%% Math 132, Spring, 2004
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\begin{center}
\textbf{College of the Holy Cross, Spring Semester, 2005} \\
\textbf{Math 132 Answers to Review Sheet for Midterm 3}
\end{center}

\be

%% ugly, fragile hackery to make sheet fit on two pages
%\item \raisebox{-26pt}{\parbox{6in}{%
%    \begin{minipage}[b]{4in} 
\item The graph at right shows either a pdf
or a cdf.  Which type of function is it, and why?  If it is a pdf,
sketch the graph of the corresponding cdf; if it is a cdf, sketch
the graph of the corresponding pdf.
%    \end{minipage}\qquad\qquad
%    \input{review3b.eepic}
%  }
%}

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%  \begin{minipage}[b]{4in} 
\soln The graph encloses more than one unit
    of area (and is non-decreasing and seems to
    approach~1 as $x\to\infty$), so it's a cdf. The corresponding
    density has one bump, centered just to the right of $x=0$ .
%  \end{minipage}\qquad\qquad
%  \input{rsoln3a.eepic}
%}

\item Let~$x$ be the size of an email message in kilobytes~(KB). An
  ISP finds that the fraction of emails between $x$~and $x+\Delta x$
  KB~in size is about $cxe^{-0.01x}\Delta x$ (that is, the density
  function for $x$ is $p(x) = c x e^{-0.01x}$).

  (a) $c=(0.01)^2=10^{-4}$
  
  (b) Fraction of email messages that are at most 100~KB
  in size: $1-\frac{2}{e}\simeq0.264$
  
  (c) Fraction that are at least 50~KB in size:
  $\frac{3}{2\sqrt{e}}\simeq0.91$. 
  
  (d) Median size of an email message: About~$167.8$~KB.
  
\item In a certain population, the height~$x$ of an individual in
  inches is normally distributed, with a mean of~68 and a standard
  deviation of~4. In other words, heights are modeled by the density
  function $p(x)=\frac{1}{4\sqrt{2\pi}}e^{-(x-68)^2/32}$.

  (a) Write down a definite integral whose value gives the fraction of
  the population whose height is between 68~and 72~inches, and perform
  the change of variables $z=(x-68)/4$ on your integral.

  \soln $\ds\frac{1}{4\sqrt{2\pi}}\int_{68}^{72} e^{-(x-68)^2/32}\,dx
  =\frac{1}{\sqrt{2\pi}}\int_0^1 e^{-z^2/2}\,dz$.

  (b) Write out the Taylor series for $e^u$ giving both the first four
  non-zero terms and the summation form.

  \soln $\ds e^u=\sum_{k=0}^\infty \frac{u^k}{k!} 
  = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$ 
 
  (c) Use your answer from (b) to find the Taylor series for
  \[
  \int_0^x e^{-z^2/2}\,dz,
  \]
  giving both the first four non-zero terms and the summation form.

  \soln $\ds\int_0^x e^{-z^2/2}\,dz
  =\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)\cdot2^k\cdot k!}
  = x - \frac{x^3}{3\cdot2} + \frac{x^5}{5\cdot2^2\cdot2!}
  - \frac{x^7}{7\cdot2^3\cdot3!} + \cdots$

  (d) Estimate the integral in part~(a) using the four-term series 
  from part~(c).
  
  \soln $\ds\frac{1}{\sqrt{2\pi}}\int_0^1 e^{-z^2/2}\,dz 
  \simeq\frac{1}{\sqrt{2\pi}}
   \sum_{k=0}^3 \frac{(-1)^k}{(2k+1)\cdot2^k\cdot k!}
  =\frac{1}{\sqrt{2\pi}}\Bigl(
  1-\frac{1}{6}+\frac{1}{40}-\frac{1}{336}\Bigr)\simeq0.34$

  (The absolute error of this estimate is at most
  $\frac{1}{\sqrt{2\pi}}\cdot\frac{1}{9\cdot2^4\cdot4!}\simeq0.0001$

\item
  
  (a) Use the Comparison Test to determine whether or not
  $\ds\sum_{n=0}^\infty\frac{n+3^n}{2^n}$ converges.

  \soln Diverges by comparison with
  $\ds\sum_{n=0}^\infty\frac{3^n}{2^n}
  =\sum_{n=0}^\infty\left(\frac{3}{2}\right)^n$.

  (b) Use the Integral Test to determine whether or not
  $\ds\sum_{k=0}^\infty\frac{k}{e^k}$ converges.

  \soln Converges (see also Question~2 above.)
  
  (c) Use the Ratio Test to determine whether or not
  $\ds\sum_{k=0}^\infty \frac{3^n}{n!}$ converges.

  \soln Converges, since
  $\ds\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
  =\lim_{n\to\infty}\left|\frac{3}{n+1}\right|=0<1$.
  
\item Determine (with justification!) whether or not the following
  series converge:
  \[
  \sum_{k=1}^\infty \frac{1}{\sqrt{k}},\qquad
  \sum_{k=3}^\infty (-1)^{k-1}\frac{1}{\ln k},\qquad
  \sum_{n=10}^\infty \frac{n^2+n}{3n^3-1000},\qquad
  \sum_{n=1}^\infty \frac{1}{n^{1.01}}.
  \]

  \soln The first and third diverge: Use the integral test for the
  first, comparison with the harmonic series for the third. The
  second and fourth converge: The second is alternating, and the terms
  decrease to~0 in absolute value, while the fourth converges by the
  integral test.

\vfil\eject
  
\item Let $f(x)=\sqrt{1+x}=(1+x)^{1/2}$. Find the 4th degree Taylor
  polynomial of~$f$ centered at $a=0$.  Find a factorial
  expression for the general term of the Taylor series.

  \soln $\ds p_4(x)
  =1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{15}{16\cdot24}x^4$.
  The first few derivatives are $\ds f'(0)=\frac{1}{2}$,
  \[
  f^{(2)}(0)=-\frac{1}{2}\cdot\frac{1}{2}\qquad
  f^{(3)}(0)=\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\qquad
  f^{(4)}(0)=-\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\qquad
  f^{(5)}(0)=\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}.
  \]
  For $k\geq2$, the general expression is
  \[
  f^{(k)}(0)=(-1)^{k-1}\frac{(2k-3)}{2}\cdot\frac{(2k-5)}{2}\cdots
  \frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}
%  =(-1)^{k-1}\frac{(2k-3)!}{2^k}\cdot
%  \frac{1}{(2k-2)(2k-4)(2k-6)\cdots6\cdot4\cdot2}
  =(-1)^{k-1}\frac{(2k-2)!}{2^k}\cdot
  \frac{1}{2^{k-1}(k-1)!}
  \]
  We do not expect you to construct such formulas, and certainly not
  under test pressure.

\item Consider the geometric series
  $f(x)=\ds\sum_{k=0}^\infty x^k=\frac{1}{1-x}$.
  
  (a) Differentiate and multiply by~$x$; the Taylor series of
  $xf'(x)=\ds\frac{x}{(1-x)^2}$ is $\ds\sum_{k=0}^\infty kx^k$.

  (b) Integrate; the Taylor series of~$-\ln(1-x)$ is
  $\ds\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}
  =\sum_{k=1}^\infty \frac{x^k}{k}$.
  
  (c) Find the radius of convergence of the series in part~(b), and
  investigate convergence at the endpoints.

  \soln The radius is~1, the interval of convergence is $-1\leq x<1$.

  (d) Set $x=1/2$ to see that 
  $\ds\sum_{k=1}^\infty\frac{1}{k\cdot 2^k}=\ln 2$,
  $\ds\sum_{k=1}^\infty\frac{k}{2^k}=2$.

\item For each of the given power series, find the interval of
  convergence.
  \[
  f(x)=\sum_{n=1}^\infty\frac{(2x)^n}{\sqrt{n}},\qquad
  g(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-5)^n}{n\cdot3^n}.
  \]
  (In particular, give the radius of convergence, and investigate
  convergence at the endpoints.)

  \soln For~$f(x)$, $-\frac{1}{2}\leq x<\frac{1}{2}$. For~$g(x)$,
  $2<x\leq 8$. (The first has radius~$1/2$, the second has radius~3.)

\vfil\eject
  
\item The second degree Taylor polynomial of~$f(x)$ at $a = 0$ is
  $p_2(x) = C_0 + C_1x + C_2x^2$.  What can you say about the signs of
  $C_0$, $C_1$, and~$C_2$ if you know the graph of~$f(x)$ is:
  \begin{center}
%    \input{review3a.eepic}
  \end{center}

  \soln $C_0<0$, $C_1$~and $C_2>0$.
  
\item Use the error bound for Taylor approximations to estimate the
  number of decimal places of accuracy if the 6th degree Taylor
  polynomial at $a = 0$ is used to approximate $\cos(0.8)$.  Do the
  same for the $n$th degree polynomial in general.  What happens to
  the error bound as $n\to\infty$?

  \soln With a polynomial of degree~$2n$, the ``standard'' error bound
  is
  \[
  |E_{2n}(0.8)|\leq\frac{(0.8)^{2n+2}}{(2n+2)!},
  \]
  the absolute value of the first term of the Taylor series that is
  omitted from the approximating polynomial. (This observation rests
  on the fact that the series for~$\cos(0.8)$ alternates and the terms
  decrease in absolute value.) 

  In particular, the degree-6 polynomial estimate is accurate to
  $\pm4.16\times10^{-6}$, or 5~decimal places. The error bounds
  decrease rapidly to~0 as $n\to\infty$: The numerators decrease
  exponentially and the denominators grow ``super-exponentially''
  (asymptotically faster than any exponential function).

\ee

\end{document}
