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\begin{document}

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\vspace*{-1cm}
\begin{center} {\Large \bf

    \textsf{MAPA 3071}\\
\textbf{Special topics in approximation theory.}\\
       {\Small 2006-} 

\medskip

  {\Large Elliptic lattices.}}



\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
       (++32)(0)(10)473157 ,
   \href{mailto:alphonse.magnus@uclouvain.be}
                  {\tt alphonse.magnus@uclouvain.be} ,
  \href{http://perso.uclouvain.be/alphonse.magnus/}{\tt http://perso.uclouvain.be/alphonse.magnus/}
\end{center}
\bigskip

Sept. 2006 -  etc.

This version: \today \qquad  (incomplete and unfinished)

The present file is http://perso.uclouvain.be/alphonse.magnus/num3/m32006.pdf

\medskip



\vspace*{-1pt}

\begin{flushright}  \textit{Ars Longa}\footnote{The author of the following thesis:
\textsl{Sur les champs stationnaires des int\'egrales multiples du
 calcul des variations. Transformation de Haar-Carath\'eodory. Espaces g\'en\'eralis\'es de  Cartan.}, 	Louvain : UCL, 1942, choose \textit{Ars Longa}
as author's name. These things happened in 1942. I really hope that he/she
had (or still has?) a \textit{Vita Longa}.}\ \ \ \ \ \ \    \end{flushright}


\tableofcontents


\section{The name of the game.}
\renewcommand{\mylmark}{The name}

\subsection{Elliptic grid, lattice, sequence}\  \\

Have a look at this, on what is called ``elliptic sequences'':

\href{http://www.research.att.com/~njas/sequences/A006721}{\tt http://www.research.att.com/\~{}njas/sequences/A006721}

  Greetings from \href{http://www.research.att.com/~njas/sequences/Seis.html}{The On-Line Encyclopedia of Integer Sequences}!

 A006721
            Somos-5 sequence: $a(n) = (a(n-1)a(n-4)+a(n-2)a(n-3))/a(n-5)$.          (Formerly M0735) 1, 1, 1, 1, 1, 2, 3, 5, 11, 37, 83, 274, 1217, 6161, 22833, 165713, 1249441, 9434290, 68570323, 1013908933, 11548470571, 142844426789, 2279343327171, 57760865728994, 979023970244321

          
REFERENCES
{\Small
          ...
          
            David Gale, "The strange and surprising saga of the Somos sequence", Math. Intelligencer 13(1) (1991), pp. 40-42.
          
           ...
          
            A. J. van der Poorten, \href{http://arXiv.org/abs/math.NT/0403225}{Elliptic curves and continued fractions http://arXiv.org/abs/math.NT/0403225}

            A. J. van der Poorten, \href{http://arXiv.org/abs/math.NT/0412293}{Recurrence relations for elliptic sequences...  http://arXiv.org/abs/math.NT/0412293}
          
            J. Propp, \href{http://www.math.wisc.edu/~propp/somos.html}{The Somos Sequence Site  http://www.math.wisc.edu/\~{}propp/somos.html}
         
           ...
         
            D. Zagier, \href{http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Problems.html}{Problems posed at the St Andrews Colloquium, 1996 http://www-groups.dcs.st-and.ac.uk/\~{}john/Zagier/Problems.html}
        
            
}

Consider (see somos.html above) recurrence relations linking products of cosines: $\cos((n+1)\theta)\cos((n-1)\theta)=
\cos^2(n\theta)\cos^2\theta -\sin^2(n\theta)\sin^2\theta$,
$\cos((n+2)\theta)\cos((n-2)\theta)=
\cos^2(n\theta)\cos^2(2\theta) -\sin^2(n\theta)\sin^2(2\theta)$, and eliminate $\sin^2(n\theta)$:
$c_{n+1}c_{n-1}/\sin^2\theta -c_{n+2}c_{n-2}/\sin^2(2\theta) = c_n^2 (\cot^2\theta-\cot^2(2\theta))$,
yielding a family of recurrence relations which showed sometimes  unsuspected integer particular
solutions. Experts made the connection with integral invariants of elliptic curves and these remarkable
sequences have been called ``elliptic''.

\includegraphics*[angle=270, scale=0.50]{ellipsol.ps}

\begin{minipage}[t]{7.2cm}
\  \par


%\parpic{
{\Small
Airfoil grid built with elliptic grid generation. \\
\includegraphics*[angle=0, scale=0.20]{nacaell.ps} \\
{\sl William D. Henshaw 2006-06-17}\\ 
http://www.llnl.gov/CASC/Overture/henshaw/overtureFigures/node5.html
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\  \par

\vspace*{-10pt}
\noindent


Nothing works! ``Elliptic grid'' means a convenient mesh for discretizing over ellipses,
``elliptic difference operator'' is a partial difference operator extending partial
differential operator of elliptic type, an ``elliptic lattice'' is a set
of complex numbers $\{m \omega_1 + n \omega_2\}$, $m$ and $n \in \mathbb{Z}$ of
periods of elliptic functions, or simply the support of a partial difference
operator, or also a special digital filter\dots
\end{minipage}

\smallskip

What is meant here is a new rule for building lattices 
$\{x_n\}, n\in\mathbb{Z}$, which may be
useful in difference calculus.

\subsection{Known lattices}  \  \\ 

\begin{enumerate}

\item Arithmetic progression $x_n=an+b$,

\item Geometric progression $x_n= a q^n +b$,

\item Double geometric progression ANSUW (Askey- Nikiforov- Suslov- Uvarov- Wilson \cite{AA,AW,Segovia,magsnul,Niki0,NS,Niki})
\begin{equation}\label{latt01} x_n= a q^n + b q^{-n} +c. \end{equation} 
\end{enumerate}

Why this latter formula? My favorite way to introduce it is that the first order
divided difference of a polynomial on such a lattice yields a
polynomial of lower degree. So, $\displaystyle \dfrac{p(x_{n+1})-p(x_n)}{x_{n+1}-x_n}$ is a polynomial of degree $k-1$ if $p$ has degree $k$. With
$p(x)=x^2$, $x_{n+1}+x_n=\alpha x_n+\beta$. No, this is too easy, we
recover the simple geometric $q-$lattice, with $q=\alpha-1$. It must have been
a central divided difference $\displaystyle \dfrac{p(x_{n+1/2})-p(x_{n-1/2})}{x_{n+1/2}-x_{n-1/2}}$.
Then, still with $p(x)=x^2$,  $x(s+1/2)+x(s-1/2)=\alpha x(s)+\beta$. Entering
\eqref{latt01}, assumed to be valid for integer as well as half-integer $s$, one
finds indeed that it fits, with $q^{1/2}+q^{-1/2} = \alpha$, certainly the fastest
derivation. 

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But why choose a difference operator built on $x-$values? A (seemingly) more
general setting uses an (apparently) independent lattice of
values: $\displaystyle \dfrac{p(y(s+1))-p(y(s))}{y(s+1)-y(s)}$
which must be a polynomial of degree lower than the degree of $p$:\\
$p(y)=y^2:\   y(s+1)+y(s)=\alpha x(s)+\beta, $ \\
$p(y)=y^3:\   y(s+1)^2+y(s+1)y(s)+y(s)^2=$ some polynomial of degree 2, from which we subtract the square of the first result, to get %
\ \ \ \ \ \ \ \ $y(s+1)y(s)=\gamma x(s)^2+\delta x(s)+\epsilon$.\\
These two equations already tell what $y(s)$ and $y(s+1)$ are, as functions of
$$
y(s), y(s+1) = \dfrac{\alpha x(s)+\beta}{2}\pm 
\sqrt{ \left(\dfrac{\alpha x(s)+\beta}{2}\right)^2 -\gamma x(s)^2-\delta x(s)-\epsilon
  },
$$
so, the $(x,y)$ are on a conic $y^2-\alpha xy +\gamma x^2 -\beta y +\delta x
  +\epsilon=0$. But $y(s+1)$ must be the ordinate of a second point
with absciss $x(s+1)$!  A rigid law rules therefore the $x-$ and $y-$sequences,
$\displaystyle y(s+2)+y(s+1)= \alpha x(s+1)+\beta =\alpha \left[\dfrac{ \alpha y(s+1)-\delta }{\gamma}
-x(s)\right] +\beta =  \alpha \dfrac{ \alpha y(s+1)-\delta }{\gamma}
-y(s)-y(s+1) +2\beta$. 

One recovers \cite{Segovia,magsnul}  $x_n$ and $y_n$ as
 combinations of $q^n$ and $q^{-n}$, where
$$
q+\dfrac{1}{q}=\dfrac{\alpha^2-2\gamma}{\gamma}.
$$
The slopes of the asymptotes of the conic above are $s_1$ and $s_2$ such that
$s_1+s_2= \alpha$ and $s_1 s_2=\gamma$, or $\displaystyle \dfrac{s_1}{s_2}+
\dfrac{s_2}{s_1}= \dfrac{\alpha^2-2\gamma}{\gamma}$: \emph{\textbf{ $\boldsymbol{q}$
is the ratio of the slopes of the asymptotes of the conic.}}

\includegraphics*[angle=0, width=6.5cm]{sti1.ps} \includegraphics*[angle=0, scale=0.50]{sti2.ps}

Figures from \cite{magsnul}


Remark also that $q^{1/2}+q^{-1/2} = \alpha/\sqrt{\gamma}$.

\medskip

\section{The new lattice, at last} \   \\
\renewcommand{\mylmark}{The new lattice}


It will be introduced through four equivalent points of view:

\subsection{Points $(x_n,y_n), (x_n,y_{n+1})$ on a biquadratic curve} \   \\

From \cite{magLuminy}:

Simplest difference equations relate two values of the
unknown fuction $f$: say, $f(\varphi(x))$ and $f(\psi(x))$.

Most instances \cite{MilneT} are $(\varphi(x),\psi(x))= (x,x+h)$, or the more
symmetric $(x-h/2,x+h/2)$, or also $(x,qx)$ in $q-$difference
equations \cite{GrH}. Recently, more complicated forms $(r(x)-\sqrt{s(x)},
r(x)+\sqrt{s(x)})$ have appeared \cite{AW,BI2,Koe,Ko,NS,Niki}, where
$r$ and $s$ are rational functions.

This latter trend will be examined here: we need, for each $x$,
two values $f(\varphi(x))$ and $f(\psi(x))$ for $f$.\\
A first-order difference equation is
$\mathcal{F}\left(x,f(\varphi(x)),f(\psi(x))\right)=0$, or
$f(\varphi(x))-f(\psi(x))= \mathcal{G}\left(x,f(\varphi(x)),f(\psi(x))\right)$
if we want to emphasize the difference of $f$. There is of course some freedom
in this latter writing. Only symmetric forms in $\varphi$ and $\psi$
will be considered here:
\begin{equation}\label{diff1}
 (\mathcal{D} f)(x) = \mathscr{F}\left(x,f(\varphi(x)),f(\psi(x))\right),
\end{equation}
where $\mathcal{D}$ is the divided difference operator
\begin{equation}\label{diffop}
 (\mathcal{D} f)(x) = \dfrac{ f(\psi(x)- f(\varphi(x) }{\psi(x)-\varphi(x) },
\end{equation}
 and where $\mathscr{F}$ is a symmetric function of its two last arguments.

For instance, a linear difference equation of  first order may be written as
$$  a(x) f(\varphi(x)) + b(x) f(\psi(x)) +c(x)=0,$$
as well as
 $$ \alpha(x) (\mathcal{D} f)(x) = \beta(x) [f(\varphi(x)) + f(\psi(x))] +\gamma(x),$$
with $\alpha(x)=[b(x)-a(x)][\psi(x)-\varphi(x)]/2$,
$\beta(x)=-[a(x)+b(x)]/2$, and $\gamma(x)=-c(x)$.

\medskip



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  The simplest choice
 for $\varphi$ and $\psi$ is to take the two determinations of an
algebraic function of degree 2, i.e., the two $y-$roots of
\begin{subequations}
\begin{equation}\label{Fy}
   F(x,y)=X_0(x)+X_1(x) y +X_2(x) y^2 =0,
\end{equation}
where $X_0, X_1$, and $X_2$ are rational functions.\\
 \    \\
 \   \\
 \   \\
 But difference equations must allow the recovery of $f$ on a whole set
 of points! An initial-value problem for a first order difference equation
 starts with a value for $f(y_0)$ at $x=x_0$, where $y_0$ is one root
 of \eqref{Fy} at $x=x_0$. The difference equation at $x=x_0$ relates then
 $f(y_0)$ to $f(y_1)$, where $y_1$ is the second root of \eqref{Fy} at $x_0$.
 We need $x_1$ such that $y_1$ is one of the two roots of \eqref{Fy} at $x_1$,
so for one of the roots of $F(x,y_1)=0$ which is not $x_0$. Here again, the
simplest case is when $F$ is of degree 2 in $x$:
 \begin{equation}\label{Fx}
   F(x,y)=Y_0(y)+Y_1(y) x +Y_2(y) x^2 =0.
\end{equation}
\end{subequations}
   \   \\
Both forms \eqref{Fy} and \eqref{Fx} hold simultaneously
when $F$ is \emph{\textbf{biquadratic}}:
\begin{equation}\label{Fbiq}
 F(x,y)= \sum_{i=0}^2 \sum_{j=0}^2 c_{i,j} x^i y^j.
\end{equation}


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We again look for a sequence $\{(x_n,y_n\}_{n\in\mathbb{Z}}$ on an algebraic
curve $F(x,y)=0$ such that 1) at $x=x_n$ correspond two ordinates $y_n$
and $y_{n+1}$ given by an equation of degree two with coefficients which are
rational functions of $x_n$; 2) at $y=y_{n+1}$ correspond two abscissae $x_n$
and $x_{n+1}$ given by an equation of degree two with coefficients which are
rational functions of $y_{n+1}$. This leads to 

\bigskip

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\psplot{2.001}{2.529}{x x mul dup 1 sub exch 4 sub mul 10 mul sqrt 4 mul -1 mul x 18 mul add 32 x x mul 5 mul sub div}
\psplot{3.2}{8}{x x mul dup 1 sub exch 4 sub mul 10 mul sqrt 4 mul x 18 mul add 32 x x mul 5 mul sub div}
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\uput[0](-8,-4){Look!  {\blue two} {\red colours} !}
\end{pspicture}
\end{center}

\bigskip

\noindent\textbf{Definition 1.} \textsl{A sequence $\{\dotsc, x_{-1}, x_0, x_1,\dotsc\}$
is an elliptic lattice if there exists a sequence $\{\dotsc, y_{-1}, y_0, y_1,\dotsc\}$
and
a biquadratic polynomial  $F$ (see \eqref{Fy}, \eqref{Fx}, \eqref{Fbiq})
 $F(x,y)= \sum_{i=0}^2 \sum_{j=0}^2 c_{i,j} x^i y^j = X_0(x)+X_1(x)y+X_2(x)y^2
   =Y_0(y)+Y_1(y)x+Y_2(y)x^2$,\\
such that $F(x_n,y_n)=0$ and $F(x_n,y_{n+1})=0$, for $n\in\mathbb{Z}$.}

As $y_n$ and $y_{n+1}$ are the two roots in $t$ of $F(x_n,t)= X_0(x_n)+X_1(x_n)t
+X_2(x_n)t^2=0$, useful identities are
\begin{equation} \label{ysump}
  y_n+y_{n+1} = -\dfrac{X_1(x_n)}{X_2(x_n)} =
  - \dfrac{c_{2,1}x_n^2 +c_{1,1}x_n+c_{0,1}  }
          {c_{2,2}x_n^2 +c_{1,2}x_n+c_{0,2}  },
y_n y_{n+1} = \dfrac{X_0(x_n)}{X_2(x_n)} =
   \dfrac{c_{2,0}x_n^2 +c_{1,0}x_n+c_{0,0}  }
          {c_{2,2}x_n^2 +c_{1,2}x_n+c_{0,2}  }.
\end{equation}
and the direct formula
\begin{equation} \label{ydirect}
y_n \  \text{and\ } y_{n+1} = \dfrac{ -X_1(x_n) \pm \sqrt{P(x_n)}  }{2 X_2(x_n) },
\end{equation}
where 
\begin{equation}\label{P} P=X_1^2-4X_0 X_2\end{equation} is a polynomial of degree 4.

Also, as $x_{n-1}$ and $x_n$ are the two roots in $t$ of $F(t,y_n)=0$,
\begin{equation} \label{xsump}
x_n+x_{n-1} = -\dfrac{Y_1(y_n)}{Y_2(y_n)} =
  - \dfrac{c_{1,2}y_n^2 +c_{1,1}y_n+c_{1,0}  }
          {c_{2,2}y_n^2 +c_{2,1}y_n+c_{2,0}  },
x_n x_{n-1} = \dfrac{Y_0(y_n)}{Y_2(y_n)} =
   \dfrac{c_{0,2}y_n^2 +c_{0,1}y_n+c_{0,0}  }
          {c_{2,2}y_n^2 +c_{2,1}y_n+c_{2,0}  }.
\end{equation}


Factorizations:

\begin{equation} \label{factori}
F(x,y)= X_2(x)(y-\varphi(x))(y-\psi(x)) \quad = Y_2(y)(x-\varphi^{-1}(y))(x-\psi^{-1}(y)).
\end{equation} $
F(x_n,y)= X_2(x_n)(y-y_n)(y-y_{n+1}),\qquad F(x,y_n)= Y_2(y_n)(x-x_n)(x-x_{n-1})$.



The construction above is called ``T-algorithm'' in \cite[Theorem~6]{SpZ2007}.

\noindent\textbf{Exercise.} Is $\{x_n\}=\{1,2,3,\dotsc\}$ an elliptic sequence?
Yes, of course, with $y_n=n$, so $F(x,y)=(y-x)(y-x-1)$. What are the
other admissible $y-$sequences? And $\{1,4,\dotsc, n^2,\dotsc\}$? Yes.
And $\{1,\sqrt{2},\sqrt{3},\dotsc\}$? No. Answers are very easy through other
equivalent definitions, see next section, \S~\ref{ss2}.

\medskip

There are 9 coefficients in \eqref{Fbiq}, actually 8 degrees of freedom.
Transformations $\displaystyle x \to \dfrac{\alpha' + \beta' x}{\gamma'+\delta' x},
                  y \to \dfrac{\alpha'' + \beta'' y}{\gamma'' +\delta'' y}$,
with $\alpha'\delta'-\beta'\gamma'$ and  $\alpha''\delta''-\beta''\gamma''\neq 0$ lead to
basically equivalent $(x_n,y_n)$ sequences.
 Two essential degrees of freedom
remain. See Spiridonov \& Zhedanov \cite{SpZ2007}

The first one is the \emph{\textbf{modulus}} $k$. It appears from the cross ratios
$\displaystyle R_{a,b,c,d}=\dfrac{\  \  \  \displaystyle\dfrac{z_a-z_c}{z_a-z_d}\ \ \ }
       {\  \  \  \displaystyle\dfrac{z_b-z_c}{z_b-z_d}\ \ \ }$, where $z_1,\dotsc, z_4$ are the
zeros of $P$, and where $(a,b,c,d)$ is a permutation of $(1,2,3,4)$. The important thing is that
the cross ratio is left unchanged under a $x \to (\alpha' + \beta' x)/(\gamma'+\delta' x)$
transformation. Among other choices, we may send the four zeros of $P$ to
a symmetric set $\{-1/k,-1,1,1/k\}$. The 6 values of $R$ under the 24 possible permutations of
the indexes  happen to be $\displaystyle \left(\dfrac{1+k}{1-k}\right)^2, 
\dfrac{(1+k)^2}{4k}, -\dfrac{(1-k)^2}{4k}$ and their inverses. If $R_1$ is one of these values,
the other ones are generated by applying one or several times $R \to 1/R$ and $R\to 1-R$ (modular
group (\textbf{?})). Then the various possible values of the modulus are $u$ such that
$\displaystyle R=\left(\dfrac{1+u}{1-u}\right)^2$ and are, if $k$ is one root,
$\pm k$, $\pm 1/k$, $\displaystyle \pm \left(\dfrac{1\pm\sqrt{k}}{1\mp \sqrt{k}}\right)^2$,


The same cross ratio would be obtained with the four zeros of $Q=Y_1^2-4Y_0Y_2$,
relevant from $F(x,y)=0 \Rightarrow x=(-Y_1(y)\pm\sqrt{Q(y)})/(2Y_2(y))$,
see \eqref{checkratio} p.~\pageref{checkratio} for a case
where the degree of $Q$ is reduced to 3, see also Appel \& Goursat p.293,
where it is stated that the four tangents to the cubic meeting a given point
of the curve define the same cross ratio.


A vanishing modulus corresponds to a polynomial $P$ with a
double zero. We then recover the ANSUW case, up to rational transformations
of coordinates.

The reduced form with two parameters which will be used here is

\begin{equation} \label{Fcanon}
\begin{split}
F(x,y)&= k^2(1-k^2 z'^2)x^2y^2 -k^2(1-z'^2)(x^2+y^2)
 -2k(1-k^2)z' xy +1-k^2z'^2\\
  &= k^2(1-k^2 z'^2)x^2y^2 -k^2(1-z'^2)(x+y)^2
 -2k(z'-k)(1+kz') xy +1-k^2z'^2
\end{split}\end{equation}

\noindent\textbf{The second parameter} $z'$ is investigated now. With \eqref{Fcanon}, \eqref{ydirect} is
$\displaystyle y=\dfrac{k(1-k^2)z' x \pm\sqrt{P(x)}  }
     {k^2[(1-k^2 z'^2)x^2 +z'^2-1]}$, with
$P(x)=k^2(1-z'^2)(1-k^2z'^2)(1-x^2)(1-k^2x^2)$.

We already see that, if $z'^2$ is close to 1 or $k^{-2}$, $P$ is almost the zero
polynomial, the curve in the $x-y$ plane has almost no width, any $x_n$ leads to $y_n$ and $y_{n+1}$ close together, and to $x_{n\pm 1}$ close together too, and close to $x_n$.\\
Well, let us take the example $z'=k^{-1}+$ a small $\varepsilon$. Then,
$y_{n+1}$ and $y_n=\displaystyle x_n \pm \sqrt{\dfrac{2k\varepsilon (1-x_n^2)(1-k^2x_n^2)}  {1-k^2}  }+O(\varepsilon)$, and
$x_{n\pm 1}=\displaystyle x_n \pm 2\sqrt{\dfrac{2k\varepsilon (1-x_n^2)(1-k^2x_n^2)}  {1-k^2}  }+O(\varepsilon)$, we see that $z'$ is linked to a \emph{\textbf{step}} of the
$x-$lattice, and that this step is small if $z'$ is close to $\pm 1$ or $\pm k^{-1}$.

\medskip

Baxter \cite[\S~15.10]{Bax} uses $x^2y^2 + c(x^2+y^2) +2dxy +1$, so that
$P(x)=4[d^2 x^2 -(c+x^2)(cx^2+1)]= -4c[x^4 -(k+k^{-1})x^2+1] = 
-4c(1-k^2(x^2/k))(1-(x^2/k))$ if  $k+k^{-1}= (d^2-c^2-1)/c$, or $(k+c)(1+kc)=kd^2$.\\
Baxter's form follows from \eqref{Fcanon} in the variables $k^{1/2}x$ and $k^{1/2}y$:
$\displaystyle (k^{1/2}x)^2(k^{1/2}y)^2 -\dfrac{k(1-z'^2)}{1-k^2z'^2}[(k^{1/2}x)^2+( k^{1/2}y)^2] 
- \dfrac{2(1-k^2)z'}{1-k^2z'^2} k^{1/2}x k^{1/2}y +1=0$. Check that
$\displaystyle \dfrac{d^2-c^2-1}{c}= -\dfrac{(1-k^2)^2z'^2 -k^2(1-z'^2)^2}{k(1-k^2z'^2)(1-z'^2)}+\dfrac{1-k^2z'^2}{k(1-z'^2)}=k+k^{-1}$ indeed.


Then, Baxter finds that $\sqrt{-1/(kc)}$ is the elliptic sine of the relevant step
\cite[eq. (15.10.8)]{Bax}. Here, $\displaystyle \sqrt{-\dfrac{1}{kc}}=
\sqrt{\dfrac{1-k^2z'^2}{k^2(1-z'^2)} }=k^{-1}$dc$(\zeta)=$sn$(\zeta+K+iK')$
if $z'$ is the elliptic sine $z'=$sn$(\zeta)$.  From \cite[p.572]{Abr}:
\includegraphics*[angle=270, scale=0.25]{dcsn.ps}

\medskip


\parpic{
%\begin{center}
\psset{unit=1.0cm}\vspace*{-3pt}\begin{pspicture}(0,-1)(4.2,3.2)
\psline{->}(0,0)(4.5,0)
\psline{->}(0,0)(0,3.5)
\psline[linestyle=dashed,dash=2mm 2mm](3.5,-0.1)(3.5,3.1)
\uput[270](3.5,-0.1){$z'$} 
\psline[linestyle=dashed,dash=2mm 2mm](0.25,-0.1)(0.25,0.6)
\uput[270](0.3,-0.1){$z_i$} 
\psline[linestyle=dashed,dash=2mm 2mm](4.25,-0.1)(4.25,2.6)
\uput[270](4.3,-0.1){$z_{i+1}$} 
\parametricplot{-135}{225}{t cos 2 mul 2.25 add t cos t sin add 1.75 add}
\end{pspicture}
}
%\end{center}
\enlargethispage{55pt}
The second parameter $z'$ is a fixed point of the relation between $x_n$
and $x_{n+1}$, i.e., $E(z',z')=0$ (see next section for $E$). In the $F(x,y)=0$
setting, the fixed points $x_n=x_{n+1}$ occur at the $y-$extrema, so when
$\partial F/\partial x=0$. \\
With \eqref{Fcanon}, at $y=\pm 1$ or $\pm 1/k$, $x=\pm 1/(kz')$ or $\pm z'$, $\partial F/\partial x=2k^2(1-k^2 z'^2)xy^2 -2k^2(1-z'^2)x
 -2k(1-k^2)z' y 
\\
= 2k^2 x[z'^2(1-k^2y^2)-(1-y^2)]-2k(1-k^2)z'y \\ = 2k(1-k^2)z'(kz'x\mp 1)\text{\ or\ } 2(1-k^2)(x\mp z')=0$.\\
The four abscissae of interest are therefore $\pm 1/(kz')$ and $\pm z'$.
A cross ratio is
$\displaystyle \dfrac{\  \  \  \displaystyle\dfrac{1/(kz')+z'}{1/(kz')+1/(kz')}\ \ \ }
       {\  \  \  \displaystyle\dfrac{z'+z'}{z'+1/(kz')}\ \ \ }
=\dfrac{(1+kz'^2)^2}{4kz'^2}$. So, calculations similar to the operations
defining the modulus lead here to $kz'^2$.

With Baxter's form, the abscissae corresponding to $y=\pm k^{1/2}$ and $y=\pm k^{-1/2}$
are $x:  x^2y^2 + c(x^2+y^2) +2dxy +1=(k+c)x^2 \pm 2 dk^{1/2} x +ck+1 = (k+c)^{-1}
[(k+c)x \pm dk^{1/2}]^2=0 \Rightarrow x=\mp dk^{1/2}/(k+c)=\mp (1+kc)/(dk^{1/2})$ and 
$(k^{-1}+c)x^2  \pm 2dk^{-1/2}x +k^{-1}c+1= k^{-1}(1+ck)^{-1}[(1+kc)x\pm d k^{1/2}]^2
\Rightarrow x=\mp dk^{1/2}/(1+kc)= \mp (k+c)/(dk^{1/2})$.
The  cross ratio is now
$\displaystyle \dfrac{\  \  \  \displaystyle\dfrac{k+c+1+kc}{k+c+k+c}\ \ \ }
       {\  \  \  \displaystyle\dfrac{1+kc+1+kc}{1+kc+k+c}\ \ \ }
= \dfrac{(1+c)^2(1+k)^2}{4(k+c)(1+kc)}$



\subsection{Symmetric biquadratic relation between $x_n$ and $x_{n\pm 1}$.\label{ss2}} \   \\

 The companion sequence $\{y_n\}$ is not
needed in the definition of an elliptic lattice, but the definition above is best
suited to the description of difference equations. 

A relation involving only $x_n$ and $x_{n+1}$ is obtained by
the elimination of $y_{n+1}$ through the resultant of the two
polynomials in $y_{n+1}$ from \eqref{xsump}
$P_1(y_{n+1})= (x_n+x_{n+1})Y_2(y_{n+1})+Y_1(y_{n+1})$ and
$P_2(y_{n+1})= x_n x_{n+1}Y_2(y_{n+1})-Y_0(y_{n+1})$.

The form of this resultant is most easily found through interpolation
at the two zeros $u$ and $v$ of $Y_2$: let $Y_2(t)=\alpha (t-u)(t-v)$,
$Y_1(t)=\beta (t-u)(t-v)+\beta' t+\beta''$, $Y_0(t)=\gamma (t-u)(t-v)+\gamma' t+\gamma''$,
then, using the basis $\{1,t,(t-u)(t-v),t(t-u)(t-v)\}$, the resultant is built with
the coefficients of $tP_1(t), P_1(t), tP_2(t)$, and $P_2(t)$:
%$\displaystyle
\begin{equation*} %\label{resul}
 E(x_n,x_{n+1})=
 \begin{vmatrix} (x_n+x_{n+1})\alpha+\beta  &  \beta'  &  (u+v)\beta'+\beta'' & -uv\beta'  \\
                     0  &(x_n+x_{n+1})\alpha+\beta  &  \beta'  &  \beta''   \\
                    x_n x_{n+1}\alpha-\gamma  &  -\gamma'  & -(u+v)\gamma' -\gamma'' & uv\gamma'  \\
                     0 & x_n x_{n+1}\alpha-\gamma  &  -\gamma'  &  -\gamma''
    \end{vmatrix}=0,
%$
\end{equation*}
which is clearly a symmetric polynomial of degree 2 in $x_n+x_{n+1}$ and $x_n x_{n+1}$

\begin{multline*}
(\beta' u+\beta'')(\beta' v+\beta '')(\alpha x_nx_{n+1}-\gamma)^2 \\
+[(\beta' u+\beta'')(\gamma' v+\gamma '')
    +(\beta' v+\beta'')(\gamma' u+\gamma '')](\alpha(x_n+x_{n+1})+\beta)(\alpha x_nx_{n+1}-\gamma) \\
+(\gamma' u+\gamma'')(\gamma' v+\gamma '')(\alpha(x_n+x_{n+1})+\beta)^2 \\
+(\beta'' \gamma' -\beta' \gamma'')[ \beta' (\alpha x_nx_{n+1}-\gamma)
   +\gamma' (\alpha(x_n+x_{n+1})+\beta)]=0,
\end{multline*}

or, with $S=x_n+x_{n+1}$ and $\Pi=x_nx_{n+1}$, \\
\begin{multline} \label{resul}
[(\gamma' v+\gamma'')(\alpha S+\beta)+(\beta' v+\beta'')(\alpha\Pi-\gamma)]
[(\gamma' u+\gamma'')(\alpha S+\beta)+(\beta' u+\beta'')(\alpha\Pi-\gamma)]\\
+( \beta''\gamma'-\beta'\gamma'')[\gamma' (\alpha S+\beta)+\beta' (\alpha\Pi-\gamma)]=0.
\end{multline}

\medskip

\noindent\textbf{\emph{Example.}} With the canonical form \eqref{Fcanon},
$\alpha=k^2(1-k^2z'^2)$, $\beta=0, \beta'=-2k(1-k^2)z', \beta''=0$,
$\displaystyle \gamma=-k^2(1-z'^2), \gamma'=0,
\gamma''= 1-k^2z'^2 -k^2\dfrac{(1-z'^2)^2}{1-k^2z'^2} =
\dfrac{(1-k^2)(1-k^2z'^4)}{1-k^2z'^2}$, so $u+v=0, uv=\displaystyle -\dfrac{1-z'^2}{1-k^2z'^2}$, 
\begin{multline*}E(x_n,x_{n+1})=
\beta'^2 u v(\alpha x_nx_{n+1}-\gamma)^2 
+\gamma''^2\alpha^2(x_n+x_{n+1})^2 
 -\beta'^2 \gamma''  (\alpha x_nx_{n+1}-\gamma)
   \\
=\alpha^2\beta'^2 uv x_n^2 x_{n+1}^2 +\gamma''^2\alpha^2(x_n+x_{n+1})^2 
-\alpha\beta'^2 \underbrace{(2\gamma uv +\gamma'')}_{[1+k^2 -4k^2z'^2 +k^2(1+k^2)z'^4]/(1-k^2z'^2)}x_nx_{n+1} 
+\beta'^2\gamma\underbrace{( uv\gamma +\gamma'')}_{1-k^2z'^2} \\
=-4k^6(1-k^2)^2z'^2(1-z'^2)(1-k^2z'^2)x_n^2 x_{n+1}^2
+k^4(1-k^2)^2 (1-k^2z'^4)^2 (x_n+x_{n+1})^2 \\
-4k^4(1-k^2)^2 z'^2[1+k^2 -4k^2z'^2 +k^2(1+k^2)z'^4]x_nx_{n+1} 
-4k^4(1-k^2)^2z'^2(1-z'^2)(1-k^2z'^2)\\
= - \displaystyle 4k^4(1-k^2)^2 z'^2(1-z'^2)(1-k^2z'^2)\left\{
 k^2 x_n^2 x_{n+1}^2 -\dfrac{(1-k^2z_0^4)^2}{4z'^2(1-k^2z'^2)(1-z'^2)}(x_n+x_{n+1})^2 \right. \\ \left.
+ \dfrac{1+k^2 -4k^2z'^2 +k^2(1+k^2)z'^4}{(1-z'^2)(1-k^2z'^2)}x_nx_{n+1} 
+1 \right\}
\end{multline*}
which is again of the form \eqref{Fcanon}, with the same $k$, but with $z'$ replaced by $z''$:
\begin{equation} \label{Ecanon1}
E(x,y)= 
 k^2(1-k^2 z''^2) x^2 y^2 -k^2(1-z''^2)(x+y)^2 
+ 2k(z''-k)(1+kz'')xy +1-k^2z''^2 =0
\end{equation} 
such that $\displaystyle \dfrac{1-z''^2}{1-k^2z''^2} = \dfrac{(1-k^2z'^4)^2}
{4k^2z'^2(1-k^2z'^2)(1-z'^2)} \  \text{and\ }
2\dfrac{z''-k}{k(1-kz'')} = -\dfrac{1+k^2 -4k^2z'^2 +k^2(1+k^2)z'^4}{k^2(1-z'^2)(1-k^2z'^2)}$,  
\begin{equation} \label{zpzs} \ \ \   \text{\ from\ which \ \ }
z''=\displaystyle -\dfrac{1-2k^2z'^2+k^2z'^4}{k(1-2z'^2+k^2z'^4)}.
\end{equation}

Interesting identities are $\displaystyle z''\pm 1 = -\dfrac{(1\mp k)(1\pm kz'^2)^2}{k(1-2z'^2+k^2z'^4)},
 1-kz''= \dfrac{2(1-z'^2)(1-k^2z'^2)}{1-2z'^2+k^2z'^4},  1+kz''=-\dfrac{2(1-k^2)z'^2}{1-2z'^2+k^2z'^4}.$

\medskip

\noindent\textbf{Definition 2.} \textsl{An elliptic lattice, or grid, is a sequence
satisfying a symmetric biquadratic relation} \cite[Theorem~5]{SpZ2007}

\begin{multline}\label{Ebiq}
 \hspace*{-12pt}E(x_n,x_{n+1})=d_{0,0} +d_{0,1}(x_n+x_{n+1}) + d_{0,2}
(x_n+x_{n+1})^2
  +d_{1,1}x_n x_{n+1} \\
 +d_{1,2}x_n x_{n+1}(x_n+x_{n+1}) +d_{2,2}x_n^2 x_{n+1}^2=0.
\end{multline}

Remark that the coefficient of the product $x_nx_{n+1}$ is $d_{1,1}+2d_{0,2}$.

From symmetry of $E$, $E(x_n,x_{n-1})= E(x_{n-1},x_n)=$ \eqref{Ebiq} at $n-1$ vanishes too.
So, $x_{n-1}$ and $x_{n+1}$ are the two $y-$roots of $E(x_n,y)=0$, and
\begin{equation} \label{Esom}
x_{n-1}+x_{n+1}= - \dfrac{d_{0,1}+(d_{1,1}+2d_{0,2})x_n+d_{1,2}x_n^2}
                       {d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2}, 
x_{n-1} x_{n+1}=  \dfrac{d_{0,0}+d_{0,1}x_n+d_{0,2}x_n^2}
                       {d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2}.
\end{equation}

\begin{equation} \label{symdirect}
x_{n-1} \  \text{and\ } x_{n+1} = \dfrac{ -(d_{0,1}+(d_{1,1}+2d_{0,2})x_n+d_{1,2}x_n^2) \pm \sqrt{P(x_n)}  }{ 2(d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2) },
\end{equation}
with the same $P$ as before (times a constant), as from \eqref{xsump} at $n+1$,
$\displaystyle x_{n+1} = -\dfrac{Y_1(y_{n+1})}{Y_2(y_{n+1})}-x_n =
 -\dfrac{Y_1}{Y_2} -x_n\text{\ at\ }  \dfrac{ -X_1(x_n) \pm \sqrt{P(x_n)}  }{2 X_2(x_n) }$,
from \eqref{ydirect}.




It becomes much easier to check if a given sequence is an elliptic one, as
$x_{n-1}+x_{n+1}$ and $x_{n-1} x_{n+1}$ must be rational functions of
degree $\leqslant 2$ of $x_n$. 

\noindent\emph{\textbf{Example.}} Let $x_n=1/\sin(n\theta)$, then \label{Exasintheta}
$\displaystyle x_{n-1}+x_{n+1} = \dfrac{1}{\cos\theta \sin(n\theta) -\sin\theta \cos(n\theta)}\\
+\dfrac{1}{\cos\theta \sin(n\theta) +\sin\theta \cos(n\theta)}=
\dfrac{2\cos\theta/x_n}{\cos^2\theta/x_n^2 - \sin^2\theta(1-1/x_n^2)} = 
\dfrac{2\cos\theta x_n}{1-\sin^2\theta x_n^2};
 x_{n-1}x_{n+1} = \dfrac{1}{[\cos\theta \sin(n\theta) -\sin\theta \cos(n\theta)]
[\cos\theta \sin(n\theta) +\sin\theta \cos(n\theta)]}=
\dfrac{1}{\cos^2\theta/x_n^2 - \sin^2\theta(1-1/x_n^2)} = 
\dfrac{x_n^2}{1-\sin^2\theta x_n^2}, \text{\ so}\\
E(x_n,x_{n\pm 1})= (1-x_n^2 \sin^2\theta)x_{n\pm 1}^2
-2x_n x_{n+1}\cos\theta +x_n^2=0, \\
E(x,y)= -\sin^2\theta\; x^2y^2 +x^2+y^2 -2\cos\theta\;xy. 
$

$E(x_n,x_{n+1})=0$ gives $x_{n+1}$ as a function of $x_n$, a procedure
quite common in numerical analysis. The \emph{\textbf{fixed points}}
$z_0: E(z_0,z_0)=0$ are especially interesting. May a sequence $x_n$ converge
towards some fixed point? Normally, no: from the symmetry of $E$,
$dx_{n+1}/dx_n\to -1$ near a fixed point. But what is ``normally''?


\subsection{\label{ss3}Coefficients of continued fraction expansion of an algebraic
 function}  \  \\ 

We consider
the algorithm for the coefficients of a continued fraction expansion of an algebraic
 function involving $\sqrt{P}$, where $P$ is a polynomial of degree $\leqslant 4$ (but a part of the algorithm is valid for any degree
[hyperelliptic case, see \S~\ref{hyperell}, p.~\pageref{hyperell}]),
and let $z_1,\dotsc, z_4$ be the zeros of $P$ ($z_4=\infty$ if degree $P<4$).

We choose a continuous $\sqrt{P(z)}$ outside a system of two cuts, say
$[z_1,z_2]$ and $[z_3,z_4]$. So, instead of
dealing with the two-sheeted Riemann surface, we will sometimes emphasize
a sign $\varepsilon$ in $\varepsilon\sqrt{P(z)}$, with $\varepsilon=+1$ or $-1$.

Let also $z_0$ different from $z_1,\dotsc, z_4$,
and  $\gamma+\delta (z-z_0)$ be the two first terms of the Taylor
expansion of $\sqrt{P(z)}$ about $z=z_0$. 

We look at continued fraction expansions about $z_0$ of functions involving $\sqrt{P}$.   

Start with $\sqrt{P}$ itself, to see how things are going:

$\displaystyle
\sqrt{P(z)} = \gamma+\delta(z-z_0) -\dfrac{(z-z_0)^2}{\text{Tayl}_1-\text{new\  cf}},
$
where ``Tayl''${}_1$ means the two first terms of the Taylor expansion of what is
needed there, i.e., of 
$\displaystyle \dfrac{(z-z_0)^2}{\gamma+\delta(z-z_0)-\sqrt{P(z)}}$,
and ``new cf'' is the remainder
Tayl${}_1(z)\displaystyle -\dfrac{(z-z_0)^2}{\gamma+\delta(z-z_0)-\sqrt{P(z)}}
= \text{Tayl}_1(z)-\dfrac{\gamma+\delta(z-z_0)+\sqrt{P(z)}}{(z-z_0)^{-2}[(\gamma+\delta(z-z_0))^2-P(z)]}$,
where the denominator is a polynomial of degree $\leqslant 2$. So, intermediate steps will
involve $\displaystyle \dfrac{S_n-\sqrt{P} }{Z_n}$, where $Z_n$ and $S_n$ are
polynomials, with $S_n(z)-\sqrt{P(z)}=O((z-z_0)^2)$ near $z_0$, so that $S_n^2(z)-P(z)=$ a polynomial multiple of $(z-z_0)^2 Z_n(z)$, say
$(z-z_0)^2 W_n(z)Z_n(z)$. We keep at each step the Taylor expansion of degree 1 and consider the
remainder (associate continued fraction, see Perron \cite[\S~20]{Per}, etc.)\\

\includegraphics*[angle=270, scale=0.50]{Perro1.ps}

...

\includegraphics*[angle=270, scale=0.50]{Perro2.ps}

{%\Small
 $\displaystyle
\dfrac{S_n(z)-\sqrt{P(z)} }{Z_n(z)} =
   \dfrac{(z-z_0)^2}{ \dfrac{(z-z_0)^2}{[S_n(z)-\sqrt{P(z)} ]/Z_n(z)  } =
                         \dfrac{  [S_n(z)+\sqrt{P(z)} ]/Z_n(z)  }
                               { (z-z_0)^{-2}[S_n^2(z)-P(z)]/Z_n^2 =
                                   W_n(z)/Z_n}    },
$}\\
and
$\displaystyle
 \dfrac{  S_n(z)+\sqrt{P(z)}   } { W_n(z)}= \text{Tayl}_1(z)-\dfrac{S_{n+1}(z)-\sqrt{P(z)}}
{Z_{n+1}(z)}$,    
so, $Z_{n+1}(z)=W_n(z)$, $S_{n+1}(z)=-S_n(z)+W_n(z)$Taylor${}_1(z)$, 
and check that $S_{n+1}^2(z)-P(z)$ is a polynomial multiple of $(z-z_0)^2Z_{n+1}(z)$:
$S_{n+1}^2(z)-P(z)= \displaystyle \underbrace{S_n^2(z)-P(z)}_{\displaystyle
  (z-z_0)^2Z_n(z)W_n(z)} -2S_n(z)W_n(z)\text{Tayl}_1(z)+W_n^2(z)\text{Tayl}^2_1(z)$,
so that $W_{n+1}(z) (=Z_{n+2}(z)) = Z_n(z) -\text{Tayl}_1(z) \displaystyle
 \dfrac{ 2S_n(z) -W_n(z)\text{Tayl}_1(z)  }{(z-z_0)^2}$
which is  still a polynomial, as $S_n(z)-\sqrt{P(z)}$, and therefore, 
$2S_n(z)/W_n(z)-$Tayl${}_1(z)$ are $O((z-z_0)^2)$. See also that
$Z_{n+2}(z) = Z_n(z) -\text{Tayl}_1(z) \displaystyle
 \dfrac{ S_n(z) -S_{n+1}(z)  }{(z-z_0)^2}$\\
Let $\text{Tayl}_1(z)$= two first Taylor terms of $(S_n+\sqrt{P})/Z_{n+1}$ be written
$[1-\beta_n(x-z_0)]/\alpha_n$, so that
$\displaystyle \dfrac{S_n(z)-\sqrt{P(z)} }{Z_n(z)} =
   \dfrac{(z-z_0)^2}{  \dfrac{  S_n(z)+\sqrt{P(z)}  }{  Z_{n+1}(z)}    }
= \dfrac{\alpha_n(x-z_0)^2 }{1-\beta_n (x-z_0) +O((x-z_0)^2) }$




\begin{equation} \label{recSZ}
\begin{bmatrix} S_{n+1}(z) \\  Z_{n+1}(z) \\  Z_{n+2}(z)\end{bmatrix} =
\begin{bmatrix} -1 & 0  & [1-\beta_n(z-z_0)]/\alpha_n  \\
                  0  &  0   &  1 \\
                \displaystyle -\dfrac{2(1-\beta_n(z-z_0)}{\alpha_n(z-z_0)^2} &  1  &  \displaystyle \dfrac{(1-\beta_n(z-z_0))^2}{\alpha_n^2(z-z_0)^2}  \end{bmatrix}
\begin{bmatrix} S_n(z) \\  Z_n(z) \\  Z_{n+1}(z) \end{bmatrix} 
\end{equation}


\noindent{\Small \textbf{Example.} $P(z)=1+z^2+z^4, z_0=0, S_0(z)=\text{Taylor\ of\ }
\sqrt{P} \equiv 1, Z_0(z)\equiv 1 \Rightarrow Z_1(z)=(S_0^2(z)-P(z))/z^2 = -1-z^2,
\text{\ new\ Taylor\ of\ } (S_0+\sqrt{P})/Z_1 = -2, S_1(z)=-S_0(z)+\text{Tayl\ }
Z_1(z)= 1+2z^2, Z_2(z)=(S_1^2(z)-P(z))/(z^2Z_1(z)) = -3, \text{\ new\ Taylor\ of\ }
(S_1+\sqrt{P})/Z_2 = -2/3, S_2(z)= 1-2z^2 \text{\ and\ }
Z_3(z)= 5/3-z^2 $ etc. Degree of $S_n=2$, degrees of $Z_n$ alternatively 0 and 2.   }

\noindent We will only consider now functions $(S_n-\sqrt{P})/Z_n$ with $S_n$ of degree 2
and $Z_n$ of degree 1.

Why the negative sign before $\sqrt{P}$? Perron uses a positive sign. But the same
Perron showed how to retrieve the measure from the Stieltjes transform
$\int_S (z-x)^{-1}d\mu(x)$: when $z$ has a small \emph{\textbf{positive}}
imaginary part $z=x^* +i\varepsilon$, $\varepsilon>0$, the imaginary part of
the Stieltjes transform is close to $-\pi i \mu'(x_0)$, see also Haydock,
Haydox \& Nex \cite{HayN}. Then, formulas with $\mu'>0$ and $Z_n(z)>0$ for large $z$
contain a negative sign before the square root.

Examples (with $z_0=\infty$, so that we consider expansions with
negative powers about $\infty$) 1: $\displaystyle \int_a^b \dfrac{\sqrt{(x-a)(b-x)}\;dx}{\pi(z-x)} = \dfrac{(b-a)^2}{8}\left(\dfrac{1}{z}+\dfrac{(a+b)/2}{z^2}+\dotsb\right)\\
\hspace*{299pt} =z-(a+b)/2 -\sqrt{(z-a)(z-b)}$.\\
2: $\displaystyle \int_a^b +\int_c^d \dfrac{\sqrt{(x-a)(x-b)(x-c)(d-x)}\;dx}{|x-x_0|(z-x)}\\ = 
\text{\ const.\ }\dfrac{S(z)=(z-x_0)^2 -(a+b+c+d-4x_0)(z-x_0)/2+S(x_0)-\sqrt{(z-a)(z-b)(z-c)(z-d)}}{z-x_0}$
if $a<b<c<d$ are real, the measure is positive if $z_0$ is in the gap $[b,c]$. The numerator
must vanish at $z=x_0$. If that does not happen, one must add to the measure a Dirac mass
at $x=x_0$.


Finally, let the continued fraction step be
\begin{equation} \label{contf}
  f_n(z):= \dfrac{S_n(z)-\sqrt{P(z)}}{(z-z_0)Z_n(z)} = \dfrac{   (z-z_0) }{\alpha_n+\beta_n(z-z_0) - (z-z_0)f_{n+1}(z)  }.
\end{equation}

The coefficients $\alpha_n$ and $\beta_n$ are immediately found from
$ \displaystyle \dfrac{  S_n(z)+\sqrt{P(z)}   } { Z_{n+1}(z)}\\= 
          \dfrac{ 2\gamma+2\delta (z-z_0)+O((z-z_0)^2) }{Z_{n+1}(z_0)
         +Z'_{n+1}(z_0)(z-z_0)+\dotsb} =
      \alpha_n+\beta_n(z-z_0)+O((z-z_0)^2)$

with
 $P$ of degree 3 or 4, asks for a sequence of polynomials
$\{S_n(x)=\gamma+\delta(x-z_0)+\xi_n (x-z_0)^2, Z_n(x)=\zeta_n(x-x_n)\}$, 
$n=0,1,\dotsc$, such that $S_n^2(x)-P(x)=\alpha_n (x-z_0)^2Z_n(x)Z_{n+1}(x)$,
see \S~\ref{cfsqrt} p.~\pageref{cfsqrt}.  



\noindent\textbf{Definition 3.} $\{x_n\}$ in the continued fraction
arrangement $\displaystyle f_n(x)=\dfrac{S_n(x)-\sqrt{P(x)}}{(x-z_0)\zeta_n(x-x_n)} \\
 = \dfrac{\alpha_n(x-z_0)}{1-\beta_n(x-z_0) -(x-z_0)f_{n+1}(x)}
 = \dfrac{\alpha_n(x-z_0)}{1-\beta_n(x-z_0) -
       \dfrac{\alpha_{n+1}(x-z_0)^2}{1-\beta_{n+1}(x-z_0)-\ddots} }$  
is an instance of
\emph{\textbf{elliptic lattice}}.


\medskip


\subsection{Values of elliptic functions with arguments in arithmetic progression} \  \\

Jacobi and Abel related the continued fraction above to a closed formula
for $x_n$ through the Jacobi inversion problem, see \eqref{Jacobiprob}
p.~\pageref{Jacobiprob}, which brings us to

\noindent\textbf{Definition 4.} \textsl{An elliptic lattice is a sequence
$x_n=\mathcal{E}(n h+u_0)$, where $\mathcal{E}$ is any elliptic function of order 2
(i.e., with 2 zeros and 2 poles in a fundamental parallelogram of periods)}.  

A first way \cite{SpZ2007} to establish this definition is to recognize \eqref{Ebiq} as an
addition formula for elliptic functions. 

One may also establish, see p.~\pageref{birational}, that the biquadratic curve $F(x,y)=0$ in \eqref{Fbiq}
has genus 1 and a parametric representation

$$ x=\mathcal{E}_1(s),\ \ y=\mathcal{E}_2(s),$$
with $\mathcal{E}_1$ and $\mathcal{E}_2$ elliptic functions of order 2,
inverting an elliptic integral of the first kind

$$ s= \int^x \dfrac{du}{\sqrt{P(u)}} = \int^y \dfrac{dv}{\sqrt{Q(v)}}.$$
with a continuous determination of the square root along the path of integration.


Now, let $s_n$ and $s'_n$ correspond to the two points $(x_n,y_n)$ and $(x_n,y_{n+1})$.
As $\mathcal{E}_1(s_n) = \mathcal{E}_1(s'_n)$ with $y_{n+1}$ normally different from
$y_n$, $s_n+s'_n=$ a constant, say $\gamma_1$ (as $s_n$ and $s'_n$ are integrals 
involving the square root of a polynomial on two paths with the same endpoints [the
second endpoint being $x_n$], the square roots are opposite on a part of the paths).
Similarly, $s'_n+s_{n+1}=$ another constant, say $\gamma_2$. Therefore, $s_{n+1}=
s_n+h$, with $h=\gamma_2-\gamma_1$.

The essential parameters in the description of an elliptic sequence is the modulus $k$
and the step $h$ (actually, the ratio $h/\omega$ one one period). The modulus is also related to the ratio $\omega_1/\omega_2$ of periods.
Finally, in a multiplicative setting, the main parameters are the nome $p$ and the multiplier $q$,
which are basically (i.e., up to multiplication by constants) the exponentials of the
periods ratio and the step.

The modulus and the step depend only on $F$ in \eqref{Fbiq} (or $E$ in \eqref{Ebiq}). For each
starting point $(x_0,y_0)$, or $s_0=h_0$, there is a different elliptic lattice with the same $k$
and $h$.

It is always possible to relate $\mathcal{E}_2$ to $\mathcal{E}_1$ through a rational
transformation of first degree $\displaystyle \mathcal{E}_2(s) =
 \dfrac{\alpha\mathcal{E}_1(s+h/2)+\beta }{\gamma\mathcal{E}_1(s+h/2)+\delta }$
\cite[p.~298]{SpZ2007}.

 

\medskip

\subsection{Relations with other problems} \  \\

and there are relations with

 Poncelet problem (see \cite{BursZhed,BursZhed2,BursZhed3})

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%\uput[180](-0.1,0.4){$y(s)$}\uput[180](-0.1,1.4){$y(s+1)$}
\parametricplot{-180}{180}{t cos 2 mul 2.25 add t cos t sin add 1.75 add}
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% \pscircle[linewidth=0.05](1,1.5){1}
\psline[linewidth=0.05](-0.2,0.8)(0.8,3)
\end{pspicture}
}
One considers two conics $C_1$ and $C_2$. From a point
$P_1$ on $C_1$, one constructs a tangent to $C_2$ meeting
$C_1$ at $P_2$, and so on. The coordinates of the $P_i$'s,
or the (central or parallel) projections of these points on any straight line,
make an elliptic lattice. \\
%{\Small
 From $P_2$ onwards, there is no double choice left  for the next point, as
one of the two possible tangents leads to  the preceding
point. At the first point, the other possible tangent simply
leads to $P_0$, and continues with $P_{-1}$, etc.
%}

From Burskii and Zhedanov: let $\{X=R_1(x), Y=R_2(x)\}$ and $\{X=S_1(y), Y=S_2(y)\}$
be parametric representations of the two conics $C_1$ and $C_2$. $R_1$, etc. are rational
functions of degree $\leqslant 2: R_i=r_i/u, S_i=s_i/v$. The line $P_n Q_n$, with parameters $x_n$ and
$y_n$, is tangent at $Q_n$, so
$$
 \dfrac{Y^{(1)}_n-Y^{(2)}_n}{X^{(1)}_n-X^{(2)}_n} = \dfrac{R_2(x_n)-S_2(y_n)}{R_1(x_n)-S_1(y_n)} 
= \dfrac{S'_2(y_n)}{S'_1(y_n)},$$
or
$$
\begin{vmatrix} R_1(x_n)-S_1(y_n)   &  R_2(x_n)-S_2(y_n)  \\
                 S'_1(y_n)            &        S'_2(y_n)           \end{vmatrix} =
\begin{vmatrix} 1  &   R_1(x_n)       &    R_2(x_n)   \\
                1  &  S_1(y_n)    &    S_2(y_n)  \\
                0   &  S'_1(y_n)     &   S'_2(y_n)   \end{vmatrix} =
\begin{vmatrix} u(x_n)  &   r_1(x_n)       &    r_2(x_n)   \\
                v(y_n)  &  s_1(y_n)    &    s_2(y_n)  \\
                v'(y_n)   &  s'_1(y_n)     &   s'_2(y_n)   \end{vmatrix} =
0,
$$  
 which is our $F(x_n,y_n)$.

 Elliptic curves.

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\end{pspicture}
}
An elliptic curve is neither an ellipse nor a curve.
It is a set of points which happens to be a group with
respect to a peculiar addition rule. The geometric
picture may seem strange: let $P$ be a polynomial
of third degree, follow the line joining two points of 
 the (cubic!) curve $y=\sqrt{P(x)}$ until it intersects a third one,
and take the symmetric point with respect to the $x-$axis.\\
\   \\

Excerpts of Appell \& Goursat   \cite{Appell}

\texttt{http://www.archive.org/details/theoriefonctions00apperich}

{\sl 
Theory of algebraic  functions and their  integrals:
 study of analytic functions on a Riemann surface.}

Contents:

{\bf 1. Two-sheeted Riemann surfaces} $u^2=z, u^2=A(z-e_1)(z-e_2)(z-e_3)(z-e_4),
u^2=A(z-e_1)\dotsb (z-e_n)$. Uniform function on a Riemann surface: zeros,
singular points, poles, orders. Rational functions of $z$ and $u$: characteristic
properties. Genus.  1--54.

{\bf 2. Hyperelliptic integrals.} Singularities. Kinds. Number of first kind integrals $=$ genus.
Third kind integrals with two logarithmic singularities. Second kind integrals with a single pole,
etc. 55--98.

{\bf 3. Connection of two-sheeted surfaces. Periods of hyperelliptic integrals. } 
Cuts. Cauchy theorem on a two-sheeted surface. Normal integrals. Periods. 99--164.

{\bf 4. Algebraic functions and the corresponding Riemann surfaces.} 
Puiseux method, $m-$sheeted surfaces. Uniform functions on a Riemann surface  165--221.

{\bf 5. Connection of Riemann surfaces. Periods of Abelian functions.} Connection order
for surfaces. Euler formula for polyhedra. Cuts on a Riemann surface. Examples. Binomial
equations. Regular Riemann surfaces. Abelian integrals: general properties, periods,
classification. 222--255.

{\bf 6. Birational transformations.} Genus conservation. Order and class of a cycle.
Halphen transformation. N\"other's theorem. Geometric definition of genus. Curves of
genus zero; one; two. 256--298.

{\bf 7. Normal integrals. Decomposition of an Abelian integral into simple elements.
Reduction cases.}  Integrals of first kind. Adjoint curves. Integrals of second and third
kind. Normal integrals of the three kinds, their periods. Exchange of parameter and argument
in integrals of third kind. Reduction in third kind integrals and $2p$ integrals
of first and second kind. Algebraic integrals. Logarithmic integrals. Integrals of first
kind reducible to elliptic integrals. 299--372.

{\bf 8. Uniform functions on a Riemann surface.} Rational function in terms of normal
integrals of second kind. Riemann-Roch's theorem. Special functions. Functions of
minimal order. Hyperelliptic curves. Relation between poles and zeros. General expression
of a uniform function with a finite number of singularities. 373--399.

{\bf 9. Abel's theorem.} General theorem. Application to integrals of first, second,
and third kind. General formula. Application to hyperelliptic integrals. Second proof.
Reduction of any sum of integrals to $p$ integrals and algebraic-logarithmic terms.
Addition theorem for first kind integrals. Solution of a system of differential
equation. Extension of Abel's theorem to algebraic skew curves. 400--434.

{\bf 10. The inversion problem.} Curves whose parametric representation is made
of uniform functions of an Abelian integral. The three forms. Inversion of an
integral of first kind related to a curve of genus one. Doubly periodic
functions. Euations $F(u,u')=0$ with uniform solutions. Method of M.~Hermite.
Application to binomial equations. Functions with algebraic addition theorems.
The Jacobi inversion problem, and extensions.  435--469.

{\bf 11. Normal curves. Modules.} Theorem of M.~Schwarz. Birational transformations
of a curve of genus one into itself. Normal curve of Clebsch. Normal curve of 
N\"other. Modules of a class of algebraic curves. Simply rational transformations. 470--487.

{\bf 12. Geometric applications of Abel's theorem.}  Intersections of an algebraic
curve and a family of algebraic curves. Applications to cubics and quartics. 
Double tangents to quartics; conics with four tangents. Applications of
Abel's theorem to areas, angles, and arcs. Skew biquadratic curves. 488--526.

\medskip

\rule{111pt}{2pt}

\medskip

From chap.~6,
one may also establish that the biquadratic curve $F(x,y)=0$ in \eqref{Fbiq}
has genus 1 and a parametric representation

$$ x=\mathcal{E}_1(s),\ \ y=\mathcal{E}_2(s),$$
with $\mathcal{E}_1$ and $\mathcal{E}_2$ elliptic functions of order 2.


Indeed, from chapter~6 of Appell \& Goursat's book \cite{Appell},   a birational \label{birational} transformation $(x,y) \leftrightarrow (\xi,\eta)$ sending the biquadratic
curve \eqref{Fbiq} $F(x,y)=0$ to the canonical elliptic curve $\eta^2=P_3(\xi)$, where $P_3$ is a
polynomial of third degree, see \cite[p.292]{Appell}: from \eqref{ydirect}
$y=(-X_1(x)+\sqrt{P(x)})/(2X_2(x))$, choose
$w=$ a square root of $P(x)$, so that $y=(-X_1(x)+w)/(2X_2(x)) \leftrightarrow
w= X_1(x)+2y X_2(x)$, and $x=z_1+1/\xi$, where $z_1$ is one of the four roots
of $P(x)=0$. Then, with $P(z_1+1/\xi)=P_3(\xi)/\xi^4$ and $\eta=w\xi^2$, $\eta^2=\xi^4 P(z_1+1/\xi)=P_3(\xi)$. 
\begin{equation} \label{xieta} \begin{split}
    x=z_1+\dfrac{1}{\xi},\qquad      &\qquad\qquad     \xi= \dfrac{1}{x-z_1}, \\
  y= \dfrac{ -X_1(z_1+1/\xi) + \eta/\xi^2 }{2X_2(z_1+1/\xi)},  &\qquad\qquad %
   \eta= [X_1(x)+2y X_2(x) ]/(x-z_1)^2 \\
    = \dfrac{ -\xi^2X_1(z_1+1/\xi) + \eta }{2\xi^2X_2(z_1+1/\xi)} &    \\
   F(x,y)= X_0(x)+X_1(x)y+X_2(x)y^2=0  & \qquad\qquad \eta^2-P_3(\xi)=0.
\end{split} \end{equation}


Then, the Weierstrass representation holds 
\begin{equation} \label{wprep} \begin{split}
\   & \xi=\wp(hu+u_0),\   \eta=\wp'(hu+u_0).  \\
\text{So,\  } &  x=z_1+1/\wp,\   y=(-X_1+\wp'/\wp^2)/(2X_2).
\end{split}\end{equation}



The birational transformation, and the Weirstrass function representation are also
described by Burskii \& Zhedanov \cite{BursZhed3}.


However, the authors of \cite{Maillard} recommend the biquadratic setting instead
of the more familiar cubic one, see \cite[pp.~300-301]{Maillard}.

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What are the $(\xi,\eta)$ corresponding to the points $(x_n,y_n)$ and $(x_n,y_{n+1})$?
Let us index them as $(\xi,\eta)_{2n}$ and $(\xi,\eta)_{2n+1}$. The $\xi$s are obviously enough
$\xi_{2n}=\xi_{2n+1}=1/(x_n-z_1)$. Then, $\eta_{2n}$ and $\eta_{2n+1}$ are the two opposite
square roots of $P_3(\xi_{2n})$.  The next point is taken on the line $y=$ constant $=y_{n+1}$, so
on the curve  $\displaystyle \dfrac{ -\xi^2X_1(z_1+1/\xi) + \eta }{2\xi^2X_2(z_1+1/\xi)} =$ constant \\$\displaystyle =\dfrac{ -\xi_{2n}^2X_1(z_1+1/\xi_{2n}) + \eta_{2n+1} }{2\xi_{2n}^2X_2(z_1+1/\xi_{2n})} $ 
$=y_{n+1}$, which is a parabola $\eta=$ a second degree polynomial in $\xi$, up to the intersection with $\eta^2=P_3(\xi)$: the equation for $\xi_{2n+2}$ is
 $\displaystyle  \xi^4[X_1(z_1+1/\xi)+2y_{n+1} X_2(z_1+1/\xi) ]^2=P_3(\xi) =\xi^4P(z_1+1/\xi)=
\xi^4[X_1(z_1+1/\xi)^2-4X_0(z_1+1/\xi)X_2(z_1+1/\xi)$, so a second degree equation
$y_{n+1}X_1(z_1+1/\xi)+y^2_{n+1}X_2(z_1+1/\xi))=-X_0(z_1+1/\xi)$, which is our $F(x,y_{n+1})=0$. The two $\xi-$roots of $\xi^2 F(z_1+1/\xi,y_{n+1})
=(1+\xi z_1)^2Y_2(y_{n+1})+\xi(1+\xi z_1)Y_1(y_{n+1})+\xi^2Y_0(y_{n+1})=0$ are therefore  
$\xi_{2n}$ and $\xi_{2n+2}$, whence $\xi_{2n}+\xi_{2n+2}=-(2z_1Y_2(y_{n+1})+Y_1(y_{n+1}))/F(z_1,y_{n+1})$ and  $\xi_{2n}\xi_{2n+2}=Y_2(y_{n+1})/F(z_1,y_{n+1})$. \\
As $F(z_1,y)=X_2(z_1)y^2+X_1(z_1)y+X_0(z_1)=0$ at the double
root $y=y(z_1)=-X_1(z_1)/(2X_2(z_1))$ when $X_1^2(z_1)-4X_0(z_1)X_2(z_1)=0$, 
$F(z_1,y)\equiv (2X_2(z_1)y+X_1(z_1))^2/(4X_2(z_1))$.\\
Remark that $(\xi_{2n+2}-\xi_{2n})^2= [(2z_1Y_2(y_{n+1})+Y_1(y_{n+1}))^2-
4Y_2(y_{n+1})F(z_1,y_{n+1})]/F^2(z_1,y_{n+1})
=[Y_1^2(y_{n+1})-4Y_0(y_{n+1})Y_2(y_{n+1})]/F^2(z_1,y_{n+1}) $.\\ 
Also, $\eta_{2n}+\eta_{2n+2}=-\eta_{2n+1}+\eta_{2n+2}=-\displaystyle [2\xi_{2n}^2X_2(z_1+1/\xi_{2n}) ]y_{n+1}
   -\xi_{2n}^2X_1(z_1+1/\xi_{2n}) +[2\xi_{2n+2}^2X_2(z_1+1/\xi_{2n+2}) ]y_{n+1}
   +\xi_{2n+2}^2X_1(z_1+1/\xi_{2n+2})=
2[(\xi_{2n+2}^2-\xi_{2n}^2)X_2(z_1)+(\xi_{2n+2}-\xi_{2n})X'_2(z_1)]y_{n+1}
+(\xi_{2n+2}^2-\xi_{2n}^2)X_1(z_1)+(\xi_{2n+2}-\xi_{2n})X'_1(z_1)$\\ 
$\xi_{2n+2}\eta_{2n}+\xi_{2n}\eta_{2n+2}=-\xi_{2n+2}\eta_{2n+1}+\xi_{2n}\eta_{2n+2}
=-\xi_{2n+2}[2\xi_{2n}^2X_2(z_1+1/\xi_{2n}) ]y_{n+1}
   -\xi_{2n+2}\xi_{2n}^2X_1(z_1+1/\xi_{2n}) +\xi_{2n}[2\xi_{2n+2}^2X_2(z_1+1/\xi_{2n+2}) ]y_{n+1}
   +\xi_{2n}\xi_{2n+2}^2X_1(z_1+1/\xi_{2n+2})=(\xi_{2n+2}-\xi_{2n})
[2\xi_{2n}\xi_{2n+2}X_2(z_1)y_{n+1}-X''_2y_{n+1} +\xi_{2n}\xi_{2n+2}X_1(z_1)-X''_1/2]
=(\xi_{2n+2}-\xi_{2n})\displaystyle \left[4X_2\dfrac{Y_2(y_{n+1})}{2X_2y_{n+1}+X_1} 
-X''_2y_{n+1} -X''_1/2\right]$.


According to the addition rule of an elliptic curve, each new point $(\xi,\eta)_{2n+2}$
is the result of adding a fixed step to $(\xi,\eta)_{2n}$, which means that the straight
line joining $(\xi,\eta)_{\text{step}}$ to $(\xi_{2n},\eta_{2n})$ must meet the cubic
$\eta^2=P_3(\xi)$ at the opposite of $(\xi_{2n+2},\eta_{2n+2})$, which happens to
be $(\xi_{2n+2},-\eta_{2n+2})=(\xi,\eta)_{2n+3}$.The  straight line 
$\displaystyle \eta = -\dfrac{\eta_{2n}+\eta_{2n+2}}{\xi_{2n+2}-\xi_{2n} }\xi+
\dfrac{\xi_{2n+2}\eta_{2n}+\xi_{2n}\eta_{2n+2}}{\xi_{2n+2}-\xi_{2n} }$
meets the cubic $\eta^2-\xi^4P(z_1+1/\xi)=\displaystyle 
\left[\dfrac{\eta_{2n}+\eta_{2n+2}}{\xi_{2n+2}-\xi_{2n} }\right]^2\xi^2+\dotsb
-P'(z_1)\xi^3 -P''(z_1)\xi^2/2-\dotsb
=\{2[(\xi_{2n}+\xi_{2n+2})X_2(z_1)+X'_2(z_1)]y_{n+1}+(\xi_{2n}+\xi_{2n+2})X_1(z_1)
+X'_1(z_1)\}^2\xi^2+\dotsb-P'(z_1)\xi^3 -P''(z_1)\xi^2/2-\dotsb$ at a third point $\xi^{(\text{step},1)}$ together with
$\xi=\xi_{2n}$ and $\xi_{2n+2}$, so that
$\displaystyle P'(z_1)\xi^{(\text{step},1)}
=\{2[(\xi_{2n}+\xi_{2n+2})X_2(z_1)+X'_2(z_1)]y_{n+1}+(\xi_{2n}+\xi_{2n+2})X_1(z_1)
+X'_1(z_1)\}^2
-P''(z_1)/2-P'(z_1)(\xi_{2n}+\xi_{2n+2})=
[(\xi_{2n}+\xi_{2n+2})(2X_2(z_1)y_{n+1}+X_1(z_1))+2X'_2(z_1)y_{n+1}
+X'_1(z_1)]^2
-P''(z_1)/2-P'(z_1)(\xi_{2n}+\xi_{2n+2})=
\left[-4X_2(z_1)\dfrac{2z_1Y_2(y_{n+1})+Y_1(y_{n+1})}{2y_{n+1}X_2(z_1)
+X_1(z_1)} +2X'_2(z_1)y_{n+1}+X'_1(z_1)\right]^2
-P''(z_1)/2+P'(z_1)4X_2(z_1)\dfrac{2z_1Y_2(y_{n+1})+Y_1(y_{n+1})}{(2y_{n+1}X_2(z_1)
+X_1(z_1))^2} 
\text{\  must\  be\  independent\  of\ }y_{n+1}$!!! Also, for all $x,y$,
$\partial F(x,y)/\partial x = X'_2 y^2+X'_1 y+X'_0 = 2Y_2(y) x +Y_1(y)$. At $x=z_1$,
$P'(z_1)=2X_1X'_1-4X'0X_2-4X_0X'_2= -4X_2[X'_2 (y(z_1))^2 +X'_1y(z_1)+X'_0]$,
where $y(z_1)=-X_1(z_1)/(2X_2(z_1))$,
as $X_0(z_1)=X_1^2(z_1)/(4X_2(z_1))$. So, $P'(z_1)=-4X_2(z_1)[2z_1 Y_2(y(z_1))
+Y_1(y(z_1))]$, and \\$\hspace*{-44pt}P'(z_1)\xi^{(\text{step},1)} =
\displaystyle 
\left\{  -2 \dfrac{X'_2y^2+X'_1y+X'_0}{y-y(z_1)}  +2X'_2y+X'_1\vphantom{\dfrac{Y}{Y}}\right\}^2-\dfrac{P''(z_1)}{2}
-4[X'_2y^2(z_1)+X'_1y(z_1)+X'_0] 
\dfrac{X'_2y^2+X'_1y+X'_0}{(y-y(z_1))^2}\\
=\left\{  -2 \dfrac{[X'_2y(z_1)+X'_1]y+X'_0}{y-y(z_1)}  +X'_1\vphantom{\dfrac{Y}{Y}}\right\}^2-\dfrac{P''(z_1)}{2}
-4[X'_2y^2(z_1)+X'_1y(z_1)+X'_0] 
\dfrac{X'_2y^2+X'_1y+X'_0}{(y-y(z_1))^2}\\
=\left\{  -2 X'_2y(z_1)-X'_1  +\dfrac{R}{y-y(z_1)}\vphantom{\dfrac{Y}{Y}}\right\}^2-\dfrac{P''(z_1)}{2}+2R
\dfrac{X'_2y^2+X'_1y+X'_0}{(y-y(z_1))^2}
$ where $R$ is the residue $-2 [X'_2y^2(z_1)+X'_1y(z_1)+X'_0]=P'(z_1)/(2X_2(z_1))$.Then,
$P'(z_1)\xi^{(\text{step},1)} =
\displaystyle 
( 2 X'_2y(z_1)+X'_1  )^2 \\-2(2 X'_2y(z_1)+X'_1)\dfrac{R}{y-y(z_1)}
+\dfrac{R^2}{(y-y(z_1))^2}
-\dfrac{P''(z_1)}{2}+2R
\dfrac{-R/2+(2X'_2y(z_1)+X'_1)(y-y(z_1))+X'_2(y-y(z_1))^2}{(y-y(z_1))^2}$\\
$=( 2 X'_2y(z_1)+X'_1  )^2 -P''(z_1)/2+2RX'_2
=X'^2_1-P''(z_1)/2-4X'_2X'_0=-X_1X''_1+2X''_0X_2+2X_0X''_2=2X_2[X''_0
+X''_1y(z_1)+X''_2y^2(z_1)]$.

$\displaystyle \xi^{(\text{step},1)} = \dfrac{ 2X_2(z_1)[X''_0
+X''_1y(z_1)+X''_2y^2(z_1)]}{P'(z_1)}=-\dfrac{1}{2}\;\dfrac{ X''_0
+X''_1y(z_1)+X''_2y^2(z_1)}{X'_0
+X'_1y(z_1)+X'_2y^2(z_1)}\\
=-\dfrac{1}{2}\;\dfrac{\partial^2 F(x,y)/\partial x^2\text{\ at\ }(z_1,y(z_1)) }{\partial F(x,y)/\partial x\text{\ at\ }(z_1,y(z_1)) }=
-\dfrac{Y_2(y(z_1))}{2z_1Y_2(y(z_1))+Y_1(z_1)}
$.

\noindent$\displaystyle \hspace*{-40pt}\eta^{(\text{step},1)} = -\dfrac{\eta_{2n}+\eta_{2n+2}}{\xi_{2n+2}-\xi_{2n} }\xi^{(\text{step},1)}+
\dfrac{\xi_{2n+2}\eta_{2n}+\xi_{2n}\eta_{2n+2}}{\xi_{2n+2}-\xi_{2n} }
= \left[ -2 X'_2y(z_1)-X'_1  +\dfrac{R=P'/(2X_2)}{y_{n+1}-y(z_1)}\right]\xi^{(\text{step},1)}\\
\hspace*{-33pt}+ \displaystyle 4X_2\dfrac{Y_2(y_{n+1})}{2X_2y_{n+1}+X_1} 
-X''_2y_{n+1} -X''_1/2=
\left[ -2 X'_2y(z_1)-X'_1  +\dfrac{P'/(2X_2)}{y_{n+1}-y(z_1)}\right]\xi^{(\text{step},1)}+ \displaystyle \dfrac{   \overbrace{2Y_2(y_{n+1})}^{\displaystyle \partial^2 F(x,y_{n+1})/\partial x^2} 
-X''_2y^2_{n+1}+X''_2y(z_1)y_{n+1}}{y_{n+1}-y(z_1)}  -X''_1/2
=
\left[ -2 X'_2y(z_1)-X'_1  +\dfrac{P'/(2X_2)}{y_{n+1}-y(z_1)}\right]\xi^{(\text{step},1)}+ \displaystyle \dfrac{  (X''_2y(z_1)+X''_1)y_{n+1}+X''_0}{y_{n+1}-y(z_1)} 
 -X''_1/2 = (2 X'_2y(z_1)+X'_1  )\xi^{\text{step},1} +\dfrac{ P'(z_1)\xi^{(\text{step},1)}/(2X_2(z_1))
 +X''_2y^2(z_1)+X''_1y(z_1)+X''_0 =0}{y_{n+1}-y(z_1) }+X''_2y(z_1)+X''_1/2\\
=(2 X'_2y(z_1)+X'_1  )\xi^{\text{step},1} +X''_2y(z_1)+X''_1/2\\
=\dfrac{1}{2}\;\dfrac{(X'_1X''_2-X''_1X'_2)y^2(z_1)+2(X'_0X''_2-X''_0X'_2)y(z_1)
+X'_0X''_1-X''_0X'_1}{X'_2y^2(z_1)
+X'_1y(z_1)+X'_0}
$

Returning to $(x,y)$, $\displaystyle x^{(\text{step},1)} = z_1-2
\dfrac{X'_0
+X'_1y(z_1)+X'_2y^2(z_1)}{X''_0
+X''_1y(z_1)+X''_2y^2(z_1)}
=z_1-\dfrac{2z_1Y_2(y(z_1))+Y_1(z_1)}{Y_2(y(z_1))}=-z_1
-\dfrac{Y_1(z_1)}{Y_2(y(z_1))}=z'_1$, the second $x-$root
of $F(x,y)=Y_2(y)x^2+Y_1(y)x+Y_0(y)=0$ at $y=y(z_1)$; \\
$\displaystyle y^{(\text{step},1)} = \dfrac{ -X_1(z_1+1/\xi^{(\text{step},1)}) + \eta^{(\text{step},1)} /(\xi^{(\text{step},1)})^2 }{2X_2(z1+1/\xi^{(\text{step},1)}) }=
\dfrac{-X_1 \xi^2 -X'_1\xi-X''_1/2+\eta}{2[X_2\xi^2+X'_2\xi+X''_2/2]} \\  =
\dfrac{-X_1 \xi^2 +2X'_2 y(z_1)\xi +X''_2 y(z_1) }{2[X_2\xi^2+X'_2\xi+X''_2/2]}=
y(z_1)$ as $-X_1(z_1)=2X_2(z_1)y(z_1)$.  

To multiples of $(\xi,\eta)_{(\text{step},1}$ according to the addition rule
of the elliptic curve correspond iterates of $(x,y)^{(\text{step},1)}$ on the
biquadratic curve $F(x,y)=0$. Let $x_n^{(\text{step},1)}$ and $y_n^{(\text{step},1)}$
be the abscissae and ordinates so constructed. To the neutral element $(\xi,\eta)=
(\infty,\infty)$ of the elliptic curve correspond the starting point
$(x,y)_0^{(\text{step},1)}=(z_1,y(z_1))$. Remark that
$x_{-n}^{(\text{step},1)}=x_n^{(\text{step},1)}, y_{-n}^{(\text{step},1)}=y_{n+1}^{(\text{step},1)}$.

Of course, the same may be done with any $z_i$, $i=1,\dotsc, 4$, not just $z_1$.

More: goto \eqref{addition}, p.~\pageref{addition}

\smallskip

{\red New notation}: let $(x_n,y_n)$ correspond to $(\xi_n,\eta_n)$, and $(x_n,y_{n+1})$
correspond to $(\xi_n,-\eta_n)$: $\displaystyle x_n=z_1+\dfrac{1}{\xi_n},
y_n=\dfrac{-X_1(z_1+1/\xi_n)+\eta_n/\xi_n^2}{2X_2(z_1+1/\xi_n)},
y_{n+1}=\dfrac{-X_1(z_1+1/\xi_n)-\eta_n/\xi_n^2}{2X_2(z_1+1/\xi_n)}$, then, $(\xi_{n+1},
\eta_{n+1})$ is the second intersection of the line $y=$ constant $=y_{n+1}$, i.e. the
parabola $\eta = \xi^2X_1(z_1+1/\xi)+2y_{n+1}\xi^2X_2(z_1+1/\xi)$ and the cubic
$\eta^2-\xi^4(X_1^2-4X_0X_2)=0$, so the equation for $\xi=\xi_{n+1}$ is 
$\displaystyle y_{n+1}X_1(z_1+1/\xi) +y_{n+1}^2 X_2(z_1+1/\xi)+X_0(z_1+1/\xi)=0$,\ \ \footnote{Of course, this is $F(z_1+1/\xi,y_{n+1})=0$.} \\ 
{\Small $\displaystyle
\dfrac{-X_1(z_1+1/\xi_n)-\eta_n/\xi_n^2}{2X_2(z_1+1/\xi_n)}X_1(z_1+1/\xi)
+\dfrac{[X_1(z_1+1/\xi_n)+\eta_n/\xi_n^2]^2}{4X_2^2(z_1+1/\xi_n)}X_2(z_1+1/\xi)
+X_0(z_1+1/\xi)=0, \\
-X_1(z_1+1/\xi_n)X_1(z_1+1/\xi) +\dfrac{X_1^2(z_1+1/\xi_n)}{2X_2(z_1+1/\xi_n)}X_2(z_1+1/\xi)
+(\eta_n/\xi_n^2) \left[
 \dfrac{X_1(z_1+1/\xi_n)}{X_2(z_1+1/\xi_n)}X_2(z_1+1/\xi)-X_1(z_1+1/\xi)\right]
+\dfrac{X_1^2(z_1+1/\xi_n)-4X_0(z_1+1/\xi_n)X_2(z_1+1/\xi_n)}{2X_2(z_1+1/\xi_n)}X_2(z_1+1/\xi)
+2X_0(z_1+1/\xi)X_2(z_1+1/\xi_n)=0\\
\left(\dfrac{\eta_n}{\xi_n^2} +X_1(z_1+1/\xi_n)\right)
\left[\dfrac{X_1(z_1+1/\xi_n)}{X_2(z_1+1/\xi_n)}X_2(z_1+1/\xi)-X_1(z_1+1/\xi)\right]
+2[X_0(z_1+1/\xi)X_2(z_1+1/\xi_n)-X_0(z_1+1/\xi_n)X_2(z_1+1/\xi)]=0, \text{\  two\  roots:\  }
\xi=\xi_n \text{\ and\  } \\ 
\left(\dfrac{\eta_n}{\xi_n^2} +X_1(z_1+1/\xi_n)\right)
\left[\dfrac{X_1(z_1+1/\xi_n)}{X_2(z_1+1/\xi_n)}X'_2(z_1+1/\xi_n)-X'_1(z_1+1/\xi_n)
 +\left(\dfrac{1}{\xi}-\dfrac{1}{\xi_n}\right)
\left[\dfrac{X_1(z_1+1/\xi_n)}{X_2(z_1+1/\xi_n)}X''_2-X''_1\right]/2
\right]
+2[X'_0(z_1+1/\xi_n)X_2(z_1+1/\xi_n)-X_0(z_1+1/\xi_n)X'_2(z_1+1/\xi_n)]
+\left(\dfrac{1}{\xi}-\dfrac{1}{\xi_n}\right)
[X''_0X_2(z_1+1/\xi_n)-X_0(z_1+1/\xi_n)X''_2]=0
 $  
}  % end Small

$\displaystyle \dfrac{1}{\xi_{n+1}}=\dfrac{1}{\xi_n}
-\dfrac{
   \left(\dfrac{\eta_n}{\xi_n^2} +X_1\right)
\left[\dfrac{X_1}{X_2}X'_2-X'_1\right] +2[X'_0X_2-X_0X'_2]
       }{
\left(\dfrac{\eta_n}{\xi_n^2} +X_1\right)
 \left[\dfrac{X_1}{X_2}X''_2-X''_1\right]/2
+X''_0X_2-X_0X''_2
}
$
at $z_1+1/\xi_n: X_j(z_1)=X_j-X'_j/\xi_n+X''_j/(2\xi_n^2)$:\\
\hspace*{-22pt}$\displaystyle \dfrac{1}{\xi_{n+1}}=
\dfrac{
   \left(\dfrac{\eta_n}{\xi_n} +X_1\xi_n\right)
\left[\dfrac{X_1}{X_2}[X_2(z_1)-X_2]-[X_1(z_1)-X_1]\right] +2\xi_n[[X_0(z_1)-X_0]X_2-X_0[X_2(z_1)-X_2]]
       }{
\left(\dfrac{\eta_n}{\xi_n^2} +X_1\right)
 \left[\dfrac{X_1}{X_2}X''_2-X''_1\right]/2
+X''_0X_2-X_0X''_2
}\\
=\dfrac{
   \left(\dfrac{\eta_n}{\xi_n} +X_1\xi_n\right)
\left[\dfrac{X_1}{X_2}X_2(z_1)-X_1(z_1)\right] +2\xi_n[X_0(z_1)X_2-X_0X_2(z_1)]
       }{
\left(\dfrac{\eta_n}{\xi_n^2} +X_1\right)
 \left[\dfrac{X_1}{X_2}X''_2-X''_1\right]/2
+X''_0X_2-X_0X''_2
}\\
\text{also,\ } \xi_{n+1}-\xi_n= 
\xi_n \dfrac{
   \left(\dfrac{\eta_n}{\xi_n^2} +X_1\right)
\left[\dfrac{X_1}{X_2}X'_2-X'_1\right] +2[X'_0X_2-X_0X'_2]
       }{\left(\dfrac{\eta_n}{\xi_n} +X_1\xi_n\right)
\left[\dfrac{X_1}{X_2}X_2(z_1)-X_1(z_1)\right] +2\xi_n[X_0(z_1)X_2-X_0X_2(z_1)]}  
\\
\eta_{n+1} = \xi_{n+1}^2X_1(z_1+1/\xi_{n+1})-2
\dfrac{\eta_n/\xi_n^2+X_1}{2X_2} 
\xi_{n+1}^2X_2(z_1+1/\xi_{n+1})\\
\hspace*{-33pt}= -\dfrac{\xi^2_{n+1}}{\xi_n^2} \eta_n
+\xi_{n+1}^2X'_1  \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)
+\xi_{n+1}^2(X''_1/2) \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)^2
-\xi_{n+1}^2\dfrac{\dfrac{\eta_n}{\xi_n^2}+X_1}{X_2}\left[
 X'_2 \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)
+(X''_2/2) \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)^2
\right]
$

where $X_1, X'_1$, etc. are the values at $z_1+1/\xi_n$.

Also, from $\displaystyle y_{n+1}=\dfrac{-X_1(z_1+1/\xi_n)-\eta_n/\xi_n^2}{2X_2(z_1+1/\xi_n)},
\eta_{n+1}=  \xi_{n+1}^2X_1(z_1+1/\xi_{n+1})+2y_{n+1}\xi_{n+1}^2X_2(z_1+1/\xi_{n+1})$.

And now, the constant difference of $(\xi,\eta)_{n+1}$ and $(\xi,\eta)_n$ 
according to the addition rule of elliptic curves: look at where the line
$\displaystyle \eta = \eta_n - \dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}(\xi-\xi_n)$
meets the cubic $\eta^2-\xi^4P(z_1+1/\xi)=\eta^2-P'(z_1)\xi^3-(P''(z_1)/2)\xi^2-\dotsb=0: \\
\xi_{\text{step}} = \dfrac{1}{P'(z_1)}\left[ \left(\dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}\right)^2 -P''(z_1)/2\right]-\xi_n-\xi_{n+1}$\\
{\Small
$\displaystyle  \hspace*{-77pt}=\dfrac{1}{P'(z_1)}\left[ \left(
\dfrac{  \left(1-\dfrac{\xi^2_{n+1}}{\xi_n^2}\right) \eta_n
+\xi_{n+1}^2X'_1  \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)
+\xi_{n+1}^2(X''_1/2) \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)^2
-\xi_{n+1}^2\dfrac{\dfrac{\eta_n}{\xi_n^2}+X_1}{X_2}\left[
 X'_2 \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)
+(X''_2/2) \left(\dfrac{1}{\xi_{n+1}}-\dfrac{1}{\xi_n}\right)^2
\right]
}{\xi_{n+1}-\xi_n}
   \right)^2
-P''(z_1)/2\right]-\xi_n-\xi_{n+1}$\\
$\displaystyle\hspace*{-44pt}=\dfrac{1}{P'(z_1)}\left[ \left( 
    -\dfrac{\xi_n+\xi_{n+1}}{\xi_n^2}\eta_n -\dfrac{\xi_{n+1}}{\xi_n}X'_1
+\dfrac{\xi_{n+1}-\xi_n}{\xi_n^2}X''_1/2
   -\xi_{n+1}^2\dfrac{\dfrac{\eta_n}{\xi_n^2}+X_1}{X_2}\left[
 -\dfrac{X'_2}{\xi_n\xi_{n+1}}
+\dfrac{(X''_2/2)}{\xi_n^2\xi_{n+1}^2}(\xi_{n+1}-\xi_n)
\right]
     \right)^2
-P''(z_1)/2\right]-\xi_n-\xi_{n+1}$\\
we isolate $\eta_n$ through $\eta_n = (\eta_n+\xi_n^2X_1) -\xi_n^2X_1 = -2X_2 \xi_n^2y_{n+1}-\xi_n^2X_1 $\\ 
$\displaystyle\hspace*{-44pt}=\dfrac{1}{P'(z_1)}\left[ \left(
  -2X_2y_{n+1} 
   \left[ \dfrac{\xi_{n+1} X'_2 }{\xi_n X_2} -\dfrac{(X''_2/2)}{\xi_n^2 X_2}(\xi_{n+1}-\xi_n)-\xi_n-\xi_{n+1} \right]
+(\xi_n+\xi_{n+1})X_1 -\dfrac{\xi_{n+1}}{\xi_n}X'_1
+\dfrac{\xi_{n+1}-\xi_n}{\xi_n^2}X''_1/2\right)^2
-P''(z_1)/2\right]-\xi_n-\xi_{n+1}\\
\hspace*{-44pt}=\dfrac{1}{P'(z_1)}\left[ \left(
   -2 y_{n+1} 
   \left[ \dfrac{(X''_2/2)}{\xi_n } -\xi_n
 -\xi_{n+1} X_2(z_1)  \right]+\xi_{n+1}X_1(z_1)
+\xi_nX_1 -\dfrac{X''_1/2}{\xi_n}\right)^2
-P''(z_1)/2\right]-\xi_n-\xi_{n+1}\\
\hspace*{-44pt}=\dfrac{1}{P'(z_1)}\left[ \left\{
   2  y_{n+1}
   \left[ X'_2(z_1) 
 +(\xi_n+\xi_{n+1}) X_2(z_1)  \right]+(\xi_n+\xi_{n+1})X_1(z_1)
+X'_1(z_1) \right\}^2
-P''(z_1)/2\right]-\xi_n-\xi_{n+1}$
}  % end Small
\\
\hspace*{-44pt}which depends only on $y_{n+1}$ as $\displaystyle \xi_n+\xi_{n+1}=\dfrac{1}{x_n-z_1}
+\dfrac{1}{x_{n+1}-z_1}=\dfrac{-Y_1(y_{n+1})/Y_2(y_{n+1})-2z_1}{[Y_0(y_{n+1})+z_1
Y_1(y_{n+1})+z_1^2Y_2(y_{n+1})]/Y_2(y_{n+1})  } =- \dfrac{ \partial F/\partial x}{F}$
at $(z_1,y_{n+1})$, so,\\
$\displaystyle\hspace*{-33pt}\xi_{\text{step}} =
\dfrac{1}{P'(z_1)}\left[ \left\{  \underbrace{
   2  y_{n+1}
   \left[ X'_2(z_1) 
 - \dfrac{ \partial F/\partial x}{F} X_2(z_1)  \right]
- \dfrac{ \partial F/\partial x}{F}X_1(z_1)
+X'_1(z_1) }_{\displaystyle F \partial^2\log F/\partial x \partial y} \right\}^2
-P''(z_1)/2\right]+ \dfrac{ \partial F/\partial x}{F}$\\
with $F(x,y)=X_2(x)(y-\varphi(x))(y-\psi(x))$, where $\varphi$ and $\psi=(-X_1
\pm\sqrt{P})/(2X_2)$, $\partial^2\log F/\partial x \partial y= \displaystyle
\dfrac{\partial}{\partial x}\left[\dfrac{1}{y-\varphi}+\dfrac{1}{y-\psi}\right] =
\dfrac{\varphi'}{(y-\varphi)^2}+\dfrac{\psi'}{(y-\psi)^2} =
\dfrac{X_2^2}{F^2} [ (\varphi'+\psi')[(y-(\varphi+\psi)/2)^2+(\psi-\varphi)^2/4]
  +2(y-(\varphi+\psi)/2)(\psi-\varphi)(\psi'-\varphi')/2]=
\dfrac{X_2^2}{F^2} \left[ -\left(\dfrac{X_1}{X_2}\right)'
\left[\left(y+\dfrac{X_1}{2X_2}\right)^2+\dfrac{P}{4X_2^2}\right] 
+ \left(y+\dfrac{X_1}{2X_2}\right) \left(\dfrac{P}{2X_2^2}\right)'  \right] $. \\
At $x=z_1$, $P=0$, $\varphi=\psi=y(z_1)=-X_1/(2X_2)$, $F(z_1,y)=X_2 (y-y(z_1))^2,
P'(z_1)=2X_1X_1'-4X_0X_2'-4X_0'X_2= -4X_2(X'_1y(z_1)+X_2'y^2(z_1)+X'_0)=-4X_2\partial F/\partial x \text{\ at\ } (z_1,y(z_1)),
P'/X_2^2=(P/X_2^2)'=(X_1^2/X_2^2)'-4(X_0/X_2)',
P''(z_1)/2=X_1X_1''+(X_1')^2 -2 X_0 X_2'' -4 X_0'X_2'-2X_0''X_2=
-2X_2 \partial^2F/\partial x^2+ (X'_1)^2 -4X'_0X'_2 $, and\\
$\displaystyle\hspace*{-44pt}\xi_{\text{step}} =
\dfrac{1}{P'(z_1)}\left[ \left\{  
\dfrac{X_2}{y-y(z_1)} \left[ -\left(\dfrac{X_1}{X_2}\right)'
\left(y-y(z_1)\right) 
+  \dfrac{P'}{2X_2^2}  \right] 
    \right\}^2
-P''(z_1)/2\right]+ \dfrac{X_2'}{X_2}+\dfrac{ (X_1/X_2)'y+(X_0/X_2)'}{(y-y(z_1))^2}\\
\hspace*{-22pt}=
\dfrac{1}{P'(z_1)}\left[ \left\{  
-X_2\left(\dfrac{X_1}{X_2}\right)'
+  X_2\dfrac{P'/(2X_2^2)}{y-y(z_1)}  \right\}^2
-P''(z_1)/2\right]+ \dfrac{X_2'}{X_2}
+\dfrac{ (X_1/X_2)'}{y-y(z_1)} -\dfrac{P'/(4X_2^2)}{(y-y(z_1))^2}\\
=
\dfrac{1}{P'(z_1)}\left[ \left\{  
-X_2\left(\dfrac{X_1}{X_2}\right)'  \right\}^2
-P''(z_1)/2\right]+ \dfrac{X_2'}{X_2}\\
\hspace*{-44pt}=-\dfrac{ 2X_2 \partial^2F/\partial x^2- (X'_1)^2 +4X'_0X'_2 
+(X'_1)^2-2\dfrac{X_1X'_1X'_2}{X_2}+\dfrac{X_1^2(X'_2)^2}{X_2^2}
+\dfrac{X'_2}{X_2}\left[2X_1X_1'-4X_0X_2'-4X_0'X_2\right]}
{4X_2\partial F/\partial x} \\
=-\dfrac{  \partial^2F/\partial x^2}{2\partial F/\partial x} 
$
at $(x,y)=(z_1,y(z_1))$.

$\displaystyle \eta_{\text{step}} = \eta_n -\left(\dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}\right)(\xi_{\text{step}}-\xi_n) =
-\eta_{n+1} -\left(\dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}\right)(\xi_{\text{step}}-\xi_{n+1})\\= 
\dfrac{\eta_n-\eta_{n+1}}{2} -\left(\dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}\right)\left(\xi_{\text{step}}-\dfrac{\xi_n+\xi_{n+1}}{2}\right) \\
= \dfrac{1}{2}\left[ \eta_n -  \xi_{n+1}^2X_1(z_1+1/\xi_{n+1})-2y_{n+1}\xi_{n+1}^2X_2(z_1+1/\xi_{n+1})\right]
-\left(\dfrac{\eta_{n+1}+\eta_n}{\xi_{n+1}-\xi_n}\right)\left(\xi_{\text{step}}-\dfrac{\xi_n+\xi_{n+1}}{2}\right) \\
\hspace*{-22pt}= -\xi_n^2 y_{n+1}X_2(z_1+1/\xi_n)-\xi_n^2X_1(z_1+1/\xi_n)/2-  \xi_{n+1}^2X_1(z_1+1/\xi_{n+1})/2-y_{n+1}\xi_{n+1}^2X_2(z_1+1/\xi_{n+1})\\-
 \{2  y_{n+1}
   \left[ X'_2(z_1) 
 +(\xi_n+\xi_{n+1}) X_2(z_1)  \right]+(\xi_n+\xi_{n+1})X_1(z_1)
+X'_1(z_1)\} \left(\xi_{\text{step}}-\dfrac{\xi_n+\xi_{n+1}}{2}\right) \\
=-y_{n+1}[(\xi_n^2+\xi_{n+1}^2)X_2(z_1)+(\xi_n+\xi_{n+1})X'_2(z_1)+X''_2]
 -(\xi_n^2+\xi_{n+1}^2)X_1(z_1)/2 \\ -(\xi_n+\xi_{n+1})X'_1(z_1)/2-X''_1/2
-\{2  y_{n+1}
   \left[ X'_2(z_1) 
 +(\xi_n+\xi_{n+1}) X_2(z_1)  \right]+(\xi_n+\xi_{n+1})X_1(z_1)
+X'_1(z_1)\} \left(\xi_{\text{step}}-\dfrac{\xi_n+\xi_{n+1}}{2}\right) \\
=(\xi_n+\xi_{n+1})^2[\underbrace{-y_{n+1}X_2-X_1/2-(2y_{n+1}X_2+X_1)(-1/2) }_{=0}]
+\xi_n\xi_{n+1}[2y_{n+1}X_2+X_1]+(\xi_n+\xi_{n+1})\{-y_{n+1}X'_2-X'_1/2
-(2y_{n+1}X_2+X_1)\xi_{\text{step}}+y_{n+1}X'_2+X'_1/2\}
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
=[2y_{n+1}X_2+X_1][  \xi_n\xi_{n+1}-(\xi_n+\xi_{n+1})\xi_{\text{step}}]
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
\hspace*{-44pt}=[2y_{n+1}X_2+X_1] \dfrac{ 
       Y_2(y_{n+1}) +(Y_1(y_{n+1})+2z_1Y_2(y_{n+1}))\xi_{\text{step}}   }
{F(z_1,y_{n+1})=X_2(y_{n+1}-y(z_1))^2    }
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
\hspace*{-66pt}=[\overbrace{2y_{n+1}X_2+X_1}^{2X_2(z_1)\,[y_{n+1}-y(z_1)]}] \dfrac{\ \ \dfrac{ 
     Y_2(y_{n+1})Y_1(y(z_1)) -Y_1(y_{n+1})Y_2(y(z_1))  }{2z_1Y_2((z_1))+Y_1(y(z_1))=-Y_2(y(z_1))/\xi_{\text{step}}  }\ \    }
{F(z_1,y_{n+1})=X_2(y_{n+1}-y(z_1))^2    }
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
= \dfrac{2}{y_{n+1}-y(z_1)} \left[  Y_1(y_{n+1})- \dfrac{ 
     Y_1(y(z_1)) }{Y_2(y(z_1)) }  
     Y_2(y_{n+1}) \right]\xi_{\text{step}}  
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
$.

Now, for any $x$ and $y$, $\partial F/\partial x = 2Y_2(y)x+Y_1(y)=X'_2(x)y^2
+X'_1(x)y+X'_0$. Of special interest is $F(x,y(z_1))$ which is a quadratic
polynomial in $x$, vanishing at $x=z_1$ and $x=z'_1$ (see p.\  ), so that
$\partial F/\partial x$ vanishes at $(x,y)=((z_1+z'_1)/2,y(z_1))$, so

\begin{subequations}

\begin{equation}  \label{xistep}
\xi_{\text{step}}  =-\dfrac{  \partial^2F/\partial x^2}{2\partial F/\partial x} (z_1,y(z_1))
= -\dfrac{Y_2(y(z_1))}{2z_1Y_2(y(z_1))+Y_1(y(z_1)) } = \dfrac{1}{z'_1-z_1}
\end{equation}

Remark that $\displaystyle x_{\text{step}} = z_1+\dfrac{1}{\xi_{\text{step}}  } =z'_1$.
Return to $\eta_{\text{step}}$:

$\displaystyle \eta_{\text{step}} 
= \dfrac{2\left[  Y_1(y_{n+1})
    +(z_1+z'_1)  Y_2(y_{n+1}) \right]\xi_{\text{step}}  }{y_{n+1}-y(z_1)} 
-y_{n+1}X''_2-X''_1/2-(2y_{n+1}X'_2+X'_1)\xi_{\text{step}}\\
=\dfrac{2[ X'_2((z_1+z'_1)/2)y^2_{n+1}+X'_1((z_1+z'_1)/2)y_{n+1}+X'_0((z_1+z'_1)/2)]
\xi_{\text{step}}  }{y_{n+1}-y(z_1)} 
-2\xi_{\text{step}} [y_{n+1}X'_2((z_1+z'_1)/2) +X'_1((z_1+z'_1)/2)/2]\\
=2\xi_{\text{step}}[  X'_1((z_1+z'_1)/2)/2 +X'_2((z_1+z'_1)/2)y(z_1) ]\\
=\xi_{\text{step}}\dfrac{ X_1(z'_1)-X_1(z_1) -[X_2(z'_1)-X_2(z_1)]\dfrac{X_1(z_1)}{X_2(z_1)}  }{z'_1-z_1}  $

\begin{equation}  \label{etastep}
\eta_{\text{step}}
 =\xi^2_{\text{step}} [ X_1(z'_1)+2y(z_1)X_2(z'_1)]
\end{equation}  

leading to $\displaystyle y_{\text{step}} =
\dfrac{ -X_1(z'_1)+ \eta_{\text{step}}/\xi^2_{\text{step}}  }{2X_2(z'_1)}
=y(z_1)$. 

\end{subequations}



\bigskip

From cubic to biquadratic, following  Appell \& Goursat \cite[p.293]{Appell}:  take $(\xi_0,\eta_0)$ on the cubic
$\eta^2=P_3(\xi)$ and intersect with the line $\eta-\eta_0 = t(\xi-\xi_0)$. For a given
$t$, the equation for $\xi$ is $[\eta_0+t(\xi-\xi_0)]^2-P_3(\xi)=0$ solved by 
$\xi
=\xi_0$ and the two roots of $\displaystyle 2\eta_0 t +t^2(\xi-\xi_0)
 -\dfrac{P_3(\xi)-P_3(\xi_0)}{\xi-\xi_0}=0$:

$$  -(P_3'''/6)(\xi-\xi_0)^2 +(t^2-P_3''(\xi_0)/2)(\xi-\xi_0) +2\eta_0 t -P_3'(\xi_0)=0,$$
$$ \xi= \xi_0 +\dfrac{ t^2-P_3''(\xi_0)/2 \pm \sqrt{ \mathcal{Q}(t):=(t^2-P_3''(\xi_0)/2)^2+2P_3'''
   (2\eta_0 t-P_3'(\xi_0)) /3  }  }{P_3'''/6}
$$
The four zeros of the quartic $\mathcal{Q}$ are the
slopes of lines leading to a double intersection, i.e., of the four tangents to the cubic issued from the point $(\xi_0,\eta_0)$.
We show that the cross ratio of these four tangents is independent of the
point $(\xi_0,\eta_0)$ of the cubic curve: let $(\xi_1,\eta_1)$ be another point
on the cubic curve, then
$ \mathcal{Q}_1(t)= (t^2-P_3''(\xi_1)/2)^2+2P_3''' (2\eta_1 t-P_3'(\xi_1)) /3$   
An interesting point is $\xi\to\infty$, then, $P_3''(\xi)\sim P_3''' \xi,
P_3'(\xi)\sim P_3''' \xi^2/2$, $\eta\sim (P_3''' \xi^3/6)^{1/2}$, and
$\mathcal{Q}_\infty(t\sqrt{6P_3'''\xi}) \sim (P_3''')^2\xi^2[ (6t^2 -1/2)^2 +2(2 t-1/2)/3]$.

\section{Proofs of equivalence. } 
\renewcommand{\mylmark}{Equivalence proofs 1 to 2}


\subsection{1. and 2.}\   \\




\subsubsection{From 1. to 2.}: as seen in \eqref{resul} in \S~\ref{ss2},
 p.~\pageref{ss2}, the elimination of $y_n$ from  \eqref{xsump} yields
 a relation of degree 2 in 
$x_n+x_{n-1}$ and $x_n x_{n-1}$, which is \eqref{Ebiq}. 

But one can also find \eqref{Esom} directly from \eqref{ysump} and \eqref{xsump}.
Indeed,  the sum of $x_n+x_{n-1}$ and $x_{n+1}+x_n$ from \eqref{xsump}
yields $x_{n-1}+2 x_n+x_{n+1}$ as $-Y_1(y_n)/Y_2(y_n)- Y_1(y_{n+1})/Y_2(y_{n+1})$,
which is a symmetric rational function of $y_n$ and $y_{n+1}$, therefore, from
\eqref{ysump}, a rational function of $x_n$, allowing to recover \eqref{Esom},
although the final rational function seems liable to be  of fourth degree,
but wait.

A expansion in simple fractions leads to $\displaystyle x_n+x_{n-1} =
-\dfrac{Y_1(y_n)}{Y_2(y_n)} =-\dfrac{c_{1,2}}{c_{2,2}}
-\dfrac{Y_1(u)/(u-v)}{c_{2,2}(y_n-u)} -\dfrac{Y_1(v)/(v-u)}{c_{2,2}(y_n-v)}, 
$
where $u$ and $v$ are the two roots of $Y_2(y)=0$ (ordinates of the
horizontal asymptotes of the curve $F(x,y)=0$). Then,
\begin{multline*}
x_{n-1}+2 x_n+x_{n+1} \\
 = -2\dfrac{c_{1,2}}{c_{2,2}}
     -\dfrac{Y_1(u)}{c_{2,2}(u-v)} \;\dfrac{y_n+y_{n+1}-2u}
            {y_n y_{n+1} -(y_n+y_{n+1})u +u^2}
     -\dfrac{Y_1(v)}{c_{2,2}(v-u)} \;\dfrac{y_n+y_{n+1}-2v}
            {y_n y_{n+1} -(y_n+y_{n+1})v +v^2} \\
 = -2\dfrac{c_{1,2}}{c_{2,2}}
     +\dfrac{Y_1(u)}{c_{2,2}(u-v)} \dfrac{X_1(x_n)+2u X_2(x_n) }
          {X_0(x_n)+X_1(x_n)u+X_2(x_n)u^2=F(x_n,u) }  \\
     +\dfrac{Y_1(v)}{c_{2,2}(v-u)} \dfrac{X_1(x_n)+2v X_2(x_n) }
          {X_0(x_n)+X_1(x_n)v+X_2(x_n)v^2=F(x_n,v) }
\end{multline*}
of small degree, as $F(x_n,u)=Y_0(u)+Y_1(u)x_n$ and
$F(x_n,v)=Y_0(v)+Y_1(v)x_n$. The numerators are $\partial F(x_n,y)/\partial y=
Y'_0(y)+Y'_1(y)x_n+Y'_2(y)x_n^2$ at $y=u$ and $y=v$, so
\begin{multline*}
x_{n-1}+x_{n+1} \\ 
= -2\dfrac{c_{1,2}}{c_{2,2}}
     +\dfrac{1}{c_{2,2}(u-v)}\left[
  \dfrac{ Y'_0(u)+Y'_1(u)x_n+Y'_2(u)x_n^2}{x_n+Y_0(u)/Y_1(u)}
 - \dfrac{ Y'_0(v)+Y'_1(v)x_n+Y'_2(v)x_n^2}{x_n+Y_0(v)/Y_1(v)}\right]-2x_n
\end{multline*}

which must reduce to
\begin{subequations}
\begin{equation}\label{recsom} 
x_{n-1}+x_{n+1} = -\dfrac{d_{0,1} +(2d_{0,2}+d_{1,1}) x_n +d_{1,2} x_n^2}
                       {d_{2,2} [x_n+Y_0(u)/Y_1(u)][x_n+Y_0(v)/Y_1(v)]}
\end{equation}
Remark that, as $Y'_2(y)=2c_{2,2}y+c_{2,1}$,
the $-2x_n$ term cancels with the contribution of 
$(Y'_2(u)-Y'_2(v))/(c_{2,2}(u-v))$.

The equation \eqref{recsom}
 is a \emph{\textbf{recurrence relation}} joining $x_{n-1}, x_n$, and $x_{n+1}$. Remark that $\rho_1$ and $\rho_2$, the two roots of $d_{2,2}\rho^2+d_{1,2}\rho+d_{0,2}=0$ must be $-Y_0(u)/Y_1(u)$ and $-Y_0(v)/Y_1(v)$.


Also, $\displaystyle x_n x_{n-1} =
\dfrac{Y_0(y_n)}{Y_2(y_n)} =\dfrac{c_{0,2}}{c_{2,2}}
+\dfrac{Y_0(u)/(u-v)}{c_{2,2}(y_n-u)} +\dfrac{Y_0(v)/(v-u)}{c_{2,2}(y_n-v)}, 
$

\begin{multline*}
(u-v)^2[c_{2,2}x_n x_{n-1}-c_{0,2}] [c_{2,2}x_n x_{n+1}-c_{0,2}] \\
 = \dfrac{Y_0^2(u)}{ (y_n-u)(y_{n+1}-u)} -\dfrac{Y_0(u)Y_0(v)}{(y_n-u)(y_{n+1}-v)} 
-\dfrac{Y_0(u)Y_0(v)}{(y_{n+1}-u)(y_n-v)} +\dfrac{Y_0^2(v)}{(y_n-v)(y_{n+1}-v)} 
\\
 \hspace*{-44pt}= \dfrac{Y_0^2(u)(y_n-v)(y_{n+1}-v)
  -Y_0(u)Y_0(v)[(y_{n+1}-u)(y_n-v)+(y_n-u)(y_{n+1}-v)]+Y_0^2(v)(y_n-u)(y_{n+1}-u) }
 {(y_n-u)(y_{n+1}-u)(y_n-v)(y_{n+1}-v)} \\
=X_2(x_n) \dfrac{ Y_0^2(u)F(x_n,v)
   -2Y_0(u)Y_0(v)[X_0(x_n)+(u+v)X_1(x_n)+uvX_2(x_n)]+ Y_0^2(v)F(x_n,u)}
 {F(x_n,u)F(x_n,v)=[Y_0(u)+Y_1(u)x_n][Y_0(v)+Y_1(v)x_n]}
\end{multline*}

\smallskip

\begin{multline*}
x_{n-1} x_{n+1} \\ 
 =\dfrac{X_2(x_n)}{(u-v)^2c_{2,2}^2 x_n^2}\; \dfrac{ Y_0^2(u)F(x_n,v)
   -2Y_0(u)Y_0(v)[X_0(x_n)+(u+v)X_1(x_n)+uvX_2(x_n)]+ Y_0^2(v)F(x_n,u)}
 {[Y_0(u)+Y_1(u)x_n][Y_0(v)+Y_1(v)x_n]} \\
\hspace*{-33pt}+ \dfrac{c_{0,2}}{c_{2,2} x_n}\; 
\dfrac{\sigma_0 +\sigma_1 x_n +\sigma_2 x_n^2}
                       { [x_n+Y_0(u)/Y_1(u)][x_n+Y_0(v)/Y_1(v)]}
-\dfrac{c_{0,2}^2}{c_{2,2}^2 x_n^2}
\end{multline*}
which must also reduce to
\begin{equation}\label{recprod} 
x_{n-1}x_{n+1} = \dfrac{d_{0,0} +d_{0,1} x_n +d_{0,2} x_n^2}
                       {d_{2,2} [x_n+Y_0(u)/Y_1(u)][x_n+Y_0(v)/Y_1(v)]},
\end{equation}
but it does not seem useful to follow this track any longer.
\end{subequations}

\renewcommand{\mylmark}{Equivalence proofs 2 to 1}

\subsubsection{From 2. to 1.}: 
From a sequence satisfying \eqref{Ebiq},  let us construct a valid sequence $\{y_n\}$.
We must construct $F(x,y)=\alpha (y-u)(y-v)x^2 +[\beta (y-u)(y-v)+\beta' y+\beta'']x
+\gamma (y-u)(y-v)+\gamma' y+\gamma''$ such that the resultant above is exactly \eqref{Ebiq}. 

With $S=x_n+x_{n+1}$ and $\Pi=x_nx_{n+1}$, \eqref{Ebiq} is
\begin{equation} \label{Rconic}
d_{0,0} +d_{0,1}S + d_{0,2}S^2   +d_{1,1}\Pi +d_{1,2}\Pi S +d_{2,2}\Pi^2=0.
\end{equation}
We see that \eqref{resul} holds if $(S,\Pi)= (-\beta/\alpha, \gamma/\alpha)$,  so,
 one chooses $(-\beta/\alpha,\gamma/\alpha)=(S,\Pi)$ as a point\footnote{The
 importance of this conic
has been stressed by A.Ronveaux \cite{Ronv}.} on the
conic \eqref{Rconic}. We now have
$$  \alpha^2E=d_{0,2}(\alpha S+\beta-\beta)^2 +d_{1,2}(\alpha \Pi-\gamma+\gamma) (\alpha S+\beta-\beta) +d_{2,2}(\alpha \Pi-\gamma+\gamma)^2
     +d_{0,1}\alpha(\alpha S+\beta-\beta)$$ 
 $$+d_{1,1}\alpha(\alpha \Pi-\gamma+\gamma)+d_{0,0}\alpha^2
=d_{0,2}(\alpha S+\beta)^2-2d_{0,2}\beta(\alpha S+\beta)+d_{0,2}\beta^2 $$
$$+d_{1,2}(\alpha \Pi-\gamma) (\alpha S+\beta)-d_{1,2}\beta(\alpha \Pi-\gamma) +d_{1,2}\gamma(\alpha S+\beta)-d_{1,2}\beta\gamma +d_{2,2}(\alpha \Pi-\gamma)^2 $$ $$
+2d_{2,2}\gamma(\alpha \Pi-\gamma) +d_{2,2}\gamma^2
     +d_{0,1}\alpha(\alpha S+\beta)-d_{0,1}\alpha\beta+d_{1,1}\alpha(\alpha \Pi-\gamma)+d_{1,1}\alpha\gamma+\alpha^2d_{0,0}
$$
$$ = d_{0,2}(\alpha S+\beta)^2 +d_{1,2}(\alpha \Pi-\gamma) (\alpha S+\beta) +d_{2,2}(\alpha \Pi-\gamma)^2
     +d'_{0,1}(\alpha S+\beta)+d'_{1,1}(\alpha \Pi-\gamma)=0,
$$
\begin{equation} \label{Rconicbg} \text{if \  } d_{0,2}\beta^2 -d_{1,2}\beta\gamma +d_{2,2}\gamma^2
     -d_{0,1}\alpha\beta+d_{1,1}\alpha\gamma+\alpha^2d_{0,0}=0,\end{equation}
with $d'_{0,1}=d_{0,1}\alpha-2d_{0,2}\beta +d_{1,2}\gamma$ and
$d'_{1,1}= d_{1,1}\alpha-d_{1,2}\beta +2 d_{2,2}\gamma $.
Let $\rho_1$ and $\rho_2$ be the two roots of $d_{2,2}\rho^2+d_{1,2}\rho+d_{0,2}=0$,
then, $d_{2,2}[\alpha\Pi-\gamma -\rho_1(\alpha S+\beta)][\alpha\Pi-\gamma -\rho_2(\alpha S+\beta)]
        +d'_{0,1}(\alpha S+\beta)+d'_{1,1}(\alpha\Pi-\gamma)=0$, and we 
 multiply by some constant $C$ and
compare with \eqref{resul}:


%\begin{multline} \label{resul}
%[(\gamma' v+\gamma'')(\alpha S+\beta)+(\beta' v+\beta'')(\alpha\Pi-\gamma)]
%[(\gamma' u+\gamma'')(\alpha S+\beta)+(\beta' u+\beta'')(\alpha\Pi-\gamma)]\\
%+( \beta''\gamma'-\beta'\gamma'')[\gamma' (\alpha S+\beta)+\beta' (\alpha\Pi-\gamma)]=0.
%\end{multline}

$$\beta' u+\beta'' = C_1,\qquad  \gamma' u+\gamma'' = -\rho_1 C_1 , $$ 
$$\beta' v+\beta'' = C_2 , \qquad \gamma' v+\gamma'' = -\rho_2 C_2 , $$ 
$$  (\beta''\gamma'-\beta'\gamma'')\beta'= C d'_{1,1},\qquad %
     (\beta''\gamma'-\beta'\gamma'')\gamma' = Cd'_{0,1},
$$ 
where $C_1C_2=Cd_{2,2}$,

so, $\displaystyle \dfrac{\gamma' u+\gamma''}{\beta' u+\beta''}$ and
$\displaystyle \dfrac{\gamma' v+\gamma''}{\beta' v+\beta''}$ are
$-\rho_1$ and $-\rho_2$ . We subtract $\gamma'/\beta'=d'_{0,1}/d'_{1,1}$:\\
$\displaystyle  \beta' u+\beta'' =\dfrac{\beta''\gamma'-\beta' \gamma''}{\beta' (\rho_1 + d'_{0,1}/d'_{1,1})},\quad  \beta' v+\beta'' = 
\dfrac{\beta''\gamma'-\beta' \gamma''}{\beta' (\rho_2 + d'_{0,1}/d'_{1,1})},
\quad\dfrac{\beta' u+\beta'' }{\beta' v+\beta'' } = \dfrac{d'_{1,1}\rho_2+d'_{0,1}}
{d'_{1,1}\rho_1+d'_{0,1}}
$  

$$
\beta'' = \beta' \dfrac{d'_{1,1}(u\rho_1-v\rho_2) +d'_{0,1}(u-v)}{d'_{1,1}(\rho_2-\rho_1)}
$$
$$
\gamma'' = \gamma' \dfrac{d'_{1,1}(v-u)\rho_1\rho_2 +d'_{0,1}(v\rho_1-u\rho_2)}{d'_{0,1}(\rho_2-\rho_1)}
$$
$$
\beta''\gamma'-\beta'\gamma'' = \dfrac{\beta'\gamma' (u-v)(d'_{1,1}\rho_1+d'_{0,1})
(d'_{1,1}\rho_2+d'_{0,1}) }{d'_{1,1}d'_{0,1}(\rho_2-\rho_1) }
$$
$$
\beta' u+\beta''= \dfrac{\gamma' (u-v)
(d'_{1,1}\rho_2+d'_{0,1}) }{d'_{0,1}(\rho_2-\rho_1) },\quad%
   \beta' v+\beta'' = \dfrac{\gamma' (u-v)(d'_{1,1}\rho_1+d'_{0,1})
}{d'_{0,1}(\rho_2-\rho_1) }
$$
$$
C= (\beta' u+\beta'')(\beta' v+\beta'')/d_{2,2} 
=\dfrac{1}{d_{2,2}}\dfrac{\gamma' (u-v)
(d'_{1,1}\rho_2+d'_{0,1}) }{d'_{0,1}(\rho_2-\rho_1)} \dfrac{\gamma' (u-v)(d'_{1,1}\rho_1+d'_{0,1})
}{d'_{0,1}(\rho_2-\rho_1) }
$$

or $\displaystyle
 \dfrac{\beta'}{d'_{1,1}}=\dfrac{\gamma'}{d'_{0,1}} =
\dfrac{C}{\beta''\gamma'-\beta'\gamma''}=
     \dfrac{u-v}{d_{2,2} (\rho_2-\rho_1)}  $

The degrees of freedom are therefore $u$, $v$, and $(\beta,\gamma)/\alpha$ on
the conic $R=0$ of \eqref{Rconic}.


\begin{equation}\label{Fnew}\begin{split}
F(x,y)&= \alpha(y-u)(y-v)x^2+[\beta(y-u)(y-v)+\beta' y+\beta'']x+\gamma(y-u)(y-v)+\gamma' y+\gamma'' \\
&= (y-u)(y-v)(\alpha x^2+\beta x+\gamma) +\dfrac{u-v}{(\rho_2-\rho_1) d_{2,2} }\left[
   d'_{1,1}x\left(y+\dfrac{\beta''}{\beta'}\right)
   +d'_{0,1} \left(y+\dfrac{\gamma''}{\gamma'}\right)
 \right]\\
= (y-u)(y-v)&(\alpha x^2+\beta x+\gamma) +\dfrac{u-v}{(\rho_2-\rho_1) d_{2,2} }\left[
  d'_{1,1}xy + \dfrac{d'_{1,1}(u\rho_1-v\rho_2) +d'_{0,1}(u-v)}{\rho_2-\rho_1}x  \right.  \\
& \qquad\hspace*{99pt} %
\left. +d'_{0,1}y+\dfrac{d'_{1,1}(v-u)\rho_1\rho_2 +d'_{0,1}(v\rho_1-u\rho_2)}{\rho_2-\rho_1}
\right]\\
= (y-u)(y-v)&(\alpha x^2+\beta x+\gamma) +\dfrac{u-v}{(\rho_2-\rho_1)^2 d_{2,2} }[
  d'_{1,1}[\rho_2(y-v)-\rho_1(y-u)]x +d'_{0,1}(u-v)x    \\
& \qquad\hspace*{99pt} %
 +d'_{0,1}[\rho_2(y-u)-\rho_1(y-v)]+d'_{1,1}\rho_1\rho_2(v-u)]
\end{split}\end{equation}

Where is is recalled that $d'_{0,1}=d_{0,1}\alpha-2d_{0,2}\beta +d_{1,2}\gamma$ and
$d'_{1,1}= d_{1,1}\alpha-d_{1,2}\beta +2 d_{2,2}\gamma $.

The two numbers $u$ and $v$ are the values of $y$ making $x$ infinite, we write
\eqref{Fnew} (divided by $y-v$) as a function of
$\displaystyle Y=\dfrac{y-u}{y-v}$, using $\displaystyle y=\dfrac{vY-u}{Y-1}$ and
$v-u = y-u-(y-v)$:
$$
\dfrac{Y}{Y-1}(\alpha x^2+\beta x+\gamma) -\dfrac{1}{(\rho_2-\rho_1)^2 d_{2,2} }[
  d'_{1,1}[\rho_2-\rho_1 Y]x +d'_{0,1}(1-Y)x    
 +d'_{0,1}[\rho_2 Y-\rho_1]+d'_{1,1}\rho_1\rho_2(Y-1)]=0
$$
Now, $u$ and $v$ have disappeared, they are completely hidden in $Y$,
very satisfactory.\\
$
(d'_{1,1}\rho_1 +d'_{0,1})(x-\rho_2)Y^2
+[(\rho_2-\rho_1)^2 d_{2,2} (\alpha x^2+\beta x+\gamma)
-(d'_{1,1}\rho_2 +d'_{0,1})(x-\rho_1)-(d'_{1,1}\rho_1 +d'_{0,1})(x-\rho_2)]Y 
+(d'_{1,1}\rho_2 +d'_{0,1})(x-\rho_1)=0
$\\
$Y=\displaystyle \dfrac{-(\rho_2-\rho_1)^2 d_{2,2} (\alpha x^2+\beta x+\gamma)
+(d'_{1,1}\rho_2 +d'_{0,1})(x-\rho_1)+(d'_{1,1}\rho_1 +d'_{0,1})(x-\rho_2)
\pm\sqrt{P(x)} }{2(d'_{1,1}\rho_1 +d'_{0,1})(x-\rho_2)}
$

with $P(x)=[(\rho_2-\rho_1)^2 d_{2,2} (\alpha x^2+\beta x+\gamma)
-(d'_{1,1}\rho_2 +d'_{0,1})(x-\rho_1)-(d'_{1,1}\rho_1 +d'_{0,1})(x-\rho_2)]^2
-(d'_{1,1}\rho_1 +d'_{0,1})(d'_{1,1}\rho_2 +d'_{0,1})(x-\rho_1)(x-\rho_2)$

With the example of p.\pageref{Exasintheta},
$\alpha^2 E(x,y)= \sin^2\theta\; x^2y^2 -x^2-y^2 +2\cos\theta\;xy
=\sin^2\theta \Pi^2 -S^2 +2(1+\cos\theta)\Pi
=[\sin\theta\, xy +x+y][\sin\theta\, xy-(x+y)]+2(1+\cos\theta)xy:
$
$\alpha=\sin\theta,  d_{2,2}=1, d_{0,2}=-1/\sin^2\theta, d_{1,1}= 2(1+\cos\theta)/\sin^2\theta, \rho_1,\rho_2=\mp 1/\sin\theta,
d'_{1,1}=2(1+\cos\theta+\gamma\sin\theta)/\sin\theta,
d'_{0,1}=2\beta/\sin^2\theta$.

Then, $F(x,y)= (y-u)(y-v)(\sin\theta \,x^2+\beta x+\gamma) +((u-v)\sin\theta)/2)\{ 
[d'_{1,1} x+d'_{0,1}](y-(u+v)/2)
 +(u-v)[d'_{0,1}x\sin\theta +d'_{1,1}/\sin\theta]/2\}
$, with
$\alpha^2\gamma^2 -\beta^2 +2(1+\cos\theta)\alpha\gamma
=(1+\cos\theta+\gamma\sin\theta)^2-(1+\cos\theta)^2-\beta^2=0$.

Also, 
\begin{equation*}\begin{split}
&\dfrac{2(Y-1)F}{(u-v)(y-v)}\\&= -2Y(\sin\theta x^2+\beta x+\gamma)
  +(Y-1)\sin\theta\{ [d'_{1,1} x+d'_{0,1}](Y+1)
 +[d'_{0,1}x\sin\theta +d'_{1,1}/\sin\theta](1-Y)\}/2\\
&=-2Y(\sin\theta x^2+\beta x+\gamma)
  +\{ [d'_{1,1}\sin\theta\; x+d'_{0,1}\sin\theta](Y^2-1)
 -[d'_{0,1}x\sin^2\theta +d'_{1,1}](Y-1)^2\}/2\\
&=[d'_{1,1}\sin\theta-d'_{0,1}\sin^2\theta](x-1/\sin\theta)Y^2/2
-[2\sin\theta\; x^2 +(2\beta -d'_{0,1}\sin^2\theta)x+2\gamma -d'_{1,1}]Y\\
&\ \ \ \ \ \ \ \ \ \ \ \ -[d'_{1,1}\sin\theta+d'_{0,1}\sin^2\theta](x+1/\sin\theta)/2\\
&=[1+\cos\theta-\beta+\gamma\sin\theta](x-1/\sin\theta)Y^2
-2[\sin\theta\; x^2  -(1+\cos\theta)/\sin\theta]Y\\
 &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  -[1+\cos\theta+\beta+\gamma\sin\theta](x+1/\sin\theta)
\end{split}\end{equation*}

$\displaystyle Y=\dfrac{ \sin\theta\; x^2  -(1+\cos\theta)/\sin\theta
\pm x\sin\theta\sqrt{x^2-1}   }{  [1+\cos\theta-\beta+\gamma\sin\theta](x-1/\sin\theta) }
=\dfrac{ (1+\cos\theta)[-x\cos\theta\pm\sqrt{x^2-1}][x\pm \sqrt{x^2-1}]   }{ \sin\theta [1+\cos\theta-\beta+\gamma\sin\theta](x-1/\sin\theta) }
$\\
$\ \ =\displaystyle \dfrac{(1+\cos\theta)[(1-\cos\theta)R^2-1-\cos\theta]}{\sin\theta[1+\cos\theta-\beta+\gamma\sin\theta]
(R+1/R-2/\sin\theta)}$ where $R=x\pm\sqrt{x^2-1}$, so that $x=(R+1/R)/2$. Remark that
$\displaystyle \dfrac{2}{\sin\theta}= \dfrac{1+\cos\theta}{\sin\theta} +\dfrac{1-\cos\theta}{\sin\theta} =
\dfrac{1+\cos\theta}{\sin\theta} +\dfrac{\sin\theta}{1+\cos\theta}\ \text{and}\ 
\dfrac{1+\cos\theta}{1-\cos\theta}=\left( \dfrac{1+\cos\theta}{\sin\theta}\right)^2=\cot^2(\theta/2) $,
$\displaystyle Y=\dfrac{ \sin\theta\; (R+\cot(\theta/2))}{[1+\cos\theta-\beta+\gamma\sin\theta](1-R^{-1}\tan(\theta/2))}.$

 
If $x=1/\sin(n\theta)$, $R=(1\pm \cos(n\theta))/\sin(n\theta),
R^{-1}=(1\mp \cos(n\theta))/\sin(n\theta),$ \\ $\displaystyle
\hspace*{-55pt}Y=\dfrac{ \sin\theta\; (1\pm\cos(n\theta)+\cot(\theta/2)\sin(n\theta))}{[1+\cos\theta-\beta+\gamma\sin\theta]
(\sin(n\theta)-(1\mp\cos(n\theta))\tan(\theta/2))}$
$\displaystyle
=\dfrac{ 2\cos^2(\theta/2)\; (\sin(\theta/2)+\sin((n\pm 1/2)\theta))}{[1+\cos\theta-\beta+\gamma\sin\theta]
(\sin((n\pm 1/2)\theta)-\cos(\theta/2))}$


Note that 
\begin{equation*}\begin{split}
E(x,y) &= d_{0,0} +d_{0,1}(x+y) +d_{0,2}(x+y)^2 +d_{1,1}xy +d_{1,2}xy(x+y)
+d_{2,2}x^2y^2\\
&= (d_{0,2}+d_{1,2}x+d_{2,2}x^2)y^2+(d_{0,1}+(2d_{0,2}+d_{1,1})x+d_{1,2}x^2)y
+d_{0,0}+d_{0,1}x+d_{0,2}x^2\\
&= d_{2,2}(x-\rho_1)(x-\rho_2)y^2 +(d_{0,1}+(2d_{0,2}+d_{1,1})x+d_{1,2}x^2)y
+d_{0,0}+d_{0,1}x+d_{0,2}x^2
\end{split}\end{equation*}

$\alpha^2E(x,y)$ is

$d_{2,2}[\alpha xy-\gamma -\rho_1(\alpha(x+y)+\beta)][\alpha xy-\gamma -\rho_2(\alpha(x+y)+\beta)]
        +[\underbrace{-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha}_{d'_{0,1}}] (\alpha(x+y)+\beta)+
[\underbrace{-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha}_{d'_{1,1}}](\alpha xy-\gamma)$



\subsection{ Half-integer indexes  (  
{\red new}  
%new
in 2011)\  {\black\  }}  \   \\

Spiridonov and Zhedanov have a very ingenious argument \cite[p.~298]{SpZ2007} amounting to
build a possible $y-$ sequence as half-integer entries of the $x-$sequence:
$y_n=x_{n-1/2}$.
They rely on elliptic functions identities, but
here is an elementary derivation: can we find a quadratic algebraic function
$r\pm \sqrt{s}$, with $r$ and $s$ rational functions, such that two
iterations of this function to $x_n$ gives $x_{n+1}$? The result must be
a quadratic algebraic functions whose two determinations are $x_{n-1}$ and
$x_{n+1}$, as explained in \eqref{symdirect}. Similarly, the two choices
of the sign of the square root in $r(x_n)\pm \sqrt{s(x_n)}$ must yield
$x_{n-1/2}$  and $x_{n+1/2}$. A new application of $r\pm\sqrt{s}$ represents
a new half-step either forward or backward, so that the four determinations
$r(r(x_n)+\sigma_1 \sqrt{s(x_n)}) +\sigma_2 \sqrt{s( r(x_n)+\sigma_1 \sqrt{s(x_n)}) }$
with independent signs $\sigma_1$ and $\sigma_2$, must be $x_{n+1}$, $x_{n-1}$, AND
$x_n$ twice. 

$$ \prod_{\sigma_1=\pm 1} \prod_{\sigma_2=\pm 1} (x-r(r(x_n)+\sigma_1 \sqrt{s(x_n)}) -\sigma_2 \sqrt{s( r(x_n)+\sigma_1 \sqrt{s(x_n)}) } \equiv (x-x_{n-1})(x-x_{n+1})(x-x_n)^2$$
$$[[x-r(r(x_n)+ \sqrt{s(x_n)})]^2-s( r(x_n)+ \sqrt{s(x_n)}) ][[x-r(r(x_n)- \sqrt{s(x_n)})]^2
 -s( r(x_n)- \sqrt{s(x_n)}) ]$$ 
$$ = x^4 -2[r(r(x_n)+ \sqrt{s(x_n)})+r(r(x_n)- \sqrt{s(x_n)})]x^3
+[r^2(r(x_n)+ \sqrt{s(x_n)})$$
$$+4r(r(x_n)+ \sqrt{s(x_n)})r(r(x_n)- \sqrt{s(x_n)})+r^2(r(x_n)- \sqrt{s(x_n)})
+s(r(x_n)+ \sqrt{s(x_n)})+s(r(x_n)- \sqrt{s(x_n)})]x^2+\dotsb$$
$$ \equiv
\left( x^2 + \dfrac{d_{0,1}+(2d_{0,2}+d_{1,1})x_n+d_{1,2}x_n^2}
                       {d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2} x
+  \dfrac{d_{0,0}+d_{0,1}x_n+d_{0,2}x_n^2}
                       {d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2}\right)(x-x_n)^2
\text{\ from\ }\eqref{Esom}$$\\
$\displaystyle  \equiv  \dfrac{E(x,x_n)}{d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2}(x-x_n)^2$.  \  \  ?


 Example: $x_n=\cos(n\theta) \Rightarrow x_{n\pm 1/2} = \cos(\theta/2)\cos(n\theta)
\mp \sin(\theta/2)\sin(n\theta)$: $r(x)=x\cos(\theta/2)$  and $s(x)=(1-x^2)\sin^2(\theta/2)$, 
$s(r(x_n)\pm\sqrt{s(x_n)})=[1-x_n^2\cos^2(\theta/2)\mp 2 x_n\cos(\theta/2)\sin(\theta/2)
\sqrt{1-x_n^2}-(1-x_n^2)\sin^2(\theta/2)]\sin^2(\theta/2)
=[x_n\sin(\theta/2)-\cos(\theta/2)\,\sqrt{1-x_n^2}]^2\sin^2(\theta/2)$ etc. So, the rational function
$s$ must contain $P$ and $s(r\pm\sqrt{s})$ must be the square of an
algebraic expression containing $\sqrt{P}$.

From $x_{n+1} = x_n+x_{n+1} -x_n = $ sum of roots of $F(x,y_{n+1})=0$ minus $x_n=
\displaystyle -\dfrac{Y_1(y_{n+1})}{Y_2(y_{n+1})}-x_n \\=
-\dfrac{  Y_1\left( \dfrac{-X_1(x_n)+\sqrt{P(x_n)} }{2X_2(x_n) }  \right)   }
      {  Y_2\left( \dfrac{-X_1(x_n)+\sqrt{P(x_n)} }{2X_2(x_n) } \right)    }-x_n
$ which must be the same as \\
$\displaystyle \dfrac{ -(d_{0,1}+(2d_{0,2}+d_{1,1})x_n+d_{1,2}x_n^2) +\text{\ const.\ }\sqrt{P(x_n)} }
                    {d_{0,2}+d_{1,2}x_n+d_{2,2}x_n^2=  d_{2,2}(x_n-\rho_1)(x_n-\rho_2)}
$   Seems to lead nowhere.



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Try something else. A half step forward from $x_n$ must be the same than a half step backward from $x_{n+1}$:\\
$F(x_n,x_{n+1/2})=0$, as well as $F(x_{n+1},x_{n+1/2})=0$, holding for half-integer
indexes as well, $F(x_{n+1/2},x_n)=F(x_{n+1/2},x_{n+1})=0$, amounting to
$F$ to be \emph{\textbf{symmetric}}, so, in \eqref{Fnew}, $u$ and $v$ are the two
roots of $\alpha x^2+\beta x +\gamma=0$, whence $E(u,v)=E(v,u)=0$, as $-\beta/\alpha$
and $\gamma/\alpha$ are the sum $S$ and the product $\Pi$ of the coordinates
of a point of the symmetric quartic \eqref{Ebiq} in the form \eqref{Rconic}. The equality of the coefficients of
$x$ and $y$ in \eqref{Fnew} yields
$ \displaystyle   \dfrac{d'_{1,1}(u\rho_1-v\rho_2) +d'_{0,1}(u-v)}{\rho_2-\rho_1} =d'_{0,1},\\
\text{or \ }
d'_{1,1}\dfrac{u+v}{2}+d'_{0,1} = (u-v)\dfrac{d'_{1,1}(\rho_2+\rho_1)/2+d'_{0,1}}{\rho_2-\rho_1}
=(u-v)\dfrac{-d'_{1,1}d_{1,2}/(2d_{2,2})+d'_{0,1}}{\rho_2-\rho_1}
$,\\

\begin{equation} \label{halfstep1}
\dfrac{d'_{1,1}\dfrac{u+v}{2}+d'_{0,1} }{u-v} =
\dfrac{d'_{1,1}\dfrac{\rho_2+\rho_1}{2}+d'_{0,1}}{\rho_2-\rho_1}
\end{equation}

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Funny symmetry between the two pairs $(u,v)$ and $(\rho_1,\rho_2)$ \dots
but do we have $E(\rho_1,\rho_2)=0$???
Of course no, $\rho_1$ and $\rho_2$ are the abscissae and ordinates of the
asymptotes of the curve $E=0$.

Now, \eqref{halfstep1} is 

\noindent$\displaystyle
\hspace*{-175pt}\dfrac{
[-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha]\dfrac{-\beta}{2\alpha}
-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha}{u-v}=
\dfrac{ [-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha]\dfrac{-d_{1,2}}{2d_{2,2}}
-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha}{\rho_2-\rho_1}$\\
$\displaystyle \hspace*{-175pt}
\dfrac{
[-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha]\dfrac{-\beta}{2\alpha}
-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha}{u-v}=
\dfrac{ [-d_{1,2}\beta  +d_{1,1}\alpha]\dfrac{-d_{1,2}}{2d_{2,2}}
-2d_{0,2}\beta +d_{0,1}\alpha}{\rho_2-\rho_1}
$


as
$d'_{0,1}=d_{0,1}\alpha-2d_{0,2}\beta +d_{1,2}\gamma$ and
$d'_{1,1}= d_{1,1}\alpha-d_{1,2}\beta +2 d_{2,2}\gamma $.

And as
 $\rho_1$ and $\rho_2$ be the two roots of $d_{2,2}\rho^2+d_{1,2}\rho+d_{0,2}=0$, remark that $(\rho_2-\rho_1)^2 = (d^2_{1,2}-4d_{2,2}d_{0,2})/d^2_{2,2}$.

$\displaystyle
\dfrac{
[-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha]\dfrac{-\beta}{2\alpha}
-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha}{u-v}=
\dfrac{ \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]\alpha
+(\rho_2-\rho_1)^2d_{2,2}\beta/2}{\rho_2-\rho_1}
$


With $u+v=-\beta/\alpha$ and $uv=\gamma/\alpha$,\\
$\displaystyle
\hspace*{-44pt}\dfrac{
[d_{1,2}(u+v) +2 d_{2,2}uv +d_{1,1}](u+v)
-4d_{0,2}(u+v) +2d_{1,2}uv+2d_{0,1}}{2(u-v)}=
\dfrac{ d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}
-(\rho_2-\rho_1)^2d_{2,2}(u+v)/2}{\rho_2-\rho_1},
$
a second equation for $(u,v)$, together with $E(u,v)=0$.





With the example above,  $E(u,v)= -\sin^2\theta\; u^2v^2 +u^2+v^2 -2\cos\theta\;uv=0$, 
 $d'_{0,1}=d_{0,1}+2d_{0,2}(u+v) +d_{1,2}uv=2(u+v)$, 
$d'_{1,1}= d_{1,1}+d_{1,2}(u+v) +2 d_{2,2}uv=-2(1+\cos\theta)-2\sin^2\theta\;uv =-4\cos^2(\theta/2)[1+2\sin^2(\theta/2)\;uv]$,\\
$F(x,y)= \displaystyle
(y-u)(y-v)( x^2-(u+v) x+uv) +\dfrac{u-v}{(\rho_2-\rho_1) d_{2,2}=-2\sin\theta }$\\
$\displaystyle\left[
  d'_{1,1}xy + \dfrac{d'_{1,1}(u\rho_1-v\rho_2) +d'_{0,1}(u-v)}{\rho_2-\rho_1}x  
    +d'_{0,1}y+\dfrac{d'_{1,1}(v-u)\rho_1\rho_2 +d'_{0,1}(v\rho_1-u\rho_2)}{\rho_2-\rho_1}
\right]$\\ $\displaystyle
=(y-u)(y-v)(x-u)(x-v) +\dfrac{u-v}{-2\sin\theta }$ \\
$\displaystyle\left[
  -2[1+\cos\theta+\sin^2\theta\;uv]xy + \dfrac{2[1+\cos\theta+\sin^2\theta\;uv](u+v)/\sin\theta +2(u+v)(u-v)}{2/\sin\theta}x  
    +2(u+v)y \right. $\\  $\displaystyle \left.
+\dfrac{2[1+\cos\theta+\sin^2\theta\;uv](v-u)/\sin^2\theta -2(u+v)^2/\sin\theta}{2/\sin\theta}
\right]$\\
 $\displaystyle
=(y-u)(y-v)(x-u)(x-v) +\dfrac{u-v}{-2\sin\theta }$\\
$\hspace*{-26pt}\left\{
  [1+\cos\theta+\sin^2\theta\;uv][-2xy + (u+v)x  +(v-u)/\sin\theta] +(u+v)(u-v)x\sin\theta   
    +2(u+v)y \displaystyle 
-(u+v)^2\right\}$

\noindent and 
$[1+\cos\theta+\sin^2\theta\;uv](u+v) +(u+v)(u-v)\sin\theta =  
    2(u+v) 
$
so, 
\\ \textbf{1.} either $u+v=0$, $u-v\neq 0$, and $E(u,-u)= -\sin^2\theta\;u^4+2(1+\cos\theta)u^2= 4\cos^2(\theta/2)
[-\sin^2(\theta/2)u^2+1]u^2=0: u=-v=1/\sin(\theta/2), F(x,y)= \displaystyle
(y^2-u^2)(x^2-u^2) $\\$+\dfrac{2u}{-2\sin\theta }%$\\
%$\hspace*{-76pt}
\left\{
  -2[1+\cos\theta-\sin^2\theta\;u^2]xy 
+2[1+\cos\theta-\sin^2\theta\;u^2]/(\sin(\theta/2)\sin\theta) \right\}$\\
$\displaystyle =(y^2-1/\sin^2(\theta/2))(x^2-1/\sin^2(\theta/2)) -\dfrac{
  [\overbrace{1+\cos\theta-\sin^2\theta\;/\sin^2(\theta/2)}^{
\displaystyle -2\cos^2(\theta/2) }][2/(\sin(\theta/2)\sin\theta)-2xy] } 
{-\sin\theta \sin(\theta/2)}\\
\hspace*{-55pt}= -\dfrac{1}{\sin^2(\theta/2)}
\left[ -\sin^2(\theta/2)x^2y^2 +x^2+y^2 -
\underbrace{\dfrac{4\sin^2(\theta/2)\cos^2(\theta/2)}{\sin\theta\sin(\theta/2)}}_{\displaystyle 2\cos(\theta/2)}xy
\underbrace{-\dfrac{1}{\sin^2(\theta/2)}+\dfrac{4\sin^2(\theta/2)\cos^2(\theta/2)}{\sin^2\theta\sin^2(\theta/2)}}_{\displaystyle =0}\right]
$
which is $E(xy)$ with parameter $\theta/2$ instead of $\theta$!
\\ \textbf{2.} or $\sin^2\theta\; uv +\cos\theta-1 = (v-u)\sin\theta$, hyperbola with asymptotes
$u=1/\sin\theta$ and $v=-1/\sin\theta$, part of the asymptotes of $E(u,v)=0$, is
that meaningful? One puts
$\displaystyle v=\dfrac{ \cos\theta-1+u\sin\theta}{\sin\theta(1-u\sin\theta)}$
 in $E(u,v)=-\sin^2\theta\;u^2v^2+u^2+v^2-2\cos\theta\;uv=0:\displaystyle
\dfrac{(1+u\sin\theta)(\cos\theta-1+u\sin\theta)^2}{\sin^2\theta(1-u\sin\theta)}
+u^2-2\dfrac{u\cos\theta(\cos\theta-1+u\sin\theta)}{
\sin\theta(1-u\sin\theta)}=
\dfrac{  (\cos\theta-1)^2}{\sin^2\theta}=0
$,
so what??

Could the TWO sides of \eqref{halfstep1} vanish simultaneously?
Alas, no: with $E(x_n,x_{n+1})=
x_{n+1}^2x_n^2 + 0.1x_{n+1}x_n(x_n+x_{n+1}) - (x_{n+1}^2+x_n^2) +  2x_{n+1}x_n + 0.25(x_n + x_{n+1}) + 1$,
one finds
\begin{tabular}{|c|c|c|c|}  \hline  
 $\beta/\alpha$    &   $\gamma/\alpha$    &   left side   &   right side  \\ \hline
-0.50000 & -0.22900 &0.11586 & -0.47566 \\
-0.40000  &-0.24788 &0.12493 & -0.37553 \\
-0.30000 & -0.26137 &0.14072 & -0.27541 \\
-0.20000 & -0.26928 &0.16199 & -0.17528 \\
-0.10000 & -0.27150 &0.18741 & -0.07516 \\ 
-0.024938          & -0.26938             &   0.20839     &    0          \\
0     & -0.26795  & 0.21560  &  0.02497 \\
0.10000 & -0.25862& 0.24511 & 0.12509 \\
0.20000  & -0.24355& 0.27442 & 0.22522 \\
0.26814            & -0.23004             & 0.29344        &  0.29344 \\
0.30000 &-0.22284 &0.30190 & 0.32534 \\
0.40000 &-0.19663 &0.32582 & 0.42547 \\
0.50000 &-0.16513 &0.34432 & 0.52559 \\
\hline
\end{tabular}

then, at equality of the two sides, $u= 0.36394 $,  $v= -0.63208$  ($\rho_1=-1.0512,  \rho_2= 0.95125$), and

$F(x,y)=x^2y^2 + 0.26814xy(x+y) -0.23004(x^2+y^2)  + 1.8193 xy + 0.31796 (x+y)  + 0.98238 $
 
\smallskip

It seems that we have to do it the hard way. From the $R-$conic \eqref{Rconic}, $\gamma$
is a root of an equation of degree 2


$ d_{2,2}\gamma^2 -(d_{1,2}\beta -d_{1,1}\alpha)\gamma +d_{0,2}\beta^2 -d_{0,1}\alpha\beta +d_{0,0}\alpha^2=0$

and also the square of \eqref{halfstep1}, using $(u-v)^2=(\beta^2-4\alpha\gamma)/\alpha^2$:

$\displaystyle
\left[[-d_{1,2}\beta +2 d_{2,2}\gamma +d_{1,1}\alpha]\dfrac{-\beta}{2\alpha}
-2d_{0,2}\beta +d_{1,2}\gamma+d_{0,1}\alpha\right]^2$ 

 $\displaystyle  \qquad=
\left[\dfrac{ \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]\alpha
+(\rho_2-\rho_1)^2d_{2,2}\beta/2}{\rho_2-\rho_1}\right]^2  (\beta^2-4\alpha\gamma)/\alpha^2
$

\noindent\hspace*{-33pt}$(d_{1,2}-d_{2,2}\beta/\alpha)^2\gamma^2 +2 (d_{1,2}-d_{2,2}\beta/\alpha)[d_{0,1}\alpha
-(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/\alpha]\gamma  
+[d_{0,1}\alpha
-(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/\alpha]^2
$ \\
$\displaystyle =\left[
     \dfrac{ \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]^2\alpha^2   }
           {  (d^2_{1,2}-4d_{2,2}d_{0,2})/d^2_{2,2}                             }
    +  \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]\alpha\beta d_{2,2}   
     +  (d^2_{1,2}-4d_{2,2}d_{0,2})\beta^2/4   \right]         (\beta^2-4\alpha\gamma)/\alpha^2
$

after elimination of $\gamma^2$:

$ (d_{1,2}-d_{2,2}\beta/\alpha)^2 [(d_{1,2}\beta -d_{1,1}\alpha)\gamma -d_{0,2}\beta^2 +d_{0,1}\alpha\beta -d_{0,0}\alpha^2]/d_{2,2}$\\
$ +2 (d_{1,2}-d_{2,2}\beta/\alpha)[d_{0,1}\alpha
-(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/(2\alpha)]\gamma  
+[d_{0,1}\alpha
-(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/(2\alpha)]^2
$ \\
$\displaystyle =\left[
     \dfrac{ \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]^2\alpha^2   }
           {  (d^2_{1,2}-4d_{2,2}d_{0,2})/d^2_{2,2}                             }
    +  \left[d_{0,1} -\dfrac{d_{1,1}d_{1,2}}{2d_{2,2}}\right]\alpha\beta d_{2,2}   
     +  (d^2_{1,2}-4d_{2,2}d_{0,2})\beta^2/4   \right]         (\beta^2-4\alpha\gamma)/\alpha^2
$

$\displaystyle \left\{  (d_{1,2}-d_{2,2}\beta/\alpha)
    [(d_{1,2}-d_{2,2}\beta/\alpha) (d_{1,2}\beta -d_{1,1}\alpha)/d_{2,2}
   +2d_{0,1}\alpha
-2(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/\alpha ]  \vphantom{  \dfrac{Y}{Y}  } \right. \\  \left.
                    +4\dfrac{ \left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]^2\alpha   }
           {  d^2_{1,2}-4d_{2,2}d_{0,2}                                               }
    +  4\left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]\beta    
     +  (d^2_{1,2}-4d_{2,2}d_{0,2})\beta^2/\alpha \right\} \gamma  \\
= (d_{1,2}-d_{2,2}\beta/\alpha)^2  
     (  d_{0,2}\beta^2 -d_{0,1}\alpha\beta +d_{0,0}\alpha^2)/d_{2,2} 
   -[d_{0,1}\alpha
-(2d_{0,2}+d_{1,1}/2)\beta+d_{1,2}\beta^2/(2\alpha)]^2
 +\dfrac{ \left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]^2\beta^2   }
           {  d^2_{1,2}-4d_{2,2}d_{0,2}                                               }
    +  \left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]\beta^3/\alpha    
     +  (d^2_{1,2}-4d_{2,2}d_{0,2})\beta^4/(4\alpha^2) 
$ \\
$\displaystyle \hspace*{-33pt}\left\{  [ d_{1,2}^3/d_{2,2} -4d_{1,2}d_{0,2} -d_{1,1}d_{1,2}
     +2d_{0,1}d_{2,2} ]\beta
+\left[    4\dfrac{ \left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]^2   }
           {  d^2_{1,2}-4d_{2,2}d_{0,2}                                               }
   -d_{1,2}^2 d_{1,1}/d_{2,2}+2d_{0,1}d_{1,2} \right]\alpha
\right\} \gamma  \\
= \left[ d_{1,2}^2 d_{0,2}/d_{2,2}+2d_{1,2}d_{0,1}+d_{2,2}d_{0,0}-d_{0,1}d_{2,2}
    -(2d_{0,2}+d_{1,1}/2)^2 
    +\dfrac{ \left[d_{0,1}d_{2,2} -d_{1,1}d_{1,2}/2\right]^2   }
           {  d^2_{1,2}-4d_{2,2}d_{0,2}                                        } \right]\beta^2
    +  \left[ -2d_{1,2}d_{0,0} -d_{1,2}^2 d_{0,1}/d_{0,0}
          +2d_{0,1} (2d_{0,2}+d_{1,1}/2)\right]\alpha\beta
   +[d_{1,2}^2d_{0,0}/d_{2,2} -d_{0,1}^2]\alpha^2
$ \\

Amazing  (13 Feb. 2011), it reduces to $\displaystyle \gamma= \dfrac{\text{degree\  2}}{\text{degree\ 1}}$!

Putting this formula for $\gamma$ in $E(u,v)=  d_{2,2}\gamma^2 -(d_{1,2}\beta -d_{1,1}\alpha)\gamma +d_{0,2}\beta^2 -d_{0,1}\alpha\beta +d_{0,0}\alpha^2=0$ amounts to a polynomial equation of degree 4 for $\beta/\alpha$. With the
example $E(u,v)=
u^2v^2  + 0.1 uv(u+v) - (u^2+v^2) +  2uv + 0.25(u+v) + 1= (\gamma/\alpha)^2
 -0.1 \beta\gamma/\alpha^2 -\beta^2/\gamma^2 +4\gamma/\alpha-0.25\beta/\gamma+1$,
one finds the four roots
\begin{tabular}{|c|c|c|c|c|c|}  \hline  
 $\beta/\alpha$    &   $\gamma/\alpha$    &  $u$ & $v$ &   left side   &   right side  \\ \hline
  -2.2172          &  -4.9068          & 3.5857 & -1.3685 &  -2.1950  &  -2.1950  \\
  -0.18928         & -0.26979          & -0.43333& 0.62261&  -0.16454 & -0.16454  \\ 
  0.0035158        &  -3.7319          & -1.9336 & 1.9301 & 0.028489 &  0.028489  \\
 0.26814          & -0.23004           & 0.36394 & -0.63208 &0.29344 &  0.29344 \\  \hline
\end{tabular}

each one leading to a valid $F$: \\
$[x^2 -2.2172 x -4.9068 ]y^2 -[2.2172 x^2 + 8.9183 x + 0.68660 ]y -4.9068 x^2 -0.68660 x + 1.2428 $\\ 
$[x^2 -0.18928 x -0.26979]y^2-[0.18928 x^2 + 1.7989 x -0.13308]y -0.26979 x^2 + 0.13308 x + 1.0303$\\
\hspace*{-66pt}$[x^2 +0.0035158 x -3.7319]y^2+[0.0035158 x^2 + 6.6839 x + 0.21100]y -3.7319 x^2 + 0.21100 x + 1.0101 $\\ 
$[x^2 +0.26814 x -0.23004]y^2+[0.26814 x^2 + 1.8193 x + 0.31796 ]y -0.23004 x^2 + 0.31796 x + 0.98238 $ 

For a given sequence $\{\cdots, x_{-1}, x_0, x_1,\cdots\}$ satisfying $E(x_n,x_{n\pm 1})=0$, we find four valid sequences $\{\cdots, x_{-3/2}, x_{-1}, x_{-1/2}, x_0, x_{1/2}, x_1,\cdots\}$
we solve $F(x_0,y)=0$ in $y$ and evaluate  $\displaystyle
 -\dfrac{Y_1(y)}{Y_2(y)}-x_0$: from \eqref{xsump}, the result must be $x_1$ (and
$y$ is then $x_{1/2}$) or $x_{-1}$ (and $y$ is then $x_{-1/2}$).  One then proceeds
with repeated use of \eqref{ysump} and \eqref{xsump}.

With the example above, starting with $x_0=0$:

\begin{tabular}{|c|c|c|c|c|c|c|}  \hline  
 $x_{-1}$    &   $x_{-1/2}$    &  $x_0$ & $x_{1/2}$ &  $x_1$ &  $x_{3/2}$ & $x_2$  \\ \hline
1.1328     & -0.57808  & 0   & 0.43815 & -0.88278 &  2.0770  &  -4.6529 \\
1.1328    & 2.2163 &   0    & -1.7231  &  -0.88278 &  -0.60265 & -4.6529 \\
1.1328     & 0.54929  &  0 & -0.49275   & -0.88278  & -1.4312 & -4.6529 \\
1.1328    & -1.4879   & 0   & 2.8701 &   -0.88278   & 0.58236  &  -4.6529   \\  \hline
\end{tabular}

 \medskip

Is it a surprise that there are 4 interpolants $\{\cdots, x_{n-1/2},x_n,x_{n+1/2},x_{n+1},\cdots\}$
to a given elliptic sequence $\{x_n\}$? If we accept that
$x_n=\mathcal{E}(an+b)$ with a periodic function of period
$p$ and integer $n$, then $\mathcal{E}(an+b+a/2)$ AND $\mathcal{E}(an+b+a/2+p/2)$
are both valid instances of $x_{n+1/2}$, as a new translation of $a/2$
(resp. $a/2+p/2$), will indeed yield $x_{n+1}$. And if the function $\mathcal{E}$
has TWO periods, say $p$ and $p'$, then there are four possible half-step
translations which are: $a/2, a/2+p/2, a/2+p'/2$, AND $a/2+(p+p')/2$.

\smallskip

So, elementary recovery of half-integer indexes gives a hint that
two-periodic functions will be somehow involved. Honestly, I knew it.

   
\medskip

More on halving-doubling step in \S~\ref{duplica}, p.~\pageref{duplica}

\medskip

Same experiment
with the form \eqref{Ecanon1} $E(x,y)= 
 k^2(1-k^2 z''^2) x^2 y^2 -k^2(1-z''^2)(x+y)^2 
+ 2k(z''-k)(1+kz'')xy +1-k^2z''^2 =0$,
$d'_{0,1}=d_{0,1}\alpha-2d_{0,2}\beta +d_{1,2}\gamma=2k^2(1-z''^2)\beta
$ and
$d'_{1,1}= d_{1,1}\alpha-d_{1,2}\beta +2 d_{2,2}\gamma=2k(z''-k)(1+kz'')\alpha+2k^2(1-k^2z''^2)\gamma $, $\rho_1, \rho_2 = \mp \sqrt{(1-z''^2)/(1-k^2z''^2)}$, \eqref{halfstep1} is

\noindent$\displaystyle \dfrac{[2k(z''-k)(1+kz'')\alpha+2k^2(1-k^2z''^2)\gamma ](-\beta/2)+2k^2(1-z''^2)\alpha\beta
}{u-v} =
\dfrac{2k^2(1-z''^2)\alpha\beta}{\rho_2-\rho_1}$, or\\
$\displaystyle \dfrac{[ -(z''-k)(1+kz'')+2k(1-z''^2)]\alpha\beta-k(1-k^2z''^2)\beta\gamma 
}{\alpha^{-1}\sqrt{\beta^2-4\alpha\gamma}} =\pm
k\sqrt{(1-z''^2)(1-k^2z''^2)}\alpha\beta$. 

Again, $\beta=0$ is an obvious solution! But we do it by the book: squaring and
eliminating $\gamma^2$ through $k^2(1-k^2 z''^2) \gamma^2  -k^2(1-z''^2)\beta^2 
+ 2k(z''-k)(1+kz'')\alpha\gamma +(1-k^2z''^2)\alpha^2 =0$,\\
$\displaystyle [ -(z''-k)(1+kz'') +2k(1-z''^2)]^2\alpha^2\beta^2 
 -2k(1-k^2z''^2)[-(z''-k)(1+kz'')+2k(1-z''^2)]\alpha\beta^2\gamma
   +(1-k^2z''^2)\beta^2[ k^2(1-z''^2)\beta^2 
- 2k(z''-k)(1+kz'')\alpha\gamma -(1-k^2z''^2)\alpha^2]= 
(\beta^2-4\alpha\gamma) 
k^2(1-z''^2)(1-k^2z''^2)\beta^2$. Amazing (again): a lot of terms cancel, and we
have $\beta=0$ as only solution!
Then, $E=k^2(1-k^2 z''^2) \gamma^2 
+ 2k(z''-k)(1+kz'')\alpha\gamma +(1-k^2z''^2)\alpha^2 =
(1-k^2z''^2)\displaystyle \left( k^2 \gamma^2 +2k\dfrac{z''-k}{1-kz''}
\alpha\gamma+\alpha^2\right)=0 \Rightarrow$ an equation for $uv=-u^2=\gamma/\alpha = \displaystyle k^{-1}\dfrac{ -z''+k\pm \sqrt{(1-k^2)(z''^2-1)} }{1-kz''}$, 
and $F$ is
from \eqref{Fnew}, $\displaystyle 
 (y^2-\gamma/\alpha)(x^2-\gamma/\alpha) +\dfrac{u-v}{(\rho_2-\rho_1)^2 d_{2,2} }[
  d'_{1,1}[\rho_2(y-v)-\rho_1(y-u)]x +d'_{0,1}(u-v)x +d'_{0,1}[\rho_2(y-u)-\rho_1(y-v)]+d'_{1,1}\rho_1\rho_2(v-u)]
=(y^2-\gamma/\alpha)(x^2-\gamma/\alpha) +\dfrac{2ud'_{1,1}[2\rho_2xy+2u\rho_2^2]}{4\rho_2^2 d_{2,2} }\\
=(y^2-\gamma/\alpha)(x^2-\gamma/\alpha) +\dfrac{d'_{1,1}\left[\pm \sqrt{
  -\dfrac{\gamma (1-z''^2)}{\alpha(1-k^2z''^2)}}xy-\dfrac{\gamma (1-z''^2)}{\alpha(1-k^2z''^2)}\right]}{((1-z''^2)/(1-k^2z''^2)) k^2(1-k^2z''^2) }
  $

now, from \eqref{zpzs}
$z''=\displaystyle -\dfrac{1-2k^2z'^2+k^2z'^4}{k(1-2z'^2+k^2z'^4)},
   z''-k=\displaystyle \dfrac{-1-k^2+4k^2z'^2-k^2(1+k^2)z'^4}{k(1-2z'^2+k^2z'^4)},\\
  z''-k^{-1}=\displaystyle \dfrac{-2+2(1+k^2)z'^2-2k^2z'^4}{k(1-2z'^2+k^2z'^4)},
 z''^2-1=\displaystyle (1-k^2)\dfrac{(1-k^2z'^4)^2}{k^2(1-2z'^2+k^2z'^4)^2},\\
  1-k^2z''^2= \displaystyle 4\dfrac{(1-k^2)z'^2(1-k^2z'^2)(1-z'^2)}{(1-2z'^2+k^2z'^4)^2},\\
  \dfrac{\gamma}{\alpha}=uv=\displaystyle
 \dfrac{(1+k^2)(1+k^2z'^4)\pm (1-k^2)(1-k^2z'^4) -4k^2z'^2}{2k^2(1-(1+k^2)z'^2+k^2z'^4)}
 =\dfrac{(1-k^2z'^2)^2  \text{\  and \  } k^2(1-z'^2)^2}{k^2(1-k^2z'^2)(1-z'^2)}
 =\dfrac{1}{k}\left(\dfrac{1-k^2z'^2}{k(1-z'^2)}\right)^{\pm 1} ,
\dfrac{\gamma (1-z''^2)}{\alpha(1-k^2z''^2)}
=-\dfrac{(1-k^2z'^4)^2}{k^2z'^2(1-z'^2)^2} \text{\  and\  }
-\dfrac{ (1-k^2z'^4)^2}{z'^2(1-k^2z'^2)^2},\\
d'_{1,1}= 2k\alpha[(z''-k)(1+kz'')+k(1-k^2z''^2)\gamma/\alpha]  \\= 
\displaystyle -2k\alpha\dfrac{2(1-k^2)z'^2}{1-2z'^2+k^2z'^4}
\left[ \displaystyle \dfrac{-1-k^2+4k^2z'^2-k^2(1+k^2)z'^4}{k(1-2z'^2+k^2z'^4)}
  +\displaystyle \dfrac{-2+2(1+k^2)z'^2-2k^2z'^4}{1-2z'^2+k^2z'^4}\dfrac{\gamma}{\alpha}
\right],
$,




\medskip



\picskip{0}

\subsection{(1. or 2.) and 3.  {\black  }} \  \\

\renewcommand{\mylmark}{Equivalence proofs 3 to 2}

\subsubsection{From 3. to 2. and 1.}

Let $  P(z)= c(z-z_1)(z-z_2)(z-z_3)(z-z_4) $,
whose square root has the Taylor expansion about $z_0$
$
\sqrt{P(z)} = \gamma +\delta (x-z_0) +\dotsb
$
which is matched by the two first Taylor  coefficients of $S_n(z)$, so that only
one unknown remains in $S_n$:
$$
S_n(x)=\gamma+\delta(x-z_0)+\xi_n(x-z_0)^2.  \    
$$

We will need further coefficients: $P(x)=\gamma^2 +2\gamma\delta (x-z_0) +P''(z_0)(x-z_0)^2/2
+P^{(3)}(z_0)(x-z_0)^3/6+P^{(4)}(z_0)(x-z_0)^4/24\Rightarrow$

$$ \sqrt{P(x)}=\gamma+\delta(x-z_0) +\underbrace{\dfrac{P''(z_0)/2-\delta^2}{2\gamma}}_{\displaystyle\epsilon} (x-z_0)^2
+\underbrace{\dfrac{P'''(z_0)/6-\dfrac{\delta}{\gamma}\left(
    \overbrace{P''(z_0)/2-\delta^2}^{\displaystyle 2\gamma\epsilon}\right)}{2\gamma}}_{\displaystyle\eta} (x-z_0)^3+\dotsb
$$

$\displaystyle \dfrac{S_n(x)+\sqrt{P(x)}}{Z_{n+1}(x)}=
\dfrac{1-\beta_n(x-z_0)}{\alpha_n}+O((x-z_0)^2)\Rightarrow  \alpha_n = \dfrac{\zeta_{n+1}(z_0-x_{n+1})}{2\gamma},
\beta_n= -\dfrac{\delta}{\gamma} + \dfrac{1}{z_0-x_{n+1}}
$, and
$\displaystyle \dfrac{S_n(z)-\sqrt{P(z)} }{Z_n(z)} =
    \dfrac{\alpha_n(x-z_0)^2 }{1-\beta_n (x-z_0) +O((x-z_0)^2) }
\Rightarrow \alpha_n = \dfrac{\xi_n-\epsilon}{\zeta_n(z_0-x_n)}, \beta_n=
-\dfrac{1}{z_0-x_n}-\dfrac{\eta}{\xi_n-\epsilon}$, so

$$\dfrac{1}{(z_0-x_n)(z_0-x_{n+1})} = \dfrac{\zeta_n\zeta_{n+1}}{2\gamma(\xi_n
-\epsilon)},\qquad
\dfrac{1}{z_0-x_n}+\dfrac{1}{z_0-x_{n+1}} = \dfrac{\delta}{\gamma}- \dfrac{\eta}{\xi_n-\epsilon}$$


We will only need $Z_n(x)=$ const. $(x-x_n)$ and $W_n(x)=Z_{n+1}(x)=$
const. $(x-x_{n+1})$.

We now expand $S_n^2(x)-P(x)=(x-z_0)^2W_n(x)Z_n(x)=$ const.
$(x-z_0)^2(x-x_n)(x-x_{n+1})=$ const. $[(x-z_0)^4 +(2z_0-x_n-x_{n+1})(x-z_0)^3
+(z_0-x_n)(z_0-x_{n+1})(x-z_0)^2]$:

$\xi_n^2 -P^{(4)}(z_0)/24$ is the constant $(=\zeta_n\zeta_{n+1})$,
 the coeff. of $(x-z_0)^4$,
\begin{equation}\label{SPxi}\begin{split}
 2\xi_n\delta -P^{(3)}(z_0)/6 =2(\xi_n\delta-\delta\epsilon-\gamma\eta)&= [\xi_n^2-P^{(4)}(0)/24](2z_0-x_n-x_{n+1}),\\
\delta^2+2\gamma \xi_n -P''(z_0)/2=2\gamma(\xi_n-\epsilon)&= [\xi_n^2-P^{(4)}(z_0)/24](z_0-x_n)(z_0-x_{n+1}),
\end{split}\end{equation}
from which $x_n+x_{n+1}$ and $x_n x_{n+1}$ are rational functions of $\xi_n$.

$x_n$ and $x_{n+1}$ are the roots of 
$(\xi_n^2-P^{(4)}(0)/24)(z_0-x)^2 -2(\xi_n\delta-\delta\epsilon-\gamma\eta)(z_0-x)
+2\gamma(\xi_n-\epsilon)=0$. Do we have $\xi_n=y_{n+1}$? Then, $\xi_{n-1}+\xi_n$ must be
the rational function $[2\delta(z_0-x_n)-2\gamma]/(z_0-x_n)^2$. 
From \eqref{recSZ}, $\xi_n+\xi_{n-1}=$ coefficient of $z^2$ of $S_n(z)+S_{n-1}(z)=
[1-\beta_{n-1}(z-z_0)]Z_n(z)/\alpha_{n-1}$, so  
$\xi_n+\xi_{n-1}=-\beta_{n-1}\zeta_n/[\zeta_n(z_0-x_n)/(2\gamma)]=
-2\beta_{n-1}\gamma/(z_0-x_n)=2[\delta - \gamma/(z_0-x_n) ]/(z_0-x_n)$ OK. \\
So $F(x,y)= (y^2-P^{(4)}(0)/24)(z_0-x)^2 -2(y\delta-\delta\epsilon-\gamma\eta)(z_0-x)
+2\gamma(y-\epsilon)$.

The elimination of $\xi_n$ yields a symmetric algebraic relation between $x_n$
and $x_{n+1}$: let $S=2z_0-x_n-x_{n+1}$ and $\Pi=(z_0-x_n)(z_0-x_{n+1})$,

$$ \xi_n=\dfrac{  \dfrac{P'''(z_0)/6}{S}+\dfrac{\delta^2-P''(z_0)/2=-2\gamma\epsilon}{\Pi}  }
               { \dfrac{2\delta}{S}-\dfrac{2\gamma}{\Pi}  },
2\delta\xi_n-P'''(z_0)/6= \dfrac{2\gamma^2\eta S}{\delta\Pi-\gamma S},
$$
$$
\dfrac{2\gamma^2\eta S}{\delta\Pi-\gamma S}= S\left[\dfrac{P'''(z_0)\Pi/6 -2\gamma\epsilon S}
  {2\delta\Pi-2\gamma S}\right]^2-S P^{(4)}(z_0)/24,
$$
$$
8\gamma^2 \eta (\delta\Pi-\gamma S)= [P'''(z_0)\Pi/6 -2\gamma\epsilon S]^2
-P^{(4)}(z_0)[\delta\Pi-\gamma S]^2/6,
$$
\begin{equation} \label{biqPS}
[(P'''(z_0)/6)^2-P^{(4)}(z_0)/6]\Pi^2 -\gamma[2\epsilon P'''(z_0)-\delta P^{(4)}(z_0)]\Pi S/3
+\gamma^2[4\epsilon^2-P^{(4)}(z_0)/6]S^2
-8\gamma^2\eta(\delta\Pi-\gamma S)=0.
\end{equation}

\bigskip
\renewcommand{\mylmark}{Equivalence proofs 3 to 1}

\subsubsection{From 3. to 1.}
There is also a direct way to construct a valid $y-$sequence, simply
from $y_{n+1}=\xi_n$ in \eqref{SPxi}:
\begin{equation}\label{Fxi}
F(x,y) = (y^2-c)(x-z_0)^2 + (2\delta y -P^{(3)}(z_0)/6)(x-z_0) +\delta^2+2\gamma y -P''(z_0)/2.
\end{equation}
From \eqref{xsump} and \eqref{SPxi},
the two $x-$roots of $F(x,y)=0$ at $y=y_{n+1}=\xi_n$ must indeed be $x_n$ and
$x_{n+1}$.
  
As a function of $y$, $X_2(x)y^2+X_1(x)y+X_0(x)$:
\begin{equation}\label{Fxi2}
F(x,y) = (x-z_0)^2y^2  + 2(\delta (x-z_0) +\gamma) y  
-c(x-z_0)^2  -(P^{(3)}(z_0)/6)(x-z_0) +\delta^2-P''(z_0)/2.
\end{equation}

Check that $X_1^2-4X_0X_2=$ const. $P$: indeed, $X_1(x)^2-4X_0(x)X_2(x)=
 4c(x-z_0)^4  +4(P^{(3)}(z_0)/6)(x-z_0)^3 +4 P''(z_0)/2(z-z_0)^2 +8\gamma\delta(x-z_0)
+4\gamma^2=4P(x)$.

Of special interest is $Q(y)=Y_1(y)^2-4Y_0(y)Y_2(y)$ from \eqref{Fxi}:
$$
Q(y)=[2\delta y -P^{(3)}(z_0)/6)]^2 -4(y^2-c)[\delta^2+2\gamma y -P''(z_0)/2]
$$
which is only of third degree
$$
-8\gamma y^3 +2P''(z_0) y^2 +4[2c+P^{(3)}(z_0)/6]y +[P^{(3)}(z_0)/6]^2
  +4c[\delta^2 -P''(z_0)/2],
$$
and happens to be basically a cubic resolvent \cite{Bew,Tignol} of
 $P(x)=c(x-z_1)\dotsb (x-z_4)$!
Indeed, with $\delta=P'(z_0)/(2\gamma)$ and $\gamma^2=P(z_0)$, we find\\
$\displaystyle
-\dfrac{Q(y)}{\gamma^4} = $\\
$$\left(\dfrac{2 y}{\gamma}\right)^3
-\dfrac{P''(z_0)}{2P(z_0)} \left(\dfrac{2 y}{\gamma}\right)^2
-\left( \dfrac{4c}{P(z_0)} -\dfrac{P'(z_0)P^{(3)}(z_0) }{6P^2(z_0)} \right)
   \dfrac{2 y}{\gamma}
 -\left( \dfrac{P^{(3)}(z_0) }{6P(z_0)}\right)^2 
  -\dfrac{c}{P(z_0)}\left(\left(\dfrac{P'(z_0)}{P(z_0)}\right)^2
        -2\dfrac{P''(z_0)}{P(z_0)}\right)
$$
whose coefficients depend only on the four values
$\rho_j:=1/(z_0-z_j)$, $j=1,\dotsc,4$, as $P'(z_0)/P(z_0)=\sum_1^4 \rho_j$;
$P''(z_0)/(2P(z_0))$ is the sum of the 6 products
$\rho_i \rho_j$ for $1\leqslant i<j\leqslant 4$;
$P^{(3)}(z_0)/(6P(z_0))$ is the sum of the 4 possible
products $\rho_i \rho_j \rho_k$, $1\leqslant i<j<k\leqslant 4$;
and $c/P(z_0)=\rho_1 \rho_2 \rho_3 \rho_4$.

Then, the coefficient of $(2y/\gamma)^2$ 
is $(-1)$ times the sum of the 6 products $\rho_i \rho_j$
above; the coefficient of $2y/\gamma$ is the sum of the 12
products  $\rho_i \rho_j \rho_k \rho_\ell$ with
one repeated factor; and the last coefficient  
is $(-1)$ times the sum of the four terms of the
form $(\rho_i \rho_j \rho_k)^2$ and the sum of the four terms
$\rho_i^3 \rho_j \rho_k\rho_\ell$.

This amounts to the zeros of $Q$ as
$$
\dfrac{2y}{\gamma} = \rho_1\rho_2+\rho_3\rho_4,
\  \rho_1\rho_3+\rho_2\rho_4,\  \text{and\ }
\rho_1\rho_4+\rho_2\rho_3,
$$
and $\infty$, if $Q$ is to be considered a fourth-degree polynomial.

Check:
\begin{equation} \label{checkratio} 
\dfrac{\rho_1\rho_2+\rho_3\rho_4 -(\rho_1\rho_3+\rho_2\rho_4)}  
{\rho_1\rho_2+\rho_3\rho_4 - (\rho_1\rho_4+\rho_2\rho_3)}=
\dfrac{ \hphantom{XX} \dfrac{z_3-z_2}{z_3-z_1} \hphantom{XX} }
         {\dfrac{z_4-z_2}{z_4-z_1}  }.
\end{equation}

The general $F$ is $\displaystyle (1+C y)^2F\left(x, \dfrac{A+B y}{1+C y}\right)$,
with the $F$ of \eqref{Fxi}, and the new $Q$ is \\
$\displaystyle (1+C y)^4 Q\left(\dfrac{A+B y}{1+C y}\right)$, whose
four zeros are given by  
$\displaystyle \dfrac{A+B y}{1+C y} = \dfrac{\gamma}{2}(\rho_j \rho_4+
  \rho_k \rho_\ell)$, and $y=-1/C$.

The choice of $A,B$, and $C$ ensuring that $Q=$ constant $P$ is such that 
$\displaystyle \dfrac{A+B z_j}{1+C z_j} = \dfrac{\gamma}{2}(\rho_j \rho_4+
  \rho_k \rho_\ell)$, $j=1,2,3$, and $z_4=-1/C$, 
in order to keep the cross-ratio in \eqref{checkratio}, so
$\displaystyle A+B z_j =  \dfrac{\gamma}{2z_4}
   (\rho_j \rho_4+
  \rho_k \rho_\ell)\left(\dfrac{1}{\rho_j}-\dfrac{1}{\rho_4}\right), j=1,2,3$, 
$\displaystyle A+Bz_1 = A+Bz_0 -\dfrac{B}{\rho_1} =
    \dfrac{\gamma}{2z_4} (\rho_1 \rho_4+
                         \rho_2 \rho_3)\left(\dfrac{1}{\rho_1}-\dfrac{1}{\rho_4}\right),
   A+Bz_0 -\dfrac{B}{\rho_2} =
    \dfrac{\gamma}{2z_4} (\rho_2 \rho_4+
                         \rho_1 \rho_3)\left(\dfrac{1}{\rho_2}-\dfrac{1}{\rho_4}\right),
  A+Bz_0 -\dfrac{B}{\rho_3} =
    \dfrac{\gamma}{2z_4} (\rho_3 \rho_4+
                         \rho_1 \rho_2)\left(\dfrac{1}{\rho_3}-\dfrac{1}{\rho_4}\right),
$
then, $\displaystyle A+B z_0 = \gamma (-\rho_1-\rho_2-\rho_3+\rho_4)/(2z_4)
\  \text{and}\   B=\gamma(-\rho_1\rho_2 -\rho_1\rho_3 -\rho_2\rho_3 
   +\rho_1\rho_2\rho_3/\rho_4)/(2z_4)$.

\medskip

For the $F$ of \eqref{Fcanon}, 
$\displaystyle x_n+x_{n+1} = \dfrac{ 2(1-k^2)z_0 y_{n+1} }
                                 {k[(1-k^2 z_0^2)y_{n+1}^2 -1+z_0^2]},
    x_n x_{n+1} = \dfrac{ 1-k^2 z_0^2 -k^2(1-z_0^2) y_{n+1}^2 }
                                 {k^2[(1-k^2 z_0^2)y_{n+1}^2 -1+z_0^2]},$

\begin{multline} \label{Ecanon}
E(x_n, x_{n+1})= \dfrac{(1-k^2 z_0^4)^2}{4z_0^2} (x_n+ x_{n+1})^2
  - k^2 (1-z_0^2)(1-k^2 z_0^2)(x_n x_{n+1} )^2  \\
  -[(1+k^2)(1+k^2 z_0^4)-4k^2 z_0^2]x_n x_{n+1} 
   -1-k^2z_0^4 +(1+k^2)z_0^2
\end{multline}

Fixed points $x_n=x_{n+1}$ satisfy 
$(1-z_0^2)(1-k^2 z_0^2)(x^2-z_0^2)(k^2 x^2 -z_0^{-2} )=0$.
  
\renewcommand{\mylmark}{Equivalence proofs 2 to 3}

\subsubsection{From 2. to 3.} 

Can we find $P$, $z_0$,etc. from \eqref{Ebiq}? If $z_0=\infty$, $d_{2,2}=0$;
when $z_0=0$, $d_{0,0}=0$. As $S=\Pi=0$ satisfies \eqref{biqPS}, so does
$x=y=z_0$, i.e., $z_0$ is one of the four roots of

$$ d_{2,2} z_0^4 +2d_{1,2} z_0^3 +(4d_{0,2}+d_{1,1})z_0^2 +2d_{0,1}z_0+d_{0,0}=0. $$

Then, with $x=z_0-(z_0-x_n)$ and $y=z_0-(z_0-x_{n+1})$,
$x+y=2z_0-S$, $xy=z_0^2-z_0 S+\Pi$, and \eqref{Ebiq} becomes

%\begin{multline}\label{Ebiq}
 %\hspace*{-12pt}E(x_n,x_{n+1})=d_{0,0} +d_{0,1}(x_n+x_{n+1}) + d_{0,2}(x_n+x_{n+1})^2
%  +d_{1,1}x_n x_{n+1} \\
% +d_{1,2}x_n x_{n+1}(x_n+x_{n+1}) +d_{2,2}x_n^2 x_{n+1}^2=0.
%\end{multline}


\begin{multline*}
d_{2,2}\Pi^2 -(2d_{2,2}z_0 +d_{1,2})S\Pi+
(d_{2,2} z_0^2+(d_{1,2}+d_{1,1})z_0 +d_{0,2})S^2 +
(2d_{2,2}z_0^2+2(d_{1,2}+d_{1,1}) z_0)\Pi \\
 -(2d_{2,2}z_0^3 +(3d_{1,2}+2d_{1,1})z_0^2+4d_{0,2}Cz_0 +d_{0,1})S=0,
\end{multline*}
must match \eqref{biqPS}
  
Another way: \eqref{Ebiq} yields $y=x_{n\pm 1}$ as a quadratic function
of $x=x_n$ as \\$\hspace*{-66pt}y=\displaystyle \dfrac{-d_{1,2}x^2 -(2d_{0,2}+d_{1,1})x-d_{0,1}
\pm\sqrt{(d_{1,2}x^2 +(2d_{0,2}+d_{1,1})x+d_{0,1})^2-4(d_{2,2}x^2+d_{1,2}x+d_{0,2})
         (d_{0,2}x^2+d_{0,1}x+d_{0,0})}}{2(d_{2,2}x^2+d_{1,2}x+d_{0,2})}$. Let us look now at the $(S_n,P,Z_n)$
construction as a way to find $x_{n+1}$ from $x_n$: if $x_n$ is known,
$S_n^2-P$ must vanish at $x=x_n\Rightarrow$ $S_n(x_n)=\gamma+
\delta(x_n-z_0)+\xi_n(x_n-z_0)^2=\pm\sqrt{P(x_n)}$, giving two possible
values for $\xi_n$. Then, we factor $S_n^2-P$:
$$
\hspace*{-66pt}\left[\gamma+\delta(x-z_0)+\dfrac{\pm\sqrt{P(x_n)}-\gamma-\delta(x_n-z_0)}{(x_n-z_0)^2}
(x-z_0)^2\right]^2-\gamma^2-2\gamma\delta(x-z_0)-\dfrac{P''(z_0)}{2}(x-z_0)^2
-\dfrac{P'''(z_0)}{6}(x-z_0)^3-\dfrac{P^{(4)}(z_0)}{24}
$$
what remains is constant times $(x-z_0)^2(x-x_n)$ times a last factor which
must be a constant times $x-x_{n+1}$, yielding for $x_{n+1}$ an expression
containing $\sqrt{P(x_n)}$, so that

$$  (d_{1,2}x^2 +(2d_{0,2}+d_{1,1})x+d_{0,1})^2-4(d_{2,2}x^2+d_{1,2}x+d_{0,2})
         (d_{0,2}x^2+d_{0,1}x+d_{0,0}) = \ \text{const.}\ P(x).$$

 Generalization.

Generalization of Pad\'e approximation and continued fraction
(recurrence relations) constructions: see \cite{Aptmul,AptSt,nutJAT,
StHer}

\subsection{3. and 4.}\  \\


\medskip


\medskip

\renewcommand{\mylmark}{Equivalence proofs 3 to 4}

\subsubsection{3. to 4.: Algorithm for square root of a polynomial\label{cfsqrt}} \  \\

Basically from Perron \cite[\S~20]{Per}

We may as well consider the general hyperelliptic case, with $P$ of degree $2m$
\cite{nutS}.

Let $P(z)=c(z-z_1)\dotsb (z-z_{2m})$, $S_0$ a polynomial of degree $m$
such that $S_0(z_0)= \gamma$ at a fixed point $z_0$, where $\gamma$ 
is the value at $z_0$ of a well defined
definition of the square root of $P$ (we will need this square
root only in a neigbourhood of $z_0$), and $Z_0$ and $W_0$  polynomials
of degree $m-1$ which are a factors of $P-S_0^2$.

As a matter of fact, this strange latter condition comes from the
consideration of a function $f$ which is the root of

$$  (x-z_0)Z_0(x) f^2(x) -2 S_0(x) f(x) +(x-z_0)W_0(x) = 0, $$

vanishing at $x=z_0$. If $z_0=\infty$, the property of $f$
is $f(x)=O(1/x)$ for large $x$. We have

$$  f(x) = \dfrac{ S_0(x) -\sqrt{P(x)} }{(x-z_0)Z_0(x)}, $$
where $P(x)=S_0(x)^2-(x-z_0)^2 Z_0(x)W_0(x)$. 

We now look at a continued fraction expansion to $f$ of the
form

$$  f(x) = \dfrac{\alpha_0(x-z_0)}{ 1-\beta_0(x-z_0) - \dfrac{\alpha_1(x-z_0)^2}{1-\beta_1(x-z_0) - \dotsb } }, $$

or $$ f_n(x) = \dfrac{\alpha_n(x-z_0)}{1-\beta_n(x-z_0) -(x-z_0)f_{n+1}(x)}\;,\;\; n=0,1,\dotsc$$
 Remark that the two first Taylor terms of the expansion of $f_n$ are
$$ f_n(x)= \alpha_n (x-z_0) +\alpha_n\beta_n (x-z_0)^2+\dotsb
$$

The form of $f$ is kept in all the $f_n$'s, as

\begin{equation}\label{fnquad}
%$$ \hspace*{-33pt}
f_n(x)=\dfrac{S_n(x)-\sqrt{P(x)}}{(x-z_0)Z_n(x)}
\end{equation}
$$ \Rightarrow
  f_{n+1}(x)=\dfrac{1}{x-z_0}-\beta_n -\dfrac{\alpha_n}{f_n(x)} =
   \dfrac{1-\beta_n(x-z_0)}{x-z_0} -\dfrac{\alpha_n (x-z_0)Z_n(x)[S_n(x)+\sqrt{P(x)}]}{S_n^2(x)-P(x)}
$$ 
which, if $f_n$ is a root of $(x-z_0)Z_n(x)f_n^2(x)-2S_n(x)f_n(x)+(x-z_0)W_n(x)=0$, 
with \\ $S_n(x)^2-(x-z_0)^2 Z_n(x)W_n(x)=P(x)$,
turns as $\displaystyle f_{n+1}(x)= \dfrac{[1-\beta_n(x-z_0)] W_n(x)/\alpha_n- S_n(x)- \sqrt{P(x)}}{(x-z_0)W_n(x)/\alpha_n}$,\\ \hspace*{-22pt} so that
\\
$ \displaystyle  Z_{n+1}(x)=W_n(x)/\alpha_n, \\
S_{n+1}(x)=[1-\beta_n(x-z_0)] Z_{n+1}(x)-S_n(x),\;  $\\
and, as we want 
\begin{equation}\label{alphaZZ}
S_{n+1}^2(x)-P(x)=(x-z_0)^2 Z_{n+1}(x)W_{n+1}(x)
    =(x-z_0)^2 \alpha_{n+1} Z_{n+1}(x)Z_{n+2}(x),
\end{equation}
remarking that $S_{n+1}^2(x)= [1-\beta_n(x-z_0)] Z_{n+1}(x)\{
  [1-\beta_n(x-z_0)] Z_{n+1}(x)-2S_n(x)\} +S_n^2(x) =
   [1-\beta_n(x-z_0)] Z_{n+1}(x)(S_{n+1}(x)-S_n(x))+S_n^2(x)$,\\
$\displaystyle
 W_{n+1}(x)=\alpha_n W_n(x) +[1-\beta_n(x-z_0)] 
       \dfrac{S_{n+1}(x)-S_n(x)}{(x-z_0)^2}$. \\
which is a polynomial, as $S_n$ and $S_{n+1}$ match
$\sqrt{P}$ up to second order, so that the degree
of the polynomial $W_{n+1}$ remains $\leqslant m-1$.

\medskip

 



\noindent\textbf{(Hyper)elliptic functions and all that.}
\renewcommand{\mylmark}{Equivalence proofs 3 to 4 (Hyper)elliptic functions}
%\pagestyle{fancy}

%  N.B. slight change in notations

The Green function. \label{Greenf} \  \\

For any $n_0$ and $p$, from \eqref{fnquad}:
$$
 f_{n_0} f_{n_0+1} \dotsc f_{n_0+p-1} =(x-z_0)^{-p}
  \frac{S_{n_0} -\sqrt{P}}{Z_{n_0}}
  \frac{S_{n_0+1} -\sqrt{P}}{Z_{n_0+1}}
   \dotsb
  \frac{S_{n_0+p-1} -\sqrt{P}}{Z_{n_0+p-1}} ,
$$
\begin{equation*} \begin{split}
 [f_{n_0} f_{n_0+1} \dotsc f_{n_0+p-1}]^2 &=(x-z_0)^{-2p}
 \frac{
       [S_{n_0} -\sqrt{P}]^2
       [S_{n_0+1} -\sqrt{P}]^2
   \dotsb
       [S_{n_0+p-1} -\sqrt{P}]^2  }
   {Z_{n_0} \; (Z_{n_0} Z_{n_0+1})\;
               (Z_{n_0+1} Z_{n_0+2})\;
          \dotsb
               (Z_{n_0+p-2} Z_{n_0+p-1})\;  Z_{n_0+p-1} } \\
 &=\frac{ (-1)^p \alpha_{n_0}\dotsc \alpha_{n_0+p-1}
       [S_{n_0} -\sqrt{P}]^2
       [S_{n_0+1} -\sqrt{P}]^2
   \dotsb
       [S_{n_0+p-1} -\sqrt{P}]^2  Z_{n_0+p}}
      { Z_{n_0}
       [S_{n_0}^2 -P]
       [S_{n_0+1}^2 -P]
   \dotsb
       [S_{n_0+p-1}^2 -P]  }   \\
    &    \text{ (from  \eqref{alphaZZ}} \\
 &=\frac{Z_{n_0+p}}{Z_{n_0}}
      (-1)^p \alpha_{n_0}\dotsc \alpha_{n_0+p-1}
     \frac{
       [S_{n_0} -\sqrt{P}]
       [S_{n_0+1} -\sqrt{P}]
   \dotsb
       [S_{n_0+p-1} -\sqrt{P}]  }
    {   [S_{n_0} +\sqrt{P}]
       [S_{n_0+1} +\sqrt{P}]
   \dotsb
       [S_{n_0+p-1} +\sqrt{P}]  }
\end{split}\end{equation*}


When $p$ is (almost) a period, $Z_{n_0+p} \approx Z_{n_0}$, we have the
limit
$$
 \hspace*{-44pt}\exp(\mathcal{G}(z)) = \exp( g(z) +ih(z)) =
  \lim_{n\to\infty}\left( \frac{a_1 \dotsb a_n}{f_0(z)\dotsb f_{n-1}(z)}
                   \right)^{1/n}
    \approx \left( \frac{(S_{n_0}+\sqrt{P})\dotsb (S_{n_0+p-1}+\sqrt{P})}
                        {(S_{n_0}-\sqrt{P})\dotsb (S_{n_0+p-1}-\sqrt{P})}
            \right)^{1/(2p)}
 $$

When $P$ has real roots, $\exp(\mathcal{G}(z))$ is the ratio of
complex conjugate numbers if $z\in$ the support  $S$, the locus where $P(z)\leqslant 0$.
$g(z)$ is the harmonic function vanishing on $S$, positive outside $S$ (minimum principle),
behaving like $-\log|z-z_0|+$ constant near $z_0$ ($\log|z|+$const.  for large $z$ if $z_0=\infty$): the \emph{\textbf{Green
function}} singular at $z_0$ of $S$. When $z_0=\infty$, the constant is the \emph{\textbf{Robin constant}}\footnote{About
Robin, see K.~Gustafson, T.~Abe: ``The third
boundary condition- was it Robin's?'', \textsl{Math. Intelligencer}
vol.~\textbf{20} (1998), nr.~1, pp.~63-71; ``(Victor) Gustave Robin: 1855-1897'',
\emph{ibid.}, vol.~\textbf{20}, nr.~2, pp.~47-53.} = -$\log$ of the
\emph{\textbf{logarithmic capacity}} of $S$.

\medskip

What if $P$ has general complex zeros $z_1,\dotsc, z_{2m}$?
$\exp{\mathcal{G}(z)}$ is already on the unit circle, i.e.,
$g(z)=0$ at the zeros of $P$. We still can think of a path $S$
where $g(z)=0$, i.e., $\mathcal{G}(z)$ is pure imaginary, and this
path is characterized by the differential condition
$$
   \mathcal{G}'(z)\, dz \text{\  pure imaginary}
$$
(or the \emph{quadratic differential} $(\mathcal{G}')^2(z) \,dz^2
\leqslant 0$). We must know more on $\mathcal{G}'$:
\begin{equation*} \begin{split}
 \mathcal{G}' &\approx \frac{1}{2p} \sum \left[
    \frac{S' +\dfrac{P'}{2\sqrt{P}}}{S+\sqrt{P}}
   -\frac{S' -\dfrac{P'}{2\sqrt{P}}}{S-\sqrt{P}}  \right]  \\
  &\approx
                       \frac{1}{2p} \sum
    \frac{-2S'\sqrt{P} +2\dfrac{P'S}{2\sqrt{P}}}
    {S^2-P = (x-z_0)^2\alpha_m  Z_m Z_{m+1}}
\end{split}\end{equation*}
is therefore a rational function divided by $\sqrt{P}$, with possible
poles at the zeros of the $Z$ polynomials. But let us look again
at
$\displaystyle{
 \exp(\mathcal{G}(z))
    \approx \left( \frac{(S_{n_0}+\sqrt{P})\dotsb (S_{n_0+p-1}+\sqrt{P})}
                        {(S_{n_0}-\sqrt{P})\dotsb (S_{n_0+p-1}-\sqrt{P})}
            \right)^{1/(2p)}
} $
if some $S_m\pm \sqrt{P}$ vanishes, so does $S_m^2 -P$, so, either
$Z_m$ or $Z_{m+1}$. If it is $Z_m$, look at
 $S_m = -S_{m-1} +[1-\beta_{m-1}(x-z_0)] Z_m$, which shows that $S_{m-1}\mp
  \sqrt{P}$ vanishes as well; and if it is $Z_{m+1}$,
 $S_{m+1} = -S_m +[1-\beta_m(x-z_0)] Z_{m+1} = -S_m$ at a zero of $Z_{m+1}$.
So, $\exp{\mathcal{G}}$ has no pole nor zero in the finite complex plane,
and we must have
\begin{equation} \label{Greenfp}
  \mathcal{G}'(z) = \frac{Q(z)}{\sqrt{P(z)}},
\end{equation}
where $Q(z) = z^{m-1} +\dotsb$ is a monic polynomial of degree $m-1$.

When $m>1$, $Q$ is completely determined by the conditions
$$
  \text{ periods } = 2\int_{z_1}^{z_k}
                    \frac{Q(z)}{\sqrt{P(z)}}\, dz
 \ \text{  pure imaginary}.
$$




\medskip



\noindent\textbf{Importance of the Green function $\mathcal{G}$ on the behaviour of
the denominator $B_n$.}\  \\


\noindent Remember: $\displaystyle \dfrac{A_1(x)}{B_1(x)} = \dfrac{\alpha_0 (x-z_0)}{1-\beta_0(x-z_0)},\\
\dfrac{A_2(x)}{B_2(x)} = \dfrac{\alpha_0 (x-z_0)}{1-\beta_0(x-z_0)
      - \dfrac{\alpha_1 (x-z_0)^2}{1-\beta_1(x-z_0)}  }=
 \dfrac{\alpha_0(x-z_0)[1-\beta_1(x-z_0)]}{
    [1-\beta_0(x-z_0)][1-\beta_1(x-z_0)] -\alpha_1 (x-z_0)^2 }$, \\
$A_{n+1}(x)=[1-\beta_n(x-z_0)]A_n(x)-\alpha_n(x-z_0)^2 A_{n-1}(x),\\
 B_{n+1}(x)=[1-\beta_n(x-z_0)]B_n(x)-\alpha_n(x-z_0)^2 B_{n-1}(x),
  A_0(x)=0, A_1(x)= \alpha_0(x-z_0), B_0(x)=1, B_1(x)=1-\beta_0(x-z_0)$.

\begin{equation} \label{recAB}
\begin{bmatrix} A_n(x) & B_n(x) \\A_{n+1}(x) &  B_{n+1}(x)\end{bmatrix} =
\begin{bmatrix} 0 & 1 \\-\alpha_n(x-z_0)^2  &1-\beta_n(x-z_0)  \end{bmatrix}
\begin{bmatrix}  A_{n-1}(x) & B_{n-1}(x) \\A_n(x)& B_n(x)  \end{bmatrix} 
\end{equation}
$\displaystyle \begin{bmatrix}  A_{-1}(x) & B_{-1}(x) \\A_0(x)& B_0(x)  \end{bmatrix} 
=\begin{bmatrix}  -(x-z_0)^{-1} & 0 \\0& 1  \end{bmatrix}$ 



The products $(x-z_0)^n f_0(x)\dotsc f_n(x)$ satisfy the same recurrence relation, from
$\displaystyle f_{n+1}(x)=\dfrac{1}{x-z_0}-\beta_n -\dfrac{\alpha_n}{f_n(x)},
  f_0(x)=\dfrac{S_0(x)-\sqrt{P(x)}}{(x-z_0)Z_0(x)},\\
 (x-z_0)f_0(x)f_1(x)=(x-z_0)\dfrac{S_0(x)-\sqrt{P(x)}}{(x-z_0)Z_0(x)}
             \dfrac{[1-\beta_0(x-z_0)]Z_1(x)-S_0(x)-\sqrt{P(x)}}{(x-z_0)Z_1(x)}\\=
   \dfrac{B_1(x)S_0(x)Z_1(x)-S_0^2(x)+P(x)-B_1(x)Z_1(x)\sqrt{P(x)}}{(x-z_0)Z_0(x)Z_1(x)}
=B_1(x)f(x)-A_1(x)$.






 Let $n$ be a quasi-period, then\\
$\displaystyle
B_{n-1}(x) f(x)-A_{n-1}(x) = \frac{B_{n-1}(x)S_0(x)-A_{n-1}(x) Z_0(x) -B_{n-1}(x)\sqrt{P(x)}}{Z_0(x)} \\=
 (z-z_0)^{n-1} f_0(x) \dotsb f_{n-1}(x) \approx
   (-1)^n a_1\dotsb a_n e^{-n\mathcal{G}}.
$
Now, take the other determination of $\sqrt{P}$. $\mathcal{G}$ becomes
$-\mathcal{G}$ (as seen form the definition of $\exp(\mathcal{G})$
through a product of $\displaystyle{\frac{X_k+\sqrt{Y}}{X_k-\sqrt{Y}}}$,
or frm the quadratic differential), and
$$
B_{n-1} f_{\text{conjugate}}-A_{n-1} = \frac{B_{n-1} S_0-A_{n-1} Z_0 +B_{n-1}\sqrt{P}}{Z_0}
  \approx
   (-1)^n a_1\dotsb a_n e^{n\mathcal{G}},
$$
whence
$$
 B_{n-1} \approx (-1)^n a_1\dotsb a_n Z \frac{e^{n\mathcal{G}}
                                             -e^{-n\mathcal{G}}   }
                               {2\sqrt{Y}}.
$$
showing that most of the zeros of $B_{n-1}$ are on the locus
$S=\{ z: \text{ Re}\, \mathcal{G}(z) = g(z)=0\}$.

\medskip



\noindent\emph{\textbf{Remark: ``Pell's equation''.}}    \label{Pell}
Make the product
\begin{equation*} \begin{split}
[B_{n-1} S_0-A_{n-1} Z_0] ^2 -B_{n-1}^2P &=  Z_0^2(x-z_0)^{2n} f_0 \dotsc f_{n-1} f_{0,\,\text{conj.}}
          \dotsc f_{n-1,\,\text{conj.}}  \\
  &= Z_0^2 \frac{S_0-\sqrt{P}}{Z_0} \dotsb   \frac{S_{n-1}-\sqrt{P}}{Z_{n-1}}
         \frac{S_0+\sqrt{P}}{Z_0} \dotsb   \frac{S_{n-1}+\sqrt{P}}{Z_{n-1}}   \\
   &= \alpha_0^2 \dotsc \alpha_{n-1}^2 (x-z_0)^{2n}Z_0 Z_n.
\end{split}\end{equation*}

\medskip

\label{Wperiod}What is (are) the period(s)?    \  \\

If $m>1$,
let us consider the integrals of\\
$\displaystyle{ \frac{t^k}
                     {2\pi i \sqrt{P(t)}} \, \log\left[
\frac{B_{n-1}(t) S_0(t)-A_{n-1}(t) Z_0(t) +B_{n-1}(t)\sqrt{P(t)}}
     {B_{n-1}(t) S_0(t)-A_{n-1}(t) Z_0(t) -B_{n-1}(t)\sqrt{P(t)}}   \right]
                \;   dt  }$,\ \  $k=0,\dotsc, m-2$,

on a big contour avoiding all the singularities, so that the value
$=0$,

\begin{minipage}[t]{7.25cm}
\  \par

\psset{unit=1cm}
\begin{pspicture}(0,2)(8,6)
  \psarc(-1,4.5){0.5}{10}{350}
  \psdots[dotsize=0.15](3.5,4.5)     \uput[0](3.5,4.5){$z_1$}
% \psdots[dotsize=0.15](3.9,7)     \uput[0](3.9,7){$z$}
  \psline{<-}(1,4.8)(1.5,4.8)
  \psline{<-}(1.5,4.2)(1.0,4.2)
  \psline(-0.5,4.65)(3.4,4.65)
  \psline(-0.5,4.35)(3.4,4.35)  \psdots[dotsize=0.15](-1,4.5) \uput[270](-1,4.5){$z_0$}
  \psdots[dotsize=0.15](5,5)
  \psdots[dotsize=0.15](5,6)
  \psdots[dotsize=0.15](6.1,3)
  \psline(3.4,4.35)(1.4,2.35)\psline(3.6,4.15)(1.6,2.15)
 \psarc(1.15,1.9){0.5}{55}{395}\psdots[dotsize=0.15](1.15,1.9)
 \psset{linewidth=0.075}
 \pscurve[linecolor=darkgray](3.5,4.5)(4,6.5)(5,5)
 \pscurve[linecolor=darkgray](6.1,3)(7,4.5)(5,6)  \uput[270](6.1,3){$z_{2m}$}
     \uput[0](3.8,6.1){$S$}
 \end{pspicture}
   \smallskip
\end{minipage}
 \begin{minipage}[t]{1mm}

 \ \par %\rule{0.2mm}{250pt}
 \end{minipage}
\begin{minipage}[t]{8.3cm}
\  \par

\vspace*{-10pt}
\noindent



On the other hand, making the contour shrink about $S$ and the zeros
of $U$ and $V=B_{n-1}(t) S_0(t)-A_{n-1}(t) Z_0(t) \pm B_{n-1}(t)\sqrt{P(t)}$ (which are $z_0$ and the zeros of $Z_0$ and $Z_n$), one finds
\end{minipage}

\begin{equation}  \label{Jacobiprob}
 0=-2n \int_{z_0}^{z_1} \frac{t^k}{\sqrt{P(t)}}\,dt+
     \sum_{\text{zeros of}\; Z_0,Z_n} \pm \int_{z_1}^{\text{zero}}
            \frac{t^k}{\sqrt{P(t)}}\,dt
+ \sum_{j} N_j\int_{z_j}^{z_{j+1}}
            \frac{t^k}{\sqrt{P(t)}}\,dt,\qquad k=0,\dotsc,m-2,
\end{equation}
where the first term accounts for the zero of multiplicity $2n$ at $z_0$
of $V=B_{n-1}(t) S_0(t)-A_{n-1}(t) Z_0(t) - B_{n-1}(t)\sqrt{P(t)}$, so that
the logarithm of $U/V$ on the lower path from $z_0$ to $z_1$ is the value
on the upper path minus $4n\pi i$; 
 the $\pm$ signs tell if the zero is a zero of $U$ or $V$; and where
$N_j$ are (unknown) integers. There are only $m-1$ integrals
$\displaystyle{ \int_{z_j}^{z_{j+1}} \frac{t^k}{\sqrt{P(t)}}\,dt }$ to
consider, in the ``gaps'' of $S$.  Integrals on the two sides of $S$
vanish, as $\sqrt{P}_+ = -\sqrt{P}_-$, so that the logarithm takes
opposite values on the sides of $S$.

\smallskip


\smallskip
It happens that, knowing $n$, $P$, and $Z_0$, \eqref{Jacobiprob}
 allows
to find the remaining unknowns, including the $\pm$ signs
(\emph{\textbf{Jacobi problem}}, \cite{akh, Apt,Mag79,nutS,nutJAT,PI}).

There is absolutely no need for $n$ to be an integer in the description
\eqref{Jacobiprob} of the Jacobi problem. Let $x^{(1)}_n,\dotsc, x^{(m-1)}_n$ be the
unknown zeros of $Z_n$. To see how these $x^{\ell}_n$'s are functions of
$n$, we \dots take the derivative of \eqref{Jacobiprob} with respect to
$n$ (!!):
$$ \nu_k = \sum_{\ell=1}^{m-1} s_\ell \frac{(x^{(\ell)}_n)^k}{\sqrt{P(x^{(\ell)}_n)}}\,
       \frac{dx^{(\ell)}_n}{dn}
     \qquad\qquad k=0,\dotsc, m-2,
$$
where $\nu_k =\displaystyle{
       2 \int_{z_0}^{z_1} \frac{t^k}{\sqrt{P(t)}}\,dt }$. We have a
system of differential equations for the $x^{(\ell)}_n$'s. A single
derivative is isolated through Lagrange interpolation polynomials:
$$
\frac{dx^{(\ell)}_n}{dn} = 2 s_\ell \sqrt{P(x^{(\ell)}_n)}
    \int_{z_0}^{z_1} \prod_{\substack{q=1 \\ q\neq \ell}}
      \frac{t-x^{(q)}_n}{x^{(\ell)}_n-x^{(q)}_n}\; \frac{dt}{\sqrt{P(t)}} .
$$
An initial condition consists of a set $x^{(1)}_n,\dotsc, x^{(m-1)}_n$
\emph{\textbf{and}} the signs $s_1,\dotsc, s_{m-1}$ (= a set of places
on the \emph{\textbf{Riemann surface}} of $\sqrt{P}$).

The vector $[x^{(1)}_n,\dotsc, x^{(m-1)}_n]$ is a well defined function
(\emph{\textbf{Jacobi-Abel function}}) of the left-hand side
$[n\nu_0,\dotsc, n\nu_{m-2}]$.

\medskip

\noindent\textbf{Periodicity:} $x^{(1)}_n,\dotsc, x^{(m-1)}_n$ are kept unchanged  if
the left-hand side of \eqref{Jacobiprob} is augmented by integers times the integrals
  $\displaystyle{2 \int_{z_j}^{z_{j+1}}
            \frac{t^k}{\sqrt{Y(t)}}\,dt }$ (\emph{\textbf{periods}})





It figures: in the canonical Jacobi setting, the $x_k$'s are $\pm 1$ and $\pm k^{-1}$.
Then, \eqref{Jacobiprob} becomes

\begin{equation}  \label{Jacobiprob2}
 2n \mathsf{K}+\int_{x_1=-k^{-1}}^{z_0}
            \frac{dt}{k\sqrt{P(t)}}
=
     \pm \int_{x_1=-k^{-1}}^{z_n}
            \frac{dt}{k\sqrt{P(t)}}
+2  N_n i \mathsf{K}',
\end{equation}

as the phases of $\sqrt{P(t)} = \sqrt{(t^2-1)(t^2-k^{-2})} = k^{-1}\sqrt{(1-t^2)(1-k^2t^2)}$ are $+1$, $-i$, $-1$, $+i$, and $+1$ on $\mathbb{R}+i\varepsilon$, and the relevant
definite integrals of $dt/\sqrt{(1-t^2)(1-k^2t^2)}$ are

\begin{tabular}{ccccc}
on $(-\infty,-k^{-1})$ &  $(-k^{-1},-1)$   &  $(-1,1)$  &  $(1,k^{-1})$  &  $(k^{-1},+\infty)$ \\
 $\mathsf{K}$  &  $-i\mathsf{K}'$ &  $-2\mathsf{K}$ & $i\mathsf{K}'$ &  $\mathsf{K}$,
\end{tabular}

where $\mathsf{K} = \displaystyle \int_0^1 \dfrac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}
 = \int_{k^{-1}}^\infty \dfrac{du}{\sqrt{(1-u^2)(1-k^2u^2)}}$ (take $t=1/ku$),
$\mathsf{K}' = \displaystyle \int_0^1 \dfrac{dt}{\sqrt{(1-t^2)(1-(1-k^2)t^2)}} =
   \int_1^{k^{-1}} \dfrac{du}{\sqrt{(u^2-1)(1-k^2u^2)}}$ (take 
$\displaystyle  t=\dfrac{1}{u} \sqrt{\dfrac{u^2-1}{1-k^2}} \Rightarrow u=[1-(1-k^2)t^2]^{-1/2}$)

(see Milne-Thomson, in Abramowitz \& Stegun:)

\includegraphics*[angle=270, scale=0.50]{ell17443.ps}

We clean a little bit \eqref{Jacobiprob2}: the integral from $-k^{-1}$ to $-1$ is
$-i\mathsf{K}'$, so that we throw in the right-hand side an \emph{even} integer multiple
of $i\mathsf{K}'$ (the periods of the Jacobi elliptic integral of first
kind are $4\mathsf{K}$ and $2i\mathsf{K}'$); the integral from $-1$ to $0$ is
$-\mathsf{K}$. We have

\begin{equation*}  
 (2n-1) \mathsf{K}+ \int_0^{z_0}
            \frac{dt}{k\sqrt{Y(t)}}
= \mp \mathsf{K}
     \pm \int_0^{z_n}
            \frac{dt}{k\sqrt{Y(t)}}
+\ \text{periods}.
\end{equation*}

We now see in an elementary way how the Jacobi problem is
solved for $z_n$, including the $\pm$ sign: when $n=0$, the
upper sign is the obvious choice; when $n=1$, take the lower sign,
the integral from $0$ to $z_1$ is the opposite of the integral 
from $0$ to $z_0$, so $z_1=-z_0$; when $n=2$, take the upper sign again,
but perform the integral to $z_2$ by adding a complete loop about
$[-1,1]$ weighting $4\mathsf{K}$ (or add $4\mathsf{K}$ to the ``periods''
term), etc.

With elliptic sine: $z_n = \sn(2n\mathsf{K}+\arg \sn z_0) = (-1)^n z_0$.

\smallskip

Now we deal with a symmetric set, BUT $x=\infty$ is sent to some value $t=-1/\gamma$,
\eqref{Jacobiprob2} is changed into

\begin{equation*}  
 2n\int_{-\infty}^{-1/\gamma}
            \frac{dt}{k\sqrt{Y(t)}}
+\int_{x_1=-k^{-1}}^{z_0}
            \frac{dt}{k\sqrt{Y(t)}}
=
     \pm \int_{x_1=-k^{-1}}^{z_n}
            \frac{dt}{k\sqrt{Y(t)}}
+2  N_n i \mathsf{K}',
\end{equation*}
and $z_n=\sn(an+b)$, with $\displaystyle a=\int_{-\infty}^{-1/\gamma}
            \frac{dt}{k\sqrt{Y(t)}}$, leading to a true periodic sequence whether
$a/\mathsf{K}$ is a rational number or not.


\ \  

\section{Further identities and formulas.}
\renewcommand{\mylmark}{Further identities and formulas}

\renewcommand{\mylmark}{Further: Hyperelliptic excursions}

\subsection{Hyperelliptic excursions\label{hyperell}}  \   \\

So, $P$ is a polynomial of degree $2m=2g+2$, of zeros $z_1,\dotsc, z_{2m}$.
We choose a continuous $\sqrt{P(z)}$ outside a system of cuts. So, instead of
dealing with the two-sheeted Riemann surface, we will sometimes emphasize
a sign $\varepsilon$ in $\varepsilon\sqrt{P(z)}$, with $\varepsilon=+1$ or $-1$.

Let also $z_0$ different from $z_1,\dotsc, z_{2m}$,
and  $\gamma+\delta (z-z_0)$ be the two first terms of the Taylor
expansion of $\sqrt{P(z)}$ about $z=z_0$. 

We look at continued fraction expansions about $z_0$ of functions involving $\sqrt{P}$.    We follow \S~\ref{ss3}, p.~\pageref{ss3}.

So, consider $\displaystyle  f_n(z):= \dfrac{S_n(z)-\sqrt{P(z)}}{(z-z_0)R_n(z)} $, with polynomials $S_n$ and $R_n$ such that
$S_n(z)=\gamma+\delta(z-z_0)+O((z-z_0)^2$, so that $f_n(z_0)=0$,
$\displaystyle \dfrac{S_n^2(z)-P(z)}{(z-z_0)^2R_n(z)}=$ the new polynomial
$R_{n+1}(z)$, and the continued fraction step
$\displaystyle  f_n(z):= \dfrac{S_n(z)-\sqrt{P(z)}}{(z-z_0)R_n(z)} = \dfrac{   (z-z_0) }{\alpha_n+\beta_n(z-z_0) - (z-z_0)f_{n+1}(z)  }$, as in \eqref{contfr}, in p.~\pageref{contfr}. We still have
$\displaystyle
 \dfrac{  S_n(z)+\sqrt{P(z)}   } { R_{n+1}(z)}= \text{Tayl}_1(z)-\dfrac{S_{n+1}(z)-\sqrt{P(z)}}{R_{n+1}(z)}$,    
so, $S_{n+1}(z)=-S_n(z)+R_{n+1}(z)$Taylor${}_1(z)$, 
$R_{n+2}(z) = R_n(z) -\text{Tayl}_1(z) \displaystyle
 \dfrac{ S_n(z) -S_{n+1}(z)  }{(z-z_0)^2}$


What about the degrees? The degree of $S_n^2-P$ can normally not be smaller than
$2m=2g+2$,
so the smallest choice for the degrees of the $R_n$s is degree$(R_n)= g$.

The $R$s and the $S$s will be the $x$s and $y$s of the hyperelliptic billiard.

Let us choose $X_0=$ such a polynomial of degree $g$, $Y_0$ of degree $g+1$, with
the two first Taylor coefficients as $\sqrt{P}$: $Y_0(z)=\gamma+\delta(z-z_0)+O((z-z_0)^2)$, and
such that $P(z)-Y_0^2(z)$ is a polynomial multiple of $(z-z_0)^2X_0(z)$,
$P(z)-Y_0^2(z)=(z-z_0)^2 X_0(z) X_1(z)$, according to what has been found above about $Q_n$.

Given $X_0$, $Y_0$ is constructed through interpolation of $\sqrt{P}$ at the zeros
$x_0^{(1)},\dotsc, x_0^{(g)}$ of $X_0$, here is how any choice of $g$ signs $\varepsilon_1,\dotsc, \varepsilon_g$
appears:
{\Small
$$\hspace*{-33pt}
Y_0(z)= \gamma+\delta(z-z_0)+(z-z_0)^2\left[\text{polynomial\ interpolant\ of\  degree\  }g-1 \text{\ of\ value\ }
   \dfrac{\varepsilon_j \sqrt{P(x_0^{(j)}}-\gamma-\delta(x_0^{(j)}-z_0)  }{(x_0^{(j)}-z_0)^2 }\text{\ at\ } x_0^{(j)}\right] 
$$
}
where each $\varepsilon_j= +1$ or $-1$. So, there are $2^g$ possible $Y_0$s for a
given $X_0$.

The continued fraction construction may proceed with
$$  \dfrac{\sqrt{P(z)}+Y_n(z) }{X_{n+1}(z) } = \text{Taylor}_1 +\dfrac{(z-z_0)^2}
          {\ \ \dfrac{\sqrt{P(z)}+Y_{n+1}(z)}{X_{n+2}(z)}\  \  },
$$
but the use of $\displaystyle f_n(z)= -\dfrac{(z-z_0)X_{n+1}(z) }{\sqrt{P(z)}+Y_n(z)} =
  \dfrac{Y_n(z)-\sqrt{P(z)} }{(z-z_0)X_n(z)}$ will be preferred here.

The algebraic function $f_n$ is the root of 
\begin{equation} \label{Hquadf}
  (z-z_0)X_n(z)f_n^2(z) -2Y_n(z) f_n(z) -(z-z_0)X_{n+1}(z)=0 
\end{equation}
vanishing at $z=z_0$. The $X$s are our object of study, they are
polynomials of degree $g$, of which $g$ elements are considered, such as  the
zeros $x_n^{(1)},\dotsc, x_n^{(g)}$ of $X_n$. $Y_n$ is a polynomial
of degree $m=g+1$ with the constraint that
\begin{equation} \label{HYP}
Y_n^2(z) + (z-z_0)^2 X_n(z) X_{n+1}(z) =P(z),
\end{equation}
constraint which is satisfied if $Y_n(z)=\gamma +\delta (z-z_0) +O((z-z_0)^2)$.
Then, $Y_n$ depends on $g$
elements too.

Finally, let the continued fraction step be
\begin{equation} \label{Hcf}
  f_n(z)= \dfrac{  \alpha_n (z-z_0) }{1+\beta_n(z-z_0) +\alpha_n (z-z_0)f_{n+1}(z)  }.
\end{equation}
The coefficients $\alpha_n$ and $\beta_n$ are immediately related
to the two first Taylor coefficients of $f_n$, which follow from
\eqref{Hquadf}, knowing that $f_n$ vanishes at $z_0$: 
$$\hspace*{-22pt}
f_n(z)= \underbrace{  \alpha_n (z-z_0) -\alpha_n \beta_n (z-z0)^2+\dotsb  }_{\text{\normalsize from\ \eqref{Hcf}}}
     \ =\   -\underbrace{  \dfrac{ X_{n+1}(z_0)}{2Y_n(z_0)=2\gamma}(z-z_0) 
          - \left(\dfrac{X_{n+1}}{2Y_n}\right)'(z_0)
          (z-z0)^2+\dotsb  }_{\text{\normalsize from\ \eqref{Hquadf}}}
$$

Billiard: first, recurrence relations from \eqref{Hcf}:
\begin{equation*}\begin{split}
f_{n+1}(z)&= \dfrac{1}{f_n(z)}+\dfrac{1}{\alpha_n(z-z_0)}+\dfrac{\beta_n}{\alpha_n} \\
      &= \dfrac{  (z-z_0)X_n(z) }{Y_n(z)-\sqrt{P(z)} }  +\dfrac{1}{\alpha_n(z-z_0)}+\dfrac{\beta_n}{\alpha_n} \\
     &= -\dfrac{Y_n(z)+\sqrt{P(z)} }{(z-z_0) X_{n+1}(z) } \text{\ (from\eqref{HYP})}
               +\dfrac{1}{\alpha_n(z-z_0)}+\dfrac{\beta_n}{\alpha_n} 
\end{split}\end{equation*}
which must be $\displaystyle \dfrac{Y_{n+1}(z)-\sqrt{P(z)} }{(z-z_0) X_{n+1}(z) }$, so
\begin{subequations}
\begin{equation} \label{HYn1}
  Y_{n+1}(z)= \dfrac{1+\beta_n(z-z_0)}{\alpha_n} X_{n+1}(z) -Y_n(z),
\end{equation}
and we must check \eqref{HYP} at the next step:
$Y_{n+1}^2(z)-P(z)= [(1+\beta_n(z-z_0))/\alpha_n]^2 X_{n+1}(z)
 -2[(1+\beta_n(z-z_0))/\alpha_n]Y_n(z) -(z-z_0)^2 X_n(z)$, so that
\begin{multline}\label{HXn1}
X_{n+2}(z)=[(1+\beta_n(z-z_0))/\alpha_n]
             \dfrac{ 2Y_n(z)-[(1+\beta_n(z-z_0))/\alpha_n]X_{n+1}(z)  }{(z-z_0)^2}
    +X_n(z) \\
 =[(1+\beta_n(z-z_0))/\alpha_n]
             \dfrac{ Y_n(z)-Y_{n+1}(z)  }{(z-z_0)^2}   +X_n(z)
\end{multline}

\end{subequations}

which is clearly a polynomial, as $Y_n$ and $Y_{n+1}$ share the two first
Taylor coefficients about $z_0$.

\medskip

So, a $x$ and a $y$ of the hyperelliptic game are vectors of
$\mathbb{C}^g$, as the zeros $(x_n^{(1)},\dotsc, x_n^{(g)})$ of
$X_n$, and the values $(\varepsilon_{n,1}\sqrt{P(x_n^{(1)})},\dotsc,
\varepsilon_{n,g}\sqrt{P(x_n^{(g)})}$ of $Y_n$ at the zeros of $X_n$.

The $g$ choices of the $\varepsilon$s have to be made only at $n=0$,
all the subsequent operations \eqref{HYn1}-\eqref{HXn1} being well
defined. What about other choices of the signs $\varepsilon$s? One
case is clear: when \textbf{\emph{all}} the $g$ signs are changed, 
\eqref{HYn1} at $n-1$ shows that $Y_n$ becomes the old $Y_{n-1}$, i.e.,
we are exploring the past! This is confirmed by
\eqref{Jacobiprob} in the form
$\displaystyle  0=-2n \int_{z_0}^{z_1} \frac{t^k}{\sqrt{P(t)}}\,dt+
     \sum_j \pm \int_{x_0^{(j)}}^{x_n^{(j)}}
            \frac{t^k}{\sqrt{P(t)}}\,dt +\text{periods}$, showing
that changing all the $\pm$s (are they the same thing as the
$\varepsilon_{n,j}$ of above?) is the same thing as replacing $n$
by $-n$.

\medskip

And here is a possible difference operator

{\Small
\begin{multline*}
(\mathcal{D}f)(x_n^{(1)},\dotsc, x_n^{(g)})=\\
\hspace*{-22pt}\left( \dfrac{f(\varepsilon_{n,1}\sqrt{P(x_n^{(1)})},0,\dotsc,0) -
              f(-\varepsilon_{n,1}\sqrt{P(x_n^{(1)})},0,\dotsc,0)}{2\varepsilon_{n,1}\sqrt{P(x_n^{(1)})}  },
      \dotsc,
    \dfrac{f(0,\dotsc,0, \varepsilon_{n,g}\sqrt{P(x_n^{(g)})}) -
              f(0,\dotsc,0,-\varepsilon_{n,g}\sqrt{P(x_n^{(g)})})}{2\varepsilon_{n,g}\sqrt{P(x_n^{(g)})}  }
\right)
\end{multline*}}
which looks like a kind of \textbf{\emph{gradient}}. Other combinations 
out of exterior algebra furnish all other difference operators, such
as \textbf{\emph{curls}}, etc. (oh no, not me, please!).

\medskip

And what may $F(X_n,Y_n)$ be here? Considering how $Y_n$ are
obtained from $X_n$, with the $\varepsilon_{n,1},\dotsc, \varepsilon_{n,g}$
degrees of freedom, $F$ must be an enormous contraption of degree $2^g$, but
such that equations may be solved by a sequence of quadratic operations,
perhaps something like a chain of squares $(a+(b+(c+\dotsb)^2)^2)^2$?


\renewcommand{\mylmark}{Further: Addition \& duplication formulas}
\subsection{Addition and duplication formulas} \label{duplica}\   \\

\parpic{
%\begin{center}
\psset{unit=1.0cm}\vspace*{-3pt}\begin{pspicture}(4.2,3.2)
\psline{->}(0,0)(4,0)
\psline{->}(0,0)(0,3.5)
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\psline[linestyle=dashed,dash=2mm 2mm](3.22,-0.1)(3.22,3.1)
\psline[linestyle=dashed,dash=2mm 2mm](-0.1,3.1)(3.9,3.1)\psline{->}(3.92,3.1)(3.92,2.7)
\psline[linestyle=dashed,dash=2mm 2mm](0.5,0.4)(-0.1,0.4)
\uput[270](0.5,-0.1){$x_n$} \uput[270](3.2,-0.1){$x_{n+1}$}
\uput[180](-0.1,0.4){$y_n$}\uput[180](-0.1,1.4){$y_{n+1}$}\uput[180](-0.1,3.1){$y_{n+2}$}
\parametricplot{-135}{225}{t cos 2 mul 2.25 add t cos t sin add 1.75 add}
\psline(0.5,0.4)(0.5,1.4)(3.22,1.4)(3.22,3.1)(3.9,3.1)
%\pscurve[linewidth=0.07,linecolor=red](0.2,0.2)(0.3,0.3)(1.3,2.2)
%\psline[linewidth=0.04](0.25,1)(0.5,0)(1,2)(1.5,0)
 %\pscircle*[linecolor=red](1.3,2.2){0.1}
 %\pscircle*[linecolor=red](0.2,0.2){0.1}
% \pscircle[linewidth=0.05](1,1.5){1}
\end{pspicture}
}
%\end{center}
\enlargethispage{55pt}
We start from the usual construction of $\{x_n\}$
and $\{y_n\}$ sequences, with $F(x_n,y_n)=0$, and 
$F(x_n,y_{n+1})=0 \Rightarrow x_n \text{\ and\ } x_{n+1}$
are the two roots in $x$ of $F(x,y_{n+1})=0$.

\

\

\

\

\



\bigskip


\parpic{
%\begin{center}
\psset{unit=1.0cm}\vspace*{-3pt}\begin{pspicture}(4.2,3.2)
\psline{->}(0,0)(4,0)
\psline{->}(0,0)(0,3.5)
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\psline[linestyle=dashed,dash=2mm 2mm](-0.1,2.4)(2.4,2.4)
\psline[linestyle=dashed,dash=2mm 2mm](2.4,-0.1)(2.4,3.2)
\psline[linestyle=dashed,dash=2mm 2mm](-0.1,3.2)(3.2,3.2)
%\psline{->}(3.92,3.1)(3.92,2.7)
%\psline[linestyle=dashed,dash=2mm 2mm](0.5,0.4)(-0.1,0.4)
\uput[270](1.5,-0.1){$x_n$} \uput[270](2.4,-0.1){$x_{n+1}$}
%\uput[180](-0.1,0.4){$y_n$}
\uput[180](-0.1,2.4){$x_{n+1}$}\uput[180](-0.1,3.2){$x_{n+2}$}
\parametricplot{-180}{180}{t cos  t sin 0.5 mul add 2.25 add t cos  t sin 0.5 mul sub  2.25 add}
\psline(0,0)(4,4)
\psline(1.5,1.1)(1.5,2.4)(2.4,2.4)(2.4,3.0)%(3.9,3.1)
%\pscurve[linewidth=0.07,linecolor=red](0.2,0.2)(0.3,0.3)(1.3,2.2)
%\psline[linewidth=0.04](0.25,1)(0.5,0)(1,2)(1.5,0)
 %\pscircle*[linecolor=red](1.3,2.2){0.1}
 %\pscircle*[linecolor=red](0.2,0.2){0.1}
% \pscircle[linewidth=0.05](1,1.5){1}
\end{pspicture}
}
We bypassed (section \ref{ss2}, p.~\pageref{ss2}) the $y$s by constructing $E$ such that $E(x_n,x_{n+1})=0$. By symmetry of $E$, the two roots of $E(x_n,y)=0$ are $y=x_{n+1}$ and
$y=x_{n-1}$. The horizontal $y=x_{n+1}$ would cut the  curve $E=0$ at a second point
of abscissa $x_{n+2}$\dots  If we want to see the familiar spiderweb showing $x_n,
x_{n+1},\dotsc$ we stop the horizontal when it meets the bisector $x=y$.

\  

\

\parpic{
%\begin{center}
\psset{unit=1.0cm}\vspace*{-3pt}\begin{pspicture}(4.2,3.2)
\psline{->}(0,0)(4,0)
\psline{->}(0,0)(0,3.5)
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\psline[linestyle=dashed,dash=2mm 2mm](-0.1,2.4)(3.2,2.4)
\psline[linestyle=dashed,dash=2mm 2mm](3.2,-0.1)(3.2,2.4)
\psline[linestyle=dashed,dash=2mm 2mm](-0.1,3.3)(3.2,3.3)
%\psline{->}(3.92,3.1)(3.92,2.7)
%\psline[linestyle=dashed,dash=2mm 2mm](0.5,0.4)(-0.1,0.4)
\uput[270](1.5,-0.1){$x_n$} \uput[270](3.2,-0.1){$x_{n+2}$}
%\uput[180](-0.1,0.4){$y_n$}
\uput[180](-0.1,2.4){$x_{n+1}$}\uput[180](-0.1,3.3){$x_{n+3}$}
\parametricplot{-180}{180}{t cos  t sin 0.5 mul add 2.25 add t cos  t sin 0.5 mul sub  2.25 add}
%\psline(0,0)(4,4)
\psline(1.5,1.1)(1.5,2.4)(3.2,2.4)(3.2,3.3)%(3.9,3.1)
%\pscurve[linewidth=0.07,linecolor=red](0.2,0.2)(0.3,0.3)(1.3,2.2)
%\psline[linewidth=0.04](0.25,1)(0.5,0)(1,2)(1.5,0)
 %\pscircle*[linecolor=red](1.3,2.2){0.1}
 %\pscircle*[linecolor=red](0.2,0.2){0.1}
% \pscircle[linewidth=0.05](1,1.5){1}
\end{pspicture}
}
But yes! Let us keep the same construction as before!


\  

\

\

\

\



Let us go further: bypassing the $x_{n\pm 1}$-step in \eqref{Ebiq},

$$ x_{n-1}+x_{n+1}= -\dfrac{ d_{1,2}x_n^2+(2d_{0,2}+d_{1,1})x_n+d_{0,1} }
                           { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  },
  x_{n-1}x_{n+1}= \dfrac{ d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0} }
                           { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  },
$$
$$ x_{n}+x_{n+2}= -\dfrac{ d_{1,2}x_{n+1}^2+(2d_{0,2}+d_{1,1})x_{n+1}+d_{0,1} }
                           { d_{2,2}x_{n+1}^2+d_{1,2}x_{n+1}+d_{0,2}  },
  x_{n}x_{n+2}= \dfrac{ d_{0,2}x_{n+1}^2+d_{0,1}x_{n+1}+d_{0,0} }
                           { d_{2,2}x_{n+1}^2+d_{1,2}x_{n+1}+d_{0,2}  },
$$
$$ x_{n-2}+x_{n}= -\dfrac{ d_{1,2}x_{n-1}^2+(2d_{0,2}+d_{1,1})x_{n-1}+d_{0,1} }
                           { d_{2,2}x_{n-1}^2+d_{1,2}x_{n-1}+d_{0,2}  },
  x_{n-2}x_{n}= \dfrac{ d_{0,2}x_{n-1}^2+d_{0,1}x_{n-1}+d_{0,0} }
                           { d_{2,2}x_{n-1}^2+d_{1,2}x_{n-1}+d_{0,2}  },
$$
\begin{equation*}\begin{split}
x_{n-2}+x_{n+2}&= -\dfrac{ d_{1,2}x_{n-1}^2+(2d_{0,2}+d_{1,1})x_{n-1}+d_{0,1} }
                           { d_{2,2}x_{n-1}^2+d_{1,2}x_{n-1}+d_{0,2}  }
                  -\dfrac{ d_{1,2}x_{n+1}^2+(2d_{0,2}+d_{1,1})x_{n+1}+d_{0,1} }
                           { d_{2,2}x_{n+1}^2+d_{1,2}x_{n+1}+d_{0,2}  }   -2x_n      \\
   &= -\{  2d_{1,2}d_{2,2}x_{n-1}^2x_{n+1}^2
              +[d_{2,2}(2d_{0,2}+d_{1,1})+d_{1,2}^2]x_{n-1}x_{n+1}(x_{n-1}+x_{n+1}) \\
         &    +(d_{0,2}d_{1,2}+d_{0,1}d_{2,2})(x_{n-1}^2+x_{n+1}^2)
              +(2d_{0,2}+d_{1,1})d_{1,2}x_{n-1}x_{n+1}  \\
         &  +[(2d_{0,2}+d_{1,1})d_{0,2}+d_{0,1}d_{1,2}](x_{n+1}+x_{n-1})+2d_{0,1}d_{0,2}
        \}  \\
           &/[ (d_{2,2}x_{n-1}^2+d_{1,2}x_{n-1}+d_{0,2})
                              (d_{2,2}x_{n+1}^2+d_{1,2}x_{n+1}+d_{0,2}  ) ]  -2x_n
\end{split}\end{equation*}
where we replace $x_{n-1}+x_{n+1}$ and $x_{n-1}x_{n+1}$ by the rational
functions of second degree in $x_n$ recalled above.

Fourth degree?

No: the denominator is the product of the four factors $(x_{n\pm 1}-\rho_i)$ rearranged
as the product of $(x_{n-1}-\rho_i)(x_{n+1}-\rho_i) = x_{n-1}x_{n+1}
-\rho_i(x_{n-1}+x_{n+1})+\rho_i^2 =\displaystyle
\dfrac{ d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0} }
                           { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  }
 +\dfrac{ d_{1,2}x_n^2+(2d_{0,2}+d_{1,1})x_n+d_{0,1} }
                           { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  }\rho_i +\rho_i^2 =
\dfrac{ E(x_n,\rho_i) } { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  }  \\  =
-\dfrac{ \overbrace{(d_{2,2}\rho_i^2+d_{1,2}\rho_i+d_{0,2})}^{\displaystyle =0} x_n^2
         +(d_{1,2}\rho_i^2+(2d_{0,2}+d_{1,1})\rho_i+d_{0,1})x_n
            +d_{0,2}\rho_i^2+d_{0,1}\rho_i+d_{0,0} }
                           { d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  }$,
so the degree of the final expression will remain 2.

Remark that the new $\rho_i$ is $\displaystyle
\rho_i^{(2)}=-\dfrac{d_{0,2}\rho_i^2+d_{0,1}\rho_i+d_{0,0} }
        {d_{1,2}\rho_i^2+(2d_{0,2}+d_{1,1})\rho_i+d_{0,1}  }$

After some struggle with a symbolic calculator, one has\\
denominator $=[d_{01}^2d_{22}^2+d_{02}d_{22}d_{11}^2+4d_{0,2}^2d_{22}d_{11} $ etc.  awful

Try again the resultant trick, from \S~\ref{ss2}: eliminate $x_{n\pm 1}$ from \eqref{Ebiq}\\
$E(x_n,x_{n\pm 1})=d_{0,0} +d_{0,1}(x_n+x_{n\pm 1}) + d_{0,2}
(x_n+x_{n\pm 1})^2
  +d_{1,1}x_n x_{n\pm 1} 
 +d_{1,2}x_n x_{n\pm 1}(x_n+x_{n\pm 1}) +d_{2,2}x_n^2 x_{n\pm 1}^2=
[d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}]x_{n\pm 1}^2 +[d_{1,2}x_n^2+(d_{1,1}+2d_{0,2})x_n
+d_{0,1}]x_{n\pm 1}+d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0}=0$,
 and $E(x_{n\pm2},x_{n\pm 1})=0$. I don't even try to be bright, the resultant is
{\Small 
\begin{equation*} %\label{resul}
  \hspace*{-55pt}\begin{vmatrix} 
d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  &   d_{1,2}x_n^2+(d_{1,1}+2d_{0,2})x_n
+d_{0,1}   &  d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0}  &   0 \\
 0 & d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  &   d_{1,2}x_n^2+(d_{1,1}+2d_{0,2})x_n
+d_{0,1}   &  d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0}      \\
 d_{2,2}x_{n\pm 2}^2+d_{1,2}x_{n\pm 2}+d_{0,2}  &   d_{1,2}x_{n\pm 2}^2+(d_{1,1}+2d_{0,2})x_{n\pm 2}
+d_{0,1}   &  d_{0,2}x_{n\pm 2}^2+d_{0,1}x_{n\pm 2}+d_{0,0}  &   0 \\
 0 & d_{2,2}x_{n\pm 2}^2+d_{1,2}x_{n\pm 2}+d_{0,2}  &   d_{1,2}x_{n\pm 2}^2+(d_{1,1}+2d_{0,2})x_{n\pm 2}
+d_{0,1}   &  d_{0,2}x_{n\pm 2}^2+d_{0,1}x_{n\pm 2}+d_{0,0}                       
    \end{vmatrix}
\end{equation*}
}\\
$=(x_{n\pm 2}-x_n)^2$ times
{\Small 
\begin{equation*} %\label{resul}
  \hspace*{-44pt}\begin{vmatrix} 
d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  &   d_{1,2}x_n^2+(d_{1,1}+2d_{0,2})x_n
+d_{0,1}   &  d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0}  &   0 \\
 0 & d_{2,2}x_n^2+d_{1,2}x_n+d_{0,2}  &   d_{1,2}x_n^2+(d_{1,1}+2d_{0,2})x_n
+d_{0,1}   &  d_{0,2}x_n^2+d_{0,1}x_n+d_{0,0}      \\
 d_{2,2}(x_{n\pm 2}+x_n)+d_{1,2}  &   d_{1,2}(x_{n\pm 2}+x_n)+d_{1,1}+2d_{0,2}   &  d_{0,2}(x_{n\pm 2}+x_n)+d_{0,1}  &   0 \\
 0 & d_{2,2}(x_{n\pm 2}+x_n)+d_{1,2}  &   d_{1,2}(x_{n\pm 2}+x_n)+d_{1,1}+2d_{0,2}   &  d_{0,2}(x_{n\pm 2}+x_n)+d_{0,1}                       
    \end{vmatrix}
\end{equation*}
\begin{equation*} %\label{resul}
  \hspace*{-74pt}=\begin{vmatrix} 
d_{2,2}(s-d)^2/4+d_{1,2}(s-d)/2+d_{0,2}  &   d_{1,2}(s-d)^2/4+(d_{1,1}+2d_{0,2})(s-d)/2
+d_{0,1}   &  d_{0,2}(s-d)^2/4+d_{0,1}(s-d)/2+d_{0,0}  &   0 \\
 0 & d_{2,2}(s-d)^2/4+d_{1,2}(s-d)/2+d_{0,2}  &   d_{1,2}(s-d)^2/4+(d_{1,1}+2d_{0,2})(s-d)/2
+d_{0,1}   &  d_{0,2}s-d)^2/4+d_{0,1}(s-d)/2+d_{0,0}      \\
 d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}  &   0 \\
 0 & d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}                       
    \end{vmatrix}
\end{equation*}
}\\ % end Small
where $s=x_{n\pm 2}+x_n$ and $d=x_{n\pm 2}-x_n$. Remark that $s^2-d^2=4p$, where
$p=x_{n\pm 2}x_n$.
{\Small
\begin{equation*} %\label{resul}
  \hspace*{-64pt}=\begin{vmatrix} 
d_{2,2}(s^2+d^2)/4+d_{1,2}s/2+d_{0,2}  &   d_{1,2}(s^2+d^2)/4+(d_{1,1}+2d_{0,2})s/2
+d_{0,1}   &  d_{0,2}(s^2+d^2)/4+d_{0,1}s/2+d_{0,0}  &   0 \\
 0 & d_{2,2}(s^2+d^2)/4+d_{1,2}s/2+d_{0,2}  &   d_{1,2}(s^2+d^2)/4+(d_{1,1}+2d_{0,2})s/2
+d_{0,1}   &  d_{0,2}(s^2+d^2)/4+d_{0,1}s/2+d_{0,0}      \\
 d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}  &   0 \\
 0 & d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}                       
    \end{vmatrix}
\end{equation*}
}   % end Small
\begin{equation*} %\label{resul}
  \hspace*{-64pt}=\begin{vmatrix} 
-d_{2,2}p+d_{0,2}  &   -d_{1,2}p
+d_{0,1}   &  -d_{0,2}p+d_{0,0}  &   0 \\
 0 & -d_{2,2}p+d_{0,2}  &   -d_{1,2}p
+d_{0,1}   &  -d_{0,2}p+d_{0,0}      \\
 d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}  &   0 \\
 0 & d_{2,2}s+d_{1,2}  &   d_{1,2}s+d_{1,1}+2d_{0,2}   &  d_{0,2}s+d_{0,1}                       
    \end{vmatrix}
\end{equation*}
\begin{equation*} %\label{resul}
  \hspace*{-64pt}=\begin{vmatrix} 
-d_{2,2}p+d_{0,2}  &   
d_{0,1} -d_{1,2}d_{0,2}/d_{2,2}  &  d_{0,0}-d_{0,2}^2/d_{2,2}  &   0 \\
 0 & d_{0,2}-d_{2,2}d_{0,0}/d_{0,2}  &   d_{0,1}-d_{1,2}d_{0,0}/d_{0,2}   &  -d_{0,2}p+d_{0,0}      \\
 d_{2,2}s+d_{1,2}  &   d_{1,1}+2d_{0,2}-d_{1,2}^2/d_{2,2}   &  d_{0,1}-d_{0,2}d_{1,2}/d_{2,2}  &   0 \\
 0 & d_{1,2}-d_{2,2}d_{0,1}/d_{0,2}  &   d_{1,1}+2d_{0,2}-d_{1,2}d_{0,1}/d_{0,2}   &  d_{0,2}s+d_{0,1}                       
    \end{vmatrix}
\end{equation*}
\begin{equation*} 
  \hspace*{-4pt}=(d_{2,2}p-d_{0,2})(d_{0,2}p-d_{0,0})\begin{vmatrix} 
   d_{1,1}+2d_{0,2}-d_{1,2}^2/d_{2,2}   &  d_{0,1}-d_{0,2}d_{1,2}/d_{2,2}   \\
 d_{1,2}-d_{2,2}d_{0,1}/d_{0,2}  &   d_{1,1}+2d_{0,2}-d_{1,2}d_{0,1}/d_{0,2}                        
    \end{vmatrix}\end{equation*}\begin{equation*}
-(d_{2,2}p-d_{0,2})(d_{0,2}s+d_{0,1})
\begin{vmatrix} 
 d_{0,2}-d_{2,2}d_{0,0}/d_{0,2}  &   d_{0,1}-d_{1,2}d_{0,0}/d_{0,2}      \\
 d_{1,1}+2d_{0,2}-d_{1,2}^2/d_{2,2}   &  d_{0,1}-d_{0,2}d_{1,2}/d_{2,2}  
    \end{vmatrix}\end{equation*}\begin{equation*}
-(d_{2,2}s+d_{1,2})(d_{0,2}p-d_{0,0})
\begin{vmatrix} 
d_{0,1} -d_{1,2}d_{0,2}/d_{2,2}  &  d_{0,0}-d_{0,2}^2/d_{2,2}   \\
 d_{1,2}-d_{2,2}d_{0,1}/d_{0,2}  &   d_{1,1}+2d_{0,2}-d_{1,2}d_{0,1}/d_{0,2}                         
    \end{vmatrix}\end{equation*}\begin{equation*}
+(d_{2,2}s+d_{1,2})(d_{0,2}s+d_{0,1})
  \begin{vmatrix} 
d_{0,1} -d_{1,2}d_{0,2}/d_{2,2}  &  d_{0,0}-d_{0,2}^2/d_{2,2}   \\
d_{0,2}-d_{2,2}d_{0,0}/d_{0,2}  &   d_{0,1}-d_{1,2}d_{0,0}/d_{0,2}  
    \end{vmatrix}
\end{equation*}

\bigskip

Hmm

With\\
$\displaystyle
f_n(z):= \dfrac{S_n(z)-\sqrt{P(z)}}{(z-z_0)Z_n(z)} = \dfrac{   (z-z_0) }{\alpha_n+\beta_n(z-z_0) - (z-z_0)f_{n+1}(z)  } \\
= \dfrac{   (z-z_0) }{\alpha_n+\beta_n(z-z_0) - 
   \dfrac{   (z-z_0)^2 }{\alpha_{n+1}+\beta_{n+1}(z-z_0) - 
         \dfrac{S_{n+2}(z)-\sqrt{P(z)}}{Z_{n+2}(z)} }
  }$

etc.


\bigskip




Multiplication formulas on a canonical form are investigated in \S~5.2  (p.307)
and appendices B and C
of  Maillard \& Boukraa's paper \cite{Maillard}.

\bigskip


\parpic{
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\psline{->}(0,2)(4,2)
\psline{->}(0,0)(0,3.5)
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\pscurve[linewidth=0.07](4,0)(3.75,1)(3.6,2)(3.75,3)(4,4)
%\psdot(1.1,2.2)
\psline(1.05,2.2)(3.8,3.1)
\psline[linestyle=dashed,dash=2mm 2mm](3.8,3.1)(3.8,0.9)
\uput[180](3.8,0.9){$P+Q$}\uput[90](1.05,2.2){$P$}\uput[90](2.1,2.6){$Q$}
\end{pspicture}
}
Finally, I return to the birational transformation to the Weierstrass cubic \eqref{xieta}
of p.~\pageref{xieta}. $(x,y)_{n+m}$ is the result of adding to $(x,y)_n$ $m$ times
the  step $(x,y)^{(\text{step},i)}$, according to the addition rule of
elliptic curves. We go to the $(\xi,\eta)$ plane with, say $(\tilde{\xi},\tilde{\eta})$
for $(x,y)_n$, and $(\xi,\eta)^*$ for $(x,y)^{(\text{step},i)}$:\\
$\displaystyle \tilde{\xi}= \dfrac{1}{x_n-z_i}, 
     \tilde{\eta}= \dfrac{X_1(x_n)+2y_n X_2(x_n)}{(x_n-z_i)^2}=
\tilde{\xi}^2[X_1(z_i+1/\tilde{\xi})+2y_nX_2(z_i+1/\tilde{\xi})], \\
 \xi^*= \dfrac{1}{x_m^{(\text{step},i)}-z_i}, 
 \eta^*= [X_1(x_m^{(\text{step},i)})+2y_m^{(\text{step},i)} X_2(x_m^{(\text{step},i)}) ]/(x_m^{(\text{step},i)}-z_i)^2=(\xi^*)^2[X_1(z_i+1/\xi^*)+2y_m^{(\text{step},i)} X_2(z_i+1/\xi^*)]. 
$
The straight line $\displaystyle  \eta = \dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }\xi+
\dfrac{\xi^*\tilde{\eta}-\tilde{\xi}\eta^*}{\xi^*-\tilde{\xi} }$
meets the cubic $\eta^2-\xi^4P(z_i+1/\xi)=
\displaystyle 
\left[ \dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }\right]^2 \xi^2+\dotsb
-P'(z_i)\xi^3 -P''(z_i)\xi^2/2-\dotsb  \\
= \left[ \dfrac{  (\xi^*)^2[X_1(x_m^{(\text{step},i)} )+2y_m^{(\text{step},i)} X_2(x_m^{(\text{step},i)} )]
                 -\tilde{\xi}^2[X_1(x_n)+2y_nX_2(x_n)]   }
   {\xi^*-\tilde{\xi} } \right]^2
\xi^2+\dotsb-P'(z_i)\xi^3 -P''(z_i)\xi^2/2-\dotsb$ at a third point $(\xi,
\eta)$: $P'(z_i)\xi = \{\  ... \}^2 -P''(z_i)/2 -P'(z_i)(\xi^*+\tilde{\xi}),\\
\eta = \dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }\xi+
\dfrac{\xi^*\tilde{\eta}-\tilde{\xi}\eta^*}{\xi^*-\tilde{\xi} }
= \dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }[\{\  ... \}^2 -P''(z_i)/2 -P'(z_i)(\xi^*+\tilde{\xi})]/P'(z_i)
+
\dfrac{\xi^*\tilde{\eta}-\tilde{\xi}\eta^*}{\xi^*-\tilde{\xi} }\\
= \dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }[\{\  ... \}^2 -P''(z_i)/2]/P'(z_i)
+
\dfrac{\tilde{\xi}\tilde{\eta}-\xi^*\eta^*}{\xi^*-\tilde{\xi} }
$

Whence an addition formula looking quite like the formula for the
Weierstrass $\wp$ function:

\begin{multline} \label{addition}
\dfrac{P'(z_i)}{x_{m+n}-z_i} =\\
\left[ \dfrac{  (x_n-z_i)[X_1(x_m^{(\text{step},i)} )+2y_m^{(\text{step},i)} X_2(x_m^{(\text{step},i)} )]}{(x_m^{(\text{step},i)}-z_i )( x_n-x_m^{(\text{step},i)} )} 
  -\dfrac{  (x_m^{(\text{step},i)}-z_i )[X_1(x_n)+2y_nX_2(x_n)]   }
   {(x_n-z_i)( x_n-x_m^{(\text{step},i)} ) }\right]^2\\
-P''(z_i)/2 -P'(z_i)\left[\dfrac{1}{ x_m^{(\text{step},i)} -z_i}
                          +\dfrac{1}{x_n-z_i  }\right]\\
y_{m+n}=\dfrac{-(x_{m+n}-z_i)^2\eta -X_1(x_{m+n})}{2X_2(x_{m+n})}=
\dfrac{-(x_{m+n}-z_i)^2\left[
\dfrac{\eta^*-\tilde{\eta}}{\xi^*-\tilde{\xi} }\xi+
\dfrac{\xi^*\tilde{\eta}-\tilde{\xi}\eta^*}{\xi^*-\tilde{\xi} } \right]
 -X_1(x_{m+n})}{2X_2(x_{m+n})}
\end{multline} 

With $2X_2y=-X_1+\sqrt{P}$,
$\displaystyle 
\dfrac{P'(z_i)}{x_{m+n}-z_i} =
\left[ \dfrac{  (x_n-z_i)\sqrt{P(x_m^{(\text{step},i)} )} }
   {(x_m^{(\text{step},i)}-z_i )( x_n-x_m^{(\text{step},i)} ) }
-\dfrac{  (x_m^{(\text{step},i)}-z_i )\sqrt{P(x_n)  } }
   {(x_n-z_i)( x_n-x_m^{(\text{step},i)} ) }\right]^2
-P''(z_i)/2 -P'(z_i)\left[\dfrac{1}{x_m^{(\text{step},i)}-z_i}
      +\dfrac{1}{ x_n-z_i  }
\right]
\\  \hspace*{-44pt}=
\dfrac{  (x_n-z_i)^2[P(x_m^{(\text{step},i)} )=P'(z_i)(x_m^{(\text{step},i)} 
   -z_i)+(P''(z_i)/2)(x_m^{(\text{step},i)} -z_i)^2+\dotsb ] }
   {(x_m^{(\text{step},i)}-z_i )^2( x_n-x_m^{(\text{step},i)} )^2 }
-2\dfrac{  \sqrt{P(x_m^{(\text{step},i)} )} \sqrt{P(x_n)}  }
   { ( x_n-x_m^{(\text{step},i)} )^2 }
 +\dfrac{  (x_m^{(\text{step},i)} -z_i)^2 [P(x_n)=P'(z_i)(x_n-z_i)+(P''(z_i)/2)(x_n-z_i)^2+\dotsb ]}
   {(x_n-z_i )^2( x_n-x_m^{(\text{step},i)} )^2 }
-P''(z_i)/2 \\-P'(z_i)\left[\dfrac{1}{ x_m^{(\text{step},i)}-z_i}
      +\dfrac{1}{ x_n-z_i  }
\right]=
$


Check: from \eqref{xsump},
$x_{n-1} = -\dfrac{Y_1(y_n)}{Y_2(y_n)} -x_n=$




\bigskip

Hmm, other construction: I want $x_{m+n}$ to be the sum of two elements
$\tilde{x}_n$ and $\tilde{x}_m$ of the \emph{\textbf{same}} sequence.
It will be more convenient to add a translation and consider $x_{m+n+j}$
with some fixed integer $j$ (which will be set to $-1$ later on). From the birational
transformation sending the biquadratic on a cubic



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\psline{->}(0,0)(0,3.5) \uput[0](0,3.5){$\eta$}
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\end{pspicture}
}
 \noindent
$\displaystyle \tilde{\xi}_n= \dfrac{1}{\tilde{x}_n-z_i}, 
     \tilde{\eta}_n= \dfrac{X_1(\tilde{x}_n)+2\tilde{y}_n X_2(\tilde{x}_n)}{(\tilde{x}_n-z_i)^2},\\
\tilde{\xi}_m= \dfrac{1}{\tilde{x}_m-z_i}, 
     \tilde{\eta}_m= \dfrac{X_1(\tilde{x}_m)+2\tilde{y}_m X_2(\tilde{x}_m)}{(\tilde{x}_m-z_i)^2}$. Remark that $(\tilde{x}_n, \tilde{y}_{n+1})$ corresponds to
$(\tilde{\xi}_n,-\tilde{\eta}_n)$. \\ %\picskip{0}
$(\xi_{m+n+j}\eta_{m+n+j})$ is (constants times) the Weierstrass elliptic functions
$\wp$ and $\wp'$ at $a(m+n+j)+b_i$. The constant term $b_i$, which we don't have to evaluate, fortunately,
depends on $i$ as the birational transformation contains $z_i$. Well, if we want $a(m+n+j)+b_i$
to be the sum of two arguments with the same constant term, the arguments must be
$am+(b_i+aj)/2$ and $an+(b_i+aj)/2$, so
$(\tilde{\xi}_n,  \tilde{\eta}_n)= $ cts $\times \ \wp$ and $\wp'$ at $an+(b_i+aj)/2$.\\
Now, $(\xi_{m+n+j},\eta_{m+n+j})$ is the sum of 
$(\tilde{\xi}_m,\tilde{\eta}_m)$ and $(\tilde{\xi}_n,\tilde{\eta}_n)$ according to the addition rule of
elliptic curves. 
 The straight line $\displaystyle  \eta = \dfrac{\tilde{\eta}_n-\tilde{\eta}_m}{\tilde{\xi}_n-\tilde{\xi}_m }\xi+
\dfrac{\tilde{\xi}_n\tilde{\eta}_m-\tilde{\xi}_m\tilde{\eta}_n}{\tilde{\xi}_n-\tilde{\xi}_m }$
meets the cubic $0=\eta^2-\xi^4P(z_i+1/\xi)=
\displaystyle 
\left[ \dfrac{\tilde{\eta}_n-\tilde{\eta}_m}{\tilde{\xi}_n-\tilde{\xi}_m }\right]^2 \xi^2+\dotsb
-P'(z_i)\xi^3 -P''(z_i)\xi^2/2-\dotsb  \\
= \left[ \dfrac{  (\tilde{x}_m-z_i)^2[X_1(\tilde{x}_n )+2\tilde{y}_n X_2(\tilde{x}_n )]
                 -(\tilde{x}_n-z_i)^2[X_1(\tilde{x}_m)+2\tilde{y}_mX_2(\tilde{x}_m)]   }
   { (\tilde{x}_m-z_i)(\tilde{x}_n-z_i) (\tilde{x}_m-\tilde{x}_n) } \right]^2
\xi^2+\dotsb-P'(z_i)\xi^3 -P''(z_i)\xi^2/2-\dotsb$ at a third point $(\xi,
\eta)$: $P'(z_i)\xi = \{\  ... \}^2 -P''(z_i)/2 -P'(z_i)(\tilde{\xi}_n+\tilde{\xi}_m),\\
\eta = \dfrac{\tilde{\eta}_n-\tilde{\eta}_m}{\tilde{\xi}_n-\tilde{\xi}_m }\xi+
\dfrac{\tilde{\xi}_n\tilde{\eta}_m-\tilde{\xi}_m\tilde{\eta}_n}{\tilde{\xi}_n-\tilde{\xi}_m }
= \dfrac{\tilde{\eta}_n-\tilde{\eta}_m}{\tilde{\xi}_n-\tilde{\xi}_m }[\{\  ... \}^2 -P''(z_i)/2 -P'(z_i) )(\tilde{\xi}_n+\tilde{\xi}_m)]/P'(z_i)
+\dfrac{\tilde{\xi}_n\tilde{\eta}_m-\tilde{\xi}_m\tilde{\eta}_n}{\tilde{\xi}_n-\tilde{\xi}_m }
$

Whence 

\begin{multline} \label{additiontilde}
\dfrac{P'(z_i)}{x_{m+n+j}-z_i} =\\
\left[ \dfrac{  (\tilde{x}_m-z_i)[X_1(\tilde{x}_n )+2\tilde{y}_n X_2(\tilde{x}_n )]}{(\tilde{x}_n-z_i )( \tilde{x}_m-\tilde{x}_n )} 
  -\dfrac{  (\tilde{x}_n-z_i )[X_1(\tilde{x}_m)+2\tilde{y}_mX_2(\tilde{x}_m)]   }
   {(\tilde{x}_m-z_i)( \tilde{x}_m-\tilde{x}_n ) }\right]^2\\
-P''(z_i)/2 -P'(z_i)\left[\dfrac{1}{ \tilde{x}_m -z_i}
                          +\dfrac{1}{\tilde{x}_n-z_i  }\right]
\end{multline} 

With $2X_2y=-X_1+\sqrt{P}$,
$\displaystyle 
\dfrac{P'(z_i)}{x_{m+n+j}-z_i} =
\left[ \dfrac{  (\tilde{x}_m-z_i)\sqrt{P(\tilde{x}_n) } }
   {(\tilde{x}_n-z_i )( \tilde{x}_m-\tilde{x}_n) }
-\dfrac{  (\tilde{x}_n-z_i )\sqrt{P(\tilde{x}_m)  } }
   {(\tilde{x}_m-z_i)( \tilde{x}_m-\tilde{x}_n) }\right]^2
-P''(z_i)/2 -P'(z_i)\left[\dfrac{1}{\tilde{x}_n-z_i}
      +\dfrac{1}{ \tilde{x}_m-z_i  }
\right]
$

Check: $x_n=\sin(\theta_0+nh)$, $y_n$ and $y_{n+1}= x_{n\mp 1/2} \Rightarrow
y=x\cos(h/2)\pm \sin(h/2)\sqrt{1-x^2}: y^2 -2\cos(h/2)xy +x^2 -\sin^2(h/2)=0, P(x)=X_1^2(x)-4X_0(x)X_2(x)=4\sin^2(h/2)(1-x^2)$. With $z_i=1$,\\
\hspace*{-44pt}$\displaystyle \xi_n=\dfrac{1}{\sin(\theta_0+nh)-1}= -\dfrac{1}{2\sin^2(\pi/4-\theta_0/2-nh/2)},
\eta_n= \dfrac{-2\cos(h/2)\sin(\theta_0+nh)+2\sin(\theta_0+(n-1/2)h)}{(\sin(\theta_0+nh)-1)^2}
=\dfrac{-2\sin(h/2)\cos(\theta_0+nh)}{(\sin(\theta_0+nh)-1)^2} =\dfrac{-\sin(h/2)\cos(\pi/4-\theta_0/2-nh/2)}{\sin^3(\pi/4-\theta_0/2-nh/2)}
$;
$\eta^2= 4\sin^2(h/2)\xi^4(1- x^2)= 4\sin^2(h/2)\xi^4[1-(1+1/\xi)^2]=
4\sin^2(h/2)\xi^2(-2\xi-1)$. Also, let
$\tilde{x}_n=\sin(\tilde{\theta}_0+nh)$.   Let $\theta=\tilde{\theta}_0+nh$
and $\varphi=\tilde{\theta}_0+mh$, then we expect $x_{m+n} =\sin(\theta+\varphi)$.
Is $\displaystyle
\dfrac{P'(1)}{x_{m+n}-1} =\dfrac{8\sin^2(h/2)}{1-\sin(\theta_0+(m+n)h)}$
the right-hand side of \eqref{additiontilde}? The big square above is\\
\hspace*{-22pt}$\displaystyle \left[ \dfrac{  (\sin\varphi -1)[-2\cos(h/2)\sin\theta+2\sin(\theta-h/2)]}{(\sin\theta-1 )(\sin\varphi-\sin\theta )} 
  -\dfrac{  (\sin\theta-1 )[-2\cos(h/2)\sin\varphi+2\sin(\varphi-h/2)]   }
   {(\sin\varphi-1)( \sin\varphi-\sin\theta ) }\right]^2
=4\sin^2(h/2)\left[\dfrac{  -\cos\theta(\sin\varphi -1)}{(\sin\theta-1 )(\sin\varphi-\sin\theta )} 
  +\dfrac{  \cos\varphi(\sin\theta-1 )  }
   {(\sin\varphi-1)( \sin\varphi-\sin\theta ) }\right]^2 \\
=\dfrac{4\sin^2(h/2)}{(\sin\varphi-\sin\theta)^2} \left[ \dfrac{  -(\sin\theta+1)(\sin\varphi -1)^2}{\sin\theta-1 } -\dfrac{  (\sin\varphi+1)(\sin\theta -1)^2}{\sin\varphi-1 } 
  -2\cos\theta\cos\varphi\right]\\
=4\sin^2(h/2)\left[ -\dfrac{  \sin\theta+1}{\sin\theta-1 } 
                   +2
                  +\dfrac{  \cos^2\theta-2\cos\theta\cos\varphi+\cos^2\varphi}{(\sin\varphi-\sin\theta )^2}
                   -\dfrac{  \sin\varphi+1}{\sin\varphi-1 } 
             \right] 
$\\
to which we add $\displaystyle -P''(z_i)/2 -P'(z_i)\left[\dfrac{1}{\tilde{x}_n-z_i}
      +\dfrac{1}{ \tilde{x}_m-z_i  }
\right]
=4\sin^2(h/2)+8\sin^2(h/2)\left[\dfrac{1}{\sin\theta-1}
      +\dfrac{1}{ \sin\varphi-1  }
\right],
$
which makes $\displaystyle 4\sin^2(h/2)\left[1
                  +\dfrac{  \cos^2\theta-2\cos\theta\cos\varphi+\cos^2\varphi}{(\sin\varphi-\sin\theta )^2}       \right] =4\sin^2(h/2)
                  \dfrac{  2-2\cos(\theta-\varphi)}{(\sin\varphi-\sin\theta )^2}    
=\dfrac{4\sin^2(h/2) }{\cos^2((\theta+\varphi)/2)}   
=\dfrac{8\sin^2(h/2) }{1+\cos(\theta+\varphi)}=\dfrac{8\sin^2(h/2) }{1-\sin(\theta+\varphi-\pi/2)}$, so that $\theta_0+(m+n)h = 2\tilde{\theta}_0+(m+n)h-\pi/2$. Where
is this $\pi/2$ coming from??

Ooh, that's because the addition formula associated to the straight line in
the $(\xi,\eta)$ plane is related to the Weierstrass function which has a double
pole at the origin. So, we must choose the parameter in $\xi=(x-z_i)^{-1}$ such that
the origin of the parameter occurs at $x=z_i$. This means here $x=\sin(\pi/2+(\theta-\pi/2))$, so that
$\theta-\pi/2$ is the parameter of interest. YES! Then,
$\tilde{x}_n=\sin(\tilde{\theta}_0+nh)=\sin(\theta_0/2 +\pi/4 +nh)=\sin(\pi/2+(\theta_0/2-\pi/4)+nh)$ and, indeed, $\theta_0+(m+n)h-\pi/2$ is the sum of $\theta_0/2+mh-\pi/4$ and
$\theta_0+nh-\pi/4$.
      
\bigskip

Return to \eqref{additiontilde} with a generic polynomial $P$.

Let $n^{(i)}$ be such that $x_{n^{(i)}}=z_i$. The natural parameter with respect
to the addition rule is then $n-n^{(i)}$, so $m+n-n^{(i)}= [m-n^{(i)}/2]
+[n-n^{(i)}/2]$, and we have $\tilde{x}_n = x_{n^{(i)}+[n-n^{(i)}/2]} =
x_{n^{(i)}/2+n}$. Actually, one must write $\tilde{x}_n^{(i)}$, as this
construction depends on $i$. But how is it done?

We don't need $n^{(i)}$, just that $\tilde{x}_0^{(i)}$ produces $x_0$
by duplication: $\tilde{x}_0 = x_{n^{(i)}+[-n^{(i)}/2]}$ duplicates into 
$ x_{n^{(i)}+2[-n^{(i)}/2]}=x_0$ of course. 
 
\bigskip

A wonderful simplification holds when $m=n$. Then, $\sqrt{ P(\tilde{x}_m) }=
 \sqrt{ P(\tilde{x}_n) } \\+(\tilde{x}_m -\tilde{x}_n) P'(\tilde{x}_n)  
/(2\sqrt{ P(\tilde{x}_n) })+\dotsb $, and 

$\displaystyle 
\dfrac{P'(z_i)}{x_{2n+j}-z_i} =
\left[ \dfrac{  2\sqrt{P(\tilde{x}_n) } }
   {\tilde{x}_n-z_i  }
-\dfrac{  P'(\tilde{x}_n )  }{2\sqrt{P(\tilde{x}_n)  } }
    \right]^2
-\dfrac{P''(z_i)}{2} -\dfrac{2P'(z_i)}{\tilde{x}_n-z_i}
=\dfrac{  4P(\tilde{x}_n)  }
   {(\tilde{x}_n-z_i)^2  }    -2\dfrac{ P'(\tilde{x}_n )  +P'(z_i)}{\tilde{x}_n-z_i}
+\dfrac{  (P'(\tilde{x}_n ))^2  }{4P(\tilde{x}_n)   }
-\dfrac{P''(z_i)}{2} $.

In the $(\xi,\eta)$ plane: the tangent  to the cubic $\eta^2=P_3(\xi)$
at $(\tilde{\xi}_n,\tilde{\eta}_n)$ meets the curve at $(\xi_{2n+j},-\eta_{2n+j})$:
$\displaystyle \eta= \tilde{\eta}_n +\dfrac{P'_3(\tilde{\xi}_n)}{2\sqrt{P_3(\tilde{\xi}_n)}= 2\tilde{\eta}_n  }(\xi-\tilde{\xi}_n)$ and $\eta^2=P_3(\xi) \Rightarrow
\displaystyle \dfrac{(P'_3(\tilde{\xi}_n))^2}{4P_3(\tilde{\xi}_n)}(\xi-\tilde{\xi}_n)^2
  +P'_3(\tilde{\xi}_n)(\xi-\tilde{\xi}_n)+\tilde{\eta}_n^2
-(P_3'''/6)(\xi-\tilde{\xi}_n)^3
-(P_3''(\tilde{\xi}_n)/2)(\xi-\tilde{\xi}_n)^2 -P_3'(\tilde{\xi}_n)(\xi-\tilde{\xi}_n)-P_3(\tilde{\xi}_n)=0 \Rightarrow
\xi_{2n+j} = \tilde{\xi}_n+\left[  \dfrac{(P'_3(\tilde{\xi}_n))^2}{4P_3(\tilde{\xi}_n)}
-(P_3''(\tilde{\xi}_n)/2)\right]/(P_3'''/6)
=\dfrac{(3\tilde{\xi}_n^2 +2\alpha \tilde{\xi}_n +\beta)^2}{4\tilde{\xi}_n^3 +4\alpha \tilde{\xi}_n^2 +4\beta\tilde{\xi}_n+4\gamma}-\alpha-2\tilde{\xi}_n
=\dfrac{\tilde{\xi}_n^4 -2\beta\tilde{\xi}_n^2-8\gamma\tilde{\xi}_n+\beta^2-4\alpha\gamma}{4\tilde{\xi}_n^3 +4\alpha \tilde{\xi}_n^2 +4\beta\tilde{\xi}_n+4\gamma = 4\tilde{\eta}_n^2}
\text{\ if\ } P_3(x)=x^3+\alpha x^2+\beta x+\gamma,$

does not look like a square, \emph{\textbf{BUT}}(2 Jan. 2012)\\
\hspace*{-22pt}$\displaystyle \xi_{2n+j} -\lambda = \dfrac{\tilde{\xi}_n^4 -2\beta\tilde{\xi}_n^2-8\gamma\tilde{\xi}_n+\beta^2-4\alpha\gamma}{4\tilde{\xi}_n^3 +4\alpha \tilde{\xi}_n^2 +4\beta\tilde{\xi}_n+4\gamma = 4\tilde{\eta}_n^2}-\lambda =
\dfrac{\tilde{\xi}_n^4 -4\lambda\tilde{\xi}_n^3 -(4\alpha \lambda+2\beta)\tilde{\xi}_n^2
-(4\beta\lambda+8\gamma)\tilde{\xi}_n+\beta^2-4\alpha\gamma-4\gamma\lambda}{4\tilde{\eta}_n^2}\\
=\dfrac{( \tilde{\xi}_n^2 -2\lambda \tilde{\xi}_n -2\lambda^2 -2\alpha\lambda-\beta)^2}{4\tilde{\eta}_n^2} \text{\  if\   } 4\lambda  (2\lambda^2 +2\alpha\lambda+\beta)=
-4\beta\lambda-8\gamma \text{\  and\  } (2\lambda^2 +2\alpha\lambda+\beta)^2=\beta^2-4\alpha\gamma-4\gamma\lambda \Leftrightarrow P_3(\lambda)=0 \Leftrightarrow \lambda=
\lambda_{i,k}= -1/(z_i-$ some other $z_k)
\Leftrightarrow  x_{2n+j}= z_i
\\
\hspace*{-22pt}+ \dfrac{1}{\xi_{2n+j}= -\dfrac{1}{z_i-z_k} +\dfrac{( \tilde{\xi}_n^2 -2\lambda \tilde{\xi}_n -2\lambda^2 -2\alpha\lambda-\beta)^2}{4\tilde{\eta}_n^2}  }
\Leftrightarrow  x_{2n+j}-z_k=  \dfrac{(z_i-z_k) ( \tilde{\xi}_n^2 -2\lambda \tilde{\xi}_n -2\lambda^2 -2\alpha\lambda-\beta)^2}{4 \xi_{2n+j} \tilde{\eta}_n^2}    $.

It figures: from \eqref{wprep} p.~\pageref{wprep}, $\xi$ is a Weierstrass $\wp$
function, and the three differences $\wp-e_k = \xi -\lambda_{i,k}$ are the squares
of the fundamental elliptic functions of Halphen (see Appendix C of 
Robert Conte  \& Micheline Musette's \textsl{
The Painlev\'e Handbook}, Springer 2008).
 

Strange: squares are recovered with $x_{2n+j}-z_k$, $k\neq i$, although $\xi_{2n+j}=
(x_{2n+j}-z_i)^{-1}$, $\tilde{\xi}_n$, and $(\alpha, \beta, \gamma) = (P''(z_i)/2, P'''(z_i)/6, P''''/24)/P'(z_i)$ depend on $i$. Incidentally, 
$2\lambda^2 +2\alpha\lambda+\beta = -\beta-2\gamma/\lambda_{i,k} = [-P'''(z_i)/6 -2 P''''/(24\lambda_{i,k})]/P'(z_i)= [ -4z_i +\sum_1^4 z_r +2(z_i-z_k)]/((z_i-z_k)(z_i-z_m)(z_i-z_p))=[ -z_i-z_k + z_m +z_p]/((z_i-z_k)(z_i-z_m)(z_i-z_p))$,
where $k,m,p$ are the three indexes $\neq i$.   

$\displaystyle  \dfrac{x_{2n+j}-z_k}{x_{2n+j}-z_i}=  \dfrac{(z_i-z_k) ( \tilde{\xi}_n^2 -2\lambda \tilde{\xi}_n -2\lambda^2 -2\alpha\lambda-\beta)^2}{4  \tilde{\eta}_n^2},
\text{\ with\ } \lambda=\lambda_{i,k}= \dfrac{1}{z_k-z_i}\\
 \dfrac{x_{2n+j}-z_k}{x_{2n+j}-z_m}=  \dfrac{(z_i-z_k) ( \tilde{\xi}_n^2 -2\lambda_{i,k} \tilde{\xi}_n -2\lambda_{i,k}^2 -2\alpha\lambda_{i,k}-\beta)^2}
{(z_i-z_m) ( \tilde{\xi}_n^2 -2\lambda_{i,m} \tilde{\xi}_n -2\lambda_{i,m}^2 -2\alpha\lambda_{i,m}-\beta)^2}
$


$\displaystyle \eta_{2n+j}= -\tilde{\eta}_n -\dfrac{P'_3(\tilde{\xi}_n)}{2\tilde{\eta}_n  }
\left[  \dfrac{(P'_3(\tilde{\xi}_n))^2}{4P_3(\tilde{\xi}_n)}
-(P_3''(\tilde{\xi}_n)/2)\right]/(P_3'''/6),
 $

From \eqref{xsump},
$\displaystyle x_{2n+j+1}= -\dfrac{Y_1(y_{2n+j+1})}{Y_2(y_{2n+j+1})} -x_{2n+j}$

For any $k\in\{1,2,3,4\}$, $\displaystyle x_{2n+j+1}-z_k= -\dfrac{Y_1(y_{2n+j+1})+2Y_2(y_{2n+j+1})z_k}{Y_2(y_{2n+j+1})} -(x_{2n+j}-z_k)
=\dfrac{Y_0(y_{2n+j+1})+Y_1(y_{2n+j+1})z_k+Y_2(y_{2n+j+1})z_k^2}{Y_2(y_{2n+j+1}) (x_{2n+j}-z_k)},
$

Hmm,  $x_{2n+j+1}$ follows from $x_{2n+j}$ by adding the
step in \eqref{xistep}-\eqref{etastep}, p.~\pageref{xistep}, so
$(\xi_{2n+j+1},\eta_{2n+j+1})= (\xi_{2n+j},\eta_{2n+j})+
(\xi_{\text{step}},\eta_{\text{step}}) =(\tilde{\xi}_n,\tilde{\eta}_n) 
+ (\tilde{\xi}_n,\tilde{eta}_n) + (\xi_{\text{step}},\eta_{\text{step}}) =
(\tilde{\xi}_n,\tilde{\eta}_n) 
+(\tilde{\xi}_{n+1},\tilde{\eta}_{n+1})\\  \displaystyle
\dfrac{\tilde{\eta}_{n+1} -\tilde{\eta}_n  }{\tilde{\xi}_{n+1}-\tilde{\xi}_n }=
\dfrac{-\eta_{2n+j+1} -\tilde{\eta}_n  }{\xi_{2n+j+1}-\tilde{\xi}_n },
(\eta_{2n+j+1})^2= P_3(\xi_{2n+j+1}) = \left[ \tilde{\eta}_n  +
(\xi_{2n+j+1}-\tilde{\xi}_n)  \dfrac{\tilde{\eta}_{n+1} -\tilde{\eta}_n  }{\tilde{\xi}_{n+1}-\tilde{\xi}_n }\right]^2,\\
\xi_{2n+j+1}= \dfrac{1}{P_3'''/6}\left[ -P_3''(0)/2 +\left(
\dfrac{\tilde{\eta}_{n+1} -\tilde{\eta}_n  }{\tilde{\xi}_{n+1}-\tilde{\xi}_n }\right)^2
\right] -\tilde{\xi}_{n+1}-\tilde{\xi}_n 
$ 

Also,  $(\tilde{\xi}_{n+1},\tilde{\eta}_{n+1})= 
 (\tilde{\xi}_n,\tilde{\eta}_n) + (\xi_{\text{step}},\eta_{\text{step}})$: 

$\displaystyle
\dfrac{ \tilde{\eta}_n-\eta_{\text{step}} } {\tilde{\xi}_n-\xi_{\text{step}}  }=
\dfrac{ -\tilde{\eta}_{n+1}-\eta_{\text{step}} }  {\tilde{\xi}_{n+1}-\xi_{\text{step}}  }$





No, there is nothing strange: $\wp-e_i$ is the square of an elliptic function
because it has a double zero AND a double pole. Now, $x_{2n+j}-z_i$ has a double zero, but\dots  Ah, when $x\to\infty$, and $F(x,y)=Y_2(y)x^2+Y_1(y)x+Y_0(y)=0\Rightarrow y=$ the
roots $u$ and $v$ of $Y_2(y)=0$. 








\bigskip


\section{\label{phipsi}General difference operator} \  \\
\renewcommand{\mylmark}{General difference operators.}

$$
\mathcal{D}f(x) = \dfrac{f(\psi(x))-f\varphi(x)) }{\psi(x)-\varphi(x))}
$$
must be a rational function of degree 2 for $f(x)=1/(x+c)$:
$$
-\dfrac{1}{(\psi(x)+c)(\varphi(x)+c)} =-\dfrac{1}{\psi(x)\varphi(x)+c(\psi(x)+\varphi(x))+c^2}
$$
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\uput[270](0.5,-0.1){$x(s)$} \uput[270](3.2,-0.1){$x(s+1)$}
\uput[180](-0.1,0.4){$\varphi(x(s))=y(s)$}\uput[180](-0.1,1.4){$\psi(x(s))=y(s+1)$}\uput[180](-0.1,3.1){$y(s+2)$}
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\end{pspicture}
}
New notation: $F(x,y)=X_2(x)y^2+X_1(x)y+X_0(x)=0$, where the $X$'s are polynomials
of degree $\leqslant 2$. So,
$\varphi+\psi=-X_1/X_2$ and $\varphi\psi=X_0/X_2\Rightarrow$ a (nonsymmetric)
biquadratic relation between $x$ and $\varphi(x)$ or $\psi(x)$. We now are
sure to have two elliptic lattices for the $x$'s and the $y$'s.\\
\   \\ \picskip{1}

\bigskip


\textbf{ Relation between $y_n$ and $y_{n+1}$:} use $y_n+y_{n+1}=-X_1(x_n)/X_2(x_n)$ and
$y_ny_{n+1}=X_0(x_n)/X_2(x_n)$, and consider the resultant of the two
polynomials in $x_n$\\
$ (y_n+y_{n+1})X_2(x_n)+X_1(x_n)\  \text{ and\ } y_ny_{n+1}X_2(x_n)-X_0(x_n),$\\ 
which is the full symmetric biquadratic relation between $y_n$ and $y_{n+1}$.

Of course, the same holds for $x_n$ and $x_{n+1}$, just by considering
$F(x,y)=Y_2(y)x^2+Y_1(y)x+Y_0(y)$. The polynomials are now\\
$ (x_n+x_{n+1})Y_2(y_{n+1})+Y_1(y_{n+1})\  \text{ and\ } x_nx_{n+1}Y_2(y_{n+1})-Y_0(y_{n+1}).$ 

Exemple: with  $\displaystyle F(x,y)=(\underbrace{x^2/2+x+1}_{X_2(x)})y^2 +(\underbrace{2x^2-2x-3}_{X_1(x)})y +x^2+x+1$,
one has

\noindent{\Small $(49/4*x_{n+1}^2 + 59/2*x_{n+1} + 71/4)*x_n^2 + (59/2*x_{n+1}^2 + 85/2*x_{n+1} + 17/2)*x_n 
+ 71/4*x_{n+1}^2 + 17/2*x_{n+1} + 1$ and
$ (37/4*y_{n+1}^2 + 3*y_{n+1} + 1/4)*y_n^2 + (3*y_{n+1}^2 - 55/2*y_{n+1} - 9/2)*y_n 
+ 1/4*y_{n+1}^2 - 9/2*y_{n+1} + 21$.}

\enlargethispage{33pt}

\textbf{Recurrence relations of the $x$'s and the $y$'s:}

$x_{n-1}+2x_n+x_{n+1} = \displaystyle -\dfrac{Y_1(y_n)}{Y_2(y_n)}
                           -\dfrac{Y_1(y_{n+1})}{Y_2(y_{n+1})} $
is a symmetric rational function of $y_n$ and $y_{n+1}$, so, a rational
function of $x_n$.

For the $y$'s:
$y_{n-1}+y_n+y_{n+1} = \displaystyle -\dfrac{X_1(x_{n-1})}{X_2(x_{n-1})}
                           -\dfrac{X_1(x_{n})}{X_2(x_{n})} $
is a symmetric rational function of $x_{n-1}$ and $x_{n}$, so, a rational
function of $y_n$.

\newpage

%\vspace*{-22pt}\hspace*{-22pt}
%\includegraphics[angle=0, scale=0.5]{ellnic02.ps}
%\includegraphics[angle=0, scale=0.5]{ellnic04.ps}
%\enlargethispage{55pt}
%\includegraphics[angle=0, scale=0.5]{ellnic01.ps}
%\includegraphics[angle=0, scale=0.5]{ellnic03.ps}



\section{Recurrences of biorthogonal rational functions.}
\renewcommand{\mylmark}{Biorthogonal rationa functions}

From excerpts of Spiridonov \& Zhedanov \cite{SpZ1}, also

A. Zhedanov, Biorthogonal rational functions and generalized eigenvalue problem, {\sl J. Approx.
Theory} {\bf 101} (1999), no. 2, 303--329, and \cite{ZhedPad}.

Also Brezinski, Iserles, Ismail, Masson, Norsett.

\subsection{Pad\'e and interpolatory continued fractions}\  \\

\subsubsection{Pad\'e.}
$\displaystyle{ \dfrac{\alpha_0}{x-\beta_0
               +\dfrac{\alpha_1}{x-\beta_1
               +\dfrac{\ddots}{\ \ \ \ \ \ +\dfrac{\alpha_{n-1}}
                                              {x-\beta_{n-1}}}}}} $
matches a given Laurent expansion $c_0/x+c_1/x^2+\dotsb$ at $\infty$
up to the $c_{2n-1}/x^{2n}$ term. Numerators and denominators satisfy
the recurrence relation $P_{n+1}(x)=(x-\beta_n)P_n(x)+\alpha_n
P_{n-1}(x)$, suggesting some kind of (formal?) orthogonality.
This is even more obvious in the matrix-vector setting

$$
\begin{bmatrix}
 \beta_0 & \sqrt{-\alpha_1}  &             &            &       \\
\sqrt{-\alpha_1}&\beta_1   & \sqrt{-\alpha_2}&         &       \\
                & \ddots  &\ddots           &\ddots    &      \\
                &         & \ddots         & \ddots  & \ddots
\end{bmatrix} \begin{bmatrix} P_0(x) \\ P_1(x) \\  \vdots \\ \vdots\end{bmatrix}
= x   \begin{bmatrix} P_0(x) \\ P_1(x) \\  \vdots \\ \vdots\end{bmatrix}.
$$

If one wants to approximate a Taylor expansion about the origin, just take
$z=1/x$ and rewrite the continued fraction as

$\displaystyle{ \dfrac{\alpha_0 z}{1-z\beta_0
               +\dfrac{\alpha_1 z^2}{1-\beta_1 z
               +\dfrac{\ddots}{\ \ \ \ \ \ +\dfrac{\alpha_{n-1} z^2}
                                              {1-\beta_{n-1}z}}}}} $
which matches a given Taylor-Maclaurin expansion up to the $z^{2n}$ term.

A slower progression is achieved with the \emph{corresponding
continued fraction} to a given expansion

$$\dfrac{\alpha'_0 z}
                                   {1+\dfrac{\alpha'_1 z}
                                      {1+\dfrac{\ddots}
                                  {\ \ \ \ \ \ \ +\alpha'_n z}}}$$

\subsubsection{Interpolation.\label{rinterp}}
Rational interpolations to a given set of values at
$x=y_0, y_1,\dotsc$ (yes, the relevant set will be a $y-$lattice) are achieved by

$$ \dfrac{q_n(x)}{p_n(x)} = \alpha'_0+\dfrac{x-y_0}
                                   {\alpha'_1+\dfrac{x-y_1}
                                      {\dfrac{\ddots}
                                  {\ \ \ \ \ \ \ \alpha'_{n-1}+\dfrac{x-y_{n-1}}{\alpha'_n}}}}$$
(Thiele) which agree with a given set up to $x=y_n$.

Remark that the degrees are $n/n$ for $q_{2n}/p_{2n}$, $(n+1)/n$ for
$q_{2n+1}/p_{2n+1}$. Contract the recurrences

$p_{2n}(x)=\alpha'_{2n}p_{2n-1}(x)+(x-y_{2n-1})p_{2n-2}(x),
p_{2n+1}(x)=\alpha'_{2n+1}p_{2n}(x)+(x-y_{2n})p_{2n-1}(x),
p_{2n+2}(x)=\alpha'_{2n+2}p_{2n+1}(x)+(x-y_{2n+1})p_{2n}(x),$

into $\displaystyle p_{2n+2}(x)=\alpha'_{2n+2}\left[\alpha'_{2n+1}p_{2n}(x)+(x-y_{2n})
        \dfrac{p_{2n}(x)-(x-y_{2n-1})p_{2n-2}(x)}{\alpha'_{2n}}\right]
   +(x-y_{2n+1})p_{2n}(x)$,

so, a recurrence relation for  $P_n=$ const. $p_{2n}$:

$$ P_{n+1}(x) = (x-\beta_n) P_n(x) +\alpha_n(x-y_{2n-1})(x-y_{2n})P_{n-1}(x).$$


Consider now rational functions $\displaystyle R_n(x)=\dfrac{P_n(x)}{(x-y_2)(x-y_4)\dotsb (x-y_{2n})}$:

$$ (x-y_{2n+2})R_{n+1}(x) = (x-\beta_n)R_n(x) +\alpha_n (x-y_{2n-1}) R_{n-1}(x),$$
so that the matrix-vector setting is now


$$
\begin{bmatrix}
 \beta_0 & -y_2            &             &            &       \\
\alpha_1y_1&\beta_1   & -y_4&         &       \\
                & \ddots  &\ddots           &\ddots    &      \\
                &         & \ddots         & \ddots  & \ddots
\end{bmatrix} \begin{bmatrix} R_0(x) \\ R_1(x) \\  \vdots \\ \vdots\end{bmatrix}
= x  
 \begin{bmatrix}
 1 & -1            &             &            &       \\
\alpha_1&  1   & -1&         &       \\
                & \ddots  &\ddots           &\ddots    &      \\
                &         & \ddots         & \ddots  & \ddots
\end{bmatrix} \begin{bmatrix} R_0(x) \\ R_1(x) \\  \vdots \\ \vdots\end{bmatrix}.
$$

So, $\{R_0,R_1,\dotsc\}$ is a right eigenvector which is in some way
biorthogonal to the set of left eigenvectors $\{T_0,T_1,\dotsc\}$
satisfying the recurrence

$$ \alpha_{n+1}(x-y_{2n+1})T_{n+1}(x) = -(x-\beta_n)T_n(x)+(x-y_{2n})T_{n-1}(x),$$
which is of the same structure that the recurrence of the $R_n$'s,
but with the odd $x$'s interchanged with the even $x$'s. Actually,
$T_n(x)$ is a constant times the same $P_n(x)$ as before, divided by $(x-y_1)(x-y_3)\dotsc (x-y_{2n-1})$.
$P_n(x)$ is a constant times the determinant

$$
\begin{vmatrix}
 x-\beta_0 & -(x-y_2)            &             &            &       \\
\alpha_1(x-y_1)&x-\beta_1   & -(x-y_4)&         &       \\
                & \ddots  &\ddots           &\ddots    &      \\
                &         &              & \alpha_{n-1}(x-y_{2n-3})  & x-\beta_{n-1}
\end{vmatrix}$$
 
\subsection{Biorthogonality and orthogonality}\  \\

But what is the bilinear form exhibiting the biorthogonality condition?
Could it be a linear form $\mathscr{L}$ applied to a product? Let us try,
for $k<n$,

\begin{equation*} \begin{split}
  0&=\mathscr{L}(R_nT_k) = \mathscr{L}\left(\dfrac{P_n(x)}{(x-y_2)\dotsc (x-y_{2n})}
                                            \dfrac{P_k(x)}{(x-y_1)\dotsc (x-y_{2k-1})}\right) \\
               &=\mathscr{L}\left(\dfrac{P_n(x) P_k(x) (x-y_{2k+1})\dotsc (x-y_{2n-1})}
                                        {(x-y_1)(x-y_2)\dotsc (x-y_{2n})}\right)
\end{split}\end{equation*}
showing that $P_n$ is plainly orthogonal with respect to the scalar product\\
%$\displaystyle \langle g_1,g_2\rangle =$
$ \mathscr{L}\left(\dfrac{g_1(x)g_2(x)}{(x-y_1)(x-y_2)\dotsc (x-y_{2n})}
\right)$. 

Confirmation: as $Q_n/P_n$ interpolates $f$ at $y_0,\dotsc, y_{2n}$,

$$  P_n(x)f(x)-Q_n(x) =0 \ \ \ \ \text{at}\ y_0,\dotsc,y_{2n}   $$

$$ P_n(x)g(x)f(x) - g(x)Q_n(x) =0 \ \ \ \ \text{at}\ y_0,\dotsc,y_{2n}$$
for any polynomial $g$ of degree $<n$. Perform the divided difference
of order $2n$ on $y_0,\dotsc, y_{2n}$:

$$ [y_0,\dotsc, y_{2n}]_{P_ngf} =0,\ \ \ \forall\ \text{polynomial}\ g \ \text{of\ degree}\ <n.$$

The divided difference is $\displaystyle  %\langle g_1,g_2\rangle=
        \sum_{j=0}^{2n} \dfrac{P_n(y_j)g(y_j) \;f(y_j)}{\omega'_{2n}(y_j)}$, where
$\omega_{2n}(x)=(x-y_0)\dotsc (x-y_{2n})$.

If $f$ is holomorphic in a domain containing $y_0,\dotsc, y_{2n}$, the
divided difference is
%$$ \langle g_1,g_2\rangle =
$$ \dfrac{1}{2\pi i}\int_C
        \dfrac{P_n(t)g(t)\;f(t)\;dt}{(t-y_0)\dotsc (t-y_{2n}) }
$$

Even simpler: if $f(x)=\displaystyle \sum_0^{N-1}\dfrac{\rho_k}{x-x'_k}$, 
$$\langle g_1, g_2\rangle = \sum_0^{N-1} \rho_k g_1(x'_k) g_2(x'_k). $$
        

\subsection{Example: the exponential function (Iserles)}\  \\

Rational interpolations to $e^{ax}$ at $x=0, h, 2h,\dotsc$

$ e^{ax} = \alpha'_0/(1+\dotsb)$ at $x=0: \alpha'_0=1$, \\
$ e^{-ax} = 1+\alpha'_1 x/(1+\dotsb)$ at $x=h: \alpha'_1=(e^{-ah}-1)/h$,
$\displaystyle \dfrac{\alpha'_1 x}{e^{-ax}-1} =1+\alpha'_2(x-h)$ at $x=2h:
  \alpha'_2=\dfrac{1-e^{-ah}}{h(1+e^{-ah})}$, \\
$\displaystyle \alpha'_{2j-1}=\dfrac{ 1-e^{ah}}{h(j-1+je^{ah})(1+e^{ah})},
               \alpha'_{2j} =\dfrac{e^{ah}(e^{ah}-1)}{h(j-1+je^{ah})(1+e^{ah})}$


\bigskip


\section{Elliptic Riccati equations.}  
\renewcommand{\mylmark}{Riccati}

An elliptic Riccati equation is

\begin{equation} \label{ellRic}
a(x)\dfrac{f(\psi(x))-f(\varphi(x))}{\psi(x)-\varphi(x)} =
b(x)f(\varphi(x))f(\psi(x)) +c(x)(f(\varphi(x))+f(\psi(x)))+d(x)
\end{equation}
with $\varphi$ and $\psi$ as in \S~\ref{phipsi}, p.~\pageref{phipsi}, and
where $a$, $b$, $c$, and $d$ are rational functions.

\smallskip

We consider now rational interpolation according to the setting
of \S~\ref{rinterp} above. Why is this relevant?

A first-order difference equation of the kind \eqref{ellRic} relates
$f(y_0)$ to $f(y_1)$ when $x=x_0$; $f(y_1)$ to $f(y_2)$ when
$x=x_1$, etc. So, we have information to introduce in the interpolation:
from $\displaystyle f_n(x)= \alpha'_n+\dfrac{x-y_n }{f_{n+1}(x)}$,
$f_n(y_n)=\alpha'_n$ and $f_n(y_{n+1})=\alpha'_n+ (y_{n+1}-y_n)/\alpha'_{n+1}$, so
$\alpha'_{n+1} = (y_{n+1}-y_n)/(f_n(y_{n+1})-f_n(y_n))$.

Now, from \eqref{ellRic} at $x=x_n$, so $\varphi(x)=y_n$ and $\psi(x)=y_{n+1}$,
$\displaystyle \dfrac{a_n(x_n)}{\alpha'_{n+1}}=b_n(x_n)\alpha'_n \left[\alpha'_n+ \dfrac{y_{n+1}-y_n}{\alpha'_{n+1}}\right] 
+c_n(x_n)\left[2\alpha'_n+ \dfrac{y_{n+1}-y_n}{\alpha'_{n+1}}\right]+d_n(x_n)$, or

\begin{equation} \label{alphapn}
  \alpha'_{n+1} = \dfrac{a_n(x_n)-(y_{n+1}-y_n)(\alpha'_n b_n(x_n)+c_n(x_n)}{b_n(x_n)\alpha'^2_n+2c_n(x_n)\alpha'_n+d_n(x_n) },\  n=0,1,\dotsc
\end{equation}
  
In \cite{magLuminy} (Luminy 2007), I considered rational approximants to
$$ f_0(x)=\dfrac{x-y_0}
                                   {\alpha_0 x+\beta_0 -\dfrac{(x-y_1)(x-y_2)}
                                      {\dfrac{\ddots}
                                  {\ \ \ \ \ \ \ \alpha_{n-2} x +\beta_{n-2}+\dfrac{(x-y_{2n-3})(x-y_{2n-2})}{\alpha_{n-1}x+\beta_{n-1}+\dotsb}}}}$$

making clear that the $n^{\text{th}}$ approximant (stopped at, and including the $\alpha_{n-1}x+\beta_{n-1}$ term) is the
rational function of degree $n$ interpolating $f_0$ at $x=y_0, y_1,\dotsc, y_{2n}$.
So, 
$\displaystyle f_n(x)= \dfrac{x-y_{2n} }{\alpha_n x +\beta_n -(x-y_{2n+1})f_{n+1}(x)}$,
$\alpha_n x+\beta_n$ is the polynomial interpolant of degree 1 to $(x-y_{2n})/f_n(x)$ at $y_{2n+1}$ and
$y_{2n+2}$.
 


Furthermore, the \emph{\textbf{Riccati}} form is well suited to 
continued fraction progression:

from the Luminy paper \cite{magLuminy}:


Let $f_0(x)=f(x)-f(y_0)$ be expanded in an interpolatory continued fraction ($R_{II}-$fraction \cite{Ism2005,IsMa,SpZ2007,zhed99}, or contracted Thiele's continued fraction \cite[Chap.~5]{MilneT})

$$ f_0(x)=\dfrac{x-y_0}
                                   {\alpha_0 x+\beta_0 -\dfrac{(x-y_1)(x-y_2)}
                                      {\dfrac{\ddots}
                                  {\ \ \ \ \ \ \ \alpha_{n-2} x +\beta_{n-2}+\dfrac{(x-y_{2n-3})(x-y_{2n-2})}{\alpha_{n-1}x+\beta_{n-1}+\dotsb}}}}$$

making clear that the $n^{\text{th}}$ approximant (stopped at, and including the $\alpha_{n-1}x+\beta_{n-1}$ term) is the
rational function of degree $n$ interpolating $f_0$ at $x=y_0, y_1,\dotsc, y_{2n}$.

Let $p_n$ be the denominator of this $n^{\text{th}}$ approximant.
If $f$ is a Stieltjes transform, it is known
\cite{IsMa,SpZ2007}\cite[\S~5]{zhed99}\cite{ZhedPad} that
$\{p_n(x)/((x-y_0)(x-y_2)\dotsc (x-y_{2n})\}$ and
$\{p_m(x)/((x-y_1)(x-y_3)\dotsc (x-y_{2m+1})\}$ are biorthogonal
sequences of rational functions.

From $\displaystyle f_n(x)= \dfrac{x-y_{2n} }{\alpha_n x +\beta_n -(x-y_{2n+1})f_{n+1}(x)}$,
$\alpha_n x+\beta_n$ is the polynomial interpolant of degree 1 to $(x-y_{2n})/f_n(x)$ at $y_{2n+1}$ and
$y_{2n+2}$, so we need
$f_n(y_{2n+1})$ and $f_n(y_{2n+2})$ in order to find $\alpha_n$ and $\beta_n$.

If $f_n$ satisfies  a difference equation of first order
$\mathcal{F}_n\left(x,f_n(\varphi(x)),f_n(\psi(x))\right)=0$, we find
$f_n(y_{2n+1})$ from the equation at $x_{2n}$, as $\varphi(x_{2n})=y_{2n}$,
$\psi(x_{2n})=y_{2n+1}$, and $f_n(y_{2n})=0$. Next, the equation at $x=x_{2n+1}$
yields $f_n(y_{2n+2})$.

As seen in the section~\ref{secPears} on the Pearson equation, a linear difference equation
of first order easily involves a weight function useful in orthogonality considerations.
However, Riccati equations are better suited to continued fraction constructions \cite{GrH,Mag84}. Of course, linear difference equations of first order are special
cases of Riccati equations, that is why the coefficients in \eqref{lindiff}
are written $a(x), c(x)$, and $d(x)$, whereas $b(x)$ is the coefficient
of the nonlinear part of a Riccati equation.

So, if $f_n$ satisfies the Riccati equation
\begin{multline} \label{ellRicn}
a_n(x)\dfrac{f_n(\psi(x))-f_n(\varphi(x))}{\psi(x)-\varphi(x)} =
b_n(x)f_n(\varphi(x))f_n(\psi(x))  \\  +c_n(x)(f_n(\varphi(x))+f_n(\psi(x)))+d_n(x),
\end{multline}
\enlargethispage{33pt}
one finds
at $x=x_{2n}$,  $\varphi(x)=y_{2n}$, $\psi(x)=y_{2n+1}$,
and knowing that $f_n(y_{2n})=0$,
$\displaystyle a_n(x_{2n})\dfrac{f_n(y_{2n+1})}{y_{2n+1}-y_{2n}}=c_n(x_{2n})f_n(y_{2n+1})+d_n(x_{2n})$ yields
$\displaystyle f_n(y_{2n+1})=\dfrac{d_n(x_{2n})}{\dfrac{a_n(x_{2n})}{y_{2n+1}-y_{2n}}-c_n(x_{2n})}$ and
\begin{equation}\label{alphanp1}
 \alpha_n y_{2n+1}+\beta_n = \dfrac{y_{2n+1}-y_{2n}}{f_n(y_{2n+1})}=
 \dfrac{a_n(x_{2n})-(y_{2n+1}-y_{2n})c_n(x_{2n})}{d_n(x_{2n})},
\end{equation}
and at $x=x_{2n+1}$,
$\displaystyle a_n(x_{2n+1}) \dfrac{
       \dfrac{y_{2n+2}-y_{2n} }{\alpha_n y_{2n+2}+\beta_n}
      -\dfrac{y_{2n+1}-y_{2n} }{\alpha_n y_{2n+1}+\beta_n}   }
               {y_{2n+2}-y_{2n+1} } = $\\
$\displaystyle b_n(x_{2n+1})  \dfrac{y_{2n+2}-y_{2n} }{\alpha_n y_{2n+2}+\beta_n}
      \dfrac{y_{2n+1}-y_{2n} }{\alpha_n y_{2n+1}+\beta_n}
+c_n(x_{2n+1})\left[ \dfrac{y_{2n+2}-y_{2n} }{\alpha_n y_{2n+2}+\beta_n}
      +\dfrac{y_{2n+1}-y_{2n} }{\alpha_n y_{2n+1}+\beta_n}  \right]
+d_n(x_{2n+1})$, or

\begin{multline}\label{alphanp2}
a_n(x_{2n+1}) (\alpha_n y_{2n}+\beta_n) = b_n(x_{2n+1})  (y_{2n+2}-y_{2n} )
         (y_{2n+1}-y_{2n}) \\
 +c_n(x_{2n+1})[(y_{2n+2}-y_{2n} )(\alpha_n y_{2n+1}+\beta_n)
      +(y_{2n+1}-y_{2n} )(\alpha_n y_{2n+2}+\beta_n)]\\
+d_n(x_{2n+1})(\alpha_n y_{2n+1}+\beta_n)(\alpha_n y_{2n+2}+\beta_n).
\end{multline}

which shows how to extract $\alpha_n$ and $\beta_n$ from $a_n,\dotsc$ at $x_{2n}$ and
$x_{2n+1}$.

Remark also that at $x=x_{2n-1}$, knowing that $f_n(\psi(x_{2n-1}))=f_n(y_{2n})=0$,
\eqref{ellRicn} yields
\begin{multline}\label{alphanp3}
\left[\dfrac{a_n(x_{2n-1})}{y_{2n}-y_{2n-1}} +c_n(x_{2n-1})\right]f_n(y_{2n-1})+d_n(x_{2n-1})=0.
\end{multline}


And here is how the \emph{\textbf{Riccati}} form is well suited to
continued fraction progression:

\subsection{\label{Ricthm}Theorem} \  \\

{\sl If $f_n$ satisfies the Riccati equation \eqref{ellRicn}
with rational coefficients $a_n$, $b_n$, $c_n$, and $d_n$,
and if $\displaystyle   f_n(x)= \dfrac{x-y_{2n} }
            {\alpha_n x +\beta_n -(x-y_{2n+1})f_{n+1}(x)}$,
then
$f_{n+1}$ satisfies an equation of same complexity (degree of the rational functions)
of its coefficients.}

(Actually, the degrees of $a_n$, etc. will at most exceed the degrees at $n=0$ by 3 units).

\smallskip

\noindent\textbf{Proof.} Let us start with \eqref{ellRicn} at $n=0$ with polynomial coefficients
$a_0, b_0, c_0$, and $d_0$. Suppose that, at the $n^{\text{th}}$ step, $a_n$, etc. are polynomials
with $b_n$ and $d_n$ containing the factors $X_2$ and $x-x_{2n-1}$, and $c_n$ containing
the factor $X_2$ (from \eqref{Fy} and \eqref{Fbiq}, $X_2$ is a polynomial of degree $\leqslant 2$).

Of course, if the initial coefficients $a_0$, etc. do not contain such factors, we may have to
multiply the four coefficients of \eqref{ellRicn} at $n=0$ by one or several factors of
$(x-x_{-1})X_2(x)$, that's why the degrees are liable to have to be augmented by up to 3 units,
but this operation has to be done only at $n=0$.

We suppose therefore that $b_n(x)=(x-x_{2n-1})X_2(x)\tilde{b}_n(x)$,
$c_n(x)=X_2(x)\tilde{c}_n(x)$, and
$d_n(x)=(x-x_{2n-1})X_2(x)\tilde{d}_n(x)$ in \eqref{ellRicn}, where $\tilde{b}_n$, $\tilde{c}_n$, and
$\tilde{d}_n$ are polynomials, so that \eqref{ellRicn} is now
\begin{multline} \label{ellRicn2}
a_n(x)\dfrac{f_n(\psi(x))-f_n(\varphi(x))}{\psi(x)-\varphi(x)} =
(x-x_{2n-1})X_2(x)\tilde{b}_n(x)f_n(\varphi(x))f_n(\psi(x))\\
+X_2(x)\tilde{c}_n(x)(f_n(\varphi(x))+f_n(\psi(x)))+(x-x_{2n-1})X_2(x)\tilde{d}_n(x),
\end{multline}

in which we enter $\displaystyle f_n(x)= \dfrac{x-y_{2n} }{\alpha_n x +\beta_n -(x-y_{2n+1})f_{n+1}(x)}$.

After multiplication by $(\alpha_n\varphi+\beta_n-(\varphi-y_{2n+1})f_{n+1}(\varphi))
(\alpha_n\psi+\beta_n-(\psi-y_{2n+1})f_{n+1}(\psi))$:

\begin{multline*}
a_n\dfrac{  (\psi-y_{2n})[\alpha_n\varphi+\beta_n-(\varphi-y_{2n+1})f_{n+1}(\varphi)]
          -(\varphi-y_{2n})[\alpha_n\psi+\beta_n-(\psi-y_{2n+1})f_{n+1}(\psi)] }
         {\psi-\varphi}  \\
=a_n\left[ \alpha_n y_{2n}+\beta_n +\left(\varphi\psi-\dfrac{(y_{2n}+y_{2n+1})
           (\varphi+\psi)}{2}+y_{2n}y_{2n+1}\right)\dfrac{f_{n+1}(\psi)-f_{n+1}(\varphi)}{\psi-\varphi}
    \right. \\
    \left. +\dfrac{y_{2n+1}-y_{2n}}{2}(f_{n+1}(\varphi)+f_{n+1}(\psi)\vphantom{\dfrac{L}{L}}\right] \\
=(x-x_{2n-1})X_2\tilde{b}_n (\varphi-y_{2n})(\psi-y_{2n}) \\
+X_2 \tilde{c}_n [(\psi-y_{2n})[\alpha_n\varphi+\beta_n-(\varphi-y_{2n+1})f_{n+1}(\varphi)]
          +(\varphi-y_{2n})[\alpha_n\psi+\beta_n-(\psi-y_{2n+1})f_{n+1}(\psi)]] \\
+(x-x_{2n-1})X_2 \tilde{d}_n [\alpha_n\varphi+\beta_n-(\varphi-y_{2n+1})f_{n+1}(\varphi)]
          [\alpha_n\psi+\beta_n-(\psi-y_{2n+1})f_{n+1}(\psi)]
\end{multline*}
which is the Riccati equation for $f_{n+1}$, $\displaystyle\hat{a}_{n+1}\dfrac{f_{n+1}(\psi)
  -f_{n+1}(\varphi)}{\psi-\varphi} = \hat{b}_{n+1} f_{n+1}(\varphi)f_{n+1}(\psi)
+\hat{c}_{n+1}(f_{n+1}(\varphi)+f_{n+1}(\psi))+\hat{d}_{n+1}$, where
$\hat{a}_{n+1}$ etc. are symmetric functions of $\varphi$ and $\psi$, so are rational functions
thanks to \eqref{ysump} (where $y_n=\varphi(x_n)$ and $y_{n+1}=\psi(x_n)$):

\enlargethispage{44pt}
\begin{subequations}
\begin{multline}\label{ahat}
\hat{a}_{n+1} = \left(\varphi\psi-\dfrac{(y_{2n}+y_{2n+1})
           (\varphi+\psi)}{2}+y_{2n}y_{2n+1}\right)a_n
     +\dfrac{y_{2n+1}-y_{2n}}{2} (\psi-\varphi)^2 X_2 \tilde{c}_n \\
    +\dfrac{\alpha_n y_{2n+1}+\beta_n}{2} (\psi-\varphi)^2 (x-x_{2n-1})X_2\tilde{d}_n
  = \dfrac{\left(X_0+\dfrac{(y_{2n}+y_{2n+1})
           X_1}{2}+y_{2n}y_{2n+1}X_2\right)}{X_2}a_n \\
     +\dfrac{[y_{2n+1}-y_{2n}]\tilde{c}_n +(\alpha_n y_{2n+1}+\beta_n)(x-x_{2n-1})\tilde{d}_n}{2} \  %
\dfrac{X_1^2-4X_0X_2}{X_2}
\end{multline}
\begin{equation}\label{bhat}
\hat{b}_{n+1} = (x-x_{2n-1})X_2 \tilde{d}_n (\varphi-y_{2n+1})(\psi-y_{2n+1})
                = (x-x_{2n-1})\tilde{d}_n (X_0 +y_{2n+1}X_1+y_{2n+1}^2X_2)
\end{equation}
\begin{multline}\label{chat}
\hat{c}_{n+1}=-(y_{2n+1}-y_{2n})a_n -\left(\varphi\psi-\dfrac{(y_{2n}+y_{2n+1})
           (\varphi+\psi)}{2}+y_{2n}y_{2n+1}\right)X_2\tilde{c}_n \\
      -\left( \alpha_n \left(\varphi\psi-y_{2n+1}\dfrac{\varphi+\psi}{2}\right)
              +\beta_n \left(\dfrac{\varphi+\psi}{2}-y_{2n+1}\right)\right)(x-x_{2n-1})X_2 \tilde{d}_n \\
 =-(y_{2n+1}-y_{2n})a_n -\left(X_0+\dfrac{(y_{2n}+y_{2n+1})
           X_1}{2}+y_{2n}y_{2n+1}X_2\right)\tilde{c}_n \\
      -\left( \alpha_n \left(X_0+y_{2n+1}\dfrac{X_1}{2}\right)
              -\beta_n \left(\dfrac{X_1}{2}+y_{2n+1}X_2\right)\right)(x-x_{2n-1}) \tilde{d}_n
\end{multline}
\begin{multline}\label{dhat}
\hat{d}_{n+1}=- (\alpha_n y_{2n}+\beta_n)a_n  +(x-x_{2n-1})X_2\tilde{b}_n (\varphi-y_{2n})(\psi-y_{2n}) \\
  +X_2 \tilde{c}_n [(\psi-y_{2n})(\alpha_n\varphi+\beta_n)
          +(\varphi-y_{2n})(\alpha_n\psi+\beta_n)]] \\
\qquad\qquad +(x-x_{2n-1})X_2 \tilde{d}_n (\alpha_n\varphi+\beta_n)(\alpha_n\psi+\beta_n) \\
\ \  =- (\alpha_n y_{2n}+\beta_n)a_n  +(x-x_{2n-1})\tilde{b}_n (X_0+y_{2n}X_1+ y_{2n}^2 X_2) \\
  + \tilde{c}_n [\alpha_n(2X_0+y_{2n}X_1)-\beta_n(X_1+2y_{2n}X_2)]
          +(x-x_{2n-1}) \tilde{d}_n (\alpha_n^2X_0-\alpha_n\beta_n X_1+\beta_n^2 X_2)
\end{multline}
\end{subequations}



 The first coefficient $\hat{a}_{n+1}$ is not a polynomial, but a rational function of
denominator $X_2$. We recover polynomials by multiplying the four coefficients
$\hat{a}_{n+1}$, $\hat{b}_{n+1}$, $\hat{c}_{n+1}$, and $\hat{d}_{n+1}$ by $X_2$. Moreover,
this already restores the factor $X_2$ in $X_2\hat{b}_{n+1}$, $X_2\hat{c}_{n+1}$, and $X_2\hat{d}_{n+1}$!

However, the degrees of the new coefficients are higher than before. The problem is settled
by seeing that the four polynomials $X_2\hat{a}_{n+1}$, $\hat{b}_{n+1}$, $\hat{c}_{n+1}$, and $\hat{d}_{n+1}$
vanish at $x=x_{2n-1}$ and at $x=x_{2n}$. Then, we will simply divide the four of them
by $(x-x_{2n-1})(x-x_{2n})$.

\begin{enumerate}
\item The most obvious case is \eqref{bhat}:
$
\hat{b}_{n+1}(x)  = (x-x_{2n-1})\tilde{d}_n(x) (X_0(x) +y_{2n+1}X_1(x)+y_{2n+1}^2X_2(x))
              = (x-x_{2n-1})\tilde{d}_n(x) F(x,y_{2n+1}) =(x-x_{2n-1})\tilde{d}_n(x) Y_2(y_{2n+1})(x-x_{2n})(x-x_{2n+1}),
$
shows indeed the factors $x-x_{2n-1}$ and $x-x_{2n}$, as well as $x-x_{2n+1}$,
so that $\displaystyle \tilde{b}_{n+1}(x) = \dfrac{ \hat{b}_{n+1}(x) }{(x-x_{2n-1})(x-x_{2n})(x-x_{2n+1}) }=
             Y_2(y_{2n+1})\tilde{d}_n(x)$.

\item Next, from \eqref{dhat}, $\hat{d}_{n+1}(x_{2n-1}) = (\alpha_n y_{2n}+\beta_n)[-a_n(x_{2n-1})-(y_{2n}-y_{2n-1})c_n(x_{2n-1})]=0$ from
   \eqref{alphanp3}, knowing that $d_n(x_{2n-1})=0$, \\
$\hat{d}_{n+1}(x_{2n}) = (\alpha_n y_{2n}+\beta_n)[-a_n(x_{2n})+(y_{2n+1}-y_{2n})c_n(x_{2n})
  +(\alpha_n y_{2n+1}+\beta_n)d_n(x_{2n})]=$\\
$\displaystyle (\alpha_n y_{2n}+\beta_n)(y_{2n+1}-y_{2n})\left[ -\dfrac{a_n(x_{2n})}{y_{2n+1}-y_{2n}} +c_n(x_{2n})
+\dfrac{d_n(x_{2n})}{f_n(y_{2n+1})}\right]=0$, as the last factor is the Riccati equation \eqref{ellRicn} at
$x=x_{2n}$ divided by $f_n(y_{2n+1})$ (see also \eqref{alphanp1}).\\
$\hat{d}_{n+1}(x_{2n+1})=- (\alpha_n y_{2n}+\beta_n)a_n(x_{2n+1})
     +b_n(x_{2n+1}) (y_{2n+1}-y_{2n})(y_{2n+2}-y_{2n})
  +c_n(x_{2n+1}) [(y_{2n+2}-y_{2n})(\alpha_n y_{2n+1}+\beta_n)
          +(y_{2n+1}-y_{2n})(\alpha_n y_{2n+2}+\beta_n)]]
+ d_n(x_{2n+1}) (\alpha_n y_{2n+1}+\beta_n)(\alpha_n y_{2n+2}+\beta_n)=0$, from
\eqref{alphanp2}.
\item In order to show that $\hat{a}_{n+1}$ and $\hat{c}_{n+1}$ both vanish at
$x=x_{2n-1}$ and $x_{2n}$, we consider  $\displaystyle \dfrac{a_n}{\psi-\varphi}\pm c_n$:
$\displaystyle \dfrac{\hat{a}_{n+1}(x)}{\psi(x)-\varphi(x)}+\hat{c}_{n+1}(x) =$\\
$\displaystyle    (\varphi(x)-y_{2n+1})\left[(\psi(x)-y_{2n})\left(\dfrac{a_n(x)}{\psi(x)-\varphi(x)}-c_n(x)\right)
-(\alpha_n\psi(x)+\beta_n)d_n(x)\right]$
vanishes at  $x=x_{2n}$, as the
big factor is $a_n(x_{2n})-(y_{2n+1}-y_{2n})c_n(x_{2n})
-(\alpha_n y_{2n+1}+\beta_n)d_n(x_{2n})=0$ from
\eqref{alphanp1}. At $x=x_{2n-1}$, we already encountered the condition
$a_n(x_{2n-1})+(y_{2n}-y_{2n-1})c_n(x_{2n-1})=0$ from \eqref{alphanp3} and $d_n(x_{2n-1})=0$.

The obvious vanishing of the first factor at $x=x_{2n+1}$ will allow the same condition
at $x_{2n+1}$: $a_{n+1}(x_{2n+1})+(y_{2n+2}-y_{2n+1})c_{n+1}(x_{2n+1})=0$, and this will imply
$f_{n+1}(y_{2n+2})=0$ at the next step.

\item
$\displaystyle
\dfrac{\hat{a}_{n+1}(x)}{\psi(x)-\varphi(x)}-\hat{c}_{n+1}(x) =$
\\ $\displaystyle
(\psi(x)-y_{2n+1})\left[(\varphi(x)-y_{2n})\left(\dfrac{a_n(x)}{\psi(x)-\varphi(x)}+c_n(x)\right)
+(\alpha_n\varphi(x)+\beta_n)d_n(x)\right]$ obviously vanishes now
at $x=x_{2n}$; at $x=x_{2n-1}$, the big factor is
$-a_n(x_{2n-1})-(y_{2n}-y_{2n-1})c_n(x_{2n-1})=0$ as already
encountered (in \eqref{alphanp3}, together with
$d_n(x_{2n-1})=0$).

\end{enumerate}

\medskip

We proceed therefore with
$a_{n+1}(x)=\dfrac{X_2(x)\hat{a}_{n+1}(x)}{(x-x_{2n-1})(x-x_{2n})}$,
$b_{n+1}(x)=\dfrac{X_2(x)\hat{b}_{n+1}(x)}{(x-x_{2n-1})(x-x_{2n})}$,
$c_{n+1}(x)=\dfrac{X_2(x)\hat{c}_{n+1}(x)}{(x-x_{2n-1})(x-x_{2n})}$,
$d_{n+1}(x)=\dfrac{X_2(x)\hat{d}_{n+1}(x)}{(x-x_{2n-1})(x-x_{2n})}$. \qed

\medskip

A very interesting identity is
$\hat{a}^2_{n+1} -(\psi-\varphi)^2\hat{c}_{n+1}^2 =
(\varphi-y_{2n})(\psi-y_{2n})(\varphi-y_{2n+1})(\psi-y_{2n+1})
(a_n^2-(\psi-\varphi)^2c_n^2)-(\psi-\varphi)^2\hat{b}_{n+1}
[\hat{d}_{n+1}-(\varphi-y_{2n})(\psi-y_{2n})b_n]$, or
$\hat{a}^2_{n+1} -(\psi-\varphi)^2(\hat{c}_{n+1}^2-\hat{b}_{n+1}\hat{d}_{n+1}) =
(\varphi-y_{2n})(\psi-y_{2n})(\varphi-y_{2n+1})(\psi-y_{2n+1})
(a_n^2-(\psi-\varphi)^2(c_n^2-b_nd_n))$,

$a^2_{n+1} -(\psi-\varphi)^2(c_{n+1}^2-b_{n+1}d_{n+1}) =
\displaystyle \left[\dfrac{X_2(x)}{(x-x_{2n-1})(x-x_{2n})}\right]^2
\dfrac{F(x,y_{2n})}{X_2(x)} \dfrac{F(x,y_{2n+1})}{X_2(x)}
(a_n^2-(\psi-\varphi)^2(c_n^2-b_nd_n))
= Y_2(y_{2n})Y_2(y_{2n+1})\dfrac{x-x_{2n+1}}{x-x_{2n-1}}
(a_n^2-(\psi-\varphi)^2(c_n^2-b_nd_n))
$,


\begin{equation} \label{cambd}
\dfrac{c_n^2(x)-b_n(x)d_n(x)}{X_2^2(x)}P(x)-a_n^2(x)
  =C_n  \dfrac{x-x_{2n-1} }{x-x_{-1}}\left[
 \dfrac{c_0^2(x)-b_0(x)d_0(x)}{X_2^2(x)}P(x)-a_0^2(x)\right],
\end{equation}
where $\displaystyle C_n
=Y_2(y_0)Y_2(y_1)\dotsb Y_2(y_{2n-2}) Y_2(y_{2n-1})$ ((probably a) mistake in \cite[p.~800, after (23)]{magLuminy},
and $P$ from \eqref{P}.



%\medskip



\section{Classical?}  
\renewcommand{\mylmark}{Classical?}

$a_n$ is a cubic polynomial, so is determined through the four
values $a_n(z_i)$, $i=1,\dotsc, 4$, at the zeros $z_1,\dotsc, z_4$ of $P$.
{\Small
\begin{verbatim}
ellhyper Tue10May2011
k= 0.5 ; P= x^4 - 5.0*x^2 + 4.0 ; z0= 2.005000 ; 
z= [-2.000000, -1.000000, 1.000000, 2.000000]~
F= (-0.001251562*y^2 + 0.7550062)*x^2 - 1.503750*y*x + (0.7550062*y^2 - 0.005006250)
X0= 0.7550062*x^2 - 0.005006250 ; X1= -1.503750*x ; X2= -0.001251562*x^2 + 0.7550062 ;
Y0= 0.7550062*y^2 - 0.005006250 ; Y1= -1.503750*y ; Y2= -0.001251562*y^2 + 0.7550062
zprime=[-2.020159, -0.9900209, 0.9900209, 2.020159] X2Y2zi=[0.5624812, 0.5681508, 0.5681508, 0.5624812]
zeros X2=[-24.56117 + 0.E-404*I, 24.56117 + 0.E-404*I]~
 x0= 0.08142935 , x-1= -0.08142935 y0=0 y-1=-0.1621851 
\end{verbatim}}

$\{x_n, y_n,  a_n(z_1),\dotsc, a_n(z_4)\}$ for $n=1,\dotsc, 50$: 

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%\begin{equation} \label{factori}
%F(x_n,y)= X_2(x_n)(y-y_n)(y-y_{n+1}),\qquad F(x,y_n)= Y_2(y_n)(x-x_n)(x-x_{n-1}).
%\end{equation}

at $x=z_i$ where $P(x)=X_1^2(x)-4X_0(x)X_2(x)=0$,
 $F(z_i,y)=X_0(z_i)+X_1(z_i)y+X_2(z_i)y^2=0$ has a double root $y=\varphi(z_i)=
\psi(z_i)= \displaystyle -\dfrac{X_1(z_i)}{2X_2(z_i)} = -\dfrac{2X_0(z_i)}{X_1(z_i)}$.

$\displaystyle
\hat{a}_{n+1}(z_i)  
  = \dfrac{\left(X_0(z_i)+\dfrac{(y_{2n}+y_{2n+1})
           X_1(z_i)}{2}+y_{2n}y_{2n+1}X_2(z_i)\right)}{X_2(z_i)}a_n(z_i)\\
= \dfrac{\left(X_0(z_i)-\dfrac{X_1(x_{2n})
           X_1(z_i)}{2X_2(x_{2n})}+\dfrac{X_0(x_{2n})}{X_2(x_{2n})}X_2(z_i)\right)}{X_2(z_i)}a_n(z_i)
= \dfrac{  F(x_{2n}, \varphi(z_i))  }{X_2(x_{2n})} a_n(z_i)\\= \dfrac{  
Y_2(\varphi(z_i))(x_{2n}-z_i)(x_{2n}-z'_i)  }{X_2(x_{2n})} a_n(z_i)
 \text{\  \  \  \  \  \  \  \  from\  \eqref{factori}\  with \ ``} y_n\text{''}=\varphi(z_i),
\text{\ so\ that\ ``} x_n \text{'' \  and\  ``} x_{n-1}\text{''\  are\  } z_i \text{\ and}\\
a_{n+1}(z_i)=\dfrac{X_2(z_i)\hat{a}_{n+1}(z_i)}{(z_i-x_{2n-1})(z_i-x_{2n})}
= \dfrac{X_2(z_i) Y_2(\varphi(z_i))(z'_i-x_{2n})   }{X_2(x_{2n})(z_i-x_{2n-1})}a_n(z_i)
 $,


The $a_n(z_i)$ of the picture above are $a_n(z_i)/\sqrt{Y_2(y_0)Y_2(y_1)\dotsb Y_2(y_{2n-1})}$


\begin{enumerate}

\item {\bf Gauss hypergeometric ratio and classical orthogonal polynomials.}

$\varphi(x)=\psi(x)=x$,

  $a_n(x)=x(1-ax)$, $b_n=b_0-n$, $d_n(x)=-\alpha_n b_n x$,\\
$c_n(x)=[-a/4+(c'_0+a/4)(-1)^n]x+b_n/2$,

$\alpha_{n+1}=[a+\alpha_n b_n +2c'_n]/b_{n+2}$.

Orthogonal polynomials (Jacobi, etc.) are $x^nP_n(1/x)$.

\item {\bf Discrete orthogonal and bi-orthogonal polynomials. }

$\varphi(x)=x$, $\psi(x)=x+h$, 

  \begin{enumerate}

    \item Orthogonal: limit case $y_0=y_1=\dotsb =\infty$,

    \item Bi-orthogonal $y_j=jh$


  \end{enumerate}

\item {\bf a\emph{S}key-\emph{N}ikiforov-s\emph{U}slov-\emph{U}varov-wi\emph{L}son.}


\item {\bf And now?}

\end{enumerate}

We try to keep the degrees as low as possible. 

\subsection{``Elliptic logarithm''} \  \\

We extend $\displaystyle f(x)=\log\dfrac{x-a}{x-b}$ which satisfies
$f'(x)= \dfrac{a-b}{(x-a)(x-b)}$ by looking for a function whose
divided difference is a rational function of low degree.

Answer: work out the difference of 

\begin{equation} \label{elllog}
 \Lambda_N(x) = \sum_0^N \dfrac{\rho_k}{x-y'_k},
\end{equation}
where $y'_k=y(s_1+k)$ is an  elliptic lattice in the same family, but with
$s_1-s_0$ not an integer.

Simplest way to look at the difference is to suppose that $x$ is some
$x_n$. Then,

$$\mathcal{D}\Lambda_N(x_n) = \dfrac{\Lambda_N(y_{n+1})-\Lambda_N(y_n)}{y_{n+1}-y_n}
 =-\sum_0^N\dfrac{1}{(y_n-y'_k)(y_{n+1}-y'_k)},$$

As $y_n$ and $y_{n+1}$ are the two ordinates associated to $x_n$,
$y_n+y_{n+1}=-X_1(x_n)/X_2(x_n)$, $y_ny_{n+1}=X_0(x_n)/X_2(x_n)$,
   $(y_n-y'_k)(y_{n+1}-y'_k)= (X_0(x_n)+X_1(x_n)y'_k +X_2(x_n)y'^2_k)/X_2(x_n)=
F(x_n,y'_k)/X_2(x_n)=Y_2(y'_k)(x_n-x'_{k-1})(x_n-x'_k)/X_2(x_n)$,

$$\mathcal{D}\Lambda_N(x_n) = -X_2(x_n)\sum_0^N\dfrac{ \rho_k /Y_2(y'_k) }{(x_n-x'_{k-1})(x_n-x'_k)} $$
  $$\hspace*{-23pt} = X_2(x_n)\left[\dfrac{ \dfrac{\rho_0}{(x'_0-x'_{-1})Y_2(y'_0)}}{x_n-x'_{-1}}
    +\sum_0^{N-1} \dfrac{\dfrac{\rho_{k+1}}{(x'_{k+1}-x'_k)Y_2(y'_{k+1})}-
           \dfrac{\rho_k}{(x'_k-x'_{k-1})Y_2(y'_k)}}{x_n-x'_k}
    -\dfrac{\dfrac{\rho_N}{(x'_N-x'_{N-1})Y_2(y'_N)}}{x_n-x'_N}\right], $$
so that, when $\rho_k=(x'_k-x'_{k-1})Y_2(y'_k)$,
$$ \mathcal{D}\Lambda_N(x) =\dfrac{(x'_{-1}-x'_N)X_2(x)}{(x-x'_{-1})(x-x'_N)}
=(x'_{-1}-x'_N)
c_{2,2}+\dfrac{X_2(x'_{-1})}{x-x'_{-1}} -\dfrac{X_2(x'_N)}{x-x'_N}
. $$

We can solve $\displaystyle  \mathcal{D}f(x) =\dfrac{1}{x-A}$ with $f(y_0)=0$, if an infinity of
$x'_n$s, starting with $x'_{-1}=A$, come as close as we want to a zero, say $\zeta$, of $X_2$, then
$$
f(x)=  -\dfrac{c_{2,2} (A-\zeta)}{X_2(A)}(x-y_0)
 +\dfrac{1}{X_2(A)}\lim_{x'_N\to \zeta} [\Lambda_N(x)-\Lambda_N(y_0)]
$$


Here is an example where $f(1)$ is estimated whenever $x'$ comes close to $\zeta$
which happens to be $-2.529822$ here:

$F(x,y)= -0.3125 x^2 y^2 + 2x^2 - 2.25 xy + 2y^2 - 1.25, A=-7$,

\begin{verbatim}

  n    x'          y'        res         f(1)

 -1 -7         -2.168439
  0  2.006819   3.351538   1.021789   -0.02471402 
  1 -19.96266   2.738246  -0.5662515   0.09425273 
  2 -2.068051  -2.371686  -0.3255921   0.1349691 
  3  4.274935  -4.641409   2.254690    0.04885997 
  4  2.433328   2.049464   0.09509338  0.004647754 
  5 -2.642949  34.53302 -141.3408      0.1267041   
 ...
 13 -2.560190  -2.024363  -0.06545824  0.1276923 
 ...
 56 -2.525475  -2.029866  -0.07278552  0.1281233 
 ...
238 -2.536399 550.1306 -35945.00       0.1279866 

\end{verbatim}

which is easy to check, as 1 is itself close to an infinity of $y$s,
so that we estimate a value of $f$ by running the difference equation
in the form  $f(y_{n+1})=f(y_n)+\displaystyle \dfrac{y_{n+1}-y_n}{x_n-A}$
:

\begin{verbatim}
  n      x           y          f
  0   0.7905694   0           0
  1   0.5167233   0.9856450   0.1265177
  2  -0.9529158  -0.3790235  -0.05503327
  3  -0.1575301  -0.8702581  -0.1362682
  4   0.9994804   0.6923469   0.09210034
  5  -0.2301987   0.6400372   0.08556120
  6  -0.9309638  -0.9011730  -0.1420984
  7   0.5766530  -0.3102076  -0.04472454
  8   0.7466518   0.9944962   0.1274760
  9  -0.8303816  -0.07436214 -0.01050081
 10  -0.4536000  -0.9726185  -0.1560943
 11   0.9706641   0.4453679   0.06051121
...
 69   0.6604197   0.9999706   0.1280656
...
431   0.6621363   0.9999845   0.1280671
...
793   0.6638488   0.9999940   0.1280681

\end{verbatim}

 
%\newpage

%\vspace*{-33pt}
%\includegraphics*[angle=0, scale=1.0]{M32T1.PS}




We therefore start the process with $a_0(x)=(x-x'_0)(x-x'_N)$, $b_0=c_0=0$,
$d_0(x)=X_2(x)$.

Managing to keep \emph{\textbf{polynomials}}, it appears that 
$b_n$, $c_n$, and $d_n$ are $X_2$ times a polynomial of degree 
1 or 2, whereas $a_n$ has degree 3 or 4.

Here is an example with 
$$  F(x,y)=(\underbrace{x^2/2+x+1}_{X_2(x)})y^2 +(\underbrace{2x^2-2x-3}_{X_1(x)})y +x^2+x+1,
$$

and starting at $x_{-1}=-0.45$:

\smallskip

{\Small

\hspace*{-44pt}\begin{tabular}{|c|r|r|r|r|r|r|r|r|r|r|r|}\hline
                 $k$ &-1&0&1&2&3&4&5&6&7&8&9 \\ \hline
          $x_k$&-0.45000 &0.36869 &-0.48027 &0.42834 &-0.50390 &0.47655 &
          -0.52104& 0.51127 &-0.53184 & 0.53096 &-0.53639 \\  \hline 
          $y_k$&2.0348&0.56784&1.8444&0.64066&1.6551&0.72731&1.4731&0.82873&1.3027&0.94572&1.1465 \\  \hline
\end{tabular}

}

The function $f$ satisfies $\mathcal{D}f(x)\displaystyle =\dfrac{X_2(x)}{(x-x'_0)(x-x'_9)}$, where $x'_0$ now starts at 0.25:

\smallskip

{\Small 

\hspace*{-44pt}\begin{tabular}{|c|r|r|r|r|r|r|r|r|r|r|r|}\hline
                 $k$ &0&1&2&3&4&5&6&7&8&9 \\ \hline
          $x'_k$ &0.25000  &-0.38410& 0.17205 & -0.33547 & 0.091968 &
            -0.28000 & 0.012074 & -0.21798 & -0.065644 & -0.14992  \\  \hline 
          $y'_k$& 2.1599 & 0.47429& 2.3339& 0.43380& 2.4852& 0.40394
              & 2.6034& 0.38414& 2.6800&0.37399\\  \hline
\end{tabular}

}
 \smallskip

The first polynomial $a_0$ is $(x-x'_0)(x-x'_9)$, then

\smallskip


Remark that $b_1$, $b_3$,\ldots are scalar multiples of $X_2$.

Remark also that $\alpha'_{2N+1}=\infty$, as it should: 
the rational function $f$ of degree $N$ is recovered exactly as $q_{2N}/p_{2N}$. 


\medskip

\section{Linear difference relations and equations for the numerators
and the denominators of the interpolants.}
\renewcommand{\mylmark}{Linear difference equations.}


\smallskip

I try to reproduce \eqref{rs}, the recurrence relations for
$p_n$ and $q_n$ being now (\S~\ref{rinterp}, p.~\pageref{rinterp})

$p_{n+1}(x)=\alpha'_{n+1}p_{n}(x)+(x-y_{n})p_{n-1}(x)$,

\emph{We now consider the \textbf{linear} recurrence satisfied by
combinations of such products}, i.e., by combinations of
$$ p_n(\varphi) p_n(\psi),
   p_n(\varphi) q_{n-1}(\psi),
    q_{n-1}(\varphi) p_n(\psi),
   \text{ and }
    q_{n-1}(\varphi) q_{n-1}(\psi) . $$
We just have to consider a product $r_n s_n$, kowing that
$ r_{n+1} = \alpha'_{n+1} r_n +(\varphi-y_n) r_{n-1}$, and
$ s_{n+1} = \alpha'_{n+1} s_n +(\psi-y_n) s_{n-1}$.
Simplest is to write it as matrix-vector recurrence (\emph{doesn't it
 ring a bell}?)


\begin{equation} \label{rs2}
  \begin{bmatrix}
 \vrz r_n s_{n+1} \\ \vrz r_{n+1} s_n  \\ \vrz r_n s_n \\ \vrz r_{n+1} s_{n+1}
  \end{bmatrix}
    =
  \begin{bmatrix}
   \vrz     & \psi-y_n   &   &  \alpha'_{n+1}\\
   \vrz  \varphi-y_n  & &    & \alpha'_{n+1}  \\
   \vrz             &         &        &    1   \\
   \vrz    \alpha'_{n+1}(\varphi-y_n)  &
            \alpha'_{n+1}(\psi-y_n)  &
             (\varphi-y_n)(\psi-y_n)  &
            \alpha'^2_{n+1}
  \end{bmatrix}
  \begin{bmatrix}
 \vrz  r_{n-1} s_n \\ \vrz r_n s_{n-1}  \\ \vrz r_{n-1} s_{n-1} \\ \vrz r_n s_n
  \end{bmatrix}
\end{equation}

slightly closer to \eqref{abcd2} if the latter is put in the form
 
{\Small
\begin{equation} \label{abcd3}
 \hspace*{-33pt}\begin{bmatrix}
 \vrz \dfrac{1}{\varphi-y_n}\left[\dfrac{\tilde{a}_{n+1}}{\psi-\varphi} -\tilde{c}_{n+1}\right]  \\
 \vrz \dfrac{1}{\psi-y_n}\left[\dfrac{\tilde{a}_{n+1}}{\psi-\varphi} +\tilde{c}_{n+1}\right]  \\
 \vrz \dfrac{\tilde{d}_{n+1}}{(\varphi-y_n)(\psi-y_n)} \\ \vrz \tilde{b}_{n+1}
  \end{bmatrix}
    =
  \begin{bmatrix}
   \vrz     & \psi-y_{n-1}   &   &  -\alpha'_n\\
   \vrz  \varphi-y_{n-1}  & &    & \alpha'_n  \\
   \vrz             &         &        &    -1   \\
   \vrz    -\alpha'_n(\varphi-y_{n-1})  & \alpha'_n(\psi-y_{n-1})  & -(\varphi-y_{n-1})(\psi-y_{n-1})  & -\alpha'^2_n
  \end{bmatrix}
  \begin{bmatrix}
 \vrz  \dfrac{1}{\varphi-y_{n-1}}\left[\dfrac{a_n}{\psi-\varphi} -c_n\right] \\
 \vrz  \dfrac{1}{\psi-y_{n-1}}\left[\dfrac{a_n}{\psi-\varphi} +c_n\right] \\
 \vrz \dfrac{d_n}{(\varphi-y_{n-1})(\psi-y_{n-1})} \\ \vrz b_n 
  \end{bmatrix}
\end{equation}
}

to do: curb sign errors etc. get difference relations \& eq. for $p_n, q_n$.

\section{Hypergeometric expansions.}  
\renewcommand{\mylmark}{Hypergeometric?}


From:

David R. Masson:
The last of the hypergeometric continued fractions,
Report-no: OP-SF 12 Sep 1994
\href{http://arxiv.org/abs/math.CA/9409229}
{\texttt{http://arxiv.org/abs/math.CA/9409229}}


Dharma P. Gupta; David R. Masson:
Contiguous relations, continued fractions and orthogonality 
\textsl{Trans. Amer. Math. Soc.} \textbf{350} (1998), 769-808. 
This article is available free of charge 
\href{http://www.ams.org/tran/1998-350-02/S0002-9947-98-01879-0/home.html}
{\texttt{http://www.ams.org/tran/1998-350-02/S0002-9947-98-01879-0/home.html}}

L. M. Milne Thomson:
\textsl{The Calculus Of Finite Differences}, Macmillan And Company., Limited, 1933

\subsection{Gauss hypergeometric ratio} \  \\

also in Perron \cite[chap.~8]{Per}, Wall, etc.
\enlargethispage{33pt}
Consider the Taylor-Maclaurin expansion about the origin of

\begin{equation} \label{Gauss1} \begin{split}
\hspace*{-44pt}f(x)=f_0(x)&=x\;\dfrac{{}_2F_1(a,b+1;c+1;x)}{{}_2F_1(a,b;c;x)}
           =\dfrac{x +\dfrac{a(b+1)}{c+1}x^2+\dfrac{a(a+1)(b+1)(b+2)}{2(c+1)(c+2)}x^3+\dotsb }
                  { 1 +\dfrac{ab}{c}x +\dfrac{a(a+1)b(b+1)}{2c(c+1)}x^2+\dotsb} \\
   &= x +\dfrac{a(c-b)}{c(c+1)}x^2 +\dfrac{a(c-b)(ac+bc+c-2ab)}{c^2(c+1)(c+2)}x^3+\dotsb\\
   &=\dfrac{x}{ 1-\dfrac{a(c-b)}{c(c+1)}x -\dfrac{a(c-b)(b+1)(c-a+1)}{c(c+1)^2(c+2)}x^2+\dotsb}\\
   &=\dfrac{x}{1- \dfrac{  \dfrac{a(c-b)}{c(c+1)}x }
                           {1 -\dfrac{   \dfrac{(b+1)(c-a+1)}{(c+1)(c+2)}x }
                                      { 1-\dotsb  }   }   }   
\end{split}\end{equation}
This nice continued fraction goes further with
$\displaystyle \alpha_{2n+1}=-\dfrac{(a+n)(c-b+n)}{(c+2n)(c+2n+1)}x $,
$\displaystyle \alpha_{2n+2}=-\dfrac{(b+n+1)(c-a+n+1)}{(c+2n+1)(c+2n+2)}x$, ($n=0,1,\dotsc$) 
The secret is in contiguity relations relating ${}_2F_1(a,b;c;x)$ to ${}_2F_1(a'+1,b';c+1;x)$, where
$(a',b')=(a,b)$ or $(b,a)$. Then, using
$\displaystyle \dfrac{a\dotsb (a+n-1) b\dotsb (b+n-1)}{c\dotsb (c+n-1)} -\dfrac{(a'+1)\dotsb (a'+n) b'\dotsb (b'+n-1)}{(c+1)\dotsb (c+n)}$\\$\displaystyle =
 (a'-c)n\dfrac{(a'+1)\dotsb (a'+n-1) b'\dotsb (b'+n-1)}{c\dotsb (c+n)} $,
$${}_2F_1(a,b;c;x)-{}_2F_1(a'+1,b';c+1;x)=\dfrac{(a'-c)b' x}{c(c+1)}
\;{}_2F_1(a+1,b+1;c+2;x), $$
or
$$
\dfrac{ {}_2F_1(a'+1,b';c+1;x) }{ {}_2F_1(a,b;c;x) }=\dfrac{1}{
  1+   \dfrac{(a'-c)b' x}{c(c+1)}\;\; \dfrac{  {}_2F_1(a+1,b+1;c+2;x) }
{ {}_2F_1(a'+1,b';c+1;x)}   }
$$
so that $\displaystyle f_n(x)=x\;\dfrac{ {}_2F_1(a_{n+1},b_{n+1};c+n+1;x) }{ {}_2F_1(a_n,b_n;c+n;x) }$, with
$a_0=a_1=a, a_2=a_3=a+1,\dotsc: a_n=a+\lfloor n/2\rfloor,
b_n=b+\lceil n/2\rceil$. 

Riccati: from Abramowitz 15.2.7, 
$(1-x)F'=a'_n F-a'_n(1-b'_n/c_n)F(a_{n+1},b_{n+1};c+n+1;x)$, where $(a'_n,b'_n)=(a_n,b_n)$ if 
$a_{n+1}> a_n$. So,
$$  a'_n (1-b'_n/c_n)\dfrac{f_n(x)}{x} = a'_n-(1-x)\dfrac{F'}{F},$$
and use the hypergeometric differential
equation $x(1-x)F''+[c_n-(a_n+b_n+1)x]F'-a_nb_nF=0$ (15.5.1):


to continue

\subsection{Elliptic hypergeometric expansions} \   \\

Let 
$$
\mathcal{Y}_n(x)=\dfrac{(x-y_0)\dotsb (x-y_{n-1}) }{(x-y'_0)\dotsb (x-y'_{n-1})}, 
 \mathcal{X}_n(x)=  \dfrac{(x-x_0)\dotsb (x-x_{n-1}) }{ (x-x'_0)\dotsb (x-x'_{n-1})},
 \widetilde{\mathcal{X}}_n(x)= \dfrac{(x-x_0)\dotsb (x-x_{n-1}) }{ (x-x'_{-1})\dotsb (x-x'_n)}.
$$

See that $\displaystyle
 \mathcal{D}\mathcal{Y}_n(x) =
      C_n X_2(x) \widetilde{\mathcal{X}}_{n-1}(x) $

  
Indeed, $(\varphi(x)-y_0)(\varphi(x)-y_1)\dotsb  (\varphi(x)-y_{n-1})$ and 
  $(\psi(x)-y_0)(\psi(x)-y_1)\dotsb  (\psi(x)-y_{n-1})$ both vanish at
$x=x_0,x_1,\dotsc, x_{n-2}$;
$(\varphi(x)-y'_0)(\varphi(x)-y'_1)\dotsb  (\varphi(x)-y'_{n-1})$ vanishes at
$x=x'_0,\dotsc, x'_{n-1}$, whereas
$(\psi(x)-y'_0)(\psi(x)-y'_1)\dotsb  (\psi(x)-y'_{n-1})$ vanishes at
$x=x'_{-1},\dotsc, x'_{n-2}$.

Simple fractions give $\mathcal{D}\dfrac{1}{x-y'_k}=
 -\dfrac{X_2(x)}{Y_2(y'_k)(x-x'_{k-1})(x-x'_k)}$, as seen earlier.

The constant $C_n$ is found through particular values of $x$, either $x_{-1}$ or $x_{n-1}$:

\begin{equation*}\begin{split}
C_n &= -\dfrac{\mathcal{Y}_n(\varphi(x_{-1})=y_{-1})}{(y_0-y_{-1})X_2(x_{-1})
                  \widetilde{\mathcal{X}}_{n-1}(x_{-1})}
                \\
   &= \dfrac{\mathcal{Y}_n(\psi(x_{n-1})=y_n)}{(y_n-y_{n-1})X_2(x_{n-1})
                  \widetilde{\mathcal{X}}_{n-1}(x_{n-1})}
\end{split}\end{equation*}
(Of course, $C_0=0$). Check $\displaystyle \mathcal{Y}_1(x)=\dfrac{x-y_0}{x-y'_0}
= 1+\dfrac{y'_0-y_0}{x-y'_0} \Rightarrow
\mathcal{D}\mathcal{Y}_1(x)= -\dfrac{(y'_0-y_0)X_2(x)}{Y_2(y'_0)(x-x'_{-1})(x-x'_0)}$ so
that $C_1=\displaystyle \dfrac{y_0-y'_0}{Y_2(y'_0)}$ to be compared with\\
$\displaystyle -\dfrac{\mathcal{Y}_1(y_{-1})}{(y_0-y_{-1})X_2(x_{-1})
                  \widetilde{\mathcal{X}}_0(x_{-1})}=
-\dfrac{(y_{-1}-y_0)(x_{-1}-x'_{-1})(x_{-1}-x'_0)}
       {(y_{-1}-y'_0)(y_0-y_{-1})X_2(x_{-1}) }
=\dfrac{F(x_{-1},y'_0)}{(y_{-1}-y'_0)Y_2(y'_0)X_2(x_{-1})}
=\dfrac{(y'_0-y_{-1})(y'_0-y_0)}{(y_{-1}-y'_0)Y_2(y'_0)}$, OK.

Also,
$$
\dfrac{(\varphi(x)-y_0)(\varphi(x)-y_1)\dotsb  (\varphi(x)-y_{n-1}) } 
       {(\varphi(x)-y'_1)(\varphi(x)-y'_2)\dotsb  (\varphi(x)-y'_n)}
   +\dfrac{ (\psi(x)-y_0)(\psi(x)-y_1)\dotsb  (\psi(x)-y_{n-1})}{
             (\psi(x)-y'_1)(\psi(x)-y'_2)\dotsb  (\psi(x)-y'_n)}
$$
$$
=  D_n(x) \dfrac{(x-x_0)(x-x_1)\dotsb  (x-x_{n-2}) }
                {(x-x'_0)(x-x'_1)\dotsb  (x-x'_n) },
$$
where $D_n$ is a polynomial of degree 2.

same with $\mathcal{D}^\dagger$

Riccati and $2^{\text{nd}}$ order linear differnce eq.

From $\displaystyle a(x_m) \dfrac{f(y_{m+1})-f(y_m)}{y_{m+1}-y_m}
 = b(x_m)f(y_{m+1})f(y_m) +c(x_m)(f(y_{m+1})+f(y_m)))+d(x_m)$,

$\displaystyle f(y_{m+1})= 
    \dfrac{   \left(\dfrac{a(x_m)}{y_{m+1}-y_m}+c(x_m)\right)f(y_m)+d(x_m) }
          {   \dfrac{a(x_m)}{y_{m+1}-y_m}-c(x_m) -b(x_m)f(y_m) }$


$\displaystyle \ \ = -\dfrac{1}{b(x_n)}\left(\dfrac{a(x_m)}{y_{m+1}-y_m}+c(x_m)\right)
      +\dfrac{1}{b(x_m)} 
        \dfrac{   \dfrac{a^2(x_m)}{(y_{m+1}-y_m)^2}+b(x_m)d(x_m)-c^2(x_m) }
          {   \dfrac{a(x_m)}{y_{m+1}-y_m}-c(x_m) -b(x_m)f(y_m) }$

$\displaystyle  \dfrac{a(x_{m+1})}{y_{m+2}-y_{m+1}}-c(x_{m+1}) -b(x_{m+1})f(y_{m+1}) =
    A_m + \dfrac{   B_m }
          {   \dfrac{a(x_m)}{y_{m+1}-y_m}-c(x_m) -b(x_m)f(y_m) }$,

\medskip

\subsection{Expansion for elliptic logarithm} \   \\

Expand \eqref{elllog}
satisfying
$$ \mathcal{D}\Lambda_N(x) = \dfrac{(x'_{-1}-x'_N)X_2(x)}{(x-x'_{-1})(x-x'_N)} $$
as $\displaystyle  \sum_1^{N+1} \gamma_n \mathcal{Y}_n(x)$:
$\displaystyle  \sum_1^{N+1} \gamma_n C_n X_2(x)\widetilde{\mathcal{X}}_{n-1}(x) = \dfrac{(x'_{-1}-x'_N)X_2(x)}{(x-x'_{-1})(x-x'_N)} $, \\
or $\displaystyle 
\sum_1^{N+1} \gamma_n C_n \dfrac{\mathcal{X}_{n-1}(x)}{x-x'_{n-1}} =
   \dfrac{x'_{-1}-x'_N}{x-x'_N} $, 

Residue at $x'_N: \displaystyle \gamma_{N+1} C_{N+1} \mathcal{X}_N(x'_N)=x'_{-1}-x'_N$;\\
at $x'_{N-1}:$   $\gamma_NC_N \mathcal{X}_{N-1}(x'_{N-1})
+\gamma_{N+1} C_{N+1}  \dfrac{(x'_{N-1}-x_0)\dotsb (x'_{N-1}-x_{N-1}) }
                   { (x'_{N-1}-x'_0)\dotsb (x'_{N-1}-x'_{N-2})(x'_{N-1}-x'_N)} =
 \gamma_NC_N \mathcal{X}_{N-1}(x'_{N-1})
+\gamma_{N+1} C_{N+1}  \mathcal{X}_{N-1}(x'_{N-1})\dfrac{x'_{N-1}-x_{N-1} }{(x'_{N-1}-x'_N)}= 0$; 
at $x'_{N-2}: \gamma_{N-1}C_{N-1} +\gamma_NC_N \dfrac{x'_{N-2}-x_{N-2}}{x'_{N-2}-x'_{N-1}}
+\gamma_{N+1}C_{N+1} \dfrac{(x'_{N-2}-x_{N-2})(x'_{N-2}-x_{N-1})}{(x'_{N-2}-x'_{N-1})(x'_{N-2}-x'_N)}=0$, etc.

$$\hspace*{-33pt} \gamma_n C_n = \dfrac{(x'_{n-1}-x_{n-1})(x'_N-x_n)(x'_N-x_{n+1})\dotsb (x'_N-x_{N-1})}
      {(x'_N-x'_{n-1})(x'_N-x'_n)\dotsb (x'_N-x'_{N-1} )}\gamma_{N+1} C_{N+1}
=\dfrac{(x'_{n-1}-x_{n-1})(x'_{-1}-x'_N) }
       {(x'_N-x_{n-1})\mathcal{X}_{n-1}(x'_N)}$$

\noindent \hspace*{-35pt}Proof: let
$\displaystyle S_i=
\sum_{i+1}^{N+1} (x'_{n-1}-x_{n-1})
\dfrac{(x-x_i)\dotsb (x-x_{n-2})}{(x-x'_i)\dotsb (x-x'_{n-1})}
\dfrac{(x'_N-x'_i)\dotsb (x'_N-x'_{n-2})}{(x'_N-x_i)\dotsb (x'_N-x_{n-1})}
=\dfrac{x'_i-x_i}{(x-x'_i)(x'_N-x_i)}+\dotsb$.
Then, $\displaystyle S_N=\dfrac{1}{x-x'_N},
S_{N-1}=\dfrac{x'_{N-1}-x_{N-1}}{(x-x'_{N-1})(x'_N-x_{N-1})} 
  +\dfrac{(x'_N-x_N)(x-x_{N-1})(x'_N-x'_{N-1})}{(x-x'_{N-1})(x-x'_N)(x'_N-x_{N-1})(x'_N-x_N)} =
\dfrac{x'_{N-1}-x_{N-1}+
  \dfrac{(x-x_{N-1})(x'_N-x'_{N-1})}{x-x'_N} 
}{(x-x'_{N-1})(x'_N-x_{N-1})} 
=\dfrac{1}{x-x'_N}$. 

Show that $S_i=S_{i+1}, i=N-1,N-2,\dotsc, 0$.

$$
S_i= \dfrac{x'_i-x_i}{(x-x'_i)(x'_N-x_i)}+
    \dfrac{(x-x_i)(x'_N-x'_i)}{(x-x'_i)(x'_N-x_i)} S_{i+1} =
  \dfrac{x'_i-x_i+ \dfrac{(x-x_i)(x'_N-x'_i)}{x-x'_N)}}{(x-x'_i)(x'_N-x_i)}
=\dfrac{1}{x-x'_N} \qed 
$$

And from $\displaystyle C_n=  -\dfrac{\mathcal{Y}_n(y_{-1})}{(y_0-y_{-1})X_2(x_{-1})
                  \widetilde{\mathcal{X}}_{n-1}(x_{-1})}$,

$$
\gamma_n= -\dfrac{(x'_{n-1}-x_{n-1})(x'_{-1}-x'_N) }
       {(x'_N-x_{n-1})\mathcal{X}_{n-1}(x'_N)}
  \dfrac {(y_0-y_{-1})X_2(x_{-1})
                  \widetilde{\mathcal{X}}_{n-1}(x_{-1})}{\mathcal{Y}_n(y_{-1})}
$$
$$
\hspace*{-44pt}=(x_{n-1}-x'_{n-1})X_2(x_{-1})\dfrac{(x'_N-x'_{-1})\dotsb (x'_N-x'_{n-2}) }
       {(x'_N-x_0)\dotsb(x'_N-x_{n-1})}
  \dfrac {(x_{-1}-x_0)\dotsb (x_{-1}-x_{n-2})(y_{-1}-y'_0)\dotsb (y_{-1}-y'_{n-1})
                  }{(x_{-1}-x'_{-1})\dotsb (x_{-1}-x'_{n-1})(y_{-1}-y_1)\dotsb (y_{-1}-y_{n-1})}
$$
$$
\dfrac{ \gamma_{n+1}/(x_n-x'_n) }{\gamma_n/(x_{n-1}-x'_{n-1})}=
 \dfrac{(x'_N-x'_{n-1})(x_{-1}-x_{n-1})(y_{-1}-y'_n)}
       {(x'_N-x_n)(x_{-1}-x'_n)(y_{-1}-y_n)}
$$

Return to
 $\displaystyle  \mathcal{D}f(x) =\dfrac{1}{x-A}$, $x'_{-1}=A$, $f(y_0)=0$,

$x'_n$s, starting with $x'_{-1}=A$, come as close as we want to a zero, say $\zeta$, of $X_2(x)=
c_{2,2}(x-\zeta)(x-\zeta')$, then
$$
f(x)=  \underbrace{-\dfrac{c_{2,2} (A-\zeta)}{X_2(A)}}_{1/(\zeta'-A)}(x-y_0)
 +\dfrac{1}{X_2(A)}\lim_{x'_N\to \zeta} [\Lambda_N(x)-\Lambda_N(y_0)]
$$


with the $\gamma_n$s above, with $x'_N$ replaced by $\zeta$.

$\displaystyle \gamma_1 = -\dfrac{(x'_0-x_0)(x'_{-1}-\zeta) }
       {\zeta-x_0}
  \dfrac {(y_0-y_{-1})X_2(x_{-1})(y_{-1}-y'_0)
                  }{(x_{-1}-A)(x_{-1}-x'_0)(y_{-1}-y_0)}
=-\dfrac{(x'_0-x_0)(A-\zeta) }
       {\zeta-x_0}
  \dfrac {Y_2(y'_0)}{y'_0-y_0}
$

So, the expansion of $f$ is 
$$
\sum_1^\infty  \left[ \dfrac{\gamma_n}{X_2(A)} -\dfrac{c_{2,2} (A-\zeta)}{X_2(A)} (y_n-y'_{n-1}) \right] \mathcal{Y}_n(x)
$$
(using the formal sum $x=y_0 +\sum_1^\infty (y_n-y'_{n-1})\mathcal{Y}_n(x)$).

Check $n=1$: 
$$
\dfrac{\gamma_1}{X_2(A)} -\dfrac{c_{2,2} (A-\zeta)}{X_2(A)} (y_1-y'_0)=
\dfrac{(y_1-y'_0)(f(y_1)-f(y_0)}{y_1-y_0} = \dfrac{y_1-y'_0}{x_0-A}
$$ 

\subsection{Convergence}  \   \\




Average behaviour: $\displaystyle \prod_1^n (x-x_k) =\prod_1^n (x-\mathcal{E}(ak+b))\approx \Phi(x)^n$. What is $\Phi(x)=\exp\mathcal{V}_+(x)$?

Let $a$ be a \textbf{\emph{real}} irrational multiple of a period $\omega$,
then the same factors reappear approximately in the product after $N$ steps
if $aN$ is close to an integer times $\omega$. $\Phi(x)$ is the limit
of the $N^{\text{th}}$ roots of such products. The various $ak$, for
$k=1,2,\dotsc, N$, modulo $\omega$, fill uniformly the segment $[0,\omega]$:

for any $j$ in $\{1,2,\dotsc, N\}$, there is a $k$ such that $ak$ is close to 
$j\omega/N$ modulo $\omega$. Indeed, let $aN$ be close to $M_N\omega$, with gcd$(N,M_N)=1$.Then,

$$
a k -\dfrac{j\omega}{N} = \omega\left(\dfrac{a}{\omega}-\dfrac{M_N}{N}\right)k
+\omega\dfrac{k M_N -j}{N},$$
to any $j$, there are  integers  $k$ and $m$ such that $k M_N -mN=j$ (Bezout).\\
So, we rearrange the product as $\displaystyle \Phi(x) \sim
 \left[ \prod_{j=1}^N  (x-\mathcal{E}(j\omega/N +b))\right]^{1/N}
   \sim \exp\left[\dfrac{1}{\omega}\int_0^\omega \log(x-\mathcal{E}(u +b))\;du\right]$. 
As $\mathcal{E}$ is the inversion of an elliptic integral of the first kind,
$\displaystyle u+b= \int^{\mathcal{E}} \dfrac{dv}{\sqrt{P(v)}}$,
we have
$\displaystyle \Phi(x)= \exp\left[\dfrac{1}{\omega}\int_{\{x_n\}}
   \dfrac{\log(x-v)\,dv}{\sqrt{P(v)}}\right]$,
where $\{x_n\}$ is the locus of the $x_n$s $=\{\mathcal{E}(u+b)\},
u\in [0,\omega]$. The constant $1/\omega$ is such that
$\Phi(x)\sim x$ for large $x$:
$\displaystyle \omega=\int_{\{x_n\}}
   \dfrac{\,dv}{\sqrt{P(v)}}$.

 So, let the complex potential $\displaystyle \mathcal{V_+}(x) =
  \dfrac{1}{\omega} \int_{\{x_n\}} \dfrac{\log(x-v)\,dv}{\sqrt{P(v)}}$,

($\mathcal{V}_-$ will be used with the $x'_n$s)

derivative: 
$\displaystyle \mathcal{V'_+}(x) =
  \dfrac{1}{\omega} \int_{\{x_n\}} \dfrac{dv}{(x-v)\sqrt{P(v)}}$,\\
$\displaystyle P(x)\mathcal{V'_+}(x) = \text{polynomial\ } +
  \dfrac{1}{\omega} \int_{\{x_n\}} \dfrac{\sqrt{P(v)}\,dv}{x-v}$,\\
$\displaystyle (P(x)\mathcal{V'_+}(x))' = \text{ (another) polynomial\ } +
  \dfrac{1}{\omega} \int_{\{x_n\}} \dfrac{P'(v)\,dv}{2(x-v)\sqrt{P(v)}}$,

Finally
$$
 P(x) \mathcal{V}''_+(x) +P'(x) \mathcal{V}'_+(x)/2 = \text{(still another) pol.}
$$
$$
   (\sqrt{P(x)}\; \mathcal{V}'_+(x) \;)' = \dfrac{\text{pol.}}{\sqrt{P(x)}}
$$
so, $\mathcal{V}'_+(x)=$ an incomplete elliptic integral of the second(?) kind divided
by $\sqrt{P(x)}$.

With $\xi$ such that $x=\mathcal{E}(\xi), dx/d\xi=\sqrt{P(x)}$:

$$
   \dfrac{d^2\mathcal{V}_+(x)}{d\xi^2} = \text{a pol. (in\ }x=\mathcal{E}(\xi))
$$

What is this polynomial, by the way?  $\mathcal{V}'_+(x)=x^{-1} +(\mu_{+,1}/\mu_{+,0})
 x^{-2}+(\mu_{+,2}/\mu_{+,0}) x^{-3}+\dotsb$,
$\sqrt{P(x)}=\pi_0 x^2+\pi_1 x+\pi_2+\dotsb$,

the pol. is
$$
(\pi_0 x^2+\pi_1 x+\pi_2+\dotsb)[\pi_0 -(\pi_2  +\pi_1(\mu_{+,1}/\mu_{+,0})
  +\pi_0(\mu_{+,2}/\mu_{+,0}))x^{-2}+\dotsb]$$
$$ =
\pi_0^2 x^2 +\pi_1\pi_0 x + \pi_2\pi_0 
   -\pi_0(\pi_2  +\pi_1(\mu_{+,1}/\mu_{+,0})
  +\pi_0(\mu_{+,2}/\mu_{+,0}))
$$
where $\displaystyle \mu_{+,k}= \int_{\{x_n\}} \dfrac{v^k\,dv}{\sqrt{P(v)}}$
is the $k^\text{th}$ moment of the contour drawn by the $x_n$s. The result is
$$
\pi_0^2 x^2 +\pi_1\pi_0 x + \pi_2\pi_0 
   -\dfrac{\pi_0}{\mu_{+,0}} \int_{\{x_n\}}
  \dfrac{\pi_2  +\pi_1 v   +\pi_0 v^2 = \sqrt{P(v)}-O(1/v)  }{\sqrt{P(v)}}\;dv
$$
The contour integrals
on the $x'_n$s are the same(??  yes, see later on), so, at last

$$
\dfrac{d^2\mathcal{V}(x)}{d\xi^2} =0,
$$
where $\mathcal{V}=\mathcal{V}_+ -\mathcal{V}_-$.

Jump of $\mathcal{V}'(x)$ when $x$ crosses the $x_n$ line:
$\displaystyle \mathcal{V}'(x)_{\text{average}} \pm \pi i 
   \dfrac{1}{\omega} \dfrac{1}{\sqrt{P(x)}}$, or in $\xi$:

$$
\left(\dfrac{d\mathcal{V}}{d\xi}\right)_{\text{average}} \pm \pi i 
  \dfrac{1}{\omega}
$$

Forget it! A much faster and more complete derivation: $\mathcal{V}_+(x)$ and $\mathcal{V}_-(x)$
are contour integrals on the locii filled by $\{x_n\}$ and $\{x'_n\}$ drawn by 
$\mathcal{E}(an+b)$ and $\mathcal{E}(an+b')$. If $x$ is between these two locii,
the two contour integrals of $\displaystyle \dfrac{dv}{(x-v)\sqrt{P(v)}}$ are the same
for $\mathcal{V}'_+(x)$ and $\mathcal{V}'_-(x)$, up to the residue at $v=x$:

$$
\mathcal{V}'(x)= \mathcal{V}'_+(x) -\mathcal{V}'_-(x)=
\dfrac{2\pi i }{\omega \sqrt{P(x)}} \Rightarrow
\dfrac{d\mathcal{V}(x)}{d\xi} = \dfrac{2\pi i}{\omega}
$$

\bigskip



\hspace*{-5pt}$\displaystyle (\gamma_n \mathcal{Y}_n(x))^{1/n} \sim$\\
$$
\hspace*{-55pt}\left( 
  \dfrac{(\zeta-x'_{-1})\dotsb (\zeta-x'_{n-2}) }
       {(\zeta-x_0)\dotsb(\zeta-x_{n-1})}
  \dfrac {(x_{-1}-x_0)\dotsb (x_{-1}-x_{n-2})(y_{-1}-y'_0)\dotsb (y_{-1}-y'_{n-1})
                  }{(x_{-1}-x'_{-1})\dotsb (x_{-1}-x'_{n-1})(y_{-1}-y_1)\dotsb (y_{-1}-y_{n-1})}
\dfrac{(x-y_0)\dotsb (x-y_{n-1}) }{(x-y'_0)\dotsb (x-y'_{n-1}) }\right)^{1/n} 
$$
$$\sim
\exp[ \mathcal{V}(x_{-1}) -\mathcal{V}(\zeta) -\mathcal{W}(y_{-1})
  +\mathcal{W}(x)]
$$


\parpic{
\psset{unit=1.0cm}\begin{pspicture}(5,5)
  \psline(0,0)(2,1)(3,5)(1,4)(0,0)
  \psline[linewidth=0.05,linecolor=red](0,0)(1,4)
  \uput[270](0,0){$b'$}    \uput[90](1,4){$b'+\omega$}
  \psline[linewidth=0.05,linecolor=blue](1.5,0.75)(2.5,4.75)
  \uput[0](2,2.75){$b$}  \pscircle*(2,2.75){0.05}
  \pscircle*(1,2){0.05}\uput[0](1,2){$\xi$}
\end{pspicture}
}
Towards a conjecture on rate of convergence:\\
 \   \\
\textsl{If the step $a$ in $x_n=\mathcal{E}_1(an+b)$, $x'_n=\mathcal{E}_1(an+b')$,
is a \emph{\textbf{real irrational}} multiple of a period $\omega$, of bounded
Lagrange-Markov constant(  see papers and book by S.Khrushchev), 
$$
|\gamma_n \mathcal{Y}_n(x)|^{1/n} \to \exp(-2\pi d(x)/|\omega|),
$$
where $d(x)$ is the distance of $\xi$ to the line} $\{an+b'\}$.

\medskip

Walsh, 1935!

\medskip

 $\displaystyle  \mathcal{D}f(x) =\dfrac{1}{x-A}$ with $f(y_0)=0$,
$\zeta=-2.529822$,
$F(x,y)= -0.3125 x^2 y^2 + 2x^2 - 2.25 xy + 2y^2 - 1.25, A=-7$,

{\small\hspace*{-22pt}
\begin{verbatim}
 k    x(k)      y(k)    x'(k-1) y'(k-1)   gk       term        f(1)               f(i)
 1  0.51672  0.98564;  2.0068  3.3515 ;-0.05542  0.02356958  0.1285033 -0.00453082 + 0.1201190*I
 2 -0.95291 -0.37902;-19.9626  2.7382 ;-0.21650 -0.00076035  0.1277430 -0.00000099 + 0.1495888*I
 3 -0.15753 -0.87025; -2.0680 -2.3716 ; 0.30473  0.00043771  0.1281807  0.01064768 + 0.1357818*I
 4  0.99948  0.69234;  4.2749 -4.6414 ;-0.23940 -0.00011400  0.1280667  0.00696292 + 0.1368071*I
 5 -0.23019  0.64003;  2.4333  2.0494 ;-0.05303  0.00000740  0.1280741  0.00664251 + 0.1371257*I
 6 -0.93096 -0.90117; -2.6429 34.5330 ;-4.06650 -0.00000609  0.1280680  0.00686082 + 0.1382964*I
 7  0.57665 -0.31020; -3.5481 -2.0143 ; 0.77233  0.00000073  0.1280687  0.00688667 + 0.1381635*I
 8  0.74665  0.99449;  2.1363  6.1416 ;-0.07636  0.00000002  0.1280687  0.00688892 + 0.1381636*I
 9 -0.83038 -0.07436;  9.3362  2.2354 ;-0.13154 -1.411 E-10  0.1280687  0.00689104 + 0.1381629*I
10 -0.45360 -0.97261; -2.0003 -3.0677 ; 0.36320  1.029 E-10  0.1280687  0.00688980 + 0.1381615*I
15  0.90479  0.92787;  2.5007121.2785 ;-0.11543  1.975 E-15  0.1280687  0.00689019 + 0.1381625*I
20 -0.98422 -0.50887; -8.1267  3.1850 ;-0.28273 -2.400 E-19  0.1280687  0.00689019 + 0.1381625*I
\end{verbatim}  }

Here, $b=0$, $\omega=4K$, $b'=iK'$, $k=1/2: K=1.68575, K'=2.15652$,

rate of convergence at $x=1$ is $\exp(-\pi K'/(2K)) = 1/7.459=10^{-0.8727}.$


\section{Interpolatory continued fraction.}


$$
\dfrac{\alpha_0(x-y_0)}
                                   {1+\beta_0(x-y_1) +\dfrac{\alpha_1(x-y_1)(x-y_2)}
                                      {\dfrac{\ddots}
                                  {\ \ \ \ \ \ \ \ 1+\beta_{n-2}(x-y_{2n-3})+\dfrac{\alpha_{n-1}(x-y_{2n-3})(x-y_{2n-2})}{1+\beta_{n-1}(x-y_{2n-1})+\dotsb}}}}$$

\begin{verbatim}
  k   a(k-1)     b(k-1)
  1  0.02342657  0.3064279 
  2  0.02442157 -0.5755228
  3  0.2808405   0.5023201 
  4  0.02875770  0.3488815 
  5 -0.002774129 0.6898141 
  6 -2.635035    3.609245 
  7  0.05390936  0.7862154 
  8  0.001055843 0.2904550 
  9  0.01638600 -0.8421168 
 10  0.4162262   0.5008208 
...
 13 -2.667423    1.473576 
 14  3.576408    0.6620672 
 16  0.01282034 -5.278312 
 17  3.286337    0.6235315 
 24 -1.406951    1.290914 
 31  2.713400   -4.291244 
 41  0.01600274  2.476500 
 42 -2.045183    0.8222617 
 49 -3.308682    4.687681 
 56 -1.128588    0.1804773 
 57  1.869775    0.6528640 
 59  0.01285615 -8.095352 
 60  5.162988    0.6383966 
\end{verbatim}




\bigskip \  


 
\bigskip \   \bigskip \  
\   
\bigskip \  



Hope to connect to the current theory of elliptic hypergeometric expansions
\cite{spir,Spir2004,Spirsupp,SpiCalo,SpZ1,SpZ2,SpZ2007,zhed99,ZhedPad,ZhedSpir99}  some day...



Remarks on this biquadratic calculus:\\
{\Small Titre du document / Document title
{\bf Modular invariance in lattice statistical mechanics}\\ 
Auteur(s) / Author(s)
MAILLARD J.-M. (1) ; BOUKRAA S. (2) ; 
Affiliation(s) du ou des auteurs / Author(s) Affiliation(s)
(1) LPTHE, Tour 16, 1er \'etage, 4 Place Jussieu, 75252 Paris, FRANCE
(2) Institut d'A\'eronautique, Universit\'e de Blida, BP 270, Blida, ALGERIE\\
R\'esum\'e / Abstract
We show that non-trivial symmetries, and structures, originating from lattice statistical mechanics, provide a quite efficient way to get interesting results on elliptic curves, algebraic varieties, and even arithmetic problems in algebraic geometry. In particular, the lattice statistical mechanics approach underlines the role played by the modular $j-$invariant, and by a reduction of elliptic curves to a symmetric biquadratic curve (instead of the well-known reduction to a canonical Weierstrass form). This representation of elliptic curves in terms of a very simple symmetric biquadratic is such that the action of a (generically infinite) discrete set of birational transformations, which corresponds to important non-trivial symmetries of lattice models, can be seen clearly. This biquadratic representation also makes a particular symmetry group of elliptic curves (group of permutation of three elements related to the modular $j-$invariant) crystal clear. Using this biquadratic representation of elliptic curves, we exhibit a remarkable polynomial representation of the multiplication of the shift of elliptic curves (associated with the group of rational points of the curve). The two expressions $g_2$ and $g_3$, occurring in the Weierstrass canonical form 
$y^2 = 4 x^3 - g_2 x - g_3$, are seen to present remarkable covariance properties with respect to this infinite set of commuting polynomial transformations (homogeneous polynomial transformations of three homogeneous variables, or rational transformations of two variables). \\
Revue / Journal Title
{\sl Annales de la Fondation Louis de Broglie (Ann. Fond. Louis Broglie)} ISSN 0182-4295 CODEN AFLBDU 
Source / Source
2001, vol. {\bf 26}, NS, no2, pp. 287-328 (73 ref.) 
Langue / Language
Anglais
Editeur / Publisher
Fondation Louis de Broglie, Paris, FRANCE (1975) (Revue)
Mots-cl\'es anglais / English Keywords
Theoretical study ; Elliptic curve ; Algebraic number ; Integer ; Dynamical systems ; Discrete systems ; Algebraic geometry ; Statistical mechanics ; Invariance ; 
Mots-cl\'es fran\c cais / French Keywords
Etude th\'eorique ; Courbe elliptique ; Nombre alg\'ebrique ; Nombre entier ; Syst\`eme dynamique ; Syst\`eme discret ; G\'eom\'etrie alg\'ebrique ; M\'ecanique statistique ; Invariance ; 
001b00e20 ; 001b00e30 ; 0520 ; 0530 ; 
Mots-cl\'es espagnols / Spanish Keywords
Curva eliptica ; N\'umero algebraico ; Entero ; Geometría algebraica ; 
Localisation / Location
INIST-CNRS, Cote INIST : 18321, 35400010080944.0070\\
Copyright 2006 INIST-CNRS. All rights reserved
Toute reproduction ou diffusion m\^eme partielle, par quelque proc\'ed\'e ou sur tout support que ce soit, ne pourra \^etre faite sans l'accord pr\'ealable \'ecrit de l'INIST-CNRS.
No part of these records may be reproduced of distributed, in any form or by any means, without the prior written permission of INIST-CNRS.
N${}^\circ$ notice refdoc (ud4) : 13559414
}

 



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\renewcommand{\mylmark}{References.}
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