91
Clearly this is smaller than
√
n
log
n
, so we will guess that
T
(
n
) = Θ(
√
n
log
n.
To complete the proof we must show that
T
(
n
)
is in
Ω(
√
n
log
n
)
, but this
turns out to be impossible.
On the other hand, the recurrence is clearly
Ω
√
n
. So, let’s guess that
T
(
n
)
is in
O
(
√
n
)
for
c
= 4
, n
0
= 2
. We prove this by induction.
Base case
:
T
(2) = 1 +
√
2
<
4
√
2
, so the hypothesis is correct.
Induction Hypothesis
: For any value less than or equal to
k
,
T
(
k
)
<
4
√
k
.
Induction Step
: For
2
k
,
T
(2
k
) =
T
(
k
) +
√
2
k.
By the induction hypothesis,
T
(
k
) +
√
2
k <
4
√
k
+
√
2
k.
For the theorem to be correct,
T
(
k
) +
√
2
k <
4
√
k
+
√
2
k <
4
√
2
k
which is true. Thus, by the principle of Mathematical Induction, the theorem
is correct.
14.11
Expanding the recurrence, we get
T
(
n
)
=
2
T
(
n/
2) +
n
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 Fall '08
 BELL,D
 Mathematical Induction, Recursion, Mathematical logic, 2k, Mathematical proof

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