\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 the exponential function} \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{$ $} %My kludge to get the first page to move down a bit \vskip.75truein \centerline{\titlefont The exponential function} \bigskip \centerline{\rm Arun Ram} %${}^\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\bigskip Define the {\bf exponential function} as $$e^x = 1+x+{x^2\over 2!}+{x^3\over 3!}+{x^4\over 4!} +{x^5\over 5!}+{x^6\over 6!}+{x^7\over 7!} +\cdots,$$ where {\bf $k$-factorial} is $$k! = k(k-1)(k-2)\cdots 3\cdot 2\cdot 1, \quad\hbox{for $k=1,2,3,\ldots.$}$$ Why would anyone be so crazy as to write down such a horrible mess?? \bigskip\noindent {\bf Example:} Is there a function $$f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots$$ that changes addition into multiplication??, $$f(x)f(y)=f(x+y).$$ If so $$\eqalign{ f(x+y) &= c_0+c_1(x+y)+c_2(x+y)^2+c_3(x+y)^3+c_4(x+y)^4+c_5(x+y)^5+\cdots \cr &=\phantom{\ \ +c_4x^4+4c_4x^3y+6c_4} c_0 \cr &\phantom{=\ \ +c_4x^4+4c_4x^3y} +c_1x+c_1y+ \cr &\phantom{=\ \ +c_4x^4+6} +c_2x^2+2c_2xy+c_2y^2+ \cr &\phantom{=\ \ +c_4x} +c_3x^3+3c_3x^2y+3c_3xy^2+c_3y^3+ \cr &\phantom{=\ \ x} +c_4x^4+4c_4x^3y+6c_4x^2y^2+4c_4xy^3+c_4y^4+ \cr &+\cdots \cr } $$ must be equal to $$\eqalign{ f(x)f(y) &= (c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+\cdots) (c_0+c_1y+c_2y^2+c_3y^3+c_4y^4+\cdots) \cr &= c_0^2+c_0c_1x+c_0c_2x^2+c_0c_3x^3+c_0c_4x^4+\cdots \cr &\qquad\quad+c_0c_1y+c_1^2xy+c_1c_2x^3y+c_1c_4x^4y+\cdots \cr &\qquad\quad+\cdots. \cr }$$ Comparing terms in these two expressions gives $$\matrix{ c_0^2=c_0, &c_0c_1=c_1, &c_0c_2=c_2, &c_0c_3=c_3, &c_0c_4=c_4, &\ldots, \cr &c_0c_1=c_1, &c_1^2=2c_2, &c_1c_2=3c_3, &c_1c_3=4c_4, &c_1c_4=5c_5, &\ldots, \cr }$$ So $$c_0=1,\quad c_2={c_1^2\over 2},\quad c_1{c_1^2\over2} = 3c_3, \quad c_1{c_1^3\over3\cdot2} = 4c_4,\quad c_1{c_1^4\over4\cdot3\cdot2} = 5c_5, \quad\ldots.$$ So $$c_0=1,\quad c_2={c_1^2\over 2},\quad c_3 = {c_1^3\over 3\cdot2\cdot1},\quad c_4 = {c_1^4\over 4\cdot3\cdot2\cdot1},\quad c_5 = {c_1^5\over 5\cdot4\cdot3\cdot2\cdot1}, \quad\ldots.$$ So $$\eqalign{ f(x) &= 1+c_1x+{c_1^2\over2} x^2 + {c_1^3\over 3!} x^3 + {c_1^4\over 4!} x^4 + {c_1^5\over 5!} x^5 +\cdots \cr &= 1+c_1x+{(c_1x)^2\over2} + {(c_1x)^3\over 3!} + {(c_1x)^4\over 4!} + {(c_1x)^5\over 5!} +\cdots \cr &=e^{c_1x}. \cr}$$ So, $$\hbox{if}\quad f(x+y)=f(x)f(y)\quad \hbox{then}\quad f(x)=e^{c_1x}.$$ \bigskip\noindent {\bf Example:} Is there a function $$f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots$$ whose derivative is itself, $${df\over dx} = f \qquad\hbox{???}$$ If so, $$ f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots$$ must be equal to $${df\over dx} = c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4+6c_6x^5+\cdots.$$ Comparing terms in these two expressions gives $$ c_1=c_0,\quad 2c_2=c_1,\quad 3c_3=c_2,\quad 4c_4=c_3,\quad 5c_5=c_4,\quad 6c_6=c_5,\quad \ldots $$ So $$ \displaystyle{ c_2={c_0\over2}}, \quad \displaystyle{ 3c_3={c_0\over2}}, \quad \displaystyle{ 4c_4={c_0\over 3\cdot 2}}, \quad \displaystyle{ 5c_5={c_0\over 4\cdot 3\cdot 2}}, \quad\ldots. $$ So $$\eqalign{ f(x) &= c_0+c_0x+{c_0\over 2}x^2+{c_0^2\over 3!}x^3 +{c_0\over 4!}x^4+{c_0\over 5!}x^5+\cdots \cr &= c_0(1+x+{x^2\over 2}+{x^3\over 3!} +{x^4\over 4!}+{x^5\over 5!}+\cdots ) \cr &= c_0e^x. \cr } $$ So,\quad if $\displaystyle{{df\over dx}=f}$\quad then\quad $f = c_0e^x$. \medskip So $f(x)=e^x$ is the ONLY function such that $$e^{x+y} = e^x e^y \quad\hbox{and}\quad {de^x\over dx} = e^x.$$ \bigskip\noindent {\bf Example:} Find $e^0$. $$e^0 = 1+0+{0^2\over 2!}+{0^3\over 3!}+\cdots = 1+0+0+0+\cdots = 1.$$ \medskip\noindent {\bf Example:} Explain why $\displaystyle{ e^{-x} = {1\over e^x} }$. $$e^x e^{-x} = e^{x+(-x)} = e^{x-x}=e^0=1.$$ Divide both sides by $e^x$. $$\hbox{So}\quad e^{-x} = {1\over e^x}.$$ \bigskip\noindent {\bf Define} $$(e^x)^y = e^{xy}.$$ \vfill\eject \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