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Unformatted text preview: a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök# 9 Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 3.4 Problem 4: We have f (0) = f (1) = 1. ( a ): f (2) = f (1) − f (0) = 0 f (3) = f (2) − f (1) = − 1 f (4) = f (3) − f (2) = − 1 f (5) = f (4) − f (3) = 0 . ( b ): f (2) = f (1) f (0) = 1 f (3) = f (2) f (1) = 1 f (4) = f (3) f (2) = 1 f (5) = f (4) f (3) = 1 . ( c ): f (2) = f (1) 2 + f (0) 3 = 2 f (3) = f (2) 2 + f (1) 3 = 5 f (4) = f (3) 2 + f (2) 3 = 33 f (5) = f (4) 2 + f (3) 3 = 1214 . ( d ): f (2) = f (1) /f (0) = 1 f (3) = f (2) /f (1) = 1 f (4) = f (3) /f (2) = 1 f (5) = f (4) /f (3) = 1 . § 3.4 Problem 5 ( a ): This is undefined for n = 1, because then f (1) = 2 f ( − 1), and f ( − 1) is not defined. ( b ): This is welldefined. We have f (1) = f (0) − 1 = 0 f (2) = f (1) − 1 = − 1 f (3) = f (2) − 1 = − 2 f (4) = f (3) − 1 = − 3 . We can see that f ( n ) = 1 − n , and can prove it by induction. For n = 0 it’s true. If it’s true for f ( n ) then f ( n + 1) = f ( n ) − 1 = (1 − n ) − 1 = 1 − ( n + 1) . ( c ): This is welldefined. We have f (0) = 2 f (1) = 3 f (2) = f (1) − 1 = 2 f (3) = f (2) − 1 = 1 f (4) = f (3) − 1 = 0 f (5) = f (4) − 1 = − 1 . The formula is f ( n ) = braceleftbigg 2 , n = 0 4 − n, n ≥ 1. 1 For n = 0 and n = 1 it’s clear. We now assume that n > 1 and that f ( n ) = 4 − n . Then f ( n + 1) = f ( n ) − 1 = (4 − n ) − 1 = 4 − ( n + 1) . ( d ): This is welldefined. We have f (0) = 1 f (1) = 2 f (2) = 2 f (0) = 2 f (3) = 2 f (1) = 4 f (4) = 2 f (2) = 4 f (5) = 2 f (3) = 8 f (6) = 2 f (4) = 8 f (7) = 2 f (5) = 16 . The pattern is f ( n ) = braceleftbigg 2 n/ 2 , n even 2 ( n +1) / 2 , n odd = 2 ⌈ n/ 2 ⌉ ....
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 Spring '06
 CALLAHAN
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