\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

