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