\documentclass{amsart}

\usepackage{graphics}

\title{Infinite Exponentiation}
\author{Math 166: Honors Calculus II}

\begin{document}

\maketitle

We can make sense of an infinite exponential, $x^{x^{x^{.^{.^{.}}}}}$, using a
recursive definition:
\[
f_1(x) = x,\qquad f_n(x) = x^{f_{n-1}(x)}
\]
The infinite exponential function $f(x)$ can then be defined to be
the limit of $f_n(x)$ as $n$ goes to infinity. This limit will not
exist for every~$x$. For example, $f_n(1) = 1$ for all $n$ so $f(1)
= 1$, but $f_n(2)$ would quickly go to infinity as $n$ increases. To
see which values might work, we'll first plot and compare a few of the
functions $f_n(x)$.

\begin{center}
\resizebox{2in}{!}{\includegraphics{infexps1.eps}}
\end{center}

There is some unusual behavior near 0 and the graphs split apart around
$x = 1.4$, but the different functions $f_n(x)$ agree closely between
these numbers. The key to understanding $f(x)$ is to realize that if this
limit exists then the recursive definition $f_n(x) = x^{f_{n-1}(x)}$
implies $f(x) = x^{f(x)}$.  Let $y = f(x)$. Then $y = x^y$ and we can
solve: $x = y^{1/y}$. The function $y^{1/y}$ is the inverse of $f(x)$, and
vice-versa, at least on intervals containing 1 where these functions are
1--1. We can easily check for such intervals for the function $y^{1/y}$.

\begin{center}
\resizebox{2in}{!}{\includegraphics{infexps2.eps}}
\end{center}

Since $y^{1/y} = e^{\log(y)/y}$, we compute the derivative to be
\[
\frac{d}{dy}[y^{1/y}] = y^{1/y}\Bigl[-\frac{\log(y)}{y^2} + \frac{1}{y^2}\Bigr]
= y^{1/y - 2}(1 - \log(y))
\]
which confirms that $y^{1/y}$  is strictly increasing for $1 - \log(y)\ge
0$, i.e., $0 < y \le e$ (this interval contains 1) and strictly decreasing
for $y > e$. Therefore the desired inverse $f(x)$ exists for $0 < x <
e^{1/e} = 1.44467$.

The plots confirm that the graphs of $f_n(x)$ split apart near $x = 1.4$:

\begin{center}
\resizebox{2in}{!}{\includegraphics{infexps3.eps}}
\end{center}

The graphs of $f_n(x)$ seem to split apart near~0. When $n$ is odd,
$f_n(x)$ goes to 0 and when $n$ is even $f_n(x)$ goes to~1. However, as
$n$ increases, these functions do approach a common limit: the ``even''
functions shrink down and the ``odd'' functions creep up.

\begin{center}
\resizebox{2in}{!}{\includegraphics{infexps4.eps}}
\end{center}

We can obtain the graph of the inverse of $y^{1/y}$ by plotting it
parametrically (in \emph{Mathematica},
\texttt{ParametricPlot[\{y\^(1/y),y\},\{y,0,E\}]}).
This plot agrees closely to the plots of $f_n(x)$ above.

\begin{center}
\resizebox{2in}{!}{\includegraphics{infexps5.eps}}
\end{center}

This analysis proves that $f(x)$ exists for $0 < x < e^{1/e}$. More
work would have to be done to prove that the limit of $f_n(x)$ as $n$
goes to infinity does not exist when $x > e^{1/e}$. Just because an
inverse to $y^{1/y}$ cannot be defined for $y > e$ does not mean that
the limit does not exist. The graphs do indicate, however, that $f_n(x)$
goes to infinity as $n$ goes to infinity if $x > e^{1/e}$.

\end{document}
