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110

answers:

2

Hi all,

I'm just trying to understand how in little o notation this is true:

f(n)/g(n) as n goes to infinity = 0?

Can someone explain that to me?

I do get the idea that f(n) = o(g(n)) means that f(n) grows no faster then cg(n) for all constants c > 0.

I just don't get the bit in bold above.

+2  A: 

http://en.wikipedia.org/wiki/Little_o_notation#Little-o_notation

You've left something out, namely your definitions for f and g.

It would appear that the precondition for the bolded statement is g(n) in o(f(n)).

According to the Wikipedia article, f(n) = o(g(n)) means that f grows slower than cg(n) for all positive constants. So f(n) is not in o(f(n)).

Potatoswatter
Sorry if I sound dumb, but not sure what that means: "So f(n) is not in o(f(n))." Can you put that in more plain english?
Tony
`o(f(n))` is a class of functions. A function is either in the class or it's not. `f(n)` is not, despite being part of the definition of that class.
Potatoswatter
so f(n) is not in o(f(n)) because f(n) cannot grow slower then itself. Would I be correct in my statement?
Tony
@Tony: You got it!
Potatoswatter
A: 

There was a great episode of BBC's Horizon (titled 'To Infinity and Beyond') recently that explained this.

Damian Powell
Darn looks like it's not on iPlayer anymore. Keep an eye out for a repeat - it was cool.
Damian Powell