1. ## little o notation

I'm on pg 288 of Apostol's Calculus volume 1. A part of the theorem at the top says that 1/(1 + g(x)) = 1 - g(x) + o(g(x)) if g(x) -> 0 as x -> a. The first subsequent example is to show tan x = x +(1/3)x ^3 + o(x^3) as x -> 0. Using the theorem and the taylor approximation to cosine, we have 1/cos x = 1/(1 - (1/2)x^2 + o(x^3)) = 1 +(1/2)x^2 +o(x^2) as x ->0. I can't see how this follows from the theorem or the definition of o notation.

2. A Taylor approximation for cosine is $\cos x = 1 - {x^2\over2!} + o(x^3)$. Letting $g(x) = - {x^2\over2!} + o(x^3)$ and using the theorem you get the approximation $\sec x = 1 - g(x) + o(g(x)) = 1 + {x^2\over2!} + o(x^3) + o(-{x^2\over2!} + o(x^3))$. Can you see that $o(x^3) + o(-{x^2\over2!} + o(x^3)) = o(x^2)$?

3. It's the last identity that i can't figure out how to justify. but since g(x) = -(x^2)/2! + o(x^3) shouldn't we have sec x = 1 + (x^2)/2! - o(x^3) + o((-x^2)/2! + o(x^3)) (with a minus sign preceding the o(x^3) term)? Either way, if you can explain how to justify the last sentence you wrote, it would be extremely helpful. i'm not good with the algebraic manipulations involving o-signs i guess.

4. Oh crap, yes, I lost the sign. It actually doesn't matter since the little o term represents only a magnitude. Anyway, basically the x^2 term is largest, so it dominates. Formally, let f and g be o(x^3) and let h be o(x^2/2! + g). We want to show that f+h is o(x^2). So compute ${f+h\over x^2} = {f\over x^2}+{h\over x^2} = x{f\over x^3} + {h\over x^2/2! + g}({1\over2!} + x{g \over x^3})$. The first term goes to zero as x does because f is o(x^3) and the second goes to zero because h is o(x^2/2! + g). (I should really have ${|f+h|\over x^2}$ I guess but I'm lazy. You get the idea.) Its easier if you have more theorems about how to manipulate little o's.