From dc4932e0e77878340d72d574dce1cab546c1cf3c Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Wed, 17 May 2023 15:38:16 -0600 Subject: [PATCH] Finish Apostol 1.22. --- Bookshelf/Apostol.tex | 173 ++++++++++++++++++++++++++++++++++++++---- 1 file changed, 158 insertions(+), 15 deletions(-) diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 253b9c8..4e6a13f 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -2670,7 +2670,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. Let $f$ be a nonnegative function, integrable on an interval $[a, b]$. Then the graph of $f$, that is, the set \begin{equation} - \label{sec:measurability-graph-nonnegative-function-eq1} + \label{sec:theorem-1.11-eq1} \{(x, y) \mid a \leq x \leq b, y = f(x)\}, \end{equation} is measurable and has area equal to $0$. @@ -2686,7 +2686,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. equal to $0$. \paragraph{(i)}% - \label{par:measurability-graph-nonnegative-function-i} + \label{par:theorem-1.11-i} By definition of integrability, there exists one and only one number $I$ such that @@ -2704,10 +2704,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. Let $Q$ denote the ordinate set of $f$. By \nameref{sec:measurability-ordinate-sets}, $Q$ is measurable with area equal to the integral $I = \int_a^b f(x) \mathop{dx}$. - By \nameref{par:measurability-graph-nonnegative-function-i}, $Q'$ is + By \nameref{par:theorem-1.11-i}, $Q'$ is measurable with area also equal to $I$. - We note the graph of $f$, - \eqref{sec:measurability-graph-nonnegative-function-eq1}, is equal to set + We note the graph of $f$, \eqref{sec:theorem-1.11-eq1}, is equal to set $Q - Q'$. By the \nameref{sec:area-difference-property}, $Q - Q'$ is measurable and $$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$ @@ -2751,7 +2750,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. on every $k$th open subinterval of $P$. Then, by \eqref{sec:theorem-1.9-eq1}, it follows \begin{equation} - \label{sec:integrability-bounded-monotonic-functions-eq1} + \label{sec:theorem-1.12-eq1} \int_a^b s(x) \mathop{dx} \leq \ubar{I}(f) \leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}. \end{equation} @@ -2770,7 +2769,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. & = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\ & = \frac{(b - a)(f(b) - f(a))}{n}. \end{align*} - By \eqref{sec:integrability-bounded-monotonic-functions-eq1}, + By \eqref{sec:theorem-1.12-eq1}, \begin{align*} \ubar{I}(f) & \leq \bar{I}(f) \\ @@ -2790,7 +2789,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. on every $k$th open subinterval of $P$. Then, by \eqref{sec:theorem-1.9-eq1}, it follows \begin{equation} - \label{sec:integrability-bounded-monotonic-functions-eq2} + \label{sec:theorem-1.12-eq2} \int_a^b s(x) \mathop{dx} \leq \ubar{I}(f) \leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}. \end{equation} @@ -2809,7 +2808,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. & = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_{k-1}) - f(x_k) \\ & = \frac{(b - a)(f(a) - f(b))}{n}. \end{align*} - By \eqref{sec:integrability-bounded-monotonic-functions-eq2}, + By \eqref{sec:theorem-1.12-eq2}, \begin{align*} \ubar{I}(f) & \leq \bar{I}(f) \\ @@ -2822,9 +2821,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. \end{proof} -\section{\unverified{% - Calculation of the Integral of a Bounded Monotonic Function}}% -\label{sec:calculation-integral-bounded-monotonic-function} +\section{\partial{% + Calculation of the Integral of a Bounded Increasing Function}}% +\label{sec:calculation-integral-bounded-increasing-function} \label{sec:theorem-1.13} \begin{theorem}[1.13] @@ -2832,13 +2831,157 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. Assume $f$ is increasing on a closed interval $[a, b]$. Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$. If $I$ is any number which satisfies the inequalities - $$\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k) - \leq I \leq - \frac{b - a}{n} \sum_{k=1}^n f(x_k)$$ + \begin{equation} + \label{sec:theorem-1.13-eq1} + \frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k) + \leq I \leq + \frac{b - a}{n} \sum_{k=1}^n f(x_k) + \end{equation} for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$. \end{theorem} +\begin{proof} + + Let $f$ be increasing on a closed interval $[a, b]$ and $I$ be a number + satisfying \eqref{sec:theorem-1.13-eq1}. + Let $s$ be the step function below $f$ with constant value $f(x_{k-1})$ + on every $k$th open subinterval of $P$. + Let $t$ be the step function above $f$ with constant value $f(x_k)$ + on every $k$th open subinterval of $P$. + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_a^b s(x) \mathop{dx} + & = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\ + & = \sum_{k=0}^{n-1} f(x_k)\left[\frac{b - a}{n}\right] \\ + \int_a^b t(x) \mathop{dx} + & = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right]. + \end{align*} + Therefore \eqref{sec:theorem-1.13-eq1} can alternatively be written as + \begin{equation} + \label{sec:theorem-1.13-eq2} + \int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}. + \end{equation} + By \nameref{sec:theorem-1.12}, $f$ is integrable. + Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies + \begin{equation} + \label{sec:theorem-1.13-eq3} + \int_a^b s(x) \mathop{dx} + \leq \int_a^b f(x) \mathop{dx} + \leq \int_a^b t(x) \mathop{dx}. + \end{equation} + Manipulating \eqref{sec:theorem-1.13-eq2} and \eqref{sec:theorem-1.13-eq3} + together yields + \begin{align*} + I - \int_a^b f(x) \mathop{dx} + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\ + \int_a^b f(x) \mathop{dx} - I + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}. + \end{align*} + Combining the above inequalities in turn yields + \begin{align*} + 0 + & \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\ + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\ + & = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] - + \sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\ + & = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\ + & = \frac{(b - a)(f(b) - f(a))}{n}. + \end{align*} + The above chain of inequalities holds for all positive integers $n \geq 1$, + meaning \nameref{sec:theorem-i.31} applies. + Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies + the desired result. + +\end{proof} + +\section{\partial{% + Calculation of the Integral of a Bounded Decreasing Function}}% +\label{sec:calculation-integral-bounded-decreasing-function} +\label{sec:theorem-1.14} + +\begin{theorem}[1.14] + + Assume $f$ is descreasing on $[a, b]$. + Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$. + If $I$ is any number which satisfies the inequalities + \begin{equation} + \label{sec:theorem-1.14-eq1} + \frac{b - a}{n} \sum_{k=1}^n f(x_k) + \leq I \leq + \frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k) + \end{equation} + for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$. + +\end{theorem} + +\begin{proof} + + Let $f$ be decreasing on a closed interval $[a, b]$ and $I$ be a number + satisfying \eqref{sec:theorem-1.14-eq1}. + Let $s$ be the step function below $f$ with constant value $f(x_k)$ + on every $k$th open subinterval of $P$. + Let $t$ be the step function above $f$ with constant value $f(x_{k-1})$ + on every $k$th open subinterval of $P$. + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_a^b s(x) \mathop{dx} + & = \sum_{k=1}^n f(x_k) \left[\frac{b - a}{n}\right] \\ + \int_a^b t(x) \mathop{dx} + & = \sum_{k=1}^n f(x_{k-1}) \left[\frac{b - a}{n}\right] \\ + & = \sum_{k=0}^{n-1} f(x) \left[\frac{b - a}{n}\right]. + \end{align*} + Therefore \eqref{sec:theorem-1.14-eq1} can alternatively be written as + \begin{equation} + \label{sec:theorem-1.14-eq2} + \int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}. + \end{equation} + By \nameref{sec:theorem-1.12}, $f$ is integrable. + Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies + \begin{equation} + \label{sec:theorem-1.14-eq3} + \int_a^b s(x) \mathop{dx} + \leq \int_a^b f(x) \mathop{dx} + \leq \int_a^b t(x) \mathop{dx}. + \end{equation} + Manipulating \eqref{sec:theorem-1.14-eq2} and \eqref{sec:theorem-1.14-eq3} + together yields + \begin{align*} + I - \int_a^b f(x) \mathop{dx} + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\ + \int_a^b f(x) \mathop{dx} - I + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}. + \end{align*} + Combining the above inequalities in turn yields + \begin{align*} + 0 + & \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\ + & \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\ + & = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] - + \sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\ + & = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\ + & = \frac{(b - a)(f(b) - f(a))}{n}. + \end{align*} + The above chain of inequalities holds for all positive integers $n \geq 1$, + meaning \nameref{sec:theorem-i.31} applies. + Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies + the desired result. + +\end{proof} + +\section{\unverified{% + Calculation of the Integral \texorpdfstring{$\int_0^b x^p \mathop{dx}$}{int-x-p} + when \texorpdfstring{$p$}{p} is a Positive Integer}}% +\label{sec:calculation-integral-int-x-p-p-positive-integer} +\label{sec:theorem-1.15} + +\begin{theorem}[1.15] + + If $p$ is a positive integer and $b > 0$, we have + $$\int_0^b x^p \mathop{dx} = \frac{b^{p+1}}{p+1}.$$ + +\end{theorem} + \begin{proof} TODO