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Consider the fixed-point problem
\[
   x_{n+1} = F(x),
\]
where 
\[
   F(x) = \left(\begin{array}{c}
              (x)_1^2 - 3(x)_1(x)_2 + 50(x)_2 \\
              (x)_1^2 + (x)_2^2
          \end{array}\right)
\]
and $x_0 = \bigl((x_0)_1, (x_0)_2\bigr)^T = (0,1/100)^T$.
Clearly, $(0,0)^T$ is a fixed-point for $F$.

The iterations are exemplied in a Matlab script available at
\begin{verbatim}
     http://www-math.bgsu.edu/~gwade/ma668/wadecode/superlinear.m
\end{verbatim}
From that script, the first six iterates produce
\begin{center}
\begin{tabular}{c|c}
    n & $\Vert x_{n+1}\Vert_2/\Vert x_n\Vert_2$ \\ \hline 
    0 & 50.0000 \\
    1 &  0.0071 \\
    2 & 35.0036 \\
    3 &  0.0018 \\
    4 & 34.6415 \\
    5 &  0.0001
\end{tabular}
\end{center}

{\bf Question:} How can we interpret this result in terms of 
 our theory of fixed-point iterations?

%% In case you're just learning LaTeX, here is how you'd write $F'(x)$
%% (of course you'll have to un-comment this line):
%%\[
%%   F'(x) = \left(\begin{array}{cc}
%%               2(x)_1 - 3(x)_2 & -3(x)_1 + 50 \\ 
%%               x(x)_1          & 2(x)_2 
%%           \end{array}\right)
%%\]

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