Mathematics for Economists and Social Sciences

Cristian Necul¼aescu

(Cristian Necul¼aescu) Academy of Economic Studies, room 2625, Calea DorobanTi nr.11-13, sector 1, BucureSti, RomâniaE-mail address, Cristian Necul¼aescu: math.fabiz.b.09.10@gmail.com

Dedicated to the memory of my Teachers and Professors: Dan Jebeleanu, Gheorghe Pântea, Aristide Halanay and StefanMiric¼a.

Contents

Part 1. Calculus 1

Chapter 1. In�nite series 31.1. Introduction 31.2. Special cases 61.3. Convergence Tests for positive series 91.4. Convergence tests for general series 111.5. Convergence tests for alternating series 121.6. Some formulas and exercises 121.7. A Macroeconomical Example 121.8. Power Series 131.9. Taylor�s expansions 14

Chapter 2. Functions of several variables (2 lectures) 172.1. Introduction 172.2. Continuity 192.3. Derivatives 212.4. Higher order derivatives 222.5. Applications in Economics 232.6. The implicit function theorem 292.7. Taylor Polynomials 312.8. Extreme points 332.9. Unconstrained Local Optimization 342.10. Constrained Optimization 392.11. Functions of several variables. Limit, continuity, partial derivatives, di¤erentiability and

di¤erential. Extremes. 422.12. Unconstrained optimization. Approximating functions by Least Square Method. 43

Chapter 3. Ordinary di¤erential equations (1/2 lecture) 47

Chapter 4. Finite di¤erence equations (1/2 lecture) 53

Chapter 5. Improper integrals. Euler functions: Gama, Beta 55

Chapter 6. Applications of Calculus to economic modelling 57

Part 2. Probabilities (7 lectures) 59

Chapter 7. Events. Probability: classic and axiomatic de�nition. Field of events. Properties ofprobability. 61

iii

iv Cristian Necul¼aescu

Chapter 8. Conditional probability. Probability of a union/intersection of events. Total probabilityformula. Bayes formulas. Classical probability schemes. 63

Chapter 9. De�nition of a random variable. Operations with random variables. Examples on thediscrete case. Cumulative distribution function: de�nition, properties. Functions ofrandom variables. 65

Chapter 10. Continuous random variables. Probability density function: de�nition, properties. 67

Chapter 11. Moments of random variables. Expectation and variance. Properties. Chebyshevinequality. 69

Chapter 12. Discrete bivariate random variables: marginal distributions, moments, conditionaldistributions, covariance, correlation. 71

Chapter 13. Discrete and classical distributions. Applications of probability theory to economicmodelling. 73

Chapter 14. Convexity 75

Appendix A. * High School Revision 77A.1. Sets 77A.2. Usual Number Sets. Countability 80A.3. Minorants, majorants 83A.4. Relations 84A.5. Functions 85A.6. Binary Logic 89A.7. Database applications for Logic, Sets, Relations and functions 92A.8. Sequences 94A.9. Symbols 99

Appendix B. Topology 101

Appendix C. Functions of one variable 105

Appendix. Bibliography 107

Part 1

Calculus

CHAPTER 1

In�nite series

"Divergent series are the invention of the devil,and it is shameful to base on themany demonstration whatsoever." Abel, 1828

The starting point for the main body of these Lecture Notes is the level of knowledge given in Mathe-matics by "High School graduate, with the maximum concentration on Mathematics". Broadly speaking,this means all "Precalculus", "Geometry and Trigonometry", "Analytic Geometry", "Linear Algebra �linear systems, matrices, determinants", "Abstract Algebra �groups, �elds, rings", "Calculus � limits,continuity, derivability, graphs of functions", "Calculus �elementary integrals". In the Appendix it maybe found a brief review of some of these topics; still, you may �nd useful to keep close appropriate high�school texts. During the lectures and seminars, each of you is welcomed to ask questions and to comment.As Murphy says, "Science advances when the student asks and the teacher doesn�t know the answer".

1.1. Introduction

Consider a sequence of real numbers denoted (an)n2N�.

1.1.1. De�nition (Formal). The symbol1Pn=1

anDef= a1 + a2 + � � �+ an + � � � is called "series" or "real

series" or "real in�nite series";the number an is called "the [general] term of the series";

the number Sn de�ned by Sn = a1 + a2 + a3 + � � �+ an =nPk=1

ak is called "the nth order partial sum

of the series";the sequence (Sn)n2N� is called "the sequence of partial sums of the series".

1.1.2. De�nition (Informal). A series is an "in�nite summation" or (more precise) a "discrete in�nitesummation" or a "countable in�nite summation".

1.1.3. Remark. Series as an abstract mathematical model may be found in "Macroeconomics" rep-resenting "discrete dynamics" or "inde�nite discrete �nancial �ows"; a typical situation describes the(expected) present value of a future accumulation process in which the accumulation will take place at aninde�nite number of future moments (e.g. dividends, insurance). The detailed study of these situationsis beyond the purpose of the present text �the interested reader may consult titles like [12] or [13]. Youmay see at the end of this chapter a little Macroeconomic model.

1.1.4. Example. A series:1Pn=0

�3 � 2

n

5n

�; The sign "

P" comes from the capital greek letter "sigma".

The general term: an = 3 �2n

5n.

3

4

!!! Pay attention at the �rst term (which is not always 0 or 1), located at the bottom of the summationsymbol:

1Xn= !!!

�rst term

an

The sequence of partial sums of the series: Sn =nPk=0

ak =nPk=0

3 � 2k

5k; in this particular case we may

obtain an explicit form for Sn:nPk=0

3 � 2k

5k= 3 �

nPk=0

�2

5

�k= 3 �

1��2

5

�n+11� 2

5

= 5

"1�

�2

5

�n+1#= Sn.

It may be seen that 9 limn!1

Sn = 5.

1.1.5. Remark. Between the sequences (an)n2N and (Sn)n2N there are certain recurrence relations:Sn+1 = Sn + an+1 (or an+1 = Sn+1 � Sn), 8n 2 N�.

1.1.6. De�nition (convergence/divergence). The symbol1Pn=1

an is called "convergent" (we say "it

converges") if the sequence (Sn)n2N is convergent (converges); only in this case is the value S = limn!1

Sn

called "the sum of the series";the sequence (Sn � S)n2N (the di¤erence between the partial sum and the sum) is "the remainder

sequence" and it vanishes (it tends toward 0);1Pn=1

an is divergent (diverges) if (Sn)n2N is divergent (diverges).

1.1.7. Example. For the previous example, since limn!1

Sn = 5 we conclude that the series1Pn=0

�3 � 2

n

5n

�converges and the sum is 5 (the value of the limit). We write

1Pn=0

�3 � 2

n

5n

�= 5.

1.1.8. Theorem. [Divergence test] limn!1

an 6= 0)Pn2N

an diverges:

IF the general term does not tend towards zero,THEN the series diverges.

Proof. By contradiction: the statement (an 6! 0 )Pn2N

an diverges) is logically equivalent with the

statement (Pn2N

an converges ) an ! 0).Pn2N

an convergesfrom de�nition) 9S = lim

n!1Sn ) an = Sn � Sn�1 !

n!1S � S = 0: �

The behavior of a series (convergent or divergent) is a qualitative information, called "the natureof the series". When convergent we may also talk about the sum of the series, which is quantitativeinformation [conditioned by the qualitative information]. A di¤erence between the two types of information(quantitative and qualitative) is that usually the algorithms embedded in software products are built uponthe claim that the qualitative part is satis�ed �and so the usage of some software products for situationswhere the qualitative part is not satis�ed may lead to unexpected results. As a general rule, it is advisableto separate the qualitative and quantitative studies.

5

1.1.9.Remark. There are some signi�cant di¤erences between �nite and in�nite addition (summation);some of them:

(1) While �nite addition always exist, this is not the case with in�nite addition.(2) While �nite addition is commutative, the rearrangement of the terms of an in�nite addition may

alter both the qualitative and the quantitative results.(3) While �nite addition is asociative, careless grouping and regrouping of the terms of an in�nite

addition is false and may lead to unexpected results.

1.1.10. Example. Consider1Pn=0

(�1)n. The series diverges because the general term doesn�t tend

towards zero. Still the following false line of reasoning "�nds the sum of the series":

S = 1� 1 + 1� 1 + 1� 1 + 1 + � � � )S � 1 = �1 + 1� 1 + 1 + � � � )S � 1 = � (1� 1 + 1� 1 + 1 + � � � ) = �S )

) 2S = 1) S =1

2:

The "result" is false and the unique mistake is "the notation" S =1Pn=0

(�1)n which implicitly and falsely

assumes that a number S exists and is equal with the abstract symbol1Pn=0

(�1)n.

1.1.11. Remark. Given a series, the inclusion/exclusion of a �nite number of terms doesn�t change thenature of the series [Because a �nite number of additions/substractions does not modify the existence ofa limit] [the series

Pn2N

an andP

n2Nnf0;1;��� ;kgan have the same nature]. Still, it may change the value of the

sum, when it exists.

1.1.12. Remark. While �nite addition is associative, in�nite addition is not always associative. Thismeans that in�nite grouping of the added objects sometimes changes the nature of the in�nite summation.Example: "0 = 1". False line of reasoning:

1 = 1 + 0 + 0 + � � �+ 0 + � � � == 1 + (�1 + 1) + (�1 + 1) + � � � (�1 + 1) + � � � == (1� 1) + (1� 1) + � � �+ (1� 1) + � � � == 0:

[The line of reasoning again makes the (hidden) false assumption that there is a number S equal to the

abstract symbol1Pn=0

(�1)n and falsely assumes that rearrangements are true for divergent series]

1.1.13. Remark. When convergent, the sum of a series is unique [because the limit of a sequence isunique].

1.1.14.Remark (Algebraic operations with series, Thms. 3.47, 3.50, 3.51 [14]). When the seriesPn2N

an

andPn2N

bn are both convergent and � 2 R, the seriesPn2N

(an + bn) andPn2N

(�an) are also convergent and

6

moreover, the following relations between the sums of series are valid:Pn2N

(an + bn) =Pn2N

an +Pn2N

bn;Pn2N

(�an) = �Pn2N

an:

1.1.15. Remark. In this result, the qualitative part is: "Pn2N

an andPn2N

bn are both convergent )

Pn2N

(an + bn) andPn2N

(�an) are also convergent" while the quantitative part is:

Pn2N

(an + bn) =Pn2N

an +Pn2N

bn;Pn2N

(�an) = �Pn2N

an:

1.1.16. Remark. The proof is based on translating the convergences in terms of ""�de�nitions".

1.2. Special cases

1.2.1. Arithmetic sequence (arithmetic progression). is a sequence of numbers so that thedi¤erence between any two consecutive terms is constant (and is called "common di¤erence") (Alternativecharacterization: For any three consecutive terms, the middle term is the arithmetic mean of boundaryterms)

an = a1 + (n� 1) d,Arithmetic series:nPk=1

ak =nPk=1

(a1 + (k � 1) d) = na1 +n (n� 1)

2d

[Used in �nance, simple interest formulas]

1.2.2. Geometric sequence (geometric progression). is a sequence of numbers such that theratio between two consecutive terms is constant (Alternative characterization: For any three consecutiveterms, the middle term is the geometric mean of the extreme terms).

an = a1 � rn�1, r 6= 1.Geometric Series:nPk=1

ak =nPk=1

a1 � rk�1 = a11� rn

1� r.P

n2Nan =

�convergent, a 2 (�1; 1)divergent, a 2 Rn (�1; 1)

In fact Sn = 1 + a+ a2 + a3 + � � �+ an =

8<: 1� an+1

1� a; a 6= 1

n; a = 1:

So limn!1

Sn =

( 1

1� a; a 2 (�1; 1)

does not exist or in�nite in rest.[Used in Finance, compounded interest formulas]

1.2.3. Harmonic sequence (harmonic progression). is a sequence of numbers such that thesequence of reciprocals is an arithmetic sequence:

an =1

a1 + (n� 1) d(such that any denominator is nonzero)

Harmonic series:

7

nPk=1

ak =nPk=1

1

a1 + (k � 1) d(no elementary formula available)

Interpretation: Given n (ordered) observations for a certain measurement (such that the observationsare comparable), say that an observation is a "record" if it is the greatest of all (up to it). Then the

expected number of records is 1 +1

2+1

3+ � � �+ 1

n.

1.2.4. The number e. e =1Pn=0

1

n!

1.2.1. Theorem (Thm. 3.31, [14]). limn!1

�1 +

1

n

�n= e.

1.2.2. Theorem (Thm. 3.32, [14]). The number e is irrational.

1.2.3. Example (Achille and the Turtle (Zenon paradox); also see Section 1.3 [3]). Achilles (A) andthe Turtle (T) race together. It is assumed that Achilles�speed is much bigger than the Turtle�s speed, socommon sense tells that even if Achille gives the Turtle an initial advantage, he will still win the race.The following line of reasoning has been known since Ancient Greece as the "Zenon paradox":Denote A�s speed vA and T�s speed vT (with vA > vT ). Consider the advance given by A in the form of

distance S0. A starts the race only when T covers S0. Then A starts and until he also covers S0 T alreadycovers another distance called S1. In the time needed by A to cover the new distance S1, T covers a newdistance S2, and so on. "Common sense" says that the distances Sn even if they are increasingly smaller,they are always strictly positive. This is interpreted in the following manner: "Achilles will never outrunthe Turtle, because the Turtle will always have a strictly positive distance in advance".

T�s total advantage is (the geometric series):1Pn=0

S0

�vTvA

�n= lim

n!1S0

1��vTvA

�n+11� vT

vA

=S0vAvA � vT

[Al-

though the Turtle�s total advantage is an in�nite sum of strictly positive distances, the total value of thesum is �nite]

The time Achille needs to cover this distance isS0

vA � vTwhich is equal with (the geometric series)

1Pn=0

S0vA

�vTvA

�n.

8

1.2.4. Example (Telescoping/collapsing series). Consider the series1Pn=1

1

n (n+ 1). It is convergent and

is a "telescopic series" in the sense that the summay be calculated "elementary", by successive cancellation:

1

k (k + 1)=1

k� 1

k + 1; so that

nPk=1

1

k (k + 1)=

nPk=1

�1

k� 1

k + 1

�=

=1

1� 12=

+1

2=

� 13=

+

+� � �=� � �=+

+1

n=

� 1

n+ 1

= 1� 1

n+ 1

) Sn = 1�1

n+ 1;

) 9 limn!1

Sn (so the series is convergent)

and limn!1

Sn = 1 (so the sum is 1) �

CAUTION: The sumnPk=1

�1

k� 1

k + 1

�has a �nite number of terms so it is not wrong to write

nPk=1

1

k�

nPk=1

1

k + 1.On the contrary, in the situation

1Pn=1

�1

n� 1

n+ 1

�, because of the in�nite number of terms, it

is wrong to write1Pn=1

1

n�

1Pn=1

1

n+ 1; both series are divergent so that actually we have:

1Xn=1

�1

n� 1

n+ 1

�" = "

1Xn=1

1

n�

1Xn=1

1

n+ 1() 1" = "1�1:

�

1.2.5. Example.1Pn=1

�pn+ 2� 2

pn+ 1 +

pn�

1Pn=1

�pn+ 2� 2

pn+ 1 +

pn�=

=1Pn=1

�pn+ 2�

pn+ 1 +

pn�

pn+ 1

�=

= limn!1

nPk=1

�pk + 2�

pk + 1 +

pk �

pk + 1

�=

= limn!1

�pn+ 2�

p2 + 1�

pn+ 1

�= 1�

p2

1.2.6. Exercise. For the following telescoping series, establish their nature and if convergent �nd thesum:

(1)1Pn=1

1pn+

pn+ 1

=1

9

(2)1Pn=1

1

n2 + 5n+ 6= 1

3

(3)1Pn=1

1

n2 + 4n+ 3= 5

12

(4)1Pn=1

lnn

n+ 1

(5)1Pn=1

3n2 + n� 1n2 � 2n+ 3

1.2.5. Various series classi�cations.

1.2.5.1. With respect to the convergence/divergence (and the type of divergence) of the sequence of

partial sums:

convergent%

series �! divergent �! sum equal to �1&

sum does not exist

1.2.5.2. With respect to the type of the general term:

general (an 2 R)%

series �! positive (an � 0)&

alternate (an = (�1)n bn, bn � 0)or an � an+1 < 0

1.3. Convergence Tests for positive series

The general term for positive series will be positive (an � 0) and strictly positive (an > 0) only whenrequired by the involved operations. The sum of these series always exists, but it may be in�nite (+1).

1.3.1. Theorem (Thm. 1.48, [3]). For a positive terms series, changing the order of the terms doesnot change the nature of the series or the value of the sum.

Proof. Consider1Pn=1

an with an > 0 for all n and1Pn=1

bn a rearrangement of the �rst series (that is,

the same terms in di¤erent order).

The sequence of the partial sums San =nPk=1

an is an increasing sequence (because San+1 = San+an+1 > San)

so it has a limit (which may be in�nite, denote it by Sa).

The sequence of the partial sums Sbn =nPk=1

bn is also an increasing sequence with limit Sb.

Consider an arbitrary �xed index n. Since b1, � � � , bn is a rearrangement of the terms an, there is pnthe biggest index for which bk = ank (pn = max fn1; � � � ; nkg). Then Sbn � Sapn � Sa so passing to limit forn!1 it follows that Sb � Sa. A similar argument leads to Sa � Sb so in fact Sa = Sb. �

10

1.3.2. Theorem (First Comparison Test; Thm. 1.49, [3]). Consider two series with positive termsPn2N

an andPn2N

bn so that there is an index n0 2 N for which 0 � an � bn, 8n � n0.

Then:(1) If

Pn2N

bn converges thenPn2N

an converges;

(2) IfPan diverges then

Pn2N

bn diverges.

Proof. Since the nature of the series does not change when substracting a �nite number of terms,it may be assumed that the inequality 0 � an � bn is valid for all n. Then between the partial sumssequences (which are increasing sequences for the present case) there is the relation San � Sbn for all nwhich means that when Sbn is bounded S

an is bounded too, and when S

an is unbounded S

bn is unbounded

too. �

Exercise: For the seriesPn2N

1

3n + 2use the inequality 3n + 2 � 3n ) 1

3n + 2� 1

3nand the �rst

comparison test to study the nature of the series.

Exercise: For the seriesPn2N

1pnuse the inequality

pn +

pn = 2

pn �

pn +

pn+ 1 ) 1p

n�

1pn+

pn+ 1

and the �rst comparison test to study the nature of the series.

1.3.3. Theorem (Ratio Comparison Test; Thm. 1.55, [3]). Consider two series with positive termsPn2N

an andPn2N

bn so that there is an index n0 2 N for whichan+1an

� bn+1bn

8n � n0.

Then:(1) If

Pn2N

bn converges thenPn2N

an also converges;

(2) IfPn2N

an diverges thenPn2N

bn also diverges.

Proof. Again consider thatan+1an

� bn+1bn

for all n. By multiplying all the inequalities from n = 0 up

to n = k� 1 it follows that aka0� bkb0so that ak �

a0b0bk for all k and The Comparison Test may be applied

to conclude the proof. �1.3.4. Theorem (Limit Comparison Test; Thm. 1.52 [3]). If 9 lim

n!1anbn= � 2 (0;1) then the seriesP

n2Nan and

Pn2N

bn have both the same nature.

1.3.5. Theorem (nth Root Test / Cauchy�s test, Thm. 1.65, [3]). For the seriesPn2N

an, an > 0.

If limn!1

npan = L 2 (0;1), then:

(1) For L < 1 the series converges;(2) For L > 1 the series diverges;(3) For L = 1 the test is inconclusive.

1.3.6. Theorem (Ratio Test / D�Alembert�s test, Thm. 1.62, [3]). For the seriesPn2N

an, an > 0.

11

If limn!1

an+1an

= L 2 (0;1), then:

(1) For L < 1 the series converges;(2) For L > 1 the series diverges;(3) For L = 1 the test is inconclusive.

1.3.7.Theorem (Integral Test, Thm. 1.57 [3]). Consider a function � (�) : [1;1)! R+ continuous anddecreasing. Then the series

Pn2N

� (n) converges if and only if the improper integralR11� (x) dx converges.

1.3.8. Theorem (Cauchy Condensation Test, Thm. 2.3 [3]). The seriesPn2N

an, an > 0 andPn2N

2n � a2n

have both the same nature.

1.3.9. Theorem. The p�seriesPn2N

1

npwith p 2 R is:

(1) Convergent if p > 1.(2) Divergent if p � 1.

1.3.10. Theorem (Schlömilch, Thm. 2.4 [3]). If an > 0 is eventually decreasing and the sequence nk

is strictly increasing such that�nk+1 � nknk � nk�1

�k

is a bounded sequence, then the seriesPn2N

an, an > 0 andPn2N

(nk+1 � nk) � ank have both the same nature.

1.3.11. Theorem (Raabe�s Test, Thm. 11, [9]). For a seriesPn2N

an with positive terms (an > 0),

suppose the limit limn!1

n

�anan+1

� 1�exist and is equal with L. Then:

(1) If L > 1 then the series converges;(2) If L < 1 then the series diverges.(3) If L = 1 then the test is inconclusive.

1.4. Convergence tests for general series

1.4.1.De�nition. The seriesPn2N

an is called absolute convergent whenPn2N

janj is convergent (the series

of absolute values).

1.4.2. Remark. For a general series (with an 2 R) the series of absolute values is a positive termsseries, so the previous section applies to it.

1.4.3. De�nition. The seriesPn2N

an is called conditionally convergent when it is convergent but not

absolute convergent.

1.4.4. Theorem. If a series converges absolute then it converges (in the ordinary sense).

Proof. Consider an absolute convergent seriesPn2N

an. ThenPn2N

janj is convergent and:

0 � an + janj � 2 janj ) the seriesPn2N

(an + janj) is with positive terms and is dominated by a

convergent series so by The Comparison Test it is convergent.

12

Then because the seriesPn2N

(an + janj) andPn2N

janj are convergent, so it is their di¤erence:Pn2N

(an + janj)�Pn2N

janj =Pn2N

(an + janj � janj) =Pn2N

an. �

1.4.5. Theorem (Abel). IfPn2N

an converges and (bn)n2N is a bounded monotone sequence thenPn2N

anbn

converges.

1.4.6. Theorem (Dirichlet). IfPn2N

an has bounded partial sums and (bn)n2N is monotone and limn!1bn =

0, thenPn2N

anbn converges.

1.5. Convergence tests for alternating series

1.5.1. De�nition. The seriesPn2N

an is called alternating when an = (�1)n bn, bn > 0.

1.5.2. Theorem (Alternating series test / Leibniz, Thm. 1.75, [3]). If:

(1) 9n0 2 N, 8n � n0, bn+1 � bn.(2) lim

n!1bn = 0.

Then the alternating seriesPn2N

(�1)n bn, bn > 0 converges.

1.6. Some formulas and exercisesnPk=1

1 = n

nPk=1

k =n (n+ 1)

2nPk=1

k2 =n (n+ 1) (2n+ 1)

6nPk=1

k3 =

�n (n+ 1)

2

�21 + x+ x2 + � � �+ xn =

8<: 1� xn+1

1� x; x 6= 1

n+ 1; x = 1

1 + x+ x2 + � � �+ xn + � � � = limn!1

(1 + x+ x2 + � � �+ xn) =

( 1

1� x; x 2 (�1; 1)

6 9 or 1 otherwise1Pn=1

2n + 2 � 3n + 5n3n + 5n

1.7. A Macroeconomical Example

Optional Macroeconomics Topic for Series: Chapter 3, Doepke, Lehnert, Sellgren MACROECO-NOMICS, 1999.

13

1.7.1. Example. A typical Macroeconomics model, called "The household�s maximization problem",may look like this:

maxfctg1t=1

1Pt=1

�t�1u (ct) ;

subject to:1Pt=1

P (yt � ct)

(1 +R)t�1= 0:

It is beyond the goal of the present text to study such models. Here we just mention the economicalinterpretations expressed by means of series:

� t 2 N� means "(discrete) time" ( 0 means "now", 1 means "a year from now" and so on); themeasurement unit for time may be "year" or a certain unspeci�ed "period of time".

� the discussion is about a "household", and not an individual; one di¤erence is that while anindividual lives a �nite number of years, the household may be considered "to live forever" (aninde�nite number of years).

� the household uses a single commodity (say bananas) measured in quantities (kilos of bananas)both for income and consumption;�yt is "the household�s income for period t" (kilos of bananas) (exogeneous)�ct is "the household�s consumption for the period t" (kilos of bananas)�P is the price of one kilo of bananas (doesn�t change over time :) );�the household has access to a "bananas market", where it may buy (at price P ), sell (at priceP ) and invest money to buy bonds on the bananas market, which bear interest R (1 USDinvested gives the next period (1 +R) USD);

�u (�) is an increasing function of consumption, called "the household�s consumption utilityfunction"

� � 2 [0; 1] is "the household�s discount factor" and it is a way to express how much the householdcares for the current consumption as oposed to future consumption�� = 0 means that the household only cares about current consumption;�� = 1 means that the household cares equally about current inde�nite future consumption;�� = 0:95 (a typical value) should mean that the household cares a little more about thepresent than the future consumption.

With the above conventions, the initial problem says: "�nd the maximum present utility and theconsumption strategy to attain this, while keeping equal the present values of all future income and allfuture consumption".

1.8. Power Series

1.8.1. De�nition. Given a sequence of real numbers (an)n2N and a 2 R the series1Pn=0

an (x� a)n is a

power series around a and the numbers an are the coe¢ cients of the power series.

1.8.2. Theorem. For the power series1Pn=0

an (x� a)n put � = limn!1

npjanj (if it exists) and R =

1

�.

Then the power series converges if jx� aj < R and diverges if jx� aj > R (R is called the radius of

convergence). A similar result is valid when � = limn!1

jan+1jjanj

(if it exists) and R =1

�).

Proof. Apply the root test (or the ratio test). �

14

1.8.3. Remark. For the values x = a�R and x = a+R there is no way to decide in advance so theyhave to be studied separately for each case. The interval j[a�R; a+R]j is called "convergence domain"and for each case it has to be decided if it is left/right open/closed.

1.9. Taylor�s expansions

1.9.1. Theorem (Taylor, Thm 5.15, [14]). Consider f (�) : [a; b]! R and n 2 N�.If:

(1) The derivatives up to nth order exist and are continuous on [a; b] (for a and b consider lateralderivatives),

(2) The (n+ 1)th derivative exist on (a; b),(3) � < � 2 [a; b]

(4) The polynomial P (�) is de�ned by: P (x) =nPk=0

f (k) (�)

k!(x� �)k [P (�) is called �The nth order

Taylor polynomial in ��; P (�) �coincides�with f (�) in �, in the sense that P (k) (�) = f (k) (�),8k = 0; n].

Then: 9� 2 (�; �) such that f (�) = P (�) +f (n+1) (�)

(n+ 1)!(� � �)n+1.

Proof. Consider the function g (x) = f (x)� P (x)� f (�)� P (�)

(� � �)n+1(x� �)n+1.

Then:

g(j) (x) = f (j) (x)� P (j) (x)� f (�)� P (�)

(� � �)n+1(n+ 1)n (n� 1) � � � (n� j + 1) (x� �)n�j, 8j = 0; n

g (�) = f (�)� P (�)� f (�)� P (�)

(� � �)n(�� �)n = 0.

g (�) = f (�)� P (�)� f (�)� P (�)

(� � �)n(� � �)n = 0.

g(j) (�) = f (j) (�)� P (j) (�)� f (�)� P (�)

(� � �)nn (n� 1) � � � (n� j + 1) (�� �)n�j = 0, 8j = 0; n

g(n+1) (x) = f (n+1) (x)� f (�)� P (�)

(� � �)n+1� (n+ 1)!.

From "The Mean Value Theorem" (TMVT) (for example Theorem 30.3 in [2]) for g (�) on [�; �] thereis �1 2 (�; �) such that g0 (�1) = 0.

From TMVT for g0 (�) on [�; �1] there is �2 2 (�; �1) such that g00 (�2) = 0.From TMVT for g(n) (�) on [�; �n] there is �n+1 2 (�; �n) such that g(n+1) (�n+1) = 0, meaning

f (n+1) (�n+1) =f (�)� P (�)

(� � �)n+1� (n+ 1)!.

For � = �n+1 it follows f (�) = P (�) +f (n+1) (�)

(n+ 1)!(� � �)n+1. �

15

1.9.2. Remark (Taylor Series). Suppose that the conditions in Taylor�s Theorem are satis�ed for any

n 2 N and that the remainder f(n+1) (�)

(n+ 1)!(h)n+1 converges to 0 as n!1 (uniformly with respect to h in

j[a�R; a+R]j. Then f (a+ h) =1Pn=0

f (n) (�)

n!hn, for h 2 j[a�R; a+R]j.

1.9.1. Some usual Taylor expansions. ex =1Pn=0

xn

n!, 8x 2 R

sin x =1Pn=0

(�1)n

(2n+ 1)!x2n+1, 8x 2 R

cosx =1Pn=0

(�1)n

(2n)!x2n, 8x 2 R

ax =1Pn=0

lnn a

n!xn, 8x 2 R

sinh x =1Pn=0

1

(2n+ 1)!x2n+1, 8x 2 R

coshx =1Pn=0

1

(2n)!x2n, 8x 2 R

(1 + x)� =1Pn=0

� (�� 1) � � � (�� n+ 1)

n!xn, jxj � 1

p1 + x = 1 +

1

2x� 1 � 1

2 � 4x2 +

1 � 1 � 32 � 4 � 6x

3 � 1 � 1 � 3 � 52 � 4 � 6 � 8x

4 + � � � , jxj � 13p1 + x = 1 +

1

3x� 1 � 2

3 � 6x2 +

1 � 2 � 53 � 6 � 9x

3 � 1 � 2 � 5 � 83 � 6 � 9 � 12x

4 + � � � , jxj � 1

ln (1 + x) =1Pn=0

(�1)n+1

nxn, 8x 2 (�1; 1]

1.9.3. Exercise.1Pn=1

n � an

1.9.4. Example.1Pn=1

(n3 + 1) an

(n+ 1)!

1.9.5. Solution.1Pn=1

(n3 + 1) an

(n+ 1)!=

1Pn=1

(n+ 1) (n2 � n+ 1) an

(n+ 1)!=

=1Pn=1

(n2 � n+ 1) an

n!=

1Pn=1

�nan

(n� 1)! �an

(n� 1)! +an

n!

�=

=1Pn=1

nan

(n� 1)! �1Pn=1

an

(n� 1)! +1Pn=1

an

n!=

= ea � 1 +1Pn=1

nan

(n� 1)! � a1Pn=1

an�1

(n� 1)! =

= ea � 1� a � ea +1Pn=1

(n� 1 + 1) an(n� 1)! =

= ea � 1� a � ea + a+1Pn=2

(n� 1 + 1) an(n� 1)! =

16

= ea � 1� a � ea + a+1Pn=2

�(n� 1) an(n� 1)! +

an

(n� 1)!

�=

= ea � 1� a � ea + a+ a21Pn=2

an�2

(n� 2)! + a1Pn=2

an�1

(n� 1)! =

= ea � 1� a � ea + a+ a2ea + a (ea � 1) = ea � 1 + a2ea