{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Sol_Prac_Final_3

# Sol_Prac_Final_3 - Solutions to the Practice Final Exam...

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

Solutions to the Practice Final Exam December 2, 2009 1. Solution: (a) a (0) = 1 , a (1) = 2 , a (2) = 7 , a (3) = 26 , a (4) = 97, and a (5) = 362. (b) Notice that the characteristic polynomial of the recurrence relation is x 2 - 4 x +1, which has roots 2 ± 3, yielding template solution a ( n ) = α (2 + 3) n + β (2 - 3) n . From the initial values, we have α + β = 1 α (2 + 3) + β (2 - 3) = 2 which has solution α = 1 / 2 and β = 1 / 2. Thus, we have a ( n ) = (2 + 3) n 2 + (2 - 3) n 2 . (c) Proof by induction. Step 1. (Base case) It is easy to verify that a (0) = (2+ 3) 0 2 + (2 - 3) 0 2 = 1 and (2+ 3) 1 2 + (2 - 3) 1 2 = 2. Step 2. (Induction) Suppose a ( n ) = (2+ 3) n 2 + (2 - 3) n 2 for some n 1. We then compute a ( n + 1) = 4 a ( n ) - a ( n - 1) = 4( (2 + 3) n 2 + (2 - 3) n 2 ) - ( (2 + 3) n - 1 2 + (2 - 3) n - 1 2 ) = (2 + 3) n - 1 2 (4(2 + 3) - 1) + (2 - 3) n - 1 2 (4(2 - 3) - 1) = (2 + 3) n - 1 2 (2 + 3) 2 + (2 - 3) n - 1 2 (2 - 3) 2 = (2 + 3) n +1 2 + (2 - 3) n +1 2 and the induction step is complete. 2. Solution: Notice that x ( n + 1) + n + 1 = 2( x ( n ) + n ) + 1. Letting y ( n ) = x ( n ) + n , we have y ( n + 1) = 2 y ( n ) + 1 , n 1 . Since x (1) = 1, then y (1) = 2. Guessing now that the general solution has the form y ( n ) = α 2 n + β , if we plug into the recurrence we get α 2 n +1 + β = 2 · α 2 n + 2 β + 1 which implies β = - 1. Now substituting into the recurrence for n = 1, we get α = 3 2 . Thus, y ( n ) = 3 · 2 n - 1 - 1 which implies that x ( n ) = 3 · 2 n - 1 - n - 1 .

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.

{[ snackBarMessage ]}