\sectionhead{A Hint of Topics (likely :-)  to Come:}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\pagedone
\subsectionhead{The Cantor Set}
     \begin{itemize}
          \item An important example (of a fractal, among other things) and tool for studying nonlinear dynamical systems
               is the {\em Cantor set}.  This set was developed by Georg Cantor in the 1880's, and has led a long and
               fruitful life.
               
               We can develop the Cantor set by starting with the unit interval, and iterating steps.  At each step, we
               will remove the middle thirds of each (remaining) piece:
                \begin{center}
               % \vspace{-0.5cm}
                     \begin{tikzpicture}[scale=16, line width=5pt]
                             \draw(0.0, 1.0) -- (1.0, 1.0);
	                     \draw[line width=2pt](0, 0.99) -- (0, 1.01);
	                     \draw[line width=2pt](1, 0.99) -- (1, 1.01);
	                     \draw(0, 0.95) -- (1/3, 0.95);
	                     \draw(2/3, 0.95) -- (1, 0.95);
	                     \draw(0, 0.90) -- (1/9, 0.90);
	                     \draw(2/9, 0.90) -- (1/3, 0.90);
	                     \draw(2/3, 0.90) -- (7/9, 0.90);
	                     \draw(8/9, 0.90) -- (1, 0.90);
	                     \draw(0, 0.85) -- (1/27, 0.85);
	                     \draw(2/27, 0.85) -- (3/27, 0.85);
	                     \draw(6/27, 0.85) -- (7/27, 0.85);
	                     \draw(8/27, 0.85) -- (9/27, 0.85);
	                     \draw(18/27, 0.85) -- (19/27, 0.85);
	                     \draw(20/27, 0.85) -- (21/27, 0.85);
	                     \draw(24/27, 0.85) -- (25/27, 0.85);
	                     \draw(26/27, 0.85) -- (27/27, 0.85);
	                     \draw(0/81, 0.80) -- (1/81, 0.80);
	                     \draw(2/81, 0.80) -- (3/81, 0.80);
	                     \draw(6/81, 0.80) -- (7/81, 0.80);
	                     \draw(8/81, 0.80) -- (9/81, 0.80);
	                     \draw(18/81, 0.80) -- (19/81, 0.80);
	                     \draw(20/81, 0.80) -- (21/81, 0.80);
	                     \draw(24/81, 0.80) -- (25/81, 0.80);
	                     \draw(26/81, 0.80) -- (27/81, 0.80);
	                     \draw(54/81, 0.80) -- (55/81, 0.80);
	                     \draw(56/81, 0.80) -- (57/81, 0.80);
	                     \draw(60/81, 0.80) -- (61/81, 0.80);
	                     \draw(62/81, 0.80) -- (63/81, 0.80);
	                     \draw(72/81, 0.80) -- (73/81, 0.80);
	                     \draw(74/81, 0.80) -- (75/81, 0.80);
	                     \draw(78/81, 0.80) -- (79/81, 0.80);
	                     \draw(80/81, 0.80) -- (81/81, 0.80);
	                     
	                     \draw(1/6, 0.75) -- (1/6 + 1/162, 0.75);
	                      \draw(5/6, 0.75) -- (5/6 + 1/162, 0.75);
	                      \draw(1/6, 0.73) -- (1/6 + 1/162, 0.73);
	                      \draw(5/6, 0.73) -- (5/6 + 1/162, 0.73);
                              \draw(1/6, 0.71) -- (1/6 + 1/162, 0.71);
	                      \draw(5/6, 0.71) -- (5/6 + 1/162, 0.71);

                \end{tikzpicture}
           \end{center}
\pagedone               
               The Cantor set is what is left after we iterate this process countably many times.
               
               We can calculate how much ``length'' is left after the process:
                    \begin{align*}
                         length &= 1 - \left( \frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \ldots \right) \\
                                     &= 1 - \left( \frac{2^0}{3^1} + \frac{2^1}{3^2} + \frac{2^2}{3^3} + \ldots \right) \\
                                     &= 1 - \frac{1}{3} \sum_{n = 0}^\infty\left(\frac{2}{3}\right)^n \\
                                     &= 1 - \frac{1}{3}\left(\frac{1}{1 - \frac{2}{3}}\right) \\
                                     &= 1 - 1 \\
                                     & = 0.
                    \end{align*}
\pagedone
               It may appear from this that there is nothing left in the set, but that is far from the case.
               
               Let's look at the set again, but labeling things with ternary (base 3) numbers.  Each point in the
               interval $[0, 1]$ can be written as a ternary number $0.d_1d_2d_3\ldots$ where each of the digits
               $d_i$ is 0, 1, or 2:
               \begin{center}
               % \vspace{-0.5cm}
                     \begin{tikzpicture}[scale=16, line width=5pt]
	                     \draw(0.0, 1.03) node{$0.0$};
	                     \draw(1.0, 1.03) node{$1.0$};
	                     \draw(0, 1) -- (1, 1);
	                     \draw[line width=2pt](0, 0.99) -- (0, 1.01);
	                     \draw[line width=2pt](1, 0.99) -- (1, 1.01);
	                     \draw(1/6, 0.972) node{$0.0d_2d_3\ldots$};
	                     \draw(3/6, 0.972) node{$0.1d_2d_3\ldots$};
	                     \draw(5/6, 0.972) node{$0.2d_2d_3\ldots$};
	                     \draw(0, 0.94) -- (1/3, 0.94);
	                     \draw(2/3, 0.94) -- (1, 0.94);
	                     \draw(1/18, 0.912) node{.00};
	                     \draw(3/18, 0.912) node{.01};
	                     \draw(5/18, 0.912) node{.02};
	                     \draw(13/18, 0.912) node{.20};
	                     \draw(15/18, 0.912) node{.21};
	                     \draw(17/18, 0.912) node{.22};
	                     \draw(0, 0.88) -- (1/9, 0.88);
	                     \draw(2/9, 0.88) -- (1/3, 0.88);
	                     \draw(2/3, 0.88) -- (7/9, 0.88);
	                     \draw(8/9, 0.88) -- (1, 0.88);
	                      \draw(1/18, 0.852) node{.001};
	                      \draw(5/18, 0.852) node{.021};
	                      \draw(13/18, 0.852) node{.201};
	                      \draw(17/18, 0.852) node{.221};
	                     \draw(0, 0.82) -- (1/27, 0.82);
	                     \draw(2/27, 0.82) -- (3/27, 0.82);
	                     \draw(6/27, 0.82) -- (7/27, 0.82);
	                     \draw(8/27, 0.82) -- (9/27, 0.82);
	                     \draw(18/27, 0.82) -- (19/27, 0.82);
	                     \draw(20/27, 0.82) -- (21/27, 0.82);
	                     \draw(24/27, 0.82) -- (25/27, 0.82);
	                     \draw(26/27, 0.82) -- (27/27, 0.82);
	                     \draw(0/81, 0.76) -- (1/81, 0.76);
	                     \draw(2/81, 0.76) -- (3/81, 0.76);
	                     \draw(6/81, 0.76) -- (7/81, 0.76);
	                     \draw(8/81, 0.76) -- (9/81, 0.76);
	                     \draw(18/81, 0.76) -- (19/81, 0.76);
	                     \draw(20/81, 0.76) -- (21/81, 0.76);
	                     \draw(24/81, 0.76) -- (25/81, 0.76);
	                     \draw(26/81, 0.76) -- (27/81, 0.76);
	                     \draw(54/81, 0.76) -- (55/81, 0.76);
	                     \draw(56/81, 0.76) -- (57/81, 0.76);
	                     \draw(60/81, 0.76) -- (61/81, 0.76);
	                     \draw(62/81, 0.76) -- (63/81, 0.76);
	                     \draw(72/81, 0.76) -- (73/81, 0.76);
	                     \draw(74/81, 0.76) -- (75/81, 0.76);
	                     \draw(78/81, 0.76) -- (79/81, 0.76);
	                     \draw(80/81, 0.76) -- (81/81, 0.76);
	                     
	                     \draw(1/6, 0.73) -- (1/6 + 1/162, 0.73);
	                      \draw(5/6, 0.73) -- (5/6 + 1/162, 0.73);
	                      \draw(1/6, 0.71) -- (1/6 + 1/162, 0.71);
	                      \draw(5/6, 0.71) -- (5/6 + 1/162, 0.71);
                              \draw(1/6, 0.69) -- (1/6 + 1/162, 0.69);
	                      \draw(5/6, 0.69) -- (5/6 + 1/162, 0.69);

                \end{tikzpicture}
           \end{center}
\pagedone
               In this form, we can see that in the first step, we remove all points of the form $.1d_2d_3d_4\ldots\ $.  In the
               second step, we remove all points of the form $.01d_3d_4\ldots\ $ and $.21d_3d_4\ldots\ $.
               In the $n$th step, we remove all points that have their first (ternary) digit $1$ in their $n$th place.  We will
               be left with (only) all points that have no $1$ in their ternary expansion.
               
               Note:  We do have the (potential) problem that some points, such as $0.222\ldots = 1.0$ have two
               representations.  We will use the representations with the repeated $2$s.  Thus, for example,
               the point $0.0222\ldots = 0.1000\ldots = \frac{1}{3}$ will not be removed.
               
               In this form, we can see that there is a $1-1$ onto function from the Cantor set to the interval $[0,1]$ --
               we simply replace every 2 in the ternary expansion with a 1.  From this, we can see that the Cantor
               set is an uncountable set.  In other words, even though we have removed ``length'' 1, we nonetheless
               have uncountably many points still in the set. 
               
               
\pagedone
          \end{itemize}   
\pagedone
\subsectionhead{Definition of chaos}
     \begin{itemize}
          \item At some point it becomes worthwhile to work with a specific definition of ``chaos,'' think some about
               to what extent it might be a useful definition, and to see the definition in action.
               
               The definition we will work with is (generally) attributed to Robert Devaney \cite{devaney}.  Our version is
               not exactly Devaney's, but is fairly close.  It contains three
               pieces, which do a reasonable job of capturing intuitive ideas of deterministic chaos.  We'll start by stating
               the definition, with some brief discussion of some aspects of the definition (e.g., redundancy among parts of the
               definition), and then apply the definition to some specific cases.
\pagedone
               \begin{mydef} Suppose $X$ is a metric space, $f: X \to X$ is continuous, and $S \subseteq X$ is an
                    infinite subset with the property that $f(S) = S$.  Then we say that $f$ is {\em chaotic} on $S$ if
                    \begin{enumerate}
                         \item $f$ has {\em sensitive dependence on initial conditions} on $S$.
                         \item Periodic points of $f$ are {\em dense} in $S$, and
                         \item $f$ exhibits {\em topological transitivity} on $S$.
                    \end{enumerate}
               \end{mydef}
               We need to clarify some terms in this definition.
\pagedone
          \item A {\em metric space} is a set of points (a {\em space}) $X$ for which we have a {\em metric function}
               $d:X \times X \to \myreal$ such that, for all $x_1, x_2, x_3 \in X$,
               \begin{enumerate}
                   \item $d(x_1, x_2) \ge 0$,
                   \item $d(x_1, x_2) = 0$ iff $x_1 = x_2$,
                   \item  $d(x_1, x_2) = d(x_2, x_1)$, and
                   \item $d(x_1, x_3) \le d(x_1, x_2) + d(x_2, x_3)$.
              \end{enumerate}
               For such a space, we will use the topology induced by the metric (on both $X$ and $S$), so that, for example,
               continuity of a function is relative to the metric topology.  The metric topology is the topology generated by
               $\epsilon$-neighborhoods  (for any $\epsilon > 0$):
               $$B(x_0, \epsilon) = \{x \in X | d(x_0, x) < \epsilon\}.$$
 \pagedone
           \item For {\em sensitive dependence on initial conditions} (SDIC), we will use the property that, for
                any $s_0 \in S$, and for any $\epsilon$-neighborhood $B(s_0, \epsilon)$ of $s_0$, there exist $s_1 \in B(s_0, \epsilon)$,
                $\delta > 0$, and $\lambda > 0$ such that
                     $$d(f^n(s_0), f^n(s_1)) \approx e^{\lambda n}\delta$$
                (at least for a while \ldots).  Here $f^n(s)$ is the $n$th iteration of $f$ on $s$ (i.e., $f^n(s) = f(f(\ldots f(s))\ldots))
                = f \circ f \circ \ldots \circ f \circ f(s)$).
                
                At this point, this is not a ``mathematical'' statement -- we would need to clarify what we mean by ``$\approx$'' and
                ``for a while''.
                
                In fact, for at least some of the examples we will work with, we can use the more specific (and in
                some sense stronger) condition that there exists a  $\lambda > 0$ such that, for any
                $s_0 \in S$, for any $\epsilon$-neighborhood $B(s_0, \epsilon)$ of $s_0$, for any $n > 0$,
                there exist $\delta > 0$ and $s_1 \in B(s_0, \epsilon)$ such that
                     $$d(f^k(s_0), f^k(s_1)) \ge e^{\lambda k}\delta \text{ for } k \le n.$$
                We use this sort of notation (and definition) to correspond with the concept of the {\twntyfrcyr{lyapun\'ov}}
                {\em (lyapunov) exponents} of continuous dynamical systems.
                
                This condition says that no matter how closely we specify initial conditions ($s_0$, but with ``error'' of a
                small $\epsilon$), our prediction error can grow exponentially \ldots
                
                There is ongoing discussion among researchers about this part of the definition (see, for example, the
                discussion in Bishop \cite{bishop}).  In particular, Devaney originally actually only required {\em linear growth}
                rather than {\em exponential growth}.
\pagedone
          \item The condition that {\em periodic points are dense} means that for any $\epsilon$-neighborhood
               $B(s_0, \epsilon)$, there exist an $n > 0$ and an $s \in B(s_0, \epsilon)$ such that $f^n(s) = s$.
               
               One way to think about this is that no matter where we look in a {\em chaotic} system, we will still see
               regularities (periodic points) -- in particular, our system is not just {\em pure noise}, but has structure
               within it.
               
               This is another area of ongoing discussion (again, see Bishop \cite{bishop}).
\pagedone
          \item The condition of topological transitivity means that for any two points $s_0, s_1 \in S$,
               and for any $\epsilon$-neighborhoods $B(s_0, \epsilon_0)$ and $B(s_1, \epsilon_1)$ of these
               two points, there is a point $s \in B(s_0, \epsilon_0)$ with $f^n(s) \in B(s_1, \epsilon_1)$ for some $n$.
               
               This condition is often called {\em mixing}.  Since the point $s_1$ can be anywhere in $S$, this means
               that points in any neighborhood of a given point $s_0$ will, over time, spread throughout all of $S$,
               and this is the case for any $s_0 \in S$.  Thus, over time, the system gets thoroughly mixed (despite
               the densely distributed periodic points).
               
               Some workers in the field take this property to be the most important property of {\em chaos}
               (i.e., the system looks ``noisy,'' in the sense ``most points get sent almost anywhere over time'' \ldots).
               
               This also is an area of active discussion among workers in the field \ldots
\pagedone
          \item It turns out that this definition (well, more precisely, Devaney's original definition) is not minimal.
               For example, Banks, et al.\cite{banks} showed that condition 1. (sensitive dependence on initial conditions)
               follows from 2. and 3. (periodic points are dense, and topological transitivity).  We could thus leave
               condition 1. out of the definition, and not change the collection of systems we would call {\em chaotic}.
               This also says that despite the fact that condition 1. (sensitive dependence) looks like a {\em metric}
               property, in fact {\em chaos} (as so defined) is really a topological property (mixing, with periodicity).
 \pagedone
          \item We can also observe that one way to show that a system is {\em chaotic} is to show that it is homeomorphic
               to a system we have already shown to be {\em chaotic}.
               In general form, if we have a commutative diagram
               \begin{center}
               % \vspace{-0.5cm}
                     \begin{tikzpicture}[scale=4, line width=2pt]
	                     \draw(-0.08, 1.0) node{$X$};
	                     \draw(1.08, 1.0) node{$X$};
	                     \draw[->](0.03, 1) -- (0.97, 1);
	                     \draw(0.5, 1.12) node{$f$};
	                     \draw(-0.08, 0.0) node{$Y$};
	                     \draw(1.08, 0.0) node{$Y$};
	                     \draw[->](0.03, 0) -- (0.97, 0);
	                     \draw(0.5, 0.1) node{$g$};
	                     \draw[->](-0.08,0.87) -- (-0.08,0.13);
	                     \draw[->](1.08,0.13) -- (1.08,0.87);
	                     \draw(-0.2,0.5) node{$h$};
	                     \draw(1.3,0.5) node{$h^{-1}$};
	             \end{tikzpicture}
                \end{center}
                where $h: X \to Y$ is a homeomorphism, and $g: Y \to Y$ is known to be chaotic on $Y$, then we can conclude
                that $f$ is also chaotic on $X$.
                
                The general idea here is known as {\em topological conjugacy}.
\pagedone
              \item It is also worth noting that this definition is far from universally accepted among mathematicians and/or physicists.
               For discussion of some of the issues, see Bishop\cite{bishop} and Knudsen\cite{knudsen}.  But, to a reasonable
               extent, this definition captures a reasonable amount of what a variety of workers in the field seem to have in
               mind when they speak of {\em chaos}.
               
               In any case, this is a useful definition.  As we will see, there are systems that are very easy to describe that are
               {\em chaotic} under this definition, and, conveniently, the proofs that they are chaotic are relatively straightforward.
               Of course, it shouldn't be any great surprise that the proofs are comparatively easy in at least some specific
               cases, because part of the motivation for choosing a specific ``definition'' is to make particular proofs relatively
               easy \ldots
               
     \end{itemize}
\pagedone
\subsectionhead{A Chaotic Map}
     \begin{itemize} 
           \item Let's look at a specific map.  Consider $f: [0,1] \to [0, 1]$ given by
                \begin{eqnarray*}
                     f(x) & = \begin{cases} 2x      & \text{if } 0 \le x < \frac{1}{2} \\  
                                                           2x - 1 & \text{if } \frac{1}{2} \le x  \le1. 
                                  \end{cases}
                \end{eqnarray*}
                A convenient way to look at the map (iterations) of the function is to use the binary representation of
                the unit interval.  Each point in the unit interval has a representation as $x = 0.d_1d_2d_3d_4\ldots$
                where each binary digit $d_i \in \{0, 1\}$.  Our function $f$ is then realized as
                $$f(x) = f(0.d_1d_2d_3d_4\ldots) = 0.d_2d_3d_4\ldots$$
                In other words, our function is just a {\em binary shift} -- it shifts all the binary digits one place to the
                left (and throws away the leading digit if it was 1).  
\pagedone
                Note:  as we did with the Cantor set, we will make the binary representations unique by using
                repeating 1's, so, for example, we will use 0.111\ldots instead of 1.000\ldots
                
                We will now claim that this function $f$ is {\em chaotic} of the entire unit interval.  Let's check
                against our 3-part definition:
                \begin{enumerate}
                     \item {\em Sensitive Dependence on Initial Conditions}:  Suppose we are given $x_0 \in [0, 1]$,
                          $\epsilon > 0$, and $n > 0$.
                          
                          Let $\lambda = \ln{(2)}$ so that $e^{\lambda k} = 2^k$.  Let $m \ge 0$ be such that
                          $2^{-(m+n)} < \epsilon$.  Let $\delta = 2^{-(m+n+1)}$.  Now, if $x_0 = 0.d_1d_2d_3d_4\ldots$,
                          we let $x_1 = 0.d_1d_2d_3d_4\ldots d_{m+n}(1-d_{m+n})d_{m+n+1}d_{m+n+2}\ldots$.
                          In other words, we change the $m+n$ digit of $x_0$ (from 0 to 1, or 1 to 0) to get $x_1$.
\pagedone
                          We then will have that $d(x_0, x_1) = 2^{-(m+n)} < \epsilon$, and $d(x_0, x_1) > 2^{-(m+n+1)}= \delta$.
                          Now, as we iterate, we get
                          $$d(f^k(x_0), f^k(x_1)) = 2^k * 2^{-(m+n)} > 2^k*\delta = e^{\lambda k}\delta$$
                          for $k \le n$, as desired.  In other words, we get local exponential spreading of iterates of (at least
                          some) nearby points.
                          
                     \item {\em Periodic Points Are Dense}:  This is easy \dots  Every rational number in $[0, 1]$ has periodic
                         binary expansion, and therefore is periodic under the shift map.
                         
                    \item {\em Topological Transitivity} (mixing):  Suppose we are given $x_0, x_1 \in [0, 1]$,  $\epsilon_0 > 0$, and
                         $\epsilon_1 > 0$.  Then, we let $n$ be such that $2^{-n} < 2^{-2}\epsilon_0$.  Now if
                         $x_0 = 0.d_1d_2d_3d_4\ldots $ and $x_1 = 0.e_1e_2e_3e_4\ldots $, we let 
                         $x = 0.d_1d_2d_3d_4\ldots d_{n}e_1e_2e_3e_4\ldots$.  In other words, $x$ has digits
                         from $x_0$ for the first $n$ positions, and then digits of $x_1$ concatenated after that.  We
                         will then have that $x \in B(x_0, \epsilon_0)$, and $f^n(x) = x_1 \in B(x_1, \epsilon_1)$,
                         as desired.
                \end{enumerate}
                From this, we see that the binary shift map is {\em chaotic} on the unit interval $[0, 1]$ (under the
                definition we have outlined above \ldots).
                
                Note:  Full disclosure -- the binary shift map isn't continuous on $[0, 1]$ -- in particular, there is a 
                 discontinuity at $1/2$.  With some care, we can handle that, so let's just move forward :-)
                
     \end{itemize}
\pagedone
\subsectionhead{Some other chaotic maps}
     \begin{itemize} 
           \item Now that we have an example of a chaotic map, we can use the "topological conjugacy" idea
                to show that some other related maps are chaotic.
                
                Let's start with a picture of the binary shift map:
                \begin{eqnarray*}
                     f(x) & = \begin{cases} 2x      & \text{if } 0 \le x < \frac{1}{2} \\  
                                                           2x - 1 & \text{if } \frac{1}{2} \le x  \le1
                                  \end{cases}
                \end{eqnarray*}
                looks like this:
                 \begin{center}
                    \vspace{-0.5cm}
                \begin{tikzpicture}[scale=5, line width=1.75pt]
                     \draw[->](0,0) -- (1.1,0);
                     \draw(1.2,0) node {$x_n$};
                     \draw[->](0,0) -- (0,1.05);
                     \draw(-0.02,1) -- (0.02,1);
                     \draw(-0.1, 1) node {$1$};
                     \draw(-0.2, 0.5) node {$x_{n+1}$};
                     \draw[->](0,0) -- (1.05,1.05);
                     \draw[thin,dashed](0,1) -- (1,1);
                     \draw[thin,dashed](1,0) -- (1,1);
                     \draw(1.45, 1) node {$x_{n+1}=x_n$};
                     \draw[color=purple](0,0) -- (0.5, 1);
                     \draw[color=purple](0.5, 0) -- (1,1);
                     \draw(2.2, 0.6) node {$x_{n+1}=\begin{cases} 2x_n      & \text{if } 0 \le x_n < \frac{1}{2} \\  
                                                           2x_n - 1 & \text{if } \frac{1}{2} \le x_n  \le1
                                  \end{cases}$};
                     \draw [->] (1.4,0.5) arc (-45:-90:1);
                \end{tikzpicture}
           \end{center}
 \pagedone
              \item  If we flip the second half, we get the {\em tent map}, which
                looks like this:
                 \begin{center}
                    \vspace{-0.5cm}
                \begin{tikzpicture}[scale=6, line width=1.75pt]
                     \draw[->](0,0) -- (1.1,0);
                     \draw(1.2,0) node {$x_n$};
                     \draw[->](0,0) -- (0,1.05);
                     \draw(-0.02,1) -- (0.02,1);
                     \draw(-0.1, 1) node {$1$};
                     \draw(-0.2, 0.5) node {$x_{n+1}$};
                     \draw[->](0,0) -- (1.05,1.05);
                     \draw[thin,dashed](0,1) -- (1,1);
                     \draw[thin,dashed](1,0) -- (1,1);
                     \draw(1.45, 1) node {$x_{n+1}=x_n$};
                     \draw[color=purple](0,0) -- (0.5, 1);
                     \draw[color=purple](0.5, 1) -- (1,0);
                     \draw(2.3, 0.6) node {$x_{n+1}=\begin{cases} 2x_n      & \text{if } 0 \le x_n < \frac{1}{2} \\  
                                                           2*(1 - x) & \text{if } \frac{1}{2} \le x  \le1 
                                  \end{cases}$};
                     \draw [->] (1.4,0.5) arc (-45:-90:0.7);
                \end{tikzpicture}
           \end{center}
           A relatively straightforward topological conjugacy argument shows that this map also is chaotic.
\pagedone
         \item  Now if we slightly modify the tent map (make is "smooth" at the top), we get the logistic map with $R = 4$,
          which looks like:
           \begin{center}
                    \vspace{-0.5cm}
                \begin{tikzpicture}[scale=7, line width=1.75pt]
                     \draw[->](0,0) -- (1.1,0);
                     \draw(1.2,0) node {$x_n$};
                     \draw[->](0,0) -- (0,1.05);
                     \draw(-0.02,1) -- (0.02,1);
                     \draw(-0.1, 1) node {$1$};
                     \draw(-0.2, 0.5) node {$x_{n+1}$};
                     \draw[->](0,0) -- (1.05,1.05);
                     \draw[thin,dashed](0,1) -- (1,1);
                     \draw[thin,dashed](1,0) -- (1,1);
                     \draw(1.28, 0.95) node {$x_{n+1}=x_n$};
                     \pgfmathsetmacro{\r}{4.0}
                     \draw[color=purple,domain=0:1,
	                  smooth,samples=600,variable=\t] plot
	                  ({\t},{\r*\t*(1-\t)})
	                  coordinate(end);
           %          \draw[color=purple](0,0) parabola[bend at end] (0.5,\r/4);
             %        \draw[color=purple](0.5,0.75) parabola[bend at start] (1,0);
                     \draw(1.58, 0.6) node {$x_{n+1}=4x_n(1-x_n)$};
                     \draw [->] (1.4,0.55) arc (-45:-90:0.7);
                \end{tikzpicture}
           \end{center}
           Again, by a relatively straightforward topological conjugacy argument, we can see that the logistics
           map for $R = 4$ is chaotic on $[0, 1]$.
\pagedone
          \item One more example:  we'll now move into the realm of the complex numbers,
          $\mycomplex = \{x + iy| x, y \in \myreal, i^2 = -1\}$.
          
          Consider the unit circle in the complex plane $C_1 = \{z \in \mycomplex | \ |z| = 1\}$, where
          $|z| = |x + iy| = (x^2 + y^2)^{\frac{1}{2}}$.
          
          $C_1$ looks like:
           \begin{center}
                \vspace{-1cm}
                \begin{tikzpicture}[scale=3,line width=1.75pt]
                     \draw[color=blue, ->] (-1.3,0) -- (1.3,0);
                     \draw[color=blue](1.4,0) node{$x$};
                     \draw[color=black](1.1,0.15) node{$1$};
                     \draw[color=red, ->](0,-1.3) -- (0,1.3);
                      \draw[color=black](0.1,1.15) node{$i$};
                     \draw[color=red](0,1.45) node{$iy$};
                     \draw[color=purple] (0,0) circle (1);
                     \draw[color=black](-1.2,0.15) node{$-1$};
                      \draw[color=black](-0.2,-1.15) node{$-i$};
                \end{tikzpicture}
           \end{center}
\pagedone
          Now consider $f: C_1 \to C_1$ given by $f(z) = z^2$.  We can observe that any point in $C_1$
          can be written as $z = x + iy = \cos(\theta) + i \sin(\theta)$ for some $\theta$, and then remember
          (from above) that $\cos(\theta) + i \sin(\theta) = e^{i \theta}$.  We then see that our function $f(z)$
          is given by
               $$f(z) = f(e^{i \theta}) = (e^{i \theta})^2 = e^{i * 2 \theta}.$$
          In pictorial form:
          \begin{center}
                \vspace{-1.2cm}
                \begin{tikzpicture}[scale=3.0,line width=1.75pt]
                     \draw[color=blue, ->] (-1.3,0) -- (1.3,0);
                     \draw[color=blue](1.4,0) node{$x$};
                     \draw[color=black](1.1,0.15) node{$1$};
                     \draw[color=red, ->](0,-1.3) -- (0,1.3);
                      \draw[color=black](0.1,1.15) node{$i$};
                     \draw[color=red](0,1.45) node{$iy$};
                     \draw[color=purple] (0,0) circle (1);
                     \draw[color=black](0,0) -- (0.5, 0.866);
                     \draw(1.0, 1.0) node{$z = e^{i \theta}$};
                     \draw [->] (0.4,0.0) arc (0:60:0.4);
                     \draw(0.45, 0.3) node{$\theta$};
                     \draw[color=black](0,0) -- (-0.5, 0.866);
                     \draw(-1.14, 1.0) node{$f(z) = e^{i *2 \theta}$};
                     \draw [->] (0.7,0.0) arc (0:120:0.7);
                     \draw(0.2, 0.8) node{$2\theta$};
                     \draw[color=black](-1.2,0.15) node{$-1$};
                      \draw[color=black](-0.2,-1.15) node{$-i$};
                \end{tikzpicture}
           \end{center}
\pagedone
          Another way to look at this (just keeping track of $\theta$) is:
          \begin{center}
                    \vspace{-0.5cm}
                \begin{tikzpicture}[scale=7, line width=1.75pt]
                     \draw[->](0,0) -- (1.1,0);
                     \draw(1.2,0) node {$\theta_n$};
                     \draw[->](0,0) -- (0,1.05);
                     \draw(-0.02,1) -- (0.02,1);
                     \draw(1, -0.1) node {$2\pi$};
                     \draw(-0.1, 1) node {$2\pi$};
                     \draw(-0.2, 0.5) node {$\theta_{n+1}$};
                     \draw[->](0,0) -- (1.05,1.05);
                     \draw[thin,dashed](0,1) -- (1,1);
                     \draw[thin,dashed](1,0) -- (1,1);
                     \draw(1.45, 1) node {$\theta_{n+1}=\theta_n$};
                     \draw[color=purple](0,0) -- (0.5, 1);
                     \draw[color=purple](0.5, 0) -- (1,1);
                     \draw(2.0, 0.6) node {$\theta_{n+1}=\begin{cases} 2\theta_n      & \text{if } 0 \le \theta_n < \pi \\  
                                                           2\theta_n - 2\pi & \text{if } \pi \le \theta_n  \le 2\pi
                                  \end{cases}$};
                     \draw [->] (1.4,0.5) arc (-45:-90:1);
                \end{tikzpicture}
           \end{center}
          
          Once again, it is easy to see that this map is chaotic on $C_1$.
          
      \end{itemize}
\pagedone
\subsectionhead{Julia Sets}
     \begin{itemize}
          \item Now we are about ready to dive in to the {\em Mandelbrot set}.  We'll start with
               Julia sets.
               
               We observed that $f(z) = z^2$ is chaotic on the unit circle in $\mycomplex$.  In particular,
               the function keeps elements of the unit circle on the unit circle.  In other words, if $|z| = 1$,
               then $|z^2| = 1$ also.
               
               On the other hand, if $|z| < 1$, then iterates of $z$ go to $0$, which is an attracting fixed point of
               the map.  For any point $z$ with $|z| < 1$,
               we have $\lim_{n \to \infty}f^n(z) = 0$.  On the other hand, if $|z| > 1$, then $\lim_{n \to \infty}f^n(z) = \infty$.
\pagedone
               The complex plane $\mycomplex$ thus breaks up into 4 pieces:
               \begin{enumerate}
                    \item An {\em attracting fixed point} at $0$,
                    \item A {\em basin of attraction} of the attracting fixed point -- in other words, a set of
                         points whose trajectories go toward the attracting fixed point
                    \item A {\em basin of attraction} of the attracting point at infinity (more on this later \ldots), and
                    \item A {\em boundary} between these two basins -- in this case, the unit circle.
               \end{enumerate}
               In the following picture, regions 2, 3, and 4 are indicated in {\color{green} green}, {\color{blue} blue},
               and {\color{red} red}.
\pagedone
                This is a view of the regions associated with $f(z) = z^2$:
                \begin{center}
                \vspace{-1.2cm}
                \begin{tikzpicture}[scale=1.9,line width=1.75pt]
                     \draw[color=black, ->] (-3.7,0) -- (4.2,0);
                     \draw[color=black](4.4,0) node{$x$};
                     \draw[color=black](1.2,0.15) node{$1$};
                     \draw[color=black, ->](0,-3.7) -- (0,3.5);
                     \draw[color=black](0.2,1.25) node{$i$};
                     \draw[color=black](0,3.7) node{$iy$};
                     \draw[color=red] (0,0) circle (1);
                   %  \draw[color=black](-1.3,0.15) node{$-1$};
                   %  \draw[color=black](-0.25,-1.2) node{$-i$};
                     \pgfmathsetmacro{\b}{0.1}
                     \draw[color=green,domain=0:15*pi,
	                  smooth,samples=300,variable=\t] plot
	                  ({(0.5)*exp((-\b/2)*\t)*2*cos(sqrt(4-\b*\b)/2*\t r)},
	                  {(0.5)*exp((-\b/2)*\t)*(\b*cos(sqrt(4-\b*\b)/2*\t r)+sqrt(4-\b*\b)*sin(sqrt(4-\b*\b)/2*\t r))}) 
	                  coordinate(end);
	             \pgfmathsetmacro{\b}{0.1}
                     \draw[color=blue,domain=0:8.25*pi,
	                  smooth,samples=300,variable=\t] plot
	                  ({(0.5)*exp((\b/2)*\t)*2*cos(sqrt(4-\b*\b)/2*\t r)},
	                  {(0.5)*exp((\b/2)*\t)*(\b*cos(sqrt(4-\b*\b)/2*\t r)+sqrt(4-\b*\b)*sin(sqrt(4-\b*\b)/2*\t r))}) 
	                  coordinate(end);
	             \pgfmathsetmacro{\tz}{8.25*pi-0.01}
	             \pgfmathsetmacro{\to}{8.25*pi+0.01}
	             \draw[color=blue,->,>=angle 60]
	                  ({(0.5)*exp((\b/2)*\tz)*2*cos(sqrt(4-\b*\b)/2*\tz r)},
	                  {(0.5)*exp((\b/2)*\tz)*(\b*cos(sqrt(4-\b*\b)/2*\tz r)+sqrt(4-\b*\b)*sin(sqrt(4-\b*\b)/2*\tz r))}) --
	                  ({(0.5)*exp((\b/2)*\to)*2*cos(sqrt(4-\b*\b)/2*\to r)},
	                  {(0.5)*exp((\b/2)*\to)*(\b*cos(sqrt(4-\b*\b)/2*\to r)+sqrt(4-\b*\b)*sin(sqrt(4-\b*\b)/2*\to r))});
                \end{tikzpicture}
           \end{center}
\pagedone
          \item The set of points that form the boundary of the basin of attraction of $\infty$ are particularly
               interesting.  In the case of the function we have been looking at ($f(z) = z^2$), the boundary
               is just the unit circle ($\{ z\ |\ |z| = 1\}$).
               
               Now let's generalize slightly, and consider the set of functions $f_c(z) = z^2 + c$, for $c \in \mycomplex$.
               We have been looking at $f_0(z)$.
               
               For these functions, we can define the {\em Julia Set} of the
               function by:
               
               \ \ $J(f_c(z)) =$ the boundary of the basin of attraction of $\infty$
               
               where the {\em basin of attraction of $\infty$} is the set
                $$B(\infty) = \{ z\ |\ \lim_{n \to \infty}f^n_c(z) = \infty\}.$$
\pagedone
                On the following pages are some examples of Julia sets for various
                values of $c$.  These are actually calculated by inverse iteration
                of the relevant function:
                $$f^{-1}_c(z) = sqrt(z - c)$$
                However, since there are two square roots, we randomly choose
                one of the square roots.  Similarly to what we did in the bifurcation
                diagrams, we discard the first few iterations, and then plot from there
                on.
                
                MatLab source code follows the Julia sets.
                
                \centerline{\includegraphics[width = 7.5in]{juliae0_0e0_0i.jpg}}
               \centerline{c = 0.0 + 0.0i}
                \centerline{\includegraphics[width = 7.5in]{juliaen0_2e0_0i.jpg}}
               \centerline{c = -0.2 + 0.0i}
                \centerline{\includegraphics[width = 7.5in]{juliaen0_6e0_4i.jpg}}
               \centerline{c = -0.6 + 0.4i}
                \centerline{\includegraphics[width = 7.5in]{juliaen1_02e0_0i.jpg}}
               \centerline{c = -1.0 + 0.0i}
                \centerline{\includegraphics[width = 7.5in]{juliae0_2e0_0i.jpg}}
               \centerline{c = 0.2 + 0.0i}
                \centerline{\includegraphics[width = 7.5in]{juliae0_3e0_0i.jpg}}
               \centerline{c = 0.3 + 0.0i}
               \centerline{\includegraphics[width = 7.5in]{juliae0_5e0_5i.jpg}}
               \centerline{c = 0.5 + 0.5i}
               \centerline{\includegraphics[width = 7.5in]{juliaen1_99e0_0i.jpg}}
               \centerline{c = -1.99 + 0.0i}
               \centerline{\includegraphics[width = 7.5in]{juliaen3_0e0_0i.jpg}}
               \centerline{c = -3.0 + 0.0i}
\pagedone
               \begin{verbatim}
			% Matlab source code:  myjulia.m
			%
			% calculating Julia sets by inverse iteration,
			%    and random choice of positive or negative
			%    square root
			%
			hold off; clf; clear
			z = 0.00001+0.0i;
			c = -0.6+0.4i
			for n=1:20
			    if (random('unid',2) == 1)
			        z = sqrt(z - c);
			    else
			        z = (-1)*sqrt(z - c);
			    end
			end
			y=zeros(1,40000);
			y(1) = z;
			for n=2:40000
			    if (random('unid',2) == 1)
			        y(n) = sqrt(y(n - 1) - c);
			    else
			        y(n) = (-1)*sqrt(y(n - 1) - c);
			    end
			end
			
			%pbaspect([1 1 1]);
			hold on;
			scatter(real(y),imag(y),4,[0 0 0],'filled');
			
	          \end{verbatim}
\pagedone
          \item  There is another way to visualize Julia sets.  Instead of doing inverse iteration,
               we look at the behavior of many different points in a region.  In particular, we
               iterate the function $f_c(z)$ for each value of $z$ in a rectangular region containing
               the Julia set.  Most of the points will run off to $\infty$, at varying rates of speed.
               
               Hence, what we will do is assign a color to each pixel in the rectangle according
               to the ``escape time'' of the corresponding $z$ value.  In particular, for each
               point $z$, we will see how many iterations of the function it takes for the value
               to have magnitude $>\ 2$.  We will also have a cutoff for total iterations, because
               points in (or inside of) the Julia set will never run off to $\infty$.  We will then
               assign a color according to the number of iterations.
               
               Following are some examples of this:
               
               \centerline{\includegraphics[width = 7in]{juliaen0_4942e0_5229i.jpg}}
               \centerline{c = -0.4942+0.5229ii}
               \centerline{\includegraphics[width = 7in]{juliaen0_7017e0_3842i.jpg}}
               \centerline{c = -0.7017 + 0.3842i}
                \centerline{\includegraphics[width = 5in]{juliae0_7322e0_2628i.png}}
               \centerline{c = 0.7322 + 0.2628i}
               \centerline{\includegraphics[width = 7in]{juliae0_2850e0_0100i.jpg}}
               \centerline{c = 0.2850 + 0.0100i}
     \end{itemize}
\pagedone
\subsectionhead{The Mandelbrot Set}
     \begin{itemize}
          \item The next step is to look at the Mandelbrot set.  Consider the functions we were looking
               at for Julia sets:
                    $$f_c(z) = z^2 + c, \text{where\ } c \in \mycomplex.$$
               We now ask, for $c \in \mycomplex$, what happens to $0$ under iterations of $f_c$ -- i.e.,
               what is the behavior of $f_c^n(0)$ as $n \to \infty$?  There are two possibilities:  either $f_c^n(0)$ stays
               bounded for all $n$, or $f_c^n(0) \to \infty$ as $n \to \infty$.  We can now define the
               {\em Mandelbrot Set}:
                    $$\text{\em Mandelbrot} = \{c \in \mycomplex : f_c^n(0) \text{\ stays bounded for all n}\}$$
               There is a property of the {\em Mandelbrot set} that is similar to the Julia sets:  if $|f^n_c(0)| > 2$
               for some $n \ge 0$, then $c \not\in \text{\em Mandelbrot}$.  We can use this to make an
               {\em escape time} picture of {\em Mandelbrot}.
\pagedone
               Escape time picture (black central area is {\em Mandelbrot}):
               \vspace{0.5cm}
               
               \centerline{\includegraphics[width = 7in]{Mandelbrot_all.png}}
               \centerline{Mandelbrot Set}
\pagedone
               In all the fancy colorful pictures you have seen of the {\em Mandelbrot set}, only the black
               parts are actually in the set -- the rest of the points are running away to $\infty$ at various rates:
               \vspace{0.5cm}
               
               \centerline{\includegraphics[width = 7in]{Mandelbroten0_742en0_147.png}}
               \centerline{Mandelbrot Set around $-0.742 - 0.147i$}
\pagedone
          \item  Another way to view this is to think about whether $0$ is ``trapped'' inside the Julia set.
               Another definition of the {\em Mandelbrot set} is this:
                    $$\text{\em Mandelbrot} = \{c \in \mycomplex : \text{the Julia set of } f_c(z) \text{is a connected set}\}$$
               For points in the {\em Mandelbrot set}, the Julia set is one connected piece.  For points outside the
               {\em Mandelbrot set}, the Julia set breaks up into a ``dust'' of isolated points.  ``Dusts'' like these are
               sometimes called ``Cantor dusts,'' because they have strong similarities to (and properties like) the
               {\em Cantor set}. 
     \end{itemize}
\pagedone
\subsectionhead{Fixed points, limit sets, stable sets, attractors}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\pagedone
\subsectionhead{Basins of attraction}
     \begin{itemize}
          \item \ldots\newline
               %\vspace{-0.5cm}
               \centerline{\includegraphics[width = 6.5in]{newton_zcubed.jpg}}
               \centerline{Newton's method, roots of $z^3 - 1$}
     \end{itemize}
\pagedone
\subsectionhead{Statistical measures}
     \begin{itemize}
          \item Suppose we have a data set.  A thing we might want to do is characterize the data
               set using a relatively small collection of numbers.  Or, at least, we might wish to use
               a relatively small collection of numbers to distinguish one data set from another.
               Each of these (hopefully characteristic) numbers is called a {\em statistic}.
               An example of a {\em statistic} is the {\em mean}.  Other examples are the {\em median},
               the {\em mode}, the {\em variance}, the {\em standard deviation}, etc.
               
               In some special cases, things can be particularly nice.  A {\em normal distribution}
               ({\em Gaussian distribution})
               $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$$
               is completely characterized by its mean ($\mu$) and standard deviation ($\sigma$).
\pagedone
               This becomes more significant in view of the (traditional) {\em central limit theorem}:
               
               If $\{X_i\}$ is a sequence of i.i.d. (independent, identically distributed) random variables,
               with mean $E[X_i] = \mu$ and variance $Var[X_i] = \sigma^2$ (in particular, finite mean
               and variance), then we have the convergence:
               $$\sqrt{n}\left(\left(\frac{1}{n}\sum_{i=1}^n X_i\right) - \mu\right)\ \xrightarrow{d}\ \mathcal{N}(0,\;\sigma^2),$$
               where $\mathcal{N}(0,\;\sigma^2)$ is the normal (Gaussian) distribution with mean $0$ and variance $\sigma$,
               and $\xrightarrow{d}$ is {\em convergence in distribution}.
\pagedone
               {\em Convergence in distribution} is the property, for a sequence of random variables $\{X_n\}$, that
               $$\lim_{n\to\infty} F_n(x) = F(x)$$
               for all $x \in \myreal$ at which $F(x)$ is continuous, where $F_n$ is the {\em cumulative distribution
               function} of $X_n$, and $F$ is the cumulative distribution function of the limit distribution.
               
               This leads to a paradigm concerning noisy data.  If we assume that {\em noise} (or {\em error})
               is primarily the result of an accumulation of independent random external influences, then it
               (often) will make sense to assume that any noisiness in our data will be normally distributed, and
               hence that we can profitably assume that the basis for statistical analysis will rest on studying
               the means and variances of our data sets.
\pagedone
               One could more or less identify this paradigm as a Gaussian / Normal Distribution world, with mean
               and variance as {\em sufficient statistics} for analysis of error / noise.
               
               If we have a data set, and we have good reason (e.g., a {\em central limit theorem}) to believe
               that a particular parametrized statistical model (e.g., a Gaussian distribution, with mean $\mu$ and
               variance $\sigma^2$ as parameters) is appropriate (correct?), then a {\em sufficient statistic}
               (or set of statistics) for the model is a statistic (or set) such that no other statistic that could be calculated
               from our data set could give additional information about the parameters of the model for that data set.
               In the case of the Gaussian / normal distribution, the sufficient statistics would be the {\em sample mean}
               and the {\em sample variance}.  In particular, there would be no reason to calculate any higher {\em moments}
               of the data set.
\pagedone
          \item If we have reason to believe our data set does not come from a Gaussian distribution, or, more
               generally that Gaussian models are inappropriate or insufficient, we may find it necessary to
               calculate additional or alternative statistics.
               
               One example of additional statistics is higher moments.  For a distribution $X$, for any
               $k \ge 1$, we have the $k$th {\em central moment}:
                    $$\mu_k = E\left[ (X - \mu)^k\right] = \left<\ (X - \mu)^k\ \right>,$$
               where $E[\ ]$ is the {\em expected value}, and $\mu$ is the mean of the distribution.
               
               The second central moment $\mu_2$ is the variance of the distribution.  The third central moment
               $\mu_3$ is zero for symmetric distributions (such as the Gaussian), and in general is a measure
               of the asymmetry of the distribution.
\pagedone
               {\em Standardized} versions of the 3rd and 4th central moments are called the {\em skew} and {\em kurtosis}
               or the distribution:
                    $$\text{skew}(X) = \gamma_1 = E\left[\left(\frac{X - \mu}{\sigma}\right)^3\right]$$
                    $$\text{kurtosis}(X) = \gamma_2 = E\left[\left(\frac{X - \mu}{\sigma}\right)^4\right] - 3$$
               where $\sigma$ is the standard deviation.  The adjustment of $-3$ is such that the {\em kurtosis}
               of the Gaussian is zero.
               
               We can also {\em standardize} the higher central moments in the same general way, as
               $\frac{\mu_k}{\sigma^k}$.
               
               One issue that needs to be kept in mind is that the moments may not exist -- the relevant sums
               (or integrals) may be unbounded.
\pagedone
               The various moments may be {\em sufficient statistics} to characterize a distribution.  More generally,
               we can use various statistics to distinguish between distributions -- in particular, if two distributions
               have at least one difference in their statistics, then we can have confidence that they are different distributions.
               
               Over time, various measures other than the {\em moments} have been developed for the analysis of
               data.  An important element of the evolving paradigm of nonlinear systems and chaos is the observation
               that what may look like random or noisy data may actually be a deterministic chaotic system.  An example of
               this, which was observed by people such as John von Neumann and Stan Ulam, is the logistic equation
               with $R = 4$ (for appropriate initial values).
               
               In the next section we'll look some at a particular statistic that has seen wide use in nonlinear systems --
               the {\em fractal dimension}.
     \end{itemize}
\pagedone
\subsectionhead{Fractal dimensions, related measures}
     \begin{itemize}
          \item There are various ways to think about {\em dimension}.  Let's look for a little while at the
               Lorenz equations:
                    \begin{align*}
                         \dot{x} & =  \sigma (y - x) \\
                         \dot{y} & =  x (\rho - z) - y \\
                         \dot{z} & =  xy - \beta z
                   \end{align*}
              where $\sigma$ is the {\em Prandtl number}, $\rho$ is the {\em Rayleigh number}, and $\beta$ is another adjustable
              parameter.  All of $\sigma$, $\rho$, and $\beta$ are positive, with typical values $\sigma = 10$, $\beta = 8 / 3$, and
              $\rho = 28$.
              
              If we pick particular values for $\sigma$, $\rho$, and $\beta$, and initial values $x_0$, $y_0$
              and $z_0$, we can use a numerical solver to follow a single trajectory.  The picture on the following page
              uses the MatLab built-in ODE45, which is a variable step Runge-Kutta (4,5) solver:
\pagedone
              \vspace{0.5cm}
          \centerline{\includegraphics[width = 7.8in]{lorenz-attractor-2}}
          \centerline{$\sigma = 10$, $\beta = 8/3$, $\rho = 28$, $x_0 = y_0 = z_0 = 10$}
\pagedone
          \item One way to think about this is that the trajectory is 1-dimensional -- as time passes, the system moves along
          a (curved) line in space.  A second way to think of it is that the system lives in 3-dimensional (x,y,z) space.
          A third way is that there are 3 adjustable parameters ($\sigma$, $\rho$, and $\beta$)
          that control the behavior of the system.  A fourth way is that any particular trajectory is determined by
          six numbers, $\sigma$, $\rho$, and $\beta$, and the initial conditions $x_0$, $y_0$, and $z_0$.
          
          There are a couple of other considerations.  One is that there is actually an {\em attractor} inside the system.
          For a comparatively large range of initial conditions, the trajectories will approach a limiting stable set.  By
          following a trajectory using a numerical solver, we get an approximation to the attractor.  We could explore
          the dimension(s) of the attractor.
\pagedone
          A second consideration is what we might think of as the ``real world'' problem.  For real systems that we might
          study, we typically don't actually know the equations of motion of the system.  We are likely just to have a
          set of measured data values extracted from the system through some instrumentation.
          
          We'll look at a relatively generic approach to assigning a number to a data set (i.e., calculating a statistic) which
          corresponds with and generalizes the idea of {\em dimension} in simple cases.  We will be developing (one
          notion of) the {\em fractal dimension} of a data set.
\pagedone
          \item  What we will do is to cover our data set with ``boxes'' (or ``balls'' -- see Carter \cite{carter2}), and see how
               the number of boxes grows as we decrease the sizes of the boxes.
               
               We begin with a line (segment) of length 1:
               \begin{center}
                    %\vspace{-0.5cm}
                     \begin{tikzpicture}[scale=10, line width=1.75pt]
                          \draw(0,0) -- (1,0);
                          \draw(0,-0.01) -- (0,0.01);
                          \draw(1,-0.01) -- (1,0.01);
                     \end{tikzpicture}
                \end{center}
                We can cover this line segment with 1 box of side $1=1/2^0$:
                \begin{center}
                    %\vspace{-0.5cm}
                     \begin{tikzpicture}[scale=5, line width=1.75pt,color=blue]
                          \draw[color=black](0,0) -- (1,0);
                          \draw(0,-0.5) -- (0,0.5);
                          \draw(0,0.5) -- (1,0.5);
                          \draw(1,0.5) -- (1,-0.5);
                          \draw(1,-0.5) -- (0,-0.5);
                     \end{tikzpicture}
                \end{center}
                We can cover the line segment with 2 boxes of side $1/2 = 1/2^1$:
                \begin{center}
                    \vspace{-0.75cm}
                     \begin{tikzpicture}[scale=10, line width=1.75pt,color=blue]
                          \draw[color=black](0,0) -- (1,0);
                          \draw(0,-0.25) -- (0,0.25);
                          \draw(0,0.25) -- (1,0.25);
                          \draw(1,0.25) -- (1,-0.25);
                          \draw(1,-0.25) -- (0,-0.25);
                          \draw(0.5,-0.25) -- (0.5,0.25);
                     \end{tikzpicture}
                \end{center}
                or 4 boxes of side $1/2^2$:
                \begin{center}
                    \vspace{-0.25cm}
                     \begin{tikzpicture}[scale=10, line width=1.75pt,color=blue]
                          \draw[color=black](0,0) -- (1,0);
                          \draw(0,-0.125) -- (0,0.125);
                          \draw(0,0.125) -- (1,0.125);
                          \draw(1,0.125) -- (1,-0.125);
                          \draw(1,-0.125) -- (0,-0.125);
                          \draw(0.25,-0.125) -- (0.25,0.125);
                          \draw(0.5,-0.125) -- (0.5,0.125);
                          \draw(0.75,-0.125) -- (0.75,0.125);
                     \end{tikzpicture}
                \end{center}
                In general, it can be covered with $2^{n}$ boxes of side $\frac{1}{2^n}$.
\pagedone
               Now consider a square:
               \begin{center}
                    %\vspace{-0.5cm}
                     \begin{tikzpicture}[scale=5, line width=1.75pt,color=black]
                          \draw(0,-0.5) -- (0,0.5);
                          \draw(0,0.5) -- (1,0.5);
                          \draw(1,0.5) -- (1,-0.5);
                          \draw(1,-0.5) -- (0,-0.5);
                     \end{tikzpicture}
                \end{center}
                We can cover this with 1 box of side $1$, or 4 of side $1/2^1$:
                \begin{center}
                    %\vspace{-0.5cm}
                     \begin{tikzpicture}[scale=5, line width=1.75pt,color=blue]
                          \draw(0,0) -- (1,0);
                          \draw(0.5,-0.5) -- (0.5,0.5);
                          \draw(0,-0.5) -- (0,0.5);
                          \draw(0,0.5) -- (1,0.5);
                          \draw(1,0.5) -- (1,-0.5);
                          \draw(1,-0.5) -- (0,-0.5);
                     \end{tikzpicture}
                \end{center}
                or 16 of side $1/2^2$:
                \begin{center}
                    %\vspace{-0.5cm}
                     \begin{tikzpicture}[scale=5, line width=1.75pt,color=blue]
                          \draw(0,0) -- (1,0);
                          \draw(0,0.5) -- (1,0.5);
                          \draw(0.5,-0.5) -- (0.5,0.5);
                          \draw(0,-0.5) -- (0,0.5);
                          \draw(0.25,-0.5) -- (0.25,0.5);
                          \draw(0.75,-0.5) -- (0.75,0.5);
                          \draw(0,-0.25) -- (1,-0.25);
                          \draw(0,0.25) -- (1,0.25);
                          \draw(1,0.5) -- (1,-0.5);
                          \draw(1,-0.5) -- (0,-0.5);
                     \end{tikzpicture}
                \end{center}
                In general, we can cover the square with $2^{2n}$ boxes of side $1/2^n$.
                
                In a similar fashion, we can cover a cube with $2^{3n}$ boxes of side $1/2^n$.
                
                We can now make a table, showing how the number of boxes needed to cover
                a ``cube'' in dimension $k$ grows as the length of the side of the boxes ($r$)
                decreases.
                
                \begin{center}
                     \begin{tabular}{ c || c | c | c | c | c | c }
                          r & $num_1$ & $num_2$ & $num_3$ & $\cdots$ & $num_k$ & $\cdots$\\
                         \hline
                           & & & & &\\
                         $\frac{1}{2^0}$ & 1 & 1 & 1 & $\cdots$ & 1 & $\cdots$ \\
                           & & & & &\\
                         \hline
                            & & & & &\\
                         $\frac{1}{2^1}$ & $\frac{1}{2^{1}}$ & $\frac{1}{2^{2}}$ & $\frac{1}{2^{3}}$
                                   & $\cdots$ & $\frac{1}{2^{k}}$ & $\cdots$ \\
                           & & & & &\\
                         \hline
                            & & & & &\\
                         $\frac{1}{2^2}$ & $\frac{1}{2^{2*1}}$ & $\frac{1}{2^{2*2}}$ & $\frac{1}{2^{2*3}}$
                                   & $\cdots$ & $\frac{1}{2^{2*k}}$ & $\cdots$ \\
                           & & & & &\\
                         \hline
                         \vdots & \vdots & \vdots & \vdots & $\cdots$& \vdots & $\cdots$ \\
                         \hline
                            & & & & &\\
                         $\frac{1}{2^n}$ & $\frac{1}{2^{n*1}}$ & $\frac{1}{2^{n*2}}$ & $\frac{1}{2^{n*3}}$
                                   & $\cdots$ & $\frac{1}{2^{n*k}}$ & $\cdots$ \\
                            & & & & &\\
                         \hline
                           & & & & &\\
                         \vdots & \vdots & \vdots & \vdots & $\cdots$& \vdots & $\cdots$

                    \end{tabular}
               \end{center}
               Putting all of this together, we can define a {\em dimension} by:
                    $$ \text{dim}(S) = \lim_{r \to 0}\frac{\log(num(r))}{\log\left(\frac{1}{r}\right)}$$
               where $r$ is the length of the side of a box, and $num(r)$ is the number of boxes
               of side $r$ needed to cover our set $S$.
               
               This {\em dimension} is often called the {\em fractal dimension} of the set $S$.
               
               For some (informal) calculations of {\em fractal dimensions} of some example sets,
               you can look at Carter \cite{carter4}.
               
               (I need to put more discussion of naming and details here \ldots)
     \end{itemize}
\pagedone
 \subsectionhead{ Strange attractors}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{About fractals}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Continuous systems and flows}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Some history (e.g., Poincar\'e)}
     \begin{itemize}
          \item \ldots
    
               \begin{tikzpicture}
                       \begin{scope}[MyCube]
		       \cube{-2}{0}{4}{2}
		       \end{scope}
		       \begin{scope}[scale=1, line width=1.75pt,color=blue]
%                         \shade[yslant=-0.6,right color=gray!10, left color=black!50]
%                                     (0,0) rectangle +(2,2);
%                         \draw[yslant=-0.6,color=gray] (0,0) grid (2,2);
%                         \shade[yslant=0.4,right color=gray!70,left color=gray!10]
%                                     (2,-2) rectangle +(2,2);
%                         \draw[yslant=0.4,color=gray] (2,-2) grid (4,0);
%                         \shade[yslant=0.4,xslant=-1,bottom color=gray!10,
%                               top color=black!80] (4,2) rectangle +(-2,-2);
%                         \draw[yslant=0.4,xslant=-1,color=gray] (2,0) grid (4,2);
%             %            \fill [black] (0,0) rectangle (2,2);
                          \draw [->] (2.0,4.5) arc (135:90:4.0);
                         % \draw(7.6,2.1) circle(0.75);    -2, +3.5
                          \draw(5.6,5.6) circle(0.75);
                          \draw[->](5.6,5.3)--(5.15,6.05);
                          \draw[->](6.4,5.6)--(7.4,5.6);
                          \draw[color=black](13,5.6) node{\ldots 0.7, 0.3, 0.8, 0.2, 0.5 \ldots};
                         % \draw[->](11,0)--(12,1);
                         \draw[<-](9,3.5) arc(180:160:4.0);
                          \draw(9,2.75) circle(0.75);
                          \draw[->](9,2.45)--(8.55,3.2);
                          \draw[<-](9,1.2)--(9,2);
                          \draw[color=black](9,0.8) node{$\mu$};
                          \draw[<-](13,3.5) arc(-20:0:4.0);
                          \draw(13,2.75) circle(0.75);
                          \draw[->](13,2.45)--(12.55,3.2);
                          \draw[<-](13,1.2)--(13,2.0);
                          \draw[color=black](13,0.8) node{$\sigma$};
                          \end{scope}
                     \end{tikzpicture}
               % \begin{center}
                    \vspace{0.5cm}
                    
                     \begin{tikzpicture}[scale=1, line width=1.75pt,color=blue]
                          \draw(0,0) node{\includegraphics[width = 2.0in]{lorenz-attractor-2}};
                          \draw [->] (1.5,0.25) arc (135:90:4.75);
                          \draw(5.6,1.6) circle(0.75);
                          \draw[->](5.6,1.3)--(5.1,2.0);
                          \draw[->](6.4,1.6)--(7.3,1.6);
                          \draw[color=black](12.75,1.6) node{\ldots 0.7, 0.3, 0.8, 0.2, 0.5 \ldots};
                     \end{tikzpicture}
              %  \end{center}
     \end{itemize}
\subsectionhead{Some other history (e.g., catastrophe theory)}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Poincar\'e sections}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Entropy and information theory}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Diagnostics and control of chaotic systems}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Other discrete systems -- e.g.:}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{\ \ \ Cellular automata}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{\ \ \ Iterated function systems}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Complex adaptive systems}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{\ \ \ and complex systems more generally}
     \begin{itemize}
          \item \ldots
     \end{itemize}
\subsectionhead{Various other topics \ldots}
     \begin{itemize}
          \item \ldots
     \end{itemize}

\pagedone