Does $f(n) \in O(g(n))$ imply $2^{f(n)} \in O(2^{g(n)})$?

Question

Is the following true: $$ f(n) \in O(g(n)) \text{ then } 2 ^ {f(n)} \in O(2^{g(n)})$$

Does this answer your question? What are you allowed to move into the big O notation for it to be still correct? — Nathaniel, Mar 15 '22 at 09:23

score 5 · Accepted Answer · edited Mar 15 '22 at 18:15

5

An obvious counterexample is $f(n) = n$ and $g(n) = \frac{n}{2}$. We know that $f(n) \in O(g(n))$, but $2^{f(n)} = 2^n \not \in O(2^{g(n)}) = O(2^{\frac{n}{2}}) = O(\sqrt{2^n})$.

edited Mar 15 '22 at 18:15

Steven

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answered Mar 15 '22 at 11:46

OmG

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score 0 · Answer 2 · answered Mar 15 '22 at 22:27

In general, you would ask: For a function h(n), if f(n) = O(g(n)), is h(f(n)) = O(h(g(n))?

We have f(n) < c * g(n) for some constant c, and for large n. The function h(n) must be such that within the range of f(n) and g(n), h(f(n)) must not grow so fast that a linear change in f(n) creates a non-linear change in h(f(n)).

Does $f(n) \in O(g(n))$ imply $2^{f(n)} \in O(2^{g(n)})$?

2 Answers2