# 55PS9 - SOLUTION SET 9—DUE 4/16/2008 Problem 1 (#7.1.18)...

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Unformatted text preview: SOLUTION SET 9—DUE 4/16/2008 Problem 1 (#7.1.18) . a) Find a recurrence relation for the number of permuta- tions of a set with n elements. b) Use this recurrence relation to find the number of permutations of a set with n elements using iteration. Solution. Let p n be the number of permutations of a set with n elements. We can make a recurrence relation by first choosing a location for the first element ( n ways), then permuting the remaining n- 1 elements. Thus, p ( n ) = np ( n- 1). Iterating, we see p ( n ) = n ( n- 1)( n- 2) ··· (2) p (1). Since there’s clearly only one way to permute a set of one element, p (1) = 1, so p ( n ) = n !. Problem 2 (#7.1.22) . a) Find a recurrence relation for the number of strictly increasing sequences of positive integers that have 1 as their first term and n as their last term, where n is a positive integer. That is, sequences a 1 ,a 2 ,...,a k , where a 1 = 1 , a k = n , and a j &amp;lt; a j +1 for j = 1 , 2 ,...,k- 1 . b) What are the initial conditions? c) How many sequences of the type described in (a) are there when n is a positive integer with n ≥ 2 ? Solution. Any sequence with n the last integer either contains n- 1 or it doesn’t. If it does, it’s a sequence counted by p n- 1 . If it doesn’t, it’s still a sequence counted by p n- 1 , but with the n- 1 itself removed. Thus, p n = 2 p n- 1 . The initial condition is clearly p 2 = 1. Thus, p n = 2 n- 2 . (You can see this more directly by noting that every number between 1 and n is a separate yes/no decision, and there are n- 2 such decisions.) Problem 3 (#7.1.44) . Show that the Fibonacci numbers satisfy the recurrence relation f n = 5 f n- 4 + 3 f n- 5 for n = 5 , 6 , 7 ,... , together with the initial conditions f = 0 ,f 1 = 1 ,f 2 = 1 ,f 3 = 2 ,f 4 = 3 . Use this recurrence relation to show that f 5 n is divisible by 5, for n = 1 , 2 , 3 ,... . Proof. For n ≥ 5, f n = f n- 1 + f n- 2 = ( f n- 2 + f n- 3 ) + ( f n- 3 + f n- 4 ) = f n- 3 + f n- 4 + 2 f n- 3 + f n- 4 = 3 f n- 3 + 2 f n- 4 = 3 f n- 4 + 3 f n- 5 + 2 f n- 4 = 5 f n- 4 + 3 f n- 5 Now, f 5 = 5. Suppose 5 | f 5 i . Now, f 5 i +5 = 5 f 5 i +1 + 3 f 5 i . The first term is a multiple of 5, while the second is divisible by 5 by assumption, so 5 | f 5 i +5 . By induction, then 5 | f 5 i for all i . 1 2 SOLUTION SET 9—DUE 4/16/2008 Problem 4 (#7.2.10) . Let c 1 and c 2 be real numbers with c 2 6 = 0 Suppose that r 2- c 1 r- c 2 = 0 has only one root r . A sequence { a n } is a solution of the recurrence relation a n = c 1 a n- 1 + c 2 a n- 2 if and only if a n = α 1 r n + α...
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## This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.

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55PS9 - SOLUTION SET 9—DUE 4/16/2008 Problem 1 (#7.1.18)...

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