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

Solutions

February 12, 2003}
\end{center}

\begin{enumerate}

\item Suppose you are given the following inequality:
	\[
	\frac{K(b-a)^5}{180n^4} < 0.1
	\]
	Suppose you know the following: $K=10$, $a=1$, $b=5$.
	\begin{enumerate}
	\item What are some possible values for $n$ that make the inequality hold?
	\item  What is the smallest value of $n$ that will make this inequality hold?
		\\\sol You can solve the inequality to get $n > 4.9$.
		We will see in class that we are interested in integers and often
		only even integers, so we will want $n\geq 6$.
	\end{enumerate}

\item  Suppose you are given the following inequality:
	\[
	E \leq \frac{K(b-a)^5}{180n^4}
	\]
	Again, you have $K=10$, $a=1$ and $b=5$ and you have control over $n$.
	You now want to ensure that $E<0.001$.
	\begin{enumerate}
	\item  What are some values of $n$ what will force $E<0.001$?
	\item  What is the smallest value of $n$ that will force $E<0.001$?
		\\\sol Solving the inequality gives: $n>15.4$ 
		(or $n \geq 16$ if we insist that $n$ be an integer).
	\item  Will all values of $n$ larger then the value you found in the previous
		part ensure that $E<0.001$?  Why?
	\end{enumerate}


\newpage

\textbf{Lecture Problems}

\item  Given the error bound formula for Simpson's rule:
	\[
	|E_S| \leq \frac{K(b-a)^5}{180n^4}
	\]
	Where $K$ is some number such that $|f^{(4)}(x)| \leq K$ for all $a\leq x \leq b$.
	We want to investigate what this means for estimating:
	\[
	\int_2^4 \ln x \; dx
	\]
	\begin{enumerate}
	\item  Find a suitable $K$.
		\\\sol Since $f(x) = \ln x$, $f^{(4)}(x) = -\frac{6}{x^4}$.
		Therefore, a suitable $K$ would be $6/16=3/8$ 
		(because for values of $x$ between $2$ and $4$, $x=2$ gives the minimum
		for $6/x^4$).
		(In fact, anything above $3/8$ is also valid, but in some sense, $3/8$ is the
		best choice for $K$.)
	\item  Determine $n$ so that the error in Simpson's rule is less then $0.00001$.
		\\\sol Solve the inequality to get:
		\[
		n^4 \geq \frac{K(b-a)^5}{(180)(0.00001)} = \frac{(3/8)2^5}{180(0.00001)} 
			\cong 6667
		\]
		Taking the $4$-th root we get $n\geq 9$.
	\end{enumerate}

\newpage

\item  Consider the following integral (and function!):
	\[
	G(t) = \int_1^t \frac{1}{x^3} \; dx
	\]
	\begin{enumerate}
	\item  Find a simple expression for $G(t)$ by using the Fundamental Theorem of 
		Calculus (the evaluation part!)
		\\\sol Find an antiderivative and plug in and subtract:
		\[
		G(t) = \frac{1}{2} - \frac{1}{2t^2}
		\]
	\item  What happens to $G(t)$ as $t \ra \infty$?
		(In other words, what is the limit: $\lim_{t\ra \infty} G(t)$?)
		\\\sol $1/2$.
	\item  What does $t \ra \infty$ correspond to in terms of area under the curve?
	\end{enumerate}


\end{enumerate}



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