big ohhh notation :P

[Sarahisme]

hey could someone please give me a bit of help with this big O notation bussiness, it makes not much sense to me, (mainly in terms of calucating limit sof indeterminate form, and how the big O's cancel etc.) anyways any help or pointers to a good website that explains the stuff would be great!

 

Thanks Guys and Gals

 

Sarah

[fuhrerkeebs]

If f(x)=O(g(x)) that just means that for large enough x and some constant c, |f(x)|<=c*g(x). You can use this to give bounds on certain limits, like f(x), if f(x)=g(x)+O(h(x)), because we know that lim _x->infinity g(x)-h(x) <= lim _x->infinity f(x) <= lim _x->infinity g(x)+h(x). I don't really know what you mean by "how the big O's cancel," so I can't help you out there.

[Sarahisme]

hang on a sec and i'll upload a pic of problem in which this "cancelling" (or looks like cancelling to me) happens....

[Sarahisme]

here it is...

[NSX]

What they mean is that there is an error term, on the order of x² and order of x, in reference to the last line ofc.

 

Since you are taking the limit as x approaches 0, x will become very small, so x² << 1, and x << 1/3.

 

Thus, they are negligible, and can be tossed away.

[Sarahisme]

? i am not quite understanding, sorry

[swansont]

If x < 1, and you're looking at a limit as x -> 0, then x >> x² >> x³, etc

 

You if you have a term of order x, you can ignore the higher order terms. They are small compared to x. If you have terms that look like x², you can ignore terms that look like x³, again, because they are small when x -> 0.

 

The O(xⁿ) just means that all the terms are that exponent, n, and the terms are small so they can be ignored. Thus it's not worth the time to actually calculate that term.

[Sarahisme]

ok thanks swansont and everyone else

[matt grime]

the importnat thing is that as x tends to zero so does O(x) or (X^n) for any positive n. just think of it as a function - if f(x) and g(x) tend to zero as x tends to zero then (1+f)/(2+g) tends to 1/3 as x tends to zero.