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\begin{document}
\chapter{Pre-requisites}

We will go over the pre-requisites for a second course on geometry in
the form of exercises which lead to the important results. If you are
willing to assume the results stated then you do not need to do the
exercises, however, it {\em is} worthwhile to go over these at
least once in your life!

\section{Introduction of co-ordinates}

Let us assume that we have the usual axiomatic framework of Euclidean
geometry. We will show that the points on a line can be given
arithmetic operations and identified with the ``usual'' decimal
numbers. Moreover, we can introduce co-ordinates in space using the
Euclidean framework. One important thing to note is that use {\em
  only} the parallel postulate---congruence (hence distance and
angle) play no role in the introduction of co-ordinates.

We are given some gadget that can draw the line joining two
points and the line parallel to a given line through a point outside
it. Such a gadget is a {\em ruler with a roller}. %, shown in the
% figure below.
(Alternatively you can use the {\tt xfig} program).
%\begin{center}
%\vspace{2cm}
%\input{rulrol.pstex_t}
%\end{center}
In the following constructions (see Figures~\ref{add} and~\ref{mul})
we have numbered the lines and points in the sequence in which they
are obtained. Assume given a pair of points $0$ and $1$. We can define
addition and multiplication for points $a$ and $b$ on the line $l$
joining $0$ and $1$ by the constructions given below (the final point
in each construction is the sum or product of the the two original
points).
\begin{figure}[htb]
\centering
%addition
\includegraphics{figures.1}
\caption{Addition}\label{add}
\end{figure}
\begin{figure}[htb]
\centering
%multiplication
\includegraphics{figures.2}
\caption{Multiplication}\label{mul}
\end{figure}
\begin{exer}
Show that the following (usual) rules of arithmetic hold; in other
words the points on a line form a field.
\begin{enumerate}
\item The commutative law for addition: $a+b=b+a$.
\item The commutative law for multiplication: $ab=ba$.
\item The associative law for addition: $(a+b)+c=a+(b+c)$.
\item The associative law for multiplication: $a(bc)=(ab)c$.
\item The distributive law: $a(b+c)=ab+ac$.
\item The identity for addition: $a+0=a$.
\item The identity for multiplication: $a\cdot 1=a$.
\item For any $a$ there is a point $(-a)$ so that $a+(-a)=0$.
\item For any non-zero $a$ there is a point $(1/a)$ so that $a(1/a)=1$.
\item If $O'$ and $1'$ are two other points then give a natural
  correspondence between the points of the line $l'$ joining $0'$ and
  $1'$ and the line $l$ so that the arithmetic structure is
  preserved.
\end{enumerate}
\end{exer}
In addition, we can use the notion of order on the points of a line to
define an order in our arithmetic by saying that a number lies between
two other numbers if the corresponding points have the same relation.
In particular, we say that $a>0$ if $a$ is between the points $1$
and $0$ or if $1$ is between $a$ and $0$ or if $a$ is $1$.
\begin{exer}
Show in addition that 
if $a>0$ and $b>0$ then $a+b>0$ and $a\cdot b>0$.
\end{exer}
The following two important axioms are due to Archimedes (but only one
carries his name):
\begin{axiom}[Archimedean Property] (Also known as ``Big step -- Little
  step'') If $x>0$ (is a Little step) and $y>0$ (is the Big step) then
  there is a natural number $n$ (the number of little steps) so that
  $y$ is less than $nx$.
\end{axiom}
The second axiom is perhaps even less ``obvious'' but is essential.
\begin{axiom}[Least Upper Bound] If $A_n$ is a sequence of points so
  that for all $n$, $A_{n+1}$ lies between $A_n$ and $D$ for some
  fixed point $D$ (i.~e.\ $A_n$ move towards $D$ but do not reach it).
  Then there is a point $B$ which is the ``limit'' of $A_n$. In other
  words, $A_{n+1}$ is between $A_n$ and $B$ for all $n$ and if $C$ is
  any other point so that $A_{n+1}$ lies between $A_n$ and $C$ then
  $B$ lies between $A_n$ and $C$ for all $n$ (see figure~\ref{lub}).
\begin{figure}[htb]
\centering
%limit
\includegraphics{figures.3}
\caption{The Least Upper Bound}\label{lub}
\end{figure}
\end{axiom}
\begin{exer}
We introduce the decimal representation of a real number as
follows.
\begin{enumerate}
\item Use the Archimedean Property to show that for any real number
  $x$ there is an integer $n$ so that $n\leq x < n+1$. This integer is
  called the {\em integer part} $\floor{x}$ of $x$.
\item Show that the sequence $x_n = \floor{10^nx}/10^n$ is a
  non-decreasing sequence.
\item Use the Least Upper Bound property to conclude that $x_n$ has a
  limit $y$.
\item Using the principle of the excluded middle show that $y=x$.
\end{enumerate}
\end{exer}
Finally, we choose four non-coplanar points in space and designate them
$o$, $e_1$, $e_2$ and $e_3$. The point $o$ is called the origin the
line through $o$ and $e_1$ (respectively $e_2$ or $e_3$) is called the
$x$-axis (respectively $y$-axis or $z$-axis). By drawing lines
parallel to the axes we can produce for any point a unique triple of points
$(x,y,z)$ one on each axis which uniquely determine the point in space.
By the above method we obtain the co-ordinates in decimals as well.
\begin{exer}
  Show that a line in the plane is the locus of all points with
  co-ordinates $(x,y)$ such that $ax+by+c=0$ for some constants $a$,
  $b$ and $c$ so that $a$ and $b$ are not both zero. Also show the
  converse.
\end{exer}

\section{Conic sections}
In the co-ordinate plane one can study more general geometric figures
that those desribed by lines. In this section we undertake a rigorous
study of conic sections. In particular, we find geometric criteria
that distinguish the different conics. We also establish Steiner's
construction of conic sections as the locus of intersection of a pair
of rotating lines.

The first equation that is more complicated that the equation of a
line as given above is one of the form
\[  a x^2 + b xy + c y^2 + d x + e y + f = 0 \]
where $a$, $b$ and $c$ are not all zero (in which case the equation
would become that of a line).  The locus of points $(x,y)$ that
satisfy this equation is called a conic or a conic section.
\begin{figure}[htb]
\begin{center}
\includegraphics{figures.4}
\caption{The Smooth Conics}
\end{center}
\end{figure}
By plotting the corresponding curves we find that we have the
following types of conics:
\begin{enumerate}
\item There are no solutions.
\item The solutions all lie on one line.
\item The solutions all lie on a pair of lines.
\item The conic lies within a bounded region of the plane
  (i.~e.\ the conic is compact). This called an ellipse (of which the
  circle is a special case).
\item The conic has two parts (i.~e.\ the conic is disconnected).This
  is called a hyperbola.
\item The conic is connected and not compact. This is called a
  parabola. 
\end{enumerate}
We note that the first three types are distinguished without reference
to order among numbers (or separation axioms in geometry) and so make
sense over other fields. We will see below how we can distinguish the
other conics in a purely algebraic way.
\begin{exer}
  Find ways of distinguishing the different conics by looking at the
  equation. (Hint: Examine the discriminant $b^2-4ac$).
\end{exer}

For any line $ax+by+c=0$ with $a$ non-zero, we can write the solutions
in terms of one parameter as $({-bt-c}/{a},t)$; similarly when $b$
is not zero. We can also ``solve'' a conic. Let us suppose that the
conic is not of type (1), (2) or (3) above. Fix a point $(x_0,y_0)$ on
the conic.
\begin{exer}
  We will find a parametric solution of a conic. (Hint: Use
  translation and scaling of co-ordinates to simplify the equations
  wherever possible).
  \begin{enumerate}
  \item Let $(y-y_0)=t(x-x_0)$ be a line through this point. Show that
    there is at most one other point of the conic that lines on this
    line.
  \item Find the co-ordinates of this point in terms of the constants
    $a$, $b$, $c$, $d$, $e$, $f$, $x_0$, $y_0$ and the parameter $t$.
  \item Show that this parametric solution is not well defined at two
    values of $t$ for a hyperbola.
  \item Show that this parametric solution misses one point in the
    case of an ellipse or circle but is well-defined at all values of
    $t$.
  \item Show that this parametric solution is not well defined for one
    value of $t$ and misses one point or is well defined and misses no
    points on a parabola.
  \end{enumerate}
\end{exer}
This can be carried further through Steiner's construction as follows.
Let $(x_1,y_1)$ another point on the conic.
\begin{exer}
  Show that there are constants $A$, $B$, $C$ and $D$ so that for any
  point $(x_2,y_2)$ of the conic we have
   \[ s = \frac{At +B}{C t + D} \text{ when }
         t = \frac{y_2 - y_0}{x_2 - x_0} \text{ and }
         s = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   Moreover, these constants are such that if we try to solve for
   $s=t$, we obtain no solutions when the conic is an ellipse (or
   circle), one solution for a parabola and two solutions for a
   hyperbola.
\end{exer}
\begin{figure}[htb]
\begin{center}
\includegraphics{figures.5}
\caption{Steiner's Construction}
\end{center}
\end{figure}
The geometric content of this is the statement that the conic is
obtained as the locus of intersection of a pair of rotating lines
based at $(x_0,y_0)$ and $(x_1,y_1)$ respectively with respective
slopes $s$ and $t$ related by $s(Ct + D) = At +D$.
\begin{exer}
  Prove the converse that such a locus is always a conic.
\end{exer}

\section{Polynomials and polynomial functions}
We now revise the definition and elementary properties of polynomials
and polynomial functions. The fundamental ideas of calculus consist of
extending these notions to a larger class of functions.

There are two ways of approaching the concept of function. The ancient
way is through formulae, while the modern approach is through the
study of functions of sets of points. Most functions that we study
arise naturally and can be defined formally (i.~e.\ by formulae or
expressions). On the other hand, many of the properties demanded of
functions are best defined by thinking of them as set functions.
Moreover, most of the formulae have a ``life of their own''; the
formal expressions have a more general validity than as functions
alone. Thus the study of formulae becomes algebra while the study of
functions becomes analysis. Calculus is thus seen differently by
algebraists and analysts.
% Changes Start Here
The fundamental example in both cases is that of a polynomial which we
study below.

\subsection{Polynomials in one variable}
A polynomial $P$ in one variable $x$ is formally defined as a follows
\[      P(x) = p_0 + p_1 x + \cdots + p_n x^n \]
where the $p_i$ are constants. If $n=0$ we identify the polynomial
with the constant $p_0$. If $p_n\neq 0$ then we say the polynomial has
degree $n$. If $p_n=0$ then we drop the corresponding term unless
$n=0$; the degree of the constant polynomial $0$ is considered
undefined.
\begin{exer}
  Define the arithmetic operations on polynomials algorithmically so
  that polynomial manipulations can be implemented on a computer.
\end{exer}
Polynomials can be ``evaluated'' to give functions; for any choice of
a of constant $b$, we can substitute $x$ by $b$ to obtain the
``value'' of the polynomial; this gives us the function associated
with a polynomial. However, note that when the field of constants is
finite (e.~g.\ the field with two elements $\bbF_2$) a non-constant
polynomial might induce a constant function.
\begin{exer}
  Give an example of a polynomial that is not constant but gives a
  constant function on the field with three elements.
\end{exer}
The points where the function associated with the polynomial
vanishes are called solutions or roots of the polynomial.
\begin{exer}
  Let $P(x)$ be a polynomial of degree $n$ in one variable. The
  constant $b$ is a root of $P$ (i.~e.\ $P(b)=0$) if and only if
  $P(x)$ can be written as a product $(x-b)Q(x)$ where $Q(x)$ has
  degree $n-1$. Hence or otherwise show that $P(x)$ has at most $n$
  roots.
\end{exer}
The polynomial $x^2+1$ has no roots over the field of real
numbers. The field of complex numbers is collection of numbers of the
form $a+b\sqrt{-1}$; where $a$ and $b$ are real numbers.
\begin{exer}
  Define the arithmetic operations on complex numbers algorithmically
  in terms of the arithmetic operations on real numbers. Show that any
  quadratic polynomial $x^2+ax+b$ is a product $(x-d)(x-c)$ over the
  field of complex numbers ($d$ and $c$ need not be distinct).
\end{exer}
The Fundamental Theorem of Algebra states that {\em any} polynomial
over the field of real numbers is a product of linear and quadratic
terms upto a non-zero constant multiple. Thus any polynomial with real
coefficients has all its roots over complex numbers.
\begin{exer}
 Assuming the fundamental theorem of algebra show that any polynomial
 with complex coefficients has all its roots over complex numbers.
\end{exer}
In the above discussion, we used the phrase ``has all its roots'' as a
synonym for writing the polynomial as a product of linear terms. Now
if some of these factors repeat then we say we have repeated or
multiple roots. In particular, we can say that $(x-b)^k$ vanishes $k$
times at $b$.
\begin{exer}
  Use the Binomial theorem to write a polynomial of $P$ degree $n$ as
  follows 
  \[      P(x) = \tilde{p}_0 + \tilde{p}_1 (x-b) +
                         \cdots + \tilde{p}_n (x-b)^n \]
  for some constants $\tilde{p}_i$.
\end{exer}
Thus we can say that a polynomial vanishes to order $k$ at $b$ if the
terms in the above expression for it satisfy $\tilde{p}_i=0$ whenever
$i<n$.
\begin{exer}
  If $P$ and $Q$ vanish to order $k$ at $b$, then so does $R\cdot P +
  Q$ for any polynomial $R$.
\end{exer}
Note that $\tilde{p}_0$ is the value $P(b)$ of $P$ at $b$. Moreover,
the term $\tilde{p}_1$ depends on $P$ {\em and} on $b$ that we will
now to determine. Suppose that we have
\begin{eqnarray*}
  P(x) & = & \tilde{p}_0 + \tilde{p}_1 (x-b) +
   \text{ terms that vanish to order 2 at $b$ } \\
  Q(x) & = & \tilde{q}_0 + \tilde{q}_1 (x-b) +
   \text{ terms that vanish to order 2 at $b$ } \\
  R(x) & = & \tilde{r}_0 + \tilde{r}_1 (x-b) +
   \text{ terms that vanish to order 2 at $b$ } \\
\end{eqnarray*}
The above algebraic property of vanishing to order $n$ shows us that
\begin{multline*} 
  R\cdot P + Q = (\tilde{p}_0\tilde{q}_0+\tilde{r}_0)
       + (\tilde{r}_0\tilde{p}_1 + \tilde{r}_1\tilde{p}_0+
            \tilde{q}_1)(x-b) \\
            + \text{ terms that vanish to order 2 at $b$ } 
\end{multline*}
Thus, if we denote the rule that associates the constant $\tilde{p}_1$ with
$P$ as $(d/dx)_{|b}$, then this satisfies
\[
(d/dx)_{|b}(R\cdot P + Q) =
        (d/dx)_{|b}(R) P(b) + R(b) (d/dx)_{|b}(P) + (d/dx)_{|b}(Q)
\]
Such a rule (which takes polynomial to constants) is called a
(constant) {\em derivation}. If we formally replace $b$ by the
variable $x$ we obtain the requirement for a polynomial derivation
(which takes polynomial to polynomial). A polynomial derivation is a
rule $D$ which associates to every polynomial $P$ another polynomial
$D(P)$ so that
\[ D(R\cdot P + Q) = R\cdot D(P) + D(R) \cdot P + D(Q). \]
and $D(c)=0$ for a constant polynomial $c$.
\begin{exer}
  For any derivation $D$ and any polynomial $P$ show that $D(P^n)=n
  P^{n-1} D(P)$. (Hint: Use induction). Hence or otherwise show that a
  derivation is determined on all polynomials once one knows what it
  does to the variable $x$.
\end{exer}
In particular, for there is a derivation which takes $x$ to 1. This is
called the derivative with respect to $x$ and is denoted by $dP/dx$.
\begin{exer} Show that the value of $dP/dx$ at $b$ is
  $(d/dx)_{|b}(P)$. 
\end{exer}
The relation between derivations and the order of vanishing is given
by
\begin{exer}
  If $P$ vanishes to order $n$ at $b$ then $D(P)$ vanishes to order
  $n-1$ (for any derivation $D$).
\end{exer}
One of the aims of calculus is to find a larger class of functions
which can be studied in a manner analogous to that given above for
polynomials. To do this we need to generalise the notion of
``vanishing to order $n$'' and derivations.

A simple way to enlarge the class is to consider ``rational
functions'', constructed from polynomials the same way as fractions are
from natural numbers. A rational function is of the form $P/Q$ where
$P$ and $Q$ are polynomials with $Q\neq 0$.
\begin{exer}
  Extend the definitions of the arithmetic operations and $d/dx$ to
  rational functions.
\end{exer}
Let $P$ and $Q$ be any polynomials. The usual division algorithm
allows us to write an expression $P=R Q + S$ , where $R$ and $S$ are
polynomials and the degree of $S$ is less than that of $Q$ or $S$ is
zero. Using this and the fundamental theorem of algebra it is not
difficult to show
\begin{exer}
  Every rational function over real numbers is the sum of terms of the
  form
  \[
              \frac{a}{(x-b)^n} \text{ and/or }
                  \frac{a x + b} { ((x-c)^2 + d^2)^n }
  \]
This the called the partial fraction expansion.
\end{exer}
Given a polynomial $P$ consider the problem of trying to find a
polynomial $Q$ so that $dQ/dx=P$. This is quite easily solved using
the fact (proved above) that $d(x^n)/dx = n x^{n-1}$. When the problem
is posed for rational functions it becomes a bit harder.
\begin{exer}
  Show that
  \begin{eqnarray*}
     \frac{d}{dx} \frac{1}{(x-b)^n} & = & \frac{-n}{(x-b)^{n+1}} \\
     \frac{d}{dx} \frac{1}{((x-c)^2 + d^2)^n} & = & 
     \frac{-2n(x-c)}{((x-c)^2 + d^2)^{n+1}} 
  \end{eqnarray*}
\end{exer}
Hence given any rational function $P/Q$ over reals the only hurdle to
solving the problem of finding a function $f$ so that $df/dx=P/Q$ is
to solve this when $P/Q$ is either $1/(x-b)$ or $(ax +
b)/((x-c)+d^2)$. In the section on integration we will see how these
problems can be solved.

\subsection{Polynomials in more than one variable}
Let us begin by considering polynomials in two variables $x$ and $y$
\begin{multline*}
P(x,y) = p_{0,0} + p_{1,0} x + p_{0,1} y + p_{2,0} x^2 +
             p_{1,1} xy + p_{0,2} y^2 + \dots + \\ 
             + \dots + p_{n,0} x^n + \dots + p_{0,n} y^n
\end{multline*}
where $p_{k,l}$ are constants. The degree of a term $p_{k,l}x^k y^l$
is defined as $k+l$ if $p_{k,l}$ is non-zero. The degree of a
polynomial is the maximum of the degrees of each of its non-zero terms
and the degree of the zero polynomial is undefined as before.
\begin{exer}
  Define the arithmetic operations on polynomials algorithmically so
  that polynomial manipulations can be implemented on a computer.
\end{exer}
Polynomials can be ``evaluated'' to give functions; for any choice of
a pair of constants $(a,b)$, we can substitute $x$ by $a$ and $y$ by
$b$ to obtain the ``value'' of the polynomial.  By identifying the
pair of constants $(a,b)$ with the corresponding point in the plane,
this gives us the function (on the plane) associated with the
polynomial. The points where the function associated with the
polynomial vanishes are called solutions of the polynomial.

In particular, a polynomial $P(x,y)$ as above has the origin $(0,0)$
as a solution only if the constant term $p_{0,0}$ is zero. More
generally, we say a polynomial vanishes to order $m$ at the origin
$(0,0)$ if all its terms have degree at least $m$.

The Binomial theorem allows us to extend this notion to points
$(a,b)$ other than the origin.
\begin{exer}
  Use the Binomial theorem to write a polynomial of degree $n$ as a
  sum of terms of the form
  \[   \tilde{p}_{k,l} (x-a)^k (y-b)^{l} \]
where $k+l$ is at most $n$.
\end{exer}
Thus we can say that a polynomial vanishes to order $m$ at $(a,b)$ if
the terms in the above expression for it satisfy $k+l\geq m$ whenever
the coefficients $\tilde{p}_{k,l}$ are non-zero.
\begin{exer}
  If $P$ and $Q$ vanish to order $m$ at a point,
  then so does $R\cdot P + Q$ for any polynomial $R$.
\end{exer}
Because of this we can work with arithmetic operations on polynomials
``modulo'' terms that vanish to order $m$ at a given point
$(a,b)$. In particular, any polynomial is like a linear
polynomial upto terms that vanish to order two.
\begin{multline*}
 P(x,y) = \tilde{p}_{0,0} + \tilde{p}_{1,0} (x-a) +
              \tilde{p}_{0,1} (x-b) + \\
   \text{ terms that vanish to order 2 at $(a,b))$ } 
\end{multline*}
As before $\tilde{p}_{0,0}$ is the value of $P$ at $(a,b)$.
\begin{exer}
If we denote the rule $P\mapsto \tilde{p}_{0,0}$ by
$(\partial/\partial x)_{|(a,b)}$
then check that this is a constant derivation.
\end{exer}
We also have polynomial derivations and as before
\begin{exer}
Any polynomial derivation is determined by what it does to the two
variables $x$ and $y$.
\end{exer}
In particular we have $(\partial/\partial x)$ which is defined as the
derivation that sends $x$ to 1 and $y$ to 0; $\partial/\partial y$ is
defined by symmetry.
\begin{exer} Repeat this subsection replacing two variables $x$ and
  $y$ with $n$ variables $(x_1,\dots,x_n)$.
\end{exer}

\section{Sequences}
In this section we revise the notion of convergence for real numbers
and prove the Bolzano-Weierstrass property.

From section~1 we have a least upper bound (greatest lower bound) for
any bounded increasing (respectively decreasing) sequence of real
numbers.
\begin{exer}
  Show that any bounded non-decreasing sequence of real numbers has a
  least upper bound; a bounded non-increasing sequence has a greatest
  lower bound.
\end{exer}
Now if $S$ is any bounded non-empty set of real numbers, let $c_1$ be an
upper bound and $s_1\in S$. Now we iteratively define a pair of
sequences $\{s_n\}$ and $\{c_n\}$ as follows. If $(s_n+c_n)/2$ is an
upper bound for $S$ then we define $s_{n+1}=s_n$ and
$c_{n+1}=(s_n+c_n)/2$; otherwise let $s_{n+1}$ be any element of $S$
so that $s_{n+1} > (s_n+c_n)/2$ and $c_{n+1}=c_n$.
\begin{exer}
  Show that $\{s_n\}$ is a non-decreasing sequence of elements of $S$
  and $\{c_n\}$ is a non-increasing sequence of upper bounds for $S$
  so that $(c_{n+1} - s_{n+1})\leq (s_n-c_n)/2$. Hence show that the
  greatest lower bound of $\{c_n\}$ is equal to the least upper bound
  of $\{s_n\}$ and this bound is the least upper bound for $S$.
\end{exer}
In particular, if $\{x_n\}$ is any bounded sequence of real numbers we
have a least upper bound and greatest lower bound for this sequence.
Let us define
\begin{eqnarray*}
  l_k & = & \text{ the greatest lower bound of }
              \{ x_n | n \geq k \} \\
  u_k & = & \text{ the least upper bound of }
              \{ x_n | n \geq k \} \\
  \liminf \{x_n\} & = & \text{ the least upper bound of } \{l_k\} \\
  \limsup \{x_n\} & = & \text{ the greatest lower bound of } \{u_k\} 
\end{eqnarray*}
Note that $\{l_k\}$ is a non-decreasing sequence and $\{u_k\}$ is a
non-increasing sequence.
\begin{exer}
  Show that $\liminf \{x_n\}\leq \limsup \{x_n\}$.
\end{exer}
We say that the sequence $\{x_n\}$ has a limit (is convergent) if
these two numbers are equal; this number $c$ is called the limit
of this sequence of numbers.
\begin{exer}
  Show that for every positive $\epsilon$ there is a index $n_0$ so
  that $\abs{x_n-c}<\epsilon$ for all $n>n_0$. Hence, there is an
  index $n_1$ so that $\abs{x_n-x_m}<\epsilon$ for all $n>n_1$; this
  called Cauchy's criterion. Conversely show that any
  sequence $\{x_n\}$ that satisfies Cauchy's criterion is convergent.
\end{exer}
Now for any sequence $\{x_n\}$ we can find subsequences
$\{y_k=x_{n_k}\}$ (with $n_1<n_2<\cdots$) so that $\{y_k\}$ is
convergent (this is called the Bolzano-Weierstrass property).
\begin{exer}
  Show that there are subsequences of $\{x_n\}$ that converge to
  $\liminf\{x_n\}$ and $\limsup\{x_n\}$.
\end{exer}
Finally, let us note some algebraic properties of convergent
sequences.
\begin{exer} 
  Show that the sum, difference and product of convergent sequences is
  limit of the sum, difference and product of the terms. If a sequence
  has a non-zero limit then show that the inverse of the limit is the
  limit of the inverses of the non-zero terms of the sequence.
\end{exer}

\section{Functions, continuity and differentiability}
In this section we will revise the definition and elementary
properties of differentiable functions ($n$ times differentiable
functions).

\subsection{Definitions}
The study of differentiable functions is the study of functions that
mimic the behaviour of polynomials ``approximately''. To begin with we
must formally define the notion of approximation.
\begin{exer}
For any real number $0<x<1$ show that $x^n$ is a decreasing sequence
with limit $0$.
\end{exer}
In particular, we see that a polynomial that vanishes to order $(n+1)$
at 0 satisfies the following condition on functions of one variable.
\begin{defn} A function $g(x)$ of one variable is said to be in
  $o(x^n)$ if for any $\epsilon >0$ there is a $\delta >0$ so that
  \[
  \abs{g(x)} < \epsilon \abs{x^n} \text{ for all $x$ so that }
            \abs{x}< \delta.
  \]
\end{defn}
An alternative notion is
\begin{defn} A function $g(x)$ one variable is said to be in $O(x^n)$
  is there is a $\delta >0$ and a constant $C$ so that 
  \[
  \abs{g(x)} < C \abs{x^n} \text{ for all $x$ so that }
            \abs{x}< \delta
  \]
\end{defn}
Clearly, any polynomial that vanishes to order $n$ is $O(x^n)$. Further,
it is clear that an function $g(x)$ that is $O(x^n)$ is $o(x^{n-1})$
and any function that is $o(x^n)$ is $O(x^n)$.

We can extend these notions to many variables as well. A function
$g(x_1,\dots,x_n)$ of $n$ variables is said to be in $o(\bldx^n)$
(respectively $O(\bldx^n)$) if for all lines
$(x_1,\dots,x_n)=(xc_1,\dots,xc_n)$ through the origin the restricted
function $f(x)=g(xc_1,\dots,xc_n)$ is in $o(x^n)$ (respectively
$O(x^n)$). We can further extend this to define $o((\bldx-\bldb)^n)$
and $O((\bldx-\bldb)^n)$ where $\bldb=(b_1,\dots,b_n)$ is some point,
as a way of approximating functions near this point.

We say that $g$ and $f$ agree upto $o((\bldx-\bldb)^n)$ (or $f$
approximates $g$ upto $o((\bldx-\bldb)^n)$) if $f-g$ is in
$o((\bldx-\bldb)^n)$. Note in particular, that $f$ and $g$ take the
same value at $\bldb$.

A function is differentiable $n$ times at the point $\bldc$ if it is
approximated upto $o((\bldx-\bldb)^n)$ by a polynomial (of degree $n$).
Clearly, a polynomial of any degree is differentiable by the results
of the previous section. In the one variable case we write this as
follows 
\[  f(x) = a_0 + a_1 x + \cdots + a_n x^n + o((x-c)^n) \]
\begin{exer}
  Show that for any two functions $f$ and $g$ in $o(\bldx^n)$ and a
  function $h$ which is differentiable $n$ times at the origin, the
  function $h\cdot f + g$ is in $o(\bldx^n)$.
\end{exer}
\begin{exer}
  Show that the numbers $a_k$ are uniquely determined by the function
  $f$.
\end{exer}
Now the number $a_1$ depends on $f$ and the point $c$. Now suppose
that $f$ is differentiable (1 times) at all points $c$ so that it can
be written as above near every point $c$. Then we can define the
derived function $f'$ by letting $f'(c)=a_1$ for each point $c$; the
function $f'$ is also called the derivative of $f$. Now it clear that
if $f$ is the function given by a polynomial $P$ then $f'$ is
$dP/dx$. Thus we also use the notation $df/dx$ for $f'$. We have the
derivation property as well.
\begin{exer}
  If $f$, $g$ and $h$ are differentiable then so is $hf+g$ and
  \[  \frac{d}{dx}(hf+g) = \frac{dh}{dx} f + h \frac{df}{dx} 
                           + \frac{dg}{dx}         
  \]
\end{exer}
However, unlike the condition of vanishing to order $n$ at
$c$, the condition $o((x-c)^n)$ is not very well behaved.
\begin{exer}
  Show that $f(x)=x^2\sin(1/x)$ is $o(x)$ but the derivative of $f'$
  is not $o(x^0)$.
\end{exer}
A function $f(\bldx)$ is called continuous at a point $\bldc$ if
$f(\bldx)-f(\bldc)$ is $o(\bldx-\bldc)$ (i.~e.\ it is differentiable 0
times!). It is called continuous it it has this property at all points.
Thus we would like to study functions $f$ which are differentiable and
in addition the derivative $f'$ is continuous. Such functions
are provided by the fundamental theorem of calculus.

\subsection{Properties}
We show the important properties called the intermediate value
property and extremal value property of continuous functions. We also
deduce the clutch of theorems called mean value theorem, Rolle's
theorem and so on for differentiable functions.

The following important property of continuous functions will be used
all the time.
\begin{exer}
  Let $f(x)$ be a continuous function and $\{x_n\}$ be a sequence
  converging to $c$, then $\{f(x_n\}$ is a sequence converging to
  $f(c)$. (Hint: Examine the condition for continuity near $c$).
\end{exer}
Let $f(x)$ be continuous for $x$ satisfying $a\leq x \leq b$. Let $c$
be a real number lying between $f(a)$ and $f(b)$ we want to show that
$c$ is a value of $f$; in other words any number intermediate to two
values is itself a value.
\begin{exer}
  Let $s$ be the least upper bound of the set
  \[  \{x| a\leq x\leq b \text{ and } f(x) \leq c \} \]
  Show that $f(s)=c$.(Hint: To show that $f(s)\geq c$ take a sequence
  of points approaching $s$ from above).
\end{exer}
Now let $C$ be the least upper bound of the values of $f(x)$, i.~e.\
it is the least upper bound of the set $\{f(x)|a\leq x\leq b\}$. The
$C$ is an extremal value for $f$.
\begin{exer}
  Show that $C=f(x)$ for some $x$ in the range $a\leq x \leq
  b$. (Hint: We have a sequence $\{x_n\}$ so that $f(x_n)$ converges
  to $C$; by the section on sequences this sequence has a convergent
  subsequence).
\end{exer}
The following property of differentiable functions is very important
\begin{exer}
  Let $f(x)$ be differentiable for $x$ satisfying $a\leq x \leq
  b$. Let $s$ be such that $f(s)$ is an extremal value for $f$. Then
  $f'(s)=0$. (Hint: Examine the condition for differentiability near
  $s$).
\end{exer}
Now suppose that $f(x)$ is differentiable in the range $a<x<b$ and
continuous at the endpoints $a$ and $b$ as well. Suppose that
$f(a)=f(b)=0$. There is a point $s$ where $f$ attains its maximal
value; similarly there is a point $t$ where $f$ attains its minimal
value. If $s$ is the point $a$ or $b$ then $f(x)\leq 0$ and if $t$ is
the point $a$ or $b$ then $f(x)\geq 0$. Thus in case both of these
occur then $f(x)=0$ for all $x$; then let $c=(a+b)/2$. Otherwise let
$c$ be any one of $s$ and $t$ which is not $a$ or $b$. Thus we have a
point where $f'(c)=0$.
\begin{exer}
  For a general function $g(x)$ which is differentiable in the range
  $a<x<b$ and continuous at the endpoints we apply this to the
  function
  \[      f(x) = g(x) + \frac{g(b) - g(a)}{b-a} \cdot (x-a) \]
  to show that there is a point $c$ where
  \[         g'(c) = \frac{g(b) - g(a)}{b-a} \]
\end{exer}

\section{New Functions from old}
There are three primary ways in which new functions can be
``constructed'' from old ones. The first and most familiar one is
integration. The second method is the inversion of a monotone
function. The third is the implicit definition by
an equation in two variables. 

Formally, the integral of a function $f$ is a function denoted by
$\int f$ which satisfies $d(\int f)/dx=f$; note that such a function
is only determined upto a constant since a constant has derivative 0.

Similarly, if $f(x)$ is a function, its formal inverse is a function
$g(y)$ so that $g\circ f = \text{identity}$. Finally, if $f(x,y)$ is a
function of two variables, we can look for a function $g(x)$ so that
$f(x,g(x))=0$ identically.

Algebraically this is all we need. We can easily compute the
values of derivatives at various orders of the new functions in terms
of those of the old functions.

Analytically, we need to show that such functions exists under certain
reasonable conditions on the given data. For exmaple, we have already
constructed an integral for a polynomial function.

Here is an example of inversion:
\begin{exer}
  Let $f(x)=x^3$ show that there is a function $g(y)$ so that
  $g(f(x))=x$ for all $x$.
\end{exer}
And an example of an implicit function:
\begin{exer}
  Let $f(x,y)=x^2+y^2-1$ show that there is a function $g(x)$ for
  $0<x<1/2$ so that $f(x,g(x))=0$.
\end{exer}
In the sections below, we shall construct such solutions in greater
generality. 

\subsection{Inverse functions}
We prove the inverse function theorem. A continuously differentiable
function which has non-zero derivative at a point has an inverse in a
neighbourhood of that point (see Figure~\ref{invfn}).
\begin{figure}[htb]
  \begin{center}
    \includegraphics{figures.6}
    \caption{The Inverse Function}
    \label{invfn}
  \end{center}
\end{figure}
A function is called {\em monotone} if it preserves or reverses order
over the domain of its definition. In other words if
$(f(x)-f(y))(x-y)$ does not change sign for all $x$ and $y$ in the
domain of definition of $f$. We say $f$ is {\em strictly} monotone if
in addition the above product is zero only when $x=y$. It is clear
that a strictly monotone function is one-to-one.
\begin{exer}
  Let $f$ be monotone and {\em continuous} in for $x$ in the interval
  $a<x<b$.  Show that there is a function $g$ on the range of $f$ so
  that $g(f(x))=x$ for all $x$ in the interval $a<x<b$. (Hint: Use the
  intermediate value theorem and extremal value theorem).
\end{exer}
One way to ensure that a function is monotone is to use Rolle's
theorem in reverse to prove:
\begin{exer}
  If $f$ is continuously differentiable and $f'(x_0)\neq 0$ then show
  that $f$ is monotonic in some interval around $x_0$. Hence show that
  $f$ has a inverse $g$ (as in the exercise above) in some small enough
  interval around $f(x_0)$.
\end{exer}
We can also compute the formal inverse
\begin{exer}
  If $f$ can be expressed as
  \[
   f(x)=f(x_0) + f_1 (x-x_0) + \dots + f_n (x-x_0)^n + o((x-x_0)^n)
  \]
  with $f_1\neq 0$, then show that the inverse function $g(y)$ has the
  following form where $y_0=f(x_0)$.
  \[
  g(y)=x_0 + \frac{1}{f_1} (y-y_0) - \frac{f_2}{f_1^3} (y-y_0)^2
                  + \dots + g_n (y-y_0)^n + o((y-y_0)^n)
  \]
  where $g_n$ is of the form $P_n(f_1,\dots,f_n)/f_1^{n+1}$, where
  $P_n$ is a polynomial function.
\end{exer}

\subsection{Implicit functions}
% Proof given by Kesavan
We will prove the implicit function theorem. 

Let $f(x,y)$ be a continuous function of two variables so that it is
continuous, and differentiable with respect to $y$ when $x$ is kept
fixed; in particular, we have expression
\[ f(x,y) = f_0(x,y_0) + f_1(x,y_0)(y-y_0) + o_x(y-y_0) \]
where the subscript in the $o_x$ denotes the dependence of the
condition on $x$. The function $f_1(x,y)$ is denoted by $\partial
f/\partial y$ and is called the partial derivative of $f$ with respect
to $y$ (we have already seen this for the case of polynomials). We
further assume that $f_1$ is continuous.
\begin{figure}[htb]
  \begin{center}
    \includegraphics{figures.7}
    \caption{The implicit function}
    \label{implfn}
  \end{center}
\end{figure}
Now suppose that $f(a,b)=0$ and $c=(\partial f/\partial y)(a,b)\neq
0$.  We want to find the implicit function $h(x)$ defined by $f=0$
(see \ref{implfn}). We do this by showing that for each $x$ near $a$
there is a unique $y$ so that $f(x,y)=0$. Equivalently, we need to
show that the function $g(x,y)=y-f(x,y)/c$ has a unique fixed point
for any chosen $x$ near $a$.

Contractions give rise to functions with a unique fixed point.
\begin{exer}
  Let $0<c<1$ be a constant. Let $g(x)$ be a function so that $g$
  send the interval $[a,b]$ to itself and $\abs{g(x)-g(y)}<c\abs{x-y}$
  for all $x$ and $y$ in $[a,b]$. Show that $g(x_0)=x_0$ for exactly
  one point $x_0$ in $[a,b]$.
\end{exer}
Consider the function $g(x,y)=y-f(x,y)/c$ as above; we have
$g(a,b)=b$ and $(\partial g/\partial y)=0$. For each fixed $x$ we
would like $g(x,y)$ to be a contraction on some interval around
$b$. As a first step:
\begin{exer}
  Using the continuity of $f_1$ show that there is an interval
  $[a-r,a+r]$ around $a$ and an interval $[b-s,b+s]$ around $b$ so
  that $\abs{(\partial g/\partial y)(x,y)}\leq 1/2$ for $x$ and $y$ in
  these respective intervals. (Hint: Write $\partial g/\partial y$ in
  terms of $f_1$ to show that it is continuous). In particular, by the
  mean value theorem show that \( \abs{g(x,y)-g(x,y')}\leq 1/2(y-y')
  \) on these intervals.
\end{exer}
Now by the continuity of $f$ (and thus of $g$) we can choose a smaller
$r$ so that $\abs{g(x,b)-g(a,b)}\leq s/2$ for $x$ in the interval
$[a-r,a+r]$. Since $g(a,b)=b$, it follows that
$g(x,y)=b+(g(x,y)-g(x,b))+(g(x,b)-g(a,b))$ lies in the interval
$[b-s,b+s]$. Applying the above exercise it follows that for every $x$
there is a unique point $y$ so that $g(x,y)=y$ or equivalently
$f(x,y)=0$. We denote this point $y$ as $h(x)$. This function $h$ is
the required implicit function.

We have the identity, 
\begin{multline*}
  h(x)-h(x')=g(x,h(x))-g(x',h(x'))= \\
       \left(g(x,h(x))-g(x',h(x))\right) +
                  \left(g(x',h(x))-g(x',h(x'))\right) 
\end{multline*}
Applying the mean value theorem to $g(x',y)$ we obtain
\[ \abs{h(x)-h(x')} \leq \abs{g(x,h(x))-g(x',h(x)} +
                         \frac{1}{2} \abs{h(x)-h(x')}
\]
\begin{exer}
  Use the above inequality and the continuity of $g(x,y_0)$
  for every fixed $y_0$ to conclude the $h(x)$ is continuous.
\end{exer}
Now if we assume in addition that $f$ is differentiable to order $k$
then it follows that $h$ is also differentiable to order $k$ by an
entirely similar reasoning to the one in the above exercise.
\begin{exer}
  If $f(x,y)$ has the form (near the point $(a,b)$)
  \begin{multline*}
    f(x,y) = f_{1,0}(x-a) + f_{0,1}(y-b) + \\
               f_{2,0}(x-a)^2 + 2 f_{1,1}(x-a)(y-b) +
               f_{0,2}(y-b)^2 + 
               o((x-a,y-b)^2)
  \end{multline*}
  with $f_{0,1}\neq 0$, then show that the implicit function $g(x)$
  has the form (near $x=a$),
  \[
    g(x) = b - \frac{f_{1,0}}{f_{0,1}}(x-a) 
      + \frac{f_{0,2}f_{1,0}^2 + 2 f_{1,1}f_{1,0}f_{0,1}
               - f_{2,0}f_{0,1}^2}{f_{0,1}^3} (x-a)^2 + o((x-a)^2)
  \]
\end{exer}

\subsection{Integration}
We develop the theory of integration of continuous functions and prove
the fundamental theorem of calculus.

Let $R$ be any (bounded) region in the plane which we want to measure
the area of. We can tile the plane with squares of unit length and
count the number of such squares that are contained in the region to
obtain an approximation to the area from below. On the other hand we
can count the number of squares that meet to region to obtain an
approximation to the area from above. We can repeat this with squares
of smaller size and appropriately scale the count it seems clear that
the approximant from below will increase and the approximant from
above will decrease. The least upper bound of the former is called the
inner measure and the greatest lower bound of the latter the outer
measure. To obtain an area for the region we must show that these two
numbers are the same; moreover, we would like these numbers to be
independent of the placement of the grid as well as rotation and/or
shearing of the grid.
\begin{exer}
  Show that the measure of any rectangle is the product of the two
  sides. More generally, show that the area of a parallelogram is the
  product of the height and the base.
\end{exer}
Let $f(x)$ be a (non-negative) continuous function for $a\leq x\leq
b$. Let $R$ be the region bounded by the lines $x=a$ and $x=b$ on the
left and right, by the $x$-axis below and the curve $y=f(x)$ at the
top. By the previous exercise we can calculate the measure by using
rectangles instead of squares. We do this in the following exercise. 
\begin{exer}
  By a partition $P$ of the interval $[a,b]$ we mean a (finite)
  collection of points $a=t_0<t_1<\cdots<t_n=b$. For any such
  partition we define $m_i$ (respectively $M_i$) to be the minimum
  (respectively maximum) value of $f(x)$ for $t_{i-1}\leq x \leq t_i$.
  \begin{enumerate}
    \item Show that the sum $L(P,f)$ (respectively $U(P,f)$)
      approximate the area of the region from below (respectively
      above), i.~e.\ are sums of areas of rectangles enclosed by
      (respectively enclosing) the region $R$. 
      \begin{eqnarray*}
          L(P,f) &=& \sum_{i=1}^n m_i (t_i-t_{i-1}) \\
          U(P,f) &=& \sum_{i=1}^n M_i (t_i-t_{i-1}) 
      \end{eqnarray*}
    \item If $P'$ is a finer partition than $P$ (i.~e.\ each point of
      $P$ is also a point of $P'$) then show that
      \[ L(P,f) \leq L(P',f) \leq U(P',f) \leq U(P,f) \]
    \item Let $P_n$ denote the partition of $[a,b]$ into $n$ equal
      parts. Show that
      \[ \sup \{ L(P,f) | P \text{ a partition } \} \geq
               \sup \{ L(P_n,f) \}
      \]
      Similarly for the infimum of the $U(P,f)$,
      \[ \inf \{ U(P,f) | P \text{ a partition } \} \leq
               \inf \{ U(P_n,f) \}
      \]
    \item Let $i=i(P,f)$ be such that the difference $M_i-m_i$ is
      maximum. Then show that
      \[ U(P,f) - L(P,f) \leq (M_i-m_i) (b-a) \]
      Let $x(P,f)$ denote the mid point of the interval
      $[i_{i-1},t_i]$ for this $i$.
    \item Let $c$ be any point of the interval $[a,b]$. For any
      positive $\epsilon$, show that there is a $\delta>0$ so that the
      difference between the maximum and minimum values of $f(x)$ on
      the interval $[c-\delta,c+\delta]$ is less than
      $\epsilon/(b-a)$. (Hint: use continuity of $f$ at $c$).
   \item The sequence $\{ x_n=x(P_n,f) \}$ has a convergent
      subsequence $\{ y_k=x_{n_k}\}$, with limit point $c$. Show that
      there is a $k_0$ so that if $k\geq k_0$ and $i=i(P_{n_k},f)$
      then the entire sub-interval $[t_{i-1},t_i]$ of the partition
      $P_{n_k}$ is contained in $[c-\delta,c+\delta]$.
   \item Deduce that $\sup\{L(P_{n_k},f)\} = \inf\{U(P_{n_k},f)\}$.
   \item Conclude that the inner and outer measure of the region $R$
     coincide. 
  \end{enumerate}
\end{exer}
This show that the area of the region $R$ is well-defined. It is
denoted by $\int_a^b f$ to denote its depends on $a$, $b$ and $f$ (we
will justify the use of the $\int$ symbol below. First of all note
that
\begin{exer}
  Let $a<c<b$ and $f(x)$ and $g(x)$ be (positive) functions that are
  continuous in the interval $[a,b]$; let $d>0$ be any positive
  constant. Then we have
  \[
      \int_a^c f + \int_c^b f  = \int_a^b f 
  \]
  and
  \[
      \int_a^b (d\cdot f+g) = d\cdot\int_a^b f + \int_a^b g
  \]
\end{exer}
Due to this we are justified in extending the definition as
follows. If $f(x)$ is continuous in the interval $[a,b]$ and $c>d$ 
are points in this interval we define $\int_c^d f := - \int_d^c
f$. Moreover, if $f$ is not everywhere positive we define
\[
   f_{+}(x) = \max\{f(x,),0\} = (f(x)+\abs{f(x)})/2 
\]
and
\[
   f_{-}(x) = \max\{-f(x,),0\} = (-f(x)+\abs{f(x)})/2 
\]
Then clearly $f_{\pm}$ are positive functions and $f=f_{+}-f_{-}$. We
then define the integral of $f$ by the formula $\int_a^b f := \int_a^b
f_{+} - \int_a^b f_{-}$. This definition too is justified by the
additive property given above.

Now if $f$ is continuous on an interval $[a,b]$ where its minimum
value is $m$ and its maximum value is $M$ it is clear that
\[ m (d-c) \leq \int_c^d f \leq M (d-c) \]
\begin{exer}
 Let $g(x)=\int_a^x f$, then $g$ is a differentiable function for all
 points $x$ so that $a<x<b$ and its derivative is $f$. (Hint: Apply
 the addition rule and use continuity of $f$ at $x$)
\end{exer}
This justifies the use of the $\int$ symbol. We have shown how to
analytically compute the function which we formally defined above.
In addition we have a way of constructing a function which is
differentiable and its derivative is continuous---a problem that was
raised in the section on continuous and differentiable functions.

\section{Curves}
Now that we have a basic understanding of differentiability of
functions we can begin the study of plane curves which can be defined
parametrically or as the locus of vanishing of a function of two
variables. At the very least we need the function to be (piecewise)
differentiable with ``good'' first order properties. We shall see
later that for any interesting (and {\ae}sthetic!) study
we will need second derivatives as well.

A plane curve is defined (locally) as the locus of points where a
``good'' function $f$ of two variables vanishes. In particular, let
$p=(a,b)$ be a point where $f$ vanishes, then we assume that 
\( f(x,y)= f_{1,0}(x-a) + f_{0,1} (y-b) + o(\bldx) \) is {\em
  continuously differentiable} at this point. The curve is said to be
{\em singular} at $p$ if both the above coefficients are 0; otherwise
we call the curve {\em non-singular} or {\em smooth}. For a smooth
curve through $p$ the line $f_{1,0}(x-a) + f_{0,1}(y-b) =0$ is called
the tangent line. It is the ``best'' linear approximation to the curve
in an obvious way:
\begin{exer}
  Consider the natural parametrisation of the line
  $\alpha(x-a)+\beta(y-b)=0$. The restriction of the function $f$ to
  this line can then be thought of as a function of one variable.
  Show that this function vanishes to order 2 if and only if the line
  is the one above {\em or} the curve $f=0$ is singular at the point
  $p=(a,b)$.
\end{exer}
In what follows we restrict our attention to smooth curves. Singular
curves are very interesting and are studied extensively in algebraic
geometry.

A different way of representing curves is to think of a curve as a
``moving point''. A curve can be given in parametric form by writing a
pair of functions $(x(t),y(t))$ so that as $t$ varies we will trace
out a curve. As before we will insist on the two functions being
continuously differentiable. We say that our curve is non-singular
at ``time'' $t=t_0$ we need at least of the pair $(x'(t_0),y'(t_0))$ to be
non-zero; otherwise we call the curve singular.
\begin{exer}
  Consider the function of $t$ given by substituting the above pair of
  functions in the linear form $l(x,y)=\alpha x + \beta y + \gamma$.
  Show that this function vanishes to order 2 at $t=t_0$ if and only
  if the curve is singular {\em or} the form $l(x,y)$ is the {\em
    tangent form} \( y'(t_0)(x-x(t_0)) - x'(t_0)(y-y(t_0)) \).
\end{exer}
We need to have some way of going from the parametric form of a curve
to the equation and vice verse. For the first problem, let us assume
(without loss of generality) that $x'(t_0)\neq 0$. Then, by the
inverse function theorem, we have $g(x)$ so that $g(x(t))=t$, so that
we can re-parametrise the curve to get $(x,y(g(x))$. The curve is the
(locally) given by the equation $y-h(x)=0$ where $h(x)=y(g(x)$.

To go from the equation to the parametric form we need to show that
lines parallel to a line which is {\em not} tangent to the curve will
meet the curve in exactly one point near the given point. This done
through the implicit function theorem. (Note to author: Exercises to
be added here).

One of the advantanges of working with ``orders'' of vanishing is that
make these theorems ``explicit'' if we only need our equations to be
satisfied upto terms of some order. For example, we say that
$(x(t),y(t))$ is a parametrisation at $t=t_0$ upto order $r$ of the
curve $f(x(t),y(t))=o((t-t_0)^r)$. Similarly, two curves, $f$ and $g$
are said to {\em osculate} upto order $r$ if $f-g=o(\bldx^r)$. In
particular, any curve osculates upto order 2 with a conic; thus it is
possible to write a parametrisation upto order two quite explicitly.

Finally, there is one distinguished parametrisatisation. Let
$(x(t),y(t))$ be a curve. Thinking of this as a moving point we have
not only a tangent line but a tangent vector (called the velocity
vector) $(x'(t),y'(t))$. It is thus natural to define the {\em speed}
of the curve as length of this vector.
We can ask for a constant speed (or more accurately constant energy)
parametrisation. In other words, can we find $t=u(s)$ so that
\[   (x'(t)^2 + y'(t)^2)_{\mid t=u(s)} u'(s)^2 = \text{constant} \]
\begin{exer}
  Use the inverse function theorem to show that the function
  $s(t)=\sqrt{x'(t)^2+y'(t)^2}$ has an inverse. Show that this inverse
  function satisfies the above equation.
\end{exer}
We will see that such a parametrisation called parametrisation by
arc-length plays an important role in geometry. Meanwhile,
\begin{exer}
  Show that to obtain such a parametrisation for the circle, we need
  to solve the equation
  \[    u'(t)^2 \frac{1}{(1+t^2)^2} = 1 \]
  (Hint: Use the following parametrisation.)
  \[
      (x(t),y(t)) =
         \left( \frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} \right)
  \]
\end{exer}
We will study the solution of this and related equations in the next
section.

\section{Elementary functions}
We are now in a position to introduce many interesting
functions. In particular, we define and obtain properties of the
exponential and trigonometric functions.

\begin{trivlist}
\item[{\bf Warning}] As a consequence of integration theory it is
clear that there are {\em more} functions that can be integrated than
those that can be differentiated. This runs contrary to the feeling
one gets in high school calculus due to the requirement there of
expressing all integrals as a ``formula'' in terms of elementary
functions.
\end{trivlist}

Recall, that the rational functions that could not be integrated in
terms of rational functions were of the form $1/(x-b)$ or $(ax +
b)/((x-c)^+d^2)$. The former is a continuous function when $x\neq b$
and the latter is continuous for all $x$. Thus we study the two
functions
\[
       A(x) = \int_1^x \frac{dt}{t} \text{ and }
       B(x) = \int_0^x \frac{dt}{1+t^2}
\]
The function $A(x)$ is called the logarithm denoted by $\log(x)$ and
the function $B(x)$ is the inverse of the tangent function denoted by
$\tan^{-1}(x)$ or $\arctan(x)$.
\begin{exer}
  Show that the function $A(x)$ is monotonic and goes to infinity as
  $x$ goes to infinity.
\end{exer}
In particular, it follows that $A$ has an inverse. This function is
called the exponential function and denoted by $\exp(x)$. We can
define the number $e$ as $\exp(1)$; it is called Napier's natural base
for the logarithm.
\begin{exer}
  Show that $e$ is between $2$ and $4$.
\end{exer}
Now the fundamental property of the logarithm is
\begin{exer}
  Show that $A(x+y)=A(x)+A(y)$. Consequently we obtain
  $\exp(xy)=\exp(x)\exp(y)$. 
\end{exer}
This property and the monotonicity of $\exp$ characterise it.
\begin{exer}
  Let $f(x)$ be a monotonic function so that $f(xy)=f(x)f(y)$ and
  $f(x)=a$. Show that $f(x)=\exp(x\log(a))$.
\end{exer}
Finally, we can define the hyperbolic functions as usual,
$\cosh(x)=1/2(\exp(x)+\exp(-x))$ and so on.

We noe study the functions related with the function $B$.
\begin{exer}
  Show that the function $B(x)$ is monotonic and remains bounded as
  $x$ goes to infinity.
\end{exer}
The least upper bound of the numbers $2B(x)$ is denoted by $\pi$. We
can thus define the function $\tan(x)$ on the range
$(-\pi/2,\pi/2)$ as the inverse of $B$.
\begin{exer}
  Show that an arclength parametrisation of the unit circle
  $x^2+y^2=1$ is given by $t\mapsto (\sin(t),\cos(t))$ where
  we define these functions by the formulae
  \begin{eqnarray*}
    \sin(x) & = & \frac{2\tan(x/2)}{1+\tan(x/2)^2} \\
    \cos(x) & = & \frac{1-\tan(x/2)^2}{1+\tan(x/2)^2} 
  \end{eqnarray*}
  Note that it follows that the perimeter of the circle $x^2+y^2=1$ is
  $2\pi$.
\end{exer}
It is thus natural to extend these functions {\em periodically} for
all $t$. We can {\em define} a multiplication on the unit circle by
\[ (x_1,y_1)\centerdot (x_2,y_2) =
           (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)
\]
\begin{exer}
  Show that the arclength parametrisation gives a group homomorphism
  from the additive group of real numbers to the circle.
\end{exer}
This exercise provides the link between the trigonometric and
exponential function which is expressed by the formula 
\[ \exp(\sqrt{-1} x ) = \cos(x) + \sqrt{-1} \sin(x) \]
which the reader can try to prove!
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