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

April 23, 2003}
\end{center}

\begin{enumerate}

\item  \label{one}
	 Find the Taylor series for $e^{-x}$ centered at $0$.
	\\\sol
	\[
	e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!} - \frac{x^5}{5!} + \cdots
	\]

\inch

\item  We know the series for $e^x$:
	\[
	e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} + \cdots
	\]
	How is this series related to your answer in question~\ref{one}?
	\\\sol
	We can plug $-x$ into our series for $e^x$ to get the series for $e^{-x}$.

\newpage

\lecture
\item  Using a Taylor polynomial for $f(x)=e^x$, find $e$, accurate to within $0.01$:
	\begin{enumerate}
	\item  For any $n$, what is the $n$-th derivative of $f(x)$?  $f^{(n)}(x)=$
	\\\sol $f^{(n)}(x)= e^x$.
\hinch

	\item  Find an integer $M$ so that $|f^{(n)}(x)|\leq M$ for all $x$ in the interval $[-1,1]$.
		(This will be your value for $M$.)
	\\\sol $|e^x|\leq 3$, so $M=3$.
\hinch

	\item  Find the smallest integer $N$ so that $\frac{M(1^{N+1})}{(N+1)!} < 0.01$
	\\\sol I found that for $N=5$ we have $\frac{3}{6!} \approx 0.0042$.
\hinch

	\item  For the value of $N$ that you found, write down the $N$-th Taylor polynomial $T_N(x)$:
	\\\sol
	\[
	T_5(x) = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} + \frac{x^5}{5!}
	\]
\hinch

	\item  For your value of $N$, compute $T_N(1)$.
	\\\sol
	\[
	T_5(1) = 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} + \frac{1}{5!} = \frac{163}{60} \approx 2.7167
	\]
\hinch

	\item  Compare $T_N(1)$ with $e$ on your calculator.
	\\\sol  $R_N(1) = e-T_N(1) \approx 0.001615161$

	\end{enumerate}


\end{enumerate}


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