Construir dos funciones y satisfacer


19

Construya dos funcionesf,g:R+R+ satisfying:

  1. f,g are continuous;
  2. f,g are monotonically increasing;
  3. fO(g) and gO(f).

2
Have you considered the possibility that such functions might not exist?
jmite

If both f,g are logarithmico-exponential, then either f=O(g) or g=O(f). Most functions encountered in practice are of this form.
Yuval Filmus

Respuestas:


16

There are many examples for such functions. Perhaps the easiest way to understand how to get such an example, is by manually constructing it.

Let's start with function over the natural numbers, as they can be continuously completed to the reals.

A good way to ensure that fO(g) and gO(f) is to alternate between their orders of magnitude. For example, we could define

f(n)={nn is oddn2n is even

Then, we could have g behave the opposite on the odds and evens. However, this doesn't work for you, because these functions are not monotonically increasing.

However, the choice of n,n2 was somewhat arbitrary, and we could simply increase the magnitudes so as to have monotonicity. This way, we may come up with:

f(n)={n2nn is oddn2n1n is even, and g(n)={n2n1n is oddn2nn is even

Clearly these are monotone functions. Also, f(n)O(g(n)), since on the odd integers, f behaves like n2n while g behaves like n2n1=n2n/n=o(n2n), and vice-versa on the evens.

Now all you need is to complete them to the reals (e.g. by adding linear parts between the integers, but this is really beside the point).

Also, now that you have this idea, you could use the trigonometric functions in order to construct ``closed formulas'' for such functions, since sin and cos are oscillating, and peak on alternating points.


Can we say that f(n)O(n2n) and g(n)O(n2n)? f(n) and g(n) are as defined in your answer.
mayank

Yes. We can even say that f(n)n2n (similarly for g), which is stronger than O.
Shaull

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