Discrete structure sequence, arithmetic progression
- 1. CSC102: Discrete Structure Credit Hours: 3(3,0) Lecture # 15 Unit #: 5 MISHAL IQBAL 1 Instructor: COMSATS University Islamabad Vehari Campus
- 2. Topic 2 Sequences Arithmetic progression Geometric progression Recurrence relation Fibonacci sequence
- 3. Sequences 3 A sequence is a discrete structure used to represent an ordered list. E.g., 1, 2, 3, 5, 8 is a sequence with five terms and 1, 3, 9, 27, 81 , . . . , 3 n , . . . is an infinite sequence. The terms of a sequence can be specified by providing a formula for each term of the sequence. To specify the terms of a sequence a recurrence relation is used, which expresses each term as a combination of the previous terms. Overview
- 4. Sequences 4 We use the notation {an} to describe the sequence and an represents an individual term of the sequence {an}. We describe sequences by listing the terms of the sequence in order of increasing subscripts Example: Consider the sequence {an}, where an = List of the terms of this sequence, beginning with a1, namely, a1, a2, a3, a4, a5, a6, . . . , starts with 1, . . . . Definition
- 5. Sequences 5 (1) an=n2 , where n= 1, 2, 3,…. What are the elements of sequence? 1, 4, 9, 16, 25,… (2) an=(-1)n , where n= 0, 1, 2, 3, …. What are the elements of sequence? 1, -1, 1, -1, 1,… (3) an=2n , where n= 0, 1, 2, 3, …. What are the elements of sequence? 1, 2, 4, 8, 16, 32, … Examples
- 6. Sequences 6 An arithmetic progression is a discrete analogue of the linear function f (x) = dx + a. The sequences {sn} with sn = −1 + 4n, is an arithmetic progression with initial term = -1, 2nd term = 3, and common difference = 4 if we start with n=0 The list of terms s0, s1, s2, s3, . . . begins with −1, 3, 7, 11, . . . Arithmetic Progression
- 7. Sequences 7 Example: The sequences {tn} with tn = 7 - 3n, is an arithmetic progression with initial term = 7, 2nd term = 4, and common difference = -3, if we start with n=0 The list of terms t0, t1, t2, t3, . . . begins with 7, 4, 1, -2, . . . Arithmetic Progression
- 8. Sequences 8 Given a sequence finding a rule for generating the sequence is not always straightforward. Example: Assume the sequence: 1,3,5,7,9, …. What is the formula for the sequence? Each term is obtained by adding 2 to the previous term. 1, 1+2=3, 3+2=5, 5+2=7, 7+2 It suggests an arithmetic progression: a+nd with a=1 and d=2 an=1+2n Arithmetic progression
- 9. Sequences 9 A geometric progression is a discrete analogue of the exponential function f (x) =ark The sequences {bn} with bn = (−1)n is a geometric progression with initial term=1, 2nd term = -1, and common ratio = -1, if we start at n = 0. The terms b0, b1, b2, b3, b4, . . . begins with 1,−1, 1,−1, 1, . . . Geometric Progression
- 10. Sequences 10 The sequences {cn} with cn = 2.5n is a geometric progression with initial term=2, 2nd term = 10, and common ratio = 5, if we start at n = 0. The terms c0, c1, c2, c3, c4, . . . begins with 2, 10, 50, 250, 1250, … The sequences {dn} with dn = 6.(1/3)n is a geometric progression with initial term=6, 2nd term = 2, and common ratio = 1/3, if we start at n = 0. The terms d0, d1, d2, d3, d4, . . . begins with 6, 2, 2/3, 2/9, 2/27, … Geometric Progression
- 11. Sequences 11 Assume the sequence: 1, 1/3, 1/9, 1/27, … What is the sequence? The denominators are powers of 3. 1, 1/3= 1/3, (1/3)/3=1/(3*3)=1/9, (1/9)/3=1/27 This suggests a geometric progression: ark with a=1 and r=1/3 is (1/3 )n Geometric progression
- 12. Sequences 12 In arithmetic and geometric progression, we specified sequences by providing explicit formulas for their terms. There are many other ways to specify a sequence. E.g., another way to specify a sequence is to provide one or more initial terms together with a rule for determining subsequent terms from those that precede them. Recurrence relation
- 13. Sequences 13 A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0, a1, . . . , an−1, for all integers n with n ≥ n0, where n0 is a nonnegative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation. (A recurrence relation is said to recursively define a sequence. Recurrence relation The n-th element of the sequence {an} is defined recursively in terms of the previous elements of the sequence and the initial elements of the sequence.
- 14. Sequences 14 The initial conditions for a recursively defined sequence specify the terms that precede the first term where the recurrence relation takes effect. Let {an} be a sequence that satisfies the recurrence relation an = an−1 + 3 for n = 1, 2, 3, . . . , and suppose that a0 = 2. What are a1, a2, and a3? Solution: We see from the recurrence relation that a1 = a0 + 3 = 2 + 3 = 5. It then follows that a2 = 5 + 3 = 8 and a3 = 8 + 3 = 11. Recurrence relation
- 15. Sequences 15 Let {an} be a sequence that satisfies the recurrence relation an = an−1 − an−2 for n =2, 3, 4, . . . , and suppose that a0 = 3 and a1 = 5. What are a2, and a3? Solution: We see from the recurrence relation that a2 = a1 − a0 = 5 − 3 = 2 and a3 = a2 −a1 = 2 − 5 = −3. We can find a4, a5, and each successive term in a similar way. Recurrence relation
- 16. Sequences 16 Example: Determine whether the sequence {an}, where an = 3n for every nonnegative integer n, is a solution of the recurrence relation an = 2an−1 − an−2 for n = 2, 3, 4, . . . . Solution: Suppose that an = 3n for every nonnegative integer n. Then, for n ≥ 2, we see that 2an−1 − an−2 = 2(3(n − 1)) − 3(n − 2) = 3n = an Therefore, {an}, where an = 3n, is a solution of the recurrence relation. Recurrence relation
- 17. Sequences 17 Example: Determine whether the sequence {an}, where an = 2n for every nonnegative integer n, is a solution of the recurrence relation an = 2an−1 − an−2 for n = 2, 3, 4, . . . . Solution: Suppose that an = 2n for every nonnegative integer n. Note that a0 = 1, a1 = 2, and a2 = 4, because a2 =2a1 − a0 = 2 * 2 − 1 = 3, we see that {an}, where an = 2n , is not a solution of the recurrence relation. Recurrence relation
- 18. Sequences 18 Example: Determine whether the sequence {an}, where an = 5 for every nonnegative integer n, is a solution of the recurrence relation an = 2an−1 − an−2 for n = 2, 3, 4, . . . . Solution: Suppose that an = 5 for every nonnegative integer n. Then for n ≥ 2, we see that an = 2an−1 − an−2 = 2 ・ 5 − 5 = 5 = an Therefore, {an}, where an = 5, is a solution of the recurrence relation. Recurrence relation
- 19. Sequences 19 Many methods have been developed for solving recurrence relations. Here, we will introduce a straightforward method known as iteration. Solve the recurrence relation and initial condition with {an} be a sequence that satisfies the recurrence relation an = an−1 + 3 for n = 1, 2, 3, . . . . Suppose that a1 = 2, and working upward until we reach an to deduce a closed formula for the sequence. (continue …) Recurrence relation
- 20. Sequences 20 The first approach is called forward substitution – we found successive terms beginning with the initial condition and ending with an. We see that a2 = 2 + 3 a3 = (2 + 3) + 3 = 2 + 3 * 2 = 2 + 2 *3 a4 = (2 + 2 * 3) + 3 = 2 + 3 * 3 … an = an−1 + 3 = (2 + 3 * (n − 2)) + 3 = 2 + 3(n − 1) (continue …) Recurrence relation
- 21. Sequences 21 The second approach backward substitution starting with the term an and working downward until we reach the initial condition a1 = 2 to deduce this same formula. The steps are an = an−1 + 3 = (an−2 + 3) + 3 = an−2 + 3 * 2 = (an−3 + 3) + 3 * 2 = an−3 + 3 * 3 ... = a + 3(n − 2) = (a + 3) + 3(n − 2) = 2 + 3(n − 1). Recurrence relation
- 22. Sequences 22 Compound Interest Suppose that a person deposits $10,000 in a savings account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years? Solution: Let Pn denote the amount in the account after n years. Because the amount in the account after n years equals the amount in the account after n−1 years plus interest for the nth year, we see that the sequence {Pn} satisfies the recurrence relation Pn = Pn−1 + 0.11Pn−1 = (1.11)Pn−1 Recurrence relation
- 23. Sequences 23 The initial condition is P0 = 10,000. We can use an iterative approach to find a formula for Pn. Note that P1 = (1.11) P0 P2 = (1.11) P1 = (1.11) 2 P0 P3 = (1.11) P2 = (1.11) 3 P0 ... Pn = (1.11)Pn−1 = (1.11)n P0 Initial condition P0 = 10,000, Pn = (1.11)n 10,000 is obtained. n = 30, Pn = (1.11) n 10,000 30 Recurrence relation
- 24. Sequences 24 A particularly useful sequence defined by a recurrence relation, known as the Fibonacci sequence. The recurrence relation for the Fibonacci sequence tells us that we find successive terms by adding the previous two terms. Fibonacci sequence
- 25. Sequences 25 Example: Find the Fibonacci numbers f2, f3, f4, f5, and f6. Solution: The initial conditions tell us that f0 = 0 and f1 = 1, using the recurrence relation in the definition we find that f2 = f1 + f0 = 1 + 0 = 1, f3 = f2 + f1 = 1 + 1 = 2, f4 = f3 + f2 = 2 + 1 = 3, f5 = f4 + f3 = 3 + 2 = 5, f6 = f5 + f4 = 5 + 3 = 8. Fibonacci sequence
- 26. Sequences 26 Suppose that {an} is the sequence of integers defined by an = n!, the value of the factorial function at the integer n, where n = 1, 2, 3, . . .. Because n! = n((n − 1)(n − 2) . . . 2 * 1) = n(n − 1)! = nan−1, we see that the sequence of factorials satisfies the recurrence relation an = nan−1, together with the initial condition a1 = 1. We have solved the recurrence relation together with the initial conditions when we find an explicit formula, called a closed formula, for the terms of the sequence. Fibonacci sequence
- 27. Sequences 27 A common problem in discrete mathematics is finding a closed formula, a recurrence relation, or some other type of general rule for constructing the terms of a sequence. Sometimes only a few terms of a sequence solving a problem are known; the goal is to identify the sequence. Even though the initial terms of a sequence do not determine the entire sequence, knowing the first few terms may help you make an educated conjecture about the identity of your sequence. Once you have made this conjecture, you can try to verify that you have the correct sequence. Special integer sequences
- 28. Sequences 28 When trying to deduce a possible formula, recurrence relation, or some other type of rule for the terms of a sequence when given the initial terms, try to find a pattern in these terms. You might also see whether you can determine how a term might have been produced from those preceding it. Are there runs of the same value? That is, does the same value occur many times in a row? Are terms obtained from previous terms by adding the same amount or an amount that depends on the position in the sequence? Are terms obtained from previous terms by multiplying by a particular amount? Are terms obtained by combining previous terms in a certain way? Are there cycles among the terms? Special integer sequences
- 29. Sequences 29 Find formulae for the sequences with the following first five terms: 1, 1/2, 1/4, 1/8, 1/16. Solution: We recognize that the denominators are powers of 2. The sequence with an = 1/2n , n = 0, 1, 2, . . . is a possible match. This proposed sequence is a geometric progression with initial a0 = 1 and common ratio r = 1/2. Special integer sequences
- 30. Sequences 30 Find formulae for the sequences with the following first five terms: 1, 3, 5, 7, 9. Solution: We note that each term is obtained by adding 2 to the previous term. The sequence with an = 2n + 1, n = 0, 1, 2, . . . is a possible match. This proposed sequence is an arithmetic progression with initial a0 = 1 and d = 2. Special integer sequences
- 31. Sequences 31 Find formulae for the sequences with the following first five terms: 1, −1, 1, −1, 1. Solution: The terms alternate between 1 and −1. The sequence with an = (−1)n , n = 0, 1, 2 . . . is a possible match. This proposed sequence is a geometric progression with initial a0 = 1 and r = −1. Special integer sequences
- 32. Sequences 32 How can we produce the terms of a sequence if the first 10 terms are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4? Solution: In this sequence, the integer 1 appears once, the integer 2 appears twice, the integer 3 appears three times, and the integer 4 appears four times. A reasonable rule for generating this sequence is that the integer n appears exactly n times, so the next five terms of the sequence would all be 5, the following six terms would all be 6, and so on. The sequence generated this way is a possible match. Special integer sequences
- 33. Sequences 33 How can we produce the terms of a sequence if the first 10 terms are 5, 11, 17, 23, 29, 35, 41, 47, 53, 59? Solution: In this sequence after the first is obtained by adding 6 to the previous term. (We could see this by noticing that the difference between consecutive terms is 6.) Consequently, the nth term could be produced by starting with 5 and adding 6 a total of n − 1 times; that is, a reasonable guess is that the nth term is 5 + 6(n − 1) = 6n − 1. (This is an arithmetic progression with a1 = 5 and d = 6.) Special integer sequences
- 34. Sequences 34 How can we produce the terms of a sequence if the first 10 terms are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123? Solution: Observe that each successive term of this sequence, starting with the third term, is the sum of the two previous terms. That is, 4 = 3 + 1, 7 = 4 + 3, 11 = 7 + 4, and so on. Consequently, if Ln is the nth term of this sequence, we guess that the sequence is determined by the recurrence relation Ln = Ln−1 + Ln−2 with initial conditions L1 = 1 and L2 = 3 This sequence is known as the Lucas sequence. Special integer sequences
- 36. Sequences 36 If the first 10 terms of the sequence {an} are 1, 7, 25, 79, 241, 727, 2185, 6559, 19681, 59047, construct a formula for an . Solution: We find that this ratio, although not a constant, is close to 3. So it is reasonable to suspect that the terms of this sequence are generated by a formula involving 3 n . Comparing these terms with the corresponding terms of the sequence {3 n }, we notice that the nth term is 2 less than the corresponding power of 3. We see that an = 3 n − 2 for 1 ≤ n ≤ 10 and this formula holds for all n. Special integer sequences
- 37. Sequences 37 A sequence is also an important data structure in computer science. Sequences of the form a1, a2, . . . , an are often used in computer science. These finite sequences are also called strings and it is also denoted by a1a2 . . . an. The length of a string is the number of terms in this string. The empty string, denoted by λ, is the string that has no terms and has length zero. The string abcd is a string of length four. strings
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