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\title{Calc Prep}
\author{JWR \& MWS}
\date{Aug 18, 2009}%
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\maketitle
 
\tableofcontents


\bigskip





Somewhere (here?) we should explain the terms
\jdef{variable}, \jdef{constant}, \jdef{unknown}, \jdef{identity},
\jdef{expand}, \jdef{simplify}, and the notations $\iff$, $\implies$, $:=$. 
  
\section{Laws of algebra}\label{sec:Laws_of_algebra}


\para\topic{Implicit Multiplication.} 
In mathematics the absence of an operation symbol usually
indicates multiplication: $ab$ mean $a\times b$. Sometimes
a dot is used to indicate multiplication and in computer languages
an asterisk is often used.
$$
    ab= a \cdot b = a*b=a\times b 
$$
\arap

\para \topic{Order of operations.}
Parentheses are used 
to indicate the order of doing the operations.
There are conventions which allow us not to write the parentheses.
For example, multiplication is done before addition
$$
ab+c\qquad\mbox{ means $(ab)+c$ and not $a(b+c)$},
$$
and powers are done before multiplication:
$$
ab^2c\qquad\mbox{ means $a(b^2)c$ and not $(ab)^2c$}.
$$
In the absence of other rules and parentheses,
the left most operations are done first. 
$$
a-b-c\qquad\mbox{ means $(a-b)-c$ and not $a-(b-c)$}.
$$
The long fraction line indicates that the division is done last:
$$
\frac{a+b}{c}\qquad\mbox{ means $(a+b)/c$ and not $a+(b/c)$}.
$$
In writing fractions the length of the fraction line indicates which fraction is evaluated first:
$$
\frac{\phantom{a}}{\phantom{\frac{b}{c}}}\!\!\!\!
\frac{\phantom{a}}{\phantom{\frac{b}{c}}}\!\!\!\!
\frac{a}{\frac{b}{c}}\!\!\!\!
\frac{\phantom{a}}{\phantom{\frac{b}{c}}}\!\!\!\!
\frac{\phantom{a}}{\phantom{\frac{b}{c}}}
 \qquad \mbox{ means $a/(b/c)$ and not $(a/b)/c$},
$$
$$
\frac{\phantom{\frac{a}{b}}}{\phantom{c}}\!\!\!\!
\frac{\phantom{\frac{a}{b}}}{\phantom{c}}\!\!\!\!
\frac{\frac{a}{b}}{c}\!\!\!\!
\frac{\phantom{\frac{a}{b}}}{\phantom{c}}\!\!\!\!
\frac{\phantom{\frac{a}{b}}}{\phantom{c}}
\qquad
\mbox{ means $(a/b)/c$ and not $a/(b/c)$}.
$$
The length of the horizontal line in the radical sign indicates the order of evaluation:
$$
\sqrt{a+b}\qquad\mbox{ means $\sqrt{(a+b)}$ and not $(\sqrt{a})+b$}.
$$
$$
\sqrt{a}+b\qquad\mbox{ means $(\sqrt{a})+b$ and not $\sqrt{(a+b)}$}.
$$
\arap

\para \topic{The Laws of Algebra.}  \label{LawsofAlgebra}
There are four fundamental operations
which can be performed on numbers.
\begin{enumerate}
\item Addition. The sum of $a$ and $b$ is denoted $a+b$.
\item Multiplication.  The product of $a$ and $b$ is denoted $ab$.
\item Reversing the sign. The negative of $a$ is denoted $-a$. 
\item Inverting.  The reciprocal of $a$ (for $a\ne0$) is denoted by $a^{-1}$
or by $\ds\frac1a$. 
\end{enumerate}
These operations satisfy the following laws. 

\begin{boxtable}{The Laws of Algebra}{1.8}{@{\qquad}ll@{\qquad\qquad}l}
Associative   & $a+(b+c)=(a+b)+c$ & $a(bc)=(ab)c$\\
Commutative & $a+(b+c)=(a+b)+c$ & $a(bc)=(ab)c$\\
Identity          &      $a+0=0+a=a$ & $a\cdot 1=1\cdot a=a$ \\
Inverse          &      $a+(-a)=(-a)+a=0$ & $a\cdot a^{-1}=a^{-1}\cdot a=1$ \\
Distributive   &  $a(b+c)=ab+ac$ & $(a+b)c=ac+bc$ 
\end{boxtable}
%
\medskipnoindent
The operations of \jdef{subtraction} and \jdef{division} are then defined by
\BOXIT{$\ds
 \quad   a-b:=a+(-b) \qquad \quad \qquad
      a\div b:= \frac{a}{b}:=a\cdot b^{-1}=a\cdot\frac{1}{b}. \quad
$}
All the rules of calculation that you learned in elementary
school follow from  the above  fundamental laws.
\arap

\para Because both addition and multiplication
satisfy the commutative, associative, identity,
and inverse laws, there are other analogies:
\medskipnoindent


\begin{boxtable}{Analogies between addition and multiplication}{1.8}{@{\qquad}ll@{\qquad\qquad}l@{\qquad\quad}}{\bf(i)} & $-(-a)=a$ & $ (a^{-1})^{-1}=a$ \\
{\bf(ii)}& $(a-b)+(c-d) = (a+c)-(b+d)$ & $\ds \frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$ \\
{\bf(iii)} &$a-b=(a+c)-(b+c)$ &$\ds \frac{a}{b}=\frac{ac}{bc}$ \\
{\bf(iv)} &$(a-b)-(c-d) = (a-b)+(c-d) $&$ \ds \frac{a/b}{c/d} =\frac{a}{b}\cdot\frac{c}{d}$\end{boxtable}

\medskipnoindent
These laws are proved in the
\href{https://webwork.math.wisc.edu/webwork2/mock_112/guided_proofs/8/}{Guided Exercises}.
%The last line explains why {\em  we invert and multiply to divide fractions}.
%$$
% \frac{a}{b}\div\frac{c}{d}:=\frac{a/b}{c/d} =\frac{a}{b}\cdot\frac{d}{c}.
%$$
\arap

\para  Here are some further laws which are proved using the  distributive law.
\medskipnoindent


\begin{boxtable}{Consequences of the distributive law}{1.8}{@{\quad}l@{\qquad}l@{\quad}}
{\bf(i)} $a\cdot0=0$  &    {\bf(ii)} $(-a)(-b)=ab$\\
{\bf(iii)} $-a=(-1)a$     & {\bf(iv)} $a(-b)=-ab$ \\
{\bf(v)} $\ds \frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$ &
{\bf(vi)} $(a+b)(c+d)=ab+ad+bc+bd$ \\
 {\bf(vii)} $(a+b)^2=a^2+2ab+b^2$& {\bf(viii)}  $(a+b)(a-b)=a^2-b^2$
\end{boxtable}
\medskipnoindent
These are also proved in the
\href{https://webwork.math.wisc.edu/webwork2/mock_112/guided_proofs/8/}{Guided Exercises}.
\arap
%\begin{proof}  Part~(f) is a 
%\href{https://webwork.math.wisc.edu/webwork2/mock_112/guided_proofs/8/}{Guided exercise}.
%% WARNING: This URL can easily change
%\end{proof}

\para 
 The following   \hypertarget{zpp}{\topic{Zero-Product Property}} will be used
to solve equations.

\BOXIT{
  $pq=0 \iff p=0\mbox{ or } q=0 \mbox{ (or both).}$
}

\noindent  We leave its proof as a Guided Exercise.
\arap


%\xrcs True or false?
%
%{\renewcommand{\arraystretch}{2.3}
%\begin{tabular}{ll}
%{\bf(i)} $(a+b)/c = (a/c)+(b/c)$\;?  & 
%  {\bf(ii)}  $ c/(a+b)= (c/a)+(c/b)\;? $\\
%{\bf(iii)} $\ds\frac{a+b}{c} = \frac{a}{c}+\frac{b}{c}$\;?&
% {\bf(iv)}   $\ds  \frac{c}{a+b}= \frac{c}{a}+\frac{c}{b}$\;?  \\
%{\bf(v)} $\ds \frac{a+b}{c+d}=\frac{a}{c}+\frac{b}{d}$\;? &
%  {\bf(vi)}   $\ds\frac{a\cdot b}{c\cdot d}=\frac{a}{c}\cdot\frac{b}{d}$\;?
%\end{tabular}
%}
%\scrx


\dfn For a natural number $n$ and any number $a$ the $n$th \jdef{power} of $a$ is
$$
   a^n:= \underbrace{a\cdot a\cdot a\cdots a}_{\mbox{ $n$ factors}}
$$
The zeroth power is
$$
   a^0:=1
$$
and negative powers are defined by
$$
  a^{-n}:= \frac1{a^n}.
$$
\nfd


\para\label{power_laws} 
The following laws are easy to understand when $m$ and $n$ are integers.
In Theorem~\ref{thm:a^x} below we will learn that these laws also hold whenever
$a$ and $b$ are positive real numbers and $m$ and $n$ are {\em any} real numbers,
not just integers.

\begin{boxtable}{Powers}{1.8}{@{\qquad}ll@{\qquad}ll@{\qquad}} 
&&&\\
{\bf (i)}& $a^ma^n=a^{m+n}$&  
e.g.  & $a^2a^3=(aa)(aaa)=a^5$ \\ 
{\bf(ii)} & $ \ds\left(a^m\right)^n = a^{mn}$ & 
e.g.  & $(a^2)^3=(aa)(aa)(aa)=a^6$\\ 
{\bf (iii)}& $ \ds\frac{a^m}{a^n}= a^{m-n}$ &
 e.g.  & $\ds \frac{a^2}{a^5}=a^{-3}=\frac1{a^3}$\\ 
{\bf(iv)} & $\ds (ab)^m= a^mb^m$ &
e.g. & $\ds(ab)^2=(ab)(ab)=(aa)(bb)=a^2b^2$ \\ 
{\bf(v)} & $\ds \left(\frac{a}{b}\right)^m=\frac{ a^m}{b^m}$ &
e.g. & $\ds\left(\frac{a}{b}\right)^2=\frac{a}{b}\cdot\frac{a}{b}=\frac{aa}{bb}=\frac{a^2}{b^2}$ \\
&&&
\end{boxtable}
\arap

\section{Kinds of Numbers}\label{sec:Kinds_of_Numbers}

\para  We distinguish the following different kinds of numbers.
\begin{itemize}
\item The \jdef{natural numbers} are  $1,2,3\ldots$.
\item  The \jdef{integers}  are  $\ldots -3,-2,-1,0,1,2,3\ldots$.
\item The \jdef{rational numbers} are ratios of integers like
$3/2$, $14/99$, $-1/2$.
\item  The \jdef{real numbers} are numbers which have an infinite decimal expansion like
$$
\frac32=1.5000\ldots,\quad  \frac{14}{99}=0.141414\ldots, \quad
\sqrt2=1.4142135623730951\ldots.
$$
\item  The \jdef{complex numbers} are those numbers of form $z=x+iy$
where $x$ and $y$ are real numbers and $i$ is a special new number called the
\jdef{imaginary unit} which has the property that
\BOXIT{
$
     i^2=-1;
$}
\end{itemize}
Every integer is a rational number (because $n=n/1$),
every rational number is a real number  
(see Remark~\ref{rmk:real-rational} below),
and every real number is a complex number (because $x=x+0i$). 
A real number which is not rational is called \jdef{irrational}.
\arap

\para
Each kind of number enables us to solve equations that the previous kind couldn't solve:
\begin{itemize}
\item 
The solution of the equation  $x+5=3$ is $x=-2$ which is an integer
but not a natural number.
\item 
The solution of the equation  $5x=3$ is $x=\frac{3}{5}$ which is a
rational number  but not an integer.
\item 
The   equation  $x^2=2$ has two solutions $x=\sqrt{2}$. The number $\sqrt{2}$ is a
real number  but not a rational number. (See Remark~\ref{rmk:sqrt2} below.)
\item 
The equation $x^2=4$ has two real solutions $x=\pm2$ but
the equation $z^2=-4$ has no real solutions because the square of a nonzero
real number is always positive. However it does have two complex solutions,
namely $z=\pm 2i$.
\end{itemize}
We will not use complex numbers until Section~\ref{sec:Complex_Numbers} 
but may refer to them implicitly as in 

\begin{center}\em
The equation $x^2=-4$ has no (real) solution.
\end{center}
\arap

 


\rmk \label{rmk:real-rational}
It will be proved in Theorem~\ref{thm:real-rational} that a
real number is rational if and only if its
decimal expansion eventually repeats periodically forever as in the following examples:
$$
\renewcommand{\arraystretch}{2.3}
\begin{array}{lcl}
\ds\frac{1}{3}=0.3333\ldots ,&\qquad&
\ds\frac{17}{6}=2.83333\ldots, \\
\ds\frac{7}{4}=1.250000\ldots, &\qquad &
\ds\frac{1}{7}=0.142857\,142857\,142857\ldots.
\end{array}
$$ 
\kmr

\rmk Unless the decimal expansion of a real number is eventually zero,
as in $\tfrac12=0.5000\ldots$,
any finite part of the decimal expansion is close to, but not exactly
equal to, the real number. For example $1.414$ is close to the square
root of two but not exactly equal:
$$
  (1.414)^2=1.999396\ne 2,\qquad (\sqrt2)^2=2.
$$
If we compute the square root to more decimal places we get a better approximation, but it still isn't exactly correct:
$$
(1.4142135623730951)^2=2.00000000000000014481069235364401.
$$
\kmr

\rmk \label{rmk:sqrt2}  Here is a proof that the square root of $2$ is irrational.
If it were rational there would be integers $m$ and $n$ with
$$
  \left(\frac{m}{n}\right)^2 =2.
$$
By canceling common factors we may assume that $m$ and $n$ have
no common factors and hence that they are not both even.
Now  $m^2=2n^2$ so $m^2$ is even so $m$ is even, say $m=2p$.
Then $4p^2=(2p)^2=m^2=2n^2$ so $2p^2=n^2$ so $n^2$ is even so $n$ is even.
This contradicts the fact   $m$ and $n$ are not both even.
\kmr

\section{Coordinates on the Line and Order}\label{sec:Coordinates_on_the_Line_and_Order}


\para\label{line-coords}
The choice of two points $O$ and $I$ on a line $\ell$
determines a correspondence between the points of the line and the real numbers as indicated in the following picture.

\input{line-coord}

\medskipnoindent 
The correspondence is called a \jdef{coordinate system} on the line $\ell$. When the point $A$ corresponds to the number $a$
 we say that the number $a$ is the \jdef{coordinate} of the point $A$.
The correspondence is characterized by the following properties:
\begin{enumerate}
\item The coordinate of the point $O$ is the number $0$.
\item The coordinate of the point $I$ is the number $1$.
\item If the coordinates of the points $A$ and $B$ are
the numbers $a$ and $b$, then the coordinate of the
 \jdef{midpoint}  $M$ of the segment $AB$ is 
 the average  $\ds \frac{a+b}{2}$.
 \item If three distinct points $A$, $B$, $C$
 have coordinates $a$, $b$, $c$ respectively,
 and $C$ lies on the line segment $AB$, then the real number
 $c$ lies between the real numbers $a$ and $b$,
 \end{enumerate}
\arap

\para The \jdef{positive numbers} are the real numbers on the same side
of $0$ as $1$ and the \jdef{negative numbers} are on the other side.
We write say $a$ is \jdef{less than} $b$ and write $a<b$  when $b-a$ is
positive; it is equivalent to say that $b$ is \jdef{greater than} $a$
and to write $b>a$. The notation  $a\le b$ means that 
$a$ is less than or equal  to $b$ i.e. either $a<b$ or else $a=b$.
Similarly,  $b\ge a$ means that 
$b$ is greater than or equal to $a$ i.e. either $b>a$ or else $b=a$.
Thus  when $a<b$,  a number
\footnote{The notation
$\iff$ is an abbreviation for ``if and only if''.}
$$
 \mbox{$c$ is between $a$ and $b$} \iff a<c<b.
$$
Sometimes we insert the word \jdef{strictly} for emphasis:
$a$ is strictly less than $b$ means that $a<b$ (not just  $a\le b$).
\arap

\para The \jdef{order relation} just described is characterized by
the following.
\begin{description}
\item[(Trichotomy)] Every real number is either positive, negative,
or zero (and no number satisfies two of these conditions).
\item[(Sum)] The sum of two positive numbers is positive.
\item[(Product)]  The product of two positive numbers is positive.
\end{description}
This characterization together with the notation explained
in the previous paragraph implies the following:

\begin{boxtable}{Order}{1.8}{@{\qquad\qquad \qquad}ll@{\qquad\qquad \qquad}}&\\
{\bf(i)} & If $a<b$ and $b<c$, then $a<c$. \\
{\bf(ii)} & If $a<b$, then $a+c<b+c$. \\
{\bf(iii)} & If $a<b$ and $c>0$, then $ac<bc$.\\
{\bf(iv)} & If $a<b$ and $c<0$, then $ac>bc$.\\
{\bf(v)} & If $0<a<b$, then $\ds0<\frac1b<\frac1a$. \\ & 
\end{boxtable}
\arap

\medskipnoindent

\para\topic{Interval Notation}. 
The \jdef{open interval} $(a,b)$ is the set of all real numbers $x$ such that $a<x<b$,
and the  \jdef{closed interval} $[a,b]$ is the set of all real numbers $x$ such that $a\le x\le b$. 
Thus
$$
   x \mbox{ is in the set } (a,b) \iff a<x<b
$$ 
and
$$
   x \mbox{ is in the set } [a,b] \iff a\le x\le b.
$$ 
These notations are extended to include \jdef{half open intervals}
and \jdef{unbounded  intervals} as in
$$
\renewcommand{\arraystretch}{1.5}
\begin{array}{lcl}
   x \mbox{ is in the set } (a,b] &\iff& a<x\le b, \\
   x \mbox{ is in the set } (a,\infty)&\iff& a<  x,\\
   x \mbox{ is in the set } (-\infty,a] &\iff& x\le a, \mbox{\qquad etc.}
\end{array}
$$
The \jdef{union} symbol $\cup$ is used to denote a set consisting of more than one interval
as in
$$
   x \mbox{ is in the set }(a,b)\cup (c,\infty) \iff 
   \mbox{ either } a<x<b \mbox{ or else } c<x.
$$ 
The symbol $\infty$ is pronounced \jdef{infinity} and is used to indicate that an interval is unbounded.
It is not a number so  we never write $(c,\infty]$.
\arap

\para Here are some pictures of sets. In these pictures 
the set is indicated by  thickening the corresponding part(s) of the line, 
parentheses are used to indicated an endpoint which is not in the interval,
brackets are used to indicated an endpoint which is in the interval,
and the arrow indicates that the interval extends to infinity.

\input{intervals}
\arap

\para  The \jdef{absolute value} of a real number $a$ is defined
by $|a|$ and defined by
$$
|a|:=\left\{\begin{array}{rl}
     a & \mbox{ if $a\ge 0$}\\
   -a & \mbox{ if $a< 0$}
   \end{array}\right.
$$
Since the $\sqrt{}$ denotes the nonnegative square root, an equivalent
definition is 

\BOXIT{ $|a|=\sqrt{a^2}$}

\noindent
The \jdef{distance}    between two numbers $a$ and $b$ is the absolute value $|a-b|$ of their difference. Thus the absolute value of a real number is its distance from $0$.
\arap


\para \label{midpoint-equidistant}
The average $c=\tfrac12(a+b)$ of two real numbers $a$ and $b$
is the midpoint of the interval  $[a,b]$.
The following calculation shows that 
the distance from $a$ to $c$ is the
same as the distance from  $b$ to $c$:
$$
  |a-c|=\left|a-\frac{a+b}2\right|=\left|\frac{2a-(a+b)}2\right|=
  \left|\frac{a-b}2\right|
$$
and
$$
   |b-c|=\left|b-\frac{a+b}2\right|=\left|\frac{2b-(a+b)}2\right|=
  \left|\frac{b-a}2\right|
$$
so $ |a-c|=\tfrac12|a-b|=\tfrac12|b-a|= |b-c|$. 
% \input{line-midpoint} 
\arap

\section{Exponents}\label{sec:Exponents}


The proof of the following theorem requires a more careful
definition of the set of real numbers than we have given
and is best left for more advanced courses. 

\begin{theorem}\label{thm:a^x}  Suppose that $a$ is a positive real number.
Then there is one and only one  way to define $a^x$ for all real  numbers $x$
such that
%%
\begin{description}
\item[(i)] $a^{x+y}=a^x\cdot a^y$,\;\;  $a^0=1$, \;\; $a^1=a$, \;\;$1^x=1$.
\item[(ii)] If $a>1$ and $x<y$ then $a^x<a^y$.
\item[(iii)] If $a<1$ and $x<y$ then $a^x>a^y$.
\end{description}
With this definition, the  power laws in Paragraph~\ref{power_laws}
continue to hold when $a$ and $b$ are positive real numbers and
$m$ and $n$ are arbitrary real numbers. The number $a^x$ is positive
(when $a$ is positive) regardless of the sign of $x$.
\end{theorem}




\para In particular by property~(v) in Paragraph~\ref{power_laws} we have $(a^x)^y=a^{xy}$ so $(a^{m/n})^n=a^m$ and $(a^m)^{1/n}=a^{m/n}$.
Hence for positive numbers $a$ and $b$ we have
$$
    b=a^{m/n}\iff  b^n=a^m.
$$
When $m=1$ and $n$ is a natural number the number $a^{1/n}$ is called the $n$th \jdef{root}
(\jdef{square root} if $n=2$ and \jdef{cube root} if $n=3$) and is sometimes denoted
$$
    \sqrt[n]{a}:= a^{1/n}.
$$
When $n$ is absent, $n=2$ is understood:
$$
    \sqrt{a}:= a^{1/2}.
$$
\arap


\rmk    
A number  $b$ is said to be an  \jdef{$n$th root} of $a$ iff $b^n=a$.  
When $n$ is odd,  every real number $a$ has exactly one  (real) $n$th root
and this is denoted by $\sqrt[n]{a}$.
When $n$ is even, a positive real number $a$ has two (real) $n$th roots
(and $\sqrt[n]{a}$ denotes the one which is positive)
but a negative number has no real $n$th roots.  
(In trigonometry it is proved that every nonzero complex number has
exactly $n$ distinct  complex $n$th roots.)  

The equation
$b^2=9$ has two solutions, namely $b=3$ and $b=-3$ and each is
``a'' square root of $9$ but only $b=3$ is ``the''   square root of $9$.
However $-2$ is the (only) real cube root of $-8$ because $(-2)^3=-8$.
The number $-9$ has no real square root (because $b^2=(-b)^2>0$
if $b\ne0$)  but does have two complex square roots
(because $(3i)^2=(-3i)^2=-9$). 
For most of this course we only use real  numbers
and we say that 
%
\begin{center}
$\sqrt{a}$  is undefined when $a<0$
\end{center}
%
and that 
\begin{center}\em
you can't take the square root of a negative number.
\end{center}

\medskipnoindent Also $\sqrt{a}$ always denotes the nonnegative square root:
thus $(-3)^2=3^2=9$ but $\sqrt{9}=3$ and  $\sqrt{9}\ne-3$.
\kmr

 
\section{Coordinates in the Plane and Graphs}\label{sec:Coordinates_in_the_Plane_and_Graphs}


\para \label{rect-coord}
In Paragraph~\ref{line-coords} we saw that a choice of two points $O$ and $I$
on a line $\ell$ determined a correspondence between the set of real numbers 
and the points of $\ell$. Similarly two perpendicular lines $\ell_1$ and $\ell_2$
intersecting in a point $O$ and points $I_1$ on $\ell_1$ and $I_2$ on $\ell_2$
determine a correspondence between pairs $(x,y)$ of real numbers and points $P$
on the plane of $\ell_1$ and $\ell_2$.  Given a point $P$ in the plane
we draw a line parallel to $\ell_1$ through $P$ intersecting $\ell_2$ in $Y$
and a line parallel to $\ell_2$ through $P$ intersecting $\ell_1$ in $X$.
The point $X$ has coordinate $x$ in $\ell_1$ and
the point $Y$ has coordinate $y$ in $\ell_2$. 
The point $P$ then corresponds to the pair $(x,y)$. 
We  say that the \jdef{coordinates} of $P$ are $(x,y)$. 
The notation $P(x,y)$ is used as an abbreviation for the more
cumbersome phrase ``the point $P$ whose coordinates are $(x,y)$.''
The lines $\ell_1$ and $\ell_2$ are called the \jdef{coordinate axes}
and the point $O$ is called the \jdef{origin}.
The point $I_1$ has coordinates $(1,0)$,
the point $I_2$ has coordinates $(0,1)$,
and the origin $O$ has coordinates $(0,0)$.
This correspondence between
points $P$ in the plane and pairs $(x,y)$ 
is called a \jdef{rectangular coordinate system}
(or sometimes a \jdef{Cartesian coordinate system} 
in honor of it's inventor Ren\'e Descartes). 

\input{rect-coord} 
%
\hspace{0.1in}
%
\input{23-32}

\medskipnoindent   
\begin{minipage}[b]{0.5\textwidth} %\the\textwidth=390pt=5.41667in
The coordinate axes divide the plane into
four parts called \jdef{quadrants}. The first
quadrant is the set of points $P(x,y)$ where
both coordinates are positive and the remaining
quadrants are numbered consecutively in the counter
clockwise direction as in the diagram.
\vspace{0.1in}
\end{minipage}
\hspace{0.1in}
%
\input{quadrants}
%
\bigskipnoindent
\arap



\dfn \label{dfn:graph-equation}
The \jdef{graph} of an equation in the variables $x$ and $y$ 
is the set of all points $P(x,y)$ such that $(x,y)$ satisfies the equation.
Here are two examples. 

\input{2line} 
%
\hspace{0.1in}
%
\input{unitcircle}
\nfd

\rmk We can use any two variables, but usually we use $x$ and $y$. 
We always label the axes and write the equation to avoid confusion.
When using the variables $x$ and $y$ we will usually call 
the first coordinate of a point the \jdef{$x$-coordinate} and 
the second coordinate the \jdef{$y$-coordinate}. For example,
the $x$-coordinate of the point $P(2,3)$ is $2$ and the $y$-coordinate is $3$.
Also we call the horizontal axis (i.e. the line $\ell_1$ in Paragraph~\ref{rect-coord})
the \jdef{$x$-axis} and the vertical axis (i.e. the line $\ell_2$) the $y$-axis.
\kmr


\rmk
When we are using coordinates to study geometry it is most natural
to make the scale on the $x$-axis the same as the scale on the $y$-axis,
i.e. to choose the points $O$, $I_1$, $I_2$ in~\ref{rect-coord} so that 
the distance from $O$ to $I_1$ is the same as
the distance from $O$ to $I_2$ and that this distance is one unit.
We will often assume this without comment.
In other problems this is not so natural.
Here is
a graph showing the  height $y$  in feet of an object $t$ seconds
after it is thrown into the air and
another graph  the showing relation between 
the temperature  $F$ in Fahrenheit and the 
temperature  $C$ in Celsius. Both graphs are graphs of functions
and the units on the vertical axis are different from the units
on the horizontal axis.

\bigskip

\input{height-time} 
%
\hspace{1in}
%
\input{fahrenheit-celsius}
\kmr


\begin{theorem}[\topic{The Distance Formula}] 
The distance $d(P_1,P_2)$
between the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$
is given by

\BOXIT{ 
   $ d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
}
\end{theorem}


\begin{proof} The point $Q(x_2,y_1)$ has the same $y$-coordinate
as $P_1$ and the same $y$-coordinate as $P_2$. The
points $P_1$, $P_2$, $Q$ are the vertices of a right triangle

\bigskip

\input{distance-formula}

\bigskip

\noindent with legs $QP_1$ and $QP_2$
of length $a=|x_1-x_2|$ and $b=|y_1-y_2|$.
The length $c$ of the hypotenuse $P_1P_2$
is the distance $d(P_1,P_2)$ from $P_1$ to $P_2$.
Hence the distance formula follows from
the Pythogorean Theorem $a^2+b^2=c^2$.
\end{proof}

\para The \jdef{circle} with center $C$ and radius $r$
is the set of all points $P$ such that the distance
$d(C,P)$ from $C$ to $P$ is exactly $r$.
If $C=C(h,k)$ and $P=P(x,y)$ the condition that
$P$ lies on this circle may be written
$$
      \sqrt{(x-h)^2+(y-k)^2}=d(C,P)=r.
$$
Squaring both sides we see that the circle is the graph of the equation

\BOXIT{ $(x-h)^2+(y-k)^2=r^2$.}
\arap

\begin{theorem}[\topic{The Midpoint Formula}]\label{thm:plane-midpoint}
The midpoint $M$ of the the line segment joining
the points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is
computed by averaging the cordinates: 
$$
    M = \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
 $$
\end{theorem}

\begin{proof} There are two ways to prove this. First we can compute
the distance 
$d(P_1,M)$ from $P_1$ to $M$ and the distance $d(P_2,M)$ from
$P_2$ to $M$ and check that the are equal:
$$
\begin{aligned}
  d(P_1,M)&=\sqrt{\left(x_1-\frac{x_1+x_2}2\right)^2+\left(y_1-\frac{y_1+y_2}2\right)^2}\\
  &=\sqrt{\left(\frac{x_1-x_2}2\right)^2+\left(\frac{y_1-y_2}2\right)^2}\\
  &= \tfrac12 d(P_1,P_2)
\end{aligned}
$$
and similarly $d(P_2,M)=\tfrac12 d(P_1,P_2)$.

\vspace*{-0.2in}


\medskipnoindent
\begin{minipage}[b]{0.5\textwidth}
\hfill The second way is to use congruent\\ triangles
as in the diagram to the right.
The point $Q(x_2,y_1)$ is the vertex of a right triangle $P_1QP_2$.
The point $A(\tfrac12(x_1+x_2),y_1)$ is the midpoint
of the line segment $P_1Q$ and 
the point $B(x_1,\tfrac12(y_1+y_2))$ is the midpoint
of the line segment $P_1Q$.   
(See Paragraphs~\ref{line-coords} and~\ref{midpoint-equidistant}.)
\hfill The point $M$ is a vertex of
\end{minipage}
%
\hspace{0.3in}
%
\input{plane-midpoint}

\noindent
the congruent right triangles $P_1AM$ and $MBP_2$
so $d(P_1M)=d(P_2M)$.
\end{proof}

\dfn   Let $\ell$ be a line and $F$ be a point not on $\ell$.
The  \jdef{parabola} with \jdef{focus} $F$ and \jdef{directrix} $\ell$
is the set of points $P$ which are equidistant from $F$ and $\ell$.
The line through $F$ and perpendicular to $\ell$ is called the
\jdef{axis} of the parabola and the point where the axis
intersects the parabola is called the \jdef{vertex} of the parabola.
%(The distance from a point $P$ to a line $\ell$ is the distance
%from $P$ to the closest point $Q$ on $\ell$. The closest point
%to $Q$ on $\ell$ is the point where the perpendicular to $\ell$ through $P$  meets $\ell$.)
\nfd

\begin{theorem}[\topic{Parabola Formula}] If the directrix is parallel to the $x$-axis
then the parabola is the graph of the equation of form

\BOXIT{$\ds y=a(x-h)^2+k.$}

\noindent The vertex is $V(h,k)$.
\end{theorem}

\begin{proof} Guided Exercise. % Make sure to emphasize reversibility
\end{proof}

\xmpl 
If the directrix is the horizontal line $y=-\tfrac14$ and
the focus is the point $F(\tfrac14,0)$, then the parabola
is the graph of the equation
$$
     y=x^2.
$$

\noindent
\begin{minipage}[b]{0.45\textwidth}  To see this assume
that the point $P(x,y)$ lies on the parabola.
The distance from the point $P(x,y)$ to the point $F$
is 
$$
 d(P,F)=\sqrt{(x^2+(y-\tfrac14)^2}.
$$
The distance from $P$ to $\ell$ is
$$
 d(P,\ell)=\left|y+\tfrac14\right|.
$$
\end{minipage}
%
\hfill
%
\input{parabola}
%

\medskipnoindent
Squaring both sides of the equation $d(P,\ell)=d(P,F)$ gives
$$
\left(y+\tfrac14\right)^2=x^2+(y-\tfrac14)^2.
$$
Expand and cancel to get $y=x^2$.  
%Conversely if $y=x^2$,
%then  we can do the steps backward to prove that $d(P,\ell)=d(P,F)$.
In this example, the vertex is at the origin.
\lpmx

\rmk A \jdef{parabolic miroir} is a mirror in the shape of a parabola.
If the sun's rays are directed perpendicular to the directrix, i.e. parallel
to the axis, then they reflect so that the angle of incidence  equals
the angle  of reflection. Using calculus it can be proved that
all the rays pass through the focus. The diagram shows the tangent line $\ell$ to the parabola at the point $P$ and the focus $F$ of the parabola. The line perpendicular to $\ell$ through $P$ is called the
{\em normal}. The ray from the sun is the vertical line through $P$.
The angle between the vertical and the normal is called the
{\em angle of incidence} and the angle between the normal and
the reflected ray is called the {\em angle of reflection}.  For the
parabola, the reflected ray goes through the focus.

\medskipnoindent
\input{parabolic-1}
%
\input{parabolic-2}

The same process works in reverse. Movie projectors used
to have a carbon arc lamp located at the focus of a parabolic mirror 
so that the light
would reflect from the mirror and pass through 
the lens in parallel supplying a uniformly illuminated image.
A whispering chamber works on the same principle. There is one
in the Museum of Science and Industry in Chicago. Parabolic
mirrors are located at opposite ends of a long hall. If you stand
at the focus of one of them and your friend stands at the focus
of the other, each of you can hear the other whisper, even though
people only a few yards away cannot hear you. (The entire room
is in the shape of an ellipse which has similar reflective properties.)
\kmr

\section{Lines}\label{sec:Lines}

\dfn  The \jdef{slope} of the line through the distinct points
$P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is

\BOXIT{
$\ds   m = \frac{y_2-y_1}{x_2-x_1}.$
}
In calculus this is sometimes written as 
$$
   m=\frac{\Delta y}{\Delta x}, \qquad 
   \Delta y= y_2-y_1,\quad \Delta x=x_2-x_1
$$
and described as the 
``change in $y$ divided by the change in $x$''.

\input{slope-defined}
%\hspace{12pt}
\hfill
\input{slope-welldefined}

\noindent
As we move from $P_1$ to $P_2$ the $x$-coordinate
``runs'' form $x_1$ to $x_2$ and the $y$-coordinate
``rises'' from $y_1$ to $y_2$, so the slope is sometimes
described as \jdef{rise over run}. If the $y$ coordinate
decreases as the $x$ coordinate increases, the slope
is negative.

The slope depends on the coordinate system but
is independent of the choice of the pair of distinct points on the
line. If $P'_1(x'_1,y'_1)$ and $P'_2(x'_2,y'_2)$ are two other points 
on the line and
$Q=Q(x_1,y_2)$,  $Q'=Q'(x'_2,y'_1)$ then
$$
    \frac{y_2-y_1}{x_2-x_1}=\frac{y'_2-y'_1}{x'_2-x'_1}
$$
as the triangles $P_1QP_2$ and $P'_1Q'P'_2$ are similar
right triangles.
\nfd

\para A \jdef{vertical line} is one which is parallel to
the $y$-axis. The $x$-coordinate is constant along such
a line so it has and equation of form $x=a$. For any
two points on  a vertical line we have $\Delta x=a-a=0$.
Hence the slope of a vertical line is not defined since
we never divide by zero. 
A \jdef{horizontal line} is one which is parallel to the $y$-axis. 
It has an equation of form $y=b$. The slope of a horizontal 
line is zero (which is defined).
\arap


\para \topic{The Point-Slope  Equation of a Line}.\label{point-slope}
Let $P_0(x_0,y_0)$ be a point on a line $\ell$ of slope $m$. 
Since any two points on the line can be used to define the slope 
we see that  a point $P(x,y)$  which llies on the line $\ell$
satisfies 
$$
        \frac{y-y_0}{x-x_0}=m.
$$
This equation has one minor flaw; 
it doesn't work when $(x,y)=(x_0,y_0)$
(never divide by zero). 
To remedy this multiply by $(x-x_0)$ and add
$y_0$ to both sides:

\BOXIT{
$y = y_0+m(x-x_0).$
}

\noindent This is the \jdef{point-slope form} of the equation for 
{\em line through $P_0(x_0,y_0)$ with slope $m$};
this form  makes it obvious that the point $P_0(x_0,y_0)$ lies on the line.
For example, the equation for the line through 
$P_0(2,3)$ and $P_1(4,11)$ is
$$
        \frac{y-3}{x-2}= \frac{11-3}{4-2}=4, \qquad \mbox{ or } \qquad
      y = 3+4(x-2),
$$
When $x_0=0$ and $y_0=b$  the point-slope form becomes
$$
   y=mx+b.
$$
This is called the \jdef{slope-intercept} form of the equation
for the line because the point $(0,b)$ is the point where
the line intercepts (intersects) the $y$-axis.
\arap

\begin{theorem}[\topic{Parallel Lines}]\label{thm:slope-parallel}
Two lines have the same slope if and only if 
they are parallel (or the same).
\end{theorem}

\begin{proof}
Assume  two lines have the same slope $m$.

\noindent
\begin{minipage}[b]{0.5\textwidth}
Then their slope intercept forms are $y=mx+b$ and $y=mx+b'$
If some point $(x,y)$ lies on
both lines then
$mx+b=y=mx+b'$ so $b=b'$ and the lines
are the same, Hence if $b\ne b'$ the lines do
not intersect, i.e. they are parallel. Conversely
if the  lines are parallel, then in the picture
to the right the two triangles are similar
so the the hypotenuses  are parallel.
\end{minipage}
%
\hspace{0.2in}
%
\input{slope-parallel}
\end{proof}

\begin{theorem}[\topic{Perpendicular Lines}]
Two lines are perpendicular if and only if the slope
of one is the negative reciprocal of the slope of the other.
\end{theorem}

\begin{proof}
If we draw the parallels to the two lines through the
origin we get two new lines with
(by Theorem~\ref{thm:slope-parallel})  the same slopes
and which are perpendicular if and only if the original lines
are. Hence we might as well assume the two lines
pass through the origin. \quad The  diagram shows two lines
with slopes $m$

\noindent
\begin{minipage}[b]{0.5\textwidth}
and $-m^{-1}$ and passing through the origin.
The point $P(a,b)$ lies on the line $y=mx$ so $b=ma$,
and hence $-a=-m^{-1}b$ so the point $(-b,a)$ lies on 
the line $y=-m^{-1}x$. Hence the two right triangles
$OAP$ and $OBQ$ are congruent. 
The two  acute angles in  a right triangle are complementary
so the angle $AOP$ is the complement of the angle $BQO$.
Since $QB$ is parallel to the $x$-axis the angle $QO$
makes with the $x$-axis is the complement to $POA$.
Hence $POQ$ is a right angle.
\end{minipage}
%
%\hspace{0.1in}
\input{slope-perpendicular}
\end{proof}


\section{Solving Equations}\label{sec:Solving_Equations}

\dfn \label{dfn:equivalent}  A number $a$ is called a \jdef{solution}\footnote{A solution of an equation
is sometimes also called a \jdef{root} of the equation.}
of an equation containing the variable $x$ if the equation becomes a true statement
when $a$ is substituted for $x$.
Two equations are said to be \jdef{equivalent}
iff they have exactly the same solutions.
We will sometimes use the symbol $\iff$ to indicate that two equations
are equivalent.
\nfd

\para \label{equiv1} Usually two equations are equivalent
 because one can be obtained from the other
by performing an operation to both sides of the equation
which can be reversed by another operation of the same kind.
 For example, the equations $3x+7=13$ and $x=2$ 
are equivalent because
$$
\begin{aligned}
    3x+7=13 &\iff& 3x =6 & \mbox{ (subtract $7$ from both sides),}\\
                   &\iff&  x =2 & \mbox{ (divide both sides by $3$).}
\end{aligned}
$$
The reasoning is reversible: we can go from $x=2$ to $3x=6$
by multiplying both sides by by $3$ and from $3x=6$ to $3x+7=13$
by adding $7$ to both sides.
\arap

\para We use the symbol $\implies$
when we want to assert that one equation \jdef{implies} another but 
do not want to assert the converse. The guiding principal here is
%
\begin{quote}\em
   If an equation $E'$ results from an equation $E$
   by performing the same operation to both sides,
   then $E\implies E'$, i.e. every solution of $E$ is a solution of $E'$.
\end{quote}
%
If the operation is not ``reversible'' as explained above,
there is the possibility that the set of solutions gets bigger
in which case the new solutions are called \jdef{extraneous solutions}.
(They do not solve the original equation.)  
The simplest example of how an extraneous solution  can arise is
$$
    x=3 \implies x^2=9 \qquad\mbox{(square both sides)}
$$
but  the  operation of squaring both sides is not  reversible:
it is incorrect
to conclude that $x^2=9$  implies that $x=3$. 
What {\em is} correct is that $x^2=9\iff x=\pm3$, 
i.e. {\em either $x=3$ or else $x=-3$}.
When solving an equation you may use operations which are not reversible provided that you
%
\BOXIT{
\em
Always check your answer!
}
%
(In addition to catching mistakes, this will show you which 
-- if any -- of the solutions you found are extraneous.)
\arap

\para Here are two ways in which extraneous solutions can arise:
\begin{description}
\item[(i)] Squaring both sides of an equation.
\item[(ii)] Multiplying both sides of of an 
equation by a quantity not known to be nonzero.
\end{description}

As an example of~(i) consider the equation
$$
  \sqrt{10-x}=-x-2.
$$
Squaring both sides gives the quadratic equation $10-x=x^2+4x+4$
which has two solutions $x=-6$ and $x=1$. Now $\sqrt{10-(-6)}=-(-6)-2$
but $\sqrt{10-1}\ne -1-2$.  ($\sqrt{}$ means the positive square root.)
Thus $x=-6$ is the only solution of the original equation and $x=1$
is an extraneous solution.

As an example of~(ii) consider
$$
    \frac{1}{x-1}= 2+\frac{1}{x-1}.
$$
This equation has no solution: if it did we would subtract $(x-1)^{-1}$ from
both sides and deduce that $0=2$ which is false.  But if we multiply both sides by
$x-1$ we get $1=2(x-1)+1$ which has the (extraneous) solution $x=1$.
\arap

\rmk The symbols $\iff$ and $\implies$ relate equations (or more generally
statements) not numbers. The notation  $P\implies Q$ means
that $Q$ is true if $P$ true. The notation   $x\implies 5$ is nonsense.
On exams always write complete sentences (e.g. $x=2$ or ``the answer is 2'') never
just numbers (nouns are neither true nor false).
\kmr
 
\para \label{equiv2} 
The definition of equivalent equations given in~\ref{dfn:equivalent}
applies to equations in two variables where the solution set is (usually) infinite:
A pair of numbers $(a,b)$ is a \jdef{solution} of an equation containing the variables
$x$ and $y$ if the equation becomes a true statement when $a$ is substituted for $x$
and $b$ is substituted for $y$. By Definition~\ref{dfn:graph-equation}
% 
   \begin{thmbox} \em
     Two equations  are equivalent if and
     only if they have the same graph.
   \end{thmbox} 
%
Again we usually show that two equations are equivalent
by transforming one to the other by operations which can be reversed
by another operation of the same kind.
As a simple example
$$
    6x+3y=12\iff 3y=-6x+12\iff y=-2x+4
$$
so the equations have the same graph, namely the line of slope $-2$
through the point $(0,4)$.
\arap

\section{Systems of Equations}\label{sec:Systems_of_Equations}

\para \label{equiv3} A \jdef{solution} to a  \jdef{system} of two equations in two
variables $x$ and $y$ is a pair $(a,b)$ of numbers which is a solution
of each of the two equations. Each equation has a graph and the solutions
to the system are the point where the two graphs intersect. Two systems
are \jdef{equivalent} iff they have exactly the same solutions. As in
Paragraphs~\ref{equiv1} and~\ref{equiv2} we usually show that two systems
are equivalent by performing an operation on one system which can be reversed
by another operation of the same kind.
\arap

\xmpl \label{xmpl:21-11} As an example we solve the system
$ 3x+3y=21$, $x+2y=11$.

\begin{eqnarray}
\begin{array}{r}
 3x+3y=21\\
x+2y=11
\end{array}
&\iff&
\begin{array}{l}
x+y=7\\
x+2y=11
\end{array}
\\ &\iff&
\begin{array}{l}
x+y=7\\
y=4
\end{array}
\\&\iff&
\begin{array}{l}
x=3\\
y=4
\end{array}
\end{eqnarray}
In step~(1) we divided the first equation by $3$; 
this can be undone by multiplying the first equation by $3$.
In step~(2) we subtracted the first equation from the second;
this can be undone by adding the first equation to the second.
In step~(3) we subtracted the second equation from the first;
this can be undone by adding the second equation to the first.
In each step we replace a system of two equations in $x$ and $y$
by an equivalent system, i.e. we replace a pair of lines through  the point $(x,y)=(3,4)$
by another pair of lines through that point. 
\lpmx

\para The process of simplifying a system of equations by multiplying
one of the equations by a nonzero number or replacing one of the equations by its difference with another is called \jdef{Gaussian Elimination}.
The guiding principle is to ``eliminate'' one of the variables from the first
equation and the other variable from the second equation
leaving us with an equivalent system of form $x=a$, $y=b$.
\arap

\para Gaussian Elimination can be used for any 
number of equations in any number of variables.
%We will study it more extensively in Section~\ref{}. 
For problems as
in the example (two equations in two unknowns)  which are intended
to be worked by hand, there is no reason to supply so much detail.
For much larger problems (100 equations in 100 unknowns) which
will be solved on a computer,  Gaussian Elimination is used because
we can prove it gets the correct answer. 
%In these larger problems  using \jdef{matrix notation}.
%This means that we don't write the unknowns but only their coefficients
%in an array. We can reconstruct the equations if we always put the coefficients
%of the first unknown in the first column, of the second unknown in the second column,
%etc. and the constant terms in the last column. 
\arap

\para Another nice feature of Gaussian Elimination is that it handles
gracefully the case where the two lines determined by the two
equations are parallel or the same. 
If the two equations correspond to
lines with the same slope, Gaussian Elimination will lead to an equivalent
system of form 
$$ 
\begin{array}{rr}
ax+by=c\\ 0=d.
\end{array}
$$
 If $d\ne 0$ then the original lines are parallel
but distinct and there is no solution. If $d=0$, then the original lines are the same
and there are infinitely many solutions (any point on the line solves both equations).
A system with no solutions is said to be \jdef{inconsistent}.
\arap

\section{Symmetry}\label{sec:Symmetry}

\dfn The \jdef{reflection} of a point $P$ in a line $\ell$
is the point $Q$ on the other side of the line at the same distance from the line.
The \jdef{reflection} of a point $P$ in a through the point $O$
is the point $Q$ on the the line $OP$ at the same distance from $O$ as is $P$.
For example:
\begin{enumerate}
\item The reflection of $(x,y)$ in the $y$-axis is $(-x,y)$.
\item The reflection of $(x,y)$ in the $x$-axis is $(x,-y)$.
\item The reflection of $(x,y)$ in the line $y=x$  is $(y,x)$.
\item The reflection of $(x,y)$ about the origin is $(-x,-y)$.
\end{enumerate}
A graph is \jdef{symmetric about the line} $\ell$ iff whenever
a point $P$ lies on the graph so does its reflection in $\ell$.
A graph is \jdef{symmetric about the point} $O$ iff whenever
a point $P$ lies on the graph so does its reflection through $O$.
\nfd

\para\topic{Symmetry tests}. 
Combining these definitions we see that
\begin{enumerate}
\item The graph of an equation is symmetric about  the $y$-axis if and and
only if the equation that results by replacing $x$ by $-x$ 
yields an equivalent equation.
\item The graph of an equation is symmetric about  the $x$-axis if and and
only if the equation that results by replacing $y$ by $-y$
yields an equivalent equation
\item The graph of an equation is symmetric about  the $y$-axis if and
only if the equation that results by replacing $x$ and $y$  by $-x$ and $-y$
yields an equivalent equation.
\end{enumerate}
For example, the graph of $y=x^2$ is symmetric about the $y$-axis
as $(-x)^2=x^2$ and similarly the graph of $x=y^2$ is symmetric about the $x$-axis.
The graph of $y=x^3$ is symmetric about the origin as $(-x)^3=-x^3$ so
$$
y=x^3\iff (-y)=(-x)^3.
$$
\medskipnoindent
\input{y=x2}\hspace{0.1in}\input{x=y2}\hspace{0.1in}\input{y=x3}
\medskipnoindent
Note that it might be better to say``replacing $x$ by $(-x)$'' rather
than ``replacing $x$ by $-x$'' as $(-x)^2=x^2\ne -x^2$.
\arap

\section{Completing the Square}\label{sec:Completing_the_Square}

\para By \jdef{completing the square} we mean the identity

\BOXIT{
$\ds
    ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}
$
}
which is easily proved by expanding the right hand side. 
In this section we use this identity for three purposes:
to solve a quadratic equation (the Quadratic Formula),
to find the center of a circle, and  
to find the vertex of a parabola.  
\arap

\begin{theorem}[\jdef{Quadratic Formula}]\label{thm:quadratic-formula}
The solutions of the quadratic equation
$$
   ax^2+bx+c=0
$$
are
\BOXIT{
$\ds   x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$
}
\end{theorem}

\begin{proof} Guided Exercise.
\end{proof}


\begin{theorem}[\jdef{Center of a Circle}] If $a\ne 0$, the equation
$$
   ax^2+ay^2+bx +cy +d=0
$$
is equivalent to the equation
$$
   (x+h)^2+(y+k)^2=R
$$
where
$$
    h=-\frac{b}{2a},\qquad k=-\frac{d}{2a},\qquad R =\frac{b^2+c^2}{4a^2} -d.
$$
Hence the graph is a circle of radius $\sqrt{R}$ if $R>0$,
the single point $(h,k)$ if $R=0$, and empty if $R<0$.
\end{theorem}

\begin{proof} Guided Exercise.
\end{proof}

\begin{theorem}[\jdef{Vertex Formula}] If $a\ne 0$, the equation
$$
   y=ax^2+bx +c
$$
is equivalent to the equation
$$
  y=a(x-h)^2+k
$$
where
$$
    h=-\frac{b}{2a},\qquad k=-\frac{b^2-4ac}{4a}.
$$
Hence the graph is a parabola with vertex $V(h,k)$.
\end{theorem}

\begin{proof} Guided Exercise.
\end{proof}


\section{Functions}\label{sec:Functions}

\dfn 
A \jdef{function} is a rule which produces
an output $f(x)$ from  an input $x$.
The set of inputs $x$ for which the function is defined 
is called the \jdef{domain}
and $f(x)$ (pronounced ``$f$ of $x$'') 
is the \jdef{value} of $f$ at $x$. The set of
all possible outputs $f(x)$ as $x$ runs over the domain 
is called the \jdef{range} of the function. 
\nfd

\para\topic{Functional Notation}. 
We usually define a function $f$ by writing an equation
$$
f(x)=\mbox{ some expression in }x.  
$$
It is then understood that $f(a)$ denotes the result of substituting
$a$  for $x$ in the expression.  If $a$ is itself an expression, it should
be surrounded by parentheses before doing the substitution. Thus
if $f(x)=x^2$, then $f(p+q)=(p+q)^2$ not $p+q^2$. 

In these notes the input to a function $f$ is usually a real number
and the output is also a real number, i.e. both the domain and the range 
of the function are sets of real numbers. To emphasize this we will say
that {\em $f(x)$ is a real valued function of the real variable $x$}.
The input to a function can as well be a pair $(x,y)$ of numbers.  
Thus an equation
in the variables $x$ and $y$ always has the form $F_1(x,y)=F_2(x,y)$.
Subtracting the right hand side from the left we see that every equation
in $x$ and $y$ is equivalent to an equation  of the form $F(x,y)=0$
where $F(x,y)=F_1(x,y)-F_2(x,y)$.
\arap

\para \label{domain}
If a function $f(x)$ is given by an expression in the variable  $x$
and the domain is not explicitly specified, then the domain is understood
to be the set of all $x$ for which the expression is meaningful.
For example, for the function $f(x)=1/x^2$ the domain is the set
of all nonzero real numbers $x$ 
(the value $f(0)$ is not defined because we don't divide by zero) 
and the range is the set of all positive real numbers 
(the square of any nonzero number is positive). 
The domain and range of the square root function $\sqrt{x}$ is the set
of all nonnegative numbers $x$. The domain of the function
$y=\sqrt{1-x^2}$ is the interval $[-1,1]$,
i.e. $\sqrt{1-x^2}$ is meaningful only if $-1\le x\le 1$
(otherwise the input to the square root function is negative).
The range of the function   is the interval $[0,1]$ as $0\le\sqrt{1-x^2}\le 1$.
\arap
 
 \dfn  
The \jdef{graph} of a function $f$ is the graph of the equation
$$
y=f(x),
$$
i.e.  the set of all points
$P(x,y)$ whose coordinates $(x,y)$ satsfy the equation $y=f(x)$.
According to Definition~\ref{dfn:graph-equation},
the graph of an equation of form
$$
     F(x,y)=0
$$
is the set  of all points
$P(x,y)$   in the $(x,y)$-plane 
whose coordinates $(x,y)$ satisfy the equation.  The graph
of a function $y=f(x)$ is a special case: take $F(x,y)=y-f(x)$.
To decide if a set is the graph of a function we apply the
%%
\begin{thmbox}\jdef{Vertical Line Test}.
\em A graph % of an equation $F(x,y)=0$ 
is the graph of a function $f$ if and only if
every vertical line $x=a$ intersects the
graph in at most one point. Then,
if the number $a$ is in the domain of $f$, the vertical line $x=a$
intersects the graph $y=f(x)$ in the point $P(a,f(a))$.
\end{thmbox}
%%
\nfd


\xmpl
The graph of the equation $x^2+y^2=1$ is a circle; it is
not the graph of a function since the vertical line $x=0$ (the $y$-axis)
intersects the graph in two points $P_1(0,1)$ and $P_2(0,-1)$.
This graph is however the union of two different graphs each of
which is the graph of a function:
$$
  x^2+y^2=1 \iff \mbox{ either }\quad 
  y=\sqrt{1-x^2} \quad\mbox{ or }\quad y=-\sqrt{1-x^2}.
$$ 

\noindent
\input{unit+semicircle}
%
\input{unit-semicircle}
\lpmx

\rmk To find the domain of a function we project its graph on the horizontal axis.
To find the range of a function we project its graph on the vertical axis.
More precisely,
\begin{thmbox}
 The domain of the function $f$ is the set of all real numbers $a$
such that the vertical line $x=a$ intersects the graph $y=f(x)$.
The range of the function $f$ is the set of all real numbers $b$
such that the horizontal line $y=b$ intersects the graph $y=f(x)$.
\end{thmbox}
\kmr
\para  You can add, subtract, multiply and divide two functions:
$$
\begin{aligned}
&(f+g)(x):=f(x)+g(x), &\qquad & 
(f-g)(x):=f(x)-g(x),\\ 
&(f\cdot g)(x):=f(x)\cdot g(x),&\qquad &
(f/g)(x):=\frac{f(x)}{g(x)}.
\end{aligned}
$$
In each case the domain is the intersection of the domains
of $f$ and $g$; in the case where the functions are divided
those $x$ where $g(x)=0$ must also be removed from the domain.
\arap

\dfn The \jdef{composition} $g\circ f$ of two functions $g$ and $f$ is defined by
$$
    (g\circ f)(x) := g(f(x)).
$$
\nfd


\rmk Composition of functions is not a commutative operation.
For example, if $f(x)=x^2$ and $g(x)=x+1$, then
$(g\circ f)(x)=x^2+1$ and $(f\circ g)(x)=(x+1)^2$.
\kmr

\para Two functions are said to be \jdef{equal} 
when they have the same domain and give the same output for every input.  
A  consequence of this definition
is that it doesn't matter what letters we use to define a function.
For example, the functions
$$
    f(x)=x^2,\qquad g(u)=u^2
$$
are equal since the domain of each is the set $(-\infty,\infty)$
of all real numbers and $f(t)=g(t)$ for all real numbers $t$.
In story problems, we usually use letters which suggest the meaning
like $t$ for time or $A$ for area.
\arap

\section{Inverse Functions}\label{sec:Inverse_Functions}

\dfn Two functions $f$ and $g$ are said to be \jdef{inverse functions}
iff the graphs the equations $y=f(x)$ and $x=g(y)$ are the same, i.e.
iff
$$
   y=f(x)\iff x=g(y).
$$
We also say  that $g$ is the \jdef{inverse} of $f$ and write
$$
       g=f^{-1}.
$$
\nfd

\xmpl Since
$$
     y=3x+7 \iff x=\frac{y-7}{3}
$$
the functions $f(x)=3x+7$ and $g(y)=(y-7)/3$ are inverse
functions.
\lpmx

\rmk Be careful not to confuse $f(x)^{-1}$ and $f^{-1}(x)$.
For example, if $f(x)=3x+7$, then
$$
   f(x)^{-1}=\frac{1}{3x+7}, \qquad\mbox{but}\qquad 
   f^{-1}(x)=\frac{x-7}{3}.
$$
\kmr

\para If $f$ and $g$ are inverse functions,
then the range of $f$ is the domain of $g$ and the
domain of $f$ is the range of $g$.
To decide if a function has an inverse we apply the
%%
\begin{thmbox}\jdef{Horizontal Line Test}.
\em A function $y=f(x)$  has an inverse if and only if
every horizontal  line $y =b$ intersects the
graph in at most one point. Then the horizontal line $y=b$
intersects the graph $y=f(x)$ in the point $P(f^{-1}(b),b)$.
\end{thmbox}
%%
\arap

\para \label{1-2-1} A function $f$ is said to be \jdef{one-to-one} iff
$$
    f(x_1)=f(x_2)\implies x_1=x_2.
$$
Saying that a function is one-to-one  
is just another way of saying its graph satisfies the horizontal line
test, 
i.e. that the function has an inverse. The function $f(x)=x^2$
is not one-to-one since $f(3)=f(-3)$ but $3\ne -3$.
\arap

\dfn A function $f$ is said to be \jdef{increasing} iff
$$
     x_1<x_2\implies f(x_1)<f(x_2).
$$
A function $f$ is said to be \jdef{decreasing} iff
$$
     x_1<x_2\implies f(x_1)>f(x_2).
$$
\nfd

\begin{theorem} If a function either increasing or decreasing 
it is one-to-one and therefore has an inverse.
\end{theorem}

\begin{proof} Assume $f$ is increasing and that $f(x_1)=f(x_2)$.
Then it is not the case that $x_1<x_2$ since that would imply $f(x_1)<f(x_2)$
and  it is not the case that $x_1>x_2$ since that would imply $f(x_1)>f(x_2)$.
The only possibility us that $x_1=x_2$.
\end{proof}

\rmk Just because a function has an inverse doesn't mean
that we can find a formula for the inverse. 
For example, the function $f(x)=x^5+x$ is increasing
and therefore has an inverse $f^{-1}$, but in graduate courses in
algebra it is proved that there is no elementary formula for $f^{-1}(y)$,
i.e. there is no expression for $f^{-1}(y)$ involving only the operations 
we have have defined in these notes. Put another way, there is no 
nice formula for the solution of the equation $3=x^5+x$.

   Even if we can't find a formula for the inverse of a function, we can still
compute its value for any given input to any degree of accuracy with (say) a computer. 
So we can give the inverse a name and compute with it using the rules of algebra. 
This is exactly what we shall do in Section~\ref{sec:Exponentials_and_Logarithms}.
with the exponential function
$$
    f(x)= 2^x.
$$
This function is increasing  so has an inverse function. and
the inverse function is denoted by $\log_2(y)$:
$$
    y=2^x\iff x=\log_2(y).
$$
\kmr

\para In Section~\ref{domain} we said that, 
unless the contrary is stated,  
the domain of a function $f(x)$ which is defined by an expression in $x$
is the set of $x$ for which the expression is meaningful.
However, we may decide to restrict the domain to a smaller set
in order to define an inverse function. For example, 
in Section~\ref{1-2-1} we saw that the function $f(x)=x^2$ is not
one-to-one since $f(3)=f(-3)$ and hence this function does not have an inverse function.
But we can define
$$
    q(x)=x^2,\qquad x\ge 0
$$
so the domain is $[0,\infty)$. Then 
$$
    y=q(x)\iff x=\sqrt{y}
$$
so $q(x)$ and $\sqrt{y}$ are inverse functions. (This device
is used in trigonometry to define the inverse trigonometric functions.)
If we were to program a computer to compute the function $q(x)$,
asking the computer to compute $q(-3)$ would produce an error message like
{\em ``function undefined for this input''}.
\arap

\para If the function $f$ has an inverse,  then 
the graph of the equation $y=f(x)$ 
is the same as 
the graph of the equation $x=f^{-1}(y)$.
Of course the graph of the equation $y=f^{-1}(x)$
is (usually) different. The graphs of the equations
$y=f(x)$ and $y=f^{-1}(x)$ are obtained from
each other by interchanging the $x$-axis and the $y$-axis, i.e.
by reflecting in the line $y=x$.

\input{y=fx}
\input{x=fy}
\arap

\section{Average Rate of Change}\label{sec:Average_Rate_of_Change}

\dfn  The \jdef{average rate of change} of the function $f$
over the interval $[a,b]$ is the slope
of the line joining the point $(a,f(a))$ to the point $(b,f(b))$.
When $y=f(x)$  the average rate of change is often written as
$$
     \frac{\Delta y}{\Delta x}= \frac{\Delta f}{\Delta x}= \frac{f(b)-f(a)}{b-a},
$$
i.e. the average rate of change is 
the change $\Delta y=f(b)-f(a)$ in $y$ 
divided by the change $\Delta x=b-a$ in $x$.
\nfd

\xmpl The average rate of change of the functions $f(x)=x^2$
on the interval $[a,b]$ is
$$
\frac{\Delta f}{\Delta x}= \frac{f(b)-f(a)}{b-a}=\frac{b^2-a^2}{b-a}
=\frac{(b+a)(b-a)}{b-a}=b+a.
$$
Note that the $b-a$ in the denominator cancels out. 
To see way this happens we will repeat the calculation using
the notation $h=b-a=\Delta x$ so that  $b=a+h$.
When we expand the numerator  the  terms which don't contain an $h$
cancel. Then the $h$ in the denominator cancels with (some of) the $h$'s 
in the numerator as follows:
$$
\frac{\Delta f}{\Delta x}= 
%\frac{f(a+h)-f(a)}{h}=
\frac{(a+h)^2-a^2}{h}
=\frac{a^2+2ah+h^2-a^2}{h}
=\frac{2ah+h^2}{h}=2a+h.
$$
This cancellation will always
happen when you are asked to simplify an average rate of change.
\lpmx

\para The average rate of change of a function $f$ over the interval $[a,b]$
is undefined when $a=b$ (zero divided by zero is nonsense)
but for $b$ very close to $a$ the average rate of change will usually be close to
a number called the \jdef{instantaneous rate of change}. This is the slope of the
tangent line to the graph $y=f(x)$ at the point $(a,f(a))$. 

\noindent

\input{average-rate}
\input{instant-rate}


\noindent In calculus 
the instantaneous rate of change is called the \jdef{derivative}.
In the previous paragraph we saw that for the function $f(x)=x^2$ 
the average rate of change over the interval $[a,b]$ is $a+b$.
This is $2a+h$ where $h=\Delta x=b-a$.
The instantaneous rate of change at the point $(a,f(a))$ is $2a$
which is obtained from the average rate of change by taking $h=0$.
\arap

\para Suppose you are traveling from Madison to Milwaukee
by automobile via the interstate highway I-94. Your position is a function 
$s=f(t)$ of the time $t$. The value $s$ is the number on the mile marker
at the side of the road. (Along much of the road there is a mile marker
every tenth of a mile, but imagine there is one every few feet.)
The change $\Delta s=f(t+\Delta t)-f(t)$ is also the change in the odometer
reading in your car. If the time interval $\Delta t$ is so short that your
speed doesn't change much in the time interval, the average rate
of change $\Delta s/\Delta t$ is your speed as shown 
on the speedometer. The speed is
the instantaneous rate of change of the position.
\arap



\section{Polynomials}\label{sec:Polynomials}

\para For any graph the points where it intersects
the $x$-axis are called the \jdef{$x$-intercepts}
and the points where it intersects the $y$-axis are
called the \jdef{$y$-intercepts}.
The equation of the $x$-axis is $y=0$ and 
the equation of the $y$-axis is $x=0$ .
Thus if the graph is the graph of the equation $F(x,y)=0$, 
the $x$-intercepts are the points $(x,0)$ such that
$F(x,0)=0$, and similarly 
the $y$-intercepts are the points $(0,y)$ such that
$F(0,y)=0$. If the graph is the graph of a function $y=f(x)$,
then (assuming that
$0$ is in the domain)
the $y$-intercept is the point $(0,f(0))$, and the
$x$-intercepts are the points $(r,0)$ such that $f(r)=0$.
The numbers $r$ such that $f(r)=0$ are called the \jdef{zeros}
of the function.
\arap

\dfn A \jdef{polynomial} is a function of form
$$
   f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0
$$
where the exponents on $x$ are nonnegative integers.
A \jdef{polynomial equation} is an equation of form
$$
 f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0=0.
$$
Its solutions are the zeros of the polynomial and are also
called the \jdef{roots}   of the polynomial.
When $a_n\ne0$ we say that the \jdef{degree}
of the polynomial  is $n$. 
The constants $a_0,a_1,\ldots,a_n$ are called
the \jdef{coefficients}.  
For any polynomial $f(0)=a_0$ (the \jdef{constant term}) 
so the $y$-intercept of the graph is the  point $(0,a_0)$.
\nfd

\para  A polynomial of degree one (or zero) is called \jdef{linear} 
(since its graph is a line), 
a polynomial of degree two is called \jdef{quadratic}, and
a polynomial of degree three is called \jdef{cubic}.
In Paragraph~\ref{point-slope} we used the notation
$f(x)=mx+b$ (slope-intercept form) rather than 
$f(x)=a_1x+a_0$ for a linear function and in
Theorem~\ref{thm:quadratic-formula}
we wrote the quadratic equation as $ax^2+bx+c=0$
rather than as $a_2x^2+a_1x+a_0=0$.
\arap

\para It is easy to graph a linear function: just find
two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$
on the graph and draw the line through them.
It is also not hard to graph a quadratic function
$f(x)=ax^2+bx+c$.  The graph is a parabola which opens up
(like $y=x^2$)  if $a>0$ and down (like $y=-x^2$)  if $a<0$.
By the Quadratic Formula in Theorem~\ref{thm:quadratic-formula}
the graph has  two $x$-intercepts  
if the \jdef{discriminant} $b^2-4ac$ is positive, 
one $x$-intercept
if the discriminant is zero, and 
no $x$-intercept if the discriminant is negative.
\arap

\para
When $a=1$ and $b$ and $c$ are integers we could try to solve 
the quadratic equation $x^2+bx+c=0$ by factoring.
If
$$
   x^2+bx+c= (x-r_1)(x-r_2)= x^2-(r_1+r_2)x+r_1r_2
$$
(for all $x$) then 
$$
b=-(r_1+r_2),\qquad c=r_1r_2. 
$$
If we suspect that the roots are integers, 
we can try all possible ways of factoring $c$. For example,
to solve $x^2-5x+6=0$ we try $(r_1,r_2)=(1,6)$,
$(-1,-6)$, $(2,3)$, $(-2,-3)$. 
Only $(r_1,r_2)=(2,3)$ gives $-(r_1+r_2)=-5$ so
$$
     x^2-5x+6=(x-2)(x-3)
$$
and  the solutions of $x^2-5x+6=0$ are $x=2$ and $x=3$. 
Of course, there is usually no reason
to suspect that the roots are integers so it is best to use the Quadratic Formula 
(Theorem~\ref{thm:quadratic-formula})  to solve a quadratic equation.
\arap

\para A polynomial inequality like 
$$
x^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0>0
$$
is easy to solve if we can factor the left hand side. Then we can write it in the form
$$
   (x-r_1)(x-r_2)\cdots(x-r_n)>0.
$$
%If the roots $r_1,\ldots,r_n$ are real and $r_1<r_2<\cdots<r_n$, then for
%$r_k<x<r_{k+1}$ we have 
%$x-r_i>0$ for $i=1,\ldots,k$ and $x-r_i<0$ for $i=k+1,\ldots,n$.
%The polynomial is positive on the interval  $(r_k,r_{k+1})$ if $n-k$ is even.
On each of the intervals $(-\infty,r_1),(r_1,r_2),\ldots,(r_{n-1},r_n),(r_n,\infty)$ the sign
of the polynomial is constant so we can compute the sign by evaluating the polynomial at some point
in the interval.
For example,
$$
  (x-1)(x-4)(x-9)>0 \iff x \mbox{ in } (1,4)\cup(9,\infty)
$$
and
$$
  (x-1)(x-4)(x-9) \ge 0 \iff x \mbox{ in } [1,4]\cup[9,\infty)
$$
As a check we evaluate \par
at $x=0$ in $(-\infty,1)$  and get $(0-1)(0-4)(0-9)=-36<0$,\par
at $x=2$ in $(1,4)$ and get  $(2-1)(2-4)(2-9)=14>0$,\par
at $x=5$ in $(4,9)$ and get  $(5-1)(5-4)(5-9)=-16<0$, and\par
at $x=10$ in $(9,\infty)$ and get  $(10-1)(10-4)(10-9)=54>0$.
\arap

\para 
To graph a polynomial $f(x)$ we use the following rules:
\begin{description}
\item[(i)] The sign of the polynomial does not change
between two adjacent roots. To determine this sign
we can evaluate the polynomial at {\em any} number 
in between.
\item[(ii)] If the polynomial can be factored so that
$$
    f(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)
$$
where $r_1<r_2<\cdots<r_n$ are the roots, then
between each pair of roots $r_k,r_{k+1}$  the graph reverses
direction exactly once, i.e. either
the function value $f(x)$ increases on some interval $[r_k,c_k]$ and then
decreases on the interval $[c_k,r_{k+1}]$  or it
decreases on some interval $[r_k,c_k]$ and then
increases on the interval $[c_k,r_{k+1}]$
\item[(iii)] The absolute value of $f(x)$ is large when the
absolute value of $x$ is large. Whether $f(x)$ is large
positive or large negative depends only on the sign of the 
coefficient of $x^n$ (where $n$ is the degree) and on whether
$n$ is odd or even. 
\end{description}
In calculus you will learn how to determine on which
intervals the function is increasing and 
you will learn enough to understand why~(ii) is true.

Below we have graphed the polynomial
$$
f(x)=x^4-2x^3-x^2+2x=(x+1)(x-0)(x-1)(x-2)
$$
which factors completely as in item~(ii). The degree
is $4$ and the roots are $-1,0,1,2$. In each of the intervals
$(-1,0)$, $(0,1)$, $(1,2)$ between roots the function
reverses direction exactly once. Next to the graph $y=f(x)$
is the graph $y=f(x)-1$. The degree is still $4$ but the graph $y=f(x)-1$
has two $x$-intercepts (not $4$ as does $y=f(x)$) and between them 
the function reverses direction three times. According to
item~(ii) this means that the polynomial $f(x)-1$ does not factor completely
(into real linear polynomials).

\medskip

\input{y=1234}
\input{y=1234+}
\arap

\section{Rational Functions}\label{sec:Rational_Functions}

\dfn A \jdef{rational function}  is a quotient 
$$
   f(x)=\frac{g(x)}{h(x)}
$$
of two polynomials. The domain is the set of all real numbers
where the denominator $h(x)$ os not zero. Since the denominator
$h(x)$ is not zero for $x$ in the domain, a zero of $f$ is the same as a zero
of the numerator $g$. A point where the denominator  is zero 
(and the numerator is not zero) is called a \jdef{pole} of $f$.
\nfd

\begin{theorem}[\jdef{Division Algorithm}] \label{thm:DivisionAlgorithm}
Every rational function may be written in the form
$$
   f(x)=\frac{g(x)}{h(x)}=q(x) + \frac{r(x)}{h(x)}
$$
where $q(x)$ and $r(x)$ are polynomials and
the degree of the remainder $r(x)$ is less than the degree of 
the divisor $h(x)$.
\end{theorem}

\begin{proof} Guided Exercise.
\end{proof}
 
 \rmk An analogous statement holds for rational numbers.
 For example,  
 $$
    \frac{22}{7}=3+\frac{1}{7}.
$$
In the case of rational numbers (as opposed to rational functions)
 the condition that the degree of $r(x)$ is less than
the degree of $h(x)$ is replaced by the condition that the remainder
is less than the divisor. In fact there is a very strong analogy between
integers and rational numbers on the one hand and polynomials
and rational functions on the other. 
\kmr

\para 
To graph a rational function 
$$
 f(x)=\frac{a_nx^n+\cdots+a_1x+a_0}{b_mx^m+\cdots+b_1x+b_0}
$$ 
we use the following rules:
\begin{description}
\item[(i)] On any interval which contains neither a zero nor a pole,
the sign of $f(x)$ does not change. To determine this sign
we can evaluate the rational function at {\em any} number 
in the interval.
\item[(ii)] If $n>m$ (and $a_n,b_m\ne0$) then
the absolute value of $f(x)$ is arbitrarily large when the
absolute value of $x$ is sufficiently large. 
\item[(iii)] If $n\le m$ (and $a_n,b_m\ne0$) then
 $f(x)$ is arbitrarily close to $c$ when the
absolute value of $x$ is sufficiently large, where $c=a_n/b_m$ if $n=m$
and $c=0$ is $n<m$.
\item{(iv)} If $p$ is a pole, then
the absolute value of $f(x)$ is arbitrarily large
when $x$ is sufficiently close to $p$.
\end{description}
In case~(iii) the horizontal line $y=c$ is called a \jdef{horizontal asymptote}
of the graph of $y=f(x)$. The vertical line $x=p$ as in case~(iv) is called
a \jdef{vertical asymptote} of the graph.


If a nonzero number is divided by a relatively small number, the
result is large. 
This is why the absolute value of $f(x)$ is large when $x$ is near a pole.
This is why~(i) is true.

The reason  why (ii)-(iv) are true is as follows. With $q(x),r(x),h(x)$
as in the Division Algorithm the term $r(x)/h(x)$ is very small when
the absolute value of $x$ is large. This is because
 the degree of $r(x)$ is less than the degree of $h(x)$ so
the absolute value of $r(x)$ is much smaller than the absolute
value of $h(x)$ when the absolute value of $x$ is large.
Hence the ratio $r(x)/h(x)$ is small and $f(x)$ is close to $q(x)$.
When $m=n$, the quotient $q(x)$ is the constant $a_n/b_n$,
when $n<m$ $q(x)$ is the zero polynomial, and when $n<m$,
the quotient $q(x)$ has positive degree and hnce has large absolute value
when the absolute value of $x$ is large.
\arap

\xmpl The $x$-axis is a horizontal asymptote of the graph
$\ds y=\frac4{1+x^2}$ and also of the graph $\ds y=\frac1{1-x^2}$.
The lines $x=\pm1$ are vertical asymptotes of the graph $\ds y=\frac1{1-x^2}$.

\bigskip

\input{y=1+x2}
\hfil
\input{y=1-x2}
\lpmx
  
  
\section{Exponentials and Logarithms}\label{sec:Exponentials_and_Logarithms}
  

\para 
Each positive number $a$ distinct from one determines a function
$$
  f(x)=a^x
$$
called the \jdef{exponential function base $a$}. It is characterized by
the following properties:
%%
\begin{description}
\item[(i)] The domain is the set of all real numbers and the range is the set of all positive real numbers.
\item[(ii)] $a^{u+v}=a^u\cdot a^v$,\;\;  $a^0=1$.
\item[(iii)] If $a>1$, the exponential function base $a$ is increasing.
\item[(iv)] If $0<a<1$, the exponential function base $a$ is decreasing.
\end{description}
This was stated as Theorem~\ref{thm:a^x}. A careful definition of
the exponential function requires concepts which are normally
taught in calculus. All the power laws in Paragraph~\ref{power_laws}
which hold hold when the exponents are integers  
continue to hold for real exponents.  Here are the graphs
$y=2^x$ and $y=\left(\frac12\right)^x$.

\medskip

\input{y=2tox}
\hfil
\input{y=halftox}

\medskipnoindent
If $a>1$ then $0<1/a=a^{-1}<1$ and $\left(1/a\right)^x\left(a^{-1}\right)^x=a^{-x}$.
Hence the graph  of $y=(1/a)^x$ is obtained from the graph of $y=a^x$ by reflection in the $y$-axis. 
\arap

\para For $a>1$ as $x$ increases through positive values the function value
$y=a^x$ increases very rapidly; one calls this \jdef{exponential growth}.
For example, $2^{10}=1024$ while $2^{20}=1,048,576$, i.e. doubling the input
from ten to twenty changes the output from about one thousand to 
about one million.
On the other hand as $x$ decreases through negative values the
value $y=a^x$ approaches the $x$-axis very rapidly,
but (since $a^x>0$) never intersects it. 
For $0<a<1$,  the situation is reversed: as $x$ increases, the function value $y=a^x$  approaches the $x$-axis very rapidly 
-- this is called \jdef{exponential decay} --
while as $x$ decreases through negative values  the function value 
becomes large very rapidly. 
Whether or not $a>1$ or $a<1$ the function value
is always positive.
\arap

\rmk We say that the $x$-axis is a \jdef{horizontal asymptote} for the graph
$y=a^x$ but the situation is not exactly the same as it was for the graph
of a rational function. When a  rational function  has a horizontal asymptote, 
the graph is close  to the asymptote whenever the absolute value of $x$ is large
whether $x$ is positive or $x$ is negative. For the exponential function,
the graph approaches the $x$-axis when $x$ is 
large positive or large negative, but not both.
\kmr

\dfn The \jdef{logarithm base $a$} is the inverse function to
the exponential function base $a$. It is denoted by $\log_a$ so

\BOXIT{$y=\log_a(x) \iff x=a^y.$}

\medskipnoindent i.e.


\BOXIT{$y=\log_a(a^y) \qquad  x=a^{\log_a(x)}.$}
\nfd



\para The exponential function satisfies the identity
$a^ua^v=a^{u+v}$ which  says that
``exponentiation changes addition into multiplication''.
Analogously, ``the logarithm changes multiplication into addition''.
If $U=a^u$ and $V=a^v$ then $UV=a^{u+v}$. But these equations also say that
$u=\log_a(U)$,  $v=\log_a(V)$, and $u+v=\log_a(UV)$. Hence
$$
\log_a(UV)=\log_a(U)+\log_a(V).
$$
The other laws of exponentiation imply in the same way 
corresponding laws for logarithms as follows:


\begin{boxtable}{Exponentials and Logarithms}{1.8}{@{\qquad}ll@{\qquad}ll@{\qquad}} 
&&&\\
{\bf (i)}& $a^ua^v=a^{u+v}$&   & $\log_a(UV)=\log_a(U)+\log_a(V)$ \\ 
{\bf (ii)}& $a^0=1$&   & $0=\log_a(1)$ \\ 
{\bf(iii)} & $ \ds\left(a^u\right)^p= a^{pu}$ & & $\log_a(U^p)=p\log_a(U)$\\ 
{\bf (iv)}& $ \ds\frac{a^u}{a^v}= a^{u-v}$ & & $\ds\log_a\left(\frac{U}{V}\right)=\log_a(U)-\log_a(V)$\\ 
&&&
\end{boxtable}
\arap

\rmk Before the advent of electronic computers,
logarithms and exponentials were used to do
arithmetic. To multiply two numbers $U$ and $V$
one would look up their logarithms in a table,
add the logarithms, and then find the number
in the table with (almost) the same logarithm as the sum.
That number was the product $UV$.
(The point is that addition of long numbers is much easier than
multiplication.)

 Scientists and engineers used
a device called a
\jdef{slide rule}   to perform these operations
quickly. Essentially it consisted of two rulers marked
with numbers spaced according to their logarithms
as in this diagram:

\noindent
\input{slide-rule}

\noindent
To compute the product of (say)
$2\times 3$ you place the $1$ on one of the logarithmic rulers
above the $2$ on the other. The number on the bottom ruler
below the $3$ on the top ruler is then at a distance of
$\log(2)+\log(3)=\log(2\times 3)$ from the $1$ on the bottom ruler
and using the markings on the ruler you can read off the product.
The above diagram was created with a computer and is quite accurate.
You can see how it works by marking off the distance between the $1$ and the $2$
on a piece of paper. You will see that this distance is exactly the same as
the distance between the $3$ and the $6$.
\kmr



\para The equations $y=\log_a(x)$ and $x=a^y$ have the same graph.
(This is always true for inverse functions.)
The graph of $y=\log_a(x)$ is obtained from the graph of $y=a^x$ by
reflecting in the line $y=x$, i.e. interchanging the $x$-axis and the $y$-axis.

\bigskip

\input{y=logx}
\input{x=logy}
\arap



\section{Exponential Growth and Decay}\label{sec:Exponential_Growth_and_Decay}

\dfn A quantity $N$ is said to obey an \jdef{exponential law}
iff its value at time $t$ is given by the formula
$$
      N = N_0 a^t
$$
where $a$ is a positive real number distinct from $1$. 
Since $a^0=1$ it follows
that $N_0$ is the value of $N$ at time $t=0$ and since 
$a^{t+1}=a^t\cdot a$ it follows that $N$ changes by a factor of $a$
in each time interval of length $1$. When $a>1$ we say that
$N$ \jdef{grows exponentialy} while if $a<1$ we say that
$N$ \jdef{decays exponentially}. 
\nfd
 
 
 \para  Many phenomena are governed by exponential
growth laws. To name a few:
\begin{enumerate}
\item 
Money invested in a bank account grows exponentially
at $6\%$ per year (or whatever the interest rate is).
If you invest $B_0$ dollars in a bank account that pays
$6\%$ per year in $t$ years your balance will be
$$
       B=B_0(1.06)^t \mbox{ dollars.}
$$
\item
The population of the world grows exponentially
at $1.5\%$ per year. (This will not continue forever.)
As long as this growth rate continues the population
$P$ of the world in $t$ years will be
$$
     P=P_0(1.015)^t
$$
where $P_0$ is the population today.  Other populations
(e.g. bacteria) obey the same law (albeit with a different growth rate).
\item 
The amount of radioactivity in a radioactive material
decays exponentially (at a rate that depends on the material).
For example carbon-14 decays at $0.012\%$ per year so
that amount $N$ of carbon-14 in a sample of organic material
is given by
$$
     N=N_0(0.988)^t
$$
where $N_0$ is the amount of carbon-14 that was in the sample $t$ years ago.
(Since the percentage of carbon-14 in living material is known, this formula
can be used to estimate how long it has been since the organic material was alive.)
%and polonium-210 decays at $0.495\%$ per day.
\end{enumerate}
\arap

\para It is easy to understand why a quantity might grow or decay exponentially.
Consider for example a population of bacteria. At any moment a certain
percentage of the population will be at the right stage of development
to subdivide. Each bacterium which subdivides adds a new bacterium to
the population. If (say) 2\% of the bacteria subdivide every hour,
then in the first hour the population grows from $N_0$ to $N_1=N_0(1.02)$.
In the second hour the population grows from $N_1$ to $N_2=N_1(1.02)$
so $N_2=N_0(1.02)^2$. In general, after $t$ hours the population will
be $N_0(1.02)^t$.
\arap

\section{The Natural Logarithm}\label{sec:The_Natural_Logarithm}

\para The irrational number
$$
e=2.7182818284590451\ldots
$$
is most often used as the base for the exponential function. 
The exponential function base $e$ is called the \jdef{natural exponential}
and the inverse function
$$
    \ln(x):=\log_e(x)
$$
is called the \jdef{natural logarithm}. We shall explain what is natural
about the natural  logarithm in two ways.
\arap

\para First we note that the average rate of change 
(see Section~\ref{sec:Average_Rate_of_Change}) for the natural exponential
on the interval $[x,x+h]$ is given by
$$
    \frac{e^{x+h}-e^x}{h}=e^x\left(\frac{e^h-1}{h}\right)
$$
It is proved in calculus that the number $e$ is 
characterized by the fact that the ratio $\ds\left(\frac{e^h-1}{h}\right)$
is very close to one when $h$ is small. This means that the  average rate
of change over a short interval $[x,x+h]$ (the {\em instantaneous rate of change})
is the value $e^x$ of the natural exponential.

Now consider a quantity $N$ governed by an exponential law
$$
     N=N_0a^t
$$
and let $r=\ln(a)$ be the natural logarithm of the base $a$. Then $a=e^r$ so
$a^t=\left(e^r\right)^t=e^{rt}$ and the exponential law takes the form
$$
     N=N_0e^{rt}.
$$
Now consider the average rate of change of $N$ over a short time
interval $[t,t+\Delta t]$ of duration $\Delta t$. Using the above formula
with $x=rt$ and $h=r\Delta t$ we get
$$
\frac{\Delta N}{\Delta t} =\frac{N_0e^{rt+r\Delta t}-e^{rt}}{\Delta  t}
=N_0e^{rt}\left( \frac{e^{r\Delta t}-1}{\Delta t}\right) =Nr\left(\frac{e^h-1}{h}\right)
$$
which says that the instantaneous rate of change is $Nr$. For example,
if $r=0.06$ this says that the instantaneous rate of change is 6\% of $N$.
(Remember: ``per'' means divide, ''of'' means times, and ''cent'' means 100
so  6 \% means 6/100 and 6\% of $N$ is $0.06N$.) In summary

\begin{thmbox}
\begin{quote} The instantaneous rate of change of a quantity
governed by the exponential law $N=N_0e^{rt}$ is 
$100r\%$ (of $N$).
\end{quote}
\end{thmbox}
\arap

\para  The second way of understanding what is natural
about the natural exponential function is to consider
a bank account which earns a nominal interest
rate of (say) $6\%$ per year. The interest rate is called
nominal because the actual amount of interest paid
depends on how often the bank pays the interest.
An interest rate  of $6\%$ per year is the same
as an interest rate of 
$6\%/12=0.5\%$ per month
but if the bank pays the interest every month,
the account will grow by slightly more than $6\%$ in a year.
Here is how a few different banks might pay interest.

 \medskipnoindent(i)  The Alaska Bank pays interest compounded annually.
This means that every year it looks at the balance in an account and adds $6\%$.
Thus an initial deposit of $B_0$ dollars grows to
$$
            B = B_0 (1.06)^t
$$
dollars after $t$ years (assuming no other deposits are made during
this period).  

 \medskipnoindent(ii)  The Minnesota Bank pays interest compounded monthly.
This means that every month it looks at the balance in an account and adds
$6/12\% = 0.5\%$.
Thus an initial deposit of $B_0$ dollars grows to
$$
            B = B_0 (1.005)^{12t}
$$
dollars after $12t$ months = $t$ years
 (assuming no other deposits are made during this period.) 

 \medskipnoindent(iii)  The Wisconsin Bank pays interest compounded weekly.
This means that every week it looks at the balance in an account and adds
$6/52\% = 0.001154\%$.
Thus an initial deposit of $B_0$ dollars grows to
$$
            B = B_0 \left(1 +\frac{0.06}{52} \right)^{52t}
$$
dollars after $52t$ weeks = $t$ years. 
 
\medskipnoindent(iv) The Delaware Bank pays interest compounded daily.
This means that every day it looks at the balance in an account and adds
$6/365\%$.
Thus an initial deposit of $B_0$ dollars grows to
$$
            B = B_0 \left(1 +\frac{0.06}{365} \right)^{365t}
$$
dollars after $365t$ days = $t$ years. 

\medskip

 The general formula for the balance $B$ in a bank account
after $t$ years if the balance is initially $B_0$, the interest rate
is $r$ per year, and the interest is \jdef{compounded} (i.e. paid) $m$ times per year
is
$$
    B=B_0\left(1+\frac{r}{m}\right)^{mt}.
$$
In all cases the formula is of the form
$$
    B=B_0 a^t
$$
where $a=(1+r/m)^m$ an  the compounding period is
$1/m$ years i.e. the interest is compounded $m$ times per year;
When $m$ becomes infinite we say that the interest
is \jdef{compounded continuously} and in calculus it is proved
that this formula works with $a=e^r$ so the balance after $t$ years is given by
$$
    B=B_0 e^{rt}.
$$
The following table shows the values of
$a^t=\left( 1+ \frac{r}{m}\right)^{mt}$
for $t=2$, $r=0.05$, $n=mt$, and various values of the
number $m$ of compounding periods per year.

\bigskip

$$\renewcommand{\arraystretch}{1.4}
\begin{array}{c|c} 
m&\ds\left(1+\frac{r}{m}\right)^{mt}=\left(1+\frac{0.05}{m}\right)^{2m}\\   &\\ \hline
1   &    1.102500000000000 \\
12  &    1.104941335558328    \\
52  &    1.105117820169223 \\
365 &    1.105163349128883\\
\infty & 1.105170918075648 
\end{array}
$$
Thus an account with a starting value of \$1000
at 5\% per year will, in two years,  earn \$102.50 in interest if the  interest is
compounded annually and \$105.17 in interest if the interest is compounded continuously.
\arap
   
   
\rmk Presumably in the old days the interest
was credited to the account at the the end of
each compounding period and if the account
was closed in the middle of a compounding
period the amount of interest earned since
the beginning of the compounding period
was forfeited. Today most banks would
use the formula $B=B_0\left( 1+ \frac{r}{m}\right)^{mt}$
for all values of $t$ not just when $mt$ is an integer.
Of course, the contract signed when the account
was opened will make this precise.
\kmr   
   

 \para
The \jdef{doubling time} of a quantity  $N=N_0e^{rt}$ which
is increasing exponentially is the time $t$ such that
$N=2N_0$. Since
$$
  2N_0=N_0e^{rt}\implies 2=e^{rt}\implies \ln 2=rt
$$
the doubling time is $t=(\ln 2)/r$. Similarly, the
\jdef{half life} of a quantity  $N=N_0e^{rt}$ which
is decreasing exponentially is the time $t$ such that
$N=N_0/2$, i.e. $t=-(\ln 2)/r$.
 \arap
 
\section{Complex Numbers}\label{sec:Complex_Numbers}


\dfn The \jdef{complex numbers} are those numbers of form 
$$
z=x+iy
$$
where $x$ and $y$ are real numbers and $i$ is a special new number called the
\jdef{imaginary unit} which has the property that
%\BOXIT{
$
     i^2=-1.
$%}
The real number $x$ is called the \jdef{real part} of $z$ and
the real number $y$ is called the \jdef{imaginary part} of $z$. 
Two complex numbers are equal iff their real parts are equal 
and their imaginary parts are equal. 
The complex number
$$
   \bar{z}:=x-iy
$$
is called the \jdef{conjugate} of $z$.  
\nfd

\para The arithmetic operations are performed by treating $i$ as
a variable and then replacing $i^2$ by $-1$ if necessary. 
This
makes clear that the commutative, associative, and distributive
laws hold, that the additive inverse is$-z=-(x_iy)=(-x)+(-y)i$ 
so that the law $z+(-z)=0$, and that the rule for multiplying
complex numbers is
$$
\begin{aligned}
zw&=(x+iy)(u+iv)\\
& =xu+(xv+yu)i+yvi^2\\
& =xu+(xv+yu)i-yv\\
&=(xu-yv)+i(xv+yu)
\end{aligned}
$$
where $z=x+iy$ and $w=u+iv$ and $x$ $y$, $u$, $v$ are real.
Using the algebraic law $(a+b)(a-b)=a^2-b^2$ we see
that the product of a complex number and its conjugate is
$$
 z\bar{z}= (x+iy)(x-iy)=x^2- i^2y^2=x^2+y^2
$$
which is a positive real number if $z\ne0$. Hence the multiplication inverse
is defined by
$$
   z^{-1}=\frac{1}{z}= \frac{\bar{z}}{z\bar{z}}= \frac{x}{x^2+y^2} -\frac{iy}{x^2+y^2}.
$$
\arap

%\begin{thmbox} 
\para Conjugation satisfies the following laws.
\begin{enumerate}
\item A complex number is real if and only if it is equal to its
conjugate:
$$
   x+iy=x-iy\iff y=0\iff z=x.
$$
\item The conjugate of the conjugate is the original number.
$$
    \bar{\bar{z}}=\overline{(x-iy)}=x+iy=z.
$$
\item Doing an arithmetic operation and then taking the
conjugate gives the same result as taking the conjugates first and then doing
the operation:
$$
   \overline{z\pm w}=\bar{z}\pm\bar{w},\qquad \overline{zw}=\bar{z}\bar{w},
   \qquad\overline{\left(\frac{z}{w}\right)}=\frac{\bar{z}}{\bar{w}}.
$$
\end{enumerate}
These are all easy to prove. For example if 
$z=x+iy$, $\bar{z}=x-iy$, $w=u+iv$, $\bar{w}=u-iv$ then
$$
\renewcommand{\arraystretch}{1.8}
\begin{aligned}
\bar{z}\bar{w} &=(x-iy)(u-iv)\\
& =(xu-yv)-(xv+yu)i\\
&=\overline{(xu-yv)+(xv+yu)i}\\
&=\overline{zw}
\end{aligned}
$$
\arap

\section{The Fundamental Theorem of Algebra}\label{sec:The_Fundamental_Theorem_of_Algebra}
 
\begin{theorem}[\jdef{Fundamental Theorem of Algebra}]
A  polynomial equation (of positive degree) has a
complex root. 
\end{theorem}

\rmk The proof of the Fundamental Theorem is rather difficult
and is not ordinarily taught in undergraduate courses. 
The theorem is true even if the coefficients in the polynomial are complex.
In this section we shall see what the theorem means for
\jdef{real polynomials}, i.e. polynomials  with real coefficients. 
\kmr

 \begin{theorem}[\jdef{Remainder Theorem}]
 If a polynomial $f(x)$ is divided by $(x-r)$
 the remainder is $f(r)$
\end{theorem}
\begin{proof}
By the Division Algorithm (Theorem~\ref{thm:DivisionAlgorithm})
we have
$$
    f(x)=(x-r)q(x)+c\eqno(*)
$$
where $c$ is a polynomial of degree less than the degree
of the divisor $(x-r)$. But the degree of the divisor $(x-r)$
is one so the degree of $c$ is zero, i.e. $c$ is a constant.
Now plug in $x=r$ to get
$
   f(r)=(r-r)q(r)+c=0+c=c
$
so~$(*)$ becomes
$
   f(x)=(x-r)q(x)+f(r).
$
\end{proof} 

\begin{corollary}[\jdef{Factor Theorem}]
If $r$ is a root of the polynomial $f(x)$,
then $(x-r)$ evenly divides $f(x)$,
i.e. $f(x)=(x-r)q(x)$.
\end{corollary}

\begin{proof}
By the Remainder Theorem just proved
we have
$$
    f(x)=(x-r)q(x)+f(r).
$$
Hence if $f(r)=0$, then $f(x)=(x-r)q(x)$. 
\end{proof} 

\begin{theorem}[\jdef{Complete Factorization}]
Let $f(x)$ be a polynomial of degree $n$.
Then 
$$
   f(x)=c(x-r_1)(x-r_2)\cdots(x-r_n)
$$
where $c$ is a constant and 
 $r_1,r_2,\ldots,r_n$ are the roots of $f(x)$, i.e. the solutions of $f(x)=0$.
\end{theorem}

\begin{proof} 
By the Fundamental Theorem of Algebra, there is a complex number $r_1$
such that $f(r_1)=0$ so by the
Factor Theorem just proved there is a polynomial $f_1(x)$ such that
$$
   f(x)=(x-r_1)f_1(x).
$$
Now $f_1(x)$ is a polynomial of degree $n-1$ so by the same argument there is a 
complex number $r_1$ and a polynomial of $f_2(x)$ with $f_1(x)=(x-r_2)f_2(x)$
and hence
$$
    f(x)=(x-r_1)f_1(x)=(x-r_1)(x-r_2)f_2(x).
$$
Repeating $n$ times gives
$$
   f(x)=(x-r_1)(x-r_2)\cdots(x-r_n)c.
$$
where $c=f_n(x)$ has degree $0$, i.e. $c$ is constant.
\end{proof} 

\begin{corollary} 
Every real polynomial  
may be written as a product of real linear polynomials 
and real quadratic polynomials with negative discriminant.
\end{corollary}

\begin{proof} If the coefficients of $f(x)$ are real and $r$ is
a complex but non real root of $f(x)$ then $0=\overline{f(r)}=f(\bar{r})$
so the conjugate $\bar{r}$ is also a root.
Let  $p$ and $q$ bet the real and imaginary parts of $r$ so  $r=p+qi$ 
and $\bar{r}=p-qi$.  The $r+\bar{r}=2p$ and $r\bar{r}=p^2+q^2$.
The quadratic polynomial
$$
    (x-r)(x-\bar{r})= x^2-(r+\bar{r})x+r\bar{r}=x^2+2px+(p^2+q^2)
$$
has negative discriminant $b^2-4ac=4p^2-4(p^2+q^2)=-q^2$.
Thus the non real complex roots occur  in pairs in the
complete factorization and may be combined to give
the desired factorization into real (quadratic and linear) polynomials.
\end{proof} 

\section{Sequences and Series}\label{sec:Sequences_and_Series}

\dfn A \jdef{sequence} is a function whose domain
is a set of integers usually either the natural numbers
$1,2,3,\ldots$ or the nonnegative integers $0,1,2,3,\ldots$.
The output of the sequence when the input is $n$ is called the
$n$th \jdef{term} of the sequence.
\nfd

\para
Usually the input is denoted by a letter like $n$ near the middle of
the alphabet and the output for input $n$ is denoted by making
$n$ a subscript rather than by surrounding $n$ by parentheses,
i.e. we write $a_n$ rather than $a(n)$. Here are some examples.
$$
    a_n=5+3n, \qquad b_n=2^n, \qquad c_n=n^2.
$$
\arap

\dfn An \jdef{arithmetic sequence} is one of form
$$
    a_n= a+nd
$$
where $a$ and $d$ are constants. An arithmetic sequence has the
property that the difference of successive terms is a constant:
$$
   a_{n+1}-a_n=(a+(n+1)d)-(a+nd) =d.
$$
\nfd

\dfn A \jdef{geometric sequence} is one of form
$$
    g_n= ar^n
$$
where $a$ and $r$ are constants. A geometric sequence has the
property that the ratio of successive terms is a constant:
$$
   \frac{g_{n+1}}{g_n}=\frac{ar^{n+1}}{ar^n}=r.
$$
\nfd

\xmpl The sequence $c_n=n^2$ is neither arithmetic nor geometric.
The sequence  is not arithmetic because the difference
$$
  c_{n+1}-c_n=(n+1)^2-n^2=2n+1
$$ 
is not constant (i.e. it depends on $n$) and the sequence
is not geometric because the ratio
$$
  \frac{c_{n+1}}{c_n}=\frac{(n+1)^2}{n^2}= 1+\frac{2}{n}=\frac{1}{n^2}
$$ 
is not constant (i.e. it also depends on $n$).
\lpmx

\dfn Every sequence determines another sequence called the
corresponding \jdef{series}. The $n$th term of the series
is the sum of the first $n$ terms of the sequence. Thus
if the $n$th term of the sequence is $a_n$, then the
$n$th term of the series is
$$
     S_n=a_1+a_2+\cdots+a_n.
$$
\nfd

\dfn A series is often written using \jdef{Sigma notation} as in
$$
    S_n=\sum_{k=1}^n a_k.
$$
More generally
$$
   \sum_{k=m}^n a_k := a_m+a_{m+1}+\cdots+a_n.
$$
\nfd

\begin{theorem}\label{thm:geometric-sum} The sum of the first $n$ terms of a geometric sequence
is given by
$$
   \sum_{k=1}^n ar^k =a\;\frac{r^{n+1}-r}{r-1}.
$$
\end{theorem}
\begin{proof}
\end{proof}

\para\topic{Mortgages}.  A family takes out a 30 year  \$100,000 mortgage
to buy a house. The interest rate is 6\% per year and they will repay
the loan in 360 equal monthly payments. To compute the monthly
payment imagine that they family has taken 360 loans
$$
100000 =L_1+L_2+\cdots+L_{360}
$$
and they will repay the $k$th loan at the end of the $k$th month
with the monthly payment of $a$ dollars. The monthly interest rate
is $6\%/12=0.005$ so
the amount owed on the
$k$th loan after $k$ months is $(1.005)^kL_k$. This is the amount
of the $k$th payment so $a=(1.005)^kL_k$ so
$L_k=a(1.005)^{-k}$ so
$$
 100000 =L_1+L_2+\cdots+L_{360}=a(1.005)^{-1}+a(1.005)^{-2}+\cdots+a(1.005)^{-360}.
$$
Using Sigma notation and Theorem~\ref{thm:geometric-sum} 
with $r=(1.005)^{-1}$
this may be written
$$
100000=\sum_{k=1}^{360}a(1.005)^{-k}= 
a\frac{(1.005)^{-361}-(1.005)^{-1}}{(1.005)^{-1}-1}
$$
To evaluate the expression on the right multiply top and bottom by $1.005$
to get
$$
\frac{(1.005)^{-361}-(1.005)^{-1}}{(1.005)^{-1}-1}=
\frac{(1.005)^{-360}-1}{1-1.005}=\frac{1-(1.005)^{-360}}{0.005}=166.79
$$
so $100000= 166.79a$ so the monthly payment is
$$
   a=100000/166.79 =599.56.
$$
\arap

\section{Infinite Series}\label{sec:Infinite_Series}

\dfn For a sequence  $a_n$   the notation
$$
    \lim_{n\to\infty} a_n=b
$$
means that 
that the numbers $a_n$ are arbitrarily close to the number $b$ when 
$n$ is sufficiently large.  The expression  on the left is called the
the \jdef{limit} of $a_n$ as $n$ becomes infinite. For a series
we also use the notation
$$
   \sum_{k=1}^\infty a_k= \lim_{n\to\infty} \sum_{k=1}^n a_k.
$$
\nfd

\begin{theorem} \label{thm:infinite-geometric-sum}
 If the ratio $r$ of successive terms in a geometric
series is less than one in absolute value, then
the sum of the infinite geometric series is
$$
  \sum_{k=1}^\infty ar^k=\frac{ar}{1-r}.
$$ 
\end{theorem}

\begin{proof} By Theorem~\ref{thm:geometric-sum} 
$$
 \sum_{k=1}^n ar^k =\frac{ar-ar^{n+1}}{1-r}=\frac{ar}{1-r}+ cr^n
 $$
 where $c=ar/(1-r)$. In Section~\ref{sec:Exponentials_and_Logarithms} we saw that
 for $|r|<1$ the graph of the exponential function $y=r^x$
 decays exponentially as $x$ becomes large positive. This implies that
 $$
    \lim_{n\to\infty}r^n=0
 $$
 so
 $$
  \sum_{k=1}^\infty ar^k=\lim_{n\to\infty} \sum_{k=1}^n ar^k
  =\frac{ar}{1-r}+ c \lim_{n\to\infty}r^n =\frac{ar}{1-r}.
$$
\end{proof}

\xmpl To travel one mile I must first travel the first half mile, then
half of the remaining distance, then half of the remaining distance, and so on.
How can I ever go whole distance? 
The answer is that the sum of all the distances is one:
$$
\frac{1}{2}+\frac{1}{4}=\frac{3}{4},\qquad \frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{3}{4},
\qquad \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{15}{16}, \qquad \ldots.
$$
The numerator  is only one less than the denominator:
$$
  \sum_{k=1}^n \left(\frac{1}{2}\right)^k=  \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^n} =\frac{2^n-1}{2^n}=1-2^{-n}.
$$
The finite sums are getting closer to one and the infinite sum is
$$
    \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k =\frac{\frac12}{1-\frac12}=1.
$$
\lpmx


\xmpl  \label{xmpl:3/22}
We use Theorem~\ref{thm:infinite-geometric-sum}
to prove that the infinite repeating decimal $0.1\,36\,36\,36\,36\,36\,36\ldots$.
is equal to $3/22$. The first step is to make
the decimal look like a geometric series.
$$
\begin{aligned}
0.1\,36\,36\,36\,36\,36\,36\ldots
&=\frac{1}{10}+36\bigl(0.0\,01\,01\,01\,01\,01\ldots\bigr)\\
&=\frac{1}{10}+\frac{36}{10}\bigl(0.01\,01\,01\,01\,01\ldots\bigr)\\
&=\frac{1}{10}+\frac{36}{10}\bigl(10^{-2}+10^{-4}+10^{-6}+10^{-8}+\cdots\bigr)\\
&=\frac{1}{10}+\frac{36}{10}\;\sum_{k=1}^\infty\left(\frac{1}{100}\right)^k
\end{aligned}
$$
Now by Theorem~\ref{thm:geometric-sum}
$$
\sum_{k=1}^\infty\left(\frac{1}{100}\right)^k=
 \frac{0.01}{1-0.01} = \frac{0.01}{0.99}= \frac{1}{99}
$$
so 
$$
0.1\,36\,36\,36\,36\ldots=
\frac{1}{10}+\frac{36}{10}\cdot \frac{1}{99} .
$$
Now we do the arithmetic:\
$$
\frac{1}{10}+\frac{36}{10}\cdot \frac{1}{99} 
=\frac{99+36}{990}=\frac{135}{990}%=\frac{9\cdot15}{9\cdot110}
=\frac{15}{110}=\frac{3}{22}.
$$
\lpmx


\para The concept of an infinite sum makes the definition of decimal
expansion more precise. If $x$ is  a real number between $0$ and $1$
it has a decimal expansion
$$
    x=\sum_{k=1}^\infty d_k \,10^{-k}
$$
where each $d_k$ is an integer between $0$ and $9$. For most real numbers
the digits $d_k$ won't follow any pattern, but
\arap
%http://www.usna.edu/Users/math/wdj/book/node43.html

\begin{theorem}\label{thm:real-rational}  
A real number $x$ is rational if and only if has a repeating
decimal expansion  like the one in Example~\ref{xmpl:3/22}.
\end{theorem}

\begin{proof}  A rational number  is a ratio  $p/q$ of two integers
$p$ and $q$.
To see why the decimal expansion eventually repeats periodically
imagine computing the decimal expansion by long division. 
At each step in the long division algorithm we compute  
the next digit in the quotient,
multiply that digit by $q$, 
subtract the product to get the next remainder,
and bring down the next digit from the dividend. 
The remainder is smaller than $q$, otherwise we would  have  used a larger 
digit in the quotient we are computing. Once we are computing digits of 
the right of the decimal point the digit we bring down from the dividend is always zero
and since the remainder is always less than $q$ we will eventually find ourselves
redoing what we have already done. 

The proof that a real number with a repeating decimal expansion
is rational is just like the computation in Example~\ref{xmpl:3/22}.
We first write the number as the sum of  a finite decimal
and a negative power of ten times a geometric series.
We then use Theorem~\ref{thm:infinite-geometric-sum}
and do the arithmetic.
\end{proof}


\appendix

\section{Where to Look in the Text}

This appendix tells you where to look in the course text
%%
\begin{quote}
David Cohen: {\em College Algebra} Fifth Edition, 
Thomson Brooks/Cole 2003.
\end{quote}
%%
for additional reading.% for the corresponding section in these notes.



\begin{description}
 \item[\ref{sec:Laws_of_algebra}](Laws of algebra) The material in this section is mostly review but some of it
is discussed in Appendices~A.2 and~B.1 of the text.
\item[\ref{sec:Kinds_of_Numbers}](Kinds of Numbers)  
Most of the material in this section is covered in Section~1.1 of the text.
See  Appendix~A.3 of the text for the proof that $\sqrt2$ is irrational.
\item[\ref{sec:Coordinates_on_the_Line_and_Order}](Coordinates on the Line and Order) 
Inteval notation is explained in Section~1,1 (Page~3). 
\item[\ref{sec:Exponents}](Exponents) See Appendices~B.2 and B.3 in the text.
\item[\ref{sec:Coordinates_in_the_Plane_and_Graphs}](Coordinates in the Plane and Graphs) See sections~1.2 and~2.5 of the text.
\item[\ref{sec:Lines}](Lines)
\item[\ref{sec:Solving_Equations}](Solving Equations)
\item[\ref{sec:Systems_of_Equations}](Systems of Equations)
\item[\ref{sec:Symmetry}](Symmetry)
\item[\ref{sec:Completing_the_Square}](Completing the Square)
\item[\ref{sec:Functions}](Functions)
\item[\ref{sec:Inverse_Functions}](Inverse Functions)
\item[\ref{sec:Average_Rate_of_Change}](Average Rate of Change)
\item[\ref{sec:Polynomials}](Polynomials)
\item[\ref{sec:Rational_Functions}](Rational Functions)
\item[\ref{sec:Exponentials_and_Logarithms}](Exponentials and Logarithms)
\item[\ref{sec:Exponential_Growth_and_Decay}](Exponential Growth and Decay)
\item[\ref{sec:The_Natural_Logarithm}](The Natural Logarithm)
\item[\ref{sec:Complex_Numbers}](Complex Numbers)
\item[\ref{sec:The_Fundamental_Theorem_of_Algebra}](The Fundamental Theorem of Algebra)
\item[\ref{sec:Sequences_and_Series}](Sequences and Series)
\item[\ref{sec:Infinite_Series}](Infinite Series)
\item[\ref{sec:Where_to_Look_in_the_Text}](Where to Look in the Text)

\end{description}






\end{document}
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