# What is the order of these functions

1. Sep 21, 2008

### Gear2d

I need now what is the order (The Big Oh notation) of these three functions:

2^x
3^x
3^(x-10)

Last edited: Sep 21, 2008
2. Sep 21, 2008

### mathman

Your question is incomplete. You need to specify x ->?.

3. Sep 21, 2008

### HallsofIvy

Staff Emeritus
Also you should specify what your definition of "order" is. The "big Oh" I am familiar with says that one function is "O" of another function. Saying that g(x) is O(f(x)) (as x -> a) means that $\lim_{x\rightarrow a} g(x)/f(x)$ is a finite number.

Often "order" is given in terms of x going to infinity but even so, your question makes no sense. A function does not just have an "order". It is or is not "O" of another function. Are you asking if 2x= O(3x) or vice-versa?

Last edited: Sep 22, 2008
4. Sep 21, 2008

### Gear2d

Good point guys. I wanted to compare two of them to each other, lets say:

f(x) = 2^x vs. g(x) = 3^x

now from this I have to determine if f=Theta(g), f<Theta(g) or f>Theta(g). Instead of using L'Hospital rule:

c=lim 2^x/3^x = lim (2/3)^x = 0 as x goes to infinite this would mean that f<Theta(g).

I wanted to do this using order where x>1. Would 2^x have O(2^x) and 3^x have O(3^x), where f(n)<g(n)?

5. Sep 22, 2008

### HallsofIvy

Staff Emeritus
Please don't shift from O(g) to "Theta(g)"!

And since that limit is 0, f(x)= o(g). "Small o" or f= o(g) means that the fraction f/g goest to 0. As I said before, most text books give: "f= O(g), as x goes to a" means that f(x)/g(x) goes to a finite limit" which includes 0 so if f(x)= o(g), it follows that f= O(g). Some text books however, require that f(x)/g(x) go to a [g]nonzero[/b] finite limit in order that f= O(g) so that does NOT include f= o(g). I recommend that you check the definition in your text book to be certain which convention it uses.

Oh, and to answer your last question, f= O(f) for any function f so saying "2x= O(2x" (which is what I guess you mean by "2^x have O(2^x)) so "2x= O(2x)" is trivially true. A function doesn't have "an order" except as compared to some other function.