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19
3.11
(a)
n
!=
n
×
(
n
−
1)
×···×
n
2
×
(
n
2
−
1)
×···×
2
×
1
≥
n
2
×
n
2
×···×
n
2
×
1
×···×
1
×
1
=(
n
2
)
n/
2
Therefore
lg
n
!
≥
lg
µ
n
2
¶
n
2
≥
1
2
(
n
lg
n
−
n
)
.
(b)
This part is easy, since clearly
1
·
2
·
3
···
n<n
·
n
·
n
···
n,
so
n
!
<n
n
yielding
log
n
!
<n
log
n
.
3.12
Clearly this recurrence is in
O
(log
n
√
n
)
since the recurrance can only be
expanded
log
n
times with each time being
√
n
or less. However, since this
series drops so quickly, it might be reasonable to guess that the closed form
solution is
O
(
√
n
)
. We can prove this to be correct quite easily with an
induction proof. We need to show that
T
(
n
)
≤
c
√
n
for a suitable constant
c
. Pick
c
=4
.
Base case
:
T
(1)
≤
4
.
Induction Hypothesis
:
T
(
n
)
≤
4
√
n
.
Induction Step
:
T
(2
n
)=
T
(
n
)+
√
2
n
≤
4
√
n
+
√
2
n
=2
√
2
√
2
n
=(2
√
2+1)
√
2
n
≤
4
√
2
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This note was uploaded on 12/27/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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