Example of Induction inequality problem
Here is a different inequality that works out, that I did NOT do
in lecture.
Define a recurrence relation g as follows:
g(0) = 2, g(1) = 4,
g(n) = 3g(n1) + 4g(n2), for all integers n>1.
Prove, using strong induction, that g(n) > 4
n
, for all n>1.
Base Case(s):
n=2: g(2) = 3g(1)+4g(0) = 3(4) + 4(2) = 20
> 4
2
n=3: g(3) = 3g(2)+4g(1) = 3(20) + 4(4) = 76 > 4
3
Assume, for some arbitrary n=k, with k>1, that g(k) > 4
k
.
We must prove under this assumption for n=k+1,
g(k+1) > 4
k+1
,
g(k+1) = 3g(k) + 4g(k1)
= 3g(k) + 4g(k1)
3(4
k
) + 4g(k1), using the inductive hypothesis
> 3(4
k
) + 4(4
k1
), using the strong inductive hypothesis
= 3(4
k
) +4
k
= (3+1)(4
k
)
= 4(4
k
)
= 4
k+1
This proves the inductive step, so we can conclude that
g(n) > 4
n
, for all n>1.
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View Full DocumentHere is the problem I ended up working out in class:
Define a recurrence relation g as follows:
g(0) = 2, g(1) = 5,
g(n) = 5g(n1)  4g(n2), for all integers n>1.
Prove, using strong induction, that g(n) = 4
n
+ 1, for all n>1.
Base case(s): n=2: g(2) = 5g(1) – 4g(0) = 5(5) – 4(2) = 17 = 4
2
+ 1
n=3: g(3) = 5g(2) – 4g(1) =5(17) – 4(5)= 65 – 4
3
+ 1
Inductive hypothesis: Assume for k
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 Fall '09
 Natural number, 2J, 2m, 4k

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