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\begin{document}

\tutprobn{12}{10/22/99}{October 21, 1999}

\begin{problems}

\problem Is a Harvard degree really worth more than an MIT degree?!  Let
us say that a person with a Harvard degree starts with \$40,000 and gets a
\$20,000 raise every year after graduation, whereas a person with an MIT
degree starts with \$30,000, but gets a 20\% raise every year.  Assume
inflation is a fixed $8\%$ every year.  That is, \$1.08 a year from now is
worth \$1.00 today.  (You'll need a calculator to get final answers; if
one is not available, it's ok to express the answer as a closed form
numerical expression.)

\bparts


\ppart How much is a Harvard degree worth today if the holder will work
for $n$ years following graduation? \hint In Lecture 8 a closed formula
for $\sum_{i=0}^{n} ir^i$ was derived.

\ppart How much is an MIT degree worth in this case?

\ppart If you plan to retire after twenty years, which degree would
be worth more?

\solution{
One dollar after year $i$ is worth $r^i$ in today's currency, where
\[
r= \frac{1}{1.08} = 0.925\, 925\, 925 \dots
\]
So 
\begin{eqnarray*}
\mbox{Hvd}_n & = & \sum_{i=0}^{n} (40000+ 20000i)r^i\\
        & = & 40000\sum_{i=0}^{n} r^i + 20000\sum_{i=0}^{n} i r^i,\\
\mbox{MIT}_n & = & 30000\sum_{i=0}^{n} {1.2}^i r^i\\
        & = & 30000\sum_{i=0}^{n} (1.2r)^i
\end{eqnarray*}

But
\[
\sum_{i=0}^{n} ir^i = \frac{{r - (n+1) r^{n+1} + n r^{n+2}}}{(1-r)^2},
\]
so 
\begin{eqnarray*}
\mbox{Hvd}_n & = & 40000 \frac{(1- r^{n+1})}{1- r}  + 
                   20000 \frac{(r - (n+1) r^{n+1} + n r^{n+2})}{(1-r)^2}\\
  & = &  \frac{20000(2(1- r^{n+1} -r +r^{n+2}) + r - (n+1) r^{n+1} + n r^{n+2})}{(1-r)^2}\\
  & = &  \frac{{20000(2 - 2r - (n-1)r^{n+1} + (n + 2)r^{n+2})}}{(1-r)^2}\\
\mbox{MIT}_n & = &  \frac{{30000(1 - (1.2r)^{n+1})}}{1 - r}
\end{eqnarray*}
and for $n=20$,
\begin{eqnarray*}
\mbox{Hvd}_20 & = &  \frac{{20000(2 - 2r - 19r^{21} + 22r^{22})}}{(1-r)^2} = 1,532,285,\\
\mbox{MIT}_21 & = &  \frac{{30000(1 - (1.2r)^{21})}}{1 - r} = 3,296,368.
\end{eqnarray*}

so the MIT degree is more than twice as valuable!}

\eparts


\problem

Regrouping or reordering the terms of an infinite series is ok for
\emph{absolutely convergent} series, namely, series for which the sum of
the absolute values of their terms converges.

In other cases, rearranging terms is not ok.
An extreme example is
\[
S \eqdef \sum_{i=0}^{\infty} (-1)^i = 1 + (-1) + 1 + (-1) + \cdots
\]
because by regrouping the terms we can deduce:
\begin{eqnarray*}
S & = & [ 1 + (-1) ] + [ 1 + (-1) ] + \cdots = 0 + 0 + \cdots = 0,\\
S & = & 1 + [(-1) + 1] + [(-1) + 1] + \cdots = 1 + 0 + 0 + \cdots = 1.
\end{eqnarray*}

To see the problem here, let $S_n \eqdef \sum_{i=0}^{n}(-1)^i$.  Then $S_n
= 1$ for odd $n$ and $S_n = 0$ for even $n$.  Hence $S_n$ does not have a
well-defined limit as $n$ tends to infinity, implying that $S$ is not a
convergent series.

Even for convergent, but not absolutely convergent, series, rearranging
terms can cause big changes.  For example, the series
\[
1 - 1/2 + 1/3 - 1/4 \dots
\]
converges, but is not absolutely convergent.

Prove that by reordering terms in $1 - 1/2 + 1/3 - 1/4 \dots$, you can
obtain three series which converge to 3, converge to 4, and don't converge.

\solution{TO BE INSERTED.}

\problem
There is a number $a$ such that $\sum_{i=1}^\infty i^p$ converges iff $p < a$.

\bparts

\ppart What is the value of $a$?

\solution{$-1$}

\ppart Prove it.

\solution{For $p=-1$, the sum is the harmonic series which we know does
not converge.  Since the term $i^p$ is increasing in $p$ for $i>1$, the
sum will be larger, and hence also diverge for $p>-1$.

By the integral method, the sum is bounded by the integral from 1 to
$\infty$ of $x^p$.  For $p<-1$, the indefinite integral is
$x^{p+1}/(p+1)$, so the integral evaluates to $1/(p+1)$.  Hence the sum
is bounded above, and since it is increasing, it has a finite limit, i.e.,
it converges.}

\eparts

\iffalse
\problem We begin with two large glasses.  The first glass contains a pint
of water, and the second contains a pint of wine.  We pour some fixed
fraction, $f$, of a pint from the first glass into the second, stir up the
wine/water mixture in the second glass, and then pour the same fraction
$f$ of a pint of the mix back into the first glass.  We now repeat this
pouring back-and-forth process a total of $n$ times.

\bparts

\ppart
Describe a closed form formula for the amount of wine in the first glass
after $n$ back-and-forth pourings.

\ppart
What is the limit of the amount of wine in each glass as $n$ approaches infinity?

\solution{
\emph{Solution with $f = 1/3$}: The state of the system of glasses/wine/water at the
beginning of a round of pouring and pouring back is determined by the
total amount of wine in the first glass.  Initially, the first glass
contains one pint of water and no wine.  Suppose the first glass contains
$w$ pints of wine, $0\leq w \leq 1$.  Since we're at the beginning of a
round, it contains one pint of liquid, and therefore contains $1-w$ pints
of water.  The second glass contains the rest of the wine and water.

Pouring 1/3 pint from first glass to second leaves 2/3 pint of liquid and
$(2/3)w$ wine in the first glass, and 4/3 pints liquid and $1 - (2/3)w$ wine
in the second glass.  Pouring 1/3 pint back from second into first
transfers a proportion of (1/3)/(4/3) of the wine in the second glass into
the first.  So the round completes with both glasses containing a pint of
liquid, and the first glass containing
\[(2/3)w + (1/4)(1- (2/3)w) = 1/4 + w/2\]
pints of wine.  After the second round, the first glass contains
\[1/4 + (1/4 + w/2)/2 = 1/4 + 1/8 + w/2^2\]
pints of wine, and after the $n$th round
\[
\begin{array}{lll}
w/2^n + \Sigma_{i=1}^n (1/2)^{i+1} &=  w/2^n + (1/2)\Sigma_{i=1}^n
(1/2)^{i}\\
& = w/2^n + (1/2)(-1 + \Sigma_{i=0}^n (1/2)^{i})\\
& = w/2^n + (1/2)(-1 + (1 - (1/2)^{n+1})/(1 - 1/2))\\
& = w/2^n - 1/2 + 1 - (1/2)^{n+1}\\
& = w/2^n + 1/2 - (1/2)^{n+1}.
\end{array}
\]
So when $w=0$ initially, the pints of wine in the first glass is
\[1/2 - (1/2)^{n+1}.\]
You would expect after a thorough mixing that the glasses would contain
the same amount of wine, viz. 1/2 pint.  Indeed, the limiting amount of
wine in the first glass approaches 1/2 from below as n approaches
infinity.  In fact, it approaches 1/2 no matter how the wine was initially
distributed.
}

\eparts
\fi



\end{problems}

\end{document}

