224
Solutions to Exercises
12. [BB] We have
an
<
b
n
for
n
2: 1 since
a
<
b,
so
an
=
O(b
n
).
We must prove
b
n
#
O(a
n
).
If
b
n
=
O(a
n
),
then there would exist a constant c such that
b
n
S;
can
for all sufficiently large
n.
Dividing by
a,
we obtain
(~)n
S;
c, for all sufficiently large
n,
but this is not true because
~
> 1
implies that
(~)n
grows without bound as
n
increases.
13. (a) [BB] We established
2
n
<
n!
for
n
2: 4 in Problem 6 of Chapter 5. Thus,
2
n
=
O(n!)
(c
=
1,
no
=
4). On the other hand,
n!
#
0(2
n
),
for consider
n!
1234
n
2n
=
2222···"2·
The product of the first three factors on the right is
~
and each of the remaining
n

3 terms is
bigger than 2. So
;l
>
~(2n3)
for
n
>
3.
If
n!
=
0(2n),
then
n!
<
c2
n
for some constant c
and all sufficiently large
n
and so
;l
< c. We have shown this is not possible. Since
2
n
=
O(n!)
but
n!
#
0(2n),
we have
2
n
<
n!
as required.
(b) Since
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 Summer '10
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 Graph Theory, Mathematical Induction, Recursion, Inductive Reasoning, Sufficiently large, Structural induction, Arbitrarily large

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