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\def\header{{\hfill Exam 3 \hfill A. Miller \hfill Fall 2004
                                       \hfill Math 210\hfill}}
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\begin{document}

\bigskip

Show all work.  

Simplify your answers.  

Circle your answer.

\bigskip

No notes, no books, no calculator, no cell phones, no pagers, no
electronic devices. 


\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\bigskip
Circle your Discussion Section:
\begin{verbatim}
     DIS 343 12:05p T B329 VAN VLECK 
     DIS 344 12:05p R B321 VAN VLECK 
     DIS 345  1:20p T  595 VAN HISE  
     DIS 346  1:20p R 3401 STERLING  
\end{verbatim}
\bigskip

\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
   \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}

\immediate\openout\examTBLFile=exam.tbl

\bigskip
 
Solutions will be posted shortly after the exam:
www.math.wisc.edu/$\sim$miller/m210

\setcounter{page}{0}

\newpage 

\prob{8} % chapter 9

Prove the geometric series formula:
$$1+x+x^2+\cdots+x^n={{1-x^{n+1}}\over{1-x^{\phantom{n+1}}}}$$ (for $x\neq 1$)

\answer Put $S=1+x+x^2+\cdots+x^n$.
Then 
$$xS=x(1+x+x^2+\cdots+x^n)=x+x^2+x^3\cdots+x^n+x^{n+1}.$$
If we subtract we get:
$$S-xS=1-x^{n+1}.$$
Hence $(1-x)S=1-x^{n+1}$ or $S={{1-x^{n+1}}\over{1-x^{\phantom{n+1}}}}$. 



\prob{9} % 9.1

\par (a) What is the effective quarterly rate $j$ which corresponds to a
nominal annual rate of 8 percent?

\vfill

\par (b) What is the effective annual rate $k$ corresponding to a nominal
annual rate of 8 percent which is compounded quarterly?

\vfill

\par (c) Suppose P=\$1000 is put in a savings account which earns a nominal
annual interest rate of 8 percent compounded quarterly.
After 18 months how much is in the account?  


\vfill

\answer \par (a) $j=.08/4=.02$
\par (b) $(1+k)=(1+j)^4$ or $k=(1.02)^4-1$.
\par (c) $18$ months is a year and a half or 6 quarters. Hence
the answer is $1000(1.02)^6$.


\prob{8} 

Sam buys a car from his sister Sally for \$3000.  He agrees to pay her \$1000
two years from now and the remaining amount $A$ three years from now.  Assuming
an effective annual interest rate of 6 percent, how much  is $A$?

\answer $$3000={1000\over{(1.06)^2}}+{A\over{(1.06)^3}}$$
Hence
$$A=(1.06)^3\left(3000 -{1000\over{(1.06)^2}}\right)$$

\prob{8} 

Harry bought 100 shares in the Inca Lost Gold Mine Corporation (ILGM) for \$90
per share on Nov 15, 2000.  On April 1, 2001 the stock price rises to \$120 per
share and the company makes a two for one split so that each share is now worth
\$60.   On Nov 15, 2002 rumors hit the market that there is no lost Inca Gold
and the stock price of ILGM plunges to \$10 per share.  However, the company
finds that tourists will pay a lot of money to be taken out into the Yucatan
Jungle to search for the Inca Lost Gold Mine.   The price of each share of ILGM
rises steadily to  \$80 on Nov 15, 2004 at which time Harry sells all his
shares of ILGM stock.   

\bigskip 
Did Harry lose money or make money on his investment? What was the
effective annual yield (or loss) on his investment?

\answer Harry made money.  On Nov 15, 2000 he paid \$9000 for his
100 shares.  The stock split 2 for 1 which means that each stock holder
doubles his number of shares.
On Nov 15, 2004 he sold his 200 shares for \$16000.
The yield is 
the effective annual interest rate which when compounded would result 
in the same amount of gain. 
His annual yield $i$ satisfies:
$$16000=9000(1+i)^4.$$
Hence 
$$i=(16/9)^{1/4}-1.$$ 
As far as I know there is no mathematical difference between
\par yield = rate of return = interest rate
\par\noindent The four year rate of return $k$ would satisfy
$$ 9000(1+k)=16000$$
since in this case there would be one four year period of time.
Hence if we set $P=9000$ and $Q=16000$ then
$$P(1+k)=Q$$
$$1+k={Q\over P}$$
$$k={Q\over P}-1={Q-P\over P}$$
$Q-P=7000$ is the gain and $P=9000$ is the amount invested.
Hence 
$$k=7/9=.7777\ldots\approx 78\%$$
The nominal annual rate of return $j$ would satisfy
$$4j=k.$$ Hence $j\approx 19{1\over 2}\%$.  The effective
annual rate of return $i$ should satisfy
$$(1+i)^4=(1+k)$$
Which is the same $i$ as above.  Using a calculator
we get that $i$ is approximately $14{1\over 2}\%.$   


\prob{9}
A loan is paid back in three annual payments of $p$ dollars 
made at the end of each
year.  The  effective annual interest rate is $i$.  

\bigskip

(a) What is the principle or amount borrowed $B$?  Assume that it is
borrowed at the beginning of the first year.  (Express $B$ in
terms of $p$ and $i$.)

\vfill

(b) What is the total amount of interest $I$ (in dollars) paid 
thru the course of the loan?  (Express $I$ in terms of $B$, $p$, and
$i$.)

\vfill

(c) Each payment $p$ in a loan is part interest and part reduction in the
principle.  What part $I_1,I_2,I_3$ of each of the three payments is
interest?

\vfill

\answer (a) $$ B={p\over {(1+i)}}+{p\over {(1+i)^2}}+{p\over {(1+i)^3}}$$
\par (b) Everything paid is either principle or interest so
$$3p=B+I \mbox{ or } I=3p-B$$
\par (c) At the beginning of the second year there are only two
payments left, so the remaining principle is 
$$C={p\over {(1+i)}}+{p\over {(1+i)^2}}$$
Thus the first payment has reduced the principle by
$$B-C={p\over {(1+i)^3}}$$
The first payment $p=I_1+(B-C)$ hence 
$$I_1=p- {p\over {(1+i)^3}}.$$
Similarly at the beginning of the third year there is only one
payment left, so the remaining principle is 
$$D={p\over {(1+i)}}$$
so the second payment has reduced the principle by
$$C-D={p\over {(1+i)^2}}.$$
Thus $p=I_2+(C-D)$ hence
$$I_2=p- {p\over {(1+i)^2}}.$$
Finally at the beginning of the fourth year there is no priniciple left
and hence the third payment reduces the principle
by $D-0={p\over {(1+i)}}$ and so $p=I_3+(D-0)$ and
so
$$I_3=p-{p\over {(1+i)}}.$$



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