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anyone correct me if i am wrong!!

- #3

HallsofIvy

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[tex]\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}[/itex]

If that limit is a finite, non-zero, number, then f= O(g) (and g= O(f)). If it is 0, then f= o(g). If it is infinite, then g= o(g).

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I agreee that's kinda stupid terminology .Get used to it.If you are given two functions and asked which function goes faster. You divide on function by the other and you use lopitals rule. If your ratio gives you a constant it is said the two fuction grow at the same rate. I can't understand why 5x and 2x would be growing at the same rate. 5=2?

It makes more sense to me to say f1(x)=5x grows "faster" than f2(x)=2x for [itex]x\in\mathbb{R^+}[/itex].

- #5

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at the last one i think halls meantIf it is 0, then f= o(g). If it is infinite, then g= o(g).

g= o(f), since at this case g(x) is said to be of a higher order than f, considering that both of them are infinitely small.

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yeah and if f, and g are infinitely large when x->infinity, than it is vice verseHallsofIvy;1285869[tex said:\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}[/itex]

If that limit is a finite, non-zero, number, then f= O(g) (and g= O(f)). If it is 0, then f= o(g). If it is infinite, then g= o(g).

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HallsofIvy

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Actually, I was assuming that f and g are unbounded so I'm not sure what you mean by "vice versa".yeah and if f, and g are infinitely large when x->infinity, than it is vice verse

- #8

Hurkyl

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You should consider "the same rate" a technical term with a technical definition that may or may not relate to a common English interpretation of those same words.

We use this definition because it's useful. Constant factors are often irrelevant. In analysis, there is no difference between "x goes to zero" and "4x goes to zero". In computer science, we choose up front to ignore the constant factors, so that we can prove things in generality without having to worry about the architectural differences between different computers, or the optimization skill of the programmer / compiler, et cetera.

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