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Unformatted text preview: Discrete Mathematics, Spring 2004 Homework 6 Sample Solutions 5.2 #22. Solve the recurrence relation 9 a n = 6 a n − 1 − a n − 2 given the initial conditions a = 6 ,a 1 = 5. Solution . To solve this 2nd order linear homogeneous recurrence relation, we first solve the associated quadratic equation: 9 t 2 = 6 t − 1 ⇒ 9 t 2 − 6 t + 1 = 0 ⇒ (3 t − 1) 2 = 0 ⇒ t = 1 / 3 . Then (as shown in class) the sequences b n = (1 / 3) n , c n = n (1 / 3) n both satisfy the same type of recurrence relation; that is, 9 b n = 6 b n − 1 − b n − 2 and 9 c n = 6 c n − 1 − c n − 2 . So Bb n + Cc n satisfies such a recurrence relation as well for any constants B and C , and hence we look for constants B and C with Bb + Cc = 6 , Bb 1 + Cc 1 = 5 . Substituting for b ,b 1 ,c ,c 1 , we obtain the equations B = 6, B/ 3 + C/ 3 = 5. Solving for C gives C = 9, so a n = Bb n + Cc n = 6 parenleftbigg 1 3 parenrightbigg n + 9 n parenleftbigg 1 3 parenrightbigg n = 2 parenleftbigg 1 3 parenrightbigg n − 1 + n parenleftbigg 1 3 parenrightbigg n − 2 . 5.2 #35. Solve the recurrence relation a n = radicalbigg a n − 2 a n − 1 with initial conditions a = 8, a 1 = 1 / (2 √ 2) by taking the logarithm of both sides and mak ing the substitution b n = lg a n . Solution . Making the substitutions as described above, we obtain the recurrence relation b n = lg a n = lg radicalbigg a n − 2 a n − 1 = 1 2 lg a n − 2 − 1 2 lg a n − 1 = − 1 2 b n − 1 + 1 2 b n − 2 . The associated quadratic equation is t 2 + 1 2 t − 1 2 = 0, which has roots t 1 = 1 2 , t 2 = − 1. Now b = lg a = 3 and b 1 = lg a 1 = − 3 / 2, so we look for constants C and D with C + D = 3 , C · 1 2 + D ( − 1) = − 3 2 ....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Math, Algebra

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