If $f$ is big or little oh of $g$, what can we say about $a^f$ and $b^g$ for $a,b>1$?

Question

I’m interested in what operations preserve asymptotic relationships.

For example, I can prove that if $f=o(g)$ (as $x\to\infty$), then $a^f=O(b^g)$, for any bases $a,b>1$.

But I think that’s the best we can say: I don’t think we can improve the big-oh in the conclusion to a little-oh. Yet I can’t find a counterexample.

Is that right? If so, what’s a good counterxample?

More generally, what’s some good intuition about how asymptotic relationships change when we apply an operator besides exponentiation?

Answer 1

Hint: $a^f/b^g=b^{\log_b(a)f}/b^g=b^{\log_b(a)f-g}$.

More generally, it depends what sort of operation you apply. Applying the exponential is very different from applying the logarithm or $\arctan$.

Answer 2

Example A

$2x = O(x)$ as $x\to\infty$, but of course $e^{2x}$ is not $O(e^x)$.

Example B

$1/x^2 = o(1/x)$ as $x\to\infty$, but $e^{1/x^2} \to 1$ and $e^{1/x} \to 1$ so $e^{1/x^2}$ is not $o(e^{1/x})$.

Example C, $a \ne b$

$x+\sqrt{x} = \Theta(x)$ but $2^{x+\sqrt{x}} = o(3^x)$

Answer 3

You might want to take a peek at Joel Spencer's "Asymptopia" (AMS, 2014), an in-depth discussion of matters asymptotical. Or take a look at Hildebrand's Short Course on Asymptotics.