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\problemgrouptitle{Taylor Series}
  
\begin{problemblock}{Find the Taylor series for $f(x)$ centered at the given value of $a$.}

\item $\displaystyle f(x)=e^{-x},\  a=\ln 3$
\answer{
$\displaystyle \frac 13-\frac 13 \left(x-\ln 3 \right)+\frac 16 \left(x-\ln 3 \right)^2-\frac 1{18} \left(x-\ln 3 \right)^3+\ldots$
}

\item $\displaystyle f(x)=\sin \pi x,\  a=\frac 12$
\answer{
$\displaystyle 1-\frac {\pi^2}{2!} \left(x-\frac 12 \right)^2+\frac {\pi^4}{4!} \left(x-\frac 12 \right)^4+\ldots$
}

\item $\displaystyle f(x)=\ln x,\  a=e$
\answer{
$\displaystyle 1+\frac {1}{e} \left(x-e \right)-\frac {1}{2e^2} \left(x-e\right)^2+\frac 1{3e^3}\left(x-e\right)^3+\ldots$
}

\item $\displaystyle f(x)=2^x,\  a=1$
\answer{
$\displaystyle 2+2\ln 2 \left(x-1 \right)+\left(\ln 2\right)^2 \left(x-1\right)^2+\frac 13 \left(\ln 2 \right)^3\left(x-1\right)^3+\ldots$
}

\item $\displaystyle f(x)=\ln (3+x),\  a=1$
\answer{
$\displaystyle \ln 4+\frac 14  \left(x-1 \right)-\frac 1{32}  \left(x-1\right)^2+\frac 1{192} \left(x-1\right)^3+\ldots$
}

\end{problemblock}

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