EC 400 Notes for Class 1.pdf

8/14/2019 EC 400 Notes for Class 1.pdf

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Chapter 6

Vectors

6.1 What is a Vector?

For most purposes economists think of vectors as an array of real numbers(1 2). The set of all vectors is written asR

, pronounced R n. Youhave to distinguish between row and column vectors when doing matrix andvector algebra. In this book all vectors are column vectors.

x=

12:

The corresponding row vector is written as:

x0 = (1 2)

The vector x0 is sometimes pronounced as "x prime". Pure mathematicianswork with a more abstract denition of a vector and a vector space. Physiciststhink of vectors as things that have both size and direction, for example velocityand force.

6.2 Vector Addition and Multiplication

6.2.1 Vector Addition

Ifxand y are two vectors in R

x + y=

12:

+

12:

=

1+12+2

:+

47


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48 CHAPTER 6. VECTORS

0 x

y

x +y

Figure 6.1: Vector Addition

Figure 6.1 illustrates vector addition with the vectors x and y shown nose totail. The alternative is to show vector addition as a parallelogram as in Figure6.2.

Vector addition is commutative, that is x + y= y + x. This is because

x + y=

1+12+2

+

=

1+12+2

+

=y + x

Ifx R and y R and and are dierent you cannot add xand y. Inmathematical language the sum ofxand y is undened.

6.2.2 Scalar Multiplication

If is a real number (called a scalar in this context)

x=

12

:

=

12

:

=x

Figure 6.3 illustrates a vector x multiplied by 2 1 and 2. Multiplying avector by a positive number keeps it pointing in the same direction but makesit longer or shorter. Multiplying a vector by a negative number makes it pointin the opposite direction.

Scalar multiplication is distributive over vector addition, that is


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6.2. VECTOR ADDITION AND MULTIPLICATION 49

0 x

y

x +y

Figure 6.2: Vector Addition Illustrated with a Parallelogram

0

x

-x

- 2x

2x

Figure 6.3: Multiplying a Vector by a Scalar


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(x + y) = x +y for all vectors xand yTo see why observe that

(x + y) =

1+12+2

+

=

(1+1) (2+2)

(+)

=

12

+

12

=x+y

The denition of vector addition also implies that if and are scalars and xa vector

(+)x=x +x

that is scalar multiplication is distributive over scalar addition.

6.2.3 Inner product

The inner product of two vectors xand y in R is

x0

y=

X=1

The inner product is sometime called the "scalar product" or "dot product"and written as x y. The properties of the inner product are:

1. The inner product is commutative, that is

x0y= y0x

for all vectors x and y This is becausex0y=P

=1 =P

=1 =y0x

2. Ifxand yare vectors and is a scalar

(x0y) = (x)0y= x0 (y)

because(x0y) = (P

=1) =P

=1()= (x)0y=

P=1() =

x0 (y)


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6.3. LENGTH AND NORM 51

3. Ift, u, vand w arevectors

t0 (u + v) = (u + v)0 t= t0u + t0v

because t0 (u + v) = (u + v)0

t=P

=1(+) =P

=1+P

=1=t0u + t0v.

4. Also

(t + w)0 (u + v) = t0u + w0u + t0v + w0v= u0t + u0w + v0t + v0w

because(t + w)0 (u + v) =P

=1(+) (+) =P

=1(+++)

5. Similarly Ift, u, v and w are vectors and are scalars

(t

+w

)

0

(u

+v

) = (t

)

0

(u

+v

) + (w

)

0

(u

+v

)= (t)0 (u) + (t)

0 (v) + (w)0 (u) + (w)

0 (v)

= t0u+t0v+w0u+w0v (6.1)

6.3 Length and Norm

6.3.1 Denition

For all vectors x, x0x=P

=12 . As

2 0 for all and

2 = 0 if and

only if= 0,

x0x 0 for all xand x0x= 0 if and only ifx= 0

where 0 is an vector with every element 0. Denekxk as

kxk= (x0x)12

so kxk 0 and kxk = 0 if and only ifx = 0. Pythagoras Theorem impliesthat in 2 and 3 dimensions kxk is the length ofx. Pure mathematicians callkxk the norm ofx. It is also called the length ofx.

6.3.2 Properties of the Length kxk

These are things you need to remember. The rst property follows directly fromthe denition. The others are proved in sections 6.5 and 6.6 of this chapter.

1. If is a scalar kxk= || kxkwhere||is the absolute value of, so||= if 0 and ||= if 0.

2. x0y= kxk kyk cos where is the angle between xand y.

3. Cauchy-Schwarz inequality:

|x0y| kxk kyk


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4. Triangle inequalitykx + yk kxk + kyk

kx yk kxk + kyk

5. Ifx0y = 0, x 6= 0 and y 6= 0 the angle between x and y is 90 becausecos90 = 0. The vectors xand yare said to be orthogonal.

6. Ifx0y = kxk kyk, x and y are parallel and point in the same directionbecause cos0 = 1

7. If x0y= kxk kyk, xand y are parallel and point in opposite directionsbecause cos 180 =1

6.4 The Least Squares Problem

6.4.1 The Least Squares Problem With Vector Notation

You will become very familiar with least squares problems in econometrics. Thissection of notes covers the simplest case. It also generates inequality 6.3 whichis an essential step on the way to the Cauchy-Schwarz inequality. The problemhere is nding the value of that minimizes

(xy) =X=1

( )2

You will come to think of this as the regression of onto a single variable

without an intercept. Supposexand yare vectors. Given the denition ofthe norm ky xk

(xy) =X=1

( )2 =ky xk2 = (y x)0(y x)

Expanding the brackets using the result in equation 6.1 on inner products gives

(xy) = (y x)0(y x)

= y0y x0y y0x +2x0x

= y0y 2x0y +2x0x

because x0y= x0y.

Assume that 6= 0 for some so x6=0. This implies thatx0x0 Thencompleting the square

(xy) = y0y2x0y +2x0x (6.2)

= x0x

(x0x)1

(x0y)2

+ y0y (x0x)1

(x0y)2

y0y (x0x)1

(x0y)2


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6.4. THE LEAST SQUARES PROBLEM 53

for all and

(xy)=y0y (x0x)1

(x0y)2

if and only if = (x0x)1 (x0y). Thus (xy)is minimized by setting

= (x0x)1

(x0y)

and has a minimum value of

y0y (x0x)1

(x0y)2

Recall that (xy) is a sum of squares, so (xy)0 for all includ-

ing = (x0x)1 (x0y) at which point (xy) =y

0

y (x0x)1 (x0y)2 Thus

y0

y (x0

x)1

(x0

y)2

Recall that by assumption x6= 0 so x0

x0. Multiply-ing by x0xand rearranging gives

(x0y)2

(x0x) (y0y) (6.3)

6.4.2 The Least Squares Problem Without Vector Nota-tion

You can also solve this simple least squares problem without using vectors. Theproblem is to nd that minimizes the sum of squares.

X=1

( )2

Expanding the brackets

X=1

( )2 =

X=1

2 2+

22

=X=1

2 2

X=1

!+2

X=1

2

!

Assume that at least one of the 0 is not zero, so

P=1

2 0 Completing

the square

X=1

( )2 =

X=1

2

!

X=1

2

!1 X

=1

!2

+X=1

2

X=1

2

!1 X

=1

!2


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which is minimized by setting

=

X=1

2

!1 X

=1

!= (x0x)

1(x0y)

and has a minimum value of

X=1

2

X=1

2

!1 X

=1

!2=y0y (x0x)

1(x0y)

2

Notice that expressions likeP

=1 take longer to write down and are harderto work with than the corresponding vector expression x0y. The algebrabecomes even worse if you are working with the general least squares problemof minimizing

X=1

X=1

!2

If you are to survive your econometrics course you must make the switch fromthe notation

P=1 to vector notation x

0y as rapidly as possible.

6.5 Inequalities for Vectors

6.5.1 The Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality says that for any vectors x and y with the samenumber of elements

|x0y| kxk kyk (6.4)

Here kxk and kyk are the norm or length of x and y so kxk2 = x0x and

kyk2 =y0y The expression |x0y| is the absolute value ofx0y. (The absolute

value of a number is if 0 and if 0. Note that this implies that2 =||2)

The Cauchy-Schwarz inequality is derived from inequality 6.3 which statesthat (x0y)

2 (x0x) (y0y). From the denitions of absolute value and norm

|x0y|2 = (x0y)2, (x0x) = kxk2 and y0y= kyk2 so the inequality (x0y)2

(x0

x) (y0

y)can be written as

|x0y|2 kxk

2 kyk2 (6.5)

which implies the Cauchy-Schwarz inequality |x0y| kxk kyk Inequality 6.3was proved under the assumption that x 6= 0, but as both sides of the Cauchy-Schwarz inequality are zero when x=0 it also holds when x=0


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6.5. INEQUALITIES FOR VECTORS 55

x

y

x +y

Figure 6.4:

6.5.2 The Triangle Inequality

The triangle inequality states that

kx + yk kxk + kyk

Geometrically you can think ofx + yx and y as three sides of a triangle. Theinequality simply states that one side of a triangle cannot be longer than thesum of the other two sides, as illustrated in Figure 6.4.

The triangle inequality is a direct consequence of the Cauchy-Schwarz in-equality. From the denition ofkx + yk

kx + yk2 = (x + y)

0 (x + y) = x0x + 2x0y + y0y


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O bx x

C

y

B

Figure 6.5:

From the Cauchy-Schwarz inequality|x0y| kxk kyk. As || for any num-ber this implies that x0y kxk kyk so

kx + yk2 = x0x + 2x0y + y0y

kxk2 +2kxk kyk + kyk2 = (kxk + kyk)2

As kx + yk kxk and kyk are all non negative this implies the triangle inequalitykx + yk kxk + kyk Replacing y by y and noting that kyk = kyk thetriangle inequality implies that

kx yk kxk + kyk

6.6 The Angle Between Two Vectors

6.6.1 Showing That x0y =kxk kyk cos

Think ofx and y as non zero vectors and as a scalar. In Figure 6.5 x isa scalar multiple of the vector x. The least squares problem is nding tominimizekyxk which is the distance between the point at the end of thevector y and the horizontal straight line from the origin 0 that goes throughthe end of the vector xGeometrically you nd the point xon the line0xthat


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6.6. THE ANGLE BETWEEN TWO VECTORS 57

0 90 180 270 360

cos

1

- 1

Figure 6.6: The Graph ofcos

minimizes the length ofyx by drawing the line from to that is at 90

to the vector x.

As = (x0x)1 (x0y) solves the least squares minimization problem the

length of the line is the length of the vector x where = (x0x)1 (x0y)

Think about the case where is positive as shown in Figure 6.5. Askxk =

(x0x)12 the length of is kxk = kxk (x0x)

1 (x0y) = (x0x)

12 (x0y) Ele-

mentary trigonometry and Figure 6.5 imply that if is the angle between thevectors xand y

cos =0

0 =

(x0x)

12 (x0y)

kyk =

x0y

kxk kyk (6.6)

wherecos is the cosine of the angle . Equation 6.6 implies that

x0y= kxk kyk cos

In words this says that the inner product of the vectors xand yis the productof their lengths and the cosine of the angle between the two vectors. Therelationship also holds when is negative, so the angle between the two vectorsis more than 90 and, as the graph ofcos in Figure 6.6 shows, cos 0.One of the properties of the cosine is that for all with 0 360

cos = cos (360 )

This has the useful implication that it does not matter which way round youthink of the angle between the two vectors, as shown in Figure 6.7 . If youthink of the angle between the two vectors as the smaller of the two anglesbetween the two vectors, so it lies between 0 and 180. The graph of thefunctioncos in Figure 6.6 shows that

0 cos 1 if 0 90


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360 -x

y

Figure 6.7:

and1 cos 0 if 90 180

which gives a result suciently useful to be stated formally.

Proposition 1 x0y 0 if the angle between x and y is between 0 and90

x0y 0 if the angle betweenx andy is between90 and180

6.6.2 Orthogonal Vectors

Suppose x and y are non zero vectors, so kxk 0 and kyk 0 Suppose

also that x0y = 0. This implies that x0y = kxk kyk cos = 0 where is theangle between the vectors x and y As kxk 0 and kyk 0 this implies thatcos = 0 Given the graph ofcos this implies that = 90 or = 270 .

As Figure 6.8 shows you can measure the angle between x and y as 90

or 270 . It is convenient to choose the smaller angle 90. In mathematicallanguage the vectors x and y are said to be orthogonal ifx0y= 0so the vectorsare at an angle of 90 to each other. In everyday English an angle of90 iscalled a "right angle".

6.6.3 Parallel vectors

If x0y= kxk kyk cos = kxk kyk

then cos = 1so = 0 The angle between xand yis 0. The two vectors areparallel and point in the same direction.

Ifx0y= kxk kyk cos = kxk kyk


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6.6. THE ANGLE BETWEEN TWO VECTORS 59

90

270x

y

Figure 6.8:


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then cos = 1so = 180 The two vectors are parallel and point in oppositedirections.

6.7 Vectors, Lines, Planes and Hyperplanes

6.7.1 Vectors and Lines in R2

The objective of this section is to introduce the idea of a hyperplane, whichis a generalization of a line in two dimensional space, and a plane in threedimensional space. Hyperplanes are important for an intuitive understandingof partial derivatives, and also for the theory of concavity and convexity thatunderlies optimization theory.

One of microeconomics favorite equations is the budget line; with twogoods1 and 2, with prices 1 and2 this is the set of points satisfying

11+22= (6.7)

where is the amount the consumer has to spend. This is straight line; if26= 0 equation 6.7 can be written as

2=

2

12

1

so the slope of the budget line is 12

. If2= 0the line is vertical. In vectornotation equation 6.7 can be written as

p0x=

Another way at looking at this equation is to choose any point x0 on the lineso p0x0= and write the equation as p

0x= p0x0, or rearranging

p0 (x x0) = 0

Recall that the inner product p0 (x x0) = kpk kx x0k cos where is theangle between the two vectors p and x x0. Thus ascos 90

= 0, if the innerproduct is 0, the angle is 90 and the two vectors are said to be "orthogonal".The line, that is the set of points satisfying p0 (x x0), is precisely the set ofpoints for which x x0 is orthogonal to p This is illustrated in Figure 6.9.The vector p orthogonal to the line is called the normal vector.

6.7.2 Vectors and Planes in R3

With three goods the budget equation becomes

11+22+33=

The vector notation appears unchanged as p0 (x x0) = 0, but now you haveto think of the vectors as living in three dimensional space R3 rather than two


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6.7. VECTORS, LINES, PLANES AND HYPERPLANES 61

x0

x1

x2

p

x - x0

p( x - x0) = 0

Figure 6.9: The vector pis orthogonal to the line p0

(x x0) = 0.

dimensional spaceR2 Geometrically, in three dimensional space, given a xedvector p the set of vectors orthogonal to p is a plane. The vector p is againcalled the normal vector to the plane.

You can perhaps imagine the plane and its normal vector p by thinking ofyour pen asp. If you put one end of the pen on the table top, and point the penvertically upwards, all the vectors in the plane of the table top are horizontal,so are orthogonal to the pen. It is possible to lift up a corner of the table sothe table top is no longer horizontal, whilst keeping the angle between pen andtable top xed at90, so the vector is still orthogonal to the plane. (Take yourcoee mug o the table before trying this.)

6.7.3 Vectors and Hyperplanes in R

With goods the budget equation becomes

11+22++=

Again the budget equation can be written as p0 (x x0) = 0, but p and x x0are now vectors. A hyperplane is dened as follows:

Denition 2 A hyperplane inR is a set of the form

{x: x Rp0 (x x0) = 0}

wherep is a non-zero vector inR, andx0 is a vector inR.

In two dimensional space a hyperplane is a straight line, and in three di-mensional space a hyperplane is a plane. For any all the vectors x x0 areorthogonal to p, and pis still called the normal vector.


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6.8 First Thoughts on Constrained Maximiza-

tion

This section is not part of the LSE Revision Maths course.Constrained maximization problems are central to economics. This section

works with a very simple problem, that of maximizing q0x subject to the con-straint that p0x . As this is the rst constrained maximization problem inthis book I will give a formal denition.

Denition 3 Ifp andq are vectors the elementx0 ofR solves the prob-

lem of maximizing q0x on R subject to the constraint that p0x if

p0x0

andq0x0 q0x for all elementsx ofR satisfyingp0x .

The results that I am about to derive are almost obvious, and not interestingin themselves, but the insight gained from thinking about these results are veryhelpful once you come to think about Lagrangians and partial derivatives. Therst result is

Proposition 4 Ifp is a non zero vectors inR, q is a vector inR, there isa real number 0 such that

q= p

and there is a vectorx0 inR such that

p0

x0=

thenx0 solves the problem of maximizing q0x on R subject to the constraint

thatp0x .

The result is pretty obvious from Figure 6.10. Ifp=qwhere 0 andp0x0 = any x that gives a higher value of the objective q

0x than x0, mustalso give a higher value ofp0xthan x0. As p

0x0= this implies that p0x ,

so x does not satisfy the constraint. The formal proof is:Proof. Suppose that there is a number 0 and a vector x0 satisfying theconditions of the proposition. Let x be any element of R satisfying theconstraint p0x As p0x and p0x0 = , p

0x p0x0. As 0 thisimplies that

(p)0 x= p0x p0x0= (p)0 x0

so as q= pq0x q0x0

Thus I have shown that ifp0x then q0x q0x0 so x0 solves the constrainedmaximization problem.

The other result is:


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6.8. FIRST THOUGHTS ON CONSTRAINED MAXIMIZATION 63

x0

x1

x2

p =q

x - x0

p( x - x0) = 0

q

Figure 6.10:

Proposition 5 Ifp is a non zero vectors inR, q is a vector inR, there isno real number 0 such thatq= p then for any vectorx0 there is anvectorx1 such that

p0x1 p

0x0

and

qx1 q0x0.

Figure 6.11 illustrates this result. The trick is to nd a vector x1 with theproperty that the angle between x1 x0 and q is less than 90

and the anglebetween x1 x0andp is more than90

so from Proposition 1 p0 (x1 x0) 0

and q (x1 x0) 0. The intuition is obvious from the diagram, but theresult is so important that it merits a formal proof.Proof. Let

r= q (p0p)1

(p0q)p (6.8)

and note thatp0r= 0 (6.9)

so as shown in Figure 6.12 r is orthogonal to p. Thus from equations 6.8 and6.9 r0r= r0q (p0p)

1 (p0q) r0p= r0qso

r0r= r0q= q0r (6.10)

Letx1

= x0

+ r p

where 0. Then

p0x1= p

0x0+ p

0r p0p= p0x0p

0p (6.11)

because from equation 6.9 p0r= 0 By assumption 0 and p6= 0, so p0p 0.Thus from equation 6.11

p0 (x1 x0) 0 (6.12)


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x1

x2

p

q

p( x - x0) = 0q( x - x0) = 0

x1

0

x0

Figure 6.11:

x1

x2

p

q

p( x - x0) = 0q( x - x0) = 0

r

r

x1

0

x0

q r =p

-p

Figure 6.12:


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6.8. FIRST THOUGHTS ON CONSTRAINED MAXIMIZATION 65

Alsoq0x

1= q0x

0+ q0r q0p= q0x

0+ r0r q0p

because from equation 6.10q0r= r0r. Thus

q0 (x1 x0) =r0r q0p (6.13)

As 0 the right hand side of equation 6.13 can be made strictly positive ifeither r 6= 0 so r0r 0 and is suciently small, or r = 0 and q0p 0. Ineither case

q0 (x1 x0) 0 (6.14)

Ifr= 0 and q0p= p0q 0 equation 6.8 implies that

q=p

where = (p0p)1 (p0q). As p0p 0 and p0q 0, 0. Thus either thereis a vector x1 satisfying conditions 6.12 and 6.14 or there is a number 0such that q=p.

Proposition 5 implies that if p 6= 0, p0x0 = and there is no number 0 such that q= p, then x0 cannot solve the problem of maximizing q

0x

subject to the constraint p0x because p0x1 p0x0 = and q

0x1 q0x0.

Proposition 4 states that ifp0x0= and there is a real number 0 such thatq= pthen x0 solves the constrained maximization problem. Taken togetherthese results imply that

Proposition 6 Ifp andx0 are vectors, p6= 0 andp0x0 = thenx0 solves

the problem of maximizingq0xsubject to p0x if and only if there is a number

0 such thatq= p.

The number is of course a Lagrange multiplier.


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Chapter 7

Open Sets, Closed Sets andContinuous Functions

7.1 Why This Matters to Economists

The mathematics in this chapter matters to economists for several reasons.There is a big result, that a function that is continuous on a closed and boundedsubset ofR has a maximum and a minimum, which is important because maxi-mization and minimization crop up very frequently in economics. You will meetthe mathematics discussed in this chapter very early in your microeconomicscourse in the form of the continuity assumptions for consumer preferences and

its implication, the existence of a continuous utility functions that captures thepreferences. Continuity is also central to theorems on the existence of generalcompetitive equilibrium.

Sections 1-4 are very useful. The rest of the chapter is harder going and lesswidely used in economics. Section 9 is the background you need to understandthe material on continuity in consumer theory. This is quite hard; if youare struggling it would be reasonable to skip this and accept that this bit ofconsumer theory will not mean much to you.

7.2 Vector Length and Open Balls

Remember from the previous chapter that in R

we have a concept of the lengthor norm of a vector

kxk=

X=1

2

! 12

In one dimensional space = 1, andkk= || the absolute value of. In twoand three dimensional space Pythagoras Theorem implies that this is indeed

67


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68CHAPTER 7. OPEN SETS, CLOSED SETS AND CONTINUOUS FUNCTIONS

the length of the vector. The norm has the property that

kxk 0 for all x (7.1)

kxk = 0 if and only ifx= 0 (7.2)

The distance between two vectors xand x0 in R is dened as

kx x0k=

X=1

( 0)2

! 12

In one dimensional space this reduces to the absolute value | 0|which is thedistance between the numbers and 0. In two and three dimensional spacePythagoras Theorem implies that kx x0k is the distance between x and x0.I therefore refer to kx x0k as distance, even when 3. Properties 7.1

and 7.2 imply that

kx x0k 0 for all x

kx x0k = 0 if and only ifx= x0

Now think about the set of points that are closer to x0 than some strictlypositive number . In one dimensional space x0 is a number and this is theopen interval (0 0+) In two dimensional space it is a circle centredon x0 with radius . In three dimensional space this is a sphere or ball. Infact we use the term ball regardless of the dimension of the space, using thefollowing denition.

Denition 7 If is a strictly positive number andx0 is an element ofR the

open ball (x0 ) centred on x0 with radius is the set of points that arecloser than to x0, that is

(x0 ) = {x: x R kx x0k }

Note that the inequality in this denition is strict. I have used theword "open" many times before when writing about "open intervals" and Iwill explain shortly what the word "open" means in general. In order to dothis I will introduce some more terms that mathematicians have borrowed fromelsewhere.

7.3 Boundaries

Imagine yourself standing on the equator. Take a tiny step to the north; youare in the northern hemisphere. Take a tiny step to the south; you are in thesouthern hemisphere. The equator is the boundary of the two hemispheres.Mathematicians have borrowed the term boundary from geographers and useit much the same sense, applying it to the dimensional space of real numberR


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7.4. OPEN AND CLOSED SETS AND BOUNDARIES 69

Denition 8 A pointx0 inR is an element of the boundaryof a subset

ofR if for any number 0 the open ball(x0

) contains both points thatare elements of and points that are in the complement of.

You may recall that the denition of complement:

Denition 9 The complement of a subset of R is the set of elementsofRthat are not in, that is

={x: x R, x }

The fact that the radius of the open ball (x0)can be a very small posi-tive number captures the idea that a step to the north, however tiny, moves youfrom the equator to the northern hemisphere. The denitions of the boundaryand the complement of a set immediately imply a result that is useful later.

Proposition 10 The boundary of a set and the boundary of its complement are the same.

The idea of a boundary may become clearer with an example.

Example 11 The boundary of the set

=

(1 2) : (1 2) R2, 1 2

is the straight line1= 2.

This example is illustrated in Figure 7.1. Any open ball such as centeredon a point on the line 1= 2 has points that are in and points that are notin , so lies in the boundary of. However any open ball such as centeredon a point that lies in does not contain any points that are not in if itsradius is small enough. Similarly an open ball such asthat is centered on apoint that does not lie in or on the line 1= 2 does not contain any pointsthat are in if its radius is small enough. Thus points that are not on theline 1= 2 are not on the boundary of.

7.4 Open and Closed Sets and Boundaries

British suburban gardens usually have something physical marking the bound-ary, a fence, a hedge or a wall; you could describe them as closed. Suburbangardens in the USA often have nothing visible to mark the boundary; you coulddescribe them as open. Mathematicians use the terms open and closed in asomewhat similar way.


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0 x1

x2

A

B

Cx1 = x2

S

Figure 7.1:

0 x1

x2

S1

Figure 7.2:


25/46

7.5. OPEN SETS 71

0 x1

x2

S2

Figure 7.3:

A subset ofR is open if none of the points in the boundary of liein .

A subsetofR isclosedifallthe points in the boundary oflie in .

For example the boundary of the set1= (1 2) : (1 2) R2 1 0 2 0

in Figure 8.3 is the set of points (1 0) with 1 0, and the set of points(0 2) with 2 0. The boundary and the set do not intersect; the set isopen.

On the other hand the boundary of the set2 =

(1 2) : (1 2) R2 1 0 2 0

in Figure 8.4 is the same as the boundary of1 that is the set of points (1 0)with 1 0, and the set of points (0 2) with 2 0. However here theboundary lies in the set; the set is closed.

Some sets are neither open nor closed. The boundary of the set 3 =(1 2) : (1 2) R2 1 0 2 0

in Figure 7.4 is the same as the bound-

ary of1 and 3 However the boundary points(1 0) with 1 0 lie in 3whereas the boundary points (0 2) with 2 0 do not lie in 3.

7.5 Open Sets

7.5.1 Denition

The denition of an open set as a set for which none of the points in the boundaryof lie in is good for an intuitive understanding of what an open set is.


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0 x1

x2

S3

Figure 7.4:

However this denition is not good for proving results about open sets, forwhich the "ocial" denition works better. This is:

Denition 12 A subset of R is open if for every elementx of there isan open ball(x ) that is a subset of.

The rst result about open sets has to be that the "ocial" denition isequivalent to the denition in terms of the set boundary. This is:

Proposition 13 A subset of R is open if and only if it does not containany points in its boundary.

Proof. As is open then for any element xofthere is an open ball (x )that is a subset of so does not contain any element of the complement

of Thus x cannot be a boundary point of, so no points in are in theboundary of.

Now suppose thatdoes not contain any of its boundary points, but is notopen. Asis not open there is an element x ofwith the property that everyopen ball(x )contains a point in . As x is inthis implies that x is inthe boundary of, contradicting the assumption that does not contain anyof its boundary points. Thus ifdoes not contain any of its boundary pointsmust be open.

7.5.2 Open Intervals, Open Balls and Open Sets


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7.5. OPEN SETS 73

I have used the term "open interval" frequently. I dened open intervals as setsof the form

( ) = {: R }

( ) = {: R }

( ) = {: R }

The next proposition, which is proved in the appendix to this chapter establishesthat this is a sensible form of words.

Proposition 14 An open interval is an open subset ofR.

The next bit of tidying up of the use of the word open is:

Proposition 15 An open ball is an open set.

Again this is proved in the appendix to this chapter.

7.5.3 Unions and Intersections of Open Sets

There are two important results here. The rst is

Proposition 16 Theunionof anite or innite number of open subsets ofR is open.

Proof. Ifx is an element of the union it is an element of one of the open sets, so there is an open ball (x ) that is a subset of, and thus a subset ofthe union of the sets. Thus the union is open.

The second result is:

Proposition 17 Theintersection of a nite number of open sets of R isopen.

Proof. Let the subsets be12and letx be a point in their intersection.As is open there is a number 0 such that the open ball (x ) is asubset of that is

(x ) (7.3)

Let= min (1 2)

Thus 0 for all so the open ball (x ) is a subset of the open ball(x ) for all that is (x ) (x ). Given relationship 7.3 thisimplies

(x ) for = 1 2implying that(x )is a subset of the intersection of12, so the inter-section is itself open.

Note that the fact that this result applies to a nite number of open setsis important. For example that sets

0 1

for = 1 2 3 are open, but

their intersection is{0} which is not an open set, because no open ball centredon 0 is a subset of{0}.


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7.6 Closed Sets

7.6.1 Formal Denition

I introduced closed sets as sets that include their boundaries. The "ocial"denition is

Denition 18 A set is closed if its complement is open.

The next proposition is that this denition is equivalent to the denition interms of the set boundary.

Proposition 19 A subset of R is closed if and only if its boundary is asubset of.

Proof. Assume the set is closed so its complement is open. FromProposition 13 as is open none of the points in its boundary lie in , soall the points in the boundary of lie in . As the boundary of and theboundary of its complement are the same (Proposition 10) this implies thatthe boundary of the closed set lies in .

Conversely if the boundary oflies inthere is no point in the boundary ofthat lies in. As the boundaries ofand are the same this implies thatno point in the boundary of lies in . From Proposition 13 this impliesthat is an open set so is a closed set.

7.6.2 Closed Sets and Innite Sequences

You may have seen a dierent denition of a closed set in terms of limits ofinnite sequences. The intuitive idea of the limit of an innite sequence is thatpoints in the innite sequence get closer and closer to the limit as you movefurther and further along the sequence. The formal denition is:

Denition 20 The innite sequence {x1x2x3} in R tends to a limit

x inR if for any number 0 there is an integersuch thatkx xk

or equivalentlyx (x ) for all .

The result linking innite sequences and closed sets is:

Proposition 21 A subset of R is closed if and only if, for any innitesequence{x1x2x3} whose points are allthat tends to a limit, that limitis in.

The proof of this proposition is in the appendix to this chapter. This resultshows that I could have dened closed sets in R in terms of innite sequences,as is sometimes done. However the denition of closed sets as sets whosecomplements are open is easier to work with.


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7.7. CONTINUOUS FUNCTIONS 75

43210

8

6

4

2

0

x

y

x

y

Figure 7.5: This function is not continuous at = 2.

7.7 Continuous Functions

7.7.1 Continuous Functions of a Single Variable

Intuitively a function is continuous if(x)is very close to (x0)whenx is veryclose to x0. Informally a function of a single variable is continuous if you candraw its graph without taking your pen o the paper. The function = for

2 and = 2 for 2 illustrated in Figure 7.5 is not continuous. Theformal denition of continuity for a function of a single variable dened on anopen1 set is:

Denition 22 Ifis an open subset ofR the function : R is continu-ous if for any0 R and any 0 there is a number 0 with the propertythat if | 0| then |() (0) |

Figure 7.1 illustrates the denition for the function () = 2, where it iseasy to check that if= 1

2 then |2 20| when | 0| .

7.7.2 Continuous Functions of Several Variables

The notation for a function ofvariables is (x) where xis an vector, thatis an element of R. The intuitive idea of continuity for functions of severalvariables is the same as the intuition for functions of a single variable; ifxis very

1 At this point I work with a function dened on an open set because then if x is in theset any open ball (x )with a small enough lies entirely in the set. If the set is not openthere are boundary points x in for which (x ) is not a subset of for any , and thefunction may not be dened at some points in the open ball.


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43210

4

3

2

1

0

x

y

x

y

1 - 1 1 +

2+

2

2 -

2x

Figure 7.6: The function () = 2 is continuous.

close tox0then(x)is very close to(x0) The formal denition of continuityfor functions of several variables requires cutting and pasting the denition forfunctions of a single variable, replacing R by R where appropriate| 0|bykx x0k and () (0)by (x) (x0).

Denition 23 (Continuity for a Function of Several Variables 1) Ifisan open subset ofR the function : R is continuous if for anyx0and any 0 there is a number 0 with the property that ifkx x0k then |(x) (x0) |

It is somewhat easier to visualize with an equivalent denition

Denition 24 (Continuity for a Function of Several Variables 2) Ifisan open subset ofR the function : R is continuous if for anyx0and any 0 there is a number 0 with the property that ifx(x0 )then |(x) (x0) |

To see that the two denitions are equivalent observe that the set of valuesofxfor which kx x0k is the open ball (x0 ).

Figure 7.7 illustrates this denition for one of economists favorite func-tions and diagrams, a utility function and indierence curves. There is anopen ball (x0 )that lies in the set between the two indierence curves with (x) = (x0) and (x) = (x0) . For all points in this open ball| (x) (x0)|


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7.8. CLOSED SETS, BOUNDED SETS, COMPACT SETS AND CONTINUOUS FUNCTIONS77

u(x) = u(x0) +

u(x) = u(x0) -

x0

B(x0,)

x1

x2

0

u(x) = u(x0)

Figure 7.7: Indierence Curves for a Continuous Utility Function

7.8 Closed Sets, Bounded Sets, Compact Sets

and Continuous Functions

For economists one of the most important results on continuous functions is:

Theorem 25 If is a closed and bounded subset of R and the function

: R is continuous the function has a maximum and a minimumon, that is there are elementsxmin andxmax ofwith the property that

(xmin) (x) (xmax)

for allx in.

The theorem is important due to economists obsession with maximizationas it gives conditions under which you can be sure that the function has amaximum and minimum. The proof is somewhat complicated and I do notgive it here.

This denition is no use without a denition of a bounded set:

Denition 26 A subsetofR isbounded if there is an element ofR withthe property that

kxk for allx in

Another way of saying this is that the distance between x and 0 is less than for all elements of. If = 1this simply means that for all in. If = 2 this means thatlies inside a circle with radius centered on theorigin. More generally it means that allxin lie in the open ball (0 ).


32/46


It is important thatbe both closed and bounded. For example the interval(0 1)is bounded but not closed, the function () = 1 is continuous on(0 1)but has no maximum, as ()grows without limit as gets closer and closer to0. The interval [0 ) is closed but not bounded, and the continuous function() = 2 grows without limit as tends to innity.

If you have done some general topology you will have come across the term"compact". There is a general denition of a compact set and then a theorem(the Heine-Borel Theorem) which states that the sets in R that are compactare the sets that are closed and bounded, so when working in R you shouldinterpret compact as meaning closed and bounded. If you have never doneany general topology ignore this comment.

7.9 Continuity for Consumer Theory

7.9.1 The Denition

Continuity is important at one point in consumer theory, showing that thecontinuity assumption on preferences implies the existence of a continuous utilityfunction that represents the preferences. In order to understand this resultyou need to know that continuity can be dened in the following way.

Denition 27 (Continuity for a Function of Several Variables 3) Ifisa subset of R the function : R is continuous if for all elements ofR the sets{x: x (x) } and{x: x (x) } are both closed.

For this to make sense this denition has to be equivalent to Denitions 23

and 24 of continuous functions. The rest of the chapter demonstrates that thisis indeed so. This is very much an exercise in pure mathematics with a numberof real analysis type arguments with and . I suggest skipping this section ifyou have never seen any real analysis.

7.9.2 Level Sets, Upper Contour Sets, Lower Contour Sets

It is useful to have some terminology.

Denition 28 If is a subset ofR and the function : R, the set

{x: x (x) =}

of elements of for which(x) = is called a level set.

This may be a new piece of vocabulary, but economists are very familiarwith level sets; we call the level set of a utility function an indierence curve,and the level set of a production function an isoquant.


33/46

7.9. CONTINUITY FOR CONSUMER THEORY 79

x1

x2

0

u(x) = y

level set

indifference curve

u(x) > y

the upper contour set

u(x) < y

the lower contour set

Figure 7.8: Level Set, Upper Countour Set and Lower Contour Set for a UtilityFunction


{x: x (x) }

of elements of for which(x) is greater than is called an upper contourset.


{x: x (x) }

of elements of for which(x) is less than is called a lower contour set.

Figure 7.8 illustrates these sets for a utility function.

7.9.3 Open Set and Closed Set Denition of Continuity

The argument that Denition 27 of continuity is equivalent to Denitions 23and 24 starts from the following proposition.

Proposition 31 Ifis an open subset ofR the function : Ris contin-uous if and only if for all elements of R the upper contour set{x: x (x) }and the lower contour set{x: x (x) } are both open.


34/46


x1

x2

0

x2B(x1,1)

B(x2,2)x1

u(x) = y

level set

indifference curve

u(x) > y

the upper contour set

u(x) < y

the lower contour set

Figure 7.9: The Upper and Lower Contour Sets are Open

Figure 7.9 illustrates the conditions of this thereon for a utility function.As the lower contour set is open, for any point x1in the lower contour set, thereis an open ball (x1 1) that is a subset of the lower contour set. As theupper contour set is open, for any point x2 in the upper contour set, there is anopen ball (x2 2) that is a subset of the upper contour set.

The proof of Proposition 31 is suggested by Figures 7.7 and 7.9.Proof. The function : R is continuous if for any 0 and any

0 there is a number 0 with the property that if x(x0 ) then|(x) (x0) | In particular if(x1) so (x1) 0 there is anopen ball(x1 1)on which|(x) (x1) | (x1). As(x) (x1)|(x) (x1) | which implies that (x) (x1) (x1) so (x) .Thus the lower contour set is open. Similarly if(x2) so (x2) 0there is an open ball (x2 2) on which |(x2) (x) | (x2) As(x2) (x) |(x2) (x)| this implies that (x2) (x) (x2) so (x). Thus the upper contour set is open. Hence the upper and lowercontour sets are open if the function is continuous.

To prove the converse result note that if the upper contour set

{x: x (x0) (x)}

and the lower contour set

{x: x (x) (x0) +}

are open then, from Proposition 17, their intersection is also open. This inter-section is the set on which (x0) (x) (x0)+or equivalently the seton which (x)(x0) , that is the set {x: x |(x) (x0)| } The point x0 is an element of this set, hence as the set is open there is an open


35/46

7.9. CONTINUITY FOR CONSUMER THEORY 81

ball (x0 ) with the property that all elements of the open ball lie in the set{x: x |(x) (x

0)| } This is what is needed to establish continu-

ity.

This proposition makes possible yet another equivalent denition of conti-nuity.

Denition 32 (Continuity for a Function of Several Variables 4) Ifisan open subset ofR the function : R is continuous if for all elements of R the upper contour set {x: x (x) } and the lower contour set{x: x (x) } are both open.

Now suppose that (x) : R is continuous and consider the comple-ment in of the open upper contour set {x: x (x) }, that is the set

{x: x (x) } This is by denition a closed set2 . Similarly the com-plement in of the open lower contour set {x: x (x) } that is theclosed set {x: x (x) } Given Denition 32 I can give yet anotherdenition of continuity.

Denition 33 (Continuity for a Function of Several Variables 5) Ifisan open subset ofR the function : Ris continuousif for all elementsofR the sets{x: x (x) } and{x: x (x) } are both closed.

I have been assuming throughout this argument that is an open set. Witha little more mathematical sophistication3 I can drop the requirement that isan open set. This is useful thing to do, because for may purposes it is necessary

to think of functions dened on R+

which is dened as the set of vectors with 0 for = 1 2. The set R+ is closed.

Denition 34 (Continuity for a Function of Several Variables) If isa subset ofR the function : R is continuous if for all elements ofR the sets{x: x (x) } and{x: x (x) } are both closed.

This is the denition used in consumer theory.

2 I am being somewhat sloppy with denitions of open and closed sets here. If you havenot noticed my sloppiness ignore this footnote. The diculty is that I am assuming that is open and is continuous, in which case the set {x: x (x) } that I am calling thecomplement of the upper contour set in is not a closed subset ofR unless = R. Ican get away with this by dening an open set in as the intersection ofand a set that is

open in R

, and a closed set in as the complement in of an open set in If is openthe open sets in are open in R, but the closed sets in are not closed in R. If isclosed some of the open sets in are not open in R, but the closed sets in are closed inR. If you know about general topology you will recognize that either way I am setting upa topology on , and know that this makes mathematical sense.

3 In order to do this I have to dene an open set in as the intersection of an open set inR, and then dene a closed subset ofas the complement in of an open subset in . Se efootnote 2.


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7.10 Appendix: Proofs

7.10.1 Proof That An Open Interval is an 0pen Subset ofR

If and = min ( ) then as

0 (7.4)

As = min ( ) and 0

1

2

so

1

2 (7.5)

As = min ( )

+ +1

2

so

+1

2 (7.6)

Taken together inequalities 7.4, 7.5 and 7.6 imply that

1

2 +

1

2 (7.7)

The set of points satisfying 12

+ 12

is the open ball

2

Inequality 7.7 then implies that 2 is a subset of the intervals( ) ( )

and ( ) Thus these intervals are open.

7.10.2 Proof That An Open Ball is an Open Subset ofR

The open ball (x0 )is the set of points satisfying kx x0k Note thatifx is an element of(x0 ) then kx x0k 0. Let z be an element ofthe open ball (x kx x0k) Then

kz xk kx x0k

From the triangle inequality (proved in the previous chapter on vectors)

kz x0k= kz x + x x0k kz xk + kx x0k

The last two inequalities imply that

kz x0k kz xk + kx x0k kx x0k + kx x0k=

Thus ifz is an element of the open ball(x kx x0k)then kz x0k , soz is an element of the open ball (x0 ). Thus the open ball (x kx x0k)is a subset of (x0 ), so the open ball (x0 ) is an open set.


37/46

7.10. APPENDIX: PROOFS 83

7.10.3 Innite Sequences and Closed Sets

The result here is

Proposition 35 A set is closed if and only if for any innite sequence{x1x2x3} whose points are all that tends to a limit, that limit is in.

Proof. Suppose thatis closed and that {x1x2x3} is an innite sequenceof points in that tend to a limit in . Asis closed is open so there isan open ball (x )that is a subset of

As{x1x2x3}tends to xthe point x is an element of(x ) and thus of for all suciently large. But by assumption all points in the sequence lie inso cannot be in .This contradiction implies that the limit of innite sequence of points a closedsetmust lie in .

Now suppose that the limit of any innite sequence of points in lies in ,but is not closed. As is not closed, is not open. This implies thatthere is a point x in for which every open ball(x )contains a point thatis not in and therefore is in regardless of how small is. Thus thereis an innite sequence of points {x1x2x3} that are not in

for whichx

x 1

This innite sequence tends to x As none of the points in

the innite sequence are in they are all in . However the limit is in.This contradicts the supposition on limits of innite sequences and not beingclosed that I started with. Thus is closed.


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39/46

Chapter 8

Introduction toMultivariate Calculus

8.1 Why Economists are Interested

We use multivariate calculus in economics because the things we are interestedin are usually functions of more than one variable. This gets suppressed inintroductory economics. You may have worked with a model where the costto a rm of producing is () = 1

22 and written

= as marginal cost.

This seems a bit special, so perhaps you generalized this to work with a costfunction () = 2At this stage you are treatingas a parameter of the cost

function. However when you think about it the cost must depend on inputprices and the technology used, so is itself a function, and changes in inputprices change . You will learn to think of cost as a function of input prices1, 2 and output (1 2 ) For example the minimum cost of producingoutput from inputs 1 and 2 with prices 1 and 2 using the technologygiven by the Cobb-Douglas production function = 1

2 is

(1 1 ) = 1

+

+

+

+

+

1

+

2 1+

(You can nd this result on page 54 of H.R.Varian, Microeconomic Analysis.)

This reduces to the cost function 2 if = = 14

and = 2212

112

2.However, having acknowledged that costs depend on 1, 2 and , you now

have to use the partial derivative

(1 1 )

=

1

+

1

+

+

+

+

+

1

+

2 1+

1

for marginal cost. When you have a function of many variables(1 2 )the partial derivative is dened analogously to the derivative of a function of a

85


40/46

86 CHAPTER 8. INTRODUCTION TO MULTIVARIATE CALCULUS

x0

y

x0

y = f(x)

y = f(x0) + f(x0) (x x0)

Figure 8.1: A function and its tangent

single variable as

(1 2 )

= lim

0

(1 2 +) (1 2 )

You calculate the partial derivative (1 2 )

in exactly the same way

as you calculate the derivative of a function of a single variable, treating all the

variables except as parameters of the function.In fact the distinction between a parameter and a variable depends entirely

on what you are interested in. Varian writes (1 1 ) treating , and as parameters. But for some purposes, for example modelling technicalchange you might want to think of, and as variables, and write costs as (1 2 )

8.2 Derivatives and Approximations

8.2.1 When Can You Approximate

Think about a function of one variable = () illustrated by the curvedline in Figure 8.1 and its tangent at 0 which is a straight line with equation= (0) +

0 (0) ( 0). The denition of the derivative0 ()implies that

close to the tangent line is a good approximation to the original function, thatis

() (0) +0 (0) ( 0) when is close to 0


41/46

8.2. DERIVATIVES AND APPROXIMATIONS 87

5 2.50 -2.5

-5

52.50-2.5

-5

120

100

80

60

x y

z

x y

z

Figure 8.2: Appoximating the function at (1 2)() = 100 21 22 at(1 2)

by the plane () = 95 2 (1 1) 4 (2 2)

Now think about the function of two variables = (1 2)illustrated in Figure8.2. The graph of the function is now a curved surface. Just as the curved linein two dimensional space corresponds to a curved surface in three dimensionalspace, the tangent line in two dimensional space corresponds to a tangent planein three dimensional space. In the chapter on vectors I showed that any planethrough the point (01 02 0)can be written in vector notation as

(1 2 3)

12

0102

0

= 0

or equivalently

1(1 01) +2(2 02) +3( 0) = 0

or if36= 0 as= 0+1(1 01) +2(2 02)

where 1 = 13

and 2 = 23

By analogy with the equation of the tangent

line as = (0) + 0 (0) ( 0)the equation of the tangent plane should be

= (01 02) + (01 02)

1(1 01) +

(01 02)

2(2 02)

and the approximation should be

(1 2) (01 02) +(01 02)

1(1 01) +

(01 02)

2(2 02)


42/46


This suggests that the extension to variables should be the approximation

(x) (x0) +X=1

(x0)

( 0) when xis close to x0

The argument I have just given is in no way rigorous, but it can be made so.This is done by assuming that the function (x) can be approximated by afunction of the form

(x) (x0) +X=1

( 0)

and then showing that must be the partial derivative (x0)

.

You have to be slightly careful with this, because unlike the situation withfunctions of a single variable it is possible that the derivatives exist but theapproximation does not work. As an example consider the function :R2 Rgiven by

(1 2) = 0 when 12= 0

(1 2) = 1 when 126= 0

This function is 0 when one or both of 1 and 2 are zero, and 1 otherwise.Thus (1 0) = 0 for all values of 1 and (0 2) = 0 for all values of 2,implying that the partial derivatives

(0 0)

1=

(0 0)

2= 0

This suggests that when (1 2) is close to (0 0)

(0 0) +(0 0)

11+

(0 0)

22 (8.1)

should be a good approximation to (1 2) However

(0 0) = (0 0)

1=

(0 0)

2= 0

so this implies that 0 is a good approximation to (1 2) when (1 2) isclose to 0. But this is very far from being a good approximation because(1 2) = 1 whenever 126= 0.

An assumption that ensures that the approximation works for a function

: R is that is open, that has partial derivatives on , and that thepartial derivatives are continuous. In order to avoid writing out the conditionsmany times I will use the following denition.

Denition 36 Ifis an open subset of R the function(x) : R is awellbehaved if it has partial derivatives, and the partial derivatives are continuous

on.


43/46

8.3. LEVEL SETS AND VECTORS OF PARTIAL DERIVATIVES 89

x0

x1

x2 Df(x0)

Df(x0)( x - x0) = 0

f(x) = f(x0)

Figure 8.3: The Level Sets, Tangent and Partial Derivative Vector for a Function

If a function is well behaved the approximation

(x) (x0) +X=1

(x0)

( 0) when xis close to x0

works. Making this argument rigorous requires a precise denition of what ismeant by approximation, and then a proof that existence and continuity of thepartial derivatives implies that the approximation is valid.

8.3 Level Sets and Vectors of Partial Derivatives8.3.1 The Level Set and Tangent for One Function

In the previous chapter I gave:


{x: x (x) = }

of elements of for which(x) = is called a level set.

A level set of a utility function is an indierence curve; a level set of a

production function is an isoquant. Now think about two pointsx and x0thatlie in the same level set so

(x) = (x0)

and assume that the function : R is well behaved so close to x0

(x) (x0) +X=1

(x0)

( 0)


44/46


As(x) = (x0)this implies that

X=1

(x0)

( 0) 0 (8.2)

If I use notation

(x0) =

(x0)1

(x0)2

(x0)

so(x0)is the vector of partial derivatives equation 8.2 becomes

(x0)0 (x x0) 0

Thus when x is very close to x0 and (x) = (x0) x very nearly satises theequation.

(x0)0 (x x0) = 0

This is an equation of the type that I discussed at some length in section 7of the chapter on vectors. If = 2 this is a straight line, if = 3 it isa plane, and for 3 it is a hyperplane. In every case the vector of partialderivatives(x0)is orthogonal (at90

) to the line, plane or hyperplane. Thisis illustrated in Figure 8.3, which shows the vector of partial derivatives (x0)which is orthogonal to the line (x0)

0 (x x0) = 0 The line is a very goodapproximation to the level set when x is close to x0, and it seems reasonablyintuitive and is in fact true that this line must be tangent to the level set.

8.3.2 Level Sets and Tangents for Two Functions

This section goes beyond revision mathematics in taking a look at a constrainedoptimization problem, and working with vectors of partial derivatives to gener-ate insights about Lagrangians. Figure 8.4 is the same as Figure 8.3 except thatI have introduced another function (x), its partial derivative vector (x0)at x0and its level set (x) = (x0) I have chosen to show the case where thetwo partial derivative vectors point in dierent directions. In Figure 8.4 thereis a point x1for which (x1) (x0) and (x1) (x0) which immediatelyimplies that if = (x0), the point x0 cannot solve the problem of maximizing(x)subject to (x) .

From Figure 8.4 it looks as if the problem is that the two partial derivative

vectors point in dierent directions. I have a result from the chapter on vectorsthat goes some way to clarifying what is happening here. This is:

Proposition 38 If p is a non zero vectors in R, q is a vector in R, andthere is no real number 0 such thatq= p then for any vectorx0 thereis an vectorx1 such that

p0x1 p0x0


45/46

8.3. LEVEL SETS AND VECTORS OF PARTIAL DERIVATIVES 91

x0

x1

x2

Df(x0)

Df(x0)( x - x0) = 0

f(x) = f(x0)

g(x) = g(x0)

Dg(x0)

Dg(x0)( x - x0) = 0

x1

Figure 8.4: Level Sets and Partial Derivative Vectors for Two Functions

and

qx1 q0x0.

Using this result with p= (x0) and q= (x0) I have the fact that ifthere is no real number 0 such that (x0) = (x0) then there is avector x1 such that

(x0)0

x1 (x0)0

x0

which implies that

(x0) + (x0)0 (x1 x0) (x0) (8.3)

and(x0)

0

x1 (x0)0

x0

which implies that

(x0) +(x0)0

(x1 x0) (x0) (8.4)

Ifx1is very close to x0the terms on the left hand side of inequality 8.3 is a verygood approximation to (x1)and the terms on the left hand side of inequality8.4 is a very good approximation to (x1) All this suggests the followingproposition

Proposition 39 If is a well behaved function and there is no number 0such that(x0) = (x0) then there is a vectorx1 such that

(x1) (x0)

and

(x1) (x0)


46/46


This proposition is true, although I certainly havent proved it. Doing thatrequires a precise statement of what how the approximation argument works.The proposition immediately implies that if there is no number 0 such that(x0) = (x0)or equivalently

(x0)

=

(x0)

for = 1 2

and (x0) = then x0 cannot solve the problem of maximizing (x) subjectto (x) . The number is of course the Lagrange multiplier. Anotherway of saying this is that if and are well behaved functions and (x0) =the rst order conditions from the Lagrangian are necessary for x0 to maximize(x)subject to (x) .

EC 400 Notes for Class 1.pdf

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