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\title{\vspace*{-1.5in}\textbf{Math 2300-007: Quiz 9}}
\date{}
\author{}


\begin{document}
\maketitle
%\underline{Math 1300-400: Quiz 1}
\lhead{Math 2300-013: Calculus II}
\rhead{March 24, 2018}
\chead{Quiz 9}
\vspace{-.5in}
\noindent
{Name: \underline{\phantom{Aalsdkjf}\textcolor{blue}{Solutions}\phantom{kjflksdjflkdsjf}}} \hfill{Score: \underline{\phantom{lksdjflkdsjf}}}\\[12pt]

\noindent
\begin{enumerate}
\item (5 points) Find the 5th degree Taylor Polynomial for $f(x) = \sin(x)$ centered at $a = 0$.\vskip12pt
\textcolor{blue}{%
\begin{tabular}{l|l|l}
$n$ & $f^{(n)}(x)$ & $f^{(n)}(a) = f^{(n)}(0)$\\\hline
$0$ & $\sin(x)$ & $0$\\
$1$ & $\cos(x)$ & $1$\\
$2$ & $-\sin(x)$ &$0$\\
$3$ & $-\cos(x)$ &$-1$\\
$4$ & $\sin(x)$ & $0$\\
$5$ & $\cos(x)$ & $1$\\
\end{tabular}\vskip12pt
Consequently, $T_5(x)$, the fifth-degree Taylor Polynomial for $f(x) = \sin(x)$, centered at $a=0$ is given by 
\begin{align*}
&T_5(x)\\
&= f(0) + \frac{f'(0)}{1!}(x-0)^1 + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 + \frac{f^{(4)}(0)}{4!}(x-0)^4 + \frac{f^{(5)}(0)}{5!}(x-0)^5\\
&= 0 + \frac{1}{1}x + \frac{0}{2!}x^2 - \frac{1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5\\
&= x - \frac{x^3}{3!} + \frac{x^5}{5!}.
\end{align*}
}
\newpage
\item (5 points) What is the interval of convergence of the series $\displaystyle \sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}(5x-1)^n$?\vskip12pt
\textcolor{blue}{%
First, we apply the ratio test:
\begin{align*}
\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to \infty} \left|\frac{(-3)^{n+1}(5x-1)^{n+1}}{\sqrt{n+1}}\cdot \frac{\sqrt{n}}{(-3)^n(5x-1)^n}\right|\\
&= \lim_{n\to \infty} \left|\frac{3(5x-1)}{1}\cdot \frac{\sqrt{n}}{\sqrt{n+1}}\right|\\
&= \left|3(5x-1)\right| \cdot \lim_{n\to \infty} \frac{\sqrt{n}}{\sqrt{n+1}}\\
&= \left|3(5x-1)\right| \cdot 1.
\end{align*}
In order for the ratio test to imply that the series converges, we need 
\begin{gather*}
-1 < 3(5x-1) < 1\\
\frac{-1}{3} < 5x-1 < \frac{1}{3}\\
\frac{2}{3} < 5x < \frac{4}{3}\\
\frac{2}{15} < x < \frac{4}{15}.
\end{gather*}
This shows that the series converges for $x$ in $\left(\frac{2}{15},\frac{4}{15}\right)$ and diverges for $x < \frac{2}{15}$ and for $x > \frac{4}{15}$. It remains to check what happens at the endpoints:\vskip24pt
\begin{itemize}
\item[\underline{$x = \frac{2}{15}$}:] We have
\[
\sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}(5\cdot\frac{2}{15}-1)^n = \sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}\left(-\frac{1}{3}\right)^n = \sum_{n=1}^\infty\frac{(1)^{n}}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{\sqrt{n}},
\]
which diverges by the $p$-test ($p = 1/2 \leq 1$). 
\item[\underline{$x = \frac{4}{15}$}:] We have
\[
\sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}(5\cdot\frac{4}{15}-1)^n = \sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}\left(\frac{1}{3}\right)^n = \sum_{n=1}^\infty\frac{(-1)^{n}}{\sqrt{n}},
\]
which converges by the Alternating Series Test.
\end{itemize}
Consequently, the interval of convergence for the series $\displaystyle \sum_{n=1}^\infty\frac{(-3)^{n}}{\sqrt{n}}(5x-1)^n$ is
\[
\left(\frac{2}{15},\frac{4}{15}\right].
\]
} 
\end{enumerate}
\end{document}
