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\noindent {\bf Question}

\noindent For each of the given functions, calculate its Taylor
series about the given point; also, determine the radius and
interval of convergence of the resulting power series whereever
possible.
\begin{enumerate}
\item $f(x) = x^3+6x^2+5x-7$ about $a =6$;
\item $f(x) = e^{3x}$ about $a=-2$;
\item $f(x) = \cosh(x)$ about $a=1$;
\end{enumerate}


\medskip

\noindent {\bf Answer}

\noindent \begin{enumerate}
\item we start by calculating the derivatives of $f$ at $a =6$:
\[ f^{(0)}(6) =f(6) = 455; f^{(1)}(6) =f'(6) = 185; f^{(2)}(6) = 48;
f^{(3)}(6) = 6; f^{(n)}(6) = 0 \mbox{ for } n\ge 4. \] Hence, the
Taylor series for $f$ centered at $a =6$ is
\[ \sum_{n=0}^\infty \frac{1}{n!} f^{(n)}(6) (x-6)^n = 455 + 185 (x-6)
+ \frac{1}{2} \: 48 (x-6)^2 + \frac{1}{6} \: 6 (x-6)^3. \] The
radius of convergence of this series is $\infty$ (using the root
test, for instance), and so the interval of convergence is ${\bf
R}$.
\item we start by calculating that $f^{(n)}(x) = 3^n e^{3x}$ for $n\ge
0$, and so $f^{(n)}(-2) = 3^n e^{-6}$.  Hence, the Taylor series
for $f$ centered at $a =-2$ is
\[ \sum_{n=0}^\infty \frac{1}{n!} f^{(n)}(-2) (x+2)^n =
e^{-6} \sum_{n=0}^\infty \frac{3^n}{n!} (x+2)^n. \] The radius of
convergence of this series is $\infty$ (using the ratio test, for
instance), and so the interval of convergence is ${\bf R}$.
\item we start here by recalling that
\[ f^{(n)}(x) = \left\{ \begin{array}{ll} \cosh(x) & \mbox{ for }
x\mbox{ even, and } \\ \sinh(x) & \mbox{ for } x \mbox{ odd.}
\end{array}\right. \]
So, we have that $f^{(n)}(1) =\cosh(1) =
\frac{1}{2}(e+\frac{1}{e})$ for $n$ even, and $f^{(n)}(1)
=\sinh(1) = \frac{1}{2}(e-\frac{1}{e})$ for $n$ odd.  Hence, the
Taylor series for $f$ centered at $a =1$ is
\begin{eqnarray*}
\sum_{n=0}^\infty \frac{1}{n!} f^{(n)}(1) (x-1)^n & = &
\sum_{k=0}^\infty \frac{1}{(2k)!} f^{(2k)}(1) (x-1)^{2k} +
\sum_{k=0}^\infty \frac{1}{(2k+1)!} f^{(2k+1)}(1) (x-1)^{2k+1} \\
 & = & \frac{e^2+1}{2e} \sum_{k=0}^\infty \frac{1}{(2k)!} (x-1)^{2k} +
\frac{e^2 -1}{2e} \sum_{k=0}^\infty \frac{1}{(2k+1)!}
(x-1)^{2k+1}.
\end{eqnarray*}
The radius of convergence of this series is $\infty$ (using the
ratio test, for instance), and so the interval of convergence is
${\bf R}$.
\end{enumerate}


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