如何显示5n = O（nlogn）

Question

I have this as a homework question and don't remember learning it in class. Can someone point me in the right direction or have documentation on how to solve these types of problems?

Answer 1

More formally...

First, we prove that if f(n) = 5n , then f ∈ O(n) . In order to show this, we must show that for some sufficiently large k and i ≥ k , f(i) ≤ ci . Fortunately, c = 5 makes this trivial.

Next, I'll prove that for all f ∈ O(n) that f ∈ O(n * log n) . Hence, we must show that for some sufficiently large k , all i ≥ k , f(i) ≤ ci * log i . Hence, if we let k be large enough that f(i) ≤ ci , and i ≥ 2 , then the result is trivial since ci ≤ ci * log i .

QED.

Answer 2

Look into the definition of big-O-notation. It means that 5n will run no slower the nlogn, which is true. nlogn is an upper bound of the number of operations to be performed.

Answer 3

I don't remember the wording of the formal definition, but what you have to show is:

c ₁ * 5 * n < c ₂ * n * logn, n>c ₃

where c ₁ and c ₂ are arbitrary constants, for some number c ₃ . Define c ₃ in terms of c ₁ and c ₂ , and you're done.

Answer 4

It's been three years since I touched big-O stuff. But I think you can try to show this:

O(5n) = O(n) = O(nlogn)

O(5n) = O(n) is very easy to show, so all you have to do now is to show O(n) = O(nlogn) which shouldn't be too hard too.

Answer 5

You can prove it by applying L'Hopitals rule to lim n-> infinity of 5n/nlogn

g(n) = 5n and f(n)=nlogn

Derivate g(n) and f(n) so you will get something like this

5/(some stuff here that will contain n)

5/infinity = 0 so 5n = O(nlogn) is true