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

Question

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

I find it difficult to understand this question. An example would be welcome. In addition, if you use the l'Hôpital rule, please show how you make differentiation.

Answer 1

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

Answer 2

The claim is incorrect.

A counterexample is as follows: we have no doubt that 2n is an element of O(n) . But we can prove that exp(2n) not an element of O(exp(n)) . It is easy to understand by calculating

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

which means that exp(2n) not in O(exp(n)) .

Given your allusion to L'Hospital: this is a rule for calculating limits using derivatives, or rather:

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

under certain circumstances (for example, both f and g tend to infinity. I don’t know which criteria should be met, 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 an element of O(g'(x)) , then we have that f(x) is an element of O(g(x)) . Another direction is not so.

Answer 3

I will 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 this to solve inf / inf or 0/0 uncertainties. But your problem is not what I think, but maybe when you try to get O (g (n)) or exp (f (n)), which are composite functions.

The chain rule for obtaining composite functions is: (fog) (x) = f '(g (x)). g '(x)

if you do, you can get any composite function.