathcal{V}_n(z)),|f(z)|\right] &\ \text{if $z$ is outside $C_n$},
 \end{aligned}
\right.
\end{equation}
as one must take into account the residue at $t=z$ in the latter case.

$\mu_{n,i}$ and $\mathcal{V}_{n,i}$ are known if one interpolates on
a given set of points. $\mathcal{V}_{n,p}$ has to be determined
from a theory of orthogonal polynomials.


\medskip

Simple example: interpolation concentrated on a single point
(Pad\'e), say 0, and $f$ analytic outside the real interval
$[a,b]$, $0\notin [a,b]$. Then $C_n$ may be deformed up to
$[a,b]$ used twice, first with limit values from above
$f_+(t)$ of $f$, and next with $-f_-(t)$. If $f_+-f_-$ has a
constant phase and basically an integrable logarithm, then
$\mathcal{V}_{n,p}(z)$ outside $[a,b]$ is close to
$\log[d(z-c) +(1-d)\sqrt{(z-a)(z-b)}]$, where $c=2/(a^{-1}+b^{-1})$
and $d=1/(1+|c|/\sqrt{ab})$ (Szeg\H o\footnote{Actually,
$t^n q(1/t)$ is orthogonal with respect to a fixed weight on
$[b^{-1},a^{-1}]$, so that $z^{-n}q(z)$ involves mainly
the logarithm of
$z^{-1}-(a^{-1}+b^{-1})/2+\sqrt{(z^{-1}-a^{-1})(z^{-1}-b^{-1})}$,
which is the ``usual'' potential related to an interval.}). Then,
$\mathcal{V}_n(z)$ is close to
$-\log[d(1-c/z) +(1-d)\sqrt{(1-a/z)(1-b/z)}]$, whose real
part is the constant $\log[(b+a)/(d(b-a))]$ on $[a,b]$, and
is less than this constant everywhere else the square root must
be taken accordingly). Also,
$c_n\sim [d(b-a)/(b+a)]^{2n}$.
For small $z$, the error behaves like
$\displaystyle{ \left[ \dfrac{d(b-a)}{b+a} \dfrac{z}{2dc}\right]^{2n}
=\left[ \dfrac{(b-a)z}{4ab}\right]^{2n} }$ ($(b-a)/(4ab)$ is the
logarithmic capacity of $[b^{-1},a^{-1}]$).

\subsection{Orthogonal polynomials behaviour} \   \\

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Remark that, as $c_n$ is the maximum of $|f \exp(-2n\mathcal{V}_n)|$
on a contour $C_n$ which may be deformed, we can as well look for
the contour yielding the smallest maximum: the smallest extimate will
be the most realistic. The point where the maximum occurs (actually,
it will be a whole subarc) is a saddle-point of
$|f \exp(-2n\mathcal{V}_n = w_n q^2|$. This is exactly what happens
with true orthogonal polynomials with respect to positive
measures,  for the $L^2$ norm on $C_n$, and we have a theory
giving $\mathcal{V}_{n,p}$ in that case (Szeg\H o, Widom).
So, for a given $C_n$, the true (monic) orthogonal polynomal
has the smallest possible $L^2$ norm on $C_n$, $w_n q^2$ has often an
almost constant absolute value (``envelope'') on a subarc $\Delta_n$, but
probably a fast varying phase there. As a consequence, the integrals
we need,
involving $w_n q^2$, without the absolute value, will be much smaller
than $L^2$ norms, and we will not get valuable estimates.

Suppose that, among all the possible $C_n$'s, a miraculous one
is such that $w_n q^2$ happens to have only a slowly varying phase
on the subarc where the absolute value is close to its maximum
$c_n$. Then, $q\sqrt{w_n}$ is almost real, we do not need complex
conjugation in the scalar products any more, and the kind of
orthogonal polynomials needed in rational interpolation look like
righteous $L^2-$orthogonal polynomials.

\subsubsection{$L^2-$orthogonal polynomials.}

Szeg\H o-Widom theory: $w_n q^2$ behaves essentially outside
$\Delta_n \bigcup E $ as $c_n \Phi^{2n}$, where $\Phi$
maps the exterior of $\Delta_n$ on $|\Phi|>1$ and behaves near
$E$ as dictated by $c_n \Phi^{2n} = f \exp(-2n\mathcal{V}_{n,i}
                                   +2n\mathcal{V}_{n,p})$.

For instance, if all the interpolation points are
concentrated on $z_0$, $c_n^{1/(2n)}\Phi$ must have a pole with unit
residue at $z_0$. The unit residue allows to compute
$c_n$. So, in the example above,
$\Phi(z)=[2ab/z-a-b +2\sqrt{ab(1-a/z)(1-b/z)}]/(b-a)$ maps
indeed\footnote{See that $\Phi+1/\Phi=2(2ab/z-a-b)$.} the exterior of $[a,b]$ on $|\Phi|>1$, and has a pole
at $z=0$ with residue $4ab/(b-a)$, whence
$c_n=[(b-a)/(4ab)]^{2n}$ as already found.

Remark also that $\log\Phi$ is the complex Green function
of $\Delta_n$ with a singularity at a given point.

``Essentially'' means that only $n^{\text{th}}$ powers are
considered for the moment.

If the interpolation points are spread on an arc $E$ with
a known distribution, $\log\Phi$ is a sum of Green fuctions
\begin{equation}   \label{logPhi}
\log\Phi(z)= \dfrac{1}{2n}\sum_{k=0}^{2n} \log\Phi(z;z_i)
  \to \int_E \log\Phi(z;t)\,d\mu_{n,i}(t).
\end{equation}
\begin{equation}   \label{cnres}
\dfrac{1}{2n}\log c_n
  \to -\int_E \log(\text{res.}\,\Phi(z;t))\,d\mu_{n,i}(t).
\end{equation}

In the example above,\\ $\displaystyle{ \Phi(z;t)=  \dfrac{
  (2t-a-b)(z-t)+2(t-a)(t-b)+2\sqrt{(t-a)(t-b)(z-a)(z-b)} }
  {(b-a)(z-t)}  }$, with the square root such that
$|\Phi|>1$ outside $[a,b]$ (there can be no doubt: the other
possibility is $1/\Phi$). The residue at the pole $z=t$ is
$4(t-a)(t-b)/(b-a)$.

The $z-$derivative is $\displaystyle{
 \dfrac{d\Phi(z;t)/dz}{\Phi(z;t)}=
 -\dfrac{1}{z-t}\sqrt{\dfrac{(t-a)(t-b)}{(z-a)(z-b)} }   }$.

Let $E$ be another interval $[c,d]$ with a uniform distribution
$d\mu_{n,i}(t)=dt/(d-c)$ (here is where point distributions are
replaced by easier smooth distributions).
So we have $\log\Phi(z)= \int_c^d \log\Phi(z;t)\,dt/(d-c)$, with
the eerie $\Phi(z;t)$ just above. But use \\
$\displaystyle{ \Phi(z;t)+\dfrac{1}{\Phi(z;t)} =
  2\dfrac{(2t-a-b)(z-t)+2(t-a)(t-b)}{(b-a)(z-t)} =
  2\dfrac{a+b-2z}{b-a} +
  4\dfrac{(z-a)(z-b)}{(b-a)(z-t)} }$,
so \\ $\displaystyle{ t=t(\Phi) = z-
 \dfrac{4(z-a)(z-b)}{(b-a)(\Phi+\Phi^{-1})-2(a+b)+4z} }$,
with poles $\Phi=\Phi^{\pm 1}(z,\infty)$ and residues  \\
$\pm\Phi^{\pm 1}\sqrt{(z-a)(z-b)}$, and
\begin{equation*}\begin{split}
\log\Phi(z)&=
  \dfrac{1}{d-c} \int_c^d \log\Phi(z;t)\,dt =
  \dfrac{1}{d-c} \int_{\Phi(z;c)}^{\Phi(z;d)} \log\Phi\,dt(\Phi) \\
   &=
  \dfrac{d\log\Phi(z;d)-c\log\Phi(z;c)}{d-c}
  -\dfrac{1}{d-c}
   \int_{\Phi(z;c)}^{\Phi(z;d)} \dfrac{t(\Phi)}{\Phi}\,d\Phi
\end{split}\end{equation*}
turning as
$\displaystyle{
  \dfrac{(d-z)\log\Phi(z;d)-(c-z)\log\Phi(z;c)}{d-c}     }$
\\$\displaystyle{
 -\dfrac{\sqrt{(z-a)(z-b)}}{d-c}
  \log \dfrac{(\Phi(z;d)-\Phi(z;\infty))
              (\Phi(z;c)-1/\Phi(z;\infty)) }
             {(\Phi(z;d)-1/\Phi(z;\infty))
              (\Phi(z;c)-\Phi(z;\infty)) }
}$
Awful. From the derivative above:
\begin{equation*}
  \dfrac{\Phi'(z)}{\Phi(z)} =-\dfrac{1}{d-c}
  \int_c^d  \dfrac{1}{z-t} \sqrt{\dfrac{(t-a)(t-b)}{(z-a)(z-b)}}\,dt
\end{equation*}

\smallskip

A more interesting interpolation points distribution is
the Chebyshev distribution on $[c,d]: d\mu_{n,i}(t)=
\pi^{-1} [(t-c)(d-t)]^{-1/2}\,dt$. Then, $\log\Phi(z)$ is the
constant term of the Chebyshev expansion of $\log\Phi(z;t)$.
The $z-$derivative is
\begin{equation*}
  \dfrac{\Phi'(z)}{\Phi(z)} =-\dfrac{1}{\pi\sqrt{(z-a)(z-b)}}
  \int_c^d  \dfrac{1}{z-t} \sqrt{\dfrac{(t-a)(t-b)}
  {(t-c)(d-t)}}\,dt,
\end{equation*}
a typical complete elliptic integral of the third kind.
``Simple examples'' do not seem much easier than the ``big''
example. But suppose that $[c,d]$ is far from $[a,b]$. Then, for
$z$ near $[c,d]$, $\Phi'/\Phi$ is not far from
$-\pi^{-1}\int_c^d (z-t)^{-1}[(t-c)(d-t)]^{-1/2}dt=
-[(z-c)(z-d)]^{-1/2}
$, which is pure imaginary on $[c,d]$, so the rational interpolant
is close to the best rational approximation.
And the error norm, from \eqref{cnres}, is about
$\displaystyle{c_n\approx \left[\dfrac{b-a}{(c+d-2a)(c+d-2b)}\right]^{2n} }$.

\subsubsection{Experiment with several poles supports.} So far,
with real intervals $[a,b]$, $[c,d]$, we discussed actual
orthogonal polynomials as true denominators of rational
interpolants. But let us keep $[c,d]$ real, and try several
arcs joining two fixed nonreal points, say $ia$ and $-ia$.

to be continued

\section{`1/9', again and again, ad nauseam.}

OK, back to `1/9'. Now, according to Trefethen {\it et al.}
\cite{SchT,TWS} recent work, I stick to rational approximation
to $e^z$ on $(-\infty,0]$.

There are still things to find! Did anybody see that the denominators
in Carpenter {\it et al.} \cite{CRV} look like
$\exp(-0.712 z)$ (after $z\leftrightarrow -z$)\footnote{Coefficient of $z$
in denominators of \cite{CRV} behave like $0.712+0.18/n$.}, and, of course,
the numerators look like $\exp(0.288 z)$. What can these numbers be??

We try to go further in investigating the distributions of poles and
interpolation points.

\subsection{The complex potential}  \   \\

\parpic{
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   \uput[90](1.5,1.5){$E$}  \uput[45](3,1.75){$\Delta_n$}
   \uput[90](2,2.5){$b_n$}\uput[270](2,0.5){$a_n$}
  \end{pspicture}
   }
  As we suspect the poles to be distributed on an single
arc $\Delta_n$ joining $a_n$ and $b_n$(still unknown),
$\mathcal{V}_n$ is a function with branch-points such that\\
 1.\   its derivative $\mathcal{V}'_n$ takes opposite
    pure imaginary values on the two sides of the negative
    real axis $=E$,\\
 2.\  $\mathcal{V}'_n +f'/(2nf) =\mathcal{V}'_n+1/(2n)$ takes opposite
    values on the two sides of $\Delta_n$.\\
 3.\  for $\mathcal{V}_n$ itself, $\mathcal{V}_n(-\infty+0i)-
       \mathcal{V}_n(-\infty-0i)= 2\pi i$. \\
 The second condition means that $[\mathcal{V}'_n(z) +1/(2n)]/\sqrt{(z-a_n)(z-b_n)}$
has no more branchpoints at $a_n$ and $b_n$, and can be recovered at any
$z\notin E$ through a Cauchy integral on a contour allowed to
stretch up to the two sides of $E$. Same experiment with a further multiplication
by $\sqrt{z}$:

\begin{equation} \label{Vp1}
 \sqrt{\dfrac{z}{(z-a_n)(z-b_n)}}\left(\mathcal{V}'_n(z)+\dfrac{1}{2n}\right) =
 \dfrac{1}{2\pi n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)(t-b_n)}}\dfrac{dt}{z-t}
\end{equation}

Explanation: there should be a numerator $\mathcal{V}'_n(t)+1/(2n)$ in the integral, but
 $\sqrt{t}\mathcal{V}'_n(t)$ has no branchpoint at 0, 
   and its contributions
 from the two sides of $E$ cancel,
 only $-1/(2n)$ remains, whose equal contributions
 on the two sides are added. 

Signs of the square roots: if the square root in $z$ at the left is
positive for positive $z$, square root inside integral is
positive. $\sqrt{z}\mathcal{V}'_n(z)$ must be positive if $z$ is a small 
positive number.


Transformation of \eqref{Vp1}: the derivative of
the left-hand side of \eqref{Vp1} is 
\begin{equation*}
\dfrac{a_nb_n-z^2}{2\sqrt{z(z-a_n)^3(z-b_n)^3}} \left(\mathcal{V}'_n(z)+\dfrac{1}{2n}\right)  
+\sqrt{\dfrac{z}{(z-a_n)(z-b_n)}}\mathcal{V}''_n(z),
\end{equation*}
and we integrate by parts (in $t$) the $z-$derivative of the 
right-hand side to get
\begin{multline*}
  \dfrac{1}{2\pi n}\int_{-\infty}^0 \dfrac{(t^2-a_nb_n)dt}{2(z-t)\sqrt{-t(t-a_n)^3(t-b_n)^3}} =
  \dfrac{1}{2\pi n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)(t-b_n)}}\dfrac{a_nb_n-t^2}
        {2(z-t)z(t-a_n)(t-b_n)}\,dt \\
   \text{(\ \ \ must\ decrease\ faster\ than\ } |z|^{-1} \text{\ for\ large\ }z, \text{\ remark\ also\ that\ } t^2-a_nb_n \\
    \text{\ over\ the\ big\ }\sqrt{\ } \text{\ is\ the\ derivative\ of\ a\ function \ vanishing \ at \ 0\ and\ }\infty\\
    \text{so,\ replace \ }1/(z-t) \text{\ by\ } 1/(z-t)-1/z= t/[z(z-t)]) \\
 =\dfrac{1}{2\pi n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)(t-b_n)}}\dfrac{a_nb_n-z^2}
        {2(z-t)z(z-a_n)(z-b_n)}\,dt  \\
 +\dfrac{1}{2\pi n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)(t-b_n)}}
    \dfrac{[2a_nb_n-(a_n+b_n)t]z+2a_nb_nt-a_nb_n(a_n+b_n)}
        {2z(z-a_n)(z-b_n)(t-a_n)(t-b_n)}\,dt 
\end{multline*}
which is
\begin{equation*}
 \dfrac{a_nb_n-z^2}{2\sqrt{z(z-a_n)^3(z-b_n)^3}} \left(\mathcal{V}'_n(z)+\dfrac{1}{2n}\right)  
 + \dfrac{\text{a\ polynomial\ of\ degree\ }\leqslant 1 \ \text{in\ }z }{z(z-a_n)(z-b_n)}
\end{equation*}
and what remains is
\begin{equation*}
 \mathcal{V}_n''(z) = \dfrac{\text{this polynomial}}{\sqrt{z^3(z-a_n)(z-b_n)}}
\end{equation*}
but, as $\mathcal{V}_n$ is the potential of the sum of two opposite charges,
$\mathcal{V}''_n(z)$ must decrease faster than $|z|^{-2}$ for large $z$, this implies
a first condition on $a_n$ and $b_n$
\begin{equation} 
  2a_nb_n \int_{-\infty}^0   \sqrt{\dfrac{-t}{(t-a_n)^3(t-b_n)^3}} \,dt =
 (a_n+b_n)\int_{-\infty}^0  \sqrt{\dfrac{-t}{(t-a_n)^3(t-b_n)^3}}  \,t\,dt,
\end{equation}
leading to (in?)famous elliptic integrals (to do: look at Carlson's forms).
Elementary change of variable $t=-u\sqrt{a_nb_n}$ and 
 $\displaystyle{ \dfrac{a_n+b_n}{2\sqrt{a_nb_n}} = -\cos\theta }$ leads to
\begin{equation}
0= \int_0^\infty  \sqrt{\dfrac{u}{(1-2u\cos\theta+u^2)^3}}\,(1-u\cos\theta)\,du =
   -\int_0^1  \dfrac{\cos\theta(1+u^2)-2u}{\sqrt{u(1-2u\cos\theta+u^2)^3}}\,du 
\end{equation}
(put $u\leftrightarrow 1/u$ in the integral from 1 to $\infty$). Apply a
crude integration formula:

\begin{center}
\begin{tabular}{cccccccc}
  0 & 0.25 & 0.5 & 0.6 & 0.65 & 0.7 & 0.75 & 1 \\  \hline
 0.8472& 0.5919& 0.2656& 0.0998&0.0045&-0.102&-0.224&$-\infty$
\end{tabular}
\end{center}

Root is $\cos\theta=0.6522295\dotsc$ (computed through elliptic
integrals \cite{Mag3}).

\parpic{
  \psset{unit=1cm}
  \begin{pspicture}(0,0)(5,8)
   \psline{->}(0,6.5)(4.5,6.5) \psline{->}(2,5.5)(2,7.7)
     \pscurve[linestyle=dashed,dash=2mm 2mm,linewidth=0.03](0,6.4)(1,6.1)(1.75,5.5)
     \pscurve[linewidth=0.03](2.25,5.4)(2.75,5.9)(3.35,6.2)
     \pscurve[linewidth=0.03](3.35,6.8)(4,6.65)(4.5,6.6)
     \uput[90](0,7){Im$\ \mathcal{V}''_n$}  \uput[270](2.5,6.5){$\mathcal{V}''_n$}
    \psline{->}(0,3.5)(4.5,3.5) \psline{->}(2,2.5)(2,4.7)
     \pscurve[linestyle=dashed,dash=2mm 2mm,linewidth=0.03](0,3.4)(1,3.1)(1.75,2.5)
     \pscurve[linewidth=0.03](2.25,4.6)(2.75,3.6)(3.35,3)
     \pscurve[linewidth=0.03](3.35,2.8)(4,3.2)(4.5,3.4)
     \uput[90](0,4){Im$\ \mathcal{V}'_n$}  \uput[270](3,4.5){$\mathcal{V}'_n$}
    \psline{->}(0,1)(4.5,1) \psline{->}(2,0)(2,2.2)
     \pscurve[linestyle=dashed,dash=2mm 2mm,linewidth=0.03](-0.2,2)(0,2)(1,1.75)(2,1.1)(2,1)
     %\pscurve[linewidth=0.03](2.25,4.6)(2.75,3.6)(3.35,3)
     %\pscurve[linewidth=0.03](3.35,2.8)(4,3.2)(4.5,3.4)
     \uput[90](0,1.5){Im$\ \mathcal{V}_n$}  \uput[270](2.5,1.8){$\mathcal{V}_n$}
   \end{pspicture}
   }
We have now
   \\
\begin{equation} \label{Vs1} 
 \mathcal{V}_n''(z) = \dfrac{A_n}{\sqrt{z^3(z-a_n)(z-b_n)}}
\end{equation}
with $\displaystyle{ A_n= 
   \dfrac{1}{4\pi n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)^3(t-b_n)^3}}
    [2a_nb_nt-a_nb_n(a_n+b_n)]\,dt }$. \\
 $\displaystyle{ \ \ = 
   -\dfrac{1}{4\pi n}\,\dfrac{a_nb_n(a_n-b_n)^2}{a_n+b_n}\int_{-\infty}^0 \sqrt{\dfrac{-t}{(t-a_n)^3(t-b_n)^3}}\,dt}<0$.
    When one crosses the line of poles, $\mathcal{V}''_n$ is replaced by its opposite. The picture at left
   shows $\mathcal{V}''_n(z)$ for positive $z$, and its imaginary part on the upper side
     of $(-\infty,0)$. \  \\
\begin{equation} \label{Vp2} 
 \mathcal{V}_n'(z) = \int_\infty^z \dfrac{A_n\,dt}{\sqrt{t^3(t-a_n)(t-b_n)}}
\end{equation}
to be sure that $\mathcal{V}'(\infty)=0$. The path of integration
in \eqref{Vp2} joins $\infty$ to $z$ by avoiding the cuts
$(-\infty,0)$ and $(a_n,b_n)$. The continuation from small positive $z$
to large $z$ would exhibit $-1/n$ as limit. However one switches to
$-1/n-\mathcal{V}'_n$ by crossing the line of poles.
 Remark that $\mathcal{V}'_n(z)\sim
 -(2/3)A_n (a_nb_n)^{-3/2} z^{-3/2}$ for large $z$.

\begin{equation} \label{V1}
  \mathcal{V}_n(z)= \ \text{const.}\ -A_n \int_\infty^z \dfrac{(z-t)\,dt}
     {\sqrt{t^3(t-a_n)(t)b_n)}}
\end{equation}


Playing with Legendre expansions:

$$
\dfrac{1}{\sqrt{1-2u\cos\theta+u^2}}=\sum_0^\infty P_m(\cos\theta)\,u^m,
 \dfrac{1}{\sqrt{(1-2u\cos\theta+u^2)^3}}=\sum_1^\infty P'_m(\cos\theta)\,u^{m-1},
$$

$$
\dfrac{1+2u\cos\theta+u^2}{\sqrt{u(1-2u\cos\theta+u^2)^3}}=\sum_1^\infty 
  [P'_m(\cos\theta)+2\cos\theta P'_{m-1}(\cos\theta)+P'_{m-2}(\cos\theta)]\,u^{m-3/2},
$$
whence an unsophisticated equation for $\theta$
$$
 \sum_1^\infty 
   \dfrac{P'_m(\cos\theta)+2\cos\theta P'_{m-1}(\cos\theta)+P'_{m-2}(\cos\theta)}{m-1/2}=0,
$$

 $\Phi$ must behave near $E$ in agreement
with this distribution. The derivative of $\log\Phi$
\begin{equation*}
   (\log\Phi(z))' = \dfrac{f'(z)}{2nf(z)}
      -\int_E \dfrac{d\mu_{n,i}(t)}{z-t} +\mathcal{V}'_{n,p}(z)
\end{equation*}
behaves on the two sides of $E$ as an average
$\pm \pi i \mu'_{n,i}(z)$
(\emph{\textbf{Sokhotskyi-Plemelj}} formulas).

\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,
english translation  in
 \textsl{Sb. Math.}  vol. \textbf{193} (2002) no. 1-2, 1-72.

\bibitem{CRV} A.J.~CARPENTER, A.~RUTTAN, and R.S.~VARGA, Extended
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\bibitem{GR} A.A.~Gonchar, E.A.~Rakhmanov, Equilibrium distribution and
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    {\sl Math.\ USSR Sbornik\/} {\bf 62\/} (1989) 305-348.




\bibitem{Mag3}
  A.P. Magnus, J. Meinguet,
The elliptic functions and integrals of the `1/9' problem,
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See in \href{http://www.math.ucl.ac.be/~magnus}
    {\texttt{http://www.math.ucl.ac.be/\~{}magnus}}

\bibitem{cafe} A.P.Magnus, J.~Nuttall, On the constructive rational
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\bibitem{Me2} J.~Meinguet, An electrostatic approach to the determination
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\bibitem{SchT} T. Schmelzer, L.N. Trefethen, Computing the Gamma
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\bibitem{Sta} H.~Stahl, Convergence of rational interpolants,
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\bibitem{TWS} L.N. Trefethen, J.A.C. Weideman, T. Schmelzer, Talbot
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\bibitem {wal} J.L.~Walsh, {\sl Interpolation and Approximation by
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\end{thebibliography}


\end{document}
