Finish Apostol 1.15.
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@ -57,15 +57,17 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}.
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Let $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let
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Let $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let
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$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{sec:def-partition} of $[a, b]$
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$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{sec:def-partition} of $[a, b]$
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such that $s$ is constant on the open subintervals of $P$.
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such that $s$ is constant on the open subintervals of $P$.
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Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval,
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Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval
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so that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k,
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of $P$, so that
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\quad k = 1, 2, \ldots, n.$$
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$$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k, \quad k
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= 1, 2, \ldots, n.$$
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The \textbf{integral of $s$ from $a$ to $b$}, denoted by the symbol
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The \textbf{integral of $s$ from $a$ to $b$}, denoted by the symbol
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$\int_a^b s(x)\mathop{dx}$, is defined by the following formula:
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$\int_a^b s(x)\mathop{dx}$, is defined by the following formula:
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$$\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}).$$
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$$\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}).$$
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Furthermore, we define
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If $a < b$, then we define
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$$\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx} \quad\text{and}\quad
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$$\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}.$$
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\int_a^a s(x) \mathop{dx} = 0.$$
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If $a = b$, then we define
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$$\int_a^b s(x) \mathop{dx} = 0.$$
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\section{\defined{Partition}}%
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\section{\defined{Partition}}%
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\label{sec:def-partition}
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\label{sec:def-partition}
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@ -1707,8 +1709,10 @@ Then
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$P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
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$P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
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$s$ and $t$ remain constant on every open subinterval of $P$.
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$s$ and $t$ remain constant on every open subinterval of $P$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
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Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
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Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$.
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$P$.
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Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
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$P$.
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By definition of the \nameref{sec:def-integral-step-function},
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\begin{align*}
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\int_a^b \left[ s(x) + t(x) \right] \mathop{dx}
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\int_a^b \left[ s(x) + t(x) \right] \mathop{dx}
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@ -1735,14 +1739,14 @@ For every real number $c$, we have
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By definition of a step function, there exists a \nameref{sec:def-partition}
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
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$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
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subinterval of $P$.
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subinterval of $P$.
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Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
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$P$.
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Then $c \cdot s$ is a step function with step partition $P$.
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Then $c \cdot s$ is a step function with step partition $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\begin{align*}
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\int_a^b c \cdot s(x) \mathop{dx}
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\int_a^b c \cdot s(x) \mathop{dx}
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& = \sum_{k=1}^n (c \cdot s(x)) \cdot (x_k - x_{k-1}) \\
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& = \sum_{k=1}^n c \cdot s_k \cdot (x_k - x_{k-1}) \\
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& = \sum_{k=1}^n c \cdot (s(x) \cdot (x_k - x_{k-1})) \\
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& = c \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\
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& = c \sum_{k=1}^n s(x) \cdot (x_k - x_{k-1}) \\
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& = c \int_a^b s(x) \mathop{dx}.
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& = c \int_a^b s(x) \mathop{dx}.
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\end{align*}
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\end{align*}
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@ -1791,8 +1795,10 @@ If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
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By construction, $P$ is a step partition for both $s$ and $t$.
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By construction, $P$ is a step partition for both $s$ and $t$.
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Thus $s$ and $t$ remain constant on every open subinterval of $P$.
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Thus $s$ and $t$ remain constant on every open subinterval of $P$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
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Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
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Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$.
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$P$.
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Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
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$P$.
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By definition of the \nameref{sec:def-integral-step-function},
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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\int_a^b s(x) \mathop{dx}
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@ -1823,7 +1829,8 @@ Then
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as a subdivision point.
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as a subdivision point.
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Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
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Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
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that $x_i = c$.
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that $x_i = c$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $Q$.
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Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
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$Q$.
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By definition of the \nameref{sec:def-integral-step-function},
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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\int_a^b s(x) \mathop{dx}
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@ -1857,7 +1864,8 @@ Then
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By definition of a step function, there exists a \nameref{sec:def-partition}
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By definition of a step function, there exists a \nameref{sec:def-partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
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$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
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subinterval of $P$.
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subinterval of $P$.
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Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$.
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Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
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$P$.
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Let $c$ be a real number.
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Let $c$ be a real number.
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Then $t(x) = s(x - c)$ is a step function on closed interval $[a + c, b + c]$
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Then $t(x) = s(x - c)$ is a step function on closed interval $[a + c, b + c]$
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@ -1938,7 +1946,7 @@ Then
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Let $s$ be a step function on closed interval $[a, b]$.
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Let $s$ be a step function on closed interval $[a, b]$.
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Then
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Then
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$$\int_a^b s(x) \mathop{dx} = - \int_{-b}^{-a} s(-x) \mathop{dx}.$$
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$$\int_a^b s(x) \mathop{dx} = -\int_{-b}^{-a} s(-x) \mathop{dx}.$$
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\begin{proof}
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\begin{proof}
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@ -1951,4 +1959,498 @@ Then
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\end{proof}
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\end{proof}
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\chapter{Exercises 1.15}%
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\label{chap:exercises-1.15}
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\section{Exercise 1.15.1}%
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\label{sec:exercise-1.15.1}
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Compute the value of each of the following integrals.
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\subsection{\partial{Exercise 1.15.1a}}%
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\label{sub:exercise-1.15.1a}
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$\int_{-1}^3 \floor{x} \mathop{dx}$.
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\begin{proof}
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Let $s(x) = \floor{x}$ with domain $[-1, 3]$.
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By construction, $s$ is a step function with partition
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$P = \{-1, 0, 1, 2, 3\} = \{x_0, x_1, x_2, x_3, x_4\}$.
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Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
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$P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_{-1}^3 \floor{x} \mathop{dx}
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& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
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& = -1 + 0 + 1 + 2 \\
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& = 2.
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\end{align*}
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\end{proof}
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\subsection{\partial{Exercise 1.15.1c}}%
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\label{sub:exercise-1.15.1c}
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$\int_{-1}^3 \left(\floor{x} + \floor{x + \frac{1}{2}}\right) \mathop{dx}$.
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\begin{proof}
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Let $s(x) = \floor{x} + \floor{x + \frac{1}{2}}$ with domain $[-1, 3]$.
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By construction, $s$ is a step function with partition
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\begin{align*}
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P
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& = \{-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2},
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3\} \\
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& = \{x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8\}.
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\end{align*}
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Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
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$P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_{-1}^3 \floor{x} + \floor{x + \frac{1}{2}}
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& = \sum_{k=1}^8 s_k \cdot (x_k - x_{k-1}) \\
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& = \frac{1}{2} \sum_{k=1}^8 s_k \\
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& = \frac{1}{2} \left( -2 - 1 + 0 + 1 + 2 + 3 + 4 + 5 \right) \\
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& = 6.
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\end{align*}
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\end{proof}
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\subsection{\partial{Exericse 1.15.1e}}%
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\label{sub:exercise-1.15.1e}
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$\int_{-1}^3 \floor{2x} \mathop{dx}$.
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\begin{proof}
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Let $s(x) = \floor{2x}$.
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By \nameref{sec:hermites-identity},
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$s(x) = \floor{x} + \floor{x + \frac{1}{2}}$.
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Thus, by \nameref{sub:exercise-1.15.1c},
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$$\int_{-1}^3 \floor{2x} \mathop{dx} = 6.$$
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\end{proof}
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\section{\partial{Exercise 1.15.3}}%
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\label{sec:exercise-1.15.3}
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Show that
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$\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx} = a - b$.
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\begin{proof}
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Let $s(x) = \floor{x}$ and $t(x) = \floor{-x}$, both with domain $[a, b]$.
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Let $x_1$, $\ldots$, $x_{n-1}$ denote the integers found in interval $(a, b)$.
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Then $P = \{x_0, x_1, \ldots, x_n\}$, $x_0 = a$ and $x_n = b$, is a step
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\nameref{sec:def-partition} of both $s$ and $t$.
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Let $s_k$ and $t_k$ denote the constant values $s$ and $t$ take on the $k$th
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open subinterval of $P$ respectively.
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By \nameref{sub:exercise-1.11.4b}, $\floor{-x} = -\floor{x} - 1$ for all $x$
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in every open subinterval of $P$.
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That is, $s_k = -t_k - 1$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx}
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& = \sum_{k=1}^n s_k (x_k - x_{k-1}) +
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\sum_{k=1}^n t_k (x_k - x_{k-1}) \\
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& = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (s_k + t_k) \\
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& = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (-t_k - 1 + t_k) \\
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& = \sum_{k=1}^n (x_{k-1} - x_k) \\
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& = x_0 - x_n \\
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& = a - b.
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\end{align*}
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\end{proof}
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\section{Exercise 1.15.5}%
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\label{sec:exercise-1.15.5}
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\subsection{\partial{Exercise 1.15.5a}}%
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\label{sub:exercise-1.15.5a}
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Prove that $\int_0^2 \floor{t^2} \mathop{dt} = 5 - \sqrt{2} - \sqrt{3}$.
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\begin{proof}
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Let $s(t) = \floor{t^2}$ with domain $[0, 2]$.
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Then $s$ is a \nameref{sec:def-step-function} with partition
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$P = \{0, 1, \sqrt{2}, \sqrt{3}, 2\} = \{x_0, x_1, \ldots, x_4\}$.
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Let $s_k$ denote the constant value that $s$ takes in the $k$th open
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subinterval of $P$.
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By the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_0^2 \floor{t^2} \mathop{dt}
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& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
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& = 0 \cdot (1 - 0) + 1 \cdot (\sqrt{2} - 1) +
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2 \cdot (\sqrt{3} - \sqrt{2}) + 3 \cdot (2 - \sqrt{3}) \\
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& = 5 - \sqrt{2} - \sqrt{3}.
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\end{align*}
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\end{proof}
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\subsection{\partial{Exercise 1.15.5b}}%
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\label{sub:exercise-1.15.5b}
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Compute $\int_{-3}^3 \floor{t^2} \mathop{dt}$.
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\begin{proof}
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Let $s(t) = \floor{t^2}$ with domain $[0, 3]$.
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Then $s$ is a \nameref{sec:def-step-function} with \nameref{sec:def-partition}
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\begin{align*}
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P
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& = \{\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5},
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\sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}\} \\
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& = \{x_0, x_1, \ldots, x_9\}.
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\end{align*}
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Let $s_k$ denote the constant value that $s$ takes in the $k$th open
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subinterval of $P$.
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By the \nameref{sec:def-integral-step-function},
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\begin{align}
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\int_0^3 \floor{t^2} \mathop{dt}
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& = \sum_{k=1}^9 s_k \cdot (x_k - x_{k-1})
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\nonumber \\
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& = \sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k}).
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\label{sub:exercise-1.15.5b-eq1}
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\end{align}
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We notice $\floor{t^2}$ is symmetric about the $y$-axis.
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Thus
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\begin{equation}
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\label{sub:exercise-1.15.5b-eq2}
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\int_{-3}^0 \floor{t^2} \mathop{dt} = \int_0^3 \floor{t^2}.
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\end{equation}
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By \nameref{sec:step-additivity-with-respect-interval-integration},
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\begin{align*}
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\int_{-3}^3 \floor{t^2} \mathop{dt}
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& = \int_{-3}^0 \floor{t^2} \mathop{dt} +
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\int_0^3 \floor{t^2} \mathop{dt} \\
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& = 2\int_0^3 \floor{t^2} \mathop{dt}
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& \eqref{sub:exercise-1.15.5b-eq2} \\
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& = 2 \left[\sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k})\right].
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& \eqref{sub:exercise-1.15.5b-eq1}
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\end{align*}
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\end{proof}
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\section{Exercise 1.15.7}%
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\label{sec:exercise-1.15.7}
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\subsection{\partial{Exercise 1.15.7a}}%
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||||||
|
\label{sub:exercise-1.15.7a}
|
||||||
|
|
||||||
|
Compute $\int_0^9 \floor{\sqrt{t}} \mathop{dt}$.
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
Let $s(t) = \floor{\sqrt{t}}$ with domain $[0, 9]$.
|
||||||
|
Then $s$ is a \nameref{sec:def-step-function} with \nameref{sec:def-partition}
|
||||||
|
$P = \{0, 1, 4, 9\} = \{x_0, x_1, x_2, x_3\}$.
|
||||||
|
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
|
||||||
|
subinterval of $P$.
|
||||||
|
By the \nameref{sec:def-integral-step-function},
|
||||||
|
\begin{align*}
|
||||||
|
\int_0^9 \floor{\sqrt{t}} \mathop{dt}
|
||||||
|
& = \sum_{k=1}^3 s_k \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = 0 \cdot (1 - 0) + 1 \cdot (4 - 1) + 2 \cdot (9 - 4) \\
|
||||||
|
& = 13.
|
||||||
|
\end{align*}
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.7b}}%
|
||||||
|
\label{sub:exercise-1.15.7b}
|
||||||
|
|
||||||
|
If $n$ is a positive integer, prove that
|
||||||
|
$$\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = n(n - 1)(4n + 1) / 6.$$
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
Define predicate $P(n)$ as
|
||||||
|
\begin{equation}
|
||||||
|
\label{sub:exercise-1.15.7b-eq1}
|
||||||
|
\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = \frac{n(n - 1)(4n + 1)}{6}.
|
||||||
|
\end{equation}
|
||||||
|
We use induction to prove $P(n)$ holds for all integers satisfying $n > 0$.
|
||||||
|
|
||||||
|
\paragraph{Base Case}%
|
||||||
|
|
||||||
|
Let $n = 1$.
|
||||||
|
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, 1]$.
|
||||||
|
Then $s$ is a \nameref{sec:def-step-function} with
|
||||||
|
\nameref{sec:def-partition} $P = \{0, 1\} = \{x_0, x_1\}$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$P$.
|
||||||
|
By definition of the \nameref{sec:def-integral-step-function}, the left-hand
|
||||||
|
side of \eqref{sub:exercise-1.15.7b-eq1} evaluates to
|
||||||
|
\begin{align*}
|
||||||
|
\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt}
|
||||||
|
& = \int_0^1 \floor{\sqrt{t}} \mathop{dt} \\
|
||||||
|
& = \sum_{k=1}^1 s_k \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = 0.
|
||||||
|
\end{align*}
|
||||||
|
The right-hand side of \eqref{sub:exercise-1.15.7b-eq1} likewise evaluates
|
||||||
|
to $0$.
|
||||||
|
Thus $P(1)$ holds.
|
||||||
|
|
||||||
|
\paragraph{Induction Step}%
|
||||||
|
|
||||||
|
Let $n > 0$ be a positive integer and suppose $P(n)$ is true.
|
||||||
|
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, (n + 1)^2]$.
|
||||||
|
Then $s$ is a \nameref{sec:def-step-function} with
|
||||||
|
\nameref{sec:def-partition}
|
||||||
|
\begin{align*}
|
||||||
|
P
|
||||||
|
& = \{0, 1, 4, \ldots, n^2, (n + 1)^2\} \\
|
||||||
|
& = \{x_0, x_1, \ldots, x_n, x_{n + 1}\}.
|
||||||
|
\end{align*}
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$P$.
|
||||||
|
By definition of the \nameref{sec:def-integral-step-function}, it follows
|
||||||
|
that
|
||||||
|
\begin{align*}
|
||||||
|
& \int_0^{(n + 1)^2} s(x) \mathop{dx} \\
|
||||||
|
& = \sum_{k=1}^{n + 1} s_k \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) +
|
||||||
|
\left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\
|
||||||
|
& = \int_0^{n^2} s(x) \mathop{dx} +
|
||||||
|
\left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\
|
||||||
|
& = \int_0^{n^2} s(x) \mathop{dx} +
|
||||||
|
\left[ n \cdot ((n + 1)^2 - n^2) \right] \\
|
||||||
|
& = \int_0^{n^2} s(x) \mathop{dx} + \left[ 2n^2 + n \right] \\
|
||||||
|
& = \frac{n(n - 1)(4n + 1)}{6} + 2n^2 + n
|
||||||
|
& \text{induction hypothesis} \\
|
||||||
|
& = \frac{n(n - 1)(4n + 1) + 12n^2 + 6n}{6} \\
|
||||||
|
& = \frac{4n^3 + 9n^2 + 5n}{6} \\
|
||||||
|
& = \frac{(n^2 + n)(4n + 5)}{6} \\
|
||||||
|
& = \frac{(n + 1)((n + 1) - 1)(4(n + 1) + 1)}{6}.
|
||||||
|
\end{align*}
|
||||||
|
Thus $P(n + 1)$ holds.
|
||||||
|
|
||||||
|
\paragraph{Conclusion}%
|
||||||
|
|
||||||
|
By mathematical induction, it follows for all positive integers $n$, $P(n)$
|
||||||
|
is true.
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\section{\partial{Exercise 1.15.9}}%
|
||||||
|
\label{sec:exercise-1.15.9}
|
||||||
|
|
||||||
|
Show that the following property is equivalent to
|
||||||
|
\nameref{sec:step-expansion-contraction-interval-integration}:
|
||||||
|
\begin{equation}
|
||||||
|
\label{sec:exercise-1.15.9-eq1}
|
||||||
|
\int_{ka}^{kb} f(x) \mathop{dx} = k \int_a^b f(kx) \mathop{dx}.
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
Let $f$ be a step function on closed interval $[a, b]$ and $k \neq 0$.
|
||||||
|
Applying \nameref{sec:step-expansion-contraction-interval-integration} to the
|
||||||
|
right-hand side of \eqref{sec:exercise-1.15.9-eq1} yields
|
||||||
|
$$k\int_{ka}^{kb} f(kx / k) \mathop{dx} =
|
||||||
|
k\left[k\int_a^b f(kx) \mathop{dx}\right].$$
|
||||||
|
Simplifying the left-hand side and dividing both sides by $k$ immediately
|
||||||
|
yields the desired result.
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\section{Exercise 1.15.11}%
|
||||||
|
\label{sec:exercise-1.15.11}
|
||||||
|
|
||||||
|
If we instead defined the integral of step functions as
|
||||||
|
\begin{equation*}
|
||||||
|
\label{sec:exercise-1.15.11-eq1}
|
||||||
|
\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}),
|
||||||
|
\end{equation*}
|
||||||
|
a new and different theory of integration would result.
|
||||||
|
Which of the following properties would remain valid in this new theory?
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.11a}}%
|
||||||
|
\label{sub:exercise-1.15.11a}
|
||||||
|
|
||||||
|
$\int_a^b s + \int_b^c s = \int_a^c s$.
|
||||||
|
|
||||||
|
\note{This property mirrors
|
||||||
|
\nameref{sec:step-additivity-with-respect-interval-integration}}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
The above property is valid.
|
||||||
|
|
||||||
|
\divider
|
||||||
|
|
||||||
|
WLOG, suppose $a < b < c$.
|
||||||
|
Let $s$ be a step function defined on closed interval $[a, c]$.
|
||||||
|
By definition of a \nameref{sec:def-step-function}, there exists a
|
||||||
|
\nameref{sec:def-partition} such that $s$ is constant on each open
|
||||||
|
subinterval of $P$.
|
||||||
|
Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $b$
|
||||||
|
as a subdivision point.
|
||||||
|
Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
|
||||||
|
that $x_i = c$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$Q$.
|
||||||
|
By \eqref{sec:exercise-1.15.11-eq1},
|
||||||
|
\begin{align*}
|
||||||
|
\int_a^c s
|
||||||
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \sum_{k=1}^i s_k^3 \cdot (x_k - x_{k-1}) +
|
||||||
|
\sum_{k=i+1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \int_a^b s + \int_b^c s.
|
||||||
|
\end{align*}
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.11b}}%
|
||||||
|
\label{sub:exercise-1.15.11b}
|
||||||
|
|
||||||
|
$\int_a^b (s + t) = \int_a^b s + \int_a^b t$.
|
||||||
|
|
||||||
|
\note{This property mirrors the \nameref{sec:step-additive-property}.}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
The above property is invalid.
|
||||||
|
|
||||||
|
\divider
|
||||||
|
|
||||||
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
||||||
|
By definition of a step function, there exists a \nameref{sec:def-partition}
|
||||||
|
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
|
||||||
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
||||||
|
open subinterval of $P_t$.
|
||||||
|
Therefore $s + t$ is a step function with step partition
|
||||||
|
$$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\},$$
|
||||||
|
the common refinement of $P_s$ and $P_t$ with subdivision points
|
||||||
|
$x_0$, $x_1$, $\ldots$, $x_n$.
|
||||||
|
|
||||||
|
$s$ and $t$ remain constant on every open subinterval of $P$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$P_s$.
|
||||||
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
||||||
|
$P_t$.
|
||||||
|
By \eqref{sec:exercise-1.15.11-eq1},
|
||||||
|
\begin{align*}
|
||||||
|
\int_a^b s + t
|
||||||
|
& = \sum_{k=1}^n (s_k + t_k)^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \sum_{k=1}^n
|
||||||
|
\left[ s_k^3 + 3s_k^2t_k + 3s_kt_k^2 + t_k^3 \right] \\
|
||||||
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \;+ \\
|
||||||
|
& \quad\qquad
|
||||||
|
\sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \;+ \\
|
||||||
|
& \quad\qquad
|
||||||
|
\sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}) \\
|
||||||
|
& = \int_a^b s + \int_a^b t +
|
||||||
|
\sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}).
|
||||||
|
\end{align*}
|
||||||
|
Since this last addend does not necessarily equal $0$, the desired property is
|
||||||
|
invalid.
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.11c}}%
|
||||||
|
\label{sub:exercise-1.15.11c}
|
||||||
|
|
||||||
|
$\int_a^b c \cdot s = c \int_a^b s$.
|
||||||
|
|
||||||
|
\note{This property mirrors the \nameref{sec:step-homogeneous-property}.}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
The above property is invalid.
|
||||||
|
|
||||||
|
\divider
|
||||||
|
|
||||||
|
Let $s$ be a step function on closed interval $[a, b]$.
|
||||||
|
By definition of a step function, there exists a \nameref{sec:def-partition}
|
||||||
|
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
||||||
|
subinterval of $P$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$P$.
|
||||||
|
Then $c \cdot s$ is a step function with step partition $P$.
|
||||||
|
By \eqref{sec:exercise-1.15.11-eq1},
|
||||||
|
\begin{align*}
|
||||||
|
\int_a^b c \cdot s
|
||||||
|
& = \sum_{k=1}^n (c \cdot s_k)^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \sum_{k=1}^n c^3 \cdot s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = c^3 \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = c^3 \int_a^b s.
|
||||||
|
\end{align*}
|
||||||
|
Since $c^3$ does not necessarily equal $c$, the desired property is invalid.
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.11d}}%
|
||||||
|
\label{sub:exercise-1.15.11d}
|
||||||
|
|
||||||
|
$\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$.
|
||||||
|
|
||||||
|
\note{This property mirrors \nameref{sec:step-invariance-under-translation}.}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
The above property is valid.
|
||||||
|
|
||||||
|
\divider
|
||||||
|
|
||||||
|
Let $s$ be a step function on closed interval $[a + c, b + c]$.
|
||||||
|
By definition of a \nameref{sec:def-step-function}, there exists a \nameref{sec:def-partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open subinterval of $P$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of $P$.
|
||||||
|
|
||||||
|
Let $c$ be a real number.
|
||||||
|
Then $t(x) = s(x + c)$ is a step function on closed interval $[a, b]$ with partition $Q = \{x_0 - c, x_1 - c, \ldots, x_n - c\}$.
|
||||||
|
Furthermore, $t$ is constant on each open subinterval of $Q$.
|
||||||
|
Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$.
|
||||||
|
By construction, $t_k = s_k$.
|
||||||
|
By \eqref{sec:exercise-1.15.11-eq1},
|
||||||
|
\begin{align*}
|
||||||
|
\int_{a+c}^{b+c} s(x) \mathop{dx}
|
||||||
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \sum_{k=1}^n s_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\
|
||||||
|
& = \sum_{k=1}^n t_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\
|
||||||
|
& = \int_a^b t(x) \mathop{dx} \\
|
||||||
|
& = \int_a^b s(x + c) \mathop{dx}.
|
||||||
|
\end{align*}
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
|
\subsection{\partial{Exercise 1.15.11e}}%
|
||||||
|
\label{sub:exercise-1.15.11e}
|
||||||
|
|
||||||
|
If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
|
||||||
|
|
||||||
|
\note{This property mirrors the \nameref{sec:step-comparison-theorem}.}
|
||||||
|
|
||||||
|
\begin{proof}
|
||||||
|
|
||||||
|
The above property is valid.
|
||||||
|
|
||||||
|
\divider
|
||||||
|
|
||||||
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
||||||
|
By definition of a \nameref{sec:def-step-function}, there exists a \nameref{sec:def-partition}
|
||||||
|
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
|
||||||
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
||||||
|
open subinterval of $P_t$.
|
||||||
|
Let $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\}$$ be the common refinement
|
||||||
|
of $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
|
||||||
|
|
||||||
|
By construction, $P$ is a step partition for both $s$ and $t$.
|
||||||
|
Thus $s$ and $t$ remain constant on every open subinterval of $P$.
|
||||||
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
||||||
|
$P$.
|
||||||
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
||||||
|
$P$.
|
||||||
|
By \eqref{sec:exercise-1.15.11-eq1},
|
||||||
|
\begin{align*}
|
||||||
|
\int_a^b s
|
||||||
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& < \sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \\
|
||||||
|
& = \int_a^b t.
|
||||||
|
\end{align*}
|
||||||
|
|
||||||
|
\end{proof}
|
||||||
|
|
||||||
\end{document}
|
\end{document}
|
||||||
|
|
|
||||||
Loading…
Reference in New Issue