Question

Is 3ⁿ = O(2ⁿ)? how about (3/2)ⁿ = O(2ⁿ)? Can you explain the answers?

I got false for the first since, 3ⁿ grows faster then 2ⁿ no matter what constant C you multiply 2ⁿ by. And same for the second?

How about log(3ⁿ) = O(log (2ⁿ) )? I think we can’t determine this because we don’t know the base of the log.

Answer 1

Let’s actually prove a stronger result: for any constants r₀ and r₁ where 1 ≤ r₀ < r₁, it is true that r₀ⁿ = O(r₁ⁿ) and it is false that r₁ⁿ = r₀ⁿ. This proves your result as a special case, since 1 < 3/2 < 2.

To prove the first part, we’ll show that r₀ⁿ = O(r₁ⁿ). To do this, we’ll use the definition of big-O and find values of n₀ and c such that for any n > n₀, we have that

r₀ⁿ ≤ c r₁ⁿ

We can choose n = n₀ and can choose c = 1. The above inequality then holds, so by definition we have that r₀ⁿ = O(r₁ⁿ).

To prove the second part, we’ll show that r₁ⁿ ≠ O(r₀ⁿ). To do this, we’ll proceed by contradiction. Assume for the sake of contradiction that there exists a choice of c and n₀ such that for any n > n₀, we have that

r₁ⁿ ≤ c r₀ⁿ

Take the log of both sides to get

n log r₁ ≤ log (c r₀ⁿ)

n log r₁ ≤ log c + n log r₀

n (log r₁ – log r₀) ≤ log c

n log(r₁ / r₀) ≤ log c

n ≤ log c / (log(r₁ / r₀))

But now we’re in trouble, since this statement should hold for any choice of n. However, if we pick any choice of n greater than log c / (log(r₁ / r₀)), the statement becomes false.

We have reached a contradiction, so our assumption must have been wrong. Thus if 1 < r₀ < r₁, we have that r₁ⁿ ≠ O(r₀ⁿ).

Hope this helps!