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diff --git a/homeworks/hw-3.tex b/homeworks/hw-3.tex new file mode 100644 index 0000000..b3d029d --- /dev/null +++ b/homeworks/hw-3.tex @@ -0,0 +1,250 @@ +% Created 2023-10-07 Sat 14:49 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt} +\date{\today} +\title{HW 03} +\hypersetup{ + pdfauthor={Elizabeth Hunt}, + pdftitle={HW 03}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} + +\section{Question One} +\label{sec:org6f2bd27} +\subsection{Three Terms} +\label{sec:orgeb827ff} +\begin{align*} +Si_3(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!}}{s} dx \\ +&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)} +\end{align*} +\subsection{Five Terms} +\label{sec:orge6a15e4} +\begin{align*} +Si_3(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!} - \frac{s^7}{7!} + \frac{s^9}{9!}}{s} dx \\ +&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)} - \frac{x^7}{(7!)(7)} + \frac{s^9}{(9!)(9)} +\end{align*} +\subsection{Ten Terms} +\label{sec:orge87e346} +\begin{align*} +Si_{10}(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!} - \frac{s^7}{7!} + \frac{s^9}{9!} - \frac{s^{11}}{11!} + \frac{s^{13}}{13!} - \frac{s^{15}}{15!} + \frac{s^{17}}{17!} - \frac{s^{19}}{19!}}{s} ds \\ +&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)} - \frac{x^7}{(7!)(7)} + \frac{s^9}{(9!)(9)} - \frac{s^{11}}{(11!)(11)} + \frac{s^{13}}{(13!)(13)} - \frac{s^{15}}{(15!)(15)} \\ +&+ \frac{s^{17}}{(17!)(17)} - \frac{s^{19}}{(19!)(19)} +\end{align*} +\section{Question Three} +\label{sec:org6e2f7fc} +For the second term in the difference quotient, we can expand the taylor series centered at x=a: + +\begin{equation*} +f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots \\ +\end{equation*} + +Which we substitute into the difference quotient: + +\begin{equation*} +\frac{f(a) - f(a - h)}{h} = \frac{f(a) - (f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots)}{h} +\end{equation*} + +And subs. \(x=a-h\): + +\begin{align*} +\frac{f(a) - (f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots)}{h} &= -f'(a)(-1) + -\frac{1}{2}f''(a)h \\ +&= f'(a) - \frac{1}{2}f''(a)h + \cdots \\ +\end{align*} + +Which we now plug into the initial \(e_{\text{abs}}\): + +\begin{align*} +e_{\text{abs}} &= |f'(a) - \frac{f(a) - f(a - h)}{h}| \\ +&= |f'(a) - (f'(a) + -\frac{f''(a)}{2}h + \cdots)| \\ +&= |- \frac{1}{2}f''(a)h + \cdots | \\ +\end{align*} + +With the Taylor Remainder theorem we can absorb the series following the second term: + +\begin{equation*} +e_{\text{abs}} = |- \frac{1}{2}f''(a)h + \cdots | = |\frac{1}{2}f''(\xi)h| \leq Ch +\end{equation*} + +Thus our error is bounded linearly with \(h\). + +\section{Question Four} +\label{sec:orga7d02a2} +For the first term in the difference quotient we know, from the given notes, + +\begin{equation*} +f(a+h) = f(a) + f'(a)h + \frac{1}{2}f''(a)h^2 + \frac{1}{6}f'''(a)(h^3) +\end{equation*} + +And from some of the work in Question Three, + +\begin{equation*} +f(a - h) = f(a) + f'(a)(-h) + \frac{1}{2}f''(a)(-h)^2 + \frac{1}{6}f'''(a)(-h^3) +\end{equation*} + +We can substitute immediately into \(e_{\text{abs}} = |f'(a) - (\frac{f(a+h) - f(a-h)}{2h})|\): + +\begin{align*} +e_{\text{abs}} &= |f'(a) - \frac{1}{2h}((f(a) + f'(a)h + \frac{1}{2}f''(a)h^2 + \cdots) - (f(a) - f'(a)h + \frac{1}{2}f''(a)h^2 + \cdots))| \\ +&= |f'(a) - \frac{1}{2h}(2f'(a)h + \frac{1}{6}f'''(a)h^3 + \cdots)| \\ +&= |f'(a) - f'(a) - \frac{1}{12}f'''(a)h^2 + \cdots| \\ +&= |-\frac{1}{12}f'''(a)h^2 + \cdots| +\end{align*} + +Finally, with the Taylor Remainder theorem we can absorb the series following the third term: + +\begin{equation*} +e_{\text{abs}} = |-\frac{1}{12}f'''(\xi)h^2| = |\frac{1}{12}f'''(\xi)h^2| \leq Ch^2 +\end{equation*} + +Meaning that as \(h\) scales linearly, our error is bounded by \(h^2\) as opposed to linearly as in Question Three. + +\section{Question Six} +\label{sec:org7b05811} +\subsection{A} +\label{sec:org8341a77} +\begin{verbatim} +(load "../lizfcm.asd") +(ql:quickload :lizfcm) + +(defun f (x) + (/ (- x 1) (+ x 1))) + +(defun fprime (x) + (/ 2 (expt (+ x 1) 2))) + +(let ((domain-values (loop for a from 0 to 2 + append + (loop for i from 0 to 9 + for h = (/ 1.0 (expt 2 i)) + collect (list a h))))) + (lizfcm.utils:table (:headers '("a" "h" "f'" "\\approx f'" "e_{\\text{abs}}") + :domain-order (a h) + :domain-values domain-values) + (fprime a) + (lizfcm.approx:fwd-derivative-at 'f a h) + (abs (- (fprime a) + (lizfcm.approx:fwd-derivative-at 'f a h))))) +\end{verbatim} + + +\section{Question Nine} +\label{sec:orgeb1839f} +\subsection{C} +\label{sec:org5691277} + +\begin{verbatim} +(load "../lizfcm.asd") +(ql:quickload :lizfcm) + +(defun factorial (n) + (if (= n 0) + 1 + (* n (factorial (- n 1))))) + +(defun taylor-term (n x) + (/ (* (expt (- 1) n) + (expt x (+ (* 2 n) 1))) + (* (factorial n) + (+ (* 2 n) 1)))) + +(defun f (x &optional (max-iterations 30)) + (let ((sum 0.0)) + (dotimes (n max-iterations) + (setq sum (+ sum (taylor-term n x)))) + (* sum (/ 2 (sqrt pi))))) + +(defun fprime (x) + (* (/ 2 (sqrt pi)) (exp (- 0 (* x x))))) + +(let ((domain-values (loop for a from 0 to 1 + append + (loop for i from 0 to 9 + for h = (/ 1.0 (expt 2 i)) + collect (list a h))))) + (lizfcm.utils:table (:headers '("a" "h" "f'" "\\approx f'" "e_{\\text{abs}}") + :domain-order (a h) + :domain-values domain-values) + (fprime a) + (lizfcm.approx:central-derivative-at 'f a h) + (abs (- (fprime a) + (lizfcm.approx:central-derivative-at 'f a h))))) +\end{verbatim} + + +\begin{center} +\begin{tabular}{rrrrr} +a & h & f' & \(\approx\) f' & e\textsubscript{\text{abs}}\\[0pt] +0 & 1.0 & 1.1283791670955126d0 & 0.8427006725464232d0 & 0.28567849454908933d0\\[0pt] +0 & 0.5 & 1.1283791670955126d0 & 1.0409997446922075d0 & 0.0873794224033051d0\\[0pt] +0 & 0.25 & 1.1283791670955126d0 & 1.1053055663206806d0 & 0.023073600774832004d0\\[0pt] +0 & 0.125 & 1.1283791670955126d0 & 1.122529655394656d0 & 0.005849511700856569d0\\[0pt] +0 & 0.0625 & 1.1283791670955126d0 & 1.1269116944798618d0 & 0.0014674726156507223d0\\[0pt] +0 & 0.03125 & 1.1283791670955126d0 & 1.1280120131008824d0 & 3.6715399463016496d-4\\[0pt] +0 & 0.015625 & 1.1283791670955126d0 & 1.1282873617826952d0 & 9.180531281738347d-5\\[0pt] +0 & 0.0078125 & 1.1283791670955126d0 & 1.128356232581468d0 & 2.293451404455915d-5\\[0pt] +0 & 0.00390625 & 1.1283791670955126d0 & 1.1283734502811613d0 & 5.71681435124205d-6\\[0pt] +0 & 0.001953125 & 1.1283791670955126d0 & 1.1283777547060847d0 & 1.4123894278572635d-6\\[0pt] +1 & 1.0 & 0.41510750774498784d0 & 0.4976611317561498d0 & 0.08255362401116195d0\\[0pt] +1 & 0.5 & 0.41510750774498784d0 & 0.44560523266293384d0 & 0.030497724917946d0\\[0pt] +1 & 0.25 & 0.41510750774498784d0 & 0.4234889628937013d0 & 0.008381455148713468d0\\[0pt] +1 & 0.125 & 0.41510750774498784d0 & 0.41725265825950153d0 & 0.002145150514513694d0\\[0pt] +1 & 0.0625 & 0.41510750774498784d0 & 0.41564710776310854d0 & 5.396000181207006d-4\\[0pt] +1 & 0.03125 & 0.41510750774498784d0 & 0.4152414157140871d0 & 1.3390796909928948d-4\\[0pt] +1 & 0.015625 & 0.41510750774498784d0 & 0.41514241394084905d0 & 3.490619586121735d-5\\[0pt] +1 & 0.0078125 & 0.41510750774498784d0 & 0.41510582632900395d0 & 1.6814159838896003d-6\\[0pt] +1 & 0.00390625 & 0.41510750774498784d0 & 0.415092913054238d0 & 1.4594690749825112d-5\\[0pt] +1 & 0.001953125 & 0.41510750774498784d0 & 0.4150670865046777d0 & 4.0421240310117845d-5\\[0pt] +\end{tabular} +\end{center} + + +\section{Question Twelve} +\label{sec:orgc55bfd1} + +First we'll place a bound on \(h\); looking at a graph of \(f\) it's pretty obvious from the asymptotes that we don't want to go much further than \(|h| = 2 - \frac{pi}{2}\). + +Following similar reasoning as Question Four, we can determine an optimal \(h\) by computing \(e_{\text{abs}}\) for the central difference, but now including a roundoff error for each time we run \(f\) +such that \(|f_{\text{machine}}(x) - f(x)| \le \epsilon_{\text{dblprec}}\) (we'll use double precision numbers, from HW 2 we know \(\epsilon_{\text{dblprec}} \approx 2.22045 (10^{-16})\)). + +We'll just assume \(|f_{\text{machine}}(x) - f(x)| = \epsilon_{\text{dblprec}}\) so our new difference quotient becomes: + +\begin{align*} +e_{\text{abs}} &= |f'(a) - (\frac{f(a+h) - f(a-h) + 2\epsilon_{\text{dblprec}}}{2h})| \\ +&= |\frac{1}{12}f'''(\xi)h^2 + \frac{\epsilon_{\text{dblprec}}}{h}| +\end{align*} + +Because we bounded our \(|h| = 2 - \frac{pi}{2}\) we'll find the maximum value of \(f'''\) between \(a - (2 - \frac{\pi}{2})\) and \(a - (2 - \frac{\pi}{3})\). Using \href{https://www.desmos.com/calculator/gen1zpohh2}{desmos} I found this to be -2. + +Thus, \(e_{\text{abs}} \leq \frac{1}{6}h^2 + \frac{\epsilon_{\text{dblprec}}}{h}\). Finding the derivative: + +\begin{equation*} +e' = \frac{1}{3}h - \frac{\epsilon_{\text{dblprec}}}{h^2} +\end{equation*} + +And solving at \(e' = 0\): + +\begin{equation*} +\frac{1}{3}h = \frac{\epsilon_{\text{dblprec}}}{h^2} \Rightarrow h^3 = 3\epsilon_{\text{dblprec}} \Rightarrow h = (3\epsilon_{\text{dblprec}})^{1/3} +\end{equation*} + +Which is \(\approx (3(2.22045 (10^{-16}))^{\frac{1}{3}} \approx 8.7335 10^{-6}\). +\end{document}
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