249 lines
7.0 KiB
TeX
249 lines
7.0 KiB
TeX
\documentclass{article}
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\input{../../preamble}
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\newcommand{\lean}[1]{\href
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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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\hyperlabel{sec:exercise-4}%
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Prove that the greatest-integer function has the properties indicated:
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\subsection*{\proceeding{Exercise 4a}}%
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\hyperlabel{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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\end{proof}
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\subsection*{\proceeding{Exercise 4b}}%
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\hyperlabel{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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\ % Force space prior to *Proof.*
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\begin{enumerate}[(a)]
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\item \lean{exercise\_4b\_1}
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\item \lean{exercise\_4b\_2}
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\end{enumerate}
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\end{proof}
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\subsection*{\proceeding{Exercise 4c}}%
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\hyperlabel{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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\end{proof}
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\subsection*{\proceeding{Exercise 4d}}%
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\hyperlabel{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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\end{proof}
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\subsection*{\proceeding{Exercise 4e}}%
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\hyperlabel{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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\end{proof}
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\section*{\proceeding{Exercise 5}}%
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\hyperlabel{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-hande 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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\hyperlabel{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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Define $S_i = \mathbb{Z} \cap \ioc{0}{f(i)}$ for all $i \in \mathbb{Z}$.
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By definition, the set of lattice points of $S$ is given by
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$$L = \{ (i, j) : i = a, \ldots, b \land j \in S_i \}.$$
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By construction, it follows $$\sum_{j \in S_i} 1 = \floor{f(i)}.$$
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Therefore $$\abs{L}
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= \sum_{i=a}^b \sum_{j \in S_i} 1
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= \sum_{i=1}^b \floor{f(i)}.$$
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\end{proof}
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\section*{Exercise 7}%
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\hyperlabel{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*{\unverified{Exercise 7a}}%
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\hyperlabel{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*{\proceeding{Exercise 7b}}%
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\hyperlabel{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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\end{proof}
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\section*{\unverified{Exercise 8}}%
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\hyperlabel{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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