1 Introduction
1.1 Numbers
•
Natural numbers
N
These are all positive integers, including 0.
•
Integers
Z
These are the elements of
N
, plus the negative integers.
•
Rational numbers
Q
These are all the numbers which can be written
p/q
, where
p
and
q
= 0 are elements of
Z
. These numbers have either a finite number of decimalsor a periodic infinite number of decimals, for example1
.
3795 3795 3795 3795
······
Q
contains
Z
, which is obvious if one takes
q
= 1.
•
Real numbers
R
These are the elements of
Q
plus all the numbers with infinite andrandom decimals. Examples of real numbers, which are not in
Q
are:
√
2
,π,e,...
.
Density property:
Between any two real numbers can be found a rational number, and viceverse.
1.2 Group
A group
G
is a set of elements
{
a
}
, together with an operation
⋆
, such that:
•
if
a,b
are two elements of
G
, then
a⋆ b
is element of
G
;
•
G
contains a unit element
u
such that, for any element
a
of
G
,
a ⋆u
=
a
;
•
for any element
a
of
G
, there is an element ˜
a
such that
a ⋆
˜
a
=
u
.Examples of groups:
{
Z
,
+
}
, or
{
Q
⋆
,
×}
, where
Q
⋆
is
Q
without 0.
1.3 Combinatorics
•
Permutations
The number of ways to choose an order for
n
elements is the factorial
n
! =
n
×
(
n
−
1)
×
(
n
−
2)
···×
3
×
2
×
1
.
Indeed, there are
n
possibilities for the first element, and for each of these possibilities,there are
n
−
1 for the second element, etc...
•
Combinations
The number of ways to choose
k
elements out of
n
, independentlyof the order, is given by the binomial coefficients
nk
=
n
!
k
!(
n
−
k
)!
.
3
Indeed, the number of possible ways to order
n
points is
n
!, and has to be dividedby the number of ways to order the
k
chosen elements, which is
k
!, and also by thenumber of ways to order the remaining
n
−
k
elements, which is (
n
−
k
)!Some simple properties are:
nn
−
k
=
nk
,
n
1
=
n,
n
0
= 1
.
•
Binomial formula
. We show here that(
a
+
b
)
n
=
k
=
n
k
=0
nk
a
n
−
k
b
k
,
(1)where
n
is an integer and
a,b
are real numbers, using a
proof by induction
:
First step:
check that eq.(1) is valid for a given value of
n
, for example
n
= 2:(
a
+
b
)
2
=
a
2
+ 2
ab
+
b
2
=
20
a
2
b
0
+
21
a
1
b
1
+
22
a
0
b
2
=
k
=2
k
=0
2
k
a
2
−
k
b
k
.
Second step:
suppose that eq.(1) is valid for
n
, and show that it is then valid for
n
+ 1:(
a
+
b
)
n
+1
= (
a
+
b
)
k
=
n
k
=0
nk
a
n
−
k
b
k
=
k
=
n
k
=0
nk
a
n
−
k
+1
b
k
+
k
=
n
k
=0
nk
a
n
−
k
b
k
+1
=
k
=
n
k
=0
nk
a
n
+1
−
k
b
k
+
k
=
n
+1
k
=1
nk
−
1
a
n
+1
−
k
b
k
=
a
n
+1
+
b
n
+1
+
k
=
n
k
=1
nk
+
nk
−
1
a
n
+1
−
k
b
k
=
a
n
+1
+
b
n
+1
+
k
=
n
k
=1
n
+ 1
k
a
n
+1
−
k
b
k
=
k
=
n
+1
k
=0
n
+ 1
k
a
n
+1
−
k
b
k
.
4
2 Functions of a real variable
A function of a real variable
f
is an operation, which to a real variable
x
associates thequantity
f
(
x
).
2.1 Continuity
Intuitively, a function
f
of the variable
x
is continuous is a small change in
x
leads to asmall change in
f
(
x
). More rigorously,
f
is continuous in
x
0
if for any
ε >
0, one canalways find a
δ >
0 such that
|
x
−
x
0
|
< δ
⇒ |
f
(
x
)
−
f
(
x
0
)
|
< ε.
2.2 Differentiation
The derivative of a function
f
at the point
x
is the slope of the tangent of the curve
y
=
f
(
x
)at
x
. In order to calculate it, let’s consider the points
M
and
M
′
with coordinates (
x,f
(
x
))and (
x
+∆
x,f
(
x
+∆
x
)) respectively, where
dx >
0 is an increment (see fig(1)). The slopeof the straight line (
MM
′
) isslope =
f
(
x
+ ∆
x
)
−
f
(
x
)(
x
+ ∆
x
)
−
x
=
f
(
x
+ ∆
x
)
−
f
(
x
)∆
x.
The slope of the tangent of the curve at
M
is obtained when ∆
x
→
0. The derivative of
f
at the point
x
is then
f
′
(
x
) = lim
∆
x
→
0
f
(
x
+ ∆
x
)
−
f
(
x
)∆
x
=
df dx,
(2)where
dx
denotes the infinitesimal increment in
x
and
df
the corresponding infinitesimalincrement in
f
(
x
).
Example
Let us calculate the derivative of
f
(
x
) =
ax
n
, where
a
is a constant and
n
is aninteger. By definition
f
′
(
x
) = lim
∆
x
→
0
a
(
x
+ ∆
x
)
n
−
ax
n
∆
x
= lim
∆
x
→
0
a
[
x
n
+
nx
n
−
1
∆
x
+
n
(
n
−
1)
x
n
−
2
(∆
x
)
2
/
2 +
···
]
−
ax
n
∆
x
=
a
lim
∆
x
→
0
nx
n
−
1
+
n
(
n
−
1)
x
n
−
2
∆
x
+
···
=
anx
n
−
1
,
where the dots represent higher orders in ∆
x
.5
xf(x)x+ xf(x+ x)
∆∆
Figure 1: The derivative is the slope of the tangent.Eq, (2) defines the “right derivative”, for ∆
x >
0. One can also define the “left derivative”bylim
∆
x
→
0
f
(
x
)
−
f
(
x
−
∆
x
)∆
x,
where ∆
x >
0. A function
f
is said to be differentiable at
x
if these two definitions leadto the same result. If these two derivatives are different, the function is singular at thepoint
x
and its derivative is not defined. An example of such a singularity is the function
f
(
x
) =
|
x
|
at
x
= 0. Indeed, for
x
= 0, the left derivative is -1 and the right derivative is1.Note that a function can be continuous but not differentiable for a given value of
x
, asshows the previous example.
Extrema of a function
Since the derivative
f
′
(
a
) of a function at the point
a
correspondsto the slope of the tangent of the curve of equation
y
=
f
(
x
), we have the followingclassification:
•
if
f
′
(
a
)
>
0, then
f
is increasing in the vicinity of
a
;
•
if
f
′
(
a
)
<
0, then
f
is decreasing in the vicinity of
a
;
•
if
f
′
(
a
) = 0 and
f
(
x
) changes sign at
x
=
a
, then
f
(
a
) is an extremum of
f
;
•
if
f
′
(
a
) = 0 and
f
(
x
) does not change sign at
x
=
a
, then the point of coordinates(
a,f
(
a
)) is called an inflexion point. At such a point, the second derivative changessign.6
Derivative of a product
If
f,g
are two functions of
x
, the derivative (
fg
)
′
is given by(
fg
)
′
(
x
) = lim
∆
x
→
0
f
(
x
+ ∆
x
)
g
(
x
+ ∆
x
)
−
f
(
x
)
g
(
x
)∆
x
= lim
∆
x
→
0
f
(
x
+ ∆
x
)
g
(
x
)
−
f
(
x
)
g
(
x
) +
f
(
x
+ ∆
x
)
g
(
x
+ ∆
x
)
−
f
(
x
+ ∆
x
)
g
(
x
)∆
x
= lim
∆
x
→
0
g
(
x
)
f
(
x
+ ∆
x
)
−
f
(
x
)∆
x
+
f
(
x
+ ∆
x
)
g
(
x
+ ∆
x
)
−
g
(
x
)∆
x
=
f
′
(
x
)
g
(
x
) +
f
(
x
)
g
′
(
x
)
.
(3)
Chain rule
Consider two functions
f
and
g
, and the function
F
defined as
F
(
x
) =
f
(
g
(
x
)).The derivative of
F
is
F
′
(
x
) = lim
∆
x
→
0
F
(
x
+ ∆
x
)
−
F
(
x
)∆
x
= lim
∆
x
→
0
f
(
g
(
x
+ ∆
x
))
−
f
(
g
(
x
))
dx
= lim
∆
x
→
0
f
(
g
(
x
+ ∆
x
))
−
f
(
g
(
x
))
g
(
x
+ ∆
x
)
−
g
(
x
)
×
g
(
x
+ ∆
x
)
−
g
(
x
)∆
x
= lim
∆
x
→
0
f
(
g
+ ∆
g
)
−
f
(
g
)∆
g
×
g
(
x
+ ∆
x
)
−
g
(
x
)∆
x
=
f
′
(
g
(
x
))
×
g
′
(
x
)
,
where ∆
g
=
g
(
x
+ ∆
x
)
−
g
(
x
) is the increment in
g
(
x
) corresponding to
x
→
x
+ ∆
x
.
Derivative of a ratio
The derivative of 1
/x
is
−
1
/x
2
, such that the derivative of thefunction 1
/f
is
1
f
(
x
)
′
=
−
1
f
2
(
x
)
×
f
′
(
x
) =
−
f
′
(
x
)
f
2
(
x
)
.
As a consequence, the derivative of the ratio of the functions
f
and
g
is
f
(
x
)
g
(
x
)
′
=
f
′
(
x
)
×
1
g
(
x
)+
f
(
x
)
×
−
g
′
(
x
)
g
2
(
x
)
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
2
(
x
)
.
Derivative of an inverse function
If
y
=
f
(
x
), the inverse function
f
−
1
, when it exists,is defined by
x
=
f
−
1
(
y
).
Do not confuse the inverse function
f
−
1
with
1
/f
!
. In order todefine the inverse of a function, one needs a one-to-one mapping between
x
and
y
. This isusually the case on a given interval for
x
at least.The derivative of the inverse is then
f
−
1
(
y
)
′
= lim
∆
y
→
0
f
−
1
(
y
+ ∆
y
)
−
f
−
1
(
y
)∆
y
7
= lim
∆
x
→
0
x
+ ∆
x
−
xf
(
x
+ ∆
x
)
−
f
(
x
)=1
f
′
(
x
)
,
where ∆
x
is defined such that
y
+ ∆
y
=
f
(
x
+ ∆
x
).
2.3 Polynomial functions
A polynomial function of
x
is of the form
P
(
x
) =
n
=
N
n
=0
a
n
x
n
,
where
a
n
are the coefficients and
N
is the degree of the polynomial.If
N
is odd, the polynomial has at least one zero. Indeed, we have thenlim
x
→−∞
P
(
x
) =
−∞
and lim
x
→
+
∞
P
(
x
) = +
∞
,
such that the line representing
y
=
P
(
x
) cuts at least once the axis
y
= 0, since thepolynomial is a continuous function.A polynomial of degree 2 can have two poles, but might not have any (real) pole:
•
P
(
x
) =
a
(
x
−
z
1
)(
x
−
z
2
) has two poles
z
1
,z
2
. The pole is double if
z
1
=
z
2
;
•
Q
(
x
) =
ax
2
+
bx
+
c
has no pole if
b
2
−
4
ac <
0.A polynomial of degree 3 has either one pole or three poles, and can be written, for all
x
,
•
P
(
x
) =
a
(
x
−
z
1
)(
x
−
z
2
)(
x
−
z
3
) if
P
has three poles;
•
Q
(
x
) = (
x
−
z
)(
ax
2
+
bx
+
c
), with
b
2
−
4
ac <
0, if
Q
has one pole
z
.In general, any polynomial function can be written
P
(
x
) = (
x
−
z
1
)
···
(
x
−
z
n
)
×
(
a
1
x
2
+
b
1
x
+
c
)
···
(
a
m
x
2
+
b
m
x
+
c
m
)
,
where
z
i
,
i
= 1
,...,n
are the poles of the polynomial,
b
2
j
−
4
a
j
c
j
<
0 for all
j
= 1
,...,m
,and
n
+ 2
m
is the degree of the polynomial.
2.4 Rational functions
A rational function is the ratio of two polynomial functions
P
and
Q
, and has the form,for each
x
,
R
(
x
) =
P
(
x
)
Q
(
x
)
.
8
xyzzz
123
Figure 2: A polynomial function of degree 5, with three poles
z
1
,z
2
,z
3
.
xyzzzz
1243
Figure 3: A polynomial function of degree 6, with four poles
z
1
,z
2
,z
3
,z
4
.9
11xcos xsin xM
Figure 4: The coordinates of
M
on the trigonometric circle are (cos
x,
sin
x
).If the degree of
P
is less than the degree of
Q
, It is always possible to reduce
R
as a sumof
irreducible rational functions
of the form
a/
(
x
−
z
) or (
ax
+
b
)
/
(
x
2
+
cx
+
d
).
Example
The fraction (
x
+ 2)
/
(
x
2
+ 5
x
+ 4) can be written
x
+ 2
x
2
+ 5
x
+ 4=
x
+ 2(
x
+ 1)(
x
+ 4)=
ax
+ 1+
bx
+ 4
,
where
a
(
x
+4)+
b
(
x
+1) =
x
+2, such that
a
+
b
= 1 and 4
a
+
b
= 2, which gives
a
= 1
/
3and
b
= 2
/
3. Finally,
x
+ 2
x
2
+ 5
x
+ 4=1
/
3
x
+ 1+2
/
3
x
+ 4
.
2.5 Trigonometric functions
For a given angle 0
≤
x
≤
2
π
, sin x and cos x are defined as the coordinates of the point
M
at the intersection of the straight line (
OM
) with the trigonometric circle (see fig(4)).
Property
Using Pythagoras’ theorem, we have sin
2
x
+ cos
2
x
= 1.
Trigonometric formula
It will be shown in the chapter on vector calculus (subsection10
6.2) that the sine and cosine of the sum of two angles are given bysin(
a
+
b
) = sin
a
cos
b
+ sin
b
cos
a
cos(
a
+
b
) = cos
a
cos
b
−
sin
a
sin
b.
(4)
Important limit
We will now show, geometrically, thatlim
x
→
0
sin
xx
= 1
,
and this limit will be very useful in deriving fundamental properties of the trigonometricfunctions.
Proof:
From the definition of sine and cos, one can see on fig.(5) thatsin
x
≤
x
≤
sin
x
+ 1
−
cos
x.
(5)But one can also see that0
≤
sin
2
x
+ (1
−
cos
x
)
2
≤
x
2
,
such that0
≤
1
−
cos
x
≤
x
2
2
.
Using this in the inequalities (5), we obtainsin
xx
≤
1
≤
sin
xx
+
x
2
,
and the only possibility for this to be valid in the limit
x
→
0 is to have
sin
xx
→
1.
Derivative of trigonometric functions
The first important consequence of the previouslimit is the calculation of the derivative of the sine. From eq.(4) we havesin
′
x
= lim
∆
x
→
0
sin(
x
+ ∆
x
)
−
sin
x
∆
x
= lim
∆
x
→
0
sin
x
cos(∆
x
) + sin(∆
x
)cos
x
−
sin
x
∆
x
= lim
∆
x
→
0
cos
x
sin(∆
x
)∆
x
+ sin
x
cos(∆
x
)
−
1∆
x
.
We have seen that 1
−
cos(∆
x
) is of order ∆
x
2
, and therefore the second term vanishes inthe limit ∆
x
→
0, whereas the first term leads to(sin
x
)
′
= cos
x.
In the same way, one can easily show that (cos
x
)
′
=
−
sin
x
. As a consequence, we alsohave (tan
x
)
′
= 1 + tan
2
x
.11
abcx
Figure 5: On the figure:
a
= sin
x
,
b
= 1
−
cos
x
and
c
2
=
a
2
+
b
12
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