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Discrete Mathematics I Lectures Chapter 7. Some material adapted from lecture notes provided by Dr. Chungsim Han and Dr. Sam Lomonaco. Dr. Adam Anthony Spring 2011. Functions. A function f from a set A to a set B is an assignment of elements in A to elements in B such that

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## Discrete Mathematics I Lectures Chapter 7

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**Discrete Mathematics ILectures Chapter 7**Some material adapted from lecture notes provided by Dr. Chungsim Han and Dr. Sam Lomonaco Dr. Adam Anthony Spring 2011**Functions**• A function f from a set A to a set B is an assignmentof elements in A to elements in B such that • Every item in A is assigned to something in B • Every item in A has only one assignment in B • 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: AB • (note: Here, ““ has nothing to do with if… then) Any function lacking these properties is not well-defined**Functions**• If f:AB, 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 an inverse imageof b. • The Complete inverse image of b B is the set {a A | f(a) = b} • there can be multiple items such that f(a) = b • The range of f:AB is the set of all images of elements of A. • We say that f:AB maps A to B.**Functions**• Let us take a look at the function f:PC 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**• Let us re-specify f as follows: • f(Linda) = Moscow • f(Max) = Boston • f(Kathy) = Hong Kong • f(Peter) = Boston • Is f still a function? yes What is its range? {Moscow, Boston, Hong Kong}**x**f(x) Linda Boston Linda Moscow Max New York Max Boston Kathy Hong Kong Kathy Hong Kong Peter Moscow Peter Boston Functions • Other ways to represent f:**Functions**• If the domain of our function f is large, it is convenient to specify f with a formula, e.g.: • f:RR • f(x) = 2x • This leads to: • f(1) = 2 • f(3) = 6 • f(-3) = -6 • …**Exercise 1**X Y 1 A 2 B 3 C 4 • Is f a function? • What are the domain and codomain of f? • Write f as a set of ordered pairs • What is the range of f? • Is 1 an inverse image of a? • Is 2 an inverse image of b? • What is the inverse image of a? • What is the inverse image of b? • What is the inverse image of c? f**Exercise 2—Is it a function?**1 2 3 2 4 No—2 is related to more than 1 element**Exercise 2—Is it a function?**1 2 3 2 4 Yes—Every element on the left is related to exactly one element on the right**Exercise 2—Is it a function?**1 2 3 2 4 No—2 is not related to anything.**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**Example Function: Floor and Ceiling**• floor(x) = x • Largest integer y such that y x • ceiling(x) = x • Smallest integer y such that y x • Useful observation—for any real number x: • floor(x) x ceiling(x)**Exercise 3**• Data stored on a computer disk or transmitted over a network is typically represented as a string of bytes. Each byte is 8 bits. How many bytes are required to encode 100 bits of data? • In asynchronous transfer mode (a communications protocol used on backbone networks), data are organized into cells of 53 bytes. How many ATM cells can be transmitted in 1 minute over a connection that transmits data at the rate of 500 kilobits per second?**Exercise 4**• Use floor and ceiling to write a new definition for n div d and n mod d**Function Example: Logarithms**• Let b 1 be a positive real number. The logarithm with base b is the logarithmic function • logb : R+ R and is defined by: • logbx = the exponent to which b must be raised to obtain x. • Logbx = y by = x**Exercise 5**• Compute the following: • log553 • log91 • log44 • log24 • log28 • log2(1/2) • log2(1/16)**Useful Logarithm Properties**• logb(xy) = logb(bsbt) = logb(bs+t) = s + t= logbbs+ logbbt= logbx + logby • Similarly, logb(x/y) = logbx – logby • Base conversion:**About log2**• log2 comes up all the time in computing • Basic fundamental problem: Given N items, divide them into two groups. How many times can you divide the groups in half until there is at most one item in each group? • Another way of thinking of it: how many times can you divide an integer by 2 before you reach 0? • Any time you see this pattern (and we will!) remember that there is a logarithmic relationship here!**2-dimensional functions**• We can use cartesian products to define a function in multiple dimensions • The function f: A x B C maps pairs of elements (a,b) to an element in C • A binary operation is defined using a single set: f:A x A Asum:Z x Z Z where sum(a,b) = a + b • A unary operation is similar, but does not use a cartesian product: f: A A**Exercise 6**• Boolean Functions • Given the set B = {T,F}, we can define a boolean function with 2 inputs as: P:B x B BWhere the definition is given by a truth table: • Draw an arrow diagram for P**Section 7.2**One-to-one and Onto functions.**Function Images**• We already know that the range of a function f:AB is the set of all images of elements aA. • If we only regard a subset SA, the set of all images of elements sS is called the image of S. • We denote the image of S by f(S): f(S) = {f(s) | sS}**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}**Properties of Functions**• A function f:AB is said to be one-to-one (or injective), if and only if • x, yA (f(x) = f(y) x = y) • In other words: f is one-to-one if and only if it does not map two distinct elements of A onto the same element of B.**Properties of Functions**g(Linda) = Moscow g(Max) = Boston g(Kathy) = Hong Kong g(Peter) = New York Is g one-to-one? Yes, each element is assigned a unique element of the image. • And again… • f(Linda) = Moscow • f(Max) = Boston • f(Kathy) = Hong Kong • f(Peter) = Boston • Is f one-to-one? • No, Max and Peter are mapped onto the same element of the image.**Properties of Functions**• How can we prove that a function f is one-to-one? • Whenever you want to prove something, first take a look at the relevant definition(s): • x, yA (f(x) = f(y) x = y) • Example: • f:RR • f(x) = x2 • Disproof by counterexample: • f(3) = f(-3), but 3 -3, so f is not one-to-one.**Properties of Functions**• … and yet another example: • f:RR • f(x) = 3x • One-to-one: x, yA (f(x) = f(y) x = y) • To show: f(x) f(y) whenever x y • x y • 3x 3y • f(x) f(y), so if x y, then f(x) f(y), that is, f is one-to-one.**Properties of Functions**• A function f:AB with A,B R is called strictly increasing, if • x,yA (x < y f(x) < f(y)), • and strictly decreasing, if • x,yA (x < y f(x) > f(y)). • Obviously, a function that is either strictly increasing or strictly decreasing is one-to-one.**Properties of Functions**• A function f:AB is called onto, or surjective, if and only if for every element bB there is an element aA with f(a) = b. • In other words, f is onto if and only if its range is its entire codomain. • A function f: AB is a one-to-one correspondence, or a bijection, if and only if it is both one-to-one and onto. • Obviously, if f is a bijection and A and B are finite sets, then |A| = |B|.**Properties of Functions**• Examples: • In the following examples, we use the arrow representation to illustrate functions f:AB. • In each example, the complete sets A and B are shown.**Linda**Boston Max New York Kathy Hong Kong Peter Moscow Properties of Functions • Is f injective? • No. • Is f surjective? • No. • Is f bijective? • No.**Linda**Boston Max New York Kathy Hong Kong Peter Moscow Properties of Functions • Is f injective? • No. • Is f surjective? • Yes. • Is f bijective? • No. Paul**Linda**Boston Max New York Kathy Hong Kong Peter Moscow Lübeck Properties of Functions • Is f injective? • Yes. • Is f surjective? • No. • Is f bijective? • No.**Linda**Boston Max New York Kathy Hong Kong Peter Moscow Lübeck Properties of Functions • Is f injective? • No! f is not evena function!**Properties of Functions**• Is f injective? • Yes. • Is f surjective? • Yes. • Is f bijective? • Yes. Linda Boston Max New York Kathy Hong Kong Peter Moscow Helena Lübeck**Exercise 1—one to one and/or onto?**• Let S by the set of all strings of 0’s and 1’s of any length. Let l:SZnonneg be the length function where l(s) = length of s. • The floor function • h:RR where h(x) = x2**Exercise 2**• Let g:RR be the function defined by: g(x) = 4x3 – 5for any real x. Prove that g is a one-to-one correspondence. • Two steps: prove it is one to one, then prove it is onto**Exercise 3**• Show that the exponential function E:RR+ given by E(x) = 2x is a one-to-one correspondence**Inversion**• An interesting property of bijections is that they have an inverse function. • The inverse functionof the bijection f:AB is the function f-1:BA with • f-1(b) = a f(a) = b, for all b B**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)**f**f-1 Inversion Linda Boston • f-1:CP is no function, because it is not defined for all elements of C and assigns two images to the inverse image New York. Max New York Kathy Hong Kong Peter Moscow Helena Lübeck**Example 4**• Find the inverse function of the exponential function E:RR+ given by E(x) = 2x**Example 5**• Find the inverse of g:RR, defined by: g(x) = 4x3 – 5for any real x.**Section 7.3**Composition of Functions**Composition**• The composition of two functions g:AB and f:BC, denoted by fg, is defined by • (fg)(a) = f(g(a)) • This means that • first, function g is applied to element aA, mapping it onto an element of B, • then, function f is applied to this element of B, mapping it onto an element of C. • Therefore, the composite function maps from A to C.**Composition**• Example: • f(x) = 7x – 4, g(x) = 3x, • f:RR, g:RR • (fg)(5) = f(g(5)) = f(15) = 105 – 4 = 101 • (fg)(x) = f(g(x)) = f(3x) = 21x - 4**Exercise 1**• Let X = {a,b,c} and Y = {1,2,3,4}. f:XY and g:YX are functions defined by: • f(a) = 2, f(b) = 1, f(c) = 3 • g(1) = c, g(2) = b, g(3) = a, g(4) = b • Draw a combined arrow diagram to represent f, g and gf • What are the domain, codomain and range of gf? • Is gf a bijection (1-1 and onto)? If so, find its inverse function.**Exercise 2**• Repeat Exercise 1, but now use f g.**Exercise 3**• Let f:RR and g:RR be defined by: f(x) = 2x g(x) = x2 + 1 for all reals x. • Find g f and f g. • Is g f = f g?

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