If f = O(g), is e^f = O(e^g)?

Question

If f = O(g), is e^f = O(e^g)?

I\'m having difficultly figuring out the above question. An example would be welcome. Also, if you use l\'Hôp

Answer 1

The claim is not correct.

A conterexample is the following: We have no doubt that 2n is element of O(n). But, we can prove that exp(2n) is not an element of O(exp(n)). This can be easily seen by computing the

                 exp(2n)
     lim        -------- = infinity
n -> infinity     exp(n)

which implies that exp(2n) is not in O(exp(n)).

Considering your hint about L'Hospital: It is a rule for computing limits using derivatives, more precisely:

                f(x)                       f'(x)
     lim       ------  =        lim     -----------
n -> infinity   g(x)      n -> infinity    g'(x)

under certain circumstances (e.g. both f and g tend towards infinity. I do not know the exact criteria to be fulfilled, so I just suggest reading this for more information.

But, what we can say about functions and their derivatives is the following:

If f'(x) is element of O(g'(x)), then we have that f(x) is element of O(g(x)). The other direction is not the case.

Answer 2

I'll try to help you with l'Hôpital's:

$\lim_{x \to a}{f(x)\over g(x)}=\lim_{x \to a}{f'(a) \over g'(a)}

We use that in order to solve inf/inf or 0/0 indetermination. But your problem is not that I think, but maybe when you try to derive the O(g(n)) or exp(f(n)) which are composite functions.

The chain rule to derive composite functions is this: (f o g)(x) = f'(g(x)).g'(x)

if you follow that, you can derive any composite function.

Answer 3

This statement is wrong, for example 2n = O(n), but exp(2n) != O(exp(n)). (The latter would mean exp(2n) <= C exp(n) for sufficiently large n, i.e. exp(n) <= C which is not true.)