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\begin{document}

\begin{problem}\textbf{Asymtotics Thugs 'n Harmony}

Prove that
\[
2\ln(n^3)+2 \sim \sum_{i=1}^{n} 6/i.
\]

\solution{
The left-hand side is $\sim 6\ln n$, while the right-hand side is $6 H_n$;
but we know that $H_n\sim\ln n$, and hence $6\ln n\sim 6H_n$, as well.
}

\end{problem}

\begin{problem} \textbf{Chicken?!}

Being a stubborn MIT student, Zardosht still won't admit that Euler is
a better Mathematician than Gauss.  Zardosht, however, learned from the
last game of chicken and decides to run.  He actually runs at a good clip,
10 miles per hour. Paul, being a good sport, gives Zardosht a 5 hour head
start before he chases him down on his bike that still goes a trusty 20
miles per hour. Eric, concerned for Zardosht's safety, decides to mediate
the race in the same way as with the game of chicken. He starts at Paul,
goes to Zardosht, and bounces back and forth until Paul catches Zardosht.
Again, Eric travels at 40 miles per hour on his speedy moped. 

This problem will deal with finding an infinite series describing the 
total distance that Eric travels and determine this distance by evaluating 
the series.

\begin{problemparts}
\problempart
If Eric starts at Paul and Zardosht is distance $d$ away, how long does
it take him to catch Zardosht? 
\solution{
Eric closes on Zardosht at a rate of 30 MPH, and catches him in $\frac{d}{30}$
hours. 
}
\problempart
If Eric starts at Zardosht and Paul is distance $p$ away, how long
does it take him to catch Paul?
\solution{
Eric will catch up to Paul at a rate of 50 MPH, and catch him in
$\frac{p}{50}$ hours.
}
\problempart
In the time it takes Eric to catch Zardosht from Paul, how much
closer will Paul and Zardosht be? How far has Eric traveled?
\solution{
Paul and Zardosht close at 10 MPH, they will be $\frac{d}{3}$ miles
closer. Eric will have traveled $\frac{4*d}{3}$ miles.
}
\problempart{
In the time it takes Eric to catch Paul from Zardosht, how much
closer will Paul and Zardosht be? How far has Eric traveled?
}
\solution{
They will be $\frac{d}{5}$ miles closer. Eric has traveled
$\frac{4*d}{5}$ miles.
}
\problempart{
If the last time that Eric traveled from Paul to Zardosht, 
\emph{and back} Paul and Zardosht were $k$ miles apart, 
how far apart will they be after the round trip? 
}
\solution{
We know that Paul and Zardosht will be $\frac{k}{3}$ miles closer
in the time it takes Eric to go from Paul to Zardosht, and in
the return trip from a distance $\frac{2*k}{3}$, and in the
return trip they close a $\frac{1}{5}$ of the miles. Thus,
the new total distance will be: $\frac{8*k}{15}$. If Eric
traveled $k$ miles on the first trip, he will travel
$\frac{8*k}{15}$ miles on the next trip from Paul to Zardosht.
}
\problempart{
What is the infinite series that describes how far Eric travels
total and its solution?
}
\solution{
First we calculate the total distance Eric traveled on his first
round trip. Zardosht is initially 50 miles away. Eric closes at 30
MPH, and catches Zardosht after $\frac{5}{3}$ hours. Paul and
Zardosht are now $\frac{100}{3}$ miles apart. Eric closes on
Paul at a rate of 50 MPH, and reaches Paul in $\frac{2}{3}$ 
hours. He travels a total of $\frac{7}{3}$ hours and thus
goes $\frac{280}{3}$ miles total. 

The infinite geometric series:
\begin{equation}\label{s}
\frac{280}{3}  \sum_{i=0}^\infty \paren{\frac{8}{15}}^i.
\end{equation}
Using the formula for the sum of a geometric series, we conclude that~\eqref{s}
equals $\frac{280}{3} * \frac{15}{7} = 200$ miles.
}
\end{problemparts}
\end{problem}

\begin{problem}
Say a particular bank
charges \$10 a month in upkeep charges, and offers offers an interest
rate of $p$ every month as well. If I invest $d$ dollars today,
and I have a monthly rate of $p$, how much money do I have after a year
(including monthly fees)? 
Specifically, interest is paid on
the balance at the end of the month \emph{after} the monthly fee is 
subtracted. Here $p$ is equal to 1 plus the percent return, so 5\%
interest would yield a $p=1.05$ for example.

Give a closed form for how much money you have at the end of a year.

\solution{
Consider that there is no monthly fee first. Then final dollar amount
is simply: $d*p^12$. However, we lose 10 dollars a month. So, on our
investment we lose 10 dollars and its associated growth for 12 months
from the first month's fee. Then we lose 10 dollars and its associated
growth for 11 months from the 2nd month's fee etc. etc. This is just
a geometric series.
\begin{equation}
\sum_{i=1}^12 {10*{p}^{i}} = 10*\paren{\frac{1-p^12}{1-p}-1}.
\end{equation
So, at the end of a year, I have: $d*p^12-10*\paren{\frac{1-p^12}{1-p}-1}$
dollars.
}
\end{problem}

\begin{problem}{More with Deserts!}

\begin{problemparts}
\problempart
Say we want to once again cross a Desert. But, this time that
darned Colonel Hogan has managed to poke a small hole the
gas tank of the truck. Say that the original truck could 
travel distance $d$ on a full tank of gas. Now, the truck loses
$k$ every time you travel the distance $k$ the original truck could
have traveled.

Assuming that we use the same strategy, and first leave bring $n$
tanks of gas out into the desert, then use 1 tank of gas to move
$n-1$ tanks of gas farther into the desert, etc. etc. what is the
series that describes the spacing between depots? Can we still cross
any dessert?
\solution{
This is the same as if the truck could only go half as far on a
tank of gas. So, we can clearly still cross a dessert. The depots
will have to be spaced half as far apart, which yields a quarter
of the harmonic numbers. On the last push into the desert you can
go $\frac{d}{2}$ miles.  Previous push you use 1 tank of gas to
make 3 trips, and thus space the depot $\frac{d}{6}$ miles previously.
Then, the previous one you make 5 trips, and space it $\frac{d}{10}$ 
etc. etc. The total series is:
\begin{equation}
\sum_{i=1}^\infty \frac{d}{4*i-2}.
\end{equation
}
\problempart
Say we cared less about how much gas we used, and more about how 
many depots we have to set up. This could be the case if depots
are expensive to set up. Describe a strategy that would allow every
depot to be spaced $\frac{d}{4}$ apart.
\solution{
To space the gas stations $\frac{d}{4}$ apart, at every step use
up half of the gas at a given station to transport the other half
of the gas to the next station. Thus, if $\frac{x}{4}=n$, you must
originally have $2^n$ tanks of gas.
}

\problempart
How many fewer gas stations would you need to cross a dessert of length $x$? 
\solution{
in this case, you will need a linear number of stations for distance
$x$, as in $\frac{x}{4}$. In the previous example, you would need
an exponential number of gas stations.
}