% revised according to Mike's comments at 11/10/08 - file name: exp2007m4.tex
%improved the graphs according to Editor's suggestions, 10/24/08 by Wang
%revised the paper according to Editor's suggestions, 10/23/08 by Ding
%received editor Henle's email suggestions on 10/23/08
%received new editor Michael Henle's email on 8/25/08 telling me that the paper'%s new number is CMJ MS #08-194 and leting me email him pdf file of the paper
%withut author names; I did it on 8/25/08.
%received revision instruction by editor Beineke in 5/08. Revised on 6/2/08,
%6/5/08, 6/9/08.
%resubmitted to CMJ on 7/27/07, 6-132
%Revised on 6/30/07,7/19/07,7/26/07
%Received two reports on 6/18/07 and editor let me revise it.
%post-submitted to CMJ's editor Lowell Beineke, 11/6/06, 6-132
%rejected by ijmest on 3/28/06
%e-submitted to ijmest on 2/16/06, m.c.harrison@lboro.ac.uk
%rejected by Monthly on 2/16/06
%resubmitted to the new editor Dan Velleman of Monthly on 1/26/06:
%mathmonthly@amherst.edu
%rejected in 11/05, but let me resubmit to the new editor
%submitted to Monthly electronically, monthly@math.utexas.edu, 11/1/2005,
%REF:  MS #05-745
%first draft, 10/2004; revised in 10/05


%\documentstyle[12pt]{article}
\documentclass[12pt]{article}
\renewcommand{\baselinestretch}{1.5}

\usepackage{graphicx,amssymb,amsfonts,amsmath,amsthm,eucal}

\begin{document}
\title{Dynamics of Exponential Functions}
\author{Jiu Ding \\ Department of
Mathematics \\ The University of Southern Mississippi \\
Hattiesburg, MS 39406-5045, USA \\ Zizhong Wang \\ Department of
Mathematics and Computer Science \\ Virginia Wesleyan College \\ Norfolk,
VA 23502, USA}
\maketitle
\newtheorem{lemma}{Lemma}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{proposition}{Proposition}[section]

Exponential functions are familiar from algebra and calculus, but
if we repeatedly iterate a given exponential function starting from
any initial point, what is the {\em eventual behavior}
of the iterates?
%their dynamical properties may not be well-known to us.
In this paper we answer this question by applying calculus
to study iterated exponential functions.
This topic has a rich history; see, for example, Knoebel \cite{kn} and the
references therein. In a more recent paper by Anderson \cite{an}, the
convergence problem for the sequence of {\em iterated
exponentials}
\begin{eqnarray}
\label{exp} a, a^a, a^{(a^a)}, a^{\left(a^{(a^a)}\right)}, \ldots
\end{eqnarray}
and the related problem of solving the equation $x^y = y^x$ were
studied along with some historical accounts. The approach there
was based on calculus applied to the function $f(x) =
x^{\frac{1}{x}}$, but did not take the modern viewpoint of
dynamical systems except for some concluding remarks. Here we
introduce a few basic concepts and techniques from one-dimensional
{\em discrete dynamical systems} while studying the asymptotic
behavior of the sequence of iterates of the one-parameter family
of exponential functions $f(x) = a^x$. Throughout, we assume $a
> 0$ and $a \neq 1$. As a by-product we obtain the results on
the convergence of the sequence (\ref{exp}) of iterated
exponentials previously obtained in \cite{an}.

%We shall use the {\em monotone convergence theorem}, which says
%that a monotone and bounded sequence of real numbers converges,
%and the {\em intermediate value theorem}, which claims that a real
%continuous function $f$ on $[c, d]$ has a zero in $(c, d)$ if
%$f(c)f(d) < 0$.

Let $f$ be a function from an interval to itself. A point $x$ in
the domain of $f$ is called a {\em fixed point} of $f$ if $f(x) =
x$. $x$ is called a {\em periodic point} of period-$k$ if $f^k(x)
= x$ and $x, f(x), \ldots, f^{k-1}(x)$ are distinct. In this case,
$(x, f(x), \ldots, f^{k-1}(x))$ is called a {\em period-$k$
orbit} of $f$. A fixed point $x$ of $f$ is {\em attracting} if the
sequence $f^n(x_0)$ of iterates converges to $x$ for $x_0$
sufficiently close to $x$ in the domain of $f$, and is {\em
repelling} if no matter how close $x_0$ is to $x$, there is an
iterate $f^n(x_0)$ of $x_0$, which is farther away from $x$ than
$x_0$. (See [2] for a general introduction to discrete dynamics.)

Suppose $f$ is differentiable on its domain. Then a fixed point
$x$ of $f$ is attracting if $|f'(x)| < 1$ and repelling if
$|f'(x)| > 1$. Such properties are easy consequences of the Mean
Value Theorem. Similarly, a period-$k$ orbit $\{x, f(x), \ldots,
f^{k-1}(x)\}$ is {\em attracting} or {\em repelling} according as
$x$ is an attracting or repelling fixed point of $f^k$. While the
derivative gives information about local properties of the
iterates, we are interested in the global picture. In the
following we divide our discussion of iterating exponential
functions $f(x) = a^x$ into the two cases of $a > 1$ and $0 < a <
1$.

%We also need the following simple and useful fact: Let $J$ be an
%interval, let $f: J \rightarrow J$ be a continuous function, and
%let $x_n = f(x_{n-1}) = f(f(x_{n-2})) = \cdots = f^n(x_0), n = 1,
%$2, \ldots$ be the sequence of the iterates of an {\em initial
%point} $x_0 \in J$ (also called the {\em orbit} of $x_0$) under
%$f$. If $\lim_{n \rightarrow \infty} x_n = x^* \in J$, then
%$f(x^*) = x^*$.
%It is well known that the graph of any exponential function $f(x)
%= a^x$ is {\em concave upward}, and $f$ is strictly increasing if
%$a > 1$ and strictly decreasing when $0 < a < 1$.

\section{Case 1: $a > 1$}

We first ask for what value
of the base $a$, does the function $f$ have a fixed point at which the tangent
line to the graph of $f$ has slope $1$. If $f$ has a fixed point $b$
such that $f'(b) = 1$, that is, if $a^b = b$ and $a^b \ln a = 1$,
then $b \ln a = \ln b$ and $b \ln a = 1$. Thus, $b = e$ and $a =
e^{\frac{1}{e}}$. This means that the graph of $f$ is tangent to
the line $y = x$ at a fixed point of $f$ if and only if $a =
e^{\frac{1}{e}}$. In this case, $e$ is the unique fixed point of
$f(x) = e^{\frac{x}{e}}$. Now we consider three subcases:

\subsection{Case 1(a): $1 < a < e^{\frac{1}{e}}$}

Figure 1 illustrates that $f$ has exactly two fixed points $x^* = x^*(a)
\in (1, e)$ and $y^* = y^*(a) \in (e, \infty)$, which can be proved by
applying the Intermediate Value Theorem to the function $f(x) - x$.

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.8]{f1.jpg}
\caption{The graph of $f(x)=a^x$ with $a=1.2$.} \label{fig1}
\end{center}
\end{figure}

Figure 1 also indicates that the graph of $f$ lies above the line
$y = x$ over $(- \infty, x^*]$ and $[y^*, \infty)$, and below that
line over $[x^*, y^*]$. If $x_0 \in (- \infty, x^*)$, then $x_0 <
f(x_0) < x^*$, and by induction $f^{n-1}(x_0) < f^n(x_0) < x^*$
for all $n$. Thus the Monotone Convergence Theorem ensures that
$f^n(x_0) \rightarrow l$ for some number $l \le x^*$. Since $l
\leftarrow f^{n+1}(x_0) = f(f^n(x_0)) \rightarrow f(l)$, we see
that $l$ is a fixed point of $f$ and so $l = x^*$. By the same
token, if $x_0 \in (x^*, y^*)$, then $f^{n-1}(x_0)
> f^n(x_0) > x^*$ for any $n$, so $f^n(x_0) \rightarrow x^*$.
Finally, the sequence $f^n(x_0) \rightarrow \infty$ if $x_0 >
y^*$, due to the fact that $f^n(x_0)$ is a monotonically
increasing sequence and there is no fixed point of $f$ in $(y^*,
\infty)$.
%Such conclusions are obviously seen by the graphic
%analysis.

Since $a^{x^*} = x^* > 1$, we have $a = (x^*)^{\frac{1}{x^*}} <
x^*$. Thus the sequence (\ref{exp}) is strictly increasing and
converges to $x^*$ when $1 < a < e^{\frac{1}{e}}$, as proved in
\cite{an}.

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.8]{f2.jpg}
\caption{The graph of $f(x)=a^x$ with $a=e^{\frac{1}{e}}$.} \label{fig2}
\end{center}
\end{figure}


\subsection{Case 1(b): $a = e^{\frac{1}{e}}$}

Since the graph of $f(x) = e^{\frac{x}{e}}$ lies above its tangent line
$y = x$ at the unique fixed point $e$ of $f$ (see Figure 2), $x < f(x) < e$
for $x < e$ and $e < x < f(x)$ for $x > e$. By the same argument as above,
if $x_0 < e$, then $f^n(x_0) \rightarrow e$, and $f^n(x_0) \rightarrow
\infty$ if $x_0 > e$. In particular, the sequence (\ref{exp}) is
monotonically increasing and converges to $e$ when $a = e^{\frac{1}{e}}$.


\subsection{Case 1(c): $a > e^{\frac{1}{e}}$}

If $a^x = x$ for some $x > 0$, then $x^{\frac{1}{x}} = a > e^{\frac{1}{e}}$.
This contradicts the fact, as shown in \cite{an}, that $t^{\frac{1}{t}}
\le e^{\frac{1}{e}}$ for all $t > 0$. Thus $f$ has no fixed point.
Since $x < f(x)$ for all real numbers $x$, the sequence $f^n(x_0)$
is monotonically increasing and diverges to infinity for any
initial point $x_0$. Consequently, the sequence (\ref{exp}) is
monotonically increasing and diverges to infinity when $a >
e^{\frac{1}{e}}$.

\subsection{Summary}

Let $x^*$ be a fixed point of a function $f$. The set of all
initial points $x_0$ for which $f^n(x_0) \rightarrow x^*$ is
called the {\em basin of attraction} for $x^*$. The basin of
attraction for a periodic orbit is defined similarly. Our analysis
shows that, if $1 < a < e^{\frac{1}{e}}$, then the basin of
attraction for the fixed point $x^* \in (1, e)$ is $(-\infty,
y^*)$, and that for the fixed point $y^* \in (e, \infty)$, it is
$\{y^*\}$; and when $a = e^{\frac{1}{e}}$, the basin of attraction
for the fixed point $x^* = e$ is $(-\infty, e]$.

The number $a^* = e^{\frac{1}{e}}$ is called the {\em bifurcation
point} for the parameter $a > 1$ in the family of exponential
functions $f(x) = a^x$, because both the number of fixed points of $f$ 
and their nature change as $a$ passes through $a^*$. Since a
pair of fixed points are born as the graph of $f$ becomes tangent
to and then crosses the line $y = x$ when $a$ decreases through
$a^*$, this kind of bifurcation is called a {\em tangent
bifurcation} (see Figure 3 below, where solid line and dashed line
represent the attracting and repelling fixed point respectively).

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.7]{f3.jpg}
\caption{The tangent bifurcation for $f(x)=a^x$ with $a > 1$.}
\label{fig3}
\end{center}
\end{figure}

\section{Case 2: $0 < a < 1$}

In this case, $f$ has a unique fixed point $x^* = x^*(a) \in (0, 1)$.
Actually $x^* \in (a, 1)$ since $a < a^{x^*} = x^* < 1$. Differentiating the
identity $x^*(a) \equiv a^{x^*(a)}$ with respect to $a$ gives
\[ \frac{dx^*}{da} = \frac{(x^*)^2}{a(1 - \ln x^*)}, \]
so the fixed point $x^*$, as a function of $a$, is strictly increasing
on $(0, 1)$. Furthermore, since $(e^{-e})^{e^{-1}} = e^{-1}$,
$x^*(e^{-e}) = e^{-1}$.

Suppose $b$ is a fixed point of $f$ at which the tangent line to
the graph of $f$ has slope $-1$. Then $a^b = b$ and $a^b \ln a = -
1$, so $b \ln a = \ln b$ and $b \ln a = -1$. Thus, $b = e^{-1}$
and $a = e^{-e}$. This means that the graph of $f$ is
perpendicular to the line $y = x$ at a fixed point of $f$ if and
only if $a = e^{-e}$. If $e^{-e} < a < 1$, then $x^*(a) > x^*(e^{-e})
= e^{-1}$, which implies that $f'(x^*) = \ln x^* > \ln e^{-1} = -1$.
Similarly, if $0 < a < e^{-e}$, then $f'(x^*) < -1$.

In order to use the Monotone Convergence Theorem again, as we did
for $a > 1$, we form a strictly increasing function $g(x) \equiv
f^2(x) = a^{\left( a^x \right)}$ for $0 < a < 1$; see Figure
\ref{fig5} with $a = 0.1$. Since $f$ maps $[0, 1]$ into itself, so
does $g$. Thus, we restrict $x \in [0, 1]$ in the following
analysis.

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.8]{f4.jpg}
\caption{The graph of $g(x)=a^{\left( a^x \right) }$ with
$a=0.1$.} \label{fig5}
\end{center}
\end{figure}

Let $h(x) = g(x) - x$. Then $h'(x) = (\ln a)^2 a^x a^{\left( a^x
\right)} - 1$ and
\begin{eqnarray}
\label{h''}
h''(x) = (\ln a)^3 (1 + a^x \ln a) a^x a^{ \left(a^x \right)}.
\end{eqnarray}
Since $(\ln t)^2 t \le 4 e^{-2}$ on $[0, 1]$,
\begin{eqnarray}
\label{h'0}
h'(0) = (\ln a)^2 a - 1 \le \frac{4}{e^2} - 1 < 0
\end{eqnarray}
and
\begin{eqnarray}
\label{h'1}
h'(1) = (\ln a)^2 a a^a - 1 \le \frac{4}{e^2} a^a - 1 < \frac{4}{e^2} - 1 < 0.
\end{eqnarray}

If $e^{-1} \le a < 1$, then $1 + a^x \ln a \ge 1 - a^x
> 0$ for all $x \in (0, 1)$. So $h''(x) < 0$ on $(0, 1)$ from
(\ref{h''}), which implies that $h'(x)$ is strictly decreasing and
is always negative by (\ref{h'0}) and (\ref{h'1}). On the other
hand, if $0 < a < e^{-1}$, then $1 + \ln a < 0$. Since $\min_{t
\in (0, 1]} t \ln t = - e^{-1}$, it follows that $1 + a \ln a \ge
1 - e^{-1} > 0$. Thus, by the Intermediate Value Theorem, there is
a unique $c \in (0, 1)$ for which
\begin{eqnarray}
\label{c} a^c = - \frac{1}{\ln a}.
\end{eqnarray}
By (\ref{h''}), $h''(c) = 0$. It follows that if $0 < a < e^{-1}$,
then $h'(x)$ is strictly increasing on $[0, c]$ and strictly
decreasing on $[c, 1]$, and so
\begin{eqnarray}
\label{max}
h'(c) = \max \{h'(x) : x \in [0, 1]\}.
\end{eqnarray}

With this preliminary analysis, we can study the dynamics of $f(x)
= a^x$ with $0 < a < 1$. Again we consider three subcases.

\subsection{Case 2(a): $e^{-e} < a < 1$}

If $e^{-1} \le a < 1$, then $h(x)$ is strictly decreasing from
$h(0) = a > 0$ to $h(1) = a^a - 1 < 0$ on $[0, 1]$. Since $h(x^*) = 0$
and $x^* \in (0, 1)$, it follows that $g(x) > x$ on $[0, x^*)$ and
$g(x) < x$ on $(x^*, 1]$. If $e^{-e} < a < e^{-1}$, then $1 < -\ln a < e$.
Using (\ref{c}), we find that
\[ h'(c) = - \frac{\ln a}{e} - 1 < 0, \]
which ensures, by (\ref{max}), that $h'(x) < 0$ on $[0, 1]$.
Therefore $h(x)$ is strictly decreasing on $[0, 1]$. Hence, $g(x)
> x$ on $[0, x^*)$ and $g(x) < x$ on $(x^*, 1]$.

In summary, for $e^{-e} < a < 1$, $x < f^2(x) < x(a)$ if $0 \le x
< x^*$ and $x > f^2(x) > x(a)$ if $x^* < x \le 1$. It follows by
induction that for $0 \le x_0 < x^*, \; x_0 < f^2(x_0) < \cdots <
x^*$ and for $x^* < x_0 \le 1, \; x_0 > f^2(x_0) > \cdots
> x^*$. The Monotone Convergence Theorem implies that
$f^{2k}(x_0) \rightarrow x^*$ for $x_0 \in [0, 1]$, because
$x^*$ is the only fixed point of $f^2$. Since $f^{2k+1}(x_0) =
f(f^{2k}(x_0)) \rightarrow f(x^*) = x^*, \; f^n(x_0) \rightarrow
x^*$ for all $x_0 \in [0, 1]$.

On the other hand, if $x_0 > 1$, then $x_1 = f(x_0) \in (0, 1)$,
and if $x_0 < 0$, then $x_1 = f(x_0) > 1$ and so $x_2 = f(x_1) \in
(0, 1)$. Therefore,
\[ f^n(x_0) \rightarrow x^* \]
for all initial points $x_0$. As a corollary, it follows that the
sequence (\ref{exp}) converges to $x^*$ for $e^{-e} < a < 1$.

\subsection{Case 2(b): $a = e^{-e}$}

When $a=e^{-e}$, $c = e^{-1}$ is the solution of (\ref{c}).
Straightforward computation shows that
\[ h'(e^{-1}) = 0, \]
so (\ref{max}) implies that $h'(x) < 0$ on $[0, e^{-1}) \cup
(e^{-1}, 1]$. Therefore $h$ is strictly decreasing on $[0, 1]$,
which guarantees that $x_0 < f^2(x_0) < e^{-1}$ for $x_0 \in [0,
e^{-1})$ and $x_0 > f^2(x_0) > e^{-1}$ for $x_0 \in (e^{-1}, 1]$.
By induction, the sequence $f^{2k}(x_0)$ is strictly increasing or
decreasing, according as $x_0 \in [0, e^{-1})$ or $x_0 \in
(e^{-1}, 1]$. Thus, $f^{2k}(x_0) \rightarrow e^{-1}$. Since
$f^{2k+1}(x_0) \rightarrow f(e^{-1}) = e^{-1}$, we have $f^n(x_0)
\rightarrow e^{-1}$ for all $x_0 \in [0, 1]$. The same argument as
in the last part of Case 2(a) shows that for all $x_0$,
\[ f^n(x_0) \rightarrow e^{-1}. \]
As a special case, the sequence (\ref{exp}) converges to $e^{-1}$
when $a = e^{-e}$.

\subsection{Case 2(c): $0 < a < e^{-e}$}

Then $h'(x^*) = g'(x^*) - 1 = [f'(x^*)]^2 - 1 > 0$
since $f'(x^*) < -1$. Since $h(x^*) = 0, \; h(x) < 0$
for $x < x^*$ and near $x^*$, and $h(x) > 0$ for
$x > x^*$ and near $x^*$. Thus, there are $u \in (0, x^*)$ and
$v \in (x^*, 1)$ such that $h(u) < 0$ and $h(v) > 0$. Since $h(0)
> 0$ and $h(1) < 0$, by the Intermediate Value Theorem, there are
$\tilde{x} = \tilde{x}(a) \in (0, u)$ and $\tilde{y} = \tilde{y}(a)
\in (v, 1)$ such that $h(\tilde{x}) = 0$ and $h(\tilde{y}) = 0$.
Since $x^*$ is the only fixed point of $f$, $\{\tilde{x}, \tilde{y}\}$
is a period-$2$ orbit for $f$, that is, $a^{\tilde{x}} = \tilde{y}$ and
$a^{\tilde{y}} = \tilde{x}$.

Now we have $g(x) > x$ on $[0, \tilde{x}), \; g(x) < x$ on
$(\tilde{x}, x^*), \; g(x) > x$ on $(x^*, \tilde{y})$, and $g(x)
< x$ on $(\tilde{y}, 1]$. The same analysis as in Case 2(a) gives
\[ x_0 < f^2(x_0) < \cdots < f^{2k}(x_0) < \cdots < \tilde{x},
\; \; \forall \; x_0 \in [0, \tilde{x}), \]
\[ x_0 > f^2(x_0) > \cdots > f^{2k}(x_0) > \cdots > \tilde{x},
\; \; \forall \; x_0 \in (\tilde{x}, x^*), \]
\[ x_0 < f^2(x_0) < \cdots < f^{2k}(x_0) < \cdots < \tilde{y},
\; \; \forall \; x_0 \in (x^*, \tilde{y}), \; \; \mbox{and} \]
\[ x_0 > f^2(x_0) > \cdots > f^{2k}(x_0) > \cdots > \tilde{y},
\; \; \forall \; x_0 \in (\tilde{y}, 1]. \] It follows that for
$x_0 \in [0, x^*)$,
\begin{eqnarray}
\label{x*} f^{2k}(x_0) \rightarrow \tilde{x}
\end{eqnarray}
and for $x_0 \in (x^*, 1]$,
\begin{eqnarray}
\label{y*} f^{2k}(x_0) \rightarrow \tilde{y}.
\end{eqnarray}

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.8]{f5copy.jpg}
\caption{Graphs of $f(x)=a^x$ and $g(x)=a^{\left( a^x \right) }$
with $a=0.03$.} \label{fig6}
\end{center}
\end{figure}

Since $x^* < f(x_0) \le 1$ if $0 \le x_0 < x^*$ and $0 \le
f(x_0) < x^*$ if $x^* < x_0 \le 1$, we immediately see that for
$x_0 \in [0, x^*)$,
\begin{eqnarray}
\label{x*y*} f^{2k+1}(x_0) \rightarrow \tilde{y}
\end{eqnarray}
and for $x_0 \in (x^*, 1]$,
\begin{eqnarray}
\label{y*x*} f^{2k+1}(x_0) \rightarrow \tilde{x}.
\end{eqnarray}

It follows from (\ref{x*})--(\ref{y*x*}) that for all $x_0 \in
[0, 1]$, the sequence $f^n(x_0)$ approaches the period-$2$ orbit
$\{\tilde{x}, \tilde{y}\}$. Since $x_0 > 1$ implies $0 < f(x_0) <
1$ and $x_0 < 0$ implies $f(x_0) > 1$, the same conclusion is true
for all real numbers $x_0$. See Figure \ref{fig6} for an
illustration. In particular, since $a \neq x^*(a)$, the sequence
(\ref{exp}) approaches the period-$2$ orbit $\{\tilde{x},
\tilde{y}\}$ for $0 < a < e^{-e}$.

\bigskip

In summary, if $e^{-e} \le a < 1$, then the basin of attraction
for the fixed point $x^*$ is $(-\infty, \infty)$, and if $0 < a <
e^{-e}$, the basin of attraction for the period-$2$ orbit
$\{\tilde{x}, \tilde{y}\}$ is $(-\infty, \infty) \setminus \{x^*\}$,
and the basin of attraction for the fixed point $x^*$
is $\{x^*\}$.

\begin{figure}[ht]
\begin{center}
\par
\includegraphics[scale=0.7]{f6.jpg}
\caption{The pitchfork bifurcation for $f(x)=a^x$ with $0 < a <
1$.} \label{fig7}
\end{center}
\end{figure}

It turns out that the number $a^{**} = e^{-e}$ is a bifurcation
point for the parameter $a$ in the family of exponential functions
$f(x) = a^x$ with $0 < a < 1$ since a periodic point appears as
$a$ decreases through $a^{**}$. Since an attracting fixed point
becomes repelling and an attracting period-$2$ orbit is born when
$a$ passes through $a^{**}$, this kind of bifurcation is often
called a {\em pitchfork bifurcation} as the shape of Figure
\ref{fig7} suggests.

\section{Global Summary}

Table 1 summarizes the dynamical
properties of exponential functions. As we have seen, this
family of elementary functions provides a good example of the application
of the calculus to discrete dynamical systems.

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|} \hline \hline
Value of $a$         &                       & Basin of Attraction  \\ \hline
$0 < a < e^{-e}$     & $x^*(a)$, fixed point & $\{x^*(a)\}$  \\ 
                     & $\{\tilde{x}(a), \tilde{y}(a)\}$, period-$2$ orbit & 
$(-\infty, \infty) \setminus \{x^*(a)\}$ \\ \hline
$e^{-e} \le a \le 1$ & $x^*(a)$, fixed point & $(-\infty, \infty)$ \\ \hline
$1 < a < e^{\frac{1}{e}}$ & $x^*(a)$, fixed point & $(-\infty, y^*(a))$ \\ 
                          & $y^*(a)$, fixed point & $\{y^*(a)\}$ \\ 
                          & $\infty$              & $(y^*(a), \infty)$ \\ \hline
$a = e^{\frac{1}{e}}$ & $e$, fixed point & $(-\infty, e]$ \\ 
                      & $\infty$         & $(e, \infty)$ \\ \hline
$e^{\frac{1}{e}} < a$ & $\infty$         & $(-\infty, \infty)$ \\ \hline
\hline
\end{tabular}
\end{center}
\caption{Dynamics of exponential functions $f(x) = a^x$.}
\end{table}

\begin{thebibliography}{7}
\bibitem{an} J. Anderson, Iterated exponentials, {\em Amer.
Math. Monthly} {\bf 111} (2004) 668-679.
\bibitem{ho} R. A. Holmgren, {\em A First Course in Discrete Dynamical
Systems}, 2nd ed., Springer, 1996.
\bibitem{kn} R. A. Knoebel, Exponentials reiterated, {\em Amer.
Math. Monthly} {\bf 88} (1981) 235-252.
\end{thebibliography}
\end{document}