% Use % to comment
\documentclass[]{article} % Delete `twocolumn' to have one column text
\usepackage[]{graphicx}
\usepackage[dvips]{color}
\usepackage{amssymb}
\usepackage{amsmath}

%\pagestyle{plain} \addtolength{\topmargin}{-.8in}
%\addtolength{\textheight}{2.0in}
%\addtolength{\evensidemargin}{-0.75in}
%\addtolength{\oddsidemargin}{-1in} \addtolength{\textwidth}{2in}
\markright{Solution Keys\hfill Page }


\newcommand{\bt}{\begin{tabular*}{\textwidth}}
\newcommand{\et}{\end{tabular*}}
%
\newcommand{\n}{\noindent}

\pagestyle{myheadings}

\begin{document}
%\baselineskip 16pt
%\thispagestyle{empty}
\n {\bf \textcolor{red}{Note:}} \textcolor{blue}{Solutions to
Sample Test 2}

%\begin{enumerate}
\n {\bf \textcolor{red}{[\#1]}} (a) Let $h$ be the water level
from the bottom, $\Delta h$ be the infinitesimal thickness of the
water slab at height $h$. The work required to pump that slab of
water over the top rim is $62.5 \times 6^2\times \Delta h\times
(2-h)$ in lb$\times$ft. A Riemann sum approximating the total work
is $\sum 62.5(36)(2-h)\Delta h$. (b) $\int_0^2
62.5(36)(2-h)dh=\lim_{\Delta h\to 0}\sum 62.5(36)(2-h)\Delta h$.
(c) Total force on one side: $\int_0^2 62.5(2-h)(6)dh$, here
$(2-h)$ is the depth measuring from the top rim.  

\n {\bf \textcolor{red}{[\#2]}} (a) $\sum \rho(x)\Delta x =\sum
(5+4\cos(2x))\Delta x$. (b) $\int_0^2(5+4\cos(2x))dx$.

\n {\bf \textcolor{red}{[\#3]}} $f(x)=\ln(1-3x)$, $f(0)=\ln 1=0,
f'(0)=\frac{1}{1-3x}(-3)|_{x=0}=-3,f''(0)=(-1)(1-3x)^{-2}(-3)^2|_{x=0}=-9,
f'''(0)=(-1)(-2)(1-3x)^{-3}(-3)^3|_{x=0}=-54$. The Taylor series
is
\[
f(0)+\frac{f'(0)}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots
=-3x-\frac{9}{2}x^2-9x^3-\cdots
\]

\n {\bf \textcolor{red}{[\#4]}}
\[
f(x)=x^3e^{-x^2}=\sum_{k=1}^\infty
x^3\frac{(-x^2)^k}{k!}=\sum_{k=1}^\infty
(-1)^k\frac{x^{2k+3}}{k!}=x^3-x^5+\frac{x^7}{2!}+\cdots
\]

\n {\bf \textcolor{red}{[\#5]}} $C_n=\frac{n}{4^n}$
$\lim_{n\to\infty}\frac{|C_n|}{|C_{n+1}|}=\lim_{n\to\infty}\frac{n}{n+1}4=4=R$.
The power series converges for $|x|<4$.

\n {\bf \textcolor{red}{[\#6]}} It is geometric series from the
second term onwards, with $3(\frac{2}{5})$ as the leading term and
$r=-\frac{2}{5}$ as the ratio. Hence
\[
20+3\left(\frac{2}{5}\right)-3\left(\frac{2}{5}\right)^2+
3\left(\frac{2}{5}\right)^3-\cdots
=20+\frac{3(2/5)}{1-(-2/5)}=\frac{146}{7}.
\]

\n {\bf \textcolor{red}{[\#7]}} $f(x)=\sum_{k=0}^\infty
(-1)^k\frac{3^k k^2}{k!}x^{4k+1}$. (a) $f^{(101)}(0)$ is the
coefficient of the $x^{101}$ times $101!$. Since the $x^{101}$
term corresponds to $k=25$, we have
$f^{(101)}(0)=(-1)^{25}\frac{3^{25}(25)^2}{25!}(101!)=-\frac{3^{25}(25)^2}{25!}(101!)$
There is no even terms in the power series, or in other words the
coefficients of all even terms are zeros. Therefore
$f^{(102)}(0)=0$. (b) $f'(x)=\sum_{k=0}^\infty (-1)^k\frac{3^k
k^2(4k+1)}{k!}x^{4k}$


\end{document}

















ddd
