\magnification=1100 \overfullrule0pt \input amssym.def \input prepictex \input pictex \input postpictex % ********************* Definitions ************************************ %\def\widetilde{\mathaccent"0365 } \def\CC{{\Bbb C}} \def\FF{{\Bbb F}} \def\HH{{\Bbb H}} \def\NN{{\Bbb N}} \def\OO{{\Bbb O}} \def\QQ{{\Bbb Q}} \def\RR{{\Bbb R}} \def\ZZ{{\Bbb Z}} \def\cA{{\cal A}} \def\cB{{\cal B}} \def\cR{{\cal R}} \def\cZ{{\cal Z}} \def\cC{{\cal C}} \def\cR{{\cal R}} \def\fb{\frak{b}} \def\fg{\frak{g}} \def\fh{\frak{h}} \def\fn{\frak{n}} \def\Card{\hbox{Card}} \def\End{\hbox{End}} \def\Hom{\hbox{Hom}} \def\Ind{\hbox{Ind}} \def\Id{\hbox{Id}} \def\codim{\hbox{codim}} \def\diag{\hbox{diag}} \def\id{\hbox{id}} \def\im{\hbox{im}} \def\tr{\hbox{tr}} \def\Tr{\hbox{Tr}} \def\fbox{} % ********************* FONTS ************************************ \font\smallcaps=cmcsc10 \font\titlefont=cmr10 scaled \magstep1 \font\titlefontbold=cmbx10 scaled \magstep1 \font\titlesubfont=cmr10 scaled \magstep1 \font\sectionfont=cmbx10 \font\tinyrm=cmr10 at 8pt % ******************** SECTION HEADERS *************************** \newcount\sectno \newcount\subsectno \newcount\resultno \def\section #1. #2\par{ \sectno=#1 \resultno=0 \bigskip\noindent{\sectionfont #1. #2}~\medbreak} \def\subsection #1\par{\bigskip\noindent{\it #1} \medbreak} %******************* MATHEMATICAL LABELS ************************** \def\prop{ \global\advance\resultno by 1 \medskip\noindent{\bf Proposition \the\sectno.\the\resultno. }\sl} \def\lemma{ \global\advance\resultno by 1 \medskip\noindent{\bf Lemma \the\sectno.\the\resultno. } \sl} \def\fact{ \global\advance\resultno by 1 \medskip\noindent{\bf Fact \the\sectno.\the\resultno. } \sl} \def\remark{ \global\advance\resultno by 1 \medskip\noindent{\bf Remark \the\sectno.\the\resultno. }} \def\example{ \global\advance\resultno by 1 \medskip\noindent{\bf Example \the\sectno.\the\resultno. }\sl} \def\cor{ \global\advance\resultno by 1 \medskip\noindent{\bf Corollary \the\sectno.\the\resultno. }\sl} \def\thm{ \global\advance\resultno by 1 \medskip\noindent{\bf Theorem \the\sectno.\the\resultno. }\sl} \def\defn{ \global\advance\resultno by 1 \medskip\noindent{\it Definition \the\sectno.\the\resultno. }\slrm} \def\endthm{\rm\medskip} \def\thmend{\rm\medskip} \def\endlemma{\rm\medskip} \def\endfact{\rm\medskip} \def\endexample{\rm\medskip} \def\endprop{\rm\medskip} \def\endcor{\rm\medskip} \def\pf{\rm\smallskip\noindent{\it Proof. }} \def\endpf{\qed\hfil\medskip} \def\pfend{\qed\hfil\medskip} \def\note{\smallbreak\noindent{Note:}} \def\enddefn{\rm\medskip} %Homemade Struts: \newbox\strutAbox \setbox\strutAbox=\hbox{\vrule height 12pt depth6pt width0pt} \def\strutA{\relax\copy\strutAbox} \newbox\strutBbox \setbox\strutBbox=\hbox{\vrule height 10pt depth5pt width0pt} \def\strutB{\relax\copy\strutBbox} \newbox\strutDbox \setbox\strutDbox=\hbox{\vrule height 11pt depth5pt width0pt} \def\strutD{\relax\copy\strutDbox} %high strut: \newbox\strutHbox \setbox\strutHbox=\hbox{\vrule height 11pt depth1pt width0pt} \def\strutH{\relax\copy\strutHbox} %low strut: \newbox\strutLbox \setbox\strutLbox=\hbox{\vrule height 1pt depth5pt width0pt} \def\strutL{\relax\copy\strutLbox} % hack to ignore lots of typed stuff.... \def\ignore#1{\relax} % ****************** QED SIGNS ********************************* \def\qed{\hbox{\hskip 1pt\vrule width4pt height 6pt depth1.5pt \hskip 1pt}} \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt \hbox{\vrule width.#2pt height#1pt \kern#1pt \vrule width.2pt} \hrule height.2pt}}}} \def\square{\mathchoice\sqr54\sqr54\sqr{3.5}3\sqr{2.5}3} \def\whiteslug{\bf $ \square $ \rm} % open square %*************** EQUATIONS WITH NUMBERS ************** \def\formula{\global\advance\resultno by 1 \eqno{(\the\sectno.\the\resultno)}} \def\formulano{\global\advance\resultno by 1 (\the\sectno.\the\resultno)} \def\tableno{\global\advance\resultno by 1 \the\sectno.\the\resultno. } \def\lformula{\global\advance\resultno by 1 \leqno(\the\sectno.\the\resultno)} %************Commutative diagrams********************** \def\mapright#1{\smash{\mathop {\longrightarrow}\limits^{#1}}} \def\mapleftright#1{\smash{\mathop {\longleftrightarrow}\limits^{#1}}} \def\mapsrightto#1{\smash{\mathop {\longmapsto}\limits^{#1}}} \def\mapleft#1{\smash{ \mathop{\longleftarrow}\limits^{#1}}} \def\mapdown#1{\Big\downarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\lmapdown#1{{\hbox{$\scriptstyle#1$}} \llap {$\vcenter{\hbox{\Big\downarrow}}$} } \def\mapup#1{\Big\uparrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapne#1{\Big\nearrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapse#1{ %{$\vcenter{ \hbox{$\scriptstyle#1$} %$} \rlap{ $\vcenter{\hbox{$\searrow$}}$ } } \def\mapnw#1{\Big\nwarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapsw#1{ %\Big \swarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} %********** DATING ****************************************** \def\monthname {\ifcase\month\or January\or February\or March\or April\or May\or June\or July\or August\or September\or October\or November\or December\fi} \newcount\mins \newcount\hours \hours=\time \mins=\time \def\now{\divide\hours by60 \multiply\hours by60 \advance\mins by-\hours \divide\hours by60 % NOTE: \divide only gives integer answers. \ifnum\hours>12 \advance\hours by-12 \number\hours:\ifnum\mins<10 0\fi\number\mins\ P.M.\else \number\hours:\ifnum\mins<10 0\fi\number\mins\ A.M.\fi} \def\today {\monthname\ \number\day, \number\year} %**************** PAGE HEADERS ************************* \nopagenumbers \def\runningtitle{\smallcaps numbers} \headline={\ifnum\pageno>1\eoheadline\else\firstheadline\fi} \def\names{\smallcaps a.\ ram} %\def\firstheadline{\noindent Preliminary Draft \hfill \today} \def\firstheadline{} \def\eoheadline{\ifodd\pageno\oddheadline\else\evenheadline\fi} \def\oddheadline{\tenrm\hfil\runningtitle\hfil\folio} \def\evenheadline{\tenrm\folio\hfil{\names}\hfil} %**************** TITLE ************************* \vphantom{$ $} %My kludge to get the first page to move down a bit \vskip.75truein \centerline{\titlefont Numbers} \bigskip \centerline{\rm Arun Ram} %${}^\ast$ \centerline{Department of Mathematics} \centerline{University of Wisconsin, Madison} \centerline{Madison, WI 53706 USA} \centerline{{\tt ram@math.wisc.edu}} \medskip \centerline{Version: \today} %\footnote{}{\tinyrm %${}^\ast$ %Research partially supported by the National Security Agency %and by EPSRC Grant GR K99015 at the Newton Institute for %Mathematical Sciences.} %\footnote{}{\tinyrm %\noindent {Keywords:} exponential functions, trigonometric functions} \bigskip %**************** ABSTRACT ************************* %\noindent{\bf Abstract.} \bigskip\noindent {\bf Calculus} is the study of \smallskip\noindent \itemitem{(1)} Derivatives \smallskip\noindent \itemitem{(2)} Integrals \smallskip\noindent \itemitem{(3)} Applications of derivatives \smallskip\noindent \itemitem{(4)} Applications of integrals \medskip\noindent A {\it derivative} is a creature you put a function into, it chews on it, and spits out a new function. \medskip\noindent A {\it function} takes in a number, chews on it, and spits out a new number. \bigskip\noindent {\ }\qquad\qquad{\bf Derivatives}\hskip2.25truein {\bf Functions} \smallskip\medskip \noindent ${\hbox{input}\atop\hbox{function}}$ $\longrightarrow$ \fbox{~$\displaystyle{d\over dx}$~} $\longrightarrow$${\hbox{output}\atop\hbox{function}}$ \hskip1.0truein ${\hbox{input}\atop \hbox{number}}$ $\longrightarrow$ \fbox{~$f$~} $\longrightarrow$${\hbox{output}\atop\hbox{number}}$ \medskip \medskip\noindent The {\it integral} is the derivative backwards: \medskip\noindent {\bf Numbers} are at the bottom of the food chain. \bigskip\bigskip At some point humankind wanted to count things and discovered the {\bf positive integers}, $$ 1,\; 2,\; 3,\; 4,\; 5,\;\ldots. $$ GREAT for counting something, \smallskip\noindent BUT what if you don't have anything? How do we talk about nothing, nulla, zilch? \smallskip\noindent $\ldots$ and so we discovered the {\bf nonnegative integers}, $$ 0,\; 1,\; 2,\; 3,\; 4,\; 5,\; \ldots. $$ GREAT for adding, $$ 5+3=8,\;\; 0+10=10,\;\; 21+37=48, $$ BUT not so great for subtraction, $$ 5-3=2,\;\; 2-0=2,\;\; 12-34=???. $$ $\ldots$ and so we discovered the {\bf integers} $$ \ldots, -3,\; -2,\; -1,\; 0,\; 1,\; 2,\; 3,\; \ldots\;. $$ GREAT for adding, subtracting and multiplying, $$ 3\cdot 6=18,\;\; -3\cdot 2=-6,\;\; 0\cdot 7=0, $$ BUT not so great if you only want part of the sausage $\ldots$, \smallskip\noindent $\ldots$ and so we discovered the {\bf rational numbers}, $$ {a\over b}\;, \qquad\hbox{$a$ an integer, $b$ an integer, $b\ne 0$}. $$ GREAT for addition, subtraction, multiplication, and division, \smallskip\noindent BUT not so great for finding $\sqrt{2} = ????$, \smallskip\noindent $\ldots$ and so we discovered the {\bf real numbers}, $$ \hbox{all decimal expansions}. $$ \smallskip\noindent Examples: $$\matrix{ \eqalign{ \pi &=3.1415926\ldots\;,\cr e &= 2.71828\ldots\;, \cr \sqrt 2&=1.414\ldots\;,\cr 10&= 10.0000\ldots\;, \cr} &\qquad\qquad &\eqalign{ {1\over 3}&=.3333\ldots\;,\cr {1\over 8}&=.125 = .125000000\ldots\;, \cr} }$$ GREAT for addition, subtraction, multiplication, and division, \smallskip\noindent BUT not so great for finding $\sqrt{-9} = ????$, \smallskip\noindent $\ldots$ and so we discovered the {\bf complex numbers}, $$ a+bi, \qquad \hbox{$a$ a real number, $b$ a real number, $i=\sqrt{-1}$}\}. $$ \smallskip \noindent {\bf Examples:} $\qquad 3+\sqrt{2}i, \qquad 6=6+0i, \qquad \pi +\sqrt{7}i,$ \medskip\noindent and $$ \sqrt{-9}=\sqrt{9(-1)}=\sqrt{9}\sqrt{-1}=3i. $$ \smallskip\noindent GREAT. \smallskip\noindent {\it Addition:} \qquad $ (3+4i)+(7+9i)=3+7+4i+9i=10+13i. $ \smallskip\noindent {\it Subtraction:} \qquad $ (3+4i)-(7+9i)=3-7+4i-9i=-4-5i. $ \smallskip\noindent {\it Multiplication:} $$\eqalign{ (3+4i)(7+9i) &=3(7+9i)+4i(7+9i)\cr &=21+27i+28i+36i^2\cr &=21+55i-36\cr &=-15+55i. \cr}$$ \smallskip\noindent {\it Division:} $$\eqalign{ {3+4i\over 7+9i} &={(3+4i)\over (7+9i)} {(7-9i)\over (7-9i)}={21-27i+28i+36\over 49-63i+63i+81}\cr &\cr &={57+i\over 130}={57\over 130}+{1\over 130}i. \cr}$$ \smallskip\noindent {\it Square Roots:} We want $\sqrt{-3+4i}$ to be some $a+bi$. $$ \hbox{If}\qquad \sqrt{-3+4i}= a+bi $$ then $$\eqalign{ -3+4i = (a+bi)^2 &= a^2+abi+abi+b^2i^2 \cr &= a^2-b^2+2abi. \cr }$$ So $$a^2-b^2=-3\qquad\hbox{and}\qquad 2ab=4.$$ \noindent Solve for $a$ and $b$. $$\eqalign{ b={4\over 2a}={2\over a}.\qquad &\hbox{So}\quad a^2-\Big({2\over a}\Big)^2 = -3. \cr &\hbox{So}\quad a^2-{4\over a^2}=-3. \cr &\hbox{So}\quad a^4-4=-3a^2. \cr &\hbox{So}\quad a^4+3a^2-4=0. \cr &\hbox{So}\quad (a^2+4)(a^2-1)=0. \cr }$$ So $a^2=-4$ or $a^2=1$. \smallskip\noindent So $a=\pm 1$,\qquad and $b={2\over \pm 1}=2$ or $-2$. \smallskip\noindent So $a+bi=1+2i$ or $a+bi=-1-2i$. \smallskip\noindent So $\sqrt{-3+4i}=\pm (1+2i)$. \medskip\noindent {\it Graphing:} \bigskip \medskip\noindent {\it Factoring:} $$\eqalign{ x^2+5 & = (x+\sqrt{5}\,i)(x-\sqrt{5}\,i), \cr x^2+x+1 &= \left(x-\Big(-\hbox{$1\over2$}+\hbox{$\sqrt{3}\over2$}i\Big)\right) \left(x-\Big(-\hbox{$1\over2$}-\hbox{$\sqrt{3}\over2$}i\Big)\right) \cr }$$ This is REALLY why we like the complex numbers. The {\bf fundamental theorem of algebra} says that ANY POLYNOMIAL (for example, $x^{12673}+2563x^{159}+\pi x^{121}+\sqrt{7}\,x^{23}+9621{1\over2}$) can be factored completely as $$(x-u_1)(x-u_2)\cdots (x-u_n)$$ where $u_1,\ldots, u_n$ are complex numbers. \vfill\eject \end \section 2. The {\it exponential\/} function \bigskip \medskip \noindent ${\hbox{input}\atop x}$$\longrightarrow$ \fbox{~f~}$\longrightarrow$${\hbox{output}\atop f(x)}$ \hskip1.25truein ${\hbox{input}\atop f}$ $\longrightarrow$ \fbox{~$\displaystyle{d\over dx}$~} $\longrightarrow$ ${\hbox{output}\atop {df\over dx}}$ \medskip {\bf Function}\hskip2.75truein {\bf Derivative} \medskip \noindent The {\it exponential function\/} is the function $e^x$ such that $$ {de^x\over dx}=e^x\quad\hbox{and}\quad e^0=1. $$ Figure out what $e^x$ is: \smallskip Suppose $e^x=a_0+a_1x+a_2x^2+a_3x^3+\cdots$ Then $e^0=a_0+0+0+\cdots =1$. So $a_0=1$. $$\eqalign{ {de^x\over dx}&=a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots\cr &\cr &=e^x=a_0+a_1x+a_2x^2+a_3x^3+\cdots \cr} $$ So $a_1=0$, $2a_2=a_1$, $4a_3=a_2$, $4a_4=a_3$, $\ldots$. So $$ a_0=1,\;\; a_1=1,\;\; a_2={1\over 2},\;\; a_3={1\over 2\cdot 3},\;\; a_4={1\over 2\cdot 3\cdot 4},\;\; a_5={1\over 2\cdot 3\cdot 4\cdot 5},\ldots $$ So $$ e^x=1+x+{1\over 2}x^2+{1\over 2\cdot 3} x^3+{1\over 2\cdot 3\cdot 4} x^4+{1\over 2\cdot 3\cdot 4\cdot 5} x^5+\cdots. $$ \medskip\noindent {\bf Factorials} $$\eqalign{ 7!&=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=5040\cr 5!&=5\cdot 4\cdot 3\cdot 2\cdot 1=120\cr 3!&=3\cdot 2\cdot 1. \cr }$$ So $$ e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+{x^4\over 4!}+{x^5\over 5!}+\cdots $$ So $$\eqalign{ e^1&=1+1+{1\over 2}+{1\over 6}+{1\over 24}+{1\over 120}+\cdots =2.781828\ldots \cr &\cr e^{-3}&=1+(-3)+{(-3)^2\over 2}+{(-3)^3\over 6}+{(-3)^4\over 24}+\cdots \cr &\cr &=1-3+{3^2\over 2}-{3^3\over 6}+{3^4\over 24}+\cdots = ????\cr }$$ \medskip \centerline{${\hbox{input}\atop x}$ $\longrightarrow$ \fbox{~$e^x$~} $\longrightarrow$ ${\hbox{output}\atop e^x}$} \medskip \noindent {\it Note:} ~By the chain rule $$ {d\over dx} e^{2+x}=e^{2+x}\cdot {d(2+x)\over dx}=e^{2+x}\cdot 1=e^{2+x} $$ and $$ e^{2+0}=e^2. $$ So $$ e^{2+x}=e^2x+{e^2x^2\over 2!}+{e^2x^3\over 3!}+\cdots $$ since, in this case $$ a_0=e^2,\;\; a_1=a_0,\;\; 2a_2=a_1,\;\; 3a_3=a_2,\ldots $$ if $$ e^{2+x}=a_0+a_1x+a_2x^2+\cdots . $$ So $$ e^{2+x}=e^2e^x. $$ Similarly, $$ e^{10+x}=e^{10}e^x\;\hbox{ and }\; e^{642+x}=e^{542}e^x $$ and $$ \hbox{\fbox{~$e^{y+x}=e^ye^x$~}}. $$ Since $e^{-x}e^x = e^{-x+x}=e^0=1$ $$e^{-x}={1\over e^x}.$$ Since $$\eqalign{ e^{10x} &= e^{x+x+x+x+x+x+x+x+x+x}\cr &=e^xe^{x+x+x+x+x+x+x+x+x}\cr &=e^xe^xe^{x+x+x+x+x+x+x+x}\cr &=e^xe^xe^xe^{x+x+x+x+x+x}\cr &=e^xe^xe^xe^xe^{x+x+x+x+x}\cr &=e^xe^xe^xe^xe^xe^{x+x+x+x}\cr &=e^xe^xe^xe^xe^xe^xe^xe^xe^xe^x=(e^x)^{10}\cr} $$ \smallskip \noindent {\bf Summary}: $e^x$ is the function such that $$ {de^x\over dx}=e^x\quad\hbox{and}\quad e^0=1. $$ Then $$\eqalign{ e^{x+y}&=e^xe^y\cr e^{-x}&={1\over e^x}\cr e^{nx}&= (e^x)^n. \cr }$$ \vfill\eject \end \magnification=1100 \overfullrule0pt \input amssym.def \input prepictex \input pictex \input postpictex % ********************* Definitions ************************************ %\def\widetilde{\mathaccent"0365 } \def\CC{{\Bbb C}} \def\FF{{\Bbb F}} \def\HH{{\Bbb H}} \def\NN{{\Bbb N}} \def\OO{{\Bbb O}} \def\QQ{{\Bbb Q}} \def\RR{{\Bbb R}} \def\ZZ{{\Bbb Z}} \def\cA{{\cal A}} \def\cB{{\cal B}} \def\cR{{\cal R}} \def\cZ{{\cal Z}} \def\cC{{\cal C}} \def\cR{{\cal R}} \def\fb{\frak{b}} \def\fg{\frak{g}} \def\fh{\frak{h}} \def\fn{\frak{n}} \def\Card{\hbox{Card}} \def\End{\hbox{End}} \def\Hom{\hbox{Hom}} \def\Ind{\hbox{Ind}} \def\Id{\hbox{Id}} \def\codim{\hbox{codim}} \def\diag{\hbox{diag}} \def\id{\hbox{id}} \def\im{\hbox{im}} \def\tr{\hbox{tr}} \def\Tr{\hbox{Tr}} % ********************* FONTS ************************************ \font\smallcaps=cmcsc10 \font\titlefont=cmr10 scaled \magstep1 \font\titlefontbold=cmbx10 scaled \magstep1 \font\titlesubfont=cmr10 scaled \magstep1 \font\sectionfont=cmbx10 \font\tinyrm=cmr10 at 8pt % ******************** SECTION HEADERS *************************** \newcount\sectno \newcount\subsectno \newcount\resultno \def\section #1. #2\par{ \sectno=#1 \resultno=0 \bigskip\noindent{\sectionfont #1. #2}~\medbreak} \def\subsection #1\par{\bigskip\noindent{\it #1} \medbreak} %******************* MATHEMATICAL LABELS ************************** \def\prop{ \global\advance\resultno by 1 \medskip\noindent{\bf Proposition \the\sectno.\the\resultno. }\sl} \def\lemma{ \global\advance\resultno by 1 \medskip\noindent{\bf Lemma \the\sectno.\the\resultno. } \sl} \def\fact{ \global\advance\resultno by 1 \medskip\noindent{\bf Fact \the\sectno.\the\resultno. } \sl} \def\remark{ \global\advance\resultno by 1 \medskip\noindent{\bf Remark \the\sectno.\the\resultno. }} \def\example{ \global\advance\resultno by 1 \medskip\noindent{\bf Example \the\sectno.\the\resultno. }\sl} \def\cor{ \global\advance\resultno by 1 \medskip\noindent{\bf Corollary \the\sectno.\the\resultno. }\sl} \def\thm{ \global\advance\resultno by 1 \medskip\noindent{\bf Theorem \the\sectno.\the\resultno. }\sl} \def\defn{ \global\advance\resultno by 1 \medskip\noindent{\it Definition \the\sectno.\the\resultno. }\slrm} \def\endthm{\rm\medskip} \def\thmend{\rm\medskip} \def\endlemma{\rm\medskip} \def\endfact{\rm\medskip} \def\endexample{\rm\medskip} \def\endprop{\rm\medskip} \def\endcor{\rm\medskip} \def\pf{\rm\smallskip\noindent{\it Proof. }} \def\endpf{\qed\hfil\medskip} \def\pfend{\qed\hfil\medskip} \def\note{\smallbreak\noindent{Note:}} \def\enddefn{\rm\medskip} %Homemade Struts: \newbox\strutAbox \setbox\strutAbox=\hbox{\vrule height 12pt depth6pt width0pt} \def\strutA{\relax\copy\strutAbox} \newbox\strutBbox \setbox\strutBbox=\hbox{\vrule height 10pt depth5pt width0pt} \def\strutB{\relax\copy\strutBbox} \newbox\strutDbox \setbox\strutDbox=\hbox{\vrule height 11pt depth5pt width0pt} \def\strutD{\relax\copy\strutDbox} %high strut: \newbox\strutHbox \setbox\strutHbox=\hbox{\vrule height 11pt depth1pt width0pt} \def\strutH{\relax\copy\strutHbox} %low strut: \newbox\strutLbox \setbox\strutLbox=\hbox{\vrule height 1pt depth5pt width0pt} \def\strutL{\relax\copy\strutLbox} % hack to ignore lots of typed stuff.... \def\ignore#1{\relax} % ****************** QED SIGNS ********************************* \def\qed{\hbox{\hskip 1pt\vrule width4pt height 6pt depth1.5pt \hskip 1pt}} \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt \hbox{\vrule width.#2pt height#1pt \kern#1pt \vrule width.2pt} \hrule height.2pt}}}} \def\square{\mathchoice\sqr54\sqr54\sqr{3.5}3\sqr{2.5}3} \def\whiteslug{\bf $ \square $ \rm} % open square %*************** EQUATIONS WITH NUMBERS ************** \def\formula{\global\advance\resultno by 1 \eqno{(\the\sectno.\the\resultno)}} \def\formulano{\global\advance\resultno by 1 (\the\sectno.\the\resultno)} \def\tableno{\global\advance\resultno by 1 \the\sectno.\the\resultno. } \def\lformula{\global\advance\resultno by 1 \leqno(\the\sectno.\the\resultno)} %************Commutative diagrams********************** \def\mapright#1{\smash{\mathop {\longrightarrow}\limits^{#1}}} \def\mapleftright#1{\smash{\mathop {\longleftrightarrow}\limits^{#1}}} \def\mapsrightto#1{\smash{\mathop {\longmapsto}\limits^{#1}}} \def\mapleft#1{\smash{ \mathop{\longleftarrow}\limits^{#1}}} \def\mapdown#1{\Big\downarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\lmapdown#1{{\hbox{$\scriptstyle#1$}} \llap {$\vcenter{\hbox{\Big\downarrow}}$} } \def\mapup#1{\Big\uparrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapne#1{\Big\nearrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapse#1{ %{$\vcenter{ \hbox{$\scriptstyle#1$} %$} \rlap{ $\vcenter{\hbox{$\searrow$}}$ } } \def\mapnw#1{\Big\nwarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapsw#1{ %\Big \swarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} %********** DATING ****************************************** \def\monthname {\ifcase\month\or January\or February\or March\or April\or May\or June\or July\or August\or September\or October\or November\or December\fi} \newcount\mins \newcount\hours \hours=\time \mins=\time \def\now{\divide\hours by60 \multiply\hours by60 \advance\mins by-\hours \divide\hours by60 % NOTE: \divide only gives integer answers. \ifnum\hours>12 \advance\hours by-12 \number\hours:\ifnum\mins<10 0\fi\number\mins\ P.M.\else \number\hours:\ifnum\mins<10 0\fi\number\mins\ A.M.\fi} \def\today {\monthname\ \number\day, \number\year} %**************** PAGE HEADERS ************************* \nopagenumbers \def\runningtitle{\smallcaps genesis} \headline={\ifnum\pageno>1\eoheadline\else\firstheadline\fi} \def\names{\smallcaps a.\ ram} %\def\firstheadline{\noindent Preliminary Draft \hfill \today} \def\firstheadline{} \def\eoheadline{\ifodd\pageno\oddheadline\else\evenheadline\fi} \def\oddheadline{\tenrm\hfil\runningtitle\hfil\folio} \def\evenheadline{\tenrm\folio\hfil{\names}\hfil} %**************** TITLE ************************* \vphantom{$ $} %My kludge to get the first page to move down a bit \vskip.75truein \centerline{\titlefont An approach to ``early trascendentals''} \bigskip \centerline{\rm Arun Ram} %${}^\ast$ \centerline{Department of Mathematics} \centerline{University of Wisconsin, Madison} \centerline{Madison, WI 53706 USA} \centerline{{\tt ram@math.wisc.edu}} \medskip \centerline{Version: \today} %\footnote{}{\tinyrm %${}^\ast$ %Research partially supported by the National Security Agency %and by EPSRC Grant GR K99015 at the Newton Institute for %Mathematical Sciences.} \footnote{}{\tinyrm \noindent {Keywords:} exponential functions, trigonometric functions} \bigskip %**************** ABSTRACT ************************* %\noindent{\bf Abstract.} \bigskip\noindent {\bf The function ${\rm god}(t)$} \bigskip\noindent There is one function that \smallskip\noindent \itemitem{(a)} in the Beginning, created something from nothing, and \smallskip\noindent \itemitem{(b)} is ``unchanging'', or rather, its change is itself. \smallskip\noindent Through the ages thinkers have contemplated this function and nowadays it is common to write (a) and (b) in abbreviated form, $$\hbox{(a$'$)}~~~~{\rm god}(0)=1, \qquad\qquad\hbox{and}\qquad\qquad \hbox{(b$'$)}~~~~{d~{\rm god}(t)\over dt} = {\rm god}(t),$$ but the meaning is still the same. \smallskip Two of the children of god are eve and adam: $${\rm god}(it) = {\rm adam}(t) + i ~{\rm eve}(t).$$ \bigskip\noindent {\bf Trying to understand ${\rm god}(t)$} \bigskip If we try to ``understand'' god in ``normal'' terms, $${\rm god}(t) = a_0 + a_1t +a_2t^2+a_3t^3+ \cdots,$$ then $$\matrix{ \hbox{since} &\qquad &{\rm god}(0)=1,\hfill &\qquad &a_0=1,\qquad \hfill&\hbox{and}\hfill \cr \cr \cr \hbox{since} &\qquad &\displaystyle{d{\rm god}(t)\over dt} = {\rm god}(t), \hfill &\qquad &a_1=a_0, \hfill &\hbox{and}\hfill\cr &&&&2a_2 = a_1, \hfill &\hbox{and}\hfill\cr &&&&3a_3 = a_2, \hfill &\hbox{and}\hfill\cr &&&&4a_4 = a_3, \hfill &\hbox{and}\hfill\cr &&&&5a_5 = a_4, \hfill &\hbox{$\ldots,$ etc.,}\hfill\cr }$$ and so $${\rm god}(t)= 1~+~t~+~{1\over 2!}t^2~+~{1\over 3!}t^3~+~{1\over 4!}t^4 ~+~ \cdots,$$ which illustrates that ${\rm god}(t)$ exists everywhere and goes on forever. \bigskip\noindent {\bf An amazing thing about ${\rm god}(t)$} \bigskip One of the amazing things about god is that $${\rm god}(t+s) = {\rm god}(t)~{\rm god}(s).$$ To see why god is this way suppose that there is a ``different'' function such that \smallskip\noindent \itemitem{(a$''$)} is ``unchanging'' \qquad $\displaystyle{ \left( \hbox{i.e.} \quad {d~\widetilde{{\rm god}}(t)\over dt} = \widetilde{{\rm god}}(t)\right) }$, \qquad and \smallskip\noindent \itemitem{(b$''$)} in the Beginning, was the way that god is after $s$ millenia \qquad $(~\hbox{i.e.}\quad \widetilde{{\rm god}}(0)={\rm god}(s)~~)$. \smallskip\noindent By the chain rule, $${d~{\rm god}(t+s)\over dt} = {\rm god}(t+s) \qquad\hbox{and}\qquad {\rm god}(0+s) = {\rm god}(s),$$ and so $${\rm god}(t+s) = \widetilde{{\rm god}}(t).$$ Also, $${d~({\rm god}(t){\rm god}(s))\over dt} = {\rm god}(t){\rm god}(s), \qquad\hbox{and}\qquad {\rm god}(0){\rm god}(s) = {\rm god}(s),$$ and so $${\rm god}(t){\rm god}(s) = \widetilde{{\rm god}}(t)= {\rm god}(t+s).$$ \bigskip\noindent {\bf What about ${\rm adam}(t)$ and ${\rm eve}(t)$?} \bigskip $$\matrix{ {\rm god}(it) &= &1 &+it &\displaystyle{ +{(it)^2\over 2!} }&\displaystyle{ +{(it)^3\over 3!} }&\displaystyle{ +{(it)^4\over 4!}} &\displaystyle{ +{(it)^5\over 5!} }&+ &\cdots \cr \cr \cr &= &1 &&\displaystyle{ +{i^2t^2\over 2!} }&&\displaystyle{ +{i^4t^4\over 4!} }&&\displaystyle{ +{i^6t^6\over 6!} }&&+\cdots \cr &&&+it &&\displaystyle{ +{i^3t^3\over 3!} }&&\displaystyle{ +{i^5t^5\over 5!} }&&\displaystyle{ +{i^7t^7\over 7!} } &&+\cdots \cr \cr \cr &= &1 &&\displaystyle{ -{t^2\over 2!} }&&\displaystyle{ +{t^4\over 4!} }&&\displaystyle{ -{t^6\over 6!} }&&+\cdots \cr &&&+it &&\displaystyle{ -{it^3\over 3!} }&&\displaystyle{ +{it^5\over 5!} }&&\displaystyle{ -{it^7\over 7!} } &&+\cdots \cr }$$ $$ = \left(1-{t^2\over 2!}+{t^4\over 4!}-{t^6\over 6!}+{t^8\over 8!} -\cdots\right) + i\left(t-{t^3\over 3!}+{t^5\over 5!}-{t^7\over 7!}+\cdots\right) \hskip.5in $$ and, since adam and eve are the children of god, $$\hbox{~~i.e.\qquad because \quad${\rm god}(it) = {\rm adam}(t) + i~{\rm eve}(t)$~,\qquad}$$ we see that $$\eqalign{ {\rm adam}(t) &= 1-{t^2\over 2!}+{t^4\over 4!}-{t^6\over 6!}+{t^8\over 8!}+\cdots, \qquad\qquad\qquad\hbox{and} \cr \cr {\rm eve}(t) &=t-{t^3\over 3!}+{t^5\over 5!}-{t^7\over 7!}+{t^9\over 9!}+\cdots, \cr }$$ from which it follows that $$\matrix{ {\rm adam}(0)=0, &\qquad\qquad &{\rm eve}(0)=1, \cr \cr {\rm adam}(-t)=-{\rm adam}(t), &\qquad\qquad &{\rm eve}(-t)={\rm eve}(t), \cr \cr \displaystyle{{d~{\rm adam}(t)\over dt}={\rm eve}(t)}, &\qquad\qquad &\displaystyle{{d~{\rm eve}(t)\over dt}=-{\rm adam}(t)}. \cr }$$ So, adam and eve are complete opposites and identical twins at the same time. \bigskip\noindent {\bf Complete opposites and identical twins at the same time, another manifestation} \medskip $$\eqalign{ 1 &= {\rm god}(0) = {\rm god}(it-it) ={\rm god}(it+i(-t)) = {\rm god}(it){\rm god}(i(-t)) \cr &= ({\rm adam}(t)+i~{\rm eve}(t))({\rm adam}(-t)+i~{\rm eve}(-t)) \cr &= ({\rm adam}(t)+i~{\rm eve}(t))({\rm adam}(t)-i~{\rm eve}(t)) \cr &= ({\rm adam}(t))^2+({\rm eve}(t))^2, \cr \cr \hbox{i.e.} \qquad\qquad 1 &= ({\rm adam}(t))^2+({\rm eve}(t))^2. \cr }$$ \bigskip\noindent {\bf Through the ages: where are we now?} \bigskip\noindent Let $x = {\rm eve}(t)$ and $y = {\rm adam}(t)$. \medskip\noindent \itemitem{(A)} In the Beginning the point $(x,y)$ was at $ ({\rm adam}(0),{\rm eve}(0))=(1,0)$,\quad and \medskip\noindent since $1={\rm adam}(t))^2+({\rm eve}(t))^2$, \qquad $x^2+y^2=1$,\qquad and \medskip\noindent \itemitem{(B)} adam and eve travel through the ages on a circle of radius 1. $$ \beginpicture \setcoordinatesystem units <2cm,2cm> % sets scale \setplotarea x from -1.7 to 1.7, y from -1.7 to 1.7 % sets plot size up \plot -1.5 0 1.5 0 / \plot 0 -1.5 0 1.5 / \put{$({\rm adam}(0),{\rm eve}(0))$}[bl] at 1.1 0.1 % \put{$\bullet$} at 1 0 % \put{$y$} at -.2 1.4 % \put{$x$} at 1.4 -.2 % %Circle \ellipticalarc axes ratio 1:1 360 degrees from 1 0 center at 0 0 \endpicture $$ \noindent Where are they after $d$ millenia? $$\eqalign{ \matrix{ \hbox{The distance traveled}\cr \hbox{after $d$ millenia}\cr} &= \int_{t=0}^{t=d} ds = \int_{t=0}^{t=d} \displaystyle{ \sqrt{\left({dx\over dt}\right)^2 +\left({dy\over dt}\right)^2} ~dt } \cr \cr &= \int_{t=0}^{t=d} \displaystyle{ \sqrt{\left({d~{\rm adam}(t)\over dt}\right)^2 +\left({d~{\rm eve}(t)\over dt}\right)^2} ~dt } \cr \cr &= \int_{t=0}^{t=d} \sqrt{({\rm eve})^2+(-{\rm adam}(t))^2} ~dt \cr \cr &= \int_{t=0}^{t=d}\sqrt{1} ~dt = \int_{t=0}^{t=d} dt = ~t~\Big\vert^{t=d}_{t=0} =d-0=d, \cr }$$ and so $$ \matrix{ {\rm adam}(t) ~~~= ~~\hbox{$x$-coordinate of the point on a circle of radius 1}\hfill \cr \phantom{{\rm adam}(t) ~~~=~~} ~~\hbox{which is distance $d$ from the point (1,0),}\hfill &\qquad\qquad\hbox{and} \cr \cr {\rm eve}(t) ~~~= ~~\hbox{$y$-coordinate of the point on a circle of radius 1} \hfill\cr \phantom{{\rm adam}(t) ~~~=~~} ~~\hbox{which is distance $d$ from the point (1,0).} \hfill\cr }$$ $$ \beginpicture \setcoordinatesystem units <2cm,2cm> % sets scale \setplotarea x from -1.7 to 1.7, y from -1.7 to 1.7 % sets plot size up \plot -1.5 0 1.5 0 / \plot 0 -1.5 0 1.5 / \plot 0 0 0.5 .866025 / \plot 0.5 0 0.5 .866025 / \put{$({\rm adam}(0),{\rm eve}(0))$}[bl] at 1.1 0.1 % \put{$({\rm adam}(d),{\rm eve}(d))$}[bl] at .6 0.9 % \put{$\bullet$} at .5 .866025 % \put{$\bullet$} at 1 0 % \put{$y$} at -.2 1.4 % \put{$x$} at 1.4 -.2 % %Circle \ellipticalarc axes ratio 1:1 360 degrees from 1 0 center at 0 0 \endpicture $$ The triangle in this picture is $$ \beginpicture \setcoordinatesystem units <2cm,2cm> % sets scale \setplotarea x from -0.3 to 1.2, y from -0.3 to 1.2 % sets plot size up \plot 0 0 0.5 0 / \plot 0 0 0.5 .866025 / \plot 0.5 0 0.5 .866025 / \put{$1$}[r] at 0.1 0.4 % \put{${\rm adam}(d)$}[t] at 0.25 -0.1 % \put{${\rm eve}(d)$}[l] at .6 0.4 % \endpicture \qquad\qquad\qquad\qquad \beginpicture \setcoordinatesystem units <2cm,2cm> % sets scale \setplotarea x from -0.3 to 1.2, y from -0.3 to 1.2 % sets plot size up \plot 0 0 0.5 0 / \plot 0 0 0.5 .866025 / \plot 0.5 0 0.5 .866025 / \put{hypotenuse}[r] at 0.1 0.4 % \put{adjacent}[t] at 0.25 -0.1 % \put{opposite}[l] at .6 0.4 % \endpicture $$ and so $$ {\rm adam}(d) = {\hbox{opposite}\over \hbox{hypotenuse}} \qquad\hbox{and}\qquad {\rm eve}(d) = {\hbox{adjacent}\over \hbox{hypotenuse}} $$ for a right triangle with angle $d$. \bigskip\noindent {\bf Some remarks on society} \bigskip\noindent {\bf 1.} It is interesting to note that our school systems like to introduce our children to ${\rm adam}(t)$ and ${\rm eve}(t)$ but prefer to hide from my child how close they really are to ${\rm god}(t)$. \bigskip\noindent {\bf 2.} Mathematicians are a cloistered group and prefer to study god, adam, and eve in anonymity. In the mathematical literature $$\matrix{ {\rm god}(t)\qquad\hfill &\hbox{is usually called}\qquad \hfill &e^t~, \hfill\cr {\rm adam}(t)\hfill &\hbox{is usually termed} \qquad &\cos t~,\hfill &\qquad\hbox{and} \hfill \cr {\rm eve}(t)\hfill &\hbox{is usually called} \hfill &\sin t~. \hfill \cr }$$ \vfill\eject \end % Last Edit 7 February 1997 % This document is written in Plain TeX % The macros: prepictex.tex, pictex.tex, and postpictex.tex are also % required for the full compilation of the document. \magnification=1000 \overfullrule0pt \nopagenumbers \input prePicTeX \input PicTeX \input postPicTeX \input amssym.def \vphantom{$ $} %My kludge to get the first page to move down a bit \vskip2truein $$ \beginpicture \setcoordinatesystem units <1cm,1cm> % sets scale \setplotarea x from -5 to 5, y from -2 to 2 % sets plot size up \put{$\bullet$} at -3 2 % \put{$\bullet$} at -2 2 % \put{$\bullet$} at -1 2 % \put{$\bullet$} at 0 2 % Top dots \put{$\bullet$} at 1 2 % \put{$\bullet$} at 2 2 % \put{$\bullet$} at 3 2 % \put{$\bullet$} at -3 .5 % \put{$\bullet$} at -2 .5 % \put{$\bullet$} at -1 .5 % \put{$\bullet$} at 0 .5 % Bottom dots \put{$\bullet$} at 1 .5 % \put{$\bullet$} at 2 .5 % \put{$\bullet$} at 3 .5 % %Border of cylinder \plot -3 .5 3 .5 / \ellipticalarc axes ratio 3:1 90 degrees from -4.5 1 center at -3 1 \ellipticalarc axes ratio 3:1 90 degrees from 3 .5 center at 3 1 \plot -3 2 3 2 / \ellipticalarc axes ratio 3:1 180 degrees from -3 3 center at -3 2.5 \plot -3 3 3 3 / %top edge of cylinder \ellipticalarc axes ratio 3:1 180 degrees from 3 2 center at 3 2.5 \plot -4.5 2.5 -4.5 1 / \plot 4.5 2.5 4.5 1 / % Vertical edges %\plot -3 2 -3 .5 / \plot -2 2 -2 .5 / \plot -1 2 -1 .5 / %\plot 0 2 0 .5 / %\plot 1 2 1 .5 / \plot 2 2 2 .5 / %\plot 3 2 3 .5 / \setquadratic % single crossing \plot 0 .5 .05 .8 .4 1.15 / \plot .6 1.35 .95 1.7 1 2 / \plot 0 2 .05 1.7 .5 1.25 .95 .8 1 .5 / %0 crossing \plot 3 .5 3.02 .8 3.2 1 / %bottom corner \plot 3 2 3.05 1.7 3.2 1.5 / %top corner \plot -3.8 1.5 -3.7 1.45 -3.5 1.25 -3.05 .8 -3 .5 / \plot -3.4 1.35 -3.05 1.7 -3 2 / \plot -3.8 1 -3.7 1.05 -3.6 1.15 / \ellipticalarc axes ratio 7:3.5 -90 degrees from -3.8 1 center at -3.8 1.35 \ellipticalarc axes ratio 13:3.5 90 degrees from 3.2 1 center at 3.2 1.35 \ellipticalarc axes ratio 7:5 -90 degrees from -3.8 1.5 center at -3.8 2 \ellipticalarc axes ratio 13:5 90 degrees from 3.2 1.5 center at 3.2 2 \setlinear \setdashes \plot -3.8 2.5 3.2 2.5 / \plot -3.8 1.7 3.2 1.7 / \ellipticalarc axes ratio 7:3.5 90 degrees from -3.8 1.7 center at -3.8 1.35 \ellipticalarc axes ratio 13:3.5 -90 degrees from 3.2 1.7 center at 3.2 1.35 \ellipticalarc axes ratio 7:5 90 degrees from -3.8 2.5 center at -3.8 2 \ellipticalarc axes ratio 13:5 -90 degrees from 3.2 2.5 center at 3.2 2 \endpicture $$ $$ \beginpicture \setcoordinatesystem units <1cm,1cm> % sets scale \setplotarea x from -5 to 5, y from -2 to 2 % sets plot size up \put{$\bullet$} at -3 2 % \put{$\bullet$} at -2 2 % \put{$\bullet$} at -1 2 % \put{$\bullet$} at 0 2 % Top dots \put{$\bullet$} at 1 2 % \put{$\bullet$} at 2 2 % \put{$\bullet$} at 3 2 % \put{$\bullet$} at -3 0.5 % \put{$\bullet$} at -2 0.5 % \put{$\bullet$} at -1 0.5 % \put{$\bullet$} at 0 0.5 % Bottom dots \put{$\bullet$} at 1 0.5 % \put{$\bullet$} at 2 0.5 % \put{$\bullet$} at 3 0.5 % %Flagpole \plot -4.5 2.5 -4.5 1.62 / \plot -4.5 1.38 -4.5 1.12 / \plot -4.5 .88 -4.5 0 / \plot -4.25 2.5 -4.25 1.62 / \plot -4.25 1.38 -4.25 1.12 / \plot -4.25 .88 -4.25 0 / \ellipticalarc axes ratio 1:1 360 degrees from -4.5 2.5 center at -4.375 2.5 \put{$*$} at -4.375 2.5 \ellipticalarc axes ratio 1:1 180 degrees from -4.5 0 center at -4.375 0 % Vertical edges %\plot -3 2 -3 .5 / \plot -2 2 -2 .5 / \plot -1 2 -1 .5 / %\plot 0 2 0 .5 / %\plot 1 2 1 .5 / \plot 2 2 2 .5 / %\plot 3 2 3 .5 / \setquadratic %0 crossing \plot 3 .5 2.98 .7 2.8 .8 / %bottom corner \plot 3 2 2.98 1.8 2.8 1.7 / %top corner \setlinear %top edge \plot 2.8 1.7 2.15 1.7 / \plot 1.85 1.7 1.15 1.7 / \plot 0.85 1.7 0.15 1.7 / \plot -0.15 1.7 -0.85 1.7 / \plot -1.15 1.7 -1.85 1.7 / \plot -2.15 1.7 -2.85 1.7 / \plot -3.15 1.7 -4.1 1.7 / \ellipticalarc axes ratio 2:1 180 degrees from -4.65 1.7 center at -4.65 1.6 \plot -4.65 1.5 -3.8 1.5 / %bottom edge \plot 2.8 .8 2.15 .8 / \plot 1.85 .8 1.15 .8 / \plot 0.85 .8 0.15 .8 / \plot -0.15 .8 -0.85 .8 / \plot -1.15 .8 -1.85 .8 / \plot -2.15 .8 -2.85 .8 / \plot -3.15 .8 -4.1 .8 / \ellipticalarc axes ratio 2:1 180 degrees from -4.65 1 center at -4.65 .9 \plot -4.65 1 -3.8 1 / \setquadratic \plot -3.8 1.5 -3.7 1.45 -3.5 1.25 -3.05 .8 -3 .5 / \plot -3.4 1.35 -3.05 1.7 -3 2 / \plot -3.8 1 -3.7 1.05 -3.6 1.15 / % single crossing \plot 0 .5 .05 .8 .4 1.15 / \plot .6 1.35 .95 1.7 1 2 / \plot 0 2 .05 1.7 .5 1.25 .95 .8 1 .5 / \endpicture $$ $$ \beginpicture \setcoordinatesystem units <1cm,1cm> % sets scale \setplotarea x from -5 to 5, y from -2 to 2 % sets plot size up \put{$\bullet$} at -3 2 % \put{$\bullet$} at -2 2 % \put{$\bullet$} at -1 2 % \put{$\bullet$} at 0 2 % Top dots \put{$\bullet$} at 1 2 % \put{$\bullet$} at 2 2 % \put{$\bullet$} at 3 2 % \put{$\bullet$} at -3 .5 % \put{$\bullet$} at -2 .5 % \put{$\bullet$} at -1 .5 % \put{$\bullet$} at 0 .5 % Bottom dots \put{$\bullet$} at 1 .5 % \put{$\bullet$} at 2 .5 % \put{$\bullet$} at 3 .5 % %Border of annulus \plot -3 .5 3 .5 / \plot -3 -.5 3 -.5 / \ellipticalarc axes ratio 2:1 180 degrees from -3 .5 center at -3 0 \ellipticalarc axes ratio 2:1 180 degrees from 3 -.5 center at 3 0 \plot -3 2 3 2 / \plot -3 -2 3 -2 / \ellipticalarc axes ratio 5:4 180 degrees from -3 2 center at -3 0 \ellipticalarc axes ratio 5:4 180 degrees from 3 -2 center at 3 0 % Vertical edges %\plot -3 2 -3 .5 / \plot -2 2 -2 .5 / \plot -1 2 -1 .5 / %\plot 0 2 0 .5 / %\plot 1 2 1 .5 / \plot 2 2 2 .5 / %\plot 3 2 3 .5 / \setquadratic %0 crossing \plot 3 .5 3.02 .8 3.2 1 / %bottom corner \plot 3 2 3.05 1.7 3.2 1.5 / %top corner \plot -3.8 1.5 -3.7 1.45 -3.5 1.25 -3.05 .8 -3 .5 / \plot -3.4 1.35 -3.05 1.7 -3 2 / \plot -3.8 1 -3.7 1.05 -3.6 1.15 / \ellipticalarc axes ratio 1.8:2 180 degrees from -3.8 1 center at -3.8 0 \ellipticalarc axes ratio 2.8:2 180 degrees from 3.2 -1 center at 3.2 0 \ellipticalarc axes ratio 2.8:3 180 degrees from -3.8 1.5 center at -3.8 0 \ellipticalarc axes ratio 3.8:3 180 degrees from 3.2 -1.5 center at 3.2 0 \setlinear \plot -3.8 -1 3.2 -1 / \plot -3.8 -1.5 3.2 -1.5 / \setquadratic % single crossing \plot 0 .5 .05 .8 .4 1.15 / \plot .6 1.35 .95 1.7 1 2 / \plot 0 2 .05 1.7 .5 1.25 .95 .8 1 .5 / \endpicture $$ \bigskip \centerline{\bf Figure 1.} \vfill\eject \end