**Financial Mathematics Prepa ration Course Notes on
Calculus**

**1 The Real Numbers**

The basis of calculus is the set of real numbers which we
shall denote by

the symbol R. Loosely speaking the real numbers is the set of all infinite

decimal expansions a_{0}.a_{1}a_{2}a_{3}a_{4} . . . where the a_{i}’s are integers and a_{i}
for

i ≥ 1 is between 0 and 9. Clearly all integers and all rational numbers

p/q where p and q are integers, q ≠ 0 are real numbers (for instance the

integer −124 can be written −124.000000 . . . ,1/3 = 0.33333333 . . . ). We

denote the set of integers . . . ,−3,−2,−1, 0, 1, 2, 3, 4, 5, . . . by the
symbol

Z and the set of all rational numbers by Q. The set of positive integers

1, 2, 3, 4, 5, . . . , n, . . . is called the set of natural numbers and is
denoted N.

We have N ⊂ Z ⊂ Q ⊂
R.

Both the natural numbers, the integers and rational
numbers have the

property of being countable i.e. we can write them all down in a sequence

z_{1}, z_{2}, z_{3}, . . . , z_{n}, . . . . For instance we can write down all the
integers in the

sequence 0, 1,−1, 2,−2, 3,−3, 4,−4, . . . . It is a little more tricky to see
that

the set of all rational numbers are countable but it can be d one . The real

numbers, however are not countable. Here is the argument: assume that

R is countable and assume we have written down all the real numbers in a

sequence:

Thus the first index indicates the position in the sequence and the second

the position in the decimal expansion. We shall show that this is impossible

by constructing a real number which is not in this sequence. Choose b_{1} ≠

a_{11}, b_{2} ≠ a_{22}, b_{3} ≠ a_{33}, b_{4} ≠ a_{44}, b_{5} ≠ a_{}, b_{6} ≠ a_{66}, . . . . Now
form the

real number 0.b_{1}b_{2}b_{4}b_{4}b_{5}b_{6}b_{7}b_{8} . . . . Assume this number is the n’th
term in

the sequence . Thus 0.b_{1}b_{2}b_{3}b_{4} . . . b_{n} · · · = a_{n0}.a_{n1}a_{n2} . . . a_{nn} . .
. , but since

b_{n} ≠ a_{nn} they are not equal and so we have a contradiction. Since the

rational numbers, Q, is countable this shows that there are strictly more

real numbers than rational numbers: Q ⊆ R. Here is
another argument,

which goes all the way back to the ancient greeks, that shows that there are

real numbers which are not rational: consider the real number
. If this is

a rational number we can write where we
assume that the fraction

p/q is in lowest terms i.e. the integers p and q have no common factors . By

squaring both sides we get 2 = p^{2}/q^{2} or by cross-multiplying 2q^{2} = p^{2}. This

show that p^{2} is an even number, now if p itself was odd p would be of the

form 2r +1 for some integer r and so p^{2} = (2r +1)^{2} = 2r^{2} +1+4r which is

also odd. Hence in order for p^{2} to be even p itself must be even so p = 2r.

But then p^{2} = 4r^{2} and we get 2q^{2} = 4r^{2}. Dividing by 2 on both sides we

get q^{2} = 2r^{2} so q^{2} is even and hence also q must be even. This shows that

2 is a common factor in p and q in contradiction with the fact that p/q is in

lowest terms.

Real numbers can be added , multiplied and divided (if the
number we

divide by is ≠ 0) just as rational numbers can. What sets the real numbers

apart from the rational numbers, and which is the discovery that forms the

foundation of calculus is the concept of a limit. This notion took some 2000

years to discover. The ancient greek mathematicians studied geometry and

certainly discovered the number π as the ratio between the circumference

of a circle and its diameter. They were, however incapable of computing

Figure 1:

π because as we now know π is not a rational number. The greeks were

certainly comfortable with some non-rational numbers as they freely operated

for instance with square roots but as we also know now π cannot be

expressed even using square roots (or any other kind of roots, third, fourth,

etc).

**Example 1.1 **We want to compute the area of the unit
disc i.e. the disc

with radius 1

Inscribing the disc in a square with sides = 2 we get that
the area

≤ 4 We divide the disc into 4 equal slices and divide the interval into

two halves . We inscribe the quarter circle in the union of two rectangles

as q shown in fig.2. The height of the smaller of the two rectangles is

Thus the combined area of the two rectangles
is

It follows that the area of the disc Next we
divide

the interval into quarters and draw four rectangles: the height of the

rectangles are The combined area is

and so we get the estimate: area

of the disc

In the next step we divide the intervals into 1/8’ths and
we get the

combined area of the eight rectangles is

From this

we get the estimate: area of disc ≤ 3.3398

We can continue this way, next we divide into 1/16’ths,
1/32’ths,. . . , 1/2^{n}, . . . .

Figure 2:

In the n’th step the combined area of the 2^{n} rectangles
is

Thus we get the estimate: area of disc We

can see from the figures that the rectangles fit better and better around the

disc and so w can continue this process and approximate the area of the disc

as closely as we want. Using MATLAB we can easily compute the terms in

this sequence (at least as many as our computing power allows)

Figure 3:

Figure 4:

Here are the 25 first terms

Type in the following code as an M-file in MATLAB and use
it to compute

some of the numbers in this sequence:

**function y=A(n)**

v=0:2^ n;

S=sqrt(1-((2^ n-v)/2^ n).^ 2);

y=4*(1/2^ n)*sum(S);

MATLAB gives the following value for π = 3.14159265358979
and as we

Figure 5:

know the area of the unit disc is π it certainly looks
likely that the terms in

this sequence approaches π as n becomes larger

The notion that a sequence of real numbers approach
another real number

is fundamental in calculus and we need a formal definition of what it means.

Before we give the formal definition let us consider a few more examples of

sequences.

**Example 1.2 **1. a_{n} = 1/n

Thus the sequence {a_{n}} is the sequence {b_{n}}
is

and {c_{n}} is the sequence

We see that the terms in both {a_{n}} and {b_{n}} ”cluster”
around 0 but while

half of the terms of {c_{n}} cluster around 0 the other half grows larger and

larger.

It is this clustering behavior we shall use to give a
strict mathematical

definition of what it means for a sequence of real numbers to ”approach” a

fixed real number a.

**Definition 1.0.1** Let {a_{n}} be a sequence of real
numbers. We say that {a_{n}}

converges to a real number a (notation: a_{n} → a for n → ∞
or

if no matter how close we are to a, all the numbers in the sequence except

for at most finitely many are even closer to a. This is the definition in

words. In mathematics we have to make precise the notion of being close.

Figure 6:

Let ε be a positive real number (you should think of ε as
a very small positive

number). The numbers which are ε close to a consists of the real numbers x

such that the distance between x and a, |a − x| is smaller than ε. Thus the

real numbers ε close to a is the set {x|a − ε < x < a + ε} = (a − ε, a + ε).

The definition of convergence means that no matter how small an ε > 0

we take all but finitely many terms in the sequence are ε close to a i.e. in

(a−ε, a+ε). In other words for each ε there will be a number N (depending

on ε) such that when n > N, |a−a_{n}| < ε i.e. from N onwards the terms of

the sequence are ε close to a. Of course as we take smaller and smaller εs

more and more of the sequence will lie outside (a − ε, a + ε) but no matter

how small an ε it will always be only a finite number.

**Example 1.3 **In our area example, consider ε =
0.0001 then we see that

from n = 19 onwards we have | π −a_{n}| < ε. Remark that this does not prove

that a_{n} → π. In order to do that we would have to show that for any ε

we can find N (depending on ε of course) such that from n = N onwards

| π − a_{n}| < ε. We have only proved it for a single ε

Let’s formally prove that
1/n → 0, n → ∞.

Consider any ε > 0 and consider
1/ε
(if ε is very small this number will

be very large). Choose a large natural number N such that N >
1/ε
. Then

for any n > N we have n >
1/ε
and hence1/n
< ε. This proves that from N

onwards |1/n
− 0| =
1/n
< ε , which is precisely what it means for
1/n → 0.

**Homework Problems (due Monday 10/6):**

1. Show that for any two real numbers a, b the following inequality holds :

|a + b| ≤ |a| + |b| (Hint: consider separately the four cases a > 0, b > 0; a <

0, b > 0; a < 0, b < 0; a > 0, b < 0)

2. Show that for any two real numbers a, b, ||a| − |b|| ≤ |a − b|

3. Show that for any real number a the sequence a +
1/n → a

4. Show that if {a_{n}} and {b_{n}} are sequences such that a_{n} → a and

b_{n} → b then the sequence {a_{n} + b_{n}} converges to a + b, and the sequence

{a_{n}b_{n}} converges to ab