Recurrence Relations and Recursion: MATH 180

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1 Recurrence Relations and Recursion: MATH 180 1: Recursively Defined Sequences Example 1: The sequence a 1,a 2,a 3,... can be defined recursively as follows: (1) For all integers k 2, a k = a k (2) a 1 = 1. Find a 2,a 3,a 4,a 5,... Conjecture a formula for a n,n N. Prove your conjecture using induction. Example 2: The sequence b 0,b 1,b 2,... can be defined recursively as follows: (1) For all integers k 2, b k = b k 1 + b k 2 recurrence relation (2) b 0 = 1,b 1 = 3 Find b 2,b 3,b 4,b 5. Example 3: The sequence c 0,c 1,c 2,... can be defined recursively as follows: (1) For all integers k 2, c k = c k 1 + k c k recurrence relation (2) c 0 = 1,c 1 = 2 Find c 2,c 3,c 4. Example 4: Let s 0,s 1,s 2,... be a sequence that satisfies the following recurrence relation: for all integers k 1,s k = 3s k 1 1 Explain why the following statement is true: for all integers k 0,s k+1 = 3s k 1 Example 5: Let a 0,a 1,a 2,...,b 0,b 1,b 2,... satisfy the recurrence relation that the k th term equals 3 times the (k 1) st term for all integers k 2: (1)a k = 3a k 1 and b k = 3b k 1 1

2 But suppose that the initial conditions for the sequences are different: (2)a 1 = 2 and b 1 = 1. Find the first four terms for each sequence. Example 6: Show that the sequence 1, -1!, 2!,-3!, 4!,...( 1) n n!,... for n 0, satisfies the recurrence relation s k = k s k 1 for all integers k 1. Example 7: The Fibonacci sequence: F 1 = 1,F 2 = 1,F n = F n 1 + F n 2. Example 8: Write the sum n k=0 k as a recurrence relation. 2

3 2 Solving Recurrence Relations by Iteration. Example 1: Let a 0,a 1,a 2,... be the sequence defined recursively as follows: for all integers k 1, a k = a k 1 + 2, where a 0 = 1. Use iteration to guess an explicit formula for the sequence. Example 2: A k = (1.055)A k 1, A 0 = Use iteration to guess an explicit formula for the sequence. Then find A 20 Example 3: M k = 2M k 1 + 1, for all integers k 2, where M 1 = 1. Use iteration to guess an explicit formula for the sequence. 3

4 3 An Important Example Let T n = 2T n/2 + n where n is any power of 2 and let T 1 = 1. Then Find a formula for T n. T 1 = 1 T 2 = 2T = 4 T 4 = 2T = 12 T 8 = 2T = First, by dividing both sides of the recurrence equation by n, we can rewrite as: T n n = T n/2 n/ Now we start listing it out, starting from 2 and going to n, doubling the value of the subscript each time: T 2 2 = T T 4 4 = T T 8 8 = T T n/2 n/2 = T n/4 n/4 + 1 T n n = T n/2 n/ How many lines have we written down as we go between 2 and 8? As we go between 2 and 16? As we go between 2 and n? 2. Add up the two columns. 3. Find a formula for T n. 4

5 4: Structural Induction When a set has been defined recursively, a version of mathematical induction, called structural induction, can be used to prove that every object in the set satisfies a given property. Structural Induction for Recursively Defined Sets: Let S be a set that has been defined recursively, and consider a property that objects in S may or may not possess. To prove that every object in S possesses the property: 1. Show that each object in the base for S satisfies the property; 2. Show that for each rule in the recursion, if the rule is applied to an object in S that satisfies the property, then the object defined by the rule also satisfies the property. Because no objects other than those obtained through the base and recursion conditions are contained in S, it must be the case that every object in S possesses the property. Example 1: (Generalized DeMorgan s Law) Prove that for all integers n 1, if A 1,A 2,...,A n are sets, then ( n ) C A i = i=1 n (A i ) C. (Now remember, we HAVE proved this for two sets: (A B) C = A C B C.) i=1 Definition 1: Let n be a positive integer. Given a finite set A, a string of length n over A is an ordered n-tuple of elements of A written without parentheses or commas. The elements of A are called the characters of the string. The null string over A is defined to be the string with no characters. It is usually denoted ε and is said to have length 0. If A = {0,1} then a string over A is called a bit string. Subexample 1: List all bit strings of length 3. Example 2: Define a set S recursively as follows: I. Base: ε S II. Recursion: If s S, then a. bs S b. sb S c. saa S d. aas S III. Restriction: Nothing is in S other than objects defined in I and II above. Use structural induction to prove that every string in S contains an even number of a s. 5

6 Example 3: Consider the alphabet A = {a,b}, all strings on this alphabet, and the inductive subset of strings S defined recursively as follows: I. Base: a S II. Recursion: a. If x S then axb S b. If x,y S then xy S. (This is the concatenation of x and y: write down x and then y.) III. Restriction: Nothing is in S other than objects defined in I and II above. Prove that for every x S, the number of a s in x is greater than the number of b s. 6

7 Homework for Recurrence Relations. 1 Problems 1. Find the first four terms of each of the recursively defined sequences below. a. a k = 2 a k 1 + k, for all integers k 2, where a 1 = 1. b. b k = k (b k 1 ) 2, for all integers k 1, where b 0 = 2. c. c k = k c k 1 c k 2, for all integers k 3, where c 1 = 1,c 2 = Show that the sequence 0,1,3,7,...2 n 1,..., for n 0, satisfies the recurrence relation for all integers k Show that the sequence 1,-1, 1 2, 1 3! relation for all integers k 1. c k = 2c k 1 + 1,,..., ( 1)n n!,... for n 0 satisfies the recurrence s k = s k 1, k 4. Use the recurrence relation and values for the Fibonacci sequence F 1,F 2,... to compute F 13 and F Problems 1. Each of these sequences are defined recursively. Use iteration to guess an explicit formula for the sequence. a. a k = k a k 1, for all integers k 1, where a 0 = 1. b. b k = 3b k 1 + 1, for all integers k 2, where b 1 = 1. c. c k = c k k, for all integers k 2, where c 1 = 1. d. d k = 2d k 1, for all integers k 1, where d 0 = Use mathematical induction to confirm your answers to problem 1 parts a-d. 3 Problems 1. Let C n = C n/2 + 1 and C 1 = 1, where n is a power of 2. Find and prove a formula for C n. 7

8 4 Problems. 1. Define a set S recursively as follows: I. Base: 1 S II. Recursion: If s S, then a. 0s S b. 1s S. III. Restriction: Nothing is in S other than objects defined in I and II above. Use structural induction to prove that every string in S ends in a For all positive integers n, if A and B 1,B 2,...,B n are sets, then prove that ( n A i=1 B i ) = n (A B i ). (Remember, we did prove the following: A (B C) = (A B) (A C).) i=1 8