lizfcm/homeworks/hw-3.org

HW 03

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

  (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)))))

a	h	f'	≈ 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

  (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)))))

a	h	f'	≈ 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 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 $\approx (3(2.22045 (10^{-16}))^{\frac{1}{3}} \approx 8.7335 10^{-6}$.