Mathematics for Economists Introduction to Functions Introduction In economics we study the relationship between variables and attempt to explain these relationships through economic theory. For instance in macroeconomics we try and understand the relationship between unemployment and economic growth. These relationships can almost always be described by mathematical functions. Functions, similar to numbers, are an essential building block in mathematics. It is very likely that you have been exposed, maybe unknowingly to functions during high school mathematics. Irrefutably the concept of functions is one of the most important mathematics and is required routinely in higher mathematics. function is simply a rule that assigns to each variable x one and only one value [f(x)], which is known as the value of the function of x. The domain of a function refers to the set of all possible values of x. The range refers to the set of all possible values for [f(x)]. Definition of a Function Let and be two sets. function f: is a binary relation from Set to Set such that for every a, there is exactly one element b such that b = f(a). When we use the notation y = f(x), we say that x is the argument of this function and y is the value of this function at x, or that y is the of this function at x. a,! b such that b = f(a) Examples of Common Functional Forms Quadratic Functions: n example of a function which assigns a number in R 1 to each number in R 1 may be quadratic such as f(x) = x 2 + 3. If for example we choose x = 2 and f(x) = x 2 + 3 then f(2) = 2 2 + 3 = 7 or if x = 3 then f(3) = 12. We may also try x = 1 2, which will give us f(1 2) = 3.25. This is an example of a linear function. Linear Functions: n example of a function which assigns a number in R 1 to each number in R 1 may be linear such as f(x) = 3x 1. If for example we choose x = 2 and f(x) = 3x 1 then f(2) = 3(2) 1 = 5 or if x = 3 then f(3) = 8. We may also try x = 1 2, which will give us f(1 2) = 1/2. Polynomial Functions: n example of a function which assigns a number in R 1 to each number in R 1 may be Polynomial such as f(x) = x 3 2x 2 + 2. If for example we choose x = 2 and f(x) = x 3 2x 2 + 2 then f(2) = (2 3 ) 2(2 2 ) + 2 = 2 or if x = 3 then f(3) = 19. We may also try x = 1 2, which will give us f(1 2) = 1.875.
Power Functions: n example of a function which assigns a number in R 1 to each number in R 1 may be Polynomial such as f(x) = 2x 3. If for example we choose x = 2 and f(x) = 2x 3 then f(2) = 2(2 3 ) = 2 or if x = 3 then f(3) = 54. We may also try x = 1 2, which will give us f(1 2) = 0.25. Note on Relations and Functions \ efore defining functions it is important to understand relations. relation is defined as a set of ordered pairs (a, b) of real numbers (R), where a represents x and b represents y. relation may also be defined as a rule a, table, equation or inequality that determines the ordered pairs. For example for the linear function f(x) = 3x 1 defines the following subset of ordered pairs, {(1,2)(2,5)(3,9)}. Likewise, the set of ordered pairs (relation) {(1,4)(2,7)(3,12)} defines the quadratic function above f(x) = x 2 + 3. The input or independent variables x comes from the set known as the domain and the output or dependent variables y are a subset of the codomain referred to as the range. Graphing Functions y using a rectangular coordinate system scaled by number lines we can represent relations. We can then graph a geometrical representation of a function. Remember that for a relation to be a function is must have only one y value for each value of x. This means that no independent input value x has more than one dependent output point y on the graph. If we were to draw a vertical line somewhere on the graph it should not intersect with the curve more than once. In other words if the relation is a function no two points of the graph will lie on the same vertical line. In the two graphs below we can see that the SX price index is a relation but not a function because there can be multiple y (prices) values for a given x (day) value because it captures the range of prices of the index for each day (Graph 1). If the graph used the SX daily average price for y then this would be a function because every x (day) has only one y (average price) value (Graph 2). Using the average, median daily prices for the y value would also allow the relation to be a function. Graph 1: SX Index Daily Price Range Price Graph 2: SX Index Daily verage Price Day The relation in Graph 3 below is not a function because there can be multiple values of y for a given x. To demonstrate that this is case we can draw a vertical line in the graph to show that x corresponds to two values of y. For instance when x = 1, y can take on either 5 or 3, because x + (1) = 5 and therefore y 2 = ±2.360. This is known as the vertical line test and is a quick simple way to verify if a graph of a relation represents a function. Graph 4 is a function because only one value of y for each value of x. Day
y 2 = 4 + x 8 7 6 y y = 2x 2 8 7 6 y 5 5 4 4 3 2 1 3 2 1-5 -4-3 -2-1 1 2 3 4 5 x -5-4 -3-2 -1 1 2 3 4 5 x { ( 3,18)( 2,8)( 1,2)(0,0)(1,2)(2,8)(3,18) } asic Definition of Functions Let and be two sets. function f: is a binary relation from Set to Set such that for every a, there is exactly one element b such that b = f(a). When we write a function such as y = f(x) we say that x is the argument of the function and y is the value of the function at x. We also refer to y as the image of the function at x. We call set the domain and set the codomain of the function. lso we call the subset of elements b the range of the function f:. We also write the range of the function as: f() = {b a, b = f(a)} lso the graph of a function f: is the subset of the Cartesian product of the set which we can denote as, {af(a)) a } Each function must have a properly defined domain, codomain and a procedure or formula which associates each point in the domain to an element of the codomain. We can think of this as taking an element from the domain and putting it through a machine which transforms the element into an element in the codomain. Remember that for every element a there exists one element b which means that each element in the set transforms into one and only one element in the codomain. This is illustrated in the following diagram where for each element a in the set there exists a b in set to which it maps to. This is why functions are often referred to as mapping an element in one domain to an element in another domain. a,! b such that b = f(a) f: a b
asic Definitions of Functions Let and be two sets. function f: is a binary relation from Set to Set such that for every a, there is exactly one element b such that b = f(a), (we often just write f(a) = b) : For f: the set is the domain : For f: the set is the codomain The argument of the function b = f(a) is a Value: b is the value of the function, and is also known as the image of the function at a The Range of the function f: f() = {b a, b = f(a)} The graph of the function f: is the subset of the Cartesian product of the set which we denote as {af(a)) a } We can also map functions from a subset of the domain. Let C be a subset of the domain, then the image of C under f:, denoted f(c), is defined as: f(c) = {f(c) c C} Remember that y is the image of the function at x. So when we write f(c) we choose elements that are part of subset C which is a subset of, and because every element of a maps to a point in b so must every point in c. This can also be expressed alternatively as: f(c) = {b c C, b = f(c)} c C,! b such that b = f(c) C c f: b Example 1: and Range of Functions Find the domain and range of the function y = 3 + x 2 The domain will be any real number {x x R}, because there are no restrictions on what can be substituted for x. The range will always be positive because even if the value of x is negative when we square the value of a negative number it will be positive. lso the minimum value y can take on is 3, when x = 0.This means that the range will be any positive real number greater than 3 (y 3).
x y 4 19 3 12 2 7 1 4 0 3 y = 3 + x 2 8 7 6 5 4 y -1 4 3-2 7 2-3 12 1-4 19-5 -4-3 -2-1 1 2 3 4 5 x Example 2: and Range of Functions Find the domain and range of the function y = 3 x For this function we know that the domain will not include all the real numbers because we cannot have 3 x < 0. Therefore the domain must include all real numbers less than or equal to 3, {x x 3}. The range will be all the real numbers greater than or equal to zero, {y y 0}. x y 4 Undefined 3 0 2 1 1 1.41421356 0 1.73205081 y = 3 x 8 7 6 5 4 y -1 2 3-2 2.23606798 2-3 2.44948974 1-4 2.64575131-5 -4-3 -2-1 1 2 3 4 5 x Example 3: and Range of Functions Find the domain and range of the function y = 1 x 3 For this function the domain is all the real numbers except for 3, {x x 33}, because the function is undefined at x = 3. Therefore there is a break point in the graph as this value of x is not part of the domain therefore the domain will be {x x 3}. To find the range we can just take the inverse of the function y 1 = 3 1 x, since the domain of y 1 will be {x x 0}. Therefore the range of the function y = 1 will be {y y 0}. x 3 X Y 7 0.25 6 0.33333333 5 0.5 4 1 3 Not Defined 2-1 1-0.5 0.5-0.4 0.25-0.3636364 0-0.3333333-0.25-0.3076923-0.5-0.2857143-1 -0.25-2 -0.2-3 -0.1666667-4 -0.1428571-5 -0.125-6 -0.1111111-7 -0.1 y = 1 x 3 as x y 1 y as x 3 y as x y 3 x as x 3 y
Properties of Functions Using the function f: where : {a 1, a 2, a 3 } and : {b 1, b 2, b 3 } f: a 1 b 1 a 2 b 2 a 3 b 3 a,! b such that b = f(a) We now turn our attention to the basic properties of functions. These properties are important when thinking about how the functions map a variable from the domain to the codomain. This section is designed to be definitional and as a short overview of the various properties of functions. Injective Functions f:, a 1, a 2, iff a 1 a 2 then f(a 1 ) f(a 2 ) f: a 1 a 2 b 1 = f(a 1 ) b 2 = f(a 2 ) Function f: is injective if and only iff whenever a 1 a 2 it is the case that f(a 1 ) f(a 2 ). This means that for every element a i in the domain must map to only one point in the codomain. nother way to express this is f: is injective iff a 1, a 2, if f(a 1 ) f(a 2 ) then a 1 a 2. So if we have the elements in the codomain that are the outcome of the functions f(a 1 ) and f(a 2 ), when we trace them back to the domain it must be the case that a 1 a 2. We can show the injective property in the following diagram. Note, that if the number of elements in is greater than the number of elements in then the function is not injective. Remember that every point in a must have one and only one arrow going to a single element in where there are no other arrows.
Surjective Functions f:, b, a such that b = f(a) a 1 f: b 1 = f(a 1 ) a 2 a 3 b 2 = f(a 2 ), f(a 3 ) function is surjective iff b, a such that b = f(a). This means that the range of f: coincides with the codomain. Therefore in the diagram every point in the codomain needs to have an arrow going to it from the domain. There can be multiple elements in the domain that map to a single element in. In the diagram below a 2 and a 3 both connect to the element b 2 in the codomain. The only requirement for surjective functions is that every element in has an element in such that b = f(a). Remember that the range f: is the subset of elements b such that a such that b = f(a). Therefore a surjective function means that the range f: must coincide with, and all points in the codomain need an arrow. Note: Whenever the number of elements in domain is less than the elements in the cdomain then the function f: is not surjective. Every point in must have an arrow from the domain and this can only happen when the number of points in is greater than or equal to the number of points in ijective Functions ijective Functions f:, a 1, a 2, iff a 1 a 2 then f(a 1 ) f(a 2 ) f:, b, a such that b = f(a) f: a 2 a 3 a 1 b 1 = f(a 1 ) b 2 = f(a 2 ) b 3 = f(a 3 ) function is bijective if and only if it is both injective and surjective. We know that for a injective function each element in the domain must map to a different point in the codomain. In addition the surjective function says that every element in the codomain must map to at least one element in the domain, (every point in the codomain needs an arrow from the domain). Therefore a bijective function implies each element in will have one arrow arriving. lternatively there is a one to one relationship between the elements in the domain and codomain where every element in domain has one and only one partner element in the codomain.
Composite Functions If f and g are functions then the composite functions g of f denoted (g o f) and f of g denoted (f o g). We write these functions as (g o f)(x) = g f(x) and (f o g)(x) = f(g(x)). The domain of (g o f) is the subset of the domain of f for which g o f is defined. Likewise the domain of (f o g) is the subset of the domain of g for which f o g is defined. lternatively we could say that if there are three non empty sets, and C. Let f: and g: C be two arbitrary functions. We define the composite function, denoted g o f: C, as (g o f)(x) = g f(x), x. Note: For the composite function the order in which we apply each function is relevant. This is demonstrated in example 5 below. Example 5: Composite Functions If f(x) = 2x + 2 and g(x) = x 2 and f: R R and g: R R we can find the composite functions. (g o f)(x) = g f(x) = (2x + 2) 2, x R. However (f o g)(x) = f g(x) = 2x 2 + 2, x R. Therefore (g o f)(x): R R and (f o g)(x): R R are different functions because they differ in at least one point of their domain. For example (g o f)(1) = 25 and (f o g)(1) = 1. This shows that the order in which we apply each function is important. Example 6: Composite Functions If f(x) = x 4 + 2 and g(x) = x 2 then find both (g o f)(x) and (f o g)(x) (g o f)(x) = g f(x) = g(x 4 + 2 )= (x 4 + 2) 2 = x 8 + 4x 4 + 4 (f o g)(x) = f(g(x)) = f(x 2 ) = (x 2 ) 4 + 2 = x 8 + 2 Example 7: Composite Functions If we have the function f(x) = x 2 1, and g(x) = x 2 we can find the following composite functions, (a) (f o f)x = f f(x) = f(x 2 1) = ((x 2 1) 2 1) = x 4 2 (b) (f o g)(x) = f g(x) = f(x 2 ) = (x 2 ) 2 1 = x 4 1 (c) (g o f)(x) = g f(x) = g(x 2 1) = (x 2 1) 2 = x 4 2x 2 + 1 (d) (f g)x = f(x) g(x) = (x 2 1)(x 2 ) = x 4 x 2 (e) (f + g)x = f(x) g(x) = (x 2 1) + (x 2 ) = 2x 2 1
Inverse Functions Inverse Functions f: g f(a) = a, a and f g(b) = b, b g(b) f b f g(b) = b a g f(a) g f(a) = a The functions f: and g: are called inverse functions if and only if g f(a) = a, a, and f g(b) = b, b. So g: is the inverse function of f:. function f: has an inverse if an only if it is bijective. lso a function is bijective if and only if it is injective and surjective at the same time. dditionally every function has at most one inverse function. These claims are proved in the appendix text box labelled the Inverse Function Proofs. Inverse functions should return the composite function back to its original point. This means that g o f: and f o g:. Example 8 Inverse Functions Consider the two functions f and g 1. f: R R, f(x) = x 2 and For the function to be injective every element a i in the domain must map to only one point in the codomain so whenever a 1 a 2 it is the case that f(a 1 ) f(a 2 ), in our example this does not hold. In our example f( 2) = (2 2 ) = 4 is the same as f(2) = (2 2 ) = 4, so we have f(a 1 ) = b = f(a 2 ) and therefore it cannot be injective. The only requirement for surjective functions is that every element in has an element in such that b = f(a). If the codomain is all the real number R then the function f is not surjective either because there are negative numbers in which have no element such that b = f(a). For instance 4 is a number in the codomain which does not have an element in that maps to it by the function f(x). From first examining the first function f it is obvious that f is neither injective nor surjective. Therefore the function is also not bijective and does not have an inverse. 2. g: [0, + ) [0, + ), g(x) = x The second function g does have an inverse and the inverse formula is similar to the definition of the function f. We must take care when defining the domain and codomain of g s inverse function. The function g is injective because whenever a 1 a 2 it is the case that f(a 1 ) f(a 2 ). This is because all the elements in the domain are positive and therefore map to a single element in the codomain. This function is also surjective because every element in codomain has one element in the domain. gain this is because the domain only includes
positive numbers. Considering the above the function g is bijective and has an inverse. Suppose we have the following function: h: [0, + ) [0, + ), h(x) = x 2 This function is bijective and has an inverse which is the function g. We can prove that functions g and h are inverses. Suppose that g: and that h:. We already know that for every b, h g(b) = b. Now choose an arbitrary a. If we let b = g(a) then by definition g(b) is the unique element of such that h g(b) = b. However b = g(a), therefore a = h(b) and a = g(h(a)). This proves that g: and h: are inverse functions. nother way to show this is by writing out the functions explicitly and choosing an arbitrary number for a such as 4 g: [0, + ) [0, + ), g(x) = x h: [0, + ) [0, + ), h(x) = x 2 g(a) = b, if a = 4 then g(a) = g(4) = 4 = 2 h(b) = a, if b = 2 then h(b) = h(2) = 2 2 = 4 h g(b) = b, if b = 2 then h g(b) = 2 2 = 2 1 = 2 g h(a) = a, if a = 4 then g h(a) = (4 2 ) = 4 1 = 4 If we use arbitrary numbers in the domain [0, + ) [0, + ) we will always be able to show that these particular functions are inverse functions. Pre-images, Extensions and Restrictions Previously we said that when we write b = f(a) we call a the argument of the function and b is the image of this function at x. To define the pre-image consider the function f: and let D. The pre-image of the subset D under f: is the subset of the domain, f 1 (D) which is defined by: f 1 (D) = {a d D, d = f(a)} The pre-image of the an element of the domain is denoted as f 1 (b) which is also defined as, f 1 (b) = {a b = f(a)} f:, b, a such that b = f(a) f: a 1 b 1 a 2 a 3 b 2 f 1 (b) = {a b = f(a)} f 1 (b 1 ) = {a 1 }
f 1 (b 2 ) = {a 2, a 3 } f 1 (b 1, b 2 ) = f 1 (b 1 ) f 1 (b 1 ) = If we let C be a nonempty subset of, C and let f: and g: C be two functions such that f(x) = g(x), x C. In this case we say that g: C is the restriction of f: and that f: is the extension of g: C. f:, b, a such that b = f(a) c C,! b such that b = f(c) f: b 1 C c g: C b 2 Restriction: g: C is the restriction of f: Extension: f: is the extension of g: C Example 9: Extension and Restrictions Suppose we have the following functions f: R [0, + ), f(x) = x 2 and g: [0, + ) [0, + ), f(x) = x 2 The function f: R [0, + ) is an extension of g: [0, + ) [0, + ). If we look at the set of non negative numbers then both functions f and g use the same formula. Likewise g: [0, + ) [0, + ) is a restriction of f: R [0, + ) The function g: [0, + ) [0, + ) is an injective function. Remember that a function f: is injective iff a 1, a 2, such that whenever f(a 1 ) f(a 2 ) then a 1 a 2. So if we have the elements in the codomain that are the outcome of the functions f(a 1 ) and f(a 2 ), when we trace them back to the domain it must be the case that a 1 a 2. s the domain and codomain are both non negative then it is the case that x 1, x 2 [0, + ), if g(x 1 ) g(x 2 ) then x 1 x 2. For x 1 x 2 and g(x 1 ) = g(x 2 ) then it must be the case that domain includes negative numbers. Since this is not the case the function must be injective. y substituting numbers in we can see that if we let g(x 1 ) = 4 = g(x 2 ) then because of the restricted domain x 1 = 2 = x 2. On the other hand the function f: R [0, + ) is not injective. This is because x 1, x 2 R, if f(x 1 ) = f(x 2 ) then x 1 x 2. For example f(x 1 ) = f(x 2 ) = 4 but x 1 x 2 because x 1 = 2 and x 2 = +2. Therefore the definition of an injective function is violated.
Definitions and Properties of Functions grument: If b = f(a) we call a the argument of the function Image: If b = f(a) then b is the image of this function at a : Given f: the set is the domain of f : Given f: the set is the codomain of f Range: f: the Range is f() = {b a, b = f(a)} Injective Function: Function f: is injective if and only iff whenever a 1 a 2 it is the case that f(a 1 ) f(a 2 ). f:, a 1, a 2, iff a 1 a 2 then f(a 1 ) f(a 2 ) Surjective Functions: function is surjective iff b, a such that b = f(a) f:, b, a such that b = f(a) ijective Functions: function is bijective iff it is both injective and surjective Composite Functions: Let f: and g: C be two arbitrary functions. We define the composite function, denoted g o f: C, as (g o f)(x) = g f(x), x, and the composite function f o g is denoted as (f o g)(x) = f g(x), x. Inverse Functions: The functions f: and g: are called inverse functions if and only if g f(a) = a, a, and f g(b) = b, b. function f: has an inverse if an only if it is bijective. lso a function is bijective if and only if it is injective and surjective at the same time. dditionally every function has at most one inverse function. Pre-Image: f: and let D. The pre-image of the subset D under f: is the subset of the domain, f 1 (D) which is defined by: f 1 (D) = {a d D, d = f(a)} Restrictions: If we let C be a nonempty subset of, C and let f: and g: C be two functions such that f(x) = g(x), x C. The restriction is g: C is the restriction of f: Extensions: If we let C be a nonempty subset of, C and let f: and g: C be two functions such that f(x) = g(x), x C. The extension is f: is the extension of g: C. Counting Functions The number of functions b a To count the number of functions we just count the number of ways the elements in, can be mapped to the elements in. Since has 3 values we can assign 3 possible value for f(a 1 ), namely, b 1, b 2 and b 3. Likewise the second element a 2 can map through the function f(a 2 ) to three points in namely, b 1, b 2 and b 3. gain f(a 3 ) can be assigned the same three possible
values. s f(a 1 ), f(a 2 ) and f(a 3 ) all have three options or ways they can map to in the codomain there are 3 3 3 = 3 3 = 27 possible functions. This means that we can count the total number of functions as b a, where b represents the number of elements in and a represents the number of elements in. Example 4: Counting Functions How many possible functions are there? f: a1b1 a1b2 a1b3 a2b1 a2b2 a2b3 a 1 b 1 a 2 b 2 b 3 a,! b such that b = f(a) We can see that there are three destinations for f(a 1 ) which are b 1, b 2 and b 3. Likewise there are three destinations for f(a 2 ) which again are b 1, b 2 and b 3. Therefore there b a = 3 2 = 9 functions. nother way to work this out is to line up the different pairs and then draw lines connecting each set of elements with the corresponding sets of elements. Then you just count the number of lines of which there will be 9. Counting the number of injective functions Suppose that : {a 1, a 2, a 3, a 4, a 5 } and : {b 1, b 2, b 3 } and we want t count the number of different injective functions. We know that there are exactly b different ways of choosing the value of the first element in, which is a 1. There are exactly b 1 different ways of choosing the value of the second element in which is a 2. Likewise there are b 2 ways of choosing the third element in. This continues until there are b (a 1) = b a + 1 ways of choosing the value of the a th element of. Therefore by using the multiplication principle, the number of different injective functions f: is simply, b! (b a)! s you may notice, in the case of where the number of elements in is equal to the number of elements in, we just use the multiplication principle b!. This is because say for instance we have : {a 1, a 2, a 3 } and : {b 1, b 2, b 3 } then we know that there are exactly b different ways of choosing the value of the first element in, which is a 1. There are exactly b 1 different ways of choosing the value of the second element in which is a 2. Likewise there are b 2 ways of choosing the third element in. Therefore b! = b (b 1) (b 2), which in this case is 3! = 3 2 1 = 6. n easy way to remember this is that if the sets are equal then we just use the top line of b! (b a)!. In the box below example 5 we work through three scenarios and show how to count the number of injective functions.
Example 5: Counting Injective Functions 1. Suppose we have diagram f: where : {a 1, a 2 } and : b 1, b 2, b 3. First is important to make sure that this function is injective. The number of elements in, a = 2 is less than the number of elements in, b = 3 so we know that the function is injective. s explained above there are exactly b different ways of choosing the value of the first element in, a 1. There are exactly b 1 different ways of choosing the value of the second element in, a 2. Therefore we can use the multiplication principle to and the equation, injective functions. b! = 3! = 3 2 1 = 6. (b a)! 1! 1 b! (b a)! to find the number of different 2. If we have f: X Z where X: {1,2,3} and Z: {1,2,3,4,5} then there are 5! 2! = 5 4 3 = 60 injective functions. 3. If we have h: J Z where J: {1,2,3,4} and Z: {1,2,3,4} then every surjective function f sends each element from X to a different element of Z. Since both sets have the same number of elements the function h: J Z is injective if an only if it is surjective and bijective. This means that overall there must be b! = 4! = 24 injective functions. Counting the number of surjective Functions If the number of elements in # = a equals the number of elements in # = b then the number of surjective functions is equal to the number of injective functions and is simply b!. However when a > b then finding the number of surjective functions is not a trivial question. In fact the answer is complicated and difficult to explain. The number of surjections from a set of a elements to a set of b elements, where a > b is given by, surj(a, b) = ( 1) i b i=0 b (b i)a i If we have : {a 1, a 2, a 3, a 4, a 5, a 6 } so # = a = 6 and : {b 1, b 2, b 3, b 4, b 5 } so # = b = 6 then there are 1800 surjective functions from a total of b a = 15265 functions. gain this is difficult to explain in simple terms. The example below demonstrates some the logic an intuition of counting surjective functions for when a > b. Suppose that # = a and # = b and a > b then the number of functions f: for which at least one element of is not an image is b 1 b 1 (b 1)a b 2 (b 2)a + ( 1) b b = ( 1)i+1 b 1 i=1 b (b i)a i s the number of total functions is = b a, then the number of functions that are surjections are those which every element of is an image, b 1 b a ( 1) i+1 b i (b i)a = ( 1) i i=1 b 1 i=0 b (b i)a i
For the case where the codomain has exactly two elements there is a special rule for counting surjective funtions. If has a elements and has 2 elements then f: is defined by the set f 1 (b 1 ) because any element in belongs to either f 1 (b 1 ) or f 1 (b 2 ). This means that f 1 (b 2 ) = f 1 (b 1 ) and the number of ways we can choose elements of to compose f 1 (b 1 ) is the number of subsets of elements in expect for the empty set and the complete set. This is required because we are requiring functions f: to be surjective. s set has 2 a subsets, the number of surjective functions is 2 a 2. Note: The mathematics used for counting surjective functions is beyond the scope of this text and for more information you may wish to consult a mathematics text for a comprehensive explanation. Example 6: Counting Surjective Functions 1. If we have h: J Z where J: {1,2,3,4} and Z: {1,2,3,4} then every surjective function f sends each element from X to a different element of Z. Since both sets have the same number of elements the function h: J Z is injective if an only if it is surjective and bijective. This means that overall there must be 4! = 24 surjective functions. The number of surjective functions is equal to the number of injective functions when the number of elements in each set are equal. 2. In the above diagram f: where : {a 1, a 2, a 3 } and : {b 1, b 2 } there are surjective functions f that send two elements from to the same element of. ecause the codomain has only 2 elements we can use the formula 2 a 2 which gives 6 different surjective functions. lternatively we could say there are 3 2 = 3 pairs {i, j} of. For a given pair in there are 2! 2! = 2 surjective functions for f such that f(i) = f(j). Therefore there must be 3 2 = 6 surjective functions. 3. If we have f: X Z where X: {1,2,3,4,5} and Z: {1,2,3,4} then every surjective function f sends two elements from X to the same element of Z. Therefore there are 5 4 = 10 possible pairs 2! {i, j}x: {1,2,3,4,5} which can be chosen. For a given pair in X there are 4! = 24 surjective functions for f such that f(i) = f(j). This means that overall there must be 10 24 = 240 surjective functions. Counting the number of bijective functions function is bijective if and only if it is both injective and surjective. To be injective we require the number of elements in the domain # = a to be less than or equal to the number of elements in the codomain # = b.therefore for injective functions we require that # = a # = b. If the number of elements in domain # = a is less than the number of elements in the codomain # = b then the function is not surjective. Therefore for surjective functions we require that # = b # = a. When we put these conditions together # = # or a = b is the requirement for the function to be bijective. This means that when we are counting bijective functions we calculate a b. In fact when the function is bijective the number of bijective functions is equal to the number of surjective and injective functions. If a = b then #bijective functions = #injective functions = #surjective functions = #functions
Example 5: Counting bijective functions If we have h: J Z where J: {1,2,3,4} and Z: {1,2,3,4} both sets have the same number of elements. This means that the function h: J Z is injective and surjective and therefore bijective. This means that overall there must be 4! = 24 bijective, injective, surjective functions. The number of bijective functions is equal to the number of injective functions and surjective functions when the number of elements in each set are equal. Example 10: Counting Functions when # < # f: a 1 b 1 a 2 b 2 b 3 a,! b such that b = f(a) For the function depicted in the above diagram: 1. Count the number of possible functions f: 2. Count the number of injective functions f: 3. Count the number of non injective functions f: 4. Count the number of surjective functions f: 1. b a = 3 2 =9 functions. 2. We can count the injective functions by seeing that there are three elements in for f(a 1 ) to map to. s f(a 1 ) maps to one element in this leaves two options for f(a 1 ) to map to. For example f(a 1 ) can map to any of the three points in so if f(a 1 ) = b 1 then f(a 2 ) can be either b 1 or b 2. Therefore f(a 1 ) has 3 choices and f(a 2 ) has 3 1 choices, and therefore there b! are = 3 2 1 = 6 injective functions. (b a)! 1 3. There are 3 non injective functions which are where both f(a 1 ) and f(a 2 ) map to b 1, both f(a 1 ) and f(a 2 ) map to b 2 and both f(a 1 ) and f(a 2 ) map to b 3 and a 1 a 2. So we can just count the number of total functions 9 and subtract the number of injective functions 6, which equals 3. 4. Whenever the number of elements in codomain b is greater than the elements in the domain a then the function f: is not surjective. In other words not every element in has a partner or originator in. This means that the function has b > a and is not surjective.
Example 11: Counting Functions when # > # f: a 1 b 1 a 2 b 2 a 3 a,! b such that b = f(a) For the function depicted in the above diagram: 1. Count the number of possible functions f: 2. Count the number of injective functions f: 3. Count the number of non injective functions f: 4. Count the number of surjective functions f: 1. b a = 2 3 = 2 2 2 = 8. 2. a > b and therefore the function f: is not injective. If the number of elements in is greater than the number of elements in then the function is not injective. For injective functions a 1 a 2 and f(a 1 ) f(a 2 ), if f(a 1 ) = b 1 then we cannot have f(a 2 ) = b 1 = f(a 3 ) because a 2 a 3. This violates the property of an injective function. This means that two of the elements in the domain must map to the same point in and this cannot happen if the function is injective. Hence there are no injective functions. 3. s there are no injective functions it must be that every possible function is a non injective function. Therefore there are 8 non injective functions. We could also list the non injective functions as: b 1 = f(a 1 ) =f(a 2 ), b 2 = f(a 1 ) =f(a 2 ) where a 1 a 2 b 1 = f(a 1 ) =f(a 3 ), b 2 = f(a 1 ) =f(a 3 ) where a 1 a 3 b 1 = f(a 2 ) =f(a 3 ), b 2 = f(a 2 ) =f(a 3 ) where a 2 a 3 b 1 = f(a 1 ) = f(a 2 ) =f(a 3 ), b 2 = f(a 1 ) = f(a 2 ) =f(a 3 ) where a 1 a 2 a 3 gain there are 8 non injective functions which is the number all possible functions. 4. a > b therefore the function is surjective. s there are 2 elements in we can use the formula 2 a 2, which gives us 6 different surjective functions. nother way of working out the surjective is to count the number of pair combinations of. In this case there are 3 2 = 3 pairs possible pairs {i, j}x: {a 1, a 2, a 3 } which can be chosen. For a given pair in X there are 2! = 2 surjective functions for f such that f(i) = f(j). This means that overall there must be 3 2 = 6 surjective functions. 2!
Implicit Functions In our examination of functions we have always presented functions where the dependent variable are explicit functions of the independent variable. In other words all the x's have been on the right hand side and y on the left hand side. Implicit functions have both the independent and dependent variables on the right hand side. Some implicit functions can be converted by simply solving for y but this is not always possible. However we often are still interested in how a one unit change in one of the independent variables affects the value of the dependent variable. To do this we use implicit differentiation. Some examples of implicit functions are below. Examples of Implicit Functions 4x + 2xy 5 = 0 8x + 5y 20 = 0 12y 4 x 2 + 34 = 0 y 2 + 3xy + x 3 = 0 Sometimes we can easily express the implicit function explicitly. For instance can be expressed as 10x + 5y 20 = 0 5y = 20 10x Note that sometimes when we transform implicit functions to be explicit functions they can be more difficult to work with. For further discussion on the implicit differentiation refer to the document Single Variable Calculus.
ppendix Proofs for Inverse Functions 1. Function has a inverse if and only if it is bijective Proof that the function is injective: If f: has an inverse, say g: we claim that f must be injective. However suppose that there are two distinct elements a 1, a 2, a 1 a 2 such that f(a 1 ) = f(a 2 ), therefore b = f(a 1 ) = f(a 2 ). Since f and g are inverse functions then a 1 = g f(a 1 ). This means that b = f(a 1 ) and that a 1 = g(b). However since f and g are inverse functions then a 2 = g f(a 2 ). This means that b = f(a 2 ) and that a 2 = g(b), but this is an absurd because a 1 and a 2 are not equal and therefore they cannot both be equal to g(b). In other words we cannot have a 1 a 2 and a 1 = g(b) = a 2. This proves that the function f is injective. Proof that the function is surjective: Let b be arbitrary. Let a be such that a = g(b). s f and g are inverse functions, then f(a) = f g(b) = b. ecause b is randomly chosen we have proved that the function f is surjective. lso since f: is surjective and injective it must be the case that the function is also bijective. Proof that the functions f: and g: are inverses: We already know that for every b, f g(b) = b. Now choose an arbitrary a. If we let b = f(a) then by definition g(b) is the g(b) is the unique element of such that f g(b) = b. However b = f(a), therefore a = g(b) and a = g(f(a)). This proves that f: and g: are inverse functions. 2. Proving the uniqueness of the inverse function Proof that every function has at most one inverse function: Suppose that we have a bijective function f: which has two inverses g: and h:. For any given y there exists an element x such that y = f(x). Therefore g(y) = g f(x) = x = h f(x) = h(y). Thus g(y) = h(y), y. This proves that g: and h: are identical.