hw6solns

# Hw6solns - 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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## 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.

### Page1 / 5

Hw6solns - 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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online