\begin{examproblem}[15]
A newlywed couple want to borrow $d$ dollars from a mortgage company so
that they can purchase a small condo.  After a grace period of $g$ years,
they will start make $n$ annual payments, after which the loan will be
considered to be paid off.  The mortgage company will immediately invest
each payment it receives at an $a$\% annual rate of return, \eg, a \$1
investment will yield \$R after one year, where $R\ eqdef 1+ a/100$.

When the laon is paid off, the mortgage company wants to wind up with at
least as much money from investing the newlywed's payments as they would
have if they had invested the whole $d$ dollar loan at $a$\% starting on
the day of the loan.

\bparts

\ppart[10] Write a closed form formula in $d,n$ and $R$ for the least annual payment
the mortgage company must ask from the newlyweds.

\solution{
Letting $p$ be the payment, the company requires
\[
p &\geq& \frac{dR^{g+n-1}(R-1)}{R^n - 1}.
\]
We derive this bound as follows:

Suppose the couple pays $p$ dollars at the end of each year.  The company
can invest the first payment for $n-1$ years, the second payment for $n-2$
years, \etc, so that after $n$ payments their total cash is:
\begin{eqnarray}
\sum_{i=1}^{n} p \cdot R^{i-1}
& = & \sum_{i=0}^{n-1} p \cdot R^i\notag\\
& = & p \frac{R^n - 1}{R - 1}.\label{pR}
\end{eqnarray}

On the other hand, the last payment is made after $g+n-1$ years.  If the
company invested the $d$ dollars today, then their total cash is $d
R^{g+n-1}$.  The amount~(\ref{pR}) must be at least as large as this, so
\begin{eqnarray*}
p \frac{R^n - 1}{R - 1} & \geq & dR^{g+n-1}\\
p &\geq& \frac{dR^{g+n-1}(R-1)}{R^n - 1}.
\end{eqnarray*}
}

\ppart[5] Suppose the number, $n$, of payments is 15.  Then the required
payment is a function, $p(g)$, of the number of years, $g$, in the grace
period.  Assuming (unrealistically) that $g$ could be an arbitrarily
large, give the least real number, $a$, such that $p(g) = O(g^a)$, or
explain why there is no such $a$.

\eparts

\end{examproblem}

