Big O Notation :|

[mathews411]

So I'm taking a summer computer science class this year, and the topic of Big O notation has surfaced. In class it seemed fairly basic, we looked at simple equations (Ex. 8log n + 7n), the answer would be O(n) as it is the term that grows the fastest. That's stuff i can do.

Now we were given an assignment of similar type questions, but they threw something in that caused me to doubt what I know. Usually, when we would encounter a log, it would simple just be log(n), now they included log(n^2), or any kind of number for the exponent.

For example I'll give a similar question that was in my assignment so if anyone can answer it they not doing my assignment for me.
3log(n^9) + 2nlog(n >2).

Me and a friend both assumed the answer is O(nlogn), as nlogn grows faster then log(n). But then we started to talk about if it might be O(n^k) from the n^9 .

In class, we never talked about how if a term is in a logarithm if it grows the fastest, or if it defaults to the original log and becomes a slower growth function.

Any help will be greatly appreciated, as Big O notation makes me die a little on the inside.

Hope I dint make this post to confusing!

[Dreadnought]

Remember that you have

code:

log(n^k) = k*log(n)

With that in mind you should be able to compare the behavior of log(n^k) with n*log(n).

[rdrake]

If you expand out out, you get .

Recall from the logarithmic rules that



where , so we get:



So

Or you can be a jerk and say it's which it technically is.