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COURSE 002: INTRODUCTORY MATHEMATICAL ECONOMICS
DELHI SCHOOL OF ECONOMICS
HANDOUT ON PRILIMINARIES
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Welcome to D’School! These notes are intended to warm-up our incoming MA students. We assume that you have taken a course on math-eco before. We hope that most of the material in these notes will be a review of what you already know, before we take you to a higher level. If you have never taken such a course before, you are strongly encouraged to go through this material presented and/or mentioned here on your own. If some of the material is unfamiliar, do not be intimidated! We hope you find these notes helpful! If not, you can consult the references listed in the syllabus, or any other textbooks of your choice for more information or another style of presentation.
Now Cheers!
Part -1: Part-2
Elementary Logic Derivative & Optimisation (1 var) Set theory Multi variate functions Functions, types, Comparative Statics (1 var) Concavity, Convexity, Contours Envelop Theorem Group theory, Ring, Field Directional Derivative
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Part- 1 1.1. Elements of logic 1.1.1. Sentential Logic
Assertions or statements have the property of being either true or false. In sentential logic we consider only those sentences which are assertions. Examples of Sentences:
(i) Sun rises in the east. (ii) Two and two make five. (iii) What time is it? (iv) One ought not to tell lies.
The !rst two are statements and the last two are not.
1.1.2. Logical Connectives (i) Negation (!!): Negation of a true statement is false and that of a false one is true.
Schematically, A !!A T F
F T (ii) Conjunction (""## $$ ##%%&&''((##))): A conjunction of two statements is true if and only if
both the constituent statements are true. Schematically, A B (A ! B) T T T T F F F T F F F F
The word ‘but’ has about the same sense as ‘and’.
(iii) Disjunction (**): In common usage ‘or’ is used both in exclusive and inclusive senses. In logic ‘or’ is always used in the inclusive sense. A disjunction of two statements is false if and only if both of its constituents are false. Schematically,
A B (A ** B) T T T T F T F T T F F F
(iv) Conditional or Implication (!): A Conditional is a statement of the form A " B (read as ‘If A then B’) where A and B are statements. A is called the antecedent and B the consequent. A conditional is false if and only if the antecedent is true and the consequent is false. Schematically,
A B (A ! B) T T T
T F F F T T F F T
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B true !"#$%&"
!"#$%"%&'("
)"#$%"%&'("
A true '"#$%&"
)"#$%"%&'("
!"#$%"%&'("
Again, consider P " Q. If the statement P is true, then the statement Q is true. But if P is false, Q may be true or false. For example, on real numbers, let P = x > 0 and Q = x2 > 0. If P is true, so is Q, but Q may be true even if P is not, e.g. x = - 2. We club these together and say P " Q.
(v) Biconditional ( " ): A # B (‘A if and only if B’) is true if and only if A and B have the same truth-value. Schematically,
A B (A " B) T T T T F F F T F F F T
A classical example of a non-truth-functional connective is that of possibility. For example, the sentence:
(1) It is possible that there is life on Mars
is true under any liberal interpretation of the notion of possibility; but then so is the sentence:
(2) It is possible that there is not any life on Mars.
On the other hand, the sentence:
(3) It is possible that 2+2=5
is ordinarily regarded as false.
1.1.3. Necessity and Sufficiency
- Necessity: A is necessary for B , !A must hold or be true in order for B to hold good
or to be true, !B is true requires that A must also be true! “A if B” or “A is implied
by B” (A!B).
If A is not true, B must be not true. But that doesn’t mean
that if B is not true, A must be not true.
! A is not true ! B must be not true ! B is not true is necessary for A is not true.
i.e. ~A ! ~B.
- Sufficiency: “A is sufficient for B” means that if A holds, B must hold. We can say “A
is true only if B is true”; or A implies B (A!B).
B is not true ! A must be not true. ! B is not true is sufficient for A is not true i.e. ~B ! ~A.
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!"'"
- Both Necessity and Sufficiency: A"B
1.1.4. Theorems and Proofs
“A ! B” : A is true! B must be true
- Here, A is called “premise” and B is call “conclusion”
- Constructive proof / Direct proof: Assume that A is true, deduce various
consequences of that, and use them to show that B must hold.
- Contrapositive proof: Assume that B does not hold, then show that A cannot hold.
~B ! ~A is the contrapositive of A ! B.
1.1.5. Theorems, Assumptions and Conclusion
- A Theorem: is simply a statement deduced from other statements and should be a
concept familiar from courses in mathematics.
- Theorems provide a compact and precise format for presenting the assumptions and
important conclusions of sometimes lengthy arguments, and so help to identify
immediately the scope and limitations of the result presented.
1.1.6. Tautology: A tautology is a statement which is true for all logically possible
combinations of truth-values of the constituent statements.
Let us call a sentence atomic if it contains no sentential connectives.
We can also define the notion of tautology as follows: A sentence is a tautology if and only if the result of replacing any of its component atomic sentences (in all occurrences) by other atomic sentences is always a true sentence.
Tautological Implication: A tautologically implies B if and only if (A " B) is a tautology.
Tautological Equivalence: A is tautologically equivalent to B if and only if (A # B) is a tautology.
Examples of tautologies:
(i) A " A
(ii) A " (A * B)
(iii) (A " B) # (B " A)
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(iv) (A * B) # (B * A)
(v) (A " (B " C)) # ((A " B) " C)
(vi) (A " (B * C)) # ((A " B) * (A " C))
(vii) (A * (B " C)) # ((A * B) " (A * C))
(viii) (!(A " B)) # (!A *!B)
(ix) (!(A * B)) # (!A "!B)
(x) (A " B) # (!B "!A)
(xi) (A # B) # ((A " B) " (B " A))
(xii) (A " C) " (B " C) " ((A * B) " C)
(xiii) (A " B) # [!(!A *!B)]
(xiv) (A " B) # [!(A "!B)]
(xv) (A * B) # [!(!A "!B)] # (!A " B)
(xvi) (A * B) # (!A " B)
(xvii) (A " B) # [!(A "!B)]
(xviii) (A " B) # (!A * B)
(xix) (A # B) # [!(A "!B) "!(B "!A)]
(xx) (A # B) #! [! (!A * B) *! (!B * A)]
(xxi) (A # B) #![(A " B) "!(B " A)].
Inference: Q logically follows from P if P ! Q is a tautology, or iff the conditional P ! Q is universally valid. In other words:
Q logically follows from P if Q is tautologically implied by P.
The consistency of a set of premises whose logical structure may be expressed by sentential connectives alone may be determined by a mechanical truth table test. The truth table for the conjunction of the premises is constructed. If every entry in the !nal column is ‘F’ then the premises are inconsistent. If at least one entry is ‘T’ the premises are consistent.
Quanti#ers: x " 5 is not a statement. In order to change x " 5 into a statement one has to replace the variable by some constant or use a quanti!er. Consider the following two statements:
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(i) For any number x, if x is divisible by 2 and 3 then x is divisible by 6. [In symbols: (! numbers x) (x is divisible by 2 and 3 " x is divisible by 6)].
(ii) For any number x, there exists a number y such that y " x. [In symbols: (! numbers x) (+ a number y) (y " x)].
In statement (i) the variable x is bound by a universal quanti!er and in statement (ii) the variable x is bound by a universal quanti!er while y is bound by an existential quanti!er.
Example: Let S be the set of natural numbers. Then, !(, x " S) (x is divisible by 5) # (#$x$" S) (x is not divisible by 5). %(#$x$" S) (x is divisible by 5) # (,#x - S) (x is not divisible by 5).
1.2. Elements of Set Theory 1.2.1. Notation and Basic Concepts
- Set: A set is a collection of objects/ elements. Elements may be numbers or vectors.
Examples: S = {1, 7, 9, 4}. A = {1, red colour}. N = {x | x is a natural number}.
The members of a set are called its elements. If x is an element of set A we say that x is in A and write x " A.
- Subset: A is a subset of B (i.e. A & B) if and only if for every x, if x " A then x " B.
- Proper Subset: Set A is a subset of set B if every element of set A is also an element
of set B. Notation: A! B.
- Empty set: S is an empty set if it contains no elements at all. Notation: S = !
- Complement Set: The complement of set S in a Universal set U is the set of all
elements in U which are not in S. Notation: Complement set of S: S# OR cS OR S c.
- Union: Notation { | }S T x x S or x T! " # # . In general: i I iS!" (I is Index set).
- Intersection: Notation { | }S T x x S and x T! " # # . Or i I iS!! (I is Index set).
- Index Set: The set of interger number starting with 1. S = (1, 2, 3, 4..., n). Notation:
...}3,2,1{!I
- Difference of Sets: A – B = { | and }x x A x B! " . Also called the complement of B relative to A. More familiar is the notion of the universal set ! , and cB B!" = .
- Set Equality: A is equal or identical to B if and only if A and B have the same elements. A = B $ [(A & B) ' (B & A)].
- n – Space: is the Set Product : 1 2... {( , ,..., ) | }nn iR R R R x x x x R= ! ! ! " # i =1,2...,n.
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- Non-Negative Orthant: 1 2{( , ,..., ) | 0}n nn iR x x x x R+ ! " # "
- The Product of two sets OR, Cartesian Product: Cartesian product of two sets A and B is defined as the set A ( B = {(a, b) | a " A ' b " B}. Or, },|),{( TtSstsTS !!"#
Example: Suppose A = {blue colour, red colour} and B = {7, 5, 2}. Then, A ( B = {(blue colour, 7), (blue colour, 5), (blue colour, 2), (red colour, 7), (red colour, 5), (red colour, 2)}.
1.2.2. Some Important Set Identities and Other Principles
(1) A . ! = A Note here: ! is null set, and ! is universal set.
(2) A $ ! = A
(3) A . B = B . A
(4) A $ B = B $ A
(5) A . (B $ C) = (A . B) $ (A . C)
(6) A $ (B . C) = (A $ B) . (A $ C)
Proof. of (6) Suppose x !LHS. So, x ! A and x ! (B $ C). So x ! A and either x ! B or x ! C or both. So, either x ! (A $ B) or x ! (A $ C) (or both). Converse?
(7) A .!A = !
(8) A $!A =!
(9) A . A = A
(10) A $ A = A
(11) A . ! = !
(12) A $ ! = !
(13) ! ! !
(14) !!A = A
(15) A = !B " B = !A
(16) A . B ! ! " A! ! * B! !
(17) A $ B ! ! " A! !
(18) A . (B . C) = (A . B) . C
(19) A $ (B $ C) = (A $ B) $ C
- Distributive Laws "
* ./00#&%&123!4%56!
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(20) A . (A $ B) = A
(21) A $ (A . B) = A
(22) !A ! A
(23) !(A . B)= !A $!B
(24) !(A $ B)= !A .!B
(25) A $ !A =!
(26) A ! (A $ B)= A ! B
(27) A $ (A ! B)= A ! B
(28) (A ! B) ! B = A ! B
(29) (A ! B) ! A =!
(30) (A ! B) . B = A . B
(31) (A . B) ! B = A ! B
1.2.3. Relations and Functions
Consider the two sets: set S (s1, s2,...) and set T (t1, t2,...).
The product of the two sets: S ! T = {(s,t)| s ! S and t ! T} , called Order pairs.
Any Collection or Set of Order pairs is said to constitute a Binary Relation (S/T) of the
sets S and T. Binary Relation: A binary relation from set S to set T is a subset of S( T.
Meaningful Relationship: (sRt) : The Set of the order pairs that are constituted by elements
of the sets S and T are meaningful.
},|),{( TSsRtandTtSstssRt !"##$
Example: },|),{(2 SySxyxSSS !!==" " " {( , ) | , , }x y x S y S x y! " = # # "
Function: A relation f from S to T , notationally f : S ! T , is called a function iff for every x
" S, there is a unique y " T such that (x, y) " f.
In other words, a relation from S to T is a function iff –
[(! x " S)[(#$y " T )((x, y) " f)] ' (! x " S)(!$y, y’" T )[(x, y) " f ' (x, y’) " f ! y = y’]].
Example: Consider sets A = {2,3,4,9} and B = {1,7,6,4,8}.
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A ( B = {(2, 1), (2, 7), (2, 6), (2, 4), (2, 8), (3, 1), (3, 7), (3, 6), (3, 4), (3, 8), (4, 1), (4, 7), (4, 6), (4, 4), (4, 8), (9, 1), (9, 7), (9, 6), (9, 4), (9, 8)}.
Let: R1 = {(2, 1), (3, 6), (9, 4)}. R2 = {(2,4), (2,6), (3,1), (4,7), (9,8)}.
R3 = {(2, 8), (3, 1), (4, 4), (9, 7)}. R4 = {(2, 8), (3, 1), (4, 4), (9, 4)}.
Among these four relations only R3 and R4 are functions. Why? Check yourself.
Image: Let f : S ! T be any function, and let A be a subset of S. De!ne f(A)= {y | for some x " A, (x, y) " f}. We call f(A) the image of A in T .
Inverse of a relation: Given any relation R from A to B, we de!ne R%1 = {(y, x) | (x, y) " R}. The relation R%1 from B to A is called the inverse of R.
1.2.4. Properties of Binary Relations: Let R be a binary relation on a set S. We de!ne:
(! x, y " S) (xPy $ xRy '$% yRx) : Preference relation
(! x, y " S) (xIy $ xRy ' yRx). : Indifference relation
• Completeness: A relation R on S is complete if, for all distinct elements x and y in S,
xRy or yRx [the order pairs (x,y) or (y,x)] are all meaningful. Ex1: x||y
• Re$exivity: R on S is re&exive i0 (! x " S)(xRx). • Symmetry: R on S is symmetric i) (! x, y " S)(xRy ! yRx). Ex: x"y • Transitive: A relation R on S is Transitive if, for all three elements x, y and z in S, xRy
and yRz implies xRz."R on S is transitive i) (! x, y, z " S)(xRy " yRz ! xRz). Ex1. • Odering: A binary relation which satisfies all properties of Completeness, Reflexivity
and Transitivity is called an Ordering.
A relation which is reflexive, transitive and symmetric is called an Equivalence relation.
Following are the some other properties of set –
- Irre$exivity: A relation R on S is reflexive if, for all elements x in S, xRx (ordered
pairs (x, x) are meaningful). R on S is irre&exive i0 (! x " S)(%xRx).
- Connected: R on S is connected i) (! x, y " S)(x = y ! xRy * yRx).
- Asymmetry: R on S is asymmetric i) (! x, y " S)(xRy !%yRx).
- Anti-symmetry: R on S is Anti-symmetric i) (! x, y " S)(xRy '$yRx ! x = y).
- Quasi-transitive: R on S is quasi-transitive i) (! x, y, z " S)(xP y " yP z ! xP z).
- Acyclic: R on S is acyclic i0 (!$n " N %{1, 2})(!$x1, x2, ... xn " S)(x1Px2 ' x2Px3 ' ... '
xn%1Pxn ! x1Rxn), where N is the set of positive integers.
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2.1. Mapping and Functions: 2.1.1. Mapping
Given two non-empty sets A and B, if it be possible to associate or match each element of the set A with a single element of B by some specified manner f, then this matching procedure is called mapping or correspondence.
This mapping or map is denoted by f: A ! B.
Consider the sets A ={1, 2, 3, 4}, B ={4, 8, 12, 16}. Here we observe that with each element of the set A, we can match element of the set B which is a number four times as large. We call this the mapping f and denote the match by writing: f: x!4x.
We define mapping (or function) in an equivalent form as below –
Let A and B be two non-empty sets. A sub-set f of A!B is called a function or mapping from A to B, if, to each a !A, exists a unique b! B such that the ordered pair (a,b) ! f. The set A is called the domain and the set B is called the co-domain of the mapping f. We see that, to each a in A, f associates an element b in B uniquely.
In general, if a !A and b !B, then we say that b is the f-image of a and write f(a) = b. a is called the pre-image of b. Note that there may be more than one pre-image of b.
A map f: A!B is well defined, if
(i) a !A ! f(a) !B, (ii) any element a !A!unique element f(a) ! B, (iii) two or more than two elements of A may have same image in B.
If A = {2, 3, 4} and B = {5, 7, 8} and if f(2) = 5, f(3) = 7, then f: x ! 2x + 1 does not define a mapping as 4!A has no image in B.
2.1.2. Different types map:
Consider the following mappings :
A = {1, 2, 3, 4, ......}, B ={ 1,1/2,1/3, ',...... }; thus f: x!1/x.
A = { 1,2,3,4, ...... }, B = {I, 4, 9, 16, ...... }; thus f: x!x2 .
In the above two mappings, one element a !A maps to one element b ! B. Such a mapping is called one-one In this case distinct element of the domain has distinct in the co-domain.
Thus a mapping f is called one-one or injective, if different elements of the set have different f-images in B. Thus, if a1, a2!A, then a1 ! a2! f(al) ! f(a2) in B, or equivalently f(a1)= f(a2) ! a1=a2.
If f be injective then each element of B has at most one image in A. The graph of f, denoted by f*, is the sub-set of A! B by {a, f(a) : a !A and f(a) ! B }.
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The range of f is the set of all images under f and is denoted by f(A) or {f(a)} for all a !A. Evidently f(A)= {b:b=j(a) for some a !A and b !R}. In general, f(A)! B. Let the domain A be {a, b, c, d} and the co-domain B be {a, b, c}· Then for the function f from A ! B as given by f = {(a, b), (b, c), (c, c), (d, b)}, the range f (A) is {b, c}.
Consider a set A = {A1, A2, A3, A4, A5}, the elements being the students of your class and a set B= {50, 55, 61}, the elements being the marks obtained by the students. It is possible that more than one student may secure the same mark; hence here more than one member of the domain may have the same image. This type of mapping is called many-one-mapping. While matching, if every element of the co-domain be an image of some element of the domain and no element of the co-domain remains unused, then the mapping is from the domain onto the co-domain. Such a mapping is called onto mapping.
On the other hand, if at least one element in the co-domain be not an image of some element of the domain, then such a mapping is not an onto mapping and is called an into mapping. If A = {1, 2, 3, 4}, B ={2, 3, 4, 5} be two sets, then f : x - x + 1 maps A onto the set B. Again, if A be the set of all even positive integers, then f: x ! x + 1, x ! A, is not onto mapping of A to the set B of all positive integers, odd or even.
Notice that for an into mapping {f(x)}! B for all x!A, while for an onto mapping {f(x)} = B for all x!A. Also notice that the range is always a sub-set, not necessarily a proper sub-set, of the co-domain. We give diagramatic examples of different types of mapping here –
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Onto mapping is also known as surjective mapping. The one-one and onto mapping is known as bijeciive mapping. In this case the two sets will contain the same number of elements.
Besides these, there are many-one onto and many-one into mappings. Let f : A!B and g : B !C be two maps. Then the two mappings f and g are said to be equal, if and f(x) = g(x) for all x!A.
In this case the domains of the mappings must be the same.
Let A be a non-empty set and f be a mapping such that, by f: A !A each element A is mapped on itself. Then f is called the identity mapping and f(x) = x for all x !A. If A={I, 2, 3}, then f= {(1, 1), (2, 2), (3, 3)} is an identity mapping of A. Identity mapping is always one-one onto. It is denoted by IA .
The mapping f is called constant, if the range of f be a singleton set, that is, f(a) = b for all a!A and b!B. Two sets A and B are said to be cardinally equiva1ent, if there exists a mapping f from A to B which is one-one and onto. We denote this by A~ B. These two sets are said to be equipotent with each other.
Let I be the set of positive integers and E be the set of all even integers. Consider the mapping f: I! E as given by f(x) = 2x for all x!I. Then I is equipotent with E for the mapping is one-one onto. The two sets I and E are cardinally equivalent. If a set A be cardinally equivalent to the set of natural numbers N={1, 2, 3, 4, ...... } then A is called a denumerable set. A set is said to be countable, if it be finite or denumerable.
The set A = { a2 : a !N} is denumerable for the mapping f: N ! A as given by f(a)=a2, a!N is one-one and onto.
2.1.3. Inverse mapping
Let f: A!B be a map and b!B be arbitrary. Then the inverse of the element b, which is denoted by f –1(b) is defined as a srt consisting of those elements of A which have b as their images. Obviously, f –1(b)!A.
Let f:A!B be a one-one onto mapping. Then the mapping f –1(b):B!A which associates, to each element b!B, the element a!A, such that f(a)=b, is called inverse mapping of the map f. Thus f –1(b)={ a!A: f(a)=b}. The map f –1(b) may be a singleton set, or a set consisting more than one element. If f be one-one onto, then f –1(b)=a ! f(a)=b.
2.2. Functions
The property of convexity is most often assumed ! analysis is mathematically tractable and
results are clear-cut and well-behaved.
Images & Preimages: Let f: A B.
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Definition: Let A0!A. The image of A0 under f, denoted f(A0), is the set {b| b=f(a), for some
a"A0}.
Let B0!B. The preimage of B0 under f, f -& (B0) = {a| f(a) "B0}.
So the image of a set is the collection of images of all its elements, and the preimage or inverse image of a set is the collection of all elements in the domain that map into this set.
Example: For the function f(x)=x', let A0=[-2,2]. Then, f(A0)=[0,4]. Let B0=[-2,9]. Then, f-&
(B0)=[-3,3].
Facts. Let f: A B, A0, A1!A, B0,B1!B. Then we claim that -
(i) B0!B1 # f-&(B0)!f-&(B1).
(ii) f-& (B0* B1) = f-& (B0) * f-&(B1), where * can be U, , -.
i.e., f -& preserves set inclusion, union, intersection and difference. f only preserves the first
two of these. The 3rd holds with ! and the 4th with $. To find counterexamples, many-to-
one functions (to which we now move) are helpful.
Proofs of the above claims are homework.
Injective & Surjective Functions:
Definition: f: A B is injective (one-to-one) if [f(a)=f(a )]#[a = a ]. It is
surjective (onto) if for every b "B, there exists a "A s.t. b=f(a). A function that is both of
these is called bijective.
For example, f: % % defined by f(x)=x( is many-to-one and not surjective. If the domain is
%+ instead, then f is injective, and further if the codomain is %+, then it is bijective.
Inverse of a Function properties:
Let f: A B, A0!A, B0!B. Then we claim that -
(i) f -& (f (A0))$A0; equality holds if f is injective.
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(ii) f (f -& (B0)) !B0; equality holds if f is surjective.
You should also show by examples that equality does not in general hold.
Now we use the f -& to mean something different, namely the inverse function.
If f is bijective, then define a function f -&: B A by f -& (b)=a if a is the unique element of A
s.t. f(a)=b. Note that f -& is also bijective.
Indeed, suppose b b and f -& (b)= f -& (b ) = a. Then f(a)=b and f(a)=b which is not
possible. So f -& is injective. Moreover, for every a"A, there is a b"B s.t. b=f(a), since f is a
function. So, there is a b"B s.t. f -& (b)=a. So f f -& is also surjective; hence it is bijective.
2.2.1. A Function is a Relation that associates each element of one set with a single, unique
element of another set. f: D ! R; where D is the Domain and R is called the Range.
Convex set in Rn : S!Rn is a conxex set iff for all x1 , x2 ! S, we have: t. x1 + (1- t). x2 ! S for all t in the interval 10 !! t . Geometrically, if x, y - Rn, then {z - Rn : z = )x + (1 % )) y for ) - [0, 1]} constitutes the straight line connecting x and y. So a convex set is any set that contains the entire line segment between any two vectors in the set.
Theorem: Intersection of Convex Sets is Convex. Can you prove this?
Theorem: The union of two convex sets is not necessarily convex. Why not?
A vector z - Rn is a convex combination of x1, ..., xm - Rn
if
1
m
j jj
z x!=
=" for some )1, …., )m* 0 with 1
1m
jj!
=
="
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
15
In the figure below –
Theorem 1: A set X 1 Rn is convex if and only if it contains any convex combination of any
vectors x1,...,xm- X.
Proof. The proof is by mathematical induction on m. For m =1, the only convex combination of vector x is x itself. So the basis statement for m =1 is true. The induction step is to suppose that the proposition is true for m % 1> 0 vectors, and then to show that this implies the proposition is true for m vectors. So consider any convex combination ( mj=1 )jxj of m vectors contained in X. Since m " 2 and each )j " 0 with ( mj=1 )j =1, we may suppose WLOG that )m < 1. Then since
1
1
1 1,1 1 1
mm jj j m
j m m m
!! !! ! !
=
=
"= = =" " "
##
the induction hypothesis implies that y +1( )1
m jjj
m
x X!!=
"#$ . Then the definition of a convex
set implies
The propoposition then follows from mathematical induction. QED.
Given any set X 1 Rn , the convex hull Co(X) is the intersection of all
convex sets that contain X. Since the intersection of any two convex
sets is convex, it follows that the convex hull is the smallest convex
set that contains X.
• If X is convex, then Co(X)=X. Why?
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
16
Observation 1: Suppose X 1 Rn . Then Co(X) is the set of all convex combinations of
vectors in X. Proof. Lemma 1 implies that any convex combination of elements x
1,..., x
m - X must be
contained in Co(X). To show that Co(X) contains only vectors that are convex combinations
of some x1,..., x
m - X, we need to show that Y + {x - R
n+: x is a convex combination of some
x1,..., x
m - X} is a convex set. So consider and y, z " Y. Then, by definition, there is a set of
vectors y1,...,ym " X and list of nonnegative numbers *1,...,*m with (*i =1 such that y = (*iyi.
Similarly, there is a set of vectors z1,...,zr" X and list of nonnegative numbers +1,...,+r with (+i
=1 such that z = 1
r iiiz!
=" . Then for any ) - [0,1],we have
)y+(1% ))z = ) 1
m
i=! ,iyi + (1% )) 1
r
i=! -izi
=1
m
i=! ),iyi + 1
r
i=! (1% ))-iz i
which, since )1
m
i=! ,i +(1% )) 1
r
i=! -i = )+(1% ))=1, implies that )y +(1% ))z is a convex
combination of y1,...,yn, z1,...,zr - X.
2.2.2. Real Valued Function
Definition of RVF: f: D!R is a real valued function if D is any set and RR ! .
We here restrict our attention to real valued functions whose domains are convex sets.
Assumption: Real Valued Function over Convex Sets.
Let f: D!R is a real valued function where D nR! is a convex set and R R! .
Increasing Functions: f: D!R is increasing function whenever f(x0)! f(x) ! xx !0 , and
xx !0 . We say f is strictly increasing function whenever f(x0) >f(x) ! x0 ! x, and xx !0 .
Note: xx !0 means that at least one of components of vector x0 is greater than the same
ordered component of vector x.
Decreasing Functions: f: D! R is decreasing function whenever f(x0)! f(x) ! xx !0 , and
xx !0 . We say f is strictly decreasing function whenever f (x0) < f(x) ! xx !0 , and xx !0 .
2.2.3. Related Sets
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
17
The graph of a function is a related set which provides an easy and intuitive way of thinking
about the function. There are some sets related to a function.
Level Sets: L(y0) is a level set of the real valued function f: D ! R iff 0 oL(y ) { | , ( ) y }x x D f x= ! = , where y0 R! . (The Contour Set)
Examples: Isoquant curve, Indiferrent curves, Isoprofit curve…
Usage of Level Sets:
- Reducing by one the number of dimesions needed to represent the function.
- Two indifferent level sets of a function can never cross
Level Sets Relative to a Point x0
L(x0) is a level sets relative to point x iff L(x0) = {x|x D! , f(x) = f(x0)}.
Superior and Inferior Sets
1. S(y0) = {x | x D! , f(x) ! y0} is call superior set for level yo.
2. I(y0) = {x | x D! , f(x) ! y0} is call inferior set for level yo.
3. S.( y0) = {x | x D! , f(x) > y0} is call strictly superior set for level yo.
4. I. (y0) = {x | x D! , f(x) < y0} is call strictly inferior set for level yo.
Theorem: Superior, Inferior, and Level Sets : For any f: D R! any Ry !0 :
1. L(y0) ! S(y0)
2. L(y0) ! I(y0)
3. L(y0) = S(y0) ! I(y0)
4. S.(y0) ! S(y0)
5. I. (y0) ! I(y0)
6. S. (y0) ! L(y0) = O
7. I. (y0) ! L(y0) = O
8. S. (y0) ! I. (y0) = O.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
18
2.2.4. Concave and Convex functions [+$&"%,("&(-./#0(&"$1"%,(2("#$%(23"4("2'55$2("
%,.%"6"1"7#"/2"."8$#9(:"2(%;"<"
Concave Function: f : X " R is concave if for any x, z - X,we have, for all ) - (0, 1) ,
f()z +(1 % )) x) * )f (z)+(1 % )) f(x).
Strictly Concave Functions: f : X " R"is strictly concave if for any x, y - X with x =z, we"have,
for all ) - (0, 1) , f()z +(1 % )) x) > )f (z)+(1 % )) f(x).
! A!constant function is concave. Why?
! A!linear function is concave. Why?
The set of points beneath concave regions is a convex set. The set of points beneath the
non-concave region is not a convex set.
For strictly concave functions, geometrically these modifications simply require the graph
of the function to lie everywhere above the the chord connecting any two points. Rule out
any flat portions of the graph of the function.
! f : X ! R!is concave if and only if f()(z % x)+ x) " ) (f(z) % f(x)) + f(x) for all x, z " X
and ) " (0, 1) . Why?
! f : X ! R!is!concave!if!and!only!if!f(),x+x) " ) (f(x + ,x) % f(x))+f(x) for!all!x, (x +
,x) " X and ) " (0, 1) . Why?
Theorem: Points on and below the graph of a Concave Fn always form a Convex Set
Let D nR! be a convex set and let R ! R. Let A })(,|),{( yxfDxyx !"# be the set of
points “on or below” the graph of f: D R! . Then f is concave function ! A is a
convex set.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
19
:"
="
Proof: We have to show that f(x) is a concave function implies A is convex and A is
convex implies f(x) is concave function.
First part: f(x) is a concave function ! A is a convex set. Let take any two points: (x1,y1)
and (x2,y2) in set A. Take convex combination of the two point: (xt, yt) such that:
xt = tx1 + (1-t) x2 and yt = ty1 + (1-t) y2 .
We need to prove that (xt, yt) is also in set A.
f(x) is a concave function ! by the definition of concave function:
f(xt) ! tf(x1) + (1-t)f(x2)
By the definition of set A, we have f(x1)! y1 and f(x2) !y2
! tf(x1) + (1-t)f(x2) ! ty1 + (1-t)y2 = yt
! f(xt) ! yt ! (xt , yt) ! A ! A is a convex set.
Second part: Proving A is convex implies f is a concave set.
Consider two any points on the graph of f(x): (x1,f(x1)) and (x2,f(x2)).
A is convex ! (xt, yt) ! A, in which :
xt = tx1 + (1-t)x2 and yt = tf(x1) + (1-t)f(x2).
We need to prove that: f(xt) ! tf(x1) + (1-t)f(x2).
According to the definition of set A:
f(xt) ! yt = t f(x1) + (1-t)f(x2). ! f(x) is a concave function.
Linear Combinations of Concave Functions
Consider a list of functions fi : X ! R for i =1,..., n, and a list of numbers *1,..., *n. The
function f +1
ni iif!
=" is called a linear combination of f1,..., fn. If each of the weights *i " 0,
then f is a nonnegative linear combination of f1,...,fn.
The next proposition establishes that any nonnegative linear combination of concave
functions is also a concave function.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
20
Theorem 2: Suppose f1,..., fn are concave functions. Then for any *1,...,*n, for which each *i "
0, f +1
ni iif!
=" is also a concave function. If, in addition, at least one fj is strictly concave and
*j > 0, then f is strictly concave.
Proof. Consider any x, y " X and ) " (0, 1) . If each fi is concave, we have
fi [)x +(1 % )) y] " ).fi(x) + (1 % )). fi(y) !
Therefore, f()x +(1% ))y) +1
ni iif!
=" ()x +(1% ))y) " 1
nii
!=" ()fi(x)+(1% ))fi(y))
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!= )1
ni iif!
=" (x)+(1% ))1
ni iif!
=" (y)!!!- )f(x)+(1% ))f(y).
This establishes that f is concave. If some fi is strictly concave and *i > 0, then the inequality
is strict. QED.
Since a constant function is concave, Theorem 2 implies
• If f is concave, then any aff"ne transformation *f ++ with * " 0!is also concave.
• If f is strictly concave, then any aff"ne transfn *f ++ with *> 0!is also strictly!concave.
2.2.5. Quasiconcavity
Quasiconcave functions: f : X ! R is quasi-concave if for any x, z " X, we have
f()z +(1% ))x) " min{f (x),f(z)} for all ) " (0, 1).
Strictly Quasiconcave fn : f is strictly quasi-concave if for any x, z " X and x =z, we have
f()z +(1% ))x)> min{f (x),f(z)} for all ) " (0, 1). !
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!#$%&"'()*(%+,!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!-).!/$%&"'()*(%+,!
- A Strictly Concave Function forbid the convex combination of two points in the same
level set also lies in that level set
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21
Theorem 3:"Concavity Implies Quasiconcavity (Concavity ! Quasiconcavity)
A concave function is always quasiconcave. A strictly concave function is always strictly quasiconcave.
Proof. The theorem follows immediately from the observation that if f is quasi-concave, then for all x, z " X, we have
f()z +(1% ))x)" )f(z)+(1% ))f(x) " min{f (x),f(z)} for all ) " (0, 1).
If f is strictly concave, we have for all x, y " X and x =y,
f()z +(1% ))x)>)f(z)+(1% ))f(x) " min{f (x),f(z)} for all ) " (0, 1). QED.
! If X * R,"then f : X ! R"is quasi-concave if and only if it is either monotonic or
f/rst non*decreasing and"then"non*increasing. Why?
Geometrically: When f(x) is an increasing function, it will be quasiconcave whenever
the level set relative to any convex combination
of two points, L(xt) is always on or above
the lowest of the level sets L(x1) and L(x2).
When f(x) is a decreasing function, it will be quasiconcave whenever the level set relative to
any convex combination of two points, L(xt) is
always on or below the highest of the level sets
L(x1), L(x2).
Theorem: Quasiconcave and the Superior Sets
f: D!R is a quasiconcave function if and only if S(x) i.e. the superior set relative to point
x: {x| x D! , f(x) ! y} is a convex set for all x D! .
Proof: We have to prove the both necessary and sufficient terms.
Sufficiency: Prove that f(x) is a quasiconcave function ! S(x) is a convex set.
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Consider any two points x1 and x2 in set S(x). We need to prove that xt made by the convex
combination of the two vectors x1 and x2 is also in S(x) ! f(xt) ! f(x).
! according to the definition of S(x), we have: f(x1) ! f(x) and f(x2) ! f(x).
By the definition of a quasiconcave function: f(xt) ! min[f(x1),f(x2)] ! f(x)
! f(xt) ! f(x).
Necessity: Prove that if S(x) is a convex set ! f(x) is a quasiconcave function.
We need to prove that f(xt) ! min[f(x1),f(x2)] under the condition that S(x) is convex. Consider
any two points x1 and x2 in S(x). Without loss of generality, assume we have: f(x1) ! f(x2).
According to the definition of S(x), we have: f(x1) ! f(x2) ! f(x).
Because f(x1) ! f(x2) ! S(x2) ! S(x1) ! x1 and x2 are both in S(x2)
! xt = tx1 + (1-t)x2 (for all t ]1,0[! ) is also in S(x2) ! f(xt) ! f(x1) ! f(x2).
Because t ]1,0[! ! f(xt) ! min[f(x1), f(x2)] ! f(x) is a quasiconcave function.
Our next theorem states that any monotone nondecreasing transformation of a quasi-concave
function is quasi-concave.
Theorem 4: Suppose f : X ! R is quasi-concave and . : f(X) ! R is nondecreasing. Then . /
f : X ! R is quasi-concave. If f is strictly quasi-concave and / is strictly increasing, then . / f
is strictly quasi-concave.
Proof. Consider any x, y " X. If f is quasi-concave, then f ()x +(1 % )) y) " min {f(x),f(y)} .
Therefore, / nondecreasing implies
.(f ()x +(1 % )) y)) " .(min {f(x),f(y)})= min {.(f(x)),.(f(y))} .
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If f is strictly quasi-concave, then for x =y, we have f()x +(1 % )) y) > min {f(x),f(y)}.
Therefore, if / is strictly increasing, we have
.(f ()x +(1 % )) y)) >.(min {f(x),f(y)})= min {.(f(x)),.(f(y))} .
!Recall that for any x - X, P (x) + {z - X : f(z) * f(x)} is called the better set of x.
Theorem 5: A function f : X ! R is quasi-concave if and only if P (x) is a convex set for each x " X.
Proof. (only if) Suppose f is quasi-concave. Choose an arbitrary x0 " X. To show that P (x
0) is
convex, consider any x, y " P (x0). Then, f(x), f(y) " f(x
0) and the quasi-concavity of f imply
that f()x +(1 % )) y) " min {f(x),f(y)} " f(x 0)
which implies that )x +(1 % )) y - P (x0) for any ) - (0, 1) .
(if) Suppose P (x0) is convex each x
0 " X. Now consider any x, y " X. WLOG, suppose that
f(x) 0 f(y). Then letting x = x0, we have that x, y " P (x0) and therefore )y +(1 % )) x " P (x
0)
for any ) " (0, 1) . It then follows from the definition of P (x0) that
f()y +(1 % )) x) " f(x)= min {f(x),f(y)} . QED. !
Theorem 5 is illustrated below for an increasing function f : R+2 ! R. Notice that all convex
combinations of vectors in P (x1) are also elements of P (x
1). Also notice that if f is strictly
concave, then the level set can contain no straight line segments.
Corollary 1: Suppose f : X ! R attains a maximum on X.
(a) If f is quasi-concave, then the set of maximizers is convex.
(b)If f is strictly quasi-concave, then the maximizer of f is unique.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
24
Proof. (a) Let x0 be a maximizer of f. Then f(x) 0 f(x
0) for all x " X implies that P (x
0) is the
set of maximizers of f. If f is quasi-concave, then Theorem 5 implies that P (x0) is convex.
(b)If f is strictly quasi-concave, suppose x, y are both maximizers of f. Then x 1 y implies
1 1( )2 2
f x y+ > min {f(x),f(y)} = f(x) which implies that x is not a maximizer of f.
Example: Let I > 0 be the income of some household and let p =(p1,..., pn) " nR+
denote
the vector of prices of the n goods. Then
B(p, I) - {x " nR+
: px 0 I }
defines its budget set – the set of all possible nonnegative bundles of goods it may
purchase within its budget. You can verify that y, z " B(p, I) implies )y +(1 % )) z "
B(p, I) for all ) " [0, 1]. Therefore, B(p, I) is a convex set.
Now suppose that the household preferences are given by some a utility function u : nR+
"
R. Then Corollary 5 implies that if u is strictly
quasi-concave, there is a unique bundle
x " B(p, I) that maximizes u : B(p, I) ! R.
The example is illustrated here.
2.2.6. Convex and Quasiconvex Functions
If we reverse the inequality sign in the definitions of concave and quasi-concave functions we obtain convex and quasi-convex functions. Convex Functions: f: D R! is a convex function iff for all x1 and x2 in D,
f(xt) ! tf(x1) + (1-t)f(x2), for all t! [0,1].
Convexity ! the region above the graph – set (x,y) or (D,R) is a convex set.
:"
="
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
25
Strictly Convex Function: f: D R! is a strictly convex function iff for all x1 and x2 in D,
f(xt) < tf(x1) + (1- t) f(x2), for all t! (0,1).
Theorem 6: (a) f is a (strictly) convex function if and only if %f is a (strictly) concave fnctn.
(b) f is a (strictly) quasi-convex function if and only if %f is a (strictly) quasi-concave fnction.
!
Theorem 6 allows us to easily translate all of our propositions for concave and quasi-concave
functions to the analogues for convex and quasi-convex functions, which are provided here
for easy reference. !
• Suppose f1,...,fn are convex functions. Then for any *1,...,*n, for which each *i " 0, then
f-(*ifi is also a convex function. If, in addition, at least one fj is strictly convex and *j > 0,
then f is strictly convex.
• A linear function is both concave and convex.
• A (strictly) convex function is (strictly) quasi-convex.
• Suppose " : X ! R is quasi-convex and / : "(X) " R is nondecreasing.
Then (. 1 " ): X " R is quasi-convex. If " is strictlyquasi-convex and / is strictly increasing,
then (. 1 "!)is strictly quasi-convex.
• A function " : X " R is quasi-convex if and only if for each x - X, W(x) is convex.
• Suppose " : X " R attains a minimum on X. (a) I0 " is quasi-convex, then the set of
minimizers is convex. (b) If " is strictly quasi-convex, then theminimizer of " is unique.
Theorem. Points On & Above the Graph of a Convex Functn Always Form a Convex Set
Let D nR! be a convex set, R ! R. Let A* = {(x,y) | x ! D, f(x) ! y} be the set of points
“on and above” the graph of f: D!R.
f is a convex function ! A* is a convex set.
Proof: Sufficiency: We need to prove that f is a convex function ! A* is a convex set.
Consider the two points (x1,y1) and (x2,y2) in set A*.
Take the convex combination of the two points, we have: (xt,yt), in which:
xt = t x1 + (1-t) x2 and yt = t f(x1) + (1-t) f(x2)
We need to prove that (xt,yt) is also in set A* ! f(xt) ! yt.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
26
According to the definition of a convex function: f(xt) ! tf(x1) + (1-t)f(x2) = yt
! f(xt) ! yt ! (xt,yt) is in set A* as well.
Necessity: We need to prove that A* is a convex set ! f is a convex function.
Consider the two points on the graph of f (x): (x1,f(x1) and (x2,f(x2)).
A* is a convex set ! f(xt) ! tf(x1) + (1-t)f(x2) ! is a convex function.
Quasiconvex and Strictly Quasiconvex Functions
1. A function f: D!R is Quasiconvex iff for all x1 and x2 in D,
f(xt) ! max[f(x1), f(x2)] for all t ! [0,1].
2. A function f: D ! R is Strictly Quasiconvex iff for all x1 and x2 in D,
f(xt) < max [f(x1), f(x2)] for all t ! (0,1).
Theorem: Quasiconvexity and the Inferior Sets
f: D ! R is a quasiconvex function iff I(x) – Inferior set is a convex set for all x!D.
Proof: Sufficiency: We need to prove that f(x) is a quasiconvex ! I(x) Inferior set is a
convex set.
We need to prove that (xt,yt) is also in the I(x) set.
Consider the two points in I(x) set (x1,y1) and (x2,y2). Without loss of generality, we
assume that f(x1) ! f(x2) ! I(x1) ! I(x2) and f(x1)-f(x2) ! 0
f(x) is quasiconvex ! f(xt) ! max[f(x1),f(x2)]
Because f(x1) ! f(x2) ! f(x2) = max[f(x1), f(x2)]
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! f(xt) ! f(x2) ! (xt,yt) ! I(x2) ! (xt,yt) ! A*.
Necessarity: We need to prove that I(x) is a convex set ! f(x) is a quasiconvex function.
We need to prove that f(xt) ! max[f(x1),f(x2)].
Consider the two points in I(x): x1(x11, x1
2) and x2(x22,x2
2)
I(x) is a convex set ! the convex combination of the two points is also in I(x).
Without loss of generality, we assume that f(x1) ! f(x2) !max[f(x1), f(x2)] = f(x2).
f(xt) = t f(x1) + (1-t) f(x2) = f(x2) + t [f(x1)-f(x2)] ! f(x2) due to f(x1) ! f(x2).
! f(xt) ! max[f(x1), f(x2)] ! f(x) is a quasiconvex function.
Theorem: Concave/Convex and Quasiconcave/Quasiconvex Funtions
1. f(x) is (strictly) concave function iff –f(x) is (strictly) convex function.
2. f(x) is (strictly) quasiconcave function iff –f(x) is (strictly) quasiconvex function.
Summaries:
1. f concave ! convex sets beneath the graph.
2. f convex ! convex sets above the graph.
3. f quasiconcave ! Superior sets are convex sets.
4. f quansiconvex ! Inferior sets are convex sets.
5. f concave ! f quasiconcave.
6. f convex ! f quasiconvex.
7. f (strictly) concave ! -f (strictly) convex.
8. f (strictly) quasiconcave ! -f (strictly) quasiconvex.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
28
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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
29
2.3. A little Topology
Topology is a study of fundamental properties of sets and mappings.
The distance between the points ),( 12
11
1 xxx and ),( 22
21
2 xxx defined as
d(x1, x2) = |x1 – x2| = 222
12
221
11 )()( xxxx !+!
Open Ball: The open ball with the center nRx !0 and radius e > 0 (a real number) is the
subset of points in nR : }),(|{)( 00 exxdRxxB ne <!" .
Closed Ball: the closed ball with center nRx !0 and the radius e > 0 (a real number) is the
subset of points in nR : }),(|{)( 00 exxdRxxB ne !"# .
Open Sets: S nR! is an Open Set if for all x ! S, there exist some e > 0 such that the open
ball )(xBe S! . It only contains interior points.
Interior Point: A point is a is an interior point of set A if there is an open ball or closed ball
about a which contains only points of set A.
Boundary Point: A point a is a boundary point of set A if for every Be(x) [regardless of how
small e>0 may be] contains both types of points which are in the set and which not.
The Theorem on Open Sets in Rn:
1. The Empty set is an Open Set
2. The entire space R n is an open set
3. The union of open sets is an Open set
4. The intersection of any finite number of open sets is an Open set.
Theorem: Every Open set is a Collection of Open Balls.
S is an open set. For every x ! S, choose some xe > 0 such that SxBxe
!)( .
)(xBSxeSx!
= !
Closed Sets: nRS ! is a closed set if and only if its complement S# is an Open set.
Theorem on Closed sets:
1. The empty set is a closed set
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
30
2. The entire space is a closed set
3. The union of closed set is a close set
4. The intersection of a finite number of closed sets is a closed set.
Theorem: Closed sets in R and the Unions of close Intervals.
Let S is any closed set in R. Then: S = ( )),[],( +!"#!$
biaiIi!
Proof:
If RS ! is closed ! S . is open. By the definition of an open set, for each x in S ., we
have: ' ( )S B x!=! .
We can rewrite as: ( , )x xx cS
S x x! !"
# = $ +! .
Let i=x, I= S ., xi xa !"= , xi xb !+= , we have ai<bi
( , )i ii I
S a b!
" = !
! ),( iiIi
bacS!
= !
Applying the De Morgan’s law, we have: ),( iiIi
bacS!
= !
),[],(),( +!"!= iiii babac !
! ( )),[],( +!"!=#
iiIi
baS !" .
Theorem: Close sets in +R and the Union of Closed Intervals:
Let S is any closed set in +R . Then: S = ( )),[],0[ +!"#
biaiIi!
Bounded Sets in nR : A subset S in nR is called bounded if and only if it is entirely
contained within some ball (an open or a closed ball). That is, S is bounded if there exists e>0
such that S )(xBe! for some x nR! .
Consider a subset S in R space:
A lower bound: A real number l is called a lower bound for S if l x! for all Sx! .
An Upper bound: A real number l is called an upper bound for S if l x! for all Sx!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
31
*+0/&5".15"60%15&5"
A subset S in R has many lower bounds and upper bounds.
The greatest lower bound (g.l.b): biggest number among those lower bounds for S.
The least upper bound (l.u.b): the smallest number among those upper bounds for S.
A close set S in R contains its upper and lower bounds. An open set S in R does not
contain its upper and lower bounds.
Theorem: Upper and Lower Bounds in Subsets of Real Numbers
1. Let S ! R is a bounded open set and let a be the g.l.b of S and b be the l.u.b of S.
Then a S! and b S! .
2. Let S is a bounded closed set in R. Let a is the g.l.b of S and b is the l.u.b of S.
Then a! S and b ! S.
Compact Sets in nR ( Heine-Borel): A set nRS ! is compact if and only
if S is closed and bounded.
nR is closed but is not bounded ! nR is not compact.
2.3.1. Continuity
- A Continuous mapping or a continuous function
- In most economic application, we will either want to assume that the functions we are
dealing with are continuous or we want to discover whether they are continuous when we are
unwilling to assume it.
- A continuous function: f: R ! R is continuous function at a point xo if for all ! > 0, there
exists ! > 0 such that d(x, xo)< ! implies d[(f(x) - f(xo)] < ! . ( ( ) ( ( ))o of B x B f x! "#
xo xo+
f(!$)
f(xo+ )
f(!$ +)
xo+ xo
f(xo) f(xo)+
f(xo+ )
f(xo+ ) (f(xo), f(xo)+ )
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
32
Continuity: Let D be a set, R be another set, and let f: D ! R. The function f is continuous at
the point xo ! D if and only if, for all ! > 0, there exists a ! > 0 such that:
( ( )) ( ( ))o of B x B f x! "# . If f is continous at all xo ! D, it is called a continuous function.
A function is continuous at a point xo if for all ! > 0, there exists ! > 0 such that any point
less than a distance ! away from xo is mapped by f into some point in the range which is less
than a distance ! away from f(xo). f(xo) is the image of xo.
Every point in o( )B x! is mapped by f into some point no father than ! from f(xo).
Basically, a function is continuous if a “small movement” in domain does not cause a “big
jump” in the range.
It is not true that a continuous function always maps an open set in the domain set into an
open set in the range, or that closed set is mapped into closed sets. Ex: y = a. (the domain is
an open set but the range (image) set may not be an open set)
Theorem: Continuity and the Inverse Image of Open Sets
Let f: D ! R be a mapping and let 1!f : R ! D be its inverse mapping from R to D. Let
RT ! be an open set in the range of f. Then f is continuous if and only if the inverse
image DTf !" )(1 is an open set in the domain.
Proof: Necessity: Let f is a continuous function and T be an open set in the range. We have
to prove that DTf !" )(1 is an open set.
T = {f(x)} is an open set ! ! some 0>! such that TxfB !))((" . The function f is a
continuous function ! according to the definition of a continous function, there exist 0>!
such that ))(())(( xfBxBf !" # ))](([)( 1 xfBfxB !"#$% ! DTf !" )(1 is an
open set because DTf !" )(1 is a set of )(xB! .
Sufficiency: Let DxfBf !" ))(((1 # is an open set. We need to prove that f(x) is a
continuous function.
If DxfBf !" ))(((1 # is an open set !" some 0>! such that DxB !)(" is
contained by ))((1 xBf !" .! !)(xB" ))((1 xBf !
" )())(( xfBxBf !" #$
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
33
for all ))}((({ 1 xfBfx !"# .
! f(x) is a continuous function for all ))}((({ 1 xfBfx !"# .
Theorem: Continuity and the Inverse Image of Closed Sets
Let f: D!R is a continuous mapping and 1!f is inverse mapping from R to D. T R! is
closed set in the range of f. Then f is continuous function if and only if DTf !" )(1 is a
closed set in the domain of f.
Proof: Let an image set T of f(x) is a closed set. We need to prove that DTf !" )(1 is a
closed set equivalent to f being a continuous function.
If T is a closed set ! cT is an open set . According to the preceding theorem, f(x) is a
continuous function if and only if )(1 cTf ! is an open set or )(1 Tcf ! is an open set !
)(1 Tf ! is a closed set.
3.1. Mathematical operation:
We are familiar with the operations of ordinary addition, multiplication of numbers. We have further studied the notions of union and intersection of sets. These are denoted by
a + b= c, a.b = d, A U B= C, A ! B= D.
In each of these situations an element (c, d, C or D) is assigned to the ordered pair of original elements as an outcome of operation considered. By an ordered pair we mean a pair of objects, say (a, b), which are considered in the order first a, then b.
Let us now consider a set S = {a, b, c, d, .. . . . } and we define an operation, given by the symbol *, (or o, or . , or any other symbol) which operating on an ordered pair of two elements of the set S gives the next element of the set as its outcome. For example, a * b= c, b * c= d, and so on. The rules or laws of such operation may be given in the form of postulates, theorems or any other definition. In general, an operation or composition is a set of rules given in the form of postulates, theorems or any definition which assigns to each ordered sub-set of a set some uniquely detennined element of that set.
Thus a mathematical system involves
(i) a non-empty set of elements ; (ii) one or more operations which associate with each ordered sub-set of the set, an
element also of the set; (iii) a set of operations or compositions, called postulates or axioms, which the
elements and the operations satisfy.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
34
3.2. Binary composition. Operations are classified according to the number of elements of the set thatare involved in it, as unary, binary, ternary, ....... , n-ary, and so on.
Extraction of square root is a unary operation in arithmetic; for, only one element of the number system is operated upon.
An operation o (or *, or " or any other symbol) which when applied to two elements of a set S gives a unique element, also of the same set, is called a binary operation. Thus, if a!S and b!S and a unique element, c, also of the same set, exists such that c= a o b then the operation o is a binary operation in S. Given a set A, any mapping of AxA into A is called a binary composition on A.
Let o be a binary composition on A. By definition, mapping of A!A into A, i.e., (a, b) ! A!A, o(a, b)!A.
Addition is a binary operation on the set of even numbers, since the Sum of two even natural numbers is an natural number; but is not a binary operation on the set of odd natural numbers; since the sum of two odd natural numbers is an even natural number. Subtraction is a binary operation the set of all integers, positive and negative, but not, if be limited' to only positive integers.
Neither addition nor multiplicationis a binary operation on the set S= {0, 1, 2, 3, 4, 5}, since 3 + 4= 7 !S and 2 x3= 6 !S.
Let M2 be the set of all 2x2 matrices whose elements are rational numbers. For any two
matrices A, B! M2 , where A= 11 12
21 22
a aa a
! "# $% &
, B= 11 12
21 22
b bb b! "# $% &
, we have the usual sum A + B=
11 11 12 12
21 21 22 22
a b a ba b a b
+ +! "# $+ +% &
, which also belongs to M2. So the usual matrix addition is a binary
composition on M2.
If o be a binary operation defined over a set S, then two elemlents a and b of S, a 0 b ! S; this is said to be the closure property of the binary operation and the set S is said to be closed with respect to the operation o. A non-empty set together with one or more than one operations is called algebraic structure. For examples, (N, +), (Z, +, -), (R, +, '). are algebraic structures.
3.3. Laws of binary composition
Associative law: A binary operation o on the elements of the set S is said to be associative, if and only if, for any three elements a, b, c!S,
a o ( b o c ) = (a 0 b ) 0 c. For example, multiplication is an associative composition in the set of integers,
since (3 x 4) x 5= 3 x (4 x 5). Let the binary operation o be defined on R by a 0 b= a + 2b; a, b!R. Now ( a 0 b ) 0 c = (a + 2b) 0 c = a + 2b + 2c; a, b, c!R.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
35
But a o (boc)= a o (b +2c)= a + 2(b+ 2c) = a +2b+4c. Therefore the binary operation o as defined here is not associative.
Commutative law : A binary operation o on the elements of the set S is commutative, if and only if, for any two ,a b!S, a o b= b o a. For example, union and intersection of sets are commutative compositions in the sets of all sub-sets of a set, since A U B= B U A and A!B= B! A. Let o be the binary operation on R, that is, 0 : R R R! "
defined by o : ( , )x y x y! " ; x, y!R. Then o (6,2) = 6 -2 = 4 but o (2, 6)= 2! 6=! 4. Hence subtraction of real numbers is
not a commutative operation. .
Distributive law : Two binary operations o and *, say, on elements of the set S are said to be distributive, if and only if, for every element a, b and c !S,
a o (b * c)= (a o b)*(a o c) [left] (b * c) o a= (b o a)*(c o a) [right]
For example, multiplication is left as well as right distributive over addition in the set of real numbers. We say that the multiplication composition ( • ) distributes addition composition (+) for the set of real numbers. But the addition composition does not distribute the multiplication position in R, since a, b, c!R,
a • (b + c)= a • b + a • c but a + (b • c)! (a + b) • (a + c). Identity element. A set S is said to possess an identity element with respect the binary operation o, if there exists an element i! S with property that
I o x = x o I = x, for every x! S.
Here (S, o) is an algebraic structure with identity element. If i o x= x, then i is called the left identity element for the operation o and if x 0 i= x, then i is called the right identity element for o. If the set contains an element which is both a right and a left identity element for an operation, then we call it an identity element for that operation.
Consider any element x of the set Q of rational numbers with respect to the binary operation addition. Obviously, zero is the identity element, since 0 + x= x + 0= x, for every x!Q.
1 is the identity element of Q for the binary operation mul-tiplication, since 1.x= x.1= x, for every x!Q.
It is easily seen that for the set N of natural numbers2 there is no identity clement for addition; but 1 is an identity element with respect to multiplication.
There is no need for an identity element to exist as we see that there is no such element for the ordinary operation of sub-traction in R, because this would have to be a number i such that
i x x i x! = ! = for all x, in the real number set R, which is impossible.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
36
3.4. GROUP THEORY:
Theory of groups is one of the most important fundamental concepts of Modem Algebra. Group is, an algebraic structure with only binary opetation.
3.4.1. Groupoid, semi-group and monoid:
Consider the algebraic structure (S, 0) in which S is a non-empty set on which the binary composition 0 is defined. Thus S is closed with the operation o. Thus, if a, b ! S, then a 0 b ! S. Such a system consisting of a non-empty set S and a binary composition defined in S is called a groupoid.
If N be the set of natural numbers, then (N, +) is a groupoid because the set N is closed under addition. But the set of integers is not a groupoid under addition composition. The set S= { -2, -1, 0, 1, 2} is not a groupoid with respect to addition, since 2 + 2 = 4 !S, that is, the set S is not closed with addition composition.
If the binary composition 0 be commutative in S, then groupoid is said to be a commutative groupoid.
If a 0 x= x for all x!S, then a!S is called the left identily element of the groupoid. Similarly a !S is a right identity element of the groupoid, if x 0 a= x for all x!S .
The groupoid (Z, -) has no left identity element but zero is a right identity element of it. If a groupoid possesses a right as well as a left identity, then they are equal.
A system consisting of a non-empty set S and an associative binary composition in S is called a semi-group.
Consider, for example, a, b, c !Z, the set of all integers. Then we have (a + b) + c = a + (b + c) and (a . b). c= a. (b. c) . Thus Z being closed with the binary operations of addition and multiplication the systems (Z, +), and (Z, .) are semi-groups as those two compositions are associative in Z. But the system (Z, -) is not a semi-group, since subtraction does not satisfy the associative law.
A system consisting of a non-empty set S and an associative binary composition with identity is called a monoid. Thus monoid is an associative groupoid with an identity element. If Z be the set of all integers, then (Z, +) is a monoid with identity element 0 and (Z, .) is a monoid with 1 as the identity element.
3.4.2. Group. A non-empty set S of elements a, b, c, ....... forms a group with respect to the binary operation *, if the following properties (axioms) hold:
(i). For every pair a and b!S, a * b is in S (closure law).
(ii). For any three elements a,b, c!S, a* (b * c)= (a * b) * c holds (associative law).
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
37
(iii) ! in S an element i, called a left identity, so that i * a= a, for every a!S.
iv) For each a in S, the equation x * a = i has a solution x !S.
This solution x is called a left inverse of a.
If, in addition to these postulates for a group; a * b= b * a (commutative law)
for all a and b in the group, the group is called a commutative or abelian group; otherwise it is called a non-abelian group.
For example, consider the set of integers (positive, negative and Zero) Z= {. ........ , -2, -1, 0, 1,2, ...... } on which the binary operation addition is applied.
For any a,b,c !Z, we have
i. a + b! Z (closure) ii. (a + b) + c= a + (b + c) (associativity), iii. 0 + a = a (0 is the left identity element) iv. (-a )+a = 0 (left inverse of a is a Z! " )
Hence the set of integers forms a group with respect to addition, since it satisfies all the group axioms. Moreover, a+ b = b +a, which shows that it is an abelian group. Thus the set Z of all integers (positive, negative zero) with additive binary operation is an abelian group.
Consider, again, the set Q+ of non-zero rational numbers on which the binary operation multiplication is applied. For any a,b,c!Q+, we have
i. a ! b!Q+ (.closure), ii. (a x b) x c = a x (b x c) (assoclativity), iii. 1 x a = a (1 is the left identity element), iv. 1 1aa! = (1 a is the left inverse of a).
Moreover a x b= b x a.
Hence the set Q+ of non-zero rational numbers is an abelian group with respect to multiplication. If Q be the set of rational numbers, then (Q, +) is an abelian group but (Q,.) is not, since 0 has no left inverse in Q.
A group with addition binary operation is known as additive group and that with multiplication binary operation is known as multiplicative group. A group (or the binary operation * is often written as (G, *) or < G, * >, where G refers to the set forming the group. If the underlying set in it group consists of a finite number of elements, then it is called a finite group; the number of elements in the underlying set determines its order. An infinite group consists o( infinite number of elements in'its underlying set. It is said to be of order zero or of infinite order.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
38
Generally the order o( the group G is denoted by a(G). Sometimes only the set symbol G is used to denote a group in context of a given operation.
3.4.3. Rings: A set of elements a, b, c, ...... forms a ring R with respect to the binary compositions --addition and multiplication, defined on R, if
i. the set forms a commutative group with respect to addition, ii. the set is closed with respect to multiplication, iii. the associative law a.(b.c) = (a.b).c holds for multiplication, and iv. the distributive laws (right as well as left)
a.(b + c) = ab + a.c and (b + c).a ... b.a + c.a hold.
This algebraic structure is often written as (R, + , • ) or < R, + , • > where R is the non-empty set of the ring. When the operations + and 2 are defined, the set R defines a ring.
A ring R in which multiplication is commutative is commutative ring. For a commutative ring R, a .b= b . a , for all a, b !R. The set of all integers with two binary operations, and multiplication forms a ring.
Cor. Since the ring is a group with respect to addition, all the group properties hold for addition ..
The unique identity of the additive group, denoted by ordinary symbol for the number zero, is called the zero element of the ring and is denoted by O. Thus a + 0 = 0 + a = a, for every a in the ring.
The unique additive inverse of an element a is denoted by (- a), such that a + (- a)= 0= (- a) + a.
For the additive operation, left as we.ll as right cancel" laws hold good in a ring; that is, for an a, b, e !R,
a + b = a + c implies b = c and b + a = c + a implies b = c.
There will be a unique solution x= b + (- a), .written as (b - a), of the equation a + x = b. 3.4.4. Fields. A ring containing at least two elements is called an integral domain, if it be commutative, has a unity element and be without zero divisor. The ring of integers (Z, +, .) is an integral domain. The ring of even integers does not contain the unity element and hence it is not an integral domain, although it is without zero divisor. The set of natural numbers N does not form an integtal domain. Note. Some authors do not demand a unity element for an integral domain. Any ring containing at least two elements is called a field,if it be commutative, has a unity element and be such that all non-zero elements ,have multiplicative inverse. Thus an integral domain is a field, if every element a (! 0) has a multiplicative inverse a -1, such that a -1. a= unity element.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
39
The ring of rational numbers is a field, since it is a commutative ring with unity and each non-zero element has a multiplicative invese. The ring of all integers is not a field, as all non-zero elements in the ring of integers do not have multiplicative inverses. Thus a set F having at least two elements which algebraic structure (F, + , .) with the binary operations of addition and multiplication is called a field if the following be satisfied :
i. a + b! F, whenever a, b!F. ii. a + (b + c)= (a + b)+ c ; a, b, c !F. iii. There exists an element 0 !F such that a + 0= 0 + a = a, for all a!F. iv. a + b = b + a. v. a + x = b is solvable, for all a, b!F. vi. a . b!F, whenever a, b!F. vii. a. (b. c)= (a. b) . c, for all a, b, c !F. viii. There exists an element 1 !F such that a . 1= 1 . a= a, for all a !F. ix. a . b = b. a, for all a, b !F. x. a . x = b is solvable, for all a, b! F, where a! 0. xi. a. (b + c) = a. b + a. c . for all a, b, c !F. xii. (b + c) . a= b . a + c . a It follows from (x) that, for every a!F, a! 0, there exists an element a-I !F such that a-I . a= a . a-I = 1.
4.1. Vector Spaces: Let S be any non-empty set. If a * b !S for all a, b!S and a * b is unique, then * is
said to be an internal composition [e.g. addition of vectors] in the set S. This is a map
S S S! " .
Let vector V and a field F be two non-empty sets. If a o * !V for all a!F and *!V
and a o * is unique, the o is said to be an external composition [e.g. multiplication of vectors
by scalars] in V over F. This is a map F V V! " .
Definition: Let (F, +, . ) be a field. Then a non-empty set V is called a vector space over
the field F, if in V there be defined an internal composition * and an external composition o
over F such that for all a, b!F for all ,, - !V,
(i) (V, *) is an abelian group.
(ii) a o [* o +] =[ a o * ] * [a o + ],
(iii) [a + b] o * =[a o * ] * [b o * ],
(iv) [a . b] o * = a o [b o * ]
(v) 1 o * = * , the unity scalar 1!F.
1 is the multiplicative identity of the field F.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
40
The vector space V over the field F is denoted by V(F) or simply V.
4.2. Linear independence and dependence of vectors: Let V be a vector space over a field F.
Any vector , is said to be a linear combination of the vectors *1, *2, ....., *n !V if
* = c1. *1 + c2. *2 + ...... + cn. *n
where the scalars c1 , c2 , ......, cn !F.
As V is a vector space, so by vector addition and scalar multiplication in V, we have ,!V.
For a fixed set of vectors *1, *2, ....., *n different linear combinations are obtained by choosing
different sets of scalars.
Thus (2, 3, 4) = 3(1, 1, 1) + (- 1, 0, 1)
and (2, 3, 4) = (-2, 1, 2) + 12
(2, 0, 3) +3(1, 2 1,3 6
)
Let V be a vector space over a field F. A finite sub-set { *1, *2, ....., *n } of vectors of V is said
to be linearly dependent, if there exist scalars c1 , c2 , ......, cn !F, not all of them zero, such
that c1. *1 + c2. *2 + ...... + cn. *n = 0 .
Let V be a vector space over a field F. A finite sub-set { *1, *2, ....., *n } of vectors of V is said
to be linearly independent, if every relation of the form
c1. *1 + c2. *2 + ...... + cn. *n = 0 , ci !F, i = 1 (1) n
implies that ci=0, for each i = 1 (1) n
An infinite set of vectors of the vector space V is said to be linearly independent, if its every
finite sub-set be linearly independent, otherwise it is linearly dependent.
A null set is assumed to be linearly independent.
The results below follow directly from the definitions.
(a) If two vectors be linearly dependent, then one of them is a scalar multiple of the other.
(b) If the vectors *1, *2, ....., *n be linearly dependent, then one of them can be expressed
as a linear combination of the others.
(c) A system consisting of a single non-zero vector is always linearly independent .
(d) Every super-set of a linearly dependent set of vectors is linearly dependent.
(e) Any sub-set of linearly independent set of vectors is linearly independent.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
41
(f) A set of vectors which contains the zero vector is necessarily linearly dependent.
4.3. Linear span. Let V be a vector space over the field F and S be any non-empty sub-set
of V. Then the linear span of S is defined as the set of all linear combinations of finite sets of
elements of S. It is denoted by L(S).
Thus we have
L(S)={ c1. *1 + c2. *2 + ...... + cn. *n : *i !S; ci !F, i=1, 2, …, n}.
L(S) is said to be generated or spanned by the set S and S is said to be the set of generators of
L(S).For example, the set L(S)= 1
2
00!
!" #$ %& &' () *& &+ ,- .
of all 2!2 diagonal matrices is the linear span of
1 00 0! "# $% &
and 0 00 1! "# $% &
.
In other words, the sub-space spanned by a non-empty sub-set S of a vector space V is the set
of all linear combination of vectors in S. Clearly S!L(S).
The null set generates the set of zero vector alone, that is, { 0 }= L(2).
Note 1. A vector V! " is in the sub-space of V generated by S, if it can be expressed as a
linear combination of a finite number of vectors in S over F.
Note 2. The sub-set containing the single element (1, 0, 0) of the space V3 in F generates the
sub-space which is the totality of the elements of the form (a, 0, 0). The sub-set {(1, 0, 0), (0,
1, 0) } of V3 in F generates sub-space which is the totality of elements of the form (a, b, 0).
4.4. Basis and dimension of a vector space.
Let V be a vector space over the field F and S be a sub-set of V(F) such that
(i) S is a set of linearly independent vectors in and
(ii) L(S) = ! "V, that is, each vector , in V is a linear combination of the finite
number of elements of S, (S generates V), then S called the basis set or simply
basis of V.
A zero vector cannot be an element of a basis set, such a set is linearly dependent;"
Consider the set B= {(1, 0, 0), (0, 1, 0), (0, 0, 1)} in V3 over the real numbers. This set is
linearly independent. Also B spans V3 because any vector (a1, a2, a3) of V3 can be written as a
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
42
linear combination of the vectors of B, viz. (a1, a2, a3) = a1(1, 0, 0)+ a2(0, 1, 0) + a3(0, 0, 1).
Hence B is a basis for V3 over the real numbers. It is called the standard basis of R3.
Consider, again, the set D = {1, 1, 0), (1, 0, 1), (0, 1, 1) } in.V3, over the real numbers. The
relation a(1, 1, 0) + b(l, 0, 1) + c(0, 1, 1) = 0 = (0, 0, 0) implies a = b = c = 0, that is, D is
linearly independent.
Furthermore any vector (a1, a2, a3) of V3 can be written as a linear combination of the vectors
of D as (a1, a2, a3) = 12
(a1 + a2 - a3) (1, 1, 0)
+ 12
(a1 + a3 – a2 ) (1, 0, 1)
+ 12
(a2 + a3 – a1) (0, 1, 1).
Thus D is also a basis of V3 over the real numbers. This shows that the basis for a vector space
need not be unique.
The vector space V is said to be finite dimensional or finitely generated, if there exists a finite
sub-set, S of V such that L(S)=V. Otherwise the vector space is infinite dimensional. The null
vector space which has no basis is of finite dimension which is zero.
The number of elements in any basis set of a finite dimensional vector space V(F) is called the
dimension of the vector space and is denoted by dimV. Vn(F) is n-dimensional, if its basis
contains n elements.
The vectot space R3 is of dimension 3 as B= {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis of R3.
Similarly for R2 it is 2 and for R4· it is 4 and so on. The vector space spanned by the vectors
(1, 0, ..... , 0), (0, 1, 0, …. , 0) .... (0, 0, ......, 0, 1) is denoted by Vn(R) or simply by Vn.
Examples.
Ex. 1. Express (3, 4, 5) as a linear combination of , = (1, 2, 3) , - = (2, 3,4) and 2 = (4,3,2) in
the vector space V, of real numbers.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
43
Let (3,4, 5) = a(1, 2, 3) + b(2, 3, 4) + c(4, 3, 2) for a. b, c !R
= (a + 2b + 4c ; 2a + 3b + 3c ; 3a+4b + 2c).
By vector treatment, we have
a + 2b +4c = 3,
2a +3b +3c = 4,
3a + 4b + 2c = 5.
Solving these equations by Cramer's rule, we have2 a=1/2, b=3/4, c=1/4
Thus (3, 4, 5)= 1/2 (1, 2, 3) + 3/4 (2, 3, 4) + 1/4 (4, 3, 2).
Ex. 2. Show that the vectors (1, 2, 1), (2,1, 0), (1,-1, 2) form a basis of the vector V3, over the
field of real numbers.
We know that if V(F) be a finite dimensional vector space of dimension n, then any set
of n linearly independent vectors in V forms a basis of V.
Now the set { (1, 0, 0), (0, 1, 0), (0, 0, 1) } forms a basis for V3 over real numbers.
Hence its dimension is 3. If we can show that the set
S = { (1, 2, 1), (2, 1, 0), (1, -1, 2) }
is linearly independent, then S will also form a basis of V3.
We have a1 (1, 2, 1) + a2 (4, 1, 0)+ a3 (1, -1, 2) = (0, 0, 0)
=> (a1 + 2a2 + a3 ; 2a1 + 2a2 – a3 ; a1 + 2a3 )= (0, 0, 0).
Therefore a1 + 2a2 + a3 = 0
2a1 + 2a2 – a3 = 0
a1 – a3 = 0
The coefficient matrix is 1 2 12 1 11 0 2
! "# $%# $# $& '
= A (say).
det A= - 9 . Thus the rank of A is 3. Hence solving these equations, we have the only solution
a1 = a2 = a3 = 0 . Therefore the set S is linearly independent.
Hence it forms a basis for V3 in the field of real numbers.
"
"
1
1
DSE 002:INTRO. MATH ECON HANDOUTS PART 2: ELEMENTARY CALCULUS SUGATA BAG July 2012.------------------------------------------------------------------------------------------------------------ 1. The Derivative of a Function of One Variable.
Let f(x) be a function of one variable x.
Definition 1: The derivative of f evaluated at x is the following limit (if the limitexists):
(1) limtÆ0 [f(x + t) - f(x)]/t ≡ f¢(x) or ≡ df (x)/dx.
Geometrically, f¢(x) may be interpreted as the slope of the line tangent to thegraph of the function through the point x, f(x) if such a tangent line exists.
As t approaches 0, the slopes of the line segments connecting (x,f(x)) to(x+t,f(x+t)) get closer to f¢(x).
Rules For Differentiation
Function Derivative
Rule 1: f(x) = axk, f¢(x) = kaxk-1for k ≠ 0Rule 2: f(x) = ex, f¢(x) = ex
Figure 1: The Derivative of a Function
f(x+t)
f(x)
x x+t
Limiting slope is f¢(x)
2
2
Rule 3: f(x) = lnx, f¢(x) = 1/x for x > 0Rule 4: f(x) = kg(x), f¢(x) = kg¢(x)Constant RuleRule 5: f(x) = g(x) + h(x), f¢(x) = g¢(x) + h¢(x)Addition RuleRule 6: f(x) = g(x)h(x), f¢(x) = g¢(x)h(x) +g(x)h¢(x) Product RuleRule 7: f(x) = g[h(x)], f¢(x) = g¢[h(x)]h¢(x)Chain RuleRule 8: f(x) = g(x)/h(x), h(x) ≠ 0f¢(x) = {g¢(x)h(x)-g(x)h¢(x)}/h(x)2 Quotient Rule
Examples
(1) f(x) ≡ a + bx + cx2 + dx3, f¢(x) = b + 2cx + 3dx2.
(2) f(x) ≡ ekx ≡ eh(x) where h(x) ≡ kx≡ g[h(x)] where g(y) ≡ ey
Therefore
f¢(x) = g¢[h(x)]h¢(x)= eh(x) h¢(x)= ekxk = kekx.
(3) f(x) ≡ [g(x) + h(x)]k, f¢(x) = k[g(x) + h(x)]k-1[g¢(x) + h¢(x)].
Higher Order Derivatives
In example (1) above, f¢(x) is a differentiable function of x for each x. Hence wemay calculate the derivative of f¢(x). [Geometric interpretation?]
Definition 2: The second derivative of f evaluated at x is (assuming that f¢(x)exists in a neighbourhood around x and that the limit below exists):
(2) f¢¢(x) ≡ limtÆ0[f¢(x + t) - f¢(x)]/t ≡ d2f(x)/dx2 (alternative notation).
Examples. Calculate f¢¢(x) for examples (1) and (2) above.
(1) f¢(x) = b + 2xc + 3dx2, f¢¢(x) = 2c + 6dx
(2) f¢(x) = kekx, f¢¢(x) = k{dekx/dx}= k {kekx}= k2ekx.
The third derivative of f evaluated at x is defined as the derivative of the function
f''(x) if it exists. It is denoted by f'''(x) or d3f(x)/dx3.
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2. Maximizing or Minimizing Differentiable Functions of One Variable
Consider the problem of maximizing or minimizing the function of one variable,f(x), over the set x > 0. Assume f¢(x) exists for x > 0.
If we are at a local maximum of f at the point x*, then we must be at the top of ahill and the slope of the function must be zero there; i.e., we must have f¢(x*) = 0.
Similarly, if f attains a local minimum at the point x*, then we must be at thebottom of a valley and the slope must be zero there; i.e., we must have f¢(x*) = 0.
Thus we have:
The First Order Necessary Condition for f to attain a local max or min at the interiorpoint x* is:
(1) f¢(x*) = 0.
Note that condition (1) above is only a necessary condition for a local min ormax; i.e., if f attains a local min or max at x*, then (1) will be satisfied. However,if condition (1) is satisfied, then we do not know whether f attains a local min ora local max at x*. In fact, f might not even attain a local min or max at x*; i.e.,condition (1) could be satisfied at a point of inflection. For example, let f(x) ≡(x - 1)3. Then f¢(x) = 3(x - 1)2 (1 + 0) so that f ¢(1) = 0. However, x* = 1 is not a localmaximizer or minimizer of f.
Second Order Sufficient Conditions for f to attain a strict local maximum at theinterior point x*:
(2) f¢(x*) = 0 and f¢¢(x*) < 0.
The second part of (2) means that the rate of change of f ¢(x) around x* is negative.This means that f¢(x) is positive for x slightly less than x*, f¢(x) = 0 for x = x*, andf¢(x) is negative for x slightly greater than x*.
Example 1:
f(x) ≡ 2 - x2 for - • < x < + •.f(x) = 0 - 2x set= 0
Æ -2x = 0Æ x* = 0
f¢¢(x) = -2 = f¢¢(0).
Therefore f attains a local maximum at x* = 0. Since there are no other pointswhere the slope is 0, x* = 0 is the global maximizer of f.
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5
Slope is + Slope is 0
Slope is -
x
f’(x) = - 2
Second Order Sufficient Conditions for f to attain a strict local minimum at theinterior point x*:
(3) f¢(x*) = 0 and f¢¢(x*) > 0.
If conditions (3) are satisfied, the slope of f is negative to the left of x*, 0 at x*, andpositive immediately to the right of x*.
Example 2:
f(x) ≡ x2 - 1 for – • < x < + •.f¢(x) = 2x+ 0 set= 0 Æ x* = 0f¢¢(x) = 2 Æ f¢¢(x*) = 2 > 0.
Therefore x* = 0 is a local minimizer. It is also the global minimizer of f (samereason as example 1).
Example 3:
f(x) ≡ x3- x2 + 2 – • < x < + •.f¢(x) = 3x2 - 2x set= 0or x(3x - 2) = 0
Æ x*1 = 0, x*2 = 2/3f¢¢(x) = 6x - 2f¢¢(x*1 ) = 6(0) - 2 = - 2 Æ local max at x = 0f¢¢(x*2 ) = 6(2/3) - 2 = 2 Æ local min at x = 2/3
However, the local max is not a global max and the local min is not a global min.(Why?)
Example 4
f(x) = x3; - • < x < + •.f¢(x) = 3x2 set= 0
Æ x* = 0f¢¢(x) = 6x
6
6
Thus f¢¢(0) = 0.
The sufficient conditions for a local max or a local min are not satisfied at x* = 0.Hence we cannot say if f attains a local max or min at x* = 0.
Sufficient Conditions for f to have an inflection point at x*.
(4) f¢(x*) = 0, f¢¢(x*) = 0 and f¢¢¢(x*) ≠ 0.
Thus the f defined in example 4 above has an inflection point at x* = 0.
Second Order Necessary Conditions for f to attain a local maximum at the interiorpoint x*:
(5) f¢(x*)= 0 and f¢¢(x*) ≤ 0.
Second Order Necessary Conditions for f to attain a local minimum at the interiorpoint x*:
(6) f¢(x*) = 0 and f¢¢(x*) ≥ 0.
Conditions (4) - (6) are not as important as conditions (2) and (3).
3. Some Consumer Theory Examples of Optimizing Behavior.
Example 1: We suppose that a consumer's preferences over nonnegative amountsof two goods, x1 ≥ 0, x2 ≥ 0, can be represented by means of the following Cobb-Douglas utility function:
(1) u(x1, x2) ≡ xa1 x1-a2 , 0 < a < 1
where a is a fixed number or parameter which characterizes the consumer'spreferences.
We suppose that the price of good 1 is some number p1 > 0 and the price of thesecond good is p2 > 0. We suppose also that the consumer has the fixed amountof income I > 0 to spend on the two goods. The consumer's budget constraint is
(2) p1x1 + p2x2 = I.
We assume that the consumer attempts to maximize the utility function (1)subject to the budget constraint (2) and the nonnegativity constraints x1 ≥ 0 andx2 ≥ 0. Mathematically, we write this constrained maximization problem asfollows:
(3) maxx1, x2 {xa1 x1-a2 : p1x1 + p2x2 = I; x1 ≥ 0; x2 ≥ 0}.
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How can we solve this rather complex looking problem (3) using the material inthe previous section? We temporarily ignore the nonnegativity restrictions x1 ≥ 0and x2 ≥ 0 and we use the budget constraint (2) to solve for x2 in terms of x1:
(4) x2 = (I - p1x1)/p2
Now substitute (4) into the utility function (1) and we have reduced the twovariable constrained maximization problem (3) into the following single variableutility maximization problem:
(5) maxx1 f(x1) ≡ xa1 [(I - p1x1)/p2]1-a; x1 ≥ 0.
f¢(x1) = axa-11 [(I - p1x1)/p2]1-a + xa
1 (1-a) [(I - p1x1)/p2]1-a-1(-p1/p2) set= 0or axa-1
1 = xa1 (1 - a) [(I - p1x1)/p2]-1(p1/p2)
or ax-11 = (1-a)(I - p1x1)-1p1or a(I - p1x1) = (1 - a)p1x1or aI = ap1x1 + (1 - a)p1x1
= p1x1
(6) or x*1 = aI/p1 > 0.
It can be verified that f¢¢(aI/p1) < 0 so that the x*1 defined by (6) does in factsolve the maximization problem (5). We may substitute (6) into (4) anddetermine the corresponding x*2 which solves (3):
(7) x*2 = (I - p1x*1 )/p2 = (I - p1(aI/p1))/p2= (1 - a)I/p2 > 0.
Since the x*1 and x*2 defined by (6) and (7) are both positive, the nonnegativityrestrictions x1 ≥ 0 and x2 ≥ 0 are satisfied. Hence we can conclude that we havesolved the consumer's constrained utility maximization problem (3). Note thatthe x1 and x2 solutions are functions of p1, p2 I and a; i.e., we have
(8) x*1 = D1(p1, p2, I, a) ≡ aI/p1 andx*2 = D2(p1, p2, I, a) ≡ (1 - a)I/p2.
The solution functions D1 and D2 are the consumer's system of utilitymaximizing demand functions. Note that the functional forms for D1 and D2 arecompletely determined by the functional form for the consumer's utility functionu; see (1).
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This example illustrates how optimizing theory is used in modernmicroeconomic theory. It also illustrates a technique that we shall use quitefrequently in dealing with constrained maximization problems: we shall use theconstraint to solve for one variable in terms of the other variables and thenreduce the constrained maximization problem into an unconstrainedmaximization problem involving one less variable.
Example 2: Notice that the parameter that characterizes the consumer'spreferences, a, occurred in both of the demand functions (8) in the previousexample. How can we determine what a is?
Suppose we collect data on the consumer's purchases of commodity 1 duringperiod t, xt
1 say, on the market price for commodity 1 during period t, pt1 say
and on the consumer's income in period t, It say. Then if equation (6) heldexactly in each period, we would have xt
1 = aIt/pt1 or pt
1 xt1 /It = a for t = 1, 2, . .
., T. It is unlikely that equation (6) will hold exactly for each period. Thus wemight have:
(9) pt1 xt
1 /It = a + et; t = 1, 2, . . ., T,
where et is an error term for period t. It is reasonable to estimate or approximatethe consumer's a parameter by choosing a to minimize the following function:
(10) f(a) ≡ S Tt=1 e2
t = S Tt=1 [(pt
1 xt1 /It) - a]2
= the sum of the squared errors.f¢(a) = S T
t=1 2[(pt1 xt
1 /It) - a]2-1(-1) set= 0 = -2 S T
t=1(pt1xt
1/It) + 2 S Tt=1 a = 0
(11) Æ a* = [S Tt=1(pt
1xt1/It) ]/T
f¢¢(a*) = 2T > 0 so we have a local (and global) minimum of f(a ) at a*. Thenumber a* defined by (11) is called the least squares estimator for a.
Now you can see how we can use observable data on a consumer's choices inorder to determine an approximation to his or her preference function. (Why arewe interested in a consumer's preferences anyway?)
4. Partial Derivatives
Each consumer and producer in a modern economy must choose betweenthousands of goods. Hence, we cannot restrict ourselves to optimizationproblems involving only one variable. However, it turns out that the singlevariable optimization techniques studied in section 2 can be modified to deal
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with the N variable case. A key concept that we will need to accomplish thisgeneralization to the N variable case is the idea of a partial derivative.
Let f be a function of N variables, x1, x2, . . ., xN.
Definition 1: The first order partial derivative of f with respect to x1 evaluated atx1, x2, . . ., xN is defined as the following limit (if it exists):
(11) limtÆ0[f(x1 + t, x2, . . ., xN) - f(x1, x2, . . ., xN)]/t≡ ∂f(x1, x2, . . ., xN)/∂x1
or ≡ fx1(x1, x2, . . ., xN)or ≡ f1(x1, x2, . . ., xN).
Note that there are three commonly used notational conventions used to denotethe concept of a partial derivative. Note that x2, . . ., xN are held constant indefinition (1). Thus if we regard f as just a function of x1, then (1) reduces tof¢(x1), the ordinary derivative of f with respect to x1. Hence in order to actuallycalculate f1(x1, x2, . . ., xN), we need only treat x2, . . ., xN as constants anddifferentiate the resulting function of x1 with respect to x1. The other N-1 firstorder partial derivatives of f may be defined in an analogous manner.
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Example 1:
f(x1, x2, x3) ≡ a0 + a1x1 + a2x2 + a3x3,f1(x1, x2, x3) ≡ a1f2(x1, x2, x3) = a2f3(x1, x2, x3) = a3.
Example 2:
f(x1, x2, x3) ≡ 3x21 + 2x1x2 + 6x
212 x
312 + ex2 + ln x3, xi > 0.
f1(x1, x2, x3) = 6x1 + 2x2
f2(x1, x2, x3) = 2x1 + 3x 2
- 12 x
312 + ex2
f3(x1, x2, x3) = 3x 212 x
2
- 12 + x-1
3 .
Example 3:
f(x1, x2) ≡ a0 + a1x1 + a2x2 + a11x21 + a12x1x2 + a22 x2
2 f1(x1, x2) = a1 + 2a11x1 + a12 x2f2(x1, x2) = a2 + a12x1 + 2a22x2
If you can differentiate functions of one variable, then you can partiallydifferentiate. It's easy; just treat the "other" variables as constants.
Question: What is the geometric interpretation of a partial derivative?
Higher Order Partial Derivatives. Once we have calculated the function ∂f(x1,. . ., xN)/∂x1 = f1(x1, . . ., xN), we can partially differentiate the resulting functionwith respect to x1, x2, . . ., or xN. The resulting function is called a second orderpartial derivative of f with respect to the variable x1 and xi say, evaluated at x1, . . ., xN:
(2) limtÆ0 f1(x1,K, xi-1, xi + t ,xi+1,K ,xN ) - f1(x1,K ,xi-1,xi ,xi+1,K ,xN )
t≡ ∂2f(x1, . . ., xN)/∂x1 ∂xi≡ fx1xi(x1, . . ., xN)≡ f1i(x1, . . ., xN).
There are three commonly used notations used to denote the concept of a secondorder partial derivative.
Example 3:
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f1(x1, x2) = a1 + 2a11x1 + a12 x2f11(x1, x2) = 2a11f12(x1, x2) = a12 f2(x1, x2) = a2 + a12 x1 + 2a22x2f21(x1, x2) = a12f 22(x1, x2) = 2a22
Note that f12(x1, x2) = f21(x1, x2). We shall show later that this is generally thecase.
Maximizing or Minimizing a Differentiable Function of N Variables
If we are at a local interior maximizing or minimizing point x*1 , . . .x*N of f, thenwe are at the top of a hill or at the bottom of a valley with respect to each variabletaken separately. Thus the following conditions must be satisfied:
First Order Necessary Conditions for an Interior Min or Max:
(3) ∂f(x*1 , . . ., x*N )/∂x1 = f1(x*1 , . . ., x*N ) = 0 M M M∂f(x*1 , . . ., x*N )/∂xN = fN(x*1 , . . ., x*N ) = 0.
Note that (3) is a system of N simultaneous equations in N unknowns, x*1 , . . ., x*N.
So far, the conditions for maximizing or minimizing a function of one variablehave generalized to the N variable case in a relatively straightforward manner.However, developing second order sufficient conditions for maximizing orminimizing a function of N variables requires that we introduce the concept ofthe directional derivative. In order to operationalize this concept, we shallrequire a few mathematical results.
5. The Mean Value Theorem
The Mean Value Theorem: Let f(x) be a continuous function over the interval a ≤ x≤ b with a < b. Suppose f¢(x) exists over this interval. Then there exists an x*such that
(1) a < x* < b and
(2) f¢(x*) = [f(b) - f(a)]/[b - a].
Proof: Define the function g(x) for a ≤ x ≤ b by
(3) g(x) ≡ f(x) - x [f(b) - f(a)]/[b - a].
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Let x = a and x = b and evaluate g(a) and g(b). We find that
(4) g(a) = f(a) - a [f(b) - f(a)]/[b - a] = [bf(a) - af(b)]/[b - a] and
(5) g(b) = f(b) - b [f(b) - f(a)]/[b - a]= [bf(a) - af(b)]/[b - a].
Since g(a) = g(b) and g is continuous, g must attain either a local maximum or alocal minimum (or both) at some point x* between a and b. The first ordernecessary condition for g to attain a local max or min at x* will be satisfied andthus we have:
(6) 0 = g¢(x*) = f¢(x*) - [f(b) - f(a)]/[b - a].
Now note that (2) is just a rearrangement of (6). Q.E.D.
Geometric Interpretation: f(x)
f(b) slope = [f(b) - f(a)]/[b - a]
slope = f’(x*)= [f(b)-f(a)]/[b - a]f(a)
a x* x** b x
Note that [f(b) - f(a)]/[b - a] is the average slope of the function over the intervala ≤ x ≤ b while f¢(x*) is the slope of the line tangent to the function at the point x*.
6. A Multivariate Function Chain Rule
Theorem: Let f(x1, x2) be a function of two variables defined over the regiona1 < x1 < b1 and a2 < x2 < b2. Suppose that the first order partial derivatives of fexist and are continuous over this region. Suppose that g1(z) and g2(z) aredifferentiable functions of z for c < z < d and for z in this region, we havea1 < g1(z) < b1 and a2 < g2(z) < b2. Finally, for c < z < d, define the (multivariatecomposite) function h(z) by:
(1) h(z) ≡ f[g1(z), g2(z)]
Then the derivative of h is given by:
(2) h¢(z) = f1[g1(z), g2(z)]g ¢1(z) + f2[g1(z), g2(z)]g ¢2(z) .
Proof: By the definition of a derivative, h¢(z) is defined as the following limit:
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(3) h¢(z) ≡ limtÆ0
[f(g1(z + t), g2(z + t)] - f[g1(z), g2(z)]]/t
(4) = limtÆ0
{f(g1(z + t), g2(z + t)]- f[g1(z), g2(z + t)] + f[g1(z), g2(z + t)] - f[g1(z), g2(z)]}/t
where we have subtracted and added the same term
(5) = limtÆ0
{f[x¢¢1 , x¢¢2 ] - f[x¢1 , x¢¢2 ] + f[x¢1 , x¢¢2 ] - f[x¢1 , x¢2 ]}/t
letting
x¢1 ≡ g1(z), x¢¢1 ≡ g1(z + t), x¢2 ≡ g2(z), x¢¢2 ≡ g2(z + t)
(6) = limtÆ0
{f[x¢¢1 , x¢¢2 ] - f[x¢1 , x¢¢2 ]}/t + limtÆ0{f[x¢1 , x¢¢2 ] - f[x¢1 , x¢2 ]}/t
(7) = limtÆ0{f1(x*1 , x¢¢2 ) (x¢¢1 - x¢1 )}/t + limtÆ0{f[x¢1 , x¢¢2 ] - f[x¢1 , x¢2 ]}/t
where x*1 is between x ¢1 = g1(z) and x ¢¢1 = g1(z + t), applying the Mean ValueTheorem to f(x, x¢¢2 ) regarded as a function of its first variable only and hence thederivative in this case is the first order partial derivative f1(x, x¢¢2 )
(8) = limtÆ0{f1(x*1 , x¢¢2 ) (x¢¢1 - x¢1 )}/t + limtÆ0{f2(x¢1 , x*2 ) (x¢¢2 - x¢2 )}/t
where x*2 is between x ¢2 = g2(z) and x¢¢2 = g2(z + t), applying the Mean ValueTheorem to f(x¢1 , x) regarded as a function of its second variable only
(9) = limtÆ0
f1(x*1 , x¢¢2 ) [g1(z%+%t)%-%g1(z)]
%t + limtÆ0
f2(x¢1 , x*2 )[g2(z%+%t)%-%g2(z)]
t
(10) = f1(g1(z), g2(z)) g ¢1(z) + f2(g1(z), g2(z)) g ¢2(z)
since as t Æ 0, x¢¢1 and x*1 tend to x ¢1 = g1(z) and x¢¢2 andx*2 tend to x ¢2 = g2(z).Q.E.D.
Example 1:
f(x1, x2) ≡ x1x2 ; g1(z) = z ; g2(z) = z2 + 1
(11) h(z) ≡ f[g1(z), g2(z)] = g1(z) g2(z) = z(z2 + 1)
(12) = z3 + z
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14
(13) Therefore h¢(z) = 3z2 + 1
where we have computed the derivative using the direct expression (12). Nowwe compute the derivative using the composite function chain rule (2) or (10):
h¢(z) = f1(x1, x2) g ¢1(z) + f2(x1, x2) g ¢2(z) = x21 + x1(2z)= z2 + 1 + z (2z)
(14) = 3z2 + 1 which agrees with (13).
The chain rule (2) seems to be a rather complex way of doing something simple.However, the following example shows how the rule may be used to deduce aformula for the slope of an indifference curve in terms of the first order partialderivatives of a consumer's utility function.
Example 2: f(x1, x2) is the consumer's utility function, x1 = g1(z) = z; x2 = g2(z) =g2(x1). Set utility equal to a constant; i.e.,
(15) h(z) ≡ f[g1(z), g2(z)] = f[z, g2(z)] = constant = u say
(16) Therefore h¢(z) = f1[z, g2(z)]• 1+ f2[z, g2(z)] g ¢2(z) = 0
(17) Therefore g¢2(z) = - f1[z, g2(z)] / f2[z, g2(z)].
Note that as z varies, x1 = z and x2 = g2(z) is the set of (x1, x2) points that give theconsumer the constant utility u. For each x1, the x2 point that gives this constantlevel of utility is x2 = g2(x1), and the slope of the indifference curve through thispoint is
(17) g ¢2(x1) = - f1(x1, x2) / f2(x1, x2) where x2 = g(x1).
As a concrete example of this formula, consider the Cobb-Douglas utilityfunction defined in section 3 with the taste parameter a = 1/2. Thus
(18) u = f(x1, x2) ≡ x 112 x
212
Using formula (17), the slope of the indifference curve through x1 > 0 and x2 > 0is
(19) -f1(x1, x2) / f2(x1, x2)= -12 x
1
- 12 x
212 / 12 x
112 x
2- 1
2 = -x2/x1
Note: In order for formula (17) to be valid, we require that f2(x1, x2) ≠ 0.
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7. Single Variable Comparative Statics Analysis and the Envelop Theorem
Recall the unconstrained maximization problem, maxxf(x), that we studied insection 2. The function f is the economic agent's objective function that he or she istrying to maximize and x is the agent's decision or choice variable. We shall nowcomplicate matters by assuming that the objective function f also depends on avariable "a" that cannot be controlled by the economic agent. Thus the objectivefunction that is to be maximized with respect to x is now f(x, a).
For an example of such a function, consider the f defined by (5) in section 3:there, "a" could be p1, p2, I or a. In general, economists are interested in knowinghow the optimal x responds to a change in the signal, "a".
Our maximization problem may now be written as:
(1) maxx f(x, a).
Assuming that f is differentiable, the first order necessary condition for solving(1) is:
(2) f1(x, a) = 0.
We suppose that for an initial "a", the solution to (1) is x* = g(a) and so we have
(3) f1[g(a), a] = 0.
We also assume that the second order sufficient condition for solving (1) issatisfied at x*:
(4) f11[g(a), a] < 0.
Equation (2) may be solved for x = g(a) and if we knew f precisely, we coulddetermine this solution function g. However, in many cases, we may not know fvery accurately, but we may know the signs of the partial derivatives of f up tosay the second order (again, recall example 1 in section 3). Under theseconditions, we can determine how x = g(a) changes as a changes; morespecifically, we can determine the slope of the response function, dx/da = g¢(a),as follows. Simply differentiate both sides of equation (3) with respect to a, usingthe composite function chain rule developed in the previous section. We obtainthe following equation:
(5) f11[g(a), a] g¢(a) + f12[g(a), a]1 = 0
(6) or g¢(a) = -f12[g(a),a] / f11[g(a), a].
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Thus the response of x to a small change in a hinges on the sign of the secondorder (cross) partial derivative of the objective function, f12[g(a), a] = f12(x*, a): ifthis derivative is positive, then the optimal x* will increase as a increases; if f12 isnegative , then the optimal x* will decrease as a increases. To deduce this rule,we used the second order condition (4).
This is a rather amazing qualitative result and we will see several concreteapplications of it later in the course. This qualitative result is an example ofcomparative statics analysis.
There is one additional result that we wish to derive in this section. Define theoptimized objective function as a function of the signal or stimulus variable "a" by
(7) h(a) ≡ maxx f(x, a)
(8) = f[g(a), a]
Now differentiate (8) with respect to "a" using our multivariate chain rule:
(9) h¢(a) = f1[g(a), a]g¢(a) + f2[g(a), a]= 0 g¢(a) + f2[g(a), a] using (3)= f2[g(a), a].
Thus to determine the slope of the optimized objective function with respect to"a", we need only partially differentiate f(x*, a) with respect to its second variable"a". The result (9) is known as the envelop theorem.
The results in this section were first derived by the famous American economist,Paul Samuelson, in his book, Foundations of Economic Analysis, 1947.
Note that the only difficult mathematical result that was required to derive all ofthis was the Theorem in section 6.
8. The Geometry of Single Variable Comparative Statics Analysis
Consider the following initial unconstrained maximization problem where thevalue of the parameter a is a1:
(1) maxx{f(x, a1): -• < x < + •}.
Suppose x1 = g(a1) solves (1) and we have
(2) f1(x1, a1) = 0;
(3) f11(x1, a1) < 0.
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Now consider problem (1) when a1 is increased by Da to a2 = a1 + Da where a2 >a1. We suppose x2 = g(a2) solves the new problem and we have:
(4) f1(x2, a2) = 0;
(5) f11(x2, a2) < 0
where (5) will follow from (3) and the continuity of f11 if Da is sufficiently small.
We can illustrate graphically the unconstrained maximization problem (1) byplotting y = f(x, a1) as a function of x, holding a1 fixed. The geometry of thismaximization problem is illustrated in Figure 4 below; see the lower of the twographs. We can also illustrate the second maximization problem when a1 isreplaced by a2 by plotting y = f(x, a2) as a function of x; see the higher graph inFigure 1 below. y
Figure 4: f12(x1, a1) > 0. Slope is f1(x2, a2) = 0
Slope is f1(x1, a1) = 0 y = f(x, a1)
Slope is f1(x1, a2) > 0
y = f(x, a2)
x1 x2 xAs we increase a from a1 to a2 holding x1 fixed, the slope f1(x1, a) increases from0 to the positive number f1 = (x1, a2). Thus f12(x1, a1) will be positive and x2 =g(a2) will be greater than x1 = g(a1); i.e., g¢(a1) = dx*(a1)/da > 0.
Figure 5 below illustrates the case where f12(x1, a1) = 0 and x1 = x2 and g¢(a1) = 0while Figure 6 illustrates the case where f12(x1, a1) < 0, x2 < x1 and g¢(a1) < 0.
Figure 5: f12(x1, a1) = 0 Figure 6: f12(x1, a1) < 0.y y
y = f ( x , a2)f1(x2,a2)=0
y=f(x,a1)
y = f ( x , a1)y=f(x,a2)
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x1 =x2 x2 x1
Thus the sign of the derivative f12(x1, a1) tells us whether the optimal x increasesor decreases as a increases.
Application to Producer Supply and Demand Functions
Suppose that the maximum output y that a producer can produce in a given timeperiod using the positive amount of input x is
(6) y = af(x)
where a > 0 is an efficiency parameter and f is a twice continuously differentiableproduction function. Given a positive price for a unit of output, p > 0, and apositive input price, w > 0, we assume that the producer solves the followingprofit maximization problem:
(7) maxy, x{py - wx: y = af(x)} ≡ p(p, w, a).
We can use the constraint to eliminate y from the objective function and weobtain the following unconstrained maximization problem that is equivalent to(7):
(8) maxx{paf(x) - wx} ≡ p(p, w, a).
We assume that the solution x* to (8) satisfies:
(9) paf¢(x*) - w = 0;
(10) paf¢¢(x*) < 0.
Now regard x* as a function of p, w and a, say x* = d(p, w, a). To determinehow the demand for input changes as the output price p increases, replace x* in(9) by d(p, w, a) and differentiate the resulting equation with respect to p. Doingthis differentiation using normal calculus rules (treating w and a as constants),we obtain the following equation:
(11) af¢[d(p, w, a)] + paf¢¢[d(p, w, a)] ∂d∂p
(p, w, a) = 0 or
(12) ∂d∂p
(p, w, a) = -af¢(x*)/paf¢¢(x*)
= - f¢(x*)/pf¢¢(x*)= (-)(+)/(+)(-) using (9) and (10)> 0.
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Thus input demand increases as the output price increases.
To obtain the response of optimal output y* to an increase in the output price, weuse the constraint in (7) to define optimal output y* in terms of the optimal inputx*; i.e., define the optimal supply function s(p, w, a) as
(13) s(p, w, a) ≡ a f[d(p, w, a)].
Now partially differentiate (13) with respect to p and use (12) to determine thederivative ∂d(p, w, a)/∂p:
(14) ∂s∂p
(p, w, a) = af¢[d(p, w, a)] ∂d∂p
(p, w, a)
= af¢(x*) (-f¢(x*)/pf¢¢(x*))= -a [f¢(x*)]2/pf¢¢(x*)= (-) (+) (+)/(+)(-) using (9) and (10)> 0.
Thus output supply increases as the output price increases.
To determine how the optimized objective function p(p, w, a) changes as pincreases, we need only differentiate p defined by (15) with respect to p:
(15) p(p, w, a) ≡ paf[d(p, w, a)] - wd(p, w, a).
Partially differentiating (15) with respect to p yields:
∂p
∂p (p, w, a) = af(x*) + [paf¢(x*) - w]
∂d∂p
(p, w, a)
(16) = af(x*) = s(p, w, a) using (9).
We could have obtained result (16) by using the Envelop Theorem: replace f(x, a)by
(17) F(x, p) ≡ paf(x) - wx;
replace x* = g(a) by x* = d(p); replace a by p; and replace h(a) by p(p) where
(18) p(p) ≡ maxx{F(x, p)}.
The Envelop Theorem tells us that
(19) p¢(p) = F2(x*, p) = af(x*)
which is (16).
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Problems
1. Adapt the above methodology to obtain formulae for the derivatives ∂d(p,w, a)/∂w, ∂d(p, w, a)/∂a, ∂s(p, w, a)/∂w and ∂s(p, w, a)/∂a. Sign thesederivatives.
2. Show that ∂p(p, w, a)/∂w = -d(p, w, a) (Hotelling's Lemma).
3. Show that ∂s(p, w, a)/∂w = -∂d(p, w, a)/∂p (Hotelling SymmetryCondition).
4. Calculate the consumer's system of demand functions D1(p1, p2, I) and D2(p1, p2, I) if the consumer's utility function is defined by:
(i) u(x1, x2) ≡ f [x112 x2
12 ]
where f is a continuously differentiable function of one variable which hasf¢(x) > 0 for all x > 0.
5. Solve maxx {f(x): x ≥ 0} for the following functions f:
(a) f(x) ≡ -x2 + 2x - 2(b) f(x) ≡ ln x - x + 1(c) f(x) ≡ -x2 - 2x.
Check the relevant second order conditions.
6. Solve maxx1, x2{(f(x1, x2)} for the following f:
(a) f(x1, x2) ≡ - x12 + x1x2 - x2
2 + 2;(b) f(x1, x2) ≡% ln x1 + ln x2 - 2x1 - 2x2 + 2.
Check the relevant second order conditions.
9. The Directional Derivative and First Order Necessary Conditions.
Let f(x1, x2, . . ., xN) be a function of N variables. In order to define thedirectional derivative of f, we first need to define a direction.
Definition 1: A direction v is defined to be N numbers v1, v2, . . ., vN whosesquared components sum to 1; i.e., v2
1 + v22 + . . . + v2
N = 1. Thus a direction v ≡
(v1, v2, . . ., vN) in N dimensional space is a point on the sphere of radius 1 withcenter at the origin.
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Definition 2: The direction derivative of the function f evaluated at the point(x1, x2, . . ., xN) in the direction v, denoted as Dvf(x1, x2, . . ., xN), is defined as thefollowing limit (if it exists):
(1) Dvf(x1, x2, . . ., xN) ≡ limtÆ0 f(x1 + tv1,K ,xN + tvN ) - f(x1,K ,xN )
t
Geometric Interpretation?
Let ei ≡ (0, . . ., 0, 1, 0, . . ., 0) denote the point in N dimensional space which hasith coordinate equal to 1 and all other coordinates equal to zero. (This is oftencalled the ith unit vector in matrix algebra).
Suppose we chose our direction v to be ei . Then using definition 1, it can be seenthat
(2) Dei f(x1, x2, . . ., xN) = fi(x1, x2, . . ., xN)
where fi denotes the ith first order partial derivative of f. Thus partial derivativesare special cases of the directional derivative: the ith partial derivative is equal tothe directional derivative in the direction given by the ith coordinate axis.
Example: Let f(x1, x2) = x21 + x2 and v = (1/ 2 , 1/ 2 ) = (2
-12 , 2
-12 )
D (2-12 ,2
-12 ) f(x1, x2) ≡ lim tÆ0 [f(x1 + t2
-12 , x2 + t2
-12 ) - f(x1, x2)] / t
= lim tÆ0 [(x1 + 2 -12 t)2 + (x2 + 2
-12 t) - (x2
1 + x2)] / t
= lim tÆ0 [x21 + 2
1- 12 tx1 + 2-1t2) + (x2 + 2
-12 t) -x2
1 - x2] / t
= lim tÆ0 [2 12 tx1 + 2-1t2 + 2
-12 t] / t
= lim tÆ0 [2 12 x1 + 2-1t + 2
-12 ]
(3) = 2 12 x1 + 2
-12
The above example shows that it is not very easy to calculate a directionalderivative in general. This contrasts to the case of partial derivatives whereordinary calculus rules for differentiation could be used. Thus we need an easierway for calculating directional derivatives, and the following Theorem does thisfor us.
First Order Directional Derivative Theorem. If the first order partial derivatives of fexist and are continuous functions in a neighbourhood around the pointx1, x2, . . ., xN, then the directional derivative of f evaluated at x1, . . ., xN for any
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direction v ≡ (v1, v2, . . ., vN) exists and is given by the following weighted sumof partial derivatives:
(4) Dvf(x1, x2, . . ., xN) = S i=1N vifi(x1, x2, . . ., xN)
= v1f1(x1, . . ., xN) + v2f2(x1, . . ., xN) + . . . + vNfN(x1, . . ., xN).
Proof: For simplicity, we prove the result for the case where N = 2. The generalcase may be proven in a similar manner.
By the definitional of the directional derivative, we have
Dvf(x1, x2) ≡ lim tÆ0 [f(x1 + tv1, x2 + tv2) - f(x1, x2)] / t
= lim tÆ0 [f(x1 + tv1, x2 + tv2) - f(x1, x2 + tv2) + f(x1, x2 + tv2) - f(x1, x2)] / t
upon adding and subtracting the term f(x1, x2 + tv2)
= lim tÆ0 [f1(x*1 , x2 + tv2) (x1 + tv1 - x1) + f2(x1, x*2 )(x2 + tv2 - x2)] / t
applying the Mean Value Theorem twice where x*1 is between x1 + tv1 and x1and x*2 is between x2 + tv2 and x2
= lim tÆ0 [f1(x*1 , x2 + tv2)v1 + f2(x1, x*2 )v2
canceling terms involving t
= f1(x1, x2)v1 + f2(x1, x2)v2
taking limits and using the continuity of f1 and f2
= v1f1(x1, x2) + v2f2(x1, x2).Q.E.D.
Example: Let f(x1, x2) = x21 + x2 and v1 = 2
-12 , v2 = 2
-12 .
Then
Dvf(x1, x2) = v1f1(x1, x2) + v2f2(x1, x2)
= 2 -12 2x1 + 2
-12 • 1
= 2 12 x1 + 2
-12 which agrees with (3).
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Now that we have introduced the concept of the directional derivative, thefollowing condition for an interior max or min should be obvious.
First Order Necessary Conditions for f to attain a local interior max or min atx1, x2, . . ., xN are:
(5) Dvf(x1, x2, . . ., xN) = 0 for every direction v = (v1, v2, . . ., vN).
If the regularity conditions for the First Order Directional Derivative Theoremare satisfied, then by (4), conditions (5) are equivalent to:
(6) S i=1N vifi(x1, x2, . . ., xN) = 0 for every v = (v1, v2, . . ., vN) such that S i=1
N v2i = 1.
At first sight, conditions (5) and (6) may not seem to be of much practical value,since as soon as N ≥ 2, we must check (5) and (6) for an infinite number ofdirections v. However, it is easy to verify that conditions (6) are equivalent toour old first order necessary conditions, (3) in section 4, which we now rewrite asconditions (7):
(7) ∂f(x1, x2, . . ., xN)/∂x1 ≡ f1(x1, x2, . . ., xN) = 0 M M M
∂f(x1, x2, . . ., xN)/∂xN ≡ fN(x1, x2, . . ., xN) = 0
It is obvious that (7) implies (6). To show that (6) implies (7), choose v to be theunit vector ei for i = 1, 2, . . ., N.
The reader may well be a bit confused at this point. Why did we bother tointroduce the concept of the directional derivative if in the end, we simply endup with conditions (7), which we derived before using only the much simplerconcept of a partial derivative?
There are two answers to this questions: (i) if the partial derivative functions arecontinuous, we now know that conditions (7) imply the seemingly muchstronger conditions (5), and (ii) in order to derive valid second order sufficientconditions, we need the concept of the directional derivative.
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10. Second Order Sufficient Conditions for a Maximum or Minimum
Before we can develop our second order conditions, we need to introduce theconcept of a second order directional derivative. Suppose we are given adirection v and the first order directional derivative in this direction exists for all(x1, x2, . . ., xN) in a neighbourhood; i.e., D vf(x1, x2, . . ., xN) exists. Now pickanother direction u ≡ (u1, u2, . . ., u2) where u2
1 + u22 + . . . + u2
N = 1.
Definition: The second order directional derivative of f in the directions v and uevaluated at the point x1, x2, . . ., xN is defined as the following limit if it exists:
(1) Dvuf(x1, x2, . . ., xN) ≡ lim tÆ0 [Dvf(x1 + tu1, . . ., xN + tuN) -Dvf(x1, . . ., xN)] / t
[Geometric Interpretation?]
Note that if v = ei and u = ej, then Deiejf(x1, . . ., xN) = fij(x1, . . . , xN), the secondorder partial derivative of f with respect to xi and xj. For example, if N = 2,v = (1, 0) and u = (0, 1), then Dvuf(x1, x2) = De1e2f(x1, x2) = f12(x1, x2) ≡ ∂2f(x1,x2)/∂x1∂x2. Recall that it is straightforward to compute second order partialderivatives using ordinary calculus rules. However, for general directions v andu, it is not easy to compute Dvuf(x1, . . . , xN).
The following theorem allows us to express a general second order directionalderivative in terms of second order partial derivatives.
Second Order Directional Derivative Theorem. If the first and second order partialderivatives of f exist and are continuous in a neighbourhood around the point x1,. . . , xN, then the second order directional derivative of f in the directions v = (v1,. . . , vN) and u = (u1, . . ., uN) evaluated at x1, . . . , xN exists and may becalculated as the following weighted sum of second order partial derivatives
(2) Dvuf(x1, . . . , xN) = S i=1N
S j=1
N vifij(x1, . . ., xN)uj.
Proof: For simplicity, we shall prove only the case where N = 2. The general casefollows in an analogous manner. By the definition of Dvuf(x1, x2), we have:
Dvuf(x1, x2) = lim tÆ0 [Dvf(x1 + tu1, x2 + tu2) - Dvf(x1, x2)] / t
= lim tÆ0 [(v1f1(x1 + tu1, x2 + tu2) + v2f2(x1 + tu1, x2 + tu2))- (v1f1(x1, x2) + v2f2(x1, x2)] / t
applying the First Order Directional Derivative Theorem to Dvf(x1 + tu1, x2 + tu2)and Dvf(x1, x2)
= limtÆ0[(v1{f1(x1 + tu1, x2 + tu2) - f1(x1, x2)}
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+ v2{f2(x1 +tu1, x2 +tu2) - f2(x1, x2)}] / t
collecting terms involving v1 and v2
= lim tÆ0 S i=12 vi{fi(x1+ tu1, x2 + tu2) - fi(x1, x2)} / t
= S i=12 viDufi(x1, x2)
by the definition of Dufi
= S i=12 vi{ S j=1
2 ujfij(x1, x2)}
applying the First Order Directional Derivative Theorem to Duf1 and Duf2
= S i=12
S j=12 vjfij(x1, x2)uj rearranging terms
= v1f11(x1, x2)u1 + v1f12(x1, x2) u2+ v2f21(x1, x2)u1 + v2f22(x1, x2)u2
Q.E.D.
Assuming that our function f(x1, . . . , xN) has continuous second order partialderivatives, we may now derive sufficient conditions for an interior local max ormin by applying our old one dimensional conditions to all possible directions.
Thus we have the following:
Second Order Sufficient Conditions for f to attain a strict local max:
(3) Dvf(x1, . . . , xN) = 0 for all directions v and
(4) Dvvf(x1, . . . , xN) = S i=1
N S j=1N vifij(x1, . . . , xN)vj < 0
for all v such that v21 + . . . + v2
N = 1
Second Order Sufficient Conditions for f to attain a strict local min at x1, . . . , xN:
(5) Dvf(x1, . . . , xN) = 0 for all directions v and
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(6) Dvvf(x1, . . . , xN) = S i=1
N S j=1N vifij(x1, . . . , xN)vj > 0
for all v such that v21 + . . . + v2
N = 1.
We indicated in the previous section, that conditions (3) or (5) are equivalent tothe following first order necessary conditions which can readily be verified:
(7) fi(x1, . . . , xN) = 0, i = 1, . . ., N
However, if N ≥ 2, then conditions (4) and (6) involve checking an infinitenumber of inequalities, which is not practical. Hence we need to convertconditions (4) and (6) into equivalent sets of conditions that can be checked. Thistask is accomplished in courses in matrix algebra and we will not attempt to do ithere for the general case. However, we shall do the case where N = 2. We first,require a preliminary result. We state it for N = 2, but it is valid for a general N.
Young's Theorem. If f1, f2 and f12 exist and are continuous functions around thepoint (x1, x2), then the second order partial derivative f21(x1, x2) exists and equals
(8) f21(x1, x2) = f12(x1, x2).
Proof: By the definition of f21(x1, x2), we have
f21(x1, x2) ≡ lim hÆ0 [f2(x1 + h, x2) - f2(x1, x2)] / h
= lim hÆ0 h-1[lim kÆ0 {f(x1 + h,x2 + k) - f(x1 + h, x2)} / k+ lim kÆ0 {f(x1, x2 + k) - f(x1, x2)} / k].
by the definition of f2 used two times
= lim hÆ0 lim kÆ0 h-1k-1[g(x1 + h) - g(x1)]
defining g(t) ≡ f(t, x2 + k) - f(t, x2)
= lim hÆ0 lim kÆ0 h-1k-1[g¢(x*1 )]
where x*1 is between x1 + h and x1, applying the Mean Value Theorem to g
= lim hÆ0 lim kÆ0 k-1[f1(x*1 , x2 + k) - f1(x*1 , x2)]
cancelling h's and using the definition of g
= lim hÆ0 f12(x*1 , x2)
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by the definition of f12
= f12(x1, x2)
using the continuity of f12.Q.E.D.
Now consider conditions (6) when N = 2. Using Young's Theorem, theseconditions become:
(9) Dvvf(x1, x2) = v21 f11(x1, x2) + 2v1v2f12(x1, x2) + v2
2 f22(x1, x2) > 0
for all v1 and v2 such that v21 + v2
2 = 1.
Our problem is to reduce the infinite number of conditions in (9) down to a finitenumber of conditions.
Suppose v1 = 0. Since v21 + v2
2 = 1, we have v22 = 1 and (9) reduces to
(10) f22(x1, x2) > 0.
Now let us fix v1≠ 0 and look at the right hand side of (9) as a function of v2. Wewant to know if
(11) g(v2) ≡ v21 f *11 + 2v1v2 f *12 + v2
2 f *22 > 0
where f*ij ≡ fij(x1, x2). Let us try to minimize g with respect to v2:
g¢(v2) = 2v1f *12 + 2 v2 f *22 set= 0 Æ v*2 = -v1 f *12 /f *22
g¢¢(v*2 ) = 2f *22
> 0 using (10).
Hence v*2 = globally minimizes g(v2). Thus if g(v*
2 ) > 0, then g(v2) ≥ g(v*2 ) > 0
for all v2. Now calculate g(v*2 ):
g(v*2 ) = v2
1 f *11 + 2v1(- v1f *12 /f *22 )f *12 + v21 f *12 2/f *22 = v2
1 [f *11 - f *12 2/f *22 ].
Thus necessary and sufficient conditions for (9) to be true are (10) andf *11 - f *12 2 / f *22 > 0 which in view of (10) is equivalent to:
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(12) f11(x1, x2) f22(x1, x2) - [f12(x1, x2)]2 > 0.
Thus the infinite number of inequalities in (9) are equivalent to the twoconditions (10) and (12).
Suppose Conditions (9) are satisfied. Then set v2 = 0 and (9) reduces to:
(13) f11(x1, x2) > 0.
Problem: Show that conditions (10) and (12) are equivalent to conditions (12) and(13).
Now consider conditions (4) when N = 2. We may modify our analysis on theprevious page and show that (4) is equivalent to the following two conditions:
(14) f11(x1, x2) < 0 and
(15) f11(x1, x2) f22(x1, x2) - [f12(x1, x2)]2 > 0.
Condition (14) may be replaced by
(16) f22(x1, x2) < 0.
(Another way to derive (14) and (15) from (13) and (12) is to observe thatmaximizing f(x1, x2) is equivalent to minimizing -f(x1, x2). Thus we require -f11(x1, x2) > 0 and [-f11(x1, x2)][-f22(x1, x2)] - [-f12(x1, x2)]2 > 0 and these twoinequalities are equivalent to (14) and (15).)
Example 1: Maximize f(x1, x2) ≡ 2x1 + 2x2 - x21 + x1x2 - x2
2 .
f1(x1, x2) = 2 - 2x1 + x2 set= 0 Solution is x*1 = x*
2 = 2.f2(x1, x2) = 2 + x1 - 2x2 set= 0
f11(x1, x2) = -2 , f12(x1, x2) = 1f21(x1, x2) = 1 , f22(x1, x2) = -2
f11(x*1 , x*
2 ) = - 2 < 0
f11(x*1 , x*
2 ) f22(*1 , x*2 ) - [f12(x*
1 , x*2 ])2
= [-2] [-2] - 12= 4 - 1= 3 > 0
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Therefore the second order sufficient conditions for a local max are satisfied atx*
1 = 2, x*2 = 2. Since there is only one solution to the first order conditions and
f(x1, x2) becomes large and negative as x21 + x%22 becomes large, we conclude
that our local max is also the global max.
Example 2: Two Input Cobb-Douglas Production Function
Suppose that a competitive firm utilizes positive amounts of two inputs, x1 andx2, in order to produce units of a single output y. The technology of the firmmay be summarized by means of a production function f; i.e., y = f(x1, x2) denotesthe maximum amount that can be produced in a certain period of time using x1units of input 1 and x2 units of input 2. Suppose that the firm can sell units ofoutput at the fixed price p0 > 0 and can purchase units of input 1 and 2 at thefixed prices p1 and p2. Then the firm's constrained profit maximization problem is
(17) maxy, x1, x2{p0y - p1x1 - p2x2: y = f(x1, x2)}.
Problem (17) is a constrained profit maximization problem involving 3 decisionvariables, y, x1, x2; 3 exogenous variables, p0, p1, p2; and one (productionfunction) constraint. Let us substitute the constraint function into the objectivefunction and reduce (17) into the following unconstrained profit maximizationproblem involving the two decision variables, x1 and x2:
(18) maxx1, x2{p0f(x1, x2) - p1x1 - p2x2}.
We cannot solve (18) until we are given a concrete functional form for theproduction function f. Suppose that
(19) y = f(x1, x2) ≡ a x1a1 x2
a2
where the technological parameters satisfy the following restrictions
(20) a > 0, a1 > 0, a2 > 0, a1 + a2 < 1.
Substituting (19) into (18), we find that the first order necessary conditions for(18) are:
(21) a1p0ax1 x1a1-1
x2a2 - p1 set= 0;
(22) a2p0ax2 x1a1
x2a2 -1- p2 set= 0.
Multiply (21) by x1, multiply (22) by x2. We get:
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(23) a1p0ax1 x1a1
x2a2 = p1x1 and
(24) a2p0ax2 x1a1
x2a2 = p2x2.
Now divide (23) by (24) and simplify. We get
(25) x1 = a1 a2-1p-1
1 p2x2.
Substitute (25) into (24) and simplify. We get
(26) x*2 = [aa2 a1a1
a2-a1 ]1/(1 - a1 - a2)p01/(1 - a1 - a2)p1- a1/(1 - a1 - a2)p2( a1 -1)/(1 - a1 - a2)
≡ k2p01/(1 - a1 - a2)p1- a1/(1 - a1 - a2)p2( a1 -1)/(1 - a1 - a2)
≡ D2(p0, p1, p2, a, a1, a2)
Now substitute (26) into (25) and get:
(27) x*1 = [a1 a2-1k2]p01/(1 - a1 - a2)p1(a2-1)(1 - a1 - a2)p2 - a2/(1 - a1 - a2)
≡ k1p01/(1 - a1 - a2)p1(a2-1)/(1 - a1 - a2)p2 - a2 /(1 - a1 - a2)
≡ D1(p0, p1, p2, a, a1, a2).
The x1 and x2 solutions to (18) defined by (27) and (26) are the firm's system ofprofit maximizing input demand functions, D1 and D2. These two functions tell ushow the firm's demands will vary as the output and input prices p0, p1 and p2change. The demand functions also depend on the technological parameters a,a1 and a2 that appeared in the production function (19).
To obtain the firm's output supply function S, substitute (26) and (27) into (19)and obtain:
(28) y* = [a k1a1 k2
a2 ]p0(a1 + a2)/(1 - a1 - a2)p1-a1/(1 - a1 - a2)p2-a2/(1 - a1 - a2)
≡ k0p0(a1 + a2)/(1 - a1 - a2)p1-a1/(1 - a1 - a2)p2-a2/(1 - a1 - a2)
≡ S(p0, p1, p2, a, a1, a2).
This example illustrates how the firm's profit maximization problem generatesoutput supply functions and input demand functions.
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Question: if we were given data on the firm's output in period t, yt, the outputprice pt
0 and the input prices pt1 and pt
2 for period t for t = 1, . . ., T, how couldwe use these data to obtain estimates for the firm's technological parameters a,a1 and a2? Hint: take the logarithm of both sides of (28) and recall example 2 insection 3 above.
We are not quite through with the above example: we have not checked whetherour solution defined by (26) and (27) satisfies the second order sufficientconditions for a local maximum. We first calculate the second order partialderivatives f*ij ≡ fij(x*1 , x*2 ) for i, j = 1, 2:
f *11 = p0aa1(a1 - 1) x1*a1-2 x2
*a2 ; f *12 = p0aa1a2 x1*a1-1x2
*a2-1= f *21
f *22 = p0aa2(a2 - 1) x1*a1 x1
*a2 -1
By (20) a1 < 1 and a2 < 1 so that
(29) f *11 < 0 and f *22 < 0.
(30) f *11 f *22 - (f *12 )2 = [p0a x1*a1-1x2
*a2-1)]2[a1(a1 - 1)a2(a2 - 1) - (a1a2)2]
= [p0a x1*a1-1x2
*a2-1)]2a1a2[1 - a1 - a2]
> 0 using (20).
Thus the second order sufficient conditions for a local max are satisfied and oursolution functions defined by (26) and (27) are valid.
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11. The Lagrange Multiplier Technique
In economics, constrained maximization problems are often solved using theLagrange multiplier technique. Thus it is necessary for us to explain what is is.
Suppose f and b are differentiable functions of two variables x1 and x2 and wewish to solve the following constrained maximization problem (e.g., recallexample 1 in section 3):
(1) maxx1,x2{f(x1, x2) : b(x1, x2) = 0}.
Suppose that the maximum to (1) occurs at a point x*1 , x*2 in the interior of thedomain of both functions. Suppose also that f2(x*1 , x*2 ) ≠ 0 and b2(x*1 , x*2 ) ≠ 0.
Consider the indifference or level curve of f through the point (x*1 , x*2 ). This isthe set {x1, x2 : f(x1, x2) = f(x*1 , x*2 )}. Consider also the constraint curve, {x1, x2 :b(x1, x2) = 0}. Obviously, if x*1 , x*2 solves (1), then x*1 , x*2 is on the constraintcurve; i.e.,
(2) b(x*1 , x*2 ) = 0.
From elementary geometrical considerations, it can be seen that if x*1 , x*2 solves(1), then the indifference curve of f through x*1 , x*2 must be tangent to theconstraint curve and the point of tangency occurs at x*1 , x*2 . Thus the slopes ofthe two curves must be equal. From formula (17) in section 6, the slope of theindifference curve through x*1 , x*2 is -f1(x*1 , x*2 )/f2(x*1 , x*2 ) while the slope of theconstraint function at x*1 , x*2 is -b1(x*1 , x*2 )/b2(x*1 , x*2 ). Equating these twoslopes yields the equation
(3) -f1(x*1 , x*2 )/f2(x*1 , x*2 ) = -b1(x*1 , x*2 )/b2(x*1 , x*2 )
Now rearrange (3) to yield the following equation:
(4) -f1(x*1 , x*2 )/b1(x*1 , x*2 ) = - f2(x*1 , x*2 )/b2(x*1 , x*2 )≡ l*
where we have defined the common ratio in (4) to be the number l*. Now wemay rearrange equations (4) to yield the following two equations involving l*:
(5) f1(x*1 , x*2 ) + l*b1(x*1 , x*2 ) = 0
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(6) f2(x*1 , x*2 ) + l*b2(x*1 , x*2 ) = 0.
Equations (2), (5) and (6) may be regarded as three equations in the threeunknowns x*1 , x*2 and l*. These are Lagrange's first order necessary conditions forx*1 , x*2 to solve the constrained maximization problem (1).
These conditions may be obtained in a simple way by defining the LagrangianL(x1, x2, l) by:
(7) L(x1, x2, l) ≡ f(x1, x2) + lb(x1, x2).
It can be verified that equations (5), (6) and (2) are equivalent to the followingfirst order conditions:
(8) ∂L(x*1 , x*2 , l*)/∂x1 = 0; ∂L(x*1 , x*2 , l*)/∂x2=0; ∂L(x*1 , x*2 , l*)/∂l = 0
The second order conditions for the Lagrangian technique are too complex for usto develop here. In practice, when using the Lagrange multiplier technique forsolving (1), one simply hopes that the point x*1, x*2 found by solving (8) is thedesired maximum.
Question: Suppose (1) was a minimization problem instead of a maximizationproblem. How could we adapt the above technique?
Problem:
Determine whether the following functions have any local minimums ormaximums. Check the relevant second order conditions. The domain ofdefinition for each function is two dimensional space.(i) f(x1,x2) ≡ x1
2 + x22 - 2x1 - 2x2 ;
(ii) f(x1,x2) ≡ - x12 + x1x2 - x2
2 + x1 - x2 ;(iii) f(x1,x2) ≡ x1
2 - 2x1x2 + x22 .
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