bookshelf/Bookshelf/Apostol/Chapter_1_11.tex

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\documentclass{article}
\input{../../preamble}
\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf]
\newcommand{\lean}[1]{\leanref
{./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
{Apostol.Chapter\_1\_11.#1}}
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\begin{document}
\header{Exercises 1.11}{Tom M. Apostol}
\section*{Exercise 4}%
\label{sec:exercise-4}
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Prove that the greatest-integer function has the properties indicated:
\subsection*{\verified{Exercise 4a}}%
\label{sub:exercise-4a}
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$\floor{x + n} = \floor{x} + n$ for every integer $n$.
\begin{proof}
\lean{exercise\_4a}
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\divider
Let $x$ be a real number and $n$ an integer.
Let $m = \floor{x + n}$.
By definition of the floor function, $m$ is the unique integer such that
$m \leq x + n < m + 1$.
Then $m - n \leq x < (m - n) + 1$.
That is, $m - n = \floor{x}$.
Rearranging terms we see that $m = \floor{x} + n$ as expected.
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\end{proof}
\subsection*{\verified{Exercise 4b}}%
\label{sub:exercise-4b}
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$\floor{-x} =
\begin{cases}
-\floor{x} & \text{if } x \text{ is an integer}, \\
-\floor{x} - 1 & \text{otherwise}.
\end{cases}$
\begin{proof}
\ \vspace{6pt}
\lean{exercise\_4b\_1}
\lean{exercise\_4b\_2}
\divider
There are two cases to consider:
\paragraph{Case 1}%
Suppose $x$ is an integer.
Then $x = \floor{x}$ and $-x = \floor{-x}$.
It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
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\paragraph{Case 2}%
Suppose $x$ is not an integer.
Let $m = \floor{-x}$.
By definition of the floor function, $m$ is the unique integer such that
$m \leq -x < m + 1$.
Equivalently, $-m - 1 < x \leq -m$.
Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
Then, by definition of the floor function, $\floor{x} = -m - 1$.
Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
\paragraph{Conclusion}%
The above two cases are exhaustive. Thus
$$\floor{-x} =
\begin{cases}
-\floor{x} & \text{if } x \text{ is an integer}, \\
-\floor{x} - 1 & \text{otherwise}.
\end{cases}$$
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\end{proof}
\subsection*{\verified{Exercise 4c}}%
\label{sub:exercise-4c}
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$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
\begin{proof}
\lean{exercise\_4c}
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\divider
Rewrite $x$ and $y$ as the sum of their floor and fractional components:
$x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
Now
\begin{align}
\floor{x + y}
& = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
& = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
& = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
& \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1}
\end{align}
There are two cases to consider:
\paragraph{Case 1}%
Suppose $\{x\} + \{y\} < 1$.
Then $\floor{\{x\} + \{y\}} = 0$.
Substituting this value into \eqref{sub:exercise-4c-eq1} yields
$$\floor{x + y} = \floor{x} + \floor{y}.$$
\paragraph{Case 2}%
Suppose $\{x\} + \{y\} \geq 1$.
Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
Thus $\floor{\{x\} + \{y\}} = 1$.
Substituting this value into \eqref{sub:exercise-4c-eq1} yields
$$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
\paragraph{Conclusion}%
Since the above two cases are exhaustive, it follows
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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\end{proof}
\subsection*{\partial{Exercise 4d}}%
\label{sub:exercise-4d}
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$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
\begin{proof}
\lean{exercise\_4d}
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\divider
This is immediately proven by applying Hermite's Identity as shown in
\nameref{sec:exercise-5}.
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\end{proof}
\subsection*{\partial{Exercise 4e}}%
\label{sub:exercise-4e}
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$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
\begin{proof}
\lean{exercise\_4e}
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\divider
This is immediately proven by applying Hermite's Identity as shown in
\nameref{sec:exercise-5}.
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\end{proof}
\section*{\partial{Exercise 5}}%
\label{sec:exercise-5}
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The formulas in Exercises 4(d) and 4(e) suggest a generalization for
$\floor{nx}$.
State and prove such a generalization.
\note{The stated generalization is known as "Hermite's Identity."}
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\begin{proof}
\lean{exercise\_5}
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\divider
We prove that for all natural numbers $n$ and real numbers $x$, the following
identity holds:
\begin{equation}
\label{sec:exercise-5-eq1}
\floor{nx} = \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}}
\end{equation}
By definition of the floor function, $x = \floor{x} + r$ for some
$r \in \ico{0}{1}$.
Define $S$ as the partition of non-overlapping subintervals
$$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots,
\ico{\frac{n-1}{n}}{1}.$$
By construction, $\cup\; S = \ico{0}{1}$.
Therefore there exists some $j \in \mathbb{N}$ such that
\begin{equation}
\label{sec:exercise-5-eq2}
r \in \ico{\frac{j}{n}}{\frac{j+1}{n}}.
\end{equation}
With these definitions established, we now show the left- and right-hand sides
of \eqref{sec:exercise-5-eq1} evaluate to the same number.
\paragraph{Left-Hand Side}%
Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}.
By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$.
Therefore $\floor{nr} = j$.
Thus
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\begin{align}
\floor{nx}
& = \floor{n(\floor{x} + r)} \nonumber \\
& = \floor{n\floor{x} + nr} \nonumber \\
& = \floor{n\floor{x}} + \floor{nr}. \nonumber
& \text{\nameref{sub:exercise-4a}} \\
& = \floor{n\floor{x}} + j \nonumber \\
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& = n\floor{x} + j. \label{sec:exercise-5-eq3}
\end{align}
\paragraph{Right-Hand Side}%
Now consider the right-hand side of identity \eqref{sec:exercise-5-eq1}.
We note each summand, by construction, is the floor of $x$ added to a
nonnegative number less than one.
Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to
the total.
Letting $z$ denote the number of summands that contribute $\floor{x} + 1$,
we have
\begin{equation}
\label{sec:exercise-5-eq4}
\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + z.
\end{equation}
The value of $z$ corresponds to the number of indices $i$ that satisfy
$$\frac{i}{n} \geq 1 - r.$$
By \eqref{sec:exercise-5-eq2}, it follows
\begin{align*}
1 - r
& \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\
& = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}.
\end{align*}
Thus we can determine the value of $z$ by instead counting the number of
indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$
Rearranging terms, we see that $i \geq n - j$ holds for
$z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands.
Substituting the value of $z$ into \eqref{sec:exercise-5-eq4} yields
\begin{equation}
\label{sec:exercise-5-eq5}
\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j.
\end{equation}
\paragraph{Conclusion}%
Since \eqref{sec:exercise-5-eq3} and \eqref{sec:exercise-5-eq5} agree with
one another, it follows identity \eqref{sec:exercise-5-eq1} holds.
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\end{proof}
\section*{\unverified{Exercise 6}}%
\label{sec:exercise-6}
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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}
Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
By construction, the number of lattice points in $S$ is
\begin{equation}
\label{sec:exercise-6-eq1}
\sum_{n = a}^b \abs{S_n}.
\end{equation}
All that remains is to show $\abs{S_i} = \floor{f(i)}$.
There are two cases to consider:
\paragraph{Case 1}%
Suppose $f(i)$ is an integer.
Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
\paragraph{Case 2}%
Suppose $f(i)$ is not an integer.
Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
$\ioc{0}{\floor{f(i)}}$.
Once again, that number is $\floor{f(i)}$.
\paragraph{Conclusion}%
By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the
proof.
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\end{proof}
\section*{Exercise 7}%
\label{sec:exercise-7}
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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 $0$.
\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
assumption $b > 1$.}
\subsection*{\unverified{Exercise 7a}}%
\label{sub:exercise-7a}
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Derive this result by a geometric argument, counting lattice points in a right
triangle.
\begin{proof}
Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
$0 < y \leq f(x)$.
By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to
the sum
\begin{equation}
\label{sub:exercise-7a-eq1}
\sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
\end{equation}
Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
By construction, $T$ does not introduce any additional lattice points.
Thus $S$ and $T$ have the same number of lattice points.
Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
lattice points.
\paragraph{(i)}%
\label{par:exercise-7a-i}
Consider the line $L$ overlapping the hypotenuse of $T$.
By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
being vertical.
Define the slope of $L$ as $$m = \frac{a}{b}.$$
$H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
$(i, i * m)$ is a lattice point.
But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
Thus $H_L = 2$.
\paragraph{(ii)}%
Next we count the number of lattice points in $T$.
Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
with bottom-left corner at $(0, 0)$.
Let $I_R$ denote the number of interior lattice points of $R$.
Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
respectively.
By \nameref{C:1:07:sub:exercise-4b-eq2},
\begin{align}
I_T
& = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
& = \frac{1}{2}(I_R - (2 - 2))
& \text{\nameref{par:exercise-7a-i}} \nonumber \\
& = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2}
\end{align}
Furthermore, since both the adjacent and opposite side of $T$ are not
included in $T$ and there exist no lattice points on $T$'s hypotenuse
besides the endpoints, it follows
\begin{equation}
\label{sub:exercise-7a-eq3}
B_T = 0.
\end{equation}
Thus the number of lattice points of $T$ equals
\begin{align}
I_T + B_T
& = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\
& = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\
& = \frac{(b - 1)(a - 1)}{2}.
& \text{\nameref{C:1:07:sub:exercise-4a}}
\label{sub:exercise-7a-eq4}
\end{align}
\paragraph{Conclusion}%
By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal
to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
But the number of lattice points of $S$ is the same as that of $T$.
By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal
to $$\frac{(b - 1)(a - 1)}{2}.$$
Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
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\end{proof}
\subsection*{\partial{Exercise 7b}}%
\label{sub:exercise-7b}
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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}
\lean{exercise\_7b}
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\divider
Let $n = 1, \ldots, b - 1$.
By hypothesis, $a$ and $b$ are coprime.
Furthermore, $n < b$ for all values of $n$.
Thus $an / b$ is not an integer.
By \nameref{sub:exercise-4b},
\begin{equation}
\label{sub:exercise-7b-eq1}
\floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
\end{equation}
Consider the following:
\begin{align*}
\sum_{n=1}^{b-1} \floor{\frac{na}{b}}
& = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
& = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
& \text{\nameref{sub:exercise-4a}} \\
& = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
& \eqref{sub:exercise-7b-eq1} \\
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
\sum_{n=1}^{b-1} a \\
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
\end{align*}
Rearranging the above yields
$$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
Dividing both sides of the above identity concludes the proof.
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\end{proof}
\section*{\partial{Exercise 8}}%
\label{sec:exercise-8}
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Let $S$ be a set of points on the real line.
The \textit{characteristic function} of $S$ is, by definition, the function
$\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and
$\mathcal{X}_S(x) = 0$ 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
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
$$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$
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This property is described by saying that every step function is a linear
combination of characteristic functions of intervals.
\begin{proof}
Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$.
Let $k \in N$ such that $x \in I_k$.
Consider an arbitrary $j \in N - \{k\}$.
By definition of a partition, $I_j \cap I_k = \emptyset$.
That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
Therefore, by definition of the characteristic function,
$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all
$j \in N - \{k\}$.
Thus
\begin{align*}
f(x)
& = c_k \\
& = (c_k)(1) + \sum\nolimits_{j \in N - \{k\}} (c_j)(0) \\
& = c_k\mathcal{X}_{I_k}(x) +
\sum\nolimits_{j \in N - \{k\}} c_j\mathcal{X}_{I_j}(x) \\
& = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).
\end{align*}
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\end{proof}
\end{document}