Recursive Definitions Rosen 3 4 Recursive or inductive

Recursive Definitions Rosen, 3. 4

Recursive (or inductive) Definitions • Sometimes easier to define an object in terms of itself. • This process is called recursion. – Sequences • {s 0, s 1, s 2, …} defined by s 0 = a and sn = 2 sn-1 + b for constants a and b and n Z+ – Sets • 3 S and x+y S if x S and y S – Functions • Example: f(n) = 2 n, f(n) = 2 f(n-1) and f(0) = 1

Recursively Defined Functions To define a function with the set of nonnegative integers as its domain 1. Specify the value of the function at zero (or sometimes, it first k terms). 2. Give a rule for finding its value at an integer from its values at smaller integers.

Examples of Recursively Defined Functions • Factorial Function n! § n! = n(n-1)(n-2)…. (1) § f(0) = 1, f(n) = n(f(n-1)) • an § f(0) = 1, f(n+1) = f(n)*a • Fibonacci Numbers § f 0=0, f 1=1, f n+1 = fn + f n-1 § {0, 1, 1, 2, 3, 5, 8, 13, . . . }

Prove that the nth term in the Fibonacci sequence is when n 2 Induction Proof: Basic Step: Let n = 2, then f 2 = 1 = =1+0 Inductive Step: Consider k 2 and assume that the expression is true for 2 n k. We must show that the expression is true for n = k+1, i. e. , that fk+1 = fk +fk-1 by definition by the inductive hypothesis

Fibonacci Proof (cont. ) Since f 2 is true and [fn is true for 2 n k fk+1] is true, then fn is true for all positive integers n 2.

Recursively Defined Set Example Let the set S be defined recursively as 3 S and x+y S if x S and y S. Prove that S is equal to the set of positive integers divisible by 3 (set A). Proof: To show that S = A, we must show that A S and that S A.

Recursive Set Example (cont. ) First we will show that A S. To do this we must show that every positive integer divisible by 3 is in S. This is the same as saying that 3 n is in S for n Z+. We will do an inductive proof. Let p(n) be the statement that 3 n S for n a positive integer. Basis Step: p(1) = 3 Inductive Step: We will show that p(n) p(n+1) Let p(n) S. Then p(n+1) = 3 n + 3. Since 3 n S and 3 S, then 3 n+3 S. Therefore every element in A must be in S.

Recursive Set Example (cont. ) Now we will show that S A. To do this we must show that every element in S is divisible by 3, i. e. , that it can be represented as 3(j) where j is a positive integer. Basis Step: 3 A since 3 = 3(1) Inductive Step: Assume x, y are in A (where x, y S). x and y are divisible by 3 since they are both in A. Therefore x = 3 j and y = 3 k for j, k Z+. Then z = x+y = 3 j+3 k = 3(j+k), which is divisible by 3, so z A (and by definition, z S). Therefore every element in S must be in A

Prove fn > n-2 where = (1+ 5)/2 whenever n 3 Inductive Proof Basis Step: First we will show fn > n-2 for f 3 and f 4. 3 -2 = 1 = (1+ 5)/2 < (1+ 9)/2 = 4/2 = f 3 4 -2 = [(1+ 5)/2]2 = (1+ 2 5 + 5)/4 = (6+ 2 5)/4 = 3/2 + 5/2 < 3/2 + 9/2 = 3 = f 4

fn > n-2 (cont. ) Inductive Step: Assume that fk > k-2 is true for all nonnegative integers 3 k n when n>4. We must show that fn+1> n-1. Lemma: 2 = [(1+ 5)/2]2 = (1+2 5 +5)/4 = 3/2 + ( 5)/2 = 1 + (1+ 5)/2 = 1+

fn > n-2 (cont. ) n-1 = 2 n-3 = ( +1)( n-3) = n-2+ n-3 By our inductive hypothesis fn-1 > n-3 and fn > n-2 fn+1= fn + fn-1 > n-2+ n-3 = n-1 Since f 3 and f 3 are true and fk is true for 3 k n fn+1 is true, then fn is true for all positive integers n 3

Find a closed form solution to the recursive equation: T(n) = T(n-1) + c 1. T(0) = c 0 One approach is to write down several terms, guess what the equation is, and then prove that your guess is correct (or not) using an induction proof. T(0) = c 0 T(1) = T(0) + c 1 = c 0 + c 1 T(2) = T(1) + c 1 = c 0 + 2 c 1 T(3) = T(2) + c 1= c 0 + 3 c 1 T(4) = T(3) + c 1= c 0 + 4 c 1 Guess that T(n) = c 0 + nc 1

Induction proof Basis Step: for n = 0, T(0) = c 0 + (0)c 1 = c 0 Inductive Step: Assume that T(n) = c 0 + nc 1, we must show that T(n+1) = c 0 + (n+1)c 1. T(n+1) = T(n) + c 1 = c 0 + nc 1 + c 1 = c 0 + (n+1)c 1.

Find a closed form solution to T(1) = c 0, T(n) = 2 T(n-1)+c 1 T(1) = c 0 T(2) = 2 T(1) + c 1 = 2 c 0 + c 1 T(3) = 2 T(2)+c 1 = 2(2 c 0+c 1)+c 1 = 4 c 0+3 c 1 T(4) = 2 T(3)+c 1 = 2(4 c 0+3 c 1)+c 1 = 8 c 0+7 c 1 T(5) = 2 T(4)+c 1 = 2(8 c 0+7 c 1)+c 1 = 16 c 0+15 c 1 Guess that T(n) = 2 n-1 c 0 + (2 n-1 -1)c 1

Prove that T(1) = c 0, T(n) = 2 T(n-1)+c 1 has closed form solution T(n) = 2 n-1 c 0 + (2 n-1 -1)c 1 Basis Step: T(1) = 21 -1 c 0 + (21 -1 -1)c 1 = c 0 Induction Step: Assume that T(n) = 2 n-1 c 0 + (2 n-1 -1)c 1. We must show that T(n+1) = 2(n+1)-1 c 0 + (2(n+1)-1 -1)c 1 = 2 nc 0 + (2 n-1)c 1. T(n+1) = 2 T(n) + c 1 = 2[2 n-1 c 0 + (2 n-1 -1)c 1]+ c 1 = 2 nc 0 + (2 n-2)c 1 + c 1 = 2 nc 0 + 2 nc 1 - c 1 = 2 nc 0 + (2 n-1)c 1.