%\tracingstats=2 % see just how much memory is left \input pictex \hsize=5.75in % Set the horizontal size to accomodate CDVI, the public domain % % viewer. Use CDVI-2 for VGA monitors with VGA cards. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Set up an "expressmode" for plots to speed up typing and TeXing % until a final draft is ready. This is done as per the PiCTeX manual. % \newif\ifexpressmode % default = \expessmodefalse % % See explanations on pages 215ff of the TeXBook. % \def\yes{y } \def\Yes{Y } \message{Type Y for EXPRESSMODE (no plotting) now: } \read-1 to\answer \ifx\answer\yes\expressmodetrue \else\ifx\answer\Yes\expressmodetrue\fi\fi % The answer is Yes (Y or y). \ifexpressmode \message {EXPRESSMODE --- No plotting} \fi %\expressmodetrue % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % \newdimen\xposition \newdimen\yposition \newdimen\dyposition \newdimen\crossbarlength \def\puterrorbar at #1 #2 with fuzz #3 {% \xposition=\Xdistance{#1} \yposition=\Ydistance{#2} \dyposition=\Ydistance{#3} \setdimensionmode \put {$\bullet$} at {\xposition} {\yposition} \dimen0 = \yposition % ** Determine the y-location of \advance \dimen0 by -\dyposition % ** the lower cross bar. \dimen2 = \yposition % ** Determine the y-location of \advance \dimen2 by \dyposition % ** the upper cross bar. \putrule from {\xposition} {\dimen0} % ** Place vertical rule. to {\xposition} {\dimen2} \dimen4 = \xposition \advance \dimen4 by -.5\crossbarlength \dimen6 = \xposition \advance \dimen6 by .5\crossbarlength \putrule from {\dimen4} {\dimen0} to {\dimen6} {\dimen0} \putrule from {\dimen4} {\dimen2} to {\dimen6} {\dimen2} \setcoordinatemode} % \def\dev{\mathop{\rm dev}\nolimits} % Define special math. function. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \nopagenumbers \hsize=5.75in \centerline{\quad} \vskip4in \centerline{\bf Mathematical Approximation and Documentation} \bigskip \centerline{Harry A. Watson, Jr.} \smallskip \centerline{Quality Assessment Directorate} \centerline{Naval Warfare Assessment Center} \centerline{Corona, California 91718-5000} \centerline{(909) 273-4787} \vfill\eject \centerline{\quad} \vskip 7.5in {\narrower\noindent This book and the associated software fit the description in the U. S. Copyright Act of a ``United States Government Work.'' It was written as a part of the author's official duties as a government employee. This means it cannot be copyrighted. This document and the software are freely available to the public for use without a copyright notice, and there are no restrictions on its use, now or subsequently.} \vfill\eject \footline={\hss\tenrm\folio\hss} % \pageno=-1 \centerline{\bf Preface} \centerline{to} \centerline{\bf Mathematical Approximations and Documentation} \bigskip There are several elementary mathematical approximations that occur on every scientific and engineering project at irregular intervals. Frequently the same approximations appear again and again as {\it ad hoc\/} report requests (and often with short notice). While the programming and documentation is straightforward, the more elaborate and expensive commercial software is often unable to deliver suitable output. This text includes the basic theory and elementary computer programs in BASIC (Beginners All-purpose Symbolic Instruction Code\footnote{$^1$}{The coding is actually done in MS-DOS QBasic (MicroSoft-Disk Operating System). There should be little problem in conversion to GW-BASIC.}) and ``{C}.''\footnote{$^2$}{The ``C'' language is written for the {\it Microsoft QuickC Compiler}.} Documentation, including graphs, is provided in \TeX\ % and \PiCTeX. A later chapter is devoted to more complex graphs done in the device-independent software {\bf METAFONT}, which can be included in \TeX\ files. No deep knowledge of programming is required, nor is the user expected to be a ``\TeX-nician;'' indeed, the user is expected to lift the coding ``as is'' and make the obvious substitutions. This book and the associated software fit the description in the U. S. Copyright Act of a ``United States Government Work.'' It was written as a part of the author's official duties as a government employee. This means it cannot be copyrighted. This document and the software are freely available to the public for use without a copyright notice, and there are no restrictions on its use, now or subsequently. Anyone using this document or these programs and files does so entirely at her/his own risk. Anyone who demands payment for the distribution of this material must make clear that the charge is for distribution only and is in no sense a license fee or purchase fee. The three topics covered in this textbook are (1) Simpson's rule for numeric integration, (2) Newton's method for approximation of roots, and (3) least-squares curve fit with a test to verify the choice of the underlying statistical distribution. Many examples are provided with hand-calculations and tables. It is expected that the user will find portions of the \TeX\ file useful. With this in mind, the macros ({\tt \char92 def}) are kept to a minimum and the displayed portions are self-contained. As often as possible, references are made to textbooks universally available, such as outline series books found in any college bookstore or encyclopedia articles. The author has attempted to locate examples of particular interest rather than pathological cases. This is to ensure a greater likelihood that an example may be copied directly into a scientific or engineering paper. \medskip % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Here is a first graph that you can cut and paste.% % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % $$\beginpicture % Draw the graph of a \setcoordinatesystem units <.78539in,.5in> % Sine curve using the \setplotarea x from -2.25 to 4.5, y from -1 to 1 % macros of PiCTeX. \plotheading {The graph of $y = \sin x$} \axis bottom shiftedto y=0 label {$\scriptstyle y = \sin x$} ticks in withvalues {$\scriptstyle -\pi$} {$-{\pi\over2}$} {\quad\ O} {$\pi\over2$} {$\scriptstyle\pi$} {${3\pi\over2}$} {$\scriptstyle2\pi$} / at -2 -1 0 1 2 3 4 / / \axis left shiftedto x=0 / \ifexpressmode \put {\bf EXPRESSMODE} at 2.00 0.5 \else \setquadratic \plot 0 0 0.2 0.30902 0.4 0.58779 0.6 0.80902 0.8 0.95106 1.0 1.00000 1.2 0.95106 1.4 0.80902 1.6 0.58779 1.8 0.30902 2.0 0.00000 2.2 -.30902 2.4 -.58779 2.6 -.80902 2.8 -.95106 3.0 -1.0000 3.2 -.95106 3.4 -.80902 3.6 -.58779 3.8 -.30902 4.0 0.00000 / % \plot 0 0 -0.2 -0.30902 -0.4 -0.58779 -0.6 -0.80902 -0.8 -0.95106 -1.0 -1.00000 -1.2 -0.95106 -1.4 -0.80902 -1.6 -0.58779 -1.8 -0.30902 -2.0 -0.00000 / % \fi \endpicture$$ \centerline{\rm F{\sevenrm IGURE} 1} \medskip % % The sine (not the "sin" or moral failing) is used over and over % in scientific and engineering texts. It is always a good "filler" % in any scientific document. % A word about fonts is in order. The only fonts employed are from the standard distribution (16 fonts) for \TeX. All computer program statements are typeset in the so-called {\tt typewriter font\/} to provide contrast and to allow alignment of the characters for keyboard input. The copyright for the program \TeX\ belongs to D.~E.~Knuth; the trademark \TeX\ belongs to the American Mathematical Society; and {\tt MS-DOS} is a trademark of the Microsoft Corporation. \vfill\eject \pageno=-2 \centerline{\bf TABLE OF CONTENTS} \bigskip {\narrower \settabs 6 \columns \+ Preface &&&&&\ i\cr \medskip \+ Table of Contents &&&&&\/ ii\cr \medskip \+ && Chapter 1---Simpson's Rule &&&\cr \+ 1.0 & Introduction &&&&\ 1\cr \+ 1.1 & Simpson's Rule &&&&\ 2\cr \+ 1.2 & Integration Computer Programs &&&&\ 4\cr \+ 1.3 & Vivasection of the Programs &&&&\ 7\cr \+ 1.4 & Numerical versus Exact Integration &&&&\ 9\cr \+ 1.5 & Truncation Error &&&&11\cr \+ 1.6 & An Example &&&&12\cr \+ 1.7 & Verifying Derivatives &&&&14\cr \+ 1.8 & Generalizations &&&&19\cr \medskip \+ && Chapter 2---Newton's Method &&&\cr \+ 2.0 & Introduction &&&& 21\cr \+ 2.1 & Theory &&&&22\cr \+ 2.2 & Applications &&&&23\cr \+ 2.3 & The Algorithm &&&&26\cr \+ 2.4 & A Second Opinion &&&&27\cr \+ 2.5 & The Quasi-Newton Method &&&&28\cr \+ 2.6 & Pathological Examples &&&&30\cr \medskip \+ && Chapter 3---Linear Least-squares Line &&&\cr \+ 3.0 & Introduction &&&& 31\cr \+ 3.1 & Dissection of a Graph &&&& 32\cr \+ 3.2 & Formula Derivation &&&& 37\cr \+ 3.3 & Goodness of Fit &&&& 39\cr \+ 3.4 & More Theory &&&& 42\cr \+ 3.5 & Problems &&&& 44\cr \medskip \+ && Chapter 4---An Adaptive Quadrature Routine &&&\cr \+ 4.0 & Introduction &&&& 47\cr \+ 4.1 & A Simple Approach &&&& 48\cr \+ 4.2 & An Adaptive Quadrature Routine &&&& 49\cr \medskip \+ && Chapter 5---{\bf METAFONT} &&&\cr \+ 5.0 & Introduction &&&& 51\cr \+ 5.1 & A Metafont Program &&&& 52\cr \medskip \+ && Chapter 6---An Important Improper Integral &&&\cr \+ 6.0 & Introduction &&&& 54\cr \+ 6.1 & Theoretical Evaluation &&&& 55\cr \+ 6.2 & Numerical Evaluation &&&& 57\cr \bigskip \+ Appendix A---Mathematical Preliminaries &&&&& 58\cr \medskip \+ Appendix B---References &&&&& 61\cr \medskip \+ Appendix C---A Geometric Persuasion &&&&& 62\cr \medskip \+ Appendix D---A Contour Integral &&&&& 64\cr \medskip \+ Appendix E---The Malgrange-Ehrenpreis Theorem &&&&& 66\cr } \vfill\eject %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pageno=1 % \headline={\tenrm\hfill Simpson's Rule} \centerline{\bf Chapter 1} \bigskip \noindent{\bf\llap{1.0\quad}Introduction.} When one finally finishes algebra and geometry and embarks into the realm of calculus, all problems seem solvable. The universe appears to be reduced to mathematical constructs and it is only a matter of time until mathematics can define everything. There is a conceit among those who first master the art and discipline of {\bf The Calculus}. The tyro\footnote{$^3$}{A beginner in learning, a novice. Don't rush to your dictionary---footnotes will be provided for many words.} feels a superiority to those who are ignornant of this powerful tool, replete with its repertoire of skills, special symbols, concise language, and geometric persuasions. The euphoria which is closely associated with the mastery of {\bf The Calculus} leaves its pupil craving a repeat experience. The classical development of calculus has evolved over several centuries; the problems and examples are thus arranged so as to direct the student along a path devoid of many of the harsh pitfalls in applied science and engineering. Embedded in this development are the two classic encounters with the dirty reality of numerical approximation: Newton's method for approximating roots and Simpson's rule for approximating integrals. These two techniques are ubiquitous\footnote{$^4$}{Being found everywhere at the same time.} in numerical analysis, calculating devices, and engineering in general. Their study and development is the subject of the first two chapters. In this, the first chapter, the topic of Simpson's rule is considered. The coding is done in several different computer languages and the text edited in different formats to ensure complete generality. Since Simpson's rule depends on properties of the parabola, a general graph of a parabola is displayed below. \bigskip % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % The graph of a parabola. % \centerline{PARABOLA} \medskip $$\beginpicture \setcoordinatesystem units <1cm,1cm> \setplotarea x from -2.5 to 5.0, y from -1.5 to 5.0 % \axis bottom shiftedto y=0.0 ticks length <0pt> withvalues {$O$} {$x$} / at -.25 5.0 / / % \axis left shiftedto x=0.0 ticks length <0pt> withvalues {$y$} / at 5.0 / / % \put {$V$} [t] <2pt,-4pt> at 1 1 % \put {$\scriptstyle\bullet$} at 1 1 % \ifexpressmode \put {\bf EXPRESSMODE} at 3 3 % \else \setquadratic \plot -2 5 1 1 4 5 / % \fi \endpicture$$ $$y = f(x) = a x^2 + bx + c,\ a>0,$$ $$\hbox{Vertex } = V = \left(-{b\over2a},\ % {b^2\over2a}\left({1\over2a}-1\right) + c\right).$$ \medskip $$\hbox{F{\sevenrm IGURE} 2}$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \vfill\eject \bigskip \noindent{\bf\llap{1.1\quad}Simpson's rule.} Most scientists and engineers have completed elementary calculus and are aware of the technique for numerical approximation by fitting arcs of parabolas; however, even someone who has only completed high school mathematics (algebra and plane geometry) will be able to apply this technique with success (and produce a professional output that will amaze even the most erudite\footnote{$^5$}{Possessing or displaying excessive knowledge acquired mainly from books.} pedant on the staff). There are pitfalls; the expert as well as the tyro can be trapped by some speciously simple-looking functions.\footnote{$^6$}{One pathological example is the so-called {\it Quadratrix of Hippias}, $y=x\tan\big(\pi y/2\big)$.} Having said all of this, let us begin the basic theoretical development of Simpson's rule.\footnote{$^7$}{George B. Thomas, Jr., {\it Calculus and Analytic Geometry}, (Reading, Massachusetts: Add\-i\-son-Wes\-ley Publishing Company, Inc., 1966), pages 385-388.} \medskip \noindent First of all, let $[\alpha,\beta]$ be a number interval, that is $$[\alpha,\beta] = \{\,x\,\mid\,\alpha \le x \le \beta\,\}.$$ We'll use the Greek letters $\alpha$ and $\beta$ because later we'll want to use $a$, $b$, and $c$ as coefficients of the quadratic equation $y=ax^2+bx+c$. % Moreover, as we develop this method, the % importance of the endpoints will diminish. We will let $f$ denote a function, $f\colon [\alpha,\beta] \to {\bf R}$, where ${\bf R}=\{\,x\,\mid\,x$ is a real number $\}$. For convenience, we will use let $y_\alpha = y_0 = f(\alpha)$, $y_\beta = y_2 = f(\beta)$, and $y_1 = f\big((\beta+\alpha)/2\big)$. This is all bookingkeeping. It is one of those mathematical concepts that causes ``math anxiety;'' however, it doesn't need to. Just consider $y$ to be the left-hand side of some expression, for instance $$ y = x^2 - 4x +2.$$ When $x=2$ we might have $y = 2\cdot2 - 4\cdot2 + 2 = -2$. This could be written as $y(2)$ or $y(x=2)$ or even $y\big|_{x=2}$, which is also written $y\big|_{2}$. We will just simplify things and write it as $y_2$. This is a fundamental idea, if the reader is confused or lost on this matter, it would be a good idea to review an algebra book (or consult appendix A). \medskip \noindent Simpson's rule follows from the so-called {\bf prismoidal formula} (for areas) $$A_p\ =\ {h\over3}\big[\,y_0+4y_1+y_2\big]\ =\ {\beta-\alpha\over6} \left[f(\alpha) + 4\cdot f\left({\beta+\alpha\over2}\right) +f(\beta)\right].$$ We will denote the left-hand side of this equation as $A_p$, which stands for the area under the arc of the parabola. This simple formula gives the exact area if the function, $f(x)$, is a polynomial of degree not higher than 3.\footnote{$^8$}{Glenn James and James, Robert C., {\it Mathematics Dictionary}, (Princeton, NJ: D. Van Nostrand Company, Inc., 1964), pages 358-359.} In particular, $A_p$, gives the area under the arch of the parabola $$y\ =\ f(x)\ =\ ax^2 + bx + c$$ between $x=-h$ and $x=+h$, as in Fig. 3.\footnote{$^9$}{Without loss of generality, we choose the interval $[a,b]$ to be $[-h,h]$. This simplifies the algebra. Everything is still completely general because we can always ``shift'' a graph back and forth along the $x$-axis without changing the area under the curve.} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % The area under an arch of a parabola. % $$\beginpicture \setcoordinatesystem units <1cm,1cm> % Use PiCTeX to \setplotarea x from -2.5 to 2.5, y from -.5 to 2.5 % construct this \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 3} % ticks length <0pt> withvalues {$\scriptstyle-h$} {$O$\quad} {$\scriptstyle h$} / at -2 0 2 / / \axis left shiftedto x=0 / \putrule from -2 0 to -2 1.5 \putrule from 2 0 to 2 0.75 \put {$\scriptstyle x$} at 2.75 0 \put {$\scriptstyle y$} at 0 2.75 \put {$\scriptstyle y_0$} at -1.75 0.75 \put {$\scriptstyle y_1$} at 0.2 0.75 \put {$\scriptstyle y_2$} at 1.75 0.375 \ifexpressmode \put {\bf EXPRESSMODE} at 1 1 \else \setquadratic \plot -2 1.5 0 1.6 2 0.75 / \fi \endpicture$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \bigskip \noindent This result is readily established as follows: in the first place, we have $$A_p = \int_{-h}^h \left(ax^2+bx+c\right)\,dx = \ {{ax^3\over3}+{bx^2\over2}+cx\,\bigg|}_{x=-h}^{x=h} \ =\ {2ah^3\over3} + 2ch.$$ Since the curve passes through the three points $(-h,y_0)$, $(0,y_1)$, and $(h,y_2)$, we also have $$y_0 = ah^2-bh+c,\quad y_1=c,\quad y_2=ah^2+bh+c,$$ from which it follows that $$\eqalign{ c\ &=\ y_1,\cr ah^2 -bh \ &=\ y_0-y_1,\cr ah^2+bh \ &=\ y_2-y_1,\cr 2ah^2 \ &=\ y_0+y_2-2y_1.\cr}$$ Hence, expressing the area $A_p$ in terms of the ordinates $y_0$, $y_1$, and $y_2$, we have $$A_p = {h\over3}\left[2ah^2 +6c\right] ={h\over3}\left[\big(y_0+y_2-2y_1\big) +6y_1\right]$$ or $$A_p = {h\over3}\left[y_0 + 4y_1 + y_2\right]\ = \ {h\over3}\left[f(-h)+4\cdot f(0) + f(h)\right].\eqno(1)$$ % \medskip \noindent Simpson's rule is derived by applying the prismoidal formula to successive subintervals of $[\alpha, \beta]$. Each separate subinterval has length $2h$. So, we may write $[\alpha,\alpha+h]$ as $[x_0, x_1]$, $[\alpha+h, \alpha+2h]$ as $[x_1, x_2]$, and so on. Now the curve $y=f(x)$ between $x=\alpha$ and $x=\beta$ is really $n/2$ curves. Each separate piece of the curve covers an $x$-subinterval of width $2h$; it is approximated by an arch of a parabola through its ends and its mid-point. The area under each parabolic arc is computed as in Eq. (1) and the resulting areas are summed together to give the rule $$\frame <5pt> {$\displaystyle A_S = {h\over3}\big[\, y_0 + 4y_1 + 2y_2 +4y_3 + \cdots + 2y_{n-2} + 4y_{n-1} + y_n \big].$}\eqno(2)$$ This result, $A_S$, is an approximate value of $\int_a^b f(x)\,dx$. Recall our bookkeeping. We let $y_0$, $y_1$, $y_2$, $\ldots$, $y_n$ stand for $f(x_0)$, $f(x_1)$, $\ldots$, $f(x_n)$. In other words, $y_0$, $y_1$, $\ldots$, $y_n$ are the ordinates of the curve $y=f(x)$ corresponding to the abscissas $x_0=a$, $x_1=a+h$, $x_2=a+2h$, $\dots$, $x_n=a+nh=b$. Each abscissa point then corresponds to an endpoint of a subdivision of the interval $\alpha \le x \le \beta$ into $n$ equal subintervals each of width $h=(\beta-\alpha)/n$. (See Fig. 4.) The number $n$ of subdivisions must be an {\it even\/} integer in order for this method to work. Not all authors use this same scheme. Some rely on subintervals of length $h$ instead of $2h$. When consulting a table book, be careful to examine the setup for this rule. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Simpson's rule over a curve (after Thomas's Calculus). % $$\beginpicture \setcoordinatesystem units <1cm,1cm> \setplotarea x from -0.5 to 10, y from -0.5 to 3.0 \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 4} % ticks length <0pt> withvalues {$O$} {$\alpha$} {$\leftarrow\! h\!\rightarrow$} {$\beta$} / at -0.25 1 3.5 7 / / % \axis left shiftedto x=0 / % \putrule from 1 0 to 1 2.8 \putrule from 2 0 to 2 2.9 \putrule from 3 0 to 3 2.65 \putrule from 4 0 to 4 2.3 \putrule from 5 0 to 5 1.85 \putrule from 6 0 to 6 1.5 \putrule from 7 0 to 7 1.4 \putrule from 3 -0.02 to 3 -0.4 \putrule from 4 -0.02 to 4 -0.4 \put {$ y=f(x)$} at 5 3 \put {$ x$} at 10.25 0 \put {$ y$} at 0 3.25 \put {$\scriptstyle 1$} <0pt,2pt> at 1.0 2.9 \put {$\scriptstyle 4$} <0pt,2pt> at 2.0 3.0 \put {$\scriptstyle 2$} <0pt,2pt> at 3.0 2.75 \put {$\scriptstyle 4$} <0pt,2pt> at 4.0 2.4 \put {$\scriptstyle 2$} <0pt,2pt> at 5.0 1.95 \put {$\scriptstyle 4$} <0pt,2pt> at 6.0 1.6 \put {$\scriptstyle 1$} <0pt,2pt> at 7.0 1.5 \put {$\scriptstyle y_0$} at 0.8 1 \put {$\scriptstyle y_1$} at 1.8 1 \put {$\scriptstyle y_2$} at 2.8 1 \put {$\scriptstyle y_{n-1}$} at 5.7 1 \put {$\scriptstyle y_{n}$} at 6.8 1 \ifexpressmode \put {\bf EXPRESSMODE} at 3 3 \else \setquadratic \plot 0.5 2.6 1.0 2.8 2.0 2.9 3.0 2.65 4.0 2.3 5.0 1.85 6.0 1.5 7.0 1.4 7.5 1.5 / % \fi \endpicture$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % It looks like the more subintervals we take, that is the smaller we let $h$ become, the more accurate the approximation will be to the area under the curve (the value of the integral). This is part of error estimation and will be done in a later paragraph. In the next paragraph we will do numerical examples and see how the method behaves numerically. \vfill\eject \noindent{\bf\llap{1.2\quad}Integration Computer Programs.} We'll list two computer programs to do the same job---one in BASIC and one in {``C''}. One integral that crops up again and again is the so-called elliptic integral. We will look at a complete elliptic integral of the second kind, usually denoted $$E\left(k,{\pi\over2}\right)\ =\ \int\nolimits_0^{\pi/2} \sqrt{1-k^2\cdot\sin^2 \theta}\,d\theta.\eqno(3)$$ Finally, we need to choose a value for $k$. If we let $e$ denote the base of the natural logarithms, that is $e \approx 2.7182828459045\dots$, then assign $k$ to be $k = \sin\big(e/2\big)$, we will have a good example to experiment with.\footnote{$^{10}$}{Harry A. Watson, Jr., ``An approximation to Planck's constant,'' {\it Abstracts\/} of the American Mathematical Society (AMS) \# 81T--181, June 1981.} \bigskip {\parindent=0pt\tt\ttraggedright\obeylines 100 DEFDBL A-H, O-Z : REM DEF FNA defines the integrand (function). 105 DEF FNA (x AS DOUBLE) = SQR(1!-(SIN(EXP(1)/2)*SIN(x)){\char94}2) 110 A = 0!: REM This is the lower endpoint of [a,b]. 120 B = 3.141592653589793\#/2!: REM This is the upper endpoint of [a,b]. 130 SUM = 0!: REM This is an accumulator. 135 REM The user is asked for a number of intervals, n, even and > 0. 140 INPUT "Enter number of intervals (must be even) "; N 145 IF N <= 0 OR (2 * FIX(N / 2) - N < 0) THEN PRINT "Input error": GOTO 140 150 H = (B - A) / N: REM The sub-interval size h = (b-a)/n. 160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times. 170 SUM = SUM + FNA(A + (I - 1) * H) 175 PRINT A + (I - 1) * H, FNA(A + (I - 1) * H) 180 SUM = SUM + 4! * FNA(A + I * H) 185 PRINT A + I * H, FNA(A + I * H) 190 SUM = SUM + FNA(A + (I + 1) * H) 195 PRINT A + (I + 1) * H, FNA(A + (I + 1) * H) 200 NEXT I: REM After loop, the sum is y\_0+4*y\_1+2*y\_2+...+4*y\_{\char123}n-1{\char125}+y\_n. 210 PRINT "Sum = "; SUM: PRINT " Value of integral = "; SUM * H / 3 220 REM This short program will integrate a function FNA(x) from x = a to b. } \medskip \noindent The ``C'' program is more complicated because it has the ``{\tt\#include}'' statements. However, the ``C'' program executes faster and is portable to other platforms. In addition, the programs below both use a math co-processor, which gives floating-point arithmetic. Minor differences in the last digit may be due to the print (format) algorithm. The BASIC program would not accept the longer value of $\pi$. If one examines the \TeX\ file for these listings, it will be noted that the control characters are input as {\tt\char92char92} for the backspace ({\tt\char92}), {\tt\char92char123} for the left brace ({\tt\char123}), {\tt\char92char125} for the right brace ({\tt\char125}), and {\tt\char92char94} for the hat or circumflex ({\tt\char94}). \medskip {\parindent=0pt\tt\ttraggedright\obeylines \#include /* Header for input/output subroutines. */ \#include /* Header for math subroutines. */ \#include /* Header for floating point subroutines. */ \medskip \#define pi 3.141592653589793238462643383279 /* Accurate value for pi. */ \medskip /* Simpson's rule for approximating integrals. \qquad a: left endpoint \qquad b: right endpoint \qquad fc: pointer to function to integrate \qquad n: number of subintervals */ double fc (double x); \medskip main() {\char123} double a,b,h,sum,x,y; /* In 'C' all variables must be assigned */ double p1, p2, p3; int i, n; printf("{\char92}007"); /* Sound bell in computer. */ a = (double) 0.0; printf("{\char92}nLeft end point = \%.16lf",a); b = (double) pi/2.0; printf("{\char92}nRight end point = \%.16lf",b); i = -1; while (i < 0){\char123} printf("{\char92}nEnter number of subintervals (must be even) "); scanf("{\char92}\%d",\&n); i = n/2; i = 2*i - n; /* Don't allow odd values of n. */ if (n<=0) i = -1; /* Don't allow zero or negative. */ {\char125} printf("{\char92}nNumber of subintervals \%d",n); h = (double) (b-a)/n; for (i=1, sum=0.0; i<=n; i = i+ 2){\char123} \quad p1 = fc((double) a+(i-1)*h); \quad p2 = fc((double) a+i*h); \quad p3 = fc((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; /* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+(i-1)*h,p1); */ /* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+i*h,p2); */ /* printf("{\char92}n x, f(x) \%.16lf \%.16lf",a+(i+1)*h,p3); */ {\char125} \medskip printf("{\char92}nValue of sum = \%.16lf", (double) sum); y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}n"); return(0); {\char125} \medskip double fc (double x) {\char123} \qquad double y; \qquad y = sqrt(1.0-sqrt(exp(1.)/2.)*sqrt(exp(1.)/2.)*sqrt(x)*sqrt(x)); \qquad return (y); {\char125} \medskip /* End of file */ } % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Parabolic curve fit for unequally spaced abscissas. % $$\beginpicture \setcoordinatesystem units <.5cm,.3cm> \setplotarea x from -0.25 to 20, y from -0.25 to 10.0 \axis bottom label {Parabolic curve fits unequally spaced abscissas} shiftedto y=0 / % \axis left shiftedto x=0 / % \putrule from 2 0 to 2 3.8 \putrule from 4 0 to 4 3.85 \putrule from 5.3 0 to 5.3 3.6 \putrule from 9 0 to 9 4.78571 \putrule from 10. 0 to 10. 4.95 \putrule from 14 0 to 14 8.35 \putrule from 16 0 to 16 8.6 \put {$O$} at 0.5 -0.75 \put {$x$} at 19 -0.5 \put {$f(x)$} [r] at -.25 9 \ifexpressmode \put {\bf EXPRESSMODE} at 10 5 \else \setquadratic \plot 1 3 3 4 7 3 / % \plot 5 3.5 8 4.5 12 5.5 / % \plot 10 5.0 13 8.0 17 8.5 / % \fi \endpicture$$ \centerline{F{\sevenrm IGURE} 5} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \vfill\eject \bigskip \noindent We'll make a table of values for various integer values of $n$, that is, various numbers of subdivisions. Much can be learned from comparing the outputs of the two programs and examining their behavior as $n$ becomes large. \smallskip $$\vbox{\offinterlineskip \hrule \halign{&\vrule#&\strut\quad\hfil#\quad&\vrule# &\quad\hfil#\quad&\vrule#&\quad\hfil#\quad&\vrule# &\quad\hfil#\quad&\vrule#\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr &$n =$\hfil&&BASIC Program&&``C'' Program\quad&&Difference\quad&\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr & 2 &&1.073441760234934 &&1.0734417602349340&&0.00000000000000&\cr & 4 &&1.057800052054470 &&1.0578000520544700&&0.00000000000000&\cr & 8 &&1.054898082319603 &&1.0548980823196030&&0.00000000000000&\cr & 10 &&1.054750510268630 &&1.0547505102686300&&0.00000000000000&\cr & 20 &&1.054686162116484 &&1.0546861621164840&&0.00000000000000&\cr & 40 &&1.054685849666779 &&1.0546858496667790&&0.00000000000001&\cr & 80 &&1.054685849645436 &&1.0546858496454370&&0.00000000000001&\cr &100 &&1.054685849645437 &&1.0546858496454360&&0.00000000000001&\cr &200 &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr &500 &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr &1,000 &&1.054685849645437 &&1.0546858496454370&&0.00000000000000&\cr &2,000 &&1.054685849645436 &&1.0546858496454370&&0.00000000000001&\cr &10,000 &&1.054685849645435 &&1.0546858496454330&&0.00000000000002&\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr} \hrule}$$ \smallskip\noindent Notice, if you will, two things are happening: (1) after $n=40$ there is no change in the first five decimal places; and, (2) after $n=80$ there is almost no change at all! So, after $n=80$, successive terms become closer and closer together and the calculated value converges to a given number. This phenomenon is called, obviously enough, convergence. It would be nice if we could tell {\sl beforehand\/} just what value of $n$ to choose to ensure the accuracy we need. Clearly, if all we want is five decimal place accuracy, $n=40$ will suffice. If we are using a double-precision calculation, then $n=80$ will do. Of course, you might say why not just take $n=10000$ to start with? Just run the program and see how long it takes to get an answer! It is possible that time is also a factor---even the fastest computers take time to do the involved mathematical operations required in numerical integration. \medskip\noindent The next item to be introduced is error estimation. It will be shown that for $n$ subintervals, there is a number $\xi\,\in\,(\alpha, \beta)$ such that the truncation error is $$-{(\beta-\alpha)^5\over 180n^4}\,f^{(4)}(\xi).$$ If the fourth derivative of $f(x)$ is bounded on $[\alpha, \beta]$, that is, if there exists a number $M$ such that $$\big| f^{(4)}(x) \big|\,\le\,M\quad\hbox{for all } x\in [\alpha,\beta],$$ then the error is bounded by $${(\beta-\alpha)^5\,M \over 180 n^4}.$$ At this point in our development, the importance of the error bound is that we will be assured that the larger the value of $n$, the closer the answer will be to the exact solution. This says that for any given degree of accuracy, there is a positive integer $N$ such that for all $n\ge N$, each approximation computed by Simpson's rule with $n$ terms will lie within the given degree of accuracy to the exact solution. \medskip\noindent This error estimation sounds like pure theory, and so it could be skipped without losing anything. That's really not the case. What we have is exactly equivalent to the mechanical ``tolerance.'' If you want something within a particular tolerance, say $\epsilon$, then you must calibrate your tools to within some accuracy, say $\delta$. It is exactly this so-called $\epsilon$-$\delta$ philosophy that is the basis for theoretical analysis. One of the great failings of pure analysis has been its inability to relate to the industrial quality efforts and statistical error estimations. \vfill\eject \noindent{\bf\llap{1.3\quad}Vivasection\footnote{$^{11}$}{\rm Disection of a living (as opposed to inoperable) computer program.} of the Programs.} The first item to note in the computer programs is the definition of the function $f(x)$\footnote{$^{12}$}{Mathematicians prefer to call a function by its single letter abbreviation, e.g. $f$, $g$, etc.; however, since the independent variable, say $x$, is required in the computer definitions, it seems better to use the convention of complex analysis and keep the notation $f(x)$, $g(x)$, etc., so that there is no ambiguity in the computer coding.} In the BASIC program the function is defined by the statement \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 105 DEF FNA (x AS DOUBLE) = SQR(1!-(SIN(EXP(1)/2)*SIN(x)){\char94}2) } \smallskip\noindent while in the ``C'' program the function is defined as a separate subroutine. ``C'' is fashioned after subroutines, so this should be no surprise. The function definition subroutine in ``C'' is thus \smallskip {\tt\parindent=0pt\ttraggedright\obeylines double fc (double x) {\char123} \qquad double y; \qquad y = sqrt(1.0-sin(exp(1.)/2.)*sin(exp(1.)/2.)*sin(x)*sin(x)); \qquad return (y); {\char125} } \smallskip\noindent Notice, further, that BASIC supports exponentiation, that is {\tt A\char94B} means $A^B$ whereas ``C'' does not (it requires a separate subroutine call to {\tt pow()}). Now we will focus on the interval endpoints, $\alpha$ and $\beta$, of the interval $[\alpha, \beta]$. Since the letters {\tt A, B} and {\tt a, b} are not used elsewhere inside the programs, they can be used for endpoint values. BASIC is not case sensitive; ``C'' is. We define the endpoints (or limits of integration) as follows: \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 110 A = 0!: REM This is the lower endpoint of [a,b]. 120 B = 3.141592653589793\#/2!: REM This is the upper endpoint of [a,b]. } \smallskip {\tt\parindent=0pt\ttraggedright\obeylines a = (double) 0.0; printf("{\char92}nLeft end point = \%.16lf",a); b = (double) pi/2.0; printf("{\char92}nRight end point = \%.16lf",b); } \smallskip\noindent Notice that in the ``C'' program the value of the endpoints is displayed via a {\tt printf()} function. It is always a good practice to display such variables. The next item of interest is the input of the number of intervals, $n$. This also determines the width (length) of the subintervals, $h = (\beta-\alpha)/n$. Each program does a check to ensure that the integer $n$ is positive and not odd. There are many ways to make such a determination, so it is not necessary to dwell on the matter. The main portion of the Simpson's rule is the iteration. Here a variable named {\tt sum} or {\tt SUM} is initialed to zero and functions as an accumulator. The values $y_{i-1} + 4\cdot y_{i}+y_{i+1}$ are added to the accumulator for $i$ between 1 and $n-1$ in steps of 2. The final value of the accumulator is multiplied by $h/3$ to get the approximation to the integral $\int_\alpha^\beta f(x)\,dx$. \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 130 SUM = 0!: REM This is an accumulator. 150 H = (B - A) / N: REM The sub-interval size h = (b-a)/n. 160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times. 170 SUM = SUM + FNA(A + (I - 1) * H) 180 SUM = SUM + 4! * FNA(A + I * H) 190 SUM = SUM + FNA(A + (I + 1) * H) 200 NEXT I: REM After loop, the sum is y\_0+4*y\_1+2*y\_2+...+4*y\_{\char123}n-1{\char125}+y\_n. } \smallskip\noindent Notice in the ``C'' coding below the statement {\tt sum += }$\dots$. This is identical to the coding {\tt sum = sum + }$\dots$. \smallskip {\tt\parindent=0pt\ttraggedright\obeylines h = (double) (b-a)/n; for (i=1, sum=0.0; i<=n; i = i+ 2){\char123} \quad p1 = fc((double) a+(i-1)*h); \quad p2 = fc((double) a+i*h); \quad p3 = fc((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; {\char125} } \smallskip\noindent The final calculation and the display is given below. First for the BASIC program and then for the ``C'' program. The BASIC program does its calculation in the {\tt PRINT} statement. \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 210 PRINT "Sum = "; SUM: PRINT " Value of integral = "; SUM * H / 3 } \smallskip\noindent Notice that the ``C'' program uses a subroutine ({\tt printf()}) to display its output on the video monitor. Notice further that in the ``C'' program the last statement is a print with the argument {\tt "{\char92}n"}. This is a carriage return/line feed that is needed after the last print statement. Otherwise, the program will type ``{\tt Hit any key to continue}'' on the same line as the output. \smallskip {\tt\parindent=0pt\ttraggedright\obeylines printf("{\char92}nValue of sum = \%.16lf", (double) sum); y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}n"); } \medskip\noindent We will examine the error estimation for this integral. Let's write down the function and its first four derivatives: $$\eqalign{f(x) &= \sqrt{1-k^2\sin^2 x}; \quad\hbox{where } k = \sin\big(e/2\big);\cr f'(x) &= -{k^2\over2}{\sin(2x)\over\sqrt{1-k^2\sin^2x} };\cr f''(x) &= -k^2\cos2x/\sqrt{X} -k^4\sin^2(2x)/\big(4\cdot X^{3/2}\big); \quad\hbox{where } X = \left(1-k^2\sin^2(x)\right)\cr f^{(3)}(x) &= 2k^2\sin(2x)/\sqrt{X} -3k^4\sin(4x)\big(4\cdot X^{3/2}\big)\cr &\qquad - 3k^6\sin^3(2x)/\big(8\cdot X^{5/2}\big);\cr f^{(4)}(x) &= 4k^2\cos(2x)/\sqrt{X} + k^4\sin^2(2x)/X^{3/2} -3k^4\cos(4x)/X^{3/2}\cr &\qquad -9k^6\sin(4x)\sin(2x)/\big(8\cdot X^{5/2}\big) -18k^6\sin^2(2x)\cos(2x)/\big(8\cdot X^{5/2}\big)\cr &\qquad - 15k^8\sin^4(2x)/\big(16\cdot X^{7/2}\big).\cr}$$ Observing that if $0.0 \le x \le 1.572$, then $-313.8 \le f^{(4)}(x) \le 85.0$, we get a bound for the fourth derivative on $[0,\pi/2]$. If $M=320$, then $$\left| f^{(4)}(x)\right| \le M,\quad\forall\,x\in\,\left[0,{\pi\over2}\right].$$ The error is thus bounded by $${\pi^5\cdot M\over 180\cdot32\cdot n^4}\ \approx\ {17\over n^4},$$ where $n$ is the number of subintervals of $[\alpha,\beta] = [0, \pi/2]$. Thus, to ensure that our answer is within say .5\% of the true answer, we would have to pick $n$ to be greater than 7 (and even!). But this is a conservative estimate. We have already seen that a much smaller value of $n$ will suffice. In the examples chosen for textbooks, the fourth derivative is generally a constant (a constant-values function, that is, a function $f^{(4)}(x) = $ constant for all $x$ under consideration). This builds the student's confidence, but does little to reflect the real-world situation. The integral we have chosen is an integral that crops up again and again in engineering problems. Is there a satisfactory resolution to this (common) problem? Indeed, there is. The idea is to examine the approximating values for increasing values of the integer $n$ which determines the step size. This technique is based on the theoretical consideration that a Cauchy sequence converges. In computer programming, it is called an {\it adaptive quadrature routine\/} (AQR). We will cover the adaptive quadrature routine in a later paragraph. \vfill\eject \noindent{\bf\llap{1.4\quad}Numerical versus Exact Integration.} The sophomore who has just completed an introduction to calculus might be surprised to discover that an integral as simple as $$\int_0^\pi \sqrt{1+\cos^2 x}\,dx\eqno(4)$$ cannot be solved by elementary techniques. (The value of this integral is the length of one ``arch'' of the sine curve, $y=\sin x$. See Fig. 1 in the preface for a graph of the sine curve.) It can be expressed in terms of a complete elliptic integral of the second kind. If we write $$E\left(k,{\pi\over2}\right) \ =\ \int_0^{\pi/2} \sqrt{1-k^2\sin^2\theta}\,d\theta,\eqno(5)$$ then we may make the simple trigonometric substitution $$\cos^2\theta = 1 - \sin^2\theta$$ to obtain $$\eqalign{\int_0^\pi \sqrt{1-\cos^2\theta}\,d\theta \ &=\ 2\cdot\sqrt{2}\cdot E(\pi/4,\pi/2)\cr \ &=\ \sqrt{2}\,\int_0^{\pi/2} \sqrt{1-\sin^2(\pi/4)\cdot\sin^2(\theta)}\,d\theta\cr &\qquad +\ \sqrt{2}\,\int_{\pi/2}^\pi \sqrt{1-\sin^2(\pi/4)\cdot\sin^2(\theta)}\,d\theta,\cr}$$ where $\sin(\pi/4) = \sqrt{2}/2$. (Recall that $\pi\over4$ radians\footnote{$^{13}$}{A radian is a unit a angular measure. Basically, $360^\circ = 2\pi$ radians. This means that 1 radian is approximately 57.29578 degrees.} is $45^\circ$. Lots of tables books give values in degrees and not in radians. You do remember radians, don't you? If not, check out a trigonometry book or look in Appendix A for a review.) We look up the value of $E(45^\circ, \pi/2)$ in a tables book\footnote{$^{14}$}{Dr. M. Fogiel, {\it Handbook of Mathematical, Scientific, and Engineering Formulas, Tables, Functions, Graphs, Transforms}, (Piscataway, NJ: Research and Education Association, 1984), page 623.} and find that $$E\left(45^\circ, {\pi\over2}\right) \approx 1.35006.$$ This says that the value of the integral in equation (6) should be approximately $$\int_0^\pi \sqrt{1+\cos^2(\theta)}\,d\theta \approx 3.82007.$$ Now we apply our integration method (Simpson's rule) and see what the results are: \smallskip $$\vbox{\offinterlineskip \hrule \halign{&\vrule#&\strut\quad\hfil#\quad&\vrule# &\quad\hfil#\quad&\vrule#&\quad\hfil#\quad&\vrule# &\quad\hfil#\quad&\vrule#\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr &$n =$\hfil&&BASIC Program&&``C'' Program\quad&&Difference\quad&\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr & 2 &&3.575356081779317 &&3.5753560817793170&&0.000000000000000&\cr & 4 &&3.829178925615088 &&3.8291789256150880&&0.000000000000000&\cr & 8 &&3.820282406120432 &&3.8202824061204320&&0.000000000000000&\cr & 16 &&3.820197813575163 &&3.8201978135751630&&0.000000000000000&\cr & 32 &&3.820197789027719 &&3.8201977890277180&&0.000000000000001&\cr & 64 &&3.820197789027712 &&3.8201977890277120&&0.000000000000000&\cr & 128 &&3.820197789027713 &&3.8201977890277120&&0.000000000000001&\cr & 256 &&3.820197789027711 &&3.8201977890277110&&0.000000000000000&\cr & 512 &&3.820197789027712 &&3.8201977890277110&&0.000000000000001&\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr} \hrule}$$ There is a lesson to be learned here. If we use a more accurate value for $E(45^\circ, \pi/2)$ than 1.3506, namely 1.3506438, then we will see that our computed answer agrees more closely with the theoretical answer. Now, lets look at the coding that was changed in the BASIC and in the ``C'' programs: \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 105 DEF FNA (x AS DOUBLE) = SQR(1!+COS(x)*COS(x)) } \smallskip {\tt\parindent=0pt\ttraggedright\obeylines double fc (double x) {\char123} \qquad double y; \qquad y = sqrt(1.0+cos(x)*cos(x)); \qquad return (y); {\char125} } \smallskip\noindent We added something else new. In the BASIC program and in the ``C'' program we added lines giving the (expected) theoretical value. \smallskip {\tt\parindent=0pt\ttraggedright\obeylines 230 PRINT "Theoretical approx = "; 2*SQR(2)*1.35064388 } \smallskip {\tt\parindent=0pt\ttraggedright\obeylines printf("Theoretical approx = %.16lf",2.0*sqrt(2.0)*1.35064388); printf("\char92n"); } \medskip\noindent We want to delve\footnote{$^{15}$}{To delve---to search carefully and painstakingly for information.} more deeply into the whole matter of numerical versus exact integration. To do this, we will evaluate an integral exactly and numerically and compare the results. Lets integrate one arch of the cosine curve: $$\int_0^{\pi/2} \cos(x)\,dx = \sin(x)\Big|_0^{\pi/2} = \sin(\pi/2) -\sin(0) = 1.\eqno(6)$$ And, numerically, setting $A_p(n)$ to be the value of Simpson's rule approximation for $n$ subintervals $$\vbox{\offinterlineskip \hrule \halign{&\vrule#&\strut\quad\hfil#\quad&\vrule# &\quad#\hfil\quad&\vrule#&\quad#\hfil\quad&\vrule# &\quad#\hfil\quad&\vrule#\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr &$n =$\hfil&&Simpson's rule&&$\big|1-A_p(n)\big|$\quad& &Error bound\quad&\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height2pt&\omit&&\omit&&\omit&&\omit&\cr & 2 &&1.00228 &&0.00228 &&0.00332 &\cr & 4 &&1.00013 &&0.00013 &&0.00021 &\cr & 6 &&1.00003 &&0.00003 &&0.00004 &\cr & 8 &&1.00001 &&0.00001 &&0.00001 &\cr & 10 &&1.0000033922209010 &&0.0000033922209010&&0.000005312 &\cr & 20 &&1.0000002115465910 &&0.0000002115465910&&0.000000332 &\cr & 50 &&1.0000000054122520 &&0.0000000054122520&&0.000000008 &\cr & 100 &&1.0000000003382360 &&0.0000000003382360&&0.0000000005 &\cr height2pt&\omit&&\omit&&\omit&&\omit&\cr} \hrule}$$ Now we see how the error estimate really works. If $f(x)=\cos(x)$, then $f^{(4)}(x) = \cos(x)$, $-1\le \cos(x)\le 1$, for all real $x$. $$\eqalign{f'(x)\ &=\ -\sin(x)\cr f''(x)\ &=\ -\cos(x)\cr f^{(3)}(x)\ &=\ \sin(x)\cr f^{(4)}(x)\ &=\ \cos(x)\cr}\eqno(7)$$ $|f^{(4)}(x)|\le 1$. We have an error bound $$\left|{(\beta-\alpha)^5\over180\cdot n^4}f^{(4)}(\xi)\right| \le {\big(\pi/2\big)^5\over180\cdot n^4}\,1 \ =\ {\pi^5\over32\cdot180\cdot n^4}.$$ \vfill\eject % \noindent{\bf\llap{1.5\quad}Truncation Error.} In this section we will derive the truncation error for Simpson's rule. This section has some theory and may be omitted by someone only interested in applications. One might note that in the previous section that the actual error was much less than the error bound. The reason for the ``looseness'' in the error bound has to do with bounding the fourth derivative over the entire interval, $[\alpha,\beta]$. \medskip\noindent We will assume that the function $f(x)$ has a Taylor series expansion. This is a reasonable assumption for theoretical purposes; however, in the real world it may present a problem. Derivation with weaker conditions may be found in standard texts. Using equation $(1)$ for $A_p$ $$\eqalign{A_p\ &=\ {1\over3}h\big(y_0+4y_1+y_2\big)\cr &=\ {1\over3}h\Bigg[ y_0 +4\left(y_0 +hy'_0 + {1\over2}h^2y^{(2)}_0 +{1\over6}h^3y_0^{(3)} + {1\over24}h^2y_0^{(4)} + \dots\right)\cr &\qquad +\left(y_0 + 2hy'_0 + 2h^2y_o^{(2)} +{4\over3}h^3y_0^{(3)} + {2\over3}h^4y_0^{(4)}+\dots\right)\,\Bigg] \cr &=\ {1\over4}h\left(6y_0 + 6hy_0' + 4h^2y_0^{(2)} +2h^3y_0^{(3)} + {5\over6}h^4y_0^{(4)} + \dots\right)\cr}\eqno(8)$$ Now do the same thing to the integral itself. $$\eqalign{\int_{x_0}^{x_2} y(x)\,dx\ &=\ F(x_2) - F(x_1)\cr &=\ 2hF'(x_0) + {1\over2}(2h)^2F^{(2)}(x_0) + {1\over6}(2h)^3F^{(3)}(x_0)\cr &\qquad+{1\over24}(2h)^4F^{(4)}(x_0) + {1\over120}(2h)^5F^{(5)}(x_0) + \dots\cr &=\ 2hy_0 + 2h^2y_0 + {4\over3}h^3y_0^2 + {2\over3}h^4y_0^{(3)} + {4\over15}h^5y_0^{(4)} + \dots.\cr}\eqno(9)$$ Subtract equation (8) from equation (9) to get $$\int_{x_0}^{x_2} y(x)\,dx - {1\over3}h\big(y_0+4y_1+y_2\big) = \left({4\over15}-{5\over18}\right)h^5y-0^{(4)} \ =\ -{h^5 y_0^{(4)}\over90}\eqno(10)$$ Equation (10) follows from the arithmetic calculation $${4\over15}-{5\over18}\ =\ {24-25\over90}\ =\ -{1\over90}.$$ If we have an interval $[\alpha,\beta]$, which can be written as the union of intervals $[x_0,x_2]$, $[x_2,x_4]$, $\ldots$, $[x_{n-2},x_n]$, then we may write the total truncation error as $$-{h^5\over90}\left(y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)}\right).$$ Now we make the assumption that the fourth derivative is continuous on the interval $[\alpha,\beta]$. $\beta-\alpha = nh$, so we have $$-{h^5\over90}\left(y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)}\right) \ =\ -{(\beta-\alpha)^5\over180\cdot n}\,y^{(4)}(\xi),$$ where $\xi\,\in\,(\alpha,\beta)$ and $$y_0^{(4)} + y_2^{(4)} + \ldots + y_{n-2}^{(4)} \ =\ {n\over2}\,\cdot\,y^{(4)}(\xi).$$ \vfill\eject % \noindent{\bf\llap{1.6\quad}An Example.} In this section we will consider an application to the particular function $$\frame <5pt> {$f(x)\ = \ (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$}$$ This remarkable function has properties as follows\footnote{$^{16}$}{The superscript before the chemical symbol gives the value of $Z$, the atomic weight; the subscript before the symbol gives the value of $A$, the atomic number; and, the superscript after the symbol gives the ionization. Values taken from {\it Quantum Physics, 2nd Ed.\/} by Eisberg and Resnick, (NY: John Wiley \& Sons, 1985), page 520.} \medskip {\settabs5\columns\thinmuskip=2mu \+Ionized &\underbar{\raise2pt\hbox{\qquad mass\qquad}} & & &relative \cr \+Isotope & electron mass &\qquad $x$ &\qquad $f(x)$ &deviation\cr \baselineskip=14pt \+\quad${}^1_1\!{\rm H}^{+}$ & {\tt\ \ 1836.152701} &\qquad $0$ & {\tt\ \ 1836.15174} & 0.00003\%\cr \+\quad${}^2_1\!{\rm H}^{+}$ & {\tt\ \ 3670.4830} &\qquad $\pi$ & {\tt\ \ 3643.34824} & 0.37\%\cr \+\quad${}^4_2{\rm He}^{++}$& {\tt\ \ 7294.2995} &\qquad $\pi+4$ & {\tt\ \ 7287.74349} & 0.01\%\cr \+\quad${}^9_4{\rm Be}^{+4}$& {\tt\ 16424.2099} &\qquad $\pi+10$ & {\tt\ 16364.47397} &0.18\%\cr \+\quad${}^{12}_{\ 6}{\rm C}^{+6}$&{\tt\ 21868.6918} &\qquad $5\pi$ &{\tt\ 21845.9892} &0.05\%\cr \+\quad${}^{63}_{29}{\rm Cu}^{+29}$&{\tt 114684.6335} &\qquad $9\pi+8$ &{\tt114237.3148} &0.19\%\cr \+\quad${}^{120}_{\ 50}{\rm Sn}^{+6}$&{\tt 218518.1598} &\qquad $14\pi+4$ &{\tt218491.3263} &0.05\%\cr} \medskip\noindent However, for our example, we will only be interested in computing the integral of the area between the curve and the $x$-axis between the roots $\alpha_1 = -4\pi$ and $\alpha_2 = -4\pi + 1/\pi$. This will involve an application of Simpson's rule. This example was chosen because it involves a polynomial (of degree three) which does not have integer coefficients. Indeed, the coefficients are not even algebraic numbers---they are transcendental numbers. We will observe how nicely our integration formula functions in this case. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Cubic graph % $$\beginpicture \setcoordinatesystem units <10cm,10000pt> \setplotarea x from -13.1 to -12.0, y from -0.015 to 0.015 \plotheading {Graph of $f(x) = (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$} \axis bottom shiftedto y=0 % label {F{\sevenrm IGURE} 6} ticks numbered from -13 to -12 by 1 / \axis left ticks in numbered from -0.015 to 0.015 by 0.005 / % \savelinesandcurves on "chap1b.t01" \put {F{\sevenrm IGURE} 6} at -12.5 -0.015 \ifexpressmode \put {\bf EXPRESSMODE} at -12.5 0.01 \else \setquadratic \inboundscheckon \plot -12.65310 -0.02541 -12.60973 -0.01066 -12.56637 0.00000 / \inboundscheckoff \plot -12.56637 0.00000 -12.53454 0.00552 -12.50271 0.00929 -12.47088 0.01151 -12.43905 0.01238 -12.40722 0.01209 -12.37538 0.01084 -12.34355 0.00880 -12.31172 0.00619 -12.27989 0.00319 -12.24806 0.00000 -12.22325 -0.00250 -12.19845 -0.00490 -12.17364 -0.00713 -12.14884 -0.00908 -12.12403 -0.01066 -12.09922 -0.01178 -12.07442 -0.01236 -12.04961 -0.01229 -12.02481 -0.01149 -12.00000 -0.00987 / \fi \putrule from -12.53454 0.0 to -12.53454 0.00552 \putrule from -12.50271 0.0 to -12.50271 0.00929 \putrule from -12.47088 0.0 to -12.47088 0.01151 \putrule from -12.43905 0.0 to -12.43905 0.01238 \putrule from -12.40722 0.0 to -12.40722 0.01209 \putrule from -12.37538 0.0 to -12.37538 0.01084 \putrule from -12.34355 0.0 to -12.34355 0.00880 \putrule from -12.31172 0.0 to -12.31172 0.00619 \putrule from -12.27989 0.0 to -12.27989 0.00319 \putrule from -12.24806 0.0 to -12.24806 0.00000 %\ifexpressmode % \put {\bf EXPRESSMODE} at -12.5 0.01 %\else % \replot "chap1b.t01" \endpicture$$ %\centerline{F{\sevenrm IGURE} 6} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \medskip\noindent The value of the integral is calculated to be approximately {\tt 0.0025664955636711}. This example points out one of the strong points of using Simpson's rule. When the answer is computed, it is in numeric form. If one obtained a so-called ``closed-form'' solution, then it would be necessary to plug in the values for $\pi$ to get a numeric result. \medskip \noindent Following this same argument, let's consider integrating the same function, $f(x) = (x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$, over the (closed) interval $\big[-4\pi+2/\pi,\,0\big]$. Because of the variation of the range of $f$, we will use a logarithmic scale on the $y$-axis. The value ``under the curve'' is approximately {\tt 5618.52714450635}. We do actually have this much accuracy because of the very precise value in our ``C'' program for the constant $\pi$. Recall the expression \medskip {\tt\obeylines\parindent=0pt\ttraggedright \#define pi 3.141592653589793238462643383279 /* Accurate value for pi. */ } \medskip\noindent The only changes necessary to the ``C'' program were \medskip {\tt\obeylines\parindent=0pt\ttraggedright a = (double) -4.0*pi+2.0/pi; printf("{\char92}nLeft end point = \%.16lf",a); b = (double) 0.0; printf("{\char92}nRight end point = \%.16lf",b); } \medskip\noindent And, now the graph. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Figure 7. % $$\beginpicture \setcoordinatesystem units <0.5cm,10cm> \setplotarea x from -13 to 2, y from 0 to .60206 % \plotheading {\lines{ The integral of\cr $f(x)=(x+4\pi)\cdot(x+4\pi-1/\pi)\cdot(x+4\pi-2/\pi)$\cr over the interval $\big[-4\pi+2/\pi,\,0\big]$\cr} } % \axis bottom shiftedto y=0 ticks withvalues {$\scriptstyle -4\pi+8/\pi$} {$-8$} {$-6$} {$-4$} {$-2$} {$0$} {$2$} / % at -10.01989 -8 -6 -4 -2 0 2 / / % \axis left label {\stack {P,o,w,e,r,s, ,of, ,$10$} } % ticks andacross logged numbered at 1 2 3 4 / % unlabeled length <0pt> at 1.25 1.5 1.75 2.25 2.5 2.75 3.5 / % from 1.5 to 3.5 by .5 from 1.25 to 2.75 by .50 / % \plot -10.01989 0.01489 -9.51890 0.11466 -9.01790 0.18282 -8.51691 0.23359 -8.01591 0.27351 -7.51492 0.30610 -7.01392 0.33343 -6.51293 0.35685 -6.01193 0.37724 -5.51094 0.39523 -5.00995 0.41129 -4.50895 0.42575 -4.00796 0.43888 -3.50696 0.45088 -3.00597 0.46192 -2.50497 0.47212 -2.00398 0.48159 -1.50298 0.49041 -1.00199 0.49868 -0.50099 0.50643 0.00000 0.51374 / \linethickness=.8pt \putrule from -10.01989 0.0 to -10.01989 0.01489 \putrule from -9.51890 0.0 to -9.51890 0.11466 \putrule from -9.01790 0.0 to -9.01790 0.18282 \putrule from -8.51691 0.0 to -8.51691 0.23359 \putrule from -8.01591 0.0 to -8.01591 0.27351 \putrule from -7.51492 0.0 to -7.51492 0.30610 \putrule from -7.01392 0.0 to -7.01392 0.33343 \putrule from -6.51293 0.0 to -6.51293 0.35685 \putrule from -6.01193 0.0 to -6.01193 0.37724 \putrule from -5.51094 0.0 to -5.51094 0.39523 \putrule from -5.00995 0.0 to -5.00995 0.41129 \putrule from -4.50895 0.0 to -4.50895 0.42575 \putrule from -4.00796 0.0 to -4.00796 0.43888 \putrule from -3.50696 0.0 to -3.50696 0.45088 \putrule from -3.00597 0.0 to -3.00597 0.46192 \putrule from -2.50497 0.0 to -2.50497 0.47212 \putrule from -2.00398 0.0 to -2.00398 0.48159 \putrule from -1.50298 0.0 to -1.50298 0.49041 \putrule from -1.00199 0.0 to -1.00199 0.49868 \putrule from -0.50099 0.0 to -0.50099 0.50643 \putrule from 0.00000 0.0 to 0.00000 0.51374 \put {$\scriptstyle\bullet$} at 0 0.51374 \put {$\longleftarrow (0,\,1836.15)$} [l] <2pt,0pt> at 0 0.51374 \put {$\mid$} at -11.92975 0.0 % \put {$\uparrow\atop -4\pi+2/\pi$} [t] <0pt,-10pt> at -11.92975 0.0 % \endpicture$$ \centerline{F{\sevenrm IGURE} 7} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \medskip\noindent We might also consider some other interesting values for the function $f(x)$, as defined above, for ratios of subatomic particles' masses to the mass of the electron. %$$f(x)=(x+4\pi)(x+4\pi-1/\pi)(x+4\pi-2/\pi)\eqno(\dagger)$$ {\settabs5\columns\thinmuskip=2mu \+Subatomic &\underbar{\raise2pt\hbox{\qquad mass\qquad}} & & &relative \cr \+Particle & electron mass &\qquad $x$ &\qquad $f(x)$ &deviation\cr \baselineskip=14pt \+\quad$n$ &{\tt\ 1838.683662} &\qquad $1/64\pi$ &{\tt\ 1838.39048447} &0.008\%\cr \+\quad$\tau^{-}$ &{\tt\ 3491.4} &\qquad $3-1/4\pi$ &{\tt\ 3488.5} & 0.039\%\cr \+\quad$\eta^{0}$ &{\tt\ 1073.97} &\qquad $-2-1/64\pi$ &{\tt\ 1073.7} & 0.014\%\cr \+\quad$\eta'$ &{\tt\ 1873.78} &\qquad $1/4\pi$ &{\tt\ 1872.2} & 0.069\%\cr \+\quad$\Sigma^{+}$ &{\tt\ 2327.5} &\qquad $1+1/64\pi$ &{\tt\ 2326.8} &0.0064\%\cr \+\quad$\Sigma^{0}$ &{\tt\ 2333.76} &\qquad $1+1/16\pi$ &{\tt\ 2334.3} &0.034\%\cr \+\quad$\Sigma^{-}$ &{\tt\ 2343.31} &\qquad $1+1/8\pi$ &{\tt\ 2344.8} &0.049\%\cr \+\quad$\Lambda$ &{\tt\ 2183.23} &\qquad $1-1/\pi$ &{\tt\ 2160.26} &0.53\%\cr } %***Other truly remarkable values not included in the tables because**** %***they either had more than two terms or did not meet the criteria.*** %***PARTICLES*** % % f(-1-1/\pi-1/16\pi)=496.79 Mev K^0 497.67 Mev % f(-2-1/\pi-1/8\pi)=493.803 Mev K^{\pm} 493.646 Mev % f(-6+1/\pi-1/4\pi)=139.1436 Mev \pi^{\pm} 139.5675 Mev % f(-6+1/\pi-1/64\pi)=134.0077 Mev \pi^0 134.9739 Mev % f(\pi/4-1/4\pi-1/64\pi)= 1115.21 Mev \Lambda 1115.63 Mev % f(\pi+1/64\pi)=1863.5536 Mev D^{0} 1864.5 \pm 0.5 Mev % f(\pi+1/16\pi)=1868.9790 Mev D^{\pm} 1869.3 \pm 0.4 Mev % f(3-1/\pi-1/4\pi)=1672.6886 Mev \Omega 1672.43 \pm 0.32 Mev % % f(8-2\pi) = 1390.93 Mev \omega 1391 \pm 18 Mev % f(-9+\pi)=132.9758 Mev \pi^{0} 134.9739 % \vfill\eject % \noindent{\bf\llap{1.7\quad}Verifying Derivatives.} Calculus is traditionally partitioned into differential and integral calculus. This partition supposes that the two concepts are independent and mutually exclusive. Then, almost as a gift from the gods, we encounter the so-called {\bf Fundamental Theorem of Calculus},\footnote{$^{17}$}{{\bf Theorem.} {\sl Let $f$ be continuous on $[\alpha,\beta\,]$ and $x\in(\alpha,\beta\,)$. If $F(x)=\int_a^x f(t)\,dt$, then $F'(x) = f(x)$.}} which unifies the two concepts (more or less---mostly less). In this era of digital computers and the future directed towards artificial intelligence, it might be a good idea to discard the artificial divisions of calculus in favor of a more pragmatic\footnote{$^{18}$}{Pragmatic---practical, especially in contrast to idealistic.} development. In this section we will examine a good technique for determining if a function, which is supposed to be the derivative of a given function, really is. The idea is simple. We have a given function and a candidate for its derivative. Looking at some interval, we choose four points and use Simpson's rule to evaluate the integral of the derivative on subintervals, ending at the four points, and compare the results with the original function. This is awkward to verbalize but easy to write ``in mathematics.'' Let $f(x)$ be a given function and let $g(x)$ be a candidate for the derivative $f'(x)$ of $f(x)$. In the interval $[\alpha,\,\beta\,]$, choose four points, $x_1$, $x_2$, $x_3$, and $x_4$. (Let $\alpha < x_1 < x_2 < x_3 < x_4 < \beta$.) We can eliminate $g(x)$ if any of the four approximations are incorrect: $$f\big(x_1\big) - f\big(\alpha\big) \ \approx \int_\alpha^{x_1} g(t)\,dt$$ $$f\big(x_2\big) - f\big(\alpha\big) \ \approx \int_\alpha^{x_2} g(t)\,dt$$ $$f\big(x_3\big) - f\big(\alpha\big) \ \approx \int_\alpha^{x_3} g(t)\,dt$$ $$f\big(x_4\big) - f\big(\alpha\big) \ \approx \int_\alpha^{x_4} g(t)\,dt$$ We want to examine this idea, using a non-trivial example. One such example comes from the four de\-riv\-a\-tives of the func\-tion $f(x)$ $=$ $\sqrt{1-k^2\sin^2(x)}$. Recall it was the fourth derivative that was used in the error estimate. This function is easy to write; however, its derivatives become more and more difficult to expand algebraically. How can we be sure that we did not make some arithmetic error? It has been shown that, on the average, 1 out of every 300 human calculations is in error. This will be done via a computer program. The program evaluates all the integrals simultaneously and outputs the results in a rectangular array of real numbers---a {\sl matrix}. By visually comparing the rows of the matrix, we can tell at once if our derivatives are correct. Without any more delay, let's write down the program and its results and then do an analysis of the coding. \medskip {\tt\parindent=0pt\ttraggedright\obeylines 100 DEFDBL A-H, O-Z 102 C2=SIN(EXP(1.)/2.)*SIN(EXP(1.)/2.) 105 DEF FNA(x AS DOUBLE)= SQR(1.-C2*SIN(x){\char94}2) 106 DEF FNB(x AS DOUBLE)= -.5*C2*SIN(2.*x)/FNA(x) 107 DEF FNC(x AS DOUBLE)= -C2*COS(2.*x)/FNA(x) \qquad\qquad -.25* C2*C2*SIN(2.*x){\char94}2/FNA(x){\char94}3 108 DEF FND(x AS DOUBLE)= 2.*C2*SIN(2.*x)/FNA(x) \qquad\qquad -.75*C2*C2*SIN(4.*x)/FNA(x){\char94}3 \qquad\qquad -.375*C2{\char94}3*SIN(2.*x){\char94}3/FNA(x){\char94}5 109 DEF FNE(x AS DOUBLE)= 4.*C2*COS(2.*x)/FNA(x) \qquad\qquad + C2{\char94}2*SIN(2.*x){\char94}2/FNA(x){\char94}3 \qquad\qquad -3.*C2{\char94}2*COS(4.*x)/FNA(x){\char94}3 \qquad\qquad -1.125*C2{\char94}3*SIN(4.*x)*SIN(2.*x)/FNA(x){\char94}5 \qquad\qquad -(18./8.)*C2{\char94}3*SIN(2.*x){\char94}2*COS(2.*x)/FNA(x){\char94}5 \qquad\qquad -(15./16.)*C2{\char94}4*SIN(2.*x){\char94}4/FNA(x){\char94}7 110 A = 0. 112 INPUT "Left endpoint = "; B 114 IF (A>B) OR (B>3.141592653589793/2.) \qquad\qquad THEN PRINT "Input out of range": GOTO 112 116 OPEN "92\_12\_14.txt" FOR APPEND AS \#1 120 REM Left endpoint <= 3.141592653589793\# / 2! 130 SUM1 = 0.: SUM2 = 0.: SUM3 = 0.: SUM4 = 0. 150 N = 512: H = (B - A) / N 160 FOR I = 1 TO N STEP 2: REM The "FOR" loop is done n/2 times. 170 SUM1 = SUM1 + FNB(A + (I - 1) * H) 172 SUM2 = SUM2 + FNC(A + (I - 1) * H) 174 SUM3 = SUM3 + FND(A + (I - 1) * H) 176 SUM4 = SUM4 + FNE(A + (I - 1) * H) 180 SUM1 = SUM1 + 4.*FNB(A + I * H) 182 SUM2 = SUM2 + 4.*FNC(A + I * H) 184 SUM3 = SUM3 + 4.*FND(A + I * H) 186 SUM4 = SUM4 + 4.*FNE(A + I * H) 190 SUM1 = SUM1 + FNB(A + (I + 1) * H) 192 SUM2 = SUM2 + FNC(A + (I + 1) * H) 194 SUM3 = SUM3 + FND(A + (I + 1) * H) 196 SUM4 = SUM4 + FNE(A + (I + 1) * H) 200 NEXT I 210 L\$ = "\#\#.\#\#\#\#\# \#\#.\#\#\#\#\# \#\#.\#\#\#\#\# \#\#.\#\#\#\#\# \#\#.\#\#\#\#\#" 300 PRINT USING L\$; FNA(B)-FNA(A); FNB(B)-FNB(A); \qquad\qquad FNC(B)-FNC(A); FND(B)-FND(A) 310 PRINT USING L\$; SUM1*H/3.; SUM2*H/3.; \qquad\qquad SUM3*H/3.; SUM4*H/3. 320 PRINT \#1, USING L\$; FNA(B)-FNA(A); FNB(B)-FNB(A); \qquad\qquad FNC(B)-FNC(A); FND(B)-FND(A) 330 PRINT \#1, USING L\$; SUM1*H/3.; SUM2*H/3.; \qquad\qquad SUM3*H/3.; SUM4*H/3. 340 CLOSE \#1 350 STOP 360 END } \medskip\noindent Of course, we want to look at the output. For input values, we used $.25$, $.50$, $1.00$, and $1.50$. Rows 1 and 2 of this 4 by 8 matrix correspond to $x=.25$, rows 3 and 4 correspond to $x=.5$, and so on. \medskip {\tt\settabs 4\columns \+\quad -0.02969 &\quad-0.23615 &\quad0.03387 &\quad\ 0.27142 \cr \+\quad -0.02969 &\quad-0.23615 &\quad0.03387 &\quad\ 0.27142 \cr \+\quad -0.11666 &\quad-0.45528 &\quad0.13655 &\quad\ 0.55428 \cr \+\quad -0.11666 &\quad-0.45528 &\quad0.13655 &\quad\ 0.55428 \cr \+\quad -0.43151 &\quad-0.76446 &\quad0.62760 &\quad\ 1.73354 \cr \+\quad -0.43151 &\quad-0.76446 &\quad0.62760 &\quad\ 1.73354 \cr \+\quad -0.77883 &\quad-0.30495 &\quad4.81402 &\quad17.17882 \cr \+\quad -0.77883 &\quad-0.30495 &\quad4.81402 &\quad17.17882 \cr} \medskip\noindent All this goes to show that the equations we coded into the computer program are {\sl probably\/} correct. This is not rigorous; however, in our days and times anything that fits this well would be a good approximation---that is, it would serve as the derivative {\sl for all practical purposes}. A word of caution: there may be more than one way to express a given equation. For instance, $y=x^2+2x+1$ and $y = (x+1)^2$ are {\sl algebraically\/} the same. Even though two expressions are algebraically they same, they may not yield the same value from a computer. This is due to several factors: (1) computers may evaluate the equations differently; (2) there may be more truncation (round-off) error with one equation than with the other (after all, computers don't have infinite accuracy); (3) a coding error may affect one equation (especially doing an integer division where the fractional part is lost); and, (4) the grouping of successive multiplications and divisions may affect the answer, especially when two nearly equal numbers are being subtracted. \medskip\noindent There are a few items worth mentioning about the computer program itself. We ``captured'' or ``saved'' the output by {\sl appending\/} to a file (in this case named {\tt 92\_12\_14.txt}. Later, we edited the file and merged it into this document. This gave two distinct advantages: (1) it was quicker and (2) there was no possibility of a human blunder in re-typing the numbers. This is always a good technique. Capturing output data and inputing the file directly into text ensures a correct copy. It may not always possible to do this. A second alternative is to compare the final text against the input via some compare routine. In Appendix B, a method of comparing text with output will be devised, developed, and discussed. In BASIC\footnote{$^{19}$}{BASIC---Beginners' All-purpose Symbolic Instruction Code. This elementary computer code is held in disdain by many programmers who prefer ``C.''} the coding line is 256 characters long, more or less (usually less, maybe 254 characters). In a text document, it is a good idea to have lines of 80 characters or less. To accommodate the document, it is necessary to ``break'' long lines into shorter segments. Whenever this is done, the line is indented. The sequence numbers are retained to aid in determining when a line has been broken. Some versions of BASIC write floating point integers using an exclamation point, {\it e.g.}, $3.00$ becomes $3!$. All floating point integers have been written in the standard mathematical format using a decimal point. QBasic\footnote{$^{20}$}{QBasic---the version of BASIC copyrighted by the Microsoft Corporation.} puts in spaces between the mathematical operators. While the spaces add to the readability of the code and aid the programmer, they do little for the documentation, which has to fit on an $8{1\over2} \times 11$ inch page. The extra spaces were removed. For the ``\TeX-nicians,'' they may note that some symbols cause the typesetting program to suffer. Before merging into a \TeX\ file, the special symbols backslash ({\tt\char92}) and hat (or circumflex) ({\tt\char94}) were replaced by their character equivalences. The pound sign, dollar sign, and underscore (\#, \$, and \_) were dealt with in the usual manner, by prefixing them with backslashes ({\tt\char92}). \medskip\noindent As usual, we will complement the BASIC program with a ``C'' program. The ``C'' program also saves its output to a file. {\tt\obeylines\parindent=0pt\ttraggedright \#include /* Header for input/output subroutines. */ \#include /* Header for math subroutines. */ \#include /* Header for floating point subroutines. */ \#define pi 3.141592653589793238462643383279 /* Accurate value for pi. */ \#define k2 0.9558669573934826 /* Simpson's rule for approximating integrals. \qquad a: left endpoint \qquad b: right endpoint \qquad fa: original function \qquad fb,fc,fd,fe: pointers to functions to integrate \qquad n: number of subintervals */ double fa (double x); double fb (double x); double fc (double x); double fd (double x); double fe (double x); main() \char123 double a,b,h,sum,x,y; /* In 'C' all variables must be assigned */ double p1, p2, p3; int i, n; FILE *fp; fp = fopen("92\_12\_13.txt", "a"); printf("{\char92}007"); /* Sound bell in computer. */ a = (double) 0.0; printf("{\char92}nLeft end point a = \%.16lf",a); printf("{\char92}nEnter right end point "); scanf("\%lf",\&b); printf("{\char92}nRight end point b = \%.16lf",b); n = 512; printf("{\char92}nNumber of subintervals \%d",n); h = (double) (b-a)/n; for (i=1, sum=0.0; i<=n; i = i+ 2)\char123 \quad p1 = fb((double) a+(i-1)*h); \quad p2 = fb((double) a+i*h); \quad p3 = fb((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; \char125 y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}nValue of f(b)-f(a) = \%.16lf", (double) fa(b)-fa(a) ); fprintf( fp, " \%.5lf \%.5lf", y, fa(b)-fa(a) ); for (i=1, sum=0.0; i<=n; i = i+ 2)\char123 \quad p1 = fc((double) a+(i-1)*h); \quad p2 = fc((double) a+i*h); \quad p3 = fc((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; \char125 y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}n f'(b) - f'(a) = \%.16lf", (double) fb(b)-fb(a) ); fprintf( fp, " \%.5lf \%.5lf", y, fb(b)-fb(a) ); for (i=1, sum=0.0; i<=n; i = i+ 2)\char123 \quad p1 = fd((double) a+(i-1)*h); \quad p2 = fd((double) a+i*h); \quad p3 = fd((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; \char125 y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}n f''(b) - f''(a) = \%.16lf", (double) fc(b)-fc(a) ); fprintf( fp, " \%.5lf \%.5lf", y, fc(b)-fc(a) ); for (i=1, sum=0.0; i<=n; i = i+ 2)\char123 \quad p1 = fe((double) a+(i-1)*h); \quad p2 = fe((double) a+i*h); \quad p3 = fe((double) a+(i+1)*h); \quad sum += p1 + 4.0 * p2 + p3; \char125 y = (double) h*sum/3.0; printf("{\char92}nValue of integral = \%.16lf", (double) y); printf("{\char92}n f'''(b)-f'''(a) = \%.16lf", (double) fd(b)-fd(a) ); fprintf( fp, " \%.5lf \%.5lf {\char92}n", y, fd(b)-fd(a) ); fclose(fp); return(0); \char125 double fa (double x) \char123 \qquad double y; \qquad y = (double) sqrt(1.0 -k2*sin(x)*sin(x)); \qquad return (y); \char125 double fb (double x) \char123 \qquad double y; \qquad y = (double) -0.5*k2*sin(2.0*x)/fa(x); \qquad return (y); \char125 double fc (double x) \char123 \qquad double y; \qquad y = (double) -k2*cos(2.0*x)/fa(x) \qquad\quad -.25*k2*k2*sin(2.0*x)*sin(2.0*x)/pow(fa(x),3.0); \qquad return (y); \char125 double fd (double x) \char123 \qquad double y; \qquad y = (double) 2.0*k2*sin(2.0*x)/fa(x) \qquad\quad -0.75*k2*k2*sin(4.0*x)/pow(fa(x),3.0) \qquad\quad -.375*k2*k2*k2*pow(sin(2.0*x),3.0)/pow(fa(x),5.0); \qquad return (y); \char125 double fe (double x) \char123 \qquad double y; \qquad y = (double) 4.0*k2*cos(2.0*x)/fa(x) \qquad\quad +k2*k2*sin(2.0*x)*sin(2.0*x)/pow(fa(x),3.0) \qquad\quad -3.0*k2*k2*cos(4.0*x)/pow(fa(x),3.0) \qquad\quad -1.125*k2*k2*k2*sin(4.0*x)*sin(2.0*x)/pow(fa(x),5.0) \qquad\quad -2.25*k2*k2*k2*sin(2.0*x)*sin(2.0*x)*cos(2.0*x)/pow(fa(x),5.0) \qquad\quad -0.9375*k2*k2*k2*k2*pow(sin(2.0*x),4.0)/pow(fa(x),7.0); \qquad return (y); \char125 /* End of file */ } \medskip\noindent Of course, there is an output file to be examined also. Here we included values of $1.51$, $1.52$, and $1.55$. \medskip {\tt\settabs 8\columns \+ -0.02969&-0.02969& -0.23615&-0.23615& 0.03387&0.03387&\ 0.27142&\ 0.27142\cr \+ -0.11666&-0.11666& -0.45528&-0.45528& 0.13655&0.13655&\ 0.55428&\ 0.55428\cr \+ -0.43151&-0.43151& -0.76446&-0.76446& 0.62760&0.62760&\ 1.73354&\ 1.73354\cr \+ -0.77883&-0.77883& -0.30495&-0.30495& 4.81402&4.81402& 17.17882& 17.17882\cr \+ -0.78168&-0.78168& -0.26553&-0.26553& 4.97895&4.97895& 15.74172& 15.74172\cr \+ -0.78414&-0.78414& -0.22454&-0.22454& 5.12756&5.12756& 13.91650& 13.91650\cr \+ -0.78894&-0.78894& -0.09416&-0.09416& 5.43883&5.43883&\ 6.37637&\ 6.37637\cr } \vfill\eject \noindent{\bf\llap{1.8\quad}Generalizations.} Mathematicians love to take a popular, useful concept and generalize it. Sometimes much can be learned by generalizing and by abstracting; more often than not, a generalization results in a more complicated, theoretical, and generally useless body of knowledge that exists solely as a requirement for a degree. This is especially true with Simpson's rule. Simpson's rule is popular and efficient. Its error calculation is straightforward and its convergence is assured. There are some examples where Simpson's rule does poorly---but these are mostly ``pathological'' in nature. (They are artificially constructed simply to demonstrate the fallability of the rule.) There are also so-called improper integrals. The improper integrals are either defined over an interval such as $[\alpha,\,+\infty]$, $[-\infty,\,\beta\,]$, or $[-\infty,\,+\infty]$, or the function is unbounded in the interval of integration (or both!) The improper integrals have to be approached with common sense. Take for example the integral $$\int_0^1 {dx\over \sqrt{x}}\ =\ 2\sqrt{x}\big|_{x=1} \ -\ \lim_{x\to0} 2\sqrt{x}\ =\ 2.$$ How could Simpson's rule be applied to such an example? The answer is simple---make the computer do the work! Choose a small number $\epsilon$, $0 < \epsilon \ll 1$. By a proper determination of the number of subintervals and a sufficiently small value of $\epsilon$, the desired answer can be obtained with the required accuracy. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % $$\beginpicture \setcoordinatesystem units <1in,.125in> \setplotarea x from -2 to 2, y from -0.25 to 18 \axis bottom shiftedto y=0 label {F{\sevenrm IGURE} 8} / % \axis left shiftedto x=0 / % \putrule from 0.0625 0.0 to 0.0625 16.0 \ifexpressmode \put {\bf EXPRESSMODE} at 0 10.0 \else \setquadratic \plot 0.0625 16. 0.125 8. 0.25 4. .375 2.66667 .5 2. .75 1.3333 1.0 1.0 1.25 0.8 1.5 0.66667 1.75 0.57142 2.0 0.5 / % \fi \endpicture$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In the case of integrals of the form $${1\over\sqrt{2\pi}}\,\int_0^{\infty} e^{-x^2/2}\,dx$$ (the classic normal curve), one needs only observe that the integral may be replaced by an integral $${1\over\sqrt{2\pi}}\,\int_0^{N} e^{-x^2/2}\,dx$$ for a suitable chosen integer $N$. We note here that $N=6$ generally suffices for most engineering work. This is called the six-sigma ($6$-$\sigma$). We can compute this estimate by considering $$\int_1^N e^{-x^2/2}\,dx \le \int_1^N e^{-x/2}\,dx.$$ If $x\ge1$, then $x^2 \ge x$ and $e^{-x^2/2} \le e^{-x/2}$. $$\int_N^\infty e^{-x/2}\,dx = 2e^{-N/2}.$$ \centerline{% \beginpicture % \setcoordinatesystem units <.5in,2.5in> % \setplotarea x from -3 to 3, y from 0 to .4 % \plotheading {\lines {% The density $\varphi(\zeta) = e^{-\zeta^2\!/2}/\sqrt{2\pi}$ of the\cr % standard normal distribution.\cr}} % \axis bottom ticks numbered from -3 to 3 by 1 % length <0pt> withvalues $\zeta$ / at 1.5 / / % \linethickness=.25pt % \putrule from 1.5 0 to 1.5 .12952 % (.12952 = density at 1.5) \setbox0 = \hbox{$swarrow$}% \put {$\swarrow$ \raise6pt\hbox{$\varphi(\zeta)$}} % [bl] at 1.5 .12952 % \ifexpressmode \put {\bf EXPRESSMODE} at 0 0.2 \else \setquadratic \plot 0.0 .39894 0.16667 .39344 0.33333 .37738 0.5 .35207 0.66667 .31945 0.83333 .28191 1. .24197 1.25 .18265 1.5 .12952 1.75 .08628 2. .05399 2.25 .03174 2.5 .01753 2.75 .00909 3.0 .00443 / \setquadratic \plot 0.0 .39894 -0.16667 .39344 -0.33333 .37738 -0.5 .35207 -0.66667 .31945 -0.83333 .28191 -1. .24197 -1.25 .18265 -1.5 .12952 -1.75 .08628 -2. .05399 -2.25 .03174 -2.5 .01753 -2.75 .00909 -3.0 .00443 / % \setshadegrid span <.025in> % \vshade 0 0 .39894 0.5 0 .35207 1 0 .24197 % 1.5 0 .12952 2 0 .05399 / % \fi \endpicture } % %\smallskip \centerline{F{\sevenrm IGURE} 9} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \medskip\noindent Having mentioned how to generalize this rule to improper integrals, it is time to examine the position of this integration scheme in mathematicians' grand scheme of things. Simpson's rule is the most popular of the so-called {\bf Newton-Cotes integration formulas}. The first three of which are given below $$\eqalign{\int_{x_0}^{x_0+h} f(x)\,dx \ &=\ {h\over2}\big(y_0+y_1\big) - {h^3\over12}f''(\xi)\cr \int_{x_0}^{x_0+2h} f(x)\,dx \ &=\ {h\over3}\big(y_0+4y_1+y_2\big) - {h^5\over90}f^{(4)}(\xi)\cr \int_{x_0}^{x_0+3h} f(x)\,dx \ &=\ {3h\over8}\big(y_0+3y_1+3y_2+y_3\big) -{3h^5\over90}f^{(4)}(\xi),\cr}$$ where $\xi$ is an intermediate value of $x$. The Newton-Cotes formulas, such as those above, are called {\sl closed} because the interval end-points are used. If the end-points are not used, but only the interior points in the interval, the formula is called open. The Newton-Cotes formulas belong to a class of formulas known as polynomial approximations; the class of polynomial approximations is contained in a class of formulas of approximation of integrals by families of analytic functions, and so on. \medskip % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Figure 10 --- Exponential curve % $$\beginpicture \setcoordinatesystem units <1in,1in> \setplotarea x from 0 to 3, y from 0 to 1 \normalgraphs \axis bottom ticks numbered from 0 to 3 by 1 length <0pt> withvalues $x$ / at .5 / / \linethickness=.25pt \putrule from .5 0 to .5 .60653 \putrule from 1 0 to 1 .36788 \putrule from 2 0 to 2 .13534 \put {$\scriptstyle\bullet$} at .5 .60653 \put {$e^{-x}$} [rt] <-4pt,-4pt> at .5 .60653 \put {$e^{-x^2}$} [lb] <4pt,4pt> at .5 .77880 \ifexpressmode \put {\bf EXPRESSMODE} at 1.5 0.5 \else \setquadratic \plot 0 1 .25 .77880 .50 .60653 .75 .47237 1.00 .36788 1.25 .28650 1.50 .22313 1.75 .17377 2.00 .13534 2.25 .10540 2.50 .08208 2.75 .06393 3.00 .04979 / % \setshadegrid span <.025in> \vshade 1 0 .36788 1.5 0 .10540 2 0 .01832 / % \linethickness=0.75pt \setquadratic \plot 0.00 1.00000 0.25 0.93941 0.50 0.77880 0.75 0.56978 1.00 0.36788 1.25 0.20961 1.50 0.10540 1.75 0.04677 2.00 0.01832 2.25 0.00633 2.50 0.00193 2.75 0.00052 3.00 0.00012 / % \fi \endpicture $$ \centerline {F{\sevenrm IGURE} 10} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \vfill\eject %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \headline={\tenrm\hfill Newton's Method} \centerline{\bf Chapter 2} \bigskip \noindent{\bf\llap{2.0\quad}Introduction.} One of the most exciting applications of differential calculus is the use of Newton's method for the determination of roots of functions. Most students of algebra think that the only roots of interest are those of polynomials. An algebra student who has done all the assigned homework might think that it's just a matter of factoring; after all, it was so clever just completing the square in the quadratic equation. (Wasn't it fun being so smart and showing that off?) That's simply not the case. In fact, there is no purely algebraic method for finding roots of polynomials of degree five or higher. The situation with the trigonometric functions is even more involved. What technique would one use to find the root or roots of $x-\cos x = 0$? $$\beginpicture \setcoordinatesystem units <1in,1in> \setplotarea x from -.5 to 2.25, y from -.5 to 3.5 \plotheading {\lines {Tangent line at $(x_0,y_0)$\cr slope $=$ $f'(x_0)$\cr}} \axis bottom shiftedto y=0 ticks unlabeled short quantity 12 / % \axis left shiftedto x=0 ticks unlabeled quantity 9 / % \ifexpressmode \put {\bf EXPRESSMODE} at 0 1 \else \setquadratic \plot -0.25 -0.2344 -0.15 -0.2844 -0.05 -0.3094 0.05 -0.3094 0.15 -0.2844 0.25 -0.2344 0.35 -0.1594 0.45 -0.0594 0.55 0.0656 0.65 0.2156 0.75 0.3906 0.85 0.5906 0.95 0.8156 1.05 1.0656 1.15 1.3406 1.25 1.6406 1.35 1.9656 1.45 2.3156 1.55 2.6906 1.65 3.0906 1.75 3.5156 / % \setlinear \plot 2.0 3.4375 0.5 -.3125 / % \fi \put {$\scriptstyle\bullet$} at 1.0 0.9375 \put {$\scriptstyle\bullet$} at .625 0.0 \put {$(x_0,y_0)$} at 1.25 .9375 \put {$\nwarrow$\lower6pt\hbox{$x$-intercept}} [lt] <0.5pt, -0.5pt> at .62 0 % \put {$f(x)$\lower6pt\hbox{$\searrow$}} [rb] at 1.45 2.3156 \put {$\nwarrow$\lower6pt\hbox{tangent line}} [lt] at 1.46 2.1 \endpicture$$ \centerline{F{\sevenrm IGURE} 11} % f(x) = y = (15/12)*(x-1/2)*(x+1/2) % f'(x) = y' = (15/6)*x % when x = 1, y(1) = f(1) = 15/16 = .9375 % when x = 1, y'(1) = f'(1) = 15/6 % when x = 2, the point on the tangent line is 3.4375 % when y = 0, the abscissa on the tangent line is 5/8 % \smallskip\noindent All of the important details of Newton's method are found in the above figure. The equation of the tangent line is determined by the so-called point-slope method from elementary algebra: $${y-y_0 \over x-x_0}\ =\ \hbox{slope } =\ {\Delta y\over\Delta x}\ =\ {\hbox{rise}\over\hbox{run}} \ =\ f'(x_0).$$ Set $y=0$ and determine the $x$-intercept of the tangent line, $\big(x_0-y_0/f'\big(x_0),0\big)$. From the above figure, we see that the point at which the tangent line crosses the $x$-axis is moving quickly towards the root---the point at which the curve $f(x)$ crosses the $x$-axis. If we only had a better first approximation $x_0$, then the $x_0-y_0/f'(x_0)$ would be even closer to the root. Why not just try again and define a new approximation? $$x_1 = x_0 - {f(x_0)\over f'(x_0)}\eqno(11)$$ \vfill\eject \noindent{\bf\llap{2.1\quad}Theory.} This paragraph is concerned with the theory behind Newton's method. The user who is interested only in results may skip this entire section. This paragraph employs the so-called Taylor series (also known as the Taylor's series) $$f(x_0) + f'(x_0)(x-x_0) + {f''(x_0)\over2}(x-x_0)^2 + \ldots + {f^{(n)}\over n!}(x-x_0)^n + \ldots.\eqno(12)$$ An alternate derivation using the Mean-Value Theorem will be presented in paragraph {\bf 2.2}. If one looks in a dictionary,\footnote{$^{21}$}{{\it Webster's Ninth New Collegiate Dictionary}, (Springfield, Massachusetts: Merriam-Webster, Inc., 1984), page 1209.} equation (12) might be written in the form $$f(x) = f(a) + {f^{[1]}\over 1!}(x-a) +{f^{[2]}\over 2!}(x-a)^2 + \ldots + {f^{n]}\over n!}(x-a)^n + \ldots, \eqno(12')$$ where $f^{[n]}(a)$ is the derivative of the $n${\raise2pt\hbox{\sevenrm th}} order of $f(x)$ evaluated at $a$. Before delving into the mathematics, a digression\footnote{$^{22}$}{Digression---A swerving away from the main subject or the {\it Leitmotif\/}; a turning aside from the main argument.} might be in order. Whenever a textbook says ``degree,'' it means the highest exponent or power. For example: $x^2$ is of second degree; the polynomial $x^3-x^2+1$ is of third degree; and, $x^2+y^2 = 1$ is a second degree relation. Whenever a textbook says ``order,'' it generally refers to a more abstract concept. In general, $x^n$ means the $n${\raise2pt\hbox{\sevenrm th}} power (degree $n$) of $x$, that is $$x^n \ =\ \underbrace{x\cdot x\cdots x}_{n\ % \hbox{\sevenrm times }}.$$ (Like every rule, this one has an exception. In tensor analysis, the $i$, $j$, and $k$ in expressions like $x^i$, $y^{jk}$, {\it etc.}, refer to contravariant indices, not powers.) On the other hand, whenever one sees such notations as $x^{(n)}$, $y^{[n]}$, and so on, a red flag ought to go up. Indeed, if a function does have a derivative, it is usually denoted by $f'(x)$, the second derivative (the derivative of the derivative) is denoted by $f''(x)$. But, the third derivative is not generally denoted by $f'''(x)$. The third derivative of $f$ at $x$ is usually written $f^{(3)}(x)$ and sometimes written as $f^{[3]}(x)$. What does $f^{(0)}(x)$ mean? Recall that (for all $x\not= 0$) $x^0\equiv1$. $f^{(0)}(x) \equiv f(x)$. The symbol made up of three horizontal lines ($\equiv$) means more that just equals ($=$); this symbol, ($\equiv$), means equivalence and that means equal for all the values under consideration. \medskip \noindent One of mathematics' favorite devices, or tricks, is to truncate\footnote{$^{23}$}{Truncate---To cut off; to make short by trimming or cutting off.} a Taylor series. In the Newton method, the series is truncated at the quadratic term, that is, at ${1\over2}f''\big(x_0\big)\cdot\big(x-x_0\big)^2$. So, the truncated series becomes $$\eqalign{f(x)\ &=\ f(x_0) + {f^{[1]}\over 1!}(x-x_0)^1\cr &=\ f\big(x_0\big) + f'\big(x_0\big)\cdot\big(x-x_0\big).\cr}\eqno(13)$$ This is the derivation. So, after all that was said and done, there's not much to do to get the result. Just pretend that $f(x_1)=0$ and get $$f\big(x_1\big)\ =\ 0\ =\ f\big(x_0\big) \ +\ f'\big(x_0\big)\big(x_1-x_0\big),$$ or $$x_1\ =\ x_0\ -\ {f(x_0)\over f'(x_0)},\eqno(14)$$ provided, of course, that $f'(x_0) \not= 0$. If $f'(x_0) = 0$, then we just choose another $x_0$, say $\tilde x_0$ where $f'\big(\tilde x_0\big) \not= 0$ and proceed. \vfill\eject \noindent{\bf\llap{2.2\quad}Applications.} One of the most important applications of Newton's method is in the calculation of square roots. Many small hand calculators have a ``hard-wired'' square root generator. The circuitry follows a simple algorithm. Since long division of decimal fractions may not be supported, we will derive an algorithm which does not involve division and finds the square root for any number $a > 0$. \medskip \noindent It is clear that if we let $h(x) = x^2 -a$ and solve this equation for the root $h(x_0)=0$, this will be the same as finding the square root of $a$. Newton's formula becomes $x_{n+1} = {1\over2}\big(x_{n} - a/x_{n}\big)$.\footnote{$^{24}$}{If the following mathematics looks too deep, try Schaum's Outline {\it Numerical Analysis, 2nd Edition}, page 334. In any case, the two BASIC computer programs ought to be useful.} \medskip\noindent The next thing that we will do is to plug the function $F(x)= a - 1/x$ into Newton's method. Then we will look at the fixed point formula. To determine the starting value, we'll require $\xi$ to satisfy the expression $|f'(\xi)| < 1$. After we've done those things, we will generate a short little BASIC program to check out some sample values. Then, to satisfy the academics, we will quote the necessary theorems and satisfy the hypotheses. \medskip \noindent $$F(x) = a - {1\over x}\eqno(15)$$ $$F'(x) = {1 \over x^2}\eqno(16)$$ Applying $f(x) = x + g\big(F(x)\big)$, where $g(y) = -y/y'$, $$\eqalign{f(x)\ &=\ x - F(x)/F'(x)\cr &=\ x - {a - 1/x \over 1/x^2}\cr &=\ x - ax^2 +x\cr &=\ 2x - ax^2\cr}\eqno(17)$$ And we observe, substituting $s = 1/a$ into $f'(s)$ that $$f'(x)\ =\ 2 - 2ax,\eqno(18)$$ $$f'(1/a)\ =\ 2 -2a(1/a)\ =\ 2-2\ =\ 0.\eqno(19)$$ Recall that $a$ $\not=$ $0$. $$f''(x)\ =\ -2a\eqno(20)$$ tell us that $f''$ does not vanish at $1/a$ so we have quadratic convergence in some (yet to be determined) neighborhood of $1/a$. So, the required function is equation $(17)$ and the iteration procedure is $$x_{n+1}\ =\ 2x_{n} - ax_{n}^2.\eqno(21)$$ All that remains is to find the necessary $x_0 = x_0(a)$ to get things started. \medskip \noindent We would like to determine an interval $[\alpha,\,\beta\,]$ on which $f$ is contractive. It isn't enough just to know that one exists, we really need to compute it. But we have an explicit function, $f$, and $f \in C^1$. By the mean value theorem ({\sl MVT for derivatives\/}), we seek values for $\xi$ such that $|f'(\xi)| < 1$. This will ensure $$\big| f(x_1) - f(x_2)\big|\ =\ |f'(\xi)|\,|x_1 - x_2|\eqno(22)$$ $$|f'(x)|\ <\ 1$$ $$|2 -2ax|\ <\ 1$$ $$-1\ <\ 2 -2ax\ <\ 1$$ $$-3\ <\ -2ax\ <\ -1$$ If we multiply an inequality by $-{1\over2}$, it changes the 'sense' of the inequality: $${3\over2}\ >\ ax\ >\ {1\over2}\eqno(23)$$ Either $a > 0$ or $a < 0$, so: $$\hbox{ interval } = \cases{{3\over 2a}\,>\,x\,>\,{1\over2a}, & if $a\,>\,0$\cr \null\cr {1\over 2a}\,>\,x\,>\,{3\over2a}, & if $a\,<\,0$\cr}\eqno(24)$$ \medskip\noindent Basic program: \medskip {\tt\obeylines\ttraggedright 10 REM ************************************************ 20 REM * Harry A. Watson, Jr. 30 REM * Math 219b 40 REM * 27 April 1991 50 REM ************************************************ 60 INPUT "a, x\_0 = "; A, X0 70 FOR I = 1 TO 100 80 X1 = 2*X0 - A*X0*X0 90 IF ABS(X1-X0) < .0001 THEN PRINT X1,1/A,I:STOP 100 X0 = X1 110 NEXT I } \bigskip\noindent Now we would like to have an iterative method for computing $\root n \of{a}$, $a > 0$, which converges locally in second order. We only want to employ addition, subtraction, multiplication and division (the four arithmetic operations of the real number field. \medskip\noindent Again, we will be successful in using Newton's method. The fixed point theorem will guarantee that the iterative method converges locally in second order, that is, quadratically. $$F(x)\ =\ a - x^n\eqno(25)$$ $$F'(x)\ =\ -nx^{n-1}\eqno(26)$$ $${F(x)\over F'(x)}\ =\ {a-x^n \over -nx^{n-1} }\ =\ -{a\over nx^{n-1}} + {x\over n}\eqno(27)$$ $$\eqalign{f(x)\ &=\ x - F(x)/F'(x)\cr &=\ x + {a\over nx^{n-1}} -{x\over n}\cr &=\ \big(1-{\textstyle{1\over n}}\big)x + {\textstyle {1\over n}}\big(a/x^{n-1}\big)\cr}\eqno(28)$$ We will ensure convergence by computing $f'$ and plugging in the value $s = \root n \of {a}$. $$f'(x)\ = (1-1/n) - [a(n-1)]/[nx^n]. \eqno(29)$$ $$\eqalign {f'\big(\root n\of {a}\big)\ &= \big(1+{\textstyle{1\over n}}\big) -{\textstyle{ n-1\over n}\,{a\over a}}\cr &=\ 1 - {\textstyle {1\over n}} - {\textstyle {n\over n}} + {\textstyle{1\over n}}\cr &=\ 0\cr}\eqno(30)$$ Theory assured us this would happen; our $f'$ has been correctly computed. $$f''(x)\ =\ [a(n-1)]/x^{n+1}\eqno(31)$$ For $a > 0$, $$f''\big(\root n\of{a}\big)\ =\ {a(n-1) \over a \root n\of{a} }\ = \ {n-1 \over \root n \of {a}}.\eqno(32)$$ $n \not= 1$ in any case. Therefore we have quadratic convergence in some (yet to be determined) interval. Again, requiring $|f'(x)| < 1$: $$-1\ <\ \left(1-{1\over n}\right) -{(n-1)a \over nx^n}\ <\ 1$$ $$-2 + {1\over n}\ <\ -{(n-1)a\over nx^n}\ <\ {1\over n}$$ $${n\over (n-1)a}\left(2-{1\over n}\right)\ >\ {1\over x^n}\ >\ {n\over (n-1)a}{1\over n}$$ $${(n-1)a\over 2n-1}\ <\ x^n\ <\ (n-1)a$$ Therefore, $$\root n \of {(n-1)a\over 2n-1}\ <\ x\ <\ \root n \of {(n-1)a}\eqno(33)$$ is an interval in which the iteration algorithm converges quadratically. Finally, we note that $x^2$ is just short-hand for $x\cdot x$ or $x\,\times\,x$, and $x^3$ stands for $x\cdot x\cdot x$ or $x\,\times\,x\,\times\,x$ and so on. For a given, arbitrary (but fixed) value of $n$, the above algorithm can be written without exponentiation, substituting $$\underbrace{x\,\cdot\,x\,\cdots\,x}_{n\ \hbox{\sevenrm times}}\ \hbox{ for } x^n.$$ \medskip\noindent BASIC computer program (fast): \medskip {\tt\obeylines\ttraggedright 10 REM ***************************************************** 20 REM * Harry A. Watson, Jr. 30 REM * Math 219b 40 REM * 27 April 1991 50 REM ***************************************************** 60 INPUT "Root n, n = "; N 70 INPUT "a, x\_0 = "; A, X0 80 FOR I = 1 TO 100 90 X1 = (1-1/N)*X0 +A/(N*X0{\char94}(N-1)) 100 IF ABS(X1-X0) < .0001 THEN PRINT X1,A{\char94}(1/N),I:STOP 110 X0 = X1 120 NEXT I 130 REM *************************************************** 140 REM * Recall that the cube root of 20 is approximately 150 REM * $e\ =\ 2.7182818\ldots$ 160 REM * Plug in $n\ =\ 3$, $a\ =\ 20$, 170 REM * and $x_0\ =\ 3$ 180 REM * Answer = 2.714418 2.714418 4 (steps) 190 REM *************************************************** } \vfill\eject \noindent{\bf\llap{2.3\quad}The Algorithm.} The algorithm associated with Newton's method may be seen graphically as follows: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Graph showing Newton's method convergence. % % $$\beginpicture \setcoordinatesystem units <1cm,1cm> \setplotarea x from 0 to 8, y from 0 to 5 \axis bottom / % \axis left / % \put {$\scriptstyle\bullet$} at 1.1 0 % \plot 6 3.5 2.75 0 / % \putrule from 2.75 0 to 2.75 0.85 % \plot 2.75 0.85 1.5 0 / % \put {$\scriptstyle\bullet$} at 2.75 0 % \put {$\scriptstyle\bullet$} at 1.5 0 % \linethickness=.75pt \ifexpressmode \put {\bf EXPRESSMODE} at 4 2.5 % \else \setquadratic \plot .25 -.25 3 1 6 3.5 / % \fi \endpicture$$ \centerline{F{\sevenrm IGURE} 12} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \medskip $$\tan\beta = f'(x_0) = {f(x_0) \over x_0 - x_1}$$ $$x_1 = x_0 -{f(x_0) \over f'(x_0)}$$ Solve the equation $f(x)=0$. We assume that $f$ has a continuous first derivative, that is $f \in C^1$, where $$C^1 = \{\,\hbox{functions with continuous } f'\,\}.$$ \medskip\noindent We will construct the following feedback diagram: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Feedback diagram %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % $$\beginpicture % \setcoordinatesystem units <1in,1in> % \setplotarea x from 0 to 4, y from 0 to 3.5 % \put {\lines{ Last estimate\cr $x_n$\cr} } at 2 3 % \put {$\displaystyle x_{n+1} = x_n - {f(x_n)\over f'(x_n)}$} at 2 2 % \put {\lines{ Next estimate\cr $x_{n+1}$\cr} } at 2 1 % \ifexpressmode \putrule from 2.0 2.75 to 2.0 2.25 % \putrule from 2.0 1.75 to 2.0 1.25 % \putrule from 3.5 3.00 to 3.0 3.00 % \else \arrow <10pt> [.2, .4] from 2.0 2.75 to 2.0 2.25 % \arrow <10pt> [.2, .4] from 2.0 1.75 to 2.0 1.25 % \arrow <10pt> [.2, .4] from 3.5 3.00 to 3.0 3.00 % \fi \putrule from 3.0 1.0 to 3.5 1.0 % \putrule from 3.5 1.0 to 3.5 3.0 % \linethickness=1pt \putrectangle corners at 1 2.75 and 3 3.25 % \putrectangle corners at 1 1.75 and 3 2.25 % \putrectangle corners at 1 0.75 and 3 1.25 % \put {$n := n+1$} [l] <5pt,0pt> at 3.5 2 % \put {$ x_0 \longrightarrow $} [r] <-2pt,0pt> at 1.00 3.00 % \put{F{\sevenrm IGURE} 13} [B] at 2.00 0.25 % \endpicture$$ % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \noindent We will see that each error is essentially proportional to the square of the previous error. This means that the number of correct decimal places roughly doubles with each successive approximation. This is called quadratic convergence. \vfill\eject \noindent{\bf\llap{2.4\quad}A Second Opinion.} We say a derivation of Newton's method for finding roots in paragraph 2.1. There was an implicit assumption made that the function in question, $f(x)$, had a Taylor series representation. This is a very strong assumption and we may find examples easily which do not satisfy this condition. One very common engineering example can be constructed as follows: $$R'(x)\ =\ \cases{ 1 & if $ 0 \le x < 1$;\cr 0 & otherwise.\cr}$$ $$R(x)\ =\ \cases{ 0 & if $x < 0$;\cr x & if $ 0 \le x < 1$;\cr 1 & if $x \ge 1$;\cr}$$ The function $R(x)$ is the so-called {\it Ramp function\/} so highly favored by electrical engineers. Clearly the derivative is discontinuous at $0$ and $1$. We can construct $f(x) = R(x) - {1\over2}$ and this function has a root, $x_0 = {1\over2}$, $f(1/2) = R(1/2)-1/2 = 1/2-1/2 = 0$. We can weaken the assumption that $f(x)$ has a Taylor series representation and keep the same basic proof if we employ the so-called {\bf Extended Law of the Mean (Mean Value Theorem).}\footnote{$^{25}$}{John M. H. Olmsted, {\it Advanced Calculus}, (NY: Appleton-Century-Crofts, Inc., 1956), pages 75-85.} \proclaim Theorem. {\bf Extended Law of the Mean (Mean Value Theorem).} If each of $f(x)$, $f'(x)$, $\ldots$, $f^{(n-1)}(x)$ is continuous on a (closed) number interval $[a,b]$, and if $f^{(n)}(x)$ exists in the (open) number interval $(a,b)$, then there exists a number $\xi\,\in\,(a,b)$ such that $$f(b) = f(a) + f'(a)(b-a) + {f''(x)\over2!}(b-a)^2 + \cdots + {f^{(n-1)}(a)\over(n-1)!}(b-a)^{n-1} + {f^{(n)}(\xi)\over n!}(b-a)^n.\eqno(34)$$ \medskip\noindent We let $n=2$ and write $$f(x) = f(x_0) + f'(x_0)(x-x_0) + {f''(\xi)\over2}(x-x_0)^2.\eqno(35)$$ We are looking for an $x_1$ such that $f(x_1) = 0$. This being the case, we have $$x_1 = x_0 - {f(x_0)\over f'(x_0)} - {f''(\xi)\over2f'(x_0)}(x_1-x_0)^2. \eqno(36)$$ \medskip\noindent We will examine other example of convergence, graphing the steps. In this case the function is not a polynomial, we will use $$f(x) = e^{-(x-1/4)} - 1/4.\eqno(37)$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Plot showing convergence of Newton's method on transcendental function. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % $$\beginpicture % \setcoordinatesystem units <5cm,5cm> % \setplotarea x from 0 to 2, y from -.25 to .75 % \axis left / % \axis bottom shiftedto y=0 / % \ifexpressmode \put {\bf EXPRESSMODE} at 1 .5 % \else % \savelinesandcurves on "chap2b.t01" % \plot .375 0 .375 .63249 1.09171 0 1.09171 .18097 1.51162 0 1.51162 0.03194 / % \linethickness=.75pt % \setquadratic \plot .25 .75 .375 .63249 .50 .52880 .625 .43728 .75 .35653 .875 .28526 1.00 .22236 1.25 .11787 1.50 .03650 1.75 -0.02686 2.00 -0.07622 / % % \replot "chap2b.t01" % \fi \put {$x_0$} <0pt,-6pt> at .375 0 % \put {$x_1$} <0pt,-6pt> at 1.09171 0 % \put {$x_2$} <0pt,-6pt> at 1.51162 0 % \put {$\swarrow$\raise6pt\hbox{$\big(x_0,f(x_0)\big)$}} % [lb] <1pt,1pt> at .375 .63249 % \put {$\swarrow$\raise6pt\hbox{$\big(x_1,f(x_1)\big)$}} % [lb] <1pt,1pt> at 1.09171 .18097 % \put {$\swarrow$\raise6pt\hbox{$\big(x_2,f(x_2)\big)$}} % [lb] <1pt,1pt> at 1.51162 .03194 % \put {$\scriptstyle\bullet$} at .375 .63249 % \put {$\scriptstyle\bullet$} at 1.09171 .18097 % \put {$\scriptstyle\bullet$} at 1.51162 .03194 % \put {$\scriptstyle\bullet$} at .375 0 % \put {$\scriptstyle\bullet$} at 1.09171 0 % \put {$\scriptstyle\bullet$} at 1.51162 0 % \put {$x \rightarrow$} [l] <2pt,0pt> at 2 0 % \put {$y=f(x)$} [r] <-2pt,0pt> at 0 0.75 % \endpicture$$ % \centerline{F{\sevenrm IGURE} 14} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \vfill\eject \noindent{\bf\llap{2.5\quad}The Quasi-Newton Method.} The quasi-Newton method, also known as the method of {\it regula falsi\/}, does not require the function for the derivative. It relies on approximation of the derivative by the ratio $${f(x_{n-1})-f(x_{n-2}) \over x_{n-1} - x_{n-2} }.$$ Making the substitution $$f'(x_{n-1}) \approx {f(x_{n-1})-f(x_{n-2}) \over x_{n-1} - x_{n-2} }, $$ we obtain $$x_n\ =\ x_{n-1} - {\big(x_{n-1}-x_{n-2}\big)\,f(x_{n-1}) \over f(x_{n-1}) - f(x_{n-2}) }.\eqno(38)$$ We will examine a problem from a standard textbook\footnote{$^{26}$}{Robert Eisberg and Resnick, Robert, {\it Quantum Physics, 2nd Edition}, (NY: John Wiley \& Sons, 1985), page 24.} on quantum physics which will employ this technique. In order to derive the Wien displacement law, $\lambda_{\hbox{max}}T = 0.2014\,hc/k$,\footnote{$^{27}$}{$h=6.626\times 10^{-34}$ joule-sec is the Planck constant; $c=2.998\times10^{8}$ m/sec is the speed of light in vacuum; $k=1.381\times10^{-23}$ joule/$^\circ$K is the Boltzmann's constant.} it is necessary to solve the transcendental equation $$e^{-x} + x/5 = 1.\eqno(39)$$ We know already that the answer is approximately 5, in fact, it is approximately 4.965. We will use the following elementary BASIC computer program to see how well this method works. \medskip {\tt\parindent=0pt\obeylines\ttraggedright 5 DEFDBL A-H, O-Z 6 OPEN "92\_12\_16.txt" FOR OUTPUT AS \#1 10 DEF fna (x) = EXP(-x) + .2 * x - 1! 20 INPUT "Initial approximation = ", x0 25 PRINT \#1, "Initial approximation = ", x0 26 PRINT \#1, "x1 = x0 + .0001 = ", x0 + .0001 30 h = .0001 40 x1 = x0 + h 50 x2 = x1 - (h * fna(x1)) / (fna(x1) - fna(x0)) 60 PRINT x2 65 PRINT \#1, "Next approximation = ", x2 70 h = x2 - x1 80 IF INKEY\$ = "" THEN GOTO 80 90 x0 = x1: x1 = x2: GOTO 50 } \medskip\noindent Now, we will look at the output (starting with an initial guess of 5): \smallskip \centerline{\bf TYPE 92\_12\_16.TXT} \medskip {\tt\parindent=0pt\obeylines Initial approximation = 5 x1 = x0 + .0001 = 5.0001 Next approximation = 4.965135680044722 Next approximation = 4.965114168528612 Next approximation = 4.965114155083961 Next approximation = 4.965114155083955 Next approximation = 4.965114155083955 } \medskip\noindent So, it didn't take but four steps to converge! Now, let's draw a graph and see just how this looks graphically. Incidently, in the real world this particular experimental value is rarely calculated with accuracy better than 10\%. There is a good approximation to 0.2014, namely, $${1\over2\pi\big(1-\cos(e/2)\big)} = {1\over4.963222169} = .20148\ldots \eqno(40)$$ (It is accurate to within 0.019\%.) \vfill\eject % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Figure 10 --- Exponential curve with intersecting straight line. % $$\beginpicture \setcoordinatesystem units <.75in,1in> \setplotarea x from 0 to 6, y from 0 to 1 \normalgraphs \axis bottom ticks numbered from 0 to 6 by 1 length <0pt> withvalues $x$ / at .75 / / \linethickness=.25pt \putrule from .75 0 to .75 .47237 \setdots \putrule from .75 .47237 to .75 .85 \setsolid \put {$\scriptstyle\bullet$} at .75 .47237 \put {$\scriptstyle\bullet$} at .75 .85 \put {$y_1=e^{-x}$} [lb] <4pt,0pt> at .75 .47236 \put {$\swarrow$\raise6pt\hbox{$y_2=1-x/5$}} [lb] <2pt,2pt> at .75 .85 \put {$\scriptstyle\bullet$} at 4.96511 0 \put {$\swarrow$\raise6pt\hbox{$y_1 = y_2$}} [lb] <2pt,2pt> at 4.96511 0 % \ifexpressmode \put {\bf EXPRESSMODE} at 1.5 0.5 \else \setquadratic \plot 0 1 .25 .77880 .50 .60653 .75 .47237 1.00 .36788 1.25 .28650 1.50 .22313 1.75 .17377 2.00 .13534 2.25 .10540 2.50 .08208 2.75 .06393 3.00 .04979 3.25 .03877 3.50 .03019 3.75 .02352 4.00 .01832 4.25 .01426 4.50 .01109 4.75 .00865 5.00 .00674 5.25 .00525 5.50 .00409 5.75 .00318 6.00 .00248 / % \setlinear \plot 0 1 5.5 -0.1 / % \fi \endpicture $$ \centerline {F{\sevenrm IGURE} 15} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \medskip\noindent Let's take a look at another example, this time an iteration for a transcendental equation. We want to find the positive solution for $2\sin(x) = x$. The first thing we will want to do is to draw a graph and estimate the root. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Graph of y_1=2*sin(x) and y_2 = x % % Figure 11. % $$\beginpicture \setcoordinatesystem units <1in,.5in> \setplotarea x from 0 to 4, y from 0 to 4 \axis left ticks numbered from 0 to 4 by 1 / \axis bottom label {F{\sevenrm IGURE} 16} ticks numbered from 0 to 3 by 1 / \ifexpressmode \put {\bf EXPRESSMODE } at 2 2 % \else \setquadratic \plot 0.0 0.00000 0.1 0.19967 0.2 0.39734 0.3 0.59104 0.4 0.77884 0.5 0.95885 0.6 1.12928 0.7 1.28844 0.8 1.43471 0.9 1.56665 1.0 1.68294 1.1 1.78241 1.2 1.86408 1.3 1.92712 1.4 1.97090 1.5 1.99499 1.6 1.99915 1.7 1.98333 1.8 1.94770 1.9 1.89260 2.0 1.81859 2.1 1.72642 2.2 1.61699 2.3 1.49141 2.4 1.35093 2.5 1.19694 2.6 1.03100 2.7 0.85476 2.8 0.66998 2.9 0.47850 3.0 0.28224 / % \setlinear \plot 0 0 4 4 / \fi \put {$\longleftarrow y_1 = 2\cdot\sin(x)$} [l] <2pt,0pt> at 2.0 1.81859 % \put {$\nwarrow$\lower6pt\hbox{$y_2=x$}} [lt] <-1pt,-1pt> at 2.5 2.5 % \put { $y_1=y_2$\lower6pt\hbox{$\downarrow$} } % [rb] <5pt,1pt> at 1.8955 1.8955 % \setdots \putrule from 1.8955 0.0 to 1.8955 1.8955 / \endpicture$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \medskip\noindent Of course, we will examine both the quasi-Newton and the Newton method. The Newton method is listed to the right and we see at once that it has converged both faster and to greater precision. The important point to note here is that with a modern (fast) computer both methods converge rapidly. We can use the two in conjunction to check each other's results. Moreover, with the Simpson's rule procedure to check the derivative function, we can be assured that the derivative $f'(x)$ of $f(x)$ is the correct function. This comparison of two different methods is similar to the adaptive quadrature routine (AQR), which will be studied in a later section. \medskip % % {\tt\parindent=0pt\obeylines Initial approximation = 2 x1 = x0 - .0001 = 1.99990 \quad 1.9009955942039090 Next approximation = 1.90099 \quad 1.8955116453795950 Next approximation = 1.89580 \quad 1.8954942672087130 Next approximation = 1.89550 \quad 1.8954952670339810 Next approximation = 1.89549 \quad 1.8954952670339810 Next approximation = 1.89549 \quad 1.8954925670339810 Next approximation = 1.89549 \quad 1.8954925670339810 } \medskip\noindent \vfill\eject \noindent{\bf\llap{2.6\quad}Pathological examples.} Newton's method will fail for certain situations. There are an excellent developments of such examples in many elementary textbooks\footnote{$^{28}$}{George B. Thomas, Jr., {\it Calculus and Analytic Geometry, 3rd Edition}, (Reading Massachusetts: Addison-Wesley Publishing Company, 1966), page 455.} We won't dwell on the analytic details of the failure of the Newton method; on the other hand, we will present graphs to illustrate the point. It has already been pointed out that the very first thing to do is to construct a graph and then estimate the root. With all the excellent graphics software available, there is no excuse for not doing so. Without further ado, let's first look at a graph where Newton's method fails to converge. $$f(x)\ =\ \cases{\sqrt{\left| x-r\right|}& if $x > r$\cr\noalign{\vskip2pt} 0 & if $x = 0$\cr -\sqrt{\left| x-r\right|}& if $x < r$\cr}\eqno(41)$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Figure 17. % $$\beginpicture \setcoordinatesystem units <2in,1in> \setplotarea x from -.125 to 2, y from -1 to 1 % \plotheading {\lines{ Successive approximants\cr oscillate back and forth.\cr} } % \axis bottom shiftedto y=0 / % \axis left shiftedto x=0 / \ifexpressmode \put {\bf EXPRESSMODE} at 2 1 % \else \linethickness=1pt \setquadratic \plot 0.35 -0.80623 0.40 -0.77460 0.45 -0.74162 0.50 -0.70711 0.55 -0.67082 0.60 -0.63246 0.65 -0.59161 0.70 -0.54772 0.75 -0.50000 0.80 -0.44721 0.85 -0.38730 0.90 -0.31623 0.95 -0.22361 0.975 -0.15811 1.00 0.00000 1.025 0.15811 1.05 0.22361 1.10 0.31623 1.15 0.38730 1.20 0.44721 1.25 0.50000 1.30 0.54772 1.35 0.59161 1.40 0.63246 1.45 0.67082 1.50 0.70711 1.55 0.74162 1.60 0.77460 1.65 0.80623 / % \fi \linethickness=0.4pt \put {$\scriptstyle\bullet$} at 0.50 -0.70711 % \put {$\scriptstyle\bullet$} at 1.50 0.70711 % \put {$\scriptstyle\bullet$} at 0.50 0.0 % \put {$\scriptstyle\bullet$} at 1.50 0.0 % \put {$x_0$} [b] <0pt,4pt> at 0.50 0.0 % \put {$x_1$} [t] <0pt,-4pt> at 1.50 0.0 % \put {$O$} [rt] <-2pt,-2pt> at 0 0 % \put {$r$} [lb] <4pt,2pt> at 1.0 0.00 % \put {$x$} at 2.1 0.0 % \put {$y$} at 0.0 1.1 % \setdashes \putrule from 0.5 0.0 to 0.5 -0.70711 % \putrule from 1.5 0.0 to 1.5 0.70711 % \setlinear \setsolid \linethickness=.25pt \plot 0.5 -0.70711 1.5 0.0 / % \plot 0.5 0.0 1.5 0.70711 / % \endpicture$$ \centerline {F{\sevenrm IGURE} 17} % \medskip\noindent Now, let's look at a graph that really gets crazy! This time the successive approximations, $x_0$ and $x_1$ don't oscillate, they get worse! $$\beginpicture \setcoordinatesystem units <2in,1in> \setplotarea x from -.125 to 2, y from -1 to 1 % \plotheading {\lines{ Successive approximants\cr diverge from root.\cr} } % \axis bottom shiftedto y=0 / % \axis left shiftedto x=0 / \ifexpressmode \put {\bf EXPRESSMODE} at 2 1 % \else \linethickness=1pt \setquadratic \plot 0.35 -0.86624 0.40 -0.84343 0.45 -0.81932 0.50 -0.79370 0.55 -0.76631 0.60 -0.73681 0.65 -0.70473 0.70 -0.66943 0.75 -0.62996 0.80 -0.58480 0.85 -0.53133 0.90 -0.46416 0.95 -0.36840 0.975 -0.29240 1.00 0.00000 1.025 0.29240 1.05 0.36840 1.10 0.46416 1.15 0.53133 1.20 0.58480 1.25 0.62996 1.30 0.66943 1.35 0.70473 1.40 0.73681 1.45 0.76631 1.50 0.79370 1.55 0.81932 1.60 0.84343 1.65 0.86624 / % \fi \linethickness=0.4pt \put {$\scriptstyle\bullet$} at 0.50 -0.79370 % \put {$\scriptstyle\bullet$} at 1.50 0.79370 % \put {$\scriptstyle\bullet$} at 0.50 0.0 % \put {$\scriptstyle\bullet$} at 1.50 0.0 % \put {$x_0$} [b] <0pt,4pt> at 0.50 0.0 % \put {$x_1$} [t] <0pt,-4pt> at 1.50 0.0 % \put {$O$} [rt] <-2pt,-2pt> at 0 0 % \put {$r$} [lb] <4pt,2pt> at 1.0 0.00 % \put {$x$} at 2.1 0.0 % \put {$y$} at 0.0 1.1 % \setdashes \putrule from 0.5 0.0 to 0.5 -0.79370 % \putrule from 1.5 0.0 to 1.5 0.79370 % % \setlinear % \setsolid % \linethickness=.25pt % \plot 0.5 -0.79370 1.5 0.0 / % % \plot 0.5 0.0 1.5 0.79370 / % \endpicture$$ \centerline {F{\sevenrm IGURE} 18} % \vfill\eject % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \headline={\tenrm\hfill Linear Least-squares Line} \centerline{\bf Chapter 3} \bigskip \noindent{\bf\llap{3.0\quad}Introduction.} One of the best introductions to the so-called methods of least-squares follows from a simple display of a least-squares linear fit graph properly done (with the error bars and all the statistical appendages) and a dissection the components of the graph. The points in the given point set are graphed as central dots ({\raise1pt\hbox{$\scriptstyle\bullet$}}) in the error bars $\big(\,{{\lower1pt\hbox{$\scriptstyle\top$}} \atop{\raise1pt\hbox{$\scriptstyle\bot$}}}\,\big)$. The length of the error bar is determined by the standard deviation, $\sigma_y$. From the central dot to the cross-bar is one unbiased standard deviation. % \newdimen\xposition \newdimen\yposition \newdimen\dyposition \newdimen\crossbarlength \def\puterrorbar at #1 #2 with fuzz #3 {% \xposition=\Xdistance{#1} \yposition=\Ydistance{#2} \dyposition=\Ydistance{#3} \setdimensionmode \put {$\bullet$} at {\xposition} {\yposition} \dimen0 = \yposition % ** Determine the y-location of \advance \dimen0 by -\dyposition % ** the lower cross bar. \dimen2 = \yposition % ** Determine the y-location of \advance \dimen2 by \dyposition % ** the upper cross bar. \putrule from {\xposition} {\dimen0} % ** Place vertical rule. to {\xposition} {\dimen2} \dimen4 = \xposition \advance \dimen4 by -.5\crossbarlength \dimen6 = \xposition \advance \dimen6 by .5\crossbarlength \putrule from {\dimen4} {\dimen0} to {\dimen6} {\dimen0} \putrule from {\dimen4} {\dimen2} to {\dimen6} {\dimen2} \setcoordinatemode} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Graph for least-squares curve fit with shading, areas, etc. % $$\beginpicture \setcoordinatesystem units <.4in,.45in> \crossbarlength=5pt \setplotarea x from -1 to 10, y from 85 to 95 \plotheading { \lines{Missile `A'\cr Warrantied in shaded area\cr} } % \linethickness=.25pt \axis bottom label { \lines{ years $\longrightarrow$\cr \noalign{\vskip3pt} % F{\sevenrm IGURE} 19\cr } } % ticks in short withvalues $1981$ $1982$ $1983$ $1984$ $1985$ $1986$ $1987$ $1988$ $1989$ $1990$ $1991$ / at 0 1 2 3 4 5 6 7 8 9 10 / / \axis left label {\lines{$\uparrow$\cr $R$\/\%\cr }} ticks in short numbered from 85 to 95 by 1 / \puterrorbar at 0 90.08 with fuzz 0.180 \puterrorbar at 1 90.57 with fuzz 0.180 \puterrorbar at 2 90.76 with fuzz 0.180 \puterrorbar at 3 91.30 with fuzz 0.180 \puterrorbar at 4 91.57 with fuzz 0.180 \puterrorbar at 5 92.44 with fuzz 0.180 \puterrorbar at 6 92.87 with fuzz 0.180 \puterrorbar at 7 93.68 with fuzz 0.180 \puterrorbar at 8 93.85 with fuzz 0.180 \puterrorbar at 9 94.49 with fuzz 0.180 \puterrorbar at 10 94.88 with fuzz 0.180 \ifexpressmode \put {\bf EXPRESSMODE} at 5 90 \else \linethickness=.4pt \plot 0. 89.9077 10. 94.9086 / \setlinear \setshadegrid span <4pt> \vshade 4 85 95 7 85 95 / \vshade 7 85 86.5 9.1 85 86.5 / \vshade 9 85 95 10 85 95 / \setlinear \setshadegrid span <4pt> \hshade 87.5 7 9.1 95 7 9.1 / \fi \put {\frame <3pt> {\lines {Unwarrantied\cr Region\cr}}} at 2 93.5 \put {\frame <4pt> {$\sigma_y = 0.180$}} at 8 87 \endpicture$$ % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % The underlying principle of the method of least squares is the determination of a curve which minimizes the squares of the deviations of a given point set to that curve. This is another way of saying that we will determine a curve such that the sum of the squares of the distances from the points in a given point set to the curve is a minimum. Since the sought-for curve is a straight line, this case is frequently referred to as a linear least-squares curve fit. The determined line is sometimes called a trend line, sometimes called a regression line, and sometimes called a line of best fit. The technique is known as the {\bf maximum likelihood criterion}. \footnote{$^{29}$}{Glenn James and James, Robert C., {\it Mathematics Dictionary}, (Princeton, NJ: D. Van Nostrand Company, Inc., 1964), page 253.} The theory involves an understanding of partial derivatives; it will be covered later in an optional paragraph. The computer programs are straightforward to easy to apply; they will be covered first. \vfill\eject \noindent{\bf\llap{3.1\quad}Dissection of a graph.} The first thing to consider in any least-squares construction is the underlying point set. This point set is sometimes called the sample space, sometimes called the sample data, and sometimes called the sample population. Confusion arises because of the various names that are attached to the data points. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Graph giving only the data points % $$\beginpicture \setcoordinatesystem units <.3in,.2in> \setplotarea x from -1 to 10, y from 85 to 95 \plotheading { Sample Data } % \linethickness=.25pt \axis bottom % label { \lines{ abscissa ($x$-axis) $\longrightarrow$\cr % \noalign{\vskip3pt} % F{\sevenrm IGURE} 20\cr} } % ticks in short withvalues {} {} {} {} {} {} {} {} {} {} {} / at 0 1 2 3 4 5 6 7 8 9 10 / / % \axis left label {\lines{$\uparrow$\cr $y(x)$\cr }} ticks in short withvalues {} {} {} {} {} {} {} {} {} {} {} / at 85 86 87 88 89 90 91 92 93 94 95 / / % \put {$\scriptstyle\bullet$} at 0 90.08 \put {$\scriptstyle\bullet$} at 1 90.57 \put {$\scriptstyle\bullet$} at 2 90.76 \put {$\scriptstyle\bullet$} at 3 91.30 \put {$\scriptstyle\bullet$} at 4 91.57 \put {$\scriptstyle\bullet$} at 5 92.44 \put {$\scriptstyle\bullet$} at 6 92.87 \put {$\scriptstyle\bullet$} at 7 93.68 \put {$\scriptstyle\bullet$} at 8 93.85 \put {$\scriptstyle\bullet$} at 9 94.49 \put {$\scriptstyle\bullet$} at 10 94.88 \ifexpressmode \put {\bf EXPRESSMODE} at 5 90 \else % \linethickness=.4pt % \plot 0. 89.9077 10. 94.9086 / \fi \endpicture$$ %\centerline{F{\sevenrm IGURE} 21} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % The next thing to consider are the arithmetic quantities associated with the data points (with the so-called point set). If we simply add up all the $x$-values (the abscissas) and divide by the count of the number of points, we will get the $x$-average, also called the $x$-mean, and usually denoted as $$\frame <5pt> {$\displaystyle \mu_x\ =\ {\sum_{i=1}^n x_i \over n}.$} \eqno(42)$$ In a similar manner, we compute the $y$-average, $\mu_y$, $$\frame <5pt> {$\displaystyle \mu_y\ =\ {\sum_{i=1}^n y_i \over n}.$} \eqno(43)$$ The averages $\mu_x$ and $\mu_y$ are familiar. Other computations are also necessary to obtain the ingredients needed to calculate the trend line, shown below. Because this line is the best possible fit to the data, it is also known as the {\sl line of best fit}. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Graph showing the trend line and data only % $$\beginpicture \setcoordinatesystem units <.3in,.2in> \setplotarea x from -1 to 10, y from 85 to 95 \plotheading { Sample Data } % \linethickness=.25pt \axis bottom % label {\lines{ abscissa ($x$-axis) $\longrightarrow$\cr % \noalign{\vskip3pt} % F{\sevenrm IGURE} 21\cr } } % ticks in short withvalues {} {} {} {} {} {} {} {} {} {} {} / at 0 1 2 3 4 5 6 7 8 9 10 / / % \axis left label {\lines{$\uparrow$\cr $y(x)$\cr }} ticks in short withvalues {} {} {} {} {} {} {} {} {} {} {} / at 85 86 87 88 89 90 91 92 93 94 95 / / % \put {$\scriptstyle\bullet$} at 0 90.08 \put {$\scriptstyle\bullet$} at 1 90.57 \put {$\scriptstyle\bullet$} at 2 90.76 \put {$\scriptstyle\bullet$} at 3 91.30 \put {$\scriptstyle\bullet$} at 4 91.57 \put {$\scriptstyle\bullet$} at 5 92.44 \put {$\scriptstyle\bullet$} at 6 92.87 \put {$\scriptstyle\bullet$} at 7 93.68 \put {$\scriptstyle\bullet$} at 8 93.85 \put {$\scriptstyle\bullet$} at 9 94.49 \put {$\scriptstyle\bullet$} at 10 94.88 \ifexpressmode \put {\bf EXPRESSMODE} at 5 90 \else \linethickness=.4pt \plot 0. 89.9077 10. 94.9086 / \fi \put {$\nwarrow\lower6pt\hbox{Line of Best Fit}$} [lt] <1pt,1pt> at 4.5 92.0 \endpicture$$ %\centerline{F{\sevenrm IGURE} 21} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \vfill\eject \noindent Now is a good time to introduce the BASIC computer program. Since we have seen several BASIC and ``C'' programs already, this listing will be provided {\it without comments}. The statistical parameters $\mu_x$ and $\mu_y$ will be denoted as {\tt XMU} and {\tt YMU}, respectively. The statistical parameters for the standard deviations, $\sigma_x$, $\sigma_y$, and $\sigma_{xy}$ (biased and unbiased), will be denoted {\tt SIGMAX}, {\tt SIGMAY}, and {\tt SIGMAXY}, respectively. \smallskip {\tt\obeylines\parindent=0pt\ttraggedright 100 CONST N = 11 110 DEFDBL A-H, O-Z: REM Define everything double precision except the indices. 120 DIM X(N), Y(N) 130 DATA 1981.,1982.,1983.,1984.,1985.,1986.,1987.,1988.,1989.,1990.,1991. 140 DATA 90.08,90.57,90.76,91.30,91.57,92.44,92.87,93.68,93.85,94.49,94.88 150 FOR I = 1 TO N: READ X(I): NEXT I 160 FOR I = 1 TO N: READ Y(I): NEXT I 170 SUMX = 0!: SUMY = 0!: SUMX2 = 0!: SUMXY = 0!: SUMX2 = 0! 180 SIGMAX = 0!: SIGMAY = 0!: SIGMAXY = 0! 190 FOR I = 1 TO N 200 SUMX = SUMX + X(I): SUMY = SUMY + Y(I): SUMX2 = SUMX2 + X(I) * X(I) 210 SUMXY = SUMXY + X(I) * Y(I): SUMY2 = SUMY2 + Y(I) * Y(I) 220 NEXT I 230 DELTA = CDBL(N) * SUMX2 - SUMX * SUMX 240 A = (SUMX2 * SUMY - SUMX * SUMXY) / DELTA 250 B = (CDBL(N) * SUMXY - SUMX * SUMY) / DELTA 260 CLS : PRINT "A (x-intercept) = "; A 270 PRINT "B (slope) = "; B 280 XMU = SUMX / CDBL(N): YMU = SUMY / CDBL(N) 290 FOR I = 1 TO N 300 SIGMAX = SIGMAX + (XMU - X(I)) * (XMU - X(I)) 310 SIGMAY = SIGMAY + (YMU - Y(I)) * (YMU - Y(I)) 320 SIGMAXY = SIGMAXY + (XMU - X(I)) * (YMU - Y(I)) 330 NEXT I 340 R = SIGMAXY / SQR(SIGMAX * SIGMAY) 350 PRINT "Correlation Coefficient r = "; R: SIGMAY = 0! 360 FOR I = 1 TO N 370 SIGMAY = SIGMAY + (Y(I) - A - B * X(I)) {\char94} 2 380 PRINT USING "\#\#\#\# \#\#.\#\#\#\#\# \#\#.\#\#\#\#\#"; X(I); Y(I); A + B * X(I) 390 NEXT I 400 SIGMAY = SQR(SIGMAY / (CDBL(N) - 2!)) 410 PRINT "sigma\_y (unbiased) = "; SIGMAY 420 PRINT "sigma\_x (unbiased) = "; SIGMAY * SQR(SIGMAX / DELTA) 430 END } \smallskip\noindent After the above BASIC computer program was executed, the following data were obtained: \smallskip {\tt\parindent=0pt\obeylines\ttraggedright A (x-intercept) = -900.77236 B (slope) = .50009 Correlation Coefficient r = .99475 Sigma\_y (unbiased) = .17980 Sigma\_x (unbiased) = .05421 } \smallskip\noindent The next important topic is the construction of the error bars. This is often the most forgotten item in the creation of a data presentation. It is the only way properly introduct statistical uncertainty into a display graph. In the hard science of Physics, such information is mandatory. The distance from the data point to the crossbars (top and bottom of the errorbar) must equal one unbiased standard deviation, $\sigma_y$. There may also be uncertainty in the abscissa. In this case, it is necessary to construct an error bar horizontal to the $x$-axis of length $\sigma_x$. The error bars tell us more than just some statistical fact about the individual data points. If we examine closely Fig. 33, we'll note that the line of best fit passes through most of the error bars. This tells us that there is a good probability that there is an underlying functional relationship between $x$ and $f(x)$, namely $f(x) \approx a\cdot x + b$. If, on the other hand, the best straight line missed a high proportion of the error bars, or if it missed any by a large distance (relative to the length of the error bar), then the relationship between $x$ and $y \approx f(x)$ would probably not be that of a straight line (if there was any relationship at all). % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % Error bar graph % $$\beginpicture % Always put % here because some word processor \setcoordinatesystem units <1cm,1cm> % truncate final space(s). \setplotarea x from -2 to 8, y from -2 to 2 % \crossbarlength = 5pt % \plotheading {Error bars with uncertainties} % \puterrorbar at 0 0 with fuzz 1.2 % \put { \lines{$\uparrow$\cr $\sigma_y$\cr $\downarrow$\cr} } % <0pt,3pt> at .5 .5 % \puterrorbar at 5 0 with fuzz 1.2 % \put { \lines{$\uparrow$\cr $\sigma_y$\cr $\downarrow$\cr} } % <0pt,3pt> at 5.5 .5 % \putrule from 3.6 0.0 to 6.4 0.0 % \putrule from 3.6 0.1 to 3.6 -0.1 % \putrule from 6.4 0.1 to 6.4 -0.1 % \put { $\leftarrow\sigma_x\rightarrow$ } at 5.7 -0.5 % \endpicture$$ % \centerline{F{\sevenrm IGURE} 22} \smallskip\noindent We'll conclude this long chapter with a another graph showing the line of best fit and the companion ``C'' program for the BASIC program above. The next graph was suggested by the textbook {\sl An Introduction to Error Analysis\/} by John R. Taylor (Mill Valley, CA: University Science Books, 1982), page 26. It indicates the uncertainties found in any physical measurement. The underlying physical law is know as {\bf Hooke's Law}, published by Robert Hooke in 1678. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % A plot of an extension of a spring $x$ versus the load $m$. % % $$\beginpicture \setcoordinatesystem units <.002in,.4in> \crossbarlength=5pt \setplotarea x from 0 to 1000, y from 0 to 5 \linethickness=.25pt \axis bottom % label {\lines{ $m$ (gm) $\longrightarrow$ \cr \noalign{\vskip2pt} % F{\sevenrm IGURE} 23\cr} } % ticks in numbered from 0 to 1000 by 500 unlabeled short quantity 11 / \axis left label {\lines{$\uparrow$\cr $x$\cr (cm)\cr}} ticks in withvalues $5$ / at 5 / quantity 6 / \plot 0 5 0 5.5 / \plot 1000 0 1050 0 / \linethickness=.4pt \plot 0 0 1000 5.5 / \puterrorbar at 200 1.1 with fuzz 0.3 \puterrorbar at 300 1.5 with fuzz 0.3 \puterrorbar at 400 1.9 with fuzz 0.3 \puterrorbar at 500 2.8 with fuzz 0.3 \puterrorbar at 600 3.4 with fuzz 0.3 \puterrorbar at 700 3.5 with fuzz 0.3 \puterrorbar at 800 4.6 with fuzz 0.3 \puterrorbar at 900 5.4 with fuzz 0.3 \endpicture$$ \vfill\eject {\tt\parindent=0pt\obeylines\ttraggedright /* Program for mean, standard deviation */ /* Program for least-squares fit to a straight line */ /* Program for computing correlation coefficient "r" */ \medskip \#include \#include \#include \medskip int main() {\char123} /* Initialize all the parameters and type variables */ \quad FILE *fp; /* file pointer */ \quad static double x[] = { 1981., 1982., 1983., 1984., 1985., \qquad 1986., 1987., 1988., 1989., 1990., \qquad 1991. }; /* x values (years) */ \quad static double y[] = { 90.08, 90.57, 90.76, 91.30, 91.57, \qquad 92.44, 92.87, 93.68, 93.85, 94.49, \qquad 94.88 }; /* y values (reliabilities) */ \quad /* The variables sum\_{something} are used as accumulators. */ \quad /* a, b are the y-intercept and slope, respectively of the \qquad least-squares line. */ \quad /* r is the correlation coefficient (the so-called Pearson's r). */ double Delta, a, b, r, sum\_x, sum\_x2, sum\_y, sum\_y2, sum\_xy; /* mu\_{something} is a mean, sigma\_{something} is a standard deviation (unbiased for least-squares = n-2 degrees of freedom). */ double mu\_x, mu\_y, sigma\_x, sigma\_y, sigma\_xy; /* The integer i is an index. (unsigned, short integer) */ int i; /* The long integer n is the count of the number of data points. */ long int n; /* Begin computations */ /* Initialize n, x[n], y[n], and give the output file a name. Then you are ready to begin execution. */ n = 11; /* Make sure the accumulators are all initialized to zero. */ sum\_x = (double) 0.0; sum\_y = (double) 0.0; sum\_y2 = (double) 0.0; sum\_x2 = (double) 0.0; sum\_xy = (double) 0.0; sigma\_x = (double) 0.0; sigma\_y = (double) 0.0; sigma\_xy = (double) 0.0; /* The so-called "for loop" computes the sum\_{something}s */ for (i=0; i < n; ++i){\char123} /* i marches from 0 to n-1. */ sum\_y += (double) y[i]; /* All the arrays in "C" */ sum\_y2 += (double) y[i]*y[i]; /* programs begin with index */ sum\_x += (double) x[i]; /* zero. */ sum\_x2 += (double) x[i]*x[i]; sum\_xy += (double) x[i]*y[i]; {\char125} /* Display the results of the computation of all the sums, Delta, \quad the slope, and the y-intercept for \quad y = a + b * x, the least-squares straight line */ printf("{\char92}n sum\_x = \%.16lf", sum\_x); printf("{\char92}n sum\_y = \%.16lf", sum\_y); printf("{\char92}n sum\_x2 = \%.16lf", sum\_x2); printf("{\char92}n sum\_y2 = \%.16lf", sum\_y2); printf("{\char92}n sum\_xy = \%.16lf", sum\_xy); Delta = (double) n*s