updates 10/9

6 files changed, 515 insertions, 43 deletions

diff --git a/homeworks/hw-2.org b/homeworks/hw-2.org
index 29502d7..f696ffb 100644
--- a/homeworks/hw-2.org
+++ b/homeworks/hw-2.org
@@ -20,35 +20,10 @@ Computing $\epsilon_{\text{mac}}$ for single precision numbers
                          :domain-values domain-values)))
 #+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)
 
 |   a |             h |           err |
+| 1.0 |           1.0 |           1.0 |
 | 1.0 |           0.5 |           0.5 |
 | 1.0 |          0.25 |          0.25 |
 | 1.0 |         0.125 |         0.125 |
@@ -76,6 +51,7 @@ Computing $\epsilon_{\text{mac}}$ for double precision numbers:
 
 (with many rows truncated)
 |     a |                      h |                    err |
+| 1.0d0 |                  1.0d0 |                  1.0d0 |
 | 1.0d0 |                  0.5d0 |                  0.5d0 |
 | 1.0d0 |                 0.25d0 |                 0.25d0 |
 | 1.0d0 |                0.125d0 |                0.125d0 |
diff --git a/homeworks/hw-2.pdf b/homeworks/hw-2.pdf
index b719796..2dc4d28 100644
--- a/homeworks/hw-2.pdf
+++ b/homeworks/hw-2.pdf
diff --git a/homeworks/hw-2.tex b/homeworks/hw-2.tex
index e0aa172..da8d3f5 100644
--- a/homeworks/hw-2.tex
+++ b/homeworks/hw-2.tex
@@ -1,4 +1,4 @@
-% Created 2023-09-27 Wed 10:09
+% Created 2023-10-07 Sat 14:51
 % Intended LaTeX compiler: pdflatex
 \documentclass[11pt]{article}
 \usepackage[utf8]{inputenc}
@@ -29,7 +29,7 @@
 \setlength\parindent{0pt}
 
 \section{Question One}
-\label{sec:orga21c813}
+\label{sec:org58b9af4}
 Computing \(\epsilon_{\text{mac}}\) for single precision numbers
 
 \begin{verbatim}
@@ -49,6 +49,7 @@ Computing \(\epsilon_{\text{mac}}\) for single precision numbers
 \begin{center}
 \begin{tabular}{rrr}
 a & h & err\\[0pt]
+1.0 & 1.0 & 1.0\\[0pt]
 1.0 & 0.5 & 0.5\\[0pt]
 1.0 & 0.25 & 0.25\\[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})
 
 \section{Question Two}
-\label{sec:org06c4a23}
+\label{sec:org27557b4}
 Computing \(\epsilon_{\text{mac}}\) for double precision numbers:
 
 \begin{verbatim}
@@ -81,6 +82,7 @@ Computing \(\epsilon_{\text{mac}}\) for double precision numbers:
 \begin{center}
 \begin{tabular}{rrr}
 a & h & err\\[0pt]
+1.0d0 & 1.0d0 & 1.0d0\\[0pt]
 1.0d0 & 0.5d0 & 0.5d0\\[0pt]
 1.0d0 & 0.25d0 & 0.25d0\\[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}
 
 \section{Question Three - |v|\textsubscript{2}}
-\label{sec:orgf181ba7}
+\label{sec:org59c6c10}
 \begin{verbatim}
 (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
       (2-norm (lizfcm.vector:p-norm 2)))
@@ -124,7 +126,7 @@ x & y & 2norm\\[0pt]
 \end{center}
 
 \section{Question Four - |v|\textsubscript{1}}
-\label{sec:org2196087}
+\label{sec:org2b67b3e}
 \begin{verbatim}
 (let ((vs '((1 1) (2 3) (4 5) (-1 2)))
       (1-norm (lizfcm.vector:p-norm 1)))
@@ -146,7 +148,7 @@ x & y & 1norm\\[0pt]
 \end{center}
 
 \section{Question Five - |v|\textsubscript{\(\infty\)}}
-\label{sec:org11b8894}
+\label{sec:org922206e}
 \begin{verbatim}
 (let ((vs '((1 1) (2 3) (4 5) (-1 2))))
   (lizfcm.utils:table (:headers '("x" "y" "max-norm")
@@ -167,11 +169,11 @@ x & y & infty-norm\\[0pt]
 \end{center}
 
 \section{Question Six - ||v - u|| via |v|\textsubscript{2}}
-\label{sec:orga2324b2}
+\label{sec:org29ec18f}
 \begin{verbatim}
-(let* ((vs '((1 1) (2 3) (4 5) (-1 2)))
-       (vs2 '((7 9) (2 2) (8 -1) (4 4)))
-       (2-norm (lizfcm.vector:p-norm 2)))
+(let ((vs '((1 1) (2 3) (4 5) (-1 2)))
+      (vs2 '((7 9) (2 2) (8 -1) (4 4)))
+      (2-norm (lizfcm.vector:p-norm 2)))
   (lizfcm.utils:table (:headers '("v1" "v2" "2-norm-d")
                        :domain-order (v1 v2)
                        :domain-values (mapcar (lambda (v1 v2)
@@ -193,11 +195,11 @@ v1 & v2 & 2-norm\\[0pt]
 \end{center}
 
 \section{Question Seven - ||v - u|| via |v|\textsubscript{1}}
-\label{sec:org388fbc7}
+\label{sec:org7a87810}
 \begin{verbatim}
-(let* ((vs '((1 1) (2 3) (4 5) (-1 2)))
-       (vs2 '((7 9) (2 2) (8 -1) (4 4)))
-       (1-norm (lizfcm.vector:p-norm 1)))
+(let ((vs '((1 1) (2 3) (4 5) (-1 2)))
+      (vs2 '((7 9) (2 2) (8 -1) (4 4)))
+      (1-norm (lizfcm.vector:p-norm 1)))
   (lizfcm.utils:table (:headers '("v1" "v2" "1-norm-d")
                        :domain-order (v1 v2)
                        :domain-values (mapcar (lambda (v1 v2)
@@ -219,10 +221,10 @@ v1 & v2 & 1-norm-d\\[0pt]
 \end{center}
 
 \section{Question Eight - ||v - u|| via |v|\textsubscript{\(\infty\)}}
-\label{sec:org6e77f76}
+\label{sec:org0f3b64f}
 \begin{verbatim}
-(let* ((vs '((1 1) (2 3) (4 5) (-1 2)))
-       (vs2 '((7 9) (2 2) (8 -1) (4 4))))
+(let ((vs '((1 1) (2 3) (4 5) (-1 2)))
+      (vs2 '((7 9) (2 2) (8 -1) (4 4))))
   (lizfcm.utils:table (:headers '("v1" "v2" "max-norm-d")
                        :domain-order (v1 v2)
                        :domain-values (mapcar (lambda (v1 v2)
diff --git a/homeworks/hw-3.org b/homeworks/hw-3.org
new file mode 100644
index 0000000..efeaa53
--- /dev/null
+++ b/homeworks/hw-3.org
@@ -0,0 +1,244 @@
+#+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 $\approx (3(2.22045 (10^{-16}))^{\frac{1}{3}} \approx 8.7335 10^{-6}$.
diff --git a/homeworks/hw-3.pdf b/homeworks/hw-3.pdf
new file mode 100644
index 0000000..1f9bac6
--- /dev/null
+++ b/homeworks/hw-3.pdf
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}
\ No newline at end of file