bookshelf/Bookshelf/Apostol/Chapter_1_11.tex

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\documentclass{article}
\usepackage{amsmath}
\usepackage[shortlabels]{enumitem}
\input{../../preamble}
\newcommand{\link}[1]{\lean{../..}
{Bookshelf/Apostol/Chapter\_1\_11}
{Apostol.Chapter\_1\_11.#1}
{Chapter\_1\_11.#1}
}
\begin{document}
\section{Exercise 4}%
\label{sec:exercise-4}
Prove that the greatest-integer function has the properties indicated:
\subsection{Exercise 4a}%
\label{sub:exercise-4a}
$\floor{x + n} = \floor{x} + n$ for every integer $n$.
\begin{proof}
\link{exercise\_4a}
\end{proof}
\subsection{Exercise 4b}%
\label{sub:exercise-4b}
$\floor{-x} =
\begin{cases}
-\floor{x} & \text{if } x \text{ is an integer}, \\
-\floor{x} - 1 & \text{otherwise}.
\end{cases}$
\begin{proof}
\ % Force space prior to *Proof.*
\begin{enumerate}[(a)]
\item \link{exercise\_4b\_1}
\item \link{exercise\_4b\_2}
\end{enumerate}
\end{proof}
\subsection{Exercise 4c}%
\label{sub:exercise-4c}
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
\begin{proof}
\link{exercise\_4c}
\end{proof}
\subsection{Exercise 4d}%
\label{sub:exercise-4d}
$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
\begin{proof}
\link{exercise\_4d}
\end{proof}
\subsection{Exercise 4e}%
\label{sub:exercise-4e}
$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
\begin{proof}
\link{exercise\_4e}
\end{proof}
\section{Exercise 5}%
\label{sec:exercise-5}
The formulas in Exercises 4(d) and 4(e) suggest a generalization for
$\floor{nx}$.
State and prove such a generalization.
\begin{proof}
\link{exercise\_5}
\end{proof}
\section{Exercise 6}%
\label{sec:exercise-6}
Recall that a lattice point $(x, y)$ in the plane is one whose coordinates are
integers.
Let $f$ be a nonnegative function whose domain is the interval $[a, b]$, where
$a$ and $b$ are integers, $a < b$.
Let $S$ denote the set of points $(x, y)$ satisfying $a \leq x \leq b$,
$0 < y \leq f(x)$.
Prove that the number of lattice points in $S$ is equal to the sum
$$\sum_{n=a}^b \floor{f(n)}.$$
\begin{proof}
TODO
\end{proof}
\section{Exercise 7}%
\label{sec:exercise-7}
If $a$ and $b$ are positive integers with no common factor, we have the formula
$$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
When $b = 1$, the sum on the left is understood to be $0$.
\subsection{Exercise 7a}%
\label{sub:exercise-7a}
Derive this result by a geometric argument, counting lattice points in a right
triangle.
\begin{proof}
TODO
\end{proof}
\subsection{Exercise 7b}%
\label{sub:exercise-7b}
Derive the result analytically as follows:
By changing the index of summation, note that
$\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$.
Now apply Exercises 4(a) and (b) to the bracket on the right.
\begin{proof}
\link{exercise\_7b}
\end{proof}
\section{Exercise 8}%
\label{sec:exercise-8}
Let $S$ be a set of points on the real line.
The \textit{characteristic function} of $S$ is, by definition, the function
$\chi_S$ such that $\chi_S(x) = 1$ for every $x$ in $S$, and $\chi_S(x) = 0$
for those $x$ not in $S$.
Let $f$ be a step function which takes the constant value $c_k$ on the $k$th
open subinterval $I_k$ of some partition of an interval $[a, b]$.
Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have
$$f(x) = \sum_{k=1}^n c_k\chi_{I_k}(x).$$
This property is described by saying that every step function is a linear
combination of characteristic functions of intervals.
\begin{proof}
TODO
\end{proof}
\end{document}