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\begin{document}

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

Solutions

April 7, 2003}
\end{center}

\begin{enumerate}

\item  Consider the probability density function below:
	\[
	f(x) = \begin{cases}
	0 & x < 0 \\
	\frac{x}{2} & 0 \leq x \leq 2 \\
	0 & x>2
	\end{cases}
	\]
	\begin{enumerate}
	\item  Show that $f(x)$ is a probablity density function.
		\\\sol The main point here is that $\int_{-\infty}^\infty f(x) \; dx = 1$.

	\item  Find the mean, $\mu$, of the random variable.
		\\\sol
		\[
		\mu = \int_{-\infty}^\infty xf(x) \; dx = \int_0^2 \frac{x^2}{2} \; dx = \frac{4}{3}
		\]

	\item  Find the median of the random variable.
	\\Hint: the median is the number $m$ so that $\int_{-\infty}^m f(x) \; dx = 0.5$.
		\\\sol  Just solve:
		\[
		\int_{-\infty}^m f(x) \; dx = \frac{m^2}{4}
		\]
		Now we solve $\frac{m^2}{4} = \frac{1}{2}$ to get $m=\sqrt{2}$.
	\end{enumerate}
	
\item  Suppose a sequence is given recursively as:
	\[
	\begin{cases}
	a_1 = 1 & \\
	a_{n+1} = \frac{1}{4-a_n} & \mbox{ if } n >1
	\end{cases}
	\]
	Suppose we know that $\lim_{n \ra \infty} =L$, find $L$.
	\\\sol
	\[
	L = \lim_{n \ra \infty} a_n = \lim_{n \ra \infty} a_{n+1} = 
	\lim_{n \ra \infty} \frac{1}{4-a_n} = 
	\frac{1}{4 - \lim_{n \ra \infty} a_n}
	= \frac{1}{4-L}
	\]
	We now solve the equation $L=\frac{1}{4-L}$ for $L$, which becomes the quadratic 
	$L^2-4L+1=0$.
	Solving gives $L=2+\sqrt{3}$ or $L=2-\sqrt{3}$.  
	A litte investigation (try computing some terms!) should convince you that 
	$L=2-\sqrt{3} \approx 0.2679492$.

\newpage
\lecture

\item  Lets find an estimate for $S=\sum_{n=1}^\infty \frac{1}{n^3}$.
	\begin{enumerate}
	\item  Using your calculator, find $s_{10}$.
		\\\sol 
		\[
		\frac{1}{1^3} + \frac{1}{2^3} + \cdots + \frac{1}{10^3} \approx 1.197531986
		\]

	\item  Find $\int_{10}^\infty \frac{1}{x^3} \; dx$.
		\\\sol
		\[
		\int_{10}^\infty \frac{1}{x^3} \; dx 
		= \lim_{t \ra \infty} \left. -\frac{1}{2x^2} \right|_{10}^t
		= \frac{1}{200}
		\]

	\item  Find $\int_{11}^\infty \frac{1}{x^3} \; dx$.
		\\\sol
		\[
		\int_{12}^\infty \frac{1}{x^3} \; dx 
		= \lim_{t \ra \infty} \left. -\frac{1}{2x^2} \right|_{11}^t
		= \frac{1}{242}
		\]

	\item  How does $R_{10}$ relate to the previous integrals?
		\\\sol  We know that 
		\[
		\int_{11}^\infty \frac{1}{x^3} \; dx \leq R_{10} \leq \int_{10}^\infty \frac{1}{x^3} \; dx
		\]

	\item  What inequalities can we set up for the sume $S=\sum_{n=1}^\infty \frac{1}{n^3}$?
		\\\sol
		\[
		s_{10} + \int_{11}^\infty \frac{1}{x^3} \; dx 
		\leq S \leq 
		s_{10} + \int_{10}^\infty \frac{1}{x^3} \; dx
		\]
		or
		\[
		1.201664217 \leq \sum_{n=1}^\infty \frac{1}{n^3} \leq 1.202531986
		\]
		
	\item  What is a good estimate for the sum and at most, how far off can you be?
		\\\sol  A good estimate would be the mid-point of the interval above, or:
		\[
		S \approx \frac{1.201664217 + 1.202531986}{2} = 1.2020981015
		\]
		At most we are off half the length of the interval or
		\[
		\mathrm{Error} \leq  \frac{1.202531986 - 1.201664217}{2} = \frac{0.000867769}{2}
		= 0.0004338845
		\]
		
	\end{enumerate}


\end{enumerate}


\end{document}