\sectionhead{Discrete Linear Systems}
\begin{itemize}
     \item The systems we have been looking at all have time ($t$) as a continuous variable (and we have been
          limiting ourselves to smooth, i.e., differentiable, systems).
          
          Now I'd like to move to {\em discrete} systems, where time proceeds in steps rather than continuously.  In
          other words, instead of looking at the variable $x(t)$, we will now take time to only take integral values,
          and we will be interested in systems with values $x_0, x_1, x_2, \dots$
          
          Instead of {\em differential} equations, we will be working with {\em difference} equations.  The general
          form of a (one variable) difference equation is
               $$x_{n+1} = f(x_n, x_{n-1}, \ldots, x_0, c).$$
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          A very simple example is the Fibonacci sequence.  This is given by the difference equation
               $$x_{n+2} = x_{n+1} + x_{n}.$$
          This is a second order difference equation.  Once we specify initial conditions (e.g., $x_0 = 0$, $x_1 = 1$),
          we have the solution sequence $0, 1, 1, 2, 3, 5, 8, 13, 21, \ldots$
          
          We can solve linear difference equations for the general solution for the $n$th term using a procedure
          similar to the approach for linear differential equations.  For the Fibonacci sequence, we rewrite the equation
          as
               $$x_{n+2} - x_{n+1} - x_{n} = 0,$$
          and then work with the characteristic polynomial
               $$r^2 - r - 1 = 0.$$
          From this, we get
               $$r = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}.$$
          The general form of the solution is then
               $$x_n = a_1\left(\frac{1 + \sqrt{5}}{2}\right)^n + a_2\left(\frac{1 - \sqrt{5}}{2}\right)^n.$$
          Once we specify the initial conditions, we can solve for $a_1$ and $a_2$.  For example,
          when $x_0 = 0$ and $x_1 = 1$, we get
               $$ a_1*1 + a_2*1 = 0 $$
               $$ a_1*\left(\frac{1 + \sqrt{5}}{2}\right) + a_2*\left(\frac{1 - \sqrt{5}}{2}\right) = 1,$$
          from which $a_2 =  - a_1$, and so
                $$ a_1*\left(\frac{1 + \sqrt{5}}{2}\right) - a_1*\left(\frac{1 - \sqrt{5}}{2}\right) = 1,$$
          and hence
               $$ a_1= \frac{1}{\sqrt{5}}$$
               $$ a_2 = - \frac{1}{\sqrt{5}}$$
          and so
               $$ x_n = \frac{1}{\sqrt{5}}*\left(\frac{1 + \sqrt{5}}{2}\right)^n
                            - \frac{1}{\sqrt{5}}*\left(\frac{1 - \sqrt{5}}{2}\right)^n .$$
                            
          This is a very general approach (with some minor subtleties when dealing with repeated roots) -- and, of
          course, there are the problems of finding the roots for higher order polynomials, but this basically gets us
          going on general linear difference equations.
          
\end{itemize}
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