Thread:

1. BIG O proof

How do I prove the following:
O(n^2 logn) is also O(n^3)?

2. Originally Posted by taurus
How do I prove the following:
O(n^2 logn) is also O(n^3)?
$f(n)\in O(n^2 \log(n))$

means there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:

$|f(n)|<k n^2 \log(n)$

Now for $n>0$ we have $n>\log(n)$ so for $n>N>0$:

$|f(n)|<k n^3$

etc.

CB

3. You will have to expand on that, I have no clue how to go on...

4. Originally Posted by CaptainBlack
$f(n)\in O(n^2 \log(n))$

means there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:

$|f(n)|<k n^2 \log(n)$

Now for $n>0$ we have $n>\log(n)$ so for $n>N>0$:

$|f(n)|<k n^3$

etc.

CB
Originally Posted by taurus
You will have to expand on that, I have no clue how to go on...
So there exists a constant $k>0$ and an $N>0$ such that for all $n>N$:

$|f(n)|<k n^3$

hence $f(n)\in O(n^3)$

CB
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