Asymptotic proof examples

Question

I came across two asymptotic function proofs.

f(n) = O(g(n)) implies 2^f(n) = O(2^g(n))

Given: f(n) ≤ C1 g(n)
     So, 2^f(n) ≤ 2^C1 g(n)                        --(i)

Now, 2^f(n) = O(2^g(n)) → 2^f(n) ≤ C2 2^g(n)                  --(ii)

From,(i) we find that (ii) will be true.
Hence 2^f(n) = O(2^g(n)) is TRUE.

Can you tell me if this proof is right? Is there any other way to solve this?

2.f(n) = O((f(n))^2)

How to prove the second example? Here I consider two cases one is if f(n)<1 and other is f(n)>1.

Note: None of them are homework questions.

right. so you are saying that f(n) = O((f(n))^2) is TRUE only for f(n) >= 1. — Andrew, Jul 06 '13 at 21:55
What does “f(n) >= 1” mean? Suppose f(n) = 1/n for odd n and 1+n for even n. — James Waldby - jwpat7, Jul 06 '13 at 22:05
what i am saying is, in your example (1/n), for n=1, it proves the condition f(n)<= c.f(n)^2 ..for any constant c ......(f(n) >= 1) but as n grows towards infinity, the f(n) grows towards zero (f(n) < 1) and in that case, it does not prove the condition f(n)<=c.f(n)^2 (big oh definition). So, the above property is true only when f(n)>=1 — Andrew, Jul 06 '13 at 22:20
It's true whenever function f is bounded below by a positive number; see answer — James Waldby - jwpat7, Jul 06 '13 at 23:35

score 0 · Accepted Answer · answered Jul 06 '13 at 23:34

The attempted-proof for example 1 looks well-intentioned but is flawed. First, “2^f(n) ≤ 2^C1 g(n)” means 2^f(n) ≤ (2^C1)*g(n), which in general is false. It should have been written 2^f(n) ≤ 2^(C1*g(n)). In the line beginning with “Now”, you should explicitly say C2 = 2^C1. The claim “(ii) will be true” is vacuous (there is no (ii)).

A function like f(n) = 1/n disproves the claim in example 2 because there are no constants N and C such that for all n > N, f(n) < C*(f(n))². Proof: Let some N and C be given. Choose n>N, n>C. f(n) = 1/n = n*(1/n²) > C*(1/n²) = C*(f(n))². Because N and C were arbitrarily chosen, this shows that there are no fixed values of N and C such that for all n > N, f(n) < C*(f(n))², QED.

Saying that “f(n) ≥ 1” is not enough to allow proving the second claim; but if you write “f(n) ≥ 1 for all n” or “f() ≥ 1” it is provable. For example, if f(n) = 1/n for odd n and 1+n for even n, we have f(n) > 1 for even n > 0, and less than 1 for odd n. To prove that f(n) = O((f(n))²) is false, use the same proof as in the previous paragraph but with the additional provision that n is even.

Actually, “f(n) ≥ 1 for all n” is stronger than necessary to ensure f(n) = O((f(n))²). Let ε be any fixed positive value. No matter how small ε is, “f(n) ≥ ε for all n > N'” ensures f(n) = O((f(n))²). To prove this, take C = max(1, 1/ε) and N=N'.

Asymptotic proof examples

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