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\begin{center}

\textbf{College of the Holy Cross, Spring Semester, 2005} \\
\textbf{Math 132, Midterm~3 Solutions} \\
\textbf{April 27, 2005}
\end{center}



\be
  
\item{} [5~points each]\quad Suppose $f(0)=0.5$, $f'(0)=-1$ and
$f''(0)=-2$.
  
  (a) Write down the Taylor polynomial of degree~$2$ for~$f$ near
  $a=0$.


 {\bf Solution:}  The polynomial is 

 $$p_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 0.5 - x - x^2$$
  
  (b) Use your answer to part~(a) to estimate~$f(0.3)$.

  {\bf Solution:} Substitute $x = 0.3$ in $p_2(x)$:
  $$p_2(0.3) = 0.5 - 0.3 - (0.3)^2 = .11$$
  
  (c) Which (if any) of the following could be the graph of~$f$?
  (Recall that $f(0)=0.5$, $f'(0)=-1$ and $f''(0)=-2$. More than one
  answer may be correct.)

  
  $\bigcirc$~(ii)\qquad
  \bigskip

  \begin{center}
    \input{midterm3_2i.eepic}\hfil
    \input{midterm3_2ii.eepic}\hfil
    \input{midterm3_2iii.eepic}\hfil
    \input{midterm3_2iv.eepic}
  \end{center}
\bigskip
{\bf Solution:}  At $a=0$, the function should be positive ($f(0) >0$), 
decreasing ($f'(0) < 0$),
and concave down ($f''(0) < 0$).  Only (ii) is all three.  

\sep
\item{} [10~points]\quad Evaluate (find the sum of) the series
  $\ds 5 - \frac{10}{3} + \frac{20}{9} - \frac{40}{27} + 
  \frac{80}{81} - \cdots$

{\bf Solution:}  This is a geometric series with $a = 5$ and $r = \frac{-2}{3}$.
Since $-1 < r < 1$, the series is convergent and the sum is 
$$\frac{a}{1-r} = \frac{5}{1+\frac{2}{3}} = 3.$$
  \sep

\item{} [10~points]\quad Use the power series for~$e^x$ about $a=0$ to
  find the power series for~$ze^{-z^2}$ about $a=0$. Express your
  answer both in summation form, and by writing out the first four
  nonzero terms.

  {\bf Solution:}  The Taylor series for $e^x$ about $a = 0$ is
  $$\sum_{k=0}^\infty \frac{x^k}{k!}$$
  To get the series for $z e^{-z^2}$, we substitute $x = -z^2$, and 
  multiply the result by $z$ to obtain:
  $$z\sum_{k=0}^\infty \frac{(-z^2)^k}{k!} = 
  z\sum_{k=0}^\infty (-1)^k \frac{z^{2k}}{k!} = 
  \sum_{k=0}^\infty (-1)^k \frac{z^{2k+1}}{k!}$$
  The first four nonzero terms are
  $$z - z^3 + \frac{z^5}{2!} - \frac{z^7}{3!}.$$


\sep

\item{} The distribution of length in a certain population of
  inchworms is described by the pdf $p(x)=c(3x^2-x^3)$ for $0\leq
x\leq 3$.

  (a) [8~points]\quad Find the value of the constant~$c$.

  {\bf Solution:}  The total area under the graph of a density
  must be 1, so 
  
  \begin{eqnarray*}
  1 &=& c \int_0^3 3x^2 - x^3\ dx \\
    &=& c \left(x^3 - \frac{x^4}{4}\right|_0^3\\
    &=& c \frac{27}{4}\\
  \Rightarrow c &=& \frac{4}{27}
  \end{eqnarray*}
  
  \sep

  (b) [7~points]\quad Find the proportion of inchworms with length at
most 2~cm.

 {\bf Solution:}  This proportion is computed by the integral $\int_0^2 p(x)\ dx$,

 which is $\frac{4}{27}\int_0^2 3x^2 - x^3\ dx = \frac{16}{27} \doteq .593$

 (about $59.3\%$ of the inchworms).
  \sep
  
\item{} [5~points each]\quad Use the indicated test to determine
  whether the given series converges or diverges. \textbf{Justify your
  answers.}
  
  (a) $\ds\sum_{n=1}^\infty\frac{n^2}{n^3+10}$; integral test

  {\bf Solution:}  We apply the substitution $u = x^3+10$ to the
  integral
  $$\int \frac{x^2\ dx}{x^3+10}
= \int \frac{\frac{1}{3}\ du}{u} = \frac{1}{3} \ln(u) 
  =\frac{1}{3} \ln(x^3+10) + C.$$
  Then $\int_1^b \frac{x^2\ dx}{x^3+10} = \ln(b^3+10) - \ln(11)$.
  Taking $\lim_{b\to\infty} \ln(b^3+10) - \ln(11)$, we do not get
  a finite value so the integral diverges, and so does the series.
  
  $\bigcirc$\quad Diverges
  \sep
  
  (b) $\ds \sum_{n=1}^\infty\frac{n}{4n^3+3n^2+5}$; comparison test

  {\bf Solution:}  For all $n\ge 1$:
  $$\frac{n}{4n^3+3n^2+5} < \frac{n}{4n^3} = \frac{1}{4n^2} < \frac{1}{n^2}$$
  The series $\sum_{n=1}^\infty \frac{1}{n^2}$ is the $p$-series with 
  $p = 2 > 1$.  Hence it is convergent.  By the comparison test,
  the given series converges too.
  
  $\bigcirc$\quad Converges\hspace*{1in}

\sep  
\item{} [15~points]\quad Let $f(x)=\sqrt{x}$. Use \textbf{the
    definition} to calculate its Taylor polynomial of degree~$3$
  at $a=1$.
 

{\bf Solution:}  We have $f(1) = 1$.  $f'(x) = \frac{1}{2}x^{-1/2}$, so
$f'(1) = \frac{1}{2}$.  $f''(x) = \frac{-1}{4}x^{-3/2}$, so
$f''(1) = \frac{-1}{4}$.  Finally, $f'''(x) = \frac{3}{8}x^{-5/2}$, so
$f'''(1) = \frac{3}{8}$.  The Taylor polynomial is
$$f(1) + f'(1)(x - 1) + \frac{f''(1)}{2}(x-1)^2 + \frac{f'''(1)}{6}(x-1)^3
= 1 + \frac{1}{2} (x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3.$$

\sep  
\item{} Both parts refer to the power series $\ds\sum_{n=1}^\infty
  \frac{(x-1)^n}{n\cdot5^n}$.
  
  (a) [10~points]\quad Use the ratio test to find the radius of
  convergence.

  {\bf Solution:}  
  $$\lim_{n\to\infty} \left|\frac{(x-1)^{n+1}}{(n+1)\cdot 5^{n+1}} 
  \cdot \frac{n\cdot5^n}{(x-1)^n}\right| = \lim_{n\to\infty} \frac{n}{n+1}\cdot\frac{1}{5}\cdot |x-1|
  =\frac{1}{5}|x-1|.$$
  So the radius of convergence is $R = \frac{1}{\frac{1}{5}} = 5$.
  \sep
  
  (b) [5~points]\quad Investigate the endpoint behavior, and determine
  the interval of convergence.

  {\bf Solution:}  At $x = 1 + R = 6$, $\ds\sum_{n=1}^\infty
  \frac{(6-1)^n}{n\cdot5^n} = \sum_{n=1}^\infty \frac{1}{n}$.  This is
  the harmonic series ($p$ series with $p = 1$).  So it diverges.  
  At $x = 1 - R = -4$, we get $\sum_{n=1}^\infty (-1)^n\frac{1}{n}$,
  which converges by the alternating series test.  The interval of
  convergence is $-4 \le x < 6$.  
  
\sep
\item{} [5~points each]\quad Suppose 
  \[
  p(x)=\frac{4}{\pi}\cdot\frac{1}{1+x^2}
  \]
  is the probability density function for the quantity~$x$, for
  $0\leq x\leq 1$.

  (a) Find the mean of~$x$.

  {\bf Solution:}  The mean is given by
  $$\int_{-\infty}^{\infty} xp(x)\ dx = 
  \frac{4}{\pi} \int_0^1\frac{x}{1+x^2}\ dx = \frac{4}{\pi}\cdot \frac{1}{2}\ln(1+x^2)|_0^1 = 
  \frac{2\ln(2)}{\pi} \doteq .4413$$
  \sep

  (b) Find the median of~$x$.

  {\bf Solution:}  The median is the $t$ such that 
  $$\frac{1}{2} = \frac{4}{\pi} \int_0^t \frac{dx}{1+x^2} = \arctan(t)$$
  This says 
  $$\arctan(t) = \frac{\pi}{8}$$
  or $t = \tan\left(\frac{\pi}{8}\right) = \doteq .4142$.
  \ee

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