\item
%% Section 2.4: Problems 8, 9, 10, 12, 14. % add 7 next time. %p85
   Section 2.4: Problems 7(a), 8, 9, 10, 12, 13, 14. % add 7 next time. %p85

Remark: Problem 13 can be derived 
by way of quadratic approximation of $x=f^{-1}(y)$
at $(y,x) = (f(x_n), x_n)$ (as shown in the slides for the linear
approximation).

% \item
% Use the code for Algorithm 2.2 to check if the standard fixed point iteration 
% $x_{n+1} = g(x_n) = 2 \cos(x_n)$ converge
% to root of $f(x) = x-2\cos(x) = 0$. 
% If not, can you explain why? 
% Can you find a constant $\beta$ such that
% the modified fixed point iteration
% $x_{n+1} = \beta x_n + (1-\beta) g(x_n) $ converges 
% (at least locally)?

\item
% Section 2.5: Problems 12(a), 14, 15, 16. %p91
Section 2.5: Problems 12(a), 14, 16. %89

\item
In Example 1 of section 2.5, the condition in Theorem 2.14 is not
satisfied. Nevertheless, one could still get faster convergence of
$\hat p_n$ than $p_n$, but of the same order. 
Analyze Aitken's $\Delta^2$ method to evaluate 
$\displaystyle{\lim_{n\to\infty}{\hat p_{n}-p \over p_{n+2} - p}}$.
Then verify your result numerically. 

\end{enumerate}
\end{document}

% next time
% \item
% Another way of computing $\pi$ is given by the
% Wallis formula
% $$
% {\pi \over 2} = \prod_{n=1}^\infty \left( {(2n)^2 \over (2n-1)(2n+1) } \right).
% $$
% There are two ways of computing the $N$-term approximation,
% namely
% $$
% {\pi_N \over 2}= 
% \left( {2\cdot 2 \over 1 \cdot 3} \right)
% \left( {4\cdot 4 \over 3 \cdot 5} \right)
% \left( {6\cdot 6 \over 5 \cdot 7} \right) \cdots
%%  \left( {2N\cdot 2N \over (2N-1) \cdot (2N+1)} \right) 
% \quad \mbox{or }
% { 4^N (N!)^2
% \over 
% \left( \displaystyle{ \prod_{n=1}^N} (2n-1) \right)
% \cdot \left( \displaystyle{ \prod_{n=1}^N} (2n+1) \right)
% }
% $$
% Which one is better in terms of the overflow issue?
% What is the rate of convergence?
% 
% Hint: The rate of convergence could be analyzed analytically
% using Stirling's formula. This is not required in this problem,
% but you can use that to help you check
% your 'numerical rate of convergence'. 
% To find out the rate of convergence
% numerically, it suffices to plot
% $|\pi - \pi_N |$ as a function of $N$.
% For example,
% the two methods in problem 4 of section 1.3 give algebraic
% rate $( N^{-k})$ and exponential rate $( q^{-N})$, respectively.
% The difference can easily be observed by plotting the results.
% The former appears as a straight line in the `loglog' plot
% on matlab and the latter appears as a straight line in 'semilogy' plot.
%