\section{(Difference Equations)}
We consider the LMM
\[y_{n+2} + 4y_{n+1} - 5y_{n} = h(4f_{n+1} + 2f_{n})\]
applied to the IVP
\[\begin{array}{ll} y' = 0,&y(0)=1 \end{array}\]
So we look at the difference equation
\[\sum_{j=0}^{k} = \alpha_{j} y_{n+j} = 0\]
where the solutions for the difference equation are
\[\rho(r) = \sum_{j=0}^{k} \alpha_{j}r^{j} = r^{2} + 4r - 5 = 0\]
So, the roots of the difference equation are $r = 1,-5$ and the general solution of the difference equation is
\[y_{n} = \alpha + \beta (-5)^{n}\]
Finding the particular solution for the starting values $y_{0} = 1$ and $y_{1} = 1 + h$, we get
\begin{eqnarray*}
	\alpha + \beta & = & 1 \\
	\alpha - 5\beta & = & 1 + h
\end{eqnarray*}
which gives us the values $\alpha = 1 + h/6$ and $\beta = -h/6$.  To show this diverges we take the limit
\[\lim_{h \rightarrow 0} (1 + \frac{h}{6}) - \frac{h}{6}(-5)^{1/h}\]
Where $n = 1/h$.  Then we see
\[\lim_{n \rightarrow \infty} (1 + \frac{1}{n6}) - \frac{1}{n6}(-5)^{n} = \lim_{n \rightarrow \infty} - \frac{(-5)^{n}}{n6}\]
Since this is an indeterminate form, we use L'Hopital's rule
\[\lim_{n \rightarrow \infty} -\frac{(-5)^{n} \ln{5}}{6}\]
This clearly diverges.  If we use the starting strategy of $y_{0} = 1$ and $y_{1} = 1 + h(-5)^{-n}$ we get a convergent method.  We can look at the effect on round off error by looking at the local trucation error.  For example, we see that 
\[C_{0} = \rho(1) = 0\]
As it should be, but
\[C_{1} = \sum_{j=1}^{k} j\alpha_{j} = 1 + 2(1 + h(-5)^{-n}) = 3 + 2\frac{h}{(-5)^{-n}}\]
And we see that as $h \rightarrow 0$, we get that $C_{1} = 3$.
