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{\bf Supplementary problems (assigned 11/24/03)}
\end{center}
\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 $N(x) = x - f(x)/f'(x)$.  Assuming that the root
$r$ of $f$ is non-degenerate (meaning $f'(r) \neq 0$), show that there exists 
$\delta > 0$
such that
\begin{itemize}
\item $x \in (r-\delta,r+\delta)$ implies that $N(x) \in (r-\delta,r+\delta)$;
\item the restriction of $N$ to the interval $(x-\delta,x+\delta)$ is a 
contraction mapping;
\item if $x_0\in (r-\delta,r+\delta)$, then $\lim_{n\to\infty} x_n = r$.
\end{itemize}
The main point here is the second one.  The other two follow fairly quickly
once you figure this one out.  
To prove the second item, you'll need the Mean Value Theorem again.
Also note that you'll definitely need the continuity of $f'$ here.

\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
$t_0\in\R$.  If such a point $t_0$ exists, what must be the 
relationship between $y_1(t)$ and $y_2(t)$ for points $t>t_0$?  $t<t_0$?

\vspace*{.2in}
\item Let $y = C(t)$ be the unique solution of the initial value problem
$$
y'' + y = 0, \qquad y(0) = 1, y'(0) = 0.
$$
It can be shown (i.e. you can assume) that $C$ is well-defined for all 
$t\in\R$.  Let $S(t) = -C'(t)$.  Prove the following about $C$ and $S$.  
\begin{enumerate}
\item $S'(t) = C(t)$ for all $t\in\R$.
\item $S$ satisfies the same differential equation as $C$ but with 
initial values $S(0) = 0$, $S'(0) = 1$.
\item $S^2(t) + C^2(t) = 1$ for all $t\in\R$.  In particular, the 
ranges of $S$ and $C$ are contained in the interval $[-1,1]$.
\item $C$ is an even function and $S$ is odd (i.e. $C(-t) = C(t)$
and $S(-t) =-S(t)$ for all $t\in \R$.
\item There exists a number $P/2 > 0$ such that $C(P/2) = 0$ but 
$C(t) > 0$ for all $t\in [0,P/2)$.  
\item $S(P/2) = 0$.  
\item $C(t + a) = C(t)C(a) - S(t)S(a)$ and $S(t+a) = S(t)C(a) + S(a)C(t)$
for all $t,a\in\R$.
\item $S$ and $C$ are periodic functions with (minimal) period $2P$.
\item The ranges of $S$ and $C$ are exactly equal to $[-1,1]$. 
\end{enumerate}

\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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