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{\bf Supplementary problems (assigned 11/24/03)}
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\vspace*{.3in}

\begin{enumerate}

\item Solve the following initial value problems
\begin{enumerate}
\item $y' = \frac{\sin t}{y}$, $y(\pi/2) = 1$.
\item $(1+t^2)y' + 4ty = (1+t^2)^{-2}$, $y(1) = 0$.
\end{enumerate}
\vspace*{.2in}

\item Disillusioned about mathematics, you descend the ivory tower and
set yourself the task of becoming a millionaire by age 50.  The plan is
simple.  You will stash away money continuously from now til then at a 
constant rate of $R$\$ per year.  You figure that in your new life as day
trader, you can make a reliable 8\% annual interest (compounded continuously,
of course) on your savings.  Unfortunatly, college has left you broke
as well as disillusioned, so you're starting from nothing.  At what rate
$R$ will you need to be saving your money?

\vspace*{.2in}
\item Remember Newton's method?  The idea is that you have an open
set $U\subset\R$ and a $C^1$ function $f:U\to\R$.  You know that
$f(r) = 0$ for some point $r\in U$, and you have a decent initial guess
$x_0$ at the location of $r$.  Beginning with this guess, you then produce
a sequence of (hopefully better) approximations of $r$ by setting
$$
x_{n+1} = N(x_n)
$$
for every $n\in\N$, where $N(x) = x - f(x)/f'(x)$.  
	Now assume that $r$ is a \emph{non-degenerate} root of $f$---i.e.
that $f'(r) \neq 0$.  Prove the following.
\begin{itemize}
\item $r$ is a fixed point of $N$.
\item There exists $\delta>0$ such that $N(N_\delta(r)) \subset N_\delta(r)$
\item The (restricted) function $N:N_\delta(r) \to N_\delta(r)$ is a 
      contraction mapping.
\end{itemize}
(The mean value theorem will be useful in the second and third items.)
What can you conclude from all this about how well Newton's method works?

\vspace*{.2in}
\item
Suppose that $f,g:\R^2\to\R$ are continuous functions and that
$f(y,t) > g(y,t)$ for all points $(y,t)\in\R^2$.  Let $y_1, y_2:\R\to\R$ 
be functions satisfying 
$$
y_1' = f(y_1,t), \quad y_2' = g(y_2,t) 
$$
for all $t\in\R$.  Show that $y_1(t_0) = y_2(t_0)$ for at most one point
$t_0\in\R$.  

\vspace*{.2in}
\item Let $f:\R\to\R$ be a function.  The \emph{support} of $f$ is the
set 
$$
K := \overline{\{x\in\R:f(x) \neq 0\}}
$$
Show that if $f$ is real analytic, then $K$ cannot be compact.  Show by
giving an example that $K$ can be compact if $f$ is merely $C^\infty$.

\vspace*{.2in}
\item Redo supplementary problem 2 from the homework assigned on 11/3/03:

As a guide, here is the solution to problem 1 from the same assignment:
\vspace*{.1in}

\noindent{\bf Proof.}
Let $\epsilon > 0$ be given. Proving that the conclusion holds is equivalent
to constructing a partition $P$ of $[a,b]$ for which 
$$
U(P,f) - L(P,f) < \epsilon.
$$
Let $M$ be an upper bound for $|f|$ on $[a,b]$.
By hypothesis we can find mutually disjoint open intervals 
$I_j$, $j=1,\dots,n$ covering the set $S$ of discontinuities of $f$ such
that 
$$
|I_1| + \dots + |I_n| < \epsilon/2M
$$
Let us write $I_j = (a_j,b_j)$.  By putting the intervals in order (and
intersecting them with $[a,b]$, if necessary)
we can suppose that 
$$
a\leq a_1 < b_1 \leq a_2 < b_2\leq \dots \leq a_n < b_n\leq b.
$$
so that $Q = \{a,a_1,b_1,\dots,a_n,b_n,b\}$ is a (not very well labeled!)
partition of $[a,b]$.  

For convenience, let us define $b_0 = a$, $a_{n+1} = b$.
Then the condition $S\subset I_1\cup\dots\cup I_n$ means that $f$ is 
continuous on the closed intervals $[b_j,a_{j+1}]$ for $0\leq j\leq n$.
Therefore $f$ is integrable on each of these intervals, and we can choose 
a partition $P_j$ of $[b_j,a_{j+1}]$ such that 
$$
U(P_j,f) - L(P_j,f) < \frac{\epsilon}{2(n+1)}.
$$
Now we define our partition $P$ to be the union of the $P_j$, $j=1,\dots,n$
(note in particular that $Q\subset P$, since every point in $Q$ is the endpoint
of one the partitions $P_j$).  Then using the upper bound $M$ for $|f|$
chosen above, we can estimate
\begin{eqnarray*}
U(P,f) & \leq & U(P_0,f) + M(b_1-a_1) + U(P_1,f) + M(b_2-a_2) + \dots
	+ M(b_n - a_n) + U(P_n,f) \\
L(P,f) & \geq & L(P_0,f) - M(b_1-a_1) + L(P_1,f) - M(b_2-a_2) + \dots
	- M(b_n - a_n) + L(P_n,f).
\end{eqnarray*}
Therefore
$$
U(P,f) - L(P,f) \leq \sum_{j=0}^n U(P_j,f) - L(P_j,f) + 
	2M \sum_{k=1}^n (b_k-a_k) < (n+1)\frac{\epsilon}{2n+1} + 
	2M\frac{\epsilon}{4M} = \epsilon.
$$
\qed




\end{enumerate}
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