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Discrete Structures
Sept 12 2011
Assignment 3:
Solutions
Prof. Hopcroft
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1
Recurrence Relations: Bottom Up
For each of the characteristic equations:
1. Find the corresponding recurrence equation.
2. How many boundary conditions are necessary for a complete solution to
f
(
n
)?
3. Show that the roots of the characteristic equations raised to the
n
th
power are solutions of the recurrence
equation. Ie
f
(
n
) =
r
n
, where
r
is a root of the corresponding characteristic equation, satisﬁes the
recurrence equation.
The characteristic equations:
1.1
x

2 = 0
1.
f
(
n
) = 2
f
(
n

1)
2. Need 1 boundary condition.
3. The roots are: 2.
2
n
= 2(2)
n

1
1.2
x
2
+ 2
x

15 = 0
1.
f
(
n
) =

2
f
(
n

1) + 15
f
(
n

2)
2. Need 2 boundary conditions.
3. The roots are: 3 and

5.
(

5)
n
=

2(

5)
n

1
+ 15(

5)
n

2
25 = 10 + 15
(1)
And:
(3)
n
=

2(3)
n

1
+ 15(3)
n

2
9 =

6 + 15
(2)
1
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View Full Document1.3
8
x
3

8
x
2
+ 4
x

1 = 0
1. 8
f
(
n
) = 8
f
(
n

1)

4
f
(
n

2) +
f
(
n

3)
2. Need 3 boundary conditions.
3. The roots are:
1
2
,
1+
√

3
4
,
1

√

3
4
.
8
±
1
2
²
n
= 8
±
1
2
²
n

1

4
±
1
2
²
n

2
+
±
1
2
²
n

3
8
2
3
=
8
2
2

4
2
+ 1
1 = 2

2 + 1
8
±
1 +
√

3
4
²
n
= 8
±
1 +
√

3
4
²
n

1

4
±
1 +
√

3
4
²
n

2
+
±
1 +
√

3
4
²
n

3
8
±
1 +
√

3
4
²
3
= 8
±
1 +
√

3
4
²
2

4
±
1 +
√

3
4
²
+ 1
(1 +
√

3)
3
8
=
(1 +
√

3)
2
2

√

3
1 + 3
√

3

9

3
√

3
8
=
1 + 2
√

3

3
2

√

3
8
8
=
2 + 2
√

3
2

√

3 = 1
In a very similar way the root
³
1

√

3
4
´
n
can be shown to satisfy the recurrence.
2
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 SELMAN

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