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updates 10/9

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Elizabeth Hunt 2023-10-09 21:08:25 -06:00
parent b35e399833
commit adda6869cb
Signed by: simponic
GPG Key ID: 52B3774857EB24B1
16 changed files with 670 additions and 54 deletions
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homeworks/hw-2.org
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@ -20,35 +20,10 @@ Computing $\epsilon_{\text{mac}}$ for single precision numbers
:domain-values domain-values))) :domain-values domain-values)))
#+END_SRC #+END_SRC
#+RESULTS:
| a | h | err |
| 1.0 | 0.5 | 0.5 |
| 1.0 | 0.25 | 0.25 |
| 1.0 | 0.125 | 0.125 |
| 1.0 | 0.0625 | 0.0625 |
| 1.0 | 0.03125 | 0.03125 |
| 1.0 | 0.015625 | 0.015625 |
| 1.0 | 0.0078125 | 0.0078125 |
| 1.0 | 0.00390625 | 0.00390625 |
| 1.0 | 0.001953125 | 0.001953125 |
| 1.0 | 0.0009765625 | 0.0009765625 |
| 1.0 | 0.00048828125 | 0.00048828125 |
| 1.0 | 0.00024414063 | 0.00024414063 |
| 1.0 | 0.00012207031 | 0.00012207031 |
| 1.0 | 6.1035156e-05 | 6.1035156e-05 |
| 1.0 | 3.0517578e-05 | 3.0517578e-05 |
| 1.0 | 1.5258789e-05 | 1.5258789e-05 |
| 1.0 | 7.6293945e-06 | 7.6293945e-06 |
| 1.0 | 3.8146973e-06 | 3.8146973e-06 |
| 1.0 | 1.9073486e-06 | 1.9073486e-06 |
| 1.0 | 9.536743e-07 | 9.536743e-07 |
| 1.0 | 4.7683716e-07 | 4.7683716e-07 |
| 1.0 | 2.3841858e-07 | 2.3841858e-07 |
| 1.0 | 1.1920929e-07 | 1.1920929e-07 |
(with many rows truncated) (with many rows truncated)
| a | h | err | | a | h | err |
| 1.0 | 1.0 | 1.0 |
| 1.0 | 0.5 | 0.5 | | 1.0 | 0.5 | 0.5 |
| 1.0 | 0.25 | 0.25 | | 1.0 | 0.25 | 0.25 |
| 1.0 | 0.125 | 0.125 | | 1.0 | 0.125 | 0.125 |
@ -76,6 +51,7 @@ Computing $\epsilon_{\text{mac}}$ for double precision numbers:
(with many rows truncated) (with many rows truncated)
| a | h | err | | a | h | err |
| 1.0d0 | 1.0d0 | 1.0d0 |
| 1.0d0 | 0.5d0 | 0.5d0 | | 1.0d0 | 0.5d0 | 0.5d0 |
| 1.0d0 | 0.25d0 | 0.25d0 | | 1.0d0 | 0.25d0 | 0.25d0 |
| 1.0d0 | 0.125d0 | 0.125d0 | | 1.0d0 | 0.125d0 | 0.125d0 |

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homeworks/hw-2.tex
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@ -1,4 +1,4 @@
% Created 2023-09-27 Wed 10:09 % Created 2023-10-07 Sat 14:51
% Intended LaTeX compiler: pdflatex % Intended LaTeX compiler: pdflatex
\documentclass[11pt]{article} \documentclass[11pt]{article}
\usepackage[utf8]{inputenc} \usepackage[utf8]{inputenc}
@ -29,7 +29,7 @@
\setlength\parindent{0pt} \setlength\parindent{0pt}
\section{Question One} \section{Question One}
\label{sec:orga21c813} \label{sec:org58b9af4}
Computing \(\epsilon_{\text{mac}}\) for single precision numbers Computing \(\epsilon_{\text{mac}}\) for single precision numbers
\begin{verbatim} \begin{verbatim}
@ -49,6 +49,7 @@ Computing \(\epsilon_{\text{mac}}\) for single precision numbers
\begin{center} \begin{center}
\begin{tabular}{rrr} \begin{tabular}{rrr}
a & h & err\\[0pt] a & h & err\\[0pt]
1.0 & 1.0 & 1.0\\[0pt]
1.0 & 0.5 & 0.5\\[0pt] 1.0 & 0.5 & 0.5\\[0pt]
1.0 & 0.25 & 0.25\\[0pt] 1.0 & 0.25 & 0.25\\[0pt]
1.0 & 0.125 & 0.125\\[0pt] 1.0 & 0.125 & 0.125\\[0pt]
@ -65,7 +66,7 @@ a & h & err\\[0pt]
\(\epsilon_{\text{mac single precision}}\) \(\approx\) 1.192(10\textsuperscript{-7}) \(\epsilon_{\text{mac single precision}}\) \(\approx\) 1.192(10\textsuperscript{-7})
\section{Question Two} \section{Question Two}
\label{sec:org06c4a23} \label{sec:org27557b4}
Computing \(\epsilon_{\text{mac}}\) for double precision numbers: Computing \(\epsilon_{\text{mac}}\) for double precision numbers:
\begin{verbatim} \begin{verbatim}
@ -81,6 +82,7 @@ Computing \(\epsilon_{\text{mac}}\) for double precision numbers:
\begin{center} \begin{center}
\begin{tabular}{rrr} \begin{tabular}{rrr}
a & h & err\\[0pt] a & h & err\\[0pt]
1.0d0 & 1.0d0 & 1.0d0\\[0pt]
1.0d0 & 0.5d0 & 0.5d0\\[0pt] 1.0d0 & 0.5d0 & 0.5d0\\[0pt]
1.0d0 & 0.25d0 & 0.25d0\\[0pt] 1.0d0 & 0.25d0 & 0.25d0\\[0pt]
1.0d0 & 0.125d0 & 0.125d0\\[0pt] 1.0d0 & 0.125d0 & 0.125d0\\[0pt]
@ -102,7 +104,7 @@ a & h & err\\[0pt]
Thus, \(\epsilon_{\text{mac double precision}}\) \(\approx\) 2.220 \(\cdot\) 10\textsuperscript{-16} Thus, \(\epsilon_{\text{mac double precision}}\) \(\approx\) 2.220 \(\cdot\) 10\textsuperscript{-16}
\section{Question Three - |v|\textsubscript{2}} \section{Question Three - |v|\textsubscript{2}}
\label{sec:orgf181ba7} \label{sec:org59c6c10}
\begin{verbatim} \begin{verbatim}
(let ((vs '((1 1) (2 3) (4 5) (-1 2))) (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
(2-norm (lizfcm.vector:p-norm 2))) (2-norm (lizfcm.vector:p-norm 2)))
@ -124,7 +126,7 @@ x & y & 2norm\\[0pt]
\end{center} \end{center}
\section{Question Four - |v|\textsubscript{1}} \section{Question Four - |v|\textsubscript{1}}
\label{sec:org2196087} \label{sec:org2b67b3e}
\begin{verbatim} \begin{verbatim}
(let ((vs '((1 1) (2 3) (4 5) (-1 2))) (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
(1-norm (lizfcm.vector:p-norm 1))) (1-norm (lizfcm.vector:p-norm 1)))
@ -146,7 +148,7 @@ x & y & 1norm\\[0pt]
\end{center} \end{center}
\section{Question Five - |v|\textsubscript{\(\infty\)}} \section{Question Five - |v|\textsubscript{\(\infty\)}}
\label{sec:org11b8894} \label{sec:org922206e}
\begin{verbatim} \begin{verbatim}
(let ((vs '((1 1) (2 3) (4 5) (-1 2)))) (let ((vs '((1 1) (2 3) (4 5) (-1 2))))
(lizfcm.utils:table (:headers '("x" "y" "max-norm") (lizfcm.utils:table (:headers '("x" "y" "max-norm")
@ -167,11 +169,11 @@ x & y & infty-norm\\[0pt]
\end{center} \end{center}
\section{Question Six - ||v - u|| via |v|\textsubscript{2}} \section{Question Six - ||v - u|| via |v|\textsubscript{2}}
\label{sec:orga2324b2} \label{sec:org29ec18f}
\begin{verbatim} \begin{verbatim}
(let* ((vs '((1 1) (2 3) (4 5) (-1 2))) (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
(vs2 '((7 9) (2 2) (8 -1) (4 4))) (vs2 '((7 9) (2 2) (8 -1) (4 4)))
(2-norm (lizfcm.vector:p-norm 2))) (2-norm (lizfcm.vector:p-norm 2)))
(lizfcm.utils:table (:headers '("v1" "v2" "2-norm-d") (lizfcm.utils:table (:headers '("v1" "v2" "2-norm-d")
:domain-order (v1 v2) :domain-order (v1 v2)
:domain-values (mapcar (lambda (v1 v2) :domain-values (mapcar (lambda (v1 v2)
@ -193,11 +195,11 @@ v1 & v2 & 2-norm\\[0pt]
\end{center} \end{center}
\section{Question Seven - ||v - u|| via |v|\textsubscript{1}} \section{Question Seven - ||v - u|| via |v|\textsubscript{1}}
\label{sec:org388fbc7} \label{sec:org7a87810}
\begin{verbatim} \begin{verbatim}
(let* ((vs '((1 1) (2 3) (4 5) (-1 2))) (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
(vs2 '((7 9) (2 2) (8 -1) (4 4))) (vs2 '((7 9) (2 2) (8 -1) (4 4)))
(1-norm (lizfcm.vector:p-norm 1))) (1-norm (lizfcm.vector:p-norm 1)))
(lizfcm.utils:table (:headers '("v1" "v2" "1-norm-d") (lizfcm.utils:table (:headers '("v1" "v2" "1-norm-d")
:domain-order (v1 v2) :domain-order (v1 v2)
:domain-values (mapcar (lambda (v1 v2) :domain-values (mapcar (lambda (v1 v2)
@ -219,10 +221,10 @@ v1 & v2 & 1-norm-d\\[0pt]
\end{center} \end{center}
\section{Question Eight - ||v - u|| via |v|\textsubscript{\(\infty\)}} \section{Question Eight - ||v - u|| via |v|\textsubscript{\(\infty\)}}
\label{sec:org6e77f76} \label{sec:org0f3b64f}
\begin{verbatim} \begin{verbatim}
(let* ((vs '((1 1) (2 3) (4 5) (-1 2))) (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
(vs2 '((7 9) (2 2) (8 -1) (4 4)))) (vs2 '((7 9) (2 2) (8 -1) (4 4))))
(lizfcm.utils:table (:headers '("v1" "v2" "max-norm-d") (lizfcm.utils:table (:headers '("v1" "v2" "max-norm-d")
:domain-order (v1 v2) :domain-order (v1 v2)
:domain-values (mapcar (lambda (v1 v2) :domain-values (mapcar (lambda (v1 v2)

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#+TITLE: HW 03
#+AUTHOR: Elizabeth Hunt
#+STARTUP: entitiespretty fold inlineimages
#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
#+LATEX: \setlength\parindent{0pt}
#+OPTIONS: toc:nil
* Question One
** Three Terms
\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*}
** Five Terms
\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*}
** Ten Terms
\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*}
* Question Three
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 $h$.
* Question Four
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.
* Question Six
** A
#+BEGIN_SRC lisp
(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_SRC
#+RESULTS:
| a | h | f' | \approx f' | e_{\text{abs}} |
| 0 | 1.0 | 2 | 1.0 | 1.0 |
| 0 | 0.5 | 2 | 1.3333333 | 0.66666675 |
| 0 | 0.25 | 2 | 1.5999999 | 0.4000001 |
| 0 | 0.125 | 2 | 1.7777777 | 0.22222233 |
| 0 | 0.0625 | 2 | 1.8823528 | 0.11764717 |
| 0 | 0.03125 | 2 | 1.939394 | 0.060606003 |
| 0 | 0.015625 | 2 | 1.9692307 | 0.030769348 |
| 0 | 0.0078125 | 2 | 1.9844971 | 0.01550293 |
| 0 | 0.00390625 | 2 | 1.992218 | 0.0077819824 |
| 0 | 0.001953125 | 2 | 1.9960938 | 0.00390625 |
| 1 | 1.0 | 1/2 | 0.33333334 | 0.16666666 |
| 1 | 0.5 | 1/2 | 0.4 | 0.099999994 |
| 1 | 0.25 | 1/2 | 0.44444445 | 0.055555552 |
| 1 | 0.125 | 1/2 | 0.47058824 | 0.029411763 |
| 1 | 0.0625 | 1/2 | 0.4848485 | 0.015151501 |
| 1 | 0.03125 | 1/2 | 0.4923077 | 0.0076923072 |
| 1 | 0.015625 | 1/2 | 0.49612403 | 0.0038759708 |
| 1 | 0.0078125 | 1/2 | 0.49805447 | 0.0019455254 |
| 1 | 0.00390625 | 1/2 | 0.49902534 | 0.00097465515 |
| 1 | 0.001953125 | 1/2 | 0.4995122 | 0.0004878044 |
| 2 | 1.0 | 2/9 | 0.16666666 | 0.055555567 |
| 2 | 0.5 | 2/9 | 0.19047618 | 0.031746045 |
| 2 | 0.25 | 2/9 | 0.2051282 | 0.017094031 |
| 2 | 0.125 | 2/9 | 0.21333337 | 0.008888856 |
| 2 | 0.0625 | 2/9 | 0.21768713 | 0.004535094 |
| 2 | 0.03125 | 2/9 | 0.21993065 | 0.002291575 |
| 2 | 0.015625 | 2/9 | 0.22106934 | 0.0011528879 |
| 2 | 0.0078125 | 2/9 | 0.22164536 | 0.00057686865 |
| 2 | 0.00390625 | 2/9 | 0.22193146 | 0.00029076636 |
| 2 | 0.001953125 | 2/9 | 0.22207642 | 0.00014580786 |
* Question Nine
** C
#+BEGIN_SRC lisp
(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_SRC
| a | h | f' | \approx f' | e_{\text{abs}} |
| 0 | 1.0 | 1.1283791670955126d0 | 0.8427006725464232d0 | 0.28567849454908933d0 |
| 0 | 0.5 | 1.1283791670955126d0 | 1.0409997446922075d0 | 0.0873794224033051d0 |
| 0 | 0.25 | 1.1283791670955126d0 | 1.1053055663206806d0 | 0.023073600774832004d0 |
| 0 | 0.125 | 1.1283791670955126d0 | 1.122529655394656d0 | 0.005849511700856569d0 |
| 0 | 0.0625 | 1.1283791670955126d0 | 1.1269116944798618d0 | 0.0014674726156507223d0 |
| 0 | 0.03125 | 1.1283791670955126d0 | 1.1280120131008824d0 | 3.6715399463016496d-4 |
| 0 | 0.015625 | 1.1283791670955126d0 | 1.1282873617826952d0 | 9.180531281738347d-5 |
| 0 | 0.0078125 | 1.1283791670955126d0 | 1.128356232581468d0 | 2.293451404455915d-5 |
| 0 | 0.00390625 | 1.1283791670955126d0 | 1.1283734502811613d0 | 5.71681435124205d-6 |
| 0 | 0.001953125 | 1.1283791670955126d0 | 1.1283777547060847d0 | 1.4123894278572635d-6 |
| 1 | 1.0 | 0.41510750774498784d0 | 0.4976611317561498d0 | 0.08255362401116195d0 |
| 1 | 0.5 | 0.41510750774498784d0 | 0.44560523266293384d0 | 0.030497724917946d0 |
| 1 | 0.25 | 0.41510750774498784d0 | 0.4234889628937013d0 | 0.008381455148713468d0 |
| 1 | 0.125 | 0.41510750774498784d0 | 0.41725265825950153d0 | 0.002145150514513694d0 |
| 1 | 0.0625 | 0.41510750774498784d0 | 0.41564710776310854d0 | 5.396000181207006d-4 |
| 1 | 0.03125 | 0.41510750774498784d0 | 0.4152414157140871d0 | 1.3390796909928948d-4 |
| 1 | 0.015625 | 0.41510750774498784d0 | 0.41514241394084905d0 | 3.490619586121735d-5 |
| 1 | 0.0078125 | 0.41510750774498784d0 | 0.41510582632900395d0 | 1.6814159838896003d-6 |
| 1 | 0.00390625 | 0.41510750774498784d0 | 0.415092913054238d0 | 1.4594690749825112d-5 |
| 1 | 0.001953125 | 0.41510750774498784d0 | 0.4150670865046777d0 | 4.0421240310117845d-5 |
* Question Twelve
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 [[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 $\approx (3(2.22045 (10^{-16}))^{\frac{1}{3}} \approx 8.7335 10^{-6}$.

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% 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}

22
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@ -0,0 +1,22 @@
[[ a_{11} a_{12} \cdots a_{1n} | b_1]
[ 0 (a_{22} - \frac{a_{}_{21}}{a_{22}}a_{11}) \cdots a_{2n} | b_2 - \frac{a_{21}}{a_{11}}b_1 ]]
#+BEGIN_SRC c
for (int i = 1; i < n; i++) {
float factor = -a[i][0] / a[0][0];
for (int j = 1; j < n; j++) {
a[i][j] = a[i][j] + factor * a[0][j];
}
b[i] = b[i] + factor * b[0];
}
for (int k = 0; k < (n - 1); k++) {
for (int i = k+1; i < n; i++) {
float factor = -a[i][k] / a[k][k];
for (int j = k+1; j < n; j++) {
a[i][j] = a[i][j] + factor * a[j][k];
}
b[i] = b[i] + factor*b[k];
}
}
#+END_SRC

13
notes/Oct-6.org Normal file
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@ -0,0 +1,13 @@
#+BEGIN_SRC c
for (int k = 0; i < (n - 1); k++) {
for (int i = k+1; i< n; i++) {
float factor = a[i][k] / a[k][k];
for (int j = k+1; j < k; j++) {
a[i][j] = a[i][j] - factor * a[k][j];
}
b[i] = b[i] - factor * b[k];
}
}
#+END_SRC

8
notes/Sep-15.org
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@ -36,11 +36,11 @@ Again, $f'(a) \approx \frac{f(a+h) - f(a)}{h}$,
$e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots$ $e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots$
$R_2 = \frac{h}{2} f''(u)$ $R_2 = \frac{h}{2} f''(\xi)$
$|\frac{h}{2} f''(u)| \leq M h^1$ $|\frac{h}{2} f''(\xi)| \leq M h^1$
$M = \frac{1}{2}|f'(u)|$ $M = \frac{1}{2}|f'(\xi)|$
*** Another approximation *** Another approximation
@ -48,5 +48,5 @@ $\text{err} = |f'(a) - \frac{f(a) - f(a - h)}{h}|$
$= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))$ $= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))$
$= |f'(a) - \frac{1}{h}(f'(a) + \frac{1}{2}f''(a)h)|$ $= |f'(a) - (f'(a) + \frac{1}{2}f''(a)h)|$

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% Created 2023-09-29 Fri 10:00
% 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}
\author{Elizabeth Hunt}
\date{\today}
\title{}
\hypersetup{
pdfauthor={Elizabeth Hunt},
pdftitle={},
pdfkeywords={},
pdfsubject={},
pdfcreator={Emacs 28.2 (Org mode 9.7-pre)},
pdflang={English}}
\begin{document}
\tableofcontents
\section{Taylor Series Approx.}
\label{sec:orgcc72ed1}
Suppose f has \(\infty\) many derivatives near a point a. Then the taylor series is given by
\(f(x) = \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)
For increment notation we can write
\(f(a + h) = f(a) + f'(a)(a+h - a) + \dots\)
\(= \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{h!} (h^n)\)
Consider the approximation
\(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |f'(a) - \frac{1}{h}(f(a + h) - f(a))|\)
Substituting\ldots{}
\(= |f'(a) - \frac{1}{h}((f(a) + f'(a) h + \frac{f''(a)}{2} h^2 + \cdots) - f(a))|\)
\(f(a) - f(a) = 0\)\ldots{} and \(distribute the h\)
\(= |-1/2 f''(a) h + \frac{1}{6}f'''(a)h^2 \cdots|\)
\subsection{With Remainder}
\label{sec:org7dfd6c7}
We can determine for some u \(f(a + h) = f(a) + f'(a)h + \frac{1}{2}f''(u)h^2\)
and so the error is \(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |\frac{h}{2}f''(u)|\)
\begin{itemize}
\item\relax [\url{https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series}]
\begin{itemize}
\item > Taylor's Theorem w/ Remainder
\end{itemize}
\end{itemize}
\subsection{Of Deriviatives}
\label{sec:org1ec7c9b}
Again, \(f'(a) \approx \frac{f(a+h) - f(a)}{h}\),
\(e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots\)
\(R_2 = \frac{h}{2} f''(u)\)
\(|\frac{h}{2} f''(u)| \leq M h^1\)
\(M = \frac{1}{2}|f'(u)|\)
\subsubsection{Another approximation}
\label{sec:org16193b9}
\(\text{err} = |f'(a) - \frac{f(a) - f(a - h)}{h}|\)
\(= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))\)
\(= |f'(a) - \frac{1}{h}(f'(a) + \frac{1}{2}f''(a)h)|\)
\end{document}

10
src/approx/derivative.lisp
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@ -1,9 +1,17 @@
(in-package :lizfcm.approx) (in-package :lizfcm.approx)
(defun derivative-at (f x &optional (delta 0.01)) (defun central-derivative-at (f x &optional (delta 0.01))
(let* ((x2 (+ x delta)) (let* ((x2 (+ x delta))
(x1 (- x delta)) (x1 (- x delta))
(y2 (apply f (list x2))) (y2 (apply f (list x2)))
(y1 (apply f (list x1)))) (y1 (apply f (list x1))))
(/ (- y2 y1) (/ (- y2 y1)
(- x2 x1)))) (- x2 x1))))
(defun fwd-derivative-at (f x &optional (delta 0.01))
(let* ((x2 (+ x delta))
(x1 x)
(y2 (apply f (list x2)))
(y1 (apply f (list x1))))
(/ (- y2 y1)
(- x2 x1))))

6
src/approx/maceps.lisp
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@ -3,10 +3,10 @@
(defun compute-maceps (f a init) (defun compute-maceps (f a init)
(let ((h init) (let ((h init)
(err init)) (err init))
(loop while (> err 0) (loop collect (list a h err)
do do
(setf h (/ h 2) (setf h (/ h 2)
err (abs (- (funcall f (+ a h)) err (abs (- (funcall f (+ a h))
(funcall f a)))) (funcall f a))))
when (> err 0) while (> err 0))))
collect (list a h err))))

3
src/approx/package.lisp
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@ -1,5 +1,6 @@
(in-package :cl-user) (in-package :cl-user)
(defpackage lizfcm.approx (defpackage lizfcm.approx
(:use :cl) (:use :cl)
(:export :derivative-at (:export :central-derivative-at
:fwd-derivative-at
:compute-maceps)) :compute-maceps))

16
tests/approx.lisp
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@ -11,7 +11,7 @@
:in lizfcm-test-suite) :in lizfcm-test-suite)
(in-suite approx-suite) (in-suite approx-suite)
(test derivative-at (test central-derivative-at
:description "derivative at is within bounds" :description "derivative at is within bounds"
(let ((f (lambda (x) (* x x))) (let ((f (lambda (x) (* x x)))
(x 2) (x 2)
@ -19,6 +19,18 @@
(f-prime-at-x 4) (f-prime-at-x 4)
(delta 0.01)) (delta 0.01))
(is (within-range-p (is (within-range-p
(derivative-at f x delta) (central-derivative-at f x delta)
f-prime-at-x
accepted-delta))))
(test fwd-derivative-at
:description "forward derivative at is within bounds"
(let ((f (lambda (x) (* x x)))
(x 2)
(accepted-delta 0.02)
(f-prime-at-x 4)
(delta 0.01))
(is (within-range-p
(forward-derivative-at f x delta)
f-prime-at-x f-prime-at-x
accepted-delta)))) accepted-delta))))