Functions

Functions A function f from a set A to a set B is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. If f is a function from A to B, we write f: AB (note: Here, ““ has nothing to do with if… then)

Functions If f:AB, we say that A is the domain of f and B is the codomain of f. If f(a) = b, we say that b is the image of a and a is the pre-image of b. The range of f:AB is the set of all images of elements of A. We say that f:AB maps A to B.

Function terminology f maps R to Z R f Z Domain Co-domain f(4.3) 4.3 4 Pre-image of 4 Image of 4.3

A string length function More functions The image of A A pre-image of 1 Domain Co-domain A B C D F Alice Bob Chris Dave Emma A class grade function 1 2 3 4 5 “a” “bb“ “cccc” “dd” “e” A string length function

Functions Let us take a look at the function f:PC with P = {Linda, Max, Kathy, Peter} C = {Boston, New York, Hong Kong, Moscow} f(Linda) = Moscow f(Max) = Boston f(Kathy) = Hong Kong f(Peter) = New York Here, the range of f is C.

Functions Other ways to represent f: x f(x) Linda Moscow Max Boston Kathy Hong Kong Peter Linda Max Kathy Peter Boston New York Hong Kong Moscow

Operations on Functions Let f1 and f2 be functions from A to R. Then the sum and the product of f1 and f2 are also functions from A to R defined by: (f1 + f2)(x) = f1(x) + f2(x) (f1f2)(x) = f1(x) f2(x) Example: f1(x) = 3x, f2(x) = x + 5 (f1 + f2)(x) = f1(x) + f2(x) = 3x + x + 5 = 4x + 5 (f1f2)(x) = f1(x) f2(x) = 3x (x + 5) = 3x2 + 15x

Functions We already know that the range of a function f:AB is the set of all images of elements aA. If we only regard a subset SA, the set of all images of elements sS is called the image of S. We denote the image of S by f(S): f(S) = {f(s) | sS}

Functions Let us look at the following well-known function: f(Linda) = Moscow f(Max) = Boston f(Kathy) = Hong Kong f(Peter) = Boston What is the image of S = {Linda, Max} ? f(S) = {Moscow, Boston} What is the image of S = {Max, Peter} ? f(S) = {Boston}

One-to-one functions A function is one-to-one if each element in the co-domain has a unique pre-image Meaning no 2 values map to the same result 1 2 3 4 5 a e i o A one-to-one function 1 2 3 4 5 a e i o A function that is not one-to-one

More on one-to-one Injective is synonymous with one-to-one “A function is injective” A function is an injection if it is one-to-one Note that there can be un-used elements in the co-domain 1 2 3 4 5 a e i o A one-to-one function

A function that is not onto Onto functions A function is onto if each element in the co-domain is an image of some pre-image Meaning all elements in the right are mapped to 1 2 3 4 a e i o u An onto function 1 2 3 4 5 a e i o A function that is not onto

More on onto Surjective is synonymous with onto “A function is surjective” A function is an surjection if it is onto Note that there can be multiply used elements in the co-domain 1 2 3 4 a e i o u An onto function

Onto vs. one-to-one Are the following functions onto, one-to-one, both, or neither? 1 2 3 4 a b c 1 2 3 4 a b c d 1 2 3 4 a b c 1-to-1, not onto Both 1-to-1 and onto Not a valid function 1 2 3 a b c d 1 2 3 4 a b c d Onto, not 1-to-1 Neither 1-to-1 nor onto

Bijections Consider a function that is both one-to-one and onto: 1 2 3 4 a b c d Consider a function that is both one-to-one and onto: Such a function is a one-to-one correspondence, or a bijection

Properties of Functions Linda Max Kathy Peter Boston New York Hong Kong Moscow Is f injective? No. Is f surjective? Is f bijective?

Properties of Functions Linda Max Kathy Peter Boston New York Hong Kong Moscow Is f injective? No. Is f surjective? Yes. Is f bijective? Paul

Properties of Functions Linda Max Kathy Peter Boston New York Hong Kong Moscow Lübeck Is f injective? Yes. Is f surjective? No. Is f bijective?

Properties of Functions Linda Max Kathy Peter Boston New York Hong Kong Moscow Lübeck Is f injective? No! f is not even a function!

Properties of Functions Linda Boston Is f injective? Yes. Is f surjective? Is f bijective? Max New York Kathy Hong Kong Peter Moscow Helena Lübeck

Identity functions A function such that the image and the pre-image are ALWAYS equal f(x) = 1*x f(x) = x + 0 The domain and the co-domain must be the same set

Inverse functions Let f(x) = 2*x R f R f-1 f(4.3) 4.3 8.6 f-1(8.6) Then f-1(x) = x/2

Inversion An interesting property of bijections is that they have an inverse function. The inverse function of the bijection f:AB is the function f-1:BA with f-1(b) = a whenever f(a) = b.

Inversion Linda Boston f Max New York f-1 Kathy Hong Kong Peter Moscow Helena Lübeck

Inversion Example: f(Linda) = Moscow f(Max) = Boston f(Kathy) = Hong Kong f(Peter) = Lübeck f(Helena) = New York Clearly, f is bijective. The inverse function f-1 is given by: f-1(Moscow) = Linda f-1(Boston) = Max f-1(Hong Kong) = Kathy f-1(Lübeck) = Peter f-1(New York) = Helena Inversion is only possible for bijections (= invertible functions)

Inversion Linda Boston f Max New York f-1 f-1:CP is no function, because it is not defined for all elements of C and assigns two images to the pre-image New York. Kathy Hong Kong Peter Moscow Helena Lübeck

Compositions of functions Let (f ○ g)(x) = f(g(x)) Let f(x) = 2x+3 Let g(x) = 3x+2 g(1) = 5, f(5) = 13 Thus, (f ○ g)(1) = f(g(1)) = 13

Compositions of functions f ○ g A B C g f g(a) f(a) a f(g(a)) g(a) (f ○ g)(a)

Compositions of functions Let f(x) = 2x+3 Let g(x) = 3x+2 f ○ g R R R g f g(1) f(5) f(g(1))=13 1 g(1)=5 (f ○ g)(1) f(g(x)) = 2(3x+2)+3 = 6x+7

Compositions of functions Does f(g(x)) = g(f(x))? Let f(x) = 2x+3 Let g(x) = 3x+2 f(g(x)) = 2(3x+2)+3 = 6x+7 g(f(x)) = 3(2x+3)+2 = 6x+11 Function composition is not commutative! Not equal!

Graphs of functions Let f(x)=2x+1 Plot (x, f(x)) This is a plot of f(x) f(x)=5 x=2

Useful functions Floor: x means take the greatest integer less than or equal to the number Ceiling: x means take the lowest integer greater than or equal to the number round(x) = floor(x+0.5)

Floor and Ceiling Functions The floor and ceiling functions map the real numbers onto the integers (RZ). The floor function assigns to rR the largest zZ with zr, denoted by r. Examples: 2.3 = 2, 2 = 2, 0.5 = 0, -3.5 = -4 The ceiling function assigns to rR the smallest zZ with zr, denoted by r. Examples: 2.3 = 3, 2 = 2, 0.5 = 1, -3.5 = -3

Sample floor/ceiling questions Find these values 1.1 1.1 -0.1 -0.1 2.99 -2.99 ½+½ ½ + ½ + ½ 1 2 -1 3 -2 ½+1 = 3/2 = 1 0 + 1 + ½ = 3/2 = 2

Ceiling and floor properties Let n be an integer (1a) x = n if and only if n ≤ x < n+1 (1b) x = n if and only if n-1 < x ≤ n (1c) x = n if and only if x-1 < n ≤ x (1d) x = n if and only if x ≤ n < x+1 (2) x-1 < x ≤ x ≤ = x < x+1 (3a) -x = - x (3b) -x = - x (4a) x+n = x+n (4b) x+n = x+n

Ceiling property proof Prove rule 4a: x+n = x+n Where n is an integer Will use rule 1a: x = n if and only if n ≤ x < n+1 Direct proof! Let m = x Thus, m ≤ x < m+1 (by rule 1a) Add n to both sides: m+n ≤ x+n < m+n+1 By rule 4a, m+n = x+n Since m = x, m+n also equals x+n Thus, x+n = m+n = x+n

Factorial Factorial is denoted by n! n! = n * (n-1) * (n-2) * … * 2 * 1 Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 Note that 0! is defined to equal 1

Proving function problems Let f be a function from A to B, and let S and T be subsets of A. Show that

Proving function problems f(SUT) = f(S) U f(T) Will show that each side is a subset of the other Two cases! Show that f(SUT)  f(S) U f(T) Let b  f(SUT). Thus, b=f(a) for some aS U T Either aS, in which case bf(S) Or aT, in which case bf(T) Thus, bf(S) U f(T) Show that f(S) U f(T)  f(S U T) Let b  f(S) U f(T) Either b  f(S) or b  f(T) (or both!) Thus, b = f(a) for some a  S or some a  T In either case, b = f(a) for some a  S U T

Proving function problems f(S∩T)  f(S) ∩ f(T) Let b  f(S∩T). Then b = f(a) for some a  S∩T This implies that a  S and a  T Thus, b  f(S) and b  f(T) Therefore, b  f(S) ∩ f(T)

Proving function problems Let f be an invertible function from Y to Z Let g be an invertible function from X to Y Show that the inverse of f○g is: (f○g)-1 = g-1 ○ f-1

Proving function problems Thus, we want to show, for all zZ and xX The second equality is similar

Graphs The graph of a function f:AB is the set of ordered pairs {(a, b) | aA and f(a) = b}. The graph is a subset of AB that can be used to visualize f in a two-dimensional coordinate system.