If f(n) = o(g(n)) , then is 2^(f(n)) = o(2^(g(n)))?

Question

Notice that I am asking for little-o here (see similar question here) - for big Oh it's clearly wrong - for little-o it feels right but can't seem to prove it...

EDIT: glad I raised a debate :) Assume f,g > 0 for simplicity

Solution 1

It is, at least if g(n) is converging to positive infinity for n to positive infinity (if g(n) isn't there are easy to find counterexamples).

Sketch of a proof:

Prerequsites: g(n) is converging to positive infinity for n to positive infinity.

f(n) in o(g(n)) means:

for every eps > 0 exists a n0 so that for all n > n0 abs(f(n)) < abs(g(n)*eps).

Form this follows:

2^abs(f(n)) < 2^abs(g(n)*eps) = (2^eps)^g(n) (for n > n0).

For eps < 1:

(2^eps)^n is in o(2^n) (as 2^eps < 2) 

it follows that for every eps2 > 0 exists a n1 so that for all n > n1

(2^eps)^n < eps2*(2^n).

Choosing eps2 = eps vields:

(2^eps)^n < eps*(2^n) for all n > n1 (n1 is dependent on eps)

Because g(n) -> pos. inf. for n -> pos. inf. there exists a n2 so that

g(n) > n1 for all n > n2

Following from there is

(2^eps)^g(n) < eps*2^g(n) for all n > n2.

So for every 0 < eps < 1 there exists a n3 >= max(n0,n2) so that

2^abs(f(n)) < 2^abs(g(n)*eps) = (2^eps)^g(n) < eps*2^g(n) for all n > n3.

For every eps3 > 1 also:

2^abs(f(n)) < eps*2^g(n) < eps3*2^g(n)

So for every eps > 0 there exists a n3 so that

2^abs(f(n)) < eps*2^g(n) for all n > n3

Becase 2^f(n) < 2^abs(f(n)) for all n, and 2^x > 0 for all real x, it follows

abs(2^f(n)) < abs(eps*2^g(n)) for all n > n3

q.e.d.

If something is unclear or wrong, please comment.

EDIT: Some thoughts on other g(n):

A subsequence of g(n) is restricted i.e. it exists some x so that there isn't an n0 with abs(g(n))>=x for all n > n0:

o(g(n)) consists only of functions that are constant 0 after some n or converge to 0. Then 2^g(n) also has a restricted subsequence, but 2^f(n) is constant 1 after some point.

There is no n0 so g(n) > 0 for all n > n0:

2^g(n) < 1 if g(n) < 0, so g(n) has a restricted subsequence meaning o(2^g(n)) consists only of functions that are constant 0 after some n or converge to 0.

Solution 2

Here's an answer. The result depends on the convergence properties of g(n).

First consider the related limit:

lim(x->inf) log_2 ((2^(f(n))) / (2^(g(n))))
=
lim(x->inf) ( log_2(2^(f(n))) - log_2(2^(g(n))) )
=
lim(x->inf) ( f(n) - g(n) ) = lim(x->inf) ( g(n) * f(n) / g(n) - g(n) )
=
lim(x->inf) ( -g(n) ) = - lim(x->inf) g(n)

... Now, to get this into the form of the original question in your post, IF it is mathematically correct to switch the limit and the log (which I think it is), then we would have:

lim(x->inf) log_2 ((2^(f(n))) / (2^(g(n))))
=
log_2 lim(x->inf) ((2^(f(n))) / (2^(g(n)))).

Now, moving on to answer the question. If it is true that

2^(f(n)) = o(2^(g(n))),

then in the limit, the right-hand side becomes

log_2 (0) = - inf

... so in this scenario it must be true that

lim(x->inf) g(n) = inf

This result does seem to make sense. Consider f(x) = exp(-x) and g(x) = 1 - exp(-x). Clearly, in this example f(x) = o(g(x)). However, 2^(f(x)) is not o(2^(g(x))) because

lim(x->inf) (2^exp(-x)) / (2^(1-exp(-x))) = 1/2.

But given f(x) = exp(x), g(x) = exp(2x), where also f(x) = o(g(x)) and where lim(x->inf) g(x) = inf, meeting the above condition, we have

lim(x->inf) (2^exp(x)) / (2^exp(2x))
=
lim(x->inf) 1 / 2^(exp(x)*(exp(x) - 1)) = 0

so 2^(f(x)) = o(2^(g(x))).

Comments

[huon]

I don't think the step from lim (g*f/g - g ) to lim -g is valid for all f and g (not without some justification/explanation at least, because g can go to infinity faster than f/g goes to zero). And it is correct to swap the limit with the log (log is continuous for positive numbers, which is all you need). I agree with your conclusion (that it depends on specific properties of g).

[Dan Nissenbaum]

Perhaps this proves it: For any eps, however small, we have lim f(n)/g(n) < eps for n>N and some N (assume the magnitude of the limit is intended). Therefore, lim (gf/g-g) < lim (g*eps-g) = lim (g(eps-1) = lim (-g). Do you think? (For the latter steps, it assumes g>0, but it should be the same form if g<0. Perhaps there is some special case if g=0 in the limit.)

[Dan Nissenbaum]

... Addendum: it seems, actually, that for the latter steps, not only does it assume that g>0, but it also washes over the sign of lim f/g, so perhaps this influences things, as well.

[Mr_and_Mrs_D]

Mistakes - please go through it again - not sure about the proof but for instance f(x) = x, g(x) = 2x, f(x) != o(g(x)) - same with the other one. Consider f,g positive

[Mr_and_Mrs_D]

first line - you mean g(n) > 0 (for all n after some point)? - this is unneeded if g->inf for n->inf (follows from it). If g(n) does not (go to infinity) can you prove it or give a counter example ? Assume g > 0 for all n. Also your definition is wrong - o means that for every c > 0 there is an n0 : f < cg for every n=>n0 - notice the strict inequality. Assume g,f >0.

[Mr_and_Mrs_D]

and what about eps > 1 ? also in line (2^eps)^g(n) <= eps*2^g(n) for all n > n3. do you mean eps2 ?

[Dan Nissenbaum]

Thanks. It was the easy case so I wasn't careful. I intended exp([2]x), not [2]x. For the other case, it was intended that f(x) != o(g(x)), as noted.

[Dan Nissenbaum]

Pardon me. For the other case it was intended that f(x) = o(g(x)) (equals sign), so I believe you're mistaken that it's not. (This case is f(x) = exp(-x), g(x) = 1 - exp(-x).)

[Ral Zarek]

@Mr_and_Mrs_D I added some sketch of a proof for the other cases I could think of (if you can think of more just drop a comment). Yes the g(n) > 0 isn't really needed. On the > vs >=: corrected it. eps > 1 usually isn't a problem, if abs(f(n)) < abs(eps*g(n)) for an eps < 1 it also is for any eps >= 1.

[Mr_and_Mrs_D]

1-exp(-x)/exp(-x) -> inf so yes but still not sure about how the definitions go for non positive functions. Consider only positive ones plz

[Mr_and_Mrs_D]

if abs(f(n)) < abs(eps*g(n)) for an eps < 1 it also is for any eps >= 1. : indeed but I need to find for every eps > 0 a suitable n0 s.t. f < eps*g for n > n0 ! Your proof seems to suggest that if there is an eps <1 s.t f < eps * g (a mere f = O(g) for an eps < 1) then 2^f = o(2^g) - correct ?

[Mr_and_Mrs_D]

Also - on g -> inf : not needed. You are just using g > 0. g may very well be bounded - as long as f goes to zero I guess. Edit : (2^eps2)^g(n) < eps2*2^g(n) for all n > n3 should read (2^eps)^g(n) < eps2*2^g(n) for all n > n3. At this point - review your question - are you really proving that a single eps1 < 1 is enough ?

[Mr_and_Mrs_D]

Aslo : There is no n0 so g(n) > 0 for all n > n0. This does not mean that g(n) is bounded. Consider g = exp( n )* abs( sin( n*pi/2 ) )

[Ral Zarek]

@Mr_and_Mrs_D It's rather common to restrict to 0 < eps < 1 in a mathematical proof, as eps is usually considered to be very small (as close to 0 as posible). In this case: if there is a n0 for an eps < 1 that n0 also works for any eps >= 1.

[Ral Zarek]

@Mr_and_Mrs_D There was a typo in the previous expression. All the eps mean any number > 0 (in one case any number between = and 1). I probably should have used eps throughout the proof, but usually that confuses non-mathematicans.

[Ral Zarek]

I didn't say g was bounded in case no n0 so g(n) > 0 for all n > n0, I said 2^g(n) has a bounded (by 1) subsequence, which has pretty much the same effect for o. I clarified this now.

[Dan Nissenbaum]

If all functions are positive, I believe the proof is straightforward as it stands. Are there incorrect steps?

[Mr_and_Mrs_D]

let us continue this discussion in chat