Real Analysis

– in a nutshell –

This book was created and used for the lecture Math-
ematical Analysis at Hamburg University of Tech-
nology in the summer term 2019 for General En-
gineering Science and Computer Science students.

Hamburg, 16th January 2022, Version 1.3

Julian P. Großmann
Hamburg University of
Technology
julian.grossmann@tuhh.de
ii

The author would like give special thanks

• to Timo Reis, Florian Bünger, Anton Schiela and Francisco Hoecker-Escut for an
excellent documentation of the Mathematical Analysis course held at Hamburg Uni-
versity of Technology before 2019,
• to Fabian Gabel and Jan Meichsner for many corrections and remarks,
• to all students who pointed out typos and other problems in this script.

Hamburg, 16th January 2022 J.P.G.

 Contents

1 Sequences and Limits 5

1.1 Just Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Convergence of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3 Subsequences and accumulation values . . . . . . . . . . . . . . . . . . . . 24
1.4 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.5 Bounded, Open, Closed and Compact Sets . . . . . . . . . . . . . . . . . . 30

2 Infinite Series 35
2.1 Basic Definitions, Convergence Criteria and Examples . . . . . . . . . . . . 36
2.2 Absolute Convergence and Criteria . . . . . . . . . . . . . . . . . . . . . . 41
2.3 The Cauchy Product of Series . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Continuous Functions 51
3.1 Bounded Functions, Pointwise and Uniform Convergence . . . . . . . . . . 51
3.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 Elementary Functions 63
4.1 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2 Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.3 Hyperbolic and Trigonometric Functions . . . . . . . . . . . . . . . . . . . 69
4.3.1 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.3.2 Area Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.3 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . 75
4.4 Arcus functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
4.5 Polynomials and Rational Functions . . . . . . . . . . . . . . . . . . . . . . 80
4.5.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.5.2 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.6 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

5 Differentiation of Functions 97
5.1 Differentiability and Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 97
5.2 Mean Value Theorems and Consequences . . . . . . . . . . . . . . . . . . . 105
5.3 Higher Derivatives, Curve Sketching . . . . . . . . . . . . . . . . . . . . . . 111
5.4 Taylor’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

iii
iv Contents

5.5 Simple methods for the numerical solution of equations . . . . . . . . . . . 120

6 The Riemann Integral 127

6.1 Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . . . . . 135
6.2 Integration Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
6.2.1 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . 138
6.2.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . 141
6.3 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . 143
6.4 Integration on Unbounded Domains and Integration of Unbounded Functions144
6.4.1 Unbounded Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
6.4.2 Unbounded Integrand . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.5 Parameter-dependent integrals . . . . . . . . . . . . . . . . . . . . . . . . . 152
6.6 Solids of revolution, path integrals . . . . . . . . . . . . . . . . . . . . . . . 154
6.7 Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Index 173
 Some words

This text should help you to understand the course Real Analysis. It goes hand in hand
with a video course you find on YouTube. Therefore, you will always find links and QR
codes that bring you to the corresponding videos.

The whole playlist can be found here::

https://jp-g.de/bsom/rab/ra-list/

To expand your knowledge even more, you can look into the following books:
• Jonathan Lewin: An Interactive Introduction to Mathematical Analysis,
• A. N. Kolmogorov: Introductory Real Analysis,
• Claudio Canute, Anita Tabacco: Mathematical Analysis I.
• Vladimir A. Zorich: Mathematical Analysis I.
Real Analysis is also known as Calculus with real numbers. It is needed for a lot of other
topics in mathematics and the foundation of every new career in mathematics or in fields
that need mathematics as a tool. We1 discuss simple examples later. Some important
bullet points are limits, continuity, derivatives and integrals. In order to describe these
things, we need a good understanding of the real numbers. They form the foundation of
a real analysis course.

complete number line

For this reason, the first step in a Real Analysis is to define the real number line. After
this, we will be able to work with these numbers and understand the field as a whole. Of
course, this is not an easy task and it will be a hiking tour that we will do together. The
1
In mathematical texts, usually, the first-person plural is used even if there is only one author. Most of
the time it simply means “we” = “I (the author) and the reader”.

1
2 Contents

summit and goal is to understand why working with real numbers is indeed a meaningful
mathematical theory.

We start in the valley of mathematics and will shortly scale the first hills. Always stay
in shape, practise and don’t hesitate to ask about the ways up. It is not an easy trip but
you can do it. Maybe the following tips can guide you:
• You will need a lot of time for this course if you really want to understand everything
you learn. Hence, make sure that you have enough time each week to do mathematics
and keep these time slots clear of everything else.
• Work in groups, solve problems together and discuss your solutions. Learning math-
ematics is not a competition.
• Explain the content of the lectures to your fellow students. Only the things you can
illustrate and explain to others are really understood by you.
• Learn the Greek letters that we use in mathematics:
α alpha β beta γ gamma Γ Gamma
δ delta  epsilon ε epsilon ζ zeta
η eta θ theta Θ Theta ϑ theta
ι iota κ kappa λ lambda Λ Lambda
µ mu ν nu ξ xi Ξ Xi
π pi Π Pi ρ rho σ sigma
Σ Sigma τ tau υ upsilon Υ Upsilon
φ phi Φ Phi ϕ phi χ chi
ψ psi Ψ Psi ω omega Ω Omega

This video may help you there:

https://jp-g.de/bsom/la/greek/

• Choosing a book is a matter of taste. Look into different ones and choose the book
that really convinces you.
Contents 3

• Keep interested, fascinated and eager to learn. However, do not expect to under-
stand everything at once.
DON’T PANIC J.P.G.
 Sequences and Limits
1
I’m a gym member. I try to go four times a week, but I’ve missed the
last twelve hundred times.
Chandler Bing

1.1 Just Numbers

Before we start with the Real Analysis course, we need to lie down some foundations. You
only need some knowledge about working with sets and maps to get started. From this,
we will introduce all the number sets we will need in this course in a quick way. Hence,
we quickly have the real numbers R we work with throughout this course.
However, if you interested in a more detailed discussion, I can recommend you my video
series about the foundations of mathematics:
Video: Start Learning Mathematics

https: // jp-g. de/ bsom/ la/ slm/

In order to construct the real number line, we need to generalise the equality sign. We
get a more abstract notion that we can use for sets to put similar elements into the same
box. In the end, we want to calculate with these boxes.
It turns out that we just need three properties from the equality sign to get the general
concept of an equivalence relation.

5
6 1 Sequences and Limits

Definition 1.1. Equivalence relation

Let X be a set. A subset R∼ ⊂ X × X is called a relation on X. We write x ∼ y
if (x, y) ∈ R∼ . A relation R∼ is called an equivalence relation if it satisfies the
following:

(a) x ∼ x for all x ∈ X. (reflexive)

(b) If x ∼ y, then also y ∼ x. (symmetric)

(c) If x ∼ y and y ∼ z, then also x ∼ z. (transitive)

By having this, we now can put equivalent elements in the corresponding boxes. These
boxes ared called equivalent classes.
Proposition & Definition 1.2. Equivalent classes
An equivalence relation ∼ on X gives a partition of X into disjoint subsets. For all
a ∈ X, we define
[a]∼ := {x ∈ X : x ∼ a}
and call it an equivalent class. We have the disjoint union:
[
X= [a]∼ .
a∈X

In the same way as we generalised the equality sign, we can also generalise the greater or
equal sign you might have seen often for numbers. It turns out that we just need some
defining properties there to get an abstract notion of such an ordering.
Definition 1.3. Ordering
Let X be a set. Let R≤ ⊂ X × X be a relation on X where we write x ≤ y if
(x, y) ∈ R≤ . A relation R≤ is called a partial order if it satisfies the following:

(a) x ≤ x for all x ∈ X. (reflexive)

(b) If x ≤ y and y ≤ x, then x = y. (antisymmetric)

(c) If x ≤ y and y ≤ z, then also x ≤ z. (transitive)

If, in addition,

(d) for all x, y ∈ X, we have either x ≤ y or y ≤ x. (total)

then ≤ is called a total order or chain.

Remark: Notation
If one has an ordering relation ≤, one usually also defines the following symbols:

x ≥ y : ⇐⇒ y ≤ x

x ≤ y and x 6= y .


x < y : ⇐⇒
1.1 Just Numbers 7

Reminder: Natural numbers

The natural numbers with zero N0 = {0, 1, 2, 3, . . .} originate from counting. There
is an operation + that is associative and commutative, called addition. Then we
can define a total order ≤. Instead of 3 + 3 + 3 + 3 + 3, one writes 5 · 3. This defines
also an associative and commutative operation on N0 , called multiplication.
The element 0 is the neutral element with respect to the addition + and the element
1 is the neutral element with respect to the multiplication ·.

Example 1.4. Sequence of numbers

An ordered infinite list of numbers is called a sequence. We will define it later in
more detail. For example:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .

Here, one would shorten that to (an )n∈N0 with an = 2n .

What can we do with this? Fibonacci numbers! Collatz conjecture!

Box 1.5. Construction of Z

We do not have an element x ∈ N0 that fulfils x + 5 = 0. This is what we would
call the inverse of 5 with respect to the addition. This element is what we recognise
inside the following elements of N0 × N0 :

(0, 5) (1, 6) (101, 106) (56, 61) (77, 82) (91, 96)

We define an equivalence relation on the set N0 × N0 by setting:

(a, b) ∼ (c, d) : ⇐⇒ a + d = b + c

The equivalent classes are excactly what we need:

[(a, b)] := {(x, y) : (x, y) ∼ (a, b)} .

We can also define an addition:

[(a, b)] + [(c, d)] := [(a + c, b + d)] .

It is well-defined, associative and commutative and [(0, 0)] defines the neutral ele-
ment. The set of all equivalence classes with this new addition is called the integers
and denoted by Z.

Also extend the multiplication to Z. Moreover, we can also define a multiplication:

[(a, b)] · [(c, d)] := [(ac + bd, ad + bc)] .

8 1 Sequences and Limits

Definition 1.6. Commutative ring

The integers Z form a so-called commutative ring. This means:

(A) Addition
(A1) associative: x + (y + z) = (x + y) + z
(A2) neutral element: There is a (unique) element 0 with x + 0 = x for all x.
(A3) inverse element: For all x there is a (unique) y with x + y = 0. We write
for this element simply −x.
(A4) commutative: x + y = y + x

(M) Multiplication
(M1) associative: x · (y · z) = (x · y) · z
(M2) neutral element: There is a (unique) element 1 with 1 · x = x for all x.
(M3) commutative: x · y = y · x

(D) Distributivity: x · (y + z) = x · y + x · z.

Box 1.7. Construction of Q

We do not have an element x ∈ Z \ {0} that fulfils x · 5 = 1. This is what we
would call the inverse of 5 with respect to the multiplication. This element is what
we recognise inside the following elements of Z \ {0} × Z \ {0}:

(1, 5) (2, 10) (101, 505) (50, 250) (11, 55) (500, 2500)

We define an equivalence relation on the set Z \ {0} × Z \ {0} by setting:

(a, b) ∼ (c, d) : ⇐⇒ a · d = b · c

The equivalent classes are excactly what we need:

[(a, b)] := {(x, y) : (x, y) ∼ (a, b)} .

We can also define a multiplication:

[(a, b)] · [(c, d)] := [(a · c, b · d)] .

It is well-defined, associative and commutative. The set of all equivalence classes

with this new multiplication is called the rational numbers (without zero).

[(a, b)] + [(c, d)] := [(ad + bc, bd)]

We can visualise these numbers in the number line:

1.1 Just Numbers 9

Definition 1.8. Real numbers

The real numbers are a non-empty set R together with the operations + : R×R → R
and · : R × R → R and an ordering relation ≤: R × R → {True, False} that fulfil
the following rules

(A) Addition
(A1) associative: x + (y + z) = (x + y) + z
(A2) neutral element: There is a (unique) element 0 with x + 0 = x for all x.
(A3) inverse element: For all x there is a (unique) y with x + y = 0. We write
for this element simply −x.
(A4) commutative: x + y = y + x

(M) Multiplication
(M1) associative: x · (y · z) = (x · y) · z
(M2) neutral element: There is a (unique) element 1 6= 0 with x·1 = x for all x.
(M3) inverse element: For all x 6= 0 there is a (unique) y with x · y = 1. We
write for this element simply x−1 .
(M4) commutative: x · y = y · x

(D) Distributivity: x · (y + z) = x · y + x · z.

(O) Ordering
(O1) x ≤ x is true for all x.
(O2) If x ≤ y and y ≤ x, then x = y.
(O3) transitive: x ≤ y and y ≤ z imply x ≤ z.
(O4) For all x, y ∈ X, we have either x ≤ y or y ≤ x.
(O5) x ≤ y implies x + z ≤ y + z for all z.
(O6) x ≤ y implies x · z ≤ y · z for all z ≥ 0.
(O7) x > 0 and ε > 0 implies x < ε + · · · + ε for sufficiently many summands.

(C) Let X, Y ⊂ R be two non-empty subsets with the property x ≤ y for all x ∈ X
and y ∈ Y . Then there is a c ∈ R with x ≤ c ≤ y for all x ∈ X and y ∈ Y .

Remark:
We will later reformulate the completeness axiom with the help of sequences. Then
it sounds like:
Completeness: Every sequence (an )n∈N with the property [For all ε > 0 there is an
N ∈ N with |an − am | < ε for all n, m > N ] has a limit.

Exercise 1.9.
Use the axioms to show:

(1) : 0 · x = 0
10 1 Sequences and Limits

(2) : −x = (−1) · x

(3) : (−1) · (−1) = 1

(4) : 1 > 0

(A2) (D) (A2)

(1): 0 · x = (0 + 0)x = 0x + 0x =⇒ 0x = 0
(∗) (A3) (D) (A2−4)
(2): −x = 0x + (−x) = (1 + (−1))x + (−x) = x + (−1)x + (−x) = (−1)x
(∗∗) (A3) (M 2)
(3): x(−1)(−1) = −(−x) = x =⇒ (−1) · (−1) = 1
(4): Exercise!

Definition 1.10. Absolute value for real numbers

The absolute value of a number x ∈ R is defined by
(
x if x ≥ 0,
|x| :=
−x if x < 0.

Proposition 1.11. Two important properties

For any two real numbers x, y ∈ R, one has

(a) |x · y| = |x| · |y|, (|·| is multiplicative),

(b) |x + y| ≤ |x| + |y|, (|·| fulfils the triangle inequality).

1.2 Convergence of Sequences 11

1.2 Convergence of Sequences

Now we start with sequences and the important definition of convergence.
Video: Start Learning Mathematics

https: // jp-g. de/ bsom/ la/ slm/

Definition 1.12.
Let M be a set. A sequence in M is a map a : N → M or a : N0 → M .

We use the following symbols for sequences:

(an )n∈N , (an ), (an )∞

n=1 , (a1 , a2 , a3 , . . .).

Remark:
M is usually a real subset (M ⊂ R), but M can also be a complex subset (M ⊂ C)
or a subset of some normed space (or M = R, M = C, M a normed space itself ).

Example 1.13. (a) an = (−1)n , then (an )n∈N = ((−1)n )n∈N = (−1, 1, −1, 1, −1, 1, . . .);
(b) an = n1 , then (an )n∈N = ( n1 )n∈N = (1, 21 , 13 , 41 , 51 , 16 , . . .);
(c) an = in (i is the imaginary unit), then (an )n∈N = (in )n∈N = (i, −1, −i, 1, i, −1, . . .);
(d) an = 1
2n
, then (an )n∈N = ( 21n )n∈N = (1, 21 , 41 , 18 , 16
1 1
, 32 , . . .);
(e) Approximation of π:
Consider a circle with radius r = 1.
The area is given by A = πr2 = π.

Now: Approximation of the circle by a regular polygon (all edges have equal length):
Area of a “piece of cake”: Acn = sin( πn ) cos( πn ) = 21 sin( 2π
n
) (the latter equation holds
true due to the general equality sin(2x) = 2 sin(x) cos(x) (will be treated later)).
The area of the polygon is therefore given by
 
n 2π
An = n · Acn = · sin .
2 n

Now consider the sequence (An )n∈N . Some values for An are listed in the following
table:
12 1 Sequences and Limits

Figure 1.1: Circle approximated by a polygon

p1
p2

Figure 1.2: “Piece of cake”

n An π − An
3 1.299038 1.84255
6 2.598076 0.54351
12 3.000000 0.14159
3072 3.14159046 0.00000219
50331648 3.141592653589 8.16 · 10−16

(f) Some linear algebra:

Consider the linear system
Ax = b, A ∈ Rn×n , b ∈ Rn .
Then o = −Ax + b and thus
x = x − Ax + b = (1 − A)x + b.
Consider the iteration
x1 = o
“Richardson iteration”
xn+1 = (1 − A)xn + b for n ≥ 1
1.2 Convergence of Sequences 13

The sequence (xn )n∈N is a sequence of approximate solutions. The method is “good”
if (xn )n∈N “converges”.

Since most (but not all) results stay the same in the real and complex case, we make the
following definition:
Definition 1.14.
The symbol F stands for either the real or complex numbers, i.e., F ∈ {R, C}.

Next we define the notions of convergence and limits:

Definition 1.15. Convergence/divergence of sequences
Let (an )n∈N be a sequence in F. We say that

– (an )n∈N is convergent to a ∈ F if for all ε > 0 there exists some N = N (ε) ∈ N
such that for all n ≥ N holds |an − a| < ε. In this case, we write

lim an = a.
n→∞

– (an )n∈N is divergent if it is not convergent, i.e., for all a ∈ F holds: There
exists some ε > 0 such that for all N there exists some n > N with |an −a| ≥ ε.

Convergence for real sequences means that if you give any small distance ε, one finds that
all sequence members an lie in the interval (a − ε, a + ε) with the exception of only finitely
many.

a1 a2 aN a
a−ε a+ε

Example 1.16. • Show: (an )n∈N with an = (1/n) is convergent with limit 0.
√
• Show: (bn )n∈N with bn = (1/ n) is convergent with limit 0.

Proof. Let ε > 0. Choose N > 1

ε2
. Then for all n ≥ N , we have

1 1
|bn − 0| = √ ≤ √ < ε
n N
This means bn is arbitrarily close to 0, eventually.

Remark:
(a) It can be shown that for a complex sequence (an )n∈N , convergence to a ∈ C holds
true if, and only if, (Re(an ))n∈N converges to Re(a) and (Im(an ))n∈N converges
to Im(a)

(b) In fact, convergence can also be defined for sequences in some arbitrary normed
14 1 Sequences and Limits

vector space V (for the definition of a normed space, e.g. consult the linear
algebra script). Then one has to replace the absolute value by the norm (e.g.,
“kan − ak < ε”.)

(c) Due to the fact that for any N ∈ N, we can find some x ∈ R with x > N , we
can equivalently reformulate the convergence definition as follows: “(an )n∈N is
convergent to a ∈ R if for all ε > 0 there exists some N = N (ε) ∈ R such that
for all n ≥ N holds |an − a| < ε”. In the following, we just write “there exists
some N ”.

What does convergence mean?

a1 a2 aN a aN +1
a−ε a+ε

Outside any ε-neighbourhood of a only finitely many elements of the sequence exist.

Example 1.17. (a) The real sequence ( n1 )n∈N converges to 0.

Proof: Let ε > 0 (be arbitrary): Choose N = 1ε + 1 = 1+ε
ε
.
Then for all n ≥ N holds
1 1 ε
≤ = < ε.
n N 1+ε
Therefore  
1
 − 0 = 1 < ε.

n  n

(b) The real sequence ((−1)n )n∈N is divergent.

Proof by contradiction:
Assume that ((−1)n )n∈N is convergent to a ∈ R. Then take e.g. ε = 1
10
. Due to
convergence, we should have some N such that for all n ≥ N holds

|(−1)n − a| < 1
10
.

Thus we have that | − 1 − a| < 1

10
and |1 − a| < 1
10
. As a consequence,
1 1
2 = |1 + a − a + 1| ≤ |1 + a| + | − a + 1| = |1 + a| + |a − 1| < 10
+ 10
= 51 .

This is a contradiction. 2
(c) For q ∈ C\{0} with |q| < 1 the complex sequence (q )n∈N converges to 0.
n

Proof: |q| < 1 gives rise to |q|

1
> 1, whence |q|
1
− 1 > 0. Therefore, we are able to apply
Bernoulli’s inequality (see tutorial) in the following way:
  n   n  
1 1 1 − |q| 1 − |q|
= 1+ −1 = 1+ ≥1+n· ,
|q|n |q| |q| |q|
and thus
1 |q|
|q|n ≤  = .
1−|q|
1 + n · |q| |q| + n · (1 − |q|)
1.2 Convergence of Sequences 15

Now let ε > 0 (be arbitrary):

Choose
|q| |q|
N= − +1
ε · (1 − |q|) 1 − |q|
Then for all n ≥ N holds
|q| |q|
n> −
ε · (1 − |q|) 1 − |q|
and thus
|q|
n · (1 − |q|) > − |q|.
ε
This leads to
|q|
|q| + n · (1 − |q|) > ,
ε
whence
|q|
< ε.
|q| + n · (1 − |q|)
The above calculations now imply
|q|
|q n − 0| = |q|n ≤ < ε.
|q| + n · (1 − |q|)
2

Remark:
The choice of the N often seems “to appear from nowhere”. However, there is
a systematic way to formulate the proof. For instance in a), we need to end up with
the equation | n1 − 0| < ε or, equivalently, n1 < ε. Inverting this expression leads to
n > 1ε . Therefore, if N is chosen as N = 1ε + 1 = 1+εε
, the desired statement follows.
If one has to formulate such a proof (for instance, in some exercise), then first
these above calculations have to be done “on some extra sheet” and then formulate
the convergence proof in the style as in a) or c).

Definition 1.18. Boundedness of sequences

Let (an )n∈N be a sequence in F. Then (an )n∈N is called

– bounded if there exists some c ∈ R such that for all n ∈ N holds |an | ≤ c;

– unbounded if it is not bounded, i.e., for all c ∈ R, there exists some n ∈ N

with |an | > c.

Theorem 1.19.
Let (an )n∈N be a convergent sequence in F. Then (an )n∈N is bounded.

Proof. Suppose that limn→∞ an = a. Take ε = 1. Then there exists some N such that for
all n ≥ N holds |an − a| < 1. Thus, for all n ≥ N holds

|an | = |an − a + a| ≤ |an − a| + |a| < 1 + |a|.

16 1 Sequences and Limits

Now choose
c = max{|a1 |, |a2 |, . . . , |aN −1 |, |a| + 1}
and consider some arbitrary sequence element ak .
If k < N , then |ak | ≤ max{|a1 |, |a2 |, . . . , |aN −1 |} ≤ c.
In the case k ≥ N , the above calculations lead to |ak | < |a| + 1 ≤ c.
Altogether, this implies that |ak | ≤ c for all k ∈ N, so (an )n∈N is bounded by c.

Remark:
For a convergent sequence (an )n∈N that is bounded by c, we can also deduce from
the above argumentation that for the limit a holds |a| ≤ c:
Suppose that limn→∞ an = a ∈ F, then for an arbitrary ε > 0 there exists an N ∈ N
such that |a−an | < ε for all n ≥ N . Hence |a| = |a−an +an | ≤ |a−an |+|an | ≤ ε+c.
Since ε > 0 can be chosen arbitrarily small this implies |a| ≤ c.

Theorem 1.20. Uniqueness of the limit of a convergent sequence

Let (an )n∈N be a convergent sequence in F. Then there exists only one limit.

Proof by contradiction: Let a, b ∈ F be two distinct (a 6= b) limits of (an )n∈N . Then we

have |a − b| > 0 and for ε = 14 |a − b| the following statements are fulfilled:
There exists some N1 such that for all n ≥ N1 holds |an − a| < ε.
There exists some N2 such that for all n ≥ N2 holds |an − b| < ε.
Let n ≥ max{N1 , N2 }. Then
|a − b| = |a − an + an − b|
1
≤ |a − an | + |an − b| ≤ ε + ε = 2ε = · |a − b|
2
Thus, |a − b| ≤ 12 |a − b|. However, this implies |a − b| ≤ 0 which is only fulfilled, if
|a − b| = 0, or, equivalently a = b. This is a contradiction to the initial assumption. 2
In the following, we present some results that allow us to determine some further limits.
Theorem 1.21. Formulae for convergent sequences
Let (an )n∈N and (bn )n∈N be a convergent sequences in F. Then the following holds
true:
(i) (an + bn )n∈N is convergent with

lim (an + bn ) = lim an + lim bn .

n→∞ n→∞ n→∞

(ii) (an · bn )n∈N is convergent with

lim (an · bn ) = lim an · lim bn .

n→∞ n→∞ n→∞

(iii) If limn→∞ bn 6= 0 and bn 6= 0 for all n ∈ N, then the sequence ( abnn )n∈N is
convergent with
an lim an
n→∞
lim = .
n→∞ bn lim bn
n→∞
1.2 Convergence of Sequences 17

Proof. Let a = lim an and b = lim bn .

n→∞ n→∞

(i) Let ε > 0 be arbitrary. Then

there exists some N1 such that for all n ≥ N1 holds |a − an | < 2ε , and
there exists some N2 such that for all n ≥ N2 holds |b − bn | < 2ε .
Now choose N = max{N1 , N2 }. Then for all n ≥ N holds
|(a + b) − (an + bn )|
ε ε
= |(a − an ) + (b − bn )| ≤ |a − an | + |b − bn | < + = ε.
2 2
(ii) Due to Theorem 1.19, both sequences (an )n∈N and (bn )n∈N are bounded. Choose
numbers c1 , c2 > 0 such that |an | < c1 and |bn | < c2 for all n ∈ N. Define
c = max{c1 , c2 }.
Let ε > 0. Convergence of (an )n∈N to a implies that there exists some N1 such
that |a − an | < 2cε for all n ≥ N1 . Furthermore, the convergence of (bn )n∈N to b
implies that there exists some N2 such that |b − bn | < 2cε for all n ≥ N2 . Now define
N = max{N1 , N2 }. Then for n ≥ N holds
|ab − an bn |
= |(ab − an b) + (an b − an bn )| ≤ |ab − an b| + |an b − an bn |
= |a − an | · |b| + |an | · |b − bn | ≤ |a − an | · c + c · |b − bn |
ε ε
< ·c+c· = ε.
2c 2c
(iii) First we specialize to the case where (an )n∈N is the constant sequence
(an )n∈N = (1, 1, 1, 1, . . .). Due to b > 0, we have the existence of some N1 such that
for all n ≥ N1 holds
|b|
|bn − b| < .
2
This just follows by an application of the “ε-criterion” to ε = |b| 2
. In particular, this
|b| |b|
leads to |b| ≤ |bn − b| + |bn | < 2 + |bn | and thus |bn | > 2 .
Let ε > 0. Let N2 such that for all n ≥ N2 holds
ε · |b|2
|bn − b| < .
2
Then for n ≥ N := max{N1 , N2 } holds
2 ε · |b|2
 
1 1  1 2
 − = · |b n − b| < · |b n − b| < · = ε.
 bn b  |bn | · |b| |b|2 |b|2 2
So far, we have shown that for some sequence (bn )n∈N with limn→∞ bn 6= 0 and bn 6= 0
for all n ∈ N holds
1 1
lim = .
n→∞ bn lim bn
n→∞
The general statement for the sequence follows by rewriting
an
( bn )n∈N
   
an 1
= an ·
bn n∈N bn n∈N
and applying the multiplication rule (ii).
18 1 Sequences and Limits

Remark:
(a) Since the constant sequence (a)n∈N = (a, a, a, . . .) is, of course, convergent to a,
statement (ii) also implies the formula

lim (a · bn ) = a · lim bn .
n→∞ n→∞

(b) For k ∈ N, a k-times application of statement (ii) yields that for some conver-
gent sequence (an )n∈N , also the sequence (akn )n∈N is convergent with
 k
lim akn = lim an .
n→∞ n→∞

Theorem 1.22. Monotonicity of limits

Let (an )n∈N and (bn )n∈N be two convergent real sequences with

lim an = a, lim bn = b.
n→∞ n→∞

Further, assume that for all n ∈ N holds an ≤ bn . Then the following holds true:

(i) a ≤ b;

(ii) If a = b and (cn )n∈N is another sequence with an ≤ cn ≤ bn for all n ∈ N, then
(cn )n∈N is convergent with
lim cn = a = b.
n→∞

(Sandwich-Theorem)

Proof. (i) Consider the sequence of differences between bn and an , i.e., (bn − an )n∈N . By
Theorem 1.21, it suffices to show that

b − a = lim (bn − an ) ≥ 0.
n→∞

Assume the converse statement, i.e., b − a < 0. Then, we have that both numbers
a − b and bn − an are positive and thus

|a − b − (an − bn )| = a − b + (bn − an ) > a − b.

In particular, there exists no n ∈ N such that |a − b − (an − bn )| < ε for ε = a − b > 0.

This is a contradiction to limn→∞ (bn − an ) = b − a.
(ii) Again consider the sequence (bn − an )n∈N which is tending to zero according to
Theorem 1.21. Further, consider the sequence (cn − an )n∈N . Then we have for all
n ∈ N that 0 ≤ cn − an ≤ bn − an . Let ε > 0. Since bn − an is tending to zero,
there exists some N such that for all n ≥ N holds |bn − an − 0| < ε. Due to
0 ≤ cn − an ≤ bn − an , we can conclude that for n ≥ N holds

|cn − an − 0| = cn − an ≤ bn − an = |bn − an − 0| < ε.

1.2 Convergence of Sequences 19

This implies that (cn − an )n∈N is convergent with limn→∞ (cn − an ) = 0. Hence
a = 0 + a = limn→∞ (cn − an ) + limn→∞ an = limn→∞ cn .

Remark:
Since the modification of finitely many sequence elements does not change the limits
(take a closer look at Definition 1.15), the statements of Theorem 1.22 can be slightly
generalised by only claiming that there exists some n0 such that for all n ≥ n0 holds
an ≤ bn (resp. for all n ≥ n0 holds an ≤ cn ≤ bn in (ii)). In the proof of (i), one
has to replace the words “there exists no n ∈ N such that” by “there exists no n ≥ n0
such that” and in the proof of (ii) the number N has to be replaced by max{N, n0 }.

Attention!
From the fact that we have the strict inequality an < bn , we cannot con-
clude that the limits satisfy a < b. To see this, consider the sequences
(an )n∈N = (0, 0, 0, . . .) and (bn )n∈N = ( n1 )n∈N . In this case, we have a = b = 0
though the strict inequality an = 0 < n1 = bn holds true for all n ∈ N.

Example 1.23. a) Consider ( n1k )n∈N for some k ∈ N. We state two alternative ways to
show that this sequence tends to zero. The first possibility is, of course, an argument-
ation as in statement (b) in Remark on page 18. The second way to treat this problem
is making use of the inequality
1 1
≥ k > 0.
n n
Since we know from Example 1.17 a) that the sequence ( n1 )n∈N tends to zero, statement
(ii) of Theorem 1.22 directly leads to the fact that ( n1k )n∈N also tends to zero.
b) Consider (an )n∈N with
2n2 + 5n − 1
an = .
−5n2 + n + 1
Rewriting
2 + n5 − n12
an = ,
−5 + n1 + n12
and using that both ( n1 )n∈N and ( n12 )n∈N tend to zero, we can apply Theorem 1.21 to
obtain that
2n2 + 5n − 1 2 + n5 − n12 2
lim an = lim 2
= lim 1 1 =− .
n→∞ n→∞ −5n + n + 1 n→∞ −5 + + n2 5
n
√
c) Consider (an )n∈N with an = n2 + 1 − n. At first glance, none of the so far presented
results seem to help to analyse convergence of this sequence. However, we can compute
√ √
√ ( n 2 + 1 − n)( n2 + 1 + n)
an = n 2 + 1 − n = √
n2 + 1 + n
n2 + 1 − n2 1 1
=√ =√ < .
n2 + 1 + n n2 + 1 + n n
By Theorem 1.22, we now get that limn→∞ an = 0.
20 1 Sequences and Limits

Next we introduce some further properties of real sequences. In particular, we declare

what the phrase “the sequence tends to infinity (∞)” means.
Definition 1.24. Monotonicity, boundedness, divergence to ±∞
A real sequence (an )n∈N is called

(a) monotonically increasing if for all n ∈ N holds an ≤ an+1 .

(b) strictly monotonically increasing if for all n ∈ N holds an < an+1 .

(c) monotonically decreasing if for all n ∈ N holds an ≥ an+1 .

(d) strictly monotonically decreasing if for all n ∈ N holds an > an+1 .

(e) bounded from above if there exists some c ∈ R with an ≤ c for all n ∈ N.

(f ) bounded from below if there exists some c ∈ R with an ≥ c for all n ∈ N.

(g) divergent to ∞ if for all c ∈ R there exists some N with an ≥ c for all n ≥ N .
In this case, we write
lim an = ∞.
n→∞

(h) divergent to −∞ if for all c ∈ R there exists some N with an ≤ c for all n ≥ N .
In this case, we write
lim an = −∞.
n→∞

Remark:
It can be readily seen from the definition that a sequence is bounded if and only if
it is both bounded from above and bounded from below.

Example 1.25. (a) For k ∈ N, the sequence ( n1k )n∈N is strongly monotonically decreasing
due to n1k > (n+1)
1
k and, moreover, both bounded from above and bounded from below.

(b) The sequence (n3 )n∈N is bounded from below and divergent to ∞.
Proof: The fact that this sequence is bounded from below directly follows from n3 ≥ 0
for all n ∈ N. To show that this sequence is divergent to ∞, let c ∈ R be arbitrary
and choose (√
3
c + 1 : if c ≥ 0,
N=
0 : else.
Then for n ≥ N , we have that n3 > c and thus, the sequence (an )n∈N tends to ∞.

Reminder:
Definition of intervals:

(a, b) := {x ∈ R | a < x < b}

[a, b) := {x ∈ R | a ≤ x < b}
(a, b] := {x ∈ R | a < x ≤ b}
1.2 Convergence of Sequences 21

[a, b] := {x ∈ R | a ≤ x ≤ b}
(−∞, b) := {x ∈ R | x < b}
(−∞, b] := {x ∈ R | x ≤ b}
(a, ∞) := {x ∈ R | a < x}
[a, ∞) := {x ∈ R | a ≤ x}

Definition 1.26. Upper and lower bounds for sets

Let M ⊂ R be any subset of real numbers. A real number b is called an upper bound
of M if x ≤ b for all x ∈ M . Analogously, a ∈ R is called a lower bound of M if
a ≤ x for all x ∈ M .

Definition 1.27. Bounded from below or above for sets

Let M ⊂ R. If there is an upper bound for M , then one calls the set bounded from
above. If there is a lower bound for M , then one calls the set bounded from below.
If both properties hold, we call the set simply bounded.

Definition 1.28. Maximal and minimal element of a set

Let M ⊂ R. An element d ∈ M is called maximal if x ≤ d for all x ∈ M . An
element c ∈ M is called minimal if x ≥ c for all x ∈ M . If these numbers exist,
one write max M = d and min M = c.

Definition 1.29. Supremum and infimum

Let M ⊂ R be a set.

(a) A real number s is called the supremum of M if:

• x ≤ s for all x ∈ M ,
• for all ε > 0 there is an x ∈ M with s − ε < x.
In this case we write s = sup M .

(b) A real number l is called the infimum of M if:

• x ≥ l for all x ∈ M ,
• for all ε > 0 there is an x ∈ M with l + ε > x.
In this case we write l = inf M .

(c) We further define

• sup M = ∞ if M is not bounded from above;
• inf M = −∞ if M is not bounded from below;
• sup ∅ = −∞;
• inf ∅ = ∞.
22 1 Sequences and Limits

To remember: Sup and Inf

The infimum is the greatest lower bound and the supremum is the lowest upper
bound.

Example 1.30. (a) sup[0, 1] = 1, inf[0, 1] = 0;

(b) sup(0, 1) = 1, inf(0, 1) = 0;
(c) sup{ n1 : n ∈ N} = 1, inf{ n1 : n ∈ N} = 0;
√ √
(d) sup{x ∈ Q : x2 < 2} = 2, inf{x ∈ Q : x2 < 2} = − 2;

Remark: Difference between sup and max (resp. inf and min)
In contrast to the maximum, the supremum does not need to belong to the respective
set. For instance, we have 1 = sup(0, 1), but max(0, 1) does not exist. The ana-
logous statement holds true for inf and min. However, we can make the following
statement: If max M (min M ) exists, then max M = sup M (min M = inf M ).

The next result concerns the special property of the real numbers that supremum and
infimum are defined for all subsets of the real numbers. This theorem goes back to
Julius Wilhelm Richard Dedekind (1831–1916). It follows from the completeness
axiom (C):
Theorem 1.31. Dedekind’s Theorem
Every non-empty bounded set M ⊂ R has a supremum and an infimum with
sup M, inf M ∈ R.

We make essential use of Dedekind’s theorem to prove the following result:

Theorem 1.32. Convergence of bounded and monotonic sequences
Let (an )n∈N be a real sequence that has one of the following properties:

– (an )n∈N is monotonically increasing and bounded from above;

– (an )n∈N is monotonically decreasing and bounded from below;

Then (an )n∈N is convergent.

Proof: Let us first assume that (an )n∈N is monotonically increasing and bounded from
above. Define the set M = {an : n ∈ N}. Since M is bounded, Dedekind’s theorem
implies that there exists some K ∈ R such that

K = sup M.

We show that K is indeed the limit of the sequence (an )n∈N .

Let ε > 0. By the definition of the supremum, we have that an ≤ K for all n ∈ N and
there exists some N ∈ N such that aN > K − ε. The monotonicity of (an )n∈N implies that
for all n ≥ N holds aN ≤ an . Altogether, we have

K − ε < aN ≤ an ≤ K
1.2 Convergence of Sequences 23

and thus |K − an | = K − an < ε for all n ≥ N . This implies convergence to K.

To prove that convergence is also guaranteed in the case where (an )n∈N is monotonically
decreasing and bounded from below, we consider the sequence (−an )n∈N , which is now
bounded from above and monotonically increasing. By the (already proven) first state-
ment of this theorem, the sequence (−an )n∈N is convergent, whence (an )n∈N is convergent
as well. 2
Remark:
By the same argumentation as in Remark from page 19, the monotone increase
(decrease) of (an )n∈N can be slightly relaxed by only claiming that an ≤ an+1 (an ≥
an+1 ) for all n ≥ n0 for some n0 in N. In such a case, the limit of the sequence is
then given by sup{an : n ≥ n0 } (resp. inf{an : n ≥ n0 }).

Example 1.33. a) Consider the sequence (an )n∈N which is recursively defined via a1 = 1
and
an + a2n
an+1 = for n ≥ 1.
2
We now prove that this sequence is convergent by showing that it is bounded from
below and for all n ≥ 2 holds an+1 ≤ an .
√
Proof: To show boundedness from below, we use the inequality xy ≤ x+y 2
for all
nonnegative x, y ∈ R. This inequality is a consequence of
√ √
( x − y)2 x+y √
0≤ = − xy.
2 2
The first inequality is a consequence of the fact that squares of real numbers cannot
be negative.
Using this inequality, we obtain for n ≥ 1
an + a2n √
r
2
an+1 = ≥ an · = 2.
2 an
Thus, (an ) is bounded from below. For showing monotonicity, we consider
2
an + an 1
an+1 − an = − an = (2 − a2n ).
2 2an
In particular, if n ≥ 2, we have that an > 0 and 2 − a2n ≤ 0. Thus, an+1 − an ≤ 0
for n ≥ 2. An application of Theorem 1.31 (resp. the slight generalisation in Remark
from above) now leads to the existence of some a ∈ R with a = limn→∞ an .

To compute the limit, we make use of the relation limn→∞ an = limn→∞ an+1 (follows
directly from Definition 1.15) and the formulae for limits in Theorem 1.21. This yields
2
an + an a + a2
a = lim an = lim an+1 = lim = .
n→∞ n→∞ n→∞ 2 2
√ √
This relation leads to the equation 2 − a2 = 0, i.e., we either have a = 2 or a = − 2.
However, the latter solution cannot be a limit since all sequence elements are positive.
Therefore, we have √
lim an = 2.
n→∞
24 1 Sequences and Limits

√
b) Let
√ x ∈ R with x > 1. Consider the sequence ( n
x)n∈N . It can be directly seen that
( x)n∈N is monotonically decreasing and bounded from below by one. Therefore, the
n

limit √
a = lim n x
n→∞

exists with a ≥ 1. To show that a = 1, we assume that a > 1 and lead this to
a contradiction.
The √equation a > 1 leads to the existence of some n ∈√ N with an > x, and thus
a > n x. On the other hand, the monotone decrease of ( n x)n∈N implies that
√ √
a = lim n x = inf{ n x : n ∈ N} < a,
n→∞

which is a contradiction.
√
c) Let x ∈ R with 0 < x < 1. Consider the sequence (an )n∈N = ( n x)n∈N . Then we have
by Example b) and Theorem 1.21 that
√
n
1 1
lim x= q = = 1.
n→∞
limn→∞ n 1 1
x

d) Let (an ) be a nonnegative sequence with an → a and k ∈ N. Then for all ε > 0 there
exists N > 0 such that |an − a| < εk . From this it follows that
√ √ p
| k an − k a| ≤ k |an − a| < ε.
√ √
Thus ( k an ) is convergent with limit k a.

1 n
e) The sequence (an )n∈N defined as an := 1 + is convergent.


n

Remark:
The limit of the sequence
 
1 n
(an )n∈N = 1+ ,
n n∈N

n
i.e. e := limn→∞ 1 + n1 is well known as Euler’s number. Later on we will define
the exponential function exp. It holds that
e = exp(1) ≈ 2.7182818... . Indeed, we
n
will show later on that ez = limn→∞ 1 + nz = exp(z).

1.3 Subsequences and accumulation values

Definition 1.34. Subsequence

Let (an )n∈N be a sequence in F. Let (nk )k∈N be a strongly monotonically increasing
sequence with nk ∈ N for all k ∈ N. Then (ank )k∈N is called a subsequence.
1.3 Subsequences and accumulation values 25

a1 a2 a3 a4 a5 a6 a7 a8 a9 a10

an1 an 2 an3 an4 an5 an6

Example 1.35. Consider the sequence (an )n∈N = ( n1 )n∈N . Then some subsequences are
given by
• (ank )k∈N = (a2k )k∈N = ( 12 , 14 , 61 , 18 , . . .);
• (ank )k∈N = (ak2 )k∈N = (1, 14 , 19 , 16
1 1
, 25 , . . .);
• (ank )k∈N = (a2k )k∈N = ( 12 , 14 , 18 , 16
1 1
, 32 , . . .);
• (ank )k∈N = (ak! )k∈N = (1, 12 , 16 , 24
1 1
, 120 1
, 720 , . . .).

Theorem 1.36. Convergence of subsequences

Let (an )n∈N be a convergent sequence in F with limn→∞ an = a. Then all sub-
sequences (ank )k∈N of (an )n∈N are also convergent with

lim ank = a.
k→∞

Proof: Since 1 ≤ n1 < n2 < n3 < . . . and nk ∈ N for all k ∈ N, we have that nk ≥ k
for all k ∈ N. Let ε > 0. By the convergence of (an )n∈N , there exists some N such that
|ak − a| < ε for all k ≥ N . Due to nk ≥ k, we thus also have that |ank − a| < ε for all
k ≥ N. 2
Attention!
The existence of a convergent subsequence (ank )k∈N does in general not imply the
convergence of (an )n∈N . For instance, consider (an )n∈N = ((−1)n )n∈N . Both sub-
sequences
(a2k )k∈N = ((−1)2k )k∈N = (1, 1, 1, 1, . . .)
(a2k+1 )k∈N = ((−1)2k+1 )k∈N = (−1, −1, −1, −1, . . .)
are convergent though (an )n∈N = ((−1)n )n∈N is divergent (see Example 1.17 b)).

However, we can “rescue” this statement by additionally claiming that (an )n∈N is mono-
tonic.
Theorem 1.37. Subsequences of monotonic sequences
Let (an )n∈N be a sequence in R. If (an )n∈N is monotonic and there exists a convergent
subsequence (ank )k∈N , then (an )n∈N is convergent with

lim an = lim ank .

n→∞ k→∞

Proof: Denote a = limk→∞ ank . We just consider the case where (an )n∈N is monotonically
increasing (the remaining part can be done analogously to the argumentations at the end
of the proof of Theorem 1.31). Since (ank )k∈N is also monotonically increasing, we have
that a = sup{ank : k ∈ N}.
Let ε > 0. Due to the convergence and monotonicity of (ank )k∈N , there exists some K ∈ N
26 1 Sequences and Limits

such that for all k ≥ K holds

a − ε < ank ≤ a.
Now assume that n ≥ N = nK . Monotonicity then implies that a − ε < anK ≤ an ≤
ann ≤ a. In particular, we have that

|a − an | = a − an < ε.

2
Next we present the famous Theorem of Bolzano-Weierstraß.
Theorem 1.38. Theorem of Bolzano-Weierstraß
Let (an )n∈N be a bounded sequence in F. Then there exists some convergent sub-
sequence (ank )k∈N .

Proof: First we consider the case F = R. Since (an )n∈N is bounded, there exist some
A, B ∈ R such that for all n ∈ N holds A ≤ an ≤ B. We will now successively construct
subintervals [An , Bn ] ⊂ [A, B] which still include infinitely many sequence elements of
(an )n∈N .
Inductively define A0 = A, B0 = B and for k ≥ 1,
a) Ak = Ak−1 , Bk = Ak−1 +B
2
k−1
, if the interval [Ak−1 , Ak−1 +B
2
k−1
] contains infinitely
many sequence elements of (an )n∈N , and
Ak−1 +Bk−1
b) Ak = 2
, Bk = Bk−1 , else.
By the construction of Ak and Bk , we have that each interval [Ak , Bk ] has infinitely many
sequence elements of (an )n∈N . We furthermore have B1 − A1 = 21 (B − A), B2 − A2 =
1
4
(B − A), . . ., Bk − Ak = 21k (B − A). Moreover, the sequence (An )n∈N is monotonically
increasing and bounded from above by B, i.e., it is convergent by Theorem 1.32. The
relation Bk − Ak = 21k (B − A) moreover implies that (Bn )n∈N is also convergent and has
the same limit as (An )n∈N . Denote

a = lim An = lim Bn .
n→∞ n→∞

Define a subsequence (ank )k∈N by n1 = 1 and nk with nk > nk−1 and ank ∈ [Ak , Bk ]
(which is possible since [Ak , Bk ] contains infinitely many elements of (an )n∈N ). Then
Ak ≤ ank ≤ Bk . Theorem 1.22 then implies that

a = lim ank .
k→∞

Finally we consider the case F = C. Write an = bn + icn where i is the imaginary unit,
bn := Re (an )p denotes the real part and cn := Im (an ) denotes the imaginary part of an .
Since |an | = b2n + c2n ≥ max{|bn |, |cn |} ≥ 0, the boundedness of the complex sequence
(an )n∈N implies the boundedness of both real sequences (bn )n∈N and (cn )n∈N . Then, by
the previous, we now that (bn )n∈N has a convergent subsequence (bnk )k∈N . Since the
subsequence (cnk )k∈N of the bounded sequence (cn )n∈N is also bounded, it also has a con-
vergent subsequence (cnkm )m∈N . The subsequence (bnkm )m∈N of the convergent sequence
(bnk )k∈N also converges. Hence (ankm )m∈N = (bnkm +icnkm )m∈N is a convergent subsequence
of (an )n∈N with limm→∞ ankm = limm→∞ bnkm + i · limm→∞ cnkm . 2
1.3 Subsequences and accumulation values 27

Definition 1.39. Accumulation value

Let (an )n∈N be a sequence in F. Then a ∈ F is called accumulation value if there
exists some subsequence (ank )k∈N with

a = lim ank .
k→∞

Attention! Names
Accumulation values are often called by other names, like accumulation points,
limits points or cluster points.

Proposition 1.40.
a ∈ F is an accumulation value if and only if in every ε-neighbourhood of a, there
are infinitely many elements of the sequence (an )n∈N .

Definition 1.41. Accumulation values ±∞

A real sequence (an )n∈N is said to have the (improper) accumulation value ∞ if it is
not bounded from above. Analogously, we define the (improper) accumulation value
−∞ if it is not bounded from below.

Definition 1.42. Limit superior - limit inferior

Let (an )n∈N be a real sequence. A number a ∈ R ∪ {∞, −∞} is called

• limit superior of (an )n∈N if a is the largest accumulation value of (an )n∈N . In
this case, we write
a = lim sup an .
n→∞

• limit inferior of (an )n∈N if a is the smallest accumulation value of (an )n∈N . In
this case, we write
a = lim inf an .
n→∞

Remark:
Almost needless to say, we define the ordering between infinity and real numbers by
−∞ < a < ∞ for all a ∈ R. It can be shown that (in contrast to the limit) the limit
superior and limit inferior always exist for any real sequence. This will follow from
the subsequent results.

Lemma 1.43.
Let (an )n∈R be a real sequence. Then the following statements hold

a) lim inf an = lim inf{ak | k ≥ n}

n→∞ n→∞

b) lim sup an = lim sup{ak | k ≥ n}

n→∞ n→∞
28 1 Sequences and Limits

c) A sequence is convergent if and only if lim inf n→∞ an = lim supn→∞ an 6∈ {±∞}.
In this case holds limn→∞ an = lim inf n→∞ an = lim supn→∞ an .

d) A sequence is divergent to ∞ if and only if lim inf n→∞ an = ∞. In this case

also holds limn→∞ an = lim supn→∞ an = ∞.

e) A sequence is divergent to −∞ if and only if lim supn→∞ an = −∞. In this case

also holds limn→∞ an = lim inf n→∞ an = −∞.

Proof:
a) If (an ) is not bounded from below, then, by Definition 1.41, −∞ is an accumula-
tion value of (an ) which necessarily must be the smallest one. By Definition 1.42
lim inf an = −∞. On the other hand, the unboundedness from below of (an ) implies
sn := inf{ak | k ≥ n} = −∞ for all n ∈ N and therefore also limn→∞ sn = −∞. Note
that formally we only defined limits for sequences with values in R and not with values
in R ∪ {−∞, ∞}. Here we implicitely used the obvious extension, namely we said that
the limit of the sequence (sn ) which is constantly −∞ has the limit −∞.
Next we consider the case where (an ) is divergent to +∞. In particular, (an ) is not
bounded from above and therefore +∞ is an accumulation value by Definition 1.41.
This is also the only accumulation value, since each subsequence of (an ) also diverges
to +∞. Hence, by Definition 1.42, lim inf an = +∞. On the other hand for each c > 0
there is an N ∈ N such that an ≥ c for all n ≥ N . Therfore sn = inf{ak | k ≥ n} ≥ c
for all n ≥ N which shows that also (sn ) diverges to +∞, i.e. limn→∞ sn = +∞.
Finally we consider the remaining case where (an ) is bounded from below and not
divergent to +∞. Then there exist constants c1 , c2 ∈ R such that c1 ≤ an for all n ∈ N
and an ≤ c2 for infinitely many n ∈ N. This implies
c1 ≤ sn = inf{ak | k ≥ n} ≤ c2
for all n ∈ N, i.e. (sn ) is bounded. Since (sn ) is also monotonically increasing as
sn+1 = inf{ak | k ≥ n + 1} ≥ min{inf{ak | k ≥ n + 1}, an } = inf{ak | k ≥ n} = sn ,
it must be convergent. Set s := limn→∞ sn . We can recursively define a subsequence
(ank ) of (an ) with n1 = 1 and nk+1 > nk such that
1
s(nk +1) = inf{am | m ≥ nk + 1} ≤ ank+1 ≤ s(nk +1) + .
k
Since the right- and left-hand sides of this inequality converge to s for k → ∞, we
also have limk→∞ ank = s which shows that s is an accumulation value of (an ). On the
other hand, if x is any other accumulation value of (an ) and if (ajk ) is a corresponding
subsequence such that limk→∞ ajk = x, then
sjk = inf{am | m ≥ jk } ≤ ajk
shows that s = limk→∞ sjk ≤ limk→∞ ajk = x which means that s is indeed the smallest
accumulation value of (an ), that is lim inf an = s.
1.4 Cauchy Sequences 29

b) Analogous to a).
c) “⇒”: Since the sequence (an ) is convergent every subsequence is convergent with the
same limit. By Definition 1.39 there exists only one accumulation value and thus
lim inf an = lim sup an .
“⇐”: Let s := lim inf an = lim sup an . Then for all ε > 0 there exists an N ∈ N such
that for all n ≥ N we have s − ε < an < s + ε. This implies convergence of (an )n∈N to
s.
d) Let sn := inf{ak : k ≥ n}.
“⇒”: We have for any c > 0 an N ∈ N such that an > c + 1 for all n ≥ N . Thus sn > c
for all n ≥ N .
“⇐”: By definition of sn we have an ≥ sn . Thus an → ∞ since sn → ∞.
e) Analogous to d).
2

Example 1.44. (a) (an )n∈N = (n)n∈N . Then ∞ is the only accumulation value and
consequently lim supn→∞ an = lim inf n→∞ an = ∞.
(b) (an )n∈N = ((−1)n n)n∈N = (−1, 2, −3, 4, −5, 6, . . .). Then ∞ and −∞ are the only
accumulation values and consequently lim sup an = ∞ and lim inf an = −∞.
(c) (an )n∈N = ((−1)n )n∈N . Then 1 and −1 are the only accumulation values and con-
sequently lim sup an = 1 and lim inf an = −1.
(d) (an )n∈N with (
(−1)n : if n is divisible by 3,
an =
n : else.
Then we have (an )n∈N = (1, 2, −1, 4, 5, 1, 7, 8, −1, 9, 10, . . .) and the set of accumula-
tion values is given by {−1, 1, ∞}. Thus, we have lim sup an = ∞ and lim inf an = −1.

1.4 Cauchy Sequences

Definition 1.45. Cauchy sequences

A sequence (an )n∈N in F is called Cauchy sequence if for all ε > 0, there exists some
N such that for all n, m ≥ N holds

|an − am | < ε.

Remark:
By the expression “n, m ≥ N ”, we mean that both n and m are greater or equal than
N , i.e., n ≥ N and m ≥ N .

Now we show that convergent sequences are indeed Cauchy sequences.

Theorem 1.46.
Let (an )n∈N be a convergent sequence. Then (an )n∈N is a Cauchy sequence.
30 1 Sequences and Limits

Proof: Let a = limn→∞ an and ε > 0. Then there exists some N such that for all k ≥ N
holds |a − ak | < 2ε . Hence, for all m, n ≥ N holds
ε ε
|an − am | = |(an − a) + (a − am )| ≤ |an − a| + |a − am | < + = ε.
2 2
2
The following theorem is closely related to Theorem 1.19.
Theorem 1.47. Cauchy sequences are bounded
Let (an )n∈N be a Cauchy sequence. Then (an )n∈N is bounded.

Proof: Take ε = 1. Then there exists some N such that for all n, m ≥ N holds |an − am | <
1. Thus, for all n ≥ N holds

|an | = |an − aN + aN | ≤ |an − aN | + |aN | < 1 + |aN |.

Now choose
c = max{|a1 |, |a2 |, . . . , |aN −1 |, |aN | + 1}
and consider some arbitrary sequence element ak .
If k < N , we have that |ak | ≤ max{|a1 |, |a2 |, . . . , |aN −1 |} ≤ c.
If k ≥ N , we have, by the above calculations, that |ak | < |aN | + 1 ≤ c.
Altogether, this implies that |ak | ≤ c for all k ∈ N, so (an )n∈N is bounded by c. 2
Now we show that Cauchy sequences in F are even convergent:
Theorem 1.48.
Every Cauchy sequence (an )n∈N in F converges.

Proof: By Theorem 1.47, (an )n∈N is bounded. By Theorem 1.38 of Bolzano-Weierstraß it

has a convergent subsequence (ank )k∈N . Set a := limk→∞ ank . For given ε > 0 there exist
N1 , N2 ∈ N such that |ank − a| < ε/2 for all k ≥ N1 and |an − am | < ε/2 for all n, m ≥ N2 .
Thus for n ≥ N := max{N1 , N2 } holds nn ≥ n ≥ N and

|an − a| ≤ |an − ann + ann − a| ≤ |an − ann | + |ann − a| < ε/2 + ε/2 = ε . 2

Theorem 1.48 is not true for arbitrary normed F-vector spaces. Those normed F-vector
spaces (V, || · ||) for which every Cauchy sequence has a limit in V are called complete or
Banach spaces (in honour of the Polish mathematician Stefan Banach). Without proof
we state that all finite dimensional normed F-vector spaces are Banach spaces.

1.5 Bounded, Open, Closed and Compact Sets

Next, we define some particular sets and special properties of sets.
Definition 1.49. ε-neighbourhood
Let x ∈ F. Then for ε > 0, the ε-neighbourhood of x is defined by the set
Bε (x) = {y ∈ F : |x − y| < ε}.
1.5 Bounded, Open, Closed and Compact Sets 31

A set M ⊂ F is called neighbourhood of x, if there exists some ε > 0 such that

Bε (x) ⊂ M.

Example 1.50. (a) If F = R, then the ε-neighbourhood of x ∈ R is given by the interval

Bε (x) = (x − ε, x + ε).

(b) If F = C, ε > 0, then the ε-neighbourhood of x ∈ C consists of all complex numbers

being in the interior of a circle in the complex plane with midpoint x and radius ε.
(c) [0, 1] is a neighbourhood of 1
2
(also of 34 , √1
2
etc.), but it is not a neighbourhood of 0
or 1.

Definition 1.51. Bounded, Open, closed, compact sets

Let M ⊂ F. Then M is called

(i) bounded if there exists some c ∈ R such that for all x ∈ M holds: |x| ≤ c.

(ii) open if for all x ∈ M holds: M is a neighbourhood of x.

(iii) closed if for all convergent sequences (an )n∈N with an ∈ M for all n ∈ N holds:
limn→∞ an = a ∈ M .

(iv) compact if for all sequences (an )n∈N with an ∈ M for all n ∈ N holds: There
exists some convergent subsequence (ank )k∈N with limk→∞ ank = a ∈ M .

Example 1.52. (a) The interval (0, 1) is open.

Proof: Consider x ∈ (0, 1). Then for ε = min{x, 1 − x} holds ε > 0 and
Bε (x) = (x − ε, x + ε) ⊂ (0, 1). 2
(b) The interval (0, 1) is not closed.
Proof: Consider the sequence (an )n∈N = ( n+1 1
)n∈N . Clearly, for all n ∈ N holds
an = n+1 ∈ (0, 1), but (an )n∈N converges to 0 ∈
1
/ (0, 1). 2
(c) The interval (0, 1) is not compact.
Proof: Again consider the sequence (an )n∈N = ( n+1
1
)n∈N in (0, 1). The convergence of
(an )n∈N to 0 ∈
/ (0, 1) also implies that this holds true for any subsequence (ank )k∈N
(see Theorem 1.36). Hence, any subsequence of the above constructed one is not
convergent to some value in (0, 1). 2
(d) The interval (0, 1] is neither open nor closed.
Proof: The closedness can be disproved by considering again the sequence
1
(an )n∈N = ( n+1 )n∈N , whereas the non-openness follows from the fact that (0, 1] is
not a neighbourhood of 1. 2
(e) The set R is open and closed but not compact.
Proof: Openness and closedness are easy to verify. To see that this set is not compact,
consider the sequence (an )n∈N = (n)n∈N (which is of course in R). It can be readily
verified that any subsequence (ank )k∈N = (nk )k∈N is unbounded, too. Therefore,
arbitrary subsequences (ank )k∈N = (nk )k∈N cannot converge. 2
32 1 Sequences and Limits

(f) The empty set ∅ is open, closed and compact.

Proof: ∅ is a neighbourhood of all x ∈ ∅ (there is none, but the statement “for all
x ∈ ∅” holds then true more than ever). By the same kind of argumentation, we
can show that this set is compact and closed. The non-existence of a sequence in ∅
implies that every statement holds true for them. In particular, all sequences (an )n∈N
in ∅ converge to some x ∈ ∅ and have a convergent subsequence with limit in ∅. 2

Next we relate these three concepts to each other.

Theorem 1.53.
For a set C ⊂ F, the following statements are equivalent:

(i) C is open;

(ii) F\C is closed.

Proof:
“(i)⇒(ii)”: Let C be open. Consider a convergent sequence (an )n∈N with an ∈ F\C. We
have to show that for a = limn→∞ an holds a ∈ F\C. Assume the converse, i.e., a ∈ C.
Since C is open, we have that Bε (a) ⊂ C for some ε > 0. By the definition of convergence,
there exists some N such that for all n ≥ N holds |a − an | < ε, i.e.,
an ∈ Bε (a) ⊂ C.
However, this is a contradiction to an ∈ F\C.

“(ii)⇒(i)”: Let F\C be closed. We have to show that C is open. Assume the converse,
i.e., C is not open. In particular, this means that there exists some a ∈ C such that for
all n ∈ N holds B 1 (a) 6⊂ C. This means that for all n ∈ N, we can find some an ∈ F\C
n
with an ∈ B 1 (a), i.e., |a − an | < n1 . As a consequence, for the sequence (an )n∈N holds that
n

lim an = a ∈ C,
n→∞

but an ∈ F\C for all n ∈ N. This is a contradiction to the closedness of F\C. 2

Now we present the connection between compactness, closedness and boundedness of

subsets of F. Note that these results hold as well in the Euclidean spaces Rn and Cn .
Theorem 1.54. Theorem of Heine-Borel
For a subset C ⊂ F, the following statements are equivalent:

(i) C is compact;

(ii) C is bounded and closed.

Proof:
“(i)⇒(ii)”: Let C be compact.
Let (an )n∈N be a convergent sequence in F with an ∈ C and a := limn→∞ an ∈ F. Since
C is compact, there is a subsequence (ank )k∈N such that b := limk→∞ ank ∈ C. By
Theorem 1.36 we have a = b ∈ C.
1.5 Bounded, Open, Closed and Compact Sets 33

Now assume that C is unbounded. Then for all n ∈ N, there exists some an ∈ C with
|an | ≥ n. Consider an arbitrary subsequence (ank )k∈N . Due to |ank | ≥ nk ≥ k, we have
that (ank )k∈N is unbounded, i.e., it cannot be convergent. This is also a contradiction to
compactness.

“(ii)⇒(i)”: Let C be closed and bounded. Let (an )n∈N be a sequence in C. The bounded-
ness of C then implies the boundedness of (an )n∈N . By the Theorem of Bolzano-Weierstraß,
there exists a convergent subsequence (ank )k∈N , i.e.,

lim ank = a
k→∞

for some a ∈ F. For compactness, we now have to show that a ∈ C. However, this is
guaranteed by the closedness of C. 2
Remark:
Taking a closer look to the proof “(i)⇒(ii)’, we did not explicitly use that we are
dealing with one of the spaces R or C. Indeed, the implication that compact sets are
bounded and closed holds true for all normed spaces. However, “(ii)⇒(i)” does not
hold true in arbitrary normed spaces. Indeed, there are examples of normed spaces
that have bounded and closed subsets which are not compact.

Definition 1.55. Interior, closure, boundary

For C ⊂ F, we define the

(i) interior of C by the set

C̊ = {x ∈ C : there exists some ε > 0 such that Bε (x) ⊂ C}.

The elements of C̊ are called inner points of C.

(ii) closure of C by the set

C = {x ∈ F : there exists a sequ. (an )n∈N in C with lim an = x}.

n→∞

The elements of C are called osculation points of C.

(iii) boundary of C by the set

∂C = C\C̊.
The elements of ∂C are called boundary points of C.

Remark:
The relation C̊ ⊂ C ⊂ C holds true for arbitrary subsets C ⊂ F. The first inclusion
holds true by definition of C̊. To verify C ⊂ C we take an arbitrary x ∈ C and
consider the constant sequence (x)n∈N . Since this sequence is completely contained
in C and converges to x ∈ C, we must have that x ∈ C.
It can be shown that for all sets C, C̊ is always open and C, ∂C are always closed
sets. In particular, if C is open (closed), then we have C̊ = C (resp. C = C)
34 1 Sequences and Limits

Example 1.56. These are examples in the case F = R:

(a) C = [0, 1], then C̊ = (0, 1), C = [0, 1] and ∂C = {0, 1};
(b) C = { n1 : n ∈ N}, then C̊ = ∅ and C = ∂C = {0} ∪ { n1 : n ∈ N}.
 Infinite Series
2
Truth hurts. Maybe not as much as jumping on a bicycle with the seat
missing, but it hurts.

Lt. Frank Drebin (Leslie Nielsen)

The topic of this part are “infinite sums” of the form

∞
X
ak
k=1

for some sequence (an )n∈N . Before we present a mathematically precise definition, we
present “a little paradoxon” that aims to show that one really has to be careful with
series.
Consider the case where (an )n∈N = ((−1)n )n∈N . On the one hand we can compute

∞
X
(−1)k = − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − . . .
k=1 = (−1 + 1) + (−1 + 1) + (−1 + 1) + (−1 + 1) + (−1 + 1) + . . .
=0 + 0 + 0 + 0 + 0 + ... = 0

and on the other hand

∞
X
(−1)k = − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − . . .
k=1 = − 1 + (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + . . .
= − 1 + 0 + 0 + 0 + 0 + 0 + . . . = −1.

This is a very dramatic contradiction! To exclude such awkward phenomena, we have to

use a precise mathematical definition of “infinite sums”.

35
36 2 Infinite Series

2.1 Basic Definitions, Convergence Criteria and

Examples

Definition 2.1. Infinite series

Let (an )n∈N be a sequence in F. Then the sequence (sn )n∈N defined by
n
X
sn := ak
k=1

is called infinite series (or just “series”). The sequence element sn is called n-th
partial sum of (an )n∈N . The series is called convergent if (sn )n∈N is convergent. In
this case, we write
X∞
ak := lim sn .
n→∞
k=1

Remark:
In the literature, the symbol ∞k=1 ak is also called series. So this symbol has a two-
P
fold meaning, namely the limit of the series (if existent) and the series itself. At
the accordant places of this manuscript, the concrete meaning will be clear from the
context.
The above definition formally does not include infinite sums of kind
∞
X ∞
X ∞
X
ak , ak , or ak for some n0 ∈ N.
k=0 k=2 k=n0

However, their meaning is straightforward to define and we will call these expres-
sions infinite series, too.

Before we give some criteria for the convergence of series, we first present the probably
most important series and analyze their convergence.

Example 2.2. (a) For q ∈ F, the geometric series

∞
X
qk
k=0

is convergent if and only if |q| < 1. Proof: We can show that the n-th partial sum is
given by
n
( n+1
X
k
1−q
1−q
: if q 6= 1,
sn = q =
k=0
n + 1 : if q = 1.

Hence, (sn )n∈N is convergent if and only if |q| < 1. In this case we have
∞
X 1 − q n+1 1
q k = lim sn = lim = .
n→∞ n→∞ 1 − q 1−q
k=0
2.1 Basic Definitions, Convergence Criteria and Examples 37

(b) The harmonic series

∞
X 1
k=1
k
is divergent to +∞.
Proof: If we construct some unbounded subsequence (snl )l∈N , the divergence of the
harmonic series is proven (since it is monotonically increasing). Indeed, we now show
the unboundedness of the subsequence (s2l )l∈N : First, observe that
l
X
s2l = s1 + (s2 − s1 ) + (s4 − s2 ) + (s8 − s4 ) + . . . + (s2l − s2l−1 ) = s1 + (s2j − s2j−1 ).
j=1

Now we take a closer look to the number s2j − s2j−1 : By definition of sn , we have
2 j j
2
X 1 X 1 j−1 1 1
s2j − s2j−1 = > = 2 = .
k 2j 2j 2
k=2j−1 +1 k=2j−1 +1

The inequality in the above formula holds true since every summand is replaced by
the smallest summand 21j . The second last equality sign then comes from the fact
that the number 21j is summed up 2j−1 -times. Now using this inequality together with
the above sum representation for s2l , we obtain
l l
X X 1 l
s2l = s1 + (s2j − s2j−1 ) > 1 + =1+ .
j=1 j=1
2 2

As a consequence, the subsequence (s2l )l∈N is unbounded. 2

(c) For α > 1, the sequence
∞
X 1
k=1
kα
is convergent.
Proof: The sequence of partial sums is strictly monotonically increasing due to
1
sn+1 − sn = ≥ 0.
(n + 1)α
Therefore, by Theorem 1.37 and Theorem 1.32, the convergence of (sn )n∈N is shown
if we find some bounded subsequence (snj )j∈N . Again we use the representation for
s2j − s2j−1 as in example b). We can estimate
2j 2 j
X 1 1 X
s2j − s2j−1 = <
kα + 1)α (2j−1
k=2j−1 +1 k=2j−1 +1
 j−1  j−1
2j−1 2j−1 2 1
= j−1 < j−1 α = = ,
(2 + 1)α (2 ) 2α 2α−1
so we have s2j − s2j−1 < q j−1 for q = 1
2α−1
and, due to α > 1, it holds that 0 < q < 1.
Using that s1 = 1 = q 0 , we obtain
l l−1
X X 1 − ql 1
s2l = s1 + (s2j − s2j−1 ) < 1 + qj = 1 + <1+ .
j=1 j=0
1−q 1−q

Hence, the sequence (s2l )l∈N is bounded. This implies the desired result. 2
38 2 Infinite Series

Remark:
Except for the first example, we have not computed the limits of the other stated
convergent series. We only proved existence or non-existence of limits. Indeed, the
computation of limits of series is, in general, a very difficult issue and is not possible
in many cases.
The function
∞
X 1
ζ(α) =
k=1
kα
is very popular in analytic number theory under the name Riemann Zeta Function.
In b) and c), we have implicitly proven that ζ(·) is defined on the interval (1, ∞)
and has a pole at 1. This function is subject of the Riemann hypothesis which is
one of the most important unsolved problems in modern mathematics. Some known
values of the Zeta function are (without proof )
∞ ∞ ∞
X 1 π2 X 1 π4 X 1 π6
2
= ζ(2) = , 4
= ζ(4) = , 6
= ζ(6) = .
k=1
k 6 k=1
k 90 k=1
k 945

Next we consider sums of convergent series and multiplication of series by some scalar
variable. The proof just consists of a straightforward application of Theorem 1.21 and is
therefore skipped.
Theorem 2.3. Formulae for convergent series
Let λ ∈ F and
∞
X ∞
X
ak , bk
k=1 k=1

be two convergent series in F. Then

(i)
∞
X ∞
X ∞
X
(ak + bk ) = ak + bk ;
k=1 k=1 k=1

(ii)
∞
X ∞
X
(λak ) = λ ak .
k=1 k=1

Theorem 2.4. Cauchy Criterion

∞
X
A series ak in F is convergent if and only if for all ε > 0, there exists some N
k=1
such that for all n ≥ m ≥ N holds
 
Xn 
ak  < ε.
 

 
k=m

Proof: By Theorem 1.46 and Theorem 1.48, a series converges if and only if the sequence
2.1 Basic Definitions, Convergence Criteria and Examples 39

(sn )n∈N of partial sums is a Cauchy sequence.

On the other hand, for n ≥ m, we have
 
Xn 
|sn − sm−1 | =  ak  .
 
 
k=m

Therefore, the Cauchy criterion is really equivalent to the fact that (sn )n∈N is a Cauchy
sequence in F. 2
Remark:
Reconsidering the example at the very beginning of this chapter, the divergence of
this sequence can be directly verified be employing the Cauchy criterion.

As a corollary, we can formulate the following criterion.

Theorem 2.5. Necessary criterion for convergence of series
Let ∞
X
ak
k=1

be a convergent series in F. Then (an )n∈N is convergent with

lim an = 0.
n→∞

Proof: Since the series converges, the Cauchy criterion implies that for all ε > 0, there
exists some N such that for all n ≥ m ≥ N holds
 
Xn 
ak  < ε.
 

 
k=m

Now considering the special case n = m, we have that for all n ≥ N holds

|an | < ε.

However, this is nothing but convergence of (an )n∈N to zero. 2

Remark:
The zero convergence of (an )n∈N is by far not sufficient for convergence. We have
already seen that the harmonic series diverges though the sequence (an )n∈N = ( n1 )n∈N
converges to zero.

For some particular cases, we nevertheless are able to show convergence.

Theorem 2.6. Leibniz Convergence Criterion
Let (an )n∈N be a monotonically decreasing real sequence with

lim an = 0.
n→∞
40 2 Infinite Series

Then the alternating sequence

∞
X
(−1)k ak
k=1
converges.

Proof: Since (an )n∈N is convergent to zero and monotonically decreasing, we have an ≥ 0
for all n ∈ N. Let (sn )n∈N be the corresponding sequence of partial sums. Then we have
for all l ∈ N that

s2l+2 − s2l = −a2l+1 + a2l+2 ≤ 0, s2l+3 − s2l+1 = a2l+2 − a2l+3 ≥ 0,

i.e., the subsequence (s2l )l∈N is monotonically decreasing and (s2l+1 )l∈N is monotonically
increasing. Furthermore, due to s2l+1 − s2l = −a2l+1 ≤ 0 holds

s2l+1 ≤ s2l .

Altogether, the (s2l )l∈N is monotonically decreasing and bounded from below, and (s2l+1 )l∈N
is monotonically increasing and bounded from above. By Theorem 1.32, both sub-
sequences are convergent. Due to

lim (s2l+1 − s2l ) = lim a2l+1 = 0,

l→∞ l→∞

an application of Theorem 1.21 yields that both subsequence have the same limit, i.e.,

lim s2l+1 = lim s2l = s

l→∞ l→∞

for some s ∈ R. Let ε > 0. Then there exists some N1 such that for all l ≥ N1
holds |s2l − s| < ε. Furthermore, there exists some N2 such that for all l ≥ N2 holds
|s2l+1 − s| < ε. Now choosing N = max{2N1 , 2N2 + 1}, we can say the following for some
m ≥ N:
In the case where m is even, we have some l ∈ N with m = 2l. By the choice of N , we
also have l ≥ N1 and thus
|sm − s| = |s2l − s| < ε.
In the case where m is odd, we have some l ∈ N with m = 2l + 1. By the choice of N , we
also have l ≥ N2 and thus
|sm − s| = |s2l+1 − s| < ε.
2

Example 2.7. (a) The alternating harmonic series

∞
X (−1)k
.
k=1
k

converges. The limit is (without proof) log(2).

(b) The alternating Leibniz series
∞
X (−1)k
.
k=1
2k + 1
converges. The limit is (without proof) π4 .
2.2 Absolute Convergence and Criteria 41

(c) The series

∞
X (−1)k
√ .
k=1
k
converges. The limit is not expressible in a closed form.

We will later treat the topic of Taylor series. Thereafter we will be able to determine
some further limits of sequences.

2.2 Absolute Convergence and Criteria

Definition 2.8. Absolute convergence

P∞
Let (an )n∈N be a sequence P in F. Then the series k=1 ak is called absolutely
convergent if the real series ∞k=1 |ak | converges.

Remark:
We will see that absolute convergence
P is really a stronger requirement than con-
vergence. However, for real series ∞ k=1 ak with ak ≥ 0 for all k ∈ N, absolute
convergence and convergence are equivalent. This is a direct consequence of the fact
that ak ≥ 0 implies |ak | = ak .

Theorem 2.9. Absolute convergence implies convergence

Let ∞ k=1 ak be an absolutely convergent series in F. Then the series is also con-
P
vergent.

Proof: Let ε > 0. By the absolute convergence of the series, the necessity of Cauchy’s
convergence criterion implies that there exists some N such that for all n ≥ m ≥ N holds
n
X
|ak | < ε.
k=m

A use of the triangular inequality gives

 n  n
X  X
ak  ≤ |ak | < ε.
 

 
k=m k=m

P∞
Then the sufficiency of Cauchy’s convergence criterion implies the convergence of k=1 ak .

Example 2.10. The alternating harmonic series is convergent as we have seen in in

Example 2.7 a), but it is not absolutely convergent, since the series of absolute values is
the harmonic series (see Example 2.2 b)).

The following criterion can be seen as a “series version” of the comparison criterion for
sequences presented in Theorem 1.22.
42 2 Infinite Series

Theorem 2.11. Majorant criterion

P∞ P∞
Let k=1 a k be a series in F. Moreover, let n 0 ∈ N and let P∞k=1 bk be a real
convergent series such that |ak | ≤ bk for all k ≥ n0 . Then k=1 ak converges
absolutely.

Proof: Let ε > 0. By the Cauchy criterion applied to ∞ k=1 bk there is an N ≥ n0 such
P
that for all n ≥ m ≥ N holds
 
n
X Xn Xn 
0≤ |ak | ≤ bk =  bk  < ε.
 
 
k=m k=m k=m

P∞
The Cauchy criterion now implies that k=1 |ak | converges. 2
Remark:
P∞
The series k=1 bk with the properties as stated in Theorem 2.11 is called a major-
ant of ∞ k=1 k .
P
a

Now we present a kind of reversed majorant criterion that gives us a sufficient criterion
for divergence.
Theorem 2.12. Minorant criterion
Let ∞
P∞
k=1 ak be a real series. Moreover, let nP
0 ∈ N and k=1 bk be a divergent series
P
∞
such that ak ≥ bk ≥ 0 for all k ≥ n0 . Then k=1 ak diverges.

Proof: We prove the result by contradiction: Let ∞ k=1 bk be divergent. Assume that
P
∞
converges. Then, due to ak ≥ bk ≥ 0, the majorant criterion implies the conver-
P
k=1 ak P
gence of ∞ k=1 bk , too. This is a contradiction to our assumption. 2
Remark:
P∞
The series k=1 bk with the properties as stated in Theorem 2.12 is called a minorant
of ∞ .
P
a
k=1 k

Theorem 2.13. Quotient criterion

Let n0 ∈ N and let ∞
k=1 ak be a series in F with the following properties:
P

– ak 6= 0 for all k ≥ n0 ;

– there exists some q ∈ (0, 1) such that for all k ≥ n0 holds

|ak+1 |
≤ q.
|ak |
P∞
Then k=1 ak converges absolutely.

Proof: We inductively obtain that for all k ≥ n0 holds

|ak | ≤ q k−n0 |an0 | .

2.2 Absolute Convergence and Criteria 43

P∞ P∞
Therefore, the series k=1 q k−n0 |an0 | is a majorant of k=1 ak . However, the majorant is
convergent due to
∞
X |an0 |
q k−n0 |an0 | = (see Example 2.2 a)).
k=1
(1 − q)q n0 −1

The majorant criterion then implies convergence. 2

Remark:
Note that the quotient criterion is different from claiming |a|ak+1
k|
|
< 1 (which is for
instance fulfilled by the divergent harmonic series). There has to exist some q < 1
such that the quotient is below q.
The quotient criterion is only sufficient for convergence and indeed, there are ex-
amples of absolutely convergent series that do not fulfill the quotient criterion. For
instance, consider the absolutely convergent series
∞
X 1
2
.
k=1
k

Observing that  2
|ak+1 |  k 
= ,
|ak | k + 1
the fact that this expression converges to 1 implies that there does not exist some
q < 1 for which the quotient criterion is fulfilled. However, this series is convergent
as we have proven in Example 2.2 c).
We could also formulate “an alternative quotient criterion”P that gives us a sufficient
criterion for divergence. Namely, consider a real series ∞ k=1 ak with positive ak
ak+1
and assume that ak ≥ 1 for all k ≥ n0 for some fixed n0 ∈ N. This gives us
0 < ak ≤ ak+1 , i.e., the sequence (an )n∈N is positivePand monotonically increasing.
Such a sequence cannot converge to zero and thus, ∞ k=1 ak is divergent.

Now we present a “limit form” of the quotient criterion:

Theorem 2.14. Quotient criterion (limit form)
Let ∞ k=1 ak be a series in F and assume that there exists some n0 ∈ N such that
P
ak 6= 0 for all k ≥ n0 . If
|ak+1 |
lim sup < 1,
k→∞ |ak |
then ∞ k=1 ak converges absolutely.
P

Proof: Set c := lim sup |a|ak+1

k|
|
< 1. Since lim sup is defined to be the largest accumulation
point of a sequence, we have for every ε > 0 that |a|ak+1
k|
|
≥ c + ε holds true for at most
finitely many k ∈ N. Hence, for ε := 2 , there exists some N ∈ N such that for all k ≥ N
1−c

holds
|ak+1 | 1−c 1+c
<c+ε=c+ = .
|ak | 2 2
Thus, the quotient criterion holds true for q := 1+c
2
which satisfies q < 1 due to c < 1.
44 2 Infinite Series

2
Remark:
|ak+1 |
Since, in case of convergence of |ak |
, the limit and limes superior coincide, the
criterion
|ak+1 |
lim <1
k→∞ |ak |

is also sufficient for absolute convergence of ∞ k=1 ak . However, this criterion re-
P
quires the convergence of the quotient sequence and is therefore weaker than the
above one.

Example 2.15. For x ∈ F, consider the series

∞
X xk
.
k=0
k!

Then an application of the limit form of the quotient criterion yields

 
 xk+1 
 (k+1)!  |x|k+1 k!
lim sup  k  = lim sup
k→∞  xk!  k→∞ |x|k (k + 1)!
1 1
= lim sup |x| = lim |x| = 0 < 1.
k→∞ k+1 k→∞ k+1

Hence, the series converges absolutely. The series ∞ xk

k=0 k! is called exponential series and
P
we will indeed define exp(x) by this expression.

Theorem 2.16.
Let x ∈ F. Then
∞
 x  n X xk
lim 1+ = .
n→∞ n k=0
k!

Proof: First we show that for fixed k ∈ N0 holds

n!
lim = 1. (2.1)
n→∞ (n − k)!nk

Note that n!
(n−k)!nk
is well defined for n ≥ k. For such an n ∈ N we have

k−1
Y n−j n! (n − k + 1)k k−1 k
1≥ = k
≥ k
= (1 − ) . (2.2)
j=0
n (n − k)!n n n

Since the right-hand side converges to 1 for n → ∞, (2.1) follows.

Now let ε > 0 and let K ∈ N be big enough such that

∞
X |x|k ε
< .
k=K
k! 3
2.2 Absolute Convergence and Criteria 45

By (2.1) there is an N ≥ K such that for all n ≥ N also

K−1
X  k
 n!  |x| ε
 (n − k)!nk − 1 k! < 3 .
 
k=0

For n ≥ N we estimate
   
∞ k n   k ∞ k
 x n X x X n x X x
 1+ − = −
   
n k! k n k k! 

  
k=0 k=0 k=0
K−1
X n xk xk  X n   ∞
n |x|k X |x|k
 
≤  k nk − k!  +
  + .
k=0 k=K
k nk k=K
k!
| {z }
< ε/3
K−1  k n ∞
|x|k X |x|k

X n!  |x| X n!
=  (n − k)!nk − 1 k! + +
 
k=0
(n − k)!nk k!
k=K | k=K
k!
| {z } {z } | {z }
≤1
< ε/3 | {z } < ε/3
< ε/3

< ε.

2
The following criterion of Raabe refines the quotient criterion.
Theorem 2.17. Raabe criterion
Let (ak )k∈N be a sequence in F.

a) If there is a k0 ∈ N and a β ∈ (1, ∞) such that ak 6= 0 and

|ak+1 | β
≤1−
|ak | k

for all k ≥ k0 , then the series ∞ k=1 ak converges absolutely.

P

b) If F = R and if there is a k0 ∈ N such that ak 6= 0 and

ak+1 1
≥1−
ak k

for all k ≥ k0 , then the series ∞k=1 ak diverges.

P

Proof: a) For k ≥ k0 holds |a|ak+1 k|

|
≤ k−β
k
. Thus k|ak+1 | ≤ (k − β)|ak | and therefore
β|ak | ≤ k|ak | − k|ak+1 |. Since β > 1 and |ak | > 0 subtracting |ak | from both sides of this
inequality yields

0 < (β − 1)|ak | ≤ (k − 1)|ak | − k|ak+1 | =: bk . (2.3)

In particular, this shows that (k|ak+1 |)k≥k0 −1 is monotonically decreasing. Since it is

bounded from below by zero,P it must converge P to some limit s := limk→∞ k|ak+1 |. Thus
also the telescoping series ∞ b
k=k0 k converges to ∞
k=k0 (k−1)|ak |−k|ak+1 | = (k0 −1)|ak0 |−
46 2 Infinite Series

P∞
s. From (2.3) we conclude that β−1
1
bk is a convergent majorant of ∞
k=k0 |ak |. By
P
P∞ k=k0
the majorant criterion, k=1 ak is absolutely convergent.

b) We may assume that k0 > 1. For k ≥ k0 holds 0 6= ak+1 ak

≥ k−1k
> 0. Since the
right-hand side is nonnegative, ak and ak+1 must have the same sign (either + or -). Since
for the same reason ak+1 and ak+2 must also have the same sign we see that all ak have
the same sign for k ≥ k0 . Without loss of generality we may therefore assume that ak > 0
for all k ≥ k0 . (Otherwise consider the sequence (−ak )k∈N .) Then kak+1 ≥ (k − 1)ak
shows thatP P∞is ak+1 ≥ k for k ≥ k0 . Con-
kak+1 ≥ (k0 − 1)ak0 =: α > 0 for all k ≥ k0 , that α
∞
sequently, k=k0 k is a divergent minorant
α
P∞ for the series k=k0 ak+1 so that the minorant
criterion implies the divergence of k=1 ak . 2

We want to remark that the assumption F = R in b) is not essential. The assertion

also holds for F = C. In this case we have to argue that for k ≥ k0 all ak = |ak |eiϕk
must have the same argument ϕk ∈ [0, 2π). Then again without loss of generality we may
assume that ϕk = 0 for all k ≥ k0 .
Theorem 2.18. Root criterion
Let n0 ∈ N and let ∞ k=1 ak be a series in F and assume that there exists some
P
q ∈ (0, 1) such that for all k ≥ n0 holds
p
k
|ak | ≤ q.

Then ∞ k=1 ak converges absolutely.

P

Proof: Taking the k-th power of the inequality k |ak | < q, we obtain that for all k ≥ n0
p

holds
|ak | < q k
Therefore, the convergent geometric series ∞
P∞
k=1 q is a majorant of k=1 ak and thus,
k
P
we have absolute convergence. 2
Theorem 2.19. Root criterion (limit form)
Let ∞k=1 ak be a series in F and assume that
P

p
lim sup k |ak | < 1.
k→∞
P∞
Then k=1 ak converges absolutely.

Proof: pThe argumentation is analogous as in the proof of Theorem 2.14. Let c :=

lim sup k |ak | < 1. Then for ε := 1−c
2
, there exists some N ∈ N such that for all k ≥ N
holds
p
k 1−c 1+c
|ak | < c + ε = c + = < 1.
2 2
The root criterion with q := 1+c
2
< 1 now implies convergence. 2

Example 2.20. Consider the series

∞
X k5
.
k=1
3k
2.2 Absolute Convergence and Criteria 47

Then we have
s  √
k 5
 5
k k  k
lim sup  k  = lim sup
k→∞ 3 k→∞ 3
√
Since we know (from the tutorial) that k
k converges to 1, the whole expression converges
to 13 < 1. Hence, the series converges.

P∞
We will now state two convergence criteria for series of the form k=1 ak bk . They are
easily deduced from the following lemma.
Lemma 2.21. Abel’s partial sums
For n ∈ N and a1 , ..., an , b1 , ..., bn+1 ∈ F holds
n
X n
X
ak bk = An bn+1 + Ak (bk − bk+1 ) ,
k=1 k=1

where Ak := ai for k ∈ {1, ..., n}.

Pk
i=1

Proof: If we additionally define A0 := 0, then

n
X n
X n
X n
X n
X n−1
X
ak b k = (Ak − Ak−1 )bk = Ak bk − Ak−1 bk = Ak bk − Ak bk+1
k=1 k=1 k=1 k=1 k=1 k=1
n
X n
X n
X
= Ak bk − Ak bk+1 + An bn+1 = Ak (bk − bk+1 ) + An bn+1 .
k=1 k=1 k=1

Theorem 2.22. Abel criterion

If the series ∞k=1 ak in F converges and if the real sequence (bk )k∈N is monotonic
P
P∞
and bounded, then the series k=1 ak bk converges.

Proof: Set Ak := ki=1 ai . By assumption both sequences (Ak )k∈N and (bk )k∈N converge
P
so that (Ak bk+1 )k∈N converges also. Since (bk )k∈N is monotonic, the telescoping series
∞
k=1 (bk − bk+1 ) converges absolutely as
P

n n
n→∞
X X
|bk − bk+1 | = | (bk − bk+1 )| = |b1 − bn+1 | −−−→ |b1 − lim bk | .
k→∞
k=1 k=1

Since (Ak )k∈N is bounded, ∞ Pb∞k+1 ) is also absolutely convergent. Summing

P
k=1 Ak (bk −
up, Lemma 2.21 implies that the series k=1 ak bk converges to the limit

∞
X ∞
X
ak bk = lim An bn+1 + Ak (bk − bk+1 ) .
n→∞
k=1 k=1

2
48 2 Infinite Series

Theorem 2.23. Dirichlet criterion

P∞
If the series k=1 ak in F is boundedP and if the real sequence (bk )k∈N converges
monotonically to zero, then the series ∞k=1 ak bk converges.

Proof: Set Ak := i=1 ai . By assumption (Ak )k∈N is bounded. Hence (Ak bk+1 )k∈N
Pk
converges
P∞ to zero. By the same argument as in the proof of Theorem 2.22, the series
P∞k=1 k k − bk+1 ) is absolutely convergent. Lemma 2.21 again implies that the series
A (b
k=1 ak bk converges to the limit
∞
X X∞
ak bk = lim An bn+1 + Ak (bk − bk+1 ) .
n→∞
k=1 | {z } k=1
=0
2

Note that the Leibniz criterion 2.6 follows from the Dirichlet criterion by taking ak :=
(−1)k .
Definition 2.24. Reordering
Let ∞ k=1 ak be a series in F and let τ : N → N be a bijective mapping. Then the
P
series ∞
X
aτ (k)
k=1
P∞
is called a reordering of k=1 ak .

Theorem 2.25. Reordering of absolutely convergent series

Let ∞
P∞
k be an absolutely convergent series in F and let k=1 aτ (k) be a reorder-
P
k=1 aP
∞
ing. Then k=1 aτ (k) is also absolutely convergent. Moreover,
∞
X ∞
X
aτ (k) = ak .
k=1 k=1

Proof: Let a = ∞ k=1 ak and let τ : N → N be bijective. Let ε > 0. Since we have absolute
P
convergence, there exists some N1 such that
∞
X ε
|ak | <
k=N
2
1

and thus    ∞ 
N1 −1 ∞
 X  X  X ε
a − ak  =  ak  ≤ |ak | < .
   

k=1
 
k=N

k=N
2
1 1

Now choose N := max{τ −1

(1), τ −1
(2), ..., τ −1
(N1 − 1)}.Then
{1, 2, . . . , N1 − 1} ⊂ {τ (1), τ (2), . . . , τ (N )}.
Then, for all n ≥ N holds
    N −1 
n N 1 −1 1 n  ε ∞
 X   X  X X X ε ε
a − aτ (k)  ≤ a − ak  +  ak − aτ (k)  < + |ak | < + = ε.
     
      2 2 2
k=1 k=1 k=1 k=1 k=N 1
2.3 The Cauchy Product of Series 49

The absolute convergence ofP ∞k=1 aτ (k) can

P∞be shown by an application of the above
P
∞
argumentation to the series k=1 |ak | and k=1 |aτ (k) |. 2

2.3 The Cauchy Product of Series

Definition 2.26. Cauchy product of sequences

Let ∞
P∞ P∞
k=0 ak , k=0 bk be two series in F. Then the Cauchy product of k=0 ak ,
P
∞
k=0 bk is given by
P
X∞ Xk
ck with ck = al bk−l .
k=0 l=0

Remark:
Note that we considered series with Plower summation index 0. The Cauchy product
can be also defined for sequences ∞
P∞
a
k=n0 k , k=n1 k with arbitrary n0 , n1 ∈ N (or
b
even n0 , n1 ∈ Z). In this case, the Cauchy product is given by
∞
X k−n
X1
ck with ck = al bk−l .
k=n0 +n1 l=n0

In order to “keep the set of indices manageable”, this is not further treated here.
Note that the following result about convergence properties of the Cauchy product
still hold true in this above mentioned more general case.

The following theorem justifies the name “product” in the above definition.
Theorem 2.27. Convergence of the Cauchy product
Let ∞
P∞ P∞
ak , k=0 bk be series in F. Assume that is absolutely conver-
P
k=0 P k=0 ak P
gent and ∞ b
k=0 k is convergent. Then the Cauchy product ∞
k=0 ck is absolutely
convergent. Moreover, the limit satisfies
∞ ∞
! ∞
!
X X X
ck = ak · bk .
k=0 k=0 k=0

Proof: Denote the sequence of partial sums of ∞

P∞ P∞
k=0 ak ,P k=0 bk and Pc∞k by (An )n∈N ,
P
k=0
∞
(Bn )n∈N and (Cn )n∈N , respectively. Moreover, set a := k=0 ak and b = k=0 bk .
Then we have
n
X n X
X k n
X k
X n
X
Cn = ck = al bk−l = an−k bl = an−k Bk .
k=0 k=0 l=0 k=0 l=0 k=0

Using this expression, we obtain

n
X n
X n
X
Cn = an−k (Bk − b) + an−k b = an−k (Bk − b) + An b.
k=0 k=0 k=0
50 2 Infinite Series

P∞
Let ε > 0. Since k=0 ak converges absolutely, there exists some N0 such that for all
n ≥ N0 holds
ε
|Bn − b| < . P∞
4( k=0 |ak | + 1)
Since (an )n∈N converges to zero (see Theorem 2.5), there exists some N1 such that for all
n ≥ N1 holds
ε
|an | < .
4N0 (sup{|b − Bl | : l ∈ N} + 1)
Also there exists some N2 such that for all n ≥ N2 holds
ε
|An − a| < .
2(|b| + 1)
Therefore, with N = max{N0 + N1 , N2 }, we have that for all n ≥ N holds
 
Xn  X n
|Cn − ab| =  an−k (Bk − b) + b(An − a) ≤ |an−k ||Bk − b| + |b||An − a|
 
 
k=0 k=0
N
X0 −1 X n
= |an−k | |Bk − b| + |an−k | |Bk − b| +|b| |An − a|
| {z } | {z } | {z }
k=0 ε k=N0 ε ε
< 4N < 4(P∞ < 2(|b|+1)
0 (sup{|b−Bl | : l∈N}+1) k=0
|ak |+1)

N0 −1 n
X ε|Bk − b| X ε|a | ε|b|
< + P∞ n−k +
k=0
4N0 (sup{|b − Bl | | l ∈ N} + 1) k=N 4( k=0 |ak | + 1) 2(|b| + 1)
0
ε ε ε
< + + =ε
4 4 2
Example 2.28. Let x, y ∈ F and consider the series
∞ ∞
X xk X yk
,
k=0
k! k=0
k!
which are absolutely convergent
P∞ as shown in Example 2.15. Then the Cauchy product of
both series is given by k=0 ck with
k k  
X xl y k−l 1 X k l k−l
ck = = · xy .
l=0
l! (k − l)! k! l=0 l
By the Binomial Theorem (see tutorial), we obtain
k  
X k l k−l
xy = (x + y)k .
l
l=0
Hence, by Theorem 2.27, we have
∞
! ∞
! ∞
X xk X yk X (x + y)k
· = .
k=0
k! k=0
k! k=0
k!
Altogether, this means that the function
∞
X xk
f (x) =
k=0
k!
fulfills f (x + y) = f (x) · f (y) for all x, y ∈ R (and even x, y ∈ C). This property is for
instance fulfilled by the exponential function. Indeed, we have that f as defined above
fulfills f (x) = ex .
 Continuous Functions
3
Now do you think you can beat the champ?
I can take him blindfolded.
What if he’s not blindfolded?
Police Squad!

The central notion of this chapter is continuity that is a property of functions. Very very
roughly speaking, it means that the function has no jumps. This is again a very abstract
concept. However, the remaining parts of the lecture are getting easier!

3.1 Bounded Functions, Pointwise and Uniform

Convergence

Definition 3.1. Bounded functions

Let I be a set. Then we call a function f : I → F bounded, if

sup{|f (x)| : x ∈ I} < ∞.

In the following, we consider sequences of functions and introduce some convergence

concepts.
Definition 3.2. Pointwise convergence
A sequence (fn )n∈N of functions fn : I → F is called pointwisely convergent to
f : I → F if for all x ∈ I holds

lim fn (x) = f (x).

n→∞

Using logical quantifiers this reads:

∀x ∈ I ∀ε > 0 ∃N ∈ N ∀n ≥ N : |fn (x) − f (x)| < ε . (3.1)

Pointwise convergence means nothing else but that for all x ∈ I, the sequence (fn (x))n∈N

51
52 3 Continuous Functions

in F converges to f (x).
We now present an alternative convergence concept for functions:
Definition 3.3. Uniform convergence
A sequence (fn )n∈N of functions fn : I → F is called uniformly convergent to
f : I → F if for all ε > 0 there exists some N such that for all n ≥ N and
x ∈ I holds
|f (x) − fn (x)| < ε.
Using logical quantifiers this reads (in contrast to (3.1)):

∀ε > 0 ∃N ∈ N ∀n ≥ N ∀x ∈ I : |fn (x) − f (x)| < ε . (3.2)

As a rule of thumb, you can think of pushing one quantifier to the right but, of course,
this will change a lot. The interpretation is that we can measure the distance between
two functions f und g as the largest distance between the two graphs, that means the
distance you can measure at a given point:

sup |f (x) − g(x)|

x∈I

We have the uniform convergence if this measured distance between fn and f is convergent
to zero. (See below.)

Look at the following example:

1 f1
f2
f3

−1

One sees that the functions fn pointwisely converges to a limit function. We also see that
we can build a jump by increasing n for the limit function. The distance between the
limit function to each member of the sequence is indeed

sup |f (x) − fn (x)| = 2

x∈R

since there is always an x ∈ R which can be chosen close enough to 0 to get an ap-
proximation of this distance. This means that the sequence of functions (fn )n∈N does not
converges uniformly in spite of being pointwisely convergent. The uniform convergence is
in fact a much stronger notion.
We will now see that uniform convergence is a stronger property than pointwise conver-
gence.
3.1 Bounded Functions, Pointwise and Uniform Convergence 53

Theorem 3.4. Uniform convergence implies pointwise convergence

Let a sequence (fn )n∈N with fn : I → F be uniformly convergent to f : I → F. Then
(fn )n∈N is also pointwisely convergent to f .

Proof: Let ε > 0. Then there exists some N such that for all n ≥ N and x ∈ I holds

|f (x) − fn (x)| < ε.

In particular, for some arbitrary x ∈ I holds

|f (x) − fn (x)| < ε.

Hence, the sequence (fn (x))n∈N in F converges to f (x). 2

Remark:
Uniform convergence to f : I → F means that for all ε > 0 holds that all (except
finitely many) functions fn are “inside some ε-stripe around f ” (see Fig. 3.1).

Theorem 3.5.
A sequence (fn )n∈N of functions fn : I → F converges uniformly to f : I → F, if,
and only if,

lim ||fn − f ||∞ = lim sup{|f (x) − fn (x)| : x ∈ I} = 0. (3.3)

n→∞ n→∞

This means that uniform convergence is nothing but convergence with respect to the
infinity norm || · ||∞ .

f+

f−

fn

Figure 3.1: Uniform convergence graphically illustrated

54 3 Continuous Functions

Proof. Assume that (fn )n∈N converges uniformly to f . Let ε > 0. Then there exists some
N such that for all n ≥ N and x ∈ I holds
ε
|f (x) − fn (x)| < .
2
Therefore, for all n ≥ N , we have
ε
sup{|f (x) − fn (x)| : x ∈ I} ≤ < ε,
2
and thus, (3.3) holds true.
Conversely, assuming that (3.3) holds true, we obtain that for ε > 0, there exists some N
with the property that for all n ≥ N holds
sup{|f (x) − fn (x)| : x ∈ I} < ε.
This means that for all n ≥ N and x ∈ I, there holds that
|f (x) − fn (x)| < ε.
However, this statement is nothing but uniform convergence of (fn )n∈N towards f .

Example 3.6. a) Let I = [0, 1] and consider the sequence fn (x) = xn . Then we have
the pointwise limit
(
0 , if x ∈ [0, 1),
f (x) = lim fn (x) =
n→∞ 1 , if x = 1.

Is (fn )n∈N also uniformly convergent√to f ?

The answer is no, since for xn = 1/ n 2, there holds
 
|f (xn ) − fn (xn )| = 0 − 12  = 21 .

b) We now consider the same sequence on the smaller interval [0, 12 ]. The pointwise limit
is now f = 0. For n ∈ N, we have
1
sup |f (x) − fn (x)| : x ∈ [0, 21 ] = sup xn : x ∈ [0, 12 ] = n .
 
2
Therefore
lim sup |f (x) − fn (x)| : x ∈ [0, 21 ] = 0

n→∞
and hence, we have uniform convergence.
c) Define the function fn : [0, 1] → R by
(
n2 x(1 − nx) , if x ∈ [0, n1 [,
fn (x) =
0 , if x ∈ [ n1 , 1].

Then for all x ∈ [0, 1] holds

lim fn (x) = 0
n→∞

since fn (0) = 0 and fn (x) = 0 if x > n1 . The sequence (fn )n∈N is however not uniformly
convergent to f = 0, since

 n
1 
sup{|fn (x) − 0| : x ∈ [0, 1]} ≥ fn 2n = .
4
3.2 Continuity 55

Theorem 3.7.
Let (fn )n∈N be a sequence of bounded functions fn : I → F. Assume that (fn )n∈N
converges uniformly to f : I → F. Then f is bounded.

Proof: For ε = 1, there exists some N such that for all n ≥ N and x ∈ I holds

|fn (x) − f (x)| < 1.

In particular, we have |fN (x) − f (x)| < 1 for all x ∈ I. This consequences that for all
x ∈ I, there holds
|f (x)| < |fN (x)| + 1.
The boundedness of fN then implies the boundedness of f . 2
Remark:
Note that the assumption of uniform convergence is essential for the boundedness of
f . For instance, consider the sequence (fn )n∈N of bounded functions fn : [0, ∞) → R
with (
x : if x < n
fn (x) =
0 : else.
First we argument that (fn )n∈N converges pointwisely to f : [0, ∞) → R with f (x) =
x: Let x ∈ [0, ∞). Then there exists some N ∈ N with x < N . Hence, for all n ≥ N ,
we have fn (x) = x. This implies convergence to f : [0, ∞) → R with f (x) = x.
Second we state that each fn is bounded: This is a consequence of the fact that, by
the definition of fn , there holds fn (x) < n for all x ∈ [0, ∞).
Altogether, we have found a sequence of bounded functions pointwisely converging
to some unbounded function. Hence, Theorem 3.7 is no longer valid, if we replace
the phrase “uniformly convergent” by “pointwisely convergent”.

3.2 Continuity
Now we begin to introduce the concept of continuity.
Definition 3.8.
Let I ⊂ F, let f : I → F be a function, and let x0 ∈ I. Then we define

(i) the limit of f as x tends to x0 by c ∈ F if for all sequences (xn )n∈N in I\{x0 }
with limn→∞ xn = x0 holds limn→∞ f (xn ) = c. In this case, we write

lim f (x) = c
x→x0

(ii) the limit of f as x tends from the left to x0 by c ∈ F if I ⊂ R and if

for all sequences (xn )n∈N in {x ∈ I : x < x0 } with limn→∞ xn = x0 holds
limn→∞ f (xn ) = c. In this case, we write

lim f (x) = c.
x%x0
56 3 Continuous Functions

(iii) the limit of f as x tends from the right to x0 by c ∈ F if I ⊂ R and if

for all sequences (xn )n∈N in {x ∈ I : x > x0 } with limn→∞ xn = x0 holds
limn→∞ f (xn ) = c. In this case, we write

lim f (x) = c.
x&x0

In all three cases we assume that at least one sequence (xn )n∈N with the stated
property exists.

Remark:
From the above definition, we can also conclude that limx→x0 f (x) exists in the case
I ⊂ R if and only if limx%x0 f (x) and limx&x0 f (x) exist and are equal. In this case,
there holds
lim f (x) = lim f (x) = lim f (x).
x%x0 x&x0 x→x0

Though not explicitly introduced in the above definition, it should be intuitively clear
what is meant by the following expressions

lim f (x) = y, lim f (x) = y, lim f (x) = ∞, lim f (x) = −∞.

x→∞ x→−∞ x→x0 x→x0

Example 3.9. a) Consider the Heaviside function H : R → R with

(
1 , if x ≥ 0,
H(x) =
0 , if x < 0.

Then we have limx%0 H(x) = 0, since for all xn ∈ R with xn < 0 holds H(xn ) = 0.
Further, limx&0 H(x) = 1, since for all xn ∈ R with xn > 0 holds H(xn ) = 1. The
n
limit limx→0 H(x) does not exist. E.g., take the sequence xn = (−1)
n
. Then
(
1 : if n is even,
H(xn ) =
0 : if n is odd.

Hence, (H(xn ))n∈N is divergent.

b) Consider the function f : R → R with
(
1 , if x = 0,
f (x) =
0 , if x =
6 0.

Then for all sequences (xn )n∈N in R\{0} holds that (f (xn ))n∈N is a constant zero
sequence. Hence, limx→0 f (x) = 0.
c) Consider a polynomial p : R → R with p(x) = an xn + . . . + a1 x + a0 for some given
a0 , . . . , an ∈ R. Let x0 ∈ R. By Theorem 1.21, we have that for all real sequences
(xn )n∈N converging to x0 holds that p(xn ) converges to p(x0 ), i.e.,

lim p(x) = p(x0 ).

x→x0
3.2 Continuity 57

Definition 3.10. Continuity

Let I ⊂ F and let f : I → F be a function. Then f is called continuous in x0 ∈ I if

lim f (x) = f (x0 ).

x→x0

Moreover, f is called continuous on I if it is continuous in x0 for all x0 ∈ I.

Remark:
Sometimes we will just say f : I → F is continuous whereby we mean it is continuous
on I.

Example 3.11. a) The constant function f : R → R with f (x) = c for some c ∈ R is

continuous on R.
b) The Heaviside function (Example 3.9 a)) is discontinuous at x0 = 0, but continuous
everywhere else.
c) The function f : R → R as in Example 3.9 b) is discontinuous at x0 = 0, but continuous
everywhere else.
d) Polynomials (Examples 3.9 d)) are continuous on R.
e) Rational functions f : I → F with f (x) = p(x)
q(x)
for some polynomials p, q (q is not the
zero polynomial) are defined on I = {x ∈ R : q(x) 6= 0} and are continuous on I (due
to Theorem 1.21).
f) The absolute value function | · | : R → R, i.e.,
(
x , if x ≥ 0,
|x| =
−x , if x < 0.

is continuous on R.
g) The function f : R → R with
(
1 , if x ∈ Q,
f (x) =
0 , if x ∈
/Q

is everywhere discontinuous.
Proof: Let x0 ∈ R:
First case: x0 ∈ Q. Then√take a sequence (xn )n∈N with limn→∞ xn = x0 and xn ∈ R\Q
(for instance, xn = x0 + n2 ). Then f (xn ) = 0 for all n ∈ N and thus limn→∞ xn = 0 6=
f (x0 ) = 1.
Second case: x0 ∈ R\Q. Then take a sequence (xn )n∈N with limn→∞ xn = x0 and
xn ∈ Q (this exists since every real number can be approximated by a rational number
in arbitrary good precision). Then f (xn ) = 1 for all n ∈ N and thus limn→∞ xn = 1 6=
f (x0 ) = 0.
58 3 Continuous Functions

Theorem 3.12. ε-δ criterion for continuity

Let I ⊂ F and f : I → F be a function. Let x0 ∈ I. Then the following two
statements are equivalent.

(i) f is continuous in x0 ;

(ii) For all ε > 0 there exists some δ > 0 such that for all x ∈ I with |x − x0 | < δ
holds
|f (x) − f (x0 )| < ε.

f+

f−

x− x+

Figure 3.2: ε-δ criterion graphically illustrated

Proof:
“(i)⇒(ii)”: Assume that (ii) is not fulfilled, i.e., there exists some ε > 0, such that for
all δ > 0, there exists some x ∈ I \ {x0 } with |x − x0 | < δ and |f (x) − f (x0 )| > ε. As
a consequence, for all n ∈ N, there exists some xn ∈ I \ {x0 } with
1
|x0 − xn | < and |f (xn ) − f (x0 )| > ε.
n
Therefore, the sequence (xn )n∈N converges to x0 , but |f (xn ) − f (x0 )| > ε, i.e., f (xn ) is
not converging to f (x0 ).
“(ii)⇒(i)”: Let (xn )n∈N be a sequence in I\{x0 } that converges to x0 . Let ε > 0. Then
there exists some δ > 0 such that for all x ∈ I with |x − x0 | < δ holds |f (x) − f (x0 )| <
ε. Since (xn )n∈N converges to x0 , there exists some N such that for all n ≥ N holds
|xn − x0 | < δ. By the ε-δ-criterion, we have then for all n ≥ N that

|f (xn ) − f (x0 )| < ε.

3.2 Continuity 59

Hence, (f (xn ))n∈N converges to f (x0 ). 2

Next we give a result on the continuity of sums, products and quotients of functions. This
looks very similar to Theorem 1.21. Indeed, those results on sums, products and quotients
of sequences are the “main ingredients” for the proof (which is therefore skipped).
Theorem 3.13. Continuity of sums, products and quotients of functions
Let I ⊂ F and let f, g : I → F be continuous in x0 ∈ I. Then also f + g and f · g
are continuous in x0 . Furthermore if g(x0 ) 6= 0, then also fg is continuous in x0 .

Now we consider the composition of functions f and g (f ◦ g) which is defined by the

formula (f ◦ g)(x) = f (g(x)).
Theorem 3.14. Continuity of compositions of functions
Let I1 , I2 ⊂ F and f : I1 → F, g : I2 → F be functions with

g(I2 ) = {g(x) : x ∈ I2 } ⊂ I1 .

Assume that g is continuous in x0 ∈ I2 and f is continuous in g(x0 ) ∈ I1 . Then

also f ◦ g is continuous in x0 .

Proof: Let (xn )n∈N be a sequence in I2 with limn→∞ xn = x0 . By the continuity of g holds
limn→∞ g(xn ) = g(x0 ). Then, by the continuity of f holds

lim f (g(xn )) = f (g(x0 )).

n→∞

2
In the following, we collect very important properties of continuous functions. The first
result is that continuous functions are bounded as far as they are defined on some compact
set. Thereafter, we present the famous Intermediate Value Theorem which basically states
that continuous functions attain every value between f (x0 ) and f (x1 ) for some arbitrary
x0 , x1 ∈ I. This result leads us to think about continuous functions as “those functions
whose graph can be drawn without putting down the pencil”.
Theorem 3.15. Continuous functions defined on a compact set
Let I ⊂ F be compact and let f : I → F be continuous. Then f (I) is compact. In
particular, by the Theorem of Heine-Borel, f (I) is bounded and closed. If further
f (I) ⊂ R, so that there exist x+ , x− ∈ I such that

f (x+ ) = max{f (x) : x ∈ I}, f (x− ) = min{f (x) : x ∈ I}.

Proof. Let (yn )n∈N be a sequence in f (I). Then for each n ∈ N there is an xn ∈ I such
that yn = f (xn ). Since I is compact, there exists a subsequence (xnk )k∈N that converges
to some x ∈ I. Now, since f is continuous, we have

lim ynk = lim f (xnk ) = f (x) =: y ∈ f (I).

k→∞ k→∞

Hence we found a subsequence (ynk )k∈N of (yn )n∈N that converges in f (I). Therefore f (I)
is compact.
60 3 Continuous Functions

Now we show that a uniformly convergent sequence of continuous functions has to converge
to a continuous function.
Theorem 3.16.
Let I ⊂ F and let (fn )n∈N be a sequence of continuous functions fn : I → F that
uniformly converges to some f : I → F. Then f is continuous.

Proof: Let ε > 0 and let x0 ∈ I. Since (fn )n∈N converges uniformly to f : I → F, there
exists some N such that for all n ≥ N and x ∈ I holds
ε
|f (x) − fn (x)| < .
3
Since fn is continuous on I, there exists some δ > 0 such that for all x ∈ I with |x−x0 | < δ
holds
ε
|fn (x0 ) − fn (x)| < .
3
Altogether, we then have

|f (x) − f (x0 )| = |f (x) − fn (x) + fn (x) − fn (x0 ) + fn (x0 ) − f (x0 )|

≤ |f (x) − fn (x)| + |fn (x) − fn (x0 )| + |fn (x0 ) − f (x0 )|
ε ε ε
< + + = ε.
3 3 3
Therefore, f is continuous by the ε-δ criterion. 2
Remark:
The above result also gives a sufficient characterisation for a sequence of pointwisely
convergent functions (fn )n∈N that are not uniformly convergent: If (fn )n∈N converges
to a discontinuous function, then this sequence cannot be uniformly convergent. For
instance, consider the sequence (fn )n∈N in Examples 3.6 a) with fn : [0, 1] → R,
x 7→ xn . The pointwise limit is given by
(
0 , if x ∈ [0, 1),
f (x) = lim fn (x) =
n→∞ 1 , if x = 1,

which is discontinuous at x = 1 though each fn is continuous. Therefore, this

sequence cannot be uniformly convergent.

Theorem 3.17. Intermediate Value Theorem

Let a, b ∈ R with a < b. Moreover, let f : [a, b] → R be continuous. Let x0 , x1 ∈ [a, b]
and let y ∈ R be between f (x0 ) and f (x1 ). Then there exists some x̂ between x0 and
x1 such that y = f (x̂).

Proof. Without loss of generality, we assume that x0 < x1 . Moreover, we can assume
without loss of generality that y = 0 (otherwise, consider the function f (x) − y instead of
f ). Furthermore, we can assume without loss of generality that f (x0 ) ≤ 0 and f (x1 ) ≥ 0
(otherwise, consider −f instead of f ).
We will construct x̂ by nested intervals (compare the proof of Theorem 1.38).
Inductively define A0 = x0 , B0 = x1 and for k ≥ 1,
3.2 Continuity 61

f+

f−

x− xh x+

Figure 3.3: Intermediate Value Theorem graphically illustrated

Ak−1 +Bk−1
a) Ak = Ak−1 , Bk = 2
, if f ( Ak−1 +B
2
k−1
) ≥ 0, and
Ak−1 +Bk−1
b) Ak = 2
, Bk = Bk−1 , if f ( Ak−1 +B
2
k−1
) ≤ 0.
Then we have
lim An = lim Bn =: x̂.
n→∞ n→∞

By the continuity of f and the fact that f (An ) ≤ 0, f (Bn ) ≥ 0 for all n ∈ N, we have
0 ≥ lim f (An ) = f (x̂) = lim f (Bn ) ≥ 0
n→∞ n→∞

and thus f (x̂) = 0.

(∗) Supplementary details: Metric spaces and topologies

We close this chapter with some remarks on generalisations. As already mentioned in Chapter 1,
almost all of the introduced concepts can be stated for general normed F-vector spaces (V, || · ||)
instead of the special one-dimensional Euclidean/unitary case (F, | · |). In those cases where
explicitly completeness properties of (F, | · |) are used, (V, || · ||) must be taken as a complete
normed F-VS, i.e. as a Banach space, and in those situations where the Theorem of Heine-Borel is
applied, (V, ||·||) can for instance be taken as a finite dimensional normed F-vector space. A further
generalization is that to metric spaces (X, d) introduced in the exercises, where the vector space V
is replaced by an arbitrary set and the norm || · || is replaced by an abstract distance measurement,
a metric d. The formal definition is: A metric on a set X is a function d : X × X → [0, ∞) that
has the following properties:
a) d(x, y) = 0 ⇔ x = y
b) d(x, y) = d(y, x) (symmetry)
c) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X (triangle inequality).
62 3 Continuous Functions

Note that a normed vector space (V, || · ||) is a special case of a metric space since we can define
X := V and d||·|| (v, w) := ||v − w|| for all v, w ∈ V to obtain a metric space (X, d||·|| ) with the
same so-called topological properties. Note also that a metric space has no algebraic structure,
that is there is no summation or scalar multiplication defined on X. In particular, this means that
generalizations of introduced concepts and theorems to metric spaces only make sense in those
cases where no algebraical properties are needed.
For example, the (ε-δ)-criterion for continuity generalizes to a function f : X1 → X2 from one
metric space (X1 , d1 ) to another metric space (X2 , d2 ) in the following way: f is continuous if
and only if for each ε > 0 there is a δ > 0 such that for all x, y ∈ X1 with d1 (x, y) < δ holds
d2 (f (x), f (y)) < ε .

Finally we mention that metric spaces are still not the end of the road for possible gener-
alizations. Even more general so-called topological spaces (X, τ ) can be considered, where X 6= ∅
is a set and τ is a set of subsets of X, i.e. each U ∈ τ is a subset of X. Then τ is called a topology
on X if the following properties are satisfied:

a) ∅, X ∈ τ
b) The union of each subset W of τ is again an element of τ , that is
∪W := {x ∈ X | ∃W ∈ W : x ∈ W } ∈ τ .
c) The intersection of each finite subset W of τ is again an element of τ , that is, if |W| ∈ N,
then ∩W := {x ∈ X | ∀W ∈ W : x ∈ W } ∈ τ .

The elements of τ are called open sets and the complements of open sets X\U where U ∈ τ are
called closed sets. A metric space (X, d) is special case of a topological space. If we set

τd := {U ⊂ X | ∀x ∈ U ∃ε > 0 : Bε (x) ⊂ U },

where Bε (x) := {y ∈ X | d(x, y) < ε} is the open d-ball with radius ε around x, then (X, τd ) is a
topological space with the same topological properties as (X, d).

Now, for example, a function f : X1 → X2 from one topological space (X1 , τ1 ) to an-
other topological space (X2 , τ2 ) is continuous, if for each U2 ∈ τ2 holds f −1 (U2 ) ∈ τ1 , which
means that each preimage of an open set in X2 is an open set in X1 . Indeed, this generalizes the
(ε-δ)-criterion for continuity for metric spaces to general topological spaces.
Good references to metric and topological spaces are the books “Mengentheoretische Topologie”
of B.v. Querenburg, and “General Topology” of R. Engelking.
 Elementary Functions
4
Her lips said no, but her eyes said ’Read my lips’ !
Niles Crane

Here we will introduce some important functions like polynomials, rational functions, exp,
sin, cos, log, sinh, cosh, power series etc. We will also consider the extension of some of
the aforementioned functions to the complex plane.

4.1 Exponential Function

Definition 4.1.
The exponential function exp : F → F is defined as
∞
X xk
exp(x) = .
k=0
k!

The number ∞
X 1
e = exp(1) =
k=0
k!
is called Euler’s number.

For Euler’s number holds

e ≈ 2.718281828459046.

We already know from Example 2.15 that the above defined series converges for all x ∈ F.
Now we present an estimate for exp(x) if the series is replaced by a finite sum.

63
64 4 Elementary Functions

Theorem 4.2.
For n ∈ N and x ∈ F with |x| ≤ 1 + n
2
holds
n
X xk |x|n+1
exp(x) = + rn (x) with |rn (x)| ≤ 2 .
k=0
k! (n + 1)!

Proof:  
∞ ∞
 X xk  X |x|k
|rn (x)| =  ≤


k=n+1
k!  k=n+1 k!
∞
|x|n+1 X |x|k
= ·
(n + 1)! k=0 (n + 2) · (n + 3) · · · (n + 1 + k)
∞
|x|n+1 X |x|k |x|n+1 1
≤ · = ·
(n + 1)! k=0 (n + 2) k (n + 1)! 1 − |x|
n+2

Since |x| ≤ n
2
+1= n+2
2
, we have

|x|n+1 1 |x|n+1
|rn (x)| ≤ · 1 =2
(n + 1)! 1 − 2
(n + 1)!

2
Theorem 4.3. Properties of the Exponential Function

(i) For all x, y ∈ C holds exp(x + y) = exp(x) exp(y).

(ii) For all x ∈ C holds exp(x) = exp(x) (y denotes the complex conjugate of
y ∈ C).

(iii) For all x ∈ C holds exp(x) 6= 0 and

1
exp(−x) = .
exp(x)

(iv) For x ∈ R with x ≥ 0 holds exp(x) ≥ 1. Where exp(x) = 1 ⇔ x = 0.

(v) For x ∈ R with x < 0 holds 0 < exp(x) < 1.

(vi) exp in continuous.

(vii) exp : R → R is strictly monotonically increasing.

(viii) limx→∞ exp(x) = ∞ and limx→−∞ exp(x) = 0.

(ix) exp : R → (0, ∞) is bijective.

(x) For all x ∈ C holds | exp(x)| = exp(Re(x)).

Proof:
(i) Already shown in Example 2.28.
4.1 Exponential Function 65

(ii) Since we have x1 + x2 = x1 + x2 and x1 x2 = x1 · x2 for all x1 , x2 ∈ C, we have

∞ ∞
X x̄k X xk
exp(x̄) = = = exp(x).
k=0
k! k=0
k!

(iii) By definition of exp, we have exp(0) = 1. From (i), we get exp(x) · exp(−x) =
exp(x − x) = exp(0) = 1. Then the statement follows.
(iv) From x ≥ 0, we get
∞ ∞
X xk X xk
exp(x) = =1+ ≥ 1.
k=0
k! k=1
k!
It is immediately clear that equality only holds for x = 0.
(v) If x < 0, we get from (iii) that exp(−x) > 1. Then, exp(x) = 1
exp(−x)
< 1.
(vi) First we show that exp is continuous at 0. Let (xn )n∈N be a sequence converging to
0. By Theorem ?? holds for (small enough) xn that | exp(xn ) − 1| = |r0 (xn )| with
|xn |
|r0 (xn )| ≤ 2 = 2|xn |.
1!
Therefore,
lim | exp(xn ) − exp(0)| = lim | exp(xn ) − 1| = lim |r0 (xn )| = 0
n→∞ n→∞ n→∞

which proves the continuity at 0. In order to show continuity on the whole real axis,
we assume that (xn )n∈N converges to x0 ∈ R and make use of
lim exp(xn ) = exp(x0 ) lim exp(xn − x0 ).
n→∞ n→∞

Since (xn − x0 )n∈N converges to 0, the above limit on the right hand side converges
to 1 (due to the continuity in 0). Therefore, we have
lim exp(xn ) = exp(x0 )
n→∞

which proves continuity of exp at x0 .

(vii) Let x1 , x2 ∈ R with x1 > x2 . Then we have x1 − x2 > 0 and thus, by (iv), exp(x1 −
x2 ) > 1. Since exp(x2 ) > 0, we have
exp(x1 ) = exp(x2 ) exp(x1 − x2 ) > exp(x2 ).

(viii) The fact limx→∞ exp(x) = ∞ follows, since we have for x > 0 that
∞ ∞
X xk X xk
exp(x) = =1+x+ > 1 + x.
k=0
k! k=2
k!

The statement (iii) then directly implies limx→−∞ exp(x) = 0.

(ix) The injectivity of exp : R → (0, ∞) follows from (vii). It remains to show surjectivity.
Let y ∈ (0, ∞).
Since limx→−∞ exp(x) = 0, there exists some x0 ∈ R such that exp(x0 ) < y.
Since limx→∞ exp(x) = ∞, there exists some x1 ∈ R such that exp(x1 ) > y.
Now, by the Mean Value Theorem, (remember that exp is continuous), there exists
some x ∈ R with exp(x) = y.
66 4 Elementary Functions

(x) Let x = x1 + ix2 with x1 = Re(x), x2 = Im(x). Then

| exp(x)| = | exp(x1 + ix2 )| = | exp(x1 ) exp(ix2 )| = | exp(x1 )|| exp(ix2 )|.

If we now show that | exp(ix2 )| = 1, the result is proven: Making use of (ii), we
obtain

| exp(ix2 )|2 = exp(ix2 ) · exp(ix2 ) = exp(ix2 − ix2 ) = exp(0) = 1.

25

20

15

10

0
−3 −2 −1 0 1 2 3

Figure 4.1: Graph of the exponential function

4.2 Logarithm
We will now define the logarithm as the inverse function of the exponential function.
As we have seen, the exponential function is bijective as a map from R to (0, ∞). This
justifies the following definition.
Definition 4.4.
The (natural) logarithm log : (0, ∞) → R is defined as the inverse function of
exp : R → (0, ∞), i.e., for all x ∈ R holds log(exp(x)) = x and for all y ∈ (0, ∞)
holds exp(log(y)) = y.

In many books, the above defined function is also denoted by logarithmus naturalis ln.
Before we collect some properties, a general result about continuity of inverse functions
is presented.
4.2 Logarithm 67

Theorem 4.5. Continuity of the inverse function

Let I, J ⊂ R be open intervals and let f : I → J be a continuous, bijective and
strictly monotonically increasing (or decreasing) function. Then the inverse func-
tion f −1 : J → I is continuous and strictly monotonically increasing (resp. decreas-
ing).

Proof: First we show strict monotonic increase. Let y1 , y2 ∈ J with y1 < y2 . Then

f (f −1 (y1 )) = y1 < y2 = f (f −1 (y2 ))

shows that, since f is strictly monotonically increasing, f −1 (y1 ) ≥ f −1 (y2 ) cannot hold,
so that f −1 (y1 ) < f −1 (y2 ). But this means that f −1 is strictly monotonically increasing.
Now we show continuity. Let ε > 0 and y0 ∈ J = f (I). Set x0 := f −1 (y0 ) ∈ I. Since I is
an open interval, there is an ε0 > 0 with ε0 < ε such that [x0 − ε0 , x0 + ε0 ] ⊂ I. Since f is
strictly monotonically increasing,

δ := min{f (x0 + ε0 ) − y0 , y0 − f (x0 − ε0 )} > 0.

Then for y ∈ J with |y − y0 | < δ holds f (x0 − ε0 ) < y < f (x0 + ε0 ) and the intermediate
value theorem yields

x := f −1 (y) ∈ [x0 − ε0 , x0 + ε0 ] ⊂ (x0 − ε, x0 + ε) = (f −1 (y0 ) − ε, f −1 (y0 ) + ε) ,

i.e. |f −1 (y) − f −1 (y0 )| < ε. By the ε-δ criterion this means that f −1 is continuous in y0 . 2

We want to remark that Theorem 4.5 also holds for intervals I and J which are not
open. In this case the proof of continuity of f −1 in y0 ∈ J has to be slightly adapted for
boundary points x0 := f −1 (y0 ). This was dropped for simplicity reasons.
Theorem 4.6. Properties of the Logarithm

(i) For all x, y ∈ (0, ∞) holds log(x · y) = log(x) + log(y).

(ii) log : (0, ∞) → R is strictly monotonically increasing.

(iii) log : (0, ∞) → R is continuous.

Proof:
(i) Define x1 = log(x) and y1 = log(y). Then, using x = exp(x1 ) and y = exp(y1 ), we
obtain

log(x · y) = log(exp(x1 ) · exp(y1 )) = log(exp(x1 + y1 )) = x1 + y1 = log(x) + log(y).

(ii) and (iii) follow from Theorem 4.5. 2

Remark:
Without further ado we introduce the logarithm for complex numbers z = r exp(iϕ)
in polar coordinates with r > 0 and ϕ ∈ [0, 2π) by
log(r exp(iϕ)) = log(r) + log(exp(iϕ)) = log(r) + iϕ .
68 4 Elementary Functions

−1

−2

−3

−4

−5
−0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Figure 4.2: Graph of the logarithm

However, formally this is a quite delicate issue and not treated in this lecture in
much detail. For further information on complex logarithms we refer to books on
Complex Analysis (German: Funktionentheorie).

By means of the exponential function, the general power ax (for a > 0, x ∈ C) can be
defined as follows:
ax := exp(log(a) · x).
This definition indeed makes sense as a0 = exp(log(a) · 0) = 1, a1 = exp(log(a) · 1) = a
and (for n ∈ N)
an = exp(log(a) + . . . + log(a)) = exp(log(a))n .
| {z }
n−times
1 √
It can further be seen that this definition implies a n = n
a. The definition of the general
power also justifies the notion exp(x) = ex .
A remaining question is how to solve the equation

ax = y

for given a > 0, y ∈ R and unknown x ∈ R. Using the definition of the general power,
this equation becomes
exp(log(a) · x) = y.
Performing the logarithm on both sides of this equation, we obtain log(a) · x = log(y) and
thus
log(y)
x= .
log(a)
4.3 Hyperbolic and Trigonometric Functions 69

In some literature, this expression is known as the logarithm of y to the basis a and
abbreviated by

log(y)
loga (y) := .
log(a)

By definition, we have log(x) = loge (x).

Remark: The pocket calculator and high school maths
In most of high school literature the logarithm (as we have defined it) is called
natural logarithm and is abreviated by ln. However, in mathematical literature the
symbol log is also commonly used for the inverse function of exp.
Furthermore, note that for many pocket calculators, the button log stands for log10 ,
the so-called decimal logarithm, whereas pushing the button ln gives the logarithm.
Note that the decimal logarithm is given by

log(x)
log10 (x) = .
log(10)

4.3 Hyperbolic and Trigonometric Functions

4.3.1 Hyperbolic Functions

Definition 4.7. Hyperbolic Sine and Hyperbolic Cosine

The hyperbolic sine (sinus hyperbolicus) and hyperbolic cosine (cosinus
hyperbolicus) sinh : C → C and cosh : C → C are defined as

1 1
sinh(x) = (exp(x) − exp(−x)) , cosh(x) = (exp(x) + exp(−x)) .
2 2
70 4 Elementary Functions

80

60

40

20

−20

sinh
−40 cosh

−60

−80
−5 −4 −3 −2 −1 0 1 2 3 4 5

Figure 4.3: Graph of the hyperbolic functions

Theorem 4.8. Properties of the Hyperbolic Functions

(i) For all x ∈ C holds sinh(x) = sinh(x), cosh(x) = cosh(x).

(ii) sinh and cosh are continuous.

(iii) For all x ∈ C holds cosh2 (x) − sinh2 (x) = 1.

(iv) For all x ∈ C holds

∞ ∞
X x2k X x2k+1
cosh(x) = , sinh(x) = .
k=0
(2k)! k=0
(2k + 1)!

(v) sinh : R → R is strictly monotonically increasing.

(vi) cosh : R → R is strictly monotonically increasing on [0, ∞) and strictly mono-

tonically decreasing on (−∞, 0].

Proof. (i) and (ii) follow from Theorem 4.3.

(iii):
1
cosh2 (x) − sinh2 (x) = (ex )2 + 2ex e−x + (e−x )2 − (ex )2 + 2ex e−x − (e−x )2


4
1
= · 4ex e−x = 1.
4
4.3 Hyperbolic and Trigonometric Functions 71

(iv): Using the series representation for exp we have

∞ ∞
!
1 1 X xk X (−x)k
sinh(x) = (ex − e−x ) = −
2 2 k=0
k! k=0
k!
∞ ∞
!
1 X xk X xk
= − (−1)k
2 k=0
k! k=0 k!
! ∞
1 X xk X x2k+1
= 2 = .
2 k=1,3,5,...
k! k=0
(2k + 1)!

The series representation for cosh can be derived analogously.

(v): Let x ∈ R and a > 0. Then ea > 1, 0 < e−a < 1 and

sinh(x + a) = 12 (ex+a − e−(x+a) ) = 12 (ea ex − e−a e−x ) > 12 (ex − e−x ) = sinh(x).

(vi): First of all it holds for x ∈ R

1 1
cosh(−x) = (e−x + ex ) = (ex + e−x ) = cosh(x).
2 2

Now let x ≥ 0. From (iii) we have cosh2 (x) = 1 + sinh2 (x) and since cosh(x) ≥ 1 and
sinh(x) ≥ 0 for x ≥ 0 it follows directly from (v) that cosh is strictly increasing on [0, ∞[.
Now let x ≤ 0. Since cosh(−x) = cosh(x) it follows from the first part that cosh is strictly
decreasing on [−∞, 0).

Remark:
The definition of the hyperbolic functions imply that sinh(−x) = − sinh(x) (resp.
cosh(−x) = cosh(x)). A function with this property is called odd (resp. even).
The monotonicity property of cosh together with the fact that cosh(0) = 1 imply
that cosh does not have any real zeros. The hyperbolic sine function has only one
zero at the origin.

Remark:
Why are these functions called hyperbolic functions?
72 4 Elementary Functions

−1

−2

−3

−4
−4 −3 −2 −1 0 1 2 3 4

Figure 4.4: Hyperbola

Since cosh2 (x) − sinh2 (x) = 1, the curve

{(− cosh(t), sinh(t)) | t ∈ R} ∪ {(cosh(t), sinh(t)) | t ∈ R}

describes a hyperbola see Figure 4.4. (In analogy the curve {(cos(t), sin(t)) | t ∈ R}
describes the unit circle.)

Definition 4.9. Hyperbolic Tangent

The hyperbolic tangent (tangens hyperbolicus) tanh : {x ∈ C : cosh(x) 6= 0} → C
is defined as
sinh(x) ex − e−x
tanh(x) = = x .
cosh(x) e + e−x
4.3 Hyperbolic and Trigonometric Functions 73

−1

−2

−3

−5 −4 −3 −2 −1 0 1 2 3 4 5

Figure 4.5: Graph of the hyperbolic tangent

Remark:
Since sinh and cosh map real numbers to real numbers and, moreover, cosh has no
zero in R, the hyperbolic tangent is defined on the whole real axis. Furthermore, it
can be seen that tanh is continuous, strictly monotonically increasing and

lim tanh(x) = 1, lim tanh(x) = −1.

x→∞ x→−∞

4.3.2 Area Functions

We already know that sinh : R → R, cosh : [0, ∞) → [1, ∞), tanh : R → (−1, 1)
are strictly monotonically increasing. They possess inverse functions defined on R (resp.
[1, ∞), (−1, 1)).
Definition 4.10. Area Functions
(i) The area hyperbolic sine or the area sinus hyperbolicus arsinh : R → R is
defined as the inverse function of sinh.

(ii) The area hyperbolic cosine or the area cosinus hyperbolicus is denoted by
arcosh : [1, ∞) → [0, ∞) is defined as the inverse function of cosh.

(iii) The area hyperbolic tangent or the area tangens hyperbolicus is denoted by
artanh : (−1, 1) → R is defined as the inverse function of tanh.
74 4 Elementary Functions

5 arsinh
arcosh
4

−1

−2

−3

−4

−5

−6 −4 −2 0 2 4 6 8

Figure 4.6: Graph of arsinh and arcosh

2.5

1.5

0.5

−0.5

−1

−1.5

−2

−2.5
−3 −2 −1 0 1 2 3

Figure 4.7: Graph of artanh

4.3 Hyperbolic and Trigonometric Functions 75

4.3.3 Trigonometric Functions

Definition 4.11. Sine and Cosine

The sine (sinus) sin : C → C and cosine (cosinus) cos : C → C are defined as
1 1
sin(x) := (exp(ix) − exp(−ix)), cos(x) := (exp(ix) + exp(−ix)).
2i 2

Theorem 4.12. Properties of sin and cos

(i) For all x ∈ C holds sin(x) = sin(x), cos(x) = cos(x).

(ii) sin and cos are continuous.

(iii) For all x ∈ C holds sin(x) = −i sinh(ix), cos(x) = cosh(ix).

(iv) For all x ∈ C holds cos2 (x) + sin2 (x) = 1.

(v) For all x ∈ C holds

∞ ∞
X x2k
k
X x2k+1
cos(x) = (−1) , sin(x) = (−1)k .
k=0
(2k)! k=0
(2k + 1)!

Proof. (i): Assume that x = x1 + ix2 with x1 , x2 ∈ R and compute

1
sin(x) = sin(x1 + ix2 ) = (exp(ix1 − x2 ) − exp(−ix1 + x2 ))
2i
1
= (exp(ix1 − x2 ) − exp(−ix1 + x2 ))
−2i
1
= (exp(−ix1 + x2 ) − exp(ix1 − x2 ))
2i
1
= (exp(ix1 + x2 ) − exp(−ix1 − x2 ))
2i
1
= (exp(ix) − exp(−ix))
2i
= sin(x)
The relation for cos(x) is analogous.
(ii): Continuity follows from that of the exponential function.
(iii): Follows by definition (and taking into account that 1i = −i).
(iv): Follows from (iii) and Theorem 4.8 (iii).
(v): The series representations can be obtained by using (iii) and the series representations
of sinh(x), cosh(x).

Remark:
By the above definition, we have that for all x ∈ C holds

exp(ix) = cos(x) + i sin(x).

76 4 Elementary Functions

For x ∈ R this gives rise to

cos(x) = Re(exp(ix)), sin(x) = Im(exp(ix)).

In particular, the equation cos2 (x) + sin2 (x) = 1 implies for x ∈ R that | sin(x)| ≤ 1
and | cos(x)| ≤ 1.

6
sin
cos

−2

−4

−6
−8 −6 −4 −2 0 2 4 6 8

Figure 4.8: Graph of sin and cos

The following result gives formulas for sine and cosine applied to sums of (complex)
numbers. These results can be readily verified by making use of Definition 4.11 and the
equation exp(x1 + x2 ) = exp(x1 ) exp(x2 ).
Theorem 4.13. Trigonometric identities
For arbitrary x, y ∈ C the trigonometric functions fulfill

sin(x + y) = sin(x) cos(y) + cos(x) sin(y),

cos(x + y) = cos(x) cos(y) − sin(x) sin(y).
4.3 Hyperbolic and Trigonometric Functions 77

Proof. This follows by

sin(x) cos(y) + cos(x) sin(y),

1 1
= (exp(ix) − exp(−ix)) · (exp(iy) + exp(−iy))
2i 2
1 1
+ (exp(ix) + exp(−ix)) · (exp(iy) − exp(−iy))
2 2i
1
= (exp(i(x + y)) − exp(i(y − x)) + exp(i(x − y)) − exp(−i(x + y))
4i
+ exp(i(x + y)) + exp(i(y − x)) − exp(i(x − y)) − exp(−i(x + y)))
1
= (2 exp(i(x + y)) − 2 exp(−i(x + y)))
4i
1
= (exp(i(x + y)) − exp(−i(x + y)))
2i
= sin(x + y)

and
cos(x) cos(y) − sin(x) sin(y),
1 1
= (exp(ix) + exp(−ix)) · (exp(iy) + exp(−iy))
2 2
1 1
− (exp(ix) − exp(−ix)) · (exp(iy) − exp(−iy))
2i 2i
1 1
= (exp(ix) + exp(−ix)) · (exp(iy) + exp(−iy))
2 2
1 1
+ (exp(ix) − exp(−ix)) · (exp(iy) − exp(−iy))
2 2
1
= (exp(i(x + y)) + exp(i(y − x)) + exp(i(x − y)) + exp(−i(x + y))
4
+ exp(i(x + y)) − exp(i(y − x)) − exp(i(x − y)) + exp(−i(x + y)))
1
= (2 exp(i(x + y)) + 2 exp(−i(x + y)))
4
1
= (exp(i(x + y)) − exp(−i(x + y)))
2
= cos(x + y).

We now define the famous number π by the double of the first positive zero of the cosine
function. The following result shows that this definition indeed makes sense.
Theorem 4.14. First positive zero of cos, definition of π
The function cos : R → R has exactly one zero in the interval [0, 2]. This zero is
called π2 .

The proof for this is not presented here. It basically consists of three steps: The first step
consists of showing that cos(2) < 0. This can be shown by using the series representation
in Theorem 4.12 (v). In the second step we have to show that cos is strictly monotonically
decreasing in the interval [0, 2]. This can be achieved by showing that sin is positive on
78 4 Elementary Functions

the interval [0, 2] and making use of the equation

   
x+y x−y
cos(x) − cos(y) = −2 sin sin ,
2 2
which follows from the trigonometric identities. Altogether, we then have that cos(0) =
1 > 0, cos(2) < 0 and cos is strictly monotonically decreasing on [0, 2]. The intermediate
value theorem implies that there exists a zero in (0, 2). The strict monotonic decrease
gives rise to the uniqueness of this zero.
Note that the positivity of sin on the interval [0, 2] implies that sin( π2 ) = 1, since cos( π2 ) = 0
and sin2 ( π2 ) + cos2 ( π2 ) = 1.
Now we present some further fundamental properties of the trigonometric functions.
These for instance include 2π-periodicity.
Theorem 4.15. Further properties of the trigonometric functions
For all x ∈ C holds:

(i) cos(x) = sin(x + π2 );

(ii) sin(x) = − cos(x + π2 );

(iii) sin(x) = − sin(x + π);

(iv) cos(x) = − cos(x + π);

(v) sin(x) = sin(x + 2π);

(vi) cos(x) = cos(x + 2π).

Proof. (i) follows from the trigonometric identity together with sin( π2 ) = 1 and cos( π2 ) = 0,
namely  π π  π 
sin x + = sin(x) cos + cos(x) sin = cos(x).
2 | {z 2 } | {z2 }
=0 =1
Item (ii) can be shown analogously.
(iii) is a consequence of
π
sin(x + π) = cos(x + ) = − sin(x),
2
(iv) is analogous.
(v) and (vi) follow by a double application of (iii) (resp. (iv)).

Definition 4.16. Tangent

The tangent tan : {x ∈ C : cos(x) 6= 0} → C is defined by

sin(x)
tan(x) = .
cos(x)

One can show that the set of zeros of the cosine function is { 2n+1
2
π : n ∈ Z}. Therefore,
the tangent is defined on
C\ 2n+1

2
π : n∈Z .
4.4 Arcus functions 79

−2

−4

−6
−4 −2 0 2 4 6

Figure 4.9: Graph of tan

4.4 Arcus functions

From the previous result, it is possible to derive that
a) sin is strictly monotonically increasing on [− π2 , π2 ] with sin(− π2 ) = −1,
sin( π2 ) = 1;
b) cos is strictly monotonically decreasing on [0, π] with cos(0) = 1, cos(π) = −1;
c) tan is strictly monotonically increasing on (− π2 , π2 ) with

lim tan(x) = −∞, lim tan(x) = ∞.

x&− π2 x% π2

Since sin, cos and tan are furthermore continuous, we can apply Theorem 4.5 to see that
the following definition makes sense:
Definition 4.17.
(i) The inverse sine or arcus sinus arcsin : [−1, 1] → R is defined as the inverse
function of sin : [− π2 , π2 ] → [−1, 1].

(ii) The inverse cosine or arcus cosinus arccos : [−1, 1] → R is defined as the
inverse function of cos : [0, π] → [−1, 1].
80 4 Elementary Functions

(iii) The inverse tangent or arcus tangens arctan : R → R is defined as the inverse
function of tan : (− π2 , π2 ) → R.

3 arcsin
arccos

2.5

1.5

0.5

−0.5

−1

−1.5
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

Figure 4.10: Graph of arcsin and arccos

4.5 Polynomials and Rational Functions

4.5.1 Polynomials

Definition 4.18. Polynomials

Let a0 , . . . , an ∈ F. Then a real (complex) polynomial is a function p : F → F with
n
X
p(x) = ak x k .
k=0

If an 6= 0, then an is called leading coefficient and n is called degree of p. We write

n =: deg p. If p is the zero polynomial, we set deg p := −∞.
The set of polynomials in F is denoted by F[x]. Moreover, we set

Fn [x] := {p ∈ F[x] | deg p ≤ n}.

4.5 Polynomials and Rational Functions 81

−1

−2

−3

−5 −4 −3 −2 −1 0 1 2 3 4 5

Figure 4.11: Graph of arctan

Remark:
Since sums and scalar multiples of polynomials are again polynomials, they form
a vector space.

Theorem 4.19. Rules for the degree

For p, q ∈ F[x] holds

deg(p · q) = deg(p) + deg(q), deg(p + q) ≤ max{deg(p), deg(q)}.

Proof. Let p(x) = nk=0 ak xk , q(x) = m

k=0 bk x with an 6= 0, bm 6= 0. The formula for
k
P P
deg(p · q) follows from
n+m
X k
X
p(x) · q(x) = ck x k
for ck = al bk−l ,
k=0 l=0

where ar := 0 =: bs for r 6∈ {0, ..., n} and s 6∈ {0, ..., m}. For the proof of the formula
deg(p + q), we assume without loss of generality that n ≥ m. Then
m
X n
X
k
p(x) + q(x) = (ak + bk )x + ak x k .
k=0 k=m+1

As a consequence, we have deg(p + q) ≤ n = max{deg(p), deg(q)}.

82 4 Elementary Functions

Remark:
As the example p(x) = x and q(x) = −x + 1 shows, it may indeed happen that
deg(p + q) < max{deg(p), deg(q)}.
Since deg(0 · p) = deg(0) = −∞ = −∞ + deg(p) and deg(0 + p) = deg(p) =
max{−∞, deg(p)}, the choice of deg p = −∞ makes indeed sense for preserving the
above formulas. However, this belongs to the “not so important facts” of mathem-
atical analysis.

Next we consider the

Pnevaluation of polynomials at some point x0 . A closer look at the
expression p(x) = k=0 ak x yields that a “stupid” determination of p(x0 ) requires n
k

summations and 1 + 2 + . . . + (n − 1) = n(n−1)

2
multiplications. With the following method,
called Horner scheme, we can evaluate polynomials with effort that is only linear in n.
The “trick” behind this method is an alternative representation, i.e we rewrite p as

p(x) = (x − x0 )(bn−1 xn−1 + . . . − b1 x + b0 ) + c.

for some constants bn−1 , . . . , b0 , c. It can be directly seen that p(x0 ) = c. Collecting powers
of x yields

p(x) = bn−1 xn + . . . + b1 x2 + b0 x − bn−1 x0 xn−1 − . . . − b1 x0 x − b0 x0 + c

= bn−1 xn + (bn−2 − bn−1 x0 )xn−1 + (bn−3 − bn−2 x0 )xn−2
+ . . . + (b0 − b1 x0 )x + (c − b0 x0 ).

By a comparison of coefficients, we obtain the system of linear equations

bn−1 = an ,
bn−2 − bn−1 x0 = an−1 ,
..
.
b 0 − b 1 x 0 = a1 ,
c − b 0 x 0 = a0 .

This system can be recursively solved as

bn−1 = an ,
bn−2 = an−1 + bn−1 x0 ,
..
.
b 0 = a1 + b 1 x 0 ,
c = a0 + b 0 x 0 .

The following schematic representation summarizes the above introduced approach.

Remark:
If a power is missing in the polynomial, then in the Horner scheme the corresponding
coefficient has to be set to 0. For instance, the polynomial x2 + 1 has to be written
as x2 + 0x + 1.
4.5 Polynomials and Rational Functions 83

an an−1 an−2 ··· a1 a0

x0 bn−1 x0 bn−2 ··· x 0 b1 x0 b 0
(mult. with x0 ) % ↓ % ↓ % ··· % ↓ % ↓
bn−1 bn−2 bn−3 b0 c

Table 4.1: Horner Scheme

Example 4.20.
Consider the polynomial p(x) = x3 − 6x2 + 7x and determine p(2).

1 −6 7 0
2 −8 −2
(mult. with 2) % ↓ % ↓ % ↓
1 −4 −1 −2

For integers, “division with remainder” is well known. The same procedure can now be
applied for polynomials and is called polynomial division. First we present an existence
result. Afterwards, we will present a method to compute the polynomials in question.
Theorem 4.21. Polynomial Division
Let p, q ∈ F[x] be given. Furthermore, assume that q is not the zero polynomial.
Then there exist polynomials g and r with deg r < deg q, such that

p = q · g + r. (4.1)

The polynomial r is called remainder. If r is the zero polynomial, we say that p is

divisible by q.

To compute the polynomials g and r as in Theorem 4.21, we use the method of polynomial
division. This method is not displayable by a mathematical theorem but only explainable
by means of concrete examples. Polynomial division is of great importance, in particular
for some special representations of rational functions.

Example 4.22. a) Let p(x) = x3 + 3x2 − 8x − 4 and q(x) = x − 2.

(x3 +3x2 −8x −4 ) : (x − 2) = x2 + 5x + 2 Remainder: 0

−(x3 −2x2 )
5x2 −8x
2
−(5x −10x)
2x −4
−(2x −4)
0
84 4 Elementary Functions

Test: We have: (x2 + 5x + 2)(x − 2) = x3 + 3x2 − 8x − 4.

This means that x3 + 3x2 − 8x − 4 is divisible by x − 2.
b) Let p(x) = x4 + 4x + 5 and q(x) = x2 + x − 1.

(x4 +4x +5 ) : (x2 + x − 1) = x2 − x + 2 Remainder: x + 7

−(x4 +x 3
−x )2

−x3 +x2 +4x

−(−x3 −x2 +x)
2x2 +3x +5
−(2x2 +2x −2)
x +7 .

Test: We have: (x2 − x + 2)(x2 + x − 1) + (x + 7) = x4 + 4x + 5.

c) Let p(x) = x4 + 4x + 5 and q(x) = x2 + 1.

(x4 +4x +5 ) : (x2 + 1) = x2 − 1 Remainder: 4x + 6

−(x4 +x )2

−x2 +4x +5
−(−x2 −1)
+4x +6 .

Test: We have: (x2 + 1)(x2 − 1) + (4x + 6) = x4 + 4x + 5.

Note that a polynomial p is divisible by x−x0 if and only if x0 is a zero of p, i.e., p(x0 ) = 0.
This follows from the fact that the division with remainder theorem implies that there
exists some polynomial q ∈ F[x] and a constant c ∈ F (i.e., a polynomial whose degree is
less than 1), such that
p(x) = (x − x0 )q(x) + c.

In particular, we have that p(x0 ) = 0 if and only if c = 0 if and only if p is divisible by

x − x0 . This leads to the following definition:
Definition 4.23.
Let p ∈ F[x] be given with zero x0 ∈ C. Then the order (also called: multiplicity)
of the zero x0 is the greatest number n ∈ N such that p is divisible by (x − x0 )n .

For sake of completeness, we say that the “order of the zero x0 is zero, if x0 is not a zero
of p”.

As the following picture shows, a zero of even order touches the x-axis, while a zero
of odd order crosses the x-axis. Finally, we present a result which is classical in poly-
nomial algebra. It states that any nonconstant complex polynomial can be represented
as a product of linear factors. Algebraists call this property the algebraic closedness of C.

Theorem 4.24. Fundamental Theorem of Algebra

Any nonconstant p ∈ C[x] with p(x) = nk=0 ak xk , n ≥ 1, an 6= 0 has a representa-
P
4.5 Polynomials and Rational Functions 85

zero of odd order

zero of even order

Figure 4.12: Qualitative behavior of even/odd order zeros of polynomials

tion n
Y
p(x) = an (x − ck )
k=1

for some c1 , . . . , cn ∈ C.

The central question about polynomials is:

How can we find (compute) zeros?

This question is by far not simple for arbitrary polynomials, since there is no “univer-
sal strategy” for this. However, for polynomials of degree at most two, we can give simple
explicit formulas:
• Polynomials of degree 1: For p(x) = a1 x + a0 with a1 6= 0, the only zero is
obviously given by x1 = − aa01 .
• Polynomials of degree 2: For p(x) = a2 x + a1 x + a0 with a2 6= 0, the zeros
x1 , x2 ∈ C are given by
p
−a1 + a21 − 4a0 a2
x1 = ,
2a2
p
−a1 − a21 − 4a0 a2
x2 = .
2a2
Note that for p(x) = x2 + px + q, the expressions for the zeros read
r
p p2
x1/2 =− ± − q.
2 4
86 4 Elementary Functions

In high school mathematics, this has the pictorial name of “pq Formula”.
Note that in the case where a21 − 4a0 a2 is a negative real number, the square root
has to be understood as the complex number(s) whose squaring gives a21 − 4a0 a2 .
For instance, the zeros of the polynomial p(x) = x2 + 1 are given by
√
x1/2 = ± −1 = ±i.

Indeed, we have the factorization

x2 + 1 = (x + i)(x − i).

• Polynomials of degree higher than 2: For polynomials of degree 3 or 4, there

exist formulas for the roots (Cardano’s formulas). However, these are really com-
plicated and hardly applicable in concrete cases. For polynomials of degree higher
than 4, it can be even shown that there do not exist any explicit formulas for the
zeros.
How can we nevertheless find zeros of polynomials of higher degree?
Let us start with an example p(x) = x3 + x2 − 6x + 4. First we just try “some
canonical candidates” for zeros, such as −1, 0, 1, 2, . . .. Plugging this into p, we see
that p(1) = 0, i.e., x1 = 1 is a zero. As a consequence, p admits a factorization
p(x) = (x − 1) · q(x) for some polynomial q. By the formulas for the degree of
products of polynomials, we get deg q = 2. The polynomial q can now be obtained
by polynomial division, i.e.,

(x3 +x2 −6x +4 ) : (x − 1) = x2 + 2x − 4 Remainder: 0

x3 −x2
2x2 −6x
2x2 −2x
−4x +4
−4x +4
0.

Altogether, we now have p(x) = (x2 + 2x − 4) · (x − 1). The zeros of p are therefore given
by x1 = 1 and the set of zeros of the polynomial q(x) = x2 + 2x − 4. The latter ones can
now be obtained by the pq formula, i.e.,
√ √
x2/3 = −1 ± 1 + 4 = −1 ± 5.

The set of zeros is therefore given by

√ √
x1 = 1, x2 = −1 + 5, x3 = −1 − 5.

The above strategy can be used for arbitrary polynomials as long as one has enough
successful guesses for zeros. In exercises (or examinations), there is oftentimes a hint
given for such guesses.

4.5.2 Rational Functions

4.5 Polynomials and Rational Functions 87

Definition 4.25.
Let p, q ∈ F[x], where q is not the constant zero polynomial. Then the function
f : D(f ) = {x ∈ F : q(x) 6= 0} → F with

p(x)
f (x) =
q(x)

is called rational function.

It can be verified that for rational functions f, g, the functions f + g, f − g, f · g, f

g
are
again rational.

Example 4.26. a) f (x) = 1

x

−1

−2

−3

−4

−5
−5 −4 −3 −2 −1 0 1 2 3 4 5

b) f (x) = 1
x2

−1
−5 −4 −3 −2 −1 0 1 2 3 4 5

We will now take a closer look at the places where f is not defined, i.e., the zeros of the
denominator polynomial.
Let f = pq be given, let x0 be a zero of q and let r ∈ N0 , s ∈ N be the multiplicities of the
zero x0 of p and q, respectively. This means that we have factorizations

p(x) = (x − x0 )r p1 (x), q(x) = (x − x0 )s q1 (x)

88 4 Elementary Functions

for some p1 , q1 ∈ F[x] with p1 (x0 ) 6= 0 and q1 (x0 ) =6 0. This means that for x ∈ D(f ) we
have
p(x) (x − x0 )r p1 (x) p1 (x)
f (x) = = s
= (x − x0 )r−s .
q(x) (x − x0 ) q1 (x) q1 (x)

Next we distinguish between several cases for s and r:

1. Case: r ≥ s:
“f has a hole at x0 ”

hole

2. Case: r < s:
f has a pole of order s − r at x0 .

Poles of even order Poles of odd order

0 0

Figure 4.13: Qualitative behavior of even/odd order poles of rational functions

x2
Example 4.27. a) The rational function f (x) = x
, defined on R\{0}, has a hole at
x0 = 0.
b) The rational function f (x) = x
x2
, defined on R\{0}, has a pole of first order at x0 = 0.
4.5 Polynomials and Rational Functions 89

Definition 4.28.
A rational function f = p
q
is called

– proper, if deg p ≤ deg q;

– strictly proper, if deg p < deg q.

Properness of a rational function can be equivalently characterized via the existence of

the limit
lim f (x).
x→∞
This limit is furthermore zero if and only if f is strictly proper.
Theorem 4.29.
Any rational function can be represented as a sum of a polynomial and a strictly
proper rational function.

Proof: Let f = pq . Applying polynomial division, we obtain that there exist g, r ∈ F[x]
with deg r < deg q and p = qg + r. Division by q gives
p(x) r(x)
f (x) = = g(x) + .
q(x) q(x)
Since the latter addend is a strictly proper rational function, the result is proven.
2

Next we consider a special representation of rational functions called partial fraction

decomposition. This will be useful in particular for integration of rational functions.
Definition 4.30.
For A, x0 ∈ C, k ∈ N, the rational function
A
(x − x0 )k

is called partial fraction.

Next we present that any strictly proper rational function has a representation as sum
of partial fractions. Note that for rational functions which are not strictly proper, we
first have to perform an additive decomposition into a polynomial and a strictly proper
rational function according to Theorem 4.29.
Theorem 4.31.
Let polynomials p(x) = an xn + . . . + a1 x + a0 , q(x) = bm xm + . . . + b1 x + b0 with
deg(p) = n < m = deg(q) be given. Assume that q has a representation
k
Y k
X
q(x) = bm (x − xj )kj , kj = m
j=0 j=0

for some pairwise distinct xj (i.e., xi 6= xj for i 6= j).

90 4 Elementary Functions

Then the rational function f (x) = p(x)

q(x)
has a representation

kj
k X
X Ajl
f (x) = .
j=0 l=1
(x − xj )l

This representation is called partial fraction decomposition.

Note that a combination of the above result with Theorem 4.29 yields that any rational
function can be represented as the sum of a polynomial and some partial fractions.

The above theorem looks more complicated then it actually is. The polynomial q has
oftentimes only simple multiplicities, i.e., kj = 1 for all j. In that case, the double sum
becomes a single sum of the form
k
X Aj1
f (x) = .
j=0
x − xj

We will give some examples. After that, we discuss how to compute the coefficients Alj .

Example 4.32. (a)

x+1
f (x) =
x2 + 1
Since x2 + 1 = (x + i)(x − i), we have a partial fraction decomposition of the form
A11 A21
f (x) = + .
x+i x−i

(b)
1
f (x) = .
x3
− 3x − 2
By the determination of the zeros of the denominator polynomial, we obtain a fac-
torization x3 − 3x − 2 = (x + 1)2 (x − 2). Therefore, there exists a partial fraction
decomposition of the form
A11 A12 A21
f (x) = + + .
x + 1 (x + 1)2 x − 2

For the computation of the coefficients, we make use of the fact that two polynomials
coincide if and only if all their coefficients coincide. This technique is called comparison
of coefficients.

Using this, we can make the following ansatz in Example a):

x+1 A11 A21
2
= + .
x +1 x+i x−i
Multiplying this equation by (x + i)(x − i) from both sides gives

x + 1 = A11 (x − i) + A21 (x + i) = (A11 + A21 )x + (−iA11 + iA21 ).

4.5 Polynomials and Rational Functions 91

A comparison of coefficients then leads to the following system of linear equations

A11 + A21 = 1,
−iA11 + iA21 = 1,

i.e.     
1 1 A11 1
= .
−i i A21 1
This leads to the solution A11 = 1+i
2
, A21 = 1−i
2
. Therefore, we have a partial fraction
decomposition
1+i 1−i
x+1 2 2
= + .
x2 + 1 x+i x−i
For the second example, we get

1 A11 A12 A21

2
= + 2
+
(x + 1) (x − 2) x + 1 (x + 1) x−2

and thus

1 =A11 (x + 1)(x − 2) + A12 (x − 2) + A21 (x + 1)2

=(A11 + A21 )x2 + (−A11 + A12 + 2A21 )x + (−2A11 − 2A12 + A21 )

and thus we get the linear system

    
1 0 1 A11 0
−1 1 2 A12  = 0 .
−2 −2 1 A21 1

Since the solution is given by A11 = − 91 , A12 = − 13 , A21 = 19 , we have

1 1 1
1 9 3 9
=− − + .
(x + 1)2 (x − 2) x + 1 (x + 1)2 x − 2

In the following, we present a nice trick to compute some of the coefficients without solving
a linear system of equations.
Theorem 4.33.
Let the assumptions of Theorem 4.31 be valid. Then the coefficients Ai,ki are given
by
p(xi )
Ai,ki = k
Y
bm (xi − xj )kj
j=0,j6=i

Again, this formula looks more complicated than it really is. The determination of Ai,ki
can be done as follows:
• Keep the factor (x − xi )ki shut.
• Plug xi into the remaining part.
92 4 Elementary Functions

In german, this method is known as “Zuhaltemethode”. A native english speaking professor

told me that there is no translation of this word.
As a consequence, we do not have to solve a linear system in the case where all zeros of the
denominator polynomial have order one. However, even the order of the zeros are higher it
is at least possible to reduce the order of the resulting linear system, since the coefficients
belonging to the partial fractions involving the biggest powers can be calculated by this
method. We now apply this method to the above two examples.

Example 4.34. (a)

x+1 x+1 A11 A21

f (x) = 2
= = + .
x +1 (x + i)(x − i) x+i x−i

The determination of A11 can be done by


x+1  −i + 1 1+i
A11 =  = =
(x+ i) (x − i) x=−i
   −i − i 2

and A21 is given by


x+1  i+1 1−i
A21 =  = = .
(x − i) x=i
(x + i)     i+i 2

This is the same result that we obtained by solving the associated linear system.
(b)
1 A11 A12 A21
f (x) = 2
= + 2
+ .
(x + 1) (x − 2) x + 1 (x + 1) x−2

We now compute the coefficients A12 and A21 by the formula given in Theorem 4.33.

1  1
A12 = 2
 =− ,
(x + 1) (x − 2) 3
  

x=−1

1  1
A21 =  = .
(x + 1)2 (x−2) x=2

9

For the coefficient A11 now we only need to solve the reduced system

1 1
1 + (x − 2) − (x + 1)2 = A11 (x + 1)(x − 2).
3 9

4.6 Power Series

Very roughly speaking, power series are “infinite polynomials”. A precise definition is the
following:
4.6 Power Series 93

Definition 4.35.
Let a sequence (ak )k∈N in F be given and let x0 ∈ F. Then the function f : D(f ) → F
defined by the series
∞
X
f (x) = ak (x − x0 )k
k=0

is called power series.

The set ( ∞
)
X
D(f ) := x∈F : ak (x − x0 )k is convergent
k=0

is called
P∞domain of convergence. The domain of convergence at least includes x0
since k=0 ak (x0 − x0 ) = a0 .
k

We have already seen several examples of power series in this chapter.

Example 4.36. a) The exponential function is defined via the power series
∞
X xk
exp(x) = ,
k=0
k!

i.e., (ak )k∈N = ( k!1 )k∈N and x0 = 0. Here D(f ) = C.

b) The sine function is defined via the power series
∞
X (−1)k x2k+1
sin(x) = ,
k=0
(2k + 1)!

i.e., (ak )k∈N = (0, 1!1 , 0, − 3!1 , 0, 5!1 , 0, − 7!1 , . . .)k∈N and x0 = 0. Again D(f ) = C.
c) cos, cosh, sinh are defined via the power series...
d) The function
∞
X (x − 1)k
f (x) =
k=1
k
is a power series.

Next we characterise the domain of convergence.

Theorem 4.37. Theorem of Cauchy-Hadamard
Let a power series
∞
X
f (x) = ak (x − x0 )k
k=0

be given. Let
1
r := p ,
lim sup k
|ak |
k→∞

where we formally define 1/∞ := 0 and 1/0 := ∞. Then for all x ∈ F with
|x − x0 | < r holds x ∈ D(f ). Furthermore, for all x ∈ F with |x − x0 | > r holds
94 4 Elementary Functions

x∈/ D(f ).
The number r as defined above is called the radius of convergence.

im

divergence

convergence

re

Figure 4.14: Domain of convergence

Proof: We have to show the following two statements:

(i) For all x ∈ F with |x − x0 | · lim supk→∞ k |ak | < 1, the power series is convergent.
p

(ii) For all x ∈ F with |x − x0 | · lim supk→∞ k |ak | > 1, the power series is divergent.
p

Statement (i) just follows from the limit form of the root criterion (Theorem 2.19), namely
p p
lim sup k |(x − x0 )k ak | = |x − x0 | · lim sup k |ak | < 1.
k→∞ k→∞

For showing (ii), we also make use of the formula

p p
lim sup k |(x − x0 )k ak | = |x − x0 | · lim sup k |ak | > 1.
k→∞ k→∞

This implies that the sequence (x−x0 )k ak does not converge to 0 and therefore, the power
series cannot converge. 2
Geometrically, the above result implies that for all x inside a circle with midpoint x0 and
radius r, the series is convergent and outside this circle, we have divergence.
The Cauchy-Hadamard Theorem characterizes convergence/divergence of the power series
in dependence of x whether it is inside or outside the circle around x0 with radius r. In
the case |x − x0 | = r, this result does not tell us anything. Indeed, we may have points
on the circle with x ∈ D(f ) and also points on the circle with x ∈ / D(f ).
4.6 Power Series 95

To see this, let us reconsider Example 4.36 d):

The radius of convergence is given by
1
r= q = 1.
lim supk→∞ k
| k1 |

So we have convergence for all x ∈ (0, 2) and divergence for all x ∈ (−∞, 0) ∪ (2, ∞). The
remaining real points which are not characterized by the Theorem of Cauchy-Hadamard
are x = 0 and x = 2. In the case x = 0, we obtain the series
∞
X (−1)k
k=0
k

which is convergent by the Leibniz criterion. Plugging in x = 2, the power series becomes
a harmonic series ∞
X 1
k=0
k
that is well-known to be divergent.

Example 4.38. a) For the power series defined by the exponential function, we have
 
1
(ak )k∈N = , x0 = 0.
k! k∈N
The radius of convergence is then given by
1 1
r= q = = ∞.
lim supk→∞ k
| k!1 | 0

As a consequence, the series converges for every x ∈ C. The same holds true for the
series of sin, cos, sinh, cosh.
b) As we have already seen above, the radius of convergence of the power series
∞
X (x − 1)k
f (x) =
k=0
k

is r = 1.
c) Consider the power series
∞
X
f (x) = k!xk .
k=0
The radius of convergence is given by
1 1
r= p = = 0.
lim sup |k!|
k ∞
k→∞

So this series is divergent for any x 6= 0.

Sometimes the computation of the radius of convergence |an | of a power

p
n
P∞ r = lim sup n→∞
series n=1 an (x − x0 ) is quite difficult. In such cases the following theorem might be
n

better suited, which follows from the quotient criterion and is stated without proof.
96 4 Elementary Functions

Theorem 4.39.
Suppose that ∞ n=1 an (x − x0 ) is a power series with coefficients an ∈ F such that
n
P

an 6= 0 for all n ≥ N with fixed N ∈ N. If limn→∞ |a|an+1

n|
|
exists in R ∪ {+∞} than
it is the radius of convergence of the power series.
 Differentiation of Functions
5
For instance, on the planet Earth, man had always assumed that he was
more intelligent than dolphins because he had achieved so much—the
wheel, New York, wars and so on—while all the dolphins had ever done
was muck about in the water having a good time. But conversely, the
dolphins had always believed that they were far more intelligent than
man—for precisely the same reasons.
Douglas Adams, The Hitchhiker’s Guide to the Galaxy

5.1 Differentiability and Derivatives

To motivate the problem, consider the functions f, g : R → R with f (x) = x2 and
g(x) = |x|. As we already know, both these functions are continuous. Let us now focus

2
f

1.8 g

1.6

1.4

1.2

1
origin
0.8

0.6

0.4

0.2

−1 −0.5 0 0.5 1

Figure 5.1: Domain of convergence

on the qualitative behavior of the functions at the origin.

97
98 5 Differentiation of Functions

f g

smooth sharp bend

there is a unique tangent no unique tangent

Table 5.1: Qualitative behavior of f and g at the origin

A straight line y(t) going through the points (x0 , f (x0 )) and (x, f (x)) is called secant 1 of
f through these points. It is given by

f (x0 ) − f (x)
y(t) = f (x0 ) + (t − x0 ).
x0 − x
In particular, the slope of y is the difference quotient

f (x0 ) − f (x) f (x) − f (x0 )

= .
x0 − x x − x0
If we now let x tend to x0 , we obtain a tangent of f at x0 . This leads to the following

fx0 f

f−f

fx
x−x

x x0

Figure 5.2: Secant of a function

definition.
Definition 5.1.
Let I ⊂ R be an interval with more than one point or an open set. Let f : I → R
be a function. Then f is called differentiable at x0 ∈ I if there exists a function

1
from the Latin word secare = “to cut”
5.1 Differentiability and Derivatives 99

∆f,x0 : I → R that is continuous in x0 and, moreover, for all x ∈ I holds

f (x) = f (x0 ) + (x − x0 ) · ∆f,x0 (x).

The number ∆f,x0 (x0 ) is called derivative of f at x0 .

The function f is called differentiable in I if it is differentiable at all x0 ∈ I.

By solving the above equation for ∆f,x0 (x), we get for x 6= x0 that

f (x) − f (x0 )
∆f,x0 (x) = ,
x − x0

i.e., it is the difference quotient. Continuity of ∆f,x0 (x) at x0 is therefore equivalent to

the existence of the limit
f (x) − f (x0 )
lim ∆f,x0 (x) = lim =: f 0 (x0 ).
x→x0 x→x0 x − x0

Also the following notation is used in the literature for f 0 (x0 ):

d
dx
∂
f (x0 ), ∂x df
f (x0 ), dx |x=x0 , ∂f |
∂x x=x0
, ∂x f (x0 ).

The next result states that differentiability is a stronger property than continuity.
Theorem 5.2.
Let f : I → R be differentiable at x0 ∈ I. Then f is continuous in x0 .

Proof: By writing
f (x) = f (x0 ) + (x − x0 ) · ∆f,x0 (x),
the continuity of ∆f,x0 at x0 implies the continuity of f at x0 . 2
As the following example shows, the opposite implication cannot be made, i.e., not every
continuous function is differentiable.

Example 5.3. Consider the absolute value function | · | : R → R. We already know that
it is continuous. For the analysis of differentiability, we distinguish between three cases:
1st Case: x0 > 0.
Then we have that |x0 | = x0 and, moreover, for x in some neighbourhood of x0 holds
|x| = x. Therefore, we have

|x| − |x0 | x − x0
lim = lim = 1.
x→x0 x − x0 x→x0 x − x0

2nd Case: x0 < 0.

Then we have that |x0 | = −x0 and, moreover, for x in some neighbourhood of x0 holds
|x| = −x. Therefore, we have

|x| − |x0 | −x + x0
lim = lim = −1.
x→x0 x − x0 x→x0 x − x0
100 5 Differentiation of Functions

3rd Case: x0 = 0.
Then the two sequences (xn )n∈N = ( n1 )n∈N , (yn )n∈N = (− n1 )n∈N both tend to x0 = 0.
However, we have
|xn | − |x0 | | 1 | − |0|
lim = lim n1 =1
n→∞ xn − x0 n→∞ −0
n
and
|yn | − |x0 | | − n1 | − |0|
lim = lim = −1.
n→∞ yn − x0 n→∞ − n1 − 0
Therefore, the limit
|x| − |0|
lim
x→0 x − 0

does not exist. Thus, | · | is not differentiable at x0 = 0.

Note that, by a substitution h := x − x0 , the difference quotient can be reformulated as

f (x) − f (x0 ) f (x0 + h) − f (x0 )

f 0 (x0 ) = lim = lim .
x→x0 x − x0 h→0 h

The derivative a := f 0 (x0 ) of f in x0 can be interpreted in the following way: The linear
mapping ϕ : R → R, x 7→ ax fulfills
 
|f (x0 + h) − (f (x0 ) + ϕ(h))|  f (x0 + h) − f (x0 ) 
lim = lim 
 − a = 0.
h→0 |h| h→0 h

This means that the affine linear mapping t(h) := f (x0 ) + ϕ(h) = f (x0 ) + ah, which
actually is the tangent of f at x0 , approximates f (x) linearly in a neighbourhood of x0 in
a best possible way.
This “local linearisation” can be generalised to functions between arbitrary normed R-
vector spaces.
Definition 5.4. Total Derivative
Let (E, || · ||E ) and (F, || · ||F ) be two normed R-vector spaces and let U be an open
subset of E. Then a function f : U → F is said to be differentiable in a point
x0 ∈ U if there is a continuous linear function ϕ : E → F such that

||f (x) − f (x0 ) − ϕ(x − x0 )||F

lim =0
x→x0 ||x − x0 ||E

or equivalently if
||f (x0 + h) − f (x0 ) − ϕ(h)||F
lim = 0.
h→0 ||h||E
In this case ϕ is called the (total) derivative of f in x0 which is denoted by f 0 (x0 ).
If f is differentiable in all points of U then f is called differentiable in U or just
differentiable and f 0 : U → L(E, F ), x 7→ f 0 (x) is called the derivative of f .

Note carefully that the total derivative f 0 (x0 ) is a linear function. If E and F are finite
dimensional R-vector spaces, say E = Rm , F = Rn for some m, n ∈ N, then f 0 (x0 ) can
be identified with its matrix representation M ∈ Rn,m with respect to the standard bases
5.1 Differentiability and Derivatives 101

of Rm and Rn . The matrix M =: Jf (x0 ) is called the Jacobian matrix of f in x0 . In the

one-dimensional case m = 1 = n, M ∈ R1,1 is a real number which is given by
f (x0 + h) − f (x0 )
M = lim = ∆f,x0 (x0 ).
h→0 h
This clearifies the connection between Definition 5.1 and Definition 5.4. Another special
case is E = R and F = C ∼ = R2 . Then, for
f : I → C, x 7→ Re(f (x)) + i Im(f (x)) ∼
= (Re(f (x)), Im(f (x))T
with I ⊂ R open, we have M = (Re(f (x))0 , Im(f (x))0 )T ∈ R2,1 and like in the one-
dimensional case we identify f 0 (x0 ) with M and therefore write

f 0 (x0 ) := Re(f (x))0 + i Im(f (x))0 ∼

= (Re(f (x))0 , Im(f (x))0 )T .
We remark that linear mappings between finite-dimensional normed R-vector spaces are
automatically continuous so that the continuity assumption of ϕ in Definition 5.4 can be
dropped in this case. Without proof we state that, if the total derivative f 0 (x0 ) of f in
x0 exists, then it is uniquely determined and its existence also implies continuity of f in
x0 like it was shown in Theorem 5.2 for the one-dimesional case.

Many of the subsequent results carried out for the cases E = R = F or E = R and
F = C can be generalised to arbitrary E and F in a straight forward manner and the
proofs are analogous and sometimes become even clearer in terms of total derivatives.
This will be part of the exercises.

Now we consider some examples of differentiable functions.

Example 5.5. a) Given is some constant c ∈ R. Consider the constant function f : R →

R with f (x) = c for all x ∈ R. Then for all x0 ∈ R holds
f (x) − f (x0 ) c−c
f 0 (x0 ) = lim = lim = 0.
x→x0 x − x0 x→x0 x − x0

b) Given is some constant c ∈ R. Consider the linear function f : R → R with f (x) = cx

for all x ∈ R. Then for all x0 ∈ R holds
f (x) − f (x0 ) cx − cx0 x − x0
f 0 (x0 ) = lim = lim = lim c = c.
x→x0 x − x0 x→x0 x − x0 x→x0 x − x0

c) For determining the derivatives of exp, sinh, cosh, sin and cos, we first determine the
following limit for λ ∈ C:
exp(λh) − 1
lim .
h→0 h
By Theorem ??, we know that for h ∈ R with |λh| < 2

exp(λh) = 1 + λh + r2 (λh)

with |r2 (λh)| ≤ |λh|2 . Therefore,

 
exp(λh) − 1 1 + λh + r2 (λh) − 1 r2 (λh)
lim = lim = lim λ + = λ.
h→0 h h→0 h h→0 h
102 5 Differentiation of Functions

We can further conclude

exp(λ(x0 + h)) − exp(λx0 ) exp(λh) − 1

lim = exp(λx0 ) · lim = λ exp(λx0 ).
h→0 h h→0 h

This has manifold consequences for the derivatives of exponential, hyperbolic and
trigonometric functions:

exp(x0 + h) − exp(x0 )
exp0 (x0 ) = lim = exp(x0 ),
h→0 h

i.e., exp0 = exp.

We can further conclude that

sinh(x0 + h) − sinh(x0 )
sinh0 (x0 ) = lim
h→0
 h 
1 exp(x0 + h) − exp(x0 ) − exp(−(x0 + h)) + exp(−x0 )
= lim +
2 h→0 h h
1
= (exp(x0 ) + exp(−x0 )) = cosh(x0 ).
2

Analogously, we can show that cosh0 = sinh. Now consider the trigonometric functions:

sin(x0 + h) − sin(x0 )
sin0 (x0 ) = lim
h→0
 h 
1 exp(i(x0 + h)) − exp(ix0 ) − exp(−i(x0 + h)) + exp(−ix0 )
= lim +
2i h→0 h h
1
= (i exp(ix0 ) + i exp(−ix0 )) = cos(x0 )
2i

and

cos(x0 + h) − cos(x0 )
cos0 (x0 ) = lim
h→0
 h 
1 exp(i(x0 + h)) − exp(ix0 ) exp(−i(x0 + h)) − exp(−ix0 )
= lim +
2 h→0 h h
1
= (i exp(ix0 ) − i exp(−ix0 ))
2
1
= − (exp(ix0 ) − exp(−ix0 )) = − sin(x0 ).
2i

Now we consider rules for the derivatives of sums, products and quotients of functions.
Theorem 5.6. Summation Rule, Product Rule, Quotient Rule
Let f, g : I → R be differentiable in x0 ∈ I.

(i) Then f + g is differentiable in x0 with (f + g)0 (x0 ) = f 0 (x0 ) + g 0 (x0 ).

(ii) Then f · g is differentiable in x0 with (f · g)0 (x0 ) = f 0 (x0 ) · g(x0 ) + f (x0 ) · g 0 (x0 ).
5.1 Differentiability and Derivatives 103

(iii) If g(x0 ) 6= 0, then fg (x0 ) is differentiable in x0 with

0
f 0 (x0 )g(x0 ) − f (x0 )g 0 (x0 )

f
(x0 ) (x0 ) = .
g g 2 (x0 )

Proof: Let f (x) = f (x0 ) + (x − x0 ) · ∆f,x0 (x), g(x) = g(x0 ) + (x − x0 ) · ∆g,x0 (x). Then
(i)
(f + g)(x) = f (x) + g(x) = (f + g)(x0 ) + (x − x0 ) · (∆f,x0 (x) + ∆g,x0 (x)).

(ii)
(f · g)(x) =(f (x0 ) + (x − x0 ) · ∆f,x0 (x0 ))(g(x0 ) + (x − x0 ) · ∆g,x0 (x))
=f (x0 )g(x0 ) + (x − x0 )(∆f,x0 (x0 )g(x0 ) + ∆g,x0 (x)f (x0 ))
+ (x − x0 )2 ∆f,x0 (x0 )∆g,x0 (x).
Thus,
f (x)g(x) − f (x0 )g(x0 )
lim
x→x0 x − x0
(x − x0 )(∆f,x0 (x)g(x0 ) + ∆g,x0 (x)f (x0 )) + (x − x0 )2 ∆f,x0 (x)∆g,x0 (x)
= lim
x→x0 x − x0
= lim ((∆f,x0 (x0 )g(x0 ) + ∆g,x0 (x)f (x0 ) + (x − x0 )∆f,x0 (x)∆g,x0 (x))
x→x0

=∆f,x0 (x)g(x0 ) + ∆g,x0 (x)f (x0 )

(iii) For convenience, we assume that f ≡ 1 (the general result follows by an application
of the product rule). Then
1 1 g(x0 ) − g(x) (x − x0 )∆g,x0 (x)
− = =−
g(x) g(x0 ) g(x0 )g(x) g(x0 )g(x)
and thus
g 0 (x0 )
 
1 1 1
lim − =− 2 .
x→x0 x − x0 g(x) g(x0 ) g (x0 )
2

Example 5.7. a) For a constant c and a differentiable function f : I → R we have

(cf )0 (x) = c0 f (x) + cf 0 (x) = cf 0 (x).

b) Let n ∈ N and f : R → R, x 7→ xn . Then f 0 (x) = nxn−1 .

Induction basis: n = 1. f 0 (x) = (1 · x)0 = 1 = 1 · x1−1 .
a)
Induction step: n > 1: Using the product rule we immediately get
f 0 (x) = (xn )0 = (x · xn−1 )0 = x0 (xn−1 ) + x(xn−1 )0 = xn−1 + x · (n − 1)xn−2 = nxn−1
Ind. Hyp.

c) f (x) = x−n = 1
xn
, n ∈ N. Then for x ∈ R\{0}
−nxn−1 1
f 0 (x) = 2n
= −n n+1 = −nx−n−1
x x
is true.
104 5 Differentiation of Functions

d) f (x) = x · exp(x). Then for x ∈ R

f 0 (x) = 1 · exp(x) + x · exp(x) = (1 + x) exp(x).

e) For a real polynomial f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 holds

f 0 (x) = nan xn−1 + (n − 1)an−1 xn−2 + . . . + 2a2 x + a1 .

Now we introduce differentiation rules for composition of functions and inverse functions.

Theorem 5.8. Differentiation of inverse functions

Let I, J ⊂ R be intervals and let f : I → J be bijective. If f is differentiable in
x0 ∈ I with f 0 (x0 ) 6= 0 and if f −1 : J → I is continuous in y0 = f (x0 ), then the
inverse function f −1 is differentiable in y0 := f (x0 ) with

0 1
f −1 (y0 ) = .
f 0 (f −1 (y0 ))

Proof: Let (yn )n∈N be a sequence in J with yn 6= y0 and limn→∞ yn = y0 . Define xn :=

f −1 (yn ). Since f −1 is continuous, we have limn→∞ xn = x0 and also xn 6= x0 by bijectivity
of f . Thus we conclude

f −1 (yn ) − f −1 (y0 ) xn − x0 1 1
lim = lim = 0 = 0 −1 .
n→∞ yn − y0 n→∞ f (xn ) − f (x0 ) f (x0 ) f (f (y0 ))
2
√ 1
Example 5.9. a) g(x) = n
x = x n is the inverse function of f : R+ → R+ , f (x) = xn .
Then for x > 0 holds
1 1 1 1 − n−1 1 1 −1
g 0 (x) = = = √ n−1 = x n = xn .
f 0 (g(x)) n(g(x))n−1 nnx n n

b) log : (0, ∞) → R is the inverse function of exp. Then for x > 0 holds
1 1 1
log0 (x) = = = .
exp0 (log(x)) exp(log(x)) x

c) arcsin : [−1, 1] → [− π2 , π2 ] is the inverse function of sin. Then for x ∈ (− π2 , π2 ) holds

1 1
arcsin0 (x) = 0 = .
sin (arcsin(x)) cos(arcsin(x))

Since sin2 (y) + cos2 (y) = 1 and cos(y) ≥ 0 for y ∈ (− π2 , π2 ), we have

q
cos(y) = 1 − sin2 (y).

Therefore
1 1
arcsin0 (x) = p 2
=√ .
1 − sin (arcsin(x)) 1 − x2
5.2 Mean Value Theorems and Consequences 105

Theorem 5.10. Chain rule

Let I, J be intervals and f : J → R and g : I → J be given. Assume that for x0 ∈ I,
g is differentiable in x0 and f is differentiable in g(x0 ). Then the composition f ◦ g
((f ◦ g)(x0 ) = f (g(x0 ))) is differentiable in x0 with

(f ◦ g)0 (x0 ) = f 0 (g(x0 )) · g 0 (x0 ).

Proof: By assumption there are functions ∆g,x0 (x) : I → R and ∆f,g(x0 ) (y) : J → R which
are continuous in x0 and g(x0 ) respectively such that

g(x) = g(x0 ) + (x − x0 )∆g,x0 (x)

f (y) = f (g(x0 )) + (y − g(x0 ))∆f,g(x0 ) (y)

for all x ∈ I and y ∈ J. Thus

(f ◦ g)(x) = f (g(x)) = f (g(x0 )) + (g(x) − g(x0 ))∆f,g(x0 ) (g(x))

= f (g(x0 )) + (x − x0 )∆g,x0 (x)∆f,g(x0 ) (g(x0 ) + (x − x0 )∆g,x0 (x))

for all x ∈ I. Hence the function

∆f ◦g (x) := ∆g,x0 (x)∆f,g(x0 ) (g(x0 ) + (x − x0 )∆g,x0 (x))

fulfills

(f ◦ g)(x) = (f ◦ g)(x0 )) + (x − x0 )∆f ◦g (x)

for all x ∈ I. As a composition of functions that are continuous in x0 it is also continuous

in x0 with ∆f ◦g (x0 ) = ∆g,x0 (x0 )∆f,g(x0 ) (g(x0 )) = g 0 (x0 )f 0 (g(x0 )). 2

Example 5.11. a) f : R → R with f (x) = sin(x2 ), then f 0 (x) = cos(x2 ) · 2x.

b) Let a ∈ R. Consider f : R → R with f (x) = ax = exp(x log(a)), then

f 0 (x) = exp(x log(a)) · log(a) = ax · log(a).

c) Let a ∈ R. Consider f : R+ → R with f (x) = xa = exp(a log(x)), then

a a
f 0 (x) = exp(a log(x)) · = xa · = a · xa−1 .
x x

d) Consider f : R+ → R with f (x) = xx = exp(x log(x)), then

x 
f 0 (x) = exp(x log(x)) · + log(x) = xx · (1 + log(x)).
x

5.2 Mean Value Theorems and Consequences

106 5 Differentiation of Functions

Definition 5.12.
Let I be an interval and f : I → R be a function. Then x0 ∈ I is called local
maximum (local minimum) if there exists some neighbourhood U of x0 such that

f (x0 ) = max{f (x) : x ∈ I ∩ U } (f (x0 ) = min{f (x) : x ∈ I ∩ U }).

A number x0 ∈ I is called local extremum if it is a local maximum or a local

minimum.

max

min

Figure 5.3: Local extrema

Theorem 5.13.
Let I be an interval and f : I → R be a function that is differentiable in x0 ∈ I.
Assume that x0 is an interior point of I and that x0 is a local extremum. Then
f 0 (x0 ) = 0.

Proof: We assume that x0 is a local maximum (the case of minimum is shown analogously).
Let U be a neighbourhood of x0 with U ⊂ I and f (x0 ) = max{f (x) : x ∈ U }. Let

f (x) = f (x0 ) + (x − x0 ) · ∆f,x0 (x).

Assume that f 0 (x0 ) = ∆f,x0 (x0 ) > 0. Since ∆f,x0 is continuous in x0 , then there exists
a neighbourhood V ⊂ U of x0 such that ∆f,x0 (x) > 0 for all x ∈ V . Then for all x1 ∈ V
with x1 > x0 holds f (x1 ) = f (x0 ) + (x1 − x0 ) · ∆f,x0 (x1 ) > f (x0 ). This is a contradiction.
On the other hand, assume that f 0 (x0 ) = ∆f,x0 (x0 ) < 0. Since ∆f,x0 is continuous in x0 ,
then there exists a neighbourhood V ⊂ U such that ∆f,x0 (x) < 0 for all x ∈ V . Then for
all x1 ∈ V with x1 < x0 holds f (x1 ) = f (x0 ) + (x1 − x0 ) · ∆f,x0 (x1 ) > f (x0 ). This is also
a contradiction. 2
As a consequence, we will formulate the following result stating that derivatives of func-
tions with equal boundary conditions have at least one zero.
5.2 Mean Value Theorems and Consequences 107

Theorem 5.14. Theorem of Rolle

Let f : [a, b] → R be differentiable in [a, b] and let f (a) = f (b). Then there exists
some x ∈ (a, b) such that f 0 (x) = 0.

Proof: If f is constant, the statement is clear (since then f 0 (x) = 0 for all x ∈ (a, b)). If
f is not constant, consider the maximum and the minimum of f on [a, b] (we know by
Theorem 3.15 that they exist). So, let x− , x+ ∈ [a, b] such that

f (x+ ) = max{f (x) : x ∈ [a, b]}, f (x− ) = min{f (x) : x ∈ [a, b]}.

Then we have that x+ ∈ (a, b) or x− ∈ (a, b) since, otherwise, f (x+ ) = f (x− ) (constant).
Then f 0 (x− ) = 0 or f 0 (x+ ) = 0. 2
As a corollary, we have that for a function f differentiable in some interval I, the following
holds: Between two zeros of f , there always exists some point x0 with f 0 (x0 ) = 0.
Now we present the famous mean value theorem.
Theorem 5.15.
Let f : [a, b] → R be differentiable. Then there exists some x̂ ∈ (a, b) such that

f (b) − f (a) = f 0 (x̂) · (b − a).

Before the proof is presented, we give some graphical interpretation: A division of the
above equation by b − a gives

f (b) − f (a)
= f 0 (x̂).
b−a

The quantity on the left hand side is equal to the slope of the secant of f through a and
b, whereas f 0 (x̂) corresponds to the slope of tangent of f at x̂. Therefore, the secant of f
through a and b is parallel to a tangent of f .
Proof: Consider the function F : [a, b] → R with

f (b) − f (a)
F (x) := f (x) − f (a) − · (x − a).
b−a

Then we have F (a) = F (b) = 0. By Rolle’s Theorem, we get that there exists some
x̂ ∈ (a, b) with
f (b) − f (a)
0 = F 0 (x̂) = f 0 (x̂) −
b−a
and thus
f (b) − f (a)
f 0 (x̂) = .
b−a
2
The mean value theorem leads us to determine monotonicity properties of a function by
means of its derivative.
108 5 Differentiation of Functions

tangent

secant

a x b

Figure 5.4: Mean value theorem

Theorem 5.16.
Let f : [a, b] → R be a differentiable function. Then the following holds true.

(i) If f 0 (x) > 0 for all x ∈ (a, b), then f is strictly monotonically increasing.

(ii) If f 0 (x) < 0 for all x ∈ (a, b), then f is strictly monotonically decreasing.

(iii) If f 0 (x) ≥ 0 for all x ∈ (a, b), then f is monotonically increasing.

(iv) If f 0 (x) ≤ 0 for all x ∈ (a, b), then f is monotonically decreasing.

Proof: (i) By the mean value theorem we have that for x1 , x2 ∈ (a, b) with x1 < x2 , there
exists some x̂ ∈ (x1 , x2 ) with

f (x2 ) − f (x1 ) = f 0 (x̂) · (x2 − x1 ) > 0.

The results (ii)-(iv) can be proven analogously. 2

Now we consider a generalisation of the mean value theorem. This gives us a completely
new tool for the determination of limits.
Theorem 5.17. Generalised mean value theorem
Let f, g : [a, b] → R be differentiable. Assume that g 0 has no zero in (a, b). Then
there exists some x̂ ∈ (a, b) such that

f (b) − f (a) f 0 (x̂)

= 0 .
g(b) − g(a) g (x̂)

Proof: By introducing the function

f (b) − f (a)
F (x) = f (x) − f (a) − · (g(x) − g(a)),
g(b) − g(a)
5.2 Mean Value Theorems and Consequences 109

we get F (a) = F (b) = 0. Now using the Theorem of Rolle, the result follows immediately.
2
Theorem 5.18. Theorem of l’Hospital
Let I be an interval and let f, g : I → R be differentiable. Let x0 ∈ I and assume
that f (x0 ) = g(x0 ) = 0 and there exists some neighbourhood U ⊂ I of x0 such that
0 (x)
g 0 (x) 6= 0 for all x ∈ U \{x0 }. Then, if limx→x0 fg0 (x) exists, then also limx→x0 fg(x)
(x)

exists and
f (x) f 0 (x)
lim = lim 0 .
x→x0 g(x) x→x0 g (x)

Proof: Let (xn )n∈N be a sequence with limn→∞ xn = x0 and xn 6= x0 for all n ∈ N. Then,
by the generalised mean value theorem, there exists a sequence (x̂n )n∈N with x̂n between
x0 and xn such that
f (xn ) f (xn ) − f (x0 ) f 0 (x̂n )
= = 0 .
g(xn ) g(xn ) − g(x0 ) g (x̂n )
In particular, since (x̂n )n∈N converges to x0 , we have

f (x) f 0 (x)
lim = lim 0 .
x→x0 g(x) x→x0 g (x)
2

Example 5.19. a) Let a ∈ R.

log(1 + ax) a
lim = lim = a.
x→0,ax>0 x x→0,ax>0 1 + ax

b)
   
1 log(1 + ax) log(1 + ax)
lim (1+ax) = x lim exp = exp lim = exp(a).
x→0,ax>0 x→0,ax>0 x x→0,ax>0 x

In particular, for x = n1 , we have

 a n
lim 1+ = exp(a).
n→∞ n

c)
1 − cos(x) sin(x) cos(x) 1
lim 2
= lim = lim = .
x→0 x x→0 2x x→0 2 2

Theorem 5.20. Generalisations of l’Hospital’s Theorem

• Expressions of type ∞ ∞
, Limit as x → x0
Let f, g : I\{x0 } → R be differentiable functions with limx→x0 f (x) = ∞,
0 (x)
limx→x0 g(x) = ∞. Then, if limx→x0 fg0 (x) exists, then also limx→x0 fg(x)
(x)
exists
110 5 Differentiation of Functions

with
f (x) f 0 (x)
lim = lim 0 .
x→x0 g(x) x→x0 g (x)

• Expressions of type 00 , Limit as x → ∞

Let f, g : [t0 , ∞) → R be differentiable functions with limx→∞ f (x) = 0,
limx→∞ g(x) = 0. Assume that there exists some x1 ∈ R such that g 0 (x) 6= 0
0 (x)
for all x > x1 . Then, if limx→∞ fg0 (x) exists, then also limx→∞ fg(x)
(x)
exists with

f (x) f 0 (x)
lim = lim 0 .
x→∞ g(x) x→∞ g (x)

• Expressions of type ∞ ∞
, Limit as x → ∞
Let f, g : [t0 , ∞) be differentiable functions with limx→∞ f (x) = ∞,
0 (x)
limx→∞ g(x) = ∞. Then, if limx→∞ fg0 (x) exists, then also limx→∞ fg(x)
(x)
ex-
ists with
f (x) f 0 (x)
lim = lim 0 .
x→∞ g(x) x→∞ g (x)

Proof: The first result follows by an application of l’Hospital’s Theorem to

1 g 0 (x)  2
f (x) g(x) (g(x))2 1 f (x)
lim = lim 1 = lim f 0 (x)
= f 0 (x)
lim .
x→x0 g(x) x→x0 x→x0 limx→x0 x→x0 g(x)
f (x) (f (x))2 g 0 (x)

The second and third statement follow by a substitution y = x1 and the consideration of
     
1 1 0 1 0 1
f (x) f y
− y 2 f y
f y f 0 (x)
lim = lim   = lim   = lim   = lim 0 .
x→∞ g(x) y→0 y→0 y→0 0 1 x→∞ g (x)
g 1 y
− 1 g0 1 y2
g y y

2
Note that we can also treat expressions of type “∞−∞” by l’Hospitals’s Theorem. Namely,
for f, g with limx→x0 f (x) = 0, limx→x0 g(x) = 0, we get that
1 1 g(x) − f (x) g 0 (x) − f 0 (x)
lim ( − ) = lim = lim 0 .
x→x0 f (x) g(x) x→x0 f (x) · g(x) x→x0 f (x) · g(x) + f (x) · g 0 (x)

Also, expressions of type “0 · ∞” can be treated by a special trick. Namely, for f, g with
limx→x0 f (x) = 0, limx→x0 g(x) = ∞, we get that
f (x) f 0 (x) f 0 (x)(g(x))2
lim f (x) · g(x) = lim 1 = lim 0 = − lim .
x→x0 x→x0
g(x)
x→x0 g (x)
− (g(x)) 2
x→x0 g 0 (x)

You do not have to keep the above two formulas in mind. These can always be derived
in concrete examples.

Example 5.21. a) Let n ∈ N and consider

 π n
lim x− · tan(x).
x→ π2 ,x< π2 2
5.3 Higher Derivatives, Curve Sketching 111

This is an expression of type “0 · ∞” and we can make use of

n
 π n x − π2
lim x− · tan(x) = lim 1
x→ π2 ,x< π2 2 x→ π2 ,x< π2
tan(x)

n−1
n · x − π2
= lim 1 1
x→ π2 ,x< π2 −
tan2 (x) cos2 (x)

n−1
n · x − π2
= lim
x→ π2 ,x< π2 − sin21(x)
(
 π n−1 2 −1 if n = 1,
= − lim n· x− sin (x) =
x→ π2 ,x< π2 2 0 else.

b)
 
1 1 x − sin(x)
lim − = lim
x→0 sin(x) x x→0 x sin(x)

1 − cos(x) sin(x)
= lim = lim = 0.
x→0 sin(x) + x cos(x) x→0 cos(x) − x sin(x) + cos(x)

5.3 Higher Derivatives, Curve Sketching

Here we consider derivatives of derivatives (of derivatives ...) and discuss consequences
for the search for local extrema of functions.
Definition 5.22.
If the derivative of a function is differentiable, we call (f 0 )0 the second derivative.
The n-th derivative of a function is inductively defined as the derivative of the
n − 1-th derivative.
d2
For the second derivative at x0 , we write f 00 (x0 ) or dx 2 f (x0 ). The n-th derivative
dn
is denoted by f (x0 ) or dxn f (x0 ).
(n)

We call a function f : I → R n-times differentiable if f (n) (x) exists for all x ∈ I.

Furthermore we call a function f : I → R n-times continuously differentiable if
f (n) : I → R exists and is continuous.

Theorem 5.23.
Let I := [a, b] and f : I → R be differentiable. Furthermore let x0 ∈ I such that
f 0 (x0 ) = 0 and f 0 is differentiable in x0 . Then

1. if f 00 (x0 ) > 0, then f has a local minimum in x0 ;

2. if f 00 (x0 ) < 0, then f has a local maximum in x0 .

Proof: We only show the case f 00 (x0 ) > 0 (the opposite case is analogous). By definition,
we have
f 0 (x) − f 0 (x0 )
f 00 (x0 ) = lim > 0.
x→x0 x − x0
112 5 Differentiation of Functions

Since f 0 is continuous in x0 , we have that there exists some ε > 0 such that for all
x ∈ I\{x0 } with |x − x0 | < ε holds
f 0 (x) − f 0 (x0 )
> 0.
x − x0
Since f 0 (x0 ) = 0, we have that
f 0 (x) < 0 for all x ∈ (x0 − ε, x0 ),
f 0 (x) > 0 for all x ∈ (x0 , x0 + ε).
Therefore, f is monotonically decreasing in (x0 − ε, x0 ) and monotonically increasing in
(x0 , x0 + ε). Therefore, f has a local minimum in x0 . 2
Remark:
Note that in the case f 00 (x0 ) = 0, we cannot make a decision whether f has a local
extremum there. For instance, consider the three functions f1 (x) = x3 , f2 (x) = x4
and f3 (x) = −x4 . We have f10 (0) = f20 (0) = f30 (0) = 0 and, furthermore, f100 (0) =
f200 (0) = f300 (0) = 0. However, f1 has no local extremum in 0, f2 has a local minimum
in 0 and f3 has a local maximum in 0.

Example 5.24. 1. Consider the rational function

x(x + 5) 36
f (x) = =x+9+ .
x−4 x−4
It can be easily seen that f has a first order pole at x0 = 4.
The first two derivatives of f are given by
x2 − 8x − 20 (x + 2)(x − 10) 72
f 0 (x) = 2
= , f 00 (x) = .
(x − 4) (x − 4)2 (x − 4)3
The zeros of f are given by x1 = 0 and x2 = −5.
Now we determine the set of local extrema: We have that f 0 (x) = 0 is only fulfilled
for x3 = −2 and x4 = 10. In this case, we have f 00 (x3 ) = − 13 < 0 and f 00 (x4 ) = 13 > 0.
As a consequence, f has a local maximum in x3 = −2 and a local minimum in
x4 = 10. We further have
a) x ∈ (−∞, −2) ⇒ f 0 (x) > 0, i.e., f is strictly monotonically increasing in
(−∞, −2);
b) x ∈ (−2, 4) ⇒ f 0 (x) < 0, i.e., f is strictly monotonically decreasing in (−2, 4);
c) x ∈ (4, 10) ⇒ f 0 (x) < 0, i.e., f is strictly monotonically decreasing in (4, 10);
d) x ∈ (10, ∞) ⇒ f 0 (x) > 0, i.e., f is strictly monotonically increasing in (10, ∞).
2. f (x) = sin(x). The zeros are given by
{0, π, −π, 2π, −2π, 3π, −3π, . . .} = {nπ | n ∈ Z}.
The first two derivatives are given by f 0 (x) = cos(x), f 00 (x) = − sin(x). The zeros
of f 0 are given by
   
π π 3π 3π 5π 5π 2n + 1
,− , ,− , , ,... = π|n∈Z .
2 2 2 2 2 2 2
5.3 Higher Derivatives, Curve Sketching 113

For xn = 2n+12
π, we have f 00 (xn ) = − sin( 2n+1
2
π) = (−1)n+1 . As a consequence, sin
has a local maximum in xn = 2n+1 2
π if n is even and a local minimum in xn = 2n+1
2
π
if n is odd.

Definition 5.25. Convexity/Concavity

A function f : [a, b] → R, a, b ∈ R, a < b, is called convex, if for all x1 < x < x2 in
[a, b] holds
f (x2 ) − f (x1 )
f (x) ≤ · (x − x1 ) + f (x1 ) . (5.1)
x2 − x1
It is called concave, if for all x1 < x < x2 in [a, b] holds

f (x2 ) − f (x1 )
f (x) ≥ · (x − x1 ) + f (x1 ) . (5.2)
x2 − x1
If the inequalities in (5.1) or (5.2) are strict then f is called strictly convex/concave.
Geometrically this means that the graph of a convex (concave) function f : [a, b] → R
restricted to any subinterval [x1 , x2 ] of [a, b] lies below (above) the secant

f (x2 ) − f (x1 )
s(x) := · (x − x1 ) + f (x1 ) .
x2 − x1

convex graph concave graph

14 14
f(x) f(x)
s(x) s(x)
12 12

10 10

8 8
y
y

6 6

4 4

2 2

0 0
0 x1= 1 2 3 4 x2= 5 6 0 x1= 1 2 3 4 x2= 5 6
x x

Figure 5.5: convex and concave graphs

Theorem 5.26.
Let f : [a, b] → R, a, b ∈ R, a < b, be 2-times differentiable.

a) If f 00 (x) ≥ 0 for all x ∈ (a, b) , then f is convex.

b) If f 00 (x) > 0 for all x ∈ (a, b) , then f is strictly convex.

c) If f 00 (x) ≤ 0 for all x ∈ (a, b) , then f is concave.

d) If f 00 (x) < 0 for all x ∈ (a, b) , then f is strictly concave.

Proof: We only prove a). The other results follow analogously. Let x1 < x < x2 in [a, b].
Since f 00 ≥ 0 we know that f 0 is monotonically increasing. By the intermediate value
114 5 Differentiation of Functions

theorem there are ξ1 ∈ (x1 , x) and ξ2 ∈ (x, x2 ) such that

f (x) − f (x1 ) f (x2 ) − f (x)

= f 0 (ξ1 ) ≤ f 0 (ξ2 ) = .
x − x1 x2 − x

This implies

(f (x) − f (x1 ))(x2 − x) ≤ (f (x2 ) − f (x))(x − x1 )

⇔ f (x)(x2 − x1 ) ≤ (f (x2 ) − f (x1 ))(x − x1 ) + f (x1 )(x2 − x1 )

f (x2 ) − f (x1 )
⇔ f (x) ≤ (x − x1 ) + f (x1 ) .
x2 − x1
2
Definition 5.27. Inflection point
Let f : [a, b] → R be a function. We say that x0 ∈ (a, b) is an inflection point of f
if there is an ε > 0 with [x0 − ε, x0 + ε] ⊂ [a, b] such that one of the following two
statements holds true:

a) f is convex on [x0 − ε, x0 ] and concave on [x0 , x0 + ε].

b) f is concave on [x0 − ε, x0 ] and convex on [x0 , x0 + ε].

Theorem 5.28.
Let f : [a, b] → R be 3-times continuously differentiable and x0 ∈ (a, b).

a) If x0 is an inflection point, then f 00 (x0 ) = 0.

b) If f 00 (x0 ) = 0 and f 000 (x0 ) 6= 0, then x0 is an inflection point.

Proof: a) follows from Theorem 5.26 and continuity of f 00 .

b) If f 00 (x0 ) = 0 and f 000 (x0 ) 6= 0, then by continuity of f 000 there is an ε > 0 with
[x0 − ε, x0 + ε] ⊂ [a, b] such that f 000 does not have a zero on [x0 − ε, x0 + ε]. This implies
that f 00 is strictly monotonic on [x0 − ε, x0 + ε]. Thus either
i) f 00 > 0 on [x0 − ε, x0 ] and f 00 < 0 on [x0 , x0 + ε] or
ii) f 00 < 0 on [x0 − ε, x0 ] and f 00 > 0 on [x0 , x0 + ε]
holds true. In case i), f is convex on [x0 − ε, x0 ] and concave on [x0 , x0 + ε] and in case
ii) the reversed behavior is given. Therefore x0 is an inflection point. 2

The aim of a so-called curve discussion of a function f : D → R, D ⊂ R, is to de-

termine its qualitative and quantitative behaviour. We give a short list of things that
have to be investigated/determined:
1. Domain of definition D
2. Symmetries
a) f is symmetrical with respect to the y-axis if f (x) = f (−x) for all x in the
domain of definition. In this case f is called an even function.
5.3 Higher Derivatives, Curve Sketching 115

b) f is point-symmetrical with respect to the origin if f (−x) = −f (x) for all x in

the domain of definition. In this case f is called an odd function.
3. Poles
4. Behaviour for x −→ ±∞, asymptotes
A straight line g(x) = ax + b, a, b ∈ R, is called an asymptote of f for x → ±∞ if
limx→±∞ (f (x) − (ax + b)) = 0. In this case the coefficients a, b can be successively
determined by

f (x)
a = lim ,
x→±∞ x
b = lim (f (x) − ax) .
x→±∞

5. Zeros
6. Extrema, monotonicity behaviour
7. Inflection points, convexity/concavity behaviour
8. Function graph

Example 5.29. We want to give a complete curve discussion for the rational function

2x2 + 3x − 4
f (x) = .
x2
1. Domain of definition: D = R\{0}
2. Symmetries: f is neither an even nor an odd function.
3. Poles: x0 = 0 is a pole of order 2, limx%0 f (x) = −∞ = limx&0 f (x)
4. Behaviour for x −→ ±∞, asymptotes: limx→±∞ f (x) x
= 0, limx→±∞ f (x) = 2. Thus
the horizontal line at y = 2 is an asymptote of f for x −→ ∞ and also for x −→ −∞.
√
5. Zeros: f (x) = 0 ⇔ 2x2 + 3x − 4 = 0 ⇔ x = x1,2 = 14 (−3 ± 41)
x1 ≈ −2.35, x2 ≈ 0.85
6. Extrema, monotonicity behaviour:
f 0 (x) = −3x+8
x3
= 0 ⇔ x = x3 = 38 , y3 := f (x3 ) ≈ 2.56
f 00 (x) 6x−24
= x4
f 00 (x3 ) < 0 ⇒ f has a local maximum at x3 .

, strictly monotonically decreasing


 < 0 , 83 < x < ∞
f 0 (x) > 0 ,0 < x < 3 8
, strictly monotonically increasing
< 0 , −∞ < x < 0 , strictly monotonically decreasing


7. Inflection points, convexity/concavity behaviour:

f 00 (x) = 0 ⇔ x = x4 = 4, y4 = f (x4 ) = 5
2
f 000 = 96−18x
x5
f 000 (x4 ) > 0 ⇒ x4 is an inflection point.
116 5 Differentiation of Functions

, strictly convex

 > 0 ,4 < x < ∞
f 00 (x) < 0 ,0 < x < 4 , strictly concave
< 0 , −∞ < x < 0 , strictly concave


8. Function graph

0
y

−1

−2

−3
−10 −8 −6 −4 −2 0 2 4 6 8 10
x

5.4 Taylor’s Formula

The aim of this part is to approximate a sufficiently smooth (that means sufficiently often
differentiable) function by a polynomial. More precisely, we will perform the approxima-
tion by
n
X f (k) (x0 ) f 00 (x0 ) 2 f 000 (x0 ) 3 f (n) (x0 ) n
f (x0 + h) ≈ hk = f (x0 ) + f 0 (x0 )h + h + h +...+ h .
k=0
k! 2 6 n!

We will also estimate the approximation error.

Theorem 5.30. Taylor’s formula
Let I be an interval and assume that f : I → R is n + 1-times differentiable. Let
x0 ∈ I and h ∈ R such that x0 + h ∈ I. Then there exists some θ ∈ (0, 1) such that
n
X f (k) (x0 ) f (n+1) (x0 + θh) n+1
f (x0 + h) = hk + h .
k! (n + 1)!
|k=0 {z } | {z }
“remainder term”
“Taylor polynomial”

The number x0 is called expansion point.

5.4 Taylor’s Formula 117

Proof: Remember the generalised mean value theorem

F (x1 ) − F (x0 ) F 0 (x0 + θ(x1 − x0 ))

= 0
g(x1 ) − g(x0 ) g (x0 + θ(x1 − x0 ))

for some θ ∈ (0, 1), which can be reformulated as (with h = x1 − x0 )

g(x0 + h) − g(x0 )
F (x0 + h) − F (x0 ) = · F 0 (x0 + θh). (5.3)
g 0 (x0 + θh)

Now consider the functions

n
X f (k) (x)
F (x) := (x0 + h − x)k , g(x) := (x0 + h − x)n+1 .
k=0
k!

Then we have
n n
0
X f (k+1) (x) X
k f (k) (x) f (n+1) (x)
F (x) = (x0 + h − x) − (x0 + h − x)k−1 = (x0 + h − x)n .
k=0
k! k=1
(k − 1)! n!

and
g 0 (x) = −(n + 1)(x0 + h − x)n .

Moreover, we have
n
X f (k) (x0 )
F (x0 ) = hk , F (x0 + h) = f (x0 + h), g(x0 ) = hn+1 , g(x0 + h) = 0.
k=0
k!

Then using (5.3), we obtain

n
X f (k) (x0 )
f (x0 + h) − hk
k=0
k!
=F (x0 + h) − F (x0 )
g(x0 + h) − g(x0 )
= · F 0 (x0 + θh)
g 0 (x0 + θh)
−hn+1 f (n+1) (x0 + θh)
= · ((1 − θ)h)n
−(n + 1)((1 − θ)h)n n!
f (n+1) (x0 + θh) n+1
= h .
(n + 1)!

This implies Taylor’s formula. 2

Note that, by the substitution h := x − x0 , Taylor’s formula can also be written as follows:

Theorem 5.31. Taylor’s formula, alternative version

Let I be an interval and assume that f : I → R is n + 1-times differentiable. Let
118 5 Differentiation of Functions

x0 , x ∈ I. Then there exists some x̂ between x0 and x such that

n
X f (k) (x0 ) f (n+1) (x̂)
f (x) = (x − x0 )k + (x − x0 )n+1 .
k! (n + 1)!
|k=0 {z } | {z }
=:Rn (x,x0 )
=:Tn (x,x0 )

The application of Taylor’s formula is twofold: First, it gives a polynomial that approx-
imates a given function quite fine in some neighbourhood. The second application is the
computation of values of “complicated functions”. We will present examples for both kinds
of application.

Example 5.32. Consider the function f (x) = sin(x). We want to determine the Taylor
polynomial of degree 6 with expansion point x0 = π2 . Since we have

sin0 (x) = cos(x), sin00 (x) = − sin(x), sin(3) (x) = − cos(x),

sin(4) (x) = sin(x), sin(5) (x) = cos(x), sin(6) (x) = − sin(x)
and π  π 
sin(x0 ) = sin = 1, cos(x0 ) = cos
= 0,
2 2
the Taylor polynomial of degree 6 is given by
1 π 2 1  π 4 1  π 6
T6 (x) = 1 − x− + x− − x− .
2 2 24 2 720 2
The remainder term reads
sin(7) (x̂)  π 7 − cos(x̂)  π 7
R6 (x, x0 ) = x− = x− .
7! 2 5040 2
Taking into account that | cos(x)| ≤ 1 for all x ∈ R, we have that
x − π 7
 
2
|R6 (x, x0 )| ≤ .
5040
This leads to the estimate
 x − π 7
 
2
| sin(x) − T6 (x)| = |R6 (x, x0 )| ≤ .
5040

Example 5.33. We want to compute log(1.2) up to 3 digit precision. A nice expansion

point is x0 = 1 since we know the precise values of log(k) (1). Consider
1 1!
log0 (x) =
, log00 (x) = − 2 ,
x x
(3) 2! (4) 3!
log (x) = 3 , log (x) = − 4
x x
and therefore, the Taylor polynomial of degree 3 is given by
1 1
T3 (x) = (x − 1) − (x − 1)2 + (x − 1)3 .
2 3
5.4 Taylor’s Formula 119

3
sin(x)
T6(x)

−1

−2

−3
−3 −2 −1 0 1 2 3 4 5 6

Figure 5.6: Taylor approximation of sin

In particular, we have
1 1 137
T3 (1.2) = 0.2 − · 0.22 + · 0.23 = .
2 3 750
Now we estimate | log(1.2) − 137
750
|: The remainder term is given by

3! (x − x0 )4 (x − x0 )4
R3 (x, x0 ) = − = −
x̂4 4! 4x̂4
for some x̂ between x and x0 . For x = 1.2, x0 = 1 we have 1 < x̂ < 1.2 and therefore

(0.2)4 1
|R3 (1.2, 1)| = 4
= 4 · 10−4 4 ≤ 4 · 10−4 .
4x̂ x̂
This leads to
| log(1.2) − 0.1826| = |R3 (1.2, 1)| ≤ 4 · 10−4 ,
so we have determined log(1.2) up to three digits.

Theorem 5.34.
Let I ⊂ R be an open interval, n ∈ N and f : I → R an n-times continuously
differentiable function. Suppose that for a ∈ I holds

f 0 (a) = f 00 (a) = ... = f (n−1) (a) = 0 and f (n) (a) 6= 0 .

If n is odd, then a is not a local extremum. If n is even and f (n) (a) > 0, then a is
a local minimum. If n is even and f (n) (a) < 0, then a is a local maximum.
120 5 Differentiation of Functions

Proof: By assumption the Taylor expansion of f of degree n − 1 in the expansion point a

reads:
f n (z)
f (x) = f (a) + (x − a)n (5.4)
n!
where z = z(x) lies between x and a. Since f (n) is continuous and f (n) (a) 6= 0, there is a
neighbourhood U := (a − ε, a + ε) ⊂ I, ε > 0, of a such that f (n) (x) 6= 0 for all x ∈ U .
This means that for all x ∈ U , f (n) (x) and f (n) (a) have the same sign. Then for any
xl ∈ (a − ε, a) and any xr ∈ (a, a + ε) Equation (5.4) implies that for zl := z(xl ) ∈ (xl , a)
and zr := z(xr ) ∈ (a, xr ) holds

f n (zl )
f (xl ) = f (a) + (xl − a)n ,
n!
f n (zr )
f (xr ) = f (a) + (xr − a)n ,
n!
and 0 6= f n (a), f n (zl ), f n (zr ) have the same sign. If n is odd, then (xl −a)n < 0 < (xr −a)n
and therefore either f (xl ) < f (a) < f (xr ) or f (xl ) > f (a) > f (xr ) so that a is not
a local extremum. If n is even and f (n) (a) > 0, then (xl − a)n , (xr − a)n > 0 and
f (xl ), f (xr ) > f (a) so that a is a local minimum. Finally, if n is even and f (n) (a) < 0,
then (xl − a)n , (xr − a)n > 0 and f (xl ), f (xr ) < f (a) so that a is a local maximum. 2

5.5 Simple methods for the numerical solution of

equations
One of the most basic concerns of mathematics is solving equations. For example, if
f : R → R is a given function and y ∈ R is a given number, then the question arises
whether or not the equation
f (x) = y (5.5)
can be solved for x. In practice the function f can be very complicated and a solution of
(5.5) can in general not be found by simple arithmetical transformations. In such cases
numerical procedures/algorithms might still be applicable to find approximate solutions
x̂ that fulfill f (x̂) ≈ y. In practical applications such approximate solutions are sufficient
if |f (x̂) − y| ≤ ε, where ε > 0 is some “acceptable” error bound. In this section we will
present two common basic numerical methods to determine such approximate solutions.
First of all note that the original Problem 5.5 can be transformed to

f˜(x) := f (x) − y + x = x (5.6)

or to
f˜(x) := f (x) − y = 0 . (5.7)

Thus solving Equation (5.6) means finding a fixed-point of f˜ and (5.7) means finding a
zero of f˜. These are in some kind the most common normalized formulations for “solving”
an equation. First we will present a simple numerical method for finding a fixed point of a
given function based on Banach’s fixed-point theorem. Afterwards we will state Newton’s
method for finding a zero of a given differentiable function.
5.5 Simple methods for the numerical solution of equations 121

Theorem 5.35. Weissinger fixed-point theorem

Let (X, d) be a complete metric space
P and f : X → X be a function on X. Assume
that there is a convergent series ∞n=0 an with an ≥ 0 such that

d(xn , yn ) ≤ an · d(x0 , y0 )

for arbitrary x0 , y0 ∈ X and recursively defined sequences

xn+1 := f (xn ) , yn+1 := f (yn ) , n ∈ N0 .

Then f possesses exactly one fixed point z ∈ X and for an arbitrary starting point
z0 ∈ X the recursively defined sequence zn+1 := f (zn ), n ∈ N0 , converges to z.
Moreover, the following error estimates hold:
∞
X
d(z, zk ) ≤ an · d(z0 , z1 ) (a priori estimate) (5.8)
n=k
∞
X
d(z, zk ) ≤ an · d(zk−1 , zk ) (a posteriori estimate) (5.9)
n=1

for all k ∈ N.

Proof: First of all we show that f is continuous. Let x ∈ X, ε > 0 and set δ := ε
a1 +1
> 0.
Then for y ∈ X with d(x, y) < δ holds by assumption
ε
d(f (x), f (y)) ≤ a1 · d(x, y) ≤ a1 · <ε.
a1 + 1
From the ε-δ-criterion it follows that f is continuous in x.

Now let x0 ∈ X be arbitrary and xn+1 := f (xn ), n ∈ N0 . Then for y0 := x1 and

yn+1 := f (yn ), n ∈ N0 , we have yn = xn+1 for all n ∈ N0 and by assumption

d(xn , xn+1 ) = d(xn , yn ) ≤ an · d(x0 , y0 ) = an · d(x0 , x1 ),

for
P∞all n ∈ N. We will show now that (xn )n∈N is a Cauchy sequence. Let ε > 0. Since
n=0 an converges, there is an N ∈ N such that for all k, m > N with k ≥ m holds

m
X ε
an < .
n=k
d(x0 , x1 ) + 1

Using the triangle inequality, for such k, m holds:

m−1
X m−1
X m−1
X
d(xk , xm ) ≤ d(xn , xn+1 ) = d(xn , yn ) ≤ an · d(x0 , x1 ) < ε .
triangle ineq.
n=k n=k n=k

Thus (xn )n∈N is a Cauchy sequence. Since (X, d) is a complete, (xn )n∈N converges to some
limit x and since f is continuous we conclude

x = lim xn = lim xn+1 = lim f (xn ) = f ( lim xn ) = f (x) ,

n→∞ n→∞ n→∞ n→∞
122 5 Differentiation of Functions

i.e. x is a fixed point of f . For any other starting point z0 ∈ X and recursively defined
zn+1 := f (zn ), n ∈ N0 , we obtain by the previous that z := limn→∞ zn is also a fixed
point of f . But the assumption applied to u0 := x, v0 := z, un+1 := f (un ) = x,
vn+1 := f (vn ) = z for n ∈ N0 yields
0 ≤ d(x, z) = d(un , vn ) ≤ an · d(u0 , v0 ) = an · d(x, z).
Since (an )n∈N0 converges to zero, the right-hand side converges to zero which implies
d(x, z) = 0, i.e. x = z. Thus f has a unique fixed point z ∈ X.
Next we derive (5.8). Let k ∈ N. For arbitrary m ≥ k holds
m−1
X
d(z, zk ) ≤ d(z, zm ) + d(zm , zk ) ≤ d(z, zm ) + d(zn , zn+1 )
n=k
m−1 ∞
m→∞
X X
≤ d(z, zm ) + an · d(z0 , z1 ) −−−→ an · d(z0 , z1 ) .
| {z }
−→ 0
m→∞ |n=k
{z } n=k
P ∞
−→ n=k an
m→∞

Finally, if we define z̃0 := zk−1 and z̃j+1 := f (z̃j ) = zk+j , j ∈ N0 , (5.9) follows from (5.8)
applied to the sequence (z̃j )j∈N0 for j := 1, namely:
∞
X ∞
X
d(z, zk ) = d(z, z̃1 ) ≤ an · d(z̃0 , z̃1 ) = an · d(zk−1 , zk ) .
n=1 n=1

The so-called Banach fixed-point theorem is a special case of Weissinger’s fixed point
theorem.
Theorem 5.36. Banach fixed-point theorem
Let (X, d) be a complete metric space and f : X → X be a function on X such that
d(f (x), f (y)) ≤ q · d(x, y) for all x, y ∈ X, where q ∈ [0, 1) is a fixed nonnegative
constant less then one a . Then f has exactly one fixed point z ∈ X and for an
arbitrary z0 ∈ X the recursively defined sequence zn+1 := f (zn ), n ∈ N0 , converges
to z. Moreover the following error estimates hold for k ∈ N:

qk
d(z, zk ) ≤ · d(z0 , z1 ) (a priori estimate) (5.10)
1−q
q
d(z, zk ) ≤ · d(zk−1 , zk ) (a posteriori estimate) (5.11)
1−q
a
Functions with this property are called contractions and q is called a contraction constant for f .

Proof: With an := q n for n ∈ N0 the asumptions of Weissinger’s fixed point theorem are
fulfilled. The estimates (5.10) and (5.11) follow directly from (5.8) and (5.9) respectively.
2

We want to reformulate Banach’s fixed-point theorem for the special case where X is
a closed subset of R and d is the Euclidean metric on X, i.e. d(x, y) := |x − y| for
x, y ∈ X. Recall that in this case (X, d) is complete.
5.5 Simple methods for the numerical solution of equations 123

Theorem 5.37.
Let X ⊂ R be closed and f : X → X be a function such that |f (x)−f (y)| ≤ q·|x−y|
for all x, y ∈ X, where q ∈ [0, 1) is a fixed nonnegative constant less then one, i.e. f
is a contraction on X with contraction constant q. Then f has exactly one fixed point
z ∈ X and for an arbitrary z0 ∈ X the recursively defined sequence zn+1 := f (zn ),
n ∈ N0 , converges to z. Moreover the following error estimates hold for k ∈ N:

qk
|z − zk | ≤ |z1 − z0 | (a priori estimate) (5.12)
1−q
q
|z − zk | ≤ |zk−1 − zk | (a posteriori estimate) (5.13)
1−q

In practical applications the function f is given but in order to apply Theorem 5.37 an
appropriate domain X with f (X) ⊂ X and a contraction constant q ∈ [0, 1) must be
determined. In practise a closed area X is guessed where a fixed-point of a given function
f might be located. Then f (X) ⊂ X must be verified and a contraction constant must
be found. If this is not possible, the guessed domain X must be changed. The following
Theorem states a standard procedure for finding an contraction constant q on a given
closed interval domain X by using an upper bound of the first derivative of f , which
requires that f is continuously differentiable on X.
Theorem 5.38.
Let X ⊂ R be a closed interval and f : X → X be continuously differentiable on
X. If q := ||f 0 ||∞ = sup{|f 0 (x)| | x ∈ X} < 1, then f is a contraction on X
with contraction constant q. In particular, by Banach’s fixed-point theorem, f has
exactly one fixed point in X and for an arbitrary z0 ∈ X the recursively defined
sequence zn+1 := f (zn ), n ∈ N0 , converges to z and the error estimates (5.12) and
(5.13) hold.

Proof: Let x, y ∈ X with x < y. By the mean value theorem there is a ξ ∈ (x, y) such
that
|f (x) − f (y)| = |f 0 (ξ) · (y − x)| = |f 0 (ξ)| · |(y − x)| ≤ q · |x − y|.

This implies that f is a contraction on X with contraction constant q and the conclusion
follows from Theorem 5.37 . 2

Now we will state Newton’s method for finding zeros of differentiable functions f . New-
ton’s method is based on the following simple iteration principle: If x0 is an approximate
solution of f (x) = 0, then f is replaced in a vicinity of x0 by the tangent of f in x0 ,
namely by
t(x) := f 0 (x0 )(x − x0 ) + f (x0 ) .

Then, if f 0 (x0 ) 6= 0, the solution x1 of t(x) = 0, which is

f (x0 )
x1 := x0 − ,
f 0 (x0 )

is taken as an improved approximate solution of f (x) = 0.

124 5 Differentiation of Functions

15

f(x)
t(x)

10

5
y

0
x1 x0

−5
−1 0 1 2 3 4 5 6 7 8
x

Figure 5.7: Newton’s method

This procedure generates a sequence

f (xn )
xn+1 := xn − , n ∈ N0
f 0 (xn )
a so-called Newton iteration, which for a general differentiable function f and an arbitrary
starting point x0 in the domain of f might not be well-defined or might not converge to
a root of f .
The following Theorem gives sufficient conditions that assure that the Newton iteration
converges to a root of f .
Theorem 5.39.
Suppose that f : [a, b] → R is two times differentiable and convex with f (a) < 0 and
f (b) > 0. Then the following holds:
a) f (ξ) = 0 for exactly one ξ ∈ (a, b) .
b) For arbitrary x0 ∈ [a, b] with f (x0 ) ≥ 0 the Newton iteration
f (xn )
xn+1 := xn − , n ∈ N0
f 0 (xn )
is well-defined, monotonically decreasing and converges to ξ.
c) If there are nonnegative constants C, K such that f 0 (ξ) ≥ C > 0 and
f 00 (x) ≤ K for all x ∈ (ξ, b) , then the following error estimate holds for
n ∈ N:
K
|xn+1 − xn | ≤ |ξ − xn | ≤ · |xn − xn−1 |2 .
2C
5.5 Simple methods for the numerical solution of equations 125

Proof: a) By the intermediate value Theorem f has a root ξ ∈ (a, b). In order to prove
uniqueness, we assume that f has two distinct roots ξ, η ∈ (a, b) with ξ < η. Since f is
convex, for x1 := a < x := ξ < x2 := η holds

f (x2 ) − f (x1 ) f (η) − f (a)

0 = f (ξ) = f (x) ≤ (x − x1 ) + f (x1 ) = (ξ − a) + f (a)
x2 − x1 η−a
−f (a) ξ−a
= (ξ − a) + f (a) = (1 − ) · f (a) < 0 ,
η−a η−a
| {z }
<1

a contradiction. Therefore, f has a unique root ξ in (a, b) .

b) By the mean value theorem there exists an η ∈ (a, ξ) such that

f (ξ) − f (a) −f (a)

f 0 (η) = = >0.
ξ−a ξ−a

Since f is convex, f 0 is monotonically increasing so that for all x ∈ [ξ, b] holds f 0 (x) ≥
f 0 (η) > 0, i.e. f 0 is positive on [ξ, b]. In particular, f is strictly monotonically increasing
on [ξ, b] so that f (x) > f (ξ) = 0 for all x ∈ (ξ, b].

Now let x0 ∈ [a, b] with f (x0 ) ≥ 0. Then necessarily x0 ≥ ξ. (Otherwise, if x0 < ξ,

then f (a) < 0 ≤ f (x0 ) would imply that f has another root in [a, x0 ] which is not
possible due to a).) By the previous f 0 (x0 ) > 0 and therefore

f (x0 )
x1 := x0 −
f 0 (x0 )

is well-defined. We know that x1 is the root of the tangent

t(x) := f 0 (x0 )(x − x0 ) + f (x0 ) .

Since t(x0 ) = f (x0 ) ≥ 0 and t0 (x0 ) = f 0 (x0 ) > 0 we immediately see that xn+1 ≤ x0 .
Moreover, (f − t)(x0 ) = 0 and (f − t)0 (x) = f 0 (x) − f 0 (x0 ) ≤ 0 for x ≤ x0 imply
f (x) ≥ t(x) for x ≤ x0 . In particular t(ξ) ≤ f (ξ) = 0 ≤ f (x0 ) = t(x0 ) and therefore
ξ ≤ x1 ≤ x0 as t(x) is a straight line. Replacing x0 by x1 we conclude inductively that
the sequence
f (xn )
xn+1 := xn − 0 , n ∈ N0 ,
f (xn )
is well-defined, monotonically decreasing and bounded from below by ξ. Thus

η := lim xn
n→∞

exists and fulfils ξ ≤ η. Since f 0 is bounded from below on [ξ, b] by some positive constant,
we also have
f (η)
η=η− 0
f (η)
which implies f (η) = 0. By a) this proves η = ξ.
126 5 Differentiation of Functions

c) Since f 0 is monotonically increasing we have f 0 (x) ≥ f 0 (ξ) ≥ C > 0 for all x ∈ [ξ, b].
Therefore, f (x) ≥ C(x − ξ) for all x ∈ [ξ, b]. In particular, this implies

f (xn )
|xn − ξ| = xn − ξ ≤ for all n ∈ N.
C
In order to estimate f (xn ) the following function g(x) is considered:

K
g(x) := f (x) − f (xn−1 ) − f 0 (xn−1 )(x − xn−1 ) − (x − xn−1 )2
2
g 0 (x) = f 0 (x) − f 0 (xn−1 ) − K(x − xn−1 )
g 00 (x) = f 00 (x) − K ≤ 0 for x ∈ (ξ, b).

Thus g 0 is monotonically decreasing on [ξ, b]. Since g 0 (xn−1 ) = 0, this implies g 0 (x) ≥ 0
for x ∈ [ξ, xn−1 ]. Since g(xn−1 ) = 0, this implies g(x) ≤ 0 for x ∈ [ξ, xn−1 ]. In particular,

K
0 ≥ g(xn ) = f (xn ) − f (xn−1 ) − f 0 (xn−1 )(xn − xn−1 ) − (xn − xn−1 )2
2
K
= f (xn ) − f (xn−1 ) + f (xn−1 ) − (xn − xn−1 )2
2
K
= f (xn ) − (xn − xn−1 )2 ,
2
that is
K
f (xn ) ≤ (xn − xn−1 )2
2
and we conclude
f (xn ) K
|xn − ξ| ≤ ≤ |xn − xn−1 |2 .
C 2C
Finally, by b) (xn )n∈N0 decreases monotonically with limit ξ which directly yields |xn+1 −
xn | ≤ |ξ − xn |.
2
Remark:
a) The error estimate given in Theorem 5.39 c) says that Newton’s method (loc-
ally) converges quadratically. If f 0 is bounded from below by some c1 > 0 and
if |f 00 | is bounded from above by some c2 , then we may choose C := c1 and
K := c2 .

b) Analog formulations of Theorem 5.39 hold if the function f is concave or if

f (a) > 0 and f (b) < 0.
 The Riemann Integral
6
Prince George: Someone said I had the wit and intellect of a donkey.
Blackadder: Oh, an absurd suggestion sir. Unless it was a particularly
stupid donkey.
Rowan Atkinson, Blackadder

Definition 6.1.
Let [a, b] ⊂ R. A set {x0 , x1 , . . . , xn } is called a decomposition or partition of [a, b]
if
a = x0 < x1 < x2 < . . . < xn−1 < xn = b.

a x1 x2 x3 x4 ··· ··· xn−1 b

Definition 6.2. Step function

f : [a, b] → R is called a step function if it is piecewisely constant, i.e. there exists
a decomposition {x0 , . . . , xn } of [a, b] and some c1 , . . . , cn ∈ R such that for all
i = 1, . . . , n holds
f (x) = ci for all x ∈ (xi−1 , xi ).
The set of step functions on [a, b] is denoted by T ([a, b]).

127
128 6 The Riemann Integral

a b

It can be readily verified that for two step functions f1 , f2 ∈ T ([a, b]) holds f1 + f2 ∈
T ([a, b]). As well, we have λf1 ∈ T ([a, b]). Hence, T ([a, b]) is a vector space. Furthermore,
since step functions only attain finitely many values, they are bounded.
Definition 6.3. Integral of step functions
Let φ ∈ T ([a, b]) and a decomposition {x0 , . . . , xn } of [a, b] be given such that for
all i = 1, . . . , n
φ(x) = ci for all x ∈ (xi−1 , xi ).
Then the integral of φ is defined as
Z b n
X
φ(x) dx := cj (xj − xj−1 ).
a j=1

However, to be sure that the integral is well-defined we need additionally that it is inde-
pendent of the special choice of a decomposition.
Lemma 6.4.
Rb Rb
a
φ(x) dx is independent of the choice of the decomposition, i.e., a φ(x) dx is well-
defined for all φ ∈ T ([a, b]).

Proof: Let
Z1 : a = x0 < x1 < x2 < . . . < xn−1 < xn = b,
Z2 : a = y0 < y1 < y2 < . . . < ym−1 < ym = b
be two decompositions of [a, b] such that φ is constant on (xi−1 , xi ) and φ is constant
on (yj−1 , yj ) for all i = 1, . . . , n, j = 1, . . . , m. We distinguish between two cases: 1st
Case: Z2 is a refinement of Z1 . This means that for all i ∈ {1, . . . , n} there exists some
j(i) ∈ {1, . . . , m} with xi = yj(i) . Then for i holds

xi = yj(i) < yj(i)+1 < . . . < yj(i+1) = xi+1

 129

and φ(x) = dk = ci for x ∈ (yk−1 , yk ), k ∈ {j(i − 1) + 1, . . . , j(i)}. This implies

m n j(i)
X X X
dj (yj − yj−1 ) = ci (yk − yk−1 )
j=1 i=1 k=j(i−1)+1

n j(i)
X X
= ci (yk − yk−1 )
i=1 k=j(i−1)+1
n
X n
X
= ci (yj(i) − yj(i−1) ) = ci (xi − xi−1 ).
i=1 i=1

2nd Case: There exists a common refinement Z3 of Z1 and Z2 , i.e., Z3 = Z1 ∪ Z2 . Then

we have (by using the 1st case)
X X X
... = ... = ... .
Z1 Z3 Z2

2
The integral can be seen as a mapping from the space T ([a, b]) to R. In the literature,
mappings from vector spaces to a field (in this case R) are called functionals. The following
result shows that the integral is a linear and monotone functional.
Theorem 6.5.
Rb
a
: T ([a, b]) → R is linear and monotonic, that is, for all φ, ψ ∈ T ([a, b]) and all
λ, µ ∈ R holds

(i)
Z b  Z b Z b
λφ(x) + µψ(x) dx = λ φ(x) dx + µ ψ(x) dx
a a a

(ii) if φ(x) ≤ ψ(x) for all x ∈ [a, b] (φ ≤ ψ) , then

Z b Z b
φ(x) dx ≤ ψ(x) dx.
a a

Proof: This directly follows by the definition of the integral. 2

Now we will define the integral for bounded functions:
Definition 6.6.
Let f : [a, b] → R be bounded. Then we define the Riemann upper integral
Z b Z b 
f (x) dx := inf φ(x) dx : φ ∈ T ([a, b]) with φ ≥ f
a a

and the Riemann lower integral by

Z b Z b 
f (x) dx := sup φ(x) dx : φ ∈ T ([a, b]) with φ ≤ f .
a a
130 6 The Riemann Integral

a b

Figure 6.1: approximation for the lower integral

a b

Figure 6.2: approximation for the upper integral

It can be seen from Fig. 6.1 and Fig. 6.2 that the integral can be seen as the “signed area”
between the function graph and the x-axis. “Signed” means that the area of the negative
parts of the function has to be counted negative.
By the above definition, we can directly deduce that for φ ∈ T ([a, b]) holds
Z b Z b Z b
φ(x) dx = φ(x) dx = φ(x) dx.
a a a

For general bounded functions f : [a, b] → R holds

Z b Z b
f (x) dx ≤ f (x) dx.
a a

Note that in general the upper integral does not coincide with the lower integral. For
instance, consider the function f : [0, 1] → R with
(
1 : x∈Q
f (x) =
0 : x ∈ R\Q
 131

Then we have
Z 1 Z 1
f (x) dx = 1 > 0 = f (x) dx.
0 0

Theorem 6.7.
Let f, g : [a, b] → R be bounded. Then

(i)
Z b Z b Z b
f (x) + g(x) dx ≤ f (x) dx + g(x) dx.
a a a

(ii)
Z b Z b Z b
f (x) + g(x) dx ≥ f (x) dx + g(x) dx.
a a a

(iii) For λ ≥ 0 holds

Z b Z b Z b Z b
λf (x) dx = λ f (x) dx, λf (x) dx = λ f (x) dx.
a a a a

(iv) For λ ≤ 0 holds

Z b Z b Z b Z b
λf (x) dx = λ f (x) dx, λf (x) dx = λ f (x) dx.
a a a a

Proof:
(i) Let ε > 0. Then there exist φ, ψ ∈ T ([a, b]) with φ ≥ f , ψ ≥ g and
Z b Z b Z b Z b
ε ε
f (x) dx + ≥ φ(x) dx, g(x) dx + ≥ ψ(x) dx.
a 2 a a 2 a

Then f + g ≤ φ + ψ and
Z b Z b 
f (x) + g(x) dx = inf ζ(x) dx : ζ ∈ T ([a, b]) with ζ ≥ f + g
a a
Z b
≤ φ(x) + ψ(x) dx
a
Z b ! Z b !
ε ε
≤ f (x) dx + + g(x) dx +
a 2 a 2
Z b Z b
= f (x) dx + g(x) dx + ε.
a a

Since this holds true for all ε > 0, we conclude

Z b Z b Z b
f (x) + g(x) dx ≤ f (x) dx + g(x) dx.
a a a
132 6 The Riemann Integral

(ii): Analogous to (i).

(iii): We only show the statement for the upper integral. The other result can be shown
analogously.
Let ε > 0. Then there exists some φ ∈ T ([a, b]) with φ ≥ f and
Z b Z b
f (x) dx + ε ≥ φ(x) dx.
a a

Then λφ ≥ λf and
Z b Z b Z b Z b
λf (x) dx ≤ λφ(x) dx = λ φ(x) dx ≤ λ f (x) dx + λε.
a a a a

Since this holds true for all ε > 0, we conclude

Z b Z b
λf (x) dx ≤ λ f (x) dx.
a a

The opposite inequality follows from the previous one applied to g(x) := λf (x) and
µ := λ1 > 0:
Z b Z b Z b Z b
1
f (x) dx = µg(x) dx ≤ µ g(x) dx = λf (x) dx.
a a a λ a

Multiplying this inequality by λ > 0 yields

Z b Z b
λ f (x) dx ≤ λf (x) dx.
a a

(iv): This result can be shown by using (iii) and the fact
Z b Z b
− f (x) dx = − f (x) dx.
a a

2
Definition 6.8.
A bounded function f : [a, b] → R is called Riemann-integrable if
Z b Z b
f (x) dx = f (x) dx.
a a

In this case, we write Z b Z b

f (x) dx = f (x) dx.
a a

The set of Riemann-integrable functions is denoted by R([a, b]).

We obviously have that T ([a, b]) ⊂ R([a, b]). In the following we state that monotonic
functions as well as continuous functions are Riemann-integrable. The proof is not presen-
ted here.
 133

Remark:
As it holds true for summation, the integration variable can be renamed without
changing the integral. That is, for f ∈ R([a, b]) holds
Z b Z b
f (x) dx = f (t) dt.
a a

Theorem 6.9.
Let f : [a, b] → R be continuous or monotonic. Then f is Riemann-integrable.

The following result is straightforward:

Theorem 6.10.
R([a, b]) is a real vector space and the mapping
Z b
: R([a, b]) → R
a

is linear and monotonic.

So far we did not compute any integrals. We will now give an example of an integral that
will be computed according to the definition. This will turn out to be really exhausting
even for this quite simple example.

RExample
1 R 1 Consider the function f : [0, 1] → R with f (x) = x. Determine
6.11.
0
f (x) dx = 0 x dx. First we consider two sequences of step functions (φn )n∈N , (ψn )n∈N
with
 
k−1 k−1 k
φn (x) = for x ∈ , , k ∈ {1, . . . , n},
n n n
 
k k−1 k
ψn (x) = for x ∈ , , k ∈ {1, . . . , n}.
n n n

Then for all n ∈ N holds φn ≤ f ≤ ψn . Now we calculate

1 n  
k−1 k k−1
Z X
φn (x) dx = · −
0 k=1
n n n
n
X k−1 1
= ·
k=1
n n
n
1 X
= 2 (k − 1)
n k=1
1 n(n − 1) 1 1
= 2
· = −
n 2 2 2n
134 6 The Riemann Integral

and n
1  
k k−1
Z X k
ψn (x) dx = · −
0 k=1
n n n
n
X k 1
= ·
k=1
n n
n
1 X
= 2 k
n k=1
1 n(n + 1) 1 1
2
· = = + .
n 2 2 2n
In particular, we have for all n ∈ N that
Z 1 Z 1 Z 1
1 1 1 1
− = φn (x) dx ≤ x dx ≤ ψn (x) dx = +
2 2n 0 0 0 2 2n
and thus Z 1
1
x dx = .
0 2

By the definition of the integral, it is not difficult to obtain that for f ∈ R([a, b]) and
c ∈ (a, b) holds
Z b Z c Z b
f (x) dx = f (x) dx + f (x) dx.
a a c
To make this formula also valid for c ≥ b or c ≤ a, we define for a ≥ b that
Z b Z a
f (x) dx := − f (x) dx.
a b

Now we present the mean value theorem of integration and its manifold consequences.
Theorem 6.12. Mean Value Theorem of Integration
Let f, g : [a, b] → R be continuous and let g(x) ≥ 0 for all x ∈ [a, b] (i.e., g ≥ 0).
Then there exists some x̂ ∈ [a, b] such that
Z b Z b
f (x)g(x) dx = f (x̂) · g(x) dx.
a a

Proof: Let

m = min{f (x) : x ∈ [a, b]}, M = max{f (x) : x ∈ [a, b]}.

Since g ≥ 0, for all x ∈ [a, b] holds mg(x) ≤ f (x)g(x) ≤ M g(x) and, by the monotonicity
of the integral holds
Z b Z b Z b Z b Z b
m g(x) dx = mg(x) dx ≤ f (x)g(x) dx ≤ M g(x) dx ≤ M g(x) dx.
a a a a a

Then there exists some m ≤ µ ≤ M such that

Z b Z b
µ g(x) dx = f (x)g(x) dx
a a
6.1 Differentiation and Integration 135

Since
min{f (x) : x ∈ [a, b]} ≤ µ ≤ max{f (x) : x ∈ [a, b]},
the intermediate value theorem implies that there exists some x̂ ∈ [a, b] with µ = f (x̂)
and thus Z b Z b
f (x̂) g(x) dx = f (x)g(x) dx.
a a
2
Corollary 6.13.
Let f : [a, b] → R be continuous. Then there exists some x̂ ∈ [a, b] such that
Z b
f (x) dx = f (x̂) · (b − a).
a

Proof: Apply the mean value theorem to g = 1. Observing that

Z b Z b
g(x) dx = 1 dx = b − a,
a a
the result then follows immediately. 2

6.1 Differentiation and Integration

Definition 6.14.
Let I be an interval and f : I → R be continuous. Then a differentiable function
F : I → R is called an antiderivative of f if F 0 = f .

Theorem 6.15.
Let I be an interval, f : I → R be continuous and a ∈ I. For x ∈ I define
Z x
F (x) = f (ξ)dξ.
a

Then F is differentiable and an antiderivative of f .

Proof: Let x ∈ I and h 6= 0 such that x + h ∈ I. Then, by using the mean value theorem
of integration we obtain
Z x+h Z x 
1 1
(F (x + h) − F (x)) = f (ξ)dξ − f (ξ)dξ
h h a a
1 x+h
Z
1
= f (ξ)dξ = · hf (x̂) = f (x̂)
h x h
for some x̂ between x and x + h. If h tends to 0 then x̂ → x and thus
1
lim (F (x + h) − F (x)) = f (x).
h→0 h

This shows the desired result. 2

Now we consider how two antiderivatives of a given continuous f : I → R differ.
136 6 The Riemann Integral

Theorem 6.16.
Let I be an interval, f : I → R be given and let F : I → R be an antiderivative of
f , i.e., F 0 = f . Then G : I → R is an antiderivative of f if and only if F − G is
constant.

Proof: “⇒”: Let G : I → R be an antiderivative of f . Then

(F − G)0 = F 0 − G0 = f − f = 0
and hence, F − G is constant due to the mean value theorem of differentiation.
“⇐”: If F − G is constant, i.e., F (x) − G(x) = c for some c ∈ R and all x ∈ I, then
0 = (F − G)0 = F 0 − G0 and thus f = F 0 = G0 . 2
Remark:
It is very important to note that the statement of Theorem 6.16 is only valid for
functions defined on intervals. For instance, consider the function f : R\{0} → R
with f (x) = x1 . An antiderivative is given by F : R\{0} → R with
(
log(x) : x > 0,
F (x) = = log(|x|).
log(−x) : x < 0

Another antiderivative is given by

(
log(x) : x > 0,
G(x) = .
log(−x) + 1 : x < 0

The difference between G and F is given by

(
0 : x > 0,
G(x) − F (x) =
1 :x<0

and therefore not constant.

The next result now states that integrals can be determined by inversion of differentiation.

Theorem 6.17. Fundamental theorem of differentiation and integration

Let I be an interval and a continuous f : I → R be given. Let F : I → R be an
antiderivative of f . Then for all a, b ∈ I holds
Z b
f (x) dx = F (b) − F (a).
a

We write Z b
f (x) dx = F (x)|x=b
x=a .
a

Proof: Consider the function F0 : I → R defined by

Z x
F0 (x) = f (ξ)dξ.
a
6.1 Differentiation and Integration 137

By Theorem
R a 6.15, we know that F0 is an antiderivative of f .R bIn particular,
R b we have that
F0 (a) = a f (ξ)dξ = 0 and thus F0 (b) − F0 (a) = F0 (b) = a f (ξ)dξ = a f (x) dx. Let
F I :→ R be an antiderivative of f . Theorem 6.16 now implies that there exists some
c ∈ R with F (x) = F0 (x) + c for all x ∈ I. Therefore
Z b
F (b) − F (a) = (F0 (b) + c) − (F0 (a) + c) = F0 (b) − F0 (a) = f (x) dx.
a

2
The above result gives rise to the following notation for an antiderivative:
Z
f (x) dx := F (x).

Based on our knowledge about differentiation, we now collect some antiderivatives of

important functions in Table 6.1.

R
f (x) f (x) dx
1
xn , n∈N xn+1
n+1
x−1 , x 6= 0 log(|x|)
1
x−n , x 6= 0, n ∈ N, n 6= 1 x1−n
1−n
exp(x) exp(x)
sinh(x) cosh(x)
cosh(x) sinh(x)
1
√ arsinh(x)
1 + x2
1
√ , x>1 arcosh(x)
x2 − 1
1
, |x| < 1 artanh(x)
1 − x2
sin(x) − cos(x)
cos(x) sin(x)
1
2
= 1 + tan2 (x) tan(x)
cos (x)
1
√ , |x| < 1 arcsin(x)
1 − x2
1
−√ , |x| < 1 arccos(x)
1 − x2
1
arctan(x)
1 + x2

Table 6.1: Some antiderivatives

138 6 The Riemann Integral

6.2 Integration Rules

We now collect some rules for the integration of more complicated functions. Unfortu-
nately, integration is not as straightforward as differentiation and one often has to have
an “inspired guess” to find out the antiderivative.

6.2.1 Integration by Substitution

Theorem 6.18. Integration by Substitution

Let I be an Interval, f : I → R be continuous and φ : [a, b] → I be continuously
differentiable. Then
Z b Z φ(b)
0
f (φ(t))φ (t) dt = f (x) dx.
a φ(a)

Proof: Let F : I → R be an antiderivative of f . Then, according to the chain rule, the

function F ◦ φ : [a, b] → R is differentiable with

(F ◦ φ)0 (t) = F 0 (φ(t))φ0 (t) = f (φ(t))φ0 (t).

Therefore,
Z b Z b Z φ(b)
x=φ(b)
0
f (φ(t))φ (t) dt = 0
(F ◦ φ) (t)dt = (F ◦ φ)(t)|t=b
t=a = F (x)|x=φ(a) = f (x) dx.
a a φ(a)

2
As a direct conclusion of this results, we can formulate the following:
Theorem 6.19. Integration by Substitution II
Let I be an interval, g : I → R be continuously differentiable and injective with
inverse function g −1 : g(I) → R. Let f : J → R with J ⊂ g(I). Then
Z b Z g −1 (b)
f (x) dx = f (g(t))g 0 (t) dt.
a g −1 (a)

Example 6.20. We can use the substitution rule to determine the area of an ellipse. The
equation of an ellipse is given by
x2 y 2
+ 2 = 1.
a2 b
This leads to r
x2
y = ±b 1 − 2 , x ∈ [−a, a].
a
As a consequence, the area of an ellipse is given by
Z a r Z ar
x2 x2
A=2 b 1 − 2 dx = 2b 1 − 2 dx
−a a −a a
6.2 Integration Rules 139

q
Now we set g(t) = a sin(t) and f (x) = 1 − xa2 . According to the substitution rule, we
2

now have
Z ar Z arcsin( a ) r
x2 a a2 sin2 (t)
1 − 2 dx = 1− 2
(a sin)0 (t) dt
−a a arcsin(− a
a)
a
Z π q Z π
2 2
= 2
1 − sin (t)a cos(t) dt = a cos2 (t) dt.
− π2 − π2

With
1 1 1 1
cos2 (t) = (exp(it) + exp(−it))2 = (exp(2it) + 2 + exp(−2it)) = cos(2t) + ,
4 4 2 2
we obtain
a
r Z π
x2
Z
2
A =2 b 1 − 2 dx = 2ab cos2 (t)dt
−a a π
−2
Z π Z π
2 1 1 2
=2ab cos(2t) + dt = ab cos(2t) + 1dt
− π2 2 2 − π2
t= π !  
1  2 1 1 π π
=ab · sin(2t) + t = ab · sin(π) − sin (−π) + + = πab.
2 t=− π 2 2 2 2
2

Note that integration

R b by substitution can also be applied by using the following formalism
for determining a f (x) dx: Consider the substitution x = g(t) ⇒ g (t) = dtd g(t) = dx
0
dt
and
“a formal multiplication with dt yields dx = g 0 (t)dt. For a formal determination of the
integration bounds, we consider the equations a = g(tl ), b = g(tu ) and thus tl = g −1 (a),
tu = g −1 (b). Integration by substitution can then be formally done by
Z b Z g −1 (b)
f (|{z}
x ) |{z}
dx = f (g(t))g 0 (t)dt.
a g −1 (a)
|{z} =g(t) =g 0 (t)dt
R g−1 (b)
g −1 (a)

Example 6.21. a) For a, b ∈ R, determine

Z b
x2 sin(x3 )dx.
a
√
Consider the “new variable” t = x3 . Then x = 3 t = t1/3 and dx dt
= 13 t−2/3 and thus
dx = 13 t−2/3 dt. The integration bounds are given by tl = a3 and tu = b3 and thus
Z b3 Z b3
1 1 1 3 1 3 x=b
sin(t) t−2/3 dt = sin(t)dt = − cos(t)|t=b
2/3

t t=a3 = − cos(x ) x=a .
a3 3 3 a3 3 3

b) For a ≥ 0, b ≥ 0, determine
Z b √
exp( x)dx.
a
140 6 The Riemann Integral

Consider the substitution x = t2 . Then dx = 2t and thus dx = 2tdt. For the integration
dt √ √
bounds, consider a = tl and b = tu which yields tl = a, tu = b. We now get
2 2

√
Z b √
Z b
exp( x)dx = √
exp(t)2tdt
a a
Z √ b
=2 √
t exp(t)dt
a
√ √ √ x=b
t=√b
= 2 exp(t)(t − 1)|t= a = 2 exp( x)( x − 1)x=a .

c) For a, b ∈ [−1, ∞), determine

b
x2 + 1
Z
√ dx.
a x+1
√
Consider the “new variable” t = x + 1. Then x = t2 − 1 and dx
dt
= 2t ⇒ dx = 2tdt
and
Z b 2 Z √b+1 2
x +1 (t − 1)2 + 1
√ dx = √ 2tdt
a x+1 a+1 t
Z √b+1
= √ 2t4 − 4t2 + 4dt
a+1
t=√b+1
2 5 4 3 
= t − t + 4t √
5 3 t= a+1
x=b
2√ 5 4√ 3 √ 
= x+1 − x + 1 + 4 x + 1 .
5 3 x=a

g 0 (x)
By using the substitution rule, we can also integrate expressions of type g(x)
.

Corollary 6.22.
For a differentiable function g : [a, b] → R with g(x) 6= 0 for all x ∈ [a, b] holds
b
g 0 (x)
Z
dx = log(|g(x)|)|x=b
x=a .
a g(x)

Proof: For f (y) = y1 , the above integral is of type

Z b
f (g(x))g 0 (x)dx
a

and thus, the result follows by the substitution rule. 2

Example 6.23. a) For a, b ∈ − π2 , π2 holds

 

b b b
− cos0 (x)
Z Z Z
sin(x)
tan(x)dx = dx = dx
a a cos(x) a cos(x)
x=b
=− log(| cos(x)|)|x=a .
6.2 Integration Rules 141

b) For a, b ∈ R holds
b
1 b 2x
Z Z
x
dx = dx
a x2 + 1 2 a x2 + 1
1 b (x2 + 1)0
Z
= dx
2 a x2 + 1
1 x=b
= log(|x2 + 1|)x=a .
2

6.2.2 Integration by Parts

In this part we integrate products of functions.
Theorem 6.24. Integration by Parts
Let I be an Interval and f, g : I → R be differentiable. Then for a, b ∈ I holds
Z b Z b
0 x=b
f (x)g(x)dx = f (x)g(x)|x=a − f (x)g 0 (x)dx.
a a

Proof: The product rule of differentiation implies

(f (x)g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)
and thus
Z b
f (x)g(x)|x=b
x=a =(f (x)g(x))0 dx
a
Z b Z b Z b
0 0 0
= f (x)g(x) + f (x)g (x)dx = f (x)g(x)dx + f (x)g 0 (x)dx.
a a a
Rb
Solving this equation for a
f 0 (x)g(x)dx, we get the desired equation. 2

Example 6.25. a) For a, b ∈ R, determine

Z b
x exp(x)dx.
a

We use integration by parts with f 0 (x) = f (x) = exp(x) and g(x) = x. Then
Z b Z b
x=b
exp(x)xdx = x exp(x)|x=a − exp(x)dx
a a
= x exp(x)|x=b
x=a − exp(x)|x=b
x=a

= (x − 1) exp(x)|x=b
x=a .

It is very important to note that an unlucky choice of f and g may be misleading. For
instance, if we choose f 0 (x) = x and g(x) = exp(x). Then integration by parts gives
Z b x=b Z b
1 2  1 2
x exp(x)dx = x exp(x) − x exp(x)dx.
a 2 x=a a 2

This formula is mathematically correct, but it does not lead to the explicit determin-
ation of the integral.
142 6 The Riemann Integral

b) Integrate sin(x) cos(x):

Z b
sin(x) cos(x) dx
a | {z } | {z }
f 0 (x)=− cos0 (x) =g(x)
x=b
Z b
2
= − cos (x)x=a − (− cos(x))(− sin(x))dx
a
x=b
Z b
2
= − cos (x)x=a − cos(x) sin(x)dx.
a
Rb
Solving this equation for a
sin(x) cos(x)dx, we obtain
Z b
x=b
1 2

sin(x) cos(x)dx = − cos (x) .
a 2 x=a

c) For a, b ∈ (0, ∞) and λ ∈ R\{−1}, determine the integral

Z b
xλ log(x)dx.
a

Defining f (x) = 1
λ+1
xλ+1 , g(x) = log(x), integration by parts leads to
Z b
xλ log(x)dx
a
x=b Z b
1 λ+1
 1 1
= x log(x) − xλ+1 dx
λ+1 x=a a λ+1 x
x=b Z b
1  1
= xλ+1 log(x) − xλ dx
λ+1 x=a a λ + 1
x=b x=b
1 λ+1
 1 
λ+1 
= x log(x) − x
λ+1 x=a (λ + 1)2 
x=a
 x=b
1 1 
= xλ+1 log(x) −  .
λ+1 λ+1  x=a

d) For a, b ∈ (0, ∞), determine the integral

Z b
log(x)
dx.
a x

Defining f (x) = g(x) = log(x), integration by parts leads to

Z b Z b
log(x) x=b log(x)
dx = log2 (x)x=a − dx.
a x a x
Rb
Solving this for a
log(x)
x
dx, we obtain
Z b
log(x) 1 x=b
dx = log2 (x)x=a .
a x 2
6.3 Integration of Rational Functions 143

6.3 Integration of Rational Functions

We know by Theorem 4.29 that every rational function f (x) possesses a representation
as the sum of some polynomial r and a strict proper rational function p(x)
q(x)
, i.e
p(x)
f (x) = r(x) + , deg(p) < deg(q).
q(x)
By using linearity of the integral, we can integrate each summand separately. Thus we
restrict our discussion here to strict proper rational functions. Furthermore by Theorem
4.31 we can write every strict proper rational function as a sum of partial fractions (x−x
A
0)
s,

with x0 being a root of q and A being some constant and s ≥ 1.

Altogether we have that it is enough to know the antiderivative of the partial fractions in
order to integrate arbitrary rational functions.
Let us first restrict to the case x0 ∈ R and s = 1. Then we have
Z
A
dx = A log |x − x0 | + const., x0 ∈ R.
x − x0
It is clear that this equality is only valid in the case x0 ∈ R since the antiderivative is a
real valued function.
Now consider x0 ∈ C \ R. Because of q having only real coefficients we know that also x0
is a root of q thus the partial fraction decomposition at least has the terms x−x
A
0
and x−x
B
0
included in the sum. Then setting x0 := α + iβ
=:a =:b
Z z }| { z }| {
(A + B)(x − α) + i (A − B) β
Z
A B
+ dx = dx.
x − x0 x − x0 β 2 + (x − α)2
Two integrals arise in this case. For the first one the antiderivative can be calculated
using the substitution t = x−α
β
, dt = dx
β
x−α
a(x − α)
Z Z Z
β dx t
2 2
dx = a x−α 2 =a dt
β + (x − α) 1+( β ) β 1 + t2
 2 !
a a x − α
= log(1 + t2 ) + const. = log 1 + + const.
2 2 β
For the second term we have again using the same substitution as above
Z Z Z
bβ 1 dx 1
2 2
dx = b x−α 2 =b dt
β + (x − α) 1+( β ) β 1 + t2
 
x−α
= b arctan(t) + const = b arctan + const.
β
Altogether we have for x0 ∈ C \ R and s = 1
 2 !  
x−α x−α
Z
A B A+B
+ dx = log 1 + + i(A − B) arctan + const.
x − x0 x − x0 2 β β
Now we want to investigate the case s > 1. Here it is not relevant if x0 is real or complex.
We have
Z
A A
s
dx = (x − x0 )1−s + const.
(x − x0 ) 1−s
144 6 The Riemann Integral

5 3 2
Example 6.26. We want to integrate f (x) = x +2x +4x −3x
(x2 +1)2
. First of all we decompose f
into a polynomial part and a strict proper function using polynomial division

x5 + 2x3 + 4x2 − 3x 4x2 − 4x 4x2 − 4x

= x + = x + .
(x2 + 1)2 (x2 + 1)2 (x + i)2 (x − i)2

The polynomial part has the antiderivative

x2
Z
x dx = + const.
2
The strict proper part has a partial fraction decomposition

4x2 − 4x A B A B
= + + + ,
(x + i)2 (x − i)2 x + i (x + i)2 x − i (x − i)2

with A = i and B = 1 − i. Thus we have

Z
i i
− dx = 2 arctan(x) + const.
x+i x−i
1−i 1−i
Z
2
dx = − + const.
(x + i) x+i
Z
1+i 1+i
2
dx = − + const.
(x − i) x−i

Altogether the antiderivative is

1−i 1+i
Z
1
f (x) dx = x2 + 2 arctan(x) − − + const.
2 x+i x−i
1 2x − 2
= x2 + 2 arctan(x) − + const.
2 x2 + 1

6.4 Integration on Unbounded Domains and

Integration of Unbounded Functions
So far, we have integrated bounded functions on compact intervals [a, b]. In this part we
skip these two assumptions by extending the integral notion to functions that may have
a pole and/or are defined on unbounded intervals.

6.4.1 Unbounded Interval

Definition 6.27. Integration on unbounded intervals
Let f : [a, ∞) → R be a function with the property that for all b > a the restriction
of f to [a, b] belongs to R([a, b]). If
Z b
lim f (x)dx
b→∞ a
6.4 Integration on Unbounded Domains and Integration of Unbounded Functions 145

exists, then we say that Z ∞

f (x)dx
a
is convergent. Otherwise, we speak of divergence.

Example 6.28. a) For integrating the function exp(−x) on the interval [0, ∞), we com-
pute Z ∞ Z b
exp(−x)dx = lim exp(−x)dx
0 b→∞ 0
= lim − exp(−x)|x=b
x=0
b→∞
=1 − lim exp(−b) = 1.
b→∞

b) For α > 0, consider Z ∞

1
dx.
1 xα
We know that for b > 1 holds
(
1 x=b
b 1

: α 6= 1,
Z
1 1−α xα−1 x=1
dx =
1 xα log(x)|x=b
x=1 : α = 1.

Since limb→∞ log(b) = ∞ and

(
1 0 : α > 1,
lim =
b→∞ xα−1 ∞ : α < 1,

we have (
Z ∞ 1
1 α−1
: α > 1,
dx =
1 xα ∞ : α ≤ 1.

It is straightforward to define the integral of a function defined on some interval unbounded

from below by Z a Z a
f (x)dx = lim f (x)dx.
−∞ b→−∞ b

We now define the integral of functions defined on the whole real axis.
Definition 6.29.
Let f : R → R be a function with the property that for all a, b ∈ R the restriction
of f to [a, b] belongs to R([a, b]). If there exists some c ∈ R such that both integrals
Z ∞ Z c
f (x)dx, f (x)dx
c −∞

exist, then we say that Z ∞

f (x)dx
−∞
146 6 The Riemann Integral

is convergent. Otherwise, we speak of divergence. In case of convergence we set

Z ∞ Z c Z ∞
f (x)dx = f (x)dx + f (x)dx.
−∞ −∞ c

We remark without proof that the above definition is independent of c.

Example 6.30. a) Consider

Z ∞ Z 0 Z ∞
1 1 1
dx = dx + dx
−∞ 1 + x2 −∞ 1 + x
2
0 1 + x2
Z 0 Z b
1 1
= lim dx + lim dx
a→−∞ a 1 + x2 b→∞ 0 1 + x2

= lim arctan(x)|x=0 x=b

x=a + lim arctan(x)|x=0
a→−∞ b→∞
= − lim arctan(a) + lim arctan(b)
a→−∞ b→∞
π π
= + = π.
2 2

b) The integral
Z ∞
xdx
−∞

diverges since both integrals

Z 0 Z 0
x=0
1 2 
xdx = lim xdx = lim x  = −∞
−∞ a→−∞ a a→−∞ 2
x=a
Z ∞ Z b x=b
1 2 
xdx = lim xdx = lim x  =∞
0 b→∞ 0 b→∞ 2
x=0
R∞
diverge. This example shows that the convergence of −∞
f (x)dx is not equivalent to
the existence of the limit Z a
lim f (x)dx.
a→∞ −a
R∞
However, in case of convergence, the integral −∞
f (x)dx coincides with the above
limit.

The majorant criterion for series says that if the absolut values of the addends of a given
series can be bounded from above by the addends of a convergent series, then (absolut)
convergence of the given series can be concluded.
Conversely, the minorant criterion for series says that if the addends of a given series can
be bounded from below by the addends of a series that diverges to +∞, then also the
given series diverges to +∞.
Analogue criteria hold true for integrals. We skip the proofs since they are totally ana-
logous to those of the majorant and minorant criteria.
6.4 Integration on Unbounded Domains and Integration of Unbounded Functions 147

Theorem 6.31.
Let f, g : [a, ∞) → R such that for all b ∈ [a, ∞), the restrictions of f and g to
[a, b] are Riemann-integrable.
R∞
(i) RIf |f (x)| ≤ g(x) for all x ∈ [a, ∞) and a g(x)dx converges, then also
∞
a
f (x)dx converges and it holds that
Z ∞  Z ∞ Z ∞
 

 f (x)dx ≤
 |f (x)|dx ≤ g(x)dx.
a a a

R∞ R∞
(ii) If g(x) ≤ f (x) for all x ∈ [a, ∞) and a
g(x)dx = +∞, then a
f (x)dx =
+∞.

Example 6.32. a) Consider Z ∞

x
dx.
1 x2 +1
We have
x
lim x · = 1.
x→∞ x2 +1
For large enough x ∈ R, we therefore have
2
 

x · x = x
 1
 x2 + 1  x2 + 1 ≥ 2

and thus
x 1
≥ .
x2 +1 2x
Since Z ∞
1
dx
1 2x
is divergent, Z ∞
x
dx
1 x2 +1
is divergent, too.
b) Consider √
Z ∞
x
dx.
1 x2 +1
We have √
3/2 x
lim x · = 1.
x→∞ x2 +1
For large enough x ∈ R, we therefore have
√ 
x2

 3/2
x · x 
= ≤2
 x2 + 1  x2 + 1

and thus √
x 2
≤ .
x2 + 1 x3/2
148 6 The Riemann Integral

Since Z ∞
2
dx
1 x3/2
is convergent, so
Z ∞ √
x
dx
1 x2 +1
is convergent, too.

Now we use integrals on unbounded domains to check whether series are convergent or
not.
Theorem 6.33. Integral Criterion for Series
Let f : [0, ∞) → R be monotonically decreasing and non-negative. Then the series
∞
X
f (k)
k=0

is convergent if and only if the integral

Z ∞
f (x)dx
0

converges. In the case of convergence, the following estimate holds true:

∞
X Z ∞
0≤ f (k) − f (x) dx ≤ f (0) (6.1)
k=0 0

Proof: Monotonicity of f implies that for k − 1 ≤ x ≤ k holds f (k) ≤ f (x) ≤ f (k − 1).

Monotonicity of the integral therefore leads to the inequality
Z k Z k Z k
f (k) = f (k)dx ≤ f (x)dx ≤ f (k − 1)dx = f (k − 1).
k−1 k−1 k−1

Therefore
Z n+2 n+1
X Z n+1 n
X
f (x)dx ≤ f (k) ≤ f (x)dx ≤ f (k).
1 k=1 0 k=0

Using this inequality, we can directly conclude that, if one of the limits (integral or sum)
as n → ∞ exists, then the other limit (sum or integral) also exists.
In case of convergence necessarily (f (k))k∈N0 is a zero sequence. Therefore

n
X Z n+1 n
X n+1
X
0≤ f (k) − f (x) dx ≤ f (k) − f (k) = f (0) − f (n + 1)
k=0 0 k=0 k=1

implies (6.1) for n → ∞.

2
6.4 Integration on Unbounded Domains and Integration of Unbounded Functions 149

Remark:
R∞
Since the convergence of a f (x)dx does not depend on a ∈ R, the aboveP result can
also be slightly generalised in a way that the Rconvergence of the series ∞ k=a f (k)
∞
for a ∈ N is equivalent to that of the integral a f (x)dx.

Example 6.34. a) Using the results of Example 6.28 b), we see that
∞
X 1
k=1
kα

converges for α > 1 and is divergent for α ≤ 1.

b) For α ∈ R consider
∞
X 1
.
n=2
n(log(n))α

We have
( (log(x))1−α x=n
∞
: α 6= 1,
Z
1 1−α 
dx = lim x=2
2 x(log(x))α n→∞
log(log(x))|x=n :α=1
x=2
( 1−α
− (log(2))
1−α
: α > 1,
=
∞ : α ≤ 1.

Therefore, the series

∞
X 1
.
n=2
n(log(n))α

is convergent if and only if α > 1.

6.4.2 Unbounded Integrand

Here we consider integrals of functions that may have a pole in the domain of integration.

Definition 6.35. Integration of unbounded functions

Let f : (a, b] → R be given with the property that for all ε > 0, the restriction of f
to the interval [a + ε, b] is Riemann-integrable. If
Z b
lim f (x)dx
ε&0 a+ε

exists, then we say that Z b

f (x)dx
a
is convergent. Otherwise, we speak of divergence.
150 6 The Riemann Integral

a b

Figure 6.3: Integral of a function with a pole

Example 6.36. a) For integrating the function log(x) from 0 to 1, we first compute
Z 1 Z 1 Z 1
1
log(x)dx = 1 · log(x)dx = x log(x)|x=1
x=ε − x · dx
ε ε ε x
= x(log(x) − 1)|x=1
x=ε .

Therefore Z 1 Z 1
log(x)dx = lim log(x)dx
0 ε&0 ε
= log(1) − 1 − lim ε(log(ε) − 1)
ε&0

= − 1 − lim ε log(ε)
ε&0
log(ε)
= − 1 − lim 1
ε&0
ε
1
ε
= − 1 − lim = −1 .
ε&0 − ε12

b) For α > 0, consider

( (
1 x=1
1 1 1
 1
: α 6= 1,
Z Z
1 1 1−α xα−1 x=ε 1−α
: α < 1,
dx = lim dx = lim =
0 xα ε&0 ε xα ε&0 log(x)|x=1
x=ε : α = 1. ∞ : α ≥ 1.

Again we can formulate a majorant and a minorant criterion for the convergence of in-
tegrals of unbounded functions.
Theorem 6.37.
Let f, g : (a, b] → R such that for all ε > 0, the restrictions of f and g to [a + ε, b]
are Riemann-integrable.
Rb Rb
(i) If |f (x)| ≤ g(x) for all x ∈ (a, b] and a g(x)dx converges, then also a f (x)dx
6.4 Integration on Unbounded Domains and Integration of Unbounded Functions 151

converges and it holds that

Z b  Z b Z b
 

 f (x)dx ≤
 |f (x)|dx ≤ g(x)dx.
a a a

Rb Rb
(ii) If g(x) ≤ f (x) for all x ∈ (a, b] and a
g(x)dx = +∞, then a
f (x)dx = +∞.

Example 6.38. a) Consider

1
cos2 (x) + 2 sin(x)
Z
√3
dx.
0 x2
This integral is convergent due to
 2 
 cos (x) + 2 sin(x) 
 √ ≤ 3 for x ∈ (0, 1].
 3
x2  x2/3

b) For integrating the function cos(x)

√
x
over the entire positive half-axis, we split
Z ∞ Z 1 Z ∞
cos(x) cos(x) cos(x)
√ dx = √ dx + √ dx.
0 x 0 x 1 x
The first addend is convergent due to
1
  Z
 cos(x) 
 √  ≤ √1 for x ∈ (0, 1],
1
√ dx = 2 < ∞.
 x  x 0 x
For the second addend, we make use of integration by parts to determine
Z b x=b
1 b sin(x)
Z
cos(x) sin(x) 
√ dx = √  + .
1 x x x=1 2 1 x3/2
R ∞ sin(x)  
The integral 1 x3/2 converges due to  x3/2  ≤ x3/2 for all x ∈ [1, ∞). Furthermore,
 sin(x)  1

x=b
sin(x)  sin(b)
lim √  = lim √ − sin(1) = − sin(1).
b→∞ x x=1 b→∞ b
R ∞ cos(x)
Therefore, the integral 0 √x dx converges.

For functions f : [a, b]\{c} → R it is straightforward to define

Z b Z ε Z b
f (x)dx = lim f (x)dx + lim f (x)dx.
a ε%c a ε&c ε

Example 6.39.
Z 1 Z ε Z 1
1 1 1
p dx = lim p dx + lim p dx
−1 |x| ε%0 −1 |x| ε&0 ε |x|
√ x=ε √ x=1
= lim −2 −x + lim 2 x
x=−1
=4 x=ε
ε%0 ε&0

The remaining parts of this Chapter on parameter-dependent integrals, solids of revolu-

tion, path integrals and Fourier series follow [?].
152 6 The Riemann Integral

6.5 Parameter-dependent integrals

In this section we consider integrals of the form
Z b
F (x) := f (x, y) dy (6.2)
a

where the integrand f additionaly depends on some free parameter x ∈ I, where I is some
real interval. Precisely, the function f : I × [a, b] → R has the property that for each fixed
x ∈ I, the function f (x, ·) : [a, b] → R, y 7→ f (x, y) is Riemann integrable.
We will investigate continuity and differentiability of the integral function
Z b
F : I → R, x 7→ f (x, y) dy .
a

For preparation we need the following definition.

Definition 6.40.
A function f : D → Rm , D ⊂ Rn , m, n ∈ N, is called uniformly continuous, if for
each ε > 0 there exits a δ > 0 such that for all x, x0 ∈ D with ||x − x0 || < δ holds
||f (x) − f (x0 )|| < ε. Here || · || denotes some norms on Rn and Rm respectively, for
example the Euclidean norm.

Note that a uniformly continuous function is continuous. In general the converse is not
true, but if the domain of a continuous function is compact it is already uniformly con-
tinuous.
Theorem 6.41.
Let D ⊂ Rn , n ∈ N, be compact. A continuous function f : D → Rm , m ∈ N, is
uniformly continuous.

Proof: Suppose that f is not uniformly continuous. Then there exists an ε > 0 such that
for each n ∈ N there are xn , yn ∈ D with ||xn − yn || < n1 and ||f (xn ) − f (yn )|| > ε. Since
D is compact, there are convergent subsequences (xnk )k∈N and (ynk )k∈N of (xn )n∈N and
(yn )n∈N which necessarily converge to the same limit. Set

z := lim xnk = lim ynk .

k→∞ k→∞

Since f is continuous, we conclude

lim (f (xnk ) − f (ynk )) = lim f (xnk ) − lim (ynk ) = f (z) − f (z) = 0,

k→∞ k→∞ k→∞

a contradiction to ||f (xnk ) − f (ynk )|| > ε for all k ∈ N. 2

Now we consider continuity of the integral function (6.2).

Theorem 6.42.
Let a, b ∈ R, a < b, and I ⊂ R be an interval. If f : I × [a, b] → R is continuous,
Rb
then F : I → R, x 7→ a f (x, y) dy is well-defined and continuous on I.
6.5 Parameter-dependent integrals 153

Proof:
Since f is continuous, for each fixed x ∈ I, the function f (x, ·) : [a, b] → R, y 7→ f (x, y) is
continuous too and therefore integrable on [a, b]. Hence F is well-defined. Now let x0 ∈ I
and I0 ⊂ I be a compact interval that contains x0 . If x0 is an inner point of I then we
can also choose I0 such that x0 is an inner point of I0 . Now I0 × [a, b] is compact in R2
and hence f is uniformly continuous on I0 × [a, b] by Theorem 6.41. This means that for
given ε > 0 there is a δ > 0 such that for all x ∈ I0 with |x − x0 | < δ and all y ∈ [a, b]
holds |f (x, y) − f (x0 , y)| < ε.
Thus for x, x0 ∈ I0 with |x − x0 | < δ and y ∈ [a, b] we have
Z b  Z b
 
|F (x) − F (x0 )| =  f (x, y) − f (x0 , y) dy  ≤ |f (x, y) − f (x0 , y)| dy ≤ ε(b − a) .
a a

Therefore F is continuous in x0 . 2

Next we consider differentiability of the integral function (6.2).

Theorem 6.43.
Let a, b ∈ R, a < b, and I ⊂ R be an interval. If f : I ×[a, b] → R is continuous and
if for each fixed y ∈ [a, b] the function f (·, y) : I → R, x 7→ f (x, y) is continuously
Rb
differentiable, then also F : I → R, x 7→ a f (x, y) dy is continuously differentiable
on I with Z b
0 ∂f
F (x) = (x, y) dy .
a ∂x

Proof:
Let x0 ∈ I. Then for x ∈ I\{x0 } and each y ∈ [a, b] the mean value theorem applied to
f (·, y) supplies a ξx,y between x and x0 such that

b b
F (x) − F (x0 ) f (x, y) − f (x0 , y)
Z Z
∂f
= dy = (ξx,y , y) dy .
x − x0 a x − x0 a ∂x

By assumption and Theorem 6.42 the function

Z b
∂f
G : I → R, x 7→ (x, y) dy
a ∂x

is well-defined and continuous on I. Since ξx,y → x0 for x → x0 this implies

b
F (x) − F (x0 )
Z
0 ∂f
F (x0 ) = lim = (x0 , y) dy .
x→x0 x − x0 a ∂x
2

Example 6.44.
Z π
sin(tx)
F (x) := dt
1 t
154 6 The Riemann Integral

Z π
0
F (x) := cos(tx) dt
1
Z π
00
F (x) := − t sin(tx) dt
1

For parameter-dependent improper intergrals analog statements for continuity and differ-
entiability hold true which we will state without proof.
Definition 6.45.
Let a ∈ R, I ⊂ R be an intervalR ∞ and f : I × [a, ∞) → R. Suppose that
R ∞for each fixed
x ∈ I the improper integral a f (x, y) dy exists. Then the integral a f (x, y) dy is
called uniformly convergent if for each ε > 0 there is a constant K > a such that
for all x ∈ I and all b1 , b2 ≥ K holds
Z b2 
 

 f (x, y) dy <ε

b1

Theorem 6.46.
Let a ∈ R and I ⊂ R be an interval. If f : I × [a, ∞) → R is continuous and
continuously differentiable with respect to the first variable x and if the integrals
Z ∞ Z ∞
∂f
f (x, y) dy and (x, y) dy
a a ∂x
converge on each compact subset of I uniformly, then
Z ∞
F : I → R, x 7→ f (x, y) dy
a

is continuously differentiable and its derivative is given by

Z ∞
0 ∂f
F (x) = (x, y) dy .
a ∂x

Example 6.47 (Gamma function).

Z ∞
Γ :(0, ∞) → R, x 7→ e−y y x−1 dy
0
Z ∞
Γ0 (x) = e−y y x−1 log(y) dy
0

6.6 Solids of revolution, path integrals

In this section we give some simple geometrical applications of one-dimensional integration
theory, for example computing volumes of certain bodies and length’s of curves in the
three-dimensional euclidean space.
A solid of revolution is a volume which is obtained by rotating a plane curve around
an axis. For example if f : [a, b] → [0, ∞) is a nonnegative continuous function and if
6.6 Solids of revolution, path integrals 155

the graph of f is rotated around the x-axis, then the enclosed volume Vrot can easily be
computed by
Z b Z b
2
Vrot = πf (x) dx = π f (x)2 dx .
a a

Example 6.48 (ellipsoid). Let us consider an ellipse given by

x2 y 2
+ 2 = 1,
a2 b
with semi-axes a, b > 0.
q
x 2
In this case f : [−a, a] → R, x 7→ b 1 − . Hence


a

Z a   x 2 Z a
2 2
Vrot = π f (x) dx = π b 1− dx
−a −a a
1 4
= πb2 (x − 2 x3 )|x=a 2
x=−a = πab .
3a 3

If a = b =: r, then the ellipsoid is a ball with radius r and volume

4 3
Vball = πr .
3

Now we will derive a formula for computing the lateral surface Mrot of solids of revolution.
First recall that the lateral surface MC of a cone with circular ground face of radius r and
lateral height l is given by
MC = πrl .

Thus the lateral surface MtC of a truncated cone with circular ground face of radius r1 ,
circular top face of radius r2 < r1 and lateral height l is given by

MtC = MC1 − MC2 = πr1 l1 − πr2 l2 = π(r1 l1 + r2 l1 − r1 l2 − r2 l2 )

= π(r1 + r2 )(l1 − l2 ) = π(r1 + r2 )l ,

where l1 is the lateral height of the complete cone and l2 that of its truncated top. (Recall
that rl11 = rl22 .)
Now let f : [a, b] → R≥0 be a non-negative function which is continuously differentiable
on (a, b). We want to approximate its lateral surface area by summing up lateral surfaces
of certain truncated cones. Precisely, consider a decomposition Z: a = x0 < x1 < ... <
xn−1 < xn = b of [a, b] and define yi := f (xi ), ∆xi = xi+1 − xi and ∆yi = yi+1 − yi . Then
the sum M (Z) of all lateral surfaces
p of the n truncated cones with circular faces of radii
yi , yi+1 and lateral heights li := (∆xi )2 + (∆yi )2 , i = 0, ..., n − 1, is given by

n−1
X p
M (Z) = π(yi + yi+1 ) (∆xi )2 + (∆yi )2
i=0
156 6 The Riemann Integral

s
n−1  2
X yi + yi+1 ∆yi
= 2π · 1+ · ∆xi
i=0
2 ∆xi

Now, roughly speaking, for ∆xi → 0 the right-hand side converges to the integral

Z b p
Mrot = 2π f (x) 1 + f 0 (x)2 dx .
a

Example 6.49 (Surface of a ball). Let us consider a ball of radius r > 0

x2 + y 2 = r 2 ,
√
The corresponding function f : [−r, r] → R, x 7→ r2 − x2 is continuously differentiable
on (−r, r) with derivative
−x
f 0 (x) = √ .
r 2 − x2
Thus the surface of the ball is
Z b Z r√ r
p x2
Mrot = 2π f (x) 1 + f 0 (x)2 dx = 2π r 2 − x2 · 1 + 2 dx
a −r r − x2
Z r√ Z r
r
= 2π r −x · √
2 2 dx = 2πr dx = 4πr2 .
2
r −x 2
−r −r

In the following we will consider curves.

Definition 6.50.

a) A continuous function c : [a, b] → Rn , x 7→ (c1 (x), ...., cn (x))T , n ∈ N, a, b ∈

R, a < b, is called a curve in Rn . The vectors c(a), c(b) ∈ Rn are called
starting point and end point of the curve. The curve is called closed if c(a) =
c(b).

b) If c : [a, b] → Rn is continuously differentiable, i.e. c ∈ C 1 ([a, b], Rn ), which

is the case if and only if each of the coordinate functions ci , i = 1, ..., n, is
continuously differentiable, then c is called a C 1 -curve. Moreover c is called
a piecewise C 1 -curve if there is a decomposition a = t0 < t1 < ... < tm = b,
m ∈ N, such that c is a C 1 -curve on each subinterval [ti , ti+1 ], i = 0, ..., m − 1.

c) A C 1 -curve c ∈ C 1 ([a, b], Rn ) is called smooth or regular if for all t ∈ [a, b].
holds
c0 (t) = (c01 (t), ...., c0n (t)) 6= 0 .

Example 6.51. a) The curve c : [0, 2π] → R2 , t 7→ (r cos(t), r sin(t))T , r > 0, de-
scribes a circle in R2 of radius r. It is a closed C 1 -curve with derivative c0 (t) =
(−r sin(t), r cos(t))T . Since cosine and sine do not have common roots, c0 (t) 6= 0 for
all t so that c is also smooth.
b) The curve c : [0, T ] → R2 , t 7→ (rt − a sin(t), r − a cos(t))T , T, a, r > 0 is called
a cycloid. It is a C 1 -curve with with derivative c0 (t) = (r − a cos(t), a sin(t))T . If
a = r, then then c0 (2πk) = 0 for all k ∈ N. Hence c is not smooth in this case.
6.6 Solids of revolution, path integrals 157

c) The curve c : [0, T ] → R3 , t 7→ (r cos(t), r sin(t), ht)T , T, r, h > 0, describes a helix

in R3 . It is a smooth C 1 -curve with derivative

c0 (t) = (−r sin(t), r cos(t), h)T 6= 0.

d) Each continuous function f : [a, b] → R can be interpreted as a curve in R2 via

c : [a, b] → R2 , t 7→ (t, f (t))T . The associated curve c is C 1 if and only if f is
continuously differentiable. In this case holds c0 (t) = (1, f 0 (t))T 6= 0 and therefore c
is smooth.

If c : [a, b] → Rn is a curve and if h : [α, β] → [a, b] is a continuous bijective and

monotonically increasing function, then the “new” curve

c̃ : [α, β] → Rn , τ 7→ c(h(τ ))

has the same shape and the same oriented direction. In this case the function h is called
a reparametrisation. In case of C 1 -curves also only C 1 -reparametrisations h with h0 > 0
are permitted.
In general curves c1 and c2 which distinguish themselves only through a reparametrisation
are considered as “equal”.
Now we want to compute the length of a C 1 -curve c : [a, b] → Rn . This is done by
approximation by polygonal paths. For a given decomposition Z := {a = t0 < t1 < .... <
tm = b}, m ∈ N, of the interval [a, b] the length L(Z) of the polygonal path with corners
c(ti ) is given by
m−1 m−1
X X c(ti+1 ) − c(ti )
L(Z) = ||c(ti+1 ) − c(ti )|| = || || · (ti+1 − ti ).
i=0 i=0
ti+1 − ti

If the right-hand side converges for ti+1 − ti → 0 then the limit is the the curve length
L(c) which is given by
Z b
L(c) = ||c0 (t)|| dt.
a

Definition 6.52.
If the set {L(Z) | Z is a decomposition of [a, b]} is bounded from above, then the
curve c : [a, b] → R is called rectifiable and

L(c) := sup{L(Z) | Z is a decomposition of [a, b]} = lim L(Z)

||Z||→0

is called the length of the curve c where ||Z|| := min{|tj+1 − tj | | j = 0, ..., m − 1}

for a decomposition Z = {a = t0 < t1 < .... < tm = b}, m ∈ N, of the interval [a, b].

Theorem 6.53.
Each C 1 -curve c is rectifiable and
Z b
L(c) = ||c0 (t)|| dt.
a
158 6 The Riemann Integral

Proof: Let Z = {a = t0 < t1 < .... < tm = b}, m ∈ N, be a decomposition of the interval
[a, b]. Using the mean value theorem for c, we have
v v
m−1
X uX
u n m−1
X uX
u n
L(Z) = t (ck (tj+1 ) − ck (tj ))2 = t (c0 (τk ))2 (tj+1 − tj )
k j
j=0 k=1 j=0 k=1

with tj ≤ τkj ≤ tj+1 . Set

v
m−1
X X
u n
u
R(Z) := t (c0k (tj ))2 (tj+1 − tj ).
j=0 k=1

We will estimate |L(Z) − R(Z)|. Let ε > 0. Since c0k is uniformly continuous on [a, b],
there is a δ > 0 such that for all t, t̃ ∈ [a, b] with |t − t̃| < δ holds |c0k (t̃) − c0k (t)| < ε for all
k = 1, ..., n. Thus if Z fulfils ||Z|| < δ, then
m−1 
X 
0 0
|L(Z) − R(Z)| =  (||c (τj )|| − ||c (tj )||) (tj+1 − tj )
 
 
j=0
m−1
X
≤ | ||c0 (τj )|| − ||c0 (tj )|| | (tj+1 − tj )
j=0
m−1
X
≤ ||c0 (τj ) − c0 (tj )|| (tj+1 − tj )
j=0
√
≤ nε(b − a) −→ 0 ,
ε&0

where τj := (τ1j , ..., τnj ) and c0 (τj ) := (c01 (τ1j ), ..., c0n (τnj )).
Rb
Since R(Z) → a ||c0 (t)|| dt for ||Z|| → 0, this also holds for L(Z).

Example 6.54. The length of a cycloid c(t) = (r(t − sin(t)), r(1 − cos(t)))T , 0 ≤ t ≤ 2π,
can be calculated as follows:

c0 (t) = (r(1 − cos(t)), r(sin(t)))T

p p
||c0 (t)|| = r (1 − cos(t))2 + sin(t)2 = r 2(1 − cos(t))
r
t t
= r 2(1 − (1 − 2 sin2 ( ))) = 2r sin( )
2 2
Z 2π
t t t=2π
L(c) = 2r sin( ) dt = −4r cos( )|t=0 = 8r .
0 2 2

The length of a curve is independent with respect to reparametrisations, because if h :

[α, β] → [a, b] is a C 1 -reparametrisation, then h0 > 0 and the substitution rule immediately
yields
Z β Z β
0
L(c ◦ h) = ||(c ◦ h) (τ )|| dτ = ||c0 (h(τ ))h0 (τ )|| dτ
α α
Z β Z b
0 0
= ||c (h(τ ))|| · h (τ ) dτ = ||c0 (t)|| · dt .
α a
6.6 Solids of revolution, path integrals 159

Definition 6.55.
Let c : [a, b] → R be a C 1 -curve. The function
Z t
S : [a, b] → R, t 7→ ||c0 (τ )|| dτ
a

is called the arc length function of the curve.

If c is a smooth C 1 -curve, then the arc length function S : [a, b] → [0, L(c)] is a C 1 -
function. In particular, the inverse function S −1 : [0, L(c)] → [a, b] exists and is a C 1 -
reparametrisation. The parametrisation c̃ := c◦S −1 is called parametrisation with respect
to the arc length.
Its derivative is given by

1
c̃0 (s) = c0 (S −1 (s)) ·
||c0 (S −1 (s))||

Obviously, this is a vector in Rn of length one. This means that the “speed” of the curve
is always constant one and that c̃0 (s) is the unit tangent vector at the curve in the point
t = S −1 (s).
Differentiation of 1 = ||c̃0 (s)||2 = hc̃0 (s), c̃0 (s)i yields

n
!0
X
0 = (hc̃0 (s), c̃0 (s)i)0 = (c˜k 0 (s))2
k=1
n
X
= 2c˜k 0 (s)c˜k 00 (s) = 2hc̃00 (s), c̃0 (s)i .
k=1

This means that the acceleration vector c̃00 (s) is perpendicular to the velocity vector c̃0 (s).
The vector
c̃00 (s)
n(s) := 00
||c̃ (s)||
is called the unit normal vector and ||c̃00 (s)|| is called the curvature of c(t) in the point
t = S −1 (s).
Finally, the plane spanned by the tangent vector c̃0 (t) and the normal vector c̃00 (t) is called
osculating plane of c(t) in the point t = S −1 (s).

Example 6.56. For a curve c : [a, b] → R, x 7→ (t, y(x))T corresponding to the graph of
a C 2 -function y : [a, b] → R holds

c0 (x) = (1, y 0 (x))T

p
||c0 (x)|| = 1 + (y 0 (x))2
Z tp
S(t) = 1 + (y 0 (x))2 dx
a
|y 00 (x)|
κ(x) = p .
( 1 + (y 0 (x))2 )3
160 6 The Riemann Integral

Now, for a given C 1 -curve c : [a, b] → R2 , we want to compute the signed area A(c)
consisting of all points “between” the curve and the origin. These are all points P = λc(t),
t ∈ [a, b], λ ∈ [0, 1]. The area A(c) is called the area enclosed by the curve c.
If Z = {a = t0 , t1 , ..., tm−1 , tm = b}, m ∈ N, is a decomposition of [a, b]. Then A(c)
can be approximated by the sum of the signed areas Ai of all triangles with corners
c(ti ), c(ti+1 ), 0R2 , i = 0, ..., m − 1. These triangle areas can easily be calculated using the
cross product:

1 1
|Ai | = ||(c1 (ti ), c2 (ti ), 0)T × (c1 (ti+1 ), c2 (ti+1 ), 0)T || = |c1 (ti )c2 (ti+1 ) − c1 (ti+1 )c2 (ti )|
2 2
1
Ai = (c1 (ti )c2 (ti+1 ) − c1 (ti+1 )c2 (ti )) .
2

Setting ∆ti := ti+1 − ti , ∆cj,i := cj (ti+1 ) − cj (ti ), j = 1, 2, i = 0, ..., m − 1, the sum A(Z)
of all signed triangle areas is
m−1
1X
A(Z) := (c1 (ti )c2 (ti+1 ) − c1 (ti+1 )c2 (ti ))
2 i=0
m−1
1 X c1 (ti )c2 (ti+1 ) − c1 (ti+1 )c2 (ti )
= · ∆ti
2 i=0 ∆ti
m−1  
1X ∆c2,i ∆c1,i
= c1 (ti ) − c2 (ti ) · ∆ti .
2 i=0 ∆ti ∆ti

For ||Z|| → 0, A(Z) converges to

Z b
1
A(c) = (c1 (t)c02 (t) − c2 (t)c01 (t)) dt . (6.3)
2 a

Example 6.57. For given a, b > 0, t1 , t2 ∈ [0, 2π], t1 < t2 we want to compute the area
At1 ,t2 of the sector of the ellipse

c : [0, 2π] → R, t 7→ (a cos(t), b sin(t))T

given by t1 ≤ t ≤ t2 . This computes as

1 t2 ab t2 ab(t2 − t1 )
Z Z
2 2
At1 ,t2 = ab(cos (t) + sin (t)) dt = 1 dt = .
2 t1 2 t1 2

In particular, if a = b =: r, then the sector of the circle has the area t2 −t1 2
2
r .

The final topic of this section are so-called curve integrals. Consider the following problem:
Given a curved wire with inhomogeneous density. By integration we want to determine
its total mass. Assume that the position of the wire is parameterised by a C 1 -curve
c : [a, b] → Rn , n := 3. The density ρ(c(t)) of the wire in the point c(t) is defined as

mass
ρ(c(t)) := .
length unit
6.6 Solids of revolution, path integrals 161

Now, in order to compute the total mass of the wire, we consider a decomposition Z =
{a = t0 , t1 , ..., tm−1 , tm = b} of the interval [a, b] and approximate the density in the
interval [ti , ti+1 ] by the constant value ρ(c(ti )) [= density in the left boundary point c(ti )].
Furthermore, also the length of the wire in the interval [ti , ti+1 ] is approximated by the
length of the straight line between c(ti ) and c(ti+1 ) which, by the mean value theorem, is
v v
u n u n
uX uX
||c(ti+1 ) − c(ti )|| = t (ck (ti+1 ) − ck (ti ))2 = t c0k (τk,i )2 (ti+1 − ti )
k=1 k=1

for suitable τk,i ∈ [ti , ti+1 ]. Thus the total mass of the wire is approximated by
v
m−1
X m−1
X
u n
uX
ρ(c(ti ))||c(ti+1 ) − c(ti )|| = ρ(c(ti ))t c0k (τk,i )2 (ti+1 − ti ).
i=0 i=0 k=1

For ||Z|| → 0 the right-hand side converges to

Z b
ρ(c(t))||c0 (t)|| dt .
a

According to this result the following notation is introduced.

Definition 6.58.
Let D ⊂ Rn , n ∈ N, f : D → R continuous and c : [a, b] → D a piecewise C 1 -curve,
a, b ∈ R, a < b. Then the path integral (or line integral, contour integral, curve
integral) of the first kind of f with respect to c is defined as
Z Z b
f (x) ds := f (c(t))||c0 (t)|| dt .
c a

If c is a closed curve then also the symbol

I
f (x) ds
c

is used instead.

The path integral of the first kind is invariant with respect to reparametrisations, since
for a C 1 -reparametrisation h : [α, β] → [a, b] with h0 > 0 holds

Z Z β
f (x) ds = f ((c ◦ h)(τ ))||(c ◦ h)0 (τ )|| dτ
c◦h α
Z β
= f (c(h(τ )))||c0 (h(τ ))h0 (τ )|| dτ
α
Z β
= f (c(h(τ )))||c0 (h(τ ))||h0 (τ ) dτ
h0 >0 α
Z b
= f (c(t))||c0 (t)|| dt (substitution t := h(τ ))
Za
= f (x) ds .
c
162 6 The Riemann Integral

Example 6.59 (center of gravity). For a system of N mass points with point masses mi
at positions xi ∈ Rn the center of gravity xs ∈ Rn is given by
PN
mi xi
xs = Pi=1 N
.
i=1 m i

For computing the center of gravity xs of a wire, the total mass of the piece of wire
between two points c(ti ) and c(ti+1 ) is approximated by ρ(c(ti ))||c(ti+1 ) − c(ti )||. Thus
the approximation of xS reads
Pm−1 c(ti+1 )−c(ti )
i=0 ρ(c(ti ))|| ∆ti
||c(ti )∆ti
xS ≈ Pm−1 c(ti+1 )−c(ti )
.
i=0 ρ(c(ti ))|| ∆ti
||∆ti

For ||Z|| → 0 this becomes

Rb
ρ(c(t))||c0 (t)||c(t) dt
R
a
ρ(x)x ds
xs = Rb = Rc .
ρ(c(t))||c 0 (t)|| dt
c
ρ(x) ds
a

The curve integral

R  R b 
ρ(x)x1 ds
c a
ρ(c(t))||c0 (t)||c1 (t) dt
.. ..
Z
ρ(x)x ds =  . = .
   

c R Rb 0
c
ρ(x)xn ds ρ(c(t))||c (t)||cn (t) dt
a

in the numerator has to be computed componentwise.

Example 6.60 (moment of inertia). If a mass point with mass m rotates around an axis
with distance r and angular velocity ω, then its kinetic energy is
1 1 1
Ekin = mv 2 = mr2 ω 2 = θω 2 .
2 2 2
The term θ := mr2 is called moment of inertia of the masspoint with respect to the given
axis of rotation. For a system of N mass points with masses mi and distances ri to the
axis of rotation, the single moments of inertia θi = mi ri2 simply add up to a total moment
of inertia
XN
θ= mi ri2 .
i=1

For a wire as considered above the moment of inertia θ is approximated by

m−1 m−1
X
2
X c(ti+1 ) − c(ti )
θ≈ ρ(c(ti ))||c(ti+1 ) − c(ti )||r(c(ti )) = ρ(c(ti )) r(c(ti ))2 ∆ti .
i=0 i=0
∆ti

Here r(c(t)) denotes the orthogonal distance of c(t) to the axis of rotation. For ||Z|| → 0
the right-hand side converges to
Z b Z
0
θ= ρ(c(t))||c (t)||r (c(t)) dt = ρ(x)r2 (x) ds .
2
a c
6.7 Fourier series 163

For example if the wire has constant density ρ and is placed along a straight line of length
l > 0 in the (x, z)-plane that encloses an angle α with the x-axis, then

c : [0, l] → R, t 7→ (t cos(α), 0, t sin(α))T

and rotation around the x-axis yields

r(c(t)) = t sin(α)
Z l
1
θx-axis = ρ · 1 · (t sin(α))2 dt = l3 ρ sin2 (α) .
0 3

6.7 Fourier series

Definition 6.61.
A function f : R → C is called periodic with period T ∈ R or shortly T -periodic if
f (t + T ) = f (t) for all t ∈ R. In this case, if T 6= 0, ν := T1 is called frequency of
f.

For example sin(t), cos(t) and exp(it) = cos(t) + i sin(t) are 2π-periodic functions. If a
function f is T -periodic, then it is also kT -periodic for each k ∈ Z. Without proof we
mention that if f is a nonconstant continuous periodic function, then there exists always
a smallest positive period T > 0 of f .
Definition 6.62.
A series of the form
∞
a0 X
f (t) = + ak cos(kωt) + bk sin(kωt)
2 k=1

with ak , bk ∈ C and ω > 0 is called Fourier series or trigonometric series. The

corresponding partial sums
n
a0 X
fn (t) = + (ak cos(kωt) + bk sin(kωt))
2 k=1

are called trigonometric polynomials.

The trigonometric polynomials fn (t) can be transformed as follows:

n
a0 X
fn (t) = + ak cos(kωt) + bk sin(kωt)
2 k=1
n
a0 X ak ikωt bk
= + (e + e−ikωt ) + (eikωt − e−ikωt )
2 k=1
2 2i
n
a0 X ak − ibk ikωt ak + ibk −ikωt
= + e + e
2 k=1
2 2
164 6 The Riemann Integral

n
X
= γk eikωt
k=−n

with
a0
γ0 := (6.4)
2
ak − ibk
γk := (6.5)
2
ak + ibk
γ−k := (6.6)
2
for k ∈ N. Conversely, if coefficients γk ∈ C, k = −n, −n + 1, ..., n, are given, then
corresponding coefficients ak ∈ C, k = 0, ..., n, bk ∈ C, k = 1, ..., n, compute as

a0 := 2γ0 (6.7)
ak := γk + γ−k (6.8)
bk := i(γk − γ−k ) . (6.9)

For Fourier series we therefore have

∞
a0 X
f (t) = + ak cos(kωt) + bk sin(kωt)
2 k=1
n
X ∞
X
= lim γk eikωt =: γk eikωt .
n→∞
k=−n k=−∞

If these series converge for each t ∈ R, then clearly the function f (t) is well-defined and
periodic with period T := 2π ω
.
Without proof we state that for two complex valued Riemann integrable functions f, g ∈
R([a, b], C), a, b ∈ R, a < b, with integrals
Z b Z b Z b
f (x) dx := Re f (x) dx + i Im f (x) dx
a a a
Z b Z b Z b
g(x) dx := Re g(x) dx + i Im g(x) dx
a a a

also their product f g : [a, b] → R, x 7→ f (x)g(x) is Riemann integrable with integral

Z b Z b

(f g)(x) dx = Re f (x) Re g(x) − Im f (x) Im g(x) dx +
a a
Z b

i Re f (x) Im g(x) + Im f (x) Re g(x) dx .
a

Definition 6.63.
Let c > 0. The mapping

h·, ··i : R([a, b], C) × R([a, b], C) → C,

6.7 Fourier series 165

Z b Z b
hf, gi = c f (x)g(x) dx = c Re f (x) Re g(x) + Im f (x) Im g(x) dx +
a a
Z b
ic Re f (x) Im g(x) − Im f (x) Re g(x) dx
a

is a so-called well-defined positive-semidefinite hermitian form on the C-vector space

R([a, b], C), which means that for all f, g, h ∈ R([a, b], C) and all λ, µ ∈ C holds

(i) hf, λg + µhi = λhf, gi + µhf, hi

(ii) hf, gi = hg, f i

Rb
(iii) hf, f i = c a |f (x)|2 dx ≥ 0 .

If f ∈ C([a, b], C)\{0} is a non-zero complex valued continuous function on [a, b], then
there exists a ξ ∈ (a, b) such that |f (ξ)| > 0 and by continuity there is an ε > 0 such that
[ξ − ε, ξ + ε] ⊂ [a, b] and such that |f (x)| ≥ 12 |f (ξ)| for all x ∈ [ξ − ε, ξ + ε]. Thus

Z b Z ξ+ε
2 1
hf, f i = c |f (x)| dx ≥ c |f (x)|2 dx ≥ cε|f (ξ)|2 > 0 .
a ξ−ε 2

This shows that h·, ··i restricted to C([a, b], C) ⊂ R([a, b], C) is positive definite, i.e. for
f ∈ C([a, b], C) holds hf, f i = 0 if and only if f = 0.
In other words h·, ··i is an inner product on the complex vector space C([a, b], C) which,
equipped with this inner product, therefore becomes a complex inner product space.
Restricted to the R-subspace C([a, b], R) consisting of all real valued continuous functions
Rb
on [a, b] the inner product becomes hf, gi = c a f (x)g(x) dx and equipped with this inner
product C([a, b], R) is a real inner product space.
The constant c is simply a scaling factor which for example can be set to the reciprocal
of the length of the integration interval, that is c := b−a
1
.
Similar arguments as given above show that h·, ··i restricted to the C/R-vector space of
piecewice complex-/real-valued continuous functions on [a, b], denoted by Cp ([a, b], C)/Cp ([a, b], R),
makes this space to a unitary/Euclidean one.
Theorem 6.64.
Let T > 0 and ω := 2π
T
.

a) The functions eikωt ∈ C([0, T ], C), k ∈ Z, form an orthonormal system with

RT
respect to the inner product hf, gi := T1 0 f (t)g(t) dt, f, g ∈ C([0, T ], C).
(Here, the inner product has the complex conjugation in the first component!)

b) The functions √12 , cos(kωt), sin(kωt) ∈ C([0, T ], R) ⊂ C([0, T ], C), k ∈

N, form an orthonormal system w.r.t. the inner product 2hf, gi =
2
RT
T 0
f (t)g(t) dt, f, g ∈ C([0, T ], C).
166 6 The Riemann Integral

Proof: a) Let k, l ∈ Z. If k = l, then

1 T −ikωt ikωt 1 T
Z Z
ikωt ilωt
he , e i = e e dt = 1 dt = 1 .
T 0 T 0
If k 6= l, then
Z T
t=T
ikωt ilωt 1 i(l−k)ωt 1 
i(l−k)ωt 
he ,e i = e dt = ·e =0.
T 0 i(l − k)ωT 
t=0

b) Let k, l ∈ N with k 6= l. Using

Z
t + sin(t) cos(t)
cos2 (t) dt =
2
t − sin(t) cos(t)
Z
sin2 (t) dt =
2
Z 2
sin (t)
sin(t) cos(t) dt =
2
immediately gives
2
2 T
Z 
1 1 1
2h √ , √ i = √ dt = 1
2 2 T 0 2
√ Z T √ t=T
1 2 2 
2h √ , cos(kωt)i = cos(kωt) dt = sin(kωt) =0

2 T 0 kωT 
t=0
√ Z T √ t=T
1 2 − 2 
2h √ , sin(kωt)i = sin(kωt) dt = cos(kωt) =0

2 T 0 kωT 
t=0
Z T t=T
2 2 kωt + sin(kωt) cos(kωt) 
2hcos(kωt), cos(kωt)i = cos (kωt) dt = =1
T 0 kωT 
t=0
t=T
2 T −
Z
kωt sin(kωt) cos(kωt) 
2hsin(kωt), sin(kωt)i = sin2 (kωt) dt =  =1
T 0 kωT 
t=0
t=T
2 T sin2 (kωt) 
Z
2hsin(kωt), cos(kωt)i = sin(kωt) cos(kωt) dt = =0
T 0 kωT t=0
2
In the sequel let T > 0 and ω := 2π
T
be fixed.
Theorem 6.65.
If the Fourier series ∞
P ikωt a0
P∞
k=−∞ γk e = 2
+ k=1 ak cos(kωt) + bk sin(kωt), γk ∈ C,
ak , bk ∈ C, converges uniformly on [0, T ] to a function f (t), then f (t) is continuous
and
1 T −ikωt
Z
ikωt
γk = he , f (t)i = e f (t) dt , k ∈ Z (6.10)
T 0
2 T
Z
ak = 2hcos(kωt), f (t)i = cos(kωt)f (t) dt , k ∈ N0 (6.11)
T 0
6.7 Fourier series 167

Z T
2
bk = 2hsin(kωt), f (t)i = sin(kωt)f (t) dt , k ∈ N. (6.12)
T 0

Proof: For n ∈ N set fn := nk=−n γk eikωt ∈ C([0, T ], C). By assumption, the sequence
P
(fn )n∈N converges uniformly to f . By Theorem 3.16 the function f must be continuous.
Furthermore, for fixed k ∈ N0 , e−ikωt fn converges uniformly to e−ikωt f as

||e−ikωt fn − e−ikωt f ||∞ = ||e−ikωt (fn − f )||∞ = ||fn − f ||∞ −→ 0 .

|e−ikωt |=1 n→∞

Thus using Theorem 6.64 a) we obtain

T
1 T 1 T −ikωt
Z Z Z
1 −ikωt −ikωt
e f (t) dt = lim (e fn (t)) dt = lim e fn (t) dt
T 0 T 0 n→∞ n→∞ T 0

= lim heikωt , fn (t)i = γk .

n→∞ | {z }
=γk for n ≥ k

Analogously using Theorem 6.64 b) we obtain the stated formulas for ak , k ≥ 0, and bk ,
k > 0.
2
Note that for arbitrary f ∈ R([0, T ], C) the coefficients γk , ak , bk defined in (6.10) to (6.12)
fulfil the relations (6.4) to (6.9). We will mainly restrict to piecewise continuous functions.
Definition 6.66. Fourier series
For a piecewise continuous function f : [0, T ] → C the Fourier coefficients (γk )k∈Z ,
(ak )k∈N0 , (bk )k∈N of f are defined by (6.10) to (6.12) and the Fourier sereies of f is
defined by
∞ ∞
X
ikωt a0 X
F (f )(t) := γk e = + ak cos(kωt) + bk sin(kωt) .
k=−∞
2 k=1

Theorem 6.67.
Let f : R → C be a piecewise continuous T -periodic function.

a) If f is even, that is f (t) = f (−t) for all t ∈ R, then

Z T
4 2
ak := f (t) cos(kωt) dt
T 0
bk = 0.

b) If f is odd, that is f (t) = −f (−t) for all t ∈ R, then

ak = 0
Z T
4 2
bk := f (t) sin(kωt) dt .
T 0
168 6 The Riemann Integral

Proof: We only prove a) since b) follows analogously.

2 T 2 −T
Z Z
bk = f (t) sin(kωt) dt = f (−τ ) sin(kωτ ) dτ
T 0 τ :=−t T 0

2 0 2 T
Z Z
= − f (τ ) sin(kωτ ) dτ = − f (τ ) sin(kωτ ) dτ = −bk
T −T T 0

This implies bk = 0. Furthermore, we compute

Z T
2 T
Z
2 2
ak = f (t) cos(kωt) dt = f (t) cos(kωt) dt
T 0 T − T2
Z 0 Z T !
2 2
= f (t) cos(kωt) dt + f (t) cos(kωt) dt
T − T2 0
Z T Z T !
2 2 2
= f (−τ ) cos(kωτ ) dτ + f (t) cos(kωt) dt
τ :=−t T 0 0
Z T
4 2
= f (t) cos(kωt) dt .
T 0

In the following we list some calculation rules for Fourier series which are easily deduced
and can be proved as an exercise.
Lemma 6.68.
Suppose that f, g : R → C are T -periodic piecewise continuous functions with
∞ ∞
X a0 X
F (f ) = F (f )(t) = γk eikωt = + ak cos(kωt) + bk sin(kωt)
k=−∞
2 k=1
X∞
F (g) = F (g)(t) = δk eikωt
k=−∞

a) linearity:
∞
X
F (αf + βg) = αF (f ) + βF (g) = (αγk + βδk )eikωt
k=−∞

b) conjugation:
∞
X
F (f ) = γ−k eikωt
k=−∞

c) time reversal:
∞
X
F (f (−t)) = γ−k eikωt
k=−∞

d) stretch:
∞
X
F (f (ct)) = γk eik(cω)t , c>0
k=−∞
6.7 Fourier series 169

e) shift:
∞
X
F (f (t + a)) = (γk eikωa )γk eikωt , a∈R
k=−∞
X∞
F (einωt f (t)) = γk−n eikωt , n∈Z
k=−∞

f ) derivation: If f is continuous and piecewise continuously differentiable, then

∞
X
F (f 0 ) = (ikωγk )eikωt
k=−∞
X∞
(kω)(bk cos(kωt) − ak sin(kωt)) .
k=1

RT
g) integration: If 0 = γ0 = 1
T 0
f (t) dt, then
Z t  Z T ∞
1 X bk ak
F f (τ ) dτ = − tf (t) dt − cos(kωt) − sin(kωt) .
0 T 0 k=1
kω kω

Without proof we state the following theorem on convergence of Fourier series.

Theorem 6.69. convergence
Let f : R → C be T -perodic and piecewise continuously differentiable with Fourier
series
∞
a0 X
F (f )(t) = + ak cos(kωt) + bk sin(kωt).
2 k=1

a) F (f )(t) converges pointwisely and for all t ∈ R holds:

1
F (f )(t) = (lim f (t) + lim f (t)) .
2 s&t s%t
| {z } | {z }
f (t+ ):= f (t− ):=

b) On each compact interval on which f is continuous, the Fourier series con-

verges uniformly.
c) In each point t of discontinuity the so-called Gibbs phenomenon can be ob-
served, which says that for δ := |f (t+ ) − f (t− )| > 0 and
1 π sin(t)
Z
1
ρ := dt − ≈ 0.0895 ≈ 9%
π 0 t 2
holds
|F (f )(t) − f (t+ )| ≈ ρ · δ ≈ |F (f )(t) − f (t− )|
i.e. the “error” between f and its Fourier series is approximately 9% of the
step size δ of the discontinuity.
170 6 The Riemann Integral

The following theorem clarifies the approximation quality of Fourier series.

Theorem 6.70.
Let f : R → C be T -periodic and piecewise continuous with Fourier series
∞ ∞
X
ikωt a0 X
F (f )(t) = γk e = + ak cos(kωt) + bk sin(kωt)
k=−∞
2 k=1

and let
n
a0 X
Sn (t) := + (ak cos(kωt) + bk sin(kωt)).
2 k=1

denote the corresponding partial sums.

a) Sn (t) is the orthogonal projection of f onto the subspace

1
Tn := span{ √ , cos(ωt), ..., cos(nωt), sin(ωt), ..., cos(n sin t)} ⊂ C([0, T ], C)
2
for n ∈ N0 with respect to the inner product
Z
2
2hu, vi = u(t)v(t) dt .
T

Setting ||u|| := 2hu, ui this means that for all g ∈ Tn holds

p

||f − Sn || ≤ ||f − g|| .

b) The so-called Bessel inequality holds:

n T
|a0 |2 X
Z
2
+ (|ak |2 + |bk |2 ) ≤ |f (t)|2 dt .
2 k=1
T 0

This especially implies the convergence of the series

∞
X ∞
X
2
|ak | , |bk |2
k=0 k=0

and therefore also limk→∞ ak = limk→∞ bk = 0

c) If f is (m − 1)-times continuously differentiable and (m + 1)-times picewise

continuously differentiable, then there is a constant C > 0 such that
C
|γk | ≤
|k|m+1

for all k ∈ Z\{0} .

Proof: a) This is clear by Theorem 6.64.

6.7 Fourier series 171

b)

0 ≤ ||f − Sn ||2 = 2hf − Sn , f − Sn i

n
a0 X
= ||f ||2 − 4 Rehf, + (ak cos(kωt) + bk sin(kωt))i + ||Sn ||2
2 k=1
n
2 |a0 |2 X
= ||f || − − |ak |2 + |bk |2 .
2 k=1

c) By using the rule for differentiation of Lemma 6.68 it is sufficient to show the assertion
for m = 0. Therefore let f be piecewise continuously differentiable. Choose a decom-
position 0 = t0 < t1 < ... < tm = T such that f |[ti ,ti+1 ] , i = 0, ..., m − 1, is continuously
differentiable. Then integration by parts yields
Z T
γk = f (t)eikωt dt
0
m−1 Z tj+1 !
1 X t=t
f (t)e−ikωt t=tj − f 0 (t)eikωt dt
 j+1
= −
ikω j=0 tj

and therefore
m−1 Z T !
1 1X 1
|γk | ≤ |f (t− +
j+1 )| + |f (tj )| + |f 0 (t)| dt .
k ω j=0 ω 0
| {z }
C:=

2
Finally, without proof, we state the following theorem on uniqueness of Fourier series:
Theorem 6.71.
If two T-periodic, piecewise continuous functions f, g : R → C have the same
Fourier series and if they fulfill
1
f (t) = (f (t− ) + f (t+ ))
2
1
g(t) = (g(t− ) + g(t+ ))
2

for all t ∈ R, then f = g.

 Index

C 1 -curve, 156 bounded, 15, 21, 31, 51

T -periodic, 163 bounded from above, 20, 21
ε-neighbourhood of x, 30 bounded from below, 20, 21
n-th derivative of a function, 111
n-th partial sum of (an )n∈N , 36 Cauchy sequence, 29
n-times continuously differentiable, 111 chain, 6
n-times differentiable, 111 closed, 31, 156
(improper) accumulation value −∞, 27 closure of C, 33
(improper) accumulation value ∞, 27 commutative ring, 8
(natural) logarithm, 66 compact, 31
concave, 113
continuous in x0 ∈ I, 57 continuous on I, 57
piecewise C 1 -curve, 156 contraction constant, 122
uniformly continuous, 152 contractions, 122
absolute value, 10 convergent, 36, 145, 146, 149
absolutely convergent, 41 convergent to a ∈ F, 13
accumulation value, 27 convex, 113
addition, 7 curvature, 159
antiderivative of f , 135 curve, 156
arc length function, 159 curve discussion, 114
arcus cosinus, 79 curve integrals, 160
arcus sinus, 79 cycloid, 156
arcus tangens, 80
area cosinus hyperbolicus, 73 decomposition, 127
area hyperbolic cosine, 73 degree of p, 80
area hyperbolic sine, 73 derivative of f at x0 , 99
area hyperbolic tangent, 73 difference quotient, 98
area sinus hyperbolicus, 73 differentiable at x0 ∈ I, 98
area tangens hyperbolicus, 73 differentiable in I, 99
divergence, 145, 146, 149
Bessel inequality, 170 divergent, 13
boundary of C, 33 divergent to −∞, 20
boundary points of C, 33 divergent to ∞, 20

173
174 INDEX

domain of convergence, 93 multiplicity, 84

enclosed, 160 natural numbers, 7

end point, 156 neighbourhood of x, 31
equivalence relation, 6 Newton iteration, 124
equivalent class, 6
Euler’s number, 63 open, 31
expansion point, 116 order, 84
exponential function, 63 osculating plane, 159
exponential series, 44 osculation points of C, 33

Fourier coefficients, 167 parametrisation with respect to the arc

Fourier sereies, 167 length, 159
Fourier series, 163 partial fraction, 89
frequency, 163 partial fraction decomposition, 90
functionals, 129 partial order, 6
partition, 127
geometric series, 36 periodic with period T ∈ R, 163
Gibbs phenomenon, 169 pointwisely convergent, 51
harmonic series, 37 polynomial, 80
hyperbolic cosine (cosinus hyperbolicus), power series, 93
69 proper, 89
hyperbolic sine (sinus hyperbolicus), 69 rational function, 87
hyperbolic tangent (tangens rectifiable, 157
hyperbolicus), 72 regular, 156
infimum of M , 21 relation, 6
infinite series, 36 reordering, 48
inflection point, 114 reparametrisation, 157
inner points of C, 33 Riemann lower integral, 129
integral, 128 Riemann upper integral, 129
interior of C, 33 Riemann Zeta Function, 38
inverse cosine, 79 Riemann-integrable, 132
inverse sine, 79
secant, 98
inverse tangent, 80
second derivative, 111
leading coefficient, 80 sequence, 7
length, 157 sequence in M , 11
limit inferior of (an )n∈N , 27 series, 36
limit superior of (an )n∈N , 27 slope, 98
local extremum, 106 smooth, 156
local maximum (local minimum), 106 starting point, 156
lower bound of M , 21 step function, 127
strictly convex/concave, 113
maximal, 21 strictly monotonically decreasing, 20
minimal, 21 strictly monotonically increasing, 20
moment of inertia, 162 strictly proper, 89
monotonically decreasing, 20 supremum of M , 21
monotonically increasing, 20
multiplication, 7 the integers, 7
INDEX 175

the limit of f as x tends from the left to trigonometric polynomials, 163

x0 , 55 trigonometric series, 163
the limit of f as x tends from the right
unbounded, 15
to x0 , 56
uniformly convergent, 52, 154
the limit of f as x tends to x0 , 55 unit normal vector, 159
the rational numbers, 8 unit tangent vector, 159
total order, 6 upper bound of M , 21