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Linear Algebra, Dave Bayer, November , [1] Solve the recurrence relation
f(0) = a, f(n + 1)
f(n) f(1) = b, −5 −4
1
0 = n f(n) = b
a = f(n) = − 5 f(n − 1) − 4 f(n − 2) (−4)n
3 4
4
−1 −1 b
a + (−1)n
3 −1 −4
1
4 b
a (−4)n
(−1)n
( − b − a) +
(b + 4a)
3
3 [2] Solve the recurrence relation
f(0) = a, f(n + 1)
f(n) = f(1) = b, n 6 −5
1
0 f(n) = f(n) = 6 f(n − 1) − 5 f(n − 2) b
a = 1
4 −1 5
−1 5 b
a + 5n
4 5 −5
1 −1 b
a 5n
1
( − b + 5a) +
(b − a)
4
4 [3] Solve the recurrence relation
f(0) = a, f(n + 1)
f(n) = −6 −8
1
0 f(1) = b, n b
a = f(n) = − 6 f(n − 1) − 8 f(n − 2) (−4)n
2 4
8
−1 −2 b
a + (−2)n
2 −2 −8
1
4 (−2)n
(−4)n
( − b − 2a) +
(b + 4a)
2
2 f(n) = [4] Solve the recurrence relation
f(0) = a, f ( n + 1)
f(n) = f(1) = b, −4 5
10 n f(n) = b
a f(n) = − 4 f(n − 1) + 5 f(n − 2) = (−5)n
6 5 −5
−1
1 b
a (−5)n
1
( − b + a) + (b + 5a)
6
6 + 1
6 15
15 b
a b
a , 2 × 2 Exercise Set J (recurrence relations), November
[5] Solve the recurrence relation
f(0) = a, f(n + 1)
f(n) = f(1) = b, n −1 6
10 b
a f(n) = − f(n − 1) + 6 f(n − 2) (−3)n
5 = 3 −6
−1
2 b
a 2n
5 + 26
13 b
a 17
17 b
a 2n
(−3)n
( − b + 2a) +
(b + 3a)
5
5 f(n) = [6] Solve the recurrence relation
f(0) = a, f(n + 1)
f(n) = f(1) = b, n −6 7
10 b
a f(n) = − 6 f(n − 1) + 7 f(n − 2) = (−7)n
8 7 −7
−1
1 b
a + 1
8 (−7)n
1
( − b + a) + (b + 7a)
8
8 f(n) = [7] Solve the recurrence re...
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This document was uploaded on 03/24/2014 for the course MATH V2010 at Barnard College.
 Spring '14
 DaveBayer
 Linear Algebra, Algebra

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