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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2008} \\
\textbf{Solutions for Math 132, Midterm Exam 3} (All Sections) \\
\textbf{Wednesday, April 23, 7 PM}
\end{center}

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%\noindent\textbf{Your Name}: $\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$
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%\noindent\textbf{Your Section}: 

%\vskip 15pt
%\leftline{Little (8:00am) $\underline{\qquad}$ Ballantine (9:00am) $\underline{\qquad}$} \leftline{DeStefano 
%(10:00am) $\underline{\qquad}$ DeStefano (noon) $\underline{\qquad}$}

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%\noindent\textbf{Instructions}:  For full credit, you must show 
%{\it all work} on the test pages and place your final answer
%in the box provided for the problem.  Use the back of the
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%next to each part of the questions are their point values.

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%\noindent\textbf{Please do not write in the space below}

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%Problem & Points/Poss \\
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%I  & \hfil/ 15\  \hfil \\
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%II  & \hfil/ 15\  \hfil \\
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%III  & \hfil/ 15\  \hfil \\
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%IV  & \hfil/ 15\  \hfil \\
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%V  & \hfil/ 15\  \hfil \\
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%VI  & \hfil/ 10\  \hfil \\
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%VII & \hfil/ 15\  \hfil \\
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%Total   & \hfil/100\hfil \\
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\noindent
I.  
\begin{enumerate}
\item[A.] [10~points] Circle the number of the plot
showing the direction field for each of the following differential equations.
(Note that there are only 3 plots, so the correct answer for one is ``None.'')
\soln
\begin{itemize}
  \item[(1)] $\ds y' = x$ is plot III (note negative slopes for all $x < 0$ and positive slopes for all $x > 0$)

    \vskip 0.1truein

  \item[(2)] $\ds y' = 1 + y^2$ matches None of the plots (by process of elimination, or by noting
that none of the direction fields has positive slope everywhere)

    \vskip 0.1truein


  \item[(3)] $\ds y' = 1 - y$ is plot I (note zero slope along the line $y = 1$)

    \vskip 0.1truein


  \item[(4)] $\ds y' = x - y$ is plot II (note zero slope along the line $y = x$)

    \vskip 0.1truein

\end{itemize}

\begin{center}
I: \includegraphics[width=2.5truein,height=2.5truein]{132P31IS.ps} 
II: \includegraphics[width=2.5truein,height=2.5truein]{132P31II.ps}
\end{center}
\begin{center}
III: \includegraphics[width=2.5truein,height=2.5truein]{132P31III.ps} 
\end{center}
\item[B.] [5~points] On the plot for the equation $y' = 1 - y$ from (3) of part A, 
give a qualitative sketch of the graph of the solution satisfying the initial
condition $y(-1) = 0$.  Show as much of the graph as you can for both positive and negative $x$.

\soln See plot I above.
\end{enumerate}

\vskip 10pt
\noindent
II.  All parts of this problem deal with the differential equation $y' = 7 - y$.
\begin{enumerate}
\item[A.]  [4~points] Use 4 steps of Euler's method to approximate the solution of this
equation with the initial condition $y(0) = 4$ at $x = 2$.  

\soln
With $x_0 = 0$ and $x_4 = 2$, we will use $\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$.  
The steps of Euler's Method are
\begin{eqnarray*}
y_1 = y_0 + (7 - y_0)\Delta x &=& 4 + (3)(.5) = 5.5\\
y_2 = y_1 + (7 - y_1)\Delta x &=& 5.5 + (1.5)(.5) = 6.25\\
y_3 = y_2 + (7 - y_2)\Delta x &=& 6.25 + (.75)(.5) = 6.625\\
y_4 = y_3 + (7 - y_3)\Delta x &=& 6.625 + (.375)(.5) = 6.8125\\
\end{eqnarray*}
The approximate value of $y(2)$ is $y(2) \doteq y_4 = 6.8125$.

\item[B.] [6~points] Find the general solution $y(x)$ of the equation 
by separating variables and integrating.

\soln
We separate variables, integrate, then exponentiate to solve for $y$:
\begin{eqnarray*}
\frac{dy}{7 - y} &=& dx\\
\int\frac{dy}{7-y} &=& \int\, dx\\
-\ln|7 - y| = x + c\\
7 - y &=& be^{-x}\quad \text{where $b = \pm e^{-c}$ is another arbitrary constant}\\
y &=& 7 - be^{-x}.
\end{eqnarray*}
\item[C.] [5~points] Find the particular solution $y(x)$ satisfying the initial 
condition $y(0) = 4$ and compute the exact value of $y(2)$.  

\soln
Substituting $x = 0$ and $y = 4$ gives $4 = 7 - b$, so $b = 3$.  The particular
solution is $y = 7 - 3e^{-x}$, and $y(2) = 7 - 3e^{-2} \doteq 6.59399$.
The approximation given by Euler's Method in part A. is an overestimate.
\end{enumerate}
\vskip 10pt
\noindent
III. All parts of this question deal with the infinite series 
\[
\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} +  \cdots\ .
\]
\begin{enumerate}
\item[A.]  [4~points] Find a general formula for the $n$th term of the series as a function
of $n$ and write the series in summation (``sigma'') notation.

\soln
If we index the first term above with $n = 1$, the second with $n = 2$, etc., then  
the general term is $a_n = \frac{n}{n+1}$, and the series is $\ds \sum_{n=1}^\infty \frac{n}{n+1}$.
\item[B.]  [4~points] Call the $n$th term in the series $a_n$.  
What is ${\displaystyle \lim_{n\to\infty} a_n}$?

\soln
We have
\begin{eqnarray*}
\lim_{n\to\infty} \frac{n}{n+1} &=& \lim_{n\to\infty} \frac{n}{n+1} \cdot \frac{\frac{1}{n}}{\frac{1}{n}}\\
&=& \lim_{n\to\infty} \frac{1}{1 + \frac{1}{n}}\\
&=& 1.
\end{eqnarray*}
(This limit could also be computed using L'Hopital's Rule.)

\item[C.]  [4~points] Write out the first 3 \emph{partial sums} of the series.

\soln
The first three partial sums are
\begin{eqnarray*}
s_1 &=& \frac{1}{2}\\
s_2 &=& \frac{1}{2} + \frac{2}{3} = \frac{7}{6}\\
s_3 &=& \frac{1}{2} + \frac{2}{3} + \frac{3}{4} = \frac{23}{12}.
\end{eqnarray*}
\item[D.]  [3~points] Does this series converge or diverge? Explain your answer.

\soln
Since $\ds \lim_{n\to\infty} \frac{n}{n+1} = 1\ne 0$, this series
\emph{diverges} by the Test for Divergence.
\end{enumerate}
\vskip 10pt
\noindent
IV. 
\begin{enumerate}
\item[A.]  [5~points] Does the series 
$\ds \sum_{n=0}^\infty \frac{(-1)^n e^n}{\pi^n}$ converge or diverge?  

\soln
This series:
\[
\sum_{n=0}^\infty \frac{(-1)^n e^n}{\pi^n} = \sum_{n=0}^\infty \left(\frac{-e}{\pi}\right)^n
\]
is a \emph{geometric series} with first term $a = 1$ and ratio $r = -e/\pi$.  Since
$|r| < 1$, the series is convergent and 
\[
\sum_{n=0}^\infty \left(\frac{-e}{\pi}\right)^n = \frac{1}{1 - \frac{-e}{\pi}} = \frac{\pi}{\pi + e}.
\]
\item[B.] [10~points] Explain why the Integral Test can be applied to the series
$\ds \sum_{n=1}^\infty \frac{n}{e^{3n}}$ and use it to determine if the series
converges or diverges.

\soln
The function $f(x) = x e^{-3x}$ gives the terms in this series when we substitute $x = n$
for $n = 1,2,3,\ldots$.  The function $f(x)$ is continuous for all $x$, and
$f(x) > 0$ for all $x > 0$.  Moreover,
\[
f'(x) = -3x e^{-3x} + e^{-3x} = e^{-3x}(1 - 3x) < 0
\]
for all $x \ge \frac{1}{3}$.  Therefore $f(x)$ is positive and decreasing for $x \ge 1$
and the Integral Test applies.  We integrate by parts with $u = x$, $dv = e^{-3x}$ to compute 
\begin{eqnarray*}
\int_1^\infty x e^{-3x}\, dx &=& \lim_{b\to\infty} \left. \frac{-1}{3} xe^{-3x} - \frac{1}{9}e^{-3x}\right|_1^b\\
&=& \lim_{b\to\infty} \frac{-b}{3e^{3b}} - \frac{1}{9e^{3b}} + \frac{1}{3e^3} + \frac{1}{9e^3}\\
&=& \frac{4}{9e^3}.
\end{eqnarray*}
Since the improper integral converges, the series also converges.
\end{enumerate}
\vskip 10pt
\noindent
V.  All parts of this question refer to the power series 
\[
\sum_{n=1}^\infty \frac{(x - 3)^n}{\sqrt{n}}.
\]
\begin{enumerate}
\item[A.] [9~points] Use the Ratio Test to determine the radius of convergence.

\soln
Applying the Ratio Test, 
\begin{eqnarray*}
\lim_{n\to\infty}\left|\frac{(x-3)^{n+1}}{\sqrt{n+1}}\cdot\frac{\sqrt{n}}{(x-3)^n}\right|
&=& \lim_{n\to\infty} \sqrt{\frac{n}{n+1}} |x - 3|\\
&=& |x - 3|.
\end{eqnarray*}
For absolute convergence, we need $|x - 3| < 1$, so $2 < x < 4$.  
The series is centered at $a = 3$, so the radius of convergence is $1$.

\item[B.] [6~points] Test convergence at the endpoints of the interval from part A to determine 
the interval of convergence.  Explain your conclusions.

\soln
At $x = 2$, we substitute and obtain
\[
\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}.
\]
This is an alternating series and $\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}$
with $\ds \lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0$.  The Alternating Series Test
implies that this is a convergent series.  

At $x = 4$, we have 
\[
\sum_{n=1}^\infty \frac{1}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{n^{1/2}}.
\]
This is a $p$-series with $p = 1/2$.  Since $p < 1$, the series diverges there.
The interval of convergence is $[2,4)$.  
\end{enumerate}
\vskip 10pt
\noindent
VI.  [10~points] Find the Taylor polynomial of degree $n = 3$ for $f(x) = 2 + 3x + x^3$ at $a = 1$.

\soln
We have
\begin{eqnarray*}
f(1) &=& 6\\
f'(x) = 3 + 3x^2 \Rightarrow f'(1) &=& 6\\
f''(x) = 6x \Rightarrow f''(1) &=& 6\\
f'''(x) = 6 \Rightarrow f'''(1) &=& 6.
\end{eqnarray*}
Therefore the Taylor polynomial is 
\[
p_3(x) = 6 + 6(x - 1) + \frac{6}{2!}(x - 1)^2 + \frac{6}{3!}(x - 1)^3 = 6 + 6(x-1) + 3(x-1)^2 + (x-1)^3.
\]
(If you expand this out and collect powers of $x$, you will see that $p_3(x) = f(x)$.)
\vskip 10pt
\noindent
VII. 
\begin{enumerate}
\item[A.] [5~points] Starting from the Taylor series for $\sin(x)$ at $a = 0$,
find a series representation for ${\displaystyle f(x) = \frac{\sin(x) - x}{x^3}}$.
Give the first three nonzero terms in your series.

\soln
Since 
\[
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots,
\]
we have 
\[
\frac{\sin(x) - x}{x^3} = \frac{1}{x^3}\left(-\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\right)
= \frac{-1}{3!} + \frac{x^2}{5!} - \frac{x^4}{7!} + \cdots .
\]
\item[B.] [5~points] 
Use your answer in part A to determine ${\displaystyle \lim_{x\to 0} \frac{\sin(x) - x}{x^3}}$.

\soln
\[
\lim_{x\to 0} \frac{\sin(x) - x}{x^3} = \lim_{x\to 0} \frac{-1}{3!} + \frac{x^2}{5!} - \frac{x^4}{7!} =
\frac{-1}{6}.
\]
\item[C.] [5~points] Use your answer in part A to determine the first three nonzero terms in a series representing the function 
\[
F(x) = \int_0^x \frac{\sin(t) - t}{t^3}\ dt.
\]

\soln
We integrate the series from part A term by term to obtain
\[
\int_0^x \frac{-1}{3!} + \frac{t^2}{5!} - \frac{t^4}{7!} + \cdots \, dt = 
\frac{-x}{3!} + \frac{x^3}{3\cdot 5!} - \frac{x^5}{5\cdot 7!} + \cdots .
\]
A summation notation form would be
\[
\sum_{n=0}^\infty (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)\cdot (2n+3)!}.
\]
\end{enumerate}
\end{document}

