Dominant term

Question

What is the dominant term ?
(N(20N^2+5+0.5N^3))^2

Answer 1

This looks like homework. So instead of the full solution, I'd only give you a hint: in a polynomial p(N), the dominant term as N tends to infinity is always the highest degree term.

Answer 2

Since they asked for "term", it should be 0.25N^8. But if they ask for big O, you can write O(N^8).

Answer 3

thank you

Answer 4

But there is no N log N term. You're dividing the second term by N, so it cancels out with the N in N/4. Right?

You can check the simplification here:
http://m.wolframalpha.com/input/?i=10MlogM%2B%5B%28N%2F4%29log%28N%2F4%29%5D%2FN%2B2N

Answer 5

I had the Q incomplete
It says that N>>M
Does this make a difference?
Kishalaya Saha

Answer 6

It might make a difference. But people mean different things when they use the "much greater than" sign >>. I mean, how much? It needs a better quantification.

If it means, for example, N > CM^2 for some constant C, then the 2N term would eventually dominate, as log x has much slower growth than x.

But without knowing what >> means in this context, it's hard to say. It's possible that theoretical computer scientists always use it to mean something specific, but I don't know.

Answer 7

Should I write n^8 or (N(0.5N^3))^2

Answer 8

I have another one it is really confusing me
Which one is the dominant term
10MlogM+[(N/4)log(N/4)]/N+2N
Where N>M
???
Kishalaya Saha

Answer 9

That simplifies to
10M log M + log(N/4)/4 + 2N,
right? Then I think it's hard to say without knowing how big N is compared to M. It could be the first term or the third. Are you sure you have written the expression correctly?

Answer 10

Yes I'm sure . It just says N>M
I did simplify it like this but some says the dominant in NlogN not 10MlogM ??

Answer 11

What I ment was someone says it does not cancel it stays the same that's why NlogN
But I'm not sure of the answer

Answer 12

Okay, I understand what you mean, but it doesn't make sense to me why they wouldn't cancel. Sorry! Maybe someone else can help you.