CS100: DISCRETE STRUCTURES

Transcription

1 1 CS100: DISCRETE STRUCTURES Computer Science Department Lecture 2: Functions, Sequences, and Sums Ch2.3, Ch2.4

2 2.3 Function introduction : 2 v Function: task, subroutine, procedure, method, mapping, v E.g. Find the grades of student x. char findgrades(string name){ } //go to grades array, //find the name, and find the corresponding grades return grades; 17-Feb-17 Computer Science Department

3 2.3 Functions 3 DEFINITION 1 Let A and B to be nonempty sets. a function f from A to 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. We can use a formula or a computer program to define a function. Example: f(x) = x + 1 described as: int increasebyone(int x){ x = x + 1; return x; }

4 2.3 Functions 4 DEFINITION 2 If f is a function from A to B, we say that A is the domain of f and B is the Co-domain of f. If f(a) = b, we say that b is the image of a and a is a preimage of b. The range of f is the set of all images of elements of A. Also, if f is a function from A to B, we say that f maps A to B. v For each function, we specify its domain, codomain and the mapping of elements of the domain to elements in the codomain v Two functions are said to be equal if they have the same domain, codomain, and the map elements of their common domain to the same elements of their common codomain v A function is differ by changing its domain, codomain or the mapping of elements

5 Exercise.. 5 What are the domain, codomain, and range of the function that assigns grades to students described in the slide 2? Solution: domain: {Adams, Chou, Goodfriend, Rodriguez, Stevens} codomain: {A, B, C, D, F} range: {A, B, C, F}

6 2.3 Functions - Example 6 v ü Let f be the function that assigns the last two bits of a bit string of length 2 or greater to that string. For example, f (11010) = 10. What is the domain, codomain and range of the function? Solution: the domain of f is the set of all bit strings of length 2 or greater.. and both the codomain and range are the set {00,01,10,11}..

7 2.3 Functions - Example 7 ü What is the domain and codomain of the function : int floor(real float){ }? Solution: domain: the set of real numbers codomain: the set of integer numbers

8 2.3 Functions 8 DEFINITION 3 If f 1 and f 2 be functions from A to R. Then f 1 + f 2 and f 1 f 2 are also functions from A to R defined by (f 1 + f 2 )(x) = f 1 (x) + f 2 (x) (f 1 f 2 ) (x) = f 1 (x) f 2 (x) v Example: Let f 1 and f 2 be functions from R to R such that f 1 (x) =x 2 and f 2 (x) = x x 2. v What are the functions f 1 + f 2 and f 1 f 2? Solution: (f 1 + f 2 )(x) = f 1 (x) + f 2 (x) = x 2 + (x x 2 ) = x (f 1 f 2 ) (x) = f 1 (x) f 2 (x) = x 2 (x x 2 ) = x 3 x 4

9 2.3 Functions One-to-One and Onto Functions 9 DEFINITION 5 A function f is said to be one-to-one, or injective, if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. A function is said to be an injection if it is one-to-one. (every element in the range is a unique image for element of A all image have at most one arrow or none) " a, b(a b f(a) f(b)) (If it s a different element, it should map to a different value.) Example: Determine whether the function f from {a,b,c,d} to {1,2,3,4,5} with f(a) = 4, f(b) = 5, f(c) = 1 and f(d) = 3 is one-to-one. Solution: Yes.

10 2.3 Functions 10 DEFINITION 7 A function f from A to B is called onto, or surjective, if and only if for every element b ÎB there is an element aîa with f(a) = b. a function f is called a surjection if it is onto. Co-domain = range Example: Let f be the function from {a,b,c,d} to {1,2,3} defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? Solution: Yes. Example: Is the function f (x) = x 2 from the set of integers to the set of integers onto? Solution: No. There is no integer x with x 2 = -1, for instance.

11 2.3 Functions 11 DEFINITION 8 The function f is a one-to-one correspondence or a bijection, if it is both oneto-one and onto. One-to-One Not onto Onto Not One-to-One One-to-One And onto bijection Not One-to-One Not onto (Neither) Not Function

12 Exercise.. 12 Determine whether each of these functions is a bijection from R to R. a) f(x) = 2x + 1 b) f(x) = x Yes No c) f(x) = x 3 Yes d) f(x) = (x 2 + I )/(x 2 + 2) No 17-Feb-17 Computer Science Department

13 Exercise.. 13 Why f is not a function from R to R? f(x)= f(0) is not defined f(x)= f(x) is not defined for x<0 f(x)= f(x) is not function because there are two values assigned to each x. 17-Feb-17 Computer Science Department

14 2.3 Functions 14 DEFINITION 9 Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a) = b. The inverse function of f is denoted by f -1. Hence, f -1 (b)=a when f(a) = b. A one-to-one correspondence is called invertible because we can define an inverse of this function. A function is not invertible if it is not a one-to-one correspondence because the inverse of such function does not exist.

15 2.3 Functions 15 Example: Let f be the function from {a, b, c} to {1, 2, 3} such that f(a) =2, f(b) = 3 and f(c) = 1. Is f invertible? And if it is, what is its inverse? Solution: Yes, it is invertible, because it is a one-to-one correspondence. f -1 (1) = c, f -1 (2) = a, f -1 (3) = b. Exercise: Let f : Z àz be such that f(x) = x+1. Is f invertible? And if it is, what is its inverse? Solution: Yes, it is invertible, because it is a one-to-one correspondence. f -1 (y) = y-1. Exercise: Let f be a function from R to R with f(x) = x 2. Is f invertible? Solution: f is not one-to-one because f(-2) = f(2) = 4. So, it is not one-toone correspondence and hence it is not invertible.

16 2.3 Functions 16 DEFINITION 10 Let g be a function from the set A to the set B, and let f be a function from the set B to the set C. The composition of the functions f and g, denoted by f o g, is defined by: (f o g)(a) = f(g(a))

17 2.3 Functions 17 o f o g is the function that assigns to the element a of A the element assigned by f to g(a). That is, to find (f o g )(a) we: 1. Apply the function g to a to obtain g(a) 2. Then we apply the function f to the result g(a) to obtain (f o g)(a) = f(g(a)) o NOTE: The composition f o g cannot be defined unless the range of g is a subset of the domain of f.

18 2.3 Functions 18 Example: Let g be the function from the set {a, b, c} to itself such that g(a) = b, g(b) = c, and g(c) = a Let f be the function from the set {a, b, c} to the set { l, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1 What is the composition of f and g, and what is the composition of g and f? Solution: The composition f o g is defined by: (f o g)(a) = f(g(a)) = f(b) =2 (f o g)(b) = f(g(b)) = f(c) =1 (f o g)(c) = f(g(c)) = f(a) = 3 g o f is not defined, because the range of f is not a subset of the domain of g

19 2.3 Functions 19 Example: Let f and g be the functions from the set of integers to the set of integers defined by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the composition of g and f? Solution: Both the compositions f o g and g o f are defined (f o g)(x) = f(g(x)) = f(3x + 2) = 2(3x + 2) + 3 = 6x + 7 (g 0 f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 2 = 6x + 11.

20 20 v 2.4 Sequences and Summations Sequences A sequence is a discrete structure used to represent an ordered list v Example: 1,2,3,5,8 1,3,9,27,81,,30, DEFINITION 1 a sequence is a function from a subset of the set of integers (usually either the set {0,1,2, } or the set {1,2,3, }) to a set S. We use the notation a n to denote the image of the integer n. We call a n a term of the sequence. We use the notation {a n } to denote the sequence. Example: Consider the sequence {a n }, where a n = 1/n. The list of the terms of this sequence, beginning with a 1, namely a 1, a 2, a 3, a 4,, starts with 1, 1/2, 1/3, 1/4,

21 2.4 Sequences and Summations 21 DEFINITION 2 a geometric progression is a sequence of the form a, ar, ar 2,, ar n, where the initial term a and the common ratio r are real numbers. Example: The following sequence are geometric progressions. {b n } with b n = (-1) n starts with 1, -1, 1, -1, 1, initial term: 1, common ratio: -1 {c n } with c n = 2*5 n starts with 2, 10, 50, 250, 1250, initial term: 2, common ratio: 5 {d n } with d n = 6 *(1/3) n starts with 6,2, 2/3, 2/9, 2/27, initial term: 6, common ratio: 1/3

22 2.4 Sequences and Summations 22 DEFINITION 3 A arithmetic progression is a sequence of the form a, a + d, a + 2d,, a + nd, where the initial term a and the common difference d are real numbers. Example: The following sequence are arithmetic progressions. {s n } with s n = n starts with -1, 3, 7, 11, initial term: -1, common difference: 4 {t n } with t n = 7 3n starts with 7, 4, 1, -2, initial term: 7, common difference: -3

23 2.4 Sequences and Summations 23 Example : Find formulae for the sequences with the following first five terms (a). 1, 1/2, 1/4, 1/8, 1/16 Solution: a n = 1/2 n (b). 1, 3, 5, 7, 9 Solution: a n = (2n )+ 1 (c). 1, -1, 1, -1, 1 Solution: a n = (-1) n

24 Exercises 24 Find the formula of this sequence and find A6, A8 : A = 10, 14, 18, 22, 26 It is arithmetic sequence, initial term a= 20, common difference d= 4, a n =10+4n A 6 = 34, A 8 = 42

25 Exercises 25 A = 4, 8, 16, 32 It is geometric sequence, initial term = 4, common ratio = 2, a n =4*2 n A 6 = 256, A 8 = 1024

26 Exercises 26 o in a geometric sequence a 0 was =3 and r =1/2 then a 3 is equal to a. 3/16 b. 3/4 c. 3/2 d. 3/8 In an arithmetic sequence a 0 was =7 and a 3 was 19 the value of a 2 is= a.14 b.11 c.15 d.23

27 2.4 Sequences and Summations Summations 27 v The sum of the terms from the sequence a m + a m+1,, a n can be expressed as n å j= m n a j å j = m, Or a å j 1 j n a j Where m is the lower limit, n is the upper limit, and j is the index of the summation Example: Express the sum of the first 100 terms of the sequence {a n }, where an = 1/n for n = 1,2,3,. Solution: 100 å j= 1 1 j

28 2.4 Sequences and Summations 28 5 å = What is the value of j 1? Solution: 5 å j = 1 j 2 = = 55 j 2 Expressed with a for loop: int sum = 0; for (int i =1; i <=5; i++) { sum = sum + i * i ; }

29 2.4 Sequences and Summations 29 What is the value of the double summation ij? Solution: 4 3 åå i= 1 j= 1 ij = 4 å i= 1 4 å ( i + 2i + 3i) = 6i = = 60 i= åå i= 1 j= 1

30 Sequences and Summations 30 Expressed with two for loops: Solution: int sum1 = 0; int sum2 = 0; for (int i =1; i <=4; i ++){ sum2 = 0; for (int j=1; j<=3; j++){ sum2 = sum2 + i *j; } sum1 = sum1 + sum2; }

31 Exercises 31 Write ( notation. ) using sigma Solution: $! k # %&' 17-Feb-17 Computer Science Department

32 ANY QUESTIONS?? 32 Refer to chapter 2 of the book for further reading