Resume
Summary Advanced Mathematics for Business & Technology
- Établissement
- Vrije Universiteit Brussel (VUB)
Summary of Advanced Mathematics for Business & Technology for BA3 Business Economics at the VUB.
[Montrer plus]
2 revues
Par: octavian-gabrielmatei • 4 mois de cela
Par: lukashuysmans • 2 semaines de cela
Aperçu du contenu
AdvancedAdvanced Mathematics
Mathematics for
for Business
Business & Technology
& Technology
1 Sequences and Series
1.1 Sequences ....... eee
1.2 InfiniteSeries 2.2... eee
1.3. The Divergence and Integral Tests 2.2... 0... 0. ee eee eee eee
1.4 Comparison Tests... 2... 2. eee
1.5 Alternating Series... 2 eee
1.6 Ratioand Root Tests 2... 0... 2. ee eee
1.7 Chapter Review .. 0... ee
2 Power Series
2.1 Power Series and Functions . 2... . 2 0. ee ee
2.2 Properties of Power Series... 2... 2 ee ees
2.3 Taylor and MacLaurin Series... 2... ee ee
2.4 Working with Taylor Series . 2... 2. ee ee
2.5 Chapter Review... 2.2... eee
3 First order equations
3.1 Integralsassolutions .. 2... 2. ee ee
3.2 Slope fields... 2... ee eee
3.3. Separable equations... 2... eee
3.4 Linear equations and the integrating factor .. 2... 2...
ee eee
3.5 Substitution . 2... ee eee
4 Higher order linear ODEs
4.1 Second orderlinear ODEs .... 1... 0... ee
4.2 Constant coefficient second order linear ODEs .............200.
4.3 Higher order linear ODEs .. 2... . ee eee
4.4 Mechanical vibrations ...... 0.0... ee ee
4.5 Nonhomogeneous equations... 6... eee
5 Systems of ODEs
5.1 Introduction to systems of ODEs .. 2... 2.2 ce eee
5.2 Matrices and linearsystems .... 2.2... 0... eee ee ee
5.3 Linear systems of ODEs .. 1... 0... ce ee eee
5.4 Eigenvaluemethod .. 2... 2... ee eee
5.5 Two-dimensional systems and their vector fields ..............0..
5.6 Second order systems and applications... .. 2... 0. ee ee ee eee
5.7 Multiple eigenvalues . 2... ee eee
5.8 Matrix exponentials... .. ee ee
5.9 Nonhomogeneous systems... 6. ee
6 Fourier series and PDEs
6.1 Boundary value problems ... 2... 0... ce ee ee
6.2 The trigonometric series... 2... ee ee
6.3 More onthe Fourier series... . ee eee
6.4 Sine and cosine series... 1... ee es
6.5 Applications of Fourier series... 2... ee eee
7 Nonlinear systems
7.1 Linearization, critical points, and equilibria... .. 2... ........040.
7.2 Stability and classification of isolated critical points ..............
7.3 Applications of nonlinear systems .... 2... 2... eee eee eee
74 Limitcycles 2... eee
75 Chaos... ee ee
1
,Part
Part 1:
1: Sequences
Sequences andSeries
and Series
Chapter
Chapter 1:
1: Sequences
Sequences
“
NTE 4
Figure 5.1 The Koch snowflake is constructed by using an iterative process. Starting with an equilateral triangle, at each step
of the process the middle third of each line segment is removed and replaced with an equilateral triangle pointing outward.
Chapter Outline
5.1 Sequences
5.2 Infinite Series
5.3 The Divergence and Integral Tests
5.4 Comparison Tests
5.5 Alternating Series
5.6 Ratio and Root Tests
Introduction
The Koch snowflake is constructed from an infinite number of nonoverlapping equilateral triangles. Consequently, we can
express its area as a sum of infinitely many terms. How do we add an infinite number of terms? Can a sum of an infinite
number of terms be finite? To answer these questions, we need to introduce the concept of an infinite series, a sum with
infinitely many terms. Having defined the necessary tools, we will be able to calculate the area of the Koch snowflake (see
Example 5.8).
The topic of infinite series may seem unrelated to differential and integral calculus. In fact, an infinite series whose terms
involve powers of a variable is a powerful tool that we can use to express functions as “infinite polynomials.” We can
use infinite series to evaluate complicated functions, approximate definite integrals, and create new functions. In addition,
infinite series are used to solve differential equations that model physical behavior, from tiny electronic circuits to Earth-
orbiting satellites.
Definition
An infinite sequence {a,,} is an ordered list of numbers of the form
Ay, AQ,---, Any...
The subscript 7 is called the index variable of the sequence. Each number a,, is a term of the sequence. Sometimes
sequences are defined by explicit formulas, in which case a, = f(n) for some function f(n) defined over the
positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one
term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the
sequence.
Explicit
Explicit Formula:
Formula: a
ann =
= 2n
2n Recurrence
Recurrence Formula:
Formula: a
a11 == 2
2
aanti
n+1 =
= an+2
ant2
2
,Arithmetic
Arithmetic sequences
sequences and
and Geometric
Geometric sequences:
sequences:
sequences are
Arithmetic sequences
Arithmetic are sequences
sequences where
where consecutive
consecutive terms
terms have
have constant
constant differences:
differences:
3,
3,5,5, 7,
7,9,9, 11,
11, …
...
> Explicit:
ð Explicit: a
a,n =
= 3
3 +
+ 2(n-1)
2(n-1)
> OR:
ð OR: an = 1 + 2n
an=1+2n
o Be
o careful with
Becareful with regards
regards to
to the
the first
first term
term of
of the
the sequence,
sequence, that’s
that’s why
why there
there are
are
multiple
multiple ways
ways of
of writing
writing it,
it, with
with n-1
n-1 or
orn.n.
= Recursive:
ð Recursive: a
a11 =
= 3
3 and
and a n+1 =
anti an + 2
=an+2
Geometric sequences are
Geometric sequences are sequences
sequences where
where the
the ratio
ratio of
of every
every pair
pair of
of consecutive
consecutive terms
terms is
is
constant:
constant:
3,
3,6,6, 12,
12, 24,
24, …
...
n-1
> Explicit:
ð Explicit: aann == 33 ** (2)
(2)™4
=> Recursive:
ð Recursive: aa11 =
= 3
3 and
and a n+1 =
anii = a
ann ** 2
2
Examples:
Examples:
Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the nth term of the sequence.
4’ 7 10’ 13’ 16°"
Solution
a. First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are
negative, and the even terms are positive. Therefore, the nth term includes a factor of (—1)”. Next,
consider the sequence of numerators {1, 2, 3,...} and the sequence of denominators {2, 3, 4,...}.
We can see that both of these sequences are arithmetic sequences. The mth term in the sequence of
numerators is m, andthe nth term in the sequence of denominators is n+ 1. Therefore, the sequence
can be described by the explicit formula
(-1)"n
an=
n+1-
b. The sequence of numerators 3, 9, 27, 81, 243,... is a geometric sequence. The numerator of the
nth term is 3” The sequence of denominators 4, 7, 10, 13, 16,... is an arithmetic sequence. The
denominator of the nth term is 4+ 3(n — 1) = 3n + 1. Therefore, we can describe the sequence by the
explicit formula a, = > T
3
, Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence
a a);=2, a,= -—3a,_ 1 forn>2
n
b. a, =5, an = 4,1 +(4) for
n>2
Solution
a. Writing out the first few terms, we have
a,=2
a, = —-3a,;=-3(2) = -6
=-6
a3 = —3a, =(-3)?2 =1212
=
a4 = —3a3 = (-3)°2. =-18
= -18
In general,
an = 2-3)".
b. Write out the first few terms:
ay= 4
ares adef—} 2
3
= 1) ~3,1_7
as=ar+(3) =f+3= 4
4
= 1) ~7, 1 15
ag=as+(3) =$+76=4
From this pattern, we derive the explicit formula
n
an = 2 =1- Fh.
Limit
Limit of
of a
a Sequence:
Sequence:
A fundamental question that arises regarding infinite sequences is the behavior of the terms as n gets larger. Since a
sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as n > oo. For
example, consider the following four sequences and their different behaviors as n — oo (see Figure 5.3):
a. {1+3n} = {4, 7, 10, 13,...}. The terms 1 + 3m become arbitrarily large as m > oo. In this case, we say that
1+3n—-ow ano.
b. {1 - (4) | = {4, 3, Z 45....}, The terms 1 — (4)’ > 1lasn-oo.
2
c. {(-1)"}={-1, 1, -1, 1,...}. The terms alternate but do not approach one single value as n > co. => Does
=> Does not
not
exist
4
Les avantages d'acheter des résumés chez Stuvia:
Qualité garantie par les avis des clients
Les clients de Stuvia ont évalués plus de 700 000 résumés. C'est comme ça que vous savez que vous achetez les meilleurs documents.
L’achat facile et rapide
Vous pouvez payer rapidement avec iDeal, carte de crédit ou Stuvia-crédit pour les résumés. Il n'y a pas d'adhésion nécessaire.
Focus sur l’essentiel
Vos camarades écrivent eux-mêmes les notes d’étude, c’est pourquoi les documents sont toujours fiables et à jour. Cela garantit que vous arrivez rapidement au coeur du matériel.
Foire aux questions
Qu'est-ce que j'obtiens en achetant ce document ?
Vous obtenez un PDF, disponible immédiatement après votre achat. Le document acheté est accessible à tout moment, n'importe où et indéfiniment via votre profil.
Garantie de remboursement : comment ça marche ?
Notre garantie de satisfaction garantit que vous trouverez toujours un document d'étude qui vous convient. Vous remplissez un formulaire et notre équipe du service client s'occupe du reste.
Auprès de qui est-ce que j'achète ce résumé ?
Stuvia est une place de marché. Alors, vous n'achetez donc pas ce document chez nous, mais auprès du vendeur Madikan. Stuvia facilite les paiements au vendeur.
Est-ce que j'aurai un abonnement?
Non, vous n'achetez ce résumé que pour €6,49. Vous n'êtes lié à rien après votre achat.
Peut-on faire confiance à Stuvia ?
4.6 étoiles sur Google & Trustpilot (+1000 avis)
52928 résumés ont été vendus ces 30 derniers jours
Fondée en 2010, la référence pour acheter des résumés depuis déjà 14 ans