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\begin{document}


\ifshowsolutions
{
\begin{center}
{\large \bf
Exam 2 Solutions}

November 22, 2000
\end{center}
\spaceoutfalse
\else
\begin{center}
{\huge \bf
6.042 Exam 2}

November 15, 2000
\end{center}

\fi

\ifshowsolutions
\else
\smallspace


\begin{tabbing} 
\hspace{1.5in}\=Name:\hspace{1in}\=\underline{\hspace{2.5in}}\\
\\
\>Tutorial Leader:\>\underline{\hspace{2.5in}}\\
\\
\>DO NOT WRITE IN THIS SPACE
\end{tabbing}
\smallspace
\begin{eqnarray*}
\hbox{Problem 1:}& \underline{\hspace{1in}} & / \;\;25\\
\hbox{Problem 2:}& \underline{\hspace{1in}} & / \;\;10\\
\hbox{Problem 3:}& \underline{\hspace{1in}} & / \;\;25\\
\hbox{Problem 4:}& \underline{\hspace{1in}} & / \;\;35\\
\hbox{Problem 5:}& \underline{\hspace{1in}} & / \;\;15\\
\hbox{Problem 6:}& \underline{\hspace{1in}} & / \;\;21\\
\hbox{Problem 7:}& \underline{\hspace{1in}} & / \;\;45\\
\hbox{Problem 8:}& \underline{\hspace{1in}} & / \;\;24\\
\hbox{Total:}& \underline{\hspace{1in}} & / \;200
\end{eqnarray*}
\nextproblem
\fi


\begin{problems}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{25} \textbf {Summations}

Find the closed form for each of the expressions listed below.  Give
your answer in simplified form.

\problempart {\bf (10 points)}: $ \sum_{i=x}^{y} 2i+1 $

\solution{ \bigskip
{\bf Solution:}

$$(y+1)^2 - x^2.$$

Grading standard:
\begin{tabbing}
3 points for knowing to break up the sum (as in $\sum_0^y -
\sum_0^{x-1}$)\\
2 points for getting the limits right when breaking up  OR\\
3 points for rearranging the limits (for instance as in $sum_0^{y-x}$)\\
2 points for simplifying correctly to an arithmetic sum \\
2 points for applying the arithmetic sum correctly\\
2 points for getting the algebra right\\
1 point for giving a simplified answer.
\end{tabbing}

There are 2 methods I'd expect to see; these correspond to the first two 
point categories and the second two point categories.  (Give points only 
for the first pair of categories or the second, not both).

Students should get at least 9 points if their answer is right.  An
example of an answer that would be right but worth only 9 points
would be

$$y(y+1) + y + 1 - x(x-1) - x.$$

}
\mediumspace
\problempart {\bf (15 points)}:
$ \sum_{i=0}^{\infty} \sum_{j=1}^n \left(\frac{j}{j+2}\right)^i $

\solution{ \bigskip
{\bf Solution:}

$\frac{1}{4}n^2 + \frac{5}{4}n$ or $\frac{n(n+5)}{4}$ or $\frac{n^2 +
5n}{4}$

Grading standard:
\begin{tabbing}
3 point for reordering the summations\\
5 points for correctly applying the formula for an infinite geometric
series\\
2 points for correctly applying the arithmetic series formula\\
3 points for getting all the algebra right\\
2 points for simplifying
\end{tabbing}

Once again, correct answers should get at least 13 points, and then only
13 if the answer is BADLY unsimplified.  An answer like

$$n + \frac{n(n+1)}{4}$$

should get 14 points.  Any of the given answers should get all 15 points.

}
\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{10} \textbf {Bounding Infinite Sums}
% not multiple choice (but could be)

Using the Integral Method, give an upper bound for 
\[\sum_{i=1}^n \frac{3}{i^{10}} \] 
 
\solution{

\textbf{Solution:}\\
\medskip
Since $f(k)=\frac{3}{i^{10}}$ is monotonically decreasing for $x>1$, we have:
    \begin{eqnarray*}
      \sum_{i=1}^n{\frac{3}{i^{10}}} &=& 1 + \sum_{i=2}^n{\frac{3}{i^{10}}} \\
      &\leq& 1+ \int_1^n{\frac{3}{x^{10}}}dx\\
      &=& 1+ 3\int_1^n{\frac{1}{x^{10}}}dx\\
      &=& 1 + 3\left [{\frac{-1}{9x^{9}}}\right ]_1^n\\
      &=& 1 + \left [{\frac{-1}{3x^{9}}}\right ]_1^n\\
      &=& 1 - \frac{1}{3n^{9}} + \frac{1}{3}\\
      &=& \frac{4}{3} - \frac{1}{3n^{9}}
    \end{eqnarray*}
    Hence, $\sum_{i=1}^n \frac{3}{x^{10}} \leq \frac{4}{3} - \frac{1}{3n^{9}}$.

\medskip
\textbf{Grading Standard:}\\
Similar to problem 5 on pset 6.\\
-10 for not using the Integral method with wrong answer\\
-7 for both limits incorrect on the integral\\
-4 for one limit incorrect on the integral\\
-2 for making algebraic mistakes\\
-6 for using the Integral method but not being tight (for example, first
saying $3/i^{10} < 3/i^2$ and using the integral method from there)
}

\nextproblem
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{25} \textbf {Asymptotic Notation}

Determine which of the eight choices below best describes each
function's asymptotic behavior.  No proofs are necessary.  {\bf (5
points each)}

\begin{tabbing}
XXX\=XXXXXXXXXXXX\=\kill
\>$\theta (n)$\>$\theta (n!)$\\
\>$\theta (n\log n)$\>$\theta (n^2)$\\
\>$\theta (1)$\>$\theta (2^n)$\\
\>$2^{\theta(n)}$\>None of the above
\end{tabbing}

\solution{ Grading Standard: Give no partial credit, except on part (d) if
they answer $2^{\theta(n)}$ instead of $\theta(2^n)$.  Give 2 points
partial credit for this answer (the function given IS $2^{\theta(n)}$ but
this is not the best answer.) }

\problempart  $g(n) = n + \ln n + (\ln n)^2$.

\solution{\bigskip
{\bf Solution:}  $\theta(n)$
}

\smallspace
\problempart $h(n) = \frac{n^2 + 2n - 3}{n^2 - 1}$

\solution{\bigskip
{\bf Solution:} $\theta(1)$
}

\smallspace
\problempart $j(n) = \sum_{i = 0}^n 2^{2i+1}$

\solution{\bigskip
{\bf Solution:} $2^{\theta(n)}$
}

\smallspace
\problempart $k(n) = (2 + \sin (n))2^{n + \sin(n)}$.

\solution{\bigskip
{\bf Solution:} $\theta(2^n)$
}

\smallspace
\problempart $f(n) = \ln ((n^2)!)$.

\solution{\bigskip
{\bf Solution:}  None of the above.
}

\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \pts{35} \textbf {3-coloring Grid Graphs}

In Exam 1, we proved that an $m$ by $n$ grid graph is always
2-colorable.  Thus, it must be 3-colorable.  However, with 3 colors,
we suddenly get a lot of choices for how to color the graph.  In this
problem we will solve the question ``How many ways are there
to 3-color a $3$ by $n$ grid graph?''

(Recall that a 3-coloring of a graph is an assignment of one of three
colors to each vertex in the graph such that if two vertices are
connected by an edge, they must have different colors.  We consider
two 3-colorings different if any vertex is colored differently between
the two.)

\problempart {\bf (8 points)} Suppose we want to color the
graph by first coloring the first $n-1$ columns, then choosing a
coloring for the last column.  However, the number of ways we can
color the last column is dependent on which coloring we choose.  There are
two cases to consider.

\begin{figure}[htbp]
\centerline{\psfig{figure=3color.eps,height=1.5in}}
\end{figure}

In the first case, the vertices $(n-1,1)$ and $(n-1,3)$ are the same color.

In the second case, the vertices $(n-1,1), (n-1,2)$, and $(n-1,3)$
are all different colors.  

\begin{itemize}
\item[1.] How many ways can we color the last column in the first
case?  \solution{5}

\item[2.] Of the ways to color the last column in the first case, how
many use three different colors? \solution{2}

\item[3.] How many ways can we color the last column in the second case?
\solution{4}

\item[4.] Of the ways to color the last column in the second case,
how many use three different colors? 
\solution{2}

\end{itemize}

\solution{
Grading:  2 points for each part, no partial credit.}

\nextproblem

\problempart {\bf (10 points)} Let $A_n$ be the number of ways to
3-color a $3$ by $n$ grid graph such that the last column is colored
with only 2 colors.  Let $B_n$ be the number of ways to 3-color a $3$
by $n$ grid graph such that the last column is colored with three
different colors.

Formulate a system of recurrence relations for $A_n$ and $B_n$, and give
the initial values $A_1$ and $B_1$.

\solution{\bigskip
{\bf Solution:}

$A_n = 3A_{n-1} + 2B_{n-1}$.

$B_n = 2A_{n-1} + 2B_{n-1}$.

$A_1 = B_1 = 6.$

Grading:
\begin{tabbing}
XXX\=XXX\=\kill
1 point for each of the two initial conditions.\\
2 points for having a system in the right form.\\
2 points for each coefficient of $A_{n-1}$.\\
1 point for each coefficient of $B_{n-1}$.
\end{tabbing}

}

\mediumspace

\problempart {\bf (7 points)} Suppose that 

$$A_n = xc_1^n + yc_2^n$$
$$B_n = xd_1c_1^n + yd_2c_2^n$$

is the solution to the recurrence relation you found in part (b).
Find the number of ways to 3-color a 3 by $n$ grid graph 
in terms of $c_1, c_2, d_1, d_2$, and $n$ and no other variables.

\solution{\bigskip
{\bf Solution:}

$$A_n+B_n = \left(\frac{-6+6d_2}{(d_2-d_1)c_1}\right)(1+d_1)c_1^n +
\left(\frac{6 - 6d_1}{(d_2-d_1)c_2}\right)(1+d_2)c_2^n$$

Grading:
\begin{tabbing}
2 points for solving for $x$ correctly ($\frac{6d_2 - 6}{(d_2-d_1)c_1}$)\\
2 points for solving for $y$ correctly ($\frac{6 - 6d_1}{(d_2-d_1)c_2}$)\\
2 points for giving a final answer by adding $A_n$ and $B_n$.\\
1 point for combining $c_1^n$ and $c_2^n$ terms in the final answer.
\end{tabbing}

}


\nextproblem

\problempart {\bf (10 points)}  Find $c_1, c_2, d_1,$ and $d_2$ such that

$$A_n = xc_1^n + yc_2^n$$
$$B_n = xd_1c_1^n + yd_2c_2^n$$

is the general form for the system of recurrence relations you found in 
part (b).  

\solution{\bigskip
{\bf Solution:}

This gives two solutions, $c_1 = \frac{5+\sqrt{17}}{2}, d_1 =
\frac{-1+\sqrt{17}}{4}$ and $c_2 = \frac{5-\sqrt{17}}{2}, d_2 =
\frac{-1-\sqrt{17}}{4}$.  

Grading:
\begin{tabbing}
3 points for using $A_n = c^n, B_n = dc^n$ to find answers for $c$ and
$d$\\
1 point for using $B_n = c^n, A_n = dc^n$, et cetera.\\
4 points for getting solutions for $c$ and $d$.\\
2 points for correctly indicating $c_1, c_2, d_1, d_2$.\\
\end{tabbing}

}
\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{15} \textbf {Nonhomogenous Recurrence}

Give a closed-form solution to the following nonhomogenous recurrence
relation,

$$G_n = 2G_{n-2} + 2^{n-1}$$

such that $G_0 = 1$ and $G_1 = 2$.

\solution{\bigskip {\bf Solution:}
$$G_n = 2^n.$$

Grading:
\begin{tabbing}
2 points for identifying the homogenous part.\\
3 points finding the general form of the solution to the homogenous 
part.\\
3 points for guessing $a 2^n$ or the like for the particular solution.\\
2 points for solving for $a$.\\
5 points for finishing their solution.\\
\end{tabbing}

If they didn't use this method to find their answer, use this standard 
instead.
\begin{tabbing}
10 points for giving $2^n$ as an answer.\\
5 points for giving correct reasoning to support their answer.\\
\end{tabbing}

So, if they just write out the first several terms and then write $2^n$ as 
their answer, they should get 10 points.  If they do this and then prove
that $2^n$ works, give them full credit.

}
\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 6
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \pts{21} \textbf {A Strange Set}
% difficulty: medium-hard

Define the function $fp(x) = x - \floor{x}$.  ($fp$ stands for
``fractional part.'')  Recall that $\floor{x}$ is the greatest integer
less than or equal to $x$.  So, $\floor{8.2} = 8$ and $\floor{-2.3} =
-3$.   Let $h = \sqrt{37}$

Let $S = \{ n \in \mathbb{Z} \ | \ n \neq 0 \wedge  fp(nh) < 0.01 \}$. 


Use the pigeonhole principle to prove that $S$ is nonempty.  Other methods
will not get credit.


\solution{\bigskip
Grading:
\begin{tabbing}
4 (mercy) points for appearing to understand the question.\\
4 points for correct ``pigeons" ($fp(nh)$ for each $n \neq 0 \in 
\mathbb{Z}$.\\
4 points for correct ``pigeonholes" (ranges of size $.01$).\\
2 points for subtracting two pigeons to get an element in $S$.\\
1 points for correctly dealing with which of the elements to subtract from
which.\\
4 points for proving their element is such that $fp(nh) < 0.01$.\\
2 points for proving their element is such that $n \neq 0$.
\end{tabbing}

The first 4 points are designed to give to people who wrote stuff that 
wasn't the start of a proof but showed some understanding of the problem 
(ie, understanding the set $S$, the function $fp$, et cetera). 
This way, if someone struggles with the problem but doesn't solve it, they
get more credit than someone who doesn't even try.

The 4 points for getting the "pigeons" right go to not only what the
pigeons *are* but also how to assign them to pigeonholes.  Thus, for
instance, saying each nonzero integer is a pigeon is only worth part of
the credit.  Saying $fp(nh)$ for each nonzero integer $n$ are the pigeons
makes it clear that we should use this value to choose the pigeonhole and
so this is worth full credit.

}
\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 7
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem \pts{45} \textbf {Rearranging the Alphabet}

For this problem you do not need to evaluate an expression fully: you
may leave it in terms of binomial coefficients, powers of numbers,
factorials, and so on.  For this problem we consider the alphabet to
consist only of capital letters.

\problempart {\bf (15 points)} How many different strings are there
of 10 letters such that no letter is used more than once?

\solution{\bigskip
{\bf Solution:}  $\frac{26!}{16!}$

Grading:
\begin{tabbing}
5 points for determining to use a permutation formula.\\
5 points for determining to use a formula that doesn't allow 
repetitions.\\
5 points for correctly applying whatever formula they used.
\end{tabbing}

If you can't tell what formula they used AND they applied whatever
formula it was incorrectly, just give them the first 5 points if they 
aren't using binomial coefficients, or no points if they are.
}
\mediumspace

\problempart {\bf (15 points)} How many ways can you rearrange the
alphabet (using each letter only once) so that none of the words ``TEST,''
``EXAM,'' and ``STUDY'' appear as consecutive letters in your arrangement?


\solution{\bigskip
{\bf Solution:}  $26! - 23! - 22! + 19!$.

Grading:
\begin{tabbing}
3 points for attempting to use inclusion/exclusion.\\
2 points for setting up inclusion/exclusion correctly.\\
2 points for using permutations without repetitions for the terms.\\
1 points for getting the 26! term correct*.\\
1 points for noticing that ``TEST'' can never appear.\\
2 points for getting the 23! term correct.\\
2 points for getting the 22! term correct.\\
2 points for getting the 19! term correct.
\end{tabbing}

* If they didn't subtract from 26! they should lose the points for both
getting the 26! term and for setting up inclusion/exclusion correctly.

}

\nextproblem

\problempart {\bf (15 points)} How many ways are there to choose a
collection of 30 letters such that none of the words ``EXAM,'' ``STUDY,''
and ``TEST'' can be formed using the letters in the collection?

\solution{\bigskip
{\bf Solution:}
$$ \binom{55}{30} - 2\binom{51}{26} - \binom{50}{25} + 2\binom{48}{23}
+ \binom{46}{21} - \binom{45}{20}.$$

Grading:
\begin{tabbing}
3 points for realizing we need to use inclusion/exclusion and setting it
up.\\
5 points for using combinations with repetitions to come up with the
terms.\\
3 points for using the formula correctly.\\
3 points for getting the size correct on each of the three two-set
intersections.\\
2 points for getting the size correct on the 3-set intersection.\\
\end{tabbing}

What I mean by getting the size correct is that they understand the
intersection issues (ie, that the `E' in EXAM and TEST need not be in the
collection twice).


}
\nextproblem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% PROBLEM 8
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem\pts{24} \textbf{Finite and Infinite Sets}

For each of the following sets, state whether it is finite, countably
infinite, or uncountably infinite.  If it is finite, give its size.
If it is countably infinite, give an injective function from the given
set to $\mathbb{N}$.  If it is uncountable, give a brief argument.

\problempart {\bf (6 points)} $R = \mathcal{P}(\{2^{-i} \ | \ i \in
\mathbb{N}\})$

\solution{\bigskip
{\bf Solution:}

Uncountable.  

Grading:
\begin{tabbing}
XXX\=\kill
2 points for saying "Uncountable."
1 point for any argument with major holes or small errors in it.\\
2 points for any argument with holes in it.\\
3 points for any argument with minor holes, or calling on results that\\
\> are correct but never done in class or in the texts.\\
4 points for any correct, complete argument.\\
\end{tabbing}

}

\smallspace

\problempart {\bf (6 points)} $S = \mathcal{P}(\mathbb{N}) \cap
\mathcal{P}(\{ 2^{-i} \ | \ i \in \mathbb{N}\})$

\solution{\bigskip
{\bf Solution:}

Finite; 2

Grading:
\begin{tabbing}
2 point for saying "finite."\\
2 points for giving 1 as the size (forgetting the empty set)\\
4 points for giving 2 as the size.
\end{tabbing}
}



\smallspace
\problempart {\bf (6 points)} $T$, the set of all finite sets of
prime numbers.

\solution{\bigskip
{\bf Solution:}

Countable.  Map a set $S \in T$ to the number $\prod_{p\in S} p$.

Grading:
\begin{tabbing}
XXX\=\kill
2 points for saying "countable."
1 point for any argument with major holes or small errors in it.\\
2 points for any argument with holes in it.\\
3 points for any argument with minor holes, or calling on results that\\
\> are correct but never done in class or in the texts.\\
4 points for any correct, complete argument.\\
\end{tabbing}

Take off 1 point if their argument doesn't involve a way to count the set 
$T$.  

}

\smallspace
\problempart {\bf (6 points)} $U$, the set of all equivalence classes
of the equivalence relation $\{ (x, y) \ | \ x, y \in \mathbb{R}$ and
$x-y \in \mathbb{Q}\}$

\solution{\bigskip
{\bf Solution:}

Uncountable.  

Grading:
\begin{tabbing}
2 points for saying "uncountable"\\\
1 point for arguing that there is a different class for every Real number.\\
2 points for setting up some incorrect diagonalization argument.\\
4 points for any correct argument.
\end{tabbing}

}


\end{problems}
\end{document}



