456 lines
14 KiB
TeX
456 lines
14 KiB
TeX
\documentclass{article}
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\input{../../preamble}
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\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf]
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\newcommand{\lean}[1]{\leanref
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{./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
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{Apostol.Chapter\_1\_11.#1}}
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\begin{document}
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\header{Exercises 1.11}{Tom M. Apostol}
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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*{\verified{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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\lean{exercise\_4a}
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\divider
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Let $x$ be a real number and $n$ an integer.
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Let $m = \floor{x + n}$.
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By definition of the floor function, $m$ is the unique integer such that
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$m \leq x + n < m + 1$.
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Then $m - n \leq x < (m - n) + 1$.
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That is, $m - n = \floor{x}$.
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Rearranging terms we see that $m = \floor{x} + n$ as expected.
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\end{proof}
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\subsection*{\verified{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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\ \vspace{6pt}
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\lean{exercise\_4b\_1}
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\lean{exercise\_4b\_2}
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\divider
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $x$ is an integer.
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Then $x = \floor{x}$ and $-x = \floor{-x}$.
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It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
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\paragraph{Case 2}%
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Suppose $x$ is not an integer.
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Let $m = \floor{-x}$.
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By definition of the floor function, $m$ is the unique integer such that
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$m \leq -x < m + 1$.
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Equivalently, $-m - 1 < x \leq -m$.
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Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
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Then, by definition of the floor function, $\floor{x} = -m - 1$.
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Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
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\paragraph{Conclusion}%
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The above two cases are exhaustive. Thus
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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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\end{proof}
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\subsection*{\verified{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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\lean{exercise\_4c}
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\divider
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Rewrite $x$ and $y$ as the sum of their floor and fractional components:
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$x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
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Now
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\begin{align}
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\floor{x + y}
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& = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
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& = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
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& = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
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& \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1}
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\end{align}
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $\{x\} + \{y\} < 1$.
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Then $\floor{\{x\} + \{y\}} = 0$.
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Substituting this value into \eqref{sub:exercise-4c-eq1} yields
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$$\floor{x + y} = \floor{x} + \floor{y}.$$
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\paragraph{Case 2}%
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Suppose $\{x\} + \{y\} \geq 1$.
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Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
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Thus $\floor{\{x\} + \{y\}} = 1$.
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Substituting this value into \eqref{sub:exercise-4c-eq1} yields
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$$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
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\paragraph{Conclusion}%
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Since the above two cases are exhaustive, it follows
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$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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\end{proof}
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\subsection*{\proceeding{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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\lean{exercise\_4d}
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-5}.
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\end{proof}
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\subsection*{\proceeding{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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\lean{exercise\_4e}
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-5}.
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\end{proof}
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\section*{\proceeding{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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\note{The stated generalization is known as "Hermite's Identity."}
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\begin{proof}
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\lean{exercise\_5}
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\divider
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We prove that for all natural numbers $n$ and real numbers $x$, the following
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identity holds:
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\begin{equation}
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\label{sec:exercise-5-eq1}
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\floor{nx} = \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}}
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\end{equation}
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By definition of the floor function, $x = \floor{x} + r$ for some
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$r \in \ico{0}{1}$.
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Define $S$ as the partition of non-overlapping subintervals
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$$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots,
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\ico{\frac{n-1}{n}}{1}.$$
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By construction, $\cup\; S = \ico{0}{1}$.
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Therefore there exists some $j \in \mathbb{N}$ such that
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\begin{equation}
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\label{sec:exercise-5-eq2}
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r \in \ico{\frac{j}{n}}{\frac{j+1}{n}}.
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\end{equation}
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With these definitions established, we now show the left- and right-hand sides
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of \eqref{sec:exercise-5-eq1} evaluate to the same number.
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\paragraph{Left-Hand Side}%
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Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}.
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By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$.
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Therefore $\floor{nr} = j$.
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Thus
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\begin{align}
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\floor{nx}
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& = \floor{n(\floor{x} + r)} \nonumber \\
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& = \floor{n\floor{x} + nr} \nonumber \\
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& = \floor{n\floor{x}} + \floor{nr}. \nonumber
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& \text{\nameref{sub:exercise-4a}} \\
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& = \floor{n\floor{x}} + j \nonumber \\
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& = n\floor{x} + j. \label{sec:exercise-5-eq3}
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\end{align}
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\paragraph{Right-Hand Side}%
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Now consider the right-hand side of identity \eqref{sec:exercise-5-eq1}.
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We note each summand, by construction, is the floor of $x$ added to a
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nonnegative number less than one.
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Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to
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the total.
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Letting $z$ denote the number of summands that contribute $\floor{x} + 1$,
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we have
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\begin{equation}
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\label{sec:exercise-5-eq4}
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\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + z.
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\end{equation}
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The value of $z$ corresponds to the number of indices $i$ that satisfy
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$$\frac{i}{n} \geq 1 - r.$$
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By \eqref{sec:exercise-5-eq2}, it follows
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\begin{align*}
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1 - r
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& \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\
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& = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}.
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\end{align*}
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Thus we can determine the value of $z$ by instead counting the number of
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indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$
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Rearranging terms, we see that $i \geq n - j$ holds for
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$z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands.
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Substituting the value of $z$ into \eqref{sec:exercise-5-eq4} yields
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\begin{equation}
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\label{sec:exercise-5-eq5}
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\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j.
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\end{equation}
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\paragraph{Conclusion}%
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Since \eqref{sec:exercise-5-eq3} and \eqref{sec:exercise-5-eq5} agree with
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one another, it follows identity \eqref{sec:exercise-5-eq1} holds.
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\end{proof}
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\section*{\unverified{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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Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
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By construction, the number of lattice points in $S$ is
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\begin{equation}
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\label{sec:exercise-6-eq1}
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\sum_{n = a}^b \abs{S_n}.
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\end{equation}
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All that remains is to show $\abs{S_i} = \floor{f(i)}$.
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $f(i)$ is an integer.
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Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
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\paragraph{Case 2}%
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Suppose $f(i)$ is not an integer.
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Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
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$\ioc{0}{\floor{f(i)}}$.
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Once again, that number is $\floor{f(i)}$.
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\paragraph{Conclusion}%
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By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
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Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the
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proof.
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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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\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
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assumption $b > 1$.}
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\subsection*{\unverified{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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Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
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Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
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$0 < y \leq f(x)$.
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By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to
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the sum
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\begin{equation}
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\label{sub:exercise-7a-eq1}
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\sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
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\end{equation}
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Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
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as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
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By construction, $T$ does not introduce any additional lattice points.
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Thus $S$ and $T$ have the same number of lattice points.
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Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
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We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
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lattice points.
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\paragraph{(i)}%
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\label{par:exercise-7a-i}
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Consider the line $L$ overlapping the hypotenuse of $T$.
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By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
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By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
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being vertical.
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Define the slope of $L$ as $$m = \frac{a}{b}.$$
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$H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
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$(i, i * m)$ is a lattice point.
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But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
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Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
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Thus $H_L = 2$.
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\paragraph{(ii)}%
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Next we count the number of lattice points in $T$.
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Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
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with bottom-left corner at $(0, 0)$.
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Let $I_R$ denote the number of interior lattice points of $R$.
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Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
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respectively.
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By \nameref{C:1:07:sub:exercise-4b-eq2},
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\begin{align}
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I_T
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& = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
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& = \frac{1}{2}(I_R - (2 - 2))
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& \text{\nameref{par:exercise-7a-i}} \nonumber \\
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& = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2}
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\end{align}
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Furthermore, since both the adjacent and opposite side of $T$ are not
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included in $T$ and there exist no lattice points on $T$'s hypotenuse
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besides the endpoints, it follows
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\begin{equation}
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\label{sub:exercise-7a-eq3}
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B_T = 0.
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\end{equation}
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Thus the number of lattice points of $T$ equals
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\begin{align}
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I_T + B_T
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& = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\
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& = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\
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& = \frac{(b - 1)(a - 1)}{2}.
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& \text{\nameref{C:1:07:sub:exercise-4a}}
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\label{sub:exercise-7a-eq4}
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\end{align}
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\paragraph{Conclusion}%
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By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal
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to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
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But the number of lattice points of $S$ is the same as that of $T$.
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By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal
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to $$\frac{(b - 1)(a - 1)}{2}.$$
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Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
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\end{proof}
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\subsection*{\proceeding{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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\lean{exercise\_7b}
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\divider
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Let $n = 1, \ldots, b - 1$.
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By hypothesis, $a$ and $b$ are coprime.
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Furthermore, $n < b$ for all values of $n$.
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Thus $an / b$ is not an integer.
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By \nameref{sub:exercise-4b},
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\begin{equation}
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\label{sub:exercise-7b-eq1}
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\floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
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\end{equation}
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Consider the following:
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\begin{align*}
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\sum_{n=1}^{b-1} \floor{\frac{na}{b}}
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& = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
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& = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
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& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
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& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
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& \text{\nameref{sub:exercise-4a}} \\
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& = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
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& \eqref{sub:exercise-7b-eq1} \\
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& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
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\sum_{n=1}^{b-1} a \\
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& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
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\end{align*}
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Rearranging the above yields
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$$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
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Dividing both sides of the above identity concludes the proof.
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\end{proof}
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\section*{\unverified{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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