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\begin{document}

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\begin{center}
\vskip 1cm{\LARGE\bf Polynomials Associated with Reciprocation
}
\vskip 1cm
\large
Clark Kimberling\\
Department of Mathematics\\
University of Evansville\\
1800 Lincoln Avenue\\
Evansville, IN 47722 \\
USA\\
\href{mailto:ck6@evansville.edu}{\tt ck6@evansville.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
Polynomials are defined recursively in various ways associated with
reciprocation; e.g., $S_{n+1}(x)/T_{n+1}(x)=S_{n}(x)/T_{n}(x)\pm
T_{n}(x)/S_{n}(x).$  Under certain conditions, the zeros of $S_{n}$
interlace those of $T_{n}.$  Identities for $S_{n},$ $T_{n},$ and related
polynomials are derived, as well as recurrence relations and infinite sums
involving roots of polynomials.
\end{abstract}

%\newtheorem{theorem}{Theorem}
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%\newtheorem{proposition}[theorem]{Proposition}
%\newtheorem{conjecture}[theorem]{Conjecture}
%\newtheorem{defin}[theorem]{Definition}
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\section{Introduction}

A well-known problem \cite{New} starts with the recurrence%
\begin{equation}
x_{n+1}=x_{n}+1/x_{n},  \label{R+}
\end{equation}%
given that $x_{0}=1.$  The sequence $(1,2,5,29,\ldots )$ thus determined is
indexed in Sloane's \textit{Online Encyclopedia of Integer Sequence}s \cite%
{Slo} as \seqnum{A073833}.  It is natural to ask what happens if the initial value $%
1 $ is replaced by an indeterminate $x.$  The purpose of this paper is to
respond to that question and related questions. For example, what if the
recurrence is replaced by 
\begin{equation}
x_{n+1}=x_{n}-1/x_{n},  \label{R-}
\end{equation}%
as in \seqnum{A127814}?  The recurrence (\ref{R-}) leads to polynomials which are
interesting because of the distribution of their zeros, as portended by the
following definition:  suppose $V=(v_{1},v_{2},\ldots ,v_{m-1})$ and $%
W=(w_{1},w_{2},\ldots ,w_{m})$ are lists of numbers satisfying 
\begin{equation*}
w_{1}<v_{1}<w_{2}<v_{2}<\cdots <v_{m-1}<w_{m};
\end{equation*}%
then $V$ \textit{interlaces} $W$.

Throughout this paper, except where otherwise stipulated, the letter $c$
denotes an arbitrary nonzero complex number.

\section{The recurrence $S_{n+1}/T_{n+1}=(1/c)(S_{n}/T_{n}-T_{n}/S_{n})$}

Define polynomials $S_{1}=S_{1}(x)=x,$ $T_{1}=T_{1}(x)=1,$ and%
\begin{equation}
\begin{tabular}{cc}
$S_{n+1}=S_{n}^{2}-T_{n}^{2},$ & $T_{n+1}=cS_{n}T_{n},$%
\end{tabular}
\label{ST}
\end{equation}%
so that%
\begin{equation}
S_{n+1}/T_{n+1}=(1/c)(S_{n}/T_{n}-T_{n}/S_{n}),  \label{STQ}
\end{equation}%
and for $n\geq 2,$%
\begin{eqnarray}
T_{n} &=&c^{n-1}S_{1}S_{2}\cdots S_{n-1},  \notag \\
S_{n+1} &=&(S_{n}-c^{n-1}S_{1}S_{2}\cdots
S_{n-1})(S_{n}+c^{n-1}S_{1}S_{2}\cdots S_{n-1}).  \label{SF}
\end{eqnarray}%
If $c=1,$ then (\ref{STQ}) is the recurrence (\ref{R-}) with $%
x_{n}=S_{n}/T_{n}.$  For $c>0$ and $n\geq 1,$ let $\mathcal{Z}_{n}$ be the
list of zeros of $S_{n}$ in increasing order, so that%
\begin{equation*}
\mathcal{Z}_{1}=(0),\text{ \ \ }\mathcal{Z}_{2}=(-1,1),
\end{equation*}%
and $\mathcal{Z}_{3}$ is the ordered list consisting of the numbers $(\pm
c\pm \sqrt{c^{2}+4})/2;$ e.g., if $c=1,$ then $\mathcal{Z}_{3}=(-\varphi
,-1/\varphi ,1/\varphi ,\varphi ),$ where $\varphi =(1+\sqrt{5})/2,$ the
golden ratio.

If $V$ and $W$ are ordered lists, we shall employ the set-union symbol $\cup 
$ for the \textit{ordered} union of the merged list formed by the numbers in 
$V$ and $W,$ thus: $\ V\cup W.$ Note that $S_{n}$ has degree $2^{n-1},$
that $T_{n}$ has degree $2^{n-1}-1,$ and that $\tbigcup\limits_{k=1}^{n}%
\mathcal{Z}_{k}$ lists the zeros of $T_{n}.$

\begin{theorem}
\label{T1}
Suppose that $c$ is a nonzero real number and $n\geq 1.$ Then $\underset{k=1}{\overset{n}{\cup }}\mathcal{Z}%
_{k}$ interlaces $\mathcal{Z}_{n+1}.$
\end{theorem}

\begin{proof}
First, suppose that $c>0.$  Clearly $\mathcal{Z}_{1}$ interlaces $\mathcal{Z}_{2}$.  It will be helpful
to denote the zeros of $S_{n}$ in increasing order by $r_{n,i}$ where $%
i=1,2,\ldots ,2^{n-1}.$  To complete a two-part first induction step, we
shall show that $\mathcal{Z}_{1}\cup \mathcal{Z}%
_{2}$ interlaces $\mathcal{Z}_{3}.$  The
function $f_{2}=S_{2}/S_{1}$ is continuous and rises strictly from $-\infty$ to $\infty$ on each
of the intervals $(-\infty ,r_{11})$ and $(r_{11},\infty ).$  Therefore
there exist unique numbers $r_{31},r_{32}$ such that%
\begin{equation}
r_{31}<r_{21}<r_{32}<r_{11},\text{ \ \ }f_{2}(r_{31})=-c,\text{ \ }%
f_{2}(r_{32})=c  \label{I1}
\end{equation}%
and $r_{33},r_{34}$ such that%
\begin{equation}
r_{11}<r_{33}<r_{22}<r_{34},\text{ \ \ }f_{2}(r_{33})=-c,\text{ \ }%
f_{2}(r_{34})=c.  \label{I2}
\end{equation}%
Thus, $r_{31},r_{33}$ are the zeros of $S_{2}+cS_{1},$ and $r_{32},r_{34}$
are the zeros of $S_{2}-cS_{1}.$ Now $S_{3}=(S_{2}+cS_{1})(S_{2}-cS_{1}),$ so
that (\ref{I1}) and (\ref{I2}) imply that $\mathcal{Z}_{1}\cup \mathcal{Z}%
_{2}$ interlaces $\mathcal{Z}_{3}.$  As a general induction hypothesis,
assume for $n\geq 3$ that $\underset{k=1}{\overset{n-2}{\cup }}\mathcal{Z}%
_{k}$ interlaces $\mathcal{Z}_{n-1}$ and that $\underset{k=1}{\overset{n-1}{%
\cup }}\mathcal{Z}_{k}$ interlaces $\mathcal{Z}_{n}.$  Write
\begin{eqnarray*}
\underset{k=1}{\overset{n-2}{\cup }}\mathcal{Z}_{k} &=&(\rho _{11},\rho
_{12},\ldots ,\rho _{1m}),\text{ \ \ where }m=2^{n-2}-1, \\
\underset{k=1}{\overset{n-1}{\cup }}\mathcal{Z}_{k} &=&(\rho _{21},\rho
_{11},\rho _{22},\rho _{12},\ldots ,\rho _{2m},\rho _{1m,}\rho _{2,m+1}).
\end{eqnarray*}%
The function $f_{n}=S_{n}/(S_{1}S_{2}\cdots S_{n-1})$ is continuous and
strictly increasing on each of the intervals%
\begin{equation}
(-\infty ,\rho _{11}),(\rho _{11},\rho _{12}),\ldots ,(\rho _{1,m-1},\rho
_{1m})  \label{Int},  \end{equation}%
with infinite limits (as in the argument above for $f_{2}.$)
Let $\rho _{31},\rho _{32}$ be the unique numbers satisfying%
\begin{equation*}
\rho _{31}<\rho _{21}<\rho _{32}<\rho _{11},\text{ \ \ }f_{n}(\rho _{31})=-c,%
\text{ \ }f_{2}(\rho _{32})=c.
\end{equation*}%
Likewise applying the intermediate value theorem to the remaining intervals
in (\ref{Int}), we conclude that $\underset{k=1}{\overset{n}{\cup }}\mathcal{%
Z}_{k}$ interlaces $\mathcal{Z}_{n+1}.$

Now suppose that $c<0.$  The recurrences (\ref{ST}) show that the only 
exponents of $c$ that occur in the polynomials $S_{n}$ are even.  Therefore,
these polynomials and their zeros are identical to those already considered.  
\end{proof}

\begin{theorem}
\label{T2}
If $n\geq 1$ and $r\in \mathcal{Z}_{n},$ then $(cr\pm \sqrt{c^{2}r^{2}+4})/2\in
\mathcal{Z}_{n+1}$.
\end{theorem}

\begin{proof}
The proposition clearly holds for $n=1.$  The zeros of $S_{2}$ are $-1$ and 
$1;$ let $r_{1}$ be either of these, and let $r_{2}$ be any number
satisfying $r_{1}=(1/c)(r_{2}-1/r_{2}),$ so that%
\begin{equation*}
r_{1}=\frac{r_{2}^{2}-1}{cr_{2}}=\frac{S_{2}(r_{2})}{T_{2}(r_{2})}.
\end{equation*}%
Taking $r_{1}=-1$ gives $S_{2}(r_{2})+T_{2}(r_{2})=0$ and taking $r_{1}=1$
gives $S_{2}(r_{2})-T_{2}(r_{2})=0.$  Consequently,%
\begin{equation*}
S_{3}(r_{2})=(S_{2}(r_{2})+T_{2}(r_{2}))(S_{2}(r_{2})-T_{2}(r_{2})),
\end{equation*}%
which implies that the $4$ numbers $r_{2}$ are the zeros of $S_{3}.$ \
Continuing with arbitrary zeros $r_{1},r_{2}$ of $S_{2},$ let $r_{3}$ be any
number satisfying $r_{2}=(1/c)(r_{3}-1/r_{3}),$ so that $%
r_{2}=S_{2}(r_{3})/T_{2}(r_{3}).$  Then%
\begin{eqnarray*}
r_{1} &=&\frac{1}{c}(r_{2}-\frac{1}{r_{2}})=\frac{1}{c}(\frac{S_{2}(r_{3})}{%
cr_{3}}-\frac{cr_{3}}{S_{2}(r_{3})}) \\
&=&\frac{S_{2}^{2}(r_{3})-c^{2}r_{3}^{2}}{cS_{2}(r_{3})T_{2}(r_{3})}=\frac{%
S_{3}(r_{3})}{T_{3}(r_{3})}.
\end{eqnarray*}%
Taking $r_{1}=-1$ gives $S_{3}(r_{3})+T_{3}(r_{3})=0$ and taking $r_{1}=1$
gives $S_{3}(r_{2})-T_{3}(r_{3})=0,$ so that the $8$ numbers $r_{3}$ are the
zeros of $S_{4}.$  This inductive procedure shows that the zeros $\rho $ of 
$S_{n+1}$ arise from the zeros $r$ of $S_{n}$ by the rule $r=(1/c)(\rho -1/\rho) $,
and solving this for $\rho $ finishes the proof.
\end{proof}

\begin{theorem}
\label{T3}
If $n\geq 1,$ then%
\begin{equation}
S_{n+1}(x)=(cx)^{2^{n-1}}S_{n}(\dfrac{x}{c}-\dfrac{1}{cx}).  \label{S1}
\end{equation}%
More generally, if $k=2,3,\ldots ,n+1,$ then%
\begin{equation}
S_{n+1}=T_{k}^{2^{n-k+1}}S_{n-k+2}(\frac{S_{k}}{T_{k}}).  \label{S2}
\end{equation}
\end{theorem}

\begin{proof}
First, we prove (\ref{S1}).  The assertion clearly holds for $n=1.$ \
Assume that $n>1$ and let $m=2^{n-1}.$  Denote the zeros of $S_{n}$ by $%
r_{k}$ for $k=1,2,\ldots ,m.$  By Theorem 2,%
\begin{eqnarray*}
S_{n+1}(x) &=&\overset{m}{\underset{h=1}{\dprod }}(x-\frac{cr_{h}-\sqrt{%
c^{2}r_{h}^{2}+4}}{2})(x-\frac{cr_{h}+\sqrt{c^{2}r_{h}^{2}+4}}{2}) \\
&=&\overset{m}{\underset{h=1}{\dprod }}(x^{2}-cr_{h}x-1) \\
&=&\overset{m}{\underset{h=1}{\dprod }}x(x-\frac{1}{x}-cr_{h}) \\
&=&(cx)^{m}S_{n}(\dfrac{x}{c}-\dfrac{1}{cx}).
\end{eqnarray*}%
In case $k=2,$ equation (\ref{S2}) is essentially a restatement of (\ref{S1}%
), just proved.  For $k>2,$ equations (\ref{S2}) will now be
proved by induction on $k.$  Assume for arbitrary $k$ satisfying $2\leq
k\leq n$ that%
\begin{equation}
S_{n+1}=T_{k-1}^{2^{n-k+2}}S_{n-k+3}(\frac{S_{k-1}}{T_{k-1}}).  \label{S3}
\end{equation}%
Substitute $S_{k}/T_{k}$ for $x$ in (\ref{S1}) with $n$ replaced by $n-k+3:$%
\begin{equation*}
S_{n-k+3}(\frac{S_{k-1}}{T_{k-1}})=(\frac{cS_{k-1}}{T_{k-1}}%
)^{2^{n-k+1}}S_{n-k+2}(\frac{S_{k}}{T_{k}}),
\end{equation*}%
so that by (\ref{S3}),%
\begin{eqnarray*}
S_{n+1} &=&T_{k-1}^{2^{n-k+2}}(\frac{cS_{k-1}}{T_{k-1}}%
)^{2^{n-k+1}}S_{n-k+2}(\frac{S_{k}}{T_{k}}) \\
&=&(cT_{k-1}S_{k-1})^{2^{n-k+1}}S_{n-k+2}(\frac{S_{k}}{T_{k}}) \\
&=&T_{k}{}^{2^{n-k+1}}S_{n-k+2}(\frac{S_{k}}{T_{k}}).
\end{eqnarray*}
\end{proof}

Using Theorems \ref{T1}-\ref{T3}, it is easy to establish the following
properties of the polynomials and zeros:\medskip

1.  $S_{n}(-x)=S_{n}(x)$ for $n\geq 2,$ and if $r\in \mathcal{Z}_{n}$ then $%
-r\in \mathcal{Z}_{n}$ for $n\geq 1.$

2.  $x^{2^{n-1}}S_{n}(1/x)=S_{n}(x)$ for $n\geq 3,$ and if $r\in \mathcal{Z}%
_{n}$ then $1/r\in \mathcal{Z}_{n}$ for $n\geq 2.$

3.  Suppose that $c>0,$ and let $r_{n}$ denote the greatest zero of $S_{n}.$
Then $(r_{n})$ is a strictly increasing sequence, and%
\begin{equation*}
\underset{n\rightarrow \infty }{\lim }r_{n}=\left\{ 
\begin{array}{cl}
(1-c)^{-1/2}, & \text{if }0<c<1; \\ 
\infty,  & \text{if }c\geq 1.%
\end{array}%
\right. 
\end{equation*}%
An outline of a proof follows.  Of course, $r_{n}=(cr_{n-1}+\sqrt{%
c^{2}r_{n-1}^{2}+4})/2,$ so that $(r_{n})$ is strictly increasing.  If $%
0<c<1,$ then, as is easily proved, $r_{n}<(1-c)^{-1/2}$ for all $n$, so that
a limit $r$ exists; since $r=(cr+\sqrt{c^{2}r^{2}+4})/2,$ we have $%
r=(1-c)^{-1/2}.$  On the other hand, supposing $c\geq 1,$ if $(r_{n})$ were
bounded above, then the previous argument would give $r=(1-c)^{-1/2},$ but
this is not a real number.  Therefore $r_{n}\rightarrow \infty .$

4.  If $n\geq 3,$ then%
\begin{equation}
S_{n}=S_{n-1}^{2}+c^{2}S_{n-1}S_{n-2}^{2}-c^{2}S_{n-2}^{4}.  \label{SS}
\end{equation}%
To prove this recurrence, from (\ref{SF}) we obtain both%
\begin{equation*}
c^{2n-4}(S_{1}S_{2}\cdots S_{n-3})^{2}S_{n-2}^{2}=S_{n-1}^{2}-S_{n}
\end{equation*}%
and%
\begin{equation*}
c^{2n-6}(S_{1}S_{2}\cdots S_{n-3})^{2}=S_{n-2}^{2}-S_{n-1},
\end{equation*}%
and (\ref{SS}) follows by eliminating $(S_{1}S_{2}\cdots S_{n-3})^{2}.$

In the case $c=1,$ the recurrence (\ref{SS}) is used (e.g., \cite{Mon}) to
define the Gor\v{s}kov-Wirsing polynomials, for which the initial
polynomials are $2x-1$ and $5x^{2}-5x+1$ rather than $x$ and $x^{2}-1.$

\section{The polynomials $V_{n}=S_{n}-c^{n-1}S_{1}S_{2}\cdots S_{n-1}$}

Equation (\ref{SF}) shows that $n\geq 2,$ the polynomial $S_{n+1}$ factors.\
We use one of factors to define a sequence of polynomials%
\begin{equation}
V_{n}=V_{n}(x)=S_{n}-c^{n-1}S_{1}S_{2}\cdots S_{n-1},  \label{V1}
\end{equation}%
so that%
\begin{eqnarray}
S_{n}(x) &=&V_{n-1}(x)V_{n-1}(-x),  \label{V2} \\
2S_{n}(x) &=&V_{n}(x)+V_{n}(-x).  \label{V3}
\end{eqnarray}%
Suppose that $n\geq 3.$  Substitute $n-1$ for $n$ in (\ref{V1}) and then
multiply both sides of the result by $cS_{n-1}$ to obtain%
\begin{equation*}
cV_{n-1}S_{n-1}=cS_{n-1}^{2}-c^{n-1}S_{1}S_{2}\cdots S_{n-1},
\end{equation*}%
so that by (\ref{V1}),%
\begin{equation*}
V_{n}=S_{n}-cS_{n-1}^{2}+cV_{n-1}S_{n-1},
\end{equation*}%
and by (\ref{V2}), and (\ref{V3}) with $n-1$ substituted for $n,$%
\begin{eqnarray*}
V_{n}(x) &=&V_{n-1}(x)V_{n-1}(-x)-c(\frac{V_{n-1}(x)+V_{n-1}(-x)}{2})^{2} \\
&&+cV_{n-1}(x)\frac{V_{n-1}(x)+V_{n-1}(-x)}{2}.
\end{eqnarray*}%
Thus, we have the following recurrence for the polynomials $V_{n}:$%
\begin{equation*}
V_{n}=V_{n-1}(x)V_{n-1}(-x)+\frac{c}{4}[V_{n-1}^{2}(x)-V_{n-1}^{2}(-x)],
\end{equation*}%
for $n\geq 3.$  Another recurrence for these polynomials stems directly
from (\ref{S1}) and (\ref{V2}):%
\begin{equation*}
V_{n+1}(x)=(cx)^{2^{n-1}}V_{n}(\frac{x}{c}-\frac{1}{cx}),
\end{equation*}%
for $n\geq 2.$

For the remainder of this section, assume that $c>0.$  For $n\geq 2,$ let $%
\mathcal{Z}_{n}^{+}$ denote the ordered list of zeros of $V_{n}$  Since $%
\mathcal{Z}_{n}^{+}\subset \mathcal{Z}_{n},$ we need only observe that in
the rule given by Theorem \ref{T2} for forming zeros, the half of the
numbers in $\mathcal{Z}_{n}$ that descend from $r=1$ in $\mathcal{Z}_{1}^{+}$
are the numbers that comprise $\mathcal{Z}_{n}^{+}.$  That is, if $n\geq 2$%
 and $r\in \mathcal{Z}_{n}^{+}$ then%
\begin{equation*}
(cr\pm \sqrt{c^{2}r^{2}+4})/2\in \mathcal{Z}_{n+1}^{+}.
\end{equation*}%
For $n\geq 3,$ no number in $\mathcal{Z}_{n}^{+}$ is rational, so that $V_{n}
$ is irreducible over the rational integers.

Let $r_{n}$ denote the greatest zero of $S_{n},$ and also of $V_{n}.$  Then%
\begin{equation*}
r_{n+1}=(cr_{n}+\sqrt{c^{2}r_{n}^{2}+4})/2,
\end{equation*}%
and from this recurrence easily follows%
\begin{equation}
r_{n+1}-1/r_{n+1}=cr_{n},  \label{V4}
\end{equation}%
of which the left-hand side is the distance from the least positive zero of $%
S_{n+1}$ to the greatest.

\section{The case $c=1$}

We turn now to the case that $c=1;$ that is, the immediate generalization of
(\ref{R-}) to the case that the initial value is $S_{1}(x)=x.$  The first
four polynomials $S_{n}$ and $V_{n}$ are as shown here:%
\begin{equation*}
\begin{tabular}{|l||l|l|}
\hline
$n$ & $S_{n}$ & $V_{n}$ \\ \hline\hline
$1$ & $x$ &  \\ \hline
$2$ & $x^{2}-1$ & $x-1$ \\ \hline
$3$ & $x^{4}-3x^{2}+1$ & $x^{2}-x-1$ \\ \hline
$4$ & $x^{8}-7x^{6}+13x^{4}-7x^{2}+1$ & $x^{4}-x^{3}-3x^{2}+x+1$ \\ \hline
\end{tabular}%
\end{equation*}%
Arrays of coefficients for $S_{n}$ are indexed \cite{Slo} as \seqnum{A147985}
and
\seqnum{A147990}, and for $T_{n}$ as \seqnum{A147986}.  The polynomials $V_{n}$ are related
by the equation $V_{n}(x)=U_{n}(-x)$ to polynomials $U_{n}$ presented at
\seqnum{A147989}.

As mentioned in Section 2, the greatest zero $r_{n}$ of $S_{n}$ grows
without bound as $n\rightarrow \infty .$  In order to discuss $r_{n}$ in
some detail, define%
\begin{equation*}
z(x)=(x+\sqrt{x^{2}+4})/2,
\end{equation*}%
so that $r_{1}=1$ and $r_{n}=z(r_{n-1})$ for $n\geq 2.$ The sequence $%
(r_{n})$ has some interesting properties arising from (\ref{V4}). For
example, if $x$ is the positive number satisfying $1+1/x=x,$ then $%
x=r_{2}=(1+\sqrt{5})/2,$ and inductively, if $x$ is the positive number
satisfying%
\begin{equation}
\frac{1}{r_{1}}+\frac{1}{r_{2}}+\cdots +\frac{1}{r_{n}}+\frac{1}{x}=x,
\label{C1}
\end{equation}%
then $x=r_{n+1}.$ Equation~(\ref{C1}) shows how the numbers $r_{n}$
arise naturally without reference to polynomials. Since%
\begin{equation*}
r_{n+1}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+\cdots +\frac{1}{r_{n}}+\frac{1}{%
r_{n+1}},
\end{equation*}%
we have $\underset{k=1}{\overset{\infty }{\sum }}\dfrac{1}{r_{k}}=\infty
.\medskip $

\begin{theorem}
\label{T4}
If $n\geq 1,$ and $r_{n}$ is the greatest zero of $S_{n},$ then%
\begin{equation*}
\sqrt{2n}-1<r_{n+1}<\sqrt{2n+1}.
\end{equation*}
\end{theorem}

\begin{proof}
Taking $c=1$ in (\ref{V4}) and squaring give%
\begin{equation*}
\frac{1}{r_{n+1}^{2}}=r_{n}^{2}-r_{n+1}^{2}+2,
\end{equation*}%
whence%
\begin{eqnarray*}
\underset{k=2}{\overset{n+1}{\sum }}\frac{1}{r_{k}^{2}}
&=&r_{1}^{2}-r_{n+1}^{2}+2n, \\
r_{n+1}^{2} &=&1+2n-\underset{k=2}{\overset{n+1}{\sum }}\frac{1}{r_{k}^{2}}
\\
&<&1+2n,
\end{eqnarray*}%
so that $r_{n+1}<\sqrt{2n+1}.$ 

We turn next to an inductive proof that
\begin{equation}
\sqrt{2n}-1<r_{n+1} \label{HH1}
\end{equation}% 
for all $n$. This is true for $n=1,$ and we assume it true for arbitrary $n$ and 
wish to prove that $r_{n+2}>\sqrt{2n+2}-1.$ We begin with the easily proved 
inequality
\begin{equation*}
4n^2+2n+1+4n\sqrt{2n}+4n+2\sqrt{2n}<(2n+2\sqrt{2n}+1)(2n+2).
\end{equation*}%
Taking the square root of both sides,
\begin{equation*}
2n+\sqrt{2n}+1<(\sqrt{2n}+1)(\sqrt{2n+2}),
\end{equation*}%
so that
\begin{equation*}
2n<(\sqrt{2n}+1)(\sqrt{2n+2}-1).
\end{equation*}%
Expanding and adding appropriate
terms to both sides,
\begin{equation*}
2n-2\sqrt{2n+2}+5>4(2n+2)+2n+1-4\sqrt{2n}\sqrt{2n+2}-4\sqrt{2n+2}+2\sqrt{2n}.
\end{equation*}%
Taking the square root of both sides,
\begin{equation*}
\sqrt{2n-2\sqrt{2n}+5}>2\sqrt{2n+2}-\sqrt{2n}-1.
\end{equation*}%
Equivalently,
\begin{equation*}   
\sqrt{2n}-1+\sqrt{2n-2\sqrt{2n}+5}>2\sqrt{2n+2}-2,
\end{equation*}%
so that by the induction hypothesis (\ref{HH1}), 
\begin{equation}
r_{n+1}+\sqrt{2n-2\sqrt{2n}+5}>2\sqrt{2n+2}-2.\label{HH2}
\end{equation}%
The inequality (\ref{HH1}), after squaring, adding $4,$ and taking square roots, gives
\begin{equation*}
\sqrt{r_{n+1}^2+4}>\sqrt{2n-2\sqrt{2n}+5.}
\end{equation*}%
In view of (\ref{HH2}), therefore,  \begin{equation*}
r_{n+2}=\frac{r_{n+1}+\sqrt{r_{n+1}^2+4}}{2}>\sqrt{2n+2}-1.
\end{equation*}%
\end{proof}

Theorem \ref{T4} implies that $\underset{n\rightarrow \infty }{\lim }r_{n}/%
\sqrt{n}=\sqrt{2}$ and that%
\begin{equation*}
\frac{1}{2n+1}<\frac{1}{r_{n+1}^{2}}<\frac{1}{(\sqrt{2n}-1)^{2}}.
\end{equation*}%
Consequently,%
\begin{equation*}
\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{r_{n}^{2}}=\infty ,\text{ \
\ }\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{nr_{n}}<\infty ,\text{ \
\ and \ \ }\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{2^{n-1}r_{n}}%
<\infty .
\end{equation*}%
The second and third sums are approximately $2.26383447$ and $1.518737247.$
For more digits of the latter, see \seqnum{A154310}.

\section{The case $c=2$}

It is easy to prove that there is exactly one choice of $c>0$ in (\ref{ST})
for which the resulting polynomial $S_{n}(x)+iT_{n}(x)$ has the form%
\begin{equation*}
(x+a+bi)^{2^{n}}
\end{equation*}%
for some real $a$ and $b$ and all $n\geq 2.$ The unique values are $c=2$
and $(a,b)=(0,1).$  In this case, the first three polynomials $S_{n},$ $%
T_{n},$ $V_{n}$ are as shown here:

\begin{equation*}
\begin{tabular}{|l||l|l|l|}
\hline
$n$ & $S_{n}$ & $T_{n}$ & $V_{n}$ \\ \hline\hline
$1$ & $x$ & $1$ &  \\ \hline
$2$ & $x^{2}-1$ & $2x$ & $x^{2}-2x-1$ \\ \hline
$3$ & $x^{4}-6x^{2}+1$ & $4x^{3}-4x$ & $x^{4}-4x^{3}-6x^{2}+4x+1$ \\ \hline
\end{tabular}%
\end{equation*}%
Arrays of coefficients for $S_{n}$ and $T_{n}$ are included as subarrays 
\cite{Slo} of arrays closely associated with Pascal's triangle. \
Specifically, for $S_{n}$ see \seqnum{A096754}, \seqnum{A135670},
and \seqnum{A141665}; for $T_{n},$
see \seqnum{A095704} and \seqnum{A135685}.  In the same way, coefficients for $V_{n}$ can be
read from A108086, modified in accord with the identity $V_{n}(x)=U_{n}(-x)$.

The fact that the zeros of $T_{n}$ interlace those of $S_{n}$ is an example
of Theorem 1.  However, in this case, one can also appeal to the
Hermite-Biehler theorem:  \textit{if }$S$\textit{\ and }$T$\textit{\ are
nonconstant polynomials with real coefficients, then the polynomials }$S$%
\textit{\ and }$T$\textit{\ have interlacing zeros if and only if all the
zeros of the polynomial }$S+iT$\textit{\ lie either in the upper half-plane
or the lower half-plane}.  For a discussion of this theorem and related
matters, see Rahman and Schmeisser \cite[pp.\ 196--209]{Rah}.

\section{The case $c=2i$}

Suppose that $S_{n}$ is a square for some $n,$ and write $%
S_{n}(x)=H_{n}^{2}(x).$  Then%
\begin{equation*}
S_{n}(\dfrac{x}{c}-\dfrac{1}{cx})=H_{n}^{2}(\dfrac{x}{c}-\dfrac{1}{cx}).
\end{equation*}%
By (\ref{S1}),%
\begin{equation}
S_{n+1}(x)=(cx)^{2^{n-1}}H_{n}^{2}(\dfrac{x}{c}-\dfrac{1}{cx}),  \label{H}
\end{equation}%
which implies that $S_{n+1}$ is a square.  It is easy to show that the only
nonzero choices of $c$ for which $S_{3}$ is a square are $\pm 2i$. \
Equation (\ref{H}) gives the recurrence%
\begin{equation*}
H_{n+1}(x)=(2ix)^{2^{n-2}}H_{n}(\dfrac{i}{2x}-\dfrac{ix}{2}),
\end{equation*}%
which implies%
\begin{equation*}
\left\vert H_{n+1}(e^{i\theta })\right\vert =2^{\deg H_{n}}\left\vert
H_{n}(\sin \theta )\right\vert
\end{equation*}%
for all real $\theta .$  Another recurrence, which follows readily from (%
\ref{SS}), is%
\begin{equation*}
H_{n}=2H_{n-2}^{4}-H_{n-1}^{2}.
\end{equation*}%
The first four of these polynomials are as follows:\medskip

$H_{3}(x)=x^{2}+1$

$H_{4}(x)=x^{4}-6x^{2}+1$

$H_{5}(x)=x^{8}+20x^{6}-26x^{4}+20x^{2}+1$

$H_{6}(x)=x^{16}-88x^{14}+92x^{12}-872x^{10}+1990x^{8}-\allowbreak
872x^{6}+92x^{4}-88x^{2}+1$\medskip

Coefficients for the polynomials $H_{7}$ and $H_{8}$ are given at 
\seqnum{A154308}.

\section{The recurrence $P_{n+1}/Q_{n+1}=(1/c)(P_{n}/Q_{n}+Q_{n}/P_{n})$}

We return now to the recurrence (\ref{R+}), with initial value $%
x_{0}=P_{1}=P_{1}(x)=x.$  Taking $Q_{1}=Q_{1}(x)=1$ leads to sequences $%
P_{n}$ and $Q_{n}$ defined by

\begin{equation*}
P_{n}=P_{n-1}^{2}+cQ_{n-1}^{2}\text{ \ \ and \ \ }Q_{n}=cP_{n-1}Q_{n-1}.
\end{equation*}%
The properties of these polynomials are analogous to those of the
polynomials $S_{n}$ and $T_{n}$ already discussed.  Indeed,%
\begin{equation}
P_{n}(x)=S_{n}(ix)  \label{PS}
\end{equation}
for $n\geq 2,$ so that the zeros of $P_{n}$ are $ir,$ where $r$ ranges
through $\mathcal{Z}_{n},$ and, if $c>0,$ we have interlaced lists of zeros
on the imaginary axis.  The recurrence (\ref{SS}) holds without change;
that is, for $n\geq 3,$ we have%
\begin{equation*}
P_{n}=P_{n-1}^{2}+P_{n-1}P_{n-2}^{2}-P_{n-2}^{4}.
\end{equation*}

Putting $x=ie^{i\theta }$ in (\ref{S1}) and applying (\ref{PS}) lead to%
\begin{equation*}
\left\vert P_{n+1}(e^{i\theta })\right\vert =\left\vert c\right\vert ^{\deg
P_{n}}\left\vert P_{n}(\frac{2}{c}\cos \theta )\right\vert
\end{equation*}%
for all real $\theta .$

For $c=1,$ coefficient arrays are given for the polynomials $P_{n}$ and $%
Q_{n}$ are indexed as \seqnum{A147987} and \seqnum{A147988}, respectively.

\section{Concluding remarks}

The author is grateful to a referee for pointing out various properties
associated with polynomials discussed in this paper --- properties which may
be worth further study.  For example, the interlacing of zeros in Theorem %
\ref{T1} implies that for fixed $n,$ the polynomials $T_{n}$ and $S_{n}$ are
consecutive members of some sequence of orthogonal polynomials.  A
consequence of Theorem \ref{T2} is that there exists Euclidean
straightedge-and-compass constructions for the zeros of $T_{n}$ and $S_{n}.$
The manner in which $S_{n+1}$ arises from the argument of $S_{n}$ in (\ref%
{S1}) is similar to the Joukowski transform. Indeed (\ref{S1}) can be
written as%
\begin{equation*}
e^{-i2^{n-1}\theta }S_{n+1}(e^{i\theta })=c^{2^{n-1}}S_{n}(\frac{2i}{c}\sin
\theta ),
\end{equation*}%
so that, apart from a constant, the modulus of $S_{n+1}$ on the unit circle
is the modulus of $S_{n}$ on a line segment. This has consequences for
estimates, such as the derivative estimates of the Bernstein-Markov type.

\begin{thebibliography}{9}
\bibitem{Mon} H. L. Montgomery, \textit{Ten Lectures on the Interface
between Analytic Number Theory and Harmonic Analysis}, CBMS Regional
Conference Series in Mathematics, 84. Published for the American
Mathematical Society, 1994, page 187.

\bibitem{New} D. J. Newman, \textit{A Problem Seminar}. \ Problem Books in
Mathematics, Springer-Verlag, 1982. \ Problem 60.

\bibitem{Rah} Q. Rahman and G. Schmeisser, \textit{Analytic Theory of
Polynomials,} Oxford University Press, 2002.

\bibitem{Slo} N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences, \\
\href{http://www.research.att.com/~njas/sequences/}{\tt http://www.research.att.com/\char'176njas/sequences/}.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 26C10; Secondary 26C15.

\noindent \emph{Keywords: } polynomial, interlacing zeros.

\bigskip
\hrule
\bigskip

\noindent 
(Concerned with sequences 
\seqnum{A073833},
\seqnum{A095704},
\seqnum{A096754},
\seqnum{A108086},
\seqnum{A127814},
\seqnum{A135670},
\seqnum{A135685},
\seqnum{A141665},
\seqnum{A147985},
\seqnum{A147986},
\seqnum{A147987},
\seqnum{A147988},
\seqnum{A154308},
\seqnum{A154310}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 11, 2009;
revised version received  March 18 2009.
Published in {\it Journal of Integer Sequences}, March 20 2009.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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