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\begin{document}
\section{First Test}
\begin{center}
{\em Mathematics 423: Test I: February 25, 1998} \end{center}

{

\noindent\large Name: }

{\large \em To receive credit you must show your work.} 

\begin{center}
\large\begin{tabular}{c|c|r}
Problem Number& Maximum Points & Points attained\\ \hline\hline 1&8\\ \hline
2&8 \\ \hline
3&16 \\ \hline
4&6 \\ \hline
5&6 \\ \hline
6&6 \\ \hline
7&6 \\ \hline
8&10 \\ \hline
9&10 \\ \hline
10&10 \\ \hline
11&6 \\ \hline
12&8\\ \hline
TOTAL&100\\
\end{tabular}
\end{center}
\section*{Some Useful Results}The following results may be of some use. You can assume them
in any argument you need to give.



\begin{theorem}[Neville's Recursion Formula]Let $x_0,\ldots,x_n$ be distinct real numbers in the interval, $[a,b]$. Let $f$ be a real valued function on $[a,b]$, and let $P(x)$ denote the
unique polynomial of degree $\le n$ such that $f(x_i)=P(x_i)$ for $i=0,\ldots,n$. For $c=0,\ldots,n$, let $P_{\widehat{c}}(x)$ denote the unique polynomial
of degree $\le n-1$ such that $P_{\widehat{c}}(x_i)=f(x_i)$ for $0\le i\le n$ with
$i\not= c$. Then for two distinct points $x_j$ and $x_k$ in the set $\{x_0,\ldots,x_n\}$:
$$P(x)=\frac{(x-x_j)P_{\widehat{j}}(x)-(x-x_k)P_{\widehat{k}}(x)}{x_k-x_j}$$ or equivalently:
$$P(x)=\frac{x-x_j}{x_k-x_j}P_{\widehat{j}}(x)+\frac{x-x_k}{x_j-x_k}P_{\wide hat{k}}(x).$$
\end{theorem}


\bigskip
\begin{theorem}[Hermite Interpolation Error Formula]\label{hermite}Let $x_0,\ldots,x_n$ be distinct real numbers in the interval, $[a,b]$. Let $f$ be an $2n+2$ times continuously differentiable function on $[a,b]$. Let $H(x)$ denote the
the unique polynomial of degree $\le 2n+1$ such that for $0\le i\le n$, $f(x_i)=H(x_i)$ and $f'(x_i)=H'(x_i)$. Then for each $x\in [a,b]$, there exists a point $\xi\in (a,b)$ such that
$$f(x)-H(x)=\frac{f^{(2n+2)}(\xi)}{(2n+2)!}(x-x_0)^2\cdots(x-x_n)^2.$$ \end{theorem}
\bigskip



\begin{theorem}[Euler Summation Formula]Let $f$ be an $2m+2$ times continuously differentiable function on $[a,b]$. Let $h:= \displaystyle\frac{b-a}{n}$ for some integer $n\ge 1$. Let $x_j:= a+jh$ for $j=0,\ldots,n$. Let $B_j$ denote the $j$-th Bernoulli number.
There exists $\xi\in (a,b)$ such that
\begin{eqnarray*}
\left(f(a)+\sum_{j=1}^{j=n-1}2f(x_j) +f(b)\right)\frac{h}{2} &=&\int_a^bf(x)dx+\sum_{j=1}^m\left[f^{(2j-1)}(b)-f^{(2j-1)}(a)\right]B_{2j} \frac{h^{2j}}{(2j)!}
\\ &+&f^{2m+2}(\xi)B_{2m+2}\frac{h^{2m+2}(b-a)}{(2m+2)!}. \end{eqnarray*}
\end{theorem}





\section*{Problems}In the following you must show your work. \bigskip

\begin{problem}[8 points total] Perform the following computations in 
a) 3 digit rounding arithmetic; and compute b) the absolute error, and
c) the relative error in each case:
\begin{enumerate}
\item $(121-0.3)-119$;
\item $(121-119)-0.3$.
\end{enumerate}
Put your answers in the table below.
\end{problem}

\vspace{0.1in}

\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
expression to &3 digit rounding answer&absolute error& relative error\\ be evaluated&{\em(2 points each)}&{\em(1 point each)}& {\em(1 point each)}\\ \hline\hline
{\rule[-2mm]{0mm}{7mm}$(121-0.3)-119$}&&&\\ \hline {\rule[-2mm]{0mm}{7mm}$(121-119)-0.3$}&&&\\ \hline \end{tabular}
\end{center}






\begin{problem}[8 points total] Perform the following computations in 
a) 3 digit chopping arithmetic; and compute b) the absolute error, and
c) the relative error in each case:
\begin{enumerate}
\item $(102-0.3)-0.7$;
\item $102-(0.3+0.7)$.
\end{enumerate}
Put your answers in the table below.
\end{problem}


\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
expression to &3 digit rounding answer&absolute error& relative error\\ be evaluated&{\bf(2 points each)}&{\bf(1 point each)}& {\bf(1 point each)}\\ \hline\hline
{\rule[-2mm]{0mm}{7mm}$(102-0.3)-0.7$}&&&\\ \hline {\rule[-2mm]{0mm}{7mm}$102-(0.3+0.7)$}&&&\\ \hline \end{tabular}
\end{center}





\begin{problem}[16 points total] A person wishes to find a zero of $f(x)=e^x-3x$ for $0\le x\le 1$. That person decides to use the bisection method to accomplish this. {\em Assume for simplicity that you are using exact arithmetic, i.e., that the errors of computer arithmetic play no role here.}
\begin{enumerate}
\item[\bf 5 pts:]State a theorem and show that it applies to guarantee 
that $f(x)$ has a zero on $[0,1]$.
\item[\bf 5 pts:]The first approximation to a solution by the bisection 
method is $0.5$ with $f(0.5)=e^{0.5}-3\cdot 0.5\approx 0.15$.
What is the second approximation to a solution by the bisection method?
\item[\bf 6 pts:]You would like to find an approximation to a zero of $f(x)$ 
on $[0,1]$ with an absolute error of no more than $0.001$. Using the bisection method as outlined in the previous part of this problem, and the error estimate for the bisection method,
which is the smallest integer $n$ for which you know that on the
$n$-th approximation you will be within $0.001$ of the correct
answer. An answer without justification by the error estimate
for the bisection method will receive no credit. \end{enumerate}
\end{problem}




\begin{problem}[6 points total] A person wishes to find a zero of $f(x)=e^x-3x$ for $0\le x\le 1$. That person decides to use the secant method
for finding a solution of the equation $f(x)=0$ on the interval, $[0,1]$. {\em Assume for simplicity that you are using exact arithmetic, i.e., that the
errors of computer arithmetic play no role here.} What is the first approximation to a solution in the open interval $(0,1)$ by the secant method? \end{problem}




\begin{problem}[6 points total] A person wishes to find a zero of $f(x)=e^x-3x$ for $0\le x\le 1$. That person decides to use Newton's method (also known as the Newton-Raphson method) for finding a solution of the equation $f(x)=0$ on the interval, $[0,1]$. {\em Assume for simplicity that you are using exact arithmetic, i.e., that the errors of computer arithmetic play no role here.} Write down iteration formula that Newton's method gives for solving $f(x)=0$, and using $1.0$ as a starting guess find the first approximation to a solution of $f(x)=0$ given by this formula.
\end{problem}









\begin{problem}[6 points total] Does $p_n=10^{-3^n}$ with $n=1,2,3,\ldots$ converge to zero of order $3$? To receive credit you must justify your answer, i.e., show it converges of order $3$ to zero or show why it does not converge of order $3$ to zero.
\end{problem}












\begin{problem}[6 points total] Write down the Lagrange form \hbox{\rm (}{\em not the Newton form}\hbox{\rm )} of the interpolating polynomial of degree $\le 2$ with the value $0$ at $x_0=0$, the value $2$ at $x_1=1$, and the value $3$ at $x_2=3$. \end{problem}








\begin{problem}[10 points total]Given the function $f(x)=x^3$:
\begin{enumerate}
\item[\bf 7 pts:]Compute the divided differences, 
$\displaystyle f[0],f[0,2],f[0,2,3];$
\item[\bf 3 pts:]Write down the interpolating polynomial, $p_2(x)$, of 
degree $\le 2$ for $f(x)$ with the node points $x_0=0$, $x_1=2$, $x_2=3$ using Newton's interpolatory 
divided-difference formula, i.e., by using the Newton polynomial built out of divided differences \hbox{\rm (} {\em not the Lagrange form}\hbox{\rm )}. \end{enumerate}
\end{problem}




\begin{problem}[10 points total]Neville's method is used to approximate $f(1)$ giving the following table:
$$
\begin{array}{llll}
x_0=0&P_0=0 \\
x_1=2&P_1=16 &P_{01}=8\\
x_2=3&P_2 &P_{12} &P_{123}=8 \\
%24,8
\end{array}
$$
Determine $P_2=f(3)$.
\end{problem}







\begin{problem}[10 points]Let $\displaystyle f(x)=\sin\left(\frac{\pi}{2} x\right)$.
Let $H_3(x)$ denote the unique polynomial of degree $\le3 $ such that $f(0)=H_3(0)$, $f'(0)=H_3'(0)$, $f(1)=H_3(1)$, $f'(1)=H_3'(1)$. \begin{enumerate}
\item[\bf 5 points:] Compute the coefficients $a_0, a_1, a_2, a_3$, of $H_3$ 
in the Newton divided difference form
$\displaystyle H_3(x) := a_0+a_1x+a_2x^2+a_3x^2(x-1).$ \item[\bf 5 points:] Use the error term in theorem (\ref{hermite}) to estimate $\displaystyle|H_3(0.5)-f(0.5)|$. 
{\em You may use the upper bound $\displaystyle\max_{[0,1]}|f^{(4)}(x)|\le 6.4$.}
\end{enumerate}
\end{problem}


\begin{problem}[6 points]Let $f(x)$ be a $2m+2$ times differentiable function
defined on the real line. Assume that $f(x)$ is {\em periodic} with period $4$,
i.e., $f(x+4)=f(x)$ for all real points $x$. Prove that the trapezoidal approximation on $[0,4]$ is an order $2m+2$ method, i.e., show that with $h=4/n$,
and $x_j=jh$ for $j=0,\ldots,n$:
$$
\left(f(0)+\sum_{j=1}^{j=n-1}2f(x_j) +f(4)\right)\frac{h}{2} 
=\int_0^4f(x)dx+O(h^{2m+2}).
$$
\end{problem}




\begin{problem}[8 points total]Consider
a four times continuously differentiable real valued function $f(x)$ on $[0,1]$.
Assume that the trapezoidal approximations for $n=1, 2$ are $2, 2.3$ respectively.
What is the Richardson extrapolation estimate to $\int_0^1f(x)dx$ that the Romberg method of integration makes using these trapezoidal approximations? \end{problem}

\setcounter{problem}{0}
\section{Second Test}
\begin{center}
{\em Mathematics 423: Test II: April 8, 1998 }
\end{center}

{

\noindent\large Name:
............................................................. 

}

{\large \em To receive credit you must show your work.} 

\begin{center}
\large\begin{tabular}{c|c|r}
Problem Number& Maximum Points & Points Lost\\ \hline\hline 1&10\\ \hline
2&10 \\ \hline
3&10 \\ \hline
4&15 \\ \hline
5&11 \\ \hline
6&12 \\ \hline
7&10 \\ \hline
8&8 \\ \hline
9&14 \\ \hline
TOTAL LOST&\\
\end{tabular}
\end{center}
\vspace{0.4in}

\noindent{\bf Runge-Kutta of Order Four} For the ordinary differential equation $y'=f(t,y)$ on $[a,b]$ with initial condition $y(a)=\alpha$ and stepsize $h$ we have
\begin{eqnarray*}
w_0&=&\alpha\\
k_1&=&hf\left(t_i,w_i\right)\\
k_2&=&hf\left(t_i+\frac{h}{2},w_i+\frac{k_1}{2}\right)\\ k_3&=&hf\left(t_i+\frac{h}{2},w_i+\frac{k_2}{2}\right)\\ k_4&=&hf\left(t_i+h,w_i+k_3\right)\\
w_{i+1}&=&w_i+\frac{1}{6}(k_1+2k_2+2k_3+k_4) \end{eqnarray*}




\section*{Problems}In the following you must show your work. \bigskip






\begin{problem}[{\bf 10 points}]\label{same}\ \ \ \begin{enumerate}
\item Use Euler's method to approximate the solution of the following initial value problem:
$$y'=ty; \ 1\le t\le 2;\ y(1)=\sqrt{e}, \mbox{with\ } h=1.0.$$ 
\item The actual solution of the above differential equation is $\displaystyle y(t)= e^\frac{t^2}{2}$.
Compute the absolute error of the approximate solution at $\displaystyle t =2.0$ found in part 1 of this problem.
\end{enumerate}\end{problem}


\begin{problem}[{\bf 10 points}]\ \ \
\begin{enumerate}
\item Use Taylor's method of order two to approximate the solution of the initial
value problem as in Problem \ref{same}:
$$y'=ty; \ 1\le t\le 2.0;\ y(1)=\sqrt{e}, \mbox{with\ } h=1.0.$$ \item The actual solution of the above differential equation is $\displaystyle y(t)= e^\frac{t^2}{2}$.
Compute the absolute error of the approximate solution at $\displaystyle t =2.0$ found in part 1 of this problem.
\end{enumerate}
\end{problem}



\begin{problem}[{\bf 10 points}]\ \ \
\begin{enumerate}
\item{\bf 5 points:} Use the Runge-Kutta method of order four to approximate the solution of the initial
value problem as in Problem \ref{same}:
$$y'=ty; \ 1\le t\le 2.0;\ y(1)=\sqrt{e}, \mbox{with\ } h=1.0.$$ (See page one for some useful formulae!) 
\item{\bf 5 points:} The actual solution of the above differential equation is $\displaystyle y(t)= e^\frac{t^2}{2}$. Compute the absolute error of the approximate solution at $\displaystyle t =2.0$ found in part 1 of this problem.
\end{enumerate}
\end{problem}






\begin{problem}[{\bf 15 points}] Consider the difference equation: $$y_n=7y_{n-1}-10y_{n-2}+4.$$
\begin{enumerate}
\item {\bf 10 points:} Solve the above difference equation, i.e., 
find $y_n$ for $n\ge 0$, subject to the initial conditions $y_0=1$, $y_1=4$. \item {\bf 5 points:} Find $\displaystyle\lim_{n\to \infty}\frac{y_{n+1}}{y_n}$.
\end{enumerate}
\end{problem}









\begin{problem}[{\bf 12 points}] Let $v=(2,2,4)$ be the vector in $\reals^3$. 
Compute $||v||_1$, $||v||_2$, and $||v||_\infty$. \end{problem}




\begin{problem}[{\bf 11 points}]
Consider the solution $y(t)$ of the initial value problem: $$y''-xy'+4y=0;\ \ -1\le x\le 1$$
with initial values $y(-1)=-1,y'(-1)=2$. 
Rewrite the ordinary differential equation as a $1$st order system of ordinary
differential equations.
\end{problem}






\begin{problem}[{\bf 8 points}] Let
$\displaystyle <f(x),g(x)> = \int_{-1}^1\frac{f(x)g(x)}{\sqrt{1-x^2}}dx$ 
be an inner product
on $V_5$, the vector space of polynomials of degree $\le5$ on $[-1,1]$. It is known that the
Chebychev polynomials
$$T_0(x), T_1(x), T_2(x), T_3(x), T_4(x), T_5(x)$$ form an orthogonal basis of $V_5$ relative to this inner product. Moreover it is known that
$\displaystyle <T_n(x),T_n(x)>=\frac{\pi}{2}$ for $n>0$ and $\displaystyle <T_0(x),T_0(x)>=\pi$ .

Assume that $f(x)=T_1(x)+4T_2(x)-4T_3(x)+T_4(x)$. Find $<f(x),f(x)>$.
\end{problem}


\begin{problem}[{\bf 10 points}] Let $\displaystyle ||g(x)|| = \int_0^1g(x)^2dx$ be the norm
associated to the inner product
$\displaystyle <f(x),g(x)> = \int_0^1f(x)g(x)dx$ on the polynomials on $[0,1]$.
\begin{enumerate}
\item {\bf 7 points:} Find the linear least squares polynomial approximation to $f(x)=x^2+2$
on $[0,1]$, i.e.,
find the polynomial $h(x)$ of degree $\le 1$ such that $||h(x)-f(x)||$ is minimum.
\item {\bf 3 points:} What is $\displaystyle\int_0^1(h(x)-f(x))(3x+7)dx$? If you do this without a computation
explain why you know the number you wrote down is the answer. \end{enumerate}
\end{problem}







\begin{problem}[{\bf 14 points}] Consider the problem of finding a solution of $f(x)=0$ where $f$ is
a system of polynomials in $x=(x_1,x_2)\in \reals^2$ $$
f(x) =\left(
\begin{array}{c}
2x_1+x_2+x_2^2-x_1x_2-1\\
x_1^2-2x_1+2x_2\\
\end{array}\right)
$$
\begin{enumerate}
\item {\bf 7 points:} Compute the Jacobian of $f$. 
\item {\bf 7 points:} Use Newton's method with $\displaystyle x^0=\left[
\begin{array}{c}
1\\
0\\
\end{array}\right]$ to compute the first approximation $x^1$ to a solution of $f(x)=0$.
\end{enumerate}
\end{problem}


\setcounter{problem}{0}

\section{Final}
\begin{center}
{\em Mathematics 423: Final Exam: May 7, 1998 }
\end{center}




\noindent{\large Name:
$\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$}
\begin{center}
\large\begin{tabular}{r|c|r}
Problem Number& Maximum Points & Points Lost\\ \hline\hline 1&12\\ \hline
2&12 \\ \hline
3&12 \\ \hline
4&12 \\ \hline
5&12 \\ \hline
6&12 \\ \hline
7&12 \\ \hline
8&12 \\ \hline
9&12 \\ \hline
10&10 \\ \hline
11&10 \\ \hline
12&10 \\ \hline
13&12 \\ \hline
TOTAL&\\
\end{tabular}
\end{center}






\vspace{0.2in}


{\bf Neville's Recursion Formula:}\ Let
$x_0,\ldots,x_n$ be distinct real numbers in the interval, $[a,b]$. Let $f$ be a real valued function on $[a,b]$, and let $P(x)$ denote the
unique polynomial of degree $\le n$ such that $f(x_i)=P(x_i)$ for $i=0,\ldots,n$.
For each $c=0,\ldots,n$, let $P_{\widehat{c}}(x)$ denote the unique polynomial of degree $\le n-1$ such that $P_{\widehat{c}}(x_i)=f(x_i)$ for $0\le i\le n$ with
$i \not= c$. Then for two distinct points $x_j$ and $x_k$ in the set $\{x_0,\ldots,x_n\}$:
$$P(x)=\frac{(x-x_j)P_{\widehat{j}}(x)-(x-x_k)P_{\widehat{k}}(x)}{x_k-x_j}$$ or equivalently:
$$P(x)=\frac{x-x_j}{x_k-x_j}P_{\widehat{j}}(x)+\frac{x-x_k}{x_j-x_k}P_{\wide hat{k}}(x).$$




\section*{Problems}{\large\em To receive credit you must show your work.} 

\bigskip

\begin{problem}[12 points total] Perform the following computations in 
a) 3 digit rounding arithmetic; and compute b) the absolute error, and
c) the relative error in each case:
\begin{enumerate}
\item $(100-0.6)+0.6$;
\item $(99.8+0.6)-0.6$.
\end{enumerate}
Put your answers in the table below.
\end{problem}

\vspace{0.1in}

\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
expression to &3 digit rounding answer&absolute error& relative error\\ be evaluated&{\em(2 points each)}&{\em(2 points each)}& {\em(2 points each)}\\ \hline\hline
{\rule[-2mm]{0mm}{7mm}$(100-0.6)+0.6$}&&&\\ \hline {\rule[-2mm]{0mm}{7mm}$(99.8+0.6)-0.6$}&&&\\ \hline \end{tabular}
\end{center}




\begin{problem}[12 points total] A person decides to use Newton's method (also known as the Newton-Raphson method) to find a zero of $f(x)=x^2-2$ for $1\le x\le 2$. {\em Assume that you are using exact arithmetic,
i.e., that the errors
of computer arithmetic play no role here.} Write down iteration formula that Newton's method gives for solving $f(x)=0$, and using $2.0$ as a starting guess find the first and the second approximation to a solution of $f(x)=0$
given by this formula.
\end{problem}





\begin{problem}[12 points total]Given a
function $f: \reals\to \reals$ such that $f(0)=0$, $f(1)=0$, $f(2)=4$, $f(3)=0$, write down in Lagrange form {\em (not the Newton form)}
of the
interpolating polynomial of degree $\le 3$ with values agreeing with $f$ at $x=0,1,2,3$.
\end{problem}






\begin{problem}[12 points total]Given a
function $f: \reals\to \reals$ such that $f(0)=0$, $f(1)=0$, $f'(1)=2$, $f(2)=4$: \begin{enumerate}
\item[\em 8 pts]Write down the divided differences, $f[0],f[0,1],f[0,1,1],f[0,1,1,2]$;
\item[\em 4 pts]Write down the Hermite interpolating polynomial of degree $\le 3$ for
$f$ at the values $0,1,1,2$ using Newton's interpolatory divided-difference formula, i.e., by using the Newton polynomial built out of divided differences.
\end{enumerate}
\end{problem}



\begin{problem}[12 points total]Let $S(x)$ be a free cubic spline defined on $[0,2]$ with nodes $x_0=0,x_1=1, x_2=2$, and $S(0)=0$, $S(1)=0$ and
$S(2)=4$.
Assume that
$S(x)$ is given by the cubic polynomials $S_0(x)=
a-x+x^3$ on $[0,1]$, and by $S_1(x)=
b(x-1)+c(x-1)^2-(x-1)^3$ on $[1,2]$.
Find the real numbers $a,b,c$, i.e.,
find the actual numbers. {\em [Recall the free splines satisfy the conditions that
$S_0''(0)=0=S_1''(2)$.]}
%0,2,3
\end{problem}




\begin{problem}[12 points total]Neville's method is used to approximate $f(2)$ giving the following table:
$$
\begin{array}{llll}
x_0=0&P_0=f(0)=1 \\
x_1=1&P_1=f(1)=2 &P_{01}=3\\
x_2=3&P_2=f(3)=\ ? &P_{12}= 15&P_{012}=11 \\ %28
\end{array}
$$
Determine $P_2=f(3)$.
\end{problem}






\begin{problem}[12 points total]You would like to integrate $\displaystyle\int_1^2[\ln(x)]^2\mbox{\rm d}x$. The trapezoidal approximation for $n=1$ is $0.240$, and for $n=2$ it is $0.202$.
What is the Richardson extrapolation estimate to $\displaystyle\int_1^2[\ln(x)]^2\mbox{\rm d}x$ that the Romberg method of integration makes using these trapezoidal approximations? \end{problem}


\begin{problem}[12 points]\ \ \
\begin{enumerate}
\item Set up the Taylor method of order $2$ to solve $y'=y^2t^2$ for $t \in [0,1]$ with $h=0.25$, $y(0)=1$.
\item Set up the Euler method to solve the same initial value problem and with $h=0.25$.
\end{enumerate}
\end{problem}







\begin{problem}[12 points]
Consider the initial value problem:
$$y'''+2y''+4y'+8y=t^4;\ \ 0\le t\le 3$$ with initial values $y(0)=1,y'(0)=2$, $y''(0)=0$. Rewrite the ordinary differential equation as a $1$st order system of ordinary differential equations. {\em [Do not forget to write down the initial conditions
for the system.]}
\end{problem}




\begin{problem}[10 points]
Give an argument based on the Gershgorin Theorem to show that the matrix 
$\displaystyle \left[\begin{array}{ccccc} 5&1&1&1&1\\
1&5&1&1&1\\
1&1&5&1&1\\
1&1&1&5&1\\
1&1&1&1&5\\
\end{array}
\right]
$ is invertible. {\em Hint: You can use the fact that a matrix $A$ is invertible if and only if $0$ is not an eigenvalue of $A$.} \end{problem}


\begin{problem}[10 points]
What is the infinity norm $||A||_\infty$ of the matrix $$A=\left[\begin{array}{ccc}
-10&1&2\\
1&0&2\\
4&2&2\end{array}
\right]
$$
\end{problem}

\begin{problem}[10 points]
What is the condition number (using the infinity norm) of the matrix $$\left[\begin{array}{cc}
4&1\\
0&0.25\\
\end{array}
\right]
$$
\end{problem}



\begin{problem}[12 points]
Consider the boundary value problem:
$$y''=y+x;\ \ 0\le x\le 2$$
with boundary values $y(0)=1,y(2)=2$.
Solve this equation using finite differences with the node points $x_0=0,x_1=1,x_2=2$.
\end{problem}



\end{document}
