Solutions to Exercises 4

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Domenic Shelton
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1 Discrete Mathematics Lent 29 MA2 Solutions to Eercises 4 () Define sequence (b n ) n by b n ( n n..., where we use n ) for > n. Verify that b, b 2 2, and that, for every n 3, we have b n b n b n 2. Solution. Using ( n ) for > n, we have ( ) b and b 2 ( ) 2 ( ) ( ) 2 ( ) 2 ( ) 2. Every b n has only finitely many non-zero summands and we can write it as n ( ) n b n. Then, ( n b n b n 2 n ( ) n n 2 ( ) n 2 ( ) n n ( ) n n 2 ( ) n 2 n ( ) n n ( ) n 2 ( ) n (( )) n n. We will use the identity ( m m m ) with m n and, also, that n ( ) and n n n ) n : n (( )) n n b n b n 2 ( ) n n ( ) n n n n

2 2 n ( ) n b n. (2) Let a n denote the number of n-digit sequences in which each digit is either or, and no two consecutive s are allowed. (a) Show that a 2 and a 2 3. What would you say a is? Solution. Both and are valid -digit sequences, hence a 2. Similarly,,, are all the valid 2-digit sequences ( is prohibited), hence a 2 3. Empty sequence (the equivalent of for sets) contains no consecutive s, so a. (b) Show that for n 3 we have a n a n a n 2. Solution. Consider any valid n-digit sequence 2... n. If, then n is a valid n -digit sequence, and we have a n of these. If, then 2 must be to avoid consecutive zeros. All the remaining n 2 digits do not contain consecutive zeros, so they form a valid (n 2)-digit sequence, and we have a n 2 of them. Thus, the total number a n of valid n-digit sequences satisfies a n a n a n 2 for n 2. (c) Give a closed form epression for a n. Solution. Let f() a n n be the generating function of (a n ) n. We shall use that a n a n a n 2 for n 2. Indeed, we have f() a n n a a a n n a a n2 (a n a n 2 ) n n2 a a a n n 2 a n 2 n 2 n2 n2 a a a n n 2 a n n a a (f() a ) 2 f() 2 ( 2 )f().

3 Hence, f() 2. Equation 2 has two solutions α ( 5)/2 and β ( 5)/2. We rewrite f() as f() (α )(β ) A α B β Aβ Bα (A B). (α )(β ) Consequently, we have AB and Aβ Bα. These two equations have solution therefore, A f() β α β β and B α 5 α β α, 5 A α A α B β A α ( ) n B α β From this we deduce that ( A α n a n B β n α ( ) n β ) n. A α B n β. n Since αβ, this can be rewritten as B β β a n ( ) n (Bα n Aβ n ) ( )n 5 (α n2 β n2 ). (3) Let f n denote the n-th Fibonacci number, i.e., f f and f n f n f n 2 for n 2. Prove that for every n 2 we have f 2 n f n f n ( ) n. Solution. We proceed by induction on n. For n 2, we loo at f 2 2 f f 3. Using the recurrence f n f n f n 2, we have f 2 f f 2 and f 3 f 2 f 3. So, f 2 2 f f ( ) 2. Suppose that f 2 n f n f n ( ) n. Then 3 f 2 n f (n) f (n) f 2 n f n (f n f n ) using f (n) f n f n f n (f n f n ) f 2 n

4 4 f n f n f 2 n using f n f n f n ( ) n using induction assumption ( ) n. Thus our assertion holds by induction. (4) Use generating functions to solve the following recurrence relation: a n 5a n 6a n 2 for n 2, a ; a 3. Solution. Using Theorem 3.5 (see the notes for Lectures 5 and 6), we need to solve 2 5 6, or ( 2)( 3). Then, the general solution is given by a n A 2 n B 3 n. Using the initial conditions a and a 3, we obtain A 2 B 3 A B and 3 A 2 B 3 2A 3B. Thus, A 3 and B 3. Consequently, a n 3(3 n 2 n ). (5) Suppose that f() generates the sequence a, a, a 2,.... Give the epressions, in terms of f, for the generating functions of the following sequences: (a), a,, a,, a 2,,... ; Solution. For f() a n n, we have f( 2 ) a n 2n and f( 2 ) a n 2n a 2 a 3 4 a Hence, f( 2 ) is the generating function of, a,, a,, a 2,,....

5 (b), a, a, a 2,... ; Solution. For f() a n n, we have f() a n () n 5 ( ) n a n n a a 2 a 2 3 a 3 4 a Hence, f() is the generating function of, a, a, a 2, a 3, a 4,.... (c) a, a, a 2, a 3, a 4,.... Solution. For f() a n n, we have f( ) a n ( ) n ( ) n a n n a a a 2 2 a 3 3 a Hence, f( ) is the generating function of a, a, a 2, a 3, a 4,.... (6a) Show that the generating function of the sequence a n n, n, is f(). ( ) 2 Solution. We now that ( ) n. By differentiating both sides, we obtain that and, therefore, f() ( ) 2 ( ) ( ) 2 n n n n n n n n. (b) Find generating functions for the sequences b n n 2, n, and c n n 3, n. Solution. Let f() is the generating function of a sequence (a n ) n, i.e., f() a n n. Then f () na n n. So, g() f () na n n na n n is the generating function of the sequence (na n ) n. For a n n, n, we proved in part (a) that its generating function is f(). We also notice that b ( ) 2 n n 2 na n for every n. Therefore, by our earlier observation, g() f () ( ) () is the generating function ( ) 2 ( ) 3 of (b n ) n. For the last part, we have that c n n 3 nb n for every n. Therefore, by our earlier observation, h() g () ( () ) (2 4) ( ) 3 ( ) 4 function of (c n ) n. is the generating

6 6 (7) Find the sequences generated by the following functions: h() (a) f() 3 ; Solution. We use that y n with y, hence y ( ) ( ) n ( ) n n. Consequently, f() 3 3 ( ) n n ( ) n n3 ( ) n 3 n. Thus, f() is the generating function of,,,,,,,.... (b) g() ; 72 2 Solution. We write g() ( 3)( 4) A 3 B 4 n3 (A B) ( 4A 3B). ( 3)( 4) Hence, A B and 4A 3B, from which we obtain A and B. Using y n with y 3 and y 4, we have y g() 4 3 (4) n (3) n (4 n 3 n ) n. Hence, g() is the generating function of (4 n 3 n ) n. (c) h() 7 ; 2 7 Solution. We see that ( From this we deduce that h() generates ( n ) n, where 2 n/7 if n and 7 divides n, n otherwise. (d) e ) n 2 n 7n. Solution. We now that e y n! yn. By taing y 2, we have () e 2 n! (2)n 2 n n! n, so () is the generating function of ( 2n ) n! n.

1 Examples of Weak Induction

1 Examples of Weak Induction More About Mathematical Induction Mathematical induction is designed for proving that a statement holds for all nonnegative integers (or integers beyond an initial one). Here are some extra examples of