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\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

April 14, 2003}
\end{center}

\begin{enumerate}

\item  Find the domain of the following functions.
	(For which values of $x$ does the series converge?)
	\\(Use the ratio test and the remember to test the ``end points.''

	\begin{enumerate}
	\item  $f(x) = \sum_{n=0}^\infty x^n$
		\\\sol $(-1,1)$
\inch

	\item  $g(x) = \sum_{n=1}^\infty \frac{(x-3)^n}{n}$
		\\\sol $[2,4)$
\inch

	\item  $h(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$
		\\\sol $(-\infty,\infty)$
\inch

	\item  $k(x) = \sum_{n=0}^\infty n!x^n$
		\\\sol $x=0$
\inch

	\item  $j(x) = \sum_{n=1}^\infty \frac{(x+2)^n}{2^n n^2}$
		\\\sol $[-4,0]$

	\end{enumerate}


\newpage

\lecture

\item  For each of the power series below, determine where the series is centered, 
	and then determine the radius of convergence and the interval of convergence.
\[
\begin{array}{|c|c|c|c|}
\hline
\mbox{Series} & \mbox{Center of Power Series} & \mbox{Radius of Convergence} & \mbox{Interval of Convergence} \\
\hline &&& \\ 
\sum_{n=0}^\infty 3^nx^n & x=0 & R=\frac{1}{3} & \left(-\frac{1}{3}, \frac{1}{3} \right) \\
& & & \\ \hline &&& \\
\sum_{n=0}^\infty 3^n(x-5)^n & x=5 & R=\frac{1}{3} & \left(5-\frac{1}{3}, 5+\frac{1}{3} \right) \\
& & & \\ \hline &&& \\
\sum_{n=0}^\infty 3^n(2x-5)^n & x=\frac{5}{2} & R=\frac{1}{6} 
	& \left(\frac{5}{2}-\frac{1}{6}, \frac{5}{2}+\frac{1}{6} \right) \\
& & & \\ \hline &&& \\
\sum_{n=1}^\infty \frac{1}{n^3}(x+3)^n & x=-3 & R=1 & [-1,1] \\
& & & \\ \hline &&& \\
\sum_{n=1}^\infty \frac{n}{(n+1)^2} (3x-4)^n & x=\frac{4}{3} & R=\frac{1}{3} & 
	\left[\frac{4}{3}-\frac{1}{3}, \frac{4}{3}+\frac{1}{3} \right) \\
& & & \\ \hline 
\end{array}
\]


\item  Find a power series representation for the following functions:
	\begin{enumerate}
	\item  $h(x) = \frac{x}{1-x}$.
		\\\sol $h(x) = x\sum_{n=0}^\infty x^n = \sum_{n=0}^\infty x^{n+1} = \sum_{n=1}^\infty x^{n}$
\inch

	\item  $k(x) = \frac{x}{1-x^3}$
		\\\sol $k(x) = x\sum_{n=0}^\infty (x^3)^n = \sum_{n=0}^\infty x^{3n+1}$
\inch

	\item  $g(x) = \frac{1}{2+x}$ 
		(Hint: you can rearrange $g$ to look like $g(x) = \frac{1}{2(1+x/2)}$
		\\\sol $g(x) = \frac{1}{2}\sum_{n=0}^\infty (-x/2)^n = \sum_{n=0}^\infty \frac{(-1)^n}{2^n}x^n$
	\end{enumerate}

\end{enumerate}


\end{document}