Right-slot division makes three elementary notions meet: reciprocal
arithmetic, a pole in an affine chart, and the projective point at infinity.
Iteration adds a fourth notion: a finite or infinite expression can be the
closure of a recurrence at one of its fixed points.  We keep the algebraic
fixed-point equation, analytic convergence, and projective continuation
distinct.

\subsection{Arithmetic reciprocals and projective poles}

For $c\ne0$, the ordinary map
\[
  J_c(z)=\frac{c}{z}
\]
is defined on $K\setminus\{0\}$.  Its matrix is
\[
  J_c\leftrightarrow
  \begin{pmatrix}0&c\\1&0\end{pmatrix},
\]
so projectively
\[
  J_c(0)=\infty,
  \qquad
  J_c(\infty)=0.
\]
This continuation does not make ordinary division by zero admissible.

The reciprocal coordinate centered at a finite point $q$ is
\begin{equation}\label{eq:p0-pole-coordinate}
  w_q(z)=-\frac1{z-q}.
\end{equation}
The sign is chosen so that $w_q$ has positive determinant and therefore maps the
upper half-plane to itself; the positive chart $1/(z-q)$ has determinant $-1$,
maps the upper half-plane to the lower one, and is not used for the horocycle
discussion below.  The coordinate sends $q$ to infinity and infinity to zero.
The rational functions whose
only projective pole is $q$---equivalently, those regular on
$\PP^1\setminus\{q\}$ and at the old point infinity---form
\[
  K[w_q]=K\!\left[\frac1{z-q}\right];
\]
the sign does not change this ring.  The usual polynomial chart $K[z]$ is the
corresponding ring when the chosen pole is $\infty$.  This is a coordinate
statement, not a claim that one pole gauge is globally preferred.

By Theorem~\ref{thm:p0-ripple-pencil-formula}, whose positivity hypothesis
$w_q$ now satisfies, the same $q$ is the common
boundary tangency point of the horocycles pulled back through $w_q$.  The
denominator zero and the ripple center are therefore two readings of one
projective datum.

\begin{warning}[Negative-determinant prefixes]
\label{warn:p0-negative-determinant}
Not every non-degenerate real Möbius prefix has positive determinant: the
elementary contexts $c/z$ with $c>0$ and $c-z$ have determinant $-c$ and $-1$
respectively.  The ripple statements of Section~\ref{sec:p0-ripples} therefore
apply to the positive-determinant prefixes; for a general prefix one either
conjugates by the reflection $z\mapsto-z$, which is the sign change of the
assignment listed in Subsection~\ref{subsec:p0-point-paths}, or extends the
action anti-holomorphically by $z\mapsto (A\overline z+B)/(C\overline z+D)$,
which preserves the upper half-plane and realizes the same fractional linear
arithmetic on the real boundary.  No ripple statement is made here for the
negative-determinant case beyond this remark.
\end{warning}

\subsection{A reciprocal from an infinite expansion}

Let
\begin{equation}\label{eq:p0-geometric-recursion}
  S_0=0,
  \qquad
  S_{n+1}=1+rS_n.
\end{equation}
Then
\[
  S_n=1+r\bigl(1+r(\cdots(1+r\cdot0)\cdots)\bigr)
      =\sum_{k=0}^{n-1}r^k.
\]
Scalar multiplication is commutative, so this example does not by itself
separate left and right slots; it shows how a reciprocal can arise as recursive
closure.

\begin{proposition}[Finite and infinite geometric expansion]
\label{prop:p0-finite-geometric-expansion}
If $r\ne1$, then
\begin{equation}\label{eq:p0-finite-geometric-quotient}
  S_n=\frac{1-r^n}{1-r}.
\end{equation}
If $K=\R$ or $\CC$ and $|r|<1$, then
\[
  S_n\longrightarrow\frac1{1-r}.
\]
Formally, $(1-r)^{-1}=\sum_{k\ge0}r^k$ in $K[[r]]$.
\end{proposition}

\begin{proof}
Multiplication of the finite sum by $1-r$ telescopes to $1-r^n$.
Under $|r|<1$, $r^n\to0$.  The formal identity follows coefficient by
coefficient.
\end{proof}

The convergence hypothesis is active.  For $r=-1$ the sequence oscillates,
and for $|r|>1$ it generally does not converge to the algebraically existing
reciprocal.

\subsection{Fixed-point closure and escape to infinity}

Put
\[
  T_r(s)=1+rs.
\]
For $r\ne1$, its finite fixed point is
\begin{equation}\label{eq:p0-geometric-fixed-point}
  q_r=\frac1{1-r}.
\end{equation}
Thus, when $|r|<1$, the geometric-series truncations converge to the fixed
point of their generating update.

For $r\ne0$, the affine update is invertible and has projective matrix
\[
  M_r=\begin{pmatrix}r&1\\0&1\end{pmatrix};
\]
it fixes $\infty$.  The exceptional value $r=0$ gives the constant update
$T_0(s)=1$, whose matrix is singular and which is not an element of
$\PGL_2$.  This degenerate case still has the finite fixed point $q_0=1$ but
carries no invertible projective dynamics.

As $r\to1$, the finite point $q_r$ approaches $\infty$ in $\PP^1$.  At
$r=1$, $T_1(s)=s+1$ has no finite fixed point and has $\infty$ as its repeated
projective fixed point.  The pole of $r\mapsto1/(1-r)$ is therefore exactly
the parameter at which finite fixed-point closure becomes parabolic closure at
infinity.

\begin{proposition}[Affine iteration]
\label{prop:p0-affine-iteration}
Let $T(z)=az+b$ with $a\ne0$.  If $a\ne1$, put
$q=b/(1-a)$.  Then
\begin{equation}\label{eq:p0-affine-centered-form}
  T(z)=q+a(z-q),
  \qquad
  T^n(z)=q+a^n(z-q).
\end{equation}
Over $\R$ or $\CC$, every finite orbit converges to $q$ when $|a|<1$.
If $a=1$ and $b\ne0$, then $T^n(z)=z+nb$ and the only projective fixed point
is $\infty$.
\end{proposition}

\begin{proof}
The relation $(1-a)q=b$ gives the centered form, and induction gives its
$n$th iterate.
\end{proof}

Algebraic existence, attraction, and projective fixed-point structure are
different statements; they coincide only under additional hypotheses.

\subsection{Right reciprocal iteration and continued fractions}

A stationary right reciprocal update is
\begin{equation}\label{eq:p0-right-reciprocal-map}
  F(z)=a+\frac{b}{z},
  \qquad b\ne0,
\end{equation}
with matrix
\[
  M_F=\begin{pmatrix}a&b\\1&0\end{pmatrix}.
\]
Ordinary iteration is defined only while every intermediate denominator is
nonzero.  Its finite truncations are generalized continued fractions; the
matrix product remains projectively meaningful when a truncation reaches
infinity.  This is the standard matrix mechanism behind continued fractions
\cite{Khinchin1964ContinuedFractions}; the arithmetic reading of the same
truncations used here is the specialization to two-step words.

\begin{proposition}[Reciprocal fixed points]
\label{prop:p0-reciprocal-fixed-points}
The finite fixed points of \eqref{eq:p0-right-reciprocal-map} solve
\begin{equation}\label{eq:p0-reciprocal-quadratic}
  z^2-az-b=0.
\end{equation}
At a nonzero fixed point $q$,
\[
  F'(q)=-\frac{b}{q^2}.
\]
Over $\CC$, $q$ is locally attracting when $|F'(q)|<1$ and locally repelling
when $|F'(q)|>1$.
\end{proposition}

\begin{proof}
Multiply $q=a+b/q$ by $q$ and differentiate $F$.
\end{proof}

\begin{example}[The golden-ratio expansion]
\label{ex:p0-golden-ratio}
For $F(z)=1+1/z$ and $z_0=1$, the orbit is
\[
  1,\quad2,\quad\frac32,\quad\frac53,\quad\frac85,\ldots.
\]
The fixed points are
\[
  \phi=\frac{1+\sqrt5}{2},
  \qquad
  -\phi^{-1}=\frac{1-\sqrt5}{2}.
\]
Since $|F'(\phi)|=\phi^{-2}<1$, the positive fixed point is attracting on the
relevant positive branch.  The infinite continued fraction
\[
  1+\cfrac1{1+\cfrac1{1+\cdots}}
\]
denotes $\phi$ only after this convergence has been established; the
fixed-point equation alone does not select an attracting root in every field
or domain.
\end{example}

For nonstationary maps
\[
  F_k(z)=a_k+\frac{b_k}{z},
  \qquad
  M_k=\begin{pmatrix}a_k&b_k\\1&0\end{pmatrix},
\]
the $n$th truncation is represented by $M_n\cdots M_1$.  The nested expression,
the recurrence, and matrix multiplication are the same finite calculation.

\subsection{General projective fixed points}

For
\[
  F(z)=\frac{Az+B}{Cz+D},
  \qquad AD-BC\ne0,
\]
a finite fixed point satisfies
\begin{equation}\label{eq:p0-general-mobius-fixed-point}
  Cz^2+(D-A)z-B=0.
\end{equation}
In homogeneous coordinates, fixed points are the eigenlines of the representing
matrix.  Infinity is fixed exactly when $C=0$.  Over an algebraic closure a
nonidentity projective map has two fixed points counted with multiplicity;
over the base field they may be distinct, repeated, or visible only after a
field extension.

The elementary chain is now complete:
\[
  \text{reciprocal}
  \longrightarrow
  \text{pole}
  \longrightarrow
  \infty
  \longrightarrow
  \text{fixed-point closure}
  \longrightarrow
  \text{matrix eigenline}.
\]