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\askaboutsolutions

\begin{document}

\ifshowsolutions
  \prsol{7}{October 17, 2001}{\stamp}
\else
  \prprob{7}{October 15, 2001}{Wednesday, Oct. 17}{\stamp}
\fi

\reading Lecture Notes 7. Rosen \S1.7 through Example 15; \S1.8

\ifshowsolutions{}\else
By~9am on the due date, send email to
\begin{center}
\texttt{6042-probs@theory.lcs.mit.edu}
\end{center}
indicating in at most three sentences one or more of the following:
the passage in the reading---\emph{including its page number}---you
\begin{itemize}
\item found most difficult,
\item found most surprising,
\item would like to have discussed in the next Lecture.
\end{itemize}
\fi

\begin{problems}

\problem Assuming you could be sure of getting a 4\% return \emph{forever}
on money you invest, what would be a proper price to pay for an annuity
that paid you \$10,000 a year forever?

\solnospace{$\$10,000 / 0.04 = \$250,000$.}

\problem
There is a number $a$ such that $\sum_{i=1}^\infty i^p$ converges iff $p < a$.

\bparts

\ppart What is the value of $a$?

\solnospace{$-1$}

\ppart Which of the following would be good approaches in a proof that
this value of $a$ is correct?
\begin{enumerate}

\item find a closed form for $\int x^p\; dx$
\item find a closed form for $\int i^x\; dx$
\item induction on $p$
\item consider $\sum_{i=1}^n i^p$ and use induction on $n$
\item compare the series term-by-term with the Harmonic series

\end{enumerate}

\solnospace{The first and last approaches make sense; the second and third
approaches are nonsense, and the fourth approach by proving something
about $\sum_{i=1}^n$ by induction would need the ideas from the good
approaches to handle the induction step anyway, at which point the
induction would be moot.

\emph{Term-by-term approach:} For $p=-1$, the sum is the harmonic series
which we know does not converge.  Since the term $i^p$ is increasing in
$p$ for $i>1$, the sum will be larger, and hence also diverge for $p>-1$.

\emph{By the integral method}, the sum is $\Theta(\int_1^\infty x^p dx)$.

For $p\neq -1$, the indefinite integral is $x^{p+1}/(p+1)$.  If $p<-1$,
then $p+1 < 0$, so $\lim_{x\rightarrow \infty} x^{p+1} =0$ and the
definite integral from 1 to $\infty$ evaluates to $-1/(p+1)$.  Hence the
sum is bounded above, and since it is increasing, it has a finite limit,
i.e., it converges. If $p>-1$, then $p+1 > 0$, so $\lim_{x\rightarrow
\infty} x^{p+1} =\infty$ and the definite integral diverges.  For $p=-1$
the indefinite integral is $\log x$ which also approaches $\infty$ as $x$
approaches $\infty$, so the the definite integral also diverges.}

\eparts

\problem Find the least integer n such that $f(x)$ is $O(x^{n})$ when $f$
is defined by each of the expressions below.  If there is none, say
``none''.

\bparts

\ppart $2x^3 + (\log x) x^2$
\solnospace{3}

\ppart $2x^2 + (\log x) x^3$
\solnospace{4}

\ppart $(1.1)^x$
\solnospace{none}

\ppart $(0.1)^x$
\solnospace{0}

\ppart $\cfrac{x^4 + x^2 + 1}{x^3+1}$
\solnospace{1}

\ppart $\cfrac{x^4 + 5\log x}{x^4 + 1}$
\solnospace{0}

\ppart
$2^{3 \log_2 x^2}$
\solnospace{6}

\iffalse
\ppart $\log x^{\log x}$

\solnospace{none; because
\[
\log x^{\log x} = e^{(\log\log x) \log x} = e^{(\log x)\log\log x} =
x^{\log\log x}
\]
}
\fi

\eparts

\end{problems}

\end{document}

