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Lecture notes for Mathematics for Economists (ECB1WIS)Bakx, T.J.Pinto, A.Rottier, M.Fall 2021Supervisor:Stefan Vukojevic, MScUtrecht University School of Economics (U.S.)Note: These lecture notes are for your own use only. It is not allowed todistribute the notes further by posting them on the Inter...

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Lecture notes for Mathematics for Economists (ECB1WIS)

Bakx, T.J.M. Pinto, A.M. Rottier, M.K. Fall 2021

Supervisor: Stefan Vukojevic, MSc Utrecht University School of Economics (U.S.E.) Note: These lecture notes are for your own use only. It is not allowed to distribute the notes further by posting them on the Internet without explicit and prior permission of the authors. c Utrecht University School of Economics 2021 

Contents Introduction

3

Prerequisites 0.1 Addition, multiplication and parentheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.1.1 Exercises with addition, multiplication and parentheses . . . . . . . . . . . . . . . . . 0.2 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.2.1 Exercises with exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.3 Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.3.1 Exercises with fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.3.2 More exercises on algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.4.1 Exercises on domain and range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5.1 The chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5.2 The product rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5.3 The quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5.4 Exercises for differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.6 Strictly Monotonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.6.1 Exercises on monotonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.7 Solving a system of linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.7.1 Exercises on linear systems of equations . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8 Solving quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.8.1 Exercises on solving quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . .

4 4 5 6 7 8 9 9 10 11 12 13 14 14 15 16 16 17 19 20 21

1 Week 1: Maxima and minima; Functions of several variables 22 1.1 Recap theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.2 Optimization with one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.2.1 The second derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.3 Differentiation of functions of several variables . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.3.1 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.3.2 The total derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.4 df , f ′ and ∂f , when to use what? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2 Week 2: Convexity/concavity; Unconstrained and constrained optimization 32 2.1 Recap theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.1.1 Convex optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.1.2 Unconstrained optimization with several variables. . . . . . . . . . . . . . . . . . . . . 34 2.1.3 Constrained optimization with several variables. . . . . . . . . . . . . . . . . . . . . . 37 2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1

3 Week 3: Implicit relations; Inverse functions 3.1 Recap theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Implicit relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44 44 44 45 47

4 Week 4: Limits and continuity; Graph sketching 4.1 Recap Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Graph Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51 51 51 56 57

5 Week 5: Midterm 5.1 Midterm Exam 2020-2021 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61 61

6 Week 6: Logarithmic and exponential functions 6.1 Recap Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Prerequisites: power functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 General logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 The natural logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.4 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.5 Constrained optimization with exponentials and logarithms . . . . . . . . . . . . . . . 6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 63 63 63 64 65 66 66

7 Week 7: Integration 7.1 Recap Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 The summation operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Integration and the fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . 7.1.3 Double integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

70 70 70 72 76 78

8 Week 8: More on integral calculus; Additional economic applications 8.1 Recap Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81 83

9 Week 9: Endterm 9.1 Endterm Exam 2020-2021 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86 86

References

81

88

2

Introduction This is a set of lecture notes for the course ‘Mathematics for economists’ (ECB1WIS) taught at Utrecht University between November and January. They are meant to supplement the material provided by the lectures, as well as the main reference ‘Mathematics for Economists’ (Pemberton & Rau, 2018). The first part of these lecture notes is devoted to discussing the prerequisites for the course. This part was meant to supplement the prerequisite videos that are provided in the course channel. Students who had unsatisfying results in the entry test are encouraged to watch the videos and read this part carefully, especially focusing on those topics for which they lost points in the test. The prerequisites will not be discussed extensively (if at all) during the regular tutorials. It is however always good to take a look at these chapters if you have forgotten some of the techniques. The second part is ordered based on the weekly schedule of the course. This is the material that will be treated in the tutorials. For each week, there is a concise reiteration of the material covered in the corresponding lecture. The corresponding sections from the book are also indicated at the start. If necessary, there are clarifications regarding the material from the book that concerns the lecture. Subsequently, there are exercises of varying difficulty. Some exercises are marked with an asterisk (⋆); these are more advanced exercises. Nevertheless these are interesting exercises where one can see if their level of understanding is sufficient. Not all of these will be discussed during the tutorials. Solutions to all of the exercises will however be made available to all students in the week after the material has been covered.

3

Prerequisites: refreshing algebra This chapter serves to refresh some of the algebra skills that are part of the prerequisites for this course. Some of these will be briefly discussed in the lecture, but you are strongly encouraged to read this chapter yourself. If you are not comfortable doing the exercises in this chapter, please ask your tutor for help at the tutorials. As a general comment: Mathematics is best learnt through practice. It’s okay if you don’t remember all of the rules immediately, but the best way to become familiar with them is by doing exercises. If you cannot solve an exercise, don’t worry. Ask your tutor for more explanation and try again. It is also important not to jump to the solution manual right away. Understanding the solution is not the same as coming up with it yourself.

0.1

Addition, multiplication and parentheses

Addition and multiplication are the two most basic operations in algebra. It’s important to remember that multiplication always comes before addition (if you don’t see any parentheses). So, 3 · 2 + 7 = 6 + 7 = 13 is correct because we should first multiply 3 by 2 and then add 7 to the result. Example 0.1. Given that 2x + 5 = 9, solve for x. Solution: when solving an equation, you can manipulate it by performing the same operation on both sides. Here, we can subtract 5 on both sides: if 2x +5 = 9, then 2x +5 − 5 = 9 − 5 = 4, so 2x = 4. Now = 42 = 2. Hence, 2x = x = 2 and we have solved the equation. we can divide both sides by 2, so 2x 2 2 Let’s look at an example with more than one variable. Example 0.2. Given that 4x + 5 = 9 − y, solve for y in terms of x. Solution: we’re trying to get y to one side of the equation and everything else to the other side. Let’s first get rid of the 9: we have (4x + 5) − 9 = (9 − y) − 9, so 4x − 4 = −y. We can now divide both sides by −1 to get y = −(4x − 4) = 4 − 4x. In this example, it was crucial that we put parentheses around (4x−4). The most important rule to remember when working with expressions with parentheses is that a(b + c) = ab + ac. The parentheses indicate that here, addition of b and c needs to be carried out before multiplying by a. So, 3 · (2 + 7) = 3 · 9 = 27, which is different from 3 · 2 + 7. As a special case, with a = −1, we see that −(b + c) = −b − c. So, if you see a minus sign in front of a parenthesis, make sure that you carry the minus sign to al l the terms inside the parenthesis!

4

Example 0.3. Clear the parentheses in the expression (x + 1)(x − 2). Solution: using the rule once with a = x + 1, b = x, c = −2, we see that (x + 1)(x − 2) = x(x + 1) + (−2)(x + 1). Using the rule again on the two terms gives x(x + 1) = x2 + x and −2(x + 1) = −2x − 2, so (x + 1)(x + 2) = x2 + x − 2x − 2 = x2 − x − 2. Example 0.4. In a similar manner to the previous example, clearing the parentheses in the expression (x + y)2 = (x + y)(x + y) gives (x + y)2 = x2 + xy + xy + y2 = x2 + 2xy + y2 . This is a useful formula to remember. Sometimes it is also useful to be able to write an expression that doesn’t contain any parentheses into a form which does. This is called factoring. Example 0.5. Given that xy + y = 4, solve for y in terms of x. Solution: Because there is more than just a linear term in y, we have to factorize first. Note that 4 xy + y = y(x + 1). Hence, y(x + 1) = 4 so y = x+1 .

0.1.1

Exercises with addition, multiplication and parentheses

Exercise 0.1. Clear the parentheses in (x + y)3 = (x + y)(x + y)(x + y). You can group some terms together. Exercise 0.2. Solve the following equations for x: a) 3x + 8 = 15

c) 3(8 + 2x) = 2(6 + 5x) + 1

b) 8 − 2x = 5 − x

d) −(8 − x) = x − 4(2 − x)

e) 5 + 5x = 6 − 2(1 − x)

Exercise 0.3. Clear the parentheses in the expression (x + y − z)2 . Exercise 0.4. Solve the following equations for x in terms of y: a) 2x + 4 = 6 − 4y b) 4(x + 1)2 − (2x − 1)2 = y + 4

c) 4 − 2(x − 3y) = 6(x − y) − y

e) (x +1)(y − 2) = (x +3)(y + 4)

d) x+6(1 −y) = y−(x− (2 −x))

Exercise 0.5. Solve the following equations for y in terms of x: a) x2 y + y + 1 = 4 b)

y x+2

+y =5

c) x =

y+1 y−3

e) x + y = x2 − 2xy

d) x2 − y = −4(1 − xy)

5

0.2

Exponents

There are a few basic rules for working with exponents of numbers. We give them here: • xa · xb = xa+b . When you multiply powers of the same number x, you can add the powers together. Example 0.6. Simplify the expression y3 · (y2 · y−8 + y5 ). Solution: this equals y3 · (y2−8 + y5 ) = y3 (y−6 + y5 ), and clearing the parenthesis gives y3 · y−6 + y3 · y5 = y3−6 + y3+5 = y−3 + y8 . • x−a =

1 . xa

A negative exponent goes into the denominator.

Example 0.7. Given that x−2 = 5, solve for x. Solution: x−2 =

1

• xn =

1 x2

= 5, so x2 = 15 . Hence, x = ±

q

1. 5

√ n x.

Example 0.8. Simplify x3 ·

√ 3 x.

Solution: this equals x3 · x1/3 = x3+1/3 = x10/3 . •

xa xb

= xa−b . When you divide powers of the same number x, you can subtract the powers. Example 0.9. Given that

x5 x3 ·x2

+2+

x4 x8

= 4, solve for x.

5

Solution: the left hand side equals xx5 + 2 + x4−8 = 1 + 2 + x−4 = 3 + x−4 . Hence, 3 + x−4 = 4, so x14 = 4 − 3 = 1. But then x4 = 1, so x = ±1. • (xa )b = xab . When taking the power of a power, you can multiply the powers. Example 0.10. Calculate

√ 3 26 .

Solution: this equals (26 )1/3 = 2(6/3) = 22 = 4.

6

• (xy)a = xa ya . The power of a product is the product of the powers. Example 0.11. Clear all the parentheses from ((2xy)3 + 5)2 . Solution: we have (2xy)3 = 23 x3 y3 = 8x3 y3 . Filling this in gives (8x3 y3 + 5)2 = (8x3 y3 )2 + 2 · 5 · 8x3 y3 + 52 = 64x6 y6 + 80x3 y3 + 25. 0.2.1

Exercises with exponents

Exercise 0.6. Simplify the following expressions: a) (xy3 )(x−2 + (x + y)2 ) b) (x + x−1 )2 − 2 c) (x2 + 2y2 − 2xy)(x2 + 2y2 + 2xy) − 4y4 √ d) (−x)2 + 4x4 √ 3 √ 3 e) xx − 2( x2 )−1 Exercise 0.7. Solve the following equations for x: a)

(x3 )2 x4

+ 3(−x)2 = 5

b) (2x)4 + 2x4 = 3 c) x−3 − 1 = (2x)−3 − 3 d) x2 · x3 = x4 e) (x−1 − x)2 = x2 Exercise 0.8. Solve the following equations for x in terms of y: a) (xy)4 + y4 = 1 b) y2 − 1 − (−xy)3 = 0 √ 4 c) x2 − 1 = y p d) xy4 − xy2 = 1 e)

x4 x2 y

=2

7

0.3

Fractions

Just like for working with exponents, we also have rules for manipulating fractions. They read as follows: •

a b

ac bc

=

Anything which is both “upstairs and downstairs”, cancels out.

Example 0.12. Simplify the expression Solution: this equals

•

a b

·

c d

ac . bd

=

xy 2 xy −3

xy 2 . y −3 x

2

= yy−3 = y5 .

You can multiply fractions by multiplying numerator and denominator.

Example 0.13. Given that

2 x+1 7 y

= 5, determine x in terms of y.

+1) Solution: the left hand side equals 2(x7y = 5. When solving an equation = 2x+2 . Hence 2x+2 7y 7y with fractions, it is often a good idea to clear the denominator by multiplying both sides by it. We get 2x + 2 = 5(7y) = 35y so x = 352 y − 1.

•

a c / b d

=

a b

·

d c

=

ad . bc

Dividing by a fraction is the same as multiplying by the opposite.

Example 0.14. Solve the equation

2x+1 3x / 3x+1 x

= 2 for x.

Solution: using the rule, we see that the left side equals 2

6x +5x+1 3x2

•

a + bc c

=

2

2

2

(2x+1)(3x+1) x(3x)

=

= 2 so 6x + 5x + 1 = 2(3x ) = 6x . This implies that 5x + 1 = 0

a+b . c

6x2 +5x+1 . 3x2 1. so x = − 5

Hence,

If two fractions have the same denominator, you can add them by adding the numerators.

Example 0.15. Write 3x/ 7x + (5x + 2)/ 7x as one fraction. Solution: this equals (3x + 5x + 2)/ 7x = •

a b

+

c d

=

ad bd

+

bc bd

=

ad+bc . bd

x(8x+2) . 7

If they don’t have the same denominator, we use rule 2 to make them equal.

Example 0.16. Solve the equation

1 x−1

−

1 x+1

= x2 + 1.

(x+1)−(x−1) x+1 x−1 x−1 Solution: the left hand side equals (x+1)(x−1) = x22−1 . − (x+1)(x−1) = xx+1 2 −1 − x2 −1 = x2 −1 2 2 2 4 2 So we get x2 −1 = x + 1. Clearing the denominator gives 2 = (x − 1)(x + 1) = x − 1. Hence √ 4 x4 = 3, so x = ± 3.

8

0.3.1

Exercises with fractions

Exercise 0.9. Write the following expressions as a single fraction: a)

x y

+

y x−1

b)

1 x

−

2x3 xy 2 z

c)

1 x−1

d)

1 (x (x+y)2

e)

y x+1

− x−2 +1+

− 1/( xy ) +

xy −x ) x 2xy 2 y

Exercise 0.10. Solve the following equations for x: a) ( x2 )2 =

1 x

b) 3/ x33 =

1 x3

c)

x x3

d)

x+1 x / x+2 x

e)

2x x2 +1

+

4 x

=

−2 x2

=1

+1=

1 x2 +1

Exercise 0.11. Solve the following equations for x in terms of y: a)

x y

=

y x

b) y = 2x+1 x−3 p 3 c) y · xx4 = 2y2

d) (x − y)(x +

1 ) 2y

= x2

=4 e) x/ 2y x 0.3.2

More exercises on algebra

Exercise 0.12. Solve the following equations for x: a)

x−2 x+1

=

x+1 x−2

b) 1 − (x23 )2 = −8 √ 1 c) x = 2/ √ 3 x

Exercise 0.13. Solve the following equations for x in terms of y: )3 a) y = ( x−1 x+5 b) x − 8 = xy + √ √ c) y = x − y

x y

9

0.4

Functions

Functions are one of the must basic and fundamental things in mathematics. But what is a function? What do we mean when we talk about these things and what can we do with them in mathematics? Definition 0.1. A function is an object that assigns to every input value x, one output value f (x). What do we mean by this? It means that if we were to pick a number, call it x, we could put it into the function and it would return a ...